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\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2013 (2013), No. 138, pp. 1--5.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2013 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2013/138\hfil Existence of solutions]
{Existence of solutions to non-local problems for parabolic-hyperbolic
 equations with three lines of type changing}

\author[E. T. Karimov, A. I. Sotvoldiev \hfil EJDE-2013/138\hfilneg]
{Erkinjon T. Karimov, Akmal I. Sotvoldiyev}  % in alphabetical order

\address{Erkinjon T. Karimov\newline
Institute of Mathematics, National University of Uzbekistan
``Mirzo Ulugbek'',
 Durmon yuli str., 29, Tashkent, 100125, Uzbekistan}
\email{erkinjon@gmail.com, erkinjon.karimov@hotmail.com}

\address{Akmal I. Sotvoldiyev \newline
Institute of Mathematics, National University of Uzbekistan
``Mirzo Ulugbek'',
 Durmon yuli str., 29, Tashkent, 100125, Uzbekistan}
\email{akmal.sotvoldiyev@mail.ru}

\thanks{Submitted  May 14, 2013. Published June 20, 2013.}
\subjclass[2000]{35M10}
\keywords{Parabolic-hyperbolic equation; non-local condition; 
\hfill\break\indent Volterra integral equation}

\begin{abstract}
 In this work, we study a boundary problem with non-local conditions,
 by relating values of the unknown function with  various characteristics.
 The parabolic-hyperbolic equation with three lines of type changing
 is equivalently reduced to a system of  Volterra integral equations of
 the second kind. 
\end{abstract}

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

\section{Introduction}

Consider an equation
\begin{equation}
\begin{gathered}
u_{xx} - u_y =0, \quad (x,y) \in \Omega _0,  \\
u_{xx} - u_{yy},\quad (x,y) \in \Omega _i\; i = 1,2,3   \\
\end{gathered} \label{e1}
\end{equation}
in the domain $\Omega  = \Omega _0 \cup \Omega _1 \cup \Omega _2 \cup \Omega _3
 \cup AB \cup A A_0  \cup B B_0$; see Figure \ref{fig1}. 


\begin{figure}[ht]
\begin{center}
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\dashline{1}(25,50)(55,50)
\put(0,20){\line(1,0){80}}
\put(79,19.1){$\rightarrow$}
\put(81,17){$x$}
\put(25,5){\line(0,1){50}}
\put(24.2,54.5){$\uparrow$}
\put(55,20){\line(0,1){30}}
\put(5,35){\line(4,3){20}}
\put(5,35){\line(4,-3){35}}
\put(75,35){\line(-4,3){20}}
\put(75,35){\line(-4,-3){35}}
\put(21,50){1}
\put(26,52){$A_0$}
\put(53,52){$B_0$}
\put(1,34){$D$}
\put(16,34){$\Omega_2$}
\put(38.5,34){$\Omega_0$}
\put(61,34){$\Omega_3$}
\put(76,34){$E$}
\put(27,22){A}
\put(51,22){B}
\put(38.5,14){$\Omega_1$}
\put(5,19){\line(0,1){2}}
\put(2,16){-1/2}
\put(21,16){0}
\put(54,16){1}
\put(75,19){\line(0,1){2}}
\put(72,16){3/2}
\put(24,10){\line(1,0){2}}
\put(16,9){-1/2}
\put(38.5,5){$C$}
\end{picture}
\end{center}
\caption{Domain $\Omega$} \label{fig1}
\end{figure}


\subsection*{Problem AS}
 Find a regular solution of  equation \eqref{e1} in the domain $\Omega $,
satisfying the following conditions:
\begin{gather}
a_1( t )u(-t,t) + a_2( t )u( {t, - t} ) = a_3( t ),\quad
0 \leqslant t \leqslant \frac{1}{2},\label{e2}\\
b_1( t )u( {t,t - 1} ) + b_2( t )u( {2 - t,1 - t} ) = b_3( t ),\quad
\frac{1}{2} \leqslant t \leqslant 1,\label{e3}\\
c_1( t )( {u_x + u_y} )(t-1,t) + c_2( t )(u_x - u_y)
( {2 - t,t} ) = c_3( t ),\quad \frac{1}{2} < t < 1.\label{e4}
\end{gather}
Here ${a_i}( t ),{b_i}( t ),{c_i}( t )$
$(i = 1,2,3 )$ are given functions, such that
\begin{gather*}
a_1( 0 ) + a_2( 0 ) \ne 0,\quad
b_1( 1 ) + b_2( 1 ) \ne 0,\quad
a_1^2( t ) + a_2^2( t ) > 0,\\
b_1^2( t ) + b_2^2( t ) > 0,  \quad
c_1^2( t ) + c_2^2( t ) > 0,\quad
a_1^2 + b_2^2 > 0,\quad
a_2^2 + b_1^2 > 0.
\end{gather*}


Note that problem AS is a generalization of the following problems:
\begin{itemize}
\item[Case A] $a_1\equiv 0$.
\begin{itemize}
\item[(1)] $a_2,b_1,b_2,c_1,c_2\neq 0$,
\item[(2)] $b_1\equiv 0, a_2,b_2,c_1,c_2\neq 0$,
\item[(3)] $c_2\equiv 0, a_2,b_1,b_2,c_1\neq 0$,
\item[(4)] $b_1\equiv 0, c_2\equiv 0, a_2, b_2, c_1\neq 0$;
\end{itemize}

\item[Case B] $a_2\equiv 0$.
\begin{itemize}
\item[(1)] $a_1,b_1,b_2,c_1,c_2\neq 0$,
\item[(2)] $b_2\equiv 0, a_1,b_1,c_1,c_2\neq 0$,
\item[(3)] $c_1\equiv 0, a_1,b_1,b_2,c_2\neq 0$,
\item[(4)] $b_2\equiv 0, c_1\equiv 0, a_1, b_1, c_2\neq 0$;
\end{itemize}

\item[Case C] $b_1\equiv 0$.
\begin{itemize}
\item[(1)] $a_1,a_2,b_2,c_1,c_2\neq 0$,
\item[(2)] $c_2\equiv 0, a_1,a_2,b_2,c_1\neq 0$;
\end{itemize}

\item[Case D] $b_2\equiv 0$.
\begin{itemize}
\item[(1)] $a_1,a_2,b_1,c_1,c_2\neq 0$,
\item[(2)] $c_1\equiv 0, a_1,a_2,b_1,c_2\neq 0$;
\end{itemize}

\item[Case E] $c_1\equiv 0$. $a_1,a_2,b_1,b_2,c_2\neq 0$;

\item[Case F] $c_2\equiv 0$. $a_1,a_2,b_1,b_2,c_1\neq 0$.
\end{itemize}

Also note that cases A4 and B4 were studied in \cite{r1}.
Other cases were not investigated, and the main result of this paper is
true for these particular cases.

Boundary problems for parabolic-hyperbolic equations with two lines
of type changing were investigated in  \cite{a1,e1,e2,n1},
and with three lines of type changing in  \cite{b1,b2}.
The main point in this present work is the
non-local condition, which relates values of the unknown function with
various characteristics. It makes very difficult the reduction
of the considered problem to a system of integral equations,
we need a special algorithm for solving this problem.

\section{Main results}

In the domain $\Omega _1$ solution of the Cauchy problem with initial data
$u(x,0) = \tau _1( x )$, $u_y(x,0) = \nu _1( x )$ can be represented, as in
\cite{b3}, by
\begin{equation}
2u(x,y) = \tau_1( {x + y} ) + \tau_1(x - y)
+ \int_{x - y}^{x + y} {\nu_1( z )dz} .\label{e5}
\end{equation}
Assuming that in condition \eqref{e2},
\begin{equation}
u(-t,t) = {\varphi _1}( t ),\quad
0 \leqslant t \leqslant \frac{1}{2}, \label{e6}
\end{equation}
from \eqref{e5}, we find that
\begin{equation}
\tau '_1( t )
= \nu_1( t ) + {\Big( {\frac{{2[ {a_3(\frac{t}{2}) - a_1
(\frac{t}{2}){\varphi _1}(\frac{t}{2})}]}}{{a_2
(\frac{t}{2})}}} \Big)' },\quad 0 < t < 1.\label{e7}
\end{equation}
In condition \eqref{e3} introduce 
\begin{equation}
u( 2 - t,1 - t ) = {\varphi _2}( t ),\quad
\frac{1}{2} \leqslant t \leqslant 1,\label{e8}
\end{equation}
and from \eqref{e5}, we obtain
\begin{equation}
\tau '_1( t ) =  - \nu _1( t ) +
\Big( \frac{2[b_3( \frac{t + 1}{2}) - b_2(\frac{t + 1}{2} )
\varphi _2 (\frac{t + }{2})]}
{b_1(\frac{t + 1}{2})}\Big)' ,\quad 0 < t < 1. \label{e9}
\end{equation}
From \eqref{e7} and \eqref{e9}, it follows that
\begin{equation}
\tau '_1( t ) = \Big( {\frac{{a_3(\frac{t}{2})
- a_1(\frac{t}{2}){\varphi _1}(\frac{t}{2})}}{{a_2(\frac{t}{2})}}} \Big)'
+ \Big( {\frac{{b_3( {\frac{{t + 1}}{2}} )
- b_2( {\frac{{t + 1}}{2}} ){\varphi _2}
( {\frac{{t + 1}}{2}} )}}{{b_1( {\frac{{t + 1}}{2}} )}}} \Big)',\quad
0 < t < 1.\label{e10}
\end{equation}

The solution of the Cauchy problem in the domain ${\Omega _2}$, with given 
data $u(0,y) = {\tau _2}( y )$, $u_x( {0,y} ) = {\nu _2}( y )$,
is written  as follows \cite{b3},
\begin{equation}
2u(x,y) = {\tau _2}( {y + x} ) + {\tau _2}( {y - x} )
+ \int_{y - x}^{y + x} {{\nu _2}( z )dz} .\label{e11}
\end{equation}
Considering \eqref{e6} from \eqref{e11} we obtain
\begin{equation}
{\tau '_2}( t ) = {\nu _2}( t ) + {\varphi '_1}(\frac{t}{2}),\quad
0 < t < 1.\label{e12}
\end{equation}
In condition \eqref{e4} introduce another designation
\begin{equation}
(u_x - u_y)( {2 - t,t} ) = {\varphi _3}( t ),\quad
\frac{1}{2} < t < 1.\label{e13}
\end{equation}
Then from \eqref{e11} we obtain
\begin{equation}
\frac{{c_3( {\frac{{t + 1}}{2}} ) - c_2( {\frac{{t + 1}}{2}} ){\varphi _3}
( {\frac{{t + 1}}{2}} )}}{{c_1( {\frac{{t + 1}}{2}} )}} = {\tau '_2}( t )
+ {\nu _2}( t ),\,\,0 < t < 1.\label{e14}
\end{equation}
From \eqref{e12} and \eqref{e14} we deduce
\begin{equation}
2{\tau '_2}( t ) = {\varphi '_1}(\frac{t}{2})
+ \frac{{c_3( {\frac{{t + 1}}{2}} )
- c_2( {\frac{{t + 1}}{2}} ){\varphi _3}
( {\frac{{t + 1}}{2}} )}}{{c_1( {\frac{{t + 1}}{2}} )}},\quad
0 < t < 1.\label{e15}
\end{equation}

The solution of the Cauchy problem with data 
$u( {1,y} ) = {\tau _3}( y ),\,u_x( {1,y} ) = {\nu _3}( y )$
in the domain ${\Omega _3}$ has a form \cite{b3}
\begin{equation}
2u(x,y) = {\tau _3}( {y + x - 1} ) + {\tau _2}( {y - x + 1} )
+ \int_{y - x + 1}^{y + x - 1} {{\nu _3}( z )dz}.\label{e16}
\end{equation}
Using \eqref{e8} and \eqref{e13} from \eqref{e16}, after some evaluations
one can get
\begin{equation}
2{\tau '_3}( t ) =  - {\varphi '_2}( {\frac{{2 - t}}{2}} )
- {\varphi _3}( {\frac{{t + 1}}{2}} ),\quad 0 < t < 1.\label{e17}
\end{equation}
Further, from the equation \eqref{e1} we pass to the limit at
 $y \to  + 0$ and considering \eqref{e7} we find
\begin{equation}
{\tau ''_1}( t ) - {\tau '_1}( t )
=  - {\Big( {\frac{{2\left[ {a_3(\frac{t}{2}) - a_1
(\frac{t}{2}){\varphi _1}(\frac{t}{2})} \right]}}{{a_2(\frac{t}{2})}}} \Big)' }.
\label{e18}
\end{equation}
The solution of  \eqref{e18}  with the conditions
\begin{equation}
\tau_1( 0 ) = \frac{{a_3( 0 )}}{{a_1( 0 )
+ a_2( 0 )}},\quad
\tau_1( 1 ) = \frac{{b_3( 1 )}}{{b_1( 1 ) + b_2( 1 )}},\label{e19}
\end{equation}
which is deduced from \eqref{e2} and \eqref{e3}, can be represented
as \cite{d1}
\begin{equation}
\begin{aligned}
  \tau_1( x )
&= \frac{{a_3( 0 )}}{{a_1( 0 ) + a_2( 0 )}}
+ x\Big[ {\frac{{b_3( 1 )}}{{b_1( 1 ) + b_2( 1 )}}
 - \frac{{a_3( 0 )}}{{a_1( 0 ) + a_2( 0 )}}} \Big]  \\
&\quad + \int_0^1 {G(x,t)} \Big[ {\frac{{b_3( 1 )}}{{b_1( 1 )
 + b_2( 1 )}} - \frac{{a_3( 0 )}}{{a_1( 0 ) + a_2( 0 )}}} \Big]dt  \\
&\quad - \int_0^1 {G(x,t)} {\Big( {\frac{{2[ {a_3(\frac{t}{2})
- a_1(\frac{t}{2}){\varphi _1}(\frac{t}{2})} ]}}{{a_2(\frac{t}{2})}}}
 \Big)'}dt,\quad 0 \leqslant x \leqslant 1,
\end{aligned} \label{e20}
\end{equation}
where $G(x,t)$ is Green's function of problem \eqref{e18}-\eqref{e19}.

Continuing to assume that the function $\varphi _1$ is known, using the formula
 \eqref{e10} we represent function ${\varphi _2}$ via ${\varphi _1}$.
Then using the solution of the first boundary problem for equation \eqref{e1}
in the domain ${\Omega _0}$ (see \cite{d1}) and functional relations between
functions ${\tau _j}$ and ${\nu _j}$ $(j = 2,3)$, we obtain
\begin{equation}
\begin{gathered}
  {{\tau '}_2}( y ) = \int_0^y {{{\tau '}_3}( \eta  )N( {0,y,1,\eta } )d\eta }
- \int_0^y {{{\tau '}_2}( \eta  )N( {0,y,0,\eta } )d\eta }  + {F_1}( y ),  \\
  {{\tau '}_3}( y ) = \int_0^y {{{\tau '}_3}( \eta  )N( {1,y,1,\eta } )d\eta }
 - \int_0^y {{{\tau '}_2}( \eta  )N( {1,y,0,\eta } )d\eta }  + {F_2}( y ),
\end{gathered} \label{e21}
\end{equation}
where
\begin{gather*}
\begin{aligned}
  {F_1}( y ) &= \int_0^1 {\tau_1( \xi  ){{\overline G }_x}
 ( {o,y,\xi ,0} )d\xi }  - \frac{{a_3( 0 )}}{{a_1( 0 )
 + a_2( 0 )}}N( {0,y,0,0} ) \\
 &\quad  + \frac{{b_3( 1 )}}{{b_1( 1 ) + b_2( 1 )}}N( {0,y,1,0} )
 + {{\varphi '}_1}( {\frac{y}{2}} ),
\end{aligned} \\
\begin{aligned}
  {F_2}( y ) &= \int_0^1 {\tau_1( \xi  ){{\overline G }_x}( {1,y,\xi ,0}
 )d\xi }  - \frac{{a_3( 0 )}}{{a_1( 0 ) + a_2( 0 )}}N( {1,y,0,0} ) \\
&\quad  + \frac{{b_3( 1 )}}{{b_1( 1 ) + b_2( 1 )}}N( {1,y,1,0} )
- {\varphi _3}( {\frac{{y + 1}}{2}} ),
\end{aligned}
\end{gather*}
and
\[
\overline G ( {x,y,\xi ,\eta } ) = \frac{1}{{2\sqrt {\pi ( {y - \eta } )} }}
\sum_{n =  - \infty }^\infty
\Big[ {{e^{ - \frac{{{{( {x - \xi  + 2n} )}^2}}}{{4( {y - \eta } )}}}}
- {e^{ - \frac{{{{( {x + \xi  + 2n} )}^2}}}{{4( {y - \eta } )}}}}} \Big]
\]
is the Green's function of the first boundary problem; see \cite{d1},
$$
N( {x,y,\xi ,\eta } ) = \frac{1}{{2\sqrt {\pi ( {y - \eta } )} }}
\sum_{n =  - \infty }^\infty
\Big[ {{e^{ - \frac{{{{( {x - \xi  + 2n} )}^2}}}{{4( {y - \eta } )}}}}
+ {e^{ - \frac{{{{( {x + \xi  + 2n} )}^2}}}{{4( {y - \eta } )}}}}} \Big].
$$

From the first equation in \eqref{e21}, we represent function ${\varphi _3}$
 via ${\varphi _1}$ and further, from the second equation of \eqref{e21},
 we find the function ${\varphi _1}$.

After the finding function ${\varphi _1}$, using appropriate formulas,
we find functions ${\varphi _2}$, ${\varphi _3}$, ${\tau _i}$, ${\nu _i}$,
$( {i = 1,2,3 } )$. Solution of the problem AS can be established
in the domain  ${\Omega _0}$ as a solution of the first boundary problem,
and in the domains ${\Omega _i}$ $(i = 1,2,3)$ as a solution
of the Cauchy problem.

\begin{theorem} \label{thm1}
If the functions $a_i,b_i,c_i$ are continuously differentiable on a segment,
and have continuous second-order derivatives on an interval, then  
problem AS has a unique regular solution.
\end{theorem}


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\end{thebibliography}
\end{document}