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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2017 (2017), No. 74, pp. 1--8.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu}
\thanks{\copyright 2017 Texas State University.}
\vspace{8mm}}

\begin{document}
\title[\hfilneg EJDE-2017/74\hfil 
 Lavrent'ev-Bitsadze equations with a fractional derivative]
{Uniqueness of solutions to Dirichlet problems
for generalized Lavrent'ev-Bitsadze equations with a fractional derivative}

\author[O. Kh. Masaeva \hfil EJDE-2017/74\hfilneg]
{Olesya Kh. Masaeva}

\address{Olesya Kh. Masaeva \newline
Institute of Applied Mathematics and Automation,
 Nalchik, Russia}
\email{olesya.masaeva@yandex.ru}

\dedicatory{Communicated by Mokhtar Kirane}

\thanks{Submitted October 26, 2016. Published March 17, 2017.}
\subjclass[2010]{35A02, 35A25, 35k10}
\keywords{Mittag-Leffler type function; differential equation of fractional order;
\hfill\break\indent Dirichlet problem; mixed type equation}

\begin{abstract}
 In this article we study the uniqueness of the solution of the Dirichlet problem
 for an equation of Lavrent'ev-Bitsadze type with a fractional derivative.
 The equation studied becomes the regular Lavrent'ev-Bitsadze equation when the
 order of the derivative is an integer.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\allowdisplaybreaks

\section{Introduction}

We consider the equation
\begin{equation}
\label{f31}
Lu\equiv\frac{\partial^2 u}{\partial x^2}-D_{0y}^{\gamma}
\frac{\partial u}{\partial y}=0, \quad 0<\gamma<1,
\end{equation}
in the domain  $\Omega = \{(x,y):0<x<r, \alpha<y<\beta \}$, $\alpha<0,\beta>0$, 
where $D_{0y}^{\gamma}$ is the Riemann-Liouville differential operator of order 
$\gamma$ \cite[p. 37]{S}.

 In  \cite{dy2012,dy2013} the Dirichlet problem  for second order partial 
differential equations with a Caputo derivative has been studied. 
The equations become the Laplace equation and a vibrating string equation
 when the order of  differentiation in the equation is an integer.
The Dirichlet problem for the Lavrent'ev-Bitsadze equation has been studied 
in \cite{Can,Col1}.

Here with the  $abc$ method a uniqueness of the solution to the Dirichlet 
problem is proved for  equation \eqref{f31} in the domain $\Omega$.
Uniqueness conditions for the  solution of the problem has been found in terms of
the upper limits for the zeros of a Mittag-Leffler type function.

Let us set  $\Omega^- =\Omega\cap\{y<0 \}$, $\Omega^+=\Omega\cap\{y>0 \}$. 
Let the function  $u=u(x,y)$ be such that  $u \in C^1(\bar \Omega)$, 
$u_{xy} \in C(\bar\Omega^+)$,
$u_{xx}, D_{0y}^{\gamma}u_y \in C(\Omega^-\cup\Omega^+)$, satisfying   
\eqref{f31} at all points $ (x,y) \in \Omega^-\cup\Omega^+$ be a regular 
solution of equation  \eqref{f31} in the domain  $\Omega$.


\section{Dirichlet problem}

We try to find a regular solution to  \eqref{f31}, satisfying the conditions
\begin{gather}\label{f67}
u(0,y)=\psi_0(y), \quad u(r,y)=\psi_r(y), \quad \alpha <y < \beta, \\
\label{f68}
u(x,\alpha)=\varphi_{\alpha}(x),\quad u(x,\beta)=\varphi_{\beta}(x), \quad 0< x< r.
\end{gather}
where  $\psi_0(y)$,  $\psi_r(y)$,  $\varphi_{\alpha}(x)$, $\varphi_{\beta}(x)$ 
are given functions.
We consider the  Mittag-Leffler type function
 \begin{equation}\label{mm7}
 E_{\rho,\, \mu} \left(z \right)=\sum_{k=0}^{\infty} \frac{z^k}{\Gamma(\rho k+\mu)}, \quad \rho>0, \,\mu \in \mathbb{C}.
 \end{equation}
It is known that this function  can have only a finite number of real zeros
for all $\rho<2$, $\mu \in \mathbb{C}$ \cite[p. 372]{Po}.
Also it is known that the set of real zeros of \eqref{mm7}
is not empty for  $1<\rho<2$,  $\mu=\rho$ and $\mu=1$; see \cite{P}.

\section{Uniqueness}

\begin{theorem} \label{thm3.1}
Let  $t_1=\max \{t\in\mathbb{R}:E_{\nu,\nu}(-t)=0\}$,
 $ t_2=\max \{t\in\mathbb{R}:E_{\nu,1}(-t)=0\}$, 
$\nu=\gamma+1$, $h=\max \{t_1, t_2\}$  and
\begin{equation}\label{f41}
\frac{\beta^\nu}{r^2} \geq \frac{h}{\pi^2}.
\end{equation}
Then the homogeneous Dirichlet problem \eqref{f31}, \eqref{f67}, \eqref{f68} 
has only the trivial solution.
 \end{theorem}


\begin{proof}
 First, we consider equation \eqref{f31} in  $\Omega^-$.
According to the definition of the Riemann-Liouville fractional derivative  in 
 $\Omega^-$  equation  \eqref{f31}   can be written as
\begin{equation}\label{fg2}
Lu=u_{xx}+\frac{\partial}{\partial y}D_{0y}^{\gamma-1}
\frac{\partial}{\partial y}u=0.
\end{equation}
Let 
$$ 
\omega^-(y)=(y-\alpha)^{ \gamma} E_{ \gamma+1, \gamma+1}
\left( \lambda_n(y-\alpha)^{ \gamma+1}\right), \quad
\lambda_n= \big( \frac{\pi n}{r}\big)^2, 
$$ 
be the solution to the Cauchy problem
\begin{gather*}
D_{\alpha y}^{\gamma}\frac{d w^-(y)}{dy}+\lambda_nw^-(y)=0,\\
\lim _{y \to \alpha} D_{\alpha y}^{\gamma-1}\frac{d w^-(y)}{dy}=1,\quad
 w^-(\alpha)=0.
\end{gather*}
We multiply  \eqref{fg2} by   $v(x,y)=\omega^-(y)\sin\left(\sqrt{\lambda_n}x\right)$ 
and rewrite it as 
$$
vLu=vu_{xx}+v\frac{\partial}{\partial y}D_{0y}^{\gamma-1}
 \frac{\partial}{\partial y}u
 =(vu_x-uv_x)_x+uv_{xx}+\big(vD_{0y}^{\gamma-1}u_y\big)_y-v_yD_{0y}^{\gamma-1}u_y.
$$
Then we consider the integral
 \begin{equation}\label{fg3}
\begin{aligned}
&\int_{\varepsilon}^{r-\varepsilon}\int_{\alpha+\varepsilon}^{-\varepsilon}vLu\,dx\,dy\\
&=\int_{\alpha+\varepsilon}^{-\varepsilon}(vu_x-uv_x)
 \big|_{x=\varepsilon}^{x=r-\varepsilon}dy
 +\int_{\varepsilon}^{r-\varepsilon}\int_{\alpha+\varepsilon}^{-\varepsilon}uv_{xx}
 \,dx\,dy\\
&\quad +\int_{\varepsilon}^{r-\varepsilon}[vD_{0y}^{\gamma-1}u_y]
 \big|_{y=\alpha+\varepsilon}^{y=-\varepsilon}dx
 -\int_{\varepsilon}^{r-\varepsilon}\int_{\alpha+\varepsilon}^{-\varepsilon}
v_yD_{0y}^{\gamma-1}u_y\,dx\,dy,
\end{aligned}
 \end{equation}
 where   $\varepsilon>0$.

Since $u(x,y)\in C^1(\bar\Omega)$, we have $D_{0y}^{\gamma-1}u_y \in C(\bar\Omega)$. 
Therefore,  in  \eqref{fg3} we can make  $\varepsilon$ tend to zero
\begin{equation}\label{g1}
\begin{aligned}
0&=\int_{\alpha}^{0}[v(r,y)u_x(r,y)-u(r,y)v_x(r,y)]dy \\
&\quad -\int_{\alpha}^{0}[v(0,y)u_x(0,y)-u(0,y)v_x(0,y)]dy
 + \int_0^r\int_{\alpha}^{0}uv_{xx}\,dx\,dy\\
&\quad +\int_0^r\left(v(x,0)[D_{0y}^{\gamma-1}u_y]_{y=0}
 -v(x,\alpha)[D_{0y}^{\gamma-1}u_y]_{y=\alpha}\right)\,dx\\
&\quad -\int_0^r\int_{\alpha}^{0}v_yD_{0y}^{\gamma-1}u_y\,dx\,dy.
\end{aligned}
\end{equation}
Since $v|_{{\{x=0\}\cup\{x=r\}\cup\{y=\alpha\}}}=0$ and 
$u|_{{\{x=0\}\cup\{x=r\}}}=0$ from \eqref{g1} we obtain 
$$
0=\int_0^r\int_{\alpha}^{0}uv_{xx}\,dx\,dy
+\int_0^rv(x,0)[D_{0y}^{\gamma-1}u_y]_{y=0}dx
-\int_0^r\int_{\alpha}^{0}v_yD_{0y}^{\gamma-1}u_ydx\,dy.
$$
 Applying here the formula of fractional integration by parts
\cite[p. 34]{S},
\begin{equation}\label{p}
\int_c^d h(t)D_{dt}^{\delta}g(t)dt
=\int_c^d g(t)D_{ct}^{\delta}h(t)dt,\quad  \delta \leq 0,
\end{equation}
we can obtain
\begin{equation}\label{g13}
0=\int_0^r\int_{\alpha}^{0}uv_{xx}\,dx\,dy
+\int_0^rv(x,0)[D_{0y}^{\gamma-1}u_y]_{y=0}dx
-\int_0^r\int_{\alpha}^{0}u_yD_{\alpha y}^{\gamma-1}v_ydx\,dy.
\end{equation}
We substitute the expression  
$$
u_yD_{\alpha y}^{\gamma-1}v_y
=(uD_{\alpha y}^{\gamma-1}v_y)_y-u\frac{\partial}{\partial y}
D_{\alpha y}^{\gamma-1}v_y
$$ 
in \eqref{g13}, then we have
\begin{align*}
0&=\int_0^r\int_{\alpha}^{0}u(v_{xx}+D_{\alpha y}^{\gamma}v_y)\,dx\,dy
+\int_0^rv(x,0)[D_{0y}^{\gamma-1}u_y]_{y=0}dx \\
&\quad -\int_0^ru(x,0)[D_{\alpha y}^{\gamma-1}v_y]_{y=0}dx
 +\int_0^ru(x,\alpha)[D_{\alpha y}^{\gamma-1}v_y]_{y=\alpha}dx.
\end{align*}
Hence, as $u(x,\alpha)=0$ and $v_{xx}+D_{\alpha y}^{\gamma}v_y=0$,  we have
\begin{equation}\label{s41}
\int_0^rv(x,0)[D_{0y}^{\gamma-1}u_y]_{y=0}dx
-\int_0^ru(x,0)[D_{\alpha y}^{\gamma-1}v_y]_{y=0}dx=0.
\end{equation}

In  $\Omega^+$ we have
\begin{equation}\label{fg4}
Lu=u_{xx}-\frac{\partial}{\partial y}D_{0y}^{\gamma-1}\frac{\partial}{\partial y}u.
\end{equation}
Denote by $\omega^+(y)=(\beta-y)^{ \gamma} E_{\gamma+1,\gamma+1}
\left(- \lambda_n(\beta-y)^{\gamma+1}\right)$ the solution of the Cauchy problem
\begin{gather*}
D_{\beta y}^{\gamma}\frac{d w^+(y)}{dy}-\lambda_nw^+(y)=0, \\
\lim _{y \to \beta} D_{\beta y}^{\gamma-1}\frac{d w^+(y)}{dy}=-1,\quad
w^+(\beta)=0.
\end{gather*}
Multiplying \eqref{fg4} by  $v(x,y)=\omega^+(y)\sin\left(\sqrt{\lambda_n}x\right)$, 
we obtain 
$$
vLu=(vu_x-uv_x)_x+uv_{xx}-\big(vD_{0y}^{\gamma-1}u_y\big)_y
+v_yD_{0y}^{\gamma-1}u_y.
$$
Then
\begin{align*}
&\int_{\varepsilon}^{r-\varepsilon}\int_{\varepsilon}^{\beta-\varepsilon}v\,Lu\,dx\,dy\\
&=\int_{\varepsilon}^{\beta-\varepsilon}(vu_x-uv_x)
 \bigl|_{x=\varepsilon}^{x=r-\varepsilon}dy 
 +\int_{\varepsilon}^{r-\varepsilon}\int_{\varepsilon}^{\beta-\varepsilon}uv_{xx}
 \,dx\,dy \\
&\quad -\int_{\varepsilon}^{r-\varepsilon}(vD_{0y}^{\gamma-1}u_y)
 \bigr|_{y=\varepsilon}^{y=\beta-\varepsilon}dx 
 +\int_{\varepsilon}^{r-\varepsilon}\int_{\varepsilon}^{\beta-\varepsilon}
 v_yD_{0y}^{\gamma-1}u_y\,dx\,dy.
\end{align*}
Making $\varepsilon$ tend to zero, then in view of $u(0,y)=u(r,y)=0$,
 $v(0,y)=v(r,y)=0$, $v(x,\beta)=0$, and formula \eqref{p}, we have
\begin{equation}\label{g14}
0=\int_0^r\int_0^{\beta}uv_{xx}\,dx\,dy+\int_0^rv(x,0)
 [D_{0y}^{\gamma-1}u_y]_{y=0}dx
+\int_0^r\int_0^{\beta}u_yD_{\beta y}^{\gamma-1}v_y\,dx\,dy.
\end{equation}
Taking into account the equality 
$u_yD_{\beta y}^{\gamma-1}v_y=(uD_{\beta y}^{\gamma-1}v_y)_y-u 
\frac {\partial}{\partial y}D_{\beta y}^{\gamma-1}v_y$, and using \eqref{g14}, 
we  obtain
\begin{align*}
&\int_0^r \int_0^{\beta}uv_{xx}\,dx\,dy
 +\int_0^rv(x,0)[D_{0y}^{\gamma-1}u_y]_{y=0}dx
 +\int_0^ru(x,\beta)[D_{\beta y}^{\gamma-1}v_y]_{y=\beta}dx \\
&-\int_0^ru(x,0)[D_{\beta y}^{\gamma-1}v_y]_{y=0}dx
 -\int_0^r \int_0^{\beta}u \frac {\partial}{\partial y}D_{\beta y}^{\gamma-1}
 v_y\,dx\,dy=0.
\end{align*}
Hence, $D_{\beta y}^{\gamma}v_y=-\frac {\partial}{\partial y}
D_{\beta y}^{\gamma-1}v_y$, $ u(x,\beta)=0$,
which lead us to
 \begin{equation}\label{fg5}
\begin{aligned}
&\int_0^r\int_0^{\beta}u(v_{xx}+ D_{\beta y}^{\gamma}v_y)\,dx\,dy
 +\int_0^rv(x,0) [D_{0y}^{\gamma-1}u_y]_{y=0}dx \\
&-\int_0^ru(x, 0)[D_{\beta y}^{\gamma-1}v_y]_{y=0}dx=0.
\end{aligned}
\end{equation}
Note that  $ v_{xx}+ D_{\beta y}^{\gamma}v_y=0$.
  Therefore, using \eqref{fg5},  we  obtain
\begin{equation}\label{3s4}
\int_0^rv(x,0)[D_{0y}^{\gamma-1}u_y]_{y=0}dx
 -\int_0^ru(x, 0)[D_{\beta y}^{\gamma-1}v_y]_{y=0}dx=0.
 \end{equation}

Considering that the function  $u(x,y)$ satisfies the conditions
$$
\lim_{y\to 0-}u(x,y)=\lim_{y\to 0+}u(x,y),\quad 
\lim_{y\to 0-}D_{0y}^{\gamma-1}u_{y}=\lim_{y\to 0+}D_{0y}^{\gamma-1}u_{y},
$$
using   \eqref{s41} and  \eqref{3s4}, one finds the values of the functions 
$u (x, 0)$ and $[D_{0y}^{\gamma-1}u_y]_{y=0}$.

Now let us consider the system of the algebraic equations
\begin{equation} \label{ss2}
\begin{gathered}
\omega^-(0) u_\gamma-[D_{\alpha y}^{\gamma-1}\frac{d}{dy}\omega^-]_{y=0} u_0=0, \\
-[D_{\beta y}^{\gamma-1}\frac{d}{dy}\omega^+]_{y=0}u_0+
\omega^+(0)u_\gamma=0,
\end{gathered}
\end{equation}
where
$$
u_0= \int_0^r u(x,0)\sin(\sqrt{\lambda_n}x)dx,\quad  
u_\gamma=\int_0^r[D_{0 y}^{\gamma-1}u_{y}]_{y=0}\, \sin(\sqrt{\lambda_n}x)dx.
$$
From the definition of the Riemann-Liouville fractional integro-differentiation, 
it follows that
$$
-\frac{d}{dy}w^+=D_{\beta y}^{1}w^+,\quad 
\frac{d}{dy}w^-=D_{\alpha y}^{1}w^-.
$$
Using the formula of fractional integro-differentiation for the Mittag-Leffler 
type function,
$$
D_{st}^\delta|t-s|^{\mu-1}E_{\rho,\,\mu}(\lambda|t-s|^{\rho})
=|t-s|^{\mu-\delta-1}E_{\rho,\,\mu-\delta}(\lambda|t-s|^{\rho}), \quad 
\delta\in\mathbb{R},
$$
$\mu>0$, if $\delta\notin\mathbb{N}\cup\{0\}$, and $\mu\in\mathbb{R}$, 
if $\delta\in\mathbb{N}\cup\{0\}$, then we find that
\begin{gather*}
D_{\beta y}^{\gamma-1}\frac{d}{dy}\omega^+
=- E_{\gamma+1, 1}\left(- \lambda_n(\beta-y)^{ \gamma+1}\right),\\
D_{\alpha y}^{\gamma-1}\frac{d}{dy}\omega^-=E_{\gamma+1,1}
\left( \lambda_n(y-\alpha)^{\gamma+1}\right). 
\end{gather*}
Then the determinant of   \eqref{ss2} has the form  
\begin{align*}
\Delta&=\beta^{\gamma} E_{\gamma+1,\gamma+1 }  
\left(-\lambda_{n}\beta^{\gamma+1} \right)
E_{\gamma+1, 1 }\left(\lambda_{n}|\alpha|^{\gamma+1}\right) \\
&\quad +|\alpha|^{\gamma}E_{\gamma+1, \gamma+1} 
 \left({\lambda_{n}|\alpha|^{\gamma+1}}
 \right)E_{\gamma+1,1 } \left(-\lambda_{n}\beta^{\gamma+1} \right), \quad
n=1,2,\dots 
\end{align*}
Let us  show that  $\Delta \not=0$.
Since
$$
E_{\gamma+1,1}(\lambda_n|\alpha|^{\gamma+1})>0, \quad
E_{\gamma+1,\gamma+1}(\lambda_n |\alpha|^{\gamma+1})>0, 
$$ 
 the existence of  roots of the equation
$\Delta=0$ depends on
 $$
 E_{\gamma+1,\gamma+1}(-\lambda_n{\beta^{\gamma+1}}),\quad 
E_{\gamma+1,1}(-\lambda_n{\beta^{\gamma+1}}).
 $$
Next, we use the asymptotic expansion  \eqref{mm7} at $\rho \in (1,2)$.
As $|z| \to \infty$,  \cite[p. 219]{Po}, the following formula holds
\begin{equation}\label{m7}
E_{\rho,\,\mu}(z)=1/\rho z^{(1-\mu)/\rho} e^{z^{1/\rho}}
-\sum_{k=1}^{m}{z^{-k}}/{\Gamma(\mu-\rho k)}+
O\left({|z|^{-m-1}}\right),
\end{equation}
for $|\arg z|\leq \pi$. 
When $z \in \mathbb{R}$ and $z \to -\infty$,
\begin{equation}\label{m8}
E_{\rho,\,\mu}(z)=-\sum_{k=1}^{m}{z^{-k}}/{\Gamma(\mu-\rho k)}
+O({|z|^{-m-1}}).
\end{equation}
From  \eqref{m7},  we obtain the expansions
\begin{gather*}
E_{\gamma+1,1}(\lambda_n|\alpha|^{\gamma+1})
=\frac{1}{\gamma+1}e^{\lambda_n^{\frac{1}{\gamma+1}}|\alpha|}
+O\left(\lambda_n^{-1}\right), \\
E_{\gamma+1,\gamma+1}(\lambda_n|\alpha|^{\gamma+1})=\frac{1}{\gamma+1}
\lambda_n^{-\frac{\gamma}{\gamma+1}} |\alpha|^{-\gamma}
e^{\lambda_n^{\frac{1}{\gamma+1}} |\alpha|}+
O\left(\lambda_n^{-2}\right). 
\end{gather*}
By  \eqref{m8} at  $m=2$ and $m=1$ we have the representations
\begin{gather*}
E_{\gamma+1, \gamma+1}(-\lambda_n{\beta^{\gamma+1}} )
=-\frac{\beta^{-2\gamma-2}}{\lambda_n^2
\Gamma(-\gamma-1)}+O\left(\lambda_n^{-3} \right),\, \\
E_{\gamma+1,1}(-\lambda_n{\beta^{\gamma+1}})
=\frac{\beta^{-\gamma-1}}{\lambda_n\Gamma(-\gamma)}
+O \left( \lambda_n^{-2} \right).
\end{gather*}
Taking into account $\Gamma(-\gamma)<0$,  $\Gamma(-\gamma-1)>0$, we obtain
\begin{align*}
\lim_{n\to\infty}E_{\gamma+1, \gamma+1}(- \lambda_n\beta^{\gamma+1})E_{\gamma+1,1}
(\lambda_n|\alpha|^{\gamma+1})=-\infty, \\
\lim_{n\to\infty}E_{\gamma+1, \gamma+1}(\lambda_n|\alpha|^{\gamma+1})E_{\gamma+1,1}
(-\lambda_n\beta^{\gamma+1})=-\infty.
\end{align*}
Therefore, 
\begin{align*}
E_{\gamma+1,\gamma+1}(-\lambda_n\beta^{\gamma+1})E_{\gamma+1,1}
(\lambda_n|\alpha|^{\gamma+1})<0,   \quad \lambda_n\beta^{\gamma+1}>t_1,\\
E_{\gamma+1,\gamma+1}(\lambda_n|\alpha|^{\gamma+1})E_{\gamma+1,1}
(-\lambda_n\beta^{\gamma+1})<0,\quad \lambda_n\beta^{\gamma+1}>t_2.
\end{align*}
Next by  \eqref{f41}, for $\lambda_n\beta^{\gamma+1}\geq h$, $ n=1,2,\dots$, 
we have
$\Delta<0$, $n=1,2,\dots$.
Thus, from  \eqref{ss2}, it follows that
\begin{equation}
\label{f5}
u_0=0,\quad u_{\gamma}=0.
\end{equation}
  Since the functions  $\{ \sin (\frac{ \pi n}{r}x) \}$ form a dense system,
using \eqref{f5} we conclude that
\begin{equation}
\label{f6}
[D_{0 y}^{\gamma-1}u_{y}]_{y=0}=0, \quad  u(x,0)=0, \quad  x \in (0,r).
\end{equation}
With this result we prove that $\Omega^-u = 0$.
We have
$$
uLu=(uu_x)_x-u_x^2+(uD_{0y}^{\gamma-1}u_{y})_y-u_yD_{0y}^{\gamma-1}u_{y}, \quad
(x,y) \in \Omega^-.
$$
We consider the integral
\begin{align*}
&\int_{\varepsilon}^{r-\varepsilon}\int_{\alpha+\varepsilon}^{-\varepsilon} 
uLu\,dx\,dy \\
&=-\int_{\varepsilon}^{r-\varepsilon}\int_{\alpha+\varepsilon}^{-\varepsilon}
 (u_x^2+u_yD_{0y}^{\gamma-1}u_{y})\,dx\,dy
+ \int_{\alpha+\varepsilon}^{-\varepsilon}(uu_x)
 \Bigl|_{x=\varepsilon}^{x=r-\varepsilon}dy\\
&\quad +\int_{\varepsilon}^{r-\varepsilon}(uD_{0y}^{\gamma-1}u_{y})
 \Bigl|_{y=\alpha+\varepsilon}^{y=-\varepsilon}\,dx, 
\end{align*}
As  $Lu=0$, we  obtain 
\begin{align*}
&-\int_{\varepsilon}^{r-\varepsilon}\int_{\alpha+\varepsilon}^{-\varepsilon}
(u_x^2+u_yD_{0y}^{\gamma-1}u_{y})\,dx\,dy
 + \int_{\alpha+\varepsilon}^{-\varepsilon}(uu_x)
 \Bigl|_{x=\varepsilon}^{x=r-\varepsilon}dy \\
&\quad +\int_{\varepsilon}^{r-\varepsilon}(uD_{0y}^{\gamma-1}u_{y})
 \Bigl|_{y=\alpha+\varepsilon}^{y=-\varepsilon}\,dx=0, 
\end{align*}
Making  $\varepsilon$ tend to zero we  get
\begin{equation} \label{g3}
\int_0^r\int_{\alpha}^{0}(u_x^2+u_yD_{0y}^
{\gamma-1}u_{y})\,dx\,dy=0.
\end{equation}
Since the fractional integral operator is positive \cite{lit_2},
$$
\int_0^r\int_{\alpha}^{0}u_yD_{0y}^{\gamma-1}u_{y}\,dx\,dy \geq 0, 
$$
then from  \eqref{g3} it follows that $u_x=0$, $u_y=0  $. 
So,   $u={\rm const}$ in  $\Omega^-$. Namely, due to  $u\in C(\bar\Omega)$, 
we can obtain $u(x,y)=0 $ for all $(x,y)\in\Omega^-$.

In  $\Omega^{+}$, we have
\begin{equation}\label{f9}
D_{0y}^{\gamma-1}u_y \cdot Lu=\frac{\partial}{\partial x}
\big(u_xD_{0y}^{\gamma-1}u_y\big)-u_x
\frac{\partial}{\partial x}D_{0y}^{\gamma-1}u_y-{2^{-1}}
\frac{\partial}{\partial y}\big(D_{0y}^{\gamma-1}u_y\big)^2.
\end{equation}
Integrating \eqref{f9}, we  obtain
\begin{equation}\label{f11}
\begin{aligned}
&\int_0^{\beta}u_x(r,y)D_{0y}^{\gamma-1}u_y(r,y)\,dy
 -\int_0^{\beta}u_x(0,y)D_{0y}^{\gamma-1}u_y(0,y)\,dy \\
&-\int_0^r\int_0^{\beta}u_xD_{0y}^{\gamma-1}u_{yx}\,dx\,dy
-{\frac{1}{2}\int_0^r}\big(D_{0y}^{\gamma-1}u_y\big)^2
 \Bigr|_{\,y=\beta}dx \\
&+{\frac{1}{2} \int_0^r}\big(D_{0y}^{\gamma-1}u_y\big)^2\Bigr|_{\,y=0}\,dx=0.
\end{aligned}
\end{equation}
Since  $u(0,y)=u(r,y)=0$, we have 
$D_{0y}^{\gamma-1}u_y(r,y)=D_{0y}^{\gamma-1}u_y(0,y)=0$.
So from  \eqref{f11} it follows that
\begin{equation}\label{f12}
-\int_0^r\int_0^{\beta}u_xD_{0y}^{\gamma-1}u_{yx}\,dx\,dy-{\frac{1}{2}
\int_0^r}\big(D_{0y}^{\gamma-1}u_y\big)^2\Bigr|_{\,y=\beta}\,dx=0.
\end{equation}
We have 
$$
D_{0y}^{\gamma-1}u_{yx}=D_{0y}^{\gamma}u_x
 -\frac{y^{-\gamma}}{\Gamma(1-\gamma)}u_x(x,0)=D_{0y}^{\gamma}u_x.
$$
Substituting the above formula in \eqref{f12},   we obtain
\begin{equation}\label{f10}
\int_0^r\int_0^{\beta}u_xD_{0y}^{\gamma}u_x\,dx\,dy
+{\frac{1}{2} \int_0^r}\big(D_{0y}^{\gamma-1}u_y\big)^2\Bigr|_{y=\beta}\,dx=0.
\end{equation}
Assume  $f=D_{0y}^{\gamma}u_x$. Then
\begin{equation}\label{f13}
u_x=D_{0y}^{-\gamma}f+\frac{y^{\gamma-1}}{\Gamma(\gamma)}
\lim_{y \to 0}D_{0y}^{\gamma-1}u_x.
\end{equation}
From \cite{Pskhu}, we know that 
$\lim_{y \to 0}D_{0y}^{\gamma-1}u_x= \Gamma(\gamma)
\lim_{y \to 0}y^{1-\gamma}u_x(x,y)$. Therefore, as 
$u_x(x,0)=0$, then $\lim_{y \to 0}D_{0y}^{\gamma-1}u_x=0$.
So from  \eqref{f13}, it follows that $u_x=D_{0y}^{-\gamma}f$.
Thereby,
$$
\int_0^r\int_0^{\beta}u_xD_{0y}^{\gamma}u_x\,dx\,dy
=\int_0^r\int_0^{\beta}fD_{0y}^{-\gamma}f\,dx\,dy \geq 0,
$$
accordingly, \eqref{f10}  makes possible the conclusion $ u_x = 0$, i.e. 
$ u = u (y)$. Then according to  \eqref{f31}, we have 
$$ 
D_{0 y}^{\gamma} u_y = 0. 
$$
Applying the operator  $D_{0y}^{-\gamma}$  to both sides of this equation, we have
$$
D_{0y}^{-\gamma}D_{0y}^{\gamma}u_y=u_{y}-\frac{y^{\gamma-1}}{\Gamma(\gamma)}
\lim_{y \to 0}D_{0y}^{\gamma-1}u_y=0.
$$
Considering the first formula of \eqref{f6}, we have  $ u_y=0$. 
Consequently,  $u(x,y)={\rm\,const}$. As  $u$ from the class $C(\bar\Omega^+)$  
and $u|_{\partial\Omega^+}=0$, then $u(x,y)=0$ $\forall (x,y) \in \Omega^+$.
Thus, $u(x,y)\equiv0$ for all points $ (x,y) \in\Omega$.
This proves the theorem.
\end{proof}

\subsection*{Acknowledgements}
 The author would like to thank the referees for their valuable comments and
suggestions which improved this article.

\begin{thebibliography}{00}


\bibitem{Can} Cannon, J.~R.;
\emph{A Dirichlet problem for an equation of mixed type with a
discontinuous coefficient.}
Ann. de Math. pura ed Appl. 1963, vol. 61, no 1. pp. 371--377. doi: 10.1007/BF02410656.

\bibitem{dy2012}  Masaeva O.~Kh.;
\emph{Dirichlet problem for the generalized Laplace
equation with the Caputo derivative.}
Differential equations. 2012, vol. 48, no 3. pp. 449-454.
 doi: 10.1134/ S0012266112030184.

\bibitem{dy2013}   Masaeva O.~Kh.;
\emph{Dirichlet problem for a nonlocal wave equation.}
 Differential equations. 2013, vol. 49, no 12. pp. 1518-1523. 
doi: 10.1134/S0012266113120069.

\bibitem{lit_2} Nakhushev, A.~M.;
\emph{On the positivity of continuous and discrete differentiation and integration 
operators that are very important in fractional calculus and in the theory of 
equations of mixed type.} Differential Equations. 1998, vol. 34, no 1. p. 103-112.

\bibitem{Po}  Popov, A.~Yu.;  Sedletskii, A.~M.;
\emph{Distribution of roots of Mittag-Leffler functions.} 
Journal of Mathematical Sciences. 2013, vol. 190, no. 2,  pp. 209-409. 
doi: 10.1007/s10958-013-1255-3.

\bibitem{P} Pskhu, A.~V.;
\emph{On the real zeros of  functions of Mittag-Leffler type}.
 Mathematical Notes, 2005, 77:4, 546-552. doi:10.1007/s11006-005-0054-7.

\bibitem{Pskhu} Pskhu, A.~V.;
\emph{Solution of a boundary value problem for a fractional partial 
differential equation. Differential Equations.} 2003. vol. 39, no 8. pp. 
1150-1158. doi: 10.1023/ B:DIEQ.0000011289.79263.02.

\bibitem{S} Samko, S.~G.;  Kilbas, A.~A.;  Marichev, O.~I.;
\emph{Fractional Integrals and Derivatives: Theory and Applications.}
 Gordon and Breach Science Publishers. Switzerland, 1993.

\bibitem{Col1} Soldatov, A.~P.;
\emph{On Dirichlet-type problems for the Lavrent'ev-Bitsadze equation.}
Proceedings of the Steklov Institute of Mathematics. 2012, vol. 278. 
no 1. pp. 233--240. doi: 10.1134/S0081543812060223.

\end{thebibliography}

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