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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2018 (2018), No. 77, pp. 1--17.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu}
\thanks{\copyright 2018 Texas State University.}
\vspace{8mm}}

\begin{document}
\title[\hfilneg EJDE-2018/77\hfil Inverse problem for a degenerate heat equation]
{Inverse problem for a two-dimensional strongly degenerate heat equation}

\author[M. Ivanchov, V. Vlasov \hfil EJDE-2018/77\hfilneg]
{Mykola Ivanchov, Vitaliy Vlasov}

\address{Mykola Ivanchov \newline
Department  of Differential Equations,
Ivan Franko National University of Lviv,
  Universytetska st., 1,   79000, Ukraine}
\email{mykola.ivanchov@lnu.edu.ua}

\address{ Vitaliy Vlasov \newline
Department  of Differential Equations,
Ivan Franko National University of Lviv,
  Universytetska st., 1, 79000, Ukraine}
\email{siphiuel@gmail.com}

\dedicatory{Communicated by Ludmila S. Pulkina}

\thanks{Submitted November 29, 2017. Published March 20, 2018.}
\subjclass[2010]{35R30, 35K10, 35A09, 35A02}
\keywords{Inverse problem; strongly degenerate heat equation; 
 Green function; 
\hfill\break\indent classical solution}

\begin{abstract}
 This article concerns the existence and uniqueness of solutions 
 in the problem  of identifying the leading coefficient in a 
 two-dimensional heat equation. We suppose that unknown coefficient 
 depends on the time variable and the  equation is strongly degenerate.
 Applying Schauder fixed-point theorem, we find conditions for 
 existence of a classical solution.
\end{abstract}

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


\section{Introduction}

We consider an inverse problem for a two-dimensional degenerate heat equation
in a rectangular domain. Direct problems of this type are  mathematical models
of various processes such as seawater desalination, movement of liquid in
porous medium, financial market behavior, etc.
\cite{Ber,Caf,Greb}. Such problems are comparatively well studied
(see, for instance, \cite{Ben}).

Inverse problems arise when certain parameters of these processes are unknown.
Various types of inverse problems for non-degenerate equations are well
investigated and some results may be found   in monographs and references
therein \cite{Imi1,Isak,Pril}. The beginning of the research of inverse problems
for one-dimensional degenerate parabolic equations was made in
\cite{Imi2, Imi3, Imi4, Loren,Sald1}.  There the conditions of existence
and uniqueness of solution for these problems were  established in the cases
of week and strong power degeneration. Later  some results were obtained
for parabolic equations with arbitrary types of degeneration
\cite{Sald2, Sald3} and for free-boundary domains \cite{Hryn1,Hryn2}.
 Inverse problem for a two-dimensional weakly degenerate heat equation in
a rectangular domain was considered in \cite{Vlas}.

In this paper we establish conditions for existence and uniqueness of
solution to an inverse problem for a two-dimensional strongly
degenerate heat equation.


\section{Statement of the problem and main result}

In the domain $Q_T:=\{(x,y,t): 0<x<h, 0<y<l, 0<t<T\} $ we consider a
two-dimensional heat equation  with unknown leading coefficient depending on
the time variable. We suppose that the equation degenerates at
the initial moment as a power with a given exponent  $\beta\ge 1$.
We choose the case of mixed Dirichlet-Neumann
boundary conditions. The additional condition (so-called
{\it overdetermination condition}) is taken accordingly to the physical
sense and it presents the value of the heat flux on the part of the boundary
of  domain. So, the problem consists of finding a pair of functions
$(a(t), u(x,y,t)),  a(t)>0, t\in[0,T]$ that satisfy the degenerate
heat equation
\begin{equation}\label{vls1}
u_t = t^{\beta}a(t)\Delta u + f(x,y,t),\quad (x,y,t) \in Q_T,
\end{equation}
the initial condition
\begin{equation}\label{vls2}
u(x,y,0) = \varphi(x,y,0),\quad (x,y) \in \overline{D}:=[0,h]\times[0,l],
\end{equation}
the boundary conditions
\begin{align}\label{vls3}
&u(0,y,t) = \mu_1(y,t),\, u(h,y,t) = \mu_2(y,t), \quad (y,t) \in [0,l]\times[0,T],\\ \label{vls4}
&u_y(x,0,t) = \nu_1(x,t),\, u_y(x,l,t) = \nu_2(x,t), \quad (x,t) \in [0,h]\times[0,T]
\end{align}
and the overdetermination condition
\begin{equation}\label{vls5}
 a(t)u_x(0,y_0,t) = \kappa(t),\quad t \in (0,T],
\end{equation}
where $y_0\in (0,l)$ is some arbitrary fixed point. Our goal is to determine
conditions of the existence and uniqueness of solution to problem
\eqref{vls1}--\eqref{vls5} in the spaces of continuously differentiable functions.
 The corresponding result is presented in a theorem,
for which we use the following assumptions:
\begin{itemize}
\item[(A1)]   $\beta \geq 1$, $\varphi \in C^2(\overline{D})$,
$\mu_i \in C^{2,1}([0,l]\times (0,T]) \cap C^{1,1}([0,l]\times [0,T])$,
$\nu_i \in C^{1,0}([0,h] \times (0,T]) \cap C([0,h]\times [0,T])$
for $i=1,2$,  $f \in C^{2,2,0}(\overline{Q}_T)$,
$\kappa(t) = \kappa_0(t)t^{\frac{1-\beta}{2}}$ with $\kappa_0 \in C[0,T]$;

\item[(A2)]  $ \varphi_x(x,y) \geq 0$ for $(x,y) \in \overline{D}$;
$\mu_{1_t}(y,t)-f(0,y,t) < 0$, $ \mu_{2_t}(y,t) - f(h,y,t) \geq 0$,
$\mu_{1_{yy}}(y,t) \geq 0$, $\mu_{2_{yy}}(y,t) \leq 0$ for
 $(y,t) \in [0,l]\times(0,T]$;
$\nu_{1_x}(x,t) \leq 0$, $\nu_{2_x}(x,t) \geq 0$ for $(x,t) \in [0,h] \times [0,T]$;
$f_x(x,y,t) \geq 0$ for $(x,y,t) \in \overline{Q}_T$;
$\kappa_0(t) > 0$ for $t \in [0,T]$;

\item[(A3)] $\mu_1(y,0) = \varphi(0,y)$, $\mu_2(y,0) = \varphi(h,y)$ for
$y \in [0,l]$;
$\nu_1(x,0) = \varphi_y(x,0)$,   $ \nu_2(x,0) = \varphi_y(x,l)$ for $x \in [0,h]$;
$\mu_{1y}(0,t) = \nu_1(0,t)$, $\mu_{1y}(l,t) = \nu_2(0,t)$,
$\mu_{2y}(0,t) = \nu_1(h,t)$,   $\mu_{2y}(l,t) = \nu_2(h,t)$ for $t \in [0,T]$.
\end{itemize}

\begin{theorem} \label{thm1}
Under assumptions {\rm (A1)--(A3)},
there exists  unique solution $(a,u)$ to \eqref{vls1}--\eqref{vls5},  which
belongs to the space $C[0,T] \times (C^{2,2,1}(Q_T)\cap C^{0,1,1}(\overline{Q}_T))$,
where $a(t) > 0$ and $t\in [0,T]$.
\end{theorem}

Here we have the usual notation \cite{Lad} for the Banach spaces
 $ C^{k,m,n}(Q_T)$ consisting
 of functions that are continuous in $Q_T$, with their derivatives of
 $k$-th order  with respect to $x$, $m$-th order  with respect to $y$,
and $n$-th order  with respect to $t$.
The spaces $C^{k,m}([0,h] \times (0,T])$ and others are defined similarly.


\section{Existence of solutions}

There are many approaches to study existence of solutions to inverse problems.
Among them, we choose the method of reducing the problem \eqref{vls1}--\eqref{vls5}
to an equation with respect to the coefficient $a(t)$ and applying the
Schauder fixed-point theorem.

  Assume temporarily that $a=a(t)>0$ from class $C[0,T] $ is known and
find the solution to the direct problem \eqref{vls1}--\eqref{vls4} \cite{Imi1}:
\begin{equation} \label{vls6}
\begin{aligned}
u(x,y,t)
&= \int_0^l \int_0^h G_{12}(x,y,t,\xi,\eta,0)\varphi(\xi,\eta)d\xi d\eta \\
&\quad + \int_0^t \int_0^l G_{12_{\xi}}(x,y,t,0,\eta,\tau)
 \tau^{\beta}a(\tau)\mu_1(\eta,\tau)\,d\eta\,d\tau \\
&\quad - \int_0^t \int_0^l G_{12_{\xi}}(x,y,t,h,\eta,\tau) \tau^{\beta}
 a(\tau) \mu_2(\eta,\tau) \,d\eta\,d\tau  \\
&\quad - \int_0^t \int_0^h G_{12}(x,y,t,\xi,0,\tau)
 \tau^{\beta}a(\tau)\nu_1(\xi,\tau)d\xi d\tau \\
&\quad + \int_0^t \int_0^h G_{12}(x,y,t,\xi,l,\tau)
  \tau^{\beta}a(\tau)\nu_2(\xi,\tau)d\xi d\tau \\
&\quad + \int_0^t \int_0^l \int_0^h G_{12}(x,y,t,\xi,\eta,\tau)f(\xi,\eta,\tau)
\,d\xi\,d\eta\,d\tau.
\end{aligned}
\end{equation}
Here we denote by $G_{ij}(x,y,t,\xi,\eta,\tau)$, $i\in\{1,2\}$ the Green
functions for the heat equation \eqref{vls1} with boundary conditions of
$i$-th kind with respect to $x$ and $j$-th kind with respect to  $y$.
 They are defined the equality \cite{Imi1}
\begin{equation} \label{vls7}
\begin{gathered}
\begin{aligned}
&G_{ij}(x,y,t,\xi,\eta,\tau)\\
& = \frac{1}{4\pi(\theta(t)-\theta(\tau))}
 \sum_{m,n=-\infty}^{\infty} \bigg(\exp\Big(-\frac{(x-\xi+2nh)^2}{4(\theta(t)
 -\theta(\tau))}\Big)   \\
&\quad + (-1)^i \exp\Big(-\frac{(x+\xi+2nh)^2}{4(\theta(t)-\theta(\tau))}\Big) \bigg)
\bigg(\exp\Big(-\frac{(y-\eta+2ml)^2}{4(\theta(t)-\theta(\tau))}\Big)    \\
&\quad + (-1)^j \exp\Big(-\frac{(y+\eta+2ml)^2}{4(\theta(t)-\theta(\tau))}\Big)\bigg),
\end{aligned}\\
\theta(t) = \int_0^t \tau^{\beta}a(\tau)d\tau,\quad i,j\in\{1,2\}.
 \end{gathered}
\end{equation}
Note that $G_{ij}$ may be expressed as a product of two Green functions for
 one-dimensional heat equations:
$G_{ij}(x,y,t,\xi,\eta,\tau)=G_i(x,t,\xi,\tau) G_j(y,t,\eta,\tau)$.

Using Green function's properties and integrating by parts, find the
 derivative $u_x(x,y,t)$:
\begin{align*}
&u_x(x,y,t) \\
&= \int_0^l \int_0^h G_{22}(x,y,t,\xi,\eta,0)\varphi_{\xi}(\xi,\eta)d\xi d\eta   \\
&\quad- \int_0^t \int_0^l G_{22}(x,y,t,0,\eta,\tau)(\mu_{1_{\tau}}(\eta,\tau)
 -\tau^{\beta}a(\tau)\mu_{1_{\eta\eta}}(\eta,\tau) - f(0,\eta,\tau))\,d\eta\,d\tau  \\
&\quad + \int_0^t \int_0^l G_{22}(x,y,t,h,\eta,\tau)(\mu_{2_{\tau}}(\eta,\tau)
 - \tau^{\beta}a(\tau)\mu_{2_{\eta\eta}}(\eta,\tau) - f(h,\eta,\tau))\,d\eta\,d\tau  \\
&\quad - \int_0^t \int_0^h G_{22}(x,y,t,\xi,0,\tau)\tau^{\beta}a(\tau)\nu_{1_{\xi}}
 (\xi,\tau)d\xi d\tau \\
&\quad + \int_0^t \int_0^h G_{22}(x,y,t,\xi,l,\tau)
 \tau^{\beta}a(\tau)\nu_{2_{\xi}}(\xi,\tau)d\xi d\tau \\
&\quad + \int_0^t \int_0^l \int_0^h G_{22}(x,y,t,\xi,\eta,\tau)f_{\xi}
(\xi,\eta,\tau)\,d\xi\,d\eta\,d\tau.
\end{align*}
Substituting expression  in \eqref{vls5} we obtain the equation 
\begin{equation} \label{vls9}
\begin{aligned}
a(t) 
& = \kappa(t)\Big(\int_0^l \int_0^h G_{22}(0,y_0,t,\xi,\eta,0)\varphi_{\xi}(\xi,\eta)
 d\xi d\eta \\
&\quad - \int_0^t \int_0^l G_{22}(0,y_0,t,0,\eta,\tau)
 (\mu_{1_{\tau}}(\eta,\tau) - \tau^{\beta}a(\tau)\mu_{1_{\eta\eta}}(\eta,\tau)\\
&\quad -f(0,\eta,\tau))\,d\eta\,d\tau \\
&\quad + \int_0^t \int_0^l G_{22}(0,y_0,t,h,\eta,\tau)
 (\mu_{2_{\tau}}(\eta,\tau)-\tau^{\beta}a(\tau)\mu_{2_{\eta\eta}}(\eta,\tau)\\
&\quad -f(h,\eta,\tau))\,d\eta\,d\tau \\
&\quad - \int_0^t \int_0^h G_{22}(0,y_0,t,\xi,0,\tau)
 \tau^{\beta}a(\tau)\nu_{1_{\xi}}(\xi,\tau)d\xi d\tau \\
&\quad + \int_0^t \int_0^h G_{22}(0,y_0,t,\xi,l,\tau)\tau^{\beta}a(\tau)
 \nu_{2_{\xi}}(\xi,\tau)d\xi d\tau   \\
&\quad + \int_0^t \int_0^l \int_0^h G_{22}(0,y_0,t,\xi,\eta,\tau)
 f_{\xi}(\xi,\eta,\tau)\,d\xi\,d\eta\,d\tau\Big)^{-1},\quad   t\in(0,T].
 \end{aligned}
\end{equation}
It is easy to see this equation  and problem \eqref{vls1}--\eqref{vls5}
are equivalent. Indeed, if $(a(t), u(x,y,t))$ is a solution to
 \eqref{vls1}--\eqref{vls5}, then the function $a(t)$ satisfies \eqref{vls9}
as it is shown above. On the other hand, if $a(t)$ is a solution to
\eqref{vls9}, then we substitute it into \eqref{vls1} and find the solution
to problem \eqref{vls1}--\eqref{vls4} under the form \eqref{vls6}.
Condition \eqref{vls5} follows from \eqref{vls9}.

To establish  estimates for solutions of \eqref{vls9},we  denote
$ a_{\rm max}:=\max_{[0,T]}a(t)$ and consider the following integral
from \eqref{vls9},
\begin{equation} \label{vls99}
\begin{aligned}
& \int_0^t \int_0^l G_{22}(0,y_0,t,0,\eta,\tau)(f(0,\eta,\tau)
 - \mu_{1\tau}(\eta,\tau))\,d\eta\,d\tau \\
&\geq \min_{[0,l]\times[0,T]}(f(0,y,t) - \mu_{1_t}(y,t))
 \int_0^t\int_0^l G_{22}(0,y_0,t,0,\eta,\tau)\,d\eta\,d\tau \\
&\geq C_1\int_0^t\frac{d\tau}{\sqrt{\theta(t)-\theta(\tau)}}\\
&\geq \frac{C_2}{\sqrt{a_{\rm max}}}\int_0^t\frac{d\tau}{\sqrt{ t^{\beta+1}
 -\tau^{\beta+1}}}
\ge \frac{C_3}{t^{\frac{\beta-1}{2}}\sqrt{a_{\rm max}}}.
\end{aligned}
\end{equation}
Here we  used the equality
\begin{equation}\label{vls10}
\int_0^l G_2( y,t,\eta,\tau)d\eta=1,
\end{equation}
that can be verified by direct computation. The constants
$C_i$, $i\in\{1,2,3\} $ depend on given data. It follows from (A2)
 that all other summands in denominator of \eqref{vls9} are non-negative,
therefore
$$
a(t)\le\frac{\kappa(t)t^{\frac{\beta-1}{2}}\sqrt{a_{\rm max}}}{C_3}.
$$
Taking into account the assumptions
$\kappa(t) = \kappa_0(t)t^{\frac{1-\beta}{2}}$, $\kappa_0(t)>0$, $t\in[0,T]$,
 we obtain
$$
a(t)\le \frac{\kappa_0(t) \sqrt{a_{\rm max}}}{C_3},\quad t\in[0,T].
$$
Hence, we get the estimate
\begin{equation}\label{vls11}
a(t)\le A_1<\infty,\quad t\in[0,T],
\end{equation}
where $A_1$ is a known constant depending on the problem data.

Now we estimate $a(t)$ from below. Denote integrals in the denominator
of \eqref{vls9} as $I_i$, $i=\overline{1,6}$.  Taking into account
\eqref{vls10}, we find
  $$
  I_1\le\max_{\overline D}\varphi_x(x,y):=C_4.
  $$
Denote   $  a_{\rm min}:=\min_{[0,T]}a(t)$. Next we have the estimate
\begin{align*}
I_2
&= \int_0^t \int_0^l G_{22}(0,y_0,t,0,\eta,\tau)(f(0,\eta,\tau)
 - \mu_{1\tau}(\eta,\tau)+\tau^{\beta}a(\tau)\mu_{1_{\eta\eta}}(\eta,\tau))
\,d\eta\,d\tau\\
&\le \max_{[0,l]\times[0,T]}(f(0,y,t)- \mu_{1_t}(y,t))
 \int_0^t\int_0^l G_{22}(0,y_0,t,0,\eta,\tau)\,d\eta\,d\tau\\
&\quad +\max_{[0,l]\times[0,T]}t^{\beta} \mu_{1_{yy}}(y,t)
 \int_0^t \int_0^l G_{22}(0,y_0,t,0,\eta,\tau)a(\tau)\,d\eta\,d\tau.
\end{align*}
Using  \eqref{vls10}, \eqref{vls11} and the known estimates  of
the Green function \cite{Imi1},
\begin{equation}\label{vls111}
G_2(0,t,0,\tau)\le\frac{C_5}{\sqrt{\theta(t)-\theta(\tau)}}+C_6,
\end{equation}
we obtain
$$
I_3\le\frac{C_7}{t^{\frac{\beta-1}{2}}\sqrt{a_{\rm min}}}+C_8.
$$
Taking into account estimates \cite{Imi1}
$$
G_2(0,t,h,\tau)\le C_7
$$
 and \eqref{vls10}, we have
$I_2\le C_9$.

To estimate $I_4$ we use  equality \eqref{vls10} and estimates
\eqref{vls11}, \eqref{vls111}:
\begin{align*}
I_4
&\le C_{10}\int_0^t G_{2}(y_0,t,0,\tau)\tau^{\beta}a(\tau)d\tau \\
&=\frac{C_{10}}{2\sqrt{\pi}}\int_0^t
\frac{1}{\sqrt{\theta(t)-\theta(\tau)}}
 \sum_{m=-\infty}^{\infty}\Big( \exp\Big(-\frac{(y_0-\eta+2ml)^2}{4(\theta(t)
-\theta(\tau))}\Big) \\
&\quad + \exp\Big(-\frac{(y_0+\eta+2ml)^2}{4(\theta(t)-\theta(\tau))}
 \Big)\Big)\tau^{\beta}a(\tau)d\tau\\
&= \frac{C_{10}}{2\sqrt{\pi}}\int_0^{\theta(t)}\frac{1}{\sqrt{z}}
 \sum_{m=-\infty}^{\infty}\Big( \exp\Big(-\frac{(y_0-\eta+2ml)^2}{4z}\Big)\\
&\quad + \exp\Big(-\frac{(y_0+\eta+2ml)^2}{4z}\Big)\Big) dz \\
&\le C_{11}\sqrt{\theta(t)}\le C_{12}.
\end{align*}
The integral $I_5$ is estimated similarly.
From \eqref{vls10}  we obtain $I_6\le C_{13}$.

The above  estimates lead us to the inequality
$$
a(t)\ge\frac{\kappa(t)}{\frac{C_{14}}{t^{\frac{\beta-1}{2}}
 \sqrt{a_{\rm min}}}+C_{15}}.
$$
Taking into account assumptions on $\kappa(t)$, we transform it to
$$
a(t)\ge\frac{\kappa_0(t)\sqrt{a_{\rm min}}}{C_{14}
+C_{15}t^{\frac{\beta-1}{2}}\sqrt{a_{\rm min}}}, \quad t\in[0,T]
$$
or
$$
\sqrt{a_{\rm min}}\ge\frac{C_{16} }{C_{14}+C_{17} \sqrt{a_{\rm min}}}.
$$
Solving this inequality we obtain an estimate for $a(t)$ from below:
\begin{equation}\label{vls12}
a(t)\ge A_0>0,\quad t\in[0,T],
\end{equation}
where $A_0$ is determined by given data. Therefore, the a priori estimates
of solutions to the equation \eqref{vls9} are established.

Now we put  equation \eqref{vls9} in the form
\begin{equation}\label{vls13}
a(t) =  Pa(t), \quad t\in[0,T].
\end{equation}
It follows from \eqref{vls11}, \eqref{vls12} that the operator $ P$
maps the set $ \mathcal N:= \{a\in C[0,T]:  A_0\leq a(t) \leq A_1 \}$
into itself. By the Arzela-Ascoli theorem and the above estimates,
it is clear that  the compactness of $P$ on $\mathcal N$ will be
established if we prove  that $\forall \varepsilon>0$ exists such
a $\delta>0$, that inequality $|Pa(t_2)-Pa(t_1)|<\varepsilon$ holds if
$|t_1-t_2|<\delta $ for all $  t_1, t_2\in[0,T]$   and for any
$a\in\mathcal N$.

Firstly,  let   show that there exists $\lim_{t\to 0}a(t)$. It follows from
the above estimates and assumptions (A2) that the limit of denominator
in \eqref{vls9} is equal to $\lim_{t\to 0} I_2(t)$.
Taking into account the equalities
\begin{equation} \label{vls14}
\begin{gathered}
\lim_{\tau\to t}\int_0^l G_{2}(y_0,t,\eta,\tau)(f(0,\eta,\tau)-\mu_{1_{\tau}}(\eta,\tau))d\eta
=f(0,y_0,t)-\mu_{1_{t}}(y_0,t) ,\\
\int_0^t\frac{d\tau}{\sqrt{t^{\beta+1}-\tau^{\beta+1}}}=t^{\frac{1-\beta}2}\int_0^1
 \frac{dz}{\sqrt{1-z^{\beta+1}}}
\end{gathered}
\end{equation}
and using the mean-value theorem, from \eqref{vls9}  we have
$$
 \lim_{t\to 0}a(t)=\sqrt{\frac{\pi}{\beta+1}}
 \frac{\kappa_0(0)\sqrt{ \lim_{t\to 0}a(t^*)}}
{(f(0,y_0,0)-\mu_{1_{t}}(y_0,0))\int_0^1 \frac{dz}{\sqrt{1-z^{\beta+1}}}},
$$
where $t^*\in[0,t]$. From this we obtain
$$
 \lim_{t\to 0}a(t)= \frac{\pi}{\beta+1}
\Big( \frac{\kappa_0(0)}{(f(0,y_0,0)-\mu_{1_{t}}(y_0,0))
\int_0^1\frac{dz}{\sqrt{1-z^{\beta+1}}}}  \Big)^2.
$$
Consider the difference
\begin{align*}
| Pa(t_1)-Pa(t_2)|
& = \big|\frac{\kappa_0(t_1)}{t_1^{\frac{\beta-1}2}u_x(0,y_0,t_1)} -
\frac{\kappa_0(t_2)}{t_2^{\frac{\beta-1}2}u_x(0,y_0,t_2)}\big|\\
&=\frac{|\kappa_0(t_1)t_2^{\frac{\beta-1}2}u_x(0,y_0,t_2)
 -\kappa_0(t_2)t_1^{\frac{\beta-1}2}u_x(0,y_0,t_1)|}
{t_1^{\frac{\beta-1}2}u_x(0,y_0,t_1)t_2^{\frac{\beta-1}2}u_x(0,y_0,t_2)}.
\end{align*}
It follows from \eqref{vls99} that
$$
t_1^{\frac{\beta-1}2}u_x(0,y_0,t_1)t_2^{\frac{\beta-1}2}u_x(0,y_0,t_2)\ge C_{18}>0.
$$
Thus, it is sufficient to estimate the difference
 $u_x(0,y_0,t_1)t_2^{\frac{\beta-1}2}-u_x(0,y_0,t_2)t_1^{\frac{\beta-1}2}$.
Consider for example the expression
\begin{align*}
\Delta:=&\big|t_2^{\frac{\beta-1}2}\int_0^{t_2}
 \int_0^l G_{22}(0,y_0,t_2,0,\eta,\tau)\mu(\eta,\tau)\,d\eta\,d\tau \\
& - t_1^{\frac{\beta-1}2}\int_0^{t_1} \int_0^l G_{22}(0,y_0,t_1,0,\eta,\tau)
 \mu(\eta,\tau)\,d\eta\,d\tau\big|,
 \end{align*}
where $\mu(y,t)$ is some continuous on $[0,l]\times[0,T]$ function.
To find the limit
$$
 \lim_{t\to 0}t^{\frac{\beta-1}2}\int_0^{t}
\int_0^l G_{22}(0,y_0,t,0,\eta,\tau)\mu(\eta,\tau)\,d\eta\,d\tau,
 $$
 we decompose the Green function into two summands
\[
G_{22}(0,y_0,t,0,\eta,\tau)=\frac1{ \pi(\theta(t)-\theta(\tau))}
\exp\Big(-\frac{(y_0-\eta)^2}{4(\theta(t)-\theta(\tau))}\Big)
  + G^0_{22}(0,y_0,t,0,\eta,\tau).
\]
It follows from the Green function estimates \eqref{vls111} that
$$
\lim_{t\to 0}t^{\frac{\beta-1}2}\int_0^{t}
\int_0^l G^0_{22}(0,y_0,t,0,\eta,\tau)\mu(\eta,\tau)\,d\eta\,d\tau=0.
 $$
On the other hand, using \eqref{vls14} we find
\begin{align*}
&\lim_{t\to 0}t^{\frac{\beta-1}2}\int_0^{t}
  \int_0^l \frac1{ \pi(\theta(t)-\theta(\tau))}
 \Big(\exp\Big(-\frac{(y_0-\eta)^2}{4(\theta(t)-\theta(\tau))}\Big) \\
&\quad + \exp\Big(-\frac{(y_0+\eta)^2}{4(\theta(t)-\theta(\tau))}\Big)\Big)
   \mu(\eta,\tau)\,d\eta\,d\tau \\
&=\sqrt{\frac{\beta+1}{\pi}}
 \frac{\mu(y_0,0)}{\sqrt{a(0)}} \int_0^1 \frac{dz}{\sqrt{1-z^{\beta+1}}}.
\end{align*}
It means that $\forall\varepsilon>0$ there exists such $t_0\in(0,T]$ that
$$
\big|t^{\frac{\beta-1}2}\int_0^{t} \int_0^l G_{22}(0,y_0,t,0,\eta,\tau)
\mu(\eta,\tau)\,d\eta\,d\tau -\sqrt{\frac{\beta+1}{\pi}}
 \frac{\mu(y_0,0)}{\sqrt{a(0)}} \int_0^1 \frac{dz}{\sqrt{1-z^{\beta+1}}}\big|
<\varepsilon,
 $$
 when $0<t<t_0$. Therefore, if $t_i< t_0, i\in\{1,2\}$,
then $\Delta<2\varepsilon$.

 Consider the case when $t_i\ge t_0, i\in\{1,2\} $ and suppose for the
 definiteness that $t_1<t_2$. Present $\Delta$ as follows:
\begin{equation} \label{vls15}
\begin{aligned}
\Delta
&=\big|t_2^{\frac{\beta-1}2}\int_{t_1}^{t_2} \int_0^l G_{22}(0,y_0,t_2,0,\eta,\tau)\mu(\eta,\tau)\,d\eta\,d\tau\\
&\quad +\int_0^{t_1} \int_0^l\bigl(t_2^{\frac{\beta-1}2}
  G_{22}(0,y_0,t_2,0,\eta,\tau)
  -t_1^{\frac{\beta-1}2} G_{22}(0,y_0,t_1,0,\eta,\tau)\bigr)\mu(\eta,\tau)
 \,d\eta\,d\tau\big|.
\end{aligned}
\end{equation}
Using  \eqref{vls111}, we obtain
 \begin{align*}
 &\big|t_2^{\frac{\beta-1}2}\int_{t_1}^{t_2}
 \int_0^l G_{22}(0,y_0,t_2,0,\eta,\tau)\mu(\eta,\tau)\,d\eta\,d\tau \big| \\
&\le t_2^{\frac{\beta-1}2}\int_{t_1}^{t_2} \int_0^l
 \Big(\frac{C_{19}}{\sqrt{\theta(t_2)-\theta(\tau)}} +C_{20}\Big)d\tau  \\
&\le C_{21}|t_2-t_1|+C_{22}t_2^{\frac{\beta-1}2}
 \int_{t_1}^{t_2}\frac{d\tau} {\sqrt{t_2^{\beta+1}  -\tau^{\beta+1}}} \\
&=C_{21}|t_2-t_1|+C_{22}\int_{\frac{t_1}{t_2}}^1\frac{dz}
 {\sqrt{1 -z^{\beta+1}}} \\
&\le C_{21}|t_2-t_1|+C_{22}\int_{\frac{t_1}{t_2}}^1\frac{dz} {\sqrt{1-z}}\\
&\le  C_{21}|t_2-t_1|+C_{23}\sqrt{t_2-t_1}.
 \end{align*}
Consider the second summand from \eqref{vls15}:
\begin{align*}
&\big|\int_0^{t_1} \int_0^l\bigl(t_2^{\frac{\beta-1}2}
 G_{22}(0,y_0,t_2,0,\eta,\tau)
  -t_1^{\frac{\beta-1}2}   G_{22}(0,y_0,t_1,0,\eta,\tau)\bigr)
 \mu(\eta,\tau)\,d\eta\,d\tau\big| \\
&\le \big|t_2^{\frac{\beta-1}2}-t_1^{\frac{\beta-1}2}\big|\,
\big|\int_0^{t_1}\int_0^l G_{22}(0,y_0,t_2,0,\eta,\tau)
 \mu(\eta,\tau)\,d\eta\,d\tau\big| \\
&\quad +t_1^{\frac{\beta-1}2}\int_0^{t_1}
 \int_0^l|G_{22}(0,y_0,t_2,0,\eta,\tau)
 -G_{22}(0,y_0,t_1,0,\eta,\tau)||\mu(\eta,\tau)|\,d\eta\,d\tau\\
&:=\Delta_1+\Delta_2.
  \end{align*}
Substituting the product of two one-dimensional Green functions instead
of the Green function $G_{22}(0,y_0,t_2,0,\eta,\tau)$  and taking into
account \eqref{vls10}, \eqref{vls111}, we find
\begin{align*}
\Delta_1
&\le C_{24}\big|t_2^{\frac{\beta-1}2}-t_1^{\frac{\beta-1}2}\big|
 \int_0^{t_1}\int_0^l G_{2}(0,t_2,0,\tau)G_{2}(y_0,t_2,\eta,\tau)
 \,d\eta\,d\tau \\
&\le C_{24}\big|t_2^{\frac{\beta-1}2}-t_1^{\frac{\beta-1}2}\big|
 \int_0^{t_1}G_{2}(0,t_2,0,\tau)d\tau \\
&\le C_{25}\big|t_2^{\frac{\beta-1}2}-t_1^{\frac{\beta-1}2}\big|
t_1^{\frac{1-\beta}2}
=C_{25}\big|\big(\frac{t_2}{t_1}\big)^{\frac{\beta-1}2}-1\big|<\epsilon\,,
\end{align*}
 when $ |t_2-t_1|<\delta_1$. By the same reasoning we obtain
\begin{align*}
\Delta_2
&\le C_{24}t_1^{\frac{\beta-1}2}
\Big(\int_0^{t_1}\int_0^l |G_{2}(0,t_2,0,\tau)-G_{2}(0,t_1,0,\tau)|
 G_{2}(y_0,t_2,\eta,\tau) \,d\eta\,d\tau\\
&\quad +\int_0^{t_1}\int_0^lG_{2}(0,t_1,0,\tau)
 |G_{2}(y_0,t_2,\eta,\tau)-G_{2}(y_0,t_1,\eta,\tau)|\,d\eta\,d\tau\Big)\\
&\le C_{24} t_1^{\frac{\beta-1}2} \Big(\int_0^{t_1}|G_{2}(0,t_2,0,\tau)
 -G_{2}(0,t_1,0,\tau)|d\tau   \\
&\quad + \int_0^{t_1}\int_0^l G_{2}(0,t_1,0,\tau)
  |G_{2}(y_0,t_2,\eta,\tau)-G_{2}(y_0,t_1,\eta,\tau)|\,d\eta\,d\tau\Big)\\
&:=\Delta_{2,1}+\Delta_{2,2}\,.
 \end{align*}
Now we put the Green function  $G_{2}(0,t,0,\tau) $ in the form
 $$
G_{2}(0,t,0,\tau)=\frac1{\sqrt{\pi(\theta(t)-\theta(\tau))}}
 +\frac2{\sqrt{\pi(\theta(t)-\theta(\tau))}}\sum_{n=1}^{\infty}
\exp\Big(-\frac{n^2h^2}{\theta(t)-\theta(\tau)}\Big).
$$
Then
\begin{equation} \label{vls16}
\begin{aligned}
\Delta_{2,1}
&\le C_{24}t_1^{\frac{\beta-1}2}\Big(\frac1{\sqrt{\pi}}
 \int_0^{t_1}\Big(\frac1{\sqrt{\theta(t_1)-\theta(\tau)}}-
\frac1{\sqrt{\theta(t_2)-\theta(\tau)}}\Big)d\tau\\
&\quad +\int_0^{t_1}\big|\sum_{n=1}^{\infty}\Big(
 \frac2{\sqrt{\pi(\theta(t_1)-\theta(\tau))}}
  \exp\Big(-\frac{n^2h^2}{\theta(t_1)-\theta(\tau)}\Big)\\
&\quad -\frac2{\sqrt{\pi(\theta(t_2)-\theta(\tau))}}
 \exp\Big(-\frac{n^2h^2}{\theta(t_2)-\theta(\tau)}\Big)\Big)\big|d\tau\Big).
 \end{aligned}
\end{equation}
Now we transform and estimate the first summand:
\begin{align*}
&C_{26}t_1^{\frac{\beta-1}2} \int_0^{t_1}
 \Big(\frac1{\sqrt{\theta(t_1)-\theta(\tau)}}-
\frac1{\sqrt{\theta(t_2)-\theta(\tau)}}\Big)d\tau \\
&=C_{26}t_1^{\frac{\beta-1}2} \int_0^{t_1}\frac{\theta(t_2)-
 \theta(t_1)}{\sqrt{(\theta(t_1)-\theta(\tau))}}
 \frac{d\tau}{(\theta(t_2)-\theta(\tau))(\sqrt{\theta(t_1)
 -\theta(\tau)}+\sqrt{\theta(t_2)-\theta(\tau)}}  \\
&\le C_{27}t_1^{\frac{\beta-1}2}
 \int_0^{t_1}\frac{(t_2^{\beta+1}-t_1^{\beta+1})d\tau}
 {\big(\sqrt{t_1^{\beta+1}-\tau^{\beta+1}}+\sqrt{t_2^{\beta+1}-
\tau^{\beta+1}}\big) \sqrt{\bigl(t_1^{\beta+1}-\tau^{\beta+1}\bigr)
 \bigl(t_2^{\beta+1}-\tau^{\beta+1}\bigr)}}  \\
&=C_{27}t_1^{\frac{\beta-1}2} \int_0^{t_1}
 \Big(\frac1{\sqrt{t_1^{\beta+1}-\tau^{\beta+1}}}
 -\frac1{\sqrt{t_2^{\beta+1}-\tau^{\beta+1}}}\Big)d\tau.
\end{align*}
After the change of variable $z=\tau/t_1$ we obtain
\begin{equation} \label{vls17}
\begin{aligned}
&C_{26}t_1^{\frac{\beta-1}2} \int_0^{t_1}
 \Big(\frac1{\sqrt{\theta(t_1)-\theta(\tau)}}-
\frac1{\sqrt{\theta(t_2)-\theta(\tau)}}\Big)d\tau \\
&\le C_{27} \int_0^1\Big(\frac1{\sqrt{1-z^{\beta+1}}}\\
&-\frac1{\sqrt{(\frac{t_2}{t_1})^{\beta+1}-z^{\beta+1}}}\Big)dz<\epsilon,
  \quad \text{when }  |\frac{t_2}{t_1}-1|<\delta_2,
 \end{aligned}
\end{equation}
 as the function
$$
I(\sigma):=\int_0^1\frac{dz}{\sqrt{\sigma-z^{\beta+1}}}
$$
is continuous for $\sigma\ge 1$.

Now we estimate the second summand from \eqref{vls16}:
\begin{align*}
&C_{26}t_1^{\frac{\beta-1}2}\int_0^{t_1}
 \big|\sum_{n=1}^{\infty}\Big(\frac2{\sqrt{\pi(\theta(t_1)-
\theta(\tau))}}\exp\Big(-\frac{n^2h^2}{\theta(t_1)-\theta(\tau)}\Big)\\
&-\frac2{\sqrt{\pi(\theta(t_2)-\theta(\tau))}}
 \exp\Big(-\frac{n^2h^2}{\theta(t_2)-\theta(\tau)} \Big)\Big)\big|d\tau \\
&=\frac{2C_{26}}{\sqrt{\pi}}t_1^{\frac{\beta-1}2}\int_0^{t_1}\big|
\int_{\theta(t_2)-\theta(\tau)}^{\theta(t_1)-\theta(\tau)}
 \frac{\partial}{\partial z} \Big(\frac1{\sqrt{z}}\sum_{n=1}^{\infty}
 \exp\Big(-\frac{n^2h^2}{z}\Big)  dz\big|d\tau \\
&\le C_{28}(\theta(t_2)-\theta(t_1))
 \le C_{29}(t_2^{\beta+1}-t_1^{\beta+1})<\epsilon,
 \quad\text{when }  |t_2-t_1|<\delta_3.
\end{align*}

To estimate $\Delta_{2,2}$, we extract the main term from the Green function,
\begin{equation} \label{vls18}
\begin{aligned}
\Delta_{2,2}
&\le C_{24}t_1^{\frac{\beta-1}2}\Big(\int_0^{t_1}G_{2}(0,t_1,0,\tau)d\tau\\
&\quad\times \int_0^l\big|\frac1{\sqrt{\theta(t_2)-\theta(\tau)}}
\exp\Big(-\frac{(\eta-y_0)^2}{4(\theta(t_2)-\theta(\tau))}\Big)\\
&\quad-\frac1{\sqrt{\theta(t_1)-\theta(\tau)}}
 \exp\Big({-\frac{(\eta-y_0)^2}{4(\theta(t_1)-\theta(\tau))}}\Big)\big|d\eta\\
&\quad +\int_0^{t_1}G_{2}(0,t_1,0,\tau)d\tau\int_0^l |G_{2}^0(y_0,t_2,\eta,\tau)
 -G_{2}^0(y_0,t_1,\eta,\tau)|d\eta\Big).
\end{aligned}
\end{equation}
Taking into account  that $G^0_{2}(y_0,t,\eta,\tau)$   has no singularity, we find
\begin{align*}
\int_0^l |G_{2}^0(y_0,t_2,\eta,\tau)-G_{2}^0(y_0,t_1,\eta,\tau)|d\eta
&\le   \int_0^l\big|\int_{t_1}^{t_2}G^0_{2_t}(y_0,t, \eta,\tau)dt\big| d\eta \\
&\le C_{30}|t_1-t_2|.
\end{align*}
We consider the  expression
\begin{equation} \label{vls19}
\begin{aligned}
&\int_0^l\big|\frac1{\sqrt{\theta(t_2)-\theta(\tau)}}
 \exp\Big(-\frac{(\eta-y_0)^2}{4(\theta(t_2)-\theta(\tau))} \Big) \\
& -\frac1{\sqrt{\theta(t_1)-\theta(\tau)}}
  \exp\Big(-\frac{(\eta-y_0)^2}{4(\theta(t_1)-\theta(\tau))} \Big)\big|d\eta \\
&\le \int_0^l\frac1{\sqrt{\theta(t_2)-\theta(\tau)}}
 \Big( \exp\Big(-\frac{(\eta-y_0)^2}{4(\theta(t_2)-\theta(\tau))} \Big)\\
&\quad -\exp\Big(-\frac{(\eta-y_0)^2}{4(\theta(t_1)-\theta(\tau))} \Big)\Big) d\eta\\
&\quad + \int_0^l \exp\Big(-\frac{(\eta-y_0)^2}{4(\theta(t_1)-\theta(\tau))} \Big) 
\Big(\frac1{\sqrt{\theta(t_1)-\theta(\tau)}}
 -\frac1{\sqrt{\theta(t_2)-\theta(\tau)}}\Big) d\eta  \,.
\end{aligned}
\end{equation}
After the change of variable $z=\frac{\eta-y_0}{2\sqrt{\theta(t_2)-\theta(\tau)}}$
in the first summand we have
\begin{align*}
&\int_0^l\big|\frac1{\sqrt{\theta(t_2)-\theta(\tau)}}
 \exp\Big(-\frac{(\eta-y_0)^2}{4(\theta(t_2)-\theta(\tau))} \Big)\\
&-\frac1{\sqrt{\theta(t_1)-\theta(\tau)}}
 \exp\Big(-\frac{(\eta-y_0)^2}{4(\theta(t_1)-\theta(\tau))} \Big)\big|d\eta\\
&=2\int_{\frac{-y_0}{2\sqrt{\theta(t_2)
 -\theta(\tau)}}}^{\frac{l-y_0}{2\sqrt{\theta(t_2)-\theta(\tau)}}}
\Big(\exp(-z^2)
 -\exp\Big(-z^2\frac{\theta(t_2)-\theta(\tau)}{\theta(t_1)-\theta(\tau)}\Big)\Big)dz\\
&\le 2\int_{-\infty}^{\infty}\Big(\exp(-z^2)
-\exp\Big(-z^2\frac{\theta(t_2)-\theta(\tau)}{\theta(t_1)-\theta(\tau)}\Big)\Big)dz\\
& =2\sqrt{\pi}\Big(1-\frac{\sqrt{\theta(t_1)-\theta(\tau)}}
 {\sqrt{\theta(t_2)-\theta(\tau)}}   \Big).
\end{align*}
The change of variable $z=\frac{\eta-y_0}{2\sqrt{\theta(t_1)-\theta(\tau)}}$
in the second summand gives
\begin{align*}
&\int_0^l \exp\Big(-\frac{(\eta-y_0)^2}{4(\theta(t_1)-\theta(\tau))}\Big)
 \Big(\frac1{\sqrt{\theta(t_1)-\theta(\tau)}}-
\frac1{\sqrt{\theta(t_2)-\theta(\tau)}}\Big) d\eta\\
&=2\Big(1-\frac{\sqrt{\theta(t_1)-\theta(\tau)}}
 {\sqrt{\theta(t_2)-\theta(\tau)}}\Big)
 \int_{\frac{-y_0}{2\sqrt{\theta(t_2)
-\theta(\tau)}}}^{\frac{l-y_0}{2\sqrt{\theta(t_2)-\theta(\tau)}}}e^{-z^2}dz \\
&\le 2\sqrt{\pi}\Big(1-\frac{\sqrt{\theta(t_1)
 -\theta(\tau)}}{\sqrt{\theta(t_2)-\theta(\tau)}}\Big).
\end{align*}

Now we return to the estimate of the first summand in \eqref{vls18},
 using \eqref{vls111} and \eqref{vls17},
\begin{align*}
&C_{24}t_1^{\frac{\beta-1}2}\int_0^{t_1}G_{2}(0,t_1,0,\tau)d\tau
\int_0^l\big|\frac1{\sqrt{\theta(t_2)-\theta(\tau)}}
\exp\Big(-\frac{(\eta-y_0)^2}{4(\theta(t_2)-\theta(\tau))} \Big)\\
&-\frac1{\sqrt{\theta(t_1)-\theta(\tau)}}\exp
 \Big(-\frac{(\eta-y_0)^2}{4(\theta(t_1)-\theta(\tau))} \Big)\big|d\eta \\
&\le C_{31}t_1^{\frac{\beta-1}2}\int_0^{t_1}
 \Big(\frac1{\sqrt{\theta(t_1)-\theta(\tau)}}
 -\frac1{\sqrt{\theta(t_2)-\theta(\tau)}}\Big)d\tau \\
&\quad +C_{32}t_1^{\frac{\beta-1}2}\int_0^{t_1}\Big(1-
\frac{\sqrt{\theta(t_1)-\theta(\tau)}}{\sqrt{\theta(t_2)-\theta(\tau)}}\Big)d\tau
<\epsilon,
\end{align*}
when $|\frac{t_2}{t_1}-1|<\delta_4$.

So, we  obtained the estimate of $\Delta$. Other expressions from 
$Pa(t)$ are estimated by a similar way. Therefore, the conditions of 
Schauder fixed-point theorem are satisfied for \eqref{vls13}, and the 
existence of the solution to the problem \eqref{vls1}--\eqref{vls4} is proved.

\section{Uniqueness of the solution}


Consider the function
$$
H(t):=\frac{\sqrt{\pi}\kappa(t)}{\sqrt{\beta+1}
 \int_0^t\frac{\mu_0(\tau)d\tau}{\sqrt{t^{\beta+1}-
\tau^{\beta+1}}}},
$$
where
\begin{equation}\label{vls51}
\mu_0(t):=\int_0^l G_2(y_0,t,\eta,\tau)(f(0,\eta,\tau)
-\mu_{1_{\tau}}(\eta,\tau))d\eta.
\end{equation}
It is easy to see that there exists the limit
$$
\lim_{t\to+0}H(t)=\frac{\sqrt{\pi}\kappa_0(0)}{\sqrt{\beta+1}\mu_0(0)\int_0^t
\frac{d\sigma}{\sqrt{1-\sigma^{\beta+1}}}}>0.
$$
It means that $H(t)$ is continuous function on $[0,T]$.

Using the function $H(t)$, we will establish the estimates of solutions 
for the equation \eqref{vls9}. Taking into account assumption  (A2),
 from  \eqref{vls9}  we have
\begin{align*}
a(t) 
&\le\frac{\kappa(t)}{\int_0^t \int_0^l G_{22}(0,y_0,t,0,\eta,\tau)
 (f(0,\eta,\tau)-\mu_{1_{\tau}}(\eta,\tau))\,d\eta\,d\tau}\\
 &=\frac{\kappa(t)}{\int_0^tG_{2}(0,t,0,\tau)\mu_0(\tau)d\tau}
 \le\frac{\kappa(t)\sqrt{\pi\tilde a_{\rm max}(t)}}{\sqrt{\beta+1}
 \int_0^t  \frac{\mu_0(\tau)d\tau}{\sqrt{t^{\beta+1}-\tau^{\beta+1}}}}\\
&\le \sqrt{\tilde a_{\rm max}(t)}H_{\rm max}(t),
 \end{align*}
where $\tilde a_{\rm max}(t):=\max_{[0,t]}a(\tau)$, 
$H_{\rm max}(t):=\max_{[0,t]}H(\tau)$.
From this follows that
\begin{equation}\label{vls20}
a(t)\le H_{\rm max}^2(t), \quad t\in[0,T].
\end{equation}


To estimate $a(t)$ from below, we put the denominator from \eqref{vls9} as a sum
$\frac1{\sqrt{\pi}}\int_0^t  \frac{\mu_0(\tau)d\tau}{\sqrt{\theta(t)
-\theta(\tau)}}+S(t)$. Using the notation
$\tilde a_{\min}(t):=\min_{[0,t]}a(\tau)$, $ H_{\min}(t):=\min_{[0,t]}H(\tau)$  
we obtain
$$
\frac1{\sqrt{\pi}}\int_0^t  \frac{\mu_0(\tau)d\tau}{\sqrt{\theta(t)-\theta(\tau)}}
\le \frac{\sqrt{\beta+1}}{\sqrt{\pi\tilde a_{\min}(t)}}
 \int_0^t  \frac{\mu_0(\tau)d\tau}{\sqrt{t^{\beta+1}-\tau^{\beta+1}}}
=\frac{\kappa(t)}{H(t)\sqrt{\tilde a_{\min}(t)}}.
$$
Repeating the reasons of  the estimation of  the denominator of \eqref{vls9} 
from   above, we establish that $ S(t)\le C_{33}$. Then
$$
a(t)\ge\frac{\kappa(t)}{\frac{\kappa(t)}{H(t)\sqrt{\tilde a_{\min}(t)}}+C_{33}}
= \frac{H(t)\sqrt{\tilde a_{\min}(t)}}{1+\frac{C_{33}H(t)
\sqrt{\tilde a_{\min}(t)}}{\kappa(t)}}.
$$
It follows from the assumptions (A2)  and the estimate \eqref{vls20} that
 $$
\frac{C_{33}H(t)\sqrt{\tilde a_{\min}(t)}}{\kappa(t)}
\le {C_{34}t^{\frac{\beta-1}2}}.
$$
Therefore,
 $$
a(t)\ge\frac{H_{\min}(t)\sqrt{\tilde a_{\min}(t)}}
{1+ C_{34}t^{\frac{\beta-1}2}},\quad t\in[0,T],
$$
or
\begin{equation}\label{vls21}
a(t)\ge\frac{H_{\min}^2(t)}{(1+ C_{34}t^{\frac{\beta-1}2})^2},\quad t\in[0,T].
\end{equation}

Having the estimates \eqref{vls20} and \eqref{vls21} of solutions of 
 equation \eqref{vls9}, we can pass to the proof of the uniqueness of solution. 
Put the equation \eqref{vls9} in the form
\begin{equation}\label{vls22}
a(t)=\frac{\kappa(t)}{\frac1{\sqrt{\pi}}\int_0^t  \frac{\mu_0(\tau)d\tau}{\sqrt{\theta(t)-\theta(\tau)}}+S(t,a(t))},\quad t\in[0,T].
\end{equation}
Suppose that this equation has two solutions $ a_1(t)$ and $ a_2(t)$. 
Denote $b(t):=a_1(t)-a_2(t)$. The function $b(t)$ satisfies the equation
\begin{equation} \label{vls23}
\begin{aligned}
b(t) &=\kappa(t)\Big(\frac1{\sqrt{\pi}}\int_0^t
 \Big(\frac1{\sqrt{\theta_2(t)-\theta_2(\tau)}}-
  \frac1{\sqrt{\theta_1(t)-\theta_1(\tau)}}\Big)\mu_0(\tau) d\tau+S(t,a_2(t)) \\
&\quad -S(t,a_1(t))\Big)
 \Big(\frac1{\sqrt{\pi}}\int_0^t  \frac{\mu_0(\tau)d\tau}
 {\sqrt{\theta_1(t)-\theta_1(\tau)}}+S(t,a_1(t))\Big)^{-1} \\
&\quad\times  \Big(\frac1{\sqrt{\pi}}\int_0^t  \frac{\mu_0(\tau)d\tau}
 {\sqrt{\theta_2(t)-\theta_2(\tau)}}+S(t,a_2(t))\Big)^{-1}.
\end{aligned}
\end{equation}
 It is clear from above that $ S(t,a_i(t))\ge 0, i\in\{1,2\}$.
Then it follows from \eqref{vls23} that
\begin{equation} \label{vls24}
\begin{aligned}
&|b(t)|\\
&\le\kappa(t)\Big(\frac1{\sqrt{\pi}}\int_0^t\big|\frac1{\sqrt{\theta_2(t)
 -\theta_2(\tau)}}-
  \frac1{\sqrt{\theta_1(t)-\theta_1(\tau)}}\big|\mu_0(\tau) d\tau+|S(t,a_2(t))\\
&\quad-S(t,a_1(t))|\Big)
 \Big(\frac1{\pi}\int_0^t  \frac{\mu_0(\tau)d\tau}{\sqrt{\theta_1(t)-\theta_1(\tau)}}
  \int_0^t  \frac{\mu_0(\tau)d\tau}{\sqrt{\theta_2(t)-\theta_2(\tau)}} \Big)^{-1}.
\end{aligned}
\end{equation}
Transform the expression
\begin{align*}
&\frac1{\sqrt{\theta_2(t)-\theta_2(\tau)}}-\frac1{\sqrt{\theta_1(t)
 -\theta_1(\tau)}} \\
&= \frac{\theta_2(t)-\theta_2(\tau)-(\theta_1(t)-\theta_1(\tau))}
 {\sqrt{(\theta_1(t)-\theta_1(\tau))(\theta_2(t)-\theta_2(\tau))}}
\frac1{\sqrt{\theta_1(t)-\theta_1(\tau)}+\sqrt{\theta_2(t)-\theta_2(\tau)}}.
\end{align*}
Using  estimate \eqref{vls21}, we find
$$
\big|\frac1{\sqrt{\theta_2(t)-\theta_2(\tau)}}
-\frac1{\sqrt{\theta_1(t)-\theta_1(\tau)}}\big|\le\frac{\sqrt{\beta+1}
(1+ C_{34}t^{\frac{\beta-1}2})^3}{2H_{\min}^3(t)
\sqrt{t^{\beta+1}-\tau^{\beta+1}}}b_{\rm max}(t),
$$
where $b_{\rm max}(t):=\max_{[0,t]}|b(\tau)|$. Therefore, we obtain
\begin{equation}
\begin{aligned}\label{vls25}
&\frac{\kappa(t)}{\sqrt{\pi}}
 \int_0^t\big|\frac1{\sqrt{\theta_2(t)-\theta_2(\tau)}}-
\frac1{\sqrt{\theta_1(t)-\theta_1(\tau)}}\big|\mu_0(\tau) d\tau\\
&\le\frac{\kappa^2(t)(1+ C_{34}t^{\frac{\beta-1}2})^3}{2H^4_{\min}(t)}
b_{\rm max}(t).
\end{aligned}
\end{equation}
Now we estimate  the difference $  S(t,a_2(t))-S(t,a_1(t))$. Consider
\begin{align*}
\Delta_1R:=
&\frac{1}{2\pi}\big|\int_0^h \int_0^l\Big( \frac{1}{\theta_2(t)}
 \sum_{m,n=-\infty}^{\infty} \exp\Big(-\frac{(\xi+2nh)^2}{4\theta_2(t)}\Big)\\
&\times\Big( \exp\Big(-\frac{(y_0-\eta+2ml)^2}{4\theta_2(t)}\Big)
+   \exp\Big(-\frac{(y_0+\eta+2ml)^2}{4\theta_2(t)}\Big) \Big)\\
&- \frac{1}{\theta_1(t)}\sum_{m,n=-\infty}^{\infty}
\exp\Big(-\frac{(\xi+2nh)^2}{4\theta_1(t)}\Big)
 \Big( \exp\Big(-\frac{(y_0-\eta+2ml)^2}{4\theta_1(t)}\Big)\\
&+  \exp\Big(-\frac{(y_0+\eta+2ml)^2}{4\theta_1(t)}\Big)\Big)\Big)
 \varphi_{\xi}(\xi,\eta)d\eta d\xi \big| \\
\le& C_{35} \int_0^h \int_0^l\Big|\int_{\theta_1(t)}^{\theta_2(t)}
\frac{\partial}{\partial z}\Big( \frac{1}{z}\sum_{m,n=-\infty}^{\infty}
 \exp\Big(-\frac{(\xi+2nh)^2}{4z}\Big) \\
&\quad\times \Big( \exp\Big(-\frac{(y_0-\eta+2ml)^2}{4z}\Big) 
 +   \exp\Big(-\frac{(y_0+\eta+2ml)^2}{4z}\Big) \Big)\Big)dz\Big|d\eta d\xi.
  \end{align*}
After differentiating   and using \eqref{vls10} we obtain
\begin{align*}
\Delta_1R
&\le C_{36}\Big|\int_{\theta_2(t)}^{\theta_1(t)} dz \int_0^h \int_0^l
\Big(\frac1{z^2}\sum_{m,n=-\infty}^{\infty}\exp\Big(-\frac{(\xi+2nh)^2}{4z}\Big)\\
&\quad\times \Big( \exp\Big(-\frac{(y_0-\eta+2ml)^2}{4z}\Big) 
 +   \exp\Big(-\frac{(y_0+\eta+2ml)^2}{4z}\Big) \Big) \\
&\quad +\frac1{z^3}\sum_{m,n=-\infty}^{\infty}(\xi+2nh)^2
 \exp\Big(-\frac{(\xi+2nh)^2}{4z}\Big) \\
&\quad\times \Big(  (y_0-\eta+2ml)^2  \exp\Big(-\frac{(y_0-\eta+2ml)^2}{4z}\Big)
 +(y_0+\eta+2ml)^2 \\
&\quad \times\exp\Big(-\frac{(y_0+\eta+2ml)^2}{4z}\Big)
\Big)\Big)d\eta d\xi\Big| \\
&\le C_{37}\big|\int_{\theta_2(t)}^{\theta_1(t)}\frac{dz}{z} \big|
\le C_{37}\frac{|\theta_1(t)-\theta_2(t)|}{\min\{\theta_1(t), \theta_2(t)\}}.
\end{align*}
Applying \eqref{vls21}, we arrive at the estimate
$$
\Delta_1R\le C_{38}\Big(\frac{(1+ C_{34}t^{\frac{\beta-1}2})^2}{t^{\beta-1}H_{\min}^2(t)}+1\Big)t^{\beta-1}b_{\rm max}(t)\le C_{39} b_{\rm max}(t).
$$

Now we estimate the expression
\begin{align*}
\Delta_2R
:=&\frac1{\sqrt{\pi}}\big|\int_0^t\sum_{n=1}^{\infty}
 \Big(\frac1{\sqrt{\theta_2(t)-\theta_2(\tau)}}
\exp\Big(-\frac{n^2h^2}{\theta_2(t)-\theta_2(\tau)}\Big)
 -\frac1{\sqrt{\theta_1(t)-\theta_1(\tau)}} \\
&\times\exp\Big(-\frac{n^2h^2}{\theta_1(t)-\theta_1(\tau)}\Big)\Big)
 \mu_0(\tau) d\tau\big| \\
&\le C_{40}\int_0^td\tau\big|\int_{\theta_1(t)-\theta_1(\tau)}^{\theta_2(t)
 -\theta_2(\tau)}\big|\frac{\partial}{\partial z}
 \Big(\frac1{z}\sum_{n=1}^{\infty}
 \exp\Big(-\frac{n^2h^2}{z}\Big)\big| dz\big| \\
&\le C_{41}|\theta_1(t)-\theta_2(t)|\le C_{43}b_{\rm max}(t).
\end{align*}

Other summands from $  S(t,a_2(t))-S(t,a_1(t))$ are estimated in a similar way.
 Hence,
\begin{equation}\label{vls26}
|S(t,a_1(t))-S(t,a_2(t))|\le C_{42}b_{\rm max}(t).
\end{equation}

To estimate the denominator in \eqref{vls24}, note that
\begin{equation}\label{vls27}
\frac{\kappa(t)}{\frac1{\sqrt{\pi}}\int_0^t  \frac{\mu_0(\tau)d\tau}
{\sqrt{\theta_i(t)-\theta_i(\tau)}}}
\le
\frac{\kappa(t)\sqrt{a_{\rm max}(t)}}{\frac{\sqrt{\beta+1}}{\sqrt{\pi}}
 \int_0^t  \frac{\mu_0(\tau)d\tau}
{\sqrt{ t^{\beta+1}-  \tau^{\beta+1}}}}
\le H_{\rm max}^2(t),
\end{equation}
for $i\in\{1,2\}$.

By applying estimates \eqref{vls25}--\eqref{vls27} to \eqref{vls24}, we  obtain
\begin{equation} \label{vls28}
\begin{aligned}
|b(t)|
&\le \Big(\frac{(1+C_{34}t^{\frac{\beta-1}2})^3}{2H^4_{\min}(t)}
 +C_{43}t^{\frac{\beta-1}2}\Big)H^4_{\rm max}(t)
b_{\rm max}(t)\\
&\le\Big(\frac{(1+C_{34}t^{\frac{\beta-1}2})^3 H^4_{\rm max}(t)}{2H^4_{\min}(t)}
 +C_{42}t^{\frac{\beta-1}2}H^4_{\rm max}(t)\Big) b_{\rm max}(t).
\end{aligned}
\end{equation}
As $\lim_{t\to 0} H_{\min}(t)=\lim_{t\to 0} H_{\rm max}(t)$,
 there exists such a number $t_0\in(0,T]$ that the following inequality holds:
\begin{align}\label{vls29}
&\Big(\frac{(1+C_{34}t^{\frac{\beta-1}2})^3 H^4_{\rm max}(t)}{2H^4_{\min}(t)}
+C_{42}t^{\frac{\beta-1}2}H^4_{\rm max}(t)\Big)<1,\quad t\in[0,t_0].
\end{align}
Then we conclude from \eqref{vls28} that $b_{\rm max}(t)\le 0$, $t\in[0,t_0]$,
 that is impossible. Therefore, $b(t)\equiv 0, t\in[0,t_0]$.
It follows from the uniqueness of solution to   problem \eqref{vls1}--\eqref{vls5}
that the function $u(x,y,t)$ is uniquely defined in $\overline Q_{t_0}$.

Now let us show that the solution to \eqref{vls1}--\eqref{vls5} is unique in 
the whole domain $\overline Q_{T}$.  Suppose that there exist two solutions
 $(a_i(t), u_i(x,y,t))$, $i\in\{1,2\}$ to the problem. For their difference 
$a:=a_1-a_2, u:=u_1-u_2$ we obtain the problem
\begin{gather}\label{vls30}
u_t = t^{\beta}a_1(t)\Delta u + t^{\beta}a(t)\Delta u_2(x,y,t),\quad (x,y,t)\in Q_T,\\
 \label{vls31}
u(x,y,0) = 0,\quad (x,y)\in\overline D,\\ 
\label{vls32}
u(0,y,t) = u(h,y,t) =0,\quad(y,t)\in[0,l]\times[0,T],  \\ 
\label{vls33}
 u_y(x,0,t) = u_y(x,l,t) = 0,\quad (x,t)\in[0,h]\times[0,T],\\ 
\label{vls133}
 a_1(t)u_x(0,y_0,t) = -a(t)u_{2x}(0,y_0,t),\quad t\in [0,T].
\end{gather}
Using the Green function $\tilde G_{12}(x,y,t,\xi,\eta,\tau)$ of the problem 
 \eqref{vls30}--\eqref{vls33}, we  find that
 \begin{equation}\label{vls34}
u(x,y,t) = \int_0^t \int_D  \tilde G_{12}(x,y,t,\xi,\eta,\tau)
\tau^{\beta}a(\tau)\Delta u_2(\xi,\eta,\tau) \,d\xi\,d\eta\,d\tau.
 \end{equation}
Substituting it into the overdetermination condition \eqref{vls133}:
\begin{equation}\label{vls35}
a(t) = -\frac{a_1(t)}{u_{2x}(0,y_0,t)}\int_0^t \int_0^l
\int_0^h \tilde G_{12_x}(0,y_0,t,\xi,\eta,\tau)
\tau^{\beta}a(\tau)\Delta u_2(\xi,\eta,\tau) \,d\xi\,d\eta\,d\tau,
\end{equation}
for $t \in [0,T]$.
Thus, we obtain a homogeneous Volterra integral equation of the second kind
for $a(t)$. Put it as
\begin{equation}\label{vls36}
a(t)=\int_0^t K(t,\tau)a(\tau) d\tau, \quad t\in[0,T].
\end{equation}
To study the behavior of the  kernel $K(t,\tau)$,  we find
\begin{equation} \label{vls37}
\begin{aligned}
u_2(x,y,t) 
&= \int_0^l \int_0^h\hat G_{12}(x,y,t,\xi,\eta,0)\varphi(\xi,\eta)d\xi d\eta \\
&\quad + \int_0^t \int_0^l \hat G_{12_{\xi}}(x,y,t,0,\eta,\tau)
 \tau^{\beta}a(\tau)\mu_1(\eta,\tau)\,d\eta\,d\tau \\
&\quad - \int_0^t \int_0^l\hat G_{12_{\xi}}(x,y,t,h,\eta,\tau) \tau^{\beta}
  a(\tau) \mu_2(\eta,\tau) \,d\eta\,d\tau  \\
&\quad - \int_0^t \int_0^h\hat G_{12}(x,y,t,\xi,0,\tau)
 \tau^{\beta}a(\tau)\nu_1(\xi,\tau)d\xi d\tau \\
&\quad + \int_0^t \int_0^h\hat G_{12}(x,y,t,\xi,l,\tau)
\tau^{\beta}a(\tau)\nu_2(\xi,\tau)d\xi d\tau \\
&\quad + \int_0^t \int_0^l \int_0^h \hat G_{12}(x,y,t,\xi,\eta,\tau)
 f(\xi,\eta,\tau)\,d\xi\,d\eta\,d\tau,
\end{aligned}
\end{equation}
where $\hat G_{12}(x,y,t,\xi,\eta,\tau)$ is the Green function of
 problem \eqref{vls2}--\eqref{vls4} for the equation
$$
u_t = t^{\beta}a_2(t)\Delta u + f(x,y,t).
$$
By differentiating \eqref{vls37} twice by  $x, y$ and integrating by parts,
we obtain

\begin{align*}
&u_{2_{xx}}(x,y,t) \\
&= \int_0^l \int_0^h\hat G_{12}(x,y,t,\xi,\eta,0)\varphi_{\xi\xi}(\xi,\eta)d\xi d\eta\\ 
&\quad + \int_0^t \int_0^l\hat G_{12_{\xi}} (x,y,t,0,\eta,\tau)
(\mu_{1_{\tau}}(\eta,\tau)- \tau^{\beta}a_2(\tau)\mu_{1_{\eta\eta}}(\eta,\tau) 
- f(0,\eta,\tau))\,d\eta\,d\tau  \\
&\quad - \int_0^t \int_0^l\hat G_{12_{\xi}} (x,y,t,h,\eta,\tau) 
 (\mu_{2_{\tau}}(\eta,\tau)-\tau^{\beta}a_2(\tau)\mu_{2_{\eta\eta}}(\eta,\tau) 
 - f(h,\eta,\tau))\,d\eta\,d\tau\\
&\quad  -\int_0^t \int_0^h\hat G_{12} (x,y,t,\xi,0,\tau)t^{\beta}  
  a_2(\tau)\nu_{1_{\xi\xi}}(\xi,\tau)d\xi d\tau \\
&\quad + \int_0^t \int_0^h\hat G_{12}(x,y,t,\xi,l,\tau)\tau^{\beta}
 a_2(\tau)\nu_{2_{\xi\xi}}(\xi,\tau)d\xi d\tau \\
&\quad + \int_0^t \int_0^l \int_0^h\hat G_{12} (x,y,t,\xi,\eta,\tau)
 f_{\xi\xi}(\xi,\eta,)\tau)\,d\xi\,d\eta\,d\tau,
\end{align*}
\begin{align*}
&u_{2yy}(x,y,t) \\
&= \int_0^l \int_0^h\hat G_{12} (x,y,t,\xi,\eta,0)\varphi_{\eta\eta}(\xi,\eta)d\xi
  d\eta \\
&\quad +\int_0^t \int_0^l\hat G_{12_{\xi}} (x,y,t,0,\eta,\tau) 
 \tau^{\beta}a_2(\tau)\mu_{1_{\eta\eta}}(\eta,\tau)\,d\eta\,d\tau \\
&\quad - \int_0^t \int_0^l \hat G_{12_{\xi}}(x,y,t,h,\eta,\tau)\tau^{\beta}
 a_2(\tau)\mu_{2_{\eta\eta}}(\eta,\tau)\,d\eta\,d\tau \\
 &\quad - \int_0^t \int_0^h \hat G_{12} (x,y,t,\xi,0,\tau)
(\nu_{1_{\tau}}(\eta,\tau)-t^{\beta}a_2(\tau)\nu_{1_{\xi\xi}}(\xi,\tau) 
 - f(\xi,0,\tau))d\xi d\tau \\
&\quad + \int_0^t \int_0^h\hat G_{12}(x,y,t,\xi,0,\tau)(\nu_{2_{\tau}}(\eta,\tau) 
  -t^{\beta}a_2(\tau)\nu_{2_{\xi\xi}}(\xi,\tau) - f(\xi,l,\tau))d\xi d\tau \\
&\quad  +\int_0^t \int_0^l \int_0^h\hat G_{12}(x,y,t,\xi,\eta,\tau)f_{\eta\eta}
 (\xi,\eta,)\tau)\,d\xi\,d\eta\,d\tau.
\end{align*}
From the above expression, we establish the following estimate for the kernel 
of \eqref{vls36}):
$$
|K(t,\tau)|\le \frac{C_{44}}{\sqrt{t(t-\tau)}}.
$$
This means that
$$
K(t,\tau)\equiv \frac{1}{\sqrt{t(t-\tau)}}K_1(t,\tau),
$$
and the equation \eqref{vls36}) can be presented as
\begin{equation}\label{vls39}
a(t)=\int_0^t \frac{1}{\sqrt{t(t-\tau)}}K_1(t,\tau)a(\tau) d\tau,
\quad t\in[0,T],
\end{equation}
 where $ |K_1(t,\tau)|\le C_{44}. $

It was proved that there exists such an interval $[0,t_0]$ where 
$a(t)\equiv 0$. Then the  equation \eqref{vls39}) can be transformed 
to the form
\begin{equation}\label{vls40}
a(t)=\int_{t_0}^t \frac{1}{\sqrt{t(t-\tau)}}K_1(t,\tau)a(\tau) d\tau, \quad 
t\in[t_0,T],
\end{equation}
and the following estimate holds:
$$
\big|\frac{1}{\sqrt{t(t-\tau)}}K_1(t,\tau)\big|
\le\frac{C_{45}}{\sqrt{t-\tau}},\quad t\in[t_0,T].
$$
Then  the properties of the integral Volterra equations of the second kind 
imply that  equation \eqref{vls40}) has only the trivial  solution.  
 Hence, $a(t)\equiv 0,\, t\in[0,T]$. Using it  in the equation \eqref{vls30}),
 we obtain $u(x,y,t)\equiv 0, (x,y,t)\in\overline Q_T$ because of the 
the uniqueness of solution of the problem \eqref{vls30}--\eqref{vls32} 
\cite{Lad}.
The proof of Theorem~\ref{thm1} is complete.



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\end{document}