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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2009(2009), No. 77, pp. 1--13.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2009 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2009/77\hfil A truncated Fourier series method]
{Remarks on a 2-D nonlinear backward heat problem
using a truncated Fourier series method}

\author[D. D. Trong, N. H. Tuan\hfil EJDE-2009/77\hfilneg]
{Dang Duc Trong, Nguyen Huy Tuan}  % in alphabetical order

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

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

\thanks{Submitted December 11, 2008. Published June 16, 2009.}
\thanks{Supported by the Council for Natural Sciences of Vietnam}
\subjclass[2000]{35K05, 35K99, 47J06, 47H10}
\keywords{Backward heat problem; nonlinearly ill-posed problem;
 \hfill\break\indent Fourier series; contraction principle}

\begin{abstract}
 The inverse conduction problem arises when experimental measurements
 are taken in the interior of a body, and it is desired to calculate
 temperature  on the surface. We consider the problem of finding,
 from the final data $u(x,y,T)=\varphi(x,y)$, the initial data
 $u(x,y,0)$ of the temperature function
 $u(x,y,t)$, $(x,y) \in U\equiv (0,\pi)\times (0,\pi)$,
 $t\in [0,T]$ satisfying the nonlinear system
 \begin{gather*}
 u_t-\Delta u= f(x,y,t,u(x,y, t)),\quad (x,y,t)\in U\times (0,T),\\
 u(0,y,t)= u(\pi,y,t)= u(x,0,t) = u(x,\pi,t) = 0,\quad
 (x,y,t) \in U\times(0,T).
 \end{gather*}
 This problem is known to be ill-posed, as the solution exhibits
 unstable dependence on the given data functions.
 Using the Fourier series method, we regularize the problem and
 to get some new error estimates.
 A numerical experiment is given.
\end{abstract}

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


\section{Introduction}

In this paper, we consider the following two dimensional problem in
an rectangle  $U=(0,\pi) \times (0,\pi) $
\begin{gather}
u_t-\Delta u =g(x,y,t,u(x,y,t))\quad (x,y,t)\in U\times(0,T),\;
 U = (0,\pi) \times (0,\pi) \label{eq1}\\
u(0,y,t)=u(\pi,y,t) = u(x,0,t)= u(x,\pi,t)= 0\quad
 (x,y,t) \in U \times [0,T]\label{eq2}\\
u(x,y,T)=\varphi(x,y)\quad x,y \in U.\label{eq3}
\end{gather}
where  we want to determine the temperature distribution
$u(.,.,t)$ for $0\le t<T$ from the data $\varphi(x,y)$ . The
problem is called the backward heat problem (BHP), the backward
Cauchy problem or the final value problem. This is a typical
ill-posed problem. In general no solution which satisfies the heat
conduction equation with final data and the boundary conditions
exists. Even if the solution exists, it will not be continuously
dependent on the final data such that the numerical simulations
are very difficult and some special regularization methods are
required. In the context of approximation method for this problem,
many approaches have been investigated. In the mathematical
literature various methods have been proposed for solving backward
Cauchy problems. Such authors  as Lattes and Lions,  Miller have
approximated BHP by quasi-reversibility methods (QR method for
short). In 1983, Showalter, in \cite{showalter1}, presented a
different method called the quasiboundary value (QBV method)
method to regularize that linear homogeneous problem which gave a
stability estimate better than the one of disscused method. The
main ideas of the method is of adding an appropriate
 ``corrector" to the final data. Using the method,
Clark and Oppenheimer, in  \cite{clark}, and Denche-Bessila, very
recently in \cite{Denche}, regularized the backward problem by
replacing the final condition by $ u(T)+\epsilon u(0)=g $ and
$u(T)-\epsilon u'(0)=g $ respectively.

Although there are many papers on the linear homogeneous case of
the backward problem, we only find a few papers on the
nonhomogeneous and nonlinear  cases of BHP. We can notably mention
the method of QBV and modified quasi-reversibility to solve the
one dimensional nonlinear backward heat problem (NBHP), such as
\cite{Trong2}.  Moreover, the two dimensional case of NBHP  is
very scarce and it is  not  considered   by   QR or QBV  methods.
To the authors's knowledge, in some  recent papers on the
nonlinear backward heat, the error estimates of most
regularization methods are the form $C \epsilon^{t/T}$. It
makes difficult  to solve  the error in the time $t=0$. To improve
this, we  develop a new regularization method which is called
Fourier method for solving the Problem \eqref{eq1}-\eqref{eq3}. As
far as we know, there are not any results of Fourier series method
for treating NBHP until now. Meanwhile, we will establish  faster
convergence results via improved error estimates.  Especially, the
convergence of the approximate solution at $t=0$ is also proved.
This is an improvement of known results in \cite{QuanPham, Trong2,
Trong4}.

Informally, problem \eqref{eq1}-\eqref{eq3} can be transformed to
an integral equation
\begin{equation}
\begin{aligned}
u(x,y,t)
&=\sum_{n,m=1}^\infty\Big(e^{-(t-T) (n^2+m^2)}\varphi_{nm}\\
&\quad -\int_t^T e^{-(t-s)
(n^2+m^2)}g_{nm}(u)(s)ds\Big)\sin (nx) \sin (my),
\end{aligned}\label{eeee1}
\end{equation}
where
\begin{gather*}
\varphi(x,y)=\sum_{n,m=1}^\infty \varphi_{nm} \sin (nx)\sin (my),
\\
g(u)(x,t)=\sum_{n,m=1}^\infty g_{nm}(u)(t)\sin (nx) \sin(my),
\end{gather*}
are the expansion of $\varphi$ and $g(u)$, respectively. Since $t<T$,
we know from  \eqref{eeee1} that, when $m^2+n^2$ becomes large,
$\exp \{(T-t)(m^2+n^2)\}$ increases rather quickly. Thus, the
term $ e^{-(t-T) (m^2+n^2)}$ is the cause of the
instability. So, we hope to recover the stability of
problem \eqref{eeee1} by filtering the high frequencies with suitable
method. The essence of our regularization method is just to eliminate
all high frequencies from the solution, and instead consider
\eqref{eeee1} only for $m^2+n^2 \le a_\beta$, where $a_\beta $
is an appropriate positive constant  satisfying $  \lim  _{\beta \to 0} a_\beta=\infty  $. We note that $a_\beta$ is a constant which will be selected appropriately as regularization parameter.
Then, we get a stable and convergent iteration scheme.
We have the following approximation problem
\begin{equation}
\begin{aligned}
v^{\beta,a_\beta }(x,y,t)
&=\sum^{m,n \geq 1}_ {m^2  + n^2  \leq a_\beta  }
\Big( {e^{(T-t) (n^2 + m^2)}} \varphi_{nm}\\
&\quad -\int_t^T {e^{(s-t) (n^2+ m^2)}} g_{nm}(v^{\beta, a_\beta})(s)ds\Big)
\sin(nx)\sin(my)
\end{aligned}\label{eeee2}
\end{equation}
where
\begin{gather*}
\varphi_{nm}=\frac{4}{\pi^2}\langle \varphi(x,y), \sin(nx) \sin(my)\rangle,\\
g_{nm}(u)(t)=\frac{4}{\pi^2}\langle g(x,y,t,u(x,y,t)),\sin(nx)
\sin(my)\rangle,
\end{gather*}
 and $\langle\cdot,\cdot\rangle$ is inner product in $L^2(U)$.

\section{Fourier regularization and the main results}

For clarity, we denote the solution of  \eqref{eq1}-\eqref{eq3} by
$u(.,.,t)$, and the solution of  \eqref{eeee2} by
$v^{\beta,a_\beta }(.,.,t)$.
The main conclusion of this article is as follows.

\begin{theorem} \label{thm1}
Let $\varphi \in L^2(U)$ and let
$g \in L^\infty(\overline U \times [0,T]\times R)$ satisfy
\[
|g(x,y,t,w) - g(x,y,t,u)| \leq k|w - u|
\]
for a $k > 0$ independent of $x,y,t,w,u$.
Then  \eqref{eeee2} has a unique solution
$v^{\beta,a_\beta } \in C([0,T];H_0^1(U)) \cap C^1((0,T);L^2(U))$.
\end{theorem}

\begin{theorem} \label{thm2}
The solution of  \eqref{eeee2} depends continuously on $\varphi$
in $L^2(U)$.
\end{theorem}


\begin{theorem} \label{thm3}
Let $\varphi, g$ be as in Theorem \ref{thm1}.
If $\frac{\partial g}{\partial z} (x,y,t,z)$ is bounded on
$U\times(0,T)\times R$ then   \eqref{eq1}-\eqref{eq3} has at most
one solution
\[
u \in C([0,T];H_0^1(U))\cap C^1((0,T);L^2(U)).
\]
\end{theorem}

\begin{theorem} \label{thm4}
 Let $\varphi, g$ be as in Theorem \ref{thm1}. Suppose that
 \eqref{eq1}-\eqref{eq3} has a unique solution $u(x,y,t)$ in
$C([0,T];H_0^1(U))\cap C^1((0,T);L^2(U))$ which satisfies
 \begin{equation}
\int_0^T \sum_{n,m=1}^\infty e^{2s (n^2+m^2)}g_{nm}^2(u)(s)ds<\infty.
\label{eeee3}
\end{equation}
Then
\begin{equation} \label{eq7}
  \norm{ u(.,.,t)- v^{\beta, a_\beta}(.,.,t)}
\le \sqrt{M}e^{k^2T(T-t)}  e^{-ta_\beta}
\end{equation}
for every $t\in[0,T]$, where
\[
M = 4\|u(0)\|^2 + \pi^2 T \int_0^T \sum_{n,m = 1}^{\infty}
e^{2s(n^2+m^2)}g_{nm}^2 (u)(s) ds,
\]
and $v^{\beta, a_\beta}$ is the unique solution of \eqref{eeee2}
 corresponding to $ \beta$.
Moreover, if $\frac{\partial u}{\partial t} \in L^2((0,T);L^2(U))$,
then there exists a $t_{\beta}$ such that
\[
\norm{u(.,.,0)- v^{\beta, a_\beta}(.,.,t_\beta)}
\le \sqrt{2}C \sqrt[4]{1/a_\beta},
\]
where
\[
N = \Big(\int _0^T \|\frac{\partial u}{\partial s}(.,.,s)\|^2 ds
\Big)^{1/2},
\quad
C = \max\{M,N\}.
\]
\end{theorem}

\begin{remark} \label{rmk1} \rm
(1) In the simple case of the function  $g(.,.,u)=0$ (it follows that k=0), we have
\[
 \norm{ u(.,.,t)- v^{\beta,a_\beta}(.,.,t)}\le   2 \|u(.,.,0)\|
  e^{-ta_\beta} .
\]
Choosing $a_\beta= \frac{1}{T}\ln(1/\beta)$,
we obtain the error estimate
\begin{eqnarray*}
 \norm{ u(.,.,t)- v^{\beta,a_\beta}(.,.,t)}\le   2 \|u(.,.,0)\|.
  \beta^{t/T}
\end{eqnarray*}
This error is given  in \cite{clark}.

(2)
In most known results, such as \cite{QuanPham,Trong2,Trong4},
 the errors between the exact
solution and approximate solution  can be calculated in the
form $ C \epsilon^{t/T}$.
Notice that the convergence estimate in this Theorem  does not give
any useful information on the continuous dependence of the solution
at $t=0$.  It is easy to see that if taking  $t=0$ in \eqref{eq7},
the error estimate is  as follows
$$
 \norm{ u(.,.,0)- v^{\beta,a_\beta}(.,.,0)}\le \sqrt{M}e^{k^2T^2}
$$
does not tend to zero when $\beta \to 0$.
So, the convergence of the approximate is large when $t \to 0$.
In next Theorem, we  will give a good estimate in  which the error
in the case $t=0$ is considered.

(3) In this  Theorem,  we  ask for a condition on the expansion
coefficient $g_{nm}$ in \eqref{eeee3}. We note that the solution $u$
depends on  the nonlinear term $g$ and therefore $g_{nm}, g_{nm}(u)$
is very difficult to be valued. Such a obscurity makes this Theorem
hard to be used for numerical computations. Hence, we ask  the
condition as follows
\begin{equation}
 \sum_{n,m=1}^\infty e^{2t(n^2+m^2)}|<u(.,.,t),\sin(nx)\sin(my)>|^2
 <\infty.\label{eeee7}
\end{equation}
 In this case, we only require the assumption of  $u$, not
need to compute the function  $g_{nm}(u)$. In the homogeneous case
of problem \eqref{eq1}-\eqref{eq3},i.e., $g=0$, then the right hand
side of \eqref{eeee7} is equal to $\|u(.,.,0)\|^2$. Hence, the
condition  \eqref{eeee7} is natural and acceptable.
\end{remark}

\begin{theorem} \label{thm5}
 Let $\varphi, g$ be as in  Theorem \ref{thm1}. Suppose
\eqref{eq1}-\eqref{eq3} has a unique solution $u(x,y,t)$
satisfying \eqref{eeee7}. Then we  have
 \begin{equation}
  \norm{ u(.,.,t)- v^{\beta,a_\beta}(.,.,t)}\le Q(\beta,t, u)
  e^{-ta_\beta}\label{eeee8}
\end{equation}
 for every $t\in[0,T]$, where
\begin{equation}
Q(\beta,t, u)=
\Big(2k^2T e^{2k^2T(T-t)}\int_0^T
P(\beta,s,u)ds+\frac{\pi^2}{2}P(\beta,t, u) \Big)^{1/2}
\end{equation}
and
\begin{equation}
\begin{aligned}
P(\beta,t, u)
&=\sum_ {m,n \geq 1,m^2  + n^2  \geq a_\beta  } \Big(e^{T (n^2+m^2)}
\varphi_{nm}-\int_t^T e^{s (n^2+m^2)}g_{nm}(u)(s)ds\Big)^2  \\
&=\sum_ {m,n \geq 1,m^2  + n^2  \geq a_\beta  } e^{2t(n^2+m^2)}
u^2_{nm}
\end{aligned}\label{eeee9}
\end{equation}
and $v^{\beta,a_\beta}$ is the unique solution of Problem \eqref{eeee2}.
\end{theorem}


\begin{remark} \label{rmk2}\rm
If we let $ t=0$ in  \eqref{eeee8},  we get the error at the original
time,
\begin{equation}
 \norm{ u(.,.,0)- v^{\beta,a_\beta }(.,.,0)}^2
\le 2k^2T e^{2k^2T^2}\int_0^T  P(\beta,s,u)ds+ \frac{\pi^2}{2}P(\beta,0, u) . \label{333}
\end{equation}
Noting that the right hand side of \eqref{333} tends to zero when
$\beta \to 0$.
\end{remark}

For  non-exact data, we have the following result.

\begin{theorem} \label{thm6}
Let $\varphi, g$ be as in Theorem \ref{thm1}. Assume that the exact solution
$u$ of \eqref{eq1}-\eqref{eq3} corresponding to $\varphi$  be defined
as in Theorem \ref{thm4}.
Let $\varphi_\beta \in L^2(U)$ be a measured data such that
$$
\norm{\varphi_\beta-\varphi}\le\beta.
$$
Suppose the problem \eqref{eq1}-\eqref{eq3} has a unique solution
$ u \in C([0,T];H_0^1(U))\cap C^1((0,T);L^2(U))$.
Let us select
$a_\beta= \ln \big((\frac{1}{\beta})^{1/T} (\ln \frac{1}{\beta})
^{-\alpha/(2T)}\big)$.

(i) If $u$ satisfies \eqref{eeee3} then for $t \in (0,T)$, there
exists a function $v^{\beta,a_\beta}$ satisfying
\begin{equation}
\norm{v^{\beta,a_\beta}(.,.,t)-u(.,.,t)}\le(M+1)
e^{k^2T(T-t)} \beta^{t/T} (\ln \frac{1}{\beta})
^{\frac{-\alpha(T-t)}{2T}}\Big( 1+ (\ln \frac{1}{\beta})
^{\frac{\alpha}{2}}\Big), \label{bi111}
\end{equation}
and
\[
\norm{v^{\beta,a_\beta}(.,.,0)-u(.,.,0)} \le
\sqrt[4]{ 1/a_\beta}
 \left(2\exp(k^2 T^2) +\sqrt{2} C \right)
\]
where
\[
M=3\norm{u(.,.,0)}^2+3\pi^2 T\int_0^T \sum_{n,m=1}^\infty
e^{2s (n^2+m^2)}g^2_{nm}(u)(s)ds
\]
and $C$ is defined in Theorem \ref{thm4}.

(ii) If $u$ such that the condition \eqref{eeee7} then for all
$t \in [0,T]$
 \begin{equation}
\begin{aligned}
&\norm{w^{\beta,a_\beta }(.,.,t)-u(.,.,t)} \\
&\le \beta^{t/T} (\ln \frac{1}{\beta})^{\frac{-\alpha(T-t)}{2T}}
\Big( \exp(k^2(T-t)^2)+ Q(\beta,t,u)(\ln \frac{1}{\beta})
^{\frac{\alpha}{2}}\Big),
\end{aligned}\label{bi1111}
\end{equation}
where $w^{\beta,a_\beta }$ be the solution of problem \eqref{eeee2}
corresponding to $\varphi_\beta$.
\end{theorem}


\begin{remark} \label{rmk3}
(1) If we let $\alpha=0$ in \eqref{bi111}, we have the simple error
\begin{equation}
\norm{v^{\beta,a_\beta}(.,.,t)-u(.,.,t)}\le(M+1) e^{k^2T(T-t)}
\beta^{t/T} ,\quad \forall t \in (0,T) \label{bi11111}.
\end{equation}
This error is similar to the one given in \cite{Trong2}. Notice that the
right hand side of \eqref{bi11111} does not converges to 0.
This is disadvantage point of the error \eqref{bi111}.

(2) In the  error \eqref{bi1111}, if we let $t=0$, we get
 \begin{equation}
\norm{w^{\beta,a_\beta }(.,.,0)-u(.,.,0)}
\le \exp(k^2T^2) (\ln \frac{1}{\beta})^{\frac{-\alpha}{2}}
+Q(\beta,0,u) . \label{bi222}
\end{equation}
Notice that if $\alpha >0$ then the right hand side of  \eqref{bi222}
converges to 0 and the Theorem \ref{thm6}(ii)  is a generalization
of the result given in \cite{Trong2}.
\end{remark}

\section{Proof of the main results}

\begin{proof}[Proof of theorem \ref{thm1}]
Put
\begin{align*}
&G(v^{\beta,a_\beta })(x,y,t) \\
&= \Psi(x,y,t) -  \!\sum_ {m,n \geq 1,m^2
+ n^2  \leq a_\beta  }
\Big(\int_t^T {e^{(s-t) (n^2+ m^2)}} g_{nm}(v^{\beta,a_\beta })(s)ds
\Big)\sin(nx)\sin(my)
\end{align*}
where
$$
\Psi (x,y,t) = \sum_ {m,n \geq 1,m^2  + n^2
\leq a_\beta  }  {e^{(T-t) (n^2 + m^2)}} \varphi_{nm}
\sin(nx)\sin(my).
$$
We claim that
\begin{equation}
\begin{aligned}
&\|G^p (v^{\beta,a_\beta })(.,.,t)-G^p(w^{\beta,a_\beta })(.,.,t)\|^2\\
&\leq k^{2p}e^{2Tp a_\beta} \frac{(T-t)^p C^p}{p!}|||v^{\beta,a_\beta }
-w^{\beta,a_\beta }|||^2
\end{aligned}\label{ff1}
\end{equation}
for every $p\geq 1$, where $C = \max \{T,1\}$ and
$|||\cdot |||$ is sup norm in $C([0,T];L^2(U))$.

We shall prove the latter inequality by induction.
For $p =1$, we have
\begin{align*}
&\|G(v^{\beta,a_\beta })(.,.,t)-G(w^{\beta,a_\beta })(.,.,t)\|_2^2 \\
&= \frac{\pi^2}{4}\sum_ {m,n \geq 1,m^2  + n^2  \leq a_\beta  }
\left[ e^{(s-t) (n^2+ m^2)} (g_{nm}(v^{\beta,a_\beta })(s)
 - g_{nm}(w^{\beta,a_\beta })(s))ds \right]^2
\\
&\leq \frac{\pi^2}{4} \sum_ {m,n \geq 1,m^2  + n^2
 \leq a_\beta } \int_t^T (e^{2(s-t) (n^2+ m^2)} ds
 \int_t^T ( g_{nm} (v^{\beta,a_\beta })(s) \\
&\quad - g_{nm} (w^{\beta,a_\beta })(s))^2 ds
\\
&\leq \frac{\pi^2}{4}\sum_ {m,n \geq 1,m^2  + n^2  \leq a_\beta  }
 e^{2Ta_\beta} (T - t) \int_t^T ( g_{nm} (v^{\beta,a_\beta })(s)
 - g_{nm} (w^{\beta,a_\beta })(s))^2 ds
\\
&= \frac{\pi^2}{4}e^{2Ta_\beta} (T - t)\int_t^T \sum_ {m,n \geq 1,m^2
 + n^2  \leq a_\beta }( g_{nm} (v^{\beta,a_\beta })(s)
 - g_{nm} (w^{\beta,a_\beta })(s))^2 ds
\\
&\le e^{2Ta_\beta} (T - t)\int_t^T  \int_0^\pi
 \int_0^\pi (g(x,y,s,v^{\beta,a_\beta }(x,y,s))\\
&\quad - g(x,y,s,w^{\beta,a_\beta }(x,y,s)))^2 \,dx\,dy\,ds
\\
&\leq k^2 e^{2Ta_\beta} (T - t) \int _t^T \int_0^\pi
 \int_0^\pi |v^{\beta,a_\beta }(x,y,s) - w^{\beta,a_\beta }(x,y,s)|^2
 \,dx\,dy\,ds
\\
&\leq C k^2 e^{2Ta_\beta} (T - t) |||v^{\beta,a_\beta }
 -w^{\beta,a_\beta }|||^2.
\end{align*}
Thus \eqref{ff1} holds.
Suppose that \eqref{ff1}  holds for $p =j$. We prove that \eqref{ff1}
holds for $p = j + 1$. We have
\begin{align*}
&\|G^{j+1}(v^{\beta,a_\beta })(.,.,t) - G^{j+1}(w^{\beta,a_\beta })(.,.,t)
\|^2
\\
&= \frac{\pi^2}{4} \sum_ {m,n \geq 1,m^2  + n^2  \leq a_\beta  }
\Big[\int_t^T (e^{2(s-t) (n^2+ m^2)} ds
 \Big(g_{nm}(G^j(v^{\beta,a_\beta }))(s) \\
&\quad - g_{nm}(G^j
 (w^{\beta,a_\beta }))(s)\Big)ds \Big]^2
\\
&\leq \frac{\pi^2}{4} e^{2Ta_\beta}  \sum_ {m,n \geq 1,m^2  + n^2
 \leq a_\beta  }   \Big[\int_t^T |g_{nm}(G^j(v^{\beta,a_\beta }))(s)
- g_{nm}(G^j (w^{\beta,a_\beta }))(s)|ds \Big]^2
\\
&\leq \frac{\pi^2}{4}  e^{2Ta_\beta} (T - t)\int_t^T
 \sum_{n,m = 1}^\infty |g_{nm}(G^j(v^{\beta,a_\beta }))(s)
 - g_{nm}(G^j (w^{\beta,a_\beta }))(s)|^2 ds
\\
&\leq  e^{2Ta_\beta} (T - t)\int_t^T \|g(.,.,s,G^j(v^{\beta,a_\beta })
 (.,.,s)) - g(.,.,s,G^j (w^{\beta,a_\beta })(.,.,s))\|^2 ds
\\
&\leq  e^{2Ta_\beta} (T - t)k^2 \int_t^T \|G^j(v^{\beta,a_\beta })
 (.,.,s) - G^j (w^{\beta,a_\beta })(.,.,s)\|^2 ds
\\
&\leq  e^{2Ta_\beta} (T - t)k^2 k^{2j}e^{2Tja_\beta}
 \int_t^T \frac{(T-s)^j}{j!} ds C^j |||v^{\beta,a_\beta }
 -w^{\beta,a_\beta } |||^2
\\
&\leq k^{2(j+1)} e^{2T(j+1)a_\beta}\frac{(T-t)^{j+1}}{(j+1)!}
 C^{j+1} |||v^{\beta,a_\beta }-w^{\beta,a_\beta } |||^2.
\end{align*}
Therefore,
\[
\|G^p (v^{\beta,a_\beta })(.,.,t)-G^p(w^{\beta,a_\beta })(.,.,t)\|^2
\leq k^{2p}e^{2Tp a_\beta} \frac{(T-t)^p C^p}{p!}|||v^{\beta,a_\beta }
-w^{\beta,a_\beta }|||^2
\]
for all $v^{\beta,a_\beta },w^{\beta,a_\beta } \in C([0,T];L^2(U))$.
We consider
$$
G: C([0,T];L^2(U))\to C([0,T];L^2(U)).
$$
Since $\lim  _{p \to \infty} k^{p}e^{Tp a_\beta}
\frac{T^{p/2}C^p}{\sqrt{p!}}=0$, there exists a positive integer
number $p_0$, such that $G^{p_0}$ is a contraction.
It follows that the equation $G^{p_0} (u) = u$ has a unique solution
$v^{\beta,a_\beta } \in C([0,T];L^2(U))$.
We claim that $G(v^{\beta,a_\beta })=v^{\beta,a_\beta }$.
In fact, one has
\[
G(G^{p_0}(v^{\beta,a_\beta }))= G(v^{\beta,a_\beta }).
\]
Hence
\[
G^{p_0}(G(v^{\beta,a_\beta }))= G(v^{\beta,a_\beta }).
\]
By the uniqueness of the fixed point of $G^{p_0}$, one has
$G(v^{\beta,a_\beta }) = v^{\beta,a_\beta }$, i.e., the equation
$G(v^{\beta,a_\beta }) =v^{\beta,a_\beta }$ has a unique solution
$v^{\beta,a_\beta }\in C([0,T];L^2(U))$.
\end{proof}

\begin{proof}[Proof of Theorem \ref{thm2}]
Let $u$ and $v$ be two solutions of \eqref{eeee2} corresponding to
the values $\varphi$ and $\omega$.
We have
\begin{align*}
 &\norm{u(.,.,t)-v(.,.,t)}^2
\\
&=\frac{\pi^2}{4} \sum_ {m,n \geq 1,m^2  + n^2  \leq a_\beta  }
\Big| e^{(T-t) (n^2+m^2)}(\varphi_{nm}-\omega_{nm})\\
&\quad -\int_t^T e^{(s-t) (n^2+m^2)} (g_{nm}(u)(s)-g_{nm}(v)(s)ds)\Big|^2
\\
&\le\frac{\pi^2}{2}  \sum_ {m,n \geq 1,m^2  + n^2  \leq a_\beta }
( e^{(T-t) (n^2+m^2)}|\varphi_{nm}-\omega_{nm}|)^2
\\
&\quad +\frac{\pi^2}{2} \sum_ {m,n \geq 1,m^2  + n^2  \leq a_\beta  }
\Big(\int_t^T e^{(s-t) (n^2+m^2)}  |g_{nm}(u)(s)-g_{nm}(v)(s)|ds\Big)^2
\end{align*}
Then, we obtain
\begin{align*}
&\|u(.,.,t) - v(.,.,t)\|^2 \leq\\
&\le 2 e^{2(T-t)a_\beta}\norm{\varphi-\omega}^2
 +2 k^2(T-t)e^{-2t a_\beta}  \int_t^Te^{2s a_\beta}
\norm{u(.,.,s)-v(.,.,s)}^2ds.
\end{align*}
Hence
\begin{align*}
&e^{2t a_\beta}  \norm{u(.,.,t)-v(.,.,t)}^2 \\
&\le e^{2T a_\beta} \norm{\varphi-\omega}^2+ 2k^2(T-t)
\int_t^Te^{2s a_\beta}  \norm{u(.,.,s)-v(.,.,s)}^2ds.
\end{align*}
Using Gronwall's inequality we have
\[
\norm{u(.,.,t)-v(.,.,t)}\le e^{(T-t) a_\beta}\exp(k^2(T-t)^2)
\norm{\varphi-\omega}.
\]
This completes the proof of  the theorem.
\end{proof}

\begin{proof}[Proof of Theorem \ref{thm3}]
Let $M>0$ be such that
$$
|\frac{\partial g}{\partial z}(x,y,t,z)|\le M
$$
for all $(x,y,t,z)\in U\times (0,T)\times R$.
Let $u_1(x,y,t)$ and $u_2(x,y,t)$ be two solutions of Problem
\eqref{eq1}-\eqref{eq3} such that
$u_1, u_2 \in C([0,T];H_0^1(U))\cap C^1((0,T);L^2(U))$.
Put $w(x,y,t)=u_1(x,y,t)-u_2(x,y,t)$.
Then $w$ satisfies the equation
\[
w_t(x,y,t)-\Delta w(x,y,t)=g(x,y,t,u_1(x,y,t))-g(x,y,t,u_2(x,y,t)).
\]
Thus
\[
w_t(x,y,t)-\Delta w(x,y,t)=\frac{\partial g}{\partial z}
(x,y,t,\overline{u}(x,y,t))w(x,y,t),
\]
for some $\overline{u}(x,y,t)$.
It follows that
$$
(w_t-\Delta w)^2\le M^2w^2.
$$
Now $w(0,y,t)=w(\pi,y,t)= w(x,0,t) = w(x,\pi,t) = 0$  and
$w(x,y,T)=0$. Hence by the Lees-Protter theorem \cite[p. 373]{ewing},
$w = 0$
which gives $u_1(x,y,t)=u_2(x,y,t)$ for all $t\in[0,T]$.
The proof is completed.
\end{proof}


\begin{proof}[Proof of Theorem \ref{thm4}]
The functions $u(.,.,t)$ can be
written in the form
\begin{align*}
u(x,y,t)
&=\sum_{n,m=1}^\infty (e^{-(t-T) (n^2+m^2)}\varphi_{nm}\\
&\quad -\int_t^T e^{-(t-s) (n^2+m^2)}g_{nm}(u)(s)ds)\sin(nx) \sin(my),
\end{align*}
and  $v^{\beta,a_\beta }(.,.,t)$ in the form
\begin{align*}
v^{\beta,a_\beta }(x,y,t)
&=\sum_ {m,n \geq 1,m^2  + n^2  \leq a_\beta }
\Big( {e^{(T-t) (n^2 + m^2)}} \varphi_{nm}\\
&\quad -\int_t^T {e^{(s-t) (n^2+ m^2)}} g_{nm}(v^{\beta, a_\beta})(s)ds\Big)
\sin(nx)\sin(my)
\end{align*}
Hence
\begin{align*}
v^{\beta,a_\beta}(x,y,t)-u(x,y,t)
&=\sum_ {m,n \geq 1,m^2  + n^2  \geq a_\beta  }
 (e^{-(t-T) (n^2+m^2)}\varphi_{nm}\\
&\quad -\int_t^T e^{-(t-s) (n^2+m^2)}
 g_{nm}(u)(s)ds)\sin(nx) \sin(my)\\
&\quad+\sum_ {m,n \geq 1,m^2  + n^2  \leq a_\beta  } \int_t^T
 \Big({e^{(s-t) (n^2+ m^2)}} (g_{nm}(v^{\beta, a_\beta})(s)\\
&\quad -g_{nm}(v)(s))ds\Big) \sin(nx)\sin(my)
\end{align*}
Using the inequality $(a+b)^2 \leq 2(a^2+b^2)$, we obtain
\begin{align*}
&\norm{ u(.,.,t)- v^{\beta,a_\beta }(.,.,t)}^2 \\
&\le\frac{\pi^2}{2}  \sum_ {m,n \geq 1,m^2  + n^2  \geq a_\beta }
\Big(e^{-(t-T) (n^2+m^2)}\varphi_{nm}-\int_t^T e^{-(t-s) (n^2+m^2)}
g_{nm}(u)(s)ds\Big)^2 \\
&\quad+ \frac{\pi^2}{2}\sum_ {m,n \geq 1,m^2  + n^2  \leq a_\beta  }
\Big(\int_t^T {e^{(s-t) (n^2+ m^2)}} |g_{nm}(u)(s)-g_{nm}
 ( v^{\beta,a_\beta })(s)|ds\Big)^2\\
&\le   2\frac{\pi^2}{2}  e^{-2t(n^2+m^2)}\sum_ {m,n \geq 1,m^2
 + n^2  \geq a_\epsilon  }
\Big(e^{T (n^2+m^2)}\varphi_{nm}-\int_0^T e^{s (n^2+m^2)}
 g_{nm}(u)(s)ds\Big)^2 \\
&\quad +2 \frac{\pi^2}{2} e^{-2t(n^2+m^2)}  \sum_ {m,n \geq 1,m^2
+ n^2  \geq a_\beta  }
 \Big(\int_0^t e^{s (n^2+m^2)}g_{nm}(u)(s)ds\Big)^2\\
&\quad + \frac{\pi^2}{2}(T-t)\int_t^T  \sum_{n,m=1}^\infty
 e^{2(s-t)a_\beta}(g_{nm}(u)(s)-g_{nm}(v^{\beta,a_\beta })(s))^2ds\\
&\le  4  e^{-2ta_\beta} \|u(.,.,0)\|^2
 + \pi^2 T e^{-2ta_\beta}\int_0^T e^{2s (n^2+m^2)}g^2_{nm}(u)(s)ds\\
&\quad+ 2(T-t)  e^{-2ta_\beta}  \int_t^T    e^{2sa_\beta}
\norm{g(.,.,s,u(.,.,s))-g(.,.,s,v^{\beta,a_\beta }(.,.,s))}^2ds\\
&\le 4  e^{-2ta_\beta} \|u(.,.,0)\|^2 + \pi^2 T e^{-2ta_\beta}
 \int_0^T e^{2s (n^2+m^2)}g^2_{nm}(u)(s)ds\\
&\quad+  e^{-2ta_\beta} 2k^2T\int_t^T e^{2sa_\beta}\norm{u(.,.,s)
- v^{\beta,a_\beta}(.,.,s)}^2ds .
\end{align*}
Then we obtain
\begin{align*}
& e^{2ta_\beta}  \norm{ u(.,.,t)- v^{\beta,a_\beta }(.,.,t)}^2\\
&\le  e^{-2ta_\beta}M+2k^2T\int_t^Te^{2sa_\beta} \norm{u(.,.,s)
  - v^{\beta,a_\beta }(.,.,s)}^2ds
\end{align*}
Using Gronwall's inequality, we obtain
\[
  e^{2ta_\beta}  \norm{ u(.,.,t)- v^{\beta,a_\beta }(.,.,t)}^2
\le M e^{2k^2T(T-t)}
\]
which implies
  \begin{equation}
 \norm{ u(.,.,t)-v^{\beta,a_\beta }(.,.,t)}\le  \sqrt{M}e^{k^2T(T-t)}
  e^{-ta_\beta} . \label{fff1}
\end{equation}
Then we have the equality
\[
u(x,y,t) - u(x,y,0) = \int_0^t \frac{\partial u}{\partial s} (x,y,s) ds\,.
\]
It follows that
\[
\|u(.,.,0) - u(.,.,t)\|^2 \leq t \int _0^t
\big\|\frac{\partial u}{\partial s}(.,.,s)\big\|^2 ds \leq N^2 t.
\]
Combining this and \eqref{fff1}, we have
\begin{align*}
\norm{u(.,.,0)- v^{\beta,a_\beta }(.,.,t)}
&\le \norm{u(.,.,0)-u(.,.,t)}+\norm{u(.,.,t)- v^{\beta,a_\beta }(.,.,t)}\\
&\le  C(\sqrt{t}+e^{-ta_\beta}).
\end{align*}

For every $\beta$, there exists a $t_\beta$ such that
$\sqrt{t_\beta}=e^{-t_\beta a_\beta}$; i.e.,
$\frac{ln t_\beta}{t_\beta}=-2 a_\beta$.
Using the inequality $\ln t > -\frac{1}{t}$ for every $t > 0$, we
obtain $t_\beta \le 1/(2 \sqrt{a_\beta})$.
Hence,
$$
\norm{u(.,.,0)- v^{\beta,a_\beta }(.,.,t_\beta)}
 \le \sqrt{2}C \sqrt[4]{1/a_\beta} .
$$
This completes the proof of Theorem \ref{thm4}.
\end{proof}

\begin{proof}[Proof of Theorem \ref{thm5}]
We recall that
\begin{equation}
\begin{aligned}
P(\beta,t, u)
&=\sum_ {m,n \geq 1,m^2  + n^2  \geq a_\beta  }
 \Big(e^{T (n^2+m^2)}\varphi_{nm}-\int_t^T e^{s (n^2+m^2)}g_{nm}(u)(s)ds
 \Big)^2 \\
&=\sum_ {m,n \geq 1,m^2  + n^2  \geq a_\beta  } e^{2t(n^2+m^2)}u^2_{nm}.
\end{aligned}\label{eeee9b}
\end{equation}
It is easy to prove that $P(\beta,t, u) \to 0$ when $\beta \to 0$.
As in the proof of Theorem \ref{thm4}, we have
\begin{align*}
&\norm{ u(.,.,t)- v^{\beta,a_\beta }(.,.,t)}^2\\
&\le\frac{\pi^2}{2}  \sum_ {m,n \geq 1,m^2  + n^2  \geq a_\beta  }
 \Big(e^{-(t-T) (n^2+m^2)}\varphi_{nm}-\int_t^T e^{-(t-s) (n^2+m^2)}
  g_{nm}(u)(s)ds\Big)^2 \\
&\quad+\frac{\pi^2}{2}\sum_ {m,n \geq 1,m^2  + n^2  \leq a_\beta  }
\Big(\int_t^T {e^{(s-t) (n^2+ m^2)}} |g_{nm}(u)(s)-g_{nm}
 ( v^{\beta,a_\beta})(s)|ds\Big)^2\\
&\le \frac{\pi^2}{2}  e^{-2ta_\beta}P(\beta,t, u)\\
&\quad +\frac{\pi^2}{2}(T-t)\int_t^T  \sum_{n,m=1}^\infty e^{2(s-t)
 a_\beta}(g_{nm}(u)(s)-g_{nm}( v^{\beta,a_\beta })(s))^2ds\\
&\le \frac{\pi^2}{2}  e^{-2ta_\beta}P(\beta,t, u) +
2(T-t)  e^{-2ta_\beta}  \int_t^T    e^{2sa_\beta}
 \big\|g(.,.,s,u(.,.,s))\\
 &\quad -g(.,.,s,v^{\beta,a_\beta }(.,.,s))\big\|^2ds\\
&\le \frac{\pi^2}{2}  e^{-2ta_\beta}P(\beta,t, u) +
 e^{-2ta_\beta} 2k^2T\int_t^T e^{2sa_\beta}\norm{u(.,.,s)
 - v^{\beta,a_\beta }(.,.,s)}^2ds .
\end{align*}
This implies that
\begin{align*}
 &e^{2ta_\beta}  \norm{ u(.,.,t)- v^{\beta,a_\beta }(.,.,t)}^2\\
&\le \frac{\pi^2}{2}  e^{-2ta_\beta}P(\beta,t, u)+2k^2T
 \int_t^Te^{2sa_\beta} \norm{u(.,.,s)- v^{\beta,a_\beta }(.,.,s)}^2ds
\end{align*}
Applying  Gronwall's inequality, we obtain
\[
  e^{2ta_\beta}  \norm{ u(.,.,t)- v^{\beta,a_\beta }(.,.,t)}^2
\le 2k^2T e^{-2k^2T t}\int_t^T  e^{2k^2T s} P(\beta,s,u)ds
+\frac{\pi^2}{2}P(\beta,t, u).
\]
Finally,
\[
 \norm{ u(.,.,t)- v^{\beta,a_\beta }(.,.,t)}^2
\le \Big(2k^2T e^{2k^2T(T-t)}\int_0^T  P(\beta,s,u)ds
 +\frac{\pi^2}{2} P(\beta,t, u) \Big)
  e^{-2ta_\beta} .
\]
This completes the proof of Theorem \ref{thm5}.
\end{proof}


\begin{proof}[Proof of Theorem \ref{thm6}]
(i) Let $v_1^{\beta,a_\beta }$ be the solution of
\eqref{eeee2} corresponding to $\varphi$ and let $w^{\beta,a_\beta }$
be the solution of \eqref{eeee2} corresponding to $\varphi_\beta$
where $\varphi, \varphi_\beta$ are defined  in Theorem \ref{thm6}.
Using Theorem \ref{thm4}, there exists a $t_\beta$ such that
$ \sqrt{t_\beta}=e^{-2t_\beta a_\beta}$
and
\begin{eqnarray}
\|v_1^{\beta,a_\beta }(.,.,t_\beta) - u(.,.,0)\| \leq
\sqrt{2}C \sqrt[4]{1/a_\beta}.
\end{eqnarray}
We denote
\[
v^{\beta,a_\beta } (.,.,t) =  \begin{cases}
 w^{\beta,a_\beta }  (.,.,t), & 0 < t < T, \\
 w^{\beta,a_\beta }  (.,.,t_\beta  ),& t = 0\,.
 \end{cases}
\]
Using Theorems \ref{thm2} and \ref{thm4}, we obtain
\begin{align*}
&\norm{v^{\beta,a_\beta }(.,.,t)-u(.,.,t)}\\
&\le \norm{w^{\beta,a_\beta }(.,.,t)-v_1^{\beta,a_\beta }(.,.,t)}
  +\norm{v_1^{\beta,a_\beta}(.,.,t)-u(.,.,t)} \\
&\le e^{(T-t) a_\beta}exp(k^2(T-t)^2)\norm{\varphi-\varphi_\beta}
  + M e^{k^2T(T-t)}e^{-ta_\beta}\\
&\le \exp(k^2(T-t)^2) \beta^{t/T} (\ln \frac{1}{\beta})^{\frac{-\alpha
  (T-t)}{2T}} + M e^{k^2T(T-t)}\beta^{t/T}(\ln \frac{1}{\beta})
  ^{\frac{\alpha t}{2T}},\\
&\le (M+1) e^{k^2T(T-t)} \beta^{t/T} (\ln \frac{1}{\beta})
  ^{\frac{-\alpha(T-t)}{2T}}
  \big( 1+ (\ln \frac{1}{\beta})^{\frac{\alpha}{2}}\big)
\end{align*}
for every $t \in (0,T)$.
Using the results in Theorem \ref{thm4}, we have the estimate
\begin{align*}
\|v^{\beta,a_\beta }(.,.,0) - u(.,.,0)\|
&\leq \|w^{\beta,a_\beta }(.,.,t_\beta) - v_1^{\beta,a_\beta }
(.,.,t_\beta)\| \\
&\quad + \| v_1^{\beta,a_\beta }(.,.,t_\beta) - u(.,.,0)\| \\
&\leq 2 e^{-t_\beta a_\beta} \exp (k^2 T^2)
 + \sqrt{2}C \sqrt[4]{1/a_\beta}
\\
&=  2\sqrt[4]{1/a_\beta} \exp (k^2 T^2)
 + \sqrt{2}C \sqrt[4]{1/a_\beta} \\
&= \sqrt[4]{1/a_\beta}\big(2\exp(k^2 T^2) +\sqrt{2} C \big).
\end{align*}
This completes the proof of part (i).

(ii) Using Theorems \ref{thm2} and \ref{thm5}, we obtain
\begin{align*}
&\norm{w^{\beta,a_\beta }(.,.,t)-u(.,.,t)}\\
&\le \norm{w^{\beta,a_\beta }(.,.,t)-v_1^{\beta,a_\beta }(.,.,t)}
 +\norm{v_1^{\beta,a_\beta}(.,.,t)-u(.,.,t)} \\
&\le e^{(T-t) a_\beta}exp(k^2(T-t)^2)\norm{\varphi-\varphi_\beta}
 +Q(\beta,t,u)e^{-ta_\beta}\\
&\le \exp(k^2(T-t)^2) \beta^{t/T} (\ln \frac{1}{\beta})
 ^{\frac{-\alpha (T-t)}{2T}} +Q(\beta,t,u)\beta^{t/T}
 (\ln \frac{1}{\beta})^{\frac{\alpha t}{2T}}  ,\\
&\le \beta^{t/T} (\ln \frac{1}{\beta})^{\frac{-\alpha(T-t)}{2T}}
 \Big( \exp(k^2(T-t)^2)+ Q(\beta,t,u)(\ln \frac{1}{\beta})
 ^{\frac{\alpha}{2}}\Big)
\end{align*}
for every $t \in [0,T]$.
\end{proof}

\section{Numerical experiments}
Let us consider the simple  two dimensional Allen-Cahn equation
\begin{gather*}
u_t -u_{xx}-u_{yy} = u-u^3+g(x,y,t),\quad
 (x,y,t)\in (0,\pi)\times (0,\pi) \times (0,1)
\\
u(0,y,t)=u(\pi,y,t)=u(x,0,t)=u(x,\pi,t)=0, \quad
 (x,y,t) \in (0,\pi)\times (0,\pi) \times [0,1],
\\
u(x,y,1)=\varphi(x,y),\,\,\, x,y \in (0,\pi)\times (0,\pi),
\end{gather*}
where
\begin{gather*}
g(x,y,t) = 2e^t \sin x \sin y + e^{3t} \sin^3 x \sin^3 y,\\
u(x,y,1) = \varphi_0 (x,y) \equiv e\sin x \sin y.
\end{gather*}
The exact solution of this equation is.
$u(x,y,t) = e^t \sin x\sin y$
In particular,
\[
u\big(x,y,\frac{39999}{40000}\big)\equiv u(x,y)
= \exp \big(\frac{39999}{40000}\big)\sin x \sin y.
\]
Let $\varphi_\beta(x,y) \equiv \varphi(x,y)
= (\beta + 1)e\sin x\sin y$.
Then we have
\[
\|\varphi_\beta -\varphi\|_2 =
\Big(\int_0^{\pi}\int_0^\pi\beta^2e^2 \sin^2(x)\sin^2 y dx dy\Big)^{1/2}
 = \beta  e \frac{\pi}{2}
\]
Choose $a_\beta= \frac{1}{\beta}$, and let $p$ be a natural number
satisfying $p=[\sqrt{\frac{1}{2}\ln(\frac{1}{\beta})  }]$.
We find the regularized solution
$v^{\beta,a_\beta}(x,y,\frac{39999}{40000}) \equiv u_\beta (x,y)$
having the  form
\[
v^{\beta,a_\beta}(x,y) = v_m(x,y) = w_{11,m}\sin x \sin y + w_{pp,m} \sin (px)\sin(py)
\]
where
\[
v_1(x,y) = (\beta + 1)e \sin x\sin y,\quad
w_{11,1} = (\beta + 1)e, \quad
w_{pp,1}= 0.
\]
and $ a =1/400000$,
$t_m  = 1 - am$ for $m = 1,2,\dots ,10$,
\begin{align*}
w_{11,m + 1}
&= e^{2( t_{m}- t_{m + 1} ) } w_{ij,m}
 - \frac{4}{\pi^2 }
 \int_{t_{m + 1} }^{t_m } {e^{2(s - t_{m + 1}) }}\\
&\quad\times \Big( \int_0^\pi \int_0^\pi \big( v_m(x,y)-v_m^3 (x,y)
 + g(x,y,s) \big)\sin x \sin y \,dx\,dy  \Big)ds\,,
\end{align*}
\begin{align*}
w_{pp,m + 1}
&= e^{2p^2(  t_{m}- t_{m + 1} ) }    w_{pp,m}
  - \frac{4}{\pi^2 }\int_{t_{m + 1} }^{t_m }
 {{e^{2p^2(s - t_{m + 1})  } }}\\
&\quad\times \Big( \int_0^\pi \int_0^\pi \big( v_m(x,y)-v_m^3 (x,y)
  + g(x,y,s) \big)\sin px \sin py dx dy  \Big)ds.
\end{align*}

Let $a_\beta = \|v^{\beta,a_\beta}- u\|$ be the error between the
regularized solution $v^{\beta,a_\beta}$  and the exact solution $u$.
Let $\beta = \beta_1 = 10^{-5}(p=2)$, $\beta = \beta_2 = 10^{-8}$,
$\beta =\beta_3 = 10^{-16}$. Then  we have
\begin{center}
\begin{tabular}{|c|c|c|}
 \hline
$\beta $& $ v^{\beta,a_\beta}$& $a_\beta $\\ \hline
$\beta_1 = 10^{-5}$
& \parbox{50mm}{$2.718241061\sin x\sin y -\\
 0.002038827910\sin (3x)
\sin (3y)$}
&0.002039009193  \\  \hline
$\beta_2 = 10^{-8}$
& \parbox{50mm}{$2.718213894 \sin x \sin y -\\
   0.0002039162480 \sin (3x) \sin (3y)$}
& 0.0002039162492 \\  \hline
$\beta_3 = 10^{-16}$
& \parbox{50mm}{$2.718220664 \sin x \sin y -\\
  0.0001835495554 \sin (3x) \sin (3y)$}
& 0.0001835495554\\ \hline
\end{tabular}
\end{center}

\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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