\documentclass[reqno]{amsart}
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\AtBeginDocument{{\noindent\small
{\em Electronic Journal of Differential Equations},
Vol. 2006(2006), No. 147, pp. 1--15.\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-2006/147\hfil Optimal regularization method]
{Optimal regularization method for ill-posed Cauchy problems}

\author[N. Boussetila, F. Rebbani\hfil EJDE-2006/147\hfilneg]
{Nadjib Boussetila, Faouzia Rebbani}  % in alphabetical order

\address{Nadjib Boussetila \newline
Applied Math Lab, University Badji Mokhtar-Annaba \\
P.O. Box 12, Annaba 23000, Algeria}
\email{naboussetila@yahoo.fr}

\address{Faouzia Rebbani \newline
Applied Math Lab, University Badji Mokhtar-Annaba \\
P.O. Box 12, Annaba 23000, Algeria}
\email{rebbani@wissal.dz}

\thanks{Submitted February 28, 2006. Published November 27, 2006.}
\subjclass[2000]{35K90, 47D06, 47A52, 35R25}
\keywords{Ill-posed Cauchy problem; quasi-reversibility method;
\hfill\break\indent nonlocal conditions; regularizing family}

\begin{abstract}
 The goal of this paper is to give an optimal regularization
 method for an ill-posed Cauchy problem associated with an
 unbounded linear operator in a Hilbert space. Key point to our
 proof is the use of Yosida approximation and nonlocal conditions
 to construct a family of regularizing operators for the considered
 problem. We show the convergence of this approach, and we estimate
 the convergence rate under a priori regularity assumptions on the
 problem data.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{remark}[theorem]{Remark}
\newtheorem{definition}[theorem]{Definition}

\section{Introduction and motivation}

Throughout this paper $H$ will denote a Hilbert space, endowed
with the inner product $(\cdot,\cdot)$ and the norm $\|\cdot\|$,
$\mathcal{L}(H)$ denotes the Banach algebra of bounded linear
operators on $H$.

Consider the backward Cauchy problem \begin{equation}
u'(t)+Au(t)=0,\quad 0<t<T,\quad u(T)=\varphi,\label{FVP}
\end{equation}
where $A$ is a positive $(A\geq\gamma >0)$ self-adjoint
$(A = A^{*})$, unbounded linear operator on $H$, and  $\varphi \in
H$.

\begin{quote}The problem is to determine $u(t)$ for $0\leq t
< T$ from the knowledge of the final value $u(T) =
\varphi$.
\end{quote}

Such problems are not well-posed in the Hadamard sense
\cite{hadamard}, that is, even if a unique solution exists on
$[0,T]$ it need not depend continuously on the final value $\varphi$.

Physically, problems of this nature arise in different contexts.
Beyond their interest in connection with standard diffusion
problems \cite{engel} (then $A$ is usually the Laplace operator
$-\Delta$), they also appear, for instance, in some deconvolution
problem, such as deblurring processes \cite{carasso1} ($A$ is
often a fractional power of $-\Delta$), material sciences
\cite{renardy}, hydrology \cite{hassanov,todd2} and also in many
other practical applications of mathematical physics and
engineering sciences.

In the mathematical literature various methods have been proposed
for solving backward Cauchy problems. We can notably mention the
method of quasi-solution (Q.S.-method) of Tikhonov
\cite{tikhonov}, the method of quasi-reversibility (Q.R.-method)
of Latt\`{e}s and Lions \cite{lattes}, the method of logarithmic
convexity \cite{agmon,carasso4,john,miller2,payne1}, the it
iterative procedures of Kozlov and Maz'ya \cite{baumeis2,koz}, the
quasi boundary value method (Q.B.V.-method)
\cite{clark,denche,ivanov,showalter2} and the C-regularized
semigroups technique
\cite{ames2,delaubenfles1,delaubenfles2,melnikova1,melnikova2,piskarev}.


In the method of quasi-reversibility, the main idea consists in
replacing $A$ in \eqref{FVP} by $A_{\alpha}=g_{\alpha}(A)$. In the
original method \cite{lattes} Latt\`{e}s and Lions have proposed
$g_{\alpha}(A)= A-\alpha A^{2}$, to obtain a well-posed problem in
the backward direction. Then, using the information from the
solution of the perturbed problem and solving the original
problem, we get another well-posed problem and this solution
sometimes can be taken to be the approximate solution of the
ill-posed problem \eqref{FVP}.


Difficulties may arise when using the method of
quasi-reversibility discussed above. The essential difficulty is
that the order of the operator is replaced by an operator of
second order, which produces serious difficulties on the numerical
implementation, in addition, the error $(e(\alpha))$ introduced by
small change in the final value $\varphi$ is of the order
$ e^{\frac{T}{4\alpha}}$.

In the Gajewski and Zaccharias quasi-reversibility method
\cite{Gajewski} (see also
\cite{boussetila,ewing,huang2,long,showalter1},
$g_{\alpha}(A) =A(I+\alpha A)^{-1}$.
The advantage of this perturbation lies in
the fact that this perturbation is bounded
$(A_{\alpha}\in\mathcal{L}(H))$, which gives a well-posedness in
the forward and backward direction for the perturbed problem, the
second advantage is that, this perturbation produces a best and
significant approximate solution by comparison with the method
proposed by Latt\`{e}s and Lions. But the amplification factor of
the error resulting from the approximated problem, remains  always
of the order $ e^{\frac{T}{\alpha}}$.

In the method developed by G.W. Clark and S.F. Oppenheimer
\cite{clark} (see also \cite{denche,ivanov,showalter2}, they
approximate problem \eqref{FVP} by
\begin{gather*}
v_t(t)+Av(t)=0,\quad 0<t<T,\\
\beta v(0)+v(T)=\varphi,
\end{gather*}
where $\beta >0$. This method is called quasi-boundary value
method (Q.B.V.-method). We note here that this method gives a
better approximation than many other quasi-reversibility type
methods and the error $(e(\beta))$ introduced by small change in
the final value $\varphi$ is of the order $\frac{1}{\beta}$.

In this paper,  We combine the nice smoothing effect of Yosida
approximation with advantages of quasi-boundary value method, to
build an optimal approximation to problem \eqref{FVP}.

\section{Preliminaries and basic results}

In this section we present the notation and the functional setting
which will be used in this paper and prepare some material which
will be used in our analysis.

 If $B\in\mathcal{L}(H)$ we denote by $\mathcal{N}(B)$
the kernel of $B$ and by $\mathcal{R}(B)$ the range of $B$. We
denote by $\{E_{\lambda},\; \lambda \geq \gamma >0\}$ the spectral
resolution of the identity associated to $A$.

 We denote by
$S(t)=e^{-tA}=\int_{\gamma}^{\infty}e^{-t\lambda}\,dE_{\lambda}
\in \mathcal{L}(H)$, $t\geq 0$, the
$C_0$-semigroup generated by $-A$. Some basic properties of
are listed in the following theorem.

\begin{theorem}[{\cite[Chap. 2, theorem 6.13]{Pazy}}]
\label{sgth}
For this family of operators we have:
\begin{enumerate}
\item $\|S(t)\|\leq 1$, for all $t \geq 0$;
\item the function $t\longmapsto S(t)$, $t>0$, is analytic;
\item for every real $r\geq 0$ and $t > 0$, the operator $S(t)\in
\mathcal{L}(H,\mathcal{D}(A^{r}))$;
\item for every integer $k \geq 0$ and $t >0$,
$\|S^{(k)}(t)\|=\|A^{k}S(t)\|\leq c(k)t^{-k}$;
\item for every $x \in \mathcal{D}(A^{r})$, $r\geq 0$ we have
$S(t)A^{r}x =A^{r}S(t)x$.
\end{enumerate}
\end{theorem}

\begin{definition} \label{def2.1} \rm
We put
$$
J_{\alpha}=(I+\alpha A)^{-1},\quad
A_{\alpha}= A(I+\alpha A)^{-1}= \frac{1}{\alpha}(I-J_{\alpha}), \quad
\alpha > 0,
$$
and call $A_{\alpha}$ the Yosida approximation of $A$.
\end{definition}

 Some basic properties of $A_{\alpha}$ are listed in the
following theorem.

\begin{theorem}[{\cite[chap. VII, p. 101-118]{brezis}}] \label{thm2.2}
We have
\begin{enumerate}
\item $A_{\alpha}$ is positive and self-adjoint;

\item $J_{\alpha}Ah=AJ_{\alpha}h$, for all
$h\in\mathcal{D}(A)$;

\item $J_{\alpha},\; A_{\alpha} \in \mathcal{L}(H)$,
 $\|J_{\alpha}\|\leq 1$, for all $\alpha > 0$;

\item $\|A_{\alpha}h\| \leq \|Ah\|$, for all $\alpha > 0$, for all
$h\in \mathcal{D}(A)$;

\item for all $h \in H$, $\lim_{\alpha \to 0}J_{\alpha}h = h$;

\item  for all $h \in \mathcal{D}(A)$,
$\lim_{\alpha \to 0}A_{\alpha}h = Ah$;

\item for all $h\in H$, for all $t\geq 0$,
$\lim_{\alpha\to 0}S_{\alpha}(t)h=\lim_{\alpha\to
0}e^{-tA_{\alpha}}h=S(t)h =e^{-tA}h$.
\end{enumerate}
\end{theorem}

\begin{theorem}\label{cl}
For $t > 0$, $S(t)$ is self-adjoint and one to one operator with
dense range ($S(t)= S(t)^{*}$, $\mathcal{N}(S(t))=\{0\}$ and
$\overline{\mathcal{R}(S(t))}=H$).
\end{theorem}

\begin{proof}
 $A$ is self-adjoint and since
$S(t)^{*} = (e^{-tA})^{*}=e^{-tA^{*}} = e^{-tA}$,
then we have $S(t)^{*} = S(t)$.

 Let $h\in\mathcal{N}(S(t_{0})),\; t_{0}>0$, then
$S(t_0)h=0$, which implies that $S(t)S(t_0)h=S(t+t_0)h=0,\; t\geq
0$. Using analyticity, one obtains that $S(t)h =0,\; t\geq 0$.
Strong continuity at $0$ now gives $h=0$. This shows that
$\mathcal{N}(S(t_0))=\{0\}$.
Thanks to
$$
\overline{\mathcal{R}(S(t_0))} = \mathcal{N}(S(t_0))^{\perp}
= \{0\}^{\perp} =H,
$$
we conclude that $\mathcal{R}(S(t_0))$ is dense in $H$.
\end{proof}

 For more details, we refer the reader to a general version of
theorem \ref{cl} in the case of analytic semigroups in Banach
spaces (Lemma 2.2, \cite{ioana}).

\begin{remark}[Smoothing effect and irreversibility] \label{rmk2.1} \rm
Thanks to Theorem \ref{sgth} and Theorem \ref{cl}, we observe that
the solution of the direct Cauchy problem
$$
u'(t)+Au(t)=0,\;\; 0< t \leq T, \quad u(0)=u_0,\label{IVP}
$$
has the following smoothing effect: admitting the initial value
$u(0)$ to belong only to $H$, then for all $t>0$,
\[
\mathcal{R}(S(t))\subset \mathcal{C}^{\infty}(A)
\overset{def}{=} \cap_{n=1}^{\infty}\mathcal{D}(A^{n})
\]
(space more regular than $H$) (see \cite{Gorbachuk}).
It follows that for the
final value problem \eqref{FVP} to have a solution, we should have
$u(T)\in \mathcal{C}(A)\subseteq \mathcal{R}(S(T))$, where
$\mathcal{C}(A)$ is an admissible class for which the \eqref{FVP} is
solvable. This shows that the \eqref{FVP} is irreversible in the
sense:
$$
S(T-t): H \to \mathcal{R}(S(T-t))\subset
\mathcal{C}^{\infty}(A)\subsetneq H,\quad 0\leq t <T,
$$
and
$\mathcal{R}(S(T-t))\neq \overline{\mathcal{R}(S(T-t))}$, in other
words $S(T-t)^{-1}=S(t-T)\notin\mathcal{L}(H)$.
\end{remark}

 For notational convenience and simplicity, we denote
$$
\mathcal{C}_{\theta}(A)=\{h\in H:\; \|h\|_{\mathcal{C}_{\theta}}^{2}
=\|e^{\theta TA}h\|^{2}
=\int_{\gamma}^{+\infty}e^{2T\theta\lambda}\,d\|E_{\lambda}h\|^{2}<
+\infty \},\quad \theta \geq 0.
$$
By the definition of
$\mathcal{C}_{\theta}(A)$ we have the following topological
inclusions:
\begin{gather*}
\mathcal{C}_{\theta_2}(A)\subseteq \mathcal{C}_{\theta_1}(A),
\quad  \theta_2 \geq \theta_1,
\\
\mathcal{C}_{\theta}(A)\subset \mathcal{D}(A^{r})\subset H,\quad
\theta >0,\; r>0,
\\
\|A^{r}h\|^{2}=\int_{\gamma}^{+\infty}\big(\frac{\lambda^{r}}
{e^{\theta T\lambda}}\big)^{2}e^{2\theta
T\lambda}\,d\|E_{\lambda}h\|^{2}\leq
c(\theta,r,T)\|h\|_{C_\theta}^{2}
\end{gather*}
where
$c(\theta,r,T)=(\frac{\theta
T}{r})^{2\theta T}e^{-2r}$.

 For $\lambda \geq \gamma$, we introduce the functions:
$$
H_{\sigma}(\lambda)=F_{\sigma}(\lambda)+G_{\sigma}(\lambda),
$$
where
\begin{gather*}
F_{\sigma}(\lambda)=\frac{\beta}
{\beta+e^{-\frac{T\lambda}{1+\alpha\lambda}}},\quad
G_{\sigma}(\lambda)=
\frac{e^{-\frac{T\lambda}{1+\alpha\lambda}}-e^{-T\lambda}}
{\beta+e^{-\frac{T\lambda}{1+\alpha\lambda}}},
\\
F_{\sigma,\theta}(\lambda)=F_{\sigma}(\lambda)e^{-T\theta\lambda},\quad
G_{\sigma,\theta}(\lambda)=G_{\sigma}(\lambda)e^{-T\theta\lambda},\quad
\theta >0,
\\
K_{\beta}(\lambda)=\frac{\beta}{\beta+e^{-T\lambda}},\quad
 M_{\theta}(\lambda)=\lambda^{2}e^{-\theta T\lambda},\quad \theta >0,
\\
F_{\sigma_1,\sigma_2}(\lambda)=\frac{|\beta_1
-\beta_2|}{\big(\beta_1 +e^{-\frac{T\lambda}{1+\alpha_1\lambda}}\big)
\big(\beta_2+e^{-\frac{T\lambda}{1+\alpha_2\lambda}}\big)},
\\
G_{\sigma_1,\sigma_2}(\lambda)=
\frac{|e^{-\frac{T\lambda}{1+\alpha_1\lambda}}-e^{-\frac{T\lambda}{1+\alpha_2\lambda}}|}
{\big(\beta_1 +e^{-\frac{T\lambda}{1+\alpha_1\lambda}}\big)
\big(\beta_2 +e^{-\frac{T\lambda}{1+\alpha_2\lambda}}\big)}.
\end{gather*}
By simple differential calculus and elementary estimates, we show
that
\begin{gather}
0<F_{\sigma}(\lambda)\leq 1,\quad F_{\sigma}(\lambda)\leq
\frac{\beta}{\beta+e^{\frac{-T}{\alpha}}},\quad
F_{\sigma}(\lambda)\leq \beta e^{T\lambda}.\label{e1}
\\
0<G_{\sigma}(\lambda)\leq 1, \nonumber\\
G_{\sigma}(\lambda) =
\frac{e^{-\frac{T\lambda}{1+\alpha\lambda}}\big(1-e^{\frac{-\alpha
T\lambda^{2}}{1+\alpha\lambda}}\big)}
{\beta+e^{-\frac{T\lambda}{1+\alpha\lambda}}}
 \leq  \big( 1-e^{\frac{-\alpha
T\lambda^{2}}{1+\alpha\lambda}}\big)
 \leq  \frac{\alpha T\lambda^{2}}{1+\alpha\lambda}
 \leq \alpha T\lambda^{2}. \label{e2}
\\
M_{\theta,\infty}(\lambda)=\sup_{\lambda\geq \gamma}M_{\theta}(\lambda)
=\big(\frac{2}{\theta Te}\big)^{2}\leq
\big(\frac{1}{\theta T}\big)^{2}.\label{e3}
\\
F_{\sigma,\theta,\infty}=
\sup_{\lambda \geq \gamma}F_{\sigma,\theta}(\lambda)\leq
K_{\beta,\infty}=
\sup_{\lambda \geq \gamma}K_{\beta}(\lambda)
\leq \begin{cases}
\beta,&\text{if } \theta \geq 1,\\
c_1(\theta)\beta^{\theta},&\text{if } 0<\theta <1,
\end{cases} \label{e4}
\end{gather}
where
$c_1(\theta)=(1-\theta)^{1-\theta}\theta^{\theta}\leq 1$.
\begin{equation}
G_{\sigma,\theta,\infty}=\sup_{\lambda \geq \gamma}G_{\sigma,\theta}(\lambda)\leq
 c_2(\theta,T)\frac{\alpha}{1+\beta}\leq
c_2(\theta,T)\alpha,\label{e5}
\end{equation}
where $c_2(\theta,T)=\frac{1}{T\theta ^{2}}$.
\begin{gather}
F_{\sigma_1,\sigma_2}(\lambda)\leq e^{T\lambda},\quad
F_{\sigma_1,\sigma_2}(\lambda)\leq |\beta_1 -\beta_2|e^{2T\lambda}.\label{e6}
\\
G_{\sigma_1,\sigma_2}(\lambda)\leq e^{T\lambda},\quad
F_{\sigma_1,\sigma_2}(\lambda)\leq |\alpha_1 -\alpha_2|T\lambda^{2}
 e^{T\lambda}.\label{e7}
\\
F_{\sigma}(\lambda)\lambda^{-1}
=\frac{\beta}{\beta\lambda+\lambda e^{\frac{-T\lambda}{1+\alpha\lambda}}}
\leq \frac{\beta}{\beta\lambda+\gamma e^{-T\lambda}}
\leq \frac{T}{1+\ln(\frac{\gamma T}{\beta})},
\quad (\beta\leq \gamma T).
\label{e8}
\end{gather}
Without loss of the generality, we suppose that
$\lambda \geq \gamma \geq 1$. By virtue of
$(1-e^{-\tau}\leq \sqrt{\tau}$, $\tau \geq 1)$, the function
$G_{\sigma}(\lambda)$ can be estimated as follows:
\begin{equation}
G_{\sigma}(\lambda)
= \frac{e^{-\frac{T\lambda}{1+\alpha\lambda}}-e^{-T\lambda}}
 {\beta+e^{-\frac{T\lambda}{1+\alpha\lambda}}}
=\frac{e^{-\frac{T\lambda}{1+\alpha\lambda}}
 (1-e^{\frac{-T\alpha\lambda^{2}} {1+\alpha\lambda}})}
 {\beta+e^{-\frac{T\lambda}{1+\alpha\lambda}}}
\leq   1-e^{\frac{-T\alpha\lambda^{2}} {1+\alpha\lambda}}
\leq \sqrt{T\alpha}\lambda.
\label{e9}
\end{equation}

\begin{remark} \label{rmk2.2}\rm
Let $u$ be a solution to the problem
\begin{equation}
u_t +Au=0,\quad 0<t<T,\quad u(T)=\varphi.\label{Pu}
\end{equation}
We set $U(t)=e^{-\nu t}u(t)$, $\nu \geq 1$, then $U$ is a solution
of the problem
\begin{equation}
U_t +A_{\nu}U=0,\;\; 0<t<T,\quad U(T)=e^{-\nu T}\varphi=\psi,\label{PU}
\end{equation}
with $A_{\nu}=A+\nu I\geq (\nu +\gamma)I\geq \nu I$. Hence,
regularizing \eqref{Pu} is equivalent to regularize \eqref{PU}.
\end{remark}

\begin{remark}\label{rmk2.3} \rm
The operational calculus for a
self-adjoint operator and estimates \eqref{e1}--\eqref{e9}
play the key role in our analysis and calculations.
\end{remark}

\section{The approximate problem}

\subsection*{Description of the method}
\textbf{Step 1} Let $v_{\sigma}$ be the solution of
the  perturbed problem
\begin{equation}
\begin{gathered}
v_{\sigma}'(t)+A_{\alpha}v_{\sigma}(t)=0,\quad 0<t<T,  \\
\beta v_{\sigma}(0)+v_{\sigma}(T)=\varphi\\
\end{gathered} \label{Psigma}
\end{equation}
 where the operator $A$ is replaced by
$A_{\alpha}=A(I+\alpha A)^{-1}$ and the final condition
$v(T)=\varphi$ is replaced by the nonlocal condition
$\beta v(0)+v(T)=\varphi$, where $\alpha >0$, $\beta >0$ and
$\sigma =(\alpha,\beta)$.

\textbf{Step 2} We use the initial value
$$
\varphi_{\sigma} = v_{\sigma}(0) =
\big(\beta+S_{\alpha}(T)\big)^{-1}\varphi,
$$
in the problem
\begin{equation}
u_{\sigma}'(t)+Au_{\sigma}(t)=0,\quad 0<t\leq T,
\quad u_{\sigma}(0)=\varphi_{\sigma}.\label{IVPsigma}
\end{equation}

\textbf{Step 3}  We show that
\begin{gather*}
\|u_{\sigma}(T)-\varphi\|\to 0,\quad \text{as } |\sigma| \to 0,\\
\|u_{\sigma}(0)-u(0)\|\to 0,\quad \text{as } |\sigma| \to 0,\\
\sup_{0\leq t \leq T}\|u_{\sigma}(t)-u(t)\|\to 0,\quad
\text{as }  |\sigma| \to 0.
\end{gather*}

\section{Analysis of the method and main results}

Now we are ready to state and prove the main results of this
paper.

\begin{definition}[\cite{showalter1}] \label{def4.1} \rm
A solution of \eqref{FVP} on the
interval $[0,T]$ is a function $u\in \mathcal{C}([0,T];H)\cap
\mathcal{C}^{1}((0,T);H)$ such that for all $t\in (0,T)$, $u(t)\in
\mathcal{D}(A)$ and \eqref{FVP} holds.
\end{definition}

 It is useful to know exactly the admissible set for
which \eqref{FVP} has a solution. The following lemma gives an answer
to this question.

\begin{lemma}[{\cite[Lemma 1]{clark}}] \label{lmmecle}
Problem \eqref{FVP} has a solution if and only if
$\varphi\in\mathcal{C}_{1}(A)$,
and its unique solution is represented by
\begin{equation}
u(t)=e^{(T-t)A}\varphi. \label{1}
\end{equation}
\end{lemma}

Using semi-groups theory and the properties of
$S_{\alpha}(t)$, we have the following theorem.

\begin{theorem}\label{thm1}
For all $\varphi\in H$, the function
\[
v_{\sigma}(t)=S_{\alpha}(t)\big( \beta + S_{\alpha}(T)\big)^{-1}\varphi
\] %\label{2}
is the unique solution of \eqref{Psigma} and it depends
continuously on $\varphi$.
\end{theorem}

\begin{proof} We consider the problem
\begin{equation}
y_{\sigma}'(t)+A_{\alpha}y_{\sigma}(t)=0,\quad 0<t\leq T,\quad
y_{\sigma}(0)=\big( \beta +
S_{\alpha}(T)\big)^{-1}\varphi.
\label{3}
\end{equation}
This problem is well-posed, and its solution is
\begin{equation}
y_{\sigma}(t)=S_{\alpha}(t)\big( \beta +
S_{\alpha}(T)\big)^{-1}\varphi.
\label{4}
\end{equation}
Observing that
\begin{equation}
\beta y_{\sigma}(0)+y_{\sigma}(T)=\big( \beta +
S_{\alpha}(T)\big) \big( \beta +
S_{\alpha}(T)\big)^{-1}\varphi=\varphi. \label{5}
\end{equation}
Thanks to \eqref{5} and the uniqueness of solution to direct problem
\eqref{3}, we deduce that the problem \eqref{Psigma} admits the
unique solution $v_{\sigma}$ given by \eqref{4}.
To show the continuous dependence of $v_{\sigma}$ on
$\varphi$, we compute
$$
\|v_{\sigma}(t)\|=\|S_{\alpha}(t)\big( \beta +
S_{\alpha}(T)\big)^{-1}\varphi\|
 \leq   \|\big( \beta + S_{\alpha}(T)\big)^{-1}\varphi\|
 \leq (\beta+e^{\frac{-T}{\alpha}})^{-1}\|\varphi\|.
$$
\end{proof}

\begin{theorem}\label{thm2}The problem \eqref{IVPsigma} is well-posed,
and its solution is
\begin{equation}
u_{\sigma}(t)=S(t)\varphi_{\alpha}=S(t)
\big(\beta+S_{\alpha}(T)\big)^{-1}\varphi. \label{6}
\end{equation}
\end{theorem}

 An easy computation shows that
\begin{equation}
\|u_{\sigma}(t)\|\leq \big( \frac{1}{\beta +
e^{-\frac{T}{\alpha}}}\big)^{\frac{T-t}{T}}\|\varphi\|.\label{7}
\end{equation}

\begin{theorem}\label{thm3}
For all $\varphi\in H$, $\|u_{\sigma}(T)-\varphi\|\to 0$, as
$|\sigma|\to 0$.
\end{theorem}

\begin{proof} We compute
\begin{equation}
\|u_{\sigma}(T)-\varphi\|^{2}=\int_{\gamma}^{+\infty}
H_{\alpha}(\lambda)^{2}\,d\|E_{\lambda}\varphi\|^{2} \leq
2(I_{1,\sigma}+I_{2,\sigma}), \label{8}
\end{equation}
where
\begin{gather*}
I_{1,\sigma}=\int_{\gamma}^{+\infty}
F_{\alpha}(\lambda)^{2}\,d\|E_{\lambda}\varphi\|^{2},\\
I_{2,\sigma}=\int_{\gamma}^{+\infty}
G_{\alpha}(\lambda)^{2}\,d\|E_{\lambda}\varphi\|^{2}.
\end{gather*}
Fix $\varepsilon >0$. Choose $N$ so that
$\int_{N}^{+\infty}d\|E_{\lambda}\varphi\|^{2} <\frac{\varepsilon}{8}$.
Thus
\begin{gather*}
I_{1,\sigma}\leq \int_{\gamma}^{N}F_{\sigma}(\lambda)^{2}
\,d\|E_{\lambda}\varphi\|^{2}+
\int_{N}^{+\infty}F_{\sigma}(\lambda)^{2}\,d\|E_{\lambda}\varphi\|^{2}, \\
I_{2,\sigma}\leq \int_{\gamma}^{N}G_{\sigma}(\lambda)^{2}
\,d\|E_{\lambda}\varphi\|^{2}+
\int_{N}^{+\infty}G_{\sigma}(\lambda)^{2}\,d\|E_{\lambda}\varphi\|^{2}.
\end{gather*}
Using inequalities \eqref{e1} and \eqref{e2}, we derive
\begin{gather*}
I_{1,\sigma}\leq \frac{\varepsilon}{8}+\beta^{2}e^{2TN}\|\varphi\|^{2},\\
I_{2,\sigma}\leq \frac{\varepsilon}{8}+\alpha^{2}T^{2}N^{4}\|\varphi\|^{2}.
\end{gather*}
So by taking $\sigma$ such that
$$
|\sigma|^{2}=\beta^{2}+\alpha^{2}\leq \frac{1}{\|\varphi\|^{2}}
\big(\frac{1}{T^{2}N^{4}}+\frac{1}{e^{2TN}}\big)\frac{\varepsilon}{4},
$$
we complete the proof.
\end{proof}

Note that we do not have a convergence rate here.


\begin{theorem}\label{thm4}
If $\varphi\in \mathcal{C}_{\theta}(A)$, $0 < \theta <1$, then we have
\begin{equation}
\|u_{\sigma}(T)-\varphi\|^{2}\leq
2\left(c_{1}^{2}(\theta)\beta^{2\theta}+c_{2}^{2}
(\theta,T)\alpha^{2}\right)\|\varphi\|_{\mathcal{C}_{\theta}}^{2}.
\label{9a}
\end{equation}
Moreover, if $\theta \geq 1$, then we have
\begin{equation}
\|u_{\sigma}(T)-\varphi\|^{2}\leq
2\left(\beta^{2}+c_{2}^{2}(\theta,T)\alpha^{2}\right)
\|\varphi\|_{\mathcal{C}_{\theta}}^{2},\label{9b}
\end{equation}
where $c_{1}(\theta)=(1-\theta)^{1-\theta}\theta^{\theta}\leq 1$
and $c_{2}(\theta,T)=T^{-1}\theta^{-2}$.
\end{theorem}

\begin{proof} By doing computation, we have
\begin{align*}
 \|u_{\sigma}(T)-\varphi\|^{2}
&= \int_{\gamma}^{+\infty}
H_{\sigma}^{2}(\lambda)e^{-2\theta T\lambda}e^{2\theta
T\lambda}\,d\|E_{\lambda}\varphi\|^{2}\\
& \leq 2\int_{\gamma}^{+\infty}
F_{\sigma,\theta}^{2}(\lambda)e^{2\theta
T\lambda}\,d\|E_{\lambda}\varphi\|^{2}+2\int_{\gamma}^{+\infty}
G_{\sigma,\theta}^{2}(\lambda)e^{2\theta
T\lambda}\,d\|E_{\lambda}\varphi\|^{2} \\
 &\leq  2\left(F_{\sigma,\theta,\infty}^{2}+
 G_{\sigma,\theta,\infty}^{2}\right)\|\varphi\|_{\mathcal{C}_{\theta}}^{2}
\end{align*}
and by virtue of inequalities \eqref{e4}, \eqref{e5} we obtain the
desired estimates.
\end{proof}

 We define
$$
\mathcal{F}: \mathbb{R}^{+}\times\mathbb{R}^{+}\to H,\quad
\sigma=(\alpha,\beta)\mapsto \mathcal{F}(\sigma)
=\begin{cases}
u_{\sigma}(0)=\varphi_{\sigma}, &  \sigma\neq (0,0),\\
u(0)=\varphi_{0}, & \sigma = (0,0).
\end{cases}
$$

\begin{theorem}\label{thm5}
For all $\varphi\in H$, \eqref{FVP} has a solution $u$ if and only if
the function $\mathcal{F}$ is continuous at $(0,0)$. Furthermore,
we have that $u_{\sigma}(t)$ converges to $u(t)$ as $|\sigma|$
tends to zero uniformly in $t$.
\end{theorem}

\begin{proof} Assume that $\lim_{|\sigma|\to 0}\varphi_{\sigma} = \varphi_{0}$
and $\|\varphi_{0}\|<+\infty$.
Let $w(t)=S(t)\varphi_{0}$. We compute
$$
\|w(t)-u_{\sigma}(t)\| =  \|S(t)\varphi_{0} - S(t)\varphi_{\sigma}\|\\
 =  \|S(t)(\varphi_{0} - \varphi_{\sigma})\| \\
\leq  \|\varphi_{0} - \varphi_{\sigma}\|.
$$
Which implies
$$
\sup_{0 \leq t \leq T}\|w(t)-u_{\sigma}(t)\| \leq
\|\varphi_{0} - \varphi_{\sigma}\| \to 0,\quad
\text{as } |\sigma| \to 0.
$$
Since $\lim_{|\sigma|\to 0}u_{\sigma}(T)=\varphi$ and
$\lim_{|\sigma|\to 0}u_{\sigma}(T)=w(T)$, we
infer that $w(T)=\varphi$. Thus, $w(t)=S(t)\varphi_{0}$ solves
\eqref{FVP} and satisfies the condition $w(T)=\varphi$.

 Now, let us assume that $u(t)$ is the solution to
\eqref{FVP}. From lemma \ref{lmmecle} we have
$u(0)=S(-T)\varphi\in H$, i.e.,
\[
\|u(0)\|^{2}=\|\varphi\|_{\mathcal{C}_1}^{2}=
\int_{\gamma}^{+\infty}e^{2T\lambda}\,d\|E_{\lambda}\varphi\|^{2}<
\infty.
\]
 Let $N>0$ and $\varepsilon >0$ such that
$\int_{N}^{+\infty}e^{2T\lambda}\,d\|E_{\lambda}\varphi\|^{2}<
\frac{\varepsilon}{8}$. Let
$\sigma_{i}=(\alpha_i,\beta_i),\; i=1,2$. Then
\begin{equation}
\begin{aligned}
\|u_{\sigma_1}(0)-u_{\sigma_2}(0)\|^{2}
&= \int_{\gamma}^{+\infty}\big(
(\beta_1+e^{\frac{-T\lambda}{1+\alpha_1
\lambda}})^{-1}-(\beta_2+e^{\frac{-T\lambda}{1+\alpha_2
\lambda}})^{-1}
\big)^{2}\,d\|E_{\lambda}\varphi\|^{2} \\
& \leq  2\int_{\gamma}^{+\infty} F_{\sigma_1,\sigma_2}^{2}
(\lambda)\,d\|E_{\lambda}\varphi\|^{2}+2\int_{\gamma}^{+\infty}
G_{\sigma_1,\sigma_2}^{2}(\lambda)\,d\|E_{\lambda}\varphi\|^{2}.
  \end{aligned}\label{10}
\end{equation}
By using \eqref{e6} and \eqref{e7}, the right hand sand of \eqref{10} can be
estimated as follows
\begin{align*}
\int_{\gamma}^{+\infty} F_{\sigma_1,\sigma_2}^{2}
(\lambda)\,d\|E_{\lambda}\varphi\|^{2}
& \leq \int_{\gamma}^{N} F_{\sigma_1,\sigma_2}^{2}
(\lambda)\,d\|E_{\lambda}\varphi\|^{2}+\int_{N}^{+\infty}
F_{\sigma_1,\sigma_2}^{2}
(\lambda)\,d\|E_{\lambda}\varphi\|^{2} \\
& \leq  (\beta_2-\beta_1)^{2}e^{2TN}\|\varphi\|_{\mathcal{C}_1}^{2}+
\frac{\varepsilon}{8},
\end{align*}
\begin{align*}
\int_{\gamma}^{+\infty} G_{\sigma_1,\sigma_2}^{2}
(\lambda)\,d\|E_{\lambda}\varphi\|^{2}
& \leq \int_{\gamma}^{N} G_{\sigma_1,\sigma_2}^{2}
(\lambda)\,d\|E_{\lambda}\varphi\|^{2}+\int_{N}^{+\infty}
G_{\sigma_1,\sigma_2}^{2}
(\lambda)\,d\|E_{\lambda}\varphi\|^{2} \\
& \leq
(\alpha_2-\alpha_1)^{2}T^{2}N^{4}\|\varphi\|_{\mathcal{C}_1}^{2}+
\frac{\varepsilon}{8}.
\end{align*}
Now if we choose $\sigma=(\alpha,\beta)$ so that
\[
|\sigma|^{2}=\alpha^{2}+\beta^{2}\leq
\frac{1}{\|\varphi\|_{\mathcal{C}_1}^{2}}
\Big(\frac{1}{T^{2}N^{4}}+\frac{1}{e^{2TN}}\Big)\frac{\varepsilon}{4}
\]
and $\sigma_{0} =(0,0)$, then  we have
$\|u_{\sigma}(0)-u_{\sigma_{0}}(0)\|^{2}=\|\varphi_{\sigma}-\varphi_{0}\|^{2}\leq
\varepsilon$. This shows that the function $\mathcal{F}$ is
continuous at $(0,0)$.
\end{proof}

\begin{remark} \label{rmk4.2} \rm
If we suppose that  $\varphi\in \mathcal{C}_{1}(A)$, then by  the equality
$$
\|u_{\sigma}(0)-u(0)\|^{2}=
\|u_{\sigma}(T)-\varphi\|_{\mathcal{C}_1}^{2}
$$
and theorem \ref{thm3}, we have
$$
\|u_{\sigma}(0)-u(0)\|^{2}\to 0,\;\; \text{as}\;\;|\sigma|\to 0.
$$
\end{remark}

\begin{theorem}\label{thm6}
If $\varphi\in \mathcal{C}_{1+\theta}(A)$, $0 < \theta <1$, we have
\begin{equation}
\|u_{\sigma}(0)-u(0)\|^{2}\leq
2\left(c_{1}^{2}(\theta)\beta^{2\theta}+c_{2}^{2}
(\theta,T)\alpha^{2}\right)\|\varphi\|_{\mathcal{C}_{1+\theta}}^{2}.
\label{11a}
\end{equation}
Moreover, if $\theta \geq 1$,  we have
\begin{equation}
\|u_{\sigma}(0)-u(0)\|^{2}\leq
2\left(\beta^{2}+c_{2}^{2}(\theta,T)\alpha^{2}\right)
\|\varphi\|_{\mathcal{C}_{1+\theta}}^{2}.\label{11b}
\end{equation}
\end{theorem}

\begin{proof}
By similar calculations to those used in Theorem \ref{thm4}
and \ref{thm5}, we have
\begin{align*}
&\|u_{\sigma}(0)-u(0)\|^{2}\\
& =  \int_{\gamma}^{+\infty}H_{\sigma}(\lambda)^{2}e^{-2\theta
T\lambda}e^{2(1+\theta)T\lambda} \,d\|E_{\lambda}\varphi\|^{2} \\
&\leq 2\int_{\gamma}^{+\infty}
F_{\sigma,\theta}(\lambda)^{2}e^{2(1+\theta)T\lambda}
\,d\|E_{\lambda}\varphi\|^{2}
+ 2\int_{\gamma}^{+\infty}G_{\sigma,\theta}(\lambda)^{2}e^{2(1+\theta)T\lambda}
\,d\|E_{\lambda}\varphi\|^{2}\\
&\leq 2\left(F_{\sigma,\theta,\infty}^{2}+
G_{\sigma,\theta,\infty}^{2}\right)\|\varphi\|_{\mathcal{C}_{1+\theta}}^{2}
\end{align*}
and by  \eqref{e4}, \eqref{e5} we obtain the
desired estimates.
\end{proof}

 From Theorem \ref{thm5} and \ref{thm6}, we have the following result.

\begin{corollary} \label{coro4.1}
If $\varphi\in \mathcal{C}_{1+\theta}(A)$, $0<\theta<1$,
then an upper bound of the rate of convergence of the method is
given by
$$
\sup_{0\leq t\leq
T}\|u_{\sigma}(t)-u(t)\|^{2}\leq\|u_{\sigma}(0)-u(0)\|^{2}\leq
2\left(c_{1}^{2}(\theta)\beta^{2\theta}+c_{2}^{2}
(\theta,T)\alpha^{2}\right)\|\varphi\|_{\mathcal{C}_{1+\theta}}^{2}.
%\label{12a}
$$
Moreover, if $\theta \geq 1$, then we have
$$
\sup_{0\leq t\leq
T}\|u_{\sigma}(t)-u(t)\|^{2}\leq\|u_{\sigma}(0)-u(0)\|^{2}\leq
2\left(\beta^{2}+c_{2}^{2}(\theta,T)\alpha^{2}\right)
\|\varphi\|_{\mathcal{C}_{1+\theta}}^{2}. %\label{12b}
$$
\end{corollary}

\begin{remark} \label{rmk4.3} \rm
 If $\varphi\in \mathcal{D}(A^{2})$, then with the help of
\eqref{e1} and \eqref{e2}, $\|u_{\sigma}(T)-\varphi\|^{2}$ can be
estimated as follows:
\begin{align*}
\|u_{\sigma}(T)-\varphi\|^{2}
&\leq 2\int_{\gamma}^{+\infty} F_{\sigma}^{2}
(\lambda)\,d\|E_{\lambda}\varphi\|^{2}+2\int_{\gamma}^{+\infty}
G_{\sigma}^{2}
(\lambda)\,d\|E_{\lambda}\varphi\|^{2} \\
&\leq  2\big(\frac{\beta}{\beta+e^{\frac{-T}{\alpha}}}
\big)^{2}\int_{\gamma}^{+\infty}
\,d\|E_{\lambda}\varphi\|^{2}+2T^{2}\alpha^{2}\int_{\gamma}^{+\infty}
\lambda^{4}\,d\|E_{\lambda}\varphi\|^{2}.
\end{align*}
Choosing $\alpha =\frac{T}{(1-r)\ln(\frac{1}{\beta})}$,
$0 < r<1$, we obtain
$$
\|u_{\sigma}(T)-\varphi\|^{2}\leq 2\Big( \beta^{2r}\|\varphi\|^{2}+
\frac{T^{4}}{(1-r)^{2}\ln^{2}
(\frac{1}{\beta})}\|A^{2}\varphi\|^{2}\Big).%\label{13}
$$
\end{remark}

\begin{theorem}\label{thm4.7}
Assuming that $\varphi\in \mathcal{D}(A)$ and letting
 $\alpha =\frac{T}{(1-r)\ln(\frac{1}{\beta})}$, $0 < r<1$,
the expression $\|u_{\sigma}(T)-\varphi\|^{2}$ can be estimated
as follows
$$
\|u_{\sigma}(T)-\varphi\|^{2}\leq \beta^{2r}\|\varphi\|^{2}+
\frac{4T}{(1-r)\ln
(\frac{1}{\beta})}\|A\varphi\|^{2}.%\label{14}
$$
\end{theorem}

\begin{proof}
We have
\begin{gather}
u_{\sigma}'(t)+Au_{\sigma}(t)=0,\label{E1}\\
v_{\sigma}'(t)+Av_{\sigma}(t)=(A-A_{\alpha})v_{\sigma}(t)=
\alpha J_{\alpha}A^{2}v_{\sigma}(t),\label{E2}\\
u_{\sigma}(0)-v_{\sigma}(0)=0, \label{C1}\\
\beta v_{\sigma}(0)+v_{\sigma}(T)=\varphi.\label{C2}
\end{gather}
If we put $x_{\sigma}(t)=v_{\sigma}(t)-u_{\sigma}(t)$, then
$x_{\sigma}(t)$ satisfies the equation
\begin{equation}
x_{\sigma}'(t)+Ax_{\sigma}(t)=
\alpha J_{\alpha}A^{2}u(t).\label{E3}
\end{equation}
Applying the operator
$\mathcal{M}(t)=e^{(t-T)A}$ to \eqref{E1}, \eqref{E2} and \eqref{E3}, we
obtain
\begin{gather}
\frac{d}{dt}\left(\mathcal{M}(t)u_{\sigma}(t)\right)=0,\label{E1'}\\
\frac{d}{dt}\left(\mathcal{M}(t)v_{\sigma}(t)\right)=
\alpha J_{\alpha}A^{2}\mathcal{M}(t)v_{\sigma}(t),\label{E2'}\\
\frac{d}{dt}\left(\mathcal{M}(t)x_{\sigma}(t)\right)=
\alpha J_{\alpha}A^{2}\mathcal{M}(t)v_{\sigma}(t).\label{E3'}
\end{gather}
Multiplying \eqref{E3'} by $\mathcal{M}(t)x_{\sigma}$ and integrating
the obtained result over $(0,\tau)$, we get
\begin{align*}
\int_{0}^{\tau}\frac{d}{dt}\|\mathcal{M}(t)x_{\sigma}(t)\|^{2}
&=  \|\mathcal{M}(\tau)x_{\sigma}(\tau)\|^{2}-
\|\mathcal{M}(0)x_{\sigma}(0)\|^{2}=\|\mathcal{M}(\tau)x_{\sigma}(\tau)\|^{2}\\
& =  2 \int_{0}^{\tau}Re(\alpha
J_{\alpha}A^{2}\mathcal{M}(t)v_{\sigma}(t),
\mathcal{M}(t)x_{\sigma})\,dt\\
&\leq  2T \int_{0}^{T}\|\alpha
J_{\alpha}A^{2}\mathcal{M}(t)v_{\sigma}(t)\|^{2}\,dt
+\frac{1}{2T}\int_{0}^{T}\|\mathcal{M}(t)x_{\sigma}(t)\|^{2}\\
&\leq  2T \int_{0}^{T}\|\alpha
J_{\alpha}A^{2}\mathcal{M}(t)v_{\sigma}(t)\|^{2}\,dt+\frac{1}{2}
\sup_{0\leq t\leq T}\|\mathcal{M}(t)x_{\sigma}(t)\|^{2}.
\end{align*}
This implies
$$
\frac{1}{2} \sup_{0\leq t\leq T}\|\mathcal{M}(t)x_{\sigma}(t)\|^{2}\leq
2T \int_{0}^{T}\|\alpha
J_{\alpha}A^{2}\mathcal{M}(t)v_{\sigma}(t)\|^{2}\,dt.
$$
In particulary for $t=T$ we have
\begin{equation}
\|\mathcal{M}(T)x_{\sigma}(T)\|^{2}=\|u_{\sigma}(T)-v_{\sigma}(T)\|^{2}\leq
4T \int_{0}^{T}\|\alpha
J_{\alpha}A^{2}\mathcal{M}(t)v_{\sigma}(t)\|^{2}\,dt.\label{15}
\end{equation}
 From \eqref{15} we can write
\begin{align*}
\|u_{\sigma}(T)-v_{\sigma}(T)\|^{2}
& = \|(u_{\sigma}(T)-\varphi)+ (\varphi-v_{\sigma}(T))\|^{2} \\
&=  \|(u_{\sigma}(T)-\varphi)+\beta v_{\sigma}(0)\|^{2}\\
&= \|u_{\sigma}(T)-\varphi\|^{2}+\|\beta v_{\sigma}(0)\|^{2}+
2Re(u_{\sigma}(T)-\varphi,\beta v_{\sigma}(0))\\
& \leq 4T \int_{0}^{T}\|\alpha
J_{\alpha}A^{2}\mathcal{M}(t)v_{\sigma}(t)\|^{2}\,dt.
\end{align*}
This last inequality implies
\begin{equation}
\|u_{\sigma}(T)-\varphi\|^{2}\leq 8T \int_{0}^{T}\|\alpha
J_{\alpha}A^{2}\mathcal{M}(t)v_{\sigma}(t)\|^{2}\,dt+2\|\beta
v_{\sigma}(0)\|^{2}.\label{16}
\end{equation}
To estimate the integral in the right-hand side, we take  the
inner product of \eqref{E2'} with $\alpha
J_{\alpha}A^{2}\mathcal{M}(t)v_{\sigma}(t)$ and integrate
over $(0,T)$:
\begin{align*}
&\int_{0}^{T}\|\alpha
J_{\alpha}A^{2}\mathcal{M}(t)v_{\sigma}(t)\|^{2}\,dt\\
& = \frac{1}{2}\int_{0}^{T}
\mathop{\rm Re}(\frac{d}{dt}\left(\mathcal{M}(t)v_{\sigma}(t)\right),
\alpha J_{\alpha}A^{2}\mathcal{M}(t)v_{\sigma}(t))\,dt \\
&= \frac{1}{2}\int_{0}^{T}\frac{d}{dt}
\|\alpha J_{\alpha}A\mathcal{M}(t)v_{\sigma}(t)\|^{2}\,dt
+\frac{1}{2}\int_{0}^{T}\frac{d}{dt}
\|\alpha^{2} J_{\alpha}A^{3/2}\mathcal{M}(t)v_{\sigma}(t)\|^{2}\,dt \\
&\leq \frac{1}{2}\left(\|\alpha
J_{\alpha}Av_{\sigma}(T)\|^{2}+\|\alpha^{2}
J_{\alpha}A^{3/2}v_{\sigma}(T)\|^{2}\right).
\end{align*}%\label{17}
By virtue of
$$
\|v_{\sigma}(T)\|=\|S_{\alpha}(T)\big(\beta+ S_{\alpha}(T)\big)^{-1}\varphi\|
\leq\|\varphi\|
$$
 and
$$
\|A\varphi\|^{2}=\|(I+\alpha A)J_{\alpha}A\varphi\|^{2}
=\|J_{\alpha}A\varphi\|^{2}+
2\alpha\|J_{\alpha}A^{3/2}\varphi\|^{2}+ \alpha^{2}\|
J_{\alpha}A^{2}\varphi\|^{2},
$$
we derive
\[
\int_{0}^{T}\|\alpha
J_{\alpha}A^{2}\mathcal{M}(t)v_{\sigma}(t)\|^{2}
 \leq \frac{1}{2}\left(\|\alpha
J_{\alpha}A\varphi\|^{2}+\|\alpha^{2}
J_{\alpha}A^{3/2}\varphi\|^{2}\right)
 \leq \frac{1}{2}\alpha\|A\varphi\|^{2}.
\] %\label{18}
Combining this inequality and \eqref{16}, we obtain
\begin{equation}
\begin{aligned}
 \|u_{\sigma}(T)-\varphi\|^{2}
& \leq  4T\alpha\|A\varphi\|^{2}+
\|\beta v_{\sigma}(0)\|^{2} \\
& \leq  4T\alpha\|A\varphi\|^{2}+
\big(\frac{\beta}{\beta+e^{\frac{-T}{\alpha}}}\big)^{2}\|\varphi\|^{2}.
\end{aligned}\label{19}
\end{equation}
If we choose  $\alpha =\frac{T}{(1-r)\ln(\frac{1}{\beta})}$, $0 < r<1$,
then \eqref{19} becomes
\[
\|u_{\sigma}(T)-\varphi\|^{2}\leq \beta^{2r}\|\varphi\|^{2}+
\frac{4T}{(1-r)\ln
(\frac{1}{\beta})}\|A\varphi\|^{2}.
\]%\label{20}
\end{proof}

\begin{theorem}\label{thm8}
Assuming that $\varphi\in \mathcal{D}(A)$ and $\gamma \geq1$, then
$\|u_{\sigma}(T)-\varphi\|^{2}$ can be estimated as follows
\[
\|u_{\sigma}(T)-\varphi\|^{2}\leq
2\Big(\big(\frac{T}{1+\ln(\frac{\gamma
T}{\beta})}\big)^{2}+T\alpha\Big)
\|A\varphi\|^{2}.
\]%\label{21}
\end{theorem}

\begin{proof} We have
\[
\|u_{\sigma}(T)-\varphi\|^{2}
 =  \int_{\gamma}^{+\infty} H_{\sigma}(\lambda)^{2}\,d
\|E_{\lambda}\varphi\|^{2}
 \leq 2(I_{1,\sigma}+I_{2,\sigma}),
\] %\label{22}
where
\begin{gather*}
I_{1,\sigma} =\int_{\gamma}^{+\infty}
F_{\sigma}(\lambda)^{2}\lambda^{-2}\lambda^{2}\,d\|E_{\lambda}\varphi\|^{2},\\
I_{2,\sigma} =\int_{\gamma}^{+\infty}
G_{\sigma}(\lambda)^{2}\,d\|E_{\lambda}\varphi\|^{2}.
\end{gather*}
 Using \eqref{e8} and \eqref{e9}, we obtain
\begin{gather}
I_{1,\sigma}\leq \Big(\sup_{\lambda\geq\gamma}F_{\sigma}(\lambda)
\lambda^{-1}\Big)^{2} \|A\varphi\|^{2}\leq
\Big(\frac{T}{1+\ln(\frac{\gamma
T}{\beta})}\Big)^{2}\|A\varphi\|^{2},\label{23} \\
I_{2,\sigma}\leq T\alpha \|A\varphi\|^{2}.\label{24}
\end{gather}
Combining \eqref{23} and \eqref{24} we obtain the desired
estimate.
\end{proof}

\begin{theorem}\label{thm9}
Assume that
$\|Au(0)\|^{2}=\int_{\gamma}^{+\infty}
\lambda^{2}e^{2T\lambda}\,d\|E_{\lambda}\varphi\|^{2} < \infty$,
i.e., $u(0)\in \mathcal{D}(A)$, and that
$\gamma \geq 1$.
  Then $\|u_{\sigma}(0)-u(0)\|^{2}$ can be
estimated as follows
\[
\|u_{\sigma}(0)-u(0)\|^{2}\leq
2\big(\big(\frac{T}{1+\ln(\frac{\gamma T}{\beta})}\big)^{2}+T\alpha\big)
\|Au(0)\|^{2}.
\]%\label{25}
\end{theorem}

\begin{proof} By a computation,
\[
\|u_{\sigma}(0)-u(0)\|^{2}
 =  \int_{\gamma}^{+\infty}
  H_{\sigma}(\lambda)^{2}e^{2T\lambda}\,d\|E_{\lambda}\varphi\|^{2} \\
\leq  2(I_{1,\sigma}+I_{2,\sigma}),\\
\] %\label{26}
where
\begin{gather*}
I_{1,\sigma} =\int_{\gamma}^{+\infty}
F_{\sigma}(\lambda)^{2}\lambda^{-2}\lambda^{2}e^{2T\lambda}
\,d\|E_{\lambda}\varphi\|^{2},\\
I_{2,\sigma} =\int_{\gamma}^{+\infty}
G_{\sigma}(\lambda)^{2}e^{2T\lambda}\,d\|E_{\lambda}\varphi\|^{2}.
\end{gather*}
 Using \eqref{e8} and \eqref{e9} we obtain
\begin{gather*}
I_{1,\sigma}\leq \Big(\sup_{\lambda\geq\gamma}F_{\sigma}(\lambda)\Big)^{2}
\|Au(0)\|^{2}\leq \Big(\frac{T}{1+\ln(\frac{\gamma
T}{\beta})}\Big)^{2}\|Au(0)\|^{2},\\
I_{2,\sigma}\leq T\alpha \|Au(0)\|^{2}.
\end{gather*}
Combining the two inequalities above,  we obtain the desired
estimate.
\end{proof}


We conclude this paper by  constructing a family of
regularizing operators to \eqref{FVP}.

\begin{definition}\label{def4.2} \rm
A family $\{R_{\sigma}(t),\; \sigma >0,\; t\in[0,T]\} \subset
\mathcal{L}(H)$ is called a family of regularizing operators for
the problem \eqref{FVP} if for each solution $u(t)$, $0\leq t\leq T$
of \eqref{FVP} with final element $\varphi$, and for any $\eta >0$,
there exists $\sigma(\eta)>0$, such that
\begin{gather}
\sigma(\eta)\to 0,\quad \eta \to 0,\label{R1}\\
\|R_{\sigma(\eta)}(t)\varphi_{\eta}-u(t)\|\to 0,\quad
\eta \to 0, \label{R2}
\end{gather}
for each $t\in[0,T]$ provided that $\varphi_{\eta}$ satisfies
$\|\varphi_{\eta}-\varphi\|\leq \eta$.
\end{definition}

 Define $R_{\sigma}(t)=
S(t)\big(\beta+S_{\alpha}(T)\big)^{-1}$,
$t \geq 0$, $\sigma > 0$; it is clear that
$R_{\sigma}(t)\in \mathcal{L}(H)$.  In the following we will show that
$R_{\sigma}(t)$ is a family of regularizing operators for \eqref{FVP}.

\begin{theorem}\label{thm7}
Under the assumption $\varphi\in \mathcal{C}_{1}(A)$, the
condition \eqref{R2} holds.
\end{theorem}

\begin{proof} We have
\[\Delta_{\sigma}(t)=\|R_{\sigma}(t)\varphi_{\eta}-u(t)\|
\leq \|R_{\sigma}(t)(\varphi_{\eta}-\varphi)\|+
\|R_{\sigma}(t)\varphi-u(t)\| =
\Delta_1(t)+\Delta_2(t),
\]%\label{29}
 where
\begin{gather*}
\Delta_1(t) = \|R_{\sigma}(t)(\varphi_{\eta}-\varphi)\|\leq
\Big(\frac{1}{\beta+e^{\frac{-T}{\alpha}}}\Big)\eta,
\\ %\label{30}
\Delta_2(t) = \|R_{\sigma}(t)f-u(t)\|.
\end{gather*} %\label{31}
We observe that
$$
\Delta_1(t)\leq\frac{\eta}{\beta},\quad
\Delta_1(t)\leq \eta e^{\frac{T}{\alpha}}.
$$
Choose $\beta =\sqrt{\eta}$ and
$\alpha =\frac{2T}{\ln(\frac{1}{\eta})}$,
then
$\sigma(\eta)=(\alpha(\eta),\beta(\eta))\to (0,0)$,
$\eta\to 0$, and
\begin{equation}
\Delta_1(t) \leq \sqrt{\eta}
\to 0,\quad \text{as } \eta \to 0.\label{32}
\end{equation}
 Now, by Theorem \ref{thm5} we have
\begin{equation}
\Delta_2(t)= \|u_{\sigma(\eta)}(t)-u(t)\|
\to 0, \quad \text{as } \eta \to
0,\label{33}
\end{equation}
uniformly in $t$. Combining \eqref{32} and \eqref{33} we
obtain
\[
\sup_{0\leq t \leq T}\|R_{\sigma}(t)\varphi_{\eta}-\varphi\|
\to 0, \;\text{as}\;\eta \to 0.
\]%\label{34}
This shows that $R_{\sigma}(t)$ is a family of regularizing
operators for \eqref{FVP}.
\end{proof}


\subsection*{Concluding remarks}
\textbf{1.} Note that the error factor $e(\sigma)$
introduced by small changes in the final value $\varphi$ is of
order $\frac{1}{\beta+e^{\frac{-T}{\alpha}}}$.

\noindent\textbf{2.} When $\alpha
=\frac{T}{(1-r)\ln(\frac{1}{\beta})}$, $0 < r<1$,
then
$$
e(\sigma)=e(\beta)=\frac{1}{\beta+\beta^{1-r}}\leq
(\frac{1}{\beta})^{1-r}.
$$
\textbf{3.} In \cite{clark} (resp.
\cite{Gajewski,lattes}) the error factor $e(\beta)$ (resp.
$e(\alpha)$ is of order $ \frac{1}{\beta}$ (resp.
$ e^{\frac{T}{\alpha}}$).

 Observe that
\[
\frac{1}{\beta+e^{\frac{-T}{\alpha}}}\leq
 \frac{1}{\beta}, \quad
\frac{1}{\beta+e^{\frac{-T}{\alpha}}}\leq
 e^{\frac{T}{\alpha}}.
\]
This shows that our approach has a nice regularizing
effect and gives a better approximation with comparison to the
methods developed in \cite{clark,Gajewski,lattes}.

In this study we have achieved a better results than those
established in \cite{clark,Gajewski,lattes}. The error resulting
from approximation and the rate of convergence of the method are
optimal.


\subsection*{Acknowledgments}
The authors give their cordial thanks to the anonymous referees
for their valuable comments and suggestions which improved the
quality of the paper.

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\end{thebibliography}
\end{document}