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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2008(2008), No. 84, pp. 1--12.\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-2008/84\hfil Stabilized quasi-reversibility method]
{Stabilized quasi-reversibility method for \\
a class of nonlinear ill-posed problems}

\author[D. D. Trong, N. H. Tuan\hfil EJDE-2008/84\hfilneg]
{Dang Duc Trong, Nguyen Huy Tuan}  % in alphabetical order

\address{Dang Duc Trong \newline
Department of Mathematics and Informatics,
Hochiminh City National University,
227 Nguyen Van Cu, Q. 5, Hochiminh City, Vietnam}
\email{ddtrong@mathdep.hcmuns.edu.vn}

\address{Nguyen Huy Tuan \newline
Department of Mathematics and Informatics,
Ton Duc Thang  University,
98 Ngo Tat To street , Binh Thanh district  Hochiminh City, Vietnam}
\email{tuanhuy\_bs@yahoo.com}


\thanks{Submitted April 28, 2008. Published June 8, 2008.}
\subjclass[2000]{35K05, 35K99, 47J06, 47H10}
\keywords{Ill-posed problem; nonlinear parabolic equation;\hfill\break\indent
quasi-reversibility methods; stabilized quasi-reversibility methods}

\begin{abstract}
 In this paper, we study a final value problem for the nonlinear
 parabolic equation
 \begin{gather*}
 u_t+Au =h(u(t),t),\quad  0<t<T\\
 u(T)= \varphi ,
 \end{gather*}
 where $A$ is a non-negative, self-adjoint operator and $h$
 is a Lipchitz function. Using the stabilized quasi-reversibility
 method presented by Miller, we  find optimal perturbations, of
 the operator $A$, depending on a small parameter $\epsilon $ to setup
 an approximate nonlocal problem. We show that the approximate problems
 are well-posed under certain conditions and that their solutions
 converges if and only if the original problem has a classical solution.
 We also obtain estimates for the solutions of the approximate problems,
 and show a convergence result. This paper extends the work by Hetrick
 and Hughes \cite{h2} to nonlinear ill-posed problems.
\end{abstract}

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

\section{Introduction}

Let $A$ be a self-adjoint operator on a Hilbert space  $H $ such that
$-A$ generates a compact contraction semi-group on $H$.
We shall consider the final value problem of finding a function
$u:[0,T] \to H$  satisfying
 \begin{gather}
u_t+Au = h(u(t),t),\quad  0<t<T, \label{e1}\\
u(T) = \varphi , \label{e2}
\end{gather}
for   some prescribed final value $\varphi$  in a Hilbert space $H$.
Such problem are not well posed, that is, even if a unique solution
exists on $[0,T]$ it need not depend continuously on the final value
$\varphi$ . Hence, a regularization is in order. We note that this
type of problems has been considered by many authors, using different
approaches. In their pioneering work Lattes and Lions \cite{l1}
presented, in a heuristic approach, the quasi-reversibility method.
In this method the main ideas are replacing $A$ by an operator
$A_\epsilon=f_\epsilon(A)$. Originally,
 $f_\epsilon(A)=A-\epsilon A^2$ which yields the
 well-posed problem, in the backward direction,
\begin{equation}
\begin{gathered}
u_t+Au-\epsilon A^{2}u=0,\quad  t\in [0,T], \\
u(T)=\varphi.
\end{gathered} \label{e3}
\end{equation}
The stability of this method is of order $e^{c\epsilon^{-1}}$.
In \cite{s3}, the problem is approximated by
\begin{equation}
\begin{gathered}
u_t+Au+\epsilon Au_t = 0,\quad  t\in [0,T],\\
u(T) = \varphi.
\end{gathered} \label{e4}
\end{equation}
 In \cite{m4}, using the method of stabilized quasi reversibility,
the author studied the general approximated problem
\begin{equation}
\begin{gathered}
u_t+f(A)u = 0,\quad  t\in [0,T],\\
u(T) = \varphi.
\end{gathered} \label{e5}
\end{equation}
It is clear  that  \eqref{e3} and \eqref{e4}
are special case of \eqref{e5} where $ f(x)=x-\epsilon x^2 $
and $f(x)=x/(1+\epsilon x)$ respectively.
Note that the solution of \eqref{e5} has the form $ e^{(T-t)f(A)}\varphi$.
And  since  these functions $f$ are bounded by $c/\epsilon$,
we know that  their  stability is of order $e^ {c/\epsilon}$. Hence,
the stability in this  case are quite large as in the original
quasi-reversibility methods.
To improve the stability result of this problem \eqref{e5},
Miller  gave some  appropriate conditions on the ``corrector''
$f(A)$ and obtain the stability  of order $c\epsilon^{-1}$.

In 1983, Showalter presented  a  method called the
quasiboundary value (QBV) method, to regularize that linear
homogeneous problem,  which gave a stability estimate better than
the one of discussed methods. The main idea of this method is
adding an appropriate ``corrector'' into the final data.
Using this method, Clark-Oppenheimer \cite{c1}, and
Denche-Bessila \cite{d2}, recently, regularized the backward problem by
replacing the final condition with
\[
u(T)+\epsilon u(0)=\varphi
\]
and
\[
u(T)-\epsilon u'(0)=\varphi,
\]
respectively.

In 2005, Ames and Hughes \cite{a2} applied semigroup theory and other
 operator-theoretic methods to prove Holder continuous dependence
 for homogeneous ill-posed Cauchy problems. The authors
consider the above problem in Banach space and give the conditions
of the function $f$, to obtained the  stability estimate
\[
\|u(t)-v(t)\| \le C \beta^{1-w(t)} M^{w(t)}
\]
where $u(t)$ is the solution of \eqref{e1}--\eqref{e2} and
$v(t)$ the solution of \eqref{e5}.

Although there are many works on the linear homogeneous case of
the backward problem (ill-posed problem), the literature on the linear
nonhomogeneous case and  the nonlinear case are quite
scarce. A conditional stability result for the Ginzburg-Landau equation
was given in [A].
In \cite{q1}, the authors used the QR method and the eigenvalue-expansion
method to regularize a 1-D linear nonhomogeneous backward problem.
In \cite{t1}, the authors used an improved version of QBV method to
regularize the latter problem.
Recently, Hetrick and Hughes \cite{h2} extended the earlier work of
Ames and Hughes \cite{a2}, by considered  nonhomogeneous ill-posed problems
and proving the continuous dependence in Banach spaces.
However, the nonlinear case of the problem in \cite{a2} in Banach space
is not given here and will be presented in future work.

Most of the above articles give better results than
the quasi-reversibility method  given by Miller.
So, it is difficult to consider the backward problem using
quasi-reversibility method. Up to the present, we can  find only
a few papers which study  \eqref{e1}-\eqref{e2}
using quasi-reversibility, such as \cite{l2}.
In fact, Long and Dinh \cite{l2} approximated \eqref{e1}-\eqref{e2}
by the  problem
\begin{gather*}
v'_\beta(t)+A_\beta v_\beta(t)=e^{-(1-t)\beta AA_\beta}h(v_\beta)\\
v_\beta\eqref{e1}=\varphi
\end{gather*}
where  $ f_{\beta}(A)= A_\beta=A(I+\beta A)^{-1}$ is the approximate
operator for $A$. Although $v^\beta$ is a good approximation of $u$,
the authors can not prove that $v^\beta$ is a regularized solution of $u$.
So, the quasi-reversibility method given in \cite{l2}, is not effective
 to regularize the backward problem with the large time.

This paper is a generalization of  Miller's paper  for the
nonlinear  right hand side.
 We prove that our method gives the same stability order  as
previous method in \cite{q1,t2}.
By replacing the operator $A$ by $f_{\epsilon}( A)$,
chosen latter under some better conditions, we approximate the
problem \eqref{e1}-\eqref{e2} as the follows:
\begin{gather}
u_t^{\epsilon}+f_{\epsilon}(A)u^\epsilon=h(u^\epsilon(t),t),\quad
 0<t<T, \label{e6} \\
u^{\epsilon}(T)=\varphi,\quad \label{e7}
\end{gather}
with $0<\epsilon<1$.

This paper is organized as follows.  In the section 2,  we
 derive conditions on the perturbation $f_\epsilon(A) $ and show
that \eqref{e6}-\eqref{e7} is well-posed.
Moreover,  the stability of this method is of order
$c\epsilon^{\frac{t}{T}-1}$.
Also, we find some  conditions on $f_\epsilon $ so that we can
get  error estimate
\begin{equation}
\| u^{\epsilon}(t)-u(t)\|\le C\beta(\epsilon)^{t/T}, \label{e8}
\end{equation}
where $\|\cdot\|$ is the norm in $ H $, $\beta(\epsilon) \to 0  $
when $\epsilon \to 0 $  and $C$ depends on $u(t)$.
 Finally, we consider the example and numerical experiment
will be given in Section 4, which show that the efficient of our method.

\section{Approximation of the non-linear problem}

We  assume that $ H $ is a separable Hilbert space and $A $ is self-adjoint
and that $  0 $  is in the resolvent set of $A$. We  also assume that
$A^{-1}$ is compact. Let $\{ \phi _n \}$ be an orthonormal eigenbasic on
$H$ corresponding to the eigenvalues $ \{\lambda _n \}$ of $A$;
 i.e., $A\phi_n=\lambda_n\phi_n $.
Without loss of generality, we shall assume that
\[
0<\lambda_{1} < \lambda_{2} < \lambda_{3} < \dots, \quad
  \lim _{n\to \infty} \lambda_{n} =\infty.
\]
For every $v$ in $H $ having the expansion
$v = \sum_{n = 1}^{ \infty } {v_n \phi _n }$,
$v_n \in \mathbb{R}$, $n=1,2,\dots$ and $g:\mathbb{R} \to \mathbb{R}$, we define
$  g(A)v = \sum_{n = 1}^{\infty } g(\lambda_n) v_n \phi _n$.
If $v \in H $, we define
$$
\mathop{\rm Dom}(g(A))= \{ v\in H:\|g(A)v\|^2
= \sum_{n = 1}^{\infty } g^2(\lambda_n) v^2_n   <\infty \}
$$

\subsection*{Definition}
Let fixed $\epsilon \in (0,1)$. Let  $f_\epsilon :[0,\infty) \to  \mathbb{R}$
be a bounded Borel function, and  assume that there exists
$\beta(\epsilon)>0 $  satisfies $\beta(\epsilon) \to 0  $ when
$\epsilon \to 0 $, and
$| f_\epsilon(\alpha) |\leq   -\frac{1}{T} \ln (\beta(\epsilon)) $
for all $\alpha \in [0,\infty) $.

1. $f$ is said to satisfy  Condition (A) if
 \[
\|(-A+f_\epsilon(A))u\| \leq  \beta(\epsilon)\|e^{TA}u\|
\]
for all $u\in \mathop{\rm Dom}(e^{TA})=\{u\in H: \|e^{TA}u\|
=\sqrt{ \sum_{n = 1}^{\infty } e^{2T\lambda_n} u^2_{n}  }<\infty \}$.

 Yongzhong Huang \cite[p.759]{h4} gave the approximate operator
 $$
A_\epsilon = -\frac{1}{pT}\ln (\epsilon+e^{-pT\epsilon})
$$
In the case $p=1$, we have
$f_\epsilon(x)= -\frac{1}{T}\ln(\epsilon +e^{-Tx})$,
where $x\in (0,\infty)$. Then, it is easy to see that
$ | f_\epsilon(\alpha) |\leq   -\frac{1}{T} \ln (\epsilon) $.
Also we have
 \begin{align*}
\|(-A+f_\epsilon(A))u\|^2
&=\sum_{n = 1}^{\infty } (-\lambda_n -\frac{1}{T}\ln(\epsilon
 +e^{-T\lambda_n}))^2 u^2_n \\
&\leq \sum_{n = 1}^{\infty } \frac{1}{T^2} \ln^2 (1+\epsilon e^{T\lambda_n})
 u^2_n\\
&\leq \sum_{n = 1}^{\infty } \frac{\epsilon^2}{T^2} e^{2T\lambda_n}u^2_n
=\frac{\epsilon^2}{T^2} \|e^{TA}u\|^2
\end{align*}
Hence, $f_\epsilon $ satisfies condition (A).


2. Let $ 0\le s\le t\le T $ and $u \in H$. Then we define the operator
$$
e^{(s-t)f_\epsilon(A)}u = \sum_{n = 1}^{\infty }e^{(s-t)
f_\epsilon(\lambda_n)}u_n \phi_n
$$

\begin{lemma} \label{lem1}
Let $\epsilon>0 $ be such that $0<\beta(\epsilon)<1$ and
$ u \in H $ has the eigen-function expansion
$u=\sum_{n = 1}^{\infty }u_n \phi_n $ where $u_n= <u,\phi_n>$. Then
$$
\|f_\epsilon(A)u\| \le  -\frac{1}{T} \ln (\beta(\epsilon)) \|u\|
$$
\end{lemma}

\begin{proof}
 Suppose that $ u \in H $ has the eigen-function expansion
$u=\sum_{n = 1}^{\infty }u_n \phi_n $ where $u_n= \langle u,\phi_n\rangle$.
Then, using  the expansion of $f_\epsilon(A)u$,
 and that $f_\epsilon$ is bounded, we obtain
\[
\|f_\epsilon(A)u\|^2=\sum_{n = 1}^{\infty } f^2_\epsilon(\lambda_n)u^2_n
\le \frac{1}{T^2} \ln^2 \frac{1}{(\beta(\epsilon))}
\sum_{n = 1}^{\infty }u^2_n=\ln^2 \frac{1}{(\beta(\epsilon))}\|u\|^2
\]
This completes the proof.
\end{proof}


\begin{lemma} \label{lem2}
 Let $\epsilon,s,t$ be as in Lemma \ref{lem1}. Then for  $ u \in H $, we have
$$
\|e^{(s-t)f_\epsilon(A)}u\| \le   (\beta(\epsilon))^\frac{t-s}{T}  \|u\|
$$
\end{lemma}

\begin{proof}
Using that $f_\epsilon$ is bounded, we have
\[
\|e^{(s-t)f_\epsilon(A)}u\|^2
=\sum_{n = 1}^{\infty } e^{2(s-t)f_\epsilon(\lambda_n)}u^2_n
\le \exp(\frac{s-t}{T} \ln \frac{1}{(\beta(\epsilon))})
\sum_{n = 1}^{\infty }u^2_n=  (\beta(\epsilon))^\frac{t-s}{T} \|u\|^2
\]
\end{proof}



\begin{theorem} \label{thm1}
Let $\epsilon$ be as in Lemma \ref{lem1}, $\varphi \in H $ and let
$ h:H \times \mathbb{R} \to H $ be a continuous operator satisfying
$ \|h(w(t),t)-h(v(t),t)\| \le k\|w-v\|$ for a $k >0$ independent
of $w(t),v(t) \in H$, $t\in \mathbb{R}$ and $f_\epsilon$ satisfies Condition (A).
Then the approximate problem \eqref{e6}-\eqref{e7} has a unique
solution $u^{\epsilon}\in C([0,T];H)$.
\end{theorem}

First, we consider two following propositions which are useful to the
proof of Theorem \ref{thm1}.

\begin{proposition} \label{prop2}
The integral equation
\begin{equation}
 u^\epsilon(t)=e^{(T-t)f_\epsilon(A)}\varphi
-\int_t^T e^{(s-t)f_\epsilon(A)}h(u^\epsilon(s),s)ds \label{e9}
\end{equation}
has a unique solution and this solution satisfies the approximate
problem \eqref{e6}-\eqref{e7}.
\end{proposition}

\begin{proof}
We put
\[
F(w)(t)=e^{(T-t)f_\epsilon(A)}\varphi
-\int_t^T e^{(s-t)f_\epsilon(A)}h(w(s),s)ds
\]
We claim that, for every $ w,v \in C([0,T];H) $ we have
\begin{equation}
\|F^m (w)(.,t)-F^m(v)(.,t)\|
\leq \Big(\frac{k(T-t)}{\beta(\epsilon)}\Big)^{m}|||w -v||| \label{e10}
\end{equation}
where $C = \max \{T,1\}$ and $|||\cdot|||$ is sup norm
in $C([0,T];H)$.
We shall prove the latter inequality by induction.

For $m =1$, we have
\begin{align*}
\|F(w)(.,t)-F(v)(.,t)\|
&= \|\int_t^T e^{(s-t)f_\epsilon(A)}(h(w(s),s)-h(v(s),s))ds\|
\\
&\leq \int_t^T \|e^{(s-t)f_\epsilon(A)}\| \|h(w(s),s)-h(v(s),s)\|ds
\\
&\leq \int_t^T \frac{k}{\beta(\epsilon)^{\frac{s-t}{T}}}
\|w(s)-v(s)\|ds
\\
&\leq \frac{k}{\beta(\epsilon)}\int_t^T
\|w(s)-v(s)\|ds
\\
&\leq \frac{k}{\beta(\epsilon)}(T-t)
|||w-v||||
\end{align*}
(We  can choose $\epsilon$ such that $0<\beta(\epsilon)<1$)

Suppose that \eqref{e10} holds for $m =j$. We prove that \eqref{e10}
holds for $m = j + 1$. We have
\begin{align*}
\|F^{j+1}(w)(.,t) - F^{j+1}(v)(.,t)\|
& = \|\int_t^T e^{(s-t)f_\epsilon(A)}(h(F^{j}w)(s)-h(F^{j}v)(s))ds\|\\
&\leq \int_t^T \|e^{(s-t)f_\epsilon(A)}\| \|h(F^{j}w)(s)-h(F^{j}v)(s)\|ds\\
&\leq \int_t^T \frac{k(T-t)}{\beta(\epsilon)^{\frac{s-t}{T}}}\|h(F^{j}w)(s)
 -h(F^{j}v)(s)\|ds\\
&\leq \frac{k(T-t)}{\beta(\epsilon)}\int_t^T k\|(F^{j}w)(s)-(F^{j}v)(s)\|ds\\
&\leq \frac{1}{\beta(\epsilon)} (T - t)k \int_t^T \|G^j(w)(.,s)
 - G^j (v)(.,s)\|^2 ds \\
&\leq \frac{1}{\beta(\epsilon)} (T - t)k
 \int_t^T \frac{k^j}{\beta(\epsilon)^j}(T-s)^j  ds|||w-v|||\\
&\leq \big(\frac{k}{\beta(\epsilon)}\big)^{(j+1)}(T-t)^{j+1}  |||w-v|||.
\end{align*}
Therefore, by the induction principle, we have \eqref{e10}.

We consider $F: C([0,T];H)\to C([0,T];H)$.
Since $\lim_{m \to \infty} \left(\frac{kT}{\beta(\epsilon)}\right)^m =0$,
there exists a positive integer number $m_0$ such that $F^{m_0}$
is a contraction. It follows that the equation $F^{m_0} (w) = w$
has a unique solution $u_\epsilon \in C([0,T];H)$.

We claim that $F(u^\epsilon) = u^\epsilon$. In fact, one has
$F(F^{m_0} (u^\epsilon)) = F(u^\epsilon)$.
Hence $F^{m_0} (F(u^\epsilon)) = F(u^\epsilon)$.
By the uniqueness of the fixed point of $F^{m_0}$, one has
$F(u^\epsilon) = u^\epsilon$, i.e., the equation $F(w) = w$ has a
unique solution $u^\epsilon \in C([0,T];H)$.

Finally, we prove the unique solution of \eqref{e9} satisfies t
\eqref{e6}-\eqref{e7}. In fact, one has in view from \eqref{e9}, we have
\[
 u^\epsilon(t)=e^{(T-t)f_\epsilon(A)}\varphi
-\int_t^T e^{(s-t)f_\epsilon(A)}h(u^\epsilon(s),s)ds
\]
This also follows that $u(T)= \varphi$, hence the condition \eqref{e7}
is satisfied.
The expansion formula of $u^\epsilon(t)$
\[
u^\epsilon(t)=\sum_{n = 1}^{\infty }
\Big(e^{(T-t)f_\epsilon(\lambda_n)}\varphi_n
-\int_t^T e^{(s-t)f_\epsilon(\lambda_n)}h_n(u^\epsilon)(s)ds \Big)\phi_n
\]
Differentiating  $u(t)$ with respect to $t$, we get
\begin{align*}
u_t^\epsilon(t)
&=-f_\epsilon(A)e^{(T-t)f_\epsilon(A)}\varphi
 +f_\epsilon(A)\int_t^T e^{(s-t)f_\epsilon(A)}h(u^\epsilon(s),s)ds
 +h(u^\epsilon(t),t)\\
&= -f_\epsilon(A)u^\epsilon+h(u^\epsilon(t),t).
\end{align*}
This completes the proof of Proposition \ref{prop2}
\end{proof}


\begin{proposition} \label{prop3}
Assume that $f_\epsilon $ satisfies condition A then
 Problem \eqref{e6}-\eqref{e7} has at most one solution in $ C([0,T];H)$.
\end{proposition}

\begin{proof}
Suppose  $ u(t) $ and $  v(t)$  are  solution in $ C([0,T];H)$
of the approximate problem \eqref{e6}-\eqref{e7}.
Putting $w(t) = e^{  m(t - T)} (u(t) - v(t))$ ($ m> 0 $),
then replacing in the equation \eqref{e6} and by direct computation,
we obtain
\begin{equation} \label{e11}
w_t^{} +f_\epsilon (A)w(t)  - mw(t)
 =  e^{m(t - T)} h(e^{ - m(t - T)} u(t),t) - h(e^{ - m(t - T)} v(t),t)
\end{equation}
Multiplying  two side of \eqref{e11} with $w$ and using global Lipchitz
properties of function  $h$ we get
\[
\frac{d}{2dt}\|w(t)\|^2+ \langle f_\epsilon(A)w,w\rangle
-m\|w\|^2+k\|w\|^2  \geq 0
\]
Using the boundedness of function $ f_\epsilon $ in Lemma \ref{lem2},  we have
\[
|\langle f_\epsilon(A)w,w\rangle | \leq
\|f_\epsilon(A)w\| \|w\|
\leq \frac{1}{T}\ln(\frac{1}{\beta(\epsilon)})\|w\|^2
\]
It follows that
\begin{equation} \label{e12}
\frac{d}{2ds}\|w(s)\|^2 \geq  m\|w\|^2-k\|w\|^2
-\frac{1}{T}\ln(\frac{1}{\beta(\epsilon)})\|w\|^2
\end{equation}
 Putting the integral with  $s$  from $t$ to $T$ in \eqref{e12},
 then choosing  $m=k+\frac{1}{T}\ln(\frac{1}{\beta(\epsilon)})$,
it can be rewritten as
\begin{equation} \label{e13}
\|w(T)\|^2-\|w(t)\|^2 \geq 2(m-k-\frac{1}{T}
\ln(\frac{1}{\beta(\epsilon)})\int_t^T \|w(s)\|^2 ds=0
\end{equation}
Using the equality $w(T)=u(T)-v(T)=0$, one has $w(t)=0$.
This completes the proof
\end{proof}

\begin{theorem} \label{thm2}
 The solution of  \eqref{e6}-\eqref{e7} depends continuously on
$\varphi$
\end{theorem}

\begin{proof}
Let $ u $ and $ v $ be two solution of \eqref{e6}-\eqref{e7}
corresponding with two final values  $\varphi $  and  $ \omega $.
By setting $w(t) = e^{  m(t - T)} (u(t) - v(t))$
(with $ m> 0 $), we have $w(T)=\varphi-\omega$.
In view of  inequality \eqref{e11} in Proposition \ref{prop2}, we get
\begin{equation} \label{e14}
\|w(T)\|^2-\|w(t)\|^2 \geq 2(m-k-\frac{1}{T}
\ln(\frac{1}{\beta(\epsilon)})\int_t^T \|w(s)\|^2 ds=0.
\end{equation}
choosing $m=k+\frac{1}{T}\ln(\frac{1}{\beta(\epsilon)})$, we have
\[
\| \varphi-\omega \|  \geq \|w(t)\| = e^{  m(t - T)}\|u(t) - v(t)\|
\]
This implies
\[
\|u(t) - v(t)\|  \leq e^{  m( T-t)}\|\varphi-\omega \|
=e^{k(T-t)}\beta(\epsilon)^{\frac{t}{T}-1}\|\varphi-\omega\|
\]
whihc proves continuity and that the stability of the solution is of order
$E \beta(\epsilon)^{\frac{t}{T}-1}$.
\end{proof}

\section{Regularization of Problem \eqref{e1}-\eqref{e2}}

\begin{theorem} \label{thm3}
 Let $\epsilon$ be as in Lemma \ref{lem1}. Suppose  problem \eqref{e1}-\eqref{e2}
has a unique solution  $ u(t)\in (C[0,T];H)$  which satisfies
$ u(t) \in \mathop{\rm Dom}(e^{TA})$.
Then for $0<t \leq T$ we have the  error estimate
$$
\|u(t) - u^\epsilon  (t)\|  \leq  M \beta(\epsilon) ^{t/T}
$$
Moreover, there exists  a $t_\epsilon \in (0,T)$ such that
\[
\|u(0) - u^\epsilon  (t_\epsilon )\| \le
2C\Big(\frac{T}{\ln \frac{1}{ \beta(\epsilon) }}\Big)^{1/2},
\]
 where
\[
M =e^{k(T-t)} \int_0^T\|e^{TA}u(s)\|ds ,\quad
C=  \max \{e^{k T}  M,  (\frac{1}{T}+k)M
+\sup_{t \in [0,T]}\|f(0,t)\| \},
\]
$u^\epsilon$ is  the unique solution of \eqref{e5},
 and $u^\epsilon  (t) $ is the unique solution of
\eqref{e6}-\eqref{e7}.
\end{theorem}

\begin{proof}
We put $ w^\epsilon(t)=u^\epsilon(t)-u(t)$ and
$g_\epsilon(A)=-A+f_\epsilon(A)$.  Then $w^\epsilon(t)$
satisfies
\begin{equation} \label{e15}
w_t^\epsilon+ f_\epsilon(A)w^\epsilon=h(u^\epsilon(t),t)-h(u(t),t)
+g_\epsilon(A)u(t)
\end{equation}
Let $ h_{1}:H \times \mathbb{R} \to H $ satisfying
$h_1(w(t),t))=h(w(t)+u(t),t)-h(u(t),t)$.
Using the Lipchitz property of  $h$ given in Theorem \ref{thm2},
we get $ \|h_1(w(t),t))\| \leq k \|w(t)\|$.
Hence, \eqref{e14} can be written as
\begin{gather*}
w_t^\epsilon+ f_\epsilon(A)w^\epsilon=h_1(w^\epsilon(t),t)
+g_\epsilon(A)u(t) \\
w^\epsilon(T)=0
\end{gather*}
It is not difficult to check   $w^\epsilon(t)$ satisfies
\begin{equation} \label{e16}
w^\epsilon(t)=-\int_t^T e^{(s-t)f_\epsilon(A)}[h_1(w^\epsilon(s),s)
+g_\epsilon(A)u(s)]ds.
\end{equation}
It follows that
% 17
\begin{align*}
\|w^\epsilon(t)\|
&=\|\int_t^T e^{(s-t)f_\epsilon(A)}[h_1(w^\epsilon(s),s)
 +g_\epsilon(A)u(s)]ds\|\\
&\leq \int_t^T e^{(s-t)\|f_\epsilon(A)\|}[\|h_1(w^\epsilon(s),s)\|+\|g_\epsilon(A)u(s)\|]ds\\
&\leq  \beta(\epsilon)^{\frac{t}{T}}k \int_t^T \beta(\epsilon)^{\frac{-s}{T}}\|w^\epsilon(s))\|ds+\beta(\epsilon)^{\frac{t}{T}}
\int_t^T \beta(\epsilon)^{\frac{-s}{T}}\|g_\epsilon(A)u\|ds\\
&\leq  \beta(\epsilon)^{\frac{t}{T}}k \int_t^T \beta(\epsilon)^{\frac{-s}{T}}\|w^\epsilon(s))\|ds+\beta(\epsilon)^{\frac{t}{T}}
\int_0^T \beta(\epsilon)^{\frac{T-s}{T}}\|e^{TA}u(s)\|ds\\
&\leq \beta(\epsilon)^{\frac{t}{T}}k \int_t^T \beta(\epsilon)^{\frac{-s}{T}}\|w^\epsilon(s))\|ds+\beta(\epsilon)^{\frac{t}{T}}T \int_0^T\|e^{TA}u(s)\|ds
\end{align*}
 From the above inequality, we have
\[
\beta(\epsilon)^{\frac{-t}{T}}\|w^\epsilon(t)\|
\leq k\int_t^T \beta(\epsilon)^{\frac{-s}{T}}
\|w^\epsilon(s)\|ds+ \int_0^T\|e^{TA}u(s)\|ds\,.
\]
Using Gronwall's inequality we obtain
\[
\|w^\epsilon(t)\|  \leq e^{k(T-t)}\beta(\epsilon)^{\frac{t}{T}}
\int_0^T\|e^{TA}u(s)\|ds,
\]
or
\[
\|u^\epsilon(t)-u(t)\|  \leq  M   \beta(\epsilon)^{\frac{t}{T}},
\]
 For   $t \in (0,T)$, considering  the function
$ h(t)=\frac{\ln t}{t}-\frac{\ln (\beta(\epsilon))}{T}$,
 we have $h(\beta(\epsilon))>0$,
$\lim_{t \to 0}h(t)=-\infty$,  $h'(t)>0$   ($0<t< \beta(\epsilon)$).
It follows that the equation $h(t)=0$ has a unique solution
$ t_\epsilon $ in $(0, \beta(\epsilon))$.
 Since  $\frac {\ln t_\epsilon}{t_\epsilon}=\frac{\ln (\beta(\epsilon))}{T}$,
the inequality $ \ln t > -\frac{1}{t}$ gives
$ t_\epsilon < \sqrt{ \frac{T}{\ln \frac{1}{\epsilon}}}$.

We  have $ u(t_\epsilon)-u(0)=\int_0^{t_\epsilon} u'(t)dt$. Hence
$\|u(0)-u(t_\epsilon)\|  \le t_\epsilon  \sup_{t \in [0,T]} \| u'(t)\| $.
On the other hand, one has
\begin{align*}
\|u'(t)\|
&\le \|Au(t)\|+\|f(u(t),t)\|\\
&\le\Big({\sum_{n = 1}^\infty  \lambda _n^2  u_n^2(t) }\Big)^{1/2}
 +k\|u(t)\|+\|f(0,t)\|\\
&\le \frac{1}{T}\Big(\sum_{n = 1}^\infty  { e^{2T\lambda _n }
 u_n^2(t) }\Big)^{1/2} +k\|u(t)\|+\|f(0,t)\|\\
&\le  (\frac{1}{T}+k)M+\|f(0,t)\| \le C.
\end{align*}
It follows that  $\|u(0)-u(t_\epsilon)\|  \le Ct_\epsilon$.
By the definition of $t_\epsilon$, we get
\begin{align*}
\|u(0)-u^\epsilon(t_\epsilon)\|
&\le \|u(0)-u(t_\epsilon)\|+\|u(t_\epsilon)-u^\epsilon(t_\epsilon)\|\\
&\le 2 C t_\epsilon \le 2C \Big(\frac{T}{\ln (\frac{1}{\beta(\epsilon)})}
\Big)^{1/2}.
\end{align*}
which completes the proof.
\end{proof}

\section{Example and applications}
First, we consider the  model problem
\begin{gather*}
u_t+Au(t)=h(u(t),t) \\
u(T)=\varphi
\end{gather*}
that is compared with the following well-posed
problem.
Taking function $f_\epsilon(x)= -\frac{1}{T}\ln(\epsilon +e^{-Tx})$
for $x\in (0,\infty)$, we have the first approximate problem
\begin{gather}
u^\epsilon_t-\frac{1}{T}\ln(\epsilon
+e^{-TA})u^\epsilon=h(u^\epsilon(t),t) \label{e18} \\
u^\epsilon(T)=\varphi\,. \label{e19}
\end{gather}
It is easy to check that $|f(x)| \leq \frac{1}{T}\ln (\frac{1}{\epsilon}) $.
Then $f$ satisfies Condition (A) with $ \beta(\epsilon)=\epsilon$.

In the Hilbert space, let $H=L^2(0,\pi)$ and let $A=-\Delta$ is the
Laplace operator. We take  $\lambda_n=n^2$,
$\phi_n=\sqrt{\frac{2}{\pi}} \sin (nx)$ are eigenvalues and
orthonormal eigenfunctions, which form a basis for $H$.
Let us consider the nonlinear backward heat problem
\begin{gather}
-u_{xx} +u_t = f(u) +g(x,t),\quad  (x,t)\in (0,\pi)\times (0,1) \label{e20}\\
u(0,t)=u(\pi,t)=0, \quad  t \in[0,1], \label{e21}\\
u(x,1)=\varphi(x),\quad x\in[0,\pi] \label{e22}
\end{gather}
where
\begin{gather*}
f(u) = \begin{cases}
 u^2 & u \in [ - e^{10} ,e^{10} ] \\
 - \frac{{e^{10} }}{{e - 1}}u + \frac{{e^{21} }}{{e - 1}}
 & u \in (e^{10} ,e^{11} ] \\
 \frac{{e^{10} }}{{e - 1}}u + \frac{{e^{21} }}{{e - 1}}
 & u \in ( - e^{11} , - e^{10} ] \\
 0 & |u| > e^{11}
 \end{cases} \\
g(x,t) = 2e^t \sin x - e^{2t} \sin^2 x,\\
u(x,1) = \varphi_0 (x) \equiv e\sin x.
\end{gather*}
The exact solution of the above equation is
$u(x,t) = e^t \sin x$.
In particular,
\[
u\big(x,\frac{999}{100}\big)\equiv u(x)
= \exp \big(\frac{999}{1000}\big)\sin x \approx 2.715564905 \sin x.
\]
Let $\varphi_\epsilon(x) \equiv \varphi(x) = (\epsilon + 1)e\sin x$.
Then
\[
\|\varphi_\epsilon -\varphi\|_2 = \Big(\int_0^{\pi}\epsilon^2e^2 \sin^2 x dx
\Big)^{1/2} = \epsilon e \sqrt{\pi/2}.
\]
Applying the method introduced in this paper,
we find the regularized solution
$u_\epsilon\left(x,\frac{999}{1000}\right) \equiv u_\epsilon (x)$
having the  form
\[
u_\epsilon(x) = v_m(x) = w_{1,m}\sin x + w_{6,m} \sin 6x\,,
\]
where
$v_1(x) = (\epsilon + 1)e \sin x$,
$w_{1,1} = (\epsilon + 1)e$,
$w_{6,1} = 0$,
and $a = \frac{1}{{5000}}$, $t_m  = 1 - am$ for $m = 1,2,\dots,5$, and
\begin{align*}
w_{i,m + 1}
&=({{\epsilon  + e^{ - t_{m} i^2 } }) ^ {\frac{t_{m+1}-t_{m}}  {t_{m}}}}
w_{i,m}\\
&\quad - \frac{2}{\pi }\int_{t_{m + 1} }^{t_m } {{e^{ (s - t_{m+1 }) i^2 }}}
\Big( {\int_0^\pi  {\left( {v_m^2 (x) + g(x,s)} \right)\sin (ix)dx} } ds\Big),
\end{align*}
for $i = 1,6$.
Table \ref{table1} shows the approximation error in this case.
\begin{table}[ht]
\caption{Error between regularized and exact solution}
\label{table1}
\begin{center}
\begin{tabular}{|c|c|c|} \hline
$\epsilon$& $u_\epsilon$ &  $\|u_\epsilon - u\|$ \\ \hline
$\epsilon_1 = 10^{-3}$ & $2.718118645\sin(x)-0.005612885749\sin(6x)$
 & 0.002585244486\\ \hline
$\epsilon_2 = 10^{-4}$ &  $2.715807105\sin(x) -0.005488275207 \sin(6 x)$
 & 0.0002723211648\\ \hline
$\epsilon_3 = 10^{-11}$ & $2.715552177\sin(x)-0.005518178192\sin(6x)$
 & 0.00004317829056\\ \hline
\end{tabular}
\end{center}
\end{table}

By applying the method in \cite{l2}, we have the approximate solution
\[
u_\epsilon(x,\frac{999}{1000}) = v_m(x) = w_{1,m}\sin x + w_{3,m} \sin 3x,
\]
where
$v_1(x) = (\epsilon + 1)e \sin x$,
$w_{1,1} = (\epsilon + 1)e$, $w_{3,1} = 0$,
$a = \frac{1}{{5000}}$, $t_m  = 1 - am$ for $m = 1,2,\dots,5$  and
\begin{align*}
 w_{i,m + 1}
&={ e^{  (t_{m}-t_{m+1}) \frac {i^2}{1+\epsilon i^2} } }  w_{i,m}\\
&\quad - \frac{2}{\pi }\int_{t_{m + 1} }^{t_m } {{e^{ s - t_{m+1 }
- \frac{ (t_{m}-t_{m+1})\epsilon i^2    }{1+\epsilon i^2}   } }}
 \Big( {\int_0^\pi  {\left( {v_m^2 (x) + g(x,s)} \right)\sin (ix)dx} } \Big)ds,
\end{align*}
for $i = 1,3$.
Table \ref{table2} shows the approximation error in this case.

\begin{table}[ht]
\caption{Error between regularized and exact solution}
\label{table2}
\begin{center}
\begin{tabular}{|c|c|c|} \hline
$\epsilon$& $u_\epsilon$ & $\|u_\epsilon - u\|$ \\ \hline
$\epsilon_1 = 10^{-3}$  & $2.718267378\sin(x)-0.005479540370\sin(3x)$
  &  0.006109723643\\ \hline
$\epsilon_2 = 10^{-4}$  & $2.715832209\sin(x)-0.005468363690\sin(3x)$
  & 0.005474892956\\ \hline
$\epsilon_3 = 10^{-11}$ & $2.715561633\sin(x)-0.005467119519\sin(3x)$
  & 0.005467120499\\ \hline
\end{tabular}
\end{center}
\end{table}

From the two tables, we see that the error in Table \ref{table1}
is smaller and increases slower than the error in Table \ref{table2}.
This indicates that in this example, our our approach has a nice
regularizing effect and give a better approximation that
the method in \cite{l2}.

\subsection*{Acknowledgments}
The authors would like to thank Professor Julio G. Dix
for his valuable help in the presentation of this paper.
 The authors are also  grateful to the anonymous referees for their valuable
comments leading to the improvement of our paper.

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\end{document}