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

\AtBeginDocument{{\noindent\small
{\em Electronic Journal of Differential Equations},
Vol. 2006(2006), No. 103, pp. 1--10.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu  (login: ftp)}
\thanks{\copyright 2006 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2006/103\hfil On a mixed nonlocal problem]
{On a mixed nonlocal problem for a wave equation}

\author[S. A. Beilin\hfil EJDE-2006/103\hfilneg]
{Sergei A. Beilin}

\address{Sergei A. Beilin \newline
Department of Mathematics, 
Samara State University, 
1, Ac. Pavlov st. 443011 Samara, Russia}
\email{sbeilin@narod.ru}

\date{}
\thanks{Submitted April 13, 2006. Published September 8, 2006.}
\subjclass[2000]{35L05, 35L20, 35L99}
\keywords{Nonlocal problem; integral condition; wave equation}

\begin{abstract}
 A nonlocal problem for a wave equation with integral condition
 is studied in a cylinder. The uniqueness and existence of a
 generalized solution is proved with the help of an a priori
 estimate and the Galerkin procedure, respectively.
\end{abstract}

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

\section{Introduction}

Certain problems of modern physics and technology can be
effectively described in terms of nonlocal problems for
partial differential equations. The history of nonlocal problems
with integral conditions for partial differential equations goes
back to \cite{Cannon:63}.
Cannon studied a problem for a heat equation, and in most papers,
devoted to nonlocal problems, parabolic and elliptic equations were
studied. Mixed problems with nonlocal integral conditions for
one-dimensional hyperbolic equations were considered in
\cite{Pulkina:MZ-1992,Bouziani:97,Gordeziani-Avalishvili:mm-2000,Beilin:EJDE-2001,Pulkina:MZ-2003,Pulkina:DU-2004}.

Nonlocal problem for a hyperbolic equation with $n$ space variables with
a different integral condition was considered in
\cite{Kozhanov-Pulkina:DAN-2005}

In this paper we consider an $n$-dimensional wave equation with
weighted integral nonlocal condition.
Consider a wave equation
    \begin{equation}\label{wRn:waveeq}
    u_{tt} - \Delta u + c(x,t) u = f(x,t)
    \end{equation}
in a cylinder $Q = \Omega \times (0,T)$, where $\Omega \in R^n$ is a bounded domain with a smooth boundary,
with Cauchy conditions
    \begin{equation}\label{wRn:cauchy_cond}
    u(x,0) = \phi(x), \ u_t(x,0) = \psi(x),
    \end{equation}
and an integral nonlocal condition
    \begin{equation}\label{wRn:int_cond}
\frac{\partial u}{\partial n} \big|_S
+ \int_0^t \int_{\Omega} K(x,\xi,\tau) u(\xi,\tau) d\xi d\tau  = 0,
\quad x\in\partial\Omega,
    \end{equation}
where $\phi(x),\psi(x),K(x,\xi,\tau)$ are given functions.

Denote $\widehat{W}_2^1(Q) = \{ v(x,t): v\in W_2^1(Q), v(x,T) = 0 \}$.
Let $u(x,t)$ be a solution of the problem.
Multiply the equation \eqref{wRn:waveeq} by the function
$v(x,t) \in \widehat{W}_2^1(Q)$ and integrate it over the cylinder $Q$.
After integrating by parts, we obtain
    \begin{align*}
    &\int_0^T \int_{\Omega} \left( \nabla u \nabla v - u_t v_t
    + c u v \right) dx\, dt \\
    &= \int_0^T \int_{\Omega} f v dx\,dt +
            \int_0^T \int_{\partial\Omega} \frac{\partial u}{\partial n}
            v ds\,dt +
                \int_{\Omega} \psi(x) v(x,0) dx.
    \end{align*}
After substituting the condition \eqref{wRn:int_cond}, we have:
    \begin{equation}
\begin{aligned}\label{wRn:gensoldef}
&\int_0^T \int_{\Omega} \left( \nabla u \nabla v
- u_t v_t + c u v \right) dx\,dt \\
&+ \int_0^T \int_{\partial\Omega} v(x,t) \int_0^t
\int_{\Omega} K(x,\xi,\tau) u(\xi,\tau) d\xi d\tau ds\,dt  \\
&= \int_0^T \int_{\Omega} f v dx\,dt + \int_{\Omega} \psi(x) v(x,0) dx.
\end{aligned}
\end{equation}

\begin{definition} \label{def1} \rm
A function $u(x,t) \in W_2^1(Q)$ is called a generalized solution
of \eqref{wRn:waveeq}-\eqref{wRn:cauchy_cond}-\eqref{wRn:int_cond}, if it
satisfies \eqref{wRn:gensoldef} for every $v \in \widehat{W}_2^1(Q)$
and $u(x,0) = \phi(x)$.
\end{definition}

The main result of this paper can be formulated as the following theorem.

\begin{theorem} \label{thm1}
If $\phi(x) \in W_2^1(\Omega)$, $\psi(x) \in L_2(\Omega)$,
$f(x,t) \in L_2(Q)$, $K(x,\xi,t) \in C(\Omega \times \Omega \times (0,T))$,
there exist $\frac{\partial K}{\partial \xi_i}$, $i=1\dots n$ and
    $$
    \max_{\bar Q}|K| \leq K_0, \quad
     \max_{\bar Q} \big| \frac{\partial K}{\partial \xi_i} \big| \leq K_1
    $$
then there exists a unique generalized solution to the problem
\eqref{wRn:waveeq}-\eqref{wRn:cauchy_cond}-\eqref{wRn:int_cond}.
\end{theorem}

The  following two sections provide the proof of the above theorem.

\section{Uniqueness}
Let $u_1 \neq u_2$ be solutions of \eqref{wRn:waveeq}-\eqref{wRn:cauchy_cond}-\eqref{wRn:int_cond};
then $u = u_1 - u_2$ is a solution of the same problem with
$f=\phi=\psi=0$; so, $u(x,t)$ satisfies
    \begin{equation} \label{wRn:gensoldef_0}
\begin{aligned}
&\int_0^T \int_{\Omega} \left( \nabla u \nabla v
- u_t v_t + c u v \right) dx\,dt  \\
&+ \int_0^T \int_{\partial\Omega} v(x,t) \int_0^t \int_{\Omega}
K(x,\xi,\tau) u(\xi,\tau) d\xi d\tau ds\,dt = 0.
    \end{aligned}
\end{equation}
Consider the function
    \begin{equation}\label{wRn:v_cases}
    v(x,t) =
        \begin{cases}
        \int_t^{\tau} u(x,\eta) d\eta,   & 0 \leq t \leq \tau, \\
        0,                          & \tau \leq t \leq T.
        \end{cases}
    \end{equation}
Note, that $v(x,t) \in \widehat{W}_2^1(Q)$, and
$v_t(x,t) = - u(x,t) \ \forall t \in [0,\tau]$.
Integration by parts in \eqref{wRn:gensoldef_0} will give us:
\[
    \int_0^T \int_{\Omega} \nabla u \nabla v dx\,dt
= - \int_0^{\tau} \int_{\Omega} \nabla v_t \nabla v dx\,dt
= \int_0^{\tau} \int_{\Omega} \nabla v \nabla v_t dx\,dt
- \int_{\Omega} |\nabla v|^2 |_0^{\tau} dx,
\]
hence,
\begin{gather*}
    \int_0^T \int_{\Omega} \nabla u \nabla v dx\,dt
= - \frac{1}{2} \int_{\Omega} |\nabla v(x,0)|^2 dx;
\\
    \int_0^T \int_{\Omega} u_t v_t dx\,dt = - \int_0^{\tau}
    \int_{\Omega} u_t u dx\,dt
=  \int_0^{\tau} \int_{\Omega} u u_t dx\,dt
   - \int_{\Omega} u^2 |_0^{\tau} dx,
\end{gather*}
hence,
\[
    \int_0^T \int_{\Omega} u_t v_t dx\,dt = \frac{1}{2} \int_{\Omega} u^2(x,\tau) dx.
\]
Thus, \eqref{wRn:gensoldef_0} will take the form:
\begin{align*}
& \frac{1}{2} \int_{\Omega} \left( |\nabla v(x,0)|^2
    + u^2(x,\tau) \right) dx\,dt \\
&= \int_0^T \int_{\partial\Omega} v(x,t) \int_0^t \int_{\Omega}
K(x,\xi,\tau) u(\xi,\tau) d\xi\, d\tau\, ds\,dt.
    \end{align*}
With the help of elementary inequalities we obtain the following
estimates:
    \begin{equation}
\begin{aligned}
&\frac{1}{2} \int_{\Omega} \left( |\nabla v(x,0)|^2 + u^2(x,\tau) \right)
  dx\,dt  \\
&\leq \int_0^T \int_{\partial\Omega} |v(x,t)| \int_0^t \int_{\Omega}
  |K(x,\xi,\tau)| |u(\xi,\tau)| d\xi d\tau ds\,dt  \\
&\leq K_0 \int_0^T \int_{\partial\Omega} |v(x,t)| \int_0^t
  \int_{\Omega} |u(\xi,\tau)| d\xi d\tau ds\,dt  \\
&\leq K_0 \int_0^T \int_{\partial\Omega} \left( v^2(x,t)
  + T |\Omega| \int_0^t \int_{\Omega} u^2(\xi,\tau) d\xi d\tau \right) ds\,dt
\\
&= K_0 \int_0^{\tau} \int_{\partial\Omega} v^2(x,t) ds\,dt + K_0 T^2
|\Omega| |\partial\Omega| \int_0^{\tau} \int_{\Omega} u^2(x,t) dx\,dt.
 \end{aligned} \end{equation}
Using the inequality
    $$
    \int_{\partial\Omega} v^2(x,t) ds \leq \int_{\Omega}
\left( \epsilon |\nabla v|^2 + c(\epsilon) v^2 \right) dx
    $$
and denoting $L = T^2 |\Omega| |\partial\Omega|$, we obtain
    \begin{equation*}
    \int_{\Omega} \left( |\nabla v(x,0)|^2 + u^2(x,\tau) \right) dx\,dt
 \leq K_0 \int_0^{\tau} \int_{\Omega}
\left( \epsilon |\nabla v|^2 + c(\epsilon) v^2 + L u^2 \right) dx\,dt.
 \end{equation*}
It's trivial that
    $$
    v^2 = \Big( \int_t^{\tau} u(x,\eta) d\eta \Big)^2
\leq \tau \int_0^{\tau} u^2 (x,\eta) d\eta,
    $$
and
    $$
    \int_0^{\tau} \int_{\Omega} v^2 dx\,dt \leq
\int_0^{\tau} \int_{\Omega} \left( \tau \int_0^{\tau} u^2 (x,\eta) d\eta
\right) dx\,dt
\leq \tau^2 \int_0^{\tau} \int_{\Omega} u^2(x,t) dx\,dt.
    $$
Using this result and denoting $C_0 = K_0 \max\{\epsilon, c(\epsilon), L \}$, we obtain the following inequality:
    \begin{equation}\label{wRn:uniq_ineq_v}
    \int_{\Omega} \left( |\nabla v(x,0)|^2 + u^2(x,\tau) \right) dx\,dt
 \leq C_0 \int_0^{\tau} \int_{\Omega} \left( |\nabla v(x,t)|^2 + u^2(x,t)
 \right) dx\,dt.
    \end{equation}
Consider the function
    $$
    w(x,t) = \int_0^t u(x,\eta) d\eta.
    $$
Then $v(x,t) = w(x,\tau) - w(x,t)$, $\nabla v(x,0) = \nabla w(x,\tau)$,
 and also
    $$
    |\nabla v|^2 = |\nabla w_(x,\tau) - \nabla w(x,t)|^2 \leq 2 |\nabla w(x,\tau)|^2 + 2 |\nabla w(x,t)|^2,
    $$
so we have
    $$
    \int_0^{\tau} \int_{\Omega} |\nabla v|^2 dx\,dt
\leq 2\tau \int_{\Omega} |\nabla w(x,\tau)|^2 dx
+ 2\int_0^{\tau} \int_{\Omega} |\nabla w|^2 dx\,dt.
    $$
Thus, the inequality \eqref{wRn:uniq_ineq_v} takes the form
    \begin{equation} \label{wRn:uniq_ineq_w}
\begin{aligned}
&\int_{\Omega} \left( |\nabla w(x,\tau)|^2 + u^2(x,\tau) \right) dx\\
&\leq 2 C_0 \tau \int_{\Omega} |\nabla w(x,\tau)|^2 dx
+ 2 C_0 \int_0^{\tau} \int_{\Omega} \left( |\nabla w|^2 + u^2 \right) dx\,dt.
    \end{aligned} \end{equation}
Because of arbitrariness of $\tau$, let $\tau$ satisfy $2 C_0 \tau < 1$.
Then it follows from \eqref{wRn:uniq_ineq_w}:
    \begin{equation} \label{wRn:uniq_ineq_w_final}
(1 - 2 C_0 \tau) \int_{\Omega} \left( |\nabla w(x,\tau)|^2 + u^2(x,\tau)
 \right) dx
\leq 2 C_0 \int_0^{\tau} \int_{\Omega} \left( |\nabla w|^2
+ u^2 \right) dx\,dt.
\end{equation}
Applying the Gronwall inequality, we conclude that
    $$
    \int_{\Omega} \left( |\nabla w(x,\tau)|^2 + u^2(x,\tau) \right) dx
 \leq 0 \quad \forall \tau \in [0,\frac{1}{2 C_0}],
    $$
whence it follows that
    $$
    u(x,\tau) = 0 \quad \forall \tau \in [0,\frac{1}{2 C_0}].
    $$
Obtaining the same inequality for the ``slices''
$\tau \in \big[\frac{(k-1)}{2 C_0}, \frac{k}{2 C_0}\big]$
to cover the $[0,T]$, we ensure that
    $$
    u(x,\tau) = 0 \quad \forall \tau \in [0,T].
    $$
Thus, the uniqueness is proved.


\section{Existence}

To prove the existence, we will apply the Galerkin method.
Let $w_k(x)$ be a fundamental system in $W_2^1(\Omega)$ such that
$(w_k,w_l)_{L_2(\Omega)} = \delta_{k,l}$.

We will try to find approximate solutions in the form
    \begin{equation}\label{wRn:gal_um}
    u^m(x,t) = \sum_{k=1}^m d_k(t) w_k(x),
    \end{equation}
where coefficients $d_k(t)$ are to be determined, from
    \begin{equation} \label{wRn:gal_equiv_1}
\begin{aligned}
&\int_{\Omega} \left( u_{tt}^m w_l + \nabla u^m \nabla w_l + c u^m w_l
\right) dx  \\
& + \int_{\partial\Omega} w_l(x) \int_0^t \int_{\Omega} K(x,\xi,\tau)
 u^m(\xi,\tau) d\xi d\tau ds
&= \int_{\Omega} f w_l dx.
    \end{aligned} \end{equation}
    \begin{equation}
    d_k(0) = \alpha_k, \quad d'_k(0) = \beta_k,
    \end{equation}
where $\alpha_k$ are coefficients of sums
$\phi^m(x) = \sum_{k=1}^m \alpha_k w_k$, approximating
the function $\phi(x)$ as $m \to \infty$ with respect
to the norm of $W_2^1(\Omega)$,
and $\beta_k = (\psi,w_k)_{L_2(\Omega)}$.

Substituting \eqref{wRn:gal_um} into \eqref{wRn:gal_equiv_1}, we obtain
\begin{align*}
\int_{\Omega} \sum_{k=1}^m \left( d''_k w_k w_l + d_k \nabla w_k \nabla w_l
 + c d_k w_k w_l \right) dx &\\
 + \int_{\partial\Omega} w_l \int_0^t \int_{\Omega} K(x,\xi,\tau)
\sum_{k=1}^m d_k(\tau) w_k(\xi) d\xi d\tau ds &= f_l(t),
\end{align*}
where $f_l(t) = (f, w_l)_{L_2(\Omega)}$. Changing the order of
 summation and integration, we have
\begin{align*}
\sum_{k=1}^m \left( d''_k(t) (w_k, w_l)_{L_2} + d_k(t)
(\nabla w_k, \nabla w_l)_{L_2} + d_k(t) (c w_k, w_l)_{L_2}) \right)& \\
 + \sum_{k=1}^m \int_0^t \left( d_k(\tau) \int_{\partial\Omega} w_l(x)
\int_{\Omega} K(x,\xi,\tau) w_k(\xi) d\xi\, ds \right) d\tau &= f_l(t).
\end{align*}
Denote, for short,
    \begin{gather}
    \gamma_{kl} (t) = (\nabla w_k, \nabla w_l) + (c w_k, w_l); \\
    \kappa_{kl} (\tau) = \int_{\partial\Omega} w_l(x) \int_{\Omega}
K(x,\xi,\tau) w_k(\xi) d\xi\, ds.
    \end{gather}
and obtain
    \begin{equation}\label{wRn:gal_syst_int_diff}
    \sum_{k=1}^m d''_k(t) \delta_{kl} + d_k (t) \gamma_{kl} (t) + \int_0^t d_k(\tau) \kappa_{kl} (\tau) d\tau = f_l(t).
    \end{equation}
The resulting system of integro--differential equations can be reduced to a system of
differential equations of the third order by differentiating
\eqref{wRn:gal_syst_int_diff}
with respect to $t$,
    \begin{equation}
    \sum_{k=1}^m d'''_k(t) \delta_{kl} + d'_k (t) \gamma_{kl} (t) + d_k(t) ( \kappa_{kl} (t) + \gamma'_{kl} (t) ) = f'_l(t).
    \end{equation}
Together with the initial conditions
    \begin{equation}
    d_k(0) = \alpha_k, \quad d'_k(0) = \beta_k, \quad
    d''_k(0) = f_l(0) - \alpha_k \gamma_k(0),
    \end{equation}
we have a Cauchy problem for the system of linear differential equations
with smooth coefficients, that is uniquely solvable.

Thus, for every $m$ there exists a unique $u^m(x,t)$, that satisfies
\eqref{wRn:gal_equiv_1}, in other words, the sequence $\{u^m\}$ is defined.
We shall investigate it's convergence. Multiply \eqref{wRn:gal_equiv_1}
 by $d'_l(t)$, sum by $l$ from $1$ to $m$ and, finally, integrate by
$t$ from $0$ to $\tau$,
\begin{align*}
& \int_0^{\tau} \int_{\Omega} \left( u_{tt}^m u_t^m
  + \nabla u^m \nabla u_t^m + c u^m u_t^m \right) dx\,dt \\
& + \int_0^{\tau} \int_{\partial\Omega} u_t^m \int_0^t
  \int_{\Omega} K(x,\xi,\eta) u^m(\xi,\eta) d\xi\,d\eta ds\,dt \\
&= \int_0^{\tau} \int_{\Omega} f u_t^m dx\,dt.
\end{align*}
Integration by parts on the left will give us
    \begin{equation}
    \int_0^{\tau} \int_{\Omega} u_{tt}^m u_t^m dx\,dt
= - \int_0^{\tau} \int_{\Omega} u_t^m u_{tt}^m dx\,dt
  + \int_{\Omega} \left( u_t^m \right)^2 |_0^{\tau} dx,
    \end{equation}
hence,
\begin{gather*}
    \int_0^{\tau} \int_{\Omega} u_{tt}^m u_t^m dx\,dt
 = \frac{1}{2} \int_{\Omega} \left( u_t^m (x,\tau) \right)^2 dx
  - \frac{1}{2} \int_{\Omega} \left( u_t^m (x,0) \right)^2 dx;
\\
    \int_0^{\tau} \int_{\Omega} \nabla u^m \nabla u_t^m dx\,dt
= - \int_0^{\tau} \int_{\Omega} \nabla u^m \nabla u_t^m dx\,dt + \int_{\Omega} \left( \nabla u^m \right)^2 |_0^{\tau} dx,
\end{gather*}
hence,
\[
    \int_0^{\tau} \int_{\Omega} \nabla u^m \nabla u_t^m dx\,dt
  = \frac{1}{2} \int_{\Omega} \left( \nabla u^m (x,\tau) \right)^2 dx
   - \frac{1}{2} \int_{\Omega} \left( \nabla u^m (x,0) \right)^2 dx;
\]
\begin{align*}
    \int_0^{\tau} \int_{\Omega} c u^m u_t^m dx\,dt
&= - \int_0^{\tau} \int_{\Omega} c u_t^m u^m dx\,dt
     - \int_0^{\tau} \int_{\Omega} c_t (u^m)^2 dx\,dt\\
&\quad + \int_{\Omega} c \left( u^m(x,t) \right)^2 |_0^{\tau} dx;
\end{align*}
hence,
\begin{align*}
    \int_0^{\tau} \int_{\Omega} c u^m u_t^m dx\,dt
&= \frac{1}{2} \int_{\Omega} c \left( u^m(x,\tau) \right)^2 dx
 -  \frac{1}{2} \int_{\Omega} c \left( u^m(x,0) \right)^2 dx  \\
&\quad - \frac{1}{2} \int_0^{\tau} \int_{\Omega} c_t (u^m)^2 dx\,dt ;
    \end{align*}
%%--
\begin{align*}
&\int_{\partial\Omega} \int_0^{\tau} u_t^m (x,t)
 \int_0^t \int_{\Omega} K(x,\xi,\eta) u^m(\xi,\eta) d\xi\,d\eta\, dt\,ds \\
&= - \int_{\partial\Omega} \int_0^{\tau} u^m (x,t)
     \int_{\Omega} K(x,\xi,t) u^m(\xi,\eta) d\xi\,dt\, ds \\
&\quad + \int_{\partial\Omega} \left( u^m (x,t) \int_0^t
  \int_{\Omega} K(x,\xi,\eta) u^m(\xi,\eta) d\xi\, d\eta \right)
  \Bigr|_0^{\tau} ds \\
&= - \int_{\partial\Omega} \int_0^{\tau} u^m (x,t)
  \int_{\Omega} K(x,\xi,t) u^m(\xi,\eta) d\xi\,dt\,ds  \\
&\quad + \int_{\partial\Omega} u^m (x,\tau) \int_0^{\tau}
  \int_{\Omega} K(x,\xi,\eta) u^m(\xi,\eta) d\xi\,d\eta\, ds.
\end{align*}
Finally, we obtain
    \begin{equation}\label{wRn:gal_equiv_4}
 \begin{aligned}
&  \frac{1}{2} \int_{\Omega} \left( \left( u_t^m(x,\tau) \right)^2
  + \left| \nabla u^m(x,\tau) \right|^2 + c(x,\tau) \left( u^m(x,\tau)
    \right)^2 \right) dx  \\
&= \frac{1}{2} \int_{\Omega} \left( \left( u_t^m(x,0) \right)^2
   + \left| \nabla u^m(x,0) \right|^2 + c(x,0) \left( u^m(x,0)
    \right)^2 \right) dx  \\
&\quad + \frac{1}{2} \int_0^{\tau} \int_{\Omega} c_t \left( u_m \right)^2
  dx\,dt + \int_0^{\tau} \int_{\Omega} f u_t^m dx\,dt  \\
&\quad + \int_0^{\tau} \int_{\partial\Omega} u^m(x,t) \int_{\Omega}
  K(x,\xi,t) u^m(\xi,t) d\xi ds\,dt \\
&\quad - \int_{\partial\Omega} u^m(x,\tau) \int_0^{\tau}
  \int_{\Omega} K(x,\xi,\eta) u^m(\xi,\eta) d\xi\,d\eta ds.
\end{aligned}
\end{equation}
Consider the right-most  integral in \eqref{wRn:gal_equiv_4}.
Applying the Cauchy inequality, we have
 \begin{align*}
&\int_0^{\tau} \int_{\partial\Omega} u^m(x,t)
 \int_{\Omega} K(x,\xi,t) u^m(\xi,t) d\xi\, ds\,dt  \\
&\leq \frac{1}{2} \int_0^{\tau} \int_{\partial\Omega}
  \left( u^m(x,t) \right)^2 ds\,dt
  + \frac{1}{2} \int_0^{\tau} \int_{\partial\Omega}
  \left( \int_{\Omega} K(x,\xi,t) u^m(\xi,t) d\xi \right)^2 ds\,dt;
\end{align*}
next, using an inequality in \cite[p.77]{Ladizhenskaja:73},
    \begin{equation}\label{ineq77}
    \int_{\partial\Omega} v^2(x,t) ds \leq \int_{\Omega}
    \left( \epsilon |\nabla v|^2 + c(\epsilon) v^2 \right) dx
    \end{equation}
we have
    \begin{align*}
&\frac{1}{2} \int_0^{\tau} \int_{\partial\Omega}
  \left( u^m(x,t) \right)^2 ds\,dt + \frac{1}{2} \int_0^{\tau}
  \int_{\partial\Omega} \Big( \int_{\Omega} K(x,\xi,t) u^m(\xi,t)
  d\xi \Big)^2 ds\,dt  \\
&\leq \frac{1}{2} \int_0^{\tau} \int_{\Omega}
  \big( \epsilon |\nabla u^m(x,t)|^2 + c(\epsilon) u^2 \big) dx\,dt\\
&\quad + \frac{1}{2} \int_0^{\tau} \int_{\Omega} \Big( \epsilon
   \left( \int_{\Omega} \nabla K(x,\xi,t) u^m(\xi,t) d\xi \right)^2 \\
&\quad +  c(\epsilon) \Big( \int_{\Omega} K(x,\xi,t) u^m(\xi,t)
  d\xi \Big)^2 \Big) dx\,dt.
\end{align*}
Thus,
    \begin{align*}
& \int_0^{\tau} \int_{\partial\Omega} u^m(x,t) \int_{\Omega}
   K(x,\xi,t) u^m(\xi,t) d\xi ds\,dt  \\
&\leq \frac{1}{2} \int_0^{\tau} \int_{\Omega} \left( \epsilon
      |\nabla u^m(x,t)|^2 + c(\epsilon) u^2 \right) dx\,dt  \\
&\quad + \frac{1}{2} \int_0^{\tau} \int_{\Omega} \epsilon |\Omega|
    \int_{\Omega} \left| \nabla K(x,\xi,t) u^m(\xi,t) \right|^2 d\xi dx\,dt \\
&\quad + \frac{1}{2} \int_0^{\tau} \int_{\Omega} c(\epsilon) |\Omega|
    \int_{\Omega} \left( K(x,\xi,t) u^m(\xi,t) \right)^2 d\xi dx\,dt \\
&\leq \frac{1}{2} \int_0^{\tau} \int_{\Omega} \left( \epsilon
    |\nabla u^m(x,t)|^2 + c(\epsilon) u^2 \right) dx\,dt
    + \frac{1}{2} \int_0^{\tau} \int_{\Omega} \epsilon
    |\Omega| K_1 \int_{\Omega} \left( u^m(\xi,t) \right)^2 d\xi dx\,dt \\
&\quad + \frac{1}{2} \int_0^{\tau} \int_{\Omega} c(\epsilon)
    |\Omega| K_0 \int_{\Omega} \left( u^m(\xi,t) \right)^2 d\xi dx\,dt,
\end{align*}
and, finally,
    \begin{equation} \label{wRn:gal_ineq_3}
\begin{aligned}
& \int_0^{\tau} \int_{\partial\Omega} u^m(x,t) \int_{\Omega}
  K(x,\xi,t) u^m(\xi,t) d\xi ds\,dt  \\
& \leq \frac{1}{2} \int_0^{\tau} \int_{\Omega} \left( \epsilon
  |\nabla u^m|^2 + \left( c(\epsilon)+\epsilon |\Omega|^2 K_1
  + c(\epsilon) |\Omega|^2 K_0 \right) \right) dx\,dt.
\end{aligned}
\end{equation}
Applying the modified Cauchy inequality to the second integral
in \eqref{wRn:gal_equiv_4}, we have
\begin{align*}
&\int_{\partial\Omega} u^m(x,\tau) \int_0^{\tau} \int_{\Omega}
   K(x,\xi,\eta) u^m(\xi,\eta) d\xi\,d\eta ds \\
&\leq \frac{\epsilon}{2} \int_{\partial\Omega} \left( u^m(x,\tau) \right)^2
  ds + \frac{1}{2\epsilon} \int_{\partial\Omega}
  \Big( \int_0^{\tau} \int_{\Omega} K(x,\xi,\eta) u^m(\xi,\eta)
   d\xi\,d\eta \Big)^2 ds.
\end{align*}
Using again the inequality \eqref{ineq77},
    \begin{equation} \label{wRn:gal_ineq_4}
\begin{aligned}
& \frac{\epsilon}{2} \int_{\partial\Omega} \left( u^m(x,\tau) \right)^2 ds
  + \frac{1}{2\epsilon} \int_{\partial\Omega} \Big( \int_0^{\tau}
    \int_{\Omega} K(x,\xi,\eta) u^m(\xi,\eta) d\xi\,d\eta \Big)^2 ds \\
&\leq \frac{\epsilon}{2} \int_{\Omega} \left( \mu |\nabla u^m(x,\tau)|^2
  + c(\mu) (u^m(x,\tau))^2 \right) dx \\
&\quad + \frac{1}{2\epsilon} \int_{\Omega} \Big( \tau \mu |\Omega|
  \int_0^{\tau} \int_{\Omega} |\nabla K(x,\xi,\eta)|^2 \left( u^m(\xi,\eta)
  \right)^2 d\xi\,d\eta   \\
&\quad + \tau c(\mu) |\Omega| \int_0^{\tau} \int_{\Omega}
  \left( K(x,\xi,\eta) \right)^2 \left( u^m(\xi,\eta)
  \right)^2 d\xi\,d\eta \Big) dx.
\end{aligned}
\end{equation}
Choose $\epsilon$ and $\mu$ so that $\epsilon\mu < 1, \ \epsilon c(\mu) < c_1$.
Inequalities~\eqref{wRn:gal_ineq_3} and~\eqref{wRn:gal_ineq_4}, give us:
\begin{align*}
&\int_{\Omega} \left( (u_t^m(x,\tau))^2 + (1-\epsilon\mu)
  |\nabla u^m (x,\tau)|^2 + (c_1 - \epsilon c(\mu)) (u^m(x,\tau))^2 \right)
  dx  \\
&\leq \int_{\Omega} \left( (u_t^m(x,0))^2 + |\nabla u^m(x,0)|^2
  + c_2 (u^m(x,0))^2 \right) dx \\
&\quad + \int_0^{\tau} \int_{\Omega} f^2 dx\,dt
   + \int_0^{\tau} \int_{\Omega} c_3 (u^m)^2 dx\,dt
      + \int_0^{\tau} \int_{\Omega} (u_t^m)^2 dx\,dt \\
&\quad + \int_0^{\tau} \int_{\Omega} \left( \epsilon |\nabla u^m|^2
      + h(\epsilon) (u^m)^2 \right) dx\,dt \\
&\quad + \frac{\mu}{\epsilon} K_1 \tau |\Omega|^2 \int_0^{\tau}
     \int_{\Omega} (u^m(x,t))^2 dx\,dt + \frac{c(\mu)}{\epsilon}
     K_0 \tau |\Omega|^2 \int_0^{\tau} \int_{\Omega} (u^m(x,t))^2 dx\,dt.
 \end{align*}
Thus, denoting
\begin{gather*}
    m = \min\{ 1, 1-\epsilon\mu, c_1-\epsilon c(\mu) \}, \\
 M = \max\{ c_3 + h(\epsilon) + \frac{\mu}{\epsilon}
     K_1 \tau |\Omega|^2 + \frac{c(\mu)}{\epsilon} K_0 \tau |\Omega|^2 \}
\end{gather*}
we obtain the  inequality
\begin{align*}
& m \int_{\Omega} \left( (u_t^m(x,\tau))^2 + |\nabla u^m(x,\tau)|^2
  + (u^m(x,\tau))^2 \right) dx  \\
&\leq \int_{\Omega} \left( (u_t^m(x,0))^2 + |\nabla u^m(x,0)|^2
  + (u^m(x,0))^2 \right) dx  \\
&\quad + M \int_0^{\tau} \int_{\Omega} \left( (u_t^m)^2 + |\nabla u^m|^2
     + (u^m)^2 \right) dx\,dt
   + \int_0^{\tau} \int_{\Omega} f^2(x,t) dx\,dt.
\end{align*}
Applying the Gronwall inequality and integrating on
$\tau$ from $0$ to $\tau$, we finally obtain
    \begin{equation}
    \| u^m \|_{W_2^1(Q_{\tau})} \leq C(T) \Big( \|f\|_{L_2(Q)}
   + \|\phi\|_{W_2^1(\Omega)} + \|\psi\|_{L_2(\Omega)} \Big).
    \end{equation}
Since $\|f\|_{L_2(Q)}, \|\phi\|_{W_2^1(\Omega)}, \|\psi\|_{L_2(\Omega)}$
are bounded, the sequence $\{u^m\}$ is bounded in $W_2^1(Q)$:
$\| u^m \|_{W_2^1(Q)} \leq \mathcal{D}(T)$; therefore, there exists
a weakly converging subsequence
(we will denote it $u^m$ again for simplicity); now we shall show,
that its limit $u \in W_2^1 (Q)$ is a desired solution.

To prove this, we multiply \eqref{wRn:gal_equiv_1} by the function
$h_l(t) \in W_2^1(0,T), h_l(T)=0$,
sum by $l$ from $1$ to $m$, and integrate on $t$ from $0$ to $T$;
denoting $\eta^m(x,t) = \sum_{l=1}^m w_l(x) h_l(t)$, we obtain
 \begin{align*}
&\int_0^T \int_{\Omega} \left( - u_t^m \eta_t^m + \nabla u^m \nabla \eta^m
 + c u^m \eta^m \right) dx\,dt \\
&+ \int_0^T \int_{\partial\Omega} \eta^m \int_0^t \int_{\Omega}
   K(x,\xi,\tau) u^m(\xi,\tau) d\xi d\tau ds\,dt \\
&= \int_{\Omega} u_t^m(x,0) \eta^m(x,0) dx + \int_0^T \int_{\Omega}
  f \eta^m dx\,dt.
\end{align*}
It is no a problem to go to the limit at $m \to \infty$ in
the latter expression but the one addend
    \begin{equation}\label{wRn:problem_int}
    \int_0^T \int_{\partial\Omega} \eta^m \int_0^t \int_{\Omega}
    K(x,\xi,\tau) u^m(\xi,\tau) d\xi d\tau ds\,dt.
    \end{equation}
Consider the integral
    $$
    \int_0^T \int_{\partial\Omega} \eta^m \int_0^t \int_{\Omega}
 K(x,\xi,\tau) \Bigl( u^m(\xi,\tau) - u(\xi,\tau) \Bigr) d\xi d\tau ds\,dt.
    $$
With the help of the Cauchy-Bunyakovski inequality, we have for the
``internal'' integral
\begin{align*}
&\int_0^t \int_{\Omega} K(x,\xi,\tau) \Bigl( u^m(\xi,\tau) - u(\xi,\tau)
\Bigr) d\xi d\tau  \\
&\leq \Big( \int_0^T \int_{\Omega} K^2(x,\xi,\tau) d\xi d\tau
\Big)^{1/2} \Big( \int_0^T \int_{\Omega} \big( u^m(\xi,\tau)
 - u(\xi,\tau) \big)^2 d\xi d\tau \Big)^{1/2} \\
&= \| K \|_{L_2(Q)} \| u^m - u \|_{L_2(Q)}.
\end{align*}
However, $\| K \|_{L_2(Q)} \leq |Q| K_0$, and
$\| u^m - u \|_{L_2(Q)} \to 0$ (?!),
therefore, we can go to the limit in \eqref{wRn:problem_int}. Hence,
the limit function $u$
satisfies \eqref{wRn:gal_equiv_1} for every
$\eta^m(x,t) = \sum_{l=1}^m w_l(x) h_l(t)$.

Denote $\mathcal{N}_m$ a set of the functions of the form
$\eta^m(x,t) = \sum_{l=1}^m w_l(x) h_l(t)$,
$h_l(t) \in W_2^1(0,T), h_l(T)=0$. While
$\bigcup_{m=1}^{\infty} \mathcal{N}_m$ is dense in ${\hat W}_2^1(Q)$,
then \eqref{wRn:gal_equiv_1} holds true for every function from
${\hat W}_2^1(Q)$.
This proves that $u(x,t)$ is a generalized solution of
\eqref{wRn:waveeq}-\eqref{wRn:cauchy_cond}-\eqref{wRn:int_cond}.

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\end{document}