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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2014 (2014), No. 176, pp. 1--19.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2014 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2014/176\hfil Weak inverse problems]
{Weak inverse problems for parabolic integro-differential equations
 containing \\ two kernels}

\author[K. Kasemets, J. Janno \hfil EJDE-2014/176\hfilneg]
{Kairi Kasemets, Jaan Janno}  % in alphabetical order

\address{Kairi Kasemets \newline
Tallinn University of Technology,
Ehitajate tee 5, Tallinn, Estonia}
\email{kairi.kasemets@ttu.ee}

\address{Jaan Janno \newline
Tallinn University of Technology,
Ehitajate tee 5, Tallinn, Estonia}
\email{janno@ioc.ee}

\thanks{Submitted April 8, 2014. Published August 15, 2014.}
\subjclass[2000]{35R30, 80A23}
\keywords{Inverse problem; parabolic integro-differential equation;
\hfill\break\indent relaxation kernel; quasi-solution}

\begin{abstract}
 An inverse problem to determine a coefficient and two kernels in a
 parabolic integro-differential equation is considered.
 A corresponding direct problem is supposed to be in the weak form.
 Existence of the quasi-solution is proved and issues related to Fr\'echet
 differentiation of the cost functional are treated.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{remark}[theorem]{Remark}
\allowdisplaybreaks


\section{Introduction}\label{s:intro}

Inverse problems to determine coefficients and kernels in integro-differential 
heat equations are well-studied in the smooth case when
the medium is continuous and corresponding direct problems hold in the 
classical sense (selection of references:
\cite{berca,Ge,gra,j1,jk1,jvw1,jk2,lome,lomo,lovra,PJ}). 
For instance, in \cite{jk1} problems to determine space-dependent
coefficients by means of final over-determination of the solution of the 
direct problem are dealt with. This paper exploits and
generalizes methods developed earlier in the usual parabolic case \cite{Bush,Isakov}.


Results are known for particular non-smooth cases,
 as well. For instance, identification problems for parabolic transmission problems 
are considered in \cite{jlo2}
under additional smoothness assumptions in neighbourhoods of observation areas. 
Several papers deal with degenerate cases
(see \cite{ILS} and references therein).
In \cite{jk3}  problems to reconstruct free terms and coefficients in a weak 
parabolic  problem
containing a single kernel (heat flux relaxation kernel) are considered.
In particular, a new method that enables to deduce formulas for
Fr\'echet derivatives for cost functionals of inverse problems  is proposed.

In the present article we consider the inverse problem of determining two 
kernels and a coefficient in a parabolic integro-differential equation.
The corresponding direct problem is posed in the weak form. We prove the 
Fr\'echet differentiability of the cost functional related
to the inverse problem and deduce a suitable form for the Fr\'echet derivative 
in terms of an adjoint problem. In this connection we
use an integrated convolutional form of the weak direct problem that enables to 
use test functions without classical time derivatives.
Finally, we prove the existence of the quasi-solution of the inverse problem 
under certain restrictions.

Inverse problems for smooth models with two kernels were formerly considered 
in \cite{gra,jvw1,PJ}.


\section{Formal statement of problems}\label{s:2}

 Let $\Omega$ be a $n$-dimensional domain, where $n\ge 1$ and 
$\Gamma=\partial\Omega$. Further, let $\Gamma= \Gamma_1\cup \Gamma_2$ with
 $\operatorname{meas} \Gamma_1\cap\Gamma_2=0$, 
$\operatorname{meas}\Gamma_2>0$ and   either $\Gamma_1=\emptyset$ or 
$\operatorname{meas}\Gamma_1>0$. In case $n\ge 2$ we assume
 $\Gamma$ to be sufficiently smooth.
Define
 $$
\Omega_t= \Omega\times (0,t),\quad
\Gamma_{1,t}= \Gamma_1\times (0,t),\quad
\Gamma_{2,t}= \Gamma_2\times (0,t)
$$
for $t\ge 0$.

Let $T>0$. We pose the  formal direct problem:
find $u(x,t): \Omega_T\to \mathbb{R}$  such that
\begin{gather} \label{e1}
u_t+(\mu*u)_t = Au - m*Au + f+\nabla \cdot \phi+\varphi_t\quad \text{in }\Omega_T,
\\
\label{e2}  u=u_0  \quad  \text{in }\Omega\times \{0\},
\\
\label{e3} u=g \quad  \text{in }\Gamma_{1,T},
\\  \label{e4}
-\nu_{A}\cdot \nabla u + m* \nu_{A}\cdot \nabla u = \vartheta
u+h +\nu\cdot \phi\quad  \text{in }\Gamma_{2,T},
\end{gather}
where
\[
Av=\sum_{i,j=1}^n \left(a_{ij}v_{x_j}\right)_{x_i}+av,\quad
\nu_{A}= \Big(\sum_{j=1}^n   a_{ij}\nu_j\bigl|_{i=1,\ldots,n}\Big),
\]
$\nu=(\nu_1,\ldots,\nu_n)$ is the outer normal of $\Gamma_2$,
$a_{ij},a,u_0 : \Omega\to \mathbb{R}$,
$f, \varphi : \Omega_T\to \mathbb{R}$,
$\phi :  \Omega_T\to \mathbb{R}^n$,
$g : \Omega_T\to \mathbb{R}$, $\vartheta : \Gamma_2\to\mathbb{R}$,
$h :\Gamma_{2,T}\to \mathbb{R}$, $\mu, m : (0,T)\to \mathbb{R}$ are given functions and
$$
z*w(t)=\int_0^t z(t-\tau)w(\tau)d\tau
$$
is the convolution with respect to the variable $t$. 
In the case $\Gamma_1=\emptyset$,
 the boundary condition \eqref{e3} is omitted.
The second and third addend of the free term of the equation \eqref{e1};
 i.e., $\nabla\cdot \phi$ and $\varphi_t$  may be singular distributions.

The problem \eqref{e1}--\eqref{e4} governs the heat conduction in the
body $\Omega$ filled with material with memory, where $\mu$ and $m$ are 
the relaxation kernels of the internal energy and the heat flux, respectively
 and $u$ is the temperature \cite{am,Ge,gra,Nu}.
 Then the condition \eqref{e4} corresponds to  the third kind boundary condition, 
namely it contains the heat flux
 to the co-normal direction $-\nu_{A}\cdot \nabla u + m* \nu_{A}\cdot \nabla u$.


Let us formulate the inverse problem:

\subsection*{IP}
 Find $a$, $m$ and $\mu$ such that the solution of \eqref{e1}--\eqref{e4} 
satisfies the following final and integral  additional conditions:
\begin{gather} \label{ip1}
u=u_T\quad \text{in }\Omega\times\{T\},\\
\label{ip2}
\int_{\Gamma_2}\kappa_j(x,\cdot)u(x,\cdot)d\Gamma= v_j \quad \text{in }(0,T),\;
 j=1,2,
\end{gather}
where $u_T: \Omega\to \mathbb{R}$, $\kappa_j: \Gamma_{2,T}\to \mathbb{R}$ and
$v_j: (0,T)\to \mathbb{R}$ are prescribed functions.

\begin{remark} \rm
 In the case $n=1$ and $\Omega=(c,d)$, the integral 
$\int_{\Gamma_2}z(x)d\Gamma$ is merely the sum
$\sum_{l=1}^L z(x_l)$, where $x_l\in \Gamma_2\subseteq \{c;d\}$ and 
$L$ is the number of points in $\Gamma_2$ (i.e $L\in \{1;2\}$). 
Then the conditions \eqref{ip2} read
\begin{equation}\label{ip2n=1}
\sum_{l=1}^L \kappa_j(x_l,\cdot)u(x_l,\cdot)
= v_j \quad \text{in $(0,T)$,}\quad  j=1,2.
\end{equation}
\end{remark}

 \section{Results concerning direct problem}\label{s:3}


Let us start by a rigorous mathematical formulation of the direct problem.
Define the following functional spaces:
\begin{gather*}
\mathcal{U}(\Omega_t)= C([0,t];L^2(\Omega))\cap L^2((0,t);W^{1}_2(\Omega)),
\\
\mathcal{U}_0(\Omega_t)=\Big\{\eta \in \mathcal{U}(\Omega_t):
 \eta|_{\Gamma_{1,t}}=0\;\;\text{in case $\Gamma_1\ne \emptyset$}\Big\},
\\
\mathcal{T}(\Omega_t)= \big\{\eta \in L^2((0,t);W_2^1(\Omega)):
\eta_t\in  L^2((0,t);L^2(\Omega))\big\},
\\
\mathcal{T}_0(\Omega_t)=\Big\{\eta \in \mathcal{T}(\Omega_t):
 \eta|_{\Gamma_{1,t}}=0\text{ in case $\Gamma_1\ne \emptyset$}\big\}
\end{gather*}
and introduce the following basic assumptions on the data of the direct problem:
\begin{gather} \label{assum1}
a_{ij}\in L^\infty(\Omega),\quad a_{ij}=a_{ji},\quad
\vartheta\in C(\overline\Gamma_{2}),\quad \vartheta\ge 0,
\\
 \label{assum4}
\sum_{i,j=1}^n a_{ij}(x)\lambda_i\lambda_i\ge \epsilon|\lambda|^2, \quad 
x\in \Omega,\; \lambda\in\mathbb{R}^n \; \text{with some }\epsilon>0,
\\ \label{assum1a}
a\in L^{q_1}(\Omega),\quad\text{where }q_1=1  \text{ if } n=1,\;
q_1>\frac{n}{2} \text{ if } n\ge 2,
\\  \label{assum2}
\mu\in L^2(0,T),\quad  m\in L^1(0,T)\,,
\\ \label{assum3aa}
 u_0\in L^2(\Omega),\quad g\in \mathcal{T}(\Omega_T),\quad 
h\in L^2(\Gamma_{2,T}),
\\ \label{assum3}
\begin{gathered} 
f\in L^2((0,T);L^{q_2}(\Omega)),\text{ where }
q_2=1  \text{ if } n=1,\\
q_2\in (1,q_1)  \text{ if } n=2,\;\;
q_2=\frac{2n}{n+2}  \text{ if } n \ge 3,
\end{gathered}
\\  \label{assum3a}
 \phi=(\phi_1,\ldots,\phi_n)\in (L^2(\Omega_T))^n\, ,
\\  \label{assumlisa}
\varphi\in \mathcal{U}(\Omega_T), \text{ in case }
\Gamma_1\ne\emptyset\\;\exists g_\varphi\in \mathcal{T}(\Omega_T) :
\varphi=g_{\varphi}\text{ in }\Gamma_{1,T}.
\end{gather}

If we assume additional conditions $a_{ij}\in W_2^1(\Omega)$,  
$\frac{\partial}{\partial x_i} \phi_{i}\in L^2(\Omega_T)$, $i=1,\ldots,n$,
$\varphi_t\in L^2(\Omega_T)$ and suppose that  \eqref{e1}--\eqref{e4} 
has a classical solution $u\in L^2(\Omega_T)$ such that 
$u_t,u_{x_i},u_{x_ix_j}\in L^2(\Omega_T)$, $i,j=1,\ldots,n$, then
multiplying \eqref{e1} with a test
function $\eta\in \mathcal{T}_0(\Omega_T)$
and integrating by parts we come to the relation
 \begin{equation}   \label{weakprobl}
\begin{aligned}
0 &=  \int_{\Omega}\left[(u+\mu*u)(x,T)\eta(x,T)  -
u_0(x)\eta(x,0)\right]dx - \iint_{\Omega_T}(u+\mu*u)\eta_t \,dx\,dt
\\
&\quad +  \iint_{\Omega_T}\Bigl[\sum_{i,j=1}^n
a_{ij}(u_{x_j}-m*u_{x_j})\eta_{x_i}-a(u-m*u)\eta\Bigl] \,dx\,dt
 \\
&\quad+\iint_{\Gamma_{2,T}} (\vartheta u+h)\eta\, d\Gamma dt   -
\iint_{\Omega_T}  (f\eta-\phi\cdot\nabla\eta) \,dx\,dt
\\
&\quad - \int_{\Omega}[\varphi(x,T)\eta(x,T)  -
\varphi(x,0)\eta(x,0)]dx+ \iint_{\Omega_T}\varphi\eta_t\,dx\,dt.
\end{aligned}
\end{equation}
This relation makes sense also in a more
general case  when $a_{ij}$, $\phi$, $\varphi$ satisfy  \eqref{assum1},
 \eqref{assum3a}, \eqref{assumlisa} and $u\in \mathcal{U}(\Omega_T)$.


 We call a \emph{weak solution} of the problem
\eqref{e1}--\eqref{e4} a function belonging to
$\mathcal{U}(\Omega_T)$
that satisfies the relation \eqref{weakprobl} for any $\eta\in
\mathcal{T}_0(\Omega_T)$ and, in case $\Gamma_1\ne \emptyset$,
that fulfills the boundary condition \eqref{e3}.


\begin{theorem} \label{thm1}
Problem \eqref{e1}--\eqref{e4} has a unique weak solution.  
 This solution satisfies the estimate
\begin{equation}\label{t1est}
\begin{aligned}
 \|u\|_{\mathcal{U}(\Omega_T)}
&\le  C_0\Bigl[\|u_0\|_{L^2(\Omega)}+\|f\|_{L^2((0,T);L^{q_2}(\Omega))}
+\|\phi\|_{(L^2(\Omega_T))^n}
\\
&\quad+\|\varphi\|_{\mathcal{U}(\Omega_T)}+\theta\{\|g\|_{\mathcal{T}(\Omega_T)}
+\|g_\varphi\|_{\mathcal{T}(\Omega_T)}\}+ \|h\|_{L^2(\Gamma_{2,T})}
\Bigl],
\end{aligned}
\end{equation}
where $\theta=0$ in case $\Gamma_1=\emptyset$ and $C_0$ is a constant independent
of $u_0,f,\phi,\varphi,g,h$.
\end{theorem}

\begin{proof}
 Since $\mu\in L^2(0,T)$, the  Volterra equation of the second kind
\begin{equation}\label{Volt}
\widehat\mu+\mu*\widehat\mu=-\mu \quad \text{in $(0,T)$}.
\end{equation}
has a unique solution $\widehat\mu\in L^2(0,T)$ \cite{Gri}.
We call $\widehat\mu$ the resolvent kernel of $\mu$.
Further, let us consider the following problem:
\begin{gather} \label{e1h}
\widehat u_t= A\widehat u - \widehat m*A\widehat u + \widehat f
+\nabla \cdot \widehat\phi\quad \text{in }\Omega_T,
\\
\label{e2h} \widehat u=\widehat u_0  \quad  \text{in }\Omega\times \{0\},
\\
\label{e3h} \widehat u=\widehat g \quad  \text{in }\Gamma_{1,T},
\\  \label{e4h}
-\nu_{A}\cdot \nabla \widehat u + \widehat m* \nu_{A}\cdot \nabla \widehat u
= \vartheta \widehat u+\vartheta\widehat\mu*\widehat u+\widehat h
 +\nu\cdot \widehat\phi\quad  \text{in }\Gamma_{2,T},
\end{gather}
where
\begin{gather*}
\widehat m= m-\widehat\mu+m*\widehat\mu\, ,\quad
\widehat f = f+a\varphi-\widehat m*a\varphi,
\\
\widehat\phi_i=\phi_i+\sum_{j=1}^n a_{ij}\varphi_{x_j}
 -\widehat m*\sum_{j=1}^n a_{ij}\varphi_{x_j},
\\
\widehat h=h+\vartheta\varphi+\vartheta\widehat\mu*\varphi\, ,\quad
\widehat g=g+\mu*g-g_\varphi\, ,\quad \widehat u_0=u_0-\varphi(\cdot,0).
\end{gather*}
By the properties of $m$ and $\widehat \mu$  we have $\widehat m\in L^1(0,T)$.
Further, \cite[Lemma 1]{jk3} yields
\begin{equation}\label{lemma1a}
\begin{gathered}
\mathcal{U}(\Omega_T)\hookrightarrow L^2((0,T);L^{q_3}(\Omega))\, ,\quad
\text{where $q_3=\infty$ if $n=1$,}
\\
q_3>\frac{q_1q_2}{q_1-q_2}\text{ if }n=2, \; q_3=\frac{2n}{n-2} \text{ if }n\ge 3
\end{gathered}
\end{equation}
and
\begin{equation}\label{lemma1b}
\begin{gathered}
av\in L^2((0,T);L^{q_2}(\Omega))\text{ if }a\in L^{q_1}(\Omega),\;
 v\in L^2((0,T);L^{q_3}(\Omega)),
\\
\|av\|_{L^2((0,T);L^{q_2}(\Omega))}
\le C\|a\|_{L^{q_1}(\Omega)}\|v\|_{L^2((0,T);L^{q_3}(\Omega))},
\end{gathered}
\end{equation}
where $C$ is a constant.  Using the relations \eqref{lemma1a}, \eqref{lemma1b},
the properties of $\widehat m$,  $\widehat\mu$, the assumptions
\eqref{assum1}--\eqref{assumlisa}, trace theorems and the Young theorem for
convolutions we obtain
\begin{align*}
&\widehat d:=(\widehat f,\widehat\phi,\widehat u_0,\widehat g,\widehat h)
\in \mathcal{X}\\
&:= L^2((0,T);L^{q_2}(\Omega))\times (L^2(\Omega_T))^n\times L^2(\Omega)\times
\mathcal{T}(\Omega_T)\times L^2(\Gamma_{2,T}),
\end{align*}
\begin{equation} \label{qq1}
\|\widehat d\|_\mathcal{X}\le \bar C\|d\|_{\bar{\mathcal{X}}}
\end{equation}
where $d=(f,\phi,u_0,g,h,\varphi,g_\varphi)$ and
\[
\bar{\mathcal{X}}=L^2((0,T);L^{q_2}(\Omega))
 \times (L^2(\Omega_T))^n\times L^2(\Omega)
\times \mathcal{T}(\Omega_T)\times L^2(\Gamma_{2,T})
\times \mathcal{U}(\Omega_T)\times \mathcal{T}(\Omega_T)
\]
and $\bar C$ is a constant.
It was proved in \cite[Theorem 1]{jk3} that
 problem \eqref{e1}--\eqref{e4} in case $\mu=0$ and $\varphi=0$
has for any $(f,\phi,u_0,g,h)\in \mathcal{X}$ a unique weak solution and the
corresponding solution operator $\mathcal{B}$ belongs
to $\mathcal{L}(\mathcal{X};\mathcal{U}(\Omega_T))$. (Here $\mathcal{L}(X,Y)$
stands for the space of linear bounded operators
from a Banach space $X$ to a Banach space $Y$.)
This implies that  problem \eqref{e1h}--\eqref{e4h} is
equivalent in $\mathcal{U}(\Omega_T)$ to the following operator equation:
\begin{equation}\label{qq2}
\widehat u= \mathcal{Q}\widehat u\quad \text{with}\quad
\mathcal{Q}\widehat u= \mathcal{B}(0,0,0,0,\vartheta\widehat\mu*\widehat u)
+\mathcal{B}\widehat d.
\end{equation}
To study this equation, we will  use the inequality
\begin{equation}\label{qq3}
\|\widehat \mu *y\|_{L^2(\Omega_t)}\le \int_0^t |\widehat \mu(t-\tau)|\,
 \|y\|_{L^2(\Omega_\tau)}d\tau\,,\quad  t\in [0,T]
\end{equation}
that holds for any $y\in L^2(\Omega_T)$.
This was proved in \cite[inequality (3.12)]{jk3}.

Let $\widehat u^1,\widehat u^2\in \mathcal{U}(\Omega_T)$, denote 
$v=\widehat u^1-\widehat u^2$ and estimate
$\mathcal{Q}\widehat u^1-\mathcal{Q}\widehat u^2
=\mathcal{B}(0,0,0,0,\vartheta\widehat\mu*v)$. To this end, fix $t\in [0,T]$
and define
\begin{equation}\label{Pt}
P_tw=\begin{cases}
w &\text{in }\Gamma_{2,t}\\
0 &\text{in }\Gamma_{2,T}\setminus\Gamma_{2,t}
\end{cases}
\end{equation}
for $w: \Gamma_{2,T}\to \mathbb{R}$.
Due to the causality, we have
$\mathcal{B}(0,0,0,P_t\vartheta\widehat\mu*v)(x,\tau)
=\mathcal{B}(0,0,0,\vartheta\widehat\mu*v)(x,\tau)$
for any $(x,\tau)\in\Omega_t$.
Since $\mathcal{B}\in \mathcal{L}(\mathcal{X};\mathcal{U}(\Omega_T))$,
the continuity of $\vartheta$, the trace theorem and
the inequality \eqref{qq3} with $y=v,v_{x_i}$, $i=1,\ldots,n$, it follows that
\begin{equation}
\begin{aligned}
\|\mathcal{Q}\widehat u^1-\mathcal{Q}\widehat u^2\|_{\mathcal{U}(\Omega_t)}
&=\|\mathcal{B}(0,0,0,0,\vartheta\widehat\mu*v)
  \|_{\mathcal{U}(\Omega_t)}
\\
&=\|\mathcal{B}(0,0,0,0,P_t\vartheta\widehat\mu*v)\|_{\mathcal{U}(\Omega_t)}\\
&\le\|\mathcal{B}(0,0,0,0,P_t\vartheta\widehat\mu*v)\|_{\mathcal{U}(\Omega_T)}
\\
&\le \|\mathcal{B}\| \|P_t\vartheta\widehat\mu*v\|_{L^2(\Gamma_{2,T})}
= \|\mathcal{B}\| \|\vartheta\widehat\mu*v\|_{L^2(\Gamma_{2,t})}
\\
&\le C_1 \|\widehat\mu*v\|_{L^2((0,t);W_2^1(\Omega))}\\
&\le C_2\int_0^t |\widehat \mu(t-\tau)|\, \|v\|_{L^2((0,\tau);W_2^1(\Omega))}d\tau
\end{aligned} \label{qq4}
\end{equation}
with some constants $C_1$ and  $C_2$.
Let us define the weighted norm in $\mathcal{U}(\Omega_T)$:
$\|v\|_\sigma=\sup_{0<t<T}e^{-\sigma t}\|v\|_{\mathcal{U}(\Omega_t)}$
where $\sigma\ge 0$. In view of \eqref{qq4}
and $\mathcal{U}(\Omega_t)\hookrightarrow L^2((0,t);W_2^1(\Omega))$  we get
\begin{align*}
\|\mathcal{Q}\widehat u^1-\mathcal{Q}\widehat u^2\|_\sigma
&\le  C_3 \sup_{0<t<T}e^{-\sigma t}\int_0^t |\widehat \mu(t-\tau)|\,
  \|v\|_{\mathcal{U}(\Omega_\tau)}d\tau
  \\
&=C_3\sup_{0<t<T}\int_0^t e^{-\sigma (t-\tau)}|\widehat \mu(t-\tau)|
 \,e^{-\sigma \tau}\, \|v\|_{\mathcal{U}(\Omega_\tau)}d\tau
\\
&\le C_3\int_0^Te^{-\sigma s}|\widehat \mu(s)|ds\sup_{0<\tau<T}e^{-\sigma \tau}\,
 \|v\|_{\mathcal{U}(\Omega_\tau)}\\
&= C_2\int_0^Te^{-\sigma s}|\widehat \mu(s)|ds\, \|v\|_\sigma
\end{align*}
with some constant $C_3$. 
By the dominated convergence theorem,
 $\int_0^Te^{-\sigma s}|\widehat \mu(s)|ds\to 0$ as $\sigma\to\infty$.
Thus, there exists  $\sigma_0$ such  that
$$
C_3\int_0^Te^{-\sigma_0 s}|\widehat \mu(s)|ds \le \frac{1}{2}.
$$
 Therefore,
$\|\mathcal{Q}\widehat u^1-\mathcal{Q}\widehat u^2\|_{\sigma_0}
\le \frac{1}{2}\|\widehat u^1-\widehat u^2\|_{\sigma_0}$.
The operator $\mathcal{Q}$ is a contraction in $\mathcal{U}(\Omega_T)$.
This implies that \eqref{e1h}--\eqref{e4h} has a unique weak solution in
$\mathcal{U}(\Omega_T)$.
Moreover, observing  \eqref{qq2} and the relation
$\mathcal{Q}0=\mathcal{B}\widehat d$, for the solution of
\eqref{e1h}--\eqref{e4h} we obtain the estimate
\[
\|\widehat u\|_{\sigma_0}
= \|\mathcal{Q}\widehat u-\mathcal{Q}0+\mathcal{Q}0\|_{\sigma_0}
\le \|\mathcal{Q}\widehat u-\mathcal{Q}0\|_{\sigma_0}
 +\|\mathcal{B}\widehat d\|_{\sigma_0}\le \frac{1}{2} \|\widehat u\|_{\sigma_0}
+\|\mathcal{B}\widehat d\|_{\sigma_0}
\]
which implies
\[
 \|\widehat u\|_{\sigma_0}\le 2\|\mathcal{B}\widehat d\|_{\sigma_0}
\le 2\|\mathcal{B}\widehat d\|_{\mathcal{U}(\Omega_T)}\le
2\|\mathcal{B}\| \|\widehat d\|_\mathcal{X}.
\]
Observing  the relation
$e^{-\sigma_0 T}\|\widehat u\|_{\mathcal{U}(\Omega_T)}\le \|\widehat u\|_{\sigma_0}$
and \eqref{qq1} we arrive at the  estimate
\begin{equation}\label{qq5}
\|\widehat u\|_{\mathcal{U}(\Omega_T)}\le C_4 \|d\|_{\bar{\mathcal{X}}}
\end{equation}
with a constant $C_4$.

Further, let us define 
\begin{equation}\label{u}
u=\widehat u+\varphi+\widehat\mu*(\widehat u+\varphi).
\end{equation}
Then $\widehat u$ is expressed in terms of $u$ as
\begin{equation}\label{widehatu}
\widehat u=u+\mu*u-\varphi.
\end{equation}
One can immediately check that the implications
$u\in \mathcal{U}(\Omega_T)\Leftrightarrow
\widehat u\in \mathcal{U}(\Omega_T)$
are valid. Moreover, it is easy to see that $\widehat u$ is a weak solution
of \eqref{e1h}--\eqref{e4h} if and only if $u$ is weak solution of
\eqref{e1}--\eqref{e4}.
In view of the above-presented arguments we can conclude that
\eqref{e1}--\eqref{e4} has a unique weak solution. From \eqref{u} we obtain
\[
\|u\|_{\mathcal{U}(\Omega_T)}\le
C_4 (\|\widehat u\|_{\mathcal{U}(\Omega_T)}+\|\varphi\|_{\mathcal{U}(\Omega_T)})
\]
 with a constant $C_5$. This with
\eqref{qq5} implies \eqref{t1est}. The proof is complete.
\end{proof}

It is possible to give an equivalent form to the relation \eqref{weakprobl} 
that does not contain the derivative of the test
function with respect to $t$. Namely,
the following theorem holds.

\begin{theorem} \label{thm2}
The function  $u\in  \mathcal{U}(\Omega_T)$ satisfies the relation
\eqref{weakprobl} for any $\eta\in \mathcal{T}_0(\Omega_T)$
if and only if it satisfies the  relation
\begin{equation}   \label{weakproblint}
\begin{split}
0 &=  \int_{\Omega} (u+\mu*u-\varphi)*\eta \,dx
- \int_{\Omega}\int_0^t (u_0(x)-\varphi(x,0))\eta(x,\tau)\,d\tau\, dx
\\
&\quad +  \int_{\Omega} 1*\Bigl[\sum_{i,j=1}^n
a_{ij}(u_{x_j}-m*u_{x_j})*\eta_{x_i}-a(u-m*u)*\eta\Bigl] dx
 \\
&\quad+\int_{\Gamma_{2}} 1* (\vartheta u+h)*\eta\, d\Gamma    -
\int_\Omega  1*\Bigl(f*\eta-\sum_{i=1}^n\phi_{i}*\eta_{x_i}\Bigl)\, dx, \quad
\end{split}
\end{equation}
for any  $t\in [0,T]$ and  $\eta\in \mathcal{U}_0(\Omega_T)$.
\end{theorem}

\begin{proof} 
It is analogous to the proof of \cite[Theorem 2]{jk3} that considers the 
case $\varphi=0$, $\mu=0$. We have
only to replace $u$ by $\widehat u=u+\mu*u-\varphi$ in the term $K_1(t)$ 
appearing in formulas \cite[(3.19), (3.20)]{jk3} to get the desired result.
\end{proof}

\begin{remark} \rm 
Theorems \ref{thm1} and \ref{thm2} remain valid also in the case $\Gamma_2=\emptyset$. 
In this case the terms $\|h\|_{\Gamma_{2,T}}$ and 
$\int_{\Gamma_{2}} 1* (\vartheta u+h)*\eta\, d\Gamma$ are missing in 
\eqref{t1est} and \eqref{weakproblint}, respectively.
\end{remark}

 \section{Quasi-solution of IP. Fr\'echet derivative of cost functional}\label{s:4}

Assume that $n\in \{1;2;3\}$. Moreover, let us
set  $q_1=2$ if $n=2$. Then any coefficient $a$ that belongs to $L^2(\Omega)$ 
satisfies \eqref{assum1a}.
For the weight functions $\kappa_j$ we assume that
\begin{equation}\label{kappaassum}
\kappa_j\in L^\infty((0,T);L^2(\Gamma_2))\,,\quad j=1,2.
\end{equation}
In the case $n=1$ this assumption is simply
$\kappa({x_l},\cdot)\in L^\infty(0,T)$, $x_l\in \Gamma_2\subseteq \{c;d\}$.
According to Theorem \ref{thm1}, $u\in \mathcal{U}(\Omega_T)$, thus
$u(\cdot,T)\in L^2(\Omega)$, and
the condition \eqref{ip1} is well-defined for $u_T\in L^2(\Omega)$.
Moreover, by a trace theorem we have $u\in L^2(\Gamma_{2,T})$. This
implies that $\int_{\Gamma_2}\kappa_j(x,\cdot)u(x,\cdot)d\Gamma \in L^2(0,T)$,
 $j=1,2$, hence the condition \eqref{ip2}
is well-defined for $v_j\in L^2(0,T)$, $j=1,2$.

Let $M\subseteq \mathcal{Z}:=L^2(\Omega)\times (L^2(0,T))^2$.
We call the \emph{quasi-solution} of IP in  the set $M$ an element
$z^* \in \arg \min_{z\in M}J(z)$,
where $J$ is the cost functional
$$
J(z)=\|u(\cdot,T;z)-u_T\|_{L^2(\Omega)}^2
 +\sum_{j=1}^2\|\int_{\Gamma_2}\kappa_j(x,\cdot)u(x,\cdot;z)d\Gamma-v_j\|_{L^2(0,T)}^2
$$
and $u(x,t;z)$ is the weak solution of the direct problem  \eqref{e1}--\eqref{e4} 
corresponding to given $z=(a,m,\mu)$. In case $n=1$ the integral 
$\int_{\Gamma_2}\kappa_j(x,t)u(x,t;z)d\Gamma$ in the definition of $J$ is 
replaced by $\sum_{l=1}^L\kappa_j(x_l,t)u(x_l,t;z)$.

\begin{theorem} \label{thm3} 
The functional $J$ is Fr\'echet differentiable in $\mathcal{Z}$ and
\begin{equation}\label{Jder}
\begin{split}
&J'(z)\Delta z\\
&= 2\int_\Omega\left[u(x,T;z)-u_T(x)\right]\Delta u(x,T)dx
\\
&\quad + 2\sum_{j=1}^2\int_0^T\Big[\int_{\Gamma_2}\kappa_j(y,t)u(y,t;z)d\Gamma
  - v_j(t)\Big]\int_{\Gamma_2}\kappa_j(x,t)\Delta u(x,t)d\Gamma dt,
\end{split}
\end{equation}
where $\Delta z=(\Delta a,\Delta m,\Delta\mu)\in \mathcal{Z}$ and
$\Delta u \in\mathcal{U}(\Omega_T)$ is the $z$- and $\Delta z$-dependent
weak solution of the following problem:
\begin{gather}\label{j3prob1}
\begin{aligned}
\Delta u_t+(\mu*\Delta u)_t
&= A\Delta u - m*A\Delta u+\Delta a [u-m*u] -\Delta m*au\\
&\quad -\nabla\cdot\bigl[\Delta m*\sum_{j=1}^na_{ij}u_{x_j}\bigl]
 -(\Delta\mu*u)_t\quad \text{in }\Omega_T,
\end{aligned}
\\
\label{j3prob2}  \Delta u=0  \quad  \text{in }\Omega\times \{0\},
\\
\label{j3prob3} \Delta u=0 \quad  \text{in } \Gamma_{1,T},
\\  \label{j3prob4}
\begin{split}
&-\nu_{A}\cdot \nabla \Delta u + m* \nu_{A}\cdot \nabla \Delta u \\
&= \vartheta \Delta u - \nu\cdot\Bigl[\Delta m*\sum_{j=1}^na_{ij}u_{x_j}\Bigl]
\quad  \text{in }\Gamma_{2,T}. \end{split}
\end{gather}
\end{theorem}

\begin{proof} 
 Denote $\widetilde\Delta u= u(x,t;z+\Delta z)-u(x,t;z)$ and define 
$\widehat \Delta u= \widetilde \Delta u-\Delta u$. 
Then we can represent the difference of $J$ as follows:
\begin{equation}\label{diff}
J(z+\Delta z)-J(z)= {\rm RHS}+\Theta,
\end{equation}
where RHS is the right-hand side of the equality \eqref{Jder} and
\begin{align*}
\Theta&=
2\int_\Omega[u(x,T)-u_T(x)]\widehat\Delta u(x,T)dx
\\
&\quad + 2\sum_{j=1}^2\int_0^T\Big[\int_{\Gamma_2}\kappa_j(y,t)u(y,t)d\Gamma
- v_j(t)\Big]\int_{\Gamma_2}\kappa_j(x,t)\widehat\Delta u(x,t)d\Gamma dt
\\
&\quad +\int_\Omega \Big\{(\Delta u+\widehat\Delta u)(x,T)\Big\}^2 dx \\
&\quad +\sum_{j=1}^2\int_0^T\Big\{\int_{\Gamma_2}\kappa_j(x,t)(\Delta u
+\widehat\Delta u)(x,t)d\Gamma\Big\}^2\, dt.
\end{align*}

Let us  study  problem \eqref{j3prob1}--\eqref{j3prob4}.
To this end we estimate the terms in the right-hand side of \eqref{j3prob1}.
Observing the relations $u\in \mathcal{U}(\Omega_T)$, \eqref{lemma1a}, 
\eqref{lemma1b}, $L^{2}(\Omega)\hookrightarrow L^{q_1}(\Omega)$
 and using the Young and Cauchy inequalities we deduce
\begin{equation}\label{jp1}
\begin{split}
&\|\Delta a [u-m*u]-\Delta m*au\|_{L^2((0,T);L^{q_2}(\Omega))}
\\
&\le c_1\|u\|_{\mathcal{U}(\Omega_T)}\left[(1+\|m\|_{L^2(0,T)})\|\Delta a\|_{L^{2}(\Omega)}+
\|a\|_{L^{2}(\Omega)}\|\Delta m\|_{L^2(0,T)}\right]
\\
&\le  c_2(z,u)\|\Delta z\|,
\end{split}
\end{equation}
where $c_1$ is a constant,
$c_2$ is a coefficient depending on $z=(a,m,\mu)$, $u$ and $\|\cdot\|$ denotes the norm in $\mathcal{Z}$. Taking the boundendness of $a_{ij}$
into account we similarly get
\begin{equation}\label{jp2}
\|\Delta m*\sum_{j=1}^na_{ij}u_{x_j}\|_{(L^2(\Omega_T))^n}\le
c_3\|u\|_{\mathcal{U}(\Omega_T)}\|\Delta m\|_{L^2(0,T)}\end{equation}
with a constant $c_3$.
Next let us estimate the term  $\Delta\mu*u$ at the right-hand side of
\eqref{j3prob1}. Since $u\in C([0,T];L^2(\Omega))$ and
$\Delta\mu\in L^2(0,T)$, it is easy to check that
$\Delta\mu*u\in C([0,T];L^2(\Omega))$ and
$\|\Delta\mu*u\|_{ C([0,T];L^2(\Omega))}
\le T^{1/2}\|u\|_{ C([0,T];L^2(\Omega))}\|\Delta \mu\|_{L^2(0,T)}$.
Similarly, $\|\Delta\mu*u\|_{L^2((0,T);W_2^1(\Omega))}\le T^{1/2}
\|u\|_{L^2((0,T);W_2^1(\Omega))}\|\Delta \mu\|_{L^2(0,T)}$.
Taking these estimates together, we have
\begin{equation}\label{jp3}
\|\Delta \mu*u\|_{\mathcal{U}(\Omega_T)}\le
T^{1/2}\|u\|_{\mathcal{U}(\Omega_T)}\|\Delta \mu\|_{L^2(0,T)}.
\end{equation}
Since $u=g$ in $\Gamma_{1,T}$, we find that
\begin{equation}\label{jp4}
\Delta \mu*u=\Delta\mu*g \quad  \text{in }\Gamma_{1,T}.
\end{equation}
Using the assumption $g\in \mathcal{T}(\Omega_T)$ and the Young and Cauchy
inequalities again, we obtain
\begin{equation}\label{jp5}
\begin{split}
\|\Delta\mu*g\|_{\mathcal{T}(\Omega_T)}
&=\|\Delta\mu*g\|_{L^2((0,T);W_2^1(\Omega))}+
\|(\Delta\mu*g)_t\|_{L^2((0,T);L^2(\Omega))}
\\
&=\|\Delta\mu*g\|_{L^2((0,T);W_2^1(\Omega))}+
\|\Delta\mu*g_t\|_{L^2((0,T);L^2(\Omega))}
\\
&\quad + \|\Delta\mu\, g(\cdot,0)\|_{L^2((0,T);L^2(\Omega))} \\
&\le c_4\|\Delta \mu\|_{L^2(0,T)}
\end{split}
\end{equation}
with a constant $c_4$.
Relations \eqref{jp1}--\eqref{jp5}  show that  Theorem \ref{thm1} holds for
problem \eqref{j3prob1}--\eqref{j3prob4}, hence it has a unique
weak solution  $\Delta u\in \mathcal{U}(\Omega_T)$.
Using the estimate \eqref{t1est} for the solution of this problem we
obtain
\begin{equation}\label{deltauest}
\begin{split}
&\|\Delta u\|_{\mathcal{U}(\Omega_T)}\\
&\le C_0\Bigl[\|\Delta a [u-m*u]+\Delta m*au\|_{L^2((0,T);L^{q_2}(\Omega))}
\\
&\quad +\|\Delta m*\sum_{j=1}^na_{ij}u_{x_j}\|_{(L^2(\Omega_T))^n}+\|\Delta \mu*u\|_{\mathcal{U}(\Omega_T)}
+\theta \|\Delta\mu*g\|_{\mathcal{T}(\Omega_T)}\Bigl]
\\
&\quad\le c_5(z,u)\|\Delta z\|
\end{split}
\end{equation}
with a coefficient $c_5$ depending on $z,u$.

The function $\widetilde\Delta u$ satisfies the problem
\begin{gather}
\label{j3prob1a}
\begin{gathered}
\widehat\Delta u_t+(\mu*\widehat\Delta u)_t
= A\widehat\Delta u - m*A\widehat\Delta u+ f+\widehat f
+\nabla\cdot\phi+\nabla\cdot\widehat \phi+\varphi_t+\widehat\varphi_t
\\
\text{in }\Omega_T,
\end{gathered}
\\
\label{j3prob2a} \widehat\Delta u=0  \quad  \text{in }\Omega\times \{0\},
\\
\label{j3prob3a} \widehat\Delta u=0 \quad  \text{in }\Gamma_{1,T},
\\  \label{j3prob4a}
-\nu_{A}\cdot \nabla \widehat\Delta u + m* \nu_{A}\cdot \nabla \widehat\Delta u 
= \vartheta \widehat\Delta u+\nu\cdot\phi+\nu\cdot\widehat\phi
\quad  \text{in }\Gamma_{2,T},
\end{gather}
where
\begin{gather*}
f=\Delta a \Delta u-(m+\Delta m)*\Delta a\Delta u-\Delta m*\Delta a u 
 -\Delta m*a\Delta u,\\
\widehat f=\Delta a \widehat\Delta u-(m+\Delta m)*\Delta a\widehat\Delta u
 -\Delta m*a\widehat\Delta u,\\
\phi=-\Delta m*\sum_{j=1}^n a_{ij}\Delta u_{x_j},\quad
\widehat\phi=-\Delta m*\sum_{j=1}^n a_{ij}\widehat\Delta u_{x_j},
\\
\varphi=-\Delta\mu*\Delta u,\quad 
\widehat\varphi=-\Delta\mu*\widehat\Delta u.
\end{gather*}
Similarly to \eqref{jp1}--\eqref{jp3} we deduce the following estimates:
\begin{gather*}
\begin{aligned}
&\|f\|_{L^2((0,T);L^{q_2}(\Omega))}\\
&\le c_6\Bigl\{(1+\|m\|_{L^2(0,T)}+\|\Delta m\|_{L^2(0,T)})\|\Delta a\|_{L^2(\Omega)}
\|\Delta u\|_{\mathcal{U}(\Omega_T)}\\
&\quad +\|u\|_{\mathcal{U}(\Omega_T)}\|\Delta m\|_{L^2(0,T)}
 \|\Delta a\|_{L^2(\Omega)}+
\|a\|_{L^2(\Omega)}\|\Delta m\|_{L^2(0,T)}
\|\Delta u\|_{\mathcal{U}(\Omega_T)}\Bigl\} \\
&\le c_7(z,u)\Big\{\big[\|\Delta z\| +\|\Delta z\|^2\big]
\|\Delta u\|_{\mathcal{U}(\Omega_T)}+\|\Delta z\|^2\Big\},
\end{aligned}\\
\|\widehat f\|_{L^2(0,T;L^{q_2}(\Omega))}\le c_8(z)
\left[\|\Delta z\| +\|\Delta z\|^2\right]\|
\widehat \Delta u\|_{\mathcal{U}(\Omega_T)},
\\
\|\phi\|_{(L^2(\Omega_T))^n}
 \le c_9 \|\Delta z\| \|\Delta u\|_{\mathcal{U}(\Omega_T)},
\\
\|\widehat\phi\|_{(L^2(\Omega_T))^n}\le c_{9} \|\Delta z\| \|\widehat\Delta u\|_{\mathcal{U}(\Omega_T)},
\\
\|\varphi\|_{\mathcal{U}(\Omega_T)}
\le T^{1/2}\|\Delta z\|\|\Delta u\|_{\mathcal{U}(\Omega_T)},
\\
\|\widehat\varphi\|_{\mathcal{U}(\Omega_T)}
\le T^{1/2}\|\Delta z\|\|\widehat\Delta u\|_{\mathcal{U}(\Omega_T)}
\end{gather*}
with some coefficients $c_6,\ldots,c_9$. Moreover, since 
$\Delta u=\widetilde \Delta u=0$ in $\Gamma_{1,T}$, we have
$\varphi=\widehat \varphi=0$ in $\Gamma_{1,T}$.
Applying the estimate \eqref{t1est} to the solution of the problem
\eqref{j3prob1a}--\eqref{j3prob4a} we get
\[
\|\widehat\Delta u\|_{\mathcal{U}(\Omega_T)}
\le c_{10}(z,u)\Big\{\left[\|\Delta z\| +\|\Delta z\|^2\right]
\Bigl\{\|\Delta u\|_{\mathcal{U}(\Omega_T)}
+\|\widehat \Delta u\|_{\mathcal{U}(\Omega_T)}\Bigl\}
+\|\Delta z\|^2\Big\}
\]
with a coefficient $c_{10}$. Provided $\|\Delta z\|$ is sufficiently small; i.e.,
$\|\Delta z\|+\|\Delta z\|^2\le \frac{1}{2 c_{10}(z,u)}$,
we have
\[
\|\widehat\Delta u\|_{\mathcal{U}(\Omega_T)}\le 2c_{10}(z,u)
\Big\{\left[\|\Delta z\| +\|\Delta z\|^2\right]\|\Delta u\|_{\mathcal{U}(\Omega_T)}
+\|\Delta z\|^2\Big\}.
\]
Due to \eqref{deltauest}, this yields
\begin{equation}\label{deltaauest1}
\|\widehat\Delta u\|_{\mathcal{U}(\Omega_T)}
\le c_{11}(z,u)\left[\|\Delta z\|^2 +\|\Delta z\|^3\right]
\end{equation}
with a coefficient $c_{11}$.

In view of \eqref{deltauest}, \eqref{deltaauest1} and the assumption
 $\kappa_j\in L^\infty((0,T);L^2(\Gamma_2))$
the right-hand side of \eqref{Jder} RHS and
the quantity $\Theta$ satisfy the estimates
\begin{equation}st
|{\rm RHS}|\le c_{12}(z,u)\|\Delta z\|,
\quad |\Theta|\le c_{13}(z,u)\sum_{l=2}^6 \|\Delta z\|^l,
\end{equation}st
where $c_{12}$ and $c_{13}$ are some coefficients. Moreover, RHS is linear
 with respect to $\Delta z$.
This with \eqref{diff} shows that $J$ is Fr\'echet differentiable in
$\mathcal{Z}$ and $J'(z)\Delta z$ equals RHS.
\end{proof}

\begin{theorem} \label{thm4} 
Assume $g=0$. Then  the Fr\'echet derivative of $J$ admits the form
\begin{equation}\label{gradi}
 J'(z)\Delta z
=\int_\Omega \gamma_1(x)\Delta a(x)dx + \int_0^T\gamma_2(t)\Delta m(t)dt
+ \int_0^T\gamma_3(t)\Delta\mu(t)dt,
\end{equation}
where
\begin{gather} \label{gamma1}
\gamma_1(x)=[(u-m*u)*\psi](x,T), \\ \label{gamma2}
\gamma_2(t)=-\int_\Omega \Bigl[au*\psi
+\sum_{i,j=1}^n a_{ij}\psi_{x_i}*u_{x_j}\Bigl](x,T-t)dx,
\\
\label{gamma3}
\begin{aligned}
&\gamma_3(t)\\
&=-\int_\Omega \Bigl[au*\psi+au*\psi*[\widehat\mu-m-m*\widehat\mu]
\\
&\quad+\sum_{i,j=1}^n a_{ij}\psi_{x_i}*u_{x_j}+\sum_{i,j=1}^n a_{ij}\psi_{x_i}*u_{x_j}*[\widehat\mu-m-m*\widehat\mu]\Bigl](x,T-t)dx
\\
&\quad-\int_{\Gamma_2}[\vartheta(u+\widehat\mu*u)*\psi](x,T-t)d\Gamma
\\
&\quad-2\int_\Omega\{u(x,T)-u_T(x)\}[u+\widehat\mu*u](x,T-t)dx
\\
&\quad-2\sum_{j=1}^2 \int_t^T
\Bigl[\int_{\Gamma_2}\kappa_j(y,\tau)u(y,\tau)d\Gamma-v_j(\tau)\Bigl]
\int_{\Gamma_2}\kappa_j(x,\tau) \big[u\\
&\quad +\widehat\mu*u\big](x,\tau-t)d\Gamma d\tau,
\end{aligned}
\end{gather}
where $\widehat\mu$ is the solution of \eqref{Volt}, $u(x,t)=u(x,t;z)$
and $\psi\in \mathcal{U}(\Omega_T)$ is the $z$-dependent weak solution of
 the following ``adjoint" problem:
\begin{gather}
\label{psiprob1}
\Delta \psi_t+(\mu*\Delta \psi)_t
= A\Delta \psi - m*A\Delta \psi\quad \text{in }\Omega_T,
\\
\label{psiprob2}
\Delta \psi=2[u(\cdot,T)-u_T]  \quad  \text{in }\Omega\times \{0\},
\\
\label{psiprob3}  \Delta \psi=0 \quad  \text{in }\Gamma_{1,T},
\\  \label{psiprob4}
-\nu_{A}\cdot \nabla \Delta \psi + m* \nu_{A}\cdot \nabla \Delta \psi
= \vartheta \Delta \psi +h^\circ
\quad  \text{in }\Gamma_{2,T},
\end{gather}
where
\begin{equation}\label{hcirc}
\begin{split}
&h^\circ(x,t)\\
&=-2\sum_{j=1}^2 \kappa_j(x,T-t)\Big[\int_{\Gamma_2}\kappa_j(y,T-t)u(y,T-t)d\Gamma
 - v_j(T-t)\Big].
\end{split}
\end{equation}
\end{theorem}

\begin{proof}
Define $\Delta w=\Delta u+\Delta\mu*u+
\widehat\mu*\Delta\mu*u$. Since $u,\Delta u\in \mathcal{U}(\Omega_T)$,
we have $\Delta w\in \mathcal{U}(\Omega_T)$. Moreover, using \eqref{Volt} 
it is easy to see that
$\Delta u+\mu*\Delta u+\Delta\mu*u=\Delta w+\mu*\Delta w$.
Using this relation for the time derivatives in \eqref{j3prob1} and 
the equality $\Delta u=\Delta w-\Delta\mu*u-
\widehat\mu*\Delta\mu*u$ for other terms containing $\Delta u$ in
\eqref{j3prob1}--\eqref{j3prob4} we see that $\Delta w$ is the weak 
solution of the  problem
\begin{gather}
\label{wprob1}
\Delta w_t+(\mu*\Delta w)_t= A\Delta w - m*A\Delta w+
f^\dagger+\nabla\cdot\phi^\dagger\quad \text{in }\Omega_T,
\\
\label{wprob2} \Delta w=0  \quad  \text{in }\Omega\times \{0\},
\\
\label{wprob3} \Delta w=0 \quad  \text{in }\Gamma_{1,T},
\\  \label{wprob4}
-\nu_{A}\cdot \nabla \Delta u + m* \nu_{A}\cdot \nabla \Delta u 
= \vartheta \Delta u +h^\dagger +\nu\cdot\phi^\dagger
\quad  \text{in $\Gamma_{2,T}$},
\end{gather}
where
\begin{gather}
\label{fdagger}
f^\dagger=\Delta a [u-m*u]-a\Delta m*u-a\Delta\mu*u
-a\Delta\mu*u*[\widehat\mu-m-m*\widehat\mu], \\
\label{phidagger}
\begin{gathered}
\phi^\dagger=(\phi_1^\dagger,\ldots,\phi_n^\dagger),\\
\phi_i^\dagger=-\Delta m*\sum_{j=1}^na_{ij}u_{x_j}
-\Delta \mu*\sum_{j=1}^na_{ij}u_{x_j}
-\Delta \mu*\sum_{j=1}^na_{ij}u_{x_j}*[\widehat\mu-m-m*\widehat\mu],
\end{gathered}
\\ \label{hdagger}
h^\dagger=-\vartheta \Delta \mu*[u+\widehat \mu*u].
\end{gather}
Let us write the weak form \eqref{weakproblint} for the problem 
for $\Delta w$ and use the test function $\eta=\psi$. Then we obtain
\begin{equation}   \label{p12}
\begin{aligned}
0 &=  \int_{\Omega} (\Delta w+\mu*\Delta w)*\psi \,dx
+\int_{\Omega}  1 * \Bigl[\sum_{i,j=1}^n
a_{ij}(\Delta w_{x_j} - m * \Delta w_{x_j})*\psi_{x_i}\\
&\quad -a(\Delta w-m * \Delta w)*\psi\Bigl] dx
+\int_{\Gamma_{2}} 1* (\vartheta \Delta w+h^\dagger)*\psi\, d\Gamma  \\
&\quad   -\int_\Omega  1*\Bigl(f^\dagger*\psi
 -\sum_{i=1}^n\phi^\dagger_{i}*\psi_{x_i}\Bigl) dx.
\end{aligned}
\end{equation}
Next we write the weak form \eqref{weakproblint} for the problem for $\psi$
and use the test function $\eta=\Delta w$ to get
\begin{equation}   \label{p13}
\begin{split}
0 &=  \int_{\Omega} (\psi+\mu*\psi)*\Delta w \,dx
- 2\int_{\Omega}\int_0^t u(x,T)-u_T(x)]\Delta w(x,\tau)d\tau dx
\\
&\quad +  \int_{\Omega}  1 * \Bigl[\sum_{i,j=1}^n
a_{ij}(\psi_{x_j} - m * \psi_{x_j})*\Delta w_{x_i}-a(\psi-m * \psi)*\Delta
w\Bigl] dx
 \\
&\quad +\int_{\Gamma_{2}} 1* (\vartheta \psi +h^\circ)*\Delta w\, d\Gamma.
\end{split}
 \end{equation}
Subtracting \eqref{p12} from \eqref{p13}, differentiating with respect to $t$
and setting $t=T$ we have
\begin{align*}
&2\int_{\Omega} [u(x,T)-u_T(x)]\Delta w(x,T)d\tau dx
 -\int_{\Gamma_{2}}  (h^\circ*\Delta w)(x,T)\, d\Gamma\\
&= \int_\Omega  \Bigl(f^\dagger*\psi-\sum_{i=1}^n\phi^\dagger_{i}*\psi_{x_i}\Bigl)(x,T)
\, dx - \int_{\Gamma_{2}} (h^\dagger*\psi)(x,T)\, d\Gamma.
\end{align*}
Observing the relations $\Delta w=\Delta u+\Delta\mu*u+
\widehat\mu*\Delta\mu*u$, \eqref{hcirc} and \eqref{Jder} we obtain the
formula
\begin{align*}
J'(z)\Delta z
&= \int_\Omega  \Bigl(f^\dagger*\psi-\sum_{i=1}^n\phi^\dagger_{i}*\psi_{x_i}
 \Bigl)(x,T) dx-\int_{\Gamma_{2}}(h^\dagger*\psi)(x,T)\, d\Gamma
\\
&\quad- 2\int_\Omega[u(x,T)-u_T(x)]\bigl\{(\Delta\mu
 +\widehat \mu*\Delta\mu)*u\bigl\}(x,T)dx
\\
&\quad- 2\sum_{j=1}^2\int_0^T\Big[\int_{\Gamma_2}\kappa_j(y,t)u(y,t;z)d\Gamma
 - v_j(t)\Big]\\
&\quad\times \int_{\Gamma_2}\kappa_j(x,t)
\bigl\{(\Delta\mu+\widehat \mu*\Delta\mu)*u\bigl\}(x,t)d\Gamma dt.
\end{align*}
Rearranging the terms yields \eqref{gradi} with \eqref{gamma1}-\eqref{gamma3}.
\end{proof}

The formula \eqref{gradi} shows that the vector $(\gamma_1,\gamma_2,\gamma_3)$ 
is a representation of $J'(z)$ in the space $\mathcal{Z}$. It can be used
in gradient-type minimization algorithms (cf. \cite{jk2,jk3}).

 \section{Existence of quasi-solutions}\label{s:5}

\begin{theorem} \label{thm5}
Let  $M$ be compact. Then IP has a quasi-solution in $M$.
\end{theorem}

\begin{proof} 
It coincides with the proof of \cite[Theorem 7 (ii)]{jk3}.
 We use the continuity of $J$ that is a consequence of
the Fr\'echet differentiability of $J$ proved in the previous section.
\end{proof}

\begin{theorem} \label{thm6}
Let $n=1$, $\Omega=(c,d)$, $\varphi=g_\varphi=0$, $g(x,0)=0$ and $M$ be bounded,
 closed and convex. Then IP has a quasi-solution in $M$.
\end{theorem}

\begin{proof} 
This theorem follows from Weierstrass existence theorem \cite{zeidler}
 provided we are able to show that $J$ is weakly sequentially
lower semi-continuous in $M$. We will prove that $J$ is in fact weakly 
sequentially continuous in $M$.

Let us choose some sequence $z_k=(a_k,m_k,\mu_k)\in M$ such that
$z_k\rightharpoonup z=(a,m,\mu)\in  M$. Then it is easy to see that 
$a_k\rightharpoonup a$ in $L^2(c,d)$ and $m_k\rightharpoonup m$,
$\mu_k\rightharpoonup \mu$ in $L^2(0,T)$.
As in the proof of Theorem \ref{thm1}, let $\widehat \mu\in L^2(0,T)$ 
be the solution of \eqref{Volt}. Similarly, let $\widehat\mu_k\in L^2(0,T)$
be the solution
of the equation $\widehat\mu_k+\mu_k*\widehat\mu_k=-\mu_k$ in $(0,T)$. 
Let us show that $\widehat\mu_k\rightharpoonup \widehat\mu$ in
$L^2(0,T)$. To this end we firstly verify the boundedness of the sequence 
$\widehat\mu_k$. Multiplying the equation of
$\widehat\mu_k$ by $e^{-\sigma t}$, $\sigma>0$,
and estimating by means of the Young and Cauchy inequalities we obtain
\begin{align*}
\|e^{-\sigma t}\widehat\mu_k\|_{L^2(0,T)}
&\le \|e^{-\sigma t}\mu_k*e^{-\sigma t}\widehat\mu_k\|_{L^2(0,T)}+
\|e^{-\sigma t}\mu_k\|_{L^2(0,T)}
\\
&\le \|e^{-\sigma t}\mu_k\|_{L^1(0,T)}\|e^{-\sigma t}\widehat\mu_k\|_{L^2(0,T)}+
\|e^{-\sigma t}\mu_k\|_{L^2(0,T)}
\\
&\le \|e^{-\sigma t}\|_{L^2(0,T)}\|\mu_k\|_{L^2(0,T)}\|e^{-\sigma t}
\widehat\mu_k\|_{L^2(0,T)}+
\|e^{-\sigma t}\mu_k\|_{L^2(0,T)}.
\end{align*}
Observing that $\|e^{-\sigma t}\|_{L^2(0,T)}\le 1/ \sqrt{2\sigma }$ and choosing 
$\sigma=\sigma_1=2[\sup\|\mu_k\|_{L^2(0,T)}]^2$ we get
$$
\|e^{-\sigma_1 t}\widehat\mu_k\|_{L^2(0,T)}\le 2 \|e^{-\sigma_1 t}\mu_k\|_{L^2(0,T)}
\,\Rightarrow\, 
\|\widehat\mu_k\|_{L^2(0,T)}\le 2 e^{\sigma_1 T}\sup \|\mu_k\|_{L^2(0,T)}.
$$
This shows that the sequence $\widehat\mu_k$ is bounded.
The difference $\widehat\mu_k-\widehat\mu$ can be expressed as
$$
\widehat\mu_k-\widehat\mu=-(\mu_k-\mu)-v_k*(\mu_k-\mu),
$$
where $v_k=\widehat\mu+\widehat\mu_k+\widehat\mu*\widehat\mu_k$ 
is a bounded sequence in $L^2(0,T)$.
Denote by $\langle\cdot,\cdot\rangle$ the inner product in $L^2(0,T)$. 
With an arbitrary $\zeta\in L^2(0,T)$ we have
\begin{equation}\label{exis1}
\langle\widehat\mu_k- \widehat\mu,\zeta\rangle=-\langle\mu_k- \mu,\zeta\rangle-N_k,\;
 N_k= \int_0^T  v_k(\tau)
\int_0^{T-\tau}    (\mu_k-\mu)(s)\zeta(\tau+s)dsd\tau.
\end{equation}
Since $\zeta(\tau+\,\cdot)\in L^2(0,T-\tau)$ for $\tau\in (0,T)$, it holds
 $\int_0^{T-\tau}(\mu_k-\mu)(s)\zeta(\tau+s)ds\to 0$ for $\tau\in (0,T)$. 
Moreover, since $\mu_k$ is bounded in $L^2(0,T)$, the sequence
of $\tau$-dependent functions $|\int_0^{T-\tau}(\mu_k-\mu)(s)\zeta(\tau+s)ds|$ 
is bounded by a constant.
By the Cauchy inequality and the dominated convergence
theorem, we find
$$
|N_k|\le \|v_k\|_{L^2(0,T)} \|\int_0^{T-\,\cdot}(\mu_k-\mu)(s)\zeta(\cdot+s)ds
\|_{L^2(0,T)}\to 0.
$$
Thus, from \eqref{exis1}, in view of $\mu_k\rightharpoonup\mu$, we obtain
 $\widehat\mu_k\rightharpoonup\widehat\mu$.

Let us define
$$
\widehat u= u+\mu*u\,,\quad \widehat u_k= u_k+\mu_k*u_k,
$$
where $u=u(x,t;z)$ and $u_k=u(x,t;z_k)$ are the weak solutions of 
\eqref{e1}--\eqref{e4} corresponding to the vectors $z$ and $z_k$, respectively.
The relations $u,u_k\in \mathcal{U}(\Omega_T)$ and $\mu,\mu_k\in L^2(0,T)$ 
imply $\widehat u,\widehat u_k\in \mathcal{U}(\Omega_T)$.
Observing the definitions of the resolvent kernels $\widehat\mu$ and 
$\widehat\mu_k$ we deduce
\begin{gather*}
u=\widehat u+\widehat\mu*\widehat u\,,\quad 
u_k= \widehat u_k+\widehat\mu_k*\widehat u_k,
\\
u_k-u=\widehat u_k-\widehat u+\widehat\mu_k*(\widehat u_k-\widehat u)
+(\widehat\mu_k-\widehat\mu)*\widehat u.
\end{gather*}
In view of the latter relation we express the difference of values of the 
functional $J$ as follows:
\begin{equation} \label{exis2}
\begin{aligned}
&J(z_k)-J(z)\\
&=\int_c^d (u_k-u)^2(x,T)dx
+ 2\int_c^d [u(x,T)-u_T(x)](u_k-u)(x,T)dx\\
&\quad +\sum_{j=1}^2\int_0^T\Big[\sum_{l=1}^L\kappa_j(x_l,t)(u_k-u)(x_l,t)\Big]^2dt
\\
&\quad +2\sum_{j=1}^2\int_0^T\Big[\sum_{l=1}^L\kappa_j(x_l,t)u(x_l,t)-v_j(t)\Big]
\Big[\sum_{l=1}^L\kappa_j(x_l,t)(u_k-u)(x_l,t)\Big]dt \\
&= I_k^1+I_k^2+I_k^3+I_k^4,
\end{aligned}
\end{equation}
where
\begin{gather*}
 I_k^1=\int_c^d \Bigl(\widehat u_k-\widehat u+\widehat\mu_k*(\widehat u_k
-\widehat u)+(\widehat\mu_k-\widehat\mu)*\widehat u\Bigl)^2(x,T)dx,
\\
 I_k^2= 2\int_c^d [u(x,T)-u_T(x)]\Bigl(\widehat u_k-\widehat u
 +\widehat\mu_k*(\widehat u_k-\widehat u)
 +(\widehat\mu_k-\widehat\mu)*\widehat u\Bigl)(x,T)dx,
\\
I_k^3=\sum_{j=1}^2\int_0^T\Big[\sum_{l=1}^L\kappa_j(x_l,t)\Bigl(\widehat u_k
-\widehat u+\widehat\mu_k*
(\widehat u_k-\widehat u)+(\widehat\mu_k-\widehat\mu)*\widehat u\Bigl)(x_l,t)\Big]^2
  dt,
\\
\begin{aligned}
 I_k^4&=2\sum_{j=1}^2\int_0^T\Big[\sum_{l=1}^L\kappa_j(x_l,t)u(x_l,t)-v_j(t)\Big]
\\
&\quad\times\Big[\sum_{l=1}^L\kappa_j(x_l,t)
\Bigl(\widehat u_k-\widehat u+\widehat\mu_k*(\widehat u_k-\widehat u)
+(\widehat\mu_k-\widehat\mu)*\widehat u\Bigl)(x_l,t)\Big]dt.
\end{aligned}
\end{gather*}
Using the  Cauchy inequality, $\widehat\mu\in L^2(0,T)$,
$\widehat u_k,\widehat u\in \mathcal{U}(\Omega_T)$ and the
boundedness of the sequence $\widehat\mu_k$ in $L^2(0,T)$ we obtain
\begin{align*}
 |I_k^1|&\le \|\bigl(\widehat u_k-\widehat u
 +\widehat\mu_k*(\widehat u_k-\widehat u)\bigl)(\cdot,T)\|_{L^2(c,d)}^2
\\
&\quad +  2\|\bigl((\widehat\mu_k-\widehat \mu)*\widehat u\bigl)(\cdot,T)
\|_{L^2(c,d)}\|\bigl(\widehat u_k-\widehat u
 +\widehat\mu_k*(\widehat u_k-\widehat u)\bigl)(\cdot,T)\|_{L^2(c,d)}+R_k^1
\\
&\le \tilde C_1\bigl(\|\widehat u_k-\widehat u\|_{\mathcal{U}(\Omega_T)}^2
 +\|\widehat u_k-\widehat u\|_{\mathcal{U}(\Omega_T)}\bigl)+R_k^1
\end{align*}
with a constant $\tilde C_1$  and
\[
R_k^1=\int_c^d \Bigl[\int_0^T (\widehat\mu_k-\widehat\mu)(\tau)
\widehat u(x,T-\tau)d\tau\Bigl]^2dx.
\]
Since $\widehat u\in \mathcal{U}(\Omega_T)\subset L^2(\Omega_T)$,
by Tonelli's theorem  it holds
$\widehat u(x,\cdot)\in L^2(0,T)$ a.e. $x\in (c,d)$ $\Rightarrow$
$\widehat u(x,T-\cdot)\in L^2(0,T)$ a.e. $x\in (c,d)$. Thus, in view of
$\mu_k\rightharpoonup\widehat\mu$ in $L^2(0,T)$ we have
$\int_0^T (\widehat\mu_k-\widehat\mu)(\tau)\widehat u(x,T-\tau)d\tau\to 0$ a.e.
$x\in (c,d)$.
Moreover, $\Bigl[\int_0^T (\widehat\mu_k-\widehat\mu)(\tau)
\widehat u(x,T-\tau)d\tau\Bigl]^2
\le \tilde C_{11} \int_0^T[\widehat u(x,\tau)]^2d\tau
\in L^1(c,d)$ with a constant $\tilde C_{11}$, because the sequence
$\widehat\mu_k$ is bounded in $L^2(0,T)$.
Therefore, by the dominated convergence theorem we obtain $R_k^1\to 0$.
Similarly for $I_k^2$ we get
\begin{align*}
|I_k^2|&\le 2\|u(\cdot,T)-u_T\|_{L^2(c,d)}\|\bigl(\widehat u_k-\widehat u
 +\widehat\mu_k*(\widehat u_k-\widehat u)\bigl)(\cdot,T)\|_{L^2(c,d)}+R_k^2
\\
&\le \tilde C_2 \|\widehat u_k-\widehat u\|_{\mathcal{U}(\Omega_T)} + R_k^2,
\end{align*}
where $\tilde C_2$ is a constant and
\begin{equation}st
R_k^2=\int_c^d [u(x,T)-u_T(x)]\int_0^T (\widehat\mu_k-\widehat\mu)(\tau)
\widehat u(x,T-\tau)d\tau dx.
\end{equation}st
By the same reasons as above, it holds $R_k^2\to 0$. Next, let us estimate $I_k^3$:
\begin{align*}
|I_k^3|
&\le L^2\sum_{j=1}^2\max_{1\le l\le L}\Bigl[\|\kappa_j(x_l,\cdot)\|_{L^\infty(0,T)}^2\|\bigl(\widehat u_k-\widehat u+
\widehat\mu_k*(\widehat u_k-\widehat u)\bigl)(x_l,\cdot)\|_{L^2(0,T)}^2\Bigl]
\\
&\quad +  2L^2\sum_{j=1}^2\max_{1\le l\le L}\Bigl[\|\kappa_j(x_l,\cdot)\|^2_{L^\infty(0,T)}\|\bigl((\widehat\mu_k-\widehat \mu)*\widehat u\bigl)(x_l,\cdot)\|_{L^2(0,T)}
\\
&\quad\times
\|\bigl(\widehat u_k-\widehat u+\widehat\mu_k*(\widehat u_k-\widehat u)\bigl)(x_l,\cdot)\|_{L^2(0,T)}\Bigl]+R_k^3
\\
&\le \tilde C_3\bigl(\|\widehat u_k-\widehat u\|_{\mathcal{U}(\Omega_T)}^2+
\|\widehat u_k-\widehat u\|_{\mathcal{U}(\Omega_T)}\bigl)+R_k^3,
\end{align*}
where $\tilde C_3$ is a constant and
\[
R_k^3=L^2\sum_{j=1}^2\max_{1\le l\le L}\Bigl\{\|\kappa_j(x_l,\cdot)
 \|_{L^\infty(0,T)}^2
\int_0^T \Bigl[\int_0^t (\widehat\mu_k-\widehat\mu)(\tau)
 \widehat u(x_l,t-\tau)d\tau\Bigl]^2dt\Bigl\}.
\]
Here we also used the embedding $\mathcal{U}(\Omega_T)\hookrightarrow
L^2((0,T);C[c,d])$ that holds in the case $n=1$.
Since $\widehat u(x_l,t-\cdot)\in L^2(0,t)$ for all $t\in (0,T)$
we get $\int_0^t (\widehat\mu_k-\widehat\mu)(\tau)\widehat u(x_l,t-\tau)d\tau\to 0$
for all $t\in (0,T)$. Moreover,
the sequence $|\int_0^t (\widehat\mu_k-\widehat\mu)(\tau)\widehat u(x_l,t-\tau)d\tau|$ is bounded by a constant. Consequently, $R_k^3\to 0$.
Analogously we deduce the estimate
\begin{gather*}
|I_k^4|
\le \tilde C_4\|\widehat u_k-\widehat u\|_{\mathcal{U}(\Omega_T)}+R_k^4,\quad
\text{where $\tilde C_4$ is a constant},
\\
\begin{aligned}
R_k^4&=2L\sum_{j=1}^2\bigl\|\sum_{l=1}^L\kappa_j(x_l,t)u(x_l,\cdot)-v_j
\bigl\|_{L^2(0,T)}\max_{1\le l\le L}\Bigl\{
\|\kappa_j(x_l,\cdot)\|_{L^\infty(0,T)}\\
&\quad\times\Bigl[ \int_0^T \Bigl[\int_0^t
(\widehat\mu_k-\widehat\mu)(\tau)\widehat u(x_l,
t-\tau)d\tau\Bigl]^2dt\Bigl]^{1/2}\Bigl\},
\end{aligned}
\end{gather*}
where $R_k^4\to 0$.

Note that if we manage to show that 
$\|\widehat u_k-\widehat u\|_{\mathcal{U}(\Omega_T)}\to 0$ 
then the proof is complete. Indeed,
in this case by virtue  of $R_k^i\to 0$, $i=1,2,3,4$, from the estimates 
of $I_k^i$ we get $I_k^i\to 0$, $i=1,2,3,4$ and due to
\eqref{exis2} we obtain $J(z_k)\to J(z)$, which implies the statement of the theorem.

As in the proof of Theorem \ref{thm1} we can show that $\widehat u$ and $\widehat u_k$ 
are the weak solutions of the following problems:
\begin{gather} \label{e1e}
\widehat u_t= A\widehat u - \widehat m*A\widehat u +  f+\phi_x\quad 
\text{in }\Omega_T,
\\
\label{e2e}  \widehat u= u_0  \quad  \text{in }\Omega\times \{0\},
\\
\label{e3e}  \widehat u=\widehat g \quad  \text{in }\Gamma_{1,T},
\\  \label{e4e}
 -\nu_{A}\cdot \nabla \widehat u + \widehat m* \nu_{A}\cdot \nabla \widehat u 
= \vartheta \widehat u+\vartheta\widehat\mu*\widehat u+ h +\nu\cdot \phi\quad  
\text{in }\Gamma_{2,T}, \\
 \label{e1ek}
\widehat u_{k,t}= A_k\widehat u_k - \widehat m_k*A_k\widehat u_k 
+  f+\phi_x\quad \text{in }\Omega_T,
\\
\label{e2ek} \widehat u_k= u_0  \quad  \text{in }\Omega\times \{0\},
\\
\label{e3ek} \widehat u_k=\widehat g_k \quad  \text{in }\Gamma_{1,T},
\\  \label{e4ek}
 -\nu_{A}\cdot \nabla \widehat u_k + \widehat m_k* \nu_{A}\cdot \nabla \widehat u_k 
= \vartheta \widehat u_k+\vartheta\widehat\mu_k*\widehat u_k+ h 
 +\nu\cdot \phi\quad  \text{in }\Gamma_{2,T}, 
\end{gather}
where $A_kv=(a_{11}v_{x})_{x}+a_kv$,
\begin{gather*}
\widehat m= m-\widehat\mu+m*\widehat\mu\, ,\quad
\widehat m_k= m_k-\widehat\mu_k+m_k*\widehat\mu_k,\\
\widehat g=g+\mu*g\, ,\quad  \widehat g_k=g+\mu_k*g.
\end{gather*}
We now show that $\widehat m_k\rightharpoonup \widehat m$.
 With any $\zeta\in L^2(0,T)$ we compute
\begin{gather*}
\langle \widehat m_k-\widehat m, \zeta\rangle
 =\langle m_k- m, \zeta\rangle  -\langle \widehat \mu_k-\widehat \mu, \zeta\rangle 
+N_k^1,\\
\begin{aligned}
N_k^1 &= \int_0^T\widehat\mu_k(\tau)\int_0^{T-\tau}(m_k-m)(s)\zeta(\tau+s)dsd\tau
\\
&\quad+ \int_0^T m(\tau)\int_0^{T-\tau}(\widehat\mu_k-\widehat\mu)(s)
\zeta(\tau+s)dsd\tau.
\end{aligned}
\end{gather*}
We use the relations $m_k\rightharpoonup m$, 
$\widehat \mu_k\rightharpoonup \widehat \mu$  and treat the term $N_k^1$ 
similarly to the term $N_k$ in
\eqref{exis1} to get $N_k^1\to 0$. As a result we get
 $\langle \widehat m_k-\widehat m, \zeta\rangle\to 0$, hence
 $\widehat m_k\rightharpoonup \widehat m$.

Subtracting the problem of $\widehat u$ from the problem of $\widehat u_k$ 
we see that $w_k:=\widehat u_k-\widehat u$ is a weak solution of the 
problem
\begin{gather} \label{e1w}
w_{k,t}= A w_k- \widehat m*A w_k +  \tilde f_k+\tilde\phi_{k,x}\quad \text{in $\Omega_T$,}
\\
\label{e2w} \widehat u= 0  \quad  \text{in }\Omega\times \{0\},
\\
\label{e3w} \widehat u=\tilde g_k \quad  \text{in }\Gamma_{1,T},
\\  \label{e4w}
-\nu_{A}\cdot \nabla w_k + \widehat m* \nu_{A}\cdot \nabla w_k = \vartheta
w_k+\tilde h_k +\nu\cdot \tilde\phi_k\quad  \text{in }\Gamma_{2,T}, 
\end{gather}
where
\begin{gather*}
\tilde f_k=(a_k-a)(\widehat u_k-\widehat m_k*\widehat u_k)
-a(\widehat m_k-\widehat m)*\widehat u_k,
\\
\tilde \phi_k=-a_{11}(\widehat m_k-\widehat m)*\widehat u_{k,x},\quad
 \tilde g_k=(\mu_k-\mu)*g,
\\
\tilde h_k=\vartheta[\widehat \mu_k*w_k+ (\widehat \mu_k-\widehat\mu)*\widehat u].
\end{gather*}
To  use the weak convergence $a_k\rightharpoonup a$ in forthcoming estimations
 we have to introduce the functions $\rho_k\in W_2^2(c,d)$ being the
solutions of the following Neumann problems:
$$
\rho_k''-\rho_k=a_k-a \quad \text{in $(c,d)$}\,,\qquad \rho_k'(c)=\rho_k'(d)=0.
$$
Then $\rho_k(x)=\int_c^d G(x,y)(a_k-a)(y)dy$, $x\in (c,d)$, where
$$
G(x,y)=\frac{1}{2(e^{c-d}-e^{d-c})}\begin{cases}
(e^{c-y}+e^{y-c})(e^{d-x}+e^{x-d}) &\text{for }y<x\\
(e^{c-x}+e^{x-c})(e^{d-y}+e^{y-d}) &\text{for }y>x
\end{cases}
$$
 is a Green function that satisfies the properties
$G,G_x\in L^\infty(\Omega_T)$. The weak convergence $a_k\rightharpoonup a$ 
in $L^2(c,d)$ implies
\begin{equation}\label{exis3}
\|\rho_k\|_{W_2^1(c,d)}\to 0.
\end{equation}
Using $\rho_k$  we rewrite the term $(a_k-a)(\widehat u_k-\widehat m_k*\widehat u_k)$
in $\tilde f_k$ as follows:
\begin{align*}
&(a_k-a)(\widehat u_k-\widehat m_k*\widehat u_k)\\
&= [\rho_k'(\widehat u_k-\widehat m_k*\widehat u_k)]_x
 -\rho_k'(\widehat u_k-\widehat m_k*\widehat u_k)_x
 -\rho_k(\widehat u_k-\widehat m_k*\widehat u_k).
\end{align*}
According to this relation we change the form of the problem for $w_k$ as follows:
\begin{gather} \label{e1ww}
w_{k,t}= A w_k- \widehat m*A w_k +  \overline f_k+\overline\phi_{k,x}\quad
\text{in }\Omega_T,
\\
\label{e2ww} \widehat u= 0  \quad  \text{in }\Omega\times \{0\},
\\
\label{e3ww} \widehat u=\tilde g_k \quad  \text{in }\Gamma_{1,T},
\\  \label{e4ww}
-\nu_{A}\cdot \nabla w_k + \widehat m* \nu_{A}\cdot \nabla w_k
= \vartheta w_k+\tilde h_k +\nu\cdot \overline\phi_k\quad
\text{in }\Gamma_{2,T},
\end{gather}
where
\begin{gather*}
\overline f_k=-\rho_k'(\widehat u_k+\widehat m_k*\widehat u_k)_x
-\rho_k(\widehat u_k+\widehat m_k*\widehat u_k)
 -a(\widehat m_k-\widehat m)*\widehat u_k,
\\
\overline \phi_k=\rho_k'(\widehat u_k+\widehat m_k*\widehat u_k)
-a_{11}(\widehat m_k-\widehat m)*\widehat u_{k,x}.
\end{gather*}

Let $t$ be an arbitrary number in $[0,T]$. 
To estimate $w_k$ we will use the projection operators $P_t$, defined in \eqref{Pt}, 
and $\overline P_tw=\begin{cases}
  w &  \text{in }\Omega_{t}\\   
 0 &  \text{in }\Omega_{T}\setminus\Omega_{t}
\end{cases}$.
for $w: \Omega_{T}\to \mathbb{R}$. Let $w^t_k$ stand for the weak solution of
 problem \eqref{e1ww}--\eqref{e4ww} with
$\overline f_k$, $\overline\phi_k$ and $\tilde h_k$ replaced by 
$\overline P_t\overline f_k$, $\overline P_t\overline\phi_k$
and $P_t\tilde h_k$, respectively. Then, due to the causality 
$w_k^t=w_k$ in $\Omega_t$. Applying \eqref{t1est} for $w_k^t$
we obtain
\begin{equation}\label{exis3a}
\begin{split}
\|w_k\|_{\mathcal{U}(\Omega_t)}
&= \|w_k^t\|_{\mathcal{U}(\Omega_t)}\le \|w_k^t\|_{\mathcal{U}(\Omega_T)}
\le \overline C_0\Bigl[\|\overline P_t\overline f_k\|_{L^2((0,T);L^{1}(c,d))}
\\
&\quad +\|\overline P_t\overline\phi_k\|_{(L^2(\Omega_T))^n}+\theta\|\tilde g_k\|_{\mathcal{T}(\Omega_T)}
+\|P_t\tilde h_k\|_{L^2(\Gamma_{2,T})}
\Bigl]
\\
& = \overline C_0\Bigl[\|\overline f_k\|_{L^2((0,t);L^{1}(c,d))}
+\|\overline\phi_k\|_{L^2(\Omega_t)}+\theta\|\tilde g_k\|_{\mathcal{T}(\Omega_T)}
+\|\tilde h_k\|_{L^2(\Gamma_{2,t})}
\Bigl]
\end{split}
\end{equation}
with a constant $\overline C_0$.
Using the relation $a\in L^2(c,d)$, Cauchy inequality, the inequality
\eqref{qq3}, $g(x,0)=0$, the embedding $W_2^1(c,d)\hookrightarrow C[c,d]$
and $\widehat u_k=w_k+\widehat u$
we estimate:
\begin{gather}  \label{exis3b}
\begin{aligned}
\|\overline f_k\|_{L^2((0,t);L^{1}(c,d))}
&\le \overline C_1\Bigl[\|(\widehat m_k-\widehat m)*\widehat u_k\|_{L^2((0,t);
 L^{2}(c,d))}\\
&\quad + \|\rho_k\|_{W_2^1(c,d)}
 (1+\|\widehat m_k\|_{L^2(0,T)})\|\widehat u_k\|_{\mathcal{U}(\Omega_t)}\Bigl]\\
&\le \overline C_1\Bigl[\int_0^t |(\widehat m_k-\widehat m)(t-\tau)|\,
 \|w_k\|_{L^2(\Omega_\tau)}d\tau\\
&\quad + \|\rho_k\|_{W_2^1(c,d)}(1+\|\widehat m_k\|_{L^2(0,T)})
\|w_k\|_{\mathcal{U}(\Omega_t)}\Bigl] 
+\overline R_k^1,
\end{aligned}\\   \label{exis3c}
\begin{aligned}
\|\overline \phi\|_{L^2(\Omega_t)}
&\le \overline C_2\Bigl[\|(\widehat m_k-\widehat m)*\widehat u_{k,x}\|_{L^2(\Omega_t)}\\
&\quad + \|\rho_k\|_{W_2^1(c,d)}(1+\|\widehat m_k\|_{L^2(0,T)})
\|\widehat u_k\|_{L^2((0,T);C[c,d])}\Bigl]
\\
&\le \overline C_2\Bigl[\int_0^t |(\widehat m_k-\widehat m)(t-\tau)|\,
\|w_{k,x}\|_{L^2(\Omega_\tau)}d\tau\\
&\quad +\|\rho_k\|_{W_2^1(c,d)}(1+\|\widehat m_k\|_{L^2(0,T)})\|w_k\|_{L^2((0,t);
C[c,d])}\Bigl]+\overline R_k^2,
\end{aligned}
\\
\label{exis3d}
\|\tilde g_k\|_{\mathcal{T}(\Omega_T)}\le \overline R_k^3,
\\  \label{exis3e}
\begin{aligned}
\|\tilde h_k\|_{L^2(\Gamma_{2,t})}
&\le \overline C_4 \Bigl[\|\widehat \mu_k*w_k\|_{L^2((0,t);W_2^1(c,d))}
 +\|(\widehat \mu_k-\widehat\mu)*\widehat u\|_{L^2((0,t);W_2^1(c,d))}\Bigl]
\\
&\le  \overline C_4\int_0^t |\widehat\mu_k(t-\tau)|\,
\|w_{k}\|_{L^2((0,\tau);W_2^1(c,d))}d\tau + \overline R_k^4,
\end{aligned}
\end{gather}
where $\overline C_1, \overline C_2, \overline C_4$ are constants and
\begin{gather*}
\overline R_k^1=\overline C_1\Bigl[\|(\widehat m_k-\widehat m)*\widehat
u\|_{L^2(\Omega_T)}
+ \|\rho_k\|_{W_2^1(c,d)}(1+\|\widehat m_k\|_{L^2(0,T)})
\|\widehat u\|_{\mathcal{U}(\Omega_T)}\Bigl],
\\
\begin{aligned}
\overline R_k^2&=\overline C_2\Bigl[\|(\widehat m_k-\widehat m)*\widehat
 u_x\|_{L^2(\Omega_T)}
 + \|\rho_k\|_{W_2^1(c,d)}\big(1 \\
&\quad +\|\widehat m_k\|_{L^2(0,T)}\big)
 \|\widehat u\|_{L^2((0,T);C[c,d])}\Bigl],
\end{aligned}\\
\overline R_k^3=\|(\mu_k-\mu)*g\|_{L^2(\Omega_T)}
 + \|(\mu_k-\mu)*g_x\|_{L^2(\Omega_T)}+\|(\mu_k-\mu)*g_t\|_{L^2(\Omega_T)},
\\
 \overline R_k^4=\overline C_4 \|(\widehat \mu_k-\widehat\mu)*\widehat u\|_{L^2((0,T);
W_2^1(c,d))}.
\end{gather*}
By  the weak convergence
$\widehat m_k\rightharpoonup \widehat m$, $\mu_k\rightharpoonup \mu$,
$\widehat \mu_k \rightharpoonup \widehat u$ in $L^2(0,T)$ and
the relation $\|\rho_k\|_{W_2^1(c,d)}\to 0$ it holds
\begin{equation}\label{exis4}
\overline R_k^j\to 0, \quad j=1,2,3,4.
\end{equation}
Indeed, to prove that $\|z_k*\hat v\|_{L^2(\Omega_T)}\to 0$, where
$z_k$ is  one of the functions $\widehat m_k-\widehat m$, $\mu_k-\mu$ or
$\widehat \mu_k-\widehat\mu$ and $\hat v\in L^2(\Omega_T)$ is one of
the functions $\widehat u$, $\widehat u_x$, $g$, $g_x$ or $g_t$,
it is possible to use the
dominated convergence theorem, again. More precisely,
\[
\|z_k*\hat v\|_{L^2(\Omega_T)}
=\Bigl\{\int_0^T\int_c^d\Bigl[\int_0^t z_k(\tau)\hat v(x,t-\tau)d\tau\Bigl]^2dx\,dt\Bigl\}^{1/2},
\]
where the component $\Bigl[\int_0^t z_k(\tau)\hat v(x,t-\tau)d\tau\Bigl]^2$
is bounded by an integrable  in $x\in (c,d)$ function
 $\sup\|z_k\|_{L^2(0,T)}^2
\|\hat v(x,\cdot)\|_{L^2(0,T)}^2$  and tends to zero
for all $t\in (0,T)$ and a.e. $x\in (c,d)$, because $z_k\rightharpoonup 0$ and
$\hat v(x,t-\cdot)\in L^2(0,T)$ for all $t\in (0,T)$ and a.e. $x\in (c,d)$.
(The latter relation follows from $\hat v\in L^2(\Omega_T)$ and Tonelli's theorem.)
Thus, $\|z_k*\hat v\|_{L^2(\Omega_T)}\to 0$.

As in proof of Theorem \ref{thm1}, we use the  norms
 $\|w\|_\sigma=\sup_{0<t<T}e^{-\sigma t}\|w\|_{\mathcal{U}(\Omega_t)}$ with
the weights $\sigma\ge 0$ in the space $\mathcal{U}(\Omega_T)$.
Then in view of \eqref{exis3b}--\eqref{exis3e} from \eqref{exis3a} we deduce
\begin{align*}
\|w_k\|_\sigma
&\le \overline C_5 \Bigl[\sup_{0<t<T}\int_0^t e^{-\sigma (t-\tau)}r_k(t-\tau)\,
 e^{-\sigma\tau}\|w_k\|_{\mathcal{U}(\Omega_\tau)}d\tau
\\
&\quad +\|\rho_k\|_{W_2^1(c,d)}(1+\|\widehat m_k\|_{L^2(0,T)})
 \|w_k\|_{\sigma}+\sum_{j=1}^4 \overline R_k^j\Bigl]
\\
&\le  \overline C_5 \Bigl[\Bigl\{\|e^{-\sigma t}\|_{L^2(0,T)}\|r_k\|_{L^2(0,T)}
 +\|\rho_k\|_{W_2^1(c,d)}(1+\|\widehat m_k\|_{L^2(0,T)})\Bigl\}\|w_k\|_{\sigma}
\\
&\quad+\sum_{j=1}^4 \overline R_k^j\Bigl],
\end{align*}
where $\overline C_5$ is a constant and 
$r_k=|\widehat m_k-\widehat m|+|\widehat\mu_k|$. Since 
$\|e^{-\sigma t}\|_{L^2(0,T)}\to 0$
as $\sigma\to\infty$, $\|\rho_k\|_{W_2^1(c,d)}\to 0$ and the sequences 
$\|r_k\|_{L^2(0,T)}$, $\|\widehat m_k\|_{L^2(0,T)}$ are bounded, there exist 
$\sigma_2>0$ and $K_2\in \mathbb{N}$
such that
$$
\|e^{-\sigma_2 t}\|_{L^2(0,T)}\|r_k\|_{L^2(0,T)}
+\|\rho_k\|_{W_2^1(c,d)}(1+\|\widehat m_k\|_{L^2(0,T)})\le \frac{1}{2\overline C_5}
$$
 for $k\ge K_2$.
This, along with the previous inequality, implies
$$
\|w_k\|_{\sigma_2}\le 2\overline C_5 \sum_{j=1}^4 \overline R_k^j \quad
\text{and hence}\quad 
\|w_k\|_{\mathcal{U}(\Omega_T)}\le 2e^{\sigma_2 T}\overline C_5 
\sum_{j=1}^4 \overline R_k^j
$$
for $k\ge K_2$.
Taking \eqref{exis4} into account we obtain the desired convergence: 
$\|\widehat u_k-\widehat u\|_{\mathcal{U}(\Omega_T)}
=\|w_k\|_{\mathcal{U}(\Omega_T)}\to 0$.
The theorem is proved. 
\end{proof}

\subsection*{Acknowledgement}
The research was supported by Estonian Ministry of Education and Science 
target financed theme SF0140011s09.

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