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     {\huge СИБИРСКИЕ \ ЭЛЕКТРОННЫЕ} \\[2mm]  %Headings in Russian
     {\huge МАТЕМАТИЧЕСКИЕ ИЗВЕСТИЯ} \\[2mm]
     {\large Siberian Electronic Mathematical Reports} \\[1mm]
     {\LARGE\tt{http://semr.math.nsc.ru}}\\[0.5mm]
%     {\small Институт математики им. С.Л.Соболева СО РАН}\\[-1mm]
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     \noindent
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     {\flushleft\it Том 16, стр. 144--144 (2019) \hspace{65mm}{\rm\small УДК \udcs}} %Volume, pages in Russian,
                                                                                    %filled by Editorial board
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         {\rm\small DOI~10.33048/semi.2019.16.xxx}\hphantom{aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa}{\rm\small MSC\ \ \mscs }
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\title[INITIAL BOUNDARY-VALUE PROBLEM...]{ INITIAL BOUNDARY-VALUE PROBLEM FOR EQUATIONS WITH FRACTIONAL DERIVATIVE WITH DEGENERATION OF THE FIRST KIND }
\author{{B.Yu. Irgashev}}%
\address{Bakhrom Yusuphanovich Irgashev
\newline\hphantom{iii} Namangan Engineering Construction Institute,
Namangan regional branch Institute of Mathematics of the Republic of Uzbekistan,
\newline\hphantom{iii} pr. I.Karimov, 12,
\newline\hphantom{iii} 160103, Namangan, uzbekistan}%
\email{bahromirgasev@gmail.com}%

\thanks{\sc Irgashev, B.Yu.
Initial boundaru-value problem for equations with fractional derivative with degeneration of the first kind}
\thanks{\copyright \ 2020 Irgashev B.Yu}
\thanks{\it Received  January, 1, 2015, published  March, 1,  2015.}%


\semrtop \vspace{1cm}
\maketitle {\small
\begin{quote}
\noindent{\sc Abstract. } The article considers an initial-boundary value problem for a higher-order equation with a fractional Caputo derivative, in a rectangular domain, degenerating in both variables. The solution to the problem is constructed explicitly, by the method of separation of variables. Uniqueness is shown by the spectral method.\medskip

\noindent{\bf Keywords:} Differential equation, high order, degeneration, fractional Caputo derivative, existence, uniqueness, series, uniform convergence.
 \end{quote}
}


\section{Introduction}


In the domain  $D = \left\{ {\left( {x,y} \right):0\, < x < 1,0 < y < 1} \right\},$ consider the equation
\[{\left( { - 1} \right)^{k + 1}}K\left( y \right){}_CD_{0x}^\alpha u\left( {x,y} \right) - {x^s}\frac{{{\partial ^{2k}}u}}{{\partial {y^{2k}}}} = 0,\eqno(1)\]
where  $0 < \alpha  < 1, s \in N \cup \{ 0\}$ , $ k \in N, D_{0x}^\alpha  - $  is the operator of Caputo fractional differentiation, of order $\alpha$
\[{}_CD_{0x}^\alpha u\left( {x,y} \right) = \frac{1}{{\Gamma \left( {1 - \alpha } \right)}}\int\limits_0^x {\frac{{\frac{{\partial u\left( {\tau ,y} \right)}}{{\partial \tau }}d\tau }}{{{{\left( {x - \tau } \right)}^\alpha }}}\,} .\]
The function $ K(y)$ satisfies the conditions:\\
1. $ K\left( y \right) > \,0 $ for $y>0$;\\
2. $K\left( 0 \right) = 0;$\\
3. $K\left( y \right) \in C\left[ {0,1} \right].$\\
For equation (1), consider the problem.

	\textbf{ Problem A.} Find a solution to equation (1) from the class
\[u\left( {x,y} \right) \in C_{x,y}^{0,2k - 1}\left( {\overline D } \right) \cap C_y^{2k}\left( D \right),{}_CD_{0x}^\alpha  \in C\left( D \right),\eqno(2)\]
satisfying the conditions
\[\frac{{{\partial ^{s}}u\left( {x,0} \right)}}{{\partial {y^{s}}}} = \frac{{{\partial ^{s}}u\left( {x,1} \right)}}{{\partial {y^{s}}}} = 0,\,\,\,0 \le x \le 1,\,s = 0,1,...,k - 1,\eqno(3)\]
\[u\left( {0,y} \right) = \varphi \left( y \right),\,\,0 \le y \le 1,\eqno(4)\]
here  $\varphi \left( y \right)- $ is sufficiently smooth . \\
Fractional differential equations arise in mathematical modeling of various physical processes and phenomena [1]. Second-order equations of the form (1) with partial derivatives of fractional order $\alpha\in\left({0,2}\right)$ were studied in [1] - [8] and others. In these papers, the Cauchy problem was considered, the first , the second and mixed boundary value problems, a fundamental solution is found, a general representation of solutions is constructed. Mixed equations and higher-order equations with a fractional derivative were studied in [9] - [12]. Degenerate fractional-order equations were studied in [1], [13]. The research will be carried out by the Fourier method. Previously, by the Fourier method, boundary value problems for equations with a fractional derivative were studied in [6] - [9], [12].
\begin{center}
\textbf{2. Existence of a solution
}\end{center}
We are looking for a solution in the form
\[u\left( {x,y} \right) = X\left( x \right)Y\left( y \right).\]
Then with respect to the variable  $y$, taking into account condition (3), we obtain the following spectral problem:
\[\left\{ \begin{array}{l}
{Y^{\left( {2k} \right)}}(y) = {( - 1)^k}\lambda K\left( y \right)Y(y),\\
{Y^{\left( s \right)}}(0) = {Y^{\left( s \right)}}\left( 1 \right) = 0,\,s = 0,1,...,k - 1.
\end{array} \right.\eqno(5)\]
First, we show that $\lambda  = 0$  is not an eigenvalue. Indeed, consider problem (5) with $\lambda  = 0$
\[\left\{ \begin{array}{l}
{Y^{\left( {2k} \right)}}(y) = 0,\\
{Y^{\left( s \right)}}(0) = {Y^{\left( s \right)}}\left( 1 \right) = 0,\,s = 0,1,...,k - 1.
\end{array} \right.\]
A solution to this problem satisfying the condition ${Y^{\left( s \right)}}(0) = 0,\,s = 0,1,...,k - 1,$  has the form
\[Y\left( y \right) = {c_k}{y^k} + {c_{k + 1}}{y^{k + 1}} + ... + {c_{2k - 1}}{y^{2k - 1}},\]
to determine the unknowns ${c_j},\,\,j = k,k + 1,...,2k - \,1,$  we obtain the system of equations
\[\left\{ \begin{array}{l}
{c_k} + {c_{k + 1}} + ... + {c_{2k - 1}} = 0,\\
k{c_k} + \left( {k + 1} \right){c_{k + 1}} + ... + \left( {2k - 1} \right){c_{2k - 1}} = 0,\\
...\\
k\left( {k - 1} \right)...2 \cdot {c_k} + \left( {k + 1} \right)k...3 \cdot {c_{k + 1}} + ... + \left( {2k - 1} \right)\left( {2k - 2} \right)...\left( {k + 1} \right){c_{2k - 1}} = 0,
\end{array} \right.\]
the main determinant of this system $\Delta $ , has the form
\[\Delta  = \left| {\begin{array}{*{20}{c}}
1&1&{...}&1\\
k&{k + 1}&{...}&{2k - 1}\\
.&.&.&.\\
{k\left( {k - 1} \right)...2}&{\left( {k + 1} \right)k...3}&{...}&{\left( {2k - 1} \right)\left( {2k - 2} \right)...\left( {k + 1} \right)}
\end{array}} \right| = \]
\[ = \prod\limits_{j = 1,j > i}^{k - 1} {\left( {j - i} \right)}  \ne 0.\]
Hence ${c_j} = 0,\,\,j = k,k + 1,...,2k - \,1,$ from here
\[Y\left( y \right) = {c_k}{y^k} + {c_{k + 1}}{y^{k + 1}} + ... + {c_{2k - 1}}{y^{2k - 1}} \equiv 0.\]
We reduce problem (6) to the integral equation using the Green function and obtain the necessary estimates for the eigenfunctions. But first, we show that $\lambda  > 0.$ Indeed, we have
\[\int\limits_0^1 {Y\left( y \right){Y^{\left( {2k} \right)}}(y)dy = {{( - 1)}^k}\lambda \int\limits_0^1 {K\left( y \right){Y^2}(y)dy} ,} \]
\[\int\limits_0^1 {{{\left( {{Y^{\left( k \right)}}} \right)}^2}dy = \lambda \int\limits_0^1 {K\left( y \right){Y^2}(y)dy} ,} \]
 because $\lambda  = 0$ is not an eigenvalue, it follows that $\lambda  > 0$. It remains to show the existence of eigenvalues and eigenfunctions of problem (5). The integral equation equivalent to problem (5) has the form
$$Y\left( y \right) = {\left( { - 1} \right)^k}\lambda \int\limits_0^1 {K\left( y \right)G\left( {y,\xi } \right)Y\left( \xi  \right)d\xi } ,\eqno(6)$$
where
$$G\left( {y,\xi } \right) =  - \frac{1}{{\left( {2k - 1} \right)!}}\left\{ \begin{array}{l}
{G_1}\left( {y,\xi } \right),\,\,\,\,0 \le y \le \xi ,\\
{G_2}\left( {y,\xi } \right),\,\,\,\,\xi  \le y \le 1,
\end{array} \right.$$
- the Green function of problem (5) (see [14]), here
$${G_1}\left( {y,\xi } \right) = {\left( {1 - \xi } \right)^k}{y^k}\sum\limits_{i = 0}^{k - 1} {\sum\limits_{j = 0}^{k - i - 1} {{{\left( { - 1} \right)}^i}C_{2k - 1}^i} } C_{k - 1 + j}^j{y^{k - i - 1}}{\xi ^{j + i}},$$
$${G_2}\left( {y,\xi } \right) = {\left( {1 - y} \right)^k}{\xi ^k}\sum\limits_{i = 0}^{k - 1} {\sum\limits_{j = 0}^{k - i - 1} {{{\left( { - 1} \right)}^i}C_{2k - 1}^i} } C_{k - 1 + j}^j{\xi ^{k - i - 1}}{y^{j + i}},$$
$$C_n^k = \frac{{n!}}{{k!\left( {n - k} \right)!}}.$$
Rewrite (6) as
$$\sqrt {K\left( y \right)} Y\left( y \right) = \lambda \int\limits_0^1 {\sqrt {K\left( \xi  \right)} \left[ {{{\left( { - 1} \right)}^k}G\left( {y,\xi } \right)} \right]\sqrt {K\left( y \right)} \left( {\sqrt {K\left( \xi  \right)} Y\left( \xi  \right)} \right)d\xi } ,$$
we introduce the notation
\[\overline Y \left( y \right) = \sqrt {K\left( y \right)} Y\left( y \right),\]
$$\overline G \left( {y,\xi } \right) = \sqrt {K\left( \xi  \right)} \left[ {{{\left( { - 1} \right)}^k}G\left( {y,\xi } \right)} \right]\sqrt {K\left( y \right)} ,$$
then we have
$$\overline Y \left( y \right) = \lambda \int\limits_0^1 {\overline G \left( {y,\xi } \right)\overline Y \left( \xi  \right)d\xi },\eqno(7)$$
(7) - there is an integral equation with a continuous, in both variables, and a symmetric kernel. According to the theory of equations with symmetric kernels, equation (7) has no more than a countable number of eigenvalues and eigenfun-ctions. So, problem (5) has eigenvalues ${\lambda _n} > 0,\,\,n = 1,2,...$, and the corresponding eigenfunctions are ${Y_n}(y)$ . Further, we assume that
\[{\left\| {{Y_n}(y)} \right\|^2} = \int\limits_0^1 {K\left( y \right)Y_n^2(y)dy}  = 1.\]
then, taking into account (7), we have the Bessel inequality
\[\sum\limits_{n = 1}^\infty  {{{\left( {\frac{{{Y_n}\left( y \right)}}{{{\lambda _n}}}} \right)}^2}}  \le \int\limits_0^1 {K\left( y \right){G^2}\left( {y,\xi } \right)} d\xi  < M < \infty .\eqno(8)\]
Now we find the conditions under which the given function  $\varphi \left( y \right)$  is expanded in a series according to the eigenfunctions ${Y_n}(y)$ . For this we use the Hilbert-Schmidt theorem.
\begin{theorem}
Let the function $\varphi \left( y \right)$ satisfy the conditions

1. $\frac{{{\varphi ^{\left( {2k} \right)}}\left( \xi  \right)}}{{\sqrt {K\left( \xi  \right)} }} \in C\left[ {0;1} \right];$

2. ${\varphi ^{\left( i \right)}}\left( 0 \right) = {\varphi ^{\left( i \right)}}\left( 1 \right) = 0,\,i = 0,1,...,k - 1.$\\
Then it can be expanded in a uniformly and absolutely converging series of the form
\[\varphi \left( y \right) = \sum\limits_{n = 1}^\infty  {{\varphi _n}{Y_n}\left( y \right)} ,\]
where
\[{\varphi _n} = \int\limits_0^1 {K\left( y \right)\varphi \left( y \right){Y_n}\left( y \right)dy\,\,} ,\]
\end{theorem}

\begin{proof}
We show the equality
\[\sqrt {K\left( y \right)} \varphi \left( y \right) = \int\limits_0^1 {\overline G \left( {y,\xi } \right)\left( {{{\left( { - 1} \right)}^k}\frac{{{\varphi ^{\left( {2k} \right)}}\left( \xi  \right)}}{{\sqrt {K\left( \xi  \right)} }}} \right)d\xi } ,\]
really
\[\int\limits_0^1 {\sqrt {K\left( \xi  \right)} \left[ {{{\left( { - 1} \right)}^k}G\left( {y,\xi } \right)} \right]\sqrt {K\left( y \right)} \left( {{{\left( { - 1} \right)}^k}\frac{{{\varphi ^{\left( {2k} \right)}}\left( \xi  \right)}}{{\sqrt {K\left( \xi  \right)} }}} \right)d\xi }  = \]
\[ = \sqrt {K\left( y \right)} \int\limits_0^1 {G\left( {y,\xi } \right){\varphi ^{\left( {2k} \right)}}\left( \xi  \right)} d\xi  = \sqrt {K\left( y \right)} \varphi \left( y \right).\]
Those for the function $\sqrt {K\left( y \right)} \varphi \left( y \right)$ the conditions of the Hilbert-Schmidt theorem are satisfied and therefore
\[\sqrt {K\left( y \right)} \varphi \left( y \right) = \sum\limits_{n = 1}^\infty  {\sqrt {K\left( y \right)} {\varphi _n}{Y_n}\left( y \right)} ,\]
dividing by  $\sqrt {K\left( y \right)}$, we have
\[\varphi \left( y \right) = \sum\limits_{n = 1}^\infty  {{\varphi _n}{Y_n}\left( y \right)}.\]
\end{proof}
We proceed to solve the equation in the variable $ x $. Taking into account condition(4), we obtain the following initial problem:
\[\left\{ \begin{array}{l}
{}_CD_{0x}^\alpha {X_n}\left( x \right) =  - {\lambda _n}{x^s}{X_n}\left( x \right),\\
{X_n}\left( 0 \right) = {\varphi _n},
\end{array} \right.\eqno(9)\]
where
\[{\varphi _n} = \int\limits_0^1 {K\left( y \right)\varphi \left( y \right){Y_n}\left( y \right)dy\,\,} ,\]
We will seek solution (9) in the form of a series
\[{X_n}\left( x \right) = \sum\limits_{j = 0}^\infty  {{c_j}{x^{\gamma j + \beta }}} ,\]
where ${c_j},\beta ,\gamma- $ are to be defined.\\
Formally, we have (the validity of the rearrangement of the series and the integral will follow below)
\[{}_CD_{0x}^\alpha {X_n}\left( x \right) = \frac{1}{{\Gamma \left( {1 - \alpha } \right)}}\int\limits_0^x {\frac{{\frac{d}{{d\tau }}\sum\limits_{j = 0}^\infty  {{c_j}{\tau ^{\gamma j + \beta }}} d\tau }}{{{{\left( {x - \tau } \right)}^\alpha }}}\, = } \]
\[=\frac{1}{{\Gamma \left( {1 - \alpha } \right)}}\sum\limits_{j = 0}^\infty  {{c_j}} \int\limits_0^x {\frac{{\left( {\gamma j + \beta } \right){\tau ^{\gamma j + \beta  - 1}}d\tau }}{{{{\left( {x - \tau } \right)}^\alpha }}}\, = } \]
\[ = \frac{1}{{\Gamma \left( {1 - \alpha } \right)}}\sum\limits_{j = 0}^\infty  {{c_j}} \left( {\gamma j + \beta } \right)\int\limits_0^1 {\frac{{{x^{\gamma j + \beta }}{z^{\gamma j + \beta  - 1}}dz}}{{{x^\alpha }{{\left( {1 - z} \right)}^\alpha }}}\, = } \]
\[ = \frac{1}{{\Gamma \left( {1 - \alpha } \right)}}\sum\limits_{j = 0}^\infty  {{c_j}} \left( {\gamma j + \beta } \right){x^{\gamma j + \beta  - \alpha }}\int\limits_0^1 {\frac{{{z^{\gamma j + \beta  - 1}}dz}}{{{{\left( {1 - z} \right)}^\alpha }}}\, = } \]
\[ = \sum\limits_{j = 0}^\infty  {{c_j}} \left( {\gamma j + \beta } \right){x^{\gamma j + \beta  - \alpha }}\frac{{\Gamma \left( {\gamma j + \beta } \right)}}{{\Gamma \left( {\gamma j + \beta  - \alpha  + 1} \right)}}.\]
Substituting the last expression in (9), we obtain
\[\sum\limits_{j = 0}^\infty  {{c_j}} \left( {\gamma j + \beta } \right){x^{\gamma j + \beta  - \alpha }}\frac{{\Gamma \left( {\gamma j + \beta } \right)}}{{\Gamma \left( {\gamma j + \beta  - \alpha  + 1} \right)}} =  - {\lambda _n}\sum\limits_{j = 0}^\infty  {{c_j}{x^{\gamma j + \beta  + s}}}  \Rightarrow \]
The solution to problem (9) has the form
\[\sum\limits_{j = 0}^\infty  {{c_j}} \left( {\gamma j + \beta } \right){x^{\gamma j - \alpha }}\frac{{\Gamma \left( {\gamma j + \beta } \right)}}{{\Gamma \left( {\gamma j + \beta  - \alpha  + 1} \right)}} =  - {\lambda _n}\sum\limits_{j = 0}^\infty  {{c_j}{x^{\gamma j + s}}}  \Rightarrow \]
\[\sum\limits_{j = 0}^\infty  {{c_j}\left( {\gamma j + \beta } \right)} {x^{\gamma \left( {j - 1} \right) + \gamma  - \alpha }}\frac{{\Gamma \left( {\gamma j + \beta } \right)}}{{\Gamma \left( {\gamma j + \beta  - \alpha  + 1} \right)}} =  - {\lambda _n}\sum\limits_{j = 0}^\infty  {{c_j}{x^{\gamma j + s}}}  \Rightarrow \]
\[\gamma  - \alpha  = s \Rightarrow \gamma  = \alpha  + s \Rightarrow \]
\[\sum\limits_{j = 0}^\infty  {{c_j}} \left( {\left( {\alpha  + s} \right)j + \beta } \right){x^{\alpha \left( {j - 1} \right)}}\frac{{\Gamma \left( {\left( {\alpha  + s} \right)j + \beta } \right)}}{{\Gamma \left( {\left( {\alpha  + s} \right)j + \beta  - \alpha  + 1} \right)}} =  - {\lambda _n}\sum\limits_{j = 0}^\infty  {{c_j}{x^{\alpha j}}}  \Rightarrow \]
\[\beta  = 0 \Rightarrow \]
\[\sum\limits_{j = 0}^\infty  {{c_j}} \left( {\left( {\alpha  + s} \right)j} \right){x^{\alpha \left( {j - 1} \right)}}\frac{{\Gamma \left( {\left( {\alpha  + s} \right)j} \right)}}{{\Gamma \left( {\left( {\alpha  + s} \right)j - \alpha  + 1} \right)}} =  - {\lambda _n}\sum\limits_{j = 0}^\infty  {{c_j}{x^{\alpha j}}}  \Rightarrow \]
\[{c_j}\left( {\alpha  + s} \right)j\frac{{\Gamma \left( {\left( {\alpha  + s} \right)j} \right)}}{{\Gamma \left( {\left( {\alpha  + s} \right)j - \alpha  + 1} \right)}} =  - {\lambda _n}{c_{j - 1}} \Rightarrow \]
\[{c_j} = \frac{{ - {\lambda _n}\Gamma \left( {\left( {\alpha  + s} \right)j - \alpha  + 1} \right)}}{{\left( {\alpha  + s} \right)j\Gamma \left( {\left( {\alpha  + s} \right)j} \right)}}{c_{j - 1}} = \frac{{ - {\lambda _n}\Gamma \left( {\left( {\alpha  + s} \right)\left( {j - 1} \right) + 1 + s} \right)}}{{\Gamma \left( {\left( {\alpha  + s} \right)j + 1} \right)}}{c_{j - 1}} \Rightarrow \]
\[{c_j} = \frac{{\left( { - {\lambda _n}} \right)\left( {\left( {\alpha  + s} \right)\left( {j - 1} \right) + s} \right)...\left( {\left( {\alpha  + s} \right)\left( {j - 1} \right) + 1} \right)\Gamma \left( {\left( {\alpha  + s} \right)\left( {j - 1} \right) + 1} \right)}}{{\Gamma \left( {\left( {\alpha  + s} \right)j + 1} \right)}}{c_{j - 1}} = \]
\[ = \frac{{\left( { - {\lambda _n}} \right){{\left( {\alpha  + s} \right)}^s}\left( {\left( {j - 1} \right) + \frac{1}{{\alpha  + s}}} \right)...\left( {\left( {j - 1} \right) + \frac{s}{{\alpha  + s}}} \right)\Gamma \left( {\left( {\alpha  + s} \right)\left( {j - 1} \right) + 1} \right)}}{{\Gamma \left( {\left( {\alpha  + s} \right)j + 1} \right)}}{c_{j - 1}} = \]
\[ = \frac{{{{\left( { - {\lambda _n}{{\left( {\alpha  + s} \right)}^s}} \right)}^j}{{\left( {\frac{1}{{\alpha  + s}}} \right)}_j}...{{\left( {\frac{s}{{\alpha  + s}}} \right)}_j}}}{{\Gamma \left( {\left( {\alpha  + s} \right)j + 1} \right)}}{c_0} = \]
\[ = \frac{{{{\left( { - {\lambda _n}{{\left( {\alpha  + s} \right)}^s}} \right)}^j}\Gamma \left( {j + \frac{1}{{\alpha  + s}}} \right)...\Gamma \left( {j + \frac{s}{{\alpha  + s}}} \right)}}{{\Gamma \left( {\left( {\alpha  + s} \right)j + 1} \right)}},\]
\[{c_0} = \Gamma \left( {\frac{1}{{\alpha  + s}}} \right)...\Gamma \left( {\frac{s}{{\alpha  + s}}} \right).\]
Hence, the solution to problem (9) has the form
\[{X_n}\left( x \right) = \frac{{{\varphi _n}}}{{\Gamma \left( {\frac{1}{{\alpha  + s}}} \right)...\Gamma \left( {\frac{s}{{\alpha  + s}}} \right)}}\sum\limits_{j = 0}^\infty  {\frac{{\Gamma \left( {j + \frac{1}{{\alpha  + s}}} \right)...\Gamma \left( {j + \frac{s}{{\alpha  + s}}} \right)\Gamma \left( {j + 1} \right)}}{{\Gamma \left( {\left( {\alpha  + s} \right)j + 1} \right)}}} \frac{{{{\left( { - {\lambda _n}{{\left( {\alpha  + s} \right)}^s}{x^{\alpha  + s}}} \right)}^j}}}{{j!}}.\]
This representation implies the uniqueness of the solution to problem (9).\\
Because $\alpha  + s + 1 > s + 1$, then the last series converges absolutely and uniformly for fixed values of ${\lambda _n}$ and for bounded values of $ x $ [see 15]. This means that the permutation of the series and the integral in above was legal. \\
In terms of special functions, the solution to problem (9) can be written in the form
\[{X_n}\left( x \right) = \frac{{{\varphi _n}}}{{\Gamma \left( {\frac{1}{{\alpha  + s}}} \right)...\Gamma \left( {\frac{s}{{\alpha  + s}}} \right)}}{}_{s + 1}{\Psi _1}\left( {\begin{array}{*{20}{c}}
{\left( {1,\frac{1}{{\alpha  + s}}} \right),...,\left( {1,\frac{s}{{\alpha  + s}}} \right),\left( {1,1} \right); - {\lambda _n}{{\left( {\alpha  + s} \right)}^s}{x^{\alpha  + s}}}\\
{\left( {\alpha  + s,1} \right)}
\end{array}} \right),\]
where
\[{}_p{\Psi _q}\left( {\begin{array}{*{20}{c}}
{\left( {{\alpha _1},{a_1}} \right),...,\left( {{\alpha _p},{a_p}} \right),z}\\
{\left( {{\beta _1},{b_1}} \right),...,\left( {{\beta _q},{b_p}} \right)}
\end{array}} \right) = \sum\limits_{m = 0}^\infty  {\frac{{\prod\limits_{i = 1}^p {\Gamma \left( {{\alpha _i}m + {a_i}} \right)} }}{{\prod\limits_{i = 1}^q {\Gamma \left( {{\beta _i}m + {b_i}} \right)} }}} \frac{{{z^m}}}{{m!}}\]
- is the generalized Wright function [see 15]. \\
Using the results obtained in [15, Theorem 4.], we obtain an asymptotic expansion of the generalized
Wright function for large values ${{\lambda _n}}$ and $x > \delta  > 0$:
\[{}_{s + 1}{\Psi _1}\left( {\begin{array}{*{20}{c}}
{\left( {1,\frac{1}{{\alpha  + s}}} \right),...,\left( {1,\frac{s}{{\alpha  + s}}} \right),\left( {1,1} \right); - {\lambda _n}{{\left( {\alpha  + s} \right)}^s}{x^{\alpha  + s}}}\\
{\left( {\alpha  + s,1} \right)}
\end{array}} \right) \sim \]
\[ \sim {H_{s + 1,1}}\left( { - {\lambda _n}{{\left( {\alpha  + s} \right)}^s}{x^{\alpha  + s}}} \right) = {H_{s + 1,1}}\left( t \right) = \]
\[ = \sum\limits_{m = 1}^s {{t^{ - \frac{m}{{\alpha  + s}}}}} {S_{s + 1,1}}\left( {t;m} \right) + {t^{ - 1}}{S_{s + 1,1}}\left( {t;s + 1} \right),\]
here
\[{S_{s + 1,1}}\left( {t;m} \right) = \sum\limits_{k = 0}^\infty  {\frac{{{{\left( { - 1} \right)}^k}}}{{k!}}\Gamma \left( {k + \frac{m}{{\alpha  + s}}} \right)} \Gamma \left( {1 - k - \frac{m}{{\alpha  + s}}} \right)\frac{{\prod\limits_{r = 1,r \ne m}^s {\Gamma \left( {\frac{r}{{\alpha  + s}} - k - \frac{m}{{\alpha  + s}}} \right)} }}{{\Gamma \left( {1 - \left( {\alpha  + s} \right)\left( {k + \frac{m}{{\alpha  + s}}} \right)} \right)}}{t^{ - k}} = \]
\[ = \sum\limits_{k = 0}^\infty  {\frac{{{{\left( { - 1} \right)}^k}}}{{k!}}\frac{\pi }{{\sin \pi \left( {k + \frac{m}{{\alpha  + s}}} \right)}}} \frac{{\prod\limits_{r = 1,r \ne m}^s {\frac{{ - \pi }}{{\Gamma \left( {k + \frac{m}{{\alpha  + s}} - \frac{r}{{\alpha  + s}} + 1} \right)\sin \pi \left( {k + \frac{m}{{\alpha  + s}} - \frac{r}{{\alpha  + s}}} \right)}}} }}{{\frac{{ - \pi }}{{\Gamma \left( {\left( {\alpha  + s} \right)k + m} \right)\sin \pi \left( {\left( {\alpha  + s} \right)k + m - 1} \right)}}}}{t^{ - k}} = \]
\[ = \frac{{{{\left( { - 1} \right)}^{m + s + 1}}{\pi ^{s - 1}}}}{{\sin \frac{{m\pi }}{{\alpha  + s}}}}\sum\limits_{k = 0}^\infty  {\frac{{{{\left( { - 1} \right)}^{k\left( {s - 1} \right)}}}}{{k!}}\frac{{\Gamma \left( {\left( {\alpha  + s} \right)k + m} \right)\sin \pi k\left( {\alpha  + s} \right)}}{{\prod\limits_{r = 1,r \ne m}^s {\Gamma \left( {k + \frac{m}{{\alpha  + s}} - \frac{r}{{\alpha  + s}} + 1} \right)\sin \pi \left( {\frac{m}{{\alpha  + s}} - \frac{r}{{\alpha  + s}}} \right)} }}} {t^{ - k}},\]
\[{S_{s + 1,1}}\left( {t;s + 1} \right) = \sum\limits_{k = 0}^\infty  {\frac{{{{\left( { - 1} \right)}^k}}}{{k!}}\Gamma \left( {k + 1} \right)\frac{{\prod\limits_{r = 1}^s {\Gamma \left( {\frac{r}{{\alpha  + s}} - k - 1} \right)} }}{{\Gamma \left( {1 - \left( {\alpha  + s} \right)\left( {k + 1} \right)} \right)}}} {t^{ - k}} = \]
\[ = \sum\limits_{k = 0}^\infty  {\frac{{{{\left( { - 1} \right)}^k}}}{1}\frac{{\prod\limits_{r = 1}^s {\frac{{ - \pi }}{{\Gamma \left( {k + 2 - \frac{r}{{\alpha  + s}}} \right)\sin \pi \left( { - \frac{r}{{\alpha  + s}} + k + 1} \right)}}} }}{{\frac{\pi }{{\Gamma \left( {\left( {\alpha  + s} \right)\left( {k + 1} \right) + 2} \right)\sin \pi \left( {\alpha  + s} \right)\left( {k + 1} \right)}}}}} {t^{ - k}} = \]
\[ = {\pi ^{s - 1}}{\left( { - 1} \right)^s}\sum\limits_{k = 0}^\infty  {{{\left( { - 1} \right)}^{k\left( {1 + s} \right)}}\frac{{\Gamma \left( {\left( {\alpha  + s} \right)\left( {k + 1} \right) + 2} \right)\sin \pi \left( {\alpha  + s} \right)\left( {k + 1} \right)}}{{\prod\limits_{r = 1}^s {\Gamma \left( {k + 2 - \frac{r}{{\alpha  + s}}} \right)\sin \frac{{\pi r}}{{\alpha  + s}}} }}} {t^{ - k}}.\]


Taking into account the latter, we have that there exists a number $ T $ such that for all $ n> T $ and for $x > \delta  > 0$ the estimate
\[\left| {{X_n}\left( x \right)} \right| \le \frac{M}{{{\lambda _n}}}\left| {{\varphi _n}} \right|{x^{\alpha  - 1}},\,\,0 < M - const.\]
So, the formal solution of the posed problem A has the form
\[u\left( {x,y} \right) = \sum\limits_{n = 0}^\infty  {{X_n}\left( x \right){Y_n}\left( y \right)}.\eqno(10)\]
Let us show that (10) is a classical solution to Eq. (1); indeed ( for $x > \delta  > 0$ )
\[\left| {{}_CD_{0x}^\alpha u\left( {x,y} \right)} \right| \le \sum\limits_{n = 0}^\infty  {\left| {{}_CD_{0x}^\alpha {X_n}\left( x \right)} \right|\left| {{Y_n}\left( y \right)} \right|}  = \sum\limits_{n = 0}^\infty  {\left| {{\lambda _n}{x^s}{X_n}\left( x \right)} \right|\left| {{Y_n}\left( y \right)} \right|}  \le \]
\[ \le {M_1}\sum\limits_{n = 0}^\infty  {\left| {{\varphi _n}} \right|} \left| {{Y_n}\left( y \right)} \right|,\,\,0 < {M_1} - const.\]
Next, we apply the Cauchy-Bunyakovsky inequality
 \[\sum\limits_{n = 0}^\infty  {\left| {{\varphi _n}} \right|\left| {{Y_n}\left( y \right)} \right|}  = \sum\limits_{n = 0}^\infty  {\left| {\lambda _n{\varphi _n}} \right|\left| {\frac{{{Y_n}\left( y \right)}}{{{\lambda _n}}}} \right|}  \le \sqrt {\sum\limits_{n = 1}^\infty  {\left| {\lambda _n^2\varphi _n^2} \right|} } \,\,\sqrt {\sum\limits_{n = 0}^\infty  {\left| {\frac{{Y_n^2\left( y \right)}}{{\lambda _n^2}}} \right|} } ,\]
 we have
\[{\varphi _n} = \int\limits_0^1 {K\left( y \right)\varphi \left( y \right){Y_n}\left( y \right)dy}  = \]
 $$= \frac{{{{\left( { - 1} \right)}^k}}}{{{\lambda _n}}}\int\limits_0^1 {\varphi \left( y \right)Y_n^{\left( {2k} \right)}(y)dy}  =$$
$$ = \frac{{{{\left( { - 1} \right)}^k}}}{{{\lambda _n}}}\int\limits_0^1 {{\varphi ^{\left( {2k} \right)}}\left( y \right)Y_n^{}(y)dy}=$$
\[ = \frac{{{{\left( { - 1} \right)}^k}}}{{{\lambda _n}}}\int\limits_0^1 {K\left( y \right)\frac{{{\varphi ^{\left( {2k} \right)}}\left( y \right)}}{{K\left( y \right)}}{Y_n}\left( y \right)dy} .\]
We apply Bessel's inequality
\[\sum\limits_{n = 0}^\infty  {\lambda _n^2{{\left| {{\varphi _n}\left( y \right)} \right|}^2}}  \le \int\limits_0^1 {K\left( y \right){{\left( {\frac{{{\varphi ^{\left( {2k} \right)}}\left( y \right)}}{{K\left( y \right)}}} \right)}^2}dy = } \int\limits_0^1 {{{\left( {\frac{{{\varphi ^{\left( {2k} \right)}}\left( y \right)}}{{\sqrt {K\left( y \right)} }}} \right)}^2}dy < \infty .}\eqno(11)\]
Taking into account (8) and (11), we obtain that the series
\[{}_CD_{0x}^\alpha u\left( {x,y} \right) = \sum\limits_{n = 0}^\infty  {{}_CD_{0x}^\alpha {X_n}\left( x \right){Y_n}\left( y \right)}  - \]
converges uniformly.
Now, in order for the calculations made above to be legal, we impose the following restrictions on the  $\varphi \left( y \right)$  function.So, the theorem is valid.
\begin{theorem}
Let the function $ \varphi \left( y \right)$   , satisfy the following conditions:
\[\frac{{{\varphi ^{\left( {2k} \right)}}\left( \xi  \right)}}{{\sqrt {K\left( \xi  \right)} }} \in C\left[ {0;1} \right],\,{\varphi ^{\left( s \right)}}\left( 0 \right) = {\varphi ^{\left( s \right)}}\left( 1 \right) = 0,\]
\[s = 0,1,...,k - 1,\]
then a solution to Problem A exists.
 \end{theorem}
The uniform convergence of the series
\[\frac{{{\partial ^{2k}}u\left( {x,y} \right)}}{{\partial {y^{2k}}}} = \sum\limits_{n = 0}^\infty  {{X_n}\left( x \right)\frac{{{\partial ^{2k}}{Y_n}\left( y \right)}}{{\partial {y^{2k}}}}}  = {\left( { - 1} \right)^k}{K(y)}\sum\limits_{n = 0}^\infty  {{\lambda _n}{X_n}\left( x \right){Y_n}\left( y \right)} .\]
\begin{center}
\textbf{3.Uniqueness
}\end{center}
Let the function $u\left( {x,y} \right)$ be a solution to Problem A with zero initial and boundary conditions. Consider its Fourier coefficients with respect to the system of eigenfunctions of problem (5)
\[{u_n}\left( x \right) = \int\limits_0^1 {K\left( y \right)u\left( {x,y} \right){Y_n}\left( y \right)dy} \]
it is easy to show that  ${u_n}\left( x \right)$   is a solution to the problem
\[\left\{ \begin{array}{l}
{}_CD_{0x}^\alpha {u_n}\left( x \right) =  - {\lambda _n}{x^s}{u_n}\left( x \right),\\
{u_n}\left( 0 \right) = 0.
\end{array} \right.\]
This problem has only a zero solution, i.e.
\[\int\limits_0^1 {K\left( y \right)u\left( {x,y} \right){Y_n}\left( y \right)dy}  = 0,\,\forall n.\]
Because\\

$\overline G \left( {y,\xi } \right) - $ symmetric, continuous,\\
\[\int\limits_0^1 {{{\overline G }^2}\left( {y,\xi } \right)d\xi }  < \infty ,\,\int\limits_0^1 {{{\overline G }^2}\left( {y,\xi } \right)dy}  < \infty ,\,\int\limits_0^1 {\int\limits_0^1 {{{\overline G }^2}\left( {y,\xi } \right)dyd\xi } }  < \infty ,\,{\lambda _n} > 0,\,\forall n,\]
then the conditions of Mercer's theorem are fulfilled and hence
\[\overline G \left( {y,\xi } \right) = \sum\limits_{n = 0}^\infty  {\frac{{\overline {{Y_n}} \left( y \right)\overline {{Y_n}} \left( \xi  \right)}}{{{\lambda _n}}}} .\]
Hence we have
\[\sqrt {K(y)} u\left( {x,y} \right) = \int\limits_0^1 {\overline G \left( {y,\xi } \right)\left( {{{\left( { - 1} \right)}^k}\frac{1}{{\sqrt {K\left( \xi  \right)} }}\frac{{{\partial ^{2k}}u\left( {x,\xi } \right)}}{{\partial {\xi ^{2k}}}}} \right)d\xi }  = \]
\[ = \int\limits_0^1 {\sum\limits_{n = 0}^\infty  {\frac{{{{\overline Y }_n}\left( y \right){{\overline Y }_n}\left( \xi  \right)}}{{{\lambda _n}}}} \left( {{{\left( { - 1} \right)}^k}\frac{1}{{\sqrt {K\left( \xi  \right)} }}\frac{{{\partial ^{2k}}u\left( {x,\xi } \right)}}{{\partial {\xi ^{2k}}}}} \right)d\xi }  = \]
\[ = \sum\limits_{n = 0}^\infty  {\frac{{\sqrt {K(y)} {Y_n}\left( y \right)}}{{{\lambda _n}}}} \int\limits_0^1 {\left( {{{\left( { - 1} \right)}^k}\frac{{\sqrt {K\left( \xi  \right)} {Y_n}\left( \xi  \right)}}{{\sqrt {K\left( \xi  \right)} }}\frac{{{\partial ^{2k}}u\left( {x,\xi } \right)}}{{\partial {\xi ^{2k}}}}} \right)d\xi }  = \]
\[ = \sum\limits_{n = 0}^\infty  {\frac{{\sqrt {K(y)} {Y_n}\left( y \right)}}{{{\lambda _n}}}} \int\limits_0^1 {\left( {{{\left( { - 1} \right)}^k}{Y_n}\left( \xi  \right)\frac{{{\partial ^{2k}}u\left( {x,\xi } \right)}}{{\partial {\xi ^{2k}}}}} \right)d\xi }  = \]
\[ = \sum\limits_{n = 0}^\infty  {\frac{{\sqrt {K(y)} {Y_n}\left( y \right)}}{{{\lambda _n}}}} \int\limits_0^1 {\left( {{{\left( { - 1} \right)}^k}Y{{_n^{(2k)}}_n}\left( \xi  \right)u\left( {x,\xi } \right)} \right)d\xi }  = \]
\[ = \sum\limits_{n = 0}^\infty  {\frac{{\sqrt {K(y)} {Y_n}\left( y \right)}}{{{\lambda _n}}}} \int\limits_0^1 {\left( {{\lambda _n}K\left( \xi  \right){Y_n}\left( \xi  \right)u\left( {x,\xi } \right)} \right)d\xi }  = 0.\]



\bigskip
\begin{thebibliography}{1}




\bibitem{1} A. M. Nakhushev ,
{\it  Drobniye ischisleniya i yego primenenie. [Fractional Calculus and its Applications]}, Moscow: FIZMATLIT, 2003.

\bibitem{2} A. V. Pskhu ,
{\it  Uravneniya v chastnykh proizvodnykh drobnogo poryadka. [Partial differential equations of fractional order]}, Moscow: NAUKA, 2005.

\bibitem{3} M.O.Mamchuev ,
{\it On the Well-Posedness of Boundary Value Problems for a Fractional Diffusion-Wave Equation and One Approach to Solving Them}, TDiff Equat , \textbf{56}:4 (2020), 756--760.https://doi.org/10.1134/S0012266120060087.

\bibitem{4} A. V. Pskhu ,
{\it The fundamental solution of a diffusion-wave equation of fractional order}, Izv. Math., \textbf{73}:2 (2009), 351--392.

\bibitem{5} A. V. Pskhu ,
{\it Solution of a Boundary Value Problem for a Fractional Partial Differential Equation},Differ. Equ., \textbf{39}:8 (2003), 1150--1158.

\bibitem{6} O. P. Agrawal  ,
{\it Solution for a fractional diffusion-wave equation defined in a bounded domain},Nonlinear Dynam., \textbf{29}:1 (2002), 145--155.

\bibitem{7}  S. Kh. Gekkieva,  M. A. Kerefov   ,
{\it Dirichlet boundary value problem for Aller–Lykov moisture transfer equation with fractional derivative in time},Ufa Math. J., \textbf{11}:2 (2019), 71--81.

\bibitem{8}  F. Mainardi ,
{\it The fundamental solutions for the fractional diffusion-wave equation},Appl. Math. Lett., \textbf{9}:6 (1996), 23--28.

\bibitem{9}  O.Kh. Masaeva ,
{\it Uniqueness of solutions to Dirichlet problems for generalized Lavrentyev – Bitsadze equations with a fractional derivative},Electron. J. Differential Eq., \textbf{2017}:74 (2017), 1--8.

\bibitem{10}  B. J. Kadirkulov ,
{\it Boundary problems for mixed parabolic-hyperbolic equations with two lines of changing type and fractional derivative},Electron. J. Differential Eq., \textbf{2014}:57 (2014), 1--7.

\bibitem{11}  A.S. Berdyshev , A.Cabada, B. J.Kadirkulov ,
{\it The Samarskii–Ionkin type problem for the fourth order parabolic equation with fractional differential operator },Computers and Mathematics with Applications, \textbf{62}:10 (2011), 3884--3893.

\bibitem{12}  A. S.Berdyshev , B. E.Eshmatov , B. J. Kadirkulov  ,
{\it Boundary value problems for fourth-order mixed type equation with fractional derivative },Electronic Journal of Differential Equations, \textbf{2016}:36 (2016), 3884--3893.

\bibitem{13}  А. N.  Аrtyushin  ,
{\it Fractional integral inequalities and their applications to degenerate differential
equations with the caputo fractional derivative },Siberian Mathematical Journal, \textbf{61}:2 (2020), 3884--3893.

\bibitem{14}  Yu. P. Apakov , B. Yu. Irgashev  ,
{\it Boundary-value problem for a degenerate high-odd-order equation },Ukrainian Mathematical Journal, \textbf{66}:10 (2015), 868--893.

\bibitem{15}  R.B. Paris ,
{\it Some remarks on the theorems of Wright and Braaksma on the Wright function ${}_p{\Psi _q}$. },rXiv:1708.04824v1 [math.CA],Aug 2017.

\end{thebibliography}

\end{document}
