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\title[On the determination of two coefficients]{\large On the determination of two coefficients in a multidimentional quasilinear parabolic equation}


\author[K.I. ISCHENKO]{{\bf K.I. Ischenko}\protect\orcidicon{0000-0002-3510-7292}}


\address{ Kira Ivanovna Ischenko % Контактные данные всех авторов и место работы указываются только на английском языке
\newline\hphantom{iii} Siberian Federal University,
\newline\hphantom{iii} 79 Svobodnyy pr.,
\newline\hphantom{iii} 660041, Krasnoyarsk, Russia}%
\email{\textcolor{blue}{ki.spirina@gmail.com}}%

\author[S.V. POLYNTSEVA]{{\bf S.V. Polyntseva}\protect\orcidicon{0000-0002-8480-6612}\\ \\
%\textit{Представлено} \textsc{П.П. Петровым}
%\textit{(заполняется редактором)}
}%
\address{Svetlana Vladimirovna Polyntseva  % Контактные данные всех авторов и место работы указываются только на английском языке
\newline\hphantom{iii} Krasnoyarsk, Russia}%
\email{\textcolor{blue}{svpolyntseva@gmail.com}}%


\thanks{\sc Ischenko, K.I., Polyntseva, S.V.
On the determination of two coefficients in a multidimentional quasilinear parabolic equation}
\thanks{\copyright \ 2024 Ischenko K.I., Polyntseva S.V.}
%\thanks{\rm The work is supported by RFFI (grant 03-01-11111)}
\thanks{\it Received January, 1, 2023, Published December, 31, 2023}%

\semrtop \vspace{1cm}
\maketitle {\small

\vspace{-12pt}
\centerline{{\it Communicated by} {\sc P.P. Petrov}
%({\it filled by Editorial board})
}

\bigskip
\bigskip

\begin{quote}
\noindent{\bf Abstract:} This article discusses the problem of determining two 
unknown coefficients in one multidimensional quasilinear parabolic 
equation. By means of overdetermination conditions, the initial 
inverse problem is reduced to the direct auxiliary Cauchy problem. 
Furthermore, using the weak approximation method, on the basis of 
sufficiently smooth input data, the solvability of the direct 
prob\-lem is established. The solution of the inverse problem 
is written out explicitly through the solution of a direct problem,
on this basis, the existence and uniqueness theorem of 
the solution of the original inverse problem in the class of smooth 
bounded functions is proved. An example of input data satisfying 
the conditions of the theorem is given.\medskip

\noindent{\bf Keywords:} inverse problem, coefficient inverse problem, Cauchy problem, quasilinear parabolic equation,  overdetermination con\-di\-tions, weak approximation method.
 \end{quote}
}

\bigskip

\section{Introduction}

\indent The class of inverse problems, that is difficult to study, is coefficient inverse problems, in which, along with the primary source, it is necessary to determine some physical properties of the process (coefficients).

Coefficient inverse problems for quasilinear parabolic equations describe, model and control nonlinear diffusion and filtration processes, are used to identify (thermophysical) characteristics (with a large change in the tempe\-re\-ture interval, the thermophysical characteristics of the medium depend on the temperature distribution) of the processes in thermophysics and mechanics of continuous media, etc. 

Note that the definition of quasilinear parabolic equations is given in \cite{1}.

The problems of determining the coefficients of quasilinear parabolic equ\-ations were investigated  by N.L. Gol'dman, K.T. Iskakov, A.D. Iskenderov, S.I. Kabanikhin, I.V. Koptyug, A. Lorenzi,  R.Z. Sagdeev, V.M. Volkov \cite{2} - \cite{9} and others.

In \cite{9}, an inverse problem is considered for a quasilinear equation in a semi-infinite strip in the case of one unknown coefficient for the second derivative with respect to the variable x, depending on the solution of the equation.

In \cite{2} - \cite{6}, \cite{8} coefficient inverse problems for quasilinear parabolic equati\-ons in bounded domains are investigated.

The problem considered in \cite{7} clearly demonstrates the practical applicati\-on of the solution of the inverse problem
 for a quasilinear parabolic equation. Namely, mass transport of liquids which obeys the equation with liquid concentration - dependent diffusivity is considered and water concentration profiles measured experimeally in the course of drying 
of water-saturated porous alumina pellets are shown to be successfully modelled assuming expo\-nential concentration 
dependence of diffusivity.

In the present paper, the problem of determining two unknown coefficients depending on time and the independent 
variable x in a multidimensional quasilinear parabolic equation is investigated for the unique solvability.

The study is conducted using the following method:

- on the basis of the overdetermination conditions, the initial inverse problem is reduced to an auxiliary direct Cauchy
 problem for a nonlinear loaded equation;

- solvability of the direct problem is proved using sufficiently smooth input data and the weak approximation method \cite{10};

- the solution of the inverse problem is written out explicitly through the solution of the direct problem, on this basis 
the theorem of existence and uniqueness of the classical solution of the inverse problem in the class of smooth
 bounded functions for $t \in [0, t^*]$ is proved.

Previously, this method was used in the case of coefficient inverse problems for linear and semilinear parabolic 
equations, see, for example, \cite{11}, \cite{12}.

Also in the present paper, the condition for the dependence of $t^*$ on the constants limiting the input data 
was written out, in \cite{11}, \cite{12} this condition was not written out. An example of input data satisfying the conditions 
of the proved theorem of existence and uniqueness of the classical solution of the inverse problem is given.

       

\section{Problem statement}

In $G_{[0,T]}=\{(t,x,z)\bigl| 0\leqslant t \leqslant T, x\in E_n, z\in E_1\}$ we have the Cauchy problem
\begin{equation}\label{uravnenie}
\frac{\partial u}{\partial
t}=L_x(u)+a(t,x)uu_{zz}+\beta_1(t,x)u_z+\beta_2(t,x)u^2+b(t,x)f(t,x,z),
\end{equation}
\begin{equation}\label{urnachdan}
u(0,x,z)=u_0(x,z), \quad (x,z)\in E_{n+1}.
\end{equation}

Here
\begin{equation*}
L_x(u)=\sum_{i,j=1}^n\alpha_{ij}\frac{\partial^2u}{\partial x_i\partial x_j}+\sum_{i=1}^n\alpha_i\frac{\partial u}{\partial x_i},
\end{equation*}
functions $f(t,x,z)$ and $u_0(x,z)$ are given in $G_{[0,T]}$ and $E_{n+1}$ respectively, coefficients $\alpha_{ij}$, $\alpha_i$, 
$i,j=\overline{1,n},$ $\beta_1(t,x)$, $\beta_2(t,x)$ are continuously differentiable, real-valued functions of $t$, and $t,x$, respectively, $0\leqslant t\leqslant T>0$, $T - const$, $E_n$ is the $n$-dimensional Euclidian space, $n\geqslant 1$, $n\in \textbf{N}.$

Let be $\alpha_{ij}(t)=\alpha_{ji}(t)$\, and the ratio 
\begin{equation*}
\sum_{i,j=1}^n \alpha_{ij}\xi_i\xi_j>0 \quad \forall \xi\in E_n \backslash \{0\}, \quad t \in[0,T],
\end{equation*}
is satisfied.

We assume that overdetermination conditions on two different hyper\-surface $z=d_1(t)$\, and  \,$z=d_2(t)$ are hold:
\begin{equation}\label{pereopr}
u(t,x,d_1(t))=\phi_1(t,x), \quad
u(t,x,d_2(t))=\phi_2(t,x),\quad (t,x)\in \Pi_{[0,T]},
\end{equation}
where $\Pi_{[0,T]}=\{(t,x)|\, 0\leqslant t \leqslant T,\, x \in
E_n\},$ $d_1(t)\neq d_2(t)$ and $\phi_1(t,x)$, $\phi_2(t,x)$ --- are given functions,
which satisfy matching conditions
\begin{equation}\label{sogl}
\begin{split}
\phi_1(0,x)=u_0(x,d_1(0)),\quad
\phi_2(0,x)=u_0(x,d_2(0)).\end{split}
\end{equation}

The solution to the inverse problem $(\ref{uravnenie})$ -- $(\ref{pereopr})$  is a triple of functions $ u(t,x,z),$ 
$a(t,x),$ $b(t,x),$ that belong to the class $Z(t^*),$ $0<t^*\leqslant T$, see (\ref{Z}), and satisfy relations $(\ref{uravnenie})$ -- $(\ref{pereopr})$.

\section{Main results}
\indent In the presented paper a following the existence and uniqueness theorem of 
the solution of the original inverse problem is proved:
\theoremstyle{plain}
\newtheorem*{teorem}{\quad\quad\,Theorem}
\begin{teorem}
Let conditions $(\ref{sogl})$, 
\begin{equation}\label{N0x}
\begin{split}
N_1(0,x)u_0(x,z)&=(P(0,x)f(0,x,d_2(0))-\\ &-Q(0,x)f(0,x,d_1(0)))u_0(x,z)\geqslant \delta_1,\\
N_2(0,x)&=\phi_1(0,x) \frac{\partial^2 u_0(x,d_1(0))}{\partial z^2}f(0,x,d_2(0))-\\
&-\phi_2(0,x)\frac{\partial^2 u_0(x,d_2(0))}{\partial z^2}f(0,x,d_1(0))\geqslant \delta_2,
\end{split}
\end{equation}
where  $\delta_1,$ $\delta_2>0$, $\delta_1,$ $\delta_2$= const,
\begin{equation*}
\begin{split}
P(0,x)=\phi_{1t}(0,x)-L_x(\phi_1(0,x))-(\beta_1(0,x)+d'_1(0))u_{0z}|_{z=d_1(0)}-\\-\beta_2(0,x)\phi_1^2(0,x),\\
Q(0,x)=\phi_{2t}(0,x)-L_x(\phi_2(0,x))-(\beta_1(0,x)+d'_2(0))u_{0z}|_{z=d_2(0)}-\\-\beta_2(0,x)\phi_2^2(0,x),
\end{split}
\end{equation*}
be fulfilled. It is assumed that the input data are smooth enough, 
have all continuous derivatives included in the following relation
\begin{equation}\label{vhoddan}
\begin{split}
|D^\gamma_x\frac{\partial^{l_1}}{\partial z^{l_1}}\frac{\partial^g}{\partial t^g}f(t,x,z)|+|D^\gamma_x\frac{\partial^g}{\partial t^g}L_x(\phi_{g_1}(t,x))|&+\\+|D^\gamma_x\frac{\partial^{l_1}}{\partial z^{l_1}}u_0(x,z)|+
|D_x^\gamma\frac{\partial^g}{\partial t^g}\beta_{g_1}(t,x)|&+\\
+|D_x^\gamma\frac{\partial^g}{\partial t^g}\phi_{g_1}(t,x)|+|\frac{d^{s_1}}{dt^{s_1}}d_{g_1}(t)|&\leqslant C,
\end{split}
\end{equation}
here $(t,x,z)\in G_{[0,T]}$, $l_1=\overline{0,10-2|\gamma|}$, $|\gamma|\leqslant 4$, $g=0,1$, $g_1=1,2$, $s_1=1,2$, $\gamma=(\gamma_1,...,\gamma_n)$\,-multi-index, $D^\gamma_x=\frac{\partial^{|\gamma|}}{\partial x_1^{\gamma_1}...\partial x_n^{\gamma_n}},$ $|\gamma|=\displaystyle \sum_{i=0}^n \gamma_i,$ $C\geqslant 1$ - constant. It should be noted that here and below $C$ are different constants.
Then there exists a unique solution  $u(t,x,z)$, $a(t,x)$, $b(t,x)$ of the problem $(\ref{uravnenie})$ - $(\ref{pereopr})$
in 
\begin{equation}\label{Z}
\begin{split}
Z(t^*)=\{u(t,x,z),a(t,x),b(t,x)|u\in C^{1,2,4}_{t,x,z}(G_{[0,t^*]}), \\ a(t,x),b(t,x)\in C^{0,2}_{t,x}(\Pi_{[0,t^*]})\},
\end{split}
\end{equation}
satisfies inequalities 
\begin{equation}\label{beta}
\sum_{|\beta|\leqslant 2}\sum_{k=0}^4 \Bigl|D_x^\beta\frac{\partial^k}{\partial z^k}u(t,x,z)\Bigr|\leqslant C,
\quad (t,x,z)\in G_{[0,t^*]},
\end{equation}
\begin{equation}\label{bet}
\sum_{|\beta|\leqslant 2} \Bigl|D_x^\beta a(t,x)\Bigr|+\sum_{|\beta|\leqslant 2} \Bigl|D_x^\beta b(t,x)\Bigr|\leqslant C,
\quad (t,x)\in \Pi_{[0,t^*]},
\end{equation}
where the class $C^{1,2,4}_{t,x,z}(G_{[0,t^*]})$ is defined as follows 
\begin{equation}\label{C124}
\begin{split}
C^{1,2,4}_{t,x,z}(G_{[0,t_*]})=\{f_1(t,x,z)|\frac{\partial^g}{\partial t^g}f_1\in C(G_{[0,t_*]})), \\
D_x^\beta \frac{\partial^k}{\partial z^k}f_1\in C(G_{[0,t_*]}),
|\beta|\leq 2, k=\overline{0,4}, g=0,1\},
\end{split}
\end{equation}
 and
\begin{equation*}
C^{0,2}_{t,x}(\Pi_{[0,t^*]})=\{a_1(t,x)|D_x^\beta a_1(t,x) \in C(\Pi_{[0,t^*]}),
|\beta|\leqslant 2\}.
\end{equation*} 
The constant is
$t^*=\min\Bigl(t_*,\frac{\delta_1}{2K_1}, \frac{\delta_2}{2K_2}\Bigr),$ where constants $K_1,$ $K_2$ 
depends on C, $\delta_1,$ $\delta_2,$ from relations  $(\ref{N0x}),$ $(\ref{vhoddan})$, $t_*$ satisfies the inequality
\begin{equation*}\label{t_*}
 e^{60Ct_*(U(0)+1)^{20}e^{Ct_*(1+U(0))^{20}}}\leqslant 2,  
\end{equation*}
here $U(0)$ from (\ref{U}) with $t=0,$ the constant $C$ depend on C, $\delta_2,$ from $(\ref{N0x})$,~$(\ref{vhoddan})$.
\end{teorem}
The proof of this theorem is given in section 4.

An example of input data satisfying the conditions of theorem is given in section 5.

\section{Proof of the theorem}
 \indent By means of overdetermination conditions $(\ref{pereopr})$ and equation $(\ref{uravnenie})$ we receive 
the system of linear algebraic equations by solving which  we get form of the coefficients 
\begin{equation}\label{ab}
\begin{split}
a(t,x)=\frac{P(t,x)f(t,x,d_2(t))-Q(t,x)f(t,x,d_1(t))}
{\phi_1(t,x)u_{zz}|_{z=d_1(t)}f|_{z=d_2(t)}-\phi_2(t,x)u_{zz}|_{z=d_2(t)}f|_{z=d_1(t)}},\\
b(t,x)=\frac{Q(t,x)\phi_1(t,x)u_{zz}|_{z=d_1(t)}-P(t,x)\phi_2(t,x)u_{zz}|_{z=d_2(t)}}
{\phi_1(t,x)u_{zz}|_{z=d_1(t)}f|_{z=d_2(t)}-\phi_2(t,x)u_{zz}|_{z=d_2(t)}f|_{z=d_1(t)}}.
\end{split}
\end{equation}
Substituting $(\ref{ab})$ into $(\ref{uravnenie})$, we arrive at the auxiliary direct problem:
\begin{equation}\label{zadN}
u_t=L_x(u)+\frac{N_1}{N_2}uu_{zz}+\beta_1(t,x)u_z+\beta_2(t,x)u^2+\frac{N_3}{N_2}f(t,x,z),
\end{equation}
\begin{equation}\label{nachdann}
u(0,x,z)=u_0(x,z),
\end{equation}
here
\begin{equation}\label{N123}
\begin{split}
N_1&=N_1(t,x)=P(t,x)f(t,x,d_2(t))-Q(t,x)f(t,x,d_1(t)),\\
N_2&=N_2(t,x)=\phi_1(t,x)u_{zz}|_{z=d_1(t)}f|_{z=d_2(t)}-\\
&-\phi_2(t,x)u_{zz}|_{z=d_2(t)}f|_{z=d_1(t)},\\
N_3&=N_3(t,x)=Q(t,x)\phi_1(t,x)u_{zz}|_{z=d_1(t)}-\\ &-
P(t,x)\phi_2(t,x)u_{zz}|_{z=d_2(t)},\\
P&=P(t,x)=\phi_{1t}(t,x)
-L_x(\phi_1(t,x))-(\beta_1(t,x)+d'_1(t))u_z|_{z=d_1(t)}-\\ &-\beta_2(t,x)\phi_1^2(t,x),\\
Q&=Q(t,x)=\phi_{2t}(t,x)-L_x(\phi_2(t,x))-(\beta_1(t,x)+d'_2(t))u_z|_{z=d_2(t)}-\\ &-\beta_2(t,x)\phi_2^2(t,x).
\end{split}
\end{equation}

In order for the coefficient of the highest derivative to be positive and the denominators 
in the coefficients do not equal zero, we introduce the cutoff function $S_\delta(y) \in C^{10}(E_1),$
with the following properties
\begin{equation}\label{srezka}
S_\delta(y)\geqslant \frac{\delta}{3}>0,\,
S_\delta(y)=\begin{cases}
y$, $y\geqslant\frac{\delta}{2},\\
\chi(y)$, $\frac{\delta}{3}<y<\frac{\delta}{2},\\
\frac{\delta}{3}$, $y\leqslant\frac{\delta}{3},
\end{cases}
\end{equation}
where $y\in E_1$, $\delta=const$, $\chi(y)\in C^{10}(E_1)$.

So we get following direct problem $(\ref{urav}),$ $(\ref{nachdann})$ 
\begin{equation}\label{urav}
\begin{split}
u_t=L_x(u)+\frac{S_{\delta_1}(N_1(t,x)u)}{S_{\delta_2}(N_2(t,x))}u_{zz}+\beta_1(t,x)u_z&+\\+
\beta_2(t,x)u^2&+\frac{N_3(t,x)}{S_{\delta_2}(N_2(t,x))}f(t,x,z).
\end{split}
\end{equation}


Let us prove the existence of a solution to the direct problem $(\ref{urav}), (\ref{nachdann})$. We apply the weak approximation method \cite{10}.
 We split the problem $(\ref{urav}), (\ref{nachdann})$ and linearize it by a time shift by $\displaystyle\frac{\tau}{3}$ in the nonlinear terms
\begin{equation}\label{uras1}
u^\tau_t=3L_x(u^\tau),  s\tau<t\leqslant (s+\frac{1}{3})\tau, 
\end{equation}
\begin{equation}\label{uras2}
\begin{split}
u^\tau_t&=3(\frac{S_{\delta_1}(N^{\tau}_1(t,x)u^{\tau}(t-\frac{\tau}{3}))}{S_{\delta_2}(N^{\tau}_2(t,x))}u^{\tau}_{zz}+\\ &+\beta_1(t,x)u^{\tau}_z),  (s+\frac{1}{3})\tau<t\leqslant (s+\frac{2}{3})\tau,
\end{split}
\end{equation}
\begin{equation}\label{uras3}
\begin{split}
u^\tau_t&=3(\beta_2(t,x)u^{\tau}u^{\tau}(t-\frac{\tau}{3})+\\ &+\frac{N^{\tau}_3(t,x)}{S_{\delta_2}(N^{\tau}_2(t,x))}f(t,x,z)),
 (s+\frac{2}{3})\tau<t \leqslant (s+1)\tau,
 \end{split}
\end{equation}
\begin{equation}\label{uras0}
u^\tau(0,x,z)=u_0(x,z), \quad x \in E_n, z \in E_1,
\end{equation}
here $s=0,1,...,N-1, \tau N=T, N>0$,\,$N \in \textbf{Z},$ $u^\tau=u^\tau(t)=u^\tau(t,x,z),$ 
$$
N^{\tau}_1=N^{\tau}_1(t,x)=P^{\tau}(t,x)f(t,x,d_2(t))-Q^{\tau}(t,x)f(t,x,d_1(t)),
$$
$$
N^{\tau}_2=N^{\tau}_2(t,x)=\phi_1(t,x)u^{\tau}_{zz}(t-\frac{\tau}{3},x,d_1(t))f|_{z=d_2(t)}-
$$
$$
-\phi_2(t,x)u^{\tau}_{zz}(t-\frac{\tau}{3},x,d_2(t))f|_{z=d_1(t)},
$$
\begin{gather*}
N^{\tau}_3=N^{\tau}_3(t,x)=Q^{\tau}(t,x)\phi_1(t,x)u^{\tau}_{zz}(t-\frac{\tau}{3},x,d_1(t))-\\-
P^{\tau}(t,x)\phi_2(t,x)u^{\tau}_{zz}(t-\frac{\tau}{3},x,d_2(t)),\\
P^{\tau}=P^{\tau}(t,x)=\phi_{1t}(t,x)-L_x(\phi_1(t,x))-(\beta_1(t,x)+d'_1(t))\cdot \\
\cdot u^{\tau}_z(t-\frac{\tau}{3},x,d_1(t))-\beta_2(t,x)\phi_1^2(t,x),\\
Q^{\tau}=Q^{\tau}(t,x)=\phi_{2t}(t,x)-L_x(\phi_2(t,x))-(\beta_1(t,x)+d'_2(t))\cdot \\
\cdot u^{\tau}_z(t-\frac{\tau}{3},x,d_2(t))-\beta_2(t,x)\phi_2^2(t,x).
\end{gather*}

We introduce the notation
\begin{equation}\label{U}
U^{\tau,t_0}(t)=\sum^{10}_{k=0}U^{\tau,t_0}_k(t),
\end{equation}
\begin{equation}\label{U1}
\begin{split}
U^{\tau,t_0}_k(t)=\sup_{t_0<\xi\leq t}\sup_{x\in E_n,z\in E_1}\left|\frac{\partial^k}{\partial z^k}u^\tau(\xi,x,z)\right|,\\
U_k(0)=\sup_{x\in E_n,z\in E_1}\left|\frac{\partial^k}{\partial z^k}u_0(x,z)\right|,
\end{split}
\end{equation}
\begin{equation}\label{U2}
\begin{split}
U^{\tau,t_0}_k(t_0)=\sup_{x\in E_n,z\in E_1}\left|\frac{\partial^k}{\partial z^k}u^\tau(t_0,x,z)\right|,\, t\in (t_0,(n+\frac{p}{3})\tau],
\, \\t_0 \in [0,(n+\frac{p}{3})\tau),\,t>t_0,\, p=1,2,3.
\end{split}
\end{equation}

The functions $U^{\tau,t_0}_k(t),$ $U^{\tau,t_0}_k(t_0),$ $U_k(0)$ are nonnegative and non-decreasing on each half-interval $(s\tau,(s+1)\tau]$.

Let us prove the a priori estimates guaranteeing the compactness of a set of solutions
$\{u^\tau(t,x,z)\}$ of the problem (\ref{uras1}) -- (\ref{uras0}).

Let the half-interval $(s\tau,(s+1)\tau]$ be s-th time step, where $s=\overline{0,N-1}.$

We consider the zero integer step $(s=0)$.

At the first fractional step $(p=1),$ we obtain the following estimate for the solution $u^\tau$ of problem (\ref{uras1}), (\ref{uras0}),  due to (\ref{vhoddan}) and the maximum principle~\cite{13}
\begin{equation}\label{ssh1}
\left|u^\tau(\xi,x,z)\right|\leqslant\sup_{x\in E_n,z\in E_1}\left|u_0(x,z)\right|, \quad 0<\xi\leqslant \frac{\tau}{3}.
\end{equation}

We obtain the following estimates using differentiating equations (\ref{uras1}), (\ref{uras0}) 
with respect to $z$ from one to ten times, respectively, due to (\ref{vhoddan}) and the maximum principle \cite{13}
\begin{equation}\label{osh1}
\left|\frac{\partial^k}{\partial z^k}u^\tau(\xi,x,z)\right|\leqslant \sup_{x\in E_n,z\in E_1}\left|\frac{\partial^k}{\partial z^k}u_0(x,z)\right|, \quad k=\overline{1,10}, \quad 0<\xi\leqslant \frac{\tau}{3}.
\end{equation}

We obtain the following estimate from (\ref{ssh1}), (\ref{osh1}) through (\ref{U}), (\ref{U1})
\begin{equation}\label{UU1}
U^{\tau,0}(t)\leqslant U(0), \qquad 0<t\leqslant \frac{\tau}{3}.
\end{equation}

At the second fractional step $(p=2),$ we obtain the following estimate for the solution of equation (\ref{uras2}) 
with initial data $u^\tau(\frac{\tau}{3},x,z)$
 due to (\ref{srezka}), (\ref{vhoddan}), (\ref{U}) -- (\ref{U2}) and the maximum principle~\cite{13}
\begin{equation}\label{UU2}
U^{\tau,\frac{\tau}{3}}(t)\leqslant (1+U^{\tau,\frac{\tau}{3}}(\frac{\tau}{3}))e^
{C\tau(1+U^{\tau,\frac{\tau}{3}}(\frac{\tau}{3}))^{20}}-1, \frac{\tau}{3}<t\leqslant\frac{2\tau}{3}.
\end{equation}

Collectively, due to (\ref{UU1}), (\ref{UU2}) on the first and second fractional steps, we get
\begin{equation}\label{sovU12}
U^{\tau,0}(t)\leqslant (1+U(0))e^
{C\tau(1+U(0))^{20}}-1,  0<t\leqslant\frac{2\tau}{3}.
\end{equation}

At the third fractional step $(p=3)$, integrating the equation (\ref{uras3}) with 
$t \in \big(\frac{2\tau}{3},\xi],$ $\frac{2\tau}{3}<\xi\leqslant\tau,$ we receive the equality
\begin{equation*}
\begin{split}
u^\tau(\xi)=u^\tau(\frac{2\tau}{3})&+3\int^{\xi}_{\frac{2\tau}{3}}(\beta_2(\eta,x)u^\tau(\eta) u^\tau(\eta-\frac{\tau}{3},x,z)
+\\ &+\frac{N_3^\tau(\eta,x)}{S_{\delta_2}(N_2^\tau(\eta,x))}f(\eta,x,z))d\eta.
\end{split}
\end{equation*}

The last relation implies the inequality
\begin{equation*}
\begin{split}
|u^\tau(\xi)| \leqslant|u^\tau(\frac{2\tau}{3})| &+3\int^{\xi}_{\frac{2\tau}{3}}(|\beta_2(\eta,x)|u^\tau(\eta)||u^\tau(\eta-\frac{\tau}{3})|+\\ &+
\frac{|N_3^\tau(\eta,x)|}{|S_{\delta_2}(N_2^\tau(\eta,x))|}|f(\eta,x,z)|)d\eta,
\end{split}
\end{equation*}
where $\frac{2\tau}{3}<\xi\leqslant t \leqslant \tau.$

Since this inequality holds for all $x,z$ we replace the functions of the integral terms by their 
exact upper bounds with respect to $x\in E_n, z\in E_1,$ and then replace the function $\left|u^\tau\right|,$ 
on the left-hand side of the inequality by $\displaystyle \sup_{x\in E_n z\in E_1}\left|u^\tau\right|$, using notations (\ref{U2}), we obtain
\begin{equation}\label{Uk0}
\begin{split}
U_0^{\tau,\frac{2\tau}{3}}(t)\leqslant U_0^{\tau,\frac{2\tau}{3}}(\frac{2\tau}{3})+C\int_{\frac{2\tau}{3}}^t
(U^{\tau,\frac{2\tau}{3}}_0(\eta)U^{\tau,\frac{2\tau}{3}}_0(\eta-\frac{\tau}{3})+\\
+U_2^{\tau,\frac{2\tau}{3}}(\eta-\frac{\tau}{3})+U_2^{\tau,\frac{2\tau}{3}}(\eta-\frac{\tau}{3})
U_1^{\tau,\frac{2\tau}{3}}(\eta-\frac{\tau}{3}))d\eta.
\end{split}
\end{equation}

Further, in the same way, differentiating equations (\ref{uras3}) with respect to $z$ from one to 10 times, 
 we get
\begin{equation}\label{Uk}
\begin{split}
U_k^{\tau,\frac{2\tau}{3}}(t)\leqslant U_k^{\tau,\frac{2\tau}{3}}(\frac{2\tau}{3})+C\int_{\frac{2\tau}{3}}^t
\sum_{q=0}^k(U^{\tau,\frac{2\tau}{3}}_{k-q}(\eta)U^{\tau,\frac{2\tau}{3}}_q(\eta-\frac{\tau}{3})+\\
+U_2^{\tau,\frac{2\tau}{3}}(\eta-\frac{\tau}{3})+U_2^{\tau,\frac{2\tau}{3}}(\eta-\frac{\tau}{3})
U_1^{\tau,\frac{2\tau}{3}}(\eta-\frac{\tau}{3}))d\eta, \quad k=\overline{1,10}.
\end{split}
\end{equation}

Adding (\ref{Uk0}) and (\ref{Uk}), by virtue of (\ref{U}) we receive
\begin{equation*}
\begin{split}
U^{\tau,\frac{2\tau}{3}}(t)\leqslant U^{\tau,\frac{2\tau}{3}}(\frac{2\tau}{3})+C\int_{\frac{2\tau}{3}}^t
(&U^{\tau,\frac{2\tau}{3}}(\eta)U^{\tau,\frac{2\tau}{3}}(\frac{2\tau}{3})+U^{\tau,\frac{2\tau}{3}}(\frac{2\tau}{3})+\\ &+
U^{\tau,\frac{2\tau}{3}}(\frac{2\tau}{3})U^{\tau,\frac{2\tau}{3}}(\frac{2\tau}{3}))d\eta,
\end{split}
\end{equation*}
where $C \geqslant 1$-constant, independent of $\tau$.

To the last inequality we apply the Gronwall lemma \cite{14},
then
\begin{equation*}
U^{\tau,\frac{2\tau}{3}}(t)\leqslant (U^{\tau,\frac{2\tau}{3}}(\frac{2\tau}{3})+1)e^{2C\tau(U^{\tau,\frac{2\tau}{3}}(\frac{2\tau}{3})
+1)}-1, \quad \frac{2\tau}{3}< t \leqslant \tau.
\end{equation*}

Consequently, due to (\ref{sovU12}) and the last inequality at the zero whole step the following estimate holds
\begin{equation*}
U^{\tau,0}(t)\leqslant (U(0)+1)e^{3C\tau(U(0)+1)^{20}e^{C\tau(1+U(0))^{20}}}-1, \quad 0< t \leqslant \tau.
\end{equation*}

Repeating similar arguments at the first whole step, we obtain
\begin{equation*}
U^{\tau,\tau}(t)\leqslant (U^{\tau,\tau}(\tau)+1)e^{3C\tau(U^{\tau,\tau}(\tau)+1)^{20}e^{C\tau(1+U^{\tau,\tau}(\tau))^{20}}}-1, \quad \tau< t \leqslant 2\tau.
\end{equation*}

Assuming that $\tau$ is sufficiently small and the inequality\\
$e^{60C\tau(U(0)+1)^{20}e^{C\tau(1+U(0))^{20}}}\leqslant 2$ holds, at the zero and first whole steps we get
\begin{equation*}
U^{\tau,0}(t)\leqslant (U(0)+1)e^{9C\tau(U(0)+1)^{20}e^{2C\tau(1+U(0))^{20}}}-1, \quad 0< t \leqslant 2\tau.
\end{equation*}
Analogous reasoning, at the $s-$th whole step $(s<N)$ we obtain
\begin{equation*}
U^{\tau,s\tau}(t)\leqslant (U^{\tau,s\tau}(s\tau)+1)e^{3C\tau(U^{\tau,s\tau}(s\tau)+1)^{20}e^{C\tau(U^{\tau,s\tau}(s\tau)+1)^{20}}}-1, 
\end{equation*}
$s\tau<t\leqslant (s+1)\tau.$

Consequently, at $s$ whole steps, we getting
\begin{equation*}
U^{\tau,0}(t)\leqslant (U(0)+1)e^{3(2s+1)C\tau(U(0)+1)^{20}e^{\psi(s)C\tau(1+U(0))^{20}}}-1,\quad 0<t\leqslant (s+1)\tau,
\end{equation*}
\begin{equation*}
\psi(s)=\begin{cases}
1$, $s=0,\\
2$, $s=1,2,...,N-1.\\
\end{cases}
\end{equation*}

Hence, the following estimate is true
\begin{equation*}
U^{\tau,0}(t)\leqslant (U(0)+1)e^{3Ct_*(U(0)+1)^{20}e^{t_*C(1+U(0))^{20}}}-1,\quad 0<t\leqslant t_*,
\end{equation*}
where $t_*$ satisfies the inequality 
\begin{equation}\label{t*}
e^{60Ct_*(U(0)+1)^{20}e^{Ct_*(1+U(0))^{20}}}\leqslant 2.
\end{equation}

And, therefore, taking into account the notation (\ref{U}), (\ref{U1}) uniformly with respect to~$\tau$
\begin{equation}\label{ocenka1}
|\frac{\partial^k}{\partial z^k} u^\tau (t,x,z)|\leqslant C, \quad k=\overline{0,10}, \quad (t,x,z)\in G_{[0,t_*]}.
\end{equation}

After differentiating problem (\ref{uras1}) -- (\ref{uras0}) with respect to $x_i,$ $x_j$, $x_l$ and $x_m$, $i, j, l ,m=\overline{1,n},$ we obtain equations that can be regarded as linear with coefficients uniformly bounded in $\tau$. Arguing by analogy, considering (\ref{ocenka1}), we obtain estimate 
\begin{equation}\label{uC}
\begin{split}
|D^\gamma_x\frac{\partial^k}{\partial z^k}u^{\tau}(t,x,z)|\leqslant C,\quad k=\overline{0,10-2|\gamma|},
\,|\gamma| \leqslant 4,
 (t,x,z)\in G_{[0,t_*]}.
\end{split}
\end{equation}

We obtain from (\ref{uC}) and (\ref{uras1}) -- (\ref{uras0}) uniformly with respect to $\tau$
\begin{equation*}
|u_t^{\tau}(t,x,z)|\leqslant C, \quad (t,x,z)\in G_{[0,t_*]}.
\end{equation*}

We differentiate equations (\ref{uras1}) -- (\ref{uras0}) once with respect to $z$. By (\ref{uC}), the right-hand 
side of the equations obtained is uniformly bounded in $\tau$, and consequently the left-hand side is also uniformly bounded in $\tau$
\begin{equation*}
|u_{tz}^{\tau}(t,x,z)|\leqslant C, \quad (t,x,z)\in G_{[0,t_*]}.
\end{equation*}

By analogy, uniformly with respect to $\tau$
\begin{equation*}
|\frac{\partial^k}{\partial z^k}D_x^\lambda u_t^{\tau}(t,x,z)|\leqslant C,\quad k=\overline{0,4},\quad|\lambda|\leqslant 2,
\quad (t,x,z)\in G_{[0,t_*]}.
\end{equation*}

Thus, the following estimate holds uniformly with respect to $\tau$  
\begin{equation}\label{utxz}
\begin{split}
|\frac{\partial}{\partial t}\frac{\partial^k}{\partial z^k}D^\lambda_x u^{\tau}(t,x,z)|+
|\frac{\partial}{\partial x_i}\frac{\partial^k}{\partial z^k}D^\lambda_x u^{\tau}(t,x,z)|+\\+
|\frac{\partial}{\partial z}\frac{\partial^k}{\partial z^k}D^\lambda_x u^{\tau}(t,x,z)|\leqslant C,
 k=\overline{0,4}, |\lambda| \leqslant 2, (t,x,z)\in G_{[0,t_*]}.
\end{split}
\end{equation}

The estimate (\ref{uC}) implies the uniform boundedness in $\tau$ of the family \\ 
$\left\{D_x^\gamma\frac{\partial^k}{\partial z^k} u^\tau\right\}$ in $G_{[0,t_*]}$, 
and from (\ref{uC}), (\ref{utxz}) their equicontinuity with respect to $t,$ $x$ and $z$ is equicontinuous 
in $G_{[0,t_*]}$. Therefore, for any fixed $\gamma,$ $k,$ $|\gamma|\leq 2,$ $k=\overline{0,4},$ 
by the Arzela theorem \cite{15} the set $\left\{D_x^\gamma\frac{\partial^k}{\partial z^k}u^\tau\right\}$ 
is compact in $C(G^M_{[0,t_*]}),$ $M>0$ is an integer, $G^M_{[0,t_*]}=\{(t,x,z)| t\in[0,T], |x|\leqslant M,$ $ |z|\leqslant~M \}.$

In a diagonal way, we choose a subsequence $\left\{u^{\tau}\right\}$ (we do not change the notation) converging together with the corresponding derivatives with respect to $x$ and $z$ to some function $u$ in $G_{[0,t_*]}$, and also uniformly in each $G^M_{[0,t_*]}.$ The function $u$ is continuous, has derivatives of the corresponding order with respect to $x$ and $z$ that are continuous in $G_{[0,t_*]}$, and satisfies the initial data (\ref{urnachdan}) and inequality
\begin{equation}\label{DC}
\left|D_x^\beta\frac{\partial^k}{\partial z^k}u(t,x,z)\right|\leqslant C,\, k=\overline{0,4},\, |\beta|\leq 2,\, (t,x,z)\in G_{[0,t_*]}.
\end{equation}

Since $D^\gamma_\alpha\frac{\partial^k}{\partial z^k} u^\tau\rightrightarrows D^\gamma_x\frac{\partial^k}{\partial z^k} u$ on $G^M_{[0,t_*]}$ $\forall M>0,$ $|\gamma|\leq 2,$ $k=\overline{0,4}$ and the inequality (\ref {DC}), it can be proved (the proof is similar to the proof of Theorem~1.4 of the weak approximation method \cite{10}) that the function $u$ is a solution of the problem (\ref{urav}), (\ref{nachdann}) in $ G^M_{[0, t_*]} $ for any fixed $ M $, and since $ M $ is arbitrary, it is also in $ G_{[0,t_*]} $.

The function $u(t,x,z)$ belongs to the class (\ref{C124}) and the estimate (\ref{DC}) is true.

In order that to prove the existence of a solution of problem (\ref{zadN}), (\ref{nachdann}), it is necessary to remove the cutoff functions in equation (\ref{urav}). For this, we prove that for $(t,x,z)\in G_{[0,t^*]}$,
\begin{gather*}
N_1(t,x)u(t,x,z)\geqslant \frac{\delta_1}{2},\quad
N_2(t,x)\geqslant \frac{\delta_2}{2}.
\end{gather*}

We differentiate $N_1(t,x)u(t,x,z)$ and $N_2(t,x)$ ($N_1(t,x)$, $N_2(t,x)$ from $(\ref{N123}))$ with respect to $t$
\begin{equation}\label{MM}
\begin{split}
M_1(t,x)&=(N_1(t,x)u(t,x,z))'_t=u(t,x,z)\bigl(P'_t f(t,x,d_2(t))+\\      
&+P\bigl(f'_t(t,x,d_2(t))+f'_z(t,x,d_2(t))d'_{2}(t)\bigr)-Q'_tf(t,x,d_1(t))-\\ &-Q\bigl(f'_t(t,x,d_1(t))+f'_z(t,x,d_1(t))d'_{1}(t)\bigr)\bigr)+\\ &+\bigl(P(t,x)f(t,x,d_2(t))-Q(t,x)f(t,x,d_1(t))\bigr)u_t(t,x,z), \\
M_2(t,x)&=(N_2(t,x))'_t=\bigl(u_{zzt}(t,x,d_1(t))+u_{zzz}(t,x,d_1(t))d'_1(t)\bigr)\cdot\\ &\cdot f(t,x,d_2(t))\phi_1(t,x)
+u_{zz}(t,x,d_1(t))\bigl(f'_t(t,x,d_2(t))+\\ &+f'_z(t,x,d_2(t))d'_{2}(t)\bigr)\phi_1(t,x)+u_{zz}(t,x,d_1(t))f(t,x,d_2(t)) \cdot\\ &\cdot\phi_{1t}(t,x)-\bigl(f'_t(t,x,d_1(t))+f'_z(t,x,d_1(t))d'_{1}(t)\bigr)\cdot\\ &\cdot u_{zz}(t,x,d_2(t)) \phi_2(t,x)-f(t,x,d_1(t))\bigl(u_{zzt}(t,x,d_2(t))+\\&+u_{zzz}(t,x,d_2(t))d'_2(t)\bigr)\phi_2(t,x)-\\ &-u_{zz}(t,x,d_2(t))f(t,x,d_1(t))\phi_{2t}(t,x), 
\end{split}
\end{equation}
where
\begin{equation*}
\begin{split}
P'_t&=\phi_{1tt}-L_{xt}(\phi_1(t,x))-(\beta_{1t}(t,x)+d''_1(t))u_z(t,x,d_1(t))-\\ &-\beta_{1}(t,x)(u_{zt}(t,x,d_1(t))+u_z(t,x,d_1(t))d'_1(t))-\\ &-\beta_{2t}(t,x)\phi_1^2(t,x)-2\beta_2(t,x)\phi_1(t,x)\phi_{1t}(t,x)-\\ &-(u_{zt}(t,x,d_1(t))+u_{z}(t,x,d_1(t))d'_1(t))d'_1(t),
\end{split}
\end{equation*}
\begin{equation*}
\begin{split}
Q'_t&=\phi_{2tt}-L_{xt}(\phi_2(t,x))-(\beta_{1t}(t,x)+d''_2(t))u_z(t,x,d_2(t))-\\ &-\beta_{1}(t,x)(u_{zt}(t,x,d_2(t))+u_z(t,x,d_2(t))d'_2(t))-\\ &-\beta_{2t}(t,x)\phi_2^2(t,x)-2\beta_2(t,x)\phi_2(t,x)\phi_{2t}(t,x)-\\
&-(u_{zt}(t,x,d_2(t))+u_{z}(t,x,d_2(t))d'_2(t))d'_2(t),
\end{split}
\end{equation*}
\begin{equation*}
\begin{split}
L_{xt}(\phi_1(t,x))&=\sum_{i,j=1}^n\Bigl((\alpha_{ij})'\frac{\partial^2 \phi_1}{\partial x_i \partial x_j}+
\alpha_{ij}\frac{\partial\phi_1}{\partial x_i \partial x_j \partial t}\Bigr)+\\
&+\sum_{i=1}^n\Bigl((\alpha_{i})'\frac{\partial\phi_1}{\partial x_i}+\alpha_{i}\frac{\partial^2\phi_1}{\partial x_i\partial t}\Bigr),
\end{split}
\end{equation*}
\begin{equation*}
\begin{split}
L_{xt}(\phi_2(t,x))&=\sum_{i,j=1}^n\Bigl((\alpha_{ij})'\frac{\partial^2 \phi_2}{\partial x_i \partial x_j}+
\alpha_{ij}\frac{\partial\phi_2}{\partial x_i \partial x_j \partial t}\Bigr)+\\  &+\sum_{i=1}^n\Bigl((\alpha_{i})'\frac{\partial\phi_2}{\partial x_i}+\alpha_{i}\frac{\partial^2\phi_2}{\partial x_i\partial t}\Bigr).
\end{split}
\end{equation*}


By virtue of  (\ref{vhoddan}), (\ref{DC})
\begin{equation}\label{K}
\begin{split}
|M_1(t,x)|\leqslant K_1,\quad
|M_2(t,x)|\leqslant K_2,
\end{split}
\end{equation}
here $K_1$, $K_2$ - are constants depending on $\delta_1$,\,$\delta_2,$\,$C.$


We integrate expressions $(\ref{MM})$ with respect to $t$ in the range from $0$ to $t$, we obtain
\begin{gather*}
N_1(t,x)u(t,x,z)=N_1(0,x)u(0,x,z)+\int_0^tM_1(\eta,x)d\eta,\\
N_2(t,x)=N_2(0,x)+\int_0^tM_2(\eta,x)d\eta.
\end{gather*}

By virtue of (\ref{N0x}), (\ref{K}) $N_1(t,x)u(t,x,z)\geqslant\delta_1-K_1t$, $N_2(t,x)\geqslant\delta_2-K_2t$
\begin{equation}\label{KK}
\begin{split}
N_1(t,x)u(t,x,z)\geqslant\frac{\delta_1}{2},\quad
N_2(t,x)\geqslant \frac{\delta_2}{2},\quad t\in [0,t^*].
\end{split}
\end{equation}

By the definition of the cutoff function (\ref{srezka}) and (\ref{KK}), we obtain \\
$S_{\delta_1}(N_1(t,x)u(t,x,z))= N_1(t,x)u(t,x,z)$, and $S_{\delta_2}(N_2(t,x))=N_2(t,x)$ with $t\in [0,t^*]$, 
$t^*=\min\Bigl(t_*,\frac{\delta_1}{2K_1}, \frac{\delta_2}{2K_2}\Bigr)$, $t_*$ see in (\ref{t*}).

Thus, in the equation $(\ref{urav})$, the cutoffs are removed. The function $u(t,x,z)$ satisfies the equation $(\ref{zadN})$.

The coefficients $a(t,x)$ and $b(t,x)$ can be written in the form $(\ref{ab})$.

Thus, we have proved the existence of a solution $u(t,x,z)$ of the direct problem $(\ref{zadN})$, $(\ref{nachdann})$ in the class $C^{1,2,4}_{t,x,z}(G_{[0,t^*]})$. 

Let us prove that the triple of functions $u(t,x,z),$\,$a(t,x),$\,$b(t,x)$ 
is the solution of the inverse problem (\ref{uravnenie}) -- (\ref{pereopr}),  
where $a(t,x)$ and $b(t,x)$ are defined in (\ref{ab}). Since $u(t,x,z)$ 
is the solution of the direct problem (\ref{zadN}), (\ref{nachdann}), 
substitu\-ting $u(t,x,z),$ $a(t,x),$ $b(t,x)$ in (\ref{uravnenie}), we obtain the correct identity.

According to (\ref{vhoddan}), (\ref{DC}) from (\ref{ab}), (\ref{zadN}), we obtain that the triple 
of functions $u(t,x,z),$ $a(t,x),$ $b(t,x)$ belongs to the class (\ref{Z})
and satisfies the inequalities (\ref{beta}), (\ref{bet}).

Using conditions (\ref{sogl}) and equation (\ref{uravnenie}), we can prove that the overdetermi\-nation conditions (\ref{pereopr}) are satisfied.

The existence in the class $Z(t^*)$ of the solution $u(t,x,z)$, $a(t,x)$, $b(t,x)$ of problem (\ref{uravnenie}) -- (\ref{pereopr}) satisfying relations (\ref{uravnenie}) -- (\ref{pereopr}) is proved.

The uniqueness of the solution is proved in the standard way. It is assumed that there are two solutions to the original problem  (\ref{uravnenie}) -- (\ref{pereopr}) that satisfy conditions (\ref{beta}), (\ref{bet}), and it is proved that the difference of these solutions is zero.

Thus the theorem is proved.

\section{Example}
In $G_{[0,1]}=\{(t,x,z)\bigl|\,0\leqslant t \leqslant 1,\, x\in E_1,\, z\in E_1\}$ 
 we have the Cashy problem
\begin{multline}\label{urprim}
\frac{\partial u}{\partial t}=(t^2+1)u_{xx}+(t+1)u_x+a(t,x)uu_{zz}+\sin(t+x)u_z+(3+\cos(x+t))u^2+\\
+b(t,x)(-10\sin(t+z)+\sin(t+x+z)),
\end{multline}
\begin{equation}\label{urnachprim}
u(0,x,z)=u_0(x,z)=\cos x+\cos z+4, \quad (x,z)\in E_{2},
\end{equation}
\begin{equation*}
\begin{split}
f(t,x,z)&=(-10\sin(t+z)+\sin(t+x+z)), \alpha_{11}(t)=t^2+1,  \alpha_1(t)=t+1,\\  
u_0(x,z)&=\cos x+\cos z+4, \beta_1(t,x)=\sin(t+x), \beta_2(t,x)=3+\cos(t+x).
\end{split}
\end{equation*}

 We assume, that overdeterminition conditions are hold on two different hyperspace
  $z=d_1(t)=-t^2$ and $z=d_2(t)=-t^2+1$:
\begin{equation}\label{per1}
\begin{split}
u(t;x;-t^2)=\phi_1(t,x)=\cos(t+x)+\cos (t-t^2)+4, \\
u(t;x;-t^2+1)=\phi_2(t,x)=\cos(t+x)+\cos(t-t^2+1)+4, 
\end{split}
\end{equation}
where $(t,x)\in \Pi_{[0,1]},$ $\Pi_{[0,1]}=\{(t,x)|\, 0\leqslant t \leqslant 1,\, x \in
E_1\}$ and $\phi_1(t,x),\, \phi_2(t,x)$ --- 
satysfied of 
\begin{equation*}
\begin{split}
\phi_1(0,x)=u_0(x,0)=\cos x+5,
\phi_2(0,x)=u_0(x,1)=\cos x+\cos 1+4, 
\end{split}
\end{equation*}
where $x\in E_1$.

The sought coefficients are
\begin{equation*}
a(t,x)=\frac{N_1(t,x)}{N_2(t,x)},\quad b(t,x)=\frac{N_3(t,x)}{N_2(t,x)},
\end{equation*}
here
\begin{equation*}
\begin{split}
N_1(t,x)&=(-\sin(t+x)-\sin (t-t^2)(1-2t)+(t^2+1)\cos(t+x)+(t+1)\cdot \\ &\cdot\sin(t+x)+
(\sin(t+x)-2t)\sin(t-t^2)-(3+\cos(t+x))\cdot \\ &\cdot(\cos(t+x)+\cos (t-t^2)+4)^2)(-10\sin(t-t^2+1)+\\ &+\sin(t+x-t^2+1))-(-\sin(t+x)-\sin(t-t^2+1)(1-2t)+\\
&+(t^2+1)\cos(t+x)+(t+1)\sin(t+x)+(\sin(t+x)-2t)\cdot \\ &\cdot\sin(t-t^2+1)
-(3+\cos(t+x))(\cos(t+x)+\cos(t-t^2+1)+\\ &+4)^2)(-10\sin(t-t^2)+\sin(t-t^2+x)),
\end{split}
\end{equation*}
\begin{equation*}
\begin{split}
N_2(t,x)&=-(\cos(t+x)+\cos(t-t^2)+4)\cos(t-t^2)(-10\sin(t-t^2+1)+\\ &+\sin(t+x-t^2+1))+(\cos(t+x)+\cos(t-t^2+1)+4)\cdot \\ &\cdot\cos(t-t^2+1)(-10\sin(t-t^2)+\sin(t+x-t^2)),
\end{split}
\end{equation*}
\begin{equation*}
\begin{split}
N_3(t,x)&=(-\cos(t+x)-\cos(t-t^2)-4)\cos(t-t^2)(-\sin(t+x)-\\ &-\sin (t-t^2+1)(1-2t)+(t^2+1)\cos(t+x)+(t+1)\sin(t+x)+\\ &+(\sin(t+x)-2t)\sin(t-t^2+1)-(3+\cos(t+x))(\cos(t+x)+\\ &+\cos(t-t^2+1)+4)^2)+(\cos(t+x)+\cos(t-t^2+1)+4)\cdot \\ &\cdot\cos(t-t^2+1)(-\sin(t+x)-\sin(t-t^2)(1-2t)+\\
&+(t^2+1)\cos(t+x)+(t+1)\sin(t+x)+(\sin(t+x)-2t)\cdot \\ &\cdot\sin(t-t^2)-(3+\cos(x+t))(\cos(t+x)+\cos (t-t^2)+4)^2).
\end{split}
\end{equation*}

The input data are fairly smooth, have all
continuous derivatives from $(\ref{vhoddan}).$


Functions $ N_1(0,x)u(0,x,z)$, $N_2(0,x)$ from $(\ref{N0x})$ in $\Pi_{[0,1]}$  have the following form
\begin{equation*}
\begin{split}
 N_1(0,x)u(0,x,z)=&((\cos(x)-(3+\cos(x))(\cos(x)+5)^2)(-10\sin(1)+\\ &+\sin(x+1))
-(-\sin(1)+\cos(x)+\sin(x) \sin(1)-\\ &-(3+\cos(x))(\cos(x)+ \cos(1)+4)^2)\sin(x))\cdot \\ &\cdot(\cos x+\cos z +4) \geqslant \delta_1\approx 600,\\
N_2(0,x)=-(&\cos(x)+5)(-10\sin(1)+\sin(x+1))+(\cos(x)+\\
&+\cos(1)+4)\cos(1)\sin(x) \geqslant \delta_2\approx 37.
\end{split}
\end{equation*}
Therefore, the conditions (\ref{vhoddan}), (\ref{N0x}) are fulfilled.

The solution of  (\ref{urprim}) -- (\ref{per1}) is a triple of functions
$u(t,x,z),$ $a(t,x),$ $b(t,x),$ which have the form
\begin{equation*}
a(t,x)=\frac{N_1(t,x)}{N_2(t,x)},\quad
b(t,x)=\frac{N_3(t,x)}{N_2(t,x)},
\end{equation*}
\begin{equation*}
u(t,x,z)=\cos(t+x)+\cos(t+z)+4.
\end{equation*}

\section{Conclusion}
 
In the presented article, a theorem of existence and uniqueness of the classical solution in the class of smooth bounded functions of the problem of determining the source function and the coefficient in the product of an unknown function and its second derivative with respect to a spatial variable in a multidimensional quasilinear parabolic equation is proved.

An example of input data satisfying the conditions of the proved theorem of existence and uniqueness of the classical solution of the posed inverse problem is given.

The obtained results are new, have theoretical value and can be used in the construction of a general theory 
of inverse problems.


\bigskip
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