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

\AtBeginDocument{{\noindent\small
{\em Electronic Journal of Differential Equations},
Vol. 2006(2006), No. 04, pp. 1--10.\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 2006 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2006/04\hfil Regularization and error estimates]
{Regularization and error estimates for nonhomogeneous backward
heat problems}
\author[D. T, Dang,  H. T. Nguyen\hfil EJDE-2006/04\hfilneg]
{Duc Trong Dang,  Huy Tuan Nguyen}  % in alphabetical order

\address{Duc Trong Dang\hfill\break
 Department of Mathematics, Hochiminh City National University,
 227 Nguyen Van Cu, Q5, HoChiMinh City, Vietnam}
\email{ddtrong@mathdep.hcmuns.edu.vn}

\address{Huy Tuan Nguyen \hfill\break
 Department of Mathematics, Hochiminh City National University,
 227 Nguyen Van Cu, Q5, HoChiMinh City, Vietnam}
\email{tuanhuy\_bs@yahoo.com}

\date{}
\thanks{Submitted November 11, 2005. Published January 11, 2006.}
\subjclass[2000]{35K05, 35K99, 47J06, 47H10}
\keywords{Backward heat  problem; ill-posed problem;
 contraction principle; \hfill\break\indent
 quasi-reversibility methods}

\begin{abstract}
 In this article, we study the inverse time problem for the
 non-homogeneous heat equation which is a severely ill-posed problem.
 We regularize this problem using the quasi-reversibility method
 and then obtain error estimates on the approximate solutions.
 Solutions are calculated by the contraction principle
 and shown in numerical experiments. We obtain also rates of
 convergence to the exact solution.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}

\section{Introduction}

For a positive real number $T$, consider the problem of finding
the temperature  $u(x,t)$,  such that
\begin{gather}
u_t-u_{xx}=f(x,t),\quad 0\leq x \leq \pi,\; 0<t<T,  \label{e1}\\
u(0,t)=u(\pi,t)=0,\quad 0<t<T, \label{e2}\\
u(x,T)=g(x), \quad 0\leq x.\leq \pi \label{e3}
\end{gather}
where $g(x), f(x,t) $ are given functions. This problem is called
the backward heat problem  (BHP), or  final-value problem. As is
known, such  problem is severely ill-posed; i.e., solutions do not
always exist, and when they exist, they do not depend continuously
on the given data. In fact, for small noise contaminating  physical
measurements, the corresponding solutions have large errors. This
makes difficult to use numerical calculations with inexact data.
Hence, a regularization is needed. When $f=0$, we have a homogenous
problem,
\begin{equation}
\begin{gathered} \label{FVP}
u_t+ Au= 0,\quad 0<t<T,\\
u(T)=\varphi.
\end{gathered}
 \end{equation}
that has been considered by several authors
in the previous four decades.  Lattes and Lions
\cite{l2}, Miller \cite{m1}, Payne \cite{p1}, Huang and Zheng  
\cite{h2},
 and Lavrentiev \cite{l3}
 have approximated \eqref{FVP} by perturbing the operator $A$. This
approach called the ``quasi-reversibility method''. The main idea
of this method is that by perturbing the equation in the ill-posed
problem, one may obtain a well-posed problem. Then use  the
solution of the well-posed problem as an approximate solutions of
the ill-posed problem.

   Lattes and Lions \cite{l2} 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*}
Alekseeva and Yurchuk \cite{a1} considered the problem
\begin{equation} \label{e4-5}
\begin{gathered}
u_t+Au+\epsilon Au_t=0,\quad  0<t<T, \\
u(T)=\varphi. \
\end{gathered}
\end{equation}
Gajewski and Zaccharias  \cite{g1} consider a problem  similar
to \eqref{e4-5}. Their error estimate for
the approximate solutions is
\begin{equation} \label{e6}
\|u^\epsilon(t)-u(t)\|^2 \leq  \frac{2}{t^2}(T-t)\|u(0)\|
\end{equation}
Note that these estimate can not be used at the time $t=0$.
Showalter \cite{s1,s2} presented a different method for regularizing
\eqref{FVP}, which is a stability estimate better than the previous 
ones.
Using Showalter's idea, Clark and Oppenheimer \cite{c1} used the
quasi-boundary method to regularize the backward problem with
 \begin{gather*}
u_t+Au(t)=0, \quad 0<t<T, \\
u(T)+\epsilon u(0)=\varphi\,.
\end{gather*}
A similar approaches known as quasiboundary method was given in
\cite{m2}. Also, we have to mention that nonstandard  conditions for
the parabolic equation have been considered in some recent papers
\cite{a2,a3}. Denche and Bessila \cite{d1} approximated this
problem  by perturbing the final condition \eqref{e2} with
a derivative of the same order as the equation:
\begin{gather*}
u_t+Au(t)=0,\quad 0<t<T, \\
u(T)-\epsilon u'(0)=\varphi\,.
\end{gather*}
   Huang and Zheng  \cite{h1} considered  problem
 \eqref{e4-5} where operator $-A$ is the generator of
an analytic semigroup in a Banach space.  However, they do not give
error estimates and effective methods of calculation.

 Although there are many publication on the backward  problem,
most of them are for the homogeneous case, and the literature of
the non-homogeneous case is quite scarce.  In this  paper, we
consider backward heat problem  in the non-homogeneous case. Our
results generalize  many results in previous papers; see for example
\cite{a1,a2,a3,a4,c1,h1,g1,s1}. We use quasi-reversibility to 
approximate
Problem \eqref{e1}--\eqref{e3} as the follows:
\begin{gather}
u_t^\epsilon   - u_{xx}^\epsilon   - \epsilon u_{xxxx}^\epsilon
 = \sum_{n = 1}^\infty  {e^{ - \epsilon n^4 (T - t)} f_n (t)\sin (nx),}
  \quad 0\leq x \leq \pi,\; 0<t<T,   \label{e7}\\
 u^\epsilon  (0,t) = u^\epsilon  (\pi ,t) = u_{xx}^\epsilon  (0,t)
 = u_{xx}^\epsilon  (\pi ,t) = 0, \quad 0<t<T,  \label{e8}\\
 u^\epsilon  (x,T) = g (x),\quad 0\leq x \leq \pi , \label{e9}
 \end{gather}
where $\epsilon$ is a positive parameter and
\[
f_n (t) = \frac{2}{\pi} \langle f(x,t),\sin (nx) \rangle
=\frac{2}{\pi}\int_{0}^{\pi}f(x,t)\sin (nx) dx\,,
    \]
where $ \langle\cdot,\cdot\rangle$ is the inner product in
$L^2(0,\pi)$. First, we shall prove that, the (unique) solution
$u^\epsilon$ of \eqref{e6}--\eqref{e8} is
\begin{equation} \label{e10}
    u^\epsilon  (x,t) = \sum_{n = 1}^{\infty } (e^{(T - t)(n^2
    - \epsilon n^4 )} g _n  - \int_t^T {e^{(s - t)(n^2
    - \epsilon n^4 )} e^{ - \epsilon n^4 (T - s)} f_n (s)ds)\sin (nx)},
\end{equation}
where $g_n=\frac{2}{\pi}\int_0^\pi g(x) \sin (nx) dx$.

In Section 2, we shall prove that \eqref{e7}--\eqref{e9} is well-posed.
In Section 3, we estimate the error between an exact solution $u$
of \eqref{e1}--\eqref{e3}) and the approximation solution
$u^\epsilon$ of  \eqref{e7}--\eqref{e9}.
In fact, we shall prove that
\begin{equation} \label{e11}
\|u(.,t) - u^\epsilon  (.,t)\|  \leq \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 }.
\end{equation}
Note that with this inequality, the error can be estimated at $t=0$.
Note also that  \eqref{e11} is similar \eqref{e6} when  $f=0$.
In Section 3, we obtain also some other results, including
converges rates.


\section{The well-posed  Problem}

In this section, we shall study the existence, uniqueness and
stability of a (weak) solution to \eqref{e7}--\eqref{e9}. In fact,
 one has the following result.

\begin{theorem} \label{thm1}
Let  $f (x,t)\in L^2 (0,T;L^2 (0,\pi ))$ and let
$g(x)  \in L^2 (0,\pi ) $. Then  \eqref{e7}--\eqref{e9}
has unique a weak solution $u^\epsilon  (x,t)$ which is
in $C([0,T];L^2 (0,\pi )) \cap L^2 (0,T;H_0^1 (0,\pi )
\cap H^2 (0,\pi ))$, and  is given by \eqref{e10}.
 Furthermore, the solution depends
continuously on $g$ in  $ C([0,T];L^2 (0,\pi ))$.
\end{theorem}

\begin{proof}
The proof is divided into three steps. In step 1, we prove that the 
function
$ u^\epsilon(t) $ given by \eqref{e10},
is a solution of  \eqref{e7}--\eqref{e9}. In Step 2,  we  prove
the  uniqueness. Finally in Step 3, we prove the stability of the 
solution.

\noindent{\bf Step 1:} Functions given by \eqref{e10} are solutions
of \eqref{e7}--\eqref{e9}.
Let $u^\epsilon(x,t)$ be given by \eqref{e10}.
Then we can verify directly that
$u^\epsilon  (x,t) \in C([0,T];L^2 (0,\pi )) \cap L^2 (0,T;H_0^1 
(0,\pi )
\cap H^2 (0,\pi ))$. In fact,
$u^\epsilon \in C^{\infty}((0,T];H_0^1(0,\pi)))$. Moreover,
\begin{align*}
u^\epsilon_t(x,t)
&= \sum_{n = 1}^{ \infty } \Big(( - n^2  + \epsilon n^4 )(e^{(T - 
t)(n^2
- \epsilon n^4 )} g _n  \\
&\quad - \int_t^T {e^{(s - t)(n^2- \epsilon n^4 )}
 e^{ - \epsilon n^4 (T - s)} f_n (s)ds)\sin (nx)} \Big) \\
&\quad + \sum_{n = 1}^{ \infty } {\int_t^T {e^{ - \epsilon n^4 (T - t)}
f_n (s)ds\sin (nx),} }
\end{align*}
\begin{align*}
&u^\epsilon_{xx}(x,t) \\
&= \sum_{n = 1}^{ \infty } ( - n^2 )(e^{(T - t)(n^2
 - \epsilon n^4 )} g _n  - \int_t^T {e^{(s - t)(n^2  - \epsilon n^4 )}
 e^{ - \epsilon n^4 (T - s)} f_n (s)ds)\sin (nx),}
\\
&u^\epsilon_{xxxx}(x,t) \\
&= \sum_{n = 1}^{ \infty } n^4 (e^{(T - t)(n^2
- \epsilon n^4 )} g _n  - \int_t^T {e^{(s - t)(n^2  - \epsilon n^4 )}
e^{ - \epsilon n^4 (T - s)} f_n (s)ds)\sin (nx).}
\end{align*}
    Hence
\[
    u_t^\epsilon(x,t)   - u_{xx}^\epsilon(x,t)
- \epsilon u_{xxxx}^\epsilon(x,t)   = \sum_{n = 1}^\infty
e^{ - \epsilon n^4 (T - t)} f_n (t)\sin (nx).
\]
We also have
\[
u^\epsilon(x,T)=\sum_{n = 1}^{ \infty }  g _n \sin (nx)= g(x) .
\]


\noindent{\bf Step 2:} Problem \eqref{e7}--\eqref{e9} has unique 
solution.
Suppose the there are  two solution
$ u(x,t)$ and $ v(x,t)$. Then we need to show that  $u(x,t)=v(x,t)$.
Let  $ w(x,t)=u(x,t)-v(x,t)$. Then
$w(x,t)$ satisfies the system
            \begin{equation} \label{e13}
\begin{gathered}
 w_t(x,t)   - w_{xx}(x,t)   - \epsilon w_{xxxx}(x,t)   = 0,
 \quad (x,t) \in (0,\pi ) \times (0,T), \\
 w (x,T) = 0, \quad x\in(0,\pi),\\
 w  (0,t) = w  (\pi ,t) = w_{xx}  (0,t) = w_{xx}  (\pi ,t) = 0.
\end{gathered}
 \end{equation}
For $ k>0 $, we  define ${\psi(x,t) = e^{k(t - T) }w(x,t)}$.
 Note that  $\psi(x,t)$ satisfies
\begin{equation} \label{e14}
\begin{gathered}
     \psi_t(x,t)   - \psi_{xx}(x,t)  - \epsilon 
\psi_{xxxx}(x,t)-k\psi(x,t)
     = 0,\quad  (x,t) \in (0,\pi ) \times (0,T),\\
     \psi (x,T) = 0, \quad x\in(0,\pi), \\
 \psi  (0,t) = \psi  (\pi ,t) = \psi_{xx}  (0,t) = \psi_{xx}  (\pi ,t) 
= 0.
\end{gathered}
\end{equation}
Multiplying   \eqref{e14} by  $\psi(x,t)$ and  integrating on $x $ from
0 to  $\pi$, we obtain
\begin{align*}
 \int_0^\pi  {\frac{d } {{dt}}} \psi (x,t)\psi (x,t)dx
 - \int_0^\pi   \psi _{xx} (x,t)\psi (x,t)dx&\\
- \int_0^\pi   \psi _{xxxx} (x,t)\psi (x,t)dx
 - \int_0^\pi  k \psi (x,t)\psi (x,t)dx &= 0.
\end{align*}
Applying  the Green formula, we have
\[
  \int_0^\pi   \psi _{xx} (x,t)\psi (x,t)dx
  =  - \int_0^\pi  {} \psi _x (x,t)\psi _x (x,t)dx
  =  - \|\nabla \psi (x,t)\|^2,
\]
\begin{align*}
  \int_0^\pi   \psi _{xxxx} (x,t)\psi (x,t)dx
&=  - \int_0^\pi  {} \psi _{xxx} (x,t)\psi _x (x,t)dx\\
&= \int_0^\pi  {} \psi _{xx} (x,t)\psi _{xx} (x,t)dx
  = \|\Delta  \psi (x,t)\|^2.
\end{align*}
It follows that
\[
\frac{d }
{{d t}}\|\psi (x,t)\|^2  + \|\nabla \psi (x,t)\|^2
- \epsilon \|\Delta  \psi (x,t)\|^2  - k\|\psi (x,t)\|^2  = 0\,.
\]
Using Schwartz inequality, we have
\begin{align*}
  \|\nabla \psi (x,t)\|^2
&= \int_0^\pi   -  \psi _{xx} (x,t)\psi (x,t)dx \\
&=  \langle - \Delta ^{} \psi (x,t),\psi (x,t)\rangle\\
&\leq \epsilon \|\Delta \psi (x,t)\|^2  + \frac{1}
{{4\epsilon }}\|\psi (x,t)\|^2.
 \end{align*}
Therefore,
\[
\frac{d }{{d t}}\|\psi (x,t)\|^2  \geq (k - \frac{1}
{{4\epsilon }})\|\psi (x,t)\|^2,
\]
Choosing   $k =  1/4\epsilon $, we have
    \begin{eqnarray*}
    \|\psi (.,T)\|^2  - \|\psi(.,t)\|^2  \geq \int_t^T {} (k - \frac{1}
    {4\epsilon })||\psi(.,s) ||^2ds  = 0.
    \end{eqnarray*}
    Since   $w(.,T)=0$ it follows that $w(.,t)=0$ and
    $\psi(.,t)=0$ therefore,  $u(x,t)=v(x,t)$.

\noindent{\bf Step 3:} The solution of \eqref{e7}--\eqref{e9}
depends continuously on $g \in L^2(0,\pi)$.
Let  $u $  and   $ v $ be  two solution of
\eqref{e7}--\eqref{e9} corresponding to the final values   $g$
and $h $, respectively. By \eqref{e10},
\begin{gather*}
    u(x,t) = \sum_{n = 1}^{ \infty } {(e^{(T - t)(n^2  - \epsilon n^4 
)}
    g _n  - \int_t^T {e^{(s - t)(n^2  - \epsilon n^4 )}
    e^{ - \epsilon n^4 (T - t)} f_n (s)ds)\sin (nx)} },\\
    v(x,t) = \sum_{n = 1}^{ \infty } {(e^{(T - t)(n^2  - \epsilon n^4 
)}
    h _n  - \int_t^T {e^{(s - t)(n^2  - \epsilon n^4 )}
    e^{ - \epsilon n^4 (T - t)} f_n (s)ds)\sin (nx)} },
\end{gather*}
    where
\begin{gather*}
    g _n  =  \frac{2}{\pi}\langle g (x),\sin (nx) \rangle, \quad
    h _n  =  \frac{2}{\pi}\langle h (x),\sin (nx) \rangle
\end{gather*}
It follows that
\[
\|u(.,t) - v(.,t)\|_H^2  = \frac{\pi}{2}\sum_{n = 1}^{ \infty }
{e^{2(n^2  - \epsilon n^4 )(T - t)} (g _n  - h _n )^2 }.
\]
In view of the  inequality  $n^2  - \epsilon n^4  \leq 1/(4\epsilon)$,
we have
\begin{align*}
\|u(.,t) - v(.,t)\|^2 &\leq \frac{\pi}{2} \sum_{n = 1}^{ \infty }
{e^{(T - t)/2\epsilon } (g _n  - h _n )^2 }\\
&= \frac{\pi}{2}e^{(T - t)/2\epsilon } \sum_{n = 1}^{ \infty }
  {(g _n  - h _n )^2 }
  = e^{(T - t)/2\epsilon } \|g  - h \|^2 .
\end{align*}
Hence
 \[
    \|u(.,t) - v(.,t)\|  \leq e^{(T - t)/4\epsilon } \|g-h \|.
\]
This completes the proof of Step 3 and the proof of the theorem.
\end{proof}

\section{Regularization of Problem \eqref{e1}--\eqref{e3}}

We first have a uniqueness result.

\begin{theorem} \label{thm2}
Let  $f (x,t)\in L^2 (0,T;L^2 (0,\pi ))$. Then
 \eqref{e1}--\eqref{e3} has at most one (weak) solution in
 $C([0,T];L^2 (0,\pi )) \cap L^2 (0,T;H_0^1 (0,\pi ) \cap H^2 (0,\pi 
))$.
\end{theorem}

The proof of the above lemma can be found in \cite{l1}.
Despite the uniqueness, Problem \eqref{e1}--\eqref{e3} is still 
ill-posed.
Hence, a regularization has to be used.

\begin{theorem} \label{thm3}
Let $f \in L^2 (0,T;L^2 (0,\pi ))$ be such that
$\frac{{\partial ^4 f(x,t)}}{{\partial x^4 }} \in L^2 (0,T;L^2 (0,\pi 
))$.
Suppose that Problem \eqref{e1}--\eqref{e3} has a weak solution $u$
in $C([0,T];L^2 (0,\pi )) \cap L^2 (0,T;H_0^1 (0,\pi ) \cap H^2 (0,\pi 
))$.
Then
\[
\|u(.,t) - u^\epsilon  (.,t)\|  \leq \epsilon (T - t)\sqrt {\frac{8}
{{t^4 }}\|u(.,0)\|^2  + t^2
\|\frac{{\partial ^4 f(x,t)}} {{\partial 
x^4}}\|_{L^2(0,T;L^2(0,\pi))}^2 } ,
\]
 for every $t \in (0,T]$, where $u^\epsilon $ is the unique solution
 of  \eqref{e7}--\eqref{e9}.
\end{theorem}

\begin{proof}
Suppose $u$ is the exact solution of \eqref{e1}--\eqref{e3}.
Then, as shown in \cite{c2},
\begin{eqnarray}
u(x,t) = \sum_{n = 1}^{ \infty } \Big(e^{ - tn^2 } u_n (0) + \int_0^t
e^{(s - t)n^2 } f_n (s)ds\Big)\sin (nx).
\end{eqnarray}
where $u_n(0)=\frac{2}{\pi}\langle u(x,0),\sin (nx)\rangle$. Then
\begin{align*}
    g (x) &= u(x,T) \\
 &= \sum_{n = 1}^{ \infty } \Big(e^{ - Tn^2 } u_n (0)
    + \int_0^T  e^{(s - T)n^2 } f_n (s)ds\Big)\sin (nx),\\
 &=\sum_{n = 1}^\infty  \varphi _n \sin (nx).
\end{align*}
Hence
$g _n  = e^{ - Tn^2 } u_n (0) + \int_0^T {e^{(s - T)n^2 } f_n (s)ds} $
and
\begin{align*}
    u_n^\epsilon  (t)
&=e^{(T - t)(n^2  - \epsilon n^4 )} g _n
    - \int_t^T {e^{(s - t)(n^2  - \epsilon n^4 )}
    e^{ - \epsilon n^4 (T - s)} f_n (s)ds,} \\
& = e^{(T - t)(n^2  - \epsilon n^4 )} (e^{ - Tn^2 } u_n (0)
    + \int_0^T {e^{(s - T)n^2 } f_n (s)ds} )\\
&\quad - \int_t^T {e^{(s - t)(n^2  - \epsilon n^4 )} e^{ - \epsilon n^4 
(T - s)} f_n (s)ds,} \\
&= e^{ - tn^2 } e^{ - \epsilon (T - t)n^4 } u_n (0)
 + \int_0^t {e^{(T - t)(n^2  - \epsilon n^4 )} e^{(s - T)n^2 } f_n 
(s)ds} )\\
&\quad + \int_t^T {e^{(T - t)(n^2  - \epsilon n^4 )} e^{(s - T)n^2 }
 f_n (s)ds} \\
&\quad - \int_t^T e^{(s - t)(n^2  - \epsilon n^4 )}
 e^{ - \epsilon n^4 (T - s)} f_n (s)\,ds.
\end{align*}
It follows that
\begin{equation} \label{e16}
u_n^\epsilon  (t) = e^{ - tn^2 } e^{ - \epsilon (T - t)n^4 } u_n (0)
+ \int_0^t e^{(s - t)n^2 } e^{ - \epsilon (T - t)n^4 } f_n (s)ds.
\end{equation}
From \eqref{e10}, \eqref{e13}, \eqref{e14} and using the inequality
$1-e^{-x} \leq  x $ for $x>0$, we have
\begin{equation} \label{e17}
\begin{aligned}
&|u_n(t) - u_n^\epsilon  (t)|\\
& \leq e^{ - tn^2 } (1 - e^{ - \epsilon n^4 (T - t)} )|u_n (0)|
  + |\int_0^t {e^{(s - t)n^2 } (1 - e^{ - \epsilon n^4 (T - t)} )
  f_n (s)ds} |  \\
&\leq e^{ - tn^2 } (1 - e^{ - \epsilon n^4 (T - t)} )|u_n (0)|
 + \int_0^t {e^{(s - t)n^2 } (1 - e^{ - \epsilon n^4 (T - t)} )|f_n 
(s)|ds}\\
&\leq e^{ - tn^2 } \epsilon n^4 (T - t)|u_n (0)| + \int_0^t {e^{(s - 
t)n^2 }
 \epsilon n^4 (T - t)|f_n (s)|ds} \\
&= \frac{\epsilon }    {{t^2 }}e^{ - tn^2 } (tn^2 )^2 (T - t)|u_n (0)|
 + \epsilon (T - t)\int_0^t {e^{(s - t)n^2 } n^4 |f_n (s)|ds}  \\
&\leq \frac{{2\epsilon }}
    {{t^2 }}(T - t)|u_n (0)| + \epsilon (T - t)\int_0^t {n^4 |f_n 
(s)|ds.}
\end{aligned}
\end{equation}
In view of  $(a + b)^2  \leq 2(a^2  + b^2 )$  and using Holder
inequality, we obtain
    \begin{align*}
      |u_n(t) - u_n^\epsilon  (t)|^2
&\leq  2 [\frac{{4\epsilon ^2 }} {{t^4 }}(T - t)^2 |u_n (0)|^2
 + \epsilon ^2 (T - t)^2 (\int_0^t {n^4 |f_n (s)|ds)^2 } ]  \\
&\leq \frac{{8\epsilon ^2 }}  {{t^4 }}(T - t)^2 |u_n (0)|^2
+ \epsilon ^2 (T - t)^2 t^2 \int_0^t n^8 |f_n (s)|^2 ds.
\end{align*}
It follows that
    \begin{align*}
&      \|u(.,t) - u^\epsilon  (.,t)\|^2\\
&= \frac{\pi}{2}\sum_{n = 1}^\infty  {|u_n } (t) - u_n^\epsilon  (t)|^2  
\\
&\leq \frac{\pi}{2}\frac{{8\epsilon ^2 }}
  {{t^4 }}(T - t)^2 \sum_{n = 1}^\infty  {} |u_n (0)|^2
  + \frac{\pi}{2}\epsilon ^2 (T - t)^2 t^2 \int_0^t {\sum_{n = 
1}^\infty
  n^8 |f_n (s)|^2 ds}   \\
&= \frac{{8\epsilon ^2 }}{{t^4 }}(T - t)^2 \|u(.,0)\|^2
 + \epsilon ^2 (T - t)^2 t^2 \int_0^t {\|\frac{{\partial ^4 f(x,s)}}
 {{\partial x^4 }}\|^2 ds.}
\end{align*}
This completes the proof.
\end{proof}

\begin{theorem} \label{thm4}
 Let  $u$ be a solution of \eqref{e1}--\eqref{e3} with
   $u\in L^\infty  (0,T;L^2 (0,\pi )) \cap L^2 (0,T;H_0^1 (0,\pi ))$
 and  such that
 $ \|\Delta ^2 u(x,t)\|  < \infty$ for all $t$ in $[0,T]$.
Then
\[
\|u(.,t) - u^\epsilon  (.,t)\| \leq  \epsilon T\|\Delta ^2 u(.,t)\|
\]
\end{theorem}

\begin{proof}
    From  \eqref{e16},  we have
\begin{align*}
u_n(t) - u_n^\epsilon  (t)
&= e^{ - tn^2 }(1 - e^{ - \epsilon n^4 (T - t)} )u_n (0) + \int_0^t 
{e^{(s - t)n^2 }
(1 - e^{ - \epsilon n^4 (T - t)} )f_n (s)ds}\\
&=(1 - e^{ - \epsilon n^4 T} )u_n (t).
\end{align*}
Hence
 \[
 \|u(.,t) - u^\epsilon  (.,t)\|^2
 =\frac{\pi}{2} \sum_{n = 1}^\infty  {|u_n } (t) - u_n^\epsilon  (t)|^2
 \leq \frac{\pi}{2}\epsilon ^2 T^2 \sum_{n = 1}^\infty  {n^8 u_n^2 (t)}
=\epsilon ^2 T^2 \|\Delta ^2 u(.,t)\|^2.
\]
This completes  Proof.
\end{proof}

\begin{theorem} \label{thm5}
 Let Problem \eqref{e1}--\eqref{e3} have exact solution
  $u\in C([0,T];L^2 (0,\pi )) \cap L^2 (0,T;H_0^1 (0,\pi ) \cap H^2 
(0,\pi ))$,
 corresponding to $g$.
Assume that
\[
\frac{{\partial ^4 f(x,t)}}
{{\partial x^4 }}, \frac{{\partial u}}
{{\partial t }}\in L^2 (0,T;L^2 (0,\pi )),\quad
\|\Delta^2 u(x,t)\|<\infty    \quad\forall t \in[0,T]\,.
\]
Let $g_\epsilon$ be the measured data such that
$ \|g_\epsilon-g\| \leq \epsilon $. Then there exist a function
$u^{\beta(\epsilon)}$ satisfying
\begin{gather*}
\|u^{\beta(\epsilon)}(.,t)-u(.,t)\| \leq
\frac{K}{\ln (1/\epsilon)}+\epsilon^{t/T}, \quad \forall  t\in (0,T],
\\
\|u^{\beta(\epsilon)}(.,0)-u(.,0)\|
\leq (1+C) \sqrt{\frac{T}{\ln(1/\epsilon)}}+
C\frac{T}{4\ln(1/\epsilon)},
\end{gather*}
where $  \beta(\epsilon)= \frac{T}{4\ln(1/\epsilon)}$
and
\begin{gather*}
K=\frac{1}{4}T(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,}\\
M=\max \big\{\sup_{0\leq t\leq T} \|u_t(x,t)\|, T \sup_{0\leq t\leq T}
\|\Delta^2 u(x,t)\| \big\}.
\end{gather*}
\end{theorem}

\begin{proof}
Let $v^{\beta(\epsilon)}(.,t)$ be a solution of \eqref{e7}--\eqref{e9}
corresponding $g$, and $w^{\beta(\epsilon)}$ be solution of
\eqref{e7}--\eqref{e9} corresponding $g_\epsilon $.
We consider the function
$h(t)=\frac{\ln t}{t}-\frac{\ln \epsilon}{T}$ for $\epsilon \in (0,T)$.
We have $h(T)>0 $ and $\lim_{t \to 0} h(t)=-\infty $ then
$h(t)=0$ has solution in $(0,T)$. We call $t_\epsilon $ is smallest
solution of it.
Apply inequality $\ln t >-\frac{1}{t}$ we get
$t_\epsilon< \sqrt{\frac{T}{\ln(1/\epsilon)}}$.
Using Lagrange Theorem for $u(.,t)$ and
$u^\epsilon(.,t)$ in $(0,t_\epsilon)$ we have
\[
\|u(0)-u(t_\epsilon)\| \leq t_\epsilon \|u'(\alpha)\| \leq Ct_\epsilon,
\quad \forall \alpha \in (0,t_\epsilon).
\]
Using Theorem \ref{thm4} we get
\begin{align*}
\|v^{\beta(\epsilon)}(t_\epsilon)-u(0)\|
&\leq  \|v^{\beta(\epsilon)}(t_\epsilon-u(t_\epsilon)\|
 +\|u(0)-u(t_\epsilon)\|\\
&\leq \beta(\epsilon)(T-t_\epsilon)\|\Delta^2 
u(t_\epsilon)\|+Ct_\epsilon\\
&\leq C\ (  
\sqrt{\frac{T}{\ln(1/\epsilon)}}+\frac{T}{4\ln(1/\epsilon)})
\end{align*}
We put
 \[
 u^{\beta(\epsilon)}(t) =  \begin{cases}
 w^{\beta(\epsilon)}  (t), \quad 0 < t \leq  T ,\\
 w^{\beta(\epsilon)}  (t_\epsilon  ), \quad t = 0.
 \end{cases}
\]
 By Step 3 of Theorem \ref{thm1},
\[
\|v^{\beta(\epsilon)}(.,t)-w^{\beta(\epsilon)}(.,t)\|
\leq  e^{\frac{T-t}{4\beta(\epsilon)}}\|g^\epsilon-g\|
= \epsilon^{t/T}\,.
\]
By Theorem \ref{thm3} and applying the triangle inequality, we have
\begin{eqnarray*}
\|u^{\beta(\epsilon)}(.,t)-u(.,t)\| &\leq 
&\|v^{\beta(\epsilon)}(.,t)-w^{\beta(\epsilon)}(.,t)\|+\|v^{\beta(\epsilon)}(.,t)-u(.,t)\|\\
&\leq&\frac{K}{\ln(1/\epsilon)}+\epsilon^{t/T}.
\end{eqnarray*}
On the other hand,
\begin{align*}
\|u^{\beta(\epsilon)}(.,0)-u(.,0)\|
 &\leq \|v^{\beta(\epsilon)}(.,t_\epsilon)-w^{\beta(\epsilon)}
 (.,t_\epsilon)\|+\|v^{\beta(\epsilon)}(.,t_\epsilon)-u(.,0)\|\\
&\leq (1+C) \sqrt{\frac{T}{\ln(1/\epsilon)}}+  
C\frac{T}{4\ln(1/\epsilon)}.
\end{align*}
This completes  the proof.
\end{proof}

\section{Numerical experiments}

Consider the problem
\begin{equation} \label{e18}
\begin{gathered}
    u_t- u_{xx}=2e^t \sin(x), \\
    u(x,1) = g (x) = e\sin(x)
\end{gathered}
\end{equation}
whose exact solution 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
\begin{eqnarray*}
    g _n  (x) = e\sin(x) + \frac{1}    {n}\sin (nx).
    \end{eqnarray*}
So that the data error, at the final time, is
    \[
    F(n) = \|g_n  - g \|_{L^2 (0,\pi )}
    = \sqrt {\int_0^\pi  {\frac{1}{{n^2 }}\sin ^2 nxdx} }
     = \frac{1} {n}\sqrt {\frac{\pi }    {2}}.
    \]
The solution of \eqref{e18}, 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
\[
 O(n):= ||u^n (.,0) - u(.,0)||_{L^2 (0,\pi )}
 = \sqrt {\int_0^\pi  {\frac{{e^{2n^2 } }}
{{n^2 }}\sin ^2 (nx)\,dx} }
 = \frac{{e^{n^2 } }}{{n }}\sqrt {\frac{\pi }{2}}.
\]
Then, we notice that
\begin{gather*}
\lim_{n \to \infty } F(n)=\mathop {\lim }_{n \to
\infty } ||\varphi _n  - \varphi _0 ||_{L^2 (0,\pi )}  =
\lim _{n \to \infty } \frac{1} {n}\sqrt {\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^{} }}\sqrt {\frac{\pi }
{2}}  = \infty.
\end{gather*}
 From the two equalities above,  we see that \eqref{e18}  is  an
ill-posed problem. Approximating the problem as in
\eqref{e7}--\eqref{e9}, the regularized solution is
\begin{align*}
u^\epsilon  (x,t)
&= \sum_{m = 1}^{ \infty }
\Big(e^{(T - t)(m^2- \epsilon m^4 )} g _m  \\
&- \int_t^T  e^{(s - t)(m^2  - \epsilon m^4 )}
e^{ - \epsilon m^4 (T - s)} f_m (s)ds\Big)\sin (mx),
\\
u^\epsilon  (x,t) &= e^{(1 - t)(1 - \epsilon ) + 1} \sin(x)\\
&- 2\Big(\int_t^1 {e^{(s - t)(1 - \epsilon )} e^{ - \epsilon (1 - s) + 
1} ds}
\Big)
\sin(x) + \frac{1}{n}e^{(1 - t)(n^2  - \epsilon n^4 )} \sin (nx).
\end{align*}
Hence
\[
u^\epsilon  (x,\frac{1} {2}) = \big[e^{\frac{{3 - \epsilon }} {2}}  -
2\int_{1/2}^1 {e^{2s - 1/2 - \epsilon /2} ds\big]\sin(x) + \frac{1}
{n}} e^{\frac{1} {2}(n^2  - \epsilon n^4 )} \sin (nx).
\]

\begin{table}[ht]
 \caption{Approximations and error estimates for several values of 
$\epsilon$}
\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}


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\end{document}