\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2008(2008), No. 33, pp. 1--14.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu \newline ftp ejde.math.txstate.edu (login: ftp)} \thanks{\copyright 2008 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2008/33\hfil Regularization and error estimates] {A nonhomogeneous backward heat problem: Regularization and error estimates} \author[D. D. Trong, N. H. Tuan\hfil EJDE-2008/33\hfilneg] {Dang Duc Trong, Nguyen Huy Tuan} \address{Dang Duc Trong \newline Department of Mathematics and Computer Sciences \\ Hochiminh City National University \\ 227 Nguyen Van Cu, Hochiminh City, Vietnam} \email{ddtrong@mathdep.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 November 9, 2007. Published March 6, 2008.} \thanks{Supported by the Council for Natural Sciences of Vietnam} \subjclass[2000]{35K05, 35K99, 47J06, 47H10} \keywords{Backward heat problem; ill-posed problem; \hfill\break\indent nonhomogeneous heat equation; contraction principle} \begin{abstract} We consider the problem of finding the initial temperature, from the final temperature, in the nonhomogeneous heat equation \begin{gather*} u_t-u_{xx}= f(x,t),\quad (x,t)\in (0,\pi)\times (0,T),\\ u(0,t)= u(\pi,t)= 0, \quad (x,t) \in (0,\pi)\times(0,T). \end{gather*} This problem is known as the backward heat problem and is severely ill-posed. Our goal is to present a simple and convenient regularization method, and sharp error estimates for its approximate solutions. We illustrate our results with a numerical example. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{remark}[theorem]{Remark} \newcommand{\norm}[1]{\|{#1}\|} \section{Introduction} For a positive number $T$, we consider the problem of finding the temperature $u(x,t)$, $(x,t)\in (0,\pi)\times [0,T]$, such that \begin{gather} u_t-u_{xx}=f(x,t),\quad (x,t)\in (0,\pi)\times(0,T), \label{e1} \\ u(0,t)=u(\pi,t) ,\quad (x,t) \in (0,\pi) \times [0,T], \label{e2}\\ u(x,T)=g(x),\quad x \in (0,\pi). \label{e3} \end{gather} where $g(x), f(x,z)$ are given. The problem is called the backward heat problem, the backward Cauchy problem, or the final value problem. As is known, the nonhomogeneous problem is severely ill-posed; i.e., solutions do not always exist, and in the case of existence, these do not depend continuously on the given data. In fact, from small noise contaminated physical measurements, the corresponding solutions have large errors. It makes difficult to do numerical calculations. Hence, a regularization is in order. Lattes and Lions, in \cite{lattes}, regularized the problem by adding a corrector'' to the main equation. They considered the problem \begin{gather*} u_t+Au-\epsilon A^*Au =0, \quad 00, we define the function $h(\lambda)=\frac{1}{\epsilon \lambda+e^{-\lambda T}}.$ Then $h(\lambda)\le h\Big(\frac{\ln(T/\epsilon)}{T}\Big) =\frac{T}{\epsilon \big(1+\ln(T/\epsilon)\big)} \quad \epsilon \in (0,eT).$ This follows that \label{e15} \begin{aligned} \norm{u(.,t)-v(.,t)}^2 &=\frac{\pi}{2}\sum_{p=1}^\infty \big|\frac{e^{-t p^2}}{\epsilon p^2 +e^{-T p^2}}(g_{p}-h_{p})\big|^2\\ &\le \frac{\pi}{2}\Big(\frac{T}{\epsilon \left(1+\ln(T/\epsilon)\right)}\Big)^2 \sum_{p=1}^\infty|g_{p}-h_{p}|^2\\ &=\Big(\frac{T}{\epsilon \left(1+\ln(T/\epsilon)\right)}\Big)^2 \|g-h\|^2. \end{aligned} Hence $\norm{u(.,t)-v(.,t)}\le\frac{T}{\epsilon \left(1+\ln(T/\epsilon)\right)}\norm{g-h}.$ This completes the proof of Step 2 and the proof of our theorem. \end{proof} \begin{remark} \label{rmk1} \rm In \cite{ewing,Gajewski,Trong}, the stability magnitude ise^{\frac{T}{\epsilon}}$(see \cite[Theorem 2.1]{Denche}, it is$\epsilon^{-1}$. One advantage of this method of regularization is that the order of the error, introduced by small changes in the final value$g$, is less than the order given in \cite{Trong}. \end{remark} \begin{theorem} \label{thm2.2} For any$g(x) \in L^2(0,\pi)$, the approximation$u^\epsilon(x,T)$converges to$g(x)$in$L^2(0,\pi)$as$\epsilon $tends to zero. \end{theorem} \begin{proof} We have$ g(x)=\sum_{p=1}^\infty g_p \sin (px)$, where$g_p $is defined in \eqref{e9}. Let$\alpha>0$, choose some$N$for which$\frac{\pi}{2}\sum_{p=N+1}^\infty g_p^2 < \alpha/2$. We have $$\label{e16} \|u^\epsilon(x,T)-g(x)\|^2=\frac{\pi}{2}\sum_{p=1}^\infty \frac{\epsilon^2 p^4 g_p^2}{ (\epsilon p^2+e^{-Tp^2})^2}\,.$$ Then $$\|u^\epsilon(x,T)-g(x)\|^2 \le \epsilon^2 \frac{\pi}{2} \sum_{p=1}^N p^4 g_p^2 e^{2Tp^2}+\frac{\alpha}{2}$$ By taking$\epsilon$such that$ \epsilon < \sqrt{\alpha}\big( {\pi \sum_{p=1}^N p^4 g_p^2 e^{2Tp^2} } \big)^{-1/2}$, we get $$\|u^\epsilon(x,T)-g(x)\|^2 < \alpha$$ which completes the proof. In the case$\frac{d^2g}{dx^2} \in L^2(0,\pi), we have the error estimate \begin{align*} \|u^\epsilon(x,T)-g(x)\|^2 &= \frac{\pi}{2} \sum_{p=1}^\infty ( \frac {{e^{-T p^2}}}{\epsilon p^2 +e^{-Tp^2}}-1)^2 g^2_p\\ &=\frac{\pi}{2}\sum_{p=1}^\infty \frac{\epsilon^2 p^4 g_p^2}{ (\epsilon p^2+e^{-Tp^2})^2}\\ &\le \frac{\pi}{2} \frac{T}{ \left(1+\ln(\frac{T^2}{\epsilon})\right)^2} \sum_{p=1}^\infty p^4 g_p^2 =\frac{T^2}{ \left(1+\ln(T/\epsilon)\right)^2}\|g_{xx}\|^2 \end{align*} Then, we get \begin{align*} \|u(x,T)-g(x)\| \le \frac{T}{ 1+\ln(T/\epsilon)}\|g_{xx}\|. \end{align*} This completes the proof. \end{proof} \begin{theorem} \label{thm2.3} Letg(x), \epsilon \in L^2(0,\pi)$be as in Theorem \ref{thm2.2}, and let$ f_{xx}$be in$L^2 (0,T; L^2(0,\pi))$. If the sequence$u^\epsilon(x,0)$converges in$L^2(0,\pi)$, then the problem \eqref{e1}--\eqref{e3} has a unique solution$u$. Furthermore,we then have that$u^\epsilon(x,t)$converges to$u(t)$as$\epsilon$tends to zero uniformly in t. \end{theorem} \begin{proof} Assume that$\lim _{\epsilon \to 0}u^\epsilon(x,0)=u_{0}(x)$exists. Let $$u(x,t)=\sum_{p=1}^\infty \Big(e^{-tp^2}u_{0p}- \int_0^t e^{(s-t)p^2}f_p(s)ds \Big)\sin (px)$$ where$u_{0p}= \frac{2}{\pi} \int_0^{\pi}u_0(x)\sin (px) dx$. It is clear to see that$u(x,t)$satisfies \eqref{e1}--\eqref{e2}. We have the formula of$u^\epsilon(x,t)$$u^\epsilon(x,t)=\sum_{p=1}^\infty \Big(e^{-tp^2}u^\epsilon_{0p}-\int_0^t \frac {e^{(s-t-T) p^2}}{\epsilon p^2+e^{-T p^2}}f_{p}(s)ds\Big)\sin (px)$ where$u^\epsilon_{0p}= \frac{2}{\pi} \int_0^{\pi}u^\epsilon(x,0)\sin (px) dx$. In view of the inequality$ (a+b)^2 \le 2(a^2+b^2), we have \begin{align*} &\|u^\epsilon(x,t)-u(x,t)\|^2 \\ &\le \frac{\pi}{2} \sum_{p=1}^\infty (u^\epsilon_{0p}-u_{0p})^2+ \frac{\pi}{2}t^2\sum_{p=1}^\infty \Big( \int_0^t e^{(2s-2t)p^2} \frac{\epsilon^2 p^4}{(\epsilon p^2 +e^{-T p^2})^2}f_{p}^2 ds\Big) \\ &\le \|u^\epsilon(x,0)-u_0(x)\|^2+ T^2 \frac{T^2}{ \left(1+\ln(T/\epsilon)\right)^2} \int_0^t \sum_{p=1}^\infty p^4 f_{p}^2 ds\\ &=\|u^\epsilon(x,0)-u_0(x)\|^2+ \frac{T^4}{ \left(1+\ln(T/\epsilon)\right)^2} \int_0^t \|f_{xx}\|^2 ds \\ &\le \|u^\epsilon(x,0)-u_0(x)\|^2 + \frac{T^4}{ \left(1+\ln(T/\epsilon)\right)^2} \|f_{xx}\|^2_{L^2 (0,T; L^2(0,\pi))} \end{align*} Hence,\lim _{\epsilon \to 0}u^\epsilon(x,t)=u(x,t)$. Thus$\lim _{\epsilon \to 0}u^\epsilon(x,T)=u(x,T)$. Using theorem \ref{thm2.2}, we have$u(x,T)=g(x)$. Hence,$u(x,t)$is the unique solution of the problem \eqref{e1}--\eqref{e3}. We also see that$u^\epsilon(x,t)$converges to$u(x,t)$uniformly in$t$. \end{proof} \begin{theorem} \label{thm2.4} Let$f(x,t) ,g(x),\epsilon$be as Theorem \ref{thm2.3}. If the sequence$u_t^\epsilon(x,0)$converges in$L^2(0,\pi)$, then the problem \eqref{e1}--\eqref{e3} has a unique solution$u$. Furthermore,we have that$u^\epsilon(x,t)$converges to$u(x,t)$as$\epsilon$tends to zero in$C^1(0,T;L^2(0,\pi))$. \end{theorem} \begin{proof} Assume that$\lim _{\epsilon \to 0}u_t^\epsilon(x,0)=v(x)$in$L^2(0,\pi)$. Let$v(x)=\sum_{p=1}^\infty v_p \sin (px)$where$v_p=\frac{2}{\pi} \int_0^{\pi}v(x)\sin (px) dx$. Denote by$w_p=-\frac{v_p}{p^2}$and$w(x)=\sum_{p=1}^\infty w_p \sin (px)$. It is easy to show that the function$u(x,t)$defined by $$u(x,t)=\sum_{p=1}^\infty \Big( e^{-tp^2}w_p - \int_0^t e^{(s-t)p^2} f_pds\Big)\sin (px)$$ is a solution of the problem \begin{gather*} u_t(x,t)-u_{xx}=f(x,t),\\ u(x,0)=w(x) \end{gather*} Since$u^\epsilon (x,t)is the solution of \eqref{e6}--\eqref{e8}, we have \begin{gather*} u^\epsilon_{tp}(t)=p^2 u^\epsilon_p(t)+\frac {{e^{-T p^2}}}{\epsilon p^2 +e^{-Tp^2}}f_p(t),\\ u_{tp}(t)=p^2 u_p (t)+f_p(t) \end{gather*} where \begin{gather*} u^\epsilon_p(t)=\frac{2}{\pi} \int_0^{\pi}u^\epsilon(x,t)\sin (px) dx,\quad u_p (t)= \frac{2}{\pi} \int_0^{\pi}u(x,t)\sin (px) dx \\ u^\epsilon_{tp}(t)=\frac{2}{\pi} \int_0^{\pi}u_t^\epsilon(x,t)\sin (px) dx, \quad u_{tp} (t)= \frac{2}{\pi} \int_0^{\pi}u_t(x,t)\sin (px) dx \end{gather*} So that $$\label{e17} u^\epsilon_p(t)-u_p (t)=\frac{1}{p^2} (u^\epsilon_{tp}(t)-u_{tp} (t)) + \frac {\epsilon}{\epsilon p^2+e^{-Tp^2}}f_p(t)$$ By a direct computation, \begin{align*} \|u^\epsilon(.,t)-u(.,t)\|^2 &=\frac{\pi}{2}\sum_{p=1}^\infty |u^\epsilon_p(t)-u_p (t)|^2 \\ &\le \sum_{p=1}^\infty \pi|u^\epsilon_{tp}(t)-u_{tp} (t)|^2 + \pi \sum_{p=1}^\infty \frac {\epsilon^2 }{(\epsilon p^2 +e^{-Tp^2})^2}f^2_p(t)\\ &\le 2\|u_t^\epsilon(x,t)-u_t(x,t)\|^2+ 2 \frac{T^2}{ (1+\ln(T/\epsilon)^2}\|f(.,t)\|^2 \end{align*} Hence $\|u^\epsilon(.,0)-u(.,0)\|^2 \le 2 \|u_t^\epsilon(x,0)-u_t(x,0)\|^2 + \frac{2T^2}{ (1+\ln(T/\epsilon))^2}\|f(.,t)\|^2$ Using\lim _{\epsilon \to 0}u_t^\epsilon(x,0)=v(x)=u_t(x,0)$, we get$\lim _{\epsilon \to 0}\|u^\epsilon(x,0)-u(x,0)\|=0. On the other hand, we have \begin{gather*} u^\epsilon(x,T)=\sum_{p=1}^\infty \Big(e^{-Tp^2}u^\epsilon_p(0) +\int_0^T \frac{e^{(s-2T)p^2}}{\epsilon p^2+e^{-Tp^2}}f_p(s)ds\Big) \sin (px)\\ u(x,T)=\sum_{p=1}^\infty \Big(e^{-Tp^2}u_p(0)+\int_0^T {e^{(s-T)p^2}} f_p(s)ds\Big) \sin (px) \end{gather*} It follows that \begin{align*} &\| u^\epsilon(.,T)-u(.,T)\|^2 \\ &\le 2 \sum_{p=1}^\infty e^{-2Tp^2}|u^\epsilon_p(0)-u_p(0)|^2 + 2 T^2 \sum_{p=1}^\infty \int_0^T \frac {\epsilon^2 p^4}{(\epsilon p^2+e^{-Tp^2})^2}f^2_p(s)ds\\ &\le \|u^\epsilon(.,0)-u(.,0)\|^2 +2\frac{T^2}{ (1+\ln(T/\epsilon))^2}\|f_{xx}(.,t)\|^2 \end{align*} Hence,\lim _{\epsilon \to 0}\|u^\epsilon(x,T)-u(x,T)\|=0$. Using the Theorem \ref{thm2.2}, we obtain$u(x,T)=g(x)$. This implies that$u(x,t)$is the unique solution of \eqref{e1}--\eqref{e3}. \end{proof} \begin{theorem} \label{thm2.5} If there exists$m\in (0,2)$so that$ \sum_{p=1}^\infty p^{2m}e^{mTp^2}g_p^2 $converges, then $$\|u^\epsilon(x,T)-g(x)\| \le \frac{\sqrt{ C_1\epsilon ^m} } {m}$$ where$C_1=4 \sum_{p=1}^\infty p^{2m}e^{mTp^2}g_p^2$. \end{theorem} \begin{proof} Let$m$be in$ (0,2)$such that$ \sum_{p=1}^\infty p^{2m}e^{mTp^2}g_p^2 $converges, and let$n$be in$(0,2) $. Fix a natural integer$p$, and define $$g_p(\epsilon)=\frac{\epsilon^n }{(\epsilon p^2 +e^{-Tp^2})^2}.$$ It can be shown that$g_p(\epsilon) \le g_p(\epsilon_{0})$, for all$\epsilon>0 $where$\epsilon_{0}=\frac{ n e^{-Tp^2} }{(2-n)p^2 } $. Furthermore, from \eqref{e16}, we have $$\label{e18} \|u^\epsilon(x,T)-g(x)\|^2 =\sum_{p=1}^\infty \frac{\epsilon^2 p^4 g_p^2}{ (\epsilon p^2+e^{-Tp^2})^2} =\epsilon^{2-n} \sum_{p=1}^\infty p^4 g^2_p g_p(\epsilon)$$ It follows that $$\label{e19} \|u^\epsilon(x,T)-g(x)\|^2 \le \epsilon^{2-n} (\frac {n}{2-n})^{n} \sum_{p=1}^\infty p^{4-2n}g_p^2e^{(2-n)Tp^2 }$$ If we choose$n=2-m$, we obtain$\|u^\epsilon(x,T)-g(x)\|^2 \le C_1\epsilon^m m^{-2}$. \end{proof} \begin{theorem} \label{thm2.6} Let$ f \in L^{2}(0,T;L^{2}(0,\pi))$and$ g \in L^{2}(0,\pi) $and$\epsilon \in (0,eT)$. Suppose that Problem \eqref{e1}--\eqref{e3} has a unique solution$u(x,t)$in$C([0,T];H_0^1(0,\pi))\cap C^1((0,T);L^2(0,\pi))$which satisfies$\|u_{xx}(.,t)\|<\infty$. Then $\norm{ u(.,t)- u^{\epsilon}(.,t)}\le \frac{C}{1+\ln(T/\epsilon)}$ for every$t\in[0,T]$, where$ C= T \sup _{t\in [0,T]} \|u_{xx}(.,t)\| $and$u^\epsilon$is the unique solution of \eqref{e6}--\eqref{e8}. \end{theorem} \begin{proof} Suppose \eqref{e1}--\eqref{e3} has an exact solution$u$in the space$C([0,T];H_0^1(I))\cap C^1((0,T);L^2(I)), we get the formula $$\label{e20} u(x,t)=\sum_{p=1}^\infty (e^{-(t-T) p^2}g_{p} -\int_t^T e^{-(t-s) p^2}f_{p}(s)ds)\sin (px)$$ From \eqref{e10} and \eqref{e20}, we obtain \label{e21} \begin{aligned} |u_{p}(t)- u^{\epsilon}_{p} (t)| &= \Big|\Big(e^{-(t-T) p^2}-\frac {{e^{-t p^2}}}{\epsilon p^2+e^{-Tp^2}} \Big)\Big(g_{p}-\int_t^T e^{-(T-s) p^2}f_{p}(s)ds)\Big)\Big| \\ &= \epsilon p^2\frac {e^{-(t-T) p^2}}{\epsilon p^2+e^{-Tp^2}} \Big|\Big(g_{p}-\int_t^T e^{-(T-s) p^2}f_{p}(s)ds)\Big)\Big| \\ & \le \frac{T}{1+\ln(T/\epsilon)} \Big|\Big(p^2e^{-(t-T) p^2}g_{p}-\int_t^T p^2e^{-(t-s) p^2}f_{p}(s)ds) \Big)\Big| \end{aligned} It follows that \begin{align*} &\norm{ u(.,.,t)- u^{\epsilon}(.,.,t)}^2\\ &=\frac{\pi}{2}\sum_{p=1}^\infty |u_{p}(t)- u^{\epsilon}_{p} (t)|^2\\ &\leq \frac{\pi}{2} \Big( \frac{T}{1+\ln(T/\epsilon)}\Big)^{2} \sum_{p=1}^\infty \Big(p^2e^{-(t-T) p^2}g_{p}-\int_t^T p^2 e^{-(t-s) p^2}f_{p}(s)ds)\Big)^{2} \\ &=\Big( \frac{T}{1+\ln(T/\epsilon)}\Big)^{2} \norm{u_{xx}(.,t)}^{2} \le \Big( \frac{C}{1+\ln(T/\epsilon)}\Big)^{2} \end{align*} Hence $\norm{ u(.,t)- u^{\epsilon}(.,t)}\le \frac{C}{1+\ln(T/\epsilon)}$ whereC= T \sup _{t\in [0,T]} \|u_{xx}(.,t)\| $. This completes the proof \end{proof} \begin{remark} \label{rmk2}\rm Note that in \cite[Theorem 3.3]{Trong}, the exact solution$u$satisfies the condition$\Delta^{2}u(x,t)\in L^2(0,\pi)$, while the condition of its in this theorem is$\Delta u \in L^2(0,\pi) $. So, this also implies that the final value$g$in our theorem is only in$L^2(0,\pi)$, not satisfying the condition (*) given in \cite{Trong2} (see Introduction). Further more, we also have the error estimate$\norm{ u_t(.,t)- u_t^{\epsilon}(.,t)}$which is not given in \cite{Trong,Trong2}. Hence, this result is an improvement of known result in \cite{Trong,Trong2}. \end{remark} \begin{theorem} \label{thm2.7} Let$ f \in L^{2}(0,T;L^{2}(0,\pi))$and$ g \in L^{2}(0,\pi) $and$\epsilon \in (0,eT)$. Suppose that Problem \eqref{e1}--\eqref{e3} has a unique solution$u(x,t)$in$C([0,T];H_0^1(0,\pi))\cap C^1((0,T);L^2(0,\pi))$which satisfies$\|u_{xxxx}(.,t)\|<\infty$. Then $\norm{ u_t(.,t)- u_t^{\epsilon}(.,t)}\le \frac{D}{1+\ln(T/\epsilon)}$ for every$t\in[0,T]$, where $D= T \Big(2 \sup _{t\in [0,T]} (\|u_{xxxx}(.,t)\|^2+\|f_{xx}(.,t)\|^2) \Big)^{1/2}$ and$u^\epsilonis the unique solution of \eqref{e6}--\eqref{e8}. \end{theorem} \begin{proof} In view of \eqref{e17}, we have \begin{align*} u^\epsilon_{tp}(t)-u_{tp} (t) &= p^2 (u^\epsilon_p(t)-u_p (t))-\frac {\epsilon p^2}{\epsilon p^2 +e^{-Tp^2}}f_p(t) \\ &= \frac {\epsilon p^4 e^{-(t-T) p^2}}{\epsilon p^2+e^{-Tp^2}} \Big(g_{p}-\int_t^T e^{-(T-s) p^2}f_{p}(s)ds)\Big) -\frac {\epsilon p^2}{\epsilon p^2+e^{-Tp^2}}f_p(t)\\ &= \frac {\epsilon p^4 }{\epsilon p^2+e^{-Tp^2}} u_p(t) -\frac {\epsilon p^2}{\epsilon p^2+e^{-Tp^2}}f_p(t)\\ &=\frac {\epsilon p^2}{\epsilon p^2+e^{-Tp^2}} (p^2 u_p(t)-f_p(t)) \end{align*} Hence, we get \begin{align*} \norm{ u_t(.,t)- u_t^{\epsilon}(.,t)}^2 &= \frac{\pi}{2}\sum_{p=1}^\infty |u^\epsilon_{tp}(t)-u_{tp} (t)|^2 \\ &\le \pi \frac {\epsilon^2 }{(\epsilon p^2+e^{-Tp^2})^2}\sum_{p=1}^\infty (p^8u^2_p(t)+ p^4f^2_p(t))\\ &= \frac{2T^2}{ \left(1+\ln(T/\epsilon)\right)^2} (\|u_{xxxx}(x,t)\|^2 + \|f_{xx}(x,t)\|^2) \end{align*} This completes the proof. \end{proof} In the case of nonexact data, one has the following result. \begin{theorem} \label{thm2.8} Letf,g,\epsilon $be as in Theorem \ref{thm2.6}. Assume that the exact solution$u$of \eqref{e1}--\eqref{e3} corresponding to$g$satisfies $$u \in C([0,T];L^2(0,\pi))\cap L^2(0,T;H_0^1(0,\pi)) \cap C^1((0,T);L^2(0,\pi)),$$ and$\|u_{xx}(.,t)\|< \infty$. Let$g_\epsilon \in L^2(0,\pi)$be a measured data such that $$\norm{g_\epsilon-g}\le\epsilon.$$ Then there exists a function$ u^{\epsilon}$satisfying $\norm{ u(.,t)- u^{\epsilon}(.,t)}\le \frac{C+T}{1+\ln(T/\epsilon)}$ for every$t\in [0,T]$and$C$is defined in Theorem \ref{thm2.6}. \end{theorem} \begin{proof} Let$v^\epsilon$be the solution of problem \eqref{e6}--\eqref{e8} corresponding to$g$and let again$u^{\epsilon}$be the solution of problem \eqref{e6}--\eqref{e8} corresponding to$g_\epsilon$where$g, g_\epsilonare in right hand side of \eqref{e6}. Using Theorem \ref{thm2.6} and Step 2 in Theorem \ref{thm2.1}, we get \begin{align*} \norm{u^\epsilon(.,t)-u(.,t)} &\le \norm{u^{\epsilon}(.,t)-v^\epsilon(.,t)}+\norm{v^\epsilon(.,t)-u(.,t)} \\ &\le \frac{T}{\epsilon (1+\ln(T/\epsilon)}\norm{g_\epsilon-g} + \frac{T}{ (1+\ln(T/\epsilon)}\|u_{xx}(.,t)\| \\ &\le \frac{C+T}{1+\ln(T/\epsilon)} \end{align*} for everyt \in (0,T)$and where$C$is defined in Theorem \ref{thm2.6}. This completed the proof. \end{proof} \section{A numerical example} We consider $$\label{e22} \begin{gathered} u_t-u_{xx} = f(x,t)\equiv 2e^{t}\sin x, \\ u(x,1) = g (x) \equiv e\sin x . \end{gathered}$$ The exact solution to this problem is $u(x,t) = e^t \sin x$ Note that$u(x,1/2)= \sqrt e \sin (x)\approx 1.648721271\sin(x)$. Let$g_n $be the measured final data $g _n (x) = e\sin(x) + \frac{1}{n} \sin (nx).$ So that the data error, at the final time, is $F(n) = \|g_n - g \|_{L^2 (0,\pi ) } = \sqrt {\int_0^\pi {\frac{1}{{n^2 }}\sin ^2 (nx)dx} } = {\frac{\pi } {2n}}.$ The solution of \eqref{e25}, corresponding the final value$g _n $, is $u^n (x,t) = e^t \sin (x) + \frac{1} {n}e^{n^2 (1 - t)} \sin (nx),$ The error at the original time is $O(n):= \|u^n (.,0) - u(.,0)\|_{L^2 (0,\pi )} = \sqrt {\int_0^\pi {\frac{{e^{2n^2 } }} {{n^2 }}\sin ^2 (nx)\,dx} } = \frac{{e^{n^2 } }}{{n }} {\frac{\pi }{2}}.$ Then, we notice that \begin{gather} \label{e23} \lim_{n \to \infty } F(n)=\mathop {\lim }_{n \to \infty } ||g _n - g ||_{L^2 (0,\pi )} = \lim _{n \to \infty } \frac{1} {n} {\frac{\pi } {2}} = 0,\\ \lim _{n \to \infty } O(n)=\lim _{n \to \infty } \|u^n (.,0) - u(.,0)\|_{L^2 (0,\pi )} = \lim_{n \to \infty } \frac{{e^{n^2 } }} {{n^{} }}{\frac{\pi } {2}} = \infty. \label{e24} \end{gather} From the two equalities above, we see that \eqref{e22} is an ill-posed problem. Approximating the problem as in \eqref{e1}--\eqref{e3}, the regularized solution is $$\label{e25} u^{\epsilon}(x,t) =\sum_{p=1}^\infty \Big(\frac {{e^{-t p^2}}}{\epsilon p^2+e^{-p^2}}g_{p} -\int_t^1 \frac {e^{(s-t-1) p^2}}{\epsilon p^2+e^{- p^2}}f_{p}(s)ds\Big) \sin (px)$$ for$0\le t \le 1$. Hence, we have $$\label{e26} u^{\epsilon}(x,t)= \frac {{e^{1-t }}}{\epsilon +e^{-1}}\sin x -2\Big(\int_t^1 \frac {e^{2s-t-1 }}{ \epsilon +e^{- 1}}ds\Big)\sin x +\frac{1}{n}\frac {{e^{-tn^{2} }}}{\epsilon n^{2} +e^{-n^{2}}} \sin (nx)\,.$$ It follows that $$\label{e27} u^{\epsilon}(x,\frac{1}{2}) =\Big(\frac {e^{1/2}}{\epsilon +e^{-1}} -2\int_{\frac{1}{2}}^1 \frac {e^{2s-\frac{3}{2} }}{\epsilon +e^{- 1}}ds \Big)\sin x +\frac{1}{n}\frac {{e^{-\frac{1}{2}n^{2} }}}{\epsilon n^{2} +e^{-n^{2}}} \sin (nx)$$ Let$a_\epsilon = \|u_\epsilon(.,\frac{1}{2}) - u(.,\frac{1}{2})\|$be the error between the regularized solution$u_\epsilon$and the exact solution$u$in the time$t=\frac{1}{2}$. Let$n=300$and \begin{gather*} \epsilon = \epsilon_1 = 10^{-2}\sqrt{\frac{\pi}{2}}, \epsilon =\epsilon_2 = 10^{-4}\sqrt{\frac{\pi}{2}},\\ \epsilon=\epsilon_3 = 10^{-10}\sqrt{\frac{\pi}{2}}, \epsilon =\epsilon_4 = 10^{-15}\sqrt{\frac{\pi}{2}}. \end{gather*} \begin{table} \caption{}$\begin{array}{|c|c|c|c|c|} \hline \epsilon& u_\epsilon& a_\epsilon \\ \hline \epsilon_1 = 10^{-2}\sqrt{\frac{\pi}{2}} & 1.59440220314355 \sin(x) & 0.06807885585\\ &+ 4.636337144 \times 10^{-39093 }\sin(300x)&\\ \hline \epsilon_2 =10^{-4}\sqrt{\frac{\pi}{2}} & 1.64815976557002 \sin(x) & 0.0007037421545\\ &+ 4.636337144 \times 10^{-39091}\sin(300x)&\\ \hline \epsilon_3 = 10^{-10} \sqrt{\frac{\pi}{2}} & 1.64872127013843\sin x & 1.253314137 \times 10^{-9} \\ &+ 4.636337144 \times 10^{-39084}\sin(300x)&\\ \hline \epsilon_4 =10^{-16}\sqrt{\frac{\pi}{2}} & 1.64872127070011\sin(x) & 5.810786885 \times 10^{-39079} \\ & +4.636337144 \times 10^{-39079}\sin(300x) &\\ \hline \end{array}$\end{table} We note that the new method in this article give a better approximation than the previous method in \cite{Trong}. To prove this, we have in view of the error table in \cite[p. 9]{Trong}. \begin{table}[ht] \caption{} \begin{tabular}{|c|c|c|} \hline$\epsilon$&$u_\epsilon$&$\|u-u_\epsilon\|$\\ \hline$10^{-2}\sqrt{\frac{\pi}{2}}$&$1.643563444\sin(x)+0.8243606355\sin200x$& 0.1462051256\\ \hline$10^{-4}\sqrt{\frac{\pi}{2}}$&$1.648617955\sin(x) + 0.1648721271\sin10000 x$& 0.02066391506\\ \hline$10^{-10}\sqrt{\frac{\pi}{2}}$&$1.648721271\big(\sin(x) +10^{-10}\sin(10^{10}x)\big)$&0.00002066365678\\ \hline$10^{-16}\sqrt{\frac{\pi}{2}}$&$1.648721271\big(\sin(x) +10^{-16}\sin(10^{16}x)\big)$&$2.066365678\times10^{-8}$\\ \hline$10^{-30}\sqrt{\frac{\pi}{2}}$&$1.648721271\big(\sin(x) +10^{-30}\sin(10^{30}x)\big)$&$2.066365678\times10^{-15}$\\ \hline \end{tabular} \end{table} Furthermore, we continue to approximate this problem by the method given in \cite{Trong}, which gives regularized solution $v^{\epsilon}(x,t) =\sum_{p=1}^\infty \Big(\frac {{e^{-t p^2}}}{\epsilon +e^{-p^2}}g_{p} -\int_t^1 \frac {e^{-t p^2}}{\epsilon ^{s}+e^{- sp^2}}f_{p}(s)ds\Big) \sin (px).$ Hence, we have $$\label{e28} v^{\epsilon}(x,t)= \frac {{e^{1-t }}}{\epsilon +e^{-1}}\sin x -2\Big(\int_t^1 \frac {e^{s-t }}{\epsilon^{s} +e^{- s}}ds\Big)\sin x +\frac{1}{n}\frac {{e^{-tn^{2} }}}{\epsilon +e^{-n^{2}}} \sin (nx)$$ for$0\le t \le 1$. It follows that $$\label{e29} v^{\epsilon}(x,\frac{1}{2})= \frac {e^{1/2}}{\epsilon +e^{-1}}\sin x -2\Big(\int_{\frac{1}{2}}^1 \frac {e^{s-\frac{1}{2} }}{\epsilon^{s} +e^{- s}}ds\Big)\sin x +\frac{1}{n}\frac {{e^{-\frac{1}{2}n^{2} }}}{\epsilon +e^{-n^{2}}} \sin (nx).$$ \begin{table} \caption{}$\begin{array}{|c|c|c|c|c|} \hline \epsilon& v_\epsilon& a_\epsilon \\ \hline \epsilon_1 = 10^{-2}\sqrt{\frac{\pi}{2}} & 1.701714206 \sin(x) & 0.06641679460\\ &+4.172703428 \times 10^{-39088 }\sin(300x) &\\ \hline \epsilon_2 =10^{-4}\sqrt{\frac{\pi}{2}} & 1.656775314\sin(x) & 0.01009424595\\ &+ 4.172703428\times 10^{-39086}\sin(300x)&\\ \hline \epsilon_3 = 10^{-10}\sqrt{\frac{\pi}{2}} & 1.648724344 \sin x & 0.000003851434344\\ &+ 4.172703428 \times 10^{-39080}\sin(300x)&\\ \hline \epsilon_4 =10^{-16}\sqrt{\frac{\pi}{2}} & 1.648721273\sin(x) & 2.506628275 \times 10^{-9}\\ &+ 4.172703428 \times 10^{-39074}\sin(300x)&\\ \hline \end{array}\$ \end{table} Looking at Tables 1,2,3, a comparison between the three methods, we can see the error results of in Table 1 are smaller than the errors in Tables 2 and 3. 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