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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2008(2008), No. 33, pp. 1--14.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or
 http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu  (login: ftp)}
\thanks{\copyright 2008 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2008/33\hfil Regularization and error estimates]
{A nonhomogeneous backward heat problem: Regularization and error estimates}

\author[D. D. Trong, N. H. Tuan\hfil EJDE-2008/33\hfilneg]
{Dang Duc Trong, Nguyen Huy Tuan} 

\address{Dang Duc Trong  \newline
Department of Mathematics and Computer Sciences \\
Hochiminh City National University \\
227 Nguyen Van Cu, Hochiminh City, Vietnam}
\email{ddtrong@mathdep.hcmuns.edu.vn}

\address{Nguyen Huy Tuan  \newline
Department of  Information Technology and Applied Mathematics \\
Ton Duc Thang University \\
98 Ngo Tat To,  Hochiminh City, Vietnam}
\email{tuanhuy\_bs@yahoo.com}

\thanks{Submitted November 9, 2007. Published March 6, 2008.}
\thanks{Supported by the Council for Natural Sciences of Vietnam}
\subjclass[2000]{35K05, 35K99, 47J06, 47H10}
\keywords{Backward heat problem; ill-posed problem; \hfill\break\indent
nonhomogeneous heat equation; contraction principle}

\begin{abstract}
 We consider the problem of finding the initial temperature,
 from the final temperature, in the nonhomogeneous heat equation
 \begin{gather*}
 u_t-u_{xx}= f(x,t),\quad  (x,t)\in (0,\pi)\times (0,T),\\
 u(0,t)= u(\pi,t)= 0, \quad (x,t) \in (0,\pi)\times(0,T).
 \end{gather*}
 This problem is known as the backward heat problem and is severely
 ill-posed. Our goal is to present a simple and convenient
 regularization method, and sharp error estimates for its approximate
 solutions. We illustrate our results with a numerical example.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{remark}[theorem]{Remark}
\newcommand{\norm}[1]{\|{#1}\|}

\section{Introduction}

For a positive number $T$, we consider the problem of finding the
temperature $u(x,t)$, $(x,t)\in (0,\pi)\times [0,T]$, such that
\begin{gather}
u_t-u_{xx}=f(x,t),\quad  (x,t)\in (0,\pi)\times(0,T), \label{e1} \\
u(0,t)=u(\pi,t) ,\quad  (x,t) \in (0,\pi) \times [0,T], \label{e2}\\
u(x,T)=g(x),\quad x \in (0,\pi). \label{e3}
\end{gather}
where $g(x), f(x,z)$ are given. The problem is called the backward
heat problem, the backward Cauchy problem, or the final value problem.
As is known, the nonhomogeneous problem is severely ill-posed;
i.e., solutions do not always exist, and in the case of existence,
these do not depend continuously on the given data.
In fact, from small noise contaminated physical measurements,
 the corresponding solutions have large errors.
It makes difficult to do numerical calculations. Hence,
a regularization is in order.
 Lattes and Lions, in \cite{lattes}, regularized the problem by adding
a ``corrector''  to the main equation. They considered the
problem
\begin{gather*}
u_t+Au-\epsilon A^*Au =0, \quad  0<t<T,\\
u(T)=\varphi.
\end{gather*}

Gajewski and Zaccharias \cite{Gajewski}  considered a  similar problem.
Their error estimate for the approximate solutions is
\[%  \label{e6}
\|u^\epsilon(t)-u(t)\|^2 \leq  \frac{2}{t^2}(T-t)\|u(0)\|.
\]
Note that these estimate can not be used at  time $t=0$.

In 1983, Showalter,  presented a different method called the
quasiboundary value (QBV) method to regularize that linear homogeneous
problem which gave a stability estimate better than the one
of discussed method. The main ideas of the method is of adding an
appropriate ``corrector'' into the final data. Using the method,
Clark and Oppenheimer, in \cite{clark}, and Denche-Bessila,
very recently in \cite{Denche},
regularized the backward problem by replacing the final condition by
\begin{equation} \label{e4}
u(T)+\epsilon u(0)=g
\end{equation}
and
\begin{equation} \label{e5}
u(T)-\epsilon u'(0)=g
\end{equation}
respectively. Although there are many papers on the linear homogeneous
case of the backward problem, we only find a few papers on the
nonhomogeneous case, such in \cite{Trong, Trong2}.

In 2006, Trong and Tuan \cite{Trong}, approximated the problem
 \eqref{e1}--\eqref{e3} by the quasi-reversibility method.
 However, the stability magnitude of the method is of order
$e^{\frac{T}{\epsilon}}$. Moreover, the error between the approximate
 problem and the exact solution is
\[
\epsilon (T - t)\sqrt {\frac{8}  {{t^4 }}\|u(.,0)\|^2  + t^2
\|\frac{{\partial ^4 f(x,t)}}
{{\partial x^4 }}\|_{L^2(0,T;L^2(0,\pi))}^2 },
\]
(see \cite[page 5]{Denche}) which is very large when $\epsilon$ fixed
 and
$t$ is small (tend to zero).

Very recently, in \cite{Trong2}, the authors used an improved version
of QBV method to regularize problem in one dimensional of
\eqref{e1}--\eqref{e3} in the nonlinear case of function $f$.
However, in \cite{Trong2}, the authors can only estimate the error
in the case which, the final value $g$ satisfies the
condition
\begin{equation} \label{e*}
\sum_{k=1}^\infty {e^{2T k^2}g_{k}^2}<\infty
\end{equation}
(see \cite[page 242]{Trong2}). The functions satisfying this condition
are quite scarce and so this method is not useful to consider
many nonhomogeneous backward problem in the another case of final
value  $g$, which the condition  \eqref{e*}
 is not satisfied for functions such as $g(x)=a $, where $a$ is
 constant.
We also note that the error between the approximate problem
and the exact solution  is $C\epsilon^{\frac{t}{T}}$, which is not
near to zero, if $\epsilon$ fixed and $t$ tend to zero.
Hence, the convergence of the approximate solution is very slow when
$t$ is near to the original time.

In the present paper, we shall regularize this problem
\eqref{e1}--\eqref{e3}by perturbing the final value $g$ with
new way, which is different the ways in \eqref{e4}and \eqref{e5}.
We approximate problem by the following problem

\begin{gather} \label{e6}
u_t^{\epsilon}-u_{xx}^{\epsilon}
=\sum_{p=1}^\infty \frac {{e^{-T p^2}}}{\epsilon p^2
+e^{-Tp^2}} f_{p}(t) \sin (px) , \quad (x,t)\in (0,\pi) \times (0,T),
\\  \label{e7}
u^{\epsilon}(0,t)=u^{\epsilon}(\pi,t)=  0 \quad (x,t) \in (0,\pi)\times
 [0,T]
\\  \label{e8}
u^{\epsilon}(x,T)=\sum_{p=1}^\infty \frac {{e^{-T p^2}}}{\epsilon p^2
+e^{-Tp^2}} g_{p} \sin (px),\quad  x\in (0,\pi)
\end{gather}
where $0<\epsilon<1$,
\begin{equation} \label{e9}
f_{p}(t)=\frac{2}{\pi}\int_0^{\pi}f(x,t)\sin (px)dx ,~~~
g_{p}=\frac{2}{\pi}\int_0^{\pi}g(x)\sin (px) dx
\end{equation}
and $\langle\cdot,\cdot\rangle$ is the inner product in $L^2((0,\pi))$.

We shall prove that, the (unique) solution $ u^{\epsilon}$ of
\eqref{e6}--\eqref{e8} satisfies the following equality
\begin{equation} \label{e10}
u^{\epsilon}(x,t)=
\sum_{p=1}^\infty \Big(\frac {{e^{-t p^2}}}{\epsilon p^2
 +e^{-Tp^2}}g_{p}
-\int_t^T \frac {e^{(s-t-T) p^2}}{\epsilon p^2+e^{-T
 p^2}}f_{p}(s)ds\Big)
\sin (px)
\end{equation}
where $0\le t \le T$.

Note that our method give a better approximation than the
quasi-reversibility method in \cite{Trong}, and the final value $g(x)$
 is
not essential to satisfy the condition (*), which only in
$L^2(0,\pi)$. Especially,the convergence of the approximate
solution at $t=0$ is also proved (In \cite{Trong2}, the error
$\|u(.,0)-u^\epsilon(.,0)\|$ is not given). This is an improvement
of many known results in
\cite{ames1,clark,Denche,ewing,Gajewski,Trong,Trong2,Yldzid,Yldzid2}.

The remainder of the paper is divided into three sections.  In
Section 1, we shall show  that \eqref{e6}--\eqref{e8} is well
posed and that the solution  $u^{\epsilon}(x,t)$ satisfies \eqref{e10}.
Then, in Section 2, we estimate the error between an exact
solution $u_0$ of Problem \eqref{e1}--\eqref{e3} and the
approximation solution $u^\epsilon$. In fact, we shall prove that
\begin{equation} \label{e11}
\| u^{\epsilon}(.,t)-u_0(.,t)\|\le \frac {C}{1+\ln
(\frac{T}{\epsilon})}
\end{equation}
where $\|\cdot\|$ is norm in
$L^2(0,\pi)$ and $C$ depends on $u_0$ and $f$.

Finally, a numerical experiment will be given in Section 3.


\section{Well-posedness of Problem \eqref{e6}--\eqref{e8}}

In this section, we shall study the existence, the uniqueness and
the stability of a (weak) solution of Problem \eqref{e6}--\eqref{e8}.


\begin{theorem} \label{thm2.1}
Let $f(x,t) \in L^2 (0,T);L^2(0,\pi))$ and $g(x)\in L^2(0,\pi) $.
Let a given $\epsilon \in (0,eT)$. Then
\eqref{e6}--\eqref{e8} has a unique weak solution
$u^\epsilon \in C([0,T];L^2(0,\pi) \cap L^2(0,T;H_0^1(0,\pi))
\cap C^1(0,T; H_0^1(0,\pi))$ satisfying \eqref{e10}.
The solution depends continuously on $g$ in $C([0,T]; L^2(0,\pi))$.
\end{theorem}

\begin{proof}
The proof is divided into two steps. In Step 1,
we prove the existence and the uniqueness of a solution
of \eqref{e6}--\eqref{e8}. In Step 2, the stability of the solution
is given.

\noindent{\bf Step 1.} The existence and the uniqueness of a solution
of \eqref{e6}--\eqref{e8}
We divide this step into two parts.

\noindent{\bf Part A} If
$u^\epsilon \in C([0,T];L^2(0,\pi)) \cap L^2(0,T;H_0^1( 0,\pi))
 \cap C^1(0,T; H_0^1(0,\pi))$  satisfies  \eqref{e11}
then $ u^{\epsilon}$ is solution of \eqref{e6}--\eqref{e8}.
We have
\begin{equation} \label{e12}
u^{\epsilon}(x,t)
=\sum_{p=1}^\infty \Big(\frac {{e^{-t p^2}}}{\epsilon
 p^2+e^{-Tp^2})}g_{p}
-\int_t^T \frac {e^{(s-t-T) p^2}}{\epsilon p^2+e^{-T
 p^2}}f_{p}(s)ds\Big)
 \sin (px)
\end{equation}
for $0\le t \le T$.
We can verify directly that
\[
u^\epsilon \in C([0,T];L^2(0,\pi)
\cap C^1 ((0,T);H_0^1(0,\pi)) \cap L^2 (0,T; H_0^1(0,\pi))).
\]
In fact, $u^\epsilon \in C^\infty ((0,T];H_0^1(0,\pi)))$.
Moreover, one has
\begin{align*}
&u^{\epsilon}_t(x,t)\\
&=\sum_{p=1}^\infty\Big(\frac{-p^2 e^{-t p^2}}{\epsilon p^2
+e^{-T p^2}}g_{p}-\int_t^T\frac{p^2 e^{(s-t-T) p^2}}{\epsilon p^2
+e^{-T p^2} } f_{p}(s) ds
+\frac{e^{-T  p^2}}{\epsilon p^2+e^{-T p^2}} f_{p}(t)\Big)\sin (px)\\
&=-\frac{2}{\pi}\sum_{p=1}^\infty  p^2
\langle  u^{\epsilon}(x,t),\sin px\rangle  \sin (px)
+\sum_{p=1}^\infty \frac{e^{-T  p^2}}{\epsilon p^2+e^{-T p^2}}
f_{p}(t)\sin (px)
\\
&= u^{\epsilon}_{xx}(x,t)+\sum_{p=1}^\infty
\frac{e^{-T  p^2}}{\epsilon p^2+e^{-T p^2}} f_{p}(t)\sin (px)
\end{align*}
and
\[
 u^{\epsilon}(x,T)=\sum_{p=1}^\infty
\frac {{e^{-T p^2}}}{\epsilon p^2+e^{-Tp^2}} g_{p} \sin (px)
\]
So $ u^{\epsilon}$ is the solution of \eqref{e6}--\eqref{e8}.

\noindent{\bf Part B}
 The Problem \eqref{e6}--\eqref{e8} has at most one solution
$C([0,T];H_0^1(0,\pi))\cap C^1((0,T);L^2(0,\pi))$.
A proof of this statement can be found in  \cite[Theorem 11]{carasso}.
Since Part A and Part B are proved, we complete the proof of Step 1.
\smallskip

\noindent {\bf  Step 2.}
The solution of the problem \eqref{e6}--\eqref{e8} depends continuously
on $g$ in $L^2(0,\pi)$.
Let $u$ and $v$ be two solutions of \eqref{e6}--\eqref{e8}
 corresponding to the final values $g$ and $h$. From we have
\begin{gather} \label{e13}
u(x,t)=\sum_{p=1}^\infty
\Big(\frac {{e^{-t p^2}}}{\epsilon p^2+e^{-Tp^2}}g_{p}
-\int_t^T \frac {e^{(s-t-T) p^2}}{\epsilon p^2+e^{-T
 p^2}}f_{p}(s)ds\Big)
\sin (px) \quad 0\le t \le T,
\\
 \label{e14}
v(x,t)=
\sum_{p=1}^\infty \Big(\frac {{e^{-t p^2}}}{\epsilon
 p^2+e^{-Tp^2}}h_{p}
-\int_t^T \frac {e^{(s-t-T) p^2}}{\epsilon p^2+e^{-T
 p^2}}f_{p}(s)ds\Big)
\sin (px)\quad 0\le t \le T,
\end{gather}
where
\[
g_{p}=\frac{2}{\pi}\int_0^{\pi}g(x)\sin (px)dx ,\quad
h_{p}=\frac{2}{\pi} \int_0^{\pi}h(x)\sin (px) dx.
\]
For $\lambda>0$, we define the function
\[
h(\lambda)=\frac{1}{\epsilon \lambda+e^{-\lambda T}}.
\]
Then
\[
h(\lambda)\le h\Big(\frac{\ln(T/\epsilon)}{T}\Big)
=\frac{T}{\epsilon \big(1+\ln(T/\epsilon)\big)}
\quad \epsilon \in (0,eT).
\]
This follows that
\begin{equation} \label{e15}
\begin{aligned}
 \norm{u(.,t)-v(.,t)}^2
&=\frac{\pi}{2}\sum_{p=1}^\infty \big|\frac{e^{-t p^2}}{\epsilon p^2
+e^{-T p^2}}(g_{p}-h_{p})\big|^2\\
&\le \frac{\pi}{2}\Big(\frac{T}{\epsilon
\left(1+\ln(T/\epsilon)\right)}\Big)^2
 \sum_{p=1}^\infty|g_{p}-h_{p}|^2\\
&=\Big(\frac{T}{\epsilon \left(1+\ln(T/\epsilon)\right)}\Big)^2
\|g-h\|^2.
\end{aligned}
\end{equation}
Hence
\[
\norm{u(.,t)-v(.,t)}\le\frac{T}{\epsilon
\left(1+\ln(T/\epsilon)\right)}\norm{g-h}.
\]
This completes the proof of Step 2 and the proof of our theorem.
\end{proof}

\begin{remark} \label{rmk1} \rm
In \cite{ewing,Gajewski,Trong}, the stability magnitude is
$e^{\frac{T}{\epsilon}}$ (see  \cite[Theorem 2.1]{Denche},
it is $\epsilon^{-1}$. One advantage of this method of regularization
is that the order of the error, introduced by small changes in the
final value $g$, is less than the order given in \cite{Trong}.
\end{remark}

\begin{theorem} \label{thm2.2}
For any $g(x) \in L^2(0,\pi)$, the approximation $u^\epsilon(x,T)$
converges to $g(x)$ in $L^2(0,\pi)$ as $\epsilon $ tends to zero.
\end{theorem}

\begin{proof}
 We have $ g(x)=\sum_{p=1}^\infty g_p \sin (px)$, where $g_p $
is defined in \eqref{e9}.
Let $\alpha>0$, choose some $N$ for which
 $\frac{\pi}{2}\sum_{p=N+1}^\infty g_p^2 < \alpha/2$. We have
\begin{equation} \label{e16}
\|u^\epsilon(x,T)-g(x)\|^2=\frac{\pi}{2}\sum_{p=1}^\infty
 \frac{\epsilon^2 p^4 g_p^2}{ (\epsilon p^2+e^{-Tp^2})^2}\,.
\end{equation}
Then
$$ \|u^\epsilon(x,T)-g(x)\|^2 \le \epsilon^2 \frac{\pi}{2}
\sum_{p=1}^N p^4 g_p^2 e^{2Tp^2}+\frac{\alpha}{2}
$$
By taking $\epsilon$  such that
$ \epsilon < \sqrt{\alpha}\big( {\pi \sum_{p=1}^N p^4 g_p^2 e^{2Tp^2} }
\big)^{-1/2}$, we get
$$
\|u^\epsilon(x,T)-g(x)\|^2 < \alpha
$$
which completes the proof.

In the case  $\frac{d^2g}{dx^2} \in L^2(0,\pi)$,
  we have the error estimate
 \begin{align*}
\|u^\epsilon(x,T)-g(x)\|^2
&= \frac{\pi}{2} \sum_{p=1}^\infty ( \frac {{e^{-T p^2}}}{\epsilon p^2
+e^{-Tp^2}}-1)^2 g^2_p\\
&=\frac{\pi}{2}\sum_{p=1}^\infty
 \frac{\epsilon^2 p^4 g_p^2}{ (\epsilon p^2+e^{-Tp^2})^2}\\
&\le  \frac{\pi}{2} \frac{T}{
 \left(1+\ln(\frac{T^2}{\epsilon})\right)^2}
\sum_{p=1}^\infty p^4 g_p^2
=\frac{T^2}{ \left(1+\ln(T/\epsilon)\right)^2}\|g_{xx}\|^2
\end{align*}
Then, we get
\begin{align*}
\|u(x,T)-g(x)\| \le \frac{T}{ 1+\ln(T/\epsilon)}\|g_{xx}\|.
\end{align*}
This completes the proof.
\end{proof}

\begin{theorem} \label{thm2.3}
 Let  $g(x), \epsilon \in  L^2(0,\pi)$ be as in Theorem \ref{thm2.2},
 and
let $ f_{xx}$ be in $L^2 (0,T; L^2(0,\pi))$.
If the sequence $u^\epsilon(x,0)$ converges in $L^2(0,\pi)$, then
the problem \eqref{e1}--\eqref{e3} has a unique solution  $u$.
Furthermore,we then have that $u^\epsilon(x,t)$ converges to
$u(t)$ as $\epsilon$ tends to zero uniformly in t.
\end{theorem}

\begin{proof}
 Assume that $\lim _{\epsilon \to 0}u^\epsilon(x,0)=u_{0}(x)$
exists.  Let
$$
u(x,t)=\sum_{p=1}^\infty \Big(e^{-tp^2}u_{0p}- \int_0^t
 e^{(s-t)p^2}f_p(s)ds
 \Big)\sin (px)
$$
where $u_{0p}=  \frac{2}{\pi} \int_0^{\pi}u_0(x)\sin (px) dx$.
It is clear to see that $u(x,t)$ satisfies \eqref{e1}--\eqref{e2}.
We have the formula of $u^\epsilon(x,t)$
\[
u^\epsilon(x,t)=\sum_{p=1}^\infty
 \Big(e^{-tp^2}u^\epsilon_{0p}-\int_0^t
 \frac {e^{(s-t-T) p^2}}{\epsilon p^2+e^{-T p^2}}f_{p}(s)ds\Big)\sin
 (px)
\]
where $u^\epsilon_{0p}=  \frac{2}{\pi} \int_0^{\pi}u^\epsilon(x,0)\sin
 (px)
dx$.
In view of the inequality $ (a+b)^2 \le 2(a^2+b^2)$, we have
\begin{align*}
&\|u^\epsilon(x,t)-u(x,t)\|^2 \\
&\le \frac{\pi}{2}  \sum_{p=1}^\infty (u^\epsilon_{0p}-u_{0p})^2+
\frac{\pi}{2}t^2\sum_{p=1}^\infty
\Big( \int_0^t e^{(2s-2t)p^2} \frac{\epsilon^2 p^4}{(\epsilon p^2
+e^{-T p^2})^2}f_{p}^2 ds\Big) \\
&\le \|u^\epsilon(x,0)-u_0(x)\|^2+ T^2
 \frac{T^2}{ \left(1+\ln(T/\epsilon)\right)^2}
\int_0^t \sum_{p=1}^\infty p^4 f_{p}^2 ds\\
&=\|u^\epsilon(x,0)-u_0(x)\|^2+
\frac{T^4}{ \left(1+\ln(T/\epsilon)\right)^2}
\int_0^t \|f_{xx}\|^2 ds \\
&\le \|u^\epsilon(x,0)-u_0(x)\|^2
+ \frac{T^4}{ \left(1+\ln(T/\epsilon)\right)^2}
 \|f_{xx}\|^2_{L^2 (0,T; L^2(0,\pi))}
\end{align*}
Hence, $\lim _{\epsilon \to 0}u^\epsilon(x,t)=u(x,t)$.
Thus $\lim _{\epsilon \to 0}u^\epsilon(x,T)=u(x,T)$.
Using  theorem \ref{thm2.2}, we have $u(x,T)=g(x)$.
Hence, $u(x,t)$ is the unique solution of the problem
\eqref{e1}--\eqref{e3}. We also see that $u^\epsilon(x,t)$
converges to $u(x,t)$ uniformly in $t$.
\end{proof}

\begin{theorem} \label{thm2.4}
Let $f(x,t) ,g(x),\epsilon$ be as Theorem \ref{thm2.3}.
If the sequence $u_t^\epsilon(x,0)$ converges in $L^2(0,\pi)$,
 then the problem \eqref{e1}--\eqref{e3} has a unique solution  $u$.
 Furthermore,we  have that $u^\epsilon(x,t)$ converges
to $u(x,t)$as $\epsilon$ tends to zero in $C^1(0,T;L^2(0,\pi))$.
\end{theorem}

\begin{proof}
 Assume that $\lim _{\epsilon \to 0}u_t^\epsilon(x,0)=v(x)$
in $L^2(0,\pi)$. Let $v(x)=\sum_{p=1}^\infty  v_p \sin (px)$
where $v_p=\frac{2}{\pi} \int_0^{\pi}v(x)\sin (px) dx$.
Denote by $w_p=-\frac{v_p}{p^2}$ and
$w(x)=\sum_{p=1}^\infty  w_p \sin (px)$.
It is easy to show that the function $u(x,t)$ defined by
$$
u(x,t)=\sum_{p=1}^\infty  \Big( e^{-tp^2}w_p - \int_0^t e^{(s-t)p^2}
f_pds\Big)\sin (px)
$$
is a solution of the problem
\begin{gather*}
u_t(x,t)-u_{xx}=f(x,t),\\
u(x,0)=w(x)
\end{gather*}
Since $u^\epsilon (x,t)$ is the solution of  \eqref{e6}--\eqref{e8},
we have
\begin{gather*}
u^\epsilon_{tp}(t)=p^2 u^\epsilon_p(t)+\frac {{e^{-T p^2}}}{\epsilon
 p^2
+e^{-Tp^2}}f_p(t),\\
u_{tp}(t)=p^2 u_p (t)+f_p(t)
\end{gather*}
where
\begin{gather*}
u^\epsilon_p(t)=\frac{2}{\pi} \int_0^{\pi}u^\epsilon(x,t)\sin (px)
 dx,\quad
u_p (t)= \frac{2}{\pi} \int_0^{\pi}u(x,t)\sin (px) dx \\
u^\epsilon_{tp}(t)=\frac{2}{\pi} \int_0^{\pi}u_t^\epsilon(x,t)\sin (px)
 dx,
\quad
u_{tp} (t)= \frac{2}{\pi} \int_0^{\pi}u_t(x,t)\sin (px) dx
\end{gather*}
So that
\begin{equation} \label{e17}
 u^\epsilon_p(t)-u_p (t)=\frac{1}{p^2} (u^\epsilon_{tp}(t)-u_{tp} (t))
+ \frac {\epsilon}{\epsilon p^2+e^{-Tp^2}}f_p(t)
\end{equation}
By a direct computation,
 \begin{align*}
\|u^\epsilon(.,t)-u(.,t)\|^2
&=\frac{\pi}{2}\sum_{p=1}^\infty  |u^\epsilon_p(t)-u_p (t)|^2 \\
&\le  \sum_{p=1}^\infty  \pi|u^\epsilon_{tp}(t)-u_{tp} (t)|^2
 + \pi \sum_{p=1}^\infty  \frac {\epsilon^2 }{(\epsilon p^2
 +e^{-Tp^2})^2}f^2_p(t)\\
&\le 2\|u_t^\epsilon(x,t)-u_t(x,t)\|^2+ 2
 \frac{T^2}{ (1+\ln(T/\epsilon)^2}\|f(.,t)\|^2
\end{align*}
Hence
\[
\|u^\epsilon(.,0)-u(.,0)\|^2 \le 2 \|u_t^\epsilon(x,0)-u_t(x,0)\|^2
+ \frac{2T^2}{ (1+\ln(T/\epsilon))^2}\|f(.,t)\|^2
\]
Using $\lim _{\epsilon \to 0}u_t^\epsilon(x,0)=v(x)=u_t(x,0)$, we get
$\lim _{\epsilon \to 0}\|u^\epsilon(x,0)-u(x,0)\|=0$.
On the other hand, we have
 \begin{gather*}
 u^\epsilon(x,T)=\sum_{p=1}^\infty \Big(e^{-Tp^2}u^\epsilon_p(0)
+\int_0^T \frac{e^{(s-2T)p^2}}{\epsilon p^2+e^{-Tp^2}}f_p(s)ds\Big)
\sin (px)\\
u(x,T)=\sum_{p=1}^\infty \Big(e^{-Tp^2}u_p(0)+\int_0^T {e^{(s-T)p^2}}
f_p(s)ds\Big) \sin (px)
\end{gather*}
It follows that
\begin{align*}
&\| u^\epsilon(.,T)-u(.,T)\|^2 \\
&\le 2 \sum_{p=1}^\infty  e^{-2Tp^2}|u^\epsilon_p(0)-u_p(0)|^2
 + 2 T^2 \sum_{p=1}^\infty  \int_0^T
 \frac {\epsilon^2 p^4}{(\epsilon p^2+e^{-Tp^2})^2}f^2_p(s)ds\\
&\le   \|u^\epsilon(.,0)-u(.,0)\|^2
 +2\frac{T^2}{ (1+\ln(T/\epsilon))^2}\|f_{xx}(.,t)\|^2
 \end{align*}
Hence, $\lim _{\epsilon \to 0}\|u^\epsilon(x,T)-u(x,T)\|=0$.
Using the Theorem \ref{thm2.2}, we obtain $u(x,T)=g(x)$. This implies
that $u(x,t)$ is the unique  solution of \eqref{e1}--\eqref{e3}.
\end{proof}

\begin{theorem} \label{thm2.5}
If there exists  $m\in (0,2)$ so that
$ \sum_{p=1}^\infty p^{2m}e^{mTp^2}g_p^2 $ converges, then
$$
\|u^\epsilon(x,T)-g(x)\| \le  \frac{\sqrt{ C_1\epsilon ^m} } {m}
$$
where $C_1=4 \sum_{p=1}^\infty p^{2m}e^{mTp^2}g_p^2$.
\end{theorem}

\begin{proof}
Let $m$ be in $ (0,2)$ such that $ \sum_{p=1}^\infty
 p^{2m}e^{mTp^2}g_p^2 $
converges, and let $n$ be in $(0,2) $. Fix a natural integer $p$,
and define
$$
g_p(\epsilon)=\frac{\epsilon^n }{(\epsilon p^2 +e^{-Tp^2})^2}.
$$
It can be shown that  $g_p(\epsilon) \le  g_p(\epsilon_{0})$, for all
$\epsilon>0 $
where $\epsilon_{0}=\frac{ n e^{-Tp^2}  }{(2-n)p^2  } $.
Furthermore, from \eqref{e16}, we have
\begin{equation} \label{e18}
\|u^\epsilon(x,T)-g(x)\|^2
=\sum_{p=1}^\infty  \frac{\epsilon^2 p^4 g_p^2}{ (\epsilon
 p^2+e^{-Tp^2})^2}
=\epsilon^{2-n} \sum_{p=1}^\infty p^4 g^2_p g_p(\epsilon)
\end{equation}
It follows that
\begin{equation} \label{e19}
\|u^\epsilon(x,T)-g(x)\|^2 \le \epsilon^{2-n} (\frac {n}{2-n})^{n}
\sum_{p=1}^\infty p^{4-2n}g_p^2e^{(2-n)Tp^2  }
\end{equation}
If we choose $n=2-m$, we obtain
$\|u^\epsilon(x,T)-g(x)\|^2 \le C_1\epsilon^m m^{-2}$.
\end{proof}

\begin{theorem} \label{thm2.6}
Let $ f \in L^{2}(0,T;L^{2}(0,\pi))$ and $ g \in L^{2}(0,\pi) $
and $\epsilon \in (0,eT)$. Suppose that Problem \eqref{e1}--\eqref{e3}
has a unique solution $u(x,t)$ in
 $C([0,T];H_0^1(0,\pi))\cap C^1((0,T);L^2(0,\pi))$ which satisfies
$\|u_{xx}(.,t)\|<\infty$. Then
\[
  \norm{ u(.,t)- u^{\epsilon}(.,t)}\le \frac{C}{1+\ln(T/\epsilon)}
\]
for every $t\in[0,T]$, where
$ C= T \sup _{t\in [0,T]} \|u_{xx}(.,t)\| $ and
$u^\epsilon$ is the unique solution of  \eqref{e6}--\eqref{e8}.
\end{theorem}

\begin{proof}
Suppose  \eqref{e1}--\eqref{e3} has an exact solution
$u$ in the space $C([0,T];H_0^1(I))\cap C^1((0,T);L^2(I))$, we get the
formula
\begin{equation} \label{e20}
 u(x,t)=\sum_{p=1}^\infty (e^{-(t-T) p^2}g_{p}
 -\int_t^T e^{-(t-s) p^2}f_{p}(s)ds)\sin (px)
\end{equation}
 From \eqref{e10} and \eqref{e20}, we obtain
\begin{equation} \label{e21}
\begin{aligned}
|u_{p}(t)- u^{\epsilon}_{p} (t)|
&= \Big|\Big(e^{-(t-T) p^2}-\frac {{e^{-t p^2}}}{\epsilon
 p^2+e^{-Tp^2}}
\Big)\Big(g_{p}-\int_t^T e^{-(T-s) p^2}f_{p}(s)ds)\Big)\Big|
\\
&= \epsilon p^2\frac {e^{-(t-T) p^2}}{\epsilon p^2+e^{-Tp^2}}
\Big|\Big(g_{p}-\int_t^T e^{-(T-s) p^2}f_{p}(s)ds)\Big)\Big|
\\
& \le  \frac{T}{1+\ln(T/\epsilon)}
\Big|\Big(p^2e^{-(t-T) p^2}g_{p}-\int_t^T p^2e^{-(t-s) p^2}f_{p}(s)ds)
\Big)\Big|
\end{aligned}
\end{equation}
It follows that
\begin{align*}
&\norm{ u(.,.,t)- u^{\epsilon}(.,.,t)}^2\\
&=\frac{\pi}{2}\sum_{p=1}^\infty |u_{p}(t)- u^{\epsilon}_{p} (t)|^2\\
&\leq \frac{\pi}{2} \Big(  \frac{T}{1+\ln(T/\epsilon)}\Big)^{2}
\sum_{p=1}^\infty \Big(p^2e^{-(t-T) p^2}g_{p}-\int_t^T p^2
e^{-(t-s) p^2}f_{p}(s)ds)\Big)^{2}
\\
&=\Big(  \frac{T}{1+\ln(T/\epsilon)}\Big)^{2} \norm{u_{xx}(.,t)}^{2}
\le \Big(  \frac{C}{1+\ln(T/\epsilon)}\Big)^{2}
\end{align*}
Hence
\[
\norm{ u(.,t)- u^{\epsilon}(.,t)}\le \frac{C}{1+\ln(T/\epsilon)}
\]
where $C= T \sup _{t\in [0,T]} \|u_{xx}(.,t)\| $.
This completes the proof
\end{proof}

\begin{remark} \label{rmk2}\rm
 Note that in \cite[Theorem 3.3]{Trong}, the exact solution
$u$ satisfies the condition $\Delta^{2}u(x,t)\in L^2(0,\pi)$,
while the condition of its in this theorem is $\Delta u \in L^2(0,\pi)
 $.
So,  this also implies that the final value $g$ in our theorem
is only in $L^2(0,\pi)$, not satisfying the condition (*) given
in \cite{Trong2} (see Introduction). Further more, we also have the
 error
estimate $\norm{ u_t(.,t)- u_t^{\epsilon}(.,t)}$ which is not given
in \cite{Trong,Trong2}. Hence, this result is an improvement of
known result in \cite{Trong,Trong2}.
\end{remark}


\begin{theorem} \label{thm2.7}
Let $ f \in L^{2}(0,T;L^{2}(0,\pi))$ and $ g \in L^{2}(0,\pi) $
and $\epsilon \in (0,eT)$. Suppose that Problem \eqref{e1}--\eqref{e3}
 has a unique solution $u(x,t)$ in
$C([0,T];H_0^1(0,\pi))\cap C^1((0,T);L^2(0,\pi))$ which satisfies
$\|u_{xxxx}(.,t)\|<\infty$. Then
 \[
  \norm{ u_t(.,t)- u_t^{\epsilon}(.,t)}\le \frac{D}{1+\ln(T/\epsilon)}
\]
for every $t\in[0,T]$, where
\[
 D= T \Big(2 \sup _{t\in [0,T]} (\|u_{xxxx}(.,t)\|^2+\|f_{xx}(.,t)\|^2)
 \Big)^{1/2}
\]
 and $u^\epsilon$ is the unique solution of  \eqref{e6}--\eqref{e8}.
\end{theorem}

\begin{proof}
In view of \eqref{e17}, we have
\begin{align*}
u^\epsilon_{tp}(t)-u_{tp} (t)
&= p^2 (u^\epsilon_p(t)-u_p (t))-\frac {\epsilon p^2}{\epsilon p^2
+e^{-Tp^2}}f_p(t) \\
&= \frac {\epsilon p^4 e^{-(t-T) p^2}}{\epsilon p^2+e^{-Tp^2}}
\Big(g_{p}-\int_t^T e^{-(T-s) p^2}f_{p}(s)ds)\Big)
-\frac {\epsilon p^2}{\epsilon p^2+e^{-Tp^2}}f_p(t)\\
&= \frac {\epsilon p^4 }{\epsilon p^2+e^{-Tp^2}} u_p(t)
 -\frac {\epsilon p^2}{\epsilon p^2+e^{-Tp^2}}f_p(t)\\
&=\frac {\epsilon p^2}{\epsilon p^2+e^{-Tp^2}} (p^2 u_p(t)-f_p(t))
\end{align*}
Hence, we get
\begin{align*}
\norm{ u_t(.,t)- u_t^{\epsilon}(.,t)}^2
&= \frac{\pi}{2}\sum_{p=1}^\infty  |u^\epsilon_{tp}(t)-u_{tp} (t)|^2 \\
&\le \pi \frac {\epsilon^2  }{(\epsilon
 p^2+e^{-Tp^2})^2}\sum_{p=1}^\infty
 (p^8u^2_p(t)+ p^4f^2_p(t))\\
&= \frac{2T^2}{ \left(1+\ln(T/\epsilon)\right)^2} (\|u_{xxxx}(x,t)\|^2
+ \|f_{xx}(x,t)\|^2)
\end{align*}
This completes the proof.
\end{proof}

In the case of nonexact data, one has the following result.

\begin{theorem} \label{thm2.8}
 Let $f,g,\epsilon $ be as in Theorem \ref{thm2.6}. Assume that the
 exact
solution $u$ of \eqref{e1}--\eqref{e3} corresponding to $g$ satisfies
$$
u \in C([0,T];L^2(0,\pi))\cap L^2(0,T;H_0^1(0,\pi))
\cap C^1((0,T);L^2(0,\pi)),
$$
and $\|u_{xx}(.,t)\|< \infty$.
Let $g_\epsilon \in L^2(0,\pi)$ be a measured data such that
$$
\norm{g_\epsilon-g}\le\epsilon.
$$
Then there exists a function $ u^{\epsilon}$satisfying
\[
  \norm{ u(.,t)- u^{\epsilon}(.,t)}\le \frac{C+T}{1+\ln(T/\epsilon)}
\]
for every $t\in [0,T]$ and $C$ is defined in Theorem \ref{thm2.6}.
\end{theorem}

\begin{proof}
Let $v^\epsilon$ be the solution of problem \eqref{e6}--\eqref{e8}
 corresponding to $g$ and let again  $u^{\epsilon}$ be the solution
of problem \eqref{e6}--\eqref{e8} corresponding to $g_\epsilon$
where $g, g_\epsilon$ are in right hand side of \eqref{e6}.
Using Theorem \ref{thm2.6} and Step 2 in Theorem \ref{thm2.1}, we get
\begin{align*}
\norm{u^\epsilon(.,t)-u(.,t)}
&\le
 \norm{u^{\epsilon}(.,t)-v^\epsilon(.,t)}+\norm{v^\epsilon(.,t)-u(.,t)} \\
&\le \frac{T}{\epsilon (1+\ln(T/\epsilon)}\norm{g_\epsilon-g}
 + \frac{T}{ (1+\ln(T/\epsilon)}\|u_{xx}(.,t)\| \\
&\le \frac{C+T}{1+\ln(T/\epsilon)}
\end{align*}
for every $t \in (0,T)$ and where $C$ is defined in Theorem
 \ref{thm2.6}.
This completed the proof.
\end{proof}

\section{A numerical example}

We consider
\begin{equation} \label{e22}
\begin{gathered}
u_t-u_{xx} = f(x,t)\equiv 2e^{t}\sin x, \\
u(x,1) = g (x) \equiv e\sin x .
\end{gathered}
\end{equation}
The exact solution to this problem is
\[
u(x,t) = e^t \sin x
\]
Note that $u(x,1/2)=  \sqrt e \sin (x)\approx 1.648721271\sin(x)$.
Let $g_n $ be the  measured final data
\[
    g _n  (x) = e\sin(x) + \frac{1}{n} \sin (nx).
\]
So that the data error, at the final time, is
\[
    F(n) = \|g_n  - g \|_{L^2 (0,\pi ) }
    = \sqrt {\int_0^\pi  {\frac{1}{{n^2 }}\sin ^2 (nx)dx} }
     = {\frac{\pi }    {2n}}.
\]
The solution of \eqref{e25}, corresponding the final value $g _n $, is
\[
 u^n (x,t) = e^t \sin (x)  + \frac{1}     {n}e^{n^2 (1 - t)} \sin
(nx),
\]
The error at the original  time is
\[
 O(n):= \|u^n (.,0) - u(.,0)\|_{L^2 (0,\pi  )}
 = \sqrt {\int_0^\pi {\frac{{e^{2n^2 } }}
{{n^2 }}\sin ^2 (nx)\,dx} }
 = \frac{{e^{n^2 } }}{{n }} {\frac{\pi }{2}}.
 \]
Then, we notice that
\begin{gather} \label{e23}
\lim_{n \to \infty } F(n)=\mathop {\lim }_{n \to
\infty } ||g _n  - g ||_{L^2 (0,\pi )}  =
\lim _{n \to \infty } \frac{1} {n} {\frac{\pi }
{2}}  = 0,\\
\lim _{n \to \infty } O(n)=\lim _{n \to
\infty } \|u^n (.,0) - u(.,0)\|_{L^2 (0,\pi )}
= \lim_{n \to \infty } \frac{{e^{n^2 } }} {{n^{} }}{\frac{\pi }
{2}}  = \infty. \label{e24}
\end{gather}
 From the two equalities above,  we see that \eqref{e22}  is  an
ill-posed problem. Approximating the problem as in
 \eqref{e1}--\eqref{e3}, the regularized solution is
\begin{equation} \label{e25}
u^{\epsilon}(x,t)
=\sum_{p=1}^\infty \Big(\frac {{e^{-t p^2}}}{\epsilon
 p^2+e^{-p^2}}g_{p}
-\int_t^1 \frac {e^{(s-t-1) p^2}}{\epsilon p^2+e^{-
 p^2}}f_{p}(s)ds\Big)
\sin (px)
\end{equation}
for $0\le t \le 1$. Hence, we have
\begin{equation} \label{e26}
u^{\epsilon}(x,t)=
\frac {{e^{1-t }}}{\epsilon +e^{-1}}\sin x
-2\Big(\int_t^1 \frac {e^{2s-t-1 }}{ \epsilon +e^{- 1}}ds\Big)\sin x
 +\frac{1}{n}\frac {{e^{-tn^{2} }}}{\epsilon n^{2}
+e^{-n^{2}}} \sin (nx)\,.
\end{equation}
It follows that
\begin{equation} \label{e27}
u^{\epsilon}(x,\frac{1}{2})
=\Big(\frac {e^{1/2}}{\epsilon +e^{-1}}
-2\int_{\frac{1}{2}}^1 \frac {e^{2s-\frac{3}{2} }}{\epsilon +e^{- 1}}ds
\Big)\sin x  +\frac{1}{n}\frac {{e^{-\frac{1}{2}n^{2} }}}{\epsilon
 n^{2}
+e^{-n^{2}}} \sin (nx)
\end{equation}
Let $a_\epsilon = \|u_\epsilon(.,\frac{1}{2}) - u(.,\frac{1}{2})\|$
be the error between the regularized solution $u_\epsilon$
 and the exact solution $u$ in the time $t=\frac{1}{2}$.
Let $n=300$ and
\begin{gather*}
 \epsilon = \epsilon_1 = 10^{-2}\sqrt{\frac{\pi}{2}}, \epsilon
=\epsilon_2 = 10^{-4}\sqrt{\frac{\pi}{2}},\\
\epsilon=\epsilon_3 = 10^{-10}\sqrt{\frac{\pi}{2}}, \epsilon
=\epsilon_4 = 10^{-15}\sqrt{\frac{\pi}{2}}.
\end{gather*}


\begin{table}
\caption{}
$\begin{array}{|c|c|c|c|c|} \hline
\epsilon& u_\epsilon&  a_\epsilon \\
\hline
\epsilon_1 = 10^{-2}\sqrt{\frac{\pi}{2}}
&   1.59440220314355 \sin(x)
 &  0.06807885585\\
 &+ 4.636337144 \times 10^{-39093 }\sin(300x)&\\
\hline
\epsilon_2 =10^{-4}\sqrt{\frac{\pi}{2}}
& 1.64815976557002 \sin(x)
& 0.0007037421545\\
    &+ 4.636337144 \times  10^{-39091}\sin(300x)&\\
\hline
\epsilon_3 = 10^{-10} \sqrt{\frac{\pi}{2}}
& 1.64872127013843\sin x
&  1.253314137 \times 10^{-9} \\
   &+ 4.636337144 \times  10^{-39084}\sin(300x)&\\
\hline
\epsilon_4 =10^{-16}\sqrt{\frac{\pi}{2}}
& 1.64872127070011\sin(x)
&  5.810786885 \times 10^{-39079}  \\
 & +4.636337144 \times 10^{-39079}\sin(300x) &\\
\hline
\end{array}$
\end{table}

We note that the new method in this  article give a  better
 approximation
than the previous method in \cite{Trong}.
To prove this, we have in view of the error table in
\cite[p. 9]{Trong}.

\begin{table}[ht]
\caption{}
\begin{tabular}{|c|c|c|}
\hline
$\epsilon$& $u_\epsilon$& $\|u-u_\epsilon\|$\\ \hline
 $10^{-2}\sqrt{\frac{\pi}{2}}$&
 $1.643563444\sin(x)+0.8243606355\sin200x$ & 0.1462051256\\ \hline
 $10^{-4}\sqrt{\frac{\pi}{2}}$&
$1.648617955\sin(x) + 0.1648721271\sin10000 x$& 0.02066391506\\
\hline
 $10^{-10}\sqrt{\frac{\pi}{2}}$
 &$1.648721271\big(\sin(x) +10^{-10}\sin(10^{10}x)\big)$
 &0.00002066365678\\
\hline
 $10^{-16}\sqrt{\frac{\pi}{2}}$
 &$1.648721271\big(\sin(x) +10^{-16}\sin(10^{16}x)\big)$
 &$2.066365678\times10^{-8}$\\
\hline
 $10^{-30}\sqrt{\frac{\pi}{2}}$
 &$1.648721271\big(\sin(x) +10^{-30}\sin(10^{30}x)\big)$
 &$2.066365678\times10^{-15}$\\
\hline
\end{tabular}
\end{table}

Furthermore, we continue to approximate this problem  by the method
given in \cite{Trong}, which gives regularized solution
\[
v^{\epsilon}(x,t)
=\sum_{p=1}^\infty \Big(\frac {{e^{-t p^2}}}{\epsilon +e^{-p^2}}g_{p}
-\int_t^1 \frac {e^{-t p^2}}{\epsilon ^{s}+e^{- sp^2}}f_{p}(s)ds\Big)
\sin (px).
\]
Hence, we have
\begin{equation} \label{e28}
v^{\epsilon}(x,t)=
\frac {{e^{1-t }}}{\epsilon +e^{-1}}\sin x
-2\Big(\int_t^1 \frac {e^{s-t }}{\epsilon^{s} +e^{- s}}ds\Big)\sin x
 +\frac{1}{n}\frac {{e^{-tn^{2} }}}{\epsilon  +e^{-n^{2}}}
\sin (nx)
\end{equation}
for $0\le t \le 1$. It follows that
\begin{equation} \label{e29}
v^{\epsilon}(x,\frac{1}{2})=
\frac {e^{1/2}}{\epsilon +e^{-1}}\sin x
-2\Big(\int_{\frac{1}{2}}^1 \frac {e^{s-\frac{1}{2} }}{\epsilon^{s}
+e^{- s}}ds\Big)\sin x
 +\frac{1}{n}\frac {{e^{-\frac{1}{2}n^{2} }}}{\epsilon
+e^{-n^{2}}} \sin (nx).
\end{equation}

\begin{table}
\caption{}
$\begin{array}{|c|c|c|c|c|} \hline
\epsilon& v_\epsilon&  a_\epsilon \\
\hline
\epsilon_1 = 10^{-2}\sqrt{\frac{\pi}{2}}
&   1.701714206 \sin(x)
 & 0.06641679460\\
   &+4.172703428 \times 10^{-39088 }\sin(300x) &\\
\hline
\epsilon_2 =10^{-4}\sqrt{\frac{\pi}{2}}
& 1.656775314\sin(x)
& 0.01009424595\\
    &+ 4.172703428\times  10^{-39086}\sin(300x)&\\
\hline
\epsilon_3 = 10^{-10}\sqrt{\frac{\pi}{2}}
& 1.648724344 \sin x
& 0.000003851434344\\
   &+ 4.172703428 \times  10^{-39080}\sin(300x)&\\
\hline
\epsilon_4 =10^{-16}\sqrt{\frac{\pi}{2}}
& 1.648721273\sin(x)
&   2.506628275 \times 10^{-9}\\
   &+ 4.172703428 \times  10^{-39074}\sin(300x)&\\
 \hline
\end{array}$
\end{table}

Looking at Tables 1,2,3,  a comparison between the three
methods, we can see the error results of in Table 1 are smaller than
the errors in  Tables 2 and 3.
This shows that our approach has a nice regularizing effect and give
 a better approximation with comparison to the  previous method in,
for example \cite{Trong,Trong2}.

 \subsection*{Acknowledgments}
The authors would like to thank the referees for their valuable
criticisms leading to the improved version of our paper.

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