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UDC 517.958
\bigskip

\centerline{\bf DETERMINATION OF NON-STATIONARY ADSORPTION COEFFICIENT}
\centerline{\bf ANALYTICAL IN PART OF SPATIAL VARIABLES}

\vskip 0.3cm
\centerline{\bf D.K. Durdiev}

\centerline{The Institute of Mathematics at the Academy of Sciences of
 the Republic of Uzbekistan,}
 
\centerline{e-mail address: d.durdiev@mathinst.uz}

\centerline{\bf Zh. D. Totieva}

\centerline{Southern Mathematical Institute of Vladikavkaz Scientific Center, Vladikavkaz, Russia}
\centerline{e-mail address: jannatuaeva@inbox.ru}

\vskip 0.3cm
\noindent{\bf }
{\small \it 
In this study the multidimensional adsorption coefficient inverse problem for the second order hyperbolic equation are considered.   It is supposed that this coefficient is  continuous with respect to the variables $t, x$ and analytic in the other spatial variables. For solving this equations the scale method of Banach spaces of analytic functions is applied. The problem are reduced to a system of nonlinear Volterra integral equations and the local existence, global uniqueness, stability estimates are established. 

\vskip 0.2cm
\noindent {\bf Key words:} inverse problem, fundamental solution, local solvability, Banach space, equivalent norm.

\noindent {\bf AMS Mathematics Subject Classification:} 35K10, 45G10}
\vskip 0.3cm

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\renewcommand{\thesection}{\large 1}

\section{ Introduction. Formulation  of problem}

 According to the terminology of [1], an inverse problem is called one-dimensional if the sought
function depends on one variable, and multidimensional if it depends on two and more variables. For multidimensional inverse
problems there are only special cases for which solvability is
established. One of such classes of functions in which local   solvability
takes place is the class of analytic functions.  The technique
used here is based on the scale method of Banach spaces of
analytic functions, developed in the works of L.V. Ovsyannikov [2,3] and L. Nirenberg [4]. This method was
first applied to the problem of solvability of multidimensional
inverse problems by V.G. Romanov [1,5,6]. The approach used in this works can be successfully applied to the study of
solvability of the problem of determining the non-stationary coefficient for
the first time derivative of the equation with string vibration operator in
a principle part.

In this paper we study the inverse problem of determining the coefficient $a(t, x, y),$ $t\in {\Bbb R},$ $(x,y)=(x, y_1,\cdots, y_m)\in {\Bbb R}^{1+m}, $ for the second order hyperbolic equation. The problem is investigated in the class of coefficients that are continuous with respect to the variables $t, x$ and analytic in the variable $y.$ 
 
In [7--13] this method was used to study of the multidimensional inverse problems of determining the convolutional kernel in parabolic and hyperbolic integro-differential equations of the second order; theorems of local unique
solvability of inverse problems in the class of functions with finite smoothness with respect to the time variable and analytic with respect to spatial
variables. This article generalizes the results of work [1] (sec. 3)  for the case of non-stationary potential.



Consider the problem of determining a pair of functions $u$ and $a$ satisfying the equations:
 $$
 u_{tt}-u_{xx}-\triangle  u-a(t,x,y)u_t=g(y)\delta(x)\delta(t-t_0),\; (t, x, y)\in {\Bbb R}^{2+m},\: t_0>0, \eqno(1)$$
$$
 u\bigl|_{t<0}\equiv0,\eqno(2)
 $$
     in which $\triangle $ is the Laplace operator on variables $(y_1,\dots,y_m)=y,$  $\delta(\cdot)$ is the Dirac delta function, $\delta'(\cdot)$
     is the derivative of the Dirac delta function,  $t_0$ is  a problem parameter, therefore $u=u(t, x,
y,t_0),$  $g(y)$ is a given smooth function so that $g(y)\neq0 $ for $y\in \mathbb{R}^m.$

 In inverse problem it is required to find the potential  $a(x, t, y)$ in (1),  if the solution of the
 problem (1)--(3) is known for  $x=0$, i.e. the condition
$$
u(t,0,y,t_0)=f(t,y,t_0),\; t>0,\   t_0>0 \eqno(3)
$$
is given.

Following the monograph [4, sec. 3], we introduce into
consideration the Banach space  $A_s(r)$, $s>0$,  of functions
$\varphi(y)$, $y\in {\Bbb R}^m$, which are analytic in the
neighborhood of the origin and satisfy the relation
 $$\|\varphi\|_s(r):=\sup\limits_{|y|<r}\sum_{|\alpha|=0}^\infty\frac{s^{|\alpha|}}{\alpha!}\left|D^\alpha\varphi(y)\right|
<\infty.$$ Here  $r>0, \ s>0$ and
 $$D^\alpha:=\frac{\partial^{|\alpha|}}
 {\partial y_1^{\alpha_1}...\partial y_m^{\alpha_m}}, \ \alpha:=(\alpha_1,...,\alpha_m),$$
$$|\alpha|:=\alpha_1+...+\alpha_m, \ \alpha!:=(\alpha_1)!\dots(\alpha_m)!.$$
In the following, the parameter $r$ will be considered fixed, while
  a parameter $s$  is  variable. Then,
it is formed  a scale of Banach spaces $A_s(r), \ s>0$ of analytic
functions. The following property is obvious: if  $\varphi(y)\in
A_s(r)$,  then $\varphi(y)\in A_{s'}(r)$ for all  $s'\in (0, \
s)$, consequently, $A_s(r)\subset A_{s'}(r)$, if   $s'\in (0, \
s)$ and the following inequality is valid: \vspace{0.1cm}
$$
\left\|D^\alpha\varphi\right\|_{s'}\leq
c_\alpha\frac{\|\varphi\|_s(r)}{\left(s-s'\right)^{|\alpha|}}
$$
for any   $\alpha$ with constant  $c_\alpha$, which depends only
on  $\alpha$.

     We present the solution of the problem (1), (2) in the
      form
       $$
       u(t,x,y,t_0)=\frac{1}{2}g(y)\theta(t-t_0-|x|)+v(t,x,y,t_0),
       $$
       where $v(t,x,y,t_0)$~--~regular function. n 
      Substituting this  expression into (1), we obtain the following problem for determining function $v$:
\vspace{0.0cm}
    $$v_{tt}-v_{xx}=\triangle v
    +\frac{1}{2}\triangle g(y)\theta(t-t_0-|x|)$$$$+a(t,x,y)\left[\frac{1}{2}g(y)\delta(t-t_0-|x|)+v_t(t,x,y,t_0)\right],\; (t,x,y)\in {\Bbb R}^{2+m},\,   t_0>0, \eqno(4)$$
            $$ v\mid_{t<0}\equiv0.\eqno (5)$$

        In the next section we will replace the inverse problem (4), (5) and (2) by the equivalent integro-differential equations. In what follows, we assume that the function $a$ is even in $x.$

     \renewcommand{\thesection}{\large 2}
\section{ Reducing the problem to the integrodifferential equations}




   According to the d'Alembert formula
the solution of problem  (4)--(5) satisfies the integral
equation
   $$v(t,x,y,t_0)=$$$$=\frac{1}{2}\int\int_{\Delta(t,x)}\bigl\{\triangle  v( \tau, \xi, y, t_0)+\frac{1}{2}\triangle g(y)\theta(\tau-t_0-|\xi|)$$$$+a(\tau, \xi, y)
   \left[\frac{1}{2}g(y)\delta(\tau-t_0-|\xi|)+v_t(\tau,\xi,y,t_0)\right]\bigr\}d\xi d\tau, \, (t,x,y)\in {\Bbb R}^{2+m},\; t_0>0, \eqno(6)$$
  in which
   $$\Delta(t,x)=\left\{(\tau,\xi)|\ 0\leq \tau\leq t-|x-\xi|,\  x-t\leq \xi \leq x+t\right\}.$$

  Let
       $$Q_{T}:=\{(t,t_0)|\ 0\leq t_0\leq t\leq T\}, \ T>0,$$

       $$ \Omega_{T}:=\{(t,x)|\ 0\leq|x|\leq t\leq T-|x|\},$$


          $$\Upsilon_{T}:=\biggl\{(t,x,t_0)\Big|\ |x|+t_0 \leq t \leq T-|x|,\; 0\leq t_0 \leq t\leq T\biggr\}.$$


      The domain  $\Upsilon_{T}$ in the space of the variables
     $x,\, t,\, t_0$ is the pyramid with the base  $\Omega_T$ and
     vertex   $(0,T,T)$.


    It follows from (6) that the function  $v(t,x,y,t_0)$ satisfies the integral equation
      $$v(t,x,y,t_0)=\frac{\triangle g(y)}{4}\tilde S(t,x,t_0)+\frac{g(y)}{4}\int_{\frac{x-(t-t_0)}{2}}^{\frac{x+(t-t_0)}{2}}a(t_0+|\xi|, \xi, y)d\xi+$$
 $$
        +\frac{1}{2}\int\int_{\Box(t,x,t_0)}\left[\triangle  v(\tau, \xi, y, t_0)+a(\tau, \xi, y)v_t(\tau, \xi, y,t_0)\right]d\tau
        d\xi,
       \ (t,x,t_0)\in \Upsilon_T,\ y\in {\Bbb R}^{m},  \eqno(7)
 $$
where $\theta(t)=1,\ t \geq 0,\ \theta(t)=0,\ t<0,$ $\tilde S(t,x,t_0)$ is area of a rectangle $\Box(x,t,t_0)$ in the plane of the variables $(\tau, \xi)$  for each fixed $t_0$, formed by characteristics passing through the points $(0, t_0)$ and $(x, t)$ of the differential operator $\partial^2/\partial t^2-\partial^2/\partial x^2:$
      $$\Box(x,t,t_0):=\biggl\{(\xi,\tau)\biggl|\ |\xi|+t_0<\tau< t-|x-\xi|,
   $$$$ \frac{x-(t-t_0)}{2}<\xi<\frac{x+t-t_0}{2},\; 0< t_0< t \biggr\}.$$


     Obviously, the equalities
     $f(t,y,t_0)=u(t,0,y,t_0)=$ $v(t,0,y,t_0),$ $t>t_0$ are true.
     Besides,  $f(t_0+0,y,t_0)=$ $v(t,0,y,t_0)\bigl|_{t=t_0+0}=0.$


 First note that if
$a(t,-x,y)=a(t,x,y),$ then
   $v(t, -x,y,t_0)=v(t,x,y,t_0)$. Calculating the derivative of the both sides of equation (7) with respect to  $t$,
   we obtain
\vspace{0.0cm}
   $$v_t(t,x,y,t_0)=\frac{\triangle g(y)}{2}\tilde S_t'(t,x,t_0)$$$$+\frac{g(y)}{8}\left[a\left(\frac{x+t+t_0}{2}, \frac{x+t-t_0}{2}, y\right)+a\left(\frac{-x+t+t_0}{2}, \frac{x-t+t_0}{2}, y\right)\right]$$
   $$+\frac{1}{2}\int_{\frac{x-(t-t_0)}{2}}^{\frac{x+(t-t_0)}{2}}\left[\triangle  v\left(t-|x-\xi|,\xi,y,t_0\right)+a\left(t-|x-\xi|, \xi, y\right)
   v_t\left(t-|x-\xi|, \xi, y,t_0\right)\right]d\xi, $$$$ (t,x,t_0)\in\Upsilon_T,\ y\in{\Bbb R}^m.$$
   Setting in this relation  $x=0$ and using  evenness of the functions $a(t,x,y),\  v(t,x,y,t_0)$ in $x,$
  we get the equality:
\vspace{0.0cm}
   $$f_t(t,y,t_0)=\frac{\triangle g(y)}{2}\tilde S_t'(t,0,t_0)+\frac{g(y)}{4}a\left(\frac{t-t_0}{2},\frac{t+t_0}{2},y\right)$$$$+\int_0^{\frac{t-t_0}{2}}\left[\triangle  v\left(t-\xi, \xi, y,t_0\right)+
   a\left(t-\xi, \xi, y\right)
   v\left(t-\xi, \xi, y,t_0\right)\right]d\xi,\  (t,t_0)\in Q_T,\ y\in {\Bbb R}^m.$$
    Rewrite this equation substituting  $(t-t_0)/2$ by $|x|$,  $(t+t_0)/2$ -
    by
    $t$ and solving with respect  $a(t,x,y).$  Then we have

    $$a(t,x,y)=\frac{2\triangle g(y)}{g(y)}\tilde S_t'(t,0,t-2|x|)+\frac{4}{g(y)}f_t\left(t+|x|,y,t-|x|\right)
    $$$$-\frac{4}{g(y)}\int_0^{|x|}\Biggl[\triangle  v\left(t+|x|-\xi, \xi, y,t-|x|\right)$$$$+a\left(t+|x|-\xi, \xi,y\right)
    v_t\left(t+|x|-\xi,\xi, y,t-|x|\right)\Biggr]d\xi, \,  (t,x)\in\Omega_T,\ y\in {\Bbb R}^m.\eqno(8)
    $$
    Thus, in order to find the value of the function  $a$ in the
    point  $(t,x,y),$ it is necessary to integrate function  $a(t,x,y)$ itself by segment with ends  $(t+|x|,0,\,y,\,0),\; ( t,|x|,\,y,\ 0)$, and
     $v(t,x,y,t_0)$~---~ by segment with ends    $(t+|x|,0,y,\ t-|x|),\; (\ t,|x|,y,\ t-|x|)$, which belong to the domain
    $\Upsilon_T\times{\Bbb R}^m$.


Note that the function $v$, even with respect to $x=0$, satisfies the condition $\partial v/\partial x|_{x=0}.$
   Taking into account this fact and considering the equations (4), (5), (3) for  $v$ in the domain $x>0$ we obtain
    $$\frac{\partial^2v}{\partial t^2}-\frac{\partial^2v}{\partial x^2}-\triangle v -a(t,x,y)v_t=0,\; 0<x<t-t_0,\  y\in\mathbb{R}^m,$$
    $$v \bigr|_{x=0}=f(t,y,t_0),\  \frac{\partial v}{\partial x}\biggr|_{x=0}=0,\;  0<t-t_0\leq T, \  y\in\mathbb{R}^m,$$
    then in accordance with the d'Alembert formula, which gives
    the Cauchy problem solution with an initial data on  $x=0,$ we find:
    $$v(t,x,y,t_0)=v_0(t,x,y,t_0)+\frac{1}{2}\iint\limits_{\triangle ^{'}(t,x)} \Biggl[\triangle  v\left(\tau,\xi,y,t_0\right)+a\left(\tau,\xi,y\right)v_t\left(\tau,\xi,y,t_0\right)\Biggr]d\xi d\tau,\eqno(9)
    $$
      where
$$
v_0(t,x,y,t_0)= \frac{1}{2}\left[f(t+x,y,t_0)+f(t-x,y,t_0)\right],
 $$  $$ \triangle'(t,x):=\left\{(\tau,\xi)\mid0< \xi< x,
    |\tau-t|< x-\xi\right\}, \, 0< x< t-t_0< T-x, \, y\in \mathbb{R}^m.$$
 Considering (8) for $x\geq 0,$ we have:
$$a(t,x,y)=a_0(t,x,y)$$$$-\frac{4}{g(y)}\int_0^{x}\Biggl[\triangle  v\left(t+x-\xi,\xi,y,t-x\right)+a(t+x-\xi,\xi,y)v_t(t+x-\xi,\xi,y,t-x)\Biggr]d\xi,  \eqno(10)$$
where
  $$
 a_0(t,x,y)=\frac{4}{g(y)}f_t(t+x,y,t-x)+\frac{2\triangle g(y)}{g(y)}\tilde S_t'(t,0,t-2x), \, 0\leq x\leq t\leq T-x, \, y\in \mathbb{R}^m.
 $$

To close the system of equations (9), (10), we use
$$v_t(t,x,y,t_0)=v_{0t}(t,x,y,t_0)+$$$$\frac{1}{2}\int\limits_{0}^{x} \Biggl[\triangle  v\left(t-x+\xi,\xi,y,t_0\right)-\triangle  v\left(t+x-\xi,\xi,y,t_0\right)$$$$+a\left(t-x+\xi,\xi,y\right)v_t\left(t-x+\xi,\xi,y,t_0\right)-a\left(t+x-\xi,\xi,y\right)v_t\left(t+x-\xi,\xi,y,t_0\right)\Biggr]d\xi d\tau,\eqno(11)
    $$

The system of equations (9), (10) is a closed integro-differential equations for the
functions $a$, $v$, $v_t$. Note that the operator $\triangle $ in the
function $v$ appears in the system only under the
integral signs. 

Let $w:=v_t$ and $w_0:=v_{0t}.$ We will henceforth consider system (9)--(11)  in the domain \vspace{0.1cm}
$$D_T=\Upsilon'_T\times \mathbb{R}^m,\  \Upsilon'_T=\left\{(t,x,t_0)|\ 0\leq x+t_0 \leq t \leq T-x\right\}.$$

   \renewcommand{\thesection}{\large 3}
\section{ The Main Results and Its Proof}

Let  $C_{(t,x,t_0)}\left(\Upsilon'_T;\ A_{s_0}\right)$ denote the class of
functions with values in $A_{s_0}$ $(s_0>0)$ which  are continuous in the
variables  $(t,x,t_0)$ in the domain $\Upsilon'_T.$  For fixed $(t,x,t_0),$ the
norm of a function $v(t,x,y,t_0)$ in $A_{s_0}$ will be denoted by
$\|v\|_{s_0}(t,x,t_0).$ The norm of a function $v$ in
$C_{(t,x,t_0)}\left(\Upsilon'_T;\,A_{s_0} \right)$ is defined  by the equality

$$
\|v\|_{C_{(t,x,t_0)}\left(\Upsilon'_T;\,A_{s_0}
\right)}=\sup\limits_{(t,x,t_0)\in \Upsilon'_T}\|v\|_{s_0}(t,x,t_0).
$$

Let  $C_{(t,x)}\left(G_T;\ A_{s_0}\right)$ be the class of
functions with values in $A_{s_0}$ which  are continuous in the
variables  $(t,x)$ in the domain $G_T=\left\{(t,x)|\ 0\leq x \leq t \leq T-x\right\}.$  For fixed $(t,x)$ the norm of a function $a(t,x,y)$ in $A_{s_0}$ will be denoted by
$\|a\|_{s_0}(t,x).$ The norm of a function $a$ in
$C_{(t,x)}\left(G_T;\, A_{s_0}\right)$ is defined  as
$$
\|a\|_{C_{(t,x)}\left(G_T;\,A_{s_0}
\right)}=\sup\limits_{(t,x)\in G_T}\|a\|_{s_0}(t,x).
$$

Denote also by $C\left(Q_{T};A_{s_0}\right)$ the   class of functions with values in $A_{s_0}$ which are continuous with respect to
 $t, \, t_0$   in domain  $Q_{T}.$



{\bf Theorem 1.} {\it Let $f\left(+t_0, y, t_0\right)=0,$ $\left|g(x)\right|\geq g_0>0,$ $g_0$ is a known number,  and
$$\left\{\frac{1}{g(y)}, \, \frac{\triangle g(y)}{g(y)} \right\}\in A_{s_0}; \,
\left\{f(t, y, t_0), \, f_t(t, y, t_0) \right\}\in C\left(Q_{T};A_{s_0}\right),$$
in addition, the relations  $$\max\left\{2\left\|\frac{\triangle g(y)}{g(y)}\right\|_{s_0},
\max\limits_{(t, t_0)\in Q_{T}}\left\|f(t, y, t_0)\right\|_{s_0}, \, \max\limits_{(t, t_0)\in Q_{T}}
\left \|\frac{4f_t(t, y, t_0)}{g(y)}\right\|_{s_0} \right\}\leq \frac{R}{2}$$ are valid   for some fixed $s_0>0,$ $R.$

Then there there is such a number $b\in\left(0, T/(2s_0)\right),$
%$bs_0<T/2,$
$b=b\left(s_0, R, T\right)$ that for each $s\in(0, s_0)$
in the domain $D_T\cap\left\{(t, x, y, t_0): 0\leq x+t_0 \leq b(s_0-s)\right\}$ there exists the unique solution to the equations (9)--(111) and $v, w\in C_{(t,x,t_0)}\left(P_{sT};\ A_{s_0}\right),$ $a(t,x,y)\in
C_{(t,x)}\left(K_{sT};\ A_{s_0}\right),$  where $P_{sT}=\Upsilon'_T\cap\left\{(t, x, t_0): 0\leq x+t_0 < b(s_0-s)\right\},
$ $K_{sT}=G_T\cap\left\{(t, x, t_0): 0\leq x+t_0 < b(s_0-s)\right\},$ moreover
$$\left\|v-v_0\right\|_{s}(t, x, t_0)\leq R, \ (t, x, t_0)\in P_{sT},$$$$
\,\left\|w-w_0\right\|_{s}(t, x)\leq\frac{ R}{s_0-s},\quad \left\|a-a_0\right\|_{s}(t, x)\leq\frac{ R}{s_0-s},\, (t, x)\in K_{sT}.$$}

\medskip


{\bf Proof.} Under the conditions of  Theorem 1 we have
$$v_0, w_0\in C_{(t,x,t_0)}\left(\Upsilon'_T;\,A_{s_0} \right), \,  a_0\in C_{(t,x)}\left(G_T;\ A_{s_0}\right),$$$$ \|v_0\|_s(t,x,t_0)\leq R,\ \|w_0\|_s(t,x,t_0)\leq R,\ (t,x,t_0) \in \Upsilon'_T, $$$$\, \|a_0\|_s(t,x)\leq R,\
 (t,x)\in G_T,\ 0<s<s_0.$$ Let  $b_n$
  denote the members  of the monotone decreasing sequence that is defined by the equalities
\vspace{0.1cm}
$$
b_{n+1}=\frac{b_n}{1+1/(n+1)^2},\ n=0,\, 1,\, 2,\,\dots.
$$
 Denote
 \vspace{0.1cm}
$$
b=\lim\limits_{n\rightarrow \infty}b_n=b_0\prod_{n=0}^{\infty}\left(1+1/(n+1)^2\right)^{-1}.
$$

The number  $b_0\in(0,T/(2s_0))$ will be chosen in an appropriate way.
For the system of equations (9)--(11) we construct the process of successive
approximations according to the following scheme:  \vspace{0.1cm}

$$v_{n+1}(t,x,y,t_0)=v_0(t,x,y,t_0)+$$
   $$
    +\frac{1}{2}\iint\limits_{\triangle ^{'}(t,x)} \Biggl[\triangle  v_n\left(\tau,\xi,y,t_0\right)+a_n\left(\tau,\xi,y\right)v_n\left(\tau,\xi,y,t_0\right)\Biggr]\,d\tau d\xi,\ 0\leq x\leq t-t_0\leq T-x,
    $$
$$w_{n+1}(t,x,y,t_0)=w_{0}(t,x,y,t_0)+$$$$\frac{1}{2}\int\limits_{0}^{x} \Biggl[\triangle  v_n\left(t-x+\xi,\xi,y,t_0\right)-\triangle  v_n\left(t+x-\xi,\xi,y,t_0\right)$$$$+a_n\left(t-x+\xi,\xi,y\right)w_n\left(t-x+\xi,\xi,y,t_0\right)-a_n\left(t+x-\xi,\xi,y\right)w_n\left(t+x-\xi,\xi,y,t_0\right)\Biggr]d\xi d\tau,
    $$
$$a_{n+1}(t,x,y)=a_0(t,x,y)$$$$-\frac{4}{g(y)}\int_0^{x}\Biggl[\triangle  v_n\left(t+x-\xi,\xi,y,t-x\right)+a_n(t+x-\xi,\xi,y)w_n(t+x-\xi,\xi,y,t-x)\Biggr]d\xi,$$$$  0\leq x\leq t\leq T-x.$$


Define the function   $\tilde s_n(x)$ by the formula \vspace{0.1cm}
$$
\tilde s_n(x)=\frac{s+\nu^n(x)}{2},\
\nu^n(x)=s_0-\frac{x}{b_n}.\eqno(12)
$$
 Introduce the notations $p_n=v_{n+1}-v_n$, $r_n=w_{n+1}-w_n$, $q_n=a_{n+1}-a_n$, $n=0,\,1,\,2,\dots .$
Then  $p_n, r_n, q_n$ satisfy the relations
\vspace{0.1cm}
$$p_0(t,x,y,t_0)=\frac{1}{2}\iint\limits_{\triangle ^{'}(t,x)} \Biggl[\triangle  v_0\left(\tau,\xi,y,t_0\right)+a_0\left(\tau,\xi,y\right)v_0\left(\tau,\xi,y,t_0\right)\Biggr]\, d\tau d\xi,
    $$
   $$r_{0}(t,x,y,t_0)=\frac{1}{2}\int\limits_{0}^{x} \Biggl[\triangle  v_0\left(t-x+\xi,\xi,y,t_0\right)-\triangle  v_0\left(t+x-\xi,\xi,y,t_0\right)$$$$+a_0\left(t-x+\xi,\xi,y\right)w_0\left(t-x+\xi,\xi,y,t_0\right)-a_0\left(t+x-\xi,\xi,y\right)w_0\left(t+x-\xi,\xi,y,t_0\right)\Biggr]d\xi d\tau,
    $$ $$
    (t,x,y,t_0)\in D_T,
    $$
 \vspace{0.1cm}
    $$q_0(t,x,y)=-\frac{4}{g(y)}\int_0^{x}\Biggl\{\triangle  v_0\left(t+x-\xi,\xi,y,t-x\right)$$$$+a_0(t+x-\xi,\xi,y)w_0(t+x-\xi,\xi,y,t-x)\Biggr\}d\xi,$$

$$p_{n+1}(t,x,y,t_0)=\frac{1}{2}\iint\limits_{\triangle ^{'}(t,x)} \Biggl\{\triangle  p_n(\tau,\xi,y,t_0)$$$$+q_n\left(\tau,\xi,y\right)w_{n+1}(\tau,\xi,y,t_0)+a_n\left(\tau,\xi,y\right)r_n\left(\tau,\xi,y,t_0\right)\Biggr\}\,d\tau d\xi , \quad (t,x,y,t_0)\in D_T,
    $$
$$r_{n+1}(t,x,y,t_0)=\frac{1}{2}\int\limits_{0}^{x} \Biggl[\triangle  p_n\left(t-x+\xi,\xi,y,t_0\right)-\triangle  p_n\left(t+x-\xi,\xi,y,t_0\right)$$$$+q_n\left(t-x+\xi,\xi,y\right)w_{n+1}\left(t-x+\xi,\xi,y,t_0\right)+a_n\left(t-x+\xi,\xi,y\right)r_{n}\left(t-x+\xi,\xi,y,t_0\right)$$$$-q_n\left(t+x-\xi,\xi,y\right)w_{n+1}\left(t+x-\xi,\xi,y,t_0\right)-a_n\left(t+x-\xi,\xi,y\right)r_{n}\left(t+x-\xi,\xi,y,t_0\right)\Biggr]d\xi d\tau,
    $$
   $$  \quad (t,x,y,t_0)\in D_T, $$
\vspace{0.1cm}
$$q_{n+1}(t,x,y)=-\frac{4}{g(y)}\int_0^{x}\Biggl\{\triangle p_n(t+x-\xi,\xi,y,t-x)$$$$+q_n(t+x-\xi,\xi,y)w_{n+1}(t+x-\xi,\xi,y,t-x)+a_n(t+x-\xi,\xi,y)r_n(t+x-\xi,\xi,y,t-x)\Biggr\}d\xi,$$$$  (t,x,y)\in G_T\times \mathbb{R}^m.$$


  Show that $b_0\in\left(0,\frac{T}{2s_0}\right)$
can be chosen so that the following inequalities be valid for all
$n=0,\,1,\,2,\dots: $
\vspace{0.1cm}
$$
\lambda_n=\max\biggl\{\sup\limits_{(t,x,s)\in
\hat F_n}\left[\|p_n\|_s(t,x,t_0)\frac{\nu^n(x)-s}{x}\right],\sup\limits_{(t,x,s)\in
F_n}\left[\|r_n\|_s(t,x,t_0)\frac{(\nu^n(x)-s)^2}{x}\right],$$
$$
\ \sup\limits_{(t,x,s)\in
F_n}\left[\|q_n\|_s(t,x)\frac{(\nu^n(x)-s)^2}{x}\right]
\biggr\}<\infty,\eqno(13)
$$
\vspace{0.1cm}
$$\|v_{n+1}-v_0\|_s(t,x,t_0)\leq R,\ \|w_{n+1}-w_0\|_s(t,x)\leq\frac{R}{s_0-s},\ 
\|a_{n+1}-a_0\|_s(t,x)\leq\frac{R}{s_0-s},\eqno(14)$$
 where
  $$\hat F_n=\left\{(t,x,t_0,s)|\ (t,x,t_0)\in \Upsilon'_T,\ 0\leq x+t_0< b_n(s_0-s),\ 0<s<s_0\right\},$$
  $$F_n=\left\{(t,x,s)|\ (t,x)\in G_T,\ 0\leq x< b_n(s_0-s),\ 0<s<s_0\right\}.$$
Indeed, using the  relations for  $p_n,\, q_n, $ we find
\vspace{0.1cm}
$$\|p_0\|_s(t,x,t_0)\leq \frac{1}{2}\iint\limits_{\triangle ^{'}(t,x)} \Biggl[\|\triangle  v_0\|_s\left(\tau,\xi,t_0\right)+\|a_0\|_s\left(\tau,\xi\right)\|w_0\|_s\left(\tau,\xi,t_0\right)\Biggr]\,d\tau d\xi
$$$$\leq \frac{1}{2}\iint\limits_{\triangle ^{'}(t,x)}\left[\frac{Rc_0}{(\tilde s_0(\xi)-s)^2}+R^2\right] d\tau d\xi.
$$

Here $c_0$ is a positive constant, such that
$$\|\triangle  v_0\|_s\leq c_0\frac{\|v_0\|_{\tilde s_n}}{\left(\tilde s_n-s\right)^2}, \
\tilde s_n>s>0,\ n=0,1,2,\dots.$$ It is easy to check $c_0=4m.$

 Taking the function  $\tilde s_n(\xi)$ from  (12),  for  $n=0$ we have  \vspace{0.1cm}
$$\|p_0\|_s(t,x,t_0)\leq\frac 12\int_0^x(x-\xi)\biggl[\frac{4Rc_0}{(\nu^0(\xi)-s)^2}+R^2\Biggr]d\xi $$$$\leq \frac 12
R\left[4c_0+s_0^2R\right]\int_0^x\frac{(x-\xi)d\xi}{(\nu^0(\xi)-s)^2}$$$$\leq
\frac 12b_0R\left[4c_0+s^2_0R\right]\frac{x}{\nu^0(x)-s},\ (t,x,s)\in
\hat F_0.$$

 \vspace{0.1cm}
Proceeding analogously, we obtain
 \vspace{0.1cm}
$$
\|r_0\|_s(t,x,t_0)\leq  \frac 12\int_0^{x}\left[\frac{4Rc_0}{(\nu^0(\xi)-s)^2}+R^2\right]d\xi\leq$$$$\leq
\frac 12R\left(4c_0+s_0^2R\right)\frac{x}{(\nu^0(x)-s)^2},\
(t,x,s)\in \hat F_0,$$
$$\|q_0\|_s(t,x)\leq  4g_0\int_0^{x}\left[\frac{4Rc_0}{(\nu^0(\xi)-s)^2}+R^2\right]d\xi\leq$$$$\leq
 4g_0R\left(4c_0+s_0^2R\right)\frac{x}{(\nu^0(x)-s)^2},\
(t,x,s)\in F_0.$$

\vspace{0.1cm}

 These estimates imply validity of inequality
(12) for $n=0.$
Moreover,we find
\vspace{0.1cm}
$$\|v_1-v_0\|_s(t,x,t_0)=\|p_0\|_s(t,x,t_0)\leq\frac{\lambda_0x}{\nu^0(x)-s}\leq\frac{\lambda_0b_1}{1-b_1/b_0}=b_0\lambda_0, \   (t,x,t_0,s)\in \hat F_1, $$
$$\|w_1-w_0\|_s(t,x,t_0)=\|w_0\|_s(t,x)\leq\frac{\lambda_0x}{(\nu^0(x)-s)^2}\leq\frac{4b_0\lambda_0}{s_0-s},\ (t,x,s)\in \hat F_1,$$
$$\|a_1-a_0\|_s(t,x)=\|a_0\|_s(t,x)\leq\frac{\lambda_0x}{(\nu^0(x)-s)^2}\leq\frac{4b_0\lambda_0}{s_0-s},\ (t,x,s)\in F_1.$$

Choosing   $b_0$ so as to have  $4b_0\lambda_0\leq R,$ we conclude
that inequalities (14)  are satisfied for   $n=0.$

By way of induction, we show  that inequalities (13) and
(14) are also valid for the other values of $n$ if $b_0$
is chosen suitably. Assume that inequalities (13) and
(14)  hold for  $n=0,\,1,\, 2 ,\dots ,i.$ Then,
$(t,x,t_0,s)\in \hat F_{i+1}$ we have\vspace{0.1cm}
$$\|p_{i+1}\|_s(t,x,t_0)\leq \frac{1}{2}\iint\limits_{\triangle ^{'}(t,x)} \Biggl\{\|\triangle  p_i\|_s\left(\tau,\xi,t_0\right)$$$$+\|q_i\|_s\left(\tau,\xi\right)\|w_{i+1}\|_s(\tau,\xi,t_0)+\|a_i\|_s\left(\tau,\xi\right)\|r_i\|_s\left(\tau,\xi,t_0\right)\Biggr\}\,d\tau d\xi
$$
$$\leq \frac{1}{2}\iint\limits_{\Delta(t,x)}\biggl[\frac{c_0\lambda_i\xi}{(s'_i(\xi)-s)^2(\nu^i(\xi)-s)}+\frac{\lambda_i\xi}{(\nu^i(\xi)-s)^2}\frac{R(1+s_0)}{(s_0-s)}
$$$$+\frac{\lambda_i
\xi}{(\nu^i(\xi)-s)^2}\frac{R(1+s_0)}{(s_0-s)}\biggr]d\tau
d\xi\leq\frac{\lambda_i}{2}\left(4c_0+2R(1+s_0)\right)\int_0^x
\frac{(x-\xi)\xi d\xi}{(\nu^{i+1}(\xi)-s)^3}$$$$\leq
\frac{\lambda_i }{2}b_0^2\left(4c_0+2R(1+s_0)\right)\frac{x}{\nu^{i+1}(x)-s}.$$
Here in the intermediate calculations, the function   $s_i'$ is
defined by equality (11) with   $n=i$ and the
inequalities \vspace{0.1cm}
$$
\|w_i\|_s(t,x,t_0)\leq R\frac{1+s_0}{s_0-s},\ \|a_i\|_s(t,x)\leq R\frac{1+s_0}{s_0-s}
$$
are used,  the
latter valid by the induction hypothesis, together with the
obvious inequalities $b_i\leq b_0$ and
$\nu^{i+1}(x)\leq\nu^{i}(x).$ Similar arguments for $r_{i+1},q_{i+1}$ lead to inequalities \vspace{0.1cm}
$$\|r_{i+1}\|_s(t,x,t_0)\leq \frac 12\int_0^{x}\biggl\{\frac{2c_0\lambda_i\xi}{(\tilde s_i(\xi)-s)^2(\nu^i(\xi)-s)}+\frac{2\lambda_i\xi}{(\nu^i(\xi)-s)^2}\frac{R(1+s_0)}{(s_0-s)}$$$$+\frac{2\lambda_i\xi}{(\nu^i(\xi)-s)^2}\frac{R(1+s_0)}{(s_0-s)}
\biggr\}d\xi\leq \lambda_i\left[4c_0+2R(1+s_0)\right]\int_0^{x}
\frac{\xi d\xi}{(\nu^{i+1}(\xi)-s)^3}$$$$\leq
\lambda_ib_0\left[4c_0+2R(1+s_0)\right]\frac{x}{(\nu^{i+1}(x)-s)^2},\
(t,x,s)\in \hat F_{i+1},$$

$$\|q_{i+1}\|_s(t,x)\leq4g_0\int_0^{x}\biggl\{\frac{c_0\lambda_i\xi}{(\tilde s_i(\xi)-s)^2(\nu^i(\xi)-s)}+\frac{\lambda_i\xi}{(\nu^i(\xi)-s)^2}\frac{R(1+s_0)}{(s_0-s)}$$$$+\frac{\lambda_i\xi}{(\nu^i(\xi)-s)^2}\frac{R(1+s_0)}{(s_0-s)}
\biggr\}d\xi\leq4\lambda_ig_0\left[4c_0+2R(1+s_0)\right]\int_0^{x}
\frac{\xi d\xi}{(\nu^{i+1}(\xi)-s)^3}$$$$\leq
4\lambda_ig_0b_0\left[4c_0+2R(1+s_0)\right]\frac{x}{(\nu^{i+1}(x)-s)^2},\
(t,x,s)\in F_{i+1}.$$

The obtained estimates yield  \vspace{0.1cm}
$$\lambda_{i+1}\leq\lambda_i\rho,\ \lambda_{i+1}<\infty,$$$$
\rho=b_0\left[4c_0+2R(1+s_0)\right]\max\left[\frac{b_0}{2},4g_0,1\right].$$
 Moreover,  we have \vspace{0.1cm}
$$\| v_{i+2}-v_0\|_s(t,x,t_0)\leq\sum_{n=0}^{i+1}\|p_n\|_s(t,x,t_0)\leq\sum_{n=0}^{i+1}\frac{\lambda_nx}{\nu^{n}(x)-s}\leq
\sum_{n=0}^{i+1}\frac{\lambda_nb_{i+2}}{1-b_{i+2}/b_n}$$
$$\leq\sum_{n=0}^{i+1}\lambda_nb_n(n+1)^2
\leq\lambda_0b_0\sum_{n=0}^{i+1}\rho^n(n+1)^2, \ (t,x,t_0,s)\in \hat F_{i+2},$$
\vspace{0.1cm}
$$\|w_{i+2}-w_0\|_s(t,x,t_0)\leq\sum_{n=0}^{i+1}\|r_n\|_s(t,x,t_0)\leq\sum_{n=0}^{i+1}\frac{\lambda_nx}{(\nu^{n}(x)-s)^2}\leq\frac{1}{s_0-s}
\sum_{n=0}^{i+1}\frac{\lambda_nb_{i+2}}{(1-b_{i+2}/b_n)^2}\leq$$$$\leq\frac{\lambda_0b_0}{s_0-s}\sum_{n=0}^{i+1}\rho^n(n+1)^4,\ (t,x,t_0,s)\in \hat F_{i+2},$$
$$\|a_{i+2}-a_0\|_s(t,x)\leq\sum_{n=0}^{i+1}\|q_n\|_s(t,x)\leq\sum_{n=0}^{i+1}\frac{\lambda_nx}{(\nu^{n}(x)-s)^2}\leq\frac{1}{s_0-s}
\sum_{n=0}^{i+1}\frac{\lambda_nb_{i+2}}{(1-b_{i+2}/b_n)^2}\leq$$$$\leq\frac{\lambda_0b_0}{s_0-s}\sum_{n=0}^{i+1}\rho^n(n+1)^4,\ (t,x,s)\in F_{i+2}.$$
\vspace{0.1cm}



Now we  choose   $b_0\in\left(0,\frac{T}{2s_0}\right)$ so as to obtain \vspace{0.1cm}
$$\rho<1,\ \lambda_0b_0\sum_{n=0}^{\infty}\rho^n(n+1)^4\leq
R.$$
 Then
\vspace{0.1cm}
$$\| v_{i+2}-v_0\|_s(t,x,t_0)\leq R,\ \|w_{i+2}-w_0\|_s(t,x,t_0)\leq\frac{R}{s_0-s},\ \ \ (t,x,t_0,s)\in \hat F_{i+2},$$$$
\|a_{i+2}-a_0\|_s(t,x)\leq\frac{R}{s_0-s},\ (t,x,s)\in F_{i+2}.$$
 Since the choice of $b_0$ is independent of the number of the approximation, all the successive approximations  $v_n$, $w_n$, $a_n$ belong
to
$$C_{(t,x,t_0)}\left(\hat F;\ A_s\right),\ \hat F=\bigcap_{n=0}^{\infty}\hat F_n,$$
and $$C_{(t,x)}\left( F;\ A_s\right),\  F=\bigcap_{n=0}^{\infty} F_n,$$
respectively. Moreover,
\vspace{0.1cm}
$$\| v_{n}- v_0\|_s(t,x,t_0)\leq R,\  \|w_{n}-w_0\|_s(t,x,t_0)\leq\frac{R}{s_0-s},(t,x,t_0,s)\in \hat F,$$
$$
\|a_{n}-a_0\|_s(t,x)\leq\frac{R}{s_0-s},\ (t,x,s)\in F.$$ For
$s\in(0,s_0)$ the series
\vspace{0.1cm}$$\sum_{n=0}^{\infty}( v_n- v_{n-1}),\ \sum_{n=0}^{\infty}( w_n- w_{n-1}),\
\sum_{n=0}^{\infty}(a_n-a_{n-1})$$ converge uniformly in the  norm
of the spaces
$$C_{(t,x,t_0)}\left(P_{sT}; \ A_s\right),\ P_{sT}= \Upsilon'_T\cap\left\{(t, x, t_0): 0\leq x+t_0 < b(s_0-s)\right\},$$
$$C_{(t,x)}\left(K_{sT}; \ A_s\right),\ K_{sT}= G_T\cap\left\{(t, x, t_0): 0\leq x+t_0 < b(s_0-s)\right\},$$

   therefore $v_n\rightarrow v $, $w_n\rightarrow w $, $a_n\rightarrow a,$ and the limit functions  $v$, $w$, $a$ are
elements of
   $C_{(t,x,t_0)}\left(P_{sT};\ A_s\right)$,  $C_{(t,x )}\left(K_{sT};\ A_s\right)$ respectively and satisfy equations (9)--(11).

   Now we prove    that this  solution  is unique.
Let  $(v,w,a)$ and  $( \widehat{v}, \widehat{w}, \widehat{a})$ be any two solutions satisfying the inequalities
$$\| v- v_0\|_s(t,x,t_0)\leq R,\ \|w-w_0\|_s(t,x,t_0)\leq\frac{R}{s_0-s}, \ (t,x,t_0,s)\in \hat F,$$$$
\|a-a_0\|_s(t,x)\leq\frac{R}{s_0-s},\ (t,x,s)\in F.
$$
Denote  $\tilde{p}=v-\widehat{v}$, $\tilde{r}=r-\widehat{r}$, $\tilde{q}=a-\widehat{a},$
$$\lambda:=\max\biggl\{\sup\limits_{(t,x,t_0,s)\in
\hat F}\left[\|\tilde{p}\|_s(t,x,t_0)\frac{\nu(x)-s}{x}\right],\ \sup\limits_{(t,x,t_0,s)\in
\hat F}\left[\|\tilde{r}\|_s(t,x,t_0)\frac{(\nu(x)-s)^2}{x}\right],$$$$
\sup\limits_{(t,x,s)\in
F}\left[\|\tilde{q}\|_s(t,x)\frac{(\nu(x)-s)^2}{x}\right]
\biggr\}<\infty,$$
 where  $\nu(x)=s_0-x/b$,
$b=b_0\prod\limits_{n=0}^{\infty}\left(1+1/(n+1)^2\right)^{-1}.$ Then,
for the functions $\tilde{p}$, $\tilde{q}$ can be
obtained the relations \vspace{0.1cm}

$$\tilde{p}(t,x,y,t_0)=
    \frac{1}{2}\iint\limits_{\triangle ^{'}(t,x)} \Biggl\{\triangle  \tilde p\left(\tau,\xi,y,t_0\right)$$$$+\tilde q\left(\tau,\xi,y\right)\widehat {w}(\tau,\xi,y,t_0)+a\left(\tau,\xi,y\right)\tilde r\left(\tau,\xi,y,t_0\right)\Biggr\}\,d\tau d\xi ,\  (t,x,y,t_0)\in D_T,
    $$
    $$
     \tilde r(t,x,y,t_0)=\frac{1}{2}\int\limits_{0}^{x} \Biggl[\triangle  \tilde p\left(t-x+\xi,\xi,y,t_0\right)-\triangle  \tilde p\left(t+x-\xi,\xi,y,t_0\right)$$$$+\tilde q\left(t-x+\xi,\xi,y\right)\widehat w\left(t-x+\xi,\xi,y,t_0\right)+a\left(t-x+\xi,\xi,y\right)\tilde r\left(t-x+\xi,\xi,y,t_0\right)$$$$-\tilde q\left(t+x-\xi,\xi,y\right)\widehat w\left(t+x-\xi,\xi,y,t_0\right)-a\left(t+x-\xi,\xi,y\right)\tilde r\left(t+x-\xi,\xi,y,t_0\right)\Biggr]d\xi d\tau,
    $$
      $$(t,x,y,t_0)\in D_T,$$
    \vspace{0.1cm}
$$\tilde q(t,x,y)=-\frac{4}{g(y)}\int_0^{x}\Biggl\{\triangle \tilde p\left(t+x-\xi,\xi,y,t-x\right)$$$$$$$$+\tilde q(t+x-\xi,\xi,y)\widehat {w}(t+x-\xi,\xi,y,t-x)+a(t+x-\xi,\xi,y)\tilde r(t+x-\xi,\xi,y,t-x)\Biggr\}d\xi,$$$$ (t,x,y)\in G_T\times \mathbb{R}^m.$$
  Show that $b_0\in\left(0,\frac{T}{2s_0}\right)$
can be chosen so that the following inequalities be valid for all
$n=0,\,1,\,2,\dots. $
Applying to these equations  the estimates given above, we find
the inequality in the form
$$\lambda\leq\lambda\rho',$$
$$\ \rho':=b\left[4c_0+2R(1+s_0)\right]\max\left[\frac{b}{2},4g_0,1\right]<\rho<1.$$
Consequently $\lambda=0.$ Therefore  $v=\widehat{v}$, $w=\widehat{w}$,$a=\widehat{a}$. Theorem 1 is proved.

\medskip

Consider the set $\Gamma$ of functions $f(t,y,t_0)$ representing the elements of $C( Q_T;\,A_{s_0}),\  s_0 > 0$,
for which the conditions of Theorem 1 are valid with $R$, $T$, $s_0$. Then we have the stability
theorem
\smallskip

{\bf Theorem 2. }{\it \  Let $f, \ \bar f \in \Gamma$. For the corresponding solutions $(v, w, a)$ and $(\bar v, \bar w, \bar a)$  to (9)--(11), we have
\vspace{0.2cm}
$$
\gathered
\|v-\bar{v}\|_s\leq cM,\  \|w-\bar{w}\|_s\leq \frac{cM}{s_0-s},\ (t,x,t_0 )\in
P_{sT},
\\
\|a-\bar{a}\|_s\leq \frac{cM}{s_0-s},\ (t,x )\in
K_{sT},\ 0<s<s_0,
\endgathered \eqno(15)
$$
where
$$
M=\max\big[\max\|f-\bar f\|_{s_0}(t,t_0),\ \max\|f_t-\bar f_{t}\|_{s_0}(t,t_0)\Big], \ (t,t_0)\in Q_T,
$$
and the constant $c$ depends on $R$, $T$, $s_0$.}

{\bf Proof.} For the differences $v-\bar v=\tilde v,\ a-\bar a=\tilde a$ and $f-\bar f= \tilde f$ the equalities follow from (9)--(10):

$$\tilde{v}(t,x,y,t_0)=\tilde v_0(t,x,y,t_0)
    +\frac{1}{2}\iint\limits_{\triangle ^{'}(t,x)} \Biggl\{\triangle  \tilde v\left(\tau,\xi,y,t_0\right)$$$$+\tilde a\left(\tau,\xi,y\right) {w}(\tau,\xi,y,t_0)+\bar a\left(\tau,\xi,y\right)\tilde w\left(\tau,\xi,y,t_0\right)\Biggr\}\,d\tau d\xi, \, (t,x,y,t_0)\in D_T, \eqno (16)
    $$
    $$
     \tilde w(t,x,y,t_0)= \tilde w_0(t,x,y,t_0)+\frac{1}{2}\int\limits_{0}^{x} \Biggl[\triangle  \tilde v\left(t-x+\xi,\xi,y,t_0\right)-\triangle  \tilde v\left(t+x-\xi,\xi,y,t_0\right)$$$$+\tilde a\left(t-x+\xi,\xi,y\right)w\left(t-x+\xi,\xi,y,t_0\right)+\bar a\left(t-x+\xi,\xi,y\right)\tilde w\left(t-x+\xi,\xi,y,t_0\right)$$$$-\tilde a\left(t+x-\xi,\xi,y\right) w\left(t+x-\xi,\xi,y,t_0\right)-\bar a\left(t+x-\xi,\xi,y\right)\tilde w\left(t+x-\xi,\xi,y,t_0\right)\Biggr]d\xi d\tau,\eqno (17)
    $$
    \vspace{0.1cm}
$$\tilde a(t,x,y)=\tilde a_0(t,x,y)-\frac{4}{g(y)}\int_0^{x}\Biggl\{\triangle \tilde v\left(t+x-\xi,\xi,y,t-x\right)$$$$$$$$+\tilde a(t+x-\xi,\xi,y){w}(t+x-\xi,\xi,y,t-x)+\bar a(t+x-\xi,\xi,y)\tilde w(t+x-\xi,\xi,y,t-x)\Biggr\}d\xi, \eqno(18)$$$$  (t,x,y)\in G_T\times\mathbb{R}^m,$$
where
$$
\tilde v_0(t,x,y,t_0)=\frac 12[\tilde f(t+x,y,t_0)+\tilde f(t-x,y,t_0)],
$$
$$
\tilde w_0(t,x,y,t_0)=\frac 12[\tilde f'_t(t+x,y,t_0)+\tilde f'_t(t-x,y,t_0)],
$$
$$
\tilde a_0(t,x,y)=\frac{4}{g(y)}\tilde f_t(t+x,y,t-x).
$$
It is obvious that
$$
\|\tilde v_0\|_{s_0}(t,x,t_0)\leq M,\ \|\tilde w_0\|_{s_0}(t,x,t_0)\leq M,\ (t,x,t_0)\in P_{sT},$$
$$\|\tilde a_0\|_{s_0}(t,x)\leq \frac{4}{\|g(y)\|_{s_0}}M,\ (t,x)\in K_{sT},\eqno (19)
$$
From Theorem 1 the estimates follow:
$$\|v\|_s \leq 2 R,\ \|w\|_s \leq \frac{R(1+s_0)}{s_0-s},\ 
\|a\|_s \leq \frac{R(1+s_0)}{s_0-s}.
$$

Applying the method of successive approximations used for the proof of Theorem 1 to the system of equations (16)--(18), which is linear with respect to $\tilde v$ and $\tilde a$, we find that,
the following inequalities are valid for solution to (16)--(18):
$$\|\tilde v-\tilde{v}_0\|_{s}(t,x,t_0)\leq c_1M,\ \|\tilde w-\tilde{w}_0\|_s(t,x,t_0) \leq \frac{c_1M}{s_0-s},(t,x,t_0)\in P_{sT},$$$$\ \|\tilde a-\tilde{a}_0\|_s(t,x) \leq \frac{c_1M}{s_0-s},\ (t,x)\in K_{sT},\ 0<s<s_0.
$$
where $c_1$ depends on $R$, $T$, $s_0$. Hence, by (19) inequalities (15) follow. Theorem 2 is proven.


   \renewcommand{\thesection}{\large 4}
\section{References}


  \vspace{0.2cm}
 \begin{flushleft}
1.  Romanov V. G. {\it Stability in Inverse Problems} [Russian], Nauchnyi Mir, Moscow (2005).
\end{flushleft}  
 \begin{flushleft}
2. Ovsyannikov L. V. A singular operator in a scale of Banach spaces, {\it Soviet Math. Dokl.}, vol. 6, no. 4, 1025--1028 (1965).
\end{flushleft}
  \begin{flushleft}
3. Ovsyannikov L. V. A nonlocal Cauchy problem in scales of Banach spaces, {\it Dokl. Akad. Nauk SSSR}, vol. 200, no. 4, 789--792 (1971).
\end{flushleft}
    \begin{flushleft}
4. Nirenberg L.  { \it Topics in Nonlinear Functional Analysis},
New York: Courant Institute Math.  Sci., New York Univ., 1974.
\end{flushleft}
     \begin{flushleft}
5.  Romanov V. G. Local solvability of some multidimensional inverse problems for hyperbolic equations, {\it Differ. Equ.}, vol. 25, no. 2, 203--209 (1989). http://www.mathnet.ru/links/7b1533bb6958f99f23cf0c9fcfb3ef57/de6759.pdf 
 \end{flushleft}
  \begin{flushleft}
6.  Romanov V. G. On solvability of inverse problems for hyperbolic equations in a class of functions analytic in part of variables,  {\it Dokl. Akad. Nauk SSSR}, vol. 304, no. 4, 807--811 (1989). https://zbmath.org/0682.35105
 \end{flushleft}
 \begin{flushleft}
7. Durdiev D. K. and Nuriddinov Z. Z. Determination of a multidimensional kernel in some parabolic integro--differential equation, {\it Journal of Siberian Federal University-Mathematics and Physics}, 14:1 (2021), 117--127. https://doi.org/10.17516/1997-1397-2021-14-1-117-127
 \end{flushleft}
  \begin{flushleft}
8. Bozorov Z. R., The problem of determining the two-dimensional kernel of a viscoelasticity equation, {\it Sib. Zh. Ind. Mat.}, vol. 23, no. 1, 28--45 (2020).  https://doi.org/10.33048/SIBJIM.2020.23.104 
\end{flushleft}
  \begin{flushleft}
9. Durdiev D. K. and Rahmonov A. A. The problem of determining the 2D kernel in a system of integro-differential equations of a viscoelastic porous medium, {\it J. Appl. Indust. Math.}, vol. 14, no. 2, 281--295 (2020). https://doi.org/10.33048/SIBJIM.2020.23.205
 \end{flushleft}
  \begin{flushleft}
10. Durdiev D. K. and Totieva Zh. D. The problem of determining the multidimensional kernel of the viscoelasticity equation, {\it Vladikavkaz. Mat. Zh.}, vol. 17, no. 4, 18--43 (2015). https://zbmath.org/?q=an:1474.45100
\end{flushleft}
  \begin{flushleft}
11.  Durdiev D. K. and Safarov Zh. Sh. Local solvability of the problem of definition of the spatial part of the multidimensional kernel in an integro-differential equation of hyperbolic type, {\it Vestnik Samarsk. Gos. Univ. Ser. Fiz.-Mat. Nauki}, vol. 29, no. 4, 37--47 (2012).
 \end{flushleft}
  \begin{flushleft}
12. Durdiev D. K. Some multidimensional inverse problems of memory determination in hyperbolic equations, Zh. Mat. Fiz., Anal. Geom., vol. 3, no. 4, 411--423 (2007). https://zbmath.org/?q=an:06137633
\end{flushleft}
  \begin{flushleft}
13.  Durdiev D. K. A multidimensional inverse problem for an equation with memory, {\it Sib. Math. J.}, vol. 35, no. 3, 514--521 (1994).
\end{flushleft}





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