DSpace at VNU: Holder stability for a class of initial inverse nonlinear heat problem in multiple dimension

Accepted Manuscript Hölder stability for a class of initial inverse nonlinear heat problem in multiple dimension Nguyen Huy Tuan, Bui Thanh Duy, Nguyen Dang Minh, Vo Anh Khoa PII: DOI: Reference: S1007-5704(14)00500-0 http://dx.doi.org/10.1016/j.cnsns.2014.10.027 CNSNS 3398 To appear in: Communications in Nonlinear Science and Numerical Simulation Received Date: Revised Date: Accepted Date: 21 June 2011 October 2014 21 October 2014 Please cite this article as: Tuan, N.H., Duy, B.T., Minh, N.D., Khoa, V.A., Hölder stability for a class of initial inverse nonlinear heat problem in multiple dimension, Communications in Nonlinear Science and Numerical Simulation (2014), doi: http://dx.doi.org/10.1016/j.cnsns.2014.10.027 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 Hă older stability for a class of initial inverse nonlinear heat problem in multiple dimension Nguyen Huy Tuan , Bui Thanh Duy , Nguyen Dang Minh 1 and Vo Anh Khoa Faculty of Mathematics and Computer Science, University of Science, Vietnam National University, HoChiMinh City, Viet Nam Faculty of Fundamental Science, Ho Chi Minh City Architecture University, Viet Nam Abstract We consider an inverse heat conduction problem for the non-linear heat equation which appears in some applied subjects The problem is severely ill-posed Using some modified regularization method, we establish a regularized solution which gives error estimate of Hăolder type for all t ∈ [0, T ] Some numerical examples show that the computational effect of these methods are all satisfactory Keywords and phrases: Backward heat problem, Nonlinearly Ill-posed problem, Quasi-boundary value methods, Quasi-Reversibility methods Mathematics subject Classification 2000: 35K05, 35K99, 47J06, 47H10 Introduction We consider the heat problem of finding a function u(x, t) such that  ∂u   − ∆u = f (x, t, u(x, t)), (x, t) ∈ Ω × (0, T )   ∂t u|∂Ω = 0, t ∈ [0, T ]     u(x, T ) = φ(x), x∈Ω (1) where Ω = (0, π)n ⊂ Rn , x = (x1 , , xn ) and φ(x), f (x, y, z) are given In practical situation, we not have exactly the value φ(x) We only know a noise datum φε (x) such that ∥φ − φε ∥ < ε where ∥.∥ is L2 (Ω)–norm From φε , we have to find a function v ε (x, t) which approximates stability the solution u of (1) The problem is severely ill-posed in the sense of Hadamard In fact, for a given final data, we are not sure that a solution of the problem exists In case a solution exists, it may not depend continuously on the final data The problem has many various application, for example in glass production [32], polymer processing The linear homogeneous case of (1) have been studied by many various methods We can notably mention the quasi-solution method (QS-method)of Tikhonov [44, 45],the quasi-reversibility method (QR method) of Lattes and Lions [11, 19], the quasi boundary value method (Q.B.V method)[5, 7, 33, 34] and the C-regularized semigroups method [1, 2, 15, 16, 25], the numerical method [14, 20, 28, 18] Previously, reconstruction of the initial temperature was considered only for the linear heat equation with source terms independent of the temperature To the author’s knowledge, there are not many papers on the initial inverse heat problem with linear and nonlinear source terms In 2006, Trong and Tuan proposed the quasi-reversibility regularization method for the nonhomogeneous backward heat, however, the retrieved time T = 0.5 is still small After that, Trong and Tuan also proposed many regularization method to improve previous stability results, for example [31, 36, 38, 39] In addition, Nam (2010) used the truncation method to studied the nonlinear backward heat and get the Hăolder error estimate However, a numerical simulation for 2-D and 3-D cases is not considered in this paper Recently, Li, Jiang and Hon (2010) [21] have proposed a meshless method based on RBFs method to solve 2-D and 3-D nonhomogeneous backward heat, but this algorithm was complex and hard to implement Without using a priori regularization, Chang and Liu [3] utilized the backward group preserving scheme (BGPS) to tackle those multi-dimensional nonlinear and nonhomogeneous backward heat and obtained good results Very recently, Chang [4] proposed a Fictitious Time Integration Method (FTIM) for solving the multi-dimensional nonhomogeneous and nonlinear backward heat The main idea of a fictitious time integration approach is transformation into a different evolutional PDE and semi-discretization which was first proposed by Liu [22] In fact, a fictitious time τ is employed to transform the dependent variable u(x, y, z, t) into a new one by (1 + τ )u(x, y, z, t) =: v(x, y, z, t, τ ), such that the original nonlinear and nonhomogeneous heat conduction equation is written as a new parabolic type partial differential equation in the space of (x, y, z, t, τ ) By employing the F T IM, Chang computed the solution and retrieve the initial data very well with a high order accuracy and showed that the efficiency of the computations for three-dimensional backward heat Motivated by these above reasons, in this paper, we propose a new modified quasi-boundary regularization method to establish the Hăolder estimates, i.e, the error is of order εk (0 < k < 1) The method of quasi-boudary value method was first introduced by Showalter [33] and then fast developed by many authors, for example Clark and Oppenheimer [5], Denche and Bessila [7], C-W Chang, C-S Liu and J-R Chang [4], etc The method of quasi-boudary value method is also called the perturbation method, whereby we modified the source term f and the final data φ The main idea of the quasi-boundary-value method is going to replace the boundary value problem with an approximate well-posed one, then to construct approximate solutions of the given boundary value problem The rest of the paper is organized as follows In Section 2, we state the linear case of the inhomogeneous backward problem We also give the estimates which are of order εk In Section 3, we extend our consideration to the nonlinear Cauchy problem, given by (1) Numerical examples are tested in Section to verify the efficacy of the our method Finally, proofs of main results will be give in Appendix The proof for the m-dimensional case is similar to the one for the one dimensional case Hence, for simplicity, in Appendix we will give the proof in the one dimensional case Inhomogeneous problem : the linear case In this section, we consider the linear inhomogeneous ill-posed problem as follows ∂u − ∆u = f (x, t), (x, t) ∈ Ω × (0, T ), ∂t u|∂Ω = 0, t ∈ [0, T ], (2) (3) (4) x ∈ Ω u(x, T ) = φ(x), Based on the ideas mentioned in paper [37], we consider the following approximation problem ∞ ∑ ∂uε,a e−T |m| ε,a − ∆u = fm (t)ϕm (x), (x, t) ∈ Ω × (0, T ), ∂t αe(a−1)T |m|2 + e−T |m|2 m=1 uε,a |∂Ω = 0, uε,a (x, T ) = ∞ ∑ m=1 t ∈ [0, T ], e−T |m| φε ϕm (x), + e−T |m|2 m (5) (6) αe(a−1)T |m|2 x∈Ω (7) where the small parametter α = α(ε) depends on ε, a ≥ is a constant and fm , φm are defined by m = (m1 , , mn ), mi ∈ Z, mi > 0, = (1, , 1) ϕm (x) = sin(m1 x1 ) sin(mn xn ), ( )n ∫ fm (t) = f (x, t)ϕm (x)dx, π Ω ( )n ∫ φm = φ(x)ϕm (x)dx π Ω If a = 1, the problem (5)–(7) is studied in [5] We shall prove that, the (unique) solution uε,a of (5)–(7) satisfies the following equality ( ) ∫ T ∞ ∑ e(T −t)|m| ε,a (s−T )|m|2 u (x, t) = φm − e fm (s)ds ϕm (x), ≤ t ≤ T (8) + αeaT |m|2 t m=1 Let v ε,a be the solution of problem (5)–(7) with noisy data φε ( ) ∫ T ∞ ∑ e(T −t)|m| ε,a ε (s−T )|m|2 v (x, t) = φm − e fm (s)ds ϕm (x), + αeaT |m|2 t m=1 where ≤ t ≤ T, (9) ( )n ∫ = φε (x)ϕm (x)dx π Ω φεm Throughout this section, we suppose that f ∈ L2 ((0, T ); L2 (Ω)) and g ∈ L2 (Ω) Now we state main results of Secion In Proposition 2.1, we have the neccessary and sufficient conditions for the existence of solution of Problem – Proposition 2.2 is devoted to the existence and the stability of regularization uε,a Finally Theorem 2.1 gives rate convergence of error estimated Proposition 2.1 The problem (2)–(4) has a unique solution u if and only if ∞ ∑  eT |m| gm − m=1 ∫T 2 es|m| fm (s)ds < ∞ (10) Proposition 2.2 Let f ∈ L2 ((0, T ); L2 (Ω)) and g ∈ L2 (Ω) The problem (5)–(7) has a unique weak solution uε,a ∈ C([0, T ]; L2 (Ω)) ∩ L2 ((0, T ); H01 (Ω)) ∩ C ([0, T ]; H01 (Ω)) satisfying (8) The solution of problem (5)–(7) depends continuously on g in L2 (Ω) Theorem 2.1 Let v ε,a (., t), defined in (9), be the unique solution of Problem (5)–(7)corresponding to the noisy data g ε Suppose that the problem (2)–(4) has a unique solution u(., t) ∈ W Let f, g be the functions satisfying the condition (10) a) Assume that the function f satisfying the condition ∫ ∞ ∑ T 2 e2s|m| fm (s)ds < ∞ m=1 Let α = εa Then one has for every t ∈ [0, T ] ∥u(., t) − v ε,a (., t)∥ ≤ t (C1 + 1)ε T (11) b) Assume that the function f satisfies the strong condition ∫ T ∞ ∑ 2 |m|4 e2s|m| fm (s)ds < ∞ m=1 and ∥uxx (., 0)∥ < ∞ Let α = ε aT T +β for β > then one has for every t ∈ [0, T ] ( ∥u(., t) − v ε,a (., t)∥ ≤ C2 ln( aT ε aT T +β )−1 ) c) Assume that there exists a positive number k ∈ [0, (a − 1)T ) such that ∞ ∑ 2 e2(T +k)|m| gm 0) is of order of Holder type for all t ∈ [0, T ] It is easy to see that the ( )−q p convergence rate ε , (0 < p) is more quickly than the logarithmic order ln( ) (q > 0) when ε → This ε proves that our method is effective Inhomogeneous problem : the nonlinear case In this section, we analyse the ill-posedness of the nonlinear backward heat problem in frequency space and explain why we introduce the approximation problem Consider the following heat equation system (1) with T = 1, f ∈ L∞ (Ω × [0, 1] × R) The inverse problem is to determine the value of u(x, t) for ≤ t < from the data φε (x) If the solution exists, then the problem has a unique solution ([36], Theorem 3.1, p 239) For system (1) we not guarantee that the solutions exist In [36], we present an simple way to check the existence of system (1) (See Theorem 3.2a, page 239) The main purpose of this section is to find a computation method for the exact solution when it exists Hence, the regularization techniques are required 3.1 Regularization and error estimate Informally, Problem (1) can be transformed the integral equation ( ) ∫ T ∞ ∑ −(t−T )|m|2 −(T −s)|m|2 u(x, t) = e φm − e fp (u)(s)ds ϕm (x) m=1 t (14) where φ(x) = ∞ ∑ φm ϕm (x), f (u)(x, t) = m=1 ∞ ∑ fm (u)(t)ϕm (x) are the expansion of φ and f (u) respectively m=1 Since t < T , we know from (14) that, when m become large, e(T −t)|m| increase rather quickly Thus, the term e−(t−T )|m| is the unstability cause Hence, to regularize the problem, we have to replace the term by different better term Let a ≥ be the fixed number We shall replace e−(t−T )|m| by the new better term e(T −t)|m| to obtain the integral equation + α|m|γ eaT |m|2 ( ) ∫ T ∞ (T −t)|m|2 ∑ e uε (x, t) = φm − e(s−T )|m| fm (uε )(s)ds ϕm (x), 0≤t≤T (15) + α|m|γ eaT |m|2 t m=1 where α is a positive regularization parameter depend on ε and γ > is a constant We denote ( )n ( )n ∫ 2 fm (u)(t) = < f (., t, u(., t)), ϕm (.) >= f (x, t, u(x, t))ϕm (x)dx, π π Ω φm = ( )n ( )n ∫ 2 < φ(.), ϕm (.) >= φ(x)ϕm (x)dx π π Ω (16) (17) and < , > is the inner product in L2 (Ω) Throught out this section, we assume φ ∈ L2 (Ω), α > and f ∈ L∞ (Ω × [0, T ] × R) satisfy |f (x, y, w) − f (x, y, v)| ≤ k|w − v| (18) for a k > independent of x, y, w, v We also denote W = C([0, T ]; L2 (Ω)) ∩ L2 ((0, T ); H01 (Ω)) ∩ C ([0, T ]; H01 (Ω)) Denote u(., t) and uε (., t) be the solution of Problem (1) and Problem (15) We also denote v ε (., t) be the solution of problem (15) corresponding to noisy data φε In the next theorem, we shall study the existence, the uniqueness and the stability of a (weak) solution of Problem (15) In fact, one has Proposition 3.1 The Problem (15) has uniquely a weak solution uε (x, t) ∈ W Furthermore, the solution depends continuously on φ in C([0, T ]; L2 (Ω)) Let w and v be two solutions of Problem (15) corresponding to the final values φ and ω, then one has ∥w(., t) − v(., t)∥ ≤ α t−T aT exp(k (T − t)2 )∥φ − ω∥ (19) From the latter proposition, we get the rate of convergence of error estimates Theorem 3.1 Assume that there exists a positive number P1 such that ∞ ∫ T ∑ 2 2∥u(., 0)∥2H γ + πT m2γ e2s|m| fm (u)(s)ds < P12 m=1 (20) Letting α = ε, one has ∥u(., t) − v ε (., t)∥ ≤ ek T (T −t) ( ) t P1 + ε1− a ε aT (21) where v ε is the solution of problem (15) corresponding to the noisy data φε We remark that If f = 0, the estimate (21) becomes ∥u(., t) − v ε (., t)∥ ≤ ek T (T −t) t ε aT ( ε1− a + ) √ 2∥u(., 0)∥ (22) This order error is the same in some paper [5, 36] The assumption of the nonlinear term f (u) in (20) is introduced in [36] ( Theorem 3.2, p.239 and Remark 3.3, page 242) Then, the assumption is acceptable The approximation error depends continuously on the measurement error for fixed < t ≤ T However, as t → 0, the accuracy of regularized solution becomes progressively lower This is a common property in the theory of ill-posed problems, if we not have additional conditions on the smoothness of the solution To retain the continuous dependence of the solution at t = 0, we introduce a stronger a priori assumption and get Theorem 3.2 Assume that there exists a positive number P2 such that ∞ ∑ m2γ e2(aT −T +t)|m| u2m (t) ≤ P22 , ≤ t ≤ T (23) m=1 Letting α = ε, one has ∥u(., t) − v ε (., t)∥ ≤ ek T (T −t) ε a−1 a t ε aT (P2 + 1) (24) Remark In t = 0, the estimate (24) becomes ∥u(., t) − v ε (., t)∥ The order of convergence of this error is ε a−1 a ≤ ek T (T −t) ε a−1 a (P2 + 1) (25) By choosing a suitable constant a, we have a better error a−1 The estimate (25) is of order of Holder type for all t ∈ [0, T ] As known, the convergence rate of ε a is ( ( ))−q more quickly than the logarithmic order ln (q > 0) which is investigated in many recent papers such ε as [2, 5, 6, 7, 9, 10, 11, 35, 36, 37, 38, 39, 40] 4.1 Numerical examples Numerical results in nonlinear 2-D case In this subsection, we will implement proposed regularization methods by mainly giving two simple examples (with choosing T = 1) We divide the section into three subsections The first one is to consider the main numerical example More precisely, we take a type of the Ginzburg-Landau equation as an example of the considered problem Then, the example in [42] is taken again in the second subsection in order to compare the stability with the present paper in the case of exact data (noise amplitude ε = 0) Finally, the last one is showing our comments and discussions 4.1.1 Example We are going to consider the following backward problem for Ginzburg-Landau equation  ( ) ∂u (x, y, t)   − ∆u (x, y, t) = 2−1 u − u2 + G (x, y, t) , (x, y, t) ∈ QT ,  ∂t , (x, y) ∈ Ω, u (x, y, 1) = g (x, y)   u (0, y, t) = u (π, y, t) = u (x, 0, t) = u (x, π, t) = , t ∈ [0, T ] , where Ω = (0, π) × (0, π) ⊂ R2 , QT = Ω × (0, 1), ∆ = (26) ∂ ∂ + is the 2-D Laplace operator Here ∂x2 ∂y G (x, y, t) = 2e− sin x + 2−1 y (π − y) e− sin3 x, t 3t (27) Figure 1: The exact solution u (x, y, t) = y (π − y) e− sin x at t = 0.9 (left) and 0.6 (right) t and g (x, y) = y (π − y) e− sin x (28) − 2t The exact solution to problem (26) is u (x, y, t) = y (π − y) e sin x The aim of the numerical examples is to observe ε = c · 10−r where c ∈ R is a positive number and r ∈ N will be given later The measured data g ε is defined by two ways below Clearly speaking, we will take perturbation, intended to define as ϵ · rand where each random term rand will be determined on [−1, 1] uniformly, in exact data g In particular, it can be chosen in one of the following g ε (.) = g ε (.) = ( ) ε · rand g (.) + , ∥g∥ ε · rand g (.) + √ π (29) (30) Besides, the regularized one is expected to be closed to the exact under a proper discretization The l2 -norm and l∞ -norm errors and the relative root mean square (RRMS) error will be considered Simultaneously, 2-D and 3-D graphs are applied and analysed The regularized solution in the problem (15) with γ = 2, a = is established as follows v ε (x, y, t) = ∫ M ∑ N ( ∑ ε Φmn (ϵ, 1, t) gmn − ) Φmn (ε, s, t) fmn (v ε ) (s) ds sin (mx) sin (ny) , (31) t m=1 n=1 ε where Φmn (ε, s, t) , gmn , fmn (v ε ) (s) is given by 2 e(s−t−1)(m +n ) Φmn (ε, s, t) = , ε (m2 + n2 ) + e−(m2 +n2 ) ε gmn ) ∫ π∫ π( ε · rand − 12 √ y (π − y) e sin x + sin (mx) sin (ny) dxdy π2 0 π  n n   e− 12 (1 − (−1) ) + 8√ ε · rand (1 − (−1) ) , m = 1,  π n3 n π2 π = m n (1 − (−1) ) (1 − (−1) )   √ ε · rand , m > 1,  mn π2 π (32) = (33) and ε fmn (v ) (s) = = ) ∫ π∫ π( ( ) ε ε v (x, y, s) − (v (x, y, s)) + G (x, y, s) sin (mx) sin (ny) dxdy π2 0 ∫ π∫ π [ ] v ε (x, y, s) − (v ε (x, y, s)) sin (mx) sin (ny) dxdy + Gmn (s) , π 0 π From (27), we calculate the Fourier coefficients Gmn in the following ∫ π∫ π( ) s 3s Gmn (s) = 2e− sin x + 2−1 y (π − y) e− sin3 x sin (mx) sin (ny) dxdy 0 ( ) n n  27 − 3s 10 − n2 π (1 − (−1) ) − 2s (1 − (−1) )   πe + πe , m = 1,   n7 (n )2 n 2 (1 − (−1) ) 3s 10 − n π = − πe− , m = 3,    n  0 , m ̸= {1; 3} (34) (35) where M, N are the truncation numbers ε In order to control the nonlinear term, choose vL (x, y) ≡ v ε (x, y, tL1 ) when dividing the time ti = i∆t, ∆t = 1 , i = 0, L1 , then our intention is to construct a iterative scheme that approximates to v ε when M becomes L1 large enough For i = 0, L1 − 1, we have v ε (x, y, ti ) = M ∑ N ∑ (Rmn (ε, ti ) − Wmn (ε, ti )) sin (mx) sin (ny) , (36) m=1 n=1 where ε Rmn (ε, ti ) = Φmn (ε, 1, ti ) gmn , Wmn (ε, ti ) = (37) ∫ π∫ π L1 ∫ th [ ] ∑ ε ε Φ (ε, s, t ) ds v (x, y) − (v (x, y)) sin (mx) sin (ny) dxdy mn i h h π2 th−1 0 h=i+1 L1 ∫ th ∑ + Φmn (ε, s, ti ) Gmn (s) ds π th−1 (38) h=i+1 In general, the whole process of computation is summarized in the following steps Step Choose L1 and L2 and L3 to have ti = i∆t, ∆t = , i = 0, L1 , L1 (39) xj = j∆x, ∆x = π , j = 0, L2 , L2 (40) yl = l∆y, ∆y = π , l = 0, L3 L3 (41) ε Step Put viε (x, y) ≡ v ε (x, y, ti ) , i = 0, L1 and set vL (x, y) = M ∑ N ∑ ε Φmn (ε, 1, 1) gmn sin (mx) sin (ny) m=1 n=1 Then, we find [ V ε (x, y) = v0ε (x, y) v1ε (x, y) ε vL (x, y) −1 ]T ε vL (x, y) ∈ RL1 +1 (42) ε Step For j = 0, L2 and l = 0, L3 , put viε (xj , yl ) = vi,j,l and u (xj , yl , ti ) = ui,j,l We construct two ε matrixes containing all discrete values of v and u at fixed time ti , denoted by Ai and Bi , respectively  ε  ε ε vi,0,0 vi,0,1 · · · vi,0,L ε ε ε  vi,1,0 vi,1,1 · · · vi,1,L3    L +1 L +1 Ai =  (43)  ∈ R × R ,   ε vi,L ,1 ε vi,L ,L3 ui,0,0  ui,1,0  Bi =   ui,0,1 ui,1,1 ··· ··· ui,0,L3 ui,1,L3 ui,L2 ,0 ui,L2 ,1 ui,L2 ,L3  ε vi,L ,0     ∈ RL2 +1 × RL3 +1  (44) Step For i = 0, L1 , compute errors ∑ ∑ |v ε (xj , yl , ti ) − u (xj , yl , ti )| , (L2 + 1) (L3 + 1) j=0 L E1 (ti ) = L (45) l=0 E2 (ti ) = max 0≤j≤L2 ,0≤l≤L3 |v ε (xj , yl , ti ) − u (xj , yl , ti )| √∑ L2 ∑L3 E3 (ti ) = (46) ε l=0 |v (xj , yl , ti ) − u (xj , yl , ti )| √∑ L2 ∑L3 |u (x , y , t )| j l i j=0 l=0 j=0 (47) Remark 4.1 In order to calculate the first term in the right-hand side of (38), we use Gauss-Legendre quadrature method (see in [47]) to estimate the term as follows ∫ π ∫ π vhε [ ] (x, y) − (vhε (x, y)) sin (mx) sin (ny) dxdy = m0 ∑ n0 ∑ [ ] wj wl vhε (xj , yl ) − (vhε (xj , yl )) j=0 l=0 × sin (mxj ) sin (nyl ) (48) where xj and yl are abscissas in [0, π] and wj and wl are associated weights, corresponding to two discrete spatial variables In general, the second term in the right-hand side of (38) can be approximated as above, but it can be clearly determined by direct computation from (32) and (35) ∫ th Φmn (ε, s, ti ) Gmn (s) ds = th−1  ( n  − 2s (1 − (−1) )   πΦ (ε, s, t ) e 1n i   2n2 + n  ) s=th  ( )  n  2  10 − n π (1 − (−1) ) 3s 27  −  e + 2n − n7 s=th−1 ( )  n  2  (1 − (−1) ) 3s 10 − n π −9π    Φ3n (ε, s, ti ) e−   2n + 15 n7     0 , m = 1, s=th , m = 3, s=th−1 , m ̸= {1; 3} (49) Moreover, we note that ∫ th Φmn (ε, s, ti ) ds = th−1 [Φmn (ε, th , ti ) − Φmn (ε, th−1 , ti )] m + n2 Therefore, Wmn (ε, ti ) in (38) can be approximately determined (50) π with t = in case Table In these figures, we have five levels of contour, respectively, u(x, 0) = 0.05, 0.1, 0.15, 0.2 and 0.25 when t decreases to Firgures 10 - Figure 12 present the solutions at z = Acknowledgments This work is supported by Vietnam National University HoChiMinh City(VNU-HCM) under Grant No.B201418-01 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 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al., Numerical recipes in Fortran 90, 2nd ed., Cambridge University Press, New York, 1996 A Lemma Lemma A.1 Let n, x, α, ε > 0, ≤ a ≤ b, then i ii iii a ena ≤ x− b nb + xe T ≤ , −αT εα + e ε ln(T /ε) Tx ≤ , ε + e−xT ε ln(T /ε) (73) ε ∈ (0, eT ) ε ∈ (0, eT ), x ≥ Proof i We have ena e−n(b−a) −a b = a a ≤ a ≤ x nb −nb + xe (x + e ) b (x + e−nb )1− b (x + e−nb ) b ii See [37], page iii We have x x T = ≤ ≤x −xT −xT −xT ε+e εx + xe εx + e ε ln(T /ε) 24 (74) (75) B Proof of Proposition 2.1 If Problem (2)–(4) has an exact solution u ∈ C([0, T ]; H01 (0, π)) ∩ C ((0, T ); L2 (0, π)), then u can be formulated in the frequency domain ∞ ∑ u(x, t) =  e−(t−T )m gm − m=1 ∫T  e−(t−s)m fm (s)ds sin(mx) (76) t This implies that um (0) = e T m2 ∫T gm − esm fm (s)ds (77) Then ∥u(., 0)∥2 = ∞ ∑  eT m gm − m=1 ∫T 2 esm fm (s)ds < ∞ If we get (10), then we put v(x) = ∞ ∑  eT m gm − m=1 ∫T  esm fm (s)ds sin mx and obtain v ∈ L2 (0, π) Consider the problem    ut − uxx = f (x, t), u(0, t) = u(π, t) = 0, t ∈ (0, T ),   u(x, 0) = v(x), x ∈ (0, π) (78) The classical problem (78) has a unique solution u (see [8]) We have u(x, t) = ∞ ∑  e−tm < v(x), sin mx > + m=1 ∫t  e(s−t)m fm (s)ds sin mx (79) Letting t = T in (79), we have u(x, T ) = ∞ ∑   e−T m eT m gm − 2 m=1 = ∞ ∑ ∫T  esm fm (s)ds + ∫T  e(s−T )m fm (s)ds sin mx gm sin mx = g(x) m=1 Hence, u is the unique solution of (2)–(4) C Proof of Proposition 2.2 The proof is divided into three step Step If uε,a ∈ C([0, T ]; L2 (0, π)) ∩ L2 (0, T ; H01 (0, π)) ∩ C (0, T ; H01 (0, π)) satisfies (8) then uε is solution of 25 (5)–(7) We have u ε,a ( ) ∫ T e(T −t)m (s−T )m2 (x, t) = gm − e fm (s)ds sin(mx) + αeaT m2 t m=1 ( ) ∫ T 2 ∞ ∑ e−tm e(s−t−T )m = gm − fm (s)ds sin(mx) e−T m2 + αe(a−1)T m2 e−T m2 + αe(a−1)T m2 t m=1 ∞ ∑ for ≤ t ≤ T We can verify directly that uε,a ∈ C([0, T ]; L2 (0, π)) ∩ C ((0, T ); H01 (0, π)) ∩ L2 (0, T ; H01 (0, π))) In fact, uε.a ∈ C ∞ ([0, T ]; H01 (0, π)) Moreover, one has uε,a t (x, t) ∞ ( ∑ = m=1 ∫ −m2 e−tm gm − −T m e + αe(a−1)T m2 T t m2 e(s−t−T )m fm (s)ds −T e m2 + αe(a−1)T m2 ) e f (t) sin(mx) m e + αe(a−1)T m ∞ ∞ ∑ ∑ ε,a e−T m =− m ⟨u (x, t), sin mx⟩ sin(mx) + fm (t) sin(mx) π m=1 e−T m2 + αe(a−1)T m2 m=1 −T m2 + −T m2 = uε,a xx (x, t) + ∞ ∑ e−T m fm (t) sin(mx) + αe(a−1)T m2 m=1 e−T m2 and ∞ ∑ uε,a (x, T ) = m=1 e−T m gm sin(mx) + αe(a−1)T m2 e−T m2 So uε,a is the solution of (5)–(7) Step The Problem (5)–(7) has at most one solution C([0, T ]; H01 (0, π)) ∩ C ((0, T ); L2 (0, π)) A proof of this statement can be found in [8] Step The solution of problem (5)–(7) given by (8) depends continuously on g in L2 (0, π) Let u and v be two solutions of (8) corresponding to the final values g and h respectively Then ∥u(., t) − v(., t)∥ ≤ α t−T aT ∥g − h∥ (80) From u and v be two solutions of (8) corresponding to the final values g and h we have ( ) ∫ T ∞ ∑ e(T −t)m (s−T )m2 u(x, t) = gm − e fm (s)ds sin(mx) ≤ t ≤ T, + αeaT m2 t m=1 ∞ ∑ e(T −t)m v(x, t) = + αeaT m2 m=1 where gm = π ∫ ( ∫ ) T hm − (81) e (s−T )m2 fm (s)ds sin(mx) ≤ t ≤ T, (82) t π g(x) sin(mx)dx, hm = π ∫ π h(x) sin(mx)dx (83) Using (75), we obtain ∥u(., t) − v(., t)∥ = ∞ π ∑ e(T −t)m (gm − hm ) , m=1 + αeaT m2 ≤ ∞ π 2t−2T ∑ α aT |gp − hp |2 p=1 = α 2t−2T aT 26 ∥g − h∥2 (84) Therefore ∥u(., t) − v(., t)∥ ≤ α t−T aT ∥g − h∥ (85) This completes the proof D Proof of Theorem 2.1 Let uε,a be the solution defined by (19) with exact data g Using the triangle inequality, we get ∥u(., t) − v ε,a (., t)∥ ≤ ∥u(., t) − uε,a (., t)∥ + ∥uε,a (., t) − v ε,a (., t)∥ (86) For the term ∥uε,a (., t) − v ε,a (., t)∥, using (85), we obtain ∥uε (., t) − v ε,a (., t)∥ ≤ α t−T aT ∥g ε (.) − g(.)∥ ≤ α t−T aT ε (87) a) Suppose the Problem (2)–(4) has an exact solution u ∈ C([0, T ]; H01 (0, π)) ∩ C ((0, T ); L2 (0, π)), then u can be formulated in the frequency domain u(x, t) = ∞ ∑ −(t−T )m2 (e ∫T e−(t−s)m fm (s)ds) sin(mx) gm − m=1 (88) t This implies that um (t) = e −(t−T )m2 ∫T e−(t−s)m fm (s)ds gm − (89) t Since (8), we get e(T −t)m = + αeaT m2 uε,a m (t) ( ∫ gm − ) T e (s−T )m2 fm (s)ds (90) t Combining (89) and (90) and using (75) give ( |um (t) − uε,a m (t)| = e−(t−T )m − (T −t)m2  ) ∫T gm − e−(T −s)m fm (s)ds) e + αeaT m  t  2 eaT m = α e(T −t)m gm − e−(T −s)m fm (s)ds) (1 + αeaT m2 ) t   ∫T 2 α = e(T −t)m gm − e−(T −s)m fm (s)ds) (α + e−aT m2 ) t ( ) ∫ T αe−tm sm2 = |um (0)| + e |fm (s)|ds (α + e−aT m2 ) ( ) ∫ T t ≤ α aT |um (0)| + esm |fm (s)|ds 27 ∫T (91) (92) It follows that ∥u(., t) − uε,a (., t)∥2 = ∞ π ∑ |um (t) − uε,a m (t)| m=1 2t ≤ α aT 2β = α aT ∞ π ∑ 2βm2 e |um (t)|2 m=1 ( ) ∫ T ∞ π ∑ 2|um (0)|2 + 2T e2sm |fm (s)|2 ds m=1 Hence t ∥u(., t) − uε,a (., t)∥ ≤ C1 α aT (93) where ∫ T 2∥u(., 0)∥2 + πT C1 = ∞ ∑ e2sm2 |fm (s)|2 ds m=1 Combining (87),(93) and α = εa gives t ∥u(., t) − v ε (., t)∥ ≤ C1 α aT + α t−T aT ε t T ≤ ε (C1 + 1) b) Since (91), we have   ∫T 2 α |um (t) − uε,a e(T −t)m gm − e−(T −s)m fm (s)ds) m (t)| = (α + e−aT m2 ) t ( ) ∫ T aT 2 sm2 ≤ α m |um (0)| + m e |fm (s)|ds α ln(aT /α) ( ) ∫ T aT = m2 |um (0)| + m2 esm |fm (s)|ds ln(aT /α) It follows that ∥u(., , t) − uε,a (., , t)∥2 ∞ π ∑ |um (t) − uε,a m (t)| m=1 ( ) ( )2 ∑ ∫ T ∞ aT π 4 2sm2 ≤ 2m |um (0)| + 2T m e |fm (s)| ds ln(aT /α) m=1 ) ( )2 ( ∞ ∫ T ∑ aT 2sm2 = 2∥uxx (., 0)∥ + πT m e |fm (s)| ds ln(aT /α) m=1 = Hence ∥u(., t) − uε,a (., t)∥ ≤ C2 where ( C2 = aT 2∥uxx (., 0)∥ + πT ∞ ∫ ∑ m=1 28 ln(aT /α) T (94) ) 2sm2 m e |fm (s)| ds aT Combining (87),(94) and α = ε T +β , we have ∥u(., t) − v ε,a (., t)∥ t−T + α aT ε ln(aT /α) t+β ≤ C2 + ε T +β ln(aT /α) ( )−1 t+β aT = C2 ln( aT ) + ε T +β T +β ε ≤ C2 c) Since (91), we have   ∫T α e(T −t)m gm − e−(T −s)m fm (s)ds) (α + e−aT m2 ) t ( ) ∫ T −(t+k)m2 αe (T +k)m2 (s+k)m2 e gm − e |fm (s)|ds (α + e−aT m2 ) t ( ) ∫ |um (t) − uε,a m (t)| = ≤ t+k T e(T +k)m gm − = α aT 2 e(s+k)m |fm (s)|ds t It implies that ∥u(., , t) − uε,a (., , t)∥2 ∞ π ∑ |um (t) − uε,a m (t)| m=1 ( ) ∫ T ∞ 2t+2k π ∑ 2 2(T +k)m 2(s+k)m ≤ α aT 2e gm + 2T e fm (s)ds m=1 ( ∞ ) ∫ T ∑ 2t+2k 2(T +k)m2 2(s+k)m2 aT = α πe gm + πT e fm (s)ds = m=1 Hence t+k ∥u(., t) − uε,a (., t)∥ ≤ C3 α aT (95) where ( C3 = ∞ ∑ ∫ πe 2(T +k)m2 + πT gm Combining (87),(95) and α = ε e 2(s+k)m2 (s)ds fm m=1 aT T +k ) T , we have ∥u(., t) − v ε,a (., t)∥ ≤ t+k t−T aT t+k T +k t+k T +k C3 α aT + α ≤ C3 ε +ε ε t+k = (C3 + 1)ε T +k This completes the proof E Proof of Proposition 3.1 Using the inequality (75), for ≤ t ≤ s ≤ T , we have t−T e(T −t)m e(T −t)m ≤ ≤ α aT + αmγ eaT m + αeaT m2 2 29 (96) and 2 t−s e(s−t)m e(s−t)m aT ≤ 2 ≤ α γ aT m aT m + αm e + αe (97) First, we have the two following inequality which be useful to the next results For w ∈ C([0, T ]; L2 (0, π)), we set 2 ∞ ∫ ∑ e(T −t)m e(s−t)m G(w)(x, t) = φ sin(mx) − fm (w)(s)ds sin(mx) m + αmγ eaT m + αmγ eaT m2 m=1 m=1 T ∞ ∑ t Denote by |||.||| the sup norm in C([0, T ]; L2 (0, π)) We shall prove by induction, for every w, v ∈ C([0, T ]; L2 (0, π)), p ≥ that ( −1 )p T p/2 √ C p |||w − v||| |||Gp (w) − Gp (v)||| ≤ kα a p! (98) For p = 1,using (96), (97) and Holder inequality, we have  T 2 ∫ ∞ (T −t)m2 ∑ π e  ∥G(w)(., t) − G(v)(., t)∥2 = (fm (w)(s) − fm (v)(s)) ds m=1 + αmγ eaT m2 ≤ π π ≤ = π ( = ∞ ∫T ∑ m=1 t ∞ ∑ ( m=1 ( ( e(T −t)m + αeaT m2 e(T −t)m + αmγ eaT m2 e(T −t)m + αmγ eaT m2 e(T −t)m + αmγ eaT m2 t )2 ∫T t )2 ∫T 2t−2T aT −2 a (fm (w)(s) − fm (v)(s)) ds t )2 (T − t) ∫T ∑ ∞ t )2 (fm (w)(s) − fm (v)(s)) ds m=1 ∫T ∫π (T − t) (f (x, s, w(x, s)) − f (x, s, v(x, s))) dxds ∫T ∫π (T − t) |w(x, s) − v(x, s)|2 dxds t ≤ Ck α (T − t) t ≤ k2 α (fm (w)(s) − fm (v)(s)) ds ds (T − t)|||w − v|||2 30 Thus (98) holds Suppose that (98) holds for p = j We prove that (11) holds for p = j + Using (96), (97) again, we have ∥Gj+1 (w)(., t) − Gj+1 (v)(., t)∥2  T 2 ∫ ∞ (T −t)m2 ∑ ( ) π e  fm (Gj (w))(s) − fm (Gj (v))(s) ds m=1 + αmγ eaT m2 ( π ≤ ( ≤ t e(T −t)m + αeaT m2 e(T −t)m + αeaT m2 )2 (T − t) ∫T ∑ ∞ t )2 |fm (Gj (w))(s) − fm (Gj (v))(s)|2 ds m=1 ∫T (T − t) ∥f (., s, Gj (w)(., s)) − f (., s, Gj (v)(., s))∥2 ds t ≤α 2t−2T aT ∫T (T − t)k ∥Gj (w)(., s) − Gj (v)(., s)∥2 ds t ≤α ( −2 a ≤ kα (T − t)k 2j+2 α −2j a ∫T (T − s)j dsC j |||w − v|||2 j! t −1 a )2j+2 (T − t)j+1 C j+1 |||w − v|||2 (j + 1)! Therefore, by the induction principle, we obtain ( −1 )p T p/2 √ C p |||w − v|||, |||Gp (w) − Gp (v)||| ≤ kα a p! for all w, v ∈ C([0, T ]; L2 (0, π)) We consider G : C([0, T ]; L2 (0, π)) → C([0, T ]; L2 (0, π)) It is easy to see that ( −1 )p T p/2 √ C p |||w − v||| = lim |||Gp (w) − Gp (v)||| ≤ kα a p→∞ p! So, there exists a positive integer number m0 such that Gm0 is a contraction It follows that the equation Gp0 (w) = w has a unique solution uε ∈ C([0, T ]; L2 (0, π)) We claim that G(uε ) = uε In fact, one has G(Gm0 (uε )) = G(uε ) Hence Gp0 (G(uε )) = G(uε ) By the uniqueness of the fixed point of Gp0 , one has G(uε ) = uε , i.e., the equation G(w) = w has a unique solution uε ∈ C([0, T ]; L2 (0, π)) Step The solution of the problem (15) depends continuously on φ in L2 (0, π) Let w and v be two solutions of (15) corresponding to the final values φ and ω From (15) and inequality (a + b)2 ≤ 2(a2 + b2 ), we have ∥w(., t) − v(., t)∥2 ∫ T 2 ∞ π ∑ e(T −t)m e(s−t)m (φ − ω ) − (fm (w)(s) − fm (v)(s)ds) m m m=1 + αmγ eaT m2 + αmγ eaT m2 t 2 ∞ ∞ ∫ T ∑ ∑ e(T −t)m e(s−t)m ≤ π ( |φm − ωm |) + π ( |fm (w)(s) − fm (v)(s)|ds))2 + αmγ eaT m2 + αmγ eaT m2 m=1 m=1 t = Using (9) and (10), we get the following inequality ∥w(., t) − v(., t)∥2 ≤ α 2t−2T aT ∥φ − ω∥2 + 2t ∫ + 2k (T − t)α aT −2s α aT ∥w(., s) − v(., s)∥2 ds t 31 T It follows that −2t α aT ∥w(., t) − v(., t)∥2 ≤ α −2 a ∥φ − ω∥2 ∫ T + 2k (T − t) −2s α aT ∥w(., s) − v(., s)∥2 ds t (99) Using Gronwall’s inequality we have ∥w(., t) − v(., t)∥ ≤ α t−T aT exp(k (T − t)2 )∥φ − ω∥ (100) This completes the proof F Proof of Theorem 3.1 Supposing the Problem (1) has an exact solution u, we get the following formula u(x, t) = ∞ ∑ (e−(t−T )m φm − ∫ T e−(t−s)m fm (u)(s)ds) sin mx (101) t m=1 Hence um (t) = (e −(t−T )m2 ∫ T φp − e−(t−s)m fm (u)(s)ds) (102) t < u(x, t), sin mx > π It follows from (15) that where um (t) = e(T −t)m = + αmγ eaT m2 uεm (t) ( ∫ ) T φm − e (s−T )m2 ε fm (u )(s)ds (103) t Hence, we deduce from (102),(103) that [( ) ( )] ∫ T ∫ T e(T −t)m ε (s−T )m2 ε (s−T )m2 um (t) − um (t) = φm − e fm (u )(s)ds − φm − e fm (u)(s)ds + αmγ eaT m2 t t ( ) ∫ T e(T −t)m (T −t)m2 (s−T )m2 + (e − ) φm − e fm (u)(s)ds + αmγ eaT m2 t ∫ T e(s−t)m = (fm (u)(s) − fm (uε )(s)) ds + αmγ eaT m2 t ( ) ∫ T αe(T −t)m aT m2 (s−T )m2 + e φm − e fm (u)(s)ds = I1 + I2 (104) + αmγ eaT m2 t where ∫ I1 = t T e(s−t)m (fm (u)(s) − fm (uε )(s)) ds + αmγ eaT m2 and αmγ e(T −t)m aT m2 I2 = e + αmγ eaT m2 ( ∫ T φm − ) e(s−T )m fm (u)(s)ds t 32 (105) (106) First, we estimate I1 Since (105) and the inequality (97), we get (∫ I12 )2 e(s−t)m ε = |fm (u (s)) − fm (u(s))|ds + αmγ eaT m2 t ∫ T e(2s−2t)m ≤ (T − t) |fm (uε (s)) − fm (u(s))|2 ds + αmγ eaT m2 t ∫ T −2s 2t ≤ (T − t)α aT α aT |fm (uε (s)) − fm (u(s))|2 ds T (107) t Next, we estimate I2 Since um (0) = e ∫ T m2 T φm − esm fm (u)(s)ds, we get ∫ φm = e−T m um (0) + T e(s−T )m fm (u)(s)ds (108) Replacing (108) into (105) and using(75) , we obtain ( I22 )2 ( )2 ∫ T ∫ T αmγ e(aT −t)m T m2 −T m2 (s−T )m2 (s−T )m2 = e e um (0) + e fm (u)(s)ds − e fm (u)(s)ds + αmγ eaT m2 t ( )2 ∫ t 2t−2aT γ sm2 aT ≤ α α m um (0) + e fm (u)(s)ds ( ) ∫ 2t ≤ 2α aT T m2γ u2m (0) + T 2 m2γ e2sm fm (u)(s)ds By the inequality (a + b)2 ≤ 2(a2 + b2 ) and (96), (97), we have |uεm (t) − um (t)|2 2t ≤ 2(T − t)α aT ( +2α 2t aT ∫ T −2s α aT |fm (uε (s)) − fm (u(s))|2 ds t ∫ m2γ u2m (0) T +T ) 2 m2γ e2sm fm (u)(s)ds Using (18), we get the estimate ∥u(., t) − uε (., t)∥2 = ≤ ∞ π ∑ |um (t) − uεm (t)|2 m=1 ∫ T ∞ −2s ∑ 2t π aT 2(T − t)α α aT |fm (uε (s)) − fm (u(s))|2 ds t m=1 ( ) ∫ T ∞ ∑ 2t 2γ 2γ 2sm + πα aT m um (0) + T m e fm (u)(s)ds 2t ∫ T −2s ≤ 2(T − t)α aT α aT ∥f (., s, u(., s)) − f (., s, uε (., s))∥2 ds t +α m=1 ( 2t aT 2t 2∥u(., 0)∥2H γ ∫ ≤ 2k (T − t)α aT + πT ∞ ∫ ∑ p=1 T −2s T ) 2 m2γ e2sp fm (u)(s)ds 2t α aT ∥u(., s) − uε (., s)∥2 ds + P12 α aT t 33 Hence α −2t aT ∫ ∥u(., t) − u (., t)∥ ≤ ε P12 T −2s α aT ∥u(., s) − uε (., s)∥2 ds + 2k T t Using Gronwall’s inequality, we get −2t α aT ∥u(., t) − uε (., t)∥2 ≤ P12 e2k T (T −t) Therefore ∥u(., t) − uε (., t)∥2 ≤ P12 e2k T (T −t) 2t α aT = P12 e2k T (T −t) 2t α aT (109) Hence ∥u(., t) − uε (., t)∥ ≤ P1 ek T (T −t) t α aT Since the condition ∥φε − φ∥ ≤ ε and (100) give ∥uε (., t) − v ε (., t)∥ ≤ α t−T aT exp(k (T − t)2 )∥φε − φ∥ ≤ exp(k (T − t)2 )α t−T aT ε (110) Putting α = ε, combining (112),(110) and using the triangle inequality, we have ∥u(., t) − v ε (., t)∥ ≤ ∥u(., t) − uε (., t)∥ + ∥uε (., t) − v ε (., t)∥ ≤ P1 ek T (T −t) t α aT + exp(k (T − t)2 )α ( ) t ≤ ek T (T −t) ε aT P1 + ε1− a G t−T aT ε Proof of Theorem 3.2 Using um (t) = e (T −t)m2 ∫ T φm − e(s−t)m fm (u)(s)ds, we estimate I2 by a different way as follows ( I22 = ≤ α αmγ e(T −t)m + αmγ eaT m2 )2 e2(aT −T +t)m u2m (t) m2γ e2(aT −T +t)m u2m (t) 2t−2T +2aT aT (111) By the inequality (a + b)2 ≤ 2(a2 + b2 ) and (104), (107),(111), we have ∫ T −2s 2t |uεm (t) − um (t)|2 ≤ 2(T − t)α aT α aT |fm (uε (s)) − fm (u(s))|2 ds t +2α 2t−2T +2aT aT m2γ e2(aT −T +t)m |um (t)|2 Using (18), we get the estimate ∥u(., t) − uε (., t)∥2 = ≤ ∞ π ∑ |um (t) − uεm (t)|2 m=1 ∫ T ∞ −2s ∑ 2t π aT aT 2(T − t)α α |fm (uε (s)) − fm (u(s))|2 ds t m=1 + πα ≤ 2(T − t)α 2t aT 2t−2T +2aT aT ∫ ∞ ∑ m2γ e2(aT −T +t)m |um (t)|2 m=1 T α −2s aT ∥f (., s, u(., s)) − f (., s, uε (., s))∥2 ds t + P22 α 2t 2t−2T +2aT aT ∫ ≤ 2k (T − t)α aT T −2s α aT ∥u(., s) − uε (., s)∥2 ds + P22 α t 34 2t−2T +2aT aT Hence α −2t aT ∥u(., t) − u (., t)∥ ≤ ε 2 P22 α2− a ∫ T −2s α aT ∥u(., s) − uε (., s)∥2 ds + 2k T t Using Gronwall’s inequality, we get −2t α aT ∥u(., t) − uε (., t)∥2 ≤ P22 α2− a e2k T (T −t) Therefore ∥u(., t) − uε (., t)∥2 ≤ P22 α2− a e2k T (T −t) 2t α aT (112) Combining (110),(111), (112) and using the triangle inequality, we have ∥u(., t) − v ε (., t)∥ ≤ ∥u(., t) − uε (., t)∥ + ∥uε (., t) − v ε (., t)∥ ≤ P2 ek T (T −t) α t−T +aT aT + exp(k (T − t)2 )α Since α = ε, we have ∥u(., t) − v ε (., t)∥ ≤ ek T (T −t) This completes the proof of Theorem 35 ε a−1 a t ε aT (P2 + 1) t−T aT ε Highlights: Up-to-dately, most of studies are focused on the homogeneous problem, andin 1-D, in this study we solve an initial inverse problem for a nonlinear heat equation in 2-D and 3-D We introduce a modified regularization methods Error estimates between the regularization solutions and the exact solution are obtained The proposed methods have been verified by numerical experiments We answer and add all comments of the reviewers ... older stability for a class of initial inverse nonlinear heat problem in multiple dimension Nguyen Huy Tuan , Bui Thanh Duy , Nguyen Dang Minh 1 and Vo Anh Khoa Faculty of Mathematics and Computer... Time Integration Method (FTIM) for solving the multi-dimensional nonhomogeneous and nonlinear backward heat The main idea of a fictitious time integration approach is transformation into a different... Trong and N.H Tuan; Regularization and error estimate for the nonlinear backward heat problem using a method of integral equation., Nonlinear Anal., 71 No.9, 4167-4176, 2009 [40] N.H Tuan, D.D