Fractional Differential Equations: Finite Difference Methods 9783110616064, 9783110615173

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Zhi-zhong Sun and Guang-hua Gao Fractional Differential Equations

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Zhi-zhong Sun and Guang-hua Gao

Fractional Differential Equations |

Finite Difference Methods

Mathematics Subject Classification 2020 65M06, 65M12, 65M15 Authors Prof. Zhi-zhong Sun Southeast University School of Mathematics No. 2, Dongnandaxue Road, Jiangning District 211189 Nanjing, Jiangsu Province People’s Republic of China [email protected]

Prof. Guang-hua Gao Nanjing University of Posts and Telecommunications College of Science No. 9, Wenyuan Road, Yadongxincheng Zone 210023 Nanjing, Jiangsu Province People’s Republic of China [email protected]

ISBN 978-3-11-061517-3 e-ISBN (PDF) 978-3-11-061606-4 e-ISBN (EPUB) 978-3-11-061530-2 Library of Congress Control Number: 2020938024 Bibliographic information published by the Deutsche Nationalbibliothek The Deutsche Nationalbibliothek lists this publication in the Deutsche Nationalbibliografie; detailed bibliographic data are available on the Internet at http://dnb.dnb.de. © 2020 China Science Publishing & Media Ltd. and Walter de Gruyter GmbH, Berlin/Boston Cover image: 00one / E+ / gettyimages.de Typesetting: VTeX UAB, Lithuania Printing and binding: CPI books GmbH, Leck www.degruyter.com

Preface As a generalization of classical calculus, fractional calculus has become an important branch of mathematics. It is popularly believed that this concept is stemmed from a letter by G. W. Leibniz (1646–1716) in the year 1695, where the one-half order of derivative was discussed. During the development of the past more than three centuries, numerous mathematicians made outstanding contributions on this field. Now the fractional differential equations (FDEs) have become one of the important tools to model complex mechanics and physical behaviors and widespread applications have been found in anomalous diffusion, viscoelasticity, fluid flow, boundary layer effect of pipeline, electromagnetism, signal processing and control, quantum economy, fractal theory, etc., whereas, it is difficult to get the analytical solutions to the FDEs, even for the linear FDEs. Hence it becomes an important task to find some effective numerical simulations in current researches. This book aims to make a systematic introduction to the finite difference method of FDEs. There are six chapters in this book. Chapter 1 serves as a mathematical introduction to fractional calculus. It commences with four basic definitions of fractional derivatives. The analytical solutions to two kinds of fractional ordinary differential equations (FODEs) are given, from which, readers can have a general idea on the behaviors of solutions to FODEs. Several numerical approximation ways to fractional derivatives are introduced together with their numerical accuracy analysis. The applications of these formulae are also illustrated by solving the FODEs. This part is the important foundation of the following numerical solutions to fractional partial differential equations (FPDEs). In Chapter 2, we study the finite difference methods for solving time-fractional subdiffusion equations. The time-fractional derivatives are approached by the G-L formula, the L1 approximation, the L2-1σ approximation, the fast L1 approximation and the fast L2-1σ approximation, respectively; The spatial derivatives are discretized by using the second-order central difference quotient or the compact approximation. For the 2D problem, several ADI difference schemes are derived. The unique solvability, stability and convergence for each scheme are proved. Chapter 3 shows the finite difference methods for solving time-fractional wave equations. The time-fractional derivatives are discretized by the L1 approximation, the fast L1 approximation, the L2-1σ approximation and the fast L2-1σ approximation, respectively. For the 1D problem, two kinds of difference schemes are developed, among which one is of order two in space and the other is of order four in space. For the 2D problem, the ADI scheme and compact ADI scheme are both mentioned. The unique solvability, stability and convergence for each scheme are proved.

https://doi.org/10.1515/9783110616064-201

VI | Preface In Chapter 4, we introduce the finite difference methods for solving the spacefractional partial differential equations. For the 1D problem, the first-order method based on the shifted G-L formula, the second-order method based on the weightedshifted G-L (WSGL) formula and the fourth-order method based on the WSGL formula are developed in turn. For the 2D problem, a fourth-order ADI method based on the WSGL formula is presented. The unique solvability, stability and convergence for each scheme are shown. Chapter 5 considers the finite difference methods for solving a class of the timespace fractional differential equations. The time Caputo derivative is treated by the L2-1σ approximation and the spatial Riesz derivatives are discretized by the secondorder fractional central difference quotient and the fourth-order weighted fractional central difference quotient formula, respectively. The second-order and the fourthorder difference schemes in space are established, respectively. The unique solvability, stability and convergence for each scheme are proved. In Chapter 6, the finite difference methods for solving a class of time distributedorder subdiffusion equations are concerned. The distributed integral is discretized using the composite trapezoid formula or composite Simpson formula and the Caputo time-fractional derivatives are approximated using the second-order WSGL formula. The second-order scheme in both time and distributed order, and another fourth-order scheme in both time and distributed order are constructed, respectively. In addition, for the 2D problem, a second-order ADI difference scheme and another fourth-order ADI difference scheme are developed, respectively. The unique solvability, stability and convergence for each scheme are analyzed. There are abundant results on the numerical method for FDEs in recent 20 years. In the last section of each chapter, we give a brief overview and only a limit part among them is listed in the references of this book, which are the resource or the referred materials of this book. The main part of this book is based on the research results from the authors and their research group. The authors express their heartfelt thanks to all the collaborators. The authors are also very grateful to Wanrong Cao, Rui Du, Ruilian Du, Xuping Wang, Renjun Qi and Xuanru Lu, who have read the manuscript and provided many valuable suggestions. The writing of this monograph was supported by the National Natural Science Foundation of China under Grant No. 11671081 and the Natural Science Foundation of Jiangsu Province under Grant No. BK20191375.

Preface | VII

As the authors’ limited scientific research work experience, they sincerely hope that scholars and colleagues will not hesitate to correct the shortcomings and omissions in the book. Thank you for sending emails to us. Prof. Zhi-zhong Sun School of Mathematics, Southeast University, Nanjing 211189, People’s Republic of China. Email box: [email protected] Prof. Guang-hua Gao College of Science, Nanjing University of Posts and Telecommunications, Nanjing 210023, People’s Republic of China. Email box: [email protected] March. 28, 2020

Contents Preface | V 1 1.1 1.1.1 1.1.2 1.1.3 1.1.4 1.1.5 1.1.6 1.2 1.3 1.3.1 1.3.2 1.4 1.5 1.6 1.6.1 1.6.2 1.6.3 1.6.4 1.6.5 1.7 1.7.1 1.7.2 1.7.3 1.8 1.8.1 1.8.2 1.8.3 1.9 1.10

2 2.1

Fractional derivatives and numerical approximations | 1 Definitions and properties of fractional derivatives | 1 Fractional integral | 1 Grünwald–Letnikov fractional derivative | 1 Riemann–Liouville fractional derivative | 2 Caputo fractional derivative | 3 Riesz fractional derivative | 4 Behaviors of fractional derivatives at the lower limit of integrals | 5 The Fourier transform of fractional derivatives | 5 Fractional ordinary differential equations | 7 Solutions to the FODEs in Riemann-Liouville type | 7 Solutions to the FODEs in Caputo type | 10 G-L approximations of Riemann-Liouville fractional derivatives | 11 Central difference quotient approximations of Riesz fractional derivatives | 25 Interpolation approximations of Caputo fractional derivatives | 31 L1 approximation | 31 L1-2 approximation | 39 L2-1σ approximation | 42 L2-1σ approximation of multi-term fractional derivatives | 49 H2N2 approximation | 57 Fast interpolation approximations of Caputo fractional derivatives | 67 Fast L1 approximation | 67 Fast L2-1σ approximation | 74 Fast H2N2 approximation | 81 Finite difference methods for FODEs | 85 Method based on G-L approximation | 85 Method based on L1 approximation | 93 Method based on L2-1σ approximation | 98 A simple classification of the fractional partial differential equations | 101 Supplementary remarks and discussions | 103 Exercises 1 | 106 Difference methods for the time-fractional subdiffusion equations | 109 The second-order method in space based on G-L approximation for 1D problem | 109

X | Contents 2.1.1 2.1.2 2.1.3 2.1.4 2.2 2.2.1 2.2.2 2.2.3 2.2.4 2.3 2.3.1 2.3.2 2.3.3 2.3.4 2.4 2.4.1 2.4.2 2.4.3 2.4.4 2.5 2.5.1 2.5.2 2.5.3 2.5.4 2.6 2.6.1 2.6.2 2.6.3 2.6.4 2.6.5 2.7 2.7.1 2.7.2 2.7.3 2.7.4

Derivation of the difference scheme | 111 Solvability of the difference scheme | 112 Stability of the difference scheme | 113 Convergence of the difference scheme | 115 The fourth-order method in space based on G-L approximation for 1D problem | 116 Derivation of the difference scheme | 116 Solvability of the difference scheme | 117 Stability of the difference scheme | 117 Convergence of the difference scheme | 118 The second-order method in space based on L1 approximation for 1D problem | 119 Derivation of the difference scheme | 119 Solvability of the difference scheme | 120 Stability of the difference scheme | 121 Convergence of the difference scheme | 122 The fast difference method based on L1 approximation for 1D problem | 123 Derivation of the difference scheme | 123 Solvability of the difference scheme | 125 Stability of the difference scheme | 126 Convergence of the difference scheme | 127 The fourth-order method in space based on L1 approximation for 1D problem | 128 Derivation of the difference scheme | 128 Solvability of the difference scheme | 129 Stability of the difference scheme | 130 Convergence of the difference scheme | 131 The difference method based on L2-1σ approximation for 1D problem | 132 Derivation of the difference scheme | 132 Solvability of the difference scheme | 133 An important lemma | 134 Stability of the difference scheme | 136 Convergence of the difference scheme | 139 The fast difference method based on L2-1σ approximation for 1D problem | 140 Derivation of the difference scheme | 140 Solvability of the difference scheme | 141 Stability of the difference scheme | 142 Convergence of the difference scheme | 144

Contents | XI

2.8 2.8.1 2.8.2 2.8.3 2.8.4 2.9 2.9.1 2.9.2 2.9.3 2.9.4 2.10 2.10.1 2.10.2 2.10.3 2.10.4 2.11 2.11.1 2.11.2 2.11.3 2.11.4 2.12

3 3.1 3.1.1 3.1.2 3.1.3 3.1.4 3.2 3.2.1 3.2.2 3.2.3 3.2.4 3.3 3.3.1 3.3.2

The difference method based on L1 approximation for the MTTFSD equations | 145 Derivation of the difference scheme | 146 Solvability of the difference scheme | 147 Stability of the difference scheme | 147 Convergence of the difference scheme | 149 The difference method based on L2-1σ approximation for the MTTFSD equations | 150 Derivation of the difference scheme | 151 Solvability of the difference scheme | 152 Stability of the difference scheme | 152 Convergence of the difference scheme | 154 The ADI method based on G-L approximation for 2D problem | 155 Derivation of the difference scheme | 156 Solvability of the difference scheme | 158 Stability of the difference scheme | 159 Convergence of the difference scheme | 161 The ADI method based on L1 approximation for 2D problem | 162 Derivation of the difference scheme | 162 Solvability of the difference scheme | 164 Stability of the difference scheme | 165 Convergence of the difference scheme | 166 Supplementary remarks and discussions | 167 Exercises 2 | 170 Difference methods for the time-fractional wave equations | 173 The second-order method in space based on L1 approximation for 1D problem | 173 Derivation of the difference scheme | 173 Solvability of the difference scheme | 174 Stability of the difference scheme | 175 Convergence of the difference scheme | 178 The fast difference method based on L1 approximation for 1D problem | 179 Derivation of the difference scheme | 179 Solvability of the difference scheme | 180 Stability of the difference scheme | 181 Convergence of the difference scheme | 186 The fourth-order method in space based on L1 approximation for 1D problem | 187 Derivation of the difference scheme | 187 Solvability of the difference scheme | 188

XII | Contents 3.3.3 3.3.4 3.4 3.4.1 3.4.2 3.4.3 3.4.4 3.5 3.5.1 3.5.2 3.5.3 3.5.4 3.6 3.6.1 3.6.2 3.6.3 3.6.4 3.7 3.7.1 3.7.2 3.7.3 3.7.4 3.8 3.8.1 3.8.2 3.8.3 3.8.4 3.9 3.9.1 3.9.2 3.9.3 3.9.4 3.10 3.10.1 3.10.2 3.10.3

Stability of the difference scheme | 189 Convergence of the difference scheme | 191 The difference method based on L2-1σ approximation for 1D problem | 192 Derivation of the difference scheme | 193 Solvability of the difference scheme | 196 Stability of the difference scheme | 198 Convergence of the difference scheme | 208 The fast difference method based on L2-1σ approximation for 1D problem | 209 Derivation of the difference scheme | 210 Solvability of the difference scheme | 212 Stability of the difference scheme | 214 Convergence of the difference scheme | 221 The difference method based on L1 approximation for the MTTFW equations | 222 Derivation of the difference scheme | 222 Solvability of the difference scheme | 223 Stability of difference scheme | 224 Convergence of the difference scheme | 226 The difference method based on L2-1σ approximation for the MTTFW equations | 227 Derivation of the difference scheme | 227 Solvability of the difference scheme | 230 Stability of the difference scheme | 232 Convergence of the difference scheme | 239 The difference method for the time-fractional mixed diffusion and wave equations | 240 Derivation of the difference scheme | 240 Solvability of the difference scheme | 242 Stability of the difference scheme | 242 Convergence of the difference scheme | 245 The ADI method based on L1 approximation for 2D problem | 246 Derivation of the difference scheme | 247 Solvability of the difference scheme | 249 Stability of the difference scheme | 250 Convergence of the difference scheme | 252 The compact ADI method based on L1 approximation for 2D problem | 253 Derivation of the difference scheme | 253 Solvability of the difference scheme | 256 Stability of the difference scheme | 257

Contents | XIII

3.10.4 3.11

4 4.1 4.1.1 4.1.2 4.1.3 4.1.4 4.2 4.2.1 4.2.2 4.2.3 4.2.4 4.3 4.3.1 4.3.2 4.3.3 4.3.4 4.4 4.4.1 4.4.2 4.4.3 4.4.4 4.4.5 4.5

5 5.1 5.1.1 5.1.2 5.1.3 5.1.4 5.1.5

Convergence of the difference scheme | 260 Supplementary remarks and discussions | 261 Exercises 3 | 263 Difference methods for the space-fractional partial differential equations | 269 The first-order method based on the shifted G-L formula for 1D problem | 269 Derivation of the difference scheme | 270 Solvability of the difference scheme | 271 Stability of the difference scheme | 272 Convergence of the difference scheme | 273 The second-order method based on the WSGL formula for 1D problem | 273 Derivation of the difference scheme | 274 Solvability of the difference scheme | 275 Stability of the difference scheme | 276 Convergence of the difference scheme | 277 The fourth-order method based on the WSGL formula for 1D problem | 278 Derivation of the difference scheme | 279 Solvability of the difference scheme | 280 Stability of the difference scheme | 281 Convergence of the difference scheme | 283 The fourth-order ADI method based on the WSGL formula for 2D problem | 284 Derivation of the difference scheme | 285 Three lemmas | 287 Solvability of the difference scheme | 289 Stability of the difference scheme | 290 Convergence of the difference scheme | 292 Supplementary remarks and discussions | 293 Exercises 4 | 294 Difference methods for the time-space-fractional differential equations | 297 The second-order method in space for 1D problem | 297 Derivation of the difference scheme | 298 Solvability of the difference scheme | 299 An important lemma | 300 Stability of the difference scheme | 302 Convergence of the difference scheme | 304

XIV | Contents 5.2 5.2.1 5.2.2 5.2.3 5.2.4 5.3 5.3.1 5.3.2 5.3.3 5.3.4 5.4 5.4.1 5.4.2 5.4.3 5.4.4 5.5

6 6.1 6.1.1 6.1.2 6.1.3 6.1.4 6.1.5 6.2 6.2.1 6.2.2 6.2.3 6.2.4 6.3 6.3.1 6.3.2 6.3.3 6.3.4 6.4 6.4.1

The fourth-order method in space for 1D problem | 305 Derivation of the difference scheme | 305 Solvability of the difference scheme | 307 Stability of the difference scheme | 308 Convergence of the difference scheme | 309 The second-order method in space for 2D problem | 310 Derivation of the difference scheme | 311 Solvability of the difference scheme | 312 Stability of the difference scheme | 313 Convergence of the difference scheme | 315 The fourth-order method in space for 2D problem | 316 Derivation of the difference scheme | 317 Solvability of the difference scheme | 318 Stability of the difference scheme | 320 Convergence of the difference scheme | 323 Supplementary remarks and discussions | 323 Exercises 5 | 325 Difference methods for the time distributed-order subdiffusion equations | 327 The second-order method in both space and distributed order for 1D problem | 327 Derivation of the difference scheme | 328 Solvability of the difference scheme | 330 Three lemmas | 330 Stability of the difference scheme | 333 Convergence of the difference scheme | 335 The fourth-order method in both space and distributed order for 1D problem | 336 Derivation of the difference scheme | 336 Solvability of the difference scheme | 338 Stability of the difference scheme | 339 Convergence of the difference scheme | 340 The second-order method in both space and distributed order for 2D problem | 342 Derivation of the difference scheme | 343 Solvability of the difference scheme | 344 Stability of the difference scheme | 345 Convergence of the difference scheme | 346 The fourth-order method in both space and distributed order for 2D problem | 347 Derivation of the difference scheme | 347

Contents | XV

6.4.2 6.4.3 6.4.4 6.5 6.5.1 6.5.2 6.5.3 6.5.4 6.6 6.6.1 6.6.2 6.6.3 6.6.4 6.7

A

Solvability of the difference scheme | 349 Stability of the difference scheme | 349 Convergence of the difference scheme | 351 The second-order ADI method in both space and distributed-order for 2D problem | 352 Derivation of the difference scheme | 352 Solvability of the difference scheme | 354 Stability of the difference scheme | 355 Convergence of the difference scheme | 356 The fourth-order ADI method in both space and distributed-order for 2D problem | 357 Derivation of the difference scheme | 358 Solvability of the difference scheme | 359 Stability of the difference scheme | 360 Convergence of the difference scheme | 361 Supplementary remarks and discussions | 362 Exercises 6 | 365 The Matlab code of sum-of-exponentials approximations for the kernel t −α in the Caputo fractional derivative | 369

Bibliography | 373 Index | 379

1 Fractional derivatives and numerical approximations In this chapter, several common definitions and properties of fractional derivatives will be introduced. The analytical solution as well its behaviors of two kinds of fractional ordinary differential equations (FODEs) will be analyzed. Several numerical approximations to the fractional derivatives and their accuracy will be introduced and applications into solving the FODEs will also be illustrated. These contents will be the important foundation of the following numerical research on fractional partial differential equations (FPDEs).

1.1 Definitions and properties of fractional derivatives 1.1.1 Fractional integral Definition 1.1.1. Suppose α is a positive real number and the function f (t) is defined on the interval [a, b]. The α-th order fractional integral of function f (t) is defined as t

−α a Dt f (t)

1 = ∫(t − τ)α−1 f (τ)dτ, Γ(α) a

where t ∈ [a, b] and Γ(z) is the Gamma function defined by ∞

Γ(z) = ∫ e−t t z−1 dt,

Re(z) > 0.

0

A direct calculation gives −α a Dt (t

− a)p =

Γ(1 + p) (t − a)p+α , Γ(1 + p + α)

p > −1.

1.1.2 Grünwald–Letnikov fractional derivative Definition 1.1.2. Suppose α is a positive real number, n − 1 ⩽ α < n with n a positive integer and the function f (t) is defined on the interval [a, b]. The α-th order Grünwald– Letnikov (G-L) fractional derivative of function f (t) is defined as α −α a Dt f (t) = lim h h→0

https://doi.org/10.1515/9783110616064-001

α ∑ (−1)j ( )f (t − jh), j j=0

[(t−a)/h]

2 | 1 Fractional derivatives and numerical approximations where t ∈ [a, b], [z] is the maximum integer no more than z and (αj ) is the binomial coefficient defined by α α(α − 1) ⋅ ⋅ ⋅ (α − j + 1) ( )= . j j! Suppose the functions f (k) (t), k = 0, 1, 2, . . . , n are all continuous on [a, b] and n is the minimum integer satisfying α < n. It can be proved that n−1

α a Dt f (t) = ∑

j=0

t

f (n) (τ)dτ 1 f (j) (a)(t − a)j−α . + ∫ Γ(1 + j − α) Γ(n − α) (t − τ)α−n+1 a

1.1.3 Riemann–Liouville fractional derivative Definition 1.1.3. Suppose α is a positive real number, n − 1 ⩽ α < n with n a positive integer and the function f (t) is defined on the interval [a, b]. The α-th order Riemann– Liouville (R-L) fractional derivative of function f (t) is defined as α a Dt f (t)

t

dn 1 f (τ)dτ = n( ), ∫ dt Γ(n − α) (t − τ)α−n+1 a

where t ∈ [a, b]. It is clear that α a Dt f (t)

=

dn [ D−(n−α) f (t)] dt n a t

and a direct calculation gives α a Dt (t

− a)p =

Γ(1 + p) (t − a)p−α , Γ(1 + p − α)

p > −1.

It can be proved that dm ( Dα f (t)) = a Dm+α f (t), t dt m a t

α > 0, m > 0, m ∈ ℤ+ .

Note that there is an equivalence relation between the R-L fractional derivative and the G-L fractional derivative: For a positive real number α, suppose n − 1 ⩽ α < n. If the function f (t), defined on [a, b], has continuous derivatives up to the (n − 1)-th order and f (n) (t) is integrable on [a, b], then the α-th order R-L fractional derivative of function f (t) is equivalent to the α-th order G-L fractional derivative.

1.1 Definitions and properties of fractional derivatives | 3

1.1.4 Caputo fractional derivative Definition 1.1.4. Suppose α is a positive real number, n − 1 < α ⩽ n with n a positive integer and the function f (t) is defined on the interval [a, b]. The α-th order Caputo fractional derivative of function f (t) is defined as C α a Dt f (t)

t

1 f (n) (τ)dτ = , ∫ Γ(n − α) (t − τ)α−n+1 a

where t ∈ [a, b]. It is easy to know that C α a Dt f (t)

= a D−(n−α) [f (n) (t)] t

and a direct calculation gives C α a Dt (t

− a)p =

Γ(1 + p) (t − a)p−α , Γ(1 + p − α)

p > n − 1 ⩾ 0.

Suppose the function f (t), defined on [a, b], has continuous derivatives up to the (n + 1)-th order, then lim

α→n−0

C α a Dt f (t)

= lim [ α→n−0

t

f (n) (a)(t − a)n−α 1 + ∫(t − τ)n−α f (n+1) (τ)dτ] Γ(n − α + 1) Γ(n − α + 1) a

t

= f (n) (a) + ∫ f (n+1) (τ)dτ = f (n) (t). a

Note that there is also an equivalence relation between the Caputo fractional derivative and the R-L fractional derivative: For a positive real number α, suppose 0 ⩽ n − 1 < α < n. If the function f (t), defined on [a, b], has continuous derivatives up to the (n − 1)-th order and f (n) (t) is integral on [a, b], then α a Dt f (t)

n−1

= Ca Dαt f (t) + ∑

j=0

f (j) (a)(t − a)j−α , Γ(1 + j − α)

a ⩽ t ⩽ b.

Particularly, when α ∈ (0, 1), it reads α a Dt f (t)

= Ca Dαt f (t) +

f (a)(t − a)−α . Γ(1 − α)

It can be found that if the function f (t) satisfies f (j) (a) = 0,

j = 0, 1, . . . , n − 1,

4 | 1 Fractional derivatives and numerical approximations the α-th order Caputo fractional derivative and the α-th order R-L fractional derivative are equivalent. It can be known from the definitions of fractional derivatives above that the value of fractional derivatives at one point t is related with all of the function values on the left-hand side of this point. Hence, they are also called the left G-L fractional derivative, the left R-L fractional derivative and the left Caputo fractional derivative, respectively. Similarly, the right G-L fractional derivative, the right R-L fractional derivative and the right Caputo fractional derivative can be defined respectively as α −α t Db f (t) = lim h h→0

α n t Db f (t) = (−1)

C α t Db f (t)

α ∑ (−1)j ( )f (t + jh), j j=0

[(b−t)/h]

b

1 f (τ)dτ dn ( ), ∫ dt n Γ(n − α) (τ − t)α−n+1

= (−1)n

t

b

f (n) (τ)dτ 1 . ∫ Γ(n − α) (τ − t)α−n+1 t

When a = −∞, the left G-L fractional derivative is defined as α −∞ Dt f (t)

∞ α = lim h−α ∑ (−1)j ( )f (t − jh); h→0 j j=0

When b = ∞, the right G-L fractional derivative is defined as α t D∞ f (t)

∞ α = lim h−α ∑ (−1)j ( )f (t + jh). h→0 j j=0

1.1.5 Riesz fractional derivative Definition 1.1.5. Suppose α is a positive real number, n − 1 ⩽ α < n with n a positive integer, α ≠ 2k + 1, k = 0, 1, . . ., and the function f (x) is defined on [a, b]. Then the α-th order Riesz fractional derivative of function f (x) is defined as 1 𝜕α f (x) =− [a Dαx f (x) + x Dαb f (x)], 𝜕|x|α 2 cos( απ ) 2 where x ∈ [a, b]. It can be known from the definition above that the Riesz fractional derivative can be regarded as a weighted sum of the left R-L fractional derivative and the right one. And the value of a Riesz fractional derivative at one point x is related with values of function f (x) on the both hand sides of x.

1.2 The Fourier transform of fractional derivatives | 5

1.1.6 Behaviors of fractional derivatives at the lower limit of integrals Here, we will describe some behaviors of the fractional derivative a Dαt f (t) at the lower limit of the integral. Suppose that for some small and positive constant ϵ, the function f (t) is analytical at least on the interval [a, a+ϵ]. By Taylor expansion, the function f (t) can be expressed as f (k) (a) (t − a)k , k! k=0 ∞

f (t) = ∑

t ∈ [a, a + ϵ].

(1.1)

Computing the R-L fractional derivative of each term in (1.1), one obtains α a Dt f (t)

f (k) (a) (t − a)k−α . Γ(k − α + 1) k=0 ∞

= ∑

(1.2)

For the function f (t) in the form of (1.1) with f (a) ≠ 0, it follows from (1.2) that α a Dt f (t)

∼

f (a) (t − a)−α , Γ(1 − α)

t → a + 0.

If the function f (t) has the integrable singularity at t = a, that is, the function f (t) can be expressed as f (t) = (t − a)q g(t),

where g(a) ≠ 0, q > −1,

and the function g(t) can be expressed into the form of Taylor series, then one has ∞ (k) g (k) (a) g (a) (t − a)k = ∑ (t − a)q+k . k! k! k=0 k=0 ∞

f (t) = (t − a)q ∑

Computing the R-L fractional derivative term by term, it follows g (k) (a) Γ(q + k + 1) (t − a)q+k−α , k! Γ(q + k − α + 1) k=0 ∞

α a Dt f (t) = ∑

which implies that α a Dt f (t)

∼

g(a)Γ(q + 1) (t − a)q−α , Γ(q − α + 1)

t → a + 0.

1.2 The Fourier transform of fractional derivatives Definition 1.2.1. Suppose the function g(t), defined on ℛ = (−∞, ∞), is piecewise continuous and absolutely integrable. The Fourier transform of function g(t), denoted by G(ω), is defined as ∞

G(ω) ≡ ℱ [g(t); ω] = ∫ g(t)eiωt dt. −∞

6 | 1 Fractional derivatives and numerical approximations Moreover, the function g(t) can be recovered by the Fourier inverse transform as g(t) = ℱ −1 [G(ω); t] ≡

∞

1 ∫ G(ω)e−iωt dω. 2π −∞

If the function g(t) is differentiable on ℛ, g 󸀠 (t) is piecewise continuous; the functions g(t) and g 󸀠 (t) are both absolutely integrable; when |t| → ∞, g(t) → 0, then ℱ [g (t); ω] = (−iω)ℱ [g(t); ω]. 󸀠

If the function g(t) has the continuous derivatives up to order n − 1 and the piecewise continuous derivative of order n on ℛ; the functions g(t), g 󸀠 (t), . . . , g (n) (t) are all absolutely integrable; the functions g(t), g 󸀠 (t), . . ., g (n−1) (t) → 0 when |t| → ∞, then ℱ [g

(n)

(t); ω] = (−iω)n ℱ [g(t); ω].

Suppose α > 0. Let t α−1

hα+ (t) = { Γ(α) 0,

t > 0,

,

t ⩽ 0,

then it follows α

ℱ [h+ (t); ω] = (−iω) . −α

Now we consider the case of fractional derivatives at lower limit a = −∞ when the function f (t) and its derivatives up to some necessary order approach to zero as t → −∞. Suppose n − 1 ⩽ α < n. It can be easily obtained from the integration by parts that the G-L fractional derivative, the R-L fractional derivative and the Caputo fractional derivative are in the same form of α −∞ Dt f (t) } } } α −∞ Dt f (t) } } } C α −∞ Dt f (t)}

t

f (n) (τ)dτ 1 = ∫ Γ(n − α) (t − τ)α+1−n −∞

t

=

dn f (τ)dτ 1 ) ≡ Dα f (t). ( ∫ dt n Γ(n − α) (t − τ)α+1−n −∞

1. 2.

Assume that: the functions f (t), f 󸀠 (t), . . . , f (n−1) (t) are existent and continuous, the function f (n) (t) is piecewise continuous and they are all absolutely integrable; the functions f (t), f 󸀠 (t), . . . , f (n−1) (t) → 0 when |t| → ∞,

then it holds α

α

ℱ [D f (t); ω] = (−iω) ℱ [f (t); ω].

1.3 Fractional ordinary differential equations | 7

1.3 Fractional ordinary differential equations In this section, two types of fractional ordinary differential equations (FODEs) will be analytically solved. 1.3.1 Solutions to the FODEs in Riemann-Liouville type Problem 1.3.1. Suppose 0 < α < 1. Consider the single-term FODEs in R-L type as follows: {

α 0 Dt y(t)

= f (t),

t > 0,

y(0) = 0,

(1.3) (1.4)

where the function f (t) is supposed to be in the Taylor series form of f (n) (0) n t , n! n=0 ∞

f (t) = ∑

with the convergence radius R, R > 0. Here, for the problem (1.3)–(1.4), we aim to seek the solution in the form of ∞

∞

n=0

n=0

y(t) = t α ∑ yn t n = ∑ yn t n+α . By means of α ν 0 Dt t

=

Γ(1 + ν) ν−α t , Γ(1 + ν − α)

the substitution of (1.5) into (1.3) gives ∞

∑ yn

n=0

∞ (n) Γ(1 + n + α) n f (0) n t = f (t) = ∑ t . Γ(1 + n) n! n=0

Comparing the coefficients of two series above arrives at yn =

f (n) (0) , Γ(1 + n + α)

n = 0, 1, 2, . . . .

Hence, it follows f (n) (0) tn, Γ(1 + n + α) n=0 ∞

y(t) = t α ∑ which can be rearranged as

f (n) (0) t n+α Γ(1 + n + α) n=0 ∞

y(t) = ∑

(1.5)

8 | 1 Fractional derivatives and numerical approximations f (n) (0) Γ(n + 1) n+α t n! Γ(1 + n + α) n=0 ∞

= ∑

f (n) (0) −α n D t n! 0 t n=0 ∞

= ∑

f (n) (0) n t ) n! n=0 ∞

= 0 D−α t (∑ = 0 D−α t f (t).

If the function f (t) has the following form: f (t) = t q g(t),

g (n) (0) n t , n! n=0 ∞

g(t) = ∑

g(0) ≠ 0, q > −1,

the method above is still valid. To seek the solution in the form of ∞

y(t) = t q+α ∑ yn t n n=0

subjected to the zero initial condition (1.4), similarly, it can be obtained that yn =

Γ(1 + n + q) g (n) (0) ⋅ , Γ(1 + n + α + q) n!

n = 0, 1, 2, . . . .

Suppose q + α ⩾ 0 and denote m = [q + α], then y(t) has the continuous derivative of order m. Problem 1.3.2. Suppose 0 < α < 1. Consider the problem {

α 0 Dt y(t)

= f (t),

y(0) = A,

t > 0,

(1.6)

A ≠ 0.

(1.7)

It can be found that a necessary condition to ensure a solution in the form of ∞

y(t) = ∑ yn t n n=0

(1.8)

is that f (t) ∼

At −α , Γ(1 − α)

t → 0.

Suppose f (t) =

∞ (n) At −α g (0) n + t 1−α ∑ t , Γ(1 − α) n! n=0

where the coefficient {g (n) (0)} is known.

(1.9)

1.3 Fractional ordinary differential equations | 9

Substituting (1.8) and (1.9) into (1.6) produces ∞

∑ yn

n=0

∞ (n) Γ(1 + n) n−α At −α g (0) n t = f (t) = + t 1−α ∑ t . Γ(1 + n − α) Γ(1 − α) n! n=0

Comparing the coefficients gives y0 = A,

yn =

Γ(1 + n − α) g (n−1) (0) ⋅ , Γ(1 + n) (n − 1)!

n = 1, 2, . . . .

Hence, Γ(1 + n − α) g (n−1) (0) n ⋅ t . (n − 1)! n=1 Γ(1 + n) ∞

y(t) = A + ∑ Let

y(t) = z(t) + A, then the problem (1.6)–(1.7) is converted to the following initial value problem with respect to z(t): {

α 0 Dt z(t)

= f ̂(t),

t > 0,

z(0) = 0,

(1.10) (1.11)

where f ̂(t) = f (t) −

∞ (n) g (0) n At −α = t 1−α ∑ t . Γ(1 − α) n! n=0

Solving (1.10)–(1.11) arrives at Γ(1 + n − α) g (n−1) (0) n ⋅ t . (n − 1)! n=1 Γ(1 + n) ∞

z(t) = ∑ Definition 1.3.1.

[54]

Suppose μ ∈ ℛ. Define

Cμ = {f | f (t) is a real-valued function and f (t) = t p g(t) with g(t) properly smooth, t > 0, g(0) ≠ 0, p ⩾ μ}. Denote N0 = N ∪ {0} and suppose m ∈ N0 . We call f ∈ Cμm if f (m) ∈ Cμ . For a reasonable large μ, when f ∈ Cμ , the existence of the solution to problem (1.3)–(1.4) is ensured and the bigger μ is, the smoother the solution is.

10 | 1 Fractional derivatives and numerical approximations 1.3.2 Solutions to the FODEs in Caputo type Problem 1.3.3. Consider the single-term FODEs in Caputo type as follows: {

C α 0 Dt y(t)

= f (t),

t > 0,

(1.12)

y(0) = A,

(1.13)

where 0 < α < 1 and the function f (t) can be expanded into the following series: g (n) (0) n t , n! n=0 ∞

f (t) = t q ∑

g (0) (0) = g(0) ≠ 0,

(1.14)

with the convergence radius R, R > 0 and q + α > 0. Suppose that the problem (1.12)–(1.13) has the solution in the form of ∞

y(t) = A + t p ∑ yn t n , n=0

y0 ≠ 0, p > 0.

(1.15)

Substituting (1.15) and (1.14) into (1.12) leads to ∞

∑ yn

n=0

Γ(1 + n + p) n+p−α ∞ g (n) (0) n+q t = ∑ t . Γ(1 + n + p − α) n! n=0

Comparing the coefficients gives p = q + α,

yn =

g (n) (0) Γ(1 + n + q) ⋅ , Γ(1 + n + q + α) n!

n = 0, 1, 2, . . . .

Hence, Γ(1 + n + q) g (n) (0) n ⋅ t . Γ(1 + n + q + α) n! n=0 ∞

y(t) = A + t q+α ∑

(1.16)

If we let y(t) = z(t) + A, then the function z(t) satisfies from (1.12)–(1.13) that {

C α 0 Dt z(t)

= f (t),

z(0) = 0.

t > 0,

(1.17) (1.18)

Comparing (1.12) and (1.17), one can get the same differential equation for y(t) and z(t). We can conclude from (1.16) that: 1. The initial value A = 0 or A ≠ 0 does not affect the smoothness of solution y(t).

1.4 G-L approximations of Riemann-Liouville fractional derivatives | 11

2. 3. 4. 5. 6. 7. 8.

The solution y(t) is smooth if q + α is a positive integer. The solution y(t) is continuous at t = 0 + 0 if q + α > 0. The solution y ∈ C 0 [0, R] if q + α ∈ (0, 1). The solution y ∈ C 1 [0, R] if q + α ∈ [1, 2). The solution y ∈ C 2 [0, R] if q + α ∈ [2, 3). The solution y ∈ C 3 [0, R] if q + α ∈ [3, 4). The solution y ∈ C 4 [0, R] if q + α ∈ [4, 5).

For a reasonable large μ, when f ∈ Cμ , the existence of the solution to problem (1.12)–(1.13) is ensured and the bigger μ is, the smoother the solution y(t) is.

1.4 G-L approximations of Riemann-Liouville fractional derivatives In this section, the G-L approximation of −∞ Dαt f (t) will be considered, where 0 ⩽ n−1 ⩽ α < n. The shifted Grünwald–Letnikov (G-L) formula is defined as ∞

Aαh,p f (t) = h−α ∑ gk(α) f (t − (k − p)h), k=0

(1.19)

where p is a constant, usually called the displacement, α gk(α) = (−1)k ( ), k

α α(α − 1) ⋅ ⋅ ⋅ (α − k + 1) . ( )= k! k

When p = 0, (1.19) is called the standard G-L formula. In fact, the term {gk(α) } here is the coefficient of power series of function (1 − z)α , namely, ∞ ∞ α (1 − z)α = ∑ (−1)k ( )z k = ∑ gk(α) z k , k k=0 k=0

−1 < z ⩽ 1.

The following recursive relation is true: g0(α) = 1,

gk(α) = (1 −

α + 1 (α) )gk−1 , k

k = 1, 2, . . . .

Several lemmas below will display some properties of the coefficient {gk(α) }. Lemma 1.4.1. The coefficient {gk(α) } in (1.19) satisfies: (I) When α = 0, g0(α) = 1,

g1(α) = g2(α) = ⋅ ⋅ ⋅ = 0;

(1.20)

12 | 1 Fractional derivatives and numerical approximations (II) When 0 < α < 1, g0(α) = 1,

g1(α) = −α,

∞

∑ gk(α) = 0,

k=0

g2(α) < g3(α) < ⋅ ⋅ ⋅ < 0,

m

∑ gk(α) > 0,

k=0

m ⩾ 1;

(III) When α = 1, g0(α) = 1,

g1(α) = −1,

g2(α) = g3(α) = ⋅ ⋅ ⋅ = 0;

g0(α) = 1,

g1(α) = −α,

g2(α) > g3(α) > ⋅ ⋅ ⋅ > 0,

(IV) When 1 < α < 2,

∞

∑ gk(α) = 0,

k=0

m

∑ gk(α) < 0,

k=0

m ⩾ 1;

(V) When α = 2, g0(α) = 1,

g1(α) = −2,

g2(α) = 1,

g3(α) = g4(α) = ⋅ ⋅ ⋅ = 0.

Lemma 1.4.2 (Inequalities on exponential functions). (I) 1 − x < e−x , (II) Lemma 1.4.3.

[13]

0 < x ⩽ 1;

−x−x2

1−x >e

,

0 0, we have n+1

k+1

n

2

k 1 1 k+1 1 ), ∑ ⩾ ∑ ∫ dx = ∫ dx = ln( n x x 2 n=2 n=2 k

thus, α+1

α2 󵄨󵄨 (α) 󵄨󵄨 −(α+1) ln( k+1 ) 2 , = 󵄨󵄨gk 󵄨󵄨 < αe (k + 1)α+1

k ⩾ 2.

In addition, it is obvious that 󵄨󵄨 (α) 󵄨󵄨 󵄨󵄨g1 󵄨󵄨 = α =

α2α+1 . (1 + 1)α+1

Therefore, α+1

α2 󵄨󵄨 (α) 󵄨󵄨 , 󵄨󵄨gk 󵄨󵄨 ⩽ (k + 1)α+1

k ⩾ 1.

(1.21)

(II) With the help of (1.20) and Lemma 1.4.2 (II), we have α + 1 󵄨󵄨 (α) 󵄨󵄨 󵄨󵄨 (α) 󵄨󵄨 )󵄨󵄨gk−1 󵄨󵄨 󵄨󵄨gk 󵄨󵄨 = (1 − k > e− > e−

α+1 −( α+1 )2 k k

󵄨󵄨 (α) 󵄨󵄨 󵄨󵄨gk−1 󵄨󵄨

α+1 −( α+1 )2 k k

α+1 α+1 2 󵄨 (α) 󵄨󵄨 e− k−1 −( k−1 ) 󵄨󵄨󵄨gk−2 󵄨󵄨

> ⋅ ⋅ ⋅ > e− =

α+1 −( α+1 )2 k k

α+1

α+1 2

e− k−1 −( k−1 ) ⋅ ⋅ ⋅ e−

α(1 − α) −(α+1) ∑kn=3 n1 −(α+1)2 ∑kn=3 e e 2

α+1 −( α+1 )2 3 3

1 n2

,

󵄨󵄨 (α) 󵄨󵄨 󵄨󵄨g2 󵄨󵄨

k ⩾ 3.

In view of k

∞ ∞ 1 1 π2 5 1 1 ⩽ = − (1 + ) = − , ∑ ∑ 2 2 2 4 6 4 n=1 n n=3 n n=3 n

∑

it follows −(α+1)2 ∑kn=3

e

1 n2

2 π2

5

⩾ e−(α+1) ( 6 − 4 ) >

1 . 5

Consequently, 󵄨󵄨 (α) 󵄨󵄨 α(1 − α) −(α+1) ∑kn=3 n1 e , 󵄨󵄨gk 󵄨󵄨 > 10

k ⩾ 3.

14 | 1 Fractional derivatives and numerical approximations Noticing that the function 1/x is monotone decreasing when x > 0, it holds k

n

k

n−1

2

k 1 1 1 k ⩽ ∑ ∫ dx = ∫ dx = ln . n x x 2 n=3 n=3

∑

Therefore, α 󵄨󵄨 (α) 󵄨󵄨 α(1 − α) −(α+1) ln k2 α(1 − α)2 = e , 󵄨󵄨gk 󵄨󵄨 > 10 5k α+1

k ⩾ 3.

In addition, α 󵄨󵄨 (α) 󵄨󵄨 α(1 − α) α(1 − α)2 , > 󵄨󵄨g2 󵄨󵄨 = α+1 2 5⋅2 (1 − α)2α 󵄨󵄨 (α) 󵄨󵄨 . 󵄨󵄨g1 󵄨󵄨 = α > α ⋅ 5 ⋅ 1α+1

Hence, α 󵄨󵄨 (α) 󵄨󵄨 α(1 − α)2 , 󵄨󵄨gk 󵄨󵄨 > α+1 5k

k ⩾ 1.

(1.22)

Combining (1.21) with (1.22) arrives at α(1 − α)2α 󵄨󵄨 (α) 󵄨󵄨 α2α+1 < 󵄨󵄨gk 󵄨󵄨 ⩽ , α+1 5k (k + 1)α+1

k ⩾ 1.

(1.23)

α(1 − α)2α ∞ 󵄨󵄨 (α) 󵄨󵄨 ∞ α2α+1 < ∑󵄨󵄨gk 󵄨󵄨 ⩽ ∑ . 5k α+1 (k + 1)α+1 k=l k=l

(1.24)

(III) Summing up for k in (1.23) gives ∞

∑ k=l

It follows from the function 1/xα+1 being monotone decreasing when x > 0 that ∞

∑ k=l

∞

∞ 1 1 1 < ∫ α+1 dx < ∑ α+1 , α+1 (k + 1) x k k=l l

namely, ∞

∑ k=l

∞ 1 1 1 < α < ∑ α+1 . α+1 αl (k + 1) k k=l

From (1.24) and the inequality above, it is easy to get α

α

∞ 1−α 2 2 󵄨 󵄨 ( ) < ∑󵄨󵄨󵄨gk(α) 󵄨󵄨󵄨 < 2( ) . 5 l l k=l

The proof ends.

1.4 G-L approximations of Riemann-Liouville fractional derivatives | 15

Lemma 1.4.4.

When 1 < α < 2, it holds α+1

α(α − 1)(2 − α)(3 − α) 4 ( ) 180 k

< gk(α) ⩽

α+1

3 α(α − 1) ( ) 2 k+1 α

(α − 1)(2 − α)(3 − α) 4 ( ) , 45 k

∞

∑ gn(α) >

n=k

,

k ⩾ 2;

k ⩾ 2.

Proof. (I) By (1.20) and Lemma 1.4.2 (I), we have gk(α) = (1 − < e− < e−

α+1 k α+1 k

α + 1 (α) )gk−1 k (α) gk−1

α+1

(α) e− k−1 gk−2

< ⋅ ⋅ ⋅ < e− =

α+1 k

α+1

α+1 3

e− k−1 ⋅ ⋅ ⋅ e−

α(α − 1) −(α+1) ∑kn=3 n1 e , 2

g2(α) k ⩾ 3.

Noticing n+1

k+1

n

3

k 1 1 k+1 1 ), ∑ ⩾ ∑ ∫ dx = ∫ dx = ln( n x x 3 n=3 n=3 k

it follows gk(α)
e− > e−

α + 1 (α) )gk−1 k

α+1 −( α+1 )2 k k α+1 −( α+1 )2 k k

> ⋅ ⋅ ⋅ > e− =

(α) gk−1

α+1

α+1 2

(α) e− k−1 −( k−1 ) gk−2

α+1 −( α+1 )2 k k

α+1

α+1 2

e− k−1 −( k−1 ) ⋅ ⋅ ⋅ e−

α+1 −( α+1 )2 5 5

g4(α)

α(α − 1)(2 − α)(3 − α) −(α+1) ∑kn=5 n1 −(α+1)2 ∑kn=5 e e 24

1 n2

,

k ⩾ 5.

(1.25)

16 | 1 Fractional derivatives and numerical approximations Noticing, when k ⩾ 5, it holds k

e−(α+1) ∑n=5

1 n

α+1

4 ⩾( ) k

(1.26)

due to k

n

k

n−1

4

k 1 1 1 k ∑ ⩽ ∑ ∫ dx = ∫ dx = ln . n x x 4 n=5 n=5

Similarly, when k ⩾ 5, it holds that −(α+1)2 ∑kn=5

e

1 n2

π2

205

⩾ e−9( 6 − 144 ) >

2 15

(1.27)

because of k

∞ ∞ 1 1 π 2 205 1 1 1 1 + + ) = − . ⩽ = − (1 + ∑ ∑ 2 2 2 4 9 16 6 144 n=5 n n=5 n n=1 n

∑

The substitution of (1.26) and (1.27) into (1.25) gives gk(α) >

α+1

α(α − 1)(2 − α)(3 − α) 4 ( ) 180 k

k ⩾ 5.

,

It is not hard to verify that the inequality above is also true for k = 4, 3, 2. (III) When k ⩾ 2, we have ∞

∞

n=k

n=k

∑ gn(α) > ∑

α+1

α(α − 1)(2 − α)(3 − α) 4 ( ) 180 n

α+1

=

α(α − 1)(2 − α)(3 − α) α ∞ 1 4 ∑( ) 45 n n=k

⩾

α(α − 1)(2 − α)(3 − α) α ∞ 1 4 ∑ ∫ α+1 dx 45 x n=k

n+1 n

∞

=

α(α − 1)(2 − α)(3 − α) α 1 4 ∫ α+1 dx 45 x k α

(α − 1)(2 − α)(3 − α) 4 = ( ) . 45 k The proof ends. Define C

n+α

1

∞

n+α 󵄨

(ℛ) = {f | f ∈ L (ℛ), ∫ (1 + |ω|) −∞

󵄨󵄨F(ω)󵄨󵄨󵄨dω < ∞}, 󵄨 󵄨

1.4 G-L approximations of Riemann-Liouville fractional derivatives | 17

where F(ω) = ∫−∞ f (t)eiωt dt is the Fourier transform of function f (t). Tuan and Gorenflo derived the asymptotic expansion of the standard Grünwald formula in [89]. As well, Tadjeran et al. established the asymptotic expansion of the shifted Grünwald formula in [86] (For the result of n = 1, see also [58]). ∞

Theorem 1.4.1. Suppose f ∈ C n+α (ℛ), then it holds n−1

l n Aαh,p f (t) = −∞ Dαt f (t) + ∑ cl(α,p) −∞ Dα+l t f (t)h + O(h ) l=1

uniformly for t ∈ ℛ, where {cl(α,p) } is the coefficient of power series of function Wα,p (z) =

( 1−ez )α epz , namely, −z

∞

Wα,p (z) = ∑ cl(α,p) z l = c0(α,p) + c1α,p z + c2(α,p) z 2 + c3(α,p) z 3 + O(|z|4 ), l=0

in particular, p2 αp α(3α + 1) α , c2(α,p) = − + , 2 2 2 24 p3 αp2 α(3α + 1)p α2 (α + 1) = − + − . 6 4 24 48

c0(α,p) = 1, c3(α,p)

c1(α,p) = p −

Proof. Let ∞

ℱ [f (t); ω] = ∫ f (t)e

iωt

dt ≡ F(ω).

−∞

According to ℱ [f (t − h); ω] = e

iωh

F(ω),

we have α

ℱ [Ah,p f (t); ω] ∞

= h−α ∑ gk(α) ℱ [f (t − (k − p)h); ω] k=0 ∞

= h−α ∑ gk(α) eiω(k−p)h F(ω) k=0

∞

= h−α [ ∑ gk(α) eikωh ]e−ipωh F(ω) k=0

α

= h (1 − eiωh ) e−ipωh F(ω) −α

18 | 1 Fractional derivatives and numerical approximations α

1 − eiωh ) e−ipωh F(ω) −iωh = (−iω)α Wα,p (−iωh)F(ω). = (−iω)α (

(1.28)

Taking into account that the function Wα,p (z) is analytical in a neighborhood of the origin, there exists a positive constant R such that ∞

Wα,p (z) = ∑ cl(α,p) z l , l=0

for all |z| ⩽ R.

Next, we aim to show that there is a constant c1 such that 󵄨󵄨 󵄨󵄨 n−1 󵄨󵄨 󵄨 󵄨󵄨Wα,p (−ix) − ∑ c(α,p) (−ix)l 󵄨󵄨󵄨 ⩽ c1 |x|n 󵄨󵄨 l 󵄨󵄨󵄨 󵄨󵄨 l=0 󵄨

(1.29)

uniformly holds for x ∈ ℛ. When |x| ⩽ R, 󵄨󵄨 󵄨󵄨 󵄨󵄨 ∞ 󵄨󵄨 n−1 󵄨󵄨 󵄨 󵄨 󵄨 󵄨󵄨Wα,p (−ix) − ∑ c(α,p) (−ix)l 󵄨󵄨󵄨 = 󵄨󵄨󵄨∑ c(α,p) (−ix)l 󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 l l 󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨l=n 󵄨󵄨 l=0 󵄨 ∞

󵄨 󵄨 ⩽ |x|n ∑󵄨󵄨󵄨cl(α,p) 󵄨󵄨󵄨 |x|l−n ⩽ c2 |x|n , l=n

(α,p) l |R < ∞. where c2 = R−n ∑∞ l=n |cl When |x| > R, on the one hand, ix α 󵄨󵄨 2α 󵄨 󵄨 󵄨󵄨 󵄨 󵄨󵄨 1 − e ) e−ipx 󵄨󵄨󵄨 ⩽ α ⩽ c3 |x|n , 󵄨󵄨Wα,p (−ix)󵄨󵄨󵄨 = 󵄨󵄨󵄨( 󵄨󵄨 R 󵄨󵄨 −ix

where c3 =

2α Rα+n

< ∞; On the other hand, 󵄨󵄨n−1 󵄨󵄨 n−1 󵄨󵄨 󵄨 󵄨󵄨 ∑ c(α,p) (−ix)l 󵄨󵄨󵄨 ⩽ |x|n ∑ 󵄨󵄨󵄨c(α,p) 󵄨󵄨󵄨 ⋅ |x|l−n ⩽ c4 |x|n , 󵄨󵄨 󵄨 󵄨 l 󵄨 󵄨󵄨 󵄨󵄨 l=0 l l=0 󵄨

(α,p) l−n where c4 = ∑n−1 |R < ∞. l=0 |cl Let c1 = max{c2 , c3 + c4 }. It is apparent that (1.29) is uniformly true for x ∈ ℛ. It follows from (1.28) that α

n−1

ℱ [Ah,p f (t); ω] = ∑ cl l=0

n−1

(α,p)

(−iω)α+l hl F(ω) + Φ(ω, h)

l = ∑ cl(α,p) ℱ [−∞ Dα+l t f (t); ω]h + Φ(ω, h), l=0

(1.30)

1.4 G-L approximations of Riemann-Liouville fractional derivatives | 19

where n−1

Φ(ω, h) = (−iω)α [Wα,p (−iωh) − ∑ cl(α,p) (−iωh)l ]F(ω). l=0

Taking the inverse Fourier transform on both hand sides of (1.30) produces n−1

l Aαh,p f (t) − ∑ cl(α,p) −∞ Dα+l t f (t)h = l=0

∞

1 ∫ Φ(ω, h)e−iωt dω. 2π −∞

Combining with (1.29) and noticing f ∈ C

n+α

(ℛ), we can get

󵄨󵄨 󵄨󵄨 n−1 󵄨󵄨 α 󵄨 󵄨󵄨A f (t) − ∑ c(α,p) −∞ Dα+l f (t)hl 󵄨󵄨󵄨 t 󵄨󵄨 h,p 󵄨󵄨 l 󵄨󵄨 󵄨󵄨 l=0 ∞

⩽

1 󵄨 󵄨 ∫ 󵄨󵄨󵄨Φ(ω, h)󵄨󵄨󵄨dω 2π −∞ ∞

⩽

1 󵄨 󵄨 ∫ c1 |ω|α |ωh|n 󵄨󵄨󵄨F(ω)󵄨󵄨󵄨dω 2π −∞

∞

⩽

c1 n n+α 󵄨 󵄨 h ∫ (1 + |ω|) 󵄨󵄨󵄨F(ω)󵄨󵄨󵄨dω 2π −∞

⩽ chn . The proof ends.

In what follows, some common approximations will be stated in detail. The first-order approximation [88] It is easy to see from Theorem 1.4.1 that Theorem 1.4.2. Suppose f ∈ C 1+α (ℛ), then it holds Aαh,p f (t) = −∞ Dαt f (t) + O(h) uniformly for t ∈ ℛ. The second-order approximation [88]

Theorem 1.4.3. Suppose f ∈ C 2+α (ℛ) and p ≠ q, then it holds λ1 Aαh,p f (t) + λ2 Aαh,q f (t) = −∞ Dαt f (t) + O(h2 ) uniformly for t ∈ ℛ, where λ1 =

α − 2q , 2(p − q)

λ2 =

2p − α . 2(p − q)

(1.31)

20 | 1 Fractional derivatives and numerical approximations Proof. It follows from Theorem 1.4.1 that λ1 Aαh,p f (t) + λ2 Aαh,q f (t)

2 = (λ1 + λ2 )−∞ Dαt f (t) + (λ1 c1(α,p) + λ2 c1(α,q) )−∞ Dα+1 t f (t)h + O(h )

uniformly holds for t ∈ ℛ. Let λ1 + λ2 = 1,

{

λ1 c1(α,p) + λ2 c1(α,q) = 0,

which implies λ1 =

α − 2q , 2(p − q)

λ2 =

2p − α 2(p − q)

in view of c1(α,p) = p −

α , 2

c1(α,q) = q −

α . 2

The proof ends. We call the left-hand side of (1.31) the weighted and shifted G-L (WSGL) formula. Some common applications of Theorem 1.4.3 can be concluded as follows. Corollary 1.4.1. When α ∈ (0, 1), taking (p, q) = (0, −1), then λ1 = 1 + α2 , λ2 = − α2 .This result is often used to solve the time-fractional PDEs [96] . At this point, λ1 Aαh,0 f (t) + λ2 Aαh,−1 f (t) = (1 +

∞ α α −α ∞ (α) )h ∑ gk f (t − kh) + (− )h−α ∑ gk(α) f (t − (k + 1)h) 2 2 k=0 k=0

∞

= h−α ∑ wk(α) f (t − kh) k=0

= −∞ Dαt f (t) + O(h2 )

(1.32)

uniformly holds for t ∈ ℛ, where α α { { w0(α) = (1 + )g0(α) = 1 + , { { 2 2 { { { { { (α) α (α) α (α) wk = (1 + )gk − gk−1 { { 2 2 { { { { { α α+1 α (α) { { = [(1 + )(1 − ) − ]gk−1 , 2 k 2 {

(1.33)

k ⩾ 1.

(1.34)

1.4 G-L approximations of Riemann-Liouville fractional derivatives | 21

It is easy to verify that α 3α + α2 α(α2 + 3α − 2) { { w0(α) = 1 + > 0, w1(α) = − < 0, w2(α) = , { { 2 2 4 { { { { { w(α) ⩽ w(α) ⩽ w(α) ⩽ ⋅ ⋅ ⋅ ⩽ 0, w(α) + w(α) > 0, 1 4 3 2 0 { { { { ∞ m { { (α) (α) { { { ∑ wk = 0, ∑ wk > 0, m ⩾ 2. k=0 k=0 { Corollary 1.4.2. When α ∈ (1, 2), taking (p, q) = (1, 0), then λ1 = α2 , λ2 = 1 − α2 . This result

is often used to solve the space-fractional PDEs [88] . At this point, λ1 Aαh,1 f (t) + λ2 Aαh,0 f (t)

=

∞ α α −α ∞ (α) h ∑ gk f (t − (k − 1)h) + (1 − )h−α ∑ gk(α) f (t − kh) 2 2 k=0 k=0 ∞

̃k(α) f (t − (k − 1)h) = h−α ∑ w =

k=0 α −∞ Dt f (t)

+ O(h2 )

uniformly holds for t ∈ ℛ, where ̃0(α) = w

α (α) g , 2 0

̃k(α) = w

α (α) α (α) g + (1 − )gk−1 , 2 k 2

k ⩾ 1.

(1.35)

It is easy to show that 2 − α − α2 α(α2 + α − 4) α { ̃1(α) = ̃2(α) = ̃0(α) = > 0, w < 0, w , w { { { 2 2 4 { { { ̃2(α) > 0, ̃0(α) ⩾ w ̃3(α) ⩾ w ̃4(α) ⩾ ⋅ ⋅ ⋅ ⩾ 0, w ̃0(α) + w 1⩾w { { { ∞ m { { { { ∑w ̃k(α) = 0, ∑ w ̃k(α) < 0, m ⩾ 2. { k=0 k=0

(1.36)

The third-order approximation [117] Theorem 1.4.4. Suppose f ∈ C 3+α (ℛ) and the constants p, q and r are distinct, then it holds λ1 Aαh,p f (t) + λ2 Aαh,q f (t) + λ3 Aαh,r f (t) = −∞ Dαt f (t) + O(h3 ) uniformly for t ∈ ℛ, where λ1 =

12qr − (6q + 6r + 1)α + 3α2 , 12(qr − pq − pr + p2 )

(1.37)

22 | 1 Fractional derivatives and numerical approximations

λ2 = λ3 =

12pr − (6p + 6r + 1)α + 3α2 , 12(pr − pq − qr + q2 )

12pq − (6p + 6q + 1)α + 3α2 . 12(pq − pr − qr + r 2 )

Proof. It follows from Theorem 1.4.1 that λ1 Aαh,p f (t) + λ2 Aαh,q f (t) + λ3 Aαh,r f (t)

= (λ1 + λ2 + λ3 )−∞ Dαt f (t) + (λ1 c1(α,p) + λ2 c1(α,q) + λ3 c1(α,r) )−∞ Dα+1 t f (t)h 2 3 + (λ1 c2(α,p) + λ2 c2(α,q) + λ3 c2(α,r) )−∞ Dα+2 t f (t)h + O(h )

uniformly holds for t ∈ ℛ. Let λ1 + λ2 + λ3 = 1, { { { (α,p) + λ2 c1(α,q) + λ3 c1(α,r) = 0, {λ1 c1 { { (α,p) (α,q) (α,r) {λ1 c2 + λ2 c2 + λ3 c2 = 0, which implies λ1 = λ2 = λ3 =

12qr − (6q + 6r + 1)α + 3α2 , 12(qr − pq − pr + p2 )

12pr − (6p + 6r + 1)α + 3α2 , 12(pr − pq − qr + q2 )

12pq − (6p + 6q + 1)α + 3α2 . 12(pq − pr − qr + r 2 )

The proof ends. We also call the left-hand side of (1.37) the weighted and shifted G-L (WSGL) formula. When α ∈ (0, 1), taking (p, q, r) = (0, −1, −2), then λ1 =

24 + 17α + 3α2 , 24

λ2 = −

11α + 3α2 , 12

λ3 =

5α + 3α2 . 24

This result is often used to solve the time-fractional differential equations[40] . When α ∈ (1, 2), taking (p, q, r) = (1, 0, −1), then λ1 =

5α + 3α2 , 24

λ2 =

12 + α − 3α2 , 12

λ3 =

−7α + 3α2 . 24

This result is often used to solve the space-fractional differential equations[88] .

1.4 G-L approximations of Riemann-Liouville fractional derivatives | 23

The fourth-order approximation [35]

Theorem 1.4.5. Suppose f ∈ C 4+α (ℛ) and denote δtα f (t) = λ1 Aαh,1 f (t) + λ0 Aαh,0 f (t) + λ−1 Aαh,−1 f (t),

(1.38)

2 4 δtα f (t) = −∞ Dαt f (t) + c2α −∞ Dα+2 t f (t)h + O(h )

(1.39)

then it holds

uniformly for t ∈ ℛ, where λ1 =

α2 + 3α + 2 , 12

λ0 =

4 − α2 , 6

λ−1 =

α2 − 3α + 2 12

(1.40)

and c2α = λ1 c2(α,1) + λ0 c2(α,0) + λ−1 c2(α,−1) =

−α2 + α + 4 . 24

(1.41)

Moreover, δtα f (t) = c2α −∞ Dαt f (t − h) + (1 − 2c2α )−∞ Dαt f (t) + c2α −∞ Dαt f (t + h) + O(h4 )

(1.42)

uniformly holds for t ∈ ℛ. Proof. It follows from Theorem 1.4.1 that Aαh,p f (t) = −∞ Dαt f (t) + c1(α,p) −∞ Dα+1 t f (t)h

(α,p) 2 α+3 3 4 + c2(α,p) −∞ Dα+2 −∞ Dt f (t)h + O(h ). t f (t)h + c3

Taking p = 1, 0 and −1 in the equality above, respectively, and then taking the weighted sum of the results arrive at δtα f (t) = (λ1 + λ0 + λ−1 )−∞ Dαt f (t)

α α+2 2 + (c1(α,1) λ1 + c1(α,0) λ0 + c1(α,−1) λ−1 )−∞ Dα+1 t f (t)h + c2 −∞ Dt f (t)h

3 4 + (c3(α,1) λ1 + c3(α,0) λ0 + c3(α,−1) λ−1 )−∞ Dα+3 t f (t)h + O(h ).

Let λ + λ + λ−1 = 1, { { 1(α,1) 0 c λ + c1(α,0) λ0 + c1(α,−1) λ−1 = 0, { { 1(α,1) 1 (α,0) (α,−1) λ−1 = 0, { c3 λ1 + c3 λ0 + c3 which can be solved to give (1.40). Furthermore, (1.41) and (1.39) are followed.

24 | 1 Fractional derivatives and numerical approximations On the other hand, noticing that the R-L operator

α+2 −∞ Dt

=

d2 ( Dα ), dt 2 −∞ t

it follows

δtα f (t) = (ℐ + c2α h2

α+2 −∞ Dt

can be written as

d2 ) Dα f (t) + O(h4 ), dt 2 −∞ t

(1.43)

where ℐ is the identity operator. In view of d2 1 v(t) = 2 [v(t + h) − 2v(t) + v(t − h)] + O(h2 ), dt 2 h we have d2 ) Dα f (t) dt 2 −∞ t 1 = −∞ Dαt f (t) + c2α h2 [ 2 (−∞ Dαt f (t − h) − 2−∞ Dαt f (t) h (ℐ + c2α h2

+ −∞ Dαt f (t + h)) + O(h2 )] = c2α −∞ Dαt f (t − h) + (1 − 2c2α ) −∞ Dαt f (t) + c2α −∞ Dαt f (t + h) + O(h4 ).

(1.44)

Then (1.42) is followed from (1.43) and (1.44). The proof ends. Substituting (1.40) into (1.38) leads to 4 − α2 α α2 − 3α + 2 α α2 + 3α + 2 α Ah,1 f (t) + Ah,0 f (t) + Ah,−1 f (t) 12 6 12 1 ∞ = α ∑ ŵ k(α) f (t − (k − 1)h), h k=0

δtα f (t) =

where α2 + 3α + 2 (α) α2 + 3α + 2 (α) 4 − α2 (α) (α) { g0 , ŵ 1(α) = g1 + g0 , { { ŵ 0 = 12 12 6 { { { (α) α2 + 3α + 2 (α) 4 − α2 (α) α2 − 3α + 2 (α) ̂ gk + gk−1 + gk−2 , k ⩾ 2. { wk = 12 6 12

(1.45)

It is easy to verify when α ∈ [1, 2] that ŵ 0(α) > 0, ŵ 1(α) ⩽ 0, ŵ k(α) ⩾ 0, k ⩾ 3, { { m { { ∞ (α) ̂ w = 0, ŵ k(α) ⩽ 0, m ⩾ 2, ∑ ∑ k { { { k=0 { k=0 (α) (α) { ŵ 0 + ŵ 2 ⩾ 0.

(1.46)

1.5 Central difference quotient approximations of Riesz fractional derivatives | 25

It is worth to mention that all results in this section are the G-L approximation of left R-L fractional derivative and the corresponding numerical differentiation formulae are called the left G-L formulae; similarly, the results for the right R-L fractional derivative can be obtained and the corresponding numerical differentiation formulae are called the right G-L formulae.

1.5 Central difference quotient approximations of Riesz fractional derivatives The Riesz fractional derivative 𝜕α f (x) = −Ψα (−∞ Dαx f (x) + x Dα+∞ f (x)), 𝜕|x|α

Ψα =

1 2 cos( απ ) 2

is a weighted sum of the left R-L fractional derivative −∞ Dαx f (x) and the right R-L fractional derivative x Dα+∞ f (x) . Denote xi = ih, i = 0, ±1, ±2, . . .. When α ∈ [1, 2], by the left G-L formula and the right one, it follows from Theorem 1.4.1 that ∞ Ψα ∞ (α) 𝜕α f (x) 󵄨󵄨󵄨󵄨 [ g f (x ) + gk(α) f (xi+k−1 )] + O(h). = − ∑ ∑ 󵄨 i−k+1 󵄨 𝜕|x|α 󵄨󵄨x=xi hα k=0 k k=0

(1.47)

Applying the left weighted and shifted G-L formula and the right weighted and shifted G-L formula, we have ∞ Ψ ∞ (α) 𝜕α f (x) 󵄨󵄨󵄨󵄨 ̃k f (xi−k+1 ) + ∑ w ̃k(α) f (xi+k−1 )] + O(h2 ), = − αα [ ∑ w 󵄨󵄨 α 𝜕|x| 󵄨󵄨x=xi h k=0 k=0

̃k(α) } is defined by (1.35). where {w

Ortigueira[62] introduced a fractional central difference operator as follows: ∞

Δαh f (x) ≡ ∑ ĝk(α) f (x − kh), k=−∞

where ĝk(α) =

(−1)k Γ(α + 1) . Γ(α/2 − k + 1)Γ(α/2 + k + 1)

As that pointed out in [62], the coefficient {ĝk(α) } satisfies α ∞ 󵄨󵄨 x 󵄨󵄨󵄨󵄨 󵄨󵄨 (α) ikx 2 sin( ) 󵄨󵄨 󵄨󵄨 = ∑ ĝk e , 2 󵄨󵄨 󵄨󵄨 k=−∞

x ∈ ℛ.

(1.48)

26 | 1 Fractional derivatives and numerical approximations When α > −1, the following recursive relations are true for {ĝk(α) }: ĝ0(α) =

Γ(α + 1) , Γ2 (α/2 + 1)

ĝk(α) = (1 −

α+1 )ĝ (α) , α/2 + k k−1

(α) g−k = gk(α) ,

k ⩾ 1;

k ⩾ 1.

(1.49) (1.50)

The following lemma has been proved by Çelik and Duman in [4]. Lemma 1.5.1. [4] Suppose f ∈ 𝒞 5 (ℛ) and its all derivatives of order up to 5 belong to L1 (ℛ). Then it holds −

Δαh f (x) 𝜕α f (x) = + O(h2 ). hα 𝜕|x|α

It is easy to see from the lemma above that lim [−

h→0

Δαh f (x) 𝜕α f (x) ] = . hα 𝜕|x|α

Moreover, we have the following conclusion. Theorem 1.5.1. Suppose f ∈ C 2n+α (ℛ), then it holds 2l

−

Δαh f (x) 𝜕α f (x) n−1 𝜕2l+α f (x) h = + ∑ (−1)l ĉlα ( ) + O(h2n ) α α 2l+α h 𝜕|x| 2 𝜕|x| l=1

󵄨 󵄨α uniformly for x ∈ ℛ, where {ĉlα } is the coefficient of power series of function 󵄨󵄨󵄨 sinz z 󵄨󵄨󵄨 , that is, 󵄨󵄨 sin z 󵄨󵄨α 󵄨󵄨 󵄨󵄨 α α 2 α 4 α 6 󵄨󵄨 󵄨󵄨 = ĉ0 + ĉ1 z + ĉ2 z + ĉ3 z + ⋅ ⋅ ⋅ , 󵄨󵄨 z 󵄨󵄨 in particular, ĉ0α = 1,

α ĉ1α = − , 6

ĉ2α =

(5α − 2)α , 360

ĉ3α = −

Proof. Making the Fourier transform of function [− ℱ {−

=− =−

(35α2 − 42α + 16)α . 45360

Δαh f (x) ] hα

Δαh f (x) ; ω} hα

gives

1 ∞ (α) ∑ ĝ ℱ {f (x − kh); ω} hα k=−∞ k 1 ∞ (α) iωkh F(ω) ∑ ĝ e hα k=−∞ k

1.5 Central difference quotient approximations of Riesz fractional derivatives | 27 α 1 󵄨󵄨󵄨󵄨 ωh 󵄨󵄨󵄨 󵄨󵄨2 sin( )󵄨󵄨󵄨 F(ω) α h 󵄨󵄨 2 󵄨󵄨 󵄨󵄨 sin ωh 󵄨󵄨α 󵄨 󵄨 = −|ω|α ⋅ 󵄨󵄨󵄨 ωh 2 󵄨󵄨󵄨 F(ω), 󵄨󵄨 󵄨󵄨 2

=−

(1.51)

where F(ω) is the Fourier transform of f (t). By the Fourier transform, we have ℱ{

𝜕α f (x) ; ω} = −Ψα [(iω)α + (−iω)α ]F(ω) = −|ω|α F(ω), 𝜕|x|α

ℱ{

𝜕α f (x) 𝜕2l+α f (x) 2l ; ω} = (iω) ℱ { ; ω} = (−1)l+1 ω2l |ω|α F(ω). 𝜕|x|α 𝜕|x|2l+α

Thus, 2l

ℱ{

𝜕α f (x) n−1 𝜕2l+α f (x) h + ∑ (−1)l ĉlα ( ) ; ω} α 𝜕|x| 𝜕|x|2l+α 2 l=1 2l

= ℱ{

n−1 𝜕2l+α f (x) 𝜕α f (x) l α h ̂ ; ω} + (−1) c ( ) ℱ { ; ω} ∑ l 𝜕|x|α 2 𝜕|x|2l+α l=1

(1.52)

2l

n−1 h = −|ω|α F(ω) − ∑ ĉlα ( ) ω2l |ω|α F(ω) 2 l=1 n−1

= −|ω|α [1 + ∑ ĉlα ( l=1

2l

ωh ) ]F(ω). 2

(1.53)

It is easy to know that for any z ∈ ℛ, there is a constant c such that 󵄨󵄨󵄨 󵄨 󵄨󵄨󵄨󵄨 sin z 󵄨󵄨󵄨α n−1 α 2l 󵄨󵄨󵄨 󵄨󵄨󵄨󵄨 󵄨󵄨 − ∑ ĉ z 󵄨󵄨 ⩽ c z 2n . l 󵄨󵄨󵄨󵄨 z 󵄨󵄨 󵄨󵄨 󵄨󵄨󵄨 󵄨 󵄨󵄨 l=0 Subtracting (1.53) from (1.51) leads to 2l

ℱ {−

Δαh f (x) 𝜕α f (x) n−1 𝜕2l+α f (x) h ( ) ; ω} ; ω} − ℱ { + ∑ (−1)l ĉlα α α h 𝜕|x| 𝜕|x|2l+α 2 l=1

󵄨󵄨 sin ωh 󵄨 = −|ω|α {󵄨󵄨󵄨 ωh 2 󵄨󵄨 2

2l n−1 󵄨󵄨α 󵄨󵄨 α ωh 󵄨󵄨 − [1 + ∑ ĉl ( ) ]}F(ω) ≡ Φ(ω, h). 󵄨󵄨 2 l=1

Taking the inverse Fourier transforms on both hand sides of the equality above gives 󵄨󵄨 α 2l 󵄨󵄨 2l+α 󵄨󵄨 Δh f (x) 󵄨󵄨 𝜕α f (x) n−1 f (x) h l α𝜕 󵄨󵄨− ̂ − [ + (−1) c ( ) ]󵄨󵄨󵄨 ∑ l 󵄨󵄨 α α 2l+α 󵄨󵄨 h 𝜕|x| 2 𝜕|x| 󵄨󵄨 l=1 󵄨 ∞ 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 1 = 󵄨󵄨󵄨 ∫ Φ(ω, h)e−iωx dω󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨 2π 󵄨 −∞ 󵄨

28 | 1 Fractional derivatives and numerical approximations ∞

⩽

1 󵄨 󵄨 ∫ 󵄨󵄨󵄨Φ(ω, h)󵄨󵄨󵄨dω 2π −∞ ∞

⩽

󵄨󵄨󵄨 󵄨󵄨󵄨󵄨 sin ωh 1 ∫ |ω|α ⋅ 󵄨󵄨󵄨󵄨󵄨󵄨 ωh 2 󵄨󵄨󵄨󵄨 2π 󵄨 2

−∞ ∞

⩽

2n

1 ωh 󵄨 󵄨 ∫ |ω|α c( ) ⋅ 󵄨󵄨󵄨F(ω)󵄨󵄨󵄨dω 2π 2 −∞

= where ĉ =

c 22n+1 π

2l 󵄨󵄨 n−1 󵄨󵄨α 󵄨󵄨 󵄨 󵄨󵄨 󵄨 α ωh 󵄨󵄨 − [1 + ∑ ĉl ( ) ]󵄨󵄨󵄨 ⋅ 󵄨󵄨󵄨F(ω)󵄨󵄨󵄨dω 󵄨󵄨 󵄨󵄨 2 l=1 󵄨

2n ∞

c h ( ) 2π 2

󵄨 󵄨 ̂ 2n , ∫ |ω|2n+α 󵄨󵄨󵄨F(ω)󵄨󵄨󵄨dω = ch

−∞

∫−∞ |ω|2n+α |F(ω)|dω, independent of h and x. The proof ends. ∞

When n = 2, the theorem says that 2

−

2+α Δαh f (x) 𝜕α f (x) f (x) h α𝜕 ̂ ( ) + O(h4 ) = − c 1 α α h 𝜕|x| 𝜕|x|2+α 2

=

α 𝜕α f (x) d2 𝜕α f (x) + h2 2 ( ) + O(h4 ) α 𝜕|x| 24 dx 𝜕|x|α

=

α 𝜕α f (x − h) 𝜕α f (x) 𝜕α f (x + h) 𝜕α f (x) + [ −2 + ] + O(h4 ) α α 𝜕|x| 24 𝜕|x| 𝜕|x|α 𝜕|x|α

=

α 𝜕α f (x) α 𝜕α f (x + h) α 𝜕α f (x − h) + (1 − ) + + O(h4 ). α α 24 𝜕|x| 12 𝜕|x| 24 𝜕|x|α

Hence the following theorem is true. Theorem 1.5.2. −

[116]

Suppose f ∈ C 4+α (ℛ), then it holds

Δαh f (x) α 𝜕α f (x − h) α 𝜕α f (x) α 𝜕α f (x + h) = + (1 − ) + + O(h4 ) α α α h 24 𝜕|x| 12 𝜕|x| 24 𝜕|x|α

uniformly for x ∈ ℛ. Lemma 1.5.2. When α ∈ [1, 2], the coefficient {ĝk(α) } satisfies ĝ0(α) = ∞

Γ(α + 1) ⩾ 0, + 1)

Γ2 (α/2

∑ ĝk(α) = 0,

k=−∞

(α) ĝ−k = ĝk(α) ⩽ 0,

i

− ∑ ĝk(α) ⩽ ĝ0(α) , k=−M+i k =0 ̸

Moreover, the next lemma can be proved.

k = 1, 2, . . . ,

1 ⩽ i ⩽ M − 1.

(1.54)

1.5 Central difference quotient approximations of Riesz fractional derivatives | 29

Lemma 1.5.3. When 1 < α ⩽ 2, it holds α Γ(α+1) (I) |ĝk(α) | ⩽ 2+α ( α/2+2 )α+1 , |k| ⩾ 1; Γ2 (α/2+1) α/2+k+1 rα , |k| (k+1)α+1 ∞ c∗(α) (III) ∑ |ĝk(α) | ⩾ , (l + 1)α |k|=l

⩾ 1;

(II) |ĝk(α) | ⩾

l ⩾ 1,

where rα = e−2 c∗(α) =

α+1

α Γ(α + 1) (4 − α)(2 − α)α (3 + ) ⋅ (6 + α)(4 + α)(2 + α) Γ2 (α/2 + 1) 2

,

2 r . α α

(1.55)

Proof. (I) From (1.49) and Lemma 1.4.2(I), we have α + 1 󵄨󵄨 (α) 󵄨󵄨 󵄨󵄨 ̂ (α) 󵄨󵄨 )󵄨ĝ 󵄨 󵄨󵄨gk 󵄨󵄨 = (1 − α/2 + k 󵄨 k−1 󵄨 α+1 󵄨 (α) 󵄨󵄨 ⩽ e− α/2+k 󵄨󵄨󵄨ĝk−1 󵄨󵄨 α+1 󵄨󵄨 ̂ (α) 󵄨󵄨 − ∑km=2 α/2+m ⩽ ⋅⋅⋅ ⩽ e 󵄨󵄨g1 󵄨󵄨,

k ⩾ 2.

Noticing m+1

k+1

m

2

k 1 1 α/2 + k + 1 1 ⩾ ∑ ∫ dx = ∫ dx = ln , α/2 + m α/2 + x α/2 + x α/2 + 2 m=2 m=2 k

∑

it follows 󵄨 (α) 󵄨󵄨 󵄨󵄨 ̂ (α) 󵄨󵄨 −(α+1) ln α/2+k+1 α/2+2 󵄨󵄨ĝ 󵄨󵄨gk 󵄨󵄨 ⩽ e 󵄨 1 󵄨󵄨 =

α+1

α Γ(α + 1) α/2 + 2 ) ( 2 + α Γ2 (α/2 + 1) α/2 + k + 1

,

k ⩾ 2.

Obviously, it is also true for k = 1. Hence combining with (1.50), the conclusion is true for all k (|k| ⩾ 1). (II) From (1.49) and Lemma 1.4.2(II), we have α + 1 󵄨󵄨 (α) 󵄨󵄨 󵄨󵄨 ̂ (α) 󵄨󵄨 )󵄨ĝ 󵄨 󵄨󵄨gk 󵄨󵄨 = (1 − α/2 + k 󵄨 k−1 󵄨 α+1 2 α+1 󵄨 (α) 󵄨󵄨 ⩾ e− α/2+k −( α/2+k ) 󵄨󵄨󵄨ĝk−1 󵄨󵄨

α+1 α+1 2 α+1 α+1 2 󵄨 (α) 󵄨󵄨 ⩾ e− α/2+k −( α/2+k ) e− α/2+k−1 −( α/2+k−1 ) 󵄨󵄨󵄨ĝk−2 󵄨󵄨 k

⩾ ⋅ ⋅ ⋅ ⩾ e−(α+1) ∑m=4

1 α/2+m

e−(α+1)

2

1 ∑km=4 ( α/2+m )2 󵄨󵄨 (α) 󵄨󵄨 󵄨󵄨ĝ3 󵄨󵄨,

Noticing ∞

∑(

m=4

2

∞ 1 1 ) ⩽ ∑ α/2 + m (α/2 + m + 1/2)(α/2 + m − 1/2) m=4

k ⩾ 4.

(1.56)

30 | 1 Fractional derivatives and numerical approximations ∞

= ∑( m=4

1 1 2 − )= , α/2 + m − 1/2 α/2 + m + 1/2 α+7

it follows 1 ∑km=4 ( α/2+m )2

2

e−(α+1)

2

⩾ e−2(α+1) /(α+7) ⩾ e−2 .

The combination with (1.56) gives k 󵄨󵄨 ̂ (α) 󵄨󵄨 −2 󵄨 (α) 󵄨 −(α+1) ∑m=4 󵄨󵄨gk 󵄨󵄨 ⩾ e 󵄨󵄨󵄨ĝ3 󵄨󵄨󵄨e

Noticing that the function k

1 α/2+x

1 α/2+m

,

k ⩾ 4.

is monotone decreasing for x > 0, we have

m

k

m−1

3

k 1 k + α/2 1 1 < ∑ ∫ dx = ∫ dx = ln( ). ∑ α/2 + m α/2 + x α/2 + x 3 + α/2 m=4 m=4

Hence, 󵄨󵄨 ̂ (α) 󵄨󵄨 ) −2 󵄨 (α) 󵄨 −(α+1) ln( k+α/2 3+α/2 󵄨󵄨gk 󵄨󵄨 ⩾ e 󵄨󵄨󵄨ĝ3 󵄨󵄨󵄨e α+1

󵄨 󵄨 3 + α/2 = e−2 󵄨󵄨󵄨ĝ3(α) 󵄨󵄨󵄨( ) k + α/2 rα = (k + α2 )α+1 rα , k ⩾ 4. ⩾ (k + 1)α+1

It can be verified that the inequality above is also fulfilled for k = 3, 2, 1. Combining with (1.50), it follows rα 󵄨󵄨 ̂ (α) 󵄨󵄨 , 󵄨󵄨gk 󵄨󵄨 ⩾ (k + 1)α+1

|k| ⩾ 1.

(III) Summing up for k in the inequality above leads to ∞

∞

k=l

k=l

󵄨 󵄨 ∑󵄨󵄨󵄨ĝk(α) 󵄨󵄨󵄨 ⩾ rα ∑

∞ 1 1 = r , ∑ α α+1 (k + 1)α+1 k k=l+1

l ⩾ 1.

Noticing that the function 1/xα+1 is monotone decreasing for x > 0, we have ∞

∑

1

k α+1 k=l+1

∞

k+1

⩾ ∑ ∫ k=l+1 k

1

∞

xα+1

dx = ∫

l+1

1 1 dx = . α(l + 1)α x α+1

Hence, ∞

󵄨 󵄨 ∑󵄨󵄨󵄨ĝk(α) 󵄨󵄨󵄨 ⩾

k=l

rα , α(l + 1)α

l⩾1

1.6 Interpolation approximations of Caputo fractional derivatives | 31

and c∗(α) 2rα = , α(l + 1)α (l + 1)α

∞

󵄨 󵄨 ∑ 󵄨󵄨󵄨ĝk(α) 󵄨󵄨󵄨 ⩾

|k|=l

l ⩾ 1.

The proof ends.

1.6 Interpolation approximations of Caputo fractional derivatives 1.6.1 L1 approximation For the Caputo fractional derivative of order α (0 < α < 1), t

C α 0 Dt f (t)

=

1 f 󸀠 (s) ds, ∫ Γ(1 − α) (t − s)α 0

one of the popular ways to discretize it is the so-called L1 approximation, which approximates the function f (s) on each small interval by a linear interpolation polynomial. We will state it here in detail. Take a positive integer N. Denote τ = NT , tk = kτ, 0 ⩽ k ⩽ N and a(α) = (l + 1)1−α − l1−α , l

l ⩾ 0.

(1.57)

Considering the Caputo fractional derivative at t = tn gives C α 0 Dt f (t)|t=tn

tn

f 󸀠 (t) 1 dt = ∫ Γ(1 − α) (tn − t)α 0

=

tk

n 1 f 󸀠 (t) dt. ∑ ∫ Γ(1 − α) k=1 (tn − t)α

(1.58)

tk−1

On the small interval [tk−1 , tk ], the linear interpolation polynomial for f (t) is L1,k (t) =

t − tk−1 tk − t f (tk−1 ) + f (tk ) τ τ

and the interpolation remainder is 1 f (t) − L1,k (t) = f 󸀠󸀠 (ξk )(t − tk−1 )(t − tk ), 2

t ∈ [tk−1 , tk ],

(1.59)

where ξk = ξk (t) ∈ (tk−1 , tk ). Approximating the function f (t) in (1.58) using L1,k (t) leads to C α 0 Dt f (t)|t=tn

tk

≈

n L󸀠1,k (t) 1 dt ∑ ∫ Γ(1 − α) k=1 (tn − t)α tk−1

32 | 1 Fractional derivatives and numerical approximations tk

n f (t ) − f (tk−1 ) 1 1 = ⋅ ∫ dt ∑ k Γ(1 − α) k=1 τ (tn − t)α tk−1

n

= =

f (t ) − f (tk−1 ) 1 1 ⋅ [(t − t )1−α − (tn − tk )1−α ] ∑ k Γ(1 − α) k=1 τ 1 − α n k−1 n τ−α ∑ [f (tk ) − f (tk−1 )] ⋅ [(n − k + 1)1−α − (n − k)1−α ] Γ(2 − α) k=1

=

n τ−α [f (tk ) − f (tk−1 )] ∑ a(α) Γ(2 − α) k=1 n−k

=

n−1 τ−α (α) (α) (α) [a(α) 0 f (tn ) − ∑ (an−k−1 − an−k )f (tk ) − an−1 f (t0 )]. Γ(2 − α) k=1

Now we have derived the numerical differentiation formula to calculate C0 Dαt f (t)|t=tn as follows: Dαt f (tn ) ≡

n−1 τ−α [a(α) f (t ) − − a(α) )f (tk ) − a(α) ∑ (a(α) n n−1 f (t0 )]. n−k−1 n−k Γ(2 − α) 0 k=1

(1.60)

It is usually called the L1 approximation or L1 formula. The coefficient {a(α) | l ⩾ 0} has the following properties. l Lemma 1.6.1. Suppose α ∈ (0, 1) and {a(α) | l = 0, 1, 2, . . . , } is defined by (1.57), then it l holds: (α) (α) (α) (I) 1 = a0(α) > a(α) 1 > a2 > ⋅ ⋅ ⋅ > al > 0; al → 0, when l → ∞; (II) (1 − α)l−α < a(α) < (1 − α)(l − 1)−α , l ⩾ 1. l−1 In [31, 47, 49, 50, 84], different techniques have been used to prove that L1 formula has the accuracy of order 2 − α. Now we will discuss the approximation error R(f (tn )) = C0 Dαt f (t)|t=tn − Dαt f (tn ) using the technique in [31]. The following theorem is true. Theorem 1.6.1. Suppose f ∈ C 2 [t0 , tn ], then it holds 1 α 1 󵄨󵄨 󵄨 󵄨 󵄨 [ + ] max 󵄨󵄨f 󸀠󸀠 (t)󵄨󵄨󵄨 τ2−α . 󵄨󵄨R(f (tn ))󵄨󵄨󵄨 ⩽ 2Γ(1 − α) 4 (1 − α)(2 − α) t0 ⩽t⩽tn 󵄨 Proof. From the definition of R(f (tn )), we have tk

R(f (tn )) =

n f 󸀠 (t) 1 dt ∑ ∫ Γ(1 − α) k=1 (tn − t)α tk−1

1.6 Interpolation approximations of Caputo fractional derivatives | 33 tk

n f (tk ) − f (tk−1 ) 1 1 − ⋅ dt ∑ ∫ Γ(1 − α) k=1 τ (tn − t)α tk−1

tk

=

n 1 1 󸀠 dt. ∑ ∫ [f (t) − L1,k (t)] Γ(1 − α) k=1 (tn − t)α tk−1

Noticing (1.59), the application of integration by parts arrives at tk

n 1 1 R(f (tn )) = − ) ∑ ∫ [f (t) − L1,k (t)]d( Γ(1 − α) k=1 (tn − t)α tk−1 tk

n 1 =− ∑ ∫ [f (t) − L1,k (t)]α(tn − t)−α−1 dt Γ(1 − α) k=1 tk−1

tk

=

n 1 1 ∑ ∫ f 󸀠󸀠 (ξk )(t − tk−1 )(tk − t)α(tn − t)−α−1 dt. Γ(1 − α) k=1 2 tk−1

Hence, tk

n 1 󵄨󵄨 󵄨 󵄨 󵄨 max 󵄨󵄨󵄨f 󸀠󸀠 (t)󵄨󵄨󵄨 ∑ ∫ (t − tk−1 )(tk − t)α(tn − t)−α−1 dt. 󵄨󵄨R(f (tn ))󵄨󵄨󵄨 ⩽ 2Γ(1 − α) t0 ⩽t⩽tn k=1

(1.61)

tk−1

Some direct calculations produce tk

n−1

∑ ∫ (t − tk−1 )(tk − t)α(tn − t)−α−1 dt

k=1 t

k−1

2 n−1

tk

τ ⩽ ∑ ∫ α(tn − t)−α−1 dt 4 k=1 tk−1

tn−1

=

τ2 ∫ α(tn − t)−α−1 dt 4 2

=

t0

τ −α −α 1 (τ − tn ) ⩽ τ2−α 4 4

and tn

∫ (t − tn−1 )(tn − t)α(tn − t)−α−1 dt tn−1

(1.62)

34 | 1 Fractional derivatives and numerical approximations tn

= α ∫ (t − tn−1 )(tn − t)−α dt tn−1 τ

= α ∫(τ − ξ )ξ −α dξ 0

=

α τ2−α . (1 − α)(2 − α)

(1.63)

Substituting (1.62) and (1.63) into (1.61), we have 1 1 α 󵄨󵄨 󵄨 󵄨 󵄨 [ + ] max 󵄨󵄨f 󸀠󸀠 (t)󵄨󵄨󵄨τ2−α . 󵄨󵄨R(f (tn ))󵄨󵄨󵄨 ⩽ 2Γ(1 − α) 4 (1 − α)(2 − α) t0 ⩽t⩽tn 󵄨 The proof ends. Now we consider the numerical approximation of the Caputo fractional derivative of order γ (1 < γ < 2), C γ 0 Dt f (t)

t

f 󸀠󸀠 (s) 1 = ds. ∫ Γ(2 − γ) (t − s)γ−1 0

Let g(t) = f 󸀠 (t),

α = γ − 1,

then C γ 0 Dt f (t)

t

g 󸀠 (s) 1 γ−1 ds = C0 Dt g(t) = C0 Dαt g(t), = ∫ Γ(1 − (γ − 1)) (t − s)γ−1 0

that is, the γ-th order derivative of function f (t) is exactly the (γ − 1)-th order derivative of function g(t). By Theorem 1.6.1, we have C α 0 Dt g(t)|t=tn

=

n−1 τ−α (α) (α) [a(α) 0 g(tn ) − ∑ (an−k−1 − an−k )g(tk ) Γ(2 − α) k=1

− a(α) n−1 g(t0 )] + R(g(tn )), where 1 1 α 󵄨󵄨 󵄨 󵄨 󵄨 [ + ] ⋅ max 󵄨󵄨g 󸀠󸀠 (t)󵄨󵄨󵄨 τ2−α 󵄨󵄨R(g(tn ))󵄨󵄨󵄨 ⩽ 2Γ(1 − α) 4 (1 − α)(2 − α) t0 ⩽t⩽tn 󵄨 =

γ−1 1 1 󵄨 󵄨 [ + ] ⋅ max 󵄨󵄨f 󸀠󸀠󸀠 (t)󵄨󵄨󵄨 τ3−γ . 2Γ(2 − γ) 4 (2 − γ)(3 − γ) t0 ⩽t⩽tn 󵄨

1.6 Interpolation approximations of Caputo fractional derivatives | 35

Denote bl

(γ)

= a(α) = (l + 1)1−α − l1−α = (l + 1)2−γ − l2−γ , l

l = 0, 1, 2, . . . .

(1.64)

Then C γ 0 Dt f (t)|t=tn

=

n−1 τ1−γ (γ) (γ) (γ) (γ) [b0 g(tn ) − ∑ (bn−k−1 − bn−k )g(tk ) − bn−1 g(t0 )] Γ(3 − γ) k=1

+ R(g(tn )).

(1.65)

Similarly, we have C γ 0 Dt f (t)|t=tn−1

=

n−2 τ1−γ (γ) (γ) (γ) (γ) [b0 g(tn−1 ) − ∑ (bn−k−2 − bn−k−1 )g(tk ) − bn−2 g(t0 )] Γ(3 − γ) k=1

+ R(g(tn−1 )).

(1.66)

Adding (1.65) and (1.66) and taking an average arrive at 1 C γ γ [ D f (t)|t=tn + C0 Dt f (t)|t=tn−1 ] 2 0 t =

n−1 τ1−γ (γ) g(t ) + g(tk−1 ) (γ) g(tn ) + g(tn−1 ) (γ) [b0 − ∑ (bn−k−1 − bn−k ) k Γ(3 − γ) 2 2 k=1

1 (γ) − bn−1 g(t0 )] + [R(g(tn )) + R(g(tn−1 ))]. 2 Noticing g(tk ) + g(tk−1 ) f 󸀠 (tk ) + f 󸀠 (tk−1 ) = 2 2 2 f (tk ) − f (tk−1 ) τ 󸀠󸀠󸀠 = + f (ηk ), ηk ∈ (tk−1 , tk ) τ 12 and denoting 1

δt f k− 2 =

f (tk ) − f (tk−1 ) , τ

it follows immediately from (1.67) that 1 C γ γ [ D f (t)|t=tn + C0 Dt f (t)|t=tn−1 ] 2 0 t =

n−1 1 1 τ1−γ (γ) (γ) (γ) (γ) [b0 δt f n− 2 − ∑ (bn−k−1 − bn−k )δt f k− 2 − bn−1 f 󸀠 (t0 )] Γ(3 − γ) k=1

+

n−1 2 2 τ1−γ (γ) τ 󸀠󸀠󸀠 (γ) (γ) τ [b0 f (ηn ) − ∑ (bn−k−1 − bn−k ) f 󸀠󸀠󸀠 (ηk )] Γ(3 − γ) 12 12 k=1

(1.67)

36 | 1 Fractional derivatives and numerical approximations 1 + [R(g(tn )) + R(g(tn−1 ))]. 2 Let 1

R̂ n− 2 =

n−1 2 2 τ1−γ (γ) τ 󸀠󸀠󸀠 (γ) (γ) τ [b0 f (ηn ) − ∑ (bn−k−1 − bn−k ) f 󸀠󸀠󸀠 (ηk )] Γ(3 − γ) 12 12 k=1

1 + [R(g(tn )) + R(g(tn−1 ))], 2

then we have γ−1 1 1 1 󵄨 󵄨 󵄨󵄨 ̂ n− 21 󵄨󵄨 + [ + ]} max 󵄨󵄨f 󸀠󸀠󸀠 (t)󵄨󵄨󵄨 τ3−γ . 󵄨󵄨R 󵄨󵄨 ⩽ { 6Γ(3 − γ) 2Γ(2 − γ) 4 (2 − γ)(3 − γ) t0 ⩽t⩽tn 󵄨

(1.68)

Therefore, the following theorem is obtained. Theorem 1.6.2. Suppose f ∈ C 3 [t0 , tn ], then it holds 1 C γ γ [ D f (t)|t=tn + C0 Dt f (t)|t=tn−1 ] 2 0 t =

n−1 1 1 1 τ1−γ (γ) (γ) (γ) (γ) [b0 δt f n− 2 − ∑ (bn−k−1 − bn−k )δt f k− 2 − bn−1 f 󸀠 (t0 )] + R̂ n− 2 , Γ(3 − γ) k=1 1

where R̂ n− 2 satisfies (1.68). Remark 1.6.1. Denote ∇t f k =

f (tk ) − f (tk−1 ) . τ

If we directly use 1

g(tk ) = f 󸀠 (tk ) = ∇t f k + τ ∫ f 󸀠󸀠 (tk − θτ)(1 − θ)dθ 0

in (1.65), we can obtain C γ 0 Dt f (t)|t=tn

=

n−1 τ1−γ (γ) (γ) (γ) (γ) [b0 ∇t f n − ∑ (bn−k−1 − bn−k )∇t f k − bn−1 f 󸀠 (t0 )] Γ(3 − γ) k=1

+ rn + R(g(tn )) =

n τ1−γ (γ) (γ) [ ∑ bn−k (∇t f k − ∇t f k−1 ) + bn−1 (∇t f 1 − f 󸀠 (t0 ))] Γ(3 − γ) k=2

+ rn + R(g(tn )) tk

=

n ∇ f k − ∇t f k−1 1 dt [∑ ∫ t ⋅ Γ(2 − γ) k=2 τ (tn − t)γ−1 tk−1

(1.69)

1.6 Interpolation approximations of Caputo fractional derivatives | 37 t1

+∫ t0

∇t f 1 − f 󸀠 (t0 ) dt ⋅ ] + rn + R(g(tn )), τ (tn − t)γ−1

where 1

τ1−γ (γ) rn = [b τ ∫ f 󸀠󸀠 (tn − θτ)(1 − θ)dθ Γ(3 − γ) 0 0

n−1

1

− ∑ (bn−k−1 − bn−k )τ ∫ f 󸀠󸀠 (tk − θτ)(1 − θ)dθ] (γ)

(γ)

k=1

2−γ

=

1

n

0

τ (γ) [ ∑ b (∫ f 󸀠󸀠 (tk − θτ)(1 − θ)dθ Γ(3 − γ) k=2 n−k 0

1

− ∫ f (tk−1 − θτ)(1 − θ)dθ) + 󸀠󸀠

0

1 (γ) bn−1 ∫ f 󸀠󸀠 (t1 0

− θτ)(1 − θ)dθ].

It is easy to know |rn | ⩽

n τ2−γ τ 1 (γ) (γ) 󵄨 󵄨 󵄨 󵄨 [ ∑ bn−k ⋅ max 󵄨󵄨󵄨f 󸀠󸀠󸀠 (t)󵄨󵄨󵄨 + bn−1 ⋅ max 󵄨󵄨󵄨f 󸀠󸀠 (t)󵄨󵄨󵄨] Γ(3 − γ) k=2 2 t0 ⩽t⩽tn 2 t0 ⩽t⩽t1 tk

n 1 dt τ 󵄨 󵄨 = [∑ ∫ ⋅ max 󵄨󵄨󵄨f 󸀠󸀠󸀠 (t)󵄨󵄨󵄨 γ−1 Γ(2 − γ) k=2 2 t0 ⩽t⩽tn (tn − t) tk−1

t1

+∫ t0

1 dt 󵄨 󵄨 ⋅ max 󵄨󵄨f 󸀠󸀠 (t)󵄨󵄨󵄨]. (tn − t)γ−1 2 t0 ⩽t⩽t1 󵄨

Noticing tk

tn

tk−1

t1

2−γ (tn − t1 )2−γ tn dt dt ⩽ , = = ∑ ∫ ∫ 2−γ 2−γ (tn − t)γ−1 (tn − t)γ−1 k=2 n

if n = 1, t1

∫ t0

dt τ2−γ = (tn − t)γ−1 2 − γ

and if n ⩾ 2, t1

∫ t0

τ τ τ dt ⩽ ⩽ = 2γ−1 γ−1 , (tn − t)γ−1 (tn − t1 )γ−1 (tn /2)γ−1 tn

(1.70)

38 | 1 Fractional derivatives and numerical approximations we have 2−γ

|rn | ⩽ =

t 1 1 τ τ 1 󵄨 󵄨 󵄨 󵄨 ⋅ max 󵄨󵄨f 󸀠󸀠 (t)󵄨󵄨󵄨] [ n ⋅ max 󵄨󵄨f 󸀠󸀠󸀠 (t)󵄨󵄨󵄨 + ⋅ Γ(2 − γ) 2 − γ 2 t0 ⩽t⩽tn 󵄨 2 − γ tnγ−1 2 t0 ⩽t⩽t1 󵄨 τ 1 󵄨 󵄨 󵄨 󵄨 [tn2−γ ⋅ τ max 󵄨󵄨󵄨f 󸀠󸀠󸀠 (t)󵄨󵄨󵄨 + γ−1 max 󵄨󵄨󵄨f 󸀠󸀠 (t)󵄨󵄨󵄨]. t ⩽t⩽t 2Γ(3 − γ) 0 n tn t0 ⩽t⩽t1

Denote γ

𝔻t f (tn ) =

n−1 τ1−γ (γ) (γ) (γ) (γ) [b0 ∇t f n − ∑ (bn−k−1 − bn−k )∇t f k − bn−1 f 󸀠 (t0 )]. Γ(3 − γ) k=1

Noticing 󵄨 󵄨 R(g(tn )) = O(τ3−γ ) max 󵄨󵄨󵄨f 󸀠󸀠󸀠 (t)󵄨󵄨󵄨, t0 ⩽t⩽tn

we obtain C γ 0 Dt f (t)|t=tn

τ γ 󵄨 󵄨 󵄨 󵄨 − 𝔻t f (tn ) = O(τ) max 󵄨󵄨󵄨f 󸀠󸀠󸀠 (t)󵄨󵄨󵄨 + O( γ−1 ) max 󵄨󵄨󵄨f 󸀠󸀠 (t)󵄨󵄨󵄨. t0 ⩽t⩽tn t ⩽t⩽t 0 1 tn

If f 󸀠󸀠 (0) = 0, we have C γ 0 Dt f (t)|t=tn

γ 󵄨 󵄨 − 𝔻t f (tn ) = O(τ) max 󵄨󵄨󵄨f 󸀠󸀠󸀠 (t)󵄨󵄨󵄨. t ⩽t⩽t 0

n

It follows from (1.70) that C γ 0 Dt f (t)|t=tn

γ

− 𝔻t f (tn ) tk

n ∇ f k − ∇t f k−1 dt 1 { ∑ ∫ [f 󸀠󸀠 (t) − t ]⋅ = Γ(2 − γ) k=2 τ (tn − t)γ−1 tk−1

t1

+ ∫[f 󸀠󸀠 (t) − t0

∇t f 1 − f 󸀠 (t0 ) dt }. ]⋅ τ (tn − t)γ−1

Notice the fact that ∇t f k − ∇t f k−1 = O(τ), t ∈ (tk−1 , tk ), 2 ⩽ k ⩽ n, τ 2[∇t f 1 − f 󸀠 (t0 )] f 󸀠󸀠 (t) − = O(τ), t ∈ (t0 , t1 ). τ

f 󸀠󸀠 (t) −

γ

If f 󸀠󸀠 (0) ≠ 0, we can modify 𝔻t f (tn ) as ̃γ f (tn ) = 𝔻 t

n−1 τ1−γ (γ) (γ) (γ) [b0 ∇t f n − ∑ (bn−k−1 − bn−k )∇t f k Γ(3 − γ) k=1

1.6 Interpolation approximations of Caputo fractional derivatives | 39

−bn−1 f 󸀠 (t0 ) + bn−1 (∇t f 1 − f 󸀠 (t0 ))] . (γ)

(γ)

Then we have C γ 0 Dt f (t)|t=tn

γ

̃ f (tn ) −𝔻 t tk

n ∇ f k − ∇t f k−1 dt 1 { ∑ ∫ [f 󸀠󸀠 (t) − t ]⋅ = Γ(2 − γ) k=2 τ (tn − t)γ−1 tk−1

t1

+ ∫[f 󸀠󸀠 (t) − t0

2(∇t f 1 − f 󸀠 (t0 )) dt ]⋅ } τ (tn − t)γ−1

󵄨 󵄨 = O(τ) max 󵄨󵄨󵄨f 󸀠󸀠󸀠 (t)󵄨󵄨󵄨. t0 ⩽t⩽tn

This is why we consider the approximation of 1 C γ γ [ D f (t)|t=tn + C0 Dt f (t)|t=tn−1 ] 2 0 t γ

instead of directly considering the approximation of C0 Dt f (t)|t=tn . 1.6.2 L1-2 approximation Gao et al. gave an L1-2 interpolation approximation formula for the fractional derivative of order α (0 < α < 1) in [31]. Make a linear interpolation polynomial L1,1 (t) for f (t) on [t0 , t1 ] by using two points (t0 , f (t0 )), (t1 , f (t1 )); make a quadratic interpolation polynomial L2,k (t) for f (t) on [tk , tk+1 ] by using three points (tk−1 , f (tk−1 )), (tk , f (tk )) and (tk+1 , f (tk+1 )). Differentiating L1,1 (t) and L2,k (t) with respect to t once produces 1

L󸀠1,1 (t) = δt f 2 ,

L󸀠2,k (t) =

tk+ 1 − t 2

τ

1

δt f k− 2 +

t − tk− 1 τ

2

1

δt f k+ 2 ,

1

where tk− 1 = (tk + tk−1 )/2, δt f k− 2 = (f (tk ) − f (tk−1 ))/τ. Using the above two equalities, 2 we can get C α 0 Dt f (t)|t=tn

t1

tk+1

t0

tk

n−1 1 [∫ f 󸀠 (t)(tn − t)−α dt + ∑ ∫ f 󸀠 (t)(tn − t)−α dt] = Γ(1 − α) k=1 t1

tk+1

t0

tk

n−1 1 ≈ [∫ L󸀠1,1 (t)(tn − t)−α dt + ∑ ∫ L󸀠2,k (t)(tn − t)−α dt] Γ(1 − α) k=1 t1

=

1 1 [∫(δt f 2 )(tn − t)−α dt Γ(1 − α)

t0

40 | 1 Fractional derivatives and numerical approximations n−1

tk+1

+ ∑ ∫(

tk+ 1 − t 2

τ

k=1 t k

t − tk− 1

1

δt f k− 2 +

2

τ

1

δt f k+ 2 )(tn − t)−α dt] t

t

k+1 1 n−1 tk+ 1 − t 1 1 1 2 [∫(δt f 2 )(tn − t)−α dt + ∑ δt f k− 2 ∫ (tn − t)−α dt = Γ(1 − α) τ k=1

tk

t0

n

tk

1

t − tk− 3

+ ∑ δt f k− 2 ∫ k=2

=

2

τ

tk−1

(tn − t)−α dt]

t

t

t0

t1

1 2 t3 − t 1 1 {[∫(tn − t)−α dt + ∫ 2 (tn − t)−α dt]δt f 2 Γ(1 − α) τ

tk+1

n−1

+ ∑[ ∫ k=2

tk+ 1 − t 2

τ

tk

tn

+[∫

t − tn− 3

tn−1

τ

2

tk

(tn − t)−α dt + ∫

t − tk− 3

tk−1

τ

2

1

(tn − t)−α dt]δt f k− 2

1

(tn − t)−α dt]δt f n− 2 }

̂αt f (tn ). ≡𝔻 The integrals in the above equality can be computed by substitution of the integral variable. When n = 1, ̂αt f (tn ) 𝔻

t1

1 1 [∫(t1 − t)−α dt]δt f 2 = Γ(1 − α)

t0

τ [f (t1 ) − f (t0 )] Γ(2 − α) τ−α ≡ c̃(1,α) [f (t1 ) − f (t0 )], Γ(2 − α) 0 =

−α

(1.71)

where c̃0(1,α) = 1. When n ⩾ 2, ̂αt f (tn ) = 𝔻

τ−α n−1 (n,α) [f (tn−k ) − f (tn−k−1 )], ∑ c̃ Γ(2 − α) k=0 k

(1.72)

where c̃0(n,α) =

1 1 + , 2 2−α

(1.73)

1.6 Interpolation approximations of Caputo fractional derivatives | 41

(k + 1)1−α − 2k 1−α + (k − 1)1−α 2 (k + 1)2−α − 2k 2−α + (k − 1)2−α , 1 ⩽ k ⩽ n − 2, + 2−α (n − 2)1−α − (n − 1)1−α + 2n1−α (n − 1)2−α − (n − 2)2−α = − . 2 2−α

c̃k(n,α) =

(n,α) c̃n−1

(1.74) (1.75)

The following results are given in [31]. ̂αt f (tn ) is defined by (1.71) and (1.72). Theorem 1.6.3. Suppose α ∈ (0, 1), f ∈ C 3 [t0 , tn ], 𝔻 Then we have the following error estimates: 󵄨󵄨C α 󵄨 α 󵄨󵄨 󸀠󸀠 󵄨󵄨 2−α 󵄨󵄨 D f (t)|t=t − 𝔻 ̂αt f (t1 )󵄨󵄨󵄨 ⩽ 󵄨󵄨f (t)󵄨󵄨τ , 󵄨󵄨0 t 󵄨󵄨 2Γ(3 − α) tmax 1 0 ⩽t⩽t1 󵄨 α 1 󵄨󵄨󵄨C α 󵄨󵄨 󸀠󸀠 󵄨󵄨 −α−1 3 ̂αt f (tn )󵄨󵄨󵄨 ⩽ τ 󵄨󵄨0 Dt f (t)|t=tn − 𝔻 󵄨󵄨f (t)󵄨󵄨(tn − t1 ) 󵄨󵄨 Γ(1 − α) { 12 tmax 󵄨 ⩽t⩽t 0 1 +[

α 1 1 1 󵄨 󵄨 + ( + )] max 󵄨󵄨f 󸀠󸀠󸀠 (t)󵄨󵄨󵄨τ3−α }, 12 3(1 − α)(2 − α) 2 3 − α t0 ⩽t⩽tn 󵄨

n ⩾ 2.

If f 󸀠󸀠 (0) = 0, it is easy to see that C α 0 Dt f (t)|t=tn

̂αt f (tn ) = O(τ3−α ), −𝔻

1 ⩽ n ⩽ N.

The following lemma provides the properties of the coefficient {c̃k(n,α) | 0 ⩽ k ⩽ n − 1}. Lemma 1.6.2. The coefficient {c̃k(n,α) | 0 ⩽ k ⩽ n − 1} defined by (1.73)–(1.75) satisfies the following properties: (I) When n = 2, c̃0(2,α) =

1 1 + , 2 2−α

1 1 c̃1(2,α) = 21−α − ( + ); 2 2−α

(II) When n ⩾ 3, 1 1 + , 2 2−α (n,α) > c̃2(n,α) > c̃3(n,α) > ⋅ ⋅ ⋅ > c̃n−1 ,

c̃0(n,α) = c̃0(n,α)

c̃1(n,α) = 2−α − 1 +

22−α − 2 , 2−α

󵄨 󵄨 c̃0(n,α) > 󵄨󵄨󵄨c̃1(n,α) 󵄨󵄨󵄨.

Let α∗ be the unique root of equation 6 − α = (4 − α)2α on the interval [0, 1] (α∗ ≈ 0.68029). Then when α ∈ (0, α∗ ), c̃1(n,α) > 0; When α ∈ (α∗ , 1), c̃1(n,α) < 0.

42 | 1 Fractional derivatives and numerical approximations 1.6.3 L2-1σ approximation Just as what was stated in last two subsections, for the α-th order (0 < α < 1) Caputo derivative, the classical L1 formula can attain the (2 − α)-th order convergence uniformly. Similarly, based on the piecewise interpolation approximation, the L1-2 approximation was established in [31]. On this basis, Alikhanov[1] discovered some superconvergent interpolation points and built L2-1σ formula, which can reach the (3 − α)-th order convergence uniformly. Let us get into it in detail. Suppose 0 < α < 1. Denote σ =1− f n = f (tn ),

α , 2

tn+σ = (n + σ)τ, 1

δt f n− 2 =

1 n (f − f n−1 ), τ

1 tn− 1 = (tn + tn−1 ), 2 2 1 1 1 δt2 f n = (δt f n+ 2 − δt f n− 2 ). τ

Reformulate the fractional derivative as C α 0 Dt f (t)|t=tn−1+σ

tk

n−1 f 󸀠 (ξ ) 1 = [∑ ∫ dξ Γ(1 − α) k=1 (tn−1+σ − ξ )α tk−1

tn−1+σ

+ ∫ tn−1

f 󸀠 (ξ ) dξ ]. (tn−1+σ − ξ )α

(1.76)

On the small interval [tk−1 , tk ], the quadratic interpolation polynomial of function f (t) using three points (tk−1 , f (tk−1 )), (tk , f (tk )) and (tk+1 , f (tk+1 )) reads 1

1

j=−1

l=−1,l=j̸

L2,k (t) = ∑ f (tk+j ) ∏

t − tk+l , tk+j − tk+l

k = 1, 2, . . . , n − 1,

the remainder for which is f (t) − L2,k (t) =

f 󸀠󸀠󸀠 (ξk ) (t − tk−1 )(t − tk )(t − tk+1 ), 6

ξk ∈ (tk−1 , tk+1 ).

Differentiating L2,k (t) with respect to t once gives L󸀠2,k (t) =

tk+ 1 − t 2

τ

1

δt f k− 2 +

t − tk− 1 τ

2

1

δt f k+ 2 .

On the interval [tn−1 , tn−1+σ ], the linear interpolation polynomial of function f (t) using two points (tn−1 , f (tn−1 )) and (tn , f (tn )) reads L1,n (t) = f (tn−1 )

t − tn−1 t − tn + f (tn ) . tn−1 − tn tn − tn−1

In addition, we have 1

L󸀠1,n (t) = δt f n− 2 .

1.6 Interpolation approximations of Caputo fractional derivatives | 43

Approximating the function f (t) on the right-hand side of (1.76) using L2,k (t) (1 ⩽ k ⩽ n − 1) and L1,n (t), respectively, leads to C α 0 Dt f (t)|t=tn−1+σ

tk

tn−1+σ

tk−1

tn−1

n−1 L󸀠2,k (t) L󸀠1,n (t) 1 ≈ [∑ ∫ dt + dt] ∫ Γ(1 − α) k=1 (tn−1+σ − t)α (tn−1+σ − t)α tk

=

n−1 t − tk− 1 tk+ 1 − t 1 1 1 2 {∑ ∫ [ 2 δt f k− 2 + δt f k+ 2 ](tn−1+σ − t)−α dt Γ(1 − α) k=1 τ τ tk−1

tn−1+σ

1

+ ∫ (δt f n− 2 )(tn−1+σ − t)−α dt} tn−1 t

1 t3 − t 1 1 = {[∫ 2 (tn−1+σ − t)−α dt]δt f 2 Γ(1 − α) τ

t0

tk

n−1

tk+ 1 − t 2

+ ∑[ ∫

τ

k=2 t k−1 tk−1

+ ∫

t − tk− 3 τ

tk−2 tn−1

+[∫

2

1

(tn−1+σ − t)−α dt]δt f k− 2

t − tn− 3 2

τ

tn−2

(tn−1+σ − t)−α dt

tn−1+σ tn−1

1

=

1 3 τ1−α {[∫( − θ)(n − 1 + σ − θ)−α dθ]δt f 2 Γ(1 − α) 2

1

0

n−1 3 + ∑ [∫( − θ)(n − k + σ − θ)−α dθ 2 k=2 0

1

1 1 + ∫(θ − )(n − k + 1 + σ − θ)−α dθ]δt f k− 2 2

0

σ

1

1 1 + [∫(θ − )(1 + σ − θ)−α dθ + ∫(σ − θ)−α dθ]δt f n− 2 } 2

0

1

1−α

=

0

1 τ 3 {[∫( − θ)(n − 1 + σ − θ)−α dθ]δt f 2 Γ(1 − α) 2

1

0

n−1 3 + ∑ [∫( − θ)(n − k + σ − θ)−α dθ 2 k=2 0

1

(tn−1+σ − t) dt + ∫ (tn−1+σ − t)−α dt]δt f n− 2 } −α

44 | 1 Fractional derivatives and numerical approximations 1 2

θ

+ ∫ θ(α ∫ (n − k + 0 1 2

−θ

1 + σ + ξ) 2

θ

−α−1

1 + [∫ θ(α ∫ ( + σ + ξ ) 2 0

−α−1

1

dξ )dθ]δt f k− 2

dξ )dθ +

−θ

1 σ 1−α ]δt f n− 2 } 1−α

≡ Δαt f (tn−1+σ ). Denote

c0(1,α) = σ 1−α .

(1.77)

If n ⩾ 2, denote 1

θ 2 { −α−1 { { 1 { (n,α) { θ(α ( + σ + ξ ) dξ )dθ + σ 1−α , c = (1 − α) ∫ ∫ { 0 { { 2 { { 0 −θ { { { { 1 { { { 3 { { ck(n,α) = (1 − α)[∫( − θ)(k + σ − θ)−α dθ { { { 2 { { 0 { 1 { { θ 2 −α−1 { { 1 { { { + θ(α (k + + σ + ξ ) dξ )dθ], 1 ⩽ k ⩽ n − 2, ∫ ∫ { { 2 { { { 0 −θ { { { { 1 { { { 3 (n,α) −α { { { { cn−1 = (1 − α) ∫( 2 − θ)(n − 1 + σ − θ) dθ. 0 {

(1.78)

(1.79)

(1.80)

Then we have Δαt f (tn−1+σ )

= = =

τ1−α n (n,α) k− 21 δf ∑c Γ(2 − α) k=1 n−k t

τ1−α n−1 (n,α) n−k− 21 δt f ∑c Γ(2 − α) k=0 k

τ−α n−1 (n,α) [f (tn−k ) − f (tn−k−1 )], ∑c Γ(2 − α) k=0 k

We call (1.81) the L2-1σ formula or L2-1σ approximation.

1 ⩽ n ⩽ N.

(1.81)

1.6 Interpolation approximations of Caputo fractional derivatives | 45

Computing yields, if n ⩾ 2, (1 + σ)2−α − σ 2−α (1 + σ)1−α − σ 1−α (n,α) { { c = − , { 0 { 2−α 2 { { { { 1 { { [(k + 1 + σ)2−α − 2(k + σ)2−α + (k − 1 + σ)2−α ] ck(n,α) = { { { 2−α { { { { 1 { − [(k + 1 + σ)1−α − 2(k + σ)1−α + (k − 1 + σ)1−α ], 2 { { { { 1 ⩽ k ⩽ n − 2, { { { { 1 { (n,α) { { cn−1 = [3(n − 1 + σ)1−α − (n − 2 + σ)1−α ] { { 2 { { { { 1 { [(n − 1 + σ)2−α − (n − 2 + σ)2−α ]. − { 2−α The computational cost is O(N 2 ) when using L2-1σ formula (1.81) to compute ⩽ n ⩽ N).

Δαt f (tn−1+σ ) (1

Theorem 1.6.4.

[1]

Suppose f ∈ C 3 [t0 , tn ], then it holds

󵄨󵄨C α 󵄨󵄨 (4σ − 1)σ −α 󵄨 󵄨 󵄨󵄨 D f (t)|t=t max 󵄨󵄨f 󸀠󸀠󸀠 (t)󵄨󵄨󵄨τ3−α , − Δαt f (tn−1+σ )󵄨󵄨󵄨 ⩽ 󵄨󵄨0 t n−1+σ 󵄨 12Γ(2 − α) t0 ⩽t⩽tn 󵄨 where σ = 1 − α2 , 0 < α < 1. Proof. Denote Rn = C0 Dαt f (t)|t=tn−1+σ − Δαt f (tn−1+σ ), Rn1

tk

n−1 f 󸀠 (t) − L󸀠2,k (t) 1 dt, = ∑ ∫ Γ(1 − α) k=1 (tn−1+σ − t)α tk−1

Rn2 =

tn−1+σ

1

f 󸀠 (t) − δt f n− 2 1 dt, ∫ Γ(1 − α) (tn−1+σ − t)α tn−1

then Rn = Rn1 + Rn2 .

(1.82)

Now the two terms Rn1 and Rn2 will be estimated, respectively. It is easy to see that Rn1 =

t

n−1 f (t) − L (t) 󵄨󵄨 k 1 2,k 󵄨󵄨 ∑[ 󵄨 Γ(1 − α) k=1 (tn−1+σ − t)α 󵄨󵄨󵄨t=tk−1 tk

− ∫ α[f (t) − L2,k (t)](tn−1+σ − t)−α−1 dt] tk−1

46 | 1 Fractional derivatives and numerical approximations tk

f 󸀠󸀠󸀠 (ξk ) −α n−1 = (t − tk−1 )(t − tk ) ∑ ∫ Γ(1 − α) k=1 6 tk−1

⋅ (t − tk+1 )(tn−1+σ − t)−α−1 dt, hence, 󵄨󵄨 n 󵄨󵄨 󵄨󵄨R1 󵄨󵄨 ⩽

α 󵄨 󵄨 max 󵄨󵄨f 󸀠󸀠󸀠 (t)󵄨󵄨󵄨 6Γ(1 − α) t0 ⩽t⩽tn 󵄨 tk

n−1

⋅ ∑ ∫ (t − tk−1 )(tk − t)(tk+1 − t)(tn−1+σ − t)−α−1 dt k=1 t

k−1

tn−1

ατ3 󵄨 󵄨 ⩽ max 󵄨󵄨f 󸀠󸀠󸀠 (t)󵄨󵄨󵄨 ∫ (tn−1+σ − t)−α−1 dt 12Γ(1 − α) t0 ⩽t⩽tn 󵄨 t0

σ −α 󵄨 󵄨 max 󵄨󵄨f 󸀠󸀠󸀠 (t)󵄨󵄨󵄨τ3−α . ⩽ 12Γ(1 − α) t0 ⩽t⩽tn 󵄨

(1.83)

Due to that 1 f 󸀠 (t) = f 󸀠 (tn− 1 ) + (t − tn− 1 )f 󸀠󸀠 (tn− 1 ) + (t − tn− 1 )2 f 󸀠󸀠󸀠 (ηn ), 2 2 2 2 2 t, ηn ∈ (tn−1 , tn−1+σ ), it follows Rn2

1

t

n−1+σ 󸀠 f (tn− 1 ) − δt f n− 2 1 2 dt = ∫ Γ(1 − α) (tn−1+σ − t)α

tn−1

t

n−1+σ (t − tn− 1 )f 󸀠󸀠 (tn− 1 ) 1 2 2 + dt ∫ Γ(1 − α) (tn−1+σ − t)α

tn−1

+

tn−1+σ 1 (t 2

1 ∫ Γ(1 − α)

tn−1

− tn− 1 )2 f 󸀠󸀠󸀠 (ηn ) 2

(tn−1+σ − t)α

dt.

Computing the three integrals in the equality above gives 1

tn−1+σ 󸀠 󵄨󵄨 f (tn− 1 ) − δt f n− 2 󵄨󵄨󵄨 󵄨󵄨 1 󵄨 2 󵄨󵄨 dt 󵄨󵄨󵄨 ∫ 󵄨󵄨 Γ(1 − α) 󵄨󵄨 (tn−1+σ − t)α 󵄨󵄨 󵄨 tn−1 tn−1+σ

⩽

1 1 󵄨󵄨 󸀠 n− 1 󵄨 dt 󵄨󵄨f (tn− 1 ) − δt f 2 󵄨󵄨󵄨 ∫ 2 Γ(1 − α) (tn−1+σ − t)α tn−1

1.6 Interpolation approximations of Caputo fractional derivatives | 47

⩽

1 1 τ2 󵄨 󵄨 ⋅ max 󵄨󵄨f 󸀠󸀠󸀠 (t)󵄨󵄨󵄨 ⋅ σ 1−α τ1−α ; Γ(1 − α) 24 tn−1 ⩽t⩽tn 󵄨 1−α t

n−1+σ (t − tn− 1 )f 󸀠󸀠 (tn− 1 ) 1 2 2 dt ∫ Γ(1 − α) (tn−1+σ − t)α

tn−1

tn−1+σ

t − tn− 1 1 2 f 󸀠󸀠 (tn− 1 ) ∫ dt = 2 Γ(1 − α) (tn−1+σ − t)α tn−1

=

1 α σ 1−α f 󸀠󸀠 (tn− 1 ) ⋅ [σ − (1 − )]τ2−α = 0 2 Γ(1 − α) (1 − α)(2 − α) 2

(1.84)

and tn−1+σ 1 󵄨󵄨 (t − tn− 1 )2 f 󸀠󸀠󸀠 (ηn ) 󵄨󵄨󵄨 󵄨󵄨 2 1 󵄨 2 󵄨󵄨 dt 󵄨󵄨󵄨 ∫ 󵄨󵄨 Γ(1 − α) 󵄨󵄨 (tn−1+σ − t)α 󵄨󵄨 󵄨 tn−1 tn−1+σ

⩽

τ2 1 󵄨 󵄨 max 󵄨󵄨f 󸀠󸀠󸀠 (t)󵄨󵄨󵄨 ⋅ ∫ dt 8Γ(1 − α) tn−1 ⩽t⩽tn 󵄨 (tn−1+σ − t)α tn−1

2

=

τ 1 󵄨 󵄨 max 󵄨󵄨f 󸀠󸀠󸀠 (t)󵄨󵄨󵄨 ⋅ σ 1−α τ1−α . 8Γ(1 − α) tn−1 ⩽t⩽tn 󵄨 1−α

Therefore, 󵄨󵄨 n 󵄨󵄨 󵄨󵄨R2 󵄨󵄨 ⩽

σ 1−α 󵄨 󵄨 max 󵄨󵄨f 󸀠󸀠󸀠 (t)󵄨󵄨󵄨τ3−α . 6Γ(2 − α) tn−1 ⩽t⩽tn 󵄨

(1.85)

The substitution of (1.83) and (1.85) into (1.82) leads to −α 󵄨 󵄨 󵄨󵄨 n 󵄨󵄨 (4σ − 1)σ max 󵄨󵄨f 󸀠󸀠󸀠 (t)󵄨󵄨󵄨τ3−α . 󵄨󵄨R 󵄨󵄨 ⩽ 12Γ(2 − α) t0 ⩽t⩽tn 󵄨

The proof ends. Remark 1.6.2. The parameter σ is chosen as 1 − tn−1+σ

∫ tn−1

t − tn− 1

2

α 2

(tn−1+σ − t)α

such that the integral in (1.84) dt

happens to be 0. If σ ≠ 1 − α2 , then C α 0 Dt f (t)|t=tn−1+σ

− Δαt f (tn−1+σ ) = O(τ2−α ).

48 | 1 Fractional derivatives and numerical approximations Lemma 1.6.3. [1] Suppose α ∈ (0, 1), σ = 1 − α2 , ck(n,α) (0 ⩽ k ⩽ n − 1, n ⩾ 1) is defined in (1.77)–(1.80), then it holds (n,α) (n,α) c0(n,α) > c1(n,α) > c2(n,α) > ⋅ ⋅ ⋅ > cn−2 > cn−1 > (1 − α)n−α ,

(2σ −

1)c0(n,α)

σc1(n,α)

−

> 0.

Proof. It is easy to know from (1.79)–(1.80) that (n,α) (n,α) c1(n,α) > c2(n,α) > ⋅ ⋅ ⋅ > cn−2 > cn−1 > (1 − α)n−α .

If n = 1, c0(n,α) = σ 1−α = (1 −

1−α

α ) 2

>1−

α > 1 − α, 2

hence, (1.86) is true. If n ⩾ 2 and (1.87) is valid, then we have c0(n,α) − c1(n,α) =

1 α (n,α) [(2σ − 1)c0(n,α) − σc1(n,α) ] + c > 0. σ 2σ 0

Consequently, we only need to prove (1.87). (I) If n = 2, we have (2σ − 1)c0(n,α) − σc1(n,α)

(1 + σ)2−α − σ 2−α (1 + σ)1−α − σ 1−α − ] 2−α 2 1 1 ((1 + σ)2−α − σ 2−α )] − σ[ (3(1 + σ)1−α − σ 1−α ) − 2 2−α 1 1 = (3 − 2σ − )(1 + σ)1−α 2 σ 1 = (2σ − 1)(1 − σ)(1 + σ)1−α 2σ >0.

= (2σ − 1)[

(II) If n ⩾ 3, we have (2σ − 1)c0(n,α) − σc1(n,α) = (2σ − 1)[ − σ{

(1 + σ)2−α − σ 2−α (1 + σ)1−α − σ 1−α − ] 2−α 2

1 [(2 + σ)2−α − 2(1 + σ)2−α + σ 2−α ] 2−α

1 − [(2 + σ)1−α − 2(1 + σ)1−α + σ 1−α ]} 2

(1.86) (1.87)

1.6 Interpolation approximations of Caputo fractional derivatives | 49

=

4σ − 1 (1 + σ)1−α − (2 + σ)1−α . 2σ

Noticing 1−α

1 ) 1+σ 1−α ⩽ (1 + σ)1−α (1 + ) 1+σ 3σ , = (1 + σ)1−α 1+σ

(2 + σ)1−α = (1 + σ)1−α (1 +

we obtain (2σ − 1)c0(n,α) − σc1(n,α) 4σ − 1 3σ ⩾ (1 + σ)1−α − (1 + σ)1−α 2σ 1+σ (2σ − 1)(1 − σ) = (1 + σ)1−α > 0. 2σ(1 + σ) This completes the proof. 1.6.4 L2-1σ approximation of multi-term fractional derivatives In [19], Gao et al. considered L2-1σ interpolation approximation of multiterm Caputo fractional derivatives m

α

Dt f (t) = ∑ λr C0 Dt r f (t),

(1.88)

r=0

where λr , r = 0, 1, . . . , m are positive constants, 0 ⩽ αm < αm−1 < ⋅ ⋅ ⋅ < α0 ⩽ 1, and at least one αr ∈ (0, 1), C0 Dαt f (t) is defined by C α 0 Dt f (t)

f (t) − f (0), α = 0, { { { 1 t = { Γ(1−α) ∫0 f 󸀠 (s)(t − s)−α ds, { { 󸀠 {f (t), α = 1.

α ∈ (0, 1),

Denote tn = nτ, a = min {1 − 0⩽r⩽m

n = 0, 1, 2, . . . ,

αr }, 2

b = max {1 − 0⩽r⩽m

αr }. 2

It is easy to know that a=1−

α0 1 ⩾ , 2 2

b=1−

αm ⩽ 1. 2

50 | 1 Fractional derivatives and numerical approximations Define m

λr α σ 1−αr [σ − (1 − r )]τ2−αr , Γ(3 − α ) 2 r r=0

F(σ) = ∑

σ > 0.

Two useful lemmas will be stated below. Lemma 1.6.4. The equation F(σ) = 0 has a unique positive root σ ∗ ∈ [a, b]. λ0 Γ(3−α0 )

Proof. When m = 0, F(σ) =

σ 1−α0 [σ − (1 − α0 . 2

α0 )]τ2−α0 . 2

It is easy to know that the

equation F(σ) = 0 has a unique root σ = 1 − Now, we suppose m ⩾ 1. When 0 ⩽ σ ⩽ a, F(σ) ⩽ 0. When σ ⩾ b, F(σ) ⩾ 0. When σ ∈ [a, b], ∗

m

λr 1 σ −αr [σ − (1 − αr )]τ2−αr > 0. Γ(2 − α ) 2 r r=0

F 󸀠 (σ) = ∑

Thus, the equation F(σ) = 0 has a unique root σ ∗ ∈ [a, b]. The proof ends. Lemma 1.6.5. Suppose m ⩾ 1. The Newton iteration sequence {σk }∞ k=0 , generated by { σk+1 = σk − { σ0 = b {

F(σk ) , F 󸀠 (σk )

k = 0, 1, 2, . . . ,

(1.89)

is monotone decreasing and convergent to σ ∗ . Proof. In view of the proof for Lemma 1.6.4, we have F(a) < 0, F(b) > 0, and when σ ∈ [a, b], F 󸀠 (σ) > 0. In addition, when σ ∈ [a, b], m

λr 1 σ −αr −1 (σ + αr )τ2−αr > 0. Γ(1 − α ) 2 r r=0

F 󸀠󸀠 (σ) = ∑ Noticing

F(σ0 )F 󸀠󸀠 (σ0 ) > 0, the Newton iteration sequence {σk }∞ k=0 generated by (1.89) is monotone decreasing and ∗ [82] convergent to σ . The proof ends. For simplicity, we denote σ = σ ∗ in this subsection, which implies this σ ∈ [ 21 , 1] satisfying F(σ) = 0. In addition, denote tn−1+σ = (n − 1 + σ)τ. When n = 1, C α 0 Dt f (t)|t=tn−1+σ

tσ

=

1 ∫ f 󸀠 (t)(tσ − t)−α dt. Γ(1 − α) t0

1.6 Interpolation approximations of Caputo fractional derivatives | 51

When n ⩾ 2, C α 0 Dt f (t)|t=tn−1+σ

tk

=

n−1 1 [ ∑ ∫ f 󸀠 (t)(tn−1+σ − t)−α dt Γ(1 − α) k=1 tk−1

tn−1+σ

+ ∫ f 󸀠 (t)(tn−1+σ − t)−α dt]. tn−1

The following theorem will give a numerical approximation formula of (1.88) at the point t = tn−1+σ and reveal its numerical accuracy. Theorem 1.6.5. Suppose f ∈ C 3 [t0 , tn ]. Let tk

n−1 λr [ ∑ ∫ L󸀠2,k (t)(tn−1+σ − t)−αr dt 𝒟t f (tn−1+σ ) ≡ ∑ Γ(1 − αr ) k=1 r=0 m

tk−1

tn−1+σ

+ ∫ L󸀠1,n (t)(tn−1+σ − t)−αr dt]. tn−1

Then we have the following error estimate: 󵄨󵄨 󵄨 󵄨󵄨Dt f (tn−1+σ ) − 𝒟t f (tn−1+σ )󵄨󵄨󵄨 m

λr 1 − αr σ −αr 3−αr 󵄨 󵄨 ⋅( + )σ τ ⋅ max 󵄨󵄨󵄨f 󸀠󸀠󸀠 (t)󵄨󵄨󵄨 = O(τ3−α0 ). t ⩽t⩽t Γ(2 − α ) 12 6 0 n r r=0

⩽∑

Proof. It is easy to know that Dt f (tn−1+σ ) − 𝒟t f (tn−1+σ ) tk

n−1 λr [ ∑ ∫ (f 󸀠 (t) − L󸀠2,k (t))(tn−1+σ − t)−αr dt =∑ Γ(1 − αr ) k=1 r=0 m

tk−1

tn−1+σ

+ ∫ (f 󸀠 (t) − L󸀠1,n (t))(tn−1+σ − t)−αr dt] tn−1

tk

n−1 λr =∑ ∑ ∫ (f 󸀠 (t) − L󸀠2,k (t))(tn−1+σ − t)−αr dt Γ(1 − α ) r r=0 k=1 m

tk−1

m

tn−1+σ

λr ∫ (f 󸀠 (t) − L󸀠1,n (t))(tn−1+σ − t)−αr dt. Γ(1 − α ) r r=0

+∑

tn−1

Denote M = max |f 󸀠󸀠󸀠 (t)|. t0 ⩽t⩽tn

(1.90)

52 | 1 Fractional derivatives and numerical approximations Since tk

n−1

Ar ≡ ∑ ∫ (f 󸀠 (t) − L󸀠2,k (t))(tn−1+σ − t)−αr dt k=1 t

k−1

n−1

t

k = ∑ [(f (t) − L2,k (t))(tn−1+σ − t)−αr |t=t k−1

k=1

tk

− ∫ (f (t) − L2,k (t))αr (tn−1+σ − t)−αr −1 dt] tk−1 tk

n−1

= − ∑ ∫ (f (t) − L2,k (t))αr (tn−1+σ − t)−αr −1 dt k=1 t

k−1

and 󵄨 󵄨 1 max 󵄨󵄨f (t) − L2,k (t)󵄨󵄨󵄨 ⩽ Mτ3 , 12

tk−1 ⩽t⩽tk 󵄨

we have tk

n−1

󵄨 󵄨 |Ar | ⩽ ∑ ∫ 󵄨󵄨󵄨f (t) − L2,k (t)󵄨󵄨󵄨αr (tn−1+σ − t)−αr −1 dt k=1 t

k−1

tk

n−1 1 ⩽ Mτ3 ∑ ∫ αr (tn−1+σ − t)−αr −1 dt 12 k=1 tk−1

tn−1

=

1 Mτ3 ∫ αr (tn−1+σ − t)−αr −1 dt 12 t0

1 = Mτ3 [(tn−1+σ − tn−1 )−αr − (tn−1+σ − t0 )−αr ] 12 1 ⩽ Mτ3 ⋅ (στ)−αr 12 1 = Mσ −αr τ3−αr . 12 For the second term of (1.90), according to 1

f 󸀠 (t) − L󸀠1,n (t) = f 󸀠 (t) − δt f n− 2

1

= [f 󸀠 (t) − f 󸀠 (tn− 1 )] + [f 󸀠 (tn− 1 ) − δt f n− 2 ] 2

2

1 1 = [f (tn− 1 )(t − tn− 1 ) + f 󸀠󸀠󸀠 (ηn )(t − tn− 1 )2 ] − τ2 f 󸀠󸀠󸀠 (η̃ n ), 2 2 2 2 24 t ∈ [tn−1 , tn ], ηn , η̃ n ∈ (tn−1 , tn ), 󸀠󸀠

(1.91)

1.6 Interpolation approximations of Caputo fractional derivatives | 53

we have m

tn−1+σ

m

tn−1+σ

λr B≡∑ ∫ [f 󸀠 (t) − L󸀠1,n (t)](tn−1+σ − t)−αr dt Γ(1 − α ) r r=0 tn−1

λr 1 ∫ [f 󸀠󸀠 (tn− 1 )(t − tn− 1 ) + f 󸀠󸀠󸀠 (ηn )(t − tn− 1 )2 2 2 2 Γ(1 − α ) 2 r r=0

=∑

tn−1

−

1 2 󸀠󸀠󸀠 τ f (η̃ n )](tn−1+σ − t)−αr dt 24 tn−1+σ

m

λr = f (tn− 1 ) ∑ ∫ (t − tn− 1 )(tn−1+σ − t)−αr dt 2 2 Γ(1 − α ) r r=0 󸀠󸀠

tn−1

tn−1+σ

m

λr 1 ∫ [ f 󸀠󸀠󸀠 (ηn )(t − tn− 1 )2 2 Γ(1 − α ) 2 r r=0

+∑

tn−1

−

1 2 󸀠󸀠󸀠 τ f (η̃ n )](tn−1+σ − t)−αr dt. 24

Noticing F(σ) = 0, we know that m

tn−1+σ

m

στ

λr ∑ ∫ (t − tn− 1 )(tn−1+σ − t)−αr dt 2 Γ(1 − αr ) r=0 tn−1

λr 1 ∫ [(σ − )τ − ξ ]ξ −αr dξ Γ(1 − α ) 2 r r=0

=∑

0

m

λr 1 (στ)1−αr (στ)2−αr =∑ [(σ − )τ − ] Γ(1 − αr ) 2 1 − αr 2 − αr r=0 m

λr α σ 1−αr [σ − (1 − r )]τ2−αr Γ(3 − α ) 2 r r=0

=∑

= F(σ) = 0. Therefore, m

tn−1+σ

λr 1 1 B= ∑ ∫ [ f 󸀠󸀠󸀠 (ηn )(t − tn− 1 )2 − τ2 f 󸀠󸀠󸀠 (η̃ n )](tn−1+σ − t)−αr dt. 2 Γ(1 − α ) 2 24 r r=0 tn−1

Furthermore, we have |B| ⩽

m λr 1 σ 1−αr 3−αr M∑ ⋅ τ . 6 r=0 Γ(1 − αr ) 1 − αr

(1.92)

54 | 1 Fractional derivatives and numerical approximations Substituting (1.91) and (1.92) into (1.90), we have 󵄨 󵄨󵄨 󵄨󵄨Dt f (tn−1+σ ) − 𝒟t f (tn−1+σ )󵄨󵄨󵄨 m λr 1 1 σ ⩽M∑ ⋅( + ⋅ )σ −αr τ3−αr . Γ(1 − α ) 12 6 1 − α r r r=0 This completes the proof. With application of (1.81), we have m

𝒟t f (tn−1+σ ) ≡ ∑ λr ⋅ r=0

n−1

τ−αr n−1 (n,αr ) [f (tn−k ) − f (tn−k−1 )] ∑c Γ(2 − αr ) k=0 k

= ∑ ĉk(n,α) [f (tn−k ) − f (tn−k−1 )], k=0

where m

ĉk(n,α) = ∑ λr ⋅ r=0

τ−αr (n,α ) c r, Γ(2 − αr ) k

0 ⩽ k ⩽ n − 1,

(1.93)

and {ck r } is defined by (1.77)–(1.80). The following two lemmas give the properties of coefficient {ĉk(n,α) }. (n,α )

Lemma 1.6.6. Given any nonnegative integer m, and positive constants λ0 , λ1 , . . . , λm , for any αr ∈ [0, 1] (0 ⩽ r ⩽ m), where at least one αr ∈ (0, 1), it holds the following inequalities: m

(n,α) (n,α) ĉ1(n,α) > ĉ2(n,α) > ⋅ ⋅ ⋅ > ĉn−2 > ĉn−1 > ∑ λr r=0

τ−αr n−αr . Γ(1 − αr )

(1.94)

Proof. When m = 0, the conclusion has been obtained in Lemma 1.6.3. Now, we suppose m ⩾ 1. For any αr ∈ (0, 1), we have (see Lemma 1.6.3) c1

(n,αr )

> c2

(n,αr )

> ⋅ ⋅ ⋅ > cn−2 r > cn−1 r > (1 − αr )n−αr . (n,α )

(n,α )

(1.95)

In particular, if αr = 0, we have c1

= c2

= ⋅ ⋅ ⋅ = cn−1 r = 1 = (1 − 0)n−0 ;

c1

= c2

= ⋅ ⋅ ⋅ = cn−1 r = 0 = (1 − 1)n−1 .

(n,αr )

(n,αr )

(n,α )

If αr = 1, we have (n,αr )

(n,αr )

(n,α )

(1.96)

Combining (1.95) with (1.96) and noticing that at least one αr ∈ (0, 1), the conclusion (1.94) holds. The proof ends.

1.6 Interpolation approximations of Caputo fractional derivatives | 55

Lemma 1.6.7. Given any nonnegative integer m, positive constants λ0 , λ1 , . . . , λm , for any αr ∈ [0, 1] (0 ⩽ r ⩽ m), where at least one αr ∈ (0, 1), then there exists a positive constant τ0 , such that, when τ ⩽ τ0 , it holds (2σ − 1)ĉ0(n,α) − σ ĉ1(n,α) > 0,

(1.97)

which implies ĉ0(n,α) > ĉ1(n,α) . Proof. If m = 0, the conclusion can be found in Lemma 1.6.3. Now we suppose m ⩾ 1, hence σ ∈ ( 21 , 1). (I) When n ⩾ 3, for every αr ∈ (0, 1), we have (see Lemma 1.6.3) (2σ − 1)c0

(n,αr )

= (2σ − 1)[ − σ{

− σc1

(n,αr )

(1 + σ)2−αr − σ 2−αr (1 + σ)1−αr − σ 1−αr − ] 2 − αr 2

1 [(2 + σ)2−αr − 2(1 + σ)2−αr + σ 2−αr ] 2 − αr

1 − [(2 + σ)1−αr − 2(1 + σ)1−αr + σ 1−αr ]} 2

(4σ 2 + 3σ − 1)sr − 4σ 2 + σ sr (2 + σ) − σ (2 + σ)1−αr + (1 + σ)1−αr 2 2σ 3σ − 1 (sr − 1)σ 1−αr , − 2

=−

where sr = 1−ασ /2 , r = 0, 1, . . . , m. r Noticing 1−αr

1 ) 1+σ 1 − αr ⩽ (1 + σ)1−αr (1 + ) 1+σ σs + 2σ = (1 + σ)1−αr r , sr (1 + σ)

(2 + σ)1−αr = (1 + σ)1−αr (1 +

we have (2σ − 1)c0

(n,αr )

− σc1

(n,αr )

1 1 2σ 2 ⩾ [(3σ 2 + 5σ + 2 − )sr + − 5σ 2 − 7σ + 1](1 + σ)−αr 2 σ sr 3σ − 1 − (sr − 1)σ 1−αr . 2

(1.98)

56 | 1 Fractional derivatives and numerical approximations For σ ∈ (1/2, 1), consider the function fσ (t) = (3σ 2 + 5σ + 2 −

1 2σ 2 )t + − 5σ 2 − 7σ + 1. σ t

When t ⩾ 1, 1 2σ 2 )− 2 σ t 1 ⩾ (3σ 2 + 5σ + 2 − ) − 2σ 2 σ

fσ󸀠 (t) = (3σ 2 + 5σ + 2 −

⩾ σ 2 + 5σ > 0.

Due to s0 > 1, we have fσ (s0 ) > fσ (1) =

(2σ − 1)(1 − σ) > 0. σ

Noticing m

∑ λr

r=0

τ−αr 2 σ 1−αr (sr − 1) = 2 F(σ) = 0, Γ(2 − αr ) τ

with the help of (1.98), we have (2σ − 1)ĉ0(n,α) − σ ĉ1(n,α) m

= ∑ λr r=0 m

⩾

τ−αr 1 f (s )(1 + σ)−αr ∑ λr 2 r=0 Γ(2 − αr ) σ r −

=

τ−αr (n,α ) (n,α ) [(2σ − 1)c0 r − σc1 r ] Γ(2 − αr )

τ−αr 3σ − 1 m σ 1−αr (sr − 1) ∑ λr 2 r=0 Γ(2 − αr )

1 m τ−αr f (s )(1 + σ)−αr ∑ λr 2 r=0 Γ(2 − αr ) σ r

τ−α0 1 f (s )(1 + σ)−α0 = λ0 2 Γ(2 − α0 ) σ 0 +

τ−αr 1 m f (s )(1 + σ)−αr ∑ λr 2 r=1 Γ(2 − αr ) σ r

1 τ−α0 > λ0 (1 + σ)−α0 [fσ (1) 2 Γ(2 − α0 ) m

+ ∑ λr r=1

τα0 −αr Γ(2 − α0 ) f (s )(1 + σ)α0 −αr ] λ0 Γ(2 − αr ) σ r

1.6 Interpolation approximations of Caputo fractional derivatives | 57

1 τ−α0 = λ0 (1 + σ)−α0 [fσ (1) + O(τα0 −α1 )]. 2 Γ(2 − α0 )

(1.99)

Since α0 − α1 > 0, there exists a positive constant τ0 such that fσ (1) + O(τα0 −α1 ) ⩾ 0 when τ ⩽ τ0 . Therefore, the inequality (1.97) holds. (II) When n = 2, we have (2σ − 1)c0

(2,αr )

= (2σ − 1)[

− σc1

(2,αr )

(1 + σ)2−αr − σ 2−αr (1 + σ)1−αr − σ 1−αr − ] 2 − αr 2

1 1 [(1 + σ)2−αr − σ 2−αr ]} − σ{ [3(1 + σ)1−αr − σ 1−αr ] − 2 2 − αr

1 (3σ − 1)(1 + σ) 1 = [ sr + 1 − 5σ](1 + σ)1−αr + (3σ − 1)(1 − sr )σ 1−αr . 2 σ 2 For σ ∈ ( 21 , 1), consider the function gσ (t) =

(3σ − 1)(1 + σ) t + 1 − 5σ. σ

It is easy to know that gσ (s0 ) > gσ (1) =

(3σ − 1)(1 + σ) 1 + 1 − 5σ = 3 − 2σ − > 0, σ σ

1 σ ∈ ( , 1). 2

Similar to the proof of (1.99), there exists a positive constant τ0 , such that the inequality (1.97) holds when τ ⩽ τ0 . This completes the proof. According to Theorem 1.6.5, we know that the accuracy is no less than second order when using 𝒟t f (tn−1+σ ) to approximate the value of the sum (1.88) of the multi-term Caputo derivatives at t = tn−1+σ . In the aftermentioned chapters, we will develop some high order accurate difference schemes for solving the multi-term time-fractional differential equations based on the approximation formula in Theorem 1.6.5.

1.6.5 H2N2 approximation In [56] and [61], the authors presented L2 method and L2C method (a variant of L2 method) to approximate the R-L fractional derivative. In [46], the authors applied L2 method and L2C method to treat the approximation of Caputo derivative C γ 0 Dt f (t)

t

1 = ∫ f 󸀠󸀠 (s)(t − s)1−γ ds, Γ(2 − γ) 0

γ ∈ (1, 2).

58 | 1 Fractional derivatives and numerical approximations L2 method: C γ 0 Dt f (t)|t=tn

tk

n 1 = ∑ ∫ f 󸀠󸀠 (s)(tn − s)1−γ ds Γ(2 − γ) k=1 tk−1

tk

≈

n f (t ) − 2f (tk ) + f (tk+1 ) 1 ⋅ ∫ (tn − s)1−γ ds. ∑ k−1 Γ(2 − γ) k=1 τ2 tk−1

L2C method: C γ 0 Dt f (t)|t=tn

tk

=

n 1 ∑ ∫ f 󸀠󸀠 (s)(tn − s)1−γ ds Γ(2 − γ) k=1 tk−1

tk

n f (t ) − f (tk−1 ) − f (tk ) + f (tk+1 ) 1 ⋅ ∫ (tn − s)1−γ ds. ≈ ∑ k−2 Γ(2 − γ) k=1 2τ2 tk−1

In this subsection, we introduce the H2N2 interpolation approximation method[71] . Consider the numerical evaluation of tn− 1

C γ 0 Dt f (t)|t=tn− 1

2

2

1 = ∫ f 󸀠󸀠 (t)(tn− 1 − t)1−γ dt, 2 Γ(2 − γ)

1 ⩽ n ⩽ N,

t0

which can be written as C γ 0 Dt f (t)|t=tn− 1

2

t1

tk+ 1

t0

tk− 1

2

2

n−1 1 [ ∫ f 󸀠󸀠 (t)(tn− 1 − t)1−γ dt + ∑ ∫ f 󸀠󸀠 (t)(tn− 1 − t)1−γ dt]. = 2 2 Γ(2 − γ) k=1 2

Using the data (t0 , f (t0 )), (t0 , f 󸀠 (t0 )) and (t1 , f (t1 )), the quadratic Hermite interpolation polynomial of f (t) reads 1 1 H2,0 (t) = f (t0 ) + f 󸀠 (t0 )(t − t0 ) + (δt f 2 − f 󸀠 (t0 ))(t − t0 )2 . τ

It is easy to know that 1 2 (δt f 2 − f 󸀠 (t0 )) τ

(1.100)

1 2 (δ f 2 − f 󸀠 (t0 )) = f 󸀠󸀠 (ζ0 ). τ t

(1.101)

󸀠󸀠 H2,0 (t) =

and there exists a ζ0 ∈ (t0 , t1 ) satisfying

1.6 Interpolation approximations of Caputo fractional derivatives | 59

Using three points (tk−1 , f (tk−1 )), (tk , f (tk )) and (tk+1 , f (tk+1 )), the quadratic Newton interpolation polynomial of f (t) reads 1 1 N2,k (t) = f (tk−1 ) + (δt f k− 2 )(t − tk−1 ) + (δt2 f k )(t − tk−1 )(t − tk ), 2

where δt2 f k =

1 1 1 (δ f k+ 2 − δt f k− 2 ). τ t

It is easy to know that 󸀠󸀠 N2,k (t) = δt2 f k

(1.102)

and there exists a ζk ∈ (tk−1 , tk+1 ) satisfying δt2 f k = f 󸀠󸀠 (ζk ).

(1.103)

From (1.100) and (1.102), we have C γ 0 Dt f (t)|t=tn− 1 t1

2

tk+ 1

2

2

n−1 1 󸀠󸀠 󸀠󸀠 [ ∫ H2,0 (t)(tn− 1 − t)1−γ dt] (t)(tn− 1 − t)1−γ dt + ∑ ∫ N2,k ≈ 2 2 Γ(2 − γ) k=1 t0

tk− 1

2

t1

2

1 1 2 = [ ∫ (δt f 2 − f 󸀠 (t0 ))(tn− 1 − t)1−γ dt 2 Γ(2 − γ) τ

t0

n−1

tk+ 1

2

+ ∑ ∫ (δt2 f k )(tn− 1 − t)1−γ dt] 2

k=1 t

k− 21

t1

2

1 2 1 [ ∫ (tn− 1 − t)1−γ dt ⋅ (δt f 2 − f 󸀠 (t0 )) = 2 Γ(2 − γ) τ

t0

n−1

tk+ 1

2

1 1 1 ∫ (tn− 1 − t)1−γ dt ⋅ (δt f k+ 2 − δt f k− 2 )] 2 τ k=1

+∑

tk− 1

2

≡ 𝒟̂ γ f (tn− 1 ). 2

Let 1−γ

{ τ2−γ [(k + 1)2−γ − k 2−γ ], 0 ⩽ k ⩽ n − 2, (n,γ) b̂ k = { 2τ 1−γ [(n − 21 )2−γ − (n − 1)2−γ ], k = n − 1. { 2−γ

60 | 1 Fractional derivatives and numerical approximations Computing arrives at tk+ 1

2

1 ∫ (tn− 1 − t)1−γ dt 2 τ tk− 1

2

1 1 = ⋅ [(t 1 − tk− 1 )2−γ − (tn− 1 − tk+ 1 )2−γ ] 2 2 2 τ 2 − γ n− 2 τ1−γ [(n − k)2−γ − (n − k − 1)2−γ ] 2−γ (n,γ) , 1⩽k ⩽n−1 = b̂ =

(1.104)

n−k−1

and t1

2

2 ∫ (tn− 1 − t)1−γ dt 2 τ t0

2 1 = ⋅ [(t 1 − t0 )2−γ − (tn− 1 − t 1 )2−γ ] 2 2 τ 2 − γ n− 2 2−γ

2τ1−γ 1 [(n − ) 2−γ 2 (n,γ) = b̂ . =

− (n − 1)2−γ ] (1.105)

n−1

γ

Thus we get the approximation formula of C0 Dt f (t)|t=t γ

𝒟̂ t f (tn− 1 ) = 2

=

n− 21

as follows:

n−1 1 1 1 1 (n,γ) (n,γ) [b̂ n−1 ⋅ (δt f 2 − f 󸀠 (t0 )) + ∑ b̂ n−k−1 ⋅ (δt f k+ 2 − δt f k− 2 )] Γ(2 − γ) k=1 n−1 1 1 1 (n,γ) (n,γ) (n,γ) (n,γ) [b̂ 0 δt f n− 2 − ∑ (b̂ n−k−1 − b̂ n−k )δt f k− 2 − b̂ n−1 f 󸀠 (t0 )]. Γ(2 − γ) k=1

We call (1.106) the H2N2 formula or H2N2 approximation. The coefficients in formula (1.106) satisfy Lemma 1.6.8. (n,γ) (n,γ) (n,γ) (n,γ) b̂ 0 > b̂ 1 > b̂ 2 > ⋅ ⋅ ⋅ > b̂ n−1 > 0.

Proof. Using (1.104), we have tk+ 1

2

(n,γ) b̂ n−k−1

1

1 = ∫ (tn− 1 − t)1−γ dt = τ1−γ ∫(n − k − ξ )1−γ dξ , 2 τ tk− 1

2

0

1 ⩽ k ⩽ n − 1,

(1.106)

1.6 Interpolation approximations of Caputo fractional derivatives | 61

that is, 1

(n,γ) b̂ k = τ1−γ ∫(k + 1 − ξ )1−γ dξ ,

0 ⩽ k ⩽ n − 2.

0

It is easy to know that (n,γ) (n,γ) (n,γ) (n,γ) b̂ 0 > b̂ 1 > b̂ 2 > ⋅ ⋅ ⋅ > b̂ n−2 .

Particularly, 1

(n,γ) b̂ n−2 = τ1−γ ∫(n − 1 − ξ )1−γ dξ . 0

Using (1.105), we have t1

1

2

(n,γ) b̂ n−1

1−γ

1+ξ 2 = ∫ (tn− 1 − t)1−γ dt = τ1−γ ∫(n − ) 2 τ 2 t0

dξ .

0

Noticing that when ξ ∈ (0, 1), (n − 1 − ξ )1−γ > (n −

1−γ

1+ξ ) 2

,

hence, 1

1

∫(n − 1 − ξ )1−γ dξ > ∫(n − 0

0

1−γ

1+ξ ) 2

dξ ,

which implies that (n,γ) (n,γ) b̂ n−2 > b̂ n−1 .

The proof is completed. Now we estimate the error. Theorem 1.6.6. Suppose f ∈ C 3 [t0 , tn ]. Denote γ

Rn = C0 Dt f (t)|t=t

n− 21

γ

− 𝒟̂ t f (tn− 1 ). 2

Then we have |Rn | ⩽ [

γ−1 1 1 󵄨 󵄨 + + ] max 󵄨󵄨f 󸀠󸀠󸀠 (t)󵄨󵄨󵄨τ3−γ . 8Γ(2 − γ) 12Γ(3 − γ) 2Γ(4 − γ) t0 ⩽t⩽tn 󵄨

62 | 1 Fractional derivatives and numerical approximations Proof. Let g(t) = f 󸀠 (t), α = γ − 1. We have C γ 0 Dt f (t)

t

t

0

0

1 1 f 󸀠󸀠 (s) g 󸀠 (s) ds = = ds = C0 Dαt g(t). ∫ ∫ γ−1 Γ(2 − γ) (t − s) Γ(1 − α) (t − s)α

(1.107)

Make the following piecewise linear interpolations for g(t) as follows: L1,0 (t) =

t − t1

2

t0 − t 1

g(t0 ) +

2

L1,k (t) =

2

t − tk+ 1

2

tk− 1 − tk+ 1 2

t − t0 g(t 1 ), 2 t 1 − t0

g(tk− 1 ) + 2

2

t ∈ [t0 , t 1 ], 2

t − tk− 1

2

tk+ 1 − tk− 1 2

2

g(tk+ 1 ),

t ∈ [tk− 1 , tk+ 1 ],

2

2

2

1 ⩽ k ⩽ n − 1.

It is obvious that there exist ξ0 ∈ (t0 , t 1 ), ξk ∈ (tk− 1 , tk+ 1 ), 1 ⩽ k ⩽ n − 1, such that 2

2

2

1 g(t) − L1,0 (t) = g 󸀠󸀠 (ξ0 )(t − t0 )(t − t 1 ), t ∈ [t0 , t 1 ], 2 2 2 1 󸀠󸀠 g(t) − L1,k (t) = g (ξk )(t − tk− 1 )(t − tk+ 1 ), t ∈ [tk− 1 , tk+ 1 ], 2 2 2 2 2

(1.108) 1 ⩽ k ⩽ n − 1.

Noticing C γ 0 Dt f (t)|t=tn− 1

2

= C0 Dαt g(t)|t=t

n− 21

t1

2

=

n−1

tk+ 1

2

g 󸀠 (t) 1 g (t) [∫ dt + ∑ ∫ dt], α Γ(1 − α) (tn− 1 − t) (tn− 1 − t)α k=1 t0

󸀠

2

tk− 1

2

γ we can rewrite 𝒟̂ t f (tn− 1 ) as 2

t1

2

γ

𝒟̂ t f (tn− 1 ) = 2

1 1 2 [ ∫ (tn− 1 − t)1−γ dt ⋅ (δt f 2 − f 󸀠 (t0 )) 2 Γ(2 − γ) τ

t0

n−1

tk+ 1

2

1 1 1 +∑ ∫ (tn− 1 − t)1−γ dt ⋅ (δt f k+ 2 − δt f k− 2 )] 2 τ k=1

tk− 1

2

t1

2

2 1 [ ∫ (tn− 1 − t)1−γ dt ⋅ (g(t 1 ) − g(t0 )) = 2 2 Γ(2 − γ) τ t0

2

(1.109)

1.6 Interpolation approximations of Caputo fractional derivatives | 63 tk+ 1

2

n−1

1 +∑ ∫ (tn− 1 − t)1−γ dt ⋅ (g(tk+ 1 ) − g(tk− 1 ))] 2 2 2 τ k=1 tk− 1

2

t1

2

1 2 1 { ∫ (tn− 1 − t)1−γ dt ⋅ (g(t 1 ) − δt f 2 ) − 2 2 Γ(2 − γ) τ

t0

tk+ 1

2

n−1

1 1 ∫ (tn− 1 − t)1−γ dt ⋅ [(g(tk+ 1 ) − δt f k+ 2 ) 2 2 τ k=1

+∑

tk− 1

2

1

− (g(tk− 1 ) − δt f k− 2 )]} 2

t1

2 g(t 1 ) − g(t0 ) 1 2 = (tn− 1 − t)−α dt [∫ τ 2 Γ(1 − α) 2

t0

n−1

tk+ 1

2

+∑ ∫

g(tk+ 1 ) − g(tk− 1 ) 2

k=1 t

k− 21

−

2

τ

(tn− 1 − t)−α dt] 2

1 1 (n,γ) (g(t 1 ) − δt f 2 ) {b̂ 2 Γ(2 − γ) n−1

n−1

1

1

(n,γ) + ∑ b̂ n−k−1 [(g(tk+ 1 ) − δt f k+ 2 ) − (g(tk− 1 ) − δt f k− 2 )]} k=1

2

2

t1

tk+ 1

t0

tk− 1

2

2

n−1 1 = [ ∫ L󸀠1,0 (t)(tn− 1 − t)−α dt + ∑ ∫ L󸀠1,k (t)(tn− 1 − t)−α dt] 2 2 Γ(1 − α) k=1

−

2

1 (n,γ) [b̂ (g(tn− 1 ) − δt f 2 Γ(2 − γ) 0 n−1

n− 21

) 1

− ∑ (b̂ n−k−1 − b̂ n−k )(g(tk− 1 ) − δt f k− 2 )]. (n,γ)

(n,γ)

2

k=1

Thus, t1

tk+ 1

2

2

n−1 g 󸀠 (t) − L󸀠1,k (t) g 󸀠 (t) − L󸀠1,0 (t) 1 [∫ dt + dt] Rn = ∑ ∫ Γ(1 − α) (tn− 1 − t)α (tn− 1 − t)α k=1 t0

+

2

tk− 1

1 1 (n,γ) (g(tn− 1 ) − δt f n− 2 ) [b̂ 2 Γ(2 − γ) 0

2

2

64 | 1 Fractional derivatives and numerical approximations n−1

1

(n,γ) (n,γ) − ∑ (b̂ n−k−1 − b̂ n−k )(g(tk− 1 ) − δt f k− 2 )] 2

k=1

≡ pn + qn ,

(1.110)

where t1

2

pn =

tk+ 1

2

g 󸀠 (t) − L󸀠1,k (t) 1 dt + dt], [∫ ∑ ∫ Γ(1 − α) (tn− 1 − t)α (tn− 1 − t)α k=1 t0

qn =

n−1

g 󸀠 (t) − L󸀠1,0 (t)

tk− 1

2

2

2

1 1 (n,γ) [b̂ (g(tn− 1 ) − δt f n− 2 ) 2 Γ(2 − γ) 0

n−1

1

(n,γ) (n,γ) − ∑ (b̂ n−k−1 − b̂ n−k )(g(tk− 1 ) − δt f k− 2 )]. 2

k=1

The estimate of pn . we obtain

Using the integration by parts and noticing (1.108)–(1.109), t1

2

1 pn = [ ∫ (g(t) − L1,0 (t))(−α)(tn− 1 − t)−α−1 dt 2 Γ(1 − α) t0

n−1

tk+ 1

2

+ ∑ ∫ (g(t) − L1,k (t))(−α)(tn− 1 − t)−α−1 dt] 2

k=1 t

k− 21

t1

2

1 α = [ ∫ g 󸀠󸀠 (ξ0 )(t − t0 )(t 1 − t)(tn− 1 − t)−α−1 dt 2 2 Γ(1 − α) 2 t0

n−1

tk+ 1

2

+∑ ∫ k=1 t

k− 21

1 󸀠󸀠 g (ξk )(t − tk− 1 )(tk+ 1 − t)(tn− 1 − t)−α−1 dt]. 2 2 2 2

When n = 1, we have t1

2

p1 =

α 1 ∫ g 󸀠󸀠 (ξ0 )(t − t0 )(t 1 − t)(t 1 − t)−α−1 dt 2 2 Γ(1 − α) 2 t0

t1

2

1 α = ∫ g 󸀠󸀠 (ξ0 )(t − t0 )(t 1 − t)−α dt 2 Γ(1 − α) 2 t0

t1

2

α 1 = ⋅ g 󸀠󸀠 (ξ0̂ ) ∫ (t − t0 )(t 1 − t)−α dt 2 Γ(1 − α) 2 t0

1.6 Interpolation approximations of Caputo fractional derivatives | 65 2−α 1

α 1 τ = ⋅ g 󸀠󸀠 (ξ0̂ )( ) Γ(1 − α) 2 2

∫ θ(1 − θ)−α dθ 0

2−α

=

1 τ α ⋅ g 󸀠󸀠 (ξ0̂ )( ) Γ(1 − α) 2 2

=

1 τ α ⋅ g 󸀠󸀠 (ξ0̂ )( ) Γ(3 − α) 2 2

( 2−α

,

1 1 − ) 1−α 2−α ξ0̂ ∈ (t0 , t 1 ). 2

Therefore, 2−α

|p1 | ⩽

α 󵄨 󵄨 τ ⋅ max 󵄨󵄨g 󸀠󸀠 (t)󵄨󵄨󵄨( ) 2Γ(3 − α) t0 ⩽t⩽t 1 󵄨 2

.

2

When n ⩾ 2, t1

2

α 1 pn = [ ∫ g 󸀠󸀠 (ξ0 )(t − t0 )(t 1 − t)(tn− 1 − t)−α−1 dt 2 2 Γ(1 − α) 2 t0

tk+ 1

2

n−2

+∑ ∫ k=1 t

k− 21

tn− 1

2

+ ∫ tn− 3

1 󸀠󸀠 g (ξk )(t − tk− 1 )(tk+ 1 − t)(tn− 1 − t)−α−1 dt 2 2 2 2

1 󸀠󸀠 g (ξn−1 )(t − tn− 3 )(tn− 1 − t)(tn− 1 − t)−α−1 dt]. 2 2 2 2

2

Hence, t1

2

1 α 󵄨 󵄨 ⋅ max 󵄨󵄨g 󸀠󸀠 (t)󵄨󵄨󵄨[ ∫ (t − t0 )(t 1 − t)(tn− 1 − t)−α−1 dt |pn | ⩽ 2 2 Γ(1 − α) 2 t0 ⩽t⩽tn− 1 󵄨 t0

2

n−2

tk+ 1

2

+ ∑ ∫ (t − tk− 1 )(tk+ 1 − t)(tn− 1 − t)−α−1 dt 2

k=1 t

2

2

k− 21

tn− 1

2

+ ∫ (t − tn− 3 )(tn− 1 − t)−α dt] tn− 3

2

2

2

t1

2

2 α 1 󵄨 󵄨 τ ⩽ ⋅ max 󵄨󵄨󵄨g 󸀠󸀠 (t)󵄨󵄨󵄨[ ∫ (tn− 1 − t)−α−1 dt 2 Γ(1 − α) 2 t0 ⩽t⩽tn− 1 16 2

t0

(1.111)

66 | 1 Fractional derivatives and numerical approximations tk+ 1

tn− 1

tk− 1

tn− 3

2

2

τ2 n−2 + ∑ ∫ (tn− 1 − t)−α−1 dt + ∫ (t − tn− 3 )(tn− 1 − t)−α dt] 2 2 2 4 k=1 2

2

α 󵄨 󵄨 ⋅ max 󵄨󵄨g 󸀠󸀠 (t)󵄨󵄨󵄨 ⩽ 2Γ(1 − α) t0 ⩽t⩽tn− 1 󵄨 2

tn− 3 2

⋅[

τ2 τ2−α ] ∫ (tn− 1 − t)−α−1 dt + 2 4 (1 − α)(2 − α) t0

⩽

2−α α τ2−α 󵄨 󵄨 τ ⋅ max 󵄨󵄨󵄨g 󸀠󸀠 (t)󵄨󵄨󵄨[ + ] 2Γ(1 − α) t0 ⩽t⩽tn− 1 4α (1 − α)(2 − α) 2

α 1 󵄨 󵄨 + ] max 󵄨󵄨g 󸀠󸀠 (t)󵄨󵄨󵄨τ2−α . =[ 8Γ(1 − α) 2Γ(3 − α) t0 ⩽t⩽tn− 1 󵄨

(1.112)

2

The estimate of qn .

Noticing 1

1

g(tk− 1 ) − δt f k− 2 = f 󸀠 (tk− 1 ) − δt f k− 2 = − 2

2

τ2 󸀠󸀠󸀠 f (ηk ), 24

ηk ∈ (tk−1 , tk ),

we have qn =

n−1 τ2 1 τ2 (n,γ) (n,γ) (n,γ) [b̂ 0 (− f 󸀠󸀠󸀠 (ηn )) − ∑ (b̂ n−k−1 − b̂ n−k )(− f 󸀠󸀠󸀠 (ηk ))]. Γ(2 − γ) 24 24 k=1

Consequently, |qn | ⩽ ⩽ = =

n−1 τ2 1 (n,γ) (n,γ) (n,γ) 󵄨 󵄨 max 󵄨󵄨󵄨f 󸀠󸀠󸀠 (t)󵄨󵄨󵄨 [b̂ 0 + ∑ (b̂ n−k−1 − b̂ n−k )] 24 t0 ⩽t⩽tn Γ(2 − γ) k=1

1 τ2 (n,γ) 󵄨 󵄨 max 󵄨󵄨f 󸀠󸀠󸀠 (t)󵄨󵄨󵄨 ⋅ 2b̂ 0 24 t0 ⩽t⩽tn 󵄨 Γ(2 − γ) τ2 τ1−γ 1 󵄨 󵄨 max 󵄨󵄨󵄨f 󸀠󸀠󸀠 (t)󵄨󵄨󵄨 12 t0 ⩽t⩽tn Γ(2 − γ) 2 − γ

τ3−γ 󵄨 󵄨 max 󵄨󵄨f 󸀠󸀠󸀠 (t)󵄨󵄨󵄨. 12Γ(3 − γ) t0 ⩽t⩽tn 󵄨

Substituting (1.111)–(1.113) into (1.110), we get |Rn | ⩽ |pn | + |qn | ⩽[

α τ3−γ 1 󵄨 󵄨 󵄨 󵄨 + ] max 󵄨󵄨󵄨g 󸀠󸀠 (t)󵄨󵄨󵄨τ2−α + max 󵄨󵄨f 󸀠󸀠󸀠 (t)󵄨󵄨󵄨 8Γ(1 − α) 2Γ(3 − α) t0 ⩽t⩽tn− 1 12Γ(3 − γ) t0 ⩽t⩽tn 󵄨 2

γ−1 1 1 󵄨 󵄨 ⩽[ + + ] max 󵄨󵄨f 󸀠󸀠󸀠 (t)󵄨󵄨󵄨τ3−γ . 8Γ(2 − γ) 12Γ(3 − γ) 2Γ(4 − γ) t0 ⩽t⩽tn 󵄨 The proof ends.

(1.113)

1.7 Fast interpolation approximations of Caputo fractional derivatives | 67

Remark 1.6.3. The formula (1.106) can also be obtained in the following way[46] . Let g(t) = f 󸀠 (t),

α = γ − 1.

Then we have C γ 0 Dt f (t)|t=tn− 1 t1

2

= C0 Dαt g(t)|t=t

n− 21

tk+ 1

2

2

n−1 1 [ ∫ g 󸀠 (t)(tn− 1 − t)−α dt + ∑ ∫ g 󸀠 (t)(tn− 1 − t)−α dt] = 2 2 Γ(1 − α) k=1 t0

tk− 1

2

t1

tk+ 1

t0

tk− 1

2

2

n−1 1 ≈ [ ∫ L󸀠1,0 (t)(tn− 1 − t)−α dt + ∑ ∫ L󸀠1,k (t)(tn− 1 − t)−α dt] 2 2 Γ(1 − α) k=1 2

t1

2

1 2 = [ (g(t 1 ) − g(t0 )) ⋅ ∫ (tn− 1 − t)−α dt 2 2 Γ(1 − α) τ t0

n−1

tk+ 1

2

1 + ∑ (g(tk+ 1 ) − g(tk− 1 )) ⋅ ∫ (tn− 1 − t)−α dt] 2 2 2 τ k=1 tk− 1

2

=

n−1 1 (n,γ) (n,γ) [b̂ n−1 ⋅ (g(t 1 ) − f 󸀠 (t0 )) + ∑ b̂ n−k−1 ⋅ (g(tk+ 1 ) − g(tk− 1 ))] 2 2 2 Γ(2 − γ) k=1

=

n−1 1 (n,γ) (n,γ) (n,γ) (n,γ) [b̂ 0 g(tn− 1 ) − ∑ (b̂ n−k−1 − b̂ n−k )g(tk− 1 ) − b̂ n−1 f 󸀠 (t0 )] 2 2 Γ(2 − γ) k=1

≈

n−1 1 1 1 (n,γ) (n,γ) (n,γ) (n,γ) [b̂ 0 δt f n− 2 − ∑ (b̂ n−k−1 − b̂ n−k )δt f k− 2 − b̂ n−1 f 󸀠 (t0 )], Γ(2 − γ) k=1

in which two approximate equalities have been used.

1.7 Fast interpolation approximations of Caputo fractional derivatives 1.7.1 Fast L1 approximation Jiang et al. gave the approximation formula of sum-of-exponentials (SOE) for the kernel function t −α in Caputo derivative. Lemma 1.7.1. [41] For given α ∈ (0, 1), ϵ > 0, τ̂ > 0 and T > 0, where τ̂ < T, there (α) exist a positive integer Nexp , positive points s(α) and corresponding positive weights ω(α) , l l

68 | 1 Fractional derivatives and numerical approximations (α) (l = 1, 2, . . . , Nexp ) satisfying (α) Nexp 󵄨 󵄨󵄨 (α) 󵄨󵄨 󵄨󵄨 −α 󵄨󵄨t − ∑ ω(α) e−sl t 󵄨󵄨󵄨 ⩽ ϵ, 󵄨󵄨 󵄨󵄨 l 󵄨󵄨 󵄨󵄨 l=1

∀ t ∈ [τ,̂ T].

In addition, the number of exponentials has the following estimate: 1 1 T 1 1 T (α) Nexp = O((log )(loglog + log ) + (log )(loglog + log )). ϵ ϵ τ̂ τ̂ ϵ τ̂ (α) It is worth to note that Nexp , ω(α) , s(α) not only depend on α but also depend on ϵ, τ̂ l l and T. (α) Without confusion, Nexp , s(α) and ω(α) will be briefly written as Nexp , sl and ωl . l l Table 1.1 lists the numbers of exponentials, Nexp needed to approximate t −α (t ∈ (τ,̂ T)) for different α, τ,̂ ϵ, with T = 1. One can find that the number of exponentials is very limited and no more than 200 in general.

Table 1.1: Nexp needed to approximate t −α (t ∈ (τ,̂ T )) for different α, τ,̂ ϵ with T = 1.

α

ϵ

0.1

10−6 10−8 10−10 10−12 10−14 10−6 10−8 10−10 10−12 10−14 10−6 10−8 10−10 10−12 10−14

0.5

0.9

10

−3

10

31 40 48 57 66 32 40 49 58 66 32 41 49 58 66

τ̂

−4

10−5

10−6

10−7

37 47 57 68 78 37 48 58 68 78 38 48 58 69 79

42 55 66 78 90 43 55 66 79 91 43 56 67 80 91

48 62 75 89 102 48 62 75 90 103 49 63 76 90 104

53 69 84 100 115 54 70 84 100 115 55 70 85 101 116

In what follows, a fast algorithm for Caputo fractional derivatives will be proposed. According to Lemma 1.7.1, we have C α 0 Dt f (t)|t=tn

=

tk

tn

tk−1

tn−1

n−1 1 [ ∑ ∫ f 󸀠 (t)(tn − t)−α dt + ∫ f 󸀠 (t)(tn − t)−α dt] Γ(1 − α) k=1

1.7 Fast interpolation approximations of Caputo fractional derivatives | 69 tk

tn

N

exp n−1 1 ≈ [ ∑ ∫ L󸀠1,k (t) ∑ ωl e−sl (tn −t) dt + ∫ L󸀠1,n (t)(tn − t)−α dt] (1.114) Γ(1 − α) k=1 l=1

tk−1

N

=

tn−1

tk

exp n−1 1 [ ∑ ωl ( ∑ ∫ L󸀠1,k (t)e−sl (tn −t) dt) Γ(1 − α) l=1 k=1

tk−1

tn

+ ∫ L󸀠1,n (t)(tn − t)−α dt] tn−1

N

=

exp τ−α 1 [ ∑ ωl Fln + (f (tn ) − f (tn−1 ))] Γ(1 − α) l=1 1−α

≡ ℱ Dαt f (tn ),

(1.115)

where tk

n−1

Fln = ∑ ∫ L󸀠1,k (t)e−sl (tn −t) dt, k=1 t

1 ⩽ l ⩽ Nexp ,

1 ⩽ n ⩽ N.

k−1

It is noted that Fln can be evaluated by the following recurrence relation: Fln

tk

n−1

= ∑ ∫ L󸀠1,k (t)e−sl (tn −t) dt k=1 t

k−1

tn−1

tk

n−2

= ∑ ∫ L󸀠1,k (t)e−sl (tn −t) dt + ∫ L󸀠1,n−1 (t)e−sl (tn −t) dt k=1 t

tn−2

k−1

=e

−sl τ

n−2

tk

∑ ∫ k=1 t

L󸀠1,k (t)e−sl (tn−1 −t) dt

+ δt f

tn−1

∫ e−sl (tn −t) dt tn−2

k−1 3

n− 32

1

= e−sl τ Fln−1 + τ δt f n− 2 ∫ e−sl (1+θ)τ dθ 0

= e−sl τ Fln−1 + Bl [f (tn−1 ) − f (tn−2 )],

2 ⩽ n ⩽ N,

where Fl1

1

= 0,

Bl = ∫ e−sl (1+θ)τ dθ, 0

1 ⩽ l ⩽ Nexp .

(1.116)

70 | 1 Fractional derivatives and numerical approximations A fast algorithm for evaluating C0 Dαt f (t)|t=tn can be obtained from (1.115)–(1.116) as

follows:

N

exp 1 τ−α { ℱ α { { Dt f (tn ) = [ ∑ ωl Fln + (f (tn ) − f (tn−1 ))], n ⩾ 1, { { Γ(1 − α) l=1 1−α { { { { Fl1 = 0, 1 ⩽ l ⩽ Nexp , { { { n −sl τ n−1 { Fl = e Fl + Bl [f (tn−1 ) − f (tn−2 )], 1 ⩽ l ⩽ Nexp n ⩾ 2.

(1.117) (1.118) (1.119)

The computational complexity for evaluating C0 Dαt f (t)|t=tn (1 ⩽ n ⩽ N) by L1 for-

mula (1.60) is O(N 2 ), and it is O(NNexp ) by the formula (1.117)–(1.119). When N is large,

O(NNexp ) a2 > ⋅ ⋅ ⋅ > an−1 > 0. If ϵ ⩽

2−21−α −α τ , 1−α

(1.124)

it holds ̂ (α) â (α) 0 ⩾ a1 .

In addition, τ−α , 1−α < tk−α + ϵ,

â (α) 0 = −α tk+1 − ϵ < â (α) k

k ⩾ 1.

Proof. According to (1.121), we have â (α) k

Nexp

1

l=1

0

= ∑ ωl ∫ e−sl (kτ+τθ) dθ,

1 ⩽ k ⩽ n − 1,

by which, the truth of (1.124) is apparent. Noticing tn

â (α) 0

1 = ∫ (tn − t)−α dt, τ tn−1

â (α) 1

tn−1 N

exp 1 = ∫ ∑ ωl e−sl (tn −t) dt, τ l=1

tn−2

we have tn−1

tn

â (α) 0

−

â (α) 1

1 1 = ∫ (tn − t)−α dt − ∫ (tn − t)−α dt τ τ tn−1

tn−2

tn−1

+

N

exp 1 ∫ [(tn − t)−α − ∑ ωl e−sl (tn −t) ]dt. τ l=1

tn−2

(1.125)

72 | 1 Fractional derivatives and numerical approximations Computing yields tn

tn−1

tn−1

tn−2

1 1 2 − 21−α −α τ . ∫ (tn − t)−α dt − ∫ (tn − t)−α dt = τ τ 1−α In addition, by Lemma 1.7.1, we have 󵄨󵄨 tn−1 󵄨󵄨 Nexp 󵄨󵄨 1 󵄨 󵄨󵄨 ∫ [(tn − t)−α − ∑ ωl e−sl (tn −t) ]dt 󵄨󵄨󵄨 󵄨󵄨 τ 󵄨󵄨 󵄨󵄨 󵄨󵄨 l=1 tn−2 tn−1󵄨 Nexp 󵄨󵄨󵄨 1 󵄨󵄨󵄨󵄨 󵄨 ⩽ ∫ 󵄨󵄨(tn − t)−α − ∑ ωl e−sl (tn −t) 󵄨󵄨󵄨dt 󵄨󵄨 τ 󵄨󵄨󵄨 l=1 󵄨 tn−2

⩽ ϵ. Hence when ϵ ⩽

2−21−α −α τ , 1−α

̂ (α) â (α) 0 − a1 ⩾

2 − 21−α −α τ − ϵ ⩾ 0. 1−α

Using (1.123), we have â (α) = n−k

tk N

exp 1 ∫ ∑ ωl e−sl (tn −t) dt τ l=1

tk−1 tk

tk

N

exp 1 1 = ∫ (tn − t)−α dt − ∫ [(tn − t)−α − ∑ ωl e−sl (tn −t) ]dt τ τ l=1

tk−1

tk−1

tk

=

N

exp 1 τ−α (α) an−k − ∫ [(tn − t)−α − ∑ ωl e−sl (tn −t) ]dt, 1−α τ l=1

tk−1

1 ⩽ k ⩽ n − 1. Further, it follows 󵄨󵄨 τ−α (α) 󵄨󵄨󵄨󵄨 󵄨󵄨 ̂ (α) a 󵄨 ⩽ ϵ, 󵄨󵄨an−k − 󵄨󵄨 1 − α n−k 󵄨󵄨󵄨

1 ⩽ k ⩽ n − 1,

namely, τ−α (α) τ−α (α) ak − ϵ ⩽ â (α) ⩽ a + ϵ, k 1−α 1−α k

1 ⩽ k ⩽ n − 1.

Combining with Lemma 1.6.1, (1.125) can be obtained. The proof ends.

(1.126)

1.7 Fast interpolation approximations of Caputo fractional derivatives | 73

1−α

For the common taken ϵ, the inequality ϵ ⩽ 2−2 τ−α always holds. 1−α ℱ α C α The truncation error of Dt f (tn ) to approximate 0 Dt f (t)|t=tn is given in the following theorem. Theorem 1.7.1. Suppose f ∈ C 2 [t0 , tn ], then we have 󵄨󵄨C α 󵄨 󵄨󵄨 D f (t)|t=t − ℱ Dα f (tn )󵄨󵄨󵄨 t 󵄨󵄨0 t 󵄨󵄨 n ϵtn 1 α 1 󵄨 󵄨 󵄨 󵄨 [ + ] max 󵄨󵄨f 󸀠󸀠 (t)󵄨󵄨󵄨 ⋅ τ2−α + max 󵄨󵄨f 󸀠 (t)󵄨󵄨󵄨. ⩽ 2Γ(1 − α) 4 (1 − α)(2 − α) t0 ⩽t⩽tn 󵄨 Γ(1 − α) t0 ⩽t⩽tn 󵄨

(1.127)

Proof. Some direct calculations yield C α 0 Dt f (t)|t=tn

− ℱ Dαt f (tn )

tn

tk

n−1 1 = [ ∑ ∫ f 󸀠 (t)(tn − t)−α dt + ∫ f 󸀠 (t)(tn − t)−α dt] Γ(1 − α) k=1 tk−1

tn−1

tk

−

N

exp n−1 1 [ ∑ ∫ L󸀠1,k (t) ∑ ωl e−sl (tn −t) dt Γ(1 − α) k=1 l=1

tk−1

tn

+ ∫ L󸀠1,n (t)(tn − t)−α dt] tn−1

tk

=

n 1 ∑ ∫ [f 󸀠 (t) − L󸀠1,k (t)](tn − t)−α dt Γ(1 − α) k=1 tk−1

tk

N

exp n−1 1 + ∑ ∫ L󸀠1,k (t)[(tn − t)−α − ∑ ωl e−sl (tn −t) ]dt Γ(1 − α) k=1 l=1

tk−1

≡ In + IIn .

(1.128)

By Theorem 1.6.1, for the first term in (1.128), we have |In | ⩽

1 1 α 󵄨 󵄨 [ + ] max 󵄨󵄨f 󸀠󸀠 (t)󵄨󵄨󵄨 ⋅ τ2−α . 2Γ(1 − α) 4 (1 − α)(2 − α) t0 ⩽t⩽tn 󵄨

(1.129)

For the second term in (1.128), we have t Nexp n−1 k 󵄨󵄨󵄨 󵄨󵄨󵄨 1 󵄨󵄨 󸀠 󵄨󵄨 󵄨 󵄨 max 󵄨󵄨f (t)󵄨󵄨 ∑ ∫ 󵄨󵄨󵄨(tn − t)−α − ∑ ωl e−sl (tn −t) 󵄨󵄨󵄨dt |IIn | ⩽ 󵄨󵄨 󵄨󵄨 Γ(1 − α) t0 ⩽t⩽tn k=1 t 󵄨 l=1 󵄨 k−1 tk

n−1 1 󵄨 󵄨 max 󵄨󵄨󵄨f 󸀠 (t)󵄨󵄨󵄨 ∑ ∫ ϵ dt ⩽ Γ(1 − α) t0 ⩽t⩽tn k=1

⩽

ϵ tn 󵄨 󵄨 max 󵄨󵄨f 󸀠 (t)󵄨󵄨󵄨. Γ(1 − α) t0 ⩽t⩽tn 󵄨

tk−1

(1.130)

74 | 1 Fractional derivatives and numerical approximations Substituting (1.129) and (1.130) into (1.128), we have (1.127) immediately. The proof ends.

1.7.2 Fast L2-1σ approximation With the help of Lemma 1.7.1, we can give a fast algorithm of L2-1σ interpolation approximation[101] . Take σ = 1 − α2 , τ̂ = στ. Table 1.2 lists the values of Nexp needed to approximate t −α with different parameters α, τ, ϵ. It shows the number of exponentials needed is very limited and no more than 200 usually. Table 1.2: Nexp needed to approximate t −α (t ∈ (τ,̂ T )) with different parameters α, τ, ϵ, when T = 1, τ ̂ = στ. τ

α

ϵ

10−3

10−4

10−5

10−6

10−7

0.1

10−6 10−8 10−10 10−12 10−14 10−6 10−8 10−10 10−12 10−14 10−6 10−8 10−10 10−12 10−14

31 40 48 58 66 32 41 50 59 68 33 43 51 61 70

37 48 57 68 78 38 49 59 70 80 39 50 60 72 82

43 55 66 79 90 43 56 68 80 92 45 57 69 82 95

48 62 75 89 103 49 63 77 91 105 50 65 78 93 107

54 69 84 100 115 55 70 85 102 117 56 72 87 104 119

0.5

0.9

Reformulating the fractional derivative as the sum of the integrals over the subintervals and approximating f (t) using L2,k (t) (1 ⩽ k ⩽ n − 1) and L1,n (t), respectively, by Lemma 1.7.1, we have C α 0 Dt f (t)|t=tn−1+σ

tn−1+σ

tk

n−1 1 f 󸀠 (t) f 󸀠 (t) = [∑ ∫ dt + ∫ dt] α Γ(1 − α) k=1 (tn−1+σ − t) (tn−1+σ − t)α tk−1 tk

≈

tn−1

Nexp

n−1 1 [ ∑ ∫ L󸀠2,k (t)( ∑ ωl e−sl (tn−1+σ −t) )dt Γ(1 − α) k=1 l=1 tk−1

1.7 Fast interpolation approximations of Caputo fractional derivatives | 75 tn−1+σ

+ ∫ L󸀠1,n (t) (tn−1+σ − t)−α dt]

(1.131)

tn−1

tk

N

exp n−1 1 = [ ∑ ωl ( ∑ ∫ L󸀠2,k (t)e−sl (tn−1+σ −t) dt) Γ(1 − α) l=1 k=1

tk−1

tn−1+σ

+ ∫ L󸀠1,n (t)(tn−1+σ − t)−α dt] tn−1

N

=

exp σ 1−α τ−α 1 [ ∑ ωl Fln + (f (tn ) − f (tn−1 ))] Γ(1 − α) l=1 1−α

≡ ℱ △ατ f (tn−1+σ ), where

Fl1 = 0, 1 ⩽ l ⩽ Nexp , { { { { { t n−1 k { {F n = ∑ ∫ L󸀠 (t)e−sl (tn−1+σ −t) dt, { { 2,k { l k=1 t { k−1

1 ⩽ l ⩽ Nexp ,

n ⩾ 2.

It is noted that Fln can be evaluated by the following recursive relation: tk

n−1

Fln = ∑ ∫ L󸀠2,k (t)e−sl (tn−1+σ −t) dt k=1 t

k−1

tn−1

tk

n−2

= ∑ ∫ L󸀠2,k (t)e−sl (tn−1+σ −t) dt + ∫ L󸀠2,n−1 (t)e−sl (tn−1+σ −t) dt k=1 t

tn−2

k−1

=e

−sl τ

tn−1

tk

n−2

∑ ∫ k=1 t

L󸀠2,k (t)e−sl (tn−2+σ −t) dt

+ ∫ L󸀠2,n−1 (t)e−sl (tn−1+σ −t) dt

k−1

tn−1

= e−sl τ Fln−1 + ∫ [ tn−2

=

e−sl τ Fln−1

tn− 1 − t 2

τ

3

δt f n− 2 +

tn−2

t − tn− 3 τ

2

1

δt f n− 2 ]e−sl (tn−1+σ −t) dt

1

3 + [f (tn−1 ) − f (tn−2 )] ∫( − ξ )e−sl (σ+1−ξ )τ dξ 2 1

0

1 + [f (tn ) − f (tn−1 )] ∫(ξ − )e−sl (σ+1−ξ )τ dξ , 2 0

2 ⩽ n ⩽ N.

76 | 1 Fractional derivatives and numerical approximations Denote 1

3 Al = ∫( − ξ )e−sl (σ+1−ξ )τ dξ , 2 0

1

1 Bl = ∫(ξ − )e−sl (σ+1−ξ )τ dξ . 2 0

Obviously, both Al and Bl are positive. Hence we obtain the following algorithm: For n = 1, 2, . . . , N, compute N

exp { σ 1−α τ−α 1 ℱ α { { [ (f (tn ) − f (tn−1 ))], ωl Fln + △ f (t ) = ∑ { n−1+σ τ { Γ(1 − α) l=1 1−α { { { { 1 Fl = 0, 1 ⩽ l ⩽ Nexp , { { { { { { Fln = e−sl τ Fln−1 + Al [f (tn−1 ) − f (tn−2 )] + Bl [f (tn ) − f (tn−1 )], { { { 1 ⩽ l ⩽ Nexp , n ⩾ 2. {

(1.132) (1.133) (1.134)

The computational complexity of the algorithm (1.132)–(1.134) is O(NNexp ). When N is very large, the computational complexity of the algorithm is much smaller compared with L2-1σ algorithm (1.81), of which the computational complexity is O(N 2 ). For this reason, we call the algorithm (1.132)–(1.134) a fast algorithm based on L2-1σ interpolation approximation, or, a fast L2-1σ approximation. Denote d0(1,α) =

σ 1−α τ−α , Γ(2 − α)

(1.135)

when n ⩾ 2, N

exp { σ 1−α τ−α 1 { (n,α) { ω B + , d = ∑ { l l 0 { { Γ(1 − α) l=1 Γ(2 − α) { { { { { Nexp { { (n,α) 1 d = ∑ ωl (e−sl tk−1 Al + e−sl tk Bl ), { k { Γ(1 − α) { { l=1 { { { Nexp { { { 1 (n,α) { { ∑ ω e−sl tn−2 Al . { dn−1 = Γ(1 − α) l=1 l {

(1.136) 1 ⩽ k ⩽ n − 2,

A direct evaluation for (1.131) yields the following results: When n = 1, ℱ

=

△ατ

tσ

L󸀠1,1 (t) 1 dt f (tn−1+σ ) = ∫ Γ(1 − α) (tσ − t)α

1−α −α

t0

σ τ [f (t1 ) − f (t0 )] = d0(1,α) [f (t1 ) − f (t0 )]. Γ(2 − α)

(1.137) (1.138)

1.7 Fast interpolation approximations of Caputo fractional derivatives | 77

When n ⩾ 2, ℱ

△ατ f (tn−1+σ )

tk

=

N

exp n−1 1 [ ∑ ∫ ( ∑ ωl e−sl (tn−1+σ −t) ) L󸀠2,k (t)dt Γ(1 − α) k=1 l=1

tk−1

tn−1+σ

L󸀠1,n (t)

+ ∫

(tn−1+σ − t)α

tn−1

dt]

t

n−1 k Nexp 1 tk+ 1 − t 1 2 = [ ∑ ∫ ( ∑ ωl e−sl (tn−1+σ −t) )(δt f k− 2 Γ(1 − α) k=1 τ l=1 tk−1

+ δt f

t − tk− 1

k+ 21

2

τ

tn−1+σ

1

δt f n− 2 dt] )dt + ∫ (tn−1+σ − t)α tn−1

t1 N

=

exp t3 − t 1 1 {[∫ ∑ ωl e−sl (tn−1+σ −t) 2 dt]δt f 2 Γ(1 − α) τ l=1

t0

n−1

tk N exp

+ ∑ [ ∫ ∑ ωl e−sl (tn−1+σ −t)

tk+ 1 − t 2

τ

k=2 t l=1 k−1 tk−1 N

exp

+ ∫ ∑ ωl e−sl (tn−1+σ −t)

t − tk− 3 τ

tk−2 l=1 tn−1 N

exp

+ [ ∫ ∑ ωl e−sl (tn−1+σ −t) tn−2

2

1

dt]δt f k− 2

t − tn− 3

l=1

tn−1+σ

2

τ

dt

dt

1

+ ∫ (tn−1+σ − t)−α dt]δt f n− 2 } tn−1 N

=

exp 1 ∑ ωl e−sl tn−2 Al [f (t1 ) − f (t0 )] Γ(1 − α) l=1

N

+

n−1 exp 1 ∑ ∑ ω (e−sl tn−k−1 Al + e−sl tn−k Bl ) [f (tk ) − f (tk−1 )] Γ(1 − α) k=2 l=1 l N

+( n

exp 1 σ 1−α τ−α )[f (tn ) − f (tn−1 )] ∑ ωl Bl + Γ(1 − α) l=1 Γ(2 − α)

(n,α) = ∑ dn−k [f (tk ) − f (tk−1 )]. k=1

78 | 1 Fractional derivatives and numerical approximations The following result is given in [101]. Theorem 1.7.2. Suppose f ∈ C 3 [t0 , tn ], then it holds 󵄨󵄨C α 󵄨󵄨 󵄨󵄨 D f (t)|t=t − ℱ △ατ f (tn−1+σ )󵄨󵄨󵄨 󵄨󵄨0 t n−1+σ 󵄨 5ϵtn (4σ − 1)σ −α 󵄨 󵄨󵄨 󸀠󸀠󸀠 󵄨󵄨 3−α 󵄨 max 󵄨f (t)󵄨󵄨τ + max 󵄨󵄨f 󸀠 (t)󵄨󵄨󵄨, ⩽ 12Γ(2 − α) t0 ⩽t⩽tn 󵄨 4Γ(1 − α) t0 ⩽t⩽tn 󵄨

(1.139)

where σ = 1 − α2 , 0 < α < 1. Proof. Denote Rn = C0 Dαt f (t)|t=tn−1+σ − ℱ △ατ f (tn−1+σ ). With the help of (1.131), we have Rn =

tn−1+σ

tk

n−1 1 f 󸀠 (t) f 󸀠 (t) [∑ ∫ dt + ∫ dt] α Γ(1 − α) k=1 (tn−1+σ − t) (tn−1+σ − t)α tk−1

tn−1

tk

−

Nexp

n−1 1 [ ∑ ∫ ( ∑ ωl e−sl (tn−1+σ −t) )L󸀠2,k (t)dt Γ(1 − α) k=1 l=1 tk−1

tn−1+σ

+ ∫ tn−1

L󸀠1,n (t)

(tn−1+σ − t)α

dt] tn−1+σ

tk

n−1 f 󸀠 (t) − L󸀠2,k (t) f 󸀠 (t) − L󸀠1,n (t) 1 [∑ ∫ dt + dt] = ∫ Γ(1 − α) k=1 (tn−1+σ − t)α (tn−1+σ − t)α tn−1

tk−1

tk

+

N

exp n−1 1 ∑ ∫ [(tn−1+σ − t)−α − ∑ ωl e−sl (tn−1+σ −t) ]L󸀠2,k (t)dt Γ(1 − α) k=1 l=1

tk−1

≡ In + IIn .

(1.140)

According to Theorem 1.6.4, we have |In | ⩽

(4σ − 1)σ −α 󵄨 󵄨 max 󵄨󵄨f 󸀠󸀠󸀠 (t)󵄨󵄨󵄨τ3−α . 12Γ(2 − α) t0 ⩽t⩽tn 󵄨

Using Lemma 1.7.1, we have t 󵄨󵄨 Nexp n−1 k 󵄨󵄨󵄨 󵄨󵄨 󵄨 1 󵄨 󵄨 |IIn | ⩽ ∑ ∫ 󵄨󵄨󵄨(tn−1+σ − t)−α − ∑ ωl e−sl (tn−1+σ −t) 󵄨󵄨󵄨 ⋅ 󵄨󵄨󵄨L󸀠2,k (t)󵄨󵄨󵄨dt 󵄨󵄨 Γ(1 − α) k=1 󵄨󵄨󵄨 l=1 󵄨 tk−1 tk

⩽

n−1 ϵ 󵄨 󵄨 ∑ ∫ 󵄨󵄨L󸀠 (t)󵄨󵄨dt Γ(1 − α) k=1 󵄨 2,k 󵄨 tk−1

(1.141)

1.7 Fast interpolation approximations of Caputo fractional derivatives | 79 t

n−1 k 󵄨󵄨 1 tk+ 1 − t 1 t − tk− 1 ϵ 󵄨 2 2 ⩽ + δt f k+ 2 ∑ ∫ 󵄨󵄨󵄨δt f k− 2 Γ(1 − α) k=1 󵄨󵄨 τ τ tk−1

t

󵄨󵄨 󵄨󵄨 󵄨󵄨dt 󵄨󵄨

n−1 k 󵄨󵄨 t 1 − t 󵄨󵄨 󵄨󵄨 t − t 1 1 k− 2 ϵ 󵄨 󵄨󵄨 󵄨󵄨 󵄨 󵄨 k+ (max 󵄨󵄨󵄨δt f k− 2 󵄨󵄨󵄨) ∑ ∫ (󵄨󵄨󵄨 2 ⩽ 󵄨󵄨 + 󵄨󵄨 󵄨 󵄨 󵄨 Γ(1 − α) 1⩽k⩽n 󵄨 τ 󵄨 󵄨 τ k=1 tk−1

󵄨󵄨 󵄨󵄨 󵄨󵄨)dt 󵄨󵄨

1 ϵ 5 󵄨 󵄨 (n − 1)τ max 󵄨󵄨󵄨δt f k− 2 󵄨󵄨󵄨 Γ(1 − α) 4 1⩽k⩽n 5ϵtn 󵄨 󵄨 ⩽ max 󵄨󵄨f 󸀠 (t)󵄨󵄨󵄨. 4Γ(1 − α) t0 ⩽t⩽tn 󵄨

=

(1.142)

Substituting (1.141) and (1.142) into (1.140), we have (1.139). This completes the proof. Lemma 1.7.3. [101] The coefficient {dk(n,α) | 0 ⩽ k ⩽ n − 1} defined by (1.135)–(1.138) satisfies the following relations: When n = 1, d0(n,α) > 0. (I)

When n ⩾ 2, (n,α) d1(n,α) > d2(n,α) > d3(n,α) > ⋅ ⋅ ⋅ > dn−1 ;

(II) If ϵ
0, d0(n,α) (n,α) dn−1

(1.144)

> 0, 1 ⩾ α . 2tn Γ(1 − α) >

d1(n,α)

(1.145) (1.146)

Proof. When n = 1, it is obvious that d0(n,α) > 0. When n ⩾ 2, (1.143) is true by using (1.137)–(1.138). In addition, (1.145) is true provided that (1.144) holds. Now we prove (1.144). τ−α As the beginning, a relation between dk(n,α) and Γ(2−α) ck(n,α) is illustrated. For n = 2, 3, . . ., we have 󵄨󵄨 󵄨󵄨 τ−α 󵄨󵄨 (n,α) 󵄨 ck(n,α) 󵄨󵄨󵄨 󵄨󵄨dk − 󵄨󵄨 󵄨󵄨 Γ(2 − α) t

|t−t

3

|

n− 1 { ∫ n−1 τ 2 dt, k = 0, { τ tn−2 { { { |t−tk− 3 | |t 1 −t| ϵ 2 ⋅ 1 ∫tk k+ 2 dt + 1 ∫tk−1 ⩽ dt, τ τ τ τ t t { Γ(1 − α) { k−1 k−2 { { { 1 t1 |t 32 −t| { τ ∫t0 τ dt, k = n − 1,

1 ⩽ k ⩽ n − 2,

80 | 1 Fractional derivatives and numerical approximations 1 , k = 0, { 4 { { ϵ 5 ⋅ 4 , 1 ⩽ k ⩽ n − 2, = { Γ(1 − α) { { {1, k = n − 1.

(1.147)

When n = 2, because of (2σ − 1)c0(2,α) − σc1(2,α) ⩾

(2σ − 1)(1 − σ) 2σ(1 + σ)α−1

and (1.147), we have (2σ − 1)d0(2,α) − σd1(2,α)

τ−α τ−α [(2σ − 1)c0(2,α) − σc1(2,α) ] + (2σ − 1)[d0(2,α) − c(2,α) ] Γ(2 − α) Γ(2 − α) 0 τ−α − σ[d1(2,α) − c(2,α) ] Γ(2 − α) 1 ϵ (2σ − 1)(1 − σ) ϵ τ−α − (2σ − 1) ⋅ −σ ⩾ Γ(2 − α) 2σ(1 + σ)α−1 4Γ(1 − α) Γ(1 − α) 1 τ−α (1 − σ) 6σ − 1 = [ − ϵ] α−1 Γ(1 − α) 2σ(1 + σ) 4

=

⩾ 0.

When n ⩾ 3, in view of (2σ − 1)c0(n,α) − σc1(n,α) ⩾

(2σ − 1)(1 − σ) 2σ(1 + σ)α

and (1.147), we have (2σ − 1)d0(n,α) − σd1(n,α)

τ−α τ−α [(2σ − 1)c0(n,α) − σc1(n,α) ] + (2σ − 1)[d0(n,α) − c(n,α) ] Γ(2 − α) Γ(2 − α) 0 τ−α − σ[d1(n,α) − c(n,α) ] Γ(2 − α) 1 (2σ − 1)(1 − σ) 5ϵ ϵ τ−α ⋅ −σ − (2σ − 1) ⩾ Γ(2 − α) 2σ(1 + σ)α 4Γ(1 − α) 4Γ(1 − α) 1 τ−α (1 − σ) 7σ − 1 = [ ϵ] − Γ(1 − α) 2σ(1 + σ)α 4

=

⩾ 0.

In addition, (n,α) dn−1 ⩾

τ−α ϵ c(n,α) − Γ(2 − α) n−1 Γ(1 − α)

1.7 Fast interpolation approximations of Caputo fractional derivatives | 81

τ−α ϵ ⋅ (1 − α)n−α − Γ(2 − α) Γ(1 − α) tn−α 1 (tn−α − ϵ) ⩾ . ⩾ Γ(1 − α) 2Γ(1 − α) ⩾

The proof ends. 1.7.3 Fast H2N2 approximation In what follows, we will give a fast algorithm for the H2N2 interpolation approximation of Caputo fractional derivative of order γ ∈ (1, 2)[71] . (γ−1) (γ−1) (γ−1) will be omitted for In this subsection, the superscript (γ − 1) in Nexp , sl , ωl brevity. Applying Lemma 1.7.1 and the quadratic interpolation polynomials H2,0 (t) and N2,k (t), we have C γ 0 Dt f (t)|t=tn− 1 t1

2

tk+ 1

2

2

n−1 1 = [ ∫ f 󸀠󸀠 (t)(tn− 1 − t)1−γ dt + ∑ ∫ f 󸀠󸀠 (t)(tn− 1 − t)1−γ dt] 2 2 Γ(2 − γ) k=1 t0

t1

2

≈

tk− 1

2

N

exp −sl (t 1 −t) 1 󸀠󸀠 [ ∫ H2,0 (t) ∑ ωl e n− 2 dt Γ(2 − γ) l=1

t0

tk+ 1

Nexp

2

n−2

󸀠󸀠 (t) ∑ ωl e + ∑ ∫ N2,k k=1 t

−sl (tn− 1 −t) 2

l=1

k− 21

dt

tn− 1

2

󸀠󸀠 + ∫ N2,n−1 (t)(tn− 1 − t)1−γ dt]

(1.148)

2

tn− 3 2

tn− 1

N

2

exp 1 = [ ∑ ωl Fln + δt2 f n−1 ∫ (tn− 1 − t)1−γ dt] 2 Γ(2 − γ) l=1

tn− 3

≡

ℱ

2

̂γ

D f (tn− 1 ), 2

2 ⩽ n ⩽ N,

(1.149)

where t1

Fln

2

=

−sl (t 1 −t) 󸀠󸀠 (t)e n− 2 dt ∫ H2,0

t0

n−2

tk+ 1

2

󸀠󸀠 + ∑ ∫ N2,k (t)e k=1 t

k− 21

−sl (tn− 1 −t) 2

dt,

82 | 1 Fractional derivatives and numerical approximations 1 ⩽ l ⩽ Nexp ,

2 ⩽ n ⩽ N.

The evaluation of Fln can be carried out using the following recursive algorithm: t1

2

󸀠󸀠 Fln = ∫ H2,0 (t)e

−sl (tn− 1 −t) 2

n−2

tk+ 1

2

󸀠󸀠 dt + ∑ ∫ N2,k (t)e

−sl (tn− 1 −t) 2

k=1 t

t0

dt

k− 21

t1

2

󸀠󸀠 = e−sl τ [ ∫ H2,0 (t)e

−sl (tn− 3 −t) 2

n−3

tk+ 1

2

󸀠󸀠 dt + ∑ ∫ N2,k (t)e

−sl (tn− 3 −t) 2

k=1 t

t0

dt]

k− 21

tn− 3 2

󸀠󸀠 (t)e + ∫ N2,n−2

−sl (tn− 1 −t) 2

dt

(1.150)

tn− 5 2

tn− 3

=

e−sl τ Fln−1

+

δt2 f n−2

2

∫ e

−sl (tn− 1 −t) 2

dt

tn− 5 2

=

e−sl τ Fln−1

+ Bl (δt f

n− 32

5

− δt f n− 2 ),

3 ⩽ n ⩽ N,

(1.151)

where t1

Fl2

2

󸀠󸀠 (t)e = ∫ H2,0

−sl (t 3 −t)

t0

2

dt =

1 3 2 −sl τ (e − e− 2 sl τ )(δt f 2 − f 󸀠 (t0 )), sl τ

(1.152)

tn− 3 2

−sl (t 1 −t) 1 1 −sl τ Bl = (e − e−2sl τ ). ∫ e n− 2 dt = τ sl τ

(1.153)

tn− 5 2

γ

From (1.149)–(1.153), the following algorithm is obtained to evaluate C0 Dt f (t)|t=t

n− 21

N

exp { 1 τ2−γ 2 n−1 n F ̂γ { { 1) = [ ω F + δ f ], 2 ⩽ n ⩽ N, D f (t ∑ { l l n− 2 { { Γ(2 − γ) l=1 2−γ t { { { { { t1 { 2 1 −sl (t 3 −t) 2 { 2 { 2 { Fl = ∫ e dt(δt f 2 − f 󸀠 (t0 )), 1 ⩽ l ⩽ Nexp , { { τ { { { t0 { { { 5 { n −sl τ n−1 n− 32 − δt f n− 2 ), 1 ⩽ l ⩽ Nexp , 3 ⩽ n ⩽ N. { Fl = e Fl + Bl (δt f

:

(1.154)

(1.155) (1.156)

In what follows, we try to analyze the truncation error by using ℱ D̂ γ f (tn− 1 ) to apγ

proximate C0 Dt f (t)|t=t

n− 21

.

2

1.7 Fast interpolation approximations of Caputo fractional derivatives | 83

Theorem 1.7.3. Let f ∈ C 3 [t0 , tn ]. Denote γ R̂ n = C0 Dt f (t)|t=t

n− 21

− ℱ D̂ γ f (tn− 1 ). 2

Then, when n ⩾ 2, γ−1 1 1 󵄨󵄨 ̂ n 󵄨󵄨 󵄨 󵄨 + + ] max 󵄨󵄨f 󸀠󸀠󸀠 (t)󵄨󵄨󵄨τ3−γ 󵄨󵄨R 󵄨󵄨 ⩽ [ 8Γ(2 − γ) 12Γ(3 − γ) 2Γ(4 − γ) t0 ⩽t⩽tn 󵄨 ϵ tn− 3 󵄨 󵄨 2 max 󵄨󵄨f 󸀠󸀠 (t)󵄨󵄨󵄨. + Γ(2 − γ) t0 ⩽t⩽tn 󵄨

(1.157)

Proof. Notice the fact that t1

tk+ 1

t0

tk− 1

2

2

n−1 1 [ ∫ f 󸀠󸀠 (t)(tn− 1 − t)1−γ dt + ∑ ∫ f 󸀠󸀠 (t)(tn− 1 − t)1−γ dt] R̂ n = 2 2 Γ(2 − γ) k=1 2

t1

N

2

−

exp −sl (t 1 −t) 1 󸀠󸀠 [ ∫ H2,0 (t) ∑ ωl e n− 2 dt Γ(2 − γ) l=1

t0

n−2

tk+ 1

Nexp

2

󸀠󸀠 + ∑ ∫ N2,k (t) ∑ ωl e k=1 t

−sl (tn− 1 −t) 2

l=1

k− 21

dt

tn− 1

2

󸀠󸀠 (t)(tn− 1 − t)1−γ dt] + ∫ N2,n−1 2

tn− 3 2

t1

2

1 󸀠󸀠 [ ∫ (f 󸀠󸀠 (t) − H2,0 (t))(tn− 1 − t)1−γ dt = 2 Γ(2 − γ) t0

n−1

tk+ 1

2

󸀠󸀠 (t))(tn− 1 − t)1−γ dt] + ∑ ∫ (f 󸀠󸀠 (t) − N2,k 2

k=1 t

k− 21

t1

N

2

exp −sl (t 1 −t) 1 󸀠󸀠 + [ ∫ H2,0 (t)((tn− 1 − t)1−γ − ∑ ωl e n− 2 )dt 2 Γ(2 − γ) l=1

t0

n−2

tk+ 1

Nexp

2

󸀠󸀠 + ∑ ∫ N2,k (t)((tn− 1 − t)1−γ − ∑ ωl e 2

k=1 t

k− 21

≡ In + IIn .

−sl (tn− 1 −t) 2

)dt]

l=1

(1.158)

84 | 1 Fractional derivatives and numerical approximations On the one hand, using Theorem 1.6.6, we have |In | ⩽ [

γ−1 1 1 󵄨 󵄨 + + ] max 󵄨󵄨f 󸀠󸀠󸀠 (t)󵄨󵄨󵄨τ3−γ . 8Γ(2 − γ) 12Γ(3 − γ) 2Γ(4 − γ) t0 ⩽t⩽tn 󵄨

(1.159)

On the other hand, combining (1.100)–(1.101) with (1.102)–(1.103), we have t1

tk+ 1

t0

tk− 1

t1

tk+ 1

t0

tk− 1

2

2

n−2 ϵ 󵄨 󸀠󸀠 󵄨󵄨 󵄨 󸀠󸀠 󵄨󵄨 |IIn | ⩽ [ ∫ 󵄨󵄨󵄨H2,0 (t)󵄨󵄨dt] (t)󵄨󵄨dt + ∑ ∫ 󵄨󵄨󵄨N2,k Γ(2 − γ) k=1 2

2

=

⩽

2

n−2 ϵ 󵄨 󵄨 󵄨 󵄨 [ ∫ 󵄨󵄨󵄨f 󸀠󸀠 (ζ0 )󵄨󵄨󵄨dt + ∑ ∫ 󵄨󵄨󵄨f 󸀠󸀠 (ζk )󵄨󵄨󵄨dt] Γ(2 − γ) k=1 2

ϵ tn− 3

󵄨 󵄨 2 max 󵄨󵄨f 󸀠󸀠 (t)󵄨󵄨󵄨, Γ(2 − γ) t0 ⩽t⩽tn 󵄨

(1.160)

where ζ0 ∈ (t0 , t1 ), ζk ∈ (tk−1 , tk+1 ), 1 ⩽ k ⩽ n−2. Using (1.158)–(1.160), we can get (1.157). The proof ends. Denote { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { {

tn− 1

2

1 τ1−γ (n,γ) , b̃ 0 = ∫ (tn− 1 − t)1−γ dt = 2 τ 2−γ

(1.161)

tn− 3 2

tk+ 1

2

1N

N

exp exp −sl (t 1 −t) 1 (n,γ) b̃ n−k−1 = ∫ ∑ ωl e n− 2 dt = ∫ ∑ ωl e−sl (n−k−ξ )τ dξ , τ l=1 l=1

tk− 1

0

2

t1

2

N

1 ⩽ k ⩽ n − 2, 1N

exp exp 1 ξ −sl (t 1 −t) 2 (n,γ) b̃ n−1 = ∫ ∑ ωl e n− 2 dt = ∫ ∑ ωl e−sl (n− 2 − 2 )τ dξ . τ l=1 l=1

t0

(1.162) (1.163)

0

Then it follows from (1.148) that ℱ

D̂ γ f (tn− 1 ) 2

n−1 1 1 1 1 (n,γ) (n,γ) = [b̃ n−1 ⋅ (δt f 2 − f 󸀠 (t0 )) + ∑ b̃ n−k−1 ⋅ (δt f k+ 2 − δt f k− 2 )] Γ(2 − γ) k=1

=

n−1 1 1 1 (n,γ) (n,γ) (n,γ) (n,γ) [b̃ 0 δt f n− 2 − ∑ (b̃ n−k−1 − b̃ n−k )δt f k− 2 − b̃ n−1 f 󸀠 (t0 )]. Γ(2 − γ) k=1

(1.164)

1.8 Finite difference methods for FODEs | 85

(n,γ) (n,γ) From the definitions of b̃ k and b̂ k , we have (n,γ) 󵄨 󵄨󵄨 ̃ (n,γ) 󵄨󵄨bk − b̂ k 󵄨󵄨󵄨 ⩽ ϵ,

(n,γ) (n,γ) b̃ 0 − b̂ 0 = 0;

1 ⩽ k ⩽ n − 1.

The coefficients in the formula (1.164) satisfy the following lemma. Lemma 1.7.4. (n,γ) (n,γ) (n,γ) (n,γ) b̃ 1 > b̃ 2 > b̃ 3 > ⋅ ⋅ ⋅ > b̃ n−1 .

When ϵ
b̃ 1 .

Proof. We rewrite (1.162) as tn−k− 1

1 (n,γ) b̃ k = ∫ τ

2

Nexp

∑ ωl e

tn−k− 3 2

−sl (tn− 1 −t) 2

l=1

dt

1 Nexp

= ∫ ∑ ωl e−sl (k+1−ξ )τ dξ ,

1 ⩽ k ⩽ n − 2.

0 l=1

(1.165)

Then the following inequalites (n,γ) (n,γ) (n,γ) b̃ 1 > b̃ 2 > ⋅ ⋅ ⋅ > b̃ n−1

are followed immediately by (1.165) and (1.163). In addition, (n,γ) (n,γ) (n,γ) (n,γ) b̃ 0 − b̃ 1 = b̂ 0 − b̃ 1

= (b̂ 0

(n,γ)

− b̂ 1

(n,γ)

) + (b̂ 1

(n,γ)

− b̃ 1

(n,γ)

)

(n,γ) (n,γ) ⩾ (b̂ 0 − b̂ 1 ) − ϵ

=

τ1−γ τ1−γ − (22−γ − 1) −ϵ 2−γ 2−γ

= (2 − 22−γ )

τ1−γ − ϵ > 0. 2−γ

This completes the proof.

1.8 Finite difference methods for FODEs 1.8.1 Method based on G-L approximation Problem 1.8.1. Solve the following initial value problem: {

α 0 Dt y(t)

= f (t),

y(0) = 0,

0 < t ⩽ T,

(1.166) (1.167)

86 | 1 Fractional derivatives and numerical approximations where α ∈ (0, 1). Define the function { { { { ̂ ={ u(t) { { { {

0, y(t), v(t), 0,

t < 0, 0 ⩽ t ⩽ T, T < t < 2T, t ⩾ 2T,

where v(t) is a smooth function satisfying v(k) (T) = y(k) (T) and v(k) (2T) = 0, k = 0, 1, 2. Suppose û ∈ C 1+α (ℛ). It should be pointed out that the aim of the introduction of function v(t) is only to provide a sufficient condition for using Theorem 1.4.2 without any practical calculation. Take a positive integer N and denote τ = NT , tk = kτ, k = 0, 1, 2, . . . , N. Considering equation (1.166) at t = tn , we have α 0 Dt y(t)|t=tn

= f (tn ),

1 ⩽ n ⩽ N.

It follows from the G-L formula (1.19) that n

τ−α ∑ gk(α) y(tn−k ) = f (tn ) + (r1 )n ,

1 ⩽ n ⩽ N,

k=0

(1.168)

where, by Theorem 1.4.2, there is a positive constant c1 such that 󵄨󵄨 n 󵄨󵄨 󵄨󵄨(r1 ) 󵄨󵄨 ⩽ c1 τ,

1 ⩽ n ⩽ N.

(1.169)

Noticing y(t0 ) = 0,

(1.170)

omitting the small term (r1 )n in (1.168) and replacing the exact solution y(tn ) with the numerical one yn produce the difference scheme for solving (1.166)–(1.167) as follows: n

−α (α) n−k { = f (tn ), { τ ∑ gk y k=0 { { { y0 = 0.

1 ⩽ n ⩽ N,

(1.171) (1.172)

The stability of the difference scheme Theorem 1.8.1. Suppose {yn | n = 0, 1, 2, . . . , N} is the solution of the difference scheme (1.171)–(1.172), then it holds 5 󵄨󵄨 k 󵄨󵄨 󵄨 󵄨 k α τα max 󵄨󵄨󵄨f (tm )󵄨󵄨󵄨, 󵄨󵄨y 󵄨󵄨 ⩽ (1 − α)2α 1⩽m⩽k

1 ⩽ k ⩽ N.

(1.173)

1.8 Finite difference methods for FODEs | 87

Proof. Noticing that when 0 < α < 1, g0(α) = 1, gk(α) < 0, k ⩾ 1, rewrite (1.171) as follows: n−1

yn = ∑ (−gk(α) )yn−k + τα f (tn ), k=1

1 ⩽ n ⩽ N.

(1.174)

Next, the induction method will be used to prove (1.173). When n = 1, it follows from (1.174) that 󵄨󵄨 1 󵄨󵄨 󵄨 α󵄨 󵄨󵄨y 󵄨󵄨 = τ 󵄨󵄨󵄨f (t1 )󵄨󵄨󵄨 ⩽

5 󵄨 󵄨 τα 󵄨󵄨f (t )󵄨󵄨. (1 − α)2α 󵄨 1 󵄨

Thus, (1.173) is true for k = 1. Now assume that (1.173) is true for k = 1, 2, . . . , n − 1, then it follows from (1.174) that n−1

󵄨󵄨 n 󵄨󵄨 󵄨 (α) 󵄨 n−k 󵄨 α󵄨 󵄨󵄨y 󵄨󵄨 ⩽ ∑ (−gk )󵄨󵄨󵄨y 󵄨󵄨󵄨 + τ 󵄨󵄨󵄨f (tn )󵄨󵄨󵄨 k=1

n−1

⩽ ∑ (−gk(α) )[ k=1

n−1

⩽ [ ∑ (−gk(α) ) k=1

5 󵄨 󵄨 󵄨 󵄨 (n − k)α τα max 󵄨󵄨󵄨f (tm )󵄨󵄨󵄨] + τα 󵄨󵄨󵄨f (tn )󵄨󵄨󵄨 (1 − α)2α 1⩽m⩽n−k 5 󵄨 󵄨 nα + 1]τα max 󵄨󵄨󵄨f (tm )󵄨󵄨󵄨 1⩽m⩽n (1 − α)2α

∞

∞

k=1

k=n α

= {[ ∑ (−gk(α) ) − ∑ (−gk(α) )]

5 󵄨 󵄨 nα + 1}τα max 󵄨󵄨󵄨f (tm )󵄨󵄨󵄨 1⩽m⩽n (1 − α)2α

1−α 2 5 󵄨 󵄨 ⩽ {[1 − ⋅( ) ] nα + 1}τα max 󵄨󵄨󵄨f (tm )󵄨󵄨󵄨 1⩽m⩽n 5 n (1 − α)2α 5 󵄨 󵄨 nα τα max 󵄨󵄨󵄨f (tm )󵄨󵄨󵄨, = 1⩽m⩽n (1 − α)2α where Lemma 1.4.3 is used in the penultimate step above. Hence, (1.173) is also true for k = n. By the principle of induction, the theorem is true. The proof ends. The convergence of the difference scheme

Theorem 1.8.2. Suppose {y(tn ) | n = 0, 1, 2, . . . , N} and {yn | n = 0, 1, 2, . . . , N} are solutions of the problem (1.166)–(1.167) and the difference scheme (1.171)–(1.172), respectively. Let en = y(tn ) − yn ,

n = 0, 1, 2, . . . , N,

then it holds 5c1 󵄨󵄨 n 󵄨󵄨 T α τ, 󵄨󵄨e 󵄨󵄨 ⩽ (1 − α)2α

1 ⩽ n ⩽ N.

88 | 1 Fractional derivatives and numerical approximations Proof. Subtracting (1.171)–(1.172) from (1.168) and (1.170), respectively, yields the system of error equations as follows: n

{ { τ−α ∑ gk(α) en−k = (r1 )n , { k=0 { { e0 = 0.

1 ⩽ n ⩽ N,

Noticing (1.169), the application of Theorem 1.8.1 produces 5c1 5 󵄨󵄨 n 󵄨󵄨 󵄨 󵄨 nα τα max 󵄨󵄨󵄨(r1 )m 󵄨󵄨󵄨 ⩽ T α τ, 󵄨󵄨e 󵄨󵄨 ⩽ α 1⩽m⩽n (1 − α)2 (1 − α)2α

1 ⩽ n ⩽ N.

The proof ends. Problem 1.8.2. Solve the initial value problem {

γ 0 Dt y(t)

= f (t),

y(0) = 0,

0 < t ⩽ T,

y (0) = 0, 󸀠

(1.175) (1.176)

where γ ∈ (1, 2). Define the function { { { { { ̂ ={ u(t) { { { { {

0, y(t),

t < 0, 0 ⩽ t ⩽ T,

v(t), 0,

T < t < 2T, t ⩾ 2T,

where v(t) is a smooth function satisfying v(k) (T) = y(k) (T), v(k) (2T) = 0, k = 0, 1, 2, 3. Suppose û ∈ C 1+γ (ℛ) and y ∈ C 3 [0, T]. Let z(t) = y󸀠 (t),

α = γ − 1,

then from (1.175)–(1.176), the differential equation for z(t) is obtained as follows: {

α 0 Dt z(t)

= f (t),

0 < t ⩽ T,

z(0) = 0.

(1.178)

Denote Y n = y(tn ),

1 1 Y n− 2 = (Y n + Y n−1 ), 2

Z n = z(tn ), 0 ⩽ n ⩽ N, 1 1 δt Y n− 2 = (Y n − Y n−1 ), 1 ⩽ n ⩽ N. τ

Considering equation (1.177) at t = tn , we have α 0 Dt z(t)|t=tn

= f (tn ),

(1.177)

1 ⩽ n ⩽ N.

1.8 Finite difference methods for FODEs | 89

It follows from the G-L formula (1.19) that n

−α (α) n−k { = f (tn ) + (r2 )n , { τ ∑ gk Z k=0 { { 0 { Z = 0,

1 ⩽ n ⩽ N,

(1.179) (1.180)

where, by Theorem 1.4.2, there is a positive constant c2 such that 󵄨󵄨 n 󵄨󵄨 󵄨󵄨(r2 ) 󵄨󵄨 ⩽ c2 τ,

1 ⩽ n ⩽ N.

(1.181)

It follows from (1.179) and (1.180) that n−1

τ−α ∑ gk(α) k=0

1 Z n−k + Z n−k−1 1 = f n− 2 + [(r2 )n + (r2 )n−1 ], 2 2

1 ⩽ n ⩽ N,

1

where (r2 )0 = 0, f n− 2 = 21 [f (tn ) + f (tn−1 )]. Substituting 1 1 δt Y n−k− 2 = (Z n−k + Z n−k−1 ) + O(τ2 ) 2

into the equality above gives n−1

τ1−γ ∑ gk

(γ−1)

k=0

1

1

1

δt Y n−k− 2 = f n− 2 + (r3 )n− 2 ,

1 ⩽ n ⩽ N,

(1.182)

and there is a positive constant c3 such that 󵄨󵄨 n− 21 󵄨󵄨 󵄨󵄨(r3 ) 󵄨󵄨 ⩽ c3 τ,

1 ⩽ n ⩽ N.

(1.183)

Noticing y(t0 ) = 0,

(1.184)

n− 21

omitting the small term (r3 ) in (1.182) and replacing the exact solution y(tn ) with its n numerical one y arrive at the difference scheme for solving the problem (1.175)–(1.176) as follows: n−1

1 1 { (γ−1) { { τ1−γ ∑ gk δt yn−k− 2 = f n− 2 , k=0 { { { 0 { y = 0.

1 ⩽ n ⩽ N,

(1.185) (1.186)

The stability of the difference scheme

Theorem 1.8.3. Suppose {yn | n = 0, 1, 2, . . . , N} is the solution of the difference scheme (1.185)–(1.186), then it holds 1 5 γ−1 󵄨󵄨 k− 21 󵄨󵄨 󵄨 󵄨 tk max 󵄨󵄨󵄨f m− 2 󵄨󵄨󵄨, 󵄨󵄨δt y 󵄨󵄨 ⩽ γ−1 1⩽m⩽k (2 − γ)2

1⩽k⩽N

(1.187)

90 | 1 Fractional derivatives and numerical approximations and 1 5 γ 󵄨 󵄨󵄨 k 󵄨󵄨 󵄨 tk max 󵄨󵄨󵄨f m− 2 󵄨󵄨󵄨, 󵄨󵄨y 󵄨󵄨 ⩽ γ−1 1⩽m⩽k (2 − γ)2

1 ⩽ k ⩽ N.

(1.188)

Proof. The inequality (1.187) can be obtained similarly to Theorem 1.8.1. Noticing k

k

1

1

yk = y0 + τ ∑ δt ym− 2 = τ ∑ δt ym− 2 , m=1

m=1

the application of the inequality (1.187) will produce (1.188). The proof ends. The convergence of the difference scheme Theorem 1.8.4. Suppose {y(tn ) | n = 0, 1, 2, . . . , N} and {yn | n = 0, 1, 2, . . . , N} are solutions of the problem (1.175)–(1.176) and the difference scheme (1.185)–(1.186), respectively. Let en = y(tn ) − yn ,

n = 0, 1, 2, . . . , N,

then it holds 5c3 󵄨󵄨 n 󵄨󵄨 T γ τ, 󵄨󵄨e 󵄨󵄨 ⩽ (2 − γ)2γ−1

1 ⩽ n ⩽ N.

Proof. Subtracting (1.185)–(1.186) from (1.182) and (1.184), respectively, yields the system of error equations as follows: n−1

1 1 { { τ1−γ ∑ gk(γ−1) δt en−k− 2 = (r3 )n− 2 , { k=0 { { e0 = 0.

1 ⩽ n ⩽ N,

Noticing (1.183), the application of Theorem 1.8.3 arrives at 1 5c3 5 󵄨 󵄨󵄨 n 󵄨󵄨 󵄨 tnγ max 󵄨󵄨󵄨(r3 )m− 2 󵄨󵄨󵄨 ⩽ T γ τ, 󵄨󵄨e 󵄨󵄨 ⩽ γ−1 1⩽m⩽n (2 − γ)2 (2 − γ)2γ−1

1 ⩽ n ⩽ N.

The proof ends. Problem 1.8.3. Solve the boundary value problem { where γ ∈ (1, 2).

γ 0 Dt y(t)

= f (t),

y(0) = 0,

0 < t < T,

y(T) = B,

(1.189) (1.190)

1.8 Finite difference methods for FODEs | 91

Define the function { { { { ̂ ={ u(t) { { { {

0, y(t), v(t), 0,

t < 0, 0 ⩽ t ⩽ T, T < t ⩽ 2T, t > 2T,

where v(t) is a smooth function satisfying v(k) (T) = y(k) (T), v(k) (2T) = 0, k = 0, 1, 2, 3 and suppose û ∈ C 1+γ (ℛ). Considering equation (1.189) at t = tn , we have γ 0 Dt y(t)|t=tn

= f (tn ),

1 ⩽ n ⩽ N − 1.

It follows from the shifted G-L formula (1.19) that n+1

τ−γ ∑ gk y(tn−k+1 ) = f (tn ) + (r4 )n , (γ)

k=0

1 ⩽ n ⩽ N − 1,

(1.191)

where, by Theorem 1.4.2, there is a positive constant c4 such that 󵄨󵄨 n󵄨 󵄨󵄨(r4 ) 󵄨󵄨󵄨 ⩽ c4 τ,

1 ⩽ n ⩽ N − 1.

(1.192)

y(tN ) = B,

(1.193)

Noticing the boundary conditions y(t0 ) = 0,

omitting the small term (r4 )n in (1.191) and replacing the exact solution y(tn ) with its numerical one yn arrive at the difference scheme for solving (1.189)–(1.190) as follows: n+1

{ (γ) { { τ−γ ∑ gk yn−k+1 = f (tn ), k=0 { { { 0 N { y = 0, y = B.

1 ⩽ n ⩽ N − 1,

(1.194) (1.195)

The stability of the difference scheme

Theorem 1.8.5. Suppose {yn | n = 0, 1, 2, . . . , N} is the solution of the difference scheme n+1

{ (γ) { { τ−γ ∑ gk yn−k+1 = f (tn ), k=0 { { { 0 N { y = 0, y = 0, then it holds

1 ⩽ n ⩽ N − 1,

(1.197) γ

‖y‖∞ ⩽

45 T ( ) ‖f ‖∞ , (γ − 1)(2 − γ)(3 − γ) 4

where 󵄨 󵄨 ‖y‖∞ = max 󵄨󵄨󵄨yn 󵄨󵄨󵄨, 1⩽n⩽N−1

(1.196)

󵄨 󵄨 ‖f ‖∞ = max 󵄨󵄨󵄨f (tn )󵄨󵄨󵄨. 1⩽n⩽N−1

(1.198)

92 | 1 Fractional derivatives and numerical approximations Proof. Noticing that g0 = 1, g1 (1.196) as follows: (γ)

(γ)

n

= −γ, g2

> g3

(γ)

(γ)

> ⋅ ⋅ ⋅ > 0 when 1 < γ < 2, rewrite

(−g1 )yn = ∑ gk yn−k+1 − τγ f (tn ), (γ)

(γ)

k=0 k =1̸

1 ⩽ n ⩽ N − 1.

(1.199)

Suppose ‖y‖∞ = |yn0 |, where n0 ∈ {1, 2, . . . , N − 1}. Letting n = n0 in (1.199) and taking the absolute values of the both hand sides, an application of the triangle inequality yields n0

(−g1 )‖y‖∞ ⩽ ∑ gk ‖y‖∞ + τγ ‖f ‖∞ (γ)

(γ)

k=0 k =1̸

N−1

⩽ ∑ gk ‖y‖∞ + τγ ‖f ‖∞ , (γ)

k=0 k =1̸

that is, N−1

(− ∑ gk )‖y‖∞ ⩽ τγ ‖f ‖∞ . (γ)

k=0

In view of ∞

∑ gk = 0, (γ)

k=0

it follows N−1

∞

− ∑ gk = ∑ gk > 0. (γ)

k=0

(γ)

k=N

Thus, ‖y‖∞ ⩽

τγ ∑∞ k=N

gk

(γ)

‖f ‖∞ .

The application of Lemma 1.4.4 gives ‖y‖∞ ⩽ = The proof ends.

τγ

(γ−1)(2−γ)(3−γ) 4 γ (N ) 45

‖f ‖∞ γ

45 T ( ) ‖f ‖∞ . (γ − 1)(2 − γ)(3 − γ) 4

1.8 Finite difference methods for FODEs | 93

The convergence of the difference scheme

Theorem 1.8.6. Suppose {y(tn ) | n = 0, 1, . . . , N} and {yn | n = 0, 1, . . . , N} are solutions of the problem (1.189)–(1.190) and the difference scheme (1.194)–(1.195), respectively. Let en = y(tn ) − yn ,

n = 0, 1, . . . , N,

then it holds γ

‖e‖∞ ⩽

45 T ( ) c4 τ. (γ − 1)(2 − γ)(3 − γ) 4

Proof. Subtracting (1.194)–(1.195) from (1.191) and (1.193), respectively, yields the system of error equations as follows: n+1

{ { τ−γ ∑ gk(γ) en−k+1 = (r4 )n , { k=0 { 0 { e = 0, eN = 0.

1 ⩽ n ⩽ N − 1,

Noticing (1.192), the application of Theorem 1.8.5 gives γ

‖e‖∞ ⩽

T 45 ( ) (γ − 1)(2 − γ)(3 − γ) 4

γ

⩽

󵄨 󵄨 max 󵄨󵄨󵄨(r4 )n 󵄨󵄨󵄨

1⩽n⩽N−1

45 T ( ) c4 τ. (γ − 1)(2 − γ)(3 − γ) 4

The proof ends. 1.8.2 Method based on L1 approximation Problem 1.8.4. Solve the initial value problem {

C α 0 Dt y(t)

= f (t),

0 < t ⩽ T,

y(0) = A,

(1.201)

where α ∈ (0, 1). Suppose y ∈ C 2 [0, T]. Considering equation (1.200) at t = tn , we have C α 0 Dt y(t)|t=tn

= f (tn ),

(1.200)

1 ⩽ n ⩽ N.

It follows from Theorem 1.6.1 that n−1 τ−α (α) (α) (α) [a(α) 0 y(tn ) − ∑ (an−k−1 − an−k )y(tk ) − an−1 y(t0 )] Γ(2 − α) k=1

94 | 1 Fractional derivatives and numerical approximations = f (tn ) + (r5 )n ,

1 ⩽ n ⩽ N,

(1.202)

and there is a positive constant c5 such that 󵄨󵄨 n 󵄨󵄨 2−α 󵄨󵄨(r5 ) 󵄨󵄨 ⩽ c5 τ ,

1 ⩽ n ⩽ N.

(1.203)

Noticing y(t0 ) = A,

(1.204)

omitting the small term (r5 )n in (1.202) and replacing the exact solution y(tn ) with the numerical one yn arrive at the difference scheme for solving (1.200)–(1.201) as follows: n−1 τ−α { n 0 { { [a(α) y − − a(α) )yk − a(α) ∑ (a(α) n−1 y ] = f (tn ), 0 n−k−1 n−k Γ(2 − α) { k=1 { { 0 y = A. {

1 ⩽ n ⩽ N, (1.205) (1.206)

The stability of the difference scheme

Theorem 1.8.7. Suppose {yn | n = 0, 1, 2, . . . , N} is the solution of the difference scheme (1.205)–(1.206), then it holds 󵄨󵄨 k 󵄨󵄨 󵄨󵄨 0 󵄨󵄨 󵄨α 󵄨 󵄨󵄨y 󵄨󵄨 ⩽ 󵄨󵄨y 󵄨󵄨 + Γ(1 − α) max󵄨󵄨󵄨tl f (tl )󵄨󵄨󵄨, 1⩽l⩽k

1 ⩽ k ⩽ N.

(1.207)

Proof. Reformulate (1.205) as follows: n−1

n (α) (α) k (α) 0 α a(α) 0 y = ∑ (an−k−1 − an−k )y + an−1 y + τ Γ(2 − α)f (tn ) k=1

n−1

0 − a(α) )yk + a(α) = ∑ (a(α) n−1 [y + n−k n−k−1 k=1

τα

a(α) n−1

Γ(2 − α)f (tn )].

Taking the absolute value on both hand sides of the equality above, the application of Lemma 1.6.1, the triangle inequality and τα

a(α) n−1

Γ(2 − α) ⩽

τα nα Γ(2 − α) = tnα Γ(1 − α) 1−α

lead to n−1

󵄨󵄨 n 󵄨󵄨 (α) (α) 󵄨󵄨 k 󵄨󵄨 a(α) 0 󵄨󵄨y 󵄨󵄨 ⩽ ∑ (an−k−1 − an−k )󵄨󵄨y 󵄨󵄨 k=1

󵄨󵄨 0 󵄨󵄨 α 󵄨󵄨 󵄨󵄨 + a(α) n−1 [󵄨󵄨y 󵄨󵄨 + tn Γ(1 − α)󵄨󵄨f (tn )󵄨󵄨],

1 ⩽ n ⩽ N.

Next the induction method will be applied to prove the truth of (1.207).

(1.208)

1.8 Finite difference methods for FODEs | 95

When n = 1, it follows from (1.208) that 󵄨󵄨 󵄨󵄨 󵄨󵄨 1 󵄨󵄨 α (α) 󵄨󵄨 0 󵄨󵄨 a(α) 0 󵄨󵄨y 󵄨󵄨 ⩽ a0 (󵄨󵄨y 󵄨󵄨 + τ Γ(1 − α)󵄨󵄨f (t1 )󵄨󵄨). Obviously, (1.207) is true for k = 1. Now suppose (1.207) is true for k = 1, 2, . . . , n − 1, then it follows from (1.208) that n−1

󵄨󵄨 n 󵄨󵄨 󵄨󵄨 α 󵄨󵄨 󵄨󵄨 0 󵄨󵄨 (α) (α) a(α) 0 󵄨󵄨y 󵄨󵄨 ⩽ ∑ (an−k−1 − an−k )[󵄨󵄨y 󵄨󵄨 + Γ(1 − α) max󵄨󵄨tl f (tl )󵄨󵄨] 1⩽l⩽k

k=1

󵄨󵄨 0 󵄨󵄨 α 󵄨󵄨 󵄨󵄨 + a(α) n−1 [󵄨󵄨y 󵄨󵄨 + tn Γ(1 − α)󵄨󵄨f (tn )󵄨󵄨] n−1

󵄨󵄨 0 󵄨󵄨 󵄨󵄨 α 󵄨󵄨 ) + a(α) − a(α) ⩽ { ∑ (a(α) n−1 }[󵄨󵄨y 󵄨󵄨 + Γ(1 − α) max󵄨󵄨tl f (tl )󵄨󵄨] n−k n−k−1 1⩽l⩽n

k=1

󵄨󵄨 0 󵄨󵄨 󵄨󵄨 α 󵄨󵄨 = a(α) 0 [󵄨󵄨y 󵄨󵄨 + Γ(1 − α) max󵄨󵄨tl f (tl )󵄨󵄨]. 1⩽l⩽n

Therefore, 󵄨󵄨 n 󵄨󵄨 󵄨󵄨 0 󵄨󵄨 󵄨α 󵄨 󵄨󵄨y 󵄨󵄨 ⩽ 󵄨󵄨y 󵄨󵄨 + Γ(1 − α) max󵄨󵄨󵄨tl f (tl )󵄨󵄨󵄨, 1⩽l⩽n hence, (1.207) is also true for k = n. By the principle of induction, the theorem is proved. The proof ends. The convergence of the difference scheme

Theorem 1.8.8. Suppose {y(tn ) | n = 0, 1, 2, . . . , N} and {yn | n = 0, 1, 2, . . . , N} are solutions of the problem (1.200)–(1.201) and the difference scheme (1.205)–(1.206), respectively. Let en = y(tn ) − yn ,

n = 0, 1, 2, . . . , N,

then it holds 󵄨󵄨 n 󵄨󵄨 α 2−α 󵄨󵄨e 󵄨󵄨 ⩽ c5 T Γ(1 − α)τ ,

1 ⩽ n ⩽ N.

Proof. Subtracting (1.205)–(1.206) from (1.202) and (1.204), respectively, yields the system of error equations as follows: n−1 τ−α 0 n n { { [a(α) e − − a(α) )ek − a(α) ∑ (a(α) n−1 e ] = (r5 ) , 0 n−k n−k−1 Γ(2 − α) { k=1 { { e0 = 0.

1 ⩽ n ⩽ N,

Noticing (1.203), the application of Theorem 1.8.7 gives 󵄨󵄨 n 󵄨󵄨 󵄨󵄨 0 󵄨󵄨 󵄨 l󵄨 α α 2−α 󵄨󵄨e 󵄨󵄨 ⩽ 󵄨󵄨e 󵄨󵄨 + T Γ(1 − α) max󵄨󵄨󵄨(r5 ) 󵄨󵄨󵄨 ⩽ c5 T Γ(1 − α)τ , 1⩽l⩽n The proof ends.

1 ⩽ n ⩽ N.

96 | 1 Fractional derivatives and numerical approximations Problem 1.8.5. Solve the initial value problem C γ 0 Dt y(t)

{

= f (t),

y(0) = A,

0 < t ⩽ T,

(1.209)

y (0) = B,

(1.210)

󸀠

where γ ∈ (1, 2). Suppose y ∈ C 3 [0, T]. Considering equation (1.209) at t = tn , we have C γ 0 Dt y(t)|t=tn

= f (tn ),

0 ⩽ n ⩽ N,

hence, 1 C γ 1 γ [ D y(t)|t=tn + C0 Dt y(t)|t=tn−1 ] = [f (tn ) + f (tn−1 )], 2 0 t 2

1 ⩽ n ⩽ N.

Denote Y n = y(tn ). It follows from Theorem 1.6.2 that n−1 1 1 τ1−γ (γ) (γ) (γ) (γ) [b0 δt Y n− 2 − ∑ (bn−k−1 − bn−k )δt Y k− 2 − bn−1 B] Γ(3 − γ) k=1

1 1 = [f (tn ) + f (tn−1 )] + (r6 )n− 2 , 2

1 ⩽ n ⩽ N,

(1.211)

and there is a positive constant c6 such that 󵄨󵄨 n− 21 󵄨󵄨 3−γ 󵄨󵄨(r6 ) 󵄨󵄨 ⩽ c6 τ ,

1 ⩽ n ⩽ N.

(1.212)

Noticing the initial value condition Y 0 = A,

(1.213)

1

omitting the small term (r6 )n− 2 in (1.211) and replacing the exact solution Y n with its numerical one yn arrive at the difference scheme for solving (1.209)–(1.210) as follows: n−1 1 1 τ1−γ (γ) (γ) (γ) (γ) n− 21 { { [b δ y − ∑ (bn−k−1 − bn−k )δt yk− 2 − bn−1 B] = f n− 2 , { t { { Γ(3 − γ) 0 k=1 { { { 1 ⩽ n ⩽ N, { { 0 { y = A,

where f

n− 21

(1.214) (1.215)

= 21 [f (tn ) + f (tn−1 )].

The stability of the difference scheme

Theorem 1.8.9. Suppose {yn | n = 0, 1, 2, . . . , N} is the solution of the difference scheme (1.214)–(1.215), then it holds 󵄨󵄨 n 󵄨󵄨 󵄨 γ−1 l− 1 󵄨 󵄨󵄨y 󵄨󵄨 ⩽ |A| + T[|B| + Γ(2 − γ) max󵄨󵄨󵄨tl f 2 󵄨󵄨󵄨], 1⩽l⩽n

1 ⩽ n ⩽ N.

1.8 Finite difference methods for FODEs | 97

Proof. Rewrite (1.214) as follows: n−1

1

1

1

b0 δt yn− 2 = ∑ (bn−k−1 − bn−k )δt yk− 2 + bn−1 B + τγ−1 Γ(3 − γ)f n− 2 (γ)

(γ)

(γ)

(γ)

k=1

n−1

1

= ∑ (bn−k−1 − bn−k )δt yk− 2 + bn−1 [B + (γ)

(γ)

(γ)

k=1

τγ−1 (γ) bn−1

1

Γ(3 − γ)f n− 2 ].

Taking the absolute value on both hand sides of the equality above and noticing τγ−1 (γ) bn−1

Γ(3 − γ) ⩽

τγ−1 nγ−1 Γ(3 − γ) = tnγ−1 Γ(2 − γ), 2−γ

the application of the triangle inequality yields n−1

1 1 (γ) 󵄨 (γ) 󵄨 (γ) 󵄨 󵄨 b0 󵄨󵄨󵄨δt yn− 2 󵄨󵄨󵄨 ⩽ ∑ (bn−k−1 − bn−k )󵄨󵄨󵄨δt yk− 2 󵄨󵄨󵄨

k=1

(γ) 󵄨 1󵄨 + bn−1 [|B| + tnγ−1 Γ(2 − γ)󵄨󵄨󵄨f n− 2 󵄨󵄨󵄨],

1 ⩽ n ⩽ N.

(1.216)

Next, the method of induction will be used to show the truth of 󵄨󵄨 k− 21 󵄨󵄨 󵄨 γ−1 l− 1 󵄨 󵄨󵄨δt y 󵄨󵄨 ⩽ |B| + Γ(2 − γ) max󵄨󵄨󵄨tl f 2 󵄨󵄨󵄨, 1⩽l⩽k

1 ⩽ k ⩽ N.

(1.217)

The result of (1.216) with n = 1 reveals that (1.217) is true for k = 1. Now suppose (1.217) is true for k = 1, 2, . . . , n − 1, it follows from (1.216) that 1 (γ) 󵄨 󵄨 b0 󵄨󵄨󵄨δt yn− 2 󵄨󵄨󵄨

n−1

(γ) (γ) 󵄨 γ−1 1 󵄨 ⩽ ∑ (bn−k−1 − bn−k )[|B| + Γ(2 − γ) max󵄨󵄨󵄨tl f l− 2 󵄨󵄨󵄨] 1⩽l⩽k

k=1

(γ) 󵄨 + bn−1 [|B| + tnγ−1 Γ(2 − γ)󵄨󵄨󵄨f

n− 21

󵄨󵄨 󵄨󵄨]

n−1

1

(γ) (γ) (γ) 󵄨 γ−1 󵄨 ⩽ [ ∑ (bn−k−1 − bn−k ) + bn−1 ] ⋅ [|B| + Γ(2 − γ) max󵄨󵄨󵄨tl f l− 2 󵄨󵄨󵄨] 1⩽l⩽n

k=1

(γ) 󵄨 γ−1 = b0 [|B| + Γ(2 − γ) max󵄨󵄨󵄨tl f 1⩽l⩽n

l− 21

󵄨󵄨 󵄨󵄨].

Therefore, 󵄨 γ−1 l− 1 󵄨 󵄨󵄨 n− 21 󵄨󵄨 󵄨󵄨δt y 󵄨󵄨 ⩽ |B| + Γ(2 − γ) max󵄨󵄨󵄨tl f 2 󵄨󵄨󵄨, 1⩽l⩽n which is precisely the result of (1.217) with k = n. Noticing n

1

yn = y0 + τ ∑ δt yk− 2 , k=1

98 | 1 Fractional derivatives and numerical approximations it follows from (1.217) that n

󵄨 k− 1 󵄨 󵄨󵄨 n 󵄨󵄨 󵄨󵄨 0 󵄨󵄨 󵄨󵄨y 󵄨󵄨 ⩽ 󵄨󵄨y 󵄨󵄨 + τ ∑ 󵄨󵄨󵄨δt y 2 󵄨󵄨󵄨 k=1 n

1

󵄨 󵄨 γ−1 ⩽ |A| + τ ∑ [|B| + Γ(2 − γ) max󵄨󵄨󵄨tl f l− 2 󵄨󵄨󵄨] 1⩽l⩽k

k=1

1

󵄨 γ−1 󵄨 ⩽ |A| + T[|B| + Γ(2 − γ) max󵄨󵄨󵄨tl f l− 2 󵄨󵄨󵄨], 1⩽l⩽n

1 ⩽ n ⩽ N.

The proof ends. The convergence of the difference scheme Theorem 1.8.10. Suppose {y(tn ) | n = 0, 1, 2, . . . , N} and {yn | n = 0, 1, 2, . . . , N} are solutions of the problem (1.209)–(1.210) and the difference scheme (1.214)–(1.215), respectively. Let en = y(tn ) − yn ,

n = 0, 1, 2, . . . , N,

then it holds 󵄨󵄨 n 󵄨󵄨 γ 3−γ 󵄨󵄨e 󵄨󵄨 ⩽ c6 T Γ(2 − γ)τ ,

1 ⩽ n ⩽ N.

Proof. Subtracting (1.214)–(1.215) from (1.211) and (1.213), respectively, produces the system of error equations as follows: n−1 1−γ 1 1 1 (γ) (γ) (γ) { { τ [b0 δt en− 2 − ∑ (bn−k−1 − bn−k )δt ek− 2 ] = (r6 )n− 2 , Γ(3 − γ) { k=1 { { e0 = 0.

1 ⩽ n ⩽ N,

Noticing (1.212), the application of Theorem 1.8.9 yields 󵄨󵄨 n 󵄨󵄨 󵄨 l− 1 󵄨 γ γ 3−γ 󵄨󵄨e 󵄨󵄨 ⩽ T Γ(2 − γ) max󵄨󵄨󵄨(r6 ) 2 󵄨󵄨󵄨 ⩽ c6 T Γ(2 − γ)τ , 1⩽l⩽n

1 ⩽ n ⩽ N.

The proof ends.

1.8.3 Method based on L2-1σ approximation Consider another numerical method for solving the Problem 1.8.4, i. e., {

C α 0 Dt y(t)

= f (t),

y(0) = A,

0 < t ⩽ T,

(1.218) (1.219)

1.8 Finite difference methods for FODEs | 99

where α ∈ (0, 1). Suppose y ∈ C 3 [0, T]. Considering equation (1.218) at t = tn−1+σ , by Theorem 1.6.4, we have τ−α n−1 (n,α) [y(tn−k ) − y(tn−k−1 )] = f (tn−1+σ ) + (r7 )n , ∑c Γ(2 − α) k=0 k

1⩽n⩽N

(1.220)

and there is a positive constant c7 such that 󵄨󵄨 n 󵄨󵄨 3−α 󵄨󵄨(r7 ) 󵄨󵄨 ⩽ c7 τ ,

1 ⩽ n ⩽ N.

(1.221)

Noticing the initial value condition (1.219) and omitting the small term (r7 )n in (1.220), a difference scheme for solving (1.218)–(1.219) can be derived as follows: τ−α n−1 (n,α) n−k { { − yn−k−1 ) = f (tn−1+σ ), ∑ ck (y { Γ(2 − α) k=0 { { { 0 y = A. {

1 ⩽ n ⩽ N,

(1.222) (1.223)

The stability of the difference scheme

Theorem 1.8.11. Suppose {yn | n = 0, 1, . . . , N} is the solution of the difference scheme τ−α n−1 (n,α) n−k { { − yn−k−1 ) = g n , ∑ ck (y { Γ(2 − α) k=0 { { { 0 { y = A,

1 ⩽ n ⩽ N,

(1.225)

then it holds 󵄨󵄨 k 󵄨󵄨 󵄨󵄨 0 󵄨󵄨 󵄨 α m󵄨 󵄨󵄨y 󵄨󵄨 ⩽ 󵄨󵄨y 󵄨󵄨 + Γ(1 − α) max 󵄨󵄨󵄨tm g 󵄨󵄨󵄨, 1⩽m⩽k

1 ⩽ k ⩽ N.

Proof. Denote s = τα Γ(2 − α). It follows from Lemma 1.6.3 that s

(n,α) cn−1

⩽

nα τα Γ(2 − α) = tnα Γ(1 − α). 1−α

Reformulate (1.224) as n−1

n−1

c0(n,α) yn = ∑ ck(n,α) yn−k−1 − ∑ ck(n,α) yn−k + sg n k=0

k=1

n

n−1

k=1

k=1

(n,α) n−k = ∑ ck−1 y − ∑ ck(n,α) yn−k + sg n n−1

(n,α) (n,α) 0 = ∑ (ck−1 − ck(n,α) )yn−k + cn−1 y + sg n , k=1

(1.224)

1 ⩽ n ⩽ N.

100 | 1 Fractional derivatives and numerical approximations Taking the absolute values on both hand sides of the equality above and noticing Lemma 1.6.3, the application of the triangle inequality yields n−1

󵄨 󵄨 󵄨 n󵄨 󵄨 󵄨 (n,α) (n,α) 󵄨󵄨 0 󵄨󵄨 c0(n,α) 󵄨󵄨󵄨yn 󵄨󵄨󵄨 ⩽ ∑ (ck−1 − ck(n,α) )󵄨󵄨󵄨yn−k 󵄨󵄨󵄨 + cn−1 󵄨󵄨y 󵄨󵄨 + s󵄨󵄨󵄨g 󵄨󵄨󵄨 k=1

n−1

󵄨 󵄨 (n,α) (n,α) 󵄨󵄨 0 󵄨󵄨 = ∑ (ck−1 − ck(n,α) )󵄨󵄨󵄨yn−k 󵄨󵄨󵄨 + cn−1 (󵄨󵄨y 󵄨󵄨 + k=1

n−1

s 󵄨󵄨 n 󵄨󵄨 󵄨󵄨g 󵄨󵄨)

(n,α) cn−1

󵄨 󵄨 󵄨 󵄨 (n,α) (n,α) 󵄨󵄨 0 󵄨󵄨 ⩽ ∑ (ck−1 − ck(n,α) )󵄨󵄨󵄨yn−k 󵄨󵄨󵄨 + cn−1 (󵄨󵄨y 󵄨󵄨 + Γ(1 − α)tnα 󵄨󵄨󵄨g n 󵄨󵄨󵄨), k=1

1 ⩽ n ⩽ N.

(1.226)

The following result 󵄨󵄨 k 󵄨󵄨 󵄨󵄨 0 󵄨󵄨 󵄨 α m󵄨 󵄨󵄨y 󵄨󵄨 ⩽ 󵄨󵄨y 󵄨󵄨 + Γ(1 − α) max 󵄨󵄨󵄨tm g 󵄨󵄨󵄨, 1⩽m⩽k

1⩽k⩽N

can be obtained from (1.226) by the method of induction. The proof ends. The convergence of the difference scheme Theorem 1.8.12. Suppose {y(tn ) | n = 0, 1, . . . , N} and {yn | n = 0, 1, . . . , N} are solutions of the problem (1.218)–(1.219) and the difference scheme (1.222)–(1.223), respectively. Let en = y(tn ) − yn ,

0 ⩽ n ⩽ N,

then it holds 󵄨󵄨 n 󵄨󵄨 α 3−α 󵄨󵄨e 󵄨󵄨 ⩽ c7 T Γ(1 − α)τ ,

1 ⩽ n ⩽ N.

Proof. The system of error equations is τ−α n−1 (n,α) n−k { { − en−k−1 ) = (r7 )n , ∑ ck (e Γ(2 − α) { k=0 { { e0 = 0.

1 ⩽ n ⩽ N,

Noticing (1.221) and applying Theorem 1.8.11 will easily produce the desired result. The proof ends. Next, we try to solve Problem 1.8.5, i. e., {

C γ 0 Dt y(t)

= f (t),

y(0) = A,

using the L2-1σ method, where γ ∈ (1, 2).

0 < t ⩽ T,

y (0) = B 󸀠

(1.227) (1.228)

1.9 A simple classification of the fractional partial differential equations | 101

Let z(t) = y󸀠 (t), α = γ − 1, then the problem (1.227)–(1.228) for y(t) can be converted to the initial value problem for z(t) as follows: {

C α 0 Dt z(t)

= f (t),

0 < t ⩽ T,

z(0) = B.

(1.229) (1.230)

It can be found that (1.229)–(1.230) is in the same form with Problem 1.8.4, which can be solved using the L2-1σ method. Once the approximate values z n of z(tn ), n = 0, 1, 2, . . . , N are obtained, with the help of the composite trapezoid formula or composite Simpson formula in the numerical integral, the approximate values yn of y(tn ), n = 0, 1, 2, . . . , N are available in view of tn

tn

y(tn ) = y(t0 ) + ∫ y󸀠 (η)dη = A + ∫ z(η)dη. t0

t0

1.9 A simple classification of the fractional partial differential equations Li et al. made a simple classification for the FPDEs in their review article[45] . First, consider the time FPDE: C α 0 Dt u(x, t)

= uxx (x, t),

where the time-fractional derivative is often defined in Caputo sense. According to the value of α, a categorization can be given in Table 1.3. Table 1.3: The classification of C0 Dαt u(x, t) = uxx (x, t) with α ∈ (0, 2]. α (0, 1) 1 (1, 2) 2

Math. type

Phys. sense

Time-fractional parabolic equation

Temporal subdiffusion

Parabolic equation

Diffusion

Time-fractional hyperbolic equation

Temporal superdiffusion or temporal fractional wave

Hyperbolic equation

Wave

Second, consider the space FPDE: ut (x, t) =

𝜕β u(x, t) , 𝜕|x|β

102 | 1 Fractional derivatives and numerical approximations β

u(x,t) where the space-fractional derivative 𝜕 𝜕|x| is often in Riemann–Liouville sense or β Riesz sense. According to the value of β, a categorization can be described as that in Table 1.4.

Table 1.4: The classification of ut (x, t) = β (0, 1) 1 (1, 2) 2

𝜕β u(x,t) 𝜕|x|β

with β ∈ (1, 2].

Math. type

Phys. sense

Space-fractional hyperbolic equation

Fractional advection

Hyperbolic equation

Advection

Space-fractional parabolic equation

Fractional diffusion

Parabolic equation

Diffusion

Finally, consider the time-space FPDE: C α 0 Dt u(x, t)

=

𝜕β u(x, t) , 𝜕|x|β

where the time derivative is in Caputo sense, and the space derivative is in Riesz sense, or in other distinct senses, such as R-L sense, fractional Laplacian sense, etc. A classification can be formed as that in Table 1.5. Table 1.5: The classification of C0 Dαt u(x, t) = α

β

(0, 1)

(1,2)

𝜕β u(x,t) 𝜕|x|β

with α ∈ (0, 2), β ∈ (0, 2).

Math. type

Phys. sense

(0, 1)

Time-space-fractional hyperbolic equation

Temporal subdiffusion and fractional advection

(1, 2)

Time-space-fractional parabolic equation

Temporal subdiffusion and fractional diffusion

(0, 1)

Space-time-fractional parabolic equation

Temporal superdiffusion and fractional advection

(1, 2)

Time-space-fractional hyperbolic equation

Temporal superdiffusion and fractional diffusion

In this book, we mainly present the finite difference methods for three types of FPDEs, namely time-fractional, space-fractional and time-space-fractional PDEs. In Chapter 2, we study the finite difference methods for the initial-boundary value problem of the time-fractional subdiffusion equation C α 0 Dt u(x, t)

= uxx (x, t) + f (x, t),

1.10 Supplementary remarks and discussions | 103

where α ∈ (0, 1); Chapter 3 shows the finite difference methods for the initial-boundary value problem of the time-fractional wave equation C γ 0 Dt u(x, t)

= uxx (x, t) + f (x, t),

where γ ∈ (1, 2); In chapter 4, we introduce the finite difference methods for the initialboundary value problem of the space-fractional partial differential equation β

ut (x, t) = K1 0 Dβx u(x, t) + K2 x DL u(x, t) + f (x, t), where β ∈ (1, 2); Chapter 5 considers the finite difference methods for the initialboundary value problem of the time-space-fractional differential equation C α 0 Dt u(x, t)

=

𝜕β u(x, t) + f (x, t), 𝜕|x|β

where α ∈ (0, 1), β ∈ (1, 2); In chapter 6, the finite difference methods are concerned for solving a class of initial-boundary value problem of time distributed-order subdiffusion equation w

𝒟t u(x, t) = uxx (x, t) + f (x, t), 1

where 𝒟tw u(x, t) = ∫0 w(α) C0 Dαt u(x, t)dα. The two dimensional problems are also considered in each chapter.

1.10 Supplementary remarks and discussions 1. In this chapter, four kinds of definitions of fractional derivatives were introduced. The analytical solutions to the linear FODEs with two types of fractional derivatives were described, which provides readers a general idea on the characteristics of solutions to fractional differential equations (FDEs). Regarding the definitions and properties of fractional derivatives, readers can refer to [63]. Fractional Laplace operators were not covered in this book. Interested readers may refer to [44]. 2. Several numerical ways to approximate the fractional derivatives were introduced in this chapter, together with their numerical accuracy and applications into solving FODEs. 3. The G-L fractional derivative is in a limit form, which is equivalent to the R-L fractional derivative. Hence it is natural to approximate the R-L fractional derivative using the G-L formula with an appropriate step size. The standard G-L formula has the accuracy of order one[58] and when it is directly used to solve the spacefractional differential equations, the resultant difference scheme is unstable[58] . Then the shifted G-L formula was proposed by Meerschaert and Tadjeran, the asymptotic

104 | 1 Fractional derivatives and numerical approximations expansion for which was derived by Tadjeran et al. in [86]. Two research teams, conducted by Deng and Sun, respectively, discussed a series of high-order numerical differentiation formulae with the aid of different weighted combinations of shifted G-L formulae[35, 88, 117] . It should be noted that for these high-order formulae, the appropriate smoothness of functions is required. Generally speaking, the possible highest order accuracy that an approximation can reach depends on the regularity of the function. For the Riesz fractional derivative, the asymptotic expansion of the central difference quotient formula[14, 15, 116] can be similarly obtained using the method in [35, 88, 117]. The asymptotic expansion of numerical differentiation formulae is the powerful tool to derive various high-order formulae. 4. The G-L approximation of Riemann-Liouville fractional derivatives is considered on the whole domain ℛ. When the G-L formula was used to solve the FDEs on bounded domains, the solutions to FDEs have to be extended to be defined on the whole domain ℛ and the extended solution is supposed to have the certain smoothness. If the condition is not satisfied, the consistency and convergence of the corresponding difference schemes cannot be ensured. Suppose p > 0. Define the function f (x) = {

xp , 0,

x ⩾ 0, x < 0.

Denote xi = ih. Noticing α 0 Dx f (x)

=

Γ(p + 1) p−α x , Γ(p + 1 − α)

a direct calculation gives α 0 Dx f (x1 ) α 0 Dx f (x1 )

1

− h−α ∑ gk(α) f (x1−k ) = [ 2

k=0

− h−α ∑ gk(α) f (x2−k ) = [ k=0

Γ(p + 1) − 1]hp−α , Γ(p + 1 − α)

Γ(p + 1) − (2p − α)]hp−α . Γ(p + 1 − α)

If p = 1, α = 0.5, then when x ⩾ 0, f (x) = x, both the G-L formula and the shifted G-L formula have only the accuracy of order O(h1/2 ). If p = 0, α = 0.5, these two formulae only have the accuracy of order O(h−1/2 ). Therefore, both the G-L formula and the shifted G-L formula are not uniformly convergent in order one. For some problems, the approximation errors may be large and a completely inconsistent difference scheme may be produced using these two formulae; hence, the solution of difference schemes may not be convergent to the solution of the differential equations. For the L1 formula, when 0 < α < 1, the second-order derivative of function f is required to be continuous. Otherwise, the expecting result cannot be obtained. 5. The Caputo fractional derivative is actually an integral with a weak singular kernel. It is a quite natural idea[61] to use the piecewise linear interpolation polynomial

1.10 Supplementary remarks and discussions | 105

to establish a weighted numerical integral and the resultant formula is called the L1 formula or L1 approximation. A rough estimation on the numerical accuracy of this formula is only the first order[2, 105, 119] . In [99], Wu and Sun proved that the L1 formula for the half-order fractional derivative has the accuracy of order 3/2. Later, in [84], they proved that for the α-th (α ∈ (0, 1)) order Caputo fractional derivative, the numerical accuracy of the L1 formula is order of 2−α, where the rigorous expression of errors was derived. Combining with the method of order reduction, the numerical differentiation formula for the γ-th (γ ∈ (1, 2)) order Caputo fractional derivative is also established, which has the numerical accuracy of order 3 − γ. Lin and Xu[50] also proved that the numerical accuracy of the L1 formula to approximate the α-th (α ∈ (0, 1)) order Caputo fractional derivative is order of 2 − α using the series method. In addition, some improved methods were investigated in [1, 31, 113]. 6. About the estimation on the coefficient {gk(α) }∞ k=0 in the G-L formula, where α ∈ 1−α 2 α (α) (0, 1), the inequality ∑∞ |g | ⩾ ( ) has been provided in Lemma 1.4.3. In fact, n=k n 5 k ∞ 1 (α) the similar result ∑n=k |gn | ⩾ kα Γ(1−α) can also be obtained and used to make the analysis on the related difference schemes. More details can be found in [9]. 7. Alikhanov[1] developed an L2-1σ method for the fractional derivative of order α (α ∈ (0, 1)), which improved L1-2 method[31] . The authors of [19] provided the L2-1σ method for the multi-term fractional derivatives. The authors of [81, 83] used the method of order reduction to give an L2-1σ method for the fractional derivative of order γ (γ ∈ (1, 2)). In [16], the authors presented the L2-1σ approximation for the variable order fractional derivatives. The advantage of the L2-1σ method is that a temporal second-order convergent difference scheme can be obtained when applied to numerically solve the time-fractional differential equations. 8. H2N2 approximation for the fractional derivative of order γ (γ ∈ (1, 2)) was obtained directly by the quadratic interpolation polynomial[71] , which avoids using linear interpolation polynomial as an intermediate transition. The application of H2N2 formula to solve the fractional wave equation is similar to that of L1 formula to solve the fractional subdiffusion equation. 9. Fractional derivatives are historically memorized. The value of the current moment depends on the value of all moments since the initial time. It is necessary to find a fast calculation method. Many fast approximations can be obtained by applying the sum of exponential functions to approach the kernel of fractional derivatives[41] . In [41] and [101], the authors presented the fast L1 approximation and fast L2-1σ approximation for the Caputo derivative of order α (α ∈ (0, 1)), respectively. In [71], a fast algorithm based on H2N2 approximation for the fractional Caputo derivative of order γ (γ ∈ (1, 2)) was investigated. Gao and Yang[32] , Sun and Sun[79] discussed the fast L2-1σ approximations for the multi-term time Caputo derivatives of orders belonging to (0, 1) and (1, 2), respectively. For the fast algorithm of Riemann-Liouville derivative, please refer to [78].

106 | 1 Fractional derivatives and numerical approximations 10. The numerical approximation of fractional derivatives with the initial singularity has been paid some attention in recent years. Stynes et al. [74] gave a numerical formula on the graded mesh. Shen et al. [72] further provided a fast approximation for this kind of fractional derivatives.

Exercises 1 1.1 Compute (1) a Dαt (t − a)−1/2 ; (2) a Dαt (t − a)1/2 ; (3) a Dαt (t − a)2 ; (4) C0 Dαt t 1/2 ; (5) C0 Dαt t 2 ; α

2

(t−t ) (6) 𝜕 𝜕|t| α . 1.2 Suppose the function f (t) can be expanded into the following Taylor series:

f (n) (0) n t , n! n=0 ∞

f (t) = ∑

with the convergence radius R, R > 0, f (0) ≠ 0. Solve the following problem: α 0 Dt y(t)

{

+ t 1−α y(t) = f (t),

t > 0,

α ∈ (0, 1),

y(0) = 0.

1.3 Suppose the function f (t) can be expanded into f (n) (0) n t , n! n=0 ∞

f (t) = t 2−γ ∑

with the convergence radius R, R > 0, f (0) ≠ 0. Solve the following problem: {

C γ 0 Dt y(t)

= f (t),

y(0) = A,

t > 0,

γ ∈ (1, 2),

y (0) = B. 󸀠

1.4 Suppose f ∈ C 2+α (ℛ). Try to prove Aαh,p f (t) = c1(α,p) −∞ Dαt f (t + h) + (1 − c1(α,p) )−∞ Dαt f (t) + O(h2 ). 1.5 Suppose f ∈ C 3 [t0 , tn ], γ ∈ (1, 2), α = γ − 1, g(t) = f 󸀠 (t). Try to estimate Pn , Q1 , Q2 , . . . , Qn , Rn in the following equalities: C γ 0 Dt f (t)|t=tn− 1 t1

2

2

= C0 Dαt g(t)|t=t

n− 21

tk+ 1

2

n−1 1 = [ ∫ g 󸀠 (t)(tn− 1 − t)−α dt + ∑ ∫ g 󸀠 (t)(tn− 1 − t)−α dt] 2 2 Γ(1 − α) k=1 t0

tk− 1

2

Exercises 1

| 107

t1

2

2 1 [ (g(t 1 ) − g(t0 )) ⋅ ∫ (tn− 1 − t)−α dt = 2 2 Γ(1 − α) τ t0

tk+ 1

2

n−1

1 + ∑ (g(tk+ 1 ) − g(tk− 1 )) ⋅ ∫ (tn− 1 − t)−α dt] + Pn 2 2 2 τ k=1 tk− 1

2

=

n−1 τ (n,γ) (n,γ) [b̂ n−1 ⋅ (g(t 1 ) − f 󸀠 (t0 )) + ∑ b̂ n−k−1 ⋅ (g(tk+ 1 ) − g(tk− 1 ))] + Pn 2 2 2 Γ(3 − γ) k=1

=

n−1 τ1−γ (n,γ) (n,γ) (n,γ) (n,γ) [b̂ 0 g(tn− 1 ) − ∑ (b̂ n−k−1 − b̂ n−k )g(tk− 1 ) − b̂ n−1 f 󸀠 (t0 )] + Pn 2 2 Γ(3 − γ) k=1

=

1 τ1−γ (n,γ) [b̂ 0 (δt f n− 2 + Qn ) Γ(3 − γ)

1−γ

n−1

1

(n,γ) (n,γ) (n,γ) − ∑ (b̂ n−k−1 − b̂ n−k )(δt f k− 2 + Qk ) − b̂ n−1 f 󸀠 (t0 )] + Pn k=1

=

n−1 1 1 τ1−γ (n,γ) (n,γ) (n,γ) (n,γ) [b̂ 0 δt f n− 2 − ∑ (b̂ n−k−1 − b̂ n−k )δt f k− 2 − b̂ n−1 f 󸀠 (t0 )] + Rn . Γ(3 − γ) k=1

2 Difference methods for the time-fractional subdiffusion equations It has been found that there are no Gauss statistics for diffusion processes in many complex systems and the second Fick’s law is no longer true. Specially, the linearity between the mean square displacement of particles in Brown motion and the time variable in classical diffusion processes is not satisfied. This kind of diffusion is often called the anomalous diffusion, whose striking feature lies in the power law dependence of the mean square displacement of particles on the time variable. The timefractional diffusion-wave equation is a typical tool to depict the phenomenon, which is derived by replacing the first-order time derivative with the α-th order fractional derivative. It has been widely used in various fields such as physics, control, signal and image processing, mechanics and dynamic systems, biology, environmental science, materials, economic and multidisciplinary in engineering fields. When 0 < α < 1, the corresponding equation is called the time-fractional subdiffusion equation; When 1 < α < 2, it is called the time-fractional wave equation; For α = 1, the classical diffusion equation is recovered. In this book, Chapter 2 and Chapter 3 will introduce the difference methods for these two kinds of time-fractional differential equations, respectively, and the unique solvability, stability together with the convergence of the presented difference schemes will be shown. In this chapter, for the 1D time-fractional subdiffusion equation, three distinct ways, G-L formula, L1 formula and L2-1σ formula, will be used to approximate the time-fractional derivative, respectively, and the spatial derivative will be handled with the second-order or compact approximation. The fast L1 approximation method and fast L2-1σ approximation method are presented. The L1 formula and L2-1σ formula are also mentioned for a class of multiterm time-fractional subdiffusion problems. Finally, for the 2D problem, the alternating direction implicit (ADI) difference method will be described. The whole chapter is divided into 12 sections.

2.1 The second-order method in space based on G-L approximation for 1D problem Consider the following initial-boundary value problem of the time-fractional subdiffusion equations: C α D u(x, t) = uxx (x, t) + f (x, t), { { { 0 t u(x, 0) = 0, x ∈ (0, L), { { { { u(0, t) = μ(t), u(L, t) = ν(t),

x ∈ (0, L), t ∈ (0, T],

(2.1) (2.2)

t ∈ [0, T],

(2.3)

where α ∈ (0, 1), the functions f , μ, ν are all given and μ(0) = ν(0) = 0. The mesh partition is firstly done. Take two positive integers M and N. Let h = L/M, τ = T/N. Denote xi = ih (0 ⩽ i ⩽ M), tk = kτ (0 ⩽ k ⩽ N), Ωh = {xi | 0 ⩽ i ⩽ M}, https://doi.org/10.1515/9783110616064-002

110 | 2 Difference methods for the time-fractional subdiffusion equations Ωτ = {tk | 0 ⩽ k ⩽ N}. Define the following mesh function spaces: 𝒰h = {u | u = (u0 , u1 , . . . , uM )},

𝒰h̊ = {u | u ∈ 𝒰h , u0 = uM = 0}.

For any mesh function u ∈ 𝒰h , denote 1 1 (ui − ui−1 ), δx2 ui = 2 (ui+1 − 2ui + ui−1 ), h h 1 { 12 (ui−1 + 10ui + ui+1 ), 1 ⩽ i ⩽ M − 1, (𝒜u)i = { i = 0, M. { ui ,

δx ui− 1 = 2

2

It is easy to know that (𝒜u)i = (ℐ + h12 δx2 )ui for 1 ⩽ i ⩽ M − 1, with ℐ an identity operator. For brevity, henceforth denote (𝒜u)i by 𝒜ui . For any mesh functions u, v ∈ 𝒰h̊ , define the following inner products and norms: M−1

(u, v) = h ∑ ui vi ,

‖u‖ = √(u, u),

i=1

M

(δx u, δx v) = h ∑(δx ui− 1 )(δx vi− 1 ), 2

i=1

2

M−1

(δx2 u, δx2 v) = h ∑ (δx2 ui )(δx2 vi ), i=1

‖δx u‖ = √(δx u, δx u), 󵄩󵄩 2 󵄩󵄩 √ 2 󵄩󵄩δx u󵄩󵄩 = (δx u, δx2 u),

‖u‖∞ = max |ui |. 0⩽i⩽M

Lemma 2.1.1.

[75]

For any mesh function u ∈ 𝒰h̊ , it holds √L L ‖δ u‖, ‖δ u‖, ‖u‖ ⩽ √6 x 2 x 2 1 ‖δx u‖ ⩽ ‖u‖, ‖u‖2 ⩽ ‖𝒜u‖2 ⩽ ‖u‖2 . h 3 ‖u‖∞ ⩽

Suppose u, v ∈ 𝒰h̊ . Let M−1

I(u, v) ≡ (𝒜u, −δx2 v) = −h ∑ (𝒜ui )δx2 vi . i=1

Then M−1

I(u, v) = −h ∑ (ui + M

i=1

h2 2 δ u )δ2 v 12 x i x i

= h ∑(δx ui− 1 )(δx vi− 1 ) − i=1

2

2

h2 M−1 2 h ∑ (δ u )(δ2 v ) 12 i=1 x i x i

2.1 The second-order method in space based on G-L approximation for 1D problem

= (δx u, δx v) − It is easy to verify

| 111

h2 2 (δ u, δx2 v). 12 x

2 ‖δ u‖2 ⩽ I(u, u) ⩽ ‖δx u‖2 . 3 x

(2.4)

Hence I(u, v) is an inner product defined on 𝒰h̊ . Denote (u, v)1,A = I(u, v),

‖δx u‖A = √(u, u)1,A .

The following lemma is easily deduced from (2.4). Lemma 2.1.2. For any mesh function u ∈ 𝒰h̊ , we have 2 ‖δ u‖2 ⩽ ‖δx u‖2A ⩽ ‖δx u‖2 . 3 x Another lemma that follows is prepared for the spatial approximation. Lemma 2.1.3. [75] (I) If function g ∈ C 4 [xi−1 , xi+1 ], then g(xi−1 ) − 2g(xi ) + g(xi+1 ) h2

g 󸀠󸀠 (xi ) =

1

h2 − ∫[g (4) (xi + λh) + g (4) (xi − λh)](1 − λ)3 dλ. 6 0

(II) Denote ζ (λ) = (1 − λ)3 [5 − 3(1 − λ)2 ]. If function g ∈ C 6 [xi−1 , xi+1 ], then g 󸀠󸀠 (xi+1 ) + 10g 󸀠󸀠 (xi ) + g 󸀠󸀠 (xi−1 ) g(xi+1 ) − 2g(xi ) + g(xi−1 ) = 12 h2 1

h4 + ∫[g (6) (xi − λh) + g (6) (xi + λh)]ζ (λ)dλ. 360 0

Define mesh functions Uin = u(xi , tn ), fin = f (xi , tn ),

0 ⩽ i ⩽ M, 0 ⩽ n ⩽ N.

2.1.1 Derivation of the difference scheme For any fixed x ∈ [0, L], define a function { { { { { ̂ t) = { u(x, { { { { {

0, u(x, t),

t < 0, 0 ⩽ t ⩽ T,

v(x, t), 0,

T < t < 2T, t ⩾ 2T,

(2.5)

112 | 2 Difference methods for the time-fractional subdiffusion equations k

k

where v(x, t) is a smooth function satisfying 𝜕 v(x,t) |t=T = 𝜕 𝜕t k 1+α 4 ̂ ⋅) ∈ C (ℛ) and u(⋅, t) ∈ C [0, L]. k = 0, 1, 2. Suppose u(x, Considering equation (2.1) at the point (xi , tn ), one has C α 0 Dt u(xi , tn )

= uxx (xi , tn ) + fin ,

k u(x,t) |t=T , 𝜕 v(x,t) |t=2T 𝜕t k 𝜕t k

1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N.

= 0,

(2.6)

Noticing the relationship between the Caputo derivative and the R-L derivative under the zero initial condition (2.2) and applying Theorem 1.4.2, one can obtain C α 0 Dt u(xi , tn )

n

= 0 Dαt u(xi , tn ) = τ−α ∑ gk(α) Uin−k + O(τ). k=0

(2.7)

By Lemma 2.1.3, it is clear that uxx (xi , tn ) = δx2 Uin + O(h2 ).

(2.8)

Substituting (2.7) and (2.8) into (2.6) gives n

τ−α ∑ gk(α) Uin−k = δx2 Uin + fin + (r1 )ni , k=0

1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N,

(2.9)

where there is a positive constant c1 such that 󵄨󵄨 n 󵄨󵄨 2 󵄨󵄨(r1 )i 󵄨󵄨 ⩽ c1 (τ + h ),

1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N.

(2.10)

Noticing the initial-boundary value conditions (2.2)–(2.3), one has {

Ui0 = 0, U0n

1 ⩽ i ⩽ M − 1,

= μ(tn ),

n UM

= ν(tn ),

(2.11) 0 ⩽ n ⩽ N.

(2.12)

Omitting the small term (r1 )ni in (2.9) and replacing the exact solution Uin with its numerical one uni , a difference scheme for solving (2.1)–(2.3) can be produced as n

{ τ−α ∑ gk(α) un−k = δx2 uni + fin , 1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N, { i { { { k=0 { 0 { u = 0, 1 ⩽ i ⩽ M − 1, { { { i n n { u0 = μ(tn ), uM = ν(tn ), 0 ⩽ n ⩽ N.

(2.13) (2.14) (2.15)

Next, the unique solvability, unconditional stability and convergence of this difference scheme will be analyzed. 2.1.2 Solvability of the difference scheme Theorem 2.1.1. The difference scheme (2.13)–(2.15) is uniquely solvable.

2.1 The second-order method in space based on G-L approximation for 1D problem

| 113

Proof. Let un = (un0 , un1 , . . . , unM ). The value of u0 is obviously determined by (2.14)–(2.15). Assume that the values of u0 , u1 , . . . , un−1 have been uniquely determined, then the linear system in un can be obtained from (2.13) and (2.15). To show its unique solvability, it suffices to verify that the corresponding homogeneous one {

τ−α uni = δx2 uni , un0

=

unM

1 ⩽ i ⩽ M − 1,

(2.16)

=0

(2.17)

has only the trivial solution. Suppose ‖un ‖∞ = |unin |, where in ∈ {1, 2, . . . , M − 1}. Rewrite (2.16) as (1 + 2

τα n τα n )u = 2 (ui−1 + uni+1 ), h2 i h

1 ⩽ i ⩽ M − 1.

Letting i = in in the equality above and taking the absolute value of both hand sides, an application of the triangle inequality yields (1 + 2

τα 󵄩 󵄩 τα 󵄩󵄩 n 󵄩󵄩 )󵄩󵄩u 󵄩󵄩∞ ⩽ 2 2 󵄩󵄩󵄩un 󵄩󵄩󵄩∞ , 2 h h

hence ‖un ‖∞ = 0, which implies un = 0. By the principle of induction, the theorem is true. The proof ends.

2.1.3 Stability of the difference scheme Theorem 2.1.2. Suppose {vin | 0 ⩽ i ⩽ M, 0 ⩽ n ⩽ N} is the solution of the difference scheme n

{ τ−α ∑ gk(α) vin−k = δx2 vin + fin , 1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N, { { { { k=0 { 0 { v = φ(xi ), 1 ⩽ i ⩽ M − 1, { { { i n n { v0 = 0, vM = 0, 0 ⩽ n ⩽ N.

(2.18) (2.19) (2.20)

Then it holds 5 󵄩󵄩 0 󵄩󵄩 5 󵄩󵄩 k 󵄩󵄩 󵄩 󵄩 k α τα max 󵄩󵄩󵄩f m 󵄩󵄩󵄩∞ , 󵄩󵄩v 󵄩󵄩∞ ⩽ 󵄩󵄩v 󵄩󵄩∞ + 1−α (1 − α)2α 1⩽m⩽k where ‖f m ‖∞ = max1⩽i⩽M−1 |fim |.

1 ⩽ k ⩽ N,

(2.21)

114 | 2 Difference methods for the time-fractional subdiffusion equations Proof. Reformulate (2.18) as follows: (1 + 2

n τα n τα n (α) n−k n )v (v + vi+1 = (−g )v ) + τα fin , + ∑ i h2 i k=1 k h2 i−1

1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N.

(2.22)

Suppose ‖vn ‖∞ = |vinn |, where in ∈ {1, 2, . . . , M − 1}. Letting i = in in (2.22) and taking

the absolute value of both hand sides, noticing −gk(α) ⩾ 0 (1 ⩽ k ⩽ n) and using the triangle inequality, one can get (1 + 2

τα 󵄩󵄩 n 󵄩󵄩 )󵄩v 󵄩 h2 󵄩 󵄩∞

n τα 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 ⩽ ∑ (−gk(α) )󵄩󵄩󵄩vn−k 󵄩󵄩󵄩∞ + 2 (󵄩󵄩󵄩vn 󵄩󵄩󵄩∞ + 󵄩󵄩󵄩vn 󵄩󵄩󵄩∞ ) + τα 󵄩󵄩󵄩f n 󵄩󵄩󵄩∞ , h k=1

which simplifies to give n

󵄩󵄩 n 󵄩󵄩 (α) 󵄩 n−k 󵄩 α󵄩 n󵄩 󵄩󵄩v 󵄩󵄩∞ ⩽ ∑ (−gk )󵄩󵄩󵄩v 󵄩󵄩󵄩∞ + τ 󵄩󵄩󵄩f 󵄩󵄩󵄩∞ , k=1

1 ⩽ n ⩽ N.

(2.23)

Next, the mathematical induction method will be used to show the truth of (2.21). Let An =

5 5 󵄩󵄩 0 󵄩󵄩 󵄩 󵄩 nα τα max 󵄩󵄩󵄩f m 󵄩󵄩󵄩∞ , 󵄩󵄩v 󵄩󵄩∞ + 1⩽m⩽n 1−α (1 − α)2α

1 ⩽ n ⩽ N.

When n = 1, it follows from (2.23) that 󵄩󵄩 1 󵄩󵄩 󵄩 0󵄩 (α) 󵄩 0 󵄩 α󵄩 1󵄩 α󵄩 1󵄩 󵄩󵄩v 󵄩󵄩∞ ⩽ (−g1 )󵄩󵄩󵄩v 󵄩󵄩󵄩∞ + τ 󵄩󵄩󵄩f 󵄩󵄩󵄩∞ = α󵄩󵄩󵄩v 󵄩󵄩󵄩∞ + τ 󵄩󵄩󵄩f 󵄩󵄩󵄩∞ ⩽ A1 , that is, (2.21) is obvious for k = 1. Suppose (2.21) is true for k = 1, 2, . . . , n − 1(n ⩾ 2) and consider the case of k = n. By (2.23), one has n−1

󵄩󵄩 n 󵄩󵄩 (α) 󵄩 n−k 󵄩 (α) 󵄩 0 󵄩 α󵄩 n󵄩 󵄩󵄩v 󵄩󵄩∞ ⩽ ∑ (−gk )󵄩󵄩󵄩v 󵄩󵄩󵄩∞ + (−gn )󵄩󵄩󵄩v 󵄩󵄩󵄩∞ + τ 󵄩󵄩󵄩f 󵄩󵄩󵄩∞ k=1

n−1

⩽ ∑ (−gk(α) )An−k + α( k=1

α+1

2 ) n+1

󵄩󵄩 0 󵄩󵄩 α󵄩 n󵄩 󵄩󵄩v 󵄩󵄩∞ + τ 󵄩󵄩󵄩f 󵄩󵄩󵄩∞

α

n−1

2 󵄩 󵄩 󵄩 󵄩 ⩽ ∑ (−gk(α) )An + α( ) 󵄩󵄩󵄩v0 󵄩󵄩󵄩∞ + τα 󵄩󵄩󵄩f n 󵄩󵄩󵄩∞ n k=1 α

2 󵄩 󵄩 󵄩 󵄩 = [ ∑ (−gk(α) ) − ∑ (−gk(α) )]An + α( ) 󵄩󵄩󵄩v0 󵄩󵄩󵄩∞ + τα 󵄩󵄩󵄩f n 󵄩󵄩󵄩∞ n k=1 k=n ∞

∞

α

⩽ [1 −

α

1−α 2 2 󵄩 󵄩 󵄩 󵄩 ( ) ]An + α( ) 󵄩󵄩󵄩v0 󵄩󵄩󵄩∞ + τα 󵄩󵄩󵄩f n 󵄩󵄩󵄩∞ 5 n n

2.1 The second-order method in space based on G-L approximation for 1D problem

= An − ⩽ An ,

α

| 115

α

1−α 2 5 5α 󵄩󵄩 0 󵄩󵄩 n 󵄩 󵄩 ( ) [An − ( ) τα 󵄩󵄩󵄩f n 󵄩󵄩󵄩∞ ] 󵄩v 󵄩 − 5 n 1 − α 󵄩 󵄩∞ 1 − α 2

2 α+1 where Lemma 1.4.1, Lemma 1.4.3 and ( n+1 ) < ( n2 )α+1 ⩽ ( n2 )α when n ⩾ 2 have been used. So (2.21) is also true for k = n. By the principle of induction, (2.21) is true for k = 1, 2, . . . , N. The proof ends.

Theorem 2.1.2 says that the difference scheme (2.13)–(2.15) is unconditionally stable with respect to both the initial value and the source term.

2.1.4 Convergence of the difference scheme We are now ready to show the convergence of the difference scheme (2.13)–(2.15). Theorem 2.1.3. Suppose {Uin | 0 ⩽ i ⩽ M, 0 ⩽ n ⩽ N} and {uni | 0 ⩽ i ⩽ M, 0 ⩽ n ⩽ N} are solutions of the problem (2.1)–(2.3) and the difference scheme (2.13)–(2.15), respectively. Let ein = Uin − uni ,

0 ⩽ i ⩽ M, 0 ⩽ n ⩽ N.

Then it holds 󵄩󵄩 n 󵄩󵄩 󵄩󵄩e 󵄩󵄩∞ ⩽

5 T α c1 (τ + h2 ), (1 − α)2α

1 ⩽ n ⩽ N.

Proof. Subtracting (2.13)–(2.15) from (2.9), (2.11)–(2.12), respectively, the system of error equations is produced as n

{ τ−α ∑ gk(α) ein−k = δx2 ein + (r1 )ni , 1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N, { { { { k=0 { 0 { e = 0, 1 ⩽ i ⩽ M − 1, { { { i n n { e0 = 0, eM = 0, 0 ⩽ n ⩽ N. Noticing (2.10), by Theorem 2.1.2, it is easy to see that 5 󵄩 󵄩 nα τα max 󵄩󵄩󵄩(r1 )m 󵄩󵄩󵄩∞ 1⩽m⩽n (1 − α)2α 5 ⩽ T α c1 (τ + h2 ), 1 ⩽ n ⩽ N. (1 − α)2α

󵄩󵄩 n 󵄩󵄩 󵄩󵄩e 󵄩󵄩∞ ⩽

The proof is completed.

116 | 2 Difference methods for the time-fractional subdiffusion equations

2.2 The fourth-order method in space based on G-L approximation for 1D problem This section will explore a new fourth-order method in space for solving the problem (2.1)–(2.3). ̂ t) like that in Section 2.1 and suppose For any x ∈ [0, L], define a function u(x, ̂ ⋅) ∈ C 1+α (ℛ) and u(⋅, t) ∈ C 6 [0, L]. u(x, 2.2.1 Derivation of the difference scheme Considering equation (2.1) at the point (xi , tn ), one has C α 0 Dt u(xi , tn )

= uxx (xi , tn ) + fin ,

0 ⩽ i ⩽ M, 1 ⩽ n ⩽ N.

Performing the operator 𝒜 to both hand sides of the equality above yields C

α

n

𝒜 0 Dt u(xi , tn ) = 𝒜uxx (xi , tn ) + 𝒜fi ,

1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N.

By Theorem 1.4.2 and Lemma 2.1.3, noticing (2.2), one can obtain 𝒜(τ

−α

n

∑ gk(α) Uin−k ) = δx2 Uin + 𝒜fin + (r2 )ni ,

k=0

1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N,

(2.24)

where there exists a positive constant c2 such that 󵄨󵄨 n 󵄨󵄨 4 󵄨󵄨(r2 )i 󵄨󵄨 ⩽ c2 (τ + h ),

1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N.

(2.25)

Noticing the initial-boundary value conditions (2.2)–(2.3), one has {

Ui0 = 0,

U0n

1 ⩽ i ⩽ M − 1,

= μ(tn ),

n UM

= ν(tn ),

(2.26) 0 ⩽ n ⩽ N.

(2.27)

Neglecting the small term (r2 )ni in (2.24) and replacing the exact solution Uin with its numerical one uni , another difference scheme for solving (2.1)–(2.3) is obtained as n

2 n n −α (α) n−k { { 𝒜(τ ∑ gk ui ) = δx ui + 𝒜fi , { { { { k=0 { { 1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N, { { { { 0 { u = 0, 1 ⩽ i ⩽ M − 1, { { { i n n { u0 = μ(tn ), uM = ν(tn ), 0 ⩽ n ⩽ N.

Next, we will analyze the difference scheme (2.28)–(2.30).

(2.28) (2.29) (2.30)

2.2 The fourth-order method in space based on G-L approximation for 1D problem

| 117

2.2.2 Solvability of the difference scheme Theorem 2.2.1. The difference scheme (2.28)–(2.30) is uniquely solvable. Proof. Denote un = (un0 , un1 , . . . , unM ). The initial value u0 is uniquely determined by (2.29)–(2.30). Suppose the values of u0 , u1 , . . . , un−1 have been uniquely determined, then the linear system in un can be obtained from (2.28) and (2.30). To prove its unique solvability, it suffices to show that the corresponding homogeneous one {

τ−α 𝒜uni = δx2 uni ,

un0

=

unM

1 ⩽ i ⩽ M − 1,

(2.31)

=0

(2.32)

has only the trivial solution. To this end, taking the inner product on both hand sides of (2.31) with −δx2 un arrives at τ−α (𝒜un , −δx2 un ) = −(δx2 un , δx2 un ), that is 󵄩 󵄩2 󵄩 󵄩2 τ−α 󵄩󵄩󵄩δx un 󵄩󵄩󵄩A = −󵄩󵄩󵄩δx2 un 󵄩󵄩󵄩 ⩽ 0. Thus ‖δx un ‖A = 0 and moreover ‖δx un ‖ = 0 by Lemma 2.1.2. Noticing (2.32), it follows un = 0. By the principle of induction, the theorem is true. The proof ends. 2.2.3 Stability of the difference scheme Theorem 2.2.2. Suppose {vin | 0 ⩽ i ⩽ M, 0 ⩽ n ⩽ N} is the solution of difference scheme n

−α (α) n−k 2 n n { { 𝒜(τ ∑ gk vi ) = δx vi + gi , { { { { k=0 { { 1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N, { { { { 0 { v = φ(xi ), 1 ⩽ i ⩽ M − 1, { { { i n n { v0 = 0, vM = 0, 0 ⩽ n ⩽ N.

(2.33) (2.34) (2.35)

Then it holds 1 15 󵄩 󵄩2 󵄩 󵄩2 󵄩󵄩 n 󵄩󵄩2 (󵄩󵄩δ v0 󵄩󵄩 + α+1 tnα max 󵄩󵄩󵄩g m 󵄩󵄩󵄩 ), 󵄩󵄩δx v 󵄩󵄩 ⩽ 1⩽m⩽n 2(1 − α) 󵄩 x 󵄩 2

1 ⩽ n ⩽ N,

118 | 2 Difference methods for the time-fractional subdiffusion equations where M−1

󵄩󵄩 m 󵄩󵄩2 m 2 󵄩󵄩g 󵄩󵄩 = h ∑ (gi ) . i=1

Proof. Making the inner product on both hand sides of (2.33) with −δx2 vn and noticing (2.5) lead to n M−1 1 󵄩 󵄩2 󵄩 󵄩2 τ−α ∑ gk(α) (vn−k , vn )1,A = −󵄩󵄩󵄩δx2 vn 󵄩󵄩󵄩 − h ∑ (δx2 vin )gin ⩽ 󵄩󵄩󵄩g n 󵄩󵄩󵄩 , 4 i=1 k=0

which can be rewritten to produce n

g0(α) (vn , vn )1,A ⩽ ∑ (−gk(α) )(vn , vn−k )1,A + k=1

1 α 󵄩󵄩 n 󵄩󵄩2 τ 󵄩g 󵄩 4 󵄩 󵄩

1 n ⩽ ∑ (−gk(α) )[(vn , vn )1,A + (vn−k , vn−k )1,A ] 2 k=1 +

1 α 󵄩󵄩 n 󵄩󵄩2 τ 󵄩g 󵄩 , 4 󵄩 󵄩

1 ⩽ n ⩽ N.

Noticing ∑nk=1 (−gk(α) ) ⩽ g0(α) = 1, it follows n 1 󵄩 󵄩2 (vn , vn )1,A ⩽ ∑ (−gk(α) )(vn−k , vn−k )1,A + τα 󵄩󵄩󵄩g n 󵄩󵄩󵄩 , 2 k=1

1 ⩽ n ⩽ N,

that is, n 1 α 󵄩 n 󵄩2 󵄩󵄩 n 󵄩󵄩2 (α) 󵄩 n−k 󵄩2 󵄩󵄩δx v 󵄩󵄩A ⩽ ∑ (−gk )󵄩󵄩󵄩δx v 󵄩󵄩󵄩A + τ 󵄩󵄩󵄩g 󵄩󵄩󵄩 , 2 k=1

1 ⩽ n ⩽ N.

By induction, similar to the proof of Theorem 2.1.2, one can get 󵄩󵄩 n 󵄩󵄩2 󵄩󵄩δx v 󵄩󵄩A ⩽

1 5 5 󵄩󵄩 0 󵄩󵄩2 󵄩 󵄩2 t α ⋅ max 󵄩󵄩g m 󵄩󵄩 , 󵄩δ v 󵄩 + 1 − α 󵄩 x 󵄩A (1 − α)2α n 2 1⩽m⩽n󵄩 󵄩

1 ⩽ n ⩽ N.

The application of Lemma 2.1.2 into the inequality above reaches the desired result. The proof is completed. Theorem 2.2.2 shows that the difference scheme (2.28)–(2.30) is unconditionally stable with respect to both the initial value and the source term. 2.2.4 Convergence of the difference scheme Theorem 2.2.3. Suppose {Uin | 0 ⩽ i ⩽ M, 0 ⩽ n ⩽ N} and {uni | 0 ⩽ i ⩽ M, 0 ⩽ n ⩽ N} are solutions of the problem (2.1)–(2.3) and the difference scheme (2.28)–(2.30), respectively. Let ein = Uin − uni ,

0 ⩽ i ⩽ M, 0 ⩽ n ⩽ N.

2.3 The second-order method in space based on L1 approximation for 1D problem

| 119

Then it holds L 15T α 󵄩󵄩 n 󵄩󵄩 c (τ + h4 ), 󵄩󵄩e 󵄩󵄩∞ ⩽ √ 4 (1 − α)2α 2

1 ⩽ n ⩽ N.

Proof. The subtraction of (2.28)–(2.30) from (2.24), (2.26)–(2.27), respectively, gives the system of error equations in the form of n

{ −α (α) n−k 2 n n { { 𝒜(τ ∑ gk ei ) = δx ei + (r2 )i , { { { k=0 { 0 { { ei = 0, 1 ⩽ i ⩽ M − 1, { { { n n { e0 = 0, eM = 0, 0 ⩽ n ⩽ N.

1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N,

Utilizing Theorem 2.2.2 and noticing (2.25) lead to 15 1 󵄩 󵄩2 󵄩󵄩 n 󵄩󵄩2 󵄩 󵄩2 (󵄩󵄩δ e0 󵄩󵄩 + α+1 tnα max 󵄩󵄩󵄩(r2 )m 󵄩󵄩󵄩 ) 󵄩󵄩δx e 󵄩󵄩 ⩽ 1⩽m⩽n 2(1 − α) 󵄩 x 󵄩 2 15 2 T α Lc22 (τ + h4 ) , 1 ⩽ n ⩽ N. ⩽ 4(1 − α)2α Moreover, if follows from Lemma 2.1.1 that α √L 󵄩 n 󵄩 L 󵄩󵄩 n 󵄩󵄩 󵄩󵄩δx e 󵄩󵄩 ⩽ √ 15T c (τ + h4 ), 󵄩󵄩e 󵄩󵄩∞ ⩽ 󵄩 󵄩 2 4 (1 − α)2α 2

1 ⩽ n ⩽ N.

The proof ends.

2.3 The second-order method in space based on L1 approximation for 1D problem Consider C α D u(x, t) = uxx (x, t) + f (x, t), { { { 0 t u(x, 0) = φ(x), x ∈ (0, L), { { { { u(0, t) = μ(t), u(L, t) = ν(t),

x ∈ (0, L), t ∈ (0, T],

(2.36) (2.37)

t ∈ [0, T],

(2.38)

where α ∈ (0, 1), the functions f , φ, μ, ν are all given and φ(0) = μ(0), φ(L) = ν(0). Take the same mesh partition and notations as those in Section 2.1. Suppose u ∈ C (4,2) ([0, L] × [0, T]). 2.3.1 Derivation of the difference scheme Considering equation (2.36) at the point (xi , tn ), one has C α 0 Dt u(xi , tn )

= uxx (xi , tn ) + fin ,

1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N.

120 | 2 Difference methods for the time-fractional subdiffusion equations Using the L1 formula (1.60) to approximate the time-fractional derivative and the second-order central difference quotient to treat the spatial derivative, by Theorem 1.6.1 and Lemma 2.1.3, one gets n−1 τ−α n 0 [a(α) U − )Uik − a(α) ∑ (a(α) − a(α) n−1 Ui ] n−k Γ(2 − α) 0 i k=1 n−k−1

= δx2 Uin + fin + (r3 )ni ,

1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N,

(2.39)

where there is a positive constant c3 such that 󵄨󵄨 n 󵄨󵄨 2−α 2 󵄨󵄨(r3 )i 󵄨󵄨 ⩽ c3 (τ + h ),

1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N.

(2.40)

Noticing the initial-boundary value conditions (2.37)–(2.38), one has {

Ui0 = φ(xi ), U0n

= μ(tn ),

1 ⩽ i ⩽ M − 1,

n UM

= ν(tn ),

(2.41) 0 ⩽ n ⩽ N.

(2.42)

Omitting the small term (r3 )ni in (2.39) and replacing the exact solution Uin with its numerical one uni , a difference scheme for solving (2.36)–(2.38) is produced as { { { { { { { { { { { { { { { { { { {

n−1 τ−α n 0 [a(α) u − )uk − a(α) ∑ (a(α) − a(α) n−1 ui ] n−k i Γ(2 − α) 0 i k=1 n−k−1

= δx2 uni + fin ,

u0i un0

= φ(xi ),

= μ(tn ),

1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N, 1 ⩽ i ⩽ M − 1,

unM

= ν(tn ),

(2.43) (2.44)

0 ⩽ n ⩽ N.

(2.45)

Denote s = τα Γ(2 − α),

λ=

s . h2

In what follows, the unique solvability, unconditional stability and convergence will be considered. 2.3.2 Solvability of the difference scheme Theorem 2.3.1. The difference scheme (2.43)–(2.45) is uniquely solvable. Proof. Denote un = (un0 , un1 , . . . , unM ). The value of u0 is apparently determined by (2.44)–(2.45). Suppose the values of u0 , u1 , . . . , un−1 have been uniquely determined, then the linear system in un can be

2.3 The second-order method in space based on L1 approximation for 1D problem

| 121

obtained from (2.43) and (2.45). To show its unique solvability, it suffices to prove the corresponding homogeneous one 1 { uni = δx2 uni , 1 ⩽ i ⩽ M − 1, s { n n { u0 = uM = 0

(2.46) (2.47)

has only the trivial solution. Suppose ‖un ‖∞ = |unin |, where in ∈ {1, 2, . . . , M − 1}. Rewrite (2.46) as (1 + 2λ)uni = λ(uni−1 + uni+1 ),

1 ⩽ i ⩽ M − 1.

Letting i = in in the above equality and taking the absolute value of both hand sides, an application of the triangle inequality yields 󵄩 󵄩 󵄩 󵄩 (1 + 2λ)󵄩󵄩󵄩un 󵄩󵄩󵄩∞ ⩽ 2λ󵄩󵄩󵄩un 󵄩󵄩󵄩∞ , so that ‖un ‖∞ = 0, which implies un = 0. By the principle of induction, the theorem is true. The proof ends.

2.3.3 Stability of the difference scheme Theorem 2.3.2. Suppose {vin | 0 ⩽ i ⩽ M, 0 ⩽ n ⩽ N} is the solution of the difference scheme { { { { { { { { { { { { { { { { { { {

1 (α) n n−1 (α) 0 [a v − ∑ (a − a(α) )vk − a(α) n−1 vi ] n−k i s 0 i k=1 n−k−1

= δx2 vin + fin ,

vi0 v0n

= φ(xi ),

= 0,

n vM

1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N, 1 ⩽ i ⩽ M − 1,

(2.49)

= 0,

(2.50)

0 ⩽ n ⩽ N.

Then it holds 󵄩󵄩 n 󵄩󵄩 󵄩 0󵄩 α 󵄩 m󵄩 󵄩󵄩v 󵄩󵄩∞ ⩽ 󵄩󵄩󵄩v 󵄩󵄩󵄩∞ + Γ(1 − α) max {tm 󵄩󵄩󵄩f 󵄩󵄩󵄩∞ }, 1⩽m⩽n where 󵄩󵄩 m 󵄩󵄩 󵄨 m󵄨 󵄩󵄩f 󵄩󵄩∞ = max 󵄨󵄨󵄨fi 󵄨󵄨󵄨. 1⩽i⩽M−1 Proof. Reformulate (2.48) as n−1

n (α) (α) k (α) 0 a(α) 0 vi = ∑ (an−k−1 − an−k )vi + an−1 vi k=1

(2.48)

1 ⩽ n ⩽ N,

122 | 2 Difference methods for the time-fractional subdiffusion equations n n + λ(vi−1 − 2vin + vi+1 ) + sfin ,

1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N,

which can be rearranged as n−1

n (α) (α) k (α) 0 (a(α) 0 + 2λ)vi = ∑ (an−k−1 − an−k )vi + an−1 vi k=1

n n + λ(vi−1 + vi+1 ) + sfin ,

1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N.

Suppose ‖vn ‖∞ = |vinn |, where in ∈ {1, 2, . . . , M − 1}. Letting i = in in the equality above and taking the absolute value of both hand sides, an application of the triangle inequality arrives at n−1

󵄩󵄩 n 󵄩󵄩 (α) (α) 󵄩 (α) 󵄩 󵄩 k 󵄩󵄩 󵄩 0 󵄩󵄩 (a(α) 󵄩v 󵄩󵄩∞ ⩽ ∑ (an−k−1 − an−k )󵄩󵄩v 󵄩󵄩∞ + an−1 󵄩󵄩v 󵄩󵄩∞ 0 + 2λ)󵄩 k=1

󵄩 󵄩 󵄩 󵄩 + 2λ󵄩󵄩󵄩vn 󵄩󵄩󵄩∞ + s󵄩󵄩󵄩f n 󵄩󵄩󵄩∞ ,

1 ⩽ n ⩽ N.

Therefore, n−1

󵄩󵄩 n 󵄩󵄩 (α) (α) 󵄩 k 󵄩 󵄩󵄩v 󵄩󵄩∞ ⩽ ∑ (an−k−1 − an−k )󵄩󵄩󵄩v 󵄩󵄩󵄩∞ k=1

󵄩󵄩 0 󵄩󵄩 + a(α) 󵄩v 󵄩󵄩∞ + n−1 (󵄩

s 󵄩󵄩 n 󵄩󵄩 󵄩f 󵄩󵄩∞ ),

󵄩 a(α) n−1

1 ⩽ n ⩽ N.

It follows from Lemma 1.6.1 that s

a(α) n−1

⩽

τα Γ(2 − α) = tnα Γ(1 − α), (1 − α)n−α

(2.51)

and then n−1

󵄩󵄩 n 󵄩󵄩 (α) (α) 󵄩 󵄩 k 󵄩󵄩 a(α) 󵄩v 󵄩󵄩∞ ⩽ ∑ (an−k−1 − an−k )󵄩󵄩v 󵄩󵄩∞ 0 󵄩 k=1

󵄩󵄩 0 󵄩󵄩 󵄩󵄩 n 󵄩󵄩 α + a(α) 󵄩v 󵄩󵄩∞ + tn Γ(1 − α)󵄩󵄩f 󵄩󵄩∞ ), n−1 (󵄩

1 ⩽ n ⩽ N.

(2.52)

An induction on n in (2.52) will yield 󵄩󵄩 n 󵄩󵄩 󵄩 0󵄩 α 󵄩 m󵄩 󵄩󵄩v 󵄩󵄩∞ ⩽ 󵄩󵄩󵄩v 󵄩󵄩󵄩∞ + Γ(1 − α) max {tm 󵄩󵄩󵄩f 󵄩󵄩󵄩∞ }, 1⩽m⩽n

1 ⩽ n ⩽ N.

The proof is completed. 2.3.4 Convergence of the difference scheme The next theorem is to describe the convergence of difference scheme (2.43)–(2.45).

2.4 The fast difference method based on L1 approximation for 1D problem

| 123

Theorem 2.3.3. Suppose {Uin | 0 ⩽ i ⩽ M, 0 ⩽ n ⩽ N} and {uni | 0 ⩽ i ⩽ M, 0 ⩽ n ⩽ N} are solutions of the problem (2.36)–(2.38) and the difference scheme (2.43)–(2.45), respectively. Let ein = Uin − uni ,

0 ⩽ i ⩽ M, 0 ⩽ n ⩽ N.

Then it holds 󵄩󵄩 n 󵄩󵄩 α 2−α 2 󵄩󵄩e 󵄩󵄩∞ ⩽ c3 T Γ(1 − α)(τ + h ),

1 ⩽ n ⩽ N.

Proof. Subtracting (2.43)–(2.45) from (2.39), (2.41)–(2.42), respectively, the system of error equations can be obtained as { { { { { { { { { { { { { { { { { { {

1 (α) n n−1 (α) 0 [a e − ∑ (a − a(α) )ek − a(α) n−1 ei ] n−k i s 0 i k=1 n−k−1

= δx2 ein + (r3 )ni , ei0 e0n

= 0,

= 0,

1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N,

1 ⩽ i ⩽ M − 1,

n eM = 0,

0 ⩽ n ⩽ N.

By Theorem 2.3.2 and the inequality (2.40), it follows 󵄩󵄩 n 󵄩󵄩 󵄩 0󵄩 󵄩 m󵄩 α 󵄩󵄩e 󵄩󵄩∞ ⩽ 󵄩󵄩󵄩e 󵄩󵄩󵄩∞ + tn Γ(1 − α) max 󵄩󵄩󵄩(r3 ) 󵄩󵄩󵄩∞ 1⩽m⩽n ⩽ tnα Γ(1 − α)c3 (τ2−α + h2 )

⩽ c3 T α Γ(1 − α)(τ2−α + h2 ),

1 ⩽ n ⩽ N.

The proof ends.

2.4 The fast difference method based on L1 approximation for 1D problem The aim of this section is to develop a difference scheme for the problem (2.36)–(2.38) by using the fast L1 approximation.

2.4.1 Derivation of the difference scheme Considering (2.36) at the point (xi , tn ), we get C α 0 Dt u(xi , tn )

= uxx (xi , tn ) + fin ,

1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N.

124 | 2 Difference methods for the time-fractional subdiffusion equations Using the fast L1 approximation (1.117)–(1.119) to discretize the time fractional derivative and second-order central difference quotient to approximate the spatial secondorder derivative, by Theorem 1.7.1, we can obtain N

exp { 1 n n n−1 { { [ ∑ ωl Fl,i + â (α) { 0 (Ui − Ui )] { { Γ(1 − α) { l=1 { { 2 n = δx Ui + fin + (r4 )ni , 1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N, { { { { n n−1 { { F 1 = 0, Fl,i = e−sl τ Fl,i + Bl (Uin−1 − Uin−2 ), 1 ⩽ l ⩽ Nexp , { { { l,i 1 ⩽ i ⩽ M − 1, 2 ⩽ n ⩽ N, {

(2.53) (2.54)

and there exists a positive constant c4 such that 󵄨󵄨 n󵄨 2−α 2 󵄨󵄨(r4 )i 󵄨󵄨󵄨 ⩽ c4 (τ + h + ϵ),

1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N.

(2.55)

Noticing the initial-boundary conditions (2.37)–(2.38), we have {

Ui0 = φ(xi ), U0n

= μ(tn ),

1 ⩽ i ⩽ M − 1,

n UM

= ν(tn ),

(2.56) 0 ⩽ n ⩽ N.

(2.57)

Omitting the small term (r4 )ni in (2.53) and using numerical solution uni to replace the exact solution Uin , we construct the difference scheme for the problem (2.36)–(2.38) as follows: { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { {

N

exp 1 n n n−1 [ ∑ ωl Fl,i + â (α) 0 (ui − ui )] Γ(1 − α) l=1

= δx2 uni + fin , 1 Fl,i

n Fl,i

= 0,

u0i un0

= φ(xi ),

= μ(tn ),

1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N, =

n−1 e−sl τ Fl,i

+

Bl (un−1 i

−

un−2 i ),

= ν(tn ),

1 ⩽ l ⩽ Nexp ,

1 ⩽ i ⩽ M − 1, 2 ⩽ n ⩽ N,

1 ⩽ i ⩽ M − 1,

unM

(2.58) (2.59) (2.60)

0 ⩽ n ⩽ N.

(2.61)

n When {uki | 0 ⩽ i ⩽ M, 0 ⩽ k ⩽ n − 1} has been known, we can get {Fl,i |1⩽l⩽ Nexp , 1 ⩽ i ⩽ M − 1} from (2.59). Substituting it into the left-hand side of (2.58), we can get a system of linear equations in {uni | 0 ⩽ i ⩽ M}, whose calculation cost is O(MNexp ). While the calculation of the system of linear equations in {uni | 0 ⩽ i ⩽ M} from (2.43)–(2.45) is O(MN). Thus, the total operations of (2.58)–(2.61) are O(MNNexp ), and those of (2.43)–(2.45) are O(MN 2 ). When N >> Nexp , we have O(MN 2 ) >> O(MNNexp ). We call (2.58)–(2.61) a fast difference scheme based on L1 approximation, or a fast L1 difference scheme.

2.4 The fast difference method based on L1 approximation for 1D problem

| 125

n Eliminating the intermediate variable {Fl,i } in (2.58)–(2.61), we can get an equivalent form

{ { { { { { { { { { { { { { { { { { {

n−1 1 n 0 ̂ (α) ̂ (α) n−k − â (α) [â (α) n−1 ui ] 0 ui − ∑ (ak−1 − ak )ui Γ(1 − α) k=1

= δx2 uni + fin ,

u0i un0

= φ(xi ),

= μ(tn ),

1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N,

(2.62)

1 ⩽ i ⩽ M − 1,

unM

= ν(tn ),

(2.63) 0 ⩽ n ⩽ N.

(2.64)

2.4.2 Solvability of the difference scheme Theorem 2.4.1. The difference scheme (2.58)–(2.61) is uniquely solvable. Proof. Denote un = (un0 , un1 , . . . , unM ). From (2.60)–(2.61), we can know u0 . Suppose u0 , u1 , . . . , un−1 have been uniquely determined, then we can get a system of linear equations in un from (2.58) and (2.61). It suffices to show that the corresponding homogeneous system 1 n 2 n â (α) { 0 ui = δx ui , Γ(1 − α) { n n { u0 = uM = 0

1 ⩽ i ⩽ M − 1,

(2.66)

has only the trivial solution. Suppose ‖un ‖∞ = |unin |, where in ∈ {1, 2, . . . , M − 1}. Denote λ = as (1 + 2λ)uni = λ(uni−1 + uni+1 ),

(2.65)

Γ(1−α) . Rewrite (2.65) 2 â (α) 0 h

1 ⩽ i ⩽ M − 1.

Let i = in in the equality above. Taking absolute value of both hand sides and using the triangle inequality, we can obtain 󵄩 󵄩 󵄩 󵄩 (1 + 2λ)󵄩󵄩󵄩un 󵄩󵄩󵄩∞ ⩽ 2λ󵄩󵄩󵄩un 󵄩󵄩󵄩∞ . Thus we have ‖un ‖∞ = 0, which implies un = 0. By induction principle, the conclusion holds. This completes the proof.

126 | 2 Difference methods for the time-fractional subdiffusion equations 2.4.3 Stability of the difference scheme Theorem 2.4.2. Suppose {vin | 0 ⩽ i ⩽ M, 0 ⩽ n ⩽ N} is the solution of { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { and ϵ
a1 > a2 > ⋅ ⋅ ⋅ > an−1 . n Eliminating the intermediate variable {Fl,i } in (2.67) using (2.68), we have n−1 1 n 0 ̂ (α) ̂ (α) n−k − â (α) [â (α) n−1 vi ] 0 vi − ∑ (ak−1 − ak )vi Γ(1 − α) k=1

= δx2 vin + gin ,

1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N,

or (

n−1 â (α) 2 1 0 0 )vk + â (α) + 2 )vin = [ ∑ (â (α) − â (α) n−1 vi ] n−k i Γ(1 − α) h Γ(1 − α) k=1 n−k−1

+

1 n n (v + vi+1 ) + gin , h2 i−1

1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N.

Suppose ‖vn ‖∞ = |vinn |, where in ∈ {1, 2, . . . , M − 1}. Letting i = in in the above equation, taking absolute value of both hand sides and using the triangle inequality, we can obtain (

â (α) 2 󵄩 󵄩 0 + )󵄩󵄩vn 󵄩󵄩 Γ(1 − α) h2 󵄩 󵄩∞

2.4 The fast difference method based on L1 approximation for 1D problem

⩽

| 127

n−1 1 󵄩 󵄩 󵄩󵄩 0 󵄩󵄩 [ ∑ (â (α) − â (α) )󵄩󵄩vk 󵄩󵄩󵄩∞ + â (α) 󵄩v 󵄩󵄩∞ ] n−1 󵄩 n−k 󵄩 Γ(1 − α) k=1 n−k−1

+

2 󵄩󵄩 n 󵄩󵄩 󵄩 n󵄩 󵄩v 󵄩 + 󵄩󵄩g 󵄩󵄩 , h2 󵄩 󵄩∞ 󵄩 󵄩∞

1 ⩽ n ⩽ N.

Thus we have n−1

󵄩󵄩 n 󵄩󵄩 (α) (α) 󵄩 󵄩 k 󵄩󵄩 â (α) 󵄩v 󵄩󵄩∞ ⩽ ∑ (â n−k−1 − â n−k )󵄩󵄩v 󵄩󵄩∞ 0 󵄩 k=1

‖g n ‖∞ 󵄩󵄩 0 󵄩󵄩 + â (α) 󵄩v 󵄩󵄩∞ + Γ(1 − α) (α) ), n−1 (󵄩 â n−1

1 ⩽ n ⩽ N.

An application of the induction method will yield ‖g m ‖ 󵄩󵄩 n 󵄩󵄩 󵄩 0󵄩 󵄩󵄩v 󵄩󵄩∞ ⩽ 󵄩󵄩󵄩v 󵄩󵄩󵄩∞ + Γ(1 − α) max (α) ∞ , 1⩽m⩽n â m−1

1 ⩽ n ⩽ N.

Noticing 1 −α −α â (α) m−1 ⩾ tm − ϵ ⩾ tm , 2 we can get (2.71). This completes the proof.

2.4.4 Convergence of the difference scheme Theorem 2.4.3. Suppose {Uin | 0 ⩽ i ⩽ M, 0 ⩽ n ⩽ N} and {uni | 0 ⩽ i ⩽ M, 0 ⩽ n ⩽ N} are the solutions of (2.36)–(2.38) and (2.58)–(2.61), respectively. Denote ein = Uin − uni ,

0 ⩽ i ⩽ M, 0 ⩽ n ⩽ N,

then we have 󵄩󵄩 n 󵄩󵄩 α 2−α 2 󵄩󵄩e 󵄩󵄩∞ ⩽ 2c4 T Γ(1 − α)(τ + h + ϵ),

1 ⩽ n ⩽ N.

n Proof. Eliminating the intermediate variable {Fl,i } in (2.53) using (2.54), we have n−1 1 n 0 [â (α) U − )Uin−k − â (α) ∑ (â (α) − â (α) n−1 Ui ] k Γ(1 − α) 0 i k=1 k−1

= δx2 Uin + fin + (r4 )ni ,

1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N.

(2.72)

128 | 2 Difference methods for the time-fractional subdiffusion equations Subtracting (2.62)–(2.64) from (2.72), (2.56)–(2.57) respectively, we get the system of error equations { { { { { { { { { { { { { { { { { { {

n−1 1 n ̂ (α) ̂ (α) k ̂ (α) 0 [â (α) 0 ei − ∑ (an−k−1 − an−k )ei − an−1 ei ] Γ(1 − α) k=1

= δx2 ein + (r4 )ni ,

ei0 e0n

= 0,

= 0,

1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N,

1 ⩽ i ⩽ M − 1,

n eM = 0,

0 ⩽ n ⩽ N.

Applying Theorem 2.4.2 and noticing (2.55), we have 󵄩󵄩 n 󵄩󵄩 󵄩 0󵄩 󵄩 α m󵄩 󵄩󵄩e 󵄩󵄩∞ ⩽ 󵄩󵄩󵄩e 󵄩󵄩󵄩∞ + 2tn Γ(1 − α) max 󵄩󵄩󵄩(r4 ) 󵄩󵄩󵄩∞ 1⩽m⩽n ⩽ 2tnα Γ(1 − α)c4 (τ2−α + h2 + ϵ)

⩽ 2c4 T α Γ(1 − α)(τ2−α + h2 + ϵ),

1 ⩽ n ⩽ N.

This completes the proof.

2.5 The fourth-order method in space based on L1 approximation for 1D problem In this section, another higher-order difference scheme for solving (2.36)–(2.38) will be developed. Suppose the exact solution u ∈ C (6,2) ([0, L] × [0, T]). 2.5.1 Derivation of the difference scheme Considering equation (2.36) at the point (xi , tn ), one has C α 0 Dt u(xi , tn )

= uxx (xi , tn ) + fin ,

0 ⩽ i ⩽ M, 1 ⩽ n ⩽ N.

Performing the operator 𝒜 to both hand sides of the equality above gives C

α

n

𝒜 0 Dt u(xi , tn ) = 𝒜uxx (xi , tn ) + 𝒜fi ,

1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N.

It follows from Theorem 1.6.1 and Lemma 2.1.3 that n−1 τ−α (α) n (α) (α) k (α) 0 𝒜[a0 Ui − ∑ (an−k−1 − an−k )Ui − an−1 Ui ] Γ(2 − α) k=1

= δx2 Uin + 𝒜fin + (r5 )ni ,

1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N,

(2.73)

where there exists a positive constant c5 such that 󵄨󵄨 n 󵄨󵄨 2−α 4 󵄨󵄨(r5 )i 󵄨󵄨 ⩽ c5 (τ + h ),

1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N.

(2.74)

2.5 The fourth-order method in space based on L1 approximation for 1D problem

| 129

Noticing the initial-boundary value conditions (2.37)–(2.38), one has {

Ui0 = φ(xi ),

U0n

= μ(tn ),

1 ⩽ i ⩽ M − 1,

n UM

= ν(tn ),

(2.75) 0 ⩽ n ⩽ N.

(2.76)

Neglecting the small term (r5 )ni in (2.73) and replacing the exact solution Uin with its numerical one uni lead to another difference scheme for solving (2.36)–(2.38) as follows: { { { { { { { { { { { { { { { { { { {

n−1 τ−α (α) n (α) (α) k (α) 0 𝒜[a0 ui − ∑ (an−k−1 − an−k )ui − an−1 ui ] Γ(2 − α) k=1

= δx2 uni + 𝒜fin , u0i un0

= φ(xi ),

= μ(tn ),

1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N,

1 ⩽ i ⩽ M − 1,

unM

= ν(tn ),

(2.77) (2.78)

0 ⩽ n ⩽ N.

(2.79)

2.5.2 Solvability of the difference scheme Let s = τα Γ(2 − α). Theorem 2.5.1. The difference scheme (2.77)–(2.79) is uniquely solvable. Proof. Denote un = (un0 , un1 , . . . , unM ). Obviously, the value of u0 is uniquely determined by (2.78) and (2.79). Suppose the values of u0 , u1 , . . . , un−1 have been uniquely determined, then the linear system in un can be obtained from (2.77) and (2.79). To prove its unique solvability, it suffices to show that its corresponding homogeneous one 1 { 𝒜uni = δx2 uni , s { n n { u0 = uM = 0

1 ⩽ i ⩽ M − 1,

(2.80) (2.81)

has only the trivial solution. Making the inner product on both hand sides of (2.80) with −δx2 un and noticing (2.5) give 1 n n 󵄩 󵄩2 (u , u )1,A = −󵄩󵄩󵄩δx2 un 󵄩󵄩󵄩 ⩽ 0, s so that ‖δx un ‖A = 0. By Lemma 2.1.2, it follows ‖δx un ‖ = 0, which implies un = 0 from (2.81). By the principle of induction, the theorem is true. The proof ends.

130 | 2 Difference methods for the time-fractional subdiffusion equations 2.5.3 Stability of the difference scheme Theorem 2.5.2. Suppose {vin | 0 ⩽ i ⩽ M, 0 ⩽ n ⩽ N} is the solution of the difference scheme { { { { { { { { { { { { { { { { { { {

n−1 1 (α) n (α) (α) k (α) 0 𝒜[a0 vi − ∑ (an−k−1 − an−k )vi − an−1 vi ] s k=1

= δx2 vin + gin , vi0 v0n

= φ(xi ),

n vM

= 0,

1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N,

(2.82)

1 ⩽ i ⩽ M − 1,

(2.83)

= 0,

(2.84)

0 ⩽ n ⩽ N.

Then it holds

󵄩󵄩 n 󵄩󵄩2 3 󵄩󵄩 0 󵄩󵄩2 3 α 󵄩 m 󵄩2 󵄩󵄩δx v 󵄩󵄩 ⩽ 󵄩󵄩δx v 󵄩󵄩 + Γ(1 − α) max {tm 󵄩󵄩󵄩g 󵄩󵄩󵄩 }, 1⩽m⩽n 2 4

1 ⩽ n ⩽ N,

(2.85)

where M−1

󵄩󵄩 m 󵄩󵄩2 m 2 󵄩󵄩g 󵄩󵄩 = h ∑ (gi ) . i=1

Proof. Taking the inner product on both hand sides of (2.82) with −δx2 vn and noticing (2.5) produce n−1 1 (α) n n 0 n [a0 (v , v )1,A − ∑ (a(α) − a(α) )(vk , vn )1,A − a(α) n−1 (v , v )1,A ] n−k−1 n−k s k=1 M−1

󵄩 󵄩2 = −󵄩󵄩󵄩δx2 vn 󵄩󵄩󵄩 + h ∑ gin (−δx2 vin ), i=1

1 ⩽ n ⩽ N.

Using the Cauchy–Schwarz inequality, one has n−1

n n (α) k n (α) 0 n (α) a(α) 0 (v , v )1,A = ∑ (an−k−1 − an−k )(v , v )1,A + an−1 (v , v )1,A k=1

M−1

󵄩 󵄩2 + s[−󵄩󵄩󵄩δx2 vn 󵄩󵄩󵄩 + h ∑ gin (−δx2 vin )] n−1

⩽

i=1

1 )[(vk , vk )1,A + (vn , vn )1,A ] ∑ (a(α) − a(α) n−k 2 k=1 n−k−1

1 󵄩 󵄩2 + a(α) [(v0 , v0 )1,A + (vn , vn )1,A ] + s[−󵄩󵄩󵄩δx2 vn 󵄩󵄩󵄩 2 n−1 󵄩 󵄩2 1 󵄩 󵄩2 + 󵄩󵄩󵄩δx2 vn 󵄩󵄩󵄩 + 󵄩󵄩󵄩g n 󵄩󵄩󵄩 ], 1 ⩽ n ⩽ N, 4

from which it follows that n−1

n n (α) (α) k k a(α) 0 (v , v )1,A ⩽ ∑ (an−k−1 − an−k )(v , v )1,A k=1

2.5 The fourth-order method in space based on L1 approximation for 1D problem

s

󵄩󵄩 n 󵄩󵄩2 󵄩󵄩g 󵄩󵄩 ],

1 ⩽ n ⩽ N.

1 α󵄩 0 0 󵄩g n 󵄩󵄩󵄩2 ], + a(α) 󵄩 󵄩 n−1 [(v , v )1,A + Γ(1 − α)tn 󵄩 2

1 ⩽ n ⩽ N.

0 0 + a(α) n−1 [(v , v )1,A +

2a(α) n−1

| 131

Moreover, it follows from (2.51) that n−1

n n (α) (α) k k a(α) 0 (v , v )1,A ⩽ ∑ (an−k−1 − an−k )(v , v )1,A k=1

(2.86)

An induction on n in (2.86) will lead to 1 α󵄩 󵄩󵄩g m 󵄩󵄩󵄩2 }, (vn , vn )1,A ⩽ (v0 , v0 )1,A + Γ(1 − α) max {tm 󵄩 󵄩 1⩽m⩽n 2

1 ⩽ n ⩽ N.

Combining with Lemma 2.1.2, it is easy to get (2.85). The proof ends. 2.5.4 Convergence of the difference scheme The rest of this section focuses on the convergence of the difference scheme (2.77)–(2.79). At this point, the following theorem will be obtained. Theorem 2.5.3. Suppose {Uin | 0 ⩽ i ⩽ M, 0 ⩽ n ⩽ N} and {uni | 0 ⩽ i ⩽ M, 0 ⩽ n ⩽ N} are solutions of the problem (2.36)–(2.38) and the difference scheme (2.77)–(2.79), respectively. Let ein = Uin − uni ,

0 ⩽ i ⩽ M, 0 ⩽ n ⩽ N.

Then it holds L 󵄩󵄩 n 󵄩󵄩 2−α 4 󵄩󵄩e 󵄩󵄩∞ ⩽ √3T α Γ(1 − α) c5 (τ + h ), 4

1 ⩽ n ⩽ N.

Proof. The subtraction of (2.77)–(2.79) from (2.73), (2.75)–(2.76), respectively, produces the system of error equations { { { { { { { { { { { { { { { { { { {

n−1 1 k (α) 0 (α) n (α) (α) 𝒜[a0 ei − ∑ (an−k−1 − an−k )ei − an−1 ei ] s k=1

= δx2 ein + (r5 )ni ,

ei0 e0n

= 0,

= 0,

1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N,

1 ⩽ i ⩽ M − 1,

n eM = 0,

0 ⩽ n ⩽ N.

By Theorem 2.5.2 and the inequality (2.74), one has 󵄩󵄩 n 󵄩󵄩2 3 󵄩󵄩 0 󵄩󵄩2 3 α󵄩 m 󵄩2 󵄩󵄩δx e 󵄩󵄩 ⩽ 󵄩󵄩δx e 󵄩󵄩 + Γ(1 − α) max {tm 󵄩󵄩󵄩(r5 ) 󵄩󵄩󵄩 } 1⩽m⩽n 2 4

132 | 2 Difference methods for the time-fractional subdiffusion equations

⩽

3 α 2 T Γ(1 − α)Lc52 (τ2−α + h4 ) , 4

1 ⩽ n ⩽ N.

Moreover, it follows from Lemma 2.1.1 that √L 󵄩 n 󵄩 L 󵄩󵄩 n 󵄩󵄩 󵄩󵄩δ e 󵄩󵄩 ⩽ √3T α Γ(1 − α) c5 (τ2−α + h4 ), 1 ⩽ n ⩽ N. 󵄩󵄩e 󵄩󵄩∞ ⩽ 2 󵄩 x 󵄩 4 This ends the proof.

2.6 The difference method based on L2-1σ approximation for 1D problem This section will still discuss the difference method for the problem (2.36)–(2.38). A new difference scheme will be built based on L2-1σ approximation for the time fractional derivatives. Suppose the exact solution u ∈ C (4,3) ([0, L] × [0, T]).

2.6.1 Derivation of the difference scheme Denote σ =1−

α , 2

tn−1+σ = (n − 1 + σ)τ,

s = τα Γ(2 − α),

fin−1+σ = f (xi , tn−1+σ ).

Considering (2.36) at the point (xi , tn−1+σ ), we have C α 0 Dt u(xi , tn−1+σ )

= uxx (xi , tn−1+σ ) + fin−1+σ ,

1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N.

(2.87)

Using L2-1σ approximation (1.81) to discretize the time fractional derivative, we have C α 0 Dt u(xi , tn−1+σ )

=

τ−α n−1 (n,α) n−k (Ui − Uin−k−1 ) + O(τ3−α ). ∑c Γ(2 − α) k=0 k

(2.88)

Using the linear interpolation and the second order central difference quotient to approximate the spatial second-order derivative, we have uxx (xi , tn−1+σ ) = σuxx (xi , tn ) + (1 − σ)uxx (xi , tn−1 ) + O(τ2 )

= σδx2 Uin + (1 − σ)δx2 Uin−1 + O(h2 ) + O(τ2 ).

Inserting (2.88) and (2.89) into (2.87), we can obtain τ−α n−1 (n,α) n−k (Ui − Uin−k−1 ) ∑c Γ(2 − α) k=0 k

(2.89)

2.6 The difference method based on L2-1σ approximation for 1D problem

= σδx2 Uin + (1 − σ)δx2 Uin−1 + fin−1+σ + (r6 )ni ,

1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N

| 133

(2.90)

and there exists a positive constant c6 such that 󵄨󵄨 n 󵄨󵄨 2 2 󵄨󵄨(r6 )i 󵄨󵄨 ⩽ c6 (τ + h ),

1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N.

(2.91)

Noticing the initial-boundary value conditions (2.37)–(2.38), we have {

Ui0 = φ(xi ), U0n

= μ(tn ),

1 ⩽ i ⩽ M − 1,

n UM

= ν(tn ),

(2.92) 0 ⩽ n ⩽ N.

(2.93)

Omitting the small term (r6 )ni in (2.90) and using numerical solution uni to replace the exact solution Uin , we construct the following difference scheme for the problem (2.36)–(2.38): { { { { { { { { { { { { { { { { { { {

τ−α n−1 (n,α) n−k (ui − un−k−1 ) = σδx2 uni ∑c i Γ(2 − α) k=0 k + (1 − σ)δx2 un−1 + fin−1+σ , i

u0i un0

= φ(xi ), = μ(tn ),

1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N,

1 ⩽ i ⩽ M − 1,

unM

= ν(tn ),

(2.94) (2.95)

0 ⩽ n ⩽ N.

(2.96)

2.6.2 Solvability of the difference scheme Theorem 2.6.1. The difference scheme (2.94)–(2.96) is uniquely solvable. Proof. Denote un = (un0 , un1 , . . . , unM ). From (2.95)–(2.96) we can know u0 . Suppose u0 , u1 , . . . , un−1 have been uniquely determined, then we can get a system of linear equations in un from (2.94) and (2.96). It suffices to show that the corresponding homogeneous system 1 { c0(n,α) uni = σδx2 uni , { sn n { u0 = uM = 0

1 ⩽ i ⩽ M − 1,

has only the trivial solution. Taking the inner product on both hand sides of (2.97) with un yields 1 (n,α) 󵄩󵄩 n 󵄩󵄩2 󵄩 n 󵄩2 n 2 n c 󵄩u 󵄩 = σ(u , δx u ) = −σ 󵄩󵄩󵄩δx u 󵄩󵄩󵄩 . s 0 󵄩 󵄩 It is easy to know that uni = 0,

0 ⩽ i ⩽ M.

By induction principle, the conclusion holds. This completes the proof.

(2.97) (2.98)

134 | 2 Difference methods for the time-fractional subdiffusion equations 2.6.3 An important lemma Lemma 2.6.1. [1] Suppose 𝒱 is an inner space, (⋅, ⋅) is an inner product in 𝒱 and ‖ ⋅ ‖ is the induced norm. In addition, assume that 0 < α < 1, and {ck(n,α) | 0 ⩽ k ⩽ n − 1, n ⩾ 1} satisfies {

(n,α) (n,α) c0(n,α) > c1(n,α) > c2(n,α) > ⋅ ⋅ ⋅ > cn−2 > cn−1 > (1 − α)n−α ,

(2σ −

1)c0(n,α)

−

σc1(n,α)

> 0.

(2.99) (2.100)

Then, for any u0 , u1 , . . . , un ∈ 𝒱 , we have n−1

∑ ck(n,α) (un−k − un−k−1 , σun + (1 − σ)un−1 )

k=0

⩾

1 n−1 (n,α) 󵄩󵄩 n−k 󵄩󵄩2 󵄩󵄩 n−k−1 󵄩󵄩2 (󵄩󵄩u 󵄩󵄩 − 󵄩󵄩u ∑c 󵄩󵄩 ), 2 k=0 k

n = 1, 2, . . . .

Proof. The proof will be carried out in three steps. (I) Prove n−1

∑ ck(n,α) (un−k − un−k−1 , un )

k=0

⩾

1 n−1 (n,α) 󵄩󵄩 n−k 󵄩󵄩2 󵄩󵄩 n−k−1 󵄩󵄩2 (󵄩󵄩u 󵄩󵄩 − 󵄩󵄩u ∑c 󵄩󵄩 ) 2 k=0 k +

󵄩󵄩 n−1 󵄩󵄩2 󵄩󵄩 󵄩 󵄩󵄩 ∑ c(n,α) (un−k − un−k−1 )󵄩󵄩󵄩 , 󵄩󵄩 k (n,α) 󵄩 󵄩 󵄩 󵄩󵄩 2c0 󵄩k=0 1

n = 1, 2, . . . .

When n = 1, we have 1 󵄩 󵄩2 󵄩 󵄩2 (un − un−1 , un ) = (󵄩󵄩󵄩un 󵄩󵄩󵄩 − 󵄩󵄩󵄩un−1 󵄩󵄩󵄩 ) + 2

1 󵄩󵄩 n n−1 󵄩2 󵄩u − u 󵄩󵄩󵄩 . 2󵄩

It is easy to know that (2.101) holds. Next, we consider n ⩾ 2. n−1

A ≡ ∑ ck(n,α) (un−k − un−k−1 , un ) − k=0 n−1

= ∑ ck(n,α) (un−k − un−k−1 , un − k=0 n−1

= ∑ ck(n,α) (un−k − un−k−1 , k=0

=

1 n−1 (n,α) 󵄩󵄩 n−k 󵄩󵄩2 󵄩󵄩 n−k−1 󵄩󵄩2 (󵄩󵄩u 󵄩󵄩 − 󵄩󵄩u ∑c 󵄩󵄩 ) 2 k=0 k

un−k + un−k−1 ) 2

k un−k − un−k−1 + ∑ (un−m+1 − un−m )) 2 m=1

1 n−1 (n,α) 󵄩󵄩 n−k 󵄩2 − un−k−1 󵄩󵄩󵄩 ∑c 󵄩u 2 k=0 k 󵄩

(2.101)

2.6 The difference method based on L2-1σ approximation for 1D problem n−1

k

k=1

m=1

| 135

+ ∑ ck(n,α) (un−k − un−k−1 , ∑ (un−m+1 − un−m )) n−1

1 󵄩 󵄩2 ∑ c(n,α) 󵄩󵄩un−k − un−k−1 󵄩󵄩󵄩 2 k=0 k 󵄩

=

n−1

n−1

m=1

k=m

+ ∑ (un−m+1 − un−m , ∑ ck(n,α) (un−k − un−k−1 )). Let n−1

wm = ∑ ck(n,α) (un−k − un−k−1 ), k=m

m = 0, 1, . . . , n − 1,

wn = 0, then we have

(n,α) n−m wm − wm+1 = cm (u − un−m−1 ),

m = 0, 1, . . . , n − 1.

Thus we get A=

n−1 1 n−1 1 1 ∑ (n,α) ‖wm − wm+1 ‖2 + ∑ (n,α) (wm−1 − wm , wm ) 2 m=0 cm c m=1 n−1

m−1

1 1 1 1 1 ‖w ‖2 + ∑ ( (n,α) − (n,α) )‖wm ‖2 2 c(n,α) 0 2 m=1 cm c m−1 0 1 1 2 ⩾ ‖w ‖ 2 c(n,α) 0 =

0

=

󵄩 󵄩󵄩2 1 1 󵄩󵄩󵄩󵄩 n−1 (n,α) n−k 󵄩󵄩 n−k−1 󵄩 c (u − u ) ∑ 󵄩 󵄩󵄩 . 󵄩󵄩 2 c(n,α) 󵄩󵄩󵄩󵄩k=0 k 󵄩 0

(II) Prove n−1

∑ ck(n,α) (un−k − un−k−1 , un−1 )

k=0

⩾

1 n−1 (n,α) 󵄩󵄩 n−k 󵄩󵄩2 󵄩󵄩 n−k−1 󵄩󵄩2 (󵄩󵄩u 󵄩󵄩 − 󵄩󵄩u ∑c 󵄩󵄩 ) 2 k=0 k

󵄩󵄩 n−1 󵄩󵄩2 󵄩󵄩 󵄩 (n,α) n−k n−k−1 󵄩 󵄩 − −u )󵄩󵄩󵄩 , 󵄩󵄩 ∑ ck (u 󵄩󵄩 2(c0(n,α) − c1(n,α) ) 󵄩󵄩󵄩k=0 󵄩 1

In fact, we have n−1

∑ ck(n,α) (un−k − un−k−1 , un−1 )

k=0

n = 1, 2, . . . .

(2.102)

136 | 2 Difference methods for the time-fractional subdiffusion equations

−

1 n−1 (n,α) 󵄩󵄩 n−k 󵄩󵄩2 󵄩󵄩 n−k−1 󵄩󵄩2 (󵄩󵄩u 󵄩󵄩 − 󵄩󵄩u ∑c 󵄩󵄩 ) 2 k=0 k

󵄩󵄩2 󵄩󵄩 n−1 󵄩󵄩 󵄩 (n,α) n−k n−k−1 󵄩 󵄩 −u )󵄩󵄩󵄩 + 󵄩 ∑ ck (u (n,α) (n,α) 󵄩 󵄩󵄩 󵄩 2(c0 − c1 ) 󵄩󵄩k=0 󵄩 1

n−1

= ∑ ck(n,α) (un−k − un−k−1 , un ) k=0

−

1 n−1 (n,α) 󵄩󵄩 n−k 󵄩󵄩2 󵄩󵄩 n−k−1 󵄩󵄩2 (󵄩󵄩u 󵄩󵄩 − 󵄩󵄩u ∑c 󵄩󵄩 ) 2 k=0 k

󵄩󵄩 n−1 󵄩󵄩2 󵄩󵄩 󵄩 (n,α) n−k n−k−1 󵄩 󵄩 −u )󵄩󵄩󵄩 + 󵄩 ∑ ck (u (n,α) (n,α) 󵄩 󵄩 󵄩󵄩 2(c0 − c1 ) 󵄩󵄩k=0 󵄩 1

n−1

− (un − un−1 , ∑ ck(n,α) (un−k − un−k−1 )) n−1

=

=

k=0

1 1 1 1 ∑ ( (n,α) − (n,α) )‖wm ‖2 + (n,α) ‖w0 ‖2 2 m=1 cm cm−1 2c0 1 1 + ‖w0 ‖2 − (n,α) (w0 − w1 , w0 ) 2(c0(n,α) − c1(n,α) ) c0

1 1 1 n−1 ∑ ( (n,α) − (n,α) )‖wm ‖2 2 m=2 cm cm−1 󵄩󵄩 c0(n,α) − c1(n,α) 󵄩󵄩󵄩2 c1(n,α) 󵄩󵄩 + (n,α) (n,α) w1 󵄩󵄩󵄩 w + 󵄩󵄩 0 󵄩󵄩 2c0 (c0 − c1(n,α) ) 󵄩󵄩 c1(n,α)

⩾ 0.

(III) Multiplying (2.101) with σ, (2.102) with (1 − σ), summing up the results and using (2.100), we have n−1

∑ ck(n,α) (un−k − un−k−1 , σun + (1 − σ)un−1 )

k=0

⩾

1 n−1 (n,α) 󵄩󵄩 n−k 󵄩󵄩2 󵄩󵄩 n−k−1 󵄩󵄩2 (󵄩󵄩u 󵄩󵄩 − 󵄩󵄩u ∑c 󵄩󵄩 ), 2 k=0 k

n = 1, 2, . . . .

This completes the proof.

2.6.4 Stability of the difference scheme Theorem 2.6.2. Suppose {vin | 0 ⩽ i ⩽ M, 0 ⩽ n ⩽ N} is the solution of the difference scheme

2.6 The difference method based on L2-1σ approximation for 1D problem

{ { { { { { { { { { { { { { { { {

1 n−1 (n,α) n−k (vi − vin−k−1 ) = σδx2 vin + (1 − σ)δx2 vin−1 + gin , ∑c s k=0 k vi0 v0n

1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N,

= φ(xi ),

= 0,

n vM

| 137

(2.103)

1 ⩽ i ⩽ M − 1,

(2.104)

= 0,

(2.105)

0 ⩽ n ⩽ N,

then we have 2 󵄩󵄩 n 󵄩󵄩2 󵄩󵄩 0 󵄩󵄩2 L α 󵄩 m 󵄩2 󵄩󵄩v 󵄩󵄩 ⩽ 󵄩󵄩v 󵄩󵄩 + Γ(1 − α) max (tm 󵄩󵄩󵄩g 󵄩󵄩󵄩 ), 1 ⩽ n ⩽ N, 1⩽m⩽n 12 󵄩󵄩 n 󵄩󵄩2 󵄩󵄩 0 󵄩󵄩2 1 α 󵄩 m 󵄩2 󵄩󵄩δx v 󵄩󵄩 ⩽ 󵄩󵄩δx v 󵄩󵄩 + Γ(1 − α) max (tm 󵄩󵄩󵄩g 󵄩󵄩󵄩 ), 1 ⩽ n ⩽ N, 1⩽m⩽n 2

(2.106) (2.107)

where M−1

󵄩󵄩 m 󵄩󵄩2 󵄨 m 󵄨2 󵄩󵄩g 󵄩󵄩 = h ∑ 󵄨󵄨󵄨gi 󵄨󵄨󵄨 . i=1

Proof. (I) Taking the inner product on both hand sides of (2.103) with σvn + (1 − σ)vn−1 , we have 1 n−1 (n,α) n−k (v − vn−k−1 , σvn + (1 − σ)vn−1 ) ∑c s k=0 k

= (δx2 (σvn + (1 − σ)vn−1 ), σvn + (1 − σ)vn−1 ) + (g n , σvn + (1 − σ)vn−1 ) 󵄩 󵄩2 = −󵄩󵄩󵄩δx (σvn + (1 − σ)vn−1 )󵄩󵄩󵄩 + (g n , σvn + (1 − σ)vn−1 ) ⩽− =

6 󵄩󵄩 n 6󵄩 n L2 󵄩 n 󵄩2 n−1 󵄩2 n−1 󵄩2 󵄩󵄩σv + (1 − σ)v 󵄩󵄩󵄩 + [ 2 󵄩󵄩󵄩σv + (1 − σ)v 󵄩󵄩󵄩 + 󵄩󵄩󵄩g 󵄩󵄩󵄩 ] 2 24 L L

L2 󵄩󵄩 n 󵄩󵄩2 󵄩g 󵄩 , 24 󵄩 󵄩

1 ⩽ n ⩽ N.

(2.108)

For the left-hand side of the inequality above, applying Lemma 2.6.1 and Lemma 1.6.3, we have 1 n−1 (n,α) n−k (v − vn−k−1 , σvn + (1 − σ)vn−1 ) ∑c s k=0 k ⩾

1 1 n−1 (n,α) 󵄩󵄩 n−k 󵄩󵄩2 󵄩󵄩 n−k−1 󵄩󵄩2 ⋅ ∑c (󵄩󵄩v 󵄩󵄩 − 󵄩󵄩v 󵄩󵄩 ). 2 s k=0 k

(2.109)

From (2.108) and (2.109), we have L2 󵄩󵄩 n 󵄩󵄩2 1 1 n−1 (n,α) 󵄩󵄩 n−k 󵄩󵄩2 󵄩󵄩 n−k−1 󵄩󵄩2 ⋅ ∑ ck (󵄩󵄩v 󵄩󵄩 − 󵄩󵄩v 󵄩󵄩 ) ⩽ 󵄩g 󵄩 , 2 s k=0 24 󵄩 󵄩

1 ⩽ n ⩽ N.

(2.110)

138 | 2 Difference methods for the time-fractional subdiffusion equations In view of Lemma 1.6.3, we have s

(n,α) cn−1

⩽

τα Γ(2 − α) ⩽ tnα Γ(1 − α). (1 − α)n−α

(2.111)

Rewrite (2.110) as n−1 2 󵄩 󵄩2 󵄩 󵄩2 (n,α) (n,α) 󵄩 󵄩󵄩v0 󵄩󵄩󵄩2 + L s󵄩󵄩󵄩g n 󵄩󵄩󵄩2 c0(n,α) 󵄩󵄩󵄩vn 󵄩󵄩󵄩 ⩽ ∑ (ck−1 − ck(n,α) )󵄩󵄩󵄩vn−k 󵄩󵄩󵄩 + cn−1 󵄩 󵄩 12 󵄩 󵄩 k=1

n−1 2 󵄩 󵄩2 󵄩2 L s 󵄩󵄩 n 󵄩󵄩2 (n,α) (n,α) 󵄩 = ∑ (ck−1 − ck(n,α) )󵄩󵄩󵄩vn−k 󵄩󵄩󵄩 + cn−1 (󵄩󵄩󵄩v0 󵄩󵄩󵄩 + 󵄩g 󵄩 ) 12 c(n,α) 󵄩 󵄩 k=1 n−1

2

n−1

󵄩 󵄩2 󵄩 󵄩2 󵄩2 L (n,α) (n,α) 󵄩 ⩽ ∑ (ck−1 − ck(n,α) )󵄩󵄩󵄩vn−k 󵄩󵄩󵄩 + cn−1 (󵄩󵄩󵄩v0 󵄩󵄩󵄩 + tnα Γ(1 − α)󵄩󵄩󵄩g n 󵄩󵄩󵄩 ), 12 k=1 1 ⩽ n ⩽ N. By mathematical induction, we can obtain (2.106). (II) Taking an inner product on both hand sides of (2.103) with −δx2 (σvn +(1−σ)vn−1 ), we have 1 n−1 (n,α) n−k (v − vn−k−1 , −δx2 (σvn + (1 − σ)vn−1 )) ∑c s k=0 k

󵄩 󵄩2 = −󵄩󵄩󵄩δx2 (σvn + (1 − σ)vn−1 )󵄩󵄩󵄩 − (g n , δx2 (σvn + (1 − σ)vn−1 ))

󵄩 󵄩2 󵄩 󵄩2 1 󵄩 󵄩2 ⩽ −󵄩󵄩󵄩δx2 (σvn + (1 − σ)vn−1 )󵄩󵄩󵄩 + 󵄩󵄩󵄩δx2 (σvn + (1 − σ)vn−1 )󵄩󵄩󵄩 + 󵄩󵄩󵄩g n 󵄩󵄩󵄩 4 1 󵄩󵄩 n 󵄩󵄩2 = 󵄩󵄩g 󵄩󵄩 , 1 ⩽ n ⩽ N. 4

(2.112)

For the left-hand side of the inequality above, applying Lemma 2.6.1, we have 1 n−1 (n,α) n−k (v − vn−k−1 , −δx2 (σvn + (1 − σ)vn−1 )) ∑c s k=0 k = ⩾

1 n−1 (n,α) (δx (vn−k − vn−k−1 ), δx (σvn + (1 − σ)vn−1 )) ∑c s k=0 k 1 1 n−1 (n,α) 󵄩󵄩 n−k 󵄩󵄩2 󵄩󵄩 n−k−1 󵄩󵄩2 ⋅ ∑c (󵄩󵄩δx v 󵄩󵄩 − 󵄩󵄩δx v 󵄩󵄩 ). 2 s k=0 k

(2.113)

From (2.112) and (2.113), we have 1 1 n−1 (n,α) 󵄩󵄩 n−k 󵄩󵄩2 󵄩󵄩 n−k−1 󵄩󵄩2 1 󵄩 n 󵄩2 ⋅ ∑c (󵄩󵄩δx v 󵄩󵄩 − 󵄩󵄩δx v 󵄩󵄩 ) ⩽ 󵄩󵄩󵄩g 󵄩󵄩󵄩 , 2 s k=0 k 4

1 ⩽ n ⩽ N.

(2.114)

2.6 The difference method based on L2-1σ approximation for 1D problem

| 139

Noticing (2.111), rewrite (2.114) as n−1 󵄩 󵄩2 󵄩 󵄩2 (n,α) (n,α) 󵄩 󵄩󵄩δx v0 󵄩󵄩󵄩2 + 1 s󵄩󵄩󵄩g n 󵄩󵄩󵄩2 c0(n,α) 󵄩󵄩󵄩δx vn 󵄩󵄩󵄩 ⩽ ∑ (ck−1 − ck(n,α) )󵄩󵄩󵄩δx vn−k 󵄩󵄩󵄩 + cn−1 󵄩 󵄩 2 󵄩 󵄩 k=1

n−1 󵄩 󵄩2 󵄩2 1 s 󵄩󵄩 n 󵄩󵄩2 (n,α) (n,α) 󵄩 = ∑ (ck−1 − ck(n,α) )󵄩󵄩󵄩δx vn−k 󵄩󵄩󵄩 + cn−1 (󵄩󵄩󵄩δx v0 󵄩󵄩󵄩 + 󵄩g 󵄩 ) 2 c(n,α) 󵄩 󵄩 k=1 n−1

n−1

󵄩 󵄩2 󵄩2 1 󵄩 󵄩2 (n,α) (n,α) 󵄩 ⩽ ∑ (ck−1 − ck(n,α) )󵄩󵄩󵄩δx vn−k 󵄩󵄩󵄩 + cn−1 (󵄩󵄩󵄩δx v0 󵄩󵄩󵄩 + Γ(1 − α)tnα 󵄩󵄩󵄩g n 󵄩󵄩󵄩 ), 2 k=1 1 ⩽ n ⩽ N. By mathematical induction, we can obtain (2.107). This completes the proof. 2.6.5 Convergence of the difference scheme Theorem 2.6.3. Suppose {Uin | 0 ⩽ i ⩽ M, 0 ⩽ n ⩽ N} and {uni | 0 ⩽ i ⩽ M, 0 ⩽ n ⩽ N} are solutions of the problem (2.36)–(2.38) and the difference scheme (2.94)–(2.96), respectively. Denote ein = Uin − uni ,

0 ⩽ i ⩽ M, 0 ⩽ n ⩽ N,

then we have 󵄩󵄩 n 󵄩󵄩 √ 1 󵄩󵄩δx e 󵄩󵄩 ⩽ 2 Γ(1 − α)tnα L c6 (τ2 + h2 ), 1 ⩽ n ⩽ N, 󵄩󵄩󵄩en 󵄩󵄩󵄩 ⩽ 1 √2Γ(1 − α)t α Lc6 (τ2 + h2 ), 1 ⩽ n ⩽ N. n 󵄩 󵄩∞ 4

(2.115) (2.116)

Proof. Subtracting (2.94)–(2.96) from (2.90), (2.92)–(2.93) respectively, we get the system of error equations τ−α n−1 (n,α) n−k { { ∑ ck (ei − ein−k−1 ) = σδx2 ein + (1 − σ)δx2 ein−1 + (r6 )ni , { { Γ(2 − α) { { k=0 { { 1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N, { { { 0 { { e = 0, 1 ⩽ i ⩽ M − 1, { { { in n { e0 = 0, eM = 0, 0 ⩽ n ⩽ N. Applying Theorem 2.6.2 and noticing (2.91), we have 󵄩󵄩 n 󵄩󵄩2 󵄩󵄩 0 󵄩󵄩2 1 α󵄩 m 󵄩2 󵄩󵄩δx e 󵄩󵄩 ⩽ 󵄩󵄩δx e 󵄩󵄩 + Γ(1 − α) max {tm 󵄩󵄩󵄩(r6 ) 󵄩󵄩󵄩 } 1⩽m⩽n 2 1 2 ⩽ Γ(1 − α)tnα L[c6 (τ2 + h2 )] , 1 ⩽ n ⩽ N. 2 Taking the square root on both hand sides of the inequality above, we can get (2.115). Noticing Lemma 2.1.1, from (2.115), we can obtain (2.116) easily. This completes the proof.

140 | 2 Difference methods for the time-fractional subdiffusion equations

2.7 The fast difference method based on L2-1σ approximation for 1D problem In this section, a fast difference method based on L2-1σ approximation will be given for the problem (2.36)–(2.38). Suppose the exact solution u ∈ C (4,3) ([0, L] × [0, T]).

2.7.1 Derivation of the difference scheme Denote σ =1−

α , 2

fin−1+σ = f (xi , tn−1+σ ).

tn−1+σ = (n − 1 + σ)τ,

Considering (2.36) at the point (xi , tn−1+σ ), we have C α 0 Dt u(xi , tn−1+σ )

= uxx (xi , tn−1+σ ) + fin−1+σ ,

1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N.

(2.117)

Using the theory in Subsection 1.7.2 to discretize the time fractional derivative, we get N

exp { 1 C α { { ω F n + d0(1,α) (Uin − Uin−1 ) D u(x , t ) = ∑ { i n−1+σ 0 t { Γ(1 − α) l=1 l l,i { { { { { { { + O(τ3−α + ϵ), 1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N, { 1 { { Fl,i = 0, 1 ⩽ l ⩽ Nexp , 1 ⩽ i ⩽ M − 1, { { { { n n−1 { { Fl,i = e−sl τ Fl,i + Al (Uin−1 − Uin−2 ) + Bl (Uin − Uin−1 ), { { { 1 ⩽ l ⩽ Nexp , 1 ⩽ i ⩽ M − 1, 2 ⩽ n ⩽ N. {

(2.118) (2.119) (2.120)

Using the linear interpolation and the second-order central difference quotient to approximate the spatial second-order derivative, we get uxx (xi , tn−1+σ ) = σuxx (xi , tn ) + (1 − σ)uxx (xi , tn−1 ) + O(τ2 )

= σδx2 Uin + (1 − σ)δx2 Uin−1 + O(h2 ) + O(τ2 ).

(2.121)

Substituting (2.118)–(2.121) into (2.117), we get N

exp { 1 n { { + d0(1,α) (Uin − Uin−1 ) = σδx2 Uin + (1 − σ)δx2 Uin−1 ∑ ωl Fl,i { { Γ(1 − α) { { l=1 { { { { { + fin−1+σ + (r7 )ni , 1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N, { 1 { { Fl,i = 0, 1 ⩽ l ⩽ Nexp , 1 ⩽ i ⩽ M − 1, { { { { n n−1 { { Fl,i = e−sl τ Fl,i + Al (Uin−1 − Uin−2 ) + Bl (Uin − Uin−1 ), { { { 1 ⩽ l ⩽ Nexp , 1 ⩽ i ⩽ M − 1, 2 ⩽ n ⩽ N {

(2.122) (2.123) (2.124)

2.7 The fast difference method based on L2-1σ approximation for 1D problem | 141

and there exists a positive constant c7 such that 󵄨󵄨 n 󵄨󵄨 2 2 󵄨󵄨(r7 )i 󵄨󵄨 ⩽ c7 (τ + h + ϵ),

1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N.

(2.125)

Substituting (2.123)–(2.124) into (2.122) and eliminating the intermediate variable n {Fl,i }, we have n−1

∑ dk(n,α) (Uin−k − Uin−k−1 ) = σδx2 Uin + (1 − σ)δx2 Uin−1

k=0

+ fin−1+σ + (r7 )ni ,

1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N.

Noticing the initial-boundary value conditions (2.37)–(2.38), we have Ui0 = φ(xi ),

{

U0n

= μ(tn ),

1 ⩽ i ⩽ M − 1,

n UM

= ν(tn ),

(2.126) 0 ⩽ n ⩽ N.

(2.127)

Omitting the small term (r7 )ni in (2.122) and using numerical solution uni to replace the exact solution Uin , we construct for the problem (2.36)–(2.38) the fast difference scheme as follows: { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { {

N

exp 1 n 2 n 2 n−1 + d0(1,α) (uni − un−1 ∑ ωl Fl,i i ) = σδx ui + (1 − σ)δx ui Γ(1 − α) l=1

1 Fl,i n Fl,i

u0i un0

= 0, =

+ fin−1+σ ,

1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N,

1 ⩽ l ⩽ Nexp , 1 ⩽ i ⩽ M − 1,

n−1 e−sl τ Fl,i

= φ(xi ),

= μ(tn ),

+

Al (un−1 i

−

un−2 i )

+

Bl (uni

−

un−1 i )

1 ⩽ l ⩽ Nexp , 1 ⩽ i ⩽ M − 1, 2 ⩽ n ⩽ N,

1 ⩽ i ⩽ M − 1,

unM

= ν(tn ),

0 ⩽ n ⩽ N.

(2.128) (2.129) (2.130) (2.131) (2.132)

Substituting (2.129)–(2.130) into (2.128), we can get the following equivalent form n−1

{ { − un−k−1 ) = σδx2 uni + (1 − σ)δx2 un−1 + fin−1+σ , ∑ dk(n,α) (un−k { i i i { { { k=0 { { 1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N, { { { 0 { { u = φ(xi ), 1 ⩽ i ⩽ M − 1, { { { i n n { u0 = μ(tn ), uM = ν(tn ), 0 ⩽ n ⩽ N.

2.7.2 Solvability of the difference scheme Theorem 2.7.1. The difference scheme (2.133)–(2.135) is uniquely solvable.

(2.133) (2.134) (2.135)

142 | 2 Difference methods for the time-fractional subdiffusion equations Proof. Denote un = (un0 , un1 , . . . , unM ). From (2.134)–(2.135), we can know u0 . Suppose u0 , u1 , . . . , un−1 have been uniquely determined, we can get the system of linear equations in un from (2.133) and (2.135). It suffices to show that the homogeneous system {

d0(n,α) uni = σδx2 uni , un0

=

unM

1 ⩽ i ⩽ M − 1,

=0

(2.136) (2.137)

has only the trivial solution. Taking an inner product on both hand sides of (2.136) with un , we have 󵄩 󵄩2 󵄩 󵄩2 d0(n,α) 󵄩󵄩󵄩un 󵄩󵄩󵄩 = σ(un , δx2 un ) = −σ 󵄩󵄩󵄩δx un 󵄩󵄩󵄩 , from which it is easy to know that uni = 0,

0 ⩽ i ⩽ M.

By induction principle, the conclusion holds. This completes the proof. 2.7.3 Stability of the difference scheme Theorem 2.7.2. Suppose {vin | 0 ⩽ i ⩽ M, 0 ⩽ n ⩽ N} is the solution of the difference scheme { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { {

N

exp 1 n + d0(1,α) (vin − vin−1 ) = σδx2 vin + (1 − σ)δx2 vin−1 + gin , ∑ ωl Fl,i Γ(1 − α) l=1

1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N,

1 Fl,i n Fl,i

= 0,

vi0 v0n

= φ(xi ),

=

1 ⩽ l ⩽ Nexp , 1 ⩽ i ⩽ M − 1,

n−1 e−sl τ Fl,i

= 0,

n vM

+

Al (vin−1

−

vin−2 )

+

Bl (vin

−

vin−1 ),

1 ⩽ l ⩽ Nexp , 1 ⩽ i ⩽ M − 1, 2 ⩽ n ⩽ N,

(2.138) (2.139) (2.140)

1 ⩽ i ⩽ M − 1,

(2.141)

= 0,

(2.142)

0 ⩽ n ⩽ N.

Then we have 2 󵄩󵄩 n 󵄩󵄩2 󵄩󵄩 0 󵄩󵄩2 L α 󵄩 m 󵄩2 󵄩󵄩v 󵄩󵄩 ⩽ 󵄩󵄩v 󵄩󵄩 + Γ(1 − α) max (tm 󵄩󵄩󵄩g 󵄩󵄩󵄩 ), 1 ⩽ n ⩽ N, 1⩽m⩽n 6 󵄩󵄩 n 󵄩󵄩2 󵄩󵄩 0 󵄩󵄩2 α󵄩 󵄩󵄩g m 󵄩󵄩󵄩2 ), 1 ⩽ n ⩽ N, max (tm 󵄩󵄩δx v 󵄩󵄩 ⩽ 󵄩󵄩δx v 󵄩󵄩 + Γ(1 − α) 1⩽m⩽n 󵄩 󵄩

where M−1

󵄩󵄩 m 󵄩󵄩2 󵄨 m 󵄨2 󵄩󵄩g 󵄩󵄩 = h ∑ 󵄨󵄨󵄨gi 󵄨󵄨󵄨 . i=1

(2.143) (2.144)

2.7 The fast difference method based on L2-1σ approximation for 1D problem | 143

Proof. Substituting (2.139)–(2.140) into (2.138) and eliminating the intermediate varin able {Fl,i }, we have n−1

∑ dk(n,α) (vin−k − vin−k−1 ) = σδx2 vin + (1 − σ)δx2 vin−1 + gin ,

k=0

1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N.

(2.145)

(I) Taking an inner product on both hand sides of (2.145) with σvn + (1 − σ)vn−1 , we have n−1

= =

∑ dk(n,α) (vn−k − vn−k−1 , σvn + (1 − σ)vn−1 ) k=0 (δx2 (σvn + (1 − σ)vn−1 ), σvn + (1 − σ)vn−1 ) + (g n , σvn 󵄩 󵄩2 −󵄩󵄩󵄩δx (σvn + (1 − σ)vn−1 )󵄩󵄩󵄩 + (g n , σvn + (1 − σ)vn−1 )

⩽− =

+ (1 − σ)vn−1 )

6 󵄩󵄩 n 6󵄩 n L2 󵄩 n 󵄩2 n−1 󵄩2 n−1 󵄩2 󵄩󵄩σv + (1 − σ)v 󵄩󵄩󵄩 + [ 2 󵄩󵄩󵄩σv + (1 − σ)v 󵄩󵄩󵄩 + 󵄩󵄩󵄩g 󵄩󵄩󵄩 ] 2 24 L L

L2 󵄩󵄩 n 󵄩󵄩2 󵄩g 󵄩 , 24 󵄩 󵄩

1 ⩽ n ⩽ N.

(2.146)

For the left-hand side of the inequality above, applying Lemma 2.6.1 and Lemma 1.7.3, we can get n−1

∑ dk(n,α) (vn−k − vn−k−1 , σvn + (1 − σ)vn−1 )

k=0

⩾

1 n−1 (n,α) 󵄩󵄩 n−k 󵄩󵄩2 󵄩󵄩 n−k−1 󵄩󵄩2 (󵄩󵄩v 󵄩󵄩 − 󵄩󵄩v ∑d 󵄩󵄩 ). 2 k=0 k

(2.147)

From (2.146) and (2.147), we get L2 󵄩󵄩 n 󵄩󵄩2 1 n−1 (n,α) 󵄩󵄩 n−k 󵄩󵄩2 󵄩󵄩 n−k−1 󵄩󵄩2 ∑ dk (󵄩󵄩v 󵄩󵄩 − 󵄩󵄩v 󵄩g 󵄩 , 󵄩󵄩 ) ⩽ 2 k=0 24 󵄩 󵄩

1 ⩽ n ⩽ N.

Noticing (1.146), rewrite (2.148) as n−1 2 󵄩 󵄩2 󵄩 󵄩2 (n,α) (n,α) 󵄩 󵄩󵄩v0 󵄩󵄩󵄩2 + L 󵄩󵄩󵄩g n 󵄩󵄩󵄩2 d0(n,α) 󵄩󵄩󵄩vn 󵄩󵄩󵄩 ⩽ ∑ (dk−1 − dk(n,α) )󵄩󵄩󵄩vn−k 󵄩󵄩󵄩 + dn−1 󵄩 󵄩 12 󵄩 󵄩 k=1

n−1 2 󵄩 󵄩2 󵄩 󵄩2 󵄩2 L (n,α) (n,α) 󵄩 ⩽ ∑ (dk−1 − dk(n,α) )󵄩󵄩󵄩vn−k 󵄩󵄩󵄩 + dn−1 [󵄩󵄩󵄩v0 󵄩󵄩󵄩 + Γ(1 − α)tnα 󵄩󵄩󵄩g n 󵄩󵄩󵄩 ], 6 k=1

1 ⩽ n ⩽ N. By induction principle, we can get (2.143).

(2.148)

144 | 2 Difference methods for the time-fractional subdiffusion equations (II) Taking an inner product on both hand sides of (2.145) with −δx2 (σvn +(1−σ)vn−1 ), we have n−1

∑ dk(n,α) (vn−k − vn−k−1 , −δx2 (σvn + (1 − σ)vn−1 ))

k=0

M−1

2

M−1

= −h ∑ [δx2 (σvin + (1 − σ)vin−1 )] − h ∑ [δx2 (σvin + (1 − σ)vin−1 )]gin i=1

1 󵄩 󵄩2 ⩽ 󵄩󵄩󵄩g n 󵄩󵄩󵄩 , 4

i=1

1 ⩽ n ⩽ N.

(2.149)

For the left-hand side of the inequality above, applying Lemma 2.6.1 and Lemma 1.7.3, we can get n−1

∑ dk(n,α) (vn−k − vn−k−1 , −δx2 (σvn + (1 − σ)vn−1 ))

k=0 n−1

= ∑ dk(n,α) (δx (vn−k − vn−k−1 ), δx (σvn + (1 − σ)vn−1 )) k=0

⩾

1 n−1 (n,α) 󵄩󵄩 n−k 󵄩󵄩2 󵄩󵄩 n−k−1 󵄩󵄩2 (󵄩󵄩δx v 󵄩󵄩 − 󵄩󵄩δx v ∑d 󵄩󵄩 ). 2 k=0 k

(2.150)

Substituting (2.150) into (2.149) gives 1 n−1 (n,α) 󵄩󵄩 n−k 󵄩󵄩2 󵄩󵄩 n−k−1 󵄩󵄩2 1 󵄩 n 󵄩2 (󵄩󵄩δx v 󵄩󵄩 − 󵄩󵄩δx v ∑d 󵄩󵄩 ) ⩽ 󵄩󵄩󵄩g 󵄩󵄩󵄩 , 2 k=0 k 4

1 ⩽ n ⩽ N.

(2.151)

Noticing (1.146), rewrite (2.151) as n−1 󵄩 󵄩2 󵄩 󵄩2 (n,α) (n,α) 󵄩 󵄩󵄩δx v0 󵄩󵄩󵄩2 + 1 󵄩󵄩󵄩g n 󵄩󵄩󵄩2 d0(n,α) 󵄩󵄩󵄩δx vn 󵄩󵄩󵄩 ⩽ ∑ (dk−1 − dk(n,α) )󵄩󵄩󵄩δx vn−k 󵄩󵄩󵄩 + dn−1 󵄩 󵄩 2󵄩 󵄩 k=1 n−1

󵄩 󵄩2 󵄩2 󵄩 󵄩2 (n,α) (n,α) 󵄩 ⩽ ∑ (dk−1 − dk(n,α) )󵄩󵄩󵄩δx vn−k 󵄩󵄩󵄩 + dn−1 (󵄩󵄩󵄩δx v0 󵄩󵄩󵄩 + Γ(1 − α)tnα 󵄩󵄩󵄩g n 󵄩󵄩󵄩 ), k=1

1 ⩽ n ⩽ N. By induction principle, we can get (2.144). This completes the proof. 2.7.4 Convergence of the difference scheme Theorem 2.7.3. Suppose {Uin | 0 ⩽ i ⩽ M, 0 ⩽ n ⩽ N} and {uni | 0 ⩽ i ⩽ M, 0 ⩽ n ⩽ N} are solutions of the problem (2.36)–(2.38) and the difference scheme (2.128)–(2.132), respectively. Denote ein = Uin − uni ,

0 ⩽ i ⩽ M, 0 ⩽ n ⩽ N,

2.8 The difference method based on L1 approximation for the MTTFSD equations | 145

then we have 󵄩󵄩 n 󵄩󵄩 󵄩󵄩δx e 󵄩󵄩 ⩽ √Γ(1 − α)tnα L c7 (τ2 + h2 + ϵ), 󵄩󵄩󵄩en 󵄩󵄩󵄩 ⩽ L √Γ(1 − α)t α c7 (τ2 + h2 + ϵ), n 󵄩 󵄩∞ 2

1 ⩽ n ⩽ N,

(2.152)

1 ⩽ n ⩽ N.

(2.153)

Proof. Subtracting (2.128)–(2.132) from (2.122)–(2.124), (2.126)–(2.127), respectively, we get the system of error equations { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { {

N

exp 1 n + d0(1,α) (ein − ein−1 ) = σδx2 ein + (1 − σ)δx2 ein−1 + (r7 )ni , ∑ ωl Fl,i Γ(1 − α) l=1

1 Fl,i n Fl,i

ei0 e0n

1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N,

= 0, =

1 ⩽ l ⩽ Nexp , 1 ⩽ i ⩽ M − 1,

n−1 e−sl τ Fl,i

= 0, = 0,

+ Al (ein−1 − ein−2 ) + Bl (ein − ein−1 ),

1 ⩽ l ⩽ Nexp , 1 ⩽ i ⩽ M − 1, 2 ⩽ n ⩽ N,

1 ⩽ i ⩽ M − 1,

n eM = 0,

0 ⩽ n ⩽ N.

Applying Theorem 2.7.2 and noticing (2.125), we have 󵄩󵄩 n 󵄩󵄩2 󵄩󵄩 0 󵄩󵄩2 α󵄩 m 󵄩2 󵄩󵄩δx e 󵄩󵄩 ⩽ 󵄩󵄩δx e 󵄩󵄩 + Γ(1 − α) max (tm 󵄩󵄩󵄩(r7 ) 󵄩󵄩󵄩 ) 1⩽m⩽n 2

⩽ Γ(1 − α)tnα L[c7 (τ2 + h2 + ϵ)] ,

1 ⩽ n ⩽ N.

Taking the square root on both hand sides of the inequality above, we can get (2.152). From (2.152) and Lemma 2.1.1, it is easy to obtain (2.153). This completes the proof.

2.8 The difference method based on L1 approximation for the MTTFSD equations This section will be devoted to the investigation of difference methods for solving a class of multiterm time-fractional subdiffusion (MTTFSD) equations. To simplify the statement, take a two-term case with the constant coefficients as an example. Consider the following two-term problem: α

C α C 1 { 0 Dt u(x, t) + 0 Dt u(x, t) = uxx (x, t) + f (x, t), { { { { { x ∈ (0, L), t ∈ (0, T], { { { u(x, 0) = φ(x), x ∈ (0, L), { { { { u(0, t) = μ(t), u(L, t) = ν(t), t ∈ [0, T],

(2.154) (2.155) (2.156)

where 0 < α1 < α < 1, the functions f , φ, μ, ν are all given and φ(0) = μ(0), φ(L) = ν(0). Take the same mesh partition and notations as those in Section 2.1. In addition, suppose u ∈ C (4,2) ([0, L] × [0, T]).

146 | 2 Difference methods for the time-fractional subdiffusion equations 2.8.1 Derivation of the difference scheme Considering equation (2.154) at the point (xi , tn ), one has C α 0 Dt u(xi , tn )

α

+ C0 Dt 1 u(xi , tn ) = uxx (xi , tn ) + fin ,

1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N.

Applying the L1 formula (1.60) to approximate the time-fractional derivatives and the second-order central difference quotient to approximate the spatial derivative in the equation above yields n−1 τ−α n 0 (a(α) U − )Uik − a(α) ∑ (a(α) − a(α) n−1 Ui ) n−k Γ(2 − α) 0 i k=1 n−k−1

+

n−1 τ−α1 (α ) (α1 ) (α1 ) (α ) (a0 1 Uin − ∑ (an−k−1 − an−k )Uik − an−11 Ui0 ) Γ(2 − α1 ) k=1

= δx2 Uin + fin + (r8 )ni ,

1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N,

(2.157)

where there is a positive constant c8 such that 󵄨󵄨 n󵄨 2−α 2 󵄨󵄨(r8 )i 󵄨󵄨󵄨 ⩽ c8 (τ + h ),

1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N.

(2.158)

Noticing the initial-boundary value conditions (2.155)–(2.156), one has {

Ui0 = φ(xi ),

U0n

= μ(tn ),

1 ⩽ i ⩽ M − 1,

n UM

= ν(tn ),

(2.159) 0 ⩽ n ⩽ N.

(2.160)

Neglecting the small term (r8 )ni in (2.157) and replacing the exact solution Uin with its numerical one uni produce a difference scheme for solving (2.154)–(2.156) as follows: { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { {

n−1 τ−α 0 n )uk − a(α) (a(α) u − ∑ (a(α) − a(α) n−1 ui ) n−k i Γ(2 − α) 0 i k=1 n−k−1

+

n−1 τ−α1 (α ) (α1 ) (α1 ) k (α ) (a0 1 uni − ∑ (an−k−1 − an−k )ui − an−11 u0i ) Γ(2 − α1 ) k=1

= δx2 uni + fin ,

u0i un0

= φ(xi ),

= μ(tn ),

1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N, 1 ⩽ i ⩽ M − 1,

unM

= ν(tn ),

(2.162) 0 ⩽ n ⩽ N.

Denote s = τα Γ(2 − α),

(2.161)

s1 = τα1 Γ(2 − α1 ).

In what follows, the difference scheme (2.161)–(2.163) will be analyzed.

(2.163)

2.8 The difference method based on L1 approximation for the MTTFSD equations | 147

2.8.2 Solvability of the difference scheme Theorem 2.8.1. The difference scheme (2.161)–(2.163) is uniquely solvable. Proof. Denote un = (un0 , un1 , . . . , unM ). The value of u0 is obviously determined by (2.162)–(2.163). Suppose that the values of u0 , u1 , . . . , un−1 have been uniquely determined. From (2.161) and (2.163), we can obtain the linear system in the unknown un . To prove its unique solvability, it suffices to show the corresponding homogeneous one a(α) a 1 { { ( 0 + 0 )un = δ2 un , i x i s s1 { { n n { u0 = uM = 0 (α )

1 ⩽ i ⩽ M − 1,

(2.164) (2.165)

has only the trivial solution. Suppose ‖un ‖∞ = |unin |, where in ∈ {1, 2, . . . , M − 1}. Rewrite (2.164) as a(α) a 1 1 2 0 + 0 + 2 )uni = 2 (uni−1 + uni+1 ), s s1 h h (α )

(

1 ⩽ i ⩽ M − 1.

Letting i = in in the equality above and taking the absolute value of both hand sides, by the triangle inequality, we have a 1 a(α) 2 󵄩 󵄩 2󵄩 󵄩 0 + 0 + 2 )󵄩󵄩󵄩un 󵄩󵄩󵄩∞ ⩽ 2 󵄩󵄩󵄩un 󵄩󵄩󵄩∞ . s s1 h h (α )

(

Thus ‖un ‖∞ = 0, which implies un = 0. By the principle of induction, the difference scheme (2.161)–(2.163) is uniquely solvable. The proof ends. 2.8.3 Stability of the difference scheme Theorem 2.8.2. Suppose {vin | 0 ⩽ i ⩽ M, 0 ⩽ n ⩽ N} is the solution of the difference scheme { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { {

1 (α) n n−1 (α) 0 (a v − ∑ (a − a(α) )vk − a(α) n−1 vi ) n−k i s 0 i k=1 n−k−1 +

n−1 1 (α ) (α1 ) (α1 ) k (α ) (a0 1 vin − ∑ (an−k−1 − an−k )vi − an−11 vi0 ) s1 k=1

= δx2 vin + fin , vi0 v0n

= φ(xi ),

= 0,

n vM

1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N,

(2.166)

1 ⩽ i ⩽ M − 1,

(2.167)

= 0,

(2.168)

0 ⩽ n ⩽ N.

148 | 2 Difference methods for the time-fractional subdiffusion equations Then it holds 󵄩 0󵄩 󵄩 m󵄩 󵄩󵄩 k 󵄩󵄩 󵄩󵄩v 󵄩󵄩∞ ⩽ 󵄩󵄩󵄩v 󵄩󵄩󵄩∞ + κ1 max 󵄩󵄩󵄩f 󵄩󵄩󵄩∞ , 1⩽m⩽k

1 ⩽ k ⩽ N,

(2.169)

where κ1 =

1 max{T α Γ(1 − α), T α1 Γ(1 − α1 )}, 2

󵄨 m󵄨 󵄩󵄩 m 󵄩󵄩 󵄩󵄩f 󵄩󵄩∞ = max 󵄨󵄨󵄨fi 󵄨󵄨󵄨. 1⩽i⩽M−1

Proof. Rewrite equation (2.166) as a 1 a(α) 2 0 + 0 + 2 )vin s s1 h (α )

(

1 n−1 0 − a(α) )vk + a(α) = ( ∑ (a(α) n−1 vi ) n−k i s k=1 n−k−1 + +

1 n−1 (α1 ) (α1 ) k (α ) ( ∑ (a − an−k )vi + an−11 vi0 ) s1 k=1 n−k−1 1 n n (v + vi+1 ) + fin , h2 i−1

1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N.

Suppose ‖vn ‖∞ = |vinn |, where in ∈ {1, 2, . . . , M − 1}. Letting i = in in the equality above and taking the absolute value of both hand sides, by the triangle inequality, we have a(α) a 1 2 󵄩 󵄩 0 + 0 + 2 )󵄩󵄩󵄩vn 󵄩󵄩󵄩∞ s s1 h (α )

(

1 n−1 󵄩 󵄩 󵄩󵄩 0 󵄩󵄩 − a(α) )󵄩󵄩vk 󵄩󵄩󵄩∞ + a(α) ⩽ [ ∑ (a(α) 󵄩v 󵄩󵄩∞ ] n−1 󵄩 n−k 󵄩 s k=1 n−k−1 +

2󵄩 󵄩 1 n−1 (α1 ) (α1 ) 󵄩 (α ) 󵄩 󵄩 󵄩 󵄩 󵄩 [ ∑ (a − an−k )󵄩󵄩󵄩vk 󵄩󵄩󵄩∞ + an−11 󵄩󵄩󵄩v0 󵄩󵄩󵄩∞ ] + 2 󵄩󵄩󵄩vn 󵄩󵄩󵄩∞ + 󵄩󵄩󵄩f n 󵄩󵄩󵄩∞ s1 k=1 n−k−1 h

n−1 1 (α1 ) 1 (α1 ) 󵄩 󵄩 − a(α) ) + (an−k−1 − an−k )]󵄩󵄩󵄩vk 󵄩󵄩󵄩∞ = ∑ [ (a(α) n−k n−k−1 s s 1 k=1

1 (α ) 󵄩 󵄩 2󵄩 󵄩 s 󵄩󵄩 n 󵄩󵄩 1 1 + a 1 ]󵄩󵄩v0 󵄩󵄩 + 󵄩󵄩vn 󵄩󵄩 + a(α) ⋅ + [ a(α) 󵄩f 󵄩 s n−1 s1 n−1 󵄩 󵄩∞ h2 󵄩 󵄩∞ s n−1 2a(α) 󵄩 󵄩∞ n−1 s1 󵄩󵄩 n 󵄩󵄩 1 (α ) + an−11 ⋅ (α 󵄩f 󵄩 , 1 ⩽ n ⩽ N. ) 󵄩 󵄩∞ s1 2a 1

(2.170)

n−1

By virtue of Lemma 1.6.1, we conclude that s

2a(α) n−1 s1

2an−11

(α )

⩽

⩽

t α Γ(1 − α) τα Γ(2 − α) = n , −α 2(1 − α)n 2 α

τα1 Γ(2 − α1 ) t 1 Γ(1 − α1 ) = n . −α 2(1 − α1 )n 1 2

(2.171) (2.172)

2.8 The difference method based on L1 approximation for the MTTFSD equations | 149

Inserting (2.171)–(2.172) into (2.170) yields a(α) a 1 󵄩 󵄩 ( 0 + 0 )󵄩󵄩󵄩vn 󵄩󵄩󵄩∞ s s1 (α )

n−1 1 1 (α1 ) (α1 ) 󵄩 󵄩 ⩽ ∑ [ (a(α) − an−k )]󵄩󵄩󵄩vk 󵄩󵄩󵄩∞ − a(α) ) + (an−k−1 n−k−1 n−k s s 1 k=1

1 1 (α ) 󵄩 󵄩 󵄩 󵄩 + ( a(α) + a 1 )(󵄩󵄩v0 󵄩󵄩 + κ1 󵄩󵄩󵄩f n 󵄩󵄩󵄩∞ ), s n−1 s1 n−1 󵄩 󵄩∞

1 ⩽ n ⩽ N.

(2.173)

Next, we proceed by the mathematical induction to prove the truth of (2.169). In view of (2.173), when n = 1, a 1 󵄩 󵄩 a(α) a 1 󵄩 󵄩 a(α) 󵄩 󵄩 0 + 0 )󵄩󵄩󵄩v1 󵄩󵄩󵄩∞ ⩽ ( 0 + 0 )(󵄩󵄩󵄩v0 󵄩󵄩󵄩∞ + κ1 󵄩󵄩󵄩f 1 󵄩󵄩󵄩∞ ), s s1 s s1 (α )

(

(α )

thus 󵄩󵄩 1 󵄩󵄩 󵄩 0󵄩 󵄩 1󵄩 󵄩󵄩v 󵄩󵄩∞ ⩽ 󵄩󵄩󵄩v 󵄩󵄩󵄩∞ + κ1 󵄩󵄩󵄩f 󵄩󵄩󵄩∞ . Obviously, (2.169) is true for k = 1. Suppose that (2.169) is true for k = 1, 2, . . . , n − 1. From (2.173), we can obtain a 1 󵄩 󵄩 a(α) 0 + 0 )󵄩󵄩󵄩vn 󵄩󵄩󵄩∞ s s1 (α )

(

n−1 1 (α1 ) 1 (α1 ) 󵄩 󵄩 − a(α) ) + (an−k−1 − an−k )](󵄩󵄩󵄩v0 󵄩󵄩󵄩∞ ⩽ ∑ [ (a(α) n−k−1 n−k s s 1 k=1

1 (α ) 󵄩 󵄩 1 󵄩 󵄩 󵄩 󵄩 + a 1 ](󵄩󵄩v0 󵄩󵄩 + κ1 󵄩󵄩󵄩f n 󵄩󵄩󵄩∞ ) + κ1 max 󵄩󵄩󵄩f m 󵄩󵄩󵄩∞ ) + [ a(α) s n−1 s1 n−1 󵄩 󵄩∞ 1⩽m⩽k

n−1 1 1 (α1 ) (α1 ) ⩽ { ∑ [ (a(α) − an−k )] − a(α) ) + (an−k−1 n−k−1 n−k s s 1 k=1

1 1 (α ) 󵄩 󵄩 󵄩 󵄩 + ( a(α) + a 1 )} ⋅ (󵄩󵄩󵄩v0 󵄩󵄩󵄩∞ + κ1 max 󵄩󵄩󵄩f m 󵄩󵄩󵄩∞ ) 1⩽m⩽n s n−1 s1 n−1 a 1 a(α) 󵄩 󵄩 󵄩 󵄩 0 + 0 )(󵄩󵄩󵄩v0 󵄩󵄩󵄩∞ + κ1 max 󵄩󵄩󵄩f m 󵄩󵄩󵄩∞ ). 1⩽m⩽n s s1 (α )

=(

Hence (2.169) is also true for k = n. The desired result can be obtained by the principle of induction. The proof is completed. 2.8.4 Convergence of the difference scheme Theorem 2.8.3. Suppose that {Uin | 0 ⩽ i ⩽ M, 0 ⩽ n ⩽ N} and {uni | 0 ⩽ i ⩽ M, 0 ⩽ n ⩽ N} are solutions of the problem (2.154)–(2.156) and the difference scheme (2.161)–(2.163),

150 | 2 Difference methods for the time-fractional subdiffusion equations respectively. Let ein = Uin − uni ,

0 ⩽ i ⩽ M, 0 ⩽ n ⩽ N.

Then it holds 󵄩󵄩 n 󵄩󵄩 2−α 2 󵄩󵄩e 󵄩󵄩∞ ⩽ c8 κ1 (τ + h ),

1 ⩽ n ⩽ N.

Proof. The subtraction of (2.161)–(2.163) from (2.157), (2.159)–(2.160), respectively, gives the system of error equations as follows: { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { {

1 (α) n n−1 (α) 0 [a e − ∑ (a − a(α) )ek − a(α) n−1 ei ] n−k i s 0 i k=1 n−k−1 +

1 (α1 ) n n−1 (α1 ) (α1 ) k (α ) [a e − ∑ (a − an−k )ei − an−11 ei0 ] s1 0 i k=1 n−k−1

= δx2 ein + (r8 )ni ,

ei0 e0n

= 0,

= 0,

1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N,

1 ⩽ i ⩽ M − 1,

n eM = 0,

0 ⩽ n ⩽ N.

Noticing (2.158), Theorem 2.8.2 immediately implies 󵄩󵄩 n 󵄩󵄩 󵄩 0󵄩 󵄩 m󵄩 2−α 2 󵄩󵄩e 󵄩󵄩∞ ⩽ 󵄩󵄩󵄩e 󵄩󵄩󵄩∞ + κ1 max 󵄩󵄩󵄩(r8 ) 󵄩󵄩󵄩∞ ⩽ c8 κ1 (τ + h ), 1⩽m⩽n

1 ⩽ n ⩽ N.

The proof ends.

2.9 The difference method based on L2-1σ approximation for the MTTFSD equations Consider the following problem of the multiterm time-fractional subdiffusion (MTTFSD) equations: m

α { { ∑ λr C D r u(x, t) = uxx (x, t) + f (x, t), x ∈ (0, L), t ∈ (0, T], { { { r=0 0 t { { u(x, 0) = φ(x), x ∈ (0, L), { { { { u(0, t) = μ(t), u(L, t) = ν(t), t ∈ [0, T],

(2.174) (2.175) (2.176)

where λ0 , λ1 , . . . , λm are positive constants, 0 ⩽ αm < αm−1 < ⋅ ⋅ ⋅ < α0 ⩽ 1, at least one αr ∈ (0, 1), functions f , φ, μ, ν are all given, and φ(0) = μ(0), φ(L) = ν(0). Suppose the exact solution u ∈ C (4,3) ([0, L] × [0, T]).

2.9 The difference method based on L2-1σ approximation for the MTTFSD equations | 151

2.9.1 Derivation of the difference scheme Applying the theory in Subsection 1.6.4, let σ be the unique root of equation F(σ) = 0, tn−1+σ = (n − 1 + σ)τ, fin−1+σ = f (xi , tn−1+σ ). Considering (2.174) at the point (xi , tn−1+σ ), we have m

α

∑ λr C0 Dt r u(xi , tn−1+σ )

r=0

= uxx (xi , tn−1+σ ) + fin−1+σ , 1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N.

(2.177)

Using the theory in Subsection 1.6.4 to discretize the time fractional derivative, we have m

n−1

α

∑ λr C0 Dt r u(xi , tn−1+σ ) = ∑ ĉk(n,α) (Uin−k − Uin−k−1 ) + O(τ3−α0 ).

r=0

k=0

(2.178)

Using the linear interpolation and the second-order central difference quotient to approximate the spatial second-order derivative, we have uxx (xi , tn−1+σ ) = σuxx (xi , tn ) + (1 − σ)uxx (xi , tn−1 ) + O(τ2 )

= σδx2 Uin + (1 − σ)δx2 Uin−1 + O(h2 ) + O(τ2 ).

(2.179)

Inserting (2.178) and (2.179) into (2.177), we have n−1

∑ ĉk(n,α) (Uin−k − Uin−k−1 )

k=0

= σδx2 Uin + (1 − σ)δx2 Uin−1 + fin−1+σ + (r9 )ni ,

1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N

(2.180)

and there exists a positive constant c9 such that 󵄨󵄨 n 󵄨󵄨 2 2 󵄨󵄨(r9 )i 󵄨󵄨 ⩽ c9 (τ + h ),

1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N.

(2.181)

Noticing the initial-boundary value conditions (2.175)–(2.176), we have {

Ui0 = φ(xi ),

U0n = μ(tn ),

1 ⩽ i ⩽ M − 1,

n UM = ν(tn ),

(2.182) 0 ⩽ n ⩽ N.

(2.183)

Omitting the small term (r9 )ni in (2.180) and using numerical solution uni to replace the exact solution Uin , we construct for the problem (2.174)–(2.176) the following difference scheme: n−1

{ { − un−k−1 ) = σδx2 uni + (1 − σ)δx2 un−1 + fin−1+σ , ∑ ĉk(n,α) (un−k { i i i { { { k=0 { { 1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N, { { { 0 { { u = φ(xi ), 1 ⩽ i ⩽ M − 1, { { { i n n { u0 = μ(tn ), uM = ν(tn ), 0 ⩽ n ⩽ N.

(2.184) (2.185) (2.186)

152 | 2 Difference methods for the time-fractional subdiffusion equations 2.9.2 Solvability of the difference scheme Theorem 2.9.1. The difference scheme (2.184)–(2.186) is uniquely solvable. Proof. Denote un = (un0 , un1 , . . . , unM ). From (2.185)–(2.186) we get u0 . Suppose u0 , u1 , . . . , un−1 have been uniquely determined, we can get the system of linear equations in un from (2.184) and (2.186). It suffices to show that the homogeneous system {

ĉ0(n,α) uni = σδx2 uni , un0

=

unM

1 ⩽ i ⩽ M − 1,

(2.187)

=0

(2.188)

has only the trivial solution. Taking an inner product on both hand sides of (2.187) with un , we have 󵄩 󵄩2 󵄩 󵄩2 ĉ0(n,α) 󵄩󵄩󵄩un 󵄩󵄩󵄩 = σ(un , δx2 un ) = −σ 󵄩󵄩󵄩δx un 󵄩󵄩󵄩 . It is easy to know that uni = 0,

0 ⩽ i ⩽ M.

By induction principle, the theorem holds. This completes the proof. 2.9.3 Stability of the difference scheme Theorem 2.9.2. Suppose {vin | 0 ⩽ i ⩽ M, 0 ⩽ n ⩽ N} is the solution of the difference scheme n−1

{ { ∑ ĉk(n,α) (vin−k − vin−k−1 ) = σδx2 vin + (1 − σ)δx2 vin−1 + gin , { { { { k=0 { { 1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N, { { { 0 { { v = φ(xi ), 1 ⩽ i ⩽ M − 1, { { { i n n { v0 = 0, vM = 0, 0 ⩽ n ⩽ N.

(2.189) (2.190) (2.191)

Then we have 󵄩󵄩 n 󵄩󵄩2 󵄩󵄩 0 󵄩󵄩2 󵄩󵄩δx v 󵄩󵄩 ⩽ 󵄩󵄩δx v 󵄩󵄩 +

2 ∑m r=0

1

󵄩 󵄩2 max󵄩󵄩g l 󵄩󵄩󵄩 ,

󵄩 λr 1⩽l⩽n T αr Γ(1−αr )

where M−1

󵄩󵄩 l 󵄩󵄩2 󵄨 l 󵄨2 󵄩󵄩g 󵄩󵄩 = h ∑ 󵄨󵄨󵄨gi 󵄨󵄨󵄨 . i=1

1 ⩽ n ⩽ N,

(2.192)

2.9 The difference method based on L2-1σ approximation for the MTTFSD equations | 153

Proof. Taking an inner product on both hand sides of (2.189) with −δx2 (σvn +(1−σ)vn−1 ), we have n−1

∑ ĉk(n,α) (vn−k − vn−k−1 , −δx2 (σvn + (1 − σ)vn−1 ))

k=0

󵄩 󵄩2 = −󵄩󵄩󵄩δx2 (σvn + (1 − σ)vn−1 )󵄩󵄩󵄩 − (δx2 (σvn + (1 − σ)vn−1 ), g n ) 1 󵄩 󵄩2 ⩽ 󵄩󵄩󵄩g n 󵄩󵄩󵄩 , 1 ⩽ n ⩽ N. 4

(2.193)

For the left-hand side of the inequality above, applying Lemmas 2.6.1, 1.6.6 and 1.6.7, we have n−1

∑ ĉk(n,α) (vn−k − vn−k−1 , −δx2 (σvn + (1 − σ)vn−1 ))

k=0 n−1

= ∑ ĉk(n,α) (δx (vn−k − vn−k−1 ), δx (σvn + (1 − σ)vn−1 )) k=0

⩾

1 n−1 (n,α) 󵄩󵄩 n−k 󵄩󵄩2 󵄩󵄩 n−k−1 󵄩󵄩2 ⋅ ∑ ĉ (󵄩󵄩δx v 󵄩󵄩 − 󵄩󵄩δx v 󵄩󵄩 ). 2 k=0 k

(2.194)

From (2.194) and (2.193), we have 1 󵄩 n 󵄩2 1 n−1 (n,α) 󵄩󵄩 n−k 󵄩󵄩2 󵄩󵄩 n−k−1 󵄩󵄩2 ⋅ ∑ ĉ (󵄩󵄩δx v 󵄩󵄩 − 󵄩󵄩δx v 󵄩󵄩 ) ⩽ 󵄩󵄩󵄩g 󵄩󵄩󵄩 , 2 k=0 k 4

1 ⩽ n ⩽ N,

that is, 2

󵄩 󵄩 ĉ0(n,α) 󵄩󵄩󵄩δx vn 󵄩󵄩󵄩

n−1 󵄩 󵄩2 (n,α) (n,α) 󵄩 󵄩󵄩δx v0 󵄩󵄩󵄩2 + 1 󵄩󵄩󵄩g n 󵄩󵄩󵄩2 , − ĉk(n,α) )󵄩󵄩󵄩δx vn−k 󵄩󵄩󵄩 + ĉn−1 ⩽ ∑ (ĉk−1 󵄩 󵄩 2󵄩 󵄩 k=1

1 ⩽ n ⩽ N.

From Lemma 1.6.6, we know m

(n,α) ĉn−1 > ∑ λr r=0

m λ τ−αr n−αr ⩾ ∑ α r . r Γ(1 − αr ) T Γ(1 − αr ) r=0

It follows that 2

󵄩 󵄩 ĉ0(n,α) 󵄩󵄩󵄩δx vn 󵄩󵄩󵄩 n−1

󵄩 󵄩2 󵄩2 (n,α) (n,α) 󵄩 ⩽ ∑ (ĉk−1 − ĉk(n,α) )󵄩󵄩󵄩δx vn−k 󵄩󵄩󵄩 + ĉn−1 (󵄩󵄩󵄩δx v0 󵄩󵄩󵄩 + k=1

n−1

󵄩 󵄩2 (n,α) ⩽ ∑ (ĉk−1 − ĉk(n,α) )󵄩󵄩󵄩δx vn−k 󵄩󵄩󵄩 k=1

1 󵄩󵄩 n 󵄩󵄩2 󵄩g 󵄩 ) (n,α) 󵄩 󵄩 ̂ 2cn−1

154 | 2 Difference methods for the time-fractional subdiffusion equations 󵄩2 (n,α) 󵄩 + ĉn−1 [󵄩󵄩󵄩δx v0 󵄩󵄩󵄩 +

1

λr 2 ∑m r=0 T αr Γ(1−αr )

󵄩󵄩 n 󵄩󵄩2 󵄩󵄩g 󵄩󵄩 ],

1 ⩽ n ⩽ N.

By mathematical induction, we can get (2.192). This completes the proof. From Theorem 2.9.2, we can know that the difference scheme (2.184)–(2.186) is stable with respect to the initial value and right-hand function.

2.9.4 Convergence of the difference scheme Theorem 2.9.3. Suppose that {Uin | 0 ⩽ i ⩽ M, 0 ⩽ n ⩽ N} and {uni | 0 ⩽ i ⩽ M, 0 ⩽ n ⩽ N} are solutions of the problem (2.174)–(2.176) and the difference scheme (2.184)–(2.186), respectively. Denote ein = Uin − uni ,

0 ⩽ i ⩽ M, 0 ⩽ n ⩽ N,

then we have 󵄩󵄩 n 󵄩󵄩 󵄩󵄩δx e 󵄩󵄩 ⩽ √

L

c9 (τ2 + h2 ),

λr 2 ∑m r=0 T αr Γ(1−αr )

1

1 󵄩󵄩 n 󵄩󵄩 󵄩󵄩e 󵄩󵄩∞ ⩽ √ m 2 2∑

λr r=0 T αr Γ(1−αr )

1 ⩽ n ⩽ N,

Lc9 (τ2 + h2 ),

1 ⩽ n ⩽ N.

(2.195) (2.196)

Proof. Subtracting (2.184)–(2.186) from (2.180), (2.182)–(2.183), we have the system of error equations n−1

{ { ∑ ĉk(n,α) (ein−k − ein−k−1 ) = σδx2 ein + (1 − σ)δx2 ein−1 + (r9 )ni , { { { { { { k=0 1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N, { { { 0 { { e = 0, 1 ⩽ i ⩽ M − 1, { { { in n { e0 = 0, eM = 0, 0 ⩽ n ⩽ N. Applying Theorem 2.9.2 and noticing (2.181), we have 󵄩󵄩 n 󵄩󵄩2 󵄩󵄩 0 󵄩󵄩2 󵄩󵄩δx e 󵄩󵄩 ⩽ 󵄩󵄩δx e 󵄩󵄩 + ⩽

1

1

󵄩󵄩

m󵄩 󵄩2

max 󵄩(r9 ) λr 1⩽m⩽n󵄩 2 ∑m r=0 T αr Γ(1−αr )

λr 2 ∑m r=0 T αr Γ(1−αr )

2

L[c9 (τ2 + h2 )] ,

󵄩󵄩

1 ⩽ n ⩽ N.

Taking the square root on both hand sides of the inequality above, we get (2.195). From (2.195) and Lemma 2.1.1, it is easy to obtain (2.196). This completes the proof.

2.10 The ADI method based on G-L approximation for 2D problem

| 155

2.10 The ADI method based on G-L approximation for 2D problem In this section, the finite difference method for solving two-dimensional time-fractional subdiffusion equations will be investigated. For this, consider the following problem: C α { 0 Dt u(x, y, t) = uxx (x, y, t) + uyy (x, y, t) + f (x, y, t), { { { { { (x, y) ∈ Ω, t ∈ (0, T], { { { u(x, y, 0) = 0, (x, y) ∈ Ω, { { { { u(x, y, t) = μ(x, y, t), (x, y) ∈ 𝜕Ω, t ∈ [0, T],

(2.197) (2.198) (2.199)

where Ω = (0, L1 )×(0, L2 ), α ∈ (0, 1), the functions f , μ are given and μ(x, y, 0)|(x,y)∈𝜕Ω = 0. For the numerical approximation, the mesh partition is made firstly. Assume that h1 = L1 /M1 , h2 = L2 /M2 and τ = T/N for some positive integers M1 , M2 and N. Let xi = ih1 (0 ⩽ i ⩽ M1 ), yj = jh2 (0 ⩽ j ⩽ M2 ) and tk = kτ (0 ⩽ k ⩽ N). Define Ω̄ h = {(xi , yj ) | 0 ⩽ i ⩽ M1 , 0 ⩽ j ⩽ M2 } as a set of mesh points on Ω,̄ Ωh = Ω̄ h ∩ Ω and 𝜕Ωh = Ω̄ h ∩ 𝜕Ω as sets of interior mesh points and boundary mesh points, respectively. Let ω = {(i, j) | (xi , yj ) ∈ Ωh }, 𝜕ω = {(i, j) | (xi , yj ) ∈ 𝜕Ωh }, ω̄ = ω ∪ 𝜕ω, Ωτ = {tk | 0 ⩽ k ⩽ N}. Define two mesh function spaces as follows: ̄ is a mesh function defined on Ω̄ h }, 𝒱h = {u | u = {uij | (i, j) ∈ ω} 𝒱̊h = {u | u ∈ 𝒱h ; uij = 0 if (i, j) ∈ 𝜕ω}.

For any mesh function v ∈ 𝒱h , introduce some notations: 1 1 (v − vi−1,j ), δx2 vij = (δx vi+ 1 ,j − δx vi− 1 ,j ), 2 2 h1 ij h1 1 1 δy vi,j− 1 = (vij − vi,j−1 ), δy2 vij = (δy vi,j+ 1 − δy vi,j− 1 ), 2 2 2 h2 h2 1 δx δy vi− 1 ,j− 1 = (δy vi,j− 1 − δy vi−1,j− 1 ), 2 2 2 2 h1 1 δx2 δy2 vij = 2 (δy2 vi−1,j − 2δy2 vij + δy2 vi+1,j ), h1

δx vi− 1 ,j = 2

δx δy2 vi− 1 ,j , δy δx2 vi,j− 1 and others can be defined similarly. 2 2 For any mesh functions u, v ∈ 𝒱̊h , define M1 −1 M2 −1

‖u‖ = √(u, u),

(u, v) = h1 h2 ∑ ∑ uij vij , i=1

j=1

M1 M2 −1

(δx u, δx v) = h1 h2 ∑ ∑ (δx ui− 1 ,j )(δx vi− 1 ,j ), i=1 j=1

2

2

M1 −1 M2

(δy u, δy v) = h1 h2 ∑ ∑(δy ui,j− 1 )(δy vi,j− 1 ), i=1 j=1

2

2

‖δx u‖ = √(δx u, δx u), ‖δy u‖ = √(δy u, δy u),

156 | 2 Difference methods for the time-fractional subdiffusion equations M1 M2

(δx δy u, δx δy v) = h1 h2 ∑ ∑(δx δy ui− 1 ,j− 1 )(δx δy vi− 1 ,j− 1 ), 2

i=1 j=1

2

2

‖∇h u‖ = √‖δx u‖2 + ‖δy u‖2 ,

‖δx δy u‖ = √(δx δy u, δx δy u), ‖u‖∞ =

2

max

1⩽i⩽M1 −1,1⩽j⩽M2 −1

|uij |.

It is easy to check that for any mesh functions u, v ∈ 𝒱̊h , it holds M1 −1 M2 −1

(−δx2 u, v) ≡ h1 h2 ∑ ∑ (−δx2 uij )vij = (δx u, δx v), i=1

(2.200)

j=1

M1 −1 M2 −1

(−δy2 u, v) ≡ h1 h2 ∑ ∑ (−δy2 uij )vij = (δy u, δy v), i=1

(2.201)

j=1

M1 −1 M2 −1

(δx2 δy2 u, v) ≡ h1 h2 ∑ ∑ (δx2 δy2 uij )vij = (δx δy u, δx δy v). i=1

(2.202)

j=1

In addition, we denote ℐ as the unit operator, or, the identity operator. The next lemma states a relationship between two different norms. Lemma 2.10.1.

[75]

For any mesh function u ∈ 𝒱̊h , we have ‖u‖2 ⩽

6 L21

1 +

6 L22

‖∇h u‖2 .

2.10.1 Derivation of the difference scheme Define mesh functions Uijn = u(xi , yj , tn ),

fijn = f (xi , yj , tn ),

(i, j) ∈ ω,̄

0 ⩽ n ⩽ N.

For any fixed (x, y) ∈ Ω,̄ define a function { { { { { ̂ y, t) = { u(x, { { { { {

0, u(x, y, t),

t < 0, 0 ⩽ t ⩽ T,

v(x, y, t), 0,

T < t < 2T, t ⩾ 2T,

k

k

with v(x, y, t) a smooth function satisfying 𝜕 v(x,y,t) |t=T = 𝜕 u(x,y,t) |t=T and 𝜕t k 𝜕t k 1+α ̄ ̂ y, ⋅) ∈ C (ℛ) and u(⋅, ⋅, t) ∈ C (4,4) (Ω). 0, k = 0, 1, 2. Assume that u(x,

𝜕k v(x,y,t) |t=2T 𝜕t k

Considering equation (2.197) at the point (xi , yj , tn ), one has C α 0 Dt u(xi , yj , tn )

= uxx (xi , yj , tn ) + uyy (xi , yj , tn ) + fijn ,

(i, j) ∈ ω, 1 ⩽ n ⩽ N.

=

2.10 The ADI method based on G-L approximation for 2D problem

| 157

Noticing the close relationship between the Caputo derivative and the R-L derivative under the zero initial value condition (2.198), by Theorem 1.4.2 and Lemma 2.1.3, one can obtain n

τ−α ∑ gk(α) Uijn−k = δx2 Uijn + δy2 Uijn + fijn + (r10 )nij , k=0

(i, j) ∈ ω, 1 ⩽ n ⩽ N,

(2.203)

where there is a positive constant c10 such that 󵄨󵄨 n󵄨 2 2 󵄨󵄨(r10 )ij 󵄨󵄨󵄨 ⩽ c10 (τ + h1 + h2 ),

(i, j) ∈ ω, 1 ⩽ n ⩽ N.

Adding a small perturbation term τ2α δx2 δy2 (τ−α ∑nk=0 gk(α) Uijn−k ) into (2.203) arrives at n

n

τ−α ∑ gk(α) Uijn−k + τ2α δx2 δy2 (τ−α ∑ gk(α) Uijn−k ) k=0

k=0

= δx2 Uijn + δy2 Uijn + fijn + (r11 )nij ,

(i, j) ∈ ω, 1 ⩽ n ⩽ N,

(2.204)

where n

(r11 )nij = (r10 )nij + τ2α (τ−α ∑ gk(α) δx2 δy2 Uijn−k ) k=0

and there is a positive constant c11 such that 󵄨󵄨 n󵄨 min{1,2α} + h21 + h22 ), 󵄨󵄨(r11 )ij 󵄨󵄨󵄨 ⩽ c11 (τ

(i, j) ∈ ω, 1 ⩽ n ⩽ N.

(2.205)

Noticing the initial-boundary value conditions (2.198)–(2.199), one has {

Uij0 = 0,

Uijn

(i, j) ∈ ω,

= μ(xi , yj , tn ),

(2.206)

(i, j) ∈ 𝜕ω, 0 ⩽ n ⩽ N.

(2.207)

Neglecting the small term (r11 )nij in (2.204) and replacing the exact solution Uijn with its numerical one unij , we get a difference scheme for solving (2.197)–(2.199) in the form of { { { { { { { { { { { { { { { { { { {

n

n

τ−α ∑ gk(α) un−k + τ2α δx2 δy2 (τ−α ∑ gk(α) un−k ij ) ij

k=0 k=0 2 n 2 n n = δx uij + δy uij + fij , (i, j) ∈ ω, 1 ⩽ n ⩽ N, u0ij = 0, (i, j) ∈ ω, unij = μ(xi , yj , tn ), (i, j) ∈ 𝜕ω, 0 ⩽ n ⩽ N.

Equation (2.208) can be rewritten as unij − τα δx2 unij − τα δy2 unij + τ2α δx2 δy2 unij

(2.208) (2.209) (2.210)

158 | 2 Difference methods for the time-fractional subdiffusion equations n

α n = ∑ (−gk(α) )(un−k + τ2α δx2 δy2 un−k ij ij ) + τ fij , k=1

namely, (ℐ − τα δx2 )(ℐ − τα δy2 )unij n

= ∑ (−gk(α) )(ℐ + τ2α δx2 δy2 )un−k + τα fijn , ij k=1

(i, j) ∈ ω, 1 ⩽ n ⩽ N.

Let u∗ij = (ℐ − τα δy2 )unij , then the difference scheme (2.208)–(2.210) can be reformulated as the following ADI form: On each time level t = tn (1 ⩽ n ⩽ N), firstly, for any fixed j from 1 to M2 − 1, solve a series of linear systems in the unknown {u∗ij | 0 ⩽ i ⩽ M1 } in x direction n

{ { (ℐ − τα δx2 )u∗ij = ∑ (−gk(α) )(ℐ + τ2α δx2 δy2 )un−k + τα fijn , ij { k=1 { ∗ α 2 n ∗ α 2 n { u0j = (ℐ − τ δy )u0j , uM1 ,j = (ℐ − τ δy )uM1 ,j

1 ⩽ i ⩽ M1 − 1,

to obtain the value of {u∗ij | 1 ⩽ i ⩽ M1 − 1} on an intermediate time level. Then, for any fixed i from 1 to M1 − 1, carry out some calculations for the unknown {unij | 0 ⩽ j ⩽ M2 } in y direction {

(ℐ − τα δy2 )unij = u∗ij , 1 ⩽ j ⩽ M2 − 1, uni0 = μ(xi , y0 , tn ), uni,M2 = μ(xi , yM2 , tn )

to get the desired value of {unij | 1 ⩽ j ⩽ M2 − 1}. 2.10.2 Solvability of the difference scheme Theorem 2.10.1. The difference scheme (2.208)–(2.210) is uniquely solvable. Proof. Denote ̄ un = {unij | (i, j) ∈ ω}.

2.10 The ADI method based on G-L approximation for 2D problem

| 159

We proceed by the mathematical induction. The values of u0 is obviously determined by (2.209)–(2.210). Now assume that the values of u0 , u1 , . . . , un−1 have been uniquely determined, then we can obtain the linear system in the unknown un from (2.208) and (2.210). To show its unique solvability, it is sufficient to verify that the corresponding homogeneous one {

τ−α g0(α) unij + τα g0(α) δx2 δy2 unij = δx2 unij + δy2 unij , unij

= 0,

(i, j) ∈ ω,

(i, j) ∈ 𝜕ω

(2.211) (2.212)

has only the trivial solution. To this end, making the inner product on both hand sides of (2.211) with un , respectively, and noticing (2.212), it follows from (2.200)–(2.202) that τ−α (un , un ) + τα (δx δy un , δx δy un ) = −[(δx un , δx un ) + (δy un , δy un )]. Thus 󵄩 󵄩2 󵄩 󵄩2 󵄩 󵄩2 τ−α 󵄩󵄩󵄩un 󵄩󵄩󵄩 + τα 󵄩󵄩󵄩δx δy un 󵄩󵄩󵄩 = −󵄩󵄩󵄩∇h un 󵄩󵄩󵄩 ⩽ 0, which implies ‖un ‖ = 0. Then noticing (2.212), un = 0 can be concluded. By the principle of induction, the difference scheme (2.208)–(2.210) is uniquely solvable. The proof ends. In what follows, the stability and convergence of the difference scheme (2.208)– (2.210) will be discussed.

2.10.3 Stability of the difference scheme Theorem 2.10.2. Suppose {vijn | (i, j) ∈ ω,̄ 0 ⩽ n ⩽ N} is the solution of the difference scheme { { { { { { { { { { { { { { { { { { {

n

n

τ−α ∑ gk(α) vijn−k + τα ∑ gk(α) δx2 δy2 vijn−k

k=0 k=0 2 n 2 n n = δx vij + δy vij + fij , (i, j) ∈ ω, 1 vij0 = φ(xi , yj ), (i, j) ∈ ω, vijn = 0, (i, j) ∈ 𝜕ω, 0 ⩽ n ⩽ N.

Then it holds 󵄩󵄩 n 󵄩󵄩2 2α 󵄩 n 󵄩2 󵄩󵄩v 󵄩󵄩 + τ 󵄩󵄩󵄩δx δy v 󵄩󵄩󵄩 5 󵄩󵄩 0 󵄩󵄩2 󵄩 󵄩2 ⩽ (󵄩v 󵄩 + τ2α 󵄩󵄩󵄩δx δy v0 󵄩󵄩󵄩 ) 1−α 󵄩 󵄩

⩽ n ⩽ N,

(2.213) (2.214) (2.215)

160 | 2 Difference methods for the time-fractional subdiffusion equations

+

L21 L22 5 󵄩 󵄩2 t α max 󵄩󵄩f m 󵄩󵄩 , ⋅ 12(L21 + L22 ) (1 − α)2α n 1⩽m⩽n󵄩 󵄩

1 ⩽ n ⩽ N,

(2.216)

where M1 −1 M2 −1

󵄩󵄩 m 󵄩󵄩2 m 2 󵄩󵄩f 󵄩󵄩 = h1 h2 ∑ ∑ (fij ) . i=1

j=1

Proof. Taking an inner product on both hand sides of (2.213) with vn , respectively, and noticing (2.215), it follows from (2.200)–(2.202) that n

n

τ−α ∑ gk(α) (vn−k , vn ) + τα ∑ gk(α) (δx δy vn−k , δx δy vn )

= =

k=0 k=0 n n n n −(δx v , δx v ) − (δy v , δy v ) + (f n , vn ) 󵄩 󵄩2 −󵄩󵄩󵄩∇h vn 󵄩󵄩󵄩 + (f n , vn ), 1 ⩽ n ⩽ N.

(2.217)

By means of the Cauchy–Schwarz inequality and Lemma 2.10.1, we get 1 󵄩 󵄩2 1 1 󵄩 󵄩 󵄩 󵄩 (f n , vn ) ⩽ 󵄩󵄩󵄩f n 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩vn 󵄩󵄩󵄩 ⩽ 6( 2 + 2 )󵄩󵄩󵄩vn 󵄩󵄩󵄩 + L1 L2 24( L12 + 󵄩 󵄩2 ⩽ 󵄩󵄩󵄩∇h vn 󵄩󵄩󵄩 +

1

24( L12 + 1

1

1 ) L22

󵄩󵄩 n 󵄩󵄩2 󵄩󵄩f 󵄩󵄩 ,

󵄩󵄩 n 󵄩󵄩2 f 󵄩󵄩

󵄩 1 󵄩 ) L22

1 ⩽ n ⩽ N.

(2.218)

The substitution of (2.218) into (2.217) arrives at n

τ−α ∑ gk(α) [(vn−k , vn ) + τ2α (δx δy vn−k , δx δy vn )] ⩽

k=0 L21 L22 󵄩󵄩 n 󵄩󵄩2 󵄩f 󵄩 , 24(L21 + L22 ) 󵄩 󵄩

1 ⩽ n ⩽ N.

Rearranging the above result and again using the Cauchy–Schwarz inequality lead to 󵄩󵄩 n 󵄩󵄩2 2α 󵄩 n 󵄩2 󵄩󵄩v 󵄩󵄩 + τ 󵄩󵄩󵄩δx δy v 󵄩󵄩󵄩 n

⩽ ∑ (−gk(α) )[(vn−k , vn ) + τ2α (δx δy vn−k , δx δy vn )] + k=1 n

L21 L22 󵄩 󵄩2 τα 󵄩󵄩f n 󵄩󵄩 24(L21 + L22 ) 󵄩 󵄩

1 󵄩 1 󵄩 󵄩2 󵄩 󵄩2 󵄩2 󵄩 󵄩2 ⩽ ∑ (−gk(α) )[ (󵄩󵄩󵄩vn−k 󵄩󵄩󵄩 + 󵄩󵄩󵄩vn 󵄩󵄩󵄩 ) + τ2α (󵄩󵄩󵄩δx δy vn−k 󵄩󵄩󵄩 + 󵄩󵄩󵄩δx δy vn 󵄩󵄩󵄩 )] 2 2 k=1 +

L21 L22 󵄩 󵄩2 τα 󵄩󵄩f n 󵄩󵄩 , 24(L21 + L22 ) 󵄩 󵄩

1 ⩽ n ⩽ N.

It follows by noticing ∑nk=1 (−gk(α) ) ⩽ g0(α) = 1 that 󵄩󵄩 n 󵄩󵄩2 2α 󵄩 n 󵄩2 󵄩󵄩v 󵄩󵄩 + τ 󵄩󵄩󵄩δx δy v 󵄩󵄩󵄩 n

󵄩 󵄩2 󵄩 󵄩2 ⩽ ∑ (−gk(α) )(󵄩󵄩󵄩vn−k 󵄩󵄩󵄩 + τ2α 󵄩󵄩󵄩δx δy vn−k 󵄩󵄩󵄩 ) k=1

2.10 The ADI method based on G-L approximation for 2D problem

+

L21 L22 󵄩 󵄩2 τα 󵄩󵄩f n 󵄩󵄩 , 12(L21 + L22 ) 󵄩 󵄩

| 161

1 ⩽ n ⩽ N.

(2.219)

Starting from (2.219), an induction method will yield the desired result (2.216). The process is quite similar to that in Theorem 2.1.2 and the details are omitted here. The proof is completed.

2.10.4 Convergence of the difference scheme Theorem 2.10.3. Suppose {Uijn | (i, j) ∈ ω,̄ 0 ⩽ n ⩽ N} and {unij | (i, j) ∈ ω,̄ 0 ⩽ n ⩽ N} are solutions of the problem (2.197)–(2.199) and the difference scheme (2.208)–(2.210), respectively. Let eijn = Uijn − unij ,

(i, j) ∈ ω,̄ 0 ⩽ n ⩽ N.

Then it holds 󵄩󵄩 n 󵄩󵄩 min{1,2α} + h21 + h22 ), 󵄩󵄩e 󵄩󵄩 ⩽ κ2 (τ

1 ⩽ n ⩽ N,

with κ2 =

α

LL L1 L2 15 T √ ( ) 2 1 2 2 c11 . 6 1 − α 2 L1 + L2

Proof. Subtracting (2.208)–(2.210) from (2.204), (2.206)–(2.207), respectively, the system of error equations is obtained as { { { { { { { { { { { { { { { { { { {

n

n

τ−α ∑ gk(α) eijn−k + τα ∑ gk(α) δx2 δy2 eijn−k

k=0 k=0 2 n 2 n n = δx eij + δy eij + (r11 )ij , (i, j) ∈ ω, eij0 = 0, (i, j) ∈ ω, eijn = 0, (i, j) ∈ 𝜕ω, 0 ⩽ n ⩽ N.

1 ⩽ n ⩽ N,

Noticing (2.205), Theorem 2.10.2 immediately implies L21 L22 5 󵄩󵄩 n 󵄩󵄩2 󵄩 󵄩2 ⋅ t α max 󵄩󵄩(r11 )m 󵄩󵄩󵄩 󵄩󵄩e 󵄩󵄩 ⩽ 12(L21 + L22 ) (1 − α)2α n 1⩽m⩽n󵄩 ⩽

2 L21 L22 5T α min{1,2α} 2 2 ⋅ L L [c (τ + h + h )] , 1 2 11 1 2 12(L21 + L22 ) (1 − α)2α

1 ⩽ n ⩽ N.

The theorem follows by taking the square root on both hand sides of the above inequality. The proof ends.

162 | 2 Difference methods for the time-fractional subdiffusion equations

2.11 The ADI method based on L1 approximation for 2D problem The aim of this section is to provide an ADI method based on L1 approximation for the 2D time-fractional subdiffusion problem C α { 0 Dt u(x, y, t) = uxx (x, y, t) + uyy (x, y, t) + f (x, y, t), { { { { { (x, y) ∈ Ω, t ∈ (0, T], { { { u(x, y, 0) = φ(x, y), (x, y) ∈ Ω, { { { { u(x, y, t) = μ(x, y, t), (x, y) ∈ 𝜕Ω, t ∈ [0, T],

(2.220) (2.221) (2.222)

where Ω = (0, L1 ) × (0, L2 ), α ∈ (0, 1), the functions f , φ, μ are all given and μ(x, y, 0)|(x,y)∈𝜕Ω = φ(x, y). Take the same mesh partition and notations as those in Section 2.10. Suppose u ∈ (4,4,2) ̄ C (Ω × [0, T]). In addition, denote s = τα Γ(2 − α). 2.11.1 Derivation of the difference scheme Considering equation (2.220) at the point (xi , yj , tn ), one has C α 0 Dt u(xi , yj , tn )

= uxx (xi , yj , tn ) + uyy (xi , yj , tn ) + fijn ,

(i, j) ∈ ω, 1 ⩽ n ⩽ N.

Using the L1 formula (1.160) to handle the Caputo derivative and the central difference approximation to discretize the spatial derivative, by Theorem 1.6.1 and Lemma 2.1.3, we get n−1 τ−α n 0 [a(α) U − )Uijk − a(α) ∑ (a(α) − a(α) n−1 Uij ] n−k Γ(2 − α) 0 ij k=1 n−k−1

= δx2 Uijn + δy2 Uijn + fijn + (r12 )nij ,

(i, j) ∈ ω, 1 ⩽ n ⩽ N,

(2.223)

where there exists a positive constant c12 such that 󵄨󵄨 n󵄨 2−α 2 2 󵄨󵄨(r12 )ij 󵄨󵄨󵄨 ⩽ c12 (τ + h1 + h2 ),

(i, j) ∈ ω, 1 ⩽ n ⩽ N.

Adding a small perturbation term s2 δx2 δy2 [

n−1 τ−α (α) k (α) 0 n (α) (a(α) 0 Uij − ∑ (an−k−1 − an−k )Uij − an−1 Uij )], Γ(2 − α) k=1

into both hand sides of (2.223) gives n−1 τ−α n 0 (ℐ + s2 δx2 δy2 )(a(α) U − − a(α) )Uijk − a(α) ∑ (a(α) ij n−1 Uij ) 0 n−k−1 n−k Γ(2 − α) k=1

= δx2 Uijn + δy2 Uijn + fijn + (r13 )nij ,

(i, j) ∈ ω, 1 ⩽ n ⩽ N,

(2.224)

2.11 The ADI method based on L1 approximation for 2D problem

| 163

where (r13 )nij = (r12 )nij + s2 [

n−1 τ−α n 0 δx2 δy2 (a(α) U − − a(α) )Uijk − a(α) ∑ (a(α) ij n−1 Uij )] 0 n−k−1 n−k Γ(2 − α) k=1

and there exists a positive constant c13 such that 󵄨󵄨 n󵄨 min{2α,2−α} + h21 + h22 ), 󵄨󵄨(r13 )ij 󵄨󵄨󵄨 ⩽ c13 (τ

(i, j) ∈ ω, 1 ⩽ n ⩽ N.

(2.225)

Noticing the initial-boundary value conditions (2.221)–(2.222), one has {

Uij0 = φ(xi , yj ),

Uijn

= μ(xi , yj , tn ),

(i, j) ∈ ω, (i, j) ∈ 𝜕ω, 0 ⩽ n ⩽ N.

(2.226) (2.227)

Omitting the small term (r13 )nij in (2.224) and replacing the exact solution Uijn with its numerical one unij , we can obtain the following difference scheme for solving (2.220)–(2.222): { { { { { { { { { { { { { { { { { { {

n−1 τ−α n (α) 0 (ℐ + s2 δx2 δy2 )(a(α) u − − a(α) )uk − a(α) ∑ (an−k−1 n−1 uij ) 0 ij n−k ij Γ(2 − α) k=1

= δx2 unij + δy2 unij + fijn , u0ij = φ(xi , yj ), unij

(i, j) ∈ ω, 1 ⩽ n ⩽ N,

(i, j) ∈ ω,

= μ(xi , yj , tn ),

(2.228) (2.229)

(i, j) ∈ 𝜕ω, 0 ⩽ n ⩽ N.

(2.230)

Rewrite equation (2.228) as (ℐ + s2 δx2 δy2 )unij − sδx2 unij − sδy2 unij n−1

0 n = (ℐ + s2 δx2 δy2 )( ∑ (a(α) − a(α) )uk + a(α) n−1 uij ) + sfij , n−k−1 n−k ij k=1

namely, (ℐ − sδx2 )(ℐ − sδy2 )unij n−1

0 n = (ℐ + s2 δx2 δy2 )( ∑ (a(α) − a(α) )uk + a(α) n−1 uij ) + sfij , n−k ij n−k−1 k=1

(i, j) ∈ ω, 1 ⩽ n ⩽ N.

Let u∗ij = (ℐ − sδy2 )unij , then the difference scheme (2.228)–(2.230) can be reformulated as the following ADI form:

164 | 2 Difference methods for the time-fractional subdiffusion equations On each time level t = tn (1 ⩽ n ⩽ N), firstly, for any fixed j from 1 to M2 − 1, solve a series of linear systems in the unknown {u∗ij | 0 ⩽ i ⩽ M1 } in x direction { { { { { { { { { { { { {

n−1

0 n (ℐ − sδx2 )u∗ij = (ℐ + s2 δx2 δy2 )( ∑ (a(α) − a(α) )uk + a(α) n−1 uij ) + sfij , n−k−1 n−k ij k=1

1 ⩽ i ⩽ M1 − 1,

u∗0j = (ℐ − sδy2 )un0j ,

u∗M1 ,j = (ℐ − sδy2 )unM1 ,j

to obtain the value of {u∗ij | 1 ⩽ i ⩽ M1 − 1} on an intermediate time level. Then, for any fixed i from 1 to M1 − 1, carry out some calculations about the unknown {unij | 0 ⩽ j ⩽ M2 } in y direction {

(ℐ − sδy2 )unij = u∗ij , uni0 = μ(xi , y0 , tn ),

1 ⩽ j ⩽ M2 − 1, uni,M2 = μ(xi , yM2 , tn )

to get the desired value of {unij | 1 ⩽ j ⩽ M2 − 1}. 2.11.2 Solvability of the difference scheme Theorem 2.11.1. The difference scheme (2.228)–(2.230) is uniquely solvable. Proof. Denote ̄ un = {unij | (i, j) ∈ ω}. The value of u0 is uniquely determined by (2.229)–(2.230). Suppose the values of u0 , u1 , . . . , un−1 have been uniquely determined, then the system in un can be obtained from (2.228) and (2.230). To show its unique solvability, it is sufficient to prove that the corresponding homogeneous one 1 { (ℐ + s2 δx2 δy2 )unij = δx2 unij + δy2 unij , s { n { uij = 0, (i, j) ∈ 𝜕ω

(i, j) ∈ ω,

(2.231) (2.232)

has only the trivial solution. Taking the inner product on both hand sides of (2.231) with un and noticing (2.232), it follows from (2.200)–(2.202) that 1 n n (u , u ) + s(δx δy un , δx δy un ) = −(δx un , δx un ) − (δy un , δy un ) s 󵄩 󵄩2 = −󵄩󵄩󵄩∇h un 󵄩󵄩󵄩 ⩽ 0,

2.11 The ADI method based on L1 approximation for 2D problem

| 165

which implies ‖un ‖ = 0. It follows un = 0. By the principle of induction, the difference scheme (2.228)–(2.230) is uniquely solvable. The proof ends.

2.11.3 Stability of the difference scheme For any mesh functions u, v ∈ 𝒱̊h , define (u, v)s ≡ (u, v) + s2 (δx δy u, δx δy v). Theorem 2.11.2. Suppose {vijn | (i, j) ∈ ω,̄ 0 ⩽ n ⩽ N} is the solution of difference scheme { { { { { { { { { { { { { { { { { { {

n−1 1 n (α) (α) k (α) 0 (ℐ + s2 δx2 δy2 )(a(α) 0 vij − ∑ (an−k−1 − an−k )vij − an−1 vij ) s k=1

= δx2 vijn + δy2 vijn + fijn , vij0 = φ(xi , yj ),

vijn

= 0,

(i, j) ∈ ω, 1 ⩽ n ⩽ N,

(2.233)

(i, j) ∈ ω,

(2.234)

(i, j) ∈ 𝜕ω, 0 ⩽ n ⩽ N.

(2.235)

Then it holds (vn , vn )s ⩽ (v0 , v0 )s +

L21 L22 α󵄩 󵄩󵄩f m 󵄩󵄩󵄩2 }, Γ(1 − α) max {tm 󵄩 󵄩 1⩽m⩽n 12(L21 + L22 )

1 ⩽ n ⩽ N,

(2.236)

where M1 −1 M2 −1

󵄩󵄩 m 󵄩󵄩2 m 2 󵄩󵄩f 󵄩󵄩 = h1 h2 ∑ ∑ (fij ) . i=1

j=1

Proof. Making the inner product on both hand sides of (2.233) with vn , respectively, it produces n−1 1 k (α) 0 n (α) n (α) ((ℐ + s2 δx2 δy2 )(a(α) 0 v − ∑ (an−k−1 − an−k )v − an−1 v ), v ) s k=1

= (δx2 vn , vn ) + (δy2 vn , vn ) + (f n , vn ),

1 ⩽ n ⩽ N.

(2.237)

Because of (2.200)–(2.202), we have by noticing (2.235) that ((ℐ + s2 δx2 δy2 )vk , vn ) = (vk , vn ) + s2 (δx δy vk , δx δy vn ) = (vk , vn )s ,

0 ⩽ k ⩽ n;

󵄩 󵄩2 (δx2 vn , vn ) + (δy2 vn , vn ) = −(δx vn , δx vn ) − (δy vn , δy vn ) = −󵄩󵄩󵄩∇h vn 󵄩󵄩󵄩 .

(2.238) (2.239)

166 | 2 Difference methods for the time-fractional subdiffusion equations By the Cauchy–Schwarz inequality and Lemma 2.10.1, it follows 1 1 1 󵄩 󵄩2 󵄩 󵄩 󵄩 󵄩 (f n , vn ) ⩽ 󵄩󵄩󵄩f n 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩vn 󵄩󵄩󵄩 ⩽ 6( 2 + 2 )󵄩󵄩󵄩vn 󵄩󵄩󵄩 + 1 L1 L2 24( L2 + 󵄩 󵄩2 ⩽ 󵄩󵄩󵄩∇h vn 󵄩󵄩󵄩 +

1

24( L12 + 1

1

1 ) L22

󵄩󵄩 n 󵄩󵄩2 󵄩󵄩f 󵄩󵄩 ,

󵄩󵄩 n 󵄩󵄩2 f 󵄩󵄩

󵄩 1 󵄩 ) L22

1 ⩽ n ⩽ N.

(2.240)

Inserting (2.238)–(2.240) into (2.237) and rearranging the result, again by the Cauchy– Schwarz inequality yield n−1

n n (α) (α) k n (α) 0 n a(α) 0 (v , v )s ⩽ ∑ (an−k−1 − an−k )(v , v )s + an−1 (v , v )s k=1

+ ⩽

L21 L22 󵄩 󵄩2 s󵄩󵄩󵄩f n 󵄩󵄩󵄩 2 2 24(L1 + L2 )

1 n−1 (α) − a(α) )[(vk , vk )s + (vn , vn )s ] ∑ (a n−k 2 k=1 n−k−1

L21 L22 1 󵄩 󵄩2 0 0 n n s󵄩󵄩f n 󵄩󵄩 , + a(α) n−1 [(v , v )s + (v , v )s ] + 2 24(L21 + L22 ) 󵄩 󵄩

which simplifies to give n−1

n n (α) (α) k k (α) 0 0 a(α) 0 (v , v )s ⩽ ∑ (an−k−1 − an−k )(v , v )s + an−1 (v , v )s k=1

+

L21 L22 󵄩 󵄩2 s󵄩󵄩f n 󵄩󵄩 , 12(L21 + L22 ) 󵄩 󵄩

1 ⩽ n ⩽ N.

Noticing (2.51), we have n−1

k k (α) 0 0 n n (α) (α) a(α) 0 (v , v )s ⩽ ∑ (an−k−1 − an−k )(v , v )s + an−1 [(v , v )s k=1

+

L21 L22 󵄩 󵄩2 tnα Γ(1 − α)󵄩󵄩󵄩f n 󵄩󵄩󵄩 ], 2 2 12(L1 + L2 )

1 ⩽ n ⩽ N.

(2.241)

Then the claimed result (2.236) can be achieved by the induction method from (2.241). The proof ends.

2.11.4 Convergence of the difference scheme We now consider the convergence of the difference scheme (2.228)–(2.230). At this point, the following theorem is true.

2.12 Supplementary remarks and discussions | 167

Theorem 2.11.3. Suppose {Uijn | (i, j) ∈ ω,̄ 0 ⩽ n ⩽ N} and {unij | (i, j) ∈ ω,̄ 0 ⩽ n ⩽ N} are solutions of the problem (2.220)-(2.222) and the difference scheme (2.228)–(2.230), respectively. Let eijn = Uijn − unij ,

(i, j) ∈ ω,̄ 0 ⩽ n ⩽ N.

Then it holds 󵄩󵄩 n 󵄩󵄩 min{2α,2−α} + h21 + h22 ), 󵄩󵄩e 󵄩󵄩 ⩽ κ3 (τ

1 ⩽ n ⩽ N,

with κ3 =

L1 L2 3L1 L2 α T Γ(1 − α) c13 . √ 2 6 L1 + L22

Proof. The substraction of (2.228)–(2.230) from (2.224), (2.226)–(2.227), respectively, produces the system of error equations as follows: { { { { { { { { { { { { { { { { { { {

n−1 1 n (α) (α) k (α) 0 (ℐ + s2 δx2 δy2 )[a(α) 0 eij − ∑ (an−k−1 − an−k )eij − an−1 eij ] s k=1

= δx2 eijn + δy2 eijn + (r13 )nij , eij0 = 0, eijn

= 0,

(i, j) ∈ ω, 1 ⩽ n ⩽ N,

(i, j) ∈ ω, (i, j) ∈ 𝜕ω, 0 ⩽ n ⩽ N.

Noticing (2.225), Theorem 2.11.2 immediately implies 󵄩󵄩 n 󵄩󵄩2 n n 󵄩󵄩e 󵄩󵄩 ⩽ (e , e )s ⩽ ⩽

L21 L22 α󵄩 󵄩󵄩(r13 )m 󵄩󵄩󵄩2 } Γ(1 − α) max {tm 󵄩 󵄩 1⩽m⩽n 12(L21 + L22 )

2 L21 L22 Γ(1 − α)T α L1 L2 [c13 (τmin{2α,2−α} + h21 + h22 )] , 2 2 12(L1 + L2 )

1 ⩽ n ⩽ N.

The desired result will be obvious from the above estimate. The proof is completed.

2.12 Supplementary remarks and discussions 1. The time-fractional subdiffusion equation mainly consists of two types, one is the Caputo type and the other is the R-L type, which are expressed in the form of C α 0 Dt u(x, t)

= uxx (x, t) + f (x, t),

and ut (x, t) = 0 D1−α t uxx (x, t) + g(x, t),

168 | 2 Difference methods for the time-fractional subdiffusion equations respectively. Under certain conditions, it is possible to convert between them. In this chapter, only for the Caputo type, the finite difference method was considered. In a similar way, the finite difference method for the R-L type can be discussed. On this topic, readers can refer to literatures, such as by Yuste et al.[108, 109] ; Langlands and Henry[42] ; Liu et al.[8, 120] ; Cui[10, 11] and Zhang et al.[112, 114] . 2. The Caputo derivative can be discretized using either the G-L approximation or the interpolation approximation. Sections 2.1, 2.2 and 2.10 reported the first kind of methods based on the G-L formula and the first-order convergent difference schemes in time have been obtained. Indeed, there are also related superconvergent results based on the G-L formula. From Theorem 1.4.1, we know that if the function f ∈ C 2+α (ℛ), Aαh,p f (t) = −∞ Dαt f (t) + (p −

α ) Dα+1 f (t)h + O(h2 ). 2 −∞ t

Let p = α2 , t = tn− α , with tn− α = (n − α2 )h, then we have 2

2

Aαh,0 f (tn ) = Aαh, α f (tn− α ) = −∞ Dαt f (tn− α ) + O(h2 ), 2

2

2

which means that the second-order accuracy can be achieved using Aαh,0 f (tn ) to approximate −∞ Dαt f (tn− α ). By the linear interpolation, further it holds that 2

Aαh,0 f (tn ) = (1 −

α α ) Dα f (t ) + −∞ Dαt f (tn−1 ) + O(h2 ), 2 −∞ t n 2

which says that the second-order accuracy can be achieved using Aαh,0 f (tn ) to approxi-

mate a linear combination of −∞ Dαt f (tn ) and −∞ Dαt f (tn−1 ). Dimitrov[13] and Gao et al.[28] have reported the related research results on the problem of Caputo type and R-L type, respectively. In addition, using the shifted and weighted G-L formula (1.32) to approximate the R-L fractional derivative, Wang and Vong[96] developed the finite difference method for two-term time-fractional subdiffusion equations of R-L type, where the second-order accuracy in time can be achieved. Ji et al.[40] applied the shifted and weighted G-L formula (1.37) in Theorem 1.4.4 to approximate the time-fractional derivative and presented a third-order convergent method in time. 3. For the Caputo time-fractional subdiffusion equations, Sections 2.3, 2.5, 2.8 and 2.11 reported the finite difference methods based on the L1 formula to approximate Caputo fractional derivatives. Indeed, there are also some superconvergent works based on the interpolation approximation. Alikhanov[1] proposed a superconvergent interpolation approximation, also called the L2-1σ approximation, for the Caputo fractional derivative based on the work in [31] and the second-order numerical method for solving the time-fractional subdiffusion equation was investigated using this approximation. The authors in [19] developed this method and applied it to solve the multiterm fractional subdiffusion equation and obtained a temporal second

2.12 Supplementary remarks and discussions | 169

order difference scheme. Du et al. investigated the L2-1σ method for the variable-order fractional subdiffusion equation in [16]. 4. This chapter mainly discussed the finite difference method for solving the timefractional subdiffusion equations with the Dirichlet boundary value conditions. For the problem with the Neumann boundary value condition, the readers can refer to the works by Langlands and Henry[42] , Zhao and Sun[115] and Ren et al.[69] . In addition, Gao et al.[22, 30] studied the 1D time-fractional subdiffusion equations on space unbounded domains. The results on the 2D problem can be found in [33]. 5. The operator 𝒜 defined in Section 2.1 is called an average operator. It is often used to construct the compact difference scheme, so that it is also called a compact operator. For the time-fractional subdiffusion equation, Gao and Sun[21] proposed a higher-order difference scheme by using the L1 approximation for the time-fractional derivative of order α(0 < α < 1) and the compact approximation for the spatial derivative, which achieved the convergence of order 2 − α in time and four in space in the maximum norm. 7. The second-order method in space was discussed for 1D multiterm time-fractional subdiffusion equations in Section 2.8 and Section 2.9. Regarding the fourthorder methods in space for the same problem and the corresponding 2D problem, readers can refer to [67]. 8. Based on the SOE approximation for the kernel function in the fractional derivative, this chapter illustrated two kinds of fast methods to solve the time-fractional subdiffusion equations[41, 101] . By taking into account of the special structure of resultant difference schemes for the time-fractional diffusion equations, Lu et al.[53] developed a different fast method. 9. Lv and Xu[55] considered a numerical method for the time-fractional subdiffusion equation based on the L1-2 method[31] . The (3−α)-order convergence of the scheme has been proved. Zhu and Xu further studied the fast L1-2 difference method for the fractional subdiffusion equations[118] . 10. Sections 2.10 and 2.11 introduced two kinds of ADI difference methods for solving 2D time-fractional subdiffusion equations based on the G-L formula and L1 approximation, respectively. The discrete energy method was used to analyze the unique solvability, stability and convergence of the resultant schemes in the L2 norm. Using the similar techniques, interested readers can try to give the estimates in the H 1 norm, or refer to the work in [111]. Besides, Cui[11, 12] reported the compact ADI difference methods for 2D problems and the Fourier method was used for the theoretical analysis. The compact ADI difference method for 2D problem in the R-L type was also studied in [112] and the discrete energy method was applied for the theoretical analysis. Moreover, another different ADI difference scheme was obtained in [111] by adding a different perturbation term into equation (2.223). 11. Stynes et al.[74] studied the numerical solution to the fractional subdiffusion equation with the initial singularity on the graded mesh. Shen et al.[72] further provided a fast difference scheme for this kind of fractional subdiffusion equations.

170 | 2 Difference methods for the time-fractional subdiffusion equations

Exercises 2 2.1 Using the discrete energy method in Section 2.2, show that the difference scheme in Section 2.1 is uniquely solvable, convergent and unconditionally stable with respect to the initial value and the source function on the right-hand side. 2.2 Using the discrete energy method in Section 2.5, show that the difference scheme in Section 2.3 is uniquely solvable, convergent and unconditionally stable with respect to the initial value and the source function on the right-hand side. 2.3 For the problem (2.154)–(2.156), when φ(x) = 0, construct the difference scheme n

n

(α ) { { τ−α ∑ gk(α) un−k + τ−α1 ∑ gk 1 un−k = δx2 uni + fin , { i i { { { k=0 k=0 { { 1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N, { { { 0 { { u = 0, 1 ⩽ i ⩽ M − 1, { { { in n { u0 = μ(tn ), uM = ν(tn ), 0 ⩽ n ⩽ N.

̂ t) like that in Section 2.1 and suppose u(x, ̂ ⋅) ∈ C 1+α (ℛ). Define the function u(x, For this difference scheme, try to (1) show the unique solvability; (2) show the stability with respect to the initial value and function f ; (3) show the convergence. 2.4 Consider the following problem of fourth-order fractional subdiffusion equation: { { { { { { { { { { { { {

C α 0 Dt u(x, t)

+ uxxxx (x, t) = f (x, t),

u(x, 0) = φ(x),

x ∈ (0, L),

u(0, t) = μ0 (t),

u(L, t) = ν0 (t),

uxx (0, t) = μ1 (t),

x ∈ (0, L), t ∈ (0, T],

(2.242) (2.243)

t ∈ [0, T],

uxx (L, t) = ν1 (t),

t ∈ [0, T],

(2.244) (2.245)

where α ∈ (0, 1), the functions f , φ, μ0 , ν0 , μ1 , ν1 are all given and φ(0) = μ0 (0), φ(L) = ν0 (0), φxx (0) = μ1 (0), φxx (L) = ν1 (0). Let v(x, t) = uxx (x, t), then the fourth-order equation (2.242) can be rewritten as a system of two second-order equations. Construct a difference scheme for (2.242)–(2.245) by using L2-1σ approximation or fast L2-1σ approximation. For this difference scheme, try to (1) show the unique solvability; (2) show the stability with respect to the initial value and function f ; (3) show the convergence. 2.5 For the problem (2.197)–(2.199), construct the difference scheme n

{ { τ−α ∑ gk(α) un−k = δx2 unij + δy2 unij + fijn , (i, j) ∈ ω, 1 ⩽ n ⩽ N, { ij { { { k=0 0 { { u = 0, (i, j) ∈ ω, { ij { { { n { uij = μ(xi , yj , tn ), (i, j) ∈ 𝜕ω, 0 ⩽ n ⩽ N.

Exercises 2

| 171

̂ y, t) like that in Section 2.10 and suppose u(x, ̂ y, ⋅) ∈ Define the function u(x,

C 1+α (ℛ). For this difference scheme, try to

(1) show the unique solvability; (2) show the stability with respect to the initial value and function f ; (3) show the convergence. 2.6 For the problem (2.220)–(2.222), construct the difference scheme: { { { { { { { { { { { { { { { { { { {

n−1 τ−α n 0 [a(α) u − )uk − a(α) ∑ (a(α) − a(α) n−1 uij ] n−k ij Γ(2 − α) 0 ij k=1 n−k−1

= δx2 unij + δy2 unij + fijn ,

u0ij unij

= φ(xi , yj ),

= μ(xi , yj , tn ),

(i, j) ∈ ω, 1 ⩽ n ⩽ N,

(i, j) ∈ ω, (i, j) ∈ 𝜕ω,

0 ⩽ n ⩽ N.

For this difference scheme, try to (1) show the unique solvability; (2) show the stability with respect to the initial value and function f ; (3) show the convergence.

3 Difference methods for the time-fractional wave equations This chapter will develop the difference methods for solving the time-fractional wave equations. The discussion on 1D problem is given in the former 8 sections. The time-fractional derivative is discretized by L1 approximation or L2-1σ approximation. The spatial derivative is discretized by the second-order central difference quotient approximation or the compact approximation. The fast L1 approximation and fast L2-1σ approximation are concerned. The difference methods for the multiterm time-fractional wave equation and the time-fractional mixed diffusion and wave equation are also investigated. The ADI and compact ADI methods for 2D problem are established. The chapter consists of 11 sections.

3.1 The second-order method in space based on L1 approximation for 1D problem Consider the following problem of the time-fractional wave equations: γ

C D u(x, t) = uxx (x, t) + f (x, t), x ∈ (0, L), t ∈ (0, T], { { { 0 t u(x, 0) = φ(x), ut (x, 0) = ψ(x), x ∈ (0, L), { { { { u(0, t) = μ(t), u(L, t) = ν(t), t ∈ [0, T],

(3.1) (3.2) (3.3)

where γ ∈ (1, 2), the functions f , φ, ψ, μ, ν are given and φ(0) = μ(0), φ(L) = ν(0), ψ(0) = μ󸀠 (0), ψ(L) = ν󸀠 (0). Suppose u ∈ C (4,3) ([0, L] × [0, T]). Take the same mesh partition and notations as those in Section 2.1. For the mesh function v = {vik | 0 ⩽ i ⩽ M, 0 ⩽ k ⩽ N} defined on Ωh × Ωτ , denote k− 21

vi

1 = (vik + vik−1 ), 2

k− 21

δt vi

=

1 k (v − vik−1 ). τ i

Define the same mesh function spaces 𝒰h and 𝒰h̊ as those in Section 2.1. Denote Uin = u(xi , tn ),

fin = f (xi , tn ),

ψi = ψ(xi ),

0 ⩽ i ⩽ M, 0 ⩽ n ⩽ N.

3.1.1 Derivation of the difference scheme Considering equation (3.1) at the point (xi , tn ), we have C γ 0 Dt u(xi , tn )

= uxx (xi , tn ) + fin ,

https://doi.org/10.1515/9783110616064-003

1 ⩽ i ⩽ M − 1, 0 ⩽ n ⩽ N.

174 | 3 Difference methods for the time-fractional wave equations Taking an average on two adjacent time levels gives 1 C γ γ [ D u(xi , tn ) + C0 Dt u(xi , tn−1 )] 2 0 t 1 n− 1 = [uxx (xi , tn ) + uxx (xi , tn−1 )] + fi 2 , 2

1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N,

n− 1

where fi 2 = 21 (fin + fin−1 ). For the approximation of the time-fractional derivative and spatial derivative in the equality above, the L1 formula (1.69) and the second-order central difference quotient are applied, respectively, and it follows from Theorem 1.6.2 and Lemma 2.1.3 that n−1 τ1−γ k− 1 n− 1 (γ) (γ) (γ) (γ) [b0 δt Ui 2 − ∑ (bn−k−1 − bn−k )δt Ui 2 − bn−1 ψi ] Γ(3 − γ) k=1 n− 21

= δx2 Ui

n− 21

+ fi

n− 21

+ (r1 )i

1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N,

,

(3.4)

and there is a positive constant c1 such that 󵄨󵄨 n− 21 󵄨󵄨 3−γ 2 󵄨󵄨(r1 )i 󵄨󵄨 ⩽ c1 (τ + h ),

1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N,

(3.5)

(γ)

where {bl } is defined in (1.64). Noticing the initial-boundary value conditions (3.2)–(3.3), we have {

Ui0 = φ(xi ),

U0n

= μ(tn ),

1 ⩽ i ⩽ M − 1,

n UM

= ν(tn ),

(3.6) 0 ⩽ n ⩽ N.

(3.7)

n− 1

Omitting the small term (r1 )i 2 in (3.4) and replacing the exact solution Uin with its numerical one uni arrives at a difference scheme for solving (3.1)–(3.3) as follows: { { { { { { { { { { { { { { { { { { {

n−1 τ1−γ k− 1 n− 1 (γ) (γ) (γ) (γ) [b0 δt ui 2 − ∑ (bn−k−1 − bn−k )δt ui 2 − bn−1 ψi ] Γ(3 − γ) k=1 n− 21

= δx2 ui

u0i un0

n− 21

+ fi

= φ(xi ), = μ(tn ),

,

1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N,

1 ⩽ i ⩽ M − 1,

unM

= ν(tn ),

(3.8) (3.9)

0 ⩽ n ⩽ N.

Denote η = τγ−1 Γ(3 − γ). 3.1.2 Solvability of the difference scheme Theorem 3.1.1. The difference scheme (3.8)–(3.10) is uniquely solvable.

(3.10)

3.1 The second-order method in space based on L1 approximation for 1D problem

| 175

Proof. Let un = (un0 , un1 , . . . , unM ). The value of u0 is determined by (3.9)–(3.10). Now assume that the values of u0 , u1 , . . . , un−1 have been uniquely determined, then the linear system in un can be obtained from (3.8) and (3.10). To show its unique solvability, it suffices to verify that the corresponding homogeneous one 1 n 1 2 n ui = δx ui , { 2 { ητ n n { u0 = uM = 0

1 ⩽ i ⩽ M − 1,

(3.11) (3.12)

has only the trivial solution. Taking the inner product on both hand sides of (3.11) with un and noticing (3.12), we have 1 1󵄩 1 n n 󵄩2 (u , u ) = (δx2 un , un ) = − 󵄩󵄩󵄩δx un 󵄩󵄩󵄩 ⩽ 0, ητ 2 2 thus ‖un ‖ = 0. It follows un = 0 from (3.12). By the principle of induction, the theorem is true. The proof ends.

3.1.3 Stability of the difference scheme The stability of the difference scheme (3.8)–(3.10) will be analyzed in this subsection. The following theorem is true. Theorem 3.1.2. Suppose {vin | 0 ⩽ i ⩽ M, 0 ⩽ n ⩽ N} is the solution of the difference scheme { { { { { { { { { { { { { { { { { { {

1 (γ) n− 21 n−1 (γ) k− 1 (γ) (γ) [b0 δt vi − ∑ (bn−k−1 − bn−k )δt vi 2 − bn−1 ψi ] η k=1 n− 21

= δx2 vi

vi0 v0n

n− 21

+ fi

= φ(xi ),

= 0,

n vM

,

1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N,

(3.13)

1 ⩽ i ⩽ M − 1,

(3.14)

= 0,

(3.15)

0 ⩽ n ⩽ N.

Then it holds 2−γ

t 󵄩󵄩 n 󵄩󵄩2 󵄩󵄩 0 󵄩󵄩2 ‖ψ‖2 󵄩󵄩δx v 󵄩󵄩 ⩽ 󵄩󵄩δx v 󵄩󵄩 + n Γ(3 − γ) n

󵄩 1 󵄩2 + Γ(2 − γ)tnγ−1 ⋅ τ ∑ 󵄩󵄩󵄩f k− 2 󵄩󵄩󵄩 , k=1

1 ⩽ n ⩽ N,

(3.16)

176 | 3 Difference methods for the time-fractional wave equations where M−1

‖ψ‖2 = h ∑ ψ2i , i=1

M−1

k− 1 2 󵄩󵄩 k− 21 󵄩󵄩2 󵄩󵄩f 󵄩󵄩 = h ∑ (fi 2 ) . i=1

1

Proof. Making the inner product on both hand sides of (3.13) with ηδt vn− 2 produces 1

n−1

1

1

1

b0 (δt vn− 2 , δt vn− 2 ) = ∑ (bn−k−1 − bn−k )(δt vk− 2 , δt vn− 2 ) (γ)

(γ)

(γ)

k=1

1

1

1

+ bn−1 (ψ, δt vn− 2 ) + η(δx2 vn− 2 , δt vn− 2 ) (γ)

1

1

+ η(f n− 2 , δt vn− 2 ),

1 ⩽ n ⩽ N.

(3.17)

Noticing (3.15), the application of the summation by parts arrives at 1

1

1

1

(δx2 vn− 2 , δt vn− 2 ) = −(δx vn− 2 , δx δt vn− 2 ) 1 󵄩 󵄩2 󵄩 󵄩2 = − (󵄩󵄩󵄩δx vn 󵄩󵄩󵄩 − 󵄩󵄩󵄩δx vn−1 󵄩󵄩󵄩 ). 2τ

(3.18)

Substituting (3.18) into (3.17), it follows from the Cauchy–Schwarz inequality that 1 2 η 󵄩 (γ) 󵄩 󵄩2 󵄩 󵄩2 󵄩 b0 󵄩󵄩󵄩δt vn− 2 󵄩󵄩󵄩 + (󵄩󵄩󵄩δx vn 󵄩󵄩󵄩 − 󵄩󵄩󵄩δx vn−1 󵄩󵄩󵄩 ) 2τ

n−1

1

1

1

= ∑ (bn−k−1 − bn−k )(δt vk− 2 , δt vn− 2 ) + bn−1 (ψ, δt vn− 2 ) (γ)

(γ)

k=1

1

(γ)

1

+ η(f n− 2 , δt vn− 2 ) ⩽

1 2 1 2 1 n−1 (γ) (γ) 󵄩 󵄩 󵄩 󵄩 ∑ (bn−k−1 − bn−k )(󵄩󵄩󵄩δt vk− 2 󵄩󵄩󵄩 + 󵄩󵄩󵄩δt vn− 2 󵄩󵄩󵄩 ) 2 k=1

1 2 1 1 1 (γ) 󵄩 󵄩 + bn−1 (‖ψ‖2 + 󵄩󵄩󵄩δt vn− 2 󵄩󵄩󵄩 ) + η(f n− 2 , δt vn− 2 ), 2

which can be simplified to 1 2 η 󵄩 (γ) 󵄩 󵄩 󵄩2 󵄩 󵄩2 b0 󵄩󵄩󵄩δt vn− 2 󵄩󵄩󵄩 + (󵄩󵄩󵄩δx vn 󵄩󵄩󵄩 − 󵄩󵄩󵄩δx vn−1 󵄩󵄩󵄩 ) τ

n−1

1 2 (γ) (γ) (γ) 󵄩 󵄩 ⩽ ∑ (bn−k−1 − bn−k )󵄩󵄩󵄩δt vk− 2 󵄩󵄩󵄩 + bn−1 ‖ψ‖2

k=1

1

1

+ 2η(f n− 2 , δt vn− 2 ),

1 ⩽ n ⩽ N.

The result can further be reformulated as n (γ) 󵄩 󵄩󵄩 n 󵄩󵄩2 τ k− 1 󵄩2 󵄩󵄩δx v 󵄩󵄩 + ∑ bn−k 󵄩󵄩󵄩δt v 2 󵄩󵄩󵄩 η k=1

3.1 The second-order method in space based on L1 approximation for 1D problem

| 177

n−1 1 2 τ (γ) (γ) 󵄩 󵄩2 τ 󵄩 󵄩 ⩽ 󵄩󵄩󵄩δx vn−1 󵄩󵄩󵄩 + ∑ bn−k−1 󵄩󵄩󵄩δt vk− 2 󵄩󵄩󵄩 + bn−1 ‖ψ‖2 η k=1 η 1

1

+ 2τ(f n− 2 , δt vn− 2 ),

1 ⩽ n ⩽ N.

Let n 1 2 (γ) 󵄩 󵄩 󵄩2 τ 󵄩 F n = 󵄩󵄩󵄩δx vn 󵄩󵄩󵄩 + ∑ bn−k 󵄩󵄩󵄩δt vk− 2 󵄩󵄩󵄩 , η k=1

󵄩 󵄩2 F 0 = 󵄩󵄩󵄩δx v0 󵄩󵄩󵄩 ,

n ⩾ 1,

then F n ⩽ F n−1 +

1 1 τ (γ) b ‖ψ‖2 + 2τ(f n− 2 , δt vn− 2 ), η n−1

1 ⩽ n ⩽ N.

The recursive process will lead to Fn ⩽ F0 + ⩽ F0 +

n 1 1 τ n−1 (γ) ∑ bk ‖ψ‖2 + 2τ ∑ (f k− 2 , δt vk− 2 ) η k=0 k=1 (γ)

n 1 2 η 󵄩 1 󵄩2 b 󵄩 τ n−1 (γ) 󵄩 ∑ bk ‖ψ‖2 + τ ∑ ( (γ) 󵄩󵄩󵄩f k− 2 󵄩󵄩󵄩 + n−k 󵄩󵄩󵄩δt vk− 2 󵄩󵄩󵄩 ), η k=0 η b k=1 n−k

1 ⩽ n ⩽ N.

Thus, n−1 n η 󵄩 k− 1 󵄩2 (γ) 󵄩󵄩 n 󵄩󵄩2 󵄩󵄩 0 󵄩󵄩2 τ 2 󵄩󵄩δx v 󵄩󵄩 ⩽ 󵄩󵄩δx v 󵄩󵄩 + ∑ bk ‖ψ‖ + τ ∑ (γ) 󵄩󵄩󵄩f 2 󵄩󵄩󵄩 , η k=0 k=1 b

1 ⩽ n ⩽ N.

(3.19)

n−k

(γ)

By the definition of bk and Lemma 1.6.1, it is easy to know that (2 − γ)(k + 1)1−γ < bk < (2 − γ)k 1−γ , (γ)

from which, we can get bn−k > (2 − γ)(n − k + 1)1−γ ⩾ (2 − γ)n1−γ , (γ)

1 ⩽ k ⩽ n.

Therefore, η

(γ) bn−k

⩽

τγ−1 Γ(3 − γ) = Γ(2 − γ)(nτ)γ−1 = Γ(2 − γ)tnγ−1 , (2 − γ)n1−γ

(3.20)

from which it follows n

τ∑

k=1

n η 󵄩󵄩 k− 21 󵄩󵄩2 󵄩󵄩 k− 21 󵄩󵄩2 γ−1 f ⩽ Γ(2 − γ)t ⋅ τ ∑ 󵄩 󵄩 󵄩󵄩f 󵄩󵄩 . n 󵄩 (γ) 󵄩

bn−k

k=1

(3.21)

178 | 3 Difference methods for the time-fractional wave equations (γ)

In addition, it follows from the definition of {bk } that n−1 τ τ n−1 (γ) ∑ bk = γ−1 ∑ [(k + 1)2−γ − k 2−γ ] η k=0 τ Γ(3 − γ) k=0

=

2−γ

tn τ2−γ 2−γ n = . Γ(3 − γ) Γ(3 − γ)

(3.22)

The substitution of (3.21) and (3.22) into (3.19) will get the assertion (3.16). The proof ends. 3.1.4 Convergence of the difference scheme Theorem 3.1.3. Suppose {Uin | 0 ⩽ i ⩽ M, 0 ⩽ n ⩽ N} and {uni | 0 ⩽ i ⩽ M, 0 ⩽ n ⩽ N} are solutions of the problem (3.1)–(3.3) and the difference scheme (3.8)–(3.10), respectively. Let ein = Uin − uni , then it holds

0 ⩽ i ⩽ M, 0 ⩽ n ⩽ N,

cL 󵄩󵄩 n 󵄩󵄩 3−γ 2 󵄩󵄩e 󵄩󵄩∞ ⩽ 1 √T γ Γ(2 − γ)(τ + h ), 2

1 ⩽ n ⩽ N.

Proof. Subtracting (3.8)–(3.10) from (3.4), (3.6)–(3.7), respectively, produces the system of error equations as follows: { { { { { { { { { { { { { { { { { { {

1 (γ) n− 21 n−1 (γ) k− 1 (γ) (γ) [b0 δt ei − ∑ (bn−k−1 − bn−k )δt ei 2 − bn−1 ⋅ 0] η k=1 n− 21

= δx2 ei

ei0 e0n

= 0,

= 0,

n− 21

+ (r1 )i

,

1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N,

1 ⩽ i ⩽ M − 1,

n eM

= 0,

(3.23) (3.24)

0 ⩽ n ⩽ N.

(3.25)

Noticing (3.5), an immediate consequence of Theorem 3.1.2 into (3.23)–(3.25) arrives at n

󵄩󵄩 n 󵄩󵄩2 󵄩 k− 1 󵄩2 γ−1 󵄩󵄩δx e 󵄩󵄩 ⩽ tn Γ(2 − γ)τ ∑ 󵄩󵄩󵄩(r1 ) 2 󵄩󵄩󵄩 k=1

γ

⩽ T Γ(2 −

γ)Lc12 (τ3−γ

2

+ h2 ) ,

1 ⩽ n ⩽ N.

Taking the square root on both hand sides of the inequality above and noticing Lemma 2.1.1, we have √L 󵄩 n 󵄩 c 1 L 󵄩󵄩 n 󵄩󵄩 󵄩󵄩δ e 󵄩󵄩 ⩽ √T γ Γ(2 − γ)(τ3−γ + h2 ), 󵄩󵄩e 󵄩󵄩∞ ⩽ 2 󵄩 x 󵄩 2 The proof ends.

1 ⩽ n ⩽ N.

3.2 The fast difference method based on L1 approximation for 1D problem

| 179

3.2 The fast difference method based on L1 approximation for 1D problem In this section, we will propose a fast difference scheme for (3.1)–(3.3) based on the fast L1 approximation.

3.2.1 Derivation of the difference scheme Denote v(x, t) = ut (x, t),

α = γ − 1.

Thus, the problem (3.1)–(3.3) is equivalent to the following one: { { { { { { { { { { { { {

C α 0 Dt v(x, t)

= uxx (x, t) + f (x, t),

ut (x, t) = v(x, t),

u(x, 0) = φ(x), u(0, t) = μ(t),

x ∈ (0, L), t ∈ (0, T],

x ∈ [0, L], t ∈ (0, T],

v(x, 0) = ψ(x), u(L, t) = ν(t),

x ∈ (0, L), t ∈ [0, T].

(3.26) (3.27) (3.28) (3.29)

Denote Uin = u(xi , tn ), φi = φ(xi ),

Vin = v(xi , tn ),

ψi = ψ(xi ),

fin = f (xi , tn ),

0 ⩽ i ⩽ M, 0 ⩽ n ⩽ N,

0 ⩽ i ⩽ M.

Considering equation (3.26) at the point (xi , tn ), we have C α 0 Dt v(xi , tn )

= uxx (xi , tn ) + fin ,

1 ⩽ i ⩽ M − 1, 0 ⩽ n ⩽ N.

Applying the fast L1 approximation and second-order central difference quotient to approximate Caputo and spatial derivatives in the equality above respectively, it follows from Theorem 1.7.1 that N

exp { 1 n n n−1 { { [ ωl Fl,i + â (α) ∑ { 0 (Vi − Vi )] { Γ(1 − α) { { l=1 { { 2 n n n = δ U + f 1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N, { x i i + (r2 )i , { { { 1 n −s τ n−1 { { Fl,i = 0, Fl,i = e l Fl,i + Bl (Vin−1 − Vin−2 ), { { { 1 ⩽ l ⩽ Nexp , 1 ⩽ i ⩽ M − 1, 2 ⩽ n ⩽ N, {

(3.30) (3.31)

and there exists a positive constant c2 such that 󵄨󵄨 n 󵄨󵄨 2−α 2 󵄨󵄨(r2 )i 󵄨󵄨 ⩽ c2 (τ + h + ϵ),

1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N.

(3.32)

180 | 3 Difference methods for the time-fractional wave equations Substituting (3.31) into (3.30), we obtain n−1 1 n 0 ̂ (α) ̂ (α) n−k − â (α) [â (α) n−1 Vi ] 0 Vi − ∑ (ak−1 − ak )Vi Γ(1 − α) k=1

= δx2 Uin + fin + (r2 )ni ,

1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N.

(3.33)

Considering equation (3.27) at the point (xi , tn− 1 ), we have 2

n− 21

δt Ui

n− 21

= Vi

+ (r3 )ni ,

0 ⩽ i ⩽ M, 1 ⩽ n ⩽ N,

(3.34)

and there exists a positive constant c3 such that 󵄨󵄨 n 󵄨󵄨 2 󵄨󵄨(r3 )i 󵄨󵄨 ⩽ c3 τ ,

0 ⩽ i ⩽ M, 1 ⩽ n ⩽ N.

(3.35)

Noticing the initial-boundary value conditions (3.28)–(3.29), we have {

Ui0 = φi , U0n

= μ(tn ),

Vi0 = ψi , n UM

0 ⩽ i ⩽ M,

= ν(tn ),

1 ⩽ n ⩽ N.

(3.36) (3.37)

Omitting the small terms (r2 )ni and (r3 )ni in (3.30) and (3.34), and replacing the exact solution {Uin , Vin } with its numerical one {uni , vin } arrive at the difference scheme for solving (3.26)–(3.29) as follows: { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { {

N

exp 1 n n n−1 [ ∑ ωl Fl,i + â (α) 0 (vi − vi )] Γ(1 − α) l=1

= δx2 uni + fin , 1 Fl,i

= 0,

n Fl,i

1 ⩽ i ⩽ M − 1, =

n−1 e−sl τ Fl,i

+

1 ⩽ n ⩽ N,

Bl (vin−1

−

vin−2 ),

1 ⩽ l ⩽ Nexp , 1 ⩽ i ⩽ M − 1, 2 ⩽ n ⩽ N,

n− 1 n− 1 δt ui 2 = vi 2 , 0 ⩽ i ⩽ M, 1 ⩽ n ⩽ u0i = φi , vi0 = ψi , 0 ⩽ i ⩽ M, un0 = μ(tn ), unM = ν(tn ), 1 ⩽ n ⩽ N.

N,

(3.38) (3.39) (3.40) (3.41) (3.42)

Substituting (3.39) into (3.38) yields n−1 1 n 0 2 n n ̂ (α) ̂ (α) n−k − â (α) [â (α) n−1 vi ] = δx ui + fi , 0 vi − ∑ (ak−1 − ak )vi Γ(1 − α) k=1

1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N.

3.2.2 Solvability of the difference scheme Theorem 3.2.1. The difference scheme (3.38)–(3.42) is uniquely solvable.

(3.43)

3.2 The fast difference method based on L1 approximation for 1D problem |

181

Proof. Let un = (un0 , un1 , . . . , unM ),

n vn = (v0n , v1n , . . . , vM ).

The value of {u0 , v0 } is determined by (3.41). Now assume that the value of {u0 , v0 , u1 , v1 , . . . , un−1 , vn−1 } has been uniquely determined. From (3.40), we have n− 21

vin = 2vi

n− 21

− vin−1 = 2δt ui

− vin−1 ,

0 ⩽ i ⩽ M, 1 ⩽ n ⩽ N.

(3.44)

Substituting (3.44) into (3.38) and noticing (3.42) give a linear system in un as follows: N

exp { 1 n− 21 n { { − 2vin−1 )] [ ωl Fl,i + â (α) ∑ { 0 (2δt ui { { Γ(1 − α) l=1 { { { = δ2 un + fin , 1 ⩽ i ⩽ M − 1, { { { n x i n { u0 = μ(tn ), uM = ν(tn ).

To show its unique solvability, it suffices to verify that the corresponding homogeneous one â (α) { 0 { Γ(1 − α) { { n { u0 = 0,

2 ⋅ uni = δx2 uni , τ unM = 0

1 ⩽ i ⩽ M − 1,

(3.45) (3.46)

has only the trivial solution. Taking the inner product on both hand sides of (3.45) with un and noticing (3.46), we have â (α) 2 󵄩 󵄩2 󵄩 󵄩2 0 ⋅ 󵄩󵄩un 󵄩󵄩 + 󵄩󵄩δ un 󵄩󵄩 = 0, Γ(1 − α) τ 󵄩 󵄩 󵄩 x 󵄩 thus ‖un ‖ = 0. It follows un = 0. When un is determined, vn will be obtained from (3.44). By the principle of induction, the theorem is true. The proof ends.

3.2.3 Stability of the difference scheme Theorem 3.2.2. Suppose {uni , vin | 0 ⩽ i ⩽ M, 0 ⩽ n ⩽ N} is the solution of the difference scheme

182 | 3 Difference methods for the time-fractional wave equations

{ { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { {

N

exp 1 n n n−1 2 n n [ ∑ ωl Fl,i + â (α) 0 (vi − vi )] = δx ui + pi , Γ(1 − α) l=1

1 Fl,i

= 0, n− 21

n− 21

δt ui u0i un0

1 ⩽ i ⩽ M − 1,

n Fl,i

= vi

vi0 unM

= φi , = 0,

=

n−1 e−sl τ Fl,i

+

1 ⩽ l ⩽ Nexp , + qin ,

= ψi ,

= 0,

Bl (vin−1

1 ⩽ n ⩽ N,

−

vin−2 ),

1 ⩽ i ⩽ M − 1,

0 ⩽ i ⩽ M,

2 ⩽ n ⩽ N,

1 ⩽ n ⩽ N,

0 ⩽ i ⩽ M,

(3.47) (3.48) (3.49) (3.50)

1 ⩽ n ⩽ N.

(3.51)

Then it holds 4 1−α t tn 󵄩󵄩 n 󵄩󵄩2 󵄩󵄩 0 󵄩󵄩2 󵄩 󵄩2 3 n δ u ⩽ δ u + [ + ϵ]󵄩󵄩v0 󵄩󵄩 󵄩󵄩 x 󵄩󵄩 󵄩󵄩 x 󵄩󵄩 Γ(2 − α) Γ(1 − α) 󵄩 󵄩

2 1 2â (α) 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 0 + 2Γ(1 − α)tn1+α [max{󵄩󵄩󵄩p1 󵄩󵄩󵄩, max 󵄩󵄩󵄩pk− 2 󵄩󵄩󵄩} + max 󵄩󵄩󵄩qk 󵄩󵄩󵄩] , Γ(1 − α) 1⩽k⩽n 2⩽k⩽n

where M−1

󵄩󵄩 n 󵄩󵄩2 n 2 󵄩󵄩w 󵄩󵄩 = h ∑ (wi ) , i=1

w = v, p, q,

and 1 = (pni + pn−1 i ), 2

n− 21

pi

1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N.

Proof. Denote 2â (α) 0 q1 , Γ(1 − α) i

Q1i = p1i + n− 21

Qni = pi

+

1 ⩽ i ⩽ M − 1,

n−1 1 n )qin−k ], [â (α) q − ∑ (â (α) − â (α) k Γ(1 − α) 0 i k=1 k−1

1 ⩽ i ⩽ M − 1, 2 ⩽ n ⩽ N. Substituting (3.48) into (3.47), we have n−1 1 n 0 2 n n [â (α) v − )vin−k − â (α) ∑ (â (α) − â (α) n−1 vi ] = δx ui + pi , k Γ(1 − α) 0 i k=1 k−1

1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N,

1 ⩽ n ⩽ N,

3.2 The fast difference method based on L1 approximation for 1D problem

| 183

which is equivalent to 1 1 0 2 1 1 2 { 1 ⩽ i ⩽ M − 1, â (α) { 0 (2vi − 2vi ) = δx ui + pi , { Γ(1 − α) { { { { n−1 { 1 1 n−k− 21 (α) n− 2 0 ̂ [ a v − − â (α) − â k(α) )vi ∑ (â (α) { n−1 vi ] 0 i k−1 { { Γ(1 − α) { k=1 { { { 1 { n− 21 2 n− 2 { = δx ui + pi , 1 ⩽ i ⩽ M − 1, 2 ⩽ n ⩽ N.

(3.52)

(3.53)

Substituting (3.49) into (3.52)–(3.53) yields 1 1 { â (α) (2δt ui2 − 2vi0 ) = δx2 u1i + Q1i , 1 ⩽ i ⩽ M − 1, { 0 { Γ(1 − α) { { { { n−1 { 1 n− 21 n−k− 21 (α) 0 (α) (α) ̂ ̂ ̂ [ a δ u − − â (α) ( a − a )δ u ∑ { t t n−1 vi ] { i i k−1 k { Γ(1 − α) 0 { k=1 { { { 1 { 2 n− 2 n { = δx ui + Qi , 1 ⩽ i ⩽ M − 1, 2 ⩽ n ⩽ N.

(3.54)

(3.55)

1

(I) Making the inner product on both hand sides of (3.54) with δt u 2 produces 1 2 󵄩󵄩 21 󵄩󵄩2 â (α) δt u 󵄩󵄩 − (δx2 u1 , δt u 2 ) 󵄩 0 󵄩 Γ(1 − α) 1 1 2 0 1 2 2 = â (α) 0 (v , δt u ) + (Q , δt u ) Γ(1 − α) 2 󵄩󵄩 0 󵄩󵄩2 3 󵄩󵄩 21 󵄩󵄩2 1 󵄩 󵄩 󵄩 1󵄩 â (α) v 󵄩󵄩 + 󵄩󵄩δt u 󵄩󵄩 ) + 󵄩󵄩󵄩Q1 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩δt u 2 󵄩󵄩󵄩. ⩽ 󵄩 0 ( 󵄩 Γ(1 − α) 3 2

Noticing (3.51), we have 1

1

−(δx2 u1 , δt u 2 ) = (δx u1 , δt δx u 2 ) =

1 󵄩󵄩 1 󵄩󵄩2 󵄩󵄩 0 󵄩󵄩2 (󵄩δ u 󵄩 − 󵄩δ u 󵄩 ) + 2τ 󵄩 x 󵄩 󵄩 x 󵄩

1 2 τ 󵄩󵄩 󵄩 󵄩󵄩δx δt u 2 󵄩󵄩󵄩 , 2

hence it follows that 2 󵄩󵄩 21 󵄩󵄩2 1 󵄩󵄩 1 󵄩󵄩2 󵄩󵄩 0 󵄩󵄩2 â (α) 󵄩δt u 󵄩󵄩 + 2τ (󵄩󵄩δx u 󵄩󵄩 − 󵄩󵄩δx u 󵄩󵄩 ) 0 󵄩 Γ(1 − α) 2 󵄩󵄩 0 󵄩󵄩2 3 󵄩󵄩 21 󵄩󵄩2 1 󵄩 󵄩 󵄩 1󵄩 â (α) v 󵄩󵄩 + 󵄩󵄩δt u 󵄩󵄩 ) + 󵄩󵄩󵄩Q1 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩δt u 2 󵄩󵄩󵄩. ⩽ 󵄩 0 ( 󵄩 Γ(1 − α) 3 2 Multiplying both hand sides of the inequality above by 2τ arrives at τ 󵄩󵄩 21 󵄩󵄩2 󵄩󵄩 1 󵄩󵄩2 â (α) 󵄩δt u 󵄩󵄩 + 󵄩󵄩δx u 󵄩󵄩 0 󵄩 Γ(1 − α) τ 󵄩󵄩 0 󵄩󵄩2 󵄩 󵄩 󵄩 1󵄩 󵄩 󵄩2 4 ⩽ 󵄩󵄩󵄩δx u0 󵄩󵄩󵄩 + ⋅ â (α) v 󵄩󵄩 + 2τ󵄩󵄩󵄩Q1 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩δt u 2 󵄩󵄩󵄩. 󵄩 0 󵄩 3 Γ(1 − α)

(3.56) 1

(II) Making the inner product on both hand sides of (3.55) with 2δt un− 2 produces 2 󵄩󵄩 n− 21 󵄩󵄩2 1 󵄩󵄩 n 󵄩󵄩2 󵄩󵄩 n−1 󵄩󵄩2 â (α) 󵄩δt u 󵄩󵄩 + τ (󵄩󵄩δx u 󵄩󵄩 − 󵄩󵄩δx u 󵄩󵄩 ) 0 󵄩 Γ(1 − α)

184 | 3 Difference methods for the time-fractional wave equations

=

n−1 1 1 2 0 n− 21 [ ∑ (â (α) − â (α) )(δt un−k− 2 , δt un− 2 ) + â (α) )] n−1 (v , δt u k−1 k Γ(1 − α) k=1 1

+ 2(Qn , δt un− 2 ) ⩽

n−1 1 2 1 2 1 󵄩 󵄩 󵄩 󵄩 [ ∑ (â (α) − â (α) )(󵄩󵄩󵄩δt un−k− 2 󵄩󵄩󵄩 + 󵄩󵄩󵄩δt un− 2 󵄩󵄩󵄩 ) k−1 k Γ(1 − α) k=1

󵄩󵄩 n 󵄩󵄩 󵄩󵄩 n− 21 󵄩󵄩 󵄩󵄩 0 󵄩󵄩2 󵄩󵄩 n− 21 󵄩󵄩2 + â (α) 󵄩v 󵄩󵄩 + 󵄩󵄩δt u 󵄩󵄩 )] + 2󵄩󵄩Q 󵄩󵄩 ⋅ 󵄩󵄩δt u 󵄩󵄩, n−1 (󵄩

2 ⩽ n ⩽ N,

that is, 1 󵄩󵄩 n− 21 󵄩󵄩2 1 󵄩󵄩 n 󵄩󵄩2 󵄩󵄩 n−1 󵄩󵄩2 â (α) 󵄩δt u 󵄩󵄩 + τ (󵄩󵄩δx u 󵄩󵄩 − 󵄩󵄩δx u 󵄩󵄩 ) 0 󵄩 Γ(1 − α)

n−1 1 2 1 1 󵄩 󵄩2 󵄩 󵄩 â (α) 󵄩󵄩v0 󵄩󵄩 )󵄩󵄩δt un−k− 2 󵄩󵄩󵄩 + ∑ (â (α) − â (α) k 󵄩 Γ(1 − α) k=1 k−1 Γ(1 − α) n−1 󵄩 󵄩

⩽

1 󵄩 󵄩 󵄩 󵄩 + 2󵄩󵄩󵄩Qn 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩δt un− 2 󵄩󵄩󵄩,

2 ⩽ n ⩽ N.

The above inequality can be rewritten as n−1 τ 󵄩󵄩 n−k− 21 󵄩󵄩2 󵄩󵄩 n 󵄩󵄩2 ∑ â (α) 󵄩󵄩δt u 󵄩󵄩 + 󵄩󵄩δx u 󵄩󵄩 k Γ(1 − α) k=0

⩽

=

n−1 τ 󵄩󵄩 n−k− 21 󵄩󵄩2 󵄩󵄩 n−1 󵄩󵄩2 ∑ â (α) 󵄩󵄩δt u 󵄩󵄩 + 󵄩󵄩δx u 󵄩󵄩 k−1 Γ(1 − α) k=1 1 τ 󵄩 󵄩2 󵄩 󵄩 󵄩 󵄩 â (α) 󵄩󵄩v0 󵄩󵄩 + 2τ󵄩󵄩󵄩Qn 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩δt un− 2 󵄩󵄩󵄩 + Γ(1 − α) n−1 󵄩 󵄩

n−2 τ 󵄩󵄩 n−1−k− 21 󵄩󵄩2 󵄩󵄩 n−1 󵄩󵄩2 ∑ â (α) 󵄩󵄩δt u 󵄩󵄩 + 󵄩󵄩δx u 󵄩󵄩 k Γ(1 − α) k=0 1 τ 󵄩 󵄩2 󵄩 󵄩 󵄩 󵄩 â (α) 󵄩󵄩v0 󵄩󵄩 + 2τ󵄩󵄩󵄩Qn 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩δt un− 2 󵄩󵄩󵄩, 2 ⩽ n ⩽ N. + Γ(1 − α) n−1 󵄩 󵄩

Replacing the superscript n with m and summing up for m from 2 to n on both hand sides of the above inequality lead to n−1 τ 󵄩󵄩 n−k− 21 󵄩󵄩2 󵄩󵄩 n 󵄩󵄩2 ∑ â (α) 󵄩󵄩δt u 󵄩󵄩 + 󵄩󵄩δx u 󵄩󵄩 k Γ(1 − α) k=0 τ 󵄩󵄩 21 󵄩󵄩2 󵄩󵄩 1 󵄩󵄩2 ⩽ â (α) 󵄩δt u 󵄩󵄩 + 󵄩󵄩δx u 󵄩󵄩 0 󵄩 Γ(1 − α)

+

n−1 n τ 󵄩󵄩 0 󵄩󵄩2 󵄩󵄩 m 󵄩󵄩 󵄩󵄩 m− 21 󵄩󵄩 v + 2τ ∑ â (α) ∑ 󵄩 󵄩 󵄩󵄩Q 󵄩󵄩 ⋅ 󵄩󵄩δt u 󵄩󵄩, m 󵄩 󵄩 Γ(1 − α) m=1 m=2

Combining (3.56) with (3.57) produces n−1 τ 󵄩󵄩 n−k− 21 󵄩󵄩2 󵄩󵄩 n 󵄩󵄩2 ∑ â (α) 󵄩󵄩δt u 󵄩󵄩 + 󵄩󵄩δx u 󵄩󵄩 k Γ(1 − α) k=0

2 ⩽ n ⩽ N.

(3.57)

3.2 The fast difference method based on L1 approximation for 1D problem

| 185

󵄩 󵄩2 ⩽ 󵄩󵄩󵄩δx u0 󵄩󵄩󵄩 +

n−1 n 1 τ 4 (α) 󵄩 󵄩󵄩v0 󵄩󵄩󵄩2 + 2τ ∑ 󵄩󵄩󵄩Qm 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩δt um− 2 󵄩󵄩󵄩 ̂ [ â (α) + a ] ∑ m 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 0 Γ(1 − α) 3 m=1 m=1

󵄩 󵄩2 = 󵄩󵄩󵄩δx u0 󵄩󵄩󵄩 +

n−1 n−1 1 4 τ 2 ̂ (α) 󵄩󵄩󵄩v0 󵄩󵄩󵄩󵄩 + 2τ ∑ 󵄩󵄩󵄩󵄩Qn−k 󵄩󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩󵄩δt un−k− 2 󵄩󵄩󵄩󵄩 [ â (α) 0 + ∑ am ]󵄩 Γ(1 − α) 3 m=1 k=0

󵄩 󵄩2 ⩽ 󵄩󵄩󵄩δx u0 󵄩󵄩󵄩 +

n−1 n−1 τ τ 4 2 󵄩󵄩 n−k− 21 󵄩󵄩2 ̂ (α) 󵄩󵄩󵄩v0 󵄩󵄩󵄩󵄩 + [ â (α) ∑ â (α) 󵄩󵄩 󵄩δt u 0 + ∑ am ]󵄩 k 󵄩 Γ(1 − α) 3 Γ(1 − α) m=1 k=0 n−1

+ τΓ(1 − α) ∑

1 󵄩󵄩 n−k 󵄩󵄩2 󵄩Q 󵄩󵄩 ,

(α) 󵄩 k=0 â k

1 ⩽ n ⩽ N,

that is, n−1 τ 4 2 󵄩󵄩 n 󵄩󵄩2 󵄩󵄩 0 󵄩󵄩2 ̂ (α) 󵄩󵄩󵄩v0 󵄩󵄩󵄩󵄩 [ â (α) 󵄩󵄩δx u 󵄩󵄩 ⩽ 󵄩󵄩δx u 󵄩󵄩 + 0 + ∑ am ]󵄩 Γ(1 − α) 3 m=1 n−1

1 󵄩󵄩 n−k 󵄩󵄩2 󵄩Q 󵄩󵄩 , (α) 󵄩 ̂ a k=0 k

+ τΓ(1 − α) ∑

1 ⩽ n ⩽ N.

(3.58)

It follows from (1.126) that n−1 4 τ ̂ (α) [ â (α) 0 + ∑ am ] Γ(1 − α) 3 m=1

⩽ = ⩽

n−1 τ 4 τ−α τ−α (α) [ ⋅ + ∑( a + ϵ)] Γ(1 − α) 3 1 − α m=1 1 − α m

τ 4 τ−α τ−α 1−α [ ⋅ + (n − 1) + (n − 1)ϵ] Γ(1 − α) 3 1 − α 1 − α 4 1−α t 3 n

Γ(2 − α)

+

tn ϵ Γ(1 − α)

(3.59)

and it follows from Lemma 1.7.2 that n−1

1 󵄩󵄩 n−k 󵄩󵄩2 󵄩Q 󵄩󵄩 (α) 󵄩 ̂ k=0 ak

τΓ(1 − α) ∑

n−1 1 󵄩 󵄩2 ⩽ τΓ(1 − α) max 󵄩󵄩󵄩Qk 󵄩󵄩󵄩 ∑ (α) 1⩽k⩽n k=0 â k n−1

󵄩 󵄩2 α ⩽ τΓ(1 − α) max 󵄩󵄩󵄩Qk 󵄩󵄩󵄩 ∑ 2 tk+1 1⩽k⩽n

k=0

󵄩 󵄩2 ⩽ 2Γ(1 − α)tn1+α max 󵄩󵄩󵄩Qk 󵄩󵄩󵄩 . 1⩽k⩽n

(3.60)

In addition, it follows from the definition of Qn that 2â (α) 󵄩󵄩 1 󵄩󵄩 󵄩󵄩 1 󵄩󵄩 󵄩󵄩 1 󵄩󵄩 0 󵄩󵄩Q 󵄩󵄩 ⩽ 󵄩󵄩p 󵄩󵄩 + 󵄩q 󵄩, Γ(1 − α) 󵄩 󵄩

(3.61)

186 | 3 Difference methods for the time-fractional wave equations n−1

1 󵄩󵄩 n 󵄩󵄩 󵄩󵄩 n− 21 󵄩󵄩 󵄩 󵄩 󵄩 󵄩 [â (α) 󵄩󵄩qn 󵄩󵄩 + ∑ (â (α) − â (α) )󵄩󵄩qn−k 󵄩󵄩󵄩] 󵄩󵄩Q 󵄩󵄩 ⩽ 󵄩󵄩p 󵄩󵄩 + k 󵄩 Γ(1 − α) 0 󵄩 󵄩 k=1 k−1 1 󵄩 󵄩 ⩽ 󵄩󵄩󵄩pn− 2 󵄩󵄩󵄩 +

1 󵄩 󵄩 󵄩󵄩 k 󵄩󵄩 ̂ (α) [â (α) 󵄩󵄩qn 󵄩󵄩 + (â (α) 󵄩q 󵄩󵄩] 0 − an−1 ) max 󵄩 Γ(1 − α) 0 󵄩 󵄩 1⩽k⩽n−1

1 2â (α) 󵄩 󵄩 󵄩 󵄩 0 ⩽ 󵄩󵄩󵄩pn− 2 󵄩󵄩󵄩 + max 󵄩󵄩qk 󵄩󵄩, Γ(1 − α) 1⩽k⩽n󵄩 󵄩

2 ⩽ n ⩽ N.

(3.62)

The substitution of (3.59)–(3.62) into (3.58) yields 4 1−α tn tn 󵄩 󵄩2 󵄩󵄩 n 󵄩󵄩2 󵄩󵄩 0 󵄩󵄩2 + ϵ]󵄩󵄩v0 󵄩󵄩 󵄩󵄩δx u 󵄩󵄩 ⩽ 󵄩󵄩δx u 󵄩󵄩 + [ 3 Γ(2 − α) Γ(1 − α) 󵄩 󵄩 1 󵄩 󵄩 󵄩 󵄩 + 2Γ(1 − α)tn1+α [max{󵄩󵄩󵄩p1 󵄩󵄩󵄩, max 󵄩󵄩󵄩pk− 2 󵄩󵄩󵄩}

+

2

2â (α) 0

󵄩 󵄩 max 󵄩󵄩qk 󵄩󵄩] , Γ(1 − α) 1⩽k⩽n󵄩 󵄩

2⩽k⩽n

1 ⩽ n ⩽ N.

The proof ends.

3.2.4 Convergence of the difference scheme Theorem 3.2.3. Suppose {Uin , Vin | 0 ⩽ i ⩽ M, 0 ⩽ n ⩽ N} and {uni , vin | 0 ⩽ i ⩽ M, 0 ⩽ n ⩽ N} are solutions of the problem (3.26)–(3.29) and the difference scheme (3.38)–(3.42), respectively. Let ein = Uin − uni ,

zin = Vin − vin ,

0 ⩽ i ⩽ M, 0 ⩽ n ⩽ N,

then it holds 󵄩󵄩 n 󵄩󵄩 √ γ 󵄩󵄩δx e 󵄩󵄩 ⩽ 2Ltn Γ(2 − γ) (c2 +

2 c )(τ3−γ + h2 + ϵ), Γ(3 − γ) 3

1 ⩽ n ⩽ N.

Proof. Subtracting (3.43), (3.40)–(3.42) from (3.33), (3.34), (3.36)–(3.37), respectively, produces the system of error equations as follows: { { { { { { { { { { { { { { { { { { { { { { { { {

n−1 1 n 0 ̂ (α) ̂ (α) n−k − â (α) [â (α) n−1 zi ] 0 zi − ∑ (ak−1 − ak )zi Γ(1 − α) k=1

= δx2 ein + (r2 )ni ,

1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N,

n− 1 n− 1 δt ei 2 = zi 2 + (r3 )ni , ei0 = 0, zi0 = 0, 0 ⩽ n e0n = 0, eM = 0, 1 ⩽

0 ⩽ i ⩽ M, 1 ⩽ n ⩽ N, i ⩽ M, n ⩽ N.

3.3 The fourth-order method in space based on L1 approximation for 1D problem

| 187

Noticing (3.32) and (3.35), an immediate consequence of Theorem 3.2.2 reads 󵄩󵄩 n 󵄩󵄩2 󵄩󵄩 0 󵄩󵄩2 󵄩󵄩δx e 󵄩󵄩 ⩽ 󵄩󵄩δx e 󵄩󵄩 + [

4 1−α t 3 n

Γ(2 − α)

+

tn 󵄩 󵄩2 ϵ]󵄩󵄩z 0 󵄩󵄩 Γ(1 − α) 󵄩 󵄩

1 󵄩 󵄩 󵄩 󵄩 + 2Γ(1 − α)tn1+α [max{󵄩󵄩󵄩(r2 )1 󵄩󵄩󵄩, max 󵄩󵄩󵄩(r2 )k− 2 󵄩󵄩󵄩}

+

2⩽k⩽n

2

2â (α) 0

󵄩 󵄩 max 󵄩󵄩(r )k 󵄩󵄩] Γ(1 − α) 1⩽k⩽n󵄩 3 󵄩

1 󵄩 󵄩 󵄩 󵄩 = 2Γ(1 − α)tn1+α [max{󵄩󵄩󵄩(r2 )1 󵄩󵄩󵄩, max 󵄩󵄩󵄩(r2 )k− 2 󵄩󵄩󵄩}

+

2

2â (α) 0

2⩽k⩽n

󵄩 󵄩 max 󵄩󵄩(r )k 󵄩󵄩] Γ(1 − α) 1⩽k⩽n󵄩 3 󵄩

⩽ 2Γ(1 − α)tn1+α [√L c2 (τ2−α + h2 + ϵ) +

2 2â (α) 0 √ L c3 τ 2 ] Γ(1 − α)

= 2Γ(1 − α)tn1+α [√L c2 (τ2−α + h2 + ϵ) +

2 √L c3 τ2−α ] Γ(2 − α)

⩽ 2Γ(1 − α)tn1+α [(c2 +

2

2 c )√L (τ2−α + h2 + ϵ)] , Γ(2 − α) 3

2

1 ⩽ n ⩽ N.

Taking the square root on both hand sides of the inequality above arrives at 2 c )(τ2−α + h2 + ϵ) Γ(2 − α) 3 2 γ c )(τ3−γ + h2 + ϵ), 1 ⩽ n ⩽ N. = √2Ltn Γ(2 − γ) (c2 + Γ(3 − γ) 3

󵄩󵄩 n 󵄩󵄩 √ 1+α 󵄩󵄩δx e 󵄩󵄩 ⩽ 2Ltn Γ(1 − α) (c2 +

The proof ends.

3.3 The fourth-order method in space based on L1 approximation for 1D problem In this section, we continue to consider the problem (3.1)–(3.3), but another high order difference scheme in space will be developed. Suppose the exact solution u ∈ C (6,3) ([0, L] × [0, T]).

3.3.1 Derivation of the difference scheme Considering equation (3.1) at the point (xi , tn ), we have C γ 0 Dt u(xi , tn )

= uxx (xi , tn ) + fin ,

0 ⩽ i ⩽ M, 0 ⩽ n ⩽ N.

188 | 3 Difference methods for the time-fractional wave equations Performing the operator 𝒜 to both hand sides of the equality above and taking an average on two adjacent time levels arrive at 1 C γ C γ 2 1 n− 1 = 𝒜{ [uxx (xi , tn ) + uxx (xi , tn−1 )]} + 𝒜fi 2 , 2 𝒜{ [0 Dt u(xi , tn ) + 0 Dt u(xi , tn−1 )]}

1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N.

Applying the L1 formula (1.69) to approximate the time-fractional derivative in the equality above, it follows from Theorem 1.6.2 and Lemma 2.1.3 that n−1 τ1−γ k− 1 n− 1 (γ) (γ) (γ) (γ) 𝒜{b0 δt Ui 2 − ∑ (bn−k−1 − bn−k )δt Ui 2 − bn−1 ψi } Γ(3 − γ) k=1 n− 21

n− 21

= δx2 Ui

n− 21

+ 𝒜fi

+ (r4 )i

1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N,

,

(3.63)

and there is a positive constant c4 such that n− 1 󵄨 󵄨󵄨 3−γ 4 󵄨󵄨(r4 )i 2 󵄨󵄨󵄨 ⩽ c4 (τ + h ),

1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N,

(3.64)

(γ)

where {bl } is defined in (1.64). Noticing the initial-boundary value conditions (3.2)–(3.3), we have {

Ui0 = φ(xi ), U0n

= μ(tn ),

1 ⩽ i ⩽ M − 1,

n UM

= ν(tn ),

(3.65) 0 ⩽ n ⩽ N.

(3.66)

n− 1

Omitting the small term (r4 )i 2 in (3.63) and replacing the exact solution Uin with its numerical one uni produce a difference scheme for (3.1)–(3.3) as follows: { { { { { { { { { { { { { { { { { { {

n−1 τ1−γ n− 1 k− 1 (γ) (γ) (γ) (γ) 𝒜{b0 δt ui 2 − ∑ (bn−k−1 − bn−k )δt ui 2 − bn−1 ψi } Γ(3 − γ) k=1 n− 21

= δx2 ui

u0i un0

n− 21

+ 𝒜fi

= φ(xi ),

= μ(tn ),

,

1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N,

1 ⩽ i ⩽ M − 1,

unM

= ν(tn ),

(3.67) (3.68)

0 ⩽ n ⩽ N.

(3.69)

Denote η = τγ−1 Γ(3 − γ). 3.3.2 Solvability of the difference scheme We now proceed to discuss the unique solvability of the difference scheme (3.67)– (3.69). The following theorem is true.

3.3 The fourth-order method in space based on L1 approximation for 1D problem

| 189

Theorem 3.3.1. The difference scheme (3.67)–(3.69) is uniquely solvable. Proof. Let un = (un0 , un1 , . . . , unM ). The value of u0 is determined by (3.68)–(3.69). Suppose the values of u0 , u1 , . . . , un−1 have been uniquely determined, then the linear system in un can be obtained from (3.67) and (3.69). To show its unique solvability, it suffices to prove the corresponding homogeneous one 1 1 2 n n 𝒜ui = δx ui , { ητ 2 { n n { u0 = uM = 0

1 ⩽ i ⩽ M − 1,

(3.70) (3.71)

has only the trivial solution. Taking the inner product on both hand sides of (3.70) with un yields 1 1 (𝒜un , un ) = (δx2 un , un ). ητ 2

(3.72)

Noticing (3.71), it follows from the summation by parts and Lemma 2.1.1 that (𝒜un , un ) = ((ℐ +

h2 2 n n δ )u , u ) 12 x

2 󵄩 󵄩2 h 󵄩 󵄩2 2 󵄩 󵄩2 = 󵄩󵄩󵄩un 󵄩󵄩󵄩 − 󵄩󵄩󵄩δx un 󵄩󵄩󵄩 ⩾ 󵄩󵄩󵄩un 󵄩󵄩󵄩 . 12 3

Substituting this result into (3.72) arrives at

2 1 󵄩󵄩 n 󵄩󵄩2 1 󵄩 n 󵄩2 ⋅ 󵄩󵄩u 󵄩󵄩 ⩽ − 󵄩󵄩󵄩δx u 󵄩󵄩󵄩 ⩽ 0, 3 ητ 2

thus ‖un ‖2 = 0. It follows un = 0 from (3.71). By the principle of induction, the theorem is true. The proof ends. 3.3.3 Stability of the difference scheme The stability analysis on the difference scheme (3.67)–(3.69) will be carried out in this subsection. Theorem 3.3.2. Suppose {vin | 0 ⩽ i ⩽ M, 0 ⩽ n ⩽ N} is the solution of the difference scheme { { { { { { { { { { { { { { { { { { {

n−1 1 n− 1 k− 1 (γ) (γ) (γ) 𝒜{b0 δt vi 2 − ∑ (bn−k−1 − bn−k )δt vi 2 − bn−1 ψi } η k=1 n− 21

= δx2 vi vi0 v0n

n− 21

+ gi

= φ(xi ),

= 0,

n vM

,

1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N,

(3.73)

1 ⩽ i ⩽ M − 1,

(3.74)

= 0,

(3.75)

0 ⩽ n ⩽ N.

190 | 3 Difference methods for the time-fractional wave equations Then it holds

2−γ

n 1 2 t 󵄩󵄩 n 󵄩󵄩2 󵄩󵄩 0 󵄩󵄩2 󵄩 󵄩 ‖𝒜ψ‖2 + Γ(2 − γ)tnγ−1 τ ∑ 󵄩󵄩󵄩g k− 2 󵄩󵄩󵄩 , 󵄩󵄩δx v 󵄩󵄩A ⩽ 󵄩󵄩δx v 󵄩󵄩A + n Γ(3 − γ) k=1

1 ⩽ n ⩽ N,

(3.76)

where M−1

M−1

k− 1 2 󵄩󵄩 k− 21 󵄩󵄩2 󵄩󵄩g 󵄩󵄩 = h ∑ (gi 2 ) .

‖𝒜ψ‖2 = h ∑ (𝒜ψi )2 , i=1

i=1

1

Proof. Making the inner product on both hand sides of (3.73) with η𝒜δt vn− 2 , it follows from the Cauchy–Schwarz inequality that 1 2 (γ) 󵄩 󵄩 b0 󵄩󵄩󵄩𝒜δt vn− 2 󵄩󵄩󵄩

n−1

1

1

= ∑ (bn−k−1 − bn−k )(𝒜δt vk− 2 , 𝒜δt vn− 2 ) (γ)

(γ)

k=1

1

1

1

1

1

+ bn−1 (𝒜ψ, 𝒜δt vn− 2 ) + η(δx2 vn− 2 , 𝒜δt vn− 2 ) + η(g n− 2 , 𝒜δt vn− 2 ) (γ)

⩽

1 2 1 2 1 n−1 (γ) (γ) 󵄩 󵄩 󵄩 󵄩 − bn−k )(󵄩󵄩󵄩𝒜δt vk− 2 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝒜δt vn− 2 󵄩󵄩󵄩 ) ∑ (b 2 k=1 n−k−1

1 2 1 1 1 (γ) 󵄩 󵄩 + bn−1 (‖𝒜ψ‖2 + 󵄩󵄩󵄩𝒜δt vn− 2 󵄩󵄩󵄩 ) + η(δx2 vn− 2 , 𝒜δt vn− 2 ) 2 1

1

+ η(g n− 2 , 𝒜δt vn− 2 ),

1 ⩽ n ⩽ N,

which can be simplified to 1 2 (γ) 󵄩 󵄩 b0 󵄩󵄩󵄩𝒜δt vn− 2 󵄩󵄩󵄩 n−1

(γ) (γ) 󵄩 (γ) 󵄩2 ⩽ ∑ (bn−k−1 − bn−k )󵄩󵄩󵄩𝒜δt vk− 2 󵄩󵄩󵄩 + bn−1 ‖𝒜ψ‖2 1

k=1

1

1

1

1

+ 2η(δx2 vn− 2 , 𝒜δt vn− 2 ) + 2η(g n− 2 , 𝒜δt vn− 2 ),

1 ⩽ n ⩽ N.

(3.77)

Applying the summation by parts and noticing (3.75), we have 1

1

1

1

−(δx2 vn− 2 , 𝒜δt vn− 2 ) = (δt vn− 2 , vn− 2 )1,A

1 [(vn , vn )1,A − (vn−1 , vn−1 )1,A ] 2τ 1 󵄩 󵄩2 󵄩 󵄩2 = (󵄩󵄩󵄩δx vn 󵄩󵄩󵄩A − 󵄩󵄩󵄩δx vn−1 󵄩󵄩󵄩A ). 2τ =

The substitution of (3.78) into (3.77) yields

n 1 2 η 󵄩 (γ) 󵄩 󵄩 󵄩2 󵄩 󵄩2 ∑ bn−k 󵄩󵄩󵄩𝒜δt vk− 2 󵄩󵄩󵄩 + (󵄩󵄩󵄩δx vn 󵄩󵄩󵄩A − 󵄩󵄩󵄩δx vn−1 󵄩󵄩󵄩A ) τ k=1

n−1

1 2 1 1 (γ) (γ) 󵄩 󵄩 ⩽ ∑ bn−k−1 󵄩󵄩󵄩𝒜δt vk− 2 󵄩󵄩󵄩 + bn−1 ‖𝒜ψ‖2 + 2η(g n− 2 , 𝒜δt vn− 2 ),

k=1

1 ⩽ n ⩽ N,

(3.78)

3.3 The fourth-order method in space based on L1 approximation for 1D problem

| 191

which can be rearranged as n (γ) 󵄩 󵄩󵄩 n 󵄩󵄩2 τ k− 1 󵄩2 󵄩󵄩δx v 󵄩󵄩A + ∑ bn−k 󵄩󵄩󵄩𝒜δt v 2 󵄩󵄩󵄩 η k=1

n−1 1 2 (γ) 󵄩 󵄩 󵄩2 τ 󵄩 ⩽ 󵄩󵄩󵄩δx vn−1 󵄩󵄩󵄩A + ∑ bn−k−1 󵄩󵄩󵄩𝒜δt vk− 2 󵄩󵄩󵄩 η k=1 1 1 τ (γ) + bn−1 ‖𝒜ψ‖2 + 2τ(g n− 2 , 𝒜δt vn− 2 ), η

1 ⩽ n ⩽ N.

(3.79)

Let n 1 2 (γ) 󵄩 󵄩 󵄩2 τ 󵄩 Gn = 󵄩󵄩󵄩δx vn 󵄩󵄩󵄩A + ∑ bn−k 󵄩󵄩󵄩𝒜δt vk− 2 󵄩󵄩󵄩 , η k=1

󵄩 󵄩2 G0 = 󵄩󵄩󵄩δx v0 󵄩󵄩󵄩A ,

1 ⩽ n ⩽ N.

It follows from (3.79) that Gn ⩽ Gn−1 +

1 1 τ (γ) b ‖𝒜ψ‖2 + 2τ(g n− 2 , 𝒜δt vn− 2 ), η n−1

1 ⩽ n ⩽ N.

The recursive process will lead to Gn ⩽ G0 + ⩽ G0 +

n 1 1 τ n−1 (γ) ∑ bk ‖𝒜ψ‖2 + 2τ ∑ (g k− 2 , 𝒜δt vk− 2 ) η k=0 k=1

τ n−1 (γ) ∑ b ‖𝒜ψ‖2 η k=0 k n

+ τ ∑( k=1

(γ)

η 󵄩󵄩 k− 21 󵄩󵄩2 bn−k 󵄩󵄩 k− 1 󵄩2 󵄩g 󵄩󵄩 + 󵄩󵄩𝒜δt v 2 󵄩󵄩󵄩 ), (γ) 󵄩 η b

1 ⩽ n ⩽ N,

n−k

that is, n−1 n η 󵄩 k− 1 󵄩2 (γ) 󵄩󵄩 n 󵄩󵄩2 󵄩󵄩 0 󵄩󵄩2 τ 2 󵄩󵄩δx v 󵄩󵄩A ⩽ 󵄩󵄩δx v 󵄩󵄩A + ∑ bk ‖𝒜ψ‖ + τ ∑ (γ) 󵄩󵄩󵄩g 2 󵄩󵄩󵄩 , η k=0 k=1 b

1 ⩽ n ⩽ N.

(3.80)

n−k

By (3.20) and (3.22), it follows (3.76) from (3.80). The proof ends. 3.3.4 Convergence of the difference scheme We now present the error analysis of the difference scheme (3.67)–(3.69). Theorem 3.3.3. Suppose {Uin | 0 ⩽ i ⩽ M, 0 ⩽ n ⩽ N} and {uni | 0 ⩽ i ⩽ M, 0 ⩽ n ⩽ N} are solutions of the problem (3.1)–(3.3) and the difference scheme (3.67)–(3.69), respectively. Let ein = Uin − uni ,

0 ⩽ i ⩽ M, 0 ⩽ n ⩽ N,

192 | 3 Difference methods for the time-fractional wave equations then it holds L 󵄩󵄩 n 󵄩󵄩 3−γ 4 󵄩󵄩e 󵄩󵄩∞ ⩽ √6T γ Γ(2 − γ) c4 (τ + h ), 4

1 ⩽ n ⩽ N.

Proof. The subtraction of (3.67)–(3.69) from (3.63), (3.65)–(3.66), respectively, produces the system of error equations as follows: { { { { { { { { { { { { { { { { { { {

n−1 1 k− 1 n− 1 (γ) (γ) (γ) (γ) 𝒜{b0 δt ei 2 − ∑ (bn−k−1 − bn−k )δt ei 2 − bn−1 ⋅ 0} η k=1 n− 21

= δx2 ei

ei0 e0n

= 0,

= 0,

n− 21

+ (r4 )i

,

1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N,

1 ⩽ i ⩽ M − 1,

n eM = 0,

0 ⩽ n ⩽ N.

Noticing (3.64), the application of Theorem 3.3.2 yields n

󵄩 󵄩󵄩 n 󵄩󵄩2 k− 1 󵄩2 γ−1 󵄩󵄩δx e 󵄩󵄩A ⩽ tn Γ(2 − γ)τ ∑ 󵄩󵄩󵄩(r4 ) 2 󵄩󵄩󵄩 k=1

γ

⩽ T Γ(2 −

γ)Lc42 (τ3−γ

2

+ h4 ) ,

1 ⩽ n ⩽ N.

Taking the square root on both hand sides of the inequality above, it follows from Lemma 2.1.1 and Lemma 2.1.2 that √L 󵄩 n 󵄩 √L 3󵄩 󵄩󵄩 n 󵄩󵄩 󵄩 󵄩󵄩δ e 󵄩󵄩 ⩽ ⋅ √ 󵄩󵄩󵄩δx en 󵄩󵄩󵄩A 󵄩󵄩e 󵄩󵄩∞ ⩽ 2 󵄩 x 󵄩 2 2 L ⩽ √6T γ Γ(2 − γ) c4 (τ3−γ + h4 ), 4

1 ⩽ n ⩽ N.

The proof ends.

3.4 The difference method based on L2-1σ approximation for 1D problem Consider the following problem of the time-fractional wave equation: γ

C D u(x, t) = uxx (x, t) + f (x, t), x ∈ (0, L), t ∈ (0, T], { { { 0 t u(x, 0) = φ(x), ut (x, 0) = ψ(x), x ∈ [0, L], { { { { u(0, t) = 0, u(L, t) = 0, t ∈ (0, T],

(3.81) (3.82) (3.83)

where γ ∈ (1, 2), the functions f , φ, ψ are all given and φ(0) = 0, φ(L) = 0, ψ(0) = 0, ψ(L) = 0. Suppose the exact solution u ∈ C (4,4) ([0, L] × [0, T]). In this section, we will use the L2-1σ approximation to establish a temporal secondorder difference scheme.

3.4 The difference method based on L2-1σ approximation for 1D problem

| 193

3.4.1 Derivation of the difference scheme At first, a useful lemma is listed. Lemma 3.4.1. [81] (I) Suppose f ∈ C 2 [tk , tk+1 ], then it holds 1 [f (tk ) + f (tk+1 )] 2 1

= f (tk+ 1 ) + 2

τ2 1 1 ∫[ftt (tk+ 1 − τs) + ftt (tk+ 1 + τs)](1 − s)ds. 2 2 8 2 2

(3.84)

0

(II) Suppose f ∈ C 2 [tk , tk+1 ] and σ ∈ (0, 1), thus we have (1 − σ)f (tk ) + σf (tk+1 ) 1

= f (tk+σ ) + τ2 ∫[σ(1 − σ)2 ftt (tk+σ + (1 − σ)τs) 0

+ (1 − σ)σ 2 ftt (tk+σ − στs)](1 − s)ds.

(3.85)

(III) Suppose f ∈ C 3 [tk , tk+1 ], then one has 1 [f (tk+1 ) − f (tk )] = f 󸀠 (tk+ 1 ) 2 τ 1

τ2 1 1 + ∫[fttt (tk+ 1 + τs) + fttt (tk+ 1 − τs)](1 − s)2 ds. 2 2 16 2 2

(3.86)

0

(IV) Suppose f ∈ C 3 [tk , tk+1 ] and σ ∈ (0, 1), then one has 1 [(2σ + 1)f (tk+1 ) − 4σf (tk ) + (2σ − 1)f (tk−1 )] 2τ = f 󸀠 (tk+σ ) +

1

τ2 [(2σ + 1)(1 − σ)3 ∫ fttt (tk+σ + (1 − σ)τs)(1 − s)2 ds 4 0

1

+ 4σ 4 ∫ fttt (tk+σ − στs)(1 − s)2 ds 0 3

1

− (2σ − 1)(1 + σ) ∫ fttt (tk+σ − (1 + σ)τs)(1 − s)2 ds].

(3.87)

0

Proof. (I) Expanding f (tk ) and f (tk+1 ) at t = tk+ 1 to the second-order derivative term 2 with the help of Taylor expansion with the integral remainder, and averaging the results yield (3.84).

194 | 3 Difference methods for the time-fractional wave equations (II) By the same process as (I), but expanding is performed at the point t = tk+σ to the second-order derivative term. Multiplying the results by 1 − σ and σ, respectively, and then adding the results together will give (3.85). (III) Expand f (tk ) and f (tk+1 ) at t = tk+ 1 with the help of Taylor expansion with 2

the integral remainder to the third-order derivative term. The equality (3.86) can be obtained. (IV) Expand f (tk+1 ), f (tk ) and f (tk−1 ) at t = tk+σ to the third-order derivative term with the help of Taylor expansion with the integral remainder. Then multiplying the obtained results by 2σ + 1, −4σ and 2σ − 1, respectively, and summing up the results will lead to (3.87).

For any {uni | 0 ⩽ i ⩽ M, 0 ⩽ n ⩽ N} defined on Ωh × Ωτ , introduce the following notation: 1 Dt ̄uni = [(2σ + 1)uni − 4σun−1 + (2σ − 1)un−2 n ⩾ 2. i i ], 2τ Let v(x, t) = ut (x, t),

α = γ − 1,

σ =1−

α , 2

s = τα Γ(2 − α).

Thus, the problem (3.81)–(3.83) is equivalent to { { { { { { { { { { {

C α 0 Dt v(x, t)

= uxx (x, t) + f (x, t),

ut (x, t) = v(x, t), u(x, 0) = φ(x), u(0, t) = 0,

x ∈ (0, L), t ∈ (0, T],

x ∈ [0, L], t ∈ (0, T],

v(x, 0) = ψ(x),

u(L, t) = 0,

x ∈ [0, L],

t ∈ (0, T].

(3.88) (3.89) (3.90) (3.91)

Denote {

Uin = u(xi , tn ), φi = φ(xi ),

Vin = v(xi , tn ),

ψi = ψ(xi ),

0 ⩽ i ⩽ M, 0 ⩽ n ⩽ N,

0 ⩽ i ⩽ M.

Considering (3.88) at the point (xi , tn−1+σ ), we have C α 0 Dt v(xi , tn−1+σ )

= uxx (xi , tn−1+σ ) + fin−1+σ ,

1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N,

(3.92)

where fin−1+σ = f (xi , tn−1+σ ). Applying the L2-1σ formula (1.81) to approximate the Caputo derivative gives C α 0 Dt v(xi , tn−1+σ )

=

1 n−1 (n,α) n−k (Vi − Vin−k−1 ) + O(τ3−α ). ∑c s k=0 k

(3.93)

Combining the linear interpolation (3.85) with the second-order central difference quotient (Lemma 2.1.3) for the spatial second-order derivative, we have uxx (xi , tn−1+σ ) = σuxx (xi , tn ) + (1 − σ)uxx (xi , tn−1 ) + O(τ2 )

3.4 The difference method based on L2-1σ approximation for 1D problem

= σδx2 Uin + (1 − σ)δx2 Uin−1 + O(h2 ) + O(τ2 ).

| 195

(3.94)

Substituting (3.93) and (3.94) into (3.92) arrives at 1 n−1 (n,α) n−k (Vi − Vin−k−1 ) = σδx2 Uin + (1 − σ)δx2 Uin−1 + fin−1+σ + (r5 )ni , ∑c s k=0 k 1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N,

(3.95)

and there exists a positive constant c5 such that 󵄨󵄨 n 󵄨󵄨 2 2 󵄨󵄨(r5 )i 󵄨󵄨 ⩽ c5 (τ + h ),

1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N.

(3.96)

Considering (3.89) at the points (xi , t 1 ) and (xi , tn−1+σ ), respectively, we have 2

{

ut (xi , t 1 ) = v(xi , t 1 ), 2

0 ⩽ i ⩽ M,

2

ut (xi , tn−1+σ ) = v(xi , tn−1+σ ),

0 ⩽ i ⩽ M,

2 ⩽ n ⩽ N.

It follows from Lemma 3.4.1 that 1

{

1

δt Ui2 = Vi2 + (r6 )1i ,

Dt ̄Uin

=

σVin

+ (1 −

0 ⩽ i ⩽ M,

σ)Vin−1

+

(r6 )ni ,

(3.97) 0 ⩽ i ⩽ M,

2 ⩽ n ⩽ N,

(3.98)

and there exists a positive constant c6 such that 󵄨󵄨 2 n󵄨 2 󵄨󵄨δx (r6 )i 󵄨󵄨󵄨 ⩽ c6 τ ,

1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N.

(3.99)

In addition, from (3.89) and (3.91), we have (r6 )n0 = 0,

(r6 )nM = 0,

1 ⩽ n ⩽ N.

(3.100)

Noticing the initial-boundary value conditions (3.90)–(3.91), we have {

Ui0 = φi ,

U0n

= 0,

Vi0 = ψi ,

n UM

= 0,

0 ⩽ i ⩽ M, 1 ⩽ n ⩽ N.

(3.101) (3.102)

Omitting the small term (r5 )ni and (r6 )ni in (3.95), (3.97) and (3.98) and replacing the exact solution {Uin , Vin } with its numerical one {uni , vin } produce the difference scheme for (3.88)–(3.91) as follows: { { { { { { { { { { { { { { { { { { { { { { { { { { { { {

1 n−1 (n,α) n−k (vi − vin−k−1 ) = σδx2 uni + (1 − σ)δx2 un−1 + fin−1+σ , ∑c i s k=0 k 1 2

1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N,

1 2

δt ui = vi ,

Dt ̄uni u0i = un0 =

=

σvin

φi ,

0,

0 ⩽ i ⩽ M, + (1 −

vi0 unM

σ)vin−1 ,

= ψi ,

= 0,

(3.103) (3.104)

0 ⩽ i ⩽ M,

0 ⩽ i ⩽ M,

1 ⩽ n ⩽ N.

2 ⩽ n ⩽ N,

(3.105) (3.106) (3.107)

On each time level, only one tridiagonal system of linear algebraic equations need be solved. See the process of the proof for Theorem 3.4.1.

196 | 3 Difference methods for the time-fractional wave equations Remark 3.4.1. By Lemma 3.4.1, it is easy to write (r6 )ni in the integral expression, and then obtain (3.99). Remark 3.4.2. For the nonhomogeneous boundary value problem: γ

C D u(x, t) = uxx (x, t) + f (x, t), x ∈ (0, L), t ∈ (0, T], { { { 0 t u(x, 0) = φ(x), ut (x, 0) = ψ(x), x ∈ [0, L], { { { { u(0, t) = μ(t), u(L, t) = ν(t), t ∈ (0, T],

the equality (3.100) is in general not valid. In this case, let v(x, t) = ut (x, t), then the problem above can be written as the following equivalent one: { { { { { { { { { { { { { { { { {

C α 0 Dt v(x, t)

= uxx (x, t) + f (x, t),

utxx (x, t) = vxx (x, t), u(x, 0) = φ(x), u(0, t) = μ(t), 󸀠

v(0, t) = μ (t),

x ∈ (0, L), t ∈ (0, T],

x ∈ (0, L),

t ∈ (0, T],

v(x, 0) = ψ(x),

x ∈ [0, L],

u(L, t) = ν(t), 󸀠

v(L, t) = ν (t),

t ∈ (0, T], t ∈ (0, T],

for which we construct the following difference scheme: { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { {

1 n−1 (n,α) n−k (vi − vin−k−1 ) = σδx2 uni + (1 − σ)δx2 un−1 + fin−1+σ , ∑c i s k=0 k 1

1

1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N,

δx2 δt ui2 = δx2 vi2 , 1 ⩽ i ⩽ M − 1, δx2 Dt ̄uni = δx2 (σvin + (1 − σ)vin−1 ), 1 ⩽ i ⩽ u0i = φi , vi0 = ψi , 0 ⩽ i ⩽ M, un0 = μ(tn ), unM = ν(tn ), 1 ⩽ n ⩽ N, n v0n = μ󸀠 (tn ), vM = ν󸀠 (tn ), 1 ⩽ n ⩽ N.

M − 1,

2 ⩽ n ⩽ N,

The interested readers can refer to [81]. Another way is firstly to make the boundary value conditions homogeneous and then construct the corresponding difference scheme. 3.4.2 Solvability of the difference scheme Theorem 3.4.1. The difference scheme (3.103)–(3.107) is uniquely solvable. Proof. Let un = (un0 , un1 , . . . , unM ),

n vn = (v0n , v1n , . . . , vM ).

(I) The value of {u0 , v0 } is determined by (3.106).

3.4 The difference method based on L2-1σ approximation for 1D problem

| 197

(II) It follows from (3.104) that 1

vi1 = 2δt ui2 − vi0 ,

0 ⩽ i ⩽ M.

(3.108)

Substituting (3.108) into (3.103), and noticing (3.107), we obtain the linear system in u1 as follows: 1 1 { c0(1,α) (2δt ui2 − 2vi0 ) = σδx2 u1i + (1 − σ)δx2 u0i + fiσ , { s1 1 { u0 = 0, uM = 0.

1 ⩽ i ⩽ M − 1,

(3.109) (3.110)

Considering its homogeneous one, we have 2 (1,α) 1 { c0 ui = σδx2 u1i , sτ { 1 1 { u0 = 0, uM = 0.

1 ⩽ i ⩽ M − 1,

(3.111) (3.112)

Making the inner product on both hand sides of (3.111) with u1 and noticing (3.112) produce 2 (1,α) 󵄩󵄩 1 󵄩󵄩2 󵄩 󵄩2 c 󵄩u 󵄩 + σ 󵄩󵄩󵄩δx u1 󵄩󵄩󵄩 = 0, sτ 0 󵄩 󵄩 which implies u1 = 0, hence the system (3.109)–(3.110) has a unique solution. Once u1 is obtained, v1 is followed from (3.108). (III) Suppose the value of {u0 , v0 , u1 , v1 , . . . , un−1 , vn−1 } has been uniquely determined, then it follows from (3.105) that vin =

1 (D ̄un − (1 − σ)vin−1 ), σ t i

0 ⩽ i ⩽ M.

(3.113)

Substituting (3.113) into (3.103), noticing (3.107), one can obtain the linear system in un as follows: { { { { { { { { { { { { {

n−1 1 (n,α) 1 {c0 [ (Dt ̄uni − (1 − σ)vin−1 ) − vin−1 ] + ∑ ck(n,α) (vin−k − vin−k−1 )} s σ k=1

= σδx2 uni + (1 − σ)δx2 un−1 + fin−1+σ , i un0

= 0,

unM

1 ⩽ i ⩽ M − 1,

= 0.

(3.114) (3.115)

To show its unique solvability, it suffices to prove the corresponding homogeneous one 1 1 2σ + 1 n { c0(n,α) ⋅ ⋅ ui = σδx2 uni , s σ 2τ { n n { u0 = 0, uM = 0 has only the trivial solution.

1 ⩽ i ⩽ M − 1,

(3.116) (3.117)

198 | 3 Difference methods for the time-fractional wave equations Taking the inner product on both hand sides of (3.116) with un yields 2σ + 1 (n,α) 󵄩󵄩 n 󵄩󵄩2 󵄩 n 󵄩2 c 󵄩u 󵄩 + σ 󵄩󵄩󵄩δx u 󵄩󵄩󵄩 = 0, 2sτσ 0 󵄩 󵄩 which implies un = 0, and the system (3.114)–(3.115) has a unique solution un . Once un is obtained, vn is followed from (3.113). By the principle of induction, the theorem is true. The proof ends.

3.4.3 Stability of the difference scheme Before analyzing the stability of the difference scheme, we give two lemmas. Lemma 3.4.2. Suppose u0 , u1 , . . . , uN ∈ 𝒰h̊ , (⋅, ⋅) is an inner product on 𝒰h̊ and ‖ ⋅ ‖ is the induced norm. Denote 󵄩 󵄩2 󵄩 󵄩2 󵄩 󵄩2 E n ≡ (2σ + 1)󵄩󵄩󵄩un 󵄩󵄩󵄩 − (2σ − 1)󵄩󵄩󵄩un−1 󵄩󵄩󵄩 + (2σ 2 + σ − 1)󵄩󵄩󵄩un − un−1 󵄩󵄩󵄩 ,

n ⩾ 1.

Then we have (Dt ̄un , σun + (1 − σ)un−1 ) ⩾

1 n (E − E n−1 ), 4τ

n⩾2

and En ⩾

1 󵄩󵄩 n 󵄩󵄩2 󵄩u 󵄩 , σ󵄩 󵄩

n ⩾ 1.

Proof. The operator Dt ̄un can be rewritten as Dt ̄un = 2σ

un − un−1 un − un−2 − (2σ − 1) , τ 2τ

or 1 un−1 − un−2 1 un − un−1 − (σ − ) . Dt ̄un = (σ + ) 2 τ 2 τ Noticing the identities (a − b)a = 21 [a2 − b2 + (a − b)2 ], (a − b)b = 21 [a2 − b2 − (a − b)2 ], we have (Dt ̄un , σun + (1 − σ)un−1 ) = σ[2σ(

un − un−1 n un − un−2 n , u ) − (2σ − 1)( , u )] τ 2τ

1 un − un−1 n−1 1 un−1 − un−2 n−1 + (1 − σ)[(σ + )( , u ) − (σ − )( , u )] 2 τ 2 τ

3.4 The difference method based on L2-1σ approximation for 1D problem

σ 󵄩 󵄩2 󵄩 󵄩2 󵄩2 󵄩 = σ[ (󵄩󵄩󵄩un 󵄩󵄩󵄩 − 󵄩󵄩󵄩un−1 󵄩󵄩󵄩 + 󵄩󵄩󵄩un − un−1 󵄩󵄩󵄩 ) τ 2σ − 1 󵄩󵄩 n 󵄩󵄩2 󵄩󵄩 n−2 󵄩󵄩2 󵄩󵄩 n 󵄩2 (󵄩󵄩u 󵄩󵄩 − 󵄩󵄩u 󵄩󵄩 + 󵄩󵄩u − un−2 󵄩󵄩󵄩 )] − 4τ + (1 − σ)[ σ−

1 2

σ+

2τ

1 2

󵄩2 󵄩2 󵄩 󵄩 󵄩2 󵄩 (󵄩󵄩󵄩un 󵄩󵄩󵄩 − 󵄩󵄩󵄩un−1 󵄩󵄩󵄩 − 󵄩󵄩󵄩un − un−1 󵄩󵄩󵄩 )

󵄩2 󵄩2 󵄩 󵄩 󵄩2 󵄩 (󵄩󵄩un−1 󵄩󵄩󵄩 − 󵄩󵄩󵄩un−2 󵄩󵄩󵄩 + 󵄩󵄩󵄩un−1 − un−2 󵄩󵄩󵄩 )] 2τ 󵄩 σ 󵄩 󵄩2 󵄩 󵄩2 󵄩 󵄩2 ⩾ σ[ (󵄩󵄩󵄩un 󵄩󵄩󵄩 − 󵄩󵄩󵄩un−1 󵄩󵄩󵄩 + 󵄩󵄩󵄩un − un−1 󵄩󵄩󵄩 ) τ 2σ − 1 󵄩󵄩 n 󵄩󵄩2 󵄩󵄩 n−2 󵄩󵄩2 󵄩 󵄩2 󵄩 󵄩2 (󵄩󵄩u 󵄩󵄩 − 󵄩󵄩u 󵄩󵄩 + 2󵄩󵄩󵄩un − un−1 󵄩󵄩󵄩 + 2󵄩󵄩󵄩un−1 − un−2 󵄩󵄩󵄩 )] − 4τ −

+ (1 − σ)[ σ−

1 2

σ+

2τ

1 2

󵄩 󵄩2 󵄩 󵄩2 󵄩 󵄩2 (󵄩󵄩󵄩un 󵄩󵄩󵄩 − 󵄩󵄩󵄩un−1 󵄩󵄩󵄩 − 󵄩󵄩󵄩un − un−1 󵄩󵄩󵄩 )

󵄩 󵄩2 󵄩 󵄩2 󵄩 󵄩2 (󵄩󵄩un−1 󵄩󵄩󵄩 − 󵄩󵄩󵄩un−2 󵄩󵄩󵄩 + 󵄩󵄩󵄩un−1 − un−2 󵄩󵄩󵄩 )] 2τ 󵄩 2σ + 1 󵄩󵄩 n 󵄩󵄩2 󵄩󵄩 n−1 󵄩󵄩2 2σ − 1 󵄩󵄩 n−1 󵄩󵄩2 󵄩󵄩 n−2 󵄩󵄩2 = (󵄩u 󵄩 − 󵄩u 󵄩󵄩 ) − (󵄩u 󵄩󵄩 − 󵄩󵄩u 󵄩󵄩 ) 4τ 󵄩 󵄩 󵄩 4τ 󵄩 2σ 2 + σ − 1 󵄩󵄩 n 󵄩2 󵄩 󵄩2 + (󵄩󵄩u − un−1 󵄩󵄩󵄩 − 󵄩󵄩󵄩un−1 − un−2 󵄩󵄩󵄩 ) 4τ 1 n (E − E n−1 ), n ⩾ 2. = 4τ −

Further, by the Cauchy–Schwarz inequality, we get 󵄩 󵄩2 󵄩 󵄩2 E n = (2σ 2 + 3σ)󵄩󵄩󵄩un 󵄩󵄩󵄩 + (2σ 2 − σ)󵄩󵄩󵄩un−1 󵄩󵄩󵄩 − 2(2σ − 1)(σ + 1)(un , un−1 ) 󵄩 󵄩2 󵄩 󵄩2 󵄩 󵄩2 ⩾ (2σ 2 + 3σ)󵄩󵄩󵄩un 󵄩󵄩󵄩 + (2σ 2 − σ)󵄩󵄩󵄩un−1 󵄩󵄩󵄩 − [(2σ − 1)σ 󵄩󵄩󵄩un−1 󵄩󵄩󵄩 (2σ − 1)(σ + 1)2 󵄩󵄩 n 󵄩󵄩2 󵄩󵄩u 󵄩󵄩 ] σ 1 󵄩 󵄩2 = 󵄩󵄩󵄩un 󵄩󵄩󵄩 . σ +

The proof ends. Lemma 3.4.3. The following two inequalities are valid: n

(m,α) (m−1,α) − cm−2 )⩽ ∑ (cm−2

m=2

n

(m,α) ⩽ ∑ cm−1

m=2

1−α α ( + 1)σ −α , 12 σ

3 (n − 1 + σ)1−α . 2

| 199

200 | 3 Difference methods for the time-fractional wave equations Proof. (I) Denote n

(m,α) (m−1,α) An ≡ ∑ (cm−2 − cm−2 ). m=2

When n = 2: An = c0(2,α) − c0(1,α) =

1 1 [(1 + σ)2−α − σ 2−α ] − [(1 + σ)1−σ + σ 1−σ ]. 2−α 2

When n ⩾ 3: n

An = ∑ [ m=2

1 ((m − 1 + σ)2−α − 2(m − 2 + σ)2−α + (m − 3 + σ)2−α ) 2−α

1 − ((m − 1 + σ)1−α − 2(m − 2 + σ)1−α + (m − 3 + σ)1−α ) 2 1 + ((m − 2 + σ)2−α − (m − 3 + σ)2−α ) 2−α 1 − (3(m − 2 + σ)1−α − (m − 3 + σ)1−α )] 2 n

= ∑[ m=2

1 ((m − 1 + σ)2−α − (m − 2 + σ)2−α ) 2−α

1 − ((m − 1 + σ)1−α + (m − 2 + σ)1−α )]. 2

It is easy to see that the equality above is also true for n = 2. Let f (x) = (x + σ)1−α , then we have n

m−1

1 An = ∑ { ∫ f (x)dx − [f (m − 1) + f (m − 2)]} 2 m=2 n

m−2

= ∑ (− m=2

1 󸀠󸀠 f (ξm )), 12

ξm ∈ (m − 2, m − 1).

A direct calculation yields −f 󸀠󸀠 (x) = (1 − α)α(x + σ)−α−1 . Then it follows that An =

n 1 [−f 󸀠󸀠 (ξ2 ) + ∑ (−f 󸀠󸀠 (ξm ))] 12 m=3

3.4 The difference method based on L2-1σ approximation for 1D problem | 201

=

n 1 (1 − α)α[(ξ2 + σ)−α−1 + ∑ (ξm + σ)−α−1 ] 12 m=3

⩽

n 1 (1 − α)α[σ −α−1 + ∑ (m − 2 + σ)−α−1 ] 12 m=3

⩽

n 1 (1 − α)α[σ −α−1 + ∑ ∫ (x + σ)−α−1 dx] 12 m=3

m−2

n−2

=

m−3

1 (1 − α)α[σ −α−1 + ∫ (x + σ)−α−1 dx] 12 0 −α

σ − (n − 2 + σ)−α 1 (1 − α)α[σ −α−1 + ] 12 α 1 σ −α ⩽ (1 − α)α[σ −α−1 + ] 12 α 1 − α −α α σ ( + 1). = 12 σ =

(II) Denote n

(m,α) Bn ≡ ∑ cm−1 . m=2

Then n 1 Bn = ∑ [ (3(m − 1 + σ)1−α − (m − 2 + σ)1−α ) 2 m=2

−

1 ((m − 1 + σ)2−α − (m − 2 + σ)2−α )] 2−α

n 1 = ∑ (m − 1 + σ)1−α + [(n − 1 + σ)1−α − σ 1−α ] 2 m=2

−

1 [(n − 1 + σ)2−α − σ 2−α ] 2−α

n−1

m

⩽ ∑ ∫ (x + σ)1−α dx + (n − 1 + σ)1−α m=2 m−1

1 1 [(n − 1 + σ)2−α − σ 2−α ] + [(n − 1 + σ)1−α − σ 1−α ] − 2 2−α (n − 1 + σ)2−α − (1 + σ)2−α 1 = + [3(n − 1 + σ)1−α − σ 1−α ] 2−α 2 1 2−α 2−α − [(n − 1 + σ) − σ ] 2−α 1 ⩽ [3(n − 1 + σ)1−α − σ 1−α ] 2

202 | 3 Difference methods for the time-fractional wave equations 3 (n − 1 + σ)1−α . 2

⩽ The proof ends.

Theorem 3.4.2. Suppose {un , vn | 0 ⩽ n ⩽ N} satisfies { { { { { { { { { { { { { { { { { { { { { { { { { { { { {

1 n−1 (n,α) n−k (vi − vin−k−1 ) = σδx2 uni + (1 − σ)δx2 un−1 + pni , ∑c i s k=0 k 1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N,

1 2

1 2

δt ui = vi + qi1 ,

Dt ̄uni u0i = un0 =

σvin

=

+ (1 −

vi0 unM

φi , 0,

0 ⩽ i ⩽ M, σ)vin−1

= ψi ,

+

(3.119)

qin ,

0 ⩽ i ⩽ M, 2 ⩽ n ⩽ N,

0 ⩽ i ⩽ M,

= 0,

(3.118) (3.120) (3.121)

1 ⩽ n ⩽ N,

(3.122)

n where q0n = qM = 0 (1 ⩽ n ⩽ N) and φ0 = φM = ψ0 = ψM = 0. Then there exists a constant C such that n

󵄩 󵄩2 󵄩 󵄩2 τ ∑ 󵄩󵄩󵄩vk 󵄩󵄩󵄩 + 󵄩󵄩󵄩δx un 󵄩󵄩󵄩 k=1

n

n

m=1

m=1

󵄩 󵄩2 󵄩 󵄩2 󵄩 󵄩2 󵄩 󵄩2 ⩽ C[τ1−α 󵄩󵄩󵄩v0 󵄩󵄩󵄩 + 󵄩󵄩󵄩δx u0 󵄩󵄩󵄩 + τ ∑ 󵄩󵄩󵄩pm 󵄩󵄩󵄩 + τ ∑ 󵄩󵄩󵄩δx2 qm 󵄩󵄩󵄩 ],

1 ⩽ n ⩽ N,

(3.123)

where M−1

󵄩󵄩 k 󵄩󵄩2 k 2 󵄩󵄩v 󵄩󵄩 = h ∑ (vi ) , i=1

M−1

󵄩󵄩 k 󵄩󵄩2 k 2 󵄩󵄩p 󵄩󵄩 = h ∑ (pi ) , i=1

M−1

󵄩󵄩 2 k 󵄩󵄩2 2 k 2 󵄩󵄩δx q 󵄩󵄩 = h ∑ (δx qi ) . i=1

n Proof. The combination of (3.119)–(3.122) with q0n = qM = 0 (1 ⩽ n ⩽ N) and φ0 = φM = ψ0 = ψM = 0 leads to

v0n = 0,

n vM = 0,

0 ⩽ n ⩽ N.

Consequently, we have 1

{

v02 = 0,

σv0n

+ (1 −

1

vM2 = 0, σ)v0n−1

= 0,

n σvM

+ (1 −

n−1 σ)vM

(3.124) = 0,

2 ⩽ n ⩽ N.

(3.125)

(I) When n = 1, (3.118) reads 1 (1,α) 1 c (vi − vi0 ) = σδx2 u1i + (1 − σ)δx2 u0i + p1i , s 0

1 ⩽ i ⩽ M − 1.

(3.126)

1

Making the inner product on both hand sides of (3.126) with v 2 and noticing (3.124), we obtain 1 1 1 (1,α) 󵄩󵄩 1 󵄩󵄩2 󵄩󵄩 0 󵄩󵄩2 c (󵄩v 󵄩 − 󵄩v 󵄩 ) = −(δx (σu1 + (1 − σ)u0 ), δx v 2 ) + (p1 , v 2 ). 2s 0 󵄩 󵄩 󵄩 󵄩

(3.127)

3.4 The difference method based on L2-1σ approximation for 1D problem | 203

It follows from (3.119) that 1

1

δt δx2 ui2 = δx2 vi2 + δx2 qi1 ,

1 ⩽ i ⩽ M − 1.

(3.128)

Making the inner product on both hand sides of (3.128) with −(σu1 + (1 − σ)u0 ) and noticing σu10 + (1 − σ)u00 = 0,

σu1M + (1 − σ)u0M = 0,

we get 1 (δ u1 − δx u0 , δx (σu1 + (1 − σ)u0 )) τ x 1

= (δx v 2 , δx (σu1 + (1 − σ)u0 )) − (δx2 q1 , σu1 + (1 − σ)u0 ). Adding (3.127) and (3.129) yields 1 (1,α) 󵄩󵄩 1 󵄩󵄩2 󵄩󵄩 0 󵄩󵄩2 1 c (󵄩v 󵄩 − 󵄩v 󵄩 ) + (δx u1 − δx u0 , δx (σu1 + (1 − σ)u0 )) 2s 0 󵄩 󵄩 󵄩 󵄩 τ 1

= (p1 , v 2 ) − (δx2 q1 , σu1 + (1 − σ)u0 ). Noticing the fact that

(δx u1 − δx u0 , δx (σu1 + (1 − σ)u0 ))

= σ(δx u1 , δx u1 ) − (2σ − 1)(δx u1 , δx u0 ) − (1 − σ)(δx u0 , δx u0 ) =

󵄩󵄩 1 󵄩󵄩2 σ 󵄩󵄩 1 󵄩󵄩2 3σ 2σ − 1 󵄩 󵄩 δx u1 − √ δx u0 󵄩󵄩󵄩 󵄩󵄩δx u 󵄩󵄩 + (2σ − 1)󵄩󵄩󵄩 √ 󵄩󵄩 2 2σ − 1 󵄩󵄩 4 3σ −

σ 2 − σ + 1 󵄩󵄩 0 󵄩󵄩2 󵄩󵄩δx u 󵄩󵄩 3σ

and c0(1,α) = σ 1−α , we arrive at 2 1 σ󵄩 σ 1−α τ−α 󵄩󵄩 1 󵄩󵄩2 󵄩󵄩 0 󵄩󵄩2 󵄩2 σ − σ + 1 󵄩󵄩 0 󵄩󵄩2 (󵄩󵄩v 󵄩󵄩 − 󵄩󵄩v 󵄩󵄩 ) + ( 󵄩󵄩󵄩δx u1 󵄩󵄩󵄩 − 󵄩󵄩δx u 󵄩󵄩 ) 2Γ(2 − α) τ 4 3σ 1

⩽ (p1 , v 2 ) − (δx2 q1 , σu1 + (1 − σ)u0 ), which can be rewritten as

2 σ 1−α τ1−α 󵄩󵄩 1 󵄩󵄩2 󵄩󵄩 0 󵄩󵄩2 σ󵄩 󵄩2 σ − σ + 1 󵄩󵄩 0 󵄩󵄩2 (󵄩󵄩v 󵄩󵄩 − 󵄩󵄩v 󵄩󵄩 ) + 󵄩󵄩󵄩δx u1 󵄩󵄩󵄩 − 󵄩󵄩δx u 󵄩󵄩 2Γ(2 − α) 4 3σ

(3.129)

204 | 3 Difference methods for the time-fractional wave equations 1

⩽ τ(p1 , v 2 ) − τ(δx2 q1 , σu1 + (1 − σ)u0 ) 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 1󵄩 ⩽ τ󵄩󵄩󵄩p1 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩v 2 󵄩󵄩󵄩 + τ󵄩󵄩󵄩δx2 q1 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩σu1 + (1 − σ)u0 󵄩󵄩󵄩 ⩽ τ[ ⩽ τ[

1󵄩 1 1 󵄩󵄩 21 󵄩󵄩2 ε0 󵄩󵄩 1 󵄩󵄩2 0 󵄩2 󵄩v 󵄩 + 󵄩p 󵄩 ] + τ[ 󵄩󵄩󵄩σu + (1 − σ)u 󵄩󵄩󵄩 + 2ε0 󵄩 󵄩 2 󵄩 󵄩 2

1 󵄩󵄩 2 1 󵄩󵄩2 󵄩δ q 󵄩 ] 2󵄩 x 󵄩

ε 󵄩 󵄩2 1 󵄩󵄩 1 󵄩󵄩2 󵄩󵄩 0 󵄩󵄩2 1 󵄩 󵄩2 󵄩 󵄩2 (󵄩v 󵄩 + 󵄩v 󵄩 ) + 0 󵄩󵄩󵄩p1 󵄩󵄩󵄩 ] + τ[ (󵄩󵄩󵄩u1 󵄩󵄩󵄩 + 󵄩󵄩󵄩u0 󵄩󵄩󵄩 ) + 4ε0 󵄩 󵄩 󵄩 󵄩 2 2

1 󵄩󵄩 2 1 󵄩󵄩2 󵄩δ q 󵄩 ]. 2󵄩 x 󵄩

Letting ε0 = Γ(2 − α)σ α−1 τα , there exists a positive constant C1 such that 󵄩 󵄩2 󵄩 󵄩2 τ1−α 󵄩󵄩󵄩v1 󵄩󵄩󵄩 + 󵄩󵄩󵄩δx u1 󵄩󵄩󵄩 󵄩 󵄩2 󵄩 󵄩2 󵄩 󵄩2 󵄩 󵄩2 󵄩 󵄩2 󵄩 󵄩2 ⩽ C1 (τ1−α 󵄩󵄩󵄩v0 󵄩󵄩󵄩 + 󵄩󵄩󵄩δx u0 󵄩󵄩󵄩 + τ󵄩󵄩󵄩u0 󵄩󵄩󵄩 + τ󵄩󵄩󵄩u1 󵄩󵄩󵄩 + τ1+α 󵄩󵄩󵄩p1 󵄩󵄩󵄩 + τ󵄩󵄩󵄩δx2 q1 󵄩󵄩󵄩 ).

(3.130)

(II) Making the inner product on both hand sides of (3.118) with σvn + (1 − σ)vn−1 and noticing (3.125), we have 1 n−1 (n,α) n−k (v − vn−k−1 , σvn + (1 − σ)vn−1 ) ∑c s k=0 k

= (δx2 (σun + (1 − σ)un−1 ), σvn + (1 − σ)vn−1 ) + (pn , σvn + (1 − σ)vn−1 ) = −(δx (σun + (1 − σ)un−1 ), δx (σvn + (1 − σ)vn−1 )) + (pn , σvn + (1 − σ)vn−1 ),

2 ⩽ n ⩽ N.

(3.131)

For the left-hand side of (3.131), by Lemma 2.6.1, we can obtain n−1

∑ ck(n,α) (vn−k − vn−k−1 , σvn + (1 − σ)vn−1 )

k=0

⩾

1 n−1 (n,α) 󵄩󵄩 n−k 󵄩󵄩2 󵄩󵄩 n−k−1 󵄩󵄩2 (󵄩󵄩v 󵄩󵄩 − 󵄩󵄩v ∑c 󵄩󵄩 ). 2 k=0 k

Therefore, 1 1 n−1 (n,α) 󵄩󵄩 n−k 󵄩󵄩2 󵄩󵄩 n−k−1 󵄩󵄩2 ⋅ ∑c (󵄩󵄩v 󵄩󵄩 − 󵄩󵄩v 󵄩󵄩 ) 2 s k=0 k

⩽ −(δx (σun + (1 − σ)un−1 ), δx (σvn + (1 − σ)vn−1 )) + (pn , σvn + (1 − σ)vn−1 ),

2 ⩽ n ⩽ N.

(3.132)

It follows from (3.120) that Dt ̄δx2 uni = δx2 (σvin + (1 − σ)vin−1 ) + δx2 qin ,

1 ⩽ i ⩽ M − 1,

2 ⩽ n ⩽ N.

Making the inner product on both hand sides of the equation above with −(σun + (1 − σ)un−1 ) and noticing σun0 + (1 − σ)un−1 0 = 0,

σunM + (1 − σ)un−1 M = 0,

2 ⩽ n ⩽ N,

3.4 The difference method based on L2-1σ approximation for 1D problem | 205

we obtain (Dt ̄δx un , δx (σun + (1 − σ)un−1 ))

= (δx (σvn + (1 − σ)vn−1 ), δx (σun + (1 − σ)un−1 )) − (δx2 qn , σun + (1 − σ)un−1 ),

2 ⩽ n ⩽ N.

(3.133)

Denote 󵄩 󵄩2 󵄩 󵄩2 F n = (2σ + 1)󵄩󵄩󵄩δx un 󵄩󵄩󵄩 − (2σ − 1)󵄩󵄩󵄩δx un−1 󵄩󵄩󵄩 󵄩 󵄩2 + (2σ 2 + σ − 1)󵄩󵄩󵄩δx (un − un−1 )󵄩󵄩󵄩 , n ⩾ 1. With the help of Lemma 3.4.2, we have Fn ⩾

1 󵄩󵄩 n 󵄩󵄩2 󵄩δ u 󵄩 , σ󵄩 x 󵄩

n⩾1

(3.134)

and (Dt ̄δx un , δx (σun + (1 − σ)un−1 )) ⩾

1 n (F − F n−1 ), 4τ

n ⩾ 2.

(3.135)

Combining (3.133) with (3.135), we get 1 n (F − F n−1 ) ⩽ (δx (σvn + (1 − σ)vn−1 ), δx (σun + (1 − σ)un−1 )) 4τ − (δx2 qn , σun + (1 − σ)un−1 ), 2 ⩽ n ⩽ N. Adding (3.132) and (3.136) arrives at 1 n 1 1 n−1 (n,α) 󵄩󵄩 n−k 󵄩󵄩2 󵄩󵄩 n−k−1 󵄩󵄩2 ⋅ ∑ ck (󵄩󵄩v 󵄩󵄩 − 󵄩󵄩v (F − F n−1 ) 󵄩󵄩 ) + 2 s k=0 4τ

⩽ (pn , σvn + (1 − σ)vn−1 ) − (δx2 qn , σun + (1 − σ)un−1 ) 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 ⩽ 󵄩󵄩󵄩pn 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩σvn + (1 − σ)vn−1 󵄩󵄩󵄩 + 󵄩󵄩󵄩δx2 qn 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩σun + (1 − σ)un−1 󵄩󵄩󵄩, Noticing the fact that n−1

󵄩 󵄩2 󵄩 󵄩2 ∑ ck(n,α) (󵄩󵄩󵄩vn−k 󵄩󵄩󵄩 − 󵄩󵄩󵄩vn−k−1 󵄩󵄩󵄩 )

k=0 n−1

n−2

k=0

k=0

󵄩 󵄩2 󵄩 󵄩2 = ∑ ck(n,α) 󵄩󵄩󵄩vn−k 󵄩󵄩󵄩 − ∑ ck(n−1,α) 󵄩󵄩󵄩vn−1−k 󵄩󵄩󵄩 n−2

󵄩 󵄩2 (n,α) 󵄩 󵄩󵄩v0 󵄩󵄩󵄩2 − ∑ (ck(n,α) − ck(n−1,α) )󵄩󵄩󵄩vn−1−k 󵄩󵄩󵄩 − cn−1 󵄩 󵄩 k=0

n−1

n−2

k=0

k=0

󵄩 󵄩2 󵄩 󵄩2 = ∑ ck(n,α) 󵄩󵄩󵄩vn−k 󵄩󵄩󵄩 − ∑ ck(n−1,α) 󵄩󵄩󵄩vn−1−k 󵄩󵄩󵄩

2 ⩽ n ⩽ N.

(3.136)

206 | 3 Difference methods for the time-fractional wave equations 󵄩2 (n,α) (n−1,α) 󵄩 (n,α) 󵄩 󵄩󵄩v0 󵄩󵄩󵄩2 , − (cn−2 − cn−2 )󵄩󵄩󵄩v1 󵄩󵄩󵄩 − cn−1 󵄩 󵄩 we have 1 n−1 (n,α) 󵄩󵄩 n−k 󵄩󵄩2 n−2 (n−1,α) 󵄩󵄩 n−1−k 󵄩󵄩2 1 n (∑ c (F − F n−1 ) 󵄩󵄩v 󵄩󵄩 ) + 󵄩v 󵄩󵄩 − ∑ ck 2s k=0 k 󵄩 4τ k=0 ⩽

1 󵄩2 (n−1,α) 󵄩 (n,α) 󵄩 󵄩󵄩v0 󵄩󵄩󵄩2 ] [(c(n,α) − cn−2 )󵄩󵄩󵄩v1 󵄩󵄩󵄩 + cn−1 󵄩 󵄩 2s n−2 󵄩󵄩 n 󵄩󵄩 󵄩󵄩 n 󵄩 󵄩 󵄩 󵄩 n−1 󵄩 + 󵄩󵄩p 󵄩󵄩 ⋅ 󵄩󵄩σv + (1 − σ)v 󵄩󵄩󵄩 + 󵄩󵄩󵄩δx2 qn 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩σun + (1 − σ)un−1 󵄩󵄩󵄩,

2 ⩽ n ⩽ N.

Replacing the superscript n with m and summing up for m from 2 to n on both hand sides of the inequality above yields 1 n−1 (n,α) 󵄩󵄩 n−k 󵄩󵄩2 1 n (1,α) 󵄩 1 󵄩2 (∑ c (F − F 1 ) 󵄩v 󵄩󵄩 − c0 󵄩󵄩󵄩v 󵄩󵄩󵄩 ) + 2s k=0 k 󵄩 4τ ⩽

n 1 n (m,α) 󵄩2 (m−1,α) 󵄩 (m,α) 󵄩 󵄩󵄩v0 󵄩󵄩󵄩2 ] [ ∑ (cm−2 − cm−2 )󵄩󵄩󵄩v1 󵄩󵄩󵄩 + ∑ cm−1 󵄩 󵄩 2s m=2 m=2 n

󵄩 󵄩 󵄩 󵄩 + ∑ 󵄩󵄩󵄩pm 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩σvm + (1 − σ)vm−1 󵄩󵄩󵄩 m=2 n

󵄩 󵄩 󵄩 󵄩 + ∑ 󵄩󵄩󵄩δx2 qm 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩σum + (1 − σ)um−1 󵄩󵄩󵄩, m=2

2 ⩽ n ⩽ N.

Using Lemma 3.4.3, we have 1 n 1 n−1 (n,α) 󵄩󵄩 n−k 󵄩󵄩2 󵄩 󵄩2 ( ∑ ck 󵄩󵄩v 󵄩󵄩 − c0(1,α) 󵄩󵄩󵄩v1 󵄩󵄩󵄩 ) + (F − F 1 ) 2s k=0 4τ ⩽

1 1−α α 󵄩 󵄩2 3 󵄩 󵄩2 [ ( + 1)σ −α 󵄩󵄩󵄩v1 󵄩󵄩󵄩 + (n − 1 + σ)1−α 󵄩󵄩󵄩v0 󵄩󵄩󵄩 ] 2s 12 σ 2 n

󵄩 󵄩 󵄩 󵄩 + ∑ 󵄩󵄩󵄩pm 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩σvm + (1 − σ)vm−1 󵄩󵄩󵄩 m=2 n

󵄩 󵄩 󵄩 󵄩 + ∑ 󵄩󵄩󵄩δx2 qm 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩σum + (1 − σ)um−1 󵄩󵄩󵄩, m=2

2 ⩽ n ⩽ N.

Using Lemma 1.6.3, we have c0(1,α) = σ 1−α and (n,α) c0(n,α) > c1(n,α) > c2(n,α) > ⋅ ⋅ ⋅ > cn−1 > (1 − α)n−α ,

n ⩾ 2.

(3.137)

3.4 The difference method based on L2-1σ approximation for 1D problem | 207

It follows from (3.137) that 1 − α −α n−1 󵄩󵄩 n−k 󵄩󵄩2 1 n n ∑ 󵄩󵄩v 󵄩󵄩 + (F − F 1 ) 2s 4τ k=0

⩽

1 1−α 󵄩󵄩 1 󵄩󵄩2 1 1 − α α 󵄩 󵄩2 3 󵄩 󵄩2 σ 󵄩󵄩v 󵄩󵄩 + [ ( + 1)σ −α 󵄩󵄩󵄩v1 󵄩󵄩󵄩 + (n − 1 + σ)1−α 󵄩󵄩󵄩v0 󵄩󵄩󵄩 ] 2s 2s 12 σ 2 n

󵄩 󵄩 󵄩 󵄩 + ∑ 󵄩󵄩󵄩pm 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩σvm + (1 − σ)vm−1 󵄩󵄩󵄩 m=2 n

󵄩 󵄩 󵄩 󵄩 + ∑ 󵄩󵄩󵄩δx2 qm 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩σum + (1 − σ)um−1 󵄩󵄩󵄩, m=2

2 ⩽ n ⩽ N.

Noticing 1 − α −α 1 − α 1 n−α ⩾ n = ⋅ , 2s 2 Γ(2 − α)τα 2T α Γ(1 − α) it yields that n−1 1 1 n 󵄩 󵄩2 (F − F 1 ) ∑ 󵄩󵄩vn−k 󵄩󵄩󵄩 + 2T α Γ(1 − α) k=0󵄩 4τ

⩽

τ−α 󵄩 󵄩2 σ 1−α 󵄩󵄩󵄩v1 󵄩󵄩󵄩 2Γ(2 − α) 1−α α τ−α 󵄩 󵄩2 3 󵄩 󵄩2 [ ( + 1)σ −α 󵄩󵄩󵄩v1 󵄩󵄩󵄩 + (n − 1 + σ)1−α 󵄩󵄩󵄩v0 󵄩󵄩󵄩 ] + 2Γ(2 − α) 12 σ 2 n

n

m=2

m=2

󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 + ∑ 󵄩󵄩󵄩pm 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩σvm + (1 − σ)vm−1 󵄩󵄩󵄩 + ∑ 󵄩󵄩󵄩δx2 qm 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩σum + (1 − σ)um−1 󵄩󵄩󵄩,

2 ⩽ n ⩽ N.

Multiplying both hand sides of the inequality above by 4τ, we get n−1 2 󵄩 󵄩2 τ ∑ 󵄩󵄩vn−k 󵄩󵄩󵄩 + F n T α Γ(1 − α) k=0󵄩

⩽ F1 + +

2τ1−α 1−α 󵄩󵄩 1 󵄩󵄩2 σ 󵄩󵄩v 󵄩󵄩 Γ(2 − α)

2τ1−α 1 − α α 󵄩 󵄩2 3 󵄩 󵄩2 [ ( + 1)σ −α 󵄩󵄩󵄩v1 󵄩󵄩󵄩 + (n − 1 + σ)1−α 󵄩󵄩󵄩v0 󵄩󵄩󵄩 ] Γ(2 − α) 12 σ 2 n

󵄩 󵄩 󵄩 󵄩 + 4τ ∑ 󵄩󵄩󵄩pm 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩σvm + (1 − σ)vm−1 󵄩󵄩󵄩 m=2 n

󵄩 󵄩 󵄩 󵄩 + 4τ ∑ 󵄩󵄩󵄩δx2 qm 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩σum + (1 − σ)um−1 󵄩󵄩󵄩 m=2

⩽ F1 +

1−α α 3 2τ1−α 󵄩 󵄩2 󵄩 󵄩2 [σ 1−α + ( + 1)σ −α ]󵄩󵄩󵄩v1 󵄩󵄩󵄩 + T 1−α 󵄩󵄩󵄩v0 󵄩󵄩󵄩 Γ(2 − α) 12 σ Γ(2 − α)

208 | 3 Difference methods for the time-fractional wave equations n

+ 2τ ∑ [ m=2 n

1 󵄩󵄩 m 󵄩 m 󵄩2 m−1 󵄩2 󵄩σv + (1 − σ)v 󵄩󵄩󵄩 + ε0 󵄩󵄩󵄩p 󵄩󵄩󵄩 ] ε0 󵄩

󵄩 󵄩2 󵄩 󵄩2 + 2τ ∑ [󵄩󵄩󵄩σum + (1 − σ)um−1 󵄩󵄩󵄩 + 󵄩󵄩󵄩δx2 qm 󵄩󵄩󵄩 ] m=2

⩽ F1 + +

3 1−α α 2τ1−α 󵄩 󵄩2 󵄩 󵄩2 [σ 1−α + ( + 1)σ −α ]󵄩󵄩󵄩v1 󵄩󵄩󵄩 + T 1−α 󵄩󵄩󵄩v0 󵄩󵄩󵄩 Γ(2 − α) 12 σ Γ(2 − α)

n 2τ n 󵄩󵄩 m 󵄩󵄩2 󵄩󵄩 m−1 󵄩󵄩2 󵄩 󵄩2 ∑ (󵄩󵄩v 󵄩󵄩 + 󵄩󵄩v 󵄩󵄩 ) + 2τε0 ∑ 󵄩󵄩󵄩pm 󵄩󵄩󵄩 ε0 m=2 m=2 n

󵄩 󵄩2 󵄩 󵄩2 󵄩 󵄩2 + 2τ ∑ (󵄩󵄩󵄩um 󵄩󵄩󵄩 + 󵄩󵄩󵄩um−1 󵄩󵄩󵄩 + 󵄩󵄩󵄩δx2 qm 󵄩󵄩󵄩 ), m=2

2 ⩽ n ⩽ N.

Taking ε0 = 4T α Γ(1 − α) in the inequality above, noticing (3.130), (3.134) and 󵄩 󵄩2 󵄩 󵄩2 󵄩 󵄩2 F 1 = (2σ + 1)󵄩󵄩󵄩δx u1 󵄩󵄩󵄩 − (2σ − 1)󵄩󵄩󵄩δx u0 󵄩󵄩󵄩 + (2σ 2 + σ − 1)󵄩󵄩󵄩δx (u1 − u0 )󵄩󵄩󵄩 󵄩 󵄩2 󵄩 󵄩2 ⩽ (2σ + 1)󵄩󵄩󵄩δx u1 󵄩󵄩󵄩 − (2σ − 1)󵄩󵄩󵄩δx u0 󵄩󵄩󵄩 󵄩 󵄩2 󵄩 󵄩2 + 2(2σ 2 + σ − 1)(󵄩󵄩󵄩δx u1 󵄩󵄩󵄩 + 󵄩󵄩󵄩δx u0 󵄩󵄩󵄩 ) 󵄩 󵄩2 󵄩 󵄩2 = (4σ 2 + 4σ − 1)󵄩󵄩󵄩δx u1 󵄩󵄩󵄩 + (4σ 2 − 1)󵄩󵄩󵄩δx u0 󵄩󵄩󵄩 , it follows that n−1

󵄩 󵄩2 󵄩 󵄩2 τ ∑ 󵄩󵄩󵄩vn−k 󵄩󵄩󵄩 + 󵄩󵄩󵄩δx un 󵄩󵄩󵄩 k=0

󵄩 󵄩2 󵄩 󵄩2 󵄩 󵄩2 ⩽ C2 (τ1−α 󵄩󵄩󵄩v0 󵄩󵄩󵄩 + 󵄩󵄩󵄩δx u0 󵄩󵄩󵄩 + τ󵄩󵄩󵄩u0 󵄩󵄩󵄩 n

n

n

m=1

m=1

m=1

󵄩 󵄩2 󵄩 󵄩2 󵄩 󵄩2 + τ ∑ 󵄩󵄩󵄩um 󵄩󵄩󵄩 + τ ∑ 󵄩󵄩󵄩pm 󵄩󵄩󵄩 + τ ∑ 󵄩󵄩󵄩δx2 qm 󵄩󵄩󵄩 ) 󵄩 󵄩2 󵄩 󵄩2 󵄩 󵄩2 ⩽ C2 (τ1−α 󵄩󵄩󵄩v0 󵄩󵄩󵄩 + 󵄩󵄩󵄩δx u0 󵄩󵄩󵄩 + τ󵄩󵄩󵄩u0 󵄩󵄩󵄩 n n L2 󵄩󵄩 m 󵄩󵄩2 󵄩 m 󵄩2 󵄩 2 m 󵄩2 󵄩󵄩δx u 󵄩󵄩 + τ ∑ 󵄩󵄩󵄩p 󵄩󵄩󵄩 + τ ∑ 󵄩󵄩󵄩δx q 󵄩󵄩󵄩 ), 6 m=1 m=1 m=1 n

+τ ∑

1 ⩽ n ⩽ N,

with C2 a constant. An application of the Gronwall inequality yields (3.123). The proof ends. 3.4.4 Convergence of the difference scheme Theorem 3.4.3. Suppose {Uin , Vin | 0 ⩽ i ⩽ M, 0 ⩽ n ⩽ N} and {uni , vin | 0 ⩽ i ⩽ M, 0 ⩽ n ⩽ N} are solutions of the problem (3.88)–(3.91) and the difference scheme (3.103)–

3.5 The fast difference method based on L2-1σ approximation for 1D problem | 209

(3.107), respectively. Let ein = Uin − uni ,

zin = Vin − vin ,

0 ⩽ i ⩽ M, 0 ⩽ n ⩽ N,

then it holds n

2 󵄩2 󵄩 󵄩2 󵄩 τ ∑ 󵄩󵄩󵄩z k 󵄩󵄩󵄩 + 󵄩󵄩󵄩δx en 󵄩󵄩󵄩 ⩽ C(c52 + c62 )LT(τ2 + h2 ) , k=1

1 ⩽ n ⩽ N,

where C is defined in Theorem 3.4.2. Proof. The subtraction of (3.103)–(3.107) from (3.95), (3.97)–(3.98), (3.101)–(3.102), respectively, produces the system of error equations as follows: { { { { { { { { { { { { { { { { { { { { { { { { { { { { {

1 n−1 (n,α) n−k (zi − zin−k−1 ) = σδx2 ein + (1 − σ)δx2 ein−1 + (r5 )ni , ∑c s k=0 k

1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N,

1 2

1 2

δt ei = zi + (r6 )1i , Dt ̄ein ei0 = e0n =

= 0,

σzin

+ (1 − σ)zin−1 + (r6 )ni ,

zi0 = 0, n eM = 0,

0,

0 ⩽ i ⩽ M,

0 ⩽ i ⩽ M, 2 ⩽ n ⩽ N,

0 ⩽ i ⩽ M,

1 ⩽ n ⩽ N.

Noticing (3.96), (3.99)–(3.100), the application of Theorem 3.4.2 yields n

󵄩 󵄩2 󵄩 󵄩2 τ ∑ 󵄩󵄩󵄩z k 󵄩󵄩󵄩 + 󵄩󵄩󵄩δx en 󵄩󵄩󵄩 k=1

n

n

m=1

m=1

󵄩 󵄩2 󵄩 󵄩2 󵄩 󵄩2 󵄩 󵄩2 ⩽ C[τ1−α 󵄩󵄩󵄩z 0 󵄩󵄩󵄩 + 󵄩󵄩󵄩δx e0 󵄩󵄩󵄩 + τ ∑ 󵄩󵄩󵄩(r5 )m 󵄩󵄩󵄩 + τ ∑ 󵄩󵄩󵄩δx2 (r6 )m 󵄩󵄩󵄩 ] n

2

n

2

⩽ C{τ ∑ L[c5 (τ2 + h2 )] + τ ∑ L(c6 τ2 ) } m=1

⩽

C(c52

+

c62 )LT(τ2

2 2

+h ) ,

m=1

1 ⩽ n ⩽ N.

The proof ends.

3.5 The fast difference method based on L2-1σ approximation for 1D problem In this section, we continue to consider the problem (3.81)–(3.83) and develop a fast temporal second-order difference scheme.

210 | 3 Difference methods for the time-fractional wave equations 3.5.1 Derivation of the difference scheme For the mesh function {uni | 0 ⩽ i ⩽ M, 0 ⩽ n ⩽ N} defined on Ωh × Ωτ , as what we did in last section, introduce the following notation: Dt ̄uni =

1 [(2σ + 1)uni − 4σun−1 + (2σ − 1)un−2 i i ]. 2τ

Let v(x, t) = ut (x, t),

α = γ − 1,

σ =1−

α , 2

s = τα Γ(2 − α).

Thus, the considered problem (3.81)–(3.83) is equivalent to the following one: { { { { { { { { { { {

C α 0 Dt v(x, t)

= uxx (x, t) + f (x, t),

ut (x, t) = v(x, t), u(x, 0) = φ(x), u(0, t) = 0,

x ∈ (0, L), t ∈ (0, T],

x ∈ [0, L], t ∈ (0, T],

v(x, 0) = ψ(x),

u(L, t) = 0,

x ∈ [0, L],

t ∈ (0, T].

(3.138) (3.139) (3.140) (3.141)

Denote U n = u(xi , tn ), Vin = v(xi , tn ), 0 ⩽ i ⩽ M, 0 ⩽ n ⩽ N, { i φi = φ(xi ), ψi = ψ(xi ), 0 ⩽ i ⩽ M. Considering (3.138) at the point (xi , tn−1+σ ), we have C α 0 Dt v(xi , tn−1+σ )

= uxx (xi , tn−1+σ ) + fin−1+σ ,

1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N.

(3.142)

Applying the theory in Subsection 1.7.2 to approximate Caputo derivative gives N

exp { 1 C α { { ω F n + d0(1,α) (Vin − Vin−1 ) D v(x , t ) = ∑ { i n−1+σ 0 t { Γ(1 − α) l=1 l l,i { { { { { { { + O(τ3−α + ϵ), 1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N, { 1 { { Fl,i = 0, 1 ⩽ l ⩽ Nexp , 1 ⩽ i ⩽ M − 1, { { { { n n−1 { { Fl,i = e−sl τ Fl,i + Al (Vin−1 − Vin−2 ) + Bl (Vin − Vin−1 ), { { { 1 ⩽ l ⩽ Nexp , 1 ⩽ i ⩽ M − 1, 2 ⩽ n ⩽ N. {

(3.143) (3.144) (3.145)

For the spatial second-order derivative, by (3.85) and Lemma 2.1.3, we have uxx (xi , tn−1+σ ) = σuxx (xi , tn ) + (1 − σ)uxx (xi , tn−1 ) + O(τ2 ) = σδx2 Uin + (1 − σ)δx2 Uin−1 + O(h2 + τ2 ).

(3.146)

3.5 The fast difference method based on L2-1σ approximation for 1D problem

| 211

Substituting (3.143) and (3.146) into (3.142) arrives at N

exp 1 n + d0(1,α) (Vin − Vin−1 ) ∑ ωl Fl,i Γ(1 − α) l=1

= σδx2 Uin + (1 − σ)δx2 Uin−1 + fin−1+σ + (r7 )ni , 1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N

(3.147)

and there exists a positive constant c7 such that 󵄨󵄨 n 󵄨󵄨 2 2 󵄨󵄨(r7 )i 󵄨󵄨 ⩽ c7 (τ + h + ϵ),

1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N.

(3.148)

n Substituting (3.144)–(3.145) into (3.147) and eliminating the intermediate variable {Fl,i } yield n−1

∑ dk(n,α) (Vin−k − Vin−k−1 ) = σδx2 Uin + (1 − σ)δx2 Uin−1 + fin−1+σ + (r7 )ni ,

k=0

1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N. Considering (3.139) at the points (xi , t 1 ) and (xi , tn−1+σ ), respectively, we have 2

{

1 2

1 2

δt Ui = Vi + (r8 )1i , 0 ⩽ i ⩽ M, Dt ̄Uin

=

σVin

+ (1 −

σ)Vin−1

+

(r8 )ni ,

(3.149) 0 ⩽ i ⩽ M, 2 ⩽ n ⩽ N,

(3.150)

and there exists a positive constant c8 such that 󵄨󵄨 2 n󵄨 2 󵄨󵄨δx (r8 )i 󵄨󵄨󵄨 ⩽ c8 τ ,

1 ⩽ i ⩽ M − 1, 2 ⩽ n ⩽ N.

(3.151)

In addition, from homogeneous boundary value conditions (3.141), we can obtain (r8 )n0 = 0,

(r8 )nM = 0,

1 ⩽ n ⩽ N.

(3.152)

Noticing the initial-boundary value conditions (3.140)–(3.141), we have {

Ui0 = φi ,

U0n

= 0,

Vi0 = ψi ,

n UM

= 0,

0 ⩽ i ⩽ M, 1 ⩽ n ⩽ N.

(3.153) (3.154)

Omitting the small term (r7 )ni and (r8 )ni in (3.147), (3.149)–(3.150) and replacing the exact solution {Uin , Vin } with its numerical one {uni , vin } produce the difference scheme for

212 | 3 Difference methods for the time-fractional wave equations (3.138)–(3.141) as follows: { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { {

N

exp 1 n + d0(1,α) (vin − vin−1 ) = σδx2 uni + (1 − σ)δx2 un−1 + fin−1+σ , ∑ ωl Fl,i i Γ(1 − α) l=1

1 Fl,i n Fl,i

1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N,

= 0, =

1 ⩽ l ⩽ Nexp , 1 ⩽ i ⩽ M − 1,

n−1 e−sl τ Fl,i 1 2

1 2

δt ui = vi , Dt ̄uni u0i = un0 =

=

σvin

φi , 0,

+

Al (vin−1

−

vin−2 )

+

Bl (vin

−

vin−1 ),

1 ⩽ l ⩽ Nexp , 1 ⩽ i ⩽ M − 1, 2 ⩽ n ⩽ N,

0 ⩽ i ⩽ M, + (1 −

vi0 unM

σ)vin−1 ,

= ψi ,

= 0,

(3.155) (3.156) (3.157) (3.158)

0 ⩽ i ⩽ M,

2 ⩽ n ⩽ N,

0 ⩽ i ⩽ M,

(3.159) (3.160)

1 ⩽ n ⩽ N.

(3.161)

n Substituting (3.156)–(3.157) into (3.155) and eliminating the intermediate variable {Fl,i } yield n−1

+ fin−1+σ , ∑ dk(n,α) (vin−k − vin−k−1 ) = σδx2 uni + (1 − σ)δx2 un−1 i

k=0

1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N.

(3.162)

3.5.2 Solvability of the difference scheme Theorem 3.5.1. The difference scheme (3.155)–(3.161) is uniquely solvable. Proof. Let un = (un0 , un1 , . . . , unM ),

n vn = (v0n , v1n , . . . , vM ).

(I) The value of {u0 , v0 } is determined by (3.160). (II) It follows from (3.158) that 1

vi1 = 2δt ui2 − vi0 ,

0 ⩽ i ⩽ M.

(3.163)

Substituting (3.163) into (3.162), and noticing (3.161), we obtain the linear system in u1 : 1

d(1,α) (2δt ui2 − 2vi0 ) = σδx2 u1i + (1 − σ)δx2 u0i + fiσ , { { { 0 1 ⩽ i ⩽ M − 1, { { { 1 1 { u0 = 0, uM = 0.

(3.164) (3.165)

3.5 The fast difference method based on L2-1σ approximation for 1D problem

| 213

Considering its homogeneous one, we have 2 { d0(1,α) u1i = σδx2 u1i , τ { 1 1 { u0 = 0, uM = 0.

1 ⩽ i ⩽ M − 1,

(3.166) (3.167)

Making the inner product on both hand sides of (3.166) with u1 produces 2 󵄩 󵄩2 󵄩 󵄩2 d0(1,α) 󵄩󵄩󵄩u1 󵄩󵄩󵄩 + σ 󵄩󵄩󵄩δx u1 󵄩󵄩󵄩 = 0, τ which implies u1 = 0, hence the system (3.164)–(3.165) has a unique solution. Once u1 is obtained, v1 is followed from (3.163). (III) Suppose the value of {u0 , v0 , u1 , v1 , . . . , un−1 , vn−1 } has been uniquely determined, then it follows from (3.159) that vin =

1 (D ̄un − (1 − σ)vin−1 ), σ t i

0 ⩽ i ⩽ M.

(3.168)

Substituting (3.168) into (3.162), noticing (3.161), one can obtain the linear system in un as follows: n−1 1 { { d0(n,α) [ (Dt ̄uni − (1 − σ)vin−1 ) − vin−1 ] + ∑ dk(n,α) (vin−k − vin−k−1 ) { { { σ { k=1 { 2 n 2 n−1 n−1+σ { { = σδ u + (1 − σ)δ u + f , 1 ⩽ i ⩽ M − 1, { x i x i i { { n n { u0 = 0, uM = 0.

(3.169) (3.170)

To show its unique solvability, it suffices to prove the corresponding homogeneous one 1 2σ + 1 n { d0(n,α) ⋅ ⋅ ui = σδx2 uni , σ 2τ { n n { u0 = 0, uM = 0

1 ⩽ i ⩽ M − 1,

(3.171) (3.172)

has only the trivial solution. Taking the inner product on both hand sides of (3.171) with un yields d0(n,α) ⋅

1 2σ + 1 󵄩󵄩 n 󵄩󵄩2 󵄩 n 󵄩2 ⋅ 󵄩󵄩u 󵄩󵄩 + σ 󵄩󵄩󵄩δx u 󵄩󵄩󵄩 = 0, σ 2τ

which implies un = 0, and the system (3.169)–(3.170) has a unique solution un . Once un is obtained, vn is followed from (3.168). By the principle of induction, the theorem is true. The proof ends.

214 | 3 Difference methods for the time-fractional wave equations 3.5.3 Stability of the difference scheme Theorem 3.5.2. Suppose {un , vn | 0 ⩽ n ⩽ N} is the solution of the difference scheme { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { {

N

exp 1 n + d0(1,α) (vin − vin−1 ) ∑ ωl Fl,i Γ(1 − α) l=1

= σδx2 uni + (1 − σ)δx2 un−1 + pni , i 1 Fl,i n Fl,i

= 0,

1 ⩽ l ⩽ Nexp , 1 ⩽ i ⩽ M − 1,

n−1 e−sl τ Fl,i

=

1 2

1 2

+

Al (vin−1

=

σvin

φi , 0,

Bl (vin

+

−

σ)vin−1

= ψi ,

= 0,

+

(3.173) (3.174)

vin−1 ),

0 ⩽ i ⩽ M,

+ (1 −

vi0 unM

−

vin−2 )

1 ⩽ l ⩽ Nexp , 1 ⩽ i ⩽ M − 1, 2 ⩽ n ⩽ N,

δt ui = vi + qi1 ,

Dt ̄uni u0i = un0 =

1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N,

(3.175) (3.176)

qin ,

0 ⩽ i ⩽ M, 2 ⩽ n ⩽ N,

0 ⩽ i ⩽ M,

(3.177) (3.178)

1 ⩽ n ⩽ N,

(3.179)

n where q0n = qM = 0 (1 ⩽ n ⩽ N) and φ0 = φM = ψ0 = ψM = 0. Then there exists a positive constant C such that n

󵄩 󵄩2 󵄩 󵄩2 τ ∑ 󵄩󵄩󵄩vk 󵄩󵄩󵄩 + 󵄩󵄩󵄩δx un 󵄩󵄩󵄩 k=1

n

n

m=1

m=1

󵄩 󵄩2 󵄩 󵄩2 󵄩 󵄩2 󵄩 󵄩2 ⩽ C[τ1−α 󵄩󵄩󵄩v0 󵄩󵄩󵄩 + ϵτα−1 󵄩󵄩󵄩δx u0 󵄩󵄩󵄩 + τ ∑ 󵄩󵄩󵄩pm 󵄩󵄩󵄩 + τ ∑ 󵄩󵄩󵄩δx2 qm 󵄩󵄩󵄩 ],

1 ⩽ n ⩽ N,

(3.180)

in which M−1

󵄩󵄩 k 󵄩󵄩2 k 2 󵄩󵄩v 󵄩󵄩 = h ∑ (vi ) , i=1

M−1

󵄩󵄩 k 󵄩󵄩2 k 2 󵄩󵄩p 󵄩󵄩 = h ∑ (pi ) , i=1

Proof. It follows from (3.176)–(3.179), ψM = 0 that v0n = 0,

q0n

=

n qM

M−1

󵄩󵄩 2 k 󵄩󵄩2 2 k 2 󵄩󵄩δx q 󵄩󵄩 = h ∑ (δx qi ) . i=1

= 0 (1 ⩽ n ⩽ N) and φ0 = φM = ψ0 =

n vM = 0,

0 ⩽ n ⩽ N.

n σvM

n−1 σ)vM

It is easy to obtain 1

{

v02 = 0, σv0n

+ (1 −

1

vM2 = 0, σ)v0n−1

= 0,

+ (1 −

(3.181) = 0,

2 ⩽ n ⩽ N.

(3.182)

Substituting (3.174)–(3.175) into (3.173) and getting rid of the intermediate variable n {Fl,i }, we have n−1

+ pni , ∑ dk(n,α) (vin−k − vin−k−1 ) = σδx2 uni + (1 − σ)δx2 un−1 i

k=0

1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N.

(3.183)

3.5 The fast difference method based on L2-1σ approximation for 1D problem

| 215

(I) When n = 1, equation (3.183) reads d0(1,α) (vi1 − vi0 ) = σδx2 u1i + (1 − σ)δx2 u0i + p1i ,

1 ⩽ i ⩽ M − 1.

(3.184)

It is easy to know from (3.176) that 1

1

δt δx2 ui2 = δx2 vi2 + δx2 qi1 ,

1 ⩽ i ⩽ M − 1.

(3.185) 1

Making the inner product on both hand sides of (3.184) with v 2 and noticing (3.181) yield that 1 1 1 (1,α) 󵄩󵄩 1 󵄩󵄩2 󵄩󵄩 0 󵄩󵄩2 d (󵄩v 󵄩 − 󵄩v 󵄩 ) = −(δx (σu1 + (1 − σ)u0 ), δx v 2 ) + (p1 , v 2 ). 2 0 󵄩 󵄩 󵄩 󵄩

(3.186)

Making the inner product on both hand sides of (3.185) with −(σu1 + (1 − σ)u0 ) and noticing σu10 + (1 − σ)u00 = 0, arrive at

σu1M + (1 − σ)u0M = 0,

1 (δ u1 − δx u0 , δx (σu1 + (1 − σ)u0 )) τ x 1

= (δx v 2 , δx (σu1 + (1 − σ)u0 )) − (δx2 q1 , σu1 + (1 − σ)u0 ). Adding (3.186) and (3.187) gives 1 1 (1,α) 󵄩󵄩 1 󵄩󵄩2 󵄩󵄩 0 󵄩󵄩2 d (󵄩v 󵄩 − 󵄩v 󵄩 ) + (δx u1 − δx u0 , δx (σu1 + (1 − σ)u0 )) 2 0 󵄩 󵄩 󵄩 󵄩 τ 1

= (p1 , v 2 ) − (δx2 q1 , σu1 + (1 − σ)u0 ). Noticing the fact that

(δx u1 − δx u0 , δx (σu1 + (1 − σ)u0 )) =

󵄩󵄩2 󵄩󵄩 1 σ 󵄩󵄩 1 󵄩󵄩2 3σ 2σ − 1 󵄩 󵄩 δx u1 − √ δx u0 󵄩󵄩󵄩 󵄩󵄩δx u 󵄩󵄩 + (2σ − 1)󵄩󵄩󵄩 √ 󵄩󵄩 󵄩󵄩 2 2σ − 1 4 3σ −

σ 2 − σ + 1 󵄩󵄩 0 󵄩󵄩2 󵄩󵄩δx u 󵄩󵄩 3σ

and d0(1,α) =

σ 1−α τ−α , Γ(2 − α)

we have 2 σ 1−α τ−α 󵄩󵄩 1 󵄩󵄩2 󵄩󵄩 0 󵄩󵄩2 1 σ󵄩 󵄩2 σ − σ + 1 󵄩󵄩 0 󵄩󵄩2 (󵄩󵄩v 󵄩󵄩 − 󵄩󵄩v 󵄩󵄩 ) + ( 󵄩󵄩󵄩δx u1 󵄩󵄩󵄩 − 󵄩󵄩δx u 󵄩󵄩 ) 2Γ(2 − α) τ 4 3σ 1

⩽ (p1 , v 2 ) − (δx2 q1 , σu1 + (1 − σ)u0 ),

(3.187)

216 | 3 Difference methods for the time-fractional wave equations which can be reduced to 2 σ 1−α τ1−α 󵄩󵄩 1 󵄩󵄩2 󵄩󵄩 0 󵄩󵄩2 σ󵄩 󵄩2 σ − σ + 1 󵄩󵄩 0 󵄩󵄩2 (󵄩󵄩v 󵄩󵄩 − 󵄩󵄩v 󵄩󵄩 ) + 󵄩󵄩󵄩δx u1 󵄩󵄩󵄩 − 󵄩󵄩δx u 󵄩󵄩 2Γ(2 − α) 4 3σ 1

⩽ τ(p1 , v 2 ) − τ(δx2 q1 , σu1 + (1 − σ)u0 ) 󵄩 󵄩 󵄩 1󵄩 󵄩 󵄩 󵄩 󵄩 ⩽ τ󵄩󵄩󵄩p1 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩v 2 󵄩󵄩󵄩 + τ󵄩󵄩󵄩δx2 q1 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩σu1 + (1 − σ)u0 󵄩󵄩󵄩 1󵄩 1 1 󵄩󵄩 21 󵄩󵄩2 ε0 󵄩󵄩 1 󵄩󵄩2 0 󵄩2 ⩽ τ[ 󵄩v 󵄩 + 󵄩p 󵄩 ] + τ[ 󵄩󵄩󵄩σu + (1 − σ)u 󵄩󵄩󵄩 + 2ε0 󵄩 󵄩 2 󵄩 󵄩 2 ⩽ τ[

1 󵄩󵄩 2 1 󵄩󵄩2 󵄩δ q 󵄩 ] 2󵄩 x 󵄩

ε 󵄩 󵄩2 1 󵄩 󵄩2 󵄩 󵄩2 1 󵄩󵄩 1 󵄩󵄩2 󵄩󵄩 0 󵄩󵄩2 (󵄩v 󵄩 + 󵄩v 󵄩 ) + 0 󵄩󵄩󵄩p1 󵄩󵄩󵄩 ] + τ[ (󵄩󵄩󵄩u1 󵄩󵄩󵄩 + 󵄩󵄩󵄩u0 󵄩󵄩󵄩 ) + 4ε0 󵄩 󵄩 󵄩 󵄩 2 2

1 󵄩󵄩 2 1 󵄩󵄩2 󵄩δ q 󵄩 ]. 2󵄩 x 󵄩

Taking ε0 = Γ(2 − α)σ α−1 τα , there exists a positive constant C1 such that 󵄩 󵄩2 󵄩 󵄩2 τ1−α 󵄩󵄩󵄩v1 󵄩󵄩󵄩 + 󵄩󵄩󵄩δx u1 󵄩󵄩󵄩 󵄩 󵄩2 󵄩 󵄩2 󵄩 󵄩2 󵄩 󵄩2 󵄩 󵄩2 󵄩 󵄩2 ⩽ C1 (τ1−α 󵄩󵄩󵄩v0 󵄩󵄩󵄩 + 󵄩󵄩󵄩δx u0 󵄩󵄩󵄩 + τ󵄩󵄩󵄩u0 󵄩󵄩󵄩 + τ󵄩󵄩󵄩u1 󵄩󵄩󵄩 + τ1+α 󵄩󵄩󵄩p1 󵄩󵄩󵄩 + τ󵄩󵄩󵄩δx2 q1 󵄩󵄩󵄩 ).

(3.188)

(II) Making the inner product on both hand sides of (3.183) with σvn + (1 − σ)vn−1 and noticing (3.182) yield n−1

∑ dk(n,α) (vn−k − vn−k−1 , σvn + (1 − σ)vn−1 )

=

k=0 (δx2 (σun

+ (1 − σ)un−1 ), σvn + (1 − σ)vn−1 ) + (pn , σvn + (1 − σ)vn−1 )

= −(δx (σun + (1 − σ)un−1 ), δx (σvn + (1 − σ)vn−1 )) + (pn , σvn + (1 − σ)vn−1 ),

2 ⩽ n ⩽ N.

(3.189)

For the term on the left-hand side of (3.189), by Lemma 1.7.3 and Lemma 2.6.1, we have n−1

∑ dk(n,α) (vn−k − vn−k−1 , σvn + (1 − σ)vn−1 )

k=0

⩾

1 n−1 (n,α) 󵄩󵄩 n−k 󵄩󵄩2 󵄩󵄩 n−k−1 󵄩󵄩2 (󵄩󵄩v 󵄩󵄩 − 󵄩󵄩v ∑d 󵄩󵄩 ). 2 k=0 k

Therefore, 1 n−1 (n,α) 󵄩󵄩 n−k 󵄩󵄩2 󵄩󵄩 n−k−1 󵄩󵄩2 (󵄩󵄩v 󵄩󵄩 − 󵄩󵄩v ∑d 󵄩󵄩 ) 2 k=0 k

⩽ −(δx (σun + (1 − σ)un−1 ), δx (σvn + (1 − σ)vn−1 )) + (pn , σvn + (1 − σ)vn−1 ),

2 ⩽ n ⩽ N.

(3.190)

It follows from (3.177) that Dt ̄δx2 uni = δx2 (σvin + (1 − σ)vin−1 ) + δx2 qin ,

1 ⩽ i ⩽ M − 1, 2 ⩽ n ⩽ N.

(3.191)

3.5 The fast difference method based on L2-1σ approximation for 1D problem

| 217

Taking the inner product on both hand sides of (3.191) with −(σun + (1 − σ)un−1 ) and noticing σun0 + (1 − σ)un−1 0 = 0,

σunM + (1 − σ)un−1 M = 0,

2⩽n⩽N

lead to (Dt ̄δx un , δx (σun + (1 − σ)un−1 ))

= (δx (σvn + (1 − σ)vn−1 ), δx (σun + (1 − σ)un−1 )) − (δx2 qn , σun + (1 − σ)un−1 ),

2 ⩽ n ⩽ N.

(3.192)

Denote 󵄩 󵄩2 󵄩 󵄩2 F n = (2σ + 1)󵄩󵄩󵄩δx un 󵄩󵄩󵄩 − (2σ − 1)󵄩󵄩󵄩δx un−1 󵄩󵄩󵄩 󵄩 󵄩2 + (2σ 2 + σ − 1)󵄩󵄩󵄩δx (un − un−1 )󵄩󵄩󵄩 , n ⩾ 1. Using Lemma 3.4.2, we have Fn ⩾

1 󵄩󵄩 n 󵄩󵄩2 󵄩δ u 󵄩 , σ󵄩 x 󵄩

n⩾1

(3.193)

and (Dt ̄δx un , δx (σun + (1 − σ)un−1 )) ⩾

1 n (F − F n−1 ), 4τ

n ⩾ 2.

(3.194)

Combining (3.192) with (3.194), we obtain 1 n (F − F n−1 ) ⩽ (δx (σvn + (1 − σ)vn−1 ), δx (σun + (1 − σ)un−1 )) 4τ − (δx2 qn , σun + (1 − σ)un−1 ), 2 ⩽ n ⩽ N. Adding (3.190) and (3.195) yields 1 n 1 n−1 (n,α) 󵄩󵄩 n−k 󵄩󵄩2 󵄩󵄩 n−k−1 󵄩󵄩2 (󵄩󵄩v 󵄩󵄩 − 󵄩󵄩v (F − F n−1 ) ∑d 󵄩󵄩 ) + 2 k=0 k 4τ

⩽ (pn , σvn + (1 − σ)vn−1 ) − (δx2 qn , σun + (1 − σ)un−1 ) 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 ⩽ 󵄩󵄩󵄩pn 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩σvn + (1 − σ)vn−1 󵄩󵄩󵄩 + 󵄩󵄩󵄩δx2 qn 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩σun + (1 − σ)un−1 󵄩󵄩󵄩, Noticing the fact that n−1

󵄩 󵄩2 󵄩 󵄩2 ∑ dk(n,α) (󵄩󵄩󵄩vn−k 󵄩󵄩󵄩 − 󵄩󵄩󵄩vn−k−1 󵄩󵄩󵄩 )

k=0 n−1

n−2

k=0

k=0

󵄩 󵄩2 󵄩 󵄩2 = ∑ dk(n,α) 󵄩󵄩󵄩vn−k 󵄩󵄩󵄩 − ∑ dk(n−1,α) 󵄩󵄩󵄩vn−1−k 󵄩󵄩󵄩

2 ⩽ n ⩽ N.

(3.195)

218 | 3 Difference methods for the time-fractional wave equations n−2

󵄩 󵄩2 (n,α) 󵄩 󵄩󵄩v0 󵄩󵄩󵄩2 − ∑ (dk(n,α) − dk(n−1,α) )󵄩󵄩󵄩vn−1−k 󵄩󵄩󵄩 − dn−1 󵄩 󵄩 k=0

n−1

n−2

k=0

k=0

󵄩 󵄩2 󵄩 󵄩2 = ∑ dk(n,α) 󵄩󵄩󵄩vn−k 󵄩󵄩󵄩 − ∑ dk(n−1,α) 󵄩󵄩󵄩vn−1−k 󵄩󵄩󵄩 󵄩2 (n,α) (n−1,α) 󵄩 (n,α) 󵄩 󵄩󵄩v0 󵄩󵄩󵄩2 , − (dn−2 − dn−2 )󵄩󵄩󵄩v1 󵄩󵄩󵄩 − dn−1 󵄩 󵄩

we have 1 n−1 (n,α) 󵄩󵄩 n−k 󵄩󵄩2 n−2 (n−1,α) 󵄩󵄩 n−1−k 󵄩󵄩2 1 n (∑ d (F − F n−1 ) 󵄩v 󵄩󵄩 − ∑ dk 󵄩󵄩v 󵄩󵄩 ) + 2 k=0 k 󵄩 4τ k=0

1 (n,α) 󵄩2 (n−1,α) 󵄩 (n,α) 󵄩 󵄩󵄩v0 󵄩󵄩󵄩2 ] ⩽ [(dn−2 − dn−2 )󵄩󵄩󵄩v1 󵄩󵄩󵄩 + dn−1 󵄩 󵄩 2 󵄩󵄩 n 󵄩󵄩 󵄩󵄩 n 󵄩 󵄩 󵄩 󵄩 n−1 󵄩 + 󵄩󵄩p 󵄩󵄩 ⋅ 󵄩󵄩σv + (1 − σ)v 󵄩󵄩󵄩 + 󵄩󵄩󵄩δx2 qn 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩σun + (1 − σ)un−1 󵄩󵄩󵄩,

2 ⩽ n ⩽ N.

(3.196)

Replacing the superscript n with m, and summing up for m from 2 to n on both hand sides of (3.196), we obtain 1 n 1 n−1 (n,α) 󵄩󵄩 n−k 󵄩󵄩2 󵄩 󵄩2 ( ∑ dk 󵄩󵄩v 󵄩󵄩 − d0(1,α) 󵄩󵄩󵄩v1 󵄩󵄩󵄩 ) + (F − F 1 ) 2 k=0 4τ n 1 n (m,α) 󵄩2 (m−1,α) 󵄩 (m,α) 󵄩 󵄩󵄩v0 󵄩󵄩󵄩2 ] − dm−2 )󵄩󵄩󵄩v1 󵄩󵄩󵄩 + ∑ dm−1 ⩽ [ ∑ (dm−2 󵄩 󵄩 2 m=2 m=2 n

󵄩 󵄩 󵄩 󵄩 + ∑ 󵄩󵄩󵄩pm 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩σvm + (1 − σ)vm−1 󵄩󵄩󵄩 m=2 n

󵄩 󵄩 󵄩 󵄩 + ∑ 󵄩󵄩󵄩δx2 qm 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩σum + (1 − σ)um−1 󵄩󵄩󵄩, m=2

2 ⩽ n ⩽ N.

Applying Lemma 3.4.3 and (1.147), it follows that n

(m,α) (m−1,α) − dm−2 ) ∑ (dm−2

m=2

⩽ ⩽

n n 1 9 τ−α (m,α) (m−1,α) − cm−2 )+ ∑ (cm−2 ∑ ϵ Γ(2 − α) m=2 Γ(1 − α) m=2 4

τ−α 1 − α α 1 9 ( + 1)σ −α + (n − 1)ϵ Γ(2 − α) 12 σ Γ(1 − α) 4

and n

(m,α) ⩽ ∑ dm−1

m=2

⩽

n n τ−α ϵ (m,α) + ∑ ∑ cm−1 Γ(2 − α) m=2 Γ(1 − α) m=2

τ−α 3 ϵ (n − 1 + σ)1−α + (n − 1) . Γ(2 − α) 2 Γ(1 − α)

3.5 The fast difference method based on L2-1σ approximation for 1D problem

| 219

Hence, we arrive at 1 n−1 (n,α) 󵄩󵄩 n−k 󵄩󵄩2 1 n (1,α) 󵄩 1 󵄩2 (∑ d (F − F 1 ) 󵄩v 󵄩󵄩 − d0 󵄩󵄩󵄩v 󵄩󵄩󵄩 ) + 2 k=0 k 󵄩 4τ

1 τ−α 1−α α 1 9 󵄩 󵄩2 ⩽ [( ⋅ ( + 1)σ −α + (n − 1)ϵ)󵄩󵄩󵄩v1 󵄩󵄩󵄩 2 Γ(2 − α) 12 σ Γ(1 − α) 4 ϵ τ−α 3 󵄩 󵄩2 (n − 1 + σ)1−α + (n − 1) )󵄩󵄩v0 󵄩󵄩 ] +( Γ(2 − α) 2 Γ(1 − α) 󵄩 󵄩 n

󵄩 󵄩 󵄩 󵄩 + ∑ 󵄩󵄩󵄩pm 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩σvm + (1 − σ)vm−1 󵄩󵄩󵄩 m=2 n

󵄩 󵄩 󵄩 󵄩 + ∑ 󵄩󵄩󵄩δx2 qm 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩σum + (1 − σ)um−1 󵄩󵄩󵄩, m=2

2 ⩽ n ⩽ N.

(3.197)

By Lemma 1.7.3, we get d0(1,α) =

σ 1−α τ−α Γ(2 − α)

and (n,α) d0(n,α) > d1(n,α) > d2(n,α) > d3(n,α) > ⋅ ⋅ ⋅ > dn−1 >

1 . 2tnα Γ(1 − α)

Thus, the inequality (3.197) can be reduced to n−1 1−α −α 1 1 1 n 󵄩 󵄩2 σ τ 󵄩󵄩 1 󵄩󵄩2 ( α (F − F 1 ) ∑ 󵄩󵄩󵄩vn−k 󵄩󵄩󵄩 − 󵄩v 󵄩 ) + 2 2tn Γ(1 − α) k=0 Γ(2 − α) 󵄩 󵄩 4τ

1 τ−α 1 − α α 1 9 󵄩 󵄩2 ⩽ [( ( + 1)σ −α + (n − 1)ϵ)󵄩󵄩󵄩v1 󵄩󵄩󵄩 2 Γ(2 − α) 12 σ Γ(1 − α) 4 ϵ τ−α 3 󵄩 󵄩2 (n − 1 + σ)1−α + (n − 1) )󵄩󵄩v0 󵄩󵄩 ] +( Γ(2 − α) 2 Γ(1 − α) 󵄩 󵄩 n

󵄩 󵄩 󵄩 󵄩 + ∑ 󵄩󵄩󵄩pm 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩σvm + (1 − σ)vm−1 󵄩󵄩󵄩 m=2 n

󵄩 󵄩 󵄩 󵄩 + ∑ 󵄩󵄩󵄩δx2 qm 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩σum + (1 − σ)um−1 󵄩󵄩󵄩, m=2

2 ⩽ n ⩽ N.

Multiplying both hand sides of (3.198) by 4τ yields n−1 1 󵄩 󵄩2 τ ∑ 󵄩󵄩vn−k 󵄩󵄩󵄩 + F n T α Γ(1 − α) k=0󵄩

⩽ F1 + [

2σ 1−α τ1−α τ1−α 1 − α α 1 9 󵄩 󵄩2 +( ( + 1)σ −α + Tϵ)]󵄩󵄩󵄩v1 󵄩󵄩󵄩 Γ(2 − α) Γ(2 − α) 6 σ Γ(1 − α) 2

(3.198)

220 | 3 Difference methods for the time-fractional wave equations

+(

3T 1−α 2ϵ T 󵄩 󵄩2 + )󵄩󵄩v0 󵄩󵄩 Γ(2 − α) Γ(1 − α) 󵄩 󵄩 n

󵄩 󵄩 󵄩 󵄩 + 4τ ∑ 󵄩󵄩󵄩pm 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩σvm + (1 − σ)vm−1 󵄩󵄩󵄩 m=2 n

󵄩 󵄩 󵄩 󵄩 + 4τ ∑ 󵄩󵄩󵄩δx2 qm 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩σum + (1 − σ)um−1 󵄩󵄩󵄩 m=2

1

⩽F +[ +(

τ1−α 1 − α α 1 9 2σ 1−α τ1−α 󵄩 󵄩2 +( ( + 1)σ −α + Tϵ)]󵄩󵄩󵄩v1 󵄩󵄩󵄩 Γ(2 − α) Γ(2 − α) 6 σ Γ(1 − α) 2

3T 1−α 2ϵ T 󵄩 󵄩2 + )󵄩󵄩v0 󵄩󵄩 Γ(2 − α) Γ(1 − α) 󵄩 󵄩 n

+ 2τ ∑ [ m=2 n

1 󵄩󵄩 m 󵄩 m 󵄩2 m−1 󵄩2 󵄩σv + (1 − σ)v 󵄩󵄩󵄩 + ε0 󵄩󵄩󵄩p 󵄩󵄩󵄩 ] ε0 󵄩

󵄩 󵄩2 󵄩 󵄩2 + 2τ ∑ [󵄩󵄩󵄩σum + (1 − σ)um−1 󵄩󵄩󵄩 + 󵄩󵄩󵄩δx2 qm 󵄩󵄩󵄩 ] m=2

1

⩽F +[ +( +

2σ 1−α τ1−α τ1−α 1 − α α 1 9 󵄩 󵄩2 +( ( + 1)σ −α + Tϵ)]󵄩󵄩󵄩v1 󵄩󵄩󵄩 Γ(1 − α) Γ(1 − α) 6 σ Γ(1 − α) 2

2ϵ T 3T 1−α 󵄩 󵄩2 + )󵄩󵄩v0 󵄩󵄩 Γ(2 − α) Γ(1 − α) 󵄩 󵄩

n 2τ n 󵄩󵄩 m 󵄩󵄩2 󵄩󵄩 m−1 󵄩󵄩2 󵄩 󵄩2 ∑ (󵄩󵄩v 󵄩󵄩 + 󵄩󵄩v 󵄩󵄩 ) + 2τε0 ∑ 󵄩󵄩󵄩pm 󵄩󵄩󵄩 ε0 m=2 m=2 n

󵄩 󵄩2 󵄩 󵄩2 󵄩 󵄩2 + 2τ ∑ (󵄩󵄩󵄩um 󵄩󵄩󵄩 + 󵄩󵄩󵄩um−1 󵄩󵄩󵄩 + 󵄩󵄩󵄩δx2 qm 󵄩󵄩󵄩 ), m=2

2 ⩽ n ⩽ N.

Taking ε0 = 8T α Γ(1 − α), noticing (3.188), (3.193) and 󵄩 󵄩2 󵄩 󵄩2 󵄩 󵄩2 F 1 = (2σ + 1)󵄩󵄩󵄩δx u1 󵄩󵄩󵄩 − (2σ − 1)󵄩󵄩󵄩δx u0 󵄩󵄩󵄩 + (2σ 2 + σ − 1)󵄩󵄩󵄩δx (u1 − u0 )󵄩󵄩󵄩 󵄩 󵄩2 󵄩 󵄩2 󵄩 󵄩2 󵄩 󵄩2 ⩽ (2σ + 1)󵄩󵄩󵄩δx u1 󵄩󵄩󵄩 − (2σ − 1)󵄩󵄩󵄩δx u0 󵄩󵄩󵄩 + 2(2σ 2 + σ − 1)(󵄩󵄩󵄩δx u1 󵄩󵄩󵄩 + 󵄩󵄩󵄩δx u0 󵄩󵄩󵄩 ) 󵄩 󵄩2 󵄩 󵄩2 = (4σ 2 + 4σ − 1)󵄩󵄩󵄩δx u1 󵄩󵄩󵄩 + (4σ 2 − 1)󵄩󵄩󵄩δx u0 󵄩󵄩󵄩 , there exists a positive constant C2 such that n−1

󵄩 󵄩2 󵄩 󵄩2 τ ∑ 󵄩󵄩󵄩vn−k 󵄩󵄩󵄩 + 󵄩󵄩󵄩δx un 󵄩󵄩󵄩 k=0

󵄩 󵄩2 󵄩 󵄩2 󵄩 󵄩2 ⩽ C2 (τ1−α 󵄩󵄩󵄩v0 󵄩󵄩󵄩 + ϵτα−1 󵄩󵄩󵄩δx u0 󵄩󵄩󵄩 + 󵄩󵄩󵄩u0 󵄩󵄩󵄩 n

n

n

m=1

m=1

m=1

󵄩 󵄩2 󵄩 󵄩2 󵄩 󵄩2 + τ ∑ 󵄩󵄩󵄩um 󵄩󵄩󵄩 + τ ∑ 󵄩󵄩󵄩pm 󵄩󵄩󵄩 + τ ∑ 󵄩󵄩󵄩δx2 qm 󵄩󵄩󵄩 ) 󵄩 󵄩2 󵄩 󵄩2 󵄩 󵄩2 ⩽ C2 (τ1−α 󵄩󵄩󵄩v0 󵄩󵄩󵄩 + ϵτα−1 󵄩󵄩󵄩δx u0 󵄩󵄩󵄩 + 󵄩󵄩󵄩u0 󵄩󵄩󵄩

3.5 The fast difference method based on L2-1σ approximation for 1D problem n n L2 󵄩󵄩 m 󵄩󵄩2 󵄩 m 󵄩2 󵄩 2 m 󵄩2 󵄩󵄩δx u 󵄩󵄩 + τ ∑ 󵄩󵄩󵄩p 󵄩󵄩󵄩 + τ ∑ 󵄩󵄩󵄩δx q 󵄩󵄩󵄩 ), 6 m=1 m=1 m=1 n

+τ ∑

| 221

1 ⩽ n ⩽ N.

The inequality (3.180) can be obtained with the help of the Gronwall inequality. The proof ends. Remark 3.5.1. In general, ϵ is taken such that ϵτα−1 ⩽ 1. 3.5.4 Convergence of the difference scheme Theorem 3.5.3. Suppose {Uin , Vin | 0 ⩽ i ⩽ M, 0 ⩽ n ⩽ N} and {uni , vin | 0 ⩽ i ⩽ M, 0 ⩽ n ⩽ N} are solutions of the problem (3.138)–(3.141) and the difference scheme (3.155)–(3.161), respectively. Let ein = Uin − uni ,

zin = Vin − vin ,

0 ⩽ i ⩽ M, 0 ⩽ n ⩽ N,

then it holds that n

2 󵄩 󵄩2 󵄩 󵄩2 τ ∑ 󵄩󵄩󵄩z k 󵄩󵄩󵄩 + 󵄩󵄩󵄩δx en 󵄩󵄩󵄩 ⩽ C(c72 + c82 )LT(τ2 + h2 + ϵ) ,

1 ⩽ n ⩽ N,

k=1

where C is defined in Theorem 3.5.2. Proof. The subtraction of (3.155)–(3.161) from (3.147), (3.144)–(3.145), (3.149)–(3.150) and (3.153)–(3.154), respectively, produces the system of error equations as follows: N

exp 1 { n { { ωl Fl,i + d0(1,α) (zin − zin−1 ) ∑ { { Γ(1 − α) { { l=1 { { 2 n { { = σδ e + (1 − σ)δx2 ein−1 + (r7 )ni , 1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N, { x i { { { 1 {F = 0, 1 ⩽ l ⩽ N , 1 ⩽ i ⩽ M − 1, { exp { l,i { { { n−1 n −sl τ n−1 { F = e F + A (z − zin−2 ) + Bl (zin − zin−1 ) { l,i l i l,i { { 1 ⩽ l ⩽ Nexp , 1 ⩽ i ⩽ M − 1, 2 ⩽ n ⩽ N, { { { 1 1 { { { δt ei2 = zi2 + (r8 )1i , 0 ⩽ i ⩽ M, { { { { { { Dt ̄ein = σzin + (1 − σ)zin−1 + (r8 )ni , 0 ⩽ i ⩽ M, 2 ⩽ n ⩽ N, { { { { { ei0 = 0, zi0 = 0, 0 ⩽ i ⩽ M, { { { { n n {e0 = 0, eM = 0, 1 ⩽ n ⩽ N.

Noticing (3.148) and (3.151)–(3.152), the application of Theorem 3.5.2 will lead to n

󵄩 󵄩2 󵄩 󵄩2 τ ∑ 󵄩󵄩󵄩z k 󵄩󵄩󵄩 + 󵄩󵄩󵄩δx en 󵄩󵄩󵄩 k=1

n

n

m=1

m=1

󵄩 󵄩2 󵄩 󵄩2 󵄩 󵄩2 󵄩 󵄩2 ⩽ C[τ1−α 󵄩󵄩󵄩z 0 󵄩󵄩󵄩 + ϵτα−1 󵄩󵄩󵄩δx e0 󵄩󵄩󵄩 + τ ∑ 󵄩󵄩󵄩(r7 )m 󵄩󵄩󵄩 + τ ∑ 󵄩󵄩󵄩δx2 (r8 )m 󵄩󵄩󵄩 ]

222 | 3 Difference methods for the time-fractional wave equations n

n

2

2

⩽ C{τ ∑ L[c7 (τ2 + h2 + ϵ)] + τ ∑ L(c8 τ2 ) } m=1

⩽

C(c72

+

c82 )LT(τ2

2

2

+ h + ϵ) ,

m=1

1 ⩽ n ⩽ N.

The proof ends.

3.6 The difference method based on L1 approximation for the MTTFW equations In this section, the finite difference method for solving a class of multiterm timefractional wave (MTTFW) equations will be introduced. For simplicity, take the twoterm case with the constant coefficients as an example. Consider the following problem of the two-term time-fractional wave equations: γ

γ

C 1 C { 0 Dt u(x, t) + 0 Dt u(x, t) = uxx (x, t) + f (x, t), { { { { { x ∈ (0, L), t ∈ (0, T], { { { u(x, 0) = φ(x), ut (x, 0) = ψ(x), x ∈ (0, L), { { { { u(0, t) = μ(t), u(L, t) = ν(t), t ∈ [0, T],

(3.199) (3.200) (3.201)

where 1 < γ1 < γ < 2, the functions f , φ, ψ, μ, ν are given, and φ(0) = μ(0), φ(L) = ν(0), ψ(0) = μ󸀠 (0), ψ(L) = ν󸀠 (0). Suppose u ∈ C (4,3) ([0, L] × [0, T]). Take the same mesh partition and notations as those in Section 3.1.

3.6.1 Derivation of the difference scheme Considering equation (3.199) at the point (xi , tn ), we have C γ1 0 Dt u(xi , tn )

γ

+ C0 Dt u(xi , tn ) = uxx (xi , tn ) + fin ,

1 ⩽ i ⩽ M − 1, 0 ⩽ n ⩽ N.

Taking an average on two adjacent time levels arrives at 1 1 C γ1 γ γ γ [ D u(xi , tn ) + C0 Dt 1 u(xi , tn−1 ) ] + [ C0 Dt u(xi , tn ) + C0 Dt u(xi , tn−1 )] 2 0 t 2 1 𝜕2 u 𝜕2 u n− 1 = [ 2 (xi , tn ) + 2 (xi , tn−1 )] + fi 2 , 1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N, 2 𝜕x 𝜕x n− 21

where fi

= 21 (fin + fin−1 ). It follows from Theorem 1.6.2 and Lemma 2.1.3 that n−1 τ1−γ1 n− 1 k− 1 (γ ) (γ1 ) (γ1 ) (γ1 ) [b0 1 δt Ui 2 − ∑ (bn−k−1 ψi ] − bn−k )δt Ui 2 − bn−1 Γ(3 − γ1 ) k=1

3.6 The difference method based on L1 approximation for the MTTFW equations | 223 n−1 τ1−γ n− 1 k− 1 (γ) (γ) (γ) (γ) [b0 δt Ui 2 − ∑ (bn−k−1 − bn−k )δt Ui 2 − bn−1 ψi ] Γ(3 − γ) k=1

+

n− 21

= δx2 Ui

n− 21

+ fi

n− 21

+ (r9 )i

1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N,

,

(3.202)

and there is a positive constant c9 such that 󵄨󵄨 n− 21 󵄨󵄨 3−γ 2 󵄨󵄨(r9 )i 󵄨󵄨 ⩽ c9 (τ + h ),

1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N.

(3.203)

Noticing the initial-boundary value conditions (3.200)–(3.201), we have {

Ui0 = φ(xi ), U0n

= μ(tn ),

1 ⩽ i ⩽ M − 1,

n UM

= ν(tn ),

(3.204) 0 ⩽ n ⩽ N.

(3.205)

n− 21

Omitting the small term (r9 )i in (3.202) and replacing the exact solution Uin with its numerical one uni produce a difference scheme for solving (3.199)–(3.201) as follows: { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { {

n−1 τ1−γ1 n− 1 k− 1 (γ ) (γ1 ) (γ1 ) (γ1 ) [b0 1 δt ui 2 − ∑ (bn−k−1 − bn−k )δt ui 2 − bn−1 ψi ] Γ(3 − γ1 ) k=1

+

n−1 τ1−γ k− 1 n− 1 (γ) (γ) (γ) (γ) [b0 δt ui 2 − ∑ (bn−k−1 − bn−k )δt ui 2 − bn−1 ψi ] Γ(3 − γ) k=1 n− 21

= δx2 ui u0i un0

n− 21

+ fi

= φ(xi ), = μ(tn ),

,

1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N,

1 ⩽ i ⩽ M − 1,

unM

= ν(tn ),

(3.206) (3.207)

0 ⩽ n ⩽ N.

(3.208)

Hereinafter, denote η = τγ−1 Γ(3 − γ),

η1 = τγ1 −1 Γ(3 − γ1 ).

3.6.2 Solvability of the difference scheme Theorem 3.6.1. The difference scheme (3.206)–(3.208) is uniquely solvable. Proof. Let un = (un0 , un1 , . . . , unM ). The value of u0 is determined by (3.207)–(3.208). Suppose that the values of u0 , u1 , . . . , un−1 have been uniquely determined. From (3.206) and (3.208), we can obtain the linear system in un . To prove its unique solvability, it suffices to show the corresponding homogeneous one 1 1 1 1 2 n n { { [ b0(γ1 ) + b(γ) 0 ]ui = 2 δx ui , τ η η { { n 1 n { u0 = uM = 0

1 ⩽ i ⩽ M − 1,

(3.209) (3.210)

224 | 3 Difference methods for the time-fractional wave equations has only the trivial solution. Making an inner product on both hand sides of (3.209) with un and noticing (3.210), we have 1 1 (γ1 ) 1 (γ) 󵄩󵄩 n 󵄩󵄩2 1 2 n n 1󵄩 󵄩2 [ b + b0 ]󵄩󵄩u 󵄩󵄩 = (δx u , u ) = − 󵄩󵄩󵄩δx un 󵄩󵄩󵄩 ⩽ 0, τ η1 0 η 2 2 thus ‖un ‖ = 0. The combination with (3.210) will arrive at un = 0. By the principle of induction, the difference scheme (3.206)–(3.208) is uniquely solvable. The proof ends.

3.6.3 Stability of difference scheme Theorem 3.6.2. Suppose {vin | 0 ⩽ i ⩽ M, 0 ⩽ n ⩽ N} is the solution of the difference scheme { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { {

n−1 1 n− 1 k− 1 (γ ) (γ1 ) (γ1 ) (γ1 ) [b0 1 δt vi 2 − ∑ (bn−k−1 − bn−k )δt vi 2 − bn−1 ψi ] η1 k=1

+

1 (γ) n− 21 n−1 (γ) k− 1 (γ) (γ) [b0 δt vi − ∑ (bn−k−1 − bn−k )δt vi 2 − bn−1 ψi ] η k=1 n− 21

= δx2 vi

vi0 v0n

n− 21

+ fi

= φ(xi ),

= 0,

n vM

1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N,

,

(3.211)

1 ⩽ i ⩽ M − 1,

(3.212)

= 0,

(3.213)

0 ⩽ n ⩽ N.

Then it holds 2−γ

2−γ

t 1 t 1 󵄩󵄩 n 󵄩󵄩2 󵄩󵄩 0 󵄩󵄩2 + n ]‖ψ‖2 + [tnγ1 −1 Γ(2 − γ1 ) 󵄩󵄩δx v 󵄩󵄩 ⩽ 󵄩󵄩δx v 󵄩󵄩 + [ n Γ(3 − γ1 ) Γ(3 − γ) 4 n

󵄩 1 󵄩2 + tnγ−1 Γ(2 − γ)] τ ∑ 󵄩󵄩󵄩f k− 2 󵄩󵄩󵄩 , k=1

1 ⩽ n ⩽ N,

(3.214)

1

where the definitions of ‖ψ‖ and ‖f k− 2 ‖ are the same as those in Theorem 3.1.2. 1

Proof. Taking an inner product on both hand sides of (3.211) with δt vn− 2 , it follows from the Cauchy–Schwarz inequality that [ =

1 1 1 (γ1 ) 1 (γ) 󵄩󵄩 n− 21 󵄩󵄩2 b0 + b0 ]󵄩󵄩δt v 󵄩󵄩 − (δx2 vn− 2 , δt vn− 2 ) η1 η

1 1 1 1 n−1 (γ1 ) (γ1 ) (γ1 ) [ ∑ (b − bn−k )(δt vk− 2 , δt vn− 2 ) + bn−1 (ψ, δt vn− 2 )] η1 k=1 n−k−1

3.6 The difference method based on L1 approximation for the MTTFW equations | 225

+

1 1 1 1 n−1 (γ) (γ) (γ) [ ∑ (bn−k−1 − bn−k )(δt vk− 2 , δt vn− 2 ) + bn−1 (ψ, δt vn− 2 )] η k=1 1

1

+ (f n− 2 , δt vn− 2 ) ⩽

1 2 1 2 1 1 n−1 (γ1 ) (γ1 ) 󵄩 󵄩 󵄩 󵄩 [ ∑ (bn−k−1 − bn−k )(󵄩󵄩󵄩δt vk− 2 󵄩󵄩󵄩 + 󵄩󵄩󵄩δt vn− 2 󵄩󵄩󵄩 ) η1 2 k=1 1 2 1 (γ1 ) 󵄩 󵄩 (‖ψ‖2 + 󵄩󵄩󵄩δt vn− 2 󵄩󵄩󵄩 )] + bn−1 2

+

1 2 1 2 1 1 n−1 (γ) (γ) 󵄩 󵄩 󵄩 󵄩 [ ∑ (b − bn−k )(󵄩󵄩󵄩δt vk− 2 󵄩󵄩󵄩 + 󵄩󵄩󵄩δt vn− 2 󵄩󵄩󵄩 ) η 2 k=1 n−k−1

1 2 1 (γ) 󵄩 󵄩 + bn−1 (‖ψ‖2 + 󵄩󵄩󵄩δt vn− 2 󵄩󵄩󵄩 )] 2 1

1

+ (f n− 2 , δt vn− 2 ),

1 ⩽ n ⩽ N,

which can be rearranged as [ ⩽

τ (γ1 ) τ (γ) 󵄩󵄩 n− 21 󵄩󵄩2 󵄩󵄩 n 󵄩󵄩2 󵄩󵄩 n−1 󵄩󵄩2 b + b0 ]󵄩󵄩δt v 󵄩󵄩 + 󵄩󵄩δx v 󵄩󵄩 − 󵄩󵄩δx v 󵄩󵄩 η1 0 η

1 2 τ n−1 (γ1 ) (γ1 ) 󵄩 (γ1 ) 󵄩 [ ∑ (bn−k−1 − bn−k )󵄩󵄩󵄩δt vk− 2 󵄩󵄩󵄩 + bn−1 ‖ψ‖2 ] η1 k=1

+

1 2 τ n−1 (γ) (γ) 󵄩 (γ) 󵄩 [ ∑ (bn−k−1 − bn−k )󵄩󵄩󵄩δt vk− 2 󵄩󵄩󵄩 + bn−1 ‖ψ‖2 ] η k=1 1

1

1

1

+ 2τ(f n− 2 , δt vn− 2 ),

1⩽n⩽N

(3.215)

by noticing that −(δx2 vn− 2 , δt vn− 2 ) =

1 󵄩󵄩 n 󵄩󵄩2 󵄩󵄩 n−1 󵄩󵄩2 (󵄩δ v 󵄩 − 󵄩δ v 󵄩󵄩 ). 2τ 󵄩 x 󵄩 󵄩 x

Let 󵄩 󵄩2 Q0 = 󵄩󵄩󵄩δx v0 󵄩󵄩󵄩 ,

(γ )

(γ)

n b b 1 1 2 󵄩 󵄩 󵄩 󵄩2 Qn = 󵄩󵄩󵄩δx vn 󵄩󵄩󵄩 + τ ∑ ( n−k + n−k )󵄩󵄩󵄩δt vk− 2 󵄩󵄩󵄩 , η η 1 k=1

1 ⩽ n ⩽ N.

Then (3.215) reads Qn ⩽ Qn−1 + τ(

(γ )

(γ)

1 1 1 bn−1 b + n−1 )‖ψ‖2 + 2τ(f n− 2 , δt vn− 2 ), η1 η

1 ⩽ n ⩽ N.

226 | 3 Difference methods for the time-fractional wave equations The application of the recursive process produces n

(γ )

bk 1

n−1

0

Q ⩽ Q + τ ∑(

η1

k=0

n−1

(γ)

+

󵄩 󵄩2 ⩽ 󵄩󵄩󵄩δx v0 󵄩󵄩󵄩 + τ ∑ (

bk

η

(γ ) bk 1

k=0 (γ )

η1

n

+

k=1

(γ) bk

b 1 󵄩 1 2 󵄩 + τ ∑ [ n−k 󵄩󵄩󵄩δt vk− 2 󵄩󵄩󵄩 + η n

k=1

1

n

(γ) bn−k 󵄩 󵄩

+ τ ∑[

󵄩δ v η 󵄩 t

k=1

1

1

)‖ψ‖2 + 2τ ∑ (f k− 2 , δt vk− 2 )

k− 21 󵄩 󵄩2

󵄩󵄩 +

η

)‖ψ‖2

η1 󵄩󵄩 k− 21 󵄩󵄩2 󵄩f 󵄩󵄩 ] (γ1 ) 󵄩 4bn−k η

󵄩󵄩 k− 21 󵄩󵄩2 󵄩󵄩 ], 󵄩f

(γ) 󵄩 4bn−k

1 ⩽ n ⩽ N.

Therefore, (γ )

(γ)

n n−1 b 1 b η 󵄩 k− 1 󵄩2 󵄩󵄩 n 󵄩󵄩2 󵄩󵄩 0 󵄩󵄩2 2 󵄩󵄩δx v 󵄩󵄩 ⩽ 󵄩󵄩δx v 󵄩󵄩 + τ ∑ ( k + k )‖ψ‖ + τ ∑ (γ1 ) 󵄩󵄩󵄩f 2 󵄩󵄩󵄩 1 η η 1 k=1 4b k=0 n

η

k=1

4bn−k

+τ∑

(γ)

n−k

󵄩󵄩 󵄩󵄩f

k− 21

󵄩2

󵄩󵄩 , 󵄩

1 ⩽ n ⩽ N.

(3.216)

By (3.20) and (3.22), the desired result (3.214) is attained from (3.216). The proof ends.

3.6.4 Convergence of the difference scheme Theorem 3.6.3. Suppose {Uin | 0 ⩽ i ⩽ M, 0 ⩽ n ⩽ N} and {uni | 0 ⩽ i ⩽ M, 0 ⩽ n ⩽ N} are solutions of the problem (3.199)–(3.201) and the difference scheme (3.206)–(3.208), respectively. Let ein = Uin − uni ,

0 ⩽ i ⩽ M, 0 ⩽ n ⩽ N,

then it holds 󵄩󵄩 n 󵄩󵄩 3−γ 2 󵄩󵄩e 󵄩󵄩∞ ⩽ κ(τ + h ),

1 ⩽ n ⩽ N,

where κ=

L √T γ1 Γ(2 − γ1 ) + T γ Γ(2 − γ) c9 . 4

3.7 The difference method based on L2-1σ approximation for the MTTFW equations | 227

Proof. The subtraction of (3.206)–(3.208) from (3.202), (3.204)–(3.205), respectively, produces the system of error equations as follows: { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { {

n−1 1 k− 1 n− 1 (γ1 ) (γ1 ) (γ1 ) (γ ) − bn−k )δt ei 2 − bn−1 ⋅ 0] [b0 1 δt ei 2 − ∑ (bn−k−1 η1 k=1

1 (γ) n− 21 n−1 (γ) k− 1 (γ) (γ) [b0 δt ei − ∑ (bn−k−1 − bn−k )δt ei 2 − bn−1 ⋅ 0] η k=1

+

n− 21

= δx2 ei ei0 e0n

= 0,

= 0,

n− 21

+ (r9 )i

,

1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N,

1 ⩽ i ⩽ M − 1,

n eM = 0,

0 ⩽ n ⩽ N.

Noticing (3.203) and applying Theorem 3.6.2 immediately arrive at n 󵄩 k− 1 󵄩2 󵄩󵄩 n 󵄩󵄩2 1 γ1 −1 γ−1 󵄩󵄩δx e 󵄩󵄩 ⩽ [tn Γ(2 − γ1 ) + tn Γ(2 − γ)]τ ∑ 󵄩󵄩󵄩(r9 ) 2 󵄩󵄩󵄩 4 k=1

⩽

1 γ1 2 [T Γ(2 − γ1 ) + T γ Γ(2 − γ)]Lc92 (τ3−γ + h2 ) , 4

1 ⩽ n ⩽ N.

Taking the square root on both hand sides of the inequality above and combining with Lemma 2.1.1 will get the desired conclusion. The proof ends.

3.7 The difference method based on L2-1σ approximation for the MTTFW equations Consider the following problem of the MTTFW equations: m

γ { { ∑ λr C D r u(x, t) = uxx (x, t) + f (x, t), x ∈ (0, L), t ∈ (0, T], { { { r=0 0 t { { u(x, 0) = φ(x), ut (x, 0) = ψ(x), x ∈ [0, L], { { { { u(0, t) = 0, u(L, t) = 0, t ∈ (0, T],

(3.217) (3.218) (3.219)

where λ0 , λ1 , . . . , λm are some positive constants, 1 ⩽ γm < γm−1 < ⋅ ⋅ ⋅ < γ0 ⩽ 2 and at least one γr belongs to (1, 2), and φ(0) = 0, φ(L) = 0, ψ(0) = 0, ψ(L) = 0. Suppose the exact solution u ∈ C (4,4) ([0, L] × [0, T]). 3.7.1 Derivation of the difference scheme For any grid function {uni | 0 ⩽ i ⩽ M, 0 ⩽ n ⩽ N} defined on Ωh × Ωτ , introduce the following notation: Dt ̄uni =

1 [(2σ + 1)uni − 4σun−1 + (2σ − 1)un−2 i i ], 2τ

0 ⩽ i ⩽ M, 2 ⩽ n ⩽ N.

228 | 3 Difference methods for the time-fractional wave equations Let v(x, t) = ut (x, t),

αr = γr − 1.

Thus, the problem (3.217)–(3.219) is equivalent to the following one: { { { { { { { { { { { { { { { { {

m

α

∑ λr C0 Dt r v(x, t) = uxx (x, t) + f (x, t),

r=0

ut (x, t) = v(x, t), u(x, 0) = φ(x), u(0, t) = 0,

x ∈ (0, L), t ∈ (0, T],

x ∈ [0, L], t ∈ (0, T], v(x, 0) = ψ(x),

u(L, t) = 0,

x ∈ [0, L],

t ∈ (0, T].

(3.220) (3.221) (3.222) (3.223)

Denote Uin = u(xi , tn ),

φi = φ(xi ),

Vin = v(xi , tn ),

ψi = ψ(xi ),

0 ⩽ i ⩽ M, 0 ⩽ n ⩽ N,

0 ⩽ i ⩽ M.

As that in Subsection 1.6.4, let tn−1+σ = (n − 1 + σ)τ, fin−1+σ = f (xi , tn−1+σ ) and σ is the root of F(σ) = 0. Considering (3.220) at the point (xi , tn−1+σ ), we have m

α

∑ λr C0 Dt r v(xi , tn−1+σ ) = uxx (xi , tn−1+σ ) + fin−1+σ ,

r=0

1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N.

(3.224)

Using the theory of Subsection 1.6.4 to approximate the Caputo time-fractional derivative, it follows that m

n−1

α

∑ λr C0 Dt r v(xi , tn−1+σ ) = ∑ ĉk(n,α) (Vin−k − Vin−k−1 ) + O(τ3−α0 ),

r=0

k=0

(3.225)

where m

ĉk(n,α) = ∑ λr ⋅ r=0

τ−αr (n,α ) c r, Γ(2 − αr ) k

0⩽k ⩽n−1

is defined by (1.93). For the spatial second-order derivative on the right-hand side of (3.224), using (3.85) and Lemma 2.1.3, we have uxx (xi , tn−1+σ ) = σuxx (xi , tn ) + (1 − σ)uxx (xi , tn−1 ) + O(τ2 )

= σδx2 Uin + (1 − σ)δx2 Uin−1 + O(h2 ) + O(τ2 ).

(3.226)

3.7 The difference method based on L2-1σ approximation for the MTTFW equations | 229

Substituting (3.225) and (3.226) into (3.224), we obtain n−1

∑ ĉk(n,α) (Vin−k − Vin−k−1 ) = σδx2 Uin + (1 − σ)δx2 Uin−1 + fin−1+σ + (r10 )ni ,

k=0

1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N,

(3.227)

and there exists a positive constant c10 such that 󵄨󵄨 n󵄨 2 2 󵄨󵄨(r10 )i 󵄨󵄨󵄨 ⩽ c10 (τ + h ),

1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N.

(3.228)

Considering (3.221) at the points (xi , t 1 ) and (xi , tn−1+σ ), respectively, we get 2

1

{

1

δt Ui2 = Vi2 + (r11 )1i ,

Dt ̄Uin

=

σVin

+ (1 −

0 ⩽ i ⩽ M,

σ)Vin−1

+

(r11 )ni ,

(3.229) 0 ⩽ i ⩽ M, 2 ⩽ n ⩽ N,

(3.230)

and there exists a positive constant c11 such that 󵄨󵄨 2 n󵄨 2 󵄨󵄨δx (r11 )i 󵄨󵄨󵄨 ⩽ c11 τ ,

1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N.

(3.231)

In addition, it is known from homogeneous boundary value condition (3.223) that (r11 )n0 = 0,

(r11 )nM = 0,

1 ⩽ n ⩽ N.

(3.232)

Noticing the initial-boundary value conditions (3.222)–(3.223), we have {

Ui0 = φi , U0n

= 0,

Vi0 = ψi ,

n UM

= 0,

0 ⩽ i ⩽ M, 1 ⩽ n ⩽ N.

(3.233) (3.234)

Omitting the small term (r10 )ni and (r11 )ni in (3.227), (3.229)–(3.230) and replacing the exact solution {Uin , Vin } with its numerical one {uni , vin } produce the difference scheme for (3.220)–(3.223) as follows: { { { { { { { { { { { { { { { { { { { { { { { { { { { { {

n−1

+ fin−1+σ , ∑ ĉk(n,α) (vin−k − vin−k−1 ) = σδx2 uni + (1 − σ)δx2 un−1 i

k=0

1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N,

1 2

1 2

δt ui = vi ,

Dt ̄uni u0i = un0 =

=

σvin

φi ,

0,

0 ⩽ i ⩽ M, + (1 −

vi0 unM

σ)vin−1 ,

= ψi ,

= 0,

(3.235) (3.236)

0 ⩽ i ⩽ M, 2 ⩽ n ⩽ N,

0 ⩽ i ⩽ M,

1 ⩽ n ⩽ N.

(3.237) (3.238) (3.239)

Remark 3.7.1. By Lemma 3.4.1, it is easy to get an integral expression of (r11 )ni , from which one can obtain (3.231).

230 | 3 Difference methods for the time-fractional wave equations Remark 3.7.2. For the nonhomogeneous boundary value problem m

γ { { ∑ λr C D r u(x, t) = uxx (x, t) + f (x, t), x ∈ (0, L), t ∈ (0, T], { { { r=0 0 t { { u(x, 0) = φ(x), ut (x, 0) = ψ(x), x ∈ [0, L], { { { { u(0, t) = μ(t), u(L, t) = ν(t), t ∈ (0, T],

let v(x, t) = ut (x, t), then one has an equivalent problem { { { { { { { { { { { { { { { { { { { { { { {

m

α

∑ λr C0 Dt r v(x, t) = uxx (x, t) + f (x, t),

r=0

uxxt (x, t) = vxx (x, t), u(x, 0) = φ(x),

u(0, t) = μ(t),

x ∈ (0, L), t ∈ (0, T],

v(x, 0) = ψ(x),

u(L, t) = ν(t),

󸀠

v(0, t) = μ (t),

x ∈ (0, L), t ∈ (0, T],

󸀠

x ∈ [0, L], t ∈ (0, T],

v(L, t) = ν (t),

t ∈ (0, T].

The following difference scheme can be obtained: { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { {

n−1

+ fin−1+σ , ∑ ĉk(n,α) (vin−k − vin−k−1 ) = σδx2 uni + (1 − σ)δx2 un−1 i

k=0

1

1

1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N,

δx2 δt ui2 = δx2 vi2 , 1 ⩽ i ⩽ M − 1, δx2 Dt ̄uni = δx2 (σvin + (1 − σ)vin−1 ), 1 ⩽ i ⩽ u0i = φi , vi0 = ψi , 0 ⩽ i ⩽ M, un0 = μ(tn ), unM = ν(tn ), 1 ⩽ n ⩽ N, n v0n = μ󸀠 (tn ), vM = ν󸀠 (tn ), 1 ⩽ n ⩽ N.

M − 1, 2 ⩽ n ⩽ N,

Interested readers are recommended to refer to [83]. Another way is firstly to make the boundary conditions homogeneous and then consider the problem along the line in this subsection. 3.7.2 Solvability of the difference scheme Theorem 3.7.1. The difference scheme (3.235)–(3.239) is uniquely solvable. Proof. Let un = (un0 , un1 , . . . , unM ),

n vn = (v0n , v1n , . . . , vM ).

3.7 The difference method based on L2-1σ approximation for the MTTFW equations | 231

(I) The value of {u0 , v0 } is determined by (3.238). (II) It follows from (3.236) that 1

1

vi1 = 2vi2 − vi0 = 2δt ui2 − vi0 ,

0 ⩽ i ⩽ M.

(3.240)

Substituting (3.240) into (3.235), and noticing (3.239), we can obtain the linear system in u1 1

{

ĉ0(1,α) (2δt ui2 − 2vi0 ) = σδx2 u1i + (1 − σ)δx2 u0i + fiσ , u10

= 0,

u1M

1 ⩽ i ⩽ M − 1,

= 0.

(3.241) (3.242)

Consider its homogeneous system: 2 { ĉ0(1,α) u1i = σδx2 u1i , { τ1 1 { u0 = 0, uM = 0.

1 ⩽ i ⩽ M − 1,

(3.243) (3.244)

Making the inner product on both hand sides of (3.243) with u1 and using (3.244), we have 2 1,α 󵄩󵄩 1 󵄩󵄩2 󵄩 󵄩2 ĉ 󵄩u 󵄩 + σ 󵄩󵄩󵄩δx u1 󵄩󵄩󵄩 = 0, τ 0 󵄩 󵄩 which implies u1 = 0. Then the system (3.241)–(3.242) has a unique solution u1 . Once u1 is obtained, v1 can be got from (3.240). (III) Now assume that the value of {u0 , v0 , u1 , v1 , . . . , un−1 , vn−1 } has been uniquely determined, then it follows from (3.237) that vin =

1 (D ̄un − (1 − σ)vin−1 ), σ t i

0 ⩽ i ⩽ M.

(3.245)

Substituting (3.245) into (3.235), and noticing (3.239), the linear system in un can be obtained as { { { { { { { { { { { { {

n−1 1 ĉ0(n,α) [ (Dt ̄uni − (1 − σ)vin−1 ) − vin−1 ] + ∑ ĉk(n,α) (vin−k − vin−k−1 ) σ k=1

= σδx2 uni + (1 − σ)δx2 un−1 + fin−1+σ , i

un0

= 0,

unM

1 ⩽ i ⩽ M − 1,

= 0.

To show its unique solvability, it suffices to verify that the corresponding homogeneous one 1 2σ + 1 n { ĉ0(n,α) ⋅ ⋅ ui = σδx2 uni , σ 2τ { n n { u0 = 0, uM = 0 has only the trivial solution.

1 ⩽ i ⩽ M − 1,

(3.246) (3.247)

232 | 3 Difference methods for the time-fractional wave equations Taking the inner product on both hand sides of (3.246) with un and noticing (3.247), we have 2σ + 1 (n,α) 󵄩󵄩 n 󵄩󵄩2 󵄩 n 󵄩2 ĉ 󵄩u 󵄩 + σ 󵄩󵄩󵄩δx u 󵄩󵄩󵄩 = 0, 2στ 0 󵄩 󵄩 thus un = 0. The system has a unique solution. Once un is obtained, vn is followed from (3.245). By the principle of induction, the theorem is true. The proof ends.

3.7.3 Stability of the difference scheme Theorem 3.7.2. Suppose {un , vn | 0 ⩽ n ⩽ N} is the solution of the difference scheme { { { { { { { { { { { { { { { { { { { { { { { { { { { { {

n−1

+ pni , ∑ ĉk(n,α) (vin−k − vin−k−1 ) = σδx2 uni + (1 − σ)δx2 un−1 i

k=0

1 2

1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N,

1 2

δt ui = vi + Dt ̄uni u0i = un0 =

=

σvin

φi ,

0,

qi1 ,

0 ⩽ i ⩽ M,

+ (1 −

vi0 unM

(3.248)

σ)vin−1

= ψi ,

+

(3.249)

qin ,

0 ⩽ i ⩽ M, 2 ⩽ n ⩽ N,

0 ⩽ i ⩽ M,

= 0,

(3.250) (3.251)

1 ⩽ n ⩽ N,

(3.252)

n where q0n = qM = 0 (1 ⩽ n ⩽ N) and φ0 = φM = ψ0 = ψM = 0. Then there exists a constant C such that n

󵄩 󵄩2 󵄩 󵄩2 τ ∑ 󵄩󵄩󵄩vk 󵄩󵄩󵄩 + 󵄩󵄩󵄩δx un 󵄩󵄩󵄩 k=1

n

n

l=1

l=1

󵄩 󵄩2 󵄩 󵄩2 󵄩 󵄩2 󵄩 󵄩2 ⩽ C[τĉ0(1,α) 󵄩󵄩󵄩v0 󵄩󵄩󵄩 + 󵄩󵄩󵄩δx u0 󵄩󵄩󵄩 + τ ∑󵄩󵄩󵄩pl 󵄩󵄩󵄩 + τ ∑󵄩󵄩󵄩δx2 ql 󵄩󵄩󵄩 ],

1 ⩽ n ⩽ N,

(3.253)

where M−1

󵄩󵄩 k 󵄩󵄩2 k 2 󵄩󵄩v 󵄩󵄩 = h ∑ (vi ) , i=1

M−1

󵄩󵄩 k 󵄩󵄩2 k 2 󵄩󵄩p 󵄩󵄩 = h ∑ (pi ) , ĉ0(1,α)

m

i=1

M−1

󵄩󵄩 2 k 󵄩󵄩2 2 k 2 󵄩󵄩δx q 󵄩󵄩 = h ∑ (δx qi ) , i=1

τ−αr = ∑ λr ⋅ σ 1−αr . Γ(2 − α ) r r=0

n Proof. It follows from (3.249)–(3.252) and q0n = qM = 0 (1 ⩽ n ⩽ N) together with φ0 = φM = ψ0 = ψM = 0 that

v0n = 0,

n vM = 0,

0 ⩽ n ⩽ N.

3.7 The difference method based on L2-1σ approximation for the MTTFW equations | 233

Consequently, 1

{

1

v02 = 0, vM2 = 0, σv0n

+ (1 −

σ)v0n−1

= 0,

n σvM

+ (1 −

n−1 σ)vM

(3.254) = 0,

2 ⩽ n ⩽ N.

(3.255)

In addition, it follows from (3.251)–(3.252) with φ0 = φM = 0 that {

σu10 + (1 − σ)u00 = 0, σu1M + (1 − σ)u0M = 0,

σun0

+ (1 −

σ)un−1 0

σunM

= 0,

+ (1 −

σ)un−1 M

(3.256)

= 0,

2 ⩽ n ⩽ N.

(3.257)

(I) When n = 1, equation (3.248) reads ĉ0(1,α) (vi1 − vi0 ) = σδx2 u1i + (1 − σ)δx2 u0i + p1i ,

1 ⩽ i ⩽ M − 1.

(3.258)

1

Making the inner product on both hand sides of (3.258) with v 2 and noticing (3.254), we arrive at 1 1 1 (1,α) 󵄩󵄩 1 󵄩󵄩2 󵄩󵄩 0 󵄩󵄩2 ĉ0 (󵄩󵄩v 󵄩󵄩 − 󵄩󵄩v 󵄩󵄩 ) = −(δx (σu1 + (1 − σ)u0 ), δx v 2 ) + (p1 , v 2 ). 2

(3.259)

It follows from (3.249) that 1

1

δt δx2 ui2 = δx2 vi2 + δx2 qi1 ,

1 ⩽ i ⩽ M − 1.

(3.260)

Making the inner product on both hand sides of (3.260) with −(σu1 + (1 − σ)u0 ) and noticing (3.256), we get 1 (δ u1 − δx u0 , δx (σu1 + (1 − σ)u0 )) τ x 1

= (δx v 2 , δx (σu1 + (1 − σ)u0 )) − (δx2 q1 , σu1 + (1 − σ)u0 ). Adding (3.259) and (3.261) yields that 1 1 (1,α) 󵄩󵄩 1 󵄩󵄩2 󵄩󵄩 0 󵄩󵄩2 ĉ (󵄩v 󵄩 − 󵄩v 󵄩 ) + (δx u1 − δx u0 , δx (σu1 + (1 − σ)u0 )) 2 0 󵄩 󵄩 󵄩 󵄩 τ 1

= (p1 , v 2 ) − (δx2 q1 , σu1 + (1 − σ)u0 ). Noticing the fact

(δx u1 − δx u0 , δx (σu1 + (1 − σ)u0 )) =

󵄩󵄩 1 󵄩󵄩󵄩2 σ 󵄩󵄩 1 󵄩󵄩2 3σ 2σ − 1 󵄩 δx u1 − √ δx u0 󵄩󵄩󵄩 󵄩󵄩δx u 󵄩󵄩 + (2σ − 1)󵄩󵄩󵄩 √ 󵄩󵄩 2 2σ − 1 󵄩󵄩 4 3σ −

σ 2 − σ + 1 󵄩󵄩 0 󵄩󵄩2 󵄩󵄩δx u 󵄩󵄩 , 3σ

(3.261)

234 | 3 Difference methods for the time-fractional wave equations we obtain 2 1 (1,α) 󵄩󵄩 1 󵄩󵄩2 󵄩󵄩 0 󵄩󵄩2 1 σ󵄩 󵄩2 σ − σ + 1 󵄩󵄩 0 󵄩󵄩2 ĉ0 (󵄩󵄩v 󵄩󵄩 − 󵄩󵄩v 󵄩󵄩 ) + ( 󵄩󵄩󵄩δx u1 󵄩󵄩󵄩 − 󵄩󵄩δx u 󵄩󵄩 ) 2 τ 4 3σ 1

⩽ (p1 , v 2 ) − (δx2 q1 , σu1 + (1 − σ)u0 ), which can be reduced to

2 τ (1,α) 󵄩󵄩 1 󵄩󵄩2 󵄩󵄩 0 󵄩󵄩2 σ󵄩 󵄩2 σ − σ + 1 󵄩󵄩 0 󵄩󵄩2 ĉ0 (󵄩󵄩v 󵄩󵄩 − 󵄩󵄩v 󵄩󵄩 ) + 󵄩󵄩󵄩δx u1 󵄩󵄩󵄩 − 󵄩󵄩δx u 󵄩󵄩 2 4 3σ 1

⩽ τ(p1 , v 2 ) − τ(δx2 q1 , σu1 + (1 − σ)u0 ) 󵄩 󵄩 󵄩 1󵄩 󵄩 󵄩 󵄩 󵄩 ⩽ τ󵄩󵄩󵄩p1 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩v 2 󵄩󵄩󵄩 + τ󵄩󵄩󵄩δx2 q1 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩σu1 + (1 − σ)u0 󵄩󵄩󵄩

1󵄩 1 1 󵄩 2 1 󵄩2 ε 󵄩 1 󵄩2 1 󵄩󵄩 1 󵄩󵄩2 0 󵄩2 ⩽ τ[ 󵄩󵄩󵄩v 2 󵄩󵄩󵄩 + 󵄩p 󵄩 ] + τ[ 󵄩󵄩󵄩σu + (1 − σ)u 󵄩󵄩󵄩 + 󵄩󵄩󵄩δx q 󵄩󵄩󵄩 ] 2 2ε 󵄩 󵄩 2 2 ε 󵄩 󵄩2 󵄩 󵄩2 1 󵄩 󵄩2 󵄩 󵄩2 1 󵄩 󵄩2 1󵄩 󵄩2 ⩽ τ[ (󵄩󵄩󵄩v1 󵄩󵄩󵄩 + 󵄩󵄩󵄩v0 󵄩󵄩󵄩 ) + 󵄩󵄩󵄩p1 󵄩󵄩󵄩 ] + τ[ (󵄩󵄩󵄩u1 󵄩󵄩󵄩 + 󵄩󵄩󵄩u0 󵄩󵄩󵄩 ) + 󵄩󵄩󵄩δx2 q1 󵄩󵄩󵄩 ]. 4 2ε 2 2

Noticing m

ĉ0(1,α) = ∑ λr ⋅ r=0

m τ−αr τ−αr (1,α ) c0 r = ∑ λr ⋅ σ 1−αr = O(τ−α0 ) Γ(2 − αr ) Γ(2 − α ) r r=0

and then taking ε = ĉ0(1,α) , there exists a constant C1 such that 󵄩 󵄩2 󵄩 󵄩2 τĉ0(1,α) 󵄩󵄩󵄩v1 󵄩󵄩󵄩 + 󵄩󵄩󵄩δx u1 󵄩󵄩󵄩 󵄩 󵄩2 󵄩 󵄩2 󵄩 󵄩2 ⩽ C1 (τĉ0(1,α) 󵄩󵄩󵄩v0 󵄩󵄩󵄩 + 󵄩󵄩󵄩δx u0 󵄩󵄩󵄩 + τ󵄩󵄩󵄩u0 󵄩󵄩󵄩 󵄩 󵄩2 󵄩 󵄩2 󵄩 󵄩2 + τ󵄩󵄩󵄩u1 󵄩󵄩󵄩 + τ1+α0 󵄩󵄩󵄩p1 󵄩󵄩󵄩 + τ󵄩󵄩󵄩δx2 q1 󵄩󵄩󵄩 ).

(3.262)

(II) Making the inner product on both hand sides of (3.248) with σvn + (1 − σ)vn−1 and noticing (3.255), we have n−1

∑ ĉk(n,α) (vn−k − vn−k−1 , σvn + (1 − σ)vn−1 )

k=0

= (δx2 (σun + (1 − σ)un−1 ), σvn + (1 − σ)vn−1 ) + (pn , σvn + (1 − σ)vn−1 ) = −(δx (σun + (1 − σ)un−1 ), δx (σvn + (1 − σ)vn−1 )) + (pn , σvn + (1 − σ)vn−1 ),

1 ⩽ n ⩽ N.

By Lemma 2.6.1, the term on the left-hand side of (3.263) can be estimated as n−1

∑ ĉk(n,α) (vn−k − vn−k−1 , σvn + (1 − σ)vn−1 )

k=0

(3.263)

3.7 The difference method based on L2-1σ approximation for the MTTFW equations | 235

⩾

1 n−1 (n,α) 󵄩󵄩 n−k 󵄩󵄩2 󵄩󵄩 n−k−1 󵄩󵄩2 (󵄩󵄩v 󵄩󵄩 − 󵄩󵄩v ∑ ĉ 󵄩󵄩 ). 2 k=0 k

Thus, 1 n−1 (n,α) 󵄩󵄩 n−k 󵄩󵄩2 󵄩󵄩 n−k−1 󵄩󵄩2 (󵄩󵄩v 󵄩󵄩 − 󵄩󵄩v ∑ ĉ 󵄩󵄩 ) 2 k=0 k

⩽ −(δx (σun + (1 − σ)un−1 ), δx (σvn + (1 − σ)vn−1 )) + (pn , σvn + (1 − σ)vn−1 ),

2 ⩽ n ⩽ N.

(3.264)

It follows from (3.250) that Dt ̄δx2 uni = δx2 (σvin + (1 − σ)vin−1 ) + δx2 qin ,

1 ⩽ i ⩽ M − 1, 2 ⩽ n ⩽ N.

Making the inner product on both hand sides of the equality above with −(σun + (1 − σ)un−1 ) and noticing (3.257) lead to (Dt ̄δx un , δx (σun + (1 − σ)un−1 ))

= (δx (σvn + (1 − σ)vn−1 ), δx (σun + (1 − σ)un−1 )) − (δx2 qn , σun + (1 − σ)un−1 ),

2 ⩽ n ⩽ N.

(3.265)

Denote 󵄩 󵄩2 󵄩 󵄩2 F n = (2σ + 1)󵄩󵄩󵄩δx un 󵄩󵄩󵄩 − (2σ − 1)󵄩󵄩󵄩δx un−1 󵄩󵄩󵄩 󵄩 󵄩2 + (2σ 2 + σ − 1)󵄩󵄩󵄩δx (un − un−1 )󵄩󵄩󵄩 , n ⩾ 1. It follows from Lemma 3.4.2 that Fn ⩾

1 󵄩󵄩 n 󵄩󵄩2 󵄩δ u 󵄩 , σ󵄩 x 󵄩

n⩾1

(3.266)

and (Dt ̄δx un , δx (σun + (1 − σ)un−1 )) ⩾

1 n (F − F n−1 ), 4τ

n ⩾ 2.

(3.267)

Combining (3.265) with (3.267) gets 1 n (F − F n−1 ) ⩽ (δx (σvn + (1 − σ)vn−1 ), δx (σun + (1 − σ)un−1 )) 4τ − (δx2 qn , σun + (1 − σ)un−1 ), 2 ⩽ n ⩽ N. Adding (3.264) and (3.268) yields 1 n−1 (n,α) 󵄩󵄩 n−k 󵄩󵄩2 󵄩󵄩 n−k−1 󵄩󵄩2 1 n (󵄩󵄩v 󵄩󵄩 − 󵄩󵄩v (F − F n−1 ) ∑ ĉ 󵄩󵄩 ) + 2 k=0 k 4τ

(3.268)

236 | 3 Difference methods for the time-fractional wave equations ⩽ (pn , σvn + (1 − σ)vn−1 ) − (δx2 qn , σun + (1 − σ)un−1 ) 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 ⩽ 󵄩󵄩󵄩pn 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩σvn + (1 − σ)vn−1 󵄩󵄩󵄩 + 󵄩󵄩󵄩δx2 qn 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩σun + (1 − σ)un−1 󵄩󵄩󵄩,

2 ⩽ n ⩽ N.

Noticing the fact that n−1

󵄩2 󵄩 󵄩2 󵄩 ∑ ĉk(n,α) (󵄩󵄩󵄩vn−k 󵄩󵄩󵄩 − 󵄩󵄩󵄩vn−k−1 󵄩󵄩󵄩 )

k=0 n−1

n−2

k=0

k=0

󵄩 󵄩2 󵄩 󵄩2 = ∑ ĉk(n,α) 󵄩󵄩󵄩vn−k 󵄩󵄩󵄩 − ∑ ĉk(n−1,α) 󵄩󵄩󵄩vn−1−k 󵄩󵄩󵄩 n−2

󵄩 󵄩2 (n,α) 󵄩 󵄩󵄩v0 󵄩󵄩󵄩2 − ∑ (ĉk(n,α) − ĉk(n−1,α) )󵄩󵄩󵄩vn−1−k 󵄩󵄩󵄩 − ĉn−1 󵄩 󵄩 k=0

n−1

n−2

󵄩 󵄩2 󵄩 󵄩2 = ∑ ĉk(n,α) 󵄩󵄩󵄩vn−k 󵄩󵄩󵄩 − ∑ ĉk(n−1,α) 󵄩󵄩󵄩vn−1−k 󵄩󵄩󵄩 k=0

−

(n,α) (ĉn−2

−

k=0 󵄩2 (n−1,α) 󵄩 ĉn−2 )󵄩󵄩󵄩v1 󵄩󵄩󵄩

(n,α) 󵄩 󵄩󵄩v0 󵄩󵄩󵄩2 , − ĉn−1 󵄩 󵄩

we have 1 n−1 (n,α) 󵄩󵄩 n−k 󵄩󵄩2 n−2 (n−1,α) 󵄩󵄩 n−1−k 󵄩󵄩2 1 n ( ∑ ĉ (F − F n−1 ) 󵄩v 󵄩󵄩 − ∑ ĉk 󵄩󵄩v 󵄩󵄩 ) + 2 k=0 k 󵄩 4τ k=0

1 (n,α) 󵄩2 (n−1,α) 󵄩 (n,α) 󵄩 󵄩󵄩v0 󵄩󵄩󵄩2 ] + 󵄩󵄩󵄩pn 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩σvn + (1 − σ)vn−1 󵄩󵄩󵄩 − ĉn−2 )󵄩󵄩󵄩v1 󵄩󵄩󵄩 + ĉn−1 ⩽ [(ĉn−2 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 2 󵄩󵄩 2 n 󵄩󵄩 󵄩󵄩 n n−1 󵄩 + 󵄩󵄩δx q 󵄩󵄩 ⋅ 󵄩󵄩σu + (1 − σ)u 󵄩󵄩󵄩, 2 ⩽ n ⩽ N.

(3.269)

Replacing the superscript n with l, and summing up for l from 2 to n on both hand sides of (3.269), we obtain 1 n−1 (n,α) 󵄩󵄩 n−k 󵄩󵄩2 1 n 󵄩 󵄩2 ( ∑ ĉk 󵄩󵄩v 󵄩󵄩 − ĉ0(1,α) 󵄩󵄩󵄩v1 󵄩󵄩󵄩 ) + (F − F 1 ) 2 k=0 4τ n 1 n (l,α) 󵄩2 (l−1,α) 󵄩 (l,α) 󵄩 󵄩󵄩v0 󵄩󵄩󵄩2 ) − ĉl−2 )󵄩󵄩󵄩v1 󵄩󵄩󵄩 + ∑ ĉl−1 ⩽ (∑(ĉl−2 󵄩 󵄩 2 l=2 l=2 n

n

l=2

l=2

󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 + ∑󵄩󵄩󵄩pl 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩σvl + (1 − σ)vl−1 󵄩󵄩󵄩 + ∑󵄩󵄩󵄩δx2 ql 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩σul + (1 − σ)ul−1 󵄩󵄩󵄩,

2 ⩽ n ⩽ N.

(3.270)

Multiplying both hand sides of (3.270) by 4τ and arranging the inequality, we have n−1

󵄩 󵄩2 2τ ∑ ĉk(n,α) 󵄩󵄩󵄩vn−k 󵄩󵄩󵄩 + F n k=0

n

n

l=2

l=2

󵄩 󵄩2 󵄩2 (l,α) (l−1,α) 󵄩 (l,α) 󵄩 󵄩󵄩v0 󵄩󵄩󵄩2 )󵄩󵄩󵄩v1 󵄩󵄩󵄩 + 2τ ∑ ĉl−1 ⩽ F 1 + 2τĉ0(1,α) 󵄩󵄩󵄩v1 󵄩󵄩󵄩 + 2τ ∑(ĉl−2 − ĉl−2 󵄩 󵄩

3.7 The difference method based on L2-1σ approximation for the MTTFW equations | 237 n

n

l=2

l=2

󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 + 4τ ∑󵄩󵄩󵄩pl 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩σvl + (1 − σ)vl−1 󵄩󵄩󵄩 + 4τ ∑󵄩󵄩󵄩δx2 ql 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩σul + (1 − σ)ul−1 󵄩󵄩󵄩,

2 ⩽ n ⩽ N. (3.271)

It follows from Lemma 1.6.6 that m

(n,α) (n,α) ĉ1(n,α) > ĉ2(n,α) > ⋅ ⋅ ⋅ > ĉn−2 > ĉn−1 > ∑ λr r=0

T −αr . Γ(1 − αr )

(3.272)

By Lemma 3.4.3, we get n

n

m

(l,α) (l−1,α) τ ∑(ĉl−2 − ĉl−2 ) = τ ∑ ∑ λr ⋅ l=2 r=0 m

l=2

= τ ∑ λr ⋅ r=0 m

⩽ ∑ λr ⋅ r=0 m

τ−αr (l,α ) (l−1,α ) (c r − cl−2 r ) Γ(2 − αr ) l−2

n τ−αr (l,α ) (l−1,α ) ∑(cl−2 r − cl−2 r ) Γ(2 − αr ) l=2

1 − αr αr τ1−αr [ ( + 1)σ −αr ] Γ(2 − αr ) 12 σ

α λr τ1−αr ⋅ ( r + 1)σ −αr 12 Γ(1 − αr ) σ r=0

= ∑

(3.273)

and n

n

m

(l,α) τ ∑ ĉl−1 = τ ∑ ∑ λr ⋅ l=2 r=0 m

l=2

= ∑ λr ⋅ r=0 m

= ∑ λr ⋅ ⩽

r=0 m

τ−αr (l,α ) c r Γ(2 − αr ) l−1

n τ1−αr (l,α ) ∑ cl−1 r Γ(2 − αr ) l=2

τ1−αr 3 (n − 1 + σ)1−αr Γ(2 − αr ) 2

3 T 1−αr . ∑ λr ⋅ 2 r=0 Γ(2 − αr )

The application of (3.272)–(3.274) in (3.271) leads to m

2 ∑ λr r=0

n−1 T −αr 󵄩 󵄩2 ⋅ τ ∑ 󵄩󵄩󵄩vn−k 󵄩󵄩󵄩 + F n Γ(1 − αr ) k=0

m α λ τ1−αr 󵄩 󵄩2 󵄩 󵄩2 ( r + 1)σ −αr 󵄩󵄩󵄩v1 󵄩󵄩󵄩 ⩽ F 1 + 2τĉ0(1,α) 󵄩󵄩󵄩v1 󵄩󵄩󵄩 + 2 ∑ r ⋅ 12 Γ(1 − αr ) σ r=0 m

+ 3 ∑ λr ⋅ r=0 n

n T 1−αr 󵄩󵄩 0 󵄩󵄩2 󵄩 l󵄩 󵄩 l l−1 󵄩 󵄩󵄩v 󵄩󵄩 + 4τ ∑󵄩󵄩󵄩p 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩σv + (1 − σ)v 󵄩󵄩󵄩 Γ(2 − αr ) l=2

󵄩 󵄩 󵄩 󵄩 + 4τ ∑󵄩󵄩󵄩δx2 ql 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩σul + (1 − σ)ul−1 󵄩󵄩󵄩 l=2

(3.274)

238 | 3 Difference methods for the time-fractional wave equations m λ α τ1−αr 󵄩 󵄩2 󵄩 󵄩2 ⩽ F 1 + 2τĉ0(1,α) 󵄩󵄩󵄩v1 󵄩󵄩󵄩 + ∑ r ⋅ ( r + 1)σ −αr 󵄩󵄩󵄩v1 󵄩󵄩󵄩 6 Γ(1 − α ) σ r r=0 m

+ 3 ∑ λr ⋅ r=0 n

T 1−αr 󵄩󵄩 0 󵄩󵄩2 󵄩v 󵄩 Γ(2 − αr ) 󵄩 󵄩

󵄩 󵄩2 1 󵄩󵄩 l 󵄩󵄩2 + 2τ ∑[ε0 󵄩󵄩󵄩σvl + (1 − σ)vl−1 󵄩󵄩󵄩 + 󵄩p 󵄩 ] ε0 󵄩 󵄩 l=2 n

󵄩2 󵄩 󵄩2 󵄩 + 2τ ∑[󵄩󵄩󵄩σul + (1 − σ)ul−1 󵄩󵄩󵄩 + 󵄩󵄩󵄩δx2 ql 󵄩󵄩󵄩 ] l=2

m λ α τ1−αr 󵄩 󵄩2 󵄩 󵄩2 ⩽ F 1 + 2τĉ0(1,α) 󵄩󵄩󵄩v1 󵄩󵄩󵄩 + ∑ r ⋅ ( r + 1)σ −αr 󵄩󵄩󵄩v1 󵄩󵄩󵄩 6 Γ(1 − α ) σ r r=0 m

+ 3 ∑ λr ⋅ r=0

n T 1−αr 󵄩󵄩 0 󵄩󵄩2 󵄩 l 󵄩2 󵄩 l−1 󵄩2 󵄩󵄩v 󵄩󵄩 + 2τε0 ∑(󵄩󵄩󵄩v 󵄩󵄩󵄩 + 󵄩󵄩󵄩v 󵄩󵄩󵄩 ) Γ(2 − αr ) l=2

n 2τ n 󵄩󵄩 l 󵄩󵄩2 󵄩 󵄩2 󵄩 󵄩2 󵄩 󵄩2 + ∑󵄩󵄩p 󵄩󵄩 + 2τ ∑(󵄩󵄩󵄩ul 󵄩󵄩󵄩 + 󵄩󵄩󵄩ul−1 󵄩󵄩󵄩 + 󵄩󵄩󵄩δx2 ql 󵄩󵄩󵄩 ), ε0 l=2 l=2

Taking ε0 =

1 4

∑m r=0 λr

T −αr , Γ(1−αr )

2 ⩽ n ⩽ N.

noticing (3.262), (3.266) and

󵄩 󵄩2 󵄩 󵄩2 F 1 = (2σ + 1)󵄩󵄩󵄩δx u1 󵄩󵄩󵄩 − (2σ − 1)󵄩󵄩󵄩δx u0 󵄩󵄩󵄩 󵄩 󵄩2 + (2σ 2 + σ − 1)󵄩󵄩󵄩δx (u1 − u0 )󵄩󵄩󵄩 󵄩 󵄩2 󵄩 󵄩2 ⩽ (2σ + 1)󵄩󵄩󵄩δx u1 󵄩󵄩󵄩 − (2σ − 1)󵄩󵄩󵄩δx u0 󵄩󵄩󵄩 󵄩 󵄩2 󵄩 󵄩2 + 2(2σ 2 + σ − 1)(󵄩󵄩󵄩δx u1 󵄩󵄩󵄩 + 󵄩󵄩󵄩δx u0 󵄩󵄩󵄩 ) 󵄩 󵄩2 󵄩 󵄩2 = (4σ 2 + 4σ − 1)󵄩󵄩󵄩δx u1 󵄩󵄩󵄩 + (4σ 2 − 1)󵄩󵄩󵄩δx u0 󵄩󵄩󵄩 , it is easy to know that there exists a positive constant C2 such that n−1

󵄩 󵄩2 󵄩 󵄩2 τ ∑ 󵄩󵄩󵄩vn−k 󵄩󵄩󵄩 + 󵄩󵄩󵄩δx un 󵄩󵄩󵄩 k=0

n

n

n

󵄩 󵄩2 󵄩 󵄩2 󵄩 󵄩2 󵄩 󵄩2 󵄩 󵄩2 ⩽ C2 (τĉ0(1,α) 󵄩󵄩󵄩v0 󵄩󵄩󵄩 + 󵄩󵄩󵄩δx u0 󵄩󵄩󵄩 + τ ∑󵄩󵄩󵄩ul 󵄩󵄩󵄩 + τ ∑󵄩󵄩󵄩pl 󵄩󵄩󵄩 + τ ∑󵄩󵄩󵄩δx2 ql 󵄩󵄩󵄩 ) l=1 n

l=1

2

l=1

n

L 󵄩 󵄩 󵄩2 󵄩 󵄩2 󵄩 󵄩2 󵄩2 ⩽ C2 (τĉ0(1,α) 󵄩󵄩󵄩v0 󵄩󵄩󵄩 + 󵄩󵄩󵄩δx u0 󵄩󵄩󵄩 + τ ∑ 󵄩󵄩󵄩δx ul 󵄩󵄩󵄩 + τ ∑󵄩󵄩󵄩pl 󵄩󵄩󵄩 6 l=1 l=1 n

󵄩 󵄩2 + τ ∑󵄩󵄩󵄩δx2 ql 󵄩󵄩󵄩 ), l=1

1 ⩽ n ⩽ N.

The application of the Gronwall inequality will lead to (3.253). The proof ends.

3.7 The difference method based on L2-1σ approximation for the MTTFW equations | 239

3.7.4 Convergence of the difference scheme Theorem 3.7.3. Suppose {Uin , Vin | 0 ⩽ i ⩽ M, 0 ⩽ n ⩽ N} and {uni , vin | 0 ⩽ i ⩽ M, 0 ⩽ n ⩽ N} are solutions of the problem (3.220)–(3.223) and the difference scheme (3.235)–(3.239), respectively. Let ein = Uin − uni ,

zin = Vin − vin ,

0 ⩽ i ⩽ M, 0 ⩽ n ⩽ N,

then it holds n

2 󵄩 󵄩2 󵄩 󵄩2 2 2 τ ∑ 󵄩󵄩󵄩z k 󵄩󵄩󵄩 + 󵄩󵄩󵄩δx en 󵄩󵄩󵄩 ⩽ C(c10 + c11 )LT(τ2 + h2 ) , k=1

1 ⩽ n ⩽ N,

where the constant C is defined in Theorem 3.7.2. Proof. Subtracting (3.235)–(3.239) from (3.227), (3.229)–(3.230), and (3.233)–(3.234) respectively, produces the system of error equations as follows: n−1

{ { ∑ ĉk(n,α) (zin−k − zin−k−1 ) = σδx2 ein + (1 − σ)δx2 ein−1 + (r10 )ni , { { { { k=0 { { { { 1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N, { { { 1 1 { 1 2 2 δt ei = zi + (r11 )i , 0 ⩽ i ⩽ M, { { { { { Dt ̄ein = σzin + (1 − σ)zin−1 + (r11 )ni , 0 ⩽ i ⩽ M, 2 ⩽ n ⩽ N, { { { { { { ei0 = 0, zi0 = 0, 0 ⩽ i ⩽ M, { { { n n { e0 = 0, eM = 0, 1 ⩽ n ⩽ N. Noticing (3.228) and (3.231)–(3.232), an immediate consequence of Theorem 3.7.2 into the system above is n

󵄩 󵄩2 󵄩 󵄩2 τ ∑ 󵄩󵄩󵄩z k 󵄩󵄩󵄩 + 󵄩󵄩󵄩δx en 󵄩󵄩󵄩 k=1

n

n

l=1

l=1

󵄩 󵄩2 󵄩 󵄩2 󵄩 󵄩2 󵄩 󵄩2 ⩽ C[τĉ0(1,α) 󵄩󵄩󵄩z 0 󵄩󵄩󵄩 + 󵄩󵄩󵄩δx e0 󵄩󵄩󵄩 + τ ∑󵄩󵄩󵄩(r10 )l 󵄩󵄩󵄩 + τ ∑󵄩󵄩󵄩δx2 (r11 )l 󵄩󵄩󵄩 ] n

2

n

2

⩽ C[τ ∑ L[c10 (τ2 + h2 )] + τ ∑ L(c11 τ2 ) ] l=1

⩽ The proof ends.

2 C(c10

+

2 c11 )LT(τ2

2 2

+h ) ,

l=1

1 ⩽ n ⩽ N.

240 | 3 Difference methods for the time-fractional wave equations

3.8 The difference method for the time-fractional mixed diffusion and wave equations Consider the difference method by using L1 approximation for the following initialboundary value problem of the time-fractional mixed diffusion and wave equation γ

C C α { 0 Dt u(x, t) + 0 Dt u(x, t) = uxx (x, t) + f (x, t), { { { { { x ∈ (0, L), t ∈ (0, T], { { { u(x, 0) = φ(x), ut (x, 0) = ψ(x), x ∈ (0, L), { { { { u(0, t) = μ(t), u(L, t) = ν(t), t ∈ [0, T],

(3.275) (3.276) (3.277)

where the functions f , φ, ψ, μ and ν are given, φ(0) = μ(0), φ(L) = ν(0), ψ(0) = μ󸀠 (0), γ ψ(L) = ν󸀠 (0), C0 Dt u(x, t) and C0 Dαt u(x, t) denote the time Caputo derivatives of u given by, respectively, C α 0 Dt u(x, t)

t

1 𝜕u(x, s) ds = , ∫ Γ(1 − α) 𝜕s (t − s)α

0 < α < 1,

0

and C γ 0 Dt u(x, t)

t

=

1 ds 𝜕2 u(x, s) , ∫ Γ(2 − γ) 𝜕s2 (t − s)γ−1

1 < γ < 2.

0

Suppose the exact solution u ∈ C (4,3) ([0, L] × [0, T]).

3.8.1 Derivation of the difference scheme Denote Uin = u(xi , tn ),

n− 21

fi

1 = [f (xi , tn ) + f (xi , tn−1 )]. 2

Considering equation (3.275) at the points (xi , tn ) and (xi , tn−1 ), respectively, and averaging the corresponding results yield 1 C γ 1 γ [ 0 Dt u(xi , tn ) + C0 Dt u(xi , tn−1 )] + [ C0 Dαt u(xi , tn ) + C0 Dαt u(xi , tn−1 )] 2 2 1 n− 21 = [uxx (xi , tn ) + uxx (xi , tn−1 )] + fi , 1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N. 2 With the help of Theorem 1.6.1 and Theorem 1.6.2, we have C α 0 Dt u(xi , tn )

(3.278)

3.8 The difference method for the time-fractional mixed diffusion and wave equations |

= =

241

n−1 τ−α n 0 2−α [a(α) U − )Uik − a(α) ) ∑ (a(α) − a(α) n−1 Ui ] + O(τ n−k Γ(2 − α) 0 i k=1 n−k−1

τ1−α n (α) k− 1 ∑ an−k δt Ui 2 + O(τ2−α ), Γ(2 − α) k=1

1 ⩽ i ⩽ M − 1, 0 ⩽ n ⩽ N

(3.279)

and 1 C γ γ [ D u(xi , tn ) + C0 Dt u(xi , tn−1 )] 2 0 t =

n−1 τ1−γ k− 1 n− 1 (γ) (γ) (γ) (γ) [b0 δt Ui 2 − ∑ (bn−k−1 − bn−k )δt Ui 2 − bn−1 ut (xi , t0 )] Γ(3 − γ) k=1

+ O(τ3−γ ),

1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N.

(3.280)

Applying (3.279) and (3.280) into (3.278), we obtain n−1 τ1−γ k− 1 n− 1 (γ) (γ) (γ) (γ) [b0 δt Ui 2 − ∑ (bn−k−1 − bn−k )δt Ui 2 − bn−1 ψi ] Γ(3 − γ) k=1

+

n−1 a(α) + a(α) a(α) τ1−α n− 1 k− 1 n−k−1 [ 0 δt Ui 2 + ∑ n−k δt Ui 2 ] Γ(2 − α) 2 2 k=1 n− 21

= δx2 Ui

n− 21

n− 21

+ fi

+ (r12 )i

,

1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N

(3.281)

and there exists a positive constant c12 such that n− 1 󵄨 󵄨󵄨 min{2−α,3−γ} + h2 ), 󵄨󵄨(r12 )i 2 󵄨󵄨󵄨 ⩽ c12 (τ

1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N.

(3.282)

For simplicity, denote sγ = τγ−1 Γ(3 − γ) and sα = τα−1 Γ(2 − α). Noticing the initial-boundary value conditions (3.276)–(3.277), we have u0i = φ(xi ),

{

un0

= μ(tn ),

1 ⩽ i ⩽ M − 1,

unM

= ν(tn ),

(3.283) 0 ⩽ n ⩽ N.

(3.284)

n− 1

Dropping (r12 )i 2 in (3.281) and replacing Uin by uni , we construct a finite difference scheme for solving the problem (3.275)–(3.277) as follows: { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { {

n−1 1 n− 1 k− 1 (γ) (γ) (γ) (γ) [b0 δt ui 2 − ∑ (bn−k−1 − bn−k )δt ui 2 − bn−1 ψi ] sγ k=1 (α) n−1 a(α) + a(α) 1 a0 n− 21 k− 1 n−1−k + [ δt ui + ∑ n−k δt ui 2 ] sα 2 2 k=1 n− 21

= δx2 ui

u0i un0

n− 21

+ fi

= φ(xi ), = μ(tn ),

,

1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N,

1 ⩽ i ⩽ M − 1,

unM

= ν(tn ),

(3.285) (3.286)

0 ⩽ n ⩽ N.

(3.287)

242 | 3 Difference methods for the time-fractional wave equations 3.8.2 Solvability of the difference scheme Theorem 3.8.1. The difference scheme (3.285)–(3.287) is uniquely solvable. Proof. Let un = (un0 , un1 , . . . , unM ). The value of u0 is determined by (3.286)–(3.287). Now assume that the values of u0 , u1 , . . . , un−1 have been uniquely determined, then the linear system in un can be obtained from (3.285) and (3.287). To show its unique solvability, it suffices to verify that the corresponding homogeneous one (γ)

b a(α) 1 1 { { ( 0 + 0 ) uni − δx2 uni = 0, s 2s τ 2 γ α { { n n { u0 = uM = 0

1 ⩽ i ⩽ M − 1,

has only the trivial solution. By the maximum principle, the conclusion is obvious. The proof ends. 3.8.3 Stability of the difference scheme Before analyzing the stability of the difference scheme, the following three lemmas are provided at first. Lemma 3.8.1. erties

[52]

Let {g0 , g1 , . . . , gn , . . .} be a sequence of real numbers with the propgn ⩾ 0,

gn − gn−1 ⩽ 0,

gn+1 − 2gn + gn−1 ⩾ 0.

Then for any positive integer m and for any mesh functions V1 , V2 , . . . , Vm ∈ 𝒰h̊ , it holds that m

n−1

∑ ( ∑ gp Vn−p , Vn ) ⩾ 0.

n=1 p=0 (γ)

Lemma 3.8.2. Let {bl } be defined in (1.64). For any mesh functions ψ, V1 , V2 , . . . , Vm ∈ 𝒰h̊ , it holds that m

n−1

∑ (b0 V n − ∑ (bn−k−1 − bn−k )V k − bn−1 ψ, V n ) (γ)

n=1

(γ)

(γ)

k=1

m

m 1 (γ) 󵄩 (γ) 󵄩2 ⩾ ( ∑ bm−k 󵄩󵄩󵄩V k 󵄩󵄩󵄩 − ∑ bn−1 ‖ψ‖2 ). 2 k=1 n=1

(γ)

3.8 The difference method for the time-fractional mixed diffusion and wave equations | 243

Proof. m

n−1

∑ (b0 V n − ∑ (bn−k−1 − bn−k )V k − bn−1 ψ, V n ) (γ)

n=1

(γ)

(γ)

(γ)

k=1

m

n−1

n=1

k=1

(γ) (γ) (γ) (γ) 󵄩 󵄩2 = ∑ [b0 󵄩󵄩󵄩V n 󵄩󵄩󵄩 − ∑ (bn−k−1 − bn−k )(V k , V n ) − bn−1 (ψ, V n )] m

n−1 (γ) 󵄩 (γ) (γ) 󵄩2 1 󵄩 󵄩2 󵄩 󵄩2 ⩾ ∑ [b0 󵄩󵄩󵄩V n 󵄩󵄩󵄩 − ∑ (bn−k−1 − bn−k )(󵄩󵄩󵄩V k 󵄩󵄩󵄩 + 󵄩󵄩󵄩V n 󵄩󵄩󵄩 ) 2 n=1 k=1

1 (γ) 󵄩 󵄩2 − bn−1 (‖ψ‖2 + 󵄩󵄩󵄩V n 󵄩󵄩󵄩 )] 2 =

1 m n (γ) 󵄩󵄩 k 󵄩󵄩2 n−1 (γ) 󵄩󵄩 k 󵄩󵄩2 (γ) 2 ∑ ( ∑ b 󵄩V 󵄩 − ∑ b 󵄩V 󵄩 − bn−1 ‖ψ‖ ) 2 n=1 k=1 n−k 󵄩 󵄩 k=1 n−k−1 󵄩 󵄩

1 m (γ) 󵄩 󵄩2 m (γ) = ( ∑ bm−k 󵄩󵄩󵄩V k 󵄩󵄩󵄩 − ∑ bn−1 ‖ψ‖2 ). 2 k=1 n=1 Lemma 3.8.3. [77] Let {a(α) } be defined in (1.57). For any mesh functions V1 , V2 , . . . , Vm ∈ l ̊ 𝒰h , it holds that m

∑(

n=1

n−1 a(α) + a(α) m a(α) 󵄩󵄩 n 󵄩󵄩2 n−1−k k 0 V n + ∑ n−k V , V n ) ⩾ −a(α) ∑ 󵄩󵄩V 󵄩󵄩 . 0 2 2 n=1 k=1

Proof. Notice the fact that m

∑(

n=1

n−1 a(α) + a(α) a(α) n−1−k k 0 V n + ∑ n−k V , V n) 2 2 k=1

m n−1 a(α) + a(α) m 3 󵄩󵄩 n 󵄩󵄩2 n n n−k n−k−1 k = ∑ ( a(α) a(α) V + V , V ) − ∑ ∑ 󵄩V 󵄩󵄩 . 0 0 󵄩 2 n=1 2 n=1 k=1

By some plain calculations, the coefficients of the first term on the right-hand side of the equality above satisfy the conditions of Lemma 3.8.1, hence it is nonnegative, which implies m

∑(

n=1

n−1 a(α) + a(α) m a(α) 󵄩󵄩 n 󵄩󵄩2 n−1−k k 0 V n + ∑ n−k V , V n ) ⩾ −a(α) 󵄩V 󵄩󵄩 . 0 ∑󵄩 2 2 n=1 k=1

The proof ends. Remark 3.8.1. In Lemma 3.8.3, a minor error in [77] has been corrected. We have the following result on the stability of the difference scheme.

244 | 3 Difference methods for the time-fractional wave equations Theorem 3.8.2. Let {uni | 0 ⩽ i ⩽ M, 0 ⩽ n ⩽ N} be the solution of the difference scheme (3.285)–(3.287), in which μ(t) ≡ 0, ν(t) ≡ 0. Denote τ0 = (

1/(1−α)

T 1−γ Γ(2 − α) ) 4Γ(2 − γ)

,

then, if τ ⩽ τ0 , it holds that T 2−γ 󵄩󵄩 n 󵄩󵄩2 󵄩󵄩 0 󵄩󵄩2 ‖ψ‖2 󵄩󵄩δx u 󵄩󵄩 ⩽ 󵄩󵄩δx u 󵄩󵄩 + Γ(3 − γ) n

󵄩2 󵄩 + 2Γ(2 − γ)T γ−1 τ ∑ 󵄩󵄩󵄩f k− 2 󵄩󵄩󵄩 , 1

k=1

1 ⩽ n ⩽ N. 1

Proof. Taking the inner product on both hand sides of (3.285) with δt un− 2 and summing up for n from 1 to m, yield that 1 1 1 m (γ) 󵄩󵄩 n− 21 󵄩󵄩2 n−1 (γ) (γ) ∑ [b 󵄩δ u 󵄩󵄩 − ∑ (bn−k−1 − bn−k )(δt uk− 2 , δt un− 2 ) sγ n=1 0 󵄩 t k=1 1

− bn−1 (ψ, δt un− 2 )] (γ)

+ + m

(α) n−1 a(α) + a(α) 1 1 1 1 m a0 n−k−1 δt un− 2 + ∑ n−k δt uk− 2 , δt un− 2 ) ∑( sα n=1 2 2 k=1

1 󵄩󵄩 m 󵄩󵄩2 󵄩󵄩 0 󵄩󵄩2 (󵄩δ u 󵄩 − 󵄩δ u 󵄩 ) 2τ 󵄩 x 󵄩 󵄩 x 󵄩 1

1

= ∑ (f n− 2 , δt un− 2 ), n=1

1 ⩽ m ⩽ N.

By Lemma 3.8.2 and Lemma 3.8.3, we can obtain m m 1 2 1 (γ) (γ) 󵄩 󵄩 ( ∑ bm−k 󵄩󵄩󵄩δt uk− 2 󵄩󵄩󵄩 − ∑ bn−1 ‖ψ‖2 ) 2sγ k=1 n=1

− m

1 m 󵄩󵄩 n− 21 󵄩󵄩2 1 󵄩󵄩 m 󵄩󵄩2 󵄩󵄩 0 󵄩󵄩2 ∑ 󵄩δ u 󵄩󵄩 + (󵄩󵄩δx u 󵄩󵄩 − 󵄩󵄩δx u 󵄩󵄩 ) sα n=1󵄩 t 2τ 1

1

⩽ ∑ (f n− 2 , δt un− 2 ), n=1

1 ⩽ m ⩽ N.

Further we obtain 1 m (γ) 󵄩󵄩 k− 21 󵄩󵄩2 ∑ b 󵄩δ u 󵄩󵄩 4sγ k=1 m−k 󵄩 t +(

1 m (γ) 󵄩󵄩 k− 21 󵄩󵄩2 1 m 󵄩󵄩 k− 21 󵄩󵄩2 ∑ b 󵄩δ u 󵄩󵄩 − ∑ 󵄩δ u 󵄩󵄩 ) 4sγ k=1 m−k 󵄩 t sα k=1󵄩 t

3.8 The difference method for the time-fractional mixed diffusion and wave equations | 245

+ ⩽

1 󵄩󵄩 m 󵄩󵄩2 󵄩󵄩 0 󵄩󵄩2 (󵄩δ u 󵄩 − 󵄩δ u 󵄩 ) 2τ 󵄩 x 󵄩 󵄩 x 󵄩

m 1 1 1 m (γ) ∑ bn−1 ‖ψ‖2 + ∑ (f n− 2 , δt un− 2 ), 2sγ n=1 n=1

1 ⩽ m ⩽ N.

(3.288)

For the second term on the left-hand side of (3.288), we have 1 m (γ) 󵄩󵄩 k− 21 󵄩󵄩2 1 m 󵄩󵄩 k− 21 󵄩󵄩2 ∑ b 󵄩δ u 󵄩󵄩 − ∑ 󵄩δ u 󵄩󵄩 4sγ k=1 m−k 󵄩 t sα k=1󵄩 t ⩾

m 1 2 T 1−γ 1 m 󵄩󵄩 k− 21 󵄩󵄩2 󵄩 󵄩 ∑ 󵄩δ u 󵄩󵄩 ∑ 󵄩󵄩󵄩δt uk− 2 󵄩󵄩󵄩 − 4Γ(2 − γ) k=1 sα k=1󵄩 t

=(

m 1 2 τ1−α T 1−γ 󵄩 󵄩 − ) ∑ 󵄩󵄩󵄩δt uk− 2 󵄩󵄩󵄩 , 4Γ(2 − γ) Γ(2 − α) k=1

1 ⩽ m ⩽ N,

(3.289)

in which the inequality (γ)

bl

= (2 − γ)(l + θl )1−γ ⩾ (2 − γ)N 1−γ ,

θl ∈ (0, 1), 0 ⩽ l ⩽ N − 1

was used. A careful observation of (3.289) shows that when τ ⩽ τ0 , the right-hand side of (3.289) is nonnegative. Thus it follows from (3.288) that m 1 2 T 1−γ 1 󵄩 󵄩 󵄩2 󵄩 󵄩2 󵄩 ∑ 󵄩󵄩󵄩δt uk− 2 󵄩󵄩󵄩 + (󵄩󵄩󵄩δx um 󵄩󵄩󵄩 − 󵄩󵄩󵄩δx u0 󵄩󵄩󵄩 ) 4Γ(2 − γ) k=1 2τ

⩽ = ⩽

m 1 1 1 m (γ) ∑ bn−1 ‖ψ‖2 + ∑ (f n− 2 , δt un− 2 ) 2sγ n=1 n=1 m 1 1 1 2−γ m ‖ψ‖2 + ∑ (f n− 2 , δt un− 2 ) 2sγ n=1

m 1 2−γ T 1−γ 󵄩󵄩 n− 21 󵄩󵄩2 γ−1 󵄩 n− 1 󵄩2 m ‖ψ‖2 + ∑ ( 󵄩󵄩δt u 󵄩󵄩 + Γ(2 − γ)T 󵄩󵄩󵄩f 2 󵄩󵄩󵄩 ), 2sγ n=1 4Γ(2 − γ)

1 ⩽ m ⩽ N.

Consequently, we have m τ2−γ 󵄩󵄩 m 󵄩󵄩2 󵄩󵄩 0 󵄩󵄩2 󵄩 1 󵄩2 m2−γ ‖ψ‖2 + 2Γ(2 − γ)T γ−1 τ ∑ 󵄩󵄩󵄩f n− 2 󵄩󵄩󵄩 󵄩󵄩δx u 󵄩󵄩 ⩽ 󵄩󵄩δx u 󵄩󵄩 + Γ(3 − γ) n=1

󵄩 󵄩2 ⩽ 󵄩󵄩󵄩δx u0 󵄩󵄩󵄩 +

m T 2−γ 󵄩 1 󵄩2 ‖ψ‖2 + 2Γ(2 − γ)T γ−1 τ ∑ 󵄩󵄩󵄩f n− 2 󵄩󵄩󵄩 , Γ(3 − γ) n=1

1 ⩽ m ⩽ N.

This completes the proof. 3.8.4 Convergence of the difference scheme With the help of Theorem 3.8.2, we can directly derive the following convergence theorem.

246 | 3 Difference methods for the time-fractional wave equations Theorem 3.8.3. Suppose that {Uin | 0 ⩽ i ⩽ M, 0 ⩽ n ⩽ N} is the solution of the problem (3.275)–(3.277) and {uni | 0 ⩽ i ⩽ M, 0 ⩽ n ⩽ N} is the solution the difference scheme (3.285)–(3.287), respectively. Denote ein = Uin − uni ,

0 ⩽ i ⩽ M, 0 ⩽ n ⩽ N.

Then, if τ ⩽ τ0 , we have 󵄩󵄩 n 󵄩󵄩 √ min{2−α,3−γ} + h2 ), 󵄩󵄩δx e 󵄩󵄩 ⩽ 2Γ(2 − γ)T γ Lc12 (τ

1 ⩽ n ⩽ N,

where c12 is defined in (3.282) and τ0 is defined in Theorem 3.8.2. Proof. Subtracting (3.285)–(3.287) from (3.281) and (3.283)–(3.284), respectively, we obtain the system of error equations as follows: { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { {

n−1 1 k− 1 n− 1 (γ) (γ) (γ) (γ) (b0 δt ei 2 − ∑ (bn−k−1 − bn−k )δt ei 2 − bn−1 ⋅ 0) sγ k=1

+

(α) n−1 a(α) + a(α) 1 a0 n− 1 k− 1 n−1−k ( δt ei 2 + ∑ n−k δt ei 2 ) sα 2 2 k=1 n− 21

= δx2 ei ei0 e0n

= 0,

= 0,

n− 21

+ (r12 )i

,

1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N,

1 ⩽ i ⩽ M − 1,

n eM = 0,

0 ⩽ n ⩽ N.

Combining (3.282) with Theorem 3.8.2, we obtain n

󵄩󵄩 n 󵄩󵄩2 󵄩󵄩 0 󵄩󵄩2 󵄩 γ−1 k− 1 󵄩2 󵄩󵄩δx e 󵄩󵄩 ⩽ 󵄩󵄩δx e 󵄩󵄩 + 2Γ(2 − γ)T τ ∑ 󵄩󵄩󵄩(r12 ) 2 󵄩󵄩󵄩 k=1

⩽ 2Γ(2 − γ)T

γ−1

nτL[c12 (τ

min{2−α,3−γ}

2

+ h2 )] 2

2 ⩽ 2Γ(2 − γ)T γ Lc12 (τmin{2−α,3−γ} + h2 ) ,

1 ⩽ n ⩽ N.

This completes the proof.

3.9 The ADI method based on L1 approximation for 2D problem Consider the following problem of the two-dimensional time-fractional wave equations: γ

C { 0 Dt u(x, y, t) = uxx (x, y, t) + uyy (x, y, t) + f (x, y, t), { { { { { (x, y) ∈ Ω, t ∈ (0, T], { { { u(x, y, 0) = φ(x, y), ut (x, y, 0) = ψ(x, y), (x, y) ∈ Ω, { { { { u(x, y, t) = μ(x, y, t), (x, y) ∈ 𝜕Ω, t ∈ [0, T],

(3.290) (3.291) (3.292)

3.9 The ADI method based on L1 approximation for 2D problem

| 247

where Ω = (0, L1 ) × (0, L2 ), γ ∈ (1, 2), the functions f , φ, ψ, μ are given and when (x, y) ∈ 𝜕Ω, μ(x, y, 0) = φ(x, y), μt (x, y, 0) = ψ(x, y). Suppose the exact solution u ∈ C (4,4,3) (Ω̄ × [0, T]). Take the same mesh partition and notations as those in Section 2.10. For any mesh function v = {vijk | (i, j) ∈ ω,̄ 0 ⩽ k ⩽ N} defined on Ωh × Ωτ , let k− 21

vij

1 = (vijk + vijk−1 ), 2

k− 21

δt vij

=

1 k (v − vijk−1 ). τ ij

Introduce the same mesh function spaces 𝒱h and 𝒱̊h as those in Section 2.10. Denote Uijn = u(xi , yj , tn ),

fijn = f (xi , yj , tn ),

ψij = ψ(xi , yj ),

(i, j) ∈ ω,̄ 0 ⩽ n ⩽ N.

3.9.1 Derivation of the difference scheme Considering equation (3.290) at the point (xi , yj , tn ), we have C γ 0 Dt u(xi , yj , tn )

= uxx (xi , yj , tn ) + uyy (xi , yj , tn ) + fijn ,

(i, j) ∈ ω, 0 ⩽ n ⩽ N.

Taking an average on two adjacent time levels and applying the L1 formula (1.69) and the second-order central difference quotient to approximate the time-fractional derivative and the spatial second-order derivative, respectively, we have from Theorem 1.6.2 and Lemma 2.1.3 that n−1 τ1−γ k− 1 n− 1 (γ) (γ) (γ) (γ) [b0 δt Uij 2 − ∑ (bn−k−1 − bn−k )δt Uij 2 − bn−1 ψij ] Γ(3 − γ) k=1 n− 21

= δx2 Uij

n− 21

+ δy2 Uij

n− 21

+ fij

n− 1

+ (r13 )ij 2 ,

(i, j) ∈ ω, 1 ⩽ n ⩽ N,

(3.293)

and there is positive constant c13 such that n− 1 󵄨 󵄨󵄨 3−γ 2 2 󵄨󵄨(r13 )ij 2 󵄨󵄨󵄨 ⩽ c13 (τ + h1 + h2 ),

(i, j) ∈ ω, 1 ⩽ n ⩽ N,

(γ)

where {bl } is defined in (1.64). Adding a small term

n− 1 Γ(3−γ) 1+γ 2 2 τ δx δy δt Uij 2 4

into both hand sides of (3.293) produces

n−1 τ1−γ n− 1 k− 1 (γ) (γ) (γ) (γ) [b0 δt Uij 2 − ∑ (bn−k−1 − bn−k )δt Uij 2 − bn−1 ψij ] Γ(3 − γ) k=1

+

Γ(3 − γ) 1+γ 2 2 n− 1 n− 1 n− 1 τ δx δy δt Uij 2 = δx2 Uij 2 + δy2 Uij 2 4

248 | 3 Difference methods for the time-fractional wave equations n− 21

+ fij

n− 1

+ (r14 )ij 2 ,

(i, j) ∈ ω, 1 ⩽ n ⩽ N,

(3.294)

where n− 21

(r14 )ij

n− 21

= (r13 )ij

+

Γ(3 − γ) 1+γ 2 2 n− 1 τ δx δy δt Uij 2 , 4

and there is a positive constant c14 such that n− 1 󵄨 󵄨󵄨 3−γ 2 2 󵄨󵄨(r14 )ij 2 󵄨󵄨󵄨 ⩽ c14 (τ + h1 + h2 ),

(i, j) ∈ ω, 1 ⩽ n ⩽ N.

(3.295)

Noticing the initial-boundary value conditions (3.291)–(3.292), we have {

Uij0 = φ(xi , yj ), Uijn

(i, j) ∈ ω,

= μ(xi , yj , tn ),

(3.296)

(i, j) ∈ 𝜕ω, 0 ⩽ n ⩽ N.

(3.297)

n− 1

Omitting the small term (r14 )ij 2 in (3.294) and replacing the exact solution Uijn with its numerical one unij give the difference scheme for (3.290)–(3.292) as follows: { { { { { { { { { { { { { { { { { { { { { { { { { { {

n−1 τ1−γ k− 1 n− 1 (γ) (γ) (γ) (γ) [b0 δt uij 2 − ∑ (bn−k−1 − bn−k )δt uij 2 − bn−1 ψij ] Γ(3 − γ) k=1

+ u0ij unij

Γ(3 − γ) 1+γ 2 2 n− 21 n− 1 n− 1 n− 1 τ δx δy δt uij = δx2 uij 2 + δy2 uij 2 + fij 2 , 4 (i, j) ∈ ω, 1 ⩽ n ⩽ N,

(3.298)

= φ(xi , yj ),

(3.299)

= μ(xi , yj , tn ),

(i, j) ∈ ω, (i, j) ∈ 𝜕ω, 0 ⩽ n ⩽ N.

(3.300)

Denote η = τγ−1 Γ(3 − γ). Equation (3.298) can be rearranged as n− 21

δt uij n−1

n− 21

− ηδx2 uij

(γ)

n− 21

− ηδy2 uij

+

k− 21

(γ)

(γ)

= ∑ (bn−k−1 − bn−k )δt uij k=1

1 2 2 2 2 n− 21 η τ δx δy δt uij 4 n− 21

+ bn−1 ψij + ηfij

,

or η 2 n η 2 n 1 2 2 2 2 n τδ u − τδ u + η τ δx δy uij 2 x ij 2 y ij 4 η 2 n−1 η 2 n−1 1 2 2 2 2 n−1 = un−1 ij + τδx uij + τδy uij + η τ δx δy uij 2 2 4 unij −

n−1

(γ)

(γ)

k− 21

+ τ ∑ (bn−k−1 − bn−k )δt uij k=1

(γ)

n− 21

+ τbn−1 ψij + τηfij

,

3.9 The ADI method based on L1 approximation for 2D problem

| 249

which is (ℐ −

η 2 η η η τδx )(ℐ − τδy2 )unij = (ℐ + τδx2 )(ℐ + τδy2 )un−1 ij 2 2 2 2

n−1

(γ)

(γ)

k− 21

+ τ ∑ (bn−k−1 − bn−k )δt uij k=1

(γ)

n− 21

+ τbn−1 ψij + τηfij

.

Let u∗ij = (ℐ −

η 2 n τδ )u , 2 y ij

then the difference scheme (3.298)–(3.300) can be written as the following ADI form: On each time level t = tn (1 ⩽ n ⩽ N), firstly, for any fixed j from 1 to M2 − 1, solve a series of linear systems in {u∗ij | 0 ⩽ i ⩽ M1 } in x direction η η η { { (ℐ − τδx2 )u∗ij = (ℐ + τδx2 )(ℐ + τδy2 )un−1 { ij { 2 2 2 { { { { n−1 { { k− 21 n− 21 (γ) (γ) (γ) )δ u − b + τ (b + τb ψ + τηf , ∑ t ij n−1 { ij ij n−k n−k−1 { { k=1 { { { { { { { u∗ = (ℐ − η τδ2 )un , u∗ = (ℐ − η τδ2 )un M1 ,j 0j 2 y 0j 2 y M1 ,j {

1 ⩽ i ⩽ M1 − 1,

to obtain the value of {u∗ij | 1 ⩽ i ⩽ M1 − 1} on the intermediate time level. Then, for any fixed i from 1 to M1 − 1, carry out some calculations for the unknown {unij | 0 ⩽ j ⩽ M2 } in y direction η { { (ℐ − τδy2 )unij = u∗ij , 1 ⩽ j ⩽ M2 − 1, 2 { { n n { ui0 = μ(xi , y0 , tn ), ui,M2 = μ(xi , yM2 , tn ) to produce the desired value of {unij | 1 ⩽ j ⩽ M2 − 1}. Next, the corresponding theoretical analysis on the difference scheme (3.298)– (3.300) will be studied. 3.9.2 Solvability of the difference scheme Theorem 3.9.1. The difference scheme (3.298)–(3.300) is uniquely solvable.

250 | 3 Difference methods for the time-fractional wave equations Proof. Let ̄ un = {unij | (i, j) ∈ ω}. The value of u0 is determined by (3.299)–(3.300). Now assume that the values of u0 , u1 , . . . , un−1 have been uniquely determined, then the linear system in un is obtained from (3.298) and (3.300). To show its unique solvability, it is sufficient to verify that the corresponding homogeneous one 1 n Γ(3 − γ) γ 2 2 n 1 2 n { u + τ δx δy uij = (δx uij + δy2 unij ), { ητ ij 4 2 { { n { uij = 0, (i, j) ∈ 𝜕ω

(i, j) ∈ ω,

(3.301) (3.302)

has only the trivial solution. Making the inner product on both hand sides of (3.301) with un , it follows from (2.200)–(2.202) that Γ(3 − γ) γ 1 n n 1 (u , u ) + τ (δx δy un , δx δy un ) = − [(δx un , δx un ) + (δy un , δy un )], ητ 4 2 that is, 1 󵄩󵄩 n 󵄩󵄩2 Γ(3 − γ) γ 󵄩󵄩 1󵄩 󵄩2 󵄩2 τ 󵄩󵄩δx δy un 󵄩󵄩󵄩 = − 󵄩󵄩󵄩∇h un 󵄩󵄩󵄩 ⩽ 0, 󵄩u 󵄩 + ητ 󵄩 󵄩 4 2 thus ‖un ‖ = 0. Noticing (3.302), we have un = 0. By the principle of induction, the difference scheme (3.298)–(3.300) is uniquely solvable. The proof ends. 3.9.3 Stability of the difference scheme Theorem 3.9.2. Suppose {vijn | (i, j) ∈ ω,̄ 0 ⩽ n ⩽ N} is the solution of the difference scheme { { { { { { { { { { { { { { { { { { { { { { { { { { {

1 (γ) n− 21 n−1 (γ) k− 1 (γ) (γ) [b0 δt vij − ∑ (bn−k−1 − bn−k )δt vij 2 − bn−1 ψij ] η k=1 + vij0 vijn

τ2 2 2 n− 21 n− 1 n− 1 n− 1 ηδx δy δt vij = δx2 vij 2 + δy2 vij 2 + fij 2 , 4 (i, j) ∈ ω, 1 ⩽ n ⩽ N, = φ(xi , yj ),

= 0,

(3.303)

(i, j) ∈ ω,

(i, j) ∈ 𝜕ω, 0 ⩽ n ⩽ N.

(3.304)

Then it holds 2−γ

n t 󵄩󵄩 n 󵄩󵄩2 󵄩󵄩 0 󵄩󵄩2 󵄩 1 󵄩2 ‖ψ‖2 + tnγ−1 Γ(2 − γ)τ ∑ 󵄩󵄩󵄩f k− 2 󵄩󵄩󵄩 , 󵄩󵄩∇h v 󵄩󵄩 ⩽ 󵄩󵄩∇h v 󵄩󵄩 + n Γ(3 − γ) k=1

1 ⩽ n ⩽ N,

(3.305)

3.9 The ADI method based on L1 approximation for 2D problem

| 251

where M1 −1 M2 −1

‖ψ‖2 = h1 h2 ∑ ∑ ψ2ij , i=1

j=1

M1 −1 M2 −1

k− 1 2 󵄩󵄩 k− 21 󵄩󵄩2 󵄩󵄩f 󵄩󵄩 = h1 h2 ∑ ∑ (fij 2 ) . i=1

j=1

1

Proof. Taking the inner product on both hand sides of (3.303) with ηδt vn− 2 yields n−1

1 1 1 1 2 (γ) (γ) (γ) (γ) 󵄩 󵄩 b0 󵄩󵄩󵄩δt vn− 2 󵄩󵄩󵄩 − ∑ (bn−k−1 − bn−k )(δt vk− 2 , δt vn− 2 ) − bn−1 (ψ, δt vn− 2 )

k=1

1 1 1 1 1 + η2 τ2 (δx2 δy2 δt vn− 2 , δt vn− 2 ) = η(δx2 vn− 2 , δt vn− 2 ) 4 1

1

1

1

+ η(δy2 vn− 2 , δt vn− 2 ) + η(f n− 2 , δt vn− 2 ),

1 ⩽ n ⩽ N.

Noticing (3.304), it follows from (2.200)–(2.202) that 1

1

1

1

(δx2 δy2 δt vn− 2 , δt vn− 2 ) = (δx δy δt vn− 2 , δx δy δt vn− 2 ) ⩾ 0,

1 󵄩󵄩 n 󵄩󵄩2 󵄩󵄩 n−1 󵄩󵄩2 (󵄩δ v 󵄩 − 󵄩δ v 󵄩󵄩 ), 2τ 󵄩 x 󵄩 󵄩 x 1 1 1 1 1 󵄩 󵄩2 󵄩 󵄩2 (δy2 vn− 2 , δt vn− 2 ) = −(δy vn− 2 , δy δt vn− 2 ) = − (󵄩󵄩󵄩δy vn 󵄩󵄩󵄩 − 󵄩󵄩󵄩δy vn−1 󵄩󵄩󵄩 ), 2τ 1

1

1

1

(δx2 vn− 2 , δt vn− 2 ) = −(δx vn− 2 , δx δt vn− 2 ) = −

hence, 1 2 η 󵄩 (γ) 󵄩 󵄩2 󵄩 󵄩2 󵄩 󵄩2 󵄩 󵄩2 󵄩 b0 󵄩󵄩󵄩δt vn− 2 󵄩󵄩󵄩 + [(󵄩󵄩󵄩δx vn 󵄩󵄩󵄩 + 󵄩󵄩󵄩δy vn 󵄩󵄩󵄩 ) − (󵄩󵄩󵄩δx vn−1 󵄩󵄩󵄩 + 󵄩󵄩󵄩δy vn−1 󵄩󵄩󵄩 )] 2τ

n−1

1

1

1

⩽ ∑ (bn−k−1 − bn−k )(δt vk− 2 , δt vn− 2 ) + bn−1 (ψ, δt vn− 2 ) k=1

(γ)

(γ)

1

1

+ η(f n− 2 , δt vn− 2 ),

(γ)

1 ⩽ n ⩽ N.

By the Cauchy–Schwarz inequality, we have 1 2 η 󵄩 (γ) 󵄩 󵄩2 󵄩 󵄩2 󵄩 b0 󵄩󵄩󵄩δt vn− 2 󵄩󵄩󵄩 + (󵄩󵄩󵄩∇h vn 󵄩󵄩󵄩 − 󵄩󵄩󵄩∇h vn−1 󵄩󵄩󵄩 ) 2τ

⩽

1 2 1 2 1 n−1 (γ) (γ) 󵄩 󵄩 󵄩 󵄩 ∑ (bn−k−1 − bn−k )(󵄩󵄩󵄩δt vk− 2 󵄩󵄩󵄩 + 󵄩󵄩󵄩δt vn− 2 󵄩󵄩󵄩 ) 2 k=1

1 2 1 1 1 (γ) 󵄩 󵄩 + bn−1 (‖ψ‖2 + 󵄩󵄩󵄩δt vn− 2 󵄩󵄩󵄩 ) + η(f n− 2 , δt vn− 2 ), 2

which can be reduced to n (γ) 󵄩 󵄩󵄩 n 󵄩󵄩2 τ k− 1 󵄩2 󵄩󵄩∇h v 󵄩󵄩 + ∑ bn−k 󵄩󵄩󵄩δt v 2 󵄩󵄩󵄩 η k=1

n−1 1 2 (γ) 󵄩 󵄩2 τ 󵄩 󵄩 ⩽ 󵄩󵄩󵄩∇h vn−1 󵄩󵄩󵄩 + ∑ bn−k−1 󵄩󵄩󵄩δt vk− 2 󵄩󵄩󵄩 η k=1

1 ⩽ n ⩽ N,

252 | 3 Difference methods for the time-fractional wave equations

+

1 1 τ (γ) bn−1 ‖ψ‖2 + 2τ(f n− 2 , δt vn− 2 ), η

1 ⩽ n ⩽ N.

The recursive process will produce n (γ) 󵄩 󵄩󵄩 n 󵄩󵄩2 τ k− 1 󵄩2 󵄩󵄩∇h v 󵄩󵄩 + ∑ bn−k 󵄩󵄩󵄩δt v 2 󵄩󵄩󵄩 η k=1

n−1 n 1 1 (γ) 󵄩 󵄩2 τ ⩽ 󵄩󵄩󵄩∇h v0 󵄩󵄩󵄩 + ∑ bk ‖ψ‖2 + 2τ ∑ (f k− 2 , δt vk− 2 ) η k=0 k=1 n−1 (γ) 󵄩 󵄩2 τ ⩽ 󵄩󵄩󵄩∇h v0 󵄩󵄩󵄩 + ∑ bk ‖ψ‖2 η k=0 n

+ τ ∑( k=1

(γ)

bn−k 󵄩 k− 1 󵄩2 1 󵄩󵄩δt v 2 󵄩󵄩 + η 󵄩󵄩󵄩f k− 2 󵄩󵄩󵄩2 ), 󵄩 󵄩 󵄩 󵄩 (γ) η bn−k

1 ⩽ n ⩽ N,

that is, n n−1 η 󵄩 k− 1 󵄩2 (γ) 󵄩󵄩 n 󵄩󵄩2 󵄩󵄩 0 󵄩󵄩2 τ 2 󵄩󵄩∇h v 󵄩󵄩 ⩽ 󵄩󵄩∇h v 󵄩󵄩 + ∑ bk ‖ψ‖ + τ ∑ (γ) 󵄩󵄩󵄩f 2 󵄩󵄩󵄩 , η k=0 k=1 b

1 ⩽ n ⩽ N.

(3.306)

n−k

By (3.20) and (3.22), it follows (3.305) from (3.306). The proof ends. 3.9.4 Convergence of the difference scheme Theorem 3.9.3. Suppose {Uijn | (i, j) ∈ ω,̄ 0 ⩽ n ⩽ N} and {unij | (i, j) ∈ ω,̄ 0 ⩽ n ⩽ N} are solutions of the problem (3.290)–(3.292) and the difference scheme (3.298)–(3.300), respectively. Let eijn = Uijn − unij ,

(i, j) ∈ ω,̄ 0 ⩽ n ⩽ N,

then it holds 󵄩󵄩 n 󵄩󵄩 √ γ 3−γ 2 2 󵄩󵄩∇h e 󵄩󵄩 ⩽ T Γ(2 − γ)L1 L2 c14 (τ + h1 + h2 ),

1 ⩽ n ⩽ N.

Proof. The subtraction of (3.298)–(3.300) from (3.294), (3.296)–(3.297), respectively, produces the system of error equations as follows: { { { { { { { { { { { { { { { { { { { { { { { { { { { { {

1 (γ) n− 21 n−1 (γ) k− 1 (γ) (γ) [b0 δt eij − ∑ (bn−k−1 − bn−k )δt eij 2 − bn−1 ⋅ 0] η k=1 +

Γ(3 − γ) 1+γ 2 2 n− 21 τ δx δy δt eij 4 n− 21

= δx2 eij

eij0 eijn

n− 21

+ δy2 eij

n− 1

+ (r14 )ij 2 ,

= 0,

(i, j) ∈ ω,

= 0,

(i, j) ∈ 𝜕ω, 0 ⩽ n ⩽ N.

(i, j) ∈ ω, 1 ⩽ n ⩽ N,

3.10 The compact ADI method based on L1 approximation for 2D problem

| 253

Noticing (3.295), the application of Theorem 3.9.2 will lead to n

󵄩󵄩 n 󵄩󵄩2 󵄩 γ−1 k− 1 󵄩2 󵄩󵄩∇h e 󵄩󵄩 ⩽ tn Γ(2 − γ)τ ∑ 󵄩󵄩󵄩(r14 ) 2 󵄩󵄩󵄩 k=1

2

γ

2 ⩽ T Γ(2 − γ)L1 L2 c14 (τ3−γ + h21 + h22 ) ,

1 ⩽ n ⩽ N.

Taking the square root on both hand sides of the inequality above will reach the desired result. The proof ends.

3.10 The compact ADI method based on L1 approximation for 2D problem In this section, the spatial compact ADI difference method will be developed for (3.290)–(3.292). Suppose u ∈ C (6,6,3) (Ω̄ × [0, T]). Take the same mesh partition, mesh function spaces and notations as those in last section. In addition, for any mesh function u ∈ 𝒱h , define the average operators as follows: { (ℐ + 𝒜x uij = { { uij , { (ℐ + 𝒜y uij = { { uij ,

h21 2 δ )u , 12 x ij

1 ⩽ i ⩽ M1 − 1,

0 ⩽ j ⩽ M2 ,

i = 0 or i = M1 , h22 2 δ )u , 12 y ij

1 ⩽ j ⩽ M2 − 1,

0 ⩽ i ⩽ M1 .

j = 0 or j = M2 ,

3.10.1 Derivation of the difference scheme Considering equation (3.290) at the point (xi , yj , tn ), we have C γ 0 Dt u(xi , yj , tn )

= uxx (xi , yj , tn ) + uyy (xi , yj , tn ) + fijn ,

(i, j) ∈ ω,̄ 0 ⩽ n ⩽ N.

Performing the operator 𝒜x 𝒜y to both hand sides of the equality above, it follows from Theorem 1.6.2 and Lemma 2.1.3 that C

γ

𝒜x 𝒜y 0 Dt u(xi , yj , tn ) = 𝒜y (𝒜x uxx (xi , yj , tn ))

+ 𝒜x (𝒜y uyy (xi , yj , tn )) + 𝒜x 𝒜y fijn

= 𝒜y δx2 Uijn + 𝒜x δy2 Uijn + 𝒜x 𝒜y fijn + O(h41 + h42 ),

(i, j) ∈ ω, 0 ⩽ n ⩽ N.

Taking an average on two adjacent time levels and applying the L1 formula (1.69) to approximate the temporal Caputo fractional derivative, it follows from Theorem 1.6.2 that 𝒜x 𝒜y

n−1 τ1−γ n− 1 k− 1 (γ) (γ) (γ) (γ) [b0 δt Uij 2 − ∑ (bn−k−1 − bn−k )δt Uij 2 − bn−1 ψij ] Γ(3 − γ) k=1

254 | 3 Difference methods for the time-fractional wave equations n− 21

= 𝒜y δx2 Uij

n− 21

+ 𝒜x δy2 Uij

n− 21

+ 𝒜x 𝒜y fij

n− 1

+ (r15 )ij 2 ,

(i, j) ∈ ω, 1 ⩽ n ⩽ N,

(3.307)

and there is a positive constant c15 such that n− 1 󵄨 󵄨󵄨 3−γ 4 4 󵄨󵄨(r15 )ij 2 󵄨󵄨󵄨 ⩽ c15 (τ + h1 + h2 ),

(i, j) ∈ ω, 1 ⩽ n ⩽ N,

(γ)

where {bl } is defined in (1.64). Adding a small term hand sides of (3.307) arrives at 𝒜x 𝒜y

+

n− 1 Γ(3−γ) 1+γ 2 2 τ δx δy δt Uij 2 4

into both

n−1 τ1−γ k− 1 n− 1 (γ) (γ) (γ) (γ) [b0 δt Uij 2 − ∑ (bn−k−1 − bn−k )δt Uij 2 − bn−1 ψij ] Γ(3 − γ) k=1

Γ(3 − γ) 1+γ 2 2 n− 1 n− 1 n− 1 τ δx δy δt Uij 2 = 𝒜y δx2 Uij 2 + 𝒜x δy2 Uij 2 4 n− 21

+ 𝒜x 𝒜y fij

n− 1

(i, j) ∈ ω, 1 ⩽ n ⩽ N,

+ (r16 )ij 2 ,

(3.308)

where n− 21

n− 21

= (r15 )ij

(r16 )ij

+

Γ(3 − γ) 1+γ 2 2 n− 1 τ δx δy δt Uij 2 , 4

and there is a positive constant c16 such that n− 1 󵄨 󵄨󵄨 3−γ 4 4 󵄨󵄨(r16 )ij 2 󵄨󵄨󵄨 ⩽ c16 (τ + h1 + h2 ),

(i, j) ∈ ω, 1 ⩽ n ⩽ N.

(3.309)

Noticing the initial-boundary value conditions (3.291)–(3.292), we have {

Uij0 = φ(xi , yj ), Uijn

= μ(xi , yj , tn ),

(i, j) ∈ ω, (i, j) ∈ 𝜕ω, 0 ⩽ n ⩽ N.

(3.310) (3.311)

n− 1

Omitting the small term (r16 )ij 2 in (3.308) and replacing the exact solution Uijn with its numerical one unij produce another difference scheme for (3.290)–(3.292) as follows: { { { { { { { { { { { { { { { { { { { { { { { { { { {

𝒜x 𝒜y

+ u0ij unij

n−1 τ1−γ n− 1 k− 1 (γ) (γ) (γ) (γ) [b0 δt uij 2 − ∑ (bn−k−1 − bn−k )δt uij 2 − bn−1 ψij ] Γ(3 − γ) k=1

Γ(3 − γ) 1+γ 2 2 n− 21 n− 1 n− 1 n− 1 τ δx δy δt uij = 𝒜y δx2 uij 2 + 𝒜x δy2 uij 2 + 𝒜x 𝒜y fij 2 , 4 (i, j) ∈ ω, 1 ⩽ n ⩽ N,

(3.312)

= φ(xi , yj ),

(3.313)

= μ(xi , yj , tn ),

(i, j) ∈ ω, (i, j) ∈ 𝜕ω, 0 ⩽ n ⩽ N.

(3.314)

3.10 The compact ADI method based on L1 approximation for 2D problem |

255

Denote η = τγ−1 Γ(3 − γ). Rewrite (3.312) as ητ 2 δ )(𝒜y − 2 x ητ 2 = (𝒜x + δ )(𝒜y + 2 x (𝒜x −

n−1

(γ)

ητ 2 n δ )u 2 y ij ητ 2 n−1 δ )u 2 y ij (γ)

k− 21

+ τ𝒜x 𝒜y [ ∑ (bn−k−1 − bn−k )δt uij k=1

(γ)

n− 21

+ bn−1 ψij ] + ητ𝒜x 𝒜y fij

.

Let u∗ij = (𝒜y −

ητ 2 n δ )u , 2 y ij

then the difference scheme (3.312)–(3.314) can be written into the following ADI form: On each time level t = tn (1 ⩽ n ⩽ N), firstly, for any fixed j from 1 to M2 − 1, solve a series of linear systems in the unknown {u∗ij | 0 ⩽ i ⩽ M1 } in x direction ητ 2 ∗ ητ 2 ητ 2 n−1 { { (𝒜x − δx )uij = (𝒜x + δx )(𝒜y + δ )u { { 2 2 2 y ij { { { { { n−1 { k− 1 { (γ) (γ) (γ) { { + τ 𝒜 𝒜 [ ∑ (bn−k−1 − bn−k )δt uij 2 + bn−1 ψij ] x y { k=1 { { { { n− 1 { { + ητ𝒜x 𝒜y fij 2 , 1 ⩽ i ⩽ M1 − 1, { { { { { { { { u∗ = (𝒜y − ητ δ2 )un , u∗ = (𝒜y − ητ δ2 )un 0j M1 ,j 2 y 0j 2 y M1 ,j { to obtain the value of {u∗ij | 1 ⩽ i ⩽ M1 − 1} on the intermediate time level. Then, for any fixed i from 1 to M1 − 1, carry out some calculations about the unknown {unij | 0 ⩽ j ⩽ M2 } in y direction ητ 2 n { { (𝒜y − δy )uij = u∗ij , 1 ⩽ j ⩽ M2 − 1, 2 { { n n { ui0 = μ(xi , y0 , tn ), ui,M2 = μ(xi , yM2 , tn ) to get the desired value of {unij | 1 ⩽ j ⩽ M2 − 1}.

256 | 3 Difference methods for the time-fractional wave equations 3.10.2 Solvability of the difference scheme Some preparation work is firstly done in order to show the unique solvability of the difference scheme (3.312)–(3.314). For any mesh functions u, v ∈ 𝒱̊h , define J(u, v) ≡ (𝒜x 𝒜y u, v). Noticing (2.200)-(2.202), it follows J(u, v) = ((ℐ +

h21 2 h2 δx )(ℐ + 2 δy2 )u, v) 12 12

= (u, v) −

h21 h2 h2 h2 (δx u, δx v) − 2 (δy u, δy v) + 1 2 (δx δy u, δx δy v). 12 12 144

It is easy to verify that 1 ‖u‖2 ⩽ J(u, u) ⩽ ‖u‖2 , 3

(3.315)

which reveals that J(u, v) is an inner product defined on 𝒱̊h . Denote (u, v)A = J(u, v),

‖u‖A = √(u, u)A .

In view of (3.315), the following lemma is true. Lemma 3.10.1. For any mesh function u ∈ 𝒱̊h , it holds 1 ‖u‖2 ⩽ ‖u‖2A ⩽ ‖u‖2 . 3 Now the existence and uniqueness of the solution to the difference scheme (3.312)–(3.314) will be proved. Theorem 3.10.1. The difference scheme (3.312)–(3.314) is uniquely solvable. Proof. Let ̄ un = {unij | (i, j) ∈ ω}. The value of u0 is determined by (3.313)–(3.314). Suppose the values of u0 , u1 , . . . , un−1 have been uniquely determined, then the system in un can be obtained from (3.312) and (3.314). To show its unique solvability, it is sufficient to prove that the corresponding homogeneous one Γ(3 − γ) γ 2 2 n 1 1 n { 𝒜 𝒜 u + τ δx δy uij = (𝒜y δx2 unij + 𝒜x δy2 unij ), { ητ x y ij 4 2 { { n u = 0, (i, j) ∈ 𝜕ω { ij

(i, j) ∈ ω, (3.316) (3.317)

3.10 The compact ADI method based on L1 approximation for 2D problem |

257

has only the trivial solution. Taking the inner product on both hand sides of (3.316) with un yields Γ(3 − γ) γ 2 2 n n 1 1 (𝒜x 𝒜y un , un ) + τ (δx δy u , u ) = [(𝒜y δx2 un , un ) + (𝒜x δy2 un , un )]. ητ 4 2

(3.318)

Noticing (3.317), it follows from Lemma 3.10.1 that 1 󵄩 󵄩2 󵄩 󵄩2 (𝒜x 𝒜y un , un ) = (un , un )A = 󵄩󵄩󵄩un 󵄩󵄩󵄩A ⩾ 󵄩󵄩󵄩un 󵄩󵄩󵄩 . 3

(3.319)

Noticing (2.200)–(2.202), we have 󵄩 󵄩2 (δx2 δy2 un , un ) = (δx δy un , δx δy un ) = 󵄩󵄩󵄩δx δy un 󵄩󵄩󵄩 , (𝒜y δx2 un , un ) = ((ℐ +

(3.320)

h22 2 2 n n δ )δ u , u ) 12 y x

2 2󵄩 󵄩2 󵄩2 󵄩 󵄩2 h 󵄩 = −󵄩󵄩󵄩δx un 󵄩󵄩󵄩 + 2 󵄩󵄩󵄩δx δy un 󵄩󵄩󵄩 ⩽ − 󵄩󵄩󵄩δx un 󵄩󵄩󵄩 , 12 3 h2 (𝒜x δy2 un , un ) = ((ℐ + 1 δx2 )δy2 un , un ) 12 2 2󵄩 󵄩 󵄩2 h 󵄩 󵄩2 󵄩2 = −󵄩󵄩󵄩δy un 󵄩󵄩󵄩 + 1 󵄩󵄩󵄩δx δy un 󵄩󵄩󵄩 ⩽ − 󵄩󵄩󵄩δy un 󵄩󵄩󵄩 . 12 3

(3.321)

(3.322)

Substituting (3.319)–(3.322) into (3.318) arrives at 1󵄩 1 󵄩󵄩 n 󵄩󵄩2 Γ(3 − γ) γ 󵄩󵄩 󵄩2 󵄩2 τ 󵄩󵄩δx δy un 󵄩󵄩󵄩 ⩽ − 󵄩󵄩󵄩∇h un 󵄩󵄩󵄩 ⩽ 0, 󵄩u 󵄩 + 3ητ 󵄩 󵄩 4 3 thus ‖un ‖ = 0. Noticing (3.317), we have un = 0. By the principle of induction, the difference scheme (3.312)–(3.314) is uniquely solvable. The proof ends.

3.10.3 Stability of the difference scheme For any mesh function u ∈ 𝒱̊h , define ‖∇h u‖2A = (‖δx u‖2 −

h2 h22 ‖δy δx u‖2 ) + (‖δy u‖2 − 1 ‖δx δy u‖2 ). 12 12

The following lemma will state the equivalence between ‖∇h ⋅ ‖A and ‖∇h ⋅ ‖. Lemma 3.10.2. For any mesh function u ∈ 𝒱̊h , it holds 2 ‖∇ u‖2 ⩽ ‖∇h u‖2A ⩽ ‖∇h u‖2 . 3 h

258 | 3 Difference methods for the time-fractional wave equations Proof. First, it follows from the inequalities of inverse estimate ‖δy δx u‖2 ⩽

4 ‖δx u‖2 , h22

‖δx δy u‖2 ⩽

4 ‖δy u‖2 h21

that ‖∇h u‖2A ⩾

2 2 (‖δ u‖2 + ‖δy u‖2 ) = ‖∇h u‖2 . 3 x 3

Second, it is easy to know that ‖∇h u‖2A ⩽ ‖δx u‖2 + ‖δy u‖2 = ‖∇h u‖2 . The proof ends. Another lemma will be needed. Lemma 3.10.3. For any mesh function u ∈ 𝒱̊h , it holds 1

1

1

(𝒜y δx2 un− 2 + 𝒜x δy2 un− 2 , δt un− 2 ) = −

1 󵄩󵄩 󵄩2 󵄩 󵄩2 (󵄩∇ un 󵄩󵄩 − 󵄩󵄩∇ un−1 󵄩󵄩󵄩A ). 2τ 󵄩 h 󵄩A 󵄩 h

Proof. Noticing (2.200)–(2.202) and applying the summation by parts arrive at 1

1

1

(𝒜y δx2 un− 2 + 𝒜x δy2 un− 2 , δt un− 2 ) = ((ℐ +

1 1 1 h22 2 2 n− 1 h2 δy )δx u 2 , δt un− 2 ) + ((ℐ + 1 δx2 )δy2 un− 2 , δt un− 2 ) 12 12

1

1

1

= (δx2 un− 2 + δy2 un− 2 , δt un− 2 )

1 1 1 1 h22 h2 (δx δy un− 2 , δx δy δt un− 2 ) + 1 (δx δy un− 2 , δx δy δt un− 2 ) 12 12 1 󵄩󵄩 n 󵄩󵄩2 󵄩󵄩 n 󵄩󵄩2 󵄩󵄩 n−1 󵄩󵄩2 󵄩󵄩 n−1 󵄩󵄩2 = − [(󵄩󵄩δx u 󵄩󵄩 + 󵄩󵄩δy u 󵄩󵄩 ) − (󵄩󵄩δx u 󵄩󵄩 + 󵄩󵄩δy u 󵄩󵄩 )] 2τ h2 1 󵄩 󵄩2 󵄩 󵄩2 + 2 ⋅ (󵄩󵄩󵄩δy δx un 󵄩󵄩󵄩 − 󵄩󵄩󵄩δy δx un−1 󵄩󵄩󵄩 ) 12 2τ h2 1 󵄩 󵄩2 󵄩 󵄩2 + 1 ⋅ (󵄩󵄩󵄩δx δy un 󵄩󵄩󵄩 − 󵄩󵄩󵄩δx δy un−1 󵄩󵄩󵄩 ) 12 2τ 2 2 1 󵄩 󵄩2 h 󵄩 󵄩2 󵄩 󵄩2 h 󵄩 󵄩2 = − {[(󵄩󵄩󵄩δx un 󵄩󵄩󵄩 − 2 󵄩󵄩󵄩δy δx un 󵄩󵄩󵄩 ) + (󵄩󵄩󵄩δy un 󵄩󵄩󵄩 − 1 󵄩󵄩󵄩δx δy un 󵄩󵄩󵄩 )] 2τ 12 12

+

2 2 󵄩2 󵄩 󵄩2 h 󵄩 󵄩2 󵄩 󵄩2 h 󵄩 − [(󵄩󵄩󵄩δx un−1 󵄩󵄩󵄩 − 2 󵄩󵄩󵄩δy δx un−1 󵄩󵄩󵄩 ) + (󵄩󵄩󵄩δy un−1 󵄩󵄩󵄩 − 1 󵄩󵄩󵄩δx δy un−1 󵄩󵄩󵄩 )]} 12 12 1 󵄩󵄩 2 2 󵄩 󵄩 n󵄩 n−1 = − (󵄩󵄩∇h u 󵄩󵄩󵄩A − 󵄩󵄩󵄩∇h u 󵄩󵄩󵄩A ). 2τ

The proof ends. Next, the stability result of the difference scheme (3.312)–(3.314) will be shown.

3.10 The compact ADI method based on L1 approximation for 2D problem

| 259

Theorem 3.10.2. Suppose {vijn | (i, j) ∈ ω,̄ 0 ⩽ n ⩽ N} is the solution of the difference scheme { { { { { { { { { { { { { { { { { { { { {

n−1 1 τ2 2 2 n− 1 n− 1 k− 1 (γ) (γ) (γ) (γ) 𝒜x 𝒜y [b0 δt vij 2 − ∑ (bn−k−1 − bn−k )δt vij 2 − bn−1 ψij ] + ηδx δy δt vij 2 η 4 k=1 n− 21

n− 21

+ 𝒜x δy2 vij

= 𝒜y δx2 vij

vij0 vijn

= φ(xi , yj ),

= 0,

n− 21

+ gij

(i, j) ∈ ω, 1 ⩽ n ⩽ N,

,

(3.323)

(i, j) ∈ ω,

(3.324)

(i, j) ∈ 𝜕ω, 0 ⩽ n ⩽ N,

(3.325)

where ψij |(i,j)∈𝜕ω = 0. Then it holds 2−γ

n 1 2 t 󵄩 󵄩 󵄩󵄩 n 󵄩󵄩2 󵄩󵄩 0 󵄩󵄩2 ‖ψ‖2 + 3tnγ−1 Γ(2 − γ)τ ∑ 󵄩󵄩󵄩g k− 2 󵄩󵄩󵄩 , 󵄩󵄩∇h v 󵄩󵄩A ⩽ 󵄩󵄩∇h v 󵄩󵄩A + n Γ(3 − γ) k=1

1 ⩽ n ⩽ N, (3.326)

with M1 −1 M2 −1

M1 −1 M2 −1

k− 1 2 󵄩󵄩 k− 21 󵄩󵄩2 󵄩󵄩g 󵄩󵄩 = h1 h2 ∑ ∑ (gij 2 ) .

‖ψ‖2 = h1 h2 ∑ ∑ (ψij )2 , i=1

i=1

j=1

j=1

1

Proof. Taking the inner product on both hand sides of (3.323) with ηδt vn− 2 yields n−1

1 2 1 1 (γ) 󵄩 (γ) (γ) 󵄩 b0 󵄩󵄩󵄩δt vn− 2 󵄩󵄩󵄩A − ∑ (bn−k−1 − bn−k )(δt vk− 2 , δt vn− 2 )A

k=1

(γ)

− bn−1 (ψ, δt v

n− 21

)A +

1

1 τ2 2 2 2 n− 21 η (δx δy δt v , δt vn− 2 ) 4 1

1

1

1

= η(𝒜y δx2 vn− 2 + 𝒜x δy2 vn− 2 , δt vn− 2 ) + η(g n− 2 , δt vn− 2 ).

(3.327)

It is easy to see that 󵄩 󵄩2 (δx2 δy2 δt vn− 2 , δt vn− 2 ) = 󵄩󵄩󵄩δx δy δt vn− 2 󵄩󵄩󵄩 ⩾ 0, 1

1

1

(3.328)

and it follows from Lemma 3.10.3 that 1

1

1

(𝒜y δx2 vn− 2 + 𝒜x δy2 vn− 2 , δt vn− 2 ) = −

1 󵄩󵄩 n 󵄩󵄩2 󵄩󵄩 n−1 󵄩󵄩2 (󵄩∇ v 󵄩 − 󵄩∇ v 󵄩󵄩A ). 2τ 󵄩 h 󵄩A 󵄩 h

(3.329)

Substituting (3.328) and (3.329) into (3.327), by the Cauchy–Schwarz inequality, we have 1 2 η 󵄩 (γ) 󵄩 󵄩 󵄩2 󵄩 󵄩2 b0 󵄩󵄩󵄩δt vn− 2 󵄩󵄩󵄩A + (󵄩󵄩󵄩∇h vn 󵄩󵄩󵄩A − 󵄩󵄩󵄩∇h vn−1 󵄩󵄩󵄩A ) 2τ

n−1

1

1

1

1

1

⩽ ∑ (bn−k−1 − bn−k )(δt vk− 2 , δt vn− 2 )A + bn−1 (ψ, δt vn− 2 )A + η(g n− 2 , δt vn− 2 ) k=1

(γ)

(γ)

(γ)

260 | 3 Difference methods for the time-fractional wave equations 1 2 1 2 1 n−1 (γ) (γ) 󵄩 󵄩 󵄩 󵄩 − bn−k )(󵄩󵄩󵄩δt vk− 2 󵄩󵄩󵄩A + 󵄩󵄩󵄩δt vn− 2 󵄩󵄩󵄩A ) ∑ (b 2 k=1 n−k−1

⩽

1 1 2 1 1 (γ) 󵄩 󵄩 + bn−1 (‖ψ‖2A + 󵄩󵄩󵄩δt vn− 2 󵄩󵄩󵄩A ) + η(g n− 2 , δt vn− 2 ), 2

1 ⩽ n ⩽ N,

which gives

1 2 η 󵄩 (γ) 󵄩 󵄩 󵄩2 󵄩2 󵄩 b0 󵄩󵄩󵄩δt vn− 2 󵄩󵄩󵄩A + (󵄩󵄩󵄩∇h vn 󵄩󵄩󵄩A − 󵄩󵄩󵄩∇h vn−1 󵄩󵄩󵄩A ) τ

n−1

1 1 1 2 (γ) (γ) 󵄩 (γ) 󵄩 ⩽ ∑ (bn−k−1 − bn−k )󵄩󵄩󵄩δt vk− 2 󵄩󵄩󵄩A + bn−1 ‖ψ‖2A + 2η(g n− 2 , δt vn− 2 ),

k=1

1 ⩽ n ⩽ N.

(3.330)

Let 󵄩 󵄩2 H 0 = 󵄩󵄩󵄩∇h v0 󵄩󵄩󵄩A ,

n 1 2 (γ) 󵄩 󵄩 󵄩2 τ 󵄩 H n = 󵄩󵄩󵄩∇h vn 󵄩󵄩󵄩A + ∑ bn−k 󵄩󵄩󵄩δt vk− 2 󵄩󵄩󵄩A , η k=1

1 ⩽ n ⩽ N,

then it follows from (3.330) that H n ⩽ H n−1 +

1 1 τ (γ) b ‖ψ‖2 + 2τ(g n− 2 , δt vn− 2 ), η n−1

1 ⩽ n ⩽ N.

The recursive process leads to Hn ⩽ H0 +

n 1 1 τ n−1 (γ) ∑ bk ‖ψ‖2 + 2τ ∑ (g k− 2 , δt vk− 2 ) η k=0 k=1 (γ)

⩽ H0 +

n b 󵄩 1 2 1 2 3η 󵄩 τ n−1 (γ) 󵄩 󵄩 ∑ bk ‖ψ‖2 + τ ∑ ( (γ) 󵄩󵄩󵄩g k− 2 󵄩󵄩󵄩 + n−k 󵄩󵄩󵄩δt vk− 2 󵄩󵄩󵄩 ) η k=0 3η b k=1

⩽ H0 +

b 󵄩 1 2 1 2 3η 󵄩 τ (γ) 󵄩 󵄩 ∑ bk ‖ψ‖2 + τ ∑ ( (γ) 󵄩󵄩󵄩g k− 2 󵄩󵄩󵄩 + n−k 󵄩󵄩󵄩δt vk− 2 󵄩󵄩󵄩A ), η k=0 η b k=1

n−k

n−1

(γ)

n

1 ⩽ n ⩽ N,

n−k

that is, n−1 n η 󵄩 k− 1 󵄩2 (γ) 󵄩󵄩 n 󵄩󵄩2 󵄩󵄩 0 󵄩󵄩2 τ 2 󵄩󵄩∇h v 󵄩󵄩A ⩽ 󵄩󵄩∇h v 󵄩󵄩A + ∑ bk ‖ψ‖ + 3τ ∑ (γ) 󵄩󵄩󵄩g 2 󵄩󵄩󵄩 , η k=0 k=1 b

1 ⩽ n ⩽ N.

(3.331)

n−k

By (3.20) and (3.22), the desired result (3.326) can be obtained from (3.331). The proof ends. 3.10.4 Convergence of the difference scheme Theorem 3.10.3. Suppose {Uijn | (i, j) ∈ ω,̄ 0 ⩽ n ⩽ N} and {unij | (i, j) ∈ ω,̄ 0 ⩽ n ⩽ N} are solutions of the problem (3.290)–(3.292) and the difference scheme (3.312)–(3.314), respectively. Let eijn = Uijn − unij ,

(i, j) ∈ ω,̄ 0 ⩽ n ⩽ N,

3.11 Supplementary remarks and discussions | 261

then it holds 󵄩󵄩 n 󵄩󵄩 3 √ γ 2T Γ(2 − γ)L1 L2 c16 (τ3−γ + h41 + h42 ), 󵄩󵄩∇h e 󵄩󵄩 ⩽ 2

1 ⩽ n ⩽ N.

Proof. Subtracting (3.312)–(3.314) from (3.308), (3.310)–(3.311), respectively, the system of error equations can be obtained as follows: { { { { { { { { { { { { { { { { { { { { { { { { { { {

n−1 1 k− 1 n− 1 (γ) (γ) (γ) (γ) 𝒜x 𝒜y [b0 δt eij 2 − ∑ (bn−k−1 − bn−k )δt eij 2 − bn−1 ⋅ 0] η k=1

+

Γ(3 − γ) 1+γ 2 2 n− 21 n− 1 n− 1 n− 1 τ δx δy δt eij = 𝒜y δx2 eij 2 + 𝒜x δy2 eij 2 + (r16 )ij 2 , 4 (i, j) ∈ ω, 1 ⩽ n ⩽ N,

eij0 = 0, eijn

= 0,

(i, j) ∈ ω, (i, j) ∈ 𝜕ω, 0 ⩽ n ⩽ N.

Noticing (3.309), the application of Theorem 3.10.2 will yield n

󵄩󵄩 n 󵄩󵄩2 󵄩 γ−1 k− 1 󵄩2 󵄩󵄩∇h e 󵄩󵄩A ⩽ 3tn Γ(2 − γ)τ ∑ 󵄩󵄩󵄩(r16 ) 2 󵄩󵄩󵄩 k=1

γ

2

2 ⩽ 3T Γ(2 − γ)L1 L2 c16 (τ3−γ + h41 + h42 ) ,

1 ⩽ n ⩽ N.

Taking the square root on both hand sides of the inequality above, it follows from Lemma 3.10.2 that 3 󵄩󵄩 n 󵄩󵄩 √ 3 󵄩󵄩 n 󵄩󵄩 3−γ 4 4 󵄩󵄩∇h e 󵄩󵄩A ⩽ √2T γ Γ(2 − γ)L1 L2 c16 (τ + h1 + h2 ), 󵄩󵄩∇h e 󵄩󵄩 ⩽ 2 2

1 ⩽ n ⩽ N.

The proof ends.

3.11 Supplementary remarks and discussions 1. In this chapter, the finite difference methods for solving the initial-boundary value problems of time-fractional wave equations were discussed. The time Caputo fractional derivative was approximated by the L1 approximation, the fast L1 approximation, the L2-1σ approximation or the fast L2-1σ approximation. The spatial second-order derivative was discretized by the second-order central difference quotient or the compact approximation. Several difference schemes were derived. The unique solvability, unconditional stability and convergence of each scheme were analyzed. For 2D problem, the ADI and compact ADI methods were mainly addressed. 2. The concerned problems in this chapter were the time-fractional wave equations in the differential form and the difference schemes were derived using the fractional numerical differentiation formulae to directly approximate the fractional

262 | 3 Difference methods for the time-fractional wave equations derivatives. It is worth to mention that in 1993, Tang[87] investigated the difference method for a class of pseudo-differential equations with the weak singularity kernel, which are precisely the integral form of the fractional wave equations of order 3/2. There are also some results on the difference methods for the integral forms of fractional differential equations. Huang et al.[38] performed the fractional integral operator 0 D1−α on both hand sides of (3.1) and obtained an equivalent form of (3.1) as t follows: 1−α ut (x, t) = ψ(x) + 0 D1−α t uxx (x, t) + 0 Dt f (x, t).

(3.332)

Then the first-order G-L formula was used to derive the difference scheme. Wang and Vong[96] also studied the difference method for solving (3.332) and the second-order convergent difference scheme was established, where the second-order G-L formula was derived to approximate the R-L integral. Huang et al.[39] continued to study the alternating direction implicit scheme for the two-dimensional time fractional nonlinear super-diffusion equations by the above transformation. 3. All problems in this chapter are ones with the Dirichlet boundary conditions. Ren and Sun[66] studied the high-order difference method for this kind of equations with the Neumann boundary conditions. Later on, Vong and Wang[90] also discussed the difference method for this kind of problem. Besides, the difference method for the time-fractional wave equations in unbounded domains was focused on by Brunner et al.[3] . 4. In Section 3.6 and Section 3.7, a class of multiterm time-fractional wave equations was briefly studied. For the numerical method solving this kind of problems, Gao and Liu [20] discussed the fourth-order multi-term time fractional wave equations.; Liu[51] also made some research results on the 1D problem. For the 2D problem, the compact ADI method was developed in [68] and the energy method was used to make some analyses. 5. Sun et al.[81] obtained a temporal second-order difference scheme for the fractional wave equation by using the method of order reduction and L2-1σ approximation. Then Sun et al.[83] combined the method of order reduction with the method in [19] to construct the temporal second-order difference scheme for the multiterm time fractional wave equation. Sun and Sun [79] developed a fast temporal second-order difference scheme for the multiterm time fractional wave equation. 6. Feng et al.[18] , Sun et al.[77] have studied the difference methods based on L1 approximation for the time fractional mixed diffusion and wave equations. Du and Sun[17] have studied the difference methods based on L2-1σ approximation with the method of order reduction for the time fractional mixed diffusion and wave equations.

Exercises 3

| 263

Exercises 3 3.1 Consider the problem γ

x ∈ (0, L), t ∈ (0, T], 0 Dt u(x, t) = uxx (x, t) + f (x, t), { { u(x, 0) = 0, ut (x, 0) = 0, x ∈ (0, L), { { { u(0, t) = μ(t), u(L, t) = ν(t), t ∈ [0, T],

(3.333) (3.334) (3.335)

where γ ∈ (1, 2), the functions f , μ, ν are given and μ(0) = ν(0) = μ󸀠 (0) = ν󸀠 (0) = 0. For any fixed x ∈ [0, L], define the function { { { { { ̂u(x, t) = { { { { { {

0, u(x, t),

t < 0, 0 ⩽ t ⩽ T,

v(x, t), 0,

T < t < 2T, t ⩾ 2T, k

k

where v(x, t) is a smooth function satisfying 𝜕 v(x,t) |t=T = 𝜕 u(x,t) |t=T , 𝜕t k 𝜕t k 1+γ ̂ ⋅) ∈ C (ℛ). 0, k = 0, 1, 2, 3. Suppose u(x, For the problem, construct the difference scheme as follows: n−1

1

1

𝜕k v(x,t) |t=2T 𝜕t k

1

n−k− 2 n− n− (γ−1) { { τ−(γ−1) ∑ gk δt ui = δx2 ui 2 + fi 2 , { { { { k=0 { { 1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N, { { { 0 { { u = 0, 1 ⩽ i ⩽ M − 1, { { { ni n { u0 = μ(tn ), uM = ν(tn ), 0 ⩽ n ⩽ N.

For this difference scheme, try to (1) analyze the truncation error; (2) show the unique solvability; (3) show the stability with respect to the function f ; (4) show the convergence. 3.2 For the problem (3.333)–(3.335), construct the following difference scheme: n−1

1 n− 21 n−k− 21 (γ−1) { 2 n− 2 −(γ−1) { ) = δ u + 𝒜 f , 𝒜 (τ g δ u ∑ { t x i i i { k { { k=0 { { 1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N, { { { { 0 { u = 0, 1 ⩽ i ⩽ M − 1, { { { i n n { u0 = μ(tn ), uM = ν(tn ), 0 ⩽ n ⩽ N.

For this difference scheme, try to (1) analyze the truncation error; (2) show the unique solvability;

=

264 | 3 Difference methods for the time-fractional wave equations (3) show the stability with respect to the function f ; (4) show the convergence. 3.3 For the problem (3.1)–(3.3), apply the H2N2 approximation (1.106) in Subsection 1.6.5 to construct the following difference scheme: { { { { { { { { { { { { { { { { { { {

n−1 1 k− 1 n− 1 (n,γ) (n,γ) (n,γ) (n,γ) [b̂ 0 δt ui 2 − ∑ (b̂ n−k−1 − b̂ n−k )δt ui 2 − b̂ n−1 ψi ] Γ(2 − γ) k=1 n− 21

n− 21

= δx2 ui u0i un0

+ fi

= φ(xi ),

= μ(tn ),

,

1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N,

1 ⩽ i ⩽ M − 1,

unM = ν(tn ),

0 ⩽ n ⩽ N.

For this difference scheme, try to (1) analyze the truncation error; (2) show the unique solvability; (3) show the stability with respect to the initial value φ and the function f ; (4) show the convergence. 3.4 For the problem (3.1)–(3.3), apply the fast H2N2 approximation (1.154)–(1.156) in Subsection 1.7.3 to construct the following difference scheme: { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { {

1 1 1 1 (1,γ) b̂ 0 (δt ui2 − ψi ) = δx2 ui2 + fi 2 , Γ(2 − γ)

1 ⩽ i ⩽ M − 1,

N

exp 1 τ2−γ 2 n−1 n− 1 n− 1 n [ ∑ ωl Fl,i δt ui ] = δx2 ui 2 + fi 2 , + Γ(2 − γ) l=1 2−γ

1 ⩽ i ⩽ M − 1, 2 ⩽ n ⩽ N,

t1

2 Fl,i =

2

1 −sl (t 3 −t) 2 2 dt(δt ui2 − ψi ), ∫e τ

1 ⩽ i ⩽ M − 1, 1 ⩽ l ⩽ Nexp ,

t0

n− 32

n− 52

n n−1 Fl,i = e−sl τ Fl,i + Bl (δt ui

− δt ui

u0i un0

0 ⩽ n ⩽ N.

1 ⩽ l ⩽ Nexp , = φ(xi ),

= μ(tn ),

),

1 ⩽ i ⩽ M − 1, 3 ⩽ n ⩽ N,

1 ⩽ i ⩽ M − 1,

unM = ν(tn ),

For this difference scheme, try to (1) analyze the truncation error; (2) show the unique solvability; (3) show the stability with respect to the initial value φ and the function f ; (4) show the convergence. 3.5 Consider the problem γ

γ

1 0 Dt u(x, t) + 0 Dt u(x, t) = uxx (x, t) + f (x, t), { { u(x, 0) = 0, ut (x, 0) = 0, x ∈ (0, L), { { { u(0, t) = μ(t), u(L, t) = ν(t), t ∈ [0, T],

x ∈ (0, L), t ∈ (0, T],

Exercises 3

| 265

where 1 < γ1 < γ < 2, the functions f , μ, ν are given and μ(0) = ν(0) = μ󸀠 (0) = ν󸀠 (0) = 0. ̂ t) like that in Exercise 3.1, and suppose u(x, ̂ ⋅) ∈ C 1+γ (ℛ). Define the function u(x, For the problem, construct the difference scheme as follows: { { { { { { { { { { { { { { { { { { {

n−1

(γ −1)

τ−(γ1 −1) ∑ gk 1 k=0

n−k− 21

δt ui

n− 1 n− 1 = δx2 ui 2 + fi 2 , 1 ⩽ i u0i = 0, 1 ⩽ i ⩽ M − 1, un0 = μ(tn ), unM = ν(tn ),

n−1

(γ−1)

+ τ−(γ−1) ∑ gk k=0

n−k− 21

δt ui

⩽ M − 1, 1 ⩽ n ⩽ N, 0 ⩽ n ⩽ N.

For this difference scheme, try to (1) analyze the truncation error; (2) show the unique solvability; (3) show the stability with respect to the function f ; (4) show the convergence. 3.6 For the following problem of the time-fractional mixed diffusion and wave equations γ

C D u(x, t) + C0 Dαt u(x, t) = uxx (x, t) + f (x, t), x ∈ (0, L), t ∈ (0, T], { { { 0 t u(x, 0) = φ(x), ut (x, 0) = ψ(x), x ∈ (0, L), { { { { u(0, t) = μ(t), u(L, t) = ν(t), t ∈ [0, T],

where 1 < γ < 2, 0 < α < 1, the functions f , φ, ψ, μ, ν are given and μ(0) = φ(0), ν(0) = φ(L), μ󸀠 (0) = ψ(0), ν󸀠 (0) = ψ(L), construct the following compact difference scheme: { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { {

𝒜{

n−1 1 n− 1 k− 1 (γ) (γ) (γ) (γ) [b0 δt ui 2 − ∑ (bn−k−1 − bn−k )δt ui 2 − bn−1 ψi ]} sγ k=1

+ 𝒜{

(α) n−1 a(α) + a(α) 1 a0 n− 1 k− 1 n−1−k [ δt ui 2 + ∑ n−k δt ui 2 ]} sα 2 2 k=1

n− 21

= δx2 ui

u0i un0

n− 21

+ 𝒜fi

= φ(xi ), = μ(tn ),

,

1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N,

1 ⩽ i ⩽ M − 1,

unM = ν(tn ),

0 ⩽ n ⩽ N.

For this difference scheme, try to (1) analyze the truncation error; (2) show the unique solvability; (3) show the stability with respect to the initial value φ and the function f ; (4) show the convergence.

266 | 3 Difference methods for the time-fractional wave equations 3.7 Consider the problem γ

0 Dt u(x, y, t) = uxx (x, y, t) + uyy (x, y, t) + f (x, y, t), { { { { { (x, y) ∈ Ω, t ∈ (0, T], { { u(x, y, 0) = 0, ut (x, y, 0) = 0, (x, y) ∈ Ω, { { { { u(x, y, t) = μ(x, y, t), (x, y) ∈ 𝜕Ω, t ∈ [0, T],

(3.336) (3.337) (3.338)

where Ω = (0, L1 ) × (0, L2 ), γ ∈ (1, 2), the functions f , μ are given and μ(x, y, 0)|(x,y)∈𝜕Ω = 0, μt (x, y, 0)|(x,y)∈𝜕Ω = 0. For any fixed (x, y) ∈ Ω,̄ define the function { { { { { ̂u(x, y, t) = { { { { { {

0, u(x, y, t),

t < 0, 0 ⩽ t ⩽ T,

v(x, y, t), 0,

T < t < 2T, t ⩾ 2T, k

k

k

|t=T = 𝜕 u(x,y,t) |t=T , 𝜕 where v(x, y, t) is a smooth function satisfying 𝜕 v(x,y,t) 𝜕t k 𝜕t k 1+γ ̂ y, ⋅) ∈ C (ℛ). 0, k = 0, 1, 2, 3. Suppose u(x, For the problem, construct the difference scheme as follows: n−1

1

1

1

v(x,y,t) |t=2T 𝜕t k

1

n−k− n− n− n− (γ−1) { { τ−(γ−1) ∑ gk δt uij 2 = δx2 uij 2 + δy2 uij 2 + fij 2 , { { { { k=0 { { (i, j) ∈ ω, 1 ⩽ n ⩽ N, { { { 0 { { uij = 0, (i, j) ∈ ω, { { { n { uij = μ(xi , yj , tn ), (i, j) ∈ 𝜕ω, 0 ⩽ n ⩽ N.

For this difference scheme, try to (1) analyze the truncation error; (2) show the unique solvability; (3) show the stability with respect to the function f ; (4) show the convergence. ̂ y, t) and suppose u(x, ̂ y, ⋅) ∈ C 1+γ (ℛ). For the 3.8 Similar to Exercise 3.7, define u(x, problem (3.336)–(3.338), construct the following difference scheme: { { { { { { { { { { { { { { { { { { { { {

𝒜x 𝒜y (τ

−(γ−1)

n− 21

= 𝒜y δx2 uij u0ij = 0, unij

n−1

(γ−1)

∑ gk

k=0

n− 21

+ 𝒜x δy2 uij

n−k− 21

δt uij

) n− 21

+ 𝒜x 𝒜y fij

,

(i, j) ∈ ω, 1 ⩽ n ⩽ N,

(i, j) ∈ ω,

= μ(xi , yj , tn ),

(i, j) ∈ 𝜕ω, 0 ⩽ n ⩽ N.

For this difference scheme, try to

=

Exercises 3

(1) (2) (3) (4)

analyze the truncation error; show the unique solvability; show the stability with respect to the function f ; show the convergence.

| 267

4 Difference methods for the space-fractional partial differential equations The space-fractional differential equation can be used to model the diffusion process related to the whole space. In mathematics, it can be obtained by replacing the spatial integer-order derivative in the classical diffusion equation with the R-L fractional derivative or the Riesz fractional derivative. In this chapter, we mainly focus on the numerical method for the space-fractional partial differential equations in R-L type. It is a natural idea to approximate the R-L fractional derivative using the standard G-L formula, whereas, the resultant difference scheme is unstable. Meerschaert and Tadjeran[57, 58, 59] applied the shifted G-L formula to study the numerical solution of the space-fractional differential equation and some stable difference schemes were derived. In this chapter, for the one-dimensional space-fractional partial differential equation, the first-order method in both time and space based on the shifted G-L formula, the second-order method in both time and space based on the WSGL formula, the method of order two in time and order four in space will be introduced successively. For the 2D case, the method of order two in time and order four in space will be developed. The whole chapter consists of 5 sections.

4.1 The first-order method based on the shifted G-L formula for 1D problem Consider the initial-boundary value problem of the space-fractional partial differential equations β

ut (x, t) = K1 0 Dβx u(x, t) + K2 x DL u(x, t) + f (x, t), { { { { { { x ∈ (0, L), t ∈ (0, T], { { { u(x, 0) = φ(x), x ∈ (0, L), { { { { u(0, t) = 0, u(L, t) = 0, t ∈ [0, T],

(4.1) (4.2) (4.3)

where K1 , K2 are two nonnegative constants, K1 + K2 > 0, β ∈ (1, 2), the functions f , φ are given and φ(0) = φ(L) = 0. Introduce the same mesh partition, notations and the mesh function spaces 𝒰h , 𝒰h̊ as those in Section 2.1 and Section 3.1. In addition, for any fixed t ∈ [0, T], define the function ̃ t) = { u(x,

u(x, t),

0 ⩽ x ⩽ L,

0,

x ∉ [0, L].

̃ t) satisfies the conditions of TheoSuppose u(x, ⋅) ∈ C 2 [0, T] and the function u(x, ̃ t) ∈ C 1+β (ℛ). rem 1.4.2, that is, u(⋅, https://doi.org/10.1515/9783110616064-004

270 | 4 Difference methods for the space-fractional partial differential equations Define the mesh functions Uin = u(xi , tn ),

fin = f (xi , tn ),

0 ⩽ i ⩽ M, 0 ⩽ n ⩽ N.

4.1.1 Derivation of the difference scheme Considering equation (4.1) at the point (xi , tn ), we have β

ut (xi , tn ) = K1 0 Dβx u(xi , tn ) + K2 x DL u(xi , tn ) + fin , 1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N.

(4.4)

For the space-fractional derivatives on the right-hand side of (4.4), by the shifted G-L formula (1.19), it follows i+1

β −β n + O(h), ∑ gk Ui−k+1 0 Dx u(xi , tn ) = h (β)

k=0

M−i+1

β

n −β + O(h). ∑ gk Ui+k−1 x DL u(xi , tn ) = h (β)

k=0

(4.5) (4.6)

For the time first-order derivative on the left-hand side of (4.4), we have ut (xi , tn ) =

1 n (U − Uin−1 ) + O(τ). τ i

(4.7)

The substitution of (4.5)–(4.7) into (4.4) gives

i+1 M−i+1 1 n (β) n (β) n (Ui − Uin−1 ) = K1 h−β ∑ gk Ui−k+1 + K2 h−β ∑ gk Ui+k−1 + fin + (r1 )ni , τ k=0 k=0

1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N,

(4.8)

and there is a positive constant c1 such that 󵄨󵄨 n 󵄨󵄨 󵄨󵄨(r1 )i 󵄨󵄨 ⩽ c1 (τ + h),

1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N.

(4.9)

Noticing the initial-boundary value conditions (4.2)–(4.3), we have {

Ui0 = φ(xi ), U0n

= 0,

n UM

1 ⩽ i ⩽ M − 1,

(4.10)

= 0,

(4.11)

0 ⩽ n ⩽ N.

Neglecting the small term (r1 )ni in (4.8) and replacing the exact solution Uin with its numerical one uni produce a difference scheme for solving (4.1)–(4.3) as follows: { { { { { { { { { { { { { { { { {

i+1 M−i+1 1 n (β) n (β) −β −β (ui − un−1 ) = K h g u + K h ∑ ∑ gk uni+k−1 + fin , 1 2 i i−k+1 k τ k=0 k=0

u0i un0

1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N,

= φ(xi ),

= 0,

unM

(4.12)

1 ⩽ i ⩽ M − 1,

(4.13)

= 0,

(4.14)

0 ⩽ n ⩽ N.

4.1 The first-order method based on the shifted G-L formula for 1D problem

| 271

Denote λ1 = K1

τ , hβ

λ2 = K2

τ . hβ

4.1.2 Solvability of the difference scheme Theorem 4.1.1. The difference scheme (4.12)–(4.14) is uniquely solvable. Proof. Let un = (un0 , un1 , . . . , unM ). The value of u0 is obviously determined by (4.13)–(4.14). Suppose that the values of u0 , u1 , . . . , un−1 have been uniquely determined. From (4.12) and (4.14), we can obtain the linear system in the unknown un . To prove its unique solvability, it suffices to show the corresponding homogeneous one M−i+1 i+1 1 n (β) (β) n −β −β { { u + K h u = K h g ∑ gk uni+k−1 , ∑ { 2 1 i−k+1 k { { τ i k=0 k=0 { { 1 ⩽ i ⩽ M − 1, { { { n n { u0 = uM = 0

(4.15) (4.16)

has only the trivial solution. When 1 < β < 2, it follows from Lemma 1.4.1 that g0 = 1, (β)

∞

g1 = −β,

∑ gk = 0, (β)

k=0

(β)

m

g2 > g3 > ⋅ ⋅ ⋅ > 0, (β)

∑ gk < 0,

(β)

m ⩾ 1.

(β)

k=0

Rewrite equation (4.15) as [1 + λ1 (−g1 ) + λ2 (−g1 )]uni (β)

i+1

(β)

M−i+1

= λ1 ∑ gk uni−k+1 + λ2 ∑ gk uni+k−1 , (β)

k=0 k =1̸

(β)

k=0 k =1̸

1 ⩽ i ⩽ M − 1.

Suppose ‖un ‖∞ = |unin |, where in ∈ {1, 2, . . . , M − 1}. Letting i = in in the equality above and taking the absolute value of both hand sides, it follows from the triangle inequality that (β) (β) 󵄩 󵄩 [1 + λ1 (−g1 ) + λ2 (−g1 )]󵄩󵄩󵄩un 󵄩󵄩󵄩∞ in +1

M−in +1

k=0 k =1̸

k=0 k =1̸

(β) 󵄨 (β) 󵄨 󵄨 󵄨 ⩽ λ1 ∑ gk 󵄨󵄨󵄨unin −k+1 󵄨󵄨󵄨 + λ2 ∑ gk 󵄨󵄨󵄨unin +k−1 󵄨󵄨󵄨

272 | 4 Difference methods for the space-fractional partial differential equations in +1

M−in +1

k=0 k =1̸

k=0 k =1̸

(β) 󵄩 󵄩 (β) 󵄩 󵄩 ⩽ λ1 ∑ gk 󵄩󵄩󵄩un 󵄩󵄩󵄩∞ + λ2 ∑ gk 󵄩󵄩󵄩un 󵄩󵄩󵄩∞ .

Noticing ∑m k=0 gk < 0 (m ⩾ 1), we have (β)

in +1

M−in +1

k=0

k=0

(β) 󵄩 n 󵄩 (β) 󵄩 n 󵄩 󵄩󵄩 n 󵄩󵄩 󵄩󵄩u 󵄩󵄩∞ ⩽ λ1 ∑ gk 󵄩󵄩󵄩u 󵄩󵄩󵄩∞ + λ2 ∑ gk 󵄩󵄩󵄩u 󵄩󵄩󵄩∞ ⩽ 0, n

thus ‖u ‖∞ = 0, which implies that (4.15)–(4.16) has only the trivial solution. By the principle of induction, the difference scheme (4.12)–(4.14) is uniquely solvable. The proof ends. 4.1.3 Stability of the difference scheme Theorem 4.1.2. The solution of the difference scheme (4.12)–(4.14) is stable with respect to both the initial value φ and the right hand function f . More precisely, it holds n

󵄩󵄩 n 󵄩󵄩 󵄩 0󵄩 󵄩 m󵄩 󵄩󵄩u 󵄩󵄩∞ ⩽ 󵄩󵄩󵄩u 󵄩󵄩󵄩∞ + τ ∑ 󵄩󵄩󵄩f 󵄩󵄩󵄩∞ ,

1 ⩽ n ⩽ N,

m=1

where 󵄩󵄩 m 󵄩󵄩 󵄨 m󵄨 󵄩󵄩f 󵄩󵄩∞ = max 󵄨󵄨󵄨fi 󵄨󵄨󵄨. 1⩽i⩽M−1 Proof. Rewrite equation (4.12) as follows: [1 + λ1 (−g1 ) + λ2 (−g1 )]uni (β)

i+1

(β)

M−i+1

= un−1 + λ1 ∑ gk uni−k+1 + λ2 ∑ gk uni+k−1 + τfin , i (β)

(β)

k=0 k =1̸

k=0 k =1̸

1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N. Suppose ‖un ‖∞ = |unin |, where in ∈ {1, 2, . . . , M − 1}. Letting i = in in the equality above and taking the absolute value of both hand sides, the application of the triangle inequality leads to (β) (β) 󵄩 󵄩 [1 + λ1 (−g1 ) + λ2 (−g1 )]󵄩󵄩󵄩un 󵄩󵄩󵄩∞ in +1

M−in +1

k=0 k =1̸

k=0 k =1̸

(β) 󵄩 󵄩 (β) 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 ⩽ 󵄩󵄩󵄩un−1 󵄩󵄩󵄩∞ + λ1 ∑ gk 󵄩󵄩󵄩un 󵄩󵄩󵄩∞ + λ2 ∑ gk 󵄩󵄩󵄩un 󵄩󵄩󵄩∞ + τ󵄩󵄩󵄩f n 󵄩󵄩󵄩∞ ,

1 ⩽ n ⩽ N,

that is, in +1

M−in +1

k=0

k=0

(β) 󵄩 n 󵄩 (β) 󵄩 n 󵄩 󵄩󵄩 n 󵄩󵄩 󵄩 n−1 󵄩 󵄩 n󵄩 󵄩󵄩u 󵄩󵄩∞ ⩽ 󵄩󵄩󵄩u 󵄩󵄩󵄩∞ + λ1 ∑ gk 󵄩󵄩󵄩u 󵄩󵄩󵄩∞ + λ2 ∑ gk 󵄩󵄩󵄩u 󵄩󵄩󵄩∞ + τ󵄩󵄩󵄩f 󵄩󵄩󵄩∞

4.2 The second-order method based on the WSGL formula for 1D problem

󵄩 󵄩 󵄩 󵄩 ⩽ 󵄩󵄩󵄩un−1 󵄩󵄩󵄩∞ + τ󵄩󵄩󵄩f n 󵄩󵄩󵄩∞ ,

| 273

1 ⩽ n ⩽ N.

The recursive process will produce n

󵄩 m󵄩 󵄩 0󵄩 󵄩󵄩 n 󵄩󵄩 󵄩󵄩u 󵄩󵄩∞ ⩽ 󵄩󵄩󵄩u 󵄩󵄩󵄩∞ + τ ∑ 󵄩󵄩󵄩f 󵄩󵄩󵄩∞ , m=1

1 ⩽ n ⩽ N.

The proof ends. 4.1.4 Convergence of the difference scheme Theorem 4.1.3. Suppose {Uin | 0 ⩽ i ⩽ M, 0 ⩽ n ⩽ N} and {uni | 0 ⩽ i ⩽ M, 0 ⩽ n ⩽ N} are solutions of the problem (4.1)–(4.3) and the difference scheme (4.12)–(4.14), respectively. Let ein = Uin − uni ,

0 ⩽ i ⩽ M, 0 ⩽ n ⩽ N,

then it holds 󵄩󵄩 n 󵄩󵄩 󵄩󵄩e 󵄩󵄩∞ ⩽ c1 T(τ + h),

1 ⩽ n ⩽ N.

Proof. The subtraction of (4.12)–(4.14) from (4.8), (4.10)–(4.11), respectively, yields the system of error equations as follows: { { { { { { { { { { { { { { { { {

M−i+1 i+1 1 n (β) n (β) n + K2 h−β ∑ gk ei+k−1 (ei − ein−1 ) = K1 h−β ∑ gk ei−k+1 τ k=0 k=0

ei0 e0n

+ (r1 )ni ,

= 0,

1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N,

1 ⩽ i ⩽ M − 1,

n eM = 0,

= 0,

0 ⩽ n ⩽ N.

Noticing (4.9), the application of Theorem 4.1.2 produces n

󵄩󵄩 n 󵄩󵄩 󵄩 m󵄩 󵄩󵄩e 󵄩󵄩∞ ⩽ τ ∑ 󵄩󵄩󵄩(r1 ) 󵄩󵄩󵄩∞ ⩽ nτc1 (τ + h) ⩽ c1 T(τ + h), m=1

1 ⩽ n ⩽ N.

The proof ends.

4.2 The second-order method based on the WSGL formula for 1D problem In this section, a difference method of order two in both time and space will be developed for the problem (4.1)–(4.3). ̃ t) defined in Section 4.1 satisfies Suppose u(x, ⋅) ∈ C 3 [0, T] and the function u(x, ̃ t) ∈ C 2+β (ℛ). the condition of Theorem 1.4.3, that is, u(⋅,

274 | 4 Difference methods for the space-fractional partial differential equations 4.2.1 Derivation of the difference scheme Considering equation (4.1) at the point (xi , tn ), we have β

ut (xi , tn ) = K1 0 Dβx u(xi , tn ) + K2 x DL u(xi , tn ) + fin , 1 ⩽ i ⩽ M − 1, 0 ⩽ n ⩽ N.

Taking an average on two adjacent time levels gives 1 1 [ ut (xi , tn ) + ut (xi , tn−1 )] = K1 [0 Dβx u(xi , tn ) + 0 Dβx u(xi , tn−1 )] 2 2 1 β β + K2 [x DL u(xi , tn ) + x DL u(xi , tn−1 )] 2 n− 21

+ fi

,

1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N.

(4.17)

For the space-fractional derivatives in (4.17), by the WSGL formula, it follows from Theorem 1.4.3 and Corollary 1.4.2 that i+1

β 0 Dx u(xi , tn )

n ̃k Ui−k+1 = h−β ∑ w + O(h2 ),

β x DL u(xi , tn )

n ̃k(β) Ui+k−1 = h−β ∑ w + O(h2 ),

(β)

k=0

M−i+1 k=0

(4.18) (4.19)

̃k } is defined by (1.35), which satisfies (1.36). where the coefficient {w For the time first-order derivative in (4.17), it follows from the Taylor’s formula that (β)

1 1 [ut (xi , tn ) + ut (xi , tn−1 )] = (Uin − Uin−1 ) + O(τ2 ). 2 τ

(4.20)

Substituting (4.18)–(4.20) into (4.17) produces i+1 M−i+1 1 n n− 21 n− 21 ̃k(β) Ui−k+1 ̃k(β) Ui+k−1 (Ui − Uin−1 ) = K1 h−β ∑ w + K2 h−β ∑ w τ k=0 k=0 n− 21

+ fi

n− 21

+ (r2 )i

,

1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N,

(4.21)

and there is a positive constant c2 such that 󵄨󵄨 n− 21 󵄨󵄨 2 2 󵄨󵄨(r2 )i 󵄨󵄨 ⩽ c2 (τ + h ),

1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N.

(4.22)

Noticing the initial-boundary value conditions (4.2)–(4.3), we have {

Ui0 = φ(xi ), U0n

= 0,

n UM

1 ⩽ i ⩽ M − 1, = 0,

0 ⩽ n ⩽ N.

(4.23) (4.24)

4.2 The second-order method based on the WSGL formula for 1D problem |

275

n− 1

Omitting the small term (r2 )i 2 in (4.21) and replacing the exact solution Uin with its numerical one uni produce another difference scheme for solving (4.1)–(4.3) as follows: { { { { { { { { { { { { { { { { {

M−i+1 i+1 1 n n− 1 n− 21 n− 21 −β ̃k(β) ui+k−1 ̃k(β) ui−k+1 + K2 h−β ∑ w + fi 2 , (ui − un−1 ∑w i ) = K1 h τ k=0 k=0

u0i un0

1 ⩽ i ⩽ M − 1,

= φ(xi ),

= 0,

unM

1 ⩽ n ⩽ N,

(4.25)

1 ⩽ i ⩽ M − 1,

(4.26)

= 0,

(4.27)

0 ⩽ n ⩽ N.

4.2.2 Solvability of the difference scheme We start our proof from the following important lemma. Lemma 4.2.1. For any mesh function u = (u0 , u1 , . . . , uM ) ∈ 𝒰h̊ , it holds M−1

i+1

i=1

k=0

̃k ui−k+1 )ui ⩽ 0, h ∑ (∑ w (β)

M−1 M−i+1

̃k(β) ui+k−1 )ui ⩽ 0. h ∑( ∑ w i=1

k=0

Proof. We only prove (4.28) here. The inequality (4.29) can be proved similarly. Noticing u0 = uM = 0 and (1.36), it follows M−1

i+1

i=1

k=0

M−1

i

̃k ui−k+1 )ui h ∑ (∑ w (β)

̃k ui−k+1 )ui = h ∑ (∑ w i=1

(β)

k=0

M−1

M−2

̃2(β) )h ∑ ui ui+1 ̃1(β) h ∑ (ui )2 + (w ̃0(β) + w =w M−1

i=1

i=1

M−1

̃k(β) h ∑ ui−k+1 ui + ∑w k=3

i=k

M−2

̃1 ‖u‖2 + (w ̃0 + w ̃2 )h ∑ ⩽w (β)

(β)

M−1

M−1

k=3

i=k

̃k(β) h ∑ + ∑w

(β)

2

i=1

(ui )2 + (ui+1 )2 2

(ui ) + (ui−k+1 )2 2

M−1

̃k(β) )‖u‖2 ⩽ 0, ⩽ (∑ w k=0

which implies the truth of (4.28). The proof ends.

(4.28) (4.29)

276 | 4 Difference methods for the space-fractional partial differential equations Next, the unique solvability of the difference scheme (4.25)–(4.27) will be proved. Theorem 4.2.1. The difference scheme (4.25)–(4.27) is uniquely solvable. Proof. Let un = (un0 , un1 , . . . , unM ). The value of u0 is uniquely determined by (4.26)–(4.27). Now suppose the values of u0 , u1 , . . . , un−1 have been uniquely determined, from (4.25) and (4.27), we can obtain the linear system in un . To show its unique solvability, it suffices to show the corresponding homogeneous one i+1 M−i+1 1 1 1 { { ̃k(β) uni−k+1 + K2 h−β ∑ w ̃k(β) uni+k−1 , { uni = K1 h−β ∑ w τ 2 2 { k=0 k=0 { { n n u = u = 0 M { 0

1 ⩽ i ⩽ M − 1, (4.30) (4.31)

has only the trivial solution. Taking the inner product on both hand sides of (4.30) with un , it follows from Lemma 4.2.1 that M−1 i+1 1 M−1 n 2 1 ̃k(β) uni−k+1 )uni h ∑ (ui ) = K1 h−β h ∑ ( ∑ w τ i=1 2 i=1 k=0 M−1 M−i+1 1 ̃k(β) uni+k−1 )uni + K2 h−β h ∑ ( ∑ w 2 i=1 k=0

⩽ 0, which implies ‖un ‖ = 0. Combining with (4.31), it follows that un = 0, namely, the homogeneous system (4.30)–(4.31) has only the trivial solution. By the principle of induction, the difference scheme (4.25)–(4.27) is uniquely solvable. The proof ends. 4.2.3 Stability of the difference scheme Theorem 4.2.2. The solution of the difference scheme (4.25)–(4.27) is stable with respect to both the initial value φ and the right-hand function f . More precisely, it holds n

󵄩󵄩 n 󵄩󵄩 󵄩󵄩 0 󵄩󵄩 󵄩 m− 1 󵄩 󵄩󵄩u 󵄩󵄩 ⩽ 󵄩󵄩u 󵄩󵄩 + τ ∑ 󵄩󵄩󵄩f 2 󵄩󵄩󵄩, m=1

1 ⩽ n ⩽ N,

where M−1

m− 1 2 󵄩󵄩 m− 21 󵄩󵄩 √ 󵄩󵄩f 󵄩󵄩 = h ∑ (fi 2 ) . i=1

4.2 The second-order method based on the WSGL formula for 1D problem

| 277

1

Proof. Taking the inner product on both hand sides of (4.25) with un− 2 yields M−1 i+1 1 󵄩󵄩 n 󵄩󵄩2 󵄩󵄩 n−1 󵄩󵄩2 n− 1 n− 21 ̃k(β) ui−k+1 )ui 2 (󵄩󵄩u 󵄩󵄩 − 󵄩󵄩u 󵄩󵄩 ) = K1 h−β h ∑ ( ∑ w 2τ i=1 k=0 M−1 M−i+1

n− 1

n− 21

2 ̃k(β) ui+k−1 + K2 h−β h ∑ ( ∑ w )ui

M−1

i=1

n− 21 n− 21 fi ,

+ h ∑ ui i=1

k=0

1 ⩽ n ⩽ N.

It follows from Lemma 4.2.1 that M−1 1 󵄩󵄩 n 󵄩󵄩2 󵄩󵄩 n−1 󵄩󵄩2 n− 1 n− 1 (󵄩󵄩u 󵄩󵄩 − 󵄩󵄩u 󵄩󵄩 ) ⩽ h ∑ ui 2 fi 2 2τ i=1

1 󵄩 󵄩 1󵄩 󵄩 ⩽ 󵄩󵄩󵄩un− 2 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩f n− 2 󵄩󵄩󵄩 1 󵄩 󵄩 󵄩 󵄩 󵄩 1󵄩 ⩽ (󵄩󵄩󵄩un 󵄩󵄩󵄩 + 󵄩󵄩󵄩un−1 󵄩󵄩󵄩)󵄩󵄩󵄩f n− 2 󵄩󵄩󵄩, 2

1 ⩽ n ⩽ N.

Hence, 1 󵄩󵄩 n 󵄩󵄩 󵄩󵄩 n−1 󵄩󵄩 󵄩 1󵄩 (󵄩󵄩u 󵄩󵄩 − 󵄩󵄩u 󵄩󵄩) ⩽ 󵄩󵄩󵄩f n− 2 󵄩󵄩󵄩, τ

1 ⩽ n ⩽ N,

or 󵄩󵄩 n 󵄩󵄩 󵄩󵄩 n−1 󵄩󵄩 󵄩 n− 1 󵄩 󵄩󵄩u 󵄩󵄩 ⩽ 󵄩󵄩u 󵄩󵄩 + τ󵄩󵄩󵄩f 2 󵄩󵄩󵄩,

1 ⩽ n ⩽ N.

Performing the recursive process gives n

󵄩 m− 1 󵄩 󵄩󵄩 n 󵄩󵄩 󵄩󵄩 0 󵄩󵄩 󵄩󵄩u 󵄩󵄩 ⩽ 󵄩󵄩u 󵄩󵄩 + τ ∑ 󵄩󵄩󵄩f 2 󵄩󵄩󵄩, m=1

1 ⩽ n ⩽ N.

The proof ends. 4.2.4 Convergence of the difference scheme Theorem 4.2.3. Suppose {Uin | 0 ⩽ i ⩽ M, 0 ⩽ n ⩽ N} and {uni | 0 ⩽ i ⩽ M, 0 ⩽ n ⩽ N} are solutions of the problem (4.1)–(4.3) and the difference scheme (4.25)–(4.27), respectively. Let ein = Uin − uni ,

0 ⩽ i ⩽ M, 0 ⩽ n ⩽ N,

then it holds 󵄩󵄩 n 󵄩󵄩 2 2 󵄩󵄩e 󵄩󵄩 ⩽ c2 T √L(τ + h ),

1 ⩽ n ⩽ N.

278 | 4 Difference methods for the space-fractional partial differential equations Proof. Subtracting (4.25)–(4.27) from (4.21), (4.23)–(4.24), respectively, yields the system of error equations as follows: i+1

M−i+1

1 n n− 21 n− 21 { { ̃k(β) ei−k+1 ̃k(β) ei+k−1 { (ei − ein−1 ) = K1 h−β ∑ w + K2 h−β ∑ w { { {τ k=0 k=0 { { { n− 21 + (r2 )i , 1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N, { { { { 0 { { {ei = 0, 1 ⩽ i ⩽ M − 1, { { n n {e0 = 0, eM = 0, 0 ⩽ n ⩽ N. Noticing (4.22), it follows from Theorem 4.2.2 that n

󵄩󵄩 n 󵄩󵄩 󵄩 m− 1 󵄩 2 2 2 2 󵄩󵄩e 󵄩󵄩 ⩽ τ ∑ 󵄩󵄩󵄩(r2 ) 2 󵄩󵄩󵄩 ⩽ nτc2 √L(τ + h ) ⩽ c2 T √L(τ + h ), m=1

1 ⩽ n ⩽ N.

The proof ends.

4.3 The fourth-order method based on the WSGL formula for 1D problem In this section, a difference method of order two in time and order four in space will be derived for (4.1)–(4.3) by the mean of the three-term WSGL formula (1.38) in Theorem 1.4.5. ̃ t) defined in Section 4.1 satisfies Suppose u(x, ⋅) ∈ C 3 [0, T] and the function u(x, ̃ t) ∈ C 4+β (ℛ). the condition of Theorem 1.4.5, that is, u(⋅, For any mesh function u = (u0 , u1 , . . . , uM ) ∈ 𝒰h , define the operator (ℬ(β) u)i = {

β

β

β

c2 ui−1 + (1 − 2c2 )ui + c2 ui+1 , ui ,

1 ⩽ i ⩽ M − 1, i = 0, M,

where β

c2 =

−β2 + β + 4 1 1 ∈ ( , ), 24 12 6

1 < β < 2.

In addition, for any mesh function u ∈ 𝒰h̊ , define β

‖u‖2B = ‖u‖2 − h2 c2 ‖δx u‖2 . The following lemma is true. Lemma 4.3.1. For any mesh function u ∈ 𝒰h̊ , it holds (ℬ(β) u, u) = ‖u‖2B ,

1 ‖u‖2 ⩽ ‖u‖2B ⩽ ‖u‖2 . 3

4.3 The fourth-order method based on the WSGL formula for 1D problem |

279

Proof. For any mesh function u ∈ 𝒰h̊ , noticing the definition of the operator ℬ(β) , a direct calculation yields β

(ℬ(β) u, u) = ((ℐ + c2 h2 δx2 )u, u) β

= (u, u) + c2 h2 (δx2 u, u) β

= ‖u‖2 − h2 c2 ‖δx u‖2 = ‖u‖2B .

β

Besides, noticing 121 < c2 < 61 , it follows from the inverse estimate inequality ‖δx u‖ ⩽ 2 ‖u‖ in Lemma 2.1.1 that h 4 β ‖u‖2 = (1 − 4c2 )‖u‖2 h2 1 1 ⩾ (1 − 4 × )‖u‖2 = ‖u‖2 . 6 3 β

‖u‖2B ⩾ ‖u‖2 − h2 c2 ⋅

It is clear that β

‖u‖2B = ‖u‖2 − h2 c2 ‖δx u‖2 ⩽ ‖u‖2 . The proof ends. 4.3.1 Derivation of the difference scheme Considering equation (4.1) at the point (xi , tn ), we have β

ut (xi , tn ) = K1 0 Dβx u(xi , tn ) + K2 x DL u(xi , tn ) + fin , 0 ⩽ i ⩽ M, 0 ⩽ n ⩽ N.

Taking an average on two adjacent time levels and performing the operator ℬ(β) to both hand sides of the resultant equality lead to 1 (β) ℬ [ut (xi , tn ) + ut (xi , tn−1 )] 2 1 = K1 [ℬ(β) 0 Dβx u(xi , tn ) + ℬ(β) 0 Dβx u(xi , tn−1 )] 2 1 β β + K2 [ℬ(β) x DL u(xi , tn ) + ℬ(β) x DL u(xi , tn−1 )] 2 n− 21

+ ℬ(β) fi

,

1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N.

(4.32)

For the space-fractional derivatives in (4.32), it follows from Theorem 1.4.5 that ℬ

(β)

β 0 Dx u(xi , tn )

i+1

n = h−β ∑ ŵ k Ui−k+1 + O(h4 ), k=0

(β)

(4.33)

280 | 4 Difference methods for the space-fractional partial differential equations

ℬ

(β)

M−i+1

β x DL u(xi , tn )

(β) n = h−β ∑ ŵ k Ui+k−1 + O(h4 ),

(4.34)

k=0

where the coefficient {ŵ k } is defined by (1.45), which satisfies (1.46). For the time first-order derivative in (4.32), it follows from the Taylor’s formula that (β)

1 1 [u (x , t ) + ut (xi , tn−1 )] = (Uin − Uin−1 ) + O(τ2 ). 2 t i n τ

(4.35)

The substitution of (4.33)–(4.35) into (4.32) produces ℬ

(β)

n− 21

δt Ui

i+1

n− 1

M−i+1

n− 1

(β) (β) 2 2 = K1 h−β ∑ ŵ k Ui−k+1 + K2 h−β ∑ ŵ k Ui+k−1 k=0

+

1 (β) n− ℬ fi 2

+

n− 1 (r3 )i 2 ,

k=0

1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N,

(4.36)

and there is a positive constant c3 such that 󵄨󵄨 n− 21 󵄨󵄨 2 4 󵄨󵄨(r3 )i 󵄨󵄨 ⩽ c3 (τ + h ),

1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N.

(4.37)

Noticing the initial-boundary value conditions (4.2)–(4.3), we have {

Ui0 = φ(xi ), U0n

= 0,

n UM

1 ⩽ i ⩽ M − 1,

(4.38)

= 0,

(4.39)

0 ⩽ n ⩽ N.

n− 21

Omitting the small term (r3 )i in (4.36) and replacing the exact solution Uin with its numerical one uni , we get another high order difference scheme for solving (4.1)–(4.3) in the form of 1

i+1

1

M−i+1

1

1

n− n− (β) n− 2 (β) n− 2 (β) −β { { ℬ δt ui 2 = K1 h ∑ ŵ k ui−k+1 + K2 h−β ∑ ŵ k ui+k−1 + ℬ(β) fi 2 , { { { { k=0 k=0 { { 1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N, (4.40) { { { { 0 { u = φ(xi ), 1 ⩽ i ⩽ M − 1, (4.41) { { { i n n (4.42) { u0 = 0, uM = 0, 0 ⩽ n ⩽ N.

4.3.2 Solvability of the difference scheme At first, there is a lemma similar to Lemma 4.2.1, which will state some properties of (β) the coefficient {ŵ k }. Lemma 4.3.2. For any mesh function u = (u0 , u1 , . . . , uM ) ∈ 𝒰h̊ , it holds M−1

i+1

i=1

k=0

h ∑ ( ∑ ŵ k ui−k+1 )ui ⩽ 0, M−1 M−i+1

(β)

(β) h ∑ ( ∑ ŵ k ui+k−1 )ui ⩽ 0. i=1

k=0

4.3 The fourth-order method based on the WSGL formula for 1D problem |

281

Next, the unique solvability of the difference scheme (4.40)–(4.42) will be shown. Theorem 4.3.1. The difference scheme (4.40)–(4.42) is uniquely solvable. Proof. Let un = (un0 , un1 , . . . , unM ). The value of u0 is uniquely determined by (4.41)–(4.42). Now assume that the values of u0 , u1 , . . . , un−1 have been uniquely determined, then we can obtain the linear system in the unknown un from (4.40) and (4.42). To show its unique solvability, it is sufficient to verify that the corresponding homogeneous one i+1 M−i+1 1 (β) n 1 1 (β) n (β) −β −β { { ℬ ui = K1 h ∑ ŵ k ui−k+1 + K2 h ∑ ŵ k uni+k−1 , { { τ 2 2 { k=0 k=0 { { 1 ⩽ i ⩽ M − 1, { { { n n u = u = 0 { 0 M

(4.43) (4.44)

has only the trivial solution. Taking the inner product on both hand sides of (4.43) with un , it follows from Lemma 4.3.2 that M−1 i+1 1 M−1 (β) n n 1 (β) h ∑ (ℬ ui )ui = K1 h−β h ∑ ( ∑ ŵ k uni−k+1 )uni τ i=1 2 i=1 k=0 M−1 M−i+1 1 (β) + K2 h−β h ∑ ( ∑ ŵ k uni+k−1 )uni 2 i=1 k=0

⩽ 0. Hence, (ℬ(β) un , un ) ⩽ 0. According to Lemma 4.3.1, we have 󵄩 󵄩2 1 󵄩 󵄩2 (ℬ(β) un , un ) = 󵄩󵄩󵄩un 󵄩󵄩󵄩B ⩾ 󵄩󵄩󵄩un 󵄩󵄩󵄩 , 3 thus ‖un ‖ = 0. It follows that un = 0 with the combination of (4.44), which reveals that the homogeneous system (4.43)–(4.44) has only the trivial solution. By the principle of induction, the difference scheme (4.40)–(4.42) is uniquely solvable. The proof ends. 4.3.3 Stability of the difference scheme Theorem 4.3.2. The solution of the difference scheme (4.40)–(4.42) is stable with respect to both the initial value φ and the right-hand function f . More precisely, it holds n

󵄩 (β) m− 1 󵄩 󵄩󵄩 n 󵄩󵄩 √ 󵄩󵄩 0 󵄩󵄩 󵄩󵄩u 󵄩󵄩 ⩽ 3󵄩󵄩u 󵄩󵄩 + 3τ ∑ 󵄩󵄩󵄩ℬ f 2 󵄩󵄩󵄩, m=1

1 ⩽ n ⩽ N,

282 | 4 Difference methods for the space-fractional partial differential equations where M−1

m− 1 2 󵄩󵄩 (β) m− 21 󵄩󵄩 √ 󵄩󵄩ℬ f 󵄩󵄩 = h ∑ (ℬ(β) fi 2 ) . i=1

Proof. Denote n− 21

gi

n− 21

= ℬ(β) fi

. 1

Taking the inner product on both hand sides of (4.40) with un− 2 , it follows from Lemma 4.3.2 that M−1

n− 21

h ∑ (ℬ(β) δt ui i=1

M−1

i+1

i=1

k=0

n− 21

)ui

n− 1

n− 21

(β) 2 = K1 h−β h ∑ ( ∑ ŵ k ui−k+1 )ui M−1 M−i+1

n− 1

n− 21

(β) 2 + K2 h−β h ∑ ( ∑ ŵ k ui+k−1 )ui M−1

⩽h ∑ i=1

i=1

M−1 i=1

k=0

n− 21 n− 21 ui

+ h ∑ gi

n− 1 n− 1 gi 2 ui 2

1 1 󵄩 󵄩 󵄩 󵄩 ⩽ 󵄩󵄩󵄩g n− 2 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩un− 2 󵄩󵄩󵄩,

1 ⩽ n ⩽ N.

Noticing M−1

n− 21

h ∑ (ℬ(β) δt ui i=1

M−1

n− 21

= h ∑ (δt ui i=1

M−1

n− 21

= h ∑ (ui i=1

n− 21

)ui

n− 21

β

+ c2 h2 δx2 δt ui n− 21

)(δt ui

β

n− 21

)ui

M

) − c2 h2 ⋅ h ∑(δt δx u i=1

n− 21

i− 21

)(δx u

n− 21 i− 21

)

1 󵄩󵄩 n 󵄩󵄩2 󵄩󵄩 n−1 󵄩󵄩2 1 󵄩 β 󵄩2 󵄩 󵄩2 (󵄩u 󵄩 − 󵄩u 󵄩󵄩 ) − c2 h2 ⋅ (󵄩󵄩󵄩δx un 󵄩󵄩󵄩 − 󵄩󵄩󵄩δx un−1 󵄩󵄩󵄩 ) 2τ 󵄩 󵄩 󵄩 2τ 1 󵄩 󵄩2 󵄩 󵄩2 = (󵄩󵄩󵄩un 󵄩󵄩󵄩B − 󵄩󵄩󵄩un−1 󵄩󵄩󵄩B ), 1 ⩽ n ⩽ N 2τ =

and Lemma 4.3.1, further we have 1 1 1 󵄩󵄩 n 󵄩󵄩2 󵄩󵄩 n−1 󵄩󵄩2 󵄩 󵄩 󵄩 󵄩 (󵄩u 󵄩 − 󵄩u 󵄩󵄩B ) ⩽ 󵄩󵄩󵄩g n− 2 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩un− 2 󵄩󵄩󵄩 2τ 󵄩 󵄩B 󵄩 1 1 󵄩 󵄩 󵄩 󵄩 ⩽ √3󵄩󵄩󵄩g n− 2 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩un− 2 󵄩󵄩󵄩

B

4.3 The fourth-order method based on the WSGL formula for 1D problem |

⩽ or

√3 󵄩 n− 1 󵄩 󵄩 n 󵄩 󵄩󵄩g 2 󵄩󵄩(󵄩󵄩u 󵄩󵄩 + 󵄩󵄩󵄩un−1 󵄩󵄩󵄩 ), 󵄩B 󵄩 󵄩 󵄩B 󵄩 2 󵄩

1 1 󵄩󵄩 n 󵄩󵄩 󵄩 󵄩 󵄩 󵄩 (󵄩󵄩u 󵄩󵄩B − 󵄩󵄩󵄩un−1 󵄩󵄩󵄩B ) ⩽ √3󵄩󵄩󵄩g n− 2 󵄩󵄩󵄩, τ

that is

󵄩 n− 1 󵄩 󵄩 n−1 󵄩 󵄩󵄩 n 󵄩󵄩 󵄩󵄩u 󵄩󵄩B ⩽ 󵄩󵄩󵄩u 󵄩󵄩󵄩B + √3 τ󵄩󵄩󵄩g 2 󵄩󵄩󵄩,

283

1 ⩽ n ⩽ N,

1 ⩽ n ⩽ N,

1 ⩽ n ⩽ N.

The recursive process will arrive at n

󵄩 m− 1 󵄩 󵄩󵄩 n 󵄩󵄩 󵄩 0󵄩 󵄩󵄩u 󵄩󵄩B ⩽ 󵄩󵄩󵄩u 󵄩󵄩󵄩B + √3 τ ∑ 󵄩󵄩󵄩g 2 󵄩󵄩󵄩, m=1

1 ⩽ n ⩽ N.

It is easy from Lemma 4.3.1 to get n 1 󵄩󵄩 n 󵄩󵄩 󵄩󵄩 0 󵄩󵄩 √ 󵄩 m− 1 󵄩 󵄩󵄩u 󵄩󵄩 ⩽ 󵄩󵄩u 󵄩󵄩 + 3τ ∑ 󵄩󵄩󵄩g 2 󵄩󵄩󵄩, √3 m=1

1 ⩽ n ⩽ N.

Hence the conclusion is true. The proof ends. 4.3.4 Convergence of the difference scheme Theorem 4.3.3. Suppose {Uin | 0 ⩽ i ⩽ M, 0 ⩽ n ⩽ N} and {uni | 0 ⩽ i ⩽ M, 0 ⩽ n ⩽ N} are solutions of the problem (4.1)–(4.3) and the difference scheme (4.40)–(4.42), respectively. Let ein = Uin − uni ,

0 ⩽ i ⩽ M,

0 ⩽ n ⩽ N,

then it holds 󵄩󵄩 n 󵄩󵄩 2 4 󵄩󵄩e 󵄩󵄩 ⩽ 3c3 T √L(τ + h ),

1 ⩽ n ⩽ N.

Proof. The subtraction of (4.40)–(4.42) from (4.36), (4.38)–(4.39), respectively, will produce the system of error equations as follows: 1

i+1

1

M−i+1

1

n− (β) n− 2 (β) n− 2 (β) −β { { ℬ δt ei 2 = K1 h ∑ ŵ k ei−k+1 + K2 h−β ∑ ŵ k ei+k−1 { { { { k=0 k=0 { { { n− 21 + (r3 )i , 1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N, { { { { 0 { { e = 0, 1 ⩽ i ⩽ M − 1, { { { i n n { e0 = 0, eM = 0, 0 ⩽ n ⩽ N.

Noticing (4.37), it follows immediately from Theorem 4.3.2 that n

󵄩󵄩 n 󵄩󵄩 󵄩 m− 1 󵄩 2 4 2 4 󵄩󵄩e 󵄩󵄩 ⩽ 3τ ∑ 󵄩󵄩󵄩(r3 ) 2 󵄩󵄩󵄩 ⩽ 3nτc3 √L(τ + h ) ⩽ 3c3 T √L(τ + h ), m=1

The proof ends.

1 ⩽ n ⩽ N.

284 | 4 Difference methods for the space-fractional partial differential equations

4.4 The fourth-order ADI method based on the WSGL formula for 2D problem The finite difference method for solving the two-dimensional space-fractional partial differential equations will be considered in this section. Consider the problem β

ut (x, y, t) = K1 0 Dβx u(x, y, t) + K2 x DL u(x, y, t) { { 1 { { { γ γ { + K D u(x, y, t) + K { 3 0 y 4 y DL2 u(x, y, t) + f (x, y, t), { { (x, y) ∈ Ω, t ∈ (0, T], { { { { { { u(x, y, 0) = φ(x, y), (x, y) ∈ Ω, { { { { u(x, y, t) = 0, (x, y) ∈ 𝜕Ω, t ∈ [0, T],

(4.45) (4.46) (4.47)

where Ω = (0, L1 ) × (0, L2 ), 𝜕Ω is the boundary of Ω, the constant Ki (i = 1, 2, 3, 4) is nonnegative, K1 + K2 > 0, K3 + K4 > 0, β, γ ∈ (1, 2), the functions f , φ are given and φ(x, y)|(x,y)∈𝜕Ω = 0. Introduce the same mesh partition, notations and mesh function spaces 𝒱h , 𝒱̊h as those in Section 2.10. Besides, for any mesh function u ∈ 𝒱h , define the average operators β

β

β

c2 ui−1,j + (1 − 2c2 )uij + c2 ui+1,j , 1 ⩽ i ⩽ M1 − 1,

β

ℬx uij = {

uij ,

i = 0 or i = M1 ,

γ

γ

γ

c2 ui,j−1 + (1 − 2c2 )uij + c2 ui,j+1 , 1 ⩽ j ⩽ M2 − 1,

γ

ℬy uij = {

uij ,

j = 0 or j = M2 ,

0 ⩽ j ⩽ M2 , 0 ⩽ i ⩽ M1 ,

where β

c2 =

−β2 + β + 4 1 1 ∈ ( , ), 24 12 6

γ

c2 =

−γ 2 + γ + 4 1 1 ∈ ( , ). 24 12 6

Introduce the following difference operators: i+1

δxβ uij = K1 h1 ∑ ŵ k ui−k+1,j + K2 h1 −β

(β)

−β

k=0 j+1

δyγ uij = K3 h2 ∑ ŵ k ui,j−k+1 + K4 h2 −γ

(γ)

−γ

k=0

M1 −i+1

(β) ∑ ŵ k ui+k−1,j ,

k=0

M2 −j+1

(γ) ∑ ŵ k ui,j+k−1 .

k=0

For any fixed y ∈ [0, L2 ], t ∈ [0, T], define the function ̂ y, t) = { v(x,

u(x, y, t),

x ∈ [0, L1 ],

0,

x ∉ [0, L1 ].

4.4 The fourth-order ADI method based on the WSGL formula for 2D problem

| 285

For any fixed x ∈ [0, L1 ], t ∈ [0, T], define the function ̂ y, t) = { w(x,

u(x, y, t),

y ∈ [0, L2 ],

0,

y ∉ [0, L2 ].

̂ y, t) ∈ C 4+β (ℛ), w(x, ̂ ⋅, t) ∈ C 4+γ (ℛ) and u(x, y, ⋅) ∈ C 3 [0, T]. Suppose v(⋅, Define the mesh functions Uijn = u(xi , yj , tn ),

fijn = f (xi , yj , tn ),

(i, j) ∈ ω,̄ 0 ⩽ n ⩽ N.

4.4.1 Derivation of the difference scheme Denote β

Dβx u(x, y, t) = K1 0 Dβx u(x, y, t) + K2 x DL u(x, y, t), 1

γ

Dγy u(x, y, t) = K3 0 Dγy u(x, y, t) + K4 y DL u(x, y, t). 2

Considering equation (4.45) at the point (xi , yj , tn ), we have ut (xi , yj , tn ) = Dβx u(xi , yj , tn ) + Dγy u(xi , yj , tn ) + fijn , (i, j) ∈ ω,̄ 0 ⩽ n ⩽ N. β γ

Taking an average on two adjacent time levels and performing the operator ℬx ℬy to both hand sides of the resultant equality lead to 1 β γ ℬ ℬ [u (x , y , t ) + ut (xi , yj , tn−1 )] 2 x y t i j n 1 1 n− 1 = ℬyγ [ℬxβ (Dβx Uijn + Dβx Uijn−1 )] + ℬxβ [ℬyγ (Dγy Uijn + Dγy Uijn−1 )] + ℬxβ ℬyγ fij 2 , 2 2 (i, j) ∈ ω, 1 ⩽ n ⩽ N,

(4.48)

n− 1

with fij 2 = 21 (fijn + fijn−1 ). For the time first-order derivative in (4.48), it follows from the Taylor’s formula that 1 1 [ut (xi , yj , tn ) + ut (xi , yj , tn−1 )] = (Uijn − Uijn−1 ) + O(τ2 ). 2 τ For the space-fractional derivatives in (4.48), it follows from Theorem 1.4.5 that β

β

n

i+1

n

ℬx Dx Uij = K1 h1 ∑ ŵ k Ui−k+1,j + K2 h1 −β

k=0

(β)

−β

M1 −i+1

(β) n + O(h41 ) ∑ ŵ k Ui+k−1,j

k=0

(4.49)

286 | 4 Difference methods for the space-fractional partial differential equations = δxβ Uijn + O(h41 ), γ

γ

n

j+1

(4.50) n

ℬy Dy Uij = K3 h2 ∑ ŵ k Ui,j−k+1 + K4 h2 −γ

=

δyγ Uijn

(γ)

−γ

k=0

+

O(h42 ).

M2 −j+1

(γ) n + O(h42 ) ∑ ŵ k Ui,j+k−1

k=0

(4.51)

Substituting (4.49)–(4.51) into (4.48), we get β γ

n− 21

ℬx ℬy δt Uij

n− 21

= ℬyγ δxβ Uij

n− 21

+ ℬxβ δyγ Uij

n− 21

+ ℬxβ ℬyγ fij

n− 1

+ (r4 )ij 2 ,

(i, j) ∈ ω, 1 ⩽ n ⩽ N,

(4.52)

and there is a positive constant c4 such that n− 1 󵄨 󵄨󵄨 2 4 4 󵄨󵄨(r4 )ij 2 󵄨󵄨󵄨 ⩽ c4 (τ + h1 + h2 ), n− 21

β γ

Adding a small term 41 τ2 δx δy δt Uij β γ

n− 21

ℬx ℬy δt Uij

n− 21

+ ℬxβ ℬyγ fij

+

(i, j) ∈ ω, 1 ⩽ n ⩽ N.

to both hand sides of (4.52) arrives at

τ2 β γ n− 1 n− 1 n− 1 δx δy δt Uij 2 = ℬyγ δxβ Uij 2 + ℬxβ δyγ Uij 2 4 n− 1

(i, j) ∈ ω, 1 ⩽ n ⩽ N,

+ (r5 )ij 2 ,

(4.53)

where n− 21

(r5 )ij

n− 21

= (r4 )ij

+

τ2 β γ n− 1 δx δy δt Uij 2 4

and there is a positive constant c5 such that 󵄨󵄨 n− 21 󵄨󵄨 2 4 4 󵄨󵄨(r5 )ij 󵄨󵄨 ⩽ c5 (τ + h1 + h2 ),

(i, j) ∈ ω, 1 ⩽ n ⩽ N.

(4.54)

Noticing the initial-boundary value conditions (4.46)–(4.47), we have {

Uij0 = φ(xi , yj ), Uijn

= 0,

(i, j) ∈ ω,

(4.55)

(i, j) ∈ 𝜕ω, 0 ⩽ n ⩽ N.

(4.56)

n− 1

Neglecting the small term (r5 )ij 2 in (4.53) and replacing the exact solution Uijn with its numerical one unij , we get a difference scheme for solving (4.45)–(4.47) as follows: 1 1 1 τ2 β γ n− 21 n− 21 β γ γ β n− 2 β γ n− 2 β γ n− 2 { ℬ ℬ δ u + δ δ δ u = ℬ δ u + ℬ δ u + ℬ ℬ f , { t t x y x y y x x y x y { ij ij ij ij ij { 4 { { { (i, j) ∈ ω, 1 ⩽ n ⩽ N, { { 0 { { uij = φ(xi , yj ), (i, j) ∈ ω, { { { n { uij = 0, (i, j) ∈ 𝜕ω, 0 ⩽ n ⩽ N.

(4.57) (4.58) (4.59)

4.4 The fourth-order ADI method based on the WSGL formula for 2D problem

| 287

Reformulate equation (4.57) as τ (ℬxβ − δxβ )(ℬyγ − 2 τ = (ℬxβ + δxβ )(ℬyγ + 2

τ γ n δ )u 2 y ij τ γ n−1 n− 1 δy )uij + τℬxβ ℬyγ fij 2 . 2

(4.60)

Let τ u∗ij = (ℬyγ − δyγ )unij , 2 then the difference scheme (4.57)–(4.59) can be decomposed into the following ADI form: On the time level t = tn (1 ⩽ n ⩽ N), at first, for any fixed j (1 ⩽ j ⩽ M2 − 1), solve the one-dimensional problem with respect to {u∗ij | 0 ⩽ i ⩽ M1 } in x direction τ { { (ℬβ − δβ )u∗ = (ℬxβ + { { { x 2 x ij { { { τ { { u∗0j = (ℬyγ − δyγ )un0j , 2 {

1 τ β τ β γ n− 2 δx )(ℬyγ + δyγ )un−1 , ij + τℬx ℬy fij 2 2

1 ⩽ i ⩽ M1 − 1,

τ u∗M1 ,j = (ℬyγ − δyγ )unM1 ,j 2

to get the value of {u∗ij | 1 ⩽ i ⩽ M1 − 1}; Then, for any fixed i (1 ⩽ i ⩽ M1 − 1), solve the one-dimensional problem with respect to {unij | 0 ⩽ j ⩽ M2 } in y direction { { { { {

τ (ℬyγ − δyγ )unij = u∗ij , 2 uni0 = 0, uni,M2 = 0

1 ⩽ j ⩽ M2 − 1,

to get the value of {unij | 1 ⩽ j ⩽ M2 − 1}. 4.4.2 Three lemmas In this subsection, three important lemmas will be listed so as to facilitate the analysis on the difference scheme presented above. β γ In view of the self-adjoint and positive definite operators ℬx and ℬy , they can be β

γ

performed the square root decomposition, that is, there are two operators 𝒬x and 𝒬y , such that β

β 2

ℬx = (𝒬x ) ,

γ

γ 2

ℬy = (𝒬y ) .

288 | 4 Difference methods for the space-fractional partial differential equations β

β

γ

γ

It is easy to see that ℬx and ℬy are commutable and so are 𝒬x and 𝒬y . Next, three useful lemmas will be introduced. Lemma 4.4.1. For any mesh function v ∈ 𝒱̊h , it holds 1 󵄩 󵄩 ‖v‖ ⩽ 󵄩󵄩󵄩𝒬βx 𝒬γy v󵄩󵄩󵄩, 3

󵄩󵄩 β −1 γ −1 󵄩󵄩 󵄩󵄩(𝒬x ) (𝒬y ) v󵄩󵄩 ⩽ 3‖v‖. β

γ

Proof. It follows from the definitions of the operators ℬx , ℬy and Lemma 4.3.1 that 1 2 ‖v‖ ⩽ (ℬxβ v, v) ⩽ ‖v‖2 , 3

1 2 ‖v‖ ⩽ (ℬyγ v, v) ⩽ ‖v‖2 . 3

On the one hand, we have 2

(ℬxβ ℬyγ v, v) = ((𝒬βx ) ℬyγ v, v) = (𝒬βx ℬyγ v, 𝒬βx v) = (ℬyγ 𝒬βx v, 𝒬βx v) ⩾

1 β 1 1 (𝒬 v, 𝒬βx v) = (ℬxβ v, v) ⩾ ‖v‖2 . 3 x 3 9

(4.61)

On the other hand, it holds 2

2

(ℬxβ ℬyγ v, v) = ((𝒬βx ) (𝒬γy ) v, v) 󵄩 󵄩2 = (𝒬βx 𝒬γy v, 𝒬βx 𝒬γy v) = 󵄩󵄩󵄩𝒬βx 𝒬γy v󵄩󵄩󵄩 .

(4.62)

The combination of (4.61) and (4.62) gives 1 󵄩 󵄩 ‖v‖ ⩽ 󵄩󵄩󵄩𝒬βx 𝒬γy v󵄩󵄩󵄩. 3

(4.63)

Moreover, it follows from (4.63) that 1 󵄩󵄩 β −1 γ −1 󵄩󵄩 󵄩󵄩 β γ β −1 γ −1 󵄩󵄩 󵄩(𝒬 ) (𝒬y ) v󵄩󵄩 ⩽ 󵄩󵄩𝒬x 𝒬y (𝒬x ) (𝒬y ) v󵄩󵄩 = ‖v‖, 3󵄩 x which implies the second inequality in this lemma. The proof ends. Lemma 4.4.2. For any mesh function v ∈ 𝒱̊h , it holds (v, δxβ v) ⩽ 0,

(v, δyγ v) ⩽ 0.

The proof can be proceeded similarly to that for Lemma 4.2.1 and the details are omitted here. Lemma 4.4.3. For any mesh function v ∈ 𝒱̊h , it holds 󵄩󵄩 󵄩󵄩 β 󵄩󵄩(𝒬x − 󵄩󵄩 󵄩󵄩 󵄩󵄩 β 󵄩󵄩(𝒬x + 󵄩󵄩

2 τ β −1 β 󵄩󵄩󵄩󵄩 󵄩 󵄩2 (𝒬x ) δx )v󵄩󵄩 ⩾ 󵄩󵄩󵄩𝒬βx v󵄩󵄩󵄩 + 󵄩󵄩 2 2 τ β −1 β 󵄩󵄩󵄩󵄩 󵄩 󵄩2 (𝒬x ) δx )v󵄩󵄩 ⩽ 󵄩󵄩󵄩𝒬βx v󵄩󵄩󵄩 + 󵄩󵄩 2

τ2 󵄩󵄩 β −1 β 󵄩󵄩2 󵄩(𝒬 ) δx v󵄩󵄩 , 4󵄩 x τ2 󵄩󵄩 β −1 β 󵄩󵄩2 󵄩(𝒬 ) δx v󵄩󵄩 , 4󵄩 x

4.4 The fourth-order ADI method based on the WSGL formula for 2D problem

󵄩󵄩 󵄩󵄩 γ 󵄩󵄩(𝒬y − 󵄩󵄩 󵄩󵄩 󵄩󵄩 γ 󵄩󵄩(𝒬y + 󵄩󵄩

2 τ γ −1 γ 󵄩󵄩󵄩󵄩 󵄩 󵄩2 (𝒬y ) δy )v󵄩󵄩 ⩾ 󵄩󵄩󵄩𝒬γy v󵄩󵄩󵄩 + 󵄩 2 󵄩 2 τ γ −1 γ 󵄩󵄩󵄩󵄩 󵄩 󵄩2 (𝒬y ) δy )v󵄩󵄩 ⩽ 󵄩󵄩󵄩𝒬γy v󵄩󵄩󵄩 + 󵄩󵄩 2

| 289

τ2 󵄩󵄩 γ −1 γ 󵄩󵄩2 󵄩(𝒬 ) δy v󵄩󵄩 , 4󵄩 y τ2 󵄩󵄩 γ −1 γ 󵄩󵄩2 󵄩(𝒬 ) δy v󵄩󵄩 . 4󵄩 y

Proof. By Lemma 4.4.2, some calculations give 󵄩2 󵄩󵄩 󵄩󵄩 β τ β −1 β 󵄩󵄩󵄩 ( 𝒬 ) δ )v ( 𝒬 − 󵄩󵄩 󵄩󵄩 x x 󵄩󵄩 󵄩󵄩 2 x τ τ −1 −1 = ((𝒬βx − (𝒬βx ) δxβ )v, (𝒬βx − (𝒬βx ) δxβ )v) 2 2 = (𝒬βx v, 𝒬βx v) +

τ2 −1 −1 −1 ((𝒬βx ) δxβ v, (𝒬βx ) δxβ v) − τ(𝒬βx v, (𝒬βx ) δxβ v) 4

2 −1 󵄩 󵄩2 τ 󵄩 󵄩2 = 󵄩󵄩󵄩𝒬βx v󵄩󵄩󵄩 + 󵄩󵄩󵄩(𝒬βx ) δxβ v󵄩󵄩󵄩 − τ(v, δxβ v) 4 2 −1 󵄩2 󵄩 󵄩2 τ 󵄩 ⩾ 󵄩󵄩󵄩𝒬βx v󵄩󵄩󵄩 + 󵄩󵄩󵄩(𝒬βx ) δxβ v󵄩󵄩󵄩 . 4

Other three inequalities in this lemma can be proved similarly. The proof ends. The following part will focus on the analysis for the difference scheme (4.57)– (4.59). 4.4.3 Solvability of the difference scheme Theorem 4.4.1. The difference scheme (4.57)–(4.59) is uniquely solvable. Proof. Let ̄ un = {unij | (i, j) ∈ ω}. The value of u0 is obviously determined by (4.58)–(4.59). Now assume that the values of u0 , u1 , . . . , un−1 have been uniquely determined, then we can obtain the linear system in the unknown un from (4.57) and (4.59). To show its unique solvability, it is sufficient to verify that the corresponding homogeneous one τ τ { { (ℬxβ − δxβ )(ℬyγ − δyγ )unij = 0, 2 2 { { n u = 0, (i, j) ∈ 𝜕ω { ij

(i, j) ∈ ω,

(4.64) (4.65)

has only the trivial solution. β γ To this end, performing the operator (𝒬x )−1 (𝒬y )−1 to both hand sides of (4.64) yields τ τ −1 −1 (𝒬βx − (𝒬βx ) δxβ )(𝒬γy − (𝒬γy ) δyγ )unij = 0, 2 2

(i, j) ∈ ω.

290 | 4 Difference methods for the space-fractional partial differential equations From Lemmas 4.4.1, 4.4.2 and 4.4.3, we obtain 󵄩󵄩 󵄩󵄩2 τ τ −1 −1 󵄩 󵄩 0 = 󵄩󵄩󵄩(𝒬βx − (𝒬βx ) δxβ )(𝒬γy − (𝒬γy ) δyγ )un 󵄩󵄩󵄩 󵄩󵄩 󵄩󵄩 2 2 2 󵄩󵄩 󵄩󵄩2 󵄩󵄩 τ2 󵄩󵄩󵄩 τ τ −1 −1 −1 󵄩 󵄩 󵄩 = 󵄩󵄩󵄩𝒬βx (𝒬γy − (𝒬γy ) δyγ )un 󵄩󵄩󵄩 + 󵄩󵄩󵄩(𝒬βx ) δxβ (𝒬γy − (𝒬γy ) δyγ )un 󵄩󵄩󵄩 󵄩󵄩 󵄩󵄩 2 4 󵄩󵄩 2 󵄩󵄩 τ τ −1 −1 −1 − τ(𝒬βx (𝒬γy − (𝒬γy ) δyγ )un , (𝒬βx ) δxβ (𝒬γy − (𝒬γy ) δyγ )un ) 2 2 󵄩󵄩󵄩 β γ τ γ −1 γ n 󵄩󵄩󵄩2 τ2 󵄩󵄩󵄩 β −1 β γ τ γ −1 γ n 󵄩󵄩󵄩2 = 󵄩󵄩󵄩𝒬x (𝒬y − (𝒬y ) δy )u 󵄩󵄩󵄩 + 󵄩󵄩󵄩(𝒬x ) δx (𝒬y − (𝒬y ) δy )u 󵄩󵄩󵄩 2 4 󵄩󵄩 2 󵄩󵄩 󵄩󵄩 󵄩󵄩 τ τ −1 −1 − τ((𝒬γy − (𝒬γy ) δyγ )un , δxβ (𝒬γy − (𝒬γy ) δyγ )un ) 2 2 󵄩󵄩󵄩 β γ τ γ −1 γ n 󵄩󵄩󵄩2 τ2 󵄩󵄩󵄩 β −1 β γ τ γ −1 γ n 󵄩󵄩󵄩2 ⩾ 󵄩󵄩󵄩𝒬x (𝒬y − (𝒬y ) δy )u 󵄩󵄩󵄩 + 󵄩󵄩󵄩(𝒬x ) δx (𝒬y − (𝒬y ) δy )u 󵄩󵄩󵄩 2 4 󵄩󵄩 2 󵄩󵄩 󵄩󵄩 󵄩󵄩 2 2󵄩 󵄩󵄩 󵄩 󵄩󵄩2 󵄩 󵄩 τ τ τ 󵄩 −1 −1 −1 󵄩 󵄩 󵄩 = 󵄩󵄩󵄩(𝒬γy − (𝒬γy ) δyγ )𝒬βx un 󵄩󵄩󵄩 + 󵄩󵄩󵄩(𝒬γy − (𝒬γy ) δyγ )(𝒬βx ) δxβ un 󵄩󵄩󵄩 2 4 󵄩󵄩 2 󵄩󵄩 󵄩󵄩 󵄩󵄩 2 −1 󵄩2 󵄩 󵄩2 τ 󵄩 ⩾ 󵄩󵄩󵄩𝒬γy 𝒬βx un 󵄩󵄩󵄩 + 󵄩󵄩󵄩(𝒬γy ) δyγ 𝒬βx un 󵄩󵄩󵄩 4 2 τ2 󵄩 −1 −1 −1 󵄩2 τ 󵄩 󵄩2 + (󵄩󵄩󵄩𝒬γy (𝒬βx ) δxβ un 󵄩󵄩󵄩 + 󵄩󵄩󵄩(𝒬γy ) δyγ (𝒬βx ) δxβ un 󵄩󵄩󵄩 ) 4 4 󵄩 󵄩2 1 󵄩 󵄩2 ⩾ 󵄩󵄩󵄩𝒬γy 𝒬βx un 󵄩󵄩󵄩 ⩾ 󵄩󵄩󵄩un 󵄩󵄩󵄩 , 9

(4.66)

thus, ‖un ‖ = 0. We conclude that un = 0 from (4.65). By the principle of induction, the difference scheme (4.57)–(4.59) is uniquely solvable. The proof ends. 4.4.4 Stability of the difference scheme Theorem 4.4.2. Suppose {unij | (i, j) ∈ ω,̄ 0 ⩽ n ⩽ N} is the solution of the difference n− 21

scheme (4.57)–(4.59) and denote sij

1 β γ n− 2

= ℬx ℬy fij

, then it holds

󵄩󵄩 n 󵄩󵄩 󵄩 γ β 0 󵄩 τ 󵄩 γ −1 γ β 0 󵄩 󵄩󵄩u 󵄩󵄩 ⩽ 3(󵄩󵄩󵄩𝒬y 𝒬x u 󵄩󵄩󵄩 + 󵄩󵄩󵄩(𝒬y ) δy 𝒬x u 󵄩󵄩󵄩 + 2 +

τ 󵄩󵄩 γ β −1 β 0 󵄩󵄩 󵄩𝒬 (𝒬 ) δx u 󵄩󵄩 2󵄩 y x

n τ2 󵄩󵄩 γ −1 γ β −1 β 0 󵄩󵄩 󵄩 m− 1 󵄩 󵄩󵄩(𝒬y ) δy (𝒬x ) δx u 󵄩󵄩) + 9τ ∑ 󵄩󵄩󵄩s 2 󵄩󵄩󵄩, 4 m=1

where M1 −1 M2 −1

m− 1 2 󵄩󵄩 m− 21 󵄩󵄩 √ 󵄩󵄩s 󵄩󵄩 = h ∑ ∑ (sij 2 ) . i=1

j=1

1 ⩽ n ⩽ N,

4.4 The fourth-order ADI method based on the WSGL formula for 2D problem

| 291

Proof. Noticing that equation (4.57) can be rewritten as (4.60) and performing the opβ γ erator (𝒬x )−1 (𝒬y )−1 to both hand sides of (4.60) arrive at τ −1 (𝒬βx − (𝒬βx ) δxβ )(𝒬γy − 2 τ −1 = (𝒬βx + (𝒬βx ) δxβ )(𝒬γy + 2 −1 n− 1

+ τ(𝒬βx ) (𝒬γy ) sij 2 , −1

τ γ −1 γ n (𝒬 ) δy )uij 2 y τ γ −1 γ n−1 (𝒬 ) δy )uij 2 y

(i, j) ∈ ω, 1 ⩽ n ⩽ N.

(4.67)

Taking the norm on both hand sides of (4.67), it follows from the triangle inequality that 󵄩󵄩 τ γ −1 γ n 󵄩󵄩󵄩 󵄩󵄩 β τ β −1 β γ 󵄩󵄩(𝒬x − (𝒬x ) δx )(𝒬y − (𝒬y ) δy )u 󵄩󵄩󵄩 󵄩󵄩 󵄩󵄩 2 2 󵄩󵄩 󵄩󵄩 τ τ −1 −1 󵄩 󵄩 ⩽ 󵄩󵄩󵄩(𝒬βx + (𝒬βx ) δxβ )(𝒬γy + (𝒬γy ) δyγ )un−1 󵄩󵄩󵄩 󵄩󵄩 󵄩󵄩 2 2 󵄩󵄩 β −1 γ −1 n− 21 󵄩󵄩 (4.68) + τ󵄩󵄩(𝒬x ) (𝒬y ) s 󵄩󵄩, 1 ⩽ n ⩽ N. In analogy to the derivation of (4.66), we have 2 󵄩󵄩 τ γ −1 γ n 󵄩󵄩󵄩 󵄩󵄩 β τ β −1 β γ 󵄩󵄩(𝒬x − (𝒬x ) δx )(𝒬y − (𝒬y ) δy )u 󵄩󵄩󵄩 󵄩󵄩 󵄩󵄩 2 2 2 2 −1 −1 󵄩2 τ 󵄩 󵄩2 󵄩 󵄩2 τ 󵄩 ⩾ 󵄩󵄩󵄩𝒬γy 𝒬βx un 󵄩󵄩󵄩 + 󵄩󵄩󵄩(𝒬γy ) δyγ 𝒬βx un 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝒬γy (𝒬βx ) δxβ un 󵄩󵄩󵄩 4 4 τ4 󵄩 −1 −1 󵄩2 + 󵄩󵄩󵄩(𝒬γy ) δyγ (𝒬βx ) δxβ un 󵄩󵄩󵄩 16

(4.69)

and 2 󵄩󵄩 τ γ −1 γ n−1 󵄩󵄩󵄩 󵄩󵄩 β τ β −1 β γ 󵄩󵄩(𝒬x + (𝒬x ) δx )(𝒬y + (𝒬y ) δy )u 󵄩󵄩󵄩 󵄩󵄩 󵄩󵄩 2 2 2 2 −1 −1 󵄩2 τ 󵄩 󵄩2 󵄩 󵄩2 τ 󵄩 ⩽ 󵄩󵄩󵄩𝒬γy 𝒬βx un−1 󵄩󵄩󵄩 + 󵄩󵄩󵄩(𝒬γy ) δyγ 𝒬βx un−1 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝒬γy (𝒬βx ) δxβ un−1 󵄩󵄩󵄩 4 4 τ4 󵄩 −1 −1 󵄩2 + 󵄩󵄩󵄩(𝒬γy ) δyγ (𝒬βx ) δxβ un−1 󵄩󵄩󵄩 . 16

Denote 2 2 −1 −1 󵄩2 τ 󵄩 󵄩2 󵄩 󵄩2 τ 󵄩 E n = (󵄩󵄩󵄩𝒬γy 𝒬βx un 󵄩󵄩󵄩 + 󵄩󵄩󵄩(𝒬γy ) δyγ 𝒬βx un 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝒬γy (𝒬βx ) δxβ un 󵄩󵄩󵄩 4 4 1/2

+

τ4 󵄩󵄩 γ −1 γ β −1 β n 󵄩󵄩2 󵄩(𝒬 ) δy (𝒬x ) δx u 󵄩󵄩 ) , 16 󵄩 y

0 ⩽ n ⩽ N.

It follows from (4.68)–(4.70) and Lemma 4.4.1 that 1 −1 −1 󵄩 󵄩 󵄩 1󵄩 E n ⩽ E n−1 + τ󵄩󵄩󵄩(𝒬βx ) (𝒬γy ) sn− 2 󵄩󵄩󵄩 ⩽ E n−1 + 3τ󵄩󵄩󵄩sn− 2 󵄩󵄩󵄩,

1 ⩽ n ⩽ N.

(4.70)

292 | 4 Difference methods for the space-fractional partial differential equations Applying the recursive process leads to n

1 󵄩 󵄩 E n ⩽ E 0 + 3τ ∑ 󵄩󵄩󵄩sm− 2 󵄩󵄩󵄩,

m=1

1 ⩽ n ⩽ N.

Noticing 󵄩 󵄩 1󵄩 󵄩 E n ⩾ 󵄩󵄩󵄩𝒬γy 𝒬βx un 󵄩󵄩󵄩 ⩾ 󵄩󵄩󵄩un 󵄩󵄩󵄩, 3 further it follows n

󵄩 m− 1 󵄩 󵄩󵄩 n 󵄩󵄩 0 󵄩󵄩u 󵄩󵄩 ⩽ 3E + 9τ ∑ 󵄩󵄩󵄩s 2 󵄩󵄩󵄩 m=1

−1 󵄩 󵄩 󵄩 τ󵄩 ⩽ 3(󵄩󵄩󵄩𝒬γy 𝒬βx u0 󵄩󵄩󵄩 + 󵄩󵄩󵄩(𝒬γy ) δyγ 𝒬βx u0 󵄩󵄩󵄩 + 2

+

τ 󵄩󵄩 γ β −1 β 0 󵄩󵄩 󵄩𝒬 (𝒬 ) δx u 󵄩󵄩 2󵄩 y x

n τ2 󵄩󵄩 γ −1 γ β −1 β 0 󵄩󵄩 󵄩 m− 1 󵄩 󵄩󵄩(𝒬y ) δy (𝒬x ) δx u 󵄩󵄩) + 9τ ∑ 󵄩󵄩󵄩s 2 󵄩󵄩󵄩, 4 m=1

1 ⩽ n ⩽ N,

which is exactly the desired result. The proof ends. 4.4.5 Convergence of the difference scheme Theorem 4.4.3. Suppose {Uijn | (i, j) ∈ ω,̄ 0 ⩽ n ⩽ N} and {unij | (i, j) ∈ ω,̄ 0 ⩽ n ⩽ N} are solutions of the problem (4.45)–(4.47) and the difference scheme (4.57)–(4.59), respectively. Let eijn = Uijn − unij ,

(i, j) ∈ ω,̄ 0 ⩽ n ⩽ N,

then it holds 󵄩󵄩 n 󵄩󵄩 2 4 4 󵄩󵄩e 󵄩󵄩 ⩽ 9T √L1 L2 c5 (τ + h1 + h2 ),

1 ⩽ n ⩽ N.

Proof. Subtracting (4.57)–(4.59) from (4.53), (4.55)–(4.56), respectively, will arrive at the system of error equations as follows: τ2 β γ n− 21 n− 21 n− 1 n− 1 n− 1 β γ { ℬ ℬ δ e + δx δy δt eij = ℬyγ δxβ eij 2 + ℬxβ δyγ eij 2 + (r5 )ij 2 , { t x y { ij { 4 { { { (i, j) ∈ ω, 1 ⩽ n ⩽ N, { { 0 { { eij = 0, (i, j) ∈ ω, { { { n { eij = 0, (i, j) ∈ 𝜕ω, 0 ⩽ n ⩽ N. Noticing (4.54), the application of Theorem 4.4.2 immediately yields n

󵄩󵄩 n 󵄩󵄩 󵄩 m− 1 󵄩 2 4 4 󵄩󵄩e 󵄩󵄩 ⩽ 9τ ∑ 󵄩󵄩󵄩(r5 ) 2 󵄩󵄩󵄩 ⩽ 9T √L1 L2 c5 (τ + h1 + h2 ), m=1

The proof ends.

1 ⩽ n ⩽ N.

4.5 Supplementary remarks and discussions | 293

4.5 Supplementary remarks and discussions 1. This chapter mainly focused on the finite difference methods for solving the initialboundary value problems of space-fractional partial differential equations in R-L type. For 1D problem (4.1)–(4.3), three difference schemes of order one in both time and space, order two in both time and space, together with order two in time and four in space, respectively, were built. The unique solvability, stability and convergence of each scheme were analyzed. For 2D problem (4.45)–(4.47), an ADI difference scheme of order two in time and four in space was mainly addressed along with the analysis on its unique solvability, stability and convergence. 2. For the R-L fractional derivative in the space-fractional partial differential equations, Meerschaert and Tadjeran pointed out that the difference scheme using the standard G-L formula to approximate the R-L fractional derivative was unstable in [58, 85], and proposed the shifted G-L formula for the approximation of the R-L fractional derivative to derive the difference scheme. 3. For the R-L fractional derivative in the space-fractional differential equations, Tian et al.[88] presented a two-term WSGL formula to establish the difference method of order two in both time and space. Besides, a three-term WSGL formula was also derived. Based on the work[88] , a series of works have been done by this group; please see [7, 117] and so on. 4. The numerical solution of the following one-dimensional space-fractional partial differential equation: ut (x, t) =

𝜕β u(x, t) + f (x, t) 𝜕|x|β

was studied in [4], where a Crank–Nicolson difference scheme of order two in both time and space was developed with the help of the central difference quotient formula (Lemma 1.5.1) to approximate the Riesz fractional derivative. In [64, 102], the numerical methods for solving the diffusion and advection-diffusion equations with Riesz fractional derivatives were also discussed. 5. A fourth-order numerical differentiation formula (1.54) to approximate the weighted value of Riesz fractional derivatives at three points was proposed by Zhao et al.[116] . On this basis, a higher-order difference method was investigated for solving the nonlinear space-fractional Schrödinger equation in the Riesz derivative type. Ding et al.[14, 15] derived a different higher-order numerical differentiation formula to approximate the Riesz fractional derivative and applied it to the numerical solution of the space-fractional differential equation in the Riesz derivative type. 6. For the difference schemes to solve the space-fractional differential equations, Wang et al.[91, 95] and Lei et al.[43] developed some fast methods in view of the special structures of difference schemes. 7. Wang et al.[97, 98] presented implicit conservative difference schemes for the space fractional nonlinear Schrödinger equations. Wang and Huang[92, 93] , Hao and

294 | 4 Difference methods for the space-fractional partial differential equations Sun[34] , He and Pan[36] have studied the difference methods for the fractional Ginzburg– Landau equation.

Exercises 4 4.1 For the problem (4.1)–(4.3), construct the following explicit difference scheme: { { { { { { { { { { { { { { { { {

i+1 M−i+1 1 n+1 (β) (β) (ui − uni ) = K1 h−β ∑ gk uni−k+1 + K2 h−β ∑ gk uni+k−1 + fin , τ k=0 k=0

u0i un0

1 ⩽ i ⩽ M − 1, 0 ⩽ n ⩽ N − 1,

= φ(xi ),

= 0,

unM

1 ⩽ i ⩽ M − 1, = 0,

0 ⩽ n ⩽ N.

̃ t) like that in Section 4.1 and suppose u(⋅, ̃ t) ∈ C 1+β (ℛ). Define the function u(x, For this difference scheme, try to (1) analyze the truncation error; (2) show the stability with respect to the initial value φ and the function f when βτ (K1 + K2 ) hβ ⩽ 1; βτ

(3) show the convergence when (K1 + K2 ) hβ ⩽ 1, and derive the error expression. 4.2 For the problem (4.1)–(4.3), construct the following implicit difference scheme: { { { { { { { { { { { { { { { { {

M−i+1 i+1 1 n (β) n −β −β ̃k(β) uni+k−1 + fin , ̃ u + K h (ui − un−1 ) = K h w ∑ w ∑ 2 1 i−k+1 i k τ k=0 k=0

u0i un0

1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N,

= φ(xi ),

= 0,

unM

1 ⩽ i ⩽ M − 1, = 0,

0 ⩽ n ⩽ N.

̃ t) like that in Section 4.1 and suppose u(⋅, ̃ t) ∈ C 2+β (ℛ). Define the function u(x, For this difference scheme, try to (1) analyze the truncation error; (2) show the unique solvability; (3) show the stability with respect to the initial value φ and the function f ; (4) show the convergence and derive the error expression. 4.3 For the problem (4.1)–(4.3), construct the following difference scheme: n n−1 i+1 M−i+1 (β) n (β) (β) ui − ui −β −β { { ̂ ℬ = K h w u + K h ∑ ∑ ŵ k uni+k−1 + ℬ(β) fin , { 1 2 i−k+1 k { { τ { k=0 k=0 { { 1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N, { { { 0 { { u = φ(xi ), 1 ⩽ i ⩽ M − 1, { { { in n { u0 = 0, uM = 0, 0 ⩽ n ⩽ N.

̃ t) like that in Section 4.1 and suppose u(⋅, ̃ t) ∈ C 4+β (ℛ). Define the function u(x,

Exercises 4

| 295

For this difference scheme, try to (1) analyze the truncation error; (2) show the unique solvability; (3) show the stability with respect to the initial value φ and the function f ; (4) show the convergence and derive the error expression. 4.4 For the problem (4.45)–(4.47), construct the following difference scheme: τ2 n− 1 n− 1 n− 1 n− 1 n− 1 { δt uij 2 + δxβ δyγ δt uij 2 = δxβ uij 2 + δyγ uij 2 + fij 2 , { { { 4 { { { (i, j) ∈ ω, 1 ⩽ n ⩽ N, { { 0 { { uij = φ(xi , yj ), (i, j) ∈ ω, { { { n { uij = 0, (i, j) ∈ 𝜕ω, 0 ⩽ n ⩽ N. ̂ y, t), w(x, ̂ y, t) like those in Section 4.4 and suppose Define the functions v(x, ̂ y, t) ∈ C 2+β (ℛ), w(x, ̂ ⋅, t) ∈ C 2+γ (ℛ). v(⋅, For this difference scheme, try to (1) analyze the truncation error; (2) show the unique solvability; (3) write the ADI form; (4) show the stability with respect to the initial value φ and the function f ; (5) show the convergence and derive the error expression. 4.5 For the problem 𝜕β u(x, t) { u (x, t) = + f (x, t), 0 < x < L, 0 < t ⩽ T, { { { t 𝜕|x|β { { u(x, 0) = φ(x), 0 < x < L, { { { u(0, t) = 0, u(L, t) = 0, 0 ⩽ t ⩽ T,

(4.71) (4.72) (4.73)

where 𝜕β u(x, t) β = −Ψβ (0 Dβx u(x, t) + x DL u(x, t)), 𝜕|x|β x

β 0 Dx u(x, t)

u(ξ , t) 1 𝜕2 = dξ , ∫ 2 Γ(2 − β) 𝜕x (x − ξ )β−1

β x DL u(x, t)

u(ξ , t) 1 𝜕2 = dξ , ∫ Γ(2 − β) 𝜕x2 (ξ − x)β−1

0

L

x

the functions f , φ are given and φ(0) = φ(L) = 0.

Ψβ =

1 2 cos(

βπ ) 2

,

296 | 4 Difference methods for the space-fractional partial differential equations Construct the following difference scheme: i uni − un−1 { (β) i { = −h−β ∑ ĝk uni−k + fin , { { { τ { k=i−M { 0 { { ui = φ(xi ), 1 ⩽ i ⩽ M − 1, { { { n n { u0 = 0, uM = 0, 0 ⩽ n ⩽ N,

1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N,

where the coefficient {ĝk } is defined by (1.48). ̃ t) like that in Section 4.1 and suppose u(⋅, ̃ t) ∈ C 2+β (ℛ). Define the function u(x, For this difference scheme, try to (1) analyze the truncation error; (2) show the unique solvability; (3) show the stability with respect to the initial value φ and the function f ; (4) show the convergence and derive the error expression. 4.6 For the problem (4.71)–(4.73), construct the following difference scheme: (β)

i 1 uni − un−1 n− 1 { (β) n− i { = −h−β ∑ ĝk ui−k2 + fi 2 , { { { τ { k=i−M { 0 { { ui = φ(xi ), 1 ⩽ i ⩽ M − 1, { { { n n { u0 = 0, uM = 0, 0 ⩽ n ⩽ N,

1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N,

where the coefficient {ĝk } is defined by (1.48). ̃ t) like that in Section 4.1 and suppose u(⋅, ̃ t) ∈ C 2+β (ℛ). Define the function u(x, For this difference scheme, try to (1) analyze the truncation error; (2) show the unique solvability; (3) show the stability with respect to the initial value φ and the function f : (4) show the convergence and derive the error expression. (β)

5 Difference methods for the time-space-fractional differential equations In this chapter, the finite difference methods for a class of time-space-fractional differential equations (Bloch–Torrey equations) will be considered. Many applications for this kind of equations can be found. It has been successfully applied to describe the diffusion image of human brain tissues and provides new insights into further investigations of tissue structures and the microenvironment. The Bloch–Torrey equation consists of both the time-fractional derivative (Caputo derivative) and the spacefractional derivative (Riesz derivative). In this chapter, for 1D problem, the method of order two in both time and space and another method of order two in time and four in space will be successively discussed. For 2D problem, the method of order two in both time and space and another method of order two in time and four in space will be developed in sequel. The unique solvability, stability and convergence of each scheme will be analyzed. The whole chapter is divided into 5 sections.

5.1 The second-order method in space for 1D problem In this section, the following 1D initial-boundary value problem of time-space-fractional differential equation 𝜕β u(x, t) C α { Dt u(x, t) = + f (x, t), 0 < x < L, 0 < t ⩽ T, { 0 { { 𝜕|x|β { { u(x, 0) = φ(x), 0 < x < L, { { { u(0, t) = 0, u(L, t) = 0, 0 ⩽ t ⩽ T

(5.1) (5.2) (5.3)

will be considered, where α ∈ (0, 1), β ∈ (1, 2), C0 Dαt u(x, t) is the α-th order Caputo frac-

tional derivative,

𝜕β u(x,t) 𝜕|x|β

is the β-th order Riesz fractional derivative, that is,

𝜕β u(x, t) β = −Ψβ (0 Dβx u(x, t) + x DL u(x, t)), 𝜕|x|β β 0 Dx u(x, t)

β

x

Ψβ =

1 2 cos(

βπ ) 2

,

u(ξ , t) 𝜕2 1 = dξ , ∫ Γ(2 − β) 𝜕x2 (x − ξ )β−1

x DL u(x, t) =

0

L

u(ξ , t) 1 𝜕2 dξ , ∫ Γ(2 − β) 𝜕x2 (ξ − x)β−1 x

the functions f , φ are given and φ(0) = φ(L) = 0. Take the same mesh partition and notations as those in Section 2.1. In addition, let σ = 1 − α2 , tn−1+σ = tn−1 + στ, s = τα Γ(2 − α). https://doi.org/10.1515/9783110616064-005

298 | 5 Difference methods for the time-space-fractional differential equations Define the mesh functions Uin = u(xi , tn ),

0 ⩽ i ⩽ M, 0 ⩽ n ⩽ N;

fin−1+σ = f (xi , tn−1+σ ),

1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N.

For any fixed t ∈ [0, T], define the function ̃ t) = { u(x,

u(x, t),

0 ⩽ x ⩽ L,

0,

x ∉ [0, L].

̃ t) ∈ C 2+β (ℛ). Suppose u(x, ⋅) ∈ C 3 [0, T] and u(⋅, 5.1.1 Derivation of the difference scheme Considering equation (5.1) at the point (xi , tn−1+σ ), we have C α 0 Dt u(xi , tn−1+σ )

=

𝜕β u(xi , tn−1+σ ) + fin−1+σ , 𝜕|x|β

1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N.

(5.4)

For the Caputo fractional derivative in equation (5.4), using L2-1σ approximation (1.81), it follows from Theorem 1.6.4 that C α 0 Dt u(xi , tn−1+σ )

=

τ−α n−1 (n,α) n−k (Ui − Uin−k−1 ) + O(τ3−α ). ∑c Γ(2 − α) k=0 k

(5.5)

For the Riesz fractional derivative in equation (5.4), the result of a linear interpolation approximation reads 𝜕β u(xi , tn−1+σ ) 𝜕β u(xi , tn−1 ) 𝜕β u(xi , tn ) = σ + (1 − σ) + O(τ2 ). 𝜕|x|β 𝜕|x|β 𝜕|x|β

(5.6)

Moreover, it follows from Theorem 1.5.1 that i 𝜕β u(xi , tn ) (β) n = −h−β ∑ ĝk Ui−k + O(h2 ). β 𝜕|x| k=i−M

(5.7)

The combination of (5.6) and (5.7) arrives at i 𝜕β u(xi , tn−1+σ ) (β) −β n n−1 = −h ĝk [σUi−k + (1 − σ)Ui−k ] + O(τ2 + h2 ). ∑ 𝜕|x|β k=i−M

Substituting (5.5) and (5.8) into (5.4), we obtain τ−α n−1 (n,α) n−k (Ui − Uin−k−1 ) ∑c Γ(2 − α) k=0 k

(5.8)

5.1 The second-order method in space for 1D problem

| 299

i

(β) n n−1 = −h−β ∑ ĝk [σUi−k + (1 − σ)Ui−k ] + fin−1+σ + (r1 )ni , k=i−M

1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N,

(5.9)

and there is a positive constant c1 such that 󵄨󵄨 n 󵄨󵄨 2 2 󵄨󵄨(r1 )i 󵄨󵄨 ⩽ c1 (τ + h ),

1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N.

(5.10)

Noticing the initial-boundary value conditions (5.2)–(5.3), we have {

Ui0 = φ(xi ), U0n

= 0,

n UM

1 ⩽ i ⩽ M − 1,

(5.11)

= 0,

(5.12)

0 ⩽ n ⩽ N.

Omitting the small term (r1 )ni in (5.9) and replacing the exact solution Uin with its numerical one uni , a difference scheme for solving (5.1)–(5.3) can be produced as { { { { { { { { { { { { { { { { { { { { { { { { { { { { {

τ−α n−1 (n,α) n−k (ui − un−k−1 ) ∑c i Γ(2 − α) k=0 k i

(β) n−1+σ = −h−β ∑ ĝk [σuni−k + (1 − σ)un−1 , i−k ] + fi k=i−M

u0i un0

= φ(xi ),

= 0,

unM

1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N,

(5.13)

1 ⩽ i ⩽ M − 1,

(5.14)

= 0,

(5.15)

0 ⩽ n ⩽ N.

Next, some analyses on the difference scheme (5.13)–(5.15) will be carried out. 5.1.2 Solvability of the difference scheme Theorem 5.1.1. The difference scheme (5.13)–(5.15) is uniquely solvable. Proof. Let un = (un0 , un1 , . . . , unM ). The value of u0 is obviously determined by (5.14)–(5.15). Now assume that the values of u0 , u1 , . . . , un−1 have been uniquely determined, then the linear system in un can be obtained from (5.13) and (5.15). To show its unique solvability, it suffices to verify that the corresponding homogeneous one i 1 { { { c0(n,α) uni = −σh−β ∑ ĝk(β) uni−k , s k=i−M { { { n n { u0 = 0, uM = 0

1 ⩽ i ⩽ M − 1,

(5.16) (5.17)

300 | 5 Difference methods for the time-space-fractional differential equations has only the trivial solution. Rewrite (5.16) as i 1 (β) (β) [ c0(n,α) + σh−β ĝ0 ]uni = σh−β ∑ (−ĝk )uni−k , s k=i−M

1 ⩽ i ⩽ M − 1.

(5.18)

k =0 ̸

Suppose ‖un ‖∞ = |unin |, where in ∈ {1, 2, . . . , M − 1}. Letting i = in in (5.18) and taking the absolute value on both hand sides of the equality, it follows from Lemma 1.5.2 that 1 (β) 󵄩 󵄩 [ c0(n,α) + σh−β ĝ0 ]󵄩󵄩󵄩un 󵄩󵄩󵄩∞ s in

(β) 󵄨 󵄨 ⩽ σh−β ∑ (−ĝk )󵄨󵄨󵄨unin −k 󵄨󵄨󵄨 k=in −M k =0 ̸ in

(β) 󵄩 󵄩 ⩽ σh−β ∑ (−ĝk )󵄩󵄩󵄩un 󵄩󵄩󵄩∞ k=in −M k =0 ̸

(β) 󵄩 󵄩 ⩽ σh−β ĝ0 󵄩󵄩󵄩un 󵄩󵄩󵄩∞ .

Hence, ‖un ‖∞ = 0, which implies that (5.16)–(5.17) has only the trivial solution. By the principle of induction, the theorem is true. The proof ends.

5.1.3 An important lemma In this subsection, we present an important lemma. Lemma 5.1.1. For any mesh function v = (v0 , v1 , . . . , vM ) ∈ 𝒰h̊ , it holds M−1

i

M−1

i=1

k=i−M

i=1

(β) −h−β h ∑ ( ∑ ĝk vi−k )vi ⩽ −c∗(β) (2L)−β h ∑ vi2 ,

where c∗(β) is defined by (1.55), 1 < β < 2. Proof. From Lemma 1.5.2 and Lemma 1.5.3, it is easy to see that (β) ĝk < 0,

|k| ⩾ 1;

∞

(β) ∑ ĝk = 0;

k=−∞

therefore, M−1

i

i=1

k=i−M

(β) A ≡ −h−β h ∑ ( ∑ ĝk vi−k )vi

∞

(β)

∑ (−ĝk ) ⩾

|k|=M

c∗(β)

(M + 1)β

,

M ⩾ 1,

5.1 The second-order method in space for 1D problem

M−1

M−1

i

| 301

= h−β [h ∑ (−ĝ0 )vi2 + h ∑ ∑ (−ĝk )vi−k vi ] (β)

i=1

(β)

i=1 k=i−M k =0 ̸

M−1 1 M−1 i (β) (β) 2 ⩽ h−β [h ∑ (−ĝ0 )vi2 + h ∑ ∑ (−ĝk )(vi−k + vi2 )] 2 i=1 i=1 k=i−M k =0 ̸

M−1

M−1

i 1 1 M−1 i (β) (β) (β) 2 = h−β [h ∑ (−ĝ0 )vi2 + h ∑ ∑ (−ĝk )vi2 + h ∑ ∑ (−ĝk )vi−k ]. 2 2 i=1 i=1 k=i−M i=1 k=i−M k =0 ̸

(5.19)

k =0 ̸

For the second term on the right-hand side in (5.19), we have M−1 1 M−1 i 1 M−1 M−1 1 M−1 (β) (β) (β) h ∑ ∑ (−ĝk )vi2 ⩽ h ∑ ∑ (−ĝk )vi2 = ∑ (−ĝk ) ⋅ h ∑ vi2 . 2 i=1 k=i−M 2 i=1 k=1−M 2 |k|=1 i=1 k =0 ̸

(5.20)

k =0 ̸

For the third term on the right-hand side in (5.19), we have 1 M−1 i (β) 2 h ∑ ∑ (−ĝk )vi−k 2 i=1 k=i−M k =0 ̸

k+M M−1 M−1 1 (β) (β) 2 2 + ∑ (−ĝk ) ∑ vi−k ] = h[ ∑ (−ĝk ) ∑ vi−k 2 k=1−M i=1 k=1 i=k −1

−1 M−1 M−1 1 (β) (β) ⩽ [ ∑ (−ĝk ) + ∑ (−ĝk )] ⋅ h ∑ vi2 2 k=1−M i=1 k=1

=

M−1 1 M−1 (β) ∑ (−ĝk ) ⋅ h ∑ vi2 . 2 |k|=1 i=1

(5.21)

Substituting (5.20) and (5.21) into (5.19), it follows from Lemma 1.5.2 and Lemma 1.5.3 that M−1

M−1

M−1

A ⩽ h−β [(−ĝ0 )h ∑ vi2 + ∑ (−ĝk ) ⋅ h ∑ vi2 ] (β)

i=1

M−1

(β)

|k|=1

M−1

(β) = h−β ∑ (−ĝk ) ⋅ h ∑ vi2 i=1

k=1−M

M−1

(β) = h−β ( ∑ ĝk ) ⋅ h ∑ vi2 |k|⩾M

⩽ −h−β

c∗(β)

(M + 1)β

= −(Mh)−β c∗(β) (

i=1

M−1

⋅ h ∑ vi2 i=1

β

M−1 M ) ⋅ h ∑ vi2 M+1 i=1

i=1

302 | 5 Difference methods for the time-space-fractional differential equations β

M−1 1 ⩽ −(Mh)−β c∗(β) ( ) ⋅ h ∑ vi2 2 i=1 M−1

= −c∗(β) (2L)−β ⋅ h ∑ vi2 . i=1

The proof ends. Similarly, the following conclusion can be proved (the only difference is to replace vi−k vi and vi2 with (vi−k , vi ) and ‖vi ‖2 , resp.). Corollary 5.1.1. Suppose 𝒱 is an inner product space, (⋅, ⋅) is an inner product in 𝒱 and ‖ ⋅ ‖ is the induced norm; For any mesh functions v 0 , v 1 , . . . , v M ∈ 𝒱 satisfying v 0 = 0, v M = 0, it holds M−1

i

M−1

i=1

k=i−M

i=1

(β) 󵄩 󵄩2 − h−β h ∑ ( ∑ ĝk v i−k , v i ) ⩽ −c∗(β) (2L)−β h ∑ 󵄩󵄩󵄩v i 󵄩󵄩󵄩 ,

where c∗(β) is defined by (1.55), 1 < β < 2. 5.1.4 Stability of the difference scheme Theorem 5.1.2. Suppose {uni | 0 ⩽ i ⩽ M, 0 ⩽ n ⩽ N} is the solution of the difference scheme (5.13)–(5.15), then it holds β 󵄩󵄩 n 󵄩󵄩2 󵄩󵄩 0 󵄩󵄩2 (2L) Γ(1 − α) α󵄩 󵄩󵄩f m−1+σ 󵄩󵄩󵄩2 }, max {tm 󵄩󵄩u 󵄩󵄩 ⩽ 󵄩󵄩u 󵄩󵄩 + 󵄩 󵄩 (β) 1⩽m⩽n 2c∗

1 ⩽ n ⩽ N,

where M−1

󵄩󵄩 m−1+σ 󵄩󵄩2 m−1+σ 2 ). 󵄩󵄩f 󵄩󵄩 = h ∑ (fi i=1

Proof. Taking the inner product on both hand sides of (5.13) with σun +(1−σ)un−1 yields 1 n−1 (n,α) M−1 n−k h ∑ (ui − un−k−1 )[σuni + (1 − σ)un−1 ∑c i i ] s k=0 k i=1 M−1

i

(β) n n−1 = −h−β h ∑ ∑ ĝk [σuni−k + (1 − σ)un−1 i−k ] ⋅ [σui + (1 − σ)ui ] i=1 k=i−M

M−1

+ h ∑ fin−1+σ [σuni + (1 − σ)un−1 i ]. i=1

Now each term in (5.22) will be estimated.

(5.22)

5.1 The second-order method in space for 1D problem

| 303

For the left-hand side, it follows from Lemma 2.6.1 that 1 n−1 (n,α) M−1 n−k h ∑ (ui − un−k−1 )[σuni + (1 − σ)un−1 ∑c i i ] s k=0 k i=1 = ⩾

1 n−1 (n,α) n−k (u − un−k−1 , σun + (1 − σ)un−1 ) ∑c s k=0 k 1 n−1 (n,α) 󵄩󵄩 n−k 󵄩󵄩2 󵄩󵄩 n−k−1 󵄩󵄩2 (󵄩󵄩u 󵄩󵄩 − 󵄩󵄩u ∑c 󵄩󵄩 ). 2s k=0 k

(5.23)

For the first term on the right-hand side, using Lemma 5.1.1, we have M−1

i

(β) n n−1 −h−β h ∑ ∑ ĝk [σuni−k + (1 − σ)un−1 i−k ] ⋅ [σui + (1 − σ)ui ] i=1 k=i−M

M−1

2

⩽ −c∗(β) (2L)−β h ∑ [σuni + (1 − σ)un−1 i ] i=1

󵄩 󵄩2 = −c∗(β) (2L)−β 󵄩󵄩󵄩σun + (1 − σ)un−1 󵄩󵄩󵄩 .

(5.24)

For the second term on the right-hand side, with the aid of the Cauchy–Schwarz inequality, we have M−1

h ∑ fin−1+σ [σuni + (1 − σ)un−1 i ] i=1

󵄩 󵄩 󵄩 󵄩 ⩽ 󵄩󵄩󵄩f n−1+σ 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩σun + (1 − σ)un−1 󵄩󵄩󵄩

β 󵄩 󵄩2 (2L) 󵄩 󵄩2 ⩽ c∗(β) (2L)−β 󵄩󵄩󵄩σun + (1 − σ)un−1 󵄩󵄩󵄩 + (β) 󵄩󵄩󵄩f n−1+σ 󵄩󵄩󵄩 . 4c∗

(5.25)

Substituting (5.23), (5.24) and (5.25) into (5.22) arrives at (2L)β 󵄩 n−1+σ 󵄩󵄩2 1 n−1 (n,α) 󵄩󵄩 n−k 󵄩󵄩2 󵄩󵄩 n−k−1 󵄩󵄩2 ∑ ck (󵄩󵄩u 󵄩󵄩 − 󵄩󵄩u 󵄩󵄩 ) ⩽ (β) 󵄩󵄩󵄩f 󵄩󵄩 , 2s k=0 4c∗

1 ⩽ n ⩽ N.

It follows from Lemma 1.6.3 that s

(n,α) cn−1

=

τα Γ(2 − α) (n,α) cn−1


0, K2 > 0, Ω = (0, L1 ) × (0, L2 ) and φ(x, y)|(x,y)∈𝜕Ω = 0. Take the same mesh partition and notations as those in Section 2.10. In addition, define σ = 1 − α2 , tn−1+σ = tn−1 + στ, s = τα Γ(2 − α). ̂ y, t) and w(x, ̂ y, t). Suppose v(⋅, ̂ y, t) ∈ Similar to Section 4.4, define functions v(x, 2+β 2+γ 3 ̂ ⋅, t) ∈ C (ℛ) and u(x, y, ⋅) ∈ C [0, T]. C (ℛ), w(x, Define the mesh functions Uijn = u(xi , yj , tn ),

fijn−1+σ

(i, j) ∈ ω,̄ 0 ⩽ n ⩽ N;

= f (xi , yj , tn−1+σ ),

(i, j) ∈ ω,̄ 1 ⩽ n ⩽ N.

5.3 The second-order method in space for 2D problem

| 311

5.3.1 Derivation of the difference scheme Considering equation (5.50) at the point (xi , yj , tn−1+σ ), we have C α 0 Dt u(xi , yj , tn−1+σ )

= K1

𝜕β u(xi , yj , tn−1+σ ) 𝜕|x|β

+K2

𝜕γ u(xi , yj , tn−1+σ ) 𝜕|y|γ

+ fijn−1+σ ,

(i, j) ∈ ω, 1 ⩽ n ⩽ N.

(5.53)

For the Caputo derivative in (5.53), using L2-1σ approximation (1.81), it follows from Theorem 1.6.4 that C α 0 Dt u(xi , yj , tn−1+σ )

=

τ−α n−1 (n,α) n−k (Uij − Uijn−k−1 ) + O(τ3−α ). ∑c Γ(2 − α) k=0 k

(5.54)

For the Riesz derivatives in (5.53), it follows from the linear interpolation and Theorem 1.5.1 that 𝜕β u(xi , yj , tn−1+σ ) 𝜕|x|β

=σ

𝜕β u(xi , yj , tn ) 𝜕|x|β

−β

𝜕β u(xi , yj , tn−1 ) 𝜕|x|β

+ O(τ2 )

i

i

k=i−M1

k=i−M1

−β (β) n−1 (β) n ] + O(τ2 + h21 ) ] + (1 − σ)[−h1 ∑ ĝk Ui−k,j ∑ ĝk Ui−k,j

−β

= σ[−h1 = −h1

+ (1 − σ)

i

(β) n n−1 + (1 − σ)Ui−k,j ] + O(τ2 + h21 ). ∑ ĝk [σUi−k,j

k=i−M1

(5.55)

Similarly, we have 𝜕γ u(xi , yj , tn−1+σ ) 𝜕|y|γ

j

(γ) n n−1 + (1 − σ)Ui,j−k ] + O(τ2 + h22 ). ∑ ĝk [σUi,j−k

−γ

= −h2

k=j−M2

(5.56)

Substituting (5.54)–(5.56) into (5.53) yields τ−α n−1 (n,α) n−k (Uij − Uijn−k−1 ) ∑c Γ(2 − α) k=0 k i

−β

= −K1 h1

k=i−M1 −γ

− K2 h2 +

(β) n n−1 + (1 − σ)Ui−k,j ] ∑ ĝk [σUi−k,j

fijn−1+σ

j

(γ) n n−1 + (1 − σ)Ui,j−k ] ∑ ĝk [σUi,j−k

k=j−M2

+ (r3 )nij ,

(i, j) ∈ ω, 1 ⩽ n ⩽ N,

(5.57)

312 | 5 Difference methods for the time-space-fractional differential equations and there is a positive constant c3 such that 󵄨󵄨 n 󵄨󵄨 2 2 2 󵄨󵄨(r3 )ij 󵄨󵄨 ⩽ c3 (τ + h1 + h2 ),

(i, j) ∈ ω, 1 ⩽ n ⩽ N.

(5.58)

Noticing the initial-boundary value conditions (5.51)–(5.52), we have {

Uij0 = φ(xi , yj ), Uijn

= 0,

(i, j) ∈ ω,

(5.59)

(i, j) ∈ 𝜕ω, 0 ⩽ n ⩽ N.

(5.60)

Omitting the small term (r3 )nij in (5.57) and replacing the exact solution Uijn with its numerical one unij , we get a difference scheme for solving (5.50)–(5.52) in the form of { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { {

τ−α n−1 (n,α) n−k (uij − un−k−1 ) ∑c ij Γ(2 − α) k=0 k i

(β) ∑ ĝk [σuni−k,j + (1 − σ)un−1 i−k,j ]

−β

= −K1 h1

k=i−M1 −γ

− K2 h2 u0ij unij

j

(γ) n−1+σ , ∑ ĝk [σuni,j−k + (1 − σ)un−1 i,j−k ] + fij

k=j−M2

(i, j) ∈ ω, 1 ⩽ n ⩽ N,

= φ(xi , yj ),

= 0,

(5.61)

(i, j) ∈ ω,

(5.62)

(i, j) ∈ 𝜕ω, 0 ⩽ n ⩽ N.

(5.63)

5.3.2 Solvability of the difference scheme Theorem 5.3.1. The difference scheme (5.61)–(5.63) is uniquely solvable. Proof. Let ̄ un = {unij | (i, j) ∈ ω}. The value of u0 is determined by (5.62)–(5.63). Now assume that the values of u0 , u1 , . . . , un−1 have been uniquely determined, then we can obtain the linear system in the unknown un from (5.61) and (5.63). To show its unique solvability, it is sufficient to verify that the corresponding homogeneous one j

i 1 (n,α) n −β (β) n −γ (γ) n { { { c0 uij = −K1 σh1 ∑ ĝk ui−k,j − K2 σh2 ∑ ĝk ui,j−k , s k=i−M1 k=j−M2 { { { n u = 0, (i, j) ∈ 𝜕ω { ij

has only the trivial solution.

(i, j) ∈ ω, (5.64) (5.65)

5.3 The second-order method in space for 2D problem

| 313

Rewrite (5.64) as 1 −β (β) −γ (γ) [ c0(n,α) + K1 σh1 ĝ0 + K2 σh2 ĝ0 ]unij s −β

= K1 σh1

i

∑ (−ĝk )uni−k,j + K2 σh2 (β)

k=i−M1 k =0 ̸

−γ

j

∑ (−ĝk )uni,j−k , (γ)

k=j−M2 k =0 ̸

(i, j) ∈ ω.

(5.66)

Suppose ‖un ‖∞ = |unin ,jn |, where (in , jn ) ∈ ω. In (5.66), letting (i, j) = (in , jn ) and taking the absolute value of both hand sides, the application of the triangle inequality gives 1 −β (β) −γ (γ) 󵄩 󵄩 [ c0(n,α) + K1 σh1 ĝ0 + K2 σh2 ĝ0 ]󵄩󵄩󵄩un 󵄩󵄩󵄩∞ s −β

⩽ K1 σh1

in

(β) 󵄩 󵄩 −γ ∑ (−ĝk )󵄩󵄩󵄩un 󵄩󵄩󵄩∞ + K2 σh2

k=in −M1 k =0 ̸

−β (β) −γ (γ) 󵄩 󵄩 ⩽ [K1 σh1 ĝ0 + K2 σh2 ĝ0 ]󵄩󵄩󵄩un 󵄩󵄩󵄩∞ ,

jn

(γ) 󵄩 󵄩 ∑ (−ĝk )󵄩󵄩󵄩un 󵄩󵄩󵄩∞

k=jn −M2 k =0 ̸

which implies ‖un ‖∞ = 0. The combination with (5.65) will reveal that (5.64)–(5.65) has only the trivial solution. By the principle of induction, the difference scheme (5.61)–(5.63) is uniquely solvable. The proof ends.

5.3.3 Stability of the difference scheme Theorem 5.3.2. Suppose {unij | (i, j) ∈ ω,̄ 0 ⩽ n ⩽ N} is the solution of (5.61)–(5.63), then it holds β (2L2 )γ 󵄩󵄩 n 󵄩󵄩2 󵄩󵄩 0 󵄩󵄩2 Γ(1 − α) (2L1 ) 󵄩 󵄩2 + ] max {t α 󵄩󵄩f m−1+σ 󵄩󵄩󵄩 }, [ 󵄩󵄩u 󵄩󵄩 ⩽ 󵄩󵄩u 󵄩󵄩 + (γ) 1⩽m⩽n m 󵄩 (β) 8 K2 c∗ K1 c∗

1 ⩽ n ⩽ N,

where M1 −1 M2 −1

󵄩󵄩 m−1+σ 󵄩󵄩2 m−1+σ 2 ). 󵄩󵄩f 󵄩󵄩 = h1 h2 ∑ ∑ (fij i=1

j=1

Proof. Taking the inner product on both hand sides of (5.61) with σun + (1 − σ)un−1 will arrive at M −1 M −1

1 2 1 n−1 (n,α) − un−k−1 )[σunij + (1 − σ)un−1 ∑ ck h1 h2 ∑ ∑ (un−k ij ij ij ] s k=0 i=1 j=1

314 | 5 Difference methods for the time-space-fractional differential equations M2 −1

M1 −1

−β

= K1 h2 ∑ {−h1 h1 ∑ j=1

i

(β) n n−1 ∑ ĝk [σuni−k,j + (1 − σ)un−1 i−k,j ][σuij + (1 − σ)uij ]}

i=1 k=i−M1

M1 −1

−γ

M2 −1

j

(γ) n n−1 ∑ ĝk [σuni,j−k + (1 − σ)un−1 i,j−k ][σuij + (1 − σ)uij ]}

+ K2 h1 ∑ {−h2 h2 ∑ i=1

j=1 k=j−M2

M1 −1 M2 −1

+ h1 h2 ∑ ∑ fijn−1+σ [σunij + (1 − σ)un−1 ij ]. i=1

(5.67)

j=1

Now each term in (5.67) will be estimated. By Lemma 2.6.1, we have M1 −1 M2 −1

n−1

− un−k−1 )[σunij + (1 − σ)un−1 ∑ ck(n,α) h1 h2 ∑ ∑ (un−k ij ij ij ] i=1

k=0

⩾

j=1

n−1

1 󵄩 󵄩2 󵄩 󵄩2 ∑ c(n,α) (󵄩󵄩󵄩un−k 󵄩󵄩󵄩 − 󵄩󵄩󵄩un−k−1 󵄩󵄩󵄩 ). 2 k=0 k

(5.68)

It follows from Lemma 5.1.1 that −β

M1 −1

−h1 h1 ∑

i

(β) n n−1 ∑ ĝk [σuni−k,j + (1 − σ)un−1 i−k,j ][σuij + (1 − σ)uij ]

i=1 k=i−M1

M1 −1

2

⩽ −c∗(β) (2L1 )−β h1 ∑ [σunij + (1 − σ)un−1 ij ]

(5.69)

i=1

and −γ

M2 −1

−h2 h2 ∑

j

(γ) n n−1 ∑ ĝk [σuni,j−k + (1 − σ)un−1 i,j−k ][σuij + (1 − σ)uij ]

j=1 k=j−M2

M2 −1

2

⩽ −c∗(γ) (2L2 )−γ h2 ∑ [σunij + (1 − σ)un−1 ij ] .

(5.70)

j=1

Substituting (5.68)–(5.70) into (5.67) gives 1 n−1 (n,α) 󵄩󵄩 n−k 󵄩󵄩2 󵄩󵄩 n−k−1 󵄩󵄩2 (󵄩󵄩u 󵄩󵄩 − 󵄩󵄩u ∑c 󵄩󵄩 ) 2s k=0 k M1 −1 M2 −1

2

⩽ −K1 c∗(β) (2L1 )−β h1 h2 ∑ ∑ [σunij + (1 − σ)un−1 ij ] i=1

j=1

M1 −1 M2 −1

2

− K2 c∗(γ) (2L2 )−γ h1 h2 ∑ ∑ [σunij + (1 − σ)un−1 ij ] i=1

M1 −1 M2 −1

j=1

+ h1 h2 ∑ ∑ fijn−1+σ [σunij + (1 − σ)un−1 ij ] i=1

j=1

5.3 The second-order method in space for 2D problem

| 315

󵄩 󵄩2 ⩽ −K1 c∗(β) (2L1 )−β 󵄩󵄩󵄩σun + (1 − σ)un−1 󵄩󵄩󵄩 󵄩 󵄩2 − K2 c∗(γ) (2L2 )−γ 󵄩󵄩󵄩σun + (1 − σ)un−1 󵄩󵄩󵄩 󵄩 󵄩 󵄩 󵄩 + 󵄩󵄩󵄩f n−1+σ 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩σun + (1 − σ)un−1 󵄩󵄩󵄩 ⩽

1 (2L1 )β (2L2 )γ 󵄩󵄩 n−1+σ 󵄩󵄩2 [ ]󵄩f + 󵄩󵄩 , 16 K c∗(β) K2 c∗(γ) 󵄩 1

1 ⩽ n ⩽ N.

(5.71)

By Lemma 1.6.3, we have s

(n,α) cn−1

⩽ tnα Γ(1 − α).

Then it follows from (5.71) that n−1

󵄩 󵄩2 󵄩 󵄩2 (n,α) 󵄩 (n,α) 󵄩󵄩u0 󵄩󵄩󵄩2 c0(n,α) 󵄩󵄩󵄩un 󵄩󵄩󵄩 ⩽ ∑ (ck−1 − ck(n,α) )󵄩󵄩󵄩un−k 󵄩󵄩󵄩 + cn−1 󵄩 󵄩 k=1

+ n−1

1 (2L1 )β (2L2 )γ 󵄩󵄩 n−1+σ 󵄩󵄩2 + [ ]s󵄩f 󵄩󵄩 8 K c∗(β) K2 c∗(γ) 󵄩 1

󵄩 󵄩2 (n,α) ⩽ ∑ (ck−1 − ck(n,α) )󵄩󵄩󵄩un−k 󵄩󵄩󵄩 k=1

β (2L2 )γ α 󵄩 󵄩2 󵄩2 1 (2L ) (n,α) 󵄩 ]t Γ(1 − α)󵄩󵄩󵄩f n−1+σ 󵄩󵄩󵄩 }, + cn−1 {󵄩󵄩󵄩u0 󵄩󵄩󵄩 + [ 1(β) + (γ) n 8 K c∗ K2 c∗ 1

1 ⩽ n ⩽ N.

The inductive process will lead to β (2L2 )γ 󵄩󵄩 n 󵄩󵄩2 󵄩󵄩 0 󵄩󵄩2 1 (2L1 ) α󵄩 󵄩󵄩f m−1+σ 󵄩󵄩󵄩2 }, ]Γ(1 − α) max {tm + 󵄩󵄩u 󵄩󵄩 ⩽ 󵄩󵄩u 󵄩󵄩 + [ 󵄩 󵄩 (γ) (β) 1⩽m⩽n 8 K c∗ K c 2 ∗ 1

1 ⩽ n ⩽ N.

The proof ends.

5.3.4 Convergence of the difference scheme Theorem 5.3.3. Suppose {Uijn | (i, j) ∈ ω,̄ 0 ⩽ n ⩽ N} and {unij | (i, j) ∈ ω,̄ 0 ⩽ n ⩽ N} are solutions of the problem (5.50)–(5.52) and the difference scheme (5.61)–(5.63), respectively. Let eijn = Uijn − unij ,

(i, j) ∈ ω,̄ 0 ⩽ n ⩽ N,

then it holds 1 (2L )β (2L2 )γ 󵄩󵄩 n 󵄩󵄩 ]Γ(1 − α)T α L1 L2 c3 (τ2 + h21 + h22 ), 󵄩󵄩e 󵄩󵄩 ⩽ √ [ 1(β) + (γ) 8 K c∗ K c 2 ∗ 1

1 ⩽ n ⩽ N.

(5.72)

316 | 5 Difference methods for the time-space-fractional differential equations Proof. The system of error equations can be obtained by the subtraction of (5.61)–(5.63) from (5.57), (5.59)–(5.60), respectively, as follows: { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { {

1 n−1 (n,α) n−k (eij − eijn−k−1 ) ∑c s k=0 k i

(β) n n−1 + (1 − σ)ei−k,j ] ∑ ĝk [σei−k,j

−β

= −K1 h1

k=i−M1 j

(γ) n n−1 + (1 − σ)ei,j−k ] + (r3 )nij , ∑ ĝk [σei,j−k

−γ

− K2 h2 eij0 = 0, eijn

= 0,

k=j−M2

(i, j) ∈ ω, 1 ⩽ n ⩽ N,

(i, j) ∈ ω, (i, j) ∈ 𝜕ω,

0 ⩽ n ⩽ N.

Noticing (5.58), the application of Theorem 5.3.2 yields β (2L2 )γ 󵄩󵄩 n 󵄩󵄩2 1 (2L1 ) α󵄩 󵄩󵄩(r3 )m 󵄩󵄩󵄩2 } + ]Γ(1 − α) max {tm 󵄩󵄩e 󵄩󵄩 ⩽ [ 󵄩 󵄩 (γ) (β) 1⩽m⩽n 8 K c∗ K c 2 ∗ 1

⩽

1 (2L1 )β (2L2 )γ 2 ]Γ(1 − α)T α L1 L2 c32 (τ2 + h21 + h22 ) , [ + (γ) (β) 8 K c∗ K2 c∗ 1

1 ⩽ n ⩽ N.

Taking the square root on both hand sides of the inequality above will produce (5.72). The proof ends.

5.4 The fourth-order method in space for 2D problem In this section, another higher-order difference scheme of order two in time and four in space for solving (5.50)–(5.52) will be developed. For any mesh function v = {vij | (i, j) ∈ ω}̄ ∈ 𝒱h , define the following average operators: β

β

ℋx vij = { 24 γ

vi−1,j + (1 −

β )v 12 ij

+

β v , 24 i+1,j

1 ⩽ i ⩽ M1 − 1,

vi,j−1 + (1 −

γ )v 12 ij

+

γ v , 24 i,j+1

1 ⩽ j ⩽ M2 − 1,

vij , i = 0, M1 , γ

ℋy vij = { 24

vij , j = 0, M2 ,

0 ⩽ j ⩽ M2 ; 0 ⩽ i ⩽ M1 .

Obviously, β 2 2 hδ v , 24 1 x ij γ 2 2 γ ℋy vij = vij + hδ v , 24 2 y ij β

ℋx vij = vij +

1 ⩽ i ⩽ M1 − 1, 0 ⩽ j ⩽ M2 ; 1 ⩽ j ⩽ M2 − 1, 0 ⩽ i ⩽ M1 .

̂ y, t) and w(x, ̂ y, t) like those in Section 4.4. Suppose Define the functions v(x, ̂ y, t) ∈ C 4+β (ℛ), w(x, ̂ ⋅, t) ∈ C 4+γ (ℛ) and u(x, y, ⋅) ∈ C 3 [0, T]. v(⋅,

5.4 The fourth-order method in space for 2D problem

| 317

5.4.1 Derivation of the difference scheme Considering equation (5.50) at the point (xi , yj , tn−1+σ ), by the mean of the linear interpolation, we have C α 0 Dt u(xi , yj , tn−1+σ ) 𝜕β u(xi , yj , tn−1+σ ) = K1 𝜕|x|β

= K1 [σ

𝜕β u(xi , yj , tn )

+ K2 [σ

+ K2

𝜕γ u(xi , yj , tn−1+σ ) 𝜕|y|γ

+ (1 − σ)

𝜕|x|β γ 𝜕 u(xi , yj , tn ) 𝜕|y|γ

+ fijn−1+σ

𝜕β u(xi , yj , tn−1 )

+ (1 − σ)

] 𝜕|x|β 𝜕γ u(xi , yj , tn−1 ) 𝜕|y|γ

] + fijn−1+σ + O(τ2 ),

(i, j) ∈ ω,̄ 1 ⩽ n ⩽ N. β

γ

Performing the operator ℋx ℋy to both hand sides of the equality above arrives at β

γC

α

ℋx ℋy 0 Dt u(xi , yj , tn−1+σ )

= K1 ℋyγ [σ ℋxβ

𝜕β u(xi , yj , tn )

+ (1 − σ)ℋxβ

𝜕|x|β γ γ 𝜕 u(xi , yj , tn )

+ K2 ℋxβ [σ ℋy

𝜕|y|γ

+ ℋxβ ℋyγ fijn−1+σ + O(τ2 ),

𝜕β u(xi , yj , tn−1 )

] 𝜕|x|β γ γ 𝜕 u(xi , yj , tn−1 )

+ (1 − σ)ℋy

𝜕|y|γ

]

(i, j) ∈ ω, 1 ⩽ n ⩽ N.

(5.73)

For the Caputo derivative in (5.73), using L2-1σ approximation (1.81), it follows from Theorem 1.6.4 that C α 0 Dt u(xi , yj , tn−1+σ )

=

τ−α n−1 (n,α) n−k (Uij − Uijn−k−1 ) + O(τ3−α ). ∑c Γ(2 − α) k=0 k

(5.74)

For the Riesz derivatives in (5.73), it follows from Theorem 1.5.2 that β

β𝜕

ℋx

𝜕|x|β

γ

γ𝜕

ℋy

u(xi , yj , tn ) u(xi , yj , tn ) 𝜕|y|γ

i

(β) n + O(h41 ), ∑ ĝk Ui−k,j

−β

= −h1

k=i−M1 j

−γ

= −h2

(γ) n + O(h42 ). ∑ ĝk Ui,j−k

k=j−M2

Substituting (5.74)–(5.76) into (5.73) gives β

γ

ℋx ℋy

τ−α n−1 (n,α) n−k (Uij − Uijn−k−1 ) ∑c Γ(2 − α) k=0 k

(5.75) (5.76)

318 | 5 Difference methods for the time-space-fractional differential equations i

−β (β) n n−1 = K1 ℋyγ (−h1 ) ∑ ĝk [σUi−k,j + (1 − σ)Ui−k,j ] k=i−M1 j

−γ (γ) n n−1 + K2 ℋxβ (−h2 ) ∑ ĝk [σUi,j−k + (1 − σ)Ui,j−k ]

+

β γ n−1+σ ℋx ℋy fij

k=j−M2

+ (r4 )nij ,

(i, j) ∈ ω, 1 ⩽ n ⩽ N,

(5.77)

(i, j) ∈ ω, 1 ⩽ n ⩽ N.

(5.78)

and there is a positive constant c4 such that 󵄨󵄨 n󵄨 2 4 4 󵄨󵄨(r4 )ij 󵄨󵄨󵄨 ⩽ c4 (τ + h1 + h2 ),

Noticing the initial-boundary value conditions (5.51)–(5.52), we have Uij0 = φ(xi , yj ),

{

Uijn

= 0,

(i, j) ∈ ω,

(i, j) ∈ 𝜕ω,

0 ⩽ n ⩽ N.

(5.79) (5.80)

Neglecting the small term (r4 )nij in (5.77) and replacing the exact solution Uijn with its numerical one unij , we get another difference scheme for solving (5.50)–(5.52) as follows: { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { {

β

γ

ℋx ℋy

τ−α n−1 (n,α) n−k (uij − un−k−1 ) ∑c ij Γ(2 − α) k=0 k i

−β (β) = K1 ℋyγ (−h1 ) ∑ ĝk [σuni−k,j + (1 − σ)un−1 i−k,j ] k=i−M1

j

−γ (γ) + K2 ℋxβ (−h2 ) ∑ ĝk [σuni,j−k + (1 − σ)un−1 i,j−k ] k=j−M2

+ u0ij unij

β γ n−1+σ ℋx ℋy fij ,

= φ(xi , yj ),

= 0,

(i, j) ∈ ω, 1 ⩽ n ⩽ N,

(5.81)

(i, j) ∈ ω,

(5.82)

(i, j) ∈ 𝜕ω, 0 ⩽ n ⩽ N.

(5.83)

In the subsequent part, the theoretical analyses on this scheme will be implemented.

5.4.2 Solvability of the difference scheme Theorem 5.4.1. The difference scheme (5.81)–(5.83) is uniquely solvable. Proof. Let ̄ un = {unij | (i, j) ∈ ω}.

5.4 The fourth-order method in space for 2D problem

| 319

The value of u0 is determined by (5.82)–(5.83). Now suppose the values of u0 , u1 , . . . , un−1 have been obtained, then the linear system in un can be determined by (5.81) and (5.83). To show its unique solvability, it suffices to prove that the corresponding homogeneous one i 1 (n,α) β γ n −β γ (β) { { c ℋ ℋ u = −σK h ℋ ĝk uni−k,j ∑ { 1 1 x y ij y 0 { s { { k=i−M1 { { { j { −γ β (γ) { − σK h ℋ ĝk uni,j−k , ∑ { 2 x { 2 { { k=j−M { 2 { { n u = 0, (i, j) ∈ 𝜕ω { ij

(i, j) ∈ ω,

(5.84) (5.85)

has only the trivial solution. Taking the inner product on both hand sides of (5.84) with un , we have M −1 M −1

1 2 1 (n,α) c0 h1 h2 ∑ ∑ (ℋxβ ℋyγ unij )unij s i=1 j=1

−β

M1 −1

= −K1 σh1 h1 ∑

M2 −1

i

(β) ∑ ĝk [h2 ∑ (ℋyγ uni−k,j )unij ]

i=1 k=i−M1

−γ

M2 −1

j=1

M1 −1

j

(γ) ∑ ĝk [h1 ∑ (ℋxβ uni,j−k )unij ].

− K2 σh2 h2 ∑

i=1

j=1 k=j−M2

M −1

(5.86)

γ

Denote v i = (0, uni1 , uni2 , . . . , uni,M2 −1 , 0)T , then the term h2 ∑j=12 (ℋy uni−k,j ) ⋅ unij can be taken as the inner product of v i−k with v i . This inner product is similar to (⋅, ⋅)A defined in Section 5.2. β M −1 Similarly, denote w j = (0, un1j , un2j , . . . , unM1 −1,j , 0)T , then the term h1 ∑i=11 (ℋx uni,j−k )unij can be taken as the inner product of w j−k with w j . This inner product is also similar to (⋅, ⋅)A defined in Section 5.2. It follows from Corollary 5.1.1 and (5.29) that −β

M1 −1

− h1 h1 ∑

M2 −1

i

(β) ∑ ĝk [h2 ∑ (ℋyγ uni−k,j )unij ]

i=1 k=i−M1

j=1

M1 −1

M2 −1

i=1

j=1

⩽ −c∗(β) (2L1 )−β h1 ∑ [h2 ∑ (ℋyγ unij )unij ] 2 󵄩 󵄩2 ⩽ − c∗(β) (2L1 )−β 󵄩󵄩󵄩un 󵄩󵄩󵄩 3

(5.87)

and −γ

M2 −1

− h2 h2 ∑

j

M1 −1

(γ) ∑ ĝk [h1 ∑ (ℋxβ uni,j−k )unij ]

j=1 k=j−M2

i=1

320 | 5 Difference methods for the time-space-fractional differential equations M2 −1

M1 −1

j=1

i=1

⩽ −c∗(γ) (2L2 )−γ h2 ∑ [h1 ∑ (ℋxβ unij )unij ] 2 󵄩 󵄩2 ⩽ − c∗(γ) (2L2 )−γ 󵄩󵄩󵄩un 󵄩󵄩󵄩 . 3

(5.88)

On the other hand, we have M1 −1 M2 −1

h1 h2 ∑ ∑ (ℋxβ ℋyγ unij )unij ⩾ i=1

j=1

1 󵄩󵄩 n 󵄩󵄩2 󵄩u 󵄩 . 3󵄩 󵄩

(5.89)

Substituting (5.87)–(5.89) into (5.86) produces ‖un ‖ = 0. Then un = 0 is followed by combining with (5.85). By the principle of induction, the difference scheme (5.81)–(5.83) is uniquely solvable. The proof ends.

5.4.3 Stability of the difference scheme Theorem 5.4.2. Suppose {unij | (i, j) ∈ ω,̄ 0 ⩽ n ⩽ N} is the solution of the difference scheme (5.81)–(5.83), then it holds β (2L2 )γ 󵄩󵄩 n 󵄩󵄩2 󵄩 0 󵄩2 9 (2L ) α󵄩 󵄩󵄩ℋβ ℋγ f m−1+σ 󵄩󵄩󵄩2 }, ]Γ(1 − α) max {tm 󵄩󵄩u 󵄩󵄩 ⩽ 3󵄩󵄩󵄩u 󵄩󵄩󵄩 + [ 1(β) + 󵄩 x y 󵄩 (γ) 1⩽m⩽n 16 K c∗ K c 2 ∗ 1

1 ⩽ n ⩽ N,

(5.90)

where M1 −1 M2 −1

󵄩󵄩 β γ m−1+σ 󵄩󵄩2 β γ m−1+σ 2 ). 󵄩󵄩ℋx ℋy f 󵄩󵄩 = h1 h2 ∑ ∑ (ℋx ℋy fij i=1

j=1

Proof. Taking the inner product on both hand sides of (5.81) with σun + (1 − σ)un−1 , we have M −1 M −1

1 2 1 n−1 (n,α) − un−k−1 )][σunij + (1 − σ)un−1 ∑ ck h1 h2 ∑ ∑ [ℋxβ ℋyγ (un−k ij ij ij ] s k=0 i=1 j=1

M2 −1

−β

M1 −1

= K1 h2 ∑ {−h1 h1 ∑ j=1

i

(β) ∑ ĝk [ℋyγ (σuni−k,j + (1 − σ)un−1 i−k,j )]

i=1 k=i−M1

⋅ [σunij + (1 − σ)un−1 ij ]} M1 −1

−γ

M2 −1

+ K2 h1 ∑ {(−h2 )h2 ∑ i=1

j

(γ) ∑ ĝk [ℋxβ (σuni,j−k + (1 − σ)un−1 i,j−k )]

j=1 k=j−M2

5.4 The fourth-order method in space for 2D problem

| 321

⋅ [σunij + (1 − σ)un−1 ij ]} M1 −1 M2 −1

+ h1 h2 ∑ ∑ (ℋxβ ℋyγ fijn−1+σ )[σunij + (1 − σ)un−1 ij ]. i=1

j=1

(5.91)

Now each term in (5.91) will be estimated. Suppose u, v ∈ 𝒱̊h . Define the inner product M1 −1 M2 −1

(u, v)H = h1 h2 ∑ ∑ (ℋxβ ℋyγ uij )vij . i=1

j=1

It follows from Lemma 2.6.1 that M1 −1 M2 −1

n−1

− un−k−1 )][σunij + (1 − σ)un−1 ∑ ck(n,α) h1 h2 ∑ ∑ [ℋxβ ℋyγ (un−k ij ij ij ] i=1

k=0

j=1

n−1

= ∑ ck(n,α) (un−k − un−k−1 , σun + (1 − σ)un−1 )H k=0

1 n−1 ⩾ ∑ ck(n,α) [(un−k , un−k )H − (un−k−1 , un−k−1 )H ]. 2 k=0

(5.92)

For the former two terms on the right-hand side of (5.91), by Corollary 5.1.1, similar to (5.87) and (5.88), we have M2 −1

−β

M1 −1

h2 ∑ {−h1 h1 ∑ j=1

i

(β) ∑ ĝk [ℋyγ (σuni−k,j + (1 − σ)un−1 i−k,j )]

i=1 k=i−M1

⋅ [σunij + (1 − σ)un−1 ij ]}

and

2 󵄩 󵄩2 ⩽ − c∗(β) (2L1 )−β 󵄩󵄩󵄩σun + (1 − σ)un−1 󵄩󵄩󵄩 3 M1 −1

−γ

M2 −1

h1 ∑ {(−h2 )h2 ∑ i=1

(5.93)

j

(γ) ∑ ĝk [ℋxβ (σuni,j−k + (1 − σ)un−1 i,j−k )]

j=1 k=j−M2

⋅ [σunij + (1 − σ)un−1 ij ]} 2 󵄩 󵄩2 ⩽ − c∗(γ) (2L2 )−γ 󵄩󵄩󵄩σun + (1 − σ)un−1 󵄩󵄩󵄩 . 3

(5.94)

For the last term on the right-hand side of (5.91), it follows from the Cauchy– Schwarz inequality that M1 −1 M2 −1

h1 h2 ∑ ∑ (ℋxβ ℋyγ fijn−1+σ )[σunij + (1 − σ)un−1 ij ] i=1

j=1

322 | 5 Difference methods for the time-space-fractional differential equations 󵄩 󵄩 󵄩 󵄩 ⩽ 󵄩󵄩󵄩ℋxβ ℋyγ f n−1+σ 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩σun + (1 − σ)un−1 󵄩󵄩󵄩

2 2 󵄩 󵄩2 ⩽ [ K1 c∗(β) (2L1 )−β + K2 c∗(γ) (2L2 )−γ ]󵄩󵄩󵄩σun + (1 − σ)un−1 󵄩󵄩󵄩 3 3 +

3 (2L1 )β (2L2 )γ 󵄩󵄩 β γ n−1+σ 󵄩󵄩2 [ ]󵄩ℋ ℋ f + 󵄩󵄩 . 32 K c∗(β) K2 c∗(γ) 󵄩 x y 1

(5.95)

Substituting (5.92)–(5.95) into (5.91) arrives at 1 n−1 (n,α) n−k n−k [(u , u )H − (un−k−1 , un−k−1 )H ] ∑c s k=0 k

⩽

3 (2L1 )β (2L2 )γ 󵄩󵄩 β γ n−1+σ 󵄩󵄩2 ]󵄩ℋ ℋ f [ + 󵄩󵄩 , 16 K c∗(β) K2 c∗(γ) 󵄩 x y 1

1 ⩽ n ⩽ N.

(5.96)

By Lemma 1.6.3, we have s

(n,α) cn−1

⩽ tnα Γ(1 − α).

Then it follows from (5.96) that c0(n,α) (un , un )H

n−1

(n,α) (n,α) 0 0 ⩽ ∑ (ck−1 − ck(n,α) )(un−k , un−k )H + cn−1 (u , u )H k=1

+ n−1

3 (2L1 )β (2L2 )γ 󵄩󵄩 β γ n−1+σ 󵄩󵄩2 ]󵄩ℋ ℋ f s[ + 󵄩󵄩 16 K c∗(β) K2 c∗(γ) 󵄩 x y 1

(n,α) (n,α) ⩽ ∑ (ck−1 − ck(n,α) )(un−k , un−k )H + cn−1 {(u0 , u0 )H k=1

+

3 (2L1 )β (2L2 )γ 󵄩 󵄩2 ]Γ(1 − α)tnα 󵄩󵄩󵄩ℋxβ ℋyγ f n−1+σ 󵄩󵄩󵄩 }, [ + 16 K c∗(β) K2 c∗(γ) 1

1 ⩽ n ⩽ N.

The application of the inductive process can yield 3 (2L1 )β (2L2 )γ + [ ]Γ(1 − α) 16 K c∗(β) K2 c∗(γ) 1 α󵄩 󵄩󵄩ℋβ ℋγ f m−1+σ 󵄩󵄩󵄩2 }, 1 ⩽ n ⩽ N. ⋅ max {tm 󵄩 x y 󵄩

(un , un )H ⩽ (u0 , u0 )H + 1⩽m⩽n

Noticing 1 󵄩󵄩 n 󵄩󵄩2 󵄩 n 󵄩2 n n 󵄩u 󵄩 ⩽ (u , u )H ⩽ 󵄩󵄩󵄩u 󵄩󵄩󵄩 , 3󵄩 󵄩 the inequality (5.90) can be obtained from (5.97). The proof ends.

(5.97)

5.5 Supplementary remarks and discussions | 323

5.4.4 Convergence of the difference scheme Theorem 5.4.3. Suppose {Uijn | (i, j) ∈ ω,̄ 0 ⩽ n ⩽ N} and {unij | (i, j) ∈ ω,̄ 0 ⩽ n ⩽ N} are solutions of the problem (5.50)–(5.52) and the difference scheme (5.81)–(5.83), respectively. Let eijn = Uijn − unij ,

(i, j) ∈ ω,̄ 0 ⩽ n ⩽ N,

then it holds (2L )β (2L2 )γ 󵄩󵄩 n 󵄩󵄩 3 ]Γ(1 − α)T α L1 L2 c4 (τ2 + h41 + h42 ), 󵄩󵄩e 󵄩󵄩 ⩽ √[ 1(β) + (γ) 4 K2 c∗ K1 c∗

1 ⩽ n ⩽ N.

(5.98)

Proof. Subtracting (5.81)–(5.83) from (5.77), (5.79)–(5.80), respectively, produces the system of error equations as follows: { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { {

1 n−1 (n,α) β γ n−k n−k−1 ℋx ℋy (eij − eij ) ∑c s k=0 k i

−β (β) n n−1 + (1 − σ)ei−k,j ] = K1 ℋyγ (−h1 ) ∑ ĝk [σei−k,j k=i−M1

j

−γ (γ) n n−1 + (1 − σ)ei,j−k ] + (r4 )nij , + K2 ℋxβ (−h2 ) ∑ ĝk [σei,j−k k=j−M2

eij0 eijn

= 0,

(i, j) ∈ ω,

= 0,

(i, j) ∈ 𝜕ω, 0 ⩽ n ⩽ N.

(i, j) ∈ ω, 1 ⩽ n ⩽ N,

Noticing (5.78), the application of Theorem 5.4.2 immediately yields 9 (2L )β (2L2 )γ 󵄩󵄩 n 󵄩󵄩2 α󵄩 󵄩󵄩(r4 )m 󵄩󵄩󵄩2 } ]Γ(1 − α) max {tm 󵄩󵄩e 󵄩󵄩 ⩽ [ 1(β) + 󵄩 󵄩 (γ) 1⩽m⩽n 16 K c∗ K2 c∗ 1 ⩽

9 (2L1 )β (2L2 )γ 2 [ ]Γ(1 − α)T α L1 L2 c42 (τ2 + h41 + h42 ) , + 16 K c∗(β) K2 c∗(γ) 1

1 ⩽ n ⩽ N.

Taking the square root on both hand sides of the inequality above will reach (5.98). The proof ends.

5.5 Supplementary remarks and discussions 1. The finite difference methods for 1D and 2D time-space-fractional Bloch–Torrey equations were discussed in this chapter. The time Caputo fractional derivative was

324 | 5 Difference methods for the time-space-fractional differential equations handled by L2-1σ approximation of order 3 − α, and the space Riesz fractional derivatives or the weighted values of space Riesz derivatives at three points were approximated by the fractional central difference quotient formula (Theorem 1.5.1 or Theorem 1.5.2). Several difference schemes were derived and for each of them, the unique solvability, stability and convergence in L2 norm were proved[80] . 2. In [110], the following fourth-order numerical differentiation formula to approximate the Riesz derivatives was established: β

(−

β

Δ f (x − h) Δ f (x) β β )[− h β ] + (1 + )[− h β ] 24 12 h h β

+ (−

Δ f (x + h) β 𝜕β f (x) ]= + O(h4 ). )[− h β 24 h 𝜕|x|β

(5.99)

Comparing (1.54) with (5.99), the former one is to use the fractional central difference quotient formula to approximate the weighted value of Riesz derivatives at three points, whereas, the latter one is to use the weighted value of the fractional central difference quotient formula to approximate the Riesz derivative at one point. 3. In [106], Yu et al. investigated the numerical solutions of 3D time-space-fractional Bloch–Torrey equations, where the time Caputo derivative was discretized by the L1 formula and the space Riesz derivatives were approximated by the shifted G-L formula (1.47) and a positive-type difference scheme was derived. The unconditional stability and convergence of the resultant scheme were proved by the maximum principle and the convergence order in the maximum norm was O(τ2−α + hx + hy + hz ). In [107], the authors studied the numerical solutions of 2D time-space-fractional Bloch–Torrey equations, where the time Caputo derivative was discretized by the L1 formula and the space Riesz derivatives were approximated by the fractional central difference quotient formula (Theorem 1.5.1, or [4]) and a positive-type difference scheme of order O(τ2−α + h2x + h2y ) was proposed. The unconditional stability and convergence of the resultant scheme were proved by the maximum principle and the convergence order in the maximum norm was O(τ2−α + h2x + h2y ). The method in [107] can also be used to solve the 3D time-space-fractional Bloch–Torrey equations by the finite difference method[73] . 4. Ran and Zhang[65] derived a two-level Crank–Nicolson difference scheme and a three-level linearized difference scheme for the nonlinear time-space-fractional Schrödinger equations. 5. Xu and Sun[100] developed a fast second-order difference scheme for the timespace fractional equation.

Exercises 5

| 325

Exercises 5 5.1 For the problem (5.1)–(5.3), construct the following difference scheme: { { { { { { { { { { { { { { { { { { { { { { { { {

n−1 τ−α n 0 [a(α) u − )uk − a(α) ∑ (a(α) − a(α) n−1 ui ] n−k i Γ(2 − α) 0 i k=1 n−k−1 i

(β) = −h−β ∑ ĝk uni−k + fin , k=i−M

u0i = φ(xi ),

un0 = 0,

1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N,

1 ⩽ i ⩽ M − 1,

unM = 0,

0 ⩽ n ⩽ N.

̃ t) like that in Section 5.1 and suppose u(⋅, ̃ t) ∈ C 2+β (ℛ). Define the function u(x, For this difference scheme, try to (1) analyze the truncation error; (2) show the unique solvability; (3) show the stability with respect to the initial value φ and the function f ; (4) show the convergence and derive the error expression. 5.2 For the problem (5.1)–(5.3), construct the following difference scheme: { { { { { { { { { { { { { { { { { { { { { { { { {

β

𝒜h

n−1 τ−α n (α) (α) k (α) 0 [a(α) 0 ui − ∑ (an−k−1 − an−k )ui − an−1 ui ] Γ(2 − α) k=1 i

β (β) = −h−β ∑ ĝk uni−k + 𝒜h fin , k=i−M

u0i = φ(xi ), un0

= 0,

1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N,

1 ⩽ i ⩽ M − 1,

unM

= 0,

0 ⩽ n ⩽ N.

̃ t) like that in Section 5.1 and suppose u(⋅, ̃ t) ∈ C 4+β (ℛ). Define the function u(x, For this difference scheme, try to (1) analyze the truncation error; (2) show the unique solvability; (3) show the stability with respect to the initial value φ and the function f ; (4) show the convergence and derive the error expression. 5.3 For the problem (5.50)–(5.52), construct the following difference scheme: { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { {

n−1 τ−α 0 n )uk − a(α) [a(α) u − ∑ (a(α) − a(α) n−1 uij ] n−k ij Γ(2 − α) 0 ij k=1 n−k−1 j

i

−β (β) −γ (γ) = K1 (−h1 ) ∑ ĝk uni−k,j + K2 (−h2 ) ∑ ĝk uni,j−k + fijn , k=i−M1

u0ij = φ(xi , yj ),

unij

= 0,

k=j−M2

(i, j) ∈ ω, 1 ⩽ n ⩽ N,

(i, j) ∈ ω,

(i, j) ∈ 𝜕ω, 0 ⩽ n ⩽ N.

326 | 5 Difference methods for the time-space-fractional differential equations ̂ y, t) and w(x, ̂ y, t) like those in Section 4.4 and suppose Define the functions v(x, ̂ y, t) ∈ C 2+β (ℛ), w(x, ̂ ⋅, t) ∈ C 2+γ (ℛ). v(⋅, For this difference scheme, try to (1) analyze the truncation error; (2) show the unique solvability; (3) show the stability with respect to the initial value φ and the function f ; (4) show the convergence and derive the error expression. 5.4 For the problem (5.50)–(5.52), construct the following difference scheme: { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { {

β

γ

ℋx ℋy

n−1 τ−α n (α) (α) k (α) 0 [a(α) 0 uij − ∑ (an−k−1 − an−k )uij − an−1 uij ] Γ(2 − α) k=1 i

j

k=i−M1

k=j−M2

−β (β) −γ (γ) = K1 (−h1 )ℋyγ ∑ ĝk uni−k,j + K2 (−h2 )ℋxβ ∑ ĝk uni,j−k + ℋxβ ℋyγ fijn ,

u0ij unij

= φ(xi , yj ),

= 0,

(i, j) ∈ ω, 1 ⩽ n ⩽ N,

(i, j) ∈ ω,

(i, j) ∈ 𝜕ω, 0 ⩽ n ⩽ N.

̂ y, t) and w(x, ̂ y, t) like those in Section 4.4 and suppose Define the functions v(x, ̂ y, t) ∈ C 4+β (ℛ), w(x, ̂ ⋅, t) ∈ C 4+γ (ℛ). v(⋅, For this difference scheme, try to (1) analyze the truncation error; (2) show the unique solvability; (3) show the stability with respect to the initial value φ and the function f ; (4) show the convergence and derive the error expression.

6 Difference methods for the time distributed-order subdiffusion equations In the previous chapters, numerical solutions of the multiterm time-fractional differential equations have been discussed. When the number of terms in the timefractional derivatives tends to the infinity, the time distributed-order differential equation is derived. It can be used to model the complex processes with the varying diffusion exponents as the time, such as the retarding subdiffusion, superslow diffusion, accelerating superdiffusion and so on. Numerous applications in polymer physics, kinetics of particles moving in the quenched random force fields, iterated map models, etc. have been found. In this chapter, the finite difference methods for solving a class of time distributed-order subdiffusion equations will be concerned. For each scheme, the unique solvability, stability and convergence will be investigated. The entire chapter consists of 7 sections.

6.1 The second-order method in both space and distributed order for 1D problem Consider the following 1D initial-boundary value problem of time distributed-order subdiffusion equation w 𝒟t u(x, t) = uxx (x, t) + f (x, t), 0 < x < L, 0 < t ⩽ T, { { u(x, 0) = 0, 0 < x < L, { { { u(0, t) = φ1 (t), u(L, t) = φ2 (t), 0 ⩽ t ⩽ T,

(6.1) (6.2) (6.3)

where φ1 (0) = 0, φ2 (0) = 0 and 1

w

C

α

𝒟t u(x, t) = ∫ w(α) 0 Dt u(x, t)dα, C α 0 Dt u(x, t)

=

0 t 1 ∫0 (t Γ(1−α) {

ut (x, t),

1

− ξ )−α uξ (x, ξ )dξ ,

0 ⩽ α < 1, α = 1,

w(α) ⩾ 0, ∫0 w(α)dα = c0 > 0, the functions f , φ1 and φ2 are given. ̂ t) like that in Section 2.1. Suppose u(x, ̂ ⋅) ∈ C 2+1 (ℛ) and Define the function u(x, 4 u(⋅, t) ∈ C [0, L].

https://doi.org/10.1515/9783110616064-006

328 | 6 Difference methods for the time distributed-order subdiffusion equations 6.1.1 Derivation of the difference scheme Take three positive integers J, M, N and denote Δα = 2J1 , h = ML , τ = NT ; αl = lΔα (0 ⩽ l ⩽ 2J); xi = ih (0 ⩽ i ⩽ M); tn = nτ (0 ⩽ n ⩽ N). Introduce the same mesh function spaces and notations like those in Section 2.1. The distributed-order integral is firstly discretized with proper quadrature formulae. The composite trapezoid formula is listed below. Lemma 6.1.1. (Composite trapezoid formula) Suppose function s ∈ C 2 [0, 1], then it holds 1

2J

∫ s(α)dα = Δα ∑ cl s(αl ) − l=0

0

Δα2 󸀠󸀠 s (ξ ), 12

ξ ∈ (0, 1),

where cl = {

1 , 2

1,

l = 0, 2J, 1 ⩽ l ⩽ 2J − 1.

Define the mesh functions Uin = u(xi , tn ),

fin = f (xi , tn ),

0 ⩽ i ⩽ M, 0 ⩽ n ⩽ N.

Considering equation (6.1) at the point (xi , tn ), we have w

n

𝒟t u(xi , tn ) = uxx (xi , tn ) + fi ,

1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N.

(6.4)

Let s(α, xi , tn ) = w(α) C0 Dαt u(xi , tn ). Suppose function s(⋅, xi , tn ) ∈ C 2 [0, 1]. It follows from Lemma 6.1.1 that w 𝒟t u(xi , tn )

1

= ∫ s(α, xi , tn )dα

2J

0

= Δα ∑ cl s(αl , xi , tn ) − l=0 2J

= Δα ∑ cl w(αl ) l=0

Δα2 𝜕2 s(α, xi , tn ) 󵄨󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨α=ξ n 12 𝜕α2 i

C αl 0 Dt u(xi , tn )

+ O(Δα2 ),

where ξin ∈ (0, 1). By Corollary 1.4.1, associated with (6.2), we have C α 0 Dt u(xi , tn )

= 0 Dαt u(xi , tn )

(6.5)

6.1 The second-order method in both space and distributed order for 1D problem n

= τ−α ∑ wk(α) Uin−k + O(τ2 ),

| 329

1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N,

k=0

(6.6)

where {wk(α) } is defined by (1.33)–(1.34). Substituting (6.6) into (6.5) produces 2J

w

𝒟t u(xi , tn ) = Δα ∑ cl w(αl )τ

−αl

l=0

n

∑ wk l Uin−k + O(τ2 + Δα2 ), (α )

k=0

1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N.

(6.7)

Notice uxx (xi , tn ) = δx2 Uin + O(h2 ),

1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N.

(6.8)

Substituting (6.7) and (6.8) into (6.4) arrives at 2J

n

l=0

k=0

Δα ∑ cl w(αl )τ−αl ∑ wk l Uin−k = δx2 Uin + fin + (r1 )ni , (α )

1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N,

(6.9)

and there is a positive constant c1 such that 󵄨󵄨 n 󵄨󵄨 2 2 2 󵄨󵄨(r1 )i 󵄨󵄨 ⩽ c1 (τ + h + Δα ),

1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N.

(6.10)

Noticing the initial-boundary value conditions (6.2)–(6.3), we have {

Ui0 = 0, U0n

1 ⩽ i ⩽ M − 1, n UM

= φ1 (tn ),

= φ2 (tn ),

(6.11) 0 ⩽ n ⩽ N.

(6.12)

Neglecting the small term (r1 )ni in (6.9) and replacing the exact solution Uin with its numerical one uni , we get a difference scheme for solving (6.1)–(6.3) as follows: 2J

n

(α ) { { Δα ∑ cl w(αl )τ−αl ∑ wk l un−k = δx2 uni + fin , { i { { { l=0 k=0 { { 1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N, { { { { 0 { u = 0, 1 ⩽ i ⩽ M − 1, { { { i n n { u0 = φ1 (tn ), uM = φ2 (tn ), 0 ⩽ n ⩽ N.

(6.13) (6.14) (6.15)

In the subsequent part, denote 2J

(α )

μ = Δα ∑ cl w(αl )τ−αl w0 l . l=0

(6.16)

330 | 6 Difference methods for the time distributed-order subdiffusion equations 6.1.2 Solvability of the difference scheme Theorem 6.1.1. The difference scheme (6.13)–(6.15) is uniquely solvable. Proof. Let un = (un0 , un1 , . . . , unM ). The value of u0 is determined by (6.14)–(6.15). Now assume that the values of u0 , u1 , . . . , un−1 have been uniquely determined, then the linear system in un can be obtained from (6.13) and (6.15). To show its unique solvability, it suffices to verify that the corresponding homogeneous one {

μ uni = δx2 uni , un0

=

unM

1 ⩽ i ⩽ M − 1,

(6.17)

=0

(6.18)

has only the trivial solution. Reformulate (6.17) as (μ +

2 1 )uni = 2 (uni−1 + uni+1 ), 2 h h

1 ⩽ i ⩽ M − 1.

Suppose ‖un ‖∞ = |unin |, where in ∈ {1, 2, . . . , M − 1}. Letting i = in in the equality above and taking the absolute value of both hand sides, the application of the triangle inequality produces (μ +

2 󵄩󵄩 n 󵄩󵄩 2󵄩 󵄩 )󵄩u 󵄩 ⩽ 󵄩󵄩un 󵄩󵄩 , h2 󵄩 󵄩∞ h2 󵄩 󵄩∞

which implies ‖un ‖∞ = 0, hence un = 0. By the principle of induction, the difference scheme (6.13)–(6.15) is uniquely solvable. The proof ends. For the analyses on the stability and convergence of the difference scheme (6.13)– (6.15), three useful lemmas are essential. 6.1.3 Three lemmas Definition 6.1.1. The matrix t0 [ t [ 1 [ [ t2 [ Tn = [ . [ .. [ [ [ tn−2 [ tn−1

t−1 t0 t1 .. . tn−3 tn−2

t−2 t−1 t0 .. . tn−4 tn−3

... ... ... ... ...

t2−n t3−n t4−n .. . t0 t1

t1−n t2−n t3−n .. . t−1 t0

] ] ] ] ] ] ] ] ] ] ]

(6.19)

6.1 The second-order method in both space and distributed order for 1D problem

| 331

is called the Toeplitz matrix. That is, the entries of an n × n Toeplitz matrix Tn are constant along each diagonal. Definition 6.1.2. [6] If the entries {tk }n−1 k=1−n of the Toeplitz matrix (6.19) are Fourier coefficients of function f (x), that is, π

1 tk = ∫ f (x)e−ikx dx, 2π −π

then the function f (x) is called the generating function of matrix Tn . Lemma 6.1.2. [5, 6] (Grenander–Szegö theorem) For the Toeplitz matrix Tn given by (6.19), if its generating function f (x) defined on [−π, π] is continuous and real-valued, then fmin ⩽ λmin (Tn ) ⩽ λmax (Tn ) ⩽ fmax , where fmin and fmax denote the minimum and maximum values of f (x), respectively, λmin (Tn ) and λmax (Tn ) denote the smallest and largest eigenvalues of Tn , respectively. Moreover, if fmin < fmax , then for n ⩾ 1, any eigenvalue λ(Tn ) of Tn satisfies fmin < λ(Tn ) < fmax . In particular, Tn is positive definite when fmin > 0. Lemma 6.1.3. [96] Let the coefficient {wk(α) } be defined by (1.33)–(1.34). Then, for any vector (v0 , v1 , . . . , vm )T ∈ ℛm+1 , it holds m

n

∑ ( ∑ wk(α) vn−k )vn ⩾ 0.

n=0 k=0

Proof. Notice the fact that the quadratic form m

n

∑ ( ∑ wk(α) vn−k )vn

n=0 k=0

is nonnegative is equivalent to that the symmetric Toeplitz matrix [ [ [ [ [ [ W =[ [ [ [ [ [ [ is positive semidefinite.

w0(α) 1 (α) w 2 1

1 (α) w 2 1

w0(α)

1 (α) w 2 2

1 (α) w 2 1

1 (α) w 2 m

1 (α) w 2 m−1

.. .

.. .

1 (α) w 2 2

...

1 (α) w 2 1

...

w0(α)

...

1 (α) w 2 m−2

...

.. .

1 (α) w 2 m

1 (α) w 2 m−1

1 (α) w 2 m−2

.. .

w0(α)

] ] ] ] ] ] ] ] ] ] ] ] ]

332 | 6 Difference methods for the time distributed-order subdiffusion equations Next, we try to show that the Toeplitz matrix W is positive semidefinite. The generating function of W is given by f (α, x) = w0(α) + = (1 + +

1 ∞ (α) ikx 1 ∞ (α) −ikx ∑ w e + ∑ wk e 2 k=1 k 2 k=1

α α (α) ikx α (α) 1 ∞ )g + ∑ [(1 + )gk(α) − gk−1 ]e 2 0 2 k=1 2 2

1 ∞ α (α) −ikx α ]e ∑ [(1 + )gk(α) − gk−1 2 k=1 2 2

1 α ∞ 1 α ∞ = (1 + ) ∑ gk(α) eikx + (1 + ) ∑ gk(α) e−ikx 2 2 k=0 2 2 k=0 −

α ∞ (α) i(k+1)x α ∞ (α) −i(k+1)x − ∑ gk e ∑g e 4 k=0 k 4 k=0

1 α 1 α α α = (1 + )(1 − eix ) + (1 + )(1 − e−ix ) 2 2 2 2 α α ix α ix −ix α −ix − (1 − e ) e − (1 − e ) e . 4 4 As we see, the function f (α, x) is an even function with respect to x with the period 2π, hence we only need to consider function f (α, x) for x ∈ [0, π]. Reformulate f (α, x) as i i i 1 α α f (α, x) = (1 + )[(e− 2 x − e 2 x )e 2 x ] 2 2 i i i α 1 α + (1 + )[(e 2 x − e− 2 x )e− 2 x ] 2 2 i i i i i i α α α α − [(e− 2 x − e 2 x )e 2 x ] eix − [(e 2 x − e− 2 x )e− 2 x ] e−ix 4 4 α α i 1 α x i α x 1 = (1 + )[2i sin(− )e 2 x ] + (1 + )[2i sin( )e− 2 x ] 2 2 2 2 2 2

α

−

α

i α α x i x [2i sin(− )e 2 x ] eix − [2i sin( )e− 2 x ] e−ix 4 2 4 2

α

x π π x x 1 α = [2 sin( )] { (1 + )[ei( 2 − 2 )α + ei( 2 − 2 )α ] 2 2 2 π x α i( x2 − π2 )α ix − [e ⋅ e + ei( 2 − 2 )α ⋅ e−ix ]} 4

α

x α α α α = [2 sin( )] {(1 + ) cos[ (π − x)] − cos[ (π − x) − x]}. 2 2 2 2 2 Let h(α, x) = (1 +

α α α α ) cos[ (π − x)] − cos[ (π − x) − x]. 2 2 2 2

6.1 The second-order method in both space and distributed order for 1D problem

| 333

One can easily check that α (1 + 2 α = (1 + 2

α α α ){sin[ (π − x)] − sin[ (π − x) − x]} 2 2 2 α α x x ){sin[( (π − x) − ) + ] 2 2 2 2 x x α − sin[( (π − x) − ) − ]} 2 2 2 α x x α = α(1 + ) cos[ (π − x) − ] sin( ). 2 2 2 2

hx (α, x) =

Therefore, hx (α, x) ⩾ 0 when x ∈ [0, π]. Hence, h(α, x) ⩾ h(α, 0) = cos απ ⩾ 0, which 2 implies that f (α, x) ⩾ 0. The lemma now follows as a result of Lemma 6.1.2. The proof ends. Lemma 6.1.4. For the constant μ defined by (6.16), it holds μτ = O(

1 ). | ln τ|

Proof. 2J

μτ = Δα ∑ cl w(αl ) ⋅ 1

l=0

∼ ∫ w(α)(1 + 0

α 1 (1 + l )τ ταl 2

α 1−α )τ dα 2 1

α∗ = w(α )(1 + ) ∫ τ1−α dα 2 ∗

0

1 α τ1−α 󵄨󵄨󵄨󵄨 ) 󵄨󵄨 2 | ln τ| 󵄨󵄨α=0 α∗ 1 − τ ) , = w(α∗ )(1 + 2 | ln τ|

= w(α∗ )(1 +

∗

where α∗ ∈ (0, 1). Hence, μτ = O( | ln1 τ| ). The proof ends. 6.1.4 Stability of the difference scheme Theorem 6.1.2. Suppose {vin | 0 ⩽ i ⩽ M, 0 ⩽ n ⩽ N} is the solution of the difference scheme

334 | 6 Difference methods for the time distributed-order subdiffusion equations 2J

n

(α ) { { Δα ∑ cl w(αl )τ−αl ∑ wk l vin−k = δx2 vin + gin , { { { { l=0 k=0 { { 1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N, { { { { 0 { v = φi , 1 ⩽ i ⩽ M − 1, { { { i n n { v0 = 0, vM = 0, 0 ⩽ n ⩽ N.

(6.20)

Then it holds m m 2 󵄩 󵄩2 󵄩 󵄩2 L 󵄩 󵄩2 τ ∑ 󵄩󵄩󵄩δx vn 󵄩󵄩󵄩 ⩽ 2μτ󵄩󵄩󵄩v0 󵄩󵄩󵄩 + τ ∑ 󵄩󵄩󵄩g n 󵄩󵄩󵄩 , 6 n=1 n=1

1 ⩽ m ⩽ N,

where M−1

󵄩󵄩 n 󵄩󵄩2 n 2 󵄩󵄩g 󵄩󵄩 = h ∑ (gi ) . i=1

Proof. Making the inner product on both hand sides of (6.20) with vn , it follows from Lemma 2.1.1 that 2J

n

Δα ∑ cl w(αl )τ−αl ∑ wk l (vn−k , vn )

= =

(α )

l=0 k=0 2 n n n n (δx v , v ) + (g , v ) 󵄩 󵄩2 −󵄩󵄩󵄩δx vn 󵄩󵄩󵄩 + (g n , vn )

2 󵄩 󵄩2 3 󵄩 󵄩2 L 󵄩 󵄩2 ⩽ −󵄩󵄩󵄩δx vn 󵄩󵄩󵄩 + 2 󵄩󵄩󵄩vn 󵄩󵄩󵄩 + 󵄩󵄩󵄩g n 󵄩󵄩󵄩 12 L 1 L2 󵄩 󵄩2 2 2 󵄩 󵄩 󵄩 󵄩 ⩽ −󵄩󵄩󵄩δx vn 󵄩󵄩󵄩 + 󵄩󵄩󵄩δx vn 󵄩󵄩󵄩 + 󵄩󵄩󵄩g n 󵄩󵄩󵄩 2 12 1 󵄩󵄩 n 󵄩󵄩2 L2 󵄩󵄩 n 󵄩󵄩2 = − 󵄩󵄩δx v 󵄩󵄩 + 󵄩󵄩g 󵄩󵄩 , 1 ⩽ n ⩽ N. 2 12

Summing up for n from 1 to m produces 2J

m

n

Δα ∑ cl w(αl )τ−αl ∑ ∑ wk l (vn−k , vn ) n=1 k=0 2 m

l=0 m

(α )

1 󵄩 󵄩2 L 󵄩 󵄩2 ⩽ − ∑ 󵄩󵄩󵄩δx vn 󵄩󵄩󵄩 + ∑ 󵄩󵄩g n 󵄩󵄩 , 2 n=1 12 n=1󵄩 󵄩

1 ⩽ m ⩽ N.

Adding the term μ(v0 , v0 ) to both hand sides of the inequality above gives 2J

m

n

Δα ∑ cl w(αl )τ−αl ∑ ∑ wk l (vn−k , vn ) l=0 m

(α )

n=0 k=0

1 󵄩 L2 m 󵄩󵄩 n 󵄩󵄩2 󵄩2 ⩽ − ∑ 󵄩󵄩󵄩δx vn 󵄩󵄩󵄩 + μ(v0 , v0 ) + ∑ 󵄩g 󵄩 , 2 n=1 12 n=1󵄩 󵄩

1 ⩽ m ⩽ N.

(6.21)

6.1 The second-order method in both space and distributed order for 1D problem

| 335

By Lemma 6.1.3, we have m

n

M−1 m

n

∑ ∑ wk l (vn−k , vn ) = h ∑ ∑ ( ∑ wk l vin−k )vin ⩾ 0. (α )

(α )

i=1 n=0 k=0

n=0 k=0

(6.22)

It is clear from (6.21) and (6.22) that m m 2 󵄩 󵄩2 L 󵄩 󵄩2 󵄩 󵄩2 τ ∑ 󵄩󵄩󵄩δx vn 󵄩󵄩󵄩 ⩽ 2μτ󵄩󵄩󵄩v0 󵄩󵄩󵄩 + τ ∑ 󵄩󵄩󵄩g n 󵄩󵄩󵄩 , 6 n=1 n=1

1 ⩽ m ⩽ N.

The proof ends.

6.1.5 Convergence of the difference scheme Theorem 6.1.3. Suppose {Uin | 0 ⩽ i ⩽ M, 0 ⩽ n ⩽ N} and {uni | 0 ⩽ i ⩽ M, 0 ⩽ n ⩽ N} are solutions of the problem (6.1)–(6.3) and the difference scheme (6.13)–(6.15), respectively. Let ein = Uin − uni ,

0 ⩽ i ⩽ M, 0 ⩽ n ⩽ N,

then it holds N √6 2 󵄩 󵄩 τ ∑ 󵄩󵄩󵄩en 󵄩󵄩󵄩∞ ⩽ L Tc1 (τ2 + h2 + Δα2 ). 12 n=1

Proof. Subtracting (6.13)–(6.15) from (6.9), (6.11)–(6.12), respectively, we get the system of error equations as follows: 2J

n

(α ) { { Δα ∑ cl w(αl )τ−αl ∑ wk l ein−k = δx2 ein + (r1 )ni , { { { { l=0 k=0 { { 1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N, { { { 0 { { e = 0, 1 ⩽ i ⩽ M − 1, { { { in n { e0 = 0, eM = 0, 0 ⩽ n ⩽ N.

Noticing (6.10), the application of Theorem 6.1.2 produces N N 2 2 2 󵄩 󵄩2 L 󵄩 󵄩2 L τ ∑ 󵄩󵄩󵄩δx en 󵄩󵄩󵄩 ⩽ τ ∑ 󵄩󵄩󵄩(r1 )n 󵄩󵄩󵄩 ⩽ TL[c1 (τ2 + h2 + Δα2 )] . 6 n=1 6 n=1

By the Cauchy–Schwarz inequality, Lemma 2.1.1 and the inequality above, we have N

2

N

N

n=1

n=1

󵄩 󵄩 󵄩 󵄩2 (τ ∑ 󵄩󵄩󵄩en 󵄩󵄩󵄩∞ ) ⩽ (τ ∑ 1)(τ ∑ 󵄩󵄩󵄩en 󵄩󵄩󵄩∞ ) n=1

336 | 6 Difference methods for the time distributed-order subdiffusion equations

⩽T⋅ ⩽

L N 󵄩󵄩 n 󵄩󵄩2 τ ∑ 󵄩δ e 󵄩 4 n=1󵄩 x 󵄩

LT L3 2 ⋅ T[c1 (τ2 + h2 + Δα2 )] , 4 6

that is, N √6 2 󵄩 󵄩 L Tc1 (τ2 + h2 + Δα2 ). τ ∑ 󵄩󵄩󵄩en 󵄩󵄩󵄩∞ ⩽ 12 n=1

The proof ends.

6.2 The fourth-order method in both space and distributed order for 1D problem In this section, we continue to consider the problem (6.1)–(6.3) and want to develop a difference method of order two in time and four in both space and distributed order. The unique solvability, stability and convergence of the proposed difference scheme are also proved. ̂ t) like that in Section 2.1. Suppose u(x, ̂ ⋅) ∈ C 2+1 (ℛ) and Define the function u(x, u(⋅, t) ∈ C 6 [0, L].

6.2.1 Derivation of the difference scheme At the beginning, two useful lemmas are presented that will be used later on. Lemma 6.2.1. (Composite Simpson formula) Suppose function s ∈ C 4 [0, 1], then it holds 1

2J

∫ s(α)dα = Δα ∑ dl s(αl ) − l=0

0

Δα4 (4) s (η), 180

η ∈ (0, 1),

where { { { dl = { { { {

1 , 3 2 , 3 4 , 3

l = 0, 2J, l = 2, 4, . . . , 2J − 4, 2J − 2, l = 1, 3, . . . , 2J − 3, 2J − 1.

Denote 2J

(α )

ν = Δα ∑ dl w(αl )τ−αl w0 l . l=0

(6.23)

6.2 The fourth-order method in both space and distributed order for 1D problem

| 337

Lemma 6.2.2. For the constant ν defined by (6.23), it holds ντ = O(

1 ). | ln τ|

Proof. The proof can be proceeded with the same trick used in the proof for Lemma 6.1.4 and the details are omitted here. Now we begin to build the difference scheme for (6.1)–(6.3). Considering equation (6.1) at the point (xi , tn ), we have w

n

𝒟t u(xi , tn ) = uxx (xi , tn ) + fi ,

0 ⩽ i ⩽ M, 1 ⩽ n ⩽ N.

Performing the operator 𝒜 to both hand sides of the equality above yields w

n

𝒜𝒟t u(xi , tn ) = 𝒜uxx (xi , tn ) + 𝒜fi ,

1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N,

(6.24)

where the operator 𝒜 is defined in Section 2.1. Let s(α, xi , tn ) = w(α) C0 Dαt u(xi , tn ). Suppose function s(⋅, xi , tn ) ∈ C 4 [0, 1]. It follows from Lemma 6.2.1 that 1

w

𝒟t u(xi , tn ) = ∫ s(α, xi , tn )dα 0

2J

= Δα ∑ dl s(αl , xi , tn ) − l=0 2J

= Δα ∑ dl w(αl ) l=0

Δα4 𝜕4 s(α, xi , tn ) 󵄨󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨α=ηn 180 𝜕α4 i

C αl 0 Dt u(xi , tn )

+ O(Δα4 ),

(6.25)

where ηni ∈ (0, 1). Noticing (6.2), Corollary 1.4.1 and (6.25), we have w

2J

𝒟t u(xi , tn ) = Δα ∑ dl w(αl )[τ

−αl

l=0

n

∑ wk l Uin−k + O(τ2 )] + O(Δα4 ), (α )

k=0

0 ⩽ i ⩽ M, 1 ⩽ n ⩽ N.

(6.26)

1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N.

(6.27)

It follows from Lemma 2.1.3 that 2

n

4

𝒜uxx (xi , tn ) = δx Ui + O(h ),

338 | 6 Difference methods for the time distributed-order subdiffusion equations Substituting (6.26) and (6.27) into (6.24) gives 2J

𝒜Δα ∑ dl w(αl )[τ

−αl

l=0

n

∑ wk l Uin−k ] = δx2 Uin + 𝒜fin + (r2 )ni , (α )

k=0

1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N,

(6.28)

and there is a positive constant c2 such that 󵄨󵄨 n 󵄨󵄨 2 4 4 󵄨󵄨(r2 )i 󵄨󵄨 ⩽ c2 (τ + h + Δα ),

1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N.

(6.29)

Noticing the initial-boundary value conditions (6.2)–(6.3), we have {

Ui0 = 0, U0n

1 ⩽ i ⩽ M − 1,

= φ1 (tn ),

n UM

= φ2 (tn ),

(6.30) 0 ⩽ n ⩽ N.

(6.31)

Omitting the small term (r2 )ni in (6.28) and replacing the exact solution Uin with its numerical one uni , another difference scheme for solving (6.1)–(6.3) is produced as 2J

n

(α ) n−k −α 2 n n { { 𝒜Δα ∑ dl w(αl )[τ l ∑ wk l ui ] = δx ui + 𝒜fi , { { { { l=0 k=0 { { 1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N, { { { { 0 { u = 0, 1 ⩽ i ⩽ M − 1, { { { i n n { u0 = φ1 (tn ), uM = φ2 (tn ), 0 ⩽ n ⩽ N.

(6.32) (6.33) (6.34)

We now turn to the analyses on this difference scheme. 6.2.2 Solvability of the difference scheme Theorem 6.2.1. The difference scheme (6.32)–(6.34) is uniquely solvable. Proof. Let un = (un0 , un1 , . . . , unM ). The value of u0 is determined by (6.33)–(6.34). Now suppose the values of u0 , u1 , . . . , un−1 have been uniquely determined, then the linear system in un can be obtained from (6.32) and (6.34). To show its unique solvability, it suffices to verify that the corresponding homogeneous one { has only the trivial solution.

ν𝒜uni = δx2 uni , un0

=

unM

=0

1 ⩽ i ⩽ M − 1,

(6.35) (6.36)

6.2 The fourth-order method in both space and distributed order for 1D problem

| 339

Making the inner product on both hand sides of (6.35) with un arrives at 󵄩 󵄩2 ν(𝒜un , un ) = (δx2 un , un ) = −󵄩󵄩󵄩δx un 󵄩󵄩󵄩 . Noticing 2 󵄩 󵄩2 h 󵄩 󵄩2 2 󵄩 󵄩2 (𝒜un , un ) = 󵄩󵄩󵄩un 󵄩󵄩󵄩 − 󵄩󵄩󵄩δx un 󵄩󵄩󵄩 ⩾ 󵄩󵄩󵄩un 󵄩󵄩󵄩 , 12 3

we have 2 󵄩󵄩 n 󵄩󵄩2 󵄩 󵄩2 ν󵄩u 󵄩 ⩽ −󵄩󵄩󵄩δx un 󵄩󵄩󵄩 ⩽ 0, 3 󵄩 󵄩 thus ‖un ‖ = 0. Then un = 0 is followed by noticing (6.36). By the principle of induction, the theorem is true. The proof ends.

6.2.3 Stability of the difference scheme Theorem 6.2.2. Suppose {vin | 0 ⩽ i ⩽ M, 0 ⩽ n ⩽ N} is the solution of the difference scheme 2J

n

(α ) n−k −α 2 n n { { 𝒜Δα ∑ dl w(αl )[τ l ∑ wk l vi ] = δx vi + gi , { { { { l=0 k=0 { { 1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N, { { { { 0 { v = φi , 1 ⩽ i ⩽ M − 1, { { { i n n { v0 = 0, vM = 0, 0 ⩽ n ⩽ N.

(6.37) (6.38) (6.39)

Then it holds m m 2 󵄩 󵄩2 󵄩 󵄩2 󵄩 󵄩2 3L τ ∑ 󵄩󵄩󵄩g n 󵄩󵄩󵄩 , τ ∑ 󵄩󵄩󵄩δx vn 󵄩󵄩󵄩 ⩽ 3ντ󵄩󵄩󵄩v0 󵄩󵄩󵄩 + 8 n=1 n=1

1 ⩽ m ⩽ N,

where M−1

󵄩󵄩 n 󵄩󵄩2 n 2 󵄩󵄩g 󵄩󵄩 = h ∑ (gi ) . i=1

Proof. Taking the inner product on both hand sides of (6.37) with 𝒜vn , it follows from Lemma 2.1.1, Lemma 2.1.2 and the Cauchy–Schwarz inequality that 2J

n

l=0

k=0

Δα ∑ dl w(αl )τ−αl ∑ wk l (𝒜vn−k , 𝒜vn ) (α )

= (δx2 vn , 𝒜vn ) + (g n , 𝒜vn )

340 | 6 Difference methods for the time distributed-order subdiffusion equations 󵄩 󵄩2 = −󵄩󵄩󵄩δx vn 󵄩󵄩󵄩A + (g n , 𝒜vn ) 2󵄩 󵄩 󵄩2 󵄩 󵄩 󵄩 ⩽ − 󵄩󵄩󵄩δx vn 󵄩󵄩󵄩 + 󵄩󵄩󵄩g n 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩𝒜vn 󵄩󵄩󵄩 3 2󵄩 󵄩2 󵄩 󵄩 󵄩 󵄩 ⩽ − 󵄩󵄩󵄩δx vn 󵄩󵄩󵄩 + 󵄩󵄩󵄩g n 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩vn 󵄩󵄩󵄩 3 2 2󵄩 󵄩2 2 󵄩 󵄩2 L 󵄩 󵄩2 ⩽ − 󵄩󵄩󵄩δx vn 󵄩󵄩󵄩 + 2 󵄩󵄩󵄩vn 󵄩󵄩󵄩 + 󵄩󵄩󵄩g n 󵄩󵄩󵄩 3 8 L 2 󵄩󵄩 n 󵄩󵄩2 1 󵄩󵄩 n 󵄩󵄩2 L2 󵄩󵄩 n 󵄩󵄩2 ⩽ − 󵄩󵄩δx v 󵄩󵄩 + 󵄩󵄩δx v 󵄩󵄩 + 󵄩󵄩g 󵄩󵄩 3 3 8 2 1󵄩 󵄩2 L 󵄩 󵄩2 = − 󵄩󵄩󵄩δx vn 󵄩󵄩󵄩 + 󵄩󵄩󵄩g n 󵄩󵄩󵄩 , 1 ⩽ n ⩽ N. 3 8 Summing up for n from 1 to m leads to 2J

m

n

Δα ∑ dl w(αl )τ−αl ∑ ∑ wk l (𝒜vn−k , 𝒜vn ) (α )

n=1 k=0 2 m

l=0 m

1 󵄩 󵄩 󵄩2 󵄩2 L ⩽ − ∑ 󵄩󵄩󵄩δx vn 󵄩󵄩󵄩 + ∑ 󵄩󵄩g n 󵄩󵄩 , 3 n=1 8 n=1󵄩 󵄩

1 ⩽ m ⩽ N.

Adding the term ν(𝒜v0 , 𝒜v0 ) to both hand sides of the inequality above yields 2J

m

n

Δα ∑ dl w(αl )τ−αl ∑ ∑ wk l (𝒜vn−k , 𝒜vn ) (α )

n=0 k=0

l=0 m

L2 m 󵄩󵄩 n 󵄩󵄩2 1 󵄩 󵄩2 ⩽ − ∑ 󵄩󵄩󵄩δx vn 󵄩󵄩󵄩 + ν(𝒜v0 , 𝒜v0 ) + ∑ 󵄩g 󵄩 , 3 n=1 8 n=1󵄩 󵄩

1 ⩽ m ⩽ N.

(6.40)

∑ ∑ wk l (𝒜vn−k , 𝒜vn ) = h ∑ ∑ [ ∑ wk l (𝒜vin−k )](𝒜vin ) ⩾ 0.

(6.41)

By Lemma 6.1.3, we have m

n

n=0 k=0

(α )

M−1 m

n

(α )

i=1 n=0 k=0

It follows from (6.40), (6.41) and Lemma 2.1.1 that m 3L2 m 󵄩󵄩 n 󵄩󵄩2 󵄩 󵄩2 τ ∑ 󵄩󵄩󵄩δx vn 󵄩󵄩󵄩 ⩽ 3ντ(𝒜v0 , 𝒜v0 ) + τ ∑ 󵄩g 󵄩 8 n=1󵄩 󵄩 n=1 m 2 󵄩 󵄩2 󵄩 󵄩2 3L τ ∑ 󵄩󵄩󵄩g n 󵄩󵄩󵄩 , ⩽ 3ντ󵄩󵄩󵄩v0 󵄩󵄩󵄩 + 8 n=1

1 ⩽ m ⩽ N.

The proof ends. 6.2.4 Convergence of the difference scheme Theorem 6.2.3. Suppose {Uin | 0 ⩽ i ⩽ M, 0 ⩽ n ⩽ N} and {uni | 0 ⩽ i ⩽ M, 0 ⩽ n ⩽ N} are solutions of the problem (6.1)–(6.3) and the difference scheme (6.32)–(6.34), respec-

6.2 The fourth-order method in both space and distributed order for 1D problem

| 341

tively. Let ein = Uin − uni ,

0 ⩽ i ⩽ M, 0 ⩽ n ⩽ N,

then it holds N √6 2 󵄩 󵄩 L Tc2 (τ2 + h4 + Δα4 ). τ ∑ 󵄩󵄩󵄩en 󵄩󵄩󵄩∞ ⩽ 8 n=1

Proof. The subtraction of (6.32)–(6.34) from (6.28), (6.30)–(6.31), respectively, produces the system of error equations as follows: 2J

n

(α ) n−k { −α 2 n n { 𝒜Δα ∑ dl w(αl )[τ l ∑ wk l ei ] = δx ei + (r2 )i , { { { { l=0 k=0 { { 1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N, { { { { 0 { e = 0, 1 ⩽ i ⩽ M − 1, { { { i n n { e0 = 0, eM = 0, 0 ⩽ n ⩽ N.

Noticing (6.29), the application of Theorem 6.2.2 immediately yields N N 2 󵄩 󵄩2 3L 󵄩 󵄩2 τ ∑ 󵄩󵄩󵄩δx en 󵄩󵄩󵄩 ⩽ τ ∑ 󵄩󵄩󵄩(r2 )n 󵄩󵄩󵄩 8 n=1 n=1

⩽ ⩽

3L2 N 2 τ ∑ L[c2 (τ2 + h4 + Δα4 )] 8 n=1

3L3 2 T[c2 (τ2 + h4 + Δα4 )] . 8

It follows from the Cauchy–Schwarz inequality, Lemma 2.1.1 and the inequality above that 2

N

N

N

󵄩 󵄩 󵄩 󵄩2 (τ ∑ 󵄩󵄩󵄩en 󵄩󵄩󵄩∞ ) ⩽ (τ ∑ 1)(τ ∑ 󵄩󵄩󵄩en 󵄩󵄩󵄩∞ ) n=1

n=1

⩽T⋅ ⩽

N

n=1

L 󵄩 󵄩2 τ ∑ 󵄩󵄩󵄩δx en 󵄩󵄩󵄩 4 n=1

LT 3L3 2 ⋅ T[c2 (τ2 + h4 + Δα4 )] . 4 8

Taking the square root on both hand sides of the inequality above gives N

󵄩 󵄩 τ ∑ 󵄩󵄩󵄩en 󵄩󵄩󵄩∞ ⩽ n=1

The proof ends.

√6 2 L Tc2 (τ2 + h4 + Δα4 ). 8

342 | 6 Difference methods for the time distributed-order subdiffusion equations

6.3 The second-order method in both space and distributed order for 2D problem Consider the following 2D initial-boundary value problem of time distributed-order subdiffusion equation w 𝒟t u(x, y, t) = uxx (x, y, t) + uyy (x, y, t) + f (x, y, t), { { { { { (x, y) ∈ Ω, t ∈ (0, T], { { u(x, y, 0) = 0, (x, y) ∈ Ω, { { { { u(x, y, t) = φ(x, y, t), (x, y) ∈ 𝜕Ω, t ∈ [0, T],

(6.42) (6.43) (6.44)

where Ω = (0, L1 ) × (0, L2 ), 𝜕Ω is the boundary of Ω; When (x, y) ∈ 𝜕Ω, φ(x, y, 0) = 0; w 𝒟t u(x, y, t)

C α 0 Dt u(x, y, t)

=

1

= ∫ w(α) C0 Dαt u(x, y, t)dα,

0 t 1 ∫0 (t Γ(1−α) {

ut (x, y, t),

− ξ )−α uξ (x, y, ξ )dξ ,

0 ⩽ α < 1, α = 1,

1

w(α) ⩾ 0, ∫0 w(α)dα = c0 > 0, the functions f and φ are given. In this section, a second-order difference scheme for solving (6.42)–(6.44) will be considered and its unique solvability, stability and convergence will also be illustrated. Take the same mesh partition and notations like those in Section 2.10 in both space and time directions. Besides, for any mesh function u ∈ 𝒱h , define Δh uij = δx2 uij + δy2 uij . For any mesh functions u, v ∈ 𝒱̊h , let M1 −1 M2 −1

(Δh u, Δh v) = h1 h2 ∑ ∑ (Δh uij )(Δh vij ), i=1

j=1

‖Δh u‖ = √(Δh u, Δh u) .

The following lemma is true. Lemma 6.3.1. that

[70, 76]

For any mesh function u ∈ 𝒱̊h , there is a positive constant c such ‖u‖∞ ⩽ c‖Δh u‖.

More precisely, we have c = √ 321 [π(L21 + L22 ) +

2L31 L32 ]. (L21 +L22 )2

̂ y, t) like that in Section 2.10. Suppose u(x, ̂ y, ⋅) ∈ C 2+1 (ℛ) Define the function u(x, (4,4) ̄ and u(⋅, ⋅, t) ∈ C (Ω).

6.3 The second-order method in both space and distributed order for 2D problem | 343

6.3.1 Derivation of the difference scheme Define the mesh functions Uijn = u(xi , yj , tn ),

fijn = f (xi , yj , tn ),

(i, j) ∈ ω,̄ 0 ⩽ n ⩽ N.

Considering equation (6.42) at the point (xi , yj , tn ), we have w

n

𝒟t u(xi , yj , tn ) = uxx (xi , yj , tn ) + uyy (xi , yj , tn ) + fij ,

(i, j) ∈ ω, 1 ⩽ n ⩽ N.

(6.45)

Let s(α, xi , yj , tn ) = w(α) C0 Dαt u(xi , yj , tn ). Suppose function s(⋅, xi , yj , tn ) ∈ C 2 [0, 1]. From Lemma 6.1.1, Corollary 1.4.1 and (6.43), we have w

𝒟t u(xi , yj , tn ) 2J

= Δα ∑ cl w(αl ) l=0

C αl 0 Dt u(xi , yj , tn )

2J

n

l=0

k=0

+ O(Δα2 )

= Δα ∑ cl w(αl )[τ−αl ∑ wk l Uijn−k + O(τ2 )] + O(Δα2 ). (α )

(6.46)

It follows from Lemma 2.1.3 that uxx (xi , yj , tn ) = δx2 Uijn + O(h21 ),

uyy (xi , yj , tn ) = δy2 Uijn + O(h22 ).

(6.47)

Substituting (6.46) and (6.47) into (6.45) gives 2J

n

l=0

k=0

Δα ∑ cl w(αl )[τ−αl ∑ wk l Uijn−k ] = Δh Uijn + fijn + (r3 )nij , (α )

(i, j) ∈ ω, 1 ⩽ n ⩽ N,

(6.48)

and there is a positive constant c3 such that 󵄨󵄨 n 󵄨󵄨 2 2 2 2 󵄨󵄨(r3 )ij 󵄨󵄨 ⩽ c3 (τ + h1 + h2 + Δα ),

(i, j) ∈ ω, 1 ⩽ n ⩽ N.

(6.49)

Noticing the initial-boundary value conditions (6.43)–(6.44), we have {

Uij0 = 0, Uijn

(i, j) ∈ ω,

= φ(xi , yj , tn ),

(6.50) (i, j) ∈ 𝜕ω, 0 ⩽ n ⩽ N.

(6.51)

344 | 6 Difference methods for the time distributed-order subdiffusion equations Neglecting the small term (r3 )nij in (6.48) and replacing the exact solution Uijn with its numerical one unij , we get a difference scheme for solving (6.42)–(6.44) in the form of 2J n { (α ) −αl n n { { Δα c w(α )[τ wk l un−k ∑ ∑ { l l ij ] = Δh uij + fij , { { { l=0 k=0 { { (i, j) ∈ ω, 1 ⩽ n ⩽ N, { { { 0 { { uij = 0, (i, j) ∈ ω, { { { { n { uij = φ(xi , yj , tn ), (i, j) ∈ 𝜕ω, 0 ⩽ n ⩽ N.

(6.52) (6.53) (6.54)

Next, we aim to make some analyses on the difference scheme (6.52)–(6.54). 6.3.2 Solvability of the difference scheme Theorem 6.3.1. The difference scheme (6.52)–(6.54) is uniquely solvable. Proof. Let ̄ un = {unij | (i, j) ∈ ω}. The value of u0 is determined by (6.53)–(6.54). Now assume that the values of u0 , u1 , . . . , un−1 have been uniquely determined, then we can obtain the linear system in the unknown un from (6.52) and (6.54). To show its unique solvability, it is sufficient to verify that the corresponding homogeneous one {

μunij = Δh unij ,

unij

= 0,

(i, j) ∈ ω,

(6.55)

(i, j) ∈ 𝜕ω

(6.56)

has only the trivial solution. Rewrite (6.55) as follows: (μ +

2 1 2 1 + )un = (un + uni+1,j ) + 2 (uni,j−1 + uni,j+1 ), h21 h22 ij h21 i−1,j h2

(i, j) ∈ ω.

Suppose ‖un ‖∞ = |unin ,jn |, where (in , jn ) ∈ ω. Letting (i, j) = (in , jn ) in the equality above and taking the absolute value on both hand sides of the result, the application of the triangle inequality gives (μ +

2 2 󵄩 󵄩 2󵄩 󵄩 2󵄩 󵄩 + )󵄩󵄩un 󵄩󵄩 ⩽ 󵄩󵄩un 󵄩󵄩 + 󵄩󵄩un 󵄩󵄩 , h21 h22 󵄩 󵄩∞ h21 󵄩 󵄩∞ h22 󵄩 󵄩∞

therefore, ‖un ‖∞ = 0. Then un = 0 is concluded from the combination with (6.56). By the principle of induction, the difference scheme (6.52)–(6.54) is uniquely solvable. The proof ends.

6.3 The second-order method in both space and distributed order for 2D problem

| 345

6.3.3 Stability of the difference scheme Theorem 6.3.2. Suppose {vijn | (i, j) ∈ ω,̄ 0 ⩽ n ⩽ N} is the solution of the difference scheme 2J

n { (α ) −αl { { Δα c w(α )τ wk l vijn−k = Δh vijn + gijn , ∑ ∑ { l l { { { l=0 k=0 { { (i, j) ∈ ω, 1 ⩽ n ⩽ N, { { { 0 { { vij = φij , (i, j) ∈ ω, { { { { n { vij = 0, (i, j) ∈ 𝜕ω, 0 ⩽ n ⩽ N.

(6.57) (6.58) (6.59)

Then it holds m

m

n=1

n=1

󵄩 󵄩2 󵄩 󵄩2 󵄩 󵄩2 τ ∑ 󵄩󵄩󵄩Δh vn 󵄩󵄩󵄩 ⩽ 2μτ󵄩󵄩󵄩∇h v0 󵄩󵄩󵄩 + τ ∑ 󵄩󵄩󵄩g n 󵄩󵄩󵄩 ,

1 ⩽ m ⩽ N,

where M1 −1 M2 −1

󵄩󵄩 n 󵄩󵄩2 n 2 󵄩󵄩g 󵄩󵄩 = h1 h2 ∑ ∑ (gij ) . i=1

j=1

Proof. Making the inner product on both hand sides of (6.57) with −Δh vn , it follows from the Cauchy–Schwarz inequality that 2J

n

Δα ∑ cl w(αl )τ−αl ∑ wk l (vn−k , −Δh vn )

= ⩽

(α )

l=0 k=0 n n −(Δh v , Δh v ) − (g n , Δh vn ) 󵄩 󵄩2 󵄩 󵄩 󵄩 󵄩 −󵄩󵄩󵄩Δh vn 󵄩󵄩󵄩 + 󵄩󵄩󵄩g n 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩Δh vn 󵄩󵄩󵄩

1 󵄩󵄩 n 󵄩󵄩2 󵄩Δ v 󵄩 + 2󵄩 h 󵄩 1󵄩 󵄩2 1 󵄩 󵄩2 = − 󵄩󵄩󵄩Δh vn 󵄩󵄩󵄩 + 󵄩󵄩󵄩g n 󵄩󵄩󵄩 , 2 2 󵄩 󵄩2 ⩽ −󵄩󵄩󵄩Δh vn 󵄩󵄩󵄩 +

1 󵄩󵄩 n 󵄩󵄩2 󵄩g 󵄩 2󵄩 󵄩

1 ⩽ n ⩽ N.

Summing up for n from 1 to m and adding the term μ(v0 , −Δh v0 ) to both hand sides of the obtained inequality arrive at 2J

m

n

Δα ∑ cl w(αl )τ−αl [ ∑ ∑ wk l (vn−k , −Δh vn )] (α )

n=0 k=0

l=0

1 m󵄩 1 m 󵄩 󵄩2 󵄩2 ⩽ − ∑ 󵄩󵄩󵄩Δh vn 󵄩󵄩󵄩 + μ(v0 , −Δh v0 ) + ∑ 󵄩󵄩󵄩g n 󵄩󵄩󵄩 , 2 n=1 2 n=1 By Lemma 6.1.3, we have m

n

∑ ∑ wk l (vn−k , −Δh vn )

n=0 k=0

(α )

1 ⩽ m ⩽ N.

(6.60)

346 | 6 Difference methods for the time distributed-order subdiffusion equations m

n

= ∑ ∑ wk l [(δx vn−k , δx vn ) + (δy vn−k , δy vn )] (α )

n=0 k=0

M1 M2 −1

m

n

n−k n = h1 h2 ∑ ∑ [ ∑ ∑ wk l (δx vi− )] 1 )(δx v i− 1 ,j ,j i=1 j=1

(α )

M1 −1 M2

m

2

2

n=0 k=0

n

n−k n )] + h1 h2 ∑ ∑[ ∑ ∑ wk l (δy vi,j− 1 )(δy v i,j− 1 (α )

2

2

i=1 j=1 n=0 k=0

⩾ 0.

(6.61)

Combining (6.60) with (6.61) leads to m

m

󵄩 󵄩2 󵄩 󵄩2 τ ∑ 󵄩󵄩󵄩Δh vn 󵄩󵄩󵄩 ⩽ 2μτ(v0 , −Δh v0 ) + τ ∑ 󵄩󵄩󵄩g n 󵄩󵄩󵄩 n=1

m

n=1

󵄩 󵄩2 󵄩 󵄩2 = 2μτ󵄩󵄩󵄩∇h v0 󵄩󵄩󵄩 + τ ∑ 󵄩󵄩󵄩g n 󵄩󵄩󵄩 , n=1

1 ⩽ m ⩽ N.

The proof ends.

6.3.4 Convergence of the difference scheme Theorem 6.3.3. Suppose {Uijn | (i, j) ∈ ω,̄ 0 ⩽ n ⩽ N} and {unij | (i, j) ∈ ω,̄ 0 ⩽ n ⩽ N} are solutions of the problem (6.42)–(6.44) and the difference scheme (6.52)–(6.54), respectively. Let eijn = Uijn − unij ,

(i, j) ∈ ω,̄ 0 ⩽ n ⩽ N,

then it holds N

󵄩 󵄩 τ ∑ 󵄩󵄩󵄩en 󵄩󵄩󵄩∞ ⩽ cT √L1 L2 c3 (τ2 + h21 + h22 + Δα2 ), n=1

(6.62)

where the constant c is defined in Lemma 6.3.1. Proof. Subtracting (6.52)–(6.54) from (6.48), (6.50)–(6.51), respectively, we get the system of error equations as follows: 2J n { (α ) −αl { { Δα c w(α )[τ wk l eijn−k ] = Δh eijn + (r3 )nij , ∑ ∑ { l l { { { l=0 k=0 { { (i, j) ∈ ω, 1 ⩽ n ⩽ N, { { { 0 { { eij = 0, (i, j) ∈ ω, { { { { n { eij = 0, (i, j) ∈ 𝜕ω, 0 ⩽ n ⩽ N.

6.4 The fourth-order method in both space and distributed order for 2D problem

| 347

Noticing (6.49), the application of Theorem 6.3.2 yields N

N

n=1

n=1

󵄩 󵄩2 󵄩 󵄩2 τ ∑ 󵄩󵄩󵄩Δh en 󵄩󵄩󵄩 ⩽ τ ∑ 󵄩󵄩󵄩(r3 )n 󵄩󵄩󵄩 N

2

⩽ τ ∑ L1 L2 [c3 (τ2 + h21 + h22 + Δα2 )] n=1

2

⩽ TL1 L2 [c3 (τ2 + h21 + h22 + Δα2 )] .

(6.63)

It follows from the Cauchy–Schwarz inequality, Lemma 6.3.1 and (6.63) that 2

N

N

N

󵄩 󵄩 󵄩 󵄩2 (τ ∑ 󵄩󵄩󵄩en 󵄩󵄩󵄩∞ ) ⩽ (τ ∑ 1)(τ ∑ 󵄩󵄩󵄩en 󵄩󵄩󵄩∞ ) n=1

n=1

N

n=1

󵄩 󵄩2 ⩽ Tc2 τ ∑ 󵄩󵄩󵄩Δh en 󵄩󵄩󵄩 n=1

2

2 2

⩽ c T L1 L2 [c3 (τ2 + h21 + h22 + Δα2 )] , which implies (6.62). The proof ends.

6.4 The fourth-order method in both space and distributed order for 2D problem This section is devoted to a difference method of order two in time and four in both space and distributed order for the problem (6.42)–(6.44). ̂ y, t) like that in Section 2.10. Suppose u(x, ̂ y, ⋅) ∈ C 2+1 (ℛ) Define the function u(x, (6,6) ̄ and u(⋅, ⋅, t) ∈ C (Ω).

6.4.1 Derivation of the difference scheme Considering equation (6.42) at the point (xi , yj , tn ), we have w

n

𝒟t u(xi , yj , tn ) = uxx (xi , yj , tn ) + uyy (xi , yj , tn ) + fij ,

(i, j) ∈ ω,̄ 1 ⩽ n ⩽ N.

Performing the operator 𝒜x 𝒜y to both hand sides of the equality above and noticing Lemma 2.1.3, we have w

𝒜x 𝒜y 𝒟t u(xi , yj , tn )

= 𝒜y (𝒜x uxx (xi , yj , tn )) + 𝒜x (𝒜y uyy (xi , yj , tn )) + 𝒜x 𝒜y fijn

348 | 6 Difference methods for the time distributed-order subdiffusion equations = 𝒜y δx2 Uijn + 𝒜x δy2 Uijn + 𝒜x 𝒜y fijn + O(h41 + h42 ), (i, j) ∈ ω, 1 ⩽ n ⩽ N,

(6.64)

where the operators 𝒜x and 𝒜y are defined in Section 3.10. Let s(α, xi , yj , tn ) = w(α) C0 Dαt u(xi , yj , tn ). Suppose function s(⋅, xi , yj , tn ) ∈ C 4 [0, 1]. By Lemma 6.2.1 and Corollary 1.4.1, we have 2J

w

C αl 0 Dt u(xi , yj , tn )

𝒟t u(xi , yj , tn ) = Δα ∑ dl w(αl ) l=0 2J

n

l=0

k=0

+ O(Δα4 )

= Δα ∑ dl w(αl )[τ−αl ∑ wk l Uijn−k + O(τ2 )] + O(Δα4 ). (α )

(6.65)

Substituting (6.65) into (6.64) yields 2J

𝒜x 𝒜y Δα ∑ dl w(αl )[τ

=

2 n 𝒜y δx Uij

l=0

+

2 n 𝒜x δy Uij

+

−αl

n

∑ wk l Uijn−k ] (α )

k=0 n 𝒜x 𝒜y fij

+ (r4 )nij ,

(i, j) ∈ ω, 1 ⩽ n ⩽ N,

(6.66)

and there is a positive constant c4 such that 󵄨󵄨 n󵄨 2 4 4 4 󵄨󵄨(r4 )ij 󵄨󵄨󵄨 ⩽ c4 (τ + h1 + h2 + Δα ),

(i, j) ∈ ω, 1 ⩽ n ⩽ N.

(6.67)

Noticing the initial-boundary value conditions (6.43)–(6.44), we have {

Uij0 = 0, Uijn

(i, j) ∈ ω,

= φ(xi , yj , tn ),

(6.68) (i, j) ∈ 𝜕ω, 0 ⩽ n ⩽ N.

(6.69)

Omitting the small term (r4 )nij in (6.66) and replacing the exact solution Uijn with its numerical one unij , we can obtain the following difference scheme for solving (6.42)–(6.44): { { { { { { { { { { { { { { { { { { {

2J

𝒜x 𝒜y Δα ∑ dl w(αl )[τ l=0

−αl

n

∑ wk l un−k ij ] (α )

k=0

= 𝒜y δx2 unij + 𝒜x δy2 unij + 𝒜x 𝒜y fijn ,

u0ij unij

= 0,

(i, j) ∈ ω, 1 ⩽ n ⩽ N,

(i, j) ∈ ω,

= φ(xi , yj , tn ),

(6.70) (6.71)

(i, j) ∈ 𝜕ω, 0 ⩽ n ⩽ N.

(6.72)

In the subsequent part, the unique solvability, stability and convergence of this difference scheme will be shown.

6.4 The fourth-order method in both space and distributed order for 2D problem | 349

6.4.2 Solvability of the difference scheme Theorem 6.4.1. The difference scheme (6.70)–(6.72) is uniquely solvable. Proof. Let ̄ un = {unij | (i, j) ∈ ω}. The value of u0 is uniquely determined by (6.71)–(6.72). Now suppose the values of u0 , u1 , . . . , un−1 have been uniquely determined, then the system in un can be obtained from (6.70) and (6.72). To show its unique solvability, it is sufficient to prove that the corresponding homogeneous one {

ν𝒜x 𝒜y unij = 𝒜y δx2 unij + 𝒜x δy2 unij ,

unij

= 0,

(i, j) ∈ ω,

(i, j) ∈ 𝜕ω

(6.73) (6.74)

has only the trivial solution. To this end, making the inner product on both hand sides of (6.73) with un produces ν(𝒜x 𝒜y un , un ) = (𝒜y δx2 un , un ) + (𝒜x δy2 un , un ).

(6.75)

It follows from Lemma 3.10.1 that (𝒜x 𝒜y un , un ) ⩾

1 󵄩󵄩 n 󵄩󵄩2 󵄩u 󵄩 . 3󵄩 󵄩

(6.76)

By (3.321) and (3.322), we have 2󵄩 󵄩2 (𝒜y δx2 un , un ) ⩽ − 󵄩󵄩󵄩δx un 󵄩󵄩󵄩 , 3

2󵄩 󵄩2 (𝒜x δy2 un , un ) ⩽ − 󵄩󵄩󵄩δy un 󵄩󵄩󵄩 . 3

Substituting (6.76) and (6.77) into (6.75) yields 2󵄩 1 󵄩󵄩 n 󵄩󵄩2 󵄩2 ν󵄩u 󵄩 ⩽ − 󵄩󵄩󵄩∇h un 󵄩󵄩󵄩 ⩽ 0, 3 󵄩 󵄩 3 hence, ‖un ‖ = 0, which implies un = 0 by noticing (6.74). By the principle of induction, the theorem is true. The proof ends. 6.4.3 Stability of the difference scheme Two useful lemmas are firstly prepared. Lemma 6.4.1.

[48]

For any mesh function v ∈ 𝒱̊h , it holds 2 ‖Δ v‖2 ⩽ (𝒜y δx2 v + 𝒜x δy2 v, Δh v) ⩽ ‖Δh v‖2 . 3 h

(6.77)

350 | 6 Difference methods for the time distributed-order subdiffusion equations Lemma 6.4.2.

[48]

For any mesh function v ∈ 𝒱̊h , it holds 1 ‖∇ v‖2 ⩽ (𝒜x 𝒜y v, −Δh v) ⩽ ‖∇h v‖2 . 3 h

Next, the stability result is given. Theorem 6.4.2. Suppose {vijn | (i, j) ∈ ω,̄ 0 ⩽ n ⩽ N} is the solution of the difference scheme { { { { { { { { { { { { { { { { { { {

2J

n

Δα ∑ dl w(αl )τ−αl ∑ wk l 𝒜x 𝒜y vijn−k (α )

l=0

k=0 2 n 2 n = 𝒜y δx vij + 𝒜x δy vij + gijn , vij0 = φij , (i, j) ∈ ω, vijn = 0, (i, j) ∈ 𝜕ω, 0 ⩽ n

(i, j) ∈ ω, 1 ⩽ n ⩽ N,

(6.78) (6.79)

⩽ N.

(6.80)

Then it holds m m 󵄩 󵄩2 󵄩 󵄩2 9 󵄩 󵄩2 τ ∑ 󵄩󵄩󵄩Δh vn 󵄩󵄩󵄩 ⩽ 3ντ󵄩󵄩󵄩∇h v0 󵄩󵄩󵄩 + τ ∑ 󵄩󵄩󵄩g n 󵄩󵄩󵄩 , 4 n=1 n=1

1 ⩽ m ⩽ N,

(6.81)

where M1 −1 M2 −1

󵄩󵄩 n 󵄩󵄩2 n 2 󵄩󵄩g 󵄩󵄩 = h1 h2 ∑ ∑ (gij ) . i=1

j=1

Proof. Taking the inner product on both hand sides of (6.78) with −Δh vn , it follows from Lemma 6.4.1 that 2J

n

l=0

k=0

Δα ∑ dl w(αl )τ−αl ∑ wk l (𝒜x 𝒜y vn−k , −Δh vn ) (α )

= (𝒜y δx2 vn + 𝒜x δy2 vn , −Δh vn ) + (g n , −Δh vn )

2󵄩 󵄩2 󵄩 󵄩 󵄩 󵄩 ⩽ − 󵄩󵄩󵄩Δh vn 󵄩󵄩󵄩 + 󵄩󵄩󵄩g n 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩Δh vn 󵄩󵄩󵄩 3 2󵄩 󵄩2 1 󵄩 󵄩2 3 󵄩 󵄩2 ⩽ − 󵄩󵄩󵄩Δh vn 󵄩󵄩󵄩 + 󵄩󵄩󵄩Δh vn 󵄩󵄩󵄩 + 󵄩󵄩󵄩g n 󵄩󵄩󵄩 3 3 4 1 󵄩󵄩 n 󵄩󵄩2 3 󵄩󵄩 n 󵄩󵄩2 = − 󵄩󵄩Δh v 󵄩󵄩 + 󵄩󵄩g 󵄩󵄩 , 1 ⩽ n ⩽ N. 3 4

Summing up over n from 1 to m and adding the term ν(𝒜x 𝒜y v0 , −Δh v0 ) to both hand sides of the result, we get 2J

m

n

Δα ∑ dl w(αl )τ−αl [ ∑ ∑ wk l (𝒜x 𝒜y vn−k , −Δh vn )] l=0

n=0 k=0

(α )

6.4 The fourth-order method in both space and distributed order for 2D problem

1 m󵄩 3 m 󵄩 󵄩2 󵄩2 ⩽ − ∑ 󵄩󵄩󵄩Δh vn 󵄩󵄩󵄩 + ν(𝒜x 𝒜y v0 , −Δh v0 ) + ∑ 󵄩󵄩󵄩g n 󵄩󵄩󵄩 , 3 n=1 4 n=1

1 ⩽ m ⩽ N.

| 351

(6.82)

Since the operators 𝒜x and 𝒜y are both positive definite, there exist two positive definite operators 𝒫x and 𝒫y , such that 𝒜x = 𝒫x2 , 𝒜y = 𝒫y2 . Therefore, (𝒜x 𝒜y vn−k , −Δh vn )

= (𝒜x 𝒜y vn−k , −δx2 vn ) + (𝒜x 𝒜y vn−k , −δy2 vn )

= (𝒫x 𝒫y vn−k , −𝒫x 𝒫y δx2 vn ) + (𝒫x 𝒫y vn−k , −𝒫x 𝒫y δy2 vn )

= (𝒫x 𝒫y δx vn−k , 𝒫x 𝒫y δx vn ) + (𝒫x 𝒫y δy vn−k , 𝒫x 𝒫y δy vn )

= (δx 𝒫x 𝒫y vn−k , δx 𝒫x 𝒫y vn ) + (δy 𝒫x 𝒫y vn−k , δy 𝒫x 𝒫y vn ).

Similar to the proof for (6.61), we have m

n

∑ ∑ wk l (𝒜x 𝒜y vn−k , −Δh vn ) ⩾ 0. (α )

n=0 k=0

(6.83)

By Lemma 6.4.2, it holds 󵄩 󵄩2 (𝒜x 𝒜y v0 , −Δh v0 ) ⩽ 󵄩󵄩󵄩∇h v0 󵄩󵄩󵄩 .

(6.84)

Substituting (6.83) and (6.84) into (6.82) will yield (6.81). The proof ends. 6.4.4 Convergence of the difference scheme Theorem 6.4.3. Suppose {Uijn | (i, j) ∈ ω,̄ 0 ⩽ n ⩽ N} and {unij | (i, j) ∈ ω,̄ 0 ⩽ n ⩽ N} are solutions of the problem (6.42)–(6.44) and the difference scheme (6.70)–(6.72), respectively. Let eijn = Uijn − unij ,

(i, j) ∈ ω,̄ 0 ⩽ n ⩽ N,

then it holds N 3 󵄩 󵄩 τ ∑ 󵄩󵄩󵄩en 󵄩󵄩󵄩∞ ⩽ cT √L1 L2 c4 (τ2 + h41 + h42 + Δα4 ), 2 n=1

(6.85)

where the constant c is defined in Lemma 6.3.1. Proof. Subtracting (6.70)–(6.72) from (6.66), (6.68)–(6.69), respectively, we get the system of error equations as follows: { { { { { { { { { { { { { { { { { { {

2J

n

Δα ∑ dl w(αl )[τ−αl ∑ wk l 𝒜x 𝒜y eijn−k ] (α )

l=0

k=0 2 n 2 n = 𝒜y δx eij + 𝒜x δy eij + (r4 )nij , eij0 = 0, (i, j) ∈ ω, eijn = 0, (i, j) ∈ 𝜕ω, 0 ⩽ n ⩽

(i, j) ∈ ω, 1 ⩽ n ⩽ N, N.

352 | 6 Difference methods for the time distributed-order subdiffusion equations Noticing (6.67), the application of Theorem 6.4.2 yields N N 󵄩 󵄩2 9 󵄩 󵄩2 τ ∑ 󵄩󵄩󵄩Δh en 󵄩󵄩󵄩 ⩽ τ ∑ 󵄩󵄩󵄩(r4 )n 󵄩󵄩󵄩 4 n=1 n=1

⩽ ⩽

9 N 2 τ ∑ L L [c (τ2 + h41 + h42 + Δα4 )] 4 n=1 1 2 4 9 2 TL L [c (τ2 + h41 + h42 + Δα4 )] . 4 1 2 4

(6.86)

By the Cauchy–Schwarz inequality, Lemma 6.3.1 and (6.86), we have 2

N

N

N

󵄩 󵄩 󵄩 󵄩2 (τ ∑ 󵄩󵄩󵄩en 󵄩󵄩󵄩∞ ) ⩽ (τ ∑ 1)(τ ∑ 󵄩󵄩󵄩en 󵄩󵄩󵄩∞ ) n=1

n=1

n=1

N

󵄩 󵄩2 ⩽ Tc2 τ ∑ 󵄩󵄩󵄩Δh en 󵄩󵄩󵄩 n=1

9 2 ⩽ c2 T 2 L1 L2 [c4 (τ2 + h41 + h42 + Δα4 )] , 4 which implies (6.85). The proof ends.

6.5 The second-order ADI method in both space and distributed-order for 2D problem In this section, an ADI difference scheme for solving the 2D problem of time distributedorder subdiffusion equation (6.42)–(6.44) will be proposed and its unique solvability, stability and convergence will also be shown. ̂ y, t) like that in Section 2.10. Suppose u(x, ̂ y, ⋅) ∈ C 2+1 (ℛ) Define the function u(x, (4,4) ̄ and u(⋅, ⋅, t) ∈ C (Ω). 6.5.1 Derivation of the difference scheme Adding the small term μτ δx2 δy2

Uijn −Uijn−1 τ

2J

to both hand sides of (6.48) arrives at n

Δα ∑ cl w(αl )[τ−αl ∑ wk l Uijn−k ] +

=

l=0 Δh Uijn + fijn

+

(r5 )nij ,

k=0

(α )

n n−1 τ 2 2 Uij − Uij δx δy μ τ

(i, j) ∈ ω, 1 ⩽ n ⩽ N,

where (r5 )nij = (r3 )nij +

n n−1 τ 2 2 Uij − Uij δx δy . μ τ

(6.87)

6.5 The second-order ADI method in both space and distributed-order for 2D problem

| 353

It is easy to see that μτ = O(τ2 | ln τ|) by Lemma 6.1.4. Hence, there exists a positive constant c5 such that 󵄨󵄨 n 󵄨󵄨 2 2 2 2 2 󵄨󵄨(r5 )ij 󵄨󵄨 ⩽ c3 (τ + h1 + h2 + Δα ) + c5 τ | ln τ|,

(i, j) ∈ ω, 1 ⩽ n ⩽ N.

(6.88)

Noticing the initial-boundary value conditions (6.43)–(6.44), we have {

Uij0 = 0,

Uijn

(i, j) ∈ ω,

= φ(xi , yj , tn ),

(6.89) (i, j) ∈ 𝜕ω, 0 ⩽ n ⩽ N.

(6.90)

Neglecting the small term (r5 )nij in (6.87) and replacing the exact solution Uijn with its numerical one unij , a difference scheme for (6.42)–(6.44) is obtained as follows: { { { { { { { { { { { { { { { { { { {

2J

n

Δα ∑ cl w(αl )[τ−αl ∑ wk l un−k ij ] + l=0

= Δh unij + fijn ,

u0ij unij

= 0,

(α )

k=0

n

n−1

τ 2 2 uij − uij δ δ μ x y τ

(i, j) ∈ ω, 1 ⩽ n ⩽ N,

(i, j) ∈ ω,

= φ(xi , yj , tn ),

(6.91) (6.92)

(i, j) ∈ 𝜕ω, 0 ⩽ n ⩽ N.

(6.93)

Rewrite (6.91) in the form of μunij − (δx2 + δy2 )unij +

1 2 2 n δ δ u μ x y ij

2J

n

l=0

k=1

= −Δα ∑ cl w(αl )τ−αl ∑ wk l un−k + ij (α )

1 2 2 n−1 δ δ u + fijn , μ x y ij

or (√μ ℐ −

1 2 1 2 n δ )(√μ ℐ − δ )u √μ x √μ y ij

2J

n

l=0

k=1

= −Δα ∑ cl w(αl )τ−αl ∑ wk l un−k + ij (α )

1 2 2 n−1 δ δ u + fijn . μ x y ij

Let u∗ij = (√μ ℐ −

1 2 n δ )u . √μ y ij

Then the difference scheme (6.91)–(6.93) can be decomposed into the following ADI form:

354 | 6 Difference methods for the time distributed-order subdiffusion equations On each time level t = tn (1 ⩽ n ⩽ N), firstly, for any fixed j from 1 to M2 − 1, solve a series of linear systems in the unknown {u∗ij | 0 ⩽ i ⩽ M1 } in x direction 2J

n

1 2 ∗ (α ) { { (√μ ℐ − δx )uij = −Δα ∑ cl w(αl )τ−αl ∑ wk l un−k { ij { μ { √ { l=0 k=1 { { { 1 n + δx2 δy2 un−1 1 ⩽ i ⩽ M1 − 1, { ij + fij , { { μ { { { { 1 2 n 1 2 n { ∗ { u0j = (√μ ℐ − δy )u0j , u∗M1 ,j = (√μ ℐ − δ )u √μ √μ y M1 ,j {

(6.94)

to get the value of {u∗ij | 1 ⩽ i ⩽ M1 − 1}; Then, for any fixed i from 1 to M1 −1, compute a series of one-dimensional problems in the unknown {unij | 0 ⩽ j ⩽ M2 } in y direction 1 2 n { { (√μ ℐ − δy )uij = u∗ij , 1 ⩽ j ⩽ M2 − 1, μ √ { { n n { ui0 = φ(xi , y0 , tn ), ui,M2 = φ(xi , yM2 , tn )

(6.95)

to get the desired value of {unij | 1 ⩽ j ⩽ M2 − 1}. The linear systems (6.94) and (6.95) are both tridiagonal, which can be easily solved using the Thomas algorithm. In the subsequent part, we aim to analyze the proposed difference scheme. 6.5.2 Solvability of the difference scheme Theorem 6.5.1. The difference scheme (6.91)–(6.93) is uniquely solvable. Proof. Let ̄ un = {unij | (i, j) ∈ ω}. The value of u0 is determined by (6.92)–(6.93). Now suppose the values of u0 , u1 , . . . , un−1 have been uniquely determined, then we can get the linear system in un from (6.91) and (6.93). To show its unique solvability, it suffices to prove that the corresponding homogeneous one 1 2 2 n n n { { μuij + δx δy uij = Δh uij , μ { { n { uij = 0, (i, j) ∈ 𝜕ω has only the trivial solution.

(i, j) ∈ ω,

(6.96) (6.97)

6.5 The second-order ADI method in both space and distributed-order for 2D problem

to

| 355

To this end, making the inner product on both hand sides of (6.96) with un leads 1 2 2 n n 󵄩 󵄩2 (δ δ u , u ) = (Δh un , un ) = −󵄩󵄩󵄩∇h un 󵄩󵄩󵄩 . μ x y

μ(un , un ) + Therefore,

󵄩 󵄩2 󵄩2 󵄩 󵄩2 1 󵄩 μ󵄩󵄩󵄩un 󵄩󵄩󵄩 + 󵄩󵄩󵄩δx δy un 󵄩󵄩󵄩 = −󵄩󵄩󵄩∇h un 󵄩󵄩󵄩 ⩽ 0, μ which implies ‖un ‖ = 0. Noticing (6.97), it is clear that un = 0. By the principle of induction, the theorem is true. The proof ends. 6.5.3 Stability of the difference scheme Theorem 6.5.2. Suppose {vijn | (i, j) ∈ ω,̄ 0 ⩽ n ⩽ N} is the solution of the difference scheme { { { { { { { { { { { { { { { { { { {

2J

n

Δα ∑ cl w(αl )τ−αl ∑ wk l vijn−k + l=0

= Δh vijn + gijn , vij0 vijn

= φij ,

(α )

k=0

n n−1 τ 2 2 vij − vij δx δy μ τ

(i, j) ∈ ω, 1 ⩽ n ⩽ N,

(6.98)

(i, j) ∈ ω,

= 0,

(6.99)

(i, j) ∈ 𝜕ω, 0 ⩽ n ⩽ N.

(6.100)

Then it holds m 󵄩 󵄩2 τ 󵄩 󵄩2 󵄩 󵄩2 τ ∑ 󵄩󵄩󵄩Δh vn 󵄩󵄩󵄩 + (󵄩󵄩󵄩δx2 δy vm 󵄩󵄩󵄩 + 󵄩󵄩󵄩δx δy2 vm 󵄩󵄩󵄩 ) μ n=1

m 󵄩2 󵄩 󵄩2 󵄩 󵄩2 󵄩 󵄩2 τ 󵄩 ⩽ 2μτ󵄩󵄩󵄩∇h v0 󵄩󵄩󵄩 + (󵄩󵄩󵄩δx2 δy v0 󵄩󵄩󵄩 + 󵄩󵄩󵄩δx δy2 v0 󵄩󵄩󵄩 ) + τ ∑ 󵄩󵄩󵄩g n 󵄩󵄩󵄩 , μ n=1

1 ⩽ m ⩽ N,

(6.101)

where M1 −1 M2 −1

󵄩󵄩 n 󵄩󵄩2 n 2 󵄩󵄩g 󵄩󵄩 = h1 h2 ∑ ∑ (gij ) . i=1

j=1

Proof. Making the inner product on both hand sides of (6.98) with −Δh vn and using the Cauchy–Schwarz inequality, it follows that 2J

n

Δα ∑ cl w(αl )τ−αl ∑ wk l (vn−k , −Δh vn ) l=0

+

n

τ 2 2v −v (δ δ μ x y τ

(α )

k=0 n−1

, −Δh vn )

356 | 6 Difference methods for the time distributed-order subdiffusion equations = −(Δh vn , Δh vn ) + (g n , −Δh vn ) 󵄩 󵄩2 1 󵄩 󵄩2 1 󵄩 󵄩2 ⩽ −󵄩󵄩󵄩Δh vn 󵄩󵄩󵄩 + 󵄩󵄩󵄩Δh vn 󵄩󵄩󵄩 + 󵄩󵄩󵄩g n 󵄩󵄩󵄩 2 2 1 󵄩󵄩 n 󵄩󵄩2 1 󵄩󵄩 n 󵄩󵄩2 (6.102) = − 󵄩󵄩Δh v 󵄩󵄩 + 󵄩󵄩g 󵄩󵄩 , 1 ⩽ n ⩽ N. 2 2 For the second term on the left-hand side of the inequality above, by noticing (6.100), we have (δx2 δy2 = (δx2 δy2

vn − vn−1 , −Δh vn ) τ

vn − vn−1 vn − vn−1 , −δx2 vn ) + (δx2 δy2 , −δy2 vn ) τ τ

vn − vn−1 2 n vn − vn−1 , δx δy v ) + (δx δy2 , δx δy2 vn ) τ τ 1 󵄩 󵄩2 󵄩 󵄩2 ⩾ (󵄩󵄩󵄩δx2 δy vn 󵄩󵄩󵄩 − 󵄩󵄩󵄩δx2 δy vn−1 󵄩󵄩󵄩 ) 2τ 1 󵄩 󵄩2 󵄩 󵄩2 + (󵄩󵄩󵄩δx δy2 vn 󵄩󵄩󵄩 − 󵄩󵄩󵄩δx δy2 vn−1 󵄩󵄩󵄩 ). 2τ

= (δx2 δy

(6.103)

Summing up for n in (6.102) from 1 to m and adding the term μ(v0 , −Δh v0 ) to both hand sides of the obtained inequality, with the aid of (6.103), we get 2J

m

n

Δα ∑ cl w(αl )τ−αl [ ∑ ∑ wk l (vn−k , −Δh vn )] (α )

n=0 k=0

l=0

1 󵄩 󵄩2 󵄩 󵄩2 󵄩 󵄩2 󵄩 󵄩2 + (󵄩󵄩󵄩δx2 δy vm 󵄩󵄩󵄩 + 󵄩󵄩󵄩δx δy2 vm 󵄩󵄩󵄩 − 󵄩󵄩󵄩δx2 δy v0 󵄩󵄩󵄩 − 󵄩󵄩󵄩δx δy2 v0 󵄩󵄩󵄩 ) 2μ

1 m󵄩 1 m 󵄩 󵄩2 󵄩2 ⩽ − ∑ 󵄩󵄩󵄩Δh vn 󵄩󵄩󵄩 + μ(v0 , −Δh v0 ) + ∑ 󵄩󵄩󵄩g n 󵄩󵄩󵄩 2 n=1 2 n=1 m 1 m󵄩 󵄩2 󵄩 󵄩2 1 󵄩 󵄩2 = − ∑ 󵄩󵄩󵄩Δh vn 󵄩󵄩󵄩 + μ󵄩󵄩󵄩∇h v0 󵄩󵄩󵄩 + ∑ 󵄩󵄩󵄩g n 󵄩󵄩󵄩 , 2 n=1 2 n=1

1 ⩽ m ⩽ N.

(6.104)

In view of (6.61), it holds m

n

∑ ∑ wk l (vn−k , −Δh vn ) ⩾ 0, (α )

n=0 k=0

hence, (6.101) is followed from (6.104). The proof ends. 6.5.4 Convergence of the difference scheme Theorem 6.5.3. Suppose {Uijn | (i, j) ∈ ω,̄ 0 ⩽ n ⩽ N} and {unij | (i, j) ∈ ω,̄ 0 ⩽ n ⩽ N} are solutions of the problem (6.42)–(6.44) and the difference scheme (6.91)–(6.93), respectively. Let eijn = Uijn − unij ,

(i, j) ∈ ω,̄ 0 ⩽ n ⩽ N,

6.6 The fourth-order ADI method in both space and distributed-order for 2D problem

| 357

then it holds N

󵄩 󵄩 τ ∑ 󵄩󵄩󵄩en 󵄩󵄩󵄩∞ ⩽ cT √L1 L2 (c3 + c5 )(τ2 | ln τ| + h21 + h22 + Δα2 ),

(6.105)

n=1

where the constant c is defined in Lemma 6.3.1. Proof. The subtraction of (6.91)–(6.93) from (6.87), (6.89)–(6.90), respectively, will produce the system of error equations as follows: { { { { { { { { { { { { { { { { { { {

2J

n

Δα ∑ cl w(αl )[τ−αl ∑ wk l eijn−k ] + l=0

k=0

= Δh eijn + (r5 )nij , eij0 = 0, eijn

= 0,

(α )

n

n−1

τ 2 2 eij − eij δ δ μ x y τ

(i, j) ∈ ω, 1 ⩽ n ⩽ N,

(i, j) ∈ ω, (i, j) ∈ 𝜕ω, 0 ⩽ n ⩽ N.

Noticing (6.88), the application of Theorem 6.5.2 immediately yields N

N

n=1

n=1

󵄩 󵄩2 󵄩 󵄩2 τ ∑ 󵄩󵄩󵄩Δh en 󵄩󵄩󵄩 ⩽ τ ∑ 󵄩󵄩󵄩(r5 )n 󵄩󵄩󵄩 N

2

⩽ τ ∑ L1 L2 [c3 (τ2 + h21 + h22 + Δα2 ) + c5 τ2 | ln τ|] n=1

2

⩽ TL1 L2 (c3 + c5 )2 (τ2 | ln τ| + h21 + h22 + Δα2 ) .

(6.106)

By the Cauchy–Schwarz inequality, Lemma 6.3.1 and (6.106), we have N

2

N

󵄩 󵄩 󵄩 󵄩2 (τ ∑ 󵄩󵄩󵄩en 󵄩󵄩󵄩∞ ) ⩽ T(τ ∑ 󵄩󵄩󵄩en 󵄩󵄩󵄩∞ ) n=1

n=1 N

󵄩 󵄩2 ⩽ Tc2 τ ∑ 󵄩󵄩󵄩Δh en 󵄩󵄩󵄩 n=1

2 2

2

⩽ c T L1 L2 (c3 + c5 )2 (τ2 | ln τ| + h21 + h22 + Δα2 ) , which implies (6.105). The proof ends.

6.6 The fourth-order ADI method in both space and distributed-order for 2D problem This section is devoted to the derivation of another high order ADI difference scheme for solving the problem (6.42)–(6.44) and the corresponding analysis on the resultant scheme. ̂ y, t) in the same way as that in Section 2.10. Suppose Define the function u(x, ̄ ̂ y, ⋅) ∈ C 2+1 (ℛ) and u(⋅, ⋅, t) ∈ C (6,6) (Ω). u(x,

358 | 6 Difference methods for the time distributed-order subdiffusion equations 6.6.1 Derivation of the difference scheme Adding the small term τν δx2 δy2

Uijn −Uijn−1 τ

2J

𝒜x 𝒜y Δα ∑ dl w(αl )[τ

−αl

l=0

to both hand sides of (6.66) gives n Uijn − Uijn−1 τ (α ) ∑ wk l Uijn−k ] + δx2 δy2 ν τ k=0

= 𝒜y δx2 Uijn + 𝒜x δy2 Uijn + 𝒜x 𝒜y fijn + (r6 )nij ,

(i, j) ∈ ω, 1 ⩽ n ⩽ N,

(6.107)

where Uijn − Uijn−1 τ (r6 )nij = (r4 )nij + δx2 δy2 , ν τ by noticing Lemma 6.2.2, there is a positive constant c6 such that 󵄨󵄨 n 󵄨󵄨 2 4 4 4 2 󵄨󵄨(r6 )ij 󵄨󵄨 ⩽ c4 (τ + h1 + h2 + Δα ) + c6 τ | ln τ|,

(i, j) ∈ ω, 1 ⩽ n ⩽ N.

(6.108)

Noticing the initial-boundary value conditions (6.43)–(6.44), we have {

Uij0 = 0, Uijn

(i, j) ∈ ω,

= φ(xi , yj , tn ),

(6.109) (i, j) ∈ 𝜕ω, 0 ⩽ n ⩽ N.

(6.110)

Omitting the small term (r6 )nij in (6.107) and replacing the exact solution Uijn with its numerical one unij , we can obtain the difference scheme for solving (6.42)–(6.44) as follows: { { { { { { { { { { { { { { { { { { {

2J

𝒜x 𝒜y Δα ∑ dl w(αl )[τ l=0

−αl

= 𝒜y δx2 unij + 𝒜x δy2 unij + 𝒜x 𝒜y fijn ,

u0ij unij

= 0,

n

n−1

n τ 2 2 uij − uij (α ) ∑ wk l un−k ij ] + δx δy ν τ k=0

(i, j) ∈ ω, 1 ⩽ n ⩽ N,

(i, j) ∈ ω,

= φ(xi , yj , tn ),

(6.112) (i, j) ∈ 𝜕ω, 0 ⩽ n ⩽ N.

Equation (6.111) can be reformulated as 1 ν𝒜x 𝒜y unij − (𝒜y δx2 unij + 𝒜x δy2 unij ) + δx2 δy2 unij ν

2J n 1 (α ) n = −Δα ∑ dl w(αl )τ−αl ∑ wk l 𝒜x 𝒜y un−k + δx2 δy2 un−1 ij ij + 𝒜x 𝒜y fij , ν l=0 k=1

or (√ν𝒜x −

(6.111)

1 2 1 2 n δ )(√ν𝒜y − δ )u √ν x √ν y ij

(6.113)

6.6 The fourth-order ADI method in both space and distributed-order for 2D problem

| 359

2J n 1 (α ) n = −Δα ∑ dl w(αl )τ−αl ∑ wk l 𝒜x 𝒜y un−k + δx2 δy2 un−1 ij ij + 𝒜x 𝒜y fij . ν l=0 k=1

Let u∗ij = (√ν𝒜y −

1 2 n δ )u . √ν y ij

Then the difference scheme (6.111)–(6.113) can be written as the following ADI form: On each time level t = tn (1 ⩽ n ⩽ N), at first, for any fixed j from 1 to M2 − 1, solve a series of one-dimensional problems in the unknown {u∗ij | 0 ⩽ i ⩽ M1 } in x direction 2J n 1 2 ∗ (α ) −αl { { √ν𝒜x − δ )u = −Δα d w(α )τ wk l 𝒜x 𝒜y un−k ( ∑ ∑ { l l ij { √ν x ij { { l=0 k=1 { { { 1 n { 1 ⩽ i ⩽ M1 − 1, + δx2 δy2 un−1 { ij + 𝒜x 𝒜y fij , { ν { { { { { { u∗ = (√ν𝒜 − 1 δ2 )un , u∗ = (√ν𝒜 − 1 δ2 )un y y M1 ,j √ν y 0j √ν y M1 ,j { 0j

(6.114)

to get the value of {u∗ij | 1 ⩽ i ⩽ M1 − 1} on an intermediate time level. Then, for any fixed i from 1 to M1 − 1, perform some calculations on a series of one-dimensional problems in the unknown {unij | 0 ⩽ j ⩽ M2 } in y direction 1 2 n { { (√ν𝒜y − δy )uij = u∗ij , 1 ⩽ j ⩽ M2 − 1, √ν { { n n { ui0 = φ(xi , y0 , tn ), ui,M2 = φ(xi , yM2 , tn )

(6.115)

to get the desired value of {unij | 1 ⩽ j ⩽ M2 − 1}. The linear systems (6.114) and (6.115) are both tridiagonal, which can be easily solved using the Thomas algorithm. 6.6.2 Solvability of the difference scheme Theorem 6.6.1. The difference scheme (6.111)–(6.113) is uniquely solvable. Proof. Let ̄ un = {unij | (i, j) ∈ ω}. The value of u0 is determined by (6.112)–(6.113).

360 | 6 Difference methods for the time distributed-order subdiffusion equations Now assume that the values of u0 , u1 , . . . , un−1 have been uniquely determined, then the system in un can be obtained from (6.111) and (6.113). To show its unique solvability, it is sufficient to prove that the corresponding homogeneous one 1 { ν𝒜x 𝒜y unij + δx2 δy2 unij = 𝒜y δx2 unij + 𝒜x δy2 unij , ν { n { uij = 0, (i, j) ∈ 𝜕ω

(i, j) ∈ ω,

(6.116) (6.117)

has only the trivial solution. To this end, making the inner product on both hand sides of (6.116) with un arrives at 1 ν(𝒜x 𝒜y un , un ) + (δx2 δy2 un , un ) = (𝒜y δx2 un , un ) + (𝒜x δy2 un , un ). ν Noticing (6.76), (6.77) and (δx2 δy2 un , un ) = ‖δx δy un ‖2 , we have 1 󵄩󵄩 n 󵄩󵄩2 ν󵄩u 󵄩 + 3 󵄩 󵄩

1 󵄩󵄩 2󵄩 n 󵄩2 n 󵄩2 󵄩δ δ u 󵄩󵄩 ⩽ − 󵄩󵄩󵄩∇h u 󵄩󵄩󵄩 ⩽ 0, ν󵄩 x y 󵄩 3

thus, ‖un ‖ = 0. Then un = 0 is followed by noticing (6.117). By the principle of induction, the theorem is true. The proof ends. 6.6.3 Stability of the difference scheme Theorem 6.6.2. Suppose {vijn | (i, j) ∈ ω,̄ 0 ⩽ n ⩽ N} is the solution of the difference scheme { { { { { { { { { { { { { { { { { { {

2J n vijn − vijn−1 τ (α ) Δα ∑ dl w(αl )τ−αl ∑ wk l 𝒜x 𝒜y vijn−k + δx2 δy2 ν τ l=0 k=0

= 𝒜y δx2 vijn + 𝒜x δy2 vijn + gijn , vij0 vijn

= φij ,

= 0,

(i, j) ∈ ω, 1 ⩽ n ⩽ N,

(i, j) ∈ ω,

(6.118) (6.119)

(i, j) ∈ 𝜕ω, 0 ⩽ n ⩽ N.

(6.120)

Then it holds m 󵄩2 󵄩 󵄩 󵄩2 3τ 󵄩 󵄩2 τ ∑ 󵄩󵄩󵄩Δh vn 󵄩󵄩󵄩 + (󵄩󵄩󵄩δx2 δy vm 󵄩󵄩󵄩 + 󵄩󵄩󵄩δx δy2 vm 󵄩󵄩󵄩 ) 2ν n=1

󵄩 󵄩2 3τ 󵄩 󵄩2 󵄩 󵄩2 ⩽ 3ντ󵄩󵄩󵄩∇h v0 󵄩󵄩󵄩 + (󵄩󵄩󵄩δx2 δy v0 󵄩󵄩󵄩 + 󵄩󵄩󵄩δx δy2 v0 󵄩󵄩󵄩 ) 2ν 9 m 󵄩 󵄩2 + τ ∑ 󵄩󵄩󵄩g n 󵄩󵄩󵄩 , 1 ⩽ m ⩽ N, 4 n=1 where M1 −1 M2 −1

󵄩󵄩 n 󵄩󵄩2 n 2 󵄩󵄩g 󵄩󵄩 = h1 h2 ∑ ∑ (gij ) . i=1

j=1

(6.121)

6.6 The fourth-order ADI method in both space and distributed-order for 2D problem | 361

Proof. Taking the inner product on both hand sides of (6.118) with −Δh vn , it follows from Lemma 6.4.1 that 2J

n

Δα ∑ dl w(αl )τ−αl ∑ wk l (𝒜x 𝒜y vn−k , −Δh vn ) l=0

n

τ v −v + (δx2 δy2 ν τ

(α )

k=0 n−1

, −Δh vn )

= (𝒜y δx2 vn + 𝒜x δy2 vn , −Δh vn ) + (g n , −Δh vn )

2󵄩 󵄩2 ⩽ − 󵄩󵄩󵄩Δh vn 󵄩󵄩󵄩 + 3 1󵄩 󵄩2 = − 󵄩󵄩󵄩Δh vn 󵄩󵄩󵄩 + 3

1 󵄩󵄩 n 󵄩󵄩2 3 󵄩󵄩 n 󵄩󵄩2 󵄩Δ v 󵄩 + 󵄩󵄩g 󵄩󵄩 3󵄩 h 󵄩 4 3 󵄩󵄩 n 󵄩󵄩2 󵄩g 󵄩 , 1 ⩽ n ⩽ N. 4󵄩 󵄩

(6.122)

By (6.103), we have vn − vn−1 , −Δh vn ) τ 1 󵄩 󵄩2 󵄩 󵄩2 󵄩 󵄩2 󵄩 󵄩2 ⩾ [(󵄩󵄩󵄩δx2 δy vn 󵄩󵄩󵄩 + 󵄩󵄩󵄩δx δy2 vn 󵄩󵄩󵄩 ) − (󵄩󵄩󵄩δx2 δy vn−1 󵄩󵄩󵄩 + 󵄩󵄩󵄩δx δy2 vn−1 󵄩󵄩󵄩 )]. 2τ (δx2 δy2

(6.123)

Summing up for n in (6.122) from 1 to m and adding the term ν(𝒜x 𝒜y v0 , −Δh v0 ) to both hand sides of the obtained inequality, it follows by noticing (6.123) that 2J

m

n

Δα ∑ dl w(αl )τ−αl [ ∑ ∑ wk l (𝒜x 𝒜y vn−k , −Δh vn )] (α )

n=0 k=0

l=0

1 󵄩 󵄩2 󵄩 󵄩2 󵄩 󵄩2 󵄩 󵄩2 + (󵄩󵄩󵄩δx2 δy vm 󵄩󵄩󵄩 + 󵄩󵄩󵄩δx δy2 vm 󵄩󵄩󵄩 − 󵄩󵄩󵄩δx2 δy v0 󵄩󵄩󵄩 − 󵄩󵄩󵄩δx δy2 v0 󵄩󵄩󵄩 ) 2ν 3 m 󵄩 󵄩2 1 m󵄩 󵄩2 ⩽ − ∑ 󵄩󵄩󵄩Δh vn 󵄩󵄩󵄩 + ν(𝒜x 𝒜y v0 , −Δh v0 ) + ∑ 󵄩󵄩󵄩g n 󵄩󵄩󵄩 , 1 ⩽ m ⩽ N. 3 n=1 4 n=1

(6.124)

It is clear from (6.83) and (6.84), respectively, that m

n

∑ ∑ wk l (𝒜x 𝒜y vn−k , −Δh vn ) ⩾ 0

(6.125)

󵄩 󵄩2 (𝒜x 𝒜y v0 , −Δh v0 ) ⩽ 󵄩󵄩󵄩∇h v0 󵄩󵄩󵄩 .

(6.126)

(α )

n=0 k=0

and

Substituting (6.125) and (6.126) into (6.124) produces (6.121). The proof ends. 6.6.4 Convergence of the difference scheme Theorem 6.6.3. Suppose {Uijn | (i, j) ∈ ω,̄ 0 ⩽ n ⩽ N} and {unij | (i, j) ∈ ω,̄ 0 ⩽ n ⩽ N} are solutions of the problem (6.42)–(6.44) and the difference scheme (6.111)–(6.113), respec-

362 | 6 Difference methods for the time distributed-order subdiffusion equations tively. Let eijn = Uijn − unij ,

(i, j) ∈ ω,̄ 0 ⩽ n ⩽ N,

then it holds N 3 󵄩 󵄩 τ ∑ 󵄩󵄩󵄩en 󵄩󵄩󵄩∞ ⩽ cT √L1 L2 (c4 + c6 )(τ2 | ln τ| + h41 + h42 + Δα4 ), 2 n=1

(6.127)

where the constant c is defined in Lemma 6.3.1. Proof. The subtraction of (6.111)–(6.113) from (6.107), (6.109)–(6.110), respectively, gives the system of error equations as follows: { { { { { { { { { { { { { { { { { { {

n

n−1

2J n eij − eij τ (α ) Δα ∑ dl w(αl )[τ−αl ∑ wk l 𝒜x 𝒜y eijn−k ] + δx2 δy2 ν τ l=0 k=0

= 𝒜y δx2 eijn + 𝒜x δy2 eijn + (r6 )nij , eij0 eijn

(i, j) ∈ ω, 1 ⩽ n ⩽ N,

= 0,

(i, j) ∈ ω,

= 0,

(i, j) ∈ 𝜕ω, 0 ⩽ n ⩽ N.

Noticing (6.108), the application of Theorem 6.6.2 yields N N 󵄩 󵄩2 󵄩 󵄩2 9 τ ∑ 󵄩󵄩󵄩Δh en 󵄩󵄩󵄩 ⩽ τ ∑ 󵄩󵄩󵄩(r6 )n 󵄩󵄩󵄩 4 n=1 n=1

⩽ ⩽

9 N 2 τ ∑ L L [c (τ2 + h41 + h42 + Δα4 ) + c6 τ2 | ln τ|] 4 n=1 1 2 4 9 2 TL L (c + c6 )2 (τ2 | ln τ| + h41 + h42 + Δα4 ) . 4 1 2 4

(6.128)

By the Cauchy–Schwarz inequality, Lemma 6.3.1 and (6.128), we have N

2

N

󵄩 󵄩 󵄩 󵄩2 (τ ∑ 󵄩󵄩󵄩en 󵄩󵄩󵄩∞ ) ⩽ T(τ ∑ 󵄩󵄩󵄩en 󵄩󵄩󵄩∞ ) n=1

n=1 N

󵄩 󵄩2 ⩽ Tc2 τ ∑ 󵄩󵄩󵄩Δh en 󵄩󵄩󵄩 n=1

9 2 ⩽ c2 T 2 L1 L2 (c4 + c6 )2 (τ2 | ln τ| + h41 + h42 + Δα4 ) , 4 which implies (6.127). The proof ends.

6.7 Supplementary remarks and discussions 1. In this chapter, the difference methods for solving 1D and 2D time distributed-order subdiffusion equations have been introduced[24, 29] . For the approximation of the distributed integral, the composite trapezoid formula or the composite Simpson formula

6.7 Supplementary remarks and discussions | 363

was used; For the discretization of time Caputo derivatives, the second-order WSGL formula was applied and several higher order difference schemes were derived. For 2D problem, two ADI difference schemes were also discussed. For each scheme, the unique solvability, stability and convergence were proved. Indeed, one can also directly apply the G-L formula to approximate the time Caputo derivatives and the firstorder difference scheme in time can be obtained, then the techniques of Richardson extrapolation can be used to improve the accuracy in time[23, 25] . 2. For the approximation of distributed integral, the composite mid-point formula can also be used. For details, see [103]. Also, the Gauss quadrature can be applied. For the approximation of time Caputo derivatives, the L1 formula can also be used, please refer to [60, 103]. 3. In this chapter, we only discussed the numerical solutions of time distributedorder subdiffusion equations. Now the numerical solutions of time distributed-order wave equations are briefly introduced. Consider the following problem: w 𝒟t u(x, t) = uxx (x, t) + f (x, t), 0 < x < L, 0 < t ⩽ T, { { u(x, 0) = 0, ut (x, 0) = 0, 0 < x < L, { { { u(0, t) = φ1 (t), u(L, t) = φ2 (t), 0 ⩽ t ⩽ T,

(6.129) (6.130) (6.131)

where φ1 (0) = φ2 (0) = 0, φ󸀠1 (0) = φ󸀠2 (0) = 0 and w 𝒟t u(x, t)

C γ 0 Dt u(x, t)

=

2

γ

= ∫ w(γ) C0 Dt u(x, t)dγ,

1 t 1 ∫ (t { Γ(2−γ) 0

utt (x, t),

− ξ )1−γ uξξ (x, ξ )dξ , 1 ⩽ γ < 2, γ = 2,

2

w(γ) ⩾ 0, ∫1 w(γ)dγ = c0 > 0, the functions f , φ1 and φ2 are all given. Denote Δγ = 2J1 , γl = 1 + lΔγ (0 ⩽ l ⩽ 2J). Considering equation (6.129) at the point (xi , tn ), we have w

𝒟t u(xi , tn ) = uxx (xi , tn ) + f (xi , tn ),

1 ⩽ i ⩽ M − 1, 0 ⩽ n ⩽ N.

Taking an average on two adjacent time levels gives 1 w 1 [𝒟 u(xi , tn ) + 𝒟tw u(xi , tn−1 )] = [ uxx (xi , tn ) + uxx (xi , tn−1 )] 2 t 2 1 + [f (xi , tn ) + f (xi , tn−1 )], 1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N. 2 Let γ

s(γ, xi , tn ) = w(γ) C0 Dt u(xi , tn ).

(6.132)

364 | 6 Difference methods for the time distributed-order subdiffusion equations Suppose function s(⋅, xi , tn ) ∈ C 2 [1, 2]. Applying the composite trapezoid formula to discretize the distributed integral, it follows from Lemma 6.1.1 that 2J

w

𝒟t u(xi , tn ) = Δγ ∑ cl w(γl ) l=0

C γl 0 Dt u(xi , tn )

+ O(Δγ 2 ).

Substituting the result above into (6.132) produces 2J 1 γ γ Δγ ∑ cl w(γl ) ⋅ [C0 Dt l u(xi , tn ) + C0 Dt l u(xi , tn−1 )] 2 l=0

1 1 = [ uxx (xi , tn ) + uxx (xi , tn−1 )] + [f (xi , tn ) + f (xi , tn−1 )] 2 2 + O(Δγ 2 ), 1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N.

(6.133)

̂ t) like that in Exercise 3.1 in Chapter 3. Suppose u(x, ̂ ⋅) ∈ Define the function u(x,

C 2+2 (ℛ) and u(⋅, t) ∈ C 4 [0, L].

Let v(x, t) = ut (x, t), then C γl 0 Dt u(xi , tn )

n

γ −1

= C0 Dt l v(xi , tn ) = τ−(γl −1) ∑ wk l

(γ −1)

k=0

v(xi , tn−k ) + O(τ2 ).

Therefore, 1 C γl γ [ D u(xi , tn ) + C0 Dt l u(xi , tn−1 )] 2 0 t

n−1 n 1 (γ −1) (γ −1) = [τ−(γl −1) ∑ wk l v(xi , tn−k ) + τ−(γl −1) ∑ wk l v(xi , tn−1−k )] 2 k=0 k=0

+ O(τ2 )

n−1

(γ −1)

= τ−(γl −1) ∑ wk l k=0 n−1

(γ −1)

= τ−(γl −1) ∑ wk l k=0

1 ⋅ [v(xi , tn−k ) + v(xi , tn−1−k )] + O(τ2 ) 2 ⋅

u(xi , tn−k ) − u(xi , tn−1−k ) + O(τ2 ). τ

Substituting (6.134) into (6.133) and denoting n− 21

fi

1 = [f (xi , tn ) + f (xi , tn−1 )], 2

we have 2J

n−1

(γ −1)

Δγ ∑ cl w(γl )τ−(γl −1) ∑ wk l l=0

=

n− 1 δx2 Ui 2

+

n− 1 fi 2

k=0

n− 21

+ (r7 )i

,

n−k− 21

δt Ui

1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N,

(6.134)

Exercises 6

| 365

and there exists a positive constant c7 such that 󵄨󵄨 n− 21 󵄨󵄨 2 2 2 󵄨󵄨(r7 )i 󵄨󵄨 ⩽ c7 (τ + h + Δγ ),

1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N.

Noticing the initial-boundary value conditions (6.130)–(6.131), we can get a difference scheme for (6.129)–(6.131) as follows: { { { { { { { { { { { { { { { { { { {

2J

n−1

l=0

k=0

(γ −1)

Δγ ∑ cl w(γl )τ−(γl −1) ∑ wk l

n− 1 n− 1 = δx2 ui 2 + fi 2 , 1 ⩽ i ⩽ M u0i = 0, 1 ⩽ i ⩽ M − 1, un0 = φ1 (tn ), unM = φ2 (tn ),

n−k− 21

δt ui

− 1, 1 ⩽ n ⩽ N,

(6.135) (6.136)

0 ⩽ n ⩽ N.

(6.137)

It can be proved that the difference scheme (6.135)–(6.137) is uniquely solvable, unconditionally stable and convergent. More details can be found in [26]. Regarding the ADI difference scheme for solving 2D time distributed-order wave equations, interested readers can refer to [27]. 4. Ye et al.[104] and Hu et al.[37] studied the difference methods for the one-dimensional and multidimensional distributed-order fractional wave equations based on the L1 approximation, respectively. 5. There are some works on the difference methods for the space distributed-order equations. The interested readers may refer to [94].

Exercises 6 6.1 For the problem (6.1)–(6.3), construct the following difference scheme: 2J

n

(α ) { { Δα ∑ cl w(αl )τ−αl ∑ gk l un−k = δx2 uni + fin , { i { { { l=0 k=0 { { 1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N, { { { 0 { { u = 0, 1 ⩽ i ⩽ M − 1, { { { ni n { u0 = φ1 (tn ), uM = φ2 (tn ), 0 ⩽ n ⩽ N.

̂ t) like that in Section 2.1 and suppose u(x, ̂ ⋅) ∈ C 1+1 (ℛ). Define the function u(x, For this difference scheme, try to (1) analyze the truncation error; (2) show the unique solvability; (3) show the stability with respect to the function f ; (4) show the convergence.

366 | 6 Difference methods for the time distributed-order subdiffusion equations 6.2 For the problem (6.1)–(6.3), construct the following difference scheme: 2J

n

(α ) { { Δα ∑ dl w(αl )τ−αl ∑ gk l 𝒜un−k = δx2 uni + 𝒜fin , { i { { { l=0 k=0 { { 1 ⩽ i ⩽ M − 1, 1 ⩽ n ⩽ N, { { { 0 { { u = 0, 1 ⩽ i ⩽ M − 1, { { { in n { u0 = φ1 (tn ), uM = φ2 (tn ), 0 ⩽ n ⩽ N.

̂ t) like that in Section 2.1 and suppose u(x, ̂ ⋅) ∈ C 1+1 (ℛ). Define the function u(x, For this difference scheme, try to (1) analyze the truncation error; (2) show the unique solvability; (3) show the stability with respect to the function f ; (4) show the convergence. 6.3 For the problem (6.42)–(6.44), construct the following difference scheme: 2J n { (α ) −αl n n { { Δα c w(α )[τ gk l un−k ∑ ∑ { l l ij ] = Δh uij + fij , { { { l=0 k=0 { { (i, j) ∈ ω, 1 ⩽ n ⩽ N, { { { 0 { { uij = 0, (i, j) ∈ ω, { { { { n { uij = φ(xi , yj , tn ), (i, j) ∈ 𝜕ω, 0 ⩽ n ⩽ N.

̂ y, t) like that in Section 2.10 and suppose u(x, ̂ y, ⋅) ∈ Define the function u(x,

C 1+1 (ℛ).

For this difference scheme, try to (1) analyze the truncation error; (2) show the unique solvability; (3) show the stability with respect to the function f ; (4) show the convergence. 6.4 For the problem (6.42)–(6.44), construct the following difference scheme: { { { { { { { { { { { { { { { { { { {

2J

n

l=0

k=0

Δα ∑ dl w(αl )[τ−αl ∑ gk l 𝒜x 𝒜y un−k ij ] (α )

= 𝒜y δx2 unij + 𝒜x δy2 unij + 𝒜x 𝒜y fijn , u0ij = 0, unij

(i, j) ∈ ω, 1 ⩽ n ⩽ N,

(i, j) ∈ ω,

= φ(xi , yj , tn ),

(i, j) ∈ 𝜕ω, 0 ⩽ n ⩽ N.

̂ y, t) like that in Section 2.10 and suppose u(x, ̂ y, ⋅) ∈ Define the function u(x, 1+1 C (ℛ). For this difference scheme, try to (1) analyze the truncation error;

Exercises 6

| 367

(2) show the unique solvability; (3) show the stability with respect to the function f ; (4) show the convergence. 6.5 Denote 2J

(α ) μ̄ = Δα ∑ cl w(αl )τ−αl g0 l . l=0

Try to show that ̄ = O( μτ

1 ). | ln τ|

For the problem (6.42)–(6.44), construct the following difference scheme: n n−1 2J n { τ 2 2 uij − uij (αl ) n−k −αl { { δ δ = Δh unij + fijn , Δα c w(α )[τ u ] + g ∑ ∑ { l l x y ij k { ̄ μ τ { { l=0 k=0 { { (i, j) ∈ ω, 1 ⩽ n ⩽ N, { { { 0 { { uij = 0, (i, j) ∈ ω, { { { { n { uij = φ(xi , yj , tn ), (i, j) ∈ 𝜕ω, 0 ⩽ n ⩽ N.

̂ y, t) like that in Section 2.10 and suppose u(x, ̂ y, ⋅) ∈ Define the function u(x, 1+1 C (ℛ). For this difference scheme, try to (1) analyze the truncation error; (2) write the ADI form; (3) show the unique solvability; (4) show the stability with respect to the function f ; (5) show the convergence. 6.6 Denote 2J

(α ) ν̄ = Δα ∑ dl w(αl )τ−αl g0 l . l=0

Try to show that ̄ = O( ντ

1 ). | ln τ|

For the problem (6.42)–(6.44), construct the following difference scheme: { { { { { { { { { { { { { { { { { { {

n

n−1

2J n τ 2 2 uij − uij (α ) Δα ∑ dl w(αl )[τ−αl ∑ gk l 𝒜x 𝒜y un−k ij ] + ̄ δx δy ν τ l=0 k=0

= 𝒜y δx2 unij + 𝒜x δy2 unij + 𝒜x 𝒜y fijn , u0ij unij

= 0,

(i, j) ∈ ω, 1 ⩽ n ⩽ N,

(i, j) ∈ ω,

= φ(xi , yj , tn ),

(i, j) ∈ 𝜕ω, 0 ⩽ n ⩽ N.

368 | 6 Difference methods for the time distributed-order subdiffusion equations ̂ y, t) like that in Section 2.10 and suppose u(x, ̂ y, ⋅) ∈ Define the function u(x,

C 1+1 (ℛ).

For this difference scheme, try to (1) analyze the truncation error; (2) write the ADI form; (3) show the unique solvability; (4) show the stability with respect to the function f ; (5) show the convergence.

A The Matlab code of sum-of-exponentials approximations for the kernel t −α in the Caputo fractional derivative function [xs,ws,nexp] = sumofexpappr2new(alpha,reps,dt,Tfinal ) %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % Copyright: all rights reserved by Shidong Jiang, Jiwei Zhang, % Qian Zhang and Zhimin Zhang. % Citation: please cite the following papers: % [1] S. Jiang, J. Zhang, Q. Zhang and Z. Zhang. Fast evaluation % of the Caputo fractional derivative and its applications to % fractional diffusion equations. Commun. Comput. Phys., 21(2017), % 650--678. % [2] G. Beylkin and L. Monzn. On approximation of functions by % exponential sums. Appl. Comput. Harmon. Anal. 19(2005), 17--48. % [3] G. Beylkin and L. Monzn. Approximation by exponential sums % revisited. Appl. Comput. Harmon. Anal., 28(2):131--149, 2010. %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % For given positive parameters: alpha, reps, dt and T, return % sum-of-exponentials approximation for 1/t^alpha for the inverval % dt1) res = norm(A*x-b)/norm(b); end end This code is provided by the authors of [41]. We remark that the parameters in Lemma 1.7.1 of this book correspond to the parameters in the above code as α = alpha, (α) T = T, ϵ = reps, τ̂ = dt, Nexp = nexp, , ω(α) = ws(l), s(α) = xs(l). l l

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Index k− 1

δt vi 2 173 δtα f (t) 23 Δαt f (tn−1+σ ) 44 (δx u, δx v) 110 δx ui− 1 110 2 i− 21 ,j

δx v

155

(δx2 u, δx2 v) 110 δx2 vij 155 δx2 δy2 vij 155 δx2 ui 110 δx δy2 vi− 1 ,j 155 2

(δx δy u, δx δy v) 156 δx δy vi− 1 ,j− 1 155 2

2

(δx u, δx v) 155 δy δx2 vi,j− 1 155 2

δy2 vij 155 δy vi,j− 1 155 2

(δy u, δy v) 155 ‖δx u‖ 110, 155 ‖δx u‖A 111 ‖δx2 u‖ 110 ‖δx δy u‖ 156 ‖δy u‖ 155 ‖∇h u‖ 156 𝜕α f (x) 4 𝜕|x|α a(α) 31 l 70 â (α) k Aαh,p f (t) 11 β 𝒜h ui 305

(𝒜u)i 110 𝒜x uij , 𝒜y uij 253 (γ) bl 35 (n,γ) b̂ k 59 (n,γ) b̃ k 84 β ℬx uij 284 γ ℬy uij 284 (ℬ(β) u)i 278 c∗(α) 29 ck(n,α) 44, 45 ck̂(n,α) 54 ck̃(n,α) 40

Cμ 9 Cμm 9 C n+α (ℛ) 16

dk(n,α) 76 Dαt f (tn ) 32 𝒟t f (tn−1+σ ) 51 γ 𝒟̂ t f (tn− 1 ) 60 γ

2

𝔻t f (tn ) 38 α −∞ Dt f (t) 4 C α a Dt f (t) 3 α a Dt f (t) 2 −α D a t f (t) 1 α a Dt f (t) 1 C α t Db f (t) 4 α t Db f (t) 4 α D t b f (t) 4 α t D∞ f (t) 4 F (σ) 50 g(α) 11 k ĝ (α) 25 k β

γ

ℋx vij , ℋy vij 316 ℐ 24 (α) , ω(α) , s(α) 68 Nexp l l ‖u‖ 110, 155 ‖u‖∞ 110, 156 (u, v) 110, 155 (u, v)1,A 111 𝒰h 110 𝒰h̊ 110 k− 1

vi 2 173 𝒱h 155 𝒱̊ h 155 wk(α) 20 ŵ k(α) 24 ̃k(α) 21 w

approximation formula of sum-of-exponentials 67 Bloch–Torrey equation 297 Caputo fractional derivative 3 fast H2N2 approximation 81 fast L1 approximation 70 fast L1 difference scheme 124 fast L2-1σ approximation 76 Fourier inverse transform 6 Fourier transform 5

380 | Index

fractional central difference operator 25 fractional integral 1 fractional ordinary differential equation (FODE) 7 Grünwald–Letnikov (G-L) fractional derivative 1 H2N2 approximation 60 H2N2 formula 60 identity operator 24 L1 approximation 32 L1 formula 32 L1-2 approximation 39 L2-1σ approximation 44 multi-term fractional derivative 49 multiterm time-fractional subdiffusion (MTTFSD) equations 145 multiterm time-fractional wave (MTTFW) equations 222

Riemann–Liouville (R-L) fractional derivative 2 Riesz fractional derivative 4 right Caputo fractional derivative 4 right G-L fractional derivative 4 right R-L fractional derivative 4 shifted Grünwald-Letnikov (G-L) formula 11 space-fractional partial differential equations 269 time distributed-order subdiffusion equations 327 time fractional mixed diffusion and wave equations 240 time-fractional subdiffusion equation 109 time-fractional wave equation 173 time-space-fractional differential equations 297 weighted and shifted G-L (WSGL) formula 20, 22