\documentclass[reqno]{amsart}
\usepackage{hyperref}

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2016 (2016), No. 164, pp. 1--10.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu}
\thanks{\copyright 2016 Texas State University.}
\vspace{8mm}}

\begin{document}
\title[\hfilneg EJDE-2016/164\hfil 
Loaded mixed type equation with fractional derivatives]
{Non-local problems with integral gluing condition for loaded mixed type
equations involving the Caputo fractional derivative}

\author[O. Kh. Abdullaev, K. S. Sadarangani \hfil EJDE-2016/164\hfilneg]
{Obidjon Kh. Abdullaev, Kishin B. Sadarangani}

\address{Obidjon Kh. Abdullaev \newline
Department of Differential equations and Mathematical Physics,
National University of Uzbekistan,
 100114 Uzbekistan, Tashkent, Uzbekistan}
 \email{obidjon.mth@gmail.com}

\address{Kishin S. Sadarangani \newline
Department of Mathematics,
University of Las-Palmas de Gran Canaria,
35017 Las Palmas, Spain}
\email{ksadaran@dma.ulpgc.es}

\thanks{Submitted April 8, 2016. Published June 28, 2016.}
\subjclass[2010]{34K37, 35M10}
\keywords{Caputo fractional derivatives; loaded equation; non-local problem;
\hfill\break\indent integral gluing condition; existence; uniqueness}

\begin{abstract}
 In this work, we study the existence and uniqueness of solutions
 to non-local boundary value problems with integral gluing condition.
 Mixed type equations (parabolic-hyperbolic) involving the Caputo
 fractional derivative have loaded parts in Riemann-Liouville integrals.
 Thus we use the method of integral energy to prove uniqueness,
 and the method of integral equations to prove existence.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\allowdisplaybreaks


\section{Introduction and formulation of a problem}

 The models of fractional-order derivatives are more adequate than the previously 
used integer-order models, because fractional-order derivatives and
 integrals enable the description of the memory and hereditary properties 
of different substances \cite{p1}. This is the most significant advantage of
 the fractional-order models in comparison with integer-order models, 
in which such effects are neglected.

 Fractional differential equations have recently been proved to be valuable tools 
in the modeling of many phenomena in various fields of science and
 engineering. Indeed, we can find numerous applications in viscoelasticity, 
neurons, electrochemistry, control, porous media, electromagnetism, etc.,
 (see \cite{d1,g1,h1,k6,l3,m1}).
There has been a significant development in fractional differential
 equations in recent years; see the monographs of  Kilbas,  Srivastava, 
 Trujillo \cite{k5}, Miller and Ross \cite{m2}, Podlubny
 \cite{p1},  Samko,  Kilbas , Marichev. \cite{s2} and the references therein.

 Very recently, some basic theory for the initial boundary value problems 
of fractional differential equations involving a Riemann-Liouville differential
operator of order $0 < \alpha \leq 1$ has been discussed by Lakshmikantham and 
 Vatsala \cite{l1, l2}. In a series of papers (see
 \cite{b1,b2}) the authors considered some classes of initial value problems
for functional differential equations involving
 Riemann-Liouville and Caputo fractional derivatives of order $0 < \alpha \leq 1$: 
For more details concerning geometric and physical interpretation of
 fractional derivatives of Riemann-Liouville and Caputo types see \cite{p2}.
Note that works \cite{a3,k3,k4,n2} are devoted
 to the studying of boundary value problems (BVP) for parabolic-hyperbolic 
equations, involving fractional derivatives. BVPs for the mixed type
 equations involving the Caputo and the Riemann-Liouville fractional differential 
operators were investigated in works \cite{k1,k2}.

 For the first time it was given the most general definition of a loaded equations 
and various loaded equations are classified in detail by
Nakhushev \cite{n1}. Note that with intensive research on problem of optimal
control of the agro-economical system, regulating the label of
 ground waters and soil moisture, it has become necessary to investigate BVPs 
for such class of partial differential equations .More results on the
 theory of BVP's for the loaded equations parabolic, parabolic-hyperbolic and 
elliptic-hyperbolic types were published in works \cite{i1,a1}.
Integral boundary conditions have various applications in thermoelasticity, 
chemical engineering, population dynamics, etc. Gluing conditions of integral 
form were used in \cite{a2,s1}.

 We consider the equation:
\begin{equation}\label{Eq1}
 \begin{gathered}
 u_{xx}-_{C} D_{0y} ^{\alpha}u + p(x,y)\int_x^1 {{{(t - x)}^{\beta - 1}}
u(t,0)dt} =0 \quad\text{for } y > 0 \\
u_{xx} - u_{yy} - q(x + y)\int_{x + y}^1 {{{(t - x - y)}^{\gamma - 1}}u(t,0)dt} =0
\quad \text{for } y < 0\,,
\end{gathered}
\end{equation}
with the operator
\begin{equation}\label{Eq2}
_{C}D_{0y} ^{\alpha}f= \frac{1}{{\Gamma (1 - \alpha )}}
\int_0^y {{{(y - t)}^{ - \alpha }}f'(t)} dt,
\end{equation}
where $0 < \alpha$, $\beta ,\gamma < 1$, $p(x,y)$ and $q(x+y)$ are given functions.
 Let  $\Omega $ be domain, bounded with segments: 
${A_1}A_2  = \{ (x,y): x = 1,\; 0 < y < h\}$, 
${B_1}B_2  = \{ (x,y): x = 0,\; 0 < y < h\} $, 
$B_2 A_2  = \{ (x,y): y = h,\; 0 < x < 1\} $ at the $y > 0$,
and characteristics : ${A_1}C: x - y = 1$; 
${B_1}C: x + y = 0$ of the equation \eqref{Eq1} at $y < 0$, where 
${A_1}( {1;0} )$,
$A_2 ( {1;h} )$, $B_1 (0;0 )$, $B_2 (0;h )$, $C(1/2; -1/2)$.

 Let us introduce: $\theta (x) = \frac{{x + 1}}{2} + i  \frac{{x - 1}}{2}$, 
${i^2} = - 1$,
\begin{equation}\label{Eq3}
D_{xa}^{ - \beta }f(x) = \frac{1}{{\Gamma (\beta )}}
\int_x^a {{{(t - x)}^{\beta - 1}}f(t)dt}, \quad 0 < \beta < 1.
\end{equation}
${\Omega ^ + } = \Omega \cap (y > 0)$, 
${\Omega ^ - } = \Omega \cap (y < 0)$, 
${I_1} = \{ {x: \frac{1}{2} < x < 1}\}$,
$I_2 =\{ {y:0 < y < h} \}$. 
In the domain $\Omega $ we study the following problem.
\smallskip

\noindent\textbf{Problem I.} 
Find a solution $u(x,y)$ of equation \eqref{Eq1} from the  class of functions:
\[
W = \{ {u(x,y): u(x,y) \in C(\bar \Omega ) \cap {C^2}({\Omega ^ - }),
\; u_{xx} \in C( {{\Omega ^ + }} ),\;_CD_{oy}^\alpha u
\in C( {{\Omega ^ + }} )} \},
\]
that satisfies the boundary conditions:
\begin{gather}\label{Eq4}
u(x,y)\big|_{A_1A_2}  = {\varphi  }(y), \quad  0 \leq y \leq h; \\
\label{Eq5}
u(x,y)\big|_{B_1B_2} = \psi (y),\quad 0 \leq y \leq h; \\
\label{Eq6}
\frac{d}{{dx}}u( {\theta (x)} ) = a(x){u_y}(x,0) + b(x){u_x}(x,0) 
+ c(x)u(x,0) + d(x),\quad x \in {I_1},
\end{gather}
and gluing condition
\begin{equation}\label{Eq7}
\lim_{y \to + 0} {y^{1 - \alpha }}{u_y}(x,y) 
= {\lambda_1}(x){u_y}(x, - 0) + {\lambda_2}(x)\int_x^1 {r(t)u(t,0)dt}, \quad 0<x<1
\end{equation}
where $\varphi (y)$, $\psi (y)$, $a(x)$, $b(x)$, $c(x)$, $d(x)$, and
 ${\lambda_j}(x)$, are given
functions, such that $\sum_{j = 1}^2 {\lambda_j^2(x)}\ne 0$.

\section{Uniqueness of solution for Problem I}

It is known that equation \eqref{Eq1} on the characteristics coordinate
 $\xi = x + y$ and $\eta = x - y$ at $y \leq 0$ has the form
\begin{equation}\label{Eq8}
{u_{\xi \eta }} = \frac{{q(\xi )}}{4}\int_\xi ^1 {{{(t - \xi )}^{\gamma - 1}}
u(t,0)dt}.
\end{equation}
We introduce the notation:
$u(x,0) = \tau (x)$, $0 \leq x \leq1$;
 ${u_y}(x, - 0) = \nu  ^ - (x)$, $0 < x < 1$;
$$
\lim_{y \to + 0} {y^{1 - \alpha }}{u_y}(x,y) = \nu  ^ + (x), \quad  0 < x < 1.
$$
A solution of the Cauchy problem for the equation \eqref{Eq1} in the domain
 ${\Omega ^ - }$ can be represented as
\begin{equation}\label{Eq9}
\begin{aligned}
u(x,y) &= \frac{{{\tau  }(x + y) + {\tau  }(x - y)}}{2} 
- \frac{1}{2}\int_{x + y}^{x - y} {\nu ^ - }(t)dt  \\
&\quad  +\frac{1}{4}\int_{x + y}^{x- y} q(\xi )d\xi \int_\xi ^{x - y} d\eta 
\int_\xi ^1 {{{(t - \xi )}^{\gamma - 1}}\tau (t)dt}.
\end{aligned}
\end{equation}
After using condition \eqref{Eq6} and taking \eqref{Eq3} into account, 
from \eqref{Eq9} we obtain
\begin{equation}\label{Eq10}
\begin{aligned}
&( {2a(x) - 1} ){\nu ^ - }(x) \\
&= \frac{{1 - x}}{2}\Gamma (\gamma )q(x)D_{x\,\,1}^{ - \gamma }\tau (x) 
+ ( {1 - 2b(x)} )\tau '(x) 
- 2c(x)\tau (x) - 2d(x).
\end{aligned}
\end{equation}
Considering above notation and gluing condition \eqref{Eq7} we have
\begin{equation}\label{Eq11}
\nu  ^ + (x) = {\lambda_1}(x)\nu  ^ - (x) + {\lambda_2}(x)\int_x^1 {r(t)\tau (t)dt},
\end{equation}
Further from \eqref{Eq1} at $y \to + 0$ considering \eqref{Eq2}, \eqref{Eq11} and
$$
\lim_{y \to 0} D_{0y}^{\alpha - 1}f(y) = \Gamma (\alpha )
\lim_{y \to 0} {y^{1 - \alpha }}f(y),
$$
we obtain k1:
\begin{equation}\label{Eq12}
{\tau '' }(x) - \Gamma (\alpha ){\lambda_1}(x)\nu  ^ - (x) 
- \Gamma (\alpha ){\lambda_2}(x)\int_x^1 {r(t)\tau (t)dt} + \Gamma (\beta
)p(x,0)D_{x1}^{ - \beta }\tau (x) = 0.
\end{equation}


\subsection*{Main results}

\begin{theorem}\label{th1}
 If the given functions satisfy conditions
\begin{gather}\label{Eq13}
 {\Big( {\frac{{{\lambda_2}(x)}}{{r(x)}}} \Big)' } \geq 0,\quad
 \frac{{{\lambda_1}(0)q(0)}}{{2a(0) - 1}} \geq 0,\quad
 p(0,0) \leq 0,\,\ p'(x,0) \leq 0,\quad
\frac{{{\lambda_2}(0)}}{{r(0)}} \geq 0; \\
\label{Eq14}
\Big(\frac{(1-x)q(x){\lambda_1}(x)}{(2a(x)-1)}\Big)'\geq0, \quad
 \frac{c(x){\lambda_1}(x)}{(2a(x)-1)}\leq0, \quad
\Big(\frac{(1-2b(x)){\lambda_1}(x)}{(2a(x)-1)}\Big)'\leq0.
\end{gather}
then, the solution $u(x,y)$ of the Problem I is unique.
\end{theorem}


\begin{proof}
Known that if homogeneous problem has only trivial solution, then we can state 
that original problem has unique solution. For this aim, we assume that
the Problem I has two solutions, then denoting difference of these as $u(x,y)$, 
we  get appropriate homogenous problem.
Equation \eqref{Eq12} we multiply to $\tau (x)$ and integrate from 0 to 1:
 %\label{Eq15}
\begin{align*}
&\int_0^1 {{{\tau ''} }(x)\tau (x)dx} 
 - \Gamma (\alpha )\int_0^1 {{\lambda_1}(x)\tau (x)\nu  ^ - (x)dx} \\
&- \Gamma (\alpha
)\int_0^1 {{\lambda_2}(x)\tau (x)dx\int_x^1 {r(t)\tau (t)} dt} 
+ \Gamma (\beta )\int_0^1 {\tau (x)p(x,0)D_{x1}^{ - \beta }\tau (x)} dx = 0.
\end{align*}
We investigate the integral
\begin{align*}
I &= \Gamma (\alpha )\int_0^1 {{\lambda_1}(x)\tau (x)\nu  ^ - (x)dx} 
 + \Gamma (\alpha )\int_0^1 {{\lambda_2}(x)\tau (x)dx\int_x^1 {r(t)\tau (t)} dt} \\
&\quad -  \Gamma (\beta )\int_0^1 {\tau (x)p(x,0)D_{x1}^{ - \beta }\tau (x)} dx.
\end{align*}
Taking \eqref{Eq10} into account at $d(x) = 0$, we obtain
\begin{equation}\label{Eq16}
\begin{aligned}
I &= \frac{{\Gamma (\alpha )\Gamma (\gamma )}}{2}
 \int_0^1 {\frac{{(1 - x)q(x)}}{{2a(x) - 1}}{\lambda_1}(x)\tau (x)
D_{x1}^{ - \gamma }\tau (x)dx} \\
&\quad + \Gamma (\alpha )\int_0^1 {\frac{{( {1 - 2b(x)} ){\lambda_1}(x)}}
 {{2a(x) - 1}}\tau (x)\tau '(x)dx} 
- 2\Gamma (\alpha )\int_0^1 {\frac{{{\lambda_1}(x)c(x)}}{{2a(x) - 1}}
 {\tau ^2}(x)dx} \\
&\quad + \Gamma (\alpha )\int_0^1 {{\lambda_2}(x)\tau (x)dx\int_0^x {r(t)\tau (t)} dt} 
 - \Gamma (\beta )\int_0^1 {\tau (x)p(x,0)D_{x1}^{ - \beta }\tau (x)} dx \\
&= \frac{{\Gamma (\alpha )}}{2}\int_0^1 {\frac{{(1 - x)q(x)}}{{2a(x) - 1}}
 {\lambda_1}(x)\tau (x)dx\int_x^1 {{{(t - x)}^{\gamma - 1}}}\tau (t)dt}\\
&\quad + \frac{{\Gamma (\alpha )}}{2}\int_0^1 {\frac{{1 - 2b(x)}}
 {{2a(x) - 1}}{\lambda_1}(x)d( {\tau  ^2(x)} )} \\
&\quad - 2\Gamma (\alpha )\int_0^1 {\frac{{{\lambda_1}(x)c(x)}}{{2a(x) - 1}}
 {\tau ^2}(x)dx} - \frac{{\Gamma (\alpha )}}{2}{\int_0^1
{\frac{{{\lambda_2}(x)}}{{r(x)}}d( {\int_x^1 {r(t)\tau (t)} dt} )} ^2} \\
&\quad - \int_0^1 {\tau (x)p(x,0)dx\int_x^1 {{{(t - x)}^{\beta - 1}}} \tau (t)} dt.
\end{aligned}
\end{equation}

Considering $\tau (1) = 0$, $\tau (0) = 0$ (which deduced from the conditions 
\eqref{Eq4}, \eqref{Eq5} in homogeneous case) and on a base of the formula
\cite{s3}:
 $$
{| {x - t} |^{ - \gamma }} = \frac{1}{{\Gamma ( \gamma )
\cos\frac{{\pi \gamma }}{2}}}\int_0^\infty
 {{z^{\gamma - 1}}cos[ {z( {x - t} )} ]dz} ,\quad 0 < \gamma < 1
$$
After some simplifications from \eqref{Eq16} we obtain
\begin{align}
I &= \frac{{\Gamma (\alpha )q(0){\lambda_1}(0)}}{{4(2a(0) - 1)
 \Gamma (1 - \gamma )\sin \frac{{\pi \gamma }}{2}}}\int_0^\infty {{z^{ - \gamma }}}
 [ {M^2(0,z) + N^2(0,z)}]dz \nonumber \\
&\quad + \frac{{\Gamma (\alpha )}}{{4\Gamma (1 - \gamma )\sin 
 \frac{{\pi \gamma }}{2}}}\int_0^\infty {{z^{ - \gamma }}dz} 
\int_0^1 {[{M^2(x,z)+N^2(x,z)} ]d{\big[ {{\lambda_1}(x)\frac{{(1 - x)q(x)}}{{2a(x) - 1}}} 
 \big]} } \nonumber\\
&\quad - \frac{{\Gamma (\alpha )}}{2}\int_0^1 {\tau  ^2(x){{
\Big( {{\lambda_1}(x)\frac{{1 - 2b(x)}}{{2a(x) - 1}}} \Big)}' }dx 
- 2} \Gamma (\alpha )\int_0^1 \frac{{{\lambda_1}(x)c(x)}}{{2a(x) - 1}}{\tau ^2}(x)dx
  \nonumber\\
&\quad + \frac{{\Gamma (\alpha )}}{2}\frac{{{\lambda_2}(0)}}{{r(0)}}
 {\Big( {\int_0^1 {r(t)\tau (t)} dt} \Big)^2}
 + \frac{{\Gamma (\alpha )}}{2}\int_0^1 
{{{\Big( {\frac{{{\lambda_2}(x)}}{{r(x)}}} \Big)}' }
 {\Big( {\int_x^1 {r(t)\tau (t)} dt} \Big)^2}dx} \nonumber\\
&\quad  - \frac{{p(0,0)}}{{2\Gamma (1 - \beta )\sin 
 \frac{{\pi \beta }}{2}}}\int_0^\infty {{z^{ - \beta }}} 
 [ {M^2(0,z) + N^2(0,z)} ]dz \nonumber\\
&\quad - \frac{1}{{2\sin \frac{{\pi \beta }}{2}\Gamma (1 - \beta )}}
\int_0^\infty {{z^{ - \beta }}dz} \int_0^1 {\frac{\partial }{{\partial
 x}}[ {p(x,0)} ][ {M^2(x,z) + N^2(x,z)} ]} dx \label{Eq17}
\end{align}
where $M(x,z)={\int_x^1 {\tau (t)\cos ztdt} } $, 
$ N(x,z)= {\int_x^1 {\tau (t)\sin ztdt} }$.

Thus, owing to \eqref{Eq13},\eqref{Eq14} from \eqref{Eq17} it is concluded, 
that $\tau (x) \equiv 0$. Hence, based on the solution of the first boundary
problem for the \eqref{Eq1} \cite{k2,p3} owing to account
\eqref{Eq4} and \eqref{Eq5} we obtain $u(x,y) \equiv 0$ in 
${\overline \Omega ^+ }$. Further, from functional relations \eqref{Eq10}, 
taking into account $\tau (x) \equiv 0$ we obtain that $\nu  ^ - (x) \equiv 0$.
 Consequently, based on the solution \eqref{Eq9} we obtain $u(x,y) \equiv 0$ 
in closed domain ${\overline \Omega ^ - }$.
\end{proof}

\section{Existence of solutions for Problem I}

\begin{theorem} \label{thm3.1}
If conditions \eqref{Eq13}, \eqref{Eq14} and
\begin{gather}\label{Eq18}
p(x,y)\in C({\overline {\Omega^+}})\cap {C^2}({\Omega^+} ),\quad
q(x+y)\in C( {\overline {\Omega^-}}) \cap {C^2}( {\Omega^-} ); \\
\label{Eq19}
\varphi (y), \psi (y) \in C( {\overline {{I_2}} } ) 
\cap {C^1}( {{I_2}} ); a(x), b(x), c(x), d(x) 
\in {C^1}( {\overline {{I_1}} } ) \cap {C^2}( {{I_1}} )
\end{gather}
hold, then the solution of the investigating problem exists.
\end{theorem}

\begin{proof}
Taking \eqref{Eq10} into account from \eqref{Eq12} we obtain
\begin{equation}\label{Eq20}
\tau ''(x) - A(x)\tau '(x) = f(x) - B(x)\tau (x)
\end{equation}
where
\begin{equation}\label{Eq21}
\begin{aligned}
f(x)& = \frac{{\Gamma (\alpha )\Gamma (\gamma )(1 - x){\lambda_1}(x)q(x)}}
{{2(2a(x) - 1)}}D_{x1}^{ - \gamma }\tau (x) - \Gamma (\beta )p(x,0)D_{x1}^{ -
\beta }\tau (x) \\
&\quad + \Gamma (\alpha ){\lambda_2}(x)\int_x^1 {r(t)\tau (t)dt}
  - \frac{{2\Gamma (\alpha ){\lambda_1}(x)d(x)}}{{2a(x) - 1}}
\end{aligned}
\end{equation}
\begin{equation}\label{Eq22}
A(x) = \frac{{\Gamma (\alpha ){\lambda_1}(x)(1 - 2b(x))}}{{2a(x) - 1}}, \quad
B(x) = \frac{{2\Gamma (\alpha ){\lambda_1}(x)c(x)}}{{1 - 2a(x)}}
\end{equation}
The solution of \eqref{Eq20}  with conditions
\begin{equation}\label{Eq23}
\tau (0) = \psi (0), \quad \tau (1) = \varphi (0)
\end{equation}
has the form
\begin{equation}\label{Eq24}
\begin{aligned}
\tau (x) 
&= \psi (0)+{A_1}(x)\Big( {\int_x^1 {( {B(t)\tau (t) - f(t)} )} {{A'}_1}(t)dt 
+ \frac{{\varphi (0) - \psi (0)}}{{{A_1}(1)}}} \Big) \\
&\quad - \frac{{{A_1}(x)}}{{{A_1}(1)}}\int_0^1 {( {B(t)\tau (t) - f(t)} )}
  \frac{{{A_1}(t)}}{{{{A'}_1}(t)}}dt  \\
&\quad + \int_0^x {( {B(t)\tau (t) - f(t)} )} \frac{{{A_1}(t)}}{{{{A'}_1}(t)}}dt
\end{aligned}
\end{equation}
where
\begin{equation}\label{Eq25}
{A_1}(x) = \int_0^x {\exp\Big( {\int_0^t {A(z)dz} } \Big)} dt.
\end{equation}
Further, considering \eqref{Eq21} and using \eqref{Eq3} from \eqref{Eq24} we obtain
\begin{align}
&\tau (x) \nonumber\\ 
&= {f_1}(x)+{A_1}(x) \int_x^1 {B(t)\tau (t)d{A_{1}(t)}} \nonumber\\
&\quad - \frac{{\Gamma (\alpha )}}{2}\int_x^1 {\frac{{(1 - t){\lambda
_1}(t)q(t)}}{{2a(t) - 1}}d{A_{1}(t)} \int_t^1 
 {\frac {\tau (s)ds}{(s - t)^{1-\gamma}}} } \nonumber\\
&\quad + {A_1}(x)\int_x^1 {{{A'}_1}(t)dt}
\int_t^1 {[ {p(t,0){{(s - t)}^{\beta - 1}} 
 - \Gamma (\alpha ){\lambda_2}(t)r(s)} ]\tau (s)ds} \nonumber\\
&\quad - \frac{{{A_1}(x)}}{{{A_1}(1)}}\int_0^1 {\frac{{{A_1}(t)}}{{{{A'}_1}(t)}}B(t)
 \tau (t)dt} + \int_0^x
{\frac{{{A_1}(t)}}{{{{A'}_1}(t)}}p(t,0)dt} \int_t^1 {{{(s - t)}^{\beta - 1}}
 \tau (s)ds} \nonumber\\
&\quad  + \Gamma (\alpha )\frac{{{A_1}(x)}}{{{A_1}(1)}}
 \int_0^1 {\frac{{{A_1}(t)}}{{{{A'}_1}(t)}}dt\int_t^1 {[ {\frac{{(1 - t){\lambda
_1}(t)q(t){{(s - t)}^{\gamma - 1}}}}{{2(2a(t) - 1)}} 
 - {\lambda_2}(t)r(s)} ]} \tau (s)} ds \nonumber\\
&\quad - \frac{{{A_1}(x)}}{{{A_1}(1)}}\int_0^1 {\frac{{{A_1}(t)}}{{{{A'}_1}(t)}}p(t,0)dt} \int_t^1 {{{(s - t)}^{\beta - 1}}\tau (s)ds} +
\int_0^x {\frac{{{A_1}(t)}}{{{{A'}_1}(t)}}B(t)\tau (t)dt} \nonumber\\
&\quad - \Gamma (\alpha )\int_0^x {\frac{{{A_1}(t)}}{{{{A'}_1}(t)}}dt
 \int_t^1 {[ {\frac{{(1 - t){\lambda_1}(t)q(t)}}{{2(2a(t) - 1)}}{{(s -
t)}^{\gamma - 1}} + {\lambda_2}(t)r(s)} ]} \tau (s)} ds \label{Eq26}
\end{align}
where
\begin{equation}\label{Eq27}
\begin{aligned}
&{f_1}(x) \\
&= ( {1 - \frac{{{A_1}(x)}}{{{A_1}(1)}}} )
 \int_0^x {\frac{{2\Gamma (\alpha )d(t){A_1}(t){\lambda_1}(t)}}{{{{A'}_1}(t)(2a(t) 
-1)}}dt} \\
&\quad +2\Gamma (\alpha ){A_1}(x)\int_x^1 
 {\frac{{d(t){{A'}_1}(t){\lambda_1}(t)}}{{2a(t) - 1}}dt} \\
&\quad - \frac{{{A_1}(x)}}{{{A_1}(1)}}\int_x^1 
 {\frac{{2\Gamma (\alpha )d(t){A_1}(t){\lambda_1}(t)}}{{{{A'}_1}(t)(2a(t) - 1)}}dt} -
\frac{{{A_1}(x)}}{{{A_1}(1)}}( {\psi (0) - \varphi (0)} ) + \psi (0).
\end{aligned}
\end{equation}
After some simplifications we rewrite \eqref{Eq26} in the form
\begin{align*}
&\tau (x) \\
&= {A_1}(x)\int_x^1 {\tau (s)ds} \int_x^s {[ {p(t,0){{(s - t)}^{\beta - 1}} 
- \Gamma (\alpha ){\lambda_2}(t)r(s)} ]{{A'}_1}(t)dt} \\
&\quad - \Gamma (\alpha ){A_1}(x)\int_x^1 {\tau (s)ds
 \int_x^s {{{(s - t)}^{\gamma - 1}}\frac{{(1 - t)\lambda (t)q(t)}}{{2( {2a(t) - 1}
)}}{{A'}_1}(t)dt} } \\
&\quad  - \Gamma (\alpha )\int_x^1 {\tau (s)ds
 \int_0^x {[ {\frac{{(1 - t){\lambda_1}(t)q(t)}}{{2(2a(t) - 1)}}
 {{(s - t)}^{\gamma - 1}} +
{\lambda_2}(t)r(s)} ]\frac{{{A_1}(t)}}{{{{A'}_1}(t)}}dt  } }\\
&\quad  + \int_x^1 {\tau (s)ds} \int_0^x {{{(s - t)}^{\beta - 1}}} 
 \frac{{{A_1}(t)}}{{{{A'}_1}(t)}}p(t,0)dt + {A_1}(x)\int_x^1
{{{A'}_1}(t)B(t)\tau (t)dt} \\
&\quad + \Gamma (\alpha )\frac{{{A_1}(x)}}{{{A_1}(1)}}
 \int_0^1 {\tau (s)ds\int_0^s {\Big[ {\frac{{(1 - t){\lambda_1}(t)q(t){{(s - t)}^{\gamma
- 1}}}}{{2(2a(t) - 1)}} - {\lambda_2}(t)r(s)} \Big]} }
  \frac{{{A_1}(t)}}{{{{A'}_1}(t)}}dt \\
&\quad - \frac{{{A_1}(x)}}{{{A_1}(1)}}\int_0^1 {\tau (s)ds} 
 \int_0^s {{{(s - t)}^{\beta - 1}}\frac{{{A_1}(t)}}{{{{A'}_1}(t)}}p(t,0)dt} -
\frac{{{A_1}(x)}}{{{A_1}(1)}}\int_0^1 {\frac{{{A_1}(t)}}{{{{A'}_1}(t)}}
 B(t)\tau (t)dt} \\
&\quad + \int_0^x {\tau (s)ds} \int_0^s {{{(s - t)}^{\beta - 1}}} 
 \frac{{{A_1}(t)}}{{{{A'}_1}(t)}}p(t,0)dt + \int_0^x
{\frac{{{A_1}(t)}}{{{{A'}_1}(t)}}B(t)\tau (t)dt} \\
&\quad - \Gamma (\alpha )\int_0^x {\tau (s)ds\int_0^s 
 {\Big[ {\frac{{(1 - t){\lambda_1}(t)q(t)}}{{2(2a(t) - 1)}}{{(s - t)}^{\gamma - 1}} 
+ {\lambda_2}(t)r(s)} \Big]\frac{{{A_1}(t)}}{{{{A'}_1}(t)}}dt} } + {f_1}(x)
\end{align*}
i.e., we have the integral equation:
\begin{equation}\label{Eq28}
\tau (x) = \int_0^1 {K(x,t)\tau (t)dt} + {f_1}(x),
\end{equation}
where
\begin{equation}\label{Eq29}
K(x,t) = \begin{cases}
 {K_1}(x,s),& 0 \leq t \leq x, \\
 {K_2}(x,s),& x \leq t \leq 1. 
\end {cases}
\end{equation}
 \begin{equation}\label{Eq30}
\begin{aligned}
 {K_1}(x,s)
& = \Big( {\frac{{{A_1}(x)}}{{{A_1}(1)}} - 1} \Big)
 {\frac{{\Gamma (\alpha )}}{2}\int_0^s {{{(s - t)}^{\gamma - 1}}\frac{{(1
 - t){A_1}(t)\lambda (t)q(t)}}{{(2a(t) - 1){{A'}_1}(t)}}dt}} \\
&\quad - \Gamma (\alpha )\Big( {\frac{{{A_1}(x)}}{{{A_1}(1)}} + 1} \Big)
r(s)\int_0^s {{\lambda_2}(t)\frac{{{A_1}(t)}}{{{{A'}_1}(t)}}dt} \\
&\quad -\Big( {\frac{{{A_1}(x)}}{{{A_1}(1)}} - 1} \Big) 
\Big[{\int_0^s {{{(s - t)}^{\beta - 1}}\frac{{{A_1}(t)}}{{{{A'}_1}(t)}}p(t,0)dt} 
+ \frac{{{A_1}(s)}}{{{{A'}_1}(s)}}B(s)} \Big]
\end{aligned}
\end{equation}
\begin{equation}\label{Eq31}
\begin{aligned}
&{K_2}(x,s)  \\
&= {A_1}(x)\Big( {{{A'}_1}(s)B(s) - \frac{{\Gamma (\alpha )}}{2}
 \int_x^s {{{(s - t)}^{\gamma - 1}}\frac{{(1 - t)\lambda
(t)q(t)}}{{2a(t) - 1}}{{A'}_1}(t)dt} } \Big) \\
&\quad - \Gamma (\alpha ){A_1}(x)\int_x^s {{\lambda_2}} (t){A'_1}(t)dt 
 + {A_1}(x)\int_x^s {{{(s - t)}^{\beta - 1}}} {A'_1}(t)p(t,0)dt \\
&\quad - \Gamma (\alpha )r(s)\int_0^x {\frac{{{A_1}(t)
 {\lambda_2}(t)}}{{{{A'}_1}(t)}}dt} \\
&\quad + \frac{{\Gamma (\alpha)}}{2}\frac{{{A_1}(x)}}{{{A_1}(1)}}
 \int_0^s {{{(s - t)}^{\gamma - 1}}\frac{{(1 - t){A_1}(t)
 \lambda (t)q(t)}}{{(2a(t) - 1){{A'}_1}(t)}}dt} \\
&\quad - \frac{{{A_1}(x)}}{{{A_1}(1)}}\frac{{{A_1}(s)}}{{{{A'}_1}(s)}}B(s)
 - \Gamma (\alpha )\frac{{{A_1}(x)}}{{{A_1}(1)}}r(s)\int_0^s
{\frac{{{A_1}(t){\lambda_2}(t)}}{{{{A'}_1}(t)}}dt} \\
&\quad - \frac{{\Gamma (\alpha )}}{2}\int_0^x {\frac{{(1 - t){A_1}(t)
 \lambda (t)q(t)}}{{{{A'}_1}(t)(2a(t) - 1)}}{{(s - t)}^{\gamma - 1}}dt} \\
&\quad +( {1 - \frac{{{A_1}(x)}}{{{A_1}(1)}}} )
\int_0^x {\frac{{{A_1}(t){{(s - t)}^{\beta - 1}}}}{{{{A'}_1}(t)}}p(t,0)dt}.
\end{aligned}
\end{equation}
Owing to class \eqref{Eq18}, \eqref{Eq19} of the given functions and 
after some evaluations from \eqref{Eq30}, \eqref{Eq31} and 
\eqref{Eq27}, \eqref{Eq29}
we  conclude that 
$$
| {K(x,t)} | \leq \text{const}, \quad | {{f_1}(x)} | \leq \text{const.}
$$
Since kernel $K(x,t)$ is continuous and function in right-side $F(x)$ 
is continuously differentiable, solution of integral equation \eqref{Eq28} we can
write via resolvent-kernel:
\begin{equation}\label{Eq32}
\tau (x) = {f_1}(x) - \int_0^1 {\Re (x,t){f_1}(t)dt},
\end{equation}
where $\Re (x,t)$ is the resolvent-kernel of $K(x,t)$.
Unknown functions ${\nu ^ - }(x)$ and ${\nu ^ + }(x)$ we found accordingly 
from \eqref{Eq10} and \eqref{Eq11}:
\begin{align*}
{\nu ^ - }(x) 
&= \frac{{(x - 1)q(x)}}{{2( {2a(x) - 1} )}}
 \int_x^1 {{{(t - x)}^{\gamma - 1}}} dt\int_0^1 {\Re (t,s){f_1}(s)ds}\\
&\quad +\frac{{(1 - x)q(x)}}{{2( {2a(x) - 1} )}}
 \int_x^1 {{{(t - x)}^{\gamma - 1}}{f_1}(t)} dt \\
&\quad + \frac{{1 - 2b(x)}}{{2a(x) - 1}}{f'_1}(x) 
 - \frac{{1 - 2b(x)}}{{2a(x) - 1}}
 \int_0^1 {\frac{{\partial \Re (x,t)}}{{\partial x}}{f_1}(t)dt} -
\frac{{2c(x)}}{{2a(x) - 1}}{f_1}(x) \\
&\quad + \frac{{2c(x)}}{{2a(x) - 1}}\int_0^1 {\Re (x,t){f_1}(t)dt} 
- \frac{{2d(x)}}{{2a(x) - 1}}
\end{align*}
and ${\nu ^ + }(x) = \lambda (x){\nu ^ - }(x)$. 

The solution of  Problem I in the domain ${\Omega ^ + }$ can be written as 
\cite{k2},
\begin{align*}
&u(x,y) \\
&= \int_0^y {{G_\xi }(x,y,0,\eta )\psi (\eta )d\eta }
 - \int_0^y {{G_\xi }(x,y,1,\eta )\varphi (\eta )d\eta } \\
&\quad + \int_0^1
{{G_0}(x - \xi ,y)\tau (\xi )d\xi }
 - \int_0^y {\int_0^1 {G(x,y,0,\eta )} p(\xi )d\xi d\eta
 \int_\xi ^1 {{{(t - \xi )}^{\beta - 1}}\tau (t)dt} } 
\end{align*}
where
\begin{gather*}
{G_0}(x - \xi ,y) = \frac{1}{{\Gamma (1 - \alpha )}}
 \int_0^y {{\eta ^{ - \alpha }}G(x,y,\xi ,\eta )} d\eta ,\\
\begin{aligned}
G(x,y,\xi ,\eta ) 
&= \frac{(y - \eta )^{\alpha /2  - 1}}{2}
\sum_{n = - \infty }^\infty \Big[ e_{1,\alpha/2}^{1,\alpha/2}
\Big( - \frac{| x - \xi + 2n |}{(y - \eta )^{\alpha/2}}\Big) \\
&\quad - e_{1,\alpha/2}^{1,\alpha/2}
\Big( - \frac{| x + \xi + 2n |}{(y - \eta )^{\alpha/2}}\Big) \Big]
\end{aligned}
\end{gather*}
Is the Green's function of the first boundary problem \eqref{Eq1} 
in the domain ${\Omega ^ + }$ with the Riemanne-Liouville fractional differential
operator instead of the Caputo ones \cite{p4},
$$
e_{1,\delta }^{1,\delta }(z) = \sum_{n = 0}^\infty 
\frac{z^n}{n!\Gamma (\delta - \delta n)}
$$
is the Wright type function \cite{p4}.
Solution of the Problem I in the domain ${\Omega ^ - }$ will be found 
by the formula \eqref{Eq9}. Hence, the proof is complete.
\end{proof}



We remark that if  $a(x)=1/2$, then from \eqref{Eq10} we find  
$\tau(x)$ as a solution Volterra type integral equation.
After that we can find $\nu^{+}(x)$ from the 
first boundary value problem problem for the \eqref{Eq1}, 
and $\nu^{-}(x)$ will be defined from the gluing condition \eqref{Eq11}.


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\end{document}
