Regularization of a Cauchy problem for the heat equation

In this paper, we study a Cauchy problem for the heat equation with linear source in the. This problem is ill-posed in the sense of Hadamard. To regularize the problem, the truncation method is proposed to solve the problem in the presence of noisy Cauchy data   and   satisfying We give some error estimates between the regularized solution and the exact solution under some different a-priori conditions of exact solution.

Science & Technology Development, Vol 5, No.T20- 2017 Regularization of a Cauchy problem for the heat equation  Vo Van Au University of Science, VNU-HCM Can Tho University of Technology  Nguyen Hoang Tuan University of Education, Ho Chi Minh (Received on 5th December, 2016, accepted on 28 th November, 2017) ABSTRACT In this paper, we study a Cauchy problem for the heat equation with linear source in the form ut ( x, t )  uxx ( x, t )  f ( x, t ), u(L, t )  (t ), ux (L, t )   (t ), ( x, t )  (0, L)  (0,2 ) This problem is ill-posed in the sense of Hadamard To regularize the problem, the truncation method is proposed to solve the problem in the presence of noisy Cauchy data and satisfying      satisfying         and that f f  ( x, )  f ( x, )   We give some error estimates between the regularized solution and the exact solution under some different a-priori conditions of exact solution Key words: elliptic equation, ill-posed problem, cauchy problem, regularization method, truncation method INTRODUCTION In this paper, the temperature u( x, t ) for ( x, t )  [0, L]  [0, 2 ] is sought from known boundary temperature u( L, t )   (t ) and heat flux ux ( L, t )   (t ) measurements satisfying the following problem:  x  L,  t  2 , ut ( x, t )  uxx ( x, t )  f ( x, t ),  u( L, t )   (t ), u ( L, t )   (t ), ,  where x  t  2 ,  t given  2 , (1) are functions (usually in L (0, 2 ) ) and f is a given linear heat source which may depend on the independent variables ( x, t ) Note that we have no initial condition prescribed at t and moreover, the Cauchy data  and  are perturbed so as to contain measurement errors in the form of the input noisy Cauchy data   and   (also in L2 (0, 2 ) ) satisfying     , (2) where denotes the L2 (0, 2 ) -norm and  is a small positive number representing the level of noise Trang 184 It is well-known that, at least in the linear case, the problem (1) has at most one solution using classical analytical sideways continuation for the parabolic heat equation The existence of solution also holds, in the case f However, the problem is still ill-posed in the sense that the solution, if it exists, does not depend continuously on the data Any small perturbation in the observation data can cause large errors in the solution u( x, t ) for x [0, L) Therefore, most classical numerical methods often fail to give an acceptable approximation of the solution Thus regularization techniques are required to stabilize the solution [3] In recent years, the homogeneous sideways heat equation, i.e., f in the first equation in (1), has been researched by many authors and various methods have been proposed, e.g the difference regularization method [8], the boundary element Tikhonov regularization method [5], the Fourier method [9], the quasi-reversibility method [1, 6], the wavelet, wavelet- TẠP CHÍ PHÁT TRIỂN KH&CN, TẬP 20, SOÁ T5- 2017 Galerkin and spectral regularization methods [2, 7], the conjugate gradient method [4], to mention only a few regularization ones and show that the solution of our regularized problem converges to the solution of the original linear problem (if such solution exists), as the regularization parameter tends to zero In the nonhomogeneous problem, we have many choises of stability terms for regularization However, in the case of non-homogeneous problem, the main solution u is complicated and is defined by a linear integral equation whose the right-hand side depends on the independent variables ( x, t ) In this paper, we develop a truncation method to solve in a stable manner this linear integral equation To the best of our knowledge, the Cauchy problem for the linear sideways heat equation has not yet been Therefore, in the present paper, we propose a new method that is based on linear integral equation to regularize problem (1) under two a priori conditions on the exact solution As will be shown in next section, for the linear sideways heat problem (1), its solution (exact solution) can be represented as an integral equation which contains some instability terms In order to restore the stability we replace these instable terms by some THE MAIN RESULTS Let denote the inner product in L2 (0, 2 ), and  represent the noise level in (2) For   (t ), exp( int ) exp(int ), where  (t ), exp( int ) have the Fourier series  (t ) n L2 (0, 2 ) -norm of 2  is  L2 (0, 2 ), we 2  (t ) exp( int )dt The 2  (t ),exp( int ) (3) n The principal value of in is (1 i) n in (1 i) n 2 , n 0, , n (4) Suppose that the solution of problem (1) is represented as a Fourier series un ( x) exp(int ), with un ( x) u( x, t ) u( x, t ), exp( int ) n 2 2 u( x, t ) exp( int ) d t From (1), we have the following systems of second-order ordinary differential equations:  d un ( x)  inun ( x)  f n ( x),  x  L,  d x2   t  (0, 2 ), un ( L)  n   (t ), exp(int ) , d u  n ( L)   n   (t ), exp( int ) , t  (0, 2 ),   dx where f n ( x)  f ( x, t ),exp(int )  2 (5) 2  f ( x, t ) exp(int ) d t for all n  Trang 185 Science & Technology Development, Vol 5, No.T20- 2017 For n  \{0}, multiplying the first equation in (5) by obtain un ( x)  cosh  ( L  x) in  un ( L)  sinh  ( L  x) in  in sinh  (  x) in  in L un ( L)   and integrating both sides from x to L, we sinh  (  x) in  in x f n ( ) d  , n  \{0} In the case n  0, multiplying the first equation in (5) by   x and integrating both sides from x to L, we obtain (6) L u0 ( x)  u0 ( L)  ( L  x)u0 ( L)   (  x) f ( ) d  (7) x From (6) - (7) the exact form of u is given by L   sinh  ( L  x) in  sinh  (  x) in  u( x, t )   cosh  ( L  x) in  n  n   f n ( ) d   exp(int ) in in  n \{0}  x   (0 , , f )( x), (8) L where (0 , , f )( x)  0  ( L  x)   (  x) f ( ) d  In a few sentences, we present a brief introduction x Fourier truncated method From equation (8), it can be observed that cosh  ( L  x) in  , sinh  ( L  x) in  in sinh  (  x) in  and are unbounded, as n tends to infinity, so in order to guarantee the convergence of the solution u in given by (8), the coefficient (n , n ) must decay rapidly But such a decay usually cannot occur for the measured data (n , n ) Hence, a natural way is to eliminate the high frequencies and consider the solution u for n  N , where N is a positive integer; this is the so-called Fourier truncated method, and N plays the role of a regularization parameter satisfying lim N   We define the following two operators:  0 QN+ ( , , f )( x, t )   QN ( , , f )( x, t )   Q n  N ( , , f )( x) exp(int )   exp  ( L  x) n  N Q n  N + N ,n   N , n   exp  ( L  x) in  in in   n  exp  (  x) in  in x exp  ( L  x) in  To approximate u, we introduce the regularized solution Trang 186 L n    f n ( ) d   exp(int ),  (9) ( , , f )( x) exp(int )  exp  ( L  x) n  N in   n  in L n   x exp  (  x) in  in  f n ( ) d   exp(int ) (10)  TẠP CHÍ PHÁT TRIỂN KH&CN, TẬP 20, SỐ T5- 2017 uNε  ( x, t )    sinh  ( L  x) in   L sinh  (  x) in   n   f n ( ) d   exp(int ) cosh  ( L  x) in  n  in in  n  N , n   x    QN (  ,  , f  )( x, t )  (0 , 0 , f 0 )( x) (11) Our these results would be applied after any necessary minor modifications have been made Lemma For n \{0} and n M , we have the following inequalities: cosh ( L x) in exp M (L sinh ( x) in exp M ( x) , (12) x) (13) Proof For n  \{0}, n  M , one has cosh ( L x) in exp ( L x) in exp ( L exp exp exp x) in M (L (L x) x) in (L x) in M (L exp exp n (L x) exp n (L x) exp n (L x) exp n (L x) x) , and sinh ( L x) in exp ( L x) in in exp ( L x) in in exp ( L x) in exp n exp M (L (L x) in n x) exp M (L x) , as required Lemma For n  N , we have u ( x)  1 QN , n ( , , f )( x)   un ( x)  n  2 in  (14) Proof Differentiating (6) with respect to x gives L cosh  ( L  x) in  cosh  (  x) in  u ( x)  n  sinh  ( L  x) in  n  n   f n ( ) d  in in in x (15) Adding (15) to (6), we infer that Trang 187 Science & Technology Development, Vol 5, No.T20- 2017 un ( x)  un ( x)  exp  ( L  x) in  n  exp  ( L  x) in  in from which complete the proof in L n   x exp  (  x) in  in f n ( ) d  , The following theorem comes from the regularization uN provides the error estimates in the L2 -norm when the exact solution belongs to new spaces Gs , (s 0) Here Gs is presented by Gs (0, 2 )  L2 (0, 2 ) : n exp  n 2s  (t ),exp( int ) , n (16) and this norm is given by  Gs ((0,2 ) n 2s exp  n  (t ),exp( int ) L2 (0,2 ) (17) For a Hilbert space X, we denote L (0, L; X )  :[0, L] X esssup  ( )  L X , (18) and  L (0, L; X ) esssup  ( ) X  L Theorem Assume that problem (1) has a weak solution u  C [0, T ]; L2 (0, 2 )  Choose N  such that   N   lim N1  lim  exp  L     0  0    (19) (20) (a) Suppose that the problem (1) has a solution u satisfying u L 0, L;G0 (0,2 )  ux L 0, L;G0 (0,2 )  E1 , L L (21) for some known constant E1  Then  N  uNε  ( x, )  u( x, )  P  2 E12 exp   x ,   where P  6 exp  L N    (22)  L exp L N  1 N (b) Suppose that the problem (1) has a solution u satisfying u L 0, L;Gr (0,2 )  ux L 0, L;Gr (0,2 )  E2 , L L (23) for r  and some known constant E2  Then  N  uNε  ( x, )  u( x, )  P  2 N2 r E22 exp   x     1 Corollary Let us choose N  ln   for   then   L   Trang 188 (24) TẠP CHÍ PHÁT TRIỂN KH&CN, TẬP 20, SOÁ T5- 2017 Estimate in (22) is calculated as follows x uNε  ( x, )  u( x, )  R  2 E12  L  , where R  6 2 L    6L   2 L  (25) 1      ln     L      Estimate in (24) is calculated as follows    uNε  ( x, )  u( x, )  R  2  ln     L      4 r x E22  L  (26) Proof of the Theorem The proof is divided into two parts  Part a Estimate the error (22) between the regularization uN and the exact solution u with a priori (21) We rewrite u as u( x, t )  L   sinh  ( L  x) in  sinh  (  x) in  n   f n ( ) d   exp(int ) cosh  ( L  x) in  n  in in  n  N , n   x  +   (0 , , f )( x)  QN ( , , f )( x, t )  QN ( , , f )( x, t )  (27) From (11) and (27), thanks to Parseval’s relation, we obtain uN  ( x, )  u( x, )  2  n  N , n  uN  , n ( x)  un ( x)  4  n  N QN ,n (  ,  , f  )( x)  QN ,n ( , , f )( x) : J1 ( x )   : J ( x )   2 (n , n , f n )( x)  (0 , , f )( x)  4 : J ( x )  n  N QN+ ,n ( , , f )( x) : J ( x ) (28) We now apply Lemma and using the Holder’s inequality, we have  2 sinh  ( L  x) in    J1 ( x)  6  cosh  ( L  x) in  n  n   n  n in n  N , n     6 L   sinh  (  x) in  in n  N , n  x  6  n  N , n   6      f n ( )  f n ( )  d      exp ( L  x) N      exp ( L  x) N      n n  n n   L  ( L  x )   x exp (  x) N n  N , n    f  n  ( )  f n ( ) d  ,  (29) where we have used the elementary inequality (a  b  c)2  3(a2  b2  c2 ) Similarly, the second equation J ( x) writes Trang 189 Science & Technology Development, Vol 5, No.T20- 2017  2 exp  ( L  x) in      J ( x)  12  exp ( L  x) in  n   n   n  n  in n  N  L    12 exp  (  x) in  in n  N x  12  exp  ( L  x) n  N   12  f n ( )  f n ( )  d      2 N  n   n  exp ( L  x) N   n  n  L      ( L  x) exp  (  x) n  N 2 N f  n x  ( )  f n ( ) d    (30) Thanks to Holder’s inequality and using the basic inequality ea  a, a  0, we deduce that L   2 J ( x)  6  0  0  ( L  x)  0   ( L  x)  (  x) f 0 ( )  f ( ) d   x       2  6 exp ( L  x) N 0  0  exp ( L  x) N   0     6 ( L  x)  exp (  x) N x  Using Lemma 2, easy calculations show that  L J ( x)  4    exp   x n  N   n  N   in   exp  x in  un ( x)  exp  x  u ( x )  in  n      in   2 exp  x N (31) exp   x in  exp  x in  un ( x )  exp  x in  un ( x) (32) in    exp  L n  un ( x)   exp  L n n  N  n  N L  0, L ;GL0 (0,2 )   ux L  0, L ;GL0 (0,2 )   u ( x ) n in      L   2  exp ( L  x) N            ( L  x)  exp (  L) N  f  ( , )  f ( , ) d  x   2    2 exp  x N u   ux L  0, L;G (0,2 )  L  L  0, L;GL (0,2 )   L    exp ( L  x) N   L  exp (  L) N d   2 exp  x N  E12 x     L  exp ( L  x) N    exp ( x  L) N    2 exp  x N  E12 , (33) N        which can be rewritten as    Trang 190  ( )  f ( ) d    n  N  2 exp  x N  u  Combining (28), (29), (30), (31) and (32) we infer uN  ( x, )  u( x, )  u ( x )  1 un ( x)  n   2 in  QN , n ( , , f )( x)  4 n  N f            (33) TAÏP CHÍ PHÁT TRIỂN KH&CN, TẬP 20, SỐ T5- 2017    2 L exp L N  1  N   2 uN ( x, )  u( x, )  6 exp L N   2 E1  exp   x       N     (34)  Part (b) Estimate the error (24) between the regularization uN and the exact solution u with a priori (23) By an argument analogous to the previous one, the estimates of J1 ( x), J ( x), J ( x) in the proof of part (a) remains valid Also, replace J ( x) by following estimate J ( x)  4   n  N  n  n r n  N  2 r un ( x) exp   x in       2 N2 r exp  x N 2 r n  N u ( x )  1 un ( x)  n   2 in  u ( x )  r  r exp   x in   n exp  x in  un ( x )  n exp  x in  n  in   n exp  x in  un ( x )  n exp  x in  r n  N  2 N  QN , n ( , , f )( x)  4 r  2r   n exp  L n  n  N exp  x N  u  L  0, L ;GLr (0,2 )   ux 2  u ( x) n  n 2r in exp  L n  u ( x ) n in n  N L  0, L ;GLr (0,2 )       (35) Combining (28), (29), (30), (31) and (35), we get We obtain uN  ( x, )  u( x, )   exp ( L  x) N        2 N 2 r exp  x N        exp ( L  x) N   exp ( L  x) N   uN  ( x, )  u( x, )  6 exp L    N      u      L  f  ( , )  f ( , ) d     ( L  x)  exp (  L) N  x L  0, L ;GLr (0,2 )   ux L  0, L ;GLr (0,2 )    L   L  exp (  L) N  d   2 N 2 r exp  x N  E22 x     L  exp ( x  L) N    2 N 2 r exp  x N  E22   N       L   exp L    exp  x N    N  N   2 N 2 r  E    2    exp   x  N    (36) (37) This completes the proof of the theorem CONCLUSION In this paper, the Cauchy problem for the heat equation has been solved by employing the truncation method for a resulting linear integral equation Convergence and stability estimates, as the regularization parameter tends to zero, are proved Trang 191 Science & Technology Development, Vol 5, No.T20- 2017 Chỉnh hóa tốn Cauchy cho phương trình nhiệt  Võ Văn Âu Trường Đại học Khoa học Tự nhiên, ĐHQG-HCM Trường Đại học Kỹ thuật Công nghệ Cần Thơ  Nguyễn Hoàng Tuấn Trường Đại học Sư phạm Thành phố Hồ Chí Minh TĨM TẮT Trong báo này, chúng tơi nghiên cứu tốn trường hợp liệu Cauchy  ,  hàm nguồn f bị Cauchy cho phương trình nhiệt với hàm nguồn tuyến nhiễu thỏa mãn  ,   f   tính thỏa phương trình:         f  ( x, )  f ( x, )   ut ( x, t )  uxx ( x, t )  f ( x, t ), u(L, t )  (t ), ux (L, t )   (t ), ( x, t )  (0, L)  (0,2 ) Chúng đưa đánh giá sai số nghiệm Đây tốn khơng chỉnh theo nghĩa chỉnh hóa nghiệm xác số tính trơn Hadamard Để chỉnh hóa tốn này, phương pháp khác nghiệm xác chặt cụt đề xuất để giải tốn Từ khóa: phương trình Eliptic, tốn khơng chỉnh, tốn Cauchy, phương pháp chỉnh hóa, phương pháp chặt cụt TÀI LIỆU THAM KHẢO [1] L Elden, Approximations for a Cauchy problem for the heat equation, Inverse Problems, 3, 263–273 (1987) [2] L Elden, F Berntsson, T Reginska, Wavelet and Fourier methods for solving the sideways heat equation, SIAM J Sci Comput., 21, 2187–2205 (2000) [3] D.N Hao, Methods for Inverse Heat Conduction Problems, Peter Lang, Frankfurt am Main (1998) [4] D.N Hao, P.X Thanh, D Lesnic, B.T Johansson, A boundary element method for a multidimensional inverse heat conduction problem, Int J Computer Math 89, 1540–1554 (2012) [5] D Lesnic, L Elliott, D.B Ingham, Application of the boundary element method to inverse heat conduction problems, Int J Heat Mass Transfer, 39, 1503–1517 (1996) Trang 192 [6] J.C Liu, T Wei, A quasi-reversibility regularization method for an inverse heat conduction problem without initial data, Appl Math Comput., 219, 10866–10881 (2013) [7] T Reginska, L Elden, Solving the sideways heat equation by a wavelet-Galerkin method, Inverse Problems, 13, 1093–1106 (1997) [8] X.T Xiong, C.L Fu, H.F Li, Central difference method of a non-standard inverse heat conduction problem for determining surface heat flux from interior observations, Appl Math Comput 173, 1265–1287 (2006) [9] X.T Xiong, C.L Fu, H.F Li, Fourier regularization method of a sideways heat equation for determining surface heat flux, J Math Anal Appl 317, 331– 348 (2006)  ... completes the proof of the theorem CONCLUSION In this paper, the Cauchy problem for the heat equation has been solved by employing the truncation method for a resulting linear integral equation Convergence... independent variables ( x, t ) In this paper, we develop a truncation method to solve in a stable manner this linear integral equation To the best of our knowledge, the Cauchy problem for the linear sideways... observations, Appl Math Comput 173, 1265–1287 (2006) [9] X.T Xiong, C.L Fu, H.F Li, Fourier regularization method of a sideways heat equation for determining surface heat flux, J Math Anal Appl