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Accepted Manuscript An improved regularization method for initial inverse problem in 2-D heat equation Nguyen Huy Tuan, Tran Thanh Binh, Nguyen Dang Minh, Truong Trong Nghia PII: DOI: Reference: S0307-904X(14)00276-5 http://dx.doi.org/10.1016/j.apm.2014.05.014 APM 10021 To appear in: Appl Math Modelling Received Date: Revised Date: Accepted Date: October 2012 18 November 2013 16 May 2014 Please cite this article as: N.H Tuan, T.T Binh, N.D Minh, T.T Nghia, An improved regularization method for initial inverse problem in 2-D heat equation, Appl Math Modelling (2014), doi: http://dx.doi.org/10.1016/j.apm 2014.05.014 This is a PDF file of an unedited manuscript that has been accepted for publication As a service to our customers we are providing this early version of the manuscript The manuscript will undergo copyediting, typesetting, and review of the resulting proof before it is published in its final form Please note that during the production process errors may be discovered which could affect the content, and all legal disclaimers that apply to the journal pertain An improved regularization method for initial inverse problem in 2-D heat equation Nguyen Huy Tuan 1,2∗ , Tran Thanh Binh Nguyen Dang Minh , Truong Trong Nghia , Department of Mathematics, University of Science, Vietnam National University - Ho Chi Minh City, VietNam Institute of Science and Technology in Ho Chi Minh City, Viet Nam Department of Mathematics and Applications, Sai Gon University, Ho Chi Minh city, VietNam School of Computing, University of Utah, USA Abstract The main purpose of this article is to present a new method to regularize the initial inverse heat problem with inhomogeneous source This problem is well known to be severely ill-posed There are many regularization methods with error estimator of logarithmic order An improved regularization method is proposed The error estimates of Hăolder type are obtained Some numerical tests illustrate that the proposed method is feasible and effective Keywords and phrases: Backward heat problem, Ill-posed problem, Quasi-boundary value methods, Quasi-Reversibility methods Mathematics subject Classification 2000: 35K05, 35K99, 47J06, 47H10 Introduction In this paper, we consider the non-homogeneous initial inverse heat problem (or called nonhomogeneous backward heat problem) in a rectangle I = (0, π) × (0, π): ut − uxx − uyy = f (x, y, t), (x, y, t) ∈ I × (0, T ) u(0, y, t) = u(π, y, t) = u(x, 0, t) = u(x, π, t) = 0, (x, y, t) ∈ I × (0, T ) (1) u(x, y, T ) = g(x, y), (x, y) ∈ I where g(x, y) and f (x, y, z) are given This problem is well known to be severely ill-posed and regularization methods for it are required (See [10]) As is known, the above problem is severely ill-posed, i.e its solutions not always exist and in the case of existence they not depend continuously on the given data In fact, from small noise contaminated from physical measurements, the corresponding solutions have large errors That makes difficult to numerical calculations Hence, a regularization is a need Authors such as K A Ames [1], R Lattes and J L Lions [13], R E Showalter [21], K Miller [17], L.E Payne [19] have approximated the Problem (1) by quasi-reversibility method and quasi-boundary value method In [32], T Schrăoter and U Tautenhahn established an optimal error estimate for the homogeneous case of (1) A mollification method has been studied by D N Hao in [11] S M Kirkup and M Wadsworth used an operator-splitting method in [15] A method of hyperbolic equation for backward heat has been considered by K Masood et al [16] Very recently, the homogeneous problem of (1) was also investigated by Y.C Hon et al [12], T.Wei et al in [20], J Rashidinia et al in [23], P Daripa et al in [24], Jin-Ru Wang in [35] ∗ Corresponding author E-mail: thnguyen2683@gmail.com Although there are many papers on the homogeneous case of the initial inverse heat problem, we only find a few results on the non-homogeneous case, especially the two-dimensional non-homogeneous case is very rare For the case, we refer the reader to some recent works of X.L Feng et al [7], M Li et al[14], P.T.Nam et al [18], Trong et al [25, 27, 30] Physically, g is measured data with an error of parameter Let u and v be the exact solution and the approximated solution of the backward heat t T problem respectively In [27, 29] , the errors are of order 1+ln T The error estimates in [26] are for t > and ln −1 for t = Very recently, Trong and Tuan [29] improved the previous stability results which is of order t T T 1+ln( T ) 1− Tt For the literature on non-homogeneous backward heat, we refer the reader to the results in Fu et al [7, 8, 9] However, the error estimates in the mentioned papers are still of logarithmic order Very recently, P.T Nam et al [18] regularized the Problem (1) by truncation method and obtained the error estimate which is of order q , < q < Using this method, N.H Tuan et al [31] considered a general version of the Problem (1) with similar results The truncation method introduced in [18, 31] is simple and effective to solve the backward heat problem with good estimates However, in practice, the computation of the approximation solution (by the truncation method) is impossible and difficult when we consider the problem in a general two-dimensional domain (See [4]) If the spectral problem of operator −∆ in this domain is unknown then the truncation method is seems to be useless This is may be a disadvantage point of papers [18, 31] Motivated by this reason, in the present paper, we provide another regularization method to established the Hă older estimates Our method is similar to quasi-boundary value method (or non-local boundary value problems method, see [3, 5, 30]) but it seems to be in a new direction In few words, we explain why this method is new By a natural way, to approximate the solution of the Problem (1), in many previous methods, we usually propose a regularized solution u for t ∈ [0, T ], then estimate the error u (., t) − u(., t) (norm in L2 ) for all t ∈ [0, T ] The method in this paper is first to compute the regularized solution for t ∈ [0, T + (h − 1)T ] where h ≥ 1, then use the resulting solution at t + (h − 1)T to approximate the exact solution at t Under some suitable conditions on the exact solution, we will introduce the error which is of order −p ln( ) for p > 0, < q < This type of error is not introduced in many related results The paper is structured as follows In Section 2, we present the solution of the 2-D initial inverse heat problem In Section 3, motivated by the idea coming from [30], we establish stability results for our problem and propose a new strategy with Hă older estimates The proofs of the main theoretical results will be given in Section In Section 5, the numerical results of our regularized method are presented, which proved the effectiveness of our method q The ill-posed initial inverse heat problem Throughout this paper, we denote < , >, by the inner product and the norm in L2 respectively Let us first make clear what a weak solution of the Problem (1) is We call a function u ∈ C [0, T ]; L2 (I) ∩ C ((0, T ); L2 (I)) to be a weak solution for the Problem (1) if d u(., , t), W − ∆u(., , t), W = f (., , t), W , dt (2) for all functions W (x, y) ∈ H (I) ∩ H01 (I) In fact, it is enough to choose W in the orthogonal basis { π2 sin(px) sin(qy)}p,q≥1 and the formula (2) is equivalent to T upq (t) = e (T −t)(p2 +q ) e(s−t)(p gpq − T ∞ p,q=1 where for e(T −t)(p2 +q2 ) gpq − upq (t) = gpq = fpq (t) = fpq (s)ds, π2 π2 π2 ∀p, q ≥ t which may also be written formally as u(x, y, t) = +q ) e(s−t)(p +q ) t fpq (s)ds sin(px) sin(qy), (3) (4) π π u(x, y, t) sin(px) sin(qy)dxdy, π π g(x, y) sin(px) sin(qy)dxdy, π π f (x, y, t) sin(px) sin(qy)dxdy 0 Let the function f ∈ L2 ((0, T ); L2 (I)) and g ∈ L2 (I) be given Note that the expression (4) is the solution of Problem (1) if it exists In the following Theorem, we provide a condition of its existence Theorem 2.1 If the Problem (1) has a solution u then we have ∞ p,q=1 2 T e T (p2 +q ) gpq − e s(p2 +q ) fpq (s)ds < ∞ (5) Else if (5) holds then the Problem (1) has a unique solution Proof Suppose the Problem (1) has a solution u ∈ C([0, T ]; H01 (I)) ∩ C ((0, T ); L2 (I)), then u is defined by T ∞ u(x, y, t) = p,q=1 This implies e−(t−T )(p2 +q2 ) gpq − e−(t−s)(p +q ) t fpq (s)ds sin(px) sin(qy) (6) T upq (0) = e T (p2 +q ) gpq − es(p +q ) fpq (s)ds (7) Then ∞ u(., , 0) = p,q=1 e T T (p2 +q ) gpq − e s(p2 +q ) 2 fpq (s)ds < ∞ (8) If (5) holds, then we define v(x, y) = ∞ π2 p,q=1 T eT (p2 +q2 ) gpq − es(p +q ) L2 (I) fpq (s)ds sin(px) sin(qy) Since (8), we see that v ∈ Consider the problem ut − uxx − uyy = f (x, y, t), u(0, y, t) = u(π, y, t) = u(x, 0, t) = u(x, π, t) = 0, t ∈ (0, T ) u(x, y, 0) = v(x, y), (x, y) ∈ (0, π) × (0, π) (9) It is clear that (9) is the direct problem so it has a unique solution u (See [6]) We have t ∞ u(x, y, t) = p,q=1 e−t(p2 +q2 ) < v(x, y), sin(px) sin(qy) > + e(s−t)(p By letting t = T in (10), we obtain u(x, y, T ) = ∞ = p,q=1 ∞ = e −T (p2 +q ) e T T (p2 +q ) gpq − e s(p2 +q ) +q ) fpq (s)ds sin(px) sin(qy).(10) T fpq (s)ds + e (s−T )(p2 +q ) fpq (s)ds sin(px) sin(qy) gpq sin(px) sin(qy) p,q=1 = g(x, y) Hence, u is a solution of (1) To prove the uniqueness of the solution to the Problem (1), we refer the readers to [30] The main theoretical results Let us recall (g, f ) ∈ L2 (I) × L2 (0, T : L2 (I)) be the exact data Assume that the noisy data (g , f ) ∈ L2 (I) × L2 (0, T : L2 (I)) satisfies T g −g ≤ ; f (., t) − f (., t) L2 (0,T :L2 (I)) = f (., t) − f (., t) dt ≤ In this paper, we establish a method to regularize the Problem (1) Infact, letting h ≥ be a fixed number, we denote Th = hT, Th−1 = (h − 1)T Let β be a positive constant (is called parameter regularization) which depends on such that lim →0 β = We consider the following well-posed problem 2 ∞ e−Th (p +q ) fpq (t) sin(px) sin(qy), (x, y, t) ∈ I × (0, Th ), ut − uxx − uyy = β(p2 + q ) + e−Th (p2 +q2 ) p,q=1 u (0, y, t) = u (π, y, t) = u (x, 0, t) = u (x, π, t) = 0, (x, y, t) ∈ I × [0, Th ] (11) 2 ∞ e−Th (p +q ) gpq sin(px) sin(qy), (x, y) ∈ I u (x, y, T ) = h β(p2 + q ) + e−Th (p2 +q2 ) p,q=1 and the regularized problem 2 ∞ e−Th (p +q ) v − v − v = f (t) sin(px) sin(qy), (x, y, t) ∈ I × (0, Th ), t xx yy β(p2 + q ) + e−Th (p2 +q2 ) pq p,q=1 v (0, y, t) = v (π, y, t) = v (x, 0, t) = v (x, π, t) = 0, (x, y, t) ∈ I × [0, Th ] (12) 2 ∞ e−Th (p +q ) g sin(px) sin(qy), (x, y) ∈ I v (x, y, T ) = h β(p2 + q ) + e−Th (p2 +q2 ) pq p,q=1 where gpq , gpq , fpq (t) are defined by gpq = gpq = fpq (t) = fpq (t) = π2 π2 π2 π2 π π g(x, y) sin(px) sin(qy)dxdy, 0 π π g (x, y) sin(px) sin(qy)dxdy, 0 π π f (x, y, t) sin(px) sin(qy)dxdy, 0 π π f (x, y, t) sin(px) sin(qy)dxdy 0 and then we take v (., , t + Th−1 ) of (12) as an approximation to u(., , t) The main purpose of our method is of considering the error v (., , t + Th−1 ) − u(., , t) Notice readers that when h = 1, Problem (11) is considered in [30] (See page 874.) However, the error estimates in [30] is only the logarithmic form (See Theorem 2.3 ,page 878,[30]) Thus, our present work significantly improves related results on the backward problem (we mention this in Remark 1) Theorem 3.1 Let g ∈ L2 (I) be the function such that gxx + gyy < ∞ (i)If the regularized solution u (x, y, Th−1 ) converges in L2 (I), then the Problem (1) has a unique solution u Furthermore, u (x, y, t + Th−1 ) converges to u(x, y, t) uniformly in t as tends to zero (ii)If there is a positive constant A1 such that uxx (., , 0) + uyy (., , 0) ≤ A1 then with β = one has v (., , Th−1 ) − u(., , 0) ≤ √ 2+T 1− h hT + ln( hT ) h + A1 hT ln( hT ) (13) (iii) If there are positive constants γ ∈ (0, hT ] and A2 such that π2 ∞ p,q=1 |(p2 + q )eγ(p +q ) < u(x, y, t), sin(px) sin(qy) > |2 ≤ A22 , (14) for all t ∈ [0, T ], then with β = hT T +γ , we have B2 v (., , t + Th−1 ) − u(., , t) ≤ B γ T +γ ln γ T +γ γ hT T −1 , < γ ≤ (h − 1)T hT T +γ −1 h T ln (15) , (h − 1)T < γ ≤ hT hT T +γ m for all t ∈ [0, T ], where the function H(m, n) is defined by H(m, n) = n1− n for all ≤ m ≤ n and B2 = H(γ, Th )A2 + √ 2+T t T +γ (hT ) h T ln t hT hT T +γ (16) Remark If h = then the error (13) is similar to the results in [30]( Theorem 2.3, page −m 878) In many results concerning the backward heat, the optimal error is of order ln T where m > The error order of logarithmic form has been investigated in many recent papers, such as [2, 3, 8, 5, 7, 9, 25, 26, 27, 28, 29, 30] The logarithmic type estimate is, in general, much worse than any Hă older type estimate, i.e q for some q > To retain the Hă older order in [0, T ], we should introduce a different priori assumption on u such as (14) As we know, the convergence rate of γ T +γ γ hT T ln −1 or hT T +γ γ T +γ T ln hT T +γ −1 h is faster than γ when → This implies that the error (15) is the Hă older type estimate k , for any < k < T +γ effective and useful The method in this paper is inherently restricted to the square domain, and does not apply to more general domain due to its reliance on the Fourier method We hope that in the future, we can derive similar estimates without resorting to the Fourier method The method can be applied to fairly general domains is very difficult and will be presented in next reports Proof of the main theoretical results Lemma For < α < en, ≤ m ≤ n, the following inequalities are true a) n ≤ −xn αx + e α + ln( αn ) m e−xm n ≤ H(m, n)α n −1 + ln( ) b) αx + e−xn α m where the function H(m, n) is defined by H(m, n) = n1− n Proof The proof of part (a) of this lemma can be found in [27] (17) m −1 n , (18) Proof of (b) We have e−xm αx + e−xn e−xm = αx + e−xn ≤ ≤ m n αx + e−xn 1− m n αx + e−xn n 1− m n n α + ln( αn ) ≤ H(m, n)α 1− m m −1 n n ln α m −1 n Lemma The Problem (12) has a unique solution v ∈ C [0, T ]; H01 (I) ∩ C ((0, T ); L2 (I)) satisfying v (x, y, t) = ∞ 2 e−t(p +q ) g − β(p2 + q ) + e−Th (p2 +q2 ) ) pq = p,q=1 sin(px) sin(qy) Th t 2 e(s−t−Th )(p +q ) fpq (s)ds β(p2 + q ) + e−Th (p2 +q2 ) ≤ t ≤ Th (19) The solution also depends continuously on g ∈ L2 (I) and we have u (., , t + Th−1 ) − v (., , t + Th−1 ) ≤ √ 2+T β hT + ln( hT β ) t−T hT T −t hT , ≤ t ≤ T (20) Proof The proof of this lemma is divided into two steps Step 1.The existence, the uniqueness and the stability of a solution of (12) can be found in the paper [30](Theorem 2.1, p 875) Step We shall prove (20) In fact, from u , v are solutions of Problem (1) corresponding to the exact data g and noisy data g respectively, we have u (x, y, t) = ∞ = p,q=1 2 e−t(p +q ) gpq − β(p2 + q ) + e−Th (p2 +q2 ) sin(px) sin(qy) Th t 2 e(s−t−Th )(p +q ) fpq (s)ds β(p2 + q ) + e−Th (p2 +q2 ) ≤ t ≤ Th (21) v (x, y, t) = ∞ = p,q=1 2 e−t(p +q ) g − β(p2 + q ) + e−Th (p2 +q2 ) pq sin(px) sin(qy) ≤ t ≤ Th Th t 2 e(s−t−Th )(p +q ) f (s)ds β(p2 + q ) + e−Th (p2 +q2 ) pq (22) Thus using Lemma 1b, we get u (., , t) − v (., , t) π2 = π2 + ≤ β ∞ p,q=1 ∞ Th 2 e(s−t−Th )(p +q ) (f (s) − fpq (s)ds) β(p2 + q ) + e−Th (p2 +q2 ) pq t p,q=1 2t −2 Th e−t(p +q ) (g − gpq ) β(p2 + q ) + e−Th (p2 +q2 ) pq 2− T2t Th h + ln( Tβh ) g −g 2 g −g 2 Th + t (fpq (s) − fpq (s)ds) By replace t with t = t + Th−1 , we obtain u (., , t + Th−1 ) − v (., , t + Th−1 ) ≤β 2t+2Th−1 −2 Th =β 2− Th + ln( Tβh ) ≤ (2 + T ) β 2t−2T hT Th + t+Th−1 (fpq (s) − fpq (s)ds) 2T −2t hT hT + ln( hT β ) 2t−2T hT 2t+2Th−1 Th g −g + T f (., t) − f (, t) L2 (0,T :L2 (I)) , 2T −2t hT hT + ln( hT β ) The lemma is proved Now we are in a position to prove the main result 4.1 Proof of Theorem 3.1 4.1.1 Proof of Part (i) Assume that lim u (x, y, Th−1 ) = w(x, y) ∈ L2 (I) exists We put →0 ∞ e−t(p u(x, y, t) = +q ) t wpq + e(s−t)(p +q ) p,q=1 fpq (s)ds sin(px) sin(qy), ≤ t ≤ T (23) where π π w(x, y) sin(px) sin(qy)dxdy π2 0 We shall prove that u(x, y, t) is the unique solution of the Problem (1) First, it is clear to see that u satisfies the system wpq = ut − uxx − uyy = f (x, y, t), (x, y, t) ∈ I × (0, T ) u(0, y, t) = u(π, y, t) = u(x, 0, t) = u(x, π, t) = 0, (x, y, t) ∈ I × (0, T ) (24) Next, we show that u(x, y, T ) = g(x, y) In fact, we have the formula of u (x, y, t + Th−1 ) u (x, y, t + Th−1 ) = ∞ e−(t+Th−1 )(p p,q=1 +q ) t+Th−1 upq (Th−1 ) − Th−1 2 e(s−t−Th−1 −Th )(p +q ) fpq (s)ds sin(px) sin(qy) β(p2 + q ) + e−Th (p2 +q2 ) (25) where upq (Th−1 ) = π2 π π u (x, y, Th−1 ) sin(px) sin(qy)dxdy 0 Combining (23) and (25), we obtain < u (x, y, t + Th−1 ) − u(x, y, t), sin(px) sin(qy) > =e −t(p2 +q ) t − e(s−t)(p t+Th−1 upq (Th−1 ) − wpq (t) + +q ) e(s−t−Th−1 −Th )(p +q ) fpq (s)ds β(p2 + q ) + e−Th (p2 +q2 ) Th−1 fpq (s)ds (26) By change variables s1 = s − Th−1 , the second term of the right hand side of (26) can be rewritten as follows t+Th−1 Th−1 2 e(s−t−Th−1 −Th )(p +q ) fpq (s)ds = β(p2 + q ) + e−Th (p2 +q2 ) e(s1 +Th−1 −t−Th−1 −Th )(p +q ) fpq (s1 )ds1 β(p2 + q ) + e−Th (p2 +q2 ) t e(s−t−Th )(p +q ) fpq (s)ds β(p2 + q ) + e−Th (p2 +q2 ) = t 2 Combining (26) and (27), we have < u (x, y, t + Th−1 ) − u(x, y, t), sin(px) sin(qy) > = e−t(p +q ) t − upq (Th−1 ) − wpq (t) βe(s−t)(p +q ) (p2 + q ) fpq (s)ds β(p2 + q ) + e−Th (p2 +q2 ) By using the inequality (17), we get upq (Th−1 ) − wpq (t) |< u (x, y, t + Th−1 ) − u(x, y, t), sin(px) sin(qy) >| ≤ t Th + ln( Tβh ) (p2 + q )|fpq |(s)ds By using the inequality (a + b)2 ≤ 2(a2 + b2 ), we arrive at u (x, y, t + Th−1 ) − u(x, y, t) ∞ ≤ π2 + π2 T p,q=1 (upq (Th−1 ) − wpq )2 ∞ p,q=1 t Th ln( Tβh ) (p + q )|fpq |(s) ≤ u (., , Th−1 ) − w(., ) T π2 + + 2T t ln( Tβh ) ln( Tβh ) ∞ p,q=1 ≤ u (., , Th−1 ) − w(., ) ds 2 Th Th 2 (p2 + q )2 fpq (s)ds 2 T fxx (., , s) + fyy (., , s) ds (27) It gives lim u (x, y, t + Th−1 ) = u(x, y, t) Thus lim u (x, y, Th ) = u(x, y, T ) On the other hand, →0 →0 we have u (., , Th ) − g(., ) = = ≤ π2 π2 ∞ p,q=1 ∞ β(p2 + q ) β(p2 + q ) + e−Th (p2 +q2 ) ) p,q=1 π2 e−Th (p +q ) −1 β(p2 + q ) + e−Th (p2 +q2 ) ) Th 2 gpq 2 gpq ∞ (p2 + q )2 gpq ln( Tβh ) p,q=1 Thus u (., , Th ) − g(., ) π2 ≤ ∞ Th = (p2 + q )2 gpq ln( Tβh ) p,q=1 Th ln( Tβh ) gxx + gyy Therefore lim u (x, y, Th ) = g(x, y) This implies u(x, y, T ) = g(x, y) Hence, u(x, t) is the unique →0 solution of the Problem (1) We also see that u (x, y, t + Th−1 ) converges to u(x, y, t) uniformly in t 4.1.2 Proof of Part (ii) Using the triangle inequality, we get u(., , 0) − v (., , Th−1 ) ≤ u(., 0) − u (., , Th−1 ) + u (., , Th−1 ) − v (., , Th−1 ) (28) First, from (20) in Theorem 2, we estimate u (., , Th−1 ) − v (., , Th−1 ) ≤ √ 2+T β −1 h h hT + ln( hT β ) (29) Next, we continue to estimate the error u(., , 0) − u (., , Th−1 ) From (21), we have u (x, y, Th−1 ) ∞ = p,q=1 2 e−Th−1 (p +q ) gpq − β(p2 + q ) + e−Th (p2 +q2 ) ) Th Th−1 2 e(s−Th−1 −Th )(p +q ) fpq (s)ds β(p2 + q ) + e−Th (p2 +q2 ) sin(px) sin(qy) ∞ = 2 e−Th−1 (p +q ) β(p2 + q ) + e−Th (p2 +q2 ) ) p,q=1 Th gpq − e(s−Th )(p +q ) fpq (s)ds sin(px) sin(qy) Th−1 (30) On the other hand, by direct calculation, we get T Th e (s−Th )(p2 +q ) e(s−T )(p fpq (s)ds = Th−1 10 +q ) fpq (s)ds and since ∞ eT (p u(x, y, 0) = +q ) p,q=1 gives us T gpq − fpq (s)ds sin(px) sin(qy) (31) < u(x, y, 0), sin(px) sin(qy) > (32) e(s−T )(p +q ) T gpq − e(s−T )(p +q ) fpq (s)ds = e−T (p +q ) Combining (30) and (32), we obtain < u (x, y, Th−1 ), sin(px) sin(qy) > 2 = e−Th−1 (p +q ) −T (p2 +q ) < u(x, y, 0), sin(px) sin(qy) > +q ) e −T (p 2 β(p + q ) + e h ) = e−Th (p +q ) < u(x, y, 0), sin(px) sin(qy) > β(p2 + q ) + e−Th (p2 +q2 ) ) 2 (33) From (31) and (33), we have < u (x, y, Th−1 ) − u(x, y, 0), sin(px) sin(qy) > −β(p2 + q ) = < u(x, y, 0), sin(px) sin(qy) > β(p2 + q ) + e−Th (p2 +q2 ) ) It follows from (17) that u (., , Th−1 ) − u(., , 0) = = ≤ π2 ∞ p,q=1 ∞ π2 | < u (x, y, Th−1 ) − u(x, y, 0), sin(px) sin(qy) > |2 p,q=1 Th ln( Tβh ) β(p2 + q ) β(p2 + q ) + e−Th (p2 +q2 ) ) | < u(x, y, 0), sin(px) sin(qy) > |2 uxx (., , 0) + uyy (., , 0) Hence u (., , Th−1 ) − u(., , 0) ≤ From β = hT A1 ln( hT ) β (34) and combining (28), (29) and (34), we conclude that u(., , 0) − v (., , Th−1 ) ≤ ≤ √ √ 2+T β 2+T 11 −1 h 1− h hT + ln( hT β ) hT + ln( hT ) h + hT A1 ln( hT β ) + hT A1 ln( hT ) h 4.1.3 Proof of Part (iii) It follows from (21) that u (x, y, t + Th−1 ) = ∞ = p,q=1 e−(t+Th−1 )(p +q ) gpq − β(p2 + q ) + e−Th (p2 +q2 ) ) ∞ )(p2 +q ) = ∞ gpq − t+Th−1 e(s−Th )(p +q ) fpq (s)ds sin(px) sin(qy) t+Th−1 T e−(t+Th−1 )(p +q ) β(p2 + q ) + e−Th (p2 +q2 ) ) p,q=1 e(s−t−Th−1 −Th )(p +q ) fpq (s)ds sin(px) sin(qy) β(p2 + q ) + e−Th (p2 +q2 ) Th e−(t+Th−1 β(p2 + q ) + e−Th (p2 +q2 ) ) p,q=1 = Th gpq − e(s−T )(p +q ) fpq (s)ds sin(px) sin(qy) t Moreover, from T gpq − e(s−T )(p +q ) fpq (s)ds = e(t−T )(p +q ) < u(x, y, t), sin(px) sin(qy) > t leads to | < u (x, y, t + Th−1 ) − u(x, y, t), sin(px) sin(qy) > | 2 β(p2 + q )e−γ(p +q ) γ(p2 +q ) = < u(x, y, t), sin(px) sin(qy) > | +q ) |e −T (p 2 h β(p + q ) + e ) ≤ H(γ, Th )β Th ln( ) β γ Th γ −1 Th |(p2 + q )eγ(p +q ) < u(x, y, t), sin(px) sin(qy) > | Therefore, we have u (., , t + Th−1 ) − u(., , t) = π2 ∞ p,q=1 | < u (x, y, t + Th−1 ) − u(x, y, t), sin(px) sin(qy) > |2 π2 γ H(γ, Th )2 β Th ≤ ≤ H(γ, Th )2 β ≤ H(γ, Th )2 β Tγ h Tγ h ln( ln( Th ln( ) β Th ) β Th ) β 2γ −2 Th 2γ −2 Th 2γ −2 Th ∞ p,q=1 π2 |(p2 + q )eγ(p +q ) < u(x, y, t), sin(px) sin(qy) > |2 ∞ p,q=1 |(p2 + q )eγ(p +q ) < u(x, y, t), sin(px) sin(qy) > |2 A22 Thus u (., , t + Th−1 ) − u(., , t) ≤ H(γ, Th )β 12 γ Th Th ln( ) β γ −1 Th A2 (35) Using the triangle inequality and (20), (35), we get u(., , t) − v (., , t + Th−1 ) ≤ u(., t) − u (., , t + Th−1 ) + u (., , t + Th−1 ) − v (., t + Th−1 ) ≤ H(γ, Th )β γ hT γ T +γ ≤ H(γ, Th ) √ t+γ T +γ + 2+T γ T +γ ≤ H(γ, Th ) γ hT hT ln( ) β ln −1 A2 + γ hT T + Tβ T −t hT hT + ln( hT β ) t−T hT −1 A2 hT T +γ T −t hT + ln( hT hT ) ln √ hT T +γ γ hT T −1 A2 + hT T +γ √ 2+T t+γ T +γ (hT ) h T ln( hT T +γ t−T hT ) (36) If < γ ≤ (h − 1)T , then ln ≤ ln T γ hT T and (36) can be written as follows hT T +γ −1 H(γ, Th )A2 + hT T +γ −1 h −1 ≥ ln hT T +γ u(., , t) − v (., , t + Th−1 ) γ T +γ γ hT T √ 2+T t T +γ (hT ) h t hT T ln hT T +γ If (h − 1)T < γ ≤ hT then (36) becomes u(., , t) − v (., , t + Th−1 ) ≤ γ T +γ ln T −1 h hT T +γ H(γ, Th )A2 + √ 2+T t T +γ (hT ) h ln T hT T +γ t hT Numerical Experiments In reality, we not know exact solution of the problem, hence, the priori assumptions (ii) and (iii) become implicit and may be skipped in practice Thus, the regularized solution is expected that it is a reasonable solution of the problem At the time, its certificate of use depends on its convergence We consider two examples of Problem (1) in the region I = (0, π) × (0, π): ut − uxx − uyy = f (x, y, t), (x, y, t) ∈ I × (0, 1) u(0, y, t) = u(π, y, t) = u(x, 0, t) = u(x, π, t) = 0, (x, y, t) ∈ I × (0, 1) (37) u(x, y, 1) = g(x, y), (x, y) ∈ I Let g ρ (x, y) be the disturbed measure data such that ||g(x, y) − g ρ (x, y)|| ≤ ρ 13 a) In the first example, we take: g(x, y) = e−xy sin(2x) sin(y), ρ g ρ (x, y) = g(x, y) + · rand() π f (x, y, t) = −e−txy xy + t2 y + t2 x2 − sin y sin 2x − − 2tx cos y sin 2x − 4ty cos 2x sin y f ρ (x, y, t) = −e−txy xy + t2 y + t2 x2 − sin y sin 2x − − 2tx cos y sin 2x − 4ty cos 2x sin y + ρ · rand() π The problem has exact solution u(x, y, t) = e−txy sin(2x) sin(y) Select h = 1.0 + 10−5 , γ = h/100 For each regularization parameter β1 = error for regularized solution is (from (ii)) and β2 = δ k, (ρ) = h 1+γ (from (iii)) we have the relative ||uk, − u|| ||u|| where uk, is the regularized solution by parameter βk Let’s observe the convergent behavior of the regularized solution regarding the changes of parameter and magnitude ρ of data error Using our proposed method with two types of regularization parameter βk , we set up the computation to find regularized solution at time t = 0, ρ = 10−r , r = 0, 1, 3, and parameter = 10−k , k = Figure visualizes the exact solution at t = The computational result is showed in type of relative error in Table and in 3D graphs in Figure We see that the second method yields more accurate result though not very significant Regularized solution converges as decreases but diverges when becomes smaller than ρ Therefore, we make an assumption that the best regularized solution for problem with perturbed data is obtained when ≈ ρ 10−1 10−2 10−3 10−4 10−5 10−6 δ 1, (ρ) 9.867E-01 8.812E-01 4.258E-01 6.884E-02 1.061E-02 6.526E-02 ρ=0 δ 2, (ρ) 9.870E-01 8.859E-01 4.426E-01 7.493E-02 1.069E-02 5.770E-02 ρ = 10−1 δ 1, (ρ) δ 2, (ρ) 9.807E-01 9.831E-01 8.306E-01 8.540E-01 7.145E-01 6.652E-01 6.735E+00 5.675E+00 6.020E+01 4.699E+01 diverged diverged ρ = 10−3 δ 1, (ρ) δ 2, (ρ) 9.866E-01 9.869E-01 8.807E-01 8.852E-01 4.233E-01 4.395E-01 8.963E-02 8.511E-02 4.754E-01 4.371E-01 3.947E+00 3.863E+00 ρ = 10−5 δ 1, (ρ) δ 2, (ρ) 9.867E-01 9.870E-01 8.812E-01 8.859E-01 4.258E-01 4.426E-01 6.879E-02 7.491E-02 1.041E-02 1.297E-02 6.572E-02 8.443E-02 Table 1: Relative error for regularized solutions at t = 0; noise amplitude ρ = (exact data), ρ = 10−r , r = 1, 3, (disturbed data) and = 10−k , k = 14 1 0.5 0.5 -0.5 U -1 -0.5 0.5 -1 3.5 1.5 2.5 2.5 1.5 X 0.5 Y 3.5 Figure 1: Example 1, exact solutions at t = b) Consider the second example with given functions: g(x, y) = x(π − x)y(π − y), ρ g ρ (x, y) = g(x, y) + · rand() π f (x, y, t) = x(π − x) y(π − y) − 2t + y(π − y) f (x, y, t) = x(π − x) y(π − y) − 2t + y(π − y) x(π − x) − 2t ρ x(π − x) − 2t + · rand() π It is easily to check that the exact solution u(x, y, t) = tx(π − x)y(π − y) Hence, we have u(x, y, 0) = We will find the regularized solution at t = as the approximation to plane z = Using presumption in the first example, we will set = ρ and choose β = β1 = With noise amplitude ρ = 10−r , r = 1, 3, the calculated relative errors for regularized solution are 0.349, 0.2172 and 0.1154 respectively We also visualize 3D graphs of these regularized solutions in Figure As expected, they converge smoothly to the exact solution z = 15 1 1 0.5 0.5 0.5 0.5 -0.5 U -0.5 U 0 -1 -1 -0.5 -0.5 0 0.5 0.5 -1 3.5 -1 3.5 1.5 2.5 0.5 X (a) u1, , 2.5 Y 2.5 1.5 1.5 0.5 X 3.5 (b) u2, , = 10−3 Y 2.5 1.5 3 3.5 = 10−3 1 1 0.5 0.5 0.5 0.5 -0.5 U -0.5 U 0 -1 -1 -0.5 -0.5 0 0.5 0.5 -1 3.5 -1 3.5 1.5 2.5 0.5 X (c) u1, , 2.5 Y 2.5 1.5 1.5 0.5 X 3.5 (d) u2, , = 10−4 Y 2.5 1.5 3 3.5 = 10−4 1.5 1.5 1.5 1.5 1 0.5 0.5 1 0 0.5 0.5 -0.5 U -0.5 U -1 -0.5 -1 -0.5 0 0.5 -1 3.5 2.5 1.5 X (e) u1, , 0.5 1.5 2.5 3.5 1.5 0.5 -1 2.5 Y 2.5 1.5 X 3.5 (f) u2, , = 10−5 0.5 Y 3.5 = 10−5 Figure 2: Example 1, convergence and divergence of regularized solutions at t = 0, noise amplitude ρ = 10−3 16 (a) = ρ = 10−1 (b) (c) = ρ = 10−3 = ρ = 10−5 Figure 3: Example 2, convergence of regularized solutions at t = 17 Conclusion In this paper, we considered an improved regularization method for initial inverse heat problem in 2D case In the theoretical results, we obtained the error estimation of Hă older type m lnn ( 1k ) for m, n, k > based on the smooth assumptions of the exact solution These estimates improve the results in many earlier works Finally, in the numerical experiments, two numerical examples are presented to show that our proposed method is effective In future work, we will consider the regularized problem for the following problem ∂u ∂t = b(x, y, t)ux x + c(x, y, t)uy y + f (x, y, t), (x, y, t) ∈ I × (0, T ) (38) u|∂Ω = 0, t ∈ (0, T ) u(x, y, T ) = g(x, y), (x, y, t) ∈ I × (0, T ) where b(x, y, t) and c(x, y, t) are functions depending on variables x, y, t Acknowledgments This project was supported by Institute for Computational Science and Technology HoChiMinh city under project named ”Inverse parabolic equation and application to groundwater polution source” The authors would like to thank the anonymous referees for their valuable suggestions and comments leading to the improvement of our manuscript We would like to thank Prof Dang Duc Trong for his most helpful comments and discussion on this paper References [1] K A Ames, R J Hughes; Structural Stability for Ill-Posed Problems in Banach Space, Semigroup Forum, Vol 70 (2005), N0 1, 127-145 [2] A S Carraso; Logarithmic convexity and the “slow 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M.Wadsworth,Solution of inverse diffusion problems by operator-splitting methods, Appl Math modelling 26 (2002) 1003–1018 [16] K Masood, F.D Zaman, An initial inverse problem of heat conduction in a plate with... 17 Conclusion In this paper, we considered an improved regularization method for initial inverse heat problem in 2D case In the theoretical results, we obtained the error estimation of Hă older