\documentclass[reqno]{amsart}

\AtBeginDocument{{\noindent\small
{\em Electronic Journal of Differential Equations},
Vol. 2002(2002), No. 85, pp. 1--7.\newline
ISSN: 1072-6691. URL: http://ejde.math.swt.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.swt.edu  (login: ftp)}
\thanks{\copyright 2002 Southwest Texas State University.}
\vspace{9mm}}

\begin{document}

\title[\hfilneg EJDE--2002/85\hfil A nonlocal problem for hyperbolic equations]
{A nonlocal problem for fourth order hyperbolic equations with
multiple characteristics}

\author[Bidzina Midodashvili\hfil EJDE--2002/85\hfilneg]
{Bidzina Midodashvili }

\address{Bidzina Midodashvili \newline
Department of Theoretical Mechanics,\newline
Georgian Technical University,\newline
Tbilisi, Georgia}
\email{bidmid@hotmail.com}

\date{}
\thanks{Submitted July 16, 2002. Published October 4, 2002.}
\subjclass[2000]{35L35}
\keywords{Goursat problem, Riemann function}

\begin{abstract}
  In this paper, we study fourth order differential equations
  with multiple characteristics and dominated low terms.
  We prove the existence and uniqueness of a Riemann function for this
  equation, and then provide an integral representation of
  the general solution of the Goursat problem. We also provide
  sufficient conditions for the solvability of a nonlocal problem.
\end{abstract}

\maketitle

\numberwithin{equation}{section}

\newtheorem{thm}{Theorem}[section]
\newtheorem{prop}[thm]{Proposition}
\newtheorem{lemm}[thm]{Lemma}
\newtheorem{rmrk}[thm]{Remark}
\allowdisplaybreaks

\section{Introduction}
 Partial differential equations of higher order with dominated
low terms are encountered when studying mathematical
models for certain natural and physical processes. As an example
of such type of equations, is the equation of moisture transfer [2]
$$
\frac {\partial w}{\partial t} = \frac {\partial}{\partial x}( D
\frac {\partial w}{\partial x} + A \frac {\partial^2 w}{\partial x
\partial t}),
$$
where $w$ is the concentration of moisture per unit, $D$ is the
coefficient of diffusivity, and $A>0$ is the varying coefficient
of Hallaire. Under the proper schematization of the process of
absorbing the soil moisture by the roots of plants, the pressure
$u(x,t)$ in the area of root absorption satisfies the equation of
form [4]
$$
(\frac {\partial}{\partial x} + \frac {1}{x})(u_{xt} + \lambda
u_{x}) = \mu u_{t}.
$$
 Obviously, the equation
$$
\frac {\partial^2 u}{\partial t^2} - \frac{\partial^2 u}{\partial
x^2} - \frac{\partial^4 u}{\partial x^2 \partial t^2} = 0,
$$
which describes the longitudinal waves in a thin elastic stem
taking into account the effects of transversal inertia, is of the
same type [5].

In the present work, a class equations with fourth order partial
derivatives and dominated lower order terms is considered.

  In the space $\mathbb{R}^3$ of the independent variables $x_1$, $x_2$
  and $x_3$ let
 $$
 \Pi := \{(x_1,x_2,x_3)\in R^{3}  : a_{i}<x_{i}<b_{i}\};
\quad \Pi_{i} := ]a_{i};b_{i}[;\quad
\Pi_{ij}:= \Pi_{i}\times\Pi_{j}
$$
for $i,j= 1,2,3$.
   For the class of functions $\varphi$, continuous in  $\bar{\Pi}$
with partial derivatives $\ D _{x_1}^{i}\varphi$,
$D_{x_2}^{j}\varphi$,  $D_{x_3}^{k}\varphi$,  $ 0\leq i\leq m$,
$0\leq j \leq n$, $0\leq k \leq l $, we use the symbol
$C ^ { i, j , k }(\bar{\Pi}) $.

Consider the Goursat problem
\begin{gather}
\frac {\partial ^{4}}{\partial x_1^{2} \partial x_2
\partial x_3}u(x) + \sum_{i,j,k}a^{i,j,k}(x) \frac {\partial
^{i+j+k}}{\partial x_1^{i}\partial x_2^{j}\partial
x_3^{k}}u(x) = f(x), \label{e1-1}
\\
\begin{gathered}
u(x_1,x_2,x_3^0) =\varphi_{12}(x_1,x_2),\quad
u(x_1,x_2^0,x_3)=\varphi_{13}(x_1,x_3), \\
u(x_1^0,x_2,x_3)=\varphi_{23}(x_2,x_3),\quad
u_{x_1}(x_1^0,x_2,x_3)=\widetilde{\varphi}_{23}(x_2,x_3),
\end{gathered} \label{e1-2}
\end{gather}
where $i=0,1,2$; $j,k=0,1$; $i+j+k\neq 4$, $x,x^0\in\bar{\Pi}$ and the
functions $\varphi_{ij}$ satisfy the
following compatibility conditions
\begin{equation}
\begin{gathered}
\varphi_1(x_1):=\varphi_{12}(x_1,x_2^0)=\varphi_{13}(x_1,x_3^0),
\quad \varphi_2(x_2):=\varphi_{12}(x_1^0,x_2)
=\varphi_{23}(x_2,x_3^0),\\
\varphi_3(x_3):=\varphi_{13}(x_1^0,x_3)
=\varphi_{23}(x_2^0,x_3),
\quad \varphi_0:=\varphi_1^0=\varphi_2(x_2^0)
=\varphi_3(x_3^0).
\end{gathered} \label{e1-3}
\end{equation}

\begin{thm}\label{thm1-1}
For any $f\in C(\bar{\Pi})$, $a^{i,j,k}\in C^{i,j,k}(\bar{\Pi})$
and $ \varphi_{12}\in C^{2,1}(\bar{\Pi}_{12})$,
 $\varphi_{13}\in C^{2,1}(\bar{\Pi}_{13})$, $ \varphi_{23}\in
C^{1,1}(\bar{\Pi}_{23})$, $\widetilde{\varphi}_{23}\in
C^{1,1}(\bar{\Pi}_{23})$ satisfying the compatibility conditions
(1.3) the Goursat problem (1.1), (1.2) has one and only one
solution $u\in C^{2,1,1}(\bar{\Pi})$.
\end{thm}

\begin{lemm}\label{lemm1-1}
Let $a(x)$ and $b(x)$ be continuous functions. An arbitrary
solution of equation
\begin{equation}
y'' + a(x)y' + b(x)y = 0,\quad x\in[\alpha,\beta]
\end{equation}
is monotonous if and only if  $b(x)= 0$ everywhere in
$[\alpha,\beta]$.
\end{lemm}

Let
\begin{gather*}
D:=\{x=(x_1,x_2,x_3)\in R^{3}: 0<x_{i}<x_{i}^0\},\\
D_{i}:=]0;x_{i}^0[;\,  D_{ij}:= D_{i}\times D_{j};\quad i,j= 1,2,3.
\end{gather*}
For equation (1.1) consider the boundary conditions
\begin{equation}
\begin{gathered}
u(x_1,x_2,0)=\varphi_{12}(x_1,x_2),\quad
u(x_1,0,x_3)=\varphi_{13}(x_1,x_3),\\
u(0,x_2,x_3)= \varphi_{23}(x_2,x_3),\quad
u(x_1^0,x_2,x_3)= \psi(x_2,x_3),
\end{gathered} \label{e1-4}
\end{equation}
where the functions $ \varphi_{ij},\, \psi $ satisfy the
compatibility conditions
\begin{equation} \begin{gathered}
\varphi_{12}(x_1,0)=\varphi_{13}(x_1,0),\quad
\varphi_{12}(0,x_2)= \varphi_{23}(x_2,0),\\
\varphi_{13}(0,x_3)= \varphi_{23}(0,x_3),\quad
\varphi_{12}(0,0)= \varphi_{13}(0,0)= \varphi_{23}(0,0),\\
\varphi_{12}(0,x_2)= \psi(x_2,0),\quad
\varphi_{13}(0,x_3)=\psi(0,x_3).
\end{gathered} \label{1-5}
\end{equation}

\begin{thm}\label{thm1-2}
Assume that $f\in C(\bar{D})$, $a^{i,j,k}\in C^{i,j,k}(\bar{D})$,
$\varphi_{12}\in C^{2,1}(\bar{D}_{12})$, $ \varphi_{13}\in
C^{2,1}(\bar{D}_{13})$, $\varphi_{23}, \psi \in
C^{1,1}(\bar{D}_{23})$. If there holds the condition
\begin{equation}
(a^{0,1,1} - a_{x_1}^{1,1,1})(x) = 0,\,\, x\in D \label{1-6}
\end{equation}
then problem (1.1), (1.5), (1.6) is uniquely solvable in the
class $C^{2,1,1}(\bar{D})$.
\end{thm}

\section{The Riemann function and the solution of (1.1)}

Following the scheme in  [1, 3],  we define the Riemann
function $v(x;\xi)$, $(x;\xi)\in\Pi\times\Pi$ as a solution
of the Goursat problem
\begin{gather}
\frac {\partial ^{4}}{\partial x_1^{2}
\partial x_2 \partial x_3}v(x) + \sum_{i,j,k}(-1)^{i+j+k}
\frac {\partial ^{i+j+k}}{\partial x_1^{i}\partial
x_2^{j}\partial x_3^{k}}(a^{i,j,k}(x)v(x)) =  0,
\label{e2-1}\\
\begin{aligned}
\big[v_{x_1x_1x_2}-(a^{2,0,1}v)_{x_1x_1}-(a^{1,1,1}v)_{x_1x_2}
+(a^{1,0,1}v)_{x_1}&\\
+(a^{0,1,1}v)_{x_2}-a^{0,0,1}v\big](x_1,x_2,\xi_3) &=0,
\quad (x_1,x_2)\in\bar{\Pi}_{12}; \end{aligned}
\nonumber\\
\begin{aligned}
\big[v_{x_1x_1x_3}-(a^{2,1,0}v)_{x_1x_1}
-(a^{1,1,1}v)_{x_1x_3}+(a^{1,1,0}v)_{x_1} \\
+(a^{0,1,1}v)_{x_3}-a^{0,1,0}v\big](x_1,\xi_2,x_3)&=0,
\quad (x_1,x_3)\in\bar{\Pi}_{13}; \end{aligned}
\nonumber \\
\big[v_{x_1x_2x_3}-(a^{2,1,0}v)_{x_1x_2}-(a^{2,0,1}v)_{x_1x_3}
+(a^{2,0,0}v)_{x_1}\big](\xi_1,x_2,x_3)=0,\;
(x_2,x_3)\in\bar{\Pi}_{23};
\nonumber \\
\big[v_{x_1x_1}-(a^{1,1,1}v)_{x_1}+a^{0,1,1}v\big]
(x_1,\xi_2,\xi_3)=0,\quad x_1\in\bar{\Pi}_1;
\label{e2-2} \\
\big[v_{x_1x_2}-(a^{2,0,1}v)_{x_1}\big](\xi_1,x_2,\xi_3)=0,
\quad x_2\in\bar{\Pi}_2;
\nonumber\\
\big[v_{x_1x_3}-(a^{2,1,0}v)_{x_1}\big](\xi_1,\xi_2,x_3)=0,
\quad x_3\in\bar{\Pi}_3;
\nonumber\\
v_{x_1}(\xi)=1;\quad
v(\xi_1,x_2,x_3)=0,\quad (x_2,x_3)\in\bar{\Pi}_{23},
\nonumber\end{gather}
where $i=0,1,2$; $j,k=0,1$; $i+j+k\neq 4$.
%\begin{rmrk}\label{1} \rm
For simplicity, we have omitted the second triplet of
arguments of the Riemann function.
%\end{rmrk}

\begin{rmrk}\label{2} \rm
The boundary conditions (2.2) for the Riemann function can be
received from the certain consideration of the integral
\begin{equation}
\int_{x^0}^{x}{(vLu-uL^{*}v)}(y)dy
\label{e2-3}
\end{equation}
\end{rmrk}

Further, by integration of equation (2.1) twice on the variable
$y_1$ and once on the variables $y_2$ and $y_3$ in corresponding
segments of integration ($y_{i}\in[\xi_{i};x_{i}], i=1,2,3$), and
taking into account the differential relations (2.2), we have
the following Volterra integral equation of the second kind,  with
respect to the first triplet of arguments of the Riemann function
$v(x;\xi)$
\begin{align*}
v(x)-\int_{\xi_1}^{x_1}[(a^{1,1,1}-(x_1-y_1)a^{0,1,1})v]
(y_1,x_2,x_3)dy_1&
\\
-\int_{\xi_2}^{x_2}(a^{2,0,1}v)(x_1,y_2,x_3)dy_2
-\int_{\xi_3}^{x_3}(a^{2,1,0}v)(x_1,x_2,y_3)dy_3&
\\
+\int_{\xi_1}^{x_1}\int_{\xi_2}^{x_2}[(a^{1,0,1}-(x_1-y_1)
a^{0,0,1})v](y_1,y_2,x_3)dy_1dy_2&
\\
+\int_{\xi_1}^{x_1}\int_{\xi_3}^{x_3}[(a^{1,1,0}-(x_1
-y_1)a^{0,1,0})v](y_1,x_2,y_3)dy_1dy_3&
\\
+\int_{\xi_2}^{x_2}\int_{\xi_3}^{x_3}(a^{2,0,0}v)
(x_1,y_2,y_3)dy_2dy_3&
\\
-\int_{\xi_1}^{x_1}\int_{\xi_2}^{x_2}\int_{\xi_3}^{x_3}
[(a^{1,0,0}-(x_1-y_1)a^{0,0,0})v](y_1,y_2,y_3)dy_1dy_2dy_3&
=x_1-\xi_1.
\end{align*}
The last equation  unconditionally has an unique solution and
therefore the existence and uniqueness of the solution of the
problem (2.1), (2.2) is proved.

 Now, integration (2.3) and taking into account the differential
relations  (2.2), for the regular
solution of problem (1.1), (1.2), (1.3), we have
\begin{multline}
u(x_1,x_2,x_3)=[v_{x_1}-a^{1,1,1}v](x_1^0,x_2^0,x_3^0;x)
\varphi_0
\\
+\int_{x_1^0}^{x_1}([v_{x_1}-a^{1,1,1}v]\varphi_1'
-a^{0,1,1}v\varphi_1)(y_1,x_2^0,x_3^0;x)dy_1
\\
+\int_{x_2^0}^{x_2}([v_{x_1}-a^{1,1,1}v]\varphi_2'+
[(a^{2,0,1}v)_{x_1}-a^{1,0,1}v]\varphi_2)(x_1^0,y_2,x_3^0;x)dy_2
\\
+\int_{x_3^0}^{x_3}([v_{x_1}-a^{1,1,1}v]\varphi_3'
+[(a^{2,1,0}v)_{x_1}
-a^{1,1,0}v]\varphi_3)(x_1^0,x_2^0,y_3;x)dy_3
\\
+\int_{x_1^0}^{x_1}\int_{x_2^0}^{x_2} \big(v_{x_1}
\big[\frac{\partial ^{2}\varphi_{12}}{\partial y_1\partial y_2}
+a^{2,0,1}\frac {\partial \varphi_{12}}{\partial y_1}\big]
\\
-v\big[a^{1,1,1}\frac {\partial ^{2}\varphi_{12}}{\partial
y_1\partial y_2}-(a_{x_1}^{2,0,1}-a^{1,0,1})\frac {\partial
\varphi_{12}}{\partial y_1} +a^{0,1,1}\frac {\partial
\varphi_{12}}{\partial y_2}+
a^{0,0,1}\varphi_{12}\big]\big)(y_1,y_2,x_3^0;x)dy_1dy_2
\\
+\int_{x_1^0}^{x_1}\int_{x_3^0}^{x_3}\big(v_{x_1}
\big[\frac{\partial ^{2}\varphi_{13}}{\partial y_1\partial y_3}+
a^{2,1,0}\frac {\partial \varphi_{13}}{\partial y_1}\big]
\\
-v\big[a^{1,1,1}\frac {\partial ^{2}\varphi_{13}}{\partial
y_1\partial y_3} -(a_{x_1}^{2,1,0}-a^{1,1,0})\frac {\partial
\varphi_{13}}{\partial y_1}+ a^{0,1,1}\frac {\partial
\varphi_{13}}{\partial y_3}
+a^{0,1,0}\varphi_{13}\big]\big)(y_1,x_2^0,y_3;x)dy_1dy_3
\\
 +\int_{x_2^0}^{x_2}\int_{x_3^0}^{x_3}\big(v_{x_1}
\big[\frac{\partial ^{2}\varphi_{23}}{\partial y_2\partial y_3}+
a^{2,1,0}\frac {\partial \varphi_{23}}{\partial y_2}  +
a^{2,0,1}\frac {\partial \varphi_{23}}{\partial y_3} +
a^{2,0,0}\varphi_{23}\big]
\\
- v\big[a^{1,1,1}\frac {\partial ^{2}\varphi_{23}}{\partial
y_2\partial y_3}- (a_{x_1}^{2,1,0}-a^{1,1,0})\frac {\partial
\varphi_{23}}{\partial y_2}- (a_{x_1}^{2,0,1}-a^{1,0,1})\frac
{\partial \varphi_{23}}{\partial y_3}
-(a_{x_1}^{2,0,0}-a^{1,0,0})\varphi_{23}
\\
+ \frac {\partial ^{2}\widetilde{\varphi}_{23}}{\partial
y_2\partial y_3}+ a^{2,1,0}\frac {\partial
\widetilde{\varphi}_{23}}{\partial y_2}+ a^{2,0,1}\frac
{\partial \widetilde{\varphi}_{23}}{\partial y_3} +
a^{2,0,0}\widetilde{\varphi}_{23}\big]\big)(x_1^0,y_2,y_3;x)dy_2dy_3
\\
-\int_{x_1^0}^{x_1}\int_{x_2^0}^{x_2}\int_{x_3^0}^{x_3}
v(y;x)f(y)dy_1dy_2dy_3.
\label{e2-4} \end{multline}
This proves the Theorem (1.1). \hfill$\square$

Let $v(x;\xi ),\,(x;\xi )\in \overline{\Pi }\times \overline{\Pi
}$ be the Riemann function for equation (1.1), and  let $x^0\in
\overline{\Pi }$ be an arbitrary point. Assuming that $u$ is the
regular solution of equation (1.1) in $\overline{\Pi }$ which
satisfies homogenous boundary conditions
$u(x_1^0,x_2,x_3)=u(x_1,x_2^0,x_3)$
$=u(x_1,x_2,x_3^0)=u_{x_1}(x_1^0,x_2,x_3)=0$, then, as it is easy
to see, from formula (2.4) it follows next representation
$$
u(x_1,x_2,x_3)=-\int_{x_1^0}^{x_1}\int_{x_2^0}^{x_2}%
\int_{x_3^0}^{x_3}v(y_1,y_2,y_3;x)f(y_1,y_2,y_3)dy_1dy_2dy_3,\,x\in
\overline{\Pi },
$$
for an arbitrary continuous function $f$.

Using the last representation and arbitrariness of the choices of
point $x^0$ and function $f$, from equation (1.1) one can get
following relations:
\begin{gather*}
\lbrack v_{\xi _1\xi _1\xi _2}+a^{2,0,1}v_{\xi _1\xi
_1}+a^{1,1,1}v_{\xi _1\xi _2}+a^{1,0,1}v_{\xi _1}+a^{0,1,1}v_{\xi
_2}+a^{0,0,1}v](x;\xi _1,\xi _2,x_3)=0, \\ \lbrack v_{\xi _1\xi
_1\xi _3}+a^{2,1,0}v_{\xi _1\xi _1}+a^{1,1,1}v_{\xi _1\xi
_3}+a^{1,1,0}v_{\xi _1}+a^{0,1,1}v_{\xi _3}+a^{0,1,0}v](x;\xi
_1,x_2,\xi _3)=0, \\ \lbrack v_{\xi _1\xi _2\xi
_3}+a^{2,1,0}v_{\xi _1\xi _2}+a^{2,0,1}v_{\xi _1\xi
_3}+a^{2,0,0}v_{\xi _1}](x;x_1,\xi _2,\xi _3)=0, \\
\lbrack v_{\xi _1\xi _1}+a^{1,1,1}v_{\xi
_1}+a^{0,1,1}v](x;\xi _1,x_2,x_3)=0, \\
\lbrack v_{\xi_1\xi _2}+a^{2,0,1}v_{\xi_1}](x;x_1,\xi
_2,x_3)=0,[v_{\xi_1\xi
_3}+a^{2,1,0}v_{\xi_1}](x;x_1,x_2,\xi _3)=0,\, \\
v_{\xi _1}(x;x)=1,\,\,v(x;x_1,\xi_2,\xi_3)=0.
\end{gather*}
 These relations are dual to relations (1.2) in the certain sense
(the left sides of (1.1) and (2.1), considered as differential
operators, are conjugated) , so, the definition of the Riemann
function as the solution of the Goursat problem (2.1),(2.2) is
logically correct.


\section{Proof of the Lemma 1.2 and the Theorem 1.3}

The if - part is obvious, therefore only the only if - part has to
be proved. Let us assume the contrary: there
exists $x_0\in[\alpha,\beta]$ satisfying   $b(x_0)\neq 0$
whereas an arbitrary solution of
\begin{equation}
y'' + a(x)y' + b(x)y = 0,\,\,\ x\in[\alpha,\beta] \label{e3-1}
\end{equation}
is monotonous. Certainly, because of  continuity of $b(x)$ there
exists the segment $[\alpha_1,\beta_1]$ such that it contains
the point $x_0$ and $b(x)\neq 0$,\,\,\ $x\in
[\alpha_1,\beta_1]$. Proceeding from the  well-known fact that
any solution of class $C^{2}[\alpha_1,\beta_1]$ can be
uniquely  prolonged till the solution of (3.1) of class
$C^{2}[\alpha,\beta]$ on whole $[\alpha,\beta]$ we shall not
restrict the generality of reasoning if assume that $b(x)\neq
0,\,\,\ x\in [\alpha,\beta]$.

Let $y(x)=c_1y_1(x)+c_2y_2(x)$ be an arbitrary solution of
equation (3.1) and $y'(x)=  c_1y'_1(x)+c_2y'_2(x)$ be a
constant-signed function where $y_1(x)$ and $y_2(x)$ form a
fundamental system of solutions of (3.1).

Consider the sets $K_{i} := \{x\in[\alpha,\beta]:  y'_{i}=0\},\,\
i=1,2$. Obviously, the sets $K_1$   and $K_2$ are closed. Let
us see that there hold the following properties
$$
A.\,\,\ K_1 \cap K_2 = \emptyset,\,\,\,\   B.\,\,\ K_1 \cup K_2 =
[\alpha,\beta].
$$

The property $A$ is obvious since assuming the opposite implies
the existence of a point $x_0\in[\alpha,\beta]$ such that
$y'_1(x_0)=y'_2(x_0)=0$ and therefore for Wronsky's
determinant we have $(W[y_1,y_2])(x_0)=0$ which contradicts
to the fundamentality of system $y_1(x),\,\ y_2(x)$.

Now suppose that the property $B$ is not true. This implies
the existence of a point $x_0\in[\alpha,\beta]$ such that
$y'_1(x_0)\neq 0$ and $y'_2(x_0)\neq 0$. Without
restriction of a reasoning generality we assume that
$y'_1(x_0)=y'_2(x_0)$ since in opposite case instead the
pair $y_1(x),\,\ y_2(x)$ one may consider the pair
$\frac{y'_2(x_0)}{y'_1(x_0)}y_1(x),\,\ y_2(x)$. It is
easy to note that $y''_1(x_0)\neq y''_2(x_0)$ because in
other case from (3.1) we would have $y_1(x_0)=y_2(x_0)$,
and according to $y'_1(x_0)=y'_2(x_0)$ and uniqueness of
Cauchy's problem solution we would get $y_1(x)=y_2(x),\,\,\
x\in[\alpha,\beta]$ contradicting to the condition of linear
independence of functions $y_1(x),\,\ y_2(x)$. Therefore
$y''_1(x_0)\neq y''_2(x_0)$ and as it is easy to verify
for $c_1=1$ and $c_2=-1$ the condition of sign-constancy of
the function $y'(x)= c_1y'_1(x)+c_2y'_2(x)$ is violated in
a neighborhood of the point $x_0$. This proves the property B.


Now, considering the segment $[\alpha,\beta]$ as a topological
space with the relative topology induced from $R$, which is
obviously connected, we have from the properties A and B that one
of the sets $K_1,\,\ K_2$ is empty, whereas another coincides
with $[\alpha,\beta]$, say $K_1=[\alpha,\beta]$. This means that
$y'_1(x) = 0,\,\ x\in[\alpha,\beta]$,  whence from  (3.1)
$b(x)y_1(x)= 0,\,\ x\in[\alpha,\beta]$. According to our
assumption $b(x)\neq 0$ and therefore $y_1(x)= 0,\,\
x\in[\alpha,\beta]$. The last contradicts to the linear
independence of the functions $y_1(x),\,\ y_2(x)$ and so the
lemma is proven.


Now, let us prove Theorem (1.3). Consider unknown function
$\tau(x_2,x_3)$ assuming that
$\tau(x_2,x_3)=u_{x_1}(0,x_2,x_3)$. Then, according to
(2.4) the regular solution of equation (1.1) with boundary
conditions
\begin{gather*}
u(x_1,x_2,0) =\varphi_{12}(x_1,x_2),\quad
u(x_1,0,x_3)=\varphi_{13}(x_1,x_3), \\
u(0,x_2,x_3)=\varphi_{23}(x_2,x_3),\quad
u_{x_1}(0,x_2,x_3)=\tau(x_2,x_3),
\end{gather*}
and the compatibility conditions
\begin{gather*}
\varphi_{12}(x_1,0)=\varphi_{13}(x_1,0)=\varphi_1(x_1),\quad
\varphi_{12}(0,x_2)=\varphi_{23}(x_2,0)=\varphi_2(x_2),
\\
\varphi_{13}(0,x_3)=\varphi_{23}(0,x_3)=\varphi_3(x_3),\quad
\varphi_1(0)=\varphi_2(0)=\varphi_3(0)=\varphi_0,
\end{gather*}
are given by formula
\begin{multline*}
u(x_1,x_2,x_3)=[v_{x_1}-a^{1,1,1}v](0,0,0;x)\varphi_0
\\
+\int_0^{x_1}([v_{x_1}-a^{1,1,1}v]\varphi_1'
-a^{0,1,1}v\varphi_1)(y_1,0,0;x)dy_1
\\
+\int_0^{x_2}([v_{x_1}-a^{1,1,1}v]\varphi_2'+
[(a^{2,0,1}v)_{x_1}-a^{1,0,1}v]\varphi_2)(0,y_2,0;x)dy_2
\\
+\int_0^{x_3}([v_{x_1}-a^{1,1,1}v]\varphi_3'+[(a^{2,1,0}v)_{x_1}
-a^{1,1,0}v]\varphi_3)(0,0,y_3;x)dy_3
\\
+\int_0^{x_1}\int_0^{x_2} \big(v_{x_1}
\big[\frac {\partial^{2}\varphi_{12}}{\partial y_1\partial y_2}
+a^{2,0,1}\frac {\partial \varphi_{12}}{\partial y_1}\big]
\\
-v\big[a^{1,1,1}\frac {\partial ^{2}\varphi_{12}}{\partial
y_1\partial y_2}-(a_{x_1}^{2,0,1}-a^{1,0,1})\frac {\partial
\varphi_{12}}{\partial y_1} +a^{0,1,1}\frac {\partial
\varphi_{12}}{\partial y_2}+
a^{0,0,1}\varphi_{12}\big]\big)(y_1,y_2,0;x)dy_1dy_2
\\
+\int_0^{x_1}\int_0^{x_3}\big(v_{x_1}
\big[\frac {\partial^{2}\varphi_{13}}{\partial y_1\partial y_3}
+ a^{2,1,0}\frac{\partial \varphi_{13}}{\partial y_1}\big]
\\
- v\big[a^{1,1,1}\frac{\partial ^{2}\varphi_{13}}{\partial y_1\partial y_3}
-(a_{x_1}^{2,1,0}-a^{1,1,0})\frac {\partial
\varphi_{13}}{\partial y_1}+ a^{0,1,1}\frac {\partial
\varphi_{13}}{\partial y_3}
+a^{0,1,0}\varphi_{13}\big]\big)(y_1,0,y_3;x)dy_1dy_3
\\
+\int_0^{x_2}\int_0^{x_3}\big(v_{x_1}
\big[\frac {\partial^{2}\varphi_{23}}{\partial y_2\partial y_3}
+ a^{2,1,0}\frac{\partial \varphi_{23}}{\partial y_2}
+ a^{2,0,1}\frac{\partial \varphi_{23}}{\partial y_3}
+ a^{2,0,0}\varphi_{23}\big]
\\
- v\big[a^{1,1,1}\frac {\partial ^{2}\varphi_{23}}{\partial
y_2\partial y_3}- (a_{x_1}^{2,1,0}-a^{1,1,0})\frac {\partial
\varphi_{23}}{\partial y_2}- (a_{x_1}^{2,0,1}-a^{1,0,1})\frac
{\partial \varphi_{23}}{\partial y_3}
-(a_{x_1}^{2,0,0}-a^{1,0,0})\varphi_{23}
\\
+ \frac {\partial ^{2}\tau}{\partial y_2\partial y_3}
+a^{2,1,0}\frac {\partial \tau}{\partial y_2}+ a^{2,0,1}\frac
{\partial \tau}{\partial y_3}
+a^{2,0,0}\tau\big]\big)(0,y_2,y_3;x)dy_2dy_3
\\
-\int_{x_1^0}^{x_1}\int_{x_2^0}^{x_2}\int_{x_3^0}^{x_3}
v(y;x)f(y)dy_1dy_2dy_3.
\end{multline*}

Now, putting  $x_1=x_1^0$ in the last expression and taking
into account that $u(x_1^0,x_2,x_3)= \psi(x_2,x_3)$ we
come after some transformations to the Volterra integral equation
with regard to the function $\tau(x_2,x_3)$:
\begin{multline}
v(0,x_2,x_3;x_1^0,x_2,x_3)\tau(x_2,x_3)+
\int_0^{x_3}\theta_1(0,x_2,y_3;x_1^0,x_2,x_3)
\tau(x_2,y_3)dy_3\\
+\int_0^{x_2}\int_0^{x_3}\theta_2(0,y_2,y_3;
x_1^0,x_2,x_3)\tau(y_2,y_3)dy_2dy_3
=\chi(x_2,x_3),  \label{e3-2}
\end{multline}
where $\theta_1,\, \theta_2$ and $\chi$ are known functions.
As it is well-known the last equation is solvable if
$$
v(0,x_2,x_3;x_1^0,x_2,x_3)\neq0,\quad
 0 \leq x_2 \leq x_2^0,\,\,0\leq x_3\leq x_3^0.
$$
 Further, according to
the fourth condition of (2.2) for the Riemann function we have
\begin{gather*}
[v_{x_1x_1}-(a^{1,1,1}v)_{x_1}+a^{0,1,1}v](x_1,x_2,
x_3;x_1^0,x_2,x_3)=0,
\\
 0\leq x_1\leq x_1^0, \,\, 0 \leq x_2 \leq
x_2^0,\,\,0\leq x_3\leq x_3^0.
\end{gather*}
Consider the last expression as an ordinary differential equation with respect
to $x_1$, for fixed $x_2$ and $x_3$, and rewrite it as
\begin{multline}
v_{x_1x_1}(x_1,x_2,x_3;x_1^0,x_2,x_3)
-a^{1,1,1}(x_1,x_2,x_3)v_{x_1}(x_1,x_2,x_3;x_1^0,x_2,x_3)
\\
+[a^{0,1,1}(x_1,x_2,x_3)-a_{x_1}^{1,1,1}(x_1,x_2,x_3)]v(x_1,
x_2,x_3;x_1^0,x_2,x_3)=0.
\label{e3-3}
\end{multline}

Now, if we assume (1.7) holds, then the solution of (3.3) is monotonous.
Taking into account that due to the last differential relations of (2.2)
$$
v(x_1^0,x_2,x_3;x_1^0,x_2,x_3) = 0,\quad
v_{x_1}(x_1^0,x_2,x_3;x_1^0,x_2,x_3) = 1
$$
we have
$$
v(0,x_2,x_3;x_1^0,x_2,x_3) \neq 0 ,\quad
0 \leq x_2 \leq x_2^0,\,\,0\leq x_3\leq x_3^0.
$$
Further, assuming (1.7) holds, (3.2) is uniquely solvable with regard to
the function $\tau(x_2,x_3)$. Replacing the last condition of (1.5) by
$u_{x_1}(0,x_2,x_3)= \tau(x_2,x_3)$ we come to the
problem (1.1), (1.2), (1.3) which solution will satisfy
conditions (1.5). This proves the Theorem (1.3).

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\end{document}