Three landweber iterative methods for solving the initial value problem of time fractional diffusion wave equation on spherically symmetric domain

Inverse Problems in Science and Engineering ISSN: (Print) (Online) Journal homepage: https://www.tandfonline.com/loi/gipe20 Three Landweber iterative methods for solving the initial value problem of time-fractional diffusionwave equation on spherically symmetric domain Fan Yang, Qiao-Xi Sun & Xiao-Xiao Li To cite this article: Fan Yang, Qiao-Xi Sun & Xiao-Xiao Li (2021): Three Landweber iterative methods for solving the initial value problem of time-fractional diffusion-wave equation on spherically symmetric domain, Inverse Problems in Science and Engineering, DOI: 10.1080/17415977.2021.1914603 To link to this article: https://doi.org/10.1080/17415977.2021.1914603 Published online: 17 Apr 2021 Submit your article to this journal View related articles View Crossmark data Full Terms & Conditions of access and use can be found at https://www.tandfonline.com/action/journalInformation?journalCode=gipe20 INVERSE PROBLEMS IN SCIENCE AND ENGINEERING https://doi.org/10.1080/17415977.2021.1914603 Three Landweber iterative methods for solving the initial value problem of time-fractional diffusion-wave equation on spherically symmetric domain Fan Yang, Qiao-Xi Sun and Xiao-Xiao Li School of Science, Lanzhou University of Technology, Lanzhou, People’s Republic of China ABSTRACT ARTICLE HISTORY In this paper, the inverse problem for identifying the initial value of time-fractional diffusion wave equation on spherically symmetric region is considered The exact solution of this problem is obtained by using the method of separating variables and the property the Mittag–Leffler functions This problem is ill-posed, i.e the solution(if exists) does not depend on the measurable data Three different kinds landweber iterative methods are used to solve this problem Under the priori and the posteriori regularization parameters choice rules, the error estimates between the exact solution and the regularization solutions are obtained Several numerical examples are given to prove the effectiveness of these regularization methods Received April 2020 Accepted 28 March 2021 KEYWORDS Time-fractional diffusion wave equation; spherically symmetric; Ill-posed problem; fractional Landweber method; inverse problem 2010 MATHEMATICS SUBJECT CLASSIFICATIONS 35R25; 47A52; 35R30 Introduction Nowadays, people find that the fractional derivative has much advantages in solving practical problem, such as in medical engineering [1], chemistry and biochemistry [2], finance and economics [3–6], inverse scattering [7] Up to now, a lot of achievements have been made in solving the direct problems [8–12] of fractional differential equations However, when solving practical problems, the initial value, or source term, or diffusion coefficient, or part of the boundary value [13,14] is unknown, and it is necessary to invert them through some measurement data, which puts forward the inverse problem of fractional diffusion equation The study of the fractional diffusion equation is still on an initial stage, the direct problem of it has been studied in [15–17] For the inverse problem of time-fractional diffusion equation as < α < 1, there are a lot of research results For identifying the unknown source, one can see [18–26] About backward heat conduction problem, one can see [27–34] About identifying the initial value problem, one can see [35–37] For an inverse unknown coefficient problem of a time-fractional equation, one can see [38,39] About identifying the source term and initial data simultaneous of time-fractional diffusion equation, one can see [40,41] About identifying some unknown parameters in time-fractional diffusion equation, one can see CONTACT Fan Yang yfggd114@163.com 730050, People’s Republic of China School of Science, Lanzhou University of Technology, Lanzhou, Gansu © 2021 Informa UK Limited, trading as Taylor & Francis Group F YANG ET AL [42] About the inverse problem for the heat equation on a columnar axis-symmetric area, one can see [43–48] In [43–45], Landweber regularization method, a simplified Tikhonov regularization method and a spectral method are used to identify source term on a columnar axis-symmetric area In [46–48], the authors considered a backward problem on a columnar axis-symmetric domain In [46], Yang et al used the quasi-boundary value regularization method to solve the inverse problem for determining the initial value of heat equation with inhomogeneous source on a columnar symmetric domain The error estimate between the regular solution and the exact solution under the corresponding regularization parameter selection rules is obtained Finally, numerical example is given to verify that the regularization method is very effective for solving this inverse problem In [47], Cheng et al used the modified Tikhonov regularization method to with the inverse time problem for an axisymmetric heat equation Finally, Hölder type error estimate between the approximate solution and exact solution is obtained In [48], Djerrar et al used standard Tikhonov regularization method to deal with an axisymmetric inverse problem for the heat equation inside the cylinder a ≤ r ≤ b, and numerical examples is used to show that this method is effective and stable About the inverse problem for the time-fractional diffusion equation as < α < on a columnar and spherical symmetric areas, one can see [49,50] In [49], Xiong proposed a backward problem model of a timefractional diffusion-heat equation on a columnar axis-symmetric domain Yang et al in [50] used Landweber iterative regularization method to solve identifying the initial value of time-fractional diffusion equation on spherically symmetric domain Compare with the inverse problem of time-fractional diffusion equation as < α < 1, there are little research result for the inverse problem of time-fractional diffusion wave equation as < α < Šišková et al in [51] used the regularization method to solve the inverse source problem of time-fractional diffusion wave equation Liao et al in [52] used conjugate gradient method combined with Morozovs discrepancy principle to identify the unknown source for the time-fractional diffusion wave equation Šišková et al in [53] used the regularization method to deal with an inverse source problem for a time-fractional wave equation Gong et al in [54] used a generalized Tikhonov to identify the time-dependent source term in a time-fractional diffusion-wave equation In recent years, in physical oceanography and global meteorology, the inversion of initial boundary value problems has always been a hot issue In order to increase the accuracy of numerical weather prediction, usually by the model combined with the observation data of the inversion of initial boundary value problems, and in numerical weather prediction model, provide a reasonable initial field At present, many domestic and foreign ocean circulation model, atmospheric general circulation model, numerical weather prediction model and torrential rain forecasting model belongs to the inversion of initial boundary value problems during initialization, so such problems of scientific research application prospect is very broad Yang et al in [55] used the truncated regularization method to solve the inverse initial value problem of the time-fractional inhomogeneous diffusion wave equation Yang et al in [56] used the Landweber iterative regularization method to solve the inverse problem for identifying the initial value problem of a space time-fractional diffusion wave equation Wei et al in [57] used the Tikhonov regularization method to solve the inverse initial value problem of time-fractional diffusion wave equation Wei et al in [58] used the conjugate gradient algorithm combined with Tikhonov regularization method to identify the initial value of time-fractional diffusion wave equation Until now, we find that there are few papers for the INVERSE PROBLEMS IN SCIENCE AND ENGINEERING inverse problem of time-fractional diffusion-wave equation on a columnar axis-symmetric domain and spherically symmetric domain In [59], the authors used the Landweber iterative method to solve an inverse source problem of time-fractional diffusion-wave equation on spherically symmetric domain It is assumed that the grain is of a spherically symmetric domain diffusion geometry as illustrated in Figure (a-b), which is actually consistent with laboratory measurements of helium diffusion from a physical point of view from apatite As a consequence of radiogenic production and diffusive loss, u(r, t) which only depends on the spherically radius r and t denotes the concentration of helium For the inverse problem of inversion initial value in spherically symmetric region, there are few research results at present Whereupon, in this paper, we consider the inverse problem to identify the initial value of time-fractional diffusion-wave equation on spherically symmetric region and give three regularization methods to deal with this inverse problem in order to find a effective regular method In this paper, we consider the following problem: ⎧ ⎪ ⎪ Dαt u(r, t) − ur (r, t) − urr (r, t) = f (r), < r < r0 , < t < T, < α < 2, ⎪ ⎪ r ⎪ ⎪ ⎪ u(r , t) = 0, ≤ t ≤ T, ⎪ ⎪ ⎪ ⎨ u(r, 0) = ϕ(r), ≤ r ≤ r0 , ⎪ ≤ r ≤ r0 , ut (r, 0) = ψ(r), ⎪ ⎪ ⎪ ⎪ ⎪ ≤ t ≤ T, lim u(r, t) bounded, ⎪ ⎪ r→0 ⎪ ⎪ ⎩ u(r, T) = g(r), ≤ r ≤ r0 , (1) where r0 is the radius, Dαt u(r, t) is the Caputo fractional derivative (1 < α < 2), it is defined as Dαt u(r, t) = (2 − α) t ∂ u(r, s) ds , ∂s2 (t − s)α−1 < α < (2) The existence and uniqueness of the direct problem solution has been proved in the [60] The inverse problem is to use the measurement data g(r) and the known function f (r) to identify the unknown initial data ϕ(r), ψ(r) The inverse initial value problem can be transformed into two cases: F YANG ET AL (IVP1): Assuming ψ(r) is known, we use the final value data g(r) and the known function f (r) to invert the initial value ϕ(r) (IVP2): Assuming ϕ(r) is known, we use the final value data g(r) and the known function f (r) to invert the initial value ψ(r) Because the measurements are error-prone, we remark the measurements with error as f δ and g δ and satisfy fδ − f L2 [0,r0 ;r2 ] g δ (r) − g(r) ≤ δ, L2 [0,r0 ;r2 ] (3) ≤ δ, (4) where δ > In this paper, L2 [0, r0 ; r2 ] represents the Hilbert space with weight r2 on the interval [0, r0 ] of the Lebesgue measurable function (·, ·) and · represent the inner product and norm of the space of [0, r0 ; r2 ], respectively · is defined as follows: · = r0 2 r | · | dr (5) This paper is organized as follows In Section 2, we recall and state some preliminary theoretical results In Section 3, we analyse the ill-posedness of the problem (IVP1) and the problem (IVP2), and give the conditional stability result In Section 4, we give the corresponding a priori error estimates and posteriori error estimates for three regularization methods In Section 5, we conduct some numerical tests to show the validity of the proposed regularization methods Since most of the solutions of fractional partial differential equations contain special functions (Mittag–Leffler functions), and the calculation of these functions is quite difficult In this paper, the difficulties are overcome through [61,62] Finally, we give some concluding remarks Preliminary results In this section, we give some important Lemmas Lemma 2.1 ([57]): If < α < 2, and β ∈ R be arbitrary Suppose μ satisfy min{π , π α} Then there exists a constant C1 = C(α, β, μ) > such that |Eα,β (z)| ≤ C1 , + |z| μ ≤ |arg(z)| ≤ π πα 0, we have Eα,β (−η) = 1 +O (β − α)η η , η → ∞ (7) (−λ T α ) Lemma 2.3: For 12 < γ < 1, a1 and a2 are relaxation factor and satisfy < a1 Eα,1 n 2 α < 1, < a2 T Eα,2 (−λn T ) < 1, m1 ≥ 1, m2 ≥ 1, we have sup [1 − (1 − a1 Eα,1 (−λn T α ))m1 ]γ λn >0 √ ≤ a1 m1 , α Eα,1 (−λn T ) (8) INVERSE PROBLEMS IN SCIENCE AND ENGINEERING sup [1 − (1 − a2 T Eα,2 (−λn T α ))m2 ]γ λn >0 √ ≤ a2 m2 α TEα,2 (−λn T ) (9) Proof: Refer to the appendix for the details of the proof 2 α Lemma 2.4: For 12 < γ < 1, m1 ≥ 1, m2 ≥ 1, λn = ( nπ r0 ) > 0, < a1 Eα,1 (−λn T ) < 1, (−λ T α ) < 1, we have < a2 T Eα,2 n p − p 2 (−λn T α ))m1 Eα,1 (−λn T α ) ≤ c(a1 , p)m1 , sup (1 − a1 Eα,1 (10) λn >0 p p − p 2 sup (1 − a2 T Eα,2 (−λn T α ))m2 T Eα,2 (−λn T α ) ≤ c(a2 , p)m2 , λn >0 (11) p p where the constant c(a1 , p) is given by c(a1 , p) = ( 4a1 ) and c(a2 , p) is given by c(a2 , p) = p p ( 4a2 ) Proof: Refer to the appendix for the details of the proof Lemma 2.5: For < α < and any fixed T > 0, there is at most a finite index set I1 = nπ α α {n1 , n2 , , nN } such that Eα,1 (−( nπ r0 ) T ) = for n ∈ I1 and Eα,1 (−( r0 ) T ) = for n∈ / I1 Meanwhile there is at most a finite index set I2 = {m1 , m2 , , mM } such that nπ α α Eα,2 (−( nπ / I2 r0 ) T ) = for n ∈ I2 and Eα,2 (−( r0 ) T ) = for n ∈ Proof: From Lemma 2.2, we know that there exists L0 > such that Eα,1 − nπ r0 Tα ≤ nπ (1 − α) r0 < 0, Tα nπ r0 T α > L0 , nπ α α for < α < 2, thus we know Eα,1 (−( nπ r0 ) T ) = only if ( r0 ) T ≤ L0 Since nπ nπ α limn→+∞ ( nπ r0 ) = +∞, there are only finite ( r0 ) satisfying ( r0 ) T ≤ L0 The proof α for Eα,2 (−( nπ r0 ) T ) is similar Remark 2.1: The index sets I1 and I2 may be empty, that means the singular values for the operators K1 and K2 are not zeros Here and below, all the results for I1 = ∅ and I2 = ∅ are regarded as the special cases Lemma 2.6 ([57]): For < α < and λn = ( nπ r0 ) , there exists positive constants C, C depending on α, T such that C C ≤ |Eα,1 (−λn T α )| ≤ , λn λn n ∈ I1 , (12) C C ≤ |Eα,2 (−λn T α )| ≤ , λn λn n ∈ I2 (13) F YANG ET AL nπ α 2 Lemma 2.7: For 12 < γ < 1, m3 ≥ 1, m4 ≥ 1, λn = ( nπ r0 ) > 0, < a1 Eα,1 (−( r0 ) T ) < (−( nπ )2 T α ) < 1, we have 1, < a2 T Eα,2 r0 a1 C2 r04 − p ) pπ T and C4 = ( α p p (1 + n2 )− ≤ C3 (m3 + 1)− , m4 nπ − r0 − a2 T Eα,2 where C3 = ( m3 nπ − r0 − a1 Eα,1 T α p (14) p (1 + n2 )− ≤ C4 (m4 + 1)− , (15) a2 T C2 r04 − p ) pπ Proof: Refer to the appendix for the details of the proof (−( nπ )2 T α ) < 1, < Lemma 2.8: For 12 < γ < 1, m5 ≥ 1, m6 ≥ 1, λn > 0, < a1 Eα,1 r0 nπ α 2 a2 T Eα,2 (−( r0 ) T ) < 1, we have γ +1 − a1 Eα,1 m5 Tα p − 2(γ +1) p where C5 = ( π 20 )− ( 2a1 (γp +1) ) m6 T α p − 2(γ +1) p (1 + n2 )− ≤ C5 m5 nπ − r0 γ +1 − a2 T γ +1 Eα,2 Cr2 nπ r0 − , p − 2(γ +1) p (1 + n2 )− ≤ C6 m6 and C6 = ( (16) , (17) CTr02 − p 2a2 (γ +1) − 2(γp+1) ) 2( p ) π2 Proof: Refer to the appendix for the details of the proof γ +1 α γ +1 Lemma 2.9: For < γ ≤ 1, m5 ≥ 1, m6 ≥ 1, < a1 Eα,1 (−( nπ r0 ) T ) < 1, < a2 T γ +1 α Eα,2 (−( nπ r0 ) T ) < 1, we have 1− γ +1 − a1 Eα,1 Eα,1 1− nπ − r0 γ +1 − a2 T γ +1 Eα,2 TEα,2 nπ − r0 m5 Tα ≤ (a1 m5 ) γ +1 , Tα nπ − r0 nπ − r0 (18) 2 m6 Tα ≤ (a2 m6 ) γ +1 (19) Tα Proof: Refer to the appendix for the details of the proof Lemma 2.10: For a1 > 0, a2 > 0, p > 0, m1 > 0, m2 > 0, we have Cr2 F(x) = 20 π a1 C r − 40 x π m1 −1 p (x)− −1 ≤ C7 (m1 + 1)− p+2 , (20) INVERSE PROBLEMS IN SCIENCE AND ENGINEERING G(x) = where C7 = TCr02 π2 1− Cr02 a1 Cr04 − p+2 ( ) π (p+2)π m2 −1 a2 T C2 r04 x2 π and C8 = p (x)− −1 ≤ C8 (m2 + 1)− p+2 , (21) TCr02 a2 TCr04 − p+2 ( ) π (p+2)π Proof: Refer to the appendix for the details of the proof Lemma 2.11: For a1 > 0, a2 > 0, p > 0, m5 ≥ 1, m6 ≥ 1, we have Cr2 F(x) = 20 π TCr02 G(x) = π2 where C9 = γ +1 m5 −1 Cr02 xπ − a1 − a2 T γ +1 m6 −1 Cr02 xπ γ +1 p p+2 Cr02 Cr02 − p −1 a1 − 2(γ +1) ( ) ( p+2 ) π2 π2 and C10 = p+2 − 2(γ +1) (x)− −1 ≤ C9 (m5 + 1) p , (22) p+2 − 2(γ +1) (x)− −1 ≤ C10 (m6 + 1) , (23) p+2 TCr02 TCr02 − p −1 a1 − 2(γ +1) ( π ) ( p+2 ) π2 Proof: Refer to the appendix for the details of the proof The ill-posedness and the conditional stability Define ∞ p p (1 + n2 ) (fn , Rn (r)) , H = f ∈ L [0, r0 ; r ]; (24) n=1 where (·, ·) is the inner product in L2 [0, r0 ; r2 ], then H p is a Hilbert space with the norm ∞ ϕ(·) Hp p (1 + n2 ) (ϕn , Rn (r)) , := n=1 and ∞ ψ(·) Hp p (1 + n2 ) (ψn , Rn (r)) := n=1 Theorem 3.1: Let ϕ(r), ψ(r) ∈ L2 [0, r0 ; r2 ], then there exists a unique weak solution and the weak solution for (1) is given by ∞ u(r, t) = nπ r0 t α−1 Eα,α − n=1 + Eα,1 − + tEα,2 − nπ r0 nπ r0 2 t α (f (r), Rn (r)) t α (ϕ(r), Rn (r)) t α (ψ(r), Rn (r)) Rn (r), (25) F YANG ET AL where (ϕ(r), Rn (r)) and (ψ(r), Rn (r)) are the Fourier coefficients Let t = T in (25), we have ∞ u(r, T) = nπ r0 T α−1 Eα,α − n=1 nπ r0 + Eα,1 − T α (f (r), Rn (r)) T α (ϕ(r), Rn (r)) nπ r0 + TEα,2 − T α (ψ(r), Rn (r)) Rn (r) = g(r) Denote ∞ T α−1 Eα,α − g1 (r) := g(r) − n=1 nπ r0 nπ r0 T α fn + TEα,2 − T α ψn Rn (r), and ∞ T α−1 Eα,α − g2 (r) := g(r) − n=1 nπ r0 T α fn + Eα,1 − nπ r0 T α ϕn Rn (r) Then we have ∞ g1 (r) = nπ r0 ϕn Eα,1 − n=1 T α Rn (r), (26) and ∞ g2 (r) = ψn TEα,2 − n=1 nπ r0 T α Rn (r) (27) Now we put the definitions of ϕn and ψn into (26) and (27), then the problem (IVP1) and the problem (IVP2) become the following integral equations (K1 ϕ)(ξ ) = r0 κ1 (r, ξ )ϕ(ξ ) dξ = g1 (r), (28) where the integral kernel is ∞ κ1 (r, ξ ) = Eα,1 − n=1 nπ r0 T α Rn (r)Rn (ξ ) (29) And (K2 ψ)(ξ ) = r0 κ2 (r, ξ )ψ(ξ ) dξ = g2 (r), (30) where the integral kernel is ∞ κ2 (r, ξ ) = TEα,2 − n=1 nπ r0 T α Rn (r)Rn (ξ ), (31) INVERSE PROBLEMS IN SCIENCE AND ENGINEERING and due to [59], we know the linear operators K1 and K2 are compact from L2 [0, r0 ; r2 ] to L2 [0, r0 ; r2 ] The problem (IVP1) and the problem (IVP2) are ill-posed √ Let K1∗ be the adjoint of K1 and K2∗ be the adjoint of K2 Since Rn (r) = standard orthogonal system with weight r2 in the L2 [0, r nπ r0 K1∗ K1 Rn (ξ ) = Eα,1 − r 2nπ sin( nπ r ) √ nπ r is a r0 ; r ], it is easy to verity r0 T α Rn (ξ ), and nπ r0 K2∗ K2 Rn (ξ ) = T Eα,2 − T α Rn (ξ ) (1) α Hence, the singular values of K1 are σ1n = |Eα,1 (−( nπ r0 ) T )| Define ψn(1) (r) = ⎧ ⎪ ⎪ ⎪ ⎪ ⎨Rn (r), ⎪ ⎪ ⎪ ⎪ ⎩−Rn (r), nπ r0 nπ − r0 Eα,1 − Eα,1 Tα ≥ 0, (32) Tα < (1) 2 It is clear that {ψn }∞ n=1 are orthonormal in L [0, r0 ; r ], we can verity (1) K1 Rn (ξ ) = σ1n ψn(1) (r) = Eα,1 − (1) K1∗ ψn(1) (r) = σ1n Rn (ξ ) = Eα,1 − nπ r0 nπ r0 T α Rn (r), (33) T α ψn(1) (ξ ) (34) (1) Therefore, the singular system of K1 is (σ1n ; Rn , ψn(1) ) (2) (2) By the similar verification, we know the singular system of K2 is (σ2n ; Rn , ψn ), where (2) α σ2n = |TEα,2 (−( nπ r0 ) T )| and ψn(2) (r) = ⎧ ⎪ ⎪ ⎪ ⎪ ⎨Rn (r), ⎪ ⎪ ⎪ ⎪ ⎩−Rn (r), nπ r0 nπ − r0 Eα,2 − Eα,2 Tα ≥ 0, (35) Tα < In the following, the integral kernels given in (29) and (31) are rewritten as ∞ κ1 (r, ξ ) = Eα,1 − n=1,n∈I /1 ∞ κ2 (r, ξ ) = TEα,2 − n=1,n∈I /2 nπ r0 nπ r0 T α Rn (r)Rn (ξ ), T α Rn (r)Rn (ξ ) (36) (37) INVERSE PROBLEMS IN SCIENCE AND ENGINEERING 37 Table The CPU time of Example 5.1 for different regularization method α 1.2 1.5 1.8 Landweber ε = 0.01 ε = 0.008 ε = 0.005 11.9304 15.4281 27.4445 5.8485 7.4576 13.4199 5.7179 7.5216 13.4263 Fractional Landweber ε = 0.01 ε = 0.008 ε = 0.005 1.7742 1.5056 2.4426 0.6786 1.4258 1.4684 0.7475 1.4617 1.5987 Modified iterative ε = 0.01 ε = 0.008 ε = 0.005 2.8694 3.4975 5.4024 1.6987 1.5797 3.4720 1.7893 1.4896 3.4510 Landweber regularization method has fewer iteration steps For α = 1.5 and α = 1.8, the same result is obtained In summary, the fractional Landweber regularization method has fewer iteration steps In Table 2, we use a computer with Intel(R) Core(TM) i5-6200U CPU @ 2.30 GHz 2.40 GHz and RAM of 4.00 GB to calculate the CPU time The specific analysis is as follows: we fix α = 1.2 For Table 2, when ε = 0.01, the CPU time of Landweber regularization method, fractional Landweber regularization method and modified iterative method are 11.9304 s, 1.7742 s and 2.8694 s When ε = 0.008, the CPU time of Landweber regularization method, fractional Landweber regularization method and modified iterative method are 15.4281 s, 1.5056 s and 3.4975 s When ε = 0.005, the CPU time of Landweber regularization method, fractional Landweber regularization method and modified iterative method are 27.4445 s, 2.4426 s and 5.4024 s We can deduced that α = 1.2 is fixed, the fractional Landweber regularization method has fewer CPU time For α = 1.5 and α = 1.8, the same result is obtained We fix ε = 0.01 When α = 1.2, the CPU time of Landweber regularization method, fractional Landweber regularization method and modified iterative method are 11.9304 s, 1.7742 s and 2.8694 s When α = 1.5, the CPU time of Landweber regularization method, fractional Landweber regularization method and modified iterative method are 5.8485 s, 0.6786 s and 1.6987 s When α = 1.8, the CPU time of Landweber regularization method, fractional Landweber regularization method and modified iterative method are 5.7179 s, 0.7475 s and 1.7893 s We can deduced that ε = 0.01 is fixed, the fractional Landweber regularization method has fewer CPU time For ε = 0.008 and ε = 0.005, the same result is obtained In summary, the fractional Landweber regularization method has fewer CPU time By Table 3, we can deduce that the errors between exact solution and approximate solution are smaller for fixed α and the smaller measurement error And we infer that the error between exact solution and approximation solution of fractional Landweber method is smaller than that obtained by Landweber regularization method and modified iterative method for fixed α and ε From Table 4, we can obtain the error between exact solution and approximation solution of fractional Landweber method is smaller than that obtained by Landweber regularization method and modified iterative method for fixed α and ε In Figures 1–3, we give numerical results of Example 5.1 under the a posteriori parameter choice rule for various noise levels ε = 0.01, 0.008, 0.005 in the case of α = 1.2, 1.5, 1.8 38 F YANG ET AL Table Error behaviour of Example 5.1 for different α with ε = 0.01, 0.005 α 1.1 1.3 1.5 1.7 1.9 Landweber ε = 0.01 ε = 0.01 ε = 0.005 ε = 0.005 Rela Abso Rela Abso 0.7320 0.6078 0.6501 0.5910 0.6001 0.4319 0.5017 0.4270 0.5101 0.3121 0.3725 0.2110 0.3923 0.1920 0.2519 0.1175 0.2431 0.1311 0.0935 0.0734 Fractional Landweber ε = 0.01 ε = 0.01 ε = 0.005 ε = 0.005 Rela Abso Rela Abso 0.4560 0.2113 0.2715 0.1091 0.3817 0.1352 0.1219 0.0836 0.3039 0.0987 0.0902 0.0430 0.2377 0.0553 0.0835 0.0259 0.2105 0.0098 0.0631 0.0047 Modified iterative ε = 0.01 ε = 0.01 ε = 0.005 ε = 0.005 Rela Abso Rela Abso 0.5734 0.3978 0.3807 0.2933 0.5180 0.2534 0.3741 0.1717 0.3945 0.2575 0.2370 0.1682 0.3030 0.1114 0.1415 0.0807 0.2509 0.0273 0.0901 0.0109 Table Error behaviour of Example 5.2 for different α with ε = 0.01, 0.005 α 1.1 1.3 1.5 1.7 1.9 Landweber ε = 0.01 ε = 0.01 ε = 0.005 ε = 0.005 Rela Abso Rela Abso 0.9845 0.7814 0.9013 0.6317 0.7516 0.5831 0.6289 0.5521 0.5322 0.4911 0.4516 0.3849 0.3303 0.2510 0.2035 0.1228 0.2730 0.1425 0.1013 0.0616 Fractional Landweber ε = 0.01 ε = 0.01 ε = 0.005 ε = 0.005 Rela Abso Rela Abso 0.4532 0.3017 0.2853 0.1350 0.3907 0.2125 0.2036 0.1049 0.3015 0.1526 0.1489 0.0731 0.2072 0.0821 0.0956 0.0232 0.1723 0.0101 0.0745 0.0059 Modified iterative ε = 0.01 ε = 0.01 ε = 0.005 ε = 0.005 Rela Abso Rela Abso 0.6711 0.3506 0.3421 0.1823 0.5332 0.3029 0.2521 0.1352 0.4136 0.2207 0.1987 0.0908 0.2871 0.1019 0.1115 0.0673 0.2020 0.0310 0.0109 0.0456 It can be seen that the numerical error also decreases when the noise is reduced And the smaller α, the better the approximate effect In Figures 4–6, we give numerical results of Example 5.2 under the a posteriori parameter choice rule for various noise levels ε = 0.01, 0.008, 0.005 in the case of α = 1.2, 1.5, 1.8 It can be seen that the numerical error also decreases when the noise is reduced And the smaller α, the better the approximate effect In Figures 7–9, we give numerical results of Example 5.3 under the a posteriori parameter choice rule for various noise levels ε = 0.01, 0.008, 0.005 in the case of α = 1.2, 1.5, 1.8 It can be seen that the numerical error also decreases when the noise is reduced And the smaller α, the better the approximate effect Example 5.3: Take function ⎧ π ⎨0, < r ≤ , ϕ(r) = ⎩1, π < r < π In Figures 10–12, we give numerical results of Example 5.4 under the a posteriori parameter choice rule for various noise levels ε = 0.01, 0.008, 0.005 in the case of α = 1.2, 1.5, 1.8 It can be seen that the numerical error also decreases when the noise is reduced And the smaller α, the better the approximate effect INVERSE PROBLEMS IN SCIENCE AND ENGINEERING 39 Figure The exact solution and regular solution of Landweber regularization method by using the a posteriori parameter choice rule for Example 5.3 (a) α = 1.2, (b) α = 1.5, (c) α = 1.8 Figure The exact solution and regular solution of fractional Landweber regularization method by using the a posteriori parameter choice rule for Example 5.3 (a) α = 1.2, (b) α = 1.5, (c) α = 1.8 40 F YANG ET AL Figure The exact solution and regular solution of modified iterative regularization method by using the a posteriori parameter choice rule for Example 5.3 (a) α = 1.2, (b) α = 1.5, (c) α = 1.8 Figure 10 The exact solution and regular solution of Landweber regularization method by using the a posteriori parameter choice rule for Example 5.4 (a) α = 1.2, (b) α = 1.5, (c) α = 1.8 INVERSE PROBLEMS IN SCIENCE AND ENGINEERING 41 Figure 11 The exact solution and regular solution of fractional Landweber regularization method by using the a posteriori parameter choice rule for Example 5.4 (a) α = 1.2, (b) α = 1.5, (c) α = 1.8 Figure 12 The exact solution and regular solution of modified iterative regularization method by using the a posteriori parameter choice rule for Example 5.4 (a) α = 1.2, (b) α = 1.5, (c) α = 1.8 42 F YANG ET AL Figure 13 The exact solution and regular solution of Landweber regularization method by using the a posteriori parameter choice rule for Example 5.5 (a) α = 1.2, (b) α = 1.5, (c) α = 1.8 In Figures 13–15, we give numerical results of Example 5.5 under the a posteriori parameter choice rule for various noise levels ε = 0.01, 0.008, 0.005 in the case of α = 1.2, 1.5, 1.8 It can be seen that the numerical error also decreases when the noise is reduced And the smaller α, the better the approximate effect In Figures 16–18, we give numerical results of Example 5.6 under the a posteriori parameter choice rule for various noise levels ε = 0.01, 0.008, 0.005 in the case of α = 1.2, 1.5, 1.8 It can be seen that the numerical error also decreases when the noise is reduced And the smaller α, the better the approximate effect Example 5.4: Take function ψ(r) = α cos(r) Example 5.5: Take function ⎧ π ⎨−2r + π , < r ≤ , ψ(r) = π ⎩ 2r − π , ≤ r < π Example 5.6: Take function ⎧ ⎪ 1, ⎪ ⎪ ⎪ ⎨ ψ(r) = 0, ⎪ ⎪ ⎪ ⎪ ⎩1, π , π 2π