\documentclass[reqno]{amsart}
\usepackage{hyperref}

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2015 (2015), No. 82, pp. 1--21.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2015 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2015/82\hfil Solvability of fractional analogues]
{Solvability of fractional analogues of the Neumann problem for a nonhomogeneous 
 biharmonic equation} 

\author[B. Kh. Turmetov \hfil EJDE-2015/82\hfilneg]
{Batirkhan Kh. Turmetov}

\address{Batirkhan Kh. Turmetov \newline
Institute of Mathematics and Mathematical Modeling,
Ministry of Education and Science Republic of Kazakhstan,
050010,Almaty, Kazakhistan. \newline
Department of Mathematics, Akhmet Yasawi International Kazakh-Turkish University,
161200, Turkistan, Kazakhistan}
\email{batirkhan.turmetov@iktu.kz}


\thanks{Submitted January 29, 2015. Published April 3, 2015.}
\subjclass[2000]{35J15, 35J25, 34B10, 26A33, 31A05, 31B05}
\keywords{Biharmonic equation; fractional derivative;  Miller-Ross operator;
\hfill\break\indent  Neumann problem}

\begin{abstract}
 In this article we study the  solvability of some boundary value problems
 for inhomogenous biharmobic equations. As a boundary operator we consider
 the differentiation operator of fractional order in the Miller-Ross sense.
 This problem is a generalization of the well known Neumann problems.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{remark}[theorem]{Remark}
\newtheorem{problem}[theorem]{Problem}
\allowdisplaybreaks


\section{Introduction}\label{intr}

Biharmonic equations appear in the study of mathematical models in several
real-life processes as, among others, radar imaging \cite{Ander} or incompressible
flows \cite {Lai-Liu}.
Omitting a huge amount of works devoted to the study of this kind of equations,
we refer some of them regarding  to their used methods. Difference schemes
and variational methods were used in the works \cite{Ali,Ehrlich}.
By using numerical and iterative methods, Dirichlet and Neumann boundary
problems for biharmonic equations were studied in the papers
\cite{Bjorstad,Dang}.
There are some works, for example \cite {Shi}, where a computational method,
based on the use of Haar wavelets was used for solving 2D and 3D
Poisson and biharmonic equations. We also point out the work made in \cite{Deng},
 where regularity of solutions for nonlinear biharmonic equations was investigated.
In \cite{Begehr} and the dissertation \cite{Wang} various problems for complex
biharmonic and polyharmonic equations were investigated.

In this article  we  refer  to  the domain $ \Omega  = \{ x \in R^n :|x| < 1\}$,
as  the  unit  ball. The dimension of the space is $n \ge 3$, and it is denoted
$\partial \Omega  = \{ x \in R^n :|x| = 1\} $  as the unit sphere.
The usual Euclidean norm is written as  
$ |x|^2  = x_1^2  + x_2^2  + \dots + x_n^2 $.
Now, for any $ u:\Omega  \to R$  smooth enough function  and a given 
$ \alpha  > 0$,
denoting by $ r = |x |$  and $ \theta  = x/|x|$, the  appropriate integral
operator  of  order $ \alpha $  in the  Riemann-Liouville sense can  be defined,
in a sense to (\cite{kilbas-book}, p.69), by the following expression
$$
J^\alpha  [u](x) = \frac{1}{{\Gamma (\alpha )}}\int_0^r {({r - \tau }
)^{\alpha  - 1} u(\tau \theta )d\tau } .
$$

In what follows, we assume that $ J^0 [u](x) = u(x) $  for all $x \in \Omega $.
Let $ m - 1 < \alpha  \le m$, $m = 1,2,\dots$. The following expressions
\begin{gather*}
{}_{RL} D^\alpha  [u](x) = \frac{{d^m }}{{dr^m }}J^{m - \alpha } [u](x),\quad
{}_C D^\alpha  [u](x) = J^{m - \alpha } [ {\frac{{d^m u}}{{dr^m }}} ](x),
\end{gather*}
are called, respectively, derivatives of  $\alpha $  order in Riemann-Liouville
and Caputo sense \cite{kilbas-book}. Here $ \frac{d}{dr}$  is a differentiation
operator of the form
\[
\frac{d}{{dr}} = \sum_{i = 1}^n {\frac{{x_i }}{r}\frac{\partial }{{\partial x_i }}},
\quad \frac{{d^k }}{{dr^k }} = \frac{d}{{dr}}({\frac{{d^{k - 1} }}{{dr^{k - 1} }}} ),
\quad k = 2,3,\dots.
\]
Let the parameter $j$ take one of the values, $j = 0,1,\dots,m$
and consider the set of operators
$$
D_j^\alpha  [u](x) = \frac{{d^{m - j} }}{{dr^{m - j} }}J^{m - \alpha }
 \frac{{d^j }}{{dr^j }}u(x).
$$
If $j \ge 1$ and $D = \frac{d}{{dr}}$, then
$$
D_j^\alpha = \underbrace {D \cdot D \cdot \dots \cdot D}_{m - j}
\cdot _C D^{\alpha  - j}.
$$
This operator is called derivative of $\alpha $ order in Miller-Ross sense
 \cite{Miller}.
Denote
\begin{gather*}
B_j^\alpha  u(x) = r^\alpha  D_j^\alpha  u(x),\\
B^{ - \alpha } u(x) = \frac{1}{{\Gamma (\alpha )}}
\int_0^1 {(1 - s)^{\alpha  - 1} s^{ - \alpha } u(sx)ds}.
\end{gather*}
Let $ 0 < \alpha  \le 2$. Consider the following problems in the domain $\Omega$.

\begin{problem} \label{probl1} \rm
Let $0 < \alpha < 2$. Find a function
$ u(x ) \in C^4 (\Omega  ) \cap C({\bar \Omega } )$ such that
$B_1^{\alpha  + k} [ u ](x ) \in C(\bar \Omega )$, $k = 0,1$
satisfying the  equation
\begin{equation}\label{e1-1}
\Delta ^2 u(x ) = g(x ),\quad x \in \Omega,
\end{equation}
and the boundary value conditions:
\begin{gather}\label{e1-2}
D_1^\alpha  [ u ](x ) = f_1 (x ),\quad x \in \partial \Omega, \\
\label{e1-3}
D_1^{\alpha  + 1} [ u ](x ) = f_2 (x ),{\rm{ }}x \in \partial \Omega.
\end{gather}
\end{problem}

\begin{problem} \label{probl2} \rm
 Let $1 < \alpha  \le 2$. Find a function
 $ u(x ) \in C^4 (\Omega  ) \cap C({\bar \Omega } )$  such that
 $ B_2^{\alpha  + k} [ u ](x ) \in C(\bar \Omega )$, $k = 0,1$
satisfying  equation \eqref{e1-1} and the boundary value condition:
\begin{gather}\label{e1-4}
D_2^\alpha  [ u ](x ) = f_1 (x ),\quad x \in \partial \Omega,  \\
\label{e1-5}
D_2^{\alpha  + 1} [ u ](x ) = f_2 (x ),\quad x \in \partial \Omega.
\end{gather}
\end{problem}

Note that the boundary value problems with boundary operators of fractional
order for elliptic equations of the second order have been studied in
\cite{berdyshev-smj,karachik-sam,kirane-smj1,kirane-smj2,muratbekova-bvp,
sadybekov-ejd,torebek-bvp,
turmetov-de,turmetov-sam,umarov-dm,umarov-fcaa}.
 Moreover, in \cite{berdyshev-ams} for the equation \eqref{e1-1} the boundary-value
problem with the conditions
\[
 D_0^\alpha [u](x) = f_1 (x),D_0^{\alpha  + 1}[u](x) = f_2 (x), \quad
\,x \in \partial \Omega,
\]
 $0<\alpha\leq1$ has been studied.

Note that for all $ x \in \partial \Omega $  we have the equality
$r\frac{{du}}{{dr}} = \frac{{du}}{{dr}} = \frac{{du}}{{d\nu }}$,
where $ \nu $ is a vector of outward normal to $ \partial \Omega $.
It is well known (see e.g. \cite{karachik-Cmat}) that for all
$ x \in \partial \Omega $  the operator
$r\frac{d}{{dr}}({r\frac{d}{{dr}} - 1} )\dots({r\frac{d}{{dr}} - k + 1} ) $
 coincides with the operator $ \frac{{d^k }}{{d\nu ^k }}$, $k = 1,2,\dots$.
Then in the case of  $ \alpha  = 1 $  for all  $ x \in \partial \Omega $  we obtain
$$
D_1^1 u(x) = \frac{{du(x)}}{{dr}}
= \frac{{du}}{{d\nu }}, r^2 D_j^2 u(x)
= r^2 \frac{{d^2 u(x)}}{{dr^2 }}
= r\frac{d}{{dr}}\Big({r\frac{d}{{dr}} - 1} \Big)u(x).
$$
Consequently, for values $ \alpha  = 1 $ or $ \alpha  = 2 $,
 problems \ref{probl1} and \ref{probl2} are analogues of the Neumann problem 
for the equation \eqref{e1-1}.
The considered problems in the case of $\alpha  = 1$  have been studied
in \cite{karachik-ijpam}, and in the case of $ \alpha  = 2 $
in \cite{turmetov-ashur}. It is proved that in the case of  $\alpha  = 1$
for solvability of the problem the following conditions are necessary
and sufficient:
\begin{equation}\label{e1-6}
\frac{1}{2}\int_\Omega  {({1 - |x|^2 } )g(x )dx}
 = \int_{\partial \Omega } {[ {f_2 (x ) - f_1 (x )} ]ds_x },
\end{equation}
and in the case of $ \alpha  = 2$,
\begin{gather}\label{e1-7}
\frac{1}{2}\int_\Omega  {({1 - |x|^2 } )\Gamma _3 [g(x)]dx}
 = \int_{\partial \Omega } {f_2 (x)dS_x }, \\
\label{e1-8}
\frac{1}{2}\int_\Omega  {x_k (1 - |x|^2 )\Gamma _4 [ g ](x)dx}
= \int_{\partial \Omega } {x_k [ {f_2 (x) - f_1 (x)} ]dS_x } ,\quad
k = 1,2,\dots,n,
\end{gather}
where
$\Gamma _c [ u ](x ) = ({r\frac{\partial }{{\partial r}} + c} )u(x )$, $c > 0$.

Note that the Neumann problem in the case of polyharmonic equation was
studied in \cite{karachik-DU,karachik-JAIM,turmetov-ashur1}.


\section{Properties of the operators $B_{j}^{\alpha}$ and $B^{-\alpha}$}
\label{POPER}

We assume that the function $u(x)$ is smooth enough in the domain $\Omega$.
The following proposition can be proved by direct calculation.

\begin{lemma}\label{l2-1}
Let $v_1 (x) = r\frac{{du(x)}}{{dr}},v_2 (x)
= r\frac{d}{{dr}}({r\frac{d}{{dr}} - 1} )u(x)$. Then the following equalities hold:
\begin{gather}\label{e2-1}
v_1 (0) = v_2 (0) = 0, \\
\label{e2-2}
\frac{{\partial v_2 }}{{\partial x_k }}(0) = 0, \quad k = 1,2,\dots,n.
\end{gather}
\end{lemma}

Similar propositions hold for the function $ B_{j}^{\alpha}[u](x)$, $j = 0,1$.

\begin{lemma}\label{l2-2}
Let $0 < \alpha  \le 2 $. Then the following equalities hold:
\begin{gather}\label{e2-3}
B_1^\alpha  [u](0) = 0, \\
\label{e2-4}
B_2^\alpha  [u](0) = 0,\frac{{\partial B_2^\alpha  [u](0)}}{{\partial x_i }} = 0,
\quad i = 1,2,\dots,n.
\end{gather}
\end{lemma}

\begin{proof}
Let $0 < \alpha  < 1$. Then by the definition of the operator $B_1^\alpha  $
 for the function $B_1^\alpha[u](x)$ we have
\begin{align*}
&B_{1}^{\alpha}[u](x)\\
&= \frac{r^\alpha}{\Gamma(1-\alpha)}
 \int_{0}^{r}(r-\tau)^{-\alpha}\frac{du}{d\tau}(\tau\theta)d\tau
=\frac{r^\alpha}{\Gamma(2-\alpha)}\frac{d}{dr}
 \int_{0}^{r}(r-\tau)^{1-\alpha}\frac{du}{d\tau}(\tau\theta)d\tau\\
&= \frac{{r^\alpha  }}{{\Gamma (1 - \alpha )}}\frac{d}{{dr}}
 \Big[ { {\frac{{(r - \tau )^{1 - \alpha } }}{{1 - \alpha }}u(\tau \theta)}
 \Big|_{\tau  = 0}^{\tau  = r}
 + \int_0^r {(r - \tau )^{ - \alpha } u(\tau \theta)d\tau } } \Big] \\
& = \frac{{r^\alpha  }}{{\Gamma (1 - \alpha )}}\frac{d}{{dr}}
 \Big[ { - \frac{{r^{1 - \alpha } }}{{1 - \alpha }}u(0)
  + r^{1 - \alpha } \int_0^1 {(1 - \xi )^{ - \alpha } u(\xi x)d\xi } } \Big] \\
&=- \frac{{u(0)}}{{\Gamma (1 - \alpha )}} + (1 - \alpha )u_1 (x)
 + r\frac{{du_1 (x)}}{{dr}},
\end{align*}
where
$$
u_1 (x) = \frac{1}{{\Gamma (1 - \alpha )}}\int_0^1 {(1 - \xi )^{ - \alpha }
u(\xi x)d\xi } .
$$
Therefore,
\begin{equation}\label{e2-5}
B_1^\alpha  [u](x) =  - \frac{{u(0)}}{{\Gamma (1 - \alpha )}}
+ (1 - \alpha )u_1 (x) + r\frac{{du_1 (x)}}{{dr}},\quad x \in \Omega .
\end{equation}
Hence,  by equality \eqref{e2-1} we obtain
\begin{align*}
\lim_{x \to 0} B_1^\alpha  [u](x)
&=  - \frac{{u(0)}}{{\Gamma (1 - \alpha )}}
+ (1 - \alpha )\lim_{x \to 0} u_1 (x)
+ \lim_{x \to 0} r\frac{{du_1 (x)}}{{dr}} \\
&= - \frac{u(0)}{\Gamma (1 - \alpha )}
 + \frac{(1 - \alpha )u(0)}{\Gamma (1 - \alpha )}
 \int_0^1 {(1 - \xi )^{ - \alpha } d\xi }  \\
&=  - \frac{u(0)}{\Gamma (1 - \alpha )}
+ \frac{(1-\alpha )u(0)}{\Gamma (2 - \alpha )}= 0.
\end{align*}
Equality \eqref{e2-3} is proved for the case $0 < \alpha  < 1$.

No let $1 < \alpha  < 2$ and $j = 1$. Then by definition of
 $B_1^\alpha $ we have
\begin{align*}
B_1^\alpha  [u](x)
&= \frac{{r^\alpha  }}{{\Gamma (2 - \alpha )}}\frac{d}{{dr}}
 \int_0^r (r - \tau )^{1 - \alpha } \frac{{du}}{{d\tau }}(\tau \theta)d\tau  \\
&= \frac{{r^\alpha  }}{{\Gamma (1 - \alpha )}}\frac{{d^2 }}{{dr^2 }}
 \int_0^r \frac{{(r - \tau )^{2 - \alpha } }}{{(2 - \alpha )}}
 \frac{{du}}{{d\tau }}(\tau \theta)d\tau    \\
& = \frac{{r^\alpha  }}{{\Gamma (2 - \alpha )}}\frac{{d^2 }}{{dr^2 }}
\Big[ { {\frac{{(r - \tau )^{2 - \alpha } }}{{2 - \alpha }}
 u(\tau \theta)} \Big|_{\tau  = 0}^{\tau  = r}
+ \int_0^r {(r - \tau )^{1 - \alpha } u(\tau \theta)d\tau } } \Big] \\
&= \frac{{r^\alpha  }}{{\Gamma (2 - \alpha )}}\frac{{d^2 }}{{dr^2 }}
\Big[ { - \frac{{r^{2 - \alpha } }}{{2 - \alpha }}u(0) + r^{2 - \alpha }
 \int_0^1 {(1 - \xi )^{1 - \alpha } u(\xi x)d\xi } } \Big] \\
& =  - \frac{{(1 - \alpha )u(0)}}{{\Gamma (2 - \alpha )}}
 + (1 - \alpha )(2 - \alpha )u_2 (x) + 2(2 - \alpha )r\frac{{du_2 (x)}}{{dr}}
 + r^2 \frac{{d^2 }}{{dr^2 }}u_2 (x),
\end{align*}
where
$$
u_2 (x) = \frac{1}{{\Gamma (2 - \alpha )}}\int_0^1 {(1 - \xi )^{1 - \alpha }
u(\xi x)d\xi } .
$$
Therefore,
\begin{equation}\label{e2-6}
\begin{split}
B_1^\alpha  [u](x)
&=  - \frac{{(1 - \alpha )}}{{\Gamma (2 - \alpha )}}u(0)
 + (1 - \alpha )(2 - \alpha )u_2 (x) \\
&\quad + 2(2 - \alpha )r\frac{{du_2 (x)}}{{dr}}
 +r\frac{d}{{dr}}({r\frac{d}{{dr}} - 1} )u_2 (x),x \in \Omega .
\end{split}
\end{equation}
Then, taking into account the equalities \eqref{e2-1} and \eqref{e2-2},
we obtain
\begin{align*}
\lim_{x \to 0} B_1^\alpha  [u](x)
&=  - \frac{{(1 - \alpha )}}{{\Gamma (2 - \alpha )}}u(0)
  + (1 - \alpha )(2 - \alpha )\lim_{x \to 0} u_2 (x) \\
&=  - \frac{{(1 - \alpha )u(0)}}{{\Gamma (2 - \alpha )}}
  + \frac{{(1 - \alpha )(2 - \alpha )u(0)}}{{\Gamma (2 - \alpha )}}
 \int_0^1 {(1 - \xi )^{1 - \alpha } d\xi } \\
& =  - \frac{{(1 - \alpha )u(0)}}{{\Gamma (2 - \alpha )}}
  + \frac{{(1 - \alpha )(2 - \alpha )u(0)}}{{\Gamma (3 - \alpha )}} = 0.
\end{align*}
Equality \eqref{e2-3} is proved for the case $1 < \alpha  < 2,j = 1$.

Now we turn to the proof of the first equality of \eqref{e2-4}.
By definition of  $B_2^\alpha  $ we have
\begin{align*}
B_2^\alpha  [u](x)
& = \frac{{r^\alpha  }}{{\Gamma (2 - \alpha )}}
  \int_0^r (r - \tau )^{1 - \alpha } \frac{{d^2 u}}{{d\tau ^2 }}(\tau \theta)d\tau \\
&=\frac{{r^\alpha  }}{{\Gamma (1 - \alpha )}}\frac{{d^2 }}{{dr^2 }}
 \int_0^r \frac{{(r - \tau )^{3 - \alpha } }}{{(2 - \alpha )(3 - \alpha )}}
 \frac{{d^2 u}}{{d\tau ^2 }}(\tau \theta)d\tau \\
&= - \frac{{(1 - \alpha )u(0)}}{{\Gamma (2 - \alpha )}}
 - \frac{r}{{\Gamma (2 - \alpha )}}\frac{{du(0)}}{{dr}}
  + (1 - \alpha )(2 - \alpha )u_2 (x) \\
&\quad + 2(2 - \alpha )r  \frac{{du_2 (x)}}{{dr}} + r^2 \frac{d^2 u_2 (x) }{dr^2}.
\end{align*}
Thus,
\begin{equation}\label{e2-7}
\begin{aligned}
B_2^\alpha  [u](x)
& =  - \frac{{(1 - \alpha )u(0)}}{{\Gamma (2 - \alpha )}}
 - \frac{r}{{\Gamma (2 - \alpha )}}\frac{{du(0)}}{{dr}}
 + (1 - \alpha )(2 - \alpha )u_2 (x)  \\
&\quad + 2(2 - \alpha )r\frac{{du_2 (x)}}{{dr}}
 + r\frac{d}{{dr}}\Big({r\frac{d}{{dr}} - 1} \Big)u_2 (x),\quad x \in \Omega.
\end{aligned}
\end{equation}
Equalities \eqref{e2-1} imply
$$
 {r\frac{{du_2 (x)}}{{dr}}} \Big|_{x = 0}  = 0,\quad
 {r\frac{d}{{dr}}\Big({r\frac{d}{{dr}} - 1}\Big)u_2 (x)} \Big|_{x = 0}  = 0.
$$
Then from  representation \eqref{e2-1} we obtain
$$
\lim_{x \to 0} B_2^\alpha  [u](x)
=  - \frac{{(1 - \alpha )u(0)}}{{\Gamma (2 - \alpha )}}
+ \frac{{(1 - \alpha )(2 - \alpha )u(0)}}{{\Gamma (2 - \alpha )}}
\frac{{\Gamma (2 - \alpha )\Gamma (1)}}{{\Gamma (3 - \alpha )}} = 0.
$$
Further, we denote $y_i  = \tau \theta _i$, $i = 1,2,\dots,n$.
 Then
$$
\frac{{du(\tau \theta )}}{{d\tau }}
= \sum_{i = 1}^n \frac{{\partial u(\tau \theta )}}{{\partial y_i }}
\frac{{dy_i }}{{d\tau }}
=  \sum_{i = 1}^n {\theta _i \frac{{\partial u(\tau \theta )}}{{\partial y_i }}} .
$$
Since $\theta  = x/r,\theta _i  = x_i /r$, it follows that
$$
\frac{r}{{\Gamma (2 - \alpha )}}\frac{{du(0)}}{{d\tau }}
= \frac{r}{{\Gamma (2 - \alpha )}}
 {\sum_{i = 1}^n {\frac{{x_i }}{r}\frac{{\partial u(\tau \theta )}}
{{\partial y_i }}} } \Big|_{\tau  = 0}
= \frac{1}{{\Gamma (2 - \alpha )}}
\sum_{i = 1}^n {x_i \frac{{\partial u(0)}}{{\partial y_i }}} .
$$
Thus, for any $k = 1,2,\dots,n$,
$$
\frac{\partial }{{\partial x_k }}
\Big[ { - \frac{r}{{\Gamma (2 - \alpha )}}
\frac{{du(0)}}{{d\tau }}} \Big]
=  - \frac{1}{{\Gamma (2 - \alpha )}}\frac{{\partial u(0)}}{{\partial y_k }}.
$$
It is obvious that
$$
\frac{\partial }{{\partial x_k }}
\big[ { - \frac{{1 - \alpha }}{{\Gamma (2 - \alpha )}}u(0)} \big] = 0.
$$
Further, for any $k = 1,2,\dots,n $,  the equality $
\frac{\partial }{{\partial x_k }}u(\xi x)
= \frac{{\partial u}}{{\partial y_k }}
\frac{{\partial y_k }}{{\partial x_k }}
= \xi \frac{{\partial u}}{{\partial y_k }}$ holds.
Hence,
$$
 {\frac{\partial }{{\partial x_k }}u(\xi x)} \big|_{x = 0}
 = \xi \frac{{\partial u(0)}}{{\partial y_k }}.
$$
Consequently,
$$
 {\frac{\partial }{{\partial x_k }}u_2 (x)} \big|_{x = 0}
 = \frac{1}{{(3 - \alpha )(2 - \alpha )\Gamma (2 - \alpha )}}
\frac{{\partial u(0)}}{{\partial y_k }}.
$$
Further, by the definition of  $r\frac{d}{{dr}}$  we have
$r\frac{{du_2 (x)}}{{dr}}
= \sum_{i = 1}^n {x_i \frac{{\partial u_2 (x)}}{{\partial x_i }}} $.
Thus,
$$
\frac{\partial }{{\partial x_k }}\big[ {r\frac{{du_2 (x)}}{{dr}}} \big]
= \sum_{i = 1}^n x_i \frac{{\partial ^2 u_2 (x)}}{{\partial x_k \partial x_i }}
 + \frac{{\partial u_2 (x)}}{{\partial x_k }}.
$$
Therefore,
\begin{align*}
\frac{\partial }{{\partial x_k }}\big[ {2(2 - \alpha )r\frac{{du_2 (x)}}{{dr}}}
\big] \big|_{x = 0}
&= 2(2 - \alpha )[ {\sum_{i = 1}^n
{x_i \frac{{\partial ^2 u_2 (x)}}{{\partial x_k \partial x_i }}
+ \frac{{\partial u_2 (x)}}{{\partial x_k }}} } \big] \big|_{x = 0}  \\
&= \frac{2}{{(3 - \alpha )\Gamma (2 - \alpha )}}
\frac{{\partial u(0)}}{{\partial y_k }}.
\end{align*}
Further, by \eqref{e2-2}, it follows that
$$
\frac{\partial }{{\partial x_i }}\big[ {r\frac{d}{{dr}}({r\frac{d}{{dr}} - 1}
)u_2 (x)} \big] \big|_{x = 0}  = 0.
$$
By using all these calculations, from the representation of the function
$B_2^\alpha  [u](x)$, we obtain
$$
\frac{\partial B_2^\alpha [u](0)}{\partial x_k}
= \frac{1}{\Gamma(2-\alpha)}
\big[ - \frac{\partial u(0)}{\partial y_k}
+\frac{1 - \alpha}{(3 - \alpha )}\frac{\partial u(0)}{\partial y_k}
+\frac{2}{(3 - \alpha )}\frac{\partial u(0)}{\partial y_k }  \big]= 0.
$$
If $\alpha  = 1$  or $\alpha  = 2$,
 then  $B_1^1 u(x) = r\frac{{du(x)}}{{dr}},B_1^2 u(x)
= r\frac{d}{{dr}}({r\frac{d}{{dr}} - 1} )u(x)$, and for these functions
the statement of the lemma follows from the lemma \ref{l2-1}.
\end{proof}

The following proposition was proved in \cite{torebek-bvp}.

\begin{lemma}\label{l2-3}
Let $ 0 < \alpha  \le 1$. Then for any $x \in \Omega $  the following
equalities hold:
\begin{equation}\label{e2-8}
B^{- \alpha } [ {B_1^\alpha  [ u ]} ](x ) = u(x ) - u(0),
\end{equation}
and if  $u(0) = 0$, then
\begin{equation}\label{e2-9}
B_1^\alpha  [ {B^{ - \alpha } [ u ]} ](x ) = u(x ).
\end{equation}
\end{lemma}

A similar statement is true in the case of $1 < \alpha  < 2$.

\begin{lemma}\label{l2-4}
 Let $ 1 < \alpha  < 2,j = 1$. Then equalities \eqref{e2-8}
and \eqref{e2-9} hold.
\end{lemma}
	
\begin{proof}
Let us prove  equality \eqref{e2-8}. Let $ x \in \Omega $
 and $t \in ({0,1} ]$.
Consider the function
$$
\Im _t [ u ](x ) = \frac{1}{{\Gamma (\alpha )}}\int_0^t {(t - \tau )^{\alpha  - 1}
 \tau ^{ - \alpha } B_1^\alpha  [ u ]({\tau x} )d\tau }.
$$
By using the definition of  $B_1^\alpha$, we have
$$
\Im _t [ u ](x ) = \frac{1}{{\Gamma (\alpha )}}
\int_0^t {(t - \tau )^{\alpha  - 1} \frac{d}{{d\tau }}J^{2 - \alpha }
 [ {\frac{d}{{d\tau }}u} ]({\tau x} )d\tau }.
$$
Integrating the above integral by parts, we obtain
\begin{align*}
\Im _t [ u ](x )
&= \frac{{\alpha  - 1}}{{\Gamma (\alpha )}}
 \int_0^t {(t - \tau )^{\alpha  - 2} J^{2 - \alpha }
 [ {\frac{d}{{d\tau }}u} ]({\tau x} )d\tau } \\
&= \frac{1}{{\Gamma (\alpha  - 1)}}\int_0^t {(t - \tau )^{\alpha  - 2}
J^{2 - \alpha } [ {\frac{d}{{d\tau }}u} ]({\tau x} )d\tau  = } u(tx) - u(0).
\end{align*}
If  we put $t = 1$, then
$$
u(x) = u(0) + \frac{1}{{\Gamma (\alpha )}}
 \int_0^1 {(1 - \tau )^{\alpha  - 1} \tau ^{ - \alpha }
  B_1^\alpha  [ u ]({\tau x} )d\tau }
= u(0) + B^{ - \alpha } [ {B_1^\alpha  [ u ]} ](x).
$$
Equality \eqref{e2-8} is proved.

	We turn to the proof of  \eqref{e2-9}. Since $u(0) = 0$, then the
operator $B^{ - \alpha } $ is determined for these functions, and, therefore,
applying the operator $B_1^\alpha  $
 to the function $B^{ - \alpha } [u](x)$, we have
\begin{align*}
B_1^\alpha  [ {B^{ - \alpha } [u]} ](x)
&= r^\alpha  \frac{d}{{dr}}J^{2 - \alpha } \frac{d}{{dr}}B^{ - \alpha } [u](x) \\
&= \frac{{r^\alpha  }}{{\Gamma (2 - \alpha )}}\frac{{d^2 }}{{dr^2 }}
 \int_0^r {\frac{{(r - \tau )^{2 - \alpha } }}{{2 - \alpha }}
 \frac{d}{{d\tau }}B^{ - \alpha } [u](\tau \theta )d\tau } .
\end{align*}
After the change of variables $\tau s = \xi $, the function
 $$
B^{ - \alpha } [u](\tau \theta )
= \frac{1}{{\Gamma (\alpha )}}\int_0^1 {(1 - s)^{\alpha  - 1}
s^{ - \alpha } u(\tau s\theta )ds}
$$
will be represented as
$$
B^{ - \alpha } [u](\tau \theta )
= \frac{1}{{\Gamma (\alpha )}}\int_0^\tau  (\tau  - \xi )^{\alpha  - 1}
\xi ^{ - \alpha } u(\xi \theta )d\xi
= J^\alpha  [ {\xi ^{ - \alpha } u} ].
$$
Then integrating by parts, we obtain 
$$
B_1^\alpha  [ {B^{ - \alpha } [u]} ](x)
= r^\alpha  \frac{{d^2 }}{{dr^2 }}[ {J^{2 - \alpha }
[ {J^\alpha  [\xi ^{ - \alpha } u]} ]} ](x)
= r^\alpha  \frac{{d^2 }}{{dr^2 }}[ {J^2 [\xi ^{ - \alpha } u]} ](x) = u(x).
$$
\end{proof}

\begin{lemma}\label{l2-5}
Let $1 < \alpha  \le 2$. Then for any $x \in \Omega$  the following equalities hold:
\begin{equation}\label{e2-10}
B^{ - \alpha } [ {B_2^\alpha  [ u ]} ](x )
= u(x ) - u(0) - \sum_{i = 1}^n {x_i \frac{{\partial u(0)}}{{\partial x_i }}},
\end{equation}
and if
$u(0) = 0$ and $\frac{{\partial u(0)}}{{\partial x_i }} = 0$ for $i = 1,2,\dots,n$,
then
\begin{equation}\label{e2-11}
B_2^\alpha  [ {B^{ - \alpha } [ u ]} ](x ) = u(x ).
\end{equation}
\end{lemma}

\begin{proof}
 Let us prove  equality \eqref{e2-10}. As in  the proof of
\eqref{e2-8} we consider the function
$$
\Im _t [ u ](x )
= \frac{1}{{\Gamma (\alpha )}}\int_0^t {(t - \tau )^{\alpha  - 1}
\tau ^{ - \alpha } B_2^\alpha  [ u ]({\tau x} )d\tau ,t \in (0,1].}
$$
By using the definition of $B_2^\alpha$, we have
$$
\Im _t [ u ](x ) = \frac{1}{{\Gamma (\alpha )}}
\int_0^t {(t - \tau )^{\alpha  - 1} J^{2 - \alpha }
[ {\frac{{d^2 }}{{d\tau ^2 }}u} ]({\tau x} )d\tau } .
$$
But this function by the definition of the fractional order integral
has the form
$$
\Im _t [ u ](x ) = \frac{{\alpha  - 1}}{{\Gamma (\alpha )}}
\int_0^t (t - \tau )^{\alpha  - 2} J^{2 - \alpha }
 [ {\frac{d}{{d\tau }}u} ]({\tau x} )d\tau
= J^\alpha  \big[ {J^{2 - \alpha } [ {\frac{{d^2 }}{{d\tau ^2 }}u} ]} \big](x).
$$
Since $J^\alpha  J^{2 - \alpha }  = J^{\alpha  + 2 - \alpha }  = J^2$,
$$
\Im _t [ u ](x )
= J^2 [ {\frac{{d^2 }}{{d\tau ^2 }}u} ]
= \int_0^t {(t - \tau )\frac{{d^2 }}{{d\tau ^2 }}u(\tau x)d\tau
=  - t\frac{d}{{d\tau }}u(0) + u(tx) - u(0)}.
$$
Further, since
$$
\frac{d}{{d\tau }}u(\tau x)
 = \sum_{i = 1}^n {\frac{{\partial u(\tau x)}}{{\partial y_i }}
\frac{{dy_i }}{{d\tau }} = }
\sum_{i = 1}^n {x_i \frac{{\partial u(\tau x)}}{{\partial y_i }}},
$$
it follows that
 $$
\frac{d}{{d\tau }}u(0)
 = \sum_{i = 1}^n {x_i \frac{{\partial u(0)}}{{\partial y_i }} \equiv }
\sum_{i = 1}^n {x_i \frac{{\partial u(0)}}{{\partial x_i }}}.
$$
If in the integral $\Im _t [ u ](x )$  we set $t = 1$, then
$$
u(x)-u(0)-\sum_{i = 1}^n {x_i \frac{\partial u(0)}{\partial x_i }}
 = \frac{1}{\Gamma (\alpha)}\int_0^1{(1-\tau)^{\alpha-1}
\tau ^{-\alpha} B_2^\alpha[u]({\tau x})d\tau }
\equiv B^{-\alpha}[{B_2^\alpha[u]}](x).
$$
Equality \eqref{e2-10} is proved.	

If $u(0) = 0$ and $\frac{{\partial u(0)}}{{\partial x_i }} = 0$, $i = 1,2,\dots,n$,
then  the operator $B^{ - \alpha }$ is defined on  these functions.
Applying  $B_2^\alpha$  we obtain
$$
B_2^\alpha  [ {B^{ - \alpha } [ u ]} ](x )
= \frac{{r^\alpha  }}{{\Gamma ({2 - \alpha } )}}
\int_0^r ({r - \tau } )^{1 - \alpha } \frac{{d^2 }}{{d\tau ^2 }}
B^{ - \alpha } [ u ]({\tau \theta } )d\tau .
$$
We represent the function $B^{ - \alpha } [ u ]({\tau \theta } )$ as
$$
B^{ - \alpha }[u]( {\tau \theta })
= \frac{1}{{\Gamma (\alpha )}}\int_0^1 {({1 - s} )^{\alpha  - 1}
 s^{ - \alpha } u} ({s\tau \theta })ds
= \frac{1}{{\Gamma (\alpha )}}\int_0^\tau  {({\tau  - \xi })^{\alpha  - 1}
\xi ^{ - \alpha } u} ({\xi \theta})d\xi\, .
$$
Since $\alpha  - 1 > 0$, the following equality holds
$$
\frac{d}{{d\tau }}B^{ - \alpha } [ u ]({\tau \theta } )
= \frac{{\alpha  - 1}}{{\Gamma (\alpha )}}
\int_0^\tau  {({\tau  - \xi } )^{\alpha  - 2} \xi ^{ - \alpha } u}
({\xi \theta } )d\xi  \equiv J^{\alpha  - 1}
 [ {\xi ^{ - \alpha } u} ](\tau \theta ).
$$
It is easy to check the following equalities:
\begin{align*}
B_2^\alpha  [ {B^{ - \alpha } [ u ]} ](x )
&= \frac{{r^\alpha  }}{{\Gamma ({2 - \alpha } )}}
 \frac{d}{{dr}}\int_0^r \frac{{({r - \tau } )^{2 - \alpha } }}{{2 - \alpha }}
 \frac{d}{{d\tau }}J^{\alpha  - 1} [ {\xi ^{ - \alpha } u} ](\tau \theta )d\tau \\
&=  r^\alpha  \frac{d}{{dr}}J[ {\xi ^{ - \alpha } u} ](x)
 = r^\alpha  \frac{d}{{dr}}\int_0^r \xi ^{ - \alpha } u(\xi \theta )d\xi \\
& =  r^\alpha  r^{ - \alpha } u(x) = u(x).
\end{align*}
\end{proof}
	
Let $0 < \alpha  \le 2$, and consider the functions:
\begin{equation} \label{e2-12}
\begin{aligned}
g_{1,\alpha } (x)
&= r^{\alpha  - 5} J^{1 - \alpha } [r^4 g](x)  \\
&\equiv \frac{{r^{\alpha  - 5} }}{{\Gamma (1 - \alpha )}}
 \int_0^r {({r - \tau } )^{ - \alpha } \tau ^4 g({\tau \theta } )d\tau },\quad
0<\alpha\leq1.
\end{aligned}
\end{equation}
\begin{equation} \label{e2-13}
\begin{aligned}
g_{2,\alpha } (x)
&= r^{\alpha  - 6} J^{2 - \alpha } [r^4 g](x) \\
&\equiv \frac{{r^{\alpha  - 6} }}{{\Gamma (2 - \alpha )}}
 \int_0^r {({r - \tau } )^{1 - \alpha } \tau ^4 g({\tau \theta } )d\tau },\quad
1<\alpha\leq2.
\end{aligned}
\end{equation}
Since $J^0 [r^4 g](x) = r^4 g(x)$, it follows that $g_{1,1} (x) = g_{2,2} (x) = g(x)$.

\begin{lemma}\label{l2-6}
Let $0 < \alpha  \le 2$,  and $ \Delta ^2 u(x ) = g(x )$ for $x \in \Omega$.
Then for any $ x \in \Omega $ and $j = 1,2$ the following statements hold:
\begin{itemize}
\item[(1)] if $0 < \alpha  \le 1$, then
\begin{equation}\label{e2-14}
\Delta ^2 B_1^\alpha  [u](x) = (1 - \alpha )g_{1,\alpha } (x)
+ \Gamma _4 [g_{1,\alpha } ](x);
\end{equation}

\item[(2)] if $ 1 < \alpha  \le 2$, $j = 1,2$, then
\begin{equation}\label{e2-15}
\Delta ^2 B_j^\alpha  [u](x)
= (1 - \alpha )(2 - \alpha )g_{2,\alpha } (x) + 2(2 - \alpha )
\Gamma _4 [g_{2,\alpha } ](x) + \Gamma _4 [ {\Gamma _3 [g_{2,\alpha } ]} ](x).
\end{equation}
\end{itemize}
\end{lemma}

\begin{proof}
Note that for $u(x)$ the following equality holds:
$$
\Delta ^2 [ {r\frac{d}{{dr}}u(x )} ]
= r\frac{d}{{dr}}\Delta ^2 u(x ) + 4\Delta ^2 u(x )
= ({r\frac{d}{{dr}} + 4} )\Delta ^2 u(x ) \equiv \Gamma _4 [\Delta ^2 u](x).
$$
Then when $\alpha  = 1$ we obtain
$$
\Delta ^2 B_1^1 [u](x) = \Gamma _4 [g_{1,1} ](x) = \Gamma _4 [g](x);
$$
and when $\alpha  = 2$ we have
 $$
\Delta ^2 B_2^2 [u](x) = \Gamma _3 [ {\Gamma _4 [g_{1,2} ]} ](x)
= \Gamma _4 [ {\Gamma _3 [g]} ](x).
$$
Consequently, in  these two values of $\alpha $
 the equalities \eqref{e2-14} and \eqref{e2-15} are proved.

	Let $0 < \alpha  < 1$. Using the representation of the function
$B_1^\alpha  [ u ](x )$  in \eqref{e2-5}, we obtain
\[
\Delta ^2 B_1^\alpha  [u](x)
= (1 - \alpha )\Delta ^2 u_1 (x) + \Gamma _4 [\Delta ^2 u_1 ](x).
\]
Since $ \Delta ^2 u(x ) = g(x )$,
$$
\Delta ^2 u_1 (x )
= \frac{1}{{\Gamma (1 - \alpha )}}\int_0^1 {(1 - \xi )^{ - \alpha }
\xi ^4 g(\xi x)d\xi }
= \frac{{r^{\alpha  - 5} }}{{\Gamma (1 - \alpha )}}
\int_0^r {(r - \tau )^{ - \alpha } \tau ^4 g(\tau \theta )d\tau } ;
$$
i.e. $ \Delta ^2 u_1 (x ) = g_{1,\alpha } (x)$. Thus, for the functions
$B_1^\alpha  [ u ](x )$ we obtain the equality \eqref{e2-14}.

 Let $1 < \alpha  < 2,j = 1$. Then the representation \eqref{e2-6} implies:
$$
\Delta ^2 B_1^\alpha  [u](x)
= (1 - \alpha )(2 - \alpha )\Delta ^2 u_2 (x) + 2(2 - \alpha )
\Gamma _4 [\Delta ^2 u_2 ](x) + \Gamma _4
[ {\Gamma _3 [\Delta ^2 u_2 ]} ](x)(x),
$$
Further, taking into account $\Delta ^2 u(x ) = g(x )$ for $\Delta ^2 u_2 (x)$,
 we obtain
\begin{align*}
\Delta ^2 u_2 (x)
&= \frac{1}{{\Gamma (2 - \alpha )}}\int_0^1 {(1 - \xi )^{1 - \alpha }
\xi ^4 g(\xi x)d\xi } \\
&=\frac{{r^{\alpha  - 6} }}{{\Gamma (2 - \alpha )}}
\int_0^r {(r - \tau )^{1 - \alpha } \tau ^4 g(\tau \theta )d\tau }
 = g_{2,\alpha } (x),
\end{align*}
i.e. for the functions $\Delta ^2 B_1^\alpha  [u](x)$
the representation \eqref{e2-15} holds.

Analogously, to the case  $1 < \alpha  < 2$ for $j = 2$,
the representation \eqref{e2-7}, by the equality
 $$
\frac{r}{{\Gamma (2 - \alpha )}}\frac{{du(0)}}{{d\tau }}
= \frac{r}{{\Gamma (2 - \alpha )}}
 {\sum_{i = 1}^n {\frac{{x_i }}{r}\frac{{\partial u(\tau \theta )}}
{{\partial y_i }}} } \big|_{\tau  = 0}
 = \frac{1}{{\Gamma (2 - \alpha )}}\sum_{i = 1}^n
{x_i \frac{{\partial u(0)}}{{\partial y_i }}} ,
$$
 yields the equality \eqref{e2-15}.
 \end{proof}

\begin{lemma}\label{l2-7}
Let $0 < \alpha  \le 2 $ and the functions $g_{1,\alpha } (y)$,
$g_{2,\alpha } (y)$  be defined by the equalities \eqref{e2-12} and
\eqref{e2-13}, respectively. Then for any $ x \in \Omega $  and $j = 1,2$
the following equalities hold:
\begin{itemize}
\item[(1)] if $ 0 < \alpha  \le 1 $, then
\begin{equation}\label{e2-16}
\Delta ^2 B_1^\alpha  [u](x) = |x|^{ - 4} B_1^\alpha  [|x|^4 g](x);
\end{equation}

\item[(2)] if $ 1 < \alpha  \le 2$, then, for $j = 1,2$,
\begin{equation}\label{e2-17}
\Delta ^2 B_j^\alpha  [u](x) = |x|^{ - 4} B_j^\alpha  [|x|^4 g](x).
\end{equation}
\end{itemize}
\end{lemma}

\begin{proof}
Since $ r\frac{d}{{dr}}[ {|x|^4 g} ] = |x|^4 \Gamma _4 [g](x)$, then we have
the equality
$$
|x|^{ - 4} r\frac{d}{{dr}}[ {|x|^4 g} ]
= \Gamma _4 [g](x) = \Gamma _4 [\Delta ^2 u](x)
= \Delta ^2 \Gamma _0 [u](x) \equiv \Delta ^2 B_1^1 [u](x).
$$
Further, if we denote $ r\frac{d}{{dr}}[ {|x|^4 g} ](x) = f(x)$,
then
\begin{align*}
\Delta ^2 B_2^2 [u]
&= \Delta ^2( {r^2 \frac{{d^2 }}{{dr^2 }}[ u ](x)})
= \Delta ^2 ( {r\frac{d}{{dr}}( {r\frac{d}{{dr}} - 1})[u](x)}) \\
&= \Gamma _4 [ {\Gamma _3 [\Delta ^2 u]} ](x)
 = \Gamma _3 [ {\Gamma _4 [g]}](x) = ({r\frac{d}{{dr}} + 3})({|x|^{ - 4} f}) \\
&= r\frac{d}{{dr}}({|x|^{ - 4} f}) + 3(|x|^{-4} f)
= |x|^{ - 4}( {r\frac{d}{{dr}} - 1})f.
\end{align*}
Thus
$$
\Delta ^2 B_2^2 [u]
=|x|^{-4} ({r\frac{d}{{dr}} - 1} )( {r\frac{d}{{dr}}[ {|x|^4 g} ]} )(x)
= |x|^{ - 4} r^2 \frac{{d^2 }}{{dr^2 }}[{|x|^4 g}]
= |x|^{ - 4} B_2^2 [ {|x|^4 g} ].
$$
Therefore,  equalities \eqref{e2-16} and \eqref{e2-17} in the case of integer
values of $\alpha $ is proved.
In the case of fractional values of $\alpha $
we use the equalities \eqref{e2-14} and \eqref{e2-15}.
To do it we transform the functions $g_{j,\alpha } (x),j = 1,2$.
After changing the variable $ \xi  = r^{ - 1} \tau
$ the integral, representing the function $g_{1,\alpha } (x)$,
can be rewritten in the following form
$$
g_{1,\alpha } (x) = \frac{{r^{\alpha  - 5} }}{{\Gamma (1 - \alpha )}}
\int_0^r {({r - \tau } )^{ - \alpha } \tau ^4 g({\tau \theta } )d\tau } .
$$
Then
\begin{align*}
(1 - \alpha )g_{1,\alpha } (x) + \Gamma _4 [g_{1,\alpha } ](x)
&= \frac{{1 - \alpha }}{{\Gamma ({1 - \alpha } )}}r^{\alpha  - 5}
\int_0^r {({r - \tau } )^{ - \alpha } \tau ^4 g({\tau \theta } )d\tau } \\
& = r^{ - 4} \frac{{r^\alpha  }}{{\Gamma ({1 - \alpha } )}}
 \frac{d}{{dr}}\int_0^r {({r - \tau } )^{ - \alpha }
\tau ^4 g({\tau \theta } )d\tau }.
\end{align*}
We transform the above integral as follows:
\begin{align*}
&\frac{{r^\alpha  }}{{\Gamma ({1 - \alpha } )}}\frac{d}{{dr}}
\int_0^r {({r - \tau } )^{ - \alpha } \tau ^4 g({\tau \theta } )d\tau }\\
&= \frac{{r^\alpha  }}{{\Gamma ({1 - \alpha } )}}\frac{d}{{dr}}
\int_0^r \tau ^4 g({\tau \theta } )\frac{{d({r - \tau } )^{1 - \alpha } }}
{{ - (1 - \alpha )}} \\
& = \frac{{r^\alpha  }}{{\Gamma ({1 - \alpha } )}}\frac{d}{{dr}}
\Big\{ {\int_0^r {\frac{{({r - \tau } )^{1 - \alpha } }}{{(1 - \alpha )}}
\frac{d}{{d\tau }}[ {\tau ^4 g({\tau \theta } )} ]d\tau } } \Big\} \\
&\quad +\frac{{r^\alpha  }}{{\Gamma ({1 - \alpha } )}}\int_0^r {({r - \tau } )
^{ - \alpha } \frac{d}{{d\tau }}[ {\tau ^4 g({\tau \theta } )} ]d\tau }
 \equiv B_1^\alpha  [ {|x|^4 g} ](x).
\end{align*}
Thus,
$$
\Delta ^2 B_1^\alpha  [u](x) = |x|^{ - 4} B_1^\alpha  [ {|x|^4 g} ](x),x \in \Omega .
$$
Let  $1 < \alpha  < 2$ and $j = 1$. Then after changing  variables $
\xi  = r^{ - 1} \tau$, for the function $g_{2,\alpha } (x)$
 we obtain
$$
g_{2,\alpha } (x)
= \frac{{r^{\alpha  - 6} }}{{\Gamma (2 - \alpha )}}
\int_0^r {({r - \tau } )^{1 - \alpha } \tau ^4 g({\tau \theta } )d\tau } .
$$
Further, if $f(x)$  is a smooth function then
$$
r\frac{d}{{dr}}[ {r^{\alpha  - 6} f} ]
= r^{\alpha  - 6} \big({r\frac{d}{{dr}} + \alpha  - 6} \big)f(x).
$$
Thus,
\begin{align*}
&(1 - \alpha )(2 - \alpha )g_{2,\alpha } (x) + 2(2 - \alpha )
\Gamma _4 [g_{2,\alpha } ](x) + \Gamma _4 [ {\Gamma _3 [g_{2,\alpha } ]} ](x) \\
& = ({r\frac{d}{{dr}} + 4} )\big[ {r^{\alpha  - 6} ({r\frac{d}{{dr}}
+ 3 + 2(2 - \alpha ) + \alpha  - 6} )J^{2 - \alpha } [\tau ^4 g]} \big](x) \\
&\quad + r^{\alpha  - 4} \frac{{d^2 }}{{dr^2 }}
\Big[ {\frac{1}{{\Gamma (2 - \alpha )}}\int_0^r {(r - \tau )^{1 - \alpha } }
\tau ^4 g(\tau \theta )d\tau } \Big].
\end{align*}
We transform the above integral as follows:
\begin{align*}
&\frac{1}{{\Gamma (2 - \alpha )}}\int_0^r {(r - \tau )^{1 - \alpha } }
\tau ^4 g(\tau \theta )d\tau  \\
&= \frac{1}{{\Gamma (2 - \alpha )}}\int_0^r \tau ^4 g(\tau \theta )
 \frac{{d(r - \tau )^{2 - \alpha } }}{{ - (2 - \alpha )}} \\
& =  { - \tau ^4 g(\tau \theta )\frac{{(r - \tau )^{2 - \alpha } }}{{(2 - \alpha )
\Gamma (2 - \alpha )}}} \Big|_{\tau  = 0}^{\tau  = r}
 + \frac{1}{{\Gamma (2 - \alpha )}}\int_0^r {(r - \tau )^{1 - \alpha }
\frac{d}{{d\tau }}[ {\tau ^4 g(\tau \theta )} ]d\tau }\\
&\equiv D_1^\alpha  [ {\tau ^4 g(\tau \theta )} ].
\end{align*}
Hence,
$$
\Delta ^2 B_1^\alpha  [u](x) = |x|^{ - 4} B_1^\alpha  [ {|x|^4 g} ].
$$
Similarly, we consider the case $1 < \alpha  < 2,j = 2$.
\end{proof}

\section{Some properties of the solution of the Dirichlet problem}\label{PDP}

Consider the Dirichlet problem
\begin{equation}\label{e3-1}
 \begin{gathered}
 \Delta ^2 v(x) = g_1 (x),\quad x \in \Omega  \\
 v(x) = \varphi _1 (x),\frac{{dv(x)}}{{d\nu }} = \varphi _2 (x),\quad
x \in \partial \Omega .
 \end{gathered}
\end{equation}
It is known that (see e.g. \cite{Agm}), if $g_1 (x),\varphi _1 (x)$ and
$\varphi _2 (x)$  are smooth functions, then the solution of  \eqref{e3-1}
 exists and is unique.
	The solution of  \eqref{e3-1} is represented as:
\begin{equation}\label{e3-2}
v(x ) = \int_\Omega  {G_{2,n} ({x,y} )g_1 (y )dy}  + w(x),
\end{equation}
where $G_{2,n} ({x,y} )$ is the Green function of \eqref{e3-1}, and $w(x) $
is a solution of  \eqref{e3-1} when $g_1 (x) = 0$; i.e.,
\begin{gather*}
\Delta ^2 w(x) = 0,\quad x \in \Omega ,\\
w(x) = \varphi _1 (x),\quad \frac{{dw(x)}}{{d\nu }} = \varphi _2 (x),\quad
x \in \partial \Omega .
\end{gather*}
Denote
$$
v_1 (x ) = \int_\Omega  {G_{2,n} ({x,y} )g_1 (y )dy} .
$$

The  explicit  form  of  the  Green's  function  for  the  Dirichlet  problem
is obtained for the cases $n \ge 2$ in \cite{Kal}.
For example, in the case when $n$ is odd or $n$ is even and $n >4$, the
Green's function of the problem \eqref{e3-1} follows from the expression
\begin{align*}
G_{2,n} ({x,y} )
&= d_{2,n} \Big[ {|{x - y} |^{4 - n}  -
\big|{x|y | - \frac{y}{{|y |}}} \big|^{4 - n} }  \\
 ({2 - \frac{n}{2}} )\big|{x|y | - \frac{y}{{|y |}}} \big|^{2 - n}
({1 - |x |^2 } )({1 - |y |^2 } ) \Big],
\end{align*}
where $d_{2,n}  = \frac{1}{{\omega _n }}\frac{1}{{2(n - 4)(n - 2)}}$ and
$\omega _n  = \frac{{2\pi ^{n/2} }}{{\Gamma (n/2)}}-$ is area of the unit sphere.
	
Furthermore, for convenience, we  consider only the case when $n$-odd or $n$-even
and $n > 4$. The following proposition was proved in [25].

\begin{lemma}\label{l3-1}
 Let $\varphi _1 (x),\varphi _2 (x)$
 be smooth functions. Then the following equalities hold:
\begin{gather}\label{e3-3}
w(0) = \frac{1}{{2\omega _n }}\int_{\partial \Omega }
{[ {2\varphi _1 (y) - \varphi _2 (y)} ]dS_y } ,\\
\label{e3-4}
\frac{{\partial w(0)}}{{\partial x_k }}
= \frac{n}{{2\omega _n }}\int_{\partial \Omega } {y_k [ {3\varphi _1 (y)
- \varphi _2 (y)} ]dS_y } ,\quad k = 1,2,\dots,n.
\end{gather}
\end{lemma}

\begin{lemma}\label{l3-2} Let $g_2 (x)$  be a smooth function. Then
	\begin{itemize}
\item[(1)] if  $g_1 (x) = \Gamma _4 [g_2 ](x)$, then
\begin{equation}\label{e3-5}
v_1 (0) = \frac{1}{{4\omega _n }}\int_\Omega  {({1 - |y|^2 } )g_2 (y)dy};
 \end{equation}

\item[(2)] if $g_1 (x ) = \Gamma _3 [ {\Gamma _4 [ {g_2 } ]} ](x )$  then
\begin{equation}\label{e3-6}
\frac{{\partial v_1 (0)}}{{\partial x_k }}
= \frac{n}{{4\omega _n }}\int_\Omega  {y_k (1 - |y|^2 )\Gamma _4 [ g ](y)dy} ,\quad
k = 1,2,\dots,n.
\end{equation}
\end{itemize}
\end{lemma}

Now we study the values of $v_1 (0)$  and
$\frac{{\partial v_1 (0)}}{{\partial x_k }}$, $k = 1,2,\dots,n$, when
\begin{gather}\label{e3-7}
g_1 (x) = (1 - \alpha )g_{1,\alpha } (x) + \Gamma _4 [g_{1,\alpha } ](x), \text{and }\\
\label{e3-8}
g_1 (x) = (1 - \alpha )(2 - \alpha )g_{2,\alpha } (x) + 2(2 - \alpha )
\Gamma _4 [g_{2,\alpha } ](x)
+ \Gamma _4 [ {\Gamma _3 [g_{2,\alpha } ]} ](x).
\end{gather}

\begin{lemma}\label{l3-3}
Let $0 < \alpha  \le 2$, $j = 1,2, g(x) $  be a smooth function, and
$g_{j,\alpha } (x ) $  be defined by \eqref{e2-12} or \eqref{e2-13}. Then

	(1) if $0 < \alpha  \le 1$  and  $g_1 (x)$ is defined by \eqref{e3-7}, then
\begin{equation} \label{e3-9}
\begin{aligned}
v_1 (0) &= \frac{1}{{2\omega _n }}\int_\Omega
{\frac{{1 - |y|^2 }}{2}g_{1,\alpha } (y)dy}
 + \frac{{1 - \alpha }}{{\omega _n 2(n - 2)(n - 4)}}
\int_\Omega  \Big[ |y|^{4 - n}  - 1 \\
&\quad + ({2 - \frac{n}{2}} )({1 - |y|^2 } ) \Big]g_{1,\alpha } (y)dy;
\end{aligned}
\end{equation}

 (2) if $1 < \alpha  < 2$, $j = 1$  and the function $g_1 (x)$ is defined by
\eqref{e3-8}, then
\begin{equation} \label{e3-10}
\begin{aligned}
v_1 (0) &= \frac{1}{{2\omega _n }}\int_\Omega  {\frac{{1 - |y|^2 }}{2}\Gamma _3
 [g_{2,\alpha } ](y)dy}
+ \frac{{2(2 - \alpha )}}{{2\omega _n }}\int_\Omega
{\frac{{1 - |y|^2 }}{2}g_{2,\alpha } (y)dy}  \\
&\quad + \frac{{(1 - \alpha )(2 - \alpha )}}{{\omega _n 2(n - 2)(n - 4)}}
\int_\Omega  [ {|y|^{4 - n}  - 1 + ({2 - \frac{n}{2}} )({1 - |y|^2 } )}
]g_{2,\alpha } (y)dy;
\end{aligned}
\end{equation}

	(3) if $1 < \alpha  \le 2$, $j = 2$  and  $g_1 (x)$ is defined by
\eqref{e3-8}, then for $v_1 (0)$  we have the equality \eqref{e3-10}, moreover
\begin{equation} \label{e3-11}
\begin{aligned}
\frac{{\partial v_1 (0)}}{{\partial x_k }}
&= \frac{n}{{4\omega _n }}\int_\Omega  {y_k ({1 - |y|^2 } )
\Gamma _4 [g_{2,\alpha } ](y)dy}  \\
&\quad + \frac{1}{{2\omega _n }} \frac{{2(2 - \alpha )}}{{(n - 2)}}
 \int_\Omega  y_k \big[ {|y|^{2 - n}  - 1 + \frac{{2 - n}}{2}(1 - |y|^2 )}
 \big]\Gamma _4 [ {g_{2,\alpha } } ](y)dy \\
&\quad  +\frac{(1 - \alpha )(2 - \alpha )}{2\omega _n(n - 2)}
 \int_\Omega {y_k [ {|y|^{2 - n}  - 1
 + \frac{2 - n}{2}(1 - |y|^2 )}]g_{2,\alpha }(y)dy} ,
\end{aligned}
\end{equation}
$k = 1,\dots,n$.
\end{lemma}

\begin{proof}
When $\alpha $ is an integer,   equalities \eqref{e3-9} and \eqref{e3-11}
follows from  Lemma \ref{l3-3}. Let $0 < \alpha  < 1$, $j = 1$ and
$g_1 (x)$ be represented in the form \eqref{e3-7}. Then
\begin{align*}
v_1 (x ) &= \int_\Omega  {G_{2,n} ({x,y} )g_1 (y )dy} \\
& = (1 - \alpha )\int_\Omega  {G_{2,n} ({x,y} )g_{1,\alpha } (y )dy}
&\quad +\int_\Omega  {G_{2,n} ({x,y} )\Gamma _4 [g_{1,\alpha } ](y)dy} .
\end{align*}
From the first statement  of Lemma \ref{l3-2}, for the second integral
of the above equality we obtain
$$
\int_\Omega  G_{2,n} ({0,y} )\Gamma _4 [g_{1,\alpha } ](y)dy
 = \frac{1}{{2\omega _n }}\int_\Omega  {\frac{{1 - |y|^2 }}{2}g_{1,\alpha } (y)dy} .
$$
For the first integral, using the representation of the functions $G_{2,n} ({x,y} )$,
 we have
$$
\int_\Omega  {G_{2,n} ({0,y} )g_{1,\alpha } (y )dy}
= d_{2,n} \int_\Omega  {[ {|y|^{4 - n}  - 1
+ ({2 - \frac{n}{2}} )({1 - |y|^2 } )} ]g_{1,\alpha } (y )dy}.
$$
Thus, for $v_1 (0)$  we obtain the equality \eqref{e3-9}.
Let $1 < \alpha  < 2,j = 1$. Then, using the equality \eqref{e3-8}, we obtain
\begin{align*}
v_1 (x ) &= \int_\Omega  {G_{2,n} ({x,y} )g_1 (y )dy} \\
& = (1 - \alpha )(2 - \alpha )\int_\Omega  {G_{2,n} ({x,y} )g_{2,\alpha } (y )dy} \\
&\quad  + 2(2 - \alpha )\int_\Omega  {G_{2,n} ({x,y} )\Gamma _4 [g_{2,\alpha } ](y)
 + \int_\Omega  {G_{2,n} ({x,y} )\Gamma _4 [ {\Gamma _3 [g_{2,\alpha } ]} ](y)dy} }
\end{align*}
By \eqref{e3-5}, for the second and third integrals of the last
equality we obtain
\begin{gather*}
\int_\Omega  G_{2,n} ({0,y} )\Gamma _4 [g_{2,\alpha } ](y)dy
=  \frac{1}{{2\omega _n }}\int_\Omega  {\frac{{1 - |y|^2 }}{2}g_{2,\alpha } (y)dy} ,\\
\int_\Omega  G_{2,n} ({0,y} )\Gamma _4 [ {\Gamma _3 [g_{2,\alpha } ]} ](y)dy
=  \frac{1}{{2\omega _n }}\int_\Omega  {\frac{{1 - |y|^2 }}{2}\Gamma _3
[g_{2,\alpha } ](y)dy} .
\end{gather*}
For the first integral we have
$$
\int_\Omega  {G_{2,n} ({0,y} )g_{1,\alpha } (y )dy}
= d_{2,n} \int_\Omega  [ {|y|^{4 - n}  - 1
 + ({2 - \frac{n}{2}} )({1 - |y|^2 } )} ]g_{2,\alpha } (y )dy.
$$
Therefore, for $v_1 (0)$  we obtain the equality \eqref{e3-10}.

No let $1 < \alpha  < 2$, $j = 2$. Since in this case $g_1 (x)$ has the form
\eqref{e3-8}, for $v_1 (0)$  again we obtain \eqref{e3-10}.
Further, we obtain
\begin{gather*}
v_{1,1} (x) = (1 - \alpha )(2 - \alpha )
\int_\Omega  {G_{2,n} ({x,y} )g_{2,\alpha } (y)dy,} \\
v_{1,2} (x) = 2(2 - \alpha )\int_\Omega  {G_{2,n} ({x,y} )
 \Gamma _4 [g_{2,\alpha } ](y)dy,} \\
v_{1,3} (x) = \int_\Omega  {G_{2,n} ({x,y} )
\Gamma _4 [ {\Gamma _3 [g_{2,\alpha } ]} ](y)dy}.
\end{gather*}
Using \eqref{e3-6}, for the function $v_{1,3} (x) $  we obtain
$$
\frac{{\partial v_{1,3} (0)}}{{\partial x_k }}
= \frac{n}{{4\omega _n }}\int_\Omega  {y_k (1 - |y|^2 )
\Gamma _4 [ {g_{2,\alpha } } ](y)dy} ,\quad k = 1,2,\dots,n.
$$
Since
\begin{gather*}
\frac{\partial }{{\partial x_k }}|{x - y} |^{4 - n}
=  {\frac{{4 - n}}{2}|{x - y} |^{2 - n} 2(x_k  - y_k )} \big|_{x = 0}
 =  - (4 - n)|y |^{2 - n} y_k , \\
\begin{aligned}
\frac{\partial }{{\partial x_k }}\big|{x|y| - \frac{y}{{|y|}}} \big|^{4 - n}
&= {\frac{{4 - n}}{2}|{x|y| - \frac{y}{{|y|}}} |^{2 - n} 2({x_k |y|
- \frac{{y_k }}{{|y|}}} )|y|} \Big|_{x = 0}  \\
&=  - (4 - n)y_k ,
\end{aligned}\\
\begin{aligned}
&\frac{\partial }{{\partial x_k }}\big[ {|{x|y| - \frac{y}{{|y|}}} |^{2 - n}
({1 - |x|^2 } )} \big] \\
&= \frac{{2 - n}}{2}\big|{x|y| - \frac{y}{{|y|}}} \big|^{ - n} 2({x_k |y|
- \frac{{y_k }}{{|y|}}} )|y| ( {1 - |x|^2 })
  + \big|{x|y| - \frac{y}{{|y|}}} \big|^{2 - n} ( - 2x_k ) \big|_{x = 0}\\
&=  - (2 - n)y_k ,
\end{aligned}
\end{gather*}
it follows that for
$$
\frac{{\partial G_{2,n} ({x,y} )}}{{\partial x_k }} \Big|_{x = 0},
$$
we obtain
$$
 \frac{{\partial G_{2,n} ({x,y} )}}{{\partial x_k }} \big|_{x = 0}
 = \frac{1}{{2\omega _n }}\frac{1}{{(n - 2)}}
\big[ {y_k |y|^{2 - n}  - y_k  + \frac{{2 - n}}{2}y_k (1 - |y|^2 )} \big]
$$
Then
\begin{gather*}
\frac{{\partial v_{1,1} (0)}}{{\partial x_k }}
 = \frac{1}{{2\omega _n }}\frac{{(1 - \alpha )(2 - \alpha )}}{{(n - 2)}}
\int_\Omega  {y_k [ {|y|^{2 - n}  - 1 + \frac{{2 - n}}{2}(1
- |y|^2 )} ]g_{2,\alpha } (y )dy} ,
\\
\frac{{\partial v_{1,2} (0)}}{{\partial x_k }}
= \frac{1}{{2\omega _n }}\frac{{2(2 - \alpha )}}{{(n - 2)}}
\int_\Omega  {y_k [ {|y|^{2 - n}  - 1 + \frac{{2 - n}}{2}(1
- |y|^2 )} ]\Gamma _4 [g_{2,\alpha } ](y)dy} .
\end{gather*}
Hence, for $\frac{{\partial v_1 (0)}}{{\partial x_k }}$
 we obtain \eqref{e3-11}.
\end{proof}

\section{Main results}\label{MR}

Let  $g_{1,\alpha } (x)$ and $g_{2,\alpha } (x), x\in R^{n}$
 be defined by  \eqref{e2-12} and  \eqref{e2-13},  and let $n$ be odd, or $n$ 
be even with $ n > 4$. 

\begin{theorem} \label{thm4.1}
Let $ 0 < \alpha  < 2$, $g(x)$, $f_1 (x) $ and $ f_2 (x)$
be smooth functions. 

	(1) If  $0 < \alpha  \le 1$ and $j = 1$, then  problem \ref{probl1} 
is solvable if and only if 
\begin{equation} \label{e4-1}
\begin{aligned}
&\int_{\partial \Omega } {[ {f_2 (y) + (\alpha  - 2)f_1 (y)} ]dS_y } \\
& = \int_\Omega  {\frac{{1 - |y|^2 }}{2}g_{1,\alpha } (y)dy}\\
&\quad + \frac{{1 - \alpha }}{{(n - 2)(n - 4)}}\int_\Omega  {[ {|y|^{4 - n}  - 1
+ ({2 - \frac{n}{2}} )({1 - |y|^2 } )} ]g_{1,\alpha } (y)dy}
\end{aligned}
\end{equation}

(2) If $1 < \alpha  < 2$ and $j = 1$, then  problem \ref{probl1} is solvable 
if and only if 
\begin{equation} \label{e4-2}
\begin{aligned}
&\int_{\partial \Omega } {[{f_2(y)+(\alpha-2)f_1 (y)}]dS_y}\\
&=\int_\Omega  {\frac{1 - |y|^2}{2}\Gamma _3[g_{2,\alpha }](y)dy}
+ 2(2-\alpha )\int_\Omega {\frac{1 - |y|^2 }{2}g_{2,\alpha } (y)dy}   \\
&\quad  + \frac{{(1 - \alpha )(2 - \alpha )}}{{(n - 2)(n - 4)}}
\int_\Omega  [ {|y|^{4 - n}  - 1 + ({2 - \frac{n}{2}} )
 ({1 - |y|^2 } )} ]g_{2,\alpha } (y)dy\,.
\end{aligned}
\end{equation}

(3) If the solution of the problem \ref{probl1} exists then it is unique 
up to a constant term and can be represented as 
\begin{equation}\label{e4-3}
u(x ) = C + B^{ - \alpha } [v](x),
\end{equation}
where $v(x)$  is a solution of  \eqref{e3-1}, satisfying the condition $v(0) = 0$,
 with the functions
\begin{gather}\label{e4-4}
  \varphi _1 (x) = f_1 (x ), \quad \varphi _2 (x) = f_2 (x ) + \alpha f_1 (x ) \\
\label{e4-5}
  g_1 (x) = |x|^{ - 4} B_j^\alpha  [ {|x|^4 g} ](x).
\end{gather}
\end{theorem}

\begin{proof}
Let $u(x )$  be a solution of  problem \ref{probl1}.
 Apply the operator  $B_1^\alpha  $  to the function $u(x )$, and denote 
$v(x ) = B_1^\alpha  [ u ](x )$. Then in the case $0 < \alpha  \le 1$,
 using \eqref{e2-16} from lemma \ref{l2-7}, we obtain
$$
\Delta ^2 v(x ) = \Delta ^2 B_1^\alpha  [ u ](x ) 
= |x|^{ - 4} B_1^\alpha  [ {|x|^4 g} ](x) \equiv g_1 (x ),\quad 0 < \alpha  \le 1.
$$
and if $1 < \alpha  < 2$, then by \eqref{e2-17}, we have
$$
\Delta ^2 v(x ) = \Delta ^2 B_1^\alpha  [ u ](x ) 
= |x|^{ - 4} B_1^\alpha  [ {|x|^4 g} ](x) \equiv g_1 (x ),\quad 1 < \alpha  < 2.
$$
Then by assumption,  $B_1^\alpha  [ u ](x ) \in C(\bar \Omega )$.
Therefore,  $v(x ) \in C(\bar \Omega )$  and 
$$
v(x ) \big|_{\partial \Omega }  = f_1 (x ) \equiv \varphi _1 (x).
$$
Further, if $0 < \alpha  \le 1$, then by the definition of  $
B_1^{\alpha  + 1}$,
\begin{align*}
B_1^{\alpha  + 1} [u](x) 
&= r^{\alpha  + 1} \frac{d}{{dr}}J^{2 - (\alpha  + 1)} [ {\frac{d}{{dr}}u} ](x)
  = r^{\alpha  + 1} \frac{d}{{dr}}J^{1 - \alpha } [ {\frac{d}{{dr}}u} ](x)\\
& = r^{\alpha  + 1} \frac{d}{{dr}}[ {r^{ - \alpha }  \cdot B_1^\alpha  [u]} ](x) 
= r\frac{d}{{dr}}B_1^\alpha  [u](x) - \alpha B_1^\alpha  [u](x).
\end{align*}
Therefore, the boundary condition \eqref{e2-3} of the problem \ref{probl1}
implies the condition
$$
\frac{{\partial v(x )}}{{\partial \nu }} \big|_{\partial \Omega }  
= f_2 (x ) + \alpha f_1 (x ) \equiv \varphi _2 (x).
$$
Similarly, in the case $1 < \alpha  < 2$, $j = 1$ from definition of  
$B_1^{\alpha  + 1}$, we have
\begin{align*}
B_1^{\alpha  + 1} [u](x)
& = r^{\alpha  + 1} \frac{{d^2 }}{{dr^2 }}J^{3 - (\alpha  + 1)}
 [ {\frac{d}{{dr}}u} ](x) = r^{\alpha  + 1} 
\frac{d}{{dr}}[ {\frac{d}{{dr}}J^{2 - \alpha } [ {\frac{d}{{dr}}u} ]} ](x) \\
& = r^{\alpha  + 1} \frac{d}{{dr}}[ {r^{ - \alpha }   B_1^\alpha  [u]} ](x) 
= r\frac{d}{{dr}}B_1^\alpha  [u](x) - \alpha B_1^\alpha  [u](x).
\end{align*}
Consequently,  in this case,
\[
 \frac{{\partial v(x )}}{{\partial \nu }}\big|_{\partial \Omega }  
= f_2 (x ) + \alpha f_1 (x ) \equiv \varphi _2 (x).
\]
Thus, if $u(x )$ is a solution of problem \ref{probl1}, then for the function 
$v(x ) = B_1^\alpha  [ u ](x ) $
 we obtain the problem \eqref{e3-1} with the functions 
 \eqref{e4-4}  and \eqref{e4-5}.

By \eqref{e2-3}, the additional condition $v(0) = 0$ holds. 
For smooth enough functions $g_1 (x ),\varphi _1 (x)$ and $\varphi _2 (x)$
the solution of \eqref{e3-1} exists, is unique and can be represented 
as in \eqref{e3-2}.

Let $0 < \alpha  \le 1$. Then, using the representation of the function $g_1 (x )$
 as \eqref{e2-14}, and from \eqref{e3-3} and \eqref{e3-9}, we obtain
\begin{align*}
v(0) &= \frac{1}{{2\omega _n }}\int_{\partial \Omega } {[ {2\varphi _1 (y) 
- \varphi _2 (y)} ]dS_y }  + \frac{1}{{2\omega _n }}
\int_\Omega  {\frac{{1 - |y|^2 }}{2}g_{1,\alpha } (y)dy}  \\
&\quad + \frac{{1 - \alpha }}{{\omega _n 2(n - 2)(n - 4)}}
\int_\Omega  {[ {|y|^{4 - n}  - 1 + ({2 - \frac{n}{2}} )
({1 - |y|^2 } )} ]g_{1,\alpha } (y)dy} .
\end{align*}
Hence, the condition $v(0) = 0$ holds if
\begin{align*}
& - \int_{\partial \Omega } {[ {2\varphi _1 (y) - \varphi _2 (y)} ]dS_y } \\
& = \int_\Omega  {\frac{{1 - |y|^2 }}{2}g_{1,\alpha } (y)dy}  \\
&\quad  + \frac{{1 - \alpha }}{{(n - 2)(n - 4)}}\int_\Omega  {[ {|y|^{4 - n}  
- 1 + ({2 - \frac{n}{2}} )({1 - |y|^2 } )} ]g_{1,\alpha } (y)dy} .
\end{align*}
Since
$$
2\varphi _1 (y) - \varphi _2 (y) = 2f_1 (y) - f_2 (y) - \alpha f_1 (y) 
=  - [ {f_2 (y) + (\alpha  - 2)f_1 (y)} ],
$$
this condition can be rewritten as \eqref{e4-1}.
 Therefore, necessity of condition \eqref{e4-1} is proved.

Applying  the equality $v(x ) = B_1^\alpha  [ u ](x )$,
the operator $B^{ - \alpha }$, by \eqref{e2-8}, yields
$$
B^{ - \alpha } [v](x) = B^{ - \alpha } [ {B_1^\alpha  [ u ]} ](x ) = u(x ) - u(0),
$$
i.e. if the solution of problem \ref{probl1} exists, and can be represented 
as in \eqref{e4-2}.
Now we show that  condition \eqref{e4-1} is also sufficient for the existence 
of solutions of problem \ref{probl1}. Indeed, if condition \eqref{e4-1} holds,
then for solutions of problem \eqref{e3-1} with functions \eqref{e4-4} 
and \eqref{e4-5}, condition $v(0) = 0$  holds.
Then for such functions the operator $B^{ - \alpha }$ is defined and 
we can consider the function $u(x ) = C + B^{ - \alpha } [v](x)$.
This function satisfies all conditions of the problem \ref{probl1}. 
Indeed, since
$\Delta ^2 v(x) = g_1 (x)$ and $g_1 (x) = (1 - \alpha )g_{1,\alpha } (x) 
+ \Gamma _4 [g_{1,\alpha } ](x)$, then, using  \eqref{e2-16} we can write 
the equalities
\begin{align*}
\Delta ^2 u(x ) 
&= \Delta ^2 [ {C + B^{ - \alpha } [v](x)} ] \\
&= \frac{1}{{\Gamma (\alpha )}}\int_0^1 {(t - \tau )^{\alpha  - 1} 
 \tau ^{ - \alpha } \Delta ^2 v({\tau x} )d\tau } \\
&= \frac{1}{{\Gamma (\alpha )}}\int_0^1 {(t - \tau )^{\alpha  - 1} 
 \tau ^{4 - \alpha } [ {|\tau x|^{ - 4} B_1^\alpha  
 [ {\tau ^4 g} ]} ]({\tau x} )d\tau }  \\
&=\frac{{|x|^{ - 4} }}{{\Gamma (\alpha )}}\int_0^1 (t - \tau )^{\alpha  - 1} 
 \tau ^{ - \alpha } B_1^\alpha  [ {\tau ^4 g} ]({\tau x} )d\tau  \\
& = |x|^{ - 4} B^{ - \alpha } [ {B_1^\alpha  [ {|x|^4 g} ]} ](x ) 
 = |x|^{ - 4} |x|^4 g(x) = g(x).
\end{align*}
Using \eqref{e2-9}, we obtain
\begin{align*}
D_1^\alpha  [ u ](x )\big|_{\partial \Omega } 
&= B_1^\alpha  [ u ](x )\big|_{\partial \Omega } 
 = B_1^\alpha  [ {C + B^{ - \alpha } [ v ]} ](x )\big|_{\partial \Omega }  \\
&= v(x )\big|_{\partial \Omega }  = \varphi _1 (x ) = f_1 (x),
\end{align*}
\begin{align*}
D_1^{\alpha  + 1} [ u ](x )\big|_{\partial \Omega } 
&= B_1^{\alpha  + 1} [u](x)\big|_{\partial \Omega } 
 = r\frac{\partial }{{\partial r}}B_1^\alpha  [u](x) 
- \alpha B_1^\alpha  [u](x)\big|_{\partial \Omega } \\
&= r\frac{\partial }{{\partial r}}v(x ) - \alpha v(x )\big|_{\partial \Omega } 
 = \varphi _2 (x) - \alpha \varphi _1 (x) \\
&= f_2 (x) + \alpha f_1 (x) - \alpha f_1 (x) = f_2 (x).
\end{align*}
Consequently, the function $u(x ) = C + B^{ - \alpha } [v](x)$ satisfies 
all conditions of the problem \ref{probl1}.

Let  $1 < \alpha  < 2$, $j = 1$. In this case $v(x ) = B_1^\alpha  [ u ](x )$
 will be a solution of problem \eqref{e3-1} with functions 
$\varphi _1 (x) = f_1 (x )$, $\varphi _2 (x) = f_2 (x ) + \alpha f_1 (x )$  and
\begin{align*}
g_1 (x)
& = |x|^{ - 4} B_1^\alpha  [ {|x|^4 g} ](x) \\
&\equiv(1 - \alpha )(2 - \alpha )g_{2,\alpha } (x) 
+ 2(2 - \alpha )\Gamma _4 [g_{2,\alpha } ](x) 
+ \Gamma _4 [ {\Gamma _3 [g_{2,\alpha } ]} ](x).
\end{align*}
By \eqref{e2-6}, the condition $v(0) = 0$,  holds additionally. 
Then, using \eqref{e3-3} and \eqref{e3-10}, we have
\begin{align*}
v(0) &= \frac{1}{{2\omega _n }}\int_{\partial \Omega } 
{[ {2\varphi _1 (y) - \varphi _2 (y)} ]dS_y }  
+ \frac{{2(2 - \alpha )}}{{2\omega _n }}
\int_\Omega  {\frac{{1 - |y|^2 }}{2}g_{1,\alpha } (y)dy} \\
&\quad + \frac{1}{{2\omega _n }}\int_\Omega  {\frac{{1 - |y|^2 }}{2}\Gamma _3 
[g_{2,\alpha } ](y)dy}  \\
&\quad +\frac{{(1 - \alpha )(2 - \alpha )}}{{\omega _n 2(n - 2)(n - 4)}}
\int_\Omega  {[ {|y|^{4 - n}  - 1 + ({2 - \frac{n}{2}} )({1 - |y|^2 } )} ]
g_{2,\alpha } (y)dy}.
\end{align*}
Thus, for the condition $v(0) = 0$,
 the following equality is necessary
\begin{align*}
&- \int_{\partial \Omega } {[{2\varphi _1 (y) - \varphi _2 (y)}]dS_y }  \\
&= 2(2 - \alpha )\int_\Omega  {\frac{{1 - |y|^2 }}{2}g_{2,\alpha } (y)dy} 
 +  \int_\Omega  {\frac{{1 - |y|^2 }}{2}\Gamma _3 [{g_{2,\alpha } }](y)dy}  \\
&\quad +\frac{{(1 - \alpha )(2 - \alpha )}}{{(n - 2)(n - 4)}}
\int_\Omega  {[ {|y|^{4 - n}  - 1 + ({2 - \frac{n}{2}} )({1 - |y|^2 } )} ]
g_{2,\alpha } (y)dy} .
\end{align*}
Since $2\varphi _1 (y) - \varphi _2 (y) =  - [ {f_2 (y) + (\alpha  - 2)f_1 (y)} ]$, 
this condition can be rewritten as \eqref{e4-3}. Therefore, necessity of the 
condition \eqref{e4-3} is proved. 
Further, by repetition of the argument in  the case $0 < \alpha  < 1$, one can show 
the rest of the theorem. 
\end{proof}

\begin{theorem} \label{thm4.2}
Let $1 < \alpha  \leq 2$, $j = 2$, $g(x)$, $f_1 (x) $ and $ f_2 (x)$ be smooth 
functions.  Then  problem \ref{probl2} is solvable if and only if:
\begin{equation} \label{e4-6}
\begin{aligned}
&\int_{\partial \Omega } {[ {f_2 (y) + (\alpha  - 2)f_1 (y)} ]dS_y }  \\
&= 2(2 - \alpha )\int_\Omega  {\frac{{1 - |y|^2 }}{2}g_{2,\alpha } (y)dy}
 + \int_\Omega  {\frac{{1 - |y|^2 }}{2}\Gamma _3 [ {g_{2,\alpha } } ](y)dy}  \\
&\quad + \frac{{(1 - \alpha )(2 - \alpha )}}{{(n - 2)(n - 4)}}
 \int_\Omega  {[ {|y|^{4 - n}  - 1
 + ({2 - \frac{n}{2}} )({1 - |y|^2 } )} ]g_{2,\alpha } (y)dy} ,
\end{aligned}
\end{equation}
 and
\begin{equation} \label{e4-7}
\begin{aligned}
&\int_{\partial \Omega } {y_k [ {f_2 (y) + (\alpha  - 3)f_1 (y)} ]dS_y }  \\
&= \frac{1}{{2}}\int_\Omega  {y_k ({1 - |y|^2 } )\Gamma _4 [g_{2,\alpha } ](y)dy}  \\
&\quad  + \frac{2(2 - \alpha )}{n(n - 2)}\int_\Omega  y_k [{|y|^{2 - n}  - 1
  + \frac{2-n}{2}(1 - |y|^2 )}]\Gamma _4[ {g_{2,\alpha } }](y)dy \\
&\quad  + \frac{{(1 - \alpha )(2 - \alpha )}}{{n(n - 2)}}
 \int_\Omega  {y_k[{|y|^{2 - n}  - 1 + \frac{{2 - n}}{2}(1 - |y|^2 )}]
 g_{2,\alpha } (y)dy} ,
\end{aligned}
\end{equation}
for $k = 1,\dots,n$.


If a solution of the problem \ref{probl2} exists, then it is unique up to 
a first order polynomial and can be represented as
\begin{equation}\label{e4-8}
  u(x ) = c_0  + \sum_{i = 1}^n {c_i x_i  + } B^{ - \alpha } [v](x),
\end{equation}
where $c_i$, $i = 0,1,\dots,n$  are arbitrary constants, and $v(x)$
 is a solution of the problem \eqref{e3-1} with functions 
$g_1 (x ) = |x|^{ - 4} B_2^\alpha  [ {|x|^4 g} ](x ),\varphi _1 (x) = f_1 (x )$ and 
 $ \varphi _2 (x) = f_2 (x ) + \alpha f_1 (x )$, and which satisfies conditions 
$v(0) = 0, \frac{{\partial v(0)}}{{\partial x_i }} = 0$, $i = 1,2,\dots,n$.
\end{theorem}

\begin{proof}
 Let $u(x )$ be a solution of problem \ref{probl2}. Apply to the function $u(x )$ 
the operator $B_2^\alpha $, and denote it by $v(x ) = B_2^\alpha  [ u ](x )$. 
Then \eqref{e2-12}  and
\begin{align*}
B_2^{\alpha  + 1} [u](x) 
&= r^{\alpha  + 1} \frac{d}{{dr}}J^{3 - (\alpha  + 1)}
 [ {\frac{{d^2 }}{{dr^2 }}u} ](x) = r^{\alpha  + 1} 
\frac{d}{{dr}}J^{2 - \alpha } [ {\frac{{d^2 }}{{dr^2 }}u} ](x) \\
& = r^{\alpha  + 1} \frac{d}{{dr}}[ {r^{ - \alpha }  
\cdot B_2^\alpha  [u]} ](x) = r\frac{d}{{dr}}B_2^\alpha  
[u](x) - \alpha B_2^\alpha  [u](x).
\end{align*}
imply that the function $v(x)$ is a solution of the problem \eqref{e3-1} 
with functions
$g_1 (x ) = |x|^{ - 4} B_2^\alpha  [ {|x|^4 g} ](x )$, 
$\varphi _1 (x) = f_1 (x )$, $\varphi _2 (x) = f_2 (x ) + \alpha f_1 (x )$.

Moreover,  by lemma \ref{l2-2}, the function $v(x ) = B_2^\alpha  [ u ](x )$ 
should satisfy conditions 
$v(0) = 0,\frac{{\partial v(0)}}{{\partial x_k }} = 0$, $k = 1,2,\dots,n$.

For enough smooth functions $g_1 (x )$, $\varphi _1 (x)$
 and $\varphi _2 (x)$  the solution of  problem \eqref{e3-1} exists, 
is unique and can be represented as \eqref{e3-2}.

	Further, using  the representation of the function $g_1 (x )$
 in the form \eqref{e2-15}, by similar arguments, as in the case 
$1 < \alpha  < 2,j = 1$, one can show that the equality $v(0) = 0$ 
 holds if the condition \eqref{e4-4} holds.

	Now we check that the equalities 
$\frac{{\partial v(0)}}{{\partial x_k }} = 0$, $k = 1,2,\dots,n$ hold if 
condition \eqref{e4-5} holds. To do it we use the representation of the
 function $v(x)$ in the form \eqref{e3-2} and the lemma \ref{l3-3}. 
Since the function $g_1 (x ) = |x|^{ - 4} B_2^\alpha  [ {|x|^4 g} ](x )$ 
can be represented as \eqref{e3-8}, then by \eqref{e3-4} and \eqref{e3-11},
 we obtain
\begin{align*}
\frac{{\partial v(0)}}{{\partial x_k }}
& = \frac{n}{{2\omega _n }}\int_{\partial \Omega } {y_k [ {3\varphi _1 (y) 
 - \varphi _2 (y)} ]dS_y }  + \frac{n}{{4\omega _n }}
 \int_\Omega  {y_k ({1 - |y|^2 } )\Gamma _4 [g_{2,\alpha } ](y)dy}  \\
&\quad + \frac{1}{{2\omega _n }}\frac{{2(2 - \alpha )}}{{(n - 2)}}
 \int_\Omega  y_k [ {|y|^{2 - n}  - 1 
 + \frac{{2 - n}}{2}(1 - |y|^2 )} ]\Gamma _4 [ {g_{2,\alpha } } ](y)dy \\ 
&\quad  + \frac{1}{{2\omega _n }}\frac{{(1 - \alpha )(2 - \alpha )}}{{(n - 2)}}
\int_\Omega  {y_k [ {|y|^{2 - n}  - 1 
+ \frac{{2 - n}}{2}(1 - |y|^2 )} ]g_{2,\alpha } (y)dy} ,
\end{align*}
for $ k = 1,\dots,n$.
Consequently, equalities $\frac{{\partial v(0)}}{{\partial x_k }} = 0$,
$k = 1,2,\dots,n$  hold if
\begin{align*}
&\int_{\partial \Omega } {y_k [ {\varphi _2 (y) - 3\varphi _1 (y)} ]dS_y } \\
& = \frac{1}{{2}}\int_\Omega  {y_k ({1 - |y|^2 } )
 \Gamma _4 [g_{2,\alpha } ](y)dy}  \\
&\quad + \frac{{2(2 - \alpha )}}{{n(n - 2)}}\int_\Omega  {y_k [ {|y|^{2 - n}  - 1 
 + \frac{{2 - n}}{2}(1 - |y|^2 )} ]\Gamma _4 [ {g_{2,\alpha } } ](y)dy  }\\
&\quad + \frac{{(1 - \alpha )(2 - \alpha )}}{{n(n - 2)}}
 \int_\Omega  {y_k [ {|y|^{2 - n}  - 1 
 + \frac{{2 - n}}{2}(1 - |y|^2 )} ]g_{2,\alpha } (y)dy} ,
\end{align*}
for $k = 1,\dots,n$.

Since $\varphi _2 (y) - 3\varphi _1 (y) = f_2 (x ) 
+ \alpha f_1 (x ) - 3f_1 (x ) = f_2 (x ) + (\alpha  - 3)f_1 (x )$, 
the above condition can be rewritten as \eqref{e4-7}.
	
Applying the operator $B^{ - \alpha } $ to the equality 
$v(x ) = B_1^\alpha  [ u ](x )$, by \eqref{e2-10}, we obtain
$$
B^{ - \alpha } [v](x) = B^{ - \alpha } [ {B_1^\alpha  [ u ]} ](x ) 
= u(x ) - u(0) - \sum_{i = 1}^n {x_i \frac{{\partial u(0)}}{{\partial x_i }}} .
$$
Denoting
 $$
c_0  = u(0),\quad c_i  = \frac{{\partial u(0)}}{{\partial x_i }},\quad
i = 1,2,\dots,n,
$$
 we obtain the representation \eqref{e4-8}. Therefore, if solution of the 
problem \ref{probl2} exists, then it can be represented as \eqref{e4-6}.

Now we show that conditions \eqref{e4-6} and \eqref{e4-7} are also sufficient 
for existence of a solution of the problem \ref{probl2}. 
Indeed, if conditions \eqref{e4-6} and \eqref{e4-7} hold, then for a solution 
of the problem \eqref{e3-1} with functions 
$$
g_1 (x ) = |x|^{ - 4} B_1^\alpha  [ {|x|^4 g} ](x ),\quad
\varphi _1 (x) = f_1 (x ), \quad
\varphi _2 (x) = f_2 (x ) + \alpha f_1 (x )
$$
the conditions
$$
v(0) = 0,\quad \frac{{\partial v(0)}}{{\partial x_i }} = 0,\quad i = 1,2,\dots,n,
$$
hold.  Then in the class of such functions the operator 
$B^{ - \alpha }$ is defined, and we can consider the function
$$
u(x ) = c_0  + \sum_{i = 1}^n {c_i x_i  + } B^{ - \alpha } [v](x).
$$
 We show that this function satisfies all conditions of the problem \ref{probl1}.
Indeed, since 
$$
\Delta ^2 v(x ) = g_1 (x) \equiv |x|^{ - 4} B_1^\alpha  [ {|x|^4 g} ](x ),
$$ 
it follows that
\begin{align*}
\Delta ^2 u(x ) 
&= \Delta ^2 \Big[ {c_0  + \sum_{i = 1}^n {c_i x_i  + } B^{ - \alpha } [v](x)} \Big]
 = \frac{1}{{\Gamma (\alpha )}}\int_0^1 {(t - \tau )^{\alpha  - 1} 
\tau ^{ - \alpha } \Delta ^2 v({\tau x} )d\tau } \\
&=  |x|^{ - 4} B^{ - \alpha } [ {B_2^\alpha  [ {|x|^4 g} ]} ](x ).
\end{align*}
The above expression, by \eqref{e2-11}, equals to $g(x)$.	
Further, using \eqref{e2-11}, we obtain
\begin{align*}
D_2^\alpha  [ u ](x )\big|_{\partial \Omega } 
& = B_2^\alpha  [ u ](x )\big|_{\partial \Omega } 
 = B_2^\alpha  [ {c_0  + \sum_{i = 1}^n {c_i x_i  + }
  B^{ - \alpha } [v]} ](x )\big|_{\partial \Omega } \\
&= v(x )\big|_{\partial \Omega }  = \varphi _1 (x ) = f_1 (x),
\end{align*}
\begin{align*}
&D_2^{\alpha  + 1} [ u ](x )\big|_{\partial \Omega } \\
&= B_2^{\alpha  + 1} [u](x)\big|_{\partial \Omega } 
 = r^{\alpha  + 1} \frac{d}{{dr}}J^{3 - (\alpha  + 1)} \frac{{d^2 }}{{dr^2 }}u(x)\\
&= r^{\alpha  + 1} \frac{d}{{dr}}J^{2 - \alpha } \frac{{d^2 }}{{dr^2 }}u(x) 
 = r\frac{d}{{dr}}B_1^\alpha  [u](x) - \alpha B_1^\alpha  [u](x)
 \big|_{\partial \Omega } \\
&= r\frac{d}{{dr}}B_1^\alpha  [ {c_0  + \sum_{i = 1}^n {c_i x_i  + } 
 B^{ - \alpha } [v]} ](x ) - \alpha B_1^\alpha  
 [ {c_0  + \sum_{i = 1}^n {c_i x_i  + } B^{ - \alpha } [v]} ](x )
\big|_{\partial \Omega } \\
& = r\frac{d}{{dr}}B_1^\alpha  [u](x) - \alpha B_1^\alpha  [u](x)\big|_{\partial \Omega }
  = r\frac{{dv(x )}}{{dr}} - \alpha v(x )|_{\partial \Omega } \\
& = \varphi _2 (x) - \alpha \varphi _1 (x) 
  = f_2 (x) + \alpha f_1 (x) - \alpha f_1 (x) = f_2 (x).
\end{align*}
Consequently, the function $ c_0  + \sum_{i = 1}^n {c_i x_i  + } B^{ - \alpha } [v]$ 
 satisfies all conditions of the problem \ref{probl2}. 
\end{proof}

\begin{remark} \label{rmk4.1} \rm
If in  \eqref{e4-1} $\alpha  = 1$, then condition on solvability of the 
problem \ref{probl1} coincides with the condition \eqref{e1-6}. 
Similarly, in the case $\alpha  = 2$  condition on solvability of the 
problem \ref{probl2} coincides with the conditions \eqref{e1-7} and \eqref{e1-8}.
\end{remark}

\subsection*{Acknowledgements} 
The author would like to thank the editor and referees  for  their  valuable  
comments  and  remarks, which  led  to  a  great  improvement of the article. 
This research is financially supported by a grant from the Ministry of Science 
and Education of the Republic of Kazakhstan (Grant No. 0819/GF4).

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\end{document}