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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2011 (2011), No. 07, pp. 1--6.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2011 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2011/07\hfil Non-local problems]
{Non-local problems for parabolic-hyperbolic equations
with deviation from the characteristics and  three
type-changing lines}

\author[A. S. Berdyshev, N. A. Rakhmatullaeva\hfil EJDE-2011/07\hfilneg]
{Abdumauvlen S. Berdyshev, Nilufar A. Rakhmatullaeva}  % in alphabetical order

\address{Abdumauvlen Suleymanovich Berdyshev \newline
Kazakh National Pedagogical University, named later Abaj,
050062 Almaty, Kazakhstan} 
\email{berdyshev@mail.ru}

\address{Nilufar Alisherovna Rakhmatullaeva \newline
Institute of Mathematics and Information Technologies,
Uzbek Academy of Sciences,
100125 Tashkent, Uzbekistan}
\email{rakhmatullaeva@mail.ru}

\thanks{Submitted November 18, 2010. Published January 15, 2011.}
\subjclass[2000]{35M10}
\keywords{Parabolic-hyperbolic equation;
deviation; type-changing line; \hfill\break\indent
integral equations}

\begin{abstract}
 We prove the existence and uniqueness of solutions
 for a partial differential equations of mixed type
 (parabolic-hyperbolic type).
 We use energy integrals and methods from integral equations 
 to study a problem that has deviation from the characteristics
 and three lines where the type changes.
\end{abstract}

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

\section{Introduction and formulation of the problem}

The need for studying boundary-value problems of
parabolic-hyperbolic type was emphasized by
Gelfand \cite{g1} in 1959.
Later Zolina \cite{z1} considered several of these problems and
gave some physical interpretations. Among the applications
of these problems, we have irrigation models 
found in the monograph by Serbina \cite{s1}.
Omitting many works for local and nonlocal problems,
we mention some recent works that
are closely related to the present investigation.
Berdyshev \cite{b1} studied the unique solvability
of Bitsadze-Samarskii type problem with deviation from
the characteristics for parabolic-hyperbolic equations with
one line where type changes
In \cite{b2,e2,k1,k2}, the unique solvability of  nonlocal
problems for parabolic-hyperbolic type equations with
continuous and special gluing conditions were studied.
Eleev and Lesev \cite{e1} studied
Parabolic-hyperbolic type equations with lines
where the type changes.
Boundary value problems with nonlocal conditions for
parabolic-hyperbolic equations with three lines where the type
changes were studied in \cite{b3}.

We use energy integrals and methods from
integral equations, to prove the unique solvability of a
boundary-value problems with nonlocal conditions.
This conditions relate values of the unknown function 
on the line where the type changes, with values of its derivatives
on curves lying inside of hyperbolic part of the domain.

Consider the equation
\begin{equation}
u_{xx}+\frac{\operatorname{sgn}(xy(1-x))-1 }{2}u_{yy}
+\frac{\operatorname{sgn}(xy(x-1))-1 }{2}u_y = 0 \label{e1}
\end{equation}
on a domain $\Omega=\Omega_0 \cup \Omega_1 \cup\Omega_2
\cup \Omega_3 \cup AB \cup AA_0 \cup BB_0$.
Here $\Omega_0 = \{(x,y): 0 < x < 1, 0 < y < 1\}$ and
$\Omega_1$, $\Omega_2$, $\Omega_3$ are characteristic triangles
with endpoints $A(0, 0), B(1,0),
C(\frac 1 2, -\frac 1 2)$;
$ D(-\frac 1 2, \frac 1 2), A_0(0,1)$;
$ E(\frac 3 2, \frac 1 2), B_0(1,1)$, respectively.

\subsection*{Problem NP}
 Find a regular solution of the \eqref{e1}, satisfying conditions
\begin{gather}
[u_x - u_y] \theta_1(t) + \mu_1(t)[u_x - u_y]\theta^*_1(t)
 = \varphi_1(t), \label{e2}\\
[u_x - u_y] \theta_2(t) + \mu_2(t)[u_x - u_y]\theta^*_2(t)
= \varphi_2(t), \label{e3}\\
[u_x + u_y] \theta_3(t) + \mu_3(t)[u_x + u_y]\theta^*_3(t)
 = \varphi_3(t), \label{e4}
\end{gather}
and  $u(A) = 0$, $u(B) = 0$.

Here $\theta_1(t)$, $\theta_2(t)$, $\theta_3(t)$,
[$\theta^*_1(t)$, $\theta^*_2(t)$, $\theta^*_3(t)$]
are affixes of intersection's points of characteristics, outgoing
from the points  $(x,0) \in AB$;
$(0,y) \in AA_0$; $(1,y) \in BB_0$ with
$AC; AD; BE$ $[AN; AK; BM]$, respectively.
$AN:y = - \gamma_1(x)$, $0\leq x\leq l_1$, $1/2\leq l_1\leq 1$,
$AK: x = - \gamma_2(y)$, $0\leq y\leq l_2$, $1/2\leq l_2\leq 1$,
$BM: x = - \gamma_3(y)$, $0\leq y\leq l_3$, $1/2\leq l_3\leq 1$;
$\mu_i(t)$ and $\varphi_i(t) (i = \overline{1,3})$ are given functions.

Regarding to curves $\gamma_i(t)$ we assume
the following conditions:
\begin{itemize}
\item $\gamma_i(0)=0$, $l_i+\gamma_i(l_i)=1$,
 $0<\gamma_i'(0)<1$, $\gamma_i(t)>0$, $t>0$;

\item $t-\gamma_i(t)$, $t+\gamma_i(t)$ are monotonically increasing;

\item  $\gamma_i(t)$ are twice continuously  differentiable functions.
\end{itemize}

\section{Unique solvability of the problem}

\begin{theorem} \label{thm1}
 If $\mu_{i}(t)\neq -1$ and $\mu_{i}(t),\varphi_{i}(t) \in C^{1}[0,1]$,
for $i=\overline{1,3}$, $0\leq t \leq 1$,
then  problem NP has unique solution.
\end{theorem}

We introduce the following notation
\begin{gather}
u(x, \pm0) = \tau_1^{\pm}(x), \quad
u_y(x, \pm0) = \nu_1^{\pm}(x), \label{e5}\\
u(\pm0,y) = \tau_2^{\pm}(y), \quad
u_x(\pm0,y) = \nu_2^{\pm}(y), \label{e6}\\
u(1\pm0,y) = \tau_3^{\pm}(y), \quad
u_x(1\pm0,y) = \nu_3^{\pm}(y)\,. \label{e7}
\end{gather}

It is known \cite{r1} that the solution of Cauchy's problem of \eqref{e1}
in the domain $\Omega_1$ has the form
\begin{equation}
u(x,y) = \frac 1 2 \{\tau_1^-(x+y) + \tau_1^-(x-y)
+ \int_{x-y}^{x+y} \nu_1^-(t)dt \}. \label{e8}
\end{equation}
Calculating  derivatives, we have
\begin{gather*}
u_x  = \frac{1}{2}\{ {\tau_1^-{'}(x + y) + \tau_1^-{'}(x - y)
     + \nu_1^- (x + y) - \nu_1^- (x - y)}\},\\
u_y  = \frac{1}{2}\{ {\tau_1^-{'}(x + y) - \tau_1^-{'}(x - y)
     + \nu_1^- (x + y) + \nu_1^- (x - y)} \},\\
u_x  - u_y  = \tau_1^-{'}(x - y) - \nu_1^-(x - y).
\end{gather*}
By the conditions on the function $\gamma_1(x)$,
an equation of the curve $AN$ in characteristic coordinates
$\xi=x+y,\,\eta=x-y$ can be given as $\xi = \lambda_1 (\eta),\,\,0 \leq \eta  \leq 1$, moreover $0 < \lambda_1{'}(0) < 1,\,\,\,\lambda _1 (\eta ) < \eta$. Then
$$
\theta _1 (t) = \big( {\frac{t}{2}; - \frac{t}{2}} \big),\quad
\theta _1^* (t) = \big( {\frac{{\lambda _1 (t) + t}}{2};
 \frac{{\lambda _1 (t) - t}}{2}} \big).
$$
Then we calculate
$$
[ {u_x  - u_y }]\theta _1 (t) = \tau _1^-{'}(t)- \nu _1^- (t),\quad
[ {u_x  - u_y }]\theta _1^{*} (t) = \tau _1^-{'}(t) - \nu _1^-  (t).
$$
Using condition \eqref{e2}, we find
\begin{equation}
\nu_1^-(t) = \tau_1^-{'}(t)-\frac{\varphi_1(t)} {1 + \mu_1(t)}, \quad
\mu_1(t) \neq -1. \label{e9}
\end{equation}

Similarly, using conditions \eqref{e3} and \eqref{e4}, we
obtain functional relations on lines $AA_0$ and $BB_0$,
reduced from the domains $\Omega_2$, $\Omega_3$, respectively:
\begin{gather}
\nu_2^-(t) = \tau_2^-{'}(t)-\frac{\varphi_2(t)} {1 + \mu_2(t)}, \quad
\mu_2(t) \neq -1, \label{e10}\\
\nu_3^+(t) = -\tau_3^+{'}(t)+\frac{\varphi_3(t)} {1 + \mu_3(t)}, \quad
\mu_3(t) \neq -1. \label{e11}
\end{gather}
On the domain $\Omega_0$ we obtain the equality
\begin{equation}
\begin{aligned}
&\iint_{\Omega _0} {u_x^2(x,y) \,dx\,dy}
+ \int_0^1\tau_2^+(y)\nu_2^+(y)dy - \int_0^1\tau_3^-(y)\nu_3^-(y)dy \\
&+ \frac 1 2 \int_0^1u^2(x,1)dx -  \frac 1 2 \int_0^1[\tau_1^+(x)]^2dx
= 0.
\end{aligned} \label{e12}
\end{equation}
To obtain this equality, first we multiplied \eqref{e1} by
 $u(x,y)$ and then integrated along the domain $\Omega_0$.
Then apply the Green's formula \cite{r1} and use the introduced notation
to obtain \eqref{e12}.

\subsection{Uniqueness of the solution}

To prove the uniqueness, as usual we suppose that the problem has
two  solutions $u_1$ and $u_2$. Taking difference of these solution
we obtain a homogeneous problem regarding for the new function
$u=u_1-u_2$. Below we prove that homogeneous problem NP
has only the trivial solution. Consequently,
 given functions $\varphi_i(t)$ are equal to zero.

Let us to prove that $u(x,\pm 0) = \tau_1^+(x) = \tau_1^-(x) =
0$. Passing to the limit in the domain $\Omega_0$, at $y \to +0$
from the equation $u_{xx} -u_y=0$, we obtain
\begin{equation}
\tau_1^+{''}(x) - \nu_1^+(x) = 0. \label{e13}
\end{equation}
Consider the integral $I_1 =\int_0^1\tau_1^{+}(x)\nu_1^+(x)dx$.
 Taking \eqref{e13} into account, we have
$$
I_1 = \int_0^1\tau_1^{+}(x)\tau_1^+{''}(x)dx
= -\int_0^1(\tau_1^+{'}(x))^2dx.
$$
It is obvious that $I_1 \leq 0$.

 From  relation \eqref{e9} we obtain
$\nu_1^-(x) = \tau_1^-{'}(x)$. Considering
$\tau_1^-(x) = \tau_1^+(x)$,
$\nu_1^-(x) = \nu_1^+(x)$,
$\varphi_1(t) = 0$, we obtain
\begin{equation}
I_1 = \int_0^1\tau_1^{+}(x){\tau_1^+}'(x)dx
= \frac 1 2 (\tau_1^{+}(x))^2\big|_0^1 = 0. \label{e14}
\end{equation}

Further, consider the integrals
$$
I_2 =\int_0^1\tau_2^{+}(y)\nu_2^+(y)dy \geq 0, \quad
I_3 = \int_0^1\tau_3^{-}(y)\nu_3^-(y)dy \leq 0.
$$
Using the functional relation  $\nu_2^+(y) ={\tau_2^+}'(y)$, we have
\begin{equation}
I_2 = \int_0^1\tau_2^{+}(y)\tau_2^+{'}(y)dy
= \frac 1 2 (\tau_2^{+}(1))^2 \geq 0, \label{e15}
\end{equation}
Similarly, we obtain
\begin{equation}
I_3 =- \int_0^1\tau_3^-(y)\tau_3^-{'}(y)dy
= -\frac 1 2 (\tau_3^{-}(1))^2 \leq 0. \label{e16}
\end{equation}
Taking \eqref{e15}, \eqref{e16} and $\tau_1(x)=0$ into account,
from \eqref{e12}, we obtain $u_x(x,y)=0$.
Since $u(x,y) \in C(\overline{\Omega}), u(x,y) \equiv 0$
in the domain $\Omega$.
The uniqueness of the solution for the problem NP is proved.

\subsection{Existence of the solution}

Excluding $\nu_1^+(x)=\nu_1^-(x)$ from \eqref{e9} and \eqref{e13},
we have
$$
\tau_1^+{''}(x) - \tau_1^+{'}(x)=- \frac{\varphi_1(x)} {1+ \mu_1(x)}.
$$
from here and considering $\tau_1^+(0) = \tau_1^+(1) = 0$, we obtain
$$
\tau _1^ + (x) = \int_0^x {\frac{{{\varphi _1}(t)
[ {1 - {e^{x - t}}}]}}{{1 + {\mu _1}(t)}}dt}
 + \frac{{{e^x} - 1}}{{e - 1}}\int_0^1 {\frac{{{\varphi _1}(t)
[ {{e^{1 - t}} - 1}]}}{{1 + {\mu _1}(t)}}dt}.
$$

By the unique solvability of the first boundary problem for the
heat equation \cite{f1}, the solution of \eqref{e1} in the domain
$\Omega_0$ is represented as
\begin{equation}
 \begin{aligned}
 u(x,y) &= \int_0^1 \tau_1^+(x)G(x,y,x_1,0)dx_1+ \int_0^y
 \tau_2^+(y_1)G_{x_1}(x,y,0,y_1)dy_1\\
 &\quad - \int_0^y \tau_3^-(y_1)G_{x_1}(x,y,1,y_1)dy_1,
 \end{aligned} \label{e17}
\end{equation}
where $G(x,y,x_1,y_1)$ is Green's function of the first boundary
problem for the heat equation \cite{f1}.

Differentiating \eqref{e17} once by $x$,
considering \eqref{e5}-\eqref{e6}, we obtain
\begin{gather}
\nu_2^{+}(y)=F_1(y) -\int_0^y \tau_2^+{'}(y_1)K_1(y,y_1)dy_1 +
\int_0^y \tau_3^-{'}(y_1)K_2(y,y_1)dy_1, \label{e18}
\\
\nu_3^{-}(y) = F_2(y) - \int_0^y \tau_2^+{'}(y_1)K_3(y,y_1)dy_1 +
\int_0^y \tau_3^-{'}(y_1)K_4(y,y_1)dy_1, \label{e19}
\end{gather}
where
\begin{gather*}
F_1(y) = -\int_0^1 \tau_1^+{'}(y_1)K_2(y,y_1)dy_1,\quad
F_2(y) = -\int_0^1 \tau_1^+{'}(y_1)K_4(y,y_1)dy_1,
\\
K_1(y,y_1)=\frac{1}{{\sqrt {\pi ({y - y_1 })} }}
\Big[ {1+ \sum_{ n =  - \infty ,\,   n \ne 0 }^{ + \infty }
{e^{ - \frac{{n^2 }}{{y - y_1 }}} } } \Big],
\\
K_2(y,y_1)=\frac{1}{{2\sqrt {\pi ({y - y_1 })} }}
\Big[ {2e^{ - \frac{1}{{4({y - y_1 })}}}
+ \sum_{ n =  - \infty,\,   n \ne 0 }^{ + \infty }
{({e^{ - \frac{{({2n - 1})^2 }}{{4({y - y_1 })}}}
+ e^{ - \frac{{({2n + 1})^2 }}{{4({y - y_1 })}}} })} } \Big],
\\
K_3(y,y_1)=\frac{1}{{\sqrt {\pi ({y - y_1 })} }}
\Big[ {e^{ - \frac{1}{{4(y - y_1) }}}
+ \sum_{ n =  - \infty,\,   n \ne 0 }^{ + \infty }
{ {e^{ - \frac{{({2n + 1})^2 }}{4({y - y_1})}} } } }\Big],
\\
K_4(y,y_1)=\frac{1}{{2\sqrt {\pi ({y - y_1 })} }}
\Big[ {1+e^{ - \frac{1}{{y - y_1 }}}  + \sum_{ n =  - \infty,\,
n \ne 0 }^{ + \infty } {({e^{ - \frac{{n^2 }}{{y - y_1 }}}
+ e^{ - \frac{{({n + 1})^2 }}{{y - y_1}}} })} } \Big].
\end{gather*}

From \eqref{e10} and \eqref{e18}, \eqref{e11} and \eqref{e19},
excluding $\nu_2^+(y), \nu_3^-(y)$, we have
\begin{gather}
{\tau_2^+}'(y) + \int_0^y {\tau_2^+}'(y_1)K_1(y,y_1)dy_1=F_3(y),
\label{e20}\\
\tau_3^-{'}(y) + \int_0^y \tau_3^-{'}(y_1)K_4(y,y_1)dy_1=F_4(y),
 \label{e21}
\end{gather}
where
\begin{gather*}
F_3(y)=F_1(y)+\frac{\varphi_2(y)}{1+\mu_2(y)}
 +\int_0^y\tau_3^-{'}(y_1)K_2(y,y_1)dy_1,\\
F_4(y)=-F_2(y)+\frac{\varphi_3(y)}{1+\mu_3(y)}
 +\int_0^y\tau_2^+{'}(y_1)K_3(y,y_1)dy_1.
\end{gather*}
Equations \eqref{e20}-\eqref{e21} can be considered as a system
of equations regarding unknown functions $\tau_2^+{'}(y)$ and
$\tau_3^-{'}(y)$. First we solve equation \eqref{e20}
 considering function $\tau_3^-{'}(y)$ as known.
Equation \eqref{e20} is a Volterra type integral equation
regarding to the function $\tau_2^+{'}(y)$ with continuous
right-hand side $F_3(y)$. Since the kernel $K_1(y,y_1)$ has a weak
singularity, one can represent solution of this equation
via the resolvent,
\begin{equation}
\tau_2^+{'}(y)=F_3(y)+\int_0^y R_1(y,y_1) F_3(y_1)dy_1, \label{e22}
\end{equation}
where $R_1(y,y_1)$ is the resolvent of the kernel $K_1(y,y_1)$.

Integrating once, from \eqref{e22}, we obtain
$$
\tau_2^+(y) = \int_0^yF_3(t)dt+\int_0^y
\Big(\int_0^t R_1(t,y_1)F_3(y_1)dy_1\Big)dt.
$$
Substituting $\tau_2^+(y)$ into \eqref{e21}, we obtain the
the second kind type Volterra integral equation regarding
$\tau_3^-{'}(y)$, which has unique solution \cite{f1}.

Since, functions $\tau_1^\pm(x)$, $\tau_2^\pm(y)$, $\tau_3^\pm(y)$
are known, using \eqref{e9}, \eqref{e10}, \eqref{e11} we
 find the functions $\nu_1^\pm(x)$, $\nu_2^\pm(y)$,
$\nu_3^\pm(y)$.

Finally, one can obtain solution to  problem NP
in the domain $\Omega_0$  by the formula \eqref{e17}, and in
domains $\Omega_i$, ($i = \overline{1,3}$) as a solution of the
Cauchy's problem, for instance \eqref{e8}.

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\end{document}