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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2017 (2017), No. 125, pp. 1--14.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu}
\thanks{\copyright 2017 Texas State University.}
\vspace{8mm}}

\begin{document}
\title[\hfilneg EJDE-2017/125\hfil 
 Solvability of nonlocal inverse BVP]
{Solvability of nonlocal inverse boundary-value problem for a second-order
parabolic equation with integral conditions}

\author[E. Azizbayov, Y. Mehraliyev \hfil EJDE-2017/125\hfilneg]
{Elvin Azizbayov, Yashar Mehraliyev}

\address{Elvin Azizbayov \newline
Department of Computational Mathematics,
Baku State University,
Z. Khalilov Str. 23,
Baku, AZ1148, Azerbaijan}
\email{eazizbayov@bsu.edu.az}

\address{Yashar Mehraliyev \newline
Department of Differential and Integral equations,
 Baku State University,
Z. Khalilov Str. 23,
Baku, AZ1148, Azerbaijan}
\email{yashar\_aze@mail.ru}

\dedicatory{Communicated by Ludmila Pulkina}

\thanks{Submitted February 17, 2017. Published May 8, 2017.}
\subjclass[2010]{35R30, 35K10, 35A09, 35A02}
\keywords{Second order parabolic equation; finite domain;
\hfill\break\indent nonlocal integral condition; inverse value problem;
classical solution}

\begin{abstract}
 This article studies a nonlocal inverse boundary-value problem for a
 second-order parabolic equation on the rectangular domain.
 First, we  introduce a definition of a classical solution, and then
 the original problem is reduced to an equivalent problem.
 Existence and  uniqueness of the solution of the equivalent problem
 is proved using a contraction mapping.
 Finally, using the equivalency, the existence and uniqueness of
 classical solution is obtained.
\end{abstract}

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


\section{Introduction}

Practical requirements often lead to the problem of determining the coefficients
or the right hand side of the differential equations for some known data about
its solutions. Such problems are called inverse problems in mathematical physics.
Inverse problems arise in various fields of human activity,
such as seismology, mineral exploration, biology, medical visualization,
computed tomography, Earth remote sensing, spectral analysis, nondestructive
control, etc. Fundamentals of the theory and practice of research of inverse
problems were established and developed in the pioneering works by
Tikhonov \cite{Tikhonov}, Lavrent'ev \cite{LRSh}, \cite{LVR}, Ivanov \cite{IVT},
Romanov \cite{Romanov}, Denisov \cite{Den}.
Subsequently, the methods developed by them were applied to investigate
a wide scale of inverse problems by their pupils and followers.
Recently, there have been  many  studies of inverse problems for parabolic
and other types of equations. A more detailed bibliography and a classification
of problems are found in
\cite{BeLyPoSoFro,HLI,Ivanchov,IVT,Kozhanov,Nakamura,PT,Pulkina}.

Recently, problems with nonlocal conditions for partial differential equations
have been of great interest.  We note that  most of the publications
about problems with  spatially nonlocal conditions and integral conditions
for partial differential equations are found in\cite{KozPul,Gordeziani,Pulkina2}.
In \cite{Kirichenko}, a problem of time nonlocal integral conditions for
hyperbolic conditions is investigated.

In this article we study an inverse boundary-value problem
for second-order parabolic equations with nonlocal conditions.
A distinctive feature of this article is the consideration of
a parabolic equation with both spatial and time
non-local conditions.

\section{Formulation of the problem}

Let $T>0$ be a fixed number and
$ D_T : = \{ (x,t):0 \le x \le 1,0 \le t \le T\}$.
We consider the  equation
\begin{equation}\label{e2.1}
c(t)u_t (x,t) = u_{xx} (x,t) + a(t)u(x,t) + b(t)g(x,t) + f(x,t)
\end{equation}
in the rectangular domain $D_T$. The inverse problem has nonlocal
initial condition
\begin{equation}\label{e2.2}
u(x,0) + \delta u(x,T) + \int_0^T {p(t)u(x,t)dt}
= \varphi (x) \quad  (0 \le x \le 1),
\end{equation}
periodic boundary condition
\begin{equation}\label{e2.3}
u(0,t) = u(1,t) \quad  (0 \le t \le T),
\end{equation}
nonlocal integral condition
\begin{equation}\label{e2.4}
\int_0^1 {u(x,t)dx}  = 0 \quad  (0 \le t \le T),
\end{equation}
and the additional conditions
\begin{equation}\label{e2.5}
u(x_i ,t) = h_i (t) \quad  (i = 1,2;\; 0 \le t \le T),
\end{equation}
where $\delta  \ge 0$, $x_i  \in (0,1)$ ($i = 1,2$; $x_1  \ne x_2$) are
fixed numbers, $0 < c(t)$, $g(x,t)$, $f(x,t)$, $0 \le p(t)$,
$\varphi (x)$, $h_i (t)$   ($i = 1,2$) are given functions,
 $u(x,t),a(t),b(t)$ are the sought functions.

\begin{definition} \label{def1} \rm
      The triplete $\{ u(x,t),a(t),b(t)\}$ is said to be a classical solution
of  problem \eqref{e2.1}--\eqref{e2.5}, if  the functions
$u(x,t),a(t)$ and $b(t)$ satisfy the following conditions:
\begin{enumerate}
  \item The function $u(x,t)$ and its derivatives
$u_t (x,t),u_x (x,t),u_{xx} (x,t)$ are continuous in the domain $D_T$;

  \item the functions $a(t)$ and $b(t)$ are continuous on the interval $[0,T]$;

  \item equation \eqref{e2.1} and conditions
\eqref{e2.2}--\eqref{e2.5} are satisfied in the classical (usual) sense.
\end{enumerate}
\end{definition}

\begin{lemma} \label{lem2.2}
		Suppose that $\delta  \ge 0$, $0 < c(t) \in C[0,T]$,
$a(t) \in C[0,T]$ and  $0 \le p(t) \in C[0,T]$ hold.
Then the  problem
\begin{gather}\label{e2.6}
c(t)y'(t) = a(t)y(t)\quad (0 \le t \le T), \\
\label{e2.7}
y(0) + \delta y(T) + \int_0^T {p(t)y(t)dt}  = 0
\end{gather}
has a unique trivial solution.
\end{lemma}

\begin{proof}
Obviously, the general solution of equation \eqref{e2.6} has the form:
\begin{equation}\label{e2.8}
y(t) = ce^{\int_0^t {\frac{{a(\tau )}}{{c(\tau )}}d\tau } } .
\end{equation}
Using \eqref{e2.7} we obtain
\[
c\Big( {1 + \delta e^{\int_0^T {\frac{{a(t)}}{{c(t)}}dt} }
+ \int_0^T {p(t)e^{\int_0^t {\frac{{a(\tau )}}{{c(\tau )}}d\tau } } dt} } \Big) = 0.
\]
By $\delta  \ge 0$, and $p(t) \ge 0$, from the latter relation it is clear that
$c=0$. Substituting the value of $c=0$ in \eqref{e2.8}, we obtain
that  problem \eqref{e2.6}, \eqref{e2.7} has only the trivial solution.
 The proof is complete.
\end{proof}

\begin{theorem} \label{thm1}
Assume the following conditions are satisfied:
$\delta  \ge 0$,  $0 < c(t) \in C[0,T]$,  $0 \le p(t) \in C[0,T]$,
$f(x,t) \in C(D_T )$, $\varphi (x) \in C[0,1]$,
$\int_0^1 {f(x,t)dx}  = 0$ $(0 \le t \le T)$,
$g(x,t) \in C(D_T )$,
\[
\int_0^1 {g(x,t)dx}  = 0  \quad (0 \le t \le T),
\]
$h_i (t) \in C^1 [0,T]$ $(i = 1,2)$,
 $h(t) \equiv h_1 (t)g(x_2 ,t) - h_2 (t)g(x_1 ,t) \ne 0$
$(0 \le t \le T)$, and the compatibility conditions
\begin{gather}\label{e2.9}
\int_0^1 {\varphi (x)dx}  = 0, \\
\label{e2.10}
h_i (0) + \delta h_i (T) + \int_0^T {p(t)h_i (t)dt}
= \varphi (x_i ) \quad (i = 1,2)\,.
\end{gather}
Then the problem of finding a classical solution of
 \eqref{e2.1}--\eqref{e2.5} is equivalent to the problem of
determining functions $u(x,t) \in C^{2,1} (D_T )$, $a(t) \in C[0,T]$, and
$b(t) \in C[0,T]$, satisfying equation \eqref{e2.1},
 conditions \eqref{e2.2} and \eqref{e2.3}, and the conditions
\begin{gather}\label{e2.11}
u_x (0,t) = u_x (1,t) \quad  (0 \le t \le T),\\
\label{e2.12}
c(t)h'_i (t) = u_{xx} (x_i ,t) + a(t)h_i (t) + b(t)g(x_i ,t)
+ f(x_i ,t)
\end{gather}
for $i = 1,2$; $0 \le t \le T$.
\end{theorem}

\begin{proof}
Suppose that $\{ u(x,t),a(t),b(t)\}$ is a classical solution of
\eqref{e2.1}--\eqref{e2.5}. Integrating both sides of
 \eqref{e2.1} with respect to $x$ from 0 to 1 gives
\begin{equation}\label{e2.13}
\begin{aligned}
c(t)\frac{d}{{dt}}\int_0^1 {u(x,t)dx}
& = u_x (1,t) - u_x (0,t) + a(t)\int_0^1 {u(x,t)dx}  \\
&\quad + b(t)\int_0^1 {g(x,t)dx}  + \int_0^1 {f(x,t)dx} \quad (0 \le t \le T).
\end{aligned}
\end{equation}
Under the assumptions $\int_0^1 {f(x,t)dx}  = 0$ and
$\int_0^1 {g(x,t)dx}  = 0$  $(0 \le t \le T)$, we obtain \eqref{e2.11}.

Setting $x = x_i$ in  \eqref{e2.1} we obtain
\begin{equation}\label{e2.14}
\begin{aligned}
c(t)u_t (x_i ,t)& = u(x_i ,t) + a(t)u(x_i ,t)+ b(t)u(x_i ,t) \\
&\quad + f(x_i ,t) \quad (i = 1,2;\; 0 \le t \le T).
\end{aligned}
\end{equation}
Further, assuming $h_i (t) \in C^1 [0,T]$ $(i = 1,2)$ and differentiating
\eqref{e2.5}, we have
\begin{equation}\label{e2.15}
u_t (x_i ,t) = h_i (t) \quad (i = 1,2).
\end{equation}
From \eqref{e2.14}, by  \eqref{e2.5} and \eqref{e2.15},
we conclude that the relation \eqref{e2.12} is fulfilled.

Now, assume that $\{ u(x,t),a(t),b(t)\}$ is the solution of
 \eqref{e2.1}--\eqref{e2.3}, \eqref{e2.11},
 \eqref{e2.12}. Then from \eqref{e2.13}, taking into account
\eqref{e2.11}, we find
\begin{equation}\label{e2.16}
c(t)\frac{d}{{dt}}\int_0^1 {u(x,t)dx}
= a(t)\int_0^1 {u(x,t)dx} \quad  (0 \le t \le T).
\end{equation}
By  \eqref{e2.2} and \eqref{e2.9}, it is easy to see that
\begin{equation}\label{e2.17}
\begin{aligned}
&\int_0^1 {u(x,0)dx}  + \delta \int_0^1 {u(x,T)dx}
 + \int_0^T {p(t)\int_0^1 {u(x,t)dx} dt}  \\
&= \int_0^1 \Big( {u(x,0) + \delta u(x,T) + \int_0^T {p(t)u(x,t)dt} } \Big)dx \\
& = \int_0^1 {\varphi (x)dx}  = 0.
\end{aligned}
\end{equation}

Since, by Lemma \ref{lem2.2},  problem \eqref{e2.16}, \eqref{e2.17} has only
a trivial solution, it follows that
\[
\int_0^1 {u(x,t)dx} = 0 \quad (0 \le t \le T),
\]
i.e. the condition \eqref{e2.4} holds.

Moreover, from \eqref{e2.12} and \eqref{e2.14} we find
\begin{equation}\label{e2.18}
c(t)\frac{d}{{dt}}(u(x_i ,t) - h_i (t))
= a(t)(u(x_i ,t) - h_i (t)) \quad (i = 1,2;\; 0 \le t \le T).
\end{equation}
Using \eqref{e2.2} and the compatibility conditions \eqref{e2.10}
we have
%\label{e2.19}
\begin{align*}
&u(x_i ,0) - h_i (0) + \delta (u(x_i ,T) - h_i (0)) + \int_0^T {p(t)(u(x_i ,t)
- h_i (t))dt}  \\
& = \Big( {u(x_i ,0) + \delta u(x_i ,T) + \int_0^T {p(t)u(x_i ,t)dt} } \Big)
- \Big( {h_i (0) + \delta h_i (0) + \int_0^T {p(t)h_i (t)dt} } \Big) \\
& = \varphi (x_i ) - \Big( {h_i (0) + \delta h_i (0) + \int_0^T {p(t)h_i (t)dt} }
\Big) = 0 \quad (i = 1,2).
\end{align*}
From this equality and \eqref{e2.18},  by Lemma \ref{lem2.2}, we conclude
that  conditions \eqref{e2.5} are satisfied. The proof is complete.
\end{proof}

\section{Solvability of inverse boundary-value problem}

In \cite{BudSamTikh}, it is known that the system
\begin{equation}\label{e3.1}
1,\cos \lambda _1 x,\sin \lambda _1 x,\dots ,\cos \lambda _k x,
\sin \lambda _k x,\dots,
\end{equation}
where $\lambda _k  = 2k\pi$  $(k = 0,1,\ldots)$,
is a basis for $L_2 (0,1)$.

Since the system \eqref{e3.1} form a basis in $L_2 (0,1)$,
 we shall seek the first component $u(x,t)$ of classical solution
$\{ u(x,t),a(t),b(t)\}$ of the problem \eqref{e2.1}--\eqref{e2.3},
\eqref{e2.11}, \eqref{e2.12} in the form
\begin{equation}\label{e3.2}
u(x,t) = \sum_{k = 0}^\infty  {u_{1k} (t)\cos \lambda _k x}
 + \sum_{k = 1}^\infty  {u_{2k} (t)\sin \lambda _k x} \quad (\lambda  = 2k\pi ),
\end{equation}
where
\begin{gather*}
u_{10} (t) = \int_0^1 {u(x,t)dx} , \\
u_{1k} (t) = 2\int_0^1 {u(x,t)\cos \lambda _k x\,dx} \quad (k = 1,2, \ldots ),\\
u_{2k} (t) = 2\int_0^1 {u(x,t)\sin \lambda _k x\,dx} \quad (k = 1,2, \ldots ).
\end{gather*}
Then applying the formal scheme of the Fourier method, for determining of
unknown coefficients $u_{1k} (t)$ $(k = 0,1,\dots )$ and
$u_{2k} (t)$  $(k = 1,2,\dots)$ of function $u(x,t)$ from \eqref{e2.1}
and \eqref{e2.2} we have
\begin{gather}\label{e3.3}
c(t)u'_{10} (t) = F_{10} (t;u,a,b) \quad (0 \le t \le T),\\
\label{e3.4}
c(t)u'_{ik} (t) + \lambda _{ik}^2 u_{ik} (t) = F_{ik} (t;u,a,b) \quad
 (i = 1,2;k = 1,2 \ldots ;\; 0 \le t \le T), \\
\label{e3.5}
u_{ik} (0) + \delta u_{ik} (T) + \int_0^T {p(t)} u_{ik} (t)dt
 = \varphi _{ik}  \quad (i = 1,2;\; k = 0,1,2 \ldots ),
\end{gather}
where
\begin{gather*}
F_{ik} (t;u,a,b) = f_{ik} (t) + b(t)g_{ik} (t) + a(t)u_{ik} (t) \quad
 (i = 1,2; \; k = 0,1,2 \ldots ), \\
f_{10} (t) = \int_0^1 {f(x,t)dx} , \quad  g_{10} (t) = \int_0^1 {g(x,t)dx} , \\
f_{1k} (t) = 2\int_0^1 {f(x,t)\cos \lambda _k x\,dx} , \quad
 f_{2k} (t) = 2\int_0^1 {f(x,t)\sin \lambda _k x\,dx} , \\
g_{1k} (t) = 2\int_0^1 {g(x,t)\cos \lambda _k x\,dx} , \quad
g_{2k} (t) = 2\int_0^1 {g(x,t)\sin \lambda _k x\,dx} , \\
\varphi _{10}  = \int_0^1 {\varphi (x)dx} ,\quad
\varphi _{1k} (t) = 2\int_0^1 {\varphi (x)\cos \lambda _k x\,dx} , \\
\varphi _{2k} (t) = 2\int_0^1 {\varphi (x)\sin \lambda _k x\,dx}
\end{gather*}
for $k = 1,2, \dots$.

Solving  problem \eqref{e3.3}--\eqref{e3.5} we obtain
\begin{gather}\label{e3.6}
\begin{aligned}
u_{10} (t) &= (1 + \delta )^{ - 1}
\Big( {\varphi _{10}  - \int_0^T {p(t)u_{10} (t)dt}
 - \delta \int_0^T {\frac{1}{{c(t)}}F_{10} (t;u,a,b)dt} } \Big) \\
&\quad + \int_0^t {\frac{1}{{c(\tau )}}F_{10} (\tau ;u,a,b)d\tau } ,
\end{aligned}\\
\label{e3.7}
\begin{aligned}
u_{ik} (t)
&= \frac{{e^{ - \int_0^t {\frac{{\lambda _k^2 }}{{c(s)}}ds} } }}
{{1 + \delta e^{ - \int_0^T {\frac{{\lambda _k^2 }}{{c(s)}}ds} } }}
\Big( {\varphi _{ik}  - \int_0^T {p(t)u_{ik} (t)dt} } \Big) \\
&\quad - \frac{{\delta e^{ - \int_0^T {\frac{{\lambda _k^2 }}{{c(s)}}ds} } }}
{{1 + \delta e^{ - \int_0^T {\frac{{\lambda _k^2 }}{{c(s)}}ds} } }}
\int_0^T {\frac{1}{{c(\tau )}}F_{ik} (\tau ;u,a,b)
 e^{ - \int_\tau ^t {\frac{{\lambda _k^2 }}{{c(s)}}ds} } d\tau }  \\
&\quad + \int_0^t \frac{1}{{c(\tau )}}F_{ik}
(\tau ;u,a,b)e^{ - \int_\tau ^t {\frac{{\lambda _k^2 }}{{c(s)}}ds} } d\tau \quad
 (i = 1,2;\; k = 1,2, \ldots ).
\end{aligned}
\end{gather}
After substituting expressions $u_{1k} (t)$ $(k = 0,1,\dots)$ and
$u_{2k} (t)$  $(k = 1,2,\dots)$ in \eqref{e3.2}, we obtain
\begin{equation}\label{e3.8}
\begin{aligned}
u(x,t) 
& = (1 + \delta )^{ - 1} \Big( {\varphi _{10}
- \int_0^T {p(t)u_{10} (t)dt}
- \delta \int_0^T {\frac{1}{{c(t)}}F_{10} (t;u,a,b)dt} } \Big) \\
&\quad  + \int_0^t {\frac{1}{{c(\tau )}}F_{10} (\tau ;u,a,b)d\tau }\\
&\quad + \sum_{k = 1}^\infty  \Big\{
\frac{{e^{ - \int_0^t {\frac{{\lambda _k^2 }}{{c(s)}}ds} } }}
{{1 + \delta e^{ - \int_0^T {\frac{{\lambda _k^2 }}{{c(s)}}ds} } }}
\Big( {\varphi _{1k}  - \int_0^T {p(t)u_{1k} (t)dt} } \Big) \\
&\quad  - \frac{{\delta e^{ - \int_0^T {\frac{{\lambda _k^2 }}{{c(s)}}ds} } }}
{{1 + \delta e^{ - \int_0^T {\frac{{\lambda _k^2 }}{{c(s)}}ds} } }}
 \int_0^T {\frac{1}{{c(\tau )}}F_{1k} (\tau ;u,a,b)
 e^{ - \int_\tau ^t {\frac{{\lambda _k^2 }}{{c(s)}}ds} } d\tau }  \\
&\quad  +  {\int_0^t {\frac{1}{{c(\tau )}}F_{1k} (\tau ;u,a,b)
 e^{ - \int_\tau ^t {\frac{{\lambda _k^2 }}{{c(s)}}ds} } d\tau } } \Big\}
 \cos \lambda _k x \\
&\quad  + \sum_{k = 1}^\infty  \Big\{ \frac{{
e^{ - \int_0^t {\frac{{\lambda _k^2 }}{{c(s)}}ds} } }}
{1 + \delta e^{ - \int_0^T {\frac{{\lambda _k^2 }}{{c(s)}}ds} } }
\Big( \varphi _{2k}  - \int_0^T p(t)u_{2k} (t)dt  \Big) \\
&\quad  - \frac{{\delta e^{ - \int_0^T {\frac{{\lambda _k^2 }}{{c(s)}}ds} } }}
{{1 + \delta e^{ - \int_0^T {\frac{{\lambda _k^2 }}{{c(s)}}ds} } }}
 \int_0^T {\frac{1}{{c(\tau )}}F_{2k} (\tau ;u,a,b)
 e^{ - \int_\tau ^t {\frac{{\lambda _k^2 }}{{c(s)}}ds} } d\tau } \\
&\quad   + \int_0^t {\frac{1}{{c(\tau )}}F_{2k}
 (\tau ;u,a,b)e^{ - \int_\tau ^t {\frac{{\lambda _k^2 }}{{c(s)}}ds} } d\tau }
 \Big\}\sin \lambda _k x.
\end{aligned}
\end{equation}

Now, using \eqref{e3.2} and \eqref{e2.12} we have
\begin{gather}\label{e3.9}
\begin{aligned}
a(t)& = [h(t)]^{ - 1} \Big\{ (c(t)h'_1 (t) - f(x_1 ,t))g(x_2 ,t) - (c(t)h'_2 (t)
 - f(x_2 ,t))g(x_1 ,t) \\
&\quad  + \sum_{k = 1}^\infty  {\lambda _k^2 u_{1k} (t)(g(x_2 ,t)
 \cos \lambda _k x_1  - g(x_1 ,t)\cos \lambda _k x_2 )} \\
&\quad  + \sum_{k = 1}^\infty  {\lambda _k^2 u_{2k} (t)(g(x_2 ,t)
 \sin \lambda _k x_1  - g(x_1 ,t)\sin \lambda _k x_2 )}  \Big\},
\end{aligned} \\
\label{e3.10}
\begin{aligned}
b(t)& = [h(t)]^{ - 1} \Big\{ h_1 (t)(c(t)h'_2 (t) - f(x_2 ,t)) - h_2 (t)(c(t)h'_1 (t)
 - f(x_1 ,t)) \\
&\quad  + \sum_{k = 1}^\infty  {\lambda _k^2 u_{1k} (t)(h_1 (t)\cos \lambda _k x_2
- h_2 (t)\cos \lambda _k x_2 )}  \\
&\quad   + \sum_{k = 1}^\infty  {\lambda _k^2 u_{2k} (t)(h_1 (t)\sin \lambda _k x_2
 - h_2 (t)\sin \lambda _k x_1 )}  \Big\},
\end{aligned}
\end{gather}
where
\begin{equation}\label{e3.11}
h(t) \equiv h_1 (t)g(x_2 ,t) - h_2 (t)g(x_1 ,t) \ne 0 \quad  (0 \le t \le T).
\end{equation}

Taking into account \eqref{e3.7}, from \eqref{e3.9} and
\eqref{e3.10} we obtain
\begin{equation}\label{e3.12}
\begin{aligned}
a(t) 
&= [h(t)]^{ - 1} \{ (c(t)h'_1 (t) - f(x_1 ,t))g(x_2 ,t)
 - (c(t)h'_2 (t) - f(x_2 ,t))g(x_2 ,t) \\
&\quad  + \sum_{k = 1}^\infty  {\lambda _k^2 }
\Big[ {\frac{{e^{ - \int_0^t {\frac{{\lambda _k^2 }}{{c(s)}}ds} } }}
{{1 + \delta e^{ - \int_0^T {\frac{{\lambda _k^2 }}{{c(s)}}ds} } }}
\Big( {\varphi _{1k}  - \int_0^T {p(t)u_{1k} (t)dt} } \Big) - } \\
&\quad  - \frac{{\delta e^{ - \int_0^T {\frac{{\lambda _k^2 }}{{c(s)}}ds} } }}
 {{1 + \delta e^{ - \int_0^T {\frac{{\lambda _k^2 }}{{c(s)}}ds} } }}
 \int_0^T {\frac{1}{{c(\tau )}}F_{1k} (\tau ;u,a,b)
 e^{ - \int_\tau ^t {\frac{{\lambda _k^2 }}{{c(s)}}ds} } d\tau }  \\
&\quad  { + \int_0^t {\frac{1}{{c(\tau )}}F_{1k} (\tau ;u,a,b)
 e^{ - \int_\tau ^t {\frac{{\lambda _k^2 }}{{c(s)}}ds} } d\tau } } \Big]
 (g(x_2 ,t)\cos \lambda _k x_1 \\
&\quad - g(x_1 ,t)\cos \lambda _k x_2 ) \\
&\quad  + \sum_{k = 1}^\infty  {\lambda _k^2 }
\Big[ \frac{{e^{ - \int_0^t {\frac{{\lambda _k^2 }}{{c(s)}}ds} } }}
 {{1 + \delta e^{ - \int_0^T {\frac{{\lambda _k^2 }}{{c(s)}}ds} } }}
\Big( {\varphi _{2k}  - \int_0^T {p(t)u_{2k} (t)dt} } \Big) \\
&\quad  - \frac{{\delta e^{ - \int_0^T {\frac{{\lambda _k^2 }}
 {{c(s)}}ds} } }}{{1 + \delta e^{ - \int_0^T {\frac{{\lambda _k^2 }}{{c(s)}}ds} } }}
 \int_0^T {\frac{1}{{c(\tau )}}F_{2k} (\tau ;u,a,b)
 e^{ - \int_\tau ^t {\frac{{\lambda _k^2 }}{{c(s)}}ds} } d\tau }  \\
&\quad   + \int_0^t {\frac{1}{{c(\tau )}}F_{2k} (\tau ;u,a,b)
 e^{ - \int_\tau ^t {\frac{{\lambda _k^2 }}{{c(s)}}ds} } d\tau }  \Big]
 (g(x_2 ,t)\sin \lambda _k x_1  \\
&\quad - g(x_1 ,t)\sin \lambda _k x_2 ) \Big\},
\end{aligned}
\end{equation}
\begin{equation}\label{e3.13}
\begin{aligned}
b(t) &= [h(t)]^{ - 1} \Big\{ h_1 (t)(c(t)h'_2 (t) - f(x_2 ,t)) - h_2 (t)(c(t)h'_1 (t)
- f(x_1 ,t)) \\
&\quad  + \sum_{k = 1}^\infty  \lambda _k^2
\Big[ \frac{{e^{ - \int_0^t {\frac{{\lambda _k^2 }}{{c(s)}}ds} } }}
{{1 + \delta e^{ - \int_0^T {\frac{{\lambda _k^2 }}{{c(s)}}ds} } }}
\Big( {\varphi _{1k}  - \int_0^T {p(t)u_{1k} (t)dt} } \Big) \\
&\quad  - \frac{{\delta e^{ - \int_0^T {\frac{{\lambda _k^2 }}{{c(s)}}ds} } }}
{{1 + \delta e^{ - \int_0^T {\frac{{\lambda _k^2 }}{{c(s)}}ds} } }}
\int_0^T {\frac{1}{{c(\tau )}}F_{1k} (\tau ;u,a,b)
e^{ - \int_\tau ^t {\frac{{\lambda _k^2 }}{{c(s)}}ds} } d\tau }  \\
&\quad { + \int_0^t {\frac{1}{{c(\tau )}}F_{1k} (\tau ;u,a,b)
 e^{ - \int_\tau ^t {\frac{{\lambda _k^2 }}{{c(s)}}ds} } d\tau } } \Big]
(h_1 (t)\cos \lambda _k x_2 \\
&\quad  - h_2 (t)\cos \lambda _k x_1 ) \\
&\quad  + \sum_{k = 1}^\infty  \lambda _k^2
\Big[ \frac{{e^{ - \int_0^t {\frac{{\lambda _k^2 }}{{c(s)}}ds} } }}
{{1 + \delta e^{ - \int_0^T {\frac{{\lambda _k^2 }}{{c(s)}}ds} } }}
\Big( {\varphi _{2k}  - \int_0^T {p(t)u_{2k} (t)dt} } \Big) \\
&\quad  - \frac{{\delta e^{ - \int_0^T {\frac{{\lambda _k^2 }}{{c(s)}}ds} } }}
{{1 + \delta e^{ - \int_0^T {\frac{{\lambda _k^2 }}{{c(s)}}ds} } }}
\int_0^T {\frac{1}{{c(\tau )}}F_{2k} (\tau ;u,a,b)
e^{ - \int_\tau ^t {\frac{{\lambda _k^2 }}{{c(s)}}ds} } d\tau }  \\
&\quad  + \int_0^t {\frac{1}{{c(\tau )}}F_{2k} (\tau ;u,a,b)
e^{ - \int_\tau ^t {\frac{{\lambda _k^2 }}{{c(s)}}ds} } d\tau }
\Big](h_1 (t)\sin \lambda _k x_2  \\
&\quad - h_2 (t)\sin \lambda _k x_1 ) \Big\}.
\end{aligned}
\end{equation}

Analogously,  the following lemma was proved in \cite{Mehraliyev}.

\begin{lemma} \label{lem3.1}
If $\{ u(x,t),a(t),b(t)\}$ is a solution of
 \eqref{e2.1}--\eqref{e2.3}, \eqref{e2.11}, \eqref{e2.12},
 then the functions
\begin{gather*}
u_{10} (t) = \int_0^1 {u(x,t)dx} , \\
u_{1k} (t) = 2\int_0^1 {u(x,t)\cos \lambda _k x\,dx} \quad (k = 1,2, \ldots ), \\
u_{2k} (t) = 2\int_0^1 {u(x,t)\sin \lambda _k x\,dx} \quad (k = 1,2, \ldots )
\end{gather*}
satisfy system \eqref{e3.6}, \eqref{e3.7} on the interval $[0,T]$.
\end{lemma}

From Lemma \ref{lem3.1} it follows that to  prove the uniqueness of the solution
of problem \eqref{e2.1}--\eqref{e2.3}, \eqref{e2.11},
\eqref{e2.12}, it suffices to prove the uniqueness of the solution of
 \eqref{e3.8}, \eqref{e3.12}, \eqref{e3.13}.

Now, consider the  space
$B_{2,T}^3$  \cite{Mehraliyev} consisting  of  functions of the form
\[
u(x,t) = \sum_{k = 0}^\infty  {u_{1k} (t)\cos \lambda _k x}
+ \sum_{k = 1}^\infty  {u_{2k} (t)\sin \lambda _k x} \quad (\lambda  = 2k\pi )
\]
in domain $D_T$, where the functions $u_{1k} (t)$  $(k = 0,1,\dots)$,
$u_{2k} (t)$ $(k = 1,2,\dots)$, are continuous on $[0,T]$ and  satisfy the
condition
\[
\| {u_{10} (t)}\|_{C[0,T]}  + \Big( {\sum_{k = 1}^\infty
{\Big( {\lambda _k^3 \| {u_{1k} (t)} \|_{C[0,T]} } \Big)^2 } }
 \Big)^{1/2}  + \Big( {\sum_{k = 1}^\infty  {\Big( {\lambda _k^3
\| {u_{2k} (t)} \|_{C[0,T]} } \Big)^2 } } \Big)^{1/2}  <  + \infty .
\]
The norm in the space $B_{2,T}^3$ is
\begin{align*}
\| {u(x,t)} \|_{B_{2,T}^3 }
& = \| {u_{10} (t)} \|_{C[0,T]}
 + \Big( {\sum_{k = 1}^\infty  {\Big( {\lambda _k^3 \| {u_{1k} (t)} \|_{C[0,T]} }
\Big)^2 } } \Big)^{1/2}  \\
&\quad + \Big( {\sum_{k = 1}^\infty  {\Big( {\lambda _k^3
\| {u_{2k} (t)} \|_{C[0,T]} } \Big)^2 } } \Big)^{1/2} .
\end{align*}
We denote by $E_T^3$, the Banach space $B_{2,T}^3  \times C[0,T] \times C[0,T]$
of vector functions $z(x,t) = \{ u(x,t),a(t),b(t)\}$ with norm
\[
\| {z(x,t)} \|_{B_{2,T}^3 }  = \| {u(x,t)} \|_{B_{2,T}^3 }
+ \| {a(t)} \|_{C[0,T]}  + \| {b(t)} \|_{C[0,T]} .
\]
It is known that $B_{2,T}^3$ and $E_T^3$ are Banach spaces

Now consider the operator
\[
\Phi (u,a,b) = \{ \Phi _1 (u,a,b),\Phi _2 (u,a,b),\Phi _3 (u,a,b)\}
\]
in the space $E_T^3$, where
\begin{gather*}
\Phi _1 (u,a,b) = \tilde u(x,t) \equiv \sum_{k = 0}^\infty
{\tilde u_{1k} (t)\cos \lambda _k x}  + \sum_{k = 1}^\infty
 {\tilde u_{2k} (t)\sin \lambda _k x} , \\
\Phi _2 (u,a,b) = \tilde a(t), \quad
\Phi _3 (u,a,b) = \tilde b(t)
\end{gather*}
and the functions $\tilde u_{10} (t)$, $\tilde u_{ik} (t)$
$(i = 1,2;\; k = 0,1,2,\dots)$, $\tilde a(t)$ and $\tilde b(t)$
are equal to the right-hand sides of \eqref{e3.6}, \eqref{e3.7}, \eqref{e3.12},
and \eqref{e3.13} respectively.

Using simple transformations from \eqref{e3.6}, \eqref{e3.7}, \eqref{e3.12},
and \eqref{e3.13} we obtain
\begin{equation}\label{e3.14}
\begin{aligned}
&\| {\tilde u_{10} (t)} \|_{C[0,T]} \\
&\le (1 + \delta )^{ - 1} \Big[ | {\varphi _{10} } |
+ T\| {p(t)} \|_{C[0,T]} \| {u_{10} (t)} \|_{C[0,T]}  \\
&\quad  + \delta \| {\frac{1}{{c(t)}}} \|_{C[0,T]}
 \Big( \sqrt T \Big( {\int_0^T {| {f_{10} (\tau )} |^2 d\tau } } \Big)^{1/2}
 +  T\| {a(t)} \|_{C[0,T]} \| {u_{10} (t)} \|_{C[0,T]}  \\
&\quad   { + \sqrt T \| {b(t)} \|_{C[0,T]}
\Big( {\int_0^T {| {g_{10} (\tau )} |^2 d\tau } } \Big)^{1/2} } \Big) \Big] \\
&\quad  + \| {\frac{1}{{c(t)}}} \|_{C[0,T]}
\Big[ \sqrt T \Big( {\int_0^T {| {f_{10} (\tau )} |^2 d\tau } } \Big)^{1/2}\\
&\quad + T\| {a(t)} \|_{C[0,T]} \| {u_{10} (t)} \|_{C[0,T]}
+  \sqrt T \| {b(t)} \|_{C[0,T]} \Big( {\int_0^T {| {g_{10} (\tau )} |^2 d\tau } }
\Big)^{1/2}  \Big],
\end{aligned}
\end{equation}

\begin{equation}\label{e3.15}
\begin{aligned}
&\Big( {\sum_{k = 1}^\infty  Big( {\lambda _k^3 \| {\tilde u_{ik} (t)} 
\|_{C[0,T]} } \Big)^2 }  \Big)^{1/2}  \\
& \le 2\sqrt 2 \Big( {\sum_{k = 1}^\infty  {\Big( 
{\lambda _k^3 | {\varphi _{ik} } |} \Big)^2 } } \Big)^{1/2} 
 + 2\sqrt 2 T\| {p(t)} \|_{C[0,T]} \Big( {\sum_{k = 1}^\infty  {
\Big( {\lambda _k^3 \| {u_{ik} } \|_{C[0,T]} } \Big)^2 } } \Big)^{1/2}  \\
&\quad  + 2\sqrt 2 (1 + \delta )\| {\frac{1}{{c(t)}}} \|_{C[0,T]} 
 \sqrt T \Big( {\int_0^T {\sum_{k = 1}^\infty  
 {\Big( {\lambda _k^3 | {f_{ik} (\tau )} |} \Big)^2 d\tau } } } \Big)^{1/2} \\
&\quad  + 2\sqrt 2 (1 + \delta )\| {\frac{1}{{c(t)}}} \|_{C[0,T]} 
 T\| {a(t)} \|_{C[0,T]} \Big( {\sum_{k = 1}^\infty  
{\Big( {\lambda _k^3 \| {u_{ik} (t)} \|_{C[0,T]} } \Big)^2 } } \Big)^{1/2}  \\
&\quad + 2\sqrt {2T} (1 + \delta )\| {\frac{1}{{c(t)}}} \|_{C[0,T]}
  \| {b(t)} \|_{C[0,T]} \Big( {\int_0^T {\sum_{k = 1}^\infty  
 {\Big( {\lambda _k^3 | {g_{ik} (\tau )} |} \Big)^2 d\tau } } } \Big)^{1/2} ,
\end{aligned}
\end{equation}

\begin{align}
&\| {\tilde a(t)} \|_{C[0,T]}  \nonumber \\
& \le \| {[h(t)]^{ - 1} } \|_{C[0,T]}  
\Big\{ \big\| (c(t)h'_1 (t) - f(x_1 ,t))g(x_2 ,t) - (c(t)h'_2 (t) \nonumber \\
&\quad - f(x_2 ,t))g(x_1 ,t) \big\|_{C[0,T]} \nonumber \\
&\quad  + \Big( {\sum_{k = 1}^\infty  {\lambda _k^{ - 2} } } \Big)^{1/2} 
\| {| {g(x_2 ,t)} | + | {g(x_1 ,t)} |} \|_{C[0,T]} 
\Big[ {\sum_{i = 1}^2 {\Big( {\sum_{k = 1}^\infty  
{\Big( {\lambda _k^3 | {\varphi _{ik} } |} \Big)^2 } } \Big)^{1/2} } } \nonumber\\
&\quad  + T\| {p(t)} \|_{C[0,T]} \sum_{i = 1}^2 
{\Big( {\sum_{k = 1}^\infty  {\Big( {\lambda _k^3 \| {u_{ik} (t)} \|_{C[0,T]} }
 \Big)^2 } } \Big)^{1/2} }   \nonumber\\
&\quad  + (1 + \delta )\| {\frac{1}{{c(t)}}} \|_{C[0,T]} 
 \sqrt T \sum_{i = 1}^2 {\Big( {\int_0^T {\sum_{k = 1}^\infty  
 {\Big( {\lambda _k^3 | {f_{ik} (\tau )} |} \Big)^2 d\tau } } } \Big)^{1/2} }
 \label{e3.16} \\
&\quad  + (1 + \delta )\| {\frac{1}{{c(t)}}} \|_{C[0,T]} 
 T\| {a(t)} \|_{C[0,T]} \sum_{i = 1}^2 
 {\Big( {\sum_{k = 1}^\infty  {\Big( {\lambda _k^3 \| {u_{ik} (t)} \|_{C[0,T]} } 
 \Big)^2 } } \Big)^{1/2} }  \nonumber\\
&\quad  + (1 + \delta )\| {\frac{1}{{c(t)}}} \|_{C[0,T]} 
\sqrt T \| {b(t)} \|_{C[0,T]} \sum_{i = 1}^2 
{\Big( {\int_0^T {\sum_{k = 1}^\infty  
 {\big( {\lambda _k^3 | {g_{ik} (\tau )d\tau } |} \big)^2 } } } \Big)^{1/2} } 
 \Big] \Big\},  \nonumber 
\end{align}

\begin{equation}\label{e3.17}
\begin{aligned}
&\| \tilde b(t) \|_{C[0,T]}  \\
&\le \| [h(t)]^{ - 1}  \|_{C[0,T]}  
 \Big\{ \| (h_1 (t)(c(t)h'_2 (t) - f(x_2 ,t)) - h_2 (t)(c(t)h'_1 (t) \\
&\quad - f(x_1 ,t)) \|_{C[0,T]}  \\
&\quad  + \Big( {\sum_{k = 1}^\infty  {\lambda _k^{ - 2} } } \Big)^{1/2}
 \| {| {h_1 (t)} | + | {h_2 (t)} |} \|_{C[0,T]} 
\Big[ {\sum_{i = 1}^2 {\Big( {\sum_{k = 1}^\infty  {\Big( {\lambda _k^3 |
 {\varphi _{ik} } |} \Big)^2 } } \Big)^{1/2} } } \\
&\quad  + T\| {p(t)} \|_{C[0,T]} \sum_{i = 1}^2 
 {\Big( {\sum_{k = 1}^\infty  {\Big( {\lambda _k^3 \| {u_{ik} (t)} \|_{C[0,T]} }
  \Big)^2 } } \Big)^{1/2} }  \\
&\quad  + (1 + \delta )\| {\frac{1}{{c(t)}}} \|_{C[0,T]} 
 \sqrt T \sum_{i = 1}^2 {\Big( {\int_0^T {\sum_{k = 1}^\infty  
 {\Big( {\lambda _k^3 | {f_{ik} (\tau )} |} \Big)^2 d\tau } } } \Big)^{1/2} }  \\
&\quad  + (1 + \delta )\| {\frac{1}{{c(t)}}} \|_{C[0,T]} 
 T\| {a(t)} \|_{C[0,T]} \sum_{i = 1}^2 
 \Big( \sum_{k = 1}^\infty  \Big( {\lambda _k^3 \| {u_{ik} (t)} \|_{C[0,T]} } 
 \Big)^2   \Big)^{1/2}   \\
&\quad  + (1 + \delta )\| {\frac{1}{{c(t)}}} \|_{C[0,T]} 
\sqrt T \| {b(t)} \|_{C[0,T]} \\
&\quad\times \sum_{i = 1}^2 
\Big( \int_0^T \sum_{k = 1}^\infty  \Big( {\lambda _k^3 | {g_{ik} 
 (\tau )d\tau } |} \Big)^2   \Big)^{1/2}  \Big] \Big\}.
\end{aligned}
\end{equation}

Assume that the data for problem \eqref{e2.1}--\eqref{e2.3},
\eqref{e2.11}  and \eqref{e2.12} satisfy the following conditions
\begin{itemize}
 \item [(A1)] 
$ \varphi (x) \in C^2 [0,1]$, $\varphi '''(x) \in L_2 (0,1)$, 
 $\varphi (0) = \varphi (1)$, 
 $ \varphi '(0) = \varphi '(1),\varphi ''(0) = \varphi ''(1)$;

 \item [(A2)] 
$f(x,t),f_x (x,t),f_{xx} (x,t) \in C^2 [0,1]$, 
$f_{xxx} (x,t) \in L_2 (D_T )$,  $f(0,t) = f(1,t)$, 
 $f_x (0,t) = f_x (1,t)$,  $f_{xx} (0,t) = f_{xx} (1,t) = 0$  $(0 \le t \le T)$;

 \item[(A3)] 
$g(x,t),g_x (x,t),g_{xx} (x,t) \in C^2 [0,1]$, 
$g_{xxx} (x,t) \in L_2 (D_T )$,  $g(0,t) = g(1,t)$, 
$g_x (0,t) = f_x (1,t)$,  $g_{xx} (0,t) = g_{xx} (1,t) = 0$ $(0 \le t \le T)$;

 \item[(A4)] 
$\delta  \ge 0$, $0 \le p(t) \in C[0,T]$, 
$h_i (t) \in C^1 [0,T]$ $(i = 1,2)$, 
$h(t) = h_1 (t)g(x_2 ,t) - h_2 (t)g(x_1 ,t) \ne 0$ $(0 \le t \le T)$.
\end{itemize}

Then from \eqref{e3.14}--\eqref{e3.17} we find that
\begin{gather} \label{e3.18}
\begin{aligned}
\| {\tilde u(x,t)} \|_{B_{2,T}^3 }  
&\le A_1 (T) + B_1 (T)\| {a(t)} \|_{C[0,T]} \| {u(x,t)} \|_{B_{2,T}^3 }  \\
&\quad + C_1 (T)\| {u(x,t)} \|_{B_{2,T}^3 }  + D_1 (T)\| {b(t)} \|_{C[0,T]} ,
\end{aligned} \\
\label{e3.19}
\begin{aligned}
\| {\tilde a(t)} \|_{C[0,T]} 
& \le A_2 (T) + B_2 (T)\| {a(t)} \|_{C[0,T]} \| {u(x,t)} \|_{B_{2,T}^3 } \\
&\quad + C_2 (T)\| {u(x,t)} \|_{B_{2,T}^3 }  + D_2 (T)\| {b(t)} \|_{C[0,T]} ,
\end{aligned} \\
\label{e3.20}
\begin{aligned}
\| {\tilde b(t)} \|_{C[0,T]}  
&\le A_3 (T) + B_3 (T)\| {a(t)} \|_{C[0,T]} \| {u(x,t)} \|_{B_{2,T}^3 }  \\
&\quad + C_3 (T)\| {u(x,t)} \|_{B_{2,T}^3 }  + D_3 (T)\| {b(t)} \|_{C[0,T]} ,
\end{aligned}
\end{gather}
where
\begin{gather*}
\begin{aligned}
A_1 (T) 
&= (1 + \delta )^{ - 1} \Big( {2\| {\varphi (x)} \|_{L_2 (0,1)}  
+ 2\delta \| {\frac{1}{{c(t)}}} \|_{C[0,T]} \| {f(x,t)} \|_{L_2 (D_T )} } \Big)\\
&\quad  + \| {\frac{1}{{c(t)}}} \|_{C[0,T]} \| {f(x,t)} \|_{L_2 (D_T )}  
+ 4\sqrt 2 \| {\varphi '''(x)} \|_{L_2 (0,1)}  \\
&\quad  + 4\sqrt {2T} \| {\frac{1}{{c(t)}}} \|_{C[0,T]}
  \| {f_{xxx} (x,t)} \|_{L_2 (D_T )} ,
\end{aligned} \\
B_1 (T) = \big( {\delta (1 + \delta )^{ - 1}  + 1} \big)
 T\| {\frac{1}{{c(t)}}} \|_{C[0,T]} , \\
C_1 (T) = \left( {(1 + \delta )^{ - 1}  + 2\sqrt 2 } \right)T\| {p(t)} \|_{C[0,T]} ,\\
D_1 (T) = \left( {\delta (1 + \delta )^{ - 1}  + 1 + 2\sqrt 2 } \right)
\sqrt T \| {\frac{1}{{c(t)}}} \|_{C[0,T]} \| g_{xxx} (x,t) \|_{L_2 (D_T )} ,\\
\begin{aligned}
A_2 (T) &= \| [h(t)]^{ - 1}  \|_{C[0,T]} \Big\{ \big\| (c(t)h'_1 (t) 
 - f(x_1 ,t))g(x_2 ,t) - (c(t)h'_2 (t) \\
&\quad - f(x_2 ,t))g(x_1 ,t) \big\|_{C[0,T]}  
 + \Big( {\sum_{k = 1}^\infty  {\lambda _k^{ - 2} } } \Big)^{1/2} 
 \| {| {g(x_1 ,t)} | + | {g(x_2 ,t)} |} \|_{C[0,T]}  \\
&\quad\times 
{ \Big[ {2\| {\varphi '''(x)} \|_{L_2 (0,1)}  + 2(1 + \delta )
\| {\frac{1}{{c(t)}}} \|_{C[0,T]} \sqrt T \| {f_{xxx} (x,t)} \|_{L_2 (D_T )} }
 \Big]} \Big\},
\end{aligned}  \\
\begin{aligned}
B_2 (T) &= \| {[h(t)]^{ - 1} } \|_{C[0,T]} 
 \Big( {\sum_{k = 1}^\infty  {(\lambda _k^{ - 2} )} } \Big)^{1/2}  \\
&\quad\times \| {| {g(x_1 ,t)} | + | {g(x_2 ,t)} |} \|_{C[0,T]} (1 + \delta )
T\| {\frac{1}{{c(t)}}} \|_{C[0,T]} ,
\end{aligned}\\
C_2 (T) = \| {[h(t)]^{ - 1} } \|_{C[0,T]} \Big( {\sum_{k = 1}^\infty 
 {(\lambda _k^{ - 2} )} } \Big)^{1/2} \| {| {g(x_1 ,t)} | 
 + | {g(x_2 ,t)} |} \|_{C[0,T]} \| {p(t)} \|_{C[0,T]} T, \\
\begin{aligned}
D_2 (T) &= \| {[h(t)]^{ - 1} } \|_{C[0,T]} 
\Big( {\sum_{k = 1}^\infty  {(\lambda _k^{ - 2} )} } \Big)^{1/2} 
\| {| {g(x_1 ,t)} | + | {g(x_2 ,t)} |} \|_{C[0,T]}  \\
&\quad \times (1 + \delta )\| {\frac{1}{{c(t)}}} \|_{C[0,T]} 
\sqrt T \| {g_{xxx} (x,t)} \|_{L_2 (D_T )} ,
\end{aligned}\\
\begin{aligned}
A_3 (T) &= \| {[h(t)]^{ - 1} } \|_{C[0,T]} 
\Big\{ \big\| h_1 (t)(c(t)h'_2 (t) - f(x_2 ,t)) - h_2 (t)(c(t)h'_1 (t) \\
&\quad - f(x_1 ,t)) \|_{C[0,T]}  
 + \Big( {\sum_{k = 1}^\infty  {\lambda _k^{ - 2} } } \Big)^{1/2} \| {| {h_1 (t)} |
 + | {h_2 (t)} |} \|_{C[0,T]}  \\
&\quad \times    \Big[ {2\| {\varphi '''(x)} \|_{L_2 (0,1)}  
+ 2(1 + \delta )\| {\frac{1}{{c(t)}}} \|_{C[0,T]} \sqrt T \| {f_{xxx} (x,t)} 
\|_{L_2 (D_T )} } \Big] \Big\},
\end{aligned} \\
B_3 (T) = \| {[h(t)]^{ - 1} } \|_{C[0,T]} 
\Big( {\sum_{k = 1}^\infty  {(\lambda _k^{ - 2} )} } \Big)^{1/2} 
\| {| {h_1 (t)} | + | {h_2 (t)} |} \|_{C[0,T]} 
 (1 + \delta )T\| {\frac{1}{{c(t)}}} \|_{C[0,T]} , 
\\
C_3 (T) = \| {[h(t)]^{ - 1} } \|_{C[0,T]} 
\Big( {\sum_{k = 1}^\infty  {(\lambda _k^{ - 2} )} } \Big)^{1/2} \| {| {h_1 (t)} | 
+ | {h_2 (t)} |} \|_{C[0,T]} \| {p(t)} \|_{C[0,T]} T, 
\\
\begin{aligned}
D_3 (T) &= \| {[h(t)]^{ - 1} } \|_{C[0,T]} 
\Big( {\sum_{k = 1}^\infty  {(\lambda _k^{ - 2} )} } \Big)^{1/2}
 \| {| {h_1 (t)} | + | {h_2 (t)} |} \|_{C[0,T]}  \\
&\quad (1 + \delta )\| \frac{1}{c(t)} \|_{C[0,T]} 
\sqrt T \| g_{xxx} (x,t) \|_{L_2 (D_T )} .
\end{aligned}
\end{gather*}

From  \eqref{e3.18}--\eqref{e3.20} we conclude that
\begin{equation}\label{e3.21}
\begin{aligned}
&\| {\tilde u(x,t)} \|_{B_{2,T}^3 }  + \| {\tilde a(t)} \|_{C[0,T]} 
  + \| {\tilde b(t)} \|_{C[0,T]} \\
& \le A(T) + B(T)\| {a(t)} \|_{C[0,T]} \| {u(x,t)} \|_{B_{2,T}^3 } \\
&\quad + C(T)\| {u(x,t)} \|_{B_{2,T}^3 }  + D(T)\| {b(t)} \|_{C[0,T]} ,
\end{aligned}
\end{equation}
where
\begin{gather*}
A(T) = A_1 (T) + A_2 (T) + A_3 (T),\ \ B(T) = B_1 (T) + B_2 (T) + B_3 (T),\\
C(T) = C_1 (T) + C_2 (T) + C_3 (T),\ \ D(T) = D_1 (T) + D_2 (T) + D_3 (T).
\end{gather*}


\begin{theorem} \label{thm2}
If conditions {\rm (A1)--(A4)} and the condition
\begin{equation}\label{e3.22}
(B(T)(A(T) + 2) + C(T) + D(T))(A(T) + 2) < 1
\end{equation}
hold, then  problem \eqref{e2.1}--\eqref{e2.3}, \eqref{e2.11},
\eqref{e2.12} has a unique solution in the ball
$K = K_R$ $(\| z \|_{E_T^3 }  \le R \le A(T) + 2)$ of the space  $E_T^3$.
\end{theorem}

\begin{proof}
In the space $E_T^3$, we consider the equation
\begin{equation}\label{e3.23}
z = \Phi z,
\end{equation}
where $z = \{ u,a,b\}$, the components  $\Phi _i (u,a,b) \ \ (i = 1,2,3)$, 
of operator  $\Phi (u,a,b)$, defined by the right side of equations 
\eqref{e3.8}, \eqref{e3.12} and \eqref{e3.13}.

Consider the operator  $\Phi (u,a,b)$, in the ball $K = K_R$ of the space $E_T^3$. 
Similarly, with the aid of \eqref{e3.21} we obtain that for any 
 $z_1 ,z_2 ,z_3  \in K_R $ the following inequalities hold 
\begin{gather} \label{e3.24}
\begin{aligned}
&\| {\Phi z} \|_{E_T^3 }  \\ 
&\le A(T) + B(T)\| {a(t)} \|_{C[0,T]} \| {u(x,t)} \|_{B_{2,T}^3 }  
+ C(T)\| {u(x,t)} \|_{B_{2,T}^3 }  + D(T)\| {b(t)} \|_{C[0,T]} \\
&\le A(T) + B(T)(A(T) + 2)^2  + C(T)(A(T) + 2) + D(T)(A(T) + 2) \\
&< A(T) + 2,
\end{aligned} \\
\label{e3.25}
\begin{aligned}
&\| {\Phi z_1  - \Phi z_2 } \|_{E_T^3 } \\
& \le B(T)R\Big( {\| {a_1 (t) - a_2 (t)} \|_{C[0,T]}  
 + \| {u_1 (x,t) - u_2 (x,t)} \|_{B_{2,T}^3 } } \Big) \\
&\quad + C(T)\| {u_1 (x,t) - u_2 (x,t)} \|_{B_{2,T}^3 }  
 + D(T)\| {b_1 (t) - b_2 (t)} \|_{C[0,T]} .
\end{aligned}
\end{gather}

Then by  \eqref{e3.22}, from \eqref{e3.24} and \eqref{e3.25} it is clear that 
the operator $\Phi$ on the set  $K = K_R$ satisfy the conditions of the 
contraction mapping principle. Therefore the operator  $\Phi$ has 
a unique fixed point $\{ z\}  = \{ u,a,b\}$, in the ball  $K = K_R$, 
which is a solution of equation \eqref{e3.23}; i.e. in the sphere 
$K = K_R$ is the unique solution of the systems 
\eqref{e3.8}, \eqref{e3.12}, \eqref{e3.13}. Then the function $u(x,t)$, 
as an element of space  $B_{2,T}^3$, is continuous and has continuous 
derivatives $u_x (x,t)$ and $u_{xx} (x,t)$ in $D_T$.

Next, from \eqref{e3.4} it follows that $u'_{ik} (t)$ 
$(i = 1,2;\; k = 1,2 \ldots )$ is continuous on $[0,T]$ and consequently we have
\begin{align*}
&\Big( {\sum_{k = 1}^\infty  {(\lambda _k \| {u'_{ik} (t)} \|_{C[0,T]} )^2 } }
 \Big)^{1/2} \\
& \le \| {\frac{1}{{c(t)}}} \|_{C[0,T]} \sqrt 2 
\Big[ \Big( {\sum_{k = 1}^\infty  {\big( {\lambda _k^3 \| {u_{ik} (t)} \|_{C[0,T]} }
 \big)^2 } } \Big)^{1/2}  \\  
&\quad + \big\| {\| {f_x (x,t) + a(t)u_x (x,t) + b(t)g_x (x,t)} \|_{C[0,T]} } 
\big\|_{L_2 (0,1)}  \Big] <  + \infty  \quad  (i=1,2).
\end{align*}
Hence we conclude that the function $u_t (x,t)$ is continuous in the domain 
 $D_T$. Further, it is possible to verify that equation \eqref{e2.1}
and conditions \eqref{e2.2}, \eqref{e2.3}, \eqref{e2.11},
\eqref{e2.12} are satisfied in the usual sense.
 Consequently, $\{ u(x,t),a(t),b(t)\}$ is a solution of 
 \eqref{e2.1}--\eqref{e2.3}, \eqref{e2.11}, \eqref{e2.12},
 and by Lemma \ref{lem3.1} it is unique in the ball $K = K_R$. The proof is complete.
\end{proof}

From Theorem \ref{thm2} and  Theorem \ref{thm1}, it follows directly
 the following assertion.

\begin{theorem} \label{thm3}
Suppose that all assumptions of Theorem \ref{thm2}, and the compatibility conditions 
\eqref{e2.9}, \eqref{e2.10} hold. If
\[
\int_0^1 {f(x,t)dx}  = 0,\quad \int_0^1 {g(x,t)dx}  = 0 \quad (0 \le t \le T)
\]
then problem \eqref{e2.1}--\eqref{e2.5} has a unique
classical solution in the ball $K = K_R$.
\end{theorem}

\subsection*{Acknowledgements}
The authors would like to express their deep gratitude to the editorial 
team and the anonymous referees for the careful reading of the manuscript 
as well as their valuable comments and suggestions which helped to 
improve the present paper.

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\end{document}