DSpace at VNU: On a Riesz-Feller space fractional backward diffusion problem with a nonlinear source

Accepted Manuscript On a Riesz–Feller space fractional backward diffusion problem with a nonlinear source Nguyen Huy Tuan, Dinh Nguyen Duy Hai, Le Dinh Long, Van Thinh Nguyen, Mokhtar Kirane PII: DOI: Reference: S0377-0427(16)00007-8 http://dx.doi.org/10.1016/j.cam.2016.01.003 CAM 10439 To appear in: Journal of Computational and Applied Mathematics Received date: September 2015 Revised date: January 2016 Please cite this article as: N.H Tuan, D.N.D Hai, L.D Long, V.T Nguyen, M Kirane, On a Riesz–Feller space fractional backward diffusion problem with a nonlinear source, Journal of Computational and Applied Mathematics (2016), http://dx.doi.org/10.1016/j.cam.2016.01.003 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 Manuscript Click here to view linked References On a Riesz - Feller space fractional backward diffusion problem with a nonlinear source Nguyen Huy Tuana , Dinh Nguyen Duy Haib , Le Dinh Longc , Van Thinh Nguyend , Mokhtar Kiranee,f,∗ a Applied Analysis Research Group, Faculty of Mathematics and Statistics, Ton Duc Thang University, Ho Chi Minh City, Viet Nam b Department of Mathematics, University of Science, Vietnam National University, 227 Nguyen Van Cu Street, District 5, Ho Chi Minh City, Viet Nam c Institute of Computational Science and Technology in Ho Chi Minh City, Viet Nam d Department of Civil and Environmental Engineering, Seoul National University, Republic of Korea e Laboratoire de Mathmatiques Pˆ ole Sciences et Technologie, Universi´ e de La Rochelle, A´ enue M Cr´ epeau, 17042 La Rochelle Cedex, France f Nonlinear Analysis and Applied Mathematics (NAAM) Research Group, Department of Mathematics, Faculty of Science, King Abdulaziz University, P.O Box 80203, Jeddah 21589, Saudi Arabia Abstract In this paper, a backward diffusion problem for a space-fractional diffusion equation with a nonlinear source in a strip is investigated This problem is obtained from the classical diffusion equation by replacing the second-order space derivative with a Riesz-Feller derivative of order α ∈ (0, 2] A nonlinear problem is severely ill-posed, therefore we propose two new modified regularization solutions to solve it We further show that the approximated problems are well-posed and their solutions converge if the original problem has a classical solution In addition, the convergence estimates are presented under a priori bounded assumption of the exact solution For estimating the error of the proposed method, a numerical example has been implemented Keywords: Space-fractional backward diffusion problem; Ill-posed problem; Regularization; Error estimate Introduction 10 15 In recent decades, fractional operators have been playing more and more important roles in science and engineering [4], for instance, in the theory of viscoelasticity and viscoplasticity (mechanics), in the modeling of polymers and proteins (biochemistry), in the transmission of ultrasound waves (electrical engineering), and in the modeling of human tissue under mechanical loads (medicine), which are referred to [6, 9, 14] These new fractional-order models are more adequate than the integer-order models, because the fractional order derivatives and integrals enable to describe the memory and hereditary properties of different substance [10] The space-fractional diffusion equation (SFDE) has been arisen from replacing the standard space partial derivative in the diffusion equation with a space fractional partial derivative It can be derived from the continuous-time random walk in statistical mechanics, and has a wide range of applications in the theory of probability distribution, especially in the modeling of the high-frequency price dynamics in financial markets [10, 12] If an initial concentration distribution and boundary conditions are given, a complete recovery of the unknown solution is attainable from solving a well-posed forward problem The well-posed forward problem, i.e initial and boundary value problem (IBVP) for the SFDE have been studied extensively in the last few years P Agarwal [13, 18, 25, 26] and his coauthors have recently developed some new methods for fractional differential equations Liu et al [8] considered stability and convergence of the difference methods for the space-time fractional advection-diffusion equation Ray et al [16] investigated application of modified ∗ Corresponding author Email address: mokhtar.kirane@univ-lr.fr (Mokhtar Kirane) Preprint submitted to Elsevier 20 25 30 decomposition method for the analytical solution of space-fractional diffusion equation Yang et al [20] studied several numerical methods for fractional partial differential equations with Riesz space-fractional derivatives The space fractional diffusion also investigated by some other authors, such as M Younis et al [17], M.Duman et al [2], F Liu [19], Z Q Chen [1], Y Povstenko [15] However, in some practical problems, the boundary data on the whole boundary cannot be obtained In situation we only may know the noisy data on a part of the boundary or at some interior points of the considered domain, which leads to an inverse problem, namely the space-fractional inverse diffusion problems (SFIDP) To the best of our knowledge, the result from the study of SFIDP is still very limited to some specific cases dealing with homogeneous problems ([[3],[21],[22],[23],[24]]) Motivated by this reason, in this work, we consider a backward problem for the following nonlinear space fractional diffusion equation  α   ut (x, t) = x Dθ u(x, t) + f (x, t, u(x, t)), (x, t) ∈ R × (0, T ), u(x, t)|x→±∞ = 0, (x, t) ∈ R × (0, T ), (1)   u(x, T ) = g(x), x ∈ R, where the space-fractional derivative x Dθα is the Riesz-Feller fractional derivative of order α(0 < α ≤ 2) and skewness θ (|θ| ≤ min{α, − α}, θ = ±1) defined in [7], as follows:  ∞  (α + θ)π f (x + s) − f (x) Γ(1 + α)  α  sin ds  x Dθ f (x) =   π s1+α      ∞ (α − θ)π f (x − s) − f (x) + sin ds , < α < 2,    s1+α         D2 f (x) = df (x) , α = x dx2 35 40 45 50 The space-fractional backward diffusion problem is to determine the distribution u(x, t) for < t < T from the final concentration distribution u(x, T ) As it is well-known, the backward diffusion problem is severely ill-posed [22], i.e., solutions not always exist, and in the case of existence, these not depend continuously on the given data In fact, from small noise contaminated physical measurements, the corresponding solutions have large errors It makes difficult to numerical calculations Hence, a regularization is in order In recent years, the homogeneous problem, i.e, f = in Equation (1) has been proposed by some authors Zheng and Wei [22] used two methods, the spectral regularization and modified equation methods, to solve this problem In [21], they developed an optimal modified method to solve this problem by an a priori and an a posteriori strategy In 2014, Zhao et al [24] applied a simplified Tikhonov regularization method to deal with this problem After then, a new regularization method of iteration type for solving this problem has been introduced by Cheng et al [3] However, in many practical engineering applications, the diffusion occur in spatially heterogeneous environments, which requires to take account of a nonlinear source term As mentioned above, up-to-date we have not found any publication dealing with the SFIDP with a nonlinear source term; because it requires a special technique to deal with the fractional terms and nonlinear term for solving this nonlinear problem, this is the most challenging task Therefore in this study, we try to develop new methods and techniques to overcome this limitation In comparison with previous studies ([[3],[21],[22],[23],[24]]) on solving the space-fractional backward diffusion problem, our paper shows a significant improvement, because it can deal with the space-fractional inverse diffusion problems with nonlinear source This paper is organized as follows In the following section we outline the main results The proofs of these results is described in Section In Section 4, a numerical example is proposed to show the effectiveness of the regularized methods Then we end up the manuscript with the conclusion at Section 55 The main results Let g(ω) denote the Fourier transform of the integrable function g(x), which defined by g(ω) := √ 2π +∞ exp(−ixω)g(x)dx, i= √ −∞ −1 In terms of the Fourier transform, we have the following properties for the Riesz-Feller space-fractional derivative (see [11]) α x Dθ (g)(ω) = −ψαθ (ω)g(ω), where ψαθ (ω) = |ω|α cos 60 θπ + isign(ω) sin πθ (2) By taking a Fourier transform to (1), we transform Problem (1) into the following differential equation ut (ω, t) = −ψαθ (ω)u(ω, t) + f (ω, t, u(ω, t)), u(ω, T ) = g(ω) Solving the latter problem, we have T u(ω, t) = exp ψαθ (ω)(T − t) g(ω) − t exp ψαθ (ω)(s − T ) f (ω, s, u(ω, s))ds (3) From (3), applying the inverse Fourier transform, we get u(x, t) = √ 2π +∞ −∞ exp ψαθ (ω)(T − t) × g(ω) − T t exp ψαθ (ω)(s − T ) f (ω, s, u(ω, s))ds exp(ixω)dω (4) Note that ψαθ (ω) has the positive real part |ω|α cos( θπ ) Since ≤ t < s < T ; from (4) that, when |ω| 65 becomes large, the terms exp ψαθ (ω)(T − t) and exp ψαθ (ω)(s − t) increases rather quickly: small errors in high-frequency components can blow up and completely destroy the solution for < t < T , therefore recovering the scalar (temperature, pollution) u(x, t) from the measured data g (x) is severely ill-posed We define a regularization as follows uβ (x, t) = √ 2π +∞ −∞ exp ψαθ (ω)(T − t) + β exp |ω|α cos πθ T T × g (ω) − t exp (ψαθ (ω)(s − T ) f ω, s, uβ (ω, s) ds exp(iωx)dω, where β is regularization parameter We introduce the first theorem for main result as following (5) Theorem Let g ∈ L2 (R) and let f ∈ L∞ (R × [0, T ] × R) satisfy f (x, t, 0) = and |f (x, t, w) − f (x, t, v)| ≤ k|w − v|, 70 for constant k > independent of x,t,w,v Then there exists a unique solution uβ ∈ C([0, T ]; L2 (R)) to Problem (5) Let g ∈ L2 (R) be measured data such that g − g L2 (R) ≤ Let us choose β such that lim →0 β −1 bounded a) Suppose that the problem (1) has a unique solution u ∈ C([0, T ], L2 (R)) that satisfies +∞ −∞ exp |ω|α cos πθ t u(ω, t) dω < ∞, for all t ∈ [0, T ], then we have the estimate uβ (., t) − u(., t) 75 t L2 (R) ≤ β T exp((T − t)2 k )K, where K= √ 2β −1  +∞ +  sup exp |ω| cos α t∈[0,T ] −∞ Moreover, if ut ∈ L2 ((0, T ); L2 (R)) then for all u(., 0) − uβ (., t ) πθ  21 t u(ω, t) dω  ∈ (0, 1) there exists a constant t > such that 1 ≤ E8 T ln L2 (R) β − 14 (6) , where    E = max     21 T us (., s)  L2 (R) ds , exp((T − t)2 k )K      b) Suppose that the problem (1) has a unique solution u ∈ C([0, T ], L2 (R)) that satisfies +∞ T −∞ exp 2ψαθ (ω)s f (ω, s, u(ω, s)) dsdω < ∞, (7) then we have the following estimate uβ (., t) − u(., t) 80 L2 (R) ≤ t β T exp (T − t)2 k D, (8) where D= √  +∞ T 2β −1 + 4||u(., 0)||2L2 (R) + 4T −∞  21 exp(2ψαθ (ω)s) f (ω, s, u(ω, s)) dsdω  Remark 1 If we choose β = , then the results in Theorem hold The error estimate (8) is no longer valid at t = To approximate u(., 0), we use the similar technique as shown in (6) In Theorem 1, we require the following condition +∞ −∞ exp |ω|α cos πθ 2 t u(ω, t) dω < ∞ (9) 85 90 95 If f (x, t, u) = then the left hand side of (9) is equal to u(., 0) 2L2 (R) Therefore, the condition (9) is natural and acceptable t If f (x, t, u) = 0, the error estimate in Theorem is of order β T It is similar to the homogeneous case in [21, 24] In Theorem 1, to obtain the error estimate, we require the strong assumptions of u as shown by (7) and (9) This is a limitation of Theorem 1, because there are not many functions u satisfied these conditions, particularly in practice these conditions are more difficult to be satisfied and checked To improve this limitation, we introduce a new regularization solution and introduce a new technique to estimate the error in the Theorem below In fact, in this Theorem we only need a weaker assumption for u which is u ∈ C([0, T ]; L2 (R)), and this condition is natural and acceptable Theorem Let g, f, β be as Theorem Suppose that < kT < and Problem (1) has a unique Then we construct a new regularized solution solution u ∈ C([0, T ]; L2 (R)) Let m ∈ 0, k21T − Uβ ∈ C([0, T ]; L2 (R)) in such a way that Uβ (., t) − u(., t) L2 (R) β −1 + u(., 0) ≤ L2 (R) 1+ m t βT , 2 − (1 + m)T k where Uβ is the function whose Fourier transform is defined by Uβ (ω, t) = T exp ψαθ (ω)(T − t) + β exp |ω|α cos t + πθ β exp |ω|α cos + β exp |ω|α T g (ω) − πθ T cos πθ t exp ψαθ (ω)(s − t) + β exp |ω|α cos πθ T f (ω, s, Uβ )ds exp (s − t)ψαθ (ω) f (ω, s, Uβ )ds T (10) Remark Using the same technique as in (6), we obtain the estimate at t = as Uβ (., t ) − u(., 0) ≤ L2 (R) H2 −1 T ln β −1 , where 100    H = max    T us (., s)  12  L2 (R) ds , β −1 + u(., 0)      In Theorem 2, our method has a drawback that the condition for k ∈ [0, T1 ) is still too strong which may restrict the class of equations We are trying to remove this constraint If k = and f = then Problem (1) becomes a homogeneous problem The error estimate in Theorem t is of order β T It is similar to the homogeneous case in [21, 24] Proof of main results 105 L2 (R) 1+ m − (1 + m)T k First, we consider the following Lemma Lemma Let t, s ∈ [0, T ] 1) If s ≥ t, then we have exp ψαθ (ω)(s − t) a) + β exp |ω|α cos T ≤β T ≤β πθ exp ψαθ (ω)(T − t) b) + β exp |ω|α cos πθ t−s T t−T T 2) If s ≤ t, then we have β exp ψαθ (ω)(s − t + T ) c) + β exp |ω|α cos θπ ≤β T t−s T Proof First, we prove (a) In fact, we have exp ψαθ (ω)(s − t) + β exp |ω|α cos( πθ )T exp |ω|α cos( πθ )(s − t − T ) = β + exp − |ω|α cos( πθ )T exp |ω|α cos( πθ )(s − t − T ) = s−t T β + exp − |ω|α cos( πθ )T ≤ 110 s−t T β + exp − |ω|α cos( πθ )T ≤ β t−s T β + exp − |ω|α cos( πθ )T Making the change s = T , we obtain (b) Next, we prove (c) In fact from (b), we have exp ψαθ (ω)(T − (t − s)) + β exp |ω|α cos( πθ )T ≤ β t−s−T T , it follows that β exp ψαθ (ω)(s − t + T ) + β exp |ω|α cos( πθ )T ≤ β t−s T 3.1 Proof of Theorem 115 Proof We divide the proof into three steps Step1 Constructing a regularized solution (5) We consider the problem uβ (ω, t) = T exp ψαθ (ω)(T − t) + β exp |ω|α cos πθ T g (ω) − t exp ψαθ (ω)(s − t) + β exp |ω|α cos πθ T f (ω, s, uβ (ω, s))ds T −s+t T or uβ (x, t) √ 2π = +∞ −∞ −√ 2π exp ψαθ (ω)(T − t) + β exp |ω|α cos +∞ T g (ω) exp(iωx)dω πθ T exp ψαθ (ω)(s − t) + β exp |ω|α cos −∞ t πθ f (ω, s, uβ (ω, s)) exp(iωx)dsdω T (11) First, we prove that Problem (11) has a unique solution uβ that belongs to C([0, T ]; L2 (R)) Denoting 1 F (w)(x, t) = √ G(x, t) − √ 2π 2π +∞ T exp ψαθ (ω)(s − t) + β exp |ω|α cos −∞ t f (ω, s, w) exp(iωx)dsdω, πθ T for all w ∈ C([0, T ]; L2 (R)) and +∞ G(x, t) = −∞ 120 exp ψαθ (ω)(T − t) + β exp |ω|α cos g (ω) exp(iωx)dω πθ T Since f (x, t, 0) = 0, and due to the Lipschitzian property of f (x, t, w) with respect to w, we get G(w) ∈ C([0, T ]; L2 (R)) for every w ∈ C([0, T ]; L2 (R)) We claim that, for every w, v ∈ C([0, T ]; L2 (R)), p ≥ 1, we have F (w)(., t) − F (v)(., t) p p L2 (R) 2p k β ≤ (T − t)p Qp w−v p! (12) , where Q = max{1, T } and is the sup norm in L2 (R) We shall prove the latter inequality by induction For p = 1, we have F (w)(., t) − F (v)(., t) L2 (R) = F (w)(., t) − F (v)(., t) +∞ T = −∞  −∞    −∞   πθ f (ω, s, w) − f (ω, s, v) ds dω T exp ψαθ (ω)(s − t) + β exp |ω|α cos t T t exp ψαθ (ω)(s − t) T +∞ = L2 (R) + β exp |ω|α cos t +∞ ≤ πθ + β exp ≤ (T − t) β2 ≤ (T − t) β2 cos πθ T t T ds t T f (., s, w(., s)) − f (., s, v(., s)) t T t k w(., s) − v(., s) L2 (R) ds ≤    f (ω, s, w) − f (ω, s, v) ds dω ds T exp |ω|α cos( πθ )(s − t) |ω|α T  f (ω, s, w) − f (ω, s, v) ds dω L2 (R) k β ds (T − t)Q w − v 125 Therefore (12) holds for p = Suppose that (12) holds for p = m (m ≥ 1) We prove that (12) holds for p = m + We have L2 (R) F m+1 (w)(., t) − F m+1 (v)(., t) T +∞ = −∞ t  T +∞ ≤ −∞   t πθ ≤ (T − t) β2 = (T − t) β2 πθ T t f (., s, F m (w)(., s)) − f (., s, F m (v)(., s)) dsdω T f (., s, F m (w)(., s)) − f (., s, F m (v)(., s)) t ≤ (T − t)k β2 ≤ (T − t)k β2   f (ω, s, F m (w)) − f (ω, s, F m (v)) ds dω ds T +∞ T −∞ t L2 (R) f (ω, s, F m (w)) − f (ω, s, F m (v)) ds dω T exp |ω|α cos( πθ )(s − t) + β exp |ω|α cos 2 exp ψαθ (ω)(s − t) + β exp |ω|α cos = F (F m (w))(., t) − F (F m (v))(., t) L2 (R) ds T F m (w)(., s) − F m (v)(., s) t k β T 2m t L2 (R) (T − s)m Qm w−v m! ds ds ≤ k β 2(m+1) (T − t)m+1 m+1 Q w−v (m + 1)! Therefore, by the induction principle, (12) holds for every integer p From (12), we get F p (w) − F p (v) L2 (R) ≤ k β p p T2 p √ Q2 w − v , p! for every w, v ∈ C([0, T ]; L2 (R)) Since lim p→∞ 130 k β p p T2 p √ Q = 0, p! there exits a positive integer number p0 such that F p0 is a contraction It follows that F p0 (w) = w has a unique solution uβ ∈ C([0, T ]; L2 (R)) We claim that F (uβ ) = uβ In fact, one has F (F p0 (uβ )) = F (uβ ) Hence, F p0 (F (uβ )) = F (uβ ) By the uniqueness of the fixed point of F p0 , one has F (uβ ) = uβ , i.e., the equation F (w) = w has a unique solution uβ ∈ C([0, T ]; L2 (R)) The main purpose of this paper is to estimate the error uβ − u 135 we proceed to the next two steps Step2 Let wβ be the function whose Fourier transform is defined by wβ (ω, t) = T exp ψαθ (ω)(T − t) + β exp |ω|α cos We estimate wβ (., t) − uβ (., t) L2 (R) πθ T g(ω) − t exp (ψαθ (ω)(s − t) + β exp |ω|α cos πθ Using the Parseval’s equality and the inequality (a + b)2 ≤ 2(a2 + b2 ), we get T L2 (R) To this end, f ω, s, wβ (ω, s) ds (13) wβ (., t) − uβ (., t) L2 (R) +∞ ≤ + β exp |ω|α cos −∞ exp ψαθ (ω)(T − t) T +∞ +2 −∞ πθ t dω exp ψαθ (ω)(s − t) + β exp |ω|α cos = A1 + A2 , g (ω) − g(ω) T πθ f (ω, s, uβ (ω, s)) − f (ω, s, wβ (ω, s)) ds dω T (14) where +∞ A1 = + β exp |ω|α cos −∞ +∞ A2 = T −∞ 140 exp ψαθ (ω)(T − t) πθ dω, exp ψαθ (ω)(s − t) + β exp |ω|α cos t g (ω) − g(ω) T πθ f (ω, s, uβ (ω, s)) − f (ω, s, wβ (ω, s)) ds dω T (15) Term (15) can be estimated as follows +∞ A1 ≤ ≤ 2β 2t−2T T ≤ 2β 2t−2T T ≤ 2β 2t−2T T −∞ exp ψαθ (ω)(T − t) + β exp g (ω) − g(ω) dω cos( πθ )T |ω|α +∞ −∞ g (ω) − g(ω) dω g −g L2 (R) (16) and +∞ A2 = T −∞ exp ψαθ (ω)(s − t) + β exp |ω|α cos t +∞ T πθ T f (ω, s, wβ (ω, s)) − f (ω, s, uβ (ω, s)) ds dω exp ψαθ (ω)(s − t) ≤ 2(T − t) + β exp |ω|α cos( πθ )T −∞ t f (ω, s, wβ (ω, s)) − f (ω, s, uβ (ω, s)) dsdω +∞ T ≤ 2(T − t) β −∞ t f (ω, s, wβ (ω, s)) − f (ω, s, uβ (ω, s) dsdω T 2t ≤ 2(T − t)β T k 2t−2s T β −2s T t wβ (., s) − uβ (., s) L2 (R) (17) ds Combining (14), (16) and (17), we have T wβ (., t) − uβ (., t) L2 (R) ≤ 2β 2t−2T T 2t T + 2(T − t)β k β t −2s T wβ (., s) − uβ (., s) L2 (R) ds or Uβ (x, t) √ 2π = +∞ +∞ T πθ g (ω) exp(iωx)dω T exp ψαθ (ω)(s − t) + β exp |ω|α cos −∞ t + √ 2π 185 + β exp |ω|α cos −∞ −√ 2π exp ψαθ (ω)(T − t) +∞ t β exp |ω|α cos πθ πθ + β exp |ω|α cos −∞ f (ω, s, Uβ ) exp(iωx)dsdω T T πθ exp (s − t)ψαθ (ω) f (ω, s, Uβ ) exp(iωx)dsdω T Let us define the norm on C([0; T ]; L2 (R)) as follows h = sup β 0≤t≤T −t T h(t) L2 (R) , for all h ∈ C([0; T ]; L2 (R)) It is easy to show that · is a norm of C([0; T ]; L2 (R)) For v ∈ C([0; T ]; L2 (R)), we consider the following function Θ(v)(x, t) = √ A(x, t) 2π −√ 2π +√ 2π +∞ T exp ψαθ (ω)(s − t) + β exp |ω|α cos −∞ t +∞ t −∞ β exp |ω|α cos πθ + β exp |ω|α cos f (ω, s, v) exp(iωx)dsdω πθ T T πθ T exp (s − t)ψαθ (ω) f (ω, s, v) exp(iωx)dsdω, where +∞ A(x, t) = −∞ exp ψαθ (ω)(T − t) + β exp |ω|α cos πθ g (ω) exp(iωx)dω T We claim that, for every v1 , v2 ∈ C([0; T ]; L2 (R)) Θ(v1 ) − Θ(v2 ) ≤ kT v1 − v2 15 190 First, we have the two following estimates for all t ∈ [0, T ]  +∞  = J1 −∞ T exp ψαθ (ω)(s − t) + β exp |ω|α cos t +∞ T πθ f (ω, s, v1 ) − f (ω, s, v2 ) ds dω T exp ψαθ (ω)(s − t) ≤ (T − t) + β exp |ω|α cos −∞ t 2 πθ f (ω, s, v1 ) − f (ω, s, v2 ) dsdω T +∞ T ≤ (T − t) 2(t−s) T β −∞ t T 2t ≤ β T k (T − t) 2t T ≤ β k (T − t) 2 f (ω, s, v1 ) − f (ω, s, v2 ) dsdω β −2s T v1 (., s) − v2 (., s) t sup β −2s T 0≤s≤T 2t ≤ β T k (T − t)2 v1 − v2 L2 (R) ds v1 (., s) − v2 (., s) L2 (R) (32) and +∞ J2 t β exp |ω|α cos = −∞ × + β exp cos T πθ f (ω, s, v1 ) − f (ω, s, v2 ) ds +∞ t β exp |ω|α cos ≤ t πθ + β exp |ω|α cos −∞ × |ω|α πθ exp (s − t)ψαθ (ω) T dω T πθ T exp (s − t)ψαθ (ω) f (ω, s, v1 ) − f (ω, s, v2 ) dsdω +∞ t ≤ t β −∞ 2t T 2 ≤ β k t f (ω, s, v1 ) − f (ω, s, v2 ) dsdω t 2t T ≤ β k t 2(t−s) T β −2s T v1 (., s) − v2 (., s) sup β −2s T L2 (R) ds v1 (., s) − v2 (., s) 0≤s≤T L2 (R) 2t (33) ≤ β T k t2 v1 − v2 21 For < t < T , using the inequality (a + b)2 ≤ (1 + m)a2 + (1 + m )b2 for all real numbers a and b and m > 0, we obtain Θ(v1 )(., t) − Θ(v2 )(., t) L2 (R) 2t ≤ (1 + m)β T k t2 v1 − v2 + By choosing m = β T −t t , −2t T 1+ m 2t β T k (T − t)2 v1 − v2 21 we have Θ(v1 )(., t) − Θ(v2 )(., t) L2 (R) ≤ k T v1 − v2 21 , for all t ∈ (0, T ) 16 (34) 195 On other hand, letting t = in (32), we have Θ(v1 )(., 0) − Θ(v2 )(., 0) L2 (R) ≤ k T v1 − v2 21 (35) By letting t = T in (33), we have β −2 Θ(v1 )(., T ) − Θ(v2 )(., T ) L2 (R) ≤ k T v1 − v2 21 (36) Combining (34), (35) and (36), we obtain β −2t T Θ(v1 )(., t) − Θ(v2 )(., t) ≤ k T v1 − v2 21 , for all t ∈ [0, T ] L2 (R) which leads to (32) Since kT < 1, Θ is a contraction It follows that the equation Θ(v) = v has a unique solution Uβ ∈ C([0; T ]; L2 (R)) 200 Step Estimate of the error Uβ (., t) − Vβ (., t) L2 (R) Let Vβ be the function whose Fourier transform is defined by exp ψαθ (ω)(T − t) Vβ (ω, t) = + β exp |ω|α cos T − exp ψαθ (ω)(s − t) t T + β exp |ω|α cos t + g(ω) πθ β exp |ω|α cos + β exp |ω|α πθ πθ cos f (ω, s, Vβ )ds T T πθ exp (s − t)ψαθ (ω) f (ω, s, Vβ )ds T From (31) and (37), we have Uβ (ω, t) − Vβ (ω, t) ≤ exp ψαθ (ω)(T − t) + β exp |ω|α cos T + exp ψαθ (ω)(s − t) t + β exp |ω|α cos + β exp g (ω) − g(ω) T + β exp |ω|α cos t × πθ |ω|α πθ πθ cos T T πθ T f (ω, s, Vβ (ω, s)) − f (ω, s, Uβ (ω, s)) ds exp (s − t)ψαθ (ω) f (ω, s, Uβ (ω, s)) − f (ω, s, Vβ (ω, s)) ds T ≤ β t−T T g (ω) − g(ω) + β t−s T t f (ω, s, Vβ (ω, s)) − f (ω, s, Uβ (ω, s)) ds t + β t−s T f (ω, s, Vβ (ω, s)) − f (ω, s, Uβ (ω, s)) ds T ≤ β t−T T g (ω) − g(ω) + β 17 t−s T f (ω, s, Vβ (ω, s)) − f (ω, s, Uβ (ω, s)) ds (37) Since m ∈ 0, k21T − , we have that < k < (a + b)2 ≤ T √1 1+m m 1+ From the inequality a2 + (1 + m)b2 (38) for all real number a, b and m > 0, we get Uβ (., t) − Vβ (., t) L2 (R) = Uβ (., t) − Vβ (., t) ≤ 1+ m β 2t−2T T  +∞ + (1 + m) −∞ 1+ m ≤ 205 β  2t−2T T L2 (R) g −g L2 (R) 2 T β t−s T f (ω, s, Vβ (ω, s)) − f (ω, s, Uβ (ω, s)) ds dω T + (1 + m)T k 2 β 2t−2s T Uβ (., s) − Vβ (., s) L2 (R) ds This leads to β −2t T Uβ (., t) − Vβ (., t) L2 (R) 1+ ≤ m β −2 T + (1 + m)T k β −2s T Uβ (., s) − Vβ (., s) −2t L2 (R) ds (39) Set Z(t) = β T Uβ (., t) − Vβ (., t) 2L2 (R) , ∀t ∈ [0, T ] Since Uβ , Vβ ∈ C([0, T ]; L2 (R)), 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 = −2t max β T Uβ (., t) − Vβ (., t) 2L2 (R) From (39), we obtain t∈[0,T ] M ≤ 1+ m β −2 + (1 + m)T k M, or − (1 + m)T k M ≤ 210 1+ m β −2 This implies that for all t ∈ [0, T ] β −2t T Uβ (., t) − Vβ (., t) L2 (R) ≤M ≤ 1+ m β −2 − (1 + m)T k Hence Uβ (., t) − Vβ (., t) L2 (R) ≤ 1+ m t−T β T − (1 + m)T k 18 Step Estimate u(., t) − Vβ (., t) L2 (R) We have u(ω, t)  + β exp ψαθ (ω)(T πθ t T g(ω) − T t − t) exp |ω| cos πθ πθ T T  T × g(ω) −  exp ψαθ (ω)(s − T ) f (ω, s, u(ω, s))ds α + β exp |ω|α cos  exp ψαθ (ω)(s − T ) f (ω, s, u(ω, s))ds  exp ψαθ (ω)(T − t) + β exp |ω|α cos  T = exp ψαθ (ω)(T − t) g(ω) − = exp ψαθ (ω)(s − T ) f (ω, s, u(ω, s))ds t On other hand, we get u(ω, T ) = g(ω) = exp − 215  T u(ω, 0) + T ψαθ (ω) This implies that sψαθ (ω) exp (40)  f (ω, s, u(ω, s))ds T g(ω) − t exp ψαθ (ω)(s − T ) f (ω, s, u(ω, s))ds t = exp − T ψαθ (ω) u(ω, 0) + exp (s − T )ψαθ (ω) f (ω, s, u(ω, s))ds (41) Combining (40) and (41), we obtain u(ω, t) =  exp ψαθ (ω)(T − t) + β exp |ω|α cos + + πθ T g(ω) − T β exp − tψαθ (ω) exp |ω|α cos + β exp |ω|α cos β exp |ω|α cos + β exp |ω|α cos πθ πθ t πθ T  − T ) f (ω, s, u(ω, s))ds u(ω, 0) T t T πθ exp ψαθ (ω)(s T exp (s − t)ψαθ (ω) f (ω, s, u(ω, s))ds It follows from (37) and (42) that u(ω, t) − Vβ (ω, t) = B1 + B2 + B3 , 19 (42) where T B1 = t B2 B3 exp ψαθ (ω)(s − t) + β exp |ω|α cos πθ f (ω, s, Vβ (ω, s)) − f (ω, s, u(ω, s)) ds, T β exp − tψαθ (ω) exp |ω|α cos = + β exp |ω|α cos β exp |ω|α cos = πθ + β exp |ω|α cos πθ T u(ω, 0), T t T πθ πθ T exp (s − t)ψαθ (ω) f (ω, s, u(ω, s)) − f (ω, s, Vβ (ω, s)) ds This leads to u(ω, t) − Vβ (ω, t) |B1 | + |B2 | + |B3 | ≤ T ≤ t−s T β t f (ω, s, u) − f (ω, s, Vβ ) ds t t + β T |u(., 0)| + β t−s T f (ω, s, u) − f (ω, s, Vβ ) ds T t β T |u(., 0)| + ≤ 220 t−s T β f (ω, s, u) − f (ω, s, Vβ ) ds (43) Using (38), we get u(., t) − Vβ (., t) L2 (R) = u(., t) − Vβ (., t) ≤ 1+ m 2t + (1 + m) ≤ L2 (R) β T u(., 0)  +∞ 1+ m −∞ β 2t T  L2 (R) 2 T β t−s T f (ω, s, u) − f (ω, s, Vβ ) ds dω u(., 0) L2 (R) T 2t + (1 + m)T k β T β −2s T u(., s) − Vβ (., s) L2 (R) ds Thus β −2t T u(., t) − Vβ (., t) L2 (R) 1+ ≤ m u(., 0) L2 (R) T × + (1 + m)T k Since u(., t) and Vβ (., t) ∈ C([0, T ]; L2 (R)), the function Therefore, there exists a positive N = max β t∈[0,T ] N≤ 1+ m −2t T β L2 (R) 20 u(., s) − Vβ (., s) u(., t) − Vβ (., t) u(., t) − Vβ (., t) u(., 0) −2s T L2 (R) L2 (R) L2 (R) is continuous on [0, T ] This implies that + (1 + m)T k N, ds that is, β −2t T u(., t) − Vβ (., t) L2 (R) ≤N ≤ Hence, we obtain the error estimate u(., t) − Vβ (., t) L2 (R) 1+ m u(., 0) − (1 + L2 (R) m)T k 1+ m u(., 0) − (1 + m)T k ≤ L2 (R) β t T From Step and Step 3, we obtain Uβ (., t) − u(., t) L2 (R) ≤ ≤ ≤ Uβ (., t) − Vβ (., t) L2 (R) + u(., t) − Vβ (., t) 1+ t−T β T + − (1 + m)T k m β −1 + u(., 0) L2 (R) L2 (R) 1+ u(., 0) − (1 + m)T k m L2 (R) β t T 1+ m t βT − (1 + m)T k Numerical experiment 225 230 To verify our proposed methods, we carry out the numerical experiment for above regularization methods The numerical example is implemented for T=1, and with the variation of α ∈ (0, 2] In order to illustrate the sensitivity of the computational accuracy to the noise of the measurement data, we use the random function to generate the noisy data similar to an observation data The perturbation was defined as rand(size()), where rand(size()) is a random number, and plays as an amplitude of the errors The approximation of the regularization solution is computed by the FFT algorithm In this example, we consider the following Riesz - Feller space fractional backward diffusion problem  α   ut (x, t) = x Dθ u(x, t) + f (x, t, u), (x, t) ∈ (−π, π) × (0, T ), u(−π, t) = u(π, t) = 0, t ∈ (0, T ), (44)   u(x, T ) = g(x), x ∈ (−π, π) Here, f (x, t, u) = −u + h(x, t), where g(x) = e sin3 (2x), h(x, t) = αtα−1 exp(tα ) sin3 (2x) − 235 240 1 exp(tα ) + α sin(2x) − + (6)α sin(6x) , with x ∈ (−π, π), subjecting to u(−π, t) = u(π, t) = 0, and u(x, 0) = sin3 (2x) The exact solution of Problem (44) is u(x, t) = exp(tα ) sin3 (2x) The regularized solution is defined by (5) with ψαθ (ω) defined by (2) The regularized solution is expected to be closed to the exact solution under a proper discretization as shown below In general, the whole numerical procedure is proceeded in the following steps: Step Choosing I and J to generate spatial and temporal discretizations as follows xi = tj = 2π , i = 0, I, I j∆t, ∆t = , j = 0, J J i∆x, ∆x = 21 Of course, the higher value of I and J will provide more stable and accurate numerical calculation, however in the following examples I = J = 201 are satisfied Step We choose the couple (g , F ) as the observed data including the noise in the manner that F (., ) 245 rand(size(g)) √ , π (rand(size(F (., )))) √ F (., ) + π = g+ g = Step Set u(., :)β( ) (xi ) = u(., :)β( ),i and u(., :) (xi ) = ui , constructing two following vectors contained all discrete values of u(., :)β and u(., :) denoted by Λβ and Ψ, respectively, Λβ = u(., :)β,0 Ψ = u(., :)0 u(., :)β,1 u(., :)1 u(., :)β,I−1 u(., :)I−1 u(., :)β,I ∈ RI+1 , u(., :)I ∈ RI+1 Step Errors between the exact and its regularized solutions are estimated by Absolute error estimation E1 = I +1 I i=0 |uβ (xi , ) − u(xi , )|2 Relative error estimation I E2 = i=0 |uβ (xi , ) − u(xi , )|2 I i=0 250 255 260 265 |u(xi , )|2 Fig shows the graphs of the exact solutions of Problem (44) corresponding to α = 0.2, 0.6, 1.4 and 1.8 In this example as shown in Fig 2, when α is small (close to zero) the concentration distribution u(x, t) does not fluctuate much once t is increased On the other hand, when t tends to zero the concentration distribution u(x, t) is more fluctuating once the value of α is decreased to zero as shown in Fig Figs and show the comparisons between the exact and its regularized solutions at t = 0.5, and for α = 0.2 and 0.8, respectively Figures and show the comparisons between the exact and its regularized solutions at t = 0.9, and for α = 0.2 and 0.8, respectively Even for α = 0.2, at the beginning the regularized solution is oscillated around the exact solution, eventually the regularized solution is converged very well to the exact solution once tends to Tables and show the estimate of absolute (E1 ) and relative (E2 ) errors between the exact and its regularized solutions with different values of α (0.2, 0.6, 1.4 and 1.8) at t = 0.5 and 0.9, respectively It clearly shows the regularized solution converges to the exact solution with different values of α Tables 3, 4, and show the estimate of absolute (E1 ) and relative (E2 ) errors between the exact and its regularized solutions at the variation of t from 0.1 to 0.8 for a lower value of α = 0.2, and a higher value of α = 1.8, respectively These tables are not including the values of t at 0.5 and 0.9, because they are already shown in Tables and above In general, the errors between the exact and its regularized solutions are slightly higher once the order α of the Riesz-Feller fractional derivative is closed to zero It also confirms again the regularized solution is converged very well to the exact solution in the range of t ∈ (0, 1), and of α ∈ (0, 2] Conclusion 270 In this study, we introduced two new modified regularization methods as shown in Theorems and for solving a Riesz-Feller space fractional backward diffusion problem with a nonlinear source In theoretical results, we overcome the strong condition as shown in (7) and (9) by a weaker condition based on Theorem 2, and the error estimates are obtained In our numerical experiment, it is shown that the above regularized solution converges nicely to the exact solution with different values of α ∈ (0, 2] 22 (a) α = 0.2 (b) α = 0.6 (c) α = 1.4 (d) α = 1.8 Figure 1: The exact solutions with different α References 275 [1] Z.Q Chen, M.M Meerschaert, E Nane, Space-time fractional diffusion on bounded domains J Math Anal Appl 393 (2012), no 2, 479-488 [2] C Celik, M Duman, Crank-Nicolson method for the fractional diffusion equation with the Riesz fractional derivative J Comput Phys 231 (2012), no 4, 17431750 280 [3] H Cheng, C.-L Fu, G H Zheng, J Gao, A regularization for a Riesz-Feller space-fractional backward diffusion problem, Inverse Probl Sci Eng 22 (2014), no 6, 860-872 [4] K Diethelm, The Analysis of Fractional Differential Equations, Springer-Verlag, Berlin, 2010 [5] L C Evans, Partial differential equations, American Mathematical Society Provi- dence, 1998 [6] Hatano Y and Hatano N 1998 Dispersive transport of ions in column experiments: an explanation of long-tailed profiles, Water Resources Res 34 1027-33 285 [7] Mainardi F, Luchko Y, Pagnini G The fundamental solution of the space-time fractional diffusion equation Fract Cacl Appl Anal 2001; 4: 153–192 [8] F Liu, P Zhuang, V Anh, I Turner, K Burrage, Stability and convergence of the difference methods for the space-time fractional advection-diffusion equation Appl Math Comput 191 (2007), no 1, 12–20 290 [9] M Ginoa, S Cerbelli, H.E Roman, Fractional diffusion equation and relaxation in complex viscoelastic materials, Phys A 191 (1992) 449–453 23 [10] R Gorenflo , F Mainardi Some recent advances in theory and simulation of fractional diffusion processes Journal of Computational and Applied Mathematics 2009; 229(2):400–415 [11] F Mainardi , Y Luchko and G Pagnini The fundamental solution of the space-time fractional diffusion equation 2001, Fract Calc Appl Anal 153-192 295 300 [12] R Gorenflo , F Mainardi , D Moretti , G Pagnini, P Paradisi Fractional diffusion: probability distributions and random walk models Physica A: Statistical Mechanics and its Applications 2002; 305(1-2):106-112 [13] X Liu, L Zhang, P Agarwal, G Wang, On some new integral inequalities of Gronwall-Bellman-Bihari type with de-lay for discontinuous functions and their applications, Indagationes Mathematicae, 2015, doi:10.1016/j.indag.2015.07.001 [14] R Metzler, J Klafter, Boundary value problems for fractional diffusion equations, Phys A 278 (2000) 107-125 [15] Y Povstenko, Theories of thermal stresses based on space-time-fractional telegraph equations Comput Math Appl 64 (2012), no 10, 3321-3328 305 [16] S.S Ray, A new approach for the application of Adomian decomposition method for the solution of fractional space diffusion equation with insulated ends Appl Math Comput 202 (2008), no 2, 544– 549 [17] M Rahmana, A Mahmooda, M Younis, Improved and more feasible numerical methods for Riesz space fractional partial differential equations, J Appl Math Comput 237 (2014), 264–273 310 [18] S Salahshour, A Ahmadian, N Senu, D Baleanu, P Agarwal, On analytical solutions of the fractional differential equation with uncertainty: application to the Basset problem Entropy 17 (2015), no 2, 885902 [19] S Shen, F Liu, V Anh, Numerical approximations and solution techniques for the space-time RieszCaputo fractional advection-diffusion equation Numer Algorithms 56 (2011), no 3, 383-403 315 [20] Q Yang, F Liu, I Turner, Numerical methods for fractional partial differential equations with Riesz space fractional derivatives Appl Math Model 34 (2010), no 1, 200–218 [21] Z.Q Zhang, T Wei, An optimal regularization method for space-fractional backward diffusion problem Math Comput Simulation 92 (2013), 14-27 320 [22] G.H Zheng, T Wei, Two regularization methods for solving a Riesz-Feller space-fractional backward diffusion problem Inverse Problems 26 (2010), no 11, 115017, 22 pp [23] C Shi, C Wang, G Zheng, and T Wei, A new a posteriori parameter choice strategy for the convolution regularization of the space-fractional backward diffusion problem J Comp and Appl Math 279 (2015), 233-248 325 [24] J Zhao, S Liu, T Liu, An inverse problem for space-fractional backward diffusion problem Math Methods Appl Sci 37 (2014), no 8, 1147-1158 [25] X Zhang, P Agarwal, Z Liu and H Peng, The general solution for impulsive differential equations with Riemann Liouville fractional-order q ∈ (1; 2), Open Math 13 (2015), Art 73 330 [26] H Zhou, L Yang, P Agarwal, Solvability for fractional p-Laplacian differential equations with multipoint boundary conditions at resonance on infinite interval, Journal of Applied Mathematics and Computing, pp 1-26, 2015 24 (a) α = 0.2 (b) α = 0.6 (c) α = 1.4 (d) α = 1.8 Figure 2: The exact solutions with different α at t=0.2, 0.5, 0.7 and 0.9 (a) t = 0.2 (b) t = 0.5 (c) t = 0.7 (d) t = 0.9 25 Figure 3: The exact solutions at different t with different values of α (α = 0.2, 0.6, 1.4 and 1.8) (a) = 10−1 (b) = 10−2 (c) = 10−3 (d) = 10−4 Figure 4: A comparison between the exact and its regularized solutions at t = 0.5 with α = 0.2 (a) = 10−1 (b) = 10−2 (c) = 10−3 (d) = 10−4 Figure 5: A comparison between the exact and its regularized solutions at t = 0.5 with α = 1.8 26 (a) = 10−1 (b) = 10−2 (c) = 10−3 (d) = 10−4 Figure 6: A comparison between the exact and its regularized solutions at t = 0.9 with α = 0.2 (a) = 10−1 (b) = 10−2 (c) = 10−3 (d) = 10−4 Figure 7: A comparison between the exact and its regularized solutions at t = 0.9 with α = 1.8 27 1E-01 1E-02 1E-03 1E-04 1E-05 α = 0.2 E1 E2 4.19E-01 6.55E-01 1.69E-01 2.65E-01 6.02E-02 9.40E-02 2.09E-02 3.27E-02 7.30E-03 1.13E-02 α = 0.6 E1 E2 3.62E-01 9.27E-01 1.24E-01 3.17E-01 4.24E-02 1.09E-01 1.47E-02 3.76E-02 5.10E-03 1.30E-02 α = 1.4 E1 E2 8.39E-01 1.55E+00 2.55E-01 4.71E-01 1.04E-01 1.92E-01 3.67E-02 6.77E-02 1.27E-02 2.35E-02 α = 1.8 E1 E2 7.69E-01 9.96E-01 7.43E-02 9.62E-02 4.36E-02 5.65E-02 3.99E-03 5.16E-03 1.24E-04 1.60E-04 Table 1: The absolute (E1 ) and relative (E2 ) error estimates between the exact and its regularized solutions at t = 0.5 1E-01 1E-02 1E-03 1E-04 1E-05 α = 0.2 E1 E2 3.32E-01 5.86E-01 1.37E-01 2.41E-01 4.85E-02 8.55E-02 1.69E-02 2.97E-02 5.90E-03 1.03E-02 α = 0.6 E1 E2 2.26E-01 4.21E-01 7.50E-02 1.40E-01 2.60E-02 4.84E-02 9.00E-03 1.68E-02 3.10E-03 5.80E-03 α = 1.4 E1 E2 9.17E-01 1.75E+00 2.69E-01 5.13E-01 8.85E-02 1.69E-01 3.06E-02 5.84E-02 1.06E-02 2.02E-02 α = 1.8 E1 E2 5.96E-01 9.97E-01 5.82E-01 9.72E-01 3.00E-02 5.01E-02 3.72E-02 6.22E-02 1.12E-03 1.86E-03 Table 2: The absolute (E1 ) and relative (E2 ) error estimates between the exact and its regularized solutions at t = 0.9 335 t 1E-01 1E-02 1E-03 1E-04 1E-05 t = 0.1 E1 E2 5.23E-01 7.44E-01 2.09E-01 2.97E-01 7.42E-02 1.05E-01 2.58E-02 3.66E-02 8.90E-03 1.27E-02 t = 0.2 E1 E2 4.94E-01 7.17E-01 1.99E-01 2.88E-01 7.04E-02 1.02E-01 2.45E-02 3.55E-02 8.50E-03 1.23E-02 t = 0.3 E1 E2 4.69E-01 6.95E-01 1.89E-01 2.80E-01 6.69E-02 9.92E-02 2.32E-02 3.45E-02 8.10E-03 1.20E-02 t = 0.4 E1 E2 4.43E-01 6.74E-01 1.79E-01 2.72E-01 6.35E-02 9.65E-02 2.21E-02 3.36E-02 7.70E-03 1.16E-02 Table 3: The absolute (E1 ) and relative (E2 ) error estimates between the exact and its regularized solutions with α = 0.2 28 t 1E-01 1E-02 1E-03 1E-04 1E-05 t = 0.6 E1 E2 3.96E-01 6.36E-01 1.61E-01 2.58E-01 5.71E-02 9.17E-02 1.99E-03 3.19E-03 6.90E-04 1.11E-03 t = 0.7 E1 E2 3.74E-01 6.19E-01 1.52E-01 2.52E-01 5.41E-02 8.95E-02 1.88E-03 3.11E-03 6.50E-04 1.08E-03 t = 0.8 E1 E2 3.53E-01 6.03E-01 1.44E-01 2.46E-01 5.13E-02 8.75E-02 1.78E-03 3.05E-03 6.20E-04 1.06E-03 Table 4: The absolute (E1 ) and relative (E2 ) error estimates between the exact and its regularized solutions with α = 0.2 t 1E-01 1E-02 1E-03 1E-04 1E-05 t = 0.1 E1 E2 5.23E+00 9.95E-01 5.08E-01 9.96E-01 9.70E-02 1.84E-02 3.01E-03 5.72E-03 9.19E-04 1.75E-04 t = 0.2 E1 E2 3.56E+00 9.96E-01 3.46E-01 9.67E-01 2.67E-02 7.46E-02 2.01E-03 5.62E-03 6.16E-04 1.72E-01 t = 0.3 E1 E2 2.32E+00 9.98E-01 2.50E-01 9.66E-01 5.98E-02 2.60E-02 2.25E-03 5.47E-03 3.93E-04 1.69E-04 t = 0.4 E1 E2 1.40E+00 9.53E-01 1.36E-01 9.62E-01 5.84E-02 4.14E-02 7.36E-03 5.21E-03 2.30E-04 1.63E-04 Table 5: The absolute (E1 ) and relative (E2 ) error estimates between the exact and its regularized solutions with α = 1.8 t 1E-01 1E-02 1E-03 1E-04 1E-05 t = 0.6 E1 E2 6.10E-01 9.97E-01 5.97E-01 9.76E-01 4.14E-02 6.77E-02 3.50E-03 5.71E-03 1.22E-04 1.99E-04 t = 0.7 E1 E2 6.54E-01 9.97E-01 6.40E-01 9.75E-01 4.28E-02 6.53E-02 4.06E-03 6.19E-03 1.27E-04 1.94E-04 t = 0.8 E1 E2 6.40E-01 9.97E-01 6.24E-01 9.73E-01 4.06E-02 6.33E-02 3.50E-03 5.46E-03 1.21E-04 1.89E-04 Table 6: The absolute (E1 ) and relative (E2 ) error estimates between the exact and its regularized solutions with α = 1.8 29 ... solution of space- fractional diffusion equation Yang et al [20] studied several numerical methods for fractional partial differential equations with Riesz space- fractional derivatives The space fractional. .. 21589, Saudi Arabia Abstract In this paper, a backward diffusion problem for a space- fractional diffusion equation with a nonlinear source in a strip is investigated This problem is obtained from... been arisen from replacing the standard space partial derivative in the diffusion equation with a space fractional partial derivative It can be derived from the continuous-time random walk in statistical