DSpace at VNU: A nonlinear parabolic equation backward in time: Regularization with new error estimates

Nonlinear Analysis 73 (2010) 1842–1852 Contents lists available at ScienceDirect Nonlinear Analysis journal homepage: www.elsevier.com/locate/na A nonlinear parabolic equation backward in time: Regularization with new error estimates Nguyen Huy Tuan a,∗ , Dang Duc Trong b a Department of Mathematics, SaiGon University, 273 AnDuongVuong, HoChiMinh city, Viet Nam b Department of Mathematics, University of Natural Science, Vietnam National University, 227 Nguyen Van Cu, Q.5, HoChiMinh City, Viet Nam article info Article history: Received 21 April 2009 Accepted May 2010 MSC: 35K05 35K99 47J06 47H10 Keywords: Backward parabolic problem Ill-posed problem Nonlinear parabolic equation Truncation method Error estimate abstract Consider a nonlinear backward parabolic problem in the form ut + Au(t ) = f (t , u(t )), u( T ) = g , where A is a positive self-adjoint unbounded operator Based on the fundamental solution to the parabolic equation, we propose to solve this problem by the Fourier truncated method, which generates a well-posed integral equation Then the well-posedness of the proposed regularizing problem and convergence property of the regularizing solution to the exact one are proven Our regularizing scheme can be considered a new regularization, with the advantage of a relatively small amount of computation compared with the quasireversibility or quasi-boundary value regularizations Error estimates for this method are provided together with a selection rule for the regularization parameter These errors show that our method works effectively © 2010 Elsevier Ltd All rights reserved Introduction We consider the backward time problem for the following nonlinear PDE: ut + Au(t ) = f (t , u(t )), < t < T, u(T ) = g , (1) (2) where A is any non-negative, self-adjoint operator which does not depend on f and g is a given vector in a Hilbert space H An example for problem (1)–(2) is the backward heat problem ut − u = f (x, t , u), (x, t ) ∈ Ω × (0, T ) u(x, t ) = 0, (x, t ) ∈ ∂ Ω × (0, T ) u(x, T ) = ϕ(x), x ∈ Ω where Ω = (0, π )N ⊂ RN This is an example for the problem (1)–(2) corresponding to H = L2 (Ω ) and A = −∆ (associated with the homogeneous Dirichlet boundary condition) which has the eigenbasis φp (x) = ( π2 )N /2 sin(p1 x1 ) sin(pN xN ) and the eigenvalue λp = p2 Here we denote p = (p1 , pN ) ∈ N N , x = (x1 , x2 , , xp ) ∈ RN and p2 = p21 + · · · p2N The backward parabolic problems can be applied to several practical areas such as image processing, mathematical finance, and physics (See [1–3]) However, the ill-posed nature of the problems requires certain types of regularization ∗ Corresponding author E-mail address: tuanhuy_bs@yahoo.com (N.H Tuan) 0362-546X/$ – see front matter © 2010 Elsevier Ltd All rights reserved doi:10.1016/j.na.2010.05.019 N.H Tuan, D.D Trong / Nonlinear Analysis 73 (2010) 1842–1852 1843 techniques One approach to regularize the ill-posed problems is based on the use of eigenfunction expansion, where the eigenpairs of the corresponding elliptic operator are available Another approach is to use the method of quasireversibility [4–15] Other approaches include the least squares method with Tikhonov-type regularization [16] and the use of heat kernels Carasso et al [17] introduced a method of transforming the problem (1)–(2) into a second-order in time problem Later Carasso [18,19] introduced the concept of a supplementary constraint such as that of slow evolution from the continuation boundary (SECB) The methods of quasi-reversibility and Tikhonov regularization introduce artificial contamination, which is not from noises, to the numerical solutions For a recent survey on backward parabolic equations (BPE), we refer the reader to [20,21] Although there are many papers on the linear homogeneous case of the backward problem, we only find a few papers on nonhomogeneous and nonlinear cases of BPE In 2007, in [22], the authors considered the problem (1)–(2) in the case A is the Laplace operator and H = L2 (0, π ) by using the quasi-boundary value method They gave the approximated problem as follows ∞ e−tn ut − uxx = t T n =1 + e−tn2 u (0, t ) = u (π , t ) = 0, fn (u )(t ) sin nx, (x, t ) ∈ (0, π ) × (0, T ), t ∈ [0, T ], (4) ∞ T u (x, 0) + u (x, T ) = ϕ(x) − n =1 (3) s T + e−sn2 fn (u )(s)ds sin nx, (5) and obtain the error estimation u (., t ) − u(., t ) ≤ C t /T (6) Very recently, in [23], we considered the problem (1)–(2) by using the method of stabilized quasi-reversibility By replacing the operator A by f (A), chosen later under some better conditions, we approximated the problem (1)–(2) as follows: ut + f (A)u = f (u (t ), t ), < t < T, u (T ) = ϕ, with < (7) (8) < We gave the following error estimate u (t ) − u(t ) ≤ C β( )t /T (9) Here u and u are solutions to the ill-posed and approximate problems, respectively, and C is a computable constant independent of β( ) where β( ) → if → It is clear that the errors (6) and (9) in the time t = 0, are not given To improve these above disadvantages, in this paper, we develop a new regularization method which is called the Fourier method for solving the problem (1)–(2) As far as we know, there have not been any results of the Fourier series method for treating nonlinear BPE until now Meanwhile, we will establish some faster convergence error estimates In particular, the convergence of the approximate solution at t = 0, is also proved This is an improvement of many recent results in [7,24,25,22,26,23,27] Namely, assume that A admits an orthonormal eigenbasis {φp }p≥1 in H, associated with the eigenvalues such that < λ1 ≤ λ2 ≤ λ3 ≤ · · · lim λp = ∞ p→∞ Informally, problem (1)–(2) can be transformed to an integral equation having the form ∞ T e(T −t )λp gp − u( t ) = e−(t −s)λp fp (u)(s)ds φp (10) t p=1 where ∞ gp φp , g = p=1 ∞ f (t , u(t )) = fp (u)(t )φp p=1 Since t < T , we know from (10) that, when p become large, exp{(T − t )λp } increase rather quickly Thus, the term e−(t −T )λp is the instability cause In [26], this term is replaced by the better term e−t λp + e−T λp In [23], we give a different better term having the form ( λp + e−T λp ) t −T T 1844 N.H Tuan, D.D Trong / Nonlinear Analysis 73 (2010) 1842–1852 So, to regularize problem (1)–(2), we hope to recover the stability of problem (10) by filtering the high frequencies with a suitable method The essence of our regularization method is just to eliminate all high frequencies from the solution, and instead consider (10) only for λp ≤ m , where m is an appropriate positive constant satisfying lim →0 m = ∞ We note that m is a constant which will be selected appropriately as the regularization parameter This regularization method is rather simple and convenient for dealing with some ill-posed problems The present paper is devoted to establishing such a method for problem (1)–(2) Fourier regularization and the main results Let g and g denote the exact and measured data at t = T , respectively, which satisfy ≤ g −g Let P = {p ≥ 1, p ∈ N , λp ≤ m }, Q = {p ≥ 1, p ∈ N , λp > m } where m is a constant which will be selected appropriately as the regularization parameter For every v in H having the ∞ expansion v = p=1 vp φp , vp ∈ R, p = 1, 2, we define the operators S (t ), S (t ) as follows ∞ e−t λp vp φp , S (t )v = p=1 et λp vp φp S (t )v = (11) p∈P We have the following approximation problem T u (t ) = S (T − t )g − S (s − t )f (s, u (s))ds t T e(T −t )λp gp − = e(s−t )λp fp (u )(s)ds φp (12) t p∈P Theorem Let g ∈ H and let f : R × H → H be a Lipschitz continuous function in both variables, i.e f (t1 , w) − f (t2 , v) ≤ k(|t1 − t2 | + w − v ), for k > independent of w, v ∈ H , t ∈ R Then problem (12) has uniquely a solution u ∈ C ([0, T ]; H ) for any positive Proof First, it is easy to prove that S (t ) ≤ etm (13) Thus, for every v in H having the expansion v = S (t )v e2t λp vp2 ≤ e2tm = p∈P ∞ p=1 vp φp , we have vp2 p∈P ∞ vp2 = e2tm v ≤ e2tm p=1 Hence, (13) is proved For w ∈ C ([0, T ]; H ), consider the function G(w)(t ) defined by T G(w)(t ) = S (T − t )g − S (s − t )f (s, w(s))ds t It is clear that t → G(w)(t ) is continuous from [0; T ] → H and G(w)(T ) = g We claim that, for every w, v ∈ C ([0, T ]; H ) we have Gn (w)(t ) − Gn (v)(t ) ≤ kn (T − t )n eTnm Cn n! |w − v|, where C = max{T , 1} and |||.||| is sup norm in C ([0, T ]; H ) We shall prove the latter inequality by induction (14) N.H Tuan, D.D Trong / Nonlinear Analysis 73 (2010) 1842–1852 1845 For n = 1, using (13) and the Lipschitz property of f , we have T G(w)(t ) − G(v)(t ) = S (s − t )( f (s, w(s))) − f (s, v(s))ds t T ≤ ke(s−t )m w(s) − v(s) ds t ≤ CkeTm (T − t )|||w − v||| Suppose that (14) holds for n = j We prove that (14) holds for n = j + In fact, we have T S (s − t )(f (s, Gj w)(s) − f (s, Gj v)(s))ds Gj+1 (w)(t ) − Gj+1 (v)(t ) = t T Gj (w)(s) − Gj (v)(s) ≤ k(T − t )eTm ds t ≤ kj+1 eTm (j+1) (T − t )j+1 j+1 C |||w − v||| (j + 1)! Therefore, by the induction principle, we have Gn (w)(t ) − Gn (v)(t ) ≤ (CkT eTm )n |||w − v|||, n! (15) for all w, v ∈ C ([0, T ]; H ) We consider G : C ([0, T ]; H ) → C ([0, T ]; H ) Since (CkT eTm )n = 0, n→∞ n! lim there exists a positive integer number n0 such that Gn0 is a contraction It follows that the equation Gn0 (w) = w has a unique solution u ∈ C ([0, T ]; H ) We claim that G(u ) = u In fact, one has G(Gn0 (u )) = G(u ) Hence Gn0 (G(u )) = G(u ) By the uniqueness of the fixed point of Gn0 , one has G(u ) = u , i.e., the equation G(w) = w has a unique solution u ∈ C ([0, T ]; H ) Remark 2.1 In the Hilbert space, let H = L2 (0, π ) and let A = −∆ is the Laplace operator We take λn = n2 , φn = π sin(nx) are eigenvalues and orthonormal eigenfunctions, which form a basis for H Let H be the norm of L2 (0, π ) The function f satisfies the condition of Theorem which is given by an example  u     e21 e10   u+ − e−1 e−1 f ( u) =  e10 e21   u+   e−1  e − u ∈ [−e10 , e10 ] u ∈ (e10 , e11 ] u ∈ (−e11 , −e10 ] |u| > e11 It is not difficult to check that the above function satisfying the Lipschitz condition of Theorem However, the present paper can not treat for example a nonlinear term, such as f (t , u) = uq for any q This is the serious disadvantage of this paper We hope the big class of function f will be studied in future reports Theorem The solution of the problem (12) depends continuously on g in H Proof of Theorem Let u and v be two solutions of (12) corresponding to the values g and h We have T u(t ) − v(t ) = S (T − t )(g − h) − S (s − t )(f (s, u(s)) − f (s, v(s))) (16) t Using (13), we obtain T u(t ) − v(t ) ≤ S (T − t ) S (s − t ) g −h + f (s, u(s)) − f (s, v(s)) ds t T ≤ e(T −t )m g − h + k t e(s−t )m u(s) − v(s) ds (17) 1846 N.H Tuan, D.D Trong / Nonlinear Analysis 73 (2010) 1842–1852 Then, we obtain T u(t ) − v(t ) ≤ eTm etm esm g −h +k u(s) − v(s) ds t Using Gronwall’s inequality we have u(t ) − v(t ) ≤ ek(T −t ) eTm etm g −h Hence, we obtain u(t ) − v(t ) ≤ ek(T −t ) e(T −t )m g −h (18) This completes the proof of the theorem Remark 2.2 (1) If we select m = , since (18), the stability of the method becomes E1 ( ) = e quasi-reversibility method in [6,8,14,15] C1 , which is the same as the t (2) If m = T1 ln( ), the stability magnitude is E2 ( , t ) = C2 T −1 , is the same as the quasi-boundary value method of some related results in [7,11,22] T T (3) If m = T1 ln( ), the stability magnitude is better than E3 ( ) = C3 , which is given in [24,26] T T (1+ln( )) (4) If m = T T ln( Te ) ln (1+ln( )) , the stability magnitude is t T E4 ( , t ) = C4 e(T −t )m = C4 −1 T −1 ln(Te −1 ) t T −1 which is similar to [27] Note that lim →0 E4 ( , t ) =0 E2 ( , t ) and lim →0 E4 ( , t ) = lim E3 ( , t ) T + ln →0 t T = 0, 1, < t < T, t = So, the error E4 ( , t ) is better than E1 ( ), E2 ( , t ), E3 ( ) To our knowledge, this is the best known result Theorem Let f , g be as in Theorem Assume the exact solution u(t) of the problem (1)–(2) such that T ∞ e2sλp fp2 (u)(s)ds < ∞ (19) p=1 Then one has √ u(t ) − u (t ) ≤ Mek T (T −t ) e−tm , (20) for every t ∈ [0, T ], where ∞ T M = u(0) e2sλp fp2 (u)(s)ds, +T p=1 and u is the unique solution of problem (12) corresponding to Proof of Theorem The functions u(t ) and u (t ) can be written as the forms ∞ T e(T −t )λp gp − u(t ) = e(s−t )λp fp (u)(s)ds φp , t p=1 T u (t ) = S (T − t )g − S (s − t )f (s, u (s))ds t T e(T −t )λp gp − = p∈P t e(s−t )λp fp (u )(s)ds φp (21) N.H Tuan, D.D Trong / Nonlinear Analysis 73 (2010) 1842–1852 1847 Hence T e(T −t )λp gp − u(t ) − u (t ) = t p∈Q T e(s−t )λp fp (u)(s)ds φp + e(s−t )λp (fp (u )(s) − fp (u)(s))ds φp t p∈P = I +J where T e(T −t )λp gp − I = e(s−t )λp fp (u)(s)ds φp t p∈Q T e(s−t )λp (fp (u )(s) − fp (u)(s))ds φp J = t p∈P T S (s − t )(f (s, u (s)) − f (s, u(s)))ds = t Since p ∈ Q , we have e−t λp ≤ e−tm , this implies that the term I can be estimated as follows I = e T (T −t )λp gp − e (s−t )λp fp (u)(s)ds t p∈Q e−2t λp ≤2 T eT λp gp − esλp fp (u)(s)ds p∈Q ∞ T eT λp gp − ≤ 2e−2tm esλp fp (u)(s)ds = 2e−2tm u(0) T + 2T e−tm u(0) T + 2T p=1 e2sλp fp2 (u)(s)ds = 2e−2tm M T ≤ (T − t ) S (s − t ) (f (s, u (s)) − f (s, u(s))) ds t T ≤ k2 T e2(s−t )m u(s) − u (s) ds t Therefore we obtain u(t ) − u (t ) ≤ 2( I + J 2) T ≤ 4e−2tm M + 2k2 T e2(s−t )m u(s) − u (s) t It follows that T e2tm u(t ) − u (t ) ≤ 4M + 2k2 T e2sm t By using Gronwall’s inequality, we obtain e2tm u(t ) − u (t ) 2 T (T −t ) ≤ e2k 4M , which implies that √ u(t ) − u (t ) ≤ Mek T (T −t ) e−tm e2sλp fp2 (u)(s)ds Using the Lipschitzian hypothesis on f and (13), we obtain J T + 2T e−2tm p=1 ∞ = e−2tm ∞ p=1 ∞ u(s) − u (s) 2 esλp fp (u)(s)ds p∈Q p=1 t e−2t λp +2 ds ds e2sλp fp2 (u)(s)ds 1848 N.H Tuan, D.D Trong / Nonlinear Analysis 73 (2010) 1842–1852 Remark 2.3 (1) Choosing m = ln( ), we obtain the error estimate T √ u(t ) − u (t ) ≤ Mek T (T −t ) t T (22) This error is given in [28,22,23] And if the function f (t , u(t )) = 0, the error is given in [7] (2) If we select m = ln ( ) T ln s √ u(t ) − u (t ) ≤ Mek T (T −t ) t (s ≥ 0), we have the another error t T ln −st (23) −st tend to when → Hence, this error is better than the error in (22) And in the case Note that if t > 0, T and ln s = 0, this error is also (22) (20) is generalization of many related errors in some previous papers such as [28,15,22,26,23] (3) In the case f (t , u(t )) = f (t ), to find the convergence of approximate solution, we don’t ask the condition (19).Thus, we have T e(T −t )λp gp − u(t ) − u (t ) = e(s−t )λp fp (s)ds φp t p∈Q Then u(t ) − u (t ) T e(T −t )λp gp − = t p∈Q u2p = u = e(s−t )λp fp (s)ds u2p − p∈Q p∈P It is easy to prove that u2p = u lim →0 p∈P Then lim u(t ) − u (t ) →0 = If there exist s > such that ∞ λsp u2p < ∞ p=1 then u(t ) − u (t ) = p∈Q λ s u2 ≤ s λsp p p m λsp u2p ≤ p∈Q ms ∞ λsp u2p p=1 (4) In most known results, such as [28,22,23], the errors between the exact solution and approximate solution are calculated in the form C estimate is t T Notice that the convergence estimate in this theorem is useless at t = Let us take t = in (11), the error u(0) − u (0) ≤ √ 2T Mek , which does not tend to zero when → In the next theorem, we will give another estimation which the error in the case t = is considered (5) As we know, m is a constant which will be easily selected appropriately as regularization parameter In [23], β( ) is also the regularization parameter However, in that paper, it is not easy to select β( ) because it depends on the operator A and approximated operator f (A) For details, the authors selected β( ) satisfying two conditions ln(β( )), α ∈ [0, ∞) T (b) (−A + f (A))u ≤ β( ) eTA u (a) |f (α)| ≤ − This proves that our method is more effective than the method in [23] (6) In this theorem, we require a condition on the expansion coefficient fp in (19) We note that the solution u depends on the nonlinear term f and therefore fp , fp (u) is very difficult to assign a value Such obscurity makes this theorem hard to use N.H Tuan, D.D Trong / Nonlinear Analysis 73 (2010) 1842–1852 1849 for numerical computations Hence, we require another condition as follows ∞ e2t λp | u(t ), φp |2 < ∞ (24) p=1 In this case, we only require the assumption of u, there is no need to compute the function fp (u) In fact, in the homogeneous case of problem (1)–(2), i.e f = 0, then the left hand side of (24) is equal to u(0) Hence, the condition (24) is natural and acceptable And we note (24) is weaker than the condition in theorem 3.1 (page 7, [23]) Theorem Suppose problem (1)–(2) has a unique solution u(t ) satisfying (24) and f , g are as in Theorem Then we obtain another estimation u(t ) − u (t ) ≤ R( , t )e−tm , (25) for every t ∈ [0, T ], where T eT λp gp − P( , t) = t p∈Q e2t λp u2p = esλp fp (u)(s)ds , (26) p∈Q T ke2k T (T −t ) R( , t ) = P ( , t )dt + P ( , t ) , (27) and u is the unique solution of problem (12) corresponding to Proof of Theorem It is easy to see that P ( , t ) → when I e−(t −T )λp gp − = e−2t λp T eT λp gp − esλp fp (u)(s)ds t p∈Q ≤e e−(t −s)λp fp (u)(s)ds t p∈Q ≤ → Since the proof of Theorem 3, we get T −2tm e T λp T gp − e sλp fp (u)(s)ds t p∈Q = e−2tm P ( , t ) This follows that u(t ) − u (t ) ≤ 2( I + J 2) T ≤ 2e−2tm P ( , t ) + 2k2 T e2(s−t )m u(s) − u (s) ds t Applying Gronwall’s inequality, we get e2tm u(t ) − u (t ) T Tt ≤ ke−2k Ts e2k P ( , s)ds + P ( , t ) t Finally u(t ) − u (t ) ≤ T ke2k T (T −t ) P ( , t )dt + P ( , t ) e−2tm = R2 ( , t )e−2tm Remark 2.4 (1) In the right hand side of (25), if t > then both term e−tm and R( , t ) tend to zero So, the convergence in (25) is better than in (20) (2) When t = 0, (25) becomes u(0) − u (0) ≤ ke2k T 2T P ( , t )dt + P ( , 0) (28) Noting that the right hand side of (28) tends to zero when → 1850 N.H Tuan, D.D Trong / Nonlinear Analysis 73 (2010) 1842–1852 (3) In the simple linear case f (t , u(t )) = 0, this means that k = 0, then the term P ( , t ) can be estimated ∞ e2T λp gp2 ≤ P( , t) = p∈Q e2T λp gp2 = u(0) (29) P ( , t )dt + P ( , t ) (30) p=1 So T R( , t ) = ke2k T (T −t ) ≤ kT e2k T (T −t ) + u(0) = u(0) (31) Theorem Let f , g be as in Theorem Assume that the exact solution u of (1)–(2) corresponding to g be defined as in Theorem Let g ∈ H be measured data such that g −g ≤ −s Let us select m = ln ( ) T (ln ) 2T and let u be the solution of problem (12) corresponding to g (a) If u(t ) satisfies (19), then u (t ) − u(t ) ≤ exp(k (T − t ) ) 2 t T ln t T ln −s(T −t ) 2T + Me k T ( T −t ) t T ln st 2T (32) (b) If u(t ) satisfies (24), then u (t ) − u(t ) ≤ exp(k2 (T − t )2 ) −s(T −t ) 2T T ( T −t ) + R( , t )ek t T ln st 2T , (33) for every t ∈ [0, T ] and M , R( , t ) is defined in Theorems and respectively Proof of Theorem We shall prove (a), and that (b) is similar to (a) Thus, let v be the solution of problem (12) corresponding to g Using Theorems and 4, we get u (t ) − u(t ) ≤ u (t ) − v (t ) + v (t ) − u(t ) √ + Mek T (T −t ) e−tm ≤ e(T −t )m exp(k2 (T − t )2 ) g − g ≤ exp(k (T − t ) ) 2 t T ln √ ≤ M + ek T (T −t ) t T −s(T −t ) √ 2T + Me ln k2 T (T −t ) −s(T −t ) 2T 1+ ln t T ln s st 2T , , for every t ∈ (0, T ) This completed the proof of the theorem Remark 2.5 (1) If the function f (t , u(t )) = 0, using (32) we obtain the following error u (t ) − u(t ) ≤ exp(k (T − t ) ) 2 t T ln −s(T −t ) 2T + u(0) e k2 T (T −t ) t T ln st 2T (34) Noting that the convergence in t = is not given here (2) When s = 0, the estimate (33) becomes u (t ) − u(t ) ≤ (R( , t ) + 1)ek T (T −t ) t T , ∀t ∈ (0, T ) (35) This error is similar to the results in [29,28,15] (3) In the above error if we let t = then u (0) − u(0) ≤ exp(k2 T ) ln −s 2T + R( , 0)ek (36) N.H Tuan, D.D Trong / Nonlinear Analysis 73 (2010) 1842–1852 1851 Since s > we get exp(k T ) ln lim →0 −s 2T + R( , 0)ek = (4) From (36) we know lim →0 R( , 0) = But the term R( , 0) is not often computed in practice, so we can’t estimate it by the term depending on This is disadvantage of the error (36) To improve this, let us select m = T T s + ln T T ln ln , and using the proof of Theorem 5(a), we obtain u (t ) − u(t ) ≤ u (t ) − v (t ) + v (t ) − u(t ) ≤ e(T −t )m exp(k2 (T − t )2 ) g − g ≤ exp(k2 T ) e(T −t )m ≤ exp(k T ) 2 where I =1+ e−Tm √ M =1+ 1+ e−Tm + ln(T −1 ) t T T √ M + ln T ln T t T √ + Mek T (T −t ) e−tm √ M , −1 ln T ( s Tt −1 ) I (37) T s If s > then lim + ln →0 ln T T s =0 t 1+ln(T −1 ) t T −1 and thus lim →0 I = Recall that in [22], the convergence of the approximated solution is C T So, the T convergence in (37) is quicker than that given in [22] We also note that the approximation in the case t = is proved Moreover, comparing (37) with the result obtained in [7,24,25,15,22,26,23] we know estimate (37) is sharp and the best known estimate Acknowledgements The authors would like to thank Professor Ravi P Agarwal 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 References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13] [14] [15] [16] [17] [18] H 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Tuan, Regularization and error estimate for the nonlinear backward heat problem using a method of integral equation, Nonlinear Anal 71 (2009) 4167–4176 P.H Quan, D.D Trong, A nonlinearly backward. .. Equations 1994 (8) (1994) 1–9 R.E Ewing, The approximation of certain parabolic equations backward in time by Sobolev equations, SIAM J Math Anal (2) (1975) 283–294 H Gajewski, K Zaccharias,... is an appropriate positive constant satisfying lim →0 m = ∞ We note that m is a constant which will be selected appropriately as the regularization parameter This regularization method is rather