\documentclass[reqno]{amsart}
\usepackage{hyperref}

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2014 (2014), No. 57, pp. 1--7.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2014 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2014/57\hfil Mixed parabolic-hyperbolic equations]
{Boundary problems for mixed parabolic-hyperbolic equations with
 two lines of changing type and fractional derivative}

\author[B. J. Kadirkulov \hfil EJDE-2014/57\hfilneg]
{Bakhtiyor J. Kadirkulov}  % in alphabetical order

\address{Bakhtiyor J. Kadirkulov \newline
Tashkent State Institute of Oriental Studies,
Tashkent, Uzbekistan}
\email{kadirkulovbj@gmail.com}

\thanks{Submitted January 21, 2014. Published February 28, 2014.}
\subjclass[2000]{35M10}
\keywords{Parabolic-hyperbolic equation;  Tricomi problem;
\hfill\break\indent   Caputo fractional derivative; diffusion equation;
 Green's function; Volterra integral equation}

\begin{abstract}
 In this article, we study a boundary value problem for a
 parabolic-hyperbolic  equation with Caputo fractional derivative.
 Under certain conditions, we prove its unique solvability
 using methods of integral equations and Green's functions.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\allowdisplaybreaks

\section{Introduction}

Gelfand started the study of mixed parabolic-hyperbolic type equations in
his work \cite{g1}. Later on, Tricomi and Gellerstedt 
studied main boundary problems, and many authors have continued these 
studies as seen in the detailed bibliographies of \cite{d1,d2}.
Son recent works \cite{e3,e3,k2,n2} have been devoted to the study of boundary
problems for parabolic-hyperbolic equations with two or more lines of changing type.
While other works have been devoted to the study of  Riemann-Liouville,
Caputo, Hadamard, Hadamard-Marchaud and other general fractional operators;
see for example \cite{a2,b1,g1,h1,k2,k1,k5,n1,p1}.

Non-local  problems for parabolic-hyperbolic equations with one or two lines
of changing type containing the Riemann-Liouville fractional derivative
were investigated in \cite{b2,b5,e1,k3,k4}.

The Caputo fractional derivative is suitable for numerical
methods in the study of fractional differential equations \cite{a1,s1}, 
and it appears in many mathematical models of real-life processes \cite{c1,h1,m1}.

In this article we study a boundary value problem for a
parabolic-hyperbolic equation with the Caputo fractional derivative,
and having two lines where it changes type.

\section{Formulation of the problem and main result}

Let $0 < \alpha$ be a real number. For a function $\varphi (t)$,
given on $(0,\ell )$, $\ell  < \infty$ an integral-differential operator
in a sense of the Riemann-Liouville starting at $0$, is defined as
follows \cite{k5,n1,p1},
\[
D_{0t}^\alpha \varphi (t) = \begin{cases}
  \frac{1}{{\Gamma ( - \alpha )}}\int_0^t
{\frac{{\varphi (\tau )d\tau }}{{{{(t - \tau )}^{\alpha  + 1}}}}} ,& \alpha  < 0, \\
 \varphi (t),& \alpha  = 0,  \\
  \frac{{{d^n}}}{{d{t^n}}}D_{0t}^{\alpha  - n}\varphi (t), &
n - 1 < \alpha  \leq n,\; n \in N.
\end{cases}
\]
The  operator
\[
{}_CD_{0t}^\alpha \varphi (t) = D_{0t}^{\alpha  - n}{\varphi ^{(n)}}(t),\quad
n - 1 < \alpha  \leq n,\; n \in N
\]
is called the Caputo fractional differential operator.

The Riemann-Liouville and the Caputo differential operators are related
by the equality
\begin{equation}
cD_{0t}^{\alpha }\varphi (t)={}_cD_{0t}^{\alpha }\varphi (t)
+\sum_{k=0}^{n}{\frac{{{\varphi }^{k}}(0+)}{\Gamma (1+k-\mu )}{{t}^{k-\alpha }}},
\quad t>0. \label{e1}
\end{equation}
Let us consider the equation
\begin{equation}
\frac{\partial^2 u}{\partial x^2}-\frac{1-\operatorname{sign}(xy)}{2}
\frac{\partial^2 u}{\partial y^2}-\frac{1+\operatorname{sign}(xy)}{2}
\cdot {}_{C}D_{0y}^{\alpha }u=f(x,y), \,\,\alpha  \in (0,1).\label{e2}
\end{equation}
This equation for $x>0, y>0$ is the fractional order diffusion equation
\[
\frac{{{\partial }^{2}}u(x,y)}{\partial {{x}^{2}}}-{}_{C}D_{0y}^{\alpha }u(x,y)
=f(x,y),
\]
which coincides at $\alpha  = 1$ with the diffusion equation \cite{p1}
\[
\frac{{{\partial ^2}u}}{{\partial {x^2}}}
- \frac{{\partial u}}{{\partial y}} = f(x,y).
\]

Consider the \eqref{e2} in a finite domain $\Omega\subset \mathbb{R}^2$,
 bounded for $x>0$, $y > 0$ by segments $A_0B_0, B_0B$ of straight lines
$y = 1$, $x = 1$; at $x>0$, $y<0$ by segments $AC, BC$ of characteristics
$x + y = 0$, $x - y = 1$ of the \eqref{e2}; at $x<0$, $y>0$ by segments
 $AD, A_0D$ of characteristics $x + y = 0$, $y - x = 1$ of the \eqref{e2}.

The parabolic part of the mixed domain $\Omega$ will be denoted by $\Omega_3$,
and the hyperbolic part by $\Omega_1$, at $x>0$ and by $\Omega_2$ at
$x<0$, respectively.

The function $u(x,y)$ is called a regular solution of the \eqref{e2},
if it has necessary continuous derivatives participating in the \eqref{e2}
and satisfies it in ${\Omega _1} \cup {\Omega _2} \cup {\Omega _3}$.

In the domain $\Omega$ we  study the following boundary problem.

\noindent\textbf{Problem DS.}
Find a function $u(x,y) \in C(\overline \Omega)$, such that:

(1) $u$ is a regular solution of the \eqref{e2} in the domain
$\Omega \backslash (AA_0 \cup AB)$;

(2) satisfies boundary conditions
\begin{equation}
u(x,y)\big|_{BB_0 \cup DC}  = 0;\label{e3}
\end{equation}

(3) on lines of type changing it satisfies the  gluing conditions
\begin{gather}
u(x, - 0) = u(x, + 0), \,u( - 0,y) = u( + 0,y),  \label{e4} \\
u_y(x,-0)=l_1(x)\cdot \lim_{y\to +0} y^{1-\alpha }u_y(x,y)
+m_1(x)\cdot u(x,0)+n_1(x),  \label{e5} \\
u_x(-0,y)=l_2(y)\cdot u_x(+0,y)+m_2(y)\cdot u(0,y)+n_2(y),\label{e6}
\end{gather}
where ${l_i}(t)$, $m_i(t)$, $n_i(t)$, $y \in [0,1]$, $i = 1,2$ are given functions.

Note that Problem DS generalizes a problem studied in \cite{e2}.

\begin{theorem} \label{thm1}
Let the following conditions be fulfilled:
$f(x,y) \in {C^\delta }(\overline \Omega  )$,
$0 < \delta  < 1$, ${l_i}(t) \in {C^1}[0,1]$, $m_i(t),\,{n_i}(t) \in C[0,1]$,
$l_i(t) \ne 0$, $l'_i(t) - 2{l_i}(t) \cdot {m_i}(t) \geq 0$ for all
$t \in [0,1]$, $l_2(1) > 0$, $i = 1,2$.
Then  problem DS has a unique regular solution.
\end{theorem}

\subsection*{Proof Theorem \ref{thm1}}
 First, we find the main functional relations on $AB$, $AA_0$ deduced
from the domains $\Omega_1$ and $\Omega_2$.
We introduce the following notation
\begin{gather}
u(x,0) = {\tau _1}(x),\quad ,u(0,y) = {\tau _2}(y),\label{e7}\\
{u_y}(x, - 0) = \nu _1^ - (x),\quad \lim _{y \to  + 0} {y^{1 - \alpha }}{u_y}(x,y)
= \nu _1^ + (x),\label{e8} \\
{u_x}( - 0,y) = \nu _2^ - (y),\quad {u_x}( + 0,y) = \nu _2^ + (x).\label{e9}
\end{gather}
The solution to problem DS in $\Omega_1$ can be represented
by the D'Alembert's formula
\begin{equation}
u(\xi ,\eta ) = \frac{1}{2}\Big[ {{\tau _1}(\xi ) + {\tau _1}(\eta )
- \int_\xi ^\eta  {\nu _1^ - (t)dt} } \Big]
- \int_\xi ^\eta  {dt\int_y^\eta  {{f_1}( {y,\tau })d\tau } }\label{e10}
\end{equation}
where
\[
\xi  = x + y,\quad \eta  = x - y, \quad
{f_1}(\xi ,\eta ) = \frac{1}{4}f\big( {\frac{{\xi  + \eta }}{2},\frac{{\xi
- \eta }}{2}} \big).
\]
Then using condition \eqref{e3} from \eqref{e10} we get the following relation,
reduced from the domain $\Omega_1$ to the segment $AB$,
\[
{\tau '_1}(x) - \nu _1^ - (x) = 2\int_0^x {{f_1}( {t,x})dt} ,\quad
x \in (0,1).\label{e11}
\]
Similarly, from the formula
\begin{equation}
u(\xi ,\eta ) = \frac{1}{2}\Big[ {{\tau _2}(\xi ) + {\tau _2}(\eta )
- \int_\xi ^\eta  {\nu _2^ - (t)dt} } \Big]
- \int_\xi ^\eta  {dt\int_y^\eta  {{f_2}( {y,\tau })d\tau } }\label{e12}
\end{equation}
by  \eqref{e3}, from the domain $\Omega_2$ one can deduce the following
 relation between functions ${\tau _2}(y)$ and $\nu _2^ - (y)$:
\begin{equation}
{\tau '_2}(y) - \nu _2^ - (y) = 2\int_0^y {{f_2}({t,y})dt} ,\quad
y \in (0,1),\label{e13}
\end{equation}
where
\[
\xi  = x + y,\quad \eta  = y - x, \quad {f_2}(\xi ,\eta )
 =  - \frac{1}{4}f\big( {\frac{{\xi  - \eta }}{2},\frac{{\xi  + \eta }}{2}} \big).
\]

First, we prove the uniqueness of the solution of problem DS.

\begin{lemma} \label{lem1}
Let the conditions of the theorem be valid. Then problem DS can not have more
than one regular solution.
\end{lemma}

\begin{proof} Suppose the opposite.
Let problem DS has two different regular solutions ${u_1}(x,y)$,
${u_2}(x,y)$ and let
\[
u(x,y) = {u_1}(x,y) - {u_2}(x,y).
\]
It is not difficult to see that $u(x,y)$ is a regular solution of the
 homogeneous problem DS ($f(x,y) = 0$, ${n_i}(t) = 0$, $i = 1,2$).
This is why one only needs to prove that homogeneous problem has
only the trivial solution.

Let $u(x,y)$ be a regular solution of the homogeneous problem DS
in the domain $\Omega$. Since $u{u_{xx}} = {(u \cdot {u_x})_x} - u_x^2$,
 then integrating the identity
$u\big( {{u}_{xx}}-{}_{C}D_{0y}^{\alpha }u(x,y) \big)=0$ along the domain
${\Omega _3}$, taking \eqref{e3}, \eqref{e7}-\eqref{e9} into account, after
some evaluations we deduce
\begin{equation}
\iint_{{{\Omega }_{3}}}{u\,\cdot {}_{C}D_{0y}^{\alpha }u(x,y)\,dx\,dy}
+\int_{0}^{1}{{{\tau }_{2}}(x)\nu _{2}^{+}(x)dx}
+\iint_{{{\Omega }_{3}}}{u_{x}^{2}\,dx\,dy}=0\label{e14}
\end{equation}
Consider the integral
\[
{I_2} = \int_0^1 {{\tau _2}(y)\nu _2^ + (y)dy}.
\]
Taking into consideration the relation
\[
\nu _2^ + (y) = \frac{1}{{{l_2}(y)}}\nu _2^ - (y)
- \frac{{{m_2}(y)}}{{{l_2}(y)}}{\tau _2}(y),
\]
which follows from gluing conditions \eqref{e5} and notation \eqref{e9},
integral ${I_2}$ is rewritten as
\[
{I_2} = \int_0^1 {\frac{1}{{{l_2}(y)}}{\tau _2}(y)\nu _2^ - (y)dy}
- \int_0^1 {\frac{{{m_2}(y)}}{{{l_2}(y)}}\tau _2^2(y)dy}.
\]
On the other hand from \eqref{e13} it follows that
\[
\int_0^1 {\frac{1}{l_2(y)}\tau_2(y)\nu_2^-(y)dy}
 = \frac{\tau_2^2(1)}{2l_2(1)}
+ \frac{1}{2}\int_0^1 {\frac{l_2'(y)}{l_2^2(y)}\tau_2^2(y)dy}.
\]
Hence
\begin{equation}
{I_2} = \frac{{\tau _2^2(1)}}{{2{l_2}(1)}}
+ \frac{1}{2}\int_0^1 {\frac{{l_2'(y)
- {l_2}(y){m_2}(y)}}{{l_2^2(y)}}\tau _2^2(y)dy}.\label{e15}
\end{equation}
Using the formula [see \cite[p. 53]{p1}
\[
\lim_{t \to 0} D_{0t}^{\beta  - 1}\varphi (t)
= \Gamma (\beta )\lim _{t \to 0} {t^{1 - \beta }}\varphi (t),
{t^{1 - \beta }}\varphi (t) \in C[0,1),0 < \beta  < 1
\]
from the \eqref{e2} passing to the limit at $y \to  + 0$, taking notations
\eqref{e7} and \eqref{e8} into account, we obtain
\begin{equation}
{\tau ''_1}(x) - \Gamma (\alpha )\nu _1^ + (x) = 0,\,\,x \in (0,1).\label{e16}
\end{equation}
Considering condition \eqref{e5}, equality \eqref{e16} can be rewritten as
\[
{\tau ''_1}(x) - \frac{{\Gamma (\alpha )}}{{{l_1}(x)}}
[ {\nu _1^ - (x) - {m_1}(x){\tau _1}(x)}] = 0.
\]
From here we obtain
\begin{equation}
 - \int_0^1 {{{[{{\tau '}_1}(x)]}^2}dx}
 - \Gamma (\alpha )\int_0^1 {\frac{1}{{{l_1}(x)}}{\tau _1}(x)\nu _1^ - (x)dx}
 + \Gamma (\alpha )\int_0^1 {\frac{{{m_1}(x)}}{{{l_1}(x)}}\tau _1^2(x)dx}
 = 0.\label{e17}
\end{equation}
On the other hand from \eqref{e11} it follows that
\[
\int_0^1 {\frac{1}{{{l_1}(x)}}{\tau _1}(x)\nu _1^ - (x)dx}
 = \int_0^1 {\frac{{l_1'(x)}}{{2l_2^2(x)}}\tau _1^2(x)dx}.
\]
Then relation \eqref{e17} will have the form
\[
\int_0^1 {{{[{{\tau '}_1}(x)]}^2}dx}
+ \Gamma (\alpha )\int_0^1 {\frac{{l_1'(x) - 2{l_1}(x){m_1}(x)}}{{2l_1^2(x)}}
\tau _1^2(x)dx}  = 0.
\]
Here considering the conditions of the theorem we have
 ${\tau '_1}(x) = 0$. Hence, ${\tau _1}(x) =$ const. Since,
 ${\tau _1}(0) = 0$, it follows that ${\tau _1}(x) = 0$, $x \in [0,1]$.

Taking \eqref{e15}, ${\tau _1}(x) = 0$ and formula \eqref{e1} into account,
\eqref{e14} is rewritten as
\begin{align*}
&\iint_{{\Omega _3}} {uD_{RL}^\alpha u(x,y)\,dx\,dy + }
\frac{{\tau _2^2(1)}}{{2{l_2}(1)}} + \frac{1}{2}\int_0^1
{\frac{{l_2'(x) - 2{l_2}(x){m_2}(x)}}{{l_2^2(x)}}\tau _2^2(x)dx}\\
&+ \iint_{{\Omega _3}} {u_x^2(x,y)}\,dx\,dy = 0.
\end{align*}

According to \cite[Theorem 1.7.1]{n1}, from the last equality
we have $u(x,y)\equiv 0$ in ${\overline \Omega_3}$. Further,
from formulas \eqref{e10} and \eqref{e12}, by virtue of the uniqueness of
the solution of the Cauchy problem we have that $u(x,y)\equiv 0$ in
 ${\overline \Omega  _1} \cup {\overline \Omega_2}$. Hence,
$u(x,y)\equiv 0$; i.e., $u_1(x,y)\equiv u_2(x,y)$ in $\bar \Omega$.
The proof is complete
\end{proof}

Now we prove the existence of the solution of problem DS.

From the \eqref{e2} we deduce the following functional relation between
functions ${\tau _1}(x)$ and $\nu _1^ + (x)$,  on $AB$
\[
{\tau ''_1}(x) - \Gamma (\alpha )\nu _1^ + (x) = f(x,0).\label{e18}
\]
Defining function $\nu _1^ + (x)$.  from conditioin \eqref{e5} and \eqref{e11}
and substituting into the above equation we obtain the following problem 
for the unknown ${\tau _1}(x)$:
\begin{gather*}
  {{\tau}_1}''(x) + p(x){{\tau}_1}'(x) + q(x){\tau _1}(x) = g(x),  \\
  {\tau _1}(0) = {\tau _1}(1) = 0,
\end{gather*}
where
\[
p(x) =  - \frac{\Gamma (\alpha )}{l_1(x)},\quad
q(x) = \frac{m_1(x)}{l_1(x)},\quad
g(x) =  - \frac{1}{l_1(x)}\big[n_1(x) + 2\int_0^x {f_1(t,x)dt } \big].
\]
The uniqueness of the solution of this problem follows from the uniqueness
of the solution of problem DS. Note that this solution can be written
by Green's function (see \cite{e2}). Since, function ${\tau _1}(x)$ is now known,
from \eqref{e11} we find $\nu _1^-(y)$. Hence, the solution of the problem
in ${\Omega_1}$ is known.

The unknown function ${\tau _2}(y)$ can be found by the formula of the solution
of the first boundary problem for the \eqref{e2} in ${\Omega _3}$ \cite{b4}:
\begin{equation}
\begin{aligned}
u(x,y) &= \int_0^y {{G_\xi }(x,y,0,\eta ){\tau _2}(\eta )d\eta }
 +\int_0^1 {\tilde G(x - \xi ,y){\tau _1}(\xi )d\xi }\\
&\quad - \iint_{{\Omega _3}} {G(x,y,\xi ,\eta )f(\xi ,\eta )d\xi d\eta },
\end{aligned}\label{e19}
\end{equation}
where $G(x,y,\xi ,\eta )$ is the Green's function of the first boundary
problem for the diffusion equation with the Riemann-Liouville fractional
differential operator (see \cite[p. 108]{p1})
\begin{gather*}
G(x,y,\xi ,\eta )
= \frac{{{{(y - \eta )}^{\beta  - 1}}}}{2}\sum_{n =  - \infty }^\infty
\Big[ {e_{1,\beta }^{1,\beta }
\Big( { - \frac{{| {x - \xi  + 2n}|}}{{{{(y - \eta )}^\beta }}}} \Big)
- e_{1,\beta }^{1,\beta }\Big( { - \frac{{| {x + \xi
+ 2n} |}}{{{{(y - \eta )}^\beta }}}} \Big)} \Big],
\\
\tilde G(x - \xi ,y) = \frac{1}{{\Gamma (1 - \alpha )}}
\int_0^y {{\eta ^{ - \alpha }}G(x,y,\xi ,\eta )d\eta }, \quad
 \beta   = \frac{\alpha }{2},
\end{gather*}
where $e_{1,\beta }^{1,\beta }( z)$ is the Wright's function, which has the form
\[
e_{1,\beta }^{1,\delta }(z)
= \sum_{n = 0}^\infty  {\frac{{{z^n}}}{{n!\,\Gamma (\delta  - \beta n)}}} .
\]

Calculating ${u_x}(x,y)$ from \eqref{e19} and letting $x$ go to zero,
bearing in mind \cite[Lemma 2.2.2]{p1}, we get a relation between
functions ${\tau _2}(y)$ and $\nu _2^ + (y)$ on $AA_0$:
\[
\nu _2^ + (y) =  - \int_0^y {K(y - t){{\tau}_2}'(t)dt}  + {\Phi _0}(y),\label{e20}
\]
where
\begin{gather*}
K(y - t) = \frac{1}{{{{(y - t)}^\beta }}}
\Big[ {\frac{1}{{\Gamma (1 - \beta )}}
+ 2\sum_{n = 1}^\infty  {e_{1,\beta }^{1,1 - \beta }
\Big( { - \frac{{2n}}{{{{(y - t)}^\beta }}}} \Big)} } \Big],
\\
{\Phi _0}(y) = \lim _{x \to  + 0}
\Big[ {\int_0^1 {{{\tilde G}_x}(x - \xi ,y){\tau _1}(\xi )d\xi }
- \iint_{{\Omega _3}} {{G_x}(x,y,\xi ,\eta )f(\xi ,\eta )d\xi d\eta }} \Big].
\end{gather*}
Considering $\nu _2^ - (y) = \nu _2^ + (x) = {\nu _2}(x)$, excluding
${\nu _2}(x)$ from \eqref{e13} and (20), we get the Volterra integral
equation of second kind regarding the unknown function ${\tau_2}'(y)$:
\begin{equation}
{\tau_2}'(y) + \int_0^y {K(y - t){{\tau}_2}'(t)dt}  = \Phi (y), \label{e21}
\end{equation}
where
\[
\Phi (y) = {\Phi _0}(y) + 2\int_0^y {{f_2}(t,y)dt}.
\]
Since a solution of the integral equation depends on the kernel $K(y - t)$,
we shall study it in detail. The function $K(y - t)$ we represent as a sum
of two kernels
\[
K(y - t) = {K_1}(y - t) + {K_2}(y - t),
\]
where
\[
{K_1}(y - t) =  - \frac{{{{(y - t)}^{ - \beta }}}}{{\Gamma (1 - \beta )}},\quad
{K_2}(y - t) =  - \frac{2}{{{{(y - t)}^\beta }}}
\sum_{n = 1}^\infty  {e_{1,\beta }^{1,1 - \beta }
\big( { - \frac{{2n}}{{{{(y - t)}^\beta }}}} \big)}.
\]
Note that ${K_1}(y - t)$ is a kernel with weak singularity.
The kernel ${K_2}(y - t)$ represented as a series of Wright's type functions.
From \cite[(2.2.5), (2.2.24)]{p1} it follows that the kernel
${K_2}(y - t)$ has also weak singularity.
Therefore, equation \eqref{e21} is a Volterra equation of second
kind with weak singularity.

Using formulas \cite[2.2.19, 2.2.25, 2.3.8, 3.3.3]{p1},
 it is not difficult to show that the right hand side of  \eqref{e21}
is a continuous function. Then, from the theory of integral equations
follows that  \eqref{e21} has a unique continuous solution
(see, for example \cite{m2}).

After finding the functions $\tau _1(x)$, $\tau _2(x)$, $\nu _1(x)$ and
$\nu _2(y)$, the solution of problem DS in the domain $\Omega_3$ will be
found by formula \eqref{e19}, and in the domains $\Omega_1$, $\Omega_2$
as a solution of the Cauchy problem by the formulas \eqref{e10}
and \eqref{e12}, respectively.

The proof of Theorem \ref{thm1} is complete.

\subsection*{Acknowledgements}
Author is grateful to Professor M. Kirane for his useful remarks and suggestions.


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