\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 $0\le t 0$ 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)}||^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. \begin{thebibliography}{00} \bibitem{Alekseeva} S. M. Alekseeva and N. 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