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In the present paper, we consider a backward problem for a space-fractional diffusion equation (SFDE) with a time-dependent coefficient. Such the problem is obtained from the classical diffusion equation by replacing the second-order spatial derivative with the Riesz-Feller derivative of order 0,2 .
Science & Technology Development, Vol 5, No.T20- 2017 Regularization for a Riesz-Feller space fractional backward diffusion problem with a time-dependent coefficient Dinh Nguyen Duy Hai University of Science, VNU-HCM Ho Chi Minh City University of Transport (Received on 5th December 2016, accepted on 28 th November 2017) ABSTRACT In the present paper, we consider a backward 0,2 This problem is ill-posed, i.e., the solution problem for a space-fractional diffusion equation (if it exists) does not depend continuously on the data (SFDE) with a time-dependent coefficient Such the Therefore, we propose one new regularization solution problem is obtained from the classical diffusion to solve it Then, the convergence estimate is obtained equation by replacing the second-order spatial under a priori bound assumptions for exact solution derivative with the Riesz-Feller derivative of order Key words: space-fractional backward diffusion problem, Ill-posed problem, Regularization, error estimate, time-dependent coefficient INTRODUCTION The fractional differential equations appear more and more frequently in physical, chemical, biology and engineering applications Nowadays, fractional diffusion equation plays important roles in modeling anomalous diffusion and subdiffusion systems [2], description of fractional random walk, unification of diffusion [3], and wave propagation phenomenon [4] It is well known that the SFDE is obtained from the classical diffusion equation in which the second-order space derivative is replaced with a space-fractional partial derivative Let : [0, T ] is a continuous function on satisfying (t )0 In this paper, we consider a [0, T ] Trang 172 backward problem for the following nonlinear SFDE with a time-dependent coefficient ut ( x, t ) (t ) x D F ( x, t , u ( x, t )), ( x, t ) (0, T ), u ( x, t ) x 0, t (0, T ), u ( x, T) G ( x ), x , (1) where the fractional spatial derivative D x is the Riesz-Feller fractional derivative of order and skewness (0 2) (| | min{ , }, 1) defined in [5], as follows: TẠP CHÍ PHÁT TRIỂN KH&CN, TẬP 20, SỐ T5- 2017 (1 ) ( ) x D f ( x) sin d f ( x) , x D0 f ( x ) dx f ( x s ) f ( x) ( ) ds sin 1 s f ( x s ) f ( x) ds , 2, s1 Here, we wish to determine the temperature u ( x, t ) from temperature measurements G ( x ) Since the measurements usually contain an error, we now could assume that the measured data function G ( x) satisfies G G , where the constant represents the noise level Moreover, assume there hold the following a L ( ) priori bound u (, 0) L ( ) E, E (2) We assume that F satisfies the Lipschitz condition F ( x, t , z1 ) F ( x, t , z2 ) L2 ( ) K F z1 z2 L2 ( ) (3) for some constant K F independent of x, t , z1 , z2 with 1 K F 0, (4) T In case of the source function F and (t ) 1, The remainder of this paper is organized as Problem (1) has been proposed by some authors Zheng follows In Section 2, we propose the regularizing and Wei [7] used two methods, the spectral scheme for Problem (1) Then, in Section 3, we show regularization and modified equation methods, to solve that the regularizing scheme of Problem (1) is wellthis problem In [6], they developed an optimal posed Finally, the convergence estimate is given in modified method to solve this problem by an a priori Section and an a posteriori strategy In 2014, Zhao et al [8] REGULARIZATION FOR PROBLEM (1) applied a simplified Tikhonov regularization method to Let G ( ) denote the Fourier transform of the deal with this problem After then, a new regularization integrable function G ( x ), which defined by method of iteration type for solving this problem has been introduced by Cheng et al [1] Although we have Gˆ ( ) : exp( ix )G ( x) dx, i 1 many works on the linear homogeneous case of the 2 backward problem, the nonlinear case of the problem is In terms of the Fourier transform, we have the following properties for the Riesz-Feller spacequite scarce For the nonlinear problem, the solution u fractional derivative [5] is complicated and defined by an integral equation such D (G )( ) ( )G ( ), that the right hand side depends on u This leads to x studying nonlinear problem is very difficult, so in this where paper we develop a new appropriate technique ( ) | | cos isign( ) sin Trang 173 (5) Science & Technology Development, Vol 5, No.T20- 2017 t We define the function k (t ) by k (t ) ds (s) By taking a Fourier transform to Problem (1), we transform Problem (1) into the following differential equation ut ( , t ) (t ) ( )u ( , t ) F ( , t , u ( , t)), u ( , T) G ( ) (6) The solution to equation (6) is given by T uˆ ( , t ) exp( ( )( k (T ) k (t )))[Gˆ ( ) exp( ( )( k ( s) k (T ))) Fˆ ( , s, u ( , s)) ds] (7) t From (7), applying the inverse Fourier transform, we get T ˆ ( ) exp( ( )( k ( s ) k (T ))) Fˆ ( , s, u (, s )) ds] exp(ix ) d u ( x, t ) exp ( ( )( k ( T ) k ( t )) ) [ G 2 t (8) From which when becomes large, the terms therefore recovering the scalar (temperature, pollution) u ( x, t ) from the measured data G ( x ) is severely illexp ( )( k (T ) k (t )) increases rather quickly: small errors in high-frequency components can blow up posed In this note, we regularize Problem (1) by the and completely destroy the solution for t T , problem U ( , t ) exp ( )( k (T ) k (t )) T G ( ) exp ( )( k ( s ) k (t )) k (T ) k (T ) t exp | | cos exp | | cos k (T ) exp | | cos exp ( k ( s ) k (t )) ( ) Fˆ ( , s, U ( , s )) ds, k (T ) exp | | cos where is regularization parameter t THE WELL POSEDNESS OF PROBLEM (9) First, we consider the following Lemma which is used in the proof of the main results Lemma Let t , s [0, T ] 1) If s t , then we have Trang 174 Fˆ ( , s, U ( , s )) ds (9) TẠP CHÍ PHÁT TRIỂN KH&CN, TẬP 20, SỐ T5- 2017 ( exp ( )( k ( s ) k (t )) a) ( exp | | cos ( k (T ) ( k ( t ) k (T ) ) k (T ) ( )k (T )) exp | | cos 2) If s t , then we have exp( ( )( k ( s ) k (t ) k (T ))) exp(| | cos( )k (T )) c) ( )k (T )) exp ( )( k (T ) k (t )) b) k (t )k ( s ) ) k (t )k ( s ) k (T ) Proof First, we prove (a) In fact, we have exp(| | cos( exp( ( )( k ( s ) k (t ))) exp(| | cos( exp( | | cos( exp(| | cos( [ exp( | | cos( ) k (T )) )( k ( s ) k (t ) k (T ))) k ( s )k (t ) ) k (T ))] [ exp( | | k (T ) cos( k (T ) k ( s ) k ( t ) ) k (T ))] k (T ) k (t )k ( s ) )( k ( s ) k (t ) k (T ))) ) k (T )) exp( | | cos( ) k (T )) k ( s )k (t ) k (T ) k (T ) As an immediate consequence of (a), making the change s T , we have (b) Next, we prove (c) In fact from (b), we obtain exp( ( )( k (T ) ( k (t ) k ( s)))) exp(| | cos( k ( t ) k ( s ) k (T ) k (T ) ) k (T )) it follows that exp( ( )( k ( s ) k (t ) k (T ))) k (t )k ( s ) ( exp | | cos( This completes the proof ) k (T ) k (T ) ) □ We are now in a position to prove the following theorem Trang 175 Science & Technology Development, Vol 5, No.T20- 2017 Theorem Suppose m 0, KFT 2 Let G L2 ( ) and F satisfies (3) then Problem (9) is well-posed Proof We divide it into two steps Step1 The existence and the uniqueness of a solution of Problem (9) Let us define the norm on C ([0; T ]; L2 ( )) as follows k (t ) ‖ h‖ sup ‖ h(t )‖ k (T ) t T , for all h C ([0; T ]; L ( )) 2 L ( ) It is easily be seen that ‖ ‖ is a norm of C ([0; T ]; L2 ( )) For v C ([0; T ]; L2 ( )) , we consider the following function A(v )( x, t ) 2 2 t 2 exp(| | cos( t exp(| | cos( exp( ( )( k ( s ) k (t ))) T B ( x, t ) exp(| | cos( )k (T )) )k (T )) Fˆ ( , s, v ) exp(i x ) dsd )k (T )) exp(( k ( s ) k (t )) ( )) Fˆ ( , s, v ) exp(i x ) dsd , where B ( x, t ) ( ) exp ( )( k (T ) k (t )) ( G ( ) exp(i x ) d ( )k (T )) exp | | cos We claim that, for every v1 , v2 C ([0; T ]; L ( )) ‖ A(v1 ) A(v2 )‖ K F T‖ v1 v2 ‖ First, by Lemma and (3), we have two following estimates for all t [0, T ] T exp( ( )( k ( s ) k (t ))) ˆ ˆ J1 F ( , s , v1 ) F ( , s, v2 ) ds d t exp(| | cos( )k (T )) 2 Trang 176 (10) TAÏP CHÍ PHÁT TRIỂN KH&CN, TẬP 20, SỐ T5- 2017 exp( ( )( k ( s ) k (t ))) T (T t ) exp(| | cos( t Fˆ ( , s , v1 ) Fˆ ( , s , v2 ) dsd )k (T )) 2 ( k ( t ) k ( s )) T (T t ) 2k (t ) Fˆ ( , s , v1 ) Fˆ ( , s , v2 ) dsd k (T ) k (T ) t k (T ) k (T ) ‖ v1 (, s ) v2 (, s )‖ 2 L ( ) ds t 2 k ( s ) 2k (t ) 2 k ( s ) T K F (T t ) K F (T t ) sup 2 k (T ) s T 2k (t ) ‖ v1 (, s ) v2 (, s )‖ ) 2 L ( k (T ) K F (T t ) ‖ v1 v2 ‖ 2 (11) and t exp(| | cos( )k (T )) J2 exp(( k ( s ) k (t )) ( )) Fˆ ( , s , v1 ) Fˆ ( , s, v2 ) ds d exp(| | cos( )k (T )) t t ( exp | | cos ( )k (T )) ( ( exp | | exp ( k ( s ) k ( t )) ( ) cos Fˆ ( , s , v1 ) Fˆ ( , s , v2 ) dsd ) ( )k (T )) t t ( k ( t ) k ( s )) 2k (t ) Fˆ ( , s , v1 ) Fˆ ( , s , v2 ) dsd k (T ) k (T ) k (T ) k (T ) ‖ v1 (, s ) v2 (, s )‖ 2 L ( ) ds (12) 2 k ( s ) 2k (t ) 2 k ( s ) t KFt K F t sup 2 k (T ) s T 2k (t ) ‖ v1 (., s ) v2 (., s )‖ 2 L ( ) k (T ) K F t ‖ v1 v2 ‖ 2 For t T , using the inequality ( a b) (1 m) a 2 1 b for all real numbers a and m b and m 0, we obtain 2k (t ) A(v1 )(, t ) A(v2 )(, t ) By choosing m T t t 2 L ( ) (1 m) k (T ) 2k (t ) 1 2 K F t ‖ v1 v2 ‖ k (T ) K F (T t ) ‖ v1 v2 ‖ m 2 , we have 2 k ( t ) k (T ) ‖ A(v1 )(, t ) A(v2 )(, t )‖ 2 L ( K F T ‖ v1 v2 ‖ , for all t (0, T ) ) 2 (13) On the other hand, letting t in (11), we have ‖ A(v1 )(, 0) A(v2 )(, 0)‖ K F T ‖ v1 v2 ‖ (14) K F T ‖ v1 v2 ‖ (15) 2 L ( ) 2 By letting t T in (12), we have ‖ A(v1 )(, T ) A(v2 )(, T )‖ 2 2 L ( ) 2 Combining (13), (14) and (15), we obtain Trang 177 Science & Technology Development, Vol 5, No.T20- 2017 2 k ( t ) k (T ) ‖ A(v1 )(, t ) A(v2 )(, t )‖ K F T ‖ v1 v2 ‖ , for all t [0, T ] 2 L ( ) 2 which leads to (10) Since K F T 1, A is a contraction It follows that the equation A(v) v has a unique solution U C ([0; T ]; L ( )) Step The solution of Problem (9) continuously depends on the data Let V , W be two solutions of Problem (9) corresponding to the final values GV and GW By straightforward computation, we write exp( ( )( k (T ) k (t ))) W ( , t ) V ( , t ) exp(| | cos( ( t ( ( ) GV ( )) ) [Fˆ ( , s, V cos( )k (T )) ( , s )) Fˆ ( , s, W ( , s )) ds ] )k (T )) W exp | | exp(| | cos( t (G exp ( )( k ( s ) k (t )) T ) k (T )) exp(| | cos( exp(( k ( s ) k (t )) ( )) [ Fˆ ( , s , W ( , s )) Fˆ ( , s , V ( , s ))]ds ) k (T )) Now applying Lemma 1, we get k ( t ) k (T ) W ( , t ) V ( , t ) k (t )k ( s ) T GW ( ) GV ( ) k (T ) k (T ) Fˆ ( , s, V ( , s )) Fˆ ( , s, W ( , s )) ds t k (t )k ( s ) t Fˆ ( , s , V ( , s )) Fˆ ( , s, W ( , s )) ds k (T ) k ( t ) k (T ) k (T ) T GW ( ) GV ( ) k (t )k ( s ) k (T ) Fˆ ( , s , V ( , s )) Fˆ ( , s , W ( , s )) ds Since m 0, KFT T 1 m , we have that K F From the inequality ( a b) for all real number a, b and m 0, we get W (, t ) V (, t ) Trang 178 2 L ( ) W (, t ) V (, t ) 2 L ( ) 1 2 a (1 m)b m (16) TẠP CHÍ PHÁT TRIỂN KH&CN, TẬP 20, SỐ T5- 2017 1 1 m 1 1 m T k ( tk)(Tk)( s ) ˆ (1 m) F ( , s , V ( , s )) Fˆ ( , s, W ( , s )) ds d ) k ( t )2 k (T ) GW GV k (T ) 2 L ( k ( t )2 k (T ) GW GV k (T ) k ( t ) 2 k ( s ) T (1 m) K F T 2 L ( ) W (, s ) V (, s ) k (T ) L ( ) ds This leads to 2 k ( t ) W (, t ) V (, t ) k (T ) L ( ) 2 GW GV m 2 k ( s ) T (1 m) K F T 2 L ( ) W (, s ) V (, s ) k (T ) 2 L ( ) ds (17) 2 k ( t ) Set Z (t ) k (T ) ‖ W (, t ) V (, t )‖ 2 L ( ) , t [0, T ] Since W , V C ([0, T ]; L ( )), we see that the function Z is continuous on [0, T ] and attains over there its maximum M at some t0 [0, T ] Let M max Z (t ) From (17), we t[0,T ] obtain M 1 2 GW GV m 2 L ( ) (1 m) K F T M , 2 or equivalently 2 2 1 (1 m) K F T M m GW GV 2 L ( ) This implies that for all t [0, T ] 2 k ( t ) k (T ) W (, t ) V (, t ) 2 L ( ) 2 G G W V m M 2 2 L ( ) (1 m) K F T Thus, we obtain 1 W (, t ) V (, t ) L ( ) k ( t ) k (T ) m 2 (1 m) K F T k (T ) GW GV L ( ) , t [0, T ] (18) This completes the proof of Step and also the proof of the theorem □ CONVERGENCE ESTIMATE Now we are ready to state the main result Theorem Let m 0, u (, 0) L ( ) KFT 2 Suppose that Problem (1) has a unique solution u C ([0, T ]; L ( )) satisfying E with E and the regularization parameter E then we have the estimate Trang 179 Science & Technology Development, Vol 5, No.T20- 2017 1 U (, t ) u (, t ) L ( ) k (t ) m 2 k (T ) E 2 (1 m) K F T 1 k (t ) k (T ) Proof Assuming that u is a solution of Problem (9) corresponding to the final values G , we shall estimate u (, t ) u (, t ) L ( ) First we have uˆ ( , t ) exp( ( )( k (T ) k (t ))) Gˆ ( ) exp( ( )( k ( s) k (T ))) Fˆ ( , s, u ( , s)) ds T exp(| | cos( ˆ ˆ G ( ) exp( ( )( k ( s ) k (T ))) F ( , s, u ( , s )) ds t )k (T )) T exp( ( )( k (T ) k (t ))) exp(| | cos( t exp( ( )( k (T ) k (t ))) exp(| | cos( (19) )k (T )) ˆ ˆ G ( ) exp( ( )( k ( s ) k (T ))) F ( , s, u ( , s )) ds t )k (T )) T On the other hand, we get uˆ ( , T ) Gˆ ( ) exp( k (T ) ( )) uˆ ( , 0) exp(k ( s ) ( )) Fˆ ( , s, u (, s)) ds T This implies that T Gˆ ( ) exp( ( )( k ( s ) k (T ))) Fˆ ( , s, u ( , s )) ds t (20) t exp( k (T ) ( ))uˆ ( , 0) exp(( k ( s ) k (T )) ( )) Fˆ ( , s, u ( , s )) ds Combining (19) and (20), we obtain uˆ ( , t ) T ˆ G ( ) exp( ( )( k ( s ) k (T ))) Fˆ ( , s, u ( , s )) ds t exp(| | cos( )k (T )) ( ) exp ( )( k (T ) k (t )) exp( k (t ) ( )) exp(| | cos( ( ( )k (T )) uˆ ( , 0) ( )k (T )) exp | | cos exp(| | cos( )k (T )) ( )k (T )) exp | | cos t exp((k (s ) k (t )) ( ) Fˆ ( , s , u ( , s ))ds ) (21) Trang 180 TAÏP CHÍ PHÁT TRIỂN KH&CN, TẬP 20, SỐ T5- 2017 It follows from (9) and (21) that uˆ ( , t ) u ( , t ) B1 B2 B3 , where ( t ) [Fˆ ( , s, u cos( )k (T )) exp ( )( k ( s ) k (t )) T B1 ( exp | | exp( k (t ) ( )) exp(| | cos( exp(| | cos( exp(| | cos( B3 ( , s )) Fˆ ( , s, u ( , s )) ds, ] B2 exp(| | cos( )k (T )) )k (T )) )k (T )) uˆ ( , 0), )k (T )) t exp((k ( s) k (t )) ( ))[ Fˆ ( , s , u ( , s )) Fˆ ( , s, u ( , s ))]ds This leads to uˆ ( , t ) u ( , t ) | B1 | | B2 | | B3 | k (t )k ( s ) T k (t ) t k (t ) k ( s ) Fˆ ( , s , u ( , s )) Fˆ ( , s , u ( , s )) ds k (T ) | uˆ (, 0) | k (T ) t Fˆ ( , s , u ( , s ) ) Fˆ ( , s , u ( , s ) ) ds k (T ) k (t ) k (t )k ( s ) T k ( T ) | uˆ (, 0) | Fˆ ( , s, u ( , s )) Fˆ ( , s , u ( , s )) ds k (T ) (22) Using this and (16), we conclude that u (, t ) u (, t ) 2 L ( ) uˆ (, t ) u (, t ) 2 L ( ) 2k (t ) T k ( tk)(Tk)( s ) ˆ k (T ) 1 ‖ u (, 0)‖ L ( ) (1 m) F ( , s, u ( , s )) Fˆ ( , s, u ( , s )) ds d m 2 1 1 m 2k (t ) k (T ) 2k (t ) T ‖ u (, 0)‖ 2 (1 m) K F T L ( ) k (T ) 2 k ( s ) k (T ) u (, s ) u (, s ) 2 L ( ) ds, and thus 2 k ( t ) k (T ) u (, t ) u (, t ) 2 L ( ) 1 1 ‖ u (, 0)‖ m T 2 L ( ) (1 m) K F T 2 k ( s ) k (T ) u (, s ) u (, s ) 2 L ( ) ds Trang 181 Science & Technology Development, Vol 5, No.T20- 2017 Since u, u C ([0, T ]; L2 ( )), the function u (, t ) u (, t ) L ( is continuous on 0, T Therefore, there exists a ) 2 k ( t ) positive N max t[ 0,T ] u (, t ) u (, t ) k (T ) N 1 2 L ( ) This implies that 1 ‖ u (, 0)‖ L ( m (1 m) K F T N , ) that is, 2 k ( t ) ‖ u (, 0)‖ L ( ) m N 2 u (, t ) u (, t ) k (T ) 2 L ( ) (1 m) K F T Hence, we obtain the error estimate 1 u (, t ) u (, t ) L ( ) E k (t ) m k (T ) 2 (1 m) K F T On the other hand, using estimate (18), we get 1 u (, t ) U (, t ) L ( ) m 2 (1 m) K F T 1 k ( t ) k (T ) k (T ) G G L ( ) m 2 (1 m) K F T k ( t ) k (T ) k (T ) From the triangle inequality and these estimates, we obtain U (, t ) u (, t ) 1 L ( ) m 2 (1 m) K F T With E U (, t ) u (, t ) ) 1 k ( t ) k (T ) k (T ) L ( u (, t ) u (, t ) L ( ) k (t ) m E k (T ) 2 (1 m) K F T , then we have the estimate 1 U (, t ) u (, t ) L ( ) 2 k (t ) m k (T ) E 2 (1 m) K F T 1 k (t ) k (T ) This completes the proof Remark If (t ) and F ( x, t , u ) then Problem (1) becomes a homogeneous problem The error estimate in t Theorem is of order T It is similar to the homogeneous case in [1, 6, 8] CONCLUSION In this paper, we use the new regularization solution to slove a Riesz-Feller space-fractional backward diffusion problem with a time-dependent Trang 182 coefficient The convergence result has been obtained under a priori bound assumptions for the exact solution and the suitable choices of the regularization parameter TẠP CHÍ PHÁT TRIỂN KH&CN, TẬP 20, SỐ T5- 2017 Acknowledgements: The author desires to thank the handling editor and anonymous referees for their most helpful comments on this paper Chỉnh hóa cho tốn khuếch tán ngược cấp phân số không gian Riesz-Feller với hệ số phụ thuộc thời gian Đinh Nguyễn Duy Hải Trường Đại học Khoa học Tự nhiên, ĐHQG-HCM Trường Đại học Giao thông Vận tải Tp Hồ Chí Minh TĨM TẮT Trong báo này, chúng tơi xét tốn chỉnh, nghĩa nghiệm (nếu tồn tại) không phụ thuộc ngược cho phương trình khuếch tán cấp phân số khơng liên tục vào liệu Vì vậy, chúng tơi đưa gian với hệ số phụ thuộc thời gian Bài toán có nghiệm chỉnh hóa để giải tốn Sau đó, từ phương trình khuếch tán cổ điển cách ước lượng hội tụ thu giả định bị chặn thay đạo hàm bậc hai biến không gian đạo hàm tiên nghiệm cho nghiệm xác Riesz-Feller với 0,2 Đây tốn khơng Từ khóa: tốn khuếch tán ngược cấp phân số khơng gian, tốn khơng chỉnh, chỉnh hóa, ước lượng lỗi, hệ số phụ thuộc thời gian TÀI LIỆU THAM KHẢO [1] H Cheng, C.L Fu, G.H Zheng, J Gao, A regularization for a Riesz-Feller 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Mainardi, Y Luchko, G Pagnini, The fundamental solution of the space- time fractional diffusion equation, Fract Cacl Appl Anal., 4, 153– 192 (2001) [6] Z.Q Zhang, T Wei, An optimal regularization. .. slove a Riesz-Feller space- fractional backward diffusion problem with a time-dependent Trang 182 coefficient The convergence result has been obtained under a priori bound assumptions for the exact