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Applied Mathematical Modelling 35 (2011) 5673–5690 Contents lists available at ScienceDirect Applied Mathematical Modelling journal homepage: www.elsevier.com/locate/apm Notes on a new approximate solution of 2-D heat equation backward in time Nguyen Huy Tuan a,d,⇑, Dang Duc Trong b, Pham Hoang Quan c a Institute for Computational Science and Technology, Quarter 6, Linh Trung Ward, Thu Duc District, HoChiMinh City, Viet Nam Department of Mathematics, University of Natural Science, Vietnam National University, 227 Nguyen Van Cu, Q.5, HoChiMinh City, Viet Nam c Department of Mathematics and Applications, Sai Gon University, 273 An Duong Vuong, Q.5, HoChiMinh City, Viet Nam d Department of Mathematics, Hochiminh University of Technology, 144/24 Dien Bien Phu, P.25, Binh Thanh Dist., HoChiMinh City, Viet Nam b a r t i c l e i n f o Article history: Received 16 September 2010 Received in revised form 29 April 2011 Accepted May 2011 Available online 20 May 2011 Keywords: Backward heat problem Nonhomogeneous heat equation Ill-posed problem Quasi-boundary value method Quasi-reversibility method Error estimate a b s t r a c t In this paper, we consider a backward heat problem that appears in many applications This problem is ill-posed The solution of the problem as the solution exhibits unstable dependence on the given data functions Using a new regularization method, we regularize the problem and get some new error estimates Some numerical tests illustrate that the proposed method is feasible and effective This work is a generalization of many recent papers, including the earlier paper [A new regularized method for two dimensional nonhomogeneous backward heat problem, Appl Math Comput 215(3) (2009) 873–880] and some other authors such as Chu-Li Fu et al [1–3], Campbell et al [4] Ó 2011 Elsevier Inc All rights reserved Introduction Let T be a positive number We consider the problem of finding the temperature u(x, y, t), (x, y, t) X  [0; T] such that ðx; y; tÞ X  ð0; TÞ; > < ut uxx uyy ẳ f x; y; tị; u0; y; tị ẳ up; y; tị ẳ ux; 0; tị ¼ uðx; p; tÞ ¼ 0; ðx; y; tÞ X 0; Tị; > : ux; y; Tị ẳ gðx; yÞ; ðx; yÞ X; ð1Þ where X = (0, p)  (0, p) and g(x, y), f(x, y, t) are given The problem is called the backward heat problem (BHP for short), or the final-boundary value problem It is known in general that the problem is ill-posed, i.e., a solution does not always exist, and in the case of existence, it does not depend continuously on the given datum In fact, from a small noise contaminated physical measurement, the corresponding solution may have a large error It makes the numerical computation difficult Hence, a regularization is in order The homogeneous backward heat problems, i.e the case f = 0, has been studied by many authors in recent years In a few words, we mention Ames and Payne [5], Lattes and Lions [6], Showalter [7], who approximated the BHP by quasireversibility method; Tautenhahn and Schroter [8] who established an optimal error estimate for a BHP; Seidman [9] ⇑ Corresponding author at: Institute for Computational Science and Technology, Quarter 6, Linh Trung Ward, Thu Duc District, HoChiMinh City, Viet Nam E-mail address: tuanhuy_bs@yahoo.com (N.H Tuan) 0307-904X/$ - see front matter Ó 2011 Elsevier Inc All rights reserved doi:10.1016/j.apm.2011.05.010 5674 N.H Tuan et al / Applied Mathematical Modelling 35 (2011) 5673–5690 who established an optimal filtering method; and Hao [10] who studied a modification method We also refer to various other works of Chang et al [11], Chu-Li Fu et al [1–3], Campbell et al [4], Lien et al [12], Murniz et al [13], Dokuchaev et al [14], and Engl et al [15] Recently, the 1-D version of the problem in a infinite strip has been considered in [2,10] Physically, this problem arises from the requirement of recovering the heat temperature at some earlier time using the knowledge about the final temperature The problem is also involved to the situation of a particle moving in a environment with constant diffusion coefficient (see [16]) when one asks to determine the particle position history from its current place The interest of backward heat equations also comes from financial mathematics, where the celebrated BlackScholes model [17] for call option can be transformed into a backward parabolic equation whose form is related closely to backward heat equations Although there are many papers on the homogeneous backward heat equation, the result on the inhomogeneous case is very scarce while the inhomogeneous case is, of course, more general and nearer to practical application than the homogeneous one Shortly, it allows the appearance of some heat source which is inevitable in nature In the present paper, a modified method is used for solving 2-D backward heat problem, which will improve the stability results in some previous papers In many earlier works, we find that only logarithmic type estimates in L2-norm are available; and estimates of Hölder type are very rate (see Remark for more detail comparisons) In our method, corresponding to different levels of the smoothness of the exact solution, the convergence rates will be improved gradually Under some suitable conditions on the exact solution u, we shall introduce the error estimate of order p, (p > 0) This is a significant improvement in comparison with [1,10,18,8,12,3] Some comments on the usefulness of this method are given in Remarks The remainder of the paper is organized as follows Section is devoted to a short presentation of a ill-posed problem (1) In Section 3, we shall construct the regularized solution and show that it works even with very weak condition on the exact solution Under some assumptions on the exact solution, some error estimates are derived Finally, a numerical experiment is given in Section to illuminate the effect of our method The ill-posed backward heat problem Throughout this paper, we denote hÁ,Ái, kÁk by the inner product and the norm in L2(X) Let us first make clear what a weak solution of the Problem (1) is We call a function u C ½0; T; H10 ðXÞ \ C ðð0; TÞ; H2 ðXÞÞ to be a weak solution for Problem (1) if d huðÁ; Á; tÞ; WiL2 ðXÞ À huðÁ; Á; tÞ; DWiL2 Xị ẳ hf ; ; tị; WiL2 Xị ; dt ð2Þ for all functions W H2 ðXÞ \ H10 ðXÞ In fact, it is enough to choose W in the orthogonal basis {sin(nx)sin(my)}n,mP1 and the formula (2) reduces to unm tị ẳ eTtịn ỵm2 ị g nm Z T estịn ỵm2 ị fnm sịds; 8m; n P 1; ð3Þ fnm ðsÞds sin nx sin my; ð4Þ t which may also be written formally as Z X 2 etTịn ỵm ị g nm ux; y; tị ẳ fnm sị ẳ ỵm2 ị etsịn t n;mẳ1 where T Z pZ p f ðx; y; sÞ sin nx sin mydx dy; Z p0 Z p0 g nm ẳ gx; yị sin nx sin mydx dy; p 0 Z pZ p unm tị ẳ ux; y; tị sin nx sin mydx dy: p p 0 Note that if the exact solution u is smooth then the exact data (f,g) is smooth also However, the real data, which come from practical measure, is often discrete and non-smooth We shall therefore always assume that f L2(0,T);L2(X)) and g L2(X) and the error of the data is given on L2 only 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 Problem (1) has a unique solution u if and only if Z X 2 eTn ỵm ị g nm n;mẳ1 T esn ỵm2 ị 2 fnm ðsÞds < 1: ð5Þ Proof Suppose the Problem (1) has a solution u Cð½0; T; H10 ðXÞÞ \ C ðð0; TÞ; L2 ðXÞÞ, then u can be formulated in the frequency domain N.H Tuan et al / Applied Mathematical Modelling 35 (2011) 5673–5690 Z X 2 etTịn ỵm ị g nm ux; y; tị ẳ T etsịn ỵm2 ị fnm sịds sin nx sin my: 5675 6ị t n;mẳ1 This implies that ỵm2 ị unm 0ị ẳ eTn g nm Z T esn ỵm2 ị fnm ðsÞds: ð7Þ Then X kuðÁ; Á; 0Þk2 ẳ eTn ỵm2 ị g nm n;mẳ1 Z T ỵm2 ị esn 2 fnm sịds < 1: If (5) holds, then define v(x,y) be as the function Z X 2 eTn ỵm ị g nm v x; yị ẳ T esn þm2 Þ fnm ðsÞds sin nx sin my L2 Xị: n;mẳ1 Consider the problem > < ut uxx uyy ẳ f x; y; tị; u0; y; tị ẳ up; y; tị ẳ ux; 0; tị ẳ ux; p; tị ẳ 0; t 0; Tị; > : ux; y; 0ị ẳ v x; yị; ðx; yÞ ð0; pÞ Â ð0; pÞ: ð8Þ It is clear that (8) is the direct problem so it has a unique solution u (see [16]) We have Z t X 2 2 eÀtðn þm Þ < v ðx; yÞ; sin nx sin my > ỵ estịn ỵm ị fnm sịds sin nx sin my: ux; y; tị ẳ 9ị n;mẳ1 Let t = T in (9), we have Z 2 2 ux; y; Tị ẳ eTn ỵm ị eTn ỵm ị g nm T ỵm2 ị esn Z fnm sịds ỵ X ẳ T esTịn ỵm2 ị fnm sịds sin nx sin my g nm sin nx sin my ¼ gx; yị: n;mẳ1 Hence, u is the unique solution of (1) h Theorem Problem (1) has at most one solution in Cẵ0; T; H10 Xịị \ C 0; TÞ; L2 ðXÞÞ Proof Let u(x, y, t), v(x, y, t) be two solutions of Problem (1) such that u; v Cẵ0; T; H10 Xịị \ C 0; TÞ; L2 ðXÞÞ Put w(x,y,t) = u(x,y,t) À v(x,y,t) Then w satisfies the equation ðx; y; tÞ X  0; Tị; X ẳ 0; pị 0; pị; wt wxx wyy ẳ 0; w0; y; tị ẳ wp; y; tị ẳ wx; 0; tị ẳ wx; p; tị ẳ 0; wx; y; Tị ẳ 0; Now, setting Gtị ẳ G0 tị ẳ Rp Rp 0 Z pZ p ðx; y; tÞ X  ½0; T: x; y X: w2 ðx; y; tÞdx dyð0 t TÞ By direct computation we get wx; y; tịwt x; y; tịdx dy ẳ Z pZ p wðx; y; tÞðwxx ðx; y; tÞ þ wyy ðx; y; tÞÞdx dy: Using the Green formula, we obtain G0 tị ẳ Z pZ p 0 w2x x; y; tị ỵ w2y x; y; tÞÞdx dy: ð10Þ Taking the derivative of G0 (t), one has G00 tị ẳ Z pZ p wx x; y; tịwxt x; y; tị ỵ wy ðx; y; tÞwyt ðx; y; tÞ dx dy: Using the technique of integration by parts, we get G00 ðtÞ ¼ Z pZ p À 0 Á wxx x; y; tịwt x; y; tị ỵ wyy x; y; tịịwt x; y; tị dx dy ẳ Z pZ p 0 wxx x; y; tị ỵ wyy x; y; tÞÞ2 dx dy: ð11Þ 5676 N.H Tuan et al / Applied Mathematical Modelling 35 (2011) 5673–5690 Now, from (10) and applying the Holder inequality, we have Z p Z p 0 Z pZ p w2x ðx; y; tị ỵ w2y x; y; tị dx dy ẳ wx; y; tịwxx x; y; tị ỵ wyy x; y; tÞÞdx dy 0 Z p Z p 0 w ðx; y; tÞdx dy 12 Z p Z p 12 wxx x; y; tị ỵ wyy ðx; y; tÞÞ dxdy : ð12Þ Thus (10)–(12) imply ðG0 ðtÞÞ2 GðtÞG00 ðtÞ: Hence Theorem 11 ([16], p 65) gives G(t) = This implies that u(x, y, t) = v(x, y, t) The proof is completed In spite of the uniqueness, the problem (1) is still ill-posed and a regularization is necessary In next section, we shall establish the approximation problem h Regularization and error estimate In this section, we introduce a regularized problem as of integral equation and investigate the error estimate between the regularization solution and the exact one Let > 0, a > 0, b > Let g be a measured data satisfying kg À gk Starting from the ideas mentioned in the paper of Clark and Oppeinheimer [19], we consider the following approximate problem X ut uxx uyy ẳ n;mẳ1 fnm tị sin nx sin my; ðx; y; tÞ X  ð0; TÞ; ỵ beaTn2 ỵm2 ị u 0; y; tị ẳ u p; y; tị ẳ u x; 0; tị ẳ u x; p; tị ẳ x; y; tị X  ½0; T; X g nm sin nx sin my; ðx; yÞ X; u ðx; y; TÞ ẳ aTn2 ỵm2 ị ỵ be n;mẳ1 13ị 14ị 15ị where fnm(t), gnm are dened by fnm tị ẳ g nm Z pZ p f ðx; y; tÞ sin nx sin mydx dy; Z p0 Z p0 ¼ gðx; yÞ sin nx sin mydx dy: p p 0 and b is a regularization parameter depending on and it is chosen latter The real number a P is a constant.The case f = 0, a = is considered in [19] The major reason to choose a P is explained in Remark Theorem Let f L2((0,T);L2 (X)) and g L2 (X) Then Problem (13)–(15) has uniquely a weak solution u Cẵ0; T; L2 Xị\ L2 0; T; H10 ðXÞÞ \ C ð0; T; H10 ðXÞÞ defined as follows: u x; y; tị ẳ X n;mẳ1 Z T 2 eTtịn ỵm ị sTịn2 þm2 Þ g À e f ðsÞds sin nx sin my: nm nm 2 ỵ beaTn ỵm ị t ð16Þ The solution depends continuously on g in C([0,T];L2(X)) Proof The proof is divided into two steps In Step 1, we prove the existence and the uniqueness of a solution of problem (13)–(15) In Step 2, the stability of the solution is given Step The existence and the uniqueness of a solution of Problem (13)–(15) We divide this step into two parts Part A If u Cð½0; T; L2 ðXÞ \ L2 ð0; T; H10 ðXÞÞ \ C ð0; T; H10 ðXÞÞ satisfies (16) then u is solution of (13)–(15) We have u ðx; y; tÞ ¼ X n;m¼1 Z T 2 eTtịn ỵm ị sTịn2 ỵm2 ị g e f sịds sin nx sin my: nm ỵ beaTn2 ỵm2 Þ nm t We can verify directly that u Cẵ0; T; L2 Xị \ C 0; Tị; H10 ðXÞÞ \ L2 ð0; T; H10 ðXÞÞÞ In fact, u C ðð0; T; H10 ðXÞÞÞ Moreover, one has N.H Tuan et al / Applied Mathematical Modelling 35 (2011) 56735690 X ut x; y; tị ẳ n;mẳ1 ẳ X p n2 ỵ m2 ịeTtịn þm Þ g nm À þ beaTðn2 þm2 Þ Z t T 2 n2 ỵ m2 ịestTịn ỵm ị fnm tị sin nx sin my fnm sịds ỵ ỵm2 ị ỵm2 ị aTn aTn ỵ be ỵ be X n2 ỵ m2 ị < u x; y; tị; sin nx sin my > sin nx sin my ỵ n;mẳ1 n;mẳ1 ẳ uxx x; y; tị ỵ uyy x; y; tị ỵ X n;mẳ1 5677 fnm tị sin nx sin my ỵ beaTn2 ỵm2 ị fnm tị sin nx sin my; ỵ beaTn2 ỵm2 ị and X u x; y; Tị ẳ n;mẳ1 g nm sin nx sin my: ỵ beaTn2 ỵm2 ị So u is the solution of (13)–(15) Part B Problem (13)–(15) has at most one solution Cẵ0; T; H10 Xịị \ C ðð0; TÞ; L2 ðXÞÞ Let u(x, y, t), v(x, y, t) be two solutions of Problem (4)–(6) such that u; v Cẵ0; T; H10 Xịị \ C ðð0; TÞ; L2 ðXÞÞ Put w(x, y, t) = u(x, y, t) À v(x, y, t) Using Theorem 2, we get u(x, y, t) = v(x, y, t) The proof is completed Since Part A and Part B, we complete the proof of Step Step The solution of the problem (13)–(15) depends continuously on g in L2(X) First, to prove Step 2, we need the following lemma h Lemma For m,n,b > 0,0 c d, we have 2 c en ỵm ịc bd : ỵ ben2 ỵm2 ịd 17ị Proof of Lemma We have 2 2 2 eðn ỵm ịc en ỵm ịc en ỵm ịc dc ẳ : c b c c6 ỵm2 ịd ỵm2 ịd d ỵm2 ịd 1d n d n n ỵ ben2 ỵm2 ịd ỵ be ị ỵ be ỵ be For t s T, if letc = T À t,d = aT in (17), then 2 tÀT eTtịn ỵm ị b aT : ỵ beaTn2 þm2 Þ ð18Þ If we let c = s À t,d = aT in (17) then 2 tÀs eðsÀtÞðn þm Þ b aT : þ beaTðn2 þm2 Þ ð19Þ Let u and v be two solutions of (13)–(15) corresponding to the final values g and h From the definitions of u and v we have uðx; y; tị ẳ X n;mẳ1 Z T 2 eTtịn ỵm ị 2 g nm esTịn þm Þ fnm ðsÞds sin nx sin my þm2 ị aTn ỵ be t and v x; y; tị ẳ X n;mẳ1 Z T 2 eTtịn ỵm ị sTịn2 ỵm2 ị h e f sịds sin nx sin my; nm nm ỵ beaTn2 ỵm2 ị t where g nm ẳ hnm ẳ p2 p2 Z pZ p Z0 p Z0 p gðx; yÞ sin nx sin my; hðx; yÞ sin nx sin mydx dy: Hence uðx; y; tÞ v x; y; tị ẳ X n;mẳ1 2 eTtịn ỵm ị g hnm ị sin nx sin my: ỵ beaTn2 ỵm2 ị nm 20ị 5678 N.H Tuan et al / Applied Mathematical Modelling 35 (2011) 5673–5690 Using (18) and (20) and the inequality (a + b)2 2(a2 + b2), we obtain 1 þ be 2 2 p2 2tÀ2T X 2t2T p2 X eTtịn ỵm ị aT ku; ; tị v ; ; tịk ẳ ðg À h Þ ðjg nm À hnm jÞ2 b aT kg À hk2 : b nm nm 2 aTn ỵm ị n;mẳ1 n;mẳ1 Therefore tT ku; ; tị v ; Á; tÞk b aT kg À hk: ð21Þ This completes the proof of theorem h Remark As shown in the Introduction, several regularization methods are established in the literature The stability of their solutions with respect to the variation of the final value (at t = T) is controlled by inequalities with coefficients of difT ferent order of magnitude For instance, in [20], the order of magnitude is e In [19], the authors give some better stability t estimates than the latter discussed methods They show that the stability estimate is of order M T À1 In [21], the authors T improve the previous results by a better estimation of the stability order, that is A ẳ C 1ỵln From (21), if we set b = ð ðTÞ then the stability order is B ¼ C Àa For a > 1, we have lim !0 B C2 T ẳ 0: ẳ lim 1a ỵ ln A C T !0 It is easy to see that the order A of the error is less than the above order B of stability estimate This proves the advantages of our method Theorem Assume that there exists a positive number P1 such that p2 X ỵm2 ị e2tn junm tịj2 < P21 ; t T: 22ị n;mẳ1 Let g L2(X) be measured data satisfying kg À gk : Let v (Á,Á,t) be the solutions of Problem (13)–(15) corresponding to the final data g Let b = a, then one has t kuðÁ; Á; tÞ À v ; ; tịk P1 ỵ 1ịT : ð23Þ Proof Suppose the problem (1) has an exact solution u Then u can be written as uðx; y; tÞ ẳ X ỵn2 ị etTịm g nm Z T ỵm2 ị eTsịn fnm sịds sin nx sin my: 24ị t m;nẳ1 Hence unm tị ẳ etTịm ỵn2 ị Z g nm T eTsịn ỵm2 ị fnm sịds ; 25ị t where unm tị ẳ p42 hux; y; tị; sin nx sin myi It follows from (16) that unm ðtÞ ẳ Z T 2 eTtịn ỵm ị 2 g nm esTịn ỵm ị fnm sịds : ỵm2 ị aTn ỵ be t 26ị Hence, we deduce from (25) and (26) that ! Z T 2 eTtịn ỵm ị sTịn2 ỵm2 ị g e À À e f ðuÞðsÞds nm nm 2 nm tị unm tị ẳ ỵ beaTn þm Þ t Z T 2 beðTÀtÞðn þm Þ aTðn2 þm2 Þ b ðsÀTÞðn2 þm2 Þ e g e f sịds ẳ u tị ẳ nm nm 2 aTn2 ỵm2 ị nm b ỵ e ỵ beaTn ỵm ị t Ttịn2 ỵm2 ị u ẳ betn ỵm ị 2 etn ỵm ị unm tị: b ỵ eaTn2 ỵm2 ị 27ị Using (18), we obtain t ỵm2 ị junm tị unm ðtÞj baT etðn unm ðtÞ: ð28Þ 5679 N.H Tuan et al / Applied Mathematical Modelling 35 (2011) 5673–5690 This implies that kuðÁ; Á; tÞ À u ðÁ; Á; tịk2 ẳ p2 X junm tị unm tịj2 n;mẳ1 p2 2t baT X þm2 Þ e2tðn 2t junm ðtÞj2 P 21 baT : 29ị n;mẳ1 Since the condition kg gk and (21) give tÀT tÀT ku ðÁ; Á; tÞ À v ðÁ; Á; tÞk b aT kg Àg k b aT : ð30Þ Combining (29), (30) and using the triangle inequality, we have t tÀT kuðÁ; Á; tÞ À v ðÁ; Á; tÞk kuðÁ; ; tị u ; ; tịk ỵ ku :; tị v ; tịk P1 baT ỵ b aT : Notice that, from b = a, then t t t kuðÁ; Á; tÞ À v ðÁ; ; tịk P T ỵ T ẳ P1 þ 1ÞT : This completes the proof of Theorem h Remark (1) If f = 0, then from (32), we get ỵm2 ị unm tị ẳ etTịn g nm : Therefore ỵm2 ị unm 0ị ẳ eTn ỵm2 ị g nm ẳ etn unm tị: This implies that p2 X e2tn ỵm2 Þ junm ðtÞj2 ¼ kuðÁ; Á; 0Þk2 : n;m¼1 The condition (22) is acceptable Moreover, the assumptions (22) can be replaced by assumptions on f and g Thus, since ỵn2 ị unm tị ẳ etTịm g nm Z T ỵm2 ị eTsịn fnm sịds 31ị t we get ỵn2 ị etm ỵn2 ị unm tị ẳ eTm g nm Z T esn ỵm2 ị fnm sịds: 32ị t R T P1 P1 2Tm2 ỵn2 ị g nm n;mẳ1 e 2 < and n;mẳ1 e2sn ỵm ị fnm ðsÞds < then (22) is holds 2.In t = 0, the error (12) does not converges to zero when ? This is of the same order as in [19] The error estimate here is not good at t = because the condition of the exact solution u is so weak In practical applications we may expect that the exact solution is smoother In these cases the explicit error estimate is available in the next Theorem If Theorem Assume that there exist positive numbers k,P2such that p2 X ỵm2 ị e2kn junm tịj2 < P22 : 33ị n;mẳ1 aT Let v(,,t) be dened in Theorem Let b ẳ Tỵk , then one has k t kuðÁ; Á; tÞ À v ; ; tịk P2 ỵ Tỵk Tỵk : ð34Þ Proof Hence, we deduce from (25) and (26) that ! Z T 2 eTtịn ỵm ị sTịn2 ỵm2 ị g e e f sịds nm nm 2 nm tị unm tị ẳ ỵ beaTn ỵm ị t Z T 2 2 Ttịn ỵm ị be beaTkịn ỵm ị kn2 ỵm2 ị aTn2 ỵm2 ị sTịn2 ỵm2 ị e g e f sịds ẳ e unm tị: ẳ nm nm ỵ beaTn2 ỵm2 ị ỵ beaTn2 ỵm2 ị t u Ttịn2 ỵm2 ị 35ị This implies that k ỵm2 ị junm tị unm ðtÞj baT ekðn unm ðtÞ: ð36Þ 5680 N.H Tuan et al / Applied Mathematical Modelling 35 (2011) 5673–5690 Using (36), we get the estimate kuðÁ; Á; tÞ À u ; ; tịk2 ẳ p2 X n; m ẳ 11 junm ðtÞ À unm ðtÞj2 p2 2k baT X ỵm2 ị e2kn 2k junm tịj2 P22 baT : 37ị n;mẳ1 Since the condition kg À gk and (21) give tÀT tÀT ku ðÁ; Á; tÞ À v ðÁ; Á; tÞk b aT kg À gk b aT : ð38Þ From (37) and (38), we have k tÀT kuðÁ; Á; tÞ À v ðÁ; Á; tÞk kuðÁ; ; tị u ; ; tịk ỵ ku ; Á; tÞ À v ðÁ; Á; tÞk P baT ỵ b aT : aT Notice that from b ẳ Tỵk , then tỵk k ku; ; tị v ; ; tịk P2 Tỵk ỵ Tỵk : This completes the proof of Theorem h Remark (1) We emphasize once more that the error (34) À(k > 0) ÁÀqis of Hölder type for all t [0;T] It is easy to see that the convergence rate of p, (0 < p) is more rapid than the that of lnð1Þ ðq > 0Þ when ? Hence, the convergence rate in this paper is better than that of the regularizations proposed in some recent papers such as Clark and Oppenheimer [19], Denche and Bessila [22], ChuLiFu et al [1–3], Tautenhahn [18,8] For t = in (34), we get k kuðÁ; Á; 0Þ v ; ; 0ịk P2 ỵ 1ịTỵk : 39ị k Tỵk The rate of convergence at t = is Since < k (a À 1)T, the fastest convergence is large constant a This proves that (39) is sharp estimate aÀ1 a Hence it can approach for the A numerical experiment In this section, we establish some numerical tests for a example To compare some different methods, we give the numerical tests with same parameter regularization We consider the problem ut À uxx À uyy ¼ f ðx; y; tị ẳ 3et sin x sin y 40ị ux; y; 1ị ẳ gx; yị ẳ e sin x sin y: ð41Þ and The exact solution of the latter equation is ux; y; tị ẳ et sin x sin y: Note that ux; y; 1=2ị ẳ p e sinxị sinyị % 1:648721271 sinðxÞ sinðyÞ Let gpbe the measured final data g p x; yị ẳ e sinxị sinyị ỵ sinðpxÞ sinðpyÞ: p So that the data error, at the final time, is sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi Z pZ p p 2 Fpị ẳ kg p gk ẳ sin pxị sin pyịdx dy ẳ : 2p 0 p The solution of (40) and (41) corresponding the final value gpis up x; y; tị ẳ et sinxị sin y ỵ ep 1tị sinpxị sinpyị: p The error at t = is sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi Z p Z p 2p2 e ep p 2 Opị :ẳ ku ; ; 0ị u; ; 0ịk ẳ sin pxị sin pyị dx dy ẳ : p2 p 0 p Then, we notice that lim Fpị ẳ lim kg p À gk ¼ lim p!1 p!1 n!1 1p ¼ 0; p2 lim OðpÞ ¼ lim kup ðÁ; ; 0ị u; ; 0ịk ẳ lim p!1 p!1 p!1 ep p ¼ 1: p 5681 N.H Tuan et al / Applied Mathematical Modelling 35 (2011) 5673–5690 From the two equalities above, we see that (40) and (41) is an ill-posed problem Let a ¼ 101 Approximating the problem as in 100 (13)–(15), the regularized solution is v x; y; tị ẳ X etn n2 ỵm2 ị 100 e n;mẳ1 Hence, we have v x; y; tị ẳ e ỵ en2 ỵm2 ị ị g nm ỵ e2 ẳ e 50 ỵ e2 Z e n2 þm2 Þ 100 e þ eÀðn2 þm2 Þ fnm ðsÞds sin nx sin my t T: ð42Þ e2tp ds sin x sin y ỵ sinðpxÞ sinðpyÞ t 1: p p e50 ỵ e2p2 e ỵ e2 ! e3s3 50 est1ịn 50 t ! ỵm2 ị e3sÀ2tÀ2 It follows that v ðx; y; Þ Z t Z sin x sin y À e1À2t 50 ỵm2 ị ỵ e2 ds sin x sin y ỵ 43ị ep sin px sin py: p2 p e50 ỵ e2p2 44ị Let a = kv À uk be the error between the regularized solution v and the exact solution u Let ¼ 1 ¼ 10À3 ¼ 3 rffiffiffiffi p ¼ 2 ¼ 10À4 ; pffiffiffiffiffiffiffi À5 ¼ 10 2p; rffiffiffiffi ¼ 4 ¼ 10À8 p ; rffiffiffiffi p ; we have Table We note that the new method in this article give a better approximation than the previous method in [24] To prove this, we continue to approximate this problem in 2-D case by the method given in [24] Infact, by reconsider the Table numerical in [24] (see p 879), we have the Table By applying that method given in [24], we have the following regularized solution v x; y; tị ẳ X n;mẳ1 2 etn ỵm ị g ỵ en2 ỵm2 ị ị nm Z t ! 2 etn ỵm ị f sịds sinnxị sinmyị: s ỵ esn2 ỵm2 ị nm Hence, we have v x; y; tị ẳ e12t sin x sin y ỵ e2 Z t es2t ds sin x sin y ỵ p s ỵ e2s e2tp sinpxị sinpyị t ỵ eÀ2p2 ð45Þ Table The error of method in this paper 1 2 3 3 pffiffiffi ¼ 10À3 p2 pffiffiffi ¼ 10À4 p2 ffiffiffi À5 pp ¼ 10 ffi À10 pffiffi p ¼ 10 v a 1.64870174813729sinxsiny + 2.706068625  10À442978sin(103x)sin(103y) 0.00003066665669 1.64872107547220sinxsiny + 4.464249273  10À44298034sin(104x)sin(104y) 3.078760801  10À7 1.64872126874784sinxsiny 3.141592652  10À9 1.64872127070013sinxsiny 1.5707963267948  10À19 Table The error of method in [24] 1 2 3 3 pffiffiffi ¼ 10À3 p2 ffi À4 pffiffi p ¼ 10 ffiffiffi À5 pp ¼ 10 pffiffiffi ¼ 10À10 p2 v a 1.618739596sin(x)sin(y) + 1.315245468  10À434301sin(103x)sin(103y) 0.04709510496 1.645673218sinxsiny + 0.2573859778  10À43429456sin(104x)sin(104y) 0.004787870457 1.648110757sinxsiny + 2.785098042  10À1085736215sin(5.104x)sin(5.104y) 0.0009589931490 1.648720965sinxsiny 4.806636760  10À7 Table The error of method in [23] 1 2 3 4 pffiffiffi ¼ 10À3 p2 ffiffiffi À4 pp ¼ 10 pffiffiffiffiffiffi ffi À5 ¼ 10 2p ffip À8 pffiffi ¼ 10 v a 1.735336020sin(x) sin(y) + 2.630490937  10À434295sin(1000x)sin(1000y) 0.1360541296 1.674100247sin(x)sin(y) + 5.147719556  10À43429449sin(104x)sin(104y) 0.03986520228 1.658588375sinxsiny + 1.392549021  10À1085736205sin(5.104x)sin(5.104y) 0.01549921072 1.648834493sin(x)sin(y) 0.0001778487018 5682 N.H Tuan et al / Applied Mathematical Modelling 35 (2011) 5673–5690 Fig The exact solution Fig The regularized solution with 1 in Table It follows that v ðx; y; 1 ịẳẳ sin x sin y ỵ e2 Then, we have the Table Z ! esÀ1 ds sin x sin y ỵ p s ỵ e2s ep sinpxị sinpyị t ỵ e2p2 46ị N.H Tuan et al / Applied Mathematical Modelling 35 (2011) 5673–5690 Fig The regularized solution with 2 in Table Fig The regularized solution with 3 in Table 5683 We have the graphics of the exact solution and the regularized solution in Tables 1–3 These are displayed on a rectangular domain (0,p)  (0,p) (See Figs 1–15) Now we consider the random measured data case Let grandom(x) = esin xsin y + e⁄randN⁄sin(randN⁄x)⁄sin(randN ⁄y) be the random measured data, where randN = ceil(NÁ⁄rand) The function ceil(NÁ⁄rand) returns the random value between and N 5684 N.H Tuan et al / Applied Mathematical Modelling 35 (2011) 5673–5690 Fig The regularized solution with 4 in Table Fig The regularized solution with 1 in Table With the random measured data grandom, we get the regularized solution v eran x; y; tị ẳ e12t 50 ee þ eÀ2 sin x sin y À  sinðrandNyÞ: Z t e3s2t2 50 ee ỵ e2 ds sin x sin y ỵ randN ee2trandNị randN ị2 50 ee ỵ e2randNị sinrandNxị N.H Tuan et al / Applied Mathematical Modelling 35 (2011) 5673–5690 Fig The regularized solution with 2 in Table Fig The regularized solution with 3 in Table It follows that e v ran ! Z 1 e3s2 eerandNị ẳ x; y; sin x sin y ds sin x sin y ỵ randN randNị2 sinrandNxị 1 ee50 ỵ e2 ee50 þ eÀ2 ee 50 þ eÀ2ðrandNÞ2  sinðrandNyÞ: We have the following error table 5685 5686 N.H Tuan et al / Applied Mathematical Modelling 35 (2011) 5673–5690 Fig The regularized solution with 4 in Table Fig 10 The regularized solution with 1 in Table Looking at three above tables with comparison between three methods, we can see the error results of the Table are smaller than many times in the Tables and This shows that our approach has a nice regularizing effect and give a better approximation with comparison to the previous method established in [23,24] (See Table 4) N.H Tuan et al / Applied Mathematical Modelling 35 (2011) 5673–5690 Fig 11 The regularized solution with 2 in Table Fig 12 The regularized solution with 3 in Table 5687 Conclusion We have considered a regularization problem for a nonhomogeneous backward heat equations in bounded domain, namely Problem (1) In many earlier works on the backward problem, while one may obtain an Hölder type error estimate 5688 N.H Tuan et al / Applied Mathematical Modelling 35 (2011) 5673–5690 Fig 13 The regularized solution with Fig 14 The random regularized solution v eran 4 in Table at t ¼ 12 and e = 10À5 at any fixed t > 0, an explicit error estimate at t = is still difficult and was given in logarithm type only The present paper proposes a regularized solution with error estimate of Hölder type for all t [0, T] Moreover, our regularization is enough for a numerical setting and the numerical results seems satisfactory In the future, we will consider the regularized problem for a nonlinear backward heat equation 5689 N.H Tuan et al / Applied Mathematical Modelling 35 (2011) 5673–5690 Fig 15 The random regularized solution v eran at t ¼ 12 and e = 10À5 Table The error of method in this paper with random measured data ¼ ¼ ¼ ¼ ¼ pffiffiffi 10À3 p2 ffi À4 pffiffi p 10 ffiffiffi À5 pp 10 pffiffiffi 10À8 p2 pffiffiffi 10À10 p2 randN 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equations Although there are many papers on the homogeneous backward heat equation, the result on the inhomogeneous case is very scarce... approximation problem h Regularization and error estimate In this section, we introduce a regularized problem as of integral equation and investigate the error estimate between the regularization solution