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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2016 (2016), No. 24, pp. 1--13.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2016 Texas State University.}
\vspace{9mm}}


\begin{document}
\title[\hfilneg EJDE-2016/24\hfil Filter regularization]
{Filter regularization for an inverse parabolic problem in several variables}

\author[T. N. Huy, M. Kirane, L. D. Le, T. V. Nguyen \hfil EJDE-2016/24\hfilneg]
{Tuan Nguyen Huy, Mokhtar Kirane, Long Dinh Le, Thinh Van Nguyen}


\address{Tuan Nguyen Huy \newline
 Applied Analysis Research Group,
Faculty of Mathematics and Statistics,
Ton Duc Thang University, Ho Chi Minh City, Vietnam}
\email{nguyenhuytuan@tdt.edu.vn}

\address{Mokhtar Kirane \newline
Laboratoire de Mathematiques P\^{o}le Sciences et Technologie,
Universi\'e de La Rochelle,
Avenue M. Cr\'epeau, 17042 La Rochelle Cedex, France. \newline
Nonlinear Analysis and Applied Mathematics (NAAM) Research Group,
Department of Mathematics,
Faculty of Science, King Abdulaziz University, P.O. Box 80203,
Jeddah 21589, Saudi Arabia.}
\email{mokhtar.kirane@univ-lr.fr}

\address{Long Dinh Le \newline
 Institute of Computational Science and Technology,
Ho Chi Minh City, Viet Nam}
\email{long04011990@gmail.com}

\address{Thinh Van Nguyen \newline
 Department of Civil and Environmental Engineering,
Seoul National University, Republic of Korea}
\email{vnguyen@snu.ac.kr}



\thanks{Submitted December 3, 2015. Published January 15, 2016.}
\subjclass[2010]{35K05, 35K99, 47J06, 47H10}
\keywords{Ill-posed problem; truncation method; heat equation; regularization}

\begin{abstract}
The backward heat problem is known to be ill possed, which has lead
to the design of several regularization methods.
In this article we apply the method of filtering out the high frequencies
from the data for a parabolic equation.
First we identify two properties that if satisfied they imply the
convergence of the approximate solution to the exact solution.
Then we provide examples of filters that satisfy the two properties,
and error estimates for their approximate solutions.
We also provide numerical experiments to illustrate our results.
\end{abstract}

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

\section{Introduction}

The forward heat conduction problem consists of predicting the temperature
of an object at a future time from the present temperature,  boundary conditions,
and heat source.
On the other hand, the backward heat problem is an
inverse problem that consists of recovering the temperature at a past
time from the present temperature.
Inverse problems are of great importance in engineering applications,
and aim to detect a previous status from  its present information.
They can be applied to several  areas such as image processing,
mathematical finance, mechanics of continuous media, etc.
The equation $u_t - b(t) \Delta u =f(x,t)$ is a simple form of
the well-known advection-convection equation that appears in
groundwater pollution  problems and have been studied
in \cite{at}.

In this article, we consider the problem of finding a function
$u(x,t)$ from the given data $u(x,T)=g(x)$ in the parabolic problem
\begin{equation}
\begin{gathered}
\frac{\partial u}{\partial t}-b(t)L[u] = f(x,t),\quad (x,t) \in \Omega
\times (0,T)  \\
u|_{\partial \Omega}=0, \quad t \in (0,T)\\
u(x,T)=g(x),\quad x \in \Omega\,.
\end{gathered} \label{prob1}
\end{equation}
Here $\Omega $ is a bounded open domain in $\mathbb{R}^n$ with
smooth boundary $\partial \Omega$; $b(t)$, $g(x)$, $f(x,t)$ are given functions;
and $L$ is a symmetric elliptic operator. As an example of operator $L$ we have
the negative Laplacian $-\Delta =-(u_{xx}+u_{yy}+\dots)$.

It is well-known that the backward problem is ill-posed; i.e.,
its solution may not exist, and if it exists, it does not depend continuously
 on the given data.  In fact, small noise on the measured data may lead to
solutions with large errors.  This makes the numerical computation difficult,
 hence a regularization  process is needed.

Many studies have been devoted to the regularization of  \eqref{prob1}.
For one dimension with $b(t)=1$ and $f(x,t)=0$, we have the following:
John \cite{john} introduced a the idea of prescribing a bound on the solution
at $t =T$ with relation on the final data $g$.
Lattes and Lions \cite{lattes}, Showalter \cite{showalter1}, and Ewing \cite{ewing}
used quasi-reversibility method.
Ames and Epperson \cite{ames1}, and Miller \cite{miller} used the least squares
 methods with Tikhonov regularization.
Lee and Sheen \cite{lee1,lee2} used a parallel method for backward parabolic
problems.
Among other researcher in this area, we have:
Clark and Oppenheimer \cite{clark}, Ames et al \cite{ames1},
 Denche and Bessila \cite{denche}, Tautenhahn et al \cite{Sch},
 Melnikova et al \cite{melkinova1,melkinova2}, Fu \cite{feng1,fu},
Yildiz et al \cite{Yldzid1,Yldzid2}.
When $f(x,t)$ is not necessarily zero, \eqref{prob1} has been regularized by
Trong et al \cite{Trong1,Trong2}.
When $b(t)$ is not necessarily constant,
\eqref{prob1} has been studied in \cite{LeTriet,Tuan1,Fu}.

All the above studies are for the one-dimensional problems.
A filter regularization for a 3-dimensional Helmholtz equation was
studied in \cite{tran}.
Here apply a filter regularization to the backward problem
of a multi-dimensional parabolic equation.
This can be seen as an extension of the work in \cite{Quan, Tuan1}.


The outline of the rest of this article is as follows.
In the next section, we establish the existence and uniqueness of
a solution to  \eqref{prob1}.
In Section 3, we present the theoretical foundations of the filter
regularization, and  state two conditions \eqref{cond1} and \eqref{cond2}
that if satisfied, approximate solutions converge to the exact solution.
Also error estimates are  presented there.
In Section 4, we consider four regularizing filters, and
present numerical experiments for two of those filters.

\section{Inverse problem}

We assume that  $b:[0,T] \to \mathbb{R}$ is a differentiable function, and that
there exist constants $b_1,b_2,c_1$ such that
\begin{equation} \label{eb}
0 < b_1 \le b(t)\le b_2,\quad   0<b'(t) \le  c_1\quad\text{for all }t \in [0, T].
\end{equation}
Also we assume that $f \in L^2 ((0,T);L^2(\Omega))$ and $g\in L^2(\Omega) $.
In the space $L^2(\Omega)$ we denote the norm by  $\|\cdot \|$, and the inner product
by $\langle\cdot,\cdot\rangle$.

First, we recall some properties of the elliptic operator $L$
on a  bounded open domain $\Omega$ with Dirichlet boundary conditions
(see \cite[Section 6.5]{evan}).
\begin{itemize}
\item Each eigenvalue of $L$  is real, and the family of eigenvalues
$\{ \lambda_p\}_{p=1}^\infty$ satisfies
$0 < \lambda_1 \le \lambda_2 \le \lambda_3 \le \dots \to \infty$
as $p \to \infty$.

\item There exists an orthonormal basis $\{X_p\}_{p=1}^\infty $ for the space
$L^2(\Omega)$, where $X_p \in H_0^1(\Omega)$ is  an eigenfunction corresponding
to $\lambda_p$; i.e., for $n\in \mathbb{N}$,
\begin{gather*}
L[X_p](x) =\lambda_p X_p(x),\quad\text{for }x \in \Omega \\
X_p(x)=0 , \quad\text{for } x\in \partial \Omega\,.
\end{gather*}
\end{itemize}
For  $0 \leq q < \infty$, let $ g_p=\int_{\Omega}g(x)X_p(x)dx$.
Then we denote by $S^q(\Omega)$  the space of
 functions $g \in L^2(\Omega)$ satisfying
\begin{equation} \label{Sq}
\sum_{p=1}^\infty (1 + \lambda)^{2q}|g_p|^2 < \infty,
\end{equation}
with the norm
 $\|g\|^{2}_{S^q(\Omega)}= \sum_{p=1}^\infty (1 + \lambda_p)^{2q}|g_p|^2$.
When $q=0$, $S^q(\Omega)=L^2(\Omega)$
 (see  \cite[Chapter V]{Brezis}, \cite[page 179]{feng1}).

As is well known, the forward problem
\begin{equation} \label{prob2}
\begin{gathered}
\frac{\partial u}{\partial t}-b(t)L[u]=f(x,t),\quad
(x,t) \in \Omega \times (0,T)  \\
u|_{\partial \Omega}=0, \quad t \in (0,T)\\
u(x,0)=g(x),\quad x \in \Omega \,,
\end{gathered}
\end{equation}
with $f \in L^2 ((0,T);L^2(\Omega))$ and
 $g\in L^2(\Omega) $, has a unique solution.
However, for the backward problem \eqref{prob1},
with $f \in L^2 ((0,T);L^2(\Omega))$ and $g\in L^2(\Omega) $,
there is no guarantee that the solution exists.

Next we obtain a solution by the Fourier series method.
For a fixed $t$, we use the eigenfunctions of the Laplacian
to write Fourier series for  $f$, $g$, and $u$.
Then \eqref{prob1} determines an ordinary first-order differential equation
for the Fourier coefficients. Then the coefficients of the solution to this equation
at time  $T$ are equated to coefficients of $g$.
This process yields the following result.

\begin{theorem} \label{thm1}
Problem \eqref{prob1} has a unique solution if and only if
 \begin{equation} \label{cond0}
 \sum_{p=1}^\infty \exp \Big(2\lambda_p \int_t^T b(\xi)d\xi\Big)
 \Big[ g_p-\int_0^T \exp \Big(-\lambda_p \int_s^T b(\xi)d\xi\Big)
 f_p(s)ds\Big]^2<\infty  ,
 \end{equation}
 where
\begin{equation} \label{efgp}
 g_p=\int_{\Omega}g(x)X_p(x)dx,\quad f_p(s)=\int_{\Omega}f(x,s)X_p(x)dx.
\end{equation}
In this case the exact solution is
 \begin{equation} \label{exactsol}
u(x,t)= \sum_{p=1}^\infty \exp \Big(\lambda_p \int_t^T b(\xi)d\xi\Big)
 \Big[ g_p-\int_0^T \exp \Big(-\lambda_p \int_s^T b(\xi)d\xi\Big)
 f_p(s)ds\Big] \,.
 \end{equation}
\end{theorem}

\begin{remark}\rm
Examples of functions $f$ and $g$ satisfying  \eqref{cond0} are given in
the section for numerical experiments.
When $b(t)=1$ and $f(x,t)=0$, problem \eqref{prob1}
has a unique solution if and only if
\[
\sum_{p=1}^\infty  e^{2T\lambda_p} |\langle g(\cdot),X_p(\cdot)\rangle|^2
< \infty,
\]
as stated in \cite[Lemma 1]{clark}.

The proof of uniqueness uses the bounds in \eqref{eb} and is similar
to the one in \cite[Corollary 2.6]{feng2}
and   \cite[page 434]{LeTriet}; so we omit it.
\end{remark}


In spite of the solution to problem \eqref{prob1} begin unique,
it is still ill-posed and some regularization methods are necessary.
In the next section, we use a regularization method for solving
the problem.

\section{Filter regularization method}

In this section, we assume that the measured data $f^\epsilon$ and
$g^\epsilon$ belong to $L^2(\Omega)$ and satisfy
\begin{equation} \label{data}
2\|g^\epsilon-g\|^2 +2\big\| \int_0^T
|f^\epsilon(\cdot,s) -f(\cdot,s))| \, ds\big\|^2  \le\epsilon^2.
\end{equation}
The main idea of the filter method is to multiply
$f^\epsilon$ and $g^\epsilon$ by functions $R_f(\alpha,p)$
and $R_g(\alpha,p)$, respectively. These two function are called
regularizing filters, and $\alpha$ a regularization parameter.
If these two functions approach zero as $p\to\infty$,
the effect that ``high frequency'' data have on the solution
will be diminished.
For simplicity we set $R(\alpha,p)=R_f(\alpha,p)=R_g(\alpha,p)$.
Therefore, the approximate solution is
\begin{equation} \label{e3.2}
\begin{aligned}
U_\alpha^\epsilon (x,t)
&= \sum_{p=1}^\infty \exp\Big(\lambda_p \int_t^T b(\xi)d\xi\Big) R(\alpha,p) \\
&\quad\times \Big[ g_p^\epsilon-\int_t^T \exp\Big(-\lambda_p \int_s^T b(\xi)d\xi\Big)
 f_p(s)ds\Big] X_p(x) \,,
\end{aligned}
\end{equation}
where $f_p$ and $g_p$ are defined by \eqref{efgp}.

\begin{theorem} \label{thm3}
 Assume that for  the exact solution $u$ of \eqref{prob1} there exist
constants $M_p$ and $E$ such that
\begin{equation}
 \sum_{p=1}^\infty M_p^2 |\langle u(x,t), X_p(x)\rangle|^2  \le E^2
\quad\forall t\in [0,T]\,. \label{priori}
 \end{equation}
Also assume that there exist functions $K_1(\alpha)$ and $K_2(\alpha)$ such that
 \begin{gather}
  \exp \Big(\lambda_p \int_t^T b(\xi)d\xi\Big) |R(\alpha,p)| \le K_1(\alpha)
 \label{cond1} \\
 |R(\alpha,p)-1| \le K_2(\alpha) M_p \label{cond2}
 \end{gather}
 for all $p\in\mathbb{N}$ and all $t \in [0,T]$.
Then $U_\alpha^\epsilon$, defined by \eqref{e3.2}, satisfies
 \begin{equation} \label{e3.8}
 \|U_\alpha^\epsilon(\cdot,t)-u(\cdot,t)\|_{L^2(\Omega)}
\le K_1(\alpha)\epsilon+ K_2(\alpha) E.
 \end{equation}
\end{theorem}

A filter  $R(\alpha,p)$ is admissible if
$\alpha(\epsilon)$, $ K_1(\alpha)$ and $K_2(\alpha)$ tend to zero as
$\epsilon$ tends to zero. By Theorem \ref{thm3} this implies the
convergence of  the approximate solution to the exact solution,
\[
\lim_{\epsilon\to 0}\|U_\alpha^\epsilon(\cdot,t)-u(\cdot,t)\|_{L^2(\Omega)}=0\,,
\]
for any $t\in [0,T]$.

\begin{proof}[Proof of Theorem \ref{thm3}]
The strategy is to define a function $U_\alpha$ and
use the triangle inequality.
Let
\begin{align*}
U_\alpha (x,t)
&= \sum_{p=1}^\infty  R(\alpha,p) \exp\Big(\lambda_p \int_t^T b(\xi)d\xi\Big) \\
&\quad\times \Big[ g_p-\int_t^T \exp\Big(-\lambda_p \int_s^T b(\xi)d\xi\Big)
f_p(s)ds\Big] X_p(x).
\end{align*}
Then we have
\begin{equation} \label{e10}
\begin{aligned}
&\|U_\alpha^\epsilon(\cdot,t)-U_\alpha(\cdot,t)\|^2\\
&=  \sum_{p=1}^\infty    |R(\alpha,p)|^2
\exp \Big(2\lambda_p \int_t^T b(\xi)d\xi\Big)\\
&\quad\times    \Big(g_p^\epsilon-g_p
  -\int_t^Te^{-\lambda_p \int_s^T b}
 \big(f_p^\epsilon(s)-f_p(s)\big) ds\Big)^2
\\
&\leq \sum_{p=1}^\infty |R(\alpha,p)|^2
\exp \Big(2\lambda_p \int_t^T b(\xi)d\xi\Big)\\
&\quad\times  2\Big((g_p^\epsilon-g_p)^2
  +\Big(\int_t^T
 |f_p^\epsilon(s)-f_p(s)| ds\Big)^2 \Big)\\
&\leq |K_1(\alpha)|^2 2\Big(\|g^\epsilon-g\|^2
+\big\| \int_t^T |f^\epsilon(\cdot,s)-f(\cdot,s)|\,ds \big\|^2\Big)
\end{aligned}
\end{equation}
Here we used that $(\alpha+\beta)^2\leq \frac12(\alpha^2+\beta^2)$ and that
$e^{-\lambda_p \int_s^T b}\leq 1$ because $\lambda_p$ and $b$ are non-negative.
The squared integral is estimated using Fubini's theorem as follows.
Let $\phi(x)=\int_0^T h(x,s)\,ds$ then the Fourier coefficients
satisfy
\[
\phi_p=\int_\Omega\int_t^T h(x,s)\,ds X_p(x)\,dx
=\int_t^T \int_\Omega h(x,s)X_p(x) \,dx\,ds
=\int_t^T h_p(s)\,ds
\]
By Parseval's equality,
\[
\|\phi\|^2= \sum_{p=1}^\infty \phi_p^2
= \sum_{p=1}^\infty \Big(\int_0^T h_p(s)\,ds \Big)^2
=\big\|\int_0^T h(\cdot,s)\,ds \|^2\,.
\]

From  \eqref{data} and \eqref{e10}, we have
\begin{equation} \label{Ue-U}
\|U_\alpha^\epsilon(\cdot,t)-U_\alpha(\cdot,t)\|
\leq  K_1(\alpha) \epsilon
\end{equation}

From the definition of $U_\alpha$, we have
\begin{align*}
\|U_\alpha(\cdot,t)-u(\cdot,t)\|^2
&=\sum_{p=1}^\infty
[ R(\alpha,p)-1]^2 \exp \Big(2\lambda_p \int_t^T b(\xi)d\xi\Big)\\
&\quad\times \Big[ g_p-\int_t^T
\exp\Big(-\lambda_p \int_s^T b(\xi)d\xi\Big) f_p(s)ds\Big]^2\\
&=\sum_{p=1}^\infty [ R(\alpha,p)-1 ]^2 |\langle u(x,t), X_p(x)\rangle| ^2 \\
&\le |K_2(\alpha)|^2  \sum_{p=1}^\infty M_p^2 |\langle u(x,t), X_p(x)\rangle |^2\\
&\le |K_2(\alpha)|^2 E^2.
\end{align*}
This inequality, \eqref{Ue-U}, and the triangle inequality
complete the proof.
\end{proof}

\begin{remark} \rm
 Assumption \eqref{priori} holds naturally  when
 $M_p=\lambda_p^k$ for any $k>0$. In this case
 \[
 \sum_{p=1}^\infty M_p^2 |\langle u(x,t), X_p(x)\rangle|^2
= \sum_{p=1}^\infty  \lambda_p^{2k} |\langle u(x,t), X_p(x)\rangle|^2
=E^2=\|u\|_{S^k(\Omega)}^2 \,,
 \]
 where $S^k (\Omega)$ is defined in Section 2.
\end{remark}

Next we present specific filters and their regularized solutions.

\begin{proposition} \label{R1}
As in \cite{LeTriet}, let
\[
R_1 (\alpha,p)= \frac{1}{1+ \alpha \exp \big(\lambda_p \int_0^T b(\xi)d\xi\big)}\,,
\]
and $\alpha(\epsilon)= \epsilon^{(1-m)b_1/b_2}$ with $m \in (0,1)$ and $b_1$ and $b_2$
as in \eqref{eb}.
Then $R_1$ satisfies \eqref{cond1} and \eqref{cond2} with
\[
K_1(\alpha)= \alpha^{-b_2/b_1},\quad
K_2(\alpha)= \frac{ \int_0^T b(\xi)d\xi  }
 { \ln \big(\lambda_1 \int_0^T b(\xi)d\xi /\alpha\big)},
\quad M_p=\lambda_p \,.
\]
\end{proposition}

From \cite[Lemma 2]{LeTriet}, we have
\begin{align*}
R_1 (\alpha,p) \exp \Big(\lambda_p \int_t^T b(\xi)d\xi\Big)
&= \frac{ \exp \big(-\lambda_p \int_0^t b(\xi)d\xi\big) }
{ \alpha+ \exp \big(-\lambda_p \int_0^T b(\xi)d\xi\big)} \\
&\le \alpha^{  \frac{b_2t}{b_1} - \frac{b_2}{b_1}  } \\
&\le \alpha^{-b_2/b_1} =K_1(\alpha) \,.
\end{align*}

To verify condition \eqref{cond2}, we apply the elementary estimate
\[
\frac{1}{\alpha z + e^{-Mz}} \le \frac{M}{\alpha \ln(M/\alpha)}
\]
for $M>0$ and $\alpha$ small enough. Therefore,
\begin{align*}
|R_1(\alpha,p)-1|
&= \frac{\alpha}{\alpha+ \exp \big(-\lambda_p \int_0^T b(\xi)d\xi\big)  } \\
&=\frac{\alpha \lambda_p}{\alpha \lambda_p+ \lambda_p
 \exp \big(-\lambda_p \int_0^T b(\xi)d\xi\big)  } \\
&\le \frac{\alpha \lambda_p}{\alpha \lambda_p
 + \lambda_1 \exp \big(-\lambda_p \int_0^T b(\xi)d\xi\big)  } \\
&\le  \frac{ \int_0^T b(\xi)d\xi  }
{ \ln \big(\lambda_1 \int_0^T b(\xi)d\xi/\alpha\big)}  \lambda_p
= K_2(\alpha) M_p \cdot
\end{align*}

\begin{proposition} \label{R2}
For  $k \ge 1$, Let
\[
R_2(\alpha,p)= \frac{1}{1+ \epsilon \lambda_p^k
\exp \big(\lambda_p \int_0^T b(\xi)d\xi\big)  } \,.
\]
Then $R_2$ satisfies \eqref{cond1} and \eqref{cond2} with  $\alpha(\epsilon)= \epsilon$,
\[
K_1(\alpha)= b_4 \epsilon^{-1}\Big(\ln\big(\frac{b_3}{\epsilon}\big)\Big)^{-k},\quad
K_2(\alpha)= b_4 \Big(\ln\big(\frac{b_3}{\epsilon}\big)\Big)^{-k}, \quad
 M_p=\lambda_p^k \,,
\]
where $b_3=(b_2T)^{k}/k$ and $b_4=(kb_2T)^{k}$.
\end{proposition}

To prove the above proposition, we need the following Lemma.

\begin{lemma} \label{lem1}
For $M, \epsilon,x>0$, $k \ge 1$, we have the inequality
\[
\frac{1}{\epsilon x^k+e^{-Mx}} \le \frac{(kM)^k}{\epsilon \ln^k(\frac{M^{k}}{k\epsilon}) } \,.
\]
\end{lemma}

\begin{proof}
Let  $f(x)=\frac{1}{\epsilon x^k+e^{-Mx}}$. Then
\[
f'(x)=\frac{\epsilon k x^{k-1}-Me^{-Mx}}{-(\epsilon x^k+e^{-Mx})^2} \,.
\]
The only critical point $x_0$  satisfies
$x_{0}^{k-1}e^{Mx_0}=\frac{M}{k\epsilon}$ and yields a maximum.
 Hence
\[
f(x)\leq \frac{1}{\epsilon x_0^k+e^{-Mx_0}}
= \frac{1}{\epsilon x_0^k+ \frac{k\epsilon}{M } x_{0}^{k-1} } \,.
\]
By using the inequality $e^{Mx_0} \ge Mx_0$, we obtain
\[
\frac{M}{k\epsilon}
=x_{0}^{k-1}e^{Mx_0}
\le \Big(\frac{e^{Mx_0}}{M}\Big)^{(k-1}e^{Mx_0}
= \frac{1}{M^{k-1}}e^{kMx_0} \cdot
\]
This gives $e^{kMx_0} \ge \frac{M^k}{k\epsilon}$ and $kM x_0 \ge \ln(\frac{M^{k}}{k\epsilon})$.
Therefore
$
x_0 \ge \frac{1}{kM}\ln(\frac{M^{k}}{k\epsilon}).
$
Hence, we obtain
\[
f(x) \le \frac{1}{\epsilon x_0^k} \le \frac{(kM)^k}{\epsilon \ln^k(\frac{M^{k}}{k\epsilon}) }\,.
\]
\end{proof}

\begin{proof}[Proof of Proposition \ref{R2}]
Condition \eqref{cond1} is obtained as follows
\begin{align*}
R_2 (\epsilon,p) \exp \Big(\lambda_p \int_t^T b(\xi)d\xi\Big)
&= \frac{ \exp \big(-\lambda_p \int_0^t b(\xi)d\xi\big) }
 { \epsilon \lambda_p^k + \exp \big(-\lambda_p \int_0^T b(\xi)d\xi\big)} \\
&\le \frac {1}{\epsilon \lambda_p^{k} +\exp\big(-b_2T\lambda_p\big)}\,.
\end{align*}
Using the inequality
\[
\frac{1}{\epsilon x^k+e^{-b_2Tx}}
\le{(kTb_2)^k} \epsilon^{-1} \Big(\ln(\frac{(b_2T)^{k}}{k\epsilon})\Big)^{-k}
=b_4 \epsilon^{-1}\Big(\ln(\frac{b_3}{\epsilon})\Big)^{-k}\,, \label{bdt1}
\]
we conclude that
\[
R_2 (\epsilon,p) \exp \Big(\lambda_p \int_t^T b(\xi)d\xi\Big)
\le b_4 \epsilon^{-1}\Big(\ln(\frac{b_3}{\epsilon})\Big)^{-k} =K_1(\epsilon) \,.
\]
We derive Condition \eqref{cond2} as follows
\begin{align*}
|R_2(\epsilon,p)-1|
&= \frac{\epsilon \lambda_p^k}{\epsilon \lambda_p^k
+ \exp \big(-\lambda_p \int_0^T b(\xi)d\xi\big)  } \\
&\le \epsilon \lambda_p^k b_4 \epsilon^{-1}\Big(\ln\big(\frac{b_3}{\epsilon}\big)\Big)^{-k} \\
&=  b_4 \Big(\ln(\frac{b_3}{\epsilon})\Big)^{-k}
=K_2(\epsilon) M_p \,.
\end{align*}
\end{proof}

\begin{proposition} \label{R3}
Let
\[
R_3 (\alpha,p)=  \begin{cases}
1, & \text{if }  \lambda_p \le 1/\alpha,  \\
0, & \text{if }  \lambda_p > 1/\alpha \,.
\end{cases}
\]
where $\alpha = b_2T/ \ln(1/\epsilon)$.
Then $R_3$ satisfies \eqref{cond1} and \eqref{cond2} with $\alpha(\epsilon)= \epsilon$,
\[
K_1(\alpha) =\epsilon^{-1},\quad K_2(\alpha)=\alpha,\quad
M_p=\lambda_p \cdot
\]
\end{proposition}

\begin{proof}
Condition \eqref{cond1} is obtained as
\begin{align*}
R_3 (\alpha,p) \exp \Big(\lambda_p \int_t^T b(\xi)d\xi\Big)
&=  \begin{cases}
\exp \Big(\lambda_p \int_t^T b(\xi)d\xi\Big),
 & \text{if }  \lambda_p \le 1/\alpha  \\
0, & \text{if }  \lambda_p > 1/\alpha
\end{cases} \\
&\le  \exp \Big(\frac{1}{\alpha} \int_t^T b(\xi)d\xi\Big)
\le \epsilon^{-1} \,.
\end{align*}
Condition \eqref{cond2}  follows from
\begin{align*}
| R_3 (\alpha,p) -1|
&= \begin{cases}
0, & \text{if }  \lambda_p \le 1/\alpha  \\
1, & \text{if }  \lambda_p >  1/\alpha
\end{cases} \\
&\le  \alpha \lambda_p= K_2(\alpha)M_p \,.
\end{align*}
\end{proof}

\begin{proposition} \label{R4}
Let
\[
R_4(\alpha,p)= \exp \Big( -\alpha \lambda_p^2  \int_0^T b(\xi)d\xi\Big).
\]
Then $R_4$ satisfies \eqref{cond1} and \eqref{cond2} with $\alpha(\epsilon)= \epsilon$,
\[
K_1(\alpha)= \exp \Big( \frac{ \int_0^T b(\xi)d\xi }{4\alpha}  \Big),\quad
K_2(\alpha)=\alpha,\quad M_p=\lambda_p^2 \,.
\]
\end{proposition}

\begin{proof}
Conditions \eqref{cond1} and \eqref{cond2}  follow from
\begin{align*}
R_4 (\alpha,p) \exp \Big(\lambda_p \int_t^T b(\xi)d\xi\Big)
&= \exp \Big(  (\lambda_p-\alpha \lambda_p^2) \int_t^T b(\xi)d\xi\Big) \\
&\le \exp \Big( \frac{\int_0^T b(\xi)d\xi}{4\alpha} \Big)=K_1(\alpha),
\end{align*}
where we used  that $\lambda_p- \alpha \lambda_p^2 \le \frac{1}{4\alpha}$.
Using the inequality $1-e^{-z} \le z$ for $z>0$, we obtain
\[
|R_4(\alpha,p)-1|= 1- \exp \Big( -\alpha \lambda_p^2  \int_0^T b(\xi)d\xi\Big)
\le \alpha \lambda_p^2  \int_0^T b(\xi)d\xi \le K_2(\alpha) M_p \,,
\]
where $M_p=\lambda_p^2$.
\end{proof}

\section{Numerical experiments}


\begin{figure}[htb]
 \begin{center}
 \includegraphics[width=0.45\textwidth]{fig1a} % u0_3.jpg
 \includegraphics[width=0.45\textwidth]{fig1b} % g_ep1.jpg
 \end{center}
\caption{Exact solution at $t=0$, and $t=1$}
\label{fig1}
\end{figure}

\begin{figure}[htb]
 \begin{center}
\includegraphics[width=0.45\textwidth]{fig2a} % u1_1_3.jpg
\includegraphics[width=0.45\textwidth]{fig2b} \\ % u2_1_3.jpg
\includegraphics[width=0.45\textwidth]{fig2c} % u1_2_3.jpg
\includegraphics[width=0.45\textwidth]{fig2d} \\ % u2_2_3.jpg
\includegraphics[width=0.45\textwidth]{fig2e} % u1_4_3.jpg
\includegraphics[width=0.45\textwidth]{fig2f} \\ % u2_4_3.jpg
 \end{center}
 \caption{Numerical solutions at $t=0$ for filters $R_2$ (left) and
$R_3$ (right) with $\epsilon=10^{-1}$, $\epsilon=10^{-2}$, $\epsilon=10^{-4}$
 (from top to bottom)}
 \label{fig2}
\end{figure}

\begin{table}[htb]
\renewcommand{\arraystretch}{1.5}
 \caption{Absolute error estimate for mesh resolution $M=N=127$,
  $\Delta x = 2.7559E-02$,
  $\Delta y= 3.1496E-02$.}
 \label{table1}
 \begin{center}  \scriptsize
  \begin{tabular}{|l|ll|ll|ll|ll|ll|}
   \hline
    &
   \multicolumn{2}{|c}{ $\epsilon = 10^{-1} $ } &
   \multicolumn{2}{|c}{ $\epsilon = 10^{-2} $ } &
   \multicolumn{2}{|c}{ $\epsilon = 10^{-3} $ } &
   \multicolumn{2}{|c|}{ $\epsilon = 10^{-4} $ }
   \\
   \hline \hline
   \multicolumn{1}{|c|}{ $t$ } &
   \multicolumn{1}{c}{ $\delta_{1,2}$ } &
   \multicolumn{1}{c|}{ $\delta_{1,3}$ } &
   \multicolumn{1}{c}{ $\delta_{1,2}$ } &
   \multicolumn{1}{c|}{ $\delta_{1,3}$ } &
   \multicolumn{1}{c}{ $\delta_{1,2}$ } &
   \multicolumn{1}{c|}{ $\delta_{1,3}$ } &
   \multicolumn{1}{c}{ $\delta_{1,2}$ } &
   \multicolumn{1}{c|}{ $\delta_{1,3}$ }
   \\
   \hline \hline
   0.00 &  5.616E-01 &  7.045E-01 &  7.263E-02 &  1.338E-01 &  2.878E-02 &  7.796E-02 &  3.954E-02 &  6.457E-02 \\ \hline
   0.11 &  2.034E-01 &  3.487E-01 &  2.678E-02 &  3.911E-02 &  9.921E-03 &  2.116E-02 &  2.078E-02 &  1.995E-02 \\ \hline
   0.22 &  1.432E-01 &  4.184E-01 &  1.976E-02 &  1.085E-01 &  6.500E-03 &  2.284E-02 &  1.299E-02 &  1.356E-02 \\ \hline
   0.33 &  1.120E-01 &  3.563E-01 &  1.635E-02 &  1.512E-01 &  4.972E-03 &  4.830E-02 &  8.450E-03 &  1.552E-02 \\ \hline
   0.44 &  9.195E-02 &  2.867E-01 &  1.418E-02 &  1.584E-01 &  4.096E-03 &  6.950E-02 &  5.649E-03 &  2.551E-02 \\ \hline
   0.55 &  7.775E-02 &  2.307E-01 &  1.263E-02 &  1.494E-01 &  3.540E-03 &  8.006E-02 &  3.905E-03 &  3.573E-02 \\ \hline
   0.66 &  6.709E-02 &  1.882E-01 &  1.144E-02 &  1.352E-01 &  3.167E-03 &  8.281E-02 &  2.855E-03 &  4.296E-02 \\ \hline
   0.77 &  5.875E-02 &  1.559E-01 &  1.047E-02 &  1.203E-01 &  2.907E-03 &  8.104E-02 &  2.279E-03 &  4.704E-02 \\ \hline
   0.88 &  5.205E-02 &  1.311E-01 &  9.669E-03 &  1.066E-01 &  2.718E-03 &  7.696E-02 &  2.011E-03 &  4.866E-02 \\ \hline
   0.99 &  4.654E-02 &  1.118E-01 &  8.978E-03 &  9.441E-02 &  2.575E-03 &  7.189E-02 &  1.917E-03 &  4.860E-02 \\ \hline
  \end{tabular}
 \end{center}
\end{table}




Since numerical experiments were implemented for filter $R_1$ in \cite{LeTriet},
we implement experiments only for $R_2$ and $R_3$.
The efficiency of the methods is observed by comparing the errors
between numerical and exact solutions.
In both examples, we choose the exponent $k=1$, and
 consider \eqref{prob1} in a two-dimensional region.
Let $\Omega = (0,a) \times (0,b)$ be an open rectangle in
$\mathbb{R}^2$, and $T > 0$.
Let us consider
\begin{gather*}
u_t - b(t) \big( u_{xx} + u_{yy} \big) = f(x,y,t),
\quad (x,y) \in \Omega, \; t \in [0,T] \\
u(x,y,t) = 0, \quad  (x,y) \in \partial \Omega, \quad t \in [0,T] \\
u(x,y,T) = g(x,y) , \quad (x,y) \in \Omega .
\end{gather*}
The eigenfunctions and eigenvalues of the Laplacian are
\begin{gather*}
\psi_{mn} (x,y)= \frac{2}{\sqrt{ab } }
\sin \big( \frac{m \pi x }{a}  \big)
\sin \big( \frac{n \pi y }{b}  \big),\\
\lambda_{mn} = \big( \frac{ m \pi }{a} \big)^2
+ \big( \frac{ n \pi }{b} \big)^2 ,
\end{gather*}
for $(m,n) \in \mathbb{N}^2$. When
\[
b(t) = \frac{1}{ 100 + \exp( t^2 ) }
\]
this problem has exact solution
\[
u(x,y,t) = e^{ -t(x^2+y^2 ) } \sin \big( \frac{xy}{a+t} \big) (a-x) (b-y)\,.
\]

For the numerical computations we use
$a = 7$, $b = 8$, and $T = 1$.
The source function $f$ and the final datum  $g(x,y) = u(x,y,T)$ are
such that $u$ is the exact solution of the problem.

For the measured data $f^\epsilon$ and $g^\epsilon$,
 we use a random number generator $\operatorname{rand()}$ in $(-1,1)$,
$$
g^{\epsilon}(x,y) = g(x,y) + \frac{\epsilon}{\pi}  \operatorname{rand()},
 \quad f^\epsilon=f\,.
$$
At a given $\epsilon$ and $t$, the absolute error  between the
exact solution and the regularized solutions is estimated by
\begin{equation} \label{d3}
\delta_{1,l}  = \Big( \frac{ \sum_{i=1}^{M}\sum_{j=1}^{N}
| u^{l,\epsilon}( x_i, y_j, t) - u( x_i, y_j, t) |^2 } {(M)(N) }\Big)^{1/2}\,.
\end{equation}
Regularized solutions by filter $R_2$ correspond to $l=2$,
and by filter $R_3$  to $l=3$.
We choose a calculation grid of $127\times 127$ interior points, with
$x_i=i\pi/I$, $y_j=j\pi/J$, and $u^{l,\epsilon}(x,y,t)$.
 See Table \ref{table1}.

Figure \ref{fig1} shows the exact solution while
Figure \ref{fig2} shows the regularized solutions at $t=0$.
From Table \ref{table1} we see that overall filter $R_2$  gives a
better approximation than filter $R_3$.
Both  regularized solutions converge to the exact solution at $t=0$.
However, when $t$ close to 1 ($t=0.99$) the solution from filter $R_3$
is strongly oscillating and slowly converges to the exact solution.
In comparison  the convergence rate of filter $R_2$ is significant better
than the convergence rate of the filter $R_3$.

\subsection*{Acknowledgements}
This research is funded by  Foundation for Science and
Technology Development of Ton Duc Thang University (FOSTECT), website: http:
//fostect.tdt.edu.vn, under Grant FOSTECT.2014.BR.03. 
The authors would like to thank Professor Julio G. Dix for his
comments and corrections that improved this article.


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\end{document}