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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2014 (2014), No. 172, pp. 1--11.\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-2014/172\hfil Inverse problem of determining coefficients]
{Inverse problem of determining the coefficients in a degenerate parabolic equation}

\author[N. Huzyk \hfil EJDE-2014/172\hfilneg]
{Nadiya Huzyk}

\address{Nadiya Huzyk \newline
Department of Differential Equations\\
Ivan Franko National University of Lviv\\
Lviv, Ukraine}
\email{hryntsiv@ukr.net}

\thanks{Submitted December 27, 2013. Published August 12, 2014.}
\subjclass[2000]{35R30, 35R35, 35K65}
\keywords{Inverse problem; determining coefficients;
parabolic equation; \hfill\break\indent weak power degeneration}

\begin{abstract}
 We consider the inverse  problem of identifying the time-dependent
 coefficients in a  degenerate  parabolic equation.
 The conditions of existence and uniqueness of the classical solution to this
 problem are established. We investigate the case of weak power degeneration.
\end{abstract}

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

\section{Introduction}\label{Sect1}

Inverse problems of  determining simultaneously several coefficients 
in parabolic equations without degeneration are studied in 
many articles. These unknown parameters can depend on spatial variables
 \cite{An1}-\cite{Bel} or on time variables \cite{Iv1}-\cite{Iv3}.

The inverse problems for the degenerate parabolic equation are 
rarely investigated. Sufficient conditions for the existence and uniqueness of 
classical solutions to inverse problems of identification for the 
time-dependent leading coefficient in a degenerate parabolic equation  
in a domain with known boundary are established in \cite{Sald2,Sald3},
and for a free boundary domain in \cite {Hr,Iv_Hr}. 
Both cases of weak and strong power degeneration are investigated. 
The conditions of  determination for the time-dependent lower coefficients 
in the parabolic equations with weak power degeneration in a fixed boundary 
domain are found in \cite{Hr_1} and in a free boundary domain in \cite{Hr_2}.

In this article we consider an inverse problem of identifying simultaneously
the two unknown time-dependent  parameters in a one-dimensional degenerate 
parabolic equation. It is known that the leading coefficient of this 
equation is the product of the power function which caused degeneration and 
an unknown function of time. Our aim is to establish the conditions of 
existence and uniqueness of the classical solution to this problem in the 
case of weak degeneration.

\section{Statement of the problem}\label{Sect2}

In a domain  $Q_T=\{(x,t): 0 < x <h, 0 < t <T \}$
we consider an inverse problem of determining the time dependent 
coefficients $a=a(t)$ and $b=b(t)$ in  the one-dimensional degenerate parabolic
 equation
\begin{equation} \label{s18_1}
u_t=a(t) t^\beta u_{xx}+b(t)u_x+c(x,t)u+f(x,t)
\end{equation}
with initial condition
\begin{equation}
\label{s18_2} u(x,0) = \varphi(x), \quad x \in [0,h],
\end{equation}
boundary conditions
\begin{equation}
\label{s18_3} u(0,t)=\mu_1(t), \quad u(h,t) = \mu_2(t),\quad
t\in[0,T]
\end{equation}
and  over-determination  conditions
\begin{gather} \label{s18_4}  
a(t) t^\beta u_x(0,t)  =\mu_3(t),\quad t\in[0,T],\\
\label{s18_5} \int_0^{h}  u(x,t) dx = \mu_4(t), \quad t\in[0,T],
\end{gather}
where $\beta$ is a known number.

\begin{definition} \label{def2.1} \rm
A triplet of functions 
$(a,b,u) \in (C[0,T])^2\times  C^{2,1}(Q_T)\cap C^{1,0}(\overline{Q}_T)$, 
with $a(t) > 0$, $t \in [0,T]$ is called a solution to the problem 
\eqref{s18_1}-\eqref{s18_5} if it verifies the equation \eqref{s18_1} 
and conditions \eqref{s18_2}-\eqref{s18_5}.
\end{definition}

We will investigate the case of weak power degeneration, when $0 < \beta < 1$. 

\section{Existence of a solution}

We use the following assumptions:
\begin{itemize}

\item[(A1)] $\varphi \in C^2[0,h]$,
$\mu_3(t) = \mu_{3,0}(t) t^\beta$, $\mu_{3,0} \in C[0,T]$, 
$\mu_i \in C^1[0,T]$, $i=1,2,4$, $c, f \in C(\overline{Q}_T)$ and satisfy 
the H\"{o}lder condition with respect to $x$ uniformly to $t$;

\item[(A2)] $\varphi'(x) > 0$, $x \in [0,h]$,  $\mu_{3,0}(t) > 0$, 
$\mu_2(t) -\mu_1(t) \neq 0$, $t\in[0,T]$;

\item[(A3)] $\varphi(0)=\mu_1(0)$, $\varphi(h)=\mu_2(0)$,
$\int_0^h \varphi(x) dx = \mu_4(0)$.
\end{itemize}

\begin{theorem}\label{th1}
Under assumptions {\rm(A1)--(A3)}, problem \eqref{s18_1}--\eqref{s18_5} 
has a solution $(a, b, u)$ for $x \in [0,h]$ and $t \in[0,T_0]$, 
where the number $T_0 $, $ 0 < T_0 \leq T$,
is defined by the data.
\end{theorem}

\begin{proof} 
First of all we reduce the problem \eqref{s18_1}--\eqref{s18_5} 
to the equivalent system of equations. Suppose temporary that $a=a(t)$ and $b=b(t)$ 
are the known functions.
Making the substitution
\begin{equation}\label{s18_6}
u(x,t)=\widetilde{u}(x,t)+\varphi(x)+\mu_1(t)-\mu_1(0)
+\frac{x}{h}\Bigl(\mu_2(t)-\mu_1(t)-\mu_2(0)+\mu_1(0)\Bigr)
\end{equation}
we reduce the direct problem \eqref{s18_1}--\eqref{s18_3} to the  problem 
with respect to function $\widetilde{u} =\widetilde{u}(x,t)$  with homogeneous 
initial and boundary conditions:
\begin{gather}\label{s18_7}
\begin{aligned}
\widetilde{u}_t
&=a(t) t^\beta \widetilde{u}_{xx}+b(t)\widetilde{u}_x+c(x,t)\widetilde{u}
 + f(x,t)-\mu'_1(t)-\frac{x}{h}(\mu'_2(t)-\mu'_1(t))\\
&\quad +a(t)t^\beta \varphi''(x)
+b(t)\Bigl(\varphi'(x)+\frac{1}{h}
(\mu_2(t)-\mu_1(t)-\mu_2(0)+\mu_1(0))\Bigr)\\
&\quad +c(x,t)\Bigl(\varphi(x)
+\mu_1(t)-\mu_1(0)+\frac{x}{h}(\mu_2(t)-\mu_1(t)-\mu_2(0)+\mu_1(0))\Bigr),\\
&\quad (x,t)\in Q_T,
\end{aligned}\\
\label{s18_8} 
\widetilde{u}(x,0) = 0,\quad x \in[0,h], \\
\label{s18_9} 
\widetilde{u}(0,t) =  \widetilde{u}(h,t) = 0, \quad t \in[0,T].
\end{gather}
For arbitrary functions $b=b(t)$ and $a=a(t) >0$ continuous on $[0,T]$ 
problem \eqref{s18_7}--\eqref{s18_9} is equivalent to the equation
\begin{equation} \label{s18_10}
\begin{aligned}
\widetilde{u}(x,t) 
&= \int_0^t \int_0^h G_1(x,t,\xi,\tau) \Bigl(b(\tau)\widetilde{u}_\xi
 +c(\xi,\tau)\widetilde{u}+ f(\xi,\tau) -\mu'_1(\tau) \\
&\quad -\frac{\xi}{h}(\mu'_2(\tau)-\mu'_1(\tau))
 +a(\tau)\tau^\beta \varphi''(\xi)+
b(\tau)\Bigl(\varphi'(\xi)+\frac{1}{h} (\mu_2(\tau)\\
&\quad -\mu_1(\tau)
 -\mu_2(0)+\mu_1(0))\Bigr) +c(\xi,\tau)\Bigl(\varphi(\xi)
+\mu_1(\tau)-\mu_1(0)\\
&\quad + \frac{\xi}{h}(\mu_2(\tau)
 -\mu_1(\tau) -\mu_2(0)+\mu_1(0))\Bigr)d\xi\, d\tau,
\end{aligned}
\end{equation}
where $G_1=G_1(x,t,\xi,\tau)$ is a Green function of the first value-boundary
 problem for the heat equation
\begin{equation}
\label{s18_11}
\widetilde{u}_t=a(t) t^\beta \widetilde{u}_{xx}.
\end{equation}
It is known \cite[p. 13]{Iv}, that  the Green functions
 $G_k=G_k(x,t,\xi,\tau)$, $k=1,2 $   of the first $(k=1)$ or the second $(k=2)$ 
value-boundary problem for \eqref{s18_11} have  the form
\begin{equation}\label{s18_12}
\begin{aligned}
G_k(x,t,\xi,\tau)
&=\frac{1}{2\sqrt{\pi(\theta(t)-\theta(\tau))}}
\sum_{n=-\infty}^{+\infty}\Big(\exp\Big({-\frac{(x-\xi+2nh)^2}{4(\theta(t)-\theta(\tau))}}\Big)
 \\
&\quad +(-1)^k \exp\Big({-\frac{(x+\xi+2nh)^2}{4(\theta(t)
 -\theta(\tau))}}\Big)\Big), \quad  k=1,2,
\end{aligned}
\end{equation}
where
$\theta(t)=\int_0^t a(\tau) \tau^\beta d\tau$.

Put $v(x,t) \equiv u_x(x,t)$. Using  \eqref{s18_6}, \eqref{s18_10},  
we reduce the direct problem \eqref{s18_1}-\eqref{s18_3} to the system of
integral equations
\begin{gather} \label{s18_13}
\begin{aligned}
u(x,t)
 &= \varphi(x)+\mu_1(t)-\mu_1(0)+\frac{x}{h}\Bigl(\mu_2(t)-\mu_1(t)
-\mu_2(0)+\mu_1(0)\Bigr)  \\ 
&\quad + \int_0^t \int_0^h G_1(x,t,\xi,\tau) \Bigl(b(\tau)v(\xi,\tau)
 +c(\xi,\tau)u(\xi,\tau) + f(\xi,\tau)\\
&\quad -\mu'_1(\tau)  - \frac{\xi}{h}(\mu'_2(\tau)-\mu'_1(\tau))
+a(\tau)\tau^\beta \varphi''(\xi)\Bigr)d\xi\, d\tau,\quad (x,t) \in \overline{Q}_T,
\end{aligned} \\
\label{s18_14}
\begin{aligned}
v(x,t) &= \varphi'(x)+\frac{1}{h}\Bigl(\mu_2(t)-\mu_1(t)-\mu_2(0)+\mu_1(0)\Bigr) \\
&\quad +\int_0^t \int_0^h G_{1x}(x,t,\xi,\tau) 
 \Bigl(b(\tau)v(\xi,\tau)+c(\xi,\tau)u(\xi,\tau)+ f(\xi,\tau)\\
&\quad -\mu'_1(\tau) -\frac{\xi}{h}(\mu'_2(\tau)-\mu'_1(\tau)) 
 +a(\tau)\tau^\beta \varphi''(\xi)\Bigr)d\xi\, d\tau,\quad
 (x,t) \in \overline{Q}_T.
\end{aligned}
\end{gather}
Note that we differentiate \eqref{s18_13} with respect to $x$ in order to 
find $v=v(x,t)$. Let us study the behavior of the integrals on the right-hand 
sides of the formulas \eqref{s18_13}, \eqref{s18_14}. Using \eqref{s18_12}, 
it is easy to verify that
\begin{equation} \label{s18_17} 
\int_0^h G_1(x,t,\xi,\tau)d\xi \leq  1, \quad
 \int_0^h |G_{1x}(x,t,\xi,\tau) |d\xi \leq \frac{C_1}{\sqrt{\theta(t)-\theta(\tau)}}.
\end{equation}
Then  from \eqref{s18_13} and \eqref{s18_14}, we deduce that
\begin{align*}
I_1 &\equiv \Bigl|\int_0^t \int_0^1 G_1(x,t,\xi,\tau) 
\Bigl(b(\tau)v(\xi,\tau)+c(\xi,\tau)u(\xi,\tau)+ f(\xi,\tau)-\mu'_1(\tau)\\ 
&\quad -\frac{\xi}{h}(\mu'_2(\tau)-\mu'_1(\tau))
+a(\tau)\tau^\beta \varphi''(\xi)\Bigr)d\xi\, d\tau\Bigr|\\
&  \leq C_2 t,
\\
I_2 &\equiv \Bigl|\int_0^t \int_0^1 G_{1x}(x,t,\xi,\tau) 
 \Bigl(b(\tau)v(\xi,\tau)+c(\xi,\tau)u(\xi,\tau)+ f(\xi,\tau)-\mu'_1(\tau)\\ 
&\quad -\frac{\xi}{h}(\mu'_2(\tau)-\mu'_1(\tau))
+a(\tau)\tau^\beta \varphi''(\xi)\Bigr)d\xi\, d\tau\Bigr|  \\
&\leq C_3 \int_0^t \frac{d\tau}{\sqrt{\theta(t)-\theta(\tau)}} 
 \leq C_4 \int_0^t \frac{d\tau}{\sqrt{t^{\beta+1}-\tau^{\beta+1}}} \\
&\leq C_5 t^{\frac{1-\beta}{2}} \int_0^1 \frac{dz}{\sqrt{1-z^{\beta+1}}} 
 \leq C_6 t^{\frac{1-\beta}{2}}.
\end{align*}
Taking into account that $0 < \beta <1$, we conclude that the integrals 
on the right-hand sides of the formulas \eqref{s18_13}, \eqref{s18_14} 
tend to zero as $t$ tends to zero.

Under the conditions of the Theorem \ref{th1} we can represent 
 equations \eqref{s18_4}, \eqref{s18_5} in the form
\begin{gather}
\label{s18_15}
a(t) = \frac{\mu_3(t)}{t^\beta v(0,t)}, \quad  t\in[0,T],\\
\label{s18_16}
\begin{aligned}
b(t) &= \frac{1}{\mu_2(t)-\mu_1(t)} \Bigl(\mu_4'(t)+\mu_3(t)-a(t) t^\beta v(h,t) \\ 
&\quad -\int_0^h (c(x,t)u(x,t) +f(x,t)) dx\Bigr), \quad t \in [0,T].
\end{aligned}
\end{gather}
To this end, it suffices to differentiate \eqref{s18_5} with respect to $t$.

Consequently, problem \eqref{s18_1}--\eqref{s18_5} is reduced to the system 
of equations \eqref{s18_13}, \eqref{s18_14}, \eqref{s18_15}, \eqref{s18_16}
 with respect to the unknowns $u=u(x,t)$, $v=v(x,t)$, $a=a(t)$, $b=b(t)$. 
It follows from the way of derivation of 
\eqref{s18_13}, \eqref{s18_14}, \eqref{s18_15}, \eqref{s18_16} that if
 $(a,b,u)$ is a solution to the problem \eqref{s18_1}-\eqref{s18_5} then 
$(u,v,a,b)$ is a continuous solution to the system of equations \eqref{s18_13}, 
\eqref{s18_14}, \eqref{s18_15}, \eqref{s18_16}. On the other hand, 
 if $(u, v, a, b) \in (C(\overline{Q}_T))^2\times (C[0,T])^2$, $a(t) > 0$,
$t\in[0,T]$ is a solution to the system \eqref{s18_13}, \eqref{s18_14}, 
\eqref{s18_15}, \eqref{s18_16}, than $(a, b, u)$ is a solution to the inverse
 problem \eqref{s18_1}-\eqref{s18_5} in the sense of the above definition. 
Indeed, using the uniqueness properties of the solutions to the system of 
Volterra integral equations of the second kind it is easy to see that
 $v(x,t) \equiv u_x(x,t)$. So  it follows from \eqref{s18_13} that function 
$u=u(x,t)$ is a solution to the equation
\begin{align*}
u(x,t) 
&= \varphi(x)+\mu_1(t)-\mu_1(0)+\frac{x}{h}\Bigl(\mu_2(t)-\mu_1(t)-\mu_2(0)
+\mu_1(0)\Bigr)  \\ 
&\quad + \int_0^t \int_0^h G_1(x,t,\xi,\tau) \Bigl(b(\tau)u_\xi(\xi,\tau)
+c(\xi,\tau)u(\xi,\tau)+ f(\xi,\tau)\\
&\quad -\mu'_1(\tau) - \frac{\xi}{h}(\mu'_2(\tau)-\mu'_1(\tau))
+a(\tau)\tau^\beta \varphi''(\xi)\Bigr)d\xi\, d\tau.
\end{align*}
This means that $u \in C^{2,1}(Q_T)\cap C^{1,0}(\overline{Q}_T)$ is a solution 
to \eqref{s18_1}--\eqref{s18_3}. Then  we can rewrite  \eqref{s18_16} in the form
$$
a(t) t^\beta (u_x(h,t)-u_x(0,t))+b(t) (\mu_2(t)-\mu_1(t))
+ \int_0^h (c(x,t)u(x,t) +f(x,t)) dx = \mu_4'(t)
$$
or , using \eqref{s18_1}, \eqref{s18_3}, in the form
$$
\int_0^h u_t (x,t) dx = \mu_4'(t).
$$
We integrate this equality with respect to time variable from 
$0$ to $t$. Taking into account compatibility conditions,
 we obtain \eqref{s18_5}. The validity of \eqref{s18_4} follows 
from \eqref{s18_15}.

Now we  study the system of equations \eqref{s18_13}, \eqref{s18_14}, 
\eqref{s18_15}, \eqref{s18_16}. We will apply the Schauder fixed point 
theorem for a compact operator to this system. For this aim we first 
establish a priori estimates for the solutions to 
\eqref{s18_13}, \eqref{s18_14}, \eqref{s18_15}, \eqref{s18_16}.

We  estimate the functions $u=u(x,t)$, $v=v(x,t)$ taking into account 
\eqref{s18_13}, \eqref{s18_14}.  We conclude  from the assumption of
 Theorem \ref{th1} that only the first terms on the right-hand side of 
 \eqref{s18_13}, \eqref{s18_14} respectively  are non equiv to zero. 
The sum of the rest terms tend to zero when $t$ tends to zero. So, there  
exists the  number $t_1, 0 < t_1 \leq T$, such that
\begin{gather} \label{s18_28}
\begin{aligned}
&\Bigl|\mu_1(t)-\mu_1(0)+\frac{x}{h}\Bigl(\mu_2(t)
 -\mu_1(t)-\mu_2(0)+\mu_1(0)\Bigr) \\
&+ \int_0^t \int_0^h G_1(x,t,\xi,\tau)  
\Bigl(b(\tau)v(\xi,\tau)+c(\xi,\tau)u(\xi,\tau)+ f(\xi,\tau)\\
&-\mu'_1(\tau)- \frac{\xi}{h}(\mu'_2(\tau)-\mu'_1(\tau)) 
+a(\tau)\tau^\beta \varphi''(\xi)\Bigr)d\xi\, d\tau\Bigr|\\
& \leq  \frac{M_0}{2}, \quad  (x,t) \in [0,h]\times [0,t_1],
\end{aligned}\\
\label{s18_19}
\begin{aligned}
&\frac{\varphi'(x)}{2} + \frac{\mu_2(t)-\mu_1(t)-\mu_2(0)+\mu_1(0)}{h} \\ 
& + \int_0^t \int_0^h G_{1x}(x,t,\xi,\tau) \Bigl(b(\tau)v(\xi,\tau)+  c(\xi,\tau)u(\xi,\tau)+ f(\xi,\tau)-\mu'_1(\tau) \\ &-
\frac{\xi}{h}(\mu'_2(\tau)-\mu'_1(\tau))
+a(\tau)\tau^\beta \varphi''(\xi)\Bigr)d\xi\, d\tau \geq 0, \quad
 (x,t) \in [0,h]\times[0,t_1],
\end{aligned}
\end{gather}
where $M_0 \equiv \max_{x \in [0,h]} |\varphi(x)| > 0$.
Then  from  \eqref{s18_13}, we obtain 
\begin{equation}
\label{s18_29}
|u(x,t)| \leq \frac{3M_0}{2} \equiv M_1, \quad (x,t) \in [0,h]\times [0,t_1].
\end{equation}
In addition,  from  \eqref{s18_14} we conclude that
\begin{equation}
\label{s18_20}
v(x,t) \geq \frac{\varphi'(x)}{2}
\geq \frac{\min_{x \in [0,h]} \varphi'(x)}{2}\equiv M_2 > 0, \quad
 (x,t) \in [0,h]\times[0,t_1]
\end{equation}
and from \eqref{s18_15},
\begin{equation} \label{s18_21}
a(t) \leq \frac{\max_{[0,T]} \mu_{3,0}(t)}{ M_2} \equiv A_1, \quad
 t\in [0,t_1].
\end{equation}
Denote $V(t) = \max_{x\in [0,h]} v(x,t)$.
Taking into account \eqref{s18_29},  from \eqref{s18_16} we obtain
\begin{equation}
\label{s18_22}
|b(t)| \leq C_7 + C_{8} a(t) t^\beta  V(t), \quad t\in[0,t_1].
\end{equation}
Taking into account \eqref{s18_29}, \eqref{s18_22}, from \eqref{s18_14}  we obtain 
 the following inequality for $V=V(t)$,
\begin{equation} \label{s18_23}
\begin{aligned}
V(t) 
&\leq C_{9} + C_{10} \int_0^t \frac{1}{\sqrt{\theta(t)-\theta(\tau)}}d\tau
  + C_{11} \int_0^t \frac{V(\tau)}{\sqrt{\theta(t)-\theta(\tau)}}d\tau \\
 &\quad + C_{12} \int_0^t \frac{a(\tau) \tau^\beta V^2(\tau)}{\sqrt{\theta(t)
-\theta(\tau)}}d\tau, \quad  t \in [0,t_1].
\end{aligned}
\end{equation}
On the other hand, from \eqref{s18_15}  we find that
\begin{equation}
\label{s18_24}
a(t)  \geq \frac{\mu_3(t)}{t^\beta V(t)}, \quad t\in[0,T],
\end{equation}
or
$$
\frac{1}{a(t) } \leq \frac{t^\beta V(t)}{\mu_3(t)}.
$$
Using this inequality, from \eqref{s18_23} we obtain
\begin{equation}
\label{s18_25}
V_1(t) \leq C_{13} + C_{14} \int_0^t \frac{a(\tau) 
\tau^\beta V_1^2(\tau)}{\mu_3(\tau) \sqrt{\theta(t)-\theta(\tau)}}d\tau, \quad
 t \in [0,t_1],
\end{equation}
where $V_1(t) = V(t) +\frac{1}{2}$.

Applying the Cauchy and Cauchy-Buniakowski inequalities to the squared 
inequality \eqref{s18_25}, we conclude
 \begin{eqnarray}\label{s18_46}
V_1^2(t) \leq 2C_{13}^2 + 2C_{14}^2  \int_0^t
\frac{V_1^4(\tau) d\tau}{\sqrt{\theta(t)-\theta(\tau)}}
 \int_0^t
\frac{a^2(\tau)\tau^{2\beta} d\tau}{\mu_3^2(\tau)
\sqrt{\theta(t)-\theta(\tau)}}.
\end{eqnarray}
Let us  consider the integral 
$$ 
J_1 =  \int_0^t \frac{a^2(\tau)\tau^{2\beta} d\tau}{\mu_3^2(\tau)
\sqrt{\theta(t)-\theta(\tau)}}.
$$
 Using \eqref{s18_21} and the definition of the function $ \theta=\theta(t)$, 
we find that
$$
\theta(t) \leq \frac{A_1}{(1+\beta)} t^{1+\beta}.
$$ 
Then
$$
t \geq (\theta(t))^{\frac{1}{1+\beta}}\Bigl(\frac{(1+\beta)}{A_1}
\Bigr)^{\frac{1}{1+\beta}}. 
$$ 
Taking into account \eqref{s18_21} and the conditions of
 Theorem \ref{th1}, we obtain
$$
J_1 \leq C_{15}
\int_0^t \frac{a(\tau)\tau^{\beta} d\tau}{\tau^{\beta}
\sqrt{\theta(t)-\theta(\tau)}} \leq C_{16} \int_0^t
\frac{a(\tau)
\tau^{\beta}d\tau}{{\theta(\tau)}^{\frac{\beta}{1+\beta}}
\sqrt{\theta(t)-\theta(\tau)}}.
$$
 Making the substitution
$z=\theta(\tau)/\theta(t)$, we obtain
\begin{equation}\label{s18_47}
J_1 \leq C_{17} {(\theta(t))}^{\frac{1-\beta}{2(1+\beta)}}
\int_0^1 \frac{dz}{z^{\frac{\beta}{1+\beta}} \sqrt{1-z}}
\leq C_{18}.
\end{equation}
Using  \eqref{s18_47} in \eqref{s18_46}, we derive the  inequality 
$$
V_1^2(t) \leq C_{19} + C_{20}  \int_0^t \frac{V_1^4(\tau)
d\tau}{ \sqrt{\theta(t)-\theta(\tau)}}.
$$ 
We replace  $t$ by $\sigma$ ,  then multiply it by
$\frac{a(\sigma)\sigma^\beta}{\mu_3(\sigma) \sqrt{\theta(t) -
\theta(\sigma)}}$, and then integrate with respect to $\sigma$ from $0$ to $t$.
Taking into account \eqref{s18_47} and 
 $$
\int_\tau^t \frac{a(\sigma)\sigma^\beta d\sigma}{\sqrt{(\theta(t) -
\theta(\sigma))(\theta(\sigma)-\theta(\tau)) }} = \pi,
$$
 we deduce that
$$ 
\int_0^t \frac{a(\sigma)\sigma^\beta V_1^2(\sigma) }{\mu_3(\sigma)
\sqrt{\theta(t)-\theta(\sigma)}} d\sigma \leq C_{21} + C_{22}
\int_0^t \frac{V_1^4(\tau)}{\tau^\beta} d\tau.
$$
Using this inequality in  \eqref{s18_25}, we derive the  inequality
$$
V_1(t) \leq C_{23} + C_{24} \int_0^t
\frac{V_1^4(\tau)}{\tau^\beta} d\tau.
$$ 
Denote $H(t) =  C_{23} + C_{24} \int_0^t
\frac{V_1^4(\tau)}{\tau^\beta} d\tau$. Differentiating, we find 
$$
H'(t) = C_{24} \frac{V_1^4(t)}{t^\beta} \leq C_{24} \frac{H^4(t)}{t^\beta}.
$$
We integrate to obtain
$$
H(t) \leq \frac{C_{23} \sqrt[3]{(1-\beta)}}{\sqrt[3]{1-\beta-3C_{23}^3
C_{24} t^{1-\beta}}} \leq M_3,\quad t\in [0,t_2],
$$ 
where the number $t_2$, $0< t_2 <T$ satisfies 
\begin{equation} \label{s18_27}
1-\beta - 3C_{23}^3 C_{24} t_2^{1-\beta} > 0.
\end{equation}
This implies
\begin{equation} \label{s18_26}
v(x,t) \leq M_3, \quad  (x,t) \in [0,h]\times[0,t_2],
\end{equation}
where  the constant $M_3$ depends on the data.

Using \eqref{s18_26},   from  \eqref{s18_22}, \eqref{s18_24}  we obtain the estimates
\begin{gather} \label{s18_30}
a(t) \geq A_2 > 0, \quad t\in [0,t_2], \\
\label{s18_31}
|b(t)| \leq M_4, \quad t\in[0,t_2],
\end{gather}
where $A_2$ and $M_4$ are  the known constants.
Thus, an a priori estimates of the solutions to 
system \eqref{s18_13}, \eqref{s18_14}, \eqref{s18_15}, \eqref{s18_16} 
are established.

Put $T_0=\min\{t_1,t_2\}$. Denote 
$\omega = (u, v, a, b)$, 
$\mathcal{N} = \{(u, v, a, b) \in (C(\overline{Q}_{T_0}))^2 \times  
 (C[0,T_0])^2: |u(x,t)| \leq M_1, M_2 \leq v(x,t) \leq M_3, A_2 \leq a(t)
 \leq A_1, |b(t)| \leq M_4\}$.
  We rewrite  system \eqref{s18_13}, \eqref{s18_14}, \eqref{s18_15}, 
\eqref{s18_16}  as the operator equation 
$$
\omega = P \omega,
$$
 where the operator  $P$ is defined by the right-hand sides of these equations. 
The obtained estimates \eqref{s18_29}, \eqref{s18_20}, \eqref{s18_21}, 
\eqref{s18_26}, \eqref{s18_30}, \eqref{s18_31} of the functions 
$(u, v, a, b)$ gua\-rantee that  $P$ maps $\mathcal{N}$ into $\mathcal{N}$. 
The compactness of operator $P$ can be established as in non degenerate 
case \cite[p. 20]{Iv}. Now the Schauder fixed-point theorem yields the existence 
of the  continuous solution to 
 system \eqref{s18_13}, \eqref{s18_14}, \eqref{s18_15}, \eqref{s18_16} on 
$[0,h]\times [0,T_0]$. This means that exists the solution $(a, b, u)$ 
to the inverse problem \eqref{s18_1}-\eqref{s18_5}.  
This completes the proof of Theorem \ref{th1}.
\end{proof}

\section{Uniqueness of a solution}

\begin{theorem} \label{th2}
Assume that the following conditions hold:
\begin{itemize}
\item[(B1)] $c, f \in C^{1,0}([0,h]\times[0,T])$, $\varphi \in C^3[0,h]$;

\item[(B2)] $\mu_{3,0}(t) \neq 0$, $\mu_2(t) -\mu_1(t) \neq 0$, $t \in [0,T]$
\end{itemize}
Then the solution to  \eqref{s18_1}-\eqref{s18_5} is unique.
\end{theorem}

\begin{proof} 
Suppose that \eqref{s18_1}-\eqref{s18_5} has two solutions
 $(a_i, b_i, u_i)$, $i=1,2$. Denote $a(t) = a_1(t)-a_2(t)$, 
$b(t) = b_1(t) -b_2(t)$, $u(x,t) = u_1(x,t)-u_2(x,t)$. 
From \eqref{s18_1}-\eqref{s18_5}, we obtain 
\begin{gather}
\label{s18_32}
u_t = a_1(t) t^\beta u_{xx} + b_1(t) u_x +c(x,t) u + a(t) t^\beta u_{2xx} 
+ b(t) u_{2x},\quad  (x,t) \in Q_T,\\
\label{s18_33}
u(x,0) =0, \quad x\in[0,h],\\
\label{s18_34}
u(0,t) = u(h,t) =0,\quad t \in [0,T],\\
\label{s18_35}
a_1(t) t^\beta u_x(0,t) + a(t) t^\beta u_{2x}(0,t) =0,\quad t\in [0,T],\\
\label{s18_36}
\int_0^h u(x,t) dx =0, \quad    t \in [0,T].
\end{gather}
With the aid of the Green function $G^*(x,t,\xi,\tau)$ of the first 
boundary-value problem for the equation
$$
u_t = a_1(t) t^\beta u_{xx} + b_1(t) u_x +c(x,t) u
$$
we represent the solution to \eqref{s18_32}-\eqref{s18_34}  in the form
\begin{equation}
\label{s18_37}
u(x,t) = \int_0^t \int_0^h G^*(x,t,\xi,\tau) (a(\tau) 
\tau^\beta u_{2\xi\xi}(\xi,\tau) + b(\tau) u_{2\xi}(\xi,\tau)) d\xi\, d\tau,
\end{equation}
$(x,t) \in \overline{Q}_T$.
Differentiating \eqref{s18_37} with respect to $x$, we find
\begin{equation}
\label{s18_38}
u_x(x,t) = \int_0^t \int_0^h G^*_x(x,t,\xi,\tau) (a(\tau) 
\tau^\beta u_{2\xi\xi}(\xi,\tau) + b(\tau) u_{2\xi}(\xi,\tau)) d\xi\, d\tau,
\end{equation}
$(x,t) \in \overline{Q}_T$.

From \eqref{s18_35}, we deduce
\begin{equation}
\label{s18_39}
a(t) = - \frac{a_1(t) }{u_{2x}(0,t)}u_x(0,t), \quad t \in [0,T].
\end{equation}
Note that the condition $u_{2x}(0,t) \neq 0$, $t\in [0,T]$ is satisfied 
because of the assumption $\mu_{3,0}(t) \neq 0$, $t\in [0,T]$ in 
Theorem \ref{th2}.

Differentiating \eqref{s18_36} with respect to $t$ and using 
\eqref{s18_32}-\eqref{s18_35}, we obtain
\begin{equation}
\label{s18_40}
b(t) = - \frac{1}{\mu_2(t)-\mu_1(t)} \Bigl(a_1(t) t^\beta u_x(h,t) + a(t) t^\beta  u_{2x}(h,t)
 + \int_0^h c(x,t) u(x,t) dx  \Bigr), 
\end{equation}
for $t\in [0,T]$.
Using \eqref{s18_37}, \eqref{s18_38}, we reduce \eqref{s18_39}, \eqref{s18_40} 
to the system of homogeneous integral Volterra equations of the second kind
\begin{gather}
\label{s18_41}
a(t) = \int_0^t (K_{11}(t,\tau) a(\tau) + K_{12} (t,\tau) b(\tau)) d\tau,
\\ \label{s18_42}
b(t) = \int_0^t (K_{21}(t,\tau) a(\tau) + K_{22} (t,\tau) b(\tau)) d\tau,
\end{gather}
where
\begin{gather*}
K_{11}(t,\tau) = -\frac{a_1(t)}{u_{2x}(0,t)} \int_0^h G^*_x(0,t,\xi,\tau) 
\tau^\beta u_{2\xi\xi} (\xi,\tau)d\xi,
\\
K_{12}(t,\tau) = -\frac{a_1(t)}{u_{2x}(0,t)} \int_0^h G^*_x(0,t,\xi,\tau) 
 u_{2\xi}(\xi,\tau) d\xi,
\\
\begin{aligned}
K_{21}(t,\tau)
& = -\frac{1}{\mu_2(t)-\mu_2(\tau)}\Bigl(a_1(t)t^\beta 
\int_0^h G^*_x(h,t,\xi,\tau) \tau^\beta u_{2\xi\xi}(\xi,\tau)d\xi
\\
&\quad +\int_0^h \int_0^h G^*(x,t,\xi,\tau) c(x,t)
 \tau^\beta u_{2\xi\xi}(\xi,\tau) d\xi dx \\
&\quad - \frac{t^\beta a_1(t) u_{2x}(h,t)}{u_{2x}(0,t)}
 \int_0^h G^*_x(0,t,\xi,\tau) \tau^\beta u_{2\xi\xi}(\xi,\tau) d\xi\Bigr),
\end{aligned} \\
\begin{aligned}
K_{22}(t,\tau) 
&= -\frac{1}{\mu_2(t)-\mu_2(\tau)}\Bigl(a_1(t)t^\beta 
 \int_0^h G^*_x(h,t,\xi,\tau) u_{2\xi}(\xi,\tau)d\xi \\
&\quad +\int_0^h \int_0^h G^*(x,t,\xi,\tau) c(x,t) u_{2\xi}(\xi,\tau) d\xi dx \\
&\quad - \frac{t^\beta a_1(t) u_{2x}(h,t)}{u_{2x}(0,t)}
 \int_0^h G^*_x(0,t,\xi,\tau) u_{2\xi}(\xi,\tau) d\xi\Bigr).
\end{aligned}
\end{gather*}
To show that the kernels $K_{11}, K_{21}$ have an integrable singularity,
 we estimate the function $u_{2xx}(x,t)$. To this end, we consider the 
corresponding problem \eqref{s18_1}-\eqref{s18_3}. For the function
 $u_2(x,t), u_{2x}(x,t)$  the equalities analogous to \eqref{s18_13}, \eqref{s18_14} 
respectively take place.
We denote by $G_1^{(2)}= G_1^{(2)}(x,t,\xi,\tau)$ the Green function 
of the first value-boundary problem for the equation
$$
u_t=a_2(t) t^\beta u_{xx}.
$$
This function defines by the formula analogous to \eqref{s18_12} with 
$\theta^{(2)}(t)= \int_0^t a_2(\tau) \tau^\beta d\tau$. 
Differentiating the corresponding formula \eqref{s18_13} twice with respect 
to $x$ and using the relationship 
$G_{1xx}^{(2)}(x,t,\xi,\tau) = G_{1\xi\xi}^{(2)}(x,t,\xi,\tau)$ 
we arrive to the equation
\begin{equation}\label{s18_43}
\begin{aligned}
u_{2xx}(x,t) 
&= \varphi''(x) + \int_0^t  G_{1\xi}^{(2)}(x,t,h,\tau) \Bigl(b(\tau)u_{2\xi}(h,\tau)
 +c(h,\tau)u_2(h,\tau)\\
&\quad + f(h,\tau)  -\mu'_2(\tau)+a_2(\tau)\tau^\beta \varphi''(h)\Bigr) d\tau \\
&\quad - \int_0^t  G_{1\xi}^{(2)}(x,t,0,\tau) \Bigl(b(\tau)u_{2\xi}(0,\tau)
+c(0,\tau)u_2(0,\tau) \\ 
&\quad + f(0,\tau)-\mu'_1(\tau)+a_2(\tau)\tau^\beta \varphi''(0)\Bigr)d\tau \\
&\quad - \int_0^t \int_0^h G_{1\xi}^{(2)}(x,t,\xi,\tau) 
 \Bigl(c(\xi,\tau)u_{2\xi}(\xi,\tau) +c_\xi(\xi,\tau)u_2(\xi,\tau)\\
&\quad + f_\xi(\xi,\tau) - \frac{1}{h}(\mu'_2(\tau)-\mu'_1(\tau))
+a_2(\tau)\tau^\beta \varphi'''(\xi)\Bigr)d\xi\, d\tau \\
&\quad - \int_0^t \int_0^h G_{1\xi}^{(2)}(x,t,\xi,\tau) 
b(\tau)u_{2\xi\xi}(\xi,\tau)d\xi\, d\tau, \quad  (x,t) \in \overline{Q}_T.
\end{aligned}
\end{equation}
Taking into account \eqref{s18_12}, we have
\begin{align*}
 J&\equiv \int_0^t |G_{1\xi}^{(2)}(x,t,0,\tau)| d\tau \\
&= \frac{1}{2\sqrt{\pi}}\int_0^t \frac{1}{(\theta^{(2)}(t) -\theta^{(2)}(\tau))^{3/2}}
 \sum_{n=-\infty}^{+\infty} |x+2nh|  
 \exp\Bigl(-\frac{{(x+2nh)}^2}{4(\theta^{(2)}(t)-\theta^{(2)}(\tau))}\Bigr) d\tau.
\end{align*}
Using the definition of  $\theta^{(2)}=\theta^{(2)}(t)$ and 
substituting $z= 1-\frac{\tau}{t} $, we obtain
$$
J \leq C_{25}t^{-\frac{3\beta+1}{2}}\int_0^1 z^{-\frac{3}{2}}
 \sum_{n= -\infty}^{+\infty}|x+2nh|
\exp\Bigl(-\frac{C_{26}{(x+2nh)}^2}{t^{1+\beta}z}\Bigr)dz.
$$
Let us change the variable  
$ \sigma = \sqrt{\frac{C_{26}}{t^{1+\beta} z }} (x+2nh)$. We get as a result 
$$
J \leq \frac{C_{27}}{t^\beta} \int_{-\infty}^
{+\infty} e^{-\sigma^2}d\sigma \leq \frac{C_{28}}{t^\beta}.
$$
We evaluate the rest integrals which the formula \eqref{s18_43} contains 
 by a similar way. So  from \eqref{s18_43}  we deduce
\begin{equation}
\label{s18_44}
|u_{2xx}(x,t)| \leq \frac{C_{29}}{t^\beta}.
\end{equation}
 This implies that the kernels of system \eqref{s18_41}, \eqref{s18_42} 
have the integrable singularities  and respectively this system have only
 trivial solution
$$
a(t) \equiv 0, \quad b(t) \equiv 0, \quad t\in[0,T].
$$  
Using this fact in the problem \eqref{s18_32}-\eqref{s18_34} we obtain that 
$$
u(x,t)\equiv 0, \; (x,t) \in \overline{Q}_T.
$$ 
The proof is complete.
\end{proof}

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\end{document}