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\usetheme{Madrid}
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\title{Inverse problems for quasiparabolic and quasihyperbolic equations}
\date{2017}
\author{Akimova Ekaterina}
\institute[Foo and Bar]{
  \begin{tabular}[h]{cc}

Novosibirsk State University

  \end{tabular}
}
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\begin{document}
	\begin{frame}
		\titlepage
	\end{frame}
\begin{frame}
		\frametitle{Quasiparabolic and quasihyperbolic equations}

                     \[
u_{ttt}+u_{xx}+\mu u=f(x,t)+q(t)h(x,t),
\]
\[
u_{tttt}+u_{xx}+\mu u=f(x,t)+q(t)h(x,t).
\]
		Case 1:
		\begin{itemize}
			\item $u(0,t)=0$ -  redefinition condition,
			\item $u(1,t)=u_x(0,t)=0$ - basic conditions.
		\end{itemize}
		Case 2:
		\begin{itemize}
			\item $u_x(0,t)=0$ - redefinition condition,
			\item $u(0,t)=u(1,t)=0$ - basic conditions.
		\end{itemize}
		Case 3:
		\begin{itemize}
			\item $u(0,t)=0$ - redefinition condition,
			\item $u_x(0,t)=u_x(1,t)=0$ - basic conditions.
		\end{itemize}
	\end{frame}


	\begin{frame}
		\frametitle{Formulation of the problem for quasiparabolic equations}

		$Q=(0,1)\times(0,T)$-rectangle,

		$\mu(x,t)$,
		$ f (x,t)$,
		$ h(x,t)$  - prescribed function, defined in $\overline{Q}$.
		\begin{rutask}
			Find function $ \ u(x,t)$ and $\ q(t)$, connected by the equation
		\[
		u_{ttt}+u_{xx}+\mu(x,t)u=f(x,t)+q(t)h(x,t)
		\]
		in area Q.
		Conditions are satisfied for the function $u(x,t)$
		\[
		u(x,0)=u_t(x,0)=u(x,T)=0,  x \in [0,1]
		\]
 and one of the group:

			1. $u(0,t)=0$ -  redefinition condition,

$u(1,t)=u_x(0,t)=0$ - basic conditions.

			2. $u_x(0,t)=0$ -redefinition condition,
			$u(0,t)=u(1,t)=0$ - basic conditions.

			3. $u(0,t)=0$ - redefinition condition,
			$u_x(0,t)=u_x(1,t)=0$ - basic conditions.
			\end{rutask}
	\end{frame}

\begin{frame}
		\frametitle{Case 1}
$u(0,t)=0$ --- redefinition condition,

$u(1,t)=u_x(0,t)=0$ --- basic conditions.
                     \[
q(t)=\frac{u_{xx}(0,t)-f(0,t)}{h(0,t)}, h(0,t)\neq 0;
\]
Notation:
\[
f_1(x,t)=f(x,t)-\frac{f(0,t)h(x,t)}{h(0,t)},
\]
\[
h_1(x,t)=\frac{h(x,t)}{h(0,t)}.
\]

	\end{frame}
	\begin{frame}
		\frametitle{Problem 1}
                    Notation: $v(x,t)=u_{xx}(x,t)$, $f_2(x,t)=f_{1xx}(x,t)$, $h_2(x,t)=h_{1xx}(x,t)$,
		 $f_1(x,t)=f(x,t)-\frac{f(0,t)h(x,t)}{h(0,t)}$, $h_1(x,t)=\frac{h(x,t)}{h(0,t)}$, $\beta(t)=-h_{1x}(1,t)$ and $\alpha(t)=h_1(1,t)$.

Consider the problem
		\[
		\begin{cases}
		 v_{ttt}(x,t)+v_{xx}(x,t)+\mu v(x,t)=f_2(x,t)+ h_2(x,t)v(0,t), \\
		v(1,t)- \alpha (t) v(0,t)=0, x\in[0,1], t\in[0,T],\\
		v_x(0,t)+\beta (t)v(0,t)=0, x\in[0,1], t\in[0,T].
		\end{cases}
		\]

	\end{frame}

	\begin{frame}
		\frametitle{Existence theorem 1}
                      $h(x,t)\in C^2(\overline{Q})$, and conditions are met when $ t\in [0,T]$, $x \in [0,1]$:
\[
h(0,t)\neq 0,
h(1,t)\equiv0,
\mu\textless 0,
-\frac{h_x(1,t)}{h(0,t)}\leqslant0,
\]
\[
\mu^3+\left(4\left(\max_{\overline {Q}}\left|\frac{h_{xx}(x,t)}{h(0,t)}\right|\right)^{2}-\mu\right)\left(\max\limits_{\overline{Q}}\left|\frac{h_{xx}(x,t)}{h(0,t)}\right|\right)^2\leqslant0,
\]
\[
h_t(0,t)h_x(x,t)-h_{xt}(x,t)h(0,t)\geqslant0.
\]
Then inverse problem I with redefinition condition $u(0,t)=0$ and with basic conditions $u(1,t)=u_x(0,t)=0$ has a solution $\{u(x,t), q(t)\}$ such that $u(x,t) \in L_2(Q)$, $u_{xx}(x,t) \in L_2(Q)$, $u_{ttt}(x,t) \in L_2(Q)$, $q(t) \in L_2([0,T])$, for any function  $f(x,t)$, such that $f(x,t), f_{xx}(x,t)\in L_2(Q)$,
\[
f(1,t)-\frac{f(0,t)h(1,t)}{h(0,t)}=0,
f_x(1,t)-\frac{f(0,t)h_x(1,t)}{h(0,t)}=0.
\]
	\end{frame}

	\begin{frame}
		\frametitle{Case 2}
$u_x(0,t)=0$ --- redefinition condition,

$u(0,t)=u(1,t)=0$ --- basic conditions.
                     \[
q(t)=\frac{u_{xxx}(0,t)-f_x(0,t)}{h_x(0,t)}, h_x(0,t)\neq 0;
\]
Notations:
\[
\tilde{f_1}(x,t)=f(x,t)-\frac{f_x(0,t)h(x,t)}{h_x(0,t)},
\]
\[
\tilde{h_1}(x,t)=\frac{h(x,t)}{h_x(0,t)}.
\]

	\end{frame}
	\begin{frame}
		\frametitle{Problem 2}
                     Notations: $v(x,t)=u_{xx}(x,t)$, $\tilde{f}_2(x,t)=\tilde{f}_{1xx}(x,t)$, $\tilde{h}_2(x,t)=\tilde{h}_{1xx}(x,t)$, 				$\tilde{f}_1(x,t)=f(x,t)-\frac{f_x(0,t)h(x,t)}{h_x(0,t)}$, $\tilde{h}_1(x,t)=\frac{h(x,t)}{h_x(0,t)}$,
		$\tilde{\alpha}(t)=\tilde{h}_1(1,t)$ и $\tilde{\beta}(t)=-\tilde{h}_{1}(0,t)$.

		Consider the problem:
		\[
		\begin{cases}
		 v_{ttt}(x,t)+v_{xx}(x,t)+\mu v(x,t)=\tilde{f_2}(x,t)+\tilde{ h_2}(x,t)v(0,t), \\
		v(1,t)-\tilde{\alpha} (t) v_x(0,t)=0, x\in[0,1], t\in[0,T],\\
		v(0,t)+\tilde{\beta }(t)v_x(0,t)=0, x\in[0,1], t\in[0,T].
		\end{cases}
		\]

	\end{frame}

	\begin{frame}
		\frametitle{Existence theorem 2}
                     $h(x,t)\in C^2(\overline{Q})$ and conditions are met when $ t\in [0,T]$, $x \in [0,1]$:
\[
h_x(0,t)\neq 0,
h(1,t)\equiv0,
\mu\textless 0,
-\frac{h(0,t)}{h_x(0,t)}\leqslant0,
\]
\[
\mu^3+\left(4\left(\max_{\overline {Q}}\left|\frac{h_{xx}(x,t)}{h_x(0,t)}\right|\right)^2-\mu\right)
\left(\max\limits_{\overline{Q}}\left|\frac{h_{xx}(x,t)}{h_x(0,t)}\right|\right)^2\leqslant0,
\]
\[
h_{xt}(0,t)h(x,t)-h_{t}(x,t)h_x(0,t)\leqslant0.
\]
Then inverse problem I with redefinition condition $u_x(0,t)=0$ and with basic conditions $u(0,t)=u(1,t)=0$ has a solution $\{u(x,t), q(t)\}$ such that $u(x,t) \in L_2(Q)$, $u_{xx}(x,t) \in L_2(Q)$, $u_{ttt}(x,t) \in L_2(Q)$, $q(t) \in L_2([0,T])$,
 for any function $f(x,t)$  such that
$f(x,t), f_{xx}(x,t)\in L_2(Q)$:
\[
f(1,t)-\frac{f_x(0,t)}{h_x(0,t)}h(1,t)=0,
f(0,t)-\frac{f_x(0,t)}{h_x(0,t)}h(0,t)=0.
\]

	\end{frame}
\begin{frame}
		\frametitle{Problem 3}
$u(0,t)=0$ --- redefinition condition,

$u_x(0,t)=u_x(1,t)=0$ --- basic conditions.
                     \[
q(t)=\frac{u_{xx}(0,t)-f(0,t)}{h(0,t)}, h(0,t)\neq 0;
\]
Notations:
\[
\hat{f_1}(x,t)=f(x,t)-\frac{f(0,t)h(x,t)}{h(0,t)},
\]
\[
\hat{h_1}(x,t)=\frac{h(x,t)}{h(0,t)}.
\]

	\end{frame}

	\begin{frame}
		\frametitle{Problem 3}
                     Notations: $v(x,t)=u_{xx}(x,t)$, $\hat{f}_2=\hat{f}_{1xx}$, $\hat{h}_2=\hat{h}_{1xx}$, $\hat{f}						_1(x,t)=f(x,t)-\frac{f(0,t)h(x,t)}{h(0,t)}$, $\hat{h}_1(x,t)=\frac{h(x,t)}{h(0,t)}$, $\hat{\alpha}(t)=\hat{h}_{1x}				(0,t)$ и $\hat{\beta}(t)=\hat{h}_{1x}(1,t)$.

		Consider the problem:
		\[
		\begin{cases}
		v_{ttt}(x,t)+v_{xx}(x,t)+\mu v(x,t)=\hat{f_2}(x,t)+\hat{ h_2}(x,t)v(0,t),\\
		v_x(0,t)-\hat{\alpha} (t)  v(0,t)=0, x\in[0,1], t\in[0,T],\\
		v_x(1,t)+\hat{\beta }(t)v(0,t)=0, x\in[0,1], t\in[0,T].
		\end{cases}
		\]

	\end{frame}

	\begin{frame}
		\frametitle{Existence theorem 3}
         $h(x,t)\in C^4(\overline{Q})$ and conditions are met when $t\in[0,T]$:
\[
h(0,t)\neq0,
h_x(1,t)\equiv 0,
\mu\textless 0,
\frac{h_x(0,t)}{h(0,t)}\geqslant0,
\]
\[
h_{xt}(0,t)h(0,t)-h_t(0,t)h_x(0,t)\geqslant0,
\]
\[
h_{xttt}(0,t)h(0,t)+h_{xtt}(0,t)h_t(0,t)-h_{ttt}(0,t)h_x(0,t)-h_{tt}(0,t)h_{xt}\geqslant0.
\]
Then inverse problem I with redefinition condition $u(0,t)=0$ with basic conditions
$u_x(0,t)=u_x(1,t)=0$ has a solution $\{u(x,t), q(t)\}$ such that $u(x,t) \in L_2(Q)$, $u_{xx}(x,t) \in L_2(Q)$,
$u_{ttt}(x,t) \in L_2(Q)$, $q(t) \in L_2([0,T])$  for any function $f(x,t)$such that $f(x,t), f_{xx}(x,t)\in L_2(Q)$:
\[
f_x(0,t)-\frac{f(0,t)}{h(0,t)}h_x(0,t)=0,
\]
\[
f_x(1,t)-\frac{f(0,t)}{h(0,t)}h_x(1,t)=0.
\]
	\end{frame}


	\begin{frame}
		\frametitle{Formulation of the problem for hyperbolic equations}

		$Q=(0,1)\times(0,T)$-rectangle,

		$\mu(x,t)$,
		$ f (x,t)$,
		$ h(x,t)$  -  prescribed function, defined in $\overline{Q}$.
		\begin{rutask}
			Find function $ \ u(x,t)$ and $\ q(t)$, connected by the equation
		\[
		u_{tttt}+u_{xx}+\mu(x,t)u=f(x,t)+q(t)h(x,t)
		\]
in area Q.
		Conditions are satisfied for the function
		 $u(x,t)$ conditions
		\[
		u(x,0)=u_t(x,0)=u_{tt}(x,0)=u_t(x,T)=0,  x \in [0,1],
		\]
and one of the group:

			1. $u(0,t)=0$ -  redefinition condition ,

$u(1,t)=u_x(0,t)=0$ -basic condition.

			2. $u_x(0,t)=0$ - redefinition,
			$u(0,t)=u(1,t)=0$ - basic.

			3. $u(0,t)=0$ - redefinition,
			$u_x(0,t)=u_x(1,t)=0$ - basic.
		\end{rutask}
	\end{frame}




	\begin{frame}
		\frametitle{Existence theorem  4}
                  $h(x,t)\in C^6(\overline{Q})$, $\mu$, $\beta(t)=-\frac{h_x(1,t)}{h(0,t)}$. Conditions are met for this functions and for
$t\in [0,T]$:
\[
h(0,t)\neq 0,
\mu\geqslant0,
\beta(T)\leqslant 0,
\beta^{''}(t)\geqslant 0,
\beta_{2}(T)=\beta_{2}^{'}(T)=0,
\]
\[
\beta^{'}(t)(A-t)-\beta(t)\geqslant 0,
-\beta^{'}(t)(A-t)-\beta(t)\leqslant 0,
\]
\[
-3\beta^{'''}(T)+(A-T)\beta^{''''}(T)\leqslant 0,
-4\beta^{''''}(t)+(A-t)\beta ^{'''''}(t)\geqslant 0.
\]
Then inverse problem II  with redefinition condition: $u(0,t)=0$ with basic conditions $u(1,t)=u_x(0,t)=0$  has a solution $\left\{
u(x,t),q(t)\right\}$  such that $u(x,t)\in{L_2(Q)}$, $q(t)\in {L_2([0,T])}$,
$u_{tttt}(x,t)\in{L_2(Q)}$, $u_{xx}(x,t)\in{L_2(Q)}$,$v(x,t)\in{L_2(Q)}$,$v_{ttt}(x,t)\in{L_2(Q)}$ and $v_{xx}(x,t)\in{L_2(Q)}$ for any function $f(x,t)$  such that $f(x,t),f_{xx}(x,t), f_{xxt}(x,t)\in L_2(Q)$:
\[
f(1,t)-\frac{f(0,t)h(1,t)}{h(0,t)}=0,
f_x(1,t)-\frac{f(0,t)h_x(1,t)}{h(0,t)}=0.
\]
	\end{frame}

           \begin{frame}
		\frametitle{Existence theorem  5}

 $h(x,t)\in C^2(\overline{Q})$, $\tilde{\beta}(t)=-\frac{h(0,t)}{h_x(0,t)}$.  Conditions are met for this functions and for
$t\in [0,T]$:
\[
h_x(0,t)\neq 0,
\mu\geqslant0,
\tilde{\beta}(T)\geqslant 0,
\tilde{\beta}^{'}(t)\geqslant 0,
\]
\[
\tilde{\beta}(t)-\tilde{\beta}^{'}(t)\leqslant 0.
\]
Then inverse problem II with redefinition condition: $u_x(0,t)=0$ with basic conditions $u(0,t)=u(1,t)=0$  has a solution $\left\{
u(x,t),q(t)\right\}$  such that $u(x,t)\in{L_2(Q)}$, $q(t)\in {L_2([0,T])}$,
$u_{tttt}(x,t)\in{L_2(Q)}$, $u_{xx}(x,t)\in{L_2(Q)}$,$v(x,t)\in{L_2(Q)}$,$v_{ttt}(x,t)\in{L_2(Q)}$ and $v_{xx}(x,t)\in{L_2(Q)}$ for any function $f(x,t)$  such that $f(x,t),f_{xx}(x,t),f_{xxt}(x,t)\in L_2(Q)$:
\[
f(0,t)-\frac{f_x(0,t)}{h_x(0,t)}h(0,t)=0,
f(1,t)-\frac{f_x(0,t)}{h_x(0,t)}h(1,t)=0.
\]
	\end{frame}



\begin{frame}
		\frametitle{Existence theorem  6}
Пусть $h(x,t)\in C^2(\overline{Q})$. Conditions are met for this functions and for
$t\in [0,T]$:
\[
h(0,t)\neq 0,
h_x(1,t)=0,
\frac{h_x(0,T)}{h(0,T)}\geqslant 0,
\]
\[
h_x(0,t)h(0,t)-(h_{xt}(0,t)h(0,t)+h_t(0,t)h_x(0,t))(A-t)\geqslant 0.
\]
Then inverse problem II  with redefinition condition: $u(0,t)=0$  with basic conditions $u_x(0,t)=u_x(1,t)=0$  has a solution $\left\{u(x,t),q(t)\right\}$  such that $u(x,t)\in{L_2(Q)}$, $q(t)\in {L_2([0,T])}$, $u_{tttt}(x,t)\in{L_2(Q)}$, $u_{xx}(x,t)\in{L_2(Q)}$,$v(x,t)\in{L_2(Q)}$,$v_{ttt}(x,t)\in{L_2(Q)}$ and $v_{xx}(x,t)\in{L_2(Q)}$ for any function $f(x,t)$  such that $f(x,t),f_{xx}(x,t), f_{xxt}(x,t)\in L_2(Q)$:
\[
f_x(0,t)-\frac{f(0,t)}{h(0,t)}h_x(0,t)=0,
f_x(1,t)-\frac{f(0,t)}{h(0,t)}h_x(1,t)=0.
\]

	\end{frame}



\begin{frame}
		\frametitle{Conclusion}
		We have studied inverse problems for quasiparabolic and quasi-hyperbolic equations
		\begin{itemize}
			\item $u_{ttt}+u_{xx}+\mu(x,t)u=f(x,t)+q(t)h(x,t)$,
			\item $u_{tttt}+u_{xx}+\mu(x,t)u=f(x,t)+q(t)h(x,t)$,
		\end{itemize}
We have proved the existence of a solution $\left\{q(t),u(x,t)\right\}$ and defined the right parts for three different cases.
	\end{frame}

\begin{frame}
		\frametitle{Conclusion}
Thank you for attention

	\end{frame}




\end{document}