Finite Difference Methods for Nonlinear Evolution Equations 311079585X, 9783110795851

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Zhi-Zhong Sun, Qifeng Zhang, and Guang-hua Gao Finite Difference Methods for Nonlinear Evolution Equations

De Gruyter Series in Applied and Numerical Mathematics

�

Edited by Rémi Abgrall, Zürich, Switzerland José Antonio Carrillo de la Plata, Oxford, United Kingdom Jean-Michel Coron, Paris, France Athanassios S. Fokas, Cambridge, United Kingdom Irene Fonseca, Pittsburgh, USA

Volume 8

Zhi-Zhong Sun, Qifeng Zhang, and Guang-hua Gao

Finite Difference Methods for Nonlinear Evolution Equations �

Mathematics Subject Classification 2020 Primary: 65Mxx; Secondary: 35Qxx Authors Prof. Zhi-Zhong Sun School of Mathematics Southeast University 210096 Nanjing People’s Republic of China [email protected] Prof. Qifeng Zhang Department of Mathematics Zhejiang Sci-Tech University 310018 Hangzhou People’s Republic of China [email protected]

Prof. Guang-hua Gao College of Science Nanjing University of Posts and Telecommunications 210023 Nanjing People’s Republic of China [email protected]

ISBN 978-3-11-079585-1 e-ISBN (PDF) 978-3-11-079601-8 e-ISBN (EPUB) 978-3-11-079611-7 ISSN 2512-1820 Library of Congress Control Number: 2022952125 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. © 2023 Walter de Gruyter GmbH, Berlin/Boston and co-pub China Science Publishing & Media Ltd, Beijing Typesetting: VTeX UAB, Lithuania Printing and binding: CPI books GmbH, Leck www.degruyter.com

Preface The study of nonlinear phenomena is concerned in the field of natural science and even social science. Since many phenomena in nature are essentially nonlinear, nonlinear problems have aroused the interest and concern of engineers, physicists, mathematicians and many others. In the mathematical and physical sciences, nonlinearity is the phenomenon in which the change in output is not proportional to that of input. A large part of nonlinear phenomena can be described by nonlinear partial differential equations, among which two typical examples are the Navier–Stokes equation in fluid mechanics and Schrödinger equation in quantum mechanics. There are more than 118 nonlinear partial differential equations listed on Wikipedia. The solution of the heat conduction equation with the Dirichlet boundary condition can be expressed as a linear combination of sinusoidal functions of different frequencies with time-dependent coefficients. The superposition principle makes it easy to solve linear problems. It is often possible to find several particular solutions for nonlinear problems, however, it is commonly very difficult to find general solutions from these particular solutions. In the process of computerization of science, as a tool, a method and a new subject, science and engineering computation has begun its new development. Numerical solutions of differential equations have also been developed in an unprecedented way. In this book, we study the difference methods to seek the numerical solutions by selecting 12 typical nonlinear partial differential equations. The 12 equations are respectively the Fisher equation, Burgers’ equation, regularized long-wave equation, Korteweg-de Vries equation, Camassa–Holm equation, Schrödinger equation, Kuramoto–Tsuzuki equation, Zakharov equation, Ginzburg–Landau equation, Cahn–Hilliard equation, epitaxial growth model and phase field crystal model. Several effective difference schemes are established for each problem. The existence, uniqueness, conservation, boundedness and convergence of the solution of each difference scheme are proved. The whole book is concise, hierarchical, gradually deepened in the level of difficulty, which is very suitable to be studied for primary scientific researchers. It is also ideal material for graduates to study and research. The main part of the book originates from a translation of the monograph “Finite difference methods for nonlinear evolution equations” in Chinese (Science Press, 2018) written by Professor Zhi-Zhong Sun with the following modifications. Difference methods of the Fisher equation are added as a new Chapter 1; In Chapter 2, L∞ error estimate of the solution to the initial-boundary value problem of the Burgers’ equation and to the two-level nonlinear implicit difference scheme is added in Section 2.1 and Section 2.2, respecitively; A new proposed compact difference scheme for the Burgers’ equation is added in Section 2.5. In Chapter 4, the convergence and unique solvability analyses of two second-order schemes for the Korteweg–de Vries equation are supplemented in Section 4.4 and Section 4.5. In Chapter 12, the proof of Theorem 12.4 is updated. In addition, https://doi.org/10.1515/9783110796018-201

VI � Preface we have supplemented and collected no more than two numerical examples by taking a difference scheme as an example in the penultimate section of each chapter. Zhi-Zhong Sun completed the main part of the book. Qifeng Zhang provided the translation of Chapters 2–9. He also supplemented and collected numerical examples in Chapters 2–12. Guang-hua Gao translated Chapters 1, 10–12 and supplemented numerical examples in Chapter 1. All of the authors have carefully checked and further polished the whole book. Before the monograph was fully published, Qifeng and Guang-hua read many parts of the contents. After more than 10 years of study and research, both authors have benefited from the analytical methods and excellent skills. Good knowledge production should be shared with the entire world. This is one of the main motivations for translating and rewriting the book. The publication of this book was supported in part by the National Natural Science Foundation of China (Grant No. 11671081) and the Natural Science Foundation of Zhejiang Province (Grant No. LZ23A010007). Most of the contents presented in this book originate from the work of the authors and collaborators. Here, we express our sincere thanks to all the collaborators. The authors are grateful to the editors of the press for their hard work. Due to the authors’ limited ability, mistakes will be inevitable. We sincerely hope that experts and readers may provide valuable advice and suggestions. November, 2022

Zhi-Zhong Sun Qifeng Zhang Guang-hua Gao

About the Authors Zhi-Zhong Sun

Born in March 1963, received his Bachelor’s degree from the Department of Mathematics, Nanjing University in 1984, Master’s degree from the Department of Mathematics, Nanjing University in 1987, and PhD from the Institute of Computational Mathematics and Scientific/Engineering Computing, Academy of Mathematics and Systems Science, Chinese Academy of Sciences in 1990. He is on the faculty at the School of Mathematics, Southeast University since 1990, and has been a full professor since April 1998 and a doctoral supervisor since July 2004. Professor Sun is an academic leader of the Jiangsu Province “Qinglan Project” and the Executive Director of the Computational Mathematics Society in the Jiangsu Province. He majors in computational mathematics and scientific/engineering computing, and is interested in the theory of difference methods in the numerical solution of partial differential equations. Professor Sun teaches computational methods, numerical analysis, numerical solutions of partial differential equations and numerical methods of nonlinear evolution equations. He has trained 32 master students, 12 doctoral students and 2 post-doctoral students. He has chaired five National Natural Science Foundation projects in China and one Natural Science Foundation project of the Jiangsu Province. Professor Sun has published 6 monographs, 3 textbooks and 5 auxiliary textbooks, and more than 160 regular research papers. He is a highly cited scholar of Elsevier in 2020 and 2021. The course of numerical analysis for engineering graduates was awarded as the outstanding graduate course of the Innovation Project for Graduate Education in the Jiangsu Province. He won the first prize of the Jiangsu Higher Education Teaching Achievement Award (Rank 6), Jiangsu Excellent Postgraduate Textbook Award, Jiangsu Science and Technology Award (Rank 2) and the title of the National Excellent Coach in Mathematical Modeling. Qifeng Zhang

Born in September 1987, received his PhD from the School of Mathematics and Statistics, Huazhong University of Science and Technology in 2014. He now is a faculty member in the Department of Mathematics, Zhejiang Sci-Tech University since 2014. He has been an associate professor since December 2017 and a master supervisor since June 2015. Professor Zhang engaged in postdoctoral research under the supervision of Professor Zhi-Zhong Sun during 2018–2021. During January 2020–January 2021, he visited Jan S. Hesthaven at the Ecole Polytechnique Federale de Lausanne. https://doi.org/10.1515/9783110796018-202

VIII � About the Authors

Professor Zhang majors in computational mathematics and scientific/engineering computing. He is now interested in the numerical solutions of partial differential equations. He teaches numerical analysis, numerical solutions of partial differential equations and linear algebra. As a project leader, Professor Zhang completed one National Natural Science Foundation project in China and chaired three Natural Science Foundation projects of the Zhejiang Province. He has coauthored one monograph and over 40 regular research papers. Guang-hua Gao

Born in November 1985, obtained her PhD from the Department of Mathematics, Southeast University in 2012, and is a faculty member at the College of Science, Nanjing University of Posts and Telecommunications since 2012. She has been an associate professor since September 2016 and a master supervisor since March 2014. During March 2014–September 2014, she visited Professor Hai-Wei Sun at the University of Macau. Professor Gao’s research interests are in the numerical solutions of partial differential equations, especially fractional differential equations in recent years. More than 30 academic papers have been published as an author or coauthor and three monographs on numerical solutions of fractional differential equations have been published as a coauthor. As a project leader, she has completed the research work of two National Natural Science Foundation projects in China and two Natural Science Foundation projects of the Jiangsu Province. Until now, Professor Gao has supervised four graduate students.

Contents Preface � V About the Authors � VII 1 1.1 1.2 1.3 1.3.1 1.3.2 1.4 1.4.1 1.4.2 1.5 1.5.1 1.5.2 1.6 1.6.1 1.6.2 1.7 1.7.1 1.7.2 1.8 1.9

Difference methods for the Fisher equation � 1 Introduction � 1 Notation and lemmas � 3 Forward Euler difference scheme � 11 Derivation of the difference scheme � 11 Solvability and convergence of the difference scheme � 12 Backward Euler difference scheme � 14 Derivation of the difference scheme � 14 Existence and convergence of the difference solution � 16 Crank–Nicolson difference scheme � 18 Derivation of the difference scheme � 19 Existence and convergence of the difference solution � 20 Fourth-order compact difference scheme � 22 Derivation of the difference scheme � 22 Existence and convergence of difference solution � 23 Three-level linearized difference scheme � 27 Derivation of the difference scheme � 27 Existence and convergence of the difference solution � 31 Numerical experiments � 36 Summary and extension � 38

2 2.1 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

Difference methods for the Burgers’ equation � 41 Introduction � 41 Two-level nonlinear difference scheme � 43 Derivation of the difference scheme � 43 Conservation and boundedness of the difference solution � 44 Existence and uniqueness of the difference solution � 46 Convergence of the difference solution � 49 Three-level linearized difference scheme � 54 Derivation of the difference scheme � 54 Existence and uniqueness of the difference solution � 55 Conservation and boundedness of the difference solution � 56 Convergence of the difference solution � 57 Hopf–Cole transformation and fourth-order difference scheme � 61 Hopf–Cole transformation � 61 Derivation of the difference scheme � 63

X � Contents 2.4.3 2.4.4 2.4.5 2.5 2.5.1 2.5.2 2.5.3 2.5.4 2.5.5 2.6 2.7

Existence and uniqueness of the difference solution � 65 Convergence of the difference solution � 67 Calculation of the solution of the original problem � 69 Fourth-order compact two-level nonlinear difference scheme � 70 Derivation of the difference scheme � 73 Conservation and boundedness of the difference solution � 74 Existence and uniqueness of the difference solution � 75 Convergence of the difference solution � 79 Stability of the difference solution � 84 Numerical experiments � 87 Summary and extension � 88

3 3.1 3.2 3.2.1 3.2.2 3.2.3 3.2.4 3.2.5 3.3 3.3.1 3.3.2 3.3.3 3.3.4 3.4 3.5

Difference methods for the regularized long-wave equation � 91 Introduction � 91 Two-level nonlinear difference scheme � 92 Derivation of the difference scheme � 92 Existence of the difference solution � 93 Conservation and boundedness of the difference solution � 94 Uniqueness of the difference solution � 95 Convergence of the difference solution � 96 Three-level linearized difference scheme � 98 Derivation of the difference scheme � 98 Conservation and boundedness of the difference solution � 99 Existence and uniqueness of the difference solution � 100 Convergence of the difference solution � 100 Numerical experiments � 103 Summary and extension � 104

4 4.1 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

Difference methods for the Korteweg–de Vries equation � 106 Introduction � 106 First-order in space two-level nonlinear difference scheme � 107 Derivation of the difference scheme � 107 Existence of the difference solution � 110 Conservation and boundedness of the difference solution � 112 Convergence of the difference solution � 113 First-order in space three-level linearized difference scheme � 115 Derivation of the difference scheme � 115 Existence and uniqueness of the difference solution � 116 Conservation and boundedness of the difference solution � 117 Convergence of the difference solution � 118 Second-order in space two-level nonlinear difference scheme � 123 Derivation of the difference scheme � 123

Contents �

4.4.2 4.4.3 4.4.4 4.5 4.5.1 4.5.2 4.5.3 4.5.4 4.6 4.7

Existence of the difference solution � 125 Conservation and boundedness of the difference solution � 127 Convergence and uniqueness of the difference solution � 128 Second-order in space three-level linearized difference scheme � 137 Derivation of the difference scheme � 137 Conservation and boundedness of the difference solution � 139 Existence and uniqueness of the difference solution � 141 Convergence of the difference solution � 143 Numerical experiments � 146 Summary and extension � 148

5 5.1 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 5.5

Difference methods for the Camassa–Holm equation � 151 Introduction � 151 Two-level nonlinear difference scheme � 152 Derivation of the difference scheme � 152 Conservation of the difference solution � 153 Existence and uniqueness of the difference solution � 154 Convergence of the difference solution � 157 Three-level linearized difference scheme � 159 Derivation of the difference scheme � 159 Conservation and boundedness of the difference solution � 160 Existence and uniqueness of the difference solution � 161 Convergence of the difference solution � 162 Numerical experiments � 168 Summary and extension � 172

6 6.1 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 6.4.2 6.4.3

Difference methods for the Schrödinger equation � 174 Introduction � 174 Two-level nonlinear difference scheme � 176 Derivation of the difference scheme � 176 Conservation and boundedness of the difference solution � 177 Existence and uniqueness of the difference solution � 180 Convergence of the difference solution � 182 Three-level linearized difference scheme � 188 Derivation of the difference scheme � 188 Conservation and boundedness of the difference solution � 189 Existence and uniqueness of the difference solution � 191 Convergence of the difference solution � 192 Fourth-order three-level linearized difference scheme � 200 Several numerical differential formulas � 200 Derivation of the difference scheme � 203 Existence and uniqueness of the difference solution � 205

XI

XII � Contents 6.4.4 6.4.5 6.5 6.6

Conservation and boundedness of the difference solution � 207 Convergence of the difference solution � 211 Numerical experiments � 216 Summary and extension � 219

7 7.1 7.2 7.2.1 7.2.2 7.2.3 7.2.4 7.2.5 7.3 7.3.1 7.3.2 7.3.3 7.3.4 7.4 7.5

Difference methods for the Kuramoto–Tsuzuki equation � 220 Introduction � 220 Two-level nonlinear difference scheme � 225 Derivation of the difference scheme � 225 Existence of the difference solution � 227 Boundedness of the difference solution � 228 Uniqueness of the difference solution � 233 Convergence of the difference solution � 234 Three-level linearized difference scheme � 237 Derivation of the difference scheme � 237 Boundedness of the difference solution � 239 Existence and uniqueness of the difference solution � 241 Convergence of the difference solution � 242 Numerical experiments � 246 Summary and extension � 247

8 8.1 8.2 8.2.1 8.2.2 8.2.3 8.2.4 8.3 8.3.1 8.3.2 8.3.3 8.3.4 8.4 8.5

Difference methods for the Zakharov equation � 249 Introduction � 249 Two-level nonlinear difference scheme � 253 Derivation of the difference scheme � 253 Existence of the difference solution � 255 Conservation and boundedness of the difference solution � 257 Convergence of the difference solution � 259 Three-level linearized locally decoupled difference scheme � 267 Derivation of the difference scheme � 267 Existence of the difference solution � 269 Conservation and boundedness of the difference solution � 270 Convergence of the difference solution � 274 Numerical experiments � 282 Summary and extension � 283

9 9.1 9.2 9.2.1 9.2.2 9.2.3

Difference methods for the Ginzburg–Landau equation � 285 Introduction � 285 Two-level nonlinear difference scheme � 286 Derivation of the difference scheme � 290 Existence of the difference solution � 291 Boundedness of the difference solution � 292

Contents �

9.2.4 9.3 9.3.1 9.3.2 9.3.3 9.3.4 9.4 9.5

Convergence of the difference solution � 294 Three-level linearized difference scheme � 298 Derivation of the difference scheme � 298 Existence of the difference solution � 299 Boundedness of the difference solution � 300 Convergence of the difference solution � 302 Numerical experiments � 307 Summary and extension � 308

10 10.1 10.2 10.2.1 10.2.2 10.2.3 10.2.4 10.3 10.3.1 10.3.2 10.3.3 10.4 10.4.1 10.4.2 10.4.3 10.5 10.6

Difference methods for the Cahn–Hilliard equation � 309 Introduction � 309 Two-level nonlinear difference scheme � 312 Derivation of the difference scheme � 315 Existence of the difference solution � 317 Boundedness of the difference solution � 319 Convergence of the difference solution � 320 Three-level linearized difference scheme � 326 Derivation of the difference scheme � 326 Existence and uniqueness of the difference solution � 327 Convergence of the difference solution � 328 Three-level linearized compact difference scheme � 336 Derivation of the difference scheme � 338 Existence and uniqueness of the difference solution � 340 Convergence of the difference solution � 342 Numerical experiments � 348 Summary and extension � 349

11 11.1 11.2 11.3 11.3.1 11.3.2 11.3.3 11.3.4 11.4 11.4.1 11.4.2 11.4.3 11.4.4 11.5 11.5.1

Difference methods for the epitaxial growth model � 351 Introduction � 351 Notation and basic lemmas � 352 Two-level nonlinear backward Euler difference scheme � 356 Derivation of the difference scheme � 356 Boundedness of the difference solution � 357 Existence and uniqueness of the difference solution � 359 Convergence of the difference solution � 362 Two-level linearized backward Euler difference scheme � 365 Derivation of the difference scheme � 366 Boundedness of the difference solution � 367 Existence of the difference solution � 367 Convergence of the difference solution � 368 Three-level linearized backward Euler difference scheme � 371 Derivation of the difference scheme � 371

XIII

XIV � Contents 11.5.2 11.5.3 11.5.4 11.6 11.7 12 12.1 12.2 12.3 12.3.1 12.3.2 12.3.3 12.3.4 12.4 12.4.1 12.4.2 12.4.3 12.5 12.6

Boundedness of the difference solution � 373 Existence of the difference solution � 376 Convergence of the difference solution � 377 Numerical experiments � 382 Summary and extension � 385 Difference methods for the phase field crystal model � 387 Introduction � 387 Notation and basic lemmas � 388 Two-level nonlinear difference scheme � 390 Derivation of the difference scheme � 390 Boundedness of the difference solution � 391 Existence and uniqueness of the difference solution � 393 Convergence of the difference solution � 396 Three-level linearized difference scheme � 399 Derivation of the difference scheme � 399 Energy stability of the difference solution � 400 Convergence of the difference solution � 402 Numerical experiments � 406 Summary and extension � 407

Bibliography � 411 Index � 415

1 Difference methods for the Fisher equation 1.1 Introduction The Fisher equation belongs to the class of reaction-diffusion equations. In fact, it is one of the simplest semilinear reaction-diffusion equations, the one which has the inhomogeneous term f (u) = λu(1 − u), which can exhibit traveling wave solutions that switch between equilibrium states given by f (u) = 0. Such an equation occurs, e. g., in ecology, physiology, combustion, crystallization, plasma physics and in general, phase transition problems. Fisher proposed this equation in 1937 to describe the spatial spread of an advantageous allele and explored its traveling wave solutions [12]. In the same year (1937) as Fisher, Kolmogorov, Petrovskii and Piskunov introduced a more general reaction-diffusion equation [18]. In this chapter, we consider the following initial and boundary value problem of a one-dimensional Fisher equation: ut − uxx = λu(1 − u), { { u(x, 0) = φ(x), { { { u(0, t) = α(t), u(L, t) = β(t),

0 < x < L, 0 < t ⩽ T, 0 ⩽ x ⩽ L, 0 < t ⩽ T,

(1.1) (1.2) (1.3)

where λ is a positive constant, functions φ(x), α(t), β(t) are all given and φ(0) = α(0), φ(L) = β(0). Suppose that the problem (1.1)–(1.3) has a smooth solution. Before introducing the difference scheme, a priori estimate on the solution of the problem (1.1)–(1.3) is given. Theorem 1.1. Let u(x, t) be the solution of the problem (1.1)–(1.3) with α(t) ≡ 0, β(t) ≡ 0. Denote L

2

E(t) = ∫ u (x, t)dx + 0

t

L

0

0

2 ∫[∫ ux2 (x, s)dx

L

L

L

+ λ ∫(u3 (x, s) − u2 (x, s))dx]ds, 0

t

L

0

0

2 F(t) = ∫ ux2 (x, t)dx + λ ∫[ u3 (x, t) − u2 (x, t)]dx + 2 ∫[∫ us2 (x, s)dx]ds. 3 0

0

Then E(t) = E(0),

F(t) = F(0),

0 < t ⩽ T.

Proof. (I) Multiplying both the right- and left-hand sides of (1.1) by u(x, t) gives u(x, t)ut (x, t) − u(x, t)uxx (x, t) + λ[u3 (x, t) − u2 (x, t)] = 0, i. e., https://doi.org/10.1515/9783110796018-001

2 � 1 Difference methods for the Fisher equation 1 d 2 [u (x, t)] − (u(x, t)ux (x, t))x + ux2 (x, t) + λ[u3 (x, t) − u2 (x, t)] = 0. 2 dt Integrating both the right- and left-hand sides with respect to x on the interval [0, L] and noticing (1.3) with α(t) = β(t) = 0, we have L

L

L

0

0

0

1 d ∫ u2 (x, t)dx + ∫ ux2 (x, t)dx + λ ∫[u3 (x, t) − u2 (x, t)]dx = 0, 2 dt which can be rewritten as L

t

L

L

0

0

0

0

d {∫ u2 (x, t)dx + 2 ∫[∫ ux2 (x, s)dx + λ ∫(u3 (x, s) − u2 (x, s))dx]ds} = 0. dt Then E(t) = E(0) is obtained. (II) Multiplying both the right- and left-hand sides of (1.1) by ut (x, t) yields ut2 (x, t) − ut (x, t)uxx (x, t) − λ[u(x, t) − u2 (x, t)]ut (x, t) = 0, i. e., 1 1 1 ut2 (x, t) − (ut (x, t)ux (x, t))x + ( ux2 (x, t)) + λ[ u3 (x, t) − u2 (x, t)] = 0. 2 3 2 t t Integrating both the right- and left-hand sides with respect to x on the interval [0, L] and noticing (1.3) with α(t) = β(t) = 0, we have L

L

L

0

0

0

1 d d 1 1 ∫ ux2 (x, t)dx + λ ∫[ u3 (x, t) − u2 (x, t)]dx + ∫ ut2 (x, t)dx = 0, 2 dt dt 3 2 which can be rewritten as L

L

t

L

0

0

0

0

d 2 [∫ ux2 (x, t)dx + λ ∫( u3 (x, t) − u2 (x, t))dx + 2 ∫(∫ us2 (x, s)dx)ds] = 0, dt 3 i. e., dF(t) = 0, dt Thus, F(t) = F(0) is followed.

0 < t ⩽ T.

1.2 Notation and lemmas

�

3

1.2 Notation and lemmas In order to derive the difference scheme, we first divide the domain [0, L] × [0, T]. Take two positive integers m, n. Divide [0, L] into m equal subintervals, and [0, T] into n subintervals. Denote h = L/m, τ = T/n; xi = ih, 0 ⩽ i ⩽ m; tk = kτ, 0 ⩽ k ⩽ n; Ωh = {xi | 0 ⩽ i ⩽ m}, Ωτ = {tk | 0 ⩽ k ⩽ n}; Ωhτ = Ωh × Ωτ . We call all of the nodes {(xi , tk ) | 0 ⩽ i ⩽ m} on the line t = tk the k-th time-level nodes. In addition, denote xi+ 1 = 21 (xi + xi+1 ), tk+ 1 = 21 (tk + tk+1 ), r = hτ2 . 2 2 Denote 𝒰h = {u | u = (u0 , u1 , . . . , um ) is the grid function defined on Ωh }, 𝒰h̊ = {u | u ∈ 𝒰h , u0 = um = 0}.

For any grid function u ∈ 𝒰h , introduce the following notation: δx ui+ 1 = 2

1 (u − ui ), h i+1

δx2 ui =

1 (ui−1 − 2ui + ui+1 ), h2

Δx ui =

1 (u − ui−1 ). 2h i+1

It follows easily that δx2 ui =

1 (δ u 1 − δx ui− 1 ), 2 h x i+ 2

1 Δx ui = (δx ui− 1 + δx ui+ 1 ). 2 2 2

Suppose u, v ∈ 𝒰h . Introduce the inner products, norms and seminorms as m−1 1 1 (u, v) = h( u0 v0 + ∑ ui vi + um vm ), 2 2 i=1 m

⟨δx u, δx v⟩ = h ∑(δx ui− 1 )(δx vi− 1 ), 2

i=1

‖u‖∞ = max |ui |,

‖u‖ = √(u, u),

0⩽i⩽m

2

|u|2 = √h ∑ (δx2 ui ) , i=1

‖δx u‖∞ = max |δx ui− 1 |, 1⩽i⩽m

2

‖u‖1 = √‖u‖2 + |u|21 ,

|u|1 = √⟨δx u, δx u⟩, m−1

2

‖u‖2 = √‖u‖2 + |u|21 + |u|22 .

If 𝒰h is a complex space, then the corresponding inner product is defined by m−1 1 1 (u, v) = h( u0 v̄0 + ∑ ui v̄i + um v̄m ), 2 2 i=1

with v̄i the conjugate of vi .

4 � 1 Difference methods for the Fisher equation Denote 0

1

n

𝒮τ = {w | w = (w , w , . . . , w ) is the grid function defined on Ωτ }.

For any w ∈ 𝒮τ , introduce the following notation: 1 ̄ 1 1 wk+ 2 = (wk + wk+1 ), wk = (wk+1 + wk−1 ), 2 2 1 1 Dt wk = (wk+1 − wk ), Dt wk = (wk − wk−1 ), τ τ 1 1 1 δt wk+ 2 = (wk+1 − wk ), Δt wk = (wk+1 − wk−1 ). τ 2τ

It is easy to know that 1 1 1 Δt wk = (δt wk− 2 + δt wk+ 2 ). 2

Suppose u = {uik | 0 ⩽ i ⩽ m, 0 ⩽ k ⩽ n} is a grid function defined on Ωhτ , then v = {uik | 0 ⩽ i ⩽ m} is a grid function defined on Ωh , w = {uik | 0 ⩽ k ⩽ n} is a grid function defined on Ωτ . Lemma 1.1 ([25, 35]). (a) Suppose u, v ∈ 𝒰h , then m−1

m

i=1

i=1

−h ∑ (δx2 ui )vi = h ∑(δx ui− 1 )(δx vi− 1 ) + (δx u 1 )v0 − (δx um− 1 )vm . 2

2

2

2

(b) Suppose u ∈ 𝒰h̊ , then m−1

−h ∑ (δx2 ui )ui = |u|21 , i=1

|u|21 ⩽ ‖u‖ ⋅ |u|2 , √L |u|1 , 2 L |u| . ‖u‖ ⩽ √6 1

‖u‖∞ ⩽

(c) Suppose u ∈ 𝒰h̊ , then ‖u‖2∞ ⩽ ‖u‖ ⋅ |u|1 , and for arbitrary ε > 0, it holds that ‖u‖∞ ⩽ ε|u|1 +

1 ‖u‖, 4ε

‖u‖2∞ ⩽ ε|u|21 +

1 ‖u‖2 . 4ε

1.2 Notation and lemmas

(d) Suppose u ∈ 𝒰h , then |u|21 ⩽

4 ‖u‖2 . h2

(e) Suppose u ∈ 𝒰h , then ‖u‖2∞ ⩽ 2‖u‖ ⋅ |u|1 +

1 ‖u‖2 , L

and for arbitrary ε > 0, it holds that 1 1 ‖u‖2∞ ⩽ ε|u|21 + ( + )‖u‖2 . ε L (f) Suppose u ∈ 𝒰h , then for arbitrary ε > 0, it holds that 1 1 ‖δx u‖2∞ ⩽ ε|u|22 + ( + )|u|21 . ε L Proof. We only prove (c) and (e). (c) Noticing that u0 = 0, when 1 ⩽ i ⩽ m − 1, we have i

i

i

l=1

l=1

l=1

2 ) = ∑(ul + ul−1 )(ul − ul−1 ) = 2h ∑ ul− 1 δx ul− 1 . ui2 = ∑(ul2 − ul−1 2

2

Hence, i

ui2 ⩽ 2h ∑ |ul− 1 | ⋅ |δx ul− 1 |. l=1

2

2

Similarly, noticing that um = 0, we have m

ui2 ⩽ 2h ∑ |ul− 1 | ⋅ |δx ul− 1 |. l=i+1

2

2

Adding the above two inequalities together, we have m

m

m

ui2 ⩽ h ∑ |ul− 1 | ⋅ |δx ul− 1 | ⩽ √h ∑ |ul− 1 |2 ⋅ √h ∑ |δx ul− 1 |2 ⩽ ‖u‖ ⋅ |u|1 . l=1

2

2

l=1

2

It follows that ‖u‖2∞ ⩽ ‖u‖ ⋅ |u|1 . For arbitrary ε > 0, then

l=1

2

�

5

6 � 1 Difference methods for the Fisher equation ‖u‖∞ ⩽ √‖u‖ ⋅ |u|1 ⩽ ε|u|1 + ‖u‖2∞ ⩽ ‖u‖ ⋅ |u|1 ⩽ ε|u|21 +

1 ‖u‖, 4ε

1 ‖u‖2 . 4ε

(e) When i > j, i

2 ui2 = uj2 + ∑ (ul2 − ul−1 ) l=j+1

i

= uj2 + 2h ∑ ul− 1 δx ul− 1 2

l=j+1

2

i

⩽ uj2 + 2h ∑ |ul− 1 | ⋅ |δx ul− 1 | 2

l=j+1

2

m

⩽ uj2 + 2h ∑ |ul− 1 | ⋅ |δx ul− 1 | ⩽

uj2

l=1

2

2

+ 2‖u‖ ⋅ |u|1 .

(1.4)

It is easy to know that the above result holds also for i ⩽ j. Denote ωj = {

1,

1 , 2

1 ⩽ j ⩽ m − 1, j = 0, m.

Multiplying (1.4) by hωj on both the right- and left-hand sides and summing up for j from 0 to m, we have m

m

m

j=0

j=0

j=0

h ∑ ωj ui2 ⩽ h ∑ ωj uj2 + 2h ∑ ωj ‖u‖ ⋅ |u|1 . It easily follows that L‖u‖2∞ ⩽ ‖u‖2 + 2L‖u‖ ⋅ |u|1 , namely, ‖u‖2∞ ⩽ 2‖u‖ ⋅ |u|1 +

1 ‖u‖2 . L

For arbitrary ε > 0, we have 1 1 ‖u‖2∞ ⩽ ε|u|21 + ( + )‖u‖2 . ε L

1.2 Notation and lemmas

� 7

Similar results hold for the continuous functions. Next, we will give several commonly used numerical differential formulas. Lemma 1.2 ([35]). Let c, h be given constants and h > 0. (a) If g(x) ∈ C 2 [c − h, c + h], then h2 1 g(c) = [g(c − h) + g(c + h)] − g ′′ (ξ0 ), 2 2

c − h < ξ0 < c + h;

(b) If g(x) ∈ C 2 [c, c + h], then g ′ (c) =

1 h [g(c + h) − g(c)] − g ′′ (ξ1 ), h 2

c < ξ1 < c + h;

(c) If g(x) ∈ C 2 [c − h, c], then g ′ (c) =

h 1 [g(c) − g(c − h)] + g ′′ (ξ2 ), h 2

c − h < ξ2 < c;

(d) If g(x) ∈ C 3 [c − h, c + h], then g ′ (c) =

1 h2 [g(c + h) − g(c − h)] − g ′′′ (ξ3 ), 2h 6

c − h < ξ3 < c + h;

(e) If g(x) ∈ C 4 [c − h, c + h], then g ′′ (c) =

1 h2 [g(c + h) − 2g(c) + g(c − h)] − g (4) (ξ4 ), 2 12 h

c − h < ξ4 < c + h;

(f) If g(x) ∈ C 3 [c, c + h], then g ′′ (c) =

2 g(c + h) − g(c) h [ − g ′ (c)] − g ′′′ (ξ5 ), h h 3

c < ξ5 < c + h;

If g(x) ∈ C 4 [c, c + h], then g ′′ (c) =

2 g(c + h) − g(c) h h2 [ − g ′ (c)] − g ′′′ (c) − g (4) (ξ6 ), h h 3 12

c < ξ6 < c + h;

(g) If g(x) ∈ C 3 [c − h, c], then g ′′ (c) =

g(c) − g(c − h) 2 ′ h [g (c) − ] + g ′′′ (ξ7 ), h h 3

c − h < ξ7 < c;

If g(x) ∈ C 4 [c − h, c], then g ′′ (c) =

g(c) − g(c − h) 2 ′ h h2 [g (c) − ] + g ′′′ (c) − g (4) (ξ8 ), h h 3 12

c − h < ξ8 < c;

8 � 1 Difference methods for the Fisher equation (h) If g(x) ∈ C 6 [c − h, c + h], then h4 (6) 1 ′′ 1 [g (c − h) + 10g ′′ (c) + g ′′ (c + h)] = 2 [g(c + h) − 2g(c) + g(c − h)] + g (ξ9 ), 12 240 h c − h < ξ9 < c + h. Now let us introduce some important Gronwall inequalities. Theorem 1.2. (a) Suppose {F k }∞ k=0 is a nonnegative sequence; c and g are two nonnegative constants satisfying F k+1 ⩽ (1 + cτ)F k + τg,

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

then F k ⩽ eckτ (F 0 +

g ), c

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

k ∞ (b) Suppose {F k }∞ k=0 and {g }k=0 are two nonnegative sequences; c is a nonnegative constant satisfying

F k+1 ⩽ (1 + cτ)F k + τg k ,

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

then k−1

F k ⩽ eckτ (F 0 + τ ∑ g l ), l=0

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

(c) Suppose {F k }∞ k=0 is a nonnegative sequence; c and g are two nonnegative constants satisfying k−1

F k ⩽ cτ ∑ F l + g,

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

l=0

then F k ⩽ eckτ g,

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

k ∞ (d) Suppose {F k }∞ k=0 is a nonnegative sequence and {g }k=0 is nonnegative monotonically increasing (allowed not strictly monotonic) sequence satisfying k−1

F k ⩽ cτ ∑ F l + g k , l=0

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

then F k ⩽ eckτ g k ,

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

1.2 Notation and lemmas

�

Proof. (a) F k+1 ⩽ (1 + cτ)F k + τg

⩽ (1 + cτ)[(1 + cτ)F k−1 + τg] + τg

= (1 + cτ)2 F k−1 + [(1 + cτ) + 1]τg

⩽ (1 + cτ)2 [(1 + cτ)F k−2 + τg] + [(1 + cτ) + 1]τg

= (1 + cτ)3 F k−2 + [(1 + cτ)2 + (1 + cτ) + 1]τg ⩽ ⋅⋅⋅

⩽ (1 + cτ)k F 1 + [(1 + cτ)k−1 + (1 + cτ)k−2 + ⋅ ⋅ ⋅ + 1]τg

⩽ (1 + cτ)k [(1 + cτ)F 0 + τg] + [(1 + cτ)k−1 + (1 + cτ)k−2 + ⋅ ⋅ ⋅ + 1]τg = (1 + cτ)k+1 F 0 + [(1 + cτ)k + (1 + cτ)k−1 + ⋅ ⋅ ⋅ + 1]τg = (1 + cτ)k+1 F 0 + ⩽ ec(k+1)τ (F 0 +

(1 + cτ)k+1 − 1 ⋅ τg cτ

g ), c

k = 0, 1, . . . .

(b) F k+1 ⩽ (1 + cτ)F k + τg k

⩽ (1 + cτ)[(1 + cτ)F k−1 + τg k−1 ] + τg k = (1 + cτ)2 F k−1 + (1 + cτ)τg k−1 + τg k

⩽ (1 + cτ)2 [(1 + cτ)F k−2 + τg k−2 ] + (1 + cτ)τg k−1 + τg k = (1 + cτ)3 F k−2 + (1 + cτ)2 τg k−2 + (1 + cτ)τg k−1 + τg k

⩽ (1 + cτ)3 [(1 + cτ)F k−3 + τg k−3 ] + (1 + cτ)2 τg k−2 + (1 + cτ)τg k−1 + τg k = (1 + cτ)4 F k−3 + (1 + cτ)3 τg k−3 + (1 + cτ)2 τg k−2 + (1 + cτ)τg k−1 + τg k ⩽ ⋅⋅⋅ k

⩽ (1 + cτ)k+1 F 0 + τ ∑(1 + cτ)k−l g l l=0

k

k

l=0

l=0

⩽ (1 + cτ)k+1 (F 0 + τ ∑ g l ) ⩽ ec(k+1)τ (F 0 + τ ∑ g l ), (c) It is easy to know that F 0 ⩽ g. Let

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

9

10 � 1 Difference methods for the Fisher equation k−1

Gk = cτ ∑ F l + g, l=0

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

Then G0 = g,

F k ⩽ Gk , k

G =G

k−1

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

+ cτF k−1 ⩽ Gk−1 + cτGk−1 = (1 + cτ)Gk−1 ,

k = 1, 2, 3, . . . ,

by recursion, we have Gk ⩽ (1 + cτ)k G0 ⩽ eckτ g,

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

so that F k ⩽ Gk ⩽ eckτ g,

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

(d) It is easy to know that F 0 ⩽ g0. Let k−1

Gk = cτ ∑ F l + g k , l=0

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

then G0 = g 0 ,

F k ⩽ Gk ,

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

k−2

Gk = cτ ∑ F l + g k−1 + cτF k−1 + (g k − g k−1 ) l=0

= Gk−1 + cτF k−1 + (g k − g k−1 ) ⩽ (1 + cτ)Gk−1 + (g k − g k−1 ),

k = 1, 2, . . . .

Applying the result of (b), we have k

F k ⩽ Gk ⩽ eckτ [G0 + ∑(g l − g l−1 )] = eckτ g k , l=1

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

1.3 Forward Euler difference scheme

� 11

1.3 Forward Euler difference scheme 1.3.1 Derivation of the difference scheme Define the grid function U = {Uik | 0 ⩽ i ⩽ m, 0 ⩽ k ⩽ n} on Ωhτ , where Uik = u(xi , tk ),

0 ⩽ i ⩽ m, 0 ⩽ k ⩽ n.

Denote 󵄨 󵄨 c0 = max󵄨󵄨󵄨u(x, t)󵄨󵄨󵄨. 0⩽x⩽L 0⩽t⩽T

Considering equation (1.1) at the point (xi , tk ), we have ut (xi , tk ) − uxx (xi , tk ) = λu(xi , tk )[1 − u(xi , tk )],

1 ⩽ i ⩽ m − 1, 0 ⩽ k ⩽ n − 1.

(1.5)

With the help of Lemma 1.2, we have 1 k+1 (U − Uik ) + O(τ) = Dt Uik + O(τ), τ i 1 k k − 2Uik + Ui−1 ) + O(h2 ) = δx2 Uik + O(h2 ), uxx (xi , tk ) = 2 (Ui+1 h u(xi , tk ) = u(xi , tk+1 ) + O(τ) = Uik+1 + O(τ). ut (xi , tk ) =

(1.6) (1.7) (1.8)

Substituting (1.6)–(1.8) into (1.5) arrives at Dt Uik − δx2 Uik = λ(Uik − Uik Uik+1 ) + (R1 )ki ,

1 ⩽ i ⩽ m − 1, 0 ⩽ k ⩽ n − 1,

(1.9)

where there is a constant c1 such that 󵄨󵄨 k󵄨 2 󵄨󵄨(R1 )i 󵄨󵄨󵄨 ⩽ c1 (τ + h ),

1 ⩽ i ⩽ m − 1, 0 ⩽ k ⩽ n − 1.

(1.10)

Noticing the initial-boundary value conditions (1.2)–(1.3), we have {

Ui0 = φ(xi ), U0k

= α(tk ),

Umk

= β(tk ),

0 ⩽ i ⩽ m, 1 ⩽ k ⩽ n.

(1.11) (1.12)

Neglecting the small term (R1 )ki in (1.9) and replacing the exact solution Uik by its numerical one uik , the following forward Euler difference scheme is obtained as Dt uk − δx2 uik = λ(uik − uik uik+1 ), { { { 0 i ui = φ(xi ), { { { k k { u0 = α(tk ), um = β(tk ), It is easy to get the following conclusion.

1 ⩽ i ⩽ m − 1, 0 ⩽ k ⩽ n − 1, 0 ⩽ i ⩽ m, 1 ⩽ k ⩽ n.

(1.13) (1.14) (1.15)

12 � 1 Difference methods for the Fisher equation Theorem 1.3 ([29]). Let {uik | 0 ⩽ i ⩽ m, 0 ⩽ k ⩽ n} be the solution of the difference scheme (1.13)–(1.15). If 0 ⩽ φ(x) ⩽ 1, 0 ⩽ α(t) ⩽ 1, 0 ⩽ β(t) ⩽ 1 and r ⩽ 21 , then it holds that 0 ⩽ uik ⩽ 1,

0 ⩽ i ⩽ m, 0 ⩽ k ⩽ n.

Proof. Reformulate (1.13) as k k (1 + λτuik )uik+1 = (1 − 2r)uik + r(ui−1 + ui+1 ) + λτuik ,

1 ⩽ i ⩽ m − 1, 0 ⩽ k ⩽ n − 1,

1 k k [(1 − 2r)uik + r(ui−1 + ui+1 ) + λτuik ], 1 + λτuik

1 ⩽ i ⩽ m − 1, 0 ⩽ k ⩽ n − 1.

or uik+1 =

If 0 ⩽ uik ⩽ 1, 0 ⩽ i ⩽ m and r ⩽ 21 , then we have uik+1 ⩾ 0,

1⩽i ⩽m−1

and uik+1 ⩽

1 [(1 − 2r) × 1 + r × (1 + 1) + λτuik ] = 1, 1 + λτuik

1 ⩽ i ⩽ m − 1.

By induction, the conclusion is true.

1.3.2 Solvability and convergence of the difference scheme Theorem 1.4. Let {Uik | 0 ⩽ i ⩽ m, 0 ⩽ k ⩽ n} and {uik | 0 ⩽ i ⩽ m, 0 ⩽ k ⩽ n} be solutions of the problem (1.1)–(1.3) and the difference scheme (1.13)–(1.15), respectively. Denote eik = Uik − uik , 0 ⩽ i ⩽ m, 0 ⩽ k ⩽ n, c1 c2 = e3λ(c0 +1)T . 2λ(c0 + 1) Then, when r ⩽ 21 , c2 (τ + h2 ) ⩽ 1 and λ(c0 + 1)τ ⩽ 31 , it holds that (I) the solution of the difference scheme (1.13)–(1.15) exists; (II) 󵄩󵄩 k 󵄩󵄩 2 󵄩󵄩e 󵄩󵄩∞ ⩽ c2 (τ + h ),

0 ⩽ k ⩽ n.

(1.16)

1.3 Forward Euler difference scheme

�

13

Proof. Subtracting (1.13)–(1.15) from (1.9), (1.11) and (1.12), respectively, the system of error equations can be produced as Dt ek − δx2 eik = λeik − λ(Uik Uik+1 − uik uik+1 ) + (R1 )ki , { { { 0 i ei = 0, { { { k k { e0 = 0, em = 0,

1 ⩽ i ⩽ m − 1, 0 ⩽ k ⩽ n − 1, (1.17) 0 ⩽ i ⩽ m,

(1.18)

1 ⩽ k ⩽ n.

(1.19)

Rewrite (1.17) as k k eik+1 = (1 − 2r)eik + r(ei−1 + ei+1 ) + λτeik − λτ(uik eik+1 + eik Uik+1 ) + τ(R1 )ki ,

1 ⩽ i ⩽ m − 1, 0 ⩽ k ⩽ n − 1.

(1.20)

When r ⩽ 21 , taking the absolute value on both the right- and left-hand sides of (1.20) and using the triangle inequality, with the help of (1.10), we have 󵄨󵄨 k+1 󵄨󵄨 󵄩 k󵄩 󵄩 k󵄩 󵄩 k󵄩 󵄩 k󵄩 󵄨󵄨ei 󵄨󵄨 ⩽ (1 − 2r)󵄩󵄩󵄩e 󵄩󵄩󵄩∞ + r(󵄩󵄩󵄩e 󵄩󵄩󵄩∞ + 󵄩󵄩󵄩e 󵄩󵄩󵄩∞ ) + λτ 󵄩󵄩󵄩e 󵄩󵄩󵄩∞ 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 + λτ(󵄩󵄩󵄩uk 󵄩󵄩󵄩∞ 󵄩󵄩󵄩ek+1 󵄩󵄩󵄩∞ + 󵄩󵄩󵄩ek 󵄩󵄩󵄩∞ 󵄩󵄩󵄩U k+1 󵄩󵄩󵄩∞ ) + c1 τ(τ + h2 ) 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 = (1 + λτ)󵄩󵄩󵄩ek 󵄩󵄩󵄩∞ + λτ(󵄩󵄩󵄩uk 󵄩󵄩󵄩∞ 󵄩󵄩󵄩ek+1 󵄩󵄩󵄩∞ + c0 󵄩󵄩󵄩ek 󵄩󵄩󵄩∞ ) + c1 τ(τ + h2 ) 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 = [1 + λ(c0 + 1)τ]󵄩󵄩󵄩ek 󵄩󵄩󵄩∞ + λτ 󵄩󵄩󵄩uk 󵄩󵄩󵄩∞ 󵄩󵄩󵄩ek+1 󵄩󵄩󵄩∞ + c1 τ(τ + h2 ), 1 ⩽ i ⩽ m − 1. It follows by noticing (1.19) that 󵄩󵄩 k+1 󵄩󵄩 󵄩 k󵄩 󵄩 k 󵄩 󵄩 k+1 󵄩 2 󵄩󵄩e 󵄩󵄩∞ ⩽ [1 + λ(c0 + 1)τ]󵄩󵄩󵄩e 󵄩󵄩󵄩∞ + λτ 󵄩󵄩󵄩u 󵄩󵄩󵄩∞ 󵄩󵄩󵄩e 󵄩󵄩󵄩∞ + c1 τ(τ + h ),

0 ⩽ k ⩽ n − 1. (1.21)

In view of (1.18), 󵄩󵄩 0 󵄩󵄩 󵄩󵄩e 󵄩󵄩∞ = 0, which implies the truth of (1.16) for k = 0. Now assume that (1.16) is true for 0 ⩽ k ⩽ l, i. e., 󵄩󵄩 k 󵄩󵄩 2 󵄩󵄩e 󵄩󵄩∞ ⩽ c2 (τ + h ),

0 ⩽ k ⩽ l.

Then noticing eik = Uik − uik , when c2 (τ + h2 ) ⩽ 1, it follows that 󵄩󵄩 k 󵄩󵄩 󵄩 k󵄩 󵄩 k󵄩 󵄩󵄩u 󵄩󵄩∞ ⩽ 󵄩󵄩󵄩U 󵄩󵄩󵄩∞ + 󵄩󵄩󵄩e 󵄩󵄩󵄩∞ ⩽ c0 + 1,

0 ⩽ k ⩽ l.

Considering (1.13) with k = l and noticing 1 + λτuil ⩾ 1 − λτ(c0 + 1) ⩾ 32 , we obtain uil+1 =

1 l l [(1 − 2r)uil + r(ui−1 + ui+1 ) + λτuil ], 1 + λτuil

1 ⩽ i ⩽ m − 1,

14 � 1 Difference methods for the Fisher equation which means that ul+1 can be solved explicitly and uniquely. In addition, by (1.21), we have 󵄩󵄩 k+1 󵄩󵄩 󵄩 k󵄩 󵄩 k+1 󵄩 2 󵄩󵄩e 󵄩󵄩∞ ⩽ [1 + λ(c0 + 1)τ]󵄩󵄩󵄩e 󵄩󵄩󵄩∞ + λτ(c0 + 1)󵄩󵄩󵄩e 󵄩󵄩󵄩∞ + c1 τ(τ + h ),

0 ⩽ k ⩽ l,

i. e., 󵄩 󵄩 󵄩 󵄩 [1 − λτ(c0 + 1)]󵄩󵄩󵄩ek+1 󵄩󵄩󵄩∞ ⩽ [1 + λ(c0 + 1)τ]󵄩󵄩󵄩ek 󵄩󵄩󵄩∞ + c1 τ(τ + h2 ),

0 ⩽ k ⩽ l.

When λτ(c0 + 1) ⩽ 31 , we have 3 󵄩󵄩 k+1 󵄩󵄩 󵄩 k󵄩 2 󵄩󵄩e 󵄩󵄩∞ ⩽ [1 + 3λ(c0 + 1)τ]󵄩󵄩󵄩e 󵄩󵄩󵄩∞ + c1 τ(τ + h ), 2

0 ⩽ k ⩽ l.

The application of the Gronwall inequality (Theorem 1.2(a)) yields c1 󵄩󵄩 l+1 󵄩󵄩 3λ(c +1)T ⋅ (τ + h2 ) = c2 (τ + h2 ), 󵄩󵄩e 󵄩󵄩∞ ⩽ e 0 2λ(c0 + 1) from which (1.16) also holds for k = l + 1. By induction, (1.16) is true for all k (0 ⩽ k ⩽ n).

1.4 Backward Euler difference scheme The forward Euler scheme requires the step size ratio r ⩽ 21 , which implies that the temporal step size must be much smaller than the spatial one. Next, an unconditionally stable difference scheme will be introduced. 1.4.1 Derivation of the difference scheme Considering equation (1.1) at the node point (xi , tk+1 ), we have ut (xi , tk+1 ) − uxx (xi , tk+1 ) = λu(xi , tk+1 )[1 − u(xi , tk+1 )],

1 ⩽ i ⩽ m − 1, 0 ⩽ k ⩽ n − 1. (1.22)

With the help of Lemma 1.2, we have 1 k+1 (U − Uik ) + O(τ) = Dt Uik+1 + O(τ), τ i 1 k+1 k+1 uxx (xi , tk+1 ) = 2 (Ui+1 − 2Uik+1 + Ui−1 ) + O(h2 ) = δx2 Uik+1 + O(h2 ), h u(xi , tk+1 ) = u(xi , tk ) + O(τ) = Uik + O(τ). ut (xi , tk+1 ) =

Substituting (1.23)–(1.25) into (1.22) arrives at

(1.23) (1.24) (1.25)

1.4 Backward Euler difference scheme

Dt Uik+1 − δx2 Uik+1 = λ(Uik − Uik Uik+1 ) + (R2 )k+1 i ,

1 ⩽ i ⩽ m − 1, 0 ⩽ k ⩽ n − 1,

�

15

(1.26)

where there is a constant c3 such that 󵄨󵄨 k+1 󵄨 2 󵄨󵄨(R2 )i 󵄨󵄨󵄨 ⩽ c3 (τ + h ),

1 ⩽ i ⩽ m − 1, 0 ⩽ k ⩽ n − 1.

(1.27)

Noticing the initial-boundary value conditions (1.2)–(1.3), we have {

Ui0 = φ(xi ), U0k

= α(tk ),

Umk

= β(tk ),

0 ⩽ i ⩽ m, 1 ⩽ k ⩽ n.

(1.28) (1.29)

Neglecting the small term (R2 )k+1 in (1.26) and replacing the exact solution Uik by its i k numerical one ui , the backward Euler difference scheme reads D uk+1 − δx2 uik+1 = λ(uik − uik uik+1 ), { { { 0t i ui = φ(xi ), { { { k k { u0 = α(tk ), um = β(tk ),

1 ⩽ i ⩽ m − 1, 0 ⩽ k ⩽ n − 1,

(1.30)

1 ⩽ k ⩽ n.

(1.32)

0 ⩽ i ⩽ m,

(1.31)

Note that (1.30)–(1.32) is a two-level linearized difference scheme. It is easy to get the following conclusion. Theorem 1.5 ([29]). Let {uik | 0 ⩽ i ⩽ m, 0 ⩽ k ⩽ n} be the solution of the difference scheme (1.30)–(1.32). If 0 ⩽ φ(x) ⩽ 1, 0 ⩽ α(t) ⩽ 1 and 0 ⩽ β(t) ⩽ 1, then it holds that 0 ⩽ uik ⩽ 1,

0 ⩽ i ⩽ m, 0 ⩽ k ⩽ n.

Proof. Rewrite (1.30) as k+1 k+1 (1 + 2r + λτuik )uik+1 = r(ui−1 + ui+1 ) + uik + λτuik ,

1 ⩽ i ⩽ m − 1, 0 ⩽ k ⩽ n − 1. (1.33)

Suppose 0 ⩽ uik ⩽ 1, 0 ⩽ i ⩽ m and notice 0 ⩽ α(tk+1 ) ⩽ 1, 0 ⩽ β(tk+1 ) ⩽ 1. Denote min uik+1 = uik+1 , ∗

0⩽i⩽m

max uik+1 = uik+1 ∗ .

0⩽i⩽m

If i∗ ≠ 0, m, letting i = i∗ in (1.33), we have (1 + 2r + λτuik∗ )uik+1 = r(uik+1 + uik+1 ) + uik∗ + λτuik∗ ⩾ 2ruik+1 + uik∗ + λτuik∗ , ∗ ∗ −1 ∗ +1 ∗ i. e., (1 + λτuik∗ )uik+1 ⩾ (1 + λτ)uik∗ , ∗ which implies

16 � 1 Difference methods for the Fisher equation uik+1 ⩾ 0. ∗ If i∗ ≠ 0, m, letting i = i∗ in (1.33), we have k+1 k k k+1 (1 + 2r + λτuik∗ )uik+1 = r(uik+1 + uik∗ + λτuik∗ , ∗ ∗ −1 + ui∗ +1 ) + ui∗ + λτui∗ ⩽ 2rui∗

i. e., (1 + λτuik∗ )uik+1 ⩽ uik∗ + λτuik∗ ⩽ 1 + λτuik∗ , ∗ which implies uik+1 ⩽ 1. ∗ The result 0 ⩽ uik+1 ⩽ 1 is followed. By induction, the proof is completed. 1.4.2 Existence and convergence of the difference solution Theorem 1.6. Let {Uik | 0 ⩽ i ⩽ m, 0 ⩽ k ⩽ n} and {uik | 0 ⩽ i ⩽ m, 0 ⩽ k ⩽ n} be solutions of the problem (1.1)–(1.3) and the difference scheme (1.30)–(1.32), respectively. Denote eik = Uik − uik , 0 ⩽ i ⩽ m, 0 ⩽ k ⩽ n, c3 c4 = e3λ(c0 +1)T . 2λ(c0 + 1) Then, when c4 (τ + h2 ) ⩽ 1 and λ(c0 + 2)τ ⩽ 31 , it holds that (I) the difference scheme (1.30)–(1.32) is uniquely solvable; (II) 󵄩󵄩 k 󵄩󵄩 2 󵄩󵄩e 󵄩󵄩∞ ⩽ c4 (τ + h ),

0 ⩽ k ⩽ n.

(1.34)

Proof. Subtracting (1.30)–(1.32) from (1.26), (1.28)–(1.29), respectively, the system of error equations is produced as Dt eik+1 − δx2 eik+1 = λeik − λ(Uik Uik+1 − uik uik+1 ) + (R2 )k+1 { i , { { { { { 1 ⩽ i ⩽ m − 1, 0 ⩽ k ⩽ n − 1, { { { ei0 = 0, 0 ⩽ i ⩽ m, { { { k k 1 ⩽ k ⩽ n. { e0 = 0, em = 0, Rewrite (1.35) as k+1 k+1 (1 + 2r)eik+1 = r(ei−1 + ei+1 ) + (1 + λτ)eik − λτ(uik eik+1 + eik Uik+1 ) + τ(R2 )k+1 i ,

1 ⩽ i ⩽ m − 1, 0 ⩽ k ⩽ n − 1.

(1.35) (1.36) (1.37)

1.4 Backward Euler difference scheme

� 17

Taking the absolute value on both the right- and left-hand sides of the equality above and using the triangle inequality, with the help of (1.27), we have 󵄨 󵄨 (1 + 2r)󵄨󵄨󵄨eik+1 󵄨󵄨󵄨 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 ⩽ 2r 󵄩󵄩󵄩ek+1 󵄩󵄩󵄩∞ + (1 + λτ)󵄩󵄩󵄩ek 󵄩󵄩󵄩∞ + λτ(󵄩󵄩󵄩uk 󵄩󵄩󵄩∞ 󵄩󵄩󵄩ek+1 󵄩󵄩󵄩∞ + 󵄩󵄩󵄩ek 󵄩󵄩󵄩∞ 󵄩󵄩󵄩U k+1 󵄩󵄩󵄩∞ ) + c3 τ(τ + h2 ), 1 ⩽ i ⩽ m − 1, 0 ⩽ k ⩽ n − 1. It follows by noticing (1.37) that 󵄩 󵄩 (1 + 2r)󵄩󵄩󵄩ek+1 󵄩󵄩󵄩∞ 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 ⩽ 2r 󵄩󵄩󵄩ek+1 󵄩󵄩󵄩∞ + (1 + λτ)󵄩󵄩󵄩ek 󵄩󵄩󵄩∞ + λτ(󵄩󵄩󵄩uk 󵄩󵄩󵄩∞ 󵄩󵄩󵄩ek+1 󵄩󵄩󵄩∞ + c0 󵄩󵄩󵄩ek 󵄩󵄩󵄩∞ ) + c3 τ(τ + h2 ), 0 ⩽ k ⩽ n − 1, i. e., 󵄩󵄩 k+1 󵄩󵄩 󵄩 k󵄩 󵄩 k 󵄩 󵄩 k+1 󵄩 2 󵄩󵄩e 󵄩󵄩∞ ⩽ [1 + λ(c0 + 1)τ]󵄩󵄩󵄩e 󵄩󵄩󵄩∞ + λτ 󵄩󵄩󵄩u 󵄩󵄩󵄩∞ 󵄩󵄩󵄩e 󵄩󵄩󵄩∞ + c3 τ(τ + h ),

0 ⩽ k ⩽ n − 1. (1.38)

From (1.36), we easily have 󵄩󵄩 0 󵄩󵄩 󵄩󵄩e 󵄩󵄩∞ = 0,

(1.39)

which means that (1.34) holds for k = 0. Now suppose that the values of u0 , u1 , . . . , ul have been obtained from (1.30)–(1.32) and the inequality (1.34) is true for 0 ⩽ k ⩽ l. Then when c4 (τ + h2 ) ⩽ 1, it follows: 󵄩󵄩 k 󵄩󵄩 2 󵄩󵄩e 󵄩󵄩∞ ⩽ c4 (τ + h ) ⩽ 1,

0⩽k⩽l

and 󵄩󵄩 k 󵄩󵄩 󵄩 k󵄩 󵄩 k󵄩 󵄩󵄩u 󵄩󵄩∞ ⩽ 󵄩󵄩󵄩U 󵄩󵄩󵄩∞ + 󵄩󵄩󵄩e 󵄩󵄩󵄩∞ ⩽ c0 + 1,

0 ⩽ k ⩽ l.

(1.40)

(I) Proof for the unique solvability. The system of linear equations in ul+1 can be obtained from (1.30) and (1.32) as {

Dt uil+1 − δx2 uil+1 = λ(uil − uil uil+1 ), u0l+1

= α(tl+1 ),

l+1 um

= β(tl+1 ).

1 ⩽ i ⩽ m − 1,

Consider its homogeneous one: { Rewrite (1.41) as

1 l+1 u − δx2 uil+1 = −λuil uil+1 , τ i l+1 u0l+1 = 0, um = 0.

1 ⩽ i ⩽ m − 1,

(1.41) (1.42)

18 � 1 Difference methods for the Fisher equation l+1 l+1 (1 + 2r)uil+1 = r(ui−1 + ui+1 ) − λτuil uil+1 ,

1 ⩽ i ⩽ m − 1.

l+1 Suppose |uil+1 ‖∞ , 1 ⩽ i∗ ⩽ m − 1. Letting i = i∗ in the equality above and taking ∗ | = ‖u the absolute value on both the right- and left-hand sides, with the help of the triangle inequality, we get

󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 (1 + 2r)󵄩󵄩󵄩ul+1 󵄩󵄩󵄩∞ ⩽ 2r 󵄩󵄩󵄩ul+1 󵄩󵄩󵄩∞ + λτ 󵄩󵄩󵄩ul 󵄩󵄩󵄩∞ 󵄩󵄩󵄩ul+1 󵄩󵄩󵄩∞ . By (1.40), it further follows: 󵄩 l+1 󵄩 󵄩󵄩 l+1 󵄩󵄩 󵄩󵄩u 󵄩󵄩∞ ⩽ λ(c0 + 1)τ 󵄩󵄩󵄩u 󵄩󵄩󵄩∞ . When λ(c0 + 1)τ < 1, it implies ‖ul+1 ‖∞ = 0. Thus, (1.30) and (1.32) are uniquely solvable in ul+1 . (II) Proof for (1.34). From (1.38) and (1.40), we have 󵄩󵄩 k+1 󵄩󵄩 󵄩 k󵄩 󵄩 k+1 󵄩 2 󵄩󵄩e 󵄩󵄩∞ ⩽ [1 + λ(c0 + 1)τ]󵄩󵄩󵄩e 󵄩󵄩󵄩∞ + λ(c0 + 1)τ 󵄩󵄩󵄩e 󵄩󵄩󵄩∞ + c3 τ(τ + h ),

0 ⩽ k ⩽ l,

i. e., 󵄩 󵄩 󵄩 󵄩 [1 − λ(c0 + 1)τ]󵄩󵄩󵄩ek+1 󵄩󵄩󵄩∞ ⩽ [1 + λ(c0 + 1)τ]󵄩󵄩󵄩ek 󵄩󵄩󵄩∞ + c3 τ(τ + h2 ),

0 ⩽ k ⩽ l.

When λ(c0 + 1)τ ⩽ 31 , it follows: 3 󵄩󵄩 k+1 󵄩󵄩 󵄩 k󵄩 2 󵄩󵄩e 󵄩󵄩∞ ⩽ [1 + 3λ(c0 + 1)τ]󵄩󵄩󵄩e 󵄩󵄩󵄩∞ + c3 τ(τ + h ), 2

0 ⩽ k ⩽ l.

Noticing (1.39), the application of the Gronwall inequality (Theorem 1.2(a)) yields c3 󵄩󵄩 l+1 󵄩󵄩 e3λ(c0 +1)T (τ + h2 ) = c4 (τ + h2 ), 󵄩󵄩e 󵄩󵄩∞ ⩽ 2λ(c0 + 1) which says that (1.34) is also true for k = l + 1. By induction, (1.34) is true for all k (0 ⩽ k ⩽ n).

1.5 Crank–Nicolson difference scheme This section is devoted to the derivation of an unconditionally convergent difference scheme with the accuracy O(τ 2 + h2 ).

1.5 Crank–Nicolson difference scheme �

19

1.5.1 Derivation of the difference scheme Considering equation (1.1) at the point (xi , tk+ 1 ), we have 2

ut (xi , tk+ 1 ) − uxx (xi , tk+ 1 ) = λ[u(xi , tk+ 1 ) − u2 (xi , tk+ 1 )], 2

2

2

2

1 ⩽ i ⩽ m − 1, 0 ⩽ k ⩽ n − 1.

(1.43)

By Lemma 1.2, we have k+ 21

ut (xi , tk+ 1 ) = δt Ui 2

+ O(τ 2 ),

(1.44)

1 uxx (xi , tk+ 1 ) = [uxx (xi , tk+1 ) + uxx (xi , tk )] + O(τ 2 ) 2 2 1 = (δx2 Uik+1 + δx2 Uik ) + O(h2 ) + O(τ 2 ) 2 k+ 21

= δx2 Ui

u(xi , tk+ 1 ) = 2

k+ 1 Ui 2

+ O(τ 2 + h2 ),

(1.45)

+ O(τ 2 ),

(1.46)

τ τ u2 (xi , tk+ 1 ) = [Uik + ut (xi , tk+ 1 ) + O(τ 2 )][Uik+1 − ut (xi , tk+ 1 ) + O(τ 2 )] 2 2 2 2 2 = Uik Uik+1 + O(τ 2 ).

(1.47)

Inserting (1.44)–(1.47) into (1.43) arrives at k+ 21

δt Ui

k+ 21

− δx2 Ui

k+ 21

= λ(Ui

k+ 21

− Uik Uik+1 ) + (R3 )i

1 ⩽ i ⩽ m − 1, 0 ⩽ k ⩽ n − 1, (1.48)

,

where there is a constant c5 such that k+ 1 󵄨 󵄨󵄨 2 2 󵄨󵄨(R3 )i 2 󵄨󵄨󵄨 ⩽ c5 (τ + h ),

1 ⩽ i ⩽ m − 1, 0 ⩽ k ⩽ n − 1.

(1.49)

Noticing the initial-boundary value conditions (1.2)–(1.3), we have {

Ui0 = φ(xi ), U0k

= α(tk ),

Umk

= β(tk ),

0 ⩽ i ⩽ m, 1 ⩽ k ⩽ n.

(1.50) (1.51)

k+ 1

Neglecting the small term (R3 )i 2 in (1.48) and replacing the exact solution Uik by its numerical one uik , the Crank–Nicolson difference scheme is derived as 1

1

1

k+ k+ k+ { δ u 2 − δx2 ui 2 = λ(ui 2 − uik uik+1 ), { { { t i ui0 = φ(xi ), { { { { k k { u0 = α(tk ), um = β(tk ),

1 ⩽ i ⩽ m − 1, 0 ⩽ k ⩽ n − 1,

(1.52)

0 ⩽ i ⩽ m,

(1.53)

1 ⩽ k ⩽ n.

The difference scheme (1.52)–(1.54) is a two-level linearized difference scheme.

(1.54)

20 � 1 Difference methods for the Fisher equation 1.5.2 Existence and convergence of the difference solution Theorem 1.7. Let {Uik | 0 ⩽ i ⩽ m, 0 ⩽ k ⩽ n} and {uik | 0 ⩽ i ⩽ m, 0 ⩽ k ⩽ n} be solutions of the problem (1.1)–(1.3) and the difference scheme (1.52)–(1.54), respectively. Denote eik = Uik − uik , c6 =

0 ⩽ i ⩽ m, 0 ⩽ k ⩽ n,

L2 2 2 c5 √ 3 e 2 λ (c0 +1) T . λ(c0 + 1) L

Then when 2L c6 (τ 2 + h2 ) ⩽ 1, L2 λ2 [1 + 2(c0 + 1)2 ]τ ⩽ 2 and λ( 32 + c0 )τ < 1, it holds that (I) the difference scheme (1.52)–(1.54) is uniquely solvable; (II) √

󵄨󵄨 k 󵄨󵄨 2 2 󵄨󵄨e 󵄨󵄨1 ⩽ c6 (τ + h ),

0 ⩽ k ⩽ n.

(1.55)

Proof. Subtracting (1.52)–(1.54) from (1.48), (1.50)–(1.51), respectively, we get the system of error equations as follows: k+ 1

k+ 1

k+ 1

k+ 1

{ δt ei 2 − δx2 ei 2 = λei 2 − λ(Uik Uik+1 − uik uik+1 ) + (R3 )i 2 , { { { { { 1 ⩽ i ⩽ m − 1, 0 ⩽ k ⩽ n − 1, { { { e0 = 0, 0 ⩽ i ⩽ m, { { { i k k 1 ⩽ k ⩽ n. { e0 = 0, em = 0,

(1.56) (1.57) (1.58)

1

Taking the inner product of (1.56) with δt ek+ 2 on both the right- and left-hand sides, and using the summation by parts, we have 󵄩󵄩 k+ 21 󵄩󵄩2 1 󵄨󵄨 k+1 󵄨󵄨2 󵄨󵄨 k 󵄨󵄨2 󵄩󵄩δt e 󵄩󵄩 + (󵄨󵄨e 󵄨󵄨1 − 󵄨󵄨e 󵄨󵄨1 ) 2τ m−1

k+ 21

= λh ∑ [ei i=1

k+ 21

− (uik eik+1 + eik Uik+1 )]δt ei

m−1

k+ 21

+ h ∑ (R3 )i i=1

k+ 21

δt ei

1 1 󵄩 1󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 ⩽ λ(󵄩󵄩󵄩ek+ 2 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩δt ek+ 2 󵄩󵄩󵄩 + 󵄩󵄩󵄩uk 󵄩󵄩󵄩∞ 󵄩󵄩󵄩ek+1 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩δt ek+ 2 󵄩󵄩󵄩 1 1 1 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 + 󵄩󵄩󵄩U k+1 󵄩󵄩󵄩∞ 󵄩󵄩󵄩ek 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩δt ek+ 2 󵄩󵄩󵄩) + 󵄩󵄩󵄩(R3 )k+ 2 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩δt ek+ 2 󵄩󵄩󵄩 1 2 1 2 1󵄩 1󵄩 󵄩 󵄩 󵄩 1 󵄩2 󵄩 󵄩2 󵄩 󵄩2 ⩽ ( 󵄩󵄩󵄩δt ek+ 2 󵄩󵄩󵄩 + λ2 󵄩󵄩󵄩ek+ 2 󵄩󵄩󵄩 ) + ( 󵄩󵄩󵄩δt ek+ 2 󵄩󵄩󵄩 + λ2 󵄩󵄩󵄩uk 󵄩󵄩󵄩∞ 󵄩󵄩󵄩ek+1 󵄩󵄩󵄩 ) 4 4 1 2 1 2 1 󵄩󵄩 k+ 21 󵄩󵄩2 1󵄩 2 󵄩 k 󵄩2 󵄩 󵄩 󵄩 󵄩 2󵄩 k+1 + ( 󵄩󵄩δt e 󵄩󵄩 + λ 󵄩󵄩󵄩U 󵄩󵄩󵄩∞ 󵄩󵄩󵄩e 󵄩󵄩󵄩 ) + ( 󵄩󵄩󵄩δt ek+ 2 󵄩󵄩󵄩 + 󵄩󵄩󵄩(R3 )k+ 2 󵄩󵄩󵄩 ), 4 4 0 ⩽ k ⩽ n − 1,

which follows by noticing (1.49) that 1 2 1 󵄨󵄨 k+1 󵄨󵄨2 󵄨󵄨 k 󵄨󵄨2 󵄩 1 󵄩2 󵄩 󵄩2 󵄩 󵄩2 󵄩 󵄩2 󵄩 󵄩2 󵄩 󵄩 (󵄨e 󵄨󵄨1 − 󵄨󵄨e 󵄨󵄨1 ) ⩽ λ2 󵄩󵄩󵄩ek+ 2 󵄩󵄩󵄩 + λ2 󵄩󵄩󵄩uk 󵄩󵄩󵄩∞ 󵄩󵄩󵄩ek+1 󵄩󵄩󵄩 + λ2 󵄩󵄩󵄩U k+1 󵄩󵄩󵄩∞ 󵄩󵄩󵄩ek 󵄩󵄩󵄩 + 󵄩󵄩󵄩(R3 )k+ 2 󵄩󵄩󵄩 2τ 󵄨

1.5 Crank–Nicolson difference scheme

1 󵄩 󵄩2 󵄩 󵄩2 󵄩2 󵄩 󵄩2 󵄩 ⩽ λ2 (󵄩󵄩󵄩ek 󵄩󵄩󵄩 + 󵄩󵄩󵄩ek+1 󵄩󵄩󵄩 ) + λ2 󵄩󵄩󵄩uk 󵄩󵄩󵄩∞ 󵄩󵄩󵄩ek+1 󵄩󵄩󵄩 2 2 󵄩 󵄩2 + λ2 c02 󵄩󵄩󵄩ek 󵄩󵄩󵄩 + Lc52 (τ 2 + h2 ) , 0 ⩽ k ⩽ n − 1.

� 21

(1.59)

In view of (1.57), we know |e0 |1 = 0, which means that (1.55) holds for k = 0. Now assume that (1.55) is true for 0 ⩽ k ⩽ l, i. e., 󵄨󵄨 k 󵄨󵄨 2 2 󵄨󵄨e 󵄨󵄨1 ⩽ c6 (τ + h ), By Lemma 1.1, when

√L c (τ 2 2 6

0 ⩽ k ⩽ l.

+ h2 ) ⩽ 1, we have

√ √L 󵄨 k 󵄨 󵄩󵄩 k 󵄩󵄩 󵄨󵄨e 󵄨󵄨 ⩽ L c6 (τ 2 + h2 ) ⩽ 1, 0 ⩽ k ⩽ l, 󵄩󵄩e 󵄩󵄩∞ ⩽ 2 󵄨 󵄨1 2 󵄩󵄩 k 󵄩󵄩 󵄩󵄩 k 󵄩󵄩 󵄩󵄩 k 󵄩󵄩 󵄩󵄩u 󵄩󵄩∞ ⩽ 󵄩󵄩U 󵄩󵄩∞ + 󵄩󵄩e 󵄩󵄩∞ ⩽ c0 + 1, 0 ⩽ k ⩽ l. (I) Proof for the unique solvability. From (1.52) and (1.54), the system of linear equations in ul+1 can be obtained as 1

1

1

{ δt ul+ 2 − δx2 ul+ 2 = λ(ul+ 2 − uil uil+1 ), i i i { l+1 l+1 u = α(t ), u = β(tl+1 ). l+1 { 0 m

1 ⩽ i ⩽ m − 1,

Consider its homogeneous one: {

1 l+1 u − 21 δx2 uil+1 = λ( 21 uil+1 τ i l+1 u0l+1 = 0, um = 0.

− uil uil+1 ),

1 ⩽ i ⩽ m − 1,

(1.60) (1.61)

Taking the inner product of (1.60) on both the right- and left-hand sides with ul+1 gives 1 󵄩󵄩 l+1 󵄩󵄩2 󵄩u 󵄩󵄩 + τ󵄩

1 󵄨󵄨 l+1 󵄨󵄨2 1 󵄩 l+1 󵄩2 󵄩 l+1 󵄩2 l 󵄨u 󵄨󵄨1 ⩽ λ( 󵄩󵄩󵄩u 󵄩󵄩󵄩 + ‖u ‖∞ 󵄩󵄩󵄩u 󵄩󵄩󵄩 ), 2󵄨 2

which further implies 1 󵄩󵄩 l+1 󵄩󵄩2 1 󵄩 l 󵄩 󵄩 l+1 󵄩2 1 󵄩 l+1 󵄩2 󵄩u 󵄩󵄩 ⩽ λ( + 󵄩󵄩󵄩u 󵄩󵄩󵄩∞ )󵄩󵄩󵄩u 󵄩󵄩󵄩 ⩽ λ[ + (c0 + 1)]󵄩󵄩󵄩u 󵄩󵄩󵄩 . τ󵄩 2 2 1 Thus, when τ < λ(3/2+c , the equality ‖ul+1 ‖ = 0 is followed. Therefore, the value of ul+1 0) is uniquely determined by (1.52) and (1.54). (II) Proof for (1.55). By (1.59) and Lemma 1.1, we have

22 � 1 Difference methods for the Fisher equation 1 󵄨󵄨 k+1 󵄨󵄨2 󵄨󵄨 k 󵄨󵄨2 (󵄨e 󵄨󵄨1 − 󵄨󵄨e 󵄨󵄨1 ) 2τ 󵄨 1 󵄩 2 󵄩 󵄩2 󵄩 󵄩2 󵄩2 󵄩 󵄩2 ⩽ λ2 (󵄩󵄩󵄩ek+1 󵄩󵄩󵄩 + 󵄩󵄩󵄩ek 󵄩󵄩󵄩 ) + λ2 (c0 + 1)2 󵄩󵄩󵄩ek+1 󵄩󵄩󵄩 + λ2 c02 󵄩󵄩󵄩ek 󵄩󵄩󵄩 + Lc52 (τ 2 + h2 ) 2 1 1 󵄨 󵄨2 󵄨 󵄨2 󵄨 󵄨2 1 󵄨 󵄨2 ⩽ λ2 L2 (󵄨󵄨󵄨ek+1 󵄨󵄨󵄨1 + 󵄨󵄨󵄨ek 󵄨󵄨󵄨1 ) + λ2 (c0 + 1)2 L2 󵄨󵄨󵄨ek+1 󵄨󵄨󵄨1 + λ2 c02 L2 󵄨󵄨󵄨ek 󵄨󵄨󵄨1 12 6 6 2

+ Lc52 (τ 2 + h2 ) ,

0 ⩽ k ⩽ l,

i. e., 1 󵄨 󵄨2 [1 − λ2 L2 (1 + 2(c0 + 1)2 )τ]󵄨󵄨󵄨ek+1 󵄨󵄨󵄨1 6 1 2 󵄨 󵄨2 ⩽ [1 + λ2 L2 (1 + 2c02 )τ]󵄨󵄨󵄨ek 󵄨󵄨󵄨1 + 2Lc52 τ(τ 2 + h2 ) , 6

0 ⩽ k ⩽ l.

When 61 λ2 L2 [1 + 2(c0 + 1)2 ]τ ⩽ 31 , it follows: 2 󵄨 k 󵄨2 󵄨󵄨 k+1 󵄨󵄨2 2 2 2 2 2 2 󵄨󵄨e 󵄨󵄨1 ⩽ [1 + λ L (c0 + 1) τ]󵄨󵄨󵄨e 󵄨󵄨󵄨1 + 3Lc5 τ(τ + h ) ,

0 ⩽ k ⩽ l.

The application of the Gronwall inequality (Theorem 1.2(a)) yields 2 2 2 3c2 3c2 2 2 󵄨󵄨 l+1 󵄨󵄨2 λ2 L2 (c0 +1)2 lτ 󵄨󵄨 0 󵄨󵄨2 [󵄨󵄨e 󵄨󵄨1 + 2 5 2 (τ 2 + h2 ) ] ⩽ 2 5 2 eλ L (c0 +1) T (τ 2 + h2 ) . 󵄨󵄨e 󵄨󵄨1 ⩽ e Lλ (c0 + 1) Lλ (c0 + 1)

Taking the square root on both the right- and left-hand sides of the inequality above produces 󵄨󵄨 l+1 󵄨󵄨 2 2 󵄨󵄨e 󵄨󵄨1 ⩽ c6 (τ + h ). By induction, the theorem is proved.

1.6 Fourth-order compact difference scheme In this section, an unconditionally convergent compact difference scheme with the accuracy O(τ 2 + h4 ) will be developed.

1.6.1 Derivation of the difference scheme For w = {wi | 0 ⩽ i ⩽ m} ∈ 𝒰h , define an averaging operator by 𝒜wi = {

1 (wi−1 12

wi ,

+ 10wi + wi+1 ),

1 ⩽ i ⩽ m − 1, i = 0, m.

1.6 Fourth-order compact difference scheme

� 23

Considering equation (1.1) at the point (xi , tk+ 1 ), we have 2

ut (xi , tk+ 1 ) − uxx (xi , tk+ 1 ) = λ[u(xi , tk+ 1 ) − u2 (xi , tk+ 1 )], 2

2

2

0 ⩽ i ⩽ m, 0 ⩽ k ⩽ n − 1.

2

By Lemma 1.2, we have 1 k+ 1 − [uxx (xi , tk+1 ) + uxx (xi , tk )] = λ(Ui 2 − Uik Uik+1 ) + O(τ 2 ), 2 0 ⩽ i ⩽ m, 0 ⩽ k ⩽ n − 1. k+ 21

δt Ui

(1.62)

Performing the operator 𝒜 on both the right- and left-hand sides of (1.62) and noticing (Lemma 1.2(h)) 2

k

4

𝒜uxx (xi , tk ) = δx Ui + O(h ),

we have k+ 21

𝒜δt Ui

k+ 21

− δx2 Ui

k+ 21

= λ𝒜(Ui

k+ 21

− Uik Uik+1 ) + (R4 )i

,

1 ⩽ i ⩽ m − 1, 0 ⩽ k ⩽ n − 1, (1.63)

where there is a constant c7 such that k+ 1 󵄨 󵄨󵄨 2 4 󵄨󵄨(R4 )i 2 󵄨󵄨󵄨 ⩽ c7 (τ + h ),

1 ⩽ i ⩽ m − 1, 0 ⩽ k ⩽ n − 1.

(1.64)

Noticing the initial-boundary value conditions (1.2)–(1.3), we have Ui0 = φ(xi ),

{

U0k

= α(tk ),

Umk

= β(tk ),

0 ⩽ i ⩽ m, 1 ⩽ k ⩽ n.

(1.65) (1.66)

k+ 1

Neglecting the small term (R4 )i 2 in (1.63) and replacing the exact solution Uik by its numerical one uik , a compact difference scheme is derived in the form of 1

1

1

k+ k+ 2 k+ k k+1 { 𝒜δt ui 2 − δx ui 2 = λ𝒜(ui 2 − ui ui ), { { { ui0 = φ(xi ), { { { { k k { u0 = α(tk ), um = β(tk ),

1 ⩽ i ⩽ m − 1, 0 ⩽ k ⩽ n − 1,

(1.67)

0 ⩽ i ⩽ m,

(1.68)

1 ⩽ k ⩽ n.

(1.69)

The difference scheme (1.67)–(1.69) is also a two-level linearized difference scheme.

1.6.2 Existence and convergence of difference solution Theorem 1.8. Let {Uik | 0 ⩽ i ⩽ m, 0 ⩽ k ⩽ n} and {uik | 0 ⩽ i ⩽ m, 0 ⩽ k ⩽ n} be solutions of the problem (1.1)–(1.3) and the difference scheme (1.67)–(1.69), respectively. Denote

24 � 1 Difference methods for the Fisher equation eik = Uik − uik , c8 =

0 ⩽ i ⩽ m, 0 ⩽ k ⩽ n,

3 2 2 2 c7 6 √ e 8 L λ [1+2(c0 +1) ]T . 2 λ L[1 + 2(c0 + 1) ] 2

Then when 2L c8 (τ 2 + h4 ) ⩽ 1, L4 λ2 [1 + 2(c0 + 1)2 ]τ ⩽ 31 and 32 ( 32 + c0 )λτ < 1, it holds that (I) the difference scheme (1.67)–(1.69) is uniquely solvable; (II) √

󵄨󵄨 k 󵄨󵄨 2 4 󵄨󵄨e 󵄨󵄨1 ⩽ c8 (τ + h ),

0 ⩽ k ⩽ n.

(1.70)

Proof. Subtracting (1.67)–(1.69) from (1.63), (1.65)–(1.66), respectively, the system of error equations is obtained as k+ 1

k+ 1

k+ 1

k+ 1

2 k k+1 k+1 k { 𝒜δt ei 2 − δx ei 2 = λ𝒜(ei 2 − ui ei − Ui ei ) + (R4 )i 2 , { { { { { 1 ⩽ i ⩽ m − 1, 0 ⩽ k ⩽ n − 1, { { { e0 = 0, 0 ⩽ i ⩽ m, { { { i k k 1 ⩽ k ⩽ n. { e0 = 0, em = 0,

(1.71) (1.72) (1.73) 1

Taking the inner product on both the right- and left-hand sides of (1.71) with δt ek+ 2 , we have 1

1

1

1

1

1

(𝒜δt ek+ 2 , δt ek+ 2 ) − (δx2 ek+ 2 , δt ek+ 2 )

1

1

= λ[(𝒜ek+ 2 , δt ek+ 2 ) − (𝒜(uk ek+1 ), δt ek+ 2 ) − (𝒜(U k+1 ek ), δt ek+ 2 )] 1

1

+ ((R4 )k+ 2 , δt ek+ 2 ),

0 ⩽ k ⩽ n − 1.

(1.74)

Now each term in (1.74) will be analyzed: 1

1

(𝒜δt ek+ 2 , δt ek+ 2 ) = ((I +

1 1 h2 2 δ )δ ek+ 2 , δt ek+ 2 ) 12 x t

1 2 1 2 h2 󵄩 󵄩 󵄩 󵄩 = 󵄩󵄩󵄩δt ek+ 2 󵄩󵄩󵄩 − 󵄩󵄩󵄩δx δt ek+ 2 󵄩󵄩󵄩 12 1 2 h2 4 󵄩󵄩 k+ 21 󵄩󵄩2 󵄩 󵄩 ⩾ 󵄩󵄩󵄩δt ek+ 2 󵄩󵄩󵄩 − ⋅ 󵄩δ e 󵄩󵄩 12 h2 󵄩 t 1 2 2󵄩 󵄩 = 󵄩󵄩󵄩δt ek+ 2 󵄩󵄩󵄩 , 3 1 1 1 󵄨 󵄨2 󵄨 󵄨2 −(δx2 ek+ 2 , δt ek+ 2 ) = (󵄨󵄨󵄨ek+1 󵄨󵄨󵄨1 − 󵄨󵄨󵄨ek 󵄨󵄨󵄨1 ), 2τ 1 2 1 2 1󵄩 3 󵄩 󵄩 󵄩 k+ 21 k+ 21 λ(𝒜e , δt e ) ⩽ 󵄩󵄩󵄩δt ek+ 2 󵄩󵄩󵄩 + λ2 󵄩󵄩󵄩𝒜ek+ 2 󵄩󵄩󵄩 6 2 1 2 1󵄩 3 󵄩 1 󵄩2 󵄩 ⩽ 󵄩󵄩󵄩δt ek+ 2 󵄩󵄩󵄩 + λ2 󵄩󵄩󵄩ek+ 2 󵄩󵄩󵄩 , 6 2

(1.75) (1.76)

(1.77)

1.6 Fourth-order compact difference scheme 1 2 1 1󵄩 󵄩 λ(𝒜(uk ek+1 ), δt ek+ 2 ) ⩽ 󵄩󵄩󵄩δt ek+ 2 󵄩󵄩󵄩 + 6 1 2 1󵄩 󵄩 ⩽ 󵄩󵄩󵄩δt ek+ 2 󵄩󵄩󵄩 + 6

1 1 2 1󵄩 󵄩 λ(𝒜(U k+1 ek ), δt ek+ 2 ) ⩽ 󵄩󵄩󵄩δt ek+ 2 󵄩󵄩󵄩 + 6 1 2 1󵄩 󵄩 ⩽ 󵄩󵄩󵄩δt ek+ 2 󵄩󵄩󵄩 + 6 1 1 2 1 1󵄩 󵄩 ((R4 )k+ 2 , δt ek+ 2 ) ⩽ 󵄩󵄩󵄩δt ek+ 2 󵄩󵄩󵄩 + 6

3 2 󵄩󵄩 󵄩2 λ 󵄩󵄩𝒜(uk ek+1 )󵄩󵄩󵄩 2 3 2 󵄩󵄩 k 󵄩󵄩2 󵄩󵄩 k+1 󵄩󵄩2 λ 󵄩u 󵄩 󵄩e 󵄩󵄩 , 2 󵄩 󵄩∞ 󵄩

3 2 󵄩󵄩 󵄩2 λ 󵄩𝒜(U k+1 ek )󵄩󵄩󵄩 2 󵄩 3 2 󵄩󵄩 k+1 󵄩󵄩2 󵄩󵄩 k 󵄩󵄩2 λ 󵄩U 󵄩󵄩∞ 󵄩󵄩e 󵄩󵄩 , 2 󵄩 3 󵄩󵄩 k+ 1 󵄩2 󵄩󵄩(R4 ) 2 󵄩󵄩󵄩 . 2

� 25

(1.78)

(1.79) (1.80)

Inserting (1.75)–(1.80) into (1.74) and noticing (1.64) lead to 1 󵄨󵄨 k+1 󵄨󵄨2 󵄨󵄨 k 󵄨󵄨2 (󵄨e 󵄨󵄨1 − 󵄨󵄨e 󵄨󵄨1 ) 2τ 󵄨 3 󵄩 1 󵄩2 3 󵄩 󵄩2 󵄩 󵄩2 󵄩 󵄩2 󵄩2 3 󵄩 ⩽ λ2 󵄩󵄩󵄩ek+ 2 󵄩󵄩󵄩 + λ2 󵄩󵄩󵄩uk 󵄩󵄩󵄩∞ 󵄩󵄩󵄩ek+1 󵄩󵄩󵄩 + λ2 󵄩󵄩󵄩U k+1 󵄩󵄩󵄩∞ 󵄩󵄩󵄩ek 󵄩󵄩󵄩 + 2 2 2 3 󵄩 󵄩2 󵄩 3 󵄩 󵄩2 󵄩 󵄩2 󵄩2 ⩽ λ2 (󵄩󵄩󵄩ek+1 󵄩󵄩󵄩 + 󵄩󵄩󵄩ek 󵄩󵄩󵄩 ) + λ2 󵄩󵄩󵄩uk 󵄩󵄩󵄩∞ 󵄩󵄩󵄩ek+1 󵄩󵄩󵄩 4 2 3 2 2 󵄩󵄩 k 󵄩󵄩2 3 2 2 2 + λ c0 󵄩󵄩e 󵄩󵄩 + Lc7 (τ + h4 ) , 0 ⩽ k ⩽ n − 1. 2 2

3 󵄩󵄩 k+ 1 󵄩2 󵄩󵄩(R4 ) 2 󵄩󵄩󵄩 2

(1.81)

In view of (1.72), we know |e0 |1 = 0, which means that (1.70) holds for k = 0. Now assume that (1.70) is true for 0 ⩽ k ⩽ l, i. e., 󵄨󵄨 k 󵄨󵄨 2 4 󵄨󵄨e 󵄨󵄨1 ⩽ c8 (τ + h ), By Lemma 1.1, when

√L c (τ 2 2 8

0 ⩽ k ⩽ l.

+ h4 ) ⩽ 1, we have

√L √L k 󵄩󵄩 k 󵄩󵄩 |e |1 ⩽ c (τ 2 + h4 ) ⩽ 1, 0 ⩽ k ⩽ l, 󵄩󵄩e 󵄩󵄩∞ ⩽ 2 2 8 󵄩󵄩 k 󵄩󵄩 󵄩󵄩 k 󵄩󵄩 󵄩󵄩 k 󵄩󵄩 󵄩󵄩u 󵄩󵄩∞ ⩽ 󵄩󵄩U 󵄩󵄩∞ + 󵄩󵄩e 󵄩󵄩∞ ⩽ c0 + 1, 0 ⩽ k ⩽ l. (I) Proof for the unique solvability. From (1.67) and (1.69), the system of linear equations in ul+1 can be obtained as 1

1

1

{ 𝒜δt ul+ 2 − δx2 ul+ 2 = λ𝒜(ul+ 2 − uil uil+1 ), i i i { l+1 l+1 { u0 = α(tl+1 ), um = β(tl+1 ).

1 ⩽ i ⩽ m − 1,

Consider its homogeneous one: {

1 𝒜uil+1 − 21 δx2 uil+1 τ l+1 u0l+1 = 0, um =

= λ𝒜( 21 uil+1 − uil uil+1 ),

0.

1 ⩽ i ⩽ m − 1,

(1.82) (1.83)

Taking the inner product on both the right- and left-hand sides of (1.82) with ul+1 gives

26 � 1 Difference methods for the Fisher equation 1 1󵄨 󵄨2 (𝒜ul+1 , ul+1 ) + 󵄨󵄨󵄨ul+1 󵄨󵄨󵄨1 τ 2 1 l+1 l l+1 = λ(𝒜( u − u u ), ul+1 ) 2 󵄩󵄩 󵄩󵄩 1 󵄩 󵄩 󵄩 󵄩 ⩽ λ󵄩󵄩󵄩𝒜( ul+1 − ul ul+1 )󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩ul+1 󵄩󵄩󵄩 󵄩󵄩 󵄩󵄩 2 󵄩󵄩 1 󵄩󵄩 󵄩 󵄩 󵄩 󵄩 ⩽ λ󵄩󵄩󵄩( + ul )ul+1 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩ul+1 󵄩󵄩󵄩 󵄩󵄩 2 󵄩󵄩 1 󵄩 󵄩2 ⩽ λ( + ‖ul ‖∞ )󵄩󵄩󵄩ul+1 󵄩󵄩󵄩 2 1 󵄩 󵄩2 ⩽ λ( + c0 + 1)󵄩󵄩󵄩ul+1 󵄩󵄩󵄩 , 2

which further implies 1 2 󵄩󵄩 l+1 󵄩󵄩2 3 󵄩 󵄩2 ⋅ 󵄩u 󵄩󵄩 ⩽ λ( + c0 )󵄩󵄩󵄩ul+1 󵄩󵄩󵄩 . τ 3󵄩 2 Thus, when 32 ( 32 + c0 )λτ < 1, the equality ‖ul+1 ‖ = 0 is followed. Therefore, the value of ul+1 is uniquely determined by (1.67) and (1.69). (II) Proof for (1.70). From (1.81), we have 1 󵄨󵄨 k+1 󵄨󵄨2 󵄨󵄨 k 󵄨󵄨2 (󵄨e 󵄨󵄨1 − 󵄨󵄨e 󵄨󵄨1 ) 2τ 󵄨 3 3 󵄩 2 󵄩2 󵄩 󵄩2 󵄩 󵄩2 3 󵄩 󵄩2 3 ⩽ λ2 (󵄩󵄩󵄩ek+1 󵄩󵄩󵄩 + 󵄩󵄩󵄩ek 󵄩󵄩󵄩 ) + λ2 (c0 + 1)2 󵄩󵄩󵄩ek+1 󵄩󵄩󵄩 + λ2 c02 󵄩󵄩󵄩ek 󵄩󵄩󵄩 + Lc72 (τ 2 + h4 ) 4 2 2 2 3 3 2 󵄩 󵄩2 󵄩 󵄩2 ⩽ λ2 [1 + 2(c0 + 1)2 ](󵄩󵄩󵄩ek+1 󵄩󵄩󵄩 + 󵄩󵄩󵄩ek 󵄩󵄩󵄩 ) + Lc72 (τ 2 + h4 ) 4 2 L2 󵄨 3 3 2 󵄨2 󵄨 󵄨2 ⩽ λ2 [1 + 2(c0 + 1)2 ] (󵄨󵄨󵄨ek+1 󵄨󵄨󵄨1 + 󵄨󵄨󵄨ek 󵄨󵄨󵄨1 ) + Lc72 (τ 2 + h4 ) , 0 ⩽ k ⩽ l, 4 6 2 i. e., [1 − ⩽ [1 + When

L2 2 λ [1 4

L2 2 󵄨 󵄨2 λ (1 + 2(c0 + 1)2 )τ]󵄨󵄨󵄨ek+1 󵄨󵄨󵄨1 4

L2 2 2 󵄨 󵄨2 λ (1 + 2(c0 + 1)2 )τ]󵄨󵄨󵄨ek 󵄨󵄨󵄨1 + 3Lc72 τ(τ 2 + h4 ) , 4

0 ⩽ k ⩽ l.

+ 2(c0 + 1)2 ]τ ⩽ 31 , it follows:

3 2 2 󵄨󵄨 k+1 󵄨󵄨2 󵄨 k 󵄨2 9 2 2 2 4 2 󵄨󵄨e 󵄨󵄨1 ⩽ [1 + L λ (1 + 2(c0 + 1) )τ]󵄨󵄨󵄨e 󵄨󵄨󵄨1 + Lc7 τ(τ + h ) , 4 2

0 ⩽ k ⩽ l.

1.7 Three-level linearized difference scheme

� 27

The application of the Gronwall inequality (Theorem 1.2(a)) leads to 3 2 2 󵄨󵄨 l+1 󵄨󵄨2 L λ [1+2(c0 +1)2 ]lτ 󵄨󵄨 0 󵄨󵄨2 {󵄨󵄨e 󵄨󵄨1 + 󵄨󵄨e 󵄨󵄨1 ⩽ e 4

⩽

3 2 2 L λ [1 4

9 Lc72 2

2

+ 2(c0 + 1)2 ]

(τ 2 + h4 ) }

3 2 2 2 6c72 2 e 4 L λ [1+2(c0 +1) ]T (τ 2 + h4 ) . 2 2 Lλ [1 + 2(c0 + 1) ]

Taking the square root on both the right- and left-hand sides of the inequality above produces 󵄨󵄨 l+1 󵄨󵄨 2 4 󵄨󵄨e 󵄨󵄨1 ⩽ c8 (τ + h ), which says that (1.70) also holds for k = l + 1. By induction, (1.70) is true for all k (0 ⩽ k ⩽ n).

1.7 Three-level linearized difference scheme This part will focus on an unconditionally convergent and conservative three-level linearized difference scheme for solving (1.1)–(1.3) with the convergence order O(τ 2 + h2 ). 1.7.1 Derivation of the difference scheme Considering equation (1.1) at the point (xi , t 1 ), we have 2

ut (xi , t 1 ) − uxx (xi , t 1 ) = λ[u(xi , t 1 ) − u2 (xi , t 1 )], 2

2

2

2

1 ⩽ i ⩽ m − 1.

By Lemma 1.2, we have 1

1

1

δt Ui2 − δx2 Ui2 = λ(Ui2 − Ui0 Ui1 ) + (R5 )0i ,

1 ⩽ i ⩽ m − 1,

(1.84)

where there is a constant c9 such that 󵄨󵄨 0󵄨 2 2 󵄨󵄨(R5 )i 󵄨󵄨󵄨 ⩽ c9 (τ + h ),

1 ⩽ i ⩽ m − 1.

(1.85)

Considering equation (1.1) at the node point (xi , tk ), we have ut (xi , tk ) − uxx (xi , tk ) = λ[u(xi , tk ) − u2 (xi , tk )],

1 ⩽ i ⩽ m − 1, 1 ⩽ k ⩽ n − 1.

By Lemma 1.2, we have ̄ ̄ 1 Δt Uik − δx2 Uik = λ[Uik − (Uik−1 + Uik + Uik+1 )Uik ] + (R5 )ki , 3

1 ⩽ i ⩽ m − 1, 1 ⩽ k ⩽ n − 1,

(1.86)

28 � 1 Difference methods for the Fisher equation where there is a constant c10 such that 󵄨󵄨 k󵄨 2 2 󵄨󵄨(R5 )i 󵄨󵄨󵄨 ⩽ c10 (τ + h ),

1 ⩽ i ⩽ m − 1, 1 ⩽ k ⩽ n − 1.

(1.87)

Noticing the initial-boundary value conditions (1.2)–(1.3), we have {

Ui0 = φ(xi ), U0k

= α(tk ),

Umk

0 ⩽ i ⩽ m,

= β(tk ),

(1.88)

1 ⩽ k ⩽ n.

(1.89)

Neglecting the small terms in (1.84) and (1.86), and replacing the exact solution Uik by its numerical one uik , the following difference scheme can be derived in the form: 1

1

̄ Δt uik − δx2 uik ui0 = φ(xi ), u0k = α(tk ),

̄ λ[uik

1

δt ui2 − δx2 ui2 = λ(ui2 − ui0 ui1 ),

{ { { { { { { { { { { { { { {

=

k um

−

1 k k−1 u (ui 3 i

= β(tk ),

+

uik

+

uik+1 )],

1 ⩽ i ⩽ m − 1,

(1.90)

1 ⩽ i ⩽ m − 1, 1 ⩽ k ⩽ n − 1,

(1.91)

0 ⩽ i ⩽ m,

(1.92)

1 ⩽ k ⩽ n.

(1.93)

The next result illustrates the conservative property of this difference scheme. Theorem 1.9. Suppose {uik | 0 ⩽ i ⩽ m, 0 ⩽ k ⩽ n} is the solution of the difference scheme (1.90)–(1.93) with α(t) ≡ 0, β(t) ≡ 0. Denote 1 1 󵄩 1 1 󵄨 1 󵄨2 k 󵄨 ̄󵄨2 󵄩 1 󵄩2 󵄩2 󵄩 󵄩2 E k = (󵄩󵄩󵄩uk+1 󵄩󵄩󵄩 + 󵄩󵄩󵄩uk 󵄩󵄩󵄩 ) + 2τ( 󵄨󵄨󵄨u 2 󵄨󵄨󵄨1 + ∑󵄨󵄨󵄨ul 󵄨󵄨󵄨1 ) + 2λτ{ [(u0 u1 , u 2 ) − 󵄩󵄩󵄩u 2 󵄩󵄩󵄩 ] 2 2 2 l=1

k ̄ 1 󵄩 ̄󵄩2 + ∑[( (ul−1 + ul + ul+1 )ul , ul ) − 󵄩󵄩󵄩ul 󵄩󵄩󵄩 ]}, 3 l=1

0 ⩽ k ⩽ n − 1,

1 󵄨 1 2 2 󵄨2 󵄨 󵄨2 F k = (󵄨󵄨󵄨uk+1 󵄨󵄨󵄨1 + 󵄨󵄨󵄨uk 󵄨󵄨󵄨1 ) + λ{ [(uk , (uk+1 ) ) + ((uk ) , uk+1 )] 2 3 k 1 2 1 󵄩 1󵄩 󵄩 󵄩 󵄩2 󵄩2 󵄩 󵄩2 − (󵄩󵄩󵄩uk+1 󵄩󵄩󵄩 + 󵄩󵄩󵄩uk 󵄩󵄩󵄩 )} + 2τ( 󵄩󵄩󵄩δt u 2 󵄩󵄩󵄩 + ∑󵄩󵄩󵄩Δt ul 󵄩󵄩󵄩 ), 2 2 l=1

0 ⩽ k ⩽ n − 1.

Then we have 󵄩 󵄩2 E k = 󵄩󵄩󵄩u0 󵄩󵄩󵄩 , 0 ⩽ k ⩽ n − 1, F k = F̂ 0 , 0 ⩽ k ⩽ n − 1, where 4 2 2 2 󵄨 󵄨2 󵄩 󵄩2 F̂ 0 = 󵄨󵄨󵄨u0 󵄨󵄨󵄨1 + λ[ ((u0 ) , u1 ) − (u0 , (u1 ) ) − 󵄩󵄩󵄩u0 󵄩󵄩󵄩 ]. 3 3

(1.94) (1.95)

1.7 Three-level linearized difference scheme

� 29

Proof. (I) Taking the inner product of (1.90) on both the right- and left-hand sides with 1 u 2 gives 1 1 1 1 1 󵄩 1 󵄩2 (δt u 2 , u 2 ) − (δx2 u 2 , u 2 ) = λ[󵄩󵄩󵄩u 2 󵄩󵄩󵄩 − (u0 u1 , u 2 )].

Noticing 1

1

(δt u 2 , u 2 ) =

1 󵄩󵄩 1 󵄩󵄩2 󵄩󵄩 0 󵄩󵄩2 (󵄩u 󵄩 − 󵄩u 󵄩 ) and 2τ 󵄩 󵄩 󵄩 󵄩

1 1 󵄨 1 󵄨2 − (δx2 u 2 , u 2 ) = 󵄨󵄨󵄨u 2 󵄨󵄨󵄨1 ,

we have 1 1 󵄩󵄩 1 󵄩󵄩2 󵄩󵄩 0 󵄩󵄩2 󵄨󵄨 21 󵄨󵄨2 󵄩 1 󵄩2 (󵄩󵄩u 󵄩󵄩 − 󵄩󵄩u 󵄩󵄩 ) + 󵄨󵄨u 󵄨󵄨1 + λ[(u0 u1 , u 2 ) − 󵄩󵄩󵄩u 2 󵄩󵄩󵄩 ] = 0, 2τ

which can be rewritten as 1 1 󵄩󵄩 1 󵄩󵄩2 󵄩󵄩 0 󵄩󵄩2 󵄨 1 󵄨2 󵄩 1 󵄩2 󵄩 󵄩2 (󵄩u 󵄩 + 󵄩u 󵄩 ) + τ 󵄨󵄨󵄨u 2 󵄨󵄨󵄨1 + λτ[(u0 u1 , u 2 ) − 󵄩󵄩󵄩u 2 󵄩󵄩󵄩 ] = 󵄩󵄩󵄩u0 󵄩󵄩󵄩 , 2 󵄩 󵄩 󵄩 󵄩

i. e., 󵄩 󵄩2 E 0 = 󵄩󵄩󵄩u0 󵄩󵄩󵄩 .

(1.96)

Taking the inner product of (1.91) on both the right- and left-hand sides with uk yields ̄

̄ ̄ ̄ ̄ 1 󵄩 ̄ 󵄩2 (Δt uk , uk ) − (δx2 uk , uk ) = λ[󵄩󵄩󵄩uk 󵄩󵄩󵄩 − ( (uk−1 + uk + uk+1 )uk , uk )], 3 1 ⩽ k ⩽ n − 1.

Noticing (Δt uk , uk ) = ̄

1 󵄩󵄩 k+1 󵄩󵄩2 󵄩󵄩 k−1 󵄩󵄩2 (󵄩u 󵄩󵄩 − 󵄩󵄩u 󵄩󵄩 ) and 4τ 󵄩

̄ ̄ 󵄨 ̄ 󵄨2 − (δx2 uk , uk ) = 󵄨󵄨󵄨uk 󵄨󵄨󵄨1 ,

we have 1 ‖uk+1 ‖2 + ‖uk ‖2 ‖uk ‖2 + ‖uk−1 ‖2 󵄨 ̄ 󵄨2 ( − ) + 󵄨󵄨󵄨uk 󵄨󵄨󵄨1 2τ 2 2 ̄ 1 󵄩 ̄ 󵄩2 + λ[( (uk−1 + uk + uk+1 )uk , uk ) − 󵄩󵄩󵄩uk 󵄩󵄩󵄩 ] = 0, 3

1 ⩽ k ⩽ n − 1.

Replacing k by l in the equality above and summing over l from 1 to k will arrive at k k ̄ 1 󵄩󵄩 k+1 󵄩󵄩2 󵄩󵄩 k 󵄩󵄩2 1 󵄨 ̄ 󵄨2 󵄩 ̄ 󵄩2 (󵄩󵄩u 󵄩󵄩 + 󵄩󵄩u 󵄩󵄩 ) + 2τ ∑󵄨󵄨󵄨ul 󵄨󵄨󵄨1 + 2λτ ∑[( (ul−1 + ul + ul+1 )ul , ul ) − 󵄩󵄩󵄩ul 󵄩󵄩󵄩 ] 2 3 l=1 l=1

1 󵄩 󵄩2 󵄩 󵄩2 = (󵄩󵄩󵄩u1 󵄩󵄩󵄩 + 󵄩󵄩󵄩u0 󵄩󵄩󵄩 ), 2

1 ⩽ k ⩽ n − 1.

30 � 1 Difference methods for the Fisher equation 1

1

1

Adding τ|u 2 |21 +λτ[(u0 u1 , u 2 )−‖u 2 ‖2 ] on both the right- and left-hand sides of the equality above yields Ek = E0 ,

1 ⩽ k ⩽ n − 1.

(1.97)

Then the equality (1.94) is followed from (1.96) and (1.97). (II) Taking the inner product of (1.90) on both the right- and left-hand sides with 1 2 δt u gives 1 1 1 1 󵄩󵄩 21 󵄩󵄩2 2 1 0 1 󵄩󵄩δt u 󵄩󵄩 − (δx u 2 , δt u 2 ) = λ[(u 2 , δt u 2 ) − (u u , δt u 2 )].

Noticing 1 󵄨󵄨 1 󵄨󵄨2 󵄨󵄨 0 󵄨󵄨2 (󵄨u 󵄨 − 󵄨u 󵄨 ), 2τ 󵄨 󵄨1 󵄨 󵄨1 1 1 1 󵄩 󵄩2 󵄩 󵄩2 (u 2 , δt u 2 ) = (󵄩󵄩󵄩u1 󵄩󵄩󵄩 − 󵄩󵄩󵄩u0 󵄩󵄩󵄩 ), 2τ 1 1 2 2 (u0 u1 , δt u 2 ) = [(u0 , (u1 ) ) − ((u0 ) , u1 )], τ 1

1

−(δx2 u 2 , δt u 2 ) =

we have 1 󵄩 1 󵄩2 󵄩 0 󵄩2 1 0 1 2 󵄩󵄩 21 󵄩󵄩2 1 󵄨󵄨 1 󵄨󵄨2 󵄨󵄨 0 󵄨󵄨2 0 2 1 󵄩󵄩δt u 󵄩󵄩 + (󵄨󵄨u 󵄨󵄨1 − 󵄨󵄨u 󵄨󵄨1 ) + λ{ [(u , (u ) ) − ((u ) , u )] − (󵄩󵄩󵄩u 󵄩󵄩󵄩 − 󵄩󵄩󵄩u 󵄩󵄩󵄩 )} = 0, 2τ τ 2τ which can be rewritten as 2 4 2 2 󵄩 󵄩2 󵄨 󵄨2 F 0 = 󵄨󵄨󵄨u0 󵄨󵄨󵄨1 + λ[ ((u0 ) , u1 ) − (u0 , (u1 ) ) − 󵄩󵄩󵄩u0 󵄩󵄩󵄩 ] ≡ F̂ 0 . 3 3

(1.98)

Taking the inner product of (1.91) with Δt uk on both the right- and left-hand sides yields 1 k−1 󵄩󵄩 k 󵄩󵄩2 k k+1 k k 2 k̄ k k̄ k 󵄩󵄩Δt u 󵄩󵄩 − (δx u , Δt u ) = λ[(u , Δt u ) − ((u + u + u )u , Δt u )], 3 1 ⩽ k ⩽ n − 1. Noticing −(δx2 uk , Δt uk ) = ̄

1 󵄨󵄨 k+1 󵄨󵄨2 󵄨󵄨 k−1 󵄨󵄨2 (󵄨u 󵄨󵄨1 − 󵄨󵄨u 󵄨󵄨1 ), 4τ 󵄨

(uk , Δt uk ) = ̄

1 󵄩󵄩 k+1 󵄩󵄩2 󵄩󵄩 k−1 󵄩󵄩2 (󵄩u 󵄩󵄩 − 󵄩󵄩u 󵄩󵄩 ) 4τ 󵄩

and 1 k−1 ((u + uk + uk+1 )uk , Δt uk ) 3 1 2 = [((uk+1 + uk−1 )uk , uk+1 − uk−1 ) + ((uk ) , uk+1 − uk−1 )] 6τ

1.7 Three-level linearized difference scheme

� 31

1 2 2 2 [(uk , (uk+1 ) − (uk−1 ) ) + ((uk ) , uk+1 − uk−1 )] 6τ 1 2 2 2 2 = [(uk , (uk+1 ) ) + ((uk ) , uk+1 ) − (uk−1 , (uk ) ) − ((uk−1 ) , uk )], 6τ =

we have |u 1 {[ 2τ

k+1 2 |1

+ |uk |21 (uk , (uk+1 )2 ) + ((uk )2 , uk+1 ) ‖uk+1 ‖2 + ‖uk ‖2 + λ( − )] 2 3 2

|uk |21 + |uk−1 |21 (uk−1 , (uk )2 ) + ((uk−1 )2 , uk ) ‖uk ‖2 + ‖uk−1 ‖2 + λ( − )]} 2 3 2 󵄩 󵄩2 + 󵄩󵄩󵄩Δt uk 󵄩󵄩󵄩 = 0, 1 ⩽ k ⩽ n − 1. −[

Replacing k by l in the equality above and summing over l from 1 to k will arrive at k 1 󵄨󵄨 k+1 󵄨󵄨2 󵄨󵄨 k 󵄨󵄨2 1 1 󵄩 2 2 󵄩2 󵄩 󵄩2 󵄩 󵄩2 (󵄨󵄨u 󵄨󵄨1 + 󵄨󵄨u 󵄨󵄨1 )+λ{ [(uk , (uk+1 ) )+((uk ) , uk+1 )]− (󵄩󵄩󵄩uk+1 󵄩󵄩󵄩 + 󵄩󵄩󵄩uk 󵄩󵄩󵄩 )} + 2τ ∑󵄩󵄩󵄩Δt ul 󵄩󵄩󵄩 2 3 2 l=1

1 󵄨 󵄨2 󵄨 󵄨2 1 1 󵄩 󵄩2 󵄩 󵄩2 2 2 = (󵄨󵄨󵄨u1 󵄨󵄨󵄨1 + 󵄨󵄨󵄨u0 󵄨󵄨󵄨1 ) + λ{ [(u0 , (u1 ) ) + ((u0 ) , u1 )] − (󵄩󵄩󵄩u1 󵄩󵄩󵄩 + 󵄩󵄩󵄩u0 󵄩󵄩󵄩 )}, 2 3 2

1 ⩽ k ⩽ n − 1. 1

Adding τ‖δt u 2 ‖2 on both the right- and left-hand sides of the equality above produces F k = F 0,

1 ⩽ k ⩽ n − 1.

(1.99)

Combining (1.98) and (1.99) arrives at (1.95).

1.7.2 Existence and convergence of the difference solution Theorem 1.10. Let {Uik | 0 ⩽ i ⩽ m, 0 ⩽ k ⩽ n} and {uik | 0 ⩽ i ⩽ m, 0 ⩽ k ⩽ n} be solutions of the problem (1.1)–(1.3) and the difference scheme (1.90)–(1.93), respectively. Denote eik = Uik − uik , c11 = √2TLc92 +

0 ⩽ i ⩽ m, 0 ⩽ k ⩽ n, 2 3c10

(c0 + 1)2 λ2 L

2 2 2

e(c0 +1) λ L T .

Then when c11 2L (τ 2 + h2 ) ⩽ 1, ( 21 + c0 )λτ ⩽ 1 and [1 + 32 (c0 + 1)2 ]λ2 L2 τ ⩽ 1, it holds that (I) the difference scheme (1.90)–(1.93) is uniquely solvable; √

32 � 1 Difference methods for the Fisher equation (II) 󵄨󵄨 k 󵄨󵄨 2 2 󵄨󵄨e 󵄨󵄨1 ⩽ c11 (τ + h ),

0 ⩽ k ⩽ n.

(1.100)

Proof. Subtracting (1.90)–(1.93) from (1.84), (1.86), (1.88)–(1.89), respectively, the system of error equations reads { { { { { { { { { { { { { { { { { { { { {

1

1

1

Δt eik

̄ δx2 eik

̄ λ[eik

δt ei2 − δx2 ei2 = λ[ei2 − (Ui0 Ui1 − ui0 ui1 )] + (R5 )0i ,

ei0 e0k

−

=

−

1 (Uik−1 3

+

1 ⩽ i ⩽ m − 1, 1 ⩽ k ⩽ n − 1, = 0,

= 0,

0 ⩽ i ⩽ m, k em

= 0,

Uik

+

Uik+1 )Uik

1 ⩽ i ⩽ m − 1, +

1 k−1 (u 3 i

+

uik

+

(1.101) uik+1 )uik ]

+

(R5 )ki ,

(1.102) (1.103)

1 ⩽ k ⩽ n.

(1.104)

The value of u0 is uniquely determined by (1.92) and the truth of (1.100) for k = 0 is obvious in view of (1.103). From (1.90) and (1.93), the system of linear equations in u1 can be obtained. Consider its homogeneous one: 1 1 u − 21 δx2 u1 = λ( 21 ui1 τ i 1 u01 = 0, um = 0.

{

− ui0 ui1 ),

1 ⩽ i ⩽ m − 1,

(1.105) (1.106)

Taking the inner product on both the right- and left-hand sides of (1.105) with u1 gives 1 󵄩󵄩 1 󵄩󵄩2 1 2 1 1 1 󵄩 1 󵄩2 0 1 1 󵄩u 󵄩 − (δx u , u ) = λ( 󵄩󵄩󵄩u 󵄩󵄩󵄩 − (u u , u )). τ󵄩 󵄩 2 2 Noticing −(δx2 u1 , u1 ) = |u1 |21 and ‖u0 ‖∞ ⩽ c0 , we have 1 󵄩󵄩 1 󵄩󵄩2 󵄩u 󵄩 + τ󵄩 󵄩

1 󵄨󵄨 1 󵄨󵄨2 1 󵄩 0󵄩 1 󵄩 1 󵄩2 󵄩 1 󵄩2 󵄨u 󵄨 ⩽ ( + 󵄩󵄩󵄩u 󵄩󵄩󵄩∞ )λ󵄩󵄩󵄩u 󵄩󵄩󵄩 ⩽ ( + c0 )λ󵄩󵄩󵄩u 󵄩󵄩󵄩 . 2 󵄨 󵄨1 2 2

When ( 21 + c0 )λτ ⩽ 1, it follows |u1 |1 = 0. Thus, (1.90) and (1.93) uniquely determine the value of u1 . 1 Taking the inner product of (1.101) on both the right- and left-hand sides with δt e 2 gives 1 󵄩󵄩 21 󵄩󵄩2 2 1 󵄩󵄩δt e 󵄩󵄩 − (δx e 2 , δt e 2 ) 1

1

1

1

= λ(e 2 , δt e 2 ) − λ(u0 e1 + e0 U 1 , δt e 2 ) + ((R5 )0 , δt e 2 ) 󵄩 1󵄩 󵄩 1󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 1󵄩 󵄩 󵄩 󵄩 1󵄩 ⩽ λ󵄩󵄩󵄩e 2 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩δt e 2 󵄩󵄩󵄩 + λ󵄩󵄩󵄩u0 󵄩󵄩󵄩∞ 󵄩󵄩󵄩e1 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩δt e 2 󵄩󵄩󵄩 + 󵄩󵄩󵄩(R5 )0 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩δt e 2 󵄩󵄩󵄩. Noticing

1.7 Three-level linearized difference scheme

� 33

1 1 󵄩󵄩 21 󵄩󵄩 󵄨󵄨 1 󵄨󵄨 󵄩 1 󵄩2 1 󵄨 1 󵄨2 󵄩󵄩 21 󵄩󵄩2 2 1 󵄩δ e 󵄩 ⋅ 󵄨e 󵄨 󵄩󵄩δt e 󵄩󵄩 − (δx e 2 , δt e 2 ) = 󵄩󵄩󵄩δt e 2 󵄩󵄩󵄩 + 󵄨󵄨󵄨e 󵄨󵄨󵄨1 ⩾ 2 √2τ 󵄩 t 󵄩 󵄨 󵄨1 2τ

and (1.85), we have 2 󵄨󵄨 1 󵄨󵄨 󵄩 0󵄩 󵄩 1󵄩 󵄩 󵄩 1󵄩 0󵄩 󵄨󵄨e 󵄨󵄨1 ⩽ λ󵄩󵄩󵄩e 2 󵄩󵄩󵄩 + λ󵄩󵄩󵄩u 󵄩󵄩󵄩∞ 󵄩󵄩󵄩e 󵄩󵄩󵄩 + 󵄩󵄩󵄩(R5 ) 󵄩󵄩󵄩 √2τ 1 󵄩 󵄩 󵄩 󵄩 ⩽ ( + c0 )λ󵄩󵄩󵄩e1 󵄩󵄩󵄩 + 󵄩󵄩󵄩(R5 )0 󵄩󵄩󵄩 2 1 L 󵄨 󵄨 ⩽ ( + c0 ) λ󵄨󵄨󵄨e1 󵄨󵄨󵄨1 + c9 √L(τ 2 + h2 ). √6 2 When ( 21 + c0 ) √L √2τλ ⩽ 1, i. e., 31 ( 21 + c0 )2 λ2 L2 τ ⩽ 1, it follows: 6

󵄨󵄨 1 󵄨󵄨 √ 2 2 2 2 󵄨󵄨e 󵄨󵄨1 ⩽ 2τc9 √L(τ + h ) ⩽ √2TLc9 (τ + h ),

(1.107)

which implies that (1.100) holds for k = 1. Now assume that the values of u0 , u1 , . . . , ul (l ⩾ 1) have been determined and (1.100) is true for 0 ⩽ k ⩽ l, i. e., 󵄨󵄨 k 󵄨󵄨 2 2 󵄨󵄨e 󵄨󵄨1 ⩽ c11 (τ + h ), Then by Lemma 1.1, when

√L c (τ 2 2 11

0 ⩽ k ⩽ l.

+ h2 ) ⩽ 1, we have

√L 󵄨 k 󵄨 √ 󵄩󵄩 k 󵄩󵄩 󵄨󵄨e 󵄨󵄨 ⩽ L c11 (τ 2 + h2 ) ⩽ 1, 0 ⩽ k ⩽ l, 󵄩󵄩e 󵄩󵄩∞ ⩽ 2 󵄨 󵄨1 2 󵄩󵄩 k 󵄩󵄩 󵄩󵄩 k 󵄩󵄩 󵄩󵄩 k 󵄩󵄩 󵄩󵄩u 󵄩󵄩∞ ⩽ 󵄩󵄩U 󵄩󵄩∞ + 󵄩󵄩e 󵄩󵄩∞ ⩽ c0 + 1, 0 ⩽ k ⩽ l. (I) Proof for the unique solvability. From (1.91) and (1.93), the system of linear equations in ul+1 can be obtained. Consider its homogeneous one: { { 2τ1 uil+1 − 21 δx2 uil+1 = λ( 21 uil+1 − 31 uil uil+1 ), { { l+1 l+1 { u0 = 0, um = 0.

1 ⩽ i ⩽ m − 1,

(1.108) (1.109)

Taking the inner product on both the right- and left-hand sides of (1.108) with ul+1 gives 1 󵄩󵄩 l+1 󵄩󵄩2 󵄩u 󵄩󵄩 + 2τ 󵄩

Noticing

1 󵄨󵄨 l+1 󵄨󵄨2 1 󵄩󵄩 l+1 󵄩󵄩2 λ 󵄩󵄩 l 󵄩󵄩 󵄩󵄩 l+1 󵄩󵄩2 󵄨u 󵄨󵄨1 ⩽ λ󵄩󵄩u 󵄩󵄩 + 󵄩󵄩u 󵄩󵄩∞ 󵄩󵄩u 󵄩󵄩 2󵄨 2 3 1 1 󵄩 󵄩2 ⩽ λ[ + (c0 + 1)]󵄩󵄩󵄩ul+1 󵄩󵄩󵄩 . 2 3

34 � 1 Difference methods for the Fisher equation 1 󵄩󵄩 l+1 󵄩󵄩2 󵄩u 󵄩󵄩 + 2τ 󵄩

1 󵄨󵄨 l+1 󵄨󵄨2 1 󵄩󵄩 l+1 󵄩󵄩 󵄨󵄨 l+1 󵄨󵄨 󵄨󵄨u 󵄨󵄨1 ⩾ 󵄩u 󵄩󵄩 ⋅ 󵄨󵄨u 󵄨󵄨1 , √τ 󵄩 2

we have 1 1 󵄨󵄨 l+1 󵄨󵄨 󵄩 l+1 󵄩 󵄨󵄨u 󵄨󵄨1 ⩽ λ[ + (c0 + 1)]√τ 󵄩󵄩󵄩u 󵄩󵄩󵄩 2 3 1 1 L 󵄨󵄨 l+1 󵄨󵄨 ⩽ λ[ + (c0 + 1)]√τ 󵄨u 󵄨󵄨1 . √6 󵄨 2 3 When 61 [ 21 + 31 (c0 + 1)]2 λ2 L2 τ < 1, it follows |ul+1 |1 = 0. Hence, (1.91) and (1.93) determine ul+1 uniquely. (II) Proof for (1.100). Taking the inner product on both the right- and left-hand sides of (1.102) with Δt ek yields 1 󵄨 k+1 󵄨2 󵄨 k−1 󵄨2 󵄩󵄩 k 󵄩󵄩2 󵄩󵄩Δt e 󵄩󵄩 + (󵄨󵄨󵄨e 󵄨󵄨󵄨1 − 󵄨󵄨󵄨e 󵄨󵄨󵄨1 ) 4τ

̄ 1 m−1 = λ(ek , Δt ek ) − λh ∑ [uik (eik−1 + eik + eik+1 ) + eik (Uik−1 + Uik + Uik+1 )]Δt eik + ((R5 )k , Δt ek ) 3 i=1

󵄩 ̄󵄩 󵄩 󵄩 1 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 ⩽ λ󵄩󵄩󵄩ek 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩Δt ek 󵄩󵄩󵄩 + λ󵄩󵄩󵄩uk 󵄩󵄩󵄩∞ ⋅ 󵄩󵄩󵄩ek−1 + ek + ek+1 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩Δt ek 󵄩󵄩󵄩 3 1 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 + λ󵄩󵄩󵄩ek 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩U k−1 + U k + U k+1 󵄩󵄩󵄩∞ ⋅ 󵄩󵄩󵄩Δt ek 󵄩󵄩󵄩 + 󵄩󵄩󵄩(R5 )k 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩Δt ek 󵄩󵄩󵄩 3 󵄩 󵄩 󵄩 󵄩 󵄩 ̄󵄩 󵄩 󵄩 1 ⩽ λ󵄩󵄩󵄩ek 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩Δt ek 󵄩󵄩󵄩 + λ(1 + c0 ) ⋅ 󵄩󵄩󵄩ek−1 + ek + ek+1 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩Δt ek 󵄩󵄩󵄩 3 1 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 + λ(3c0 )󵄩󵄩󵄩ek 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩Δt ek 󵄩󵄩󵄩 + 󵄩󵄩󵄩(R5 )k 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩Δt ek 󵄩󵄩󵄩 3

1󵄩 1󵄩 󵄩2 󵄩2 1 󵄩 󵄩 ̄ 󵄩2 󵄩2 ⩽ ( 󵄩󵄩󵄩Δt ek 󵄩󵄩󵄩 + λ2 󵄩󵄩󵄩ek 󵄩󵄩󵄩 ) + ( 󵄩󵄩󵄩Δt ek 󵄩󵄩󵄩 + λ2 (c0 + 1)2 󵄩󵄩󵄩ek−1 + ek + ek+1 󵄩󵄩󵄩 ) 4 4 9 1󵄩 1󵄩 󵄩 󵄩2 󵄩2 󵄩2 󵄩 󵄩2 + ( 󵄩󵄩󵄩Δt ek 󵄩󵄩󵄩 + λ2 c02 󵄩󵄩󵄩ek 󵄩󵄩󵄩 ) + 󵄩󵄩󵄩Δt ek 󵄩󵄩󵄩 + 󵄩󵄩󵄩(R5 )k 󵄩󵄩󵄩 , 1 ⩽ k ⩽ l. 4 4 Hence, noticing (1.87), we have 1 󵄨󵄨 k+1 󵄨󵄨2 󵄨󵄨 k−1 󵄨󵄨2 (󵄨e 󵄨󵄨1 − 󵄨󵄨e 󵄨󵄨1 ) 4τ 󵄨 λ2 󵄩 λ2 󵄩2 󵄩 󵄩2 󵄩 󵄩2 󵄩 󵄩2 󵄩 󵄩2 ⩽ (󵄩󵄩󵄩ek+1 󵄩󵄩󵄩 + 󵄩󵄩󵄩ek−1 󵄩󵄩󵄩 ) + (c0 + 1)2 (󵄩󵄩󵄩ek−1 󵄩󵄩󵄩 + 󵄩󵄩󵄩ek 󵄩󵄩󵄩 + 󵄩󵄩󵄩ek+1 󵄩󵄩󵄩 ) 2 3 2 󵄩 󵄩2 2 + λ2 c02 󵄩󵄩󵄩ek 󵄩󵄩󵄩 + c10 L(τ 2 + h2 ) λ2 λ2 λ2 λ2 󵄩 󵄩2 󵄩 󵄩2 = [ + (c0 + 1)2 ]󵄩󵄩󵄩ek+1 󵄩󵄩󵄩 + [ + (c0 + 1)2 ]󵄩󵄩󵄩ek−1 󵄩󵄩󵄩 2 3 2 3 +[

λ2 2 󵄩 󵄩2 2 (c + 1)2 + λ2 c02 ]󵄩󵄩󵄩ek 󵄩󵄩󵄩 + c10 L(τ 2 + h2 ) 3 0

1.7 Three-level linearized difference scheme

� 35

1 1 1 1 L2 󵄨 L2 󵄨 󵄨2 󵄨2 ⩽ [ + (c0 + 1)2 ]λ2 󵄨󵄨󵄨ek+1 󵄨󵄨󵄨1 + [ + (c0 + 1)2 ]λ2 󵄨󵄨󵄨ek−1 󵄨󵄨󵄨1 2 3 6 2 3 6 1 L2 󵄨 󵄨2 2 2 L(τ 2 + h2 ) , + [ (c0 + 1)2 + c02 ]λ2 󵄨󵄨󵄨ek 󵄨󵄨󵄨1 + c10 3 6

1 ⩽ k ⩽ l,

i. e., 2 L2 󵄨 󵄨2 {1 − [1 + (c0 + 1)2 ]λ2 τ}󵄨󵄨󵄨ek+1 󵄨󵄨󵄨1 3 3

2 L2 󵄨 L2 󵄨 󵄨2 2 2 󵄨2 2 ⩽ {1 + [1 + (c0 + 1)2 ]λ2 τ}󵄨󵄨󵄨ek−1 󵄨󵄨󵄨1 + [ (c0 + 1)2 + 2c02 ]λ2 τ 󵄨󵄨󵄨ek 󵄨󵄨󵄨1 + 4c10 Lτ(τ 2 + h2 ) , 3 3 3 3 1 ⩽ k ⩽ l. 2

When [1 + 32 (c0 + 1)2 ]λ2 L3 τ ⩽ 31 , it follows: 2 󵄨󵄨 k+1 󵄨󵄨2 󵄨 k−1 󵄨2 2 2 2 󵄨󵄨e 󵄨󵄨1 ⩽ {1 + [1 + (c0 + 1) ]λ L τ}󵄨󵄨󵄨e 󵄨󵄨󵄨1 3 1 2 󵄨 󵄨2 2 + [ (c0 + 1)2 + c02 ]λ2 L2 τ 󵄨󵄨󵄨ek 󵄨󵄨󵄨1 + 6c10 Lτ(τ 2 + h2 ) 3 2 󵄨 󵄨2 󵄨 󵄨2 2 Lτ(τ 2 + h2 ) , ⩽ [1 + 2(c0 + 1)2 λ2 L2 τ] max{󵄨󵄨󵄨ek 󵄨󵄨󵄨1 , 󵄨󵄨󵄨ek−1 󵄨󵄨󵄨1 } + 6c10

1 ⩽ k ⩽ l,

so that 2 󵄨 󵄨2 󵄨 󵄨2 󵄨 󵄨2 󵄨 󵄨2 2 max{󵄨󵄨󵄨ek+1 󵄨󵄨󵄨1 , 󵄨󵄨󵄨ek 󵄨󵄨󵄨1 } ⩽ [1 + 2(c0 + 1)2 λ2 L2 τ] max{󵄨󵄨󵄨ek 󵄨󵄨󵄨1 , 󵄨󵄨󵄨ek−1 󵄨󵄨󵄨1 } + 6c10 Lτ(τ 2 + h2 ) , 1 ⩽ k ⩽ l.

Applying the Gronwall inequality (Theorem 1.2(a)) and (1.107), we get 2 2 2 3c2 (τ 2 + h2 )2 󵄨 󵄨2 󵄨 󵄨2 󵄨 󵄨2 󵄨 󵄨2 ] max{󵄨󵄨󵄨el+1 󵄨󵄨󵄨1 , 󵄨󵄨󵄨el 󵄨󵄨󵄨1 } ⩽ e2(c0 +1) λ L lτ [max{󵄨󵄨󵄨e1 󵄨󵄨󵄨1 , 󵄨󵄨󵄨e0 󵄨󵄨󵄨1 } + 10 (c0 + 1)2 λ2 L 2 2 2

⩽ e2(c0 +1) λ L T [2TLc92 +

2 3c10 2 ](τ 2 + h2 ) . (c0 + 1)2 λ2 L

Then 2 3c10 󵄨󵄨 l+1 󵄨󵄨 (c +1)2 λ2 L2 T √ 2TLc92 + (τ 2 + h2 ) 󵄨󵄨e 󵄨󵄨1 ⩽ e 0 (c0 + 1)2 λ2 L

= c11 (τ 2 + h2 ), which implies that (1.100) is true for k = l + 1. By induction, the theorem is true. In view of Lemma 1.1(b), the difference scheme (1.90)–(1.93) is unconditionally convergent in the maximum norm and the convergence order is also O(τ 2 + h2 ).

36 � 1 Difference methods for the Fisher equation

1.8 Numerical experiments In this part, we are concerned with the numerical test for the above difference schemes. We take the difference scheme (1.90)–(1.93) as an example. Two numerical examples are used to test the computational efficiency of the difference scheme (1.90)–(1.93). Denote 󵄨 󵄨 E∞ (h, τ) = max󵄨󵄨󵄨Uik (h, τ) − uik (h, τ)󵄨󵄨󵄨, 0⩽i⩽m 0⩽k⩽n

Orderh = log2

E∞ (2h, τ) , E∞ (h, τ)

Orderτ = log2

󵄨󵄨 τ 󵄨󵄨󵄨 󵄨 F∞ (h, τ) = max󵄨󵄨󵄨uik (h, τ) − ui2k (h, )󵄨󵄨󵄨, 0⩽i⩽m 󵄨 2 󵄨󵄨 0⩽k⩽n 󵄨 󵄨󵄨 󵄨󵄨󵄨 󵄨 k h G∞ (h, τ) = max󵄨󵄨󵄨uik (h, τ) − u2i ( , τ)󵄨󵄨󵄨, 0⩽i⩽m 󵄨 󵄨󵄨 2 󵄨 0⩽k⩽n

E∞ (h, 2τ) ; E∞ (h, τ)

Ordτ = log2

F∞ (h, 2τ) ; F∞ (h, τ)

Ordh = log2

G∞ (2h, τ) . G∞ (h, τ)

Example 1.1 ([26]). In the problem (1.1)–(1.3), take L = 1, T = 1, λ = 6, φ(x) = (1 + ex )−2 , α(t) = (1 + e−5t )−2 , β(t) = (1 + e1−5t )−2 . The exact solution is given by u(x, t) = (1 + ex−5t )−2 . 1 and vary τ to observe the temporal convergence, and fix τ = 1/10000 We fix h = 400 and vary h to observe the spatial convergence. The maximum absolute error E∞ (h, τ) and related convergence orders are presented in Tables 1.1 and 1.2, respectively. From these two tables, it is clear that we obtain approximate second-order accuracy in both the temporal and the spatial directions, which is consistent with our theoretical results in Section 1.7.2. Table 1.1: (Example 1.1) The temporal convergence orders in maximum norm with h = 1/400. τ

E∞ (h, τ)

Orderτ

�/�� �/�� �/�� �/��� �/���

�.������e − � �.������e − � �.������e − � �.������e − � �.������e − �

�.�� �.�� �.�� �.��

Table 1.2: (Example 1.1) The spatial convergence orders in maximum norm with τ = 1/10000. h

E∞ (h, τ)

Orderh

�/�� �/�� �/�� �/�� �/���

�.������e − � �.������e − � �.������e − � �.������e − � �.������e − �

�.�� �.�� �.�� �.��

1.8 Numerical experiments

� 37

Example 1.2. In the problem (1.1)–(1.3), take L = 1, T = 1, λ = π 2 , φ(x) = sin(πx), α(t) = β(t) = 0. The exact solution is unknown. 1 1 1 1 1 1 1 and different temporal step sizes 80 , 160 , 320 , 640 , 1280 , 2560 . Table 1.3 Take h = 1000 presents the numerical error F∞ (h, τ) and temporal convergence order Ordτ in the maximum norm, demonstrating that the difference scheme (1.90)–(1.93) generates the temporal convergence of order two. Table 1.3: (Example 1.2) The temporal convergence orders in maximum norm with h = 1/1000. τ

F∞ (h, τ)

Ordτ

�/�� �/��� �/��� �/��� �/���� �/����

�.������e − � �.������e − � �.������e − � �.������e − � �.������e − �

�.�� �.�� �.�� �.��

1 1 1 1 1 1 Fix τ = 1000 and take different spatial step sizes 201 , 40 , 80 , 160 , 320 , 640 . Table 1.4 records the numerical error G∞ (h, τ) and spatial convergence order Ordh in the maximum norm, verifying that the difference scheme (1.90)–(1.93) generates the spatial convergence of order two.

Table 1.4: (Example 1.2) The spatial convergence orders in maximum norm with τ = 1/1000. h

G∞ (h, τ)

Ordh

�/�� �/�� �/�� �/��� �/��� �/���

�.������e − � �.������e − � �.������e − � �.������e − � �.������e − �

�.�� �.�� �.�� �.��

In order to verify the conservation of the difference scheme (1.90)–(1.93), we compute the discrete energy E k and F k defined in Theorem 1.9. Figure 1.1 collects the energy curves under different step sizes, where the exact energy E(t) = E(0) = ‖u(⋅, 0)‖2 = 1 1 1 ∫0 sin2 (πx)dx = 21 , F(t) = F(0) = π 2 ∫0 cos2 (πx)dx + π 2 ∫0 [ 32 sin3 (πx) − sin2 (πx)]dx = 8 π 9

≈ 2.79253. The conservation of the discrete energy E k and F k of the difference scheme (1.90)–(1.93) is numerically verified.

38 � 1 Difference methods for the Fisher equation

Figure 1.1: The conservation of energy E k and F k .

1.9 Summary and extension In this chapter, several difference schemes are introduced for solving the Fisher equation. The conservative property of the problem (1.1)–(1.3) is first proved and then the forward Euler difference scheme is discussed, which is explicit and conditionally convergent. The induction method is used to show the conditional convergence. Next, the backward Euler difference scheme, the Crank–Nicolson difference scheme, a fourthorder compact difference scheme and a three-level linearized difference scheme are introduced, respectively, where the former three are all two-level linearized ones, while the last one is three-level linearized implicit. The coefficient matrices of these implicit difference schemes are all tri-diagonal, which can be solved using the Thomas algorithm. The unique solvability and unconditional convergence of these difference schemes are proved by induction. In [29], Sun, Wu and Wang proposed the forward Euler difference scheme (1.13)– (1.15) and the backward Euler difference scheme (1.30)–(1.32), and then proved Theorem 1.3 and Theorem 1.5. In addition, they also derived a difference scheme similar to the Crank–Nicolson one. Considering equation (1.1) at the point (xi , tk+ 1 ), one has 2

ut (xi , tk+ 1 ) − uxx (xi , tk+ 1 ) = λ[u(xi , tk+ 1 ) − u(xi , tk )u(xi , tk+1 )] + O(τ 2 ). 2

2

2

Noticing τ u(xi , tk ) = u(xi , tk+ 1 ) − ut (xi , tk+ 1 ) + O(τ 2 ), 2 2 2 it follows: λ (1 − τ)ut (xi , tk+ 1 ) − uxx (xi , tk+ 1 ) = λ[u(xi , tk ) − u(xi , tk )u(xi , tk+1 )] + O(τ 2 ). 2 2 2

1.9 Summary and extension

� 39

Furthermore, one can get 1

1 + λ2 τ

ut (xi , tk+ 1 ) − uxx (xi , tk+ 1 ) = λu(xi , tk )[1 − u(xi , tk+1 )] + O(τ 2 ), 2

2

or λ λ ut (xi , tk+ 1 ) − (1 + τ)uxx (xi , tk+ 1 ) = λ(1 + τ)u(xi , tk )[1 − u(xi , tk+1 )] + O(τ 2 ). 2 2 2 2 Hence, a difference scheme was built in the form of k+ 1

k+ 1

{ δt ui 2 − (1 + λ2 τ)δx2 ui 2 = λ(1 + λ2 τ)uik (1 − uik+1 ), { { { { { 1 ⩽ i ⩽ m − 1, 0 ⩽ k ⩽ n − 1, { { { u0 = φ(xi ), 0 ⩽ i ⩽ m, { { { i k k 1 ⩽ k ⩽ n. { u0 = α(tk ), um = β(tk ),

(1.110) (1.111) (1.112)

For the difference scheme (1.110)–(1.112), the following result is true. Theorem 1.11 ([29]). Let {uik | 0 ⩽ i ⩽ m, 0 ⩽ k ⩽ n} be the solution of the difference scheme (1.110)–(1.112). If 0 ⩽ φ(x) ⩽ 1, 0 ⩽ α(t) ⩽ 1, 0 ⩽ β(t) ⩽ 1 and r ⩽ 1, then it holds that 0 ⩽ uik ⩽ 1,

0 ⩽ i ⩽ m, 0 ⩽ k ⩽ n.

It can be proved that the difference scheme (1.110)–(1.112) is uniquely solvable and unconditionally convergent with the convergence order O(τ 2 + h2 ) in the maximum norm by using the similar method in the proof for Theorem 1.7. For the problem (1.1)–(1.3), the following two-level nonlinear difference scheme can be established: 1

1

1

k+ k+ k+ { δt ui 2 − δx2 ui 2 = λ{ui 2 − 31 [(uik+1 )2 + uik uik+1 + (uik )2 ]}, { { { { { { 1 ⩽ i ⩽ m − 1, 0 ⩽ k ⩽ n − 1, { { 0 { { u = φ(xi ), 0 ⩽ i ⩽ m, { { { ik k 1 ⩽ k ⩽ n. { u0 = α(tk ), um = β(tk ),

(1.113) (1.114) (1.115)

The difference scheme satisfies the following conservative law: k 2 k+1 3 2 0 3 󵄨󵄨 k+1 󵄨󵄨2 󵄩 k+1 󵄩2 󵄩 l+ 1 󵄩2 󵄨 0 󵄨2 󵄩 0 󵄩2 󵄨󵄨u 󵄨󵄨1 + λ[ ((u ) , 1) − 󵄩󵄩󵄩u 󵄩󵄩󵄩 ] + 2τ ∑󵄩󵄩󵄩δt u 2 󵄩󵄩󵄩 = 󵄨󵄨󵄨u 󵄨󵄨󵄨1 + λ[ ((u ) , 1) − 󵄩󵄩󵄩u 󵄩󵄩󵄩 ], 3 3 l=0

0 ⩽ k ⩽ n − 1. Using the cut-off function method, which will be introduced in Chapter 10, one can show that this difference scheme is unconditionally convergent.

40 � 1 Difference methods for the Fisher equation Finally, we would like to mention that the equation ut (x, t) − uxx (x, t) = λu(x, t)[1 − u(x, t)] is usually called the Fisher-KPP equation or KPP equation [12]. The equation ut (x, t) − uxx (x, t) = λu(x, t)[1 − u2 (x, t)] is usually called the Allen–Cahn equation [3]. Many studies on the initial and boundary value problem of the later equation can be found.

2 Difference methods for the Burgers’ equation 2.1 Introduction The Burgers’ equation is a model equation that describes many physical phenomena, such as fluid dynamics, nonlinear acoustics, gas dynamics and traffic flow dynamics. The Burgers’ equation can also be used as a simplified model of the Navier–Stokes equation for fluid dynamics. In recent years, researchers have paid more and more attention to the numerical methods for the Burgers’ equation. Consider the initial and boundary value problem of a one-dimensional nonlinear Burgers’ equation: ut + uux = νuxx , { { u(x, 0) = φ(x), { { { u(0, t) = 0, u(L, t) = 0,

0 < x < L, 0 < t ⩽ T,

(2.1)

0 < x < L,

(2.2)

0 ⩽ t ⩽ T,

(2.3)

where ν > 0 is the dynamic viscosity coefficient, φ(x) is a given function satisfying φ(0) = φ(L) = 0. Before introducing the difference scheme, we first use the energy method to give a priori estimate on the solution of the problem (2.1)–(2.3). Theorem 2.1. Let u(x, t) be the solution of the problem (2.1)–(2.3). Denote L

t

L

0

0

E(t) = ∫ u (x, t)dx + 2ν ∫[∫ ux2 (x, s)dx]ds. 2

0

Then E(t) = E(0),

0 < t ⩽ T.

Proof. Multiplying (2.1) by u(x, t) on both the right- and left-hand sides, we have 1 1 ( u2 ) + ( u3 ) = ν[(uux )x − ux2 ]. 2 3 t x Integrating the above equality on both the right- and left-hand sides with respect to x on the interval [0, L] and applying (2.3), we have L

L

0

0

1 d ∫ u2 (x, t)dx + ν ∫ ux2 (x, t)dx = 0, 2 dt which can be written as https://doi.org/10.1515/9783110796018-002

42 � 2 Difference methods for the Burgers’ equation L

t

L

0

0

0

1 d {∫ u2 (x, t)dx + 2ν ∫[∫ ux2 (x, s)dx]ds} = 0, 2 dt that is to say dE(t) = 0, dt

0 < t ⩽ T.

E(t) = E(0),

0 < t ⩽ T.

Hence,

Obviously, we have 󵄩󵄩 󵄩2 2 󵄩󵄩u(⋅, t)󵄩󵄩󵄩 ⩽ E(t) = E(0) = ‖φ‖ .

(2.4)

Denote c0 = ‖φ‖. In the following, we will use the energy method to estimate the analytical solution in L∞ -norm. Theorem 2.2. Let u(x, t) be the solution of the problem (2.1)–(2.3). Then we have c2

c2

0 ( 1 + 0 )t 󵄨󵄨 󵄨 ′ 󵄨󵄨u(⋅, t)󵄨󵄨󵄨1 ⩽ ‖φ ‖e 2ν L ν2 ,

0 < t ⩽ T.

Proof. Multiplying by −uxx (x, t) on both the right- and left-hand sides of (2.1) and integrating the obtained equality with respect to x from 0 to L, we have 1 d 󵄩󵄩 󵄩2 󵄩 󵄩2 ⋅ 󵄩u (⋅, t)󵄩󵄩󵄩 + ν󵄩󵄩󵄩uxx (⋅, t)󵄩󵄩󵄩 2 dt 󵄩 x L

= ∫ u(x, t)ux (x, t)uxx (x, t)dx 0

L

󵄩 󵄩 󵄨 󵄨 ⩽ 󵄩󵄩󵄩ux (⋅, t)󵄩󵄩󵄩∞ ∫󵄨󵄨󵄨u(x, t)uxx (x, t)󵄨󵄨󵄨dx 0

󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 ⩽ 󵄩󵄩󵄩ux (⋅, t)󵄩󵄩󵄩∞ ⋅ 󵄩󵄩󵄩u(⋅, t)󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩uxx (⋅, t)󵄩󵄩󵄩 󵄩 󵄩 󵄩 󵄩 ⩽ c0 󵄩󵄩󵄩ux (⋅, t)󵄩󵄩󵄩∞ ⋅ 󵄩󵄩󵄩uxx (⋅, t)󵄩󵄩󵄩

ν 󵄩󵄩 󵄩2 󵄩u (⋅, t)󵄩󵄩󵄩 + 2 󵄩 xx ν󵄩 󵄩2 ⩽ 󵄩󵄩󵄩uxx (⋅, t)󵄩󵄩󵄩 + 2 ⩽

c02 󵄩󵄩 󵄩2 󵄩u (⋅, t)󵄩󵄩󵄩∞ 2ν 󵄩 x c02 󵄩󵄩 1 1 󵄩 󵄩2 󵄩2 [ε󵄩u (⋅, t)󵄩󵄩󵄩 + ( + )󵄩󵄩󵄩ux (⋅, t)󵄩󵄩󵄩 ], 2ν 󵄩 xx ε L

where the third inequality is obtained by using (2.4), and the last inequality is obtained by using the continuous counterpart of Lemma 1.1(f). Taking ε = we have

ν2 c02

such that

c02 ε 2ν

= ν2 ,

2.2 Two-level nonlinear difference scheme 2 2 d 󵄩󵄩 󵄩2 c 1 c 󵄩 󵄩2 󵄩󵄩ux (⋅, t)󵄩󵄩󵄩 ⩽ 0 ( + 02 )󵄩󵄩󵄩ux (⋅, t)󵄩󵄩󵄩 , dt ν L ν

� 43

0 < t ⩽ T.

Using the Gronwall inequality, we have c2

c2

c2

c2

1 󵄩2 0 ( 1 + 0 )t 󵄩2 󵄩 󵄨2 󵄩 󵄨󵄨 ′ 2 0 ( + 0 )t 󵄨󵄨u(⋅, t)󵄨󵄨󵄨1 = 󵄩󵄩󵄩ux (⋅, t)󵄩󵄩󵄩 ⩽ 󵄩󵄩󵄩ux (⋅, 0)󵄩󵄩󵄩 e ν L ν2 = ‖φ ‖ e ν L ν2 ,

0 < t ⩽ T.

This completes the proof. From Theorem 2.2 and the continuous counterpart of Lemma 1.1(b), we can obtain the following corollary. Corollary 2.1. Let u(x, t) be the solution of the problem (2.1)–(2.3). Then we have √L ′ c02 ( 1 + c022 )t 󵄩󵄩 󵄩 ‖φ ‖e 2ν L ν , 󵄩󵄩u(⋅, t)󵄩󵄩󵄩∞ ⩽ 2

0 < t ⩽ T.

2.2 Two-level nonlinear difference scheme 2.2.1 Derivation of the difference scheme Define the grid function U = {Uik | 0 ⩽ i ⩽ m, 0 ⩽ k ⩽ n} on Ωhτ , where Uik = u(xi , tk ),

0 ⩽ i ⩽ m, 0 ⩽ k ⩽ n.

Considering equation (2.1) at the point (xi , tk+ 1 ), we have 2

ut (xi , tk+ 1 ) + u(xi , tk+ 1 )ux (xi , tk+ 1 ) = νuxx (xi , tk+ 1 ), 2

2

2

2

1 ⩽ i ⩽ m − 1, 0 ⩽ k ⩽ n − 1. (2.5)

With the help of Lemma 1.2, we have k+ 21

ut (xi , tk+ 1 ) = δt Ui 2

+ O(τ 2 ),

1 u(xi , tk+ 1 ) = [u(xi−1 , tk+ 1 ) + u(xi , tk+ 1 ) + u(xi+1 , tk+ 1 )] + O(h2 ) 2 2 2 2 3 1 k+ 21 k+ 21 k+ 21 2 2 = (Ui−1 + Ui + Ui+1 ) + O(τ + h ), 3 1 ux (xi , tk+ 1 ) = [ux (xi , tk ) + ux (xi , tk+1 )] + O(τ 2 ) 2 2 1 = (Δx Uik + Δx Uik+1 ) + O(τ 2 + h2 ) 2 k+ 21

= Δx Ui

+ O(τ 2 + h2 ),

1 uxx (xi , tk+ 1 ) = [uxx (xi , tk ) + uxx (xi , tk+1 )] + O(τ 2 ) 2 2

(2.6)

(2.7)

(2.8)

44 � 2 Difference methods for the Burgers’ equation 1 = (δx2 Uik + δx2 Uik+1 ) + O(τ 2 + h2 ) 2 k+ 21

= δx2 Ui

+ O(τ 2 + h2 ).

(2.9)

Substituting (2.6)–(2.9) into (2.5), we have 1 k+ 1 k+ 1 k+ 1 k+ 1 k+ 1 k+ 1 + (Ui−1 2 + Ui 2 + Ui+1 2 )Δx Ui 2 = νδx2 Ui 2 + Ri 2 , 3 1 ⩽ i ⩽ m − 1, 0 ⩽ k ⩽ n − 1. k+ 21

δt Ui

(2.10)

There is a positive constant c1 such that 󵄨󵄨 k+ 21 󵄨󵄨 2 2 󵄨󵄨Ri 󵄨󵄨 ⩽ c1 (τ + h ),

1 ⩽ i ⩽ m − 1, 0 ⩽ k ⩽ n − 1.

(2.11)

Noticing the initial-boundary value conditions (2.2)–(2.3), we have {

Ui0 = φ(xi ), U0k

k+ 21

Omitting the small term Ri (2.1)–(2.3) reads

= 0,

Umk

= 0,

1 ⩽ i ⩽ m − 1,

(2.12)

0 ⩽ k ⩽ n.

(2.13)

in (2.10) and replacing Uik by uik , a difference scheme for

1 1 k+ 21 k+ 21 k+ 21 k+ 21 k+ 21 2 k+ 2 { δ u + + u + u )Δ u = νδ u , (u { t x x i { i i i+1 i { 3 i−1 { { { 1 ⩽ i ⩽ m − 1, 0 ⩽ k ⩽ n − 1, { { { ui0 = φ(xi ), 1 ⩽ i ⩽ m − 1, { { { { k k 0 ⩽ k ⩽ n. { u0 = 0, um = 0,

The difference scheme (2.14)–(2.16) is a two-level nonlinear difference scheme. 2.2.2 Conservation and boundedness of the difference solution The nonlinear term in (2.14) can be rewritten as follows: 1 k+ 21 k+ 1 k+ 1 k+ 1 (ui−1 + ui 2 + ui+1 2 )Δx ui 2 3 1 k+ 1 k+ 1 k+ 1 k+ 1 k+ 1 = [ui 2 Δx ui 2 + (ui+1 2 + ui−1 2 )Δx ui 2 ] 3 1 1 1 k+ 1 k+ 1 = [ui 2 Δx ui 2 + Δx (uk+ 2 uk+ 2 )i ]. 3 For any v, w ∈ 𝒰h , define 1 ψ(v, w)i = [vi Δx wi + Δx (vw)i ], 3

1 ⩽ i ⩽ m − 1.

(2.14) (2.15) (2.16)

2.2 Two-level nonlinear difference scheme

� 45

Then 1 1 1 k+ 21 k+ 1 k+ 1 k+ 1 (ui−1 + ui 2 + ui+1 2 )Δx ui 2 = ψ(uk+ 2 , uk+ 2 )i , 3

1 1 1 k+ 21 k+ 1 k+ 1 k+ 1 (Ui−1 + Ui 2 + Ui+1 2 )Δx Ui 2 = ψ(U k+ 2 , U k+ 2 )i . 3

And (2.14) can be rewritten as k+ 21

δt ui

1

k+ 21

1

+ ψ(uk+ 2 , uk+ 2 )i = νδx2 ui

,

1 ⩽ i ⩽ m − 1, 0 ⩽ k ⩽ n − 1. 1

1

Reformulate uux as 31 [uux + (u2 )x ]. Then ψ(U k+ 2 , U k+ 2 )i can be viewed as the discretization of 31 [uux + (u2 )x ] at the point (xi , tk+ 1 ). 2 The operator ψ has the following property. Lemma 2.1. Suppose v ∈ 𝒰h , w ∈ 𝒰h̊ , then (ψ(v, w), w) = 0. Proof.

(ψ(v, w), w) 1 = (vΔx w + Δx (vw), w) 3 1 = [(vΔx w, w) + (Δx (vw), w)] 3 1 = [(Δx w, vw) + (Δx (vw), w)] 3 = 0.

Theorem 2.3. Let {uik | 0 ⩽ i ⩽ m, 0 ⩽ k ⩽ n} be the solution of the difference scheme (2.14)–(2.16). Denote k−1

󵄩 󵄩2 󵄨 1 󵄨2 E k = 󵄩󵄩󵄩uk 󵄩󵄩󵄩 + 2ντ ∑ 󵄨󵄨󵄨ul+ 2 󵄨󵄨󵄨1 , l=0

0 ⩽ k ⩽ n,

then Ek = E0 ,

1 ⩽ k ⩽ n.

(2.17)

Proof. Notice that the difference equation (2.14) can be written as k+ 21

δt ui

k+ 21

1

1

+ ψ(uk+ 2 , uk+ 2 )i − νδx2 ui

= 0,

1 ⩽ i ⩽ m − 1, 0 ⩽ k ⩽ n − 1.

Taking the inner product on both the right- and left-hand sides of the above equality 1 with uk+ 2 , we have 1

1

1

1

1

1

1

(δt uk+ 2 , uk+ 2 ) + (ψ(uk+ 2 , uk+ 2 ), uk+ 2 ) − ν(δx2 uk+ 2 , uk+ 2 ) = 0.

46 � 2 Difference methods for the Burgers’ equation 1

Noticing that uk+ 2 ∈ 𝒰h̊ , we have 1

1 󵄩󵄩 k+1 󵄩󵄩2 󵄩󵄩 k 󵄩󵄩2 (󵄩u 󵄩󵄩 − 󵄩󵄩u 󵄩󵄩 ), 2τ 󵄩

1

(δt uk+ 2 , uk+ 2 ) = 1

1

1

(ψ(uk+ 2 , uk+ 2 ), uk+ 2 ) = 0, 1 2 1 1 󵄨 󵄨 −(δx2 uk+ 2 , uk+ 2 ) = 󵄨󵄨󵄨uk+ 2 󵄨󵄨󵄨1 . Therefore, 1 2 1 󵄩󵄩 k+1 󵄩󵄩2 󵄩󵄩 k 󵄩󵄩2 󵄨 󵄨 (󵄩󵄩u 󵄩󵄩 − 󵄩󵄩u 󵄩󵄩 ) + ν󵄨󵄨󵄨uk+ 2 󵄨󵄨󵄨1 = 0, 2τ

0 ⩽ k ⩽ n − 1.

Replacing k by l in the above equality and summing over l from 0 to k − 1, we have k−1 1 󵄩󵄩 k 󵄩󵄩2 󵄩󵄩 0 󵄩󵄩2 󵄨 1 󵄨2 (󵄩󵄩u 󵄩󵄩 − 󵄩󵄩u 󵄩󵄩 ) + ν ∑ 󵄨󵄨󵄨ul+ 2 󵄨󵄨󵄨1 = 0, 2τ l=0

1 ⩽ k ⩽ n.

It yields (2.17) by rewriting the above equality. From Theorem 2.3, it is easy to know that 󵄩󵄩 k 󵄩󵄩 󵄩󵄩 0 󵄩󵄩 󵄩󵄩u 󵄩󵄩 ⩽ 󵄩󵄩u 󵄩󵄩 ≡ c2 ,

1 ⩽ k ⩽ n.

(2.18)

2.2.3 Existence and uniqueness of the difference solution We will prove the existence of the solution of the difference scheme by means of the following Browder theorem. Theorem 2.4 (Browder theorem [2, 5]). Let (H, (⋅, ⋅)) be a finite-dimensional inner product space, with ‖ ⋅ ‖ the associated norm, and Π : H → H be continuous. Assume moreover that there exists an α > 0, for any z ∈ H and ‖z‖ = α, it holds that Re(Π(z), z) ⩾ 0. Then there is a z∗ ∈ H satisfying ‖z∗ ‖ ⩽ α such that Π(z∗ ) = 0. Theorem 2.5. The difference scheme (2.14)–(2.16) has a solution. Proof. From (2.15)–(2.16), we get u0 . Suppose that the numerical solution at the k-th time level has been obtained. Let k+ 21

wi = ui we then have

,

0 ⩽ i ⩽ m,

2.2 Two-level nonlinear difference scheme

uik+1 = 2wi − uik ,

� 47

0 ⩽ i ⩽ m.

According to (2.14) and (2.16), we have the system of nonlinear equations in w = (w0 , w1 , . . . , wm ) as { 2 (wi − uik ) + ψ(w, w)i − νδx2 wi = 0, { τ { w0 = 0, wm = 0.

1 ⩽ i ⩽ m − 1,

(2.19) (2.20)

For any u, v ∈ 𝒰h̊ , define the inner product m−1

(u, v) = h ∑ ui vi , i=1

then 𝒰h̊ is an inner product space with the induced norm ‖u‖ = √(u, u). Define the operator Π : 𝒰h̊ → 𝒰h̊ by 2 (wi − uik ) + ψ(w, w)i − νδx2 wi , Π(w)i = { τ 0,

1 ⩽ i ⩽ m − 1, i = 0, m.

Then Π(w) is a continuous function in 𝒰h̊ . Applying Lemma 2.1, we have 2 [(w, w) − (uk , w)] + (ψ(w, w), w) − ν(δx2 w, w) τ 2 = [‖w‖2 − (uk , w)] + ν|w|21 τ 2 ⩾ (‖w‖2 − ‖uk ‖ ⋅ ‖w‖) τ 2 = (‖w‖ − ‖uk ‖) ⋅ ‖w‖. τ

(Π(w), w) =

Therefore, (Π(w), w) ⩾ 0 when ‖w‖ = ‖uk ‖. By Theorem 2.4, there is a w∗ ∈ 𝒰h̊ satisfying ‖w∗ ‖ ⩽ ‖uk ‖ such that Π(w∗ ) = 0; namely, the system (2.19)–(2.20) has a solution w∗ . Theorem 2.6. If τ
0, we have ‖z‖∞ ⩽ ε|z|1 +

1 ‖z‖. 4ε

Hence, 1 2 2 ‖z‖ + ν|z|21 ⩽ c2 (ε|z|1 + ‖z‖)|z|1 τ 4ε c = c2 ε|z|21 + 2 ‖z‖ ⋅ |z|1 4ε ⩽ c2 ε|z|21 + c2 ε|z|21 + = 2c2 ε|z|21 + in which ε =

ν 2c2

2

c 1 ( 2 ) ‖z‖2 4c2 ε 4ε

c2 ‖z‖2 , 64ε3

gives c4 2 2 ‖z‖ ⩽ 23 ‖z‖2 . τ 8ν

When τ < unique.

16ν3 , c24

we have ‖z‖ = 0, which reveals that the solution of (2.19)–(2.20) is

2.2.4 Convergence of the difference solution Denote c3 =

max

0⩽x⩽L,0⩽t⩽T

󵄨 󵄨󵄨 󵄨 {󵄨󵄨󵄨u(x, t)󵄨󵄨󵄨, 󵄨󵄨󵄨ux (x, t)󵄨󵄨󵄨}.

(2.28)

Theorem 2.7. Let {Uik | 0 ⩽ i ⩽ m, 0 ⩽ k ⩽ n} be the solution of (2.1)–(2.3) and {uik | 0 ⩽ i ⩽ m, 0 ⩽ k ⩽ n} be the solution of the difference scheme (2.14)–(2.16), and denote eik = Uik − uik ,

0 ⩽ i ⩽ m, 0 ⩽ k ⩽ n.

Then there is a constant c4 such that 󵄩󵄩 k 󵄩󵄩 2 2 󵄩󵄩e 󵄩󵄩 ⩽ c4 (τ + h ),

0 ⩽ k ⩽ n.

50 � 2 Difference methods for the Burgers’ equation Proof. Subtracting (2.14)–(2.16) from (2.10), (2.12) and (2.13), respectively, we have the system of error equations: k+ 1

1

1

1

k+ 1

k+ 1

1

{ δt ei 2 + ψ(U k+ 2 , U k+ 2 )i − ψ(uk+ 2 , uk+ 2 )i = νδx2 ei 2 + Ri 2 , { { { { { 1 ⩽ i ⩽ m − 1, 0 ⩽ k ⩽ n − 1, { { { e0 = 0, 1 ⩽ i ⩽ m − 1, { { { i k k 0 ⩽ k ⩽ n. { e0 = 0, em = 0,

(2.29) (2.30) (2.31) 1

Taking the inner product of (2.29) on both the right- and left-hand sides with ek+ 2 , we have 1

1

1

1

1

1

1

1

(δt ek+ 2 , ek+ 2 ) + (ψ(U k+ 2 , U k+ 2 ) − ψ(uk+ 2 , uk+ 2 ), ek+ 2 ) + ν|ek+ 2 |21 1

1

= (Rk+ 2 , ek+ 2 ),

0 ⩽ k ⩽ n − 1.

(2.32)

It is easy to know that 1

1

(δt ek+ 2 , ek+ 2 ) =

1 󵄩󵄩 k+1 󵄩󵄩2 (󵄩e 󵄩󵄩 − ‖ek ‖2 ). 2τ 󵄩

(2.33)

Noticing that 1

1

1

1

1

1

ψ(U k+ 2 , U k+ 2 ) − ψ(uk+ 2 , uk+ 2 ) 1

1

1

1

= ψ(U k+ 2 , U k+ 2 ) − ψ(U k+ 2 − ek+ 2 , U k+ 2 − ek+ 2 ) 1

1

1

1

1

1

= ψ(ek+ 2 , U k+ 2 ) + ψ(U k+ 2 , ek+ 2 ) − ψ(ek+ 2 , ek+ 2 ) and using Lemma 2.1 again, we have 1

1

1

1

1

−(ψ(U k+ 2 , U k+ 2 ) − ψ(uk+ 2 , uk+ 2 ), ek+ 2 ) 1

1

1

= −(ψ(ek+ 2 , U k+ 2 ), ek+ 2 ) 1 1 1 m−1 k+ 1 k+ 1 k+ 1 = − h ∑ [ei 2 Δx Ui 2 + Δx (ek+ 2 U k+ 2 )i ]ei 2 3 i=1

1 m−1 k+ 1 k+ 1 1 m−1 k+ 1 2 k+ 1 k+ 1 = − h ∑ (ei 2 ) Δx Ui 2 + h ∑ ei 2 Ui 2 Δx ei 2 3 i=1 3 i=1 k+ 1

k+ 21

2 − Ui 1 m−1 k+ 1 2 1 m−1 U k+ 1 = − h ∑ (ei 2 ) Δx Ui 2 − h ∑ i+1 3 i=1 6 i=0 h

1 󵄩 1 󵄩2 ⩽ c3 󵄩󵄩󵄩ek+ 2 󵄩󵄩󵄩 . 2

Substituting (2.33) and (2.34) into (2.32), we have

k+ 21 k+ 21 ei+1

ei

(2.34)

2.2 Two-level nonlinear difference scheme

� 51

1 󵄩󵄩 k+1 󵄩󵄩2 󵄩󵄩 k 󵄩󵄩2 (󵄩e 󵄩󵄩 − 󵄩󵄩e 󵄩󵄩 ) 2τ 󵄩 1 1 󵄩 1 󵄩2 󵄩 󵄩 󵄩 1󵄩 ⩽ c3 󵄩󵄩󵄩ek+ 2 󵄩󵄩󵄩 + 󵄩󵄩󵄩Rk+ 2 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩ek+ 2 󵄩󵄩󵄩 2 2

k k+1 1 1 ‖ek ‖ + ‖ek+1 ‖ 󵄩 󵄩 ‖e ‖ + ‖e ‖ ⩽ c3 ( ) + 󵄩󵄩󵄩Rk+ 2 󵄩󵄩󵄩 ⋅ , 2 2 2

It follows after eliminating

‖ek+1 ‖+‖ek ‖ 2

0 ⩽ k ⩽ n − 1.

on both the right- and left-hand sides that

1 c 󵄩 󵄩 󵄩 1 󵄩󵄩 k+1 󵄩󵄩 󵄩󵄩 k 󵄩󵄩 󵄩 󵄩 󵄩 (󵄩e 󵄩󵄩 − 󵄩󵄩e 󵄩󵄩) ⩽ 3 (󵄩󵄩󵄩ek 󵄩󵄩󵄩 + 󵄩󵄩󵄩ek+1 󵄩󵄩󵄩) + 󵄩󵄩󵄩Rk+ 2 󵄩󵄩󵄩, τ 󵄩 4

0 ⩽ k ⩽ n − 1,

i. e., (1 − When

c3 τ 4

1 c3 󵄩󵄩 k+1 󵄩󵄩 cτ 󵄩 󵄩 󵄩 󵄩 τ)󵄩e 󵄩󵄩 ⩽ (1 + 3 )󵄩󵄩󵄩ek 󵄩󵄩󵄩 + τ 󵄩󵄩󵄩Rk+ 2 󵄩󵄩󵄩, 4 󵄩 4

0 ⩽ k ⩽ n − 1.

⩽ 31 , in view of (2.11), we have 3c τ 󵄩 k 󵄩 󵄩󵄩 k+1 󵄩󵄩 󵄩󵄩e 󵄩󵄩 ⩽ (1 + 3 )󵄩󵄩󵄩e 󵄩󵄩󵄩 + 4 3c 󵄩 󵄩 ⩽ (1 + 3 τ)󵄩󵄩󵄩ek 󵄩󵄩󵄩 + 4

3 󵄩󵄩 k+ 21 󵄩󵄩 τ 󵄩R 󵄩󵄩 2 󵄩 3 √ c Lτ(τ 2 + h2 ), 2 1

0 ⩽ k ⩽ n − 1.

Using the Gronwall inequality (Theorem 1.2(a)), and noticing that ‖e0 ‖ = 0, we have 3c3 󵄩󵄩 k+1 󵄩󵄩 kτ 2c √L 2 2 2 2 󵄩󵄩e 󵄩󵄩 ⩽ e 4 ⋅ 1 (τ + h ) ⩽ c4 (τ + h ), c3

where c4 = e

3c3 4

T

⋅

0 ⩽ k ⩽ n − 1,

2c1 √L . c3

Theorem 2.8. Let {Uik | 0 ⩽ i ⩽ m, 0 ⩽ k ⩽ n} be the solution of (2.1)–(2.3) and {uik | 0 ⩽ i ⩽ m, 0 ⩽ k ⩽ n} be the solution of the difference scheme (2.14)–(2.16), respectively. Denote eik = Uik − uik ,

0 ⩽ i ⩽ m, 0 ⩽ k ⩽ n

and c3̂ =

125(c3 √L + c2 )4 5(c3 √L + c2 )2 5c32 L2 + + (1 + ), 4νL 4ν 6 16ν3

c4̂ =

c1 32 c3̂ T 5L √ e . 2 νc3̂

Then we have 󵄨󵄨 k 󵄨󵄨 2 2 󵄨󵄨e 󵄨󵄨1 ⩽ c4̂ (τ + h ),

0 ⩽ k ⩽ n.

52 � 2 Difference methods for the Burgers’ equation Proof. Noticing (2.18) and (2.28), we have 󵄩 k󵄩 󵄩 k󵄩 󵄩󵄩 k 󵄩󵄩 󵄩󵄩 k k󵄩 󵄩󵄩e 󵄩󵄩 = 󵄩󵄩U − u 󵄩󵄩󵄩 ⩽ 󵄩󵄩󵄩U 󵄩󵄩󵄩 + 󵄩󵄩󵄩u 󵄩󵄩󵄩 ⩽ c3 √L + c2 ,

0 ⩽ k ⩽ n.

(2.35)

From (2.31), we have k+ 21

e0

k+ 21

= 0,

em

= 0,

0 ⩽ k ⩽ n − 1.

(2.36) 1

Taking the inner product of (2.29) on both the right- and left-hand sides with −δx2 ek+ 2 , we obtain 1 1 󵄨 1 󵄨2 (δt ek+ 2 , −δx2 ek+ 2 ) + ν󵄨󵄨󵄨ek+ 2 󵄨󵄨󵄨2 1

1

1

1

1

1

1

= (ψ(U k+ 2 , U k+ 2 ) − ψ(uk+ 2 , uk+ 2 ), δx2 ek+ 2 ) − (Rk+ 2 , δx2 ek+ 2 ) 1

1

1

1

1

1

1

1

= (ψ(U k+ 2 , ek+ 2 ) + ψ(ek+ 2 , U k+ 2 ), δx2 ek+ 2 ) − (ψ(ek+ 2 , ek+ 2 ), δx2 ek+ 2 ) 1

1

−(Rk+ 2 , δx2 ek+ 2 ),

0 ⩽ k ⩽ n − 1.

(2.37)

For the first term on the left-hand side of (2.37), using the summation by parts and noticing (2.36), we have 1

1

(δt ek+ 2 , −δx2 ek+ 2 ) =

1 󵄨󵄨 k+1 󵄨󵄨2 󵄨󵄨 k 󵄨󵄨2 (󵄨e 󵄨󵄨1 − 󵄨󵄨e 󵄨󵄨1 ). 2τ 󵄨

(2.38)

For the first term on the right-hand side of (2.37), using the Cauchy–Schwarz inequality and Lemma 1.1(b), we have 1

1

1

1

1

(ψ(U k+ 2 , ek+ 2 ) + ψ(ek+ 2 , U k+ 2 ), δx2 ek+ 2 )

= =

h m−1 k+ 21 k+ 1 k+ 1 k+ 1 k+ 1 k+ 1 ∑ [Ui Δx ei 2 + ei 2 Δx Ui 2 + 2Δx (Ue)i 2 ] ⋅ δx2 ei 2 3 i=1

h m−1 k+ 21 k+ 1 k+ 1 k+ 1 k+ 1 k+ 1 k+ 1 k+ 1 k+ 1 ∑ (Ui Δx ei 2 + 3ei 2 Δx Ui 2 + Ui+1 2 δx e 12 + Ui−1 2 δx e 12 ) ⋅ δx2 ei 2 i+ 2 i− 2 3 i=1

󵄨 1 󵄨 󵄩 1 󵄩 󵄨 1 󵄨2 ⩽ c3 (󵄨󵄨󵄨ek+ 2 󵄨󵄨󵄨1 + 󵄩󵄩󵄩ek+ 2 󵄩󵄩󵄩) ⋅ 󵄨󵄨󵄨ek+ 2 󵄨󵄨󵄨2 󵄩 1 󵄩 󵄨 1 󵄨2 󵄨 1 󵄨 󵄨 1 󵄨2 = c3 󵄨󵄨󵄨ek+ 2 󵄨󵄨󵄨1 ⋅ 󵄨󵄨󵄨ek+ 2 󵄨󵄨󵄨2 + c3 󵄩󵄩󵄩ek+ 2 󵄩󵄩󵄩 ⋅ 󵄨󵄨󵄨ek+ 2 󵄨󵄨󵄨2 2 2 ν 󵄨 1 󵄨2 5c 󵄨 1 󵄨2 ν 󵄨 1 󵄨2 5c 󵄩 1 󵄩2 ⩽ ( 󵄨󵄨󵄨ek+ 2 󵄨󵄨󵄨2 + 3 󵄨󵄨󵄨ek+ 2 󵄨󵄨󵄨1 ) + ( 󵄨󵄨󵄨ek+ 2 󵄨󵄨󵄨2 + 3 󵄩󵄩󵄩ek+ 2 󵄩󵄩󵄩 ) 5 4ν 5 4ν ⩽

2 2ν 󵄨󵄨 k+ 21 󵄨󵄨2 5c3 L2 󵄨 1 󵄨2 (1 + )󵄨󵄨󵄨ek+ 2 󵄨󵄨󵄨1 . 󵄨󵄨e 󵄨󵄨2 + 5 4ν 6

For the second term on the right-hand side of (2.37), we have 1

1

1

−(ψ(ek+ 2 , ek+ 2 ), δx2 ek+ 2 )

(2.39)

2.2 Two-level nonlinear difference scheme

=− ⩽

� 53

h m−1 k+ 21 k+ 1 k+ 1 k+ 1 k+ 1 ∑ (ei+1 + ei 2 + ei−1 2 )Δx ei 2 ⋅ δx2 ei 2 3 i=1

h m−1 󵄨󵄨 k+ 21 󵄨󵄨 󵄨󵄨 k+ 21 󵄨󵄨 󵄨󵄨 k+ 21 󵄨󵄨 󵄨󵄨 k+ 21 󵄨󵄨 󵄨󵄨 2 k+ 21 󵄨󵄨 ∑ (󵄨e 󵄨 + 󵄨e 󵄨 + 󵄨e 󵄨)󵄨Δ e 󵄨 ⋅ 󵄨δ e 󵄨 3 i=1 󵄨 i+1 󵄨 󵄨 i 󵄨 󵄨 i−1 󵄨 󵄨 x i 󵄨 󵄨 x i 󵄨

1 h m−1 󵄨 k+ 1 󵄨 󵄨 k+ 1 󵄨 󵄨 k+ 1 󵄨 󵄨 k+ 1 󵄨 󵄩 󵄩 ⩽ 󵄩󵄩󵄩δx ek+ 2 󵄩󵄩󵄩∞ ⋅ ∑ (󵄨󵄨󵄨ei+1 2 󵄨󵄨󵄨 + 󵄨󵄨󵄨ei 2 󵄨󵄨󵄨 + 󵄨󵄨󵄨ei−1 2 󵄨󵄨󵄨) ⋅ 󵄨󵄨󵄨δx2 ei 2 󵄨󵄨󵄨 3 i=1

1 󵄩 󵄩 󵄩 1 󵄩 󵄨 1 󵄨2 ⩽ 󵄩󵄩󵄩δx ek+ 2 󵄩󵄩󵄩∞ ⋅ 󵄩󵄩󵄩ek+ 2 󵄩󵄩󵄩 ⋅ 󵄨󵄨󵄨ek+ 2 󵄨󵄨󵄨2 1 󵄩 󵄩 󵄨 1 󵄨2 ⩽ (c3 √L + c2 )󵄩󵄩󵄩δx ek+ 2 󵄩󵄩󵄩∞ 󵄨󵄨󵄨ek+ 2 󵄨󵄨󵄨2 1 2 ν 󵄨 1 󵄨2 5 󵄩 󵄩 ⩽ 󵄨󵄨󵄨ek+ 2 󵄨󵄨󵄨2 + (c3 √L + c2 )2 󵄩󵄩󵄩δx ek+ 2 󵄩󵄩󵄩∞ 5 4ν ν 󵄨 1 󵄨2 5 1 1 󵄨 1 󵄨2 󵄨 1 󵄨2 ⩽ 󵄨󵄨󵄨ek+ 2 󵄨󵄨󵄨2 + (c3 √L + c2 )2 [ε󵄨󵄨󵄨ek+ 2 󵄨󵄨󵄨2 + ( + )󵄨󵄨󵄨ek+ 2 󵄨󵄨󵄨1 ] 5 4ν ε L 25(c3 √L + c2 )2 1 󵄨󵄨 k+ 21 󵄨󵄨2 2ν 󵄨 1 󵄨2 5 + )󵄨󵄨e 󵄨󵄨1 , = 󵄨󵄨󵄨ek+ 2 󵄨󵄨󵄨2 + (c3 √L + c2 )2 ( 5 4ν L 4ν2

(2.40)

where in the fourth inequality, we have used (2.35); In the last inequality, we have used Lemma 1.1(f); In the last equality, we have taken ε=

4ν2 . 25(c3 √L + c2 )2

For the third term on the right-hand side of (2.37), using the Cauchy–Schwarz inequality and noticing (2.11), we have 1

1

− (Rk+ 2 , δx2 ek+ 2 ) ⩽

ν 󵄨󵄨 k+ 21 󵄨󵄨2 5 󵄩󵄩 k+ 21 󵄩󵄩2 ν 󵄨󵄨 k+ 21 󵄨󵄨2 5 2 2 2 2 󵄨e 󵄨󵄨2 + 󵄩󵄩R 󵄩󵄩 ⩽ 󵄨󵄨e 󵄨󵄨2 + Lc1 (τ + h ) . 5󵄨 4ν 5 4ν

Substituting (2.38)–(2.41) into (2.37), we get 1 󵄨󵄨 k+1 󵄨󵄨2 󵄨󵄨 k 󵄨󵄨2 (󵄨e 󵄨󵄨1 − 󵄨󵄨e 󵄨󵄨1 ) 2τ 󵄨 125(c3 √L + c2 )4 5(c3 √L + c2 )2 5c32 L2 󵄨󵄨 k+ 21 󵄨󵄨2 + + (1 + )]󵄨e 󵄨󵄨1 ⩽[ 4νL 4ν 6 󵄨 16ν3 5 2 + Lc12 (τ 2 + h2 ) , 0 ⩽ k ⩽ n − 1, 4ν or 2 2 󵄨 󵄨2 󵄨 󵄨2 5Lc1 (1 − c3̂ τ)󵄨󵄨󵄨ek+1 󵄨󵄨󵄨1 ⩽ (1 + c3̂ τ)󵄨󵄨󵄨ek 󵄨󵄨󵄨1 + τ(τ 2 + h2 ) , 2ν

0 ⩽ k ⩽ n − 1.

When c3̂ τ ⩽ 31 , we have 2 2 󵄨󵄨 k+1 󵄨󵄨2 󵄨 k 󵄨2 15Lc1 τ(τ 2 + h2 ) , 󵄨󵄨e 󵄨󵄨1 ⩽ (1 + 3c3̂ τ)󵄨󵄨󵄨e 󵄨󵄨󵄨1 + 4ν

0 ⩽ k ⩽ n − 1.

(2.41)

54 � 2 Difference methods for the Burgers’ equation Using the Gronwall inequality (Theorem 1.2(a)), we have 2 2 2 2 ̂ 5Lc1 󵄨󵄨 k 󵄨󵄨2 3ĉ kτ 5Lc1 (τ 2 + h2 ) ⩽ e3c3 T (τ 2 + h2 ) , 󵄨󵄨e 󵄨󵄨1 ⩽ e 3 4νĉ 4νĉ 3

3

0 ⩽ k ⩽ n.

The conclusion is true.

2.3 Three-level linearized difference scheme 2.3.1 Derivation of the difference scheme Considering equation (2.1) at the node point (xi , t0 ), and noticing (2.2), we have ut (xi , 0) = νφ′′ (xi ) − φ(xi )φ′ (xi ). Denote τ û i = φ(xi ) + [νφ′′ (xi ) − φ(xi )φ′ (xi )], 2 get

0 ⩽ i ⩽ m.

Considering equation (2.1) at the point (xi , t 1 ) and using the Taylor expansion, we 2

1

1

1

δt Ui2 + ψ(u,̂ U 2 )i = νδx2 Ui2 + R0i ,

1 ⩽ i ⩽ m − 1,

(2.42)

where there is a constant c5 satisfying 󵄨󵄨 0 󵄨󵄨 2 2 󵄨󵄨Ri 󵄨󵄨 ⩽ c5 (τ + h ),

1 ⩽ i ⩽ m − 1.

(2.43)

Considering equation (2.1) at the node point (xi , tk ) and using the Taylor expansion, we get Δt Uik + ψ(U k , U k )i = νδx2 Uik + Rki , ̄

̄

1 ⩽ i ⩽ m − 1, 1 ⩽ k ⩽ n − 1,

(2.44)

where there is a constant c6 satisfying 󵄨󵄨 k 󵄨󵄨 2 2 󵄨󵄨Ri 󵄨󵄨 ⩽ c6 (τ + h ),

1 ⩽ i ⩽ m − 1, 1 ⩽ k ⩽ n − 1.

(2.45)

Noticing the initial-boundary value conditions (2.2)–(2.3), we have {

Ui0 = φ(xi ), U0k

= 0,

Umk

= 0,

1 ⩽ i ⩽ m − 1,

(2.46)

0 ⩽ k ⩽ n.

(2.47)

2.3 Three-level linearized difference scheme

� 55

Omitting the small terms in (2.42) and (2.44), we construct a difference scheme for (2.1)–(2.3) as follows: { { { { { { { { { { { { { { {

1

1

1

δt ui2 + ψ(u,̂ u 2 )i = νδx2 ui2 ,

̄ Δt uik + ψ(uk , uk )i = ui0 = φ(xi ), k u0k = 0, um = 0,

̄ νδx2 uik ,

1 ⩽ i ⩽ m − 1,

(2.48)

1 ⩽ i ⩽ m − 1, 1 ⩽ k ⩽ n − 1,

(2.49)

1 ⩽ i ⩽ m − 1,

(2.50)

0 ⩽ k ⩽ n.

(2.51)

2.3.2 Existence and uniqueness of the difference solution Theorem 2.9. The difference scheme (2.48)–(2.51) is uniquely solvable. Proof. From (2.50) and (2.51), u0 has been given. From (2.48) and (2.51), we have the system of linear equations in u1 at the 1st time level. Consider its homogeneous one: 1 1 1 { ui1 + ψ(u,̂ u1 )i = νδx2 ui1 , 2 2 { τ1 1 { u0 = 0, um = 0.

1 ⩽ i ⩽ m − 1,

(2.52) (2.53)

Taking the inner product of (2.52) on both the right- and left-hand sides with u1 , we have 1 1 󵄩󵄩 1 󵄩󵄩2 1 1 1 2 1 1 󵄩u 󵄩 + (ψ(u,̂ u ), u ) = ν(δx u , u ). τ󵄩 󵄩 2 2 Combining (ψ(u,̂ u1 ), u1 ) = 0 with (δx2 u1 , u1 ) = −|u1 |21 , it follows that 1 󵄩󵄩 1 󵄩󵄩2 1 󵄨󵄨 1 󵄨󵄨2 󵄩u 󵄩 + ν󵄨󵄨u 󵄨󵄨1 = 0. τ󵄩 󵄩 2 Hence, ‖u1 ‖ = 0, which implies that the system of linear equations (2.52)–(2.53) has only the trivial solution. Therefore, the value of u1 is uniquely determined by (2.48) and (2.51). Suppose now the numerical solutions uk−1 at the (k − 1)-th time level and uk at the k-th time level have been determined, then we have a system of linear equations in uk+1 from (2.49) and (2.51). Consider its homogeneous one: 1 k+1 1 1 { ui + ψ(uk , uk+1 )i = νδx2 uik+1 , 2τ 2 2 { k+1 k+1 { u0 = 0, um = 0.

1 ⩽ i ⩽ m − 1,

(2.54) (2.55)

Taking the inner product on both the right- and left-hand sides of (2.54) with uk+1 , we have 1 󵄩󵄩 k+1 󵄩󵄩2 1 1 k k+1 k+1 2 k+1 k+1 󵄩u 󵄩󵄩 + (ψ(u , u ), u ) = ν(δx u , u ). 2τ 󵄩 2 2

56 � 2 Difference methods for the Burgers’ equation With the application of (ψ(uk , uk+1 ), uk+1 ) = 0 and (δx2 uk+1 , uk+1 ) = −|uk+1 |21 , we have 1 󵄩󵄩 k+1 󵄩󵄩2 1 󵄨󵄨 k+1 󵄨󵄨2 󵄩u 󵄩󵄩 + ν󵄨󵄨u 󵄨󵄨1 = 0. 2τ 󵄩 2 Hence, ‖uk+1 ‖ = 0. It implies that the system of linear equations (2.54)–(2.55) has only the trivial solution. Therefore, (2.49) and (2.51) determine uk+1 uniquely. By induction, this theorem is true.

2.3.3 Conservation and boundedness of the difference solution Theorem 2.10. Let {uik | 0 ⩽ i ⩽ m, 0 ⩽ k ⩽ n} be the solution of the difference scheme (2.48)–(2.51). Denote 1 󵄩 1 󵄨 1 󵄨2 k 󵄨 ̄󵄨2 󵄩2 󵄩 󵄩2 E(uk+1 , uk ) = (󵄩󵄩󵄩uk+1 󵄩󵄩󵄩 + 󵄩󵄩󵄩uk 󵄩󵄩󵄩 ) + 2ντ( 󵄨󵄨󵄨u 2 󵄨󵄨󵄨1 + ∑󵄨󵄨󵄨ul 󵄨󵄨󵄨1 ), 2 2 l=1

k = 0, 1, . . . , n − 1.

Then we have 󵄩 󵄩2 E(uk+1 , uk ) = 󵄩󵄩󵄩u0 󵄩󵄩󵄩 ,

k = 0, 1, 2, . . . , n − 1.

(2.56)

Proof. (I) Taking the inner product on both the right- and left-hand sides of (2.48) with 1 u 2 , we have 1

1

1

1

1

1

(δt u 2 , u 2 ) + (ψ(u,̂ u 2 ), u 2 ) = ν(δx2 u 2 , u 2 ). Using 1

1

(δt u 2 , u 2 ) = 1

1

1 󵄩󵄩 1 󵄩󵄩2 󵄩󵄩 0 󵄩󵄩2 (󵄩u 󵄩 − 󵄩u 󵄩 ), 2τ 󵄩 󵄩 󵄩 󵄩

(ψ(u,̂ u 2 ), u 2 ) = 0, 1 1 󵄨 1 󵄨2 (δx2 u 2 , u 2 ) = −󵄨󵄨󵄨u 2 󵄨󵄨󵄨1 , we have 1 󵄩󵄩 1 󵄩󵄩2 󵄩󵄩 0 󵄩󵄩2 󵄨 1 󵄨2 (󵄩󵄩u 󵄩󵄩 − 󵄩󵄩u 󵄩󵄩 ) + ν󵄨󵄨󵄨u 2 󵄨󵄨󵄨1 = 0; 2τ

(2.57)

namely, 󵄩 󵄩2 E(u1 , u0 ) = 󵄩󵄩󵄩u0 󵄩󵄩󵄩 .

(2.58)

(II) Taking the inner product on both the right- and left-hand sides of (2.49) with uk , we have ̄

2.3 Three-level linearized difference scheme

(Δt uk , uk ) + (ψ(uk , uk ), uk ) = ν(δx2 uk , uk ), ̄

̄

̄

̄

̄

� 57

1 ⩽ k ⩽ n − 1.

It is easy to get 1 󵄩󵄩 k+1 󵄩󵄩2 󵄩󵄩 k−1 󵄩󵄩2 󵄨 ̄ 󵄨2 (󵄩u 󵄩󵄩 − 󵄩󵄩u 󵄩󵄩 ) + ν󵄨󵄨󵄨uk 󵄨󵄨󵄨1 = 0, 4τ 󵄩

1 ⩽ k ⩽ n − 1,

(2.59)

or 1 ‖uk+1 ‖2 + ‖uk ‖2 ‖uk ‖2 + ‖uk−1 ‖2 󵄨 ̄ 󵄨2 ( − ) + ν󵄨󵄨󵄨uk 󵄨󵄨󵄨1 = 0, 2τ 2 2

1 ⩽ k ⩽ n − 1.

Replacing k by l and summing over l from 1 to k, we obtain k 1 󵄩󵄩 k+1 󵄩󵄩2 󵄩󵄩 k 󵄩󵄩2 󵄨 ̄󵄨2 1 󵄩 󵄩2 󵄩 󵄩2 (󵄩󵄩u 󵄩󵄩 + 󵄩󵄩u 󵄩󵄩 ) + 2ντ ∑󵄨󵄨󵄨ul 󵄨󵄨󵄨1 = (󵄩󵄩󵄩u1 󵄩󵄩󵄩 + 󵄩󵄩󵄩u0 󵄩󵄩󵄩 ), 2 2 l=1

0 ⩽ k ⩽ n − 1.

1

Adding ντ|u 2 |21 on both the right- and left-hand sides of the above equality, we have E(uk+1 , uk ) = E(u1 , u0 ),

1 ⩽ k ⩽ n − 1.

(2.60)

Combining (2.58) and (2.60) yields (2.56). Remark 2.1. E(uk+1 , uk ) can be regarded as the discrete counterpart of the energy E(tk+ 1 ). According to (2.57) and (2.59), we have 2

󵄩󵄩 k 󵄩󵄩 󵄩󵄩 0 󵄩󵄩 󵄩󵄩u 󵄩󵄩 ⩽ 󵄩󵄩u 󵄩󵄩,

1 ⩽ k ⩽ n.

2.3.4 Convergence of the difference solution Theorem 2.11. Let {Uik | 0 ⩽ i ⩽ m, 0 ⩽ k ⩽ n} be the solution of the problem (2.1)–(2.3) and {uik | 0 ⩽ i ⩽ m, 0 ⩽ k ⩽ n} be the solution of the difference scheme (2.48)–(2.51). Denote eik = Uik − uik , Suppose that

τ 2ν

⩽ 1 and

L(√Lc3 +1)2 τ 2ν

0 ⩽ i ⩽ m, 0 ⩽ k ⩽ n.

⩽ 1. Then there is a constant c7 , when τ 2 + h2 ⩽

󵄨󵄨 k 󵄨󵄨 2 2 󵄨󵄨e 󵄨󵄨1 ⩽ c7 (τ + h ), 0 ⩽ k ⩽ n, √ 󵄩󵄩󵄩ek 󵄩󵄩󵄩 ⩽ L c7 (τ 2 + h2 ), 0 ⩽ k ⩽ n. 󵄩 󵄩∞ 2

1 , c7

such

(2.61) (2.62)

58 � 2 Difference methods for the Burgers’ equation Proof. Subtracting (2.48)–(2.51) from (2.42), (2.44), (2.46)–(2.47), respectively, we arrive at the system of error equations { { { { { { { { { { { { { { {

1

1

1

δt ei2 + ψ(u,̂ e 2 )i = νδx2 ei2 + R0i , k̄

k̄

1 ⩽ i ⩽ m − 1,

Δt eik + ψ(U k , U )i − ψ(uk , u )i = ei0 e0k

= 0,

= 0,

k em

̄ νδx2 eik

+ Rki ,

= 0,

(2.63)

1 ⩽ i ⩽ m − 1, 1 ⩽ k ⩽ n − 1, (2.64) 1 ⩽ i ⩽ m − 1,

(2.65)

0 ⩽ k ⩽ n.

(2.66)

We will prove the result by induction. It follows from (2.65) and (2.66) that 󵄨󵄨 0 󵄨󵄨 󵄨󵄨e 󵄨󵄨1 = 0,

󵄩󵄩 0 󵄩󵄩 󵄩󵄩e 󵄩󵄩∞ = 0.

(2.67)

Hence, (2.61) and (2.62) hold for k = 0. 1 (I) Taking the inner product on both the right- and left-hand sides of (2.63) with δt e 2 , we have 1 1 1 1 󵄩󵄩 21 󵄩󵄩2 2 1 0 󵄩󵄩δt e 󵄩󵄩 + (ψ(u,̂ e 2 ), δt e 2 ) = ν(δx e 2 , δt e 2 ) + (R , δt e 2 ).

Noticing ei0 = 0,

0 ⩽ i ⩽ m,

we have ν 󵄨 1 󵄨2 1 0 1 1 󵄩󵄩 1 󵄩󵄩2 1 1 1 󵄩󵄩e 󵄩󵄩 + (ψ(u,̂ e ), e ) = − 󵄨󵄨󵄨e 󵄨󵄨󵄨1 + (R , e ). 2 2τ 2τ τ τ With the help of (ψ(u,̂ e1 ), e1 ) = 0, we have 1 󵄩󵄩 1 󵄩󵄩2 ν 1 2 1 0 1 1 󵄩 1 󵄩2 󵄩e 󵄩 + |e |1 = (R , e ) ⩽ 2 󵄩󵄩󵄩e 󵄩󵄩󵄩 + 2τ τ τ2 󵄩 󵄩 τ

1 󵄩󵄩 0 󵄩󵄩2 󵄩R 󵄩 . 4󵄩 󵄩

By using (2.43), it follows that 󵄨󵄨 1 󵄨󵄨2 2τ ⋅ 󵄨󵄨e 󵄨󵄨1 ⩽ ν

1 󵄩󵄩 0 󵄩󵄩2 τ 2 2 2 2 󵄩R 󵄩 ⩽ Lc5 (τ + h ) . 4󵄩 󵄩 2ν

When τ ⩽ 2ν, 󵄨󵄨 1 󵄨󵄨2 2 2 2 2 󵄨󵄨e 󵄨󵄨1 ⩽ Lc5 (τ + h ) ,

2.3 Three-level linearized difference scheme

� 59

or 󵄨󵄨 1 󵄨󵄨 √ 2 2 󵄨󵄨e 󵄨󵄨1 ⩽ Lc5 (τ + h ).

(2.68)

(II) Taking the inner product on both the right- and left-hand sides of (2.64) with Δt ek , we have 󵄩󵄩 k 󵄩󵄩2 k 2 k̄ k k k k̄ k k̄ k 󵄩󵄩Δt e 󵄩󵄩 + (ψ(U , U ) − ψ(u , u ), Δt e ) = ν(δx e , Δt e ) + (R , Δt e ), 1 ⩽ k ⩽ n − 1,

or 󵄩󵄩 k 󵄩󵄩2 ν 󵄨󵄨 k+1 󵄨󵄨2 󵄨󵄨 k−1 󵄨󵄨2 󵄩󵄩Δt e 󵄩󵄩 + (󵄨󵄨e 󵄨󵄨1 − 󵄨󵄨e 󵄨󵄨1 ) 4τ

= −(ψ(U k , U k ) − ψ(uk , uk ), Δt ek ) + (Rk , Δt ek ), ̄

̄

1 ⩽ k ⩽ n − 1.

(2.69)

It is easy to know that 󵄩󵄩 k 󵄩󵄩 󵄩󵄩U 󵄩󵄩∞ ⩽ c3 ,

󵄨󵄨 k 󵄨󵄨 √ 󵄨󵄨U 󵄨󵄨1 ⩽ Lc3 ,

0 ⩽ k ⩽ n.

(2.70)

Suppose (2.61) holds for k = 1, 2, . . . , l. Then when c7 (τ 2 + h2 ) ⩽ 1, we have 󵄨󵄨 k 󵄨󵄨 󵄨󵄨 k 󵄨󵄨 󵄨󵄨 k 󵄨󵄨 √ 󵄨󵄨u 󵄨󵄨1 ⩽ 󵄨󵄨U 󵄨󵄨1 + 󵄨󵄨e 󵄨󵄨1 ⩽ Lc3 + 1, 1 ⩽ k ⩽ l, √L 󵄨 k 󵄨 √ 󵄩󵄩 k 󵄩󵄩 󵄨󵄨u 󵄨󵄨 ⩽ L (√Lc3 + 1), 1 ⩽ k ⩽ l. 󵄩󵄩u 󵄩󵄩∞ ⩽ 󵄨 󵄨 1 2 2 Noticing that ψ(U k , U k )i − ψ(uk , uk )i ̄

̄

= ψ(ek , U k )i + ψ(uk , ek )i ̄

̄

̄ ̄ ̄ ̄ 1 1 = [eik Δx Uik + Δx (ek U k )i ] + [uik Δx eik + Δx (uk ek )i ] 3 3 ̄ 1 1 1 k k̄ k k̄ k k̄ = [eik Δx Uik + (δx ei+ (δx ei− 1 )Ui+1 + ei Δx Ui + 1 )Ui−1 ] 3 2 2 2 2 ̄k ̄ ̄ ̄ 1 k 1 1 k k k k (δ uk 1 )ek ], + [ui Δx ei + (δx ui+ 1 )ei+1 + ui Δx ei + 3 2 2 x i− 2 i−1 2

and using (2.70)–(2.72) together with Lemma 1.1(b), we have −(ψ(U k , U k ) − ψ(uk , uk ), Δt ek ) 1 󵄩 󵄩 󵄨 ̄󵄨 󵄩 ̄󵄩 󵄨 󵄨 󵄩 󵄩 󵄨 ̄󵄨 󵄩 󵄩 ⩽ (󵄩󵄩󵄩ek 󵄩󵄩󵄩∞ 󵄨󵄨󵄨U k 󵄨󵄨󵄨1 + 󵄩󵄩󵄩U k 󵄩󵄩󵄩∞ 󵄨󵄨󵄨ek 󵄨󵄨󵄨1 + 󵄩󵄩󵄩ek 󵄩󵄩󵄩∞ 󵄨󵄨󵄨U k 󵄨󵄨󵄨1 )󵄩󵄩󵄩Δt ek 󵄩󵄩󵄩 3 1 󵄩 󵄩 󵄨 ̄󵄨 󵄩 ̄󵄩 󵄨 󵄨 󵄩 󵄩 󵄨 ̄󵄨 󵄩 󵄩 + (󵄩󵄩󵄩uk 󵄩󵄩󵄩∞ 󵄨󵄨󵄨ek 󵄨󵄨󵄨1 + 󵄩󵄩󵄩ek 󵄩󵄩󵄩∞ 󵄨󵄨󵄨uk 󵄨󵄨󵄨1 + 󵄩󵄩󵄩uk 󵄩󵄩󵄩∞ 󵄨󵄨󵄨ek 󵄨󵄨󵄨1 )󵄩󵄩󵄩Δt ek 󵄩󵄩󵄩 3 ̄

̄

(2.71) (2.72)

60 � 2 Difference methods for the Burgers’ equation 1 ⩽ (2√Lc3 ‖ek ‖∞ + c3 |ek |1 )‖Δt ek ‖ 3 √L ̄ ̄ 1 󵄩 󵄩 + (2 ⋅ (√Lc3 + 1)|ek |1 + (√Lc3 + 1)‖ek ‖∞ )󵄩󵄩󵄩Δt ek 󵄩󵄩󵄩 3 2 √L 󵄨 k 󵄨 1 󵄨󵄨e 󵄨󵄨 + c3 󵄨󵄨󵄨ek 󵄨󵄨󵄨 )󵄩󵄩󵄩Δt ek 󵄩󵄩󵄩 ⩽ (2√Lc3 󵄩 󵄨 󵄨1 󵄩 3 2 󵄨 󵄨1 √L 󵄨 k̄ 󵄨 󵄩 k 󵄩 1 󵄨 ̄󵄨 󵄨󵄨e 󵄨󵄨 ]󵄩󵄩Δ e 󵄩󵄩 + [√L(√Lc3 + 1)󵄨󵄨󵄨ek 󵄨󵄨󵄨1 + (√Lc3 + 1) 3 2 󵄨 󵄨1 󵄩 t 󵄩 1 󵄨 󵄨 󵄩 󵄩 1 󵄨 ̄󵄨 󵄩 󵄩 = (L + 1)c3 󵄨󵄨󵄨ek 󵄨󵄨󵄨1 ⋅ 󵄩󵄩󵄩Δt ek 󵄩󵄩󵄩 + √L(√Lc3 + 1)󵄨󵄨󵄨ek 󵄨󵄨󵄨1 ⋅ 󵄩󵄩󵄩Δt ek 󵄩󵄩󵄩 3 2 2 2 2 1󵄩 󵄩2 (L + 1) c3 󵄨󵄨 k 󵄨󵄨2 1 󵄩󵄩 k 󵄩󵄩2 L(√Lc3 + 1) 󵄨󵄨 k̄ 󵄨󵄨2 ⩽ 󵄩󵄩󵄩Δt ek 󵄩󵄩󵄩 + 󵄨󵄨e 󵄨󵄨1 + 󵄩󵄩Δt e 󵄩󵄩 + 󵄨󵄨e 󵄨󵄨1 , 4 9 4 4

1 ⩽ k ⩽ l.

Moreover, we have 1󵄩 󵄩2 (Rk , Δt ek ) ⩽ 󵄩󵄩󵄩Δt ek 󵄩󵄩󵄩 + 2

1 󵄩󵄩 k 󵄩󵄩2 󵄩R 󵄩 . 2󵄩 󵄩

Substituting the above two inequalities into (2.69) and noticing (2.45), we have ν 󵄨󵄨 k+1 󵄨󵄨2 󵄨󵄨 k−1 󵄨󵄨2 (󵄨e 󵄨󵄨1 − 󵄨󵄨e 󵄨󵄨1 ) 4τ 󵄨 (L + 1)2 c32 󵄨󵄨 k 󵄨󵄨2 L(√Lc3 + 1)2 󵄨󵄨 k̄ 󵄨󵄨2 1 󵄩󵄩 k 󵄩󵄩2 ⩽ 󵄨󵄨e 󵄨󵄨1 + 󵄨󵄨e 󵄨󵄨1 + 󵄩󵄩R 󵄩󵄩 9 4 2 (L + 1)2 c32 󵄨󵄨 k 󵄨󵄨2 L(√Lc3 + 1)2 |ek+1 |21 + |ek−1 |21 1 2 2 2 ⩽ ⋅ + Lc6 (τ + h2 ) , 󵄨󵄨e 󵄨󵄨1 + 9 4 2 2 Multiplying by have

4τ ν

1 ⩽ k ⩽ l.

on both the right- and left-hand sides and rearranging the result, we

2 2 󵄨 󵄨󵄨 k+1 󵄨󵄨2 󵄨󵄨 k−1 󵄨󵄨2 4(L + 1) c3 󵄨󵄨 k 󵄨󵄨2 1 √ 󵄨2 󵄨 󵄨2 τ 󵄨󵄨e 󵄨󵄨1 + L( Lc3 + 1)2 τ(󵄨󵄨󵄨ek+1 󵄨󵄨󵄨1 + 󵄨󵄨󵄨ek−1 󵄨󵄨󵄨1 ) 󵄨󵄨e 󵄨󵄨1 ⩽ 󵄨󵄨e 󵄨󵄨1 + 9ν 2ν 2 2 + Lc62 τ(τ 2 + h2 ) , 1 ⩽ k ⩽ l; ν

namely, [1 − ⩽ When

L(√Lc3 + 1)2 󵄨󵄨 k+1 󵄨󵄨2 τ]󵄨󵄨e 󵄨󵄨1 2ν

4(L + 1)2 c32 󵄨󵄨 k 󵄨󵄨2 L(√Lc3 + 1)2 󵄨󵄨 k−1 󵄨󵄨2 2 2 2 2 τ 󵄨󵄨e 󵄨󵄨1 + [1 + τ]󵄨󵄨e 󵄨󵄨1 + Lc6 τ(τ + h2 ) , 9ν 2ν ν L(√Lc3 +1)2 τ 2ν

1 ⩽ k ⩽ l.

⩽ 31 , it follows:

2 2 3L(√Lc3 + 1)2 󵄨󵄨 k−1 󵄨󵄨2 3 2 2 2 󵄨󵄨 k+1 󵄨󵄨2 2(L + 1) c3 󵄨󵄨 k 󵄨󵄨2 τ 󵄨󵄨e 󵄨󵄨1 + [1 + τ]󵄨󵄨e 󵄨󵄨1 + Lc6 τ(τ + h2 ) , 󵄨󵄨e 󵄨󵄨1 ⩽ 3ν 2ν ν

1 ⩽ k ⩽ l.

� 61

2.4 Hopf–Cole transformation and fourth-order difference scheme

It is easy to know that 󵄨2 󵄨 󵄨2 󵄨 max{󵄨󵄨󵄨ek 󵄨󵄨󵄨1 , 󵄨󵄨󵄨ek+1 󵄨󵄨󵄨1 } 4(L + 1)2 c32 + 9L(√Lc3 + 1)2 3 2 󵄨 󵄨2 󵄨 󵄨2 ⩽ [1 + τ] max{󵄨󵄨󵄨ek−1 󵄨󵄨󵄨1 , 󵄨󵄨󵄨ek 󵄨󵄨󵄨1 } + Lc62 τ(τ 2 + h2 ) , 6ν ν

1 ⩽ k ⩽ l.

With the help of the Gronwall inequality (Theorem 1.2(a)), we have 󵄨 󵄨2 󵄨 󵄨2 max{󵄨󵄨󵄨el 󵄨󵄨󵄨1 , 󵄨󵄨󵄨el+1 󵄨󵄨󵄨1 } ⩽e

4(L+1)2 c32 +9L(√Lc3 +1)2 6ν

T

󵄨 󵄨2 󵄨 󵄨2 [max{󵄨󵄨󵄨e0 󵄨󵄨󵄨1 , 󵄨󵄨󵄨e1 󵄨󵄨󵄨1 } +

18Lc62

4(L + 1)2 c32 + 9L(√Lc3 + 1)2

2

(τ 2 + h2 ) ].

Noticing (2.67) and (2.68), we have 2 2

√Lc3 +1)2

4(L+1) c3 +9L( 󵄨󵄨 l+1 󵄨󵄨2 6ν 󵄨󵄨e 󵄨󵄨1 ⩽ e

≡

c72 (τ 2

T

[Lc52 +

4(L +

2 2

1)2 c32

18Lc62

+ 9L(√Lc3 + 1)2

](τ 2 + h2 )

2

+h ) ,

where c7 = e

4(L+1)2 c32 +9L(√Lc3 +1)2 12ν

T

[Lc52

+

18Lc62

4(L + 1)2 c32 + 9L(√Lc3 + 1)2

1 2

] .

Thus, (2.61) also holds for k = l + 1. By induction, it is concluded that (2.61) holds for k = 0, 1, 2, . . . , n. Combining (2.61) with Lemma 1.1(b), it yields (2.62).

2.4 Hopf–Cole transformation and fourth-order difference scheme 2.4.1 Hopf–Cole transformation Let u(x, t) = −2ν

wx (x, t) , w(x, t)

then w w − wx wt w wx ) = −2ν xt = −2ν( t ) , w t w x w2 w ux = −2ν( x ) , w x w uxx = −2ν( x ) . w xx ut = −2ν(

(2.73)

62 � 2 Difference methods for the Burgers’ equation Substituting the above equalities into (2.1), we have −2ν(

wt w w w ) + (−2ν x )(−2ν( x ) ) = ν[−2ν( x ) ]; w x w w x w xx

namely, 2

(

wt w w ) − ν[( x ) ] = ν( x ) , w x w w xx x

or 2

[

wt w w − ν( x ) − ν( x ) ] = 0. w w w x x

The above result can be rewritten as (

wt − νwxx ) = 0. w x

Hence, wt − νwxx = q(t), w which can be rearranged as wt − q(t)w = νwxx . t

Multiplying on both the right- and left-hand sides by e− ∫0 q(s)ds , we have t

t

[we− ∫0 q(s)ds ]t = ν[we− ∫0 q(s)ds ]xx . Let t

̃ (x, t) = w(x, t)e− ∫0 q(s)ds , w then −2ν

̃x w w = −2ν x = u(x, t). ̃ w w

That is, for arbitrary function q(t), u(x, t) will not be affected. Therefore, we could take q(t) = 0, which exports the following equivalent problem: wt − νwxx = 0, { { ̃ (x), w(x, 0) = φ { { { wx (0, t) = 0, wx (L, t) = 0,

0 < x < L, 0 < t ⩽ T,

(2.74)

0 ⩽ x ⩽ L,

(2.75)

0 ⩽ t ⩽ T,

(2.76)

2.4 Hopf–Cole transformation and fourth-order difference scheme

� 63

where 1

x

̃ (x) = e− 2ν ∫0 φ

φ(s)ds

(2.77)

.

The transformation (2.73) is called the Hopf–Cole transformation.

2.4.2 Derivation of the difference scheme Several numerical differential formulas with an integral remainder will be given below. Lemma 2.2 ([21]). Denote α(s) = (1 − s)3 [5 − 3(1 − s)2 ]. (a) If g(x) ∈ C 6 [x0 , x1 ], we have 5 ′′ 1 2 g(x1 ) − g(x0 ) g (x0 ) + g ′′ (x1 ) − [ − g ′ (x0 )] 6 6 h h 1

h h3 h4 = − g ′′′ (x0 ) + g (5) (x0 ) + ∫ g (6) (x0 + sh)α(s)ds 6 90 180 3

0

4

h h (6) h g (x0 + θ0 h), = − g ′′′ (x0 ) + g (5) (x0 ) + 6 90 240

θ0 ∈ (0, 1).

(b) If g(x) ∈ C 6 [xm−1 , xm ], we have g(xm ) − g(xm−1 ) 1 ′′ 5 2 g (xm−1 ) + g ′′ (xm ) − [g ′ (xm ) − ] 6 6 h h 1

h h3 h4 = g ′′′ (xm ) − g (5) (xm ) + ∫ g (6) (xm − sh)α(s)ds 6 90 180 3

=

4

0

h ′′′ h h (6) g (xm ) − g (5) (xm ) + g (xm − θm h), 6 90 240

θm ∈ (0, 1).

(c) If g(x) ∈ C 6 [xi−1 , xi+1 ], we have 1 ′′ 1 [g (xi−1 ) + 10g ′′ (xi ) + g ′′ (xi+1 )] − 2 [g(xi+1 ) − 2g(xi ) + g(xi−1 )] 12 h 1

=

h4 ∫[g (6) (xi + sh) + g (6) (xi − sh)]α(s)ds 360 0

h4 (6) = g (xi + θi h), 240

θi ∈ (−1, 1).

64 � 2 Difference methods for the Burgers’ equation For any v ∈ 𝒰h , define the averaging operator 𝒜: { { { 𝒜vi = { { { {

5 v + 61 v1 , 6 0 1 (v + 10vi + vi+1 ), 12 i−1 5 v + 61 vm−1 , 6 m

i = 0, 1 ⩽ i ⩽ m − 1, i = m.

Suppose the problem (2.74)–(2.76) has a solution w(x, t) ∈ C 6,4 ([0, L] × [0, T]). Define the grid functions Uik = u(xi , tk ),

Wik = w(xi , tk ),

0 ⩽ i ⩽ m, 0 ⩽ k ⩽ n.

With the help of (2.74) and (2.76), we have wxxx (0, t) = 0,

wxxxxx (0, t) = 0,

wxxx (L, t) = 0,

0 ⩽ t ⩽ T,

wxxxxx (L, t) = 0,

0 ⩽ t ⩽ T.

(2.78) (2.79)

Considering (2.74) at the point (xi , tk+ 1 ), we have 2

wt (xi , tk+ 1 ) − νwxx (xi , tk+ 1 ) = 0, 2

2

0 ⩽ i ⩽ m, 0 ⩽ k ⩽ n − 1.

Applying Lemma 1.2, we have ν k+ 1 wt (xi , tk+ 1 ) − [wxx (xi , tk ) + wxx (xi , tk+1 )] = (Rxt w)i 2 , 2 2 0 ⩽ i ⩽ m, 0 ⩽ k ⩽ n − 1, where k+ 21

(Rxt w)i

=−

ντ 2 𝜕4 w (x , t + ηki τ), 8 𝜕x 2 𝜕t 2 i k

ηki ∈ (0, 1), 0 ⩽ i ⩽ m.

Performing the operator 𝒜 on both the right- and left-hand sides of the above equality, we have ν k+ 1 [𝒜wxx (xi , tk ) + 𝒜wxx (xi , tk+1 )] = 𝒜(Rxt w)i 2 , 2 2 0 ⩽ i ⩽ m, 0 ⩽ k ⩽ n − 1.

𝒜wt (xi , tk+ 1 ) −

Applying Lemma 2.2 and noticing (2.76), (2.78)–(2.79), we have 2 k+ 21 k+ 21 k+ 21 { 𝒜 δ W − ν( δ W ) = R , { 1 t x 0 0 { h { 2 { { { 1 1 1 k+ k+ k+ 2 𝒜δt Wi 2 − ν δx Wi 2 = Ri 2 , { { { { { 1 1 1 { { 𝒜δ W k+ 2 − ν(− 2 δ W k+ 2 ) = Rk+ 2 , m t m x m− 1 { h 2

0 ⩽ k ⩽ n − 1,

(2.80)

1 ⩽ i ⩽ m − 1, 0 ⩽ k ⩽ n − 1,

(2.81)

0 ⩽ k ⩽ n − 1,

(2.82)

2.4 Hopf–Cole transformation and fourth-order difference scheme

� 65

where there is a constant c8 satisfying 󵄨󵄨 k+ 21 󵄨󵄨 2 4 󵄨󵄨Ri 󵄨󵄨 ⩽ c8 (τ + h ),

0 ⩽ i ⩽ m, 0 ⩽ k ⩽ n − 1.

(2.83)

Noticing the initial condition ̃ (xi ), Wi0 = φ

0 ⩽ i ⩽ m,

(2.84)

and omitting the small terms in (2.80)–(2.82), a difference scheme for (2.74)–(2.76) reads { { { { { { { { { { { { { { { { { { { { {

k+ 21

𝒜δt w0

k+ 21

𝒜δt wi

k+ 21

𝒜δt wm

2 k+ 1 − ν( δx w 1 2 ) = 0, h 2

0 ⩽ k ⩽ n − 1,

(2.85)

− νδx2 wi

1 ⩽ i ⩽ m − 1, 0 ⩽ k ⩽ n − 1,

(2.86)

0 ⩽ k ⩽ n − 1,

(2.87)

0 ⩽ i ⩽ m.

(2.88)

k+ 21

= 0,

2 − ν(− δx w h

̃ (xi ), wi0 = φ

k+ 21 m− 21

) = 0,

2.4.3 Existence and uniqueness of the difference solution Lemma 2.3. For any v = (v0 , v1 , . . . , vm ) ∈ 𝒰h , we have (𝒜v, v) = ‖v‖2 −

h2 2 |v| , 12 1

2 2 ‖v‖ ⩽ (𝒜v, v) ⩽ ‖v‖2 . 3 Proof. By using

i = 0, v0 + h6 δx v 1 , { { 2 { { { 2 1 ⩽ i ⩽ m − 1, (𝒜v)i = { vi + h12 δx2 vi , { { { { v − h δ v 1 , i = m, { m 6 x m− 2 we have m−1 1 1 (𝒜v, v) = h[ (𝒜v0 )v0 + ∑ (𝒜vi )vi + (𝒜vm )vm ] 2 2 i=1 m−1 1 h h2 1 h = h[ (v0 + δx v 1 )v0 + ∑ (vi + δx2 vi )vi + (vm − δx vm− 1 )vm ] 2 2 2 6 12 2 6 i=1 m−1 m−1 1 1 h2 = h( v20 + ∑ v2i + v2m ) + [(δx v 1 )v0 + h ∑ (δx2 vi )vi − (δx vm− 1 )vm ] 2 2 2 2 12 i=1 i=1

66 � 2 Difference methods for the Burgers’ equation

= ‖v‖2 − = ‖v‖2 −

m−1 h2 ⋅ h ∑ (δx vi+ 1 )2 2 12 i=0

h2 2 |v| . 12 1

It is easy to see (𝒜v, v) ⩽ ‖v‖2 . By Lemma 1.1(d), we easily have (𝒜v, v) ⩾ ‖v‖2 −

2 h2 4 ⋅ ‖v‖2 = ‖v‖2 . 12 h2 3

Define the norm ‖v‖𝒜 = √(𝒜v, v) on 𝒰h . With the help of Lemma 2.3, it follows that ‖v‖𝒜 and ‖v‖ are equivalent. Theorem 2.12. The difference scheme (2.85)–(2.88) has a unique solution. Proof. The value of w0 at the 0-th time level has been determined by (2.88). Suppose the value of wk at the k-th time level has been obtained, then from (2.85)–(2.87), we have the system of linear equations in wk+1 at the (k + 1)-th time level. Consider its homogeneous one: { { { { { { { { { { { { { { {

1 1 k+1 k+1 𝒜w0 − ν δx w 1 = 0, τ h 2 1 2 k+1 1 k+1 𝒜wi − ν δx wi = 0, 1 ⩽ i ⩽ m − 1, τ 2 1 1 k+1 k+1 𝒜wm − ν(− δx wm− 1 ) = 0. τ h 2

(2.89) (2.90) (2.91)

k+1 Multiplying (2.89) by 21 hw0k+1 , (2.90) by hwik+1 and (2.91) by 21 hwm on both the rightand left-hand sides, respectively, then adding the results together, we have m−1 1 1 k+1 k+1 (𝒜wk+1 , wk+1 ) − ν[w0k+1 δx wk+1 + h ∑ wik+1 δx2 wik+1 − wm δx wm− 1 ] = 0. 1 τ 2 2 2 i=1

It follows easily from the above result that 1 󵄩󵄩 k+1 󵄩󵄩2 1 󵄨 k+1 󵄨2 󵄩󵄩w 󵄩󵄩𝒜 + ν󵄨󵄨󵄨w 󵄨󵄨󵄨1 = 0. τ 2

2.4 Hopf–Cole transformation and fourth-order difference scheme

� 67

Hence, wk+1 = 0; namely, (2.89)–(2.91) has only the trivial solution. Therefore, (2.85)–(2.87) determine wk+1 uniquely. By induction, the conclusion is true. 2.4.4 Convergence of the difference solution Theorem 2.13. Let {Wik | 0 ⩽ i ⩽ m, 0 ⩽ k ⩽ n} be the solution of the problem (2.74)–(2.76) and {wik | 0 ⩽ i ⩽ m, 0 ⩽ k ⩽ n} be the solution of the difference scheme (2.85)–(2.88). Denote eik = Wik − wik ,

0 ⩽ i ⩽ m, 0 ⩽ k ⩽ n.

Then we have 󵄩󵄩 k 󵄩󵄩 2 4 󵄩󵄩e 󵄩󵄩 ⩽ c9 (τ + h ), 0 ⩽ k ⩽ n, 󵄨󵄨 k 󵄨󵄨 2 4 󵄨󵄨e 󵄨󵄨1 ⩽ c10 (τ + h ), 0 ⩽ k ⩽ n, 1 󵄩󵄩 k 󵄩󵄩 2 4 󵄩󵄩e 󵄩󵄩∞ ⩽ √2c9 c10 + c92 (τ + h ), 0 ⩽ k ⩽ n, L

(2.92) (2.93) (2.94)

where c9 = 32 T √Lc8 , c10 = 21 √ ν3 TLc8 . Proof. Subtracting (2.85)–(2.88) from (2.80)–(2.82), (2.84), respectively, we have the system of error equations as follows: { { { { { { { { { { { { { { { { { { { { {

k+ 21

𝒜δt e0

k+ 21

𝒜δt ei

k+ 21

𝒜δt em

ei0 = 0,

2 k+ 1 k+ 1 − ν( δx e 1 2 ) = R0 2 , h 2

0 ⩽ k ⩽ n − 1,

(2.95)

− νδx2 ei

1 ⩽ i ⩽ m − 1, 0 ⩽ k ⩽ n − 1,

(2.96)

0 ⩽ k ⩽ n − 1,

(2.97)

0 ⩽ i ⩽ m.

(2.98)

k+ 21

k+ 21

= Ri

2 − ν(− δx e h

k+ 21 m− 21

, k+ 21

) = Rm ,

k+ 1

k+ 1

k+ 1

(I) Multiplying (2.95) by 21 he0 2 , (2.96) by hei 2 , (2.97) by 21 hem 2 on both the rightand left-hand sides, respectively, and adding the results together, we arrive at m−1 1 󵄩󵄩 k+1 󵄩󵄩2 󵄩󵄩 k 󵄩󵄩2 k+ 1 k+ 1 k+ 1 k+ 1 k+ 1 k+ 1 (󵄩󵄩e 󵄩󵄩𝒜 − 󵄩󵄩e 󵄩󵄩𝒜 ) − ν(e0 2 δx e 1 2 + h ∑ ei 2 δx2 ei 2 − em 2 δx e 21 ) m− 2 2τ 2 i=1 1

1

= (Rk+ 2 , ek+ 2 ),

0 ⩽ k ⩽ n − 1;

68 � 2 Difference methods for the Burgers’ equation namely, 1 1 1 󵄩󵄩 k+1 󵄩󵄩2 󵄩󵄩 k 󵄩󵄩2 󵄨 1 󵄨2 (󵄩e 󵄩󵄩𝒜 − 󵄩󵄩e 󵄩󵄩𝒜 ) + ν󵄨󵄨󵄨ek+ 2 󵄨󵄨󵄨1 = (Rk+ 2 , ek+ 2 ), 2τ 󵄩

0 ⩽ k ⩽ n − 1.

Applying the Cauchy–Schwarz inequality to the right-hand side of the above equality and using Lemma 2.3, we have 1 󵄩󵄩 k+1 󵄩󵄩2 󵄩󵄩 k 󵄩󵄩2 (󵄩e 󵄩󵄩𝒜 − 󵄩󵄩e 󵄩󵄩𝒜 ) 2τ 󵄩 1 󵄩 󵄩 󵄩 1󵄩 ⩽ 󵄩󵄩󵄩Rk+ 2 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩ek+ 2 󵄩󵄩󵄩 1 3󵄩 󵄩 󵄩 1󵄩 ⩽ √ 󵄩󵄩󵄩Rk+ 2 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩ek+ 2 󵄩󵄩󵄩𝒜 2 k+1 k 1 3 󵄩 󵄩 ‖e ‖𝒜 + ‖e ‖𝒜 , ⩽ √ 󵄩󵄩󵄩Rk+ 2 󵄩󵄩󵄩 ⋅ 2 2 which follows: 1 1 󵄩󵄩 k+1 󵄩󵄩 3󵄩 󵄩 󵄩 󵄩 (󵄩󵄩e 󵄩󵄩𝒜 − 󵄩󵄩󵄩ek 󵄩󵄩󵄩𝒜 ) ⩽ √ 󵄩󵄩󵄩Rk+ 2 󵄩󵄩󵄩, τ 2

0 ⩽ k ⩽ n − 1,

after eliminating the term 21 (‖ek+1 ‖𝒜 + ‖ek ‖𝒜 ) on both the right- and left-hand sides. Hence, 3 k 󵄩 l+ 1 󵄩 󵄩󵄩 k+1 󵄩󵄩 󵄩 0󵄩 󵄩󵄩e 󵄩󵄩𝒜 ⩽ 󵄩󵄩󵄩e 󵄩󵄩󵄩𝒜 + √ τ ∑󵄩󵄩󵄩R 2 󵄩󵄩󵄩, 2 l=0

0 ⩽ k ⩽ n − 1.

With the help of (2.83) and (2.98), we arrive at 3 3 󵄩󵄩 k+1 󵄩󵄩 2 4 2 4 󵄩󵄩e 󵄩󵄩𝒜 ⩽ √ (k + 1)τ √Lc8 (τ + h ) ⩽ √ T √Lc8 (τ + h ), 2 2

0 ⩽ k ⩽ n − 1.

Applying Lemma 2.3 again, we have 3 󵄩󵄩 k 󵄩󵄩 √ 3 󵄩󵄩 k 󵄩󵄩 2 4 󵄩󵄩e 󵄩󵄩 ⩽ 󵄩e 󵄩 ⩽ T √Lc8 (τ + h ), 2 󵄩 󵄩𝒜 2 k+ 1

1 ⩽ k ⩽ n.

k+ 1

k+ 1

(II) Multiplying (2.95) by 21 hδt e0 2 , (2.96) by hδt ei 2 , (2.97) by 21 hδt em 2 on both the right- and left-hand sides, respectively, and adding the results together, we have m−1

1 k+ 1 k+ 1 k+ 1 k+ 1 k+ 1 󵄩󵄩 k+ 21 󵄩󵄩2 2 k+ 󵄩󵄩δt e 󵄩󵄩𝒜 − ν[(δx e 1 2 )δt e0 2 + h ∑ (δx ei 2 )δt ei 2 − (δx e 21 )δt em 2 ] m− 2

k+ 21

= (R namely,

, δt e

k+ 21

),

0 ⩽ k ⩽ n − 1;

i=1

2

2.4 Hopf–Cole transformation and fourth-order difference scheme

� 69

1 󵄨 k+1 󵄨2 󵄨 k 󵄨2 󵄩󵄩 k+ 21 󵄩󵄩2 󵄩󵄩δt e 󵄩󵄩𝒜 + ν ⋅ (󵄨󵄨󵄨e 󵄨󵄨󵄨1 − 󵄨󵄨󵄨e 󵄨󵄨󵄨1 ) 2τ 1

1

= (Rk+ 2 , δt ek+ 2 ) 1 1 󵄩 󵄩 󵄩 󵄩 ⩽ 󵄩󵄩󵄩Rk+ 2 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩δt ek+ 2 󵄩󵄩󵄩 1 2 1 2 2󵄩 3󵄩 󵄩 󵄩 ⩽ 󵄩󵄩󵄩δt ek+ 2 󵄩󵄩󵄩 + 󵄩󵄩󵄩Rk+ 2 󵄩󵄩󵄩 3 8 1 2 1 2 3󵄩 󵄩 󵄩 󵄩 ⩽ 󵄩󵄩󵄩δt ek+ 2 󵄩󵄩󵄩𝒜 + 󵄩󵄩󵄩Rk+ 2 󵄩󵄩󵄩 , 8

0 ⩽ k ⩽ n − 1.

Hence, 1 󵄨󵄨 k+1 󵄨󵄨2 󵄨󵄨 k 󵄨󵄨2 3 󵄩󵄩 k+ 21 󵄩󵄩2 (󵄨e 󵄨󵄨1 − 󵄨󵄨e 󵄨󵄨1 ) ⩽ 󵄩R 󵄩󵄩 , 2τ 󵄨 8ν 󵄩

0 ⩽ k ⩽ n − 1.

By recursion, we have k 3 2 󵄨󵄨 k+1 󵄨󵄨2 󵄨󵄨 0 󵄨󵄨2 3 󵄩 l+ 1 󵄩2 (k + 1)τLc82 (τ 2 + h4 ) , 󵄨󵄨e 󵄨󵄨1 ⩽ 󵄨󵄨e 󵄨󵄨1 + τ ∑󵄩󵄩󵄩R 2 󵄩󵄩󵄩 ⩽ 4ν l=0 4ν

0 ⩽ k ⩽ n − 1.

Hence, 1 3 󵄨󵄨 k 󵄨󵄨 2 4 󵄨󵄨e 󵄨󵄨1 ⩽ √ TL c8 (τ + h ), 2 ν

1 ⩽ k ⩽ n.

(III) Combining (2.92)–(2.93) with Lemma 1.1(e), it follows: 1 󵄩 󵄩2 1 󵄩󵄩 k 󵄩󵄩 󵄩 󵄩 󵄨 󵄨 2 4 󵄩󵄩e 󵄩󵄩∞ ⩽ √2󵄩󵄩󵄩ek 󵄩󵄩󵄩 ⋅ 󵄨󵄨󵄨ek 󵄨󵄨󵄨1 + 󵄩󵄩󵄩ek 󵄩󵄩󵄩 ⩽ √2c9 c10 + c92 (τ + h ), L L

1 ⩽ k ⩽ n.

2.4.5 Calculation of the solution of the original problem Let g(x) ∈ C 5 [x0 , xm ], then there are constants c1̂ , c2̂ , . . . , c6̂ satisfying g ′ (x1 ) = c1̂ δx g 1 + c2̂ δx g 3 + c3̂ δx g 5 + c4̂ δx g 7 + O(h4 ), 2

2

2

(2.99)

2

g ′ (xi ) = c5̂ δx gi− 3 + c6̂ δx gi− 1 + c6̂ δx gi+ 1 + c5̂ δx gi+ 3 + O(h4 ), 2

2

2

2

2 ⩽ i ⩽ m − 2, (2.100) 4

g (xm−1 ) = c1̂ δx gm− 1 + c2̂ δx gm− 3 + c3̂ δx gm− 5 + c4̂ δx gm− 7 + O(h ). ′

2

2

2

2

By the transformation (2.73), we have u(xi , tk ) = −2ν

wx (xi , tk ) . w(xi , tk )

(2.101)

70 � 2 Difference methods for the Burgers’ equation Using (2.99)–(2.101), we have 2ν { U1k = − k (c1̂ δx W k1 + c2̂ δx W k3 + c3̂ δx W k5 + c4̂ δx W k7 ) + R̂ k1 , 1 ⩽ k ⩽ n, { { { 2 2 2 2 W { 1 { { { { 2ν { k { ̂ δx W k 1 + c6̂ δx W k 1 + c5̂ δx W k 3 ) + R̂ ki , U k = − k (c5̂ δx Wi− { 3 + c6 { i− 2 i+ 2 i+ 2 { i 2 Wi { { 2 ⩽ i ⩽ m − 2, 1 ⩽ k ⩽ n. { { { { { 2ν k k { { ̂ δx W k 5 + c2̂ δx W k 3 + c1̂ δx W k 1 ) + R̂ km−1 , Um−1 = − k (c4̂ δx Wm− 7 + c3 { m− 2 m− 2 m− 2 { 2 { W { m−1 { { 1 ⩽ k ⩽ n,

(2.102)

(2.103)

(2.104)

where there is a constant c11 satisfying 󵄨󵄨 ̂ k 󵄨󵄨 4 󵄨󵄨Ri 󵄨󵄨 ⩽ c11 h ,

1 ⩽ i ⩽ m − 1, 1 ⩽ k ⩽ n.

Omitting the small terms in (2.102)–(2.104), we arrive at the following numerical scheme for calculations: 2ν { u1k = − k (c1̂ δx wk1 + c2̂ δx wk3 + c3̂ δx wk5 + c4̂ δx wk7 ), 1 ⩽ k ⩽ n, { { 2 2 2 2 w1 { { { { { 2ν { k k ̂ δx wk 1 + c6̂ δx wk 1 + c5̂ δx wk 3 ), 2 ⩽ i ⩽ m − 2, 1 ⩽ k ⩽ n, ui = − k (c5̂ δx wi− 3 + c6 i− 2 i+ 2 i+ 2 { 2 w { { i { { { { 2ν k k { { um−1 ̂ δx wk 5 + c2̂ δx wk 3 + c1̂ δx wk 1 ), 1 ⩽ k ⩽ n. = − k (c4̂ δx wm− 7 + c3 m− 2 m− 2 m− 2 2 wm−1 { In view of Theorem 2.13, it can be proved that m−1

2

√h ∑ (Uik − uik ) ⩽ c12 (τ 2 + h4 ), i=1

1⩽k⩽n

with c12 a constant.

2.5 Fourth-order compact two-level nonlinear difference scheme Before deriving the compact difference scheme, we introduce several lemmas. Lemma 2.4. Let g(x) ∈ C 5 [xi−1 , xi+1 ] and G(x) = g ′′ (x), we have g(xi )g ′ (xi ) = ψ(g, g)i −

h2 ψ(G, g)i + O(h4 ). 2

Proof. Based on the Taylor expansion, we have 1 ψ(g, g)i = [g(xi−1 ) + g(xi ) + g(xi+1 )]Δx g(xi ) 3

2.5 Fourth-order compact two-level nonlinear difference scheme

= [g(xi ) +

� 71

h2 ′′ h2 g (xi ) + O(h4 )] × [g ′ (xi ) + g ′′′ (xi ) + O(h4 )] 3 6

h2 [2g ′′ (xi )g ′ (xi ) + g(xi )g ′′′ (xi )] + O(h4 ) 6 h2 = g(xi )g ′ (xi ) + [G(xi )g ′ (xi ) + (Gg)′ (xi )] + O(h4 ) 6 h2 ′ = g(xi )g (xi ) + ψ(G, g)i + O(h4 ). 2 = g(xi )g ′ (xi ) +

Lemma 2.5. For any u, v ∈ 𝒰h̊ and S ∈ 𝒰h satisfying vi = δx2 ui −

h2 2 δ v + Si , 12 x i

1 ⩽ i ⩽ m − 1,

(2.105)

we have h2 2 ‖v‖ + 12 h2 (v, u) ⩽ −|u|21 − ‖v‖2 + 18

(v, u) = −|u|21 −

h4 2 h2 |v| + (S, v) + (S, u), 144 1 12 h2 (S, v) + (S, u). 12

(2.106) (2.107)

Proof. Taking an inner product of (2.105) with u on both the right- and left-hand sides, we have (v, u) = (δx2 u, u) −

h2 2 (δ v, u) + (S, u) 12 x

h2 (v, δx2 u) + (S, u) 12 h2 h2 = − |u|21 − (v, v + δx2 v − S) + (S, u) 12 12 = − |u|21 −

= − |u|21 −

h2 2 h4 2 h2 ‖v‖ + |v| + (S, v) + (S, u). 12 144 1 12

Together with the inverse estimate (Lemma 1.1(d)), we have h2 2 ‖v‖ + 12 h2 = − |u|21 − ‖v‖2 + 18

(v, u) ⩽ − |u|21 −

h4 4 h2 ⋅ 2 ‖v‖2 + (S, v) + (S, u) 144 h 12 h2 (S, v) + (S, u). 12

Lemma 2.6. For any u, v, S defined on Ωhτ satisfying h2 2 k k 2 k { { v = δ u − δ v + Sik , { i x i { { 12 x i k { u0k = um = 0, { { { { k k { v0 = vm = 0,

1 ⩽ i ⩽ m − 1, 0 ⩽ k ⩽ n,

(2.108)

0 ⩽ k ⩽ n,

(2.109)

0 ⩽ k ⩽ n,

(2.110)

72 � 2 Difference methods for the Burgers’ equation we have 1

1

(vk+ 2 , δt uk+ 2 ) =−

h2 󵄩 h4 󵄨󵄨 k+1 󵄨󵄨2 󵄨󵄨 k 󵄨󵄨2 1 󵄨󵄨 k+1 󵄨󵄨2 󵄨󵄨 k 󵄨󵄨2 󵄩2 󵄩 󵄩2 [(󵄨󵄨u 󵄨󵄨1 − 󵄨󵄨u 󵄨󵄨1 ) + (󵄩󵄩󵄩vk+1 󵄩󵄩󵄩 − 󵄩󵄩󵄩vk 󵄩󵄩󵄩 ) − (󵄨v 󵄨󵄨1 − 󵄨󵄨v 󵄨󵄨1 )] 2τ 12 144 󵄨

+

1 1 1 h2 k+ 21 (v , δt S k+ 2 ) + (S k+ 2 , δt uk+ 2 ), 12

0 ⩽ k ⩽ n − 1.

Proof. It follows from (2.108) that k+ 21

k+ 21

vi

= δx2 ui

k+ 21

= δt δx2 ui

−

h2 2 k+ 21 k+ 1 δx vi + Si 2 , 12

−

h2 k+ 1 k+ 1 δt δx2 vi 2 + δt Si 2 , 12

1 ⩽ i ⩽ m − 1, 0 ⩽ k ⩽ n − 1

(2.111)

and k+ 21

δt vi

1 ⩽ i ⩽ m − 1, 0 ⩽ k ⩽ n − 1. (2.112)

It follows from (2.109) and (2.110) that k+ 21

δt u0

k+ 21

= δt um

= 0,

0⩽k ⩽n−1

(2.113)

and k+ 21

v0

k+ 21

= vm

= 0,

0 ⩽ k ⩽ n − 1.

(2.114) 1

Taking the inner product of (2.111) on both the right- and left-hand sides with δt uk+ 2 , we have 1

1

(vk+ 2 , δt uk+ 2 ) 1

= (δx2 uk+ 2 −

1 1 h2 2 k+ 21 δx v + S k+ 2 , δt uk+ 2 ) 12

=−

1 1 1 1 1 󵄨󵄨 k+1 󵄨󵄨2 󵄨󵄨 k 󵄨󵄨2 h2 (󵄨󵄨u 󵄨󵄨1 − 󵄨󵄨u 󵄨󵄨1 ) − (vk+ 2 , δt δx2 uk+ 2 ) + (S k+ 2 , δt uk+ 2 ) 2τ 12

=−

1 1 󵄨󵄨 k+1 󵄨󵄨2 󵄨󵄨 k 󵄨󵄨2 h2 h2 (󵄨󵄨u 󵄨󵄨1 − 󵄨󵄨u 󵄨󵄨1 ) − (vk+ 2 , δt (v + δx2 v − S) 2τ 12 12

k+ 21

=−

1

1

) + (S k+ 2 , δt uk+ 2 )

1 󵄨󵄨 k+1 󵄨󵄨2 󵄨󵄨 k 󵄨󵄨2 h2 󵄩 h4 󵄨󵄨 k+1 󵄨󵄨2 󵄨󵄨 k 󵄨󵄨2 󵄩2 󵄩 󵄩2 [(󵄨󵄨u 󵄨󵄨1 − 󵄨󵄨u 󵄨󵄨1 ) + (󵄩󵄩󵄩vk+1 󵄩󵄩󵄩 − 󵄩󵄩󵄩vk 󵄩󵄩󵄩 ) − (󵄨v 󵄨󵄨1 − 󵄨󵄨v 󵄨󵄨1 )] 2τ 12 144 󵄨

+

1 1 1 h2 k+ 21 (v , δt S k+ 2 ) + (S k+ 2 , δt uk+ 2 ), 12

0 ⩽ k ⩽ n − 1.

In the second equality, we have used (2.113) and (2.114). In the third equality, we have used (2.112).

2.5 Fourth-order compact two-level nonlinear difference scheme

�

73

2.5.1 Derivation of the difference scheme Let v = uxx , then the problem (2.1)–(2.3) is equivalent to ut + uux − ν v = 0, { { { { { v = uxx , { { u(x, 0) = φ(x), { { { { u(0, t) = 0, u(L, t) = 0,

0 < x < L, 0 < t ⩽ T,

(2.115)

0 < x < L, 0 ⩽ t ⩽ T,

(2.116)

0 < x < L,

(2.117)

0 ⩽ t ⩽ T.

(2.118)

According to (2.115) and (2.118), we easily get v(L, t) = 0,

v(0, t) = 0,

0 ⩽ t ⩽ T.

(2.119)

Define the grid functions: U = {Uik | 0 ⩽ i ⩽ m, 0 ⩽ k ⩽ n} and

V = {Vik | 0 ⩽ i ⩽ m, 0 ⩽ k ⩽ n},

where Uik = u(xi , tk ) and Vik = v(xi , tk ). Considering (2.115) at the point (xi , tk+ 1 ) and (2.116) at the point (xi , tk ), respectively, 2 we have {

ut (xi , tk+ 1 ) + u(xi , tk+ 1 )ux (xi , tk+ 1 ) − ν v(xi , tk+ 1 ) = 0, 2

v(xi , tk ) = uxx (xi , tk ),

2

2

2

1 ⩽ i ⩽ m − 1, 0 ⩽ k ⩽ n − 1, 1 ⩽ i ⩽ m − 1, 0 ⩽ k ⩽ n.

Using the Taylor expansion and Lemma 2.4, we have 1 1 h2 k+ 21 k+ 1 k+ 1 k+ 21 k+ 21 { { δ U + ψ(U , U ) − ψ(V k+ 2 , U k+ 2 )i − νVi 2 = Pi 2 , { t i i { { 2 { 1 ⩽ i ⩽ m − 1, 0 ⩽ k ⩽ n − 1, { { { 2 { { { V k = δ2 U k − h δ2 V k + Qk , 1 ⩽ i ⩽ m − 1, 0 ⩽ k ⩽ n, i i x i 12 x i {

(2.120) (2.121)

and there is a constant c13 such that { { { { { { { { {

󵄨󵄨 k+ 21 󵄨󵄨 2 4 󵄨󵄨Pi 󵄨󵄨 ⩽ c13 (τ + h ), 󵄨󵄨 k 󵄨󵄨 4 󵄨󵄨Qi 󵄨󵄨 ⩽ c13 h , 1 󵄨󵄨󵄨δt Qk+ 2 󵄨󵄨󵄨 ⩽ c13 h4 , 󵄨 i 󵄨

1 ⩽ i ⩽ m − 1, 0 ⩽ k ⩽ n − 1,

(2.122)

1 ⩽ i ⩽ m − 1, 0 ⩽ k ⩽ n,

(2.123)

1 ⩽ i ⩽ m − 1, 0 ⩽ k ⩽ n − 1.

(2.124)

Noticing the initial-boundary value conditions (2.117)–(2.119), we have U 0 = φ(xi ), { { { ik U0 = 0, Umk = 0, { { { k k { V0 = 0, Vm = 0,

1 ⩽ i ⩽ m − 1,

(2.125)

0 ⩽ k ⩽ n,

(2.126)

0 ⩽ k ⩽ n.

(2.127)

74 � 2 Difference methods for the Burgers’ equation Omitting the small terms in (2.120) and (2.121) and combining with (2.125)–(2.127), we construct a difference scheme for (2.115)–(2.118) as follows: { { { { { { { { { { { { { { { { { { { { { { { { { { { { {

1 1 h2 k+ 1 ψ(vk+ 2 , uk+ 2 )i − νvi 2 = 0, 2 1 ⩽ i ⩽ m − 1, 0 ⩽ k ⩽ n − 1, h2 vki = δx2 uik − δx2 vki , 1 ⩽ i ⩽ m − 1, 0 ⩽ k ⩽ n, 12 ui0 = φ(xi ), 1 ⩽ i ⩽ m − 1,

k+ 21

δt ui

u0k vk0

= 0,

= 0,

1

1

+ ψ(uk+ 2 , uk+ 2 )i −

k um = 0, k vm = 0,

(2.128) (2.129) (2.130)

0 ⩽ k ⩽ n,

(2.131)

0 ⩽ k ⩽ n.

(2.132)

2.5.2 Conservation and boundedness of the difference solution Theorem 2.14. Let {uik , vki | 0 ⩽ i ⩽ M, 0 ⩽ k ⩽ n} be the solution of (2.128)–(2.132). Denote k−1 2 4 󵄩 󵄩2 󵄨 1 󵄨2 h 󵄩 1 󵄩2 h 󵄨󵄨 l+ 21 󵄨󵄨2 E k = 󵄩󵄩󵄩uk 󵄩󵄩󵄩 + 2ντ ∑ (󵄨󵄨󵄨ul+ 2 󵄨󵄨󵄨1 + 󵄩󵄩󵄩vl+ 2 󵄩󵄩󵄩 − 󵄨v 󵄨󵄨1 ), 12 144 󵄨 l=0

0 ⩽ k ⩽ n.

Then we have Ek = E0 ,

1 ⩽ k ⩽ n. 1

Proof. Taking an inner product of (2.128) on both the right- and left-hand sides with uk+ 2 , we have 1

1

1

1

1

(δt uk+ 2 , uk+ 2 ) + (ψ(uk+ 2 , uk+ 2 ), uk+ 2 ) − 0 ⩽ k ⩽ n − 1.

1 1 1 1 1 h2 (ψ(vk+ 2 , uk+ 2 ), uk+ 2 ) − ν(vk+ 2 , uk+ 2 ) = 0, 2

From Lemma 2.1 and taking S = 0 in (2.106), we can easily get 1 2 1 󵄩󵄩 k+1 󵄩󵄩2 󵄩󵄩 k 󵄩󵄩2 h2 󵄩 1 󵄩2 h4 󵄨󵄨 k+ 21 󵄨󵄨2 󵄨 󵄨 (󵄩󵄩u 󵄩󵄩 − 󵄩󵄩u 󵄩󵄩 ) + ν(󵄨󵄨󵄨uk+ 2 󵄨󵄨󵄨1 + 󵄩󵄩󵄩vk+ 2 󵄩󵄩󵄩 − 󵄨v 󵄨󵄨1 ) = 0, 2τ 12 144 󵄨 0 ⩽ k ⩽ n − 1.

Replacing the superscript k with l and summing over l from 0 to k − 1, we have k−1 2 4 1 󵄩󵄩 k 󵄩󵄩2 󵄩󵄩 0 󵄩󵄩2 󵄨 1 󵄨2 h 󵄩 1 󵄩2 h 󵄨󵄨 l+ 21 󵄨󵄨2 (󵄩󵄩u 󵄩󵄩 − 󵄩󵄩u 󵄩󵄩 ) + ν ∑ (󵄨󵄨󵄨ul+ 2 󵄨󵄨󵄨1 + 󵄩󵄩󵄩vl+ 2 󵄩󵄩󵄩 − 󵄨v 󵄨󵄨1 ) = 0, 2τ 12 144 󵄨 l=0

1 ⩽ k ⩽ n,

2.5 Fourth-order compact two-level nonlinear difference scheme

�

75

or 󵄩 󵄩2 E k = 󵄩󵄩󵄩u0 󵄩󵄩󵄩 = E 0 ,

1 ⩽ k ⩽ n.

Remark 2.2. Thanks to Lemma 1.1(d), we have k−1 2 4 4 󵄩 1 󵄩2 󵄨 1 󵄨2 h 󵄩 1 󵄩2 h 󵄩 󵄩2 E k ⩾ 󵄩󵄩󵄩uk 󵄩󵄩󵄩 + 2ντ ∑ (󵄨󵄨󵄨ul+ 2 󵄨󵄨󵄨1 + 󵄩󵄩󵄩vl+ 2 󵄩󵄩󵄩 − ⋅ 2 󵄩󵄩󵄩vl+ 2 󵄩󵄩󵄩 ) 12 144 h l=0 k−1 2 󵄩 󵄩2 󵄨 1 󵄨2 h 󵄩 1 󵄩2 = 󵄩󵄩󵄩uk 󵄩󵄩󵄩 + 2ντ ∑ (󵄨󵄨󵄨ul+ 2 󵄨󵄨󵄨1 + 󵄩󵄩󵄩vl+ 2 󵄩󵄩󵄩 ), 18 l=0

0 ⩽ k ⩽ n.

Together with Theorem 2.14, it is easy to know that 󵄩󵄩 k 󵄩󵄩 √ k √ 0 󵄩󵄩 0 󵄩󵄩 󵄩󵄩u 󵄩󵄩 ⩽ E = E = 󵄩󵄩u 󵄩󵄩,

1 ⩽ k ⩽ n.

It is obvious that there exists a constant c14 such that 󵄩󵄩 k 󵄩󵄩 󵄩󵄩u 󵄩󵄩 ⩽ c14 ,

0 ⩽ k ⩽ n.

(2.133)

2.5.3 Existence and uniqueness of the difference solution Theorem 2.15. The difference scheme (2.128)–(2.132) has at least a solution. Proof. Denote k uk = (u0k , u1k , . . . , um ),

vk = (vk0 , vk1 , . . . , vkm ).

It is easy to get u0 from (2.130) and (2.131). From (2.129) and (2.132), we can get v0 by computing a system of linear equations whose coefficient matrix is strictly diagonally 1 1 dominant. Suppose {uk , vk } has been determined, then we can regard {uk+ 2 , vk+ 2 } as unknowns and k+ 21

uik+1 = 2ui

− uik ,

k+ 21

vk+1 = 2vi i

− vki ,

0 ⩽ i ⩽ m.

Denote k+ 21

ui = ui

,

k+ 21

vi = vi

,

0 ⩽ i ⩽ m.

76 � 2 Difference methods for the Burgers’ equation From (2.128), (2.129), (2.131) and (2.132), we can get a system of equations in {ui , vi | 0 ⩽ i ⩽ m}: { { { { { { { { { { { { { { { { { { {

2 h2 (ui − uik ) + ψ(u, u)i − ψ(v, u)i − νvi = 0, τ 2 2 h vi = δx2 ui − δx2 vi , 12 u0 = um = 0,

1 ⩽ i ⩽ m − 1,

(2.134)

1 ⩽ i ⩽ m − 1,

(2.135) (2.136)

v0 = vm = 0.

(2.137)

Define the operator Π(u) : 𝒰h̊ → 𝒰h̊ by 2 { { 2 (ui − uik ) + ψ(u, u)i − h ψ(v, u)i − νvi , Π(u)i = { τ 2 { 0, {

1 ⩽ i ⩽ m − 1, i = 0, m,

where (v0 , v1 , . . . , vm ) is determined by (2.135) and (2.137). Then Π(u) is a continuous function in 𝒰h̊ . Taking an inner product of Π(u) with u, using Lemma 2.1 and taking S = 0 in (2.107), we have 2 h2 [‖u‖2 − (uk , u)] + (ψ(u, u), u) − (ψ(v, u), u) − ν(v, u) τ 2 2 2 h 󵄩 2 󵄩 k 2 ⩾ (‖u‖ − 󵄩󵄩󵄩u 󵄩󵄩󵄩 ⋅ ‖u‖) + ν(|u|1 + ‖v‖2 ) τ 18 2 󵄩󵄩 k 󵄩󵄩 ⩾ (‖u‖ − 󵄩󵄩u 󵄩󵄩)‖u‖. τ

(Π(u), u) =

Thus, when ‖u‖ = ‖uk ‖, it follows (Π(u), u) ⩾ 0. By Theorem 2.4, there exists a u∗ ∈ 𝒰h̊ satisfying ‖u∗ ‖ ⩽ ‖uk ‖ such that Π(u∗ ) = 0. Consequently, the difference scheme (2.128), (2.129), (2.131) and (2.132) has at least a solution uk+1 = 2u∗ − uk . ∗ Observing (2.135) and (2.137), when (u1∗ , u2∗ , . . . , um−1 ) is known, (v∗1 , v∗2 , . . . , v∗m−1 ) can be determined by (2.135) and (2.137) uniquely. Thus, vk+1 = 2v∗i − vki , i

1 ⩽ i ⩽ m − 1.

By induction, the conclusion is true. Now we are going to verify the uniqueness of the solution of the difference scheme. We have the following result. Theorem 2.16. When τ < unique.

16ν3 4 , 89c14

the solution of the difference scheme (2.128)–(2.132) is

Proof. According to Theorem 2.15, we just need to prove that (2.134)–(2.137) has a unique solution. Suppose that both {u(1) , v(1) ∈ 𝒰h̊ } and {u(2) , v(2) ∈ 𝒰h̊ } are solutions

2.5 Fourth-order compact two-level nonlinear difference scheme

� 77

of (2.134)–(2.137), i. e., they satisfy { { { { { { { { { { { { { { { { { { {

2 (1) h2 (ui − uik ) + ψ(u(1) , u(1) )i − ψ(v(1) , u(1) )i − νv(1) = 0, i τ 2 h2 v(1) = δx2 ui(1) − δx2 v(1) , i i 12 (1) u0(1) = um = 0, v(1) 0

=

v(1) m

1 ⩽ i ⩽ m − 1,

(2.138)

1 ⩽ i ⩽ m − 1,

(2.139) (2.140)

=0

(2.141)

and { { { { { { { { { { { { { { { { { { {

h2 2 (2) = 0, (ui − uik ) + ψ(u(2) , u(2) )i − ψ(v(2) , u(2) )i − νv(2) i τ 2 h2 2 (2) 2 (2) v(2) = δ u − δ v , x i i 12 x i (2) u0(2) = um = 0, v(2) 0

=

v(2) m

1 ⩽ i ⩽ m − 1, (2.142) 1 ⩽ i ⩽ m − 1, (2.143)

= 0.

(2.144) (2.145)

Let ui = ui(1) − ui(2) ,

vi = v(1) − v(2) , i i

0 ⩽ i ⩽ m.

Subtracting (2.142)–(2.145) from (2.138)–(2.141), we have { { { { { { { { { { { { { { { { { { { { { { {

2 h2 ui + [ψ(u(1) , u(1) )i − ψ(u(2) , u(2) )i ] − [ψ(v(1) , u(1) )i − ψ(v(2) , u(2) )i ] − νvi τ 2 1 ⩽ i ⩽ m − 1, h2 vi = δx2 ui − δx2 vi , 1 ⩽ i ⩽ m − 1, 12 u0 = um = 0, v0 = vm = 0.

= 0, (2.146) (2.147) (2.148) (2.149)

Taking an inner product of (2.146) on both the right- and left-hand sides with u, we have 2 h2 ‖u‖2 + (ψ(u(1) , u(1) ) − ψ(u(2) , u(2) ), u) − (ψ(v(1) , u(1) ) − ψ(v(2) , u(2) ), u) − ν(v, u) = 0. τ 2 (2.150) From (2.133), we have 󵄩󵄩 (1) 󵄩󵄩 󵄩󵄩u 󵄩󵄩 ⩽ c14 . Noticing the definition of u and v and Lemma 2.1, we have

78 � 2 Difference methods for the Burgers’ equation − (ψ(u(1) , u(1) ) − ψ(u(2) , u(2) ), u) = − (ψ(u(1) , u(1) ) − ψ(u(1) − u, u(1) − u), u) = − (ψ(u, u(1) ), u) and (ψ(v(1) , u(1) ) − ψ(v(2) , u(2) ), u) = (ψ(v(1) , u(1) ) − ψ(v(1) − v, u(1) − u), u) = (ψ(v, u(1) ), u). Together with the definition of ψ(⋅, ⋅), computing yields that − (ψ(u(1) , u(1) ) − ψ(u(2) , u(2) ), u) =− =

h m−1 ∑ [u Δ u(1) + Δx (uu(1) )i ]ui 3 i=1 i x i

h m−1 (1) ∑ [u Δ (u2 )i + (uu(1) )i Δx ui ] 3 i=1 i x

1 󵄩 󵄩 󵄩 󵄩 ⩽ (2‖u‖∞ ⋅ 󵄩󵄩󵄩u(1) 󵄩󵄩󵄩 ⋅ |u|1 + ‖u‖∞ ⋅ 󵄩󵄩󵄩u(1) 󵄩󵄩󵄩 ⋅ |u|1 ) 3 ⩽ c14 ‖u‖∞ ⋅ |u|1 . Taking ε =

ν 4c14

in Lemma 1.1(c), then we have −(ψ(u(1) , u(1) ) − ψ(u(2) , u(2) ), u) ⩽ c14 (

c4 c ν ν |u|1 + 14 ‖u‖)|u|1 ⩽ |u|21 + 143 ‖u‖2 . 4c14 ν 2 ν

Similarly, we have h2 (ψ(v(1) , u(1) ) − ψ(v(2) , u(2) ), u) 2

=

h3 m−1 ∑ [v Δ u(1) + Δx (vu(1) )i ]ui 6 i=1 i x i

=− =− ⩽

h3 m−1 (1) ∑ [u Δ (uv)i + (vu(1) )i Δx ui ] 6 i=1 i x

h3 m−1 (1) 1 1 ∑ [ui (ui Δx vi + vi+1 δx ui+ 1 + vi−1 δx ui− 1 ) + (vu(1) )i Δx ui ] 2 2 6 i=1 2 2

h2 󵄩 󵄩 (|v|1 ⋅ ‖u‖∞ + 2‖v‖∞ ⋅ |u|1 )󵄩󵄩󵄩u(1) 󵄩󵄩󵄩 6

(2.151)

2.5 Fourth-order compact two-level nonlinear difference scheme

⩽

79

c14 h2 (|v|1 ⋅ ‖u‖∞ + 2‖v‖∞ ⋅ |u|1 ). 6

ν 3√2c14

Taking ε =

�

in Lemma 1.1(c) and using the Cauchy–Schwarz inequality, then we get

h2 (ψ(v(1) , u(1) ) − ψ(v(2) , u(2) ), u) 2 3√2c14 3√2c14 c h2 c h2 ν ν |u|1 + |v|1 + ⩽ 14 |v|1 ( ‖u‖) + 14 |u|1 ( ‖v‖) 6 4ν 3 4ν 3√2c14 3√2c14 2 2 2 2 √2c14 √2c14 h h νh2 |u|1 ⋅ |v|1 + |u|1 ⋅ ‖v‖ + |v|1 ⋅ ‖u‖ 4ν 8ν 6√2 2 3√2c14 h νh ⩽ |u|1 ⋅ ‖v‖ + ‖u‖ ⋅ ‖v‖ 4ν 3√2

=

⩽

4 ν 2 νh2 2 81c14 |u|1 + ‖v‖ + ‖u‖2 . 2 18 8ν3

(2.152)

Substituting (2.151) and (2.152) into (2.150) and taking S = 0 in (2.107), we can obtain 4 c4 h2 ν νh2 2 81c14 2 ν ‖u‖2 + ν(|u|21 + ‖v‖2 ) ⩽ ( |u|21 + 143 ‖u‖2 ) + ( |u|21 + ‖v‖ + ‖u‖2 ), 3 τ 18 2 2 18 ν 8ν

or 4 89c14 2 ‖u‖2 ⩽ ‖u‖2 . τ 8ν3

When τ
0 such that 󵄨󵄨 ̌ 0 󵄨󵄨 2 󵄨󵄨Pi 󵄨󵄨 ⩽ c11 (τ + h ),

󵄨󵄨 ̌ k 󵄨󵄨 2 2 󵄨󵄨Pi 󵄨󵄨 ⩽ c11 (τ + h ),

1 ⩽ i ⩽ m − 1, 1 ⩽ k ⩽ n − 1.

(4.131)

Considering equation (4.68) at the point (xi+ 1 , tk ) and using the Taylor expansion, we 2 have k k k Vi+ + Qi+ 1 = δx U 1, i+ 1 2

2

2

0 ⩽ i ⩽ m − 1, 0 ⩽ k ⩽ n,

(4.132)

which is the same as (4.72). In addition, Lemma 4.3 holds. Noticing the initial-boundary value conditions (4.69)–(4.70), we have U 0 = φ(xi ), { { { ik U0 = 0, Umk = 0, { { { k { Vm = 0,

0 ⩽ i ⩽ m,

(4.133)

1 ⩽ k ⩽ n,

(4.134)

0 ⩽ k ⩽ n.

(4.135)

Omitting the small terms in (4.129), (4.130) and (4.132) and combining with (4.133)–(4.135), we construct a three-level linearized difference scheme for the problem (4.67)–(4.70) as follows: 1

{ { { { { { { { { { { { { { { { { { { { { { { { { { {

1

1

δt ui2 + γψ(u0 , u 2 )i + δx2 vi2 = 0, ̄

̄

Δt uik + γψ(uk , uk )i + δx2 vki = 0, vki+ 1 2

=

k δx ui+ 1, 2

ui0 = φ(xi ), u0k vkm

= 0,

= 0,

k um

= 0,

1 ⩽ i ⩽ m − 1,

(4.136)

1 ⩽ i ⩽ m − 1, 1 ⩽ k ⩽ n − 1,

(4.137)

0 ⩽ i ⩽ m − 1, 0 ⩽ k ⩽ n,

(4.138)

0 ⩽ i ⩽ m,

(4.139)

1 ⩽ k ⩽ n,

(4.140)

0 ⩽ k ⩽ n.

(4.141)

Combining (4.138) with (4.141), we have 1

1

2 2 { { vi+ 21 = δx ui+ 21 , { { k̄ k̄ { vi+ 21 = δx ui+ 21 ,

1

0 ⩽ i ⩽ m − 1,

vm2 = 0,

0 ⩽ i ⩽ m − 1,

vkm = 0,

̄

1 ⩽ k ⩽ n − 1.

4.5 Second-order in space three-level linearized difference scheme

� 139

Theorem 4.13. The difference scheme (4.136)–(4.141) is equivalent to { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { {

1 1 1 γ + [ψ(u0 , u 2 )i + ψ(u0 , u 2 )i+1 ] + δx2 (δx u 2 1 ) = 0, i+ 2 2 1 ⩽ i ⩽ m − 2, 1 1 1 1 2 2 + γψ(u0 , u 2 )m−1 + 2 (δx u 2 3 − 3δx u 2 1 ) = 0, δt um−1 m− m− h 2 2 γ 2 k̄ k k k̄ k k̄ Δt ui+ 1 + [ψ(u , u )i + ψ(u , u )i+1 ] + δx (δx ui+ 1 ) = 0, 2 2 2 1 ⩽ i ⩽ m − 2, 1 ⩽ k ⩽ n − 1, ̄ 2 k̄ k̄ k Δt um−1 + γψ(uk , uk )m−1 + 2 (δx um− 1 ) = 0, 3 − 3δx u m− 2 2 h 1 ⩽ k ⩽ n − 1, 1

δt u 2

i+ 21

ui0 u0k

= φ(xi ), = 0,

k um

(4.142) (4.143)

(4.144) (4.145)

0 ⩽ i ⩽ m,

(4.146)

= 0,

(4.147)

1⩽k⩽n

and vkm = 0,

k k vki = 2δx ui+ 1 − vi+1 , 2

i = m − 1, m − 2, . . . , 0, 0 ⩽ k ⩽ n.

Proof. The proof is omitted. Using Theorem 4.13, we construct the difference scheme (4.142)–(4.147) for solving (4.1)–(4.3). We see that (4.142)–(4.143) is a system of linear equations in u1 ; when the values of uk−1 and uk have been determined, (4.144)–(4.145) is a system of linear equations in uk+1 . We can prove that there is a unique solution for the difference scheme (4.136)–(4.141). By Theorem 4.13, we know that the solution of (4.142)–(4.147) exists and is unique.

4.5.2 Conservation and boundedness of the difference solution First, we give a lemma that will be used later. Lemma 4.4. Let v ∈ 𝒰h and u ∈ 𝒰h̊ satisfy vm = 0,

vi+ 1 = δx ui+ 1 , 2

2

0 ⩽ i ⩽ m − 1.

We have 1 (δx2 v, u) = (v0 )2 . 2 The difference scheme (4.136)–(4.141) has the following conservation law.

140 � 4 Difference methods for the Korteweg–de Vries equation Theorem 4.14. Suppose {uik , vki | 0 ⩽ i ⩽ m, 0 ⩽ k ⩽ n} is the solution of (4.136)–(4.141). Denote Ek =

1 1 2 k−1 ̄ 2 ‖uk ‖2 + ‖uk−1 ‖2 + τ[ (v02 ) + ∑ (vl0 ) ], 2 2 l=1

1 ⩽ k ⩽ n.

Then we have 󵄩 󵄩2 E k = 󵄩󵄩󵄩u0 󵄩󵄩󵄩 ,

1 ⩽ k ⩽ n.

Proof. (I) Taking the inner product on both the right- and left-hand sides of (4.136) 1 with u 2 , we have 1

1

1

1

1

1

(δt u 2 , u 2 ) + γ(ψ(u0 , u 2 ), u 2 ) + (δx2 v 2 , u 2 ) = 0. Together with Lemma 2.1 and Lemma 4.4, we can get ‖u1 ‖2 − ‖u0 ‖2 1 21 2 + (v0 ) = 0. 2τ 2 That is, ‖u1 ‖2 + ‖u0 ‖2 τ 21 2 󵄩󵄩 0 󵄩󵄩2 + (v0 ) = 󵄩󵄩u 󵄩󵄩 . 2 2

(4.148) ̄

(II) Taking the inner product on both the right- and left-hand sides of (4.137) with 2uk , we have ̄

̄

̄

̄

̄

2(Δt uk , uk ) + 2γ(ψ(uk , uk ), uk ) + 2(δx2 vk , uk ) = 0,

1 ⩽ k ⩽ n − 1.

Together with Lemma 2.1 and Lemma 4.4, we can get ̄ 2 1 ‖uk+1 ‖2 + ‖uk ‖2 ‖uk ‖2 + ‖uk−1 ‖2 ( − ) + (vk0 ) = 0, τ 2 2

1 ⩽ k ⩽ n − 1,

which leads to k ̄ 2 ‖uk+1 ‖2 + ‖uk ‖2 ‖u1 ‖2 + ‖u0 ‖2 + τ ∑(vl0 ) = , 2 2 l=1

1 ⩽ k ⩽ n − 1.

Adding (4.148) and the above equality, we can obtain ‖uk+1 ‖2 + ‖uk ‖2 1 1 2 k ̄ 2 󵄩 󵄩2 + τ[ (v02 ) + ∑(vl0 ) ] = 󵄩󵄩󵄩u0 󵄩󵄩󵄩 , 2 2 l=1 This completes this proof.

0 ⩽ k ⩽ n − 1.

(4.149)

4.5 Second-order in space three-level linearized difference scheme

� 141

From (4.148), we have 󵄩󵄩 1 󵄩󵄩 󵄩󵄩 0 󵄩󵄩 󵄩󵄩u 󵄩󵄩 ⩽ 󵄩󵄩u 󵄩󵄩. From (4.149), we have 󵄩󵄩 k+1 󵄩󵄩 󵄩󵄩 k−1 󵄩󵄩 󵄩󵄩u 󵄩󵄩 ⩽ 󵄩󵄩u 󵄩󵄩,

1 ⩽ k ⩽ n − 1.

Combining the above two inequalities produces 󵄩󵄩 k 󵄩󵄩 󵄩󵄩 0 󵄩󵄩 󵄩󵄩u 󵄩󵄩 ⩽ 󵄩󵄩u 󵄩󵄩,

1 ⩽ k ⩽ n.

4.5.3 Existence and uniqueness of the difference solution Next, we prove the unique solvability. Theorem 4.15. The difference scheme (4.136)–(4.141) has a unique solution. Proof. We can get u0 from (4.139) and v0 from (4.138) and (4.141). From (4.136), (4.138) and (4.140)–(4.141), we can get the system of linear equations in {u1 , v1 }. Consider its homogeneous one: { { { { { { { { { { { { { { { { {

1 1 γ 1 ui + ψ(u0 , u1 )i + δx2 v1i = 0, τ 2 2 1 v1i+ 1 = δx ui+ 1, 2

u01 = 0,

v1m

= 0.

2

1 ⩽ i ⩽ m − 1,

(4.150)

0 ⩽ i ⩽ m − 1,

(4.151)

1 um = 0,

(4.152) (4.153)

Taking the inner product on both the right- and left-hand sides of (4.150) with u1 , we have 1 󵄩󵄩 1 󵄩󵄩2 γ 1 2 1 1 0 1 1 󵄩u 󵄩 + (ψ(u , u ), u ) + (δx v , u ) = 0. τ󵄩 󵄩 2 2 Together with Lemma 2.1 and Lemma 4.4, we get 1 󵄩󵄩 1 󵄩󵄩2 1 1 2 󵄩u 󵄩 + (v0 ) = 0. τ󵄩 󵄩 4 Then we have 󵄩󵄩 1 󵄩󵄩 󵄩󵄩u 󵄩󵄩 = 0, which follows:

142 � 4 Difference methods for the Korteweg–de Vries equation ui1 = 0,

0 ⩽ i ⩽ m.

Then from (4.151) and (4.153), we can get v1i = 0,

0 ⩽ i ⩽ m.

That is, (4.136), (4.138) and (4.140)–(4.141) have a unique solution {u1 , v1 }. Suppose that {uk , uk−1 } and {vk , vk−1 } are known. From (4.137)–(4.138) and (4.140)– (4.141), we can get the system of linear equations in {uk+1 , vk+1 }. Consider its homogeneous one: { { { { { { { { { { { { { { { { {

1 1 k+1 γ ui + ψ(uk , uk+1 )i + δx2 vk+1 = 0, i 2τ 2 2 k+1 = δx ui+ vk+1 1 , i+ 1 2

u0k+1 = 0, vk+1 m

= 0.

2

1 ⩽ i ⩽ m − 1,

(4.154)

0 ⩽ i ⩽ m − 1,

(4.155)

k+1 um = 0,

(4.156) (4.157)

Taking the inner product on both the right- and left-hand sides of (4.154) with 2uk+1 , we have 1 󵄩󵄩 k+1 󵄩󵄩2 k k+1 k+1 2 k+1 k+1 󵄩u 󵄩󵄩 + γ(ψ(u , u ), u ) + (δx v , u ) = 0. τ󵄩 Using Lemma 2.1 and Lemma 4.4, we can get 1 󵄩󵄩 k+1 󵄩󵄩2 1 k+1 2 󵄩u 󵄩󵄩 + (v0 ) = 0. τ󵄩 2 Then we have 󵄩󵄩 k+1 󵄩󵄩 󵄩󵄩u 󵄩󵄩 = 0, which follows: uik+1 = 0,

0 ⩽ i ⩽ m.

Then from (4.155) and (4.157), we can get vk+1 = 0, i

0 ⩽ i ⩽ m.

That is, (4.137)–(4.138) and (4.140)–(4.141) have a unique solution {uk+1 , vk+1 }. By induction, the conclusion is true.

4.5 Second-order in space three-level linearized difference scheme

� 143

4.5.4 Convergence of the difference solution Assume that {u(x, t), v(x, t)} is the solution of the problem (4.67)–(4.70) and {uik , vki | 0 ⩽ i ⩽ m, 0 ⩽ k ⩽ n} is the solution of the difference scheme (4.136)–(4.141), respectively. Denote eik = u(xi , tk ) − uik ,

fik = v(xi , tk ) − vki ,

0 ⩽ i ⩽ m, 0 ⩽ k ⩽ n.

Subtracting (4.136)–(4.141) from (4.129), (4.130), (4.132)–(4.135), respectively, we can get the system of error equations: 1

{ { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { {

1

1

δt ei2 + γψ(u0 , e 2 )i + δx2 fi 2 = P̌ i0 , k̄

1 ⩽ i ⩽ m − 1,

k̄

Δt eik + γ[ψ(U k , U )i − ψ(uk , u )i ] + 1 ⩽ i ⩽ m − 1, 1 ⩽ k ⩽ n − 1,

fi+k 1 2

=

k δx ei+ 1 2

+

k Qi+ 1, 2

ei0 = 0,

0 ⩽ i ⩽ m,

= 0,

0 ⩽ k ⩽ n.

e0k fmk

k em

= 0,

Theorem 4.16. Denote c0 =

= 0,

̄ δx2 fik

= P̌ ik ,

(4.159)

0 ⩽ i ⩽ m − 1, 0 ⩽ k ⩽ n,

(4.160) (4.161)

1 ⩽ k ⩽ n,

(4.162) (4.163)

󵄨 󵄨 {󵄨󵄨󵄨u(x, t)󵄨󵄨󵄨} and λ =

max

(4.158)

(x,t)∈[0,L]×[0,T]

there is a constant c12 such that

󵄩󵄩 k 󵄩󵄩 2 2 󵄩󵄩e 󵄩󵄩 ⩽ c12 (τ + h ),

c0 |γ|τ . 3h

Suppose λ < 1, then

0 ⩽ k ⩽ n.

Proof. (I) It is easy to know from (4.161) that 󵄩󵄩 0 󵄩󵄩 󵄩󵄩e 󵄩󵄩 = 0.

(4.164) 1

Taking the inner product on both the right- and left-hand sides of (4.158) with e 2 , we have 1

1

1

1

1

1

1

(δt e 2 , e 2 ) + γ(ψ(u0 , e 2 ), e 2 ) + (δx2 f 2 , e 2 ) = (P̌ 0 , e 2 ). That is, 1 1 󵄩󵄩 1 󵄩󵄩2 1 0 1 2 1 󵄩󵄩e 󵄩󵄩 + (δx f 2 , e 2 ) = (P̌ , e ). 2τ 2

From (4.160), we have f

1 2

i+ 21

1

= δx e 2

i+ 21

1

+ Q2 1 , i+ 2

0 ⩽ i ⩽ m − 1.

(4.165)

144 � 4 Difference methods for the Korteweg–de Vries equation Similar to the proof for (4.122), we can get 1

1

−(δx2 f 2 , e 2 )

m−2 1 m−1 1 1 1 1 1 1 1 1 1 2 = − (f02 ) + f02 Q 21 − hf02 R02 + 2h ∑ Q 2 1 Rj2 − 2h ∑ ej2 (δx R 2 1 ) j+ 2 j− 2 2 2 j=0 j=1

󵄩 1 󵄩2 ⩽ 󵄩󵄩󵄩e 2 󵄩󵄩󵄩 + (c82 + c92 + 2Lc8 c9 + Lc92 )h4 . Substituting the above inequality into (4.165), we get 1 󵄩󵄩 1 󵄩󵄩2 󵄩e 󵄩 2τ 󵄩 󵄩 1 ⩽ ‖e1 ‖2 + (c82 + c92 + 2Lc8 c9 + Lc92 )h4 + 4 1 ⩽ ‖e1 ‖2 + (c82 + c92 + 2Lc8 c9 + Lc92 )h4 + 4

1 ̌0 1 (P , e ) 2 1 󵄩󵄩 1 󵄩󵄩2 󵄩e 󵄩 + 4τ 󵄩 󵄩

τ 󵄩󵄩 ̌ 0 󵄩󵄩2 󵄩P 󵄩 . 4󵄩 󵄩

That is, 󵄩 󵄩2 󵄩 󵄩2 (1 − τ)󵄩󵄩󵄩e1 󵄩󵄩󵄩 ⩽ 4τ(c82 + c92 + 2Lc8 c9 + Lc92 )h4 + τ 2 󵄩󵄩󵄩P̌ 0 󵄩󵄩󵄩 . When τ ⩽ 21 , noticing (4.131) we obtain 󵄩󵄩 1 󵄩󵄩2 2 2 2 4 2 󵄩 0 󵄩2 󵄩󵄩e 󵄩󵄩 ⩽ 8τ(c8 + c9 + 2Lc8 c9 + Lc9 )h + 2τ 󵄩󵄩󵄩P̌ 󵄩󵄩󵄩

2 ⩽ 8τ(c82 + c92 + 2Lc8 c9 + Lc92 )h4 + 2τ 2 Lc11 (τ + h2 )

2

2

2 ⩽ (8(c82 + c92 + 2Lc8 c9 + Lc92 ) + 2Lc11 )(τ 2 + h2 ) 2

≡ c13 (τ 2 + h2 ) .

(4.166)

(II) Taking the inner product on both the right- and left-hand sides of (4.159) with ̄ ek , we have ̄

̄

̄

̄

̄

̄

̄

(Δt ek , ek ) + γ(ψ(U k , U k ) − ψ(uk , uk ), ek ) + (δx2 f k , ek ) = (P̌ k , ek ),

1 ⩽ k ⩽ n − 1.

That is, ̄ ̄ 1 ‖ek+1 ‖2 + ‖ek ‖2 ‖ek ‖2 + ‖ek−1 ‖2 ( − ) + (δx2 f k , ek ) 2τ 2 2

̄ ̄ ̄ ̄ = (P̌ k , ek ) − γ(ψ(U k , U k ) − ψ(uk , uk ), ek ),

1 ⩽ k ⩽ n − 1.

From (4.160), we have ̄

̄

̄

k k fi+k 1 = δx ei+ , 1 + Q i+ 1 2

2

2

0 ⩽ i ⩽ m − 1, 1 ⩽ k ⩽ n − 1.

(4.167)

4.5 Second-order in space three-level linearized difference scheme

� 145

Similar to the proof for (4.122), we have ̄

̄

−(δx2 f k , ek )

m−2 m−1 ̄ ̄ ̄ ̄ 1 ̄ 2 k̄ k̄ k̄ k̄ = − (f0k ) + f0k Qk1 − hf0k Rk0 + 2h ∑ Qj+ ) 1 Rj − 2h ∑ ej (δx R j− 21 2 2 2 j=0 j=1

󵄩 ̄ 󵄩2 ⩽ 󵄩󵄩󵄩ek 󵄩󵄩󵄩 + (c82 + c92 + 2Lc8 c9 + Lc92 )h4 .

(4.168)

Denote 1

Gk− 2 =

γτ m−2 k− 21 k k−1 k ). ∑ U (ei ei+1 − eik−1 ei+1 6 i=1 i

Substituting (4.168) into (4.167) and noticing (4.51) with (4.131), we have 1 1 1 ‖ek+1 ‖2 + ‖ek ‖2 ‖ek ‖2 + ‖ek−1 ‖2 [( + Gk+ 2 ) − ( + Gk− 2 )] 2τ 2 2

⩽ −γ{

+

̄ ̄ h m−1 h m−2 k̄ k k̄ ∑ (Δx Uik )eik eik + ∑ (δx Ui+ 1 )ei+1 ei 3 i=1 6 i=1 2

̄ 1 m−2 k̄ k+ 1 k− 1 k k+1 k−1 k − eik ei+1 ) + (Ui 2 − Uik )(eik ei+1 − eik−1 ei+1 )]} ∑ [(Ui − Ui 2 )(eik+1 ei+1 12 i=1

̄ 󵄩 ̄ 󵄩2 + 󵄩󵄩󵄩ek 󵄩󵄩󵄩 + (c82 + c92 + 2Lc8 c9 + Lc92 )h4 + (P̌ k , ek )

1 󵄩 󵄩 󵄩 ̄󵄩 1 τ 󵄩 󵄩 󵄩 󵄩 1 τ󵄩 󵄩 󵄩 󵄩 ⩽ |γ|[ c3 󵄩󵄩󵄩ek 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩ek 󵄩󵄩󵄩 + c6 󵄩󵄩󵄩ek+1 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩ek 󵄩󵄩󵄩 + c6 󵄩󵄩󵄩ek 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩ek−1 󵄩󵄩󵄩] 2 12 h 12 h 1 󵄩 ̄ 󵄩2 1 󵄩 ̄ 󵄩2 + 󵄩󵄩󵄩ek 󵄩󵄩󵄩 + (c82 + c92 + 2Lc8 c9 + Lc92 )h4 + ‖P̌ k ‖2 + 󵄩󵄩󵄩ek 󵄩󵄩󵄩 2 2 c |γ|τ 󵄩󵄩 k+1 󵄩󵄩2 |γ| 󵄩󵄩 k 󵄩󵄩2 󵄩󵄩 k̄ 󵄩󵄩2 󵄩 󵄩2 󵄩 󵄩2 c3 (󵄩󵄩e 󵄩󵄩 + 󵄩󵄩e 󵄩󵄩 ) + 6 (󵄩󵄩e 󵄩󵄩 + 2󵄩󵄩󵄩ek 󵄩󵄩󵄩 + 󵄩󵄩󵄩ek−1 󵄩󵄩󵄩 ) ⩽ 4 24h 1 2 2 3 󵄩󵄩 k̄ 󵄩󵄩2 2 2 2 (τ + h2 ) . + 󵄩󵄩e 󵄩󵄩 + (c8 + c9 + 2Lc8 c9 + Lc92 )h4 + Lc11 2 2 Let Ek =

1 ‖ek ‖2 + ‖ek−1 ‖2 + Gk− 2 . 2

It is easy to know that ‖ek ‖2 + ‖ek−1 ‖2 󵄨󵄨 k− 21 󵄨󵄨 . 󵄨󵄨G 󵄨󵄨 ⩽ λ ⋅ 2 If λ < 1, we have (1 − λ)

‖ek ‖2 + ‖ek−1 ‖2 ‖ek ‖2 + ‖ek−1 ‖2 󵄩󵄩 k 󵄩󵄩2 󵄩󵄩 k−1 󵄩󵄩2 ⩽ E k ⩽ (1 + λ) ⩽ 󵄩󵄩e 󵄩󵄩 + 󵄩󵄩e 󵄩󵄩 . 2 2

(4.169)

146 � 4 Difference methods for the Korteweg–de Vries equation It follows from (4.169) that there exists a constant c14 such that 1 k+1 2 (E − E k ) ⩽ c14 (E k+1 + E k ) + c14 (τ 2 + h2 ) , τ

1 ⩽ k ⩽ n − 1.

That is, 2

(1 − c14 τ)E k+1 ⩽ (1 + c14 τ)E k + c14 τ(τ 2 + h2 ) ,

1 ⩽ k ⩽ n − 1.

When c14 τ ⩽ 31 , we have 3 2 E k+1 ⩽ (1 + 3c14 τ)E k + c14 τ(τ 2 + h2 ) , 2

1 ⩽ k ⩽ n − 1.

Using the Gronwall inequality (Theorem 1.2(a)), we can get 1 2 E k ⩽ e3c14 (k−1)τ [E 1 + (τ 2 + h2 ) ] 2 2 󵄩 󵄩2 󵄩 󵄩2 1 ⩽ e3c14 T [󵄩󵄩󵄩e1 󵄩󵄩󵄩 + 󵄩󵄩󵄩e0 󵄩󵄩󵄩 + (τ 2 + h2 ) ], 2

1 ⩽ k ⩽ n.

Substituting (4.164) and (4.166) into the above inequality, we can get 1 2 E k ⩽ e3c14 T (c13 + )(τ 2 + h2 ) , 2

1 ⩽ k ⩽ n.

It is easy to know that 1 3c14 T 1 ‖ek ‖2 + ‖ek−1 ‖2 2 ⩽ e (c13 + )(τ 2 + h2 ) , 2 1−λ 2

1 ⩽ k ⩽ n.

Consequently, 3 󵄩󵄩 k 󵄩󵄩 c T 2c + 1 2 (τ + h2 ), 󵄩󵄩e 󵄩󵄩 ⩽ e 2 14 √ 13 1−λ

1 ⩽ k ⩽ n.

This completes the proof.

4.6 Numerical experiments In this section, we present a numerical example to illustrate the accuracy and conservation of the second-order difference schemes (4.83)–(4.86) and (4.142)–(4.147). The nonlinear difference scheme (4.83)–(4.86) is solved by using the Newton iterative method.

4.6 Numerical experiments

� 147

Let the numerical solution corresponding to the step sizes h and τ be {ujk (h, τ) | 0 ⩽ j ⩽ m, 0 ⩽ k ⩽ n}. Denote the numerical errors by 2

m−1 k ( h , τ)) }, E(h, τ) = max {√h ∑ (ujk (h, τ) − u2j 2 1⩽k⩽n j=1 2

m−1 τ F(h, τ) = √h ∑ (ujn (h, τ) − uj2n (h, )) . 2 j=1

When τ is sufficiently small, the spatial convergence order is defined by Orderh = log2 (

E(2h, τ) ). E(h, τ)

When h is sufficiently small, the temporal convergence order is defined by Orderτ = log2 (

F(h, 2τ) ). F(h, τ)

Example 4.1. In (4.1)–(4.3), take T = 1, L = 1, γ = −6, φ(x) = x(x − 1)2 (x 3 − 2x 2 + 2). The exact solution is unknown. Taking various step size h with the sufficiently small step size τ = 1/12800 and varying step size τ with the sufficiently small step size h = 1/12800, the numerical errors and convergence orders of difference schemes (4.83)–(4.86) and (4.142)–(4.147) are listed in Table 4.1 and Table 4.2. From these tables, we know that the numerical convergence orders of both schemes can achieve O(h2 + τ 2 ), which are in good agreement with the theoretical results. Furthermore, we can see that the difference scheme (4.142)–(4.147) is more computationally efficient than the nonlinear difference scheme (4.83)–(4.86). Figure 4.1 indicates that the energy of the schemes (4.83)–(4.86) and (4.142)–(4.147) is both conservative. Table 4.1: Errors and convergence orders in space of Example 4.1 (τ = 1/12800). h �/�� �/��� �/��� �/��� �/����

Scheme (4.142)–(4.147) E(h, τ) Orderh 8.250008e-7 2.062408e-7 5.156014e-8 1.289100e-8 3.215215e-9

�.��� �.��� �.��� �.���

CPU 0.143 s 0.189 s 0.284 s 0.484 s 0.943 s

Scheme (4.83)–(4.86) E(h, τ) Orderh 8.249301e-7 2.062284e-7 5.155684e-8 1.288886e-8 3.210015e-9

�.��� �.��� �.��� �.���

CPU 0.416 s 0.436 s 0.631 s 1.063 s 8.395 s

148 � 4 Difference methods for the Korteweg–de Vries equation Table 4.2: Errors and convergence orders in time of Example 4.1 (h = 1/12800). τ �/�� �/��� �/��� �/��� �/����

Scheme (4.142)–(4.147) F(h, τ) Orderτ 1.410467e-2 4.362537e-3 1.004853e-3 2.451153e-4 7.068438e-5

�.��� �.��� �.��� �.���

CPU 0.086 s 0.161 s 0.316 s 0.631 s 1.235 s

Scheme (4.83)–(4.86) F(h, τ) Orderτ 4.445480e-3 1.030215e-3 2.507916e-4 7.133256e-5 1.842235e-5

�.��� �.��� �.��� �.���

CPU 4.601 s 9.113 s 17.354 s 32.648 s 63.290 s

Figure 4.1: The energy conservation of the difference schemes (4.83)–(4.86) (left) and (4.142)–(4.147) (right).

4.7 Summary and extension The KdV equation is a third order in space evolution equation. The solution of the initialboundary value problem (4.1)–(4.3) of the KdV equation satisfies the conservation law (4.4). In this chapter, four difference schemes are derived. The first two schemes are firstorder accurate in space, and the last two schemes are second-order accurate in space. Introducing a new variable v = ux , the original equation (4.1) could be rewritten as an equivalent system (4.67)–(4.68), where (4.67) is a second-order equation in space and (4.68) is a first-order one in space. Then we derive a difference scheme (4.78)–(4.82) for solving (4.67)–(4.70). It is a natural way to derive a difference scheme like this. The introduced intermediate variable {vki } does not have to actually participate in the calculation. Making the variable separation for the difference scheme (4.78)–(4.82), we obtain a difference scheme (4.83)–(4.86) only with respect to the variable {uik }. We have proved the unique existence, boundedness and convergence of the solution for the difference scheme (4.83)–(4.86). The Browder theorem is utilized to prove the existence of the solution of the difference scheme (4.78)–(4.82), and thereby indirectly prove the existence of the solution of the difference scheme (4.83)–(4.86). In order to directly prove the existence of the

4.7 Summary and extension

� 149

solution of the difference scheme (4.83)–(4.86), one needs to prove the existence of the solution of nonlinear system (4.100)–(4.102). Introducing a new variable z and defining the operator Π(w), the existence result can be proved by the Browder theorem. Equations (4.67)–(4.68) can be discretized as follows: k+ 21

δt Ui

1

k+ 21

1

+ γψ(U k+ 2 , U k+ 2 )i + δx2 Vi

Vik

1

k+ = P̃i 2 ,

=

k δx Ui+ 1 2

1 ⩽ i ⩽ m − 1, 0 ⩽ k ⩽ n − 1, ̃k, +Q i

0 ⩽ i ⩽ m − 1, 0 ⩽ k ⩽ n,

where there is a constant c15 such that 󵄨󵄨 ̃ k+ 21 󵄨󵄨 2 2 󵄨󵄨Pi 󵄨󵄨 ⩽ c15 (τ + h ), 1 ⩽ i ⩽ m − 1, 0 ⩽ k ⩽ n − 1, 󵄨󵄨 ̃ k 󵄨󵄨 󵄨󵄨Qi 󵄨󵄨 ⩽ c15 h, 0 ⩽ i ⩽ m − 1, 0 ⩽ k ⩽ n. Omitting the small terms, a first order in space difference scheme for solving (4.67)–(4.70) reads { { { { { { { { { { { { { { {

k+ 21

δt ui

vki

=

1

= 0,

1 ⩽ i ⩽ m − 1, 0 ⩽ k ⩽ n − 1, (4.170)

k δx ui+ 1, 2

ui0 = φ(xi ),

u0k

k+ 21

1

+ γψ(uk+ 2 , uk+ 2 )i + δx2 vi

k um

= 0,

vkm

= 0,

= 0,

0 ⩽ i ⩽ m − 1, 0 ⩽ k ⩽ n,

(4.171)

1 ⩽ i ⩽ m − 1,

(4.172)

0 ⩽ k ⩽ n.

(4.173)

We can prove that the difference scheme (4.170)–(4.173) is equivalent to { { { { { { { { { { { { { { { { { { { { { { { { { { {

k+ 21

δt ui

1

1

+ γψ(uk+ 2 , uk+ 2 )i + δx2 (δx u

1 ⩽ i ⩽ m − 2, 0 ⩽ k ⩽ n − 1,

k+ 1 δt um−12

ui0 u0k

+ γψ(u

k+ 21

,u

k+ 21

0 ⩽ k ⩽ n − 1, = φ(xi ), = 0,

k um

k+ 21

i+ 21

) = 0,

1 k+ 1 k+ 1 )m−1 + 2 (δx u 23 − 2δx u 21 ) = 0, m− 2 m− 2 h

(4.174) (4.175)

1 ⩽ i ⩽ m − 1,

(4.176)

= 0,

(4.177)

0⩽k⩽n

and k k { vi = δx ui+ 1 , 2 { k { vm = 0,

0 ⩽ i ⩽ m − 1, 0 ⩽ k ⩽ n,

(4.178)

0 ⩽ k ⩽ n.

(4.179)

We can prove the existence, boundedness and conservation of the solution of the difference scheme (4.174)–(4.177). Comparing (4.175) with (4.14), one may find that they are different.

150 � 4 Difference methods for the Korteweg–de Vries equation Noticing uxxx (L, t) = 0, the local truncation error of (4.14) is O(τ 2 + h2 ), and the local truncation error of (4.175) is O(τ 2 + h2 ) + O(h−1 )uxx (L, tk+ 1 ). 2 Numerical examples show that the condition λ < 1 in Theorems 4.7 and 4.16 may be unnecessary. We conjecture that this condition may be removed for the three-level linearized difference scheme (4.36)–(4.41) and (4.136)–(4.141). However, our analytic technique is limited at present. The results in Sections 4.2–4.5 have been published in [27], [44] and [45], respectively.

5 Difference methods for the Camassa–Holm equation 5.1 Introduction The Camassa–Holm equation is one of the important research subjects in the shallow water wave theory since it owns the peakon. Many scholars have deeply investigated on various properties of the Camassa–Holm equation, such as solitary wave solutions, the bihamiltonian structure, the complete symmetry, etc. In this chapter, we consider the initial and boundary value problem of the Camassa– Holm equation: ut − uxxt + 3uux = 2ux uxx + uuxxx , { { u(x, 0) = φ(x), { { { u(0, t) = 0, u(L, t) = 0,

0 < x < L, 0 < t ⩽ T,

(5.1)

0 < x < L,

(5.2)

0 ⩽ t ⩽ T,

(5.3)

where φ(0) = φ(L) = 0. Equation (5.1) can be written as ut − uxxt + 3uux = ux uxx + (uuxx )x .

(5.4)

Before introducing the difference method, we first give a priori estimate on the solution of the problem (5.1)–(5.3) by using the energy method. Theorem 5.1. Let u(x, t) be the solution of the problem (5.1)–(5.3). Denote L

L

E(t) = ∫ u2 (x, t)dx + ∫ ux2 (x, t)dx, 0

(5.5)

0

then E(t) = E(0),

0 < t ⩽ T.

(5.6)

Proof. Multiplying (5.1) by u(x, t) on both the right- and left-hand sides and then integrating the result with respect to x from 0 to L, we have L

L

L

0

0

0

∫ u(x, t)ut (x, t)dx − ∫ u(x, t)uxxt (x, t)dx + 3 ∫ u2 (x, t)ux (x, t)dx L

= ∫(2uux uxx + u2 uxxx )(x, t)dx. 0

Applying the integration by parts to the above equality, we have https://doi.org/10.1515/9783110796018-005

152 � 5 Difference methods for the Camassa–Holm equation L

L

0

0

1 d 󵄨L 1 d 󵄨L ∫ u2 (x, t)dx − u(x, t)uxt (x, t)󵄨󵄨󵄨0 + ∫ ux2 (x, t)dx + u3 (x, t)󵄨󵄨󵄨x=0 2 dt 2 dt 󵄨L = (u uxx )(x, t)󵄨󵄨󵄨x=0 . 2

It follows from (5.3) that d E(t) = 0, dt

0 < t ⩽ T.

Hence, (5.6) holds. From (5.5), we know that the solution of the problem (5.1)–(5.3) is conservative in the H 1 -norm. In this chapter, we denote c0 = c2 =

max

󵄨󵄨 󵄨 󵄨u(x, t)󵄨󵄨󵄨,

max

󵄨󵄨 󵄨 󵄨uxx (x, t)󵄨󵄨󵄨,

c1 =

0⩽x⩽L,0⩽t⩽T 󵄨

c3 =

0⩽x⩽L,0⩽t⩽T 󵄨

󵄨󵄨 󵄨 󵄨ux (x, t)󵄨󵄨󵄨,

max

0⩽x⩽L,0⩽t⩽T 󵄨

󵄨󵄨 󵄨 󵄨ut (x, t)󵄨󵄨󵄨.

max

0⩽x⩽L,0⩽t⩽T 󵄨

5.2 Two-level nonlinear difference scheme 5.2.1 Derivation of the difference scheme Considering equation (5.4) at the point (xi , tk+ 1 ), we have 2

ut (xi , tk+ 1 ) − uxxt (xi , tk+ 1 ) + 3u(xi , tk+ 1 )ux (xi , tk+ 1 ) 2

2

2

2

= ux (xi , tk+ 1 )uxx (xi , tk+ 1 ) + (uuxx )x (xi , tk+ 1 ). 2

2

2

With the application of the numerical differential formula, we have k+ 21

δt Ui

k+ 21

− δt δx2 Ui

k+ 21

= (Δx Ui

k+ 21

)δx2 Ui

1

1

+ 3ψ(U k+ 2 , U k+ 2 )i k+ 21 2 k+ 21 δx Ui )

+ Δx (Ui

1 ⩽ i ⩽ m − 1, 0 ⩽ k ⩽ n − 1,

k+ 21

+ Ri

, (5.7)

where there is a constant c4 such that 󵄨󵄨 k+ 21 󵄨󵄨 2 2 󵄨󵄨Ri 󵄨󵄨 ⩽ c4 (τ + h ),

1 ⩽ i ⩽ m − 1, 0 ⩽ k ⩽ n − 1.

(5.8)

Noticing the initial-boundary conditions (5.2)–(5.3), we have {

Ui0 = φ(xi ), U0k

= 0,

Umk

1 ⩽ i ⩽ m − 1, = 0,

0 ⩽ k ⩽ n.

(5.9) (5.10)

5.2 Two-level nonlinear difference scheme k+ 21

in (5.7), a difference scheme for solving (5.1)–(5.3) reads

Omitting the small term Ri k+ 1

� 153

k+ 1

1

k+ 21

1

{ δt ui 2 − δt δx2 ui 2 + 3ψ(uk+ 2 , uk+ 2 )i = (Δx ui { { { { { 1 ⩽ i ⩽ m − 1, 0 ⩽ k ⩽ n − 1, { { { u0 = φ(xi ), 1 ⩽ i ⩽ m − 1, { { { i k k { u0 = 0, um = 0, 0 ⩽ k ⩽ n.

k+ 21 2 k+ 21 δx ui ),

k+ 21

)δx2 ui

+ Δx (ui

(5.11) (5.12) (5.13)

The right-hand side of the difference equation (5.11) can be rewritten as k+ 1

k+ 21

− δx2 ui−1 2

2 1 k+ 1 δx ui + [ui−1 2 2

k+ 1 k+ 1 2(Δx ui 2 )δx2 ui 2

h

+

k+ 1 ui+1 2

k+ 21

k+ 1

δx2 ui+1 2 − δx2 ui h

].

5.2.2 Conservation of the difference solution Theorem 5.2. Let {uik | 0 ⩽ i ⩽ m, 0 ⩽ k ⩽ n} be the solution of the difference scheme (5.11)–(5.13). Denote 󵄩 󵄩2 󵄨 󵄨2 E k = 󵄩󵄩󵄩uk 󵄩󵄩󵄩 + 󵄨󵄨󵄨uk 󵄨󵄨󵄨1 ,

0 ⩽ k ⩽ n,

then Ek = E0 ,

0 ⩽ k ⩽ n. 1

Proof. Taking the inner product on both the right- and left-hand sides of (5.11) with uk+ 2 , we have 1

1

1

1

1

1

1

(δt uk+ 2 , uk+ 2 ) − (δt δx2 uk+ 2 , uk+ 2 ) + 3(ψ(uk+ 2 , uk+ 2 ), uk+ 2 ) 1

1

1

1

1

1

= ((Δx uk+ 2 )δx2 uk+ 2 , uk+ 2 ) + (Δx (uk+ 2 δx2 uk+ 2 ), uk+ 2 ). Now we analyze each term in the above equality: 1

1

(δt uk+ 2 , uk+ 2 ) = 1

1

1 󵄩󵄩 k+1 󵄩󵄩2 󵄩󵄩 k 󵄩󵄩2 (󵄩u 󵄩󵄩 − 󵄩󵄩u 󵄩󵄩 ); 2τ 󵄩 1

1

−(δt δx2 uk+ 2 , uk+ 2 ) = (δt δx uk+ 2 , δx uk+ 2 ) = 1

1

1

(ψ(uk+ 2 , uk+ 2 ), uk+ 2 ) = 0; 1

1

1

1

1 󵄨󵄨 k+1 󵄨󵄨2 󵄨󵄨 k 󵄨󵄨2 (󵄨u 󵄨󵄨1 − 󵄨󵄨u 󵄨󵄨1 ); 2τ 󵄨

1

1

((Δx uk+ 2 )δx2 uk+ 2 , uk+ 2 ) + (Δx (uk+ 2 δx2 uk+ 2 ), uk+ 2 ) 1

1

1

1

1

1

= (Δx uk+ 2 , uk+ 2 δx2 uk+ 2 ) + (uk+ 2 , Δx (uk+ 2 δx2 uk+ 2 )) = 0.

(5.14)

154 � 5 Difference methods for the Camassa–Holm equation Substituting the above four equalities into (5.14), we have 1 k+1 (E − E k ) = 0, 2τ

0 ⩽ k ⩽ n − 1.

Hence, Ek = E0 ,

0 ⩽ k ⩽ n.

5.2.3 Existence and uniqueness of the difference solution Theorem 5.3. The difference scheme (5.11)–(5.13) has a solution. Proof. Suppose the value of uk at the k-th time-level has been determined, then combining (5.11) with (5.13), we have 1 1 2 k+ 21 2 k+ 1 { (ui − uik ) − (δx2 ui 2 − δx2 uik ) + 3ψ(uk+ 2 , uk+ 2 )i { { { τ τ { { 1 k+ 21 k+ 1 k+ 1 2 k+ 2 { = (Δx ui )δx ui + Δx (ui 2 δx2 ui 2 ), 1 ⩽ i ⩽ m − 1, { { { { { k+ 21 k+ 21 { u0 = 0, um = 0.

k+ 21

Once ui

has been determined, it is easy to get uik+1 from k+ 21

uik+1 = 2ui

− uik .

Denote k+ 21

wi = ui

,

0 ⩽ i ⩽ m,

then 2 2 { (wi − uik ) − (δx2 wi − δx2 uik ) + 3ψ(w, w)i { { τ { τ 2 2 { − [(Δ w )(δ 1 ⩽ i ⩽ m − 1, { x i x wi ) + Δx (wi δx wi )] = 0, { { { w0 = 0, wm = 0. Define the operator Π : 𝒰h̊ → 𝒰h̊ by 2 2 { (wi − uik ) − (δx2 wi − δx2 uik ) + 3ψ(w, w)i { { τ {τ Π(w)i = { − [(Δ w )(δ2 w ) + Δ (w δ2 w )], 1 ⩽ i ⩽ m − 1, x i x i x i x i { { { 0, i = 0, m. {

(5.15) (5.16)

5.2 Two-level nonlinear difference scheme

� 155

Taking the inner product of Π(w) with w, we have 2 2 (Π(w), w) = [(w, w) − (uk , w)] − [(δx2 w, w) − (δx2 uk , w)] + 3(ψ(w, w), w) τ τ − ((Δx w)(δx2 w) + Δx (wδx2 w), w) 2 2 = [‖w‖2 − (uk , w)] + [(δx w, δx w) − (δx uk , δx w)] τ τ 2 2 1 󵄩 󵄩2 1󵄩 󵄩2 ⩾ [‖w‖2 − (󵄩󵄩󵄩uk 󵄩󵄩󵄩 + ‖w‖2 )] + [‖δx w‖2 − (‖δx w‖2 + 󵄩󵄩󵄩δx uk 󵄩󵄩󵄩 )] τ 2 τ 4 1 󵄩 󵄩2 1 󵄨 󵄨2 = [‖w‖2 − (󵄩󵄩󵄩uk 󵄩󵄩󵄩 + 󵄨󵄨󵄨uk 󵄨󵄨󵄨1 )]. τ 2 When ‖w‖2 = ‖uk ‖2 + 21 |uk |21 , it follows that (Π(w), w) ⩾ 0. By the Browder theorem (Theorem 2.4), there is a solution for (5.15)–(5.16). Theorem 5.4. There is a constant c5 such that the solution of the difference scheme (5.11)–(5.13) is unique when hτ < c2 . 5

Proof. According to the proof of Theorem 5.3, it is sufficient to prove that the solution of (5.15)–(5.16) is unique. Suppose (5.15)–(5.16) has another solution {vi | 0 ⩽ i ⩽ m} satisfying 2 2 { (v − uik ) − (δx2 vi − δx2 uik ) + 3ψ(v, v)i = (Δx vi )δx2 vi + Δx (vi δx2 vi ), { { { τ i τ { 1 ⩽ i ⩽ m − 1, { { { { v0 = 0, vm = 0.

(5.17) (5.18)

Let zi = wi − vi ,

0 ⩽ i ⩽ m.

Subtracting (5.17)–(5.18) from (5.15)–(5.16), respectively, we have 2 2 { zi − δx2 zi + 3[ψ(w, w)i − ψ(v, v)i ] = [(Δx wi )δx2 wi + Δx (wi δx2 wi )] { { τ { τ { − [(Δx vi )δx2 vi + Δx (vi δx2 vi )], 1 ⩽ i ⩽ m − 1, { { { { z0 = 0, zm = 0.

(5.19) (5.20)

Taking the inner product on both the right- and left-hand sides of (5.19) with z, we have 2 2 (z, z) − (δx2 z, z) + 3(ψ(w, w) − ψ(v, v), z) τ τ = ((Δx w)δx2 w + Δx (wδx2 w) − (Δx v)δx2 v − Δx (vδx2 v), z). Noticing

(5.21)

156 � 5 Difference methods for the Camassa–Holm equation ψ(w, w)i − ψ(v, v)i

= ψ(w, w)i − ψ(w − z, w − z)i

= ψ(w, z)i + ψ(z, w)i − ψ(z, z)i , we have (ψ(w, w) − ψ(v, v), z) = (ψ(w, z), z) + (ψ(z, w), z) − (ψ(z, z), z) = (ψ(z, w), z).

(5.22)

Noticing [(Δx wi )(δx2 wi ) + Δx (wi δx2 wi )] − [(Δx vi )δx2 vi + Δx (vi δx2 vi )]

= [(Δx wi )(δx2 wi ) + Δx (wi δx2 wi )] − [(Δx (wi − zi ))δx2 (wi − zi ) + Δx ((wi − zi )δx2 (wi − zi ))]

= (Δx zi )δx2 wi + (Δx wi )δx2 zi − (Δx zi )δx2 zi + Δx (zi δx2 wi + wi δx2 zi − zi δx2 zi )

= [(Δx zi )δx2 wi + Δx (zi δx2 wi )] + [(Δx wi )δx2 zi + Δx (wi δx2 zi )] − [(Δx zi )δx2 zi + Δx (zi δx2 zi )], we have ((Δx w)δx2 w + Δx (wδx2 w) − (Δx v)δx2 v − Δx (vδx2 v), z)

= ((Δx z)δx2 w + Δx (zδx2 w), z) + ((Δx w)δx2 z + Δx (wδx2 z), z) − ((Δx z)δx2 z + Δx (zδx2 z), z) = ((Δx w)δx2 z, z) + (Δx (wδx2 z), z) = ((Δx w)δx2 z, z) − (wδx2 z, Δx z).

(5.23)

Substituting (5.22) and (5.23) into (5.21), we have 2 (‖z‖2 + |z|21 ) = −3(ψ(z, w), z) + ((Δx w)δx2 z, z) − (wδx2 z, Δx z). τ

(5.24)

Now we estimate each term on the right-hand side of (5.24): 1 (−ψ(z, w), z) ⩽ (‖z‖∞ ⋅ ‖z‖ ⋅ |w|1 + ‖z‖∞ |z|1 ⋅ ‖w‖); 3 󵄩 󵄩 2 ((Δx w)δx2 z, z) ⩽ ‖z‖∞ |w|1 ⋅ 󵄩󵄩󵄩δx2 z󵄩󵄩󵄩 ⩽ ‖z‖∞ |w|1 |z|1 ; h 2 󵄩 2 󵄩󵄩 2 󵄩 −(wδx z, Δx z) ⩽ ‖w‖∞ 󵄩󵄩δx z󵄩󵄩 ⋅ |z|1 ⩽ ‖w‖∞ |z|21 . h Substituting the above three inequalities into (5.24) and using Theorem 5.2, there is a constant c5 satisfying 2 2 c5 2 |z| ⩽ |z| . τ 1 h 1 It follows that |z|21 = 0 when

τ h


0. In combination of (6.4) and Lemma 1.1(c), we have L

󵄩2 󵄩2 󵄩 󵄩 󵄨4 󵄨 ∫󵄨󵄨󵄨u(x, t)󵄨󵄨󵄨 dx ⩽ 󵄩󵄩󵄩u(⋅, t)󵄩󵄩󵄩∞ 󵄩󵄩󵄩u(⋅, t)󵄩󵄩󵄩 0

󵄨2 1 󵄩 󵄨 󵄩2 󵄩2 󵄩 ⩽ (ε󵄨󵄨󵄨u(⋅, t)󵄨󵄨󵄨1 + 󵄩󵄩󵄩u(⋅, t)󵄩󵄩󵄩 )󵄩󵄩󵄩u(⋅, t)󵄩󵄩󵄩 4ε 󵄨2 1 󵄨 = [ε󵄨󵄨󵄨u(⋅, t)󵄨󵄨󵄨1 + Q(0)]Q(0). 4ε

According to (6.5), we have L

q 󵄨 󵄨4 󵄨2 1 󵄨󵄨 󵄨2 q 󵄨 󵄨󵄨u(⋅, t)󵄨󵄨󵄨1 = ∫󵄨󵄨󵄨u(x, t)󵄨󵄨󵄨 dx + E(0) ⩽ [ε󵄨󵄨󵄨u(x, t)󵄨󵄨󵄨1 + Q(0)]Q(0) + E(0). 2 2 4ε 0

When Q(0) = 0, it follows that E(0) = 0 and hence, 󵄨󵄨 󵄨2 󵄨󵄨u(⋅, t)󵄨󵄨󵄨1 = 0. When Q(0) ≠ 0, taking ε =

1 , qQ(0)

(6.9)

one gives that

q 2 1 2 3 󵄨󵄨 󵄨2 󵄨󵄨u(⋅, t)󵄨󵄨󵄨1 ⩽ Q (0) + 2E(0) = q Q (0) + 2E(0). 4ε 4 Combining Lemma 1.1(b) with (6.8)–(6.10), it follows that L󵄨 󵄩󵄩 󵄩2 󵄨2 L 1 2 3 󵄩󵄩u(⋅, t)󵄩󵄩󵄩∞ ⩽ 󵄨󵄨󵄨u(⋅, t)󵄨󵄨󵄨1 ⩽ ( q Q (0) + 2E(0)). 4 4 4 Throughout this chapter, we denote 󵄩 󵄩 c0 = max 󵄩󵄩󵄩u(⋅, t)󵄩󵄩󵄩∞ . 0⩽t⩽T

6.2 Two-level nonlinear difference scheme 6.2.1 Derivation of the difference scheme Considering equation (6.1) at the point (xj , tk+ 1 ), we have 2

󵄨 󵄨2 iut (xj , tk+ 1 ) + uxx (xj , tk+ 1 ) + q󵄨󵄨󵄨u(xj , tk+ 1 )󵄨󵄨󵄨 u(xj , tk+ 1 ) = 0, 2

2

1 ⩽ j ⩽ m − 1, 0 ⩽ k ⩽ n − 1.

2

2

Using the numerical differential formula and the Taylor expansion, we arrive at

(6.10)

6.2 Two-level nonlinear difference scheme

� 177

q 󵄨 󵄨2 󵄨 󵄨2 k+ 1 + (󵄨󵄨󵄨Ujk 󵄨󵄨󵄨 + 󵄨󵄨󵄨Ujk+1 󵄨󵄨󵄨 )Uj 2 = Rkj , 2 1 ⩽ j ⩽ m − 1, 0 ⩽ k ⩽ n − 1, k+ 21

iδt Uj

k+ 21

+ δx2 Uj

(6.11)

where there is a positive constant c1 such that 󵄨󵄨 k 󵄨󵄨 2 2 󵄨󵄨Rj 󵄨󵄨 ⩽ c1 (τ + h ), 󵄨󵄨 k+ 21 󵄨󵄨 2 2 󵄨󵄨δt Rj 󵄨󵄨 ⩽ c1 (τ + h ),

1 ⩽ j ⩽ m − 1, 0 ⩽ k ⩽ n − 1,

(6.12)

1 ⩽ j ⩽ m − 1, 0 ⩽ k ⩽ n − 2.

(6.13)

The Taylor expansion with the integral remainder is necessary to obtain (6.13). Noticing the initial-boundary value conditions (6.2)–(6.3), we have {

Uj0 = φ(xj ), U0k

Umk

= 0,

= 0,

1 ⩽ j ⩽ m − 1,

(6.14)

0 ⩽ k ⩽ n.

(6.15)

Omitting the small term in (6.11) and replacing Ujk by ujk , a difference scheme reads q 󵄨 󵄨2 󵄨 k+ 1 k+ 1 󵄨2 k+ 1 { iδt uj 2 + δx2 uj 2 + (󵄨󵄨󵄨ujk 󵄨󵄨󵄨 + 󵄨󵄨󵄨ujk+1 󵄨󵄨󵄨 )uj 2 = 0, { { 2 { { { { 1 ⩽ j ⩽ m − 1, 0 ⩽ k ⩽ n − 1, { { { uj0 = φ(xj ), 1 ⩽ j ⩽ m − 1, { { { { k k { u0 = 0, um = 0, 0 ⩽ k ⩽ n.

(6.16) (6.17) (6.18)

6.2.2 Conservation and boundedness of the difference solution Denote m−1

m−1 m−1 󵄨 k 󵄨󵄨2 q 󵄨 󵄨4 E k = h ∑ 󵄨󵄨󵄨δx uj+ h ∑ 󵄨󵄨󵄨ujk 󵄨󵄨󵄨 . 1 󵄨󵄨 − 2 2 j=0 j=1

󵄨 󵄨2 Qk = h ∑ 󵄨󵄨󵄨ujk 󵄨󵄨󵄨 , j=1

Theorem 6.2. Suppose {ujk | 0 ⩽ j ⩽ m, 0 ⩽ k ⩽ n} is the solution of the difference scheme (6.16)–(6.18), then we have Qk = Q0 ,

1 ⩽ k ⩽ n,

(6.19)

E =E ,

1 ⩽ k ⩽ n.

(6.20)

k

0

k+ 21

Proof. (I) Multiplying (6.16) on both the right- and left-hand sides by hū j over j from 1 to m − 1, we have m−1

k+ 21

ih ∑ (δt uj j=1

k+ 21

)ū j

0 ⩽ k ⩽ n − 1.

m−1

k+ 21

+ h ∑ (δx2 uj j=1

k+ 21

)ū j

and summing

q m−1 󵄨 󵄨2 󵄨 󵄨2 󵄨 k+ 1 󵄨2 + h ∑ (󵄨󵄨󵄨ujk 󵄨󵄨󵄨 + 󵄨󵄨󵄨ujk+1 󵄨󵄨󵄨 )󵄨󵄨󵄨uj 2 󵄨󵄨󵄨 = 0, 2 j=1 (6.21)

178 � 6 Difference methods for the Schrödinger equation k+ 21

k+ 21

= 0 and um

By using u0

= 0, we have

m−1

k+ 21

h ∑ (δx2 uj j=1

m−1

k+ 21

1

󵄨 k+ 󵄨2 = −h ∑ 󵄨󵄨󵄨δx u 12 󵄨󵄨󵄨 . j+

)ū j

2

j=0

Noticing m−1

k+ 21

Re{h ∑ (δt uj j=1

k+ 21

)ū j

}=

1 󵄩󵄩 k+1 󵄩󵄩2 󵄩󵄩 k 󵄩󵄩2 (󵄩u 󵄩󵄩 − 󵄩󵄩u 󵄩󵄩 ), 2τ 󵄩

and taking the imaginary parts on both the right- and left-hand sides of (6.21), we have 1 󵄩󵄩 k+1 󵄩󵄩2 󵄩󵄩 k 󵄩󵄩2 (󵄩u 󵄩󵄩 − 󵄩󵄩u 󵄩󵄩 ) = 0, 2τ 󵄩

0 ⩽ k ⩽ n − 1,

which implies 󵄩󵄩 k+1 󵄩󵄩2 󵄩󵄩 k 󵄩󵄩2 󵄩󵄩u 󵄩󵄩 = 󵄩󵄩u 󵄩󵄩 ,

0 ⩽ k ⩽ n − 1.

Therefore, (6.19) is valid.

k+ 21

(II) Multiplying (6.16) on both the right- and left-hand sides by −hδt ū j ming over j from 1 to m − 1, we have m−1

m−1

1

1

and sum-

1

k+ k+ 󵄨 k+ 󵄨2 −ih ∑ 󵄨󵄨󵄨δt uj 2 󵄨󵄨󵄨 − h ∑ (δx2 uj 2 )δt ū j 2 j=1

j=1

m−1

q k+ 1 󵄨 󵄨2 󵄨 󵄨2 k+ 1 − h ∑ (󵄨󵄨󵄨ujk 󵄨󵄨󵄨 + 󵄨󵄨󵄨ujk+1 󵄨󵄨󵄨 )uj 2 δt ū j 2 = 0. 2 j=1 k+ 21

Noticing δt u0

k+ 21

= 0, δt um m−1

= 0, it follows: k+ 21

−h ∑ (δx2 uj j=1

(6.22)

k+ 21

)δt ū j

m−1

= h ∑ (δx u j=0

k+ 21

j+ 21

)(δx δt ū

k+ 21

j+ 21

).

Taking the real parts on both the right- and left-hand sides of (6.22), we have m−1 m−1 q 1 1 󵄨󵄨 k+1 󵄨󵄨2 󵄨󵄨 k 󵄨󵄨2 󵄨 󵄨4 󵄨 󵄨4 (󵄨󵄨u 󵄨󵄨1 − 󵄨󵄨u 󵄨󵄨1 ) − ⋅ (h ∑ 󵄨󵄨󵄨ujk+1 󵄨󵄨󵄨 − h ∑ 󵄨󵄨󵄨ujk 󵄨󵄨󵄨 ) = 0, 2τ 2 2τ j=1 j=1

0 ⩽ k ⩽ n − 1,

which implies m−1 m−1 󵄨󵄨 k+1 󵄨󵄨2 q 󵄨 k+1 󵄨4 󵄨 k 󵄨2 q 󵄨 k 󵄨4 󵄨󵄨u 󵄨󵄨1 − h ∑ 󵄨󵄨󵄨uj 󵄨󵄨󵄨 = 󵄨󵄨󵄨u 󵄨󵄨󵄨1 − h ∑ 󵄨󵄨󵄨uj 󵄨󵄨󵄨 , 2 j=1 2 j=1

0 ⩽ k ⩽ n − 1.

6.2 Two-level nonlinear difference scheme

� 179

Hence, (6.20) is valid. For arbitrary grid function u = {uj | 0 ⩽ j ⩽ m} in 𝒰h , we define the following norm: m−1 p 1 1 ‖u‖p = √h( |u0 |p + ∑ |uj |p + |um |p ). 2 2 j=1

Theorem 6.3. Suppose {ujk | 0 ⩽ j ⩽ m, 0 ⩽ k ⩽ n} is the solution of the difference scheme (6.16)–(6.18). Then we have L 1 2 󵄩 0 󵄩6 󵄨 0 󵄨2 󵄩󵄩 k 󵄩󵄩2 󵄩󵄩u 󵄩󵄩∞ ⩽ ( q 󵄩󵄩󵄩u 󵄩󵄩󵄩 + 󵄨󵄨󵄨u 󵄨󵄨󵄨1 − 2 8

q 󵄩󵄩 0 󵄩󵄩4 󵄩u 󵄩 ), 2 󵄩 󵄩4

1 ⩽ k ⩽ n.

(6.23)

Proof. By Theorem 6.2, we have 󵄩󵄩 k 󵄩󵄩2 󵄩󵄩 0 󵄩󵄩2 󵄩󵄩u 󵄩󵄩 = 󵄩󵄩u 󵄩󵄩 , 1 ⩽ k ⩽ n, q 2 󵄨󵄨 k 󵄨󵄨 󵄩 k 󵄩4 󵄨 0 󵄨2 q 󵄩 0 󵄩4 󵄨󵄨u 󵄨󵄨1 − 󵄩󵄩󵄩u 󵄩󵄩󵄩4 = 󵄨󵄨󵄨u 󵄨󵄨󵄨1 − 󵄩󵄩󵄩u 󵄩󵄩󵄩4 , 1 ⩽ k ⩽ n. 2 2

(6.24) (6.25)

󵄨 󵄨2 󵄩 󵄩4 When q ⩽ 0, it follows that |uk |21 ⩽ |uk |21 − q2 ‖uk ‖44 = 󵄨󵄨󵄨u0 󵄨󵄨󵄨1 − q2 󵄩󵄩󵄩u0 󵄩󵄩󵄩4 in the light of (6.25). Therefore, L 󵄨 k 󵄨2 L 󵄨 0 󵄨2 󵄩󵄩 k 󵄩󵄩2 󵄩󵄩u 󵄩󵄩∞ ⩽ 󵄨󵄨󵄨u 󵄨󵄨󵄨1 ⩽ (󵄨󵄨󵄨u 󵄨󵄨󵄨1 − 4 4

q 󵄩󵄩 0 󵄩󵄩4 󵄩u 󵄩 ). 2 󵄩 󵄩4

Hence, (6.23) is valid. When q > 0, it follows from (6.25) that 󵄨󵄨 k 󵄨󵄨2 󵄨󵄨 0 󵄨󵄨2 󵄨󵄨u 󵄨󵄨1 = 󵄨󵄨u 󵄨󵄨1 −

q 󵄩󵄩 0 󵄩󵄩4 󵄩u 󵄩 + 2 󵄩 󵄩4

q 󵄩󵄩 k 󵄩󵄩4 󵄩u 󵄩 , 2 󵄩 󵄩4

1 ⩽ k ⩽ n.

(6.26)

It is easy to know that m−1

󵄩󵄩 k 󵄩󵄩4 󵄨 k 󵄨4 󵄩 k 󵄩2 󵄩 k 󵄩2 󵄩󵄩u 󵄩󵄩4 = h ∑ 󵄨󵄨󵄨uj 󵄨󵄨󵄨 ⩽ 󵄩󵄩󵄩u 󵄩󵄩󵄩∞ 󵄩󵄩󵄩u 󵄩󵄩󵄩 . j=1

Thus, 󵄩󵄩 k 󵄩󵄩4 󵄨 k 󵄨2 1 󵄩 k 󵄩2 󵄩 k 󵄩2 󵄩󵄩u 󵄩󵄩4 ⩽ (ε󵄨󵄨󵄨u 󵄨󵄨󵄨1 + 󵄩󵄩󵄩u 󵄩󵄩󵄩 )󵄩󵄩󵄩u 󵄩󵄩󵄩 . 4ε Using (6.27) on the right-hand side of (6.26), we have 󵄨󵄨 k 󵄨󵄨2 󵄨󵄨 0 󵄨󵄨2 󵄨󵄨u 󵄨󵄨1 ⩽ 󵄨󵄨u 󵄨󵄨1 −

q 󵄩󵄩 0 󵄩󵄩4 q 󵄨󵄨 k 󵄨󵄨2 1 󵄩󵄩 k 󵄩󵄩2 󵄩󵄩 k 󵄩󵄩2 󵄩u 󵄩 + (ε󵄨u 󵄨 + 󵄩u 󵄩 )󵄩u 󵄩 2 󵄩 󵄩4 2 󵄨 󵄨1 4ε 󵄩 󵄩 󵄩 󵄩

(6.27)

180 � 6 Difference methods for the Schrödinger equation 󵄨 󵄨2 q 󵄩 󵄩4 q 󵄨 󵄨2 1 󵄩 󵄩2 󵄩 󵄩2 = 󵄨󵄨󵄨u0 󵄨󵄨󵄨1 − 󵄩󵄩󵄩u0 󵄩󵄩󵄩4 + (ε󵄨󵄨󵄨uk 󵄨󵄨󵄨1 + 󵄩󵄩󵄩u0 󵄩󵄩󵄩 )󵄩󵄩󵄩u0 󵄩󵄩󵄩 . 2 2 4ε When ‖u0 ‖2 = 0, it follows that 󵄨󵄨 k 󵄨󵄨2 󵄨󵄨u 󵄨󵄨1 = 0. When ‖u0 ‖2 ≠ 0, it follows by taking ε = 1/(q ⋅ ‖u0 ‖2 ) that 1 2 󵄩 0 󵄩6 󵄨 0 󵄨2 󵄨󵄨 k 󵄨󵄨2 󵄨󵄨u 󵄨󵄨1 ⩽ 2( q 󵄩󵄩󵄩u 󵄩󵄩󵄩 + 󵄨󵄨󵄨u 󵄨󵄨󵄨1 − 8

q 󵄩󵄩 0 󵄩󵄩4 󵄩u 󵄩 ). 2 󵄩 󵄩4

L 󵄨 k 󵄨2 L 1 2 󵄩 0 󵄩6 󵄨 0 󵄨2 󵄩󵄩 k 󵄩󵄩2 󵄩󵄩u 󵄩󵄩∞ ⩽ 󵄨󵄨󵄨u 󵄨󵄨󵄨1 ⩽ ( q 󵄩󵄩󵄩u 󵄩󵄩󵄩 + 󵄨󵄨󵄨u 󵄨󵄨󵄨1 − 4 2 8

q 󵄩󵄩 0 󵄩󵄩4 󵄩u 󵄩 ), 2 󵄩 󵄩4

Therefore, 1 ⩽ k ⩽ n.

Denote the right-hand side of (6.23) by c22 , then we have 󵄩󵄩 k 󵄩󵄩 󵄩󵄩u 󵄩󵄩∞ ⩽ c2 ,

1 ⩽ k ⩽ n.

6.2.3 Existence and uniqueness of the difference solution Theorem 6.4. (I) There is a solution to the difference scheme (6.16)–(6.18). (II) When τ < 1/(3c22 |q|), the solution of the difference scheme (6.16)–(6.18) is unique. Proof. (I) Suppose {ujk | 0 ⩽ j ⩽ m} has been obtained. Rewrite (6.16) and (6.18) as q 󵄨 󵄨2 2 k+ 1 k+ 1 k+ 1 k+ 1 i (uj 2 − ujk ) + δx2 uj 2 + (󵄨󵄨󵄨ujk 󵄨󵄨󵄨 + |2uj 2 − ujk |2 )uj 2 = 0, { { { 2 { τ 1 ⩽ j ⩽ m − 1, { { { { k+ 1 k+ 21 2 { u0 = 0, um = 0. k+ 21

Once {uj

| 0 ⩽ j ⩽ m} is obtained, then we have k+ 21

ujk+1 = 2uj

− ujk ,

0 ⩽ j ⩽ m.

Let k+ 21

wj = uj

,

0 ⩽ j ⩽ m.

(6.28) (6.29)

6.2 Two-level nonlinear difference scheme

� 181

Then (6.28)–(6.29) can be rewritten as { { {

q 󵄨 󵄨2 󵄨 2 󵄨2 i (wj − ujk ) + δx2 wj + (󵄨󵄨󵄨ujk 󵄨󵄨󵄨 + 󵄨󵄨󵄨2wj − ujk 󵄨󵄨󵄨 )wj = 0, τ 2 w0 = 0, wm = 0.

1 ⩽ j ⩽ m − 1,

(6.30) (6.31)

Rewrite (6.30) as q 󵄨 󵄨2 󵄨 2 󵄨2 (wj − ujk ) − iδx2 wj − i (󵄨󵄨󵄨ujk 󵄨󵄨󵄨 + 󵄨󵄨󵄨2wj − ujk 󵄨󵄨󵄨 )wj = 0, τ 2

1 ⩽ j ⩽ m − 1.

Define the operator Π : 𝒰h̊ → 𝒰h̊ by q 2 2 2 { (wj − ujk ) − iδx2 wj − i (󵄨󵄨󵄨󵄨ujk 󵄨󵄨󵄨󵄨 + 󵄨󵄨󵄨󵄨2wj − ujk 󵄨󵄨󵄨󵄨 )wj , 1 ⩽ j ⩽ m − 1, 2 Π(w)j = { τ j = 0, m. { 0, Taking the inner product of Π(w) with w, we have (Π(w), w) = =

q m−1 󵄨 󵄨2 󵄨 2 󵄨2 ((w, w) − (uk , w)) − i(δx2 w, w) − i h ∑ (󵄨󵄨󵄨ujk 󵄨󵄨󵄨 + 󵄨󵄨󵄨2wj − ujk 󵄨󵄨󵄨 )|wj |2 τ 2 j=1 q m−1 󵄨 󵄨2 󵄨 2 󵄨2 (‖w‖2 − (uk , w)) + i|w|21 − i h ∑ (󵄨󵄨󵄨ujk 󵄨󵄨󵄨 + 󵄨󵄨󵄨2wj − ujk 󵄨󵄨󵄨 )|wj |2 . τ 2 j=1

Therefore, Re{(Π(w), w)} =

2 2 󵄩 󵄩 (‖w‖2 − Re(uk , w)) ⩾ ‖w‖(‖w‖ − 󵄩󵄩󵄩uk 󵄩󵄩󵄩). τ τ

When ‖w‖ = ‖uk ‖, it follows Re{(Π(w), w)} ⩾ 0. By the Browder theorem (Theorem 2.4), there is a w∗ ∈ 𝒰h̊ satisfying ‖w∗ ‖ ⩽ ‖uk ‖ such that Π(w∗ ) = 0. Therefore, there exists a solution to the difference scheme (6.16)–(6.18). (II) Suppose (6.30)–(6.31) has another solution v = {vj | 0 ⩽ j ⩽ m} ∈ 𝒰h̊ satisfying { { {

q 󵄨 󵄨2 󵄨 2 󵄨2 i (vj − ujk ) + δx2 vj + (󵄨󵄨󵄨ujk 󵄨󵄨󵄨 + 󵄨󵄨󵄨2vj − ujk 󵄨󵄨󵄨 )vj = 0, τ 2 v0 = 0, vm = 0.

1 ⩽ j ⩽ m − 1,

(6.32) (6.33)

Denote θj = wj − vj ,

0 ⩽ j ⩽ m.

Subtracting (6.32)–(6.33) from (6.30)–(6.31), it follows: { { {

q 󵄨 󵄨2 2 󵄨 󵄨2 󵄨 󵄨2 i θj + δx2 θj + (󵄨󵄨󵄨ujk 󵄨󵄨󵄨 θj + 󵄨󵄨󵄨2wj − ujk 󵄨󵄨󵄨 wj − 󵄨󵄨󵄨2vj − ujk 󵄨󵄨󵄨 vj ) = 0, τ 2 θ0 = 0, θm = 0.

1 ⩽ j ⩽ m − 1,

(6.34) (6.35)

182 � 6 Difference methods for the Schrödinger equation Multiplying (6.34) on both the right- and left-hand sides by −ihθ̄j and summing over j from 1 to m − 1, it follows: m−1 q m−1󵄨 󵄨2 2 m−1 2 h ∑ |θj | − ih ∑ (δx2 θj )θ̄j − i h ∑ 󵄨󵄨󵄨ujk 󵄨󵄨󵄨 ⋅ |θj |2 τ j=1 2 j=1 j=1

q m−1 󵄨 󵄨 󵄨2 󵄨2 − i h ∑ (󵄨󵄨󵄨2wj − ujk 󵄨󵄨󵄨 wj − 󵄨󵄨󵄨2vj − ujk 󵄨󵄨󵄨 vj )θ̄j = 0. 2 j=1

(6.36)

Noticing 󵄨󵄨 󵄨 k 󵄨2 k 󵄨2 󵄨󵄨2wj − uj 󵄨󵄨󵄨 wj − 󵄨󵄨󵄨2vj − uj 󵄨󵄨󵄨 vj 󵄨 󵄨2 󵄨 󵄨2 󵄨 󵄨2 = 󵄨󵄨󵄨2wj − ujk 󵄨󵄨󵄨 (wj − vj ) + (󵄨󵄨󵄨2wj − ujk 󵄨󵄨󵄨 − 󵄨󵄨󵄨2vj − ujk 󵄨󵄨󵄨 )vj 󵄨 󵄨2 = 󵄨󵄨󵄨2wj − ujk 󵄨󵄨󵄨 θj + [(2wj − ujk )2wj − ujk − (2vj − ujk )2vj − ujk ]vj 󵄨 󵄨2 = 󵄨󵄨󵄨2wj − ujk 󵄨󵄨󵄨 θj + [2(wj − vj )2wj − ujk + 2(2vj − ujk )wj − vj ]vj 󵄨 󵄨2 = 󵄨󵄨󵄨2wj − ujk 󵄨󵄨󵄨 θj + 2[θj 2wj − ujk + (2vj − ujk )θ̄j ]vj and then taking the real parts on both the right- and left-hand sides of (6.36), we have m−1 2 2 󵄨 󵄨 󵄨 󵄨 ‖θ‖ ⩽ |q|h ∑ (|θj |2 󵄨󵄨󵄨2wj − ujk 󵄨󵄨󵄨 + 󵄨󵄨󵄨2vj − ujk 󵄨󵄨󵄨|θj |2 )|vj | τ j=1

⩽ 6c22 |q| ⋅ ‖θ‖2 . When τ < 1/(3c22 |q|), it follows that ‖θ‖2 = 0, which implies θj = 0, 0 ⩽ j ⩽ m. Therefore, the difference scheme (6.16)–(6.18) is uniquely solvable.

6.2.4 Convergence of the difference solution Theorem 6.5. Suppose {Ujk | 0 ⩽ j ⩽ m, 0 ⩽ k ⩽ n} is the solution of the problem (6.1)–(6.3) and {ujk | 0 ⩽ j ⩽ m, 0 ⩽ k ⩽ n} is the solution of the difference scheme (6.16)–(6.18). Denote ejk = Ujk − ujk ,

0 ⩽ j ⩽ m, 0 ⩽ k ⩽ n.

There is a positive constant c3 such that 󵄩󵄩 k 󵄩󵄩 2 2 󵄩󵄩e 󵄩󵄩 ⩽ c3 (τ + h ),

1 ⩽ k ⩽ n.

6.2 Two-level nonlinear difference scheme

� 183

Proof. Subtracting (6.16)–(6.18) from (6.11) and (6.14)–(6.15), respectively, one obtains the system of error equations as 1 q 󵄨󵄨 k 󵄨󵄨2 󵄨󵄨 k+1 󵄨󵄨2 k+ 21 k+ 21 󵄨󵄨 k 󵄨󵄨2 󵄨󵄨 k+1 󵄨󵄨2 k+ 21 2 k+ 2 k { )U U + U iδ e + δ e + [( − ( 󵄨 󵄨 󵄨 󵄨 { t j j x 󵄨 󵄨 󵄨 󵄨 󵄨󵄨uj 󵄨󵄨 + 󵄨󵄨uj 󵄨󵄨 )uj ] = Rj , j j j { 2 { { { { 1 ⩽ j ⩽ m − 1, 0 ⩽ k ⩽ n − 1, (6.37) { 0 { { ej = 0, 1 ⩽ j ⩽ m − 1, (6.38) { { { { k k (6.39) { e0 = 0, em = 0, 0 ⩽ k ⩽ n.

Denote 󵄨 󵄨2 󵄨 󵄨2 k+ 1 G(U k , U k+1 )j = (󵄨󵄨󵄨Ujk 󵄨󵄨󵄨 + 󵄨󵄨󵄨Ujk+1 󵄨󵄨󵄨 )Uj 2 . k+ 21

Multiplying (6.37) on both the right- and left-hand sides by hēj from 1 to m − 1, we have m−1

k+ 21

ih ∑ (δt ej j=1

k+ 21

)ēj

m−1

k+ 21

+ h ∑ (δx2 ej j=1

and summing over j

k+ 21

)ēj

m−1

q k+ 1 + h ∑ [G(U k , U k+1 )j − G(uk , uk+1 )j ]ēj 2 2 j=1 m−1

k+ 21

= h ∑ Rkj ēj j=1

,

0 ⩽ k ⩽ n − 1.

(6.40)

Noticing m−1

k+ 21

h ∑ (δx2 ej j=1

k+ 21

)ēj

m−1

󵄨 k+ 1 󵄨2 = −h ∑ 󵄨󵄨󵄨δx e 12 󵄨󵄨󵄨 , j+ j=0

2

G(U k , U k+1 )j − G(uk , uk+1 )j k+ 1 k+ 1 󵄨 󵄨2 󵄨 󵄨 󵄨2 󵄨 󵄨2 󵄨2 󵄨 󵄨2 󵄨 󵄨2 k+ 1 = (󵄨󵄨󵄨Ujk 󵄨󵄨󵄨 + 󵄨󵄨󵄨Ujk+1 󵄨󵄨󵄨 )(Uj 2 − uj 2 ) + (󵄨󵄨󵄨Ujk 󵄨󵄨󵄨 + 󵄨󵄨󵄨Ujk+1 󵄨󵄨󵄨 − 󵄨󵄨󵄨ujk 󵄨󵄨󵄨 − 󵄨󵄨󵄨ujk+1 󵄨󵄨󵄨 )uj 2 k+ 1 󵄨 󵄨2 󵄨 󵄨2 k+ 1 = (󵄨󵄨󵄨Ujk 󵄨󵄨󵄨 + 󵄨󵄨󵄨Ujk+1 󵄨󵄨󵄨 )ej 2 + (ejk Ū jk + ujk ējk + ejk+1 Ū jk+1 + ujk+1 ējk+1 )uj 2

(6.41)

and then taking the imaginary parts on both the right- and left-hand sides of (6.40), we obtain 1 󵄩󵄩 k+1 󵄩󵄩2 󵄩󵄩 k 󵄩󵄩2 (󵄩e 󵄩󵄩 − 󵄩󵄩e 󵄩󵄩 ) 2τ 󵄩

m−1 q m−1 k+ 1 k+ 1 k+ 1 = Im{− h ∑ (ejk Ū jk + ujk ējk + ejk+1 Ū jk+1 + ujk+1 ējk+1 )uj 2 ēj 2 + h ∑ Rkj ēj 2 } 2 j=1 j=1

184 � 6 Difference methods for the Schrödinger equation

⩽

m−1 |q| m−1 󵄨󵄨 k 󵄨󵄨 󵄨 󵄨 󵄨 k+ 1 󵄨 󵄨 󵄨 󵄨 󵄨 󵄨 󵄨 k+ 1 h ∑ (c0 󵄨󵄨ej 󵄨󵄨 + c2 󵄨󵄨󵄨ejk 󵄨󵄨󵄨 + c0 󵄨󵄨󵄨ejk+1 󵄨󵄨󵄨 + c2 󵄨󵄨󵄨ejk+1 󵄨󵄨󵄨)c2 󵄨󵄨󵄨ēj 2 󵄨󵄨󵄨 + h ∑ 󵄨󵄨󵄨Rkj 󵄨󵄨󵄨 ⋅ |ēj 2 | 2 j=1 j=1

1 |q|(c0 + c2 )c2 󵄩󵄩 k 󵄩󵄩 󵄩󵄩 k+1 󵄩󵄩 k+ 21 󵄩 󵄩 (󵄩󵄩e 󵄩󵄩 + 󵄩󵄩e 󵄩󵄩)‖e ‖ + 󵄩󵄩󵄩Rk 󵄩󵄩󵄩 ⋅ ‖ek+ 2 ‖ 2 󵄩 󵄩 󵄩 󵄩 |q|(c0 + c2 )c2 󵄩󵄩 k 󵄩󵄩 󵄩󵄩 k+1 󵄩󵄩 ‖ek ‖ + 󵄩󵄩󵄩ek+1 󵄩󵄩󵄩 󵄩󵄩 k 󵄩󵄩 ‖ek ‖ + 󵄩󵄩󵄩ek+1 󵄩󵄩󵄩 (󵄩󵄩e 󵄩󵄩 + 󵄩󵄩e 󵄩󵄩) ⋅ + 󵄩󵄩R 󵄩󵄩 ⋅ , ⩽ 2 2 2 0 ⩽ k ⩽ n − 1.

⩽

When 21 (‖ek+1 ‖ + ‖ek ‖) ≠ 0, dividing the above inequality on both the right- and left-hand sides by 21 (‖ek+1 ‖ + ‖ek ‖), we have |q| 1 󵄩󵄩 k+1 󵄩󵄩 󵄩󵄩 k 󵄩󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 (󵄩󵄩e 󵄩󵄩 − 󵄩󵄩e 󵄩󵄩) ⩽ (c + c2 )c2 (󵄩󵄩󵄩ek 󵄩󵄩󵄩 + 󵄩󵄩󵄩ek+1 󵄩󵄩󵄩) + 󵄩󵄩󵄩Rk 󵄩󵄩󵄩. τ 2 0 When 21 (‖ek+1 ‖ + ‖ek ‖) = 0, the above result also holds. Therefore, we have |q| |q| 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 (c + c2 )c2 τ]󵄩󵄩󵄩ek+1 󵄩󵄩󵄩 ⩽ [1 + (c + c2 )c2 τ]󵄩󵄩󵄩ek 󵄩󵄩󵄩 + τ 󵄩󵄩󵄩Rk 󵄩󵄩󵄩, 2 0 2 0 0 ⩽ k ⩽ n − 1.

[1 −

When

|q| (c0 2

(6.42)

+ c2 )c2 τ ⩽ 31 , using (6.12), we have

3|q| 󵄩󵄩 k+1 󵄩󵄩 󵄩 󵄩 (c + c2 )c2 τ]󵄩󵄩󵄩ek 󵄩󵄩󵄩 + 󵄩󵄩e 󵄩󵄩 ⩽ [1 + 2 0 3|q| 󵄩 󵄩 (c + c2 )c2 τ]󵄩󵄩󵄩ek 󵄩󵄩󵄩 + ⩽ [1 + 2 0

3 󵄩󵄩 k 󵄩󵄩 τ 󵄩R 󵄩 2 󵄩 󵄩 3√ Lc1 τ(τ 2 + h2 ), 2

0 ⩽ k ⩽ n − 1.

When q = 0, it follows from (6.42) that 󵄩󵄩 k 󵄩󵄩 󵄩󵄩 0 󵄩󵄩 √ 2 2 2 2 󵄩󵄩e 󵄩󵄩 ⩽ 󵄩󵄩e 󵄩󵄩 + Lc1 kτ(τ + h ) ⩽ √Lc1 T(τ + h ),

1⩽k⩽n

by recursion of 󵄩󵄩 k+1 󵄩󵄩 󵄩󵄩 k 󵄩󵄩 √ 2 2 󵄩󵄩e 󵄩󵄩 ⩽ 󵄩󵄩e 󵄩󵄩 + Lc1 τ(τ + h ),

0 ⩽ k ⩽ n − 1.

When q ≠ 0, it follows from the Gronwall inequality (Theorem 1.2(a)) that 3|q| 󵄩󵄩 k 󵄩󵄩 (c +c )c T 󵄩󵄩e 󵄩󵄩 ⩽ e 2 0 2 2

√Lc1 (τ 2 + h2 ), |q|(c0 + c2 )c2

1 ⩽ k ⩽ n.

Theorem 6.6. Suppose {Ujk | 0 ⩽ j ⩽ m, 0 ⩽ k ⩽ n} is the solution of the problem

(6.1)–(6.3) and {ujk | 0 ⩽ j ⩽ m, 0 ⩽ k ⩽ n} is the solution of the difference scheme (6.16)–(6.18). Denote ejk = Ujk − ujk ,

0 ⩽ j ⩽ m, 0 ⩽ k ⩽ n.

6.2 Two-level nonlinear difference scheme

� 185

Then there is a constant c4 such that 󵄨󵄨 k 󵄨󵄨 2 2 󵄨󵄨e 󵄨󵄨1 ⩽ c4 (τ + h ),

1 ⩽ k ⩽ n.

(6.43) k+ 21

Proof. Multiplying (6.37) on both the right- and left-hand sides by −hδt ēj ming over j from 1 to m − 1, we have m−1

m−1

1

1

and sum-

1

k+ k+ 󵄨 k+ 󵄨2 − ih ∑ 󵄨󵄨󵄨δt ej 2 󵄨󵄨󵄨 − h ∑ (δx2 ej 2 )δt ēj 2 j=1

j=1

m−1

q k+ 1 − h ∑ [G(U k+1 , U k )j − G(uk+1 , uk )j ]δt ēj 2 2 j=1 m−1

k+ 21

= − h ∑ Rkj δt ēj j=1

0 ⩽ k ⩽ n − 1.

,

Taking the real parts on both the right- and left-hand sides of the above equality, we have 1 󵄨󵄨 k+1 󵄨󵄨2 󵄨󵄨 k 󵄨󵄨2 (󵄨e 󵄨󵄨1 − 󵄨󵄨e 󵄨󵄨1 ) 2τ 󵄨

q m−1 k+ 1 = Re{ h ∑ [G(U k+1 , U k )j − G(uk+1 , uk )j ]δt ēj 2 } 2 j=1 m−1

k+ 21

+ Re{−h ∑ Rkj δt ēj j=1

},

0 ⩽ k ⩽ n − 1.

(6.44)

It follows from (6.37) that k+ 21

δt ej

k+ 21

= iδx2 ej

q + i [G(U k+1 , U k )j − G(uk+1 , uk )j ] − iRkj . 2

Then taking the conjugate on both the right- and left-hand sides of the above equality, we have k+ 21

δt ēj

k+ 21

= −iδx2 ēj

q − i ⋅ G(U k+1 , U k )j − G(uk+1 , uk )j + iR̄ kj . 2

Therefore, q m−1 k+ 1 Re{ h ∑ [G(U k+1 , U k )j − G(uk+1 , uk )j ]δt ēj 2 } 2 j=1 =

m−1 q k+ 1 Re{h ∑ [G(U k+1 , U k )j − G(uk+1 , uk )j ][−iδx2 ēj 2 2 j=1

186 � 6 Difference methods for the Schrödinger equation q − i ⋅ G(U k+1 , U k )j − G(uk+1 , uk )j + iR̄ kj ]} 2 =

m−1 q k+ 1 Re{−ih ∑ [G(U k+1 , U k )j − G(uk+1 , uk )j ](δx2 ēj 2 ) 2 j=1 m−1

+ ih ∑ [G(U k+1 , U k )j − G(uk+1 , uk )j ]R̄ kj } j=1

=

m−1 q k+ 1 Re{ih ∑ [δx (G(U k+1 , U k ) − G(uk+1 , uk ))j+ 1 ](δx ē 12 ) j+ 2 2 2 j=0 m−1

+ ih ∑ [G(U k+1 , U k )j − G(uk+1 , uk )j ]R̄ kj } j=1

|q| 󵄨󵄨 󵄨 󵄨 1󵄨 ⩽ [󵄨G(U k+1 , U k ) − G(uk+1 , uk )󵄨󵄨󵄨1 ⋅ 󵄨󵄨󵄨ek+ 2 󵄨󵄨󵄨1 2 󵄨 󵄩 󵄩 󵄩 󵄩 + 󵄩󵄩󵄩G(U k+1 , U k ) − G(uk+1 , uk )󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩Rk 󵄩󵄩󵄩].

(6.45)

With the help of (6.41), there are two constants c5 and c6 satisfying 󵄩󵄩 󵄩 k 󵄩 󵄩 k+1 󵄩 k+1 k k+1 k 󵄩 󵄩󵄩G(U , U ) − G(u , u )󵄩󵄩󵄩 ⩽ c5 (󵄩󵄩󵄩e 󵄩󵄩󵄩 + 󵄩󵄩󵄩e 󵄩󵄩󵄩), 󵄨󵄨 󵄩 k 󵄩 󵄩 k+1 󵄩 󵄨 k 󵄨 󵄨 k+1 󵄨 k+1 k k+1 k 󵄨 󵄨󵄨G(U , U ) − G(u , u )󵄨󵄨󵄨1 ⩽ c6 (󵄩󵄩󵄩e 󵄩󵄩󵄩 + 󵄩󵄩󵄩e 󵄩󵄩󵄩 + 󵄨󵄨󵄨e 󵄨󵄨󵄨1 + 󵄨󵄨󵄨e 󵄨󵄨󵄨1 ).

(6.46) (6.47)

Substituting (6.46)–(6.47) into (6.45), we have q m−1 k+ 1 Re{ h ∑ [G(U k+1 , U k )j − G(uk+1 , uk )j ]δt ēj 2 } 2 j=1 ⩽

|q| 1 󵄨 󵄨 󵄨 󵄩 󵄩 󵄩 󵄩 󵄨 󵄨 󵄨 󵄨 󵄨 ⋅ [c6 (󵄩󵄩󵄩ek 󵄩󵄩󵄩 + 󵄩󵄩󵄩ek+1 󵄩󵄩󵄩 + 󵄨󵄨󵄨ek 󵄨󵄨󵄨1 + 󵄨󵄨󵄨ek+1 󵄨󵄨󵄨1 ) ⋅ (󵄨󵄨󵄨ek 󵄨󵄨󵄨1 + 󵄨󵄨󵄨ek+1 󵄨󵄨󵄨1 ) 2 2 󵄩 󵄩 󵄩 󵄩󵄩 󵄩 + c5 (󵄩󵄩󵄩ek 󵄩󵄩󵄩 + 󵄩󵄩󵄩ek+1 󵄩󵄩󵄩)󵄩󵄩󵄩Rk 󵄩󵄩󵄩].

(6.48)

Substituting (6.48) into (6.44), replacing k by l and summing over l from 0 to k, we have k |q| 1 󵄨󵄨 k+1 󵄨󵄨2 󵄩 l 󵄩 󵄩 l+1 󵄩 󵄨 l 󵄨 󵄨 l+1 󵄨 󵄨 l 󵄨 󵄨 l+1 󵄨 󵄨󵄨e 󵄨󵄨1 ⩽ ∑ [c6 (󵄩󵄩󵄩e 󵄩󵄩󵄩 + 󵄩󵄩󵄩e 󵄩󵄩󵄩 + 󵄨󵄨󵄨e 󵄨󵄨󵄨1 + 󵄨󵄨󵄨e 󵄨󵄨󵄨1 )(󵄨󵄨󵄨e 󵄨󵄨󵄨1 + 󵄨󵄨󵄨e 󵄨󵄨󵄨1 ) 2τ 4 l=0 m−1 k

l+ 1 󵄩 󵄩 󵄩 󵄩󵄩 󵄩 + 2c5 (󵄩󵄩󵄩el 󵄩󵄩󵄩 + 󵄩󵄩󵄩el+1 󵄩󵄩󵄩)󵄩󵄩󵄩Rl 󵄩󵄩󵄩] + Re{−h ∑ ∑ Rlj δt ēj 2 }, j=1 l=0

0 ⩽ k ⩽ n − 1.

(6.49)

6.2 Two-level nonlinear difference scheme

� 187

Now we analyze the last term in the above inequality. Noticing that k

l+ 21

∑ Rlj δt ēj

l=0 k

= ∑ Rlj

ējl+1 − ējl τ

l=0

k−1 1 k ̄l+1 = (∑ Rlj ējl+1 − ∑ Rl+1 j ej ) τ l=0 l=−1 k−1 1 l l+1 0 0 ̄ ̄ = [Rkj ējk+1 − ∑ (Rl+1 j − Rj )ej − Rj ej ] τ l=0 k−1 Rl+1 − Rl 1 j l+1 j = Rkj ējk+1 − ∑ ēj , τ τ l=0

we have m−1 k

l+ 21

Re{−h ∑ ∑ Rlj δt ēj j=1 l=0

}

m−1 k−1 Rl+1 − Rl 1 j l+1 j = Re{−h ∑ [ Rkj ējk+1 − ∑ ēj ]} τ τ j=1 l=0 k−1

1 1󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 ⩽ 󵄩󵄩󵄩Rk 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩ek+1 󵄩󵄩󵄩 + ∑ 󵄩󵄩󵄩δt Rl+ 2 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩el+1 󵄩󵄩󵄩. τ l=0

(6.50)

Substituting (6.50) into (6.49) and multiplying both the right- and left-hand sides of the resulting inequality by 2τ, we have k |q| 󵄨󵄨 k+1 󵄨󵄨2 󵄩 l 󵄩 󵄩 l+1 󵄩 󵄨 l 󵄨 󵄨 l+1 󵄨 󵄨 l 󵄨 󵄨 l+1 󵄨 󵄨󵄨e 󵄨󵄨1 ⩽ 2τ ∑ [c6 (󵄩󵄩󵄩e 󵄩󵄩󵄩 + 󵄩󵄩󵄩e 󵄩󵄩󵄩 + 󵄨󵄨󵄨e 󵄨󵄨󵄨1 + 󵄨󵄨󵄨e 󵄨󵄨󵄨1 )(󵄨󵄨󵄨e 󵄨󵄨󵄨1 + 󵄨󵄨󵄨e 󵄨󵄨󵄨1 ) 4 l=0 k−1

1 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 + 2c5 (󵄩󵄩󵄩el 󵄩󵄩󵄩 + 󵄩󵄩󵄩el+1 󵄩󵄩󵄩)󵄩󵄩󵄩Rl 󵄩󵄩󵄩] + 2󵄩󵄩󵄩Rk 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩ek+1 󵄩󵄩󵄩 + 2τ ∑ 󵄩󵄩󵄩δt Rl+ 2 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩el+1 󵄩󵄩󵄩.

l=0

With the application of (6.12)–(6.13), Lemma 1.1(b) and Theorem 6.5, we have k |q| L 󵄨󵄨 l 󵄨󵄨 󵄨󵄨 l+1 󵄨󵄨 2 2 󵄨󵄨 k+1 󵄨󵄨2 )(󵄨󵄨e 󵄨󵄨1 + 󵄨󵄨e 󵄨󵄨1 ) + 4√Lc1 c3 c5 (τ 2 + h2 ) ] 󵄨󵄨e 󵄨󵄨1 ⩽ 2τ ∑ [c6 (1 + √ 4 6 l=0

+ 2√Lc1 (τ 2 + h2 ) 0 ⩽ k ⩽ n − 1.

k−1 L 󵄨󵄨 k+1 󵄨󵄨 L 󵄨󵄨 l+1 󵄨󵄨 2 2 󵄨󵄨e 󵄨󵄨1 + 2τ ∑ √Lc1 (τ + h ) ⋅ 󵄨e 󵄨 , √6 √6 󵄨 󵄨1 l=0

188 � 6 Difference methods for the Schrödinger equation Hence, there is a constant c7 satisfying k

󵄨 l 󵄨2 󵄨󵄨 k+1 󵄨󵄨2 2 2 2 󵄨󵄨e 󵄨󵄨1 ⩽ c7 τ ∑󵄨󵄨󵄨e 󵄨󵄨󵄨1 + c7 (τ + h ) ,

0 ⩽ k ⩽ n − 1.

l=1

It reveals that (6.43) is valid by using the Gronwall inequality (Theorem 1.2(c)).

6.3 Three-level linearized difference scheme 6.3.1 Derivation of the difference scheme Considering equation (6.1) at the point (xj , t 1 ), we have 2

󵄨 󵄨2 iut (xj , t 1 ) + uxx (xj , t 1 ) + q󵄨󵄨󵄨u(xj , t 1 )󵄨󵄨󵄨 u(xj , t 1 ) = 0, 2

2

2

2

1 ⩽ j ⩽ m − 1,

which results in 1

1

1

iδt Uj2 + δx2 Uj2 + q|û j |2 Uj2 = Pj0 ,

1 ⩽ j ⩽ m − 1,

(6.51)

from the numerical differential formula with

τ û j = u(xj , 0) + ut (xj , 0), 2

1 ⩽ j ⩽ m − 1.

There is a constant c8 satisfying 󵄨󵄨 0 󵄨󵄨 2 2 󵄨󵄨Pj 󵄨󵄨 ⩽ c8 (τ + h ),

1 ⩽ j ⩽ m − 1.

(6.52)

Considering equation (6.1) at the point (xj , tk ), we have 󵄨 󵄨2 iut (xj , tk ) + uxx (xj , tk ) + q󵄨󵄨󵄨u(xj , tk )󵄨󵄨󵄨 u(xj , tk ) = 0,

1 ⩽ j ⩽ m − 1, 1 ⩽ k ⩽ n − 1.

It follows from the numerical differential formula that ̄ 󵄨 󵄨2 ̄ iΔt Ujk + δx2 Ujk + q󵄨󵄨󵄨Ujk 󵄨󵄨󵄨 Ujk = Pjk ,

1 ⩽ j ⩽ m − 1, 1 ⩽ k ⩽ n − 1.

(6.53)

There is a constant c9 satisfying 󵄨󵄨 k 󵄨󵄨 2 2 󵄨󵄨Pj 󵄨󵄨 ⩽ c9 (τ + h ), { 󵄨 k󵄨 󵄨󵄨Δt P 󵄨󵄨 ⩽ c9 (τ 2 + h2 ), 󵄨 j󵄨

1 ⩽ j ⩽ m − 1, 1 ⩽ k ⩽ n − 1,

(6.54)

1 ⩽ j ⩽ m − 1, 2 ⩽ k ⩽ n − 2.

(6.55)

Noticing the initial-boundary value conditions (6.2)–(6.3), we have {

Uj0 = φ(xj ), U0k = 0,

Umk = 0,

1 ⩽ j ⩽ m − 1,

(6.56)

0 ⩽ k ⩽ n.

(6.57)

6.3 Three-level linearized difference scheme

� 189

Omitting the small terms in (6.51) and (6.53), a linearized difference scheme for the problem (6.1)–(6.3) reads { { { { { { { { { { { { { { { { {

1

1

1

̄ δx2 ujk

󵄨 󵄨2 ̄ q󵄨󵄨󵄨ujk 󵄨󵄨󵄨 ujk

iδt uj2 + δx2 uj2 + q|û j |2 uj2 = 0, iΔt ujk +

uj0

+

= 0,

= φ(xj ),

u0k = 0,

k um = 0,

1 ⩽ j ⩽ m − 1,

(6.58)

1 ⩽ j ⩽ m − 1, 1 ⩽ k ⩽ n − 1,

(6.59)

1 ⩽ j ⩽ m − 1,

(6.60)

0 ⩽ k ⩽ n.

(6.61)

6.3.2 Conservation and boundedness of the difference solution Theorem 6.7. Suppose {ujk | 0 ⩽ j ⩽ m, 0 ⩽ k ⩽ n} is the solution of the difference scheme (6.58)–(6.61). Denote q m−1󵄨 󵄨2 󵄨 1 󵄨 󵄨2 󵄨 󵄨2 󵄨2 E k = (󵄨󵄨󵄨uk+1 󵄨󵄨󵄨1 + 󵄨󵄨󵄨uk 󵄨󵄨󵄨1 ) − h ∑ 󵄨󵄨󵄨ujk 󵄨󵄨󵄨 ⋅ 󵄨󵄨󵄨ujk+1 󵄨󵄨󵄨 , 2 2 j=1

0 ⩽ k ⩽ n − 1.

Then we have 󵄩󵄩 k 󵄩󵄩2 󵄩󵄩 0 󵄩󵄩2 󵄩󵄩u 󵄩󵄩 = 󵄩󵄩u 󵄩󵄩 , 1 ⩽ k ⩽ n, m−1 q m−1 1 󵄨󵄨 1 󵄨󵄨2 󵄨󵄨 0 󵄨󵄨2 󵄨 󵄨2 󵄨 󵄨2 q 󵄨 󵄨2 (󵄨󵄨u 󵄨󵄨1 + 󵄨󵄨u 󵄨󵄨1 ) − h ∑ |û j |2 󵄨󵄨󵄨uj1 󵄨󵄨󵄨 = 󵄨󵄨󵄨u0 󵄨󵄨󵄨1 − h ∑ |û j |2 󵄨󵄨󵄨uj0 󵄨󵄨󵄨 , 2 2 j=1 2 j=1 Ek = E0 ,

1 ⩽ k ⩽ n − 1.

(6.62) (6.63) (6.64)

1

Proof. (I) Multiplying both the right- and left-hand sides of (6.58) by hū j2 and summing over j from 1 to m − 1, we obtain m−1

m−1

m−1

j=1

j=1

j=1

1 1 1 1 󵄨 1 󵄨2 ih ∑ (δt uj2 )ū j2 + h ∑ (δx2 uj2 )ū j2 + qh ∑ |û j |2 󵄨󵄨󵄨uj2 󵄨󵄨󵄨 = 0.

1

(6.65)

1

Noticing u02 = 0, um2 = 0, we have m−1

1

1

󵄨 󵄨2 h ∑ (δx2 uj2 )ū j2 = −󵄨󵄨󵄨u 2 󵄨󵄨󵄨1 . 1

j=1

Taking the imaginary parts on both the right- and left-hand sides of (6.65), we have 1 󵄩󵄩 1 󵄩󵄩2 󵄩󵄩 0 󵄩󵄩2 (󵄩u 󵄩 − 󵄩u 󵄩 ) = 0, 2τ 󵄩 󵄩 󵄩 󵄩 which implies

190 � 6 Difference methods for the Schrödinger equation 󵄩󵄩 1 󵄩󵄩2 󵄩󵄩 0 󵄩󵄩2 󵄩󵄩u 󵄩󵄩 = 󵄩󵄩u 󵄩󵄩 .

(6.66) ̄

Multiplying both the right- and left-hand sides of (6.59) by hū jk and summing over j from 1 to m − 1, we have m−1

m−1

m−1

j=1

j=1

j=1

̄ ̄ ̄ 󵄨 󵄨2 󵄨 ̄ 󵄨2 ih ∑ (Δt ujk )ū jk + h ∑ (δx2 ujk )ū jk + qh ∑ 󵄨󵄨󵄨ujk 󵄨󵄨󵄨 󵄨󵄨󵄨ujk 󵄨󵄨󵄨 = 0.

̄

(6.67)

̄

k Noticing u0k = 0, um = 0, we have m−1

̄ ̄ 󵄨 ̄ 󵄨2 h ∑ (δx2 ujk )ū jk = −󵄨󵄨󵄨uk 󵄨󵄨󵄨1 . j=1

Taking the imaginary parts on both the right- and left-hand sides of (6.67), we have 1 󵄩󵄩 k+1 󵄩󵄩2 󵄩󵄩 k−1 󵄩󵄩2 (󵄩u 󵄩󵄩 − 󵄩󵄩u 󵄩󵄩 ) = 0, 4τ 󵄩

1 ⩽ k ⩽ n − 1,

which implies 󵄩󵄩 k+1 󵄩󵄩2 󵄩󵄩 k−1 󵄩󵄩2 󵄩󵄩u 󵄩󵄩 = 󵄩󵄩u 󵄩󵄩 ,

1 ⩽ k ⩽ n − 1.

(6.68)

Combining (6.66) with (6.68), it follows (6.62).

1

(II) Multiplying both the right- and left-hand sides of (6.58) by −hδt ū j2 and summing over j from 1 to m − 1, we have m−1

m−1

m−1

j=1

j=1

j=1

1 1 1 1 1 󵄨 󵄨2 − ih ∑ 󵄨󵄨󵄨δt uj2 󵄨󵄨󵄨 − h ∑ (δx2 uj2 )(δt ū j2 ) − qh ∑ |û j |2 uj2 δt ū j2 = 0.

1

1

Noticing that δt u02 = 0, δt um2 = 0, we have m−1

1

1

m−1

1

1

j+ 2

j+ 2

−h ∑ (δx2 uj2 )δt ū j2 = h ∑ (δx u 2 1 )(δt δx ū 2 1 ). j=1

j=0

Taking the real parts on both the right- and left-hand sides of (6.69), we have m−1 1 󵄨󵄨 1 󵄨󵄨2 󵄨󵄨 0 󵄨󵄨2 1 󵄨 󵄨2 󵄨 󵄨2 (󵄨󵄨u 󵄨󵄨1 − 󵄨󵄨u 󵄨󵄨1 ) − qh ∑ |û j |2 ⋅ (󵄨󵄨󵄨uj1 󵄨󵄨󵄨 − 󵄨󵄨󵄨uj0 󵄨󵄨󵄨 ) = 0, 2τ 2τ j=1

which implies m−1 q m−1 1 󵄨󵄨 1 󵄨󵄨2 󵄨󵄨 0 󵄨󵄨2 󵄨 󵄨2 󵄨 󵄨2 q 󵄨 󵄨2 (󵄨󵄨u 󵄨󵄨1 + 󵄨󵄨u 󵄨󵄨1 ) − h ∑ |û j |2 󵄨󵄨󵄨uj1 󵄨󵄨󵄨 = 󵄨󵄨󵄨u0 󵄨󵄨󵄨1 − h ∑ |û j |2 󵄨󵄨󵄨uj0 󵄨󵄨󵄨 . 2 2 j=1 2 j=1

(6.69)

6.3 Three-level linearized difference scheme

� 191

Therefore, (6.63) is valid. Multiplying both the right- and left-hand sides of (6.59) by −hΔt ū jk and summing over j from 1 to m − 1, we have m−1

m−1

m−1

j=1

j=1

j=1

̄ 󵄨 󵄨2 󵄨 󵄨2 ̄ −ih ∑ 󵄨󵄨󵄨Δt ujk 󵄨󵄨󵄨 − h ∑ (δx2 ujk )(Δt ū jk ) − qh ∑ 󵄨󵄨󵄨ujk 󵄨󵄨󵄨 ujk Δt ū jk = 0.

Taking the real parts on both the right- and left-hand sides of the above equality, we get m−1 1 󵄨󵄨 k+1 󵄨󵄨2 󵄨󵄨 k−1 󵄨󵄨2 󵄨2 󵄨 󵄨2 󵄨 󵄨2 1 󵄨 (󵄨󵄨u 󵄨󵄨1 − 󵄨󵄨u 󵄨󵄨1 ) − qh ∑ 󵄨󵄨󵄨ujk 󵄨󵄨󵄨 ⋅ (󵄨󵄨󵄨ujk+1 󵄨󵄨󵄨 − 󵄨󵄨󵄨ujk−1 󵄨󵄨󵄨 ) = 0, 4τ 4τ j=1

which implies q m−1󵄨 󵄨2 󵄨 1 󵄨󵄨 k+1 󵄨󵄨2 󵄨󵄨 k 󵄨󵄨2 󵄨2 (󵄨󵄨u 󵄨󵄨1 + 󵄨󵄨u 󵄨󵄨1 ) − h ∑ 󵄨󵄨󵄨ujk 󵄨󵄨󵄨 ⋅ 󵄨󵄨󵄨ujk+1 󵄨󵄨󵄨 2 2 j=1 q m−1󵄨 1 󵄨 󵄨2 󵄨 󵄨2 󵄨2 󵄨 󵄨2 = (󵄨󵄨󵄨uk 󵄨󵄨󵄨1 + 󵄨󵄨󵄨uk−1 󵄨󵄨󵄨1 ) − h ∑ 󵄨󵄨󵄨ujk−1 󵄨󵄨󵄨 ⋅ 󵄨󵄨󵄨ujk 󵄨󵄨󵄨 , 2 2 j=1

1 ⩽ k ⩽ n − 1.

Therefore, (6.64) is valid. According to Theorem 6.7, there is a constant c10 satisfying 󵄩󵄩 k 󵄩󵄩 󵄩󵄩u 󵄩󵄩∞ ⩽ c10 ,

1 ⩽ k ⩽ n.

Remark 6.1. If one takes û j = uj0 ,

0⩽j⩽m

in (6.58), then we have m−1 󵄨 󵄨4 󵄨 󵄨2 q E k = 󵄨󵄨󵄨u0 󵄨󵄨󵄨1 − h ∑ 󵄨󵄨󵄨uj0 󵄨󵄨󵄨 , 2 j=1

0 ⩽ k ⩽ n − 1.

The corresponding difference scheme is second-order convergent in the L2 -norm.

6.3.3 Existence and uniqueness of the difference solution Theorem 6.8. The difference scheme (6.58)–(6.61) is uniquely solvable. Proof. (I) From (6.60)–(6.61), we know that u0 has been determined. From (6.58) and (6.61), a system of linear equations in {uj1 | 0 ⩽ j ⩽ m} is obtained. Consider its homoge-

192 � 6 Difference methods for the Schrödinger equation neous one: 1 1 1 { i uj1 + δx2 uj1 + q|û j |2 uj1 = 0, 2 2 { τ1 1 { u0 = 0, um = 0.

1 ⩽ j ⩽ m − 1,

(6.70) (6.71)

Multiplying both the right- and left-hand sides of (6.70) by hū j1 , summing over j from 1 to m − 1, and then using (6.71) and taking the imaginary parts of the result, it follows: 󵄩󵄩 1 󵄩󵄩2 󵄩󵄩u 󵄩󵄩 = 0, which implies uj1 = 0,

0 ⩽ j ⩽ m.

Thus, the difference scheme uniquely determines the value of u1 at the first time level. (II) Suppose {ujk−1 | 0 ⩽ j ⩽ m} and {ujk | 0 ⩽ j ⩽ m} have been obtained, then we

have the system of linear equations in {ujk+1 | 0 ⩽ j ⩽ m} from (6.59) and (6.61). Consider its homogeneous one: 1 1 { i ⋅ ujk+1 + δx2 ujk+1 + 2τ 2 { k+1 k+1 { u0 = 0, um = 0.

q 󵄨󵄨 k 󵄨󵄨2 k+1 󵄨u 󵄨 u = 0, 2󵄨 j 󵄨 j

1 ⩽ j ⩽ m − 1,

(6.72) (6.73)

Multiplying both the right- and left-hand sides of (6.72) by hū jk+1 , summing over j from 1 to m − 1 and then taking the imaginary parts of the result, we have 󵄩󵄩 k+1 󵄩󵄩2 󵄩󵄩u 󵄩󵄩 = 0, which implies ujk+1 = 0,

0 ⩽ j ⩽ m.

Therefore, the difference equations (6.59) and (6.61) admit a unique solution {ujk+1 | 0 ⩽ j ⩽ m}. 6.3.4 Convergence of the difference solution Theorem 6.9. Suppose {Ujk | 0 ⩽ j ⩽ m, 0 ⩽ k ⩽ n} is the solution of the problem

(6.1)–(6.3) and {ujk | 0 ⩽ j ⩽ m, 0 ⩽ k ⩽ n} is the solution of the difference scheme (6.58)–(6.61). Denote ejk = Ujk − ujk ,

0 ⩽ j ⩽ m, 0 ⩽ k ⩽ n.

6.3 Three-level linearized difference scheme

� 193

Then there is a constant c11 satisfying 󵄩󵄩 k 󵄩󵄩 2 2 󵄩󵄩e 󵄩󵄩 ⩽ c11 (τ + h ),

0 ⩽ k ⩽ n.

(6.74)

Proof. Subtracting (6.58)–(6.61) from (6.51), (6.53), (6.56) and (6.57), respectively, we have the system of error equations: { { { { { { { { { { { { { { { { {

1

1

1

iδt ej2 + δx2 ej2 + q|û j |2 ej2 = Pj0 , iΔt ejk + ej0

̄ δx2 ejk

+

󵄨 󵄨2 ̄ q(󵄨󵄨󵄨Ujk 󵄨󵄨󵄨 Ujk

−

󵄨󵄨 k 󵄨󵄨2 k̄ 󵄨󵄨uj 󵄨󵄨 uj )

= Pjk ,

= 0,

e0k = 0,

k em = 0,

1 ⩽ j ⩽ m − 1,

(6.75)

1 ⩽ j ⩽ m − 1, 1 ⩽ k ⩽ n − 1,

(6.76)

1 ⩽ j ⩽ m − 1,

(6.77)

0 ⩽ k ⩽ n.

(6.78)

It follows from (6.78) that 1

{ e02 = 0, { k̄ { e0 = 0,

1

em2 = 0, k̄ em

(6.79)

= 0,

1 ⩽ k ⩽ n − 1.

(6.80) 1

(I) Multiplying both the right- and left-hand sides of (6.75) by hēj2 and summing over j from 1 to m − 1, we have m−1

m−1

m−1

m−1

j=1

j=1

j=1

j=1

1 1 1 1 1 󵄨 1 󵄨2 ih ∑ (δt ej2 )ēj2 + h ∑ (δx2 ej2 )ēj2 + qh ∑ |û j |2 󵄨󵄨󵄨ej2 󵄨󵄨󵄨 = h ∑ Pj0 ēj2 .

(6.81)

Noticing m−1

m−1

j=1

j=0

1 1 1 󵄨2 󵄨 h ∑ (δx2 ej2 )ēj2 = −h ∑ 󵄨󵄨󵄨δx e 2 1 󵄨󵄨󵄨 j+ 2

and taking the imaginary parts on both the right- and left-hand sides of (6.81), we have m−1 1 1 󵄩󵄩 1 󵄩󵄩2 󵄩󵄩 0 󵄩󵄩2 󵄩 󵄩 󵄩 1󵄩 (󵄩󵄩e 󵄩󵄩 − 󵄩󵄩e 󵄩󵄩 ) = Im{h ∑ Pj0 ēj2 } ⩽ 󵄩󵄩󵄩P0 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩e 2 󵄩󵄩󵄩. 2τ j=1

Noticing e0 = 0, we have 1 󵄩󵄩 1 󵄩󵄩2 󵄩󵄩 0 󵄩󵄩 󵄩e 󵄩 ⩽ 󵄩󵄩P 󵄩󵄩 ⋅ 2τ 󵄩 󵄩

1 󵄩󵄩 1 󵄩󵄩 󵄩e 󵄩. 2󵄩 󵄩

Therefore, 󵄩󵄩 1 󵄩󵄩 󵄩 0󵄩 2 2 2 2 󵄩󵄩e 󵄩󵄩 ⩽ τ 󵄩󵄩󵄩P 󵄩󵄩󵄩 ⩽ τc8 √L(τ + h ) ⩽ c8 √L(τ + h ).

(6.82)

194 � 6 Difference methods for the Schrödinger equation ̄

(II) Multiplying both the right- and left-hand sides of (6.76) by hējk and summing over j from 1 to m − 1, we have m−1

m−1

m−1

m−1

j=1

j=1

j=1

j=1

̄ ̄ ̄ ̄ 󵄨 󵄨2 ̄ 󵄨 󵄨2 ̄ ̄ ih ∑ (Δt ejk )ējk + h ∑ (δx2 ejk )ējk + qh ∑ (󵄨󵄨󵄨Ujk 󵄨󵄨󵄨 Ujk − 󵄨󵄨󵄨ujk 󵄨󵄨󵄨 ujk )ējk = h ∑ Pjk ējk .

(6.83)

Noticing m−1

m−1

j=1

j=0

̄ ̄ 󵄨 k̄ 󵄨󵄨2 h ∑ (δx2 ejk )ējk = −h ∑ 󵄨󵄨󵄨δx ej+ 1 󵄨󵄨 2

and 󵄨 󵄨2 ̄ 󵄨 󵄨2 ̄ ̄ 󵄨 󵄨2 ̄ 󵄨 󵄨2 󵄨 󵄨2 ̄ ̄ (󵄨󵄨󵄨Ujk 󵄨󵄨󵄨 Ujk − 󵄨󵄨󵄨ujk 󵄨󵄨󵄨 ujk )ējk = [󵄨󵄨󵄨Ujk 󵄨󵄨󵄨 ejk + (󵄨󵄨󵄨Ujk 󵄨󵄨󵄨 − 󵄨󵄨󵄨ujk 󵄨󵄨󵄨 )ujk ]ējk ̄ ̄ 󵄨 󵄨2 󵄨 ̄ 󵄨2 = 󵄨󵄨󵄨Ujk 󵄨󵄨󵄨 󵄨󵄨󵄨ejk 󵄨󵄨󵄨 + (ejk Ū jk + ujk ējk )ujk ējk , and then taking the imaginary parts on both the right- and left-hand sides of (6.83), we have 1 󵄩󵄩 k+1 󵄩󵄩2 󵄩󵄩 k−1 󵄩󵄩2 (󵄩e 󵄩󵄩 − 󵄩󵄩e 󵄩󵄩 ) 4τ 󵄩 m−1

̄

m−1

̄

̄

= −q Im{h ∑ (ejk Ū jk + ujk ējk )ujk ējk } + Im{h ∑ Pjk ējk } j=1

j=1

󵄩 󵄩 󵄩 ̄󵄩 󵄩 󵄩 󵄩 ̄󵄩 ⩽ |q|(c0 + c10 )c10 󵄩󵄩󵄩ek 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩ek 󵄩󵄩󵄩 + 󵄩󵄩󵄩Pk 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩ek 󵄩󵄩󵄩 󵄩 󵄩 󵄩 󵄩 ‖e ⩽ (|q|(c0 + c10 )c10 󵄩󵄩󵄩ek 󵄩󵄩󵄩 + 󵄩󵄩󵄩Pk 󵄩󵄩󵄩)

k+1

‖ + ‖ek−1 ‖ , 2

1 ⩽ k ⩽ n − 1,

which implies 1 󵄩󵄩 k+1 󵄩󵄩 󵄩󵄩 k−1 󵄩󵄩 (󵄩e 󵄩󵄩 − 󵄩󵄩e 󵄩󵄩) 2τ 󵄩 󵄩 󵄩 󵄩 󵄩 ⩽ |q|(c0 + c10 )c10 󵄩󵄩󵄩ek 󵄩󵄩󵄩 + 󵄩󵄩󵄩Pk 󵄩󵄩󵄩 󵄩 󵄩 ⩽ |q|(c0 + c10 )c10 󵄩󵄩󵄩ek 󵄩󵄩󵄩 + √Lc9 (τ 2 + h2 ),

1 ⩽ k ⩽ n − 1,

or 󵄩󵄩 k+1 󵄩󵄩 󵄩󵄩 k−1 󵄩󵄩 󵄩 k󵄩 2 2 󵄩󵄩e 󵄩󵄩 ⩽ 󵄩󵄩e 󵄩󵄩 + 2|q|(c0 + c10 )c10 τ 󵄩󵄩󵄩e 󵄩󵄩󵄩 + 2√Lc9 τ(τ + h ),

1 ⩽ k ⩽ n − 1,

where we have used (6.54). It follows from the above inequality that 󵄩 󵄩󵄩 󵄩 󵄩 󵄩󵄩 󵄩 max{󵄩󵄩󵄩ek+1 󵄩󵄩󵄩, 󵄩󵄩󵄩ek 󵄩󵄩󵄩} ⩽ [1 + 2|q|(c0 + c10 )c10 τ] max{󵄩󵄩󵄩ek 󵄩󵄩󵄩, 󵄩󵄩󵄩ek−1 󵄩󵄩󵄩} + 2√Lc9 τ(τ 2 + h2 ), 1 ⩽ k ⩽ n − 1.

6.3 Three-level linearized difference scheme

� 195

When q = 0, it follows that 󵄩 󵄩󵄩 󵄩 󵄩 󵄩󵄩 󵄩 max{󵄩󵄩󵄩ek+1 󵄩󵄩󵄩, 󵄩󵄩󵄩ek 󵄩󵄩󵄩} ⩽ max{󵄩󵄩󵄩e1 󵄩󵄩󵄩, 󵄩󵄩󵄩e0 󵄩󵄩󵄩} + 2√Lc9 kτ(τ 2 + h2 ) ⩽ (c8 + 2c9 T)√L(τ 2 + h2 ),

1 ⩽ k ⩽ n − 1.

When q ≠ 0, it follows from the Gronwall inequality (Theorem 1.2(a)) that 󵄩 󵄩󵄩 󵄩 max{󵄩󵄩󵄩ek+1 󵄩󵄩󵄩, 󵄩󵄩󵄩ek 󵄩󵄩󵄩} 󵄩 󵄩󵄩 󵄩 ⩽ e2|q|(c0 +c10 )c10 T [max{󵄩󵄩󵄩e1 󵄩󵄩󵄩, 󵄩󵄩󵄩e0 󵄩󵄩󵄩} + ⩽ e2|q|(c0 +c10 )c10 T [c8 +

√Lc9 (τ 2 + h2 )] |q|(c0 + c10 )c10

c9 ]√L(τ 2 + h2 ), |q|(c0 + c10 )c10

0 ⩽ k ⩽ n − 1.

Theorem 6.10. Suppose {Ujk | 0 ⩽ j ⩽ m, 0 ⩽ k ⩽ n} is the solution of the problem

(6.1)–(6.3) and {ujk | 0 ⩽ j ⩽ m, 0 ⩽ k ⩽ n} is the solution of the difference scheme (6.58)–(6.61). Denote ejk = Ujk − ujk ,

0 ⩽ j ⩽ m, 0 ⩽ k ⩽ n.

Then there is a constant c12 satisfying 󵄨󵄨 k 󵄨󵄨 2 2 󵄨󵄨e 󵄨󵄨1 ⩽ c12 (τ + h ),

0 ⩽ k ⩽ n. 1

Proof. (I) Multiplying both the right- and left-hand sides of (6.75) by −hδt ēj2 and summing over j from 1 to m − 1, we have m−1

m−1

m−1

m−1

j=1

j=1

j=1

j=1

1 1 1 1 1 󵄨 1 󵄨2 −ih ∑ 󵄨󵄨󵄨δt ej2 󵄨󵄨󵄨 − h ∑ (δx2 ej2 )(δt ēj2 ) − qh ∑ |û j |2 ej2 δt ēj2 = −h ∑ Pj0 δt ēj2 .

(6.84)

Noticing m−1

1

1

m−1

1

1

j+ 2

j+ 2

−h ∑ (δx2 ej2 )δt ēj2 = h ∑ (δx e 2 1 )(δt δx ē 2 1 ) j=1

j=0

and then taking the real parts on both the right- and left-hand sides of (6.84), it follows: m−1 m−1 1 |ej1 |2 − |ej0 |2 1 󵄨󵄨 1 󵄨󵄨2 󵄨󵄨 0 󵄨󵄨2 (󵄨󵄨e 󵄨󵄨1 − 󵄨󵄨e 󵄨󵄨1 ) − qh ∑ |û j |2 = Re{−h ∑ Pj0 δt ēj2 } 2τ 2τ j=1 j=1

󵄩 󵄩 󵄩 1󵄩 ⩽ 󵄩󵄩󵄩P0 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩δt e 2 󵄩󵄩󵄩. Noticing

196 � 6 Difference methods for the Schrödinger equation ej0 = 0,

0 ⩽ j ⩽ m,

it follows: 1 󵄨󵄨 1 󵄨󵄨2 q m−1 2 󵄨󵄨 1 󵄨󵄨2 1 󵄩󵄩 0 󵄩󵄩 󵄩󵄩 1 󵄩󵄩 󵄨e 󵄨 − h ∑ |û | 󵄨e 󵄨 ⩽ 󵄩󵄩P 󵄩󵄩 ⋅ 󵄩󵄩e 󵄩󵄩. 2τ 󵄨 󵄨1 2τ j=1 j 󵄨 j 󵄨 τ Multiplying by 2τ on both the right- and left-hand sides gives m−1

󵄨󵄨 1 󵄨󵄨2 󵄩 0󵄩 󵄩 1󵄩 2 󵄨 1 󵄨2 󵄨󵄨e 󵄨󵄨1 ⩽ |q|h ∑ |û j | 󵄨󵄨󵄨ej 󵄨󵄨󵄨 + 2󵄩󵄩󵄩P 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩e 󵄩󵄩󵄩.

(6.85)

j=1

Denote 󵄨 󵄨 c13 = max 󵄨󵄨󵄨ut (x, 0)󵄨󵄨󵄨. 0⩽x⩽L

Then we have τ |û j | ⩽ c0 + c13 ⩽ c0 + c13 , 2

1 ⩽ j ⩽ m − 1.

It follows from (6.85), (6.52) and (6.82) that 󵄨󵄨 1 󵄨󵄨2 󵄩 0󵄩 󵄩 1󵄩 2 󵄩 1 󵄩2 󵄨󵄨e 󵄨󵄨1 ⩽ |q|(c0 + c13 ) 󵄩󵄩󵄩e 󵄩󵄩󵄩 + 2󵄩󵄩󵄩P 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩e 󵄩󵄩󵄩 2 ⩽ |q|(c0 + c13 )2 [c8 √L(τ 2 + h2 )] + 2√Lc8 (τ 2 + h2 )c8 √L(τ 2 + h2 ) 2

= [|q|(c0 + c13 )2 + 2][c8 √L(τ 2 + h2 )] , which implies that 󵄨󵄨 1 󵄨󵄨 √ 2 2 󵄨󵄨e 󵄨󵄨1 ⩽ |q|(c0 + c13 )2 + 2 c8 √L(τ + h ).

(6.86)

(II) Multiplying both the right- and left-hand sides of (6.76) by −Δt ējk and summing over j from 1 to m − 1, we have m−1

m−1

j=1

j=1

̄ 󵄨 󵄨2 − ih ∑ 󵄨󵄨󵄨Δt ejk 󵄨󵄨󵄨 − h ∑ (δx2 ejk )(Δt ējk ) m−1

m−1

j=1

j=1

󵄨 󵄨2 ̄ 󵄨 󵄨2 ̄ = qh ∑ (󵄨󵄨󵄨Ujk 󵄨󵄨󵄨 Ujk − 󵄨󵄨󵄨ujk 󵄨󵄨󵄨 ujk )Δt ējk − h ∑ Pjk Δt ējk ,

1 ⩽ k ⩽ n − 1.

For the second term on the left-hand side of (6.87), we have m−1

̄

m−1

̄

k k . −h ∑ (δx2 ejk )Δt ējk = h ∑ (δx ej+ 1 )Δt δx ē j+ 1 j=1

j=0

2

2

(6.87)

6.3 Three-level linearized difference scheme

� 197

Then taking the real parts of the above equality results in m−1

̄

Re{−h ∑ (δx2 ejk )Δt ējk } = j=1

1 󵄨󵄨 k+1 󵄨󵄨2 󵄨󵄨 k−1 󵄨󵄨2 (󵄨e 󵄨󵄨1 − 󵄨󵄨e 󵄨󵄨1 ). 4τ 󵄨

(6.88)

Now we analyze the first term on the right-hand side of (6.87). It follows from (6.76) that ̄ 󵄨 󵄨2 ̄ 󵄨 󵄨2 ̄ Δt ejk = iδx2 ejk + iq(󵄨󵄨󵄨Ujk 󵄨󵄨󵄨 Ujk − 󵄨󵄨󵄨ujk 󵄨󵄨󵄨 ujk ) − iPjk .

Thus, we have m−1

󵄨 󵄨2 ̄ 󵄨 󵄨2 ̄ Re{qh ∑ (󵄨󵄨󵄨Ujk 󵄨󵄨󵄨 Ujk − 󵄨󵄨󵄨ujk 󵄨󵄨󵄨 ujk )Δt ējk } j=1

m−1

󵄨 󵄨2 ̄ 󵄨 󵄨2 ̄ = Re{qh ∑ (󵄨󵄨󵄨Ujk 󵄨󵄨󵄨 Ū jk − 󵄨󵄨󵄨ujk 󵄨󵄨󵄨 ū jk )Δt ejk } j=1

m−1

̄ 󵄨 󵄨2 ̄ 󵄨 󵄨2 ̄ 󵄨 󵄨2 ̄ 󵄨 󵄨2 ̄ = Re{qh ∑ (󵄨󵄨󵄨Ujk 󵄨󵄨󵄨 Ū jk − 󵄨󵄨󵄨ujk 󵄨󵄨󵄨 ū jk )[iδx2 ejk + iq(󵄨󵄨󵄨Ujk 󵄨󵄨󵄨 Ujk − 󵄨󵄨󵄨ujk 󵄨󵄨󵄨 ujk ) − iPjk ]}. j=1

̂ satisfying There is a constant c14 m−1

󵄨 󵄨2 ̄ 󵄨 󵄨2 ̄ Re{qh ∑ (󵄨󵄨󵄨Ujk 󵄨󵄨󵄨 Ujk − 󵄨󵄨󵄨ujk 󵄨󵄨󵄨 ujk )Δt ējk } j=1

m−1

̄ 󵄨 󵄨2 ̄ 󵄨 󵄨2 ̄ = Re{qh ∑ (󵄨󵄨󵄨Ujk 󵄨󵄨󵄨 Ū jk − 󵄨󵄨󵄨ujk 󵄨󵄨󵄨 ū jk )(iδx2 ejk − iPjk )} j=1

1 2 2 2 2 ̂ (󵄩󵄩󵄩󵄩ek−1 󵄩󵄩󵄩󵄩 + 󵄩󵄩󵄩󵄩ek 󵄩󵄩󵄩󵄩 + 󵄩󵄩󵄩󵄩ek+1 󵄩󵄩󵄩󵄩 + 󵄨󵄨󵄨󵄨ek−1 󵄨󵄨󵄨󵄨1 ⩽ c14 2 󵄨 󵄨2 󵄨 󵄨2 󵄩 󵄩2 + 󵄨󵄨󵄨ek 󵄨󵄨󵄨1 + 󵄨󵄨󵄨ek+1 󵄨󵄨󵄨1 + 󵄩󵄩󵄩Pk 󵄩󵄩󵄩 ).

(6.89)

Taking the real parts on both the right- and left-hand sides of (6.87) and combining (6.88) with (6.89), it follows that 1 󵄨󵄨 k+1 󵄨󵄨2 󵄨󵄨 k−1 󵄨󵄨2 (󵄨e 󵄨󵄨1 − 󵄨󵄨e 󵄨󵄨1 ) 4τ 󵄨 1 2 2 2 2 2 2 2 ̂ (󵄩󵄩󵄩󵄩ek−1 󵄩󵄩󵄩󵄩 + 󵄩󵄩󵄩󵄩ek 󵄩󵄩󵄩󵄩 + 󵄩󵄩󵄩󵄩ek+1 󵄩󵄩󵄩󵄩 + 󵄨󵄨󵄨󵄨ek−1 󵄨󵄨󵄨󵄨1 + 󵄨󵄨󵄨󵄨ek 󵄨󵄨󵄨󵄨1 + 󵄨󵄨󵄨󵄨ek+1 󵄨󵄨󵄨󵄨1 + 󵄩󵄩󵄩󵄩Pk 󵄩󵄩󵄩󵄩 ) ⩽ c14 2 m−1

+ Re{−h ∑ Pjk Δt ējk }, j=1

1 ⩽ k ⩽ n − 1.

Replacing k with l and summing over l from 1 to k, we have

198 � 6 Difference methods for the Schrödinger equation 1 󵄨󵄨 k+1 󵄨󵄨2 󵄨󵄨 k 󵄨󵄨2 󵄨󵄨 1 󵄨󵄨2 󵄨󵄨 0 󵄨󵄨2 (󵄨e 󵄨󵄨1 + 󵄨󵄨e 󵄨󵄨1 − 󵄨󵄨e 󵄨󵄨1 − 󵄨󵄨e 󵄨󵄨1 ) 4τ 󵄨

k 1 2 2 2 2 2 2 2 ̂ ∑(󵄩󵄩󵄩󵄩el−1 󵄩󵄩󵄩󵄩 + 󵄩󵄩󵄩󵄩el 󵄩󵄩󵄩󵄩 + 󵄩󵄩󵄩󵄩el+1 󵄩󵄩󵄩󵄩 + 󵄨󵄨󵄨󵄨el−1 󵄨󵄨󵄨󵄨1 + 󵄨󵄨󵄨󵄨el 󵄨󵄨󵄨󵄨1 + 󵄨󵄨󵄨󵄨el+1 󵄨󵄨󵄨󵄨1 + 󵄩󵄩󵄩󵄩Pl 󵄩󵄩󵄩󵄩 ) ⩽ c14 2 l=1 󵄨󵄨 󵄨󵄨 m−1 k 󵄨󵄨 󵄨󵄨 + 󵄨󵄨󵄨h ∑ ∑ Pjl Δt ējl 󵄨󵄨󵄨, 1 ⩽ k ⩽ n − 1. 󵄨󵄨 󵄨󵄨 󵄨 󵄨 j=1 l=1

(6.90)

Noticing k

∑ Pjl Δt ējl l=1

=

1 k l l+1 ∑ P (ē − ējl−1 ) 2τ l=1 j j

=

1 k+1 l−1 l k−1 l+1 l ( ∑ P ē − ∑ P ē ) 2τ l=2 j j l=0 j j

=

k−1 1 [Pjk ējk+1 + Pjk−1 ējk − ∑ (Pjl+1 − Pjl−1 )ējl − Pj2 ēj1 ], 2τ l=2

it follows that 󵄨󵄨 m−1 k 󵄨󵄨 󵄨󵄨 󵄨 󵄨󵄨h ∑ ∑ Pl Δt ēl 󵄨󵄨󵄨 j 󵄨󵄨 j 󵄨󵄨 󵄨󵄨 j=1 l=1 󵄨󵄨 󵄨󵄨 󵄨󵄨 1 m−1 1 m−1 = 󵄨󵄨󵄨 h ∑ Pjk ējk+1 + h ∑ Pjk−1 ējk 󵄨󵄨 2τ 2τ j=1 j=1 󵄨 k−1

m−1

l=2

j=1

−∑h ∑ ⩽

Pjl+1 − Pjl−1 2τ

ējl −

󵄨󵄨 m−1 󵄨󵄨 1 ⋅ h ∑ Pj2 ēj1 󵄨󵄨󵄨 󵄨󵄨 2τ j=1 󵄨

1 󵄩󵄩 k 󵄩󵄩 󵄩󵄩 k+1 󵄩󵄩 1 󵄩󵄩 k−1 󵄩󵄩 󵄩󵄩 k 󵄩󵄩 󵄩P 󵄩 ⋅ 󵄩e 󵄩󵄩 + 󵄩󵄩P 󵄩󵄩 ⋅ 󵄩󵄩e 󵄩󵄩 2τ 󵄩 󵄩 󵄩 2τ

k−1 󵄩 󵄩 1󵄩 󵄩 󵄩 󵄩 + ∑ ‖Δt Pl ‖ ⋅ 󵄩󵄩󵄩el 󵄩󵄩󵄩 + 󵄩󵄩󵄩P2 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩e1 󵄩󵄩󵄩. 2τ l=2

(6.91)

Substituting (6.91) into (6.90) and multiplying the result on both the right- and left- hand sides by 4τ, it follows: 󵄨󵄨 k+1 󵄨󵄨2 󵄨󵄨 k 󵄨󵄨2 󵄨󵄨 1 󵄨󵄨2 󵄨󵄨 0 󵄨󵄨2 󵄨󵄨e 󵄨󵄨1 + 󵄨󵄨e 󵄨󵄨1 − 󵄨󵄨e 󵄨󵄨1 − 󵄨󵄨e 󵄨󵄨1 k

2 2 2 2 2 2 2 ̂ τ ∑(󵄩󵄩󵄩󵄩el−1 󵄩󵄩󵄩󵄩 + 󵄩󵄩󵄩󵄩el 󵄩󵄩󵄩󵄩 + 󵄩󵄩󵄩󵄩el+1 󵄩󵄩󵄩󵄩 + 󵄨󵄨󵄨󵄨el−1 󵄨󵄨󵄨󵄨1 + 󵄨󵄨󵄨󵄨el 󵄨󵄨󵄨󵄨1 + 󵄨󵄨󵄨󵄨el+1 󵄨󵄨󵄨󵄨1 + 󵄩󵄩󵄩󵄩Pl 󵄩󵄩󵄩󵄩 ) ⩽ 2c14 l=1

6.3 Three-level linearized difference scheme

� 199

k−1

󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 + 2(󵄩󵄩󵄩Pk 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩ek+1 󵄩󵄩󵄩 + 󵄩󵄩󵄩Pk−1 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩ek 󵄩󵄩󵄩 + 2τ ∑ 󵄩󵄩󵄩Δt Pl 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩el 󵄩󵄩󵄩 + 󵄩󵄩󵄩P2 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩e1 󵄩󵄩󵄩), 1 ⩽ k ⩽ n − 1.

l=2

Noticing 3 󵄩 k+1 󵄩2 L2 󵄩 k 󵄩2 1 󵄨 k+1 󵄨2 L2 󵄩 k 󵄩2 󵄩󵄩 k 󵄩󵄩 󵄩󵄩 k+1 󵄩󵄩 󵄩󵄩P 󵄩󵄩 ⋅ 󵄩󵄩e 󵄩󵄩 ⩽ 2 󵄩󵄩󵄩e 󵄩󵄩󵄩 + 󵄩󵄩󵄩P 󵄩󵄩󵄩 ⩽ 󵄨󵄨󵄨e 󵄨󵄨󵄨1 + 󵄩󵄩󵄩P 󵄩󵄩󵄩 , 12 2 12 L 3 󵄩󵄩 k 󵄩󵄩2 L2 󵄩󵄩 k−1 󵄩󵄩2 1 󵄨󵄨 k 󵄨󵄨2 L2 󵄩󵄩 k−1 󵄩󵄩2 󵄩󵄩 k−1 󵄩󵄩 󵄩󵄩 k 󵄩󵄩 󵄩󵄩P 󵄩󵄩 ⋅ 󵄩󵄩e 󵄩󵄩 ⩽ 2 󵄩󵄩e 󵄩󵄩 + 󵄩󵄩P 󵄩󵄩 ⩽ 󵄨󵄨e 󵄨󵄨1 + 󵄩󵄩P 󵄩󵄩 , 12 2 12 L 󵄩󵄩 2 󵄩󵄩 󵄩󵄩 1 󵄩󵄩 1 󵄩󵄩 2 󵄩󵄩2 1 󵄩󵄩 1 󵄩󵄩2 󵄩󵄩P 󵄩󵄩 ⋅ 󵄩󵄩e 󵄩󵄩 ⩽ 󵄩󵄩P 󵄩󵄩 + 󵄩󵄩e 󵄩󵄩 2 2 and in combination of (6.74), (6.86), (6.54) and (6.55), there is a constant c14 satisfying 󵄨󵄨 k+1 󵄨󵄨2 󵄨󵄨 k 󵄨󵄨2 󵄨󵄨e 󵄨󵄨1 + 󵄨󵄨e 󵄨󵄨1 k

2 󵄨 󵄨2 󵄨 󵄨2 󵄨 󵄨2 ⩽ c14 τ ∑(󵄨󵄨󵄨el+1 󵄨󵄨󵄨1 + 2󵄨󵄨󵄨el 󵄨󵄨󵄨1 + 󵄨󵄨󵄨el−1 󵄨󵄨󵄨1 ) + c14 (τ 2 + h2 ) , l=1

1 ⩽ k ⩽ n − 1.

In other words, 󵄨 󵄨2 󵄨 󵄨2 (1 − c14 τ)(󵄨󵄨󵄨ek+1 󵄨󵄨󵄨1 + 󵄨󵄨󵄨ek 󵄨󵄨󵄨1 ) k−1

2 󵄨 󵄨2 󵄨 󵄨2 ⩽ 2c14 τ ∑ (󵄨󵄨󵄨el+1 󵄨󵄨󵄨1 + 󵄨󵄨󵄨el 󵄨󵄨󵄨1 ) + c14 (τ 2 + h2 ) , l=0

1 ⩽ k ⩽ n − 1.

When c14 τ ⩽ 31 , it follows: k−1 3 󵄨󵄨 k+1 󵄨󵄨2 󵄨󵄨 k 󵄨󵄨2 󵄨 l+1 󵄨2 󵄨 l 󵄨2 2 2 2 󵄨󵄨e 󵄨󵄨1 + 󵄨󵄨e 󵄨󵄨1 ⩽ 3c14 τ ∑ (󵄨󵄨󵄨e 󵄨󵄨󵄨1 + 󵄨󵄨󵄨e 󵄨󵄨󵄨1 ) + c14 (τ + h ) , 2 l=0

1 ⩽ k ⩽ n − 1.

Using the Gronwall inequality (Theorem 1.2(c)) and noticing (6.86), we have 󵄨󵄨 k+1 󵄨󵄨2 󵄨󵄨 k 󵄨󵄨2 󵄨󵄨e 󵄨󵄨1 + 󵄨󵄨e 󵄨󵄨1

2 󵄨 󵄨2 󵄨 󵄨2 3 ⩽ e3c14 kτ [󵄨󵄨󵄨e1 󵄨󵄨󵄨1 + 󵄨󵄨󵄨e0 󵄨󵄨󵄨1 + c14 (τ 2 + h2 ) ] 2 3 2 ⩽ e3c14 T [(|q|(c0 + c13 )2 + 2)c82 L + c14 ](τ 2 + h2 ) , 2

1 ⩽ k ⩽ n − 1.

200 � 6 Difference methods for the Schrödinger equation

6.4 Fourth-order three-level linearized difference scheme 6.4.1 Several numerical differential formulas Lemma 6.1. Let f (x) ∈ C 6 [xj−2 , xj+2 ]. Then we have f ′′ (xj ) =

4 f (xj+1 ) − 2f (xj ) + f (xj−1 ) 1 f (xj+2 ) − 2f (xj ) + f (xj−2 ) × − × 3 3 h2 (2h)2 1

h4 − ∫[f (6) (xj + sh) + f (6) (xj − sh)](1 − s)5 ds 90 0

1

2h4 + ∫[f (6) (xj + 2sh) + f (6) (xj − 2sh)](1 − s)5 ds. 45 0

Proof. Using the Taylor expansion with the integral remainder, we have 1

k

hl hk+1 f (xj + h) = ∑ f (l) (xj ) + ∫ f (k+1) (xj + sh)(1 − s)k ds, l! k! l=0

(6.92)

0

which implies f (xj + h) = f (xj ) + hf ′ (xj ) +

h2 ′′ h3 h4 f (xj ) + f ′′′ (xj ) + f (4) (xj ) 2 6 24 1

+

h6 h5 (5) f (xj ) + ∫ f (6) (xj + sh)(1 − s)5 ds 120 120 0

and f (xj − h) = f (xj ) − hf ′ (xj ) +

h2 ′′ h3 h4 f (xj ) − f ′′′ (xj ) + f (4) (xj ) 2 6 24 1

−

h5 (5) h6 f (xj ) + ∫ f (6) (xj − sh)(1 − s)5 ds. 120 120 0

Adding the above two equalities together, we have f ′′ (xj ) =

f (xj+1 ) − 2f (xj ) + f (xj−1 ) 1

−

h2 (4) f (xj ) 12

h4 ∫[f (6) (xj + sh) + f (6) (xj − sh)](1 − s)5 ds. 120 0

Similarly,

h2

−

(6.93)

6.4 Fourth-order three-level linearized difference scheme

f ′′ (xj ) =

f (xj+2 ) − 2f (xj ) + f (xj−2 ) (2h)2

1

−

� 201

(2h)2 (4) f (xj ) 12

(2h)4 − ∫[f (6) (xj + 2sh) + f (6) (xj − 2sh)](1 − s)5 ds. 120

(6.94)

0

Multiplying (6.93) by 43 and (6.94) by 31 on both the right- and left-hand sides, respectively, and then subtracting the results, the conclusion is valid. Lemma 6.2. Let f (x) ∈ C 6 [x0 , x3 ]. Then we have f ′′ (x1 ) =

f (x3 ) − 2f (x2 ) + f (x1 ) 7 f (x2 ) − f (x1 ) + f (x0 ) 1 × − × 6 12 h2 h2 2 1 h (4) 1 ̃1 − R ̃2 − R ̃3 − R ̃ 4 ), − f ′′ (x0 ) − f (x0 ) − (14R 12 144 12

where 1

4 ̃ 1 = h ∫[f (6) (x1 + sh) + f (6) (x1 − sh)](1 − s)5 ds, R 120 0

1

4 ̃ 2 = h ∫[f (6) (x2 + sh) + f (6) (x2 − sh)](1 − s)5 ds, R 120 0

1

4 ̃ 3 = h ∫[f (6) (x1 + sh) + f (6) (x1 − sh)](1 − s)3 ds, R 6 0

4

1

̃ 4 = h ∫[f (6) (x1 + sh) + f (6) (x1 − sh)](1 − s)ds. R 12 0

Proof. Combining (6.92) with (6.93), it follows: f (x2 ) − 2f (x1 ) + f (x0 ) h2 (4) ̃1, − f (x1 ) − R 12 h2 f (x3 ) − 2f (x2 ) + f (x1 ) h2 (4) ̃2, f ′′ (x2 ) = − f (x2 ) − R 12 h2 f ′′ (x2 ) − 2f ′′ (x1 ) + f ′′ (x0 ) 1 ̃ f (4) (x1 ) = − 2R 3, h2 h f ′′ (x1 ) =

1

(6.95) (6.96) (6.97)

f (4) (x2 ) − 2f (4) (x1 ) + f (4) (x0 ) 12 ̃ = ∫[f (6) (x1 + sh) + f (6) (x1 − sh)](1 − s)ds = 4 R . (6.98) 2 h 4 h 0

Substituting (6.95)–(6.96) into (6.97), we have

202 � 6 Difference methods for the Schrödinger equation 1 f (x3 ) − 2f (x2 ) + f (x1 ) h2 (4) ̃2 [ − f (x2 ) − R 12 h2 h2 f (x ) − 2f (x1 ) + f (x0 ) h2 (4) ̃ 1 ) + f ′′ (x0 )] − 1 R ̃3 − 2( 2 − f (x1 ) − R 12 h2 h2 f (x2 ) − 2f (x1 ) + f (x0 ) 1 f (x ) − 2f (x2 ) + f (x1 ) −2× ] = 2[ 3 2 h h h2 1 1 ̃ ̃ ̃ + 2 f ′′ (x0 ) + 2 (2R 1 − R2 − R3 ) h h h2 f (4) (x2 ) − 2f (4) (x1 ) + f (4) (x0 ) 1 − + f (4) (x0 ) 2 12 12 h f (x2 ) − 2f (x1 ) + f (x0 ) 1 f (x3 ) − 2f (x2 ) + f (x1 ) −2× ] = 2[ h h2 h2 1 1 1 ̃ ̃ ̃ ̃ + 2 f ′′ (x0 ) + f (4) (x0 ) + 2 (2R 1 − R2 − R3 − R4 ). 12 h h

f (4) (x1 ) =

Further substituting the above equality into (6.95), we have f (x2 ) − 2f (x1 ) + f (x0 ) h2 f (x2 ) − 2f (x1 ) + f (x0 ) 1 f (x ) − 2f (x2 ) + f (x1 ) − [ 3 −2× ] 12 h2 h2 1 h2 (4) 1 ̃ ̃ ̃3 − R ̃4) − R ̃1 − f ′′ (x0 ) − f (x0 ) − (2R − R2 − R 12 144 12 1 f (x3 ) − 2f (x2 ) + f (x1 ) 7 f (x2 ) − 2f (x1 ) + f (x0 ) 1 = × − × 6 12 h2 h2 2 1 h (4) 1 ̃1 − R ̃2 − R ̃3 − R ̃ 4 ). − f ′′ (x0 ) − f (x0 ) − (14R 12 144 12

f ′′ (x1 ) =

Lemma 6.3. Let f (x) ∈ C 6 [xm−3 , xm ]. Then we have f ′′ (xm−1 ) =

7 f (xm ) − 2f (xm−1 ) + f (xm−2 ) × 6 h2 f (xm−1 ) − 2f (xm−2 ) + f (xm−3 ) 1 × − 12 h2 2 1 h (4) 1 f (xm ) − (14R̂ 1 − R̂ 2 − R̂ 3 − R̂ 4 ). − f ′′ (xm ) − 12 144 12

Proof. The proof is similar to Lemma 6.2, which is omitted for brevity.

6.4 Fourth-order three-level linearized difference scheme

� 203

6.4.2 Derivation of the difference scheme Taking x = 0 and x = L in equation (6.1), respectively, and using the boundary value condition (6.3), we have uxx (x0 , t) = 0,

uxx (xm , t) = 0,

0 ⩽ t ⩽ T.

(6.99)

Differentiating both the right- and left-hand sides of (6.1) with respect to x twice, it follows: ̄ x) = 0 iuxt + uxxx + q(ū x u2 + 2uuu and ̄ x2 + 2uuu ̄ xx ) = 0. iuxxt + uxxxx + q(ū xx u2 + 2ū x uux + 2ū x uux + 2uu Taking x = 0 and x = L in the above equation and combining (6.3) with (6.99), it follows: uxxxx (x0 , t) = 0,

uxxxx (xm , t) = 0,

0 ⩽ t ⩽ T.

Denote Δx ujk =

1 k k (u − uj−1 ), 2h j+1

1 k (uk − 2ujk + uj−2 ) (2h)2 j+2

Δ2x ujk =

and τ û j = u(xj , 0) + ut (xj , 0), 2

1 ⩽ j ⩽ m − 1.

Considering equation (6.1) at the point (xj , t 1 ), we have 2

󵄨 󵄨2 iut (xj , t 1 ) + uxx (xj , t 1 ) + q󵄨󵄨󵄨u(xj , t 1 )󵄨󵄨󵄨 u(xj , t 1 ) = 0, 2

2

2

2

1 ⩽ j ⩽ m − 1.

Considering equation (6.1) at the node point (xj , tk ), we have 󵄨 󵄨2 iut (xj , tk ) + uxx (xj , tk ) + q󵄨󵄨󵄨u(xj , tk )󵄨󵄨󵄨 u(xj , tk ) = 0,

1 ⩽ j ⩽ m − 1, 1 ⩽ k ⩽ n − 1.

Applying Lemma 6.1 for 2 ⩽ j ⩽ m − 2, Lemma 6.2 for j = 1 and Lemma 6.3 for j = m − 1, respectively, we have 1 1 1 1 7 1 { iδt U12 + δx2 U12 − δx2 U22 + q|û 1 |2 U12 = R01 , { { 6 12 { { { 1 1 { 4 2 21 1 2 21 2 2 2 ̂ iδ U + δ U − Δ U + q| u | U = R0j , 2 ⩽ j ⩽ m − 2, j t { x j x j j j { 3 3 { { { 1 1 1 1 { { iδ U 2 + 7 δ2 U 2 − 1 δ2 U 2 + q|û |2 U 2 = R0 m−1 t m−1 x m−1 x m−2 m−1 m−1 { 6 12

(6.100) (6.101) (6.102)

204 � 6 Difference methods for the Schrödinger equation and ̄ ̄ ̄ 7 1 iΔt U1k + δx2 U1k − δx2 U2k + q|U1k |2 U1k = Rk1 , 1 ⩽ k ⩽ n − 1, { { { 6 12 { { { { 4 2 k̄ 1 2 k̄ 󵄨󵄨 k 󵄨󵄨2 k̄ k k { { { { iΔt Uj + 3 δx Uj − 3 Δx Uj + q󵄨󵄨Uj 󵄨󵄨 Uj = Rj , { 2 ⩽ j ⩽ m − 2, 1 ⩽ k ⩽ n − 1, { { { { { 1 7 󵄨 k 󵄨󵄨2 k̄ { k k̄ k̄ k { − δx2 Um−2 + q󵄨󵄨󵄨Um−1 iΔt Um−1 + δx2 Um−1 󵄨󵄨 Um−1 = Rm−1 , { { { 6 12 { 1 ⩽ k ⩽ n − 1.

(6.103)

(6.104) (6.105)

There is a constant c15 satisfying 󵄨󵄨 k 󵄨󵄨 2 4 󵄨󵄨Rj 󵄨󵄨 ⩽ c15 (τ + h ), { 󵄨 k󵄨 󵄨󵄨Δt R 󵄨󵄨 ⩽ c15 (τ 2 + h4 ), 󵄨 j󵄨

1 ⩽ j ⩽ m − 1, 0 ⩽ k ⩽ n − 1,

(6.106)

1 ⩽ j ⩽ m − 1, 2 ⩽ k ⩽ n − 2.

(6.107)

Noticing the initial-boundary value conditions (6.2)–(6.3), it follows: Uj0 = φ(xj ),

{

U0k

= 0,

Umk

= 0,

1 ⩽ j ⩽ m − 1,

(6.108)

0 ⩽ k ⩽ n.

(6.109)

Omitting the small terms in (6.100)–(6.105), another three-level linearized difference scheme for solving (6.1)–(6.3) reads { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { {

1 1 1 1 7 1 iδt u12 + δx2 u12 − δx2 u22 + q|û 1 |2 u12 = 0, 6 12 1 1 4 2 21 1 2 21 2 iδt uj + δx uj − Δx uj + q|û j |2 uj2 = 0, 2 ⩽ j ⩽ m − 2, 3 3 1 1 1 7 2 21 1 2 2 2 = 0, iδt um−1 + δx um−1 − δx2 um−2 + q|û m−1 |2 um−1 6 12 ̄ ̄ 7 1 󵄨 󵄨2 ̄ iΔt u1k + δx2 u1k − δx2 u2k + q󵄨󵄨󵄨u1k 󵄨󵄨󵄨 u1k = 0, 1 ⩽ k ⩽ n − 1, 6 12 ̄ ̄ 1 4 󵄨 󵄨2 ̄ iΔt ujk + δx2 ujk − Δ2x ujk + q󵄨󵄨󵄨ujk 󵄨󵄨󵄨 ujk = 0, 3 3 2 ⩽ j ⩽ m − 2, 1 ⩽ k ⩽ n − 1, 1 7 󵄨 k 󵄨󵄨2 k̄ k k̄ k̄ − δx2 um−2 + q󵄨󵄨󵄨um−1 iΔt um−1 + δx2 um−1 󵄨󵄨 um−1 = 0, 6 12

uj0

(6.110) (6.111) (6.112) (6.113)

(6.114)

1 ⩽ k ⩽ n − 1,

(6.115)

= φ(xj ),

(6.116)

u0k = 0,

1 ⩽ j ⩽ m − 1,

k um = 0,

0 ⩽ k ⩽ n.

(6.117)

6.4 Fourth-order three-level linearized difference scheme

� 205

6.4.3 Existence and uniqueness of the difference solution Lemma 6.4. Let w ∈ 𝒰h̊ , 7 6

2

𝒜(w) ≡ ( δx w1 −

m−2 1 2 4 1 δx w2 )w̄ 1 + ∑ ( δx2 wj − Δ2x wj )w̄ j 12 3 3 j=2

7 1 + ( δx2 wm−1 − δx2 wm−2 )w̄ m−1 . 6 12 Then 𝒜(w) is real and satisfies h𝒜(w) = −|w|21 −

4 h2 2 |w| ⩾ − |w|21 . 12 2 3

Proof. Noticing Δ2x wj =

1 2 (δ w + 2δx2 wj + δx2 wj+1 ), 4 x j−1

we have 1 1 7 1 4 2 δ w − Δ2 w = − δx2 wj−1 + δx2 wj − δx2 wj+1 . 3 x j 3 x j 12 6 12 Then 7 6

2

𝒜(w) = ( δx w1 −

m−2 1 2 1 7 1 δx w2 )w̄ 1 + ∑ (− δx2 wj−1 + δx2 wj − δx2 wj+1 )w̄ j 12 12 6 12 j=2

7 1 + ( δx2 wm−1 − δx2 wm−2 )w̄ m−1 6 12

=

7 m−1 2 1 m−2 2 1 m−1 2 ∑ (δx wj )w̄ j − ∑ (δx wj )w̄ j+1 − ∑ (δ w )w̄ 6 j=1 12 j=1 12 j=2 x j j−1 m−1

= ∑ (δx2 wj )w̄ j − j=1

1 m−1 2 ∑ (δ w )(w̄ j+1 − 2w̄ j + w̄ j−1 ) 12 j=1 x j

m−1

= − ∑ |δx wj+ 1 |2 − j=0

2

h2 m−1󵄨󵄨 2 󵄨󵄨2 ∑ 󵄨δ w 󵄨 , 12 j=1 󵄨 x j 󵄨

and h𝒜(w) = −|w|21 −

h2 2 h2 4 4 |w|2 ⩾ −|w|21 − × |w|2 = − |w|21 . 12 12 h2 1 3

Theorem 6.11. The difference scheme (6.110)–(6.117) is uniquely solvable.

206 � 6 Difference methods for the Schrödinger equation Proof. (I) From (6.116)–(6.117), we know that u0 has been determined. From (6.110)– (6.112) and (6.117), we obtain the system of linear equations in u1 . Consider its homogeneous one: { { { { { { { { { { { { { { { { { { { { { { { { {

q 1 1 7 1 i u11 + ( δx2 u11 − δx2 u21 ) + |û 1 |2 u11 = 0, τ 2 6 12 2

q 1 1 4 1 i uj1 + ( δx2 uj1 − Δ2x uj1 ) + |û j |2 uj1 = 0, τ 2 3 3 2

(6.118) 2 ⩽ j ⩽ m − 2,

(6.119)

q 1 7 1 1 1 1 1 1 + × ( δx2 um−1 − δx2 um−2 ) + |û m−1 |2 um−1 = 0, i um−1 τ 2 6 12 2

(6.120)

u01 = 0,

1 um = 0.

(6.121)

1 Multiplying (6.118) by 2hū 11 , (6.119) by 2hū j1 , (6.120) by 2hū m−1 on both the right- and lefthand sides respectively, and then adding the results together, we have m−1 2 󵄩 󵄩2 󵄨 󵄨2 i 󵄩󵄩󵄩u1 󵄩󵄩󵄩 + h𝒜(u1 ) + qh ∑ |û j |2 󵄨󵄨󵄨uj1 󵄨󵄨󵄨 = 0. τ j=1

Taking the imaginary parts of the above equality on both the right- and left-hand sides and using Lemma 6.4, we have 󵄩󵄩 1 󵄩󵄩2 󵄩󵄩u 󵄩󵄩 = 0. Therefore, u1 is uniquely determined by (6.110)–(6.112) and (6.117). (II) Now suppose uk−1 and uk have been uniquely determined. Then from (6.113)– (6.115) and (6.117), we have the system of linear equations in uk+1 at the (k + 1)-th time level. Consider its homogeneous one: { { { { { { { { { { { { { { { { { { { { { { { { { { { { {

1 k+1 1 7 2 k+1 u + ( δx u1 − 2τ 1 2 6 1 k+1 1 4 2 k+1 i uj + ( δx uj − 2τ 2 3 2 ⩽ j ⩽ m − 2,

i

q 󵄨 󵄨2 1 2 k+1 δ u ) + 󵄨󵄨󵄨u1k 󵄨󵄨󵄨 u1k+1 = 0, 12 x 2 2 q 󵄨󵄨 k 󵄨󵄨2 k+1 1 2 k+1 Δ u ) + 󵄨󵄨uj 󵄨󵄨 uj = 0, 3 x j 2

1 1 k+1 1 7 2 k+1 k+1 + ( δx um−1 − δx2 um−2 )+ i um−1 2τ 2 6 12 u0k+1 = 0,

k+1 um = 0.

q 󵄨󵄨 k 󵄨󵄨2 k+1 = 0, 󵄨u 󵄨 u 2 󵄨 m−1 󵄨 m−1

(6.122)

(6.123) (6.124) (6.125)

Multiplying both the right- and left-hand sides of (6.122) by 2hū 1k+1 , of (6.123) by 2hū jk+1 , k+1 of (6.124) by 2hū m−1 , respectively, and then adding the results together, we have m−1 1󵄩 󵄩2 󵄨 󵄨2 󵄨 󵄨2 i 󵄩󵄩󵄩uk+1 󵄩󵄩󵄩 + h𝒜(uk+1 ) + qh ∑ 󵄨󵄨󵄨ujk 󵄨󵄨󵄨 󵄨󵄨󵄨ujk+1 󵄨󵄨󵄨 = 0. τ j=1

6.4 Fourth-order three-level linearized difference scheme

Taking the imaginary parts and using Lemma 6.4, it follows that 󵄩󵄩 k+1 󵄩󵄩2 󵄩󵄩u 󵄩󵄩 = 0. Therefore, uk+1 is also uniquely determined by (6.113)–(6.115) and (6.117). By induction, the conclusion is true.

6.4.4 Conservation and boundedness of the difference solution k Lemma 6.5. Let uk = (u0k , u1k , . . . , um ) ∈ 𝒰h̊ , k

ℬ(u , u

k+1

1 7 k+ 1 k+ 1 k+ 1 ) ≡ ( δx2 u1 2 − δx2 u2 2 )(−δt ū 1 2 ) 6 12

m−2 4 1 k+ 1 k+ 1 k+ 1 + ∑ ( δx2 uj 2 − Δ2x uj 2 )(−δt ū j 2 ) 3 3 j=2

7 1 k+ 1 k+ 1 k+ 1 + ( δx2 um−12 − δx2 um−22 )(−δt ū m−12 ). 6 12 Then we have Re{hℬ(uk , uk+1 )} =

2 2 1 󵄨 󵄨2 h 󵄨 󵄨2 󵄨 󵄨2 h 󵄨 󵄨2 [(󵄨󵄨󵄨uk+1 󵄨󵄨󵄨1 + 󵄨󵄨󵄨uk+1 󵄨󵄨󵄨2 ) − (󵄨󵄨󵄨uk 󵄨󵄨󵄨1 + 󵄨󵄨󵄨uk 󵄨󵄨󵄨2 )]. 2τ 12 12

Proof. Direct calculation yields k

ℬ(u , u

k+1

7 1 k+ 1 k+ 1 k+ 1 ) = ( δx2 u1 2 − δx2 u2 2 )(−δt ū 1 2 ) 6 12 m−2

+ ∑ (− j=2

1 2 k+ 21 7 2 k+ 21 1 k+ 1 k+ 1 δx uj−1 + δx uj − δx2 uj+1 2 )(−δt ū j 2 ) 12 6 12

7 1 k+ 1 k+ 1 k+ 1 + ( δx2 um−12 − δx2 um−22 )(−δt ū m−12 ) 6 12

m−1

k+ 21

= ∑ (δx2 uj j=1

−

k+ 21

)(−δt ū j

)−

1 m−1 2 k+ 21 k+ 1 ∑ (δx uj−1 )(−δt ū j 2 ) 12 j=2

2 m−1 2 k+ 21 1 m−2 2 k+ 21 k+ 1 k+ 1 ∑ (δx uj+1 )(−δt ū j 2 ) + ∑ (δx uj )(−δt ū j 2 ) 12 j=1 12 j=1

m−1

k+ 21

= ∑ (δx2 uj j=1

m−1

k+ 21

)(−δt ū j k+ 21

− 2 ∑ (δx2 uj j=1

)− k+ 21

)(−δt ū j

1 m−2 2 k+ 21 k+ 1 { ∑ (δx uj )(−δt ū j+1 2 ) 12 j=1 m−1

k+ 21

) + ∑ (δx2 uj j=2

k+ 1

)(−δt ū j−1 2 )}

� 207

208 � 6 Difference methods for the Schrödinger equation m−1

k+ 21

= ∑ (δx2 uj j=1

m−1

= ∑ (δx u j=0

k+ 21

j+ 21

k+ 21

)(−δt ū j

)(δt δx ū

)+

k+ 21

j+ 21

1 m−1 2 k+ 21 k+ 1 k+ 1 k+ 1 ∑ (δx uj )δt (ū j+1 2 − 2ū j 2 + ū j−1 2 ) 12 j=1

)+

h2 m−1 2 k+ 21 k+ 1 ∑ (δx uj )δt (δx2 ū j 2 ). 12 j=1

Multiplying both the right- and left-hand sides of the above equality by h, and then taking the real parts yield Re{hℬ(uk , uk+1 )} 1 󵄨󵄨 k+1 󵄨󵄨2 󵄨󵄨 k 󵄨󵄨2 h2 1 󵄨󵄨 k+1 󵄨󵄨2 󵄨󵄨 k 󵄨󵄨2 (󵄨󵄨u 󵄨󵄨1 − 󵄨󵄨u 󵄨󵄨1 ) + ⋅ (󵄨u 󵄨󵄨2 − 󵄨󵄨u 󵄨󵄨2 ) 2τ 12 2τ 󵄨 2 2 1 󵄨 󵄨2 h 󵄨 󵄨2 󵄨 󵄨2 h 󵄨 󵄨2 = [(󵄨󵄨󵄨uk+1 󵄨󵄨󵄨1 + 󵄨󵄨󵄨uk+1 󵄨󵄨󵄨2 ) − (󵄨󵄨󵄨uk 󵄨󵄨󵄨1 + 󵄨󵄨󵄨uk 󵄨󵄨󵄨2 )]. 2τ 12 12 =

Similarly, the following conclusion is obtained. k Lemma 6.6. Let uk = (u0k , u1k , . . . , um ) ∈ 𝒰h̊ ,

𝒞 (u

k−1

̄ ̄ 1 7 , uk+1 ) ≡ ( δx2 u1k − δx2 u2k )(−Δt ū 1k ) 6 12

m−2 ̄ ̄ 1 4 + ∑ ( δx2 ujk − Δ2x ujk )(−Δt ū jk ) 3 3 j=2

7 1 k k̄ k̄ + ( δx2 um−1 ). − δx2 um−2 )(−Δt ū m−1 6 12 Then we have Re{h𝒞 (uk−1 , uk+1 )}

=

2 2 1 󵄨 󵄨2 h 󵄨 󵄨2 󵄨2 󵄨 󵄨2 h 󵄨 [(󵄨󵄨󵄨uk+1 󵄨󵄨󵄨1 + 󵄨󵄨󵄨uk+1 󵄨󵄨󵄨2 ) − (󵄨󵄨󵄨uk−1 󵄨󵄨󵄨1 + 󵄨󵄨󵄨uk−1 󵄨󵄨󵄨2 )]. 4τ 12 12

Theorem 6.12. Suppose {ujk | 0 ⩽ j ⩽ m, 0 ⩽ k ⩽ n} is the solution of the difference scheme (6.110)–(6.117). Denote 2 2 q m−1󵄨 󵄨2 󵄨 1 󵄨 󵄨2 󵄨2 h 󵄨 󵄨2 󵄨 󵄨2 h 󵄨 󵄨2 E k = (󵄨󵄨󵄨uk+1 󵄨󵄨󵄨1 + 󵄨󵄨󵄨uk+1 󵄨󵄨󵄨2 + 󵄨󵄨󵄨uk 󵄨󵄨󵄨1 + 󵄨󵄨󵄨uk 󵄨󵄨󵄨2 ) − h ∑ 󵄨󵄨󵄨ujk 󵄨󵄨󵄨 󵄨󵄨󵄨ujk+1 󵄨󵄨󵄨 . 2 12 12 2 j=1

Then we have 󵄩󵄩 k 󵄩󵄩 󵄩󵄩 0 󵄩󵄩 󵄩󵄩u 󵄩󵄩 = 󵄩󵄩u 󵄩󵄩, 0 ⩽ k ⩽ n, q m−1 1 󵄨󵄨 1 󵄨󵄨2 h2 󵄨󵄨 1 󵄨󵄨2 󵄨󵄨 0 󵄨󵄨2 h2 󵄨󵄨 0 󵄨󵄨2 󵄨 󵄨2 (󵄨󵄨u 󵄨󵄨1 + 󵄨󵄨u 󵄨󵄨2 + 󵄨󵄨u 󵄨󵄨1 + 󵄨󵄨u 󵄨󵄨2 ) − h ∑ |û j |2 󵄨󵄨󵄨uj1 󵄨󵄨󵄨 2 12 12 2 j=1

(6.126)

� 209

6.4 Fourth-order three-level linearized difference scheme m−1 2 󵄨 󵄨2 h 󵄨 󵄨2 q 󵄨 󵄨2 = 󵄨󵄨󵄨u0 󵄨󵄨󵄨1 + 󵄨󵄨󵄨u0 󵄨󵄨󵄨2 − h ∑ |û j |2 󵄨󵄨󵄨uj0 󵄨󵄨󵄨 , 12 2 j=1

Ek = E0 ,

(6.127)

1 ⩽ k ⩽ n − 1.

(6.128) 1

Proof. (I) Multiplying both the right- and left-hand sides of (6.110) by hū 12 , of (6.111) by 1

1

2 hū j2 , of (6.112) by hū m−1 , respectively, and then adding the results together, we have

m−1

1 1 1 󵄨 1 󵄨2 i(δt u 2 , u 2 ) + h𝒜(u 2 ) + qh ∑ |û j |2 󵄨󵄨󵄨uj2 󵄨󵄨󵄨 = 0.

j=1

Taking the imaginary parts on both the right- and left-hand sides of the above equality and utilizing Lemma 6.4, we have 1 󵄩󵄩 1 󵄩󵄩2 󵄩󵄩 0 󵄩󵄩2 (󵄩u 󵄩 − 󵄩u 󵄩 ) = 0, 2τ 󵄩 󵄩 󵄩 󵄩 which implies 󵄩󵄩 1 󵄩󵄩 󵄩󵄩 0 󵄩󵄩 󵄩󵄩u 󵄩󵄩 = 󵄩󵄩u 󵄩󵄩.

(6.129) ̄

̄

Multiplying both the right- and left-hand sides of (6.113) by hū 1k , of (6.114) by hū jk , of ̄

k (6.115) by hū m−1 , respectively, and then adding the results together, we have m−1

̄ ̄ 󵄨 󵄨2 󵄨 ̄ 󵄨2 i(Δt uk , uk ) + h𝒜(uk ) + qh ∑ 󵄨󵄨󵄨ujk 󵄨󵄨󵄨 󵄨󵄨󵄨ujk 󵄨󵄨󵄨 = 0. j=1

Taking the imaginary parts on both the right- and left-hand sides of the above equality and using Lemma 6.4 again, we have 1 󵄩󵄩 k+1 󵄩󵄩2 󵄩󵄩 k−1 󵄩󵄩2 (󵄩u 󵄩󵄩 − 󵄩󵄩u 󵄩󵄩 ) = 0, 4τ 󵄩

1 ⩽ k ⩽ n − 1.

Combining (6.129) with (6.130), one yields (6.126).

(6.130) 1

(II) Multiplying both the right- and left-hand sides of (6.110) by −hδt ū 12 , of (6.111) 1

1

2 by −hδt ū j2 , of (6.112) by −hδt ū m−1 , respectively, and then adding the results together, we have

m−1

1 1 1 2 󵄩 󵄩 −i󵄩󵄩󵄩δt u 2 󵄩󵄩󵄩 + hℬ(u0 , u1 ) + qh ∑ |û j |2 uj2 (−δt ū j2 ) = 0.

j=1

Taking the real parts of the above equality and using Lemma 6.5, we have m−1 2 2 |uj1 |2 − |uj0 |2 1 󵄨 󵄨2 h 󵄨 󵄨2 󵄨 󵄨2 h 󵄨 󵄨2 [(󵄨󵄨󵄨u1 󵄨󵄨󵄨1 + 󵄨󵄨󵄨u1 󵄨󵄨󵄨2 ) − (󵄨󵄨󵄨u0 󵄨󵄨󵄨1 + 󵄨󵄨󵄨u0 󵄨󵄨󵄨2 )] − qh ∑ |û j |2 = 0. 2τ 12 12 2τ j=1

210 � 6 Difference methods for the Schrödinger equation Therefore, it follows that m−1 m−1 2 2 󵄨 0 󵄨2 h 󵄨 0 󵄨2 󵄨󵄨 1 󵄨󵄨2 h 󵄨󵄨 1 󵄨󵄨2 2 󵄨 1 󵄨2 2 󵄨 0 󵄨2 󵄨󵄨u 󵄨󵄨1 + 󵄨󵄨u 󵄨󵄨2 − qh ∑ |û j | 󵄨󵄨󵄨uj 󵄨󵄨󵄨 = 󵄨󵄨󵄨u 󵄨󵄨󵄨1 + 󵄨󵄨󵄨u 󵄨󵄨󵄨2 − qh ∑ |û j | 󵄨󵄨󵄨uj 󵄨󵄨󵄨 , 12 12 j=1 j=1

which is exactly (6.127). (III) Multiplying both the right- and left-hand sides of (6.113) by −hΔt ū 1k , of (6.114) k , respectively, and then adding the results together, we by −hΔt ū jk , of (6.115) by −hΔt ū m−1 have m−1

󵄩 󵄩2 󵄨 󵄨2 ̄ −i󵄩󵄩󵄩Δt uk 󵄩󵄩󵄩 + h𝒞 (uk−1 , uk+1 ) − qh ∑ 󵄨󵄨󵄨ujk 󵄨󵄨󵄨 ujk Δt ū jk = 0, j=1

1 ⩽ k ⩽ n − 1.

Taking the real parts on both the right- and left-hand sides of the above equality and using Lemma 6.6, we have 2 2 1 󵄨 󵄨2 󵄨2 󵄨2 h 󵄨 󵄨 󵄨2 h 󵄨 [(󵄨󵄨󵄨uk+1 󵄨󵄨󵄨1 + 󵄨󵄨󵄨uk+1 󵄨󵄨󵄨2 ) − (󵄨󵄨󵄨uk−1 󵄨󵄨󵄨1 + 󵄨󵄨󵄨uk−1 󵄨󵄨󵄨2 )] 4τ 12 12

−

m−1 1 󵄨 󵄨2 󵄨 󵄨2 󵄨 󵄨2 qh ∑ 󵄨󵄨󵄨ujk 󵄨󵄨󵄨 (󵄨󵄨󵄨ujk+1 󵄨󵄨󵄨 − 󵄨󵄨󵄨ujk−1 󵄨󵄨󵄨 ) = 0, 4τ j=1

1 ⩽ k ⩽ n − 1.

Rearranging the above equality leads to |u 1 [( 2τ −(

k+1 2 |1

h2 k+1 2 |u |2 12

+

|uk |21 +

2

2

h |uk |22 12

+ |uk |21 +

+ |uk−1 |21 + 2

h2 k 2 |u |2 12

h2 k−1 2 |u |2 12

q m−1󵄨 󵄨2 󵄨 󵄨2 − h ∑ 󵄨󵄨󵄨ujk 󵄨󵄨󵄨 󵄨󵄨󵄨ujk+1 󵄨󵄨󵄨 ) 2 j=1 q m−1󵄨 󵄨2 󵄨 󵄨2 − h ∑ 󵄨󵄨󵄨ujk−1 󵄨󵄨󵄨 󵄨󵄨󵄨ujk 󵄨󵄨󵄨 )] = 0, 2 j=1

1 ⩽ k ⩽ n − 1, which implies (6.128). Applying Theorem 6.12, there is a constant c16 satisfying 󵄩󵄩 k 󵄩󵄩 󵄩󵄩u 󵄩󵄩∞ ⩽ c16 ,

0 ⩽ k ⩽ n.

Remark 6.2. If one takes û j = uj0 ,

0⩽j⩽m

in (6.110)–(6.112), then we have m−1 2 󵄨 󵄨2 h 󵄨 󵄨2 q 󵄨 󵄨4 E k = 󵄨󵄨󵄨u0 󵄨󵄨󵄨1 + 󵄨󵄨󵄨u0 󵄨󵄨󵄨2 − h ∑ 󵄨󵄨󵄨uj0 󵄨󵄨󵄨 , 12 2 j=1

0 ⩽ k ⩽ n − 1.

The corresponding difference scheme is second-order convergent in the L2 -norm.

6.4 Fourth-order three-level linearized difference scheme

� 211

6.4.5 Convergence of the difference solution Theorem 6.13. Suppose {Ujk | 0 ⩽ j ⩽ m, 0 ⩽ k ⩽ n} is the solution of the problem (6.1)–(6.3) and {ujk | 0 ⩽ j ⩽ m, 0 ⩽ k ⩽ n} is the solution of the difference scheme (6.110)–(6.117). Denote ejk = Ujk − ujk ,

0 ⩽ j ⩽ m, 0 ⩽ k ⩽ n.

Then there is a constant c17 satisfying 󵄩󵄩 k 󵄩󵄩 2 4 󵄩󵄩e 󵄩󵄩 ⩽ c17 (τ + h ),

0 ⩽ k ⩽ n.

(6.131)

Proof. Subtracting (6.110)–(6.117) from (6.100)–(6.105), (6.108)–(6.109), respectively, we have the system of error equations: { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { {

1 1 1 1 7 1 iδt e12 + δx2 e12 − δx2 e22 + q|û 1 |2 e12 = R01 , 6 12 1 1 1 1 4 1 2 iδt ej2 + δx ej2 − Δ2x ej2 + q|û j |2 ej2 = R0j , 2 ⩽ j ⩽ m − 2, 3 3 1 1 1 1 7 1 2 2 2 2 2 + δx em−1 − δx2 em−2 = R0m−1 , + q|û m−1 |2 em−1 iδt em−1 6 12 ̄ ̄ 7 1 󵄨 󵄨2 ̄ 󵄨 󵄨2 ̄ iΔt e1k + δx2 e1k − δx2 e2k + q(󵄨󵄨󵄨U1k 󵄨󵄨󵄨 U1k − 󵄨󵄨󵄨u1k 󵄨󵄨󵄨 u1k ) = Rk1 , 6 12 1 ⩽ k ⩽ n − 1, ̄ ̄ 1 4 󵄨 󵄨2 ̄ 󵄨 󵄨2 ̄ iΔt ejk + δx2 ejk − Δ2x ejk + q(󵄨󵄨󵄨Ujk 󵄨󵄨󵄨 Ujk − 󵄨󵄨󵄨ujk 󵄨󵄨󵄨 ujk ) = Rkj , 3 3 2 ⩽ j ⩽ m − 2, 1 ⩽ k ⩽ n − 1, 1 7 k̄ k̄ k − δx2 em−2 iΔt em−1 + δx2 em−1 6 12 󵄨 k 󵄨󵄨2 k̄ 󵄨 k 󵄨2 k̄ k + q(󵄨󵄨󵄨Um−1 󵄨󵄨 Um−1 − 󵄨󵄨󵄨um−1 󵄨󵄨󵄨 um−1 ) = Rm−1 , 1 ⩽ k ⩽ n − 1, ej0 = 0, 1 ⩽ j ⩽ m − 1,

e0k

= 0,

k em

= 0,

0 ⩽ k ⩽ n.

(6.132) (6.133) (6.134)

(6.135) (6.136)

(6.137) (6.138) (6.139)

Using the preceding notation and conclusions, we have c0 =

max

󵄨󵄨 󵄨 󵄨u(x, t)󵄨󵄨󵄨,

0⩽x⩽L,0⩽t⩽T 󵄨

󵄨 󵄨 c13 = max 󵄨󵄨󵄨ut (x, 0)󵄨󵄨󵄨, 0⩽x⩽L

τ ‖u‖̂ ∞ ⩽ c0 + c13 ⩽ c0 + c13 , 2 󵄩󵄩 k 󵄩󵄩 󵄩󵄩u 󵄩󵄩∞ ⩽ c16 .

1

1

(I) Multiplying both the right- and left-hand sides of (6.132) by hē12 , of (6.133) by hēj2 , 1

2 of (6.134) by hēm−1 , respectively, and then adding the results together, we have

212 � 6 Difference methods for the Schrödinger equation m−1

1 1 1 1 󵄨 1 󵄨2 i(δt e 2 , e 2 ) + h𝒜(e 2 ) + qh ∑ |û j |2 󵄨󵄨󵄨ej2 󵄨󵄨󵄨 = (R0 , e 2 ).

j=1

Taking the imaginary parts on both the right- and left-hand sides of the above equality and using Lemma 6.4, it follows that 1 1 󵄩󵄩 1 󵄩󵄩2 󵄩󵄩 0 󵄩󵄩2 󵄩 󵄩 󵄩 1󵄩 (󵄩e 󵄩 − 󵄩e 󵄩 ) = Im{(R0 , e 2 )} ⩽ 󵄩󵄩󵄩R0 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩e 2 󵄩󵄩󵄩, 2τ 󵄩 󵄩 󵄩 󵄩

which implies 1 󵄩󵄩 1 󵄩󵄩2 󵄩󵄩 0 󵄩󵄩 󵄩e 󵄩 ⩽ 󵄩󵄩R 󵄩󵄩 ⋅ 2τ 󵄩 󵄩

1 󵄩󵄩 1 󵄩󵄩 󵄩e 󵄩 2󵄩 󵄩

by noticing e0 = 0. Therefore, we have 󵄩󵄩 1 󵄩󵄩 󵄩 0󵄩 2 4 󵄩󵄩e 󵄩󵄩 ⩽ τ 󵄩󵄩󵄩R 󵄩󵄩󵄩 ⩽ τ √Lc15 (τ + h ), where (6.106) has been used. ̄ ̄ (II) Multiplying both the right- and left-hand sides of (6.135) by hē1k , of (6.136) by hējk , ̄

k of (6.137) by hēm−1 , respectively, and then adding the results together, we have ̄ ̄ ̄ 󵄨 󵄨2 ̄ 󵄨 󵄨2 ̄ ̄ i(Δt ek , ek ) + h𝒜(ek ) + q(󵄨󵄨󵄨U k 󵄨󵄨󵄨 U k − 󵄨󵄨󵄨uk 󵄨󵄨󵄨 uk , ek ) = (Rk , ek ),

1 ⩽ k ⩽ n − 1.

(6.140)

Noticing 󵄨󵄨 k 󵄨󵄨2 k̄ 󵄨󵄨 k 󵄨󵄨2 k̄ 󵄨󵄨Uj 󵄨󵄨 Uj − 󵄨󵄨uj 󵄨󵄨 uj ̄ ̄ ̄ 󵄨 󵄨2 󵄨 󵄨2 󵄨 󵄨2 = 󵄨󵄨󵄨ujk 󵄨󵄨󵄨 (Ujk − ujk ) + (󵄨󵄨󵄨Ujk 󵄨󵄨󵄨 − 󵄨󵄨󵄨ujk 󵄨󵄨󵄨 )Ujk ̄ 󵄨 󵄨2 ̄ = 󵄨󵄨󵄨ujk 󵄨󵄨󵄨 ejk + (ejk Ū jk + ujk ējk )Ujk , then taking the imaginary parts on both the right- and left-hand sides of (6.140) and using Lemma 6.4 again, we have m−1 ̄ ̄ 1 󵄩󵄩 k+1 󵄩󵄩2 󵄩󵄩 k−1 󵄩󵄩2 (󵄩󵄩e 󵄩󵄩 − 󵄩󵄩e 󵄩󵄩 ) + Im{qh ∑ (ejk Ū jk + ujk ējk )Ujk ējk } 4τ j=1 ̄

= Im(Rk , ek ),

1 ⩽ k ⩽ n − 1.

The above equality further implies 1 󵄩󵄩 k+1 󵄩󵄩2 󵄩󵄩 k−1 󵄩󵄩2 (󵄩e 󵄩󵄩 − 󵄩󵄩e 󵄩󵄩 ) 4τ 󵄩 󵄩 󵄩 󵄩 ̄󵄩 󵄩 󵄩 󵄩 ̄󵄩 ⩽ |q|c0 (c0 + c16 )󵄩󵄩󵄩ek 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩ek 󵄩󵄩󵄩 + 󵄩󵄩󵄩Rk 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩ek 󵄩󵄩󵄩 k+1 k−1 󵄩 󵄩 󵄩 󵄩 ‖e ‖ + ‖e ‖ ⩽ (|q|c0 (c0 + c16 )󵄩󵄩󵄩ek 󵄩󵄩󵄩 + 󵄩󵄩󵄩Rk 󵄩󵄩󵄩) , 2

1 ⩽ k ⩽ n − 1.

� 213

6.4 Fourth-order three-level linearized difference scheme

Dividing the above inequality by 21 (‖ek+1 ‖ + ‖ek−1 ‖) on both the right- and left-hand sides, it follows: 1 󵄩󵄩 k+1 󵄩󵄩 󵄩󵄩 k−1 󵄩󵄩 󵄩 󵄩 󵄩 󵄩 (󵄩e 󵄩󵄩 − 󵄩󵄩e 󵄩󵄩) ⩽ |q|c0 (c0 + c16 )󵄩󵄩󵄩ek 󵄩󵄩󵄩 + 󵄩󵄩󵄩Rk 󵄩󵄩󵄩, 2τ 󵄩

1 ⩽ k ⩽ n − 1,

󵄩 k󵄩 󵄩 k󵄩 󵄩󵄩 k+1 󵄩󵄩 󵄩󵄩 k−1 󵄩󵄩 󵄩󵄩e 󵄩󵄩 ⩽ 󵄩󵄩e 󵄩󵄩 + 2|q|c0 (c0 + c16 )τ 󵄩󵄩󵄩e 󵄩󵄩󵄩 + 2τ 󵄩󵄩󵄩R 󵄩󵄩󵄩,

1 ⩽ k ⩽ n − 1,

or

which implies by noticing (6.106) that 󵄩 󵄩󵄩 󵄩 max{󵄩󵄩󵄩ek+1 󵄩󵄩󵄩, 󵄩󵄩󵄩ek 󵄩󵄩󵄩} 󵄩 󵄩󵄩 󵄩 󵄩 󵄩 ⩽ [1 + 2|q|c0 (c0 + c16 )τ] max{󵄩󵄩󵄩ek 󵄩󵄩󵄩, 󵄩󵄩󵄩ek−1 󵄩󵄩󵄩} + 2τ 󵄩󵄩󵄩Rk 󵄩󵄩󵄩 󵄩 󵄩󵄩 󵄩 ⩽ [1 + 2|q|c0 (c0 + c16 )τ] max{󵄩󵄩󵄩ek 󵄩󵄩󵄩, 󵄩󵄩󵄩ek−1 󵄩󵄩󵄩} + 2√Lc15 τ(τ 2 + h4 ),

1 ⩽ k ⩽ n − 1.

When q = 0, using the recursion of 󵄩 󵄩󵄩 󵄩 󵄩 󵄩󵄩 󵄩 max{󵄩󵄩󵄩ek+1 󵄩󵄩󵄩, 󵄩󵄩󵄩ek 󵄩󵄩󵄩} ⩽ max{󵄩󵄩󵄩ek 󵄩󵄩󵄩, 󵄩󵄩󵄩ek−1 󵄩󵄩󵄩} + 2√Lc15 τ(τ 2 + h4 ),

1 ⩽ k ⩽ n − 1,

one yields 󵄩 󵄩󵄩 󵄩 max{󵄩󵄩󵄩ek+1 󵄩󵄩󵄩, 󵄩󵄩󵄩ek 󵄩󵄩󵄩} ⩽ 2√Lc15 T(τ 2 + h4 ) ⩽ c17 (τ 2 + h4 ),

0 ⩽ k ⩽ n − 1.

When q ≠ 0, it follows from the Gronwall inequality (Theorem 1.2(a)) that √Lc15 (τ 2 + h4 ) 󵄩 󵄩󵄩 󵄩 󵄩 󵄩󵄩 󵄩 max{󵄩󵄩󵄩ek+1 󵄩󵄩󵄩, 󵄩󵄩󵄩ek 󵄩󵄩󵄩} ⩽ e2|q|c0 (c0 +c16 )T [max{󵄩󵄩󵄩e1 󵄩󵄩󵄩, 󵄩󵄩󵄩e0 󵄩󵄩󵄩} + ] |q|c0 (c0 + c16 ) ⩽ c17 (τ 2 + h4 ),

0 ⩽ k ⩽ n − 1.

Hence, (6.131) is true. Theorem 6.14. Suppose {Ujk | 0 ⩽ j ⩽ m, 0 ⩽ k ⩽ n} is the solution of the problem

(6.1)–(6.3) and {ujk | 0 ⩽ j ⩽ m, 0 ⩽ k ⩽ n} is the solution of the difference scheme (6.110)–(6.117). Denote ejk = Ujk − ujk ,

0 ⩽ j ⩽ m, 0 ⩽ k ⩽ n.

Then there is a constant c18 satisfying 󵄨󵄨 k 󵄨󵄨 2 4 󵄨󵄨e 󵄨󵄨1 ⩽ c18 (τ + h ),

0 ⩽ k ⩽ n. 1

Proof. (I) Multiplying both the right- and left-hand sides of (6.132) by −hδt ē12 , of (6.133) 1

1

2 by −hδt ēj2 , of (6.134) by −hδt ēm−1 , respectively, and then adding the results together, we have

214 � 6 Difference methods for the Schrödinger equation m−1

1 1 1 󵄩 1 󵄩2 −i󵄩󵄩󵄩δt e 2 󵄩󵄩󵄩 + hℬ(e0 , e1 ) − h ∑ |û j |2 ej2 δt ēj2 = −h(R0 , δt e 2 ).

j=1

Taking the real parts on both the right- and left-hand sides of the above equality and with the application of Lemma 6.5, we have m−1 2 2 |e1 |2 − |ej0 |2 1 󵄨󵄨 0 󵄨󵄨2 h 󵄨󵄨 0 󵄨󵄨2 󵄨󵄨 1 󵄨󵄨2 h 󵄨󵄨 1 󵄨󵄨2 2 j ̂ [(󵄨e 󵄨 + 󵄨e 󵄨 ) − (󵄨󵄨e 󵄨󵄨1 + 󵄨󵄨e 󵄨󵄨2 )] − h ∑ |uj | 2τ 󵄨 󵄨1 12 󵄨 󵄨2 12 2τ j=1 1

= Re{−h(R0 , δt e 2 )} 󵄩 󵄩 󵄩 1󵄩 ⩽ 󵄩󵄩󵄩R0 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩δt e 2 󵄩󵄩󵄩. Noticing e0 = 0, it follows: 1 󵄨󵄨 1 󵄨󵄨2 h2 󵄨󵄨 1 󵄨󵄨2 1 m−1 2 󵄨󵄨 1 󵄨󵄨2 (󵄨󵄨e 󵄨󵄨1 + 󵄨󵄨e 󵄨󵄨2 ) ⩽ h ⋅ ∑ |û | 󵄨e 󵄨 + 2τ 12 2τ j=1 j 󵄨 j 󵄨

1 󵄩󵄩 0 󵄩󵄩 󵄩󵄩 1 󵄩󵄩 󵄩R 󵄩 ⋅ 󵄩e 󵄩, τ󵄩 󵄩 󵄩 󵄩

which implies m−1

󵄨󵄨 1 󵄨󵄨2 󵄩 0󵄩 󵄩 1󵄩 2 󵄨 1 󵄨2 󵄨󵄨e 󵄨󵄨1 ⩽ h ∑ |û j | 󵄨󵄨󵄨ej 󵄨󵄨󵄨 + 2󵄩󵄩󵄩R 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩e 󵄩󵄩󵄩 j=1

󵄩 󵄩2 󵄩 󵄩 󵄩 󵄩 ⩽ (c0 + c13 )2 󵄩󵄩󵄩e1 󵄩󵄩󵄩 + 2󵄩󵄩󵄩R0 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩e1 󵄩󵄩󵄩 2

2 ⩽ (c0 + c13 )2 c17 (τ 2 + h4 ) + 2c15 √L(τ 2 + h4 )c17 (τ 2 + h4 ) 2

2 = [(c0 + c13 )2 c17 + 2√Lc15 c17 ](τ 2 + h4 ) .

(6.141)

In the third inequality, we have used (6.106) and (6.131). (II) Multiplying both the right- and left-hand sides of (6.135) by −hΔt ē1k , of (6.136) by k −hΔt ējk , of (6.137) by −hΔt ēm−1 , respectively, and then adding the results together, we have m−1

󵄩 󵄨 󵄨2 ̄ 󵄨 󵄨2 ̄ 󵄩2 − i󵄩󵄩󵄩Δt ek 󵄩󵄩󵄩 + h𝒞 (ek−1 , ek+1 ) − qh ∑ (󵄨󵄨󵄨Ujk 󵄨󵄨󵄨 Ujk − 󵄨󵄨󵄨ujk 󵄨󵄨󵄨 ujk )Δt ējk j=1

k

k

= − (R , Δt e ),

1 ⩽ k ⩽ n − 1.

Taking the real parts of the above equality and using Lemma 6.6, we have 2 2 1 󵄨 󵄨2 h 󵄨 󵄨2 󵄨 󵄨2 󵄨2 h 󵄨 [(󵄨󵄨󵄨ek+1 󵄨󵄨󵄨1 + 󵄨󵄨󵄨ek+1 󵄨󵄨󵄨2 ) − (󵄨󵄨󵄨ek−1 󵄨󵄨󵄨1 + 󵄨󵄨󵄨ek−1 󵄨󵄨󵄨2 )] 4τ 12 12 m−1

󵄨 󵄨2 ̄ 󵄨 󵄨2 ̄ = q Re{h ∑ (󵄨󵄨󵄨Ujk 󵄨󵄨󵄨 Ujk − 󵄨󵄨󵄨ujk 󵄨󵄨󵄨 ujk )Δt ējk } + Re{−(Rk , Δt ek )}, j=1

With the help of (6.135)–(6.137), we have

1 ⩽ k ⩽ n − 1. (6.142)

6.4 Fourth-order three-level linearized difference scheme

� 215

m−1

󵄨 󵄨2 ̄ 󵄨 󵄨2 ̄ q Re{h ∑ (󵄨󵄨󵄨Ujk 󵄨󵄨󵄨 Ujk − 󵄨󵄨󵄨ujk 󵄨󵄨󵄨 ujk )Δt ējk } j=1

m−1

󵄨 󵄨2 ̄ 󵄨 󵄨2 ̄ = q Re{h ∑ (󵄨󵄨󵄨Ujk 󵄨󵄨󵄨 Ū jk − 󵄨󵄨󵄨ujk 󵄨󵄨󵄨 ū jk )Δt ejk } j=1

m−1

󵄨 󵄨2 ̄ 󵄨 󵄨2 ̄ = q Im{h ∑ (󵄨󵄨󵄨Ujk 󵄨󵄨󵄨 Ū jk − 󵄨󵄨󵄨ujk 󵄨󵄨󵄨 ū jk )iΔt ejk } j=1

m−2

󵄨 󵄨2 ̄ 󵄨 󵄨2 ̄ 󵄨 󵄨2 ̄ 󵄨 󵄨2 ̄ = qh Im{(󵄨󵄨󵄨U1k 󵄨󵄨󵄨 Ū 1k − 󵄨󵄨󵄨u1k 󵄨󵄨󵄨 ū 1k )iΔt e1k + ∑ (󵄨󵄨󵄨Ujk 󵄨󵄨󵄨 Ū jk − 󵄨󵄨󵄨ujk 󵄨󵄨󵄨 ū jk )iΔt ejk j=2

󵄨 k 󵄨󵄨2 ̄ k̄ 󵄨 k 󵄨2 k̄ k + (󵄨󵄨󵄨Um−1 󵄨󵄨 Um−1 − 󵄨󵄨󵄨um−1 󵄨󵄨󵄨 ū m−1 )iΔt em−1 } ̄ ̄ 7 1 󵄨 󵄨2 ̄ 󵄨 󵄨2 ̄ 󵄨 󵄨2 ̄ 󵄨 󵄨2 ̄ = qh Im{(󵄨󵄨󵄨U1k 󵄨󵄨󵄨 Ū 1k − 󵄨󵄨󵄨u1k 󵄨󵄨󵄨 ū 1k )[Rk1 − ( δx2 e1k − δx2 e2k + q(󵄨󵄨󵄨U1k 󵄨󵄨󵄨 U1k − 󵄨󵄨󵄨u1k 󵄨󵄨󵄨 u1k ))] 6 12 m−2 ̄ ̄ 1 4 󵄨 󵄨2 ̄ 󵄨 󵄨2 ̄ 󵄨 󵄨2 ̄ 󵄨 󵄨2 ̄ + ∑ (󵄨󵄨󵄨Ujk 󵄨󵄨󵄨 Ū jk − 󵄨󵄨󵄨ujk 󵄨󵄨󵄨 ū jk )[Rkj − ( δx2 ejk − Δ2x ejk + q(󵄨󵄨󵄨Ujk 󵄨󵄨󵄨 Ujk − 󵄨󵄨󵄨ujk 󵄨󵄨󵄨 ujk ))] 3 3 j=2

󵄨 k 󵄨󵄨2 ̄ k̄ 󵄨 k 󵄨2 k̄ + (󵄨󵄨󵄨Um−1 󵄨󵄨 Um−1 − 󵄨󵄨󵄨um−1 󵄨󵄨󵄨 ū m−1 ) 1 7 󵄨 k 󵄨󵄨2 k̄ 󵄨 k 󵄨2 k̄ k̄ k̄ − δx2 em−2 + q(󵄨󵄨󵄨Um−1 ⋅ [Rkm−1 − ( δx2 em−1 󵄨󵄨 Um−1 − 󵄨󵄨󵄨um−1 󵄨󵄨󵄨 um−1 ))]} 6 12 ̄ ̄ 1 7 󵄨 󵄨2 ̄ 󵄨 󵄨2 ̄ = qh Im{(󵄨󵄨󵄨U1k 󵄨󵄨󵄨 Ū 1k − 󵄨󵄨󵄨u1k 󵄨󵄨󵄨 ū 1k )[Rk1 − ( δx2 e1k − δx2 e2k )] 6 12 m−2 ̄ ̄ 1 4 󵄨 󵄨2 ̄ 󵄨 󵄨2 ̄ + ∑ (󵄨󵄨󵄨Ujk 󵄨󵄨󵄨 Ū jk − 󵄨󵄨󵄨ujk 󵄨󵄨󵄨 ū jk )[Rkj − ( δx2 ejk − Δ2x ejk )] 3 3 j=2

1 2 k̄ 7 2 k̄ 󵄨 k 󵄨󵄨2 ̄ k̄ 󵄨 k 󵄨2 k̄ k + (󵄨󵄨󵄨Um−1 󵄨󵄨 Um−1 − 󵄨󵄨󵄨um−1 󵄨󵄨󵄨 ū m−1 )[Rm−1 − ( δx em−1 − δx em−2 )]} 6 12 󵄩 󵄩2 󵄩 󵄩2 󵄩 󵄩2 󵄨 󵄨2 󵄨 󵄨2 󵄨 󵄨2 󵄩 󵄩2 ⩽ c19 (󵄩󵄩󵄩ek+1 󵄩󵄩󵄩 + 󵄩󵄩󵄩ek 󵄩󵄩󵄩 + 󵄩󵄩󵄩ek−1 󵄩󵄩󵄩 + 󵄨󵄨󵄨ek+1 󵄨󵄨󵄨1 + 󵄨󵄨󵄨ek 󵄨󵄨󵄨1 + 󵄨󵄨󵄨ek−1 󵄨󵄨󵄨1 + 󵄩󵄩󵄩Rk 󵄩󵄩󵄩 ), 1 ⩽ k ⩽ n − 1.

(6.143)

Similar to the derivation of (6.91), we have 󵄨󵄨 k 󵄨󵄨 󵄨󵄨 󵄨 󵄨󵄨∑(Rl , Δt el )󵄨󵄨󵄨 ⩽ 1 󵄩󵄩󵄩Rk 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩ek+1 󵄩󵄩󵄩 + 1 󵄩󵄩󵄩Rk−1 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩ek 󵄩󵄩󵄩 󵄨󵄨 2τ 󵄩 󵄩 󵄩 󵄩 2τ 󵄩 󵄩 󵄩 󵄩 󵄨󵄨󵄨 󵄨󵄨 󵄨l=1 k−1 󵄩 󵄩 󵄩 󵄩 1󵄩 󵄩 󵄩 󵄩 + ∑ 󵄩󵄩󵄩Δt Rl 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩el 󵄩󵄩󵄩 + 󵄩󵄩󵄩R2 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩e1 󵄩󵄩󵄩. 2τ l=2

(6.144)

Replacing k by l in (6.142), summing over l from 1 to k, and then combining (6.143) with (6.144), we have

216 � 6 Difference methods for the Schrödinger equation 2 2 1 󵄨 󵄨2 h 󵄨 󵄨2 󵄨 󵄨2 h 󵄨 󵄨2 [(󵄨󵄨󵄨ek+1 󵄨󵄨󵄨1 + 󵄨󵄨󵄨ek+1 󵄨󵄨󵄨2 + 󵄨󵄨󵄨ek 󵄨󵄨󵄨1 + 󵄨󵄨󵄨ek 󵄨󵄨󵄨2 ) 4τ 12 12 2 2 󵄨 󵄨2 h 󵄨 󵄨2 󵄨 󵄨2 h 󵄨 󵄨2 − (󵄨󵄨󵄨e1 󵄨󵄨󵄨1 + 󵄨󵄨󵄨e1 󵄨󵄨󵄨2 + 󵄨󵄨󵄨e0 󵄨󵄨󵄨1 + 󵄨󵄨󵄨e0 󵄨󵄨󵄨2 )] 12 12 k

󵄩 󵄩2 󵄩 󵄩2 󵄩 󵄩2 󵄨 󵄨2 󵄨 󵄨2 󵄨 󵄨2 󵄩 󵄩2 ⩽ ∑ c19 (󵄩󵄩󵄩el+1 󵄩󵄩󵄩 + 󵄩󵄩󵄩el 󵄩󵄩󵄩 + 󵄩󵄩󵄩el−1 󵄩󵄩󵄩 + 󵄨󵄨󵄨el+1 󵄨󵄨󵄨1 + 󵄨󵄨󵄨el 󵄨󵄨󵄨1 + 󵄨󵄨󵄨el−1 󵄨󵄨󵄨1 + 󵄩󵄩󵄩Rl 󵄩󵄩󵄩 ) l=1

+

1 󵄩󵄩 k 󵄩󵄩 󵄩󵄩 k+1 󵄩󵄩 1 󵄩󵄩 k−1 󵄩󵄩 󵄩󵄩 k 󵄩󵄩 󵄩R 󵄩 ⋅ 󵄩e 󵄩󵄩 + 󵄩󵄩R 󵄩󵄩 ⋅ 󵄩󵄩e 󵄩󵄩 2τ 󵄩 󵄩 󵄩 2τ

k−1 󵄩 󵄩 󵄩 󵄩 1󵄩 󵄩 󵄩 󵄩 + ∑ 󵄩󵄩󵄩Δt Rl 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩el 󵄩󵄩󵄩 + 󵄩󵄩󵄩R2 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩e1 󵄩󵄩󵄩, 2τ l=2

1 ⩽ k ⩽ n − 1.

Noticing 2 2 3󵄩 󵄩2 L 󵄩 󵄩2 1 󵄨 󵄨2 L 󵄩 󵄩2 󵄩 󵄩 󵄩 󵄩 2󵄩󵄩󵄩Rk 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩ek+1 󵄩󵄩󵄩 ⩽ 2 󵄩󵄩󵄩ek+1 󵄩󵄩󵄩 + 󵄩󵄩󵄩Rk 󵄩󵄩󵄩 ⩽ 󵄨󵄨󵄨ek+1 󵄨󵄨󵄨1 + 󵄩󵄩󵄩Rk 󵄩󵄩󵄩 , 3 2 3 L 2 2 1 󵄨 󵄨2 L 3 󵄩 󵄩2 L 󵄩 󵄩 2‖Rk−1 ‖ ⋅ 󵄩󵄩󵄩ek 󵄩󵄩󵄩 ⩽ 2 󵄩󵄩󵄩ek 󵄩󵄩󵄩 + ‖Rk−1 ‖2 ⩽ 󵄨󵄨󵄨ek 󵄨󵄨󵄨1 + ‖Rk−1 ‖2 , 3 2 3 L

and (6.106)–(6.107), (6.131), (6.141), there is a constant c20 satisfying k

󵄨󵄨 k+1 󵄨󵄨2 󵄨 l 󵄨2 2 4 2 󵄨󵄨e 󵄨󵄨1 ⩽ c20 τ ∑󵄨󵄨󵄨e 󵄨󵄨󵄨1 + c20 (τ + h ) ,

0 ⩽ k ⩽ n − 1.

l=1

With the application of the Gronwall inequality (Theorem 1.2(c)), it follows: 󵄨󵄨 k 󵄨󵄨2 c T 2 4 2 󵄨󵄨e 󵄨󵄨1 ⩽ e 20 ⋅ c20 (τ + h ) ,

0 ⩽ k ⩽ n.

Taking the square root on both the right- and left-hand sides of the above inequality, one has 1 󵄨󵄨 k 󵄨󵄨 c T 2 4 2 4 󵄨󵄨e 󵄨󵄨1 ⩽ e 2 20 √c20 (τ + h ) ≡ c18 (τ + h ),

0 ⩽ k ⩽ n.

Using Lemma 1.1(b), we have √L 󵄨 k 󵄨 √ 󵄩󵄩 k 󵄩󵄩 󵄨󵄨e 󵄨󵄨 ⩽ L c18 (τ 2 + h4 ), 󵄩󵄩e 󵄩󵄩∞ ⩽ 󵄨 󵄨 1 2 2

0 ⩽ k ⩽ n.

6.5 Numerical experiments In the section, we take the difference schemes (6.16)–(6.18) and (6.58)–(6.61) as examples to show the accuracy and conservation. Define the maximum norm error and convergence orders, respectively, as

6.5 Numerical experiments

E∞ (h, τ) =

max

0⩽k⩽n,0⩽j⩽m

Orderh = log2

� 217

󵄨󵄨 k󵄨 󵄨󵄨u(xj , tk ) − uj 󵄨󵄨󵄨,

E∞ (2h, τ) , E∞ (h, τ)

Orderτ = log2

E∞ (h, 2τ) . E∞ (h, τ)

Example 6.1. We consider the initial and boundary value problem (6.1)–(6.3) with q = 2 and φ(x) = sech(x − 40)e2i(x−40) , L = 80 and T = 1. The exact solution of the problem is u(x, t) = sech(x − 4t − 40)e2i(x−40)−3it . Since the difference scheme (6.16)–(6.18) is a system of nonlinear equations, we will use a simple iteration to solve it: { { { { { { { { {

q 󵄨 󵄨2 󵄨 2 󵄨2 i (wj(l+1) − ujk ) + δx2 wj(l+1) + (󵄨󵄨󵄨ujk 󵄨󵄨󵄨 + 󵄨󵄨󵄨2wj(l) − ujk 󵄨󵄨󵄨 )wj(l+1) = 0, τ 2 1 ⩽ j ⩽ m − 1, w0(l+1) = 0,

(l+1) wm = 0,

until ‖w(l+1) − w(l) ‖∞ ⩽ 1e − 8, where l denotes the iterative index. Taking the initial iteration value as wj(0) = ujk ,

0 ⩽ j ⩽ m.

When {wj | 1 ⩽ j ⩽ m − 1} is obtained, let ujk+1 = 2wj − ujk ,

1 ⩽ j ⩽ m − 1.

The numerical results are listed in Tables 6.1–6.4 and Figure 6.1. Table 6.1 and Table 6.2 give the maximum errors of the numerical solutions with different step sizes for the difference scheme (6.16)–(6.18). We see that when the step size decreases by half, the maximum norm error is reduced to 1/4. Similar results are obtained from Table 6.3 and Table 6.4 for the linearized difference scheme (6.58)–(6.61). The energy curves are shown in Figure 6.1, which confirms the conservation of both difference schemes. Table 6.1: Maximum errors and spatial convergence orders with different spatial step sizes (τ = 1/1600) by the difference scheme (6.16)–(6.18). h

E∞ (h, τ)

Orderh

�/� �/�� �/�� �/��

�.������e − � �.������e − � �.������e − � �.������e − �

�.���� �.���� �.����

218 � 6 Difference methods for the Schrödinger equation Table 6.2: Maximum errors and temporal convergence orders with different temporal step sizes (h = 1/400) by the difference scheme (6.16)–(6.18). τ

E∞ (h, τ)

Orderτ

�/�� �/�� �/�� �/���

�.������e − � �.������e − � �.������e − � �.������e − �

�.���� �.���� �.����

Table 6.3: Maximum errors and spatial convergence orders with different spatial step sizes (τ = 1/1600) by the difference scheme (6.58)–(6.61). h

E∞ (h, τ)

Orderh

�/� �/�� �/�� �/��

�.������e − � �.������e − � �.������e − � �.������e − �

�.���� �.���� �.����

Table 6.4: Maximum errors and temporal convergence orders with different temporal step sizes (h = 1/400) by the difference scheme (6.58)–(6.61). τ

E∞ (h, τ)

Orderτ

�/�� �/�� �/�� �/���

�.������e − � �.������e − � �.������e − � �.������e − �

�.���� �.���� �.����

Figure 6.1: The conservative energy of the difference scheme (6.16)–(6.18) (left) and (6.58)–(6.61) (right).

6.6 Summary and extension

� 219

6.6 Summary and extension In this chapter, three difference schemes for the initial and boundary value problem of the one-dimensional nonlinear Schrödinger equation are established and analyzed theoretically. The first difference scheme is a two-level nonlinear one. It is proved that the solution of the difference scheme satisfies two conservation laws, and the error estimate of the solution in the maximum norm is given. The Browder theorem is utilized to prove the existence of the solution of the difference scheme. It is proved that the solution of the difference scheme is unique and second-order unconditionally convergent in both time and space in the maximum norm. This work is referred to [2] and [36]. The second and third difference schemes are both three-level linearized. It is proved that each difference scheme satisfies the conservation of energy, and the error estimates of the difference solution in the maximum norm are given. The unique solvability and convergence of both difference schemes are proved. The third difference scheme is developed based on the works [9] and [53] where a two-level nonlinear difference scheme is established. Here, we construct a three-level linearized difference scheme. In [41], Wang and Guo established a two-level nonlinear compact difference scheme and a three-level linearized compact difference scheme for the Schrödinger equation. In [28, 36, 38], the authors considered the difference methods for solving the coupled systems of the Schrödinger equations. In [43], the authors studied the periodic boundary value problem of the twodimensional Schrödinger equation. The numerical experiments in this chapter originate from [35].

7 Difference methods for the Kuramoto–Tsuzuki equation The Kuramoto–Tsuzuki equation [19] describes the behavior of two branches near the bifurcation point. It can be regarded as the one-dimensional form of the Ginzburg– Landau equation. In this chapter, the difference methods for the Neumann boundary value problem of the one-dimensional Kuramoto–Tsuzuki equation are studied. In Chapter 9, the difference methods for the Dirichlet boundary value problem of the two-dimensional Ginzburg–Landau equations will be discussed.

7.1 Introduction In this chapter, we study the difference methods for the following initial and boundary value problem of the Kuramoto–Tsuzuki equation: ut = (1 + ic1 )uxx + u − (1 + ic2 )|u|2 u, { { { u(x, 0) = φ(x), { { { { ux (0, t) = 0, ux (L, t) = 0,

0 < x < L, 0 < t ⩽ T,

(7.1)

0 ⩽ x ⩽ L,

(7.2)

0 < t ⩽ T,

(7.3)

where c1 and c2 are real constants, φ(x) and u(x, t) are complex functions and φx (0) = φx (L) = 0. We first give an estimate of the solution to the problem (7.1)–(7.3). Let f (x) ∈ C[0, L]. Denote L

p 󵄨 󵄨p ‖f ‖p = √∫󵄨󵄨󵄨f (x)󵄨󵄨󵄨 dx,

‖f ‖ = ‖f ‖2 .

0

In addition, if f (x) ∈ C 1 [0, L], we denote L

󵄨 󵄨2 |f |1 = √∫󵄨󵄨󵄨f ′ (x)󵄨󵄨󵄨 dx. 0

Lemma 7.1. Let s1 and s2 be two nonnegative constants, f (x) ∈ C[0, L], s1 < s2 and θ ∈ (0, 1). Then we have θs +(1−θ)s

(1−θ)s2 1 ‖f ‖θs1 +(1−θ)s2 ⩽ ‖f ‖θs . s1 ⋅ ‖f ‖s2 1

2

Proof. With the application of Hölder’s inequality of the integral type, we have θs +(1−θ)s ‖f ‖θs1 +(1−θ)s2 1 2 https://doi.org/10.1515/9783110796018-007

L

󵄨 󵄨θs +(1−θ)s2 = ∫󵄨󵄨󵄨f (x)󵄨󵄨󵄨 1 dx 0

7.1 Introduction θ

L

1−θ

L

1 󵄨θs 1 󵄨 󵄨(1−θ)s2 1−θ 󵄨 ) dx] ⩽ [∫(󵄨󵄨󵄨f (x)󵄨󵄨󵄨 1 ) θ dx] ⋅ [∫(󵄨󵄨󵄨f (x)󵄨󵄨󵄨

0

θ

L

0

L

1−θ

󵄨s 󵄨 󵄨s 󵄨 = [∫󵄨󵄨󵄨f (x)󵄨󵄨󵄨 1 dx] ⋅ [∫󵄨󵄨󵄨f (x)󵄨󵄨󵄨 2 dx] 0

0

=

1 ‖f ‖θs s1

⋅

2 ‖f ‖(1−θ)s . s2

Corollary 7.1. For arbitrary f (x) ∈ C[0, L] and p > 4, we have p−2

4

p

p

p(p−4)

‖f ‖p−2 ⩽ ‖f ‖ p−2 ⋅ ‖f ‖p p−2 , p(p−4)

‖f ‖ p2 ⩽ ‖f ‖ p−2 ⋅ ‖f ‖p2(p−2) . 2

Lemma 7.2. For arbitrary f (x) ∈ C 1 [0, L] and p > 2, we have p

2 ‖f ‖∞ ⩽

p p p 1 −1 2 ‖f ‖p−2 ⋅ |f |1 + ‖f ‖ p2 . 2 L 2

Proof. Let 0 ⩽ y < x ⩽ L. Then we have 󵄨󵄨 󵄨p 󵄨 󵄨p 󵄨󵄨f (x)󵄨󵄨󵄨 2 − 󵄨󵄨󵄨f (y)󵄨󵄨󵄨 2 x

=∫ y

d 󵄨󵄨 󵄨p (󵄨󵄨f (s)󵄨󵄨󵄨 2 )ds ds x

=

p 󵄨󵄨 󵄨 p −1 d(|f (s)|) ds ∫󵄨󵄨f (s)󵄨󵄨󵄨 2 2 ds y

L

p 󵄨 󵄨 p −1 d(|f (s)|) ⩽ ∫󵄨󵄨󵄨f (s)󵄨󵄨󵄨 2 | |ds 2 ds 0

L

1 2

L

2

p 󵄨 d(|f (s)|) 󵄨p−2 ⩽ [∫󵄨󵄨󵄨f (s)󵄨󵄨󵄨 ds] ⋅ [∫( ) ds] 2 ds 0

L

1 2

0

L

p 󵄨 󵄨 󵄨p−2 󵄨2 = [∫󵄨󵄨󵄨f (s)󵄨󵄨󵄨 ds] ⋅ [∫󵄨󵄨󵄨f ′ (s)󵄨󵄨󵄨 ds] 2 0

p−2 p 2 = ‖f ‖p−2 ⋅ |f |1 , 2

1 2

0

which implies p−2 󵄨 󵄨p 󵄨󵄨 󵄨p p 2 ⋅ |f |1 + 󵄨󵄨󵄨f (y)󵄨󵄨󵄨 2 . 󵄨󵄨f (x)󵄨󵄨󵄨 2 ⩽ ‖f ‖p−2 2

1 2

� 221

222 � 7 Difference methods for the Kuramoto–Tsuzuki equation The above inequality is also valid for 0 ⩽ x ⩽ y ⩽ L. Integrating this inequality with respect to y on [0, L], we have p p−2 󵄨p p 󵄨 2 ⋅ |f |1 + ‖f ‖ p2 , L󵄨󵄨󵄨f (x)󵄨󵄨󵄨 2 ⩽ L‖f ‖p−2 2 2

which gives p

2 ‖f ‖∞ ⩽

p−2 p p 1 2 ‖f ‖p−2 ⋅ |f |1 + ‖f ‖ p2 . 2 L 2

Remark 7.1. In the proof of Lemma 7.2, we admit the following equality: L

L

2

d 󵄨 󵄨 󵄨2 󵄨 ∫[ (󵄨󵄨󵄨f (s)󵄨󵄨󵄨)] ds = ∫󵄨󵄨󵄨f ′ (s)󵄨󵄨󵄨 ds. ds 0

0

If f (x) ∈ C 1 [0, L] and it has only finite zero points on [0, L], then the above equality is obviously valid. Lemma 7.3. For arbitrary f ∈ C 1 [0, L] and p ⩾ 2, we have ‖f ‖p ⩽ κ(‖f ‖1−α ⋅ |f |α1 + ‖f ‖), Proof.

L

α=

α

p 1 1 − , κ = max{( ) , L−α }. 2 p 2 p

L

p

p

󵄨p 󵄨 󵄨p 󵄨 󵄨 2 󵄨 2 ⋅ ‖f ‖ p2 . ‖f ‖pp = ∫󵄨󵄨󵄨f (x)󵄨󵄨󵄨 dx ⩽ ( max 󵄨󵄨󵄨f (x)󵄨󵄨󵄨) ∫󵄨󵄨󵄨f (x)󵄨󵄨󵄨 2 dx = ‖f ‖∞ 0⩽x⩽L 2 0

0

Applying Lemma 7.2, it follows: p p p p 1 −1 2 ‖f ‖pp ⩽ ( ‖f ‖p−2 ⋅ |f |1 + ‖f ‖ p2 )‖f ‖ p2 . 2 L 2 2

Further using Corollary 7.1, it results in p(p−4) p(p−4) p(p−4) p p 2 p 1 ‖f ‖pp ⩽ ( ‖f ‖ p−2 ⋅ ‖f ‖p2(p−2) ⋅ |f |1 + ‖f ‖ p−2 ⋅ ‖f ‖p2(p−2) )‖f ‖ p−2 ⋅ ‖f ‖p2(p−2) , 2 L

which implies 2p

‖f ‖pp−2 ⩽

p+2 2p p 1 ‖f ‖ p−2 ⋅ |f |1 + ‖f ‖ p−2 . 2 L

Notice that the following inequality (a + b)r ⩽ ar + br holds when a ⩾ 0, b ⩾ 0, r ∈ (0, 1). It follows from (7.4) that

(7.4)

7.1 Introduction p−2

� 223

α

2p p+2 2p p p 1 ‖f ‖p ⩽ [ ‖f ‖ p−2 ⋅ |f |1 + ‖f ‖ p−2 ] ⩽ ( ) ‖f ‖1−α ⋅ |f |α1 + L−α ‖f ‖. 2 L 2

Theorem 7.1. Suppose u(x, t) is the solution of (7.1)–(7.3), then we have L

󵄨2 󵄨 ∫󵄨󵄨󵄨u(x, t)󵄨󵄨󵄨 dx ⩽ e2t ‖φ‖2 ,

0 < t ⩽ T,

0

L

c 󵄨 󵄨2 ∫󵄨󵄨󵄨ux (x, t)󵄨󵄨󵄨 dx ⩽ ec3 t (|φ|21 + 4 ), c3

0 < t ⩽ T,

0

where c3 = (1 + c22 )κ6 25 e4T ‖φ‖4 ,

c4 = e2T ‖φ‖2 + (1 + c22 )κ6 25 e6T ‖φ‖6 .

̄ t) on both the right- and left-hand sides and integratProof. (I) Multiplying (7.1) by u(x, ing the result with respect to x from 0 to L, we have L

L

̄ t)dx ̄ t)dx − (1 + ic1 ) ∫ uxx (x, t)u(x, ∫ ut (x, t)u(x, 0

L

L

0

0

0

󵄨 󵄨2 󵄨 󵄨4 = ∫󵄨󵄨󵄨u(x, t)󵄨󵄨󵄨 dx − (1 + ic2 ) ∫󵄨󵄨󵄨u(x, t)󵄨󵄨󵄨 dx.

(7.5)

Using the integration by parts and (7.3), it follows: L

̄ t)dx − ∫ uxx (x, t)u(x, 0

L

L

L 2 2 ̄ t)󵄨󵄨󵄨󵄨x=0 + ∫󵄨󵄨󵄨󵄨ux (x, t)󵄨󵄨󵄨󵄨 dx = ∫󵄨󵄨󵄨󵄨ux (x, t)󵄨󵄨󵄨󵄨 dx. = −ux (x, t)u(x, 0

(7.6)

0

Taking the real parts on both the right- and left-hand sides of (7.5) and using (7.6), we have L

L

L

L

0

0

0

0

1 d 󵄨󵄨 󵄨2 󵄨 󵄨2 󵄨 󵄨2 󵄨 󵄨4 ∫󵄨u(x, t)󵄨󵄨󵄨 dx + ∫󵄨󵄨󵄨ux (x, t)󵄨󵄨󵄨 dx = ∫󵄨󵄨󵄨u(x, t)󵄨󵄨󵄨 dx − ∫󵄨󵄨󵄨u(x, t)󵄨󵄨󵄨 dx, 2 dt 󵄨 which implies L

L

0

0

1 d 󵄨󵄨 󵄨2 󵄨 󵄨2 ∫󵄨u(x, t)󵄨󵄨󵄨 dx ⩽ ∫󵄨󵄨󵄨u(x, t)󵄨󵄨󵄨 dx, 2 dt 󵄨

0 < t ⩽ T.

224 � 7 Difference methods for the Kuramoto–Tsuzuki equation By the Gronwall inequality, we have L

L

L

󵄨2 󵄨 󵄨2 󵄨 󵄨2 󵄨 ∫󵄨󵄨󵄨u(x, t)󵄨󵄨󵄨 dx ⩽ e2t ∫󵄨󵄨󵄨u(x, 0)󵄨󵄨󵄨 dx = e2t ∫󵄨󵄨󵄨φ(x)󵄨󵄨󵄨 dx, 0

0

0

0 < t ⩽ T.

(II) Multiplying both the right- and left-hand sides of (7.1) by the result with respect to x from 0 to L, we have L

L

0

0

1 ū 1+ic1 t

and integrating

1 󵄨 󵄨2 ∫󵄨󵄨u (x, t)󵄨󵄨󵄨 dx − ∫ uxx (x, t)ū t (x, t)dx 1 + ic1 󵄨 t L

L

1 + ic2 󵄨󵄨 1 󵄨2 ∫ u(x, t)ū t (x, t)dx − ∫󵄨u(x, t)󵄨󵄨󵄨 u(x, t)ū t (x, t)dx. 1 + ic1 1 + ic1 󵄨

=

0

(7.7)

0

By the integration by parts and using (7.3), we have L

L

󵄨L − ∫ uxx (x, t)ū t (x, t)dx = −ux (x, t)ū t (x, t)󵄨󵄨󵄨x=0 + ∫ ux (x, t)ū xt (x, t)dx 0

0

L

= ∫ ux (x, t)ū xt (x, t)dx. 0

Taking the real parts on both the right- and left-hand sides of (7.7) and using the above equality, we have L

L

1 1 d 󵄨󵄨 󵄨2 󵄨 󵄨2 ∫󵄨u (x, t)󵄨󵄨󵄨 dx ∫󵄨󵄨ut (x, t)󵄨󵄨󵄨 dx + ⋅ 2 dt 󵄨 x 1 + c12 󵄨 0

0

L

L

1 + ic2 󵄨󵄨 1 󵄨2 = Re{ ∫ u(x, t)ū t (x, t)dx − ∫󵄨󵄨u(x, t)󵄨󵄨󵄨 u(x, t)ū t (x, t)dx} 1 + ic1 1 + ic1 0

⩽

⩽

1

0

󵄩󵄩 󵄩 󵄩 󵄩 󵄩󵄩u(⋅, t)󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩ut (⋅, t)󵄩󵄩󵄩 + 2

√1 + c1

1 󵄩󵄩 󵄩2 󵄩ut (⋅, t)󵄩󵄩󵄩 + 2(1 + c12 ) 󵄩

√1 + c22 󵄩 󵄩󵄩 󵄩 󵄩󵄩󵄨󵄨 󵄨2 󵄩 󵄩󵄩󵄨󵄨u(⋅, t)󵄨󵄨󵄨 u(⋅, t)󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩ut (⋅, t)󵄩󵄩󵄩 󵄩 󵄩 √1 + c12

2 1 󵄩󵄩 1 󵄩󵄩 󵄩2 󵄩󵄩2 1 + c2 󵄩󵄩 󵄩6 u (⋅, t) + 󵄩󵄩u(⋅, t)󵄩󵄩󵄩 + 󵄩 󵄩 󵄩u(⋅, t)󵄩󵄩󵄩6 , t 󵄩 󵄩 2 2 󵄩 2(1 + c12 )

which implies L

d 󵄨󵄨 󵄨2 󵄩 󵄩2 󵄩 󵄩6 ∫󵄨u (x, t)󵄨󵄨󵄨 dx ⩽ 󵄩󵄩󵄩u(⋅, t)󵄩󵄩󵄩 + (1 + c22 )󵄩󵄩󵄩u(⋅, t)󵄩󵄩󵄩6 , dt 󵄨 x 0

0 < t ⩽ T.

7.2 Two-level nonlinear difference scheme

� 225

Combining Lemma 7.3 with the inequality (x+y)6 ⩽ 25 (x 6 +y6 ) (where x, y ⩾ 0), it follows: d 󵄨󵄨 󵄩2 󵄨 󵄩 󵄩2 󵄨2 󵄩 󵄨1 󵄩 󵄩 6 2 󵄨󵄨u(⋅, t)󵄨󵄨󵄨1 ⩽ 󵄩󵄩󵄩u(⋅, t)󵄩󵄩󵄩 + (1 + c2 )[κ(󵄩󵄩󵄩u(⋅, t)󵄩󵄩󵄩 3 󵄨󵄨󵄨u(⋅, t)󵄨󵄨󵄨13 + 󵄩󵄩󵄩u(⋅, t)󵄩󵄩󵄩)] dt 󵄩6 󵄨2 󵄩 󵄩4 󵄨 󵄩 󵄩2 󵄩 ⩽ 󵄩󵄩󵄩u(⋅, t)󵄩󵄩󵄩 + (1 + c22 )κ6 25 (󵄩󵄩󵄩u(⋅, t)󵄩󵄩󵄩 󵄨󵄨󵄨u(⋅, t)󵄨󵄨󵄨1 + 󵄩󵄩󵄩u(⋅, t)󵄩󵄩󵄩 ) 󵄨2 󵄨 ⩽ e2t ‖φ‖2 + (1 + c22 )κ6 25 (e4t ‖φ‖4 󵄨󵄨󵄨u(⋅, t)󵄨󵄨󵄨1 + e6t ‖φ‖6 ) 󵄨2 󵄨 = (1 + c22 )κ6 25 ‖φ‖4 e4t 󵄨󵄨󵄨u(⋅, t)󵄨󵄨󵄨1 + e2t ‖φ‖2 + (1 + c22 )κ6 25 e6t ‖φ‖6 󵄨 󵄨2 ⩽ (1 + c22 )k 6 25 ‖φ‖4 e4T 󵄨󵄨󵄨u(⋅, t)󵄨󵄨󵄨1 + e2T ‖φ‖2 + (1 + c22 )κ6 25 e6T ‖φ‖6 󵄨 󵄨2 = c3 󵄨󵄨󵄨u(⋅, t)󵄨󵄨󵄨1 + c4 , 0 < t ⩽ T. By the Gronwall inequality, we have 󵄨󵄨 󵄨2 󵄨2 c ct 󵄨 󵄨󵄨u(⋅, t)󵄨󵄨󵄨1 ⩽ e 3 [󵄨󵄨󵄨u(⋅, 0)󵄨󵄨󵄨1 + 4 ], c3

0 < t ⩽ T.

According to the above theorem, there is a constant c5 such that 󵄩󵄩 󵄩 󵄩󵄩u(⋅, t)󵄩󵄩󵄩∞ ⩽ c5 ,

0 ⩽ t ⩽ T.

(7.8)

7.2 Two-level nonlinear difference scheme 7.2.1 Derivation of the difference scheme Considering equation (7.1) at the point (xj , tk+ 1 ), we have 2

󵄨 󵄨2 ut (xj , tk+ 1 ) = (1 + ic1 )uxx (xj , tk+ 1 ) + u(xj , tk+ 1 ) − (1 + ic2 )󵄨󵄨󵄨u(xj , tk+ 1 )󵄨󵄨󵄨 u(xj , tk+ 1 ), 2

2

0 ⩽ j ⩽ m, 0 ⩽ k ⩽ n − 1.

2

2

2

(7.9)

With the help of the numerical differential formula, it follows: k+ 21

δt Uj

k+ 21

= (1 + ic1 )δx2 Uj

k+ 21

+ Uj

k+ 1 󵄨 k+ 1 󵄨2 k+ 1 − (1 + ic2 )󵄨󵄨󵄨Uj 2 󵄨󵄨󵄨 Uj 2 + Rj 2 ,

1 ⩽ j ⩽ m − 1, 0 ⩽ k ⩽ n − 1,

(7.10)

where there is a constant c6 satisfying 󵄨󵄨 k+ 21 󵄨󵄨 2 2 󵄨󵄨Rj 󵄨󵄨 ⩽ c6 (τ + h ),

1 ⩽ j ⩽ m − 1, 0 ⩽ k ⩽ n − 1.

Differentiating (7.1) once with respect to x, it follows: uxt = (1 + ic1 )uxxx + ux − (1 + ic2 )(2|u|2 ux + u2 ū x ).

(7.11)

226 � 7 Difference methods for the Kuramoto–Tsuzuki equation Applying the boundary value condition (7.3), one has uxxx (0, t) = 0,

uxxx (L, t) = 0,

0 < t ⩽ T.

(7.12)

Combining (7.3) with (7.12) and using Lemma 1.2, we have 2 [u(x1 , t) − u(x0 , t)] + O(h2 ), h2 2 uxx (L, t) = − 2 [u(xm , t) − u(xm−1 , t)] + O(h2 ). h

uxx (0, t) =

(7.13) (7.14)

Thus, it holds that 1 uxx (0, tk+ 1 ) = [uxx (0, tk ) + uxx (0, tk+1 )] + O(τ 2 ) 2 2 2 1 2 = [ 2 (U1k − U0k ) + 2 (U1k+1 − U0k+1 )] + O(τ 2 + h2 ) 2 h h 2 k+ 1 = δx U 1 2 + O(τ 2 + h2 ), h 2 1 uxx (xm , tk+ 1 ) = [uxx (xm , tk ) + uxx (xm , tk+1 )] + O(τ 2 ) 2 2 1 2 2 k k+1 = [− 2 (Umk − Um−1 ) − 2 (Umk+1 − Um−1 )] + O(τ 2 + h2 ) 2 h h 2 k+ 1 = − δx U 21 + O(τ 2 + h2 ). m− 2 h

(7.15)

(7.16)

With the application of (7.15)–(7.16) into (7.9) for j = 0 and j = m, respectively, we have k+ 21

δt U0

2 k+ 1 k+ 1 k+ 1 󵄨 k+ 1 󵄨2 k+ 1 = (1 + ic1 ) δx U 1 2 + U0 2 − (1 + ic2 )󵄨󵄨󵄨U0 2 󵄨󵄨󵄨 U0 2 + R0 2 , h 2

0 ⩽ k ⩽ n − 1, k+ 21

δt Um

2 = (1 + ic1 )(− δx U h

k+ 21 m− 21

k+ 21

) + Um

k+ 21

k+ 21

󵄨 󵄨2 − (1 + ic2 )󵄨󵄨󵄨Um 󵄨󵄨󵄨 Um

0 ⩽ k ⩽ n − 1,

(7.17) k+ 21

+ Rm , (7.18)

and there is a constant c7 satisfying 󵄨󵄨 k+ 21 󵄨󵄨 2 2 󵄨󵄨R0 󵄨󵄨 ⩽ c7 (τ + h ), 󵄨󵄨 k+ 21 󵄨󵄨 2 2 󵄨󵄨Rm 󵄨󵄨 ⩽ c7 (τ + h ), tion

0 ⩽ k ⩽ n − 1,

(7.19)

0 ⩽ k ⩽ n − 1.

(7.20)

Omitting the small terms in (7.10), (7.17)–(7.18) and noticing the initial value condiUj0 = φ(xj ),

0 ⩽ j ⩽ m,

(7.21)

7.2 Two-level nonlinear difference scheme

� 227

a difference scheme for solving the problem (7.1)–(7.3) reads { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { {

2 k+ 1 k+ 1 󵄨 k+ 1 󵄨2 k+ 1 = (1 + ic1 ) δx u 1 2 + u0 2 − (1 + ic2 )󵄨󵄨󵄨u0 2 󵄨󵄨󵄨 u0 2 , h 2 0 ⩽ k ⩽ n − 1, k+ 1 k+ 1 k+ 1 󵄨 k+ 1 󵄨2 k+ 1 δt uj 2 = (1 + ic1 )δx2 uj 2 + uj 2 − (1 + ic2 )󵄨󵄨󵄨uj 2 󵄨󵄨󵄨 uj 2 , 1 ⩽ j ⩽ m − 1, 0 ⩽ k ⩽ n − 1, k+ 21

δt u0

2 = (1 + ic1 )(− δx u h 0 ⩽ k ⩽ n − 1, k+ 21

δt um uj0

k+ 21 m− 21

k+ 21

) + um

k+ 21

(7.22) (7.23) k+ 21

󵄨 󵄨2 − (1 + ic2 )󵄨󵄨󵄨um 󵄨󵄨󵄨 um , (7.24)

= φ(xj ), 0 ⩽ j ⩽ m.

(7.25)

7.2.2 Existence of the difference solution The difference scheme (7.22)–(7.25) is two-level nonlinear. When uk at the k-th time level k+ 21

has been determined, it can be viewed as a system of nonlinear equations in {uj

j ⩽ m}. When

k+ 1 {uj 2

|0 ⩽

| 0 ⩽ j ⩽ m} is obtained, it results in k+ 21

ujk+1 = 2uj

− ujk ,

0 ⩽ j ⩽ m.

Theorem 7.2. There is a solution to the difference scheme (7.22)–(7.25). Proof. The difference equations (7.22)–(7.24) can be uniformly written as k+ 21

δt uj

k+ 21

= (1 + ic1 )δx2 uj

k+ 21

+ uj

0 ⩽ j ⩽ m, 0 ⩽ k ⩽ n − 1,

󵄨 k+ 1 󵄨2 k+ 1 − (1 + ic2 )󵄨󵄨󵄨uj 2 󵄨󵄨󵄨 uj 2 ,

(7.26)

where { { { { { { { 2 δx uj = { { { { { { { {

2 δ u1 , h x 2 1 (δ u 1 − δx uj− 1 ), 2 h x j+ 2 2 − δx um− 1 , 2 h

j = 0, 1 ⩽ j ⩽ m − 1, j = m.

Suppose uk has been determined and let k+ 21

wj = uj Then rewrite (7.26) as

,

0 ⩽ j ⩽ m.

(7.27)

228 � 7 Difference methods for the Kuramoto–Tsuzuki equation 2 (w − ujk ) − (1 + ic1 )δx2 wj − wj + (1 + ic2 )|wj |2 wj = 0, τ j

0 ⩽ j ⩽ m.

(7.28)

Define the operator Π : 𝒰h → 𝒰h by Π(w)j =

2 (w − ujk ) − (1 + ic1 )δx2 wj − wj + (1 + ic2 )|wj |2 wj , τ j

0 ⩽ j ⩽ m.

Now we calculate (Π(w), w) as (Π(w), w) =

2 [(w, w) − (uk , w)] − (1 + ic1 )(δx2 w, w) − (w, w) + (1 + ic2 )(|w|2 w, w). τ

Noticing m−1 1 1 −(δx2 w, w) = −h[ (δx2 w0 )w̄ 0 + ∑ (δx2 wj )w̄j + (δx2 wm )w̄ m ] 2 2 j=1 m−1

= −[(δx w 1 )w̄ 0 + ∑ (δx wj+ 1 − δx wj− 1 )w̄ j − (δx wm− 1 )w̄ m ] 2

2

j=1

2

2

m−1

= h ∑ |δx wj+ 1 |2 , 2

j=0

it follows: 2 [(w, w) − Re(uk , w)] + |w|21 − ‖w‖2 + (|w|2 w, w) τ 2 󵄩 󵄩 ⩾ (‖w‖2 − 󵄩󵄩󵄩uk 󵄩󵄩󵄩 ⋅ ‖w‖) − ‖w‖2 τ 2 󵄩󵄩 k 󵄩󵄩 2−τ (‖w‖ − = 󵄩u 󵄩)‖w‖. τ 2−τ󵄩 󵄩

Re(Π(w), w) =

When τ < 2 and ‖w‖ =

2 ‖uk ‖, 2−τ

one yields Re(Π(w), w) ⩾ 0.

Therefore, there is a solution to (7.28) with the help of the Browder theorem (Theorem 2.4).

7.2.3 Boundedness of the difference solution Denote { 1, ωj = { 1 , { 2

1 ⩽ j ⩽ m − 1, j = 0, m.

7.2 Two-level nonlinear difference scheme

� 229

Lemma 7.4. Suppose s1 and s2 are two nonnegative constants and θ ∈ (0, 1). For arbitrary u ∈ 𝒰h , we have θs +(1−θ)s

(1−θ)s2 1 . ‖u‖θs1 +(1−θ)s2 ⩽ ‖u‖θs s1 ⋅ ‖u‖s2 1

2

Proof. Applying the discrete Hölder’s inequality, it follows: θs +(1−θ)s

‖u‖θs1 +(1−θ)s2 1

m

2

θs1 +(1−θ)s2

= h ∑ ωj |uj | j=0

m

θs1

θ

1 θ

m

⩽ [h ∑ ωj (|uj | ) ] ⋅ [h ∑ ωj (|uj | j=0

(1−θ)s2

j=0

θ

m

m

= [h ∑ ωj (|uj |s1 )] ⋅ [h ∑ ωj (|uj |s2 )] j=0

=

1 ‖u‖θs s1

)

1 1−θ

1−θ

]

1−θ

j=0

⋅

2 . ‖u‖(1−θ)s s2

Corollary 7.2. For arbitrary u ∈ 𝒰h and p > 4, we have p−2

4

p

p

p(p−4)

‖u‖p−2 ⩽ ‖u‖ p−2 ⋅ ‖u‖p p−2 , p(p−4)

‖u‖ p2 ⩽ ‖u‖ p−2 ⋅ ‖u‖p2(p−2) . 2

Lemma 7.5. For arbitrary u ∈ 𝒰h and p > 2, we have p

p

2 2 ‖u‖∞ ⩽ p‖u‖p−2 ⋅ |u|1 +

−1

p 1 ‖u‖ p2 . L 2

Proof. For arbitrary l(0 ⩽ l ⩽ m − 1), there is a ξl between |ul | and |ul+1 | such that p

p

|ul+1 | 2 − |ul | 2 =

p p2 −1 ξ (|ul+1 | − |ul |). 2 l

Therefore, it follows: p p p p 󵄨 p 󵄨󵄨 󵄨 −1 −1 󵄨 󵄨󵄨|ul+1 | 2 − |ul | 2 󵄨󵄨󵄨 ⩽ (|ul+1 | 2 + |ul | 2 )󵄨󵄨󵄨|ul+1 | − |ul |󵄨󵄨󵄨 2 p p p ⩽ (|ul+1 | 2 −1 + |ul | 2 −1 )|ul+1 − ul |. 2

Let s > j and using (7.29), we have p

p

|us | 2 − |uj | 2

(7.29)

230 � 7 Difference methods for the Kuramoto–Tsuzuki equation s−1

p

p

= ∑(|ul+1 | 2 − |ul | 2 ) l=j

s−1 p p p ⩽ ∑ [|ul+1 | 2 −1 + |ul | 2 −1 ] ⋅ |ul+1 − ul | 2 l=j s−1 p p p = h ∑ [|ul+1 | 2 −1 + |ul | 2 −1 ] ⋅ |δx ul+ 1 | 2 2 l=j p−2

2 ⩽ p‖u‖p−2 ⋅ |u|1 .

The above inequality is also valid for s ⩽ j. Thus, p−2

p

p

2 ⋅ |u|1 + |uj | 2 , |us | 2 ⩽ p‖u‖p−2

0 ⩽ s, j ⩽ m.

Multiplying both the right- and left-hand sides of the above inequality by hωj and summing over j from 0 to m, it follows: p−2

p

p

2 L|us | 2 ⩽ Lp‖u‖p−2 ⋅ |u|1 + ‖u‖ p2 , 2

0 ⩽ s ⩽ m,

which implies p

p

2 2 ⩽ p‖u‖p−2 ⋅ |u|1 + ‖u‖∞

−1

p 1 ‖u‖ p2 . L 2

Lemma 7.6. For arbitrary u ∈ 𝒰h and p ⩾ 2, we have ‖u‖p ⩽ κ(‖u‖1−α |u|α1 + ‖u‖), Proof.

m

α=

1 1 − , κ = max{pα , L−α }. 2 p

p

m

p

p

p

2 ‖u‖pp = h ∑ ωj |uj |p ⩽ max |uj | 2 ⋅ h ∑ ωj |uj | 2 = ‖u‖∞ ‖u‖ p2 .

0⩽j⩽m

j=0

j=0

2

With the help of Lemma 7.5, it follows: p

2 ‖u‖pp ⩽ (p‖u‖p−2 ⋅ |u|1 +

−1

p p 1 ‖u‖ p2 )‖u‖ p2 . L 2 2

Using Corollary 7.2, one has 2

p(p−4)

‖u‖pp ⩽ [p‖u‖ p−2 ⋅ ‖u‖p2(p−2) ⋅ |u|1 + p+2

= [p‖u‖ p−2 ⋅ |u|1 + which implies

p(p−4) p(p−4) p p 1 ‖u‖ p−2 ⋅ ‖u‖p2(p−2) ]‖u‖ p−2 ⋅ ‖u‖p2(p−2) L

p(p−4) 2p 1 ‖u‖ p−2 ]‖u‖p p−2 , L

7.2 Two-level nonlinear difference scheme 2p

p+2

‖u‖pp−2 ⩽ p‖u‖ p−2 ⋅ |u|1 +

2p 1 ‖u‖ p−2 . L

� 231

(7.30)

Noticing that when a ⩾ 0, b ⩾ 0, r ∈ (0, 1), it holds that (a + b)r ⩽ ar + br . In combination of (7.30), one has ‖u‖p ⩽ [p‖u‖

p+2 p−2

p−2

2p 2p 1 ⋅ |u|1 + ‖u‖ p−2 ] ⩽ pα ‖u‖1−α |u|α1 + L−α ‖u‖. L

Similar to Theorem 7.1, we have the following conclusion for the difference scheme (7.22)–(7.25). Theorem 7.3. Suppose {ujk | 0 ⩽ j ⩽ m, 0 ⩽ k ⩽ n} is the solution of the difference scheme (7.22)–(7.25), then it holds that 3 󵄩󵄩 k 󵄩󵄩 T 󵄩 0󵄩 󵄩󵄩u 󵄩󵄩 ⩽ e 2 󵄩󵄩󵄩u 󵄩󵄩󵄩, 1 ⩽ k ⩽ n, c 󵄨󵄨 k 󵄨󵄨2 3c T 󵄨 0 󵄨2 󵄨󵄨u 󵄨󵄨1 ⩽ e 8 (󵄨󵄨󵄨u 󵄨󵄨󵄨1 + 9 ), 1 ⩽ k ⩽ n, c

(7.31) (7.32)

8

where c8 = (1 + c22 )κ6 24 e6T ‖u0 ‖4 , c9 = 21 e3T ‖u0 ‖2 + (1 + c22 )κ6 24 e9T ‖u0 ‖6 . Proof. (I) Taking the inner product on both the right- and left-hand sides of (7.26) with 1 uk+ 2 , it holds that 1

1

1

1

1

1

(δt uk+ 2 , uk+ 2 ) = (1 + ic1 )(δx2 uk+ 2 , uk+ 2 ) + (uk+ 2 , uk+ 2 ) 1 2 1 1 󵄨 󵄨 − (1 + ic2 )(󵄨󵄨󵄨uk+ 2 󵄨󵄨󵄨 uk+ 2 , uk+ 2 ) 1 2 1 2 1 4 󵄨 󵄨 󵄩 󵄩 󵄨 󵄨 = −(1 + ic1 )󵄨󵄨󵄨uk+ 2 󵄨󵄨󵄨1 + 󵄩󵄩󵄩uk+ 2 󵄩󵄩󵄩 − (1 + ic2 )(󵄨󵄨󵄨uk+ 2 󵄨󵄨󵄨 , 1). Taking the real parts on both the right- and left-hand sides of the above equality, we have 1 2 1 4 1 2 1 󵄩󵄩 k+1 󵄩󵄩2 󵄩󵄩 k 󵄩󵄩2 󵄨 󵄩 󵄩 󵄨 󵄨 󵄨 (󵄩󵄩u 󵄩󵄩 − 󵄩󵄩u 󵄩󵄩 ) = −󵄨󵄨󵄨uk+ 2 󵄨󵄨󵄨1 + 󵄩󵄩󵄩uk+ 2 󵄩󵄩󵄩 − (󵄨󵄨󵄨uk+ 2 󵄨󵄨󵄨 , 1) 2τ 󵄩 󵄩 2 1 2 ‖uk+1 ‖ + 󵄩󵄩󵄩uk 󵄩󵄩󵄩 󵄩 󵄩 ⩽ 󵄩󵄩󵄩uk+ 2 󵄩󵄩󵄩 ⩽ ( ) , 0 ⩽ k ⩽ n − 1. 2

Rearranging the above inequality, it follows: τ 󵄩 τ 󵄩 󵄩 󵄩 (1 − )󵄩󵄩󵄩uk+1 󵄩󵄩󵄩 ⩽ (1 + )󵄩󵄩󵄩uk 󵄩󵄩󵄩, 2 2

0 ⩽ k ⩽ n − 1.

232 � 7 Difference methods for the Kuramoto–Tsuzuki equation When τ ⩽ 32 , it holds that 3 󵄩 k󵄩 󵄩󵄩 k+1 󵄩󵄩 󵄩󵄩u 󵄩󵄩 ⩽ (1 + τ)󵄩󵄩󵄩u 󵄩󵄩󵄩, 2

0 ⩽ k ⩽ n − 1,

which implies 3 󵄩󵄩 k+1 󵄩󵄩 T 󵄩 0󵄩 󵄩󵄩u 󵄩󵄩 ⩽ e 2 󵄩󵄩󵄩u 󵄩󵄩󵄩,

0⩽k ⩽n−1

by recursion. (II) Taking the inner product on both the right- and left-hand sides of (7.26) with 1 1 δ uk+ 2 , we have 1−ic t 1

1 1 1 1 1 (δt uk+ 2 , δt uk+ 2 ) − (δx2 uk+ 2 , δt uk+ 2 ) 1 + ic1 1 1 1 1 + ic2 󵄨󵄨 k+ 21 󵄨󵄨2 k+ 21 1 (uk+ 2 , δt uk+ 2 ) − (󵄨u 󵄨󵄨 u , δt uk+ 2 ). = 1 + ic1 1 + ic1 󵄨

Taking the real parts on both the right- and left-hand sides of the above equality results in 1 󵄩󵄩 k+ 21 󵄩󵄩2 1 󵄨󵄨 k+1 󵄨󵄨2 󵄨󵄨 k 󵄨󵄨2 󵄩δt u 󵄩󵄩 + (󵄨󵄨u 󵄨󵄨1 − 󵄨󵄨u 󵄨󵄨1 ) 2τ 1 + c12 󵄩 ⩽

⩽

1

󵄩󵄩 k+ 21 󵄩󵄩 󵄩󵄩 k+ 21 󵄩󵄩 󵄩󵄩u 󵄩󵄩 ⋅ 󵄩󵄩δt u 󵄩󵄩 + 2

√1 + c1

1 󵄩󵄩 k+ 21 󵄩󵄩2 󵄩δt u 󵄩󵄩 + 2 2(1 + c1 ) 󵄩 0 ⩽ k ⩽ n − 1.

√1 + c22

󵄩󵄩 k+ 21 󵄩󵄩3 󵄩󵄩 k+ 21 󵄩󵄩 󵄩󵄩u 󵄩󵄩6 ⋅ 󵄩󵄩δt u 󵄩󵄩 2

√1 + c1

2 1 󵄩󵄩 k+ 21 󵄩󵄩2 1 󵄩󵄩 k+ 21 󵄩󵄩2 1 + c2 󵄩󵄩 k+ 21 󵄩󵄩6 δ u + 󵄩󵄩u 󵄩󵄩 + 󵄩 󵄩u 󵄩󵄩6 , 󵄩 t 󵄩 2 2 󵄩 2(1 + c12 ) 󵄩

With the help of Lemma 7.6, it follows: 1 󵄨󵄨 k+1 󵄨󵄨2 󵄨󵄨 k 󵄨󵄨2 (󵄨u 󵄨󵄨1 − 󵄨󵄨u 󵄨󵄨1 ) 2τ 󵄨 1 2 1 + c22 󵄩󵄩 k+ 1 󵄩󵄩6 1󵄩 󵄩 ⩽ 󵄩󵄩󵄩uk+ 2 󵄩󵄩󵄩 + 󵄩u 2 󵄩󵄩6 2 2 󵄩 1 2 1 + c22 󵄨󵄨 k+ 1 󵄨󵄨 31 k+ 1 2 󵄩󵄩 k+ 1 󵄩󵄩 6 1󵄩 󵄩 ⩽ 󵄩󵄩󵄩uk+ 2 󵄩󵄩󵄩 + [κ(󵄨󵄨u 2 󵄨󵄨1 ‖u 2 ‖ 3 + 󵄩󵄩u 2 󵄩󵄩)] 2 2 1 2 1 + c22 6 5 󵄨󵄨 k+ 1 󵄨󵄨2 󵄩󵄩 k+ 1 󵄩󵄩4 󵄩󵄩 k+ 1 󵄩󵄩6 1󵄩 󵄩 ⩽ 󵄩󵄩󵄩uk+ 2 󵄩󵄩󵄩 + κ 2 (󵄨󵄨u 2 󵄨󵄨1 󵄩󵄩u 2 󵄩󵄩 + 󵄩󵄩u 2 󵄩󵄩 ). 2 2

(7.33)

With the help of (7.31), we know 3 󵄩󵄩 k+ 21 󵄩󵄩 T 󵄩 0󵄩 󵄩󵄩u 󵄩󵄩 ⩽ e 2 󵄩󵄩󵄩u 󵄩󵄩󵄩,

0 ⩽ k ⩽ n − 1.

(7.34)

7.2 Two-level nonlinear difference scheme

� 233

Substituting (7.34) into (7.33), we have 1 2 1 󵄨󵄨 k+1 󵄨󵄨2 󵄨󵄨 k 󵄨󵄨2 󵄨 󵄨 (󵄨u 󵄨󵄨1 − 󵄨󵄨u 󵄨󵄨1 ) ⩽ c8 󵄨󵄨󵄨uk+ 2 󵄨󵄨󵄨1 + c9 2τ 󵄨 c 󵄨 󵄨2 󵄨 󵄨2 ⩽ 8 (󵄨󵄨󵄨uk+1 󵄨󵄨󵄨1 + 󵄨󵄨󵄨uk 󵄨󵄨󵄨1 ) + c9 , 0 ⩽ k ⩽ n − 1, 2

which implies 󵄨 󵄨2 󵄨 󵄨2 (1 − c8 τ)󵄨󵄨󵄨uk+1 󵄨󵄨󵄨1 ⩽ (1 + c8 τ)󵄨󵄨󵄨uk 󵄨󵄨󵄨1 + 2c9 τ,

0 ⩽ k ⩽ n − 1.

When c8 τ ⩽ 31 , 󵄨󵄨 k+1 󵄨󵄨2 󵄨 k 󵄨2 󵄨󵄨u 󵄨󵄨1 ⩽ (1 + 3c8 τ)󵄨󵄨󵄨u 󵄨󵄨󵄨1 + 3c9 τ,

0 ⩽ k ⩽ n − 1.

It follows that c 󵄨󵄨 k+1 󵄨󵄨2 3c T 󵄨 0 󵄨2 󵄨󵄨u 󵄨󵄨1 ⩽ e 8 (󵄨󵄨󵄨u 󵄨󵄨󵄨1 + 9 ), c8

0⩽k ⩽n−1

by using the Gronwall inequality (Theorem 1.2(a)). Combining Theorem 7.3 with Lemma 1.1(e), there is a constant c10 such that 󵄩󵄩 k 󵄩󵄩 󵄩󵄩u 󵄩󵄩∞ ⩽ c10 ,

0 ⩽ k ⩽ n.

(7.35)

7.2.4 Uniqueness of the difference solution 2 ), the solution of the difference scheme Theorem 7.4. When τ < 2/(1 + 3√1 + c22 c10 (7.22)–(7.25) is unique.

Proof. From the arguments in Theorem 7.2, we know that it suffices to prove that (7.28) has a unique solution. Suppose (7.28) has another solution {vj | 0 ⩽ j ⩽ m}, which satisfies 2 (v − ujk ) = (1 + ic1 )δx2 vj + vj − (1 + ic2 )|vj |2 vj , τ j Noticing (7.35), we know ‖w‖∞ ⩽ c10 ,

‖v‖∞ ⩽ c10 .

zj = wj − vj ,

0 ⩽ j ⩽ m.

Let

Subtracting (7.36) from (7.28), it follows:

0 ⩽ j ⩽ m.

(7.36)

234 � 7 Difference methods for the Kuramoto–Tsuzuki equation 2 z = (1 + ic1 )δx2 zj + zj − (1 + ic2 )(|wj |2 wj − |vj |2 vj ), τ j

0 ⩽ j ⩽ m.

Taking the inner product on both the right- and left-hand sides of the above equality with z, one yields 2 2 ‖z‖ = (1 + ic1 )(δx2 z, z) + (z, z) − (1 + ic2 )(|w|2 w − |v|2 v, z). τ

(7.37)

Noticing 󵄨󵄨 2 2 󵄨 2 2 2 󵄨󵄨|wj | wj − |vj | vj 󵄨󵄨󵄨 ⩽ (|wj | + |wj ||vj | + |vj | )|wj − vj | ⩽ 3c10 |zj |, we obtain 󵄨󵄨 󵄨 2 2 2 2 󵄨󵄨(|w| w − |v| v, z)󵄨󵄨󵄨 ⩽ 3c10 ‖z‖ .

(7.38)

(δx2 z, z) = −|z|21 .

(7.39)

In addition,

Taking the real parts on both the right- and left-hand sides of (7.37) and combining (7.38) with (7.39), we have 2 2 2 ‖z‖ ⩽ −|z|21 + ‖z‖2 + √1 + c22 ⋅ 3c10 ‖z‖2 τ 2 ⩽ (1 + 3√1 + c22 c10 )‖z‖2 ,

2 which implies ‖z‖2 = 0 when τ < 2/(1 + 3√1 + c22 c10 ).

7.2.5 Convergence of the difference solution Theorem 7.5. Suppose {Ujk | 0 ⩽ j ⩽ m, 0 ⩽ k ⩽ n} is the solution of the problem (7.1)–(7.3) and {ujk | 0 ⩽ j ⩽ m, 0 ⩽ k ⩽ n} is the solution of the difference scheme (7.22)–(7.25). Denote ejk = Ujk − ujk ,

0 ⩽ j ⩽ m, 0 ⩽ k ⩽ n.

There is a constant c11 satisfying 󵄨󵄨 k 󵄨󵄨 2 2 󵄨󵄨e 󵄨󵄨1 ⩽ c11 (τ + h ),

0 ⩽ k ⩽ n.

7.2 Two-level nonlinear difference scheme

� 235

Proof. Subtracting (7.22)–(7.25) from (7.17), (7.10), (7.18), (7.21), respectively, we have the system of error equations: k+ 1 k+ 1 k+ 1 k+ 1 󵄨 k+ 1 󵄨2 k+ 1 󵄨 k+ 1 󵄨2 k+ 1 { δt ej 2 = (1 + ic1 )δx2 ej 2 + ej 2 − (1 + ic2 )(󵄨󵄨󵄨Uj 2 󵄨󵄨󵄨 Uj 2 − 󵄨󵄨󵄨uj 2 󵄨󵄨󵄨 uj 2 ) + Rj 2 , { { { 0 ⩽ j ⩽ m, 0 ⩽ k ⩽ n − 1, (7.40) { { { { 0 (7.41) { ej = 0, 0 ⩽ j ⩽ m.

By (7.8), there is a constant c5 such that 󵄨󵄨 k 󵄨󵄨 󵄨󵄨Uj 󵄨󵄨 ⩽ c5 ,

0 ⩽ j ⩽ m, 0 ⩽ k ⩽ n.

(7.42)

Combining (7.35) with (7.42), we know 󵄨󵄨󵄨󵄨 k+ 21 󵄨󵄨2 k+ 21 󵄨󵄨 k+ 21 󵄨󵄨2 k+ 21 󵄨󵄨 − 󵄨󵄨uj 󵄨󵄨 uj 󵄨󵄨 󵄨󵄨󵄨󵄨Uj 󵄨󵄨 Uj 1 k+ 1 󵄨 󵄨 k+ 󵄨 󵄨 k+ 1 󵄨 2 󵄨 k+ 1 ⩽ (󵄨󵄨󵄨Uj 2 󵄨󵄨󵄨 + 󵄨󵄨󵄨uj 2 󵄨󵄨󵄨) 󵄨󵄨󵄨Uj 2 − uj 2 󵄨󵄨󵄨 󵄨 k+ 1 󵄨 ⩽ (c5 + c10 )2 󵄨󵄨󵄨ej 2 󵄨󵄨󵄨, 0 ⩽ j ⩽ m, 0 ⩽ k ⩽ n − 1.

(7.43) 1

(I) Taking the inner product on both the right- and left-hand sides of (7.40) with ek+ 2 and taking the real parts of the result, we have 1 󵄩󵄩 k+1 󵄩󵄩2 󵄩󵄩 k 󵄩󵄩2 (󵄩e 󵄩󵄩 − 󵄩󵄩e 󵄩󵄩 ) 2τ 󵄩 1 󵄩 1 󵄩2 󵄩 󵄩 󵄩 1󵄩 󵄨 1 󵄨2 󵄩 1 󵄩2 ⩽ −󵄨󵄨󵄨ek+ 2 󵄨󵄨󵄨1 + 󵄩󵄩󵄩ek+ 2 󵄩󵄩󵄩 + √1 + c22 (c5 + c10 )2 󵄩󵄩󵄩ek+ 2 󵄩󵄩󵄩 + 󵄩󵄩󵄩Rk+ 2 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩ek+ 2 󵄩󵄩󵄩 1 󵄩 1 󵄩2 󵄩 󵄩 󵄩 1󵄩 ⩽ [1 + √1 + c22 (c5 + c10 )2 ]󵄩󵄩󵄩ek+ 2 󵄩󵄩󵄩 + 󵄩󵄩󵄩Rk+ 2 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩ek+ 2 󵄩󵄩󵄩, 0 ⩽ k ⩽ n − 1. It follows from the above inequality that 1 1 󵄩󵄩 k+1 󵄩󵄩 󵄩󵄩 k 󵄩󵄩 󵄩 󵄩 1󵄩 󵄩 (󵄩󵄩e 󵄩󵄩 − 󵄩󵄩e 󵄩󵄩) ⩽ [1 + √1 + c22 (c5 + c10 )2 ]󵄩󵄩󵄩ek+ 2 󵄩󵄩󵄩 + 󵄩󵄩󵄩Rk+ 2 󵄩󵄩󵄩, τ

0 ⩽ k ⩽ n − 1. (7.44)

From (7.11), (7.19) and (7.20), one knows 󵄩󵄩 k+ 21 󵄩󵄩 √ 2 2 󵄩󵄩R 󵄩󵄩 ⩽ L max{c6 , c7 }(τ + h ),

0 ⩽ k ⩽ n − 1.

Combining (7.44) with (7.45), we have 1 󵄩 󵄩 {1 − [1 + √1 + c22 (c5 + c10 )2 ]τ}󵄩󵄩󵄩ek+1 󵄩󵄩󵄩 2 1 󵄩 󵄩 ⩽ {1 + [1 + √1 + c22 (c5 + c10 )2 ]τ}󵄩󵄩󵄩ek 󵄩󵄩󵄩 2 + √L max{c6 , c7 }τ(τ 2 + h2 ), 0 ⩽ k ⩽ n − 1.

(7.45)

236 � 7 Difference methods for the Kuramoto–Tsuzuki equation When 21 [1 + √1 + c22 (c5 + c10 )2 ]τ ⩽ 31 , it follows that 3 󵄩󵄩 k+1 󵄩󵄩 󵄩 k󵄩 2 󵄩󵄩e 󵄩󵄩 ⩽ {1 + [1 + √1 + c22 (c5 + c10 ) ]τ}󵄩󵄩󵄩e 󵄩󵄩󵄩 2 3 + √L max{c6 , c7 }τ(τ 2 + h2 ), 0 ⩽ k ⩽ n − 1. 2 By the Gronwall inequality (Theorem 1.2(a)), we have 3 󵄩󵄩 k 󵄩󵄩 [1+√1+c22 (c5 +c10 )2 ]T ⋅ 󵄩󵄩e 󵄩󵄩 ⩽ e 2

≡ c12 (τ 2 + h2 ),

√L max{c6 , c7 } 1 + √1 + c22 (c5 + c10 )2

(τ 2 + h2 )

0 ⩽ k ⩽ n.

(7.46)

(II) Taking the inner product on both the right- and left-hand sides of (7.40) with it follows:

1 1 δ ek+ 2 , 1−ic1 t

1 󵄩󵄩 k+ 21 󵄩󵄩2 2 k+ 1 k+ 1 󵄩󵄩δt e 󵄩󵄩 − (δx e 2 , δt e 2 ) 1 + ic1 1 1 1 1 + ic2 󵄨󵄨 k+ 21 󵄨󵄨2 k+ 21 󵄨󵄨 k+ 21 󵄨󵄨2 k+ 21 1 (ek+ 2 , δt ek+ 2 ) − (󵄨󵄨U 󵄨󵄨 U = − 󵄨󵄨u 󵄨󵄨 u , δt ek+ 2 ) 1 + ic1 1 + ic1 1 k+ 21 k+ 21 (R , δt e ). + 1 + ic1 Taking the real parts on both the right- and left-hand sides of the above equality, we have 1 󵄩󵄩 k+ 21 󵄩󵄩2 1 󵄨󵄨 k+1 󵄨󵄨2 󵄨󵄨 k 󵄨󵄨2 󵄩δt e 󵄩󵄩 + (󵄨󵄨e 󵄨󵄨1 − 󵄨󵄨e 󵄨󵄨1 ) 2τ 1 + c12 󵄩 = Re{

1 1 1 (ek+ 2 , δt ek+ 2 )} 1 + ic1

+ Re{− + Re{ ⩽

1

1 1 1 (Rk+ 2 , δt ek+ 2 )} 1 + ic1

󵄩󵄩 k+ 21 󵄩󵄩 󵄩󵄩 k+ 21 󵄩󵄩 󵄩󵄩e 󵄩󵄩 ⋅ 󵄩󵄩δt e 󵄩󵄩 + 2

√1 + c1 +

1

√1 + ⩽

1 1 + ic2 󵄨󵄨 k+ 21 󵄨󵄨2 k+ 21 󵄨󵄨 k+ 21 󵄨󵄨 k+ 21 (󵄨󵄨U 󵄨󵄨 U − 󵄨󵄨u 󵄨󵄨u , δt ek+ 2 )} 1 + ic1

c12

√1 + c22 √1 +

c12

1 󵄩 1󵄩 󵄩 󵄩 (c5 + c10 )2 󵄩󵄩󵄩ek+ 2 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩δt ek+ 2 󵄩󵄩󵄩

󵄩󵄩 k+ 21 󵄩󵄩 󵄩󵄩 k+ 21 󵄩󵄩 󵄩󵄩R 󵄩󵄩 ⋅ 󵄩󵄩δt e 󵄩󵄩

1 1 󵄩󵄩 k+ 21 󵄩󵄩2 󵄩󵄩 k+ 21 󵄩󵄩2 󵄩󵄩 k+ 21 󵄩󵄩2 2 4 󵄩 k+ 1 󵄩2 󵄩󵄩δt e 󵄩󵄩 + 󵄩󵄩e 󵄩󵄩 + 󵄩󵄩δt e 󵄩󵄩 + (1 + c2 )(c5 + c10 ) 󵄩󵄩󵄩e 2 󵄩󵄩󵄩 2 2 4(1 + c1 ) 4(1 + c1 )

7.3 Three-level linearized difference scheme

+

1 󵄩󵄩 k+ 21 󵄩󵄩2 󵄩δt e 󵄩󵄩 + 2 2(1 + c1 ) 󵄩

1 󵄩󵄩 k+ 21 󵄩󵄩2 󵄩R 󵄩󵄩 , 2󵄩

� 237

0 ⩽ k ⩽ n − 1,

i. e., 1 󵄨󵄨 k+1 󵄨󵄨2 󵄨󵄨 k 󵄨󵄨2 󵄩 1 󵄩2 (󵄨󵄨e 󵄨󵄨1 − 󵄨󵄨e 󵄨󵄨1 ) ⩽ [1 + (1 + c22 )(c5 + c10 )4 ]󵄩󵄩󵄩ek+ 2 󵄩󵄩󵄩 + 2τ 0 ⩽ k ⩽ n − 1.

1 󵄩󵄩 k+ 21 󵄩󵄩2 󵄩R 󵄩󵄩 , 2󵄩

Combining (7.45) with (7.46), it follows: 1 󵄨󵄨 k+1 󵄨󵄨2 󵄨󵄨 k 󵄨󵄨2 L 2 2 2 (󵄨e 󵄨󵄨1 − 󵄨󵄨e 󵄨󵄨1 ) ⩽ [1 + (1 + c22 )(c5 + c10 )4 ]c12 (τ 2 + h2 ) + max{c62 , c72 }(τ 2 + h2 ) , 2τ 󵄨 2 0 ⩽ k ⩽ n − 1, which implies 󵄨󵄨 k+1 󵄨󵄨2 󵄨󵄨 k 󵄨󵄨2 2 4 2 2 2 2 2 2 󵄨󵄨e 󵄨󵄨1 ⩽ 󵄨󵄨e 󵄨󵄨1 + {2[1 + (1 + c2 )(c5 + c10 ) ]c12 + L max{c6 , c7 }}τ(τ + h ) , 0 ⩽ k ⩽ n − 1. With the application of recursion to the above inequality, it follows: 󵄨󵄨 k 󵄨󵄨2 2 4 2 2 2 2 2 2 󵄨󵄨e 󵄨󵄨1 ⩽ {2[1 + (1 + c2 )(c5 + c10 ) ]c12 + L max{c6 , c7 }}T(τ + h ) , 1 ⩽ k ⩽ n. This completes the proof.

7.3 Three-level linearized difference scheme 7.3.1 Derivation of the difference scheme Considering equation (7.1) at the node point (xj , tk ), we have 󵄨 󵄨2 ut (xj , tk ) = (1 + ic1 )uxx (xj , tk ) + u(xj , tk ) − (1 + ic2 )󵄨󵄨󵄨u(xj , tk )󵄨󵄨󵄨 u(xj , tk ), 0 ⩽ j ⩽ m, 1 ⩽ k ⩽ n − 1.

(7.47)

With the application of the numerical differential formula, it is easy to know ̄ ̄ 󵄨 󵄨2 ̄ Δt Ujk = (1 + ic1 )δx2 Ujk + Ujk − (1 + ic2 )󵄨󵄨󵄨Ujk 󵄨󵄨󵄨 Ujk + Pjk ,

1 ⩽ j ⩽ m − 1, 1 ⩽ k ⩽ n − 1 and there is a constant c13 satisfying

(7.48)

238 � 7 Difference methods for the Kuramoto–Tsuzuki equation 󵄨󵄨 k 󵄨󵄨 2 2 󵄨󵄨Pj 󵄨󵄨 ⩽ c13 (τ + h ),

1 ⩽ j ⩽ m − 1, 1 ⩽ k ⩽ n − 1.

(7.49)

Combining (7.13) with (7.14), it follows: 1 uxx (x0 , tk ) = [uxx (x0 , tk−1 ) + uxx (x0 , tk+1 )] + O(τ 2 ) 2 2 1 2 + O(h2 ) + δx U k+1 + O(h2 )] + O(τ 2 ) = [ δx U k−1 1 1 2 h h 2 2 ̄ 2 = δx U k1 + O(τ 2 + h2 ), h 2 1 uxx (xm , tk ) = [uxx (xm , tk−1 ) + uxx (xm , tk+1 )] + O(τ 2 ) 2 1 2 2 k−1 2 = [(− δx Um− δ U k+11 + O(h2 ))] + O(τ 2 ) 1 + O(h )) + (− 2 h h x m− 2 2 2 2 2 k̄ = − δx Um− 1 + O(τ + h ). h 2

(7.50)

(7.51)

Applying (7.50) and (7.51) into (7.47) with j = 0 and j = m, respectively, we have ̄ ̄ 2 󵄨 󵄨2 ̄ Δt U0k = (1 + ic1 ) δx U k1 + U0k − (1 + ic2 )󵄨󵄨󵄨U0k 󵄨󵄨󵄨 U0k + P0k , h 2 1 ⩽ k ⩽ n − 1, 2 󵄨󵄨 k 󵄨󵄨2 k̄ k̄ k̄ k Δt Umk = (1 + ic1 )(− δx Um− 1 ) + Um − (1 + ic2 )󵄨󵄨Um 󵄨󵄨 Um + Pm , h 2 1 ⩽ k ⩽ n − 1,

(7.52) (7.53)

and there is a constant c14 satisfying 󵄨󵄨 k 󵄨󵄨 2 2 󵄨󵄨P0 󵄨󵄨 ⩽ c14 (τ + h ),

󵄨󵄨 k 󵄨󵄨 2 2 󵄨󵄨Pm 󵄨󵄨 ⩽ c14 (τ + h ),

1 ⩽ k ⩽ n − 1.

(7.54)

Combining (7.1) with (7.2), it follows: 󵄨 󵄨2 ut (x, 0) = (1 + ic1 )φxx (x) + φ(x) − (1 + ic2 )󵄨󵄨󵄨φ(x)󵄨󵄨󵄨 φ(x). Therefore, we have Uj1 = φ(xj ) + τut (xj , 0) + Pj0 ,

0 ⩽ j ⩽ m.

(7.55)

There is a constant c15 satisfying 󵄨󵄨 0 󵄨󵄨 2 󵄨󵄨Pj 󵄨󵄨 ⩽ c15 τ , 󵄨󵄨 2 0 󵄨 󵄨󵄨δx Pj+ 1 󵄨󵄨󵄨 ⩽ c15 τ , 2

0 ⩽ j ⩽ m,

(7.56)

0 ⩽ j ⩽ m − 1.

(7.57)

7.3 Three-level linearized difference scheme

� 239

Noticing the initial value condition Uj0 = φ(xj ),

0 ⩽ j ⩽ m,

(7.58)

and omitting the small terms in (7.48), (7.52)–(7.53), (7.55), a three-level linearized difference scheme for solving the problem (7.1)–(7.3) reads { { { { { { { { { { { { { { { { { { { { { { { { { { {

̄ ̄ 2 󵄨 󵄨2 ̄ Δt u0k = (1 + ic1 ) δx uk1 + u0k − (1 + ic2 )󵄨󵄨󵄨u0k 󵄨󵄨󵄨 u0k , 1 ⩽ k ⩽ n − 1, h 2 ̄ ̄ 󵄨 󵄨2 ̄ Δt ujk = (1 + ic1 )δx2 ujk + ujk − (1 + ic2 )󵄨󵄨󵄨ujk 󵄨󵄨󵄨 ujk , 1 ⩽ j ⩽ m − 1, 1 ⩽ k ⩽ n − 1, 2 󵄨󵄨 k 󵄨󵄨2 k̄ k k̄ k̄ Δt um = (1 + ic1 )(− δx um− 1 ⩽ k ⩽ n − 1, 1 ) + um − (1 + ic2 )󵄨󵄨um 󵄨󵄨 um , h 2

(7.60)

uj1 = φ(xj ) + τut (xj , 0),

(7.63)

uj0 = φ(xj ),

0 ⩽ j ⩽ m,

0 ⩽ j ⩽ m.

(7.59)

(7.61) (7.62)

7.3.2 Boundedness of the difference solution Theorem 7.6. Suppose {ujk | 0 ⩽ j ⩽ m, 0 ⩽ k ⩽ n} is the solution of the difference scheme (7.59)–(7.63). There is a constant c16 satisfying 󵄩󵄩 k 󵄩󵄩 󵄩󵄩u 󵄩󵄩∞ ⩽ c16 ,

0 ⩽ k ⩽ n.

(7.64)

Proof. From (7.62)–(7.63), there is a constant c17 such that 󵄩󵄩 0 󵄩󵄩 󵄩󵄩u 󵄩󵄩 ⩽ c17 , 󵄩󵄩 1 󵄩󵄩 󵄩󵄩u 󵄩󵄩 ⩽ c17 ,

󵄨󵄨 0 󵄨󵄨 󵄨󵄨u 󵄨󵄨1 ⩽ c17 , 󵄨󵄨 1 󵄨󵄨 󵄨󵄨u 󵄨󵄨1 ⩽ c17 .

(7.65) (7.66)

Rewrite (7.59)–(7.61) in a unified form as ̄ ̄ 󵄨 󵄨2 ̄ Δt ujk = (1 + ic1 )δx2 ujk + ujk − (1 + ic2 )󵄨󵄨󵄨ujk 󵄨󵄨󵄨 ujk ,

0 ⩽ j ⩽ m, 1 ⩽ k ⩽ n − 1.

(7.67)

(I) Taking the inner product on both the right- and left-hand sides of (7.67) with uk and then taking the real parts of the result, we have ̄

1 󵄩󵄩 k+1 󵄩󵄩2 󵄩󵄩 k−1 󵄩󵄩2 (󵄩u 󵄩󵄩 − 󵄩󵄩u 󵄩󵄩 ) 4τ 󵄩 󵄨 ̄ 󵄨2 󵄩 ̄ 󵄩2 󵄨 󵄨2 ̄ ̄ = −󵄨󵄨󵄨uk 󵄨󵄨󵄨1 + 󵄩󵄩󵄩uk 󵄩󵄩󵄩 − (󵄨󵄨󵄨uk 󵄨󵄨󵄨 uk , uk ) 󵄩 ̄ 󵄩2 ⩽ 󵄩󵄩󵄩uk 󵄩󵄩󵄩 2

⩽( which implies

‖uk+1 ‖ + ‖uk−1 ‖ ), 2

1 ⩽ k ⩽ n − 1,

240 � 7 Difference methods for the Kuramoto–Tsuzuki equation 󵄩 k+1 󵄩 󵄩 k−1 󵄩 󵄩󵄩 k+1 󵄩󵄩 󵄩󵄩 k−1 󵄩󵄩 󵄩󵄩u 󵄩󵄩 − 󵄩󵄩u 󵄩󵄩 ⩽ τ(󵄩󵄩󵄩u 󵄩󵄩󵄩 + 󵄩󵄩󵄩u 󵄩󵄩󵄩),

1 ⩽ k ⩽ n − 1.

When τ ⩽ 31 , we have 󵄩 k−1 󵄩 󵄩󵄩 k+1 󵄩󵄩 󵄩󵄩u 󵄩󵄩 ⩽ (1 + 3τ)󵄩󵄩󵄩u 󵄩󵄩󵄩,

1 ⩽ k ⩽ n − 1.

By recursion, it follows: 3T 󵄩󵄩 k 󵄩󵄩 󵄩 1󵄩 󵄩 0󵄩 󵄩 1󵄩 󵄩 0󵄩 [n] 󵄩󵄩u 󵄩󵄩 ⩽ (1 + 3τ) 2 max{󵄩󵄩󵄩u 󵄩󵄩󵄩, 󵄩󵄩󵄩u 󵄩󵄩󵄩} ⩽ e 2 max{󵄩󵄩󵄩u 󵄩󵄩󵄩, 󵄩󵄩󵄩u 󵄩󵄩󵄩},

1 ⩽ k ⩽ n.

Noticing (7.65) and (7.66), it follows: 3 3T 󵄩 0󵄩 󵄩 1󵄩 󵄩󵄩 k 󵄩󵄩 T 󵄩󵄩u 󵄩󵄩 ⩽ e 2 ⋅ max{󵄩󵄩󵄩u 󵄩󵄩󵄩, 󵄩󵄩󵄩u 󵄩󵄩󵄩} ⩽ c17 e 2 ,

1 ⩽ k ⩽ n.

(7.68)

(II) Taking the inner product on both the right- and left-hand sides of (7.67) with we have

1 Δ uk , 1−ic1 t

̄ 1 + ic2 󵄨󵄨 k 󵄨󵄨2 k̄ 1 1 󵄩󵄩 k 󵄩󵄩2 2 k̄ k (uk , Δt uk ) − (󵄨u 󵄨 u , Δt uk ). 󵄩󵄩Δt u 󵄩󵄩 − (δx u , Δt u ) = 1 + ic1 1 + ic1 1 + ic1 󵄨 󵄨

Taking the real parts on both the right- and left-hand sides of the above equality, we have 1 󵄩󵄩 k 󵄩󵄩2 1 󵄨 k+1 󵄨2 󵄨 k−1 󵄨2 󵄩󵄩Δt u 󵄩󵄩 + (󵄨󵄨󵄨u 󵄨󵄨󵄨1 − 󵄨󵄨󵄨u 󵄨󵄨󵄨1 ) 2 4τ 1 + c1 = Re{ ⩽

⩽

̄ 1 + ic2 󵄨󵄨 k 󵄨󵄨2 k̄ 1 (uk , Δt uk )} + Re{− (󵄨u 󵄨 u , Δt uk )} 1 + ic1 1 + ic1 󵄨 󵄨

1

󵄩󵄩 k̄ 󵄩󵄩 󵄩󵄩 k 󵄩󵄩 󵄩󵄩u 󵄩󵄩 ⋅ 󵄩󵄩Δt u 󵄩󵄩 + 2

√1 + c1

1 󵄩󵄩 k 󵄩󵄩2 󵄩Δt u 󵄩󵄩 + 2(1 + c12 ) 󵄩

√1 + c22 󵄩 󵄩󵄩󵄨󵄨 k 󵄨󵄨2 k̄ 󵄩󵄩󵄩 󵄩󵄩 k 󵄩󵄩 󵄩󵄩󵄨󵄨u 󵄨󵄨 u 󵄩󵄩 ⋅ 󵄩󵄩Δt u 󵄩󵄩 󵄩 2 √1 + c1 󵄩

2 1 󵄩󵄩 k̄ 󵄩󵄩2 1 󵄩󵄩 k 󵄩󵄩2 1 + c2 󵄩󵄩󵄩󵄨󵄨 k 󵄨󵄨2 k̄ 󵄩󵄩󵄩2 󵄩󵄩u 󵄩󵄩 + 󵄩󵄩Δt u 󵄩󵄩 + 󵄩󵄨u 󵄨 u 󵄩 , 2 2 2 󵄩󵄩󵄨 󵄨 󵄩󵄩 2(1 + c1 )

which implies 1 󵄨󵄨 k+1 󵄨󵄨2 󵄨󵄨 k−1 󵄨󵄨2 (󵄨u 󵄨󵄨1 − 󵄨󵄨u 󵄨󵄨1 ) ⩽ 4τ 󵄨

2 1 󵄩󵄩 k̄ 󵄩󵄩2 1 + c2 󵄩󵄩󵄩󵄨󵄨 k 󵄨󵄨2 k̄ 󵄩󵄩󵄩2 󵄩󵄩u 󵄩󵄩 + 󵄩󵄨u 󵄨 u 󵄩 , 2 2 󵄩󵄩󵄨 󵄨 󵄩󵄩

By Young’s inequality, it follows: 󵄨󵄨 k 󵄨󵄨4 󵄨󵄨 k̄ 󵄨󵄨2 2 󵄨󵄨 k 󵄨󵄨6 󵄨󵄨uj 󵄨󵄨 󵄨󵄨uj 󵄨󵄨 ⩽ 󵄨󵄨uj 󵄨󵄨 + 3 Again using Lemma 7.6, we have

1 󵄨󵄨 k̄ 󵄨󵄨6 󵄨u 󵄨 . 3󵄨 j 󵄨

1 ⩽ k ⩽ n − 1.

(7.69)

7.3 Three-level linearized difference scheme

� 241

󵄩󵄩󵄨 k 󵄨2 k̄ 󵄩󵄩2 󵄩󵄩󵄨󵄨u 󵄨󵄨 u 󵄩󵄩 󵄩󵄩󵄨 󵄨 󵄩󵄩 2 󵄩󵄩 k 󵄩󵄩6 1 󵄩󵄩 k̄ 󵄩󵄩6 ⩽ 󵄩󵄩u 󵄩󵄩6 + 󵄩󵄩u 󵄩󵄩6 3 3 2 6 󵄨󵄨 k 󵄨󵄨 31 󵄩󵄩 k 󵄩󵄩 32 󵄩󵄩 k 󵄩󵄩 6 1 6 󵄨󵄨 k̄ 󵄨󵄨 31 󵄩󵄩 k̄ 󵄩󵄩 32 󵄩󵄩 k̄ 󵄩󵄩 6 ⩽ κ (󵄨󵄨u 󵄨󵄨1 ⋅ 󵄩󵄩u 󵄩󵄩 + 󵄩󵄩u 󵄩󵄩) + κ (󵄨󵄨u 󵄨󵄨1 ⋅ 󵄩󵄩u 󵄩󵄩 + 󵄩󵄩u 󵄩󵄩) 3 3 1 2 6 5 󵄨󵄨 k 󵄨󵄨2 󵄩󵄩 k 󵄩󵄩4 󵄩󵄩 k 󵄩󵄩6 󵄨 ̄ 󵄨2 󵄩 ̄ 󵄩4 󵄩 ̄ 󵄩6 ⩽ κ 2 (󵄨󵄨u 󵄨󵄨1 ⋅ 󵄩󵄩u 󵄩󵄩 + 󵄩󵄩u 󵄩󵄩 ) + κ6 25 (󵄨󵄨󵄨uk 󵄨󵄨󵄨1 ⋅ 󵄩󵄩󵄩uk 󵄩󵄩󵄩 + 󵄩󵄩󵄩uk 󵄩󵄩󵄩 ). 3 3

(7.70)

Substituting (7.70) into (7.69) and further using (7.68), there is a constant c18 such that 1 󵄨󵄨 k+1 󵄨󵄨2 󵄨󵄨 k−1 󵄨󵄨2 (󵄨u 󵄨󵄨1 − 󵄨󵄨u 󵄨󵄨1 ) 4τ 󵄨 󵄨 󵄨2 󵄨 ̄ 󵄨2 ⩽ c18 (󵄨󵄨󵄨uk 󵄨󵄨󵄨1 + 󵄨󵄨󵄨uk 󵄨󵄨󵄨1 ) + c18 ⩽ c18 (

󵄨 󵄨2 |uk |21 + |uk+1 |21 |uk−1 |21 + 󵄨󵄨󵄨uk 󵄨󵄨󵄨1 + ) + c18 , 2 2

1 ⩽ k ⩽ n − 1;

namely, (1 − 2c18 τ)

|uk |21 + |uk+1 |21 |uk−1 |21 + |uk |21 ⩽ (1 + 2c18 τ) + 2c18 τ, 2 2

1 ⩽ k ⩽ n − 1.

When 2c18 τ ⩽ 31 , it follows: |uk |21 + |uk+1 |21 |uk−1 |21 + |uk |21 ⩽ (1 + 6c18 τ) + 3c18 τ, 2 2

1 ⩽ k ⩽ n − 1.

Using the Gronwall inequality (Theorem 1.2(a)) and noting (7.65)–(7.66), we have |u0 |21 + |u1 |21 1 |uk |21 + |uk+1 |21 1 2 ⩽ e6c18 T ( + ) ⩽ e6c18 T (c17 + ), 2 2 2 2

0 ⩽ k ⩽ n − 1.

(7.71)

(III) Combining (7.68), (7.71) with Lemma 1.1(e), we know that (7.64) is valid.

7.3.3 Existence and uniqueness of the difference solution Theorem 7.7. The difference scheme (7.59)–(7.63) is uniquely solvable. Proof. From (7.62)–(7.63), we know that u0 and u1 are uniquely determined. Suppose

uk−1 and uk have been determined. Then from (7.59)–(7.61) we have the system of linear

242 � 7 Difference methods for the Kuramoto–Tsuzuki equation equations in uk+1 . Consider its homogeneous one: { { { { { { { { { { { { { { {

1 k+1 1 1 1 󵄨 󵄨2 u = (1 + ic1 ) δx uk+1 + u0k+1 − (1 + ic2 )󵄨󵄨󵄨u0k 󵄨󵄨󵄨 u0k+1 , 1 2τ 0 h 2 2 2 1 k+1 1 1 1 󵄨 󵄨2 u = (1 + ic1 )δx2 ujk+1 + ujk+1 − (1 + ic2 )󵄨󵄨󵄨ujk 󵄨󵄨󵄨 ujk+1 , 1 ⩽ j ⩽ m − 1, 2τ j 2 2 2 1 k+1 1 k+1 1 1 󵄨 k 󵄨󵄨2 k+1 k+1 u = (1 + ic1 )(− δx um− u − (1 + ic2 )󵄨󵄨󵄨um 󵄨󵄨 um . 1) + 2τ m h 2 m 2 2

(7.72) (7.73) (7.74)

Multiplying both the right- and left-hand sides of (7.72) by 21 hū 0k+1 , of (7.73) by hū jk+1 , k+1 , respectively, and then adding these results together, it follows: of (7.74) by 21 hū m 1 󵄩󵄩 k+1 󵄩󵄩2 󵄩u 󵄩󵄩 2τ 󵄩 1 󵄨 󵄨2 = − (1 + ic1 )󵄨󵄨󵄨uk+1 󵄨󵄨󵄨1 + 2

1 󵄩󵄩 k+1 󵄩󵄩2 󵄩u 󵄩󵄩 2󵄩

m−1 1 󵄨 󵄨2 󵄨 1 󵄨2 󵄨 󵄨2 󵄨 󵄨2 − (1 + ic2 )h[ 󵄨󵄨󵄨u0k 󵄨󵄨󵄨 󵄨󵄨󵄨u0k+1 󵄨󵄨󵄨 + ∑ 󵄨󵄨󵄨ujk 󵄨󵄨󵄨 󵄨󵄨󵄨ujk+1 󵄨󵄨󵄨 + 2 2 j=1

1 󵄨󵄨 k 󵄨󵄨2 󵄨󵄨 k+1 󵄨󵄨2 󵄨u 󵄨 󵄨u 󵄨 ]. 2 󵄨 m󵄨 󵄨 m 󵄨

Taking the real parts on both the right- and left-hand sides of the above equality, we have 1 󵄩󵄩 k+1 󵄩󵄩2 󵄩u 󵄩󵄩 = − 2τ 󵄩

1 󵄨󵄨 k+1 󵄨󵄨2 󵄨u 󵄨󵄨1 + 2󵄨

1 󵄩󵄩 k+1 󵄩󵄩2 󵄩u 󵄩󵄩 2󵄩

m−1

1 1 󵄨 󵄨2 󵄨 󵄨2 󵄨 󵄨2 󵄨 󵄨2 − h[ 󵄨󵄨󵄨u0k 󵄨󵄨󵄨 󵄨󵄨󵄨u0k+1 󵄨󵄨󵄨 + ∑ 󵄨󵄨󵄨ujk 󵄨󵄨󵄨 󵄨󵄨󵄨ujk+1 󵄨󵄨󵄨 + 2 2 j=1

1 󵄨󵄨 k 󵄨󵄨2 󵄨󵄨 k+1 󵄨󵄨2 󵄨u 󵄨 󵄨u 󵄨 ] 2 󵄨 m󵄨 󵄨 m 󵄨

1󵄩 󵄩2 ⩽ 󵄩󵄩󵄩uk+1 󵄩󵄩󵄩 . 2 When τ < 1, we have

󵄩󵄩 k+1 󵄩󵄩 󵄩󵄩u 󵄩󵄩 = 0, which implies uk+1 = 0. Therefore, (7.59)–(7.61) is uniquely solvable with respect to uk+1 . By induction, the conclusion is true. 7.3.4 Convergence of the difference solution Theorem 7.8. Suppose {Ujk | 0 ⩽ j ⩽ m, 0 ⩽ k ⩽ n} is the solution of the problem (7.1)–(7.3) and {ujk | 0 ⩽ j ⩽ m, 0 ⩽ k ⩽ n} is the solution of the difference scheme (7.59)–(7.63). Denote

7.3 Three-level linearized difference scheme

ejk = Ujk − ujk ,

� 243

0 ⩽ j ⩽ m, 0 ⩽ k ⩽ n.

Then there is a constant c19 satisfying 󵄨󵄨 k 󵄨󵄨 2 2 󵄨󵄨e 󵄨󵄨1 ⩽ c19 (τ + h ),

0 ⩽ k ⩽ n.

(7.75)

Proof. Subtracting (7.59)–(7.63) from (7.52), (7.48), (7.53), (7.58), (7.55), respectively, we have the system of error equations: { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { {

̄ ̄ 2 󵄨 󵄨2 ̄ 󵄨 󵄨2 ̄ Δt e0k = (1 + ic1 ) δx ek1 + e0k − (1 + ic2 )(󵄨󵄨󵄨U0k 󵄨󵄨󵄨 U0k − 󵄨󵄨󵄨u0k 󵄨󵄨󵄨 u0k ) + P0k , h 2 1 ⩽ k ⩽ n − 1, ̄ ̄ 󵄨 󵄨2 ̄ 󵄨 󵄨2 ̄ Δt ejk = (1 + ic1 )δx2 ejk + ejk − (1 + ic2 )(󵄨󵄨󵄨Ujk 󵄨󵄨󵄨 Ujk − 󵄨󵄨󵄨ujk 󵄨󵄨󵄨 ujk ) + Pjk , 1 ⩽ j ⩽ m − 1, 1 ⩽ k ⩽ n − 1,

(7.76) (7.77)

2 󵄨󵄨 k 󵄨󵄨2 k̄ 󵄨󵄨 k 󵄨󵄨2 k̄ k̄ k̄ k k Δt em = (1 + ic1 )(− δx em− 1 ) + em − (1 + ic2 )(󵄨󵄨Um 󵄨󵄨 Um − 󵄨󵄨um 󵄨󵄨 um ) + Pm , h 2 1 ⩽ k ⩽ n − 1, (7.78) ej0 = 0, ej1

=

0 ⩽ j ⩽ m,

Pj0 ,

(7.79)

0 ⩽ j ⩽ m.

(7.80)

Note that (7.76)–(7.78) can be rewritten uniformly as ̄ ̄ 󵄨 󵄨2 ̄ 󵄨 󵄨2 ̄ Δt ejk = (1 + ic1 )δx2 ejk + ejk − (1 + ic2 )(󵄨󵄨󵄨Ujk 󵄨󵄨󵄨 Ujk − 󵄨󵄨󵄨ujk 󵄨󵄨󵄨 ujk ) + Pjk ,

0 ⩽ j ⩽ m, 1 ⩽ k ⩽ n − 1.

(7.81)

Combining (7.79)–(7.80) with (7.56)–(7.57), it follows: 󵄩󵄩 0 󵄩󵄩 󵄩󵄩e 󵄩󵄩 = 0, 󵄨󵄨 0 󵄨󵄨 󵄨󵄨e 󵄨󵄨1 = 0,

󵄩󵄩 1 󵄩󵄩 √ 2 󵄩󵄩e 󵄩󵄩 ⩽ Lc15 τ , 󵄨󵄨 1 󵄨󵄨 √ 2 󵄨󵄨e 󵄨󵄨1 ⩽ Lc15 τ .

(7.82) (7.83)

According to (7.8) and (7.64), we know that 󵄨󵄨 k 󵄨󵄨 󵄨󵄨Uj 󵄨󵄨 ⩽ c5 ,

󵄨󵄨 k 󵄨󵄨 󵄨󵄨uj 󵄨󵄨 ⩽ c16 ,

0 ⩽ j ⩽ m, 0 ⩽ k ⩽ n.

(7.84)

Moreover, it holds 󵄨󵄨 k 󵄨󵄨2 k̄ 󵄨󵄨 k 󵄨󵄨2 k̄ 󵄨󵄨 k 󵄨󵄨2 k̄ k k k k k̄ 󵄨󵄨Uj 󵄨󵄨 Uj − 󵄨󵄨uj 󵄨󵄨 uj = 󵄨󵄨uj 󵄨󵄨 ej + (uj ēj + ej Ū j )Uj . (I) Taking the inner product of (7.81) on both the right- and left-hand sides with ek , it follows: ̄

(Δt ek , ek ) =(1 + ic1 )(δx2 ek , ek ) + (ek , ek ) ̄

̄

̄

̄

̄

244 � 7 Difference methods for the Kuramoto–Tsuzuki equation − (1 + ic2 )(|U k |2 U k − |uk |2 uk , ek ) + (Pk , ek ) ̄ 󵄩 ̄ 󵄩2 = − (1 + ic1 )|ek |21 + 󵄩󵄩󵄩ek 󵄩󵄩󵄩 ̄ ̄ ̄ 󵄨 󵄨2 ̄ − (1 + ic2 )(󵄨󵄨󵄨uk 󵄨󵄨󵄨 ek + (uk ēk + ek Ū k )U k , ek ) + (Pk , ek ). ̄

̄

̄

̄

Taking the real parts of the above equality on both the right- and left-hand sides and using (7.84), we have 1 󵄩󵄩 k+1 󵄩󵄩2 󵄩󵄩 k−1 󵄩󵄩2 (󵄩e 󵄩󵄩 − 󵄩󵄩e 󵄩󵄩 ) 4τ 󵄩 ̄ ̄ ̄ 󵄩 ̄ 󵄩2 ⩽ 󵄩󵄩󵄩ek 󵄩󵄩󵄩 + Re{−(1 + ic2 )((uk ēk + ek Ū k )U k , ek )} + Re{(Pk , ek )} ̄ 󵄩 ̄󵄩 󵄩 󵄩 󵄩 ̄󵄩 ⩽ ‖ek ‖2 + √1 + c22 (c5 + c16 )c5 ‖ek ‖ ⋅ 󵄩󵄩󵄩ek 󵄩󵄩󵄩 + 󵄩󵄩󵄩Pk 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩ek 󵄩󵄩󵄩

k+1 k−1 ‖ek+1 ‖ + ‖ek−1 ‖ 󵄩 󵄩 󵄩 󵄩 ‖e ‖ + ‖e ‖ + √1 + c22 (c5 + c16 )c5 󵄩󵄩󵄩ek 󵄩󵄩󵄩 + 󵄩󵄩󵄩Pk 󵄩󵄩󵄩] ⋅ , 2 2 1 ⩽ k ⩽ n − 1,

⩽[

which implies 1 󵄩󵄩 k+1 󵄩󵄩 󵄩󵄩 k−1 󵄩󵄩 ‖ek+1 ‖ + ‖ek−1 ‖ 󵄩 󵄩 󵄩 󵄩 (󵄩󵄩e 󵄩󵄩 − 󵄩󵄩e 󵄩󵄩) ⩽ + √1 + c22 (c5 + c16 )c5 󵄩󵄩󵄩ek 󵄩󵄩󵄩 + 󵄩󵄩󵄩Pk 󵄩󵄩󵄩, 2τ 2 1 ⩽ k ⩽ n − 1. Noticing (7.49) and (7.54), we have 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 (1 − τ)󵄩󵄩󵄩ek+1 󵄩󵄩󵄩 ⩽ (1 + τ)󵄩󵄩󵄩ek−1 󵄩󵄩󵄩 + 2√1 + c22 (c5 + c16 )c5 τ 󵄩󵄩󵄩ek 󵄩󵄩󵄩 + 2τ √L max{c13 , c14 }(τ 2 + h2 ),

1 ⩽ k ⩽ n − 1.

When τ ⩽ 31 , we have 󵄩󵄩 k+1 󵄩󵄩 󵄩 k−1 󵄩 󵄩 k󵄩 2 2 󵄩󵄩e 󵄩󵄩 ⩽ (1 + 3τ)󵄩󵄩󵄩e 󵄩󵄩󵄩 + 3√1 + c22 (c5 + c16 )c5 τ 󵄩󵄩󵄩e 󵄩󵄩󵄩 + 3√L max{c13 , c14 }τ(τ + h ) 󵄩 󵄩󵄩 󵄩 ⩽ {1 + 3[1 + √1 + c22 (c5 + c16 )c5 ]τ} max{󵄩󵄩󵄩ek−1 󵄩󵄩󵄩, 󵄩󵄩󵄩ek 󵄩󵄩󵄩} + 3√L max{c13 , c14 }τ(τ 2 + h2 ), 1 ⩽ k ⩽ n − 1, or 󵄩 󵄩󵄩 󵄩 󵄩󵄩 󵄩 󵄩 max{󵄩󵄩󵄩ek 󵄩󵄩󵄩, 󵄩󵄩󵄩ek+1 󵄩󵄩󵄩} ⩽ {1 + 3[1 + √1 + c22 (c5 + c16 )c5 ]τ} max{󵄩󵄩󵄩ek−1 󵄩󵄩󵄩, 󵄩󵄩󵄩ek 󵄩󵄩󵄩} + 3√L max{c13 , c14 }τ(τ 2 + h2 ), 1 ⩽ k ⩽ n − 1. Using the Gronwall inequality (Theorem 1.2(a)) and noticing (7.82), we have 󵄩 󵄩󵄩 󵄩 max{󵄩󵄩󵄩ek 󵄩󵄩󵄩, 󵄩󵄩󵄩ek+1 󵄩󵄩󵄩}

7.3 Three-level linearized difference scheme

2 󵄩 󵄩󵄩 󵄩 ⩽ e3[1+√1+c2 (c5 +c16 )c5 ]T [max{󵄩󵄩󵄩e0 󵄩󵄩󵄩, 󵄩󵄩󵄩e1 󵄩󵄩󵄩} +

2

2

≡ c20 (τ + h ),

1 + √1 +

c22 (c5

√L max{c13 , c14 }

2

⩽ e3[1+√1+c2 (c5 +c16 )c5 ]T √L[c15 +

√L max{c13 , c14 }

1 + √1 +

c22 (c5

+ c16 )c5

+ c16 )c5

� 245

(τ 2 + h2 )]

](τ 2 + h2 )

0 ⩽ k ⩽ n − 1.

(7.85)

(II) Taking the inner product of (7.81) on both the right- and left-hand sides with one has

1 Δ ek , 1−ic1 t

̄ 1 1 󵄩󵄩 k 󵄩󵄩 2 k̄ k (ek , Δt ek ) 󵄩Δ e 󵄩 = (δx e , Δt e ) + 1 + ic1 󵄩 t 󵄩 1 + ic1 1 + ic2 󵄨󵄨 k 󵄨󵄨2 k̄ 󵄨󵄨 k 󵄨󵄨2 k̄ 1 − (󵄨U 󵄨 U − 󵄨󵄨u 󵄨󵄨 u , Δt ek ) + (Pk , Δt ek ). 1 + ic1 󵄨 󵄨 1 + ic1

Taking the real parts on both the right- and left-hand sides of the above equality, it follows: 1 󵄩󵄩 k 󵄩󵄩2 1 󵄨 k+1 󵄨2 󵄨 k−1 󵄨2 󵄩󵄩Δt e 󵄩󵄩 + (󵄨󵄨󵄨e 󵄨󵄨󵄨1 − 󵄨󵄨󵄨e 󵄨󵄨󵄨1 ) 2 4τ 1 + c1 = Re{

̄ ̄ 1 + ic2 󵄨󵄨 k 󵄨󵄨2 k̄ 1 (ek , Δt ek )} + Re{− (󵄨󵄨u 󵄨󵄨 e + (uk ēk + ek Ū k )U k , Δt ek )} 1 + ic1 1 + ic1

+ Re{ ⩽

1

1 (Pk , Δt ek )} 1 + ic1

󵄩󵄩 k̄ 󵄩󵄩 󵄩󵄩 k 󵄩󵄩 󵄩󵄩e 󵄩󵄩 ⋅ 󵄩󵄩Δt e 󵄩󵄩 + 2

√1 + c1 1

+

√1 + ⩽[

c12

√1 + c22 √1 +

c12

󵄩 k󵄩 󵄩 k󵄩 2 󵄩 k̄ 󵄩 (c16 󵄩󵄩󵄩e 󵄩󵄩󵄩 + (c5 + c16 )c5 󵄩󵄩󵄩e 󵄩󵄩󵄩)󵄩󵄩󵄩Δt e 󵄩󵄩󵄩

󵄩󵄩 k 󵄩󵄩 󵄩󵄩 k 󵄩󵄩 󵄩󵄩P 󵄩󵄩 ⋅ 󵄩󵄩Δt e 󵄩󵄩

1 1 󵄩󵄩 k 󵄩󵄩2 󵄩󵄩 k̄ 󵄩󵄩2 4 󵄩 󵄩󵄩ek̄ 󵄩󵄩󵄩2 ] ‖Δt ek ‖2 + (1 + c22 )c16 󵄩󵄩Δt e 󵄩󵄩 + 󵄩󵄩e 󵄩󵄩 ] + [ 󵄩 󵄩 2 2 4(1 + c1 ) 4(1 + c1 )

+[ +[

1 󵄩󵄩 k 󵄩󵄩2 2 2 2 󵄩 k 󵄩2 󵄩Δt e 󵄩󵄩 + (1 + c2 )(c5 + c16 ) c5 󵄩󵄩󵄩e 󵄩󵄩󵄩 ] 4(1 + c12 ) 󵄩 1 󵄩󵄩 k 󵄩󵄩2 󵄩󵄩 k 󵄩󵄩2 󵄩Δt e 󵄩󵄩 + 󵄩󵄩P 󵄩󵄩 ], 4(1 + c12 ) 󵄩

1 ⩽ k ⩽ n − 1,

which implies 1 󵄨󵄨 k+1 󵄨󵄨2 󵄨󵄨 k−1 󵄨󵄨2 (󵄨e 󵄨󵄨1 − 󵄨󵄨e 󵄨󵄨1 ) 4τ 󵄨 ̄ 󵄩2 󵄩 󵄩2 󵄩 󵄩2 4 󵄩 ⩽ [1 + (1 + c22 )c16 ]󵄩󵄩󵄩ek 󵄩󵄩󵄩 + (1 + c22 )(c5 + c16 )2 c52 󵄩󵄩󵄩ek 󵄩󵄩󵄩 + 󵄩󵄩󵄩Pk 󵄩󵄩󵄩 ,

1 ⩽ k ⩽ n − 1.

246 � 7 Difference methods for the Kuramoto–Tsuzuki equation Combining (7.85) with (7.49) and (7.54), it follows that 1 󵄨󵄨 k+1 󵄨󵄨2 󵄨󵄨 k−1 󵄨󵄨2 (󵄨e 󵄨󵄨1 − 󵄨󵄨e 󵄨󵄨1 ) 4τ 󵄨

4 2 ⩽ [1 + (1 + c22 )c16 ]c20 (τ 2 + h2 )

2 2

2

2 2 2 + (1 + c22 )(c5 + c16 )2 c52 c20 (τ 2 + h2 ) + L max{c13 , c14 }(τ 2 + h2 )

4 2 2 2 2 = {[1 + (1 + c22 )c16 ]c20 + (1 + c22 )(c5 + c16 )2 c52 c20 + L max{c13 , c14 }}(τ 2 + h2 ) 2

≡ c21 (τ 2 + h2 ) ,

2

1 ⩽ k ⩽ n − 1,

which leads to 󵄨󵄨 k+1 󵄨󵄨2 󵄨󵄨 k−1 󵄨󵄨2 2 2 2 󵄨󵄨e 󵄨󵄨1 ⩽ 󵄨󵄨e 󵄨󵄨1 + 4c21 τ(τ + h ) ,

1 ⩽ k ⩽ n − 1.

By recursion, we have 󵄨󵄨 k 󵄨󵄨2 󵄨 1 󵄨2 󵄨 0 󵄨2 2 2 2 󵄨󵄨e 󵄨󵄨1 ⩽ max{󵄨󵄨󵄨e 󵄨󵄨󵄨1 , 󵄨󵄨󵄨e 󵄨󵄨󵄨1 } + 2c21 kτ(τ + h ) ,

1 ⩽ k ⩽ n.

Noticing (7.83), it follows: 󵄨󵄨 k 󵄨󵄨2 2 2 2 2 󵄨󵄨e 󵄨󵄨1 ⩽ (Lc15 + 2c21 T)(τ + h ) ,

1 ⩽ k ⩽ n.

7.4 Numerical experiments In this section, we will take the difference scheme (7.59)–(7.63) as an example to test the theoretical results. Our theoretical results in this chapter can be easily extended to the spatial domain [−1, 1]. Take h = m2 , xi = −1 + ih, 0 ⩽ i ⩽ m. We define the maximum norm errors between the numerical solution and the exact solution as E∞ (h, τ) = max0⩽k⩽n ‖U k − uk ‖∞ . The convergence orders in time and in space are respectively defined by Orderτ = log2 (E∞ (h, 2τ)/E∞ (h, τ)),

Orderh = log2 (E∞ (2h, τ)/E∞ (h, τ)).

Example 7.1. In the problem (7.1)–(7.3), take c1 = c2 = 1, φ(x) = cos(πx) and replace 0 on the right-hand side of (7.1) by −ie−it cos(πx) + (1 + i)π 2 e−it cos(πx) − e−it cos(πx) + (1 + i)e−it cos3 (πx). The exact solution is u(x, t) = e−it cos(πx). The numerical results are listed in Table 7.1 and Table 7.2. As we see from these data, all of them confirm that the numerical convergence orders are two in both space and time. The real part and imaginary part of numerical solutions and errors are respectively shown in Figure 7.1

7.5 Summary and extension

� 247

Table 7.1: Maximum errors and the temporal convergence orders with h = 1/600. τ

E∞ (h, τ)

Orderτ

1/4 1/8 1/16 1/32 1/64

4.924042e-2 1.221859e-2 3.048770e-3 7.627343e-4 1.916350e-4

2.0108 2.0028 1.9990 1.9928

Table 7.2: Maximum errors and the spatial convergence orders with τ = 1/600. h

E∞ (h, τ)

Orderh

1/4 1/8 1/16 1/32 1/64

4.282604e-2 9.017158e-3 2.038037e-3 4.882680e-4 1.220695e-4

2.2477 2.1455 2.0614 2.0000

Figure 7.1: The numerical solution and the error surface with h = 1/40, τ = 1/40; left: the real part of the numerical solution and its numerical error; right: imaginary part of the numerical solution and its numerical error.

7.5 Summary and extension In this chapter, a two-level nonlinear difference scheme and a three-level linearized difference scheme for the initial-boundary value problem of the Kuramoto–Tsuzuki equa-

248 � 7 Difference methods for the Kuramoto–Tsuzuki equation tion are derived. The existence, boundedness and convergence of the solution of both difference schemes are proved. The results of this chapter are taken from [32, 33, 34]. The Kuramoto–Tsuzuki equation and the Schrödinger equation are both complex semilinear evolution equations. Boundary value conditions are formulated in different ways. Hu et al. [16] established a compact difference scheme for the Kuramoto–Tsuzuki equation and proved the convergence of the solution of the compact difference scheme.

8 Difference methods for the Zakharov equation 8.1 Introduction The quantum plays an important role in ultrasmall electronic devices, dense astrophysics, plasma systems and laser plasmas. In 1972, Vladimir Zakharov first established the system of nonlinear coupled wave equations, which describes the interaction between the high frequency Langmuir waves and the low frequency ion acoustic waves [49]. Since then, the Zakharov system has attracted extensive research. In this chapter, we consider the finite difference methods for solving the following initial and boundary value problem of the Zakharov equation: iut + uxx − uv = 0, { { { { { vtt − vxx − (|u|2 )xx = 0, { { u(x, 0) = φ(x), v(x, 0) = ψ(x), vt (x, 0) = ψ1 (x), { { { { u(0, t) = 0, u(L, t) = 0, v(0, t) = 0, v(L, t) = 0,

0 < x < L, 0 < t ⩽ T, (8.1) 0 < x < L, 0 < t ⩽ T, (8.2) 0 < x < L,

(8.3)

0 ⩽ t ⩽ T,

(8.4)

where φ(0) = φ(L) = ψ(0) = ψ(L) = ψ1 (0) = ψ1 (L) = 0, u(x, t) and φ(x) are complex functions; v(x, t), ψ(x) and ψ1 (x) are real functions. For any fixed time t, let w(x, t) be the solution of the following problem: {

wxx (x, t) = vt (x, t), w(0, t) = 0,

0 < x < L,

w(L, t) = 0.

(8.5) (8.6)

Then w(x, t) exists uniquely according to the theory of the two-point boundary value problem of the second-order ordinary differential equation. Moreover, we easily have the following estimates: √6L 󵄩 󵄨󵄨 󵄨 󵄩󵄩v (⋅, t)󵄩󵄩󵄩, 󵄨󵄨w(⋅, t)󵄨󵄨󵄨1 ⩽ 󵄩 6 󵄩 t L√6L 󵄩󵄩 󵄩 󵄩󵄩 󵄩 󵄩v (⋅, t)󵄩󵄩󵄩. 󵄩󵄩w(⋅, t)󵄩󵄩󵄩∞ ⩽ 12 󵄩 t In particular, when t = 0, it holds that {

wxx (x, 0) = ψ1 (x), w(0, 0) = 0,

0 < x < L,

w(L, 0) = 0.

Solving the above problem yields L

x

0

0

x w(x, 0) = − ∫(L − s)ψ1 (s)ds + ∫(x − s)ψ1 (s)ds. L https://doi.org/10.1515/9783110796018-008

250 � 8 Difference methods for the Zakharov equation Denote L

x

0

0

x F(ψ1 )(x) = − ∫(L − s)ψ1 (s)ds + ∫(x − s)ψ1 (s)ds. L

(8.7)

Then we have w(x, 0) = F(ψ1 )(x),

0 ⩽ x ⩽ L.

Substituting (8.5) into (8.2), it follows: (wt − v − |u|2 )xx = 0,

0 < x < L.

Since (wt − v − |u|2 )(0, t) = 0,

(wt − v − |u|2 )(L, t) = 0,

one has wt − v − |u|2 = 0,

0 ⩽ x ⩽ L.

Therefore, the problem (8.1)–(8.4) is equivalent to the following one: { { { { { { { { { { { { { { { { {

iut + uxx − uv = 0, vt − wxx = 0, 2

wt − v − |u| = 0, u(x, 0) = φ(x),

u(0, t) = 0,

v(x, 0) = ψ(x),

u(L, t) = 0,

w(x, 0) = F(ψ1 )(x),

w(0, t) = 0,

w(L, t) = 0,

0 < x < L, 0 < t ⩽ T,

(8.8)

0 < x < L, 0 < t ⩽ T,

(8.9)

0 ⩽ x ⩽ L, 0 < t ⩽ T,

(8.10)

0 < x < L,

(8.11)

0 ⩽ t ⩽ T,

(8.12)

where F(ψ1 )(x) is defined by (8.7). Theorem 8.1. Suppose {u(x, t), v(x, t), w(x, t)} is the solution of (8.8)–(8.12). Denote L

󵄨 󵄨2 Q(t) = ∫󵄨󵄨󵄨u(x, t)󵄨󵄨󵄨 dx, 0

L

1 󵄨 󵄨2 1 󵄨 󵄨2 E(t) = ∫[󵄨󵄨󵄨ux (x, t)󵄨󵄨󵄨 + wx2 (x, t) + v2 (x, t) + 󵄨󵄨󵄨u(x, t)󵄨󵄨󵄨 v(x, t)]dx. 2 2 0

Then we have Q(t) = Q(0),

0 ⩽ t ⩽ T,

(8.13)

E(t) = E(0),

0 ⩽ t ⩽ T.

(8.14)

8.1 Introduction

� 251

̄ t) and integrating Proof. (I) Multiplying both the right- and left-hand sides of (8.8) by u(x, the result with respect to x from 0 to L, we have L

L

L

0

0

0

2 ̄ t)dx + ∫ uxx (x, t)u(x, ̄ t)dx − ∫󵄨󵄨󵄨󵄨u(x, t)󵄨󵄨󵄨󵄨 v(x, t)dx = 0. i ∫ ut (x, t)u(x,

Taking the imaginary parts on both the right- and left-hand sides, one yields L

1 d 󵄨󵄨 󵄨2 ∫󵄨u(x, t)󵄨󵄨󵄨 dx = 0, 2 dt 󵄨 0

which implies that (8.13) is valid. (II) Multiplying both the right- and left-hand sides of (8.8) by −2ū t (x, t) and integrating the result with respect to x from 0 to L, we have L

L

L

0

0

0

󵄨 󵄨2 −2i ∫󵄨󵄨󵄨ut (x, t)󵄨󵄨󵄨 dx − 2 ∫ uxx (x, t)ū t (x, t)dx + 2 ∫ v(x, t)u(x, t)ū t (x, t)dx = 0. Taking the real parts on both the right- and left-hand sides of the above equality, we have L

L

0

0

d 󵄨󵄨 󵄨2 󵄨 󵄨2 ∫󵄨u (x, t)󵄨󵄨󵄨 dx + ∫ v(x, t)(󵄨󵄨󵄨u(x, t)󵄨󵄨󵄨 )t dx = 0. dt 󵄨 x

(8.15)

Multiplying both the right- and left-hand sides of (8.9) by wt (x, t) and integrating the result with respect to x from 0 to L, we have L

L

∫ vt (x, t)wt (x, t)dx − ∫ wt (x, t)wxx (x, t)dx = 0. 0

(8.16)

0

Multiplying both the right- and left-hand sides of (8.10) by −vt (x, t) and integrating the result with respect to x from 0 to L, we have L

L

L

0

0

0

󵄨 󵄨2 − ∫ wt (x, t)vt (x, t)dx + ∫ v(x, t)vt (x, t)dx + ∫󵄨󵄨󵄨u(x, t)󵄨󵄨󵄨 vt (x, t)dx = 0.

(8.17)

A combination of (8.16) with (8.17) leads to L

L

L

0

0

0

1 d 1 d 󵄨 󵄨2 ⋅ ∫ wx2 (x, t)dx + ⋅ ∫ v2 (x, t)dx + ∫󵄨󵄨󵄨u(x, t)󵄨󵄨󵄨 vt (x, t)dx = 0. 2 dt 2 dt

(8.18)

252 � 8 Difference methods for the Zakharov equation Adding (8.15) and (8.18) together, we have L

d 1 󵄨2 1 󵄨2 󵄨 󵄨 ∫[󵄨󵄨󵄨ux (x, t)󵄨󵄨󵄨 + wx2 (x, t) + v2 (x, t) + 󵄨󵄨󵄨u(x, t)󵄨󵄨󵄨 v(x, t)]dx = 0, dt 2 2 0

which implies that (8.14) is valid. According to the above theorem, we have L

󵄨 󵄨2 1 ∫[󵄨󵄨󵄨ux (x, t)󵄨󵄨󵄨 + v2 (x, t)]dx 2 0

L

󵄨 󵄨2 ⩽ − ∫󵄨󵄨󵄨u(x, t)󵄨󵄨󵄨 v(x, t)dx + E(0) 0

L

L

0

0

1 󵄨 󵄨4 ⩽ ∫ v2 (x, t)dx + ∫󵄨󵄨󵄨u(x, t)󵄨󵄨󵄨 dx + E(0) 4 L

⩽

1 󵄩 󵄩2 󵄩 󵄩2 ∫ v2 (x, t)dx + 󵄩󵄩󵄩u(⋅, t)󵄩󵄩󵄩∞ 󵄩󵄩󵄩u(⋅, t)󵄩󵄩󵄩 + E(0) 4 0

L

=

1 󵄩 󵄩2 ∫ v2 (x, t)dx + 󵄩󵄩󵄩u(⋅, t)󵄩󵄩󵄩∞ Q(0) + E(0) 4 0

L

1 󵄨 󵄩2 󵄨2 1 󵄩 ⩽ ∫ v2 (x, t)dx + (ε󵄨󵄨󵄨u(⋅, t)󵄨󵄨󵄨1 + 󵄩󵄩󵄩u(⋅, t)󵄩󵄩󵄩 )Q(0) + E(0). 4 4ε 0

Taking ε =

1 , 2Q(0)

we have 󵄨󵄨 󵄨2 󵄨󵄨u(⋅, t)󵄨󵄨󵄨1 +

1 󵄩󵄩 󵄩2 3 󵄩v(⋅, t)󵄩󵄩󵄩 ⩽ Q (0) + 2E(0), 2󵄩

0 ⩽ t ⩽ T,

which further implies 4 󵄩󵄩 󵄩2 󵄩u(⋅, t)󵄩󵄩󵄩∞ + L󵄩

1 󵄩󵄩 󵄩2 3 󵄩v(⋅, t)󵄩󵄩󵄩 ⩽ Q (0) + 2E(0), 2󵄩

0 ⩽ t ⩽ T.

Therefore, there is a constant c1 satisfying 󵄩󵄩 󵄩 󵄩󵄩u(⋅, t)󵄩󵄩󵄩∞ ⩽ c1 ,

󵄩󵄩 󵄩 󵄩󵄩v(⋅, t)󵄩󵄩󵄩 ⩽ c1 ,

0 ⩽ t ⩽ T.

8.2 Two-level nonlinear difference scheme

� 253

8.2 Two-level nonlinear difference scheme 8.2.1 Derivation of the difference scheme Considering equations (8.8)–(8.10) at the point (xj , tk+ 1 ) and then using the numerical 2 differential formula, we have k+ 21

iδt Uj

k+ 21

k+ 21

+ δx2 Uj

− Uj

k+ 21

δt Vj k+ 21

δt Wj

k+ 21

− Vj

−

k+ 21

k+ 21

Vj

= Pj

k+ 21

k+ 21

− δx2 Wj

= Qj

|Ujk |2 + |Ujk+1 |2

k+ 21

= Rj

2

,

1 ⩽ j ⩽ m − 1, 0 ⩽ k ⩽ n − 1,

(8.19)

,

1 ⩽ j ⩽ m − 1, 0 ⩽ k ⩽ n − 1,

(8.20)

,

0 ⩽ j ⩽ m, 0 ⩽ k ⩽ n − 1,

(8.21)

where there is a constant c2 satisfying 󵄨󵄨 k+ 21 󵄨󵄨 2 2 󵄨󵄨Pj 󵄨󵄨 ⩽ c2 (τ + h ), 1 ⩽ j ⩽ m − 1, 0 ⩽ k ⩽ n − 1, 󵄨󵄨 k+ 21 󵄨󵄨 2 2 󵄨󵄨Qj 󵄨󵄨 ⩽ c2 (τ + h ), 1 ⩽ j ⩽ m − 1, 0 ⩽ k ⩽ n − 1, 1 󵄨󵄨󵄨Rk+ 2 󵄨󵄨󵄨 ⩽ c2 τ 2 , 0 ⩽ j ⩽ m, 0 ⩽ k ⩽ n − 1, 󵄨 j 󵄨 k+ 21

k− 21

󵄨󵄨 Pj 󵄨󵄨 󵄨󵄨 󵄨󵄨

− Pj

󵄨󵄨 Qj 󵄨󵄨 󵄨󵄨 󵄨󵄨

− Qj

󵄨󵄨 Rj 󵄨󵄨 󵄨󵄨 󵄨󵄨

− Rj

k+ 21

k+ 21

τ

k− 21

τ

k− 21

(8.23) (8.24)

󵄨󵄨 󵄨󵄨 2 2 󵄨󵄨 ⩽ c2 (τ + h ), 󵄨󵄨

1 ⩽ j ⩽ m − 1, 1 ⩽ k ⩽ n − 1,

(8.25)

󵄨󵄨 󵄨󵄨 2 2 󵄨󵄨 ⩽ c2 (τ + h ), 󵄨󵄨

1 ⩽ j ⩽ m − 1, 1 ⩽ k ⩽ n − 1,

(8.26)

󵄨󵄨 󵄨󵄨 2 󵄨󵄨 ⩽ c2 τ , 󵄨󵄨

τ

(8.22)

0 ⩽ j ⩽ m, 1 ⩽ k ⩽ n − 1.

(8.27)

By the initial-boundary value conditions (8.11)–(8.12), it follows: Uj0 = φ(xj ), U0k = 0,

Vj0 = ψ(xj ),

Umk = 0,

Wj0 = F(ψ1 )(xj ),

W0k = 0,

Wmk = 0,

0 ⩽ j ⩽ m, 1 ⩽ k ⩽ n.

(8.28) (8.29)

Omitting the small terms in (8.19)–(8.21), a difference scheme for solving the problem (8.8)–(8.12) reads { { { { { { { { { { { { { { { { { { { { { { { { { { {

k+ 21

iδt uj

k+ 21

δ t vj

k+ 21

δt wj

k+ 21

k+ 21

− δx2 wj

k+ 21

− vj

uj0 = φ(xj ), u0k = 0,

k+ 21 k+ 21 vj

+ δx2 uj

−

− uj

= 0,

= 0, |ujk |2 + |ujk+1 |2

2 v0j = ψ(xj ),

k um = 0,

= 0,

wj0 = G(ψ1 )j ,

w0k = 0,

k wm = 0,

1 ⩽ j ⩽ m − 1, 0 ⩽ k ⩽ n − 1,

(8.30)

1 ⩽ j ⩽ m − 1, 0 ⩽ k ⩽ n − 1,

(8.31)

0 ⩽ j ⩽ m, 0 ⩽ k ⩽ n − 1,

(8.32)

0 ⩽ j ⩽ m,

(8.33)

1 ⩽ k ⩽ n,

(8.34)

254 � 8 Difference methods for the Zakharov equation where wj0 = G(ψ1 )j (0 ⩽ j ⩽ m) is determined by δx2 wj0 = ψ1 (xj ),

{

w00 = 0,

1 ⩽ j ⩽ m − 1,

0 wm = 0.

(8.35) (8.36)

From the consistency of the initial and boundary value conditions, we know 0 0 u00 = um = w00 = wm = v00 = v0m = 0.

Taking j = 0 and j = m in (8.32), it follows: vk0 = 0,

vkm = 0,

1 ⩽ k ⩽ n.

In other words, (8.32) is equivalent to 1 1 |uk |2 + |ujk+1 |2 { { δ wk+ 2 − vk+ 2 − j = 0, t j j 2 { { k k { v0 = 0, vm = 0,

1 ⩽ j ⩽ m − 1, 0 ⩽ k ⩽ n − 1, 1 ⩽ k ⩽ n.

Combining (8.31)–(8.32) with (8.35), we have 󵄨 󵄨2 1 1 |uj0 |2 + 󵄨󵄨󵄨uj1 󵄨󵄨󵄨 2 (δt vj2 − ψ1 (xj )) − δx2 vj2 − δx2 = 0, 1 ⩽ j ⩽ m − 1, τ 2 |uk+1 |2 + 2|ujk |2 + |ujk−1 |2 vk+1 + 2vkj + vk−1 j 2 j 2 k 2 j − δx = 0, δt vj − δx 4 4 1 ⩽ j ⩽ m − 1, 1 ⩽ k ⩽ n − 1. Thus, the solution {ujk , vkj | 0 ⩽ j ⩽ m, 0 ⩽ k ⩽ n} of the difference scheme (8.30)–(8.34) satisfies { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { {

k+ 21

iδt uj

k+ 21

+ δx2 uj

k+ 21 k+ 21 vj

− uj

= 0,

1 ⩽ j ⩽ m − 1, 0 ⩽ k ⩽ n − 1,

1 1 |uj0 |2 + |uj1 |2 2 (δt vj2 − ψ1 (xj )) − δx2 vj2 − δx2 = 0, 1 ⩽ j ⩽ m − 1, τ 2 vk+1 + 2vkj + vk−1 |ujk+1 |2 + 2|ujk |2 + |ujk−1 |2 j j δt2 vkj − δx2 − δx2 = 0, 4 4 1 ⩽ j ⩽ m − 1, 1 ⩽ k ⩽ n − 1,

uj0 u0k

= φ(xj ),

v0j

= 0,

= 0,

k um

= ψ(xj ), vk0

0 ⩽ j ⩽ m,

= 0,

vkm

= 0,

(8.37) (8.38)

(8.39) (8.40)

1 ⩽ k ⩽ n.

(8.41)

Therefore, we can construct the difference scheme (8.37)–(8.41) for the original problem (8.1)–(8.4). It is easy to see that the truncation errors of (8.37) and (8.39) are O(τ 2 + h2 ), but that of (8.38) is O(τ + h2 ).

8.2 Two-level nonlinear difference scheme

� 255

8.2.2 Existence of the difference solution When the value of {uk , vk , wk } at the k-th time level is known, the average value 1 1 1 {uk+ 2 , vk+ 2 , wk+ 2 } can be viewed as unknowns. It follows from (8.30)–(8.34) that { { { { { { { { { { { { { { { { { { { { {

2 k+ 21 k+ 1 k+ 1 k+ 1 (uj − ujk ) − iδx2 uj 2 + iuj 2 vj 2 = 0, τ 2 k+ 21 k+ 1 (vj − vkj ) − δx2 wj 2 = 0, τ 2 k+ 21 1 󵄨 󵄨2 󵄨 k+ 1 k+ 1 󵄨2 (wj − wjk ) − vj 2 − (󵄨󵄨󵄨ujk 󵄨󵄨󵄨 + 󵄨󵄨󵄨2uj 2 − ujk 󵄨󵄨󵄨 ) = 0, τ 2 k+ 21

u0

k+ 21

= um

k+ 21

= v0

k+ 21

= vm

k+ 21

= w0

k+ 21

= wm

1 ⩽ j ⩽ m − 1,

(8.42)

1 ⩽ j ⩽ m − 1,

(8.43)

1 ⩽ j ⩽ m − 1,

(8.44)

= 0.

(8.45)

1 ⩽ j ⩽ m − 1.

(8.46)

By (8.43), we have k+ 21

vj

τ k+ 1 = vkj + δx2 wj 2 , 2

Substituting (8.46) into (8.42) and (8.44), respectively, we have the system of nonlinear equations: { { { { { { { { { { { { { { { { { { { { { { { { {

2 k+ 21 τ k+ 1 k+ 1 k+ 1 (uj − ujk ) − iδx2 uj 2 + iuj 2 (vkj + δx2 wj 2 ) = 0, τ 2 1 ⩽ j ⩽ m − 1, 2 τ (w − wjk ) − (vkj + δx2 wj τ j 2 1 ⩽ j ⩽ m − 1, k+ 21

k+ 1 u0 2

=

k+ 21

k+ 1 um 2

=

k+ 1 w0 2

=

k+ 1 wm 2

1 󵄨 󵄨2 󵄨 ) − (󵄨󵄨󵄨ujk 󵄨󵄨󵄨 + 󵄨󵄨󵄨2uj 2

k+ 21

= 0.

(8.47)

󵄨2 − ujk 󵄨󵄨󵄨 ) = 0, (8.48) (8.49)

Denote k+ 21

uj = uj

,

k+ 21

wj = wj

,

0 ⩽ j ⩽ m.

Rewrite (8.47)–(8.49) as { { { { { { { { { { { { {

2 τ (uj − ujk ) − iδx2 uj + iuj (vkj + δx2 wj ) = 0, τ 2 2 τ 1 󵄨 󵄨2 (w − wjk ) − (vkj + δx2 wj ) − (󵄨󵄨󵄨ujk 󵄨󵄨󵄨 + |2uj − ujk |2 ) = 0, τ j 2 2 u0 = um = w0 = wm = 0.

1 ⩽ j ⩽ m − 1, (8.50) 1 ⩽ j ⩽ m − 1, (8.51) (8.52)

If (u1 , u2 , . . . , um−1 ) is known, (w1 , w2 , . . . , wm−1 ) is uniquely determined in view of (8.51) and (8.52). Therefore, we have wj = wj (u1 , u2 , . . . , um−1 ),

j = 1, 2, . . . , m − 1.

256 � 8 Difference methods for the Zakharov equation In this way, (8.50) can be viewed as a system of nonlinear equations in (u1 , u2 , . . . , um−1 ). Define the operator Π : 𝒰h̊ → 𝒰h̊ by 2 τ { { (uj − ujk ) − iδx2 uj + iuj (vkj + δx2 wj ), 2 Π(u)j = { τ { 0, {

1 ⩽ j ⩽ m − 1, j = 0, m,

where (w1 , w2 , . . . , wm−1 ) is determined by { 2 (wj − wjk ) − (vkj + τ δx2 wj ) − 1 (󵄨󵄨󵄨ujk 󵄨󵄨󵄨2 + 󵄨󵄨󵄨2uj − ujk 󵄨󵄨󵄨2 ) = 0, 󵄨 2 2 󵄨 󵄨 󵄨 { τ w = 0, w = 0. { 0 m

1 ⩽ j ⩽ m − 1,

Calculation shows Re(Π(u), u) =

2 2 2 󵄩 󵄩 Re(u − uk , u) = [‖u‖2 − Re(uk , u)] ⩾ ‖u‖(‖u‖ − 󵄩󵄩󵄩uk 󵄩󵄩󵄩). τ τ τ

When ‖u‖ = ‖uk ‖, it follows that Re(Π(u), u) ⩾ 0. By the Browder theorem (Theorem 2.4), there is a solution to the system of equations (8.50)–(8.52). Naturally, we obtain the following theorem. Theorem 8.2. There is a solution to the difference scheme (8.30)–(8.34). We can solve (8.50)–(8.52) by the following iterative algorithm. When {uj(l−1) | 1 ⩽ j ⩽ m − 1} is known, we have {wj(l) | 1 ⩽ j ⩽ m − 1} by solving the tri-diagonal linear system: τ 1 󵄨 󵄨2 2 { { (wj(l) − wjk ) − (vkj + δx2 wj(l) ) − (󵄨󵄨󵄨ujk 󵄨󵄨󵄨 + |2uj(l−1) − ujk |2 ) = 0, τ 2 2 { { (l) (l) { w0 = 0, wm = 0.

1 ⩽ j ⩽ m − 1,

Then we have {uj(l) | 1 ⩽ j ⩽ m − 1} by solving the tri-diagonal linear system: τ 2 { { (uj(l) − ujk ) − iδx2 uj(l) + iuj(l) (vkj + δx2 wj(l) ) = 0, τ 2 { { (l) (l) { u0 = 0, um = 0. We take uj = uj(l) ,

wj = wj(l) ,

1⩽j ⩽m−1

until 󵄩󵄩 (l) (l−1) 󵄩 󵄩󵄩 ⩽ ε. 󵄩󵄩u − u 󵄩∞

1 ⩽ j ⩽ m − 1,

8.2 Two-level nonlinear difference scheme

� 257

According to ujk+1 = 2uj − ujk ,

wjk+1 = 2wj − wjk ,

1 ⩽ j ⩽ m − 1,

we have {uk+1 , wk+1 }. With the help of (8.31), we finally get vk+1 = vkj + τδx2 wj , j

1 ⩽ j ⩽ m − 1.

8.2.3 Conservation and boundedness of the difference solution Theorem 8.3. Suppose {ujk , vkj , wjk | 0 ⩽ j ⩽ m, 0 ⩽ k ⩽ n} is the solution of the difference scheme (8.30)–(8.34). Denote 󵄩 󵄩2 Qk = 󵄩󵄩󵄩uk 󵄩󵄩󵄩 , 󵄨 󵄨2 E k = 󵄨󵄨󵄨uk 󵄨󵄨󵄨1 +

1 󵄨󵄨 k 󵄨󵄨2 󵄨w 󵄨 + 2 󵄨 󵄨1

0 ⩽ k ⩽ n,

m−1 1 󵄩󵄩 k 󵄩󵄩2 󵄨 k 󵄨2 k 󵄩󵄩v 󵄩󵄩 + h ∑ 󵄨󵄨󵄨uj 󵄨󵄨󵄨 vj , 2 j=1

0 ⩽ k ⩽ n.

Then we have Qk = Q0 ,

0 ⩽ k ⩽ n,

(8.53)

E =E ,

0 ⩽ k ⩽ n.

(8.54)

k

0

k+ 21

Proof. (I) Multiplying both the right- and left-hand sides of (8.30) by hū j over j from 1 to m − 1, we have m−1

k+ 21

ih ∑ (δt uj j=1

k+ 21

)ū j

m−1

k+ 21

+ h ∑ (δx2 uj j=1

k+ 21

)ū j

m−1

k+ 21 󵄨 k+ 21 󵄨2 󵄨󵄨u 󵄨󵄨 󵄨 j 󵄨

− h ∑ vj j=1

and summing

= 0.

Taking the imaginary parts on both the right- and left-hand sides, it follows: 1 󵄩󵄩 k+1 󵄩󵄩2 󵄩󵄩 k 󵄩󵄩2 (󵄩u 󵄩󵄩 − 󵄩󵄩u 󵄩󵄩 ) = 0, 2τ 󵄩

0 ⩽ k ⩽ n − 1,

which implies Qk+1 = Qk , Therefore, (8.53) holds.

0 ⩽ k ⩽ n − 1. k+ 21

(II) Multiplying both the right- and left-hand sides of (8.30) by −2hδt ū j ming over j from 1 to m − 1, we have

and sum-

258 � 8 Difference methods for the Zakharov equation m−1

m−1

1

m−1

1

1

1

1

1

k+ k+ k+ k+ k+ 󵄨 k+ 󵄨2 −2ih ∑ 󵄨󵄨󵄨δt uj 2 󵄨󵄨󵄨 − 2h ∑ (δx2 uj 2 )(δt ū j 2 ) + 2h ∑ vj 2 uj 2 δt ū j 2 = 0, j=1

j=1

0 ⩽ k ⩽ n − 1.

j=1

Taking the real parts on both the right- and left-hand sides of the above equality, we have k+1 2 k 2 m−1 1 󵄨󵄨 k+1 󵄨󵄨2 󵄨󵄨 k 󵄨󵄨2 k+ 1 |uj | − |uj | (󵄨󵄨u 󵄨󵄨1 − 󵄨󵄨u 󵄨󵄨1 ) + h ∑ vj 2 = 0, τ τ j=1

0 ⩽ k ⩽ n − 1. k+ 21

Multiplying both the right- and left-hand sides of (8.31) by hδt wj over j from 1 to m − 1, we have m−1

k+ 21

h ∑ (δt vj j=1

k+ 21

)(δt wj

m−1

k+ 21

) − h ∑ (δx2 wj j=1

k+ 21

)(δt wj

k+ 21

m−1

k+ 21

− h ∑ (δt wj j=1

k+ 21

)δt vj

m−1

k+ 21

+ h ∑ vj j=1

k+ 21

δt vj

m−1

+h ∑ j=1

|ujk |2 + |ujk+1 |2 2

and summing

) = 0.

Multiplying both the right- and left-hand sides of (8.32) by −hδt vj j from 1 to m − 1, we have

(8.55)

(8.56)

and summing over

k+ 21

δt vj

= 0.

(8.57)

Adding (8.56) with (8.57) together, it follows: m−1 |uk |2 + |uk+1 |2 1 󵄨󵄨 k+1 󵄨󵄨2 󵄨󵄨 k 󵄨󵄨2 1 󵄩 k+ 1 j j 󵄩2 󵄩 󵄩2 (󵄨󵄨w 󵄨󵄨1 − 󵄨󵄨w 󵄨󵄨1 ) + (󵄩󵄩󵄩vk+1 󵄩󵄩󵄩 − 󵄩󵄩󵄩vk 󵄩󵄩󵄩 ) + h ∑ δt vj 2 = 0, 2τ 2τ 2 j=1

0 ⩽ k ⩽ n − 1.

Then adding (8.55) and (8.58) together, it follows: 1 󵄨 󵄨2 [(󵄨󵄨󵄨uk+1 󵄨󵄨󵄨1 + τ

1 󵄨󵄨 k+1 󵄨󵄨2 󵄨w 󵄨󵄨1 + 2󵄨

󵄨 󵄨2 1 󵄨 󵄨2 − (󵄨󵄨󵄨uk 󵄨󵄨󵄨1 + 󵄨󵄨󵄨wk 󵄨󵄨󵄨1 + 2

m−1 1 󵄩󵄩 k+1 󵄩󵄩2 󵄨 k+1 󵄨2 k+1 󵄩󵄩v 󵄩󵄩 + h ∑ 󵄨󵄨󵄨uj 󵄨󵄨󵄨 vj ) 2 j=1

m−1 1 󵄩󵄩 k 󵄩󵄩2 󵄨 k 󵄨2 k 󵄩󵄩v 󵄩󵄩 + h ∑ 󵄨󵄨󵄨uj 󵄨󵄨󵄨 vj )] = 0, 2 j=1

which implies 1 k+1 (E − E k ) = 0, τ Therefore, (8.54) is valid.

0 ⩽ k ⩽ n − 1.

0 ⩽ k ⩽ n − 1,

(8.58)

8.2 Two-level nonlinear difference scheme

� 259

With the help of the above theorem, we know 󵄨󵄨 k 󵄨󵄨2 󵄨󵄨u 󵄨󵄨1 + m−1

1 󵄩󵄩 k 󵄩󵄩2 󵄩v 󵄩 2󵄩 󵄩

󵄨 󵄨2 ⩽ −h ∑ 󵄨󵄨󵄨ujk 󵄨󵄨󵄨 vkj + E 0 j=1

m−1

󵄨 󵄨4 ⩽ h ∑ (󵄨󵄨󵄨ujk 󵄨󵄨󵄨 + j=1

1 󵄨󵄨 k 󵄨󵄨2 0 󵄨v 󵄨 ) + E 4󵄨 j 󵄨

󵄩 󵄩2 󵄩 󵄩2 1 󵄩 󵄩2 ⩽ 󵄩󵄩󵄩uk 󵄩󵄩󵄩∞ 󵄩󵄩󵄩uk 󵄩󵄩󵄩 + 󵄩󵄩󵄩vk 󵄩󵄩󵄩 + E 0 4 󵄩󵄩 k 󵄩󵄩2 0 1 󵄩󵄩 k 󵄩󵄩2 = 󵄩󵄩u 󵄩󵄩∞ Q + 󵄩󵄩v 󵄩󵄩 + E 0 4 󵄨󵄨 k 󵄨󵄨2 1 󵄩󵄩 k 󵄩󵄩2 0 1 󵄩󵄩 k 󵄩󵄩2 ⩽ (ε󵄨󵄨u 󵄨󵄨1 + 󵄩󵄩u 󵄩󵄩 )Q + 󵄩󵄩v 󵄩󵄩 + E 0 4ε 4 1 󵄩 󵄩2 󵄨 󵄨2 1 = (ε󵄨󵄨󵄨uk 󵄨󵄨󵄨1 + Q0 )Q0 + 󵄩󵄩󵄩vk 󵄩󵄩󵄩 + E 0 . 4ε 4 Taking ε =

1 , 2Q0

it follows: 1 󵄨󵄨 k 󵄨󵄨2 󵄨u 󵄨 + 2 󵄨 󵄨1

1 󵄩󵄩 k 󵄩󵄩2 1 0 3 0 󵄩v 󵄩 ⩽ (Q ) + E , 4󵄩 󵄩 2

0 ⩽ k ⩽ n.

It is obtained from Lemma 1.1(b) that 2 󵄩󵄩 k 󵄩󵄩2 󵄩u 󵄩 + L 󵄩 󵄩∞

1 󵄩󵄩 k 󵄩󵄩2 1 0 3 0 󵄩v 󵄩 ⩽ (Q ) + E , 4󵄩 󵄩 2

0 ⩽ k ⩽ n.

Therefore, there is a constant c3 satisfying 󵄩󵄩 k 󵄩󵄩 󵄩󵄩u 󵄩󵄩∞ ⩽ c3 ,

󵄩󵄩 k 󵄩󵄩 󵄩󵄩v 󵄩󵄩 ⩽ c3 ,

0 ⩽ k ⩽ n.

8.2.4 Convergence of the difference solution Theorem 8.4. Suppose {Ujk , Vjk , Wjk | 0 ⩽ j ⩽ m, 0 ⩽ k ⩽ n} is the solution of the problem (8.8)–(8.12) and {ujk , vkj , wjk | 0 ⩽ j ⩽ m, 0 ⩽ k ⩽ n} is the solution of the difference scheme (8.30)–(8.34). Denote ejk = Ujk − ujk ,

fjk

gjk

= =

Vjk

−

Wjk

vkj ,

−

wjk ,

Then there is a constant c4 such that

0 ⩽ j ⩽ m, 0 ⩽ k ⩽ n, 0 ⩽ j ⩽ m, 0 ⩽ k ⩽ n, 0 ⩽ j ⩽ m, 0 ⩽ k ⩽ n.

260 � 8 Difference methods for the Zakharov equation 󵄩󵄩 k 󵄩󵄩 󵄨󵄨 k 󵄨󵄨 󵄩󵄩 k 󵄩󵄩 󵄨󵄨 k 󵄨󵄨 2 2 󵄩󵄩e 󵄩󵄩 + 󵄨󵄨e 󵄨󵄨1 + 󵄩󵄩f 󵄩󵄩 + 󵄨󵄨g 󵄨󵄨1 ⩽ c4 (τ + h ),

0 ⩽ k ⩽ n.

(8.59)

Proof. Subtracting (8.30)–(8.34) from (8.19)–(8.21), (8.28)–(8.29), respectively, we have the system of error equations: { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { {

k+ 21

iδt ej

k+ 21

k+ 21

+ δx2 ej

− (Uj

k+ 21

Vj

1 ⩽ j ⩽ m − 1, 0 ⩽ k ⩽ n − 1,

k+ 1 δt fj 2

−

k+ 21

k+ 21

δt gj ej0 e0k

k+ 1 δx2 gj 2

− fj

=

−(

k+ 1 Qj 2 ,

k+ 21 k+ 21 vj )

− uj

|Ujk |2 + |Ujk+1 |2

= 0,

= 0,

k em

= 0,

gj0

g0k

= 0,

, (8.60)

1 ⩽ j ⩽ m − 1, 0 ⩽ k ⩽ n − 1, |ujk |2 + |ujk+1 |2

(8.61) k+ 21

2

) = Rj

= F(ψ1 )(xj ) − G(ψ1 )j ,

0 ⩽ j ⩽ m,

2 0 ⩽ j ⩽ m, 0 ⩽ k ⩽ n − 1, fj0

k+ 21

= Pj

k gm

= 0,

−

= 0,

, (8.62)

1 ⩽ k ⩽ n.

(8.63) (8.64)

By the numerical analysis of the two-point boundary value problem of the secondorder ordinary differential equation, there is a constant c such that √L 2 󵄩󵄩 0 󵄩󵄩 ch . 󵄩󵄩g 󵄩󵄩∞ ⩽ 2

󵄨󵄨 0 󵄨󵄨 2 󵄨󵄨g 󵄨󵄨1 ⩽ ch ,

(8.65) k+ 21

(I) Multiplying both the right- and left-hand sides of (8.60) by hēj over j from 1 to m − 1, we have m−1

k+ 21

k+ 21

ih ∑ (δt ej

)ēj

j=1

m−1

k+ 21

= h ∑ (Uj j=1

m−1

k+ 21

Vj

k+ 21 k+ 21 fj

= h ∑ (Uj j=1

m−1

k+ 21

+ h ∑ (δx2 ej j=1

k+ 21

− uj

k+ 21

vj

k+ 21 k+ 21 k+ 21 vj )ēj

+ ej

k+ 21

)ēj

m−1

k+ 21

)ēj

and summing

k+ 21 k+ 21 ēj

+ h ∑ Pj j=1

m−1

k+ 21 k+ 21 ēj .

+ h ∑ Pj j=1

Taking the imaginary parts on both the right- and left-hand sides of the above equality, it follows: 1 󵄩󵄩 k+1 󵄩󵄩2 󵄩󵄩 k 󵄩󵄩2 (󵄩e 󵄩󵄩 − 󵄩󵄩e 󵄩󵄩 ) 2τ 󵄩 m−1

k+ 21 k+ 21 k+ 21 fj ēj

= Im{h ∑ Uj j=1

m−1

k+ 21 k+ 21 ēj }

+ h ∑ Pj j=1

1 󵄩 󵄩 1󵄩 󵄩 1󵄩 󵄩 1󵄩 󵄩 ⩽ c1 󵄩󵄩󵄩f k+ 2 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩ek+ 2 󵄩󵄩󵄩 + 󵄩󵄩󵄩Pk+ 2 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩ek+ 2 󵄩󵄩󵄩,

0 ⩽ k ⩽ n − 1.

(8.66)

8.2 Two-level nonlinear difference scheme k+ 21

(II) Multiplying both the right- and left-hand sides of (8.60) by −2hδt ēj ming over j from 1 to m − 1, we have m−1

m−1

j=1

j=1

� 261

and sum-

k+ 1 k+ 1 󵄨 k+ 1 󵄨2 − 2ih ∑ 󵄨󵄨󵄨δt ej 2 󵄨󵄨󵄨 − 2h ∑ (δx2 ej 2 )δt ēj 2 m−1

k+ 21

= − 2h ∑ (Uj j=1

k+ 21

Vj

m−1

k+ 21 k+ 21 k+ 1 vj )δt ēj 2

− uj

k+ 21

− 2h ∑ Pj j=1

k+ 21

δt ēj

0 ⩽ k ⩽ n − 1.

,

Taking the real parts on both the right- and left-hand sides of the above equality, it follows: m−1 1 󵄨󵄨 k+1 󵄨󵄨2 󵄨󵄨 k 󵄨󵄨2 k+ 1 k+ 1 k+ 1 k+ 1 k+ 1 (󵄨󵄨e 󵄨󵄨1 − 󵄨󵄨e 󵄨󵄨1 ) = Re{−2h ∑ (Uj 2 Vj 2 − uj 2 vj 2 )δt ēj 2 } τ j=1 m−1

k+ 21

+ Re{−2h ∑ Pj j=1

k+ 21

δt ēj

},

0 ⩽ k ⩽ n − 1. k+ 21

Multiplying both the right- and left-hand sides of (8.61) by hδt gj over j from 1 to m − 1, we have m−1

k+ 21

h ∑ (δt fj j=1

k+ 21

)(δt gj

m−1

k+ 21

) − h ∑ (δx2 gj j=1

0 ⩽ k ⩽ n − 1.

k+ 21

)δt gj

m−1

k+ 21

= h ∑ Qj j=1

k+ 21

Multiplying both the right- and left-hand sides of (8.62) by −hδt fj j from 1 to m − 1, we have m−1

k+ 21

− h ∑ (δt gj j=1

m−1 (|U k+1 |2 j

=−h ∑ j=1

k+ 21

)(δt fj −

m−1

k+ 21

) + h ∑ fj j=1

|ujk+1 |2 ) 2

(8.67)

and summing

k+ 21

δt gj

,

and summing over

k+ 21

δt fj

+ (|Ujk |2 − |ujk |2 )

k+ 21

δt fj

m−1

k+ 21

− h ∑ Rj j=1

0 ⩽ k ⩽ n − 1. Adding the above two equalities together results in 1 󵄩 1 󵄨󵄨 k+1 󵄨󵄨2 󵄨󵄨 k 󵄨󵄨2 󵄩2 󵄩 󵄩2 (󵄨󵄨g 󵄨󵄨1 − 󵄨󵄨g 󵄨󵄨1 ) + (󵄩󵄩󵄩f k+1 󵄩󵄩󵄩 − 󵄩󵄩󵄩f k 󵄩󵄩󵄩 ) 2τ 2τ m−1 (|U k+1 |2 − |uk+1 |2 )f k+1 − (|U k |2 − |uk |2 )f k j j j j j j +h ∑ τ j=1

k+ 21

δt fj

,

262 � 8 Difference methods for the Zakharov equation m−1

=h ∑

(|Ujk+1 |2 − |ujk+1 |2 ) − (|Ujk |2 − |ujk |2 ) τ

j=1

m−1

k+ 21

+ h ∑ Qj j=1

m−1

k+ 21

δt gj

k+ 21

− h ∑ Rj j=1

k+ 21

δt fj

k+ 21

fj

0 ⩽ k ⩽ n − 1.

,

(8.68)

Then adding (8.67) and (8.68) together, it follows: 1 󵄨󵄨 k+1 󵄨󵄨2 {[󵄨󵄨e 󵄨󵄨1 + τ

󵄨 󵄨2 1 󵄨 󵄨2 − [󵄨󵄨󵄨ek 󵄨󵄨󵄨1 + 󵄨󵄨󵄨g k 󵄨󵄨󵄨1 + 2 m−1

k+ 21

= 2 Re{−h ∑ (Uj j=1

m−1

+h ∑

m−1 1 󵄩󵄩 k+1 󵄩󵄩2 k+1 2 󵄨 k+1 󵄨2 k+1 󵄩󵄩f 󵄩󵄩 + h ∑ (|Uj | − 󵄨󵄨󵄨uj 󵄨󵄨󵄨 )fj ] 2 j=1

1 󵄨󵄨 k+1 󵄨󵄨2 󵄨g 󵄨󵄨1 + 2󵄨

m−1 1 󵄩󵄩 k 󵄩󵄩2 󵄨 k 󵄨2 󵄨 k 󵄨2 k 󵄩󵄩f 󵄩󵄩 + h ∑ (󵄨󵄨󵄨Uj 󵄨󵄨󵄨 − 󵄨󵄨󵄨uj 󵄨󵄨󵄨 )fj ]} 2 j=1 k+ 1 k+ 21 k+ 21 vj )δt ēj 2 }

k+ 21

Vj

− uj

(|Ujk+1 |2 − |ujk+1 |2 ) − (|Ujk |2 − |ujk |2 ) τ

j=1

m−1

k+ 21

δt ēj

j=1

m−1

k+ 21

− h ∑ Rj j=1

k+ 21

δt fj

m−1

k+ 21

+ 2 Re{−h ∑ Pj

k+ 21

} + h ∑ Qj j=1

k+ 21

fj

k+ 21

δt gj

0 ⩽ k ⩽ n − 1.

,

(8.69)

Denote m−1

k+ 21

Ak = 2 Re{−h ∑ (Uj j=1

m−1

Bk = h ∑

(|Ujk+1 |2

k+ 21

Vj

k+ 21 k+ 21 k+ 1 vj )δt ēj 2 },

− uj

− |ujk+1 |2 ) − (|Ujk |2 − |ujk |2 ) τ

j=1

k+ 21

fj

.

Then we have m−1

k+ 21

Ak = 2 Re{−h ∑ (Uj j=1

m−1

k+ 21

= −2 Re{h ∑ (Uj j=1

k+ 21

− uj

k+ 21

vj

k+ 21

Vj

k+ 21

Vj 1

k+ 21 k+ 21 k+ 1 vj )δt (Ū j 2

− uj

1

1

1

k+ 21

− ū j

1

k+ k+ k+ k+ δt Ū j 2 − Uj 2 Vj 2 δt ū j 2 1

1

1

k+ k+ k+ k+ δt Ū j 2 + uj 2 vj 2 δt ū j 2 )}

)}

8.2 Two-level nonlinear difference scheme

m−1

k+ 21

= − 2h ∑ (Vj

|Ujk+1 |2 − |Ujk |2 2τ

j=1

m−1

k+ 21

+ 2 Re{h ∑ (Uj j=1

m−1

k

B = h ∑(

|Ujk+1 |2

k+ 21

Vj

− |Ujk |2

τ

j=1

k+ 21

+ vj

−

k+ 21

|ujk+1 |2 − |ujk |2 2τ

� 263

)

k+ 21 k+ 21 k+ 1 vj δt Ū j 2 )},

δt ū j

+ uj

|ujk+1 |2 − |ujk |2

k+ 21

)(Vj

τ

k+ 21

− vj

).

Adding the above two equalities together, we have Ak + Bk

m−1

k+ 21

= 2 Re{h ∑ (Uj j=1

m−1

+ h ∑ (−

k+ 21

Vj

k+ 21

δt ū j

|Ujk+1 |2 − |Ujk |2 τ

j=1

m−1

k+ 21

= 2 Re{h ∑ (Uj j=1

m−1

k+ 21

Vj

k+ 21

− 2 Re{h ∑ Uj j=1

m−1

k+ 21

vj

k+ 21

δt ū j

m−1

k+ 21

j=1

m−1 j=1

m−1

k+ 21

= 2 Re{h ∑ fj j=1

k+ 21

δt ū j

k+ 21

= 2 Re{h ∑ [(Vj

τ

k+ 21

Vj

)

k+ 1 k+ 21 k+ 21 vj δt Ū j 2 )}

+ uj

1

1

k+ 1 k+ 21 k+ 21 vj δt Ū j 2

= 2 Re{h ∑ (Vj

|ujk+1 |2 − |ujk |2

−

1

1

1

k+ k+ k+ k+ k+ (δt Ū j 2 )vj 2 + uj 2 (δt ū j 2 )Vj 2 }

= 2 Re{h ∑ (−ej j=1

k+ 1 k+ 21 k+ 21 vj δt Ū j 2 )}

+ uj

k+ 21

− vj

k+ 21

− vj

k+ 21

(δt Ū j

k+ 21

+ ej

k+ 21

Vj 1

k+ 21

δt ū j

)}

1

k+ k+ δt Ū j 2 )ej 2 } 1

1

1

1

k+ 21

)ej

m−1

k+ 21 k+ 21 k+ 1 ēj δt ej 2 }.

} + 2 Re{−h ∑ Vj j=1

With the help of (8.60), it follows: k+ 21

δt ej Therefore, we have m−1

k+ 21 k+ 21 k+ 1 ēj δt ej 2

−h ∑ Vj j=1

k+ 21

= iδx2 ej

1

k+ k+ k+ k+ k+ )δt Ū j 2 + Vj 2 (δt ū j 2 − δt Ū j 2 )]ej 2 }

k+ 21

− i(ej

k+ 21

Vj

k+ 21 k+ 21 fj )

+ uj

k+ 21

− iPj

.

(8.70)

264 � 8 Difference methods for the Zakharov equation m−1

k+ 21 k+ 21 k+ 1 ēj [iδx2 ej 2

= −h ∑ Vj j=1

m−1

1

1

= ih ∑ [δx (V k+ 2 ēk+ 2 )j+ 1 ]δx e 2

j=0

m−1

k+ 21

+ ih ∑ Vj j=1

k+ 21

ēj

k+ 21

Pj

k+ 21

− i(ej

k+ 21

k+ 21 k+ 21 fj )

Vj

+ uj

m−1

k+ 21

k+ 21

− iPj

]

k+ 21 k+ 21 k+ 21 k+ 21 ēj (ej Vj

+ ih ∑ Vj

j+ 21

j=1

k+ 21 k+ 21 fj )

+ uj

(8.71)

.

Substituting (8.71) into (8.70) and using the Cauchy–Schwarz inequality, there is a constant c5 such that 1 2 󵄩 󵄩 1 󵄩2 󵄨 1 󵄨2 󵄩 1 󵄩2 󵄩 Ak + Bk ⩽ c5 (󵄩󵄩󵄩ek+ 2 󵄩󵄩󵄩 + 󵄨󵄨󵄨ek+ 2 󵄨󵄨󵄨1 + 󵄩󵄩󵄩f k+ 2 󵄩󵄩󵄩 + 󵄩󵄩󵄩Pk+ 2 󵄩󵄩󵄩 ).

(8.72)

Denote 󵄩 󵄩2 󵄨 󵄨2 Ê k = 2(c1 + c3 )2 󵄩󵄩󵄩ek 󵄩󵄩󵄩 + 󵄨󵄨󵄨ek 󵄨󵄨󵄨1 +

m−1 1 󵄨󵄨 k 󵄨󵄨2 󵄨 k 󵄨2 󵄨 k 󵄨2 k 󵄨󵄨g 󵄨󵄨1 + h ∑ (󵄨󵄨󵄨Uj 󵄨󵄨󵄨 − 󵄨󵄨󵄨uj 󵄨󵄨󵄨 )fj . 2 j=1

1 󵄩󵄩 k 󵄩󵄩2 󵄩f 󵄩 + 2󵄩 󵄩

One has 1 󵄨󵄨 k 󵄨󵄨2 󵄨g 󵄨 . 2 󵄨 󵄨1

󵄩 󵄩2 󵄨 󵄨2 1 󵄩 󵄩2 Ê k ⩾ (c1 + c3 )2 󵄩󵄩󵄩ek 󵄩󵄩󵄩 + 󵄨󵄨󵄨ek 󵄨󵄨󵄨1 + 󵄩󵄩󵄩f k 󵄩󵄩󵄩 + 4

(8.73)

Multiplying both the right- and left-hand sides of (8.66) by 4(c1 + c3 )2 , then adding the result with (8.69) and noticing (8.72), there is a constant c6 such that m−1 1 2 1 ̂ k+1 k+ 1 k+ 1 󵄩 󵄩 (E − Ê k ) ⩽ c6 (Ê k+1 + Ê k ) + c6 󵄩󵄩󵄩Pk+ 2 󵄩󵄩󵄩 + 2 Re{−h ∑ Pj 2 δt ēj 2 } τ j=1 m−1

k+ 21

+ h ∑ Qj j=1

k+ 21

δt gj

m−1

k+ 21

− h ∑ Rj j=1

k+ 21

δt fj

,

0 ⩽ k ⩽ n − 1.

Replacing k by l in the above inequality and summing over l from 0 to k, it follows: k k 1 ̂ k+1 󵄩 1 󵄩2 (E − Ê 0 ) ⩽ c6 ∑(Ê l+1 + Ê l ) + c6 ∑󵄩󵄩󵄩Pl+ 2 󵄩󵄩󵄩 τ l=0 l=0 m−1 k

l+ 21

+ 2 Re{−h ∑ ∑ Pj j=1 l=0

m−1 k

l+ 21

− h ∑ ∑ Rj j=1 l=0

Noticing

l+ 21

δt fj

,

l+ 21

δt ēj

m−1 k

l+ 21

} + h ∑ ∑ Qj j=1 l=0

0 ⩽ k ⩽ n − 1.

l+ 21

δt gj

(8.74)

8.2 Two-level nonlinear difference scheme

k

l+ 1 l+ 1 ∑ Pj 2 δt ēj 2 l=0 k

l+ 1 l+ 1 ∑ Qj 2 δt gj 2 l=0 k

l+ 21

∑ Rj

l=0

l+ 21

δt fj

l+ 21

k P 1 1 k+ 1 j = (Pj 2 ējk+1 − Pj2 ēj0 ) − ∑ τ l=1

l− 21

− Pj τ

l+ 21

k Q 1 1 k+ 1 j = (Qj 2 gjk+1 − Qj2 gj0 ) − ∑ τ l=1

l− 21

− Qj τ

l− 21

l+ 21

− Rj

k R 1 1 k+ 1 j = (Rj 2 fjk+1 − Rj2 fj0 ) − ∑ τ l=1

ējl ,

τ

gjl ,

fjl

and (8.63), it follows from (8.74) that 1 ̂ k+1 (E − Ê 0 ) τ

k

k

l=1

l=0

1

⩽ c6 Ê k+1 + 2c6 ∑ Ê l + c6 Ê 0 + c6 ∑ ‖Pl+ 2 ‖2 l+ 21

k 󵄨󵄨 P 1 1 󵄨 k+ 1 󵄨 󵄨 j + 2h ∑ { 󵄨󵄨󵄨Pj 2 ējk+1 − Pj2 ēj0 󵄨󵄨󵄨 + ∑󵄨󵄨󵄨 󵄨 τ j=1 l=1 󵄨 m−1

l+ 21

k 󵄨󵄨 Q 1 1 󵄨 k+ 1 󵄨 󵄨 j + h ∑ { 󵄨󵄨󵄨Qj 2 gjk+1 − Qj2 gj0 󵄨󵄨󵄨 + ∑󵄨󵄨󵄨 󵄨 τ j=1 l=1 󵄨 m−1

l+ 21

k 󵄨󵄨 R 1 1 󵄨 k+ 1 󵄨 󵄨 j + h ∑ { 󵄨󵄨󵄨Rj 2 fjk+1 − Rj2 fj0 󵄨󵄨󵄨 + ∑󵄨󵄨󵄨 󵄨 τ j=1 l=1 󵄨 m−1

k

k

l=1

l=0

l− 21

− Pj τ

l− 21

− Qj τ

l− 21

− Rj τ

󵄨󵄨 󵄨󵄨 󵄨󵄨 l 󵄨󵄨 󵄨󵄨 ⋅ 󵄨󵄨ej 󵄨󵄨} 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 l 󵄨󵄨 󵄨󵄨 ⋅ 󵄨󵄨gj 󵄨󵄨} 󵄨󵄨

󵄨󵄨 󵄨󵄨 󵄨󵄨 l 󵄨󵄨 󵄨󵄨 ⋅ 󵄨󵄨fj 󵄨󵄨} 󵄨󵄨

󵄩 1 󵄩2 ⩽ c6 Ê k+1 + 2c6 ∑ Ê l + c6 Ê 0 + c6 ∑󵄩󵄩󵄩Pl+ 2 󵄩󵄩󵄩 1 2󵄩 󵄩 󵄩 󵄩 + 󵄩󵄩󵄩Pk+ 2 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩ek+1 󵄩󵄩󵄩 + τ

1 󵄩󵄩 k+ 21 󵄩󵄩 󵄩󵄩 k+1 󵄩󵄩 1 󵄩󵄩 21 󵄩󵄩 󵄩󵄩 0 󵄩󵄩 󵄩Q 󵄩󵄩 ⋅ 󵄩󵄩g 󵄩󵄩 + 󵄩󵄩Q 󵄩󵄩 ⋅ 󵄩󵄩g 󵄩󵄩 τ󵄩 τ 1 1 k 󵄩 󵄩󵄩 Pl+ 2 − Pl− 2 󵄩󵄩󵄩 󵄩 l 󵄩 1 󵄩󵄩 k+ 21 󵄩󵄩 󵄩󵄩 k+1 󵄩󵄩 󵄩󵄩 ⋅ 󵄩󵄩e 󵄩󵄩 + 󵄩󵄩R 󵄩󵄩 ⋅ 󵄩󵄩f 󵄩󵄩 + 2 ∑󵄩󵄩󵄩 󵄩󵄩 󵄩 󵄩 󵄩󵄩 τ τ 󵄩 l=1 1

1

k 󵄩 󵄩󵄩 Ql+ 2 − Ql− 2 + ∑󵄩󵄩󵄩 󵄩 τ l=1 󵄩

0 ⩽ k ⩽ n − 1.

1 1 k 󵄩 󵄩󵄩 󵄩󵄩 Rl+ 2 − Rl− 2 󵄩󵄩 󵄩󵄩 l 󵄩󵄩 󵄩󵄩 ⋅ 󵄩󵄩g 󵄩󵄩 + ∑󵄩󵄩󵄩 󵄩󵄩 󵄩 τ l=1 󵄩

Furthermore, it holds that 1 ̂ k+1 (E − Ê 0 ) τ

k

k

l=1

l=0

󵄩 1 󵄩2 ⩽ c6 Ê k+1 + 2c6 ∑ Ê l + c6 Ê 0 + c6 ∑󵄩󵄩󵄩Pl+ 2 󵄩󵄩󵄩

󵄩󵄩 󵄩󵄩 󵄩󵄩 l 󵄩󵄩 󵄩󵄩 ⋅ 󵄩󵄩f 󵄩󵄩, 󵄩󵄩

� 265

266 � 8 Difference methods for the Zakharov equation 1 2 1 1 6󵄩 L2 󵄩 󵄩 󵄩2 + ( ⋅ 2 󵄩󵄩󵄩ek+1 󵄩󵄩󵄩 + 2 ⋅ 󵄩󵄩󵄩Pk+ 2 󵄩󵄩󵄩 ) τ 2 L 6 2 1 2 1 1 1 6󵄩 󵄩 󵄩2 L 󵄩 + ( ⋅ ⋅ 2 󵄩󵄩󵄩g k+1 󵄩󵄩󵄩 + 󵄩󵄩󵄩Qk+ 2 󵄩󵄩󵄩 ) τ 2 2 L 6 1 2 1 1 󵄩 1 󵄩2 1 󵄩 󵄩2 1 1 1󵄩 󵄩 󵄩2 󵄩 + ( 󵄩󵄩󵄩Q 2 󵄩󵄩󵄩 + 󵄩󵄩󵄩g 0 󵄩󵄩󵄩 ) + ( ⋅ 󵄩󵄩󵄩f k+1 󵄩󵄩󵄩 + 2󵄩󵄩󵄩Rk+ 2 󵄩󵄩󵄩 ) τ 2 2 τ 2 4 1 1 2 k 6 󵄩 󵄩2 L2 󵄩󵄩󵄩 Pl+ 2 − Pl− 2 󵄩󵄩󵄩󵄩 + ∑( 2 󵄩󵄩󵄩el 󵄩󵄩󵄩 + 󵄩󵄩󵄩 󵄩󵄩 ) 6 󵄩󵄩 τ L 󵄩󵄩 l=1 1 1 1 6 󵄩󵄩 l 󵄩󵄩2 1 L2 󵄩󵄩󵄩󵄩 Ql+ 2 − Ql− 2 + ∑( ⋅ 2 󵄩󵄩g 󵄩󵄩 + ⋅ 󵄩󵄩 2 L 2 6 󵄩󵄩 τ l=1

k

1 1 1 󵄩󵄩 l 󵄩󵄩2 󵄩󵄩󵄩󵄩 Rl+ 2 − Rl− 2 + ∑( 󵄩󵄩f 󵄩󵄩 + 󵄩󵄩 󵄩󵄩 4 τ l=1

k

󵄩󵄩2 󵄩󵄩 󵄩󵄩 ) 󵄩󵄩

󵄩󵄩2 󵄩󵄩 󵄩󵄩 ) 󵄩󵄩 k

k

󵄩 1 󵄩2 ⩽ c6 Ê k+1 + 2c6 ∑ Ê l + c6 Ê 0 + c6 ∑󵄩󵄩󵄩Pl+ 2 󵄩󵄩󵄩 l=1

1 󵄨 󵄨2 + (󵄨󵄨󵄨ek+1 󵄨󵄨󵄨1 + 2τ

1 󵄨󵄨 k+1 󵄨󵄨2 󵄨g 󵄨󵄨1 + 2󵄨

l=0

1 󵄩󵄩 k+1 󵄩󵄩2 󵄩f 󵄩󵄩 ) 4󵄩

1 2 1 2 1 2 1 1 󵄩 L2 󵄩 󵄩 󵄩 󵄩 󵄩 + ( L2 󵄩󵄩󵄩Pk+ 2 󵄩󵄩󵄩 + 󵄩󵄩󵄩Qk+ 2 󵄩󵄩󵄩 + 2󵄩󵄩󵄩Rk+ 2 󵄩󵄩󵄩 + τ 3 6

k 󵄨 󵄨2 1 󵄨 󵄨2 + ∑(󵄨󵄨󵄨el 󵄨󵄨󵄨1 + 󵄨󵄨󵄨g l 󵄨󵄨󵄨1 + 2 l=1 1

k

1 󵄩󵄩 21 󵄩󵄩2 󵄩Q 󵄩 + 2󵄩 󵄩

1 󵄩󵄩 0 󵄩󵄩2 󵄩g 󵄩 ) 2󵄩 󵄩

1 󵄩󵄩 l 󵄩󵄩2 󵄩f 󵄩 ) 4󵄩 󵄩 1

L2 󵄩󵄩󵄩 Pl+ 2 − Pl− 2 + ∑( 󵄩󵄩󵄩 6 󵄩󵄩 τ l=1

󵄩󵄩2 L2 󵄩󵄩 Ql+ 21 − Ql− 21 󵄩󵄩 󵄩 󵄩󵄩 + 󵄩󵄩󵄩 󵄩󵄩 12 󵄩󵄩 τ

󵄩󵄩2 󵄩󵄩 Rl+ 21 − Rl− 21 󵄩󵄩 󵄩󵄩 󵄩󵄩 + 󵄩󵄩 󵄩󵄩 󵄩󵄩 τ

󵄩󵄩2 󵄩󵄩 󵄩󵄩 ), 󵄩󵄩

0 ⩽ k ⩽ n − 1.

Multiplying both the right- and left-hand sides of the above inequality by 2τ and rearranging the result in view of (8.73), it follows: (1 − 2c6 τ)Ê k+1

k 1 2 2 󵄩 󵄩 ⩽ 2(1 + 2c6 )τ ∑ Ê l + (2 + 2c6 τ)Ê 0 + L2 󵄩󵄩󵄩Pk+ 2 󵄩󵄩󵄩 3 l=1

L2 󵄩󵄩 k+ 21 󵄩󵄩2 󵄩 k+ 1 󵄩2 󵄩 1 󵄩2 󵄩 0 󵄩2 󵄩󵄩Q 󵄩󵄩 + 4󵄩󵄩󵄩R 2 󵄩󵄩󵄩 + 󵄩󵄩󵄩Q 2 󵄩󵄩󵄩 + 󵄩󵄩󵄩g 󵄩󵄩󵄩 3 1 1 2 1 1 k L2 󵄩󵄩󵄩 Pl+ 2 − Pl− 2 󵄩󵄩󵄩󵄩 L2 󵄩󵄩󵄩 Ql+ 2 − Ql− 2 + τ ∑( 󵄩󵄩󵄩 󵄩󵄩 + 󵄩󵄩󵄩 󵄩󵄩 3 󵄩󵄩 τ 6 󵄩󵄩 τ l=1 +

󵄩󵄩 Rl+ 21 − Rl− 21 󵄩 + 2󵄩󵄩󵄩 󵄩󵄩 τ

k 󵄩󵄩2 󵄩󵄩 󵄩 l+ 1 󵄩2 󵄩󵄩 ) + 2c6 τ ∑󵄩󵄩󵄩P 2 󵄩󵄩󵄩 , 󵄩󵄩 l=0

Noticing (8.22)–(8.27), there is a constant c7 such that

󵄩󵄩2 󵄩󵄩 󵄩󵄩 󵄩󵄩

0 ⩽ k ⩽ n − 1.

8.3 Three-level linearized locally decoupled difference scheme k

2 Ê k+1 ⩽ c7 τ ∑ Ê l + c7 (τ 2 + h2 ) , l=1

� 267

0⩽k ⩽n−1

when 2c6 τ ⩽ 21 . By the Gronwall inequality (Theorem 1.2(c)), it follows: 2 2 Ê k+1 ⩽ c7 ec7 kτ (τ 2 + h2 ) ⩽ c7 ec7 T (τ 2 + h2 ) ,

0 ⩽ k ⩽ n − 1.

Then (8.59) follows by noticing (8.73).

8.3 Three-level linearized locally decoupled difference scheme 8.3.1 Derivation of the difference scheme Considering equations (8.8)–(8.10) at the point (xj , tk ) and then using the numerical differential formula, it follows: ̄

̄

{ iΔt Ujk + δx2 Ujk − Ujk Vjk = P̂ jk , { { { { ̄ Δt Vjk − δx2 Wjk = Q̂ jk , { { { { { k k̄ k 2 ̂k { Δt Wj − Vj − |Uj | = Rj ,

1 ⩽ j ⩽ m − 1, 1 ⩽ k ⩽ n − 1,

(8.75)

1 ⩽ j ⩽ m − 1, 1 ⩽ k ⩽ n − 1,

(8.76)

0 ⩽ j ⩽ m, 1 ⩽ k ⩽ n − 1,

(8.77)

where there is a constant c8 satisfying { { { { { { { { {

󵄨󵄨 ̂ k 󵄨󵄨 2 2 󵄨󵄨Pj 󵄨󵄨 ⩽ c8 (τ + h ), 󵄨󵄨 ̂ k 󵄨󵄨 2 2 󵄨󵄨Qj 󵄨󵄨 ⩽ c8 (τ + h ), 󵄨󵄨 ̂ k 󵄨󵄨 2 󵄨󵄨Rj 󵄨󵄨 ⩽ c8 τ ,

1 ⩽ j ⩽ m − 1, 1 ⩽ k ⩽ n − 1,

(8.78)

1 ⩽ j ⩽ m − 1, 1 ⩽ k ⩽ n − 1,

(8.79)

0 ⩽ j ⩽ m, 1 ⩽ k ⩽ n − 1

(8.80)

and { { { { { { { { { { { { { { { { { { { { {

󵄨󵄨 P̂ jk+1 − P̂ jk−1 󵄨󵄨 󵄨󵄨 󵄨󵄨 2 2 󵄨󵄨 ⩽ c8 (τ + h ), 󵄨󵄨 󵄨󵄨 󵄨󵄨 2τ 󵄨󵄨 Q̂ jk+1 − Q̂ jk−1 󵄨󵄨 󵄨󵄨 󵄨󵄨 2 2 󵄨󵄨 ⩽ c8 (τ + h ), 󵄨󵄨 󵄨󵄨 󵄨󵄨 2τ 󵄨󵄨 󵄨󵄨 R̂ jk+1 − R̂ k−1 j 󵄨󵄨 󵄨󵄨 2 󵄨󵄨 󵄨󵄨 ⩽ c8 τ , 󵄨󵄨 󵄨󵄨 2τ

1 ⩽ j ⩽ m − 1, 2 ⩽ k ⩽ n − 2,

(8.81)

1 ⩽ j ⩽ m − 1, 2 ⩽ k ⩽ n − 2,

(8.82)

0 ⩽ j ⩽ m, 2 ⩽ k ⩽ n − 2.

(8.83)

It is easily known that R̂ k0 = 0,

R̂ km = 0,

1 ⩽ k ⩽ n − 1.

Combining equations (8.8)–(8.10) with the initial value condition (8.11), we can get the values of ut (x, 0), vt (x, 0), wt (x, 0).

268 � 8 Difference methods for the Zakharov equation Considering equations (8.8)–(8.10) at the point (xj , t 1 ) and then using the numerical 2 differential formula, it follows: 1

1

1

1

{ iδt Uj2 + δx2 Uj2 − Uj2 v̂j2 = P̂ j0 , { { { { 1 1 { δt Vj 2 − δx2 Wj 2 = Q̂ j0 , { { { { 1 1 1 { { 2 2 2 2 ̂ δ W − V − | u | = R̂ 0j , t j j j {

1 ⩽ j ⩽ m − 1,

(8.84)

1 ⩽ j ⩽ m − 1,

(8.85)

0 ⩽ j ⩽ m,

(8.86)

where 1 τ v̂j2 = v(xj , 0) + vt (xj , 0), 2

1 τ û j2 = u(xj , 0) + ut (xj , 0), 2

0 ⩽ j ⩽ m,

(8.87)

and there is a constant c9 satisfying 󵄨󵄨 ̂ 0 󵄨󵄨 2 2 󵄨󵄨Pj 󵄨󵄨 ⩽ c9 (τ + h ), 󵄨󵄨 ̂ 0 󵄨󵄨 2 2 󵄨󵄨Qj 󵄨󵄨 ⩽ c9 (τ + h ), 󵄨󵄨 ̂ 0 󵄨󵄨 2 󵄨󵄨Rj 󵄨󵄨 ⩽ c9 τ ,

{ { { { { { { { {

1 ⩽ j ⩽ m − 1,

(8.88)

1 ⩽ j ⩽ m − 1,

(8.89)

0 ⩽ j ⩽ m.

(8.90)

Noticing the initial-boundary value conditions (8.11)–(8.12), we have

{

Uj0 = φ(xj ), U0k

Umk

= 0,

Vj0 = ψ(xj ), = 0,

W0k

Wj0 = F(ψ1 )(xj ), = 0,

Wmk

= 0,

0 ⩽ j ⩽ m,

(8.91)

1 ⩽ k ⩽ n,

(8.92)

where F(ψ1 )(x) is determined by (8.7). Omitting the small terms in (8.75)–(8.77) and (8.84)–(8.86), another difference scheme for (8.8)–(8.12) reads { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { {

1

1

1

1

iδt uj2 + δx2 uj2 − uj2 v̂j2 = 0,

1 ⩽ j ⩽ m − 1,

(8.93)

1 ⩽ j ⩽ m − 1,

(8.94)

󵄨 δt wj − vj − 󵄨󵄨󵄨û j 󵄨󵄨󵄨 = 0,

0 ⩽ j ⩽ m,

(8.95)

̄ ̄ iΔt ujk + δx2 ujk − ujk vkj = 0, ̄ Δt vkj − δx2 wjk = 0, ̄ 󵄨 󵄨2 Δt wjk − vkj − 󵄨󵄨󵄨ujk 󵄨󵄨󵄨 = 0, uj0 = φ(xj ), v0j = ψ(xj ),

1 ⩽ j ⩽ m − 1, 1 ⩽ k ⩽ n − 1,

(8.96)

1 ⩽ j ⩽ m − 1, 1 ⩽ k ⩽ n − 1,

(8.97)

0 ⩽ j ⩽ m, 1 ⩽ k ⩽ n − 1,

(8.98)

0 ⩽ j ⩽ m,

(8.99)

1 ⩽ k ⩽ n,

(8.100)

1 2

1 2

δt vj − δx2 wj = 0, 1 2

u0k = 0,

1 2

1 2

󵄨2

k um = 0,

wj0

w0k = 0,

= G(ψ1 )j , k wm = 0,

8.3 Three-level linearized locally decoupled difference scheme

� 269

1

1

where û j2 and v̂j2 are determined by (8.87), and wj0 = G(ψ1 )j (1 ⩽ j ⩽ m − 1) is determined by δx2 wj0 = ψ1 (xj ),

{

w00 = 0,

1 ⩽ j ⩽ m − 1,

0 wm = 0.

(8.101) (8.102)

It follows from (8.95) and (8.98)–(8.100) that vk0 = 0,

vkm = 0,

0 ⩽ k ⩽ n.

8.3.2 Existence of the difference solution It follows from (8.95) that 1

δx2

wj2 − wj0 τ/2

1 󵄨 1 󵄨2 − δx2 vj2 − δx2 (󵄨󵄨󵄨û j2 󵄨󵄨󵄨 ) = 0,

1 ⩽ j ⩽ m − 1.

Substituting (8.101) into the above equation and then noticing (8.94), we have 1 1 2 󵄨 1 󵄨2 [δt vj2 − ψ1 (xj )] − δx2 vj2 − δx2 (󵄨󵄨󵄨û j2 󵄨󵄨󵄨 ) = 0, τ

1 ⩽ j ⩽ m − 1.

(8.103)

From (8.98), it follows that δx2

̄

wjk − wjk−1 τ

̄ 󵄨 󵄨2 − δx2 vkj − δx2 (󵄨󵄨󵄨ujk 󵄨󵄨󵄨 ) = 0,

1 ⩽ j ⩽ m − 1, 1 ⩽ k ⩽ n − 1.

With the help of (8.97), we have ̄ 1 󵄨 󵄨2 (Δt vkj − δx2 wjk−1 ) − δx2 vkj − δx2 (󵄨󵄨󵄨ujk 󵄨󵄨󵄨 ) = 0, τ

1

1 ⩽ j ⩽ m − 1, 1 ⩽ k ⩽ n − 1.

(8.104)

We can solve the difference scheme (8.93)–(8.102) as follows: (1) {wj0 | 0 ⩽ j ⩽ m} is obtained by solving the tri-diagonal system (8.101)–(8.102); 1

{û j2 | 0 ⩽ j ⩽ m} and {v̂j2 | 0 ⩽ j ⩽ m} are determined by (8.87).

(2) {uj1 | 0 ⩽ j ⩽ m} is obtained by solving the tri-diagonal system (8.93) with the

1 boundary value conditions u01 = 0, um = 0; {v1j | 0 ⩽ j ⩽ m} is obtained by solving the

tri-diagonal system (8.103) with the boundary value conditions v10 = 0, v1m = 0; Then {wj1 | 0 ⩽ j ⩽ m} is obtained by (8.95).

k k−1 k k+1 (3) When {ujk−1 , ujk , vk−1 | 0 ⩽ j ⩽ m} is j , vj , wj , wj | 0 ⩽ j ⩽ m} are known, {uj obtained by solving the tri-diagonal system (8.96) with the boundary value conditions k+1 u0k+1 = 0, um = 0; {vk+1 | 0 ⩽ j ⩽ m} is obtained by solving the tri-diagonal system j

270 � 8 Difference methods for the Zakharov equation k+1 (8.104) with the boundary value conditions vk+1 = 0, vk+1 | 0 ⩽ j ⩽ m} is 0 m = 0; Then {wj obtained by (8.98). It is observed from the preceding process that only two tri-diagonal systems need to be solved independently at each time level. The coefficient matrix of each tri-diagonal system is strictly diagonally dominant. Thus, the solution of the difference scheme exists uniquely.

8.3.3 Conservation and boundedness of the difference solution Theorem 8.5. Suppose {ujk , vkj , wjk | 0 ⩽ j ⩽ m, 0 ⩽ k ⩽ n} is the solution of the difference scheme (8.93)–(8.102). Denote 󵄩 󵄩2 Qk = 󵄩󵄩󵄩uk 󵄩󵄩󵄩 , 0 ⩽ k ⩽ n, 1 󵄨 1 󵄩 1 󵄨 󵄨2 󵄨 󵄨2 󵄨2 󵄨 󵄨2 󵄩2 󵄩 󵄩2 E k = (󵄨󵄨󵄨uk+1 󵄨󵄨󵄨1 + 󵄨󵄨󵄨uk 󵄨󵄨󵄨1 ) + (󵄨󵄨󵄨wk+1 󵄨󵄨󵄨1 + 󵄨󵄨󵄨wk 󵄨󵄨󵄨1 ) + (󵄩󵄩󵄩vk+1 󵄩󵄩󵄩 + 󵄩󵄩󵄩vk 󵄩󵄩󵄩 ) 2 4 4 1 m−1 󵄨 󵄨2 󵄨󵄨 k 󵄨󵄨2 + h ∑ (vkj 󵄨󵄨󵄨ujk+1 󵄨󵄨󵄨 + vk+1 j 󵄨󵄨uj 󵄨󵄨 ), 2 j=1

0 ⩽ k ⩽ n − 1.

Then we have Qk = Q0 ,

0 ⩽ k ⩽ n;

(8.105)

m−1 1 m−1 1 1 1 󵄨󵄨 1 󵄨󵄨2 󵄨󵄨 0 󵄨󵄨2 1 󵄨 󵄨2 󵄨 󵄨2 1 󵄩 󵄩2 󵄩 󵄩2 󵄨 󵄨2 󵄨 󵄨2 (󵄨󵄨u 󵄨󵄨1 + 󵄨󵄨u 󵄨󵄨1 ) + (󵄨󵄨󵄨w1 󵄨󵄨󵄨1 + 󵄨󵄨󵄨w0 󵄨󵄨󵄨1 ) + (󵄩󵄩󵄩v1 󵄩󵄩󵄩 + 󵄩󵄩󵄩v0 󵄩󵄩󵄩 ) + (h ∑ v̂j2 󵄨󵄨󵄨uj1 󵄨󵄨󵄨 + h ∑ 󵄨󵄨󵄨û j2 󵄨󵄨󵄨 v1j ) 2 4 4 2 j=1 j=1

󵄨 󵄨2 = 󵄨󵄨󵄨u0 󵄨󵄨󵄨1 + Ek = E0 ,

1 󵄨󵄨 0 󵄨󵄨2 󵄨w 󵄨 + 2 󵄨 󵄨1

m−1 1 m−1 1 1 󵄩󵄩 0 󵄩󵄩2 1 󵄨 0 󵄨2 󵄨 󵄨2 0 󵄩󵄩v 󵄩󵄩 + (h ∑ v̂j2 󵄨󵄨󵄨uj 󵄨󵄨󵄨 + h ∑ 󵄨󵄨󵄨û j2 󵄨󵄨󵄨 vj ); 2 2 j=1 j=1

(8.106)

1 ⩽ k ⩽ n − 1.

(8.107) 1

Proof. (I) Multiplying both the right- and left-hand sides of (8.93) by hū j2 and summing over j from 1 to m − 1, it follows: m−1

1

1

m−1

1

1

m−1

1

1

󵄨 󵄨2 ih ∑ (δt uj2 )ū j2 + h ∑ (δx2 uj2 )ū j2 − h ∑ 󵄨󵄨󵄨uj2 󵄨󵄨󵄨 v̂j2 = 0. j=1

j=1

j=1

Taking the imaginary parts on both the right- and left-hand sides, we have 1 󵄩󵄩 1 󵄩󵄩2 󵄩󵄩 0 󵄩󵄩2 (󵄩u 󵄩 − 󵄩u 󵄩 ) = 0, 2τ 󵄩 󵄩 󵄩 󵄩 which implies Q1 = Q0 .

(8.108)

8.3 Three-level linearized locally decoupled difference scheme

� 271

̄

Multiplying both the right- and left-hand sides of (8.96) by hū jk and summing over j from 1 to m − 1, it follows: m−1

m−1

m−1

j=1

j=1

j=1

̄ ̄ ̄ 󵄨 ̄ 󵄨2 ih ∑ (Δt ujk )ū jk + h ∑ (δx2 ujk )ū jk − h ∑ 󵄨󵄨󵄨ujk 󵄨󵄨󵄨 vkj = 0,

1 ⩽ k ⩽ n − 1.

Taking the imaginary parts on both the right- and left-hand sides, we have 1 󵄩󵄩 k+1 󵄩󵄩2 󵄩󵄩 k−1 󵄩󵄩2 (󵄩u 󵄩󵄩 − 󵄩󵄩u 󵄩󵄩 ) = 0, 4τ 󵄩

1 ⩽ k ⩽ n − 1,

which implies 󵄩󵄩 k+1 󵄩󵄩2 󵄩󵄩 k−1 󵄩󵄩2 󵄩󵄩u 󵄩󵄩 = 󵄩󵄩u 󵄩󵄩 ,

1 ⩽ k ⩽ n − 1,

or Qk+1 = Qk−1 ,

1 ⩽ k ⩽ n − 1.

(8.109)

Combining (8.108) with (8.109), we have (8.105).

1

(II) Multiplying both the right- and left-hand sides of (8.93) by −2hδt ū j2 and summing over j from 1 to m − 1, we have m−1

m−1

m−1

j=1

j=1

j=1

1 1 1 1 1 1 󵄨 󵄨2 −2ih ∑ 󵄨󵄨󵄨δt uj2 󵄨󵄨󵄨 − 2h ∑ (δx2 uj2 )δt ū j2 + 2h ∑ v̂j2 uj2 δt ū j2 = 0.

Taking the real parts on both the right- and left-hand sides, it follows: m−1 1 |u1 |2 − |u0 |2 1 󵄨󵄨 1 󵄨󵄨2 󵄨󵄨 0 󵄨󵄨2 j j (󵄨󵄨u 󵄨󵄨1 − 󵄨󵄨u 󵄨󵄨1 ) + h ∑ v̂j2 = 0, τ τ j=1

which implies m−1

m−1

j=1

j=1

1 1 󵄨 1 󵄨2 󵄨 0 󵄨2 󵄨 0 󵄨2 󵄨󵄨 1 󵄨󵄨2 󵄨󵄨u 󵄨󵄨1 + h ∑ v̂j2 󵄨󵄨󵄨uj 󵄨󵄨󵄨 = 󵄨󵄨󵄨u 󵄨󵄨󵄨1 + h ∑ v̂j2 󵄨󵄨󵄨uj 󵄨󵄨󵄨 .

(8.110) 1

Multiplying both the right- and left-hand sides of (8.94) by hδt wj2 and summing over j from 1 to m − 1, it follows: m−1

1

1

m−1

1

1

h ∑ (δt vj2 )δt wj2 − h ∑ (δx2 wj2 )δt wj2 = 0. j=1

j=1

1

Multiplying both the right- and left-hand sides of (8.95) by −hδt vj2 and summing over j from 1 to m − 1, it follows:

272 � 8 Difference methods for the Zakharov equation m−1

1

m−1

1

1

1

m−1

1

1

󵄨 󵄨2 −h ∑ (δt wj2 )δt vj2 + h ∑ vj2 δt vj2 + h ∑ 󵄨󵄨󵄨û j2 󵄨󵄨󵄨 δt vj2 = 0. j=1

j=1

j=1

Adding the above two equalities together results in 1 󵄩 󵄩2 󵄩 󵄩2 1 m−1󵄨 1 󵄨2 1 󵄨󵄨 1 󵄨󵄨2 󵄨󵄨 0 󵄨󵄨2 (󵄨󵄨w 󵄨󵄨1 − 󵄨󵄨w 󵄨󵄨1 ) + (󵄩󵄩󵄩v1 󵄩󵄩󵄩 − 󵄩󵄩󵄩v0 󵄩󵄩󵄩 ) + h ∑ 󵄨󵄨󵄨û j2 󵄨󵄨󵄨 (v1j − v0j ) = 0, 2τ 2τ τ j=1 which implies 1 󵄨󵄨 1 󵄨󵄨2 󵄨w 󵄨 + 2 󵄨 󵄨1

m−1 1 1 󵄩󵄩 1 󵄩󵄩2 󵄨 󵄨2 1 1 󵄨 0 󵄨2 󵄩󵄩v 󵄩󵄩 + h ∑ 󵄨󵄨󵄨û j2 󵄨󵄨󵄨 vj = 󵄨󵄨󵄨w 󵄨󵄨󵄨1 + 2 2 j=1

m−1 1 1 󵄩󵄩 0 󵄩󵄩2 󵄨 󵄨2 0 󵄩󵄩v 󵄩󵄩 + h ∑ 󵄨󵄨󵄨û j2 󵄨󵄨󵄨 vj . 2 j=1

(8.111)

Adding (8.110) and (8.111) together gives 󵄨󵄨 1 󵄨󵄨2 󵄨󵄨u 󵄨󵄨1 +

1 󵄨󵄨 1 󵄨󵄨2 󵄨w 󵄨 + 2 󵄨 󵄨1

󵄨 󵄨2 1 󵄨 󵄨2 = 󵄨󵄨󵄨u0 󵄨󵄨󵄨1 + 󵄨󵄨󵄨w0 󵄨󵄨󵄨1 + 2

m−1 1 m−1 1 1 󵄩󵄩 1 󵄩󵄩2 󵄨 1 󵄨2 2 1 󵄩󵄩v 󵄩󵄩 + h ∑ v̂j2 󵄨󵄨󵄨uj 󵄨󵄨󵄨 + h ∑ |û j2 | vj 2 j=1 j=1 m−1 1 m−1 1 1 󵄩󵄩 0 󵄩󵄩2 󵄨 0 󵄨2 󵄨 󵄨2 0 󵄩󵄩v 󵄩󵄩 + h ∑ v̂j2 󵄨󵄨󵄨uj 󵄨󵄨󵄨 + h ∑ 󵄨󵄨󵄨û j2 󵄨󵄨󵄨 vj , 2 j=1 j=1

which implies 1 󵄨 󵄨2 󵄨 󵄨2 1 󵄩 󵄩2 󵄩 󵄩2 1 󵄨󵄨 1 󵄨󵄨2 󵄨󵄨 0 󵄨󵄨2 (󵄨u 󵄨 + 󵄨u 󵄨 ) + (󵄨󵄨󵄨w1 󵄨󵄨󵄨1 + 󵄨󵄨󵄨w0 󵄨󵄨󵄨1 ) + (󵄩󵄩󵄩v1 󵄩󵄩󵄩 + 󵄩󵄩󵄩v0 󵄩󵄩󵄩 ) 2 󵄨 󵄨1 󵄨 󵄨1 4 4 m−1 1 m−1 1 1 󵄨 󵄨2 󵄨 󵄨2 + (h ∑ v̂j2 󵄨󵄨󵄨uj1 󵄨󵄨󵄨 + h ∑ 󵄨󵄨󵄨û j2 󵄨󵄨󵄨 v1j ) 2 j=1 j=1

󵄨 󵄨2 = 󵄨󵄨󵄨u0 󵄨󵄨󵄨1 +

1 󵄨󵄨 0 󵄨󵄨2 󵄨w 󵄨 + 2 󵄨 󵄨1

m−1 1 m−1 1 1 󵄩󵄩 0 󵄩󵄩2 1 󵄨 0 󵄨2 󵄨 󵄨2 0 󵄩󵄩v 󵄩󵄩 + (h ∑ v̂j2 󵄨󵄨󵄨uj 󵄨󵄨󵄨 + h ∑ 󵄨󵄨󵄨û j2 󵄨󵄨󵄨 vj ). 2 2 j=1 j=1

Therefore, (8.106) is valid. (III) Multiplying both the right- and left-hand sides of (8.96) by −2hΔt ū jk and summing over j from 1 to m − 1, it follows that: m−1

m−1

m−1

j=1

j=1

j=1

̄ ̄ 󵄨 󵄨2 −2ih ∑ 󵄨󵄨󵄨Δt ujk 󵄨󵄨󵄨 − 2h ∑ (δx2 ujk )Δt ū jk + 2h ∑ vkj ujk Δt ū jk = 0.

Taking the real parts on both the right- and left-hand sides leads to k+1 2

m−1 |uj 1 󵄨󵄨 k+1 󵄨󵄨2 󵄨󵄨 k−1 󵄨󵄨2 (󵄨󵄨u 󵄨󵄨1 − 󵄨󵄨u 󵄨󵄨1 ) + h ∑ vkj 2τ j=1

| − |ujk−1 |2 2τ

= 0,

1 ⩽ k ⩽ n − 1.

(8.112)

8.3 Three-level linearized locally decoupled difference scheme

� 273

Multiplying both the right- and left-hand sides of (8.97) by hΔt wjk and summing over j from 1 to m − 1, it follows: m−1

m−1

j=1

j=1

̄

h ∑ (Δt vkj )Δt wjk − h ∑ (δx2 wjk )Δt wjk = 0,

1 ⩽ k ⩽ n − 1.

(8.113)

Multiplying both the right- and left-hand sides of (8.98) by −hΔt vkj and summing over j from 1 to m − 1, it follows: m−1

m−1

m−1

j=1

j=1

j=1

̄ 󵄨 󵄨2 − h ∑ (Δt wjk )(Δt vkj ) + h ∑ vkj Δt vkj + h ∑ 󵄨󵄨󵄨ujk 󵄨󵄨󵄨 Δt vkj = 0,

1 ⩽ k ⩽ n − 1.

(8.114)

Adding (8.113) and (8.114) together, it follows: m−1 1 󵄩 1 󵄨󵄨 k+1 󵄨󵄨2 󵄨󵄨 k−1 󵄨󵄨2 󵄩2 󵄩 󵄩2 󵄨 󵄨2 (󵄨󵄨w 󵄨󵄨1 − 󵄨󵄨w 󵄨󵄨1 ) + (󵄩󵄩󵄩vk+1 󵄩󵄩󵄩 − 󵄩󵄩󵄩vk−1 󵄩󵄩󵄩 ) + h ∑ 󵄨󵄨󵄨ujk 󵄨󵄨󵄨 Δt vkj = 0, 4τ 4τ j=1

1 ⩽ k ⩽ n − 1. (8.115)

Then adding (8.112) and (8.115) together produces 1 󵄨󵄨 k+1 󵄨󵄨2 󵄨󵄨 k−1 󵄨󵄨2 1 󵄨 1 󵄩 󵄨2 󵄨 󵄩2 󵄩 󵄨2 󵄩2 (󵄨u 󵄨󵄨1 − 󵄨󵄨u 󵄨󵄨1 ) + (󵄨󵄨󵄨wk+1 󵄨󵄨󵄨1 − 󵄨󵄨󵄨wk−1 󵄨󵄨󵄨1 ) + (󵄩󵄩󵄩vk+1 󵄩󵄩󵄩 − 󵄩󵄩󵄩vk−1 󵄩󵄩󵄩 ) 2τ 󵄨 4τ 4τ m−1

+h ∑ j=1

1 k 󵄨󵄨 k+1 󵄨󵄨2 󵄨󵄨 k−1 󵄨󵄨2 󵄨󵄨 k 󵄨󵄨2 k+1 [v (󵄨u 󵄨 − 󵄨u 󵄨 ) + 󵄨󵄨uj 󵄨󵄨 (vj − vk−1 j )] = 0, 2τ j 󵄨 j 󵄨 󵄨 j 󵄨

1 ⩽ k ⩽ n − 1,

which implies 1 󵄨 1 󵄩 1 󵄨󵄨 k+1 󵄨󵄨2 󵄨󵄨 k 󵄨󵄨2 󵄨2 󵄨 󵄨2 󵄩2 󵄩 󵄩2 (󵄨u 󵄨󵄨1 + 󵄨󵄨u 󵄨󵄨1 ) + (󵄨󵄨󵄨wk+1 󵄨󵄨󵄨1 + 󵄨󵄨󵄨wk 󵄨󵄨󵄨1 ) + (󵄩󵄩󵄩vk+1 󵄩󵄩󵄩 + 󵄩󵄩󵄩vk 󵄩󵄩󵄩 ) 2 󵄨 4 4 1 m−1 󵄨 󵄨2 󵄨󵄨 k 󵄨󵄨2 + h ∑ (vkj 󵄨󵄨󵄨ujk+1 󵄨󵄨󵄨 + vk+1 j 󵄨󵄨uj 󵄨󵄨 ) 2 j=1

1 󵄨 󵄨2 󵄨 1 󵄩 󵄩2 󵄩 1 󵄨 󵄨2 󵄨 󵄨2 󵄨2 󵄩2 = (󵄨󵄨󵄨uk 󵄨󵄨󵄨1 + 󵄨󵄨󵄨uk−1 󵄨󵄨󵄨1 ) + (󵄨󵄨󵄨wk 󵄨󵄨󵄨1 + 󵄨󵄨󵄨wk−1 󵄨󵄨󵄨1 ) + (󵄩󵄩󵄩vk 󵄩󵄩󵄩 + 󵄩󵄩󵄩vk−1 󵄩󵄩󵄩 ) 2 4 4 1 m−1 󵄨󵄨 k 󵄨󵄨2 k 󵄨 k−1 󵄨2 + h ∑ (vk−1 󵄨u 󵄨 + vj 󵄨󵄨󵄨uj 󵄨󵄨󵄨 ), 2 j=1 j 󵄨 j 󵄨

1 ⩽ k ⩽ n − 1,

or E k = E k−1 , Therefore, (8.107) holds.

1 ⩽ k ⩽ n − 1.

274 � 8 Difference methods for the Zakharov equation By Theorem 8.5, there is a constant c10 such that 󵄩󵄩 k 󵄩󵄩 󵄩󵄩u 󵄩󵄩∞ ⩽ c10 ,

󵄩󵄩 k 󵄩󵄩 󵄩󵄩w 󵄩󵄩∞ ⩽ c10 ,

󵄩󵄩 k 󵄩󵄩 󵄩󵄩v 󵄩󵄩 ⩽ c10 ,

0 ⩽ k ⩽ n.

8.3.4 Convergence of the difference solution Theorem 8.6. Suppose {Ujk , Vjk , Wjk | 0 ⩽ j ⩽ m, 0 ⩽ k ⩽ n} is the solution of the problem (8.8)–(8.12) and {ujk , vkj , wjk | 0 ⩽ j ⩽ m, 0 ⩽ k ⩽ n} is the solution of the difference scheme (8.93)–(8.102). Denote ejk = Ujk − ujk ,

fjk = Vjk − vkj ,

gjk = Wjk − wjk ,

0 ⩽ j ⩽ m, 0 ⩽ k ⩽ n.

Then there is a constant c11 satisfying 󵄩󵄩 k 󵄩󵄩 󵄨󵄨 k 󵄨󵄨 󵄩󵄩 k 󵄩󵄩 󵄨󵄨 k 󵄨󵄨 2 2 󵄩󵄩e 󵄩󵄩 + 󵄨󵄨e 󵄨󵄨1 + 󵄩󵄩f 󵄩󵄩 + 󵄨󵄨g 󵄨󵄨1 ⩽ c11 (τ + h ),

0 ⩽ k ⩽ n.

(8.116)

Proof. Subtracting (8.93)–(8.100) from (8.84)–(8.86), (8.75)–(8.77), (8.91)–(8.92), respectively, we have the system of error equations: { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { {

1

1

1

1

iδt ej2 + δx2 ej2 − ej2 v̂j2 = P̂ j0 , 1 2

1 2

δt fj − δx2 gj = Q̂ j0 , 1 2

1 2

δt gj − fj = R̂ 0j , ̄

̄

̄

iΔt ejk + δx2 ejk − (Ujk Vjk − ujk vkj ) = P̂ jk , Δt fjk Δt gjk ej0

=

̄ − δx2 gjk = Q̂ jk , ̄ 󵄨 󵄨2 − fjk − (󵄨󵄨󵄨Ujk 󵄨󵄨󵄨 0, fj0 = 0,

e0k = 0,

k em = 0,

1 ⩽ j ⩽ m − 1,

(8.117)

1 ⩽ j ⩽ m − 1,

(8.118)

0 ⩽ j ⩽ m,

(8.119)

1 ⩽ j ⩽ m − 1, 1 ⩽ k ⩽ n − 1, (8.120) 1 ⩽ j ⩽ m − 1, 1 ⩽ k ⩽ n − 1, (8.121)

󵄨 󵄨2 − 󵄨󵄨󵄨ujk 󵄨󵄨󵄨 ) = R̂ kj , gj0 = F(ψ1 )(xj ) g0k = 0,

− G(ψ1 )j ,

k gm = 0,

0 ⩽ j ⩽ m, 1 ⩽ k ⩽ n − 1,

(8.122)

0 ⩽ j ⩽ m,

(8.123)

1 ⩽ k ⩽ n.

(8.124)

It concludes from (8.123)–(8.124) that 󵄩󵄩 0 󵄩󵄩 󵄩󵄩e 󵄩󵄩 = 0,

󵄨󵄨 0 󵄨󵄨 󵄨󵄨e 󵄨󵄨1 = 0,

󵄩󵄩 0 󵄩󵄩 󵄩󵄩f 󵄩󵄩 = 0.

(8.125)

With the help of (8.65), there is a constant c such that 󵄨󵄨 0 󵄨󵄨 2 󵄨󵄨g 󵄨󵄨1 ⩽ ch ,

√L 2 󵄩󵄩 0 󵄩󵄩 ch . 󵄩󵄩g 󵄩󵄩∞ ⩽ 2

(8.126)

In addition, there is a constant c12 such that 󵄨󵄨 ̂ 21 󵄨󵄨 󵄨󵄨uj 󵄨󵄨 ⩽ c12 ,

󵄨󵄨 ̂ 21 󵄨󵄨 󵄨󵄨vj 󵄨󵄨 ⩽ c12 ,

1 ⩽ j ⩽ m − 1.

(8.127)

� 275

8.3 Three-level linearized locally decoupled difference scheme 1

(I) Multiplying both the right- and left-hand sides of (8.117) by hēj2 and summing over j from 1 to m − 1, we have m−1

m−1

m−1

m−1

j=1

j=1

j=1

j=1

1 1 1 1 1 󵄨 1 󵄨2 1 ih ∑ (δt ej2 )ēj2 + h ∑ (δx2 ej2 )ēj2 − h ∑ 󵄨󵄨󵄨ej2 󵄨󵄨󵄨 v̂j2 = h ∑ P̂ j0 ej2 .

Taking the imaginary parts on both the right- and left-hand sides, we have m−1 1 1 󵄩󵄩 1 󵄩󵄩2 󵄩󵄩 0 󵄩󵄩2 󵄩 󵄩 󵄩 1󵄩 (󵄩󵄩e 󵄩󵄩 − 󵄩󵄩e 󵄩󵄩 ) = Im{h ∑ P̂ j0 ej2 } ⩽ 󵄩󵄩󵄩P̂ 0 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩e 2 󵄩󵄩󵄩. 2τ j=1

Noting e0 = 0, it follows: 1 󵄩󵄩 1 󵄩󵄩2 󵄩󵄩 ̂ 0 󵄩󵄩 󵄩e 󵄩 ⩽ 󵄩󵄩P 󵄩󵄩 ⋅ 2τ 󵄩 󵄩

1 󵄩󵄩 1 󵄩󵄩 󵄩e 󵄩. 2󵄩 󵄩

Again using (8.88), we have 󵄩󵄩 1 󵄩󵄩 󵄩 0󵄩 2 2 2 2 󵄩󵄩e 󵄩󵄩 ⩽ τ 󵄩󵄩󵄩P̂ 󵄩󵄩󵄩 ⩽ c9 √Lτ(τ + h ) ⩽ c9 √L(τ + h ).

(8.128) 1

(II) Multiplying both the right- and left-hand sides of (8.117) by −2hδt ēj2 and summing over j from 1 to m − 1, it follows: m−1

m−1

1

1

1

m−1

1

1

1

m−1

1

󵄨 󵄨2 −2ih ∑ 󵄨󵄨󵄨δt ej2 󵄨󵄨󵄨 − 2h ∑ (δx2 ej2 )(δt ēj2 ) + 2h ∑ v̂j2 ej2 δt ēj2 = −2h ∑ P̂ j0 δt ēj2 . j=1

j=1

j=1

j=1

Taking the real parts on both the right- and left-hand sides, we have m−1 m−1 1 |e1 |2 − |e0 |2 1 1 󵄨󵄨 1 󵄨󵄨2 󵄨󵄨 0 󵄨󵄨2 j j (󵄨󵄨e 󵄨󵄨1 − 󵄨󵄨e 󵄨󵄨1 ) + 2h ∑ v̂j2 = −2 Re{h ∑ P̂ j0 δt ēj2 }. τ 2τ j=1 j=1

Noting e0 = 0, it holds that m−1

m−1

j=1

j=1

1 󵄨󵄨 1 󵄨󵄨2 󵄨 1 󵄨2 󵄩 0 󵄩 󵄩 1 󵄩 󵄩 0 󵄩2 󵄩 1 󵄩2 0 1 󵄨󵄨e 󵄨󵄨1 + h ∑ v̂j2 󵄨󵄨󵄨ej 󵄨󵄨󵄨 = −2 Re{h ∑ P̂ j ēj } ⩽ 2󵄩󵄩󵄩P̂ 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩e 󵄩󵄩󵄩 ⩽ 󵄩󵄩󵄩P̂ 󵄩󵄩󵄩 + 󵄩󵄩󵄩e 󵄩󵄩󵄩 ,

which implies 󵄨󵄨 1 󵄨󵄨2 󵄩 1 󵄩2 󵄩 0 󵄩2 󵄨󵄨e 󵄨󵄨1 ⩽ (c12 + 1)󵄩󵄩󵄩e 󵄩󵄩󵄩 + 󵄩󵄩󵄩P̂ 󵄩󵄩󵄩 by using (8.127). Combining (8.88) with (8.128) yields 󵄨󵄨 1 󵄨󵄨2 2 2 2 2 2 2 2 2 󵄨󵄨e 󵄨󵄨1 ⩽ (c12 + 1)c9 L(τ + h ) + c9 L(τ + h ) ,

276 � 8 Difference methods for the Zakharov equation which implies 󵄨󵄨 1 󵄨󵄨 √ 2 2 󵄨󵄨e 󵄨󵄨1 ⩽ (c12 + 2)Lc9 (τ + h ).

(8.129) 1

(III) Multiplying both the right- and left-hand sides of (8.118) by hδt gj2 and summing over j from 1 to m − 1, it follows: m−1

1

m−1

1

1

1

m−1

1

h ∑ (δt fj 2 )δt gj2 − h ∑ (δx2 gj2 )δt gj2 = h ∑ Q̂ j0 δt gj2 . j=1

j=1

j=1

1

Multiplying both the right- and left-hand sides of (8.119) by −hδt fj 2 and summing over j from 1 to m − 1, it follows: m−1

1

1

m−1

1

1

m−1

1

−h ∑ (δt gj2 )δt fj 2 + h ∑ fj 2 δt fj 2 = −h ∑ R̂ 0j δt fj 2 . j=1

j=1

j=1

Then adding the above two equalities together, we have m−1 m−1 1 1 1 󵄨󵄨 1 󵄨󵄨2 󵄨󵄨 0 󵄨󵄨2 1 󵄩 󵄩2 󵄩 󵄩2 (󵄨󵄨g 󵄨󵄨1 − 󵄨󵄨g 󵄨󵄨1 ) + (󵄩󵄩󵄩f 1 󵄩󵄩󵄩 − 󵄩󵄩󵄩f 0 󵄩󵄩󵄩 ) = h ∑ Q̂ j0 δt gj2 − h ∑ R̂ 0j δt fj 2 . 2τ 2τ j=1 j=1

In view of f 0 = 0, we have 󵄨󵄨 1 󵄨󵄨2 󵄩󵄩 1 󵄩󵄩2 󵄨󵄨g 󵄨󵄨1 + 󵄩󵄩f 󵄩󵄩

m−1

m−1

j=1

j=1

󵄨 󵄨2 = 󵄨󵄨󵄨g 0 󵄨󵄨󵄨1 + 2h ∑ Q̂ j0 (gj1 − gj0 ) − 2h ∑ R̂ 0j fj1 󵄨 󵄨2 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 ⩽ 󵄨󵄨󵄨g 0 󵄨󵄨󵄨1 + 2󵄩󵄩󵄩Q̂ 0 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩g 1 󵄩󵄩󵄩 + 2󵄩󵄩󵄩Q̂ 0 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩g 0 󵄩󵄩󵄩 + 2󵄩󵄩󵄩R̂ 0 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩f 1 󵄩󵄩󵄩 1 󵄩 󵄩2 3 󵄩 󵄩2 L2 󵄩 󵄩2 󵄩 󵄩2 󵄩 󵄩2 󵄩 󵄩2 󵄨 󵄨2 ⩽ 󵄨󵄨󵄨g 0 󵄨󵄨󵄨1 + ( 2 󵄩󵄩󵄩g 1 󵄩󵄩󵄩 + 󵄩󵄩󵄩Q̂ 0 󵄩󵄩󵄩 ) + (󵄩󵄩󵄩Q̂ 0 󵄩󵄩󵄩 + 󵄩󵄩󵄩g 0 󵄩󵄩󵄩 ) + ( 󵄩󵄩󵄩f 1 󵄩󵄩󵄩 + 2󵄩󵄩󵄩R̂ 0 󵄩󵄩󵄩 ) 3 2 L L2 󵄨 󵄨2 L2 󵄩 󵄩2 1 󵄨 󵄨2 1 󵄩 󵄩2 󵄩 󵄩2 ⩽ 󵄨󵄨󵄨g 1 󵄨󵄨󵄨1 + 󵄩󵄩󵄩f 1 󵄩󵄩󵄩 + (1 + )󵄨󵄨󵄨g 0 󵄨󵄨󵄨1 + (1 + )󵄩󵄩󵄩Q̂ 0 󵄩󵄩󵄩 + 2󵄩󵄩󵄩R̂ 0 󵄩󵄩󵄩 . 2 2 6 3 Noticing (8.89)–(8.90) and (8.126), we have L2 󵄩 0 󵄩2 L2 󵄨 0 󵄨2 󵄨󵄨 1 󵄨󵄨2 󵄩󵄩 1 󵄩󵄩2 󵄩 0 󵄩2 󵄨󵄨g 󵄨󵄨1 + 󵄩󵄩f 󵄩󵄩 ⩽ 2(1 + )󵄨󵄨󵄨g 󵄨󵄨󵄨1 + 2(1 + )󵄩󵄩󵄩Q̂ 󵄩󵄩󵄩 + 4󵄩󵄩󵄩R̂ 󵄩󵄩󵄩 6 3 ⩽ 2(1 +

L2 L2 2 4 2 )c h + 2(3 + )c92 L(τ 2 + h2 ) 6 3

⩽ [2(1 +

L2 2 L2 2 )c + 2(3 + )c92 L](τ 2 + h2 ) . 6 3

(8.130)

Thus, (8.116) is valid for k = 0 and k = 1 by using (8.125), (8.126) and (8.128)–(8.130).

8.3 Three-level linearized locally decoupled difference scheme

� 277

̄

(IV) Multiplying both the right- and left-hand sides of (8.120) by hējk and summing over j from 1 to m − 1, it follows that m−1

m−1

m−1

m−1

j=1

j=1

j=1

j=1

̄ ̄ ̄ ̄ ̄ ̄ ̄ ih ∑ (Δt ejk )ējk + ∑ (δx2 ejk )ējk − h ∑ (ejk vkj + Ujk fjk )ējk = h ∑ P̂ jk ējk .

Taking the imaginary parts on both the right- and left-hand sides, we have m−1 m−1 ̄ ̄ ̄ 1 󵄩󵄩 k+1 󵄩󵄩2 󵄩󵄩 k−1 󵄩󵄩2 (󵄩󵄩e 󵄩󵄩 − 󵄩󵄩e 󵄩󵄩 ) = Im{h ∑ Ujk fjk ējk + h ∑ P̂ jk ējk } 4τ j=1 j=1

󵄩 󵄩 󵄩 ̄󵄩 󵄩 󵄩 󵄩 ̄󵄩 ⩽ c1 󵄩󵄩󵄩f k 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩ek 󵄩󵄩󵄩 + 󵄩󵄩󵄩P̂ k 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩ek 󵄩󵄩󵄩.

(8.131)

(V) Multiplying both the right- and left-hand sides of (8.120) by −2hΔt ējk and summing over j from 1 to m − 1, it follows: m−1

m−1

j=1

j=1

m−1

̄

̄

̄

− 2ih ∑ |Δt ejk |2 − 2h ∑ (δx2 ejk )Δt ējk + 2h ∑ (Ujk Vjk − ujk vkj )Δt ējk m−1

= − 2h ∑ P̂ jk Δt ējk , j=1

j=1

1 ⩽ k ⩽ n − 1.

Taking the real parts on both the right- and left-hand sides, we have m−1 ̄ ̄ 1 󵄨󵄨 k+1 󵄨󵄨2 󵄨󵄨 k−1 󵄨󵄨2 (󵄨󵄨e 󵄨󵄨1 − 󵄨󵄨e 󵄨󵄨1 ) = −2 Re{h ∑ (Ujk Vjk − ujk vkj )Δt ējk } 2τ j=1 m−1

−2 Re{h ∑ P̂ jk Δt ējk }, j=1

1 ⩽ k ⩽ n − 1.

(8.132)

(VI) Multiplying both the right- and left-hand sides of (8.121) by hΔt gjk and summing over j from 1 to m − 1, we have m−1

m−1

m−1

j=1

j=1

j=1

̄ h ∑ (Δt fjk )Δt gjk − h ∑ (δx2 gjk )Δt gjk = h ∑ Q̂ jk Δt gjk ,

1 ⩽ k ⩽ n − 1.

(8.133)

Multiplying both the right- and left-hand sides of (8.122) by −hΔt fjk and summing over j from 1 to m − 1, we have m−1

m−1

m−1

m−1

j=1

j=1

j=1

j=1

̄ 󵄨 󵄨2 󵄨 󵄨2 −h ∑ (Δt gjk )Δt fjk + h ∑ fjk Δt fjk + h ∑ (󵄨󵄨󵄨Ujk 󵄨󵄨󵄨 − 󵄨󵄨󵄨ujk 󵄨󵄨󵄨 )Δt fjk = −h ∑ R̂ kj Δt fjk ,

1 ⩽ k ⩽ n − 1.

(8.134)

278 � 8 Difference methods for the Zakharov equation Adding (8.133) and (8.134) together, it follows: 1 󵄩 1 󵄨󵄨 k+1 󵄨󵄨2 󵄨󵄨 k−1 󵄨󵄨2 󵄩2 󵄩2 󵄩 (󵄨g 󵄨󵄨1 − 󵄨󵄨g 󵄨󵄨1 ) + (󵄩󵄩󵄩f k+1 󵄩󵄩󵄩 − 󵄩󵄩󵄩f k−1 󵄩󵄩󵄩 ) 4τ 󵄨 4τ k+1 2 k+1 2 k k 2 k 2 k+1 1 m−1 (|Uj | − |uj | )fj + (|Uj | − |uj | )fj + [h ∑ τ 2 j=1 m−1

−h ∑

(|Ujk |2 − |ujk |2 )fjk−1 + (|Ujk−1 |2 − |ujk−1 |2 )fjk 2

j=1

m−1

= h ∑(

|Ujk+1 |2 − |Ujk−1 |2 2τ

j=1

m−1

m−1

j=1

j=1

|ujk+1 |2 − |ujk−1 |2

−

2τ

]

)fjk

+ h ∑ Q̂ jk Δt gjk − h ∑ R̂ kj Δt fjk .

(8.135)

Denote m−1

̄

̄

C k = −2 Re{h ∑ (Ujk Vjk − ujk vkj )Δt ējk }, j=1

|Ujk+1 |2

m−1

Dk = h ∑ ( j=1

− |Ujk−1 |2

2τ

−

|ujk+1 |2 − |ujk−1 |2 2τ

)fjk .

It follows by calculation that m−1

̄

̄

C k = −2 Re{h ∑ (Ujk Vjk − ujk vkj )(Δt Ū jk − Δt ū jk )} j=1

m−1

̄ ̄ ̄ ̄ = −2 Re{h ∑ (Ujk Vjk Δt Ū jk + ujk vkj Δt ū jk − Ujk Vjk Δt ū jk − ujk vkj Δt Ū jk )}; j=1

m−1

̄ ̄ Dk = 2 Re{h ∑ (Ujk Δt Ū jk − ujk Δt ū jk )(Vjk − vkj )} j=1

m−1

̄

̄

̄

̄

= 2 Re{h ∑ (Ujk Vjk Δt Ū jk + ujk vkj Δt ū jk − ujk Vjk Δt ū jk − Ujk vkj Δt Ū jk )}. j=1

Adding the above two equalities together, we have m−1

̄

̄

̄

̄

C k + Dk = 2 Re{h ∑ [(Ujk − ujk )Vjk Δt ū jk − (Ujk − ujk )vkj Δt Ū jk ]} j=1

m−1

̄

= 2 Re{h ∑ (Vjk Δt ū jk − vkj Δt Ū jk )ejk } j=1

8.3 Three-level linearized locally decoupled difference scheme

� 279

m−1

̄ = 2 Re{h ∑ [Vjk (Δt ū jk − Δt Ū jk ) + (Vjk − vkj )Δt Ū jk ]ejk } j=1

m−1

̄ = 2 Re{h ∑ (−Vjk Δt ējk + fjk Δt Ū jk )ejk } j=1

m−1

m−1

̄

̄

= 2 Re{−h ∑ Vjk ējk Δt ejk } + 2 Re{h ∑ fjk (Δt Ū jk )ejk }. j=1

j=1

(8.136)

From (8.120), we have ̄

̄

̄

Δt ejk = iδx2 ejk − i(Ujk Vjk − ujk vkj ) − iP̂ jk ̄

̄

̄

= iδx2 ejk − i(Ujk fjk + ejk vkj ) − iP̂ jk .

(8.137)

Substituting (8.137) into (8.136), there is a constant c13 satisfying 󵄩 ̄ 󵄩2 󵄨 ̄ 󵄨2 󵄩 󵄩2 󵄩 󵄩2 C k + Dk ⩽ c13 (󵄩󵄩󵄩ek 󵄩󵄩󵄩 + 󵄨󵄨󵄨ek 󵄨󵄨󵄨1 + 󵄩󵄩󵄩f k 󵄩󵄩󵄩 + 󵄩󵄩󵄩P̂ k 󵄩󵄩󵄩 ).

(8.138)

Denote k+1 2 k 2 ‖ek+1 ‖2 + ‖ek ‖2 |e |1 + |e |1 ‖f k+1 ‖2 + ‖f k ‖2 + + 2 2 4 k+1 2 k 2 m−1 (|U k+1 |2 − |uk+1 |2 )f k + (|U k |2 − |uk |2 )f k+1 |g |1 + |g |1 j j j j j j + +h ∑ . 4 2 j=1

F k = 2(c1 + c10 )2

Then we have F k ⩾ (c1 + c10 )2

k+1 2 k 2 k+1 2 k 2 ‖ek+1 ‖2 + ‖ek ‖2 |e |1 + |e |1 ‖f k+1 ‖2 + ‖f k ‖2 |g |1 + |g |1 + + + . 2 2 8 4

Multiplying both the right- and left-hand sides of (8.131) by 4(c1 + c10 )2 , then adding the result with (8.132) and (8.135), there are two constants c14 and c15 such that 1 k 󵄩 ̄ 󵄩2 󵄨 ̄ 󵄨2 󵄩 󵄩2 󵄩 󵄩2 (F − F k−1 ) ⩽ c14 (󵄩󵄩󵄩ek 󵄩󵄩󵄩 + 󵄨󵄨󵄨ek 󵄨󵄨󵄨1 + 󵄩󵄩󵄩f k 󵄩󵄩󵄩 + 󵄩󵄩󵄩P̂ k 󵄩󵄩󵄩 ) τ m−1

m−1

j=1

j=1

m−1

+ 2 Re{−h ∑ P̂ jk Δt ējk } + h ∑ Q̂ jk Δt gjk − h ∑ R̂ kj Δt fjk j=1

m−1

󵄩 󵄩2 ⩽ c15 (F k + F k−1 ) + c14 󵄩󵄩󵄩P̂ k 󵄩󵄩󵄩 + 2 Re{−h ∑ P̂ jk Δt ējk } j=1

m−1

m−1

j=1

j=1

+ h ∑ Q̂ jk Δt gjk − h ∑ R̂ kj Δt fjk ,

1 ⩽ k ⩽ n − 1,

280 � 8 Difference methods for the Zakharov equation where (8.138) is utilized in the first inequality. Replacing k by l in the above inequality and summing over l from 1 to k, it follows: k−1 k 1 k 󵄩 󵄩2 (F − F 0 ) ⩽ c15 (F k + F 0 ) + c14 ∑󵄩󵄩󵄩P̂ l 󵄩󵄩󵄩 + 2c15 ∑ F l τ l=1 l=1 m−1 k

m−1 k

m−1 k

j=1 l=1

j=1 l=1

j=1 l=1

+ 2 Re{−h ∑ ∑ P̂ jl Δt ējl } + h ∑ ∑ Q̂ jl Δt gjl − h ∑ ∑ R̂ lj Δt fjl , 1 ⩽ k ⩽ n − 1.

(8.139)

We have by careful calculations that k

∑ P̂ jl Δt ējl = l=1

k−1 1 k+1 1 k ̂ l l+1 ∑ Pj (ēj − ējl−1 ) = ( ∑ P̂ jl−1 ējl − ∑ P̂ jl+1 ējl ) 2τ l=1 2τ l=2 l=0

k−1 P̂ l+1 − P̂ l−1 1 ̂ k−1 k j j k k+1 1 0 2 1 ̂ ̂ ̂ ̄ ̄ ̄ ̄ = (Pj ej + Pj ej − Pj ej − Pj ej ) − ∑ ējl . 2τ 2τ l=2

Therefore, 󵄨󵄨 󵄨󵄨 m−1 k 󵄨󵄨 󵄨 󵄨󵄨2 Re{−h ∑ ∑ P̂ l Δt ēl }󵄨󵄨󵄨 j j 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 j=1 l=1 k−1󵄩 󵄩󵄩 P̂ l+1 − P̂ l−1 󵄩󵄩󵄩 󵄩 l 󵄩 1 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩󵄩 ⋅ 󵄩󵄩e 󵄩󵄩 ⩽ (󵄩󵄩󵄩P̂ k−1 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩ek 󵄩󵄩󵄩 + 󵄩󵄩󵄩P̂ k 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩ek+1 󵄩󵄩󵄩 + 󵄩󵄩󵄩P̂ 1 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩e0 󵄩󵄩󵄩 + 󵄩󵄩󵄩P̂ 2 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩e1 󵄩󵄩󵄩) + 2 ∑ 󵄩󵄩󵄩 󵄩󵄩 󵄩 󵄩 󵄩󵄩 τ 2τ 󵄩 l=2

⩽

⩽

2 3 󵄩 󵄩2 L2 󵄩 3󵄩 3 󵄩 󵄩2 L2 󵄩 󵄩2 1 󵄩2 󵄩2 L 󵄩 󵄩2 [( 2 󵄩󵄩󵄩ek 󵄩󵄩󵄩 + 󵄩󵄩󵄩P̂ k−1 󵄩󵄩󵄩 ) + ( 2 󵄩󵄩󵄩ek+1 󵄩󵄩󵄩 + 󵄩󵄩󵄩P̂ k 󵄩󵄩󵄩 ) + 2 󵄩󵄩󵄩e1 󵄩󵄩󵄩 + 󵄩󵄩󵄩P̂ 2 󵄩󵄩󵄩 ] 2τ L 3 3 3 L L 2 k−1 2󵄩 l+1 l−1 3 󵄩 󵄩2 L 󵄩󵄩 P̂ − P̂ 󵄩󵄩󵄩󵄩 + ∑ ( 2 󵄩󵄩󵄩el 󵄩󵄩󵄩 + 󵄩󵄩󵄩 󵄩󵄩 ) 󵄩󵄩 3 󵄩󵄩 2τ L l=2 k 2 k+1 2 1 2 1 |e |1 + |e |1 |e |1 L2 󵄩󵄩 ̂ k−1 󵄩󵄩2 󵄩󵄩 ̂ k 󵄩󵄩2 󵄩󵄩 ̂ 2 󵄩󵄩2 [ + + (󵄩󵄩P 󵄩󵄩 + 󵄩󵄩P 󵄩󵄩 + 󵄩󵄩P 󵄩󵄩 )] 2τ 2 2 3 k−1 2 k−1󵄩 l+1 l−1 󵄩2 󵄩 ̂ 1 󵄨 󵄨2 L 󵄩 P − P̂ 󵄩󵄩󵄩 + ∑ 󵄨󵄨󵄨el 󵄨󵄨󵄨1 + ∑ 󵄩󵄩󵄩 󵄩󵄩 , 1 ⩽ k ⩽ n − 1. 󵄩󵄩 2 3 2τ 󵄩󵄩 l=2

l=2

Similarly, by using k−1 Q̂ l+1 − Q̂ l−1 1 ̂ k−1 k j j k k+1 1 0 2 1 l l ̂ ̂ ̂ ̂ gjl , ∑ Qj Δt gj = (Qj gj + Qj gj − Qj gj − Qj gj ) − ∑ 2τ 2τ l=2 l=1 k

it follows: 󵄨󵄨 m−1 k 󵄨 󵄨󵄨 l l 󵄨󵄨 󵄨󵄨h ∑ ∑ Q̂ j Δt gj 󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨 j=1 l=1

(8.140)

8.3 Three-level linearized locally decoupled difference scheme

⩽ ⩽

⩽

� 281

1 󵄩󵄩 ̂ k−1 󵄩󵄩 󵄩󵄩 k 󵄩󵄩 󵄩󵄩 ̂ k 󵄩󵄩 󵄩󵄩 k+1 󵄩󵄩 󵄩󵄩 ̂ 1 󵄩󵄩 󵄩󵄩 0 󵄩󵄩 󵄩󵄩 ̂ 2 󵄩󵄩 󵄩󵄩 1 󵄩󵄩 k−1󵄩󵄩󵄩󵄩 Q̂ l+1 − Q̂ l−1 󵄩󵄩󵄩󵄩 󵄩󵄩 l 󵄩󵄩 (󵄩Q 󵄩󵄩 ⋅ 󵄩󵄩g 󵄩󵄩 + 󵄩󵄩Q 󵄩󵄩 ⋅ 󵄩󵄩g 󵄩󵄩 + 󵄩󵄩Q 󵄩󵄩 ⋅ 󵄩󵄩g 󵄩󵄩 + 󵄩󵄩Q 󵄩󵄩 ⋅ 󵄩󵄩g 󵄩󵄩) + ∑ 󵄩󵄩 󵄩󵄩 ⋅ 󵄩󵄩g 󵄩󵄩 󵄩 󵄩󵄩 2τ 󵄩 2τ l=2 󵄩 2 1 1 6󵄩 1 6 󵄩 󵄩2 L2 󵄩 󵄩2 󵄩2 L 󵄩 󵄩2 [( ⋅ 2 󵄩󵄩󵄩g k 󵄩󵄩󵄩 + 󵄩󵄩󵄩Q̂ k−1 󵄩󵄩󵄩 ) + ( ⋅ 2 󵄩󵄩󵄩g k+1 󵄩󵄩󵄩 + 󵄩󵄩󵄩Q̂ k 󵄩󵄩󵄩 ) 2τ 4 L 6 4 L 6 2 2 1 6 󵄩 󵄩2 L 󵄩 󵄩2 1 6 󵄩 󵄩2 L 󵄩 󵄩2 + ( ⋅ 2 󵄩󵄩󵄩g 0 󵄩󵄩󵄩 + 󵄩󵄩󵄩Q̂ 1 󵄩󵄩󵄩 ) + ( ⋅ 2 󵄩󵄩󵄩g 1 󵄩󵄩󵄩 + 󵄩󵄩󵄩Q̂ 2 󵄩󵄩󵄩 )] 4 L 6 4 L 6 2 k−1 2󵄩 l+1 l−1 󵄩 1 6 󵄩 󵄩2 L 󵄩󵄩 Q̂ − Q̂ 󵄩󵄩󵄩 + ∑ ( ⋅ 2 󵄩󵄩󵄩g l 󵄩󵄩󵄩 + 󵄩󵄩󵄩 󵄩󵄩 ) 󵄩󵄩 4 6 󵄩󵄩 2τ L l=2

1 1 󵄨󵄨 k 󵄨󵄨2 󵄨󵄨 k+1 󵄨󵄨2 1 󵄨 󵄨2 󵄨 󵄨2 [ (󵄨g 󵄨 + 󵄨g 󵄨󵄨1 ) + (󵄨󵄨󵄨g 0 󵄨󵄨󵄨1 + 󵄨󵄨󵄨g 1 󵄨󵄨󵄨1 ) 2τ 4 󵄨 󵄨1 󵄨 4

L2 󵄩󵄩 ̂ k−1 󵄩󵄩2 󵄩󵄩 ̂ k 󵄩󵄩2 󵄩󵄩 ̂ 1 󵄩󵄩2 󵄩󵄩 ̂ 2 󵄩󵄩2 (󵄩Q 󵄩󵄩 + 󵄩󵄩Q 󵄩󵄩 + 󵄩󵄩Q 󵄩󵄩 + 󵄩󵄩Q 󵄩󵄩 )] 6 󵄩 2 1 k−1󵄨 󵄨2 L2 k−1󵄩󵄩󵄩󵄩 Q̂ l+1 − Q̂ l−1 󵄩󵄩󵄩󵄩 + ∑ 󵄨󵄨󵄨g l 󵄨󵄨󵄨1 + ∑ 󵄩󵄩 󵄩󵄩 , 1 ⩽ k ⩽ n − 1. 󵄩󵄩 󵄩󵄩 4 6 2τ

+

l=2

l=2

(8.141)

By using k

∑ R̂ lj Δt fjl = l=1

k−1 R̂ l+1 − R̂ l−1 1 ̂ k−1 k j j (Rj fj + R̂ kj fjk+1 − R̂ 1j fj0 − R̂ 2j fj1 ) − ∑ fjl , 2τ 2τ l=2

it follows: 󵄨󵄨 m−1 k 󵄨 󵄨󵄨 l l 󵄨󵄨 󵄨󵄨−h ∑ ∑ R̂ j Δt fj 󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨 j=1 l=1 ⩽ ⩽

1 󵄩󵄩 ̂ k−1 󵄩󵄩 󵄩󵄩 k 󵄩󵄩 󵄩󵄩 ̂ k 󵄩󵄩 󵄩󵄩 k+1 󵄩󵄩 󵄩󵄩 ̂ 1 󵄩󵄩 󵄩󵄩 0 󵄩󵄩 󵄩󵄩 ̂ 2 󵄩󵄩 󵄩󵄩 1 󵄩󵄩 k−1󵄩󵄩󵄩󵄩 R̂ l+1 − R̂ l−1 󵄩󵄩󵄩󵄩 󵄩󵄩 l 󵄩󵄩 (󵄩R 󵄩󵄩 ⋅ 󵄩󵄩f 󵄩󵄩 + 󵄩󵄩R 󵄩󵄩 ⋅ 󵄩󵄩f 󵄩󵄩 + 󵄩󵄩R 󵄩󵄩 ⋅ 󵄩󵄩f 󵄩󵄩 + 󵄩󵄩R 󵄩󵄩 ⋅ 󵄩󵄩f 󵄩󵄩) + ∑ 󵄩󵄩 󵄩󵄩 ⋅ 󵄩󵄩f 󵄩󵄩 󵄩 󵄩󵄩 2τ 󵄩 2τ l=2 󵄩 1 󵄩 󵄩2 1󵄩 1 󵄩 󵄩2 1 󵄩 󵄩2 󵄩2 󵄩 󵄩2 󵄩 󵄩2 [( 󵄩󵄩f k 󵄩󵄩 + 2󵄩󵄩󵄩R̂ k−1 󵄩󵄩󵄩 ) + ( 󵄩󵄩󵄩f k+1 󵄩󵄩󵄩 + 2󵄩󵄩󵄩R̂ k 󵄩󵄩󵄩 ) + ( 󵄩󵄩󵄩f 1 󵄩󵄩󵄩 + 2󵄩󵄩󵄩R̂ 2 󵄩󵄩󵄩 )] 2τ 8 󵄩 󵄩 8 8 k−1 󵄩󵄩 R̂ l+1 − R̂ l−1 󵄩󵄩2 1 󵄩 󵄩2 󵄩 󵄩󵄩 + ∑ ( 󵄩󵄩󵄩f l 󵄩󵄩󵄩 + 2󵄩󵄩󵄩 󵄩󵄩 ). 󵄩󵄩 󵄩󵄩 8 2τ l=2

(8.142)

Substituting (8.140)–(8.142) into (8.139) and using (8.78)–(8.83), (8.125), (8.128)–(8.130), there is a constant c16 such that k−1

2

F k ⩽ c16 τ ∑ F l + c16 (τ 2 + h2 ) , l=1

1⩽k ⩽n−1

when 2τc15 ⩽ 31 . By the Gronwall inequality (Theorem 1.2(c)), we have 2

F k ⩽ ec16 T c16 (τ 2 + h2 ) , Therefore, (8.116) is valid.

1 ⩽ k ⩽ n − 1.

282 � 8 Difference methods for the Zakharov equation

8.4 Numerical experiments In this section, we will take the three-level linearized difference scheme (8.93)–(8.102) as an example to test the accuracy and conservation. Our theoretical results in this chapx −x ter can be easily extended to the spatial domain [xl , xr ]. Take h = rm l , xi = xl + ih, 0 ⩽ i ⩽ m. Denote the numerical errors and the convergence orders, respectively, as 󵄩 󵄩 Eu (h, τ) = max 󵄩󵄩󵄩ek 󵄩󵄩󵄩∞ , 1⩽k⩽n

Orduh = log2 (Eu (2h, τ)/Eu (h, τ)),

Ordvh = log2 (Ev (2h, τ)/Ev (h, τ)),

󵄩 󵄩 Ev (h, τ) = max 󵄩󵄩󵄩f k 󵄩󵄩󵄩, 1⩽k⩽n

Orduτ = log2 (Eu (h, 2τ)/Eu (h, τ)), Ordvτ = log2 (Ev (h, 2τ)/Ev (h, τ)).

Example 8.1 ([4]). In the problem (8.1)–(8.4), we take the left and right boundaries as xl = −32 and xr = 32 and T = 4. At this point, the problem has the solitary wave solution as 2

2

u(x, t) = √2B2 (1 − C 2 ) sech(B(x − Ct))ei[(C/2)x−((C/2) −B )t] , v(x, t) = −2B2 sech2 (B(x − Ct)), where B and C are two constants. Here, we choose B = 1, C = 0.5. The initial and boundary conditions are taken from the above exact solutions. The numerical results calculated by the difference scheme (8.93)–(8.102) are listed in Tables 8.1–8.2 and displayed in Figures 8.1–8.3. From the numerical results in Table 8.1 and Table 8.2, we know that the convergence orders are both two in temporal and spatial directions. Figures 8.1 and 8.2 show that the difference scheme (8.93)–(8.102) is accurate. Figure 8.3 displays the conservation in Theorem 8.5. All of the numerical results manifest that the difference scheme (8.93)–(8.102) is accurate and practical. Table 8.1: The temporal convergence orders with h = 8/625. τ

Eu (h, τ)

Orduτ

Ev (h, τ)

Ordvτ

�/� �/� �/�� �/�� �/��

�.����e − � �.����e − � �.����e − � �.����e − � �.����e − �

�.���� �.���� �.���� �.����

�.����e − � �.����e − � �.����e − � �.����e − � �.����e − �

�.���� �.���� �.���� �.����

8.5 Summary and extension

� 283

Table 8.2: The spatial convergence orders with τ = 1/500. h

Eu (h, τ)

Orduh

Ev (h, τ)

Ordvh

�/� �/�� �/�� �/�� �/���

�.����e − � �.����e − � �.����e − � �.����e − � �.����e − �

�.���� �.���� �.���� �.����

�.����e − � �.����e − � �.����e − � �.����e − � �.����e − �

�.���� �.���� �.���� �.����

Figure 8.1: The numerical profiles for |u| and v, and their exact solutions |U| and V , m = 2560, n = 640.

Figure 8.2: The numerical error surfaces for |u| and v, m = 2560, n = 640.

8.5 Summary and extension In this chapter, we first introduce a new auxiliary function w(x, t) to convert the initial and boundary value problem (8.1)–(8.4) into an equivalent problem (8.8)–(8.12). It is proved that the solution of the initial and boundary value problem (8.8)–(8.12) satisfies two conservation laws. Then two difference schemes are studied in Section 8.2 and Section 8.3, respectively.

284 � 8 Difference methods for the Zakharov equation

Figure 8.3: The invariant curves of Qk and E k and their error curves; drawing a grid every five points with m = 320, n = 400.

A two-level nonlinear difference scheme is given in Section 8.2. The existence and uniqueness of the solution of the difference scheme are proved. Then it is shown that the solution satisfies two conservation laws. Finally, the convergence of the solution of the difference scheme is proved. A three-level linearized locally decoupled difference scheme is established in Section 8.3, in which only two independent tri-diagonal linear system need to be solved at each time level. The existence and uniqueness of solutions of the difference scheme are proved. It is also proved that the solution of the difference scheme satisfies two conservation laws and the convergence of the solution of the difference scheme is given. The Chinese scholars, Guo and Chang, studied the finite difference method for solving the Zakharov equation earlier [8, 13, 14]. The content in this chapter is developed based on their work. There is another class of nonlinear equations called the Klein–Gordon–Zakharov equation: {

utt − uxx + u + uv + |u|2 u = 0, vtt − vxx = (|u|2 )xx ,

which is similar to the Zakharov equation. In [39], Wang et al. established a three-level nonlinear difference scheme for the above problem. They proved that the solution of the difference scheme satisfies the discrete conservation law. Then the existence and convergence of the solution of the difference scheme were also given. In [42], Wang and Guo established a three-level linearized difference scheme for the above problem. The multidimensional Zakharov equation is in the form of {

iut + Δu − uv = 0,

vtt − Δv − Δ(|u|2 ) = 0.

9 Difference methods for the Ginzburg–Landau equation The Ginzburg–Landau equation as a model of low-temperature superconductivity was developed by famous physicists, Ginzburg and Landau, in the 1950s. This model was awarded the Nobel prize in physics in 2003. It is also widely used in the non-equilibrium fluid dynamics system and physical phase transition process.

9.1 Introduction In this chapter, the initial and boundary value problem of the two-dimensional Ginzburg–Landau equation ut − (ν + iα)Δu + (κ + iβ)|u|2 u − γu = 0, { { { u(x, y, t) = 0, { { { { u(x, y, 0) = φ(x, y),

(x, y) ∈ Ω, 0 < t ⩽ T,

(9.1)

(x, y) ∈ 𝜕Ω, 0 < t ⩽ T, (x, y) ∈ Ω̄

(9.2) (9.3)

will be considered, where Ω = (0, L1 ) × (0, L2 ), 𝜕Ω is the boundary of Ω, Δ is the Laplace operator and ν > 0, κ > 0, α, β, γ are given real constants. When (x, y) ∈ 𝜕Ω, φ(x, y) = 0. Before introducing the difference scheme, we first give a priori estimate on the solution of the problem (9.1)–(9.3) based on the energy method. Theorem 9.1. Suppose u(x, y, t) is the solution of the problem (9.1)–(9.3). Then we have 󵄩󵄩 󵄩 󵄩 γt 󵄩 󵄩󵄩u(⋅, ⋅, t)󵄩󵄩󵄩 ⩽ e 󵄩󵄩󵄩u(⋅, ⋅, 0)󵄩󵄩󵄩,

0 ⩽ t ⩽ T.

Proof. Taking the inner product on both the right- and the left-hand sides of (9.1) with u, we have (ut , u) − (ν + iα)(Δu, u) + (κ + iβ)‖u‖44 − γ‖u‖2 = 0,

0 < t ⩽ T.

(9.4)

Noting (9.2) and using the integration by parts, we have − (Δu, u) = ‖∇u‖2 ,

0 ⩽ t ⩽ T.

(9.5)

Substituting (9.5) into (9.4) and taking the real parts on both the right- and left-hand sides, it follows: 1 󵄩󵄩 󵄩2 󵄩 󵄩2 󵄩 󵄩4 󵄩 󵄩2 (󵄩u(⋅, ⋅, t)󵄩󵄩󵄩 )t + ν󵄩󵄩󵄩∇u(⋅, ⋅, t)󵄩󵄩󵄩 + κ󵄩󵄩󵄩u(⋅, ⋅, t)󵄩󵄩󵄩4 − γ󵄩󵄩󵄩u(⋅, ⋅, t)󵄩󵄩󵄩 = 0, 2 󵄩 which implies 󵄩 󵄩2 󵄩 󵄩2 (󵄩󵄩󵄩u(⋅, ⋅, t)󵄩󵄩󵄩 )t ⩽ 2γ󵄩󵄩󵄩u(⋅, ⋅, t)󵄩󵄩󵄩 , https://doi.org/10.1515/9783110796018-009

0 < t ⩽ T.

286 � 9 Difference methods for the Ginzburg–Landau equation With the help of the Gronwall inequality, we obtain 󵄩2 󵄩2 󵄩󵄩 2γt 󵄩 󵄩󵄩u(⋅, ⋅, t)󵄩󵄩󵄩 ⩽ e 󵄩󵄩󵄩u(⋅, ⋅, 0)󵄩󵄩󵄩 ,

0 ⩽ t ⩽ T.

Taking the square root on both the right- and left-hand sides gives 󵄩 󵄩 󵄩󵄩 γt 󵄩 󵄩󵄩u(⋅, ⋅, t)󵄩󵄩󵄩 ⩽ e 󵄩󵄩󵄩u(⋅, ⋅, 0)󵄩󵄩󵄩,

0 ⩽ t ⩽ T.

Denote c0 =

󵄨󵄨 󵄨 󵄨󵄨u(x, y, t)󵄨󵄨󵄨.

max

(x,y,t)∈Ω×[0,T]

9.2 Two-level nonlinear difference scheme Take three positive integers m1 , m2 and n. Let h1 =

h2 =

L2 , m2

τ = Tn . Denote

1 tk+ 1 = (tk + tk+1 ), 2 2 = {(xi , yj ) | 0 ⩽ i ⩽ m1 , 0 ⩽ j ⩽ m2 }, Ωτ = {tk | 0 ⩽ k ⩽ n},

xi = ih1 , Ωh1 ,h2

L1 , m1

yj = jh2 ,

tk = kτ,

ω = {(i, j) | 0 ⩽ i ⩽ m1 , 0 ⩽ j ⩽ m2 },

ω = {(i, j) | 1 ⩽ i ⩽ m1 − 1, 1 ⩽ j ⩽ m2 − 1}, ωi = {

1, 1 , 2

1 ⩽ i ⩽ m1 − 1, i = 0, m1 ,

ω̂ j = {

1, 1 , 2

𝜕ω = ω\ω, 1 ⩽ j ⩽ m2 − 1, j = 0, m2 .

Define 𝒱h = {v | v = {vij | 0 ⩽ i ⩽ m1 , 0 ⩽ j ⩽ m2 } is the mesh function on Ωh1 ,h2 }, 𝒱̊h = {v | v ∈ 𝒱h , vij = 0 when (i, j) ∈ 𝜕ω}.

Let v ∈ 𝒱h . Introduce the following notation: δx vi+ 1 ,j = 2

δy vi,j+ 1 = 2

1 (v − vij ), h1 i+1,j

1 (v − vij ), h2 i,j+1

Δh vij = δx2 vij + δy2 vij .

1 (vi−1,j − 2vij + vi+1,j ), h12 1 δy2 vij = 2 (vi,j−1 − 2vij + vi,j+1 ), h2 δx2 vij =

It is easy to know that δx2 vij =

1 (δ v 1 − δx vi− 1 ,j ), 2 h1 x i+ 2 ,j

δy2 vij =

1 (δ v 1 − δy vi,j− 1 ). 2 h2 y i,j+ 2

9.2 Two-level nonlinear difference scheme

� 287

Let u, v ∈ 𝒱h . Introduce the following inner product and norms (or seminorms): m1 m2

‖u‖ = √(u, u),

(u, v) = h1 h2 ∑ ∑ ωi ω̂ j uij v̄ij , i=0 j=0

m1 m2

m1 m2

‖δx u‖ = √h1 h2 ∑ ∑ ω̂ j |δx ui− 1 ,j |2 , 2

i=1 j=0

i=0 j=1

2

m1 −1 m2 −1

|u|1 = √‖δx u‖2 + ‖δy u‖2 , 4

‖δy u‖ = √h1 h2 ∑ ∑ ωi |δy ui,j− 1 |2 ,

‖Δh u‖ = √h1 h2 ∑ ∑ (Δh uij )2 , i=1

m1 m2

‖u‖4 = √h1 h2 ∑ ∑ ωi ω̂ j |uij |4 ,

‖u‖∞ =

i=0 j=0

j=1

max

0⩽i⩽m1 ,0⩽j⩽m2

|uij |.

Denote 0

1

n

𝒮τ = {w | w = (w , w , . . . , w ) is the mesh function on Ωτ }.

Let w ∈ 𝒮τ . Introduce the following notation: 1 ̄ 1 1 wk+ 2 = (wk + wk+1 ), wk = (wk+1 + wk−1 ), 2 2 1 1 1 δt wk+ 2 = (wk+1 − wk ), Δt wk = (wk+1 − wk−1 ), τ 2τ 1 1 ∇τ wk = (wk − wk−1 ), ∇2τ wk = (3wk − 4wk−1 + wk−2 ). τ 2τ

It is easy to know that 1 1 1 Δt wk = (δt wk− 2 + δt wk+ 2 ). 2

Suppose u = {uijk | 0 ⩽ i ⩽ m1 , 0 ⩽ j ⩽ m2 , 0 ⩽ k ⩽ n} is a mesh function defined on

Ωh1 ,h2 × Ωτ . Then v = {uijk | 0 ⩽ i ⩽ m1 , 0 ⩽ j ⩽ m2 } ∈ 𝒱h and w = {uijk | 0 ⩽ k ⩽ n} ∈ 𝒮τ . Lemma 9.1 ([35]). Let u ∈ 𝒱̊h . Then we have ‖u‖ ⩽

1 6 √ L2 + 1

6 L22

|u|1 .

Proof. With the help of Lemma 1.1(b), it follows: m1 −1

h1 ∑ |uij |2 ⩽ i=1

L21 m1 h ∑ |δ u 1 |2 . 6 1 i=1 x i− 2 ,j

Multiplying both the right- and left-hand sides by h2 and summing over j, it follows:

288 � 9 Difference methods for the Ginzburg–Landau equation m1 −1 m2 −1

h1 h2 ∑ ∑ |uij |2 ⩽ i=1

j=1

m1 m2 −1 L21 h1 h2 ∑ ∑ |δx ui− 1 ,j |2 . 2 6 i=1 j=1

In other words, 6 ‖u‖2 ⩽ ‖δx u‖2 . L21 Similarly, we can get 6 ‖u‖2 ⩽ ‖δy u‖2 . L22 Adding the above two inequalities together, we have (

6 6 + )‖u‖2 ⩽ ‖δx u‖2 + ‖δy u‖2 = |u|21 . L21 L22

Therefore, it holds that ‖u‖ ⩽

1 6 √ L2 1

+

6 L22

|u|1 .

Lemma 9.2. Let u ∈ 𝒱̊h . Then we have (Δh u, u) = −|u|21 ,

|u|21 ⩽ ‖u‖ ⋅ ‖Δh u‖.

Proof. In combination of the summation by parts, it follows: m1 −1 m2 −1

(Δh u, u) = h1 h2 ∑ ∑ (δx2 uij + δy2 uij )ū ij i=1

j=1

m1 −1 m2 −1

m1 −1 m2 −1

= h1 h2 ∑ ∑ (δx2 uij )ū ij + h1 h2 ∑ ∑ (δy2 uij )ū ij i=1

j=1

i=1

j=1

m2 −1

m1 −1

m1 −1

m2 −1

j=1

i=1

i=1

j=1

= h2 ∑ [h1 ∑ (δx2 uij )ū ij ] + h1 ∑ [h2 ∑ (δy2 uij )ū ij ] m2 −1

m1

j=1

i=1

m1 −1

m2

i=1

j=1

= h2 ∑ [−h1 ∑ |δx ui− 1 ,j |2 ] + h1 ∑ [−h2 ∑ |δy ui,j− 1 |2 ] 2

= −|u|21 . And it is given by the Cauchy–Schwarz inequality that |u|21 = −(Δh u, u) ⩽ ‖u‖ ⋅ ‖Δh u‖.

2

9.2 Two-level nonlinear difference scheme

� 289

Lemma 9.3 ([22]). Let u ∈ 𝒱̊h . Then we have |u|1 ⩽

1 6 √ L2 1

+

6 L22

‖Δh u‖.

Proof. In combination of Lemmas 9.1 and 9.2, it follows: |u|21 ⩽ ‖u‖ ⋅ ‖Δh u‖ ⩽

1 6 √ L2 + 1

6 L22

|u|1 ⋅ ‖Δh u‖.

Dividing the above inequality by |u|1 on both the right- and left-hand sides, we obtain |u|1 ⩽

1 6 √ L2 1

+

6 L22

‖Δh u‖.

Lemma 9.4 ([22]). Let u ∈ 𝒱̊h . Then we have ‖u‖∞ ⩽

1 √2L1 L2 ‖Δh u‖. 8

Proof. Noting ‖Δh u‖2

m1 −1 m2 −1

= h1 h2 ∑ ∑ (δx2 uij + δy2 uij )(δx2 ū ij + δy2 ū ij ) i=1

j=1

m1 −1 m2 −1

m1 −1 m2 −1

m1 −1 m2 −1

󵄨 󵄨2 󵄨 󵄨2 = h1 h2 ∑ ∑ 󵄨󵄨󵄨δx2 uij 󵄨󵄨󵄨 +h1 h2 ∑ ∑ [(δx2 uij )(δy2 ū ij )+(δy2 uij )(δx2 ū ij )]+h1 h2 ∑ ∑ 󵄨󵄨󵄨δy2 uij 󵄨󵄨󵄨 i=1

j=1

i=1

m1 −1 m2 −1

j=1

i=1

j=1

m1 −1 m2 −1

m1 m2

󵄨 󵄨2 󵄨 󵄨2 = h1 h2 ∑ ∑ 󵄨󵄨󵄨δx2 uij 󵄨󵄨󵄨 + 2h1 h2 ∑ ∑ |δx δy ui− 1 ,j− 1 |2 + h1 h2 ∑ ∑ 󵄨󵄨󵄨δy2 uij 󵄨󵄨󵄨 2 2 i=1

=

󵄩󵄩 2 󵄩󵄩2 󵄩󵄩δx u󵄩󵄩

j=1

+ 2‖δx δy u‖2 +

i=1

i=1 j=1

j=1

󵄩󵄩 2 󵄩󵄩2 󵄩󵄩δy u󵄩󵄩 ,

it follows: 1 ‖δx δy u‖2 ⩽ ‖Δh u‖2 . 2 For arbitrary (i, j) ∈ ω, we have uij2 ⩽

m

1 L1 h1 ∑(δx ul− 1 ,j )2 2 4 l=1

m

⩽

m

1 2 L1 L h1 ∑[ 2 h2 ∑(δy δx ul− 1 ,s− 1 )2 ] 2 2 4 l=1 4 s=1

(9.6)

290 � 9 Difference methods for the Ginzburg–Landau equation

=

L1 L2 ‖δy δx u‖2 . 16

This further implies L1 L2 ‖δy δx u‖2 . 16

‖u‖2∞ ⩽ With the application of (9.6), it follows:

‖u‖2∞ ⩽

L1 L2 ‖Δh u‖2 . 32

Taking the square root on both the right- and left-hand sides, we have ‖u‖∞ ⩽

1 √2L1 L2 ‖Δh u‖. 8

9.2.1 Derivation of the difference scheme Define the mesh function U on Ωh1 ,h2 × Ωτ : Uijk = u(xi , yj , tk ),

(i, j) ∈ ω, 0 ⩽ k ⩽ n.

Considering equation (9.1) at the point (xi , yj , tk+ 1 ), we have 2

ut (xi , yj , tk+ 1 ) − (ν + iα)Δu(xi , yj , tk+ 1 ) + (κ + iβ)(|u|2 u)(xi , yj , tk+ 1 ) − γu(xi , yj , tk+ 1 ) = 0, 2

2

(i, j) ∈ ω, 0 ⩽ k ⩽ n − 1.

2

2

Applying the numerical differential formula results in k+ 21

δt Uij

k+ 21

− (ν + iα)Δh Uij

k+ 1 k+ 1 󵄨 k+ 1 󵄨2 k+ 1 + (κ + iβ)󵄨󵄨󵄨Uij 2 󵄨󵄨󵄨 Uij 2 − γUij 2 = Pij 2 ,

(i, j) ∈ ω, 0 ⩽ k ⩽ n − 1,

(9.7)

where there is a constant c1 satisfying 󵄨󵄨 k+ 21 󵄨󵄨 2 2 2 󵄨󵄨Pij 󵄨󵄨 ⩽ c1 (τ + h1 + h2 ), k+ 21

Omitting the small term Pij tions {

(9.8)

in (9.7) and noticing the initial-boundary value condi-

Uij0 = φ(xi , yj ),

Uijk

(i, j) ∈ ω, 0 ⩽ k ⩽ n − 1.

= 0,

(i, j) ∈ ω, (i, j) ∈ 𝜕ω, 0 ⩽ k ⩽ n,

(9.9) (9.10)

9.2 Two-level nonlinear difference scheme

� 291

a difference scheme for solving (9.1)–(9.3) reads k+ 21 k+ 21 k+ 21 󵄨󵄨 k+ 21 󵄨󵄨2 k+ 21 { u u δ u − (ν + iα)Δ u + (κ + iβ) − γu = 0, 󵄨 󵄨 { t h 󵄨 󵄨 ij ij ij ij ij { { { { { (i, j) ∈ ω, 0 ⩽ k ⩽ n − 1, { { uij0 = φ(xi , yj ), (i, j) ∈ ω, { { { { { k { uij = 0, (i, j) ∈ 𝜕ω, 0 ⩽ k ⩽ n.

(9.11) (9.12) (9.13)

9.2.2 Existence of the difference solution The difference scheme (9.11)–(9.13) is two-level nonlinear. We call uk = {uijk | (i, j) ∈ ω} the numerical solution at the k-th time level. From (9.12)–(9.13), u0 at the 0-th time level is already known. Now suppose uk at the k-th time level has been obtained. Then (9.11) 1 and (9.13) can be viewed as a nonlinear system in the average value uk+ 2 = 21 (uk + uk+1 ): 2 k+ 21 k+ 21 k+ 21 󵄨󵄨 k+ 21 󵄨󵄨2 k+ 21 k { { τ (uij − uij ) − (ν + iα)Δh uij + (κ + iβ)󵄨󵄨uij 󵄨󵄨 uij − γuij = 0, { { k+ 21 { uij = 0, k+ 21

Once {uij

(i, j) ∈ ω,

(9.14)

(i, j) ∈ 𝜕ω. (9.15)

| (i, j) ∈ ω}̄ has been obtained, then we have k+ 21

uijk+1 = 2uij

− uijk ,

(i, j) ∈ ω.

Denote k+ 21

wij = uij

,

(i, j) ∈ ω.

Then rewrite (9.14)–(9.15) as 2 { (wij − uijk ) − (ν + iα)Δh wij + (κ + iβ)|wij |2 wij − γwij = 0, τ { { wij = 0,

(i, j) ∈ ω,

(9.16)

(i, j) ∈ 𝜕ω.

(9.17)

Define the operator Π : 𝒱̊h → 𝒱̊h by 2 { (wij − uijk ) − (ν + iα)Δh wij + (κ + iβ)|wij |2 wij − γwij , (i, j) ∈ ω, Π(w)ij = { τ (i, j) ∈ 𝜕ω. { 0, Then taking the inner product of Π(w) with w gives

292 � 9 Difference methods for the Ginzburg–Landau equation 2 (‖w‖2 − (uk , w)) − (ν + iα)(Δh w, w) + (κ + iβ)‖w‖44 − γ‖w‖2 τ 2 = (‖w‖2 − (uk , w)) + (ν + iα)|w|21 + (κ + iβ)‖w‖44 − γ‖w‖2 . τ

(Π(w), w) =

Hence, we have 2 (‖w‖2 − Re(uk , w)) + ν|w|21 + κ‖w‖44 − γ‖w‖2 τ 2 󵄩 󵄩 ⩾ (‖w‖2 − 󵄩󵄩󵄩uk 󵄩󵄩󵄩 ⋅ ‖w‖) − γ‖w‖2 τ 1 2 󵄩 󵄩 = [(1 − γτ)‖w‖ − 󵄩󵄩󵄩uk 󵄩󵄩󵄩] ⋅ ‖w‖. τ 2

Re(Π(w), w) =

When 21 γτ < 1 and ‖w‖ =

‖uk ‖ , 1− 21 γτ

it follows: Re(Π(w), w) ⩾ 0.

With the help of the Browder theorem (Theorem 2.4), there is a w∗ satisfying ‖w∗ ‖ ⩽ ‖uk ‖ such that 1 1− 2 γτ

Π(w∗ ) = 0. Therefore, there is a solution to (9.16)–(9.17). Subsequently, we have the following existence theorem. Theorem 9.2. When 21 γτ < 1, the difference scheme (9.11)–(9.13) has a solution. 9.2.3 Boundedness of the difference solution Theorem 9.3. Suppose {uijk | (i, j) ∈ ω, 0 ⩽ k ⩽ n} is the solution of the difference scheme

(9.11)–(9.13). Then when γτ ⩽ 32 , we have

3 󵄩󵄩 k 󵄩󵄩 T max{0,γ} 󵄩 󵄩󵄩u0 󵄩󵄩󵄩, 󵄩󵄩u 󵄩󵄩 ⩽ e 2 󵄩 󵄩

1 ⩽ k ⩽ n. 1

Proof. Taking the inner product on both the right- and left-hand sides of (9.11) with uk+ 2 , we have 1 1 1 1 1 1 2 1 1 1 󵄨 󵄨 (δt uk+ 2 , uk+ 2 ) − (ν + iα)(Δh uk+ 2 , uk+ 2 ) + (κ + iβ)(󵄨󵄨󵄨uk+ 2 󵄨󵄨󵄨 uk+ 2 , uk+ 2 ) − γ(uk+ 2 , uk+ 2 ) = 0. (9.18) Noting the fact that 1 1 1 2 󵄨 󵄨 −(Δh uk+ 2 , uk+ 2 ) = 󵄨󵄨󵄨uk+ 2 󵄨󵄨󵄨1 ,

9.2 Two-level nonlinear difference scheme

� 293

and taking the real parts on both the right- and left-hand sides of (9.18), we have 1 2 1 4 1 2 1 󵄩󵄩 k+1 󵄩󵄩2 󵄩󵄩 k 󵄩󵄩2 󵄨 󵄩 󵄩 󵄨 󵄩 󵄩 (󵄩󵄩u 󵄩󵄩 − 󵄩󵄩u 󵄩󵄩 ) + ν󵄨󵄨󵄨uk+ 2 󵄨󵄨󵄨1 + κ󵄩󵄩󵄩uk+ 2 󵄩󵄩󵄩4 − γ󵄩󵄩󵄩uk+ 2 󵄩󵄩󵄩 = 0, 2τ

0 ⩽ k ⩽ n − 1.

(9.19)

When γ ⩽ 0, it follows: 1 󵄩󵄩 k+1 󵄩󵄩2 󵄩󵄩 k 󵄩󵄩2 (󵄩u 󵄩󵄩 − 󵄩󵄩u 󵄩󵄩 ) ⩽ 0, 2τ 󵄩

0 ⩽ k ⩽ n − 1,

which implies 󵄩󵄩 k 󵄩󵄩 󵄩󵄩 0 󵄩󵄩 󵄩󵄩u 󵄩󵄩 ⩽ 󵄩󵄩u 󵄩󵄩,

1 ⩽ k ⩽ n.

When γ > 0, it follows from (9.19) that 2

1 2 1 󵄩󵄩 k+1 󵄩󵄩2 󵄩󵄩 k 󵄩󵄩2 ‖uk ‖ + ‖uk+1 ‖ 󵄩 󵄩 (󵄩󵄩u 󵄩󵄩 − 󵄩󵄩u 󵄩󵄩 ) ⩽ γ ⋅ 󵄩󵄩󵄩uk+ 2 󵄩󵄩󵄩 ⩽ γ( ), 2τ 2

0 ⩽ k ⩽ n − 1.

Dividing both the right- and left-hand sides of the above inequality by 21 (‖uk ‖ + ‖uk+1 ‖), it follows: γ 󵄩 󵄩 󵄩 1 󵄩󵄩 k+1 󵄩󵄩 󵄩󵄩 k 󵄩󵄩 󵄩 (󵄩u 󵄩󵄩 − 󵄩󵄩u 󵄩󵄩) ⩽ (󵄩󵄩󵄩uk 󵄩󵄩󵄩 + 󵄩󵄩󵄩uk+1 󵄩󵄩󵄩), τ 󵄩 2 When τ ⩽

2 , 3γ

0 ⩽ k ⩽ n − 1.

(9.20)

it follows from (9.20) that

γ 3γτ 󵄩󵄩 k 󵄩󵄩 󵄩󵄩 k+1 󵄩󵄩 1 + 2 τ 󵄩󵄩 k 󵄩󵄩 )󵄩u 󵄩, 󵄩u 󵄩 ⩽ (1 + 󵄩󵄩u 󵄩󵄩 ⩽ 2 󵄩 󵄩 1 − γ2 τ 󵄩 󵄩

0 ⩽ k ⩽ n − 1,

which by recursion deduces to 3γ 3γ 󵄩󵄩 k 󵄩󵄩 kτ 󵄩 0 󵄩 T 󵄩 0󵄩 󵄩󵄩u 󵄩󵄩 ⩽ e 2 󵄩󵄩󵄩u 󵄩󵄩󵄩 ⩽ e 2 󵄩󵄩󵄩u 󵄩󵄩󵄩,

1 ⩽ k ⩽ n.

Remark 9.1. According to the above theorem, there is a constant c2 such that 󵄩󵄩 k 󵄩󵄩 󵄩󵄩u 󵄩󵄩 ⩽ c2 ,

0 ⩽ k ⩽ n.

(9.21)

Remark 9.2. Replacing k by l in (9.19), then summing over l from 0 to k, and using (9.21), it follows: k

k

l=0

l=0

󵄩󵄩 k+1 󵄩󵄩2 󵄨 l+ 1 󵄨2 󵄩 l+ 1 󵄩4 2 󵄩󵄩u 󵄩󵄩 + 2τν ∑󵄨󵄨󵄨u 2 󵄨󵄨󵄨1 + 2κτ ∑󵄩󵄩󵄩u 2 󵄩󵄩󵄩4 ⩽ (1 + 2 max{0, γ}T)c2 ,

0 ⩽ k ⩽ n − 1.

294 � 9 Difference methods for the Ginzburg–Landau equation 9.2.4 Convergence of the difference solution Lemma 9.5 ([54]). Let u ∈ 𝒱h . Then we have ‖u‖44 ⩽ [4|u|21 + (

1 1 + 2 )‖u‖2 ]‖u‖2 . 2 L1 L2

Proof. When 0 ⩽ s ⩽ m ⩽ m1 , we have m−1

|umj |2 − |usj |2 = ∑ (|ui+1,j |2 − |uij |2 ) i=s

m−1

= ∑ (|ui+1,j | − |uij |)(|ui+1,j | + |uij |) i=s

m−1󵄨󵄨 u 󵄨 󵄨 i+1,j − uij 󵄨󵄨󵄨 ⩽ h1 ∑ 󵄨󵄨󵄨 󵄨󵄨(|ui+1,j | + |uij |) 󵄨 󵄨󵄨 h1 i=s 󵄨 1 2

m−1

m−1

2

⩽ (h1 ∑ |δx ui+ 1 ,j |2 ) [h1 ∑ (|ui+1,j | + |uij |) ] 2

i=s

m1 −1

2

i=s

1 2

m1

2

1 2

1 2

⩽ 2(h1 ∑ |δx ui+ 1 ,j | ) (h1 ∑ ωi |uij | ) . 2

i=0

i=0

It is also true for 0 ⩽ m ⩽ s ⩽ m1 . Therefore, 2

m1 −1

1 2

2

m1

1 2

|umj | ⩽ 2(h1 ∑ |δx ui+ 1 ,j | ) (h1 ∑ ωi |uij | ) + |usj |2 , 2

i=0

2

i=0

0 ⩽ m, s ⩽ m1 .

Multiplying both the right- and left-hand sides of the above inequality by ωs h1 and summing over s from 0 to m1 , it follows: 2

m1 −1

2

1 2

m1

1 2

m1

L1 |umj | ⩽ 2L1 (h1 ∑ |δx ui+ 1 ,j | ) (h1 ∑ ωi |uij | ) + h1 ∑ ωs |usj |2 , 2

i=0

2

s=0

i=0

0 ⩽ m ⩽ m1 .

Dividing the above inequality by L1 on both the right- and left-hand sides and taking the maximum value for m from 0 to m1 , we have 1 2

m1 −1

m1

1 2

max |uij |2 ⩽ 2(h1 ∑ |δx ui+ 1 ,j |2 ) (h1 ∑ ωi |uij |2 ) +

0⩽i⩽m1

i=0

2

i=0

m

1 1 h1 ∑ ωs |usj |2 . L1 s=0

Then multiplying the above inequality by h2 ω̂ j and summing over j from 0 to m2 , we have

9.2 Two-level nonlinear difference scheme

� 295

m2

h2 ∑ ω̂ j max |uij |2 0⩽i⩽m1

j=0

m2

m1 −1

2

1 2

m1

2

1 2

⩽ 2h2 ∑ ω̂ j (h1 ∑ |δx ui+ 1 ,j | ) (h1 ∑ ωi |uij | ) + j=0

2

i=0

m2

m1 −1

i=0

2

1 2

m2

m1

j=0

i=0

1 ‖u‖2 L1 2

1 2

⩽ 2(h2 ∑ ω̂ j h1 ∑ |δx ui+ 1 ,j | ) (h2 ∑ ω̂ j h1 ∑ ωi |uij | ) + j=0

2

i=0

1 = 2‖δx u‖ ⋅ ‖u‖ + ‖u‖2 . L1

1 ‖u‖2 L1 (9.22)

Analogously, we have m1

h1 ∑ ωi max |uij |2 ⩽ 2‖δy u‖ ⋅ ‖u‖ + 0⩽j⩽m2

i=0

1 ‖u‖2 . L2

(9.23)

Now we estimate ‖u‖44 . m1 m2

m1

m2

i=0 j=0

i=0

j=0

‖u‖44 = h1 h2 ∑ ∑ ωi ω̂ j |uij |4 = h1 ∑ ωi [h2 ∑ ω̂ j |uij |4 ] m1

m2

⩽ h1 ∑ ωi [( max |uij |2 )(h2 ∑ ω̂ j |uij |2 )] 0⩽j⩽m2

i=0

m1

j=0 m2

⩽ (h1 ∑ ωi max |uij |2 )(h2 ∑ ω̂ j max |uij |2 ). 0⩽j⩽m2

i=0

j=0

0⩽i⩽m1

Combining (9.22) with (9.23), it follows: ‖u‖44 ⩽ (2‖δy u‖ ⋅ ‖u‖ + = (2‖δy u‖ +

1 1 ‖u‖)(2‖δx u‖ + ‖u‖)‖u‖2 L2 L1

⩽ (4‖δy u‖2 + = [4|u|21 + (

1 1 ‖u‖2 )(2‖δx u‖ ⋅ ‖u‖ + ‖u‖2 ) L2 L1

1 1 ‖u‖2 + 4‖δx u‖2 + 2 ‖u‖2 )‖u‖2 L22 L1

1 1 + )‖u‖2 ]‖u‖2 . L21 L22

Theorem 9.4. Suppose {Uijk | (i, j) ∈ ω, 0 ⩽ k ⩽ n} is the solution of the problem (9.1)–(9.3)

and {uijk | (i, j) ∈ ω, 0 ⩽ k ⩽ n} is the solution of the difference scheme (9.11)–(9.13). Denote eijk = Uijk − uijk , Then there is a constant c3 satisfying

(i, j) ∈ ω, 0 ⩽ k ⩽ n.

296 � 9 Difference methods for the Ginzburg–Landau equation 󵄩󵄩 k 󵄩󵄩 2 2 2 󵄩󵄩e 󵄩󵄩 ⩽ c3 (τ + h1 + h2 ),

0 ⩽ k ⩽ n.

Proof. Subtracting (9.11)–(9.13) from (9.7), (9.9)–(9.10), respectively, we have the system of error equations: k+ 21 k+ 21 k+ 21 k+ 21 󵄨󵄨 k+ 21 󵄨󵄨2 k+ 21 󵄨󵄨 k+ 21 󵄨󵄨2 k+ 21 { U u U u − (ν + iα)Δ e + (κ + iβ)( − ) − γe = P , δ e 󵄨 󵄨 󵄨 󵄨 { h t 󵄨 ij 󵄨 ij 󵄨 ij 󵄨 ij ij ij ij ij { { { { { (i, j) ∈ ω, 0 ⩽ k ⩽ n − 1, (9.24) { 0 { eij = 0, (i, j) ∈ ω, (9.25) { { { { { k (9.26) { eij = 0, (i, j) ∈ 𝜕ω, 0 ⩽ k ⩽ n. 1

Taking the inner product of (9.24) on both the right- and left-hand sides with ek+ 2 and taking the real parts, we have 1 󵄩󵄩 k+1 󵄩󵄩2 󵄩󵄩 k 󵄩󵄩2 󵄩 1 󵄩2 󵄨 1 󵄨2 (󵄩󵄩e 󵄩󵄩 − 󵄩󵄩e 󵄩󵄩 ) + ν󵄨󵄨󵄨ek+ 2 󵄨󵄨󵄨1 − γ󵄩󵄩󵄩ek+ 2 󵄩󵄩󵄩 2τ 1 2 1 1 2 1 1 1 1 󵄨 󵄨 󵄨 󵄨 = − Re{(κ + iβ)(󵄨󵄨󵄨U k+ 2 󵄨󵄨󵄨 U k+ 2 − 󵄨󵄨󵄨uk+ 2 󵄨󵄨󵄨 uk+ 2 , ek+ 2 )} + Re{(Pk+ 2 , ek+ 2 )}, 0 ⩽ k ⩽ n − 1.

(9.27)

Noting 󵄨󵄨 k+ 21 󵄨󵄨2 k+ 21 󵄨󵄨 k+ 21 󵄨󵄨2 k+ 21 󵄨󵄨Uij 󵄨󵄨 Uij − 󵄨󵄨uij 󵄨󵄨 uij k+ 1 󵄨2 k+ 1 k+ 1 󵄨 k+ 1 󵄨2 k+ 1 󵄨 k+ 1 = 󵄨󵄨󵄨Uij 2 󵄨󵄨󵄨 Uij 2 − 󵄨󵄨󵄨Uij 2 − eij 2 󵄨󵄨󵄨 (Uij 2 − eij 2 ) k+ 1 k+ 1 2 k+ 1 2 k+ 1 k+ 1 󵄨 k+ 1 󵄨2 󵄨 k+ 1 󵄨2 k+ 1 󵄨 k+ 1 󵄨2 k+ 1 = 2󵄨󵄨󵄨Uij 2 󵄨󵄨󵄨 eij 2 + (Uij 2 ) eij 2 − U ij 2 (eij 2 ) − 2Uij 2 󵄨󵄨󵄨eij 2 󵄨󵄨󵄨 + 󵄨󵄨󵄨eij 2 󵄨󵄨󵄨 eij 2 ,

and with the help of Lemma 9.5, we have 1 1 1 1 2 1 2 󵄨 󵄨 󵄨 󵄨 − Re{(κ + iβ)(󵄨󵄨󵄨U k+ 2 󵄨󵄨󵄨 U k+ 2 − 󵄨󵄨󵄨uk+ 2 󵄨󵄨󵄨 uk+ 2 , ek+ 2 )}

m1 m2

󵄨 k+ 1 󵄨2 󵄨 k+ 1 󵄨3 ⩽ √κ2 + β2 h1 h2 ∑ ∑ ωi ω̂ j (3c02 󵄨󵄨󵄨eij 2 󵄨󵄨󵄨 + 3c0 󵄨󵄨󵄨eij 2 󵄨󵄨󵄨 ) i=0 j=0

⩽ √κ 2 +

󵄩 1 󵄩2 β2 (3c02 󵄩󵄩󵄩ek+ 2 󵄩󵄩󵄩

󵄩 1 󵄩2 󵄩 1 󵄩 + 3c0 󵄩󵄩󵄩ek+ 2 󵄩󵄩󵄩4 ⋅ 󵄩󵄩󵄩ek+ 2 󵄩󵄩󵄩)

1/2

1 1 󵄩 1 󵄩2 󵄨 1 󵄨2 󵄩 1 󵄩2 󵄩 1 󵄩2 ⩽ √κ2 + β2 {3c02 󵄩󵄩󵄩ek+ 2 󵄩󵄩󵄩 + 3c0 [4󵄨󵄨󵄨ek+ 2 󵄨󵄨󵄨1 + ( 2 + 2 )󵄩󵄩󵄩ek+ 2 󵄩󵄩󵄩 ] 󵄩󵄩󵄩ek+ 2 󵄩󵄩󵄩 } L1 L2

9c2 󵄩 1 󵄩4 1 1 󵄩 1 󵄩2 󵄩 1 󵄩2 󵄨 1 󵄨2 ⩽ √κ2 + β2 [3c02 󵄩󵄩󵄩ek+ 2 󵄩󵄩󵄩 + ε(4󵄨󵄨󵄨ek+ 2 󵄨󵄨󵄨1 + ( 2 + 2 )󵄩󵄩󵄩ek+ 2 󵄩󵄩󵄩 ) + 0 󵄩󵄩󵄩ek+ 2 󵄩󵄩󵄩 ]. 4ε L1 L2 1

1

1

Taking 4√κ2 + β2 ε = ν and noticing ‖ek+ 2 ‖ = ‖U k+ 2 − uk+ 2 ‖ ⩽ c0 √L1 L2 + c2 , we have 1 2 1 1 2 1 1 󵄨 󵄨 󵄨 󵄨 − Re{(κ + iβ)(󵄨󵄨󵄨U k+ 2 󵄨󵄨󵄨 U k+ 2 − 󵄨󵄨󵄨uk+ 2 󵄨󵄨󵄨 uk+ 2 , ek+ 2 )}

9.2 Two-level nonlinear difference scheme

⩽ √κ2 + β2 [3c02 +

+

ν 4√κ2

+

β2

(

� 297

1 1 + ) L21 L22

2 2 1 2 9c02 4√κ + β 2 󵄩 󵄨 1 󵄨2 󵄩 ⋅ (c0 √L1 L2 + c2 ) ]󵄩󵄩󵄩ek+ 2 󵄩󵄩󵄩 + ν󵄨󵄨󵄨ek+ 2 󵄨󵄨󵄨1 4 ν

1 2 9c2 (κ2 + β2 ) ν 1 1 2 󵄩 󵄨 1 󵄨2 󵄩 = [3c02 √κ2 + β2 + ( 2 + 2 ) + 0 (c0 √L1 L2 + c2 ) ]󵄩󵄩󵄩ek+ 2 󵄩󵄩󵄩 + ν󵄨󵄨󵄨ek+ 2 󵄨󵄨󵄨1 . 4 L1 L2 ν

Substituting the above inequality into (9.27), we have 1 󵄩󵄩 k+1 󵄩󵄩2 󵄩󵄩 k 󵄩󵄩2 (󵄩e 󵄩󵄩 − 󵄩󵄩e 󵄩󵄩 ) 2τ 󵄩 9c2 (κ2 + β2 ) 1 ν 1 2 󵄩 1 󵄩2 ⩽ [3c02 √κ2 + β2 + ( 2 + 2 ) + 0 (c0 √L1 L2 + c2 ) + max{0, γ}]󵄩󵄩󵄩ek+ 2 󵄩󵄩󵄩 4 L1 L2 ν 󵄩󵄩 k+ 21 󵄩󵄩 󵄩󵄩 k+ 21 󵄩󵄩 + 󵄩󵄩e 󵄩󵄩 ⋅ 󵄩󵄩P 󵄩󵄩, 0 ⩽ k ⩽ n − 1. 1

Noticing ‖ek+ 2 ‖ ⩽

‖ek+1 ‖+‖ek ‖ 2

and (9.8), we have

1 󵄩󵄩 k+1 󵄩󵄩 󵄩󵄩 k 󵄩󵄩 (󵄩e 󵄩󵄩 − 󵄩󵄩e 󵄩󵄩) τ 󵄩 9c2 (κ2 + β2 ) 1 ν 1 ‖ek+1 ‖ + ‖ek ‖ 2 (c0 √L1 L2 + c2 ) + max{0, γ}] ⩽ [3c02 √κ2 + β2 + ( 2 + 2 ) + 0 4 L1 L2 ν 2 + c1 √L1 L2 (τ 2 + h12 + h22 ),

0 ⩽ k ⩽ n − 1.

Denote 9c2 (κ2 + β2 ) ν 1 1 1 2 c4 = [3c02 √κ2 + β2 + ( 2 + 2 ) + 0 (c0 √L1 L2 + c2 ) + max{0, γ}]. 2 4 L1 L2 ν We know 󵄩 󵄩 󵄩 󵄩 (1 − c4 τ)󵄩󵄩󵄩ek+1 󵄩󵄩󵄩 ⩽ (1 + c4 τ)󵄩󵄩󵄩ek 󵄩󵄩󵄩 + c1 √L1 L2 τ(τ 2 + h12 + h22 ),

0 ⩽ k ⩽ n − 1.

When c4 τ ⩽ 31 , it follows: 󵄩󵄩 k+1 󵄩󵄩 󵄩 k󵄩 3 2 2 2 󵄩󵄩e 󵄩󵄩 ⩽ (1 + 3c4 τ)󵄩󵄩󵄩e 󵄩󵄩󵄩 + c1 √L1 L2 τ(τ + h1 + h2 ), 2

0 ⩽ k ⩽ n − 1.

By the Gronwall inequality (Theorem 1.2(a)), we have c1 √L1 L2 2 󵄩󵄩 k+1 󵄩󵄩 3c kτ 󵄩 0 󵄩 (τ + h12 + h22 )] 󵄩󵄩e 󵄩󵄩 ⩽ e 4 [󵄩󵄩󵄩e 󵄩󵄩󵄩 + 2c4 ⩽ e3c4 T ⋅

c1 √L1 L2 2 (τ + h12 + h22 ), 2c4

0 ⩽ k ⩽ n − 1.

298 � 9 Difference methods for the Ginzburg–Landau equation

9.3 Three-level linearized difference scheme 9.3.1 Derivation of the difference scheme Considering equation (9.1) at the node point (xi , yj , tk ), we have ut (xi , yj , tk ) − (ν + iα)Δu(xi , yj , tk ) + (κ + iβ)(|u|2 u)(xi , yj , tk ) − γu(xi , yj , tk ) = 0, (i, j) ∈ ω, 1 ⩽ k ⩽ n.

(9.28)

In view of (9.28) with k = 1, we have 󵄨 󵄨2 ∇τ Uij1 − (ν + iα)Δh Uij1 + (κ + iβ)󵄨󵄨󵄨Uij0 󵄨󵄨󵄨 Uij1 − γUij1 = R1ij ,

(i, j) ∈ ω,

(9.29)

where there is a constant c5 satisfying 󵄨󵄨 1 󵄨󵄨 2 2 󵄨󵄨Rij 󵄨󵄨 ⩽ c5 (τ + h1 + h2 ),

(i, j) ∈ ω.

(9.30)

With the application of the numerical differential formula, we have 1 (3Uijk − 4Uijk−1 + Uijk−2 ) + O(τ 2 ) 2τ = ∇2τ Uijk + O(τ 2 ),

ut (xi , yj , tk ) =

(9.31) 2

u(xi , yj , tk ) = 2u(xi , yj , tk−1 ) − u(xi , yj , tk−2 ) + O(τ ) = 2Uijk−1 − Uijk−2 + O(τ 2 ).

(9.32)

Substituting the above two equalities into (9.28), we have ∇2τ Uijk − (ν + iα)Δh Uijk + (κ + iβ)|2Uijk−1 − Uijk−2 |2 Uijk − γUijk = Rkij , (i, j) ∈ ω, 2 ⩽ k ⩽ n,

(9.33)

where there is a constant c6 such that 󵄨󵄨 k 󵄨󵄨 2 2 2 󵄨󵄨Rij 󵄨󵄨 ⩽ c6 (τ + h1 + h2 ),

(i, j) ∈ ω, 2 ⩽ k ⩽ n.

(9.34)

Omitting the small terms in (9.29) and (9.33) and noting the initial-boundary value conditions {

Uij0 = φ(xi , yj ), Uijk

= 0,

(i, j) ∈ ω,

(9.35)

(i, j) ∈ 𝜕ω, 0 ⩽ k ⩽ n,

(9.36)

9.3 Three-level linearized difference scheme

� 299

a linearized difference scheme for solving (9.1)–(9.3) reads { { { { { { { { { { { { { { { { { { { { {

󵄨 󵄨2 ∇τ uij1 − (ν + iα)Δh uij1 + (κ + iβ)󵄨󵄨󵄨uij0 󵄨󵄨󵄨 uij1 − γuij1 = 0, (i, j) ∈ ω, 󵄨 󵄨2 ∇2τ uijk − (ν + iα)Δh uijk + (κ + iβ)󵄨󵄨󵄨2uijk−1 − uijk−2 󵄨󵄨󵄨 uijk − γuijk = 0, (i, j) ∈ ω, 2 ⩽ k ⩽ n, uij0

= φ(xi , yj ),

uijk = 0,

(9.37) (9.38)

(i, j) ∈ ω,

(9.39)

(i, j) ∈ 𝜕ω, 0 ⩽ k ⩽ n.

(9.40)

9.3.2 Existence of the difference solution Theorem 9.5. When max{0, γ}τ ⩽ 1, the difference scheme (9.37)–(9.40) has a unique solution. Proof. From (9.39)–(9.40), we know that u0 has been given. Combining (9.37) with (9.40), we obtain the system of linear equations in u1 . Consider its homogeneous one: 1 1 󵄨 0 󵄨2 1 1 1 { uij − (ν + iα)Δh uij + (κ + iβ)󵄨󵄨󵄨uij 󵄨󵄨󵄨 uij − γuij = 0, τ { 1 { uij = 0,

(i, j) ∈ ω,

(9.41)

(i, j) ∈ 𝜕ω.

(9.42)

Taking the inner product on both the right- and left-hand sides of (9.41) with u1 , we obtain 1 󵄩󵄩 1 󵄩󵄩2 󵄨 0 󵄨2 1 1 󵄩 1 󵄩2 1 1 󵄩u 󵄩 − (ν + iα)(Δh u , u ) + (κ + iβ)(󵄨󵄨󵄨u 󵄨󵄨󵄨 u , u ) − γ󵄩󵄩󵄩u 󵄩󵄩󵄩 = 0. τ󵄩 󵄩 Noticing −(Δh u1 , u1 ) = |u1 |21 and taking the real parts of the above equality, we have 1 󵄩󵄩 1 󵄩󵄩2 󵄨 1 󵄨2 󵄨 0 󵄨2 1 1 󵄩 1 󵄩2 󵄩u 󵄩 + ν󵄨󵄨󵄨u 󵄨󵄨󵄨1 + κ(󵄨󵄨󵄨u 󵄨󵄨󵄨 u , u ) − γ󵄩󵄩󵄩u 󵄩󵄩󵄩 = 0. τ󵄩 󵄩 When γ ⩽ 0, or γ > 0 with τ < 1/γ, it holds that ‖u1 ‖ = 0. Hence, the difference scheme is uniquely solvable with respect to u1 . Suppose that uk−2 and uk−1 have been uniquely determined. Then we have the system of linear equations (9.38) and (9.40) in uk . Consider its homogeneous one: 3 k 󵄨 󵄨2 u − (ν + iα)Δh uijk + (κ + iβ)󵄨󵄨󵄨2uijk−1 − uijk−2 󵄨󵄨󵄨 uijk − γuijk = 0, { 2τ ij { k { uij = 0,

(i, j) ∈ ω,

(9.43)

(i, j) ∈ 𝜕ω. (9.44)

Taking the inner product on both the right- and left-hand sides of (9.43) with uk , we have

300 � 9 Difference methods for the Ginzburg–Landau equation 3 󵄩󵄩 k 󵄩󵄩2 󵄩 k 󵄩2 󵄨 k−1 󵄨 k 󵄨2 k−2 󵄨2 k k 󵄩󵄩u 󵄩󵄩 + (ν + iα)󵄨󵄨󵄨u 󵄨󵄨󵄨1 + (κ + iβ)(󵄨󵄨󵄨2u − u 󵄨󵄨󵄨 u , u ) − γ󵄩󵄩󵄩u 󵄩󵄩󵄩 = 0. 2τ Taking the real parts on both the right- and left-hand sides, we have 3 󵄩󵄩 k 󵄩󵄩2 󵄩 k 󵄩2 󵄨 k−1 󵄨 k 󵄨2 k−2 󵄨2 k k 󵄩u 󵄩 + ν󵄨󵄨󵄨u 󵄨󵄨󵄨1 + κ(󵄨󵄨󵄨2u − u 󵄨󵄨󵄨 u , u ) − γ󵄩󵄩󵄩u 󵄩󵄩󵄩 = 0. 2τ 󵄩 󵄩 3 , 2γ

When γ ⩽ 0, or γ > 0 with τ
0, with the help of (9.46), it follows: 󵄩 󵄩2 󵄩 󵄩2 (1 − 2γτ)󵄩󵄩󵄩u1 󵄩󵄩󵄩 ⩽ 󵄩󵄩󵄩u0 󵄩󵄩󵄩 .

(9.46)

9.3 Three-level linearized difference scheme

� 301

When 2γτ ⩽ 41 , it holds that 1 󵄩󵄩 0 󵄩󵄩2 4 󵄩󵄩 0 󵄩󵄩2 󵄩󵄩 1 󵄩󵄩2 󵄩󵄩u 󵄩󵄩 ⩽ 󵄩u 󵄩 ⩽ 󵄩󵄩u 󵄩󵄩 . 1 − 2γτ 󵄩 󵄩 3

(9.47)

(II) Taking the inner product on both the right- and left-hand sides of (9.38) with uk , we have 󵄨 󵄨2 󵄩 󵄩2 (∇2τ uk , uk ) − (ν + iα)(Δh uk , uk ) + (κ + iβ)(󵄨󵄨󵄨2uk−1 − uk−2 󵄨󵄨󵄨 uk , uk ) − γ󵄩󵄩󵄩uk 󵄩󵄩󵄩 = 0.

(9.48)

Noticing Re(∇2τ uk , uk ) 1 󵄩 󵄩2 󵄩 󵄩2 󵄩 󵄩2 󵄩 󵄩2 = [(󵄩󵄩󵄩uk 󵄩󵄩󵄩 + 󵄩󵄩󵄩2uk − uk−1 󵄩󵄩󵄩 ) − (󵄩󵄩󵄩uk−1 󵄩󵄩󵄩 + 󵄩󵄩󵄩2uk−1 − uk−2 󵄩󵄩󵄩 ) 4τ 󵄩 󵄩2 + 󵄩󵄩󵄩uk − 2uk−1 + uk−2 󵄩󵄩󵄩 ],

(9.49)

and taking the real parts on both the right- and left-hand sides of (9.48), we have 1 󵄩󵄩 k 󵄩󵄩2 󵄩󵄩 k 󵄩2 󵄩 󵄩2 󵄩 󵄩2 [(󵄩󵄩u 󵄩󵄩 + 󵄩󵄩2u − uk−1 󵄩󵄩󵄩 ) − (󵄩󵄩󵄩uk−1 󵄩󵄩󵄩 + 󵄩󵄩󵄩2uk−1 − uk−2 󵄩󵄩󵄩 ) 4τ 󵄩 󵄩2 󵄨 󵄨2 󵄨 󵄨2 󵄩 󵄩2 + 󵄩󵄩󵄩uk − 2uk−1 + uk−2 󵄩󵄩󵄩 ] + ν󵄨󵄨󵄨uk 󵄨󵄨󵄨1 + κ(󵄨󵄨󵄨2uk−1 − uk−2 󵄨󵄨󵄨 uk , uk ) = γ󵄩󵄩󵄩uk 󵄩󵄩󵄩 . With the help of the above equality, we have 1 󵄩󵄩 k 󵄩󵄩2 󵄩󵄩 k 󵄩2 󵄩 󵄩2 󵄩 󵄩2 󵄩 󵄩2 [(󵄩u 󵄩 + 󵄩2u − uk−1 󵄩󵄩󵄩 ) − (󵄩󵄩󵄩uk−1 󵄩󵄩󵄩 + 󵄩󵄩󵄩2uk−1 − uk−2 󵄩󵄩󵄩 )] ⩽ γ󵄩󵄩󵄩uk 󵄩󵄩󵄩 , 4τ 󵄩 󵄩 󵄩

2 ⩽ k ⩽ n. (9.50)

When γ ⩽ 0, one easily has 󵄩󵄩 k 󵄩󵄩2 󵄩󵄩 k 󵄩 k−1 󵄩2 󵄩 k−1 k−1 󵄩2 k−2 󵄩2 󵄩󵄩u 󵄩󵄩 + 󵄩󵄩2u − u 󵄩󵄩󵄩 ⩽ 󵄩󵄩󵄩u 󵄩󵄩󵄩 + 󵄩󵄩󵄩2u − u 󵄩󵄩󵄩 ,

2 ⩽ k ⩽ n.

(9.51)

When γ > 0, it follows from (9.50) that 󵄩 󵄩2 󵄩 󵄩2 󵄩 󵄩2 󵄩 󵄩2 (1 − 4γτ)(󵄩󵄩󵄩uk 󵄩󵄩󵄩 + 󵄩󵄩󵄩2uk − uk−1 󵄩󵄩󵄩 ) ⩽ 󵄩󵄩󵄩uk−1 󵄩󵄩󵄩 + 󵄩󵄩󵄩2uk−1 − uk−2 󵄩󵄩󵄩 ,

2 ⩽ k ⩽ n.

Furthermore, when 4γτ ⩽ 21 , we have 󵄩󵄩 k 󵄩󵄩2 󵄩󵄩 k k−1 󵄩2 󵄩󵄩u 󵄩󵄩 + 󵄩󵄩2u − u 󵄩󵄩󵄩 1 󵄩 󵄩2 󵄩 󵄩2 ⩽ (󵄩󵄩uk−1 󵄩󵄩󵄩 + 󵄩󵄩󵄩2uk−1 − uk−2 󵄩󵄩󵄩 ) 1 − 4γτ 󵄩 󵄩 󵄩2 󵄩 󵄩2 ⩽ (1 + 8γτ)(󵄩󵄩󵄩uk−1 󵄩󵄩󵄩 + 󵄩󵄩󵄩2uk−1 − uk−2 󵄩󵄩󵄩 ),

2 ⩽ k ⩽ n.

(9.52)

302 � 9 Difference methods for the Ginzburg–Landau equation Combining (9.51) with (9.52), we have 󵄩 k−1 󵄩2 󵄩 k−1 󵄩󵄩 k 󵄩󵄩2 󵄩󵄩 k k−2 󵄩2 k−1 󵄩 󵄩󵄩u 󵄩󵄩 + 󵄩󵄩2u − u 󵄩󵄩󵄩 ⩽ (1 + 8 max{0, γ}τ)(󵄩󵄩󵄩u 󵄩󵄩󵄩 + 󵄩󵄩󵄩2u − u 󵄩󵄩󵄩 ),

2 ⩽ k ⩽ n.

By recursion, we have 󵄩󵄩 k 󵄩󵄩2 󵄩󵄩 k 󵄩2 󵄩 󵄩2 8 max{0,γ}kτ 󵄩 k−1 󵄩2 (󵄩󵄩󵄩u1 󵄩󵄩󵄩 + 󵄩󵄩󵄩2u1 − u0 󵄩󵄩󵄩 ), 󵄩󵄩u 󵄩󵄩 + 󵄩󵄩2u − u 󵄩󵄩󵄩 ⩽ e

2 ⩽ k ⩽ n.

In combination with (9.47), it follows: 󵄩2 󵄩 󵄩2 󵄩 󵄩2 󵄩󵄩 k 󵄩󵄩2 8 max{0,γ}T 󵄩 [󵄩󵄩󵄩u1 󵄩󵄩󵄩 + 4󵄩󵄩󵄩u1 󵄩󵄩󵄩 − 4 Re(u1 , u0 ) + 󵄩󵄩󵄩u0 󵄩󵄩󵄩 ] 󵄩󵄩u 󵄩󵄩 ⩽ e 23 + 8√3 8 max{0,γ}T 󵄩󵄩 0 󵄩󵄩2 ⩽ e 󵄩󵄩u 󵄩󵄩 , 1 ⩽ k ⩽ n. 3 9.3.4 Convergence of the difference solution Theorem 9.7. Suppose {Uijk | (i, j) ∈ ω, 0 ⩽ k ⩽ n} is the solution of the problem (9.1)–(9.3)

and {uijk | (i, j) ∈ ω, 0 ⩽ k ⩽ n} is the solution of the difference scheme (9.37)–(9.40). Denote eijk = Uijk − uijk ,

(i, j) ∈ ω, 0 ⩽ k ⩽ n.

Then when τ is appropriately small, there is a constant c7 satisfying 󵄩󵄩 k 󵄩󵄩 2 2 2 󵄩󵄩e 󵄩󵄩 ⩽ c7 (τ + h1 + h2 ),

0 ⩽ k ⩽ n.

Proof. Subtracting (9.37)–(9.40) from (9.29), (9.33), (9.35)–(9.36), respectively, we have { { { { { { { { { { { { { { { { { { { { { { { { { { {

󵄨 󵄨2 ∇τ eij1 − (ν + iα)Δh eij1 + (κ + iβ)󵄨󵄨󵄨Uij0 󵄨󵄨󵄨 eij1 − γeij1 = R1ij , ∇2τ eijk

eij0 eijk

− (ν +

(i, j) ∈ ω,

(9.53)

iα)Δh eijk

󵄨 󵄨2 󵄨 󵄨2 + (κ + iβ)(󵄨󵄨󵄨2Uijk−1 − Uijk−2 󵄨󵄨󵄨 Uijk − 󵄨󵄨󵄨2uijk−1 − uijk−2 󵄨󵄨󵄨 uijk ) − γeijk = Rkij , (i, j) ∈ ω, 2 ⩽ k ⩽ n,

(9.54)

= 0,

(i, j) ∈ ω,

(9.55)

= 0,

(i, j) ∈ 𝜕ω, 0 ⩽ k ⩽ n.

(9.56)

(I) Taking the inner product on both the right- and left-hand sides of (9.53) with e1 and noting (9.55)–(9.56), we have 1 󵄩󵄩 1 󵄩󵄩2 󵄨 1 󵄨2 󵄨 0 󵄨2 1 1 󵄩 1 󵄩2 1 1 󵄩e 󵄩 + (ν + iα)󵄨󵄨󵄨e 󵄨󵄨󵄨1 + (κ + iβ)(󵄨󵄨󵄨U 󵄨󵄨󵄨 e , e ) − γ󵄩󵄩󵄩e 󵄩󵄩󵄩 = (R , e ). τ󵄩 󵄩 Taking the real parts on both the right- and left-hand sides of the equality above, we have

9.3 Three-level linearized difference scheme

� 303

1 󵄩󵄩 1 󵄩󵄩2 󵄩 1 󵄩2 󵄨 0 󵄨2 1 1 󵄨 1 󵄨2 1 1 󵄩󵄩e 󵄩󵄩 + ν󵄨󵄨󵄨e 󵄨󵄨󵄨1 + κ(󵄨󵄨󵄨U 󵄨󵄨󵄨 e , e ) = γ󵄩󵄩󵄩e 󵄩󵄩󵄩 + Re(R , e ). τ

(9.57)

Therefore, we have 1 󵄩󵄩 1 󵄩󵄩2 󵄩 1 󵄩2 󵄩 1 󵄩 󵄩 1 󵄩 󵄩e 󵄩 ⩽ γ󵄩󵄩󵄩e 󵄩󵄩󵄩 + 󵄩󵄩󵄩R 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩e 󵄩󵄩󵄩. τ󵄩 󵄩 Dividing the above inequality on both the right- and left-hand sides by ‖e1 ‖, we have 1 󵄩󵄩 1 󵄩󵄩 󵄩 1󵄩 󵄩 1󵄩 󵄩e 󵄩 ⩽ γ󵄩󵄩󵄩e 󵄩󵄩󵄩 + 󵄩󵄩󵄩R 󵄩󵄩󵄩. τ󵄩 󵄩 When γτ ⩽ 21 , we have τ 󵄩󵄩 1 󵄩󵄩 󵄩 1󵄩 󵄩󵄩 1 󵄩󵄩 󵄩R 󵄩 ⩽ 2τ 󵄩󵄩󵄩R 󵄩󵄩󵄩, 󵄩󵄩e 󵄩󵄩 ⩽ 1 − γτ 󵄩 󵄩 which implies 󵄩󵄩 1 󵄩󵄩 2 2 2 2 2 󵄩󵄩e 󵄩󵄩 ⩽ 2τc5 √L1 L2 (τ + h1 + h2 ) ⩽ 2c5 √L1 L2 (τ + h1 + h2 )

(9.58)

by noticing (9.30). (II) Taking the inner product on both the right- and left-hand sides of (9.54) with ek , it follows: 󵄨 󵄨2 󵄨 󵄨2 (∇2τ ek , ek ) + (ν + iα)󵄨󵄨󵄨ek 󵄨󵄨󵄨1 + (κ + iβ)(󵄨󵄨󵄨2U k−1 − U k−2 󵄨󵄨󵄨 U k 󵄨 󵄨2 󵄩 󵄩2 − 󵄨󵄨󵄨2uk−1 − uk−2 󵄨󵄨󵄨 uk , ek ) − γ󵄩󵄩󵄩ek 󵄩󵄩󵄩 = (Rk , ek ).

(9.59)

Combining 󵄨󵄨 k−1 󵄨2 󵄨 󵄨2 − U k−2 󵄨󵄨󵄨 U k − 󵄨󵄨󵄨2uk−1 − uk−2 󵄨󵄨󵄨 uk 󵄨󵄨2U 󵄨 󵄨2 󵄨 󵄨2 󵄨 󵄨2 = 󵄨󵄨󵄨2uk−1 − uk−2 󵄨󵄨󵄨 (U k − uk ) + (󵄨󵄨󵄨2U k−1 − U k−2 󵄨󵄨󵄨 − 󵄨󵄨󵄨2uk−1 − uk−2 󵄨󵄨󵄨 )U k 󵄨 󵄨2 = 󵄨󵄨󵄨2uk−1 − uk−2 󵄨󵄨󵄨 ek + {(2uk−1 − uk−2 )[(2Ū k−1 − Ū k−2 ) − (2ū k−1 − ū k−2 )] + [(2U k−1 − U k−2 ) − (2uk−1 − uk−2 )](2Ū k−1 − Ū k−2 )}U k 󵄨 󵄨2 = 󵄨󵄨󵄨2uk−1 − uk−2 󵄨󵄨󵄨 ek + [(2uk−1 − uk−2 )(2ēk−1 − ēk−2 ) + (2ek−1 − ek−2 )(2Ū k−1 − Ū k−2 )]U k

with (9.49) and taking the real parts of (9.59) on both the right- and left-hand sides, we have Ak ≡

1 󵄩󵄩 k 󵄩󵄩2 󵄩󵄩 k 󵄩2 󵄩 󵄩2 󵄩 󵄩2 [(󵄩e 󵄩 + 󵄩2e − ek−1 󵄩󵄩󵄩 ) − (󵄩󵄩󵄩ek−1 󵄩󵄩󵄩 + 󵄩󵄩󵄩2ek−1 − ek−2 󵄩󵄩󵄩 ) 4τ 󵄩 󵄩 󵄩 󵄩 󵄩2 󵄨 󵄨2 󵄩 󵄩2 + 󵄩󵄩󵄩ek − 2ek−1 + ek−2 󵄩󵄩󵄩 ] + ν󵄨󵄨󵄨ek 󵄨󵄨󵄨1 − γ󵄩󵄩󵄩ek 󵄩󵄩󵄩

304 � 9 Difference methods for the Ginzburg–Landau equation ⩽ − Re{(κ + iβ)[((2uk−1 − uk−2 )(2ēk−1 − ēk−2 )U k , ek ) 󵄩 󵄩 󵄩 󵄩 + ((2ek−1 − ek−2 )(2Ū k−1 − Ū k−2 )U k , ek )]} + 󵄩󵄩󵄩Rk 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩ek 󵄩󵄩󵄩. Noticing 󵄩󵄩 k 󵄩󵄩 󵄩󵄩U 󵄩󵄩∞ ⩽ c0 , 󵄩󵄩 k 󵄩󵄩 󵄩󵄩u 󵄩󵄩 ⩽

√3(23 + 8√3) 3

󵄩 󵄩 e4 max{0,γ}T 󵄩󵄩󵄩u0 󵄩󵄩󵄩 ≡ c8 ,

we have 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 Ak ⩽ √κ2 + β2 (c0 󵄩󵄩󵄩2uk−1 − uk−2 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩2ek−1 − ek−2 󵄩󵄩󵄩4 󵄩󵄩󵄩ek 󵄩󵄩󵄩4 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 + 3c02 󵄩󵄩󵄩2ek−1 − ek−2 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩ek 󵄩󵄩󵄩) + 󵄩󵄩󵄩Rk 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩ek 󵄩󵄩󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 ⩽ √κ2 + β2 c0 ⋅ 3c8 (2󵄩󵄩󵄩ek−1 󵄩󵄩󵄩4 + 󵄩󵄩󵄩ek−2 󵄩󵄩󵄩4 )󵄩󵄩󵄩ek 󵄩󵄩󵄩4 󵄩 󵄩 󵄩 󵄩󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 + √κ2 + β2 ⋅ 3c02 (2󵄩󵄩󵄩ek−1 󵄩󵄩󵄩 + 󵄩󵄩󵄩ek−2 󵄩󵄩󵄩)󵄩󵄩󵄩ek 󵄩󵄩󵄩 + 󵄩󵄩󵄩Rk 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩ek 󵄩󵄩󵄩 √5 󵄩 k 󵄩2 󵄩 k−1 󵄩2 󵄩 k−2 󵄩2 (󵄩󵄩e 󵄩󵄩 + 󵄩󵄩e 󵄩󵄩󵄩4 + 󵄩󵄩󵄩e 󵄩󵄩󵄩4 ) ⩽ √κ2 + β2 3c0 c8 2 󵄩 󵄩4 󵄩 √5 󵄩 k 󵄩2 󵄩 k−1 󵄩2 󵄩 k−2 󵄩2 󵄩 k 󵄩 󵄩 k 󵄩 + √κ2 + β2 ⋅ 3c02 ⋅ (󵄩󵄩e 󵄩󵄩 + 󵄩󵄩e 󵄩󵄩󵄩 + 󵄩󵄩󵄩e 󵄩󵄩󵄩 ) + 󵄩󵄩󵄩R 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩e 󵄩󵄩󵄩. 2 󵄩 󵄩 󵄩 With the application of Lemma 9.5, it follows: Ak ⩽ √κ2 + β2

1/2

3√5c0 c8 1 1 󵄩 󵄩2 󵄨 󵄨2 󵄩 󵄩 {[4󵄨󵄨󵄨ek 󵄨󵄨󵄨1 + ( 2 + 2 )󵄩󵄩󵄩ek 󵄩󵄩󵄩 ] 󵄩󵄩󵄩ek 󵄩󵄩󵄩 2 L1 L2 1/2

1 󵄩 1 󵄩2 󵄩 󵄨 󵄩 󵄨2 + [4󵄨󵄨󵄨ek−1 󵄨󵄨󵄨1 + ( 2 + 2 )󵄩󵄩󵄩ek−1 󵄩󵄩󵄩 ] 󵄩󵄩󵄩ek−1 󵄩󵄩󵄩 L1 L2 1/2

1 󵄩 1 󵄩2 󵄩 󵄩 󵄨 󵄨2 + [4󵄨󵄨󵄨ek−2 󵄨󵄨󵄨1 + ( 2 + 2 )󵄩󵄩󵄩ek−2 󵄩󵄩󵄩 ] 󵄩󵄩󵄩ek−2 󵄩󵄩󵄩} L1 L2

3√5c02 󵄩󵄩 k 󵄩󵄩2 󵄩󵄩 k−1 󵄩󵄩2 󵄩󵄩 k−2 󵄩󵄩2 󵄩󵄩 k 󵄩󵄩 󵄩󵄩 k 󵄩󵄩 (󵄩󵄩e 󵄩󵄩 + 󵄩󵄩e 󵄩󵄩 + 󵄩󵄩e 󵄩󵄩 ) + 󵄩󵄩R 󵄩󵄩 ⋅ 󵄩󵄩e 󵄩󵄩 2 3√5c0 c8 1 1 󵄩 󵄩2 1 󵄩 󵄩2 󵄨 󵄨2 [ε(4󵄨󵄨󵄨ek 󵄨󵄨󵄨1 + ( 2 + 2 )󵄩󵄩󵄩ek 󵄩󵄩󵄩 ) + 󵄩󵄩󵄩ek 󵄩󵄩󵄩 ⩽ √κ 2 + β 2 ⋅ 2 4ε L1 L2 + √κ2 + β2 ⋅

1 󵄩 1 1󵄩 󵄨 󵄩2 󵄩2 󵄨2 + ε(4󵄨󵄨󵄨ek−1 󵄨󵄨󵄨1 + ( 2 + 2 )󵄩󵄩󵄩ek−1 󵄩󵄩󵄩 ) + 󵄩󵄩󵄩ek−1 󵄩󵄩󵄩 4ε L1 L2

1󵄩 1 1 󵄩 󵄩2 󵄩2 󵄨 󵄨2 + ε(4󵄨󵄨󵄨ek−2 󵄨󵄨󵄨1 + ( 2 + 2 )󵄩󵄩󵄩ek−2 󵄩󵄩󵄩 ) + 󵄩󵄩󵄩ek−2 󵄩󵄩󵄩 ] 4ε L1 L2 + √κ2 + β2 ⋅

3√5c02 󵄩󵄩 k 󵄩󵄩2 󵄩󵄩 k−1 󵄩󵄩2 󵄩󵄩 k−2 󵄩󵄩2 󵄩󵄩 k 󵄩󵄩 󵄩󵄩 k 󵄩󵄩 (󵄩󵄩e 󵄩󵄩 + 󵄩󵄩e 󵄩󵄩 + 󵄩󵄩e 󵄩󵄩 ) + 󵄩󵄩R 󵄩󵄩 ⋅ 󵄩󵄩e 󵄩󵄩. 2

(9.60)

9.3 Three-level linearized difference scheme

� 305

Taking 6√5√κ2 + β2 c0 c8 ε = ν3 , then we have Ak ⩽

ν 󵄨󵄨 k 󵄨󵄨2 󵄨󵄨 k−1 󵄨󵄨2 󵄨󵄨 k−2 󵄨󵄨2 (󵄨e 󵄨 + 󵄨e 󵄨󵄨1 + 󵄨󵄨e 󵄨󵄨1 ) 3 󵄨 󵄨1 󵄨 3 2 󵄩 󵄩2 󵄩 󵄩2 󵄩2 󵄩 + (√κ2 + β2 ⋅ 3√5c0 c8 ) (󵄩󵄩󵄩ek 󵄩󵄩󵄩 + 󵄩󵄩󵄩ek−1 󵄩󵄩󵄩 + 󵄩󵄩󵄩ek−2 󵄩󵄩󵄩 ) 4ν ν 1 1 󵄩 󵄩2 󵄩 󵄩2 󵄩2 󵄩 + ( 2 + 2 )(󵄩󵄩󵄩ek 󵄩󵄩󵄩 + 󵄩󵄩󵄩ek−1 󵄩󵄩󵄩 + 󵄩󵄩󵄩ek−2 󵄩󵄩󵄩 ) 12 L1 L2 + √κ2 + β2 ⋅

3√5c02 󵄩󵄩 k 󵄩󵄩2 󵄩󵄩 k−1 󵄩󵄩2 󵄩󵄩 k−2 󵄩󵄩2 󵄩󵄩 k 󵄩󵄩 󵄩󵄩 k 󵄩󵄩 (󵄩󵄩e 󵄩󵄩 + 󵄩󵄩e 󵄩󵄩 + 󵄩󵄩e 󵄩󵄩 ) + 󵄩󵄩R 󵄩󵄩 ⋅ 󵄩󵄩e 󵄩󵄩. 2

Denote c9 =

3√5c02 1 3 2 ν 1 (κ + β2 ) ⋅ 45c02 c82 + ( 2 + 2 ) + √κ2 + β2 ⋅ . 4ν 12 L1 L2 2

Then we have ν 󵄨 󵄨2 󵄨 󵄨2 󵄨 󵄨2 󵄩 󵄩2 󵄩 󵄩2 󵄩 󵄩2 󵄩 󵄩 󵄩 󵄩 Ak ⩽ (󵄨󵄨󵄨ek 󵄨󵄨󵄨1 + 󵄨󵄨󵄨ek−1 󵄨󵄨󵄨1 + 󵄨󵄨󵄨ek−2 󵄨󵄨󵄨1 ) + c9 (󵄩󵄩󵄩ek 󵄩󵄩󵄩 + 󵄩󵄩󵄩ek−1 󵄩󵄩󵄩 + 󵄩󵄩󵄩ek−2 󵄩󵄩󵄩 ) + 󵄩󵄩󵄩Rk 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩ek 󵄩󵄩󵄩. 3 Substituting the above inequality into (9.60), we have 1 󵄩󵄩 k 󵄩󵄩2 󵄩󵄩 k 2ν 󵄨 󵄨2 󵄩2 󵄩 󵄩2 󵄩 󵄩2 [(󵄩󵄩e 󵄩󵄩 + 󵄩󵄩2e − ek−1 󵄩󵄩󵄩 ) − (󵄩󵄩󵄩ek−1 󵄩󵄩󵄩 + 󵄩󵄩󵄩2ek−1 − ek−2 󵄩󵄩󵄩 )] + 󵄨󵄨󵄨ek 󵄨󵄨󵄨1 4τ 3 ν 󵄨 󵄨2 󵄨 󵄨2 󵄩 󵄩2 󵄩 󵄩2 󵄩 󵄩2 ⩽ (󵄨󵄨󵄨ek−1 󵄨󵄨󵄨1 + 󵄨󵄨󵄨ek−2 󵄨󵄨󵄨1 ) + c9 (󵄩󵄩󵄩ek 󵄩󵄩󵄩 + 󵄩󵄩󵄩ek−1 󵄩󵄩󵄩 + 󵄩󵄩󵄩ek−2 󵄩󵄩󵄩 ) 3 󵄩 󵄩2 󵄩 󵄩 󵄩 󵄩 + γ󵄩󵄩󵄩ek 󵄩󵄩󵄩 + 󵄩󵄩󵄩Rk 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩ek 󵄩󵄩󵄩, 2 ⩽ k ⩽ n.

(9.61)

Denote 󵄩 󵄩2 󵄩 󵄩2 F k = 󵄩󵄩󵄩ek 󵄩󵄩󵄩 + 󵄩󵄩󵄩2ek − ek−1 󵄩󵄩󵄩 ,

1 ⩽ k ⩽ n.

Replacing k by l in (9.61) and summing over l from 2 to k, we have 2ν k 󵄨󵄨 l 󵄨󵄨2 1 k (F − F 1 ) + ∑󵄨e 󵄨 4τ 3 l=2󵄨 󵄨1 ⩽

k ν k 󵄨󵄨 l−1 󵄨󵄨2 󵄨󵄨 l−2 󵄨󵄨2 󵄩 󵄩2 󵄩 󵄩2 󵄩 󵄩2 ∑(󵄨󵄨e 󵄨󵄨1 + 󵄨󵄨e 󵄨󵄨1 ) + c9 ∑(󵄩󵄩󵄩el 󵄩󵄩󵄩 + 󵄩󵄩󵄩el−1 󵄩󵄩󵄩 + 󵄩󵄩󵄩el−2 󵄩󵄩󵄩 ) 3 l=2 l=2 k k 󵄩 󵄩2 󵄩 󵄩2 󵄩 󵄩2 1 + γ ∑󵄩󵄩󵄩el 󵄩󵄩󵄩 + ∑(󵄩󵄩󵄩Rl 󵄩󵄩󵄩 + 󵄩󵄩󵄩el 󵄩󵄩󵄩 ), 2 l=2 l=2

2 ⩽ k ⩽ n.

(9.62)

1 󵄩󵄩 1 󵄩󵄩2 󵄨 1 󵄨2 󵄩 1 󵄩2 1 󵄩 1 󵄩2 󵄩 1 󵄩2 󵄩e 󵄩 + ν󵄨󵄨󵄨e 󵄨󵄨󵄨1 ⩽ γ󵄩󵄩󵄩e 󵄩󵄩󵄩 + (󵄩󵄩󵄩R 󵄩󵄩󵄩 + 󵄩󵄩󵄩e 󵄩󵄩󵄩 ). τ󵄩 󵄩 2

(9.63)

According to (9.57), we have

306 � 9 Difference methods for the Ginzburg–Landau equation Noticing ‖e0 ‖ = 0 and F 1 = 5‖e1 ‖2 , and then combining (9.62) with (9.63), we have k 1 k 󵄩 󵄩2 󵄩 󵄩2 󵄩 󵄩2 F ⩽ c9 ∑(󵄩󵄩󵄩el 󵄩󵄩󵄩 + 󵄩󵄩󵄩el−1 󵄩󵄩󵄩 + 󵄩󵄩󵄩el−2 󵄩󵄩󵄩 ) 4τ l=2 k k 󵄩 󵄩2 1 󵄩 󵄩2 󵄩 󵄩2 + γ ∑󵄩󵄩󵄩el 󵄩󵄩󵄩 + ∑(󵄩󵄩󵄩Rl 󵄩󵄩󵄩 + 󵄩󵄩󵄩el 󵄩󵄩󵄩 ) 2 l=2 l=2

5 󵄩 󵄩2 1 󵄩 󵄩2 󵄩 󵄩2 + [γ󵄩󵄩󵄩e1 󵄩󵄩󵄩 + (󵄩󵄩󵄩R1 󵄩󵄩󵄩 + 󵄩󵄩󵄩e1 󵄩󵄩󵄩 )]. 4 2 In other words, 1 󵄩󵄩 k 󵄩󵄩2 󵄩e 󵄩 4τ 󵄩 󵄩

1 󵄩 󵄩2 1 k−1󵄩 󵄩2 ⩽ (c9 + γ + )󵄩󵄩󵄩ek 󵄩󵄩󵄩 + (3c9 + γ + ) ∑ 󵄩󵄩󵄩el 󵄩󵄩󵄩 2 2 l=2 5 󵄩 󵄩2 1 k 󵄩 󵄩2 5 + (2c9 + γ + )󵄩󵄩󵄩e1 󵄩󵄩󵄩 + ∑󵄩󵄩󵄩Rl 󵄩󵄩󵄩 + 4 8 2 l=2

5 󵄩󵄩 1 󵄩󵄩2 󵄩R 󵄩 8󵄩 󵄩

1 󵄩 󵄩2 1 k−1󵄩 󵄩2 ⩽ (c9 + max{0, γ} + )󵄩󵄩󵄩ek 󵄩󵄩󵄩 + (3c9 + max{0, γ} + ) ∑ 󵄩󵄩󵄩el 󵄩󵄩󵄩 2 2 l=2 + (2c9 +

5 5 󵄩 󵄩2 1 k 󵄩 󵄩2 max{0, γ} + )󵄩󵄩󵄩e1 󵄩󵄩󵄩 + ∑󵄩󵄩󵄩Rl 󵄩󵄩󵄩 + 4 8 2 l=2

5 󵄩󵄩 1 󵄩󵄩2 󵄩R 󵄩 , 8󵄩 󵄩

2 ⩽ k ⩽ n.

(9.64)

We know that (9.64) also holds for k = 1 by observing (9.63). From (9.64), we have 1 󵄩 󵄩2 [1 − 4(c9 + max{0, γ} + )τ]󵄩󵄩󵄩ek 󵄩󵄩󵄩 2 ⩽ 4(3c9 +

5 k−1󵄩 󵄩2 1 k 󵄩 󵄩2 5 max{0, γ} + )τ ∑ 󵄩󵄩󵄩el 󵄩󵄩󵄩 + 4τ( ∑󵄩󵄩󵄩Rl 󵄩󵄩󵄩 + 4 8 l=1 2 l=2

5 󵄩󵄩 1 󵄩󵄩2 󵄩R 󵄩 ), 8󵄩 󵄩

1 ⩽ k ⩽ n.

When 4(c9 + max{0, γ} + 21 )τ ⩽ 21 , it holds that k−1

k

l=1

l=2

󵄩󵄩 k 󵄩󵄩2 󵄩 l 󵄩2 󵄩 l 󵄩2 󵄩󵄩e 󵄩󵄩 ⩽ (24c9 + 10 max{0, γ} + 5)τ ∑ 󵄩󵄩󵄩e 󵄩󵄩󵄩 + 4τ(∑󵄩󵄩󵄩R 󵄩󵄩󵄩 +

5 󵄩󵄩 1 󵄩󵄩2 󵄩R 󵄩 ), 4󵄩 󵄩

1 ⩽ k ⩽ n.

With the application of the Gronwall inequality (Theorem 1.2(c)) and noting (9.34) and (9.30), we have 2 󵄩󵄩 k 󵄩󵄩2 (24c +10 max{0,γ}+5)T ⋅ (4Tc62 + 5c52 )L1 L2 (τ 2 + h12 + h22 ) , 󵄩󵄩e 󵄩󵄩 ⩽ e 9

1 ⩽ k ⩽ n,

which implies 󵄩󵄩 k 󵄩󵄩 (12c +5 max{0,γ}+ 52 )T √ (4Tc62 + 5c52 )L1 L2 (τ 2 + h12 + h22 ), 󵄩󵄩e 󵄩󵄩 ⩽ e 9

1 ⩽ k ⩽ n.

9.4 Numerical experiments

� 307

9.4 Numerical experiments In this section, we take the difference scheme (9.37)–(9.40) as an example to test the accuracy and convergence. Our theoretical results in this chapter can be easily extended to the spatial domain [−1, 1] × [−1, 1]. Take h1 = m2 , xi = −1 + ih1 , 0 ⩽ i ⩽ m1 ; h2 = m2 , 1 2 yj = −1 + jh2 , 0 ⩽ j ⩽ m2 . Example 9.1. Consider the Ginzburg–Landau equation as follows: ut − (ν + iη)Δu + (κ + iζ )|u|2 u − γu = f (x, y, t),

(x, y) ∈ (−1, 1) × (−1, 1), t ∈ (0, 1].

The exact solution is u(x, y, t) = (x + 1)4 (x − 1)4 (y + 1)4 (y − 1)4 eit ,

(9.65)

where ν = η = κ = 1, ζ = 2, γ = 3. The source term f (x, y, t) is determined by (9.65). In the runs, we use the same spatial step size h = m2 in each direction, and compute the L2 -norm error of the solution at t = 1. Denote 󵄩 󵄩 E(τ, h) = 󵄩󵄩󵄩U n − un 󵄩󵄩󵄩,

Orderh = log2 (

E(τ, 2h) ), E(τ, h)

Orderτ = log2 (

E(2τ, h) ). E(τ, h)

First, we test the spatial error and convergence order by varying h and fixing the temporal step size τ sufficiently small to avoid contamination of the temporal error. Table 9.1 gives the L2 -norm error and spatial convergence order. Second, we investigate the temporal error and convergence order. In this test, we fix h = 1/120, a value small enough such that the spatial error is negligible as compared with the temporal error. Tables 9.2 presents the L2 -norm error and the temporal convergence order. It is clear to see that they confirm the theoretical result. Table 9.1: The L2 -norm errors and spatial convergence orders with the temporal step size τ = 1/400. m

E(τ, h)

Orderh

4 8 16 32 64

2.5123e-1 5.8616e-2 1.4184e-2 3.5148e-3 8.7615e-4

2.4912 2.1801 2.0428 2.0151

308 � 9 Difference methods for the Ginzburg–Landau equation Table 9.2: The L2 -errors and temporal convergence orders with the spatial step size h = 1/120. n

E(τ, h)

Orderτ

5 10 20 40 80

7.3622e-2 2.2083e-2 6.2192e-3 1.6348e-3 4.2525e-4

1.8448 1.9019 2.0265 2.1050

9.5 Summary and extension In this chapter, the finite difference methods for solving the Ginzburg–Landau equation are studied. A two-level nonlinear difference scheme and a three-level linearized difference scheme are introduced, respectively. The existence and boundedness of solutions to difference schemes are proved. By using the boundedness of the solution of the difference scheme in the L2 -norm, we prove that the difference scheme is unconditionally convergent in the L2 -norm. The present schemes are developed based on the work [40] and [54]. The compact difference scheme and compact alternating direction implicit scheme are considered for the Ginzburg–Landau equation in [15], and the unconditional convergence of both difference schemes are proved. The numerical experiment section in this chapter originates from [54]. In this chapter, the second-order backward difference formula (9.31) is used to approximate the temporal derivative, and the extrapolation formula (9.32) is used to deal with the nonlinear term in the three-level linearized difference scheme. For the problem (9.1)–(9.3), a three-level Crank–Nicolson-type linearized difference scheme can be established as follows: { { { { { { { { { { { { { { {

̄ ̄ 󵄨 󵄨2 ̄ Δt uijk − (ν + iα)Δh uijk + (κ + iβ)󵄨󵄨󵄨uijk 󵄨󵄨󵄨 uijk − γuijk = 0, 󵄨 󵄨2 ∇τ uij1 − (ν + iα)Δh uij1 + (κ + iβ)󵄨󵄨󵄨uij0 󵄨󵄨󵄨 uij1 − γuij1 = 0,

uij0

= φ(xi , yj ),

uijk = 0,

(i, j) ∈ ω, 1 ⩽ k ⩽ n − 1, (i, j) ∈ ω, (i, j) ∈ ω, (i, j) ∈ 𝜕ω, 0 ⩽ k ⩽ n.

10 Difference methods for the Cahn–Hilliard equation The Cahn–Hilliard equation is a typical fourth-order nonlinear reaction-diffusion equation. It was originally introduced in 1958 by Cahn and Hilliard to describe the mutual diffusion phenomenon in the phase separation of binary mixtures such as alloys, polymers, etc. Later, it was widely used to study species competition and exclusion, migration process of the river bed, diffusion of droplets on solid surface and so on.

10.1 Introduction Consider the initial and boundary value problem of the Cahn–Hilliard equation as follows: ut = Δ(ϕ(u) − αΔu), { { { 𝜕u 𝜕(ϕ(u) − αΔu) = 0, = 0, { { 𝜕ν 𝜕ν { { u(x, y, 0) = φ(x, y),

(x, y) ∈ Ω, 0 < t ⩽ T,

(10.1)

(x, y) ∈ 𝜕Ω, 0 < t ⩽ T,

(10.2)

(x, y) ∈ Ω,̄

(10.3)

where Ω = (0, L1 ) × (0, L2 ), ν is the unit outer normal vector on the boundary 𝜕Ω of the domain Ω, ϕ(u) = ψ′ (u), ψ(u) = γ(u2 − β2 )2 /4, α, β and γ are three positive constants. The solution of the problem (10.1)–(10.3) satisfies the following conservation law. Theorem 10.1. Let u(x, y, t) be the solution of the problem (10.1)–(10.3) and denote t

F(t) =

1 󵄨 󵄨2 ∬ u2 (x, y, t)dxdy + ∫[∬ ϕ′ (u(x, y, s))󵄨󵄨󵄨∇u(x, y, s)󵄨󵄨󵄨 dxdy 2 0

Ω

Ω

2

+ α ∬[Δu(x, y, s)] dxdy]ds. Ω

Then it holds that F(t) = F(0),

0 < t ⩽ T.

Proof. Multiplying both the right- and left-hand sides of (10.1) with u(x, y, t) and integrating the result with respect to (x, y) on Ω, we have ∬ u(x, y, t)ut (x, y, t)dxdy − ∬ u(x, y, t) ⋅ [Δ(ϕ(u) − αΔu)](x, y, t)dxdy = 0. Ω

Ω

Now each term on the left-hand side of the above equality will be analyzed. https://doi.org/10.1515/9783110796018-010

(10.4)

310 � 10 Difference methods for the Cahn–Hilliard equation For the first term, it is easy to obtain ∬ u(x, y, t)ut (x, y, t)dxdy =

1 d ∬ u2 (x, y, t)dxdy. 2 dt

(10.5)

Ω

Ω

For the second term, the application of the integration by parts leads to − ∬[Δ(ϕ(u(x, y, t)) − αΔu(x, y, t))]u(x, y, t)dxdy Ω

L2 L1

= − ∫[∫(ϕ(u(x, y, t)) − αΔu(x, y, t))xx u(x, y, t)dx]dy 0

0

L1 L2

− ∫[∫(ϕ(u(x, y, t)) − αΔu(x, y, t))yy u(x, y, t)dy]dx L2

0

0

󵄨L1 = ∫[−(ϕ(u(x, y, t)) − αΔu(x, y, t))x u(x, y, t)󵄨󵄨󵄨x=0 0

L1

+ ∫(ϕ(u(x, y, t)) − αΔu(x, y, t))x ux (x, y, t)dx]dy 0

L1

󵄨L2 + ∫[−(ϕ(u(x, y, t)) − αΔu(x, y, t))y u(x, y, t)󵄨󵄨󵄨y=0 0

L2

+ ∫(ϕ(u(x, y, t)) − αΔu(x, y, t))y uy (x, y, t)dy]dx 0

L2 L1

󵄨L1 = ∫[∫ ϕ′ (u(x, y, t))ux2 (x, y, t)dx − α(Δu(x, y, t))ux (x, y, t)󵄨󵄨󵄨x=0 0

0

L1

+ α ∫(Δu(x, y, t))uxx (x, y, t)dx]dy 0

L1 L2

󵄨L2 + ∫[∫ ϕ′ (u(x, y, t))uy2 (x, y, t)dy − α(Δu(x, y, t))uy (x, y, t)󵄨󵄨󵄨y=0 0

0

L2

+ α ∫(Δu(x, y, t))uyy (x, y, t)dy]dx 0 2 󵄨 󵄨2 = ∬ ϕ′ (u(x, y, t))󵄨󵄨󵄨∇u(x, y, t)󵄨󵄨󵄨 dxdy + α ∬(Δu(x, y, t)) dxdy. Ω

Ω

(10.6)

10.1 Introduction

� 311

Substituting (10.5)–(10.6) into (10.4) produces 1 d 󵄨2 󵄨 ∬ u2 (x, y, t)dxdy + ∬ ϕ′ (u(x, y, t))󵄨󵄨󵄨∇u(x, y, t)󵄨󵄨󵄨 dxdy 2 dt Ω

Ω

2

+ α ∬(Δu(x, y, t)) dxdy = 0,

0 < t ⩽ T,

(10.7)

Ω

i. e., d F(t) = 0, dt

0 < t ⩽ T.

Then the desired result is followed. Remark 10.1. Noticing ϕ′ (u) = γ(3u2 − β2 ) ⩾ −γβ2 and |u|21 ⩽ |u|2 ⋅ ‖u‖, it follows from (10.7) that 1 d 󵄨 󵄨2 󵄨 󵄨2 ∬ u2 (x, y, t)dxdy + α󵄨󵄨󵄨u(⋅, ⋅, t)󵄨󵄨󵄨2 = − ∬ ϕ′ (u(x, y, t))󵄨󵄨󵄨∇u(x, y, t)󵄨󵄨󵄨 dxdy 2 dt Ω

Ω

󵄨 󵄨2 󵄨 󵄨 󵄩 󵄩 ⩽ γβ2 󵄨󵄨󵄨u(⋅, ⋅, t)󵄨󵄨󵄨1 ⩽ γβ2 󵄨󵄨󵄨u(⋅, ⋅, t)󵄨󵄨󵄨2 ⋅ 󵄩󵄩󵄩u(⋅, ⋅, t)󵄩󵄩󵄩

2 4 󵄩2 󵄨 󵄨2 γ β 󵄩󵄩 ⩽ α󵄨󵄨󵄨u(⋅, ⋅, t)󵄨󵄨󵄨2 + 󵄩u(⋅, ⋅, t)󵄩󵄩󵄩 , 4α 󵄩

which gives 2 4 1 d 󵄩󵄩 󵄩2 γ β 󵄩󵄩 󵄩2 󵄩󵄩u(⋅, ⋅, t)󵄩󵄩󵄩 ⩽ 󵄩u(⋅, ⋅, t)󵄩󵄩󵄩 . 2 dt 4α 󵄩

Hence, 2 4

󵄩󵄩 󵄩2 󵄩 󵄩2 γ β t 󵄩󵄩u(⋅, ⋅, t)󵄩󵄩󵄩 ⩽ 󵄩󵄩󵄩u(⋅, ⋅, 0)󵄩󵄩󵄩 e 2α ,

0 < t ⩽ T,

or 2 4

γ β 󵄩󵄩 󵄩 󵄩 T󵄩 󵄩󵄩u(⋅, ⋅, t)󵄩󵄩󵄩 ⩽ e 4α 󵄩󵄩󵄩u(⋅, ⋅, 0)󵄩󵄩󵄩,

0 ⩽ t ⩽ T.

Theorem 10.2. Let u(x, y, t) be the solution of the problem (10.1)–(10.3) and denote v = ϕ(u) − αΔu,

t

E(t) = ∬ ψ(u(x, y, t))dxdy + Ω

α 󵄨󵄨 󵄨2 󵄨 󵄨2 ∬󵄨∇u(x, y, t)󵄨󵄨󵄨 dxdy + ∫[∬󵄨󵄨󵄨∇v(x, y, s)󵄨󵄨󵄨 dxdy]ds. 2 󵄨 0

Ω

Then E(t) = E(0),

0 < t ⩽ T.

Ω

312 � 10 Difference methods for the Cahn–Hilliard equation Proof. Multiplying both the right- and left-hand sides of (10.1) by ϕ(u) − αΔu, we have ut [ϕ(u) − αΔu] = [Δ(ϕ(u) − αΔu)] ⋅ [ϕ(u) − αΔu] = (Δv)v. Integrating the above equality with respect to (x, y) on Ω gives ∬ ϕ(u)ut dxdy − α ∬(Δu)ut dxdy − ∬(Δv)vdxdy = 0. Ω

Ω

(10.8)

Ω

It follows from the definition of ϕ(u) that ∬ ϕ(u)ut dxdy = ∬ ψ′ (u)ut dxdy = Ω

Ω

d ∬ ψ(u)dxdy. dt

(10.9)

Ω

Noticing the first equation in (10.2), it follows from the integration by parts that −α ∬(Δu)ut dxdy = α ∬ ∇u ⋅ ∇ut dxdy = Ω

Ω

α d ∬ |∇u|2 dxdy. 2 dt

(10.10)

Ω

Noticing the second one in (10.2), in a similar manner, we obtain − ∬(Δv)vdxdy = ∬ |∇v|2 dxdy. Ω

(10.11)

Ω

Substituting (10.9), (10.10) and (10.11) into (10.8) arrives at d α d 󵄨 󵄨2 ∬ ψ(u(x, y, t))dxdy + ∬󵄨󵄨∇u(x, y, t)󵄨󵄨󵄨 dxdy dt 2 dt 󵄨 Ω

Ω

󵄨 󵄨2 + ∬󵄨󵄨󵄨∇v(x, y, t)󵄨󵄨󵄨 dxdy = 0,

0 < t ⩽ T,

Ω

which can be rewritten as d E(t) = 0, dt

0 < t ⩽ T.

Then the desired result is followed.

10.2 Two-level nonlinear difference scheme This section is devoted to the development of a two-level nonlinear difference scheme for solving (10.1)–(10.3).

10.2 Two-level nonlinear difference scheme

For any mesh function v ∈ 𝒱h , introduce notation as follows: 2 δ v1 , h1 x 2 ,j

{ { { { 2 δx vij = { { { { {

1 (δ v 1 h1 x i+ 2 ,j

i = 0, − δx vi− 1 ,j ),

2 (−δx vm − 1 ,j ), h1 1 2 2 δ v 1, h2 y i, 2

{ { { { 2 δy vij = { { { { {

1 (δ v 1 h2 y i,j+ 2

2

1 ⩽ i ⩽ m1 − 1, i = m1 , j = 0,

− δy vi,j− 1 ), 2

2 (−δy vi,m − 1 ), h2 2 2

1 ⩽ j ⩽ m2 − 1, j = m2 .

At the beginning, two useful lemmas will be brought here. Lemma 10.1. For any u ∈ 𝒱h , it holds that −(Δh u, u) = |u|21 . Proof. It follows from the summation by parts that −(Δh u, u)

m1 m2

= −h1 h2 ∑ ∑ ωi ω̂ j (δx2 uij + δy2 uij )uij i=0 j=0

m2

m1

m1

m2

j=0

i=0

i=1

j=0

= h2 ∑ ω̂ j [−h1 ∑ ωi (δx2 uij )uij ] + h1 ∑ ωi [−h2 ∑ ω̂ j (δy2 uij )uij ] m2

m1

j=0

i=1

m1

m2

i=0

j=1

= h2 ∑ ω̂ j [h1 ∑(δx ui− 1 ,j )2 ] + h1 ∑ ωi [h2 ∑(δy ui,j− 1 )2 ] 2

2

= |u|21 . Lemma 10.2. For any u ∈ 𝒱h and any ε > 0, it holds that 1 1 1 1 ‖u‖∞ ⩽ ε‖Δh u‖ + √3[ + ( + )]‖u‖. ε 2 L1 L2 Proof. Let ‖u‖∞ = |ui0 ,j0 |. By Lemma 1.1(e), we have m

m

1 1 1 1 2 ui20 ,j0 ⩽ εh1 ∑(δx ui− 1 ,j )2 + ( + )h1 ∑ ωi ui,j 0 2 0 ε L 1 i=1 i=0

� 313

314 � 10 Difference methods for the Cahn–Hilliard equation m

m

m

2 1 2 1 1 ⩽ εh1 ∑[εh2 ∑(δy δx ui− 1 ,j− 1 )2 + ( + )h2 ∑ ω̂ j (δx ui− 1 ,j )2 ] 2 2 2 ε L2 j=0 i=1 j=1

m

m

m

1 2 2 1 1 1 1 + ( + )h1 ∑ ωi [εh2 ∑(δy ui,j− 1 )2 + ( + )h2 ∑ ω̂ j uij2 ] 2 ε L1 ε L2 i=0 j=1 j=0

⩽ ε2 ‖δy δx u‖2 + [ε(

1 1 + ) + 1]|u|21 L1 L2

1 1 1 1 + ( + )( + )‖u‖2 . ε L1 ε L2

(10.12)

In addition, it follows from m1 m2

2

‖Δh u‖2 = h1 h2 ∑ ∑ ωi ω̂ j (δx2 uij + δy2 uij ) i=0 j=0 m1 m2

m1 m2

2

= h1 h2 ∑ ∑ ωi ω̂ j (δx2 uij ) + 2h1 h2 ∑ ∑ ωi ω̂ j (δx2 uij )(δy2 uij ) i=0 j=0

i=0 j=0

m1 m2

+ h1 h2 ∑ ∑ ωi ω̂ j (δy2 uij )

2

i=0 j=0

and m1 m2

h1 h2 ∑ ∑ ωi ω̂ j (δx2 uij )(δy2 uij ) i=0 j=0

m2

m1

j=0

i=0

= h2 ∑ ω̂ j [h1 ∑ ωi (δx2 uij )(δy2 uij )] m2

m1 −1

j=0

i=0

= h2 ∑ ω̂ j [−h1 ∑ (δx ui+ 1 ,j )(δx δy2 ui+ 1 ,j )] m1 −1

m2

i=0

j=0

2

2

= −h1 ∑ [h2 ∑ ω̂ j (δy2 δx ui+ 1 ,j )(δx ui+ 1 ,j )] m1 −1

m2 −1

i=0

j=0

2

2

= −h1 ∑ [−h2 ∑ (δy δx ui+ 1 ,j+ 1 )2 ] 2

2

m1 −1 m2 −1

= h1 h2 ∑ ∑ (δy δx ui+ 1 ,j+ 1 )2 i=0 j=0

2

2

= ‖δy δx u‖2 that 󵄩 󵄩2 󵄩 󵄩2 ‖Δh u‖2 = 󵄩󵄩󵄩δx2 u󵄩󵄩󵄩 + 2‖δy δx u‖2 + 󵄩󵄩󵄩δy2 u󵄩󵄩󵄩 .

(10.13)

� 315

10.2 Two-level nonlinear difference scheme

Obviously, 1 ‖δy δx u‖2 ⩽ ‖Δh u‖2 . 2

(10.14)

By Lemma 10.1, for any ε1 > 0, we have |u|21 = −(Δh u, u) ⩽ ε1 ‖Δh u‖2 +

1 ‖u‖2 . 4ε1

(10.15)

Substituting (10.14) and (10.15) into (10.12) produces 1 1 1 1 1 1 1 1 ‖u‖2 ) + ( + )( + )‖u‖2 . ui20 ,j0 ⩽ ε2 ‖Δh u‖2 + [ε( + ) + 1](ε1 ‖Δh u‖2 + 2 L1 L2 4ε1 ε L1 ε L2 Taking ε1 such that [ε( L1 + 1

1 ) L2

+ 1]ε1 = 21 ε2 and noticing 2

4 1 1 ⩽( + ) , L1 L2 L1 L2 we get ui20 ,j0

2

2

⩽ ε ‖Δh u‖ + { = ε2 ‖Δh u‖2 + [

[ε( L1 + 1

1 ) L2 2ε2

+ 1]2

1 1 1 1 + ( + )( + )}‖u‖2 ε L1 ε L2 2

3 2 1 1 1 1 1 1 + ( + )+ ( + ) + ]‖u‖2 2 L1 L2 L1 L2 2ε2 ε L1 L2 2

1 1 1 1 ⩽ ε2 ‖Δh u‖2 + 3[ + ( + )] ‖u‖2 , ε 2 L1 L2 which implies 1 1 1 1 ‖u‖∞ ⩽ ε‖Δh u‖ + √3[ + ( + )]‖u‖. ε 2 L1 L2 Remark 10.2. For any u ∈ 𝒱̊h and arbitrary ε > 0, we have ‖u‖∞ ⩽ ε‖Δh u‖ +

√2 ‖u‖. 8ε

10.2.1 Derivation of the difference scheme This subsection is devoted to the establishment of a finite difference scheme for solving the problem (10.1)–(10.3). Let v = ϕ(u) − αΔu.

316 � 10 Difference methods for the Cahn–Hilliard equation Then the original problem (10.1)–(10.3) is equivalent to { { { { { { { { { { { { {

ut = Δv,

v = ϕ(u) − αΔu, 𝜕u 𝜕v = 0, = 0, 𝜕ν 𝜕ν u(x, y, 0) = φ(x, y),

(x, y) ∈ Ω, 0 < t ⩽ T,

(10.16)

(x, y) ∈ Ω, 0 < t ⩽ T,

(10.17)

(x, y) ∈ 𝜕Ω, 0 < t ⩽ T,

(10.18)

(x, y) ∈ Ω.̄

(10.19)

Differentiating (10.16) with respect to x once gives (ux )t = vxxx + (vx )yy . Taking x = 0, L1 , respectively, and noticing (10.18), we have vxxx |x=0 = 0,

vxxx |x=L1 = 0.

(10.20)

Differentiating (10.16) with respect to y once and then taking y = 0, L2 , respectively, by (10.18), we have vyyy |y=0 = 0,

vyyy |y=L2 = 0.

(10.21)

Similarly, differentiating (10.17) with respect to x and y, respectively, by (10.18), it follows: uxxx |x=0 = 0,

uxxx |x=L1 = 0,

uyyy |y=0 = 0,

uyyy |y=L2 = 0.

(10.22) (10.23)

Define the mesh functions: Uijk = u(xi , yj , tk ),

Vijk = v(xi , yj , tk ),

0 ⩽ i ⩽ m1 , 0 ⩽ j ⩽ m2 , 0 ⩽ k ⩽ n.

Considering equations (10.16)–(10.17) at the point (xi , yj , tk+ 1 ), by Lemma 1.2, (10.18), 2 (10.20)–(10.23), we obtain k+ 1

k+ 1

k+ 1

2 2 2 { { δt Uij = Δh Vij + Pij , { { k+ 21 k+ 21 k+ 21 k+ 21 V = ϕ(U ) − αΔ U + Q , h ij ij ij { ij

0 ⩽ i ⩽ m1 , 0 ⩽ j ⩽ m2 , 0 ⩽ k ⩽ n − 1, (10.24) 0 ⩽ i ⩽ m1 , 0 ⩽ j ⩽ m2 , 0 ⩽ k ⩽ n − 1,

(10.25)

where there is a positive constant c1 such that { { { { { { { { { { { { {

󵄨󵄨 k+ 21 󵄨󵄨 2 2 2 󵄨󵄨Pij 󵄨󵄨 ⩽ c1 (τ + h1 + h2 ), 󵄨󵄨 k+ 21 󵄨󵄨 2 2 2 󵄨󵄨Qij 󵄨󵄨 ⩽ c1 (τ + h1 + h2 ), 1 󵄨󵄨 k+ 21 k− 1 󵄨 2 2 2 󵄨󵄨Qij − Qij 2 󵄨󵄨󵄨 ⩽ c1 (τ + h1 + h2 ), τ

0 ⩽ i ⩽ m1 , 0 ⩽ j ⩽ m2 , 0 ⩽ k ⩽ n − 1, (10.26) 0 ⩽ i ⩽ m1 , 0 ⩽ j ⩽ m2 , 0 ⩽ k ⩽ n − 1, (10.27) 0 ⩽ i ⩽ m1 , 0 ⩽ j ⩽ m2 , 1 ⩽ k ⩽ n − 1. (10.28)

10.2 Two-level nonlinear difference scheme

� 317

Omitting small terms in (10.24)–(10.25) and noticing the initial value condition Uij0 = φ(xi , yj ),

0 ⩽ i ⩽ m1 , 0 ⩽ j ⩽ m2 ,

(10.29)

a difference scheme for solving (10.16)–(10.19) can be established as k+ 1

k+ 1

{ δt uij 2 = Δh vij 2 , { { { { k+ 1 k+ 1 k+ 1 { vij 2 = ϕ(uij 2 ) − αΔh uij 2 , { { { { 0 { uij = φ(xi , yj ),

0 ⩽ i ⩽ m1 , 0 ⩽ j ⩽ m2 , 0 ⩽ k ⩽ n − 1,

(10.30)

0 ⩽ i ⩽ m1 , 0 ⩽ j ⩽ m2 , 0 ⩽ k ⩽ n − 1,

(10.31)

0 ⩽ i ⩽ m1 , 0 ⩽ j ⩽ m2 .

(10.32)

Inserting (10.31) into (10.30) arrives at k+ 1

k+ 21

{ δt uij 2 = Δh (ϕ(uij { 0 { uij = φ(xi , yj ),

k+ 21

) − αΔh uij

),

0 ⩽ i ⩽ m1 , 0 ⩽ j ⩽ m2 , 0 ⩽ k ⩽ n − 1, (10.33) 0 ⩽ i ⩽ m1 , 0 ⩽ j ⩽ m2 .

(10.34)

10.2.2 Existence of the difference solution The following theorem states the existence of the solution of the proposed difference scheme above. Theorem 10.3. If

γ2 β 4 τ 8α

< 1, the difference scheme (10.33)–(10.34) has a solution.

Proof. The value of u0 is determined by (10.34). Assume that the value of uk at the k-th time level has been obtained. It is noted that the nonlinear system in the unknown uk+1 can be determined by (10.33). Let k+ 21

wij = uij

,

0 ⩽ i ⩽ m1 , 0 ⩽ j ⩽ m2 .

It follows from (10.33) that 2 (w − uijk ) = Δh (ϕ(wij ) − αΔh wij ), τ ij

0 ⩽ i ⩽ m1 , 0 ⩽ j ⩽ m2 .

Once the value of w is obtained, the value of uk+1 can be followed by uijk+1 = 2wij − uijk ,

0 ⩽ i ⩽ m1 , 0 ⩽ j ⩽ m2 .

Hence, it suffices to verify that the solution of (10.35) exists. Define the operator Π : 𝒱h → 𝒱h by Π(w)ij =

2 (w − uijk ) − Δh (ϕ(wij ) − αΔh wij ), τ ij

0 ⩽ i ⩽ m1 , 0 ⩽ j ⩽ m2 .

(10.35)

318 � 10 Difference methods for the Cahn–Hilliard equation Then 2 (Π(w), w) = [(w, w) − (uk , w)] − (Δh (ϕ(w) − αΔh w), w) τ 2 = [‖w‖2 − (uk , w)] − (Δh ϕ(w), w) + α‖Δh w‖2 . τ

(10.36)

Noticing that there is a ξi+ 1 ,j lying between wij and wi+1,j , and an ηi,j+ 1 lying between wij 2 2 and wi,j+1 such that −(Δh ϕ(w), w) = (δx ϕ(w), δx w) + (δy ϕ(w), δy w) m1 −1 m2

= h1 h2 ∑ ∑ ω̂ j (δx ϕ(w)i+ 1 ,j )(δx wi+ 1 ,j ) 2

i=0 j=0

2

m1 m2 −1

+ h1 h2 ∑ ∑ ωi (δy ϕ(w)i,j+ 1 )(δy wi,j+ 1 ) 2

i=0 j=0

2

m1 −1 m2

= h1 h2 ∑ ∑ ω̂ j ϕ′ (ξi+ 1 ,j )(δx wi+ 1 ,j )2 2

i=0 j=0

2

m1 m2 −1

+ h1 h2 ∑ ∑ ωi ϕ′ (ηi,j+ 1 )(δy wi,j+ 1 )2 , 2

i=0 j=0

2

as well as ϕ′ (u) = γ(3u2 − β2 ) ⩾ −γβ2 , it follows: − (Δh ϕ(w), w) ⩾ −γβ2 |w|21 ⩾ −γβ2 ‖Δh w‖ ⋅ ‖w‖ ⩾ −(α‖Δh w‖2 +

γ2 β 4 ‖w‖2 ). 4α

(10.37)

Plugging (10.37) into (10.36) leads to 2 1 2 4 󵄩 󵄩 γ β ‖w‖2 (Π(w), w) ⩾ (‖w‖2 − 󵄩󵄩󵄩uk 󵄩󵄩󵄩 ⋅ ‖w‖) − τ 4α γ2 β 4 2 󵄩 󵄩 = ‖w‖[(1 − τ)‖w‖ − 󵄩󵄩󵄩uk 󵄩󵄩󵄩], τ 8α from which, one can find that (Π(w), w) ⩾ 0 if

γ2 β 4 τ 8α

< 1 and ‖w‖ =

has a solution by the Browder theorem (Theorem 2.4). By induction, the conclusion is true.

‖uk ‖ 1−

γ2 β 4 τ 8α

. Then (10.35)

10.2 Two-level nonlinear difference scheme

� 319

10.2.3 Boundedness of the difference solution The boundedness of the solution of the difference scheme (10.33)–(10.34) will be discussed here. Theorem 10.4. Suppose {uijk | 0 ⩽ i ⩽ m1 , 0 ⩽ j ⩽ m2 , 0 ⩽ k ⩽ n} is the solution of the

difference scheme (10.33)–(10.34). Then if

3γ2 β4 τ 8α

⩽ 1, it holds that

2 4

3γ β 󵄩󵄩 k 󵄩󵄩 T 󵄩 0󵄩 󵄩󵄩u 󵄩󵄩 ⩽ e 8α 󵄩󵄩󵄩u 󵄩󵄩󵄩,

1 ⩽ k ⩽ n.

Proof. Taking the inner product on both the right- and left-hand sides of (10.33) with 1 uk+ 2 gives 1

1

1

1

1

(δt uk+ 2 , uk+ 2 ) = (Δh (ϕ(uk+ 2 ) − αΔh uk+ 2 ), uk+ 2 ) 1

1

1

1

= (Δh ϕ(uk+ 2 ), uk+ 2 ) − α(Δh uk+ 2 , Δh uk+ 2 ). By (10.37), we have 1 2 γ2 β4 󵄩󵄩 k+ 21 󵄩󵄩2 1 󵄩󵄩 k+1 󵄩󵄩2 󵄩󵄩 k 󵄩󵄩2 󵄩 󵄩 󵄩 k+ 1 󵄩2 (󵄩󵄩u 󵄩󵄩 − 󵄩󵄩u 󵄩󵄩 ) ⩽ (α󵄩󵄩󵄩Δh uk+ 2 󵄩󵄩󵄩 + 󵄩󵄩u 󵄩󵄩 ) − α󵄩󵄩󵄩Δh u 2 󵄩󵄩󵄩 2τ 4α

2

⩽

γ2 β4 ‖uk+1 ‖ + ‖uk ‖ ( ), 4α 2

0 ⩽ k ⩽ n − 1,

which implies γ2 β4 󵄩󵄩 k+1 󵄩󵄩 󵄩󵄩 k 󵄩󵄩 1 󵄩󵄩 k+1 󵄩󵄩 󵄩󵄩 k 󵄩󵄩 (󵄩󵄩u 󵄩󵄩 − 󵄩󵄩u 󵄩󵄩) ⩽ (󵄩u 󵄩󵄩 + 󵄩󵄩u 󵄩󵄩), τ 8α 󵄩

0 ⩽ k ⩽ n − 1,

γ2 β4 󵄩󵄩 k 󵄩󵄩 γ2 β4 󵄩󵄩 k+1 󵄩󵄩 τ)󵄩󵄩u 󵄩󵄩 ⩽ (1 + τ)󵄩󵄩u 󵄩󵄩, 8α 8α

0 ⩽ k ⩽ n − 1.

i. e., (1 − If

γ2 β 4 τ 8α

⩽ 31 , it follows: 3γ2 β4 󵄩󵄩 k 󵄩󵄩 󵄩󵄩 k+1 󵄩󵄩 τ)󵄩󵄩u 󵄩󵄩, 󵄩󵄩u 󵄩󵄩 ⩽ (1 + 8α

0 ⩽ k ⩽ n − 1.

By recursion, one gets 2 4

3γ β 󵄩󵄩 k 󵄩󵄩 T 󵄩 0󵄩 󵄩󵄩u 󵄩󵄩 ⩽ e 8α 󵄩󵄩󵄩u 󵄩󵄩󵄩,

This completes the proof.

1 ⩽ k ⩽ n.

320 � 10 Difference methods for the Cahn–Hilliard equation 10.2.4 Convergence of the difference solution Denote c0 =

max

󵄨 󵄨󵄨 󵄨u(x, y, t)󵄨󵄨󵄨.

󵄨 ̄ (x,y)∈Ω,0⩽t⩽T

Introduce a second-order smooth function: Φ(u) = {

ϕ(u), 0,

|u| ⩽ c0 + 1, |u| ⩾ c0 + 2.

Then there is a constant c2 such that 󵄨󵄨 ′ 󵄨󵄨 󵄨󵄨Φ (u)󵄨󵄨 ⩽ c2 ,

󵄨󵄨 ′′ 󵄨󵄨 󵄨󵄨Φ (u)󵄨󵄨 ⩽ c2 .

Remark 10.3. When u ∈ [c0 +1, c0 +2], the function Φ(u) can be taken as the interpolation polynomial of degree 5 satisfying Φ(l) (c0 + 1) = ϕ(l) (c0 + 1),

Φ(l) (c0 + 2) = 0,

l = 0, 1, 2.

When u ∈ [−c0 − 2, −c0 − 1], the function Φ(u) can be taken as the interpolation polynomial of degree 5 satisfying Φ(l) (−c0 − 2) = 0,

Φ(l) (−c0 − 1) = ϕ(l) (−c0 − 1),

l = 0, 1, 2.

Remark 10.4. The auxiliary function Φ(u) is usually called a cut-off function, and the method to prove the convergence of difference schemes with the help of the cut off function is called the cut-off function method. Denote c3 =

max

󵄨󵄨 󵄨 󵄨ut (x, y, t)󵄨󵄨󵄨.

󵄨 ̄ (x,y)∈Ω,0⩽t⩽T

Consider the following difference scheme: k+ 1

k+ 1

{ δt uij 2 = Δh vij 2 , { { { { k+ 1 k+ 1 k+ 1 { vij 2 = Φ(uij 2 ) − αΔh uij 2 , { { { { 0 { uij = φ(xi , yj ),

0 ⩽ i ⩽ m1 , 0 ⩽ j ⩽ m2 , 0 ⩽ k ⩽ n − 1,

(10.38)

0 ⩽ i ⩽ m1 , 0 ⩽ j ⩽ m2 , 0 ⩽ k ⩽ n − 1,

(10.39)

0 ⩽ i ⩽ m1 , 0 ⩽ j ⩽ m2 .

(10.40)

Theorem 10.5. Suppose {Uijk , Vijk | 0 ⩽ i ⩽ m1 , 0 ⩽ j ⩽ m2 , 0 ⩽ k ⩽ n} and {uijk , vkij | 0 ⩽ i ⩽ m1 , 0 ⩽ j ⩽ m2 , 0 ⩽ k ⩽ n} are solutions of the problem (10.16)–(10.19) and the difference scheme (10.38)–(10.40), respectively. Let

10.2 Two-level nonlinear difference scheme

eijk = Uijk − uijk ,

� 321

0 ⩽ i ⩽ m1 , 0 ⩽ j ⩽ m2 , 0 ⩽ k ⩽ n.

Then there is a positive constant c5 such that 󵄩󵄩 k 󵄩󵄩 2 2 2 󵄩󵄩e 󵄩󵄩∞ ⩽ c5 (τ + h1 + h2 ),

1 ⩽ k ⩽ n.

(10.41)

Proof. Denote fijk = Vijk − vkij ,

0 ⩽ i ⩽ m1 , 0 ⩽ j ⩽ m2 , 0 ⩽ k ⩽ n.

k+ 1

k+ 1

Noticing Φ(Uij 2 ) = ϕ(Uij 2 ) and subtracting (10.38)–(10.40) from (10.24)–(10.25), (10.29), respectively, the system of error equations can be obtained as k+ 1

k+ 1

k+ 1

{ δt eij 2 = Δh fij 2 + Pij 2 , { { { { { { 0 ⩽ i ⩽ m1 , 0 ⩽ j ⩽ m2 , 0 ⩽ k ⩽ n − 1, { { { k+ 21 k+ 1 k+ 1 k+ 1 k+ 1 fij = Φ(Uij 2 ) − Φ(uij 2 ) − αΔh eij 2 + Qij 2 , { { { { { { 0 ⩽ i ⩽ m1 , 0 ⩽ j ⩽ m2 , 0 ⩽ k ⩽ n − 1, { { { { 0 { eij = 0, 0 ⩽ i ⩽ m1 , 0 ⩽ j ⩽ m2 . e

k+ 21

(10.42) (10.43) (10.44)

(I) Taking the inner product on both the right- and left-hand sides of (10.42) with , we have 1 1 1 1 1 󵄩󵄩 k+1 󵄩󵄩2 󵄩󵄩 k 󵄩󵄩2 (󵄩󵄩e 󵄩󵄩 − 󵄩󵄩e 󵄩󵄩 ) = (Δh f k+ 2 , ek+ 2 ) + (Pk+ 2 , ek+ 2 ) 2τ 1

1

1

1

= (f k+ 2 , Δh ek+ 2 ) + (Pk+ 2 , ek+ 2 ).

(10.45) 1

Taking the inner product on both the right- and left-hand sides of (10.43) with α1 f k+ 2 produces 1 󵄩󵄩 k+ 21 󵄩󵄩2 1 k+ 1 k+ 1 k+ 1 󵄩󵄩 = (Φ(U 2 ) − Φ(u 2 ), f 2 ) 󵄩󵄩f α α 1 1 1 1 1 − (Δh ek+ 2 , f k+ 2 ) + (Qk+ 2 , f k+ 2 ). α

(10.46)

Adding (10.45) with (10.46) leads to 1 󵄩󵄩 k+1 󵄩󵄩2 󵄩󵄩 k 󵄩󵄩2 1 󵄩 1 󵄩2 (󵄩󵄩e 󵄩󵄩 − 󵄩󵄩e 󵄩󵄩 ) + 󵄩󵄩󵄩f k+ 2 󵄩󵄩󵄩 2τ α 1 1 1 1 1 1 1 k+ 21 k+ 21 = (Φ(U ) − Φ(u ), f k+ 2 ) + (Pk+ 2 , ek+ 2 ) + (Qk+ 2 , f k+ 2 ) α α 1 1 1 󵄩 1󵄩 󵄩 1󵄩 󵄩 󵄩 󵄩 1󵄩 1󵄩 󵄩 󵄩 1󵄩 ⩽ c2 󵄩󵄩󵄩ek+ 2 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩f k+ 2 󵄩󵄩󵄩 + 󵄩󵄩󵄩Pk+ 2 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩ek+ 2 󵄩󵄩󵄩 + 󵄩󵄩󵄩Qk+ 2 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩f k+ 2 󵄩󵄩󵄩 α α 2 1 2 1 2 1 󵄩 1 󵄩2 c 󵄩 1 󵄩2 1󵄩 1 󵄩 1 󵄩2 1 󵄩 1 󵄩2 1 󵄩 󵄩 󵄩 ⩽ ( 󵄩󵄩󵄩f k+ 2 󵄩󵄩󵄩 + 2 󵄩󵄩󵄩ek+ 2 󵄩󵄩󵄩 ) + ( 󵄩󵄩󵄩Pk+ 2 󵄩󵄩󵄩 + 󵄩󵄩󵄩ek+ 2 󵄩󵄩󵄩 ) + ( 󵄩󵄩󵄩f k+ 2 󵄩󵄩󵄩 + 󵄩󵄩󵄩Qk+ 2 󵄩󵄩󵄩 ). 2α 2α 2 2 2α 2α

322 � 10 Difference methods for the Cahn–Hilliard equation Noticing (10.26) and (10.27), it follows: 1 󵄩󵄩 k+1 󵄩󵄩2 󵄩󵄩 k 󵄩󵄩2 (󵄩e 󵄩󵄩 − 󵄩󵄩e 󵄩󵄩 ) 2τ 󵄩 c2 󵄩 1 󵄩2 1 ⩽ (1 + 2 )󵄩󵄩󵄩ek+ 2 󵄩󵄩󵄩 + 2 α

1 󵄩󵄩 k+ 21 󵄩󵄩2 1 󵄩 k+ 1 󵄩2 󵄩󵄩P 󵄩󵄩 + 󵄩󵄩󵄩Q 2 󵄩󵄩󵄩 2 2α

c2 󵄩 1 󵄩2 1 1 1 2 ⩽ (1 + 2 )󵄩󵄩󵄩ek+ 2 󵄩󵄩󵄩 + (1 + )L1 L2 c12 (τ 2 + h12 + h22 ) , 2 α 2 α

When 21 (1 +

c22 )τ α

0 ⩽ k ⩽ n − 1.

⩽ 31 , it is not hard to get

c2 1 3 󵄩 k 󵄩2 3 󵄩󵄩 k+1 󵄩󵄩2 2 2 2 2 2 󵄩󵄩e 󵄩󵄩 ⩽ [1 + (1 + 2 )τ]󵄩󵄩󵄩e 󵄩󵄩󵄩 + (1 + )L1 L2 c1 τ(τ + h1 + h2 ) , 2 α 2 α

0 ⩽ k ⩽ n − 1.

The application of the Gronwall inequality (Theorem 1.2(a)) can yield 1 2 c2 3 2 󵄩󵄩 k 󵄩󵄩2 (1+ 2 )T (1 + α )L1 L2 c1 (τ 2 + h12 + h22 ) , 󵄩󵄩e 󵄩󵄩 ⩽ e 2 α c22 1+ α

1 ⩽ k ⩽ n.

(10.47)

(II) Taking the inner product on both the right- and left-hand sides of (10.42) with 1 δt ek+ 2 , we have 󵄩󵄩 k+ 21 󵄩󵄩2 k+ 1 k+ 1 k+ 1 k+ 1 󵄩󵄩δt e 󵄩󵄩 = (Δh f 2 , δt e 2 ) + (P 2 , δt e 2 ),

0 ⩽ k ⩽ n − 1. 1

Taking the inner product on both the right- and left-hand sides of (10.43) with Δh δt ek+ 2 produces 1

1

1

1

1

(f k+ 2 , Δh δt ek+ 2 ) = (Φ(U k+ 2 ) − Φ(uk+ 2 ), Δh δt ek+ 2 ) 1

1

− α(Δh ek+ 2 , Δh δt ek+ 2 ) 1

1

+ (Qk+ 2 , Δh δt ek+ 2 ),

0 ⩽ k ⩽ n − 1.

Adding the above two equalities together arrives at α 󵄩󵄩 k+1 󵄩󵄩2 󵄩󵄩 k 󵄩󵄩2 󵄩󵄩 k+ 21 󵄩󵄩2 (󵄩Δ e 󵄩󵄩 − 󵄩󵄩Δh e 󵄩󵄩 ) + 󵄩󵄩δt e 󵄩󵄩 2τ 󵄩 h 1

1

1

1

1

= (Φ(U k+ 2 ) − Φ(uk+ 2 ), Δh δt ek+ 2 ) + (Qk+ 2 , Δh δt ek+ 2 ) 1

1

+ (Pk+ 2 , δt ek+ 2 ),

0 ⩽ k ⩽ n − 1.

Replacing k with l in the equality above and summing over l from 0 to k lead to k 1 2 α 󵄩󵄩 k+1 󵄩󵄩2 󵄩󵄩 0 󵄩󵄩2 󵄩 󵄩 (󵄩󵄩Δh e 󵄩󵄩 − 󵄩󵄩Δh e 󵄩󵄩 ) + ∑󵄩󵄩󵄩δt el+ 2 󵄩󵄩󵄩 2τ l=0

10.2 Two-level nonlinear difference scheme k

1

1

� 323

1

= ∑(Φ(U l+ 2 ) − Φ(ul− 2 ), Δh δt el+ 2 ) l=0

k

k

1

1

1

1

+ ∑(Ql+ 2 , Δh δt el+ 2 ) + ∑(Pl+ 2 , δt el+ 2 ), l=0

l=0

0 ⩽ k ⩽ n − 1.

(10.48)

For the second term on the right-hand side of (10.48), one has k

1

1

∑(Ql+ 2 , Δh δt el+ 2 )

l=0

k 1 1 1 k = [∑(Ql+ 2 , Δh el+1 ) − ∑(Ql+ 2 , Δh el )] τ l=0 l=0

1

1

k 1 1 Ql+ 2 − Ql− 2 1 = [(Qk+ 2 , Δh ek+1 ) − (Q 2 , Δh e0 )] − ∑( , Δh el ) τ τ l=1 1 1 k 󵄩 1 󵄩󵄩 Ql+ 2 − Ql− 2 1 󵄩 󵄩 󵄩 󵄩 󵄩 1󵄩 󵄩 󵄩 ⩽ (󵄩󵄩󵄩Qk+ 2 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩Δh ek+1 󵄩󵄩󵄩 + 󵄩󵄩󵄩Q 2 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩Δh e0 󵄩󵄩󵄩) + ∑󵄩󵄩󵄩 󵄩 τ τ l=1 󵄩

󵄩󵄩 󵄩󵄩 󵄩󵄩 l 󵄩󵄩 󵄩󵄩 ⋅ 󵄩󵄩Δh e 󵄩󵄩. 󵄩󵄩

For the first term on the right-hand side of (10.48), one has k

1

1

1

∑(Φ(U l+ 2 ) − Φ(ul+ 2 ), Δh δt el+ 2 )

l=0

1 1 1 1 1 = [(Φ(U k+ 2 ) − Φ(uk+ 2 ), Δh ek+1 ) − (Φ(U 2 ) − Φ(u 2 ), Δh e0 )] τ

k

− ∑( l=1

1

1

1

1

[Φ(U l+ 2 ) − Φ(ul+ 2 )] − [Φ(U l− 2 ) − Φ(ul− 2 )] , Δh el ) τ

1 1 1 1 1 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 ⩽ (󵄩󵄩󵄩Φ(U k+ 2 ) − Φ(uk+ 2 )󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩Δh ek+1 󵄩󵄩󵄩 + 󵄩󵄩󵄩Φ(U 2 ) − Φ(u 2 )󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩Δh e0 󵄩󵄩󵄩) τ 1 1 1 1 k 󵄩 󵄩󵄩 [Φ(U l+ 2 ) − Φ(ul+ 2 )] − [Φ(U l− 2 ) − Φ(ul− 2 )] 󵄩󵄩󵄩 󵄩 l 󵄩 󵄩󵄩 ⋅ 󵄩󵄩Δh e 󵄩󵄩. + ∑󵄩󵄩󵄩 󵄩󵄩 󵄩 󵄩 τ 󵄩󵄩 󵄩

l=1

Noticing that there exist ρ ∈ (0, 1) and θ ∈ (0, 1) such that 1

1

1

1

[Φ(U l+ 2 ) − Φ(ul+ 2 )] − [Φ(U l− 2 ) − Φ(ul− 2 )] 1

1

1

1

= [Φ(U l− 2 + τΔt U l ) − Φ(ul− 2 + τΔt ul )] − [Φ(U l− 2 ) − Φ(ul− 2 )] 1

1

= Φ′ (U l− 2 + τρΔt U l )τΔt U l − Φ′ (ul− 2 + τρΔt ul )τΔt ul 1

= Φ′ (ul− 2 + τρΔt ul )τ(Δt U l − Δt ul ) 1

1

+ [Φ′ (U l− 2 + τρΔt U l ) − Φ′ (ul− 2 + τρΔt ul )]τΔt U l 1

= Φ′ (ul− 2 + τρΔt ul )τΔt el 1

1

1

+ Φ′′ (θ(U l− 2 + τρΔt U l ) + (1 − θ)(ul− 2 + τρΔt ul ))(el− 2 + τρΔt el )τΔt U l ,

(10.49)

324 � 10 Difference methods for the Cahn–Hilliard equation we get 󵄩󵄩 [Φ(U l+ 21 ) − Φ(ul+ 21 )] − [Φ(U l− 21 ) − Φ(ul− 21 )] 󵄩󵄩 󵄩󵄩 󵄩󵄩 󵄩󵄩 󵄩󵄩 󵄩󵄩 󵄩󵄩 τ 󵄩󵄩 l 󵄩󵄩 1 󵄩󵄩 l−1 󵄩󵄩 󵄩󵄩 l 󵄩󵄩 󵄩󵄩 l+1 󵄩󵄩 ⩽ c2 󵄩󵄩Δt e 󵄩󵄩 + c2 c3 (󵄩󵄩e 󵄩󵄩 + 󵄩󵄩e 󵄩󵄩 + 󵄩󵄩e 󵄩󵄩). 2 Therefore, k

1

1

1

∑(Φ(U l+ 2 ) − Φ(ul+ 2 ), Δh δt el+ 2 )

l=0

1 1 1 1 1 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 ⩽ (󵄩󵄩󵄩Φ(U k+ 2 ) − Φ(uk+ 2 )󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩Δh ek+1 󵄩󵄩󵄩 + 󵄩󵄩󵄩Φ(U 2 ) − Φ(u 2 )󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩Δh e0 󵄩󵄩󵄩) τ

k 󵄩 󵄩 1 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 + ∑[c2 󵄩󵄩󵄩Δt el 󵄩󵄩󵄩 + c2 c3 (󵄩󵄩󵄩el−1 󵄩󵄩󵄩 + 󵄩󵄩󵄩el 󵄩󵄩󵄩 + 󵄩󵄩󵄩el+1 󵄩󵄩󵄩)]󵄩󵄩󵄩Δh el 󵄩󵄩󵄩 2 l=1

k 1 󵄩 1󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 ⩽ c2 󵄩󵄩󵄩ek+ 2 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩Δh ek+1 󵄩󵄩󵄩 + c2 ∑󵄩󵄩󵄩Δt el 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩Δh el 󵄩󵄩󵄩 τ l=1 k 1 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩󵄩 󵄩 + c2 c3 ∑(󵄩󵄩󵄩el−1 󵄩󵄩󵄩 + 󵄩󵄩󵄩el 󵄩󵄩󵄩 + 󵄩󵄩󵄩el+1 󵄩󵄩󵄩)󵄩󵄩󵄩Δh el 󵄩󵄩󵄩. 2 l=1

Substituting (10.49) and (10.50) into (10.48) arrives at α 󵄩󵄩 k+1 󵄩󵄩2 k 󵄩󵄩 l+ 21 󵄩󵄩2 󵄩Δ e 󵄩󵄩 + ∑󵄩󵄩δt e 󵄩󵄩 2τ 󵄩 h l=0

1

1

k 󵄩 1 󵄩󵄩 Ql+ 2 − Ql− 2 1󵄩 󵄩 󵄩 󵄩 ⩽ 󵄩󵄩󵄩Qk+ 2 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩Δh ek+1 󵄩󵄩󵄩 + ∑󵄩󵄩󵄩 󵄩 τ τ l=1 󵄩

󵄩󵄩 󵄩󵄩 󵄩󵄩 l 󵄩󵄩 󵄩󵄩 ⋅ 󵄩󵄩Δh e 󵄩󵄩 󵄩󵄩

k

1 󵄩 1󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 + c2 󵄩󵄩󵄩ek+ 2 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩Δh ek+1 󵄩󵄩󵄩 + c2 ∑󵄩󵄩󵄩Δt el 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩Δh el 󵄩󵄩󵄩 τ l=1 k k 1 1 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩󵄩 󵄩 󵄩 󵄩 1󵄩 󵄩 + c2 c3 ∑(󵄩󵄩󵄩el−1 󵄩󵄩󵄩 + 󵄩󵄩󵄩el 󵄩󵄩󵄩 + 󵄩󵄩󵄩el+1 󵄩󵄩󵄩)󵄩󵄩󵄩Δh el 󵄩󵄩󵄩 + ∑󵄩󵄩󵄩Pl+ 2 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩δt el+ 2 󵄩󵄩󵄩 2 l=1 l=0

1 2 1 α󵄩 󵄩2 2 󵄩 󵄩 ⩽ [ 󵄩󵄩󵄩Δh ek+1 󵄩󵄩󵄩 + 󵄩󵄩󵄩Qk+ 2 󵄩󵄩󵄩 ] τ 8 α k 󵄩 l+ 1 l− 1 1 k󵄩 󵄩2 1 󵄩󵄩 Q 2 − Q 2 + ∑󵄩󵄩󵄩Δh el 󵄩󵄩󵄩 + ∑󵄩󵄩󵄩 2 l=1 2 l=1󵄩󵄩 τ

2 1 α󵄩 󵄩2 2c 󵄩 1 󵄩2 + [ 󵄩󵄩󵄩Δh ek+1 󵄩󵄩󵄩 + 2 󵄩󵄩󵄩ek+ 2 󵄩󵄩󵄩 ] τ 8 α

+

1 k 󵄩󵄩 l 󵄩󵄩2 1 2 k 󵄩󵄩 l 󵄩󵄩2 ∑󵄩Δ e 󵄩 + c2 ∑󵄩󵄩Δh e 󵄩󵄩 2 l=1󵄩 t 󵄩 2 l=1

󵄩󵄩2 󵄩󵄩 󵄩󵄩 󵄩󵄩

(10.50)

10.2 Two-level nonlinear difference scheme

+

cc k󵄩 3c2 c3 k 󵄩󵄩 l−1 󵄩󵄩2 󵄩󵄩 l 󵄩󵄩2 󵄩󵄩 l+1 󵄩󵄩2 󵄩2 ∑(󵄩󵄩e 󵄩󵄩 + 󵄩󵄩e 󵄩󵄩 + 󵄩󵄩e 󵄩󵄩 ) + 2 3 ∑󵄩󵄩󵄩Δh el 󵄩󵄩󵄩 8 l=1 2 l=1

+

1 k 󵄩󵄩 l+ 21 󵄩󵄩2 1 k 󵄩󵄩 l+ 21 󵄩󵄩2 ∑󵄩δ e 󵄩󵄩 + ∑󵄩󵄩P 󵄩󵄩 , 2 l=0󵄩 t 2 l=0

� 325

which implies 2 󵄩󵄩 k+1 󵄩󵄩2 8c2 󵄩󵄩 k+ 21 󵄩󵄩2 8 󵄩󵄩 k+ 21 󵄩󵄩2 󵄩󵄩Δh e 󵄩󵄩 ⩽ 2 󵄩󵄩e 󵄩󵄩 + 2 󵄩󵄩Q 󵄩󵄩 α α

+

k k 2τ 󵄩 󵄩 󵄩2 󵄩 󵄩2 󵄩 󵄩2 󵄩2 3c c (1 + c2 c3 + c22 ) ∑󵄩󵄩󵄩Δh el 󵄩󵄩󵄩 + 2 3 τ ∑(󵄩󵄩󵄩el−1 󵄩󵄩󵄩 + 󵄩󵄩󵄩el 󵄩󵄩󵄩 + 󵄩󵄩󵄩el+1 󵄩󵄩󵄩 ) α 2α l=1 l=1

+

2τ k 󵄩󵄩 l+ 21 󵄩󵄩2 k 󵄩󵄩󵄩󵄩 Ql+ 2 − Ql− 2 (∑󵄩P 󵄩󵄩 + ∑󵄩󵄩 󵄩 α l=0󵄩 τ l=1 󵄩

1

1

󵄩󵄩2 󵄩󵄩 󵄩󵄩 ), 󵄩󵄩

0 ⩽ k ⩽ n − 1.

Substituting (10.47) into the right-hand side of the inequality above and noticing (10.26)– (10.28), by the Gronwall inequality (Theorem 1.2(c)), we know that there is a constant c4 such that 󵄩󵄩 k 󵄩󵄩 2 2 2 󵄩󵄩Δh e 󵄩󵄩 ⩽ c4 (τ + h1 + h2 ),

1 ⩽ k ⩽ n.

(10.51)

The combination of (10.47), (10.51) and Lemma 10.2 shows that there is a constant c5 such that 󵄩󵄩 k 󵄩󵄩 2 2 2 󵄩󵄩e 󵄩󵄩∞ ⩽ c5 (τ + h1 + h2 ),

1 ⩽ k ⩽ n,

which reveals that the solution of the difference scheme (10.38)–(10.40) is convergent to the solution of (10.16)–(10.19), and the estimation (10.41) is true. It is noted that when c5 (τ 2 + h12 + h22 ) ⩽ 1, it follows: 󵄨󵄨 k 󵄨󵄨 󵄨󵄨 k 󵄨󵄨 󵄨󵄨 k 󵄨󵄨 󵄨󵄨uij 󵄨󵄨 ⩽ 󵄨󵄨Uij 󵄨󵄨 + 󵄨󵄨eij 󵄨󵄨 ⩽ c0 + 1,

0 ⩽ i ⩽ m1 , 0 ⩽ j ⩽ m2 , 0 ⩽ k ⩽ n.

Hence, k+ 21

Φ(uij

k+ 21

) = ϕ(uij

),

0 ⩽ i ⩽ m1 , 0 ⩽ j ⩽ m2 , 0 ⩽ k ⩽ n − 1,

which says that the difference scheme (10.38)–(10.40) is precisely the scheme (10.30)– (10.32). Therefore, the solution of the difference scheme (10.30)–(10.32) is also convergent to the solution of (10.16)–(10.19) with the estimation (10.41). The following theorem states this result. Theorem 10.6. Suppose {Uijk , Vijk | 0 ⩽ i ⩽ m1 , 0 ⩽ j ⩽ m2 , 0 ⩽ k ⩽ n} and {uijk , vkij | 0 ⩽ i ⩽ m1 , 0 ⩽ j ⩽ m2 , 0 ⩽ k ⩽ n} are solutions of the problem (10.16)–(10.19) and the difference scheme (10.30)–(10.32), respectively. Let

326 � 10 Difference methods for the Cahn–Hilliard equation eijk = Uijk − uijk ,

0 ⩽ i ⩽ m1 , 0 ⩽ j ⩽ m2 , 0 ⩽ k ⩽ n.

Then there is a constant c5 such that 󵄩󵄩 k 󵄩󵄩 2 2 2 󵄩󵄩e 󵄩󵄩∞ ⩽ c5 (τ + h1 + h2 ),

1⩽k⩽n

when τ 2 + h12 + h22 ⩽ 1/c5 .

10.3 Three-level linearized difference scheme Another linearized difference scheme for solving (10.16)–(10.19) will be developed in this section. 10.3.1 Derivation of the difference scheme Considering equations (10.16)–(10.17) at the point (xi , yj , t 1 ) and using Lemma 1.2 with 2 (10.18), (10.20)–(10.23), we obtain 1

1

0 2 2 { { δt Uij = Δh Vij + pij , 1 { { 21 0 2 ̂ { Vij = ϕ(uij ) − αΔh Uij + qij ,

0 ⩽ i ⩽ m1 , 0 ⩽ j ⩽ m2 ,

(10.52)

0 ⩽ i ⩽ m1 , 0 ⩽ j ⩽ m2 ,

(10.53)

where τ û ij = u(xi , yj , 0) + ut (xi , yj , 0), 2

0 ⩽ i ⩽ m1 , 0 ⩽ j ⩽ m2 ,

and there is a constant c6 such that 󵄨󵄨 0 󵄨󵄨 2 2 2 󵄨󵄨pij 󵄨󵄨 ⩽ c6 (τ + h1 + h2 ), { 󵄨 0󵄨 󵄨󵄨q 󵄨󵄨 ⩽ c6 (τ 2 + h2 + h2 ), 1 2 󵄨 ij 󵄨

0 ⩽ i ⩽ m1 , 0 ⩽ j ⩽ m2 ,

(10.54)

0 ⩽ i ⩽ m1 , 0 ⩽ j ⩽ m2 .

(10.55)

Considering equations (10.16)–(10.17) at the point (xi , yj , tk ) and using Lemma 1.2 with (10.18), (10.20)–(10.23), we obtain k k k { Δt Uij = Δh Vij + pij , { k k k̄ k { Vij = ϕ(Uij ) − αΔh Uij + qij ,

0 ⩽ i ⩽ m1 , 0 ⩽ j ⩽ m2 , 1 ⩽ k ⩽ n − 1,

(10.56)

0 ⩽ i ⩽ m1 , 0 ⩽ j ⩽ m2 , 1 ⩽ k ⩽ n − 1,

(10.57)

where there is a positive constant c7 such that { { { { { { { { {

󵄨󵄨 k 󵄨󵄨 2 2 2 󵄨󵄨pij 󵄨󵄨 ⩽ c7 (τ + h1 + h2 ), 󵄨󵄨 k 󵄨󵄨 2 2 2 󵄨󵄨qij 󵄨󵄨 ⩽ c7 (τ + h1 + h2 ), 󵄨󵄨 k 󵄨󵄨 2 2 2 󵄨󵄨Δt qij 󵄨󵄨 ⩽ c7 (τ + h1 + h2 ),

0 ⩽ i ⩽ m1 , 0 ⩽ j ⩽ m2 , 1 ⩽ k ⩽ n − 1,

(10.58)

0 ⩽ i ⩽ m1 , 0 ⩽ j ⩽ m2 , 1 ⩽ k ⩽ n − 1,

(10.59)

0 ⩽ i ⩽ m1 , 0 ⩽ j ⩽ m2 , 2 ⩽ k ⩽ n − 2.

(10.60)

10.3 Three-level linearized difference scheme

� 327

Noticing the initial value condition (10.19), we have Uij0 = φ(xi , yj ),

0 ⩽ i ⩽ m1 , 0 ⩽ j ⩽ m2 .

(10.61)

Omitting small terms in (10.52)–(10.53), (10.56)–(10.57) and noticing (10.61), the following difference scheme for solving (10.16)–(10.19) can be derived: { { { { { { { { { { { { { { { { { { { { { { { { {

1

1

δt uij2 = Δh vij2 , 1 2

1 2

vij = ϕ(û ij ) − αΔh uij ,

Δt uijk = Δh vkij ,

vkij = ϕ(uijk ) − uij0

̄ αΔh uijk ,

= φ(xi , yj ),

0 ⩽ i ⩽ m1 , 0 ⩽ j ⩽ m2 ,

(10.62)

0 ⩽ i ⩽ m1 , 0 ⩽ j ⩽ m2 ,

(10.63)

0 ⩽ i ⩽ m1 , 0 ⩽ j ⩽ m2 , 1 ⩽ k ⩽ n − 1,

(10.64)

0 ⩽ i ⩽ m1 , 0 ⩽ j ⩽ m2 , 1 ⩽ k ⩽ n − 1,

(10.65)

0 ⩽ i ⩽ m1 , 0 ⩽ j ⩽ m2 .

(10.66)

Substituting (10.63) and (10.65) into (10.62) and (10.64), respectively, yields 1

1

{ δt uij2 = Δh (ϕ(û ij ) − αΔh uij2 ), { { { { ̄ Δt uijk = Δh (ϕ(uijk ) − αΔh uijk ), { { { { { 0 { uij = φ(xi , yj ),

0 ⩽ i ⩽ m1 , 0 ⩽ j ⩽ m2 ,

(10.67)

0 ⩽ i ⩽ m1 , 0 ⩽ j ⩽ m2 , 1 ⩽ k ⩽ n − 1,

(10.68)

0 ⩽ i ⩽ m1 , 0 ⩽ j ⩽ m2 .

(10.69)

It is easy to see that (10.67)–(10.69) is a three-level linearized difference scheme. Now we have derived a new difference scheme (10.67)–(10.69) for solving (10.1)–(10.3).

10.3.2 Existence and uniqueness of the difference solution Theorem 10.7. The difference scheme (10.67)–(10.69) is uniquely solvable. Proof. The value of u0 is uniquely determined by (10.69) clearly. From (10.67), the linear system in u1 can be obtained. Consider its homogeneous one: α 1 1 u = − Δ2h uij1 , τ ij 2

0 ⩽ i ⩽ m1 , 0 ⩽ j ⩽ m2 .

(10.70)

Taking the inner product on both the right- and left-hand sides of (10.70) with u1 gives 1 󵄩󵄩 1 󵄩󵄩2 α 󵄩󵄩 1 󵄩2 󵄩u 󵄩 + 󵄩󵄩Δh u 󵄩󵄩󵄩 = 0, τ󵄩 󵄩 2 which implies ‖u1 ‖ = 0, i. e., (10.70) has only the trivial solution. Hence, the value of u1 is uniquely determined by (10.67).

328 � 10 Difference methods for the Cahn–Hilliard equation Now assume that the values of uk−1 and uk have been uniquely determined, then the linear system in uk+1 can be obtained from (10.68). Its homogeneous system reads 1 k+1 α u = − Δ2h uijk+1 , 2τ ij 2

0 ⩽ i ⩽ m1 , 0 ⩽ j ⩽ m2 .

Then ‖uk+1 ‖ = 0 can be shown in a similar manner. Thus, the value of uk+1 can be uniquely determined by (10.68). By induction, the solution of (10.67)–(10.69) is uniquely determined.

10.3.3 Convergence of the difference solution To discuss whether the solution of the difference scheme (10.67)–(10.69) is convergent or not, several lemmas need to be brought here. Lemma 10.3. Suppose u = (u0 , u1 , . . . , un ) and U = (U 0 , U 1 , . . . , U n ) are two mesh functions defined on Ωτ , then there exist ρ ∈ (0, 1) and θ ∈ (0, 1) such that Δt [ϕ(U k ) − ϕ(uk )] = ϕ′ (ρuk+1 + (1 − ρ)uk−1 )Δt (U k − uk )

+ ϕ′′ (θak + (1 − θbk ))[ρ(U k+1 − uk+1 ) + (1 − ρ)(U k−1 − uk−1 )]Δt U k ,

where ak = ρU k+1 + (1 − ρ)U k−1 ,

bk = ρuk+1 + (1 − ρ)uk−1 .

Proof. Some direct calculation yields Δt (ϕ(U k ) − ϕ(uk )) 1 {[ϕ(U k+1 ) − ϕ(uk+1 )] − [ϕ(U k−1 ) − ϕ(uk−1 )]} = 2τ 1 = {[ϕ(U k−1 + 2τΔt U k ) − ϕ(uk−1 + 2τΔt uk )] − [ϕ(U k−1 ) − ϕ(uk−1 )]} 2τ = ϕ′ (U k−1 + 2ρτΔt U k )Δt U k − ϕ′ (uk−1 + 2ρτΔt uk )Δt uk = ϕ (u ′

k−1

k

k

k

(10.71)

+ 2ρτΔt u )Δt (U − u )

+ [ϕ (U k−1 + 2ρτΔt U k ) − ϕ′ (uk−1 + 2ρτΔt uk )]Δt U k ′

= ϕ′ (ρuk+1 + (1 − ρ)uk−1 )Δt (U k − uk )

+ [ϕ′ (ρU k+1 + (1 − ρ)U k−1 ) − ϕ′ (ρuk+1 + (1 − ρ)uk−1 )]Δt U k

= ϕ′ (ρuk+1 + (1 − ρ)uk−1 )Δt (U k − uk )

+ ϕ′′ (θak + (1 − θ)bk )[ρ(U k+1 − uk+1 ) + (1 − ρ)(U k−1 − uk−1 )]Δt U k .

(10.72)

10.3 Three-level linearized difference scheme

� 329

It is worth to mention that the term ϕ(U k−1 + 2τρΔt U k ) − ϕ(uk−1 + 2τρΔt uk ) can be regarded as a function of ρ ∈ [0, 1] and the mean value theorem of differentiation is applied to get (10.71). The application again of the mean value theorem of differentiation can produce (10.72). Lemma 10.4. For any mesh functions u = (u0 , u1 , . . . , un ) and v = (v0 , v1 , . . . , vn ) defined on Ωτ , it holds that k

∑ ul Δt vl = l=1 k

∑ ul Δt vl = l=1

k−1 1 k k+1 (u v + uk−1 vk − u0 v1 − u1 v0 ) − ∑ (Δt ul )vl , 2τ l=1

1 ⩽ k ⩽ n − 1;

k−1 1 k k+1 (u v + uk−1 vk − u2 v1 − u1 v0 ) − ∑ (Δt ul )vl , 2τ l=2

2 ⩽ k ⩽ n − 1.

Proof. Noticing k

∑ ul Δt vl = l=1

=

1 k l l+1 ∑ u (v − vl−1 ) 2τ l=1 k−1 1 k l l+1 k l l−1 1 k+1 (∑ u v − ∑ u v ) = ( ∑ ul−1 vl − ∑ ul+1 vl ), 2τ l=1 2τ l=2 l=1 l=0

we have k

∑ ul Δt vl = l=1

=

k−1 1 k k+1 (u v + uk−1 vk − u0 v1 − u1 v0 ) − ∑ (Δt ul )vl 2τ l=1 k−1 1 k k+1 (u v + uk−1 vk − u2 v1 − u1 v0 ) − ∑ (Δt ul )vl . 2τ l=2

Now the convergence of the difference scheme (10.67)–(10.69) is concerned. Theorem 10.8. Suppose {Uijk | 0 ⩽ i ⩽ m1 , 0 ⩽ j ⩽ m2 , 0 ⩽ k ⩽ n} and {uijk | 0 ⩽ i ⩽ m1 , 0 ⩽ j ⩽ m2 , 0 ⩽ k ⩽ n} are solutions of the problem (10.1)–(10.3) and the difference scheme (10.67)–(10.69), respectively. Let eijk = Uijk − uijk ,

0 ⩽ i ⩽ m1 , 0 ⩽ j ⩽ m2 , 0 ⩽ k ⩽ n.

Then there is a constant c8 such that when τ 2 + h12 + h22 ⩽ 󵄩󵄩 k 󵄩󵄩 2 2 2 󵄩󵄩e 󵄩󵄩∞ ⩽ c8 (τ + h1 + h2 ),

1 , c8

we have

0 ⩽ k ⩽ n.

(10.73)

Proof. Denote

c0 =

fijk = Vijk − vkij , 0 ⩽ i ⩽ m1 , 0 ⩽ j ⩽ m2 , 0 ⩽ k ⩽ n. 󵄨 󵄨 󵄨 󵄨 󵄨 󵄨 max 󵄨󵄨󵄨u(x, y, t)󵄨󵄨󵄨, c9 = max 󵄨󵄨󵄨ϕ′ (u)󵄨󵄨󵄨, c10 = max 󵄨󵄨󵄨ϕ′′ (u)󵄨󵄨󵄨. ̄ |u|⩽c +1 |u|⩽c +1

(x,y)∈Ω,t∈[0,T]

0

0

330 � 10 Difference methods for the Cahn–Hilliard equation Subtracting (10.62)–(10.66) from (10.52)–(10.53), (10.56)–(10.57), (10.61), respectively, the system of error equations can be obtained as { { { { { { { { { { { { { { { { { { { { { { { { {

1

1

δt eij2 = Δh fij2 + p0ij , 1 2

1 2

fij = −αΔh eij + qij0 , Δt eijk = Δh fijk + pkij ,

(10.74)

0 ⩽ i ⩽ m1 , 0 ⩽ j ⩽ m2 ,

(10.75)

0 ⩽ i ⩽ m1 , 0 ⩽ j ⩽ m2 , 1 ⩽ k ⩽ n − 1, (10.76)

fijk = ϕ(Uijk ) − ϕ(uijk ) − eij0

0 ⩽ i ⩽ m1 , 0 ⩽ j ⩽ m2 ,

̄ αΔh eijk

+ qijk ,

= 0,

0 ⩽ i ⩽ m1 , 0 ⩽ j ⩽ m2 , 1 ⩽ k ⩽ n − 1, (10.77) 0 ⩽ i ⩽ m1 , 0 ⩽ j ⩽ m2 .

(10.78)

It is easy from (10.78) to find that 󵄩󵄩 0 󵄩󵄩 󵄩󵄩e 󵄩󵄩∞ = 0. (I) Estimation of ‖e1 ‖ and ‖Δh e1 ‖. 1 (a) Taking the inner product on both the right- and left-hand sides of (10.74) with e 2 gives 1

1

1

1

1

(δt e 2 , e 2 ) = (Δh f 2 , e 2 ) + (p0 , e 2 ); 1

Taking the inner product on both the right- and left-hand sides of (10.75) with α1 f 2 produces 1 1 1 󵄩󵄩 21 󵄩󵄩2 1 0 1 󵄩󵄩f 󵄩󵄩 = −(Δh e 2 , f 2 ) + (q , f 2 ). α α

Adding the above two equalities together arrives at 1 1 1 󵄩󵄩 1 󵄩󵄩2 󵄩󵄩 0 󵄩󵄩2 1 1 󵄩 1 󵄩2 (󵄩󵄩e 󵄩󵄩 − 󵄩󵄩e 󵄩󵄩 ) + 󵄩󵄩󵄩f 2 󵄩󵄩󵄩 = (p0 , e 2 ) + (q0 , f 2 ) 2τ α α 1 󵄩󵄩 0 󵄩󵄩2 󵄩󵄩 0 󵄩󵄩 󵄩󵄩 21 󵄩󵄩 1 󵄩󵄩 21 󵄩󵄩2 ⩽ 󵄩󵄩p 󵄩󵄩 ⋅ 󵄩󵄩e 󵄩󵄩 + 󵄩󵄩f 󵄩󵄩 + 󵄩q 󵄩 . α 4α 󵄩 󵄩

Noticing e0 = 0, it follows: 1 󵄩󵄩 1 󵄩󵄩2 󵄩󵄩 0 󵄩󵄩 1 󵄩󵄩 1 󵄩󵄩 1 󵄩󵄩 0 󵄩󵄩2 󵄩e 󵄩 ⩽ 󵄩󵄩p 󵄩󵄩 ⋅ 󵄩󵄩e 󵄩󵄩 + 󵄩q 󵄩 2τ 󵄩 󵄩 2 4α 󵄩 󵄩 1 󵄩󵄩 1 󵄩󵄩2 1 󵄩󵄩 0 󵄩󵄩2 1 󵄩󵄩 0 󵄩󵄩2 ⩽ 󵄩e 󵄩 + τ 󵄩󵄩p 󵄩󵄩 + 󵄩q 󵄩 . 4τ 󵄩 󵄩 4 4α 󵄩 󵄩 Hence, it follows from (10.54)–(10.55) that 󵄩󵄩 1 󵄩󵄩2 󵄩 0 󵄩2 1 󵄩 0 󵄩2 󵄩󵄩e 󵄩󵄩 ⩽ τ(τ 󵄩󵄩󵄩p 󵄩󵄩󵄩 + 󵄩󵄩󵄩q 󵄩󵄩󵄩 ) α 1 2 ⩽ τ(τ + )c62 L1 L2 (τ 2 + h12 + h22 ) . α

(10.79)

10.3 Three-level linearized difference scheme

� 331

(b) Taking the inner product on both the right- and left-hand sides of (10.74) with 1 δt e 2 yields 1 1 1 󵄩󵄩 21 󵄩󵄩2 0 󵄩󵄩δt e 󵄩󵄩 = (Δh f 2 , δt e 2 ) + (p , δt e 2 ). 1

Taking the inner product on both the right- and left-hand sides of (10.75) with Δh δt e 2 gives 1

1

1

1

1

(f 2 , Δh δt e 2 ) = −α(Δh e 2 , Δh δt e 2 ) + (q0 , Δh δt e 2 ). Adding the above two equalities together arrives at 1 1 1 1 󵄩󵄩 21 󵄩󵄩2 0 0 󵄩󵄩δt e 󵄩󵄩 + α(Δh e 2 , Δh δt e 2 ) = (p , δt e 2 ) + (q , Δh δt e 2 ).

Noticing e0 = 0, it follows: 1 󵄩󵄩 1 󵄩󵄩2 α 󵄩󵄩 1 󵄩󵄩2 1 0 1 1 0 1 󵄩e 󵄩 + 󵄩󵄩Δh e 󵄩󵄩 = (p , e ) + (q , Δh e ) 2τ τ τ τ2 󵄩 󵄩 1 󵄩 󵄩2 1 󵄩 󵄩2 1 󵄩 󵄩2 α 󵄩 󵄩2 ⩽ 2 󵄩󵄩󵄩e1 󵄩󵄩󵄩 + 󵄩󵄩󵄩p0 󵄩󵄩󵄩 + 󵄩󵄩󵄩Δh e1 󵄩󵄩󵄩 + 󵄩󵄩󵄩q0 󵄩󵄩󵄩 , 4 4τ ατ τ which can be reformulated as 1 󵄩 0 󵄩2 󵄩󵄩 1 󵄩󵄩2 4τ 1 󵄩󵄩 0 󵄩󵄩2 󵄩󵄩Δh e 󵄩󵄩 ⩽ ( 󵄩󵄩p 󵄩󵄩 + 󵄩󵄩󵄩q 󵄩󵄩󵄩 ) α 4 ατ τ 󵄩 󵄩2 4 󵄩 󵄩2 = 󵄩󵄩󵄩p0 󵄩󵄩󵄩 + 2 󵄩󵄩󵄩q0 󵄩󵄩󵄩 α α 4 2 τ 2 ⩽ ( + 2 )c6 L1 L2 (τ 2 + h12 + h22 ) . α α (II) Assume that (10.73) is true for 1 ⩽ k ⩽ m. Then when τ 2 +h12 +h22 ⩽ 󵄩󵄩 k 󵄩󵄩 󵄩󵄩e 󵄩󵄩∞ ⩽ 1,

(10.80) 1 , it holds that c8

1 ⩽ k ⩽ m.

Thus, 󵄩󵄩 k 󵄩󵄩 󵄩 k󵄩 󵄩 k󵄩 󵄩󵄩u 󵄩󵄩∞ ⩽ 󵄩󵄩󵄩U 󵄩󵄩󵄩∞ + 󵄩󵄩󵄩e 󵄩󵄩󵄩∞ ⩽ c0 + 1,

1⩽k⩽m

and 󵄨󵄨 󵄨 k󵄨 k k 󵄨 󵄨󵄨ϕ(Uij ) − ϕ(uij )󵄨󵄨󵄨 ⩽ c9 󵄨󵄨󵄨eij 󵄨󵄨󵄨,

0 ⩽ i ⩽ m1 , 0 ⩽ j ⩽ m2 , 1 ⩽ k ⩽ m.

(10.81)

By Lemma 10.3, we have 󵄨󵄨 󵄨 k󵄨 󵄨 k+1 󵄨 󵄨 k−1 󵄨 k k 󵄨 󵄨󵄨Δt [ϕ(Uij ) − ϕ(uij )]󵄨󵄨󵄨 ⩽ c9 󵄨󵄨󵄨Δt eij 󵄨󵄨󵄨 + c3 c10 (󵄨󵄨󵄨eij 󵄨󵄨󵄨 + 󵄨󵄨󵄨eij 󵄨󵄨󵄨), 0 ⩽ i ⩽ m1 , 0 ⩽ j ⩽ m2 , 1 ⩽ k ⩽ m − 1. Now we will show that (10.73) is also true for k = m + 1.

(10.82)

332 � 10 Difference methods for the Cahn–Hilliard equation (a) Taking the inner product on both the right- and left-hand sides of (10.76) with ek gives ̄

(Δt ek , ek ) = (Δh f k , ek ) + (pk , ek ), ̄

̄

̄

1 ⩽ k ⩽ m.

Taking the inner product on both the right- and left-hand sides of (10.77) with produces 1 󵄩󵄩 k 󵄩󵄩2 1 1 k k k k k k̄ k 󵄩󵄩f 󵄩󵄩 = (ϕ(U ) − ϕ(u ), f ) − (Δh e , f ) + (q , f ), α α α

1 ⩽ k ⩽ m.

Adding the above two equalities together arrives at 1 󵄩󵄩 k+1 󵄩󵄩2 󵄩󵄩 k−1 󵄩󵄩2 1 󵄩 󵄩2 (󵄩󵄩e 󵄩󵄩 − 󵄩󵄩e 󵄩󵄩 ) + 󵄩󵄩󵄩f k 󵄩󵄩󵄩 4τ α ̄ 1 1 k k k = (ϕ(U ) − ϕ(u ), f ) + (pk , ek ) + (qk , f k ), α α

1 ⩽ k ⩽ m.

It follows from (10.81) that 1 󵄩󵄩 k+1 󵄩󵄩2 󵄩󵄩 k−1 󵄩󵄩2 1 󵄩 󵄩2 (󵄩e 󵄩󵄩 − 󵄩󵄩e 󵄩󵄩 ) + 󵄩󵄩󵄩f k 󵄩󵄩󵄩 4τ 󵄩 α c 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 ̄󵄩 1 󵄩 󵄩 󵄩 󵄩 ⩽ 9 󵄩󵄩󵄩ek 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩f k 󵄩󵄩󵄩 + 󵄩󵄩󵄩pk 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩ek 󵄩󵄩󵄩 + 󵄩󵄩󵄩qk 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩f k 󵄩󵄩󵄩 α α 1 󵄩󵄩 k 󵄩󵄩2 1 2 󵄩󵄩 k 󵄩󵄩2 1 󵄩󵄩 k 󵄩󵄩2 ⩽ 󵄩󵄩f 󵄩󵄩 + c9 󵄩󵄩e 󵄩󵄩 + 󵄩󵄩p 󵄩󵄩 2α 2α 2 1 󵄩󵄩 k 󵄩󵄩2 1 󵄩󵄩 k 󵄩󵄩2 1 󵄩󵄩 k̄ 󵄩󵄩2 + 󵄩󵄩e 󵄩󵄩 + 󵄩󵄩f 󵄩󵄩 + 󵄩󵄩q 󵄩󵄩 , 1 ⩽ k ⩽ m, 2 2α 2α i. e., 1 󵄩 󵄩2 1 󵄩 ̄ 󵄩2 1 󵄩 󵄩2 1 󵄩 󵄩2 1 󵄩󵄩 k+1 󵄩󵄩2 󵄩󵄩 k−1 󵄩󵄩2 (󵄩e 󵄩󵄩 − 󵄩󵄩e 󵄩󵄩 ) ⩽ c92 󵄩󵄩󵄩ek 󵄩󵄩󵄩 + 󵄩󵄩󵄩ek 󵄩󵄩󵄩 + 󵄩󵄩󵄩pk 󵄩󵄩󵄩 + 󵄩󵄩󵄩qk 󵄩󵄩󵄩 4τ 󵄩 2α 2 2 2α 1 󵄩 󵄩2 1 󵄩 󵄩2 󵄩 󵄩2 ⩽ c92 󵄩󵄩󵄩ek 󵄩󵄩󵄩 + (󵄩󵄩󵄩ek+1 󵄩󵄩󵄩 + 󵄩󵄩󵄩ek−1 󵄩󵄩󵄩 ) 2α 4 1 󵄩 󵄩2 1 󵄩 󵄩2 + 󵄩󵄩󵄩pk 󵄩󵄩󵄩 + 󵄩󵄩󵄩qk 󵄩󵄩󵄩 , 1 ⩽ k ⩽ m. 2 2α Reformulating the above inequality gives 2 󵄩 󵄩2 󵄩 󵄩2 2c 󵄩 󵄩2 (1 − τ)󵄩󵄩󵄩ek+1 󵄩󵄩󵄩 ⩽ (1 + τ)󵄩󵄩󵄩ek−1 󵄩󵄩󵄩 + 9 τ 󵄩󵄩󵄩ek 󵄩󵄩󵄩 α 1 2 󵄩 󵄩2 󵄩 󵄩 + 2τ(󵄩󵄩󵄩pk 󵄩󵄩󵄩 + 󵄩󵄩󵄩qk 󵄩󵄩󵄩 ), 1 ⩽ k ⩽ m. α

When τ ⩽ 31 , it follows: 2 󵄩󵄩 k+1 󵄩󵄩2 󵄩 k−1 󵄩2 3c 󵄩 k 󵄩2 󵄩 k 󵄩2 1 󵄩 k 󵄩2 󵄩󵄩e 󵄩󵄩 ⩽(1 + 3τ)󵄩󵄩󵄩e 󵄩󵄩󵄩 + 9 τ 󵄩󵄩󵄩e 󵄩󵄩󵄩 + 3τ(󵄩󵄩󵄩p 󵄩󵄩󵄩 + 󵄩󵄩󵄩q 󵄩󵄩󵄩 ), α α

1 ⩽ k ⩽ m.

1 k f α

10.3 Three-level linearized difference scheme

� 333

It is not hard to get c2 󵄩2 󵄩 󵄩2 󵄩 󵄩2 󵄩 󵄩2 󵄩 max{󵄩󵄩󵄩ek+1 󵄩󵄩󵄩 , 󵄩󵄩󵄩ek 󵄩󵄩󵄩 } ⩽ [1 + 3(1 + 9 )τ] max{󵄩󵄩󵄩ek 󵄩󵄩󵄩 , 󵄩󵄩󵄩ek−1 󵄩󵄩󵄩 } α 󵄩󵄩 k 󵄩󵄩2 1 󵄩󵄩 k 󵄩󵄩2 + 3τ(󵄩󵄩p 󵄩󵄩 + 󵄩󵄩q 󵄩󵄩 ), 1 ⩽ k ⩽ m. α The application of (10.58) and (10.59) into the above inequality produces c2 󵄩 󵄩2 󵄩 󵄩2 󵄩 󵄩2 󵄩 󵄩2 max{󵄩󵄩󵄩ek+1 󵄩󵄩󵄩 , 󵄩󵄩󵄩ek 󵄩󵄩󵄩 } ⩽ [1 + 3(1 + 9 )τ] max{󵄩󵄩󵄩ek 󵄩󵄩󵄩 , 󵄩󵄩󵄩ek−1 󵄩󵄩󵄩 } α 1 2 + 3(1 + )L1 L2 c72 τ(τ 2 + h12 + h22 ) , 1 ⩽ k ⩽ m. α By the Gronwall inequality (Theorem 1.2(a)), (10.78) and (10.79), there is a positive constant c11 such that 2 󵄩 󵄩2 󵄩 󵄩2 2 max{󵄩󵄩󵄩ek+1 󵄩󵄩󵄩 , 󵄩󵄩󵄩ek 󵄩󵄩󵄩 } ⩽ c11 (τ 2 + h12 + h22 ) ,

1 ⩽ k ⩽ m,

which implies 󵄩󵄩 k 󵄩󵄩 2 2 2 󵄩󵄩e 󵄩󵄩 ⩽ c11 (τ + h1 + h2 ),

1 ⩽ k ⩽ m + 1.

(10.83)

(b) Taking the inner product on both the right- and left-hand sides of (10.76) with Δt ek gives 󵄩󵄩 k 󵄩󵄩2 k k k k 󵄩󵄩Δt e 󵄩󵄩 = (Δh f , Δt e ) + (p , Δt e ),

1 ⩽ k ⩽ m.

Taking the inner product on both the right- and left-hand sides of (10.77) with Δh Δt ek yields (f k , Δh Δt ek ) =(ϕ(U k ) − ϕ(uk ), Δh Δt ek ) − α(Δh ek , Δh Δt ek ) + (qk , Δh Δt ek ), ̄

1 ⩽ k ⩽ m.

Adding the above two equalities together arrives at 󵄩󵄩 k 󵄩󵄩2 k̄ k 󵄩󵄩Δt e 󵄩󵄩 + α(Δh e , Δh Δt e ) = (ϕ(U k ) − ϕ(uk ), Δh Δt ek ) + (qk , Δh Δt ek ) + (pk , Δt ek ),

1 ⩽ k ⩽ m.

Hence, 󵄩󵄩 k 󵄩󵄩2 α 󵄩󵄩 k+1 󵄩󵄩2 󵄩󵄩 k−1 󵄩󵄩2 󵄩󵄩Δt e 󵄩󵄩 + (󵄩󵄩Δh e 󵄩󵄩 − 󵄩󵄩Δh e 󵄩󵄩 ) 4τ k = (ϕ(U ) − ϕ(uk ), Δt Δh ek ) + (qk , Δt Δh ek ) + (pk , Δt ek ) ⩽ (ϕ(U k ) − ϕ(uk ), Δt Δh ek ) + (qk , Δt Δh ek ) 1󵄩 󵄩2 1 󵄩 󵄩2 + 󵄩󵄩󵄩Δt ek 󵄩󵄩󵄩 + 󵄩󵄩󵄩pk 󵄩󵄩󵄩 , 1 ⩽ k ⩽ m. 2 2

334 � 10 Difference methods for the Cahn–Hilliard equation Replacing k in the above result with l and summing over l from 1 to k, it follows: 1 k 󵄩󵄩 l 󵄩󵄩2 α 󵄩󵄩 k+1 󵄩󵄩2 󵄩󵄩 k 󵄩󵄩2 󵄩󵄩 1 󵄩󵄩2 󵄩󵄩 0 󵄩󵄩2 ∑󵄩Δ e 󵄩 + (󵄩󵄩Δh e 󵄩󵄩 + 󵄩󵄩Δh e 󵄩󵄩 − 󵄩󵄩Δh e 󵄩󵄩 − 󵄩󵄩Δh e 󵄩󵄩 ) 2 l=1󵄩 t 󵄩 4τ k

k

l=1

l=1

⩽ ∑(ϕ(U l ) − ϕ(ul ), Δt Δh el ) + ∑(ql , Δt Δh el ) + 1 ⩽ k ⩽ m.

1 k 󵄩󵄩 l 󵄩󵄩2 ∑󵄩p 󵄩 , 2 l=1󵄩 󵄩

(10.84)

Using the first equality in Lemma 10.4 and (10.78), we have k

∑(ϕ(U l ) − ϕ(ul ), Δt Δh el ) l=1

1 [(ϕ(U k−1 ) − ϕ(uk−1 ), Δh ek ) + (ϕ(U k ) − ϕ(uk ), Δh ek+1 )] 2τ

=

k−1

− ∑ (Δt (ϕ(U l ) − ϕ(ul )), Δh el ), l=1

1 ⩽ k ⩽ m.

By (10.81)–(10.82), we obtain k

∑(ϕ(U l ) − ϕ(ul ), Δt Δh el ) l=1

⩽

1 󵄩󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 (󵄩ϕ(U k−1 ) − ϕ(uk−1 )󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩Δh ek 󵄩󵄩󵄩 + 󵄩󵄩󵄩ϕ(U k ) − ϕ(uk )󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩Δh ek+1 󵄩󵄩󵄩) 2τ 󵄩 k−1

󵄨 󵄨 + ∑ 󵄨󵄨󵄨(Δt (ϕ(U l ) − ϕ(ul )), Δh el )󵄨󵄨󵄨 l=1

1 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 ⩽ (c9 󵄩󵄩󵄩ek−1 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩Δh ek 󵄩󵄩󵄩 + c9 󵄩󵄩󵄩ek 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩Δh ek+1 󵄩󵄩󵄩) 2τ k−1

󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 + ∑ (c9 󵄩󵄩󵄩Δt el 󵄩󵄩󵄩 + c3 c10 (󵄩󵄩󵄩el+1 󵄩󵄩󵄩 + 󵄩󵄩󵄩el−1 󵄩󵄩󵄩))󵄩󵄩󵄩Δh el 󵄩󵄩󵄩, l=1

1 ⩽ k ⩽ m.

(10.85)

It follows from the second equality in Lemma 10.4 that k

∑(ql , Δt Δh el ) l=1

= ⩽

k−1 1 [(qk , Δh ek+1 ) + (qk−1 , Δh ek ) − (q2 , Δh e1 )] − ∑ (Δt ql , Δh el ) 2τ l=2

1 󵄩󵄩 k 󵄩󵄩 󵄩󵄩 k+1 󵄩󵄩 󵄩󵄩 k−1 󵄩󵄩 󵄩󵄩 k 󵄩󵄩 󵄩󵄩 2 󵄩󵄩 󵄩󵄩 1 󵄩󵄩 (󵄩q 󵄩 ⋅ 󵄩Δ e 󵄩󵄩 + 󵄩󵄩q 󵄩󵄩 ⋅ 󵄩󵄩Δh e 󵄩󵄩 + 󵄩󵄩q 󵄩󵄩 ⋅ 󵄩󵄩Δh e 󵄩󵄩) 2τ 󵄩 󵄩 󵄩 h k−1

󵄩 󵄩 󵄩 󵄩 + ∑ 󵄩󵄩󵄩Δt ql 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩Δh el 󵄩󵄩󵄩, l=2

1 ⩽ k ⩽ m.

(10.86)

10.3 Three-level linearized difference scheme

� 335

Inserting (10.85) and (10.86) into (10.84) arrives at 1 k 󵄩󵄩 l 󵄩󵄩2 α 󵄩󵄩 k+1 󵄩󵄩2 󵄩󵄩 k 󵄩󵄩2 󵄩󵄩 1 󵄩󵄩2 ∑󵄩Δ e 󵄩 + (󵄩󵄩Δh e 󵄩󵄩 + 󵄩󵄩Δh e 󵄩󵄩 − 󵄩󵄩Δh e 󵄩󵄩 ) 2 l=1󵄩 t 󵄩 4τ ⩽

1 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 (c 󵄩󵄩ek−1 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩Δh ek 󵄩󵄩󵄩 + c9 󵄩󵄩󵄩ek 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩Δh ek+1 󵄩󵄩󵄩) 2τ 9 󵄩 k−1

󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 + ∑ (c9 󵄩󵄩󵄩Δt el 󵄩󵄩󵄩 + c3 c10 (󵄩󵄩󵄩el+1 󵄩󵄩󵄩 + 󵄩󵄩󵄩el−1 󵄩󵄩󵄩))󵄩󵄩󵄩Δh el 󵄩󵄩󵄩 l=1

+

1 󵄩󵄩 k 󵄩󵄩 󵄩󵄩 k+1 󵄩󵄩 󵄩󵄩 k−1 󵄩󵄩 󵄩󵄩 k 󵄩󵄩 󵄩󵄩 2 󵄩󵄩 󵄩󵄩 1 󵄩󵄩 (󵄩q 󵄩 ⋅ 󵄩Δ e 󵄩󵄩 + 󵄩󵄩q 󵄩󵄩 ⋅ 󵄩󵄩Δh e 󵄩󵄩 + 󵄩󵄩q 󵄩󵄩 ⋅ 󵄩󵄩Δh e 󵄩󵄩) 2τ 󵄩 󵄩 󵄩 h

k−1 k 󵄩 󵄩 󵄩 󵄩 1 󵄩 󵄩2 + ∑ 󵄩󵄩󵄩Δt ql 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩Δh el 󵄩󵄩󵄩 + ∑󵄩󵄩󵄩pl 󵄩󵄩󵄩 2 l=1 l=2

⩽

2 2 1 α 󵄩󵄩 k 󵄩󵄩2 2c9 󵄩󵄩 k−1 󵄩󵄩2 α 󵄩󵄩 k+1 󵄩󵄩2 2c9 󵄩󵄩 k 󵄩󵄩2 ( 󵄩󵄩Δh e 󵄩󵄩 + 󵄩󵄩e 󵄩󵄩 + 󵄩󵄩Δh e 󵄩󵄩 + 󵄩e 󵄩 ) 2τ 8 α 8 α 󵄩 󵄩 2 k−1 1󵄩 󵄩2 c 󵄩 󵄩2 + ∑ ( 󵄩󵄩󵄩Δt el 󵄩󵄩󵄩 + 9 󵄩󵄩󵄩Δh el 󵄩󵄩󵄩 + 2 2 l=1

1 󵄩󵄩 l+1 󵄩󵄩2 󵄩e 󵄩 2󵄩 󵄩

2 2 2 c32 c10 󵄩󵄩 l 󵄩󵄩2 1 󵄩󵄩 l−1 󵄩󵄩2 c3 c10 󵄩󵄩 l 󵄩󵄩2 󵄩󵄩Δh e 󵄩󵄩 + 󵄩󵄩e 󵄩󵄩 + 󵄩Δ e 󵄩 ) 2 2 2 󵄩 h 󵄩 1 α 󵄩󵄩 k+1 󵄩󵄩2 2 󵄩󵄩 k 󵄩󵄩2 α 󵄩󵄩 k 󵄩󵄩2 ( 󵄩Δ e 󵄩󵄩 + 󵄩󵄩q 󵄩󵄩 + 󵄩󵄩Δh e 󵄩󵄩 + 2τ 8 󵄩 h α 8

+

+

2 󵄩󵄩 k−1 󵄩󵄩2 α 󵄩󵄩 1 󵄩󵄩2 1 󵄩 2 󵄩2 1 k−1 󵄩 l 󵄩2 󵄩 l 󵄩2 󵄩󵄩q 󵄩󵄩 + 󵄩󵄩Δh e 󵄩󵄩 + 󵄩󵄩󵄩q 󵄩󵄩󵄩 ) + ∑ (󵄩󵄩󵄩Δt q 󵄩󵄩󵄩 + 󵄩󵄩󵄩Δh e 󵄩󵄩󵄩 ) α 2 2α 2 l=2

+

1 k 󵄩󵄩 l 󵄩󵄩2 ∑󵄩p 󵄩 , 2 l=1󵄩 󵄩

1 ⩽ k ⩽ m.

It can be obtained from the above inequality that α 󵄩󵄩 k+1 󵄩󵄩2 󵄩󵄩 k 󵄩󵄩2 (󵄩Δ e 󵄩󵄩 + 󵄩󵄩Δh e 󵄩󵄩 ) 8τ 󵄩 h 2 2 α󵄩 󵄩2 1 2c 󵄩 󵄩2 2c 󵄩 󵄩2 2 󵄩 󵄩2 2 󵄩 󵄩2 ⩽ 󵄩󵄩󵄩Δh e1 󵄩󵄩󵄩 + ( 9 󵄩󵄩󵄩ek−1 󵄩󵄩󵄩 + 9 󵄩󵄩󵄩ek 󵄩󵄩󵄩 + 󵄩󵄩󵄩qk 󵄩󵄩󵄩 + 󵄩󵄩󵄩qk−1 󵄩󵄩󵄩 4τ 2τ α α α α +

k−1 α 󵄩󵄩 1 󵄩󵄩2 1 󵄩 2 󵄩2 1 2 2 2 󵄩 l 󵄩2 󵄩󵄩Δh e 󵄩󵄩 + 󵄩󵄩󵄩q 󵄩󵄩󵄩 ) + ∑ [( c9 + c3 c10 )󵄩󵄩󵄩Δh e 󵄩󵄩󵄩 + 2 2α 2 l=1

+

1 k−1 󵄩󵄩 l 󵄩󵄩2 󵄩󵄩 l 󵄩󵄩2 1 k 󵄩 󵄩2 ∑ (󵄩󵄩Δt q 󵄩󵄩 + 󵄩󵄩Δh e 󵄩󵄩 ) + ∑󵄩󵄩󵄩pl 󵄩󵄩󵄩 , 2 l=2 2 l=1

which can be reformulated as 󵄩󵄩 k+1 󵄩󵄩2 󵄩󵄩 k 󵄩󵄩2 󵄩󵄩Δh e 󵄩󵄩 + 󵄩󵄩Δh e 󵄩󵄩

1 ⩽ k ⩽ m,

1 󵄩󵄩 l+1 󵄩󵄩2 󵄩e 󵄩 + 2󵄩 󵄩

1 󵄩󵄩 l−1 󵄩󵄩2 󵄩e 󵄩 ] 2󵄩 󵄩

336 � 10 Difference methods for the Cahn–Hilliard equation

⩽

8c92 󵄩󵄩 k−1 󵄩󵄩2 󵄩󵄩 k 󵄩󵄩2 (󵄩e 󵄩󵄩 + 󵄩󵄩e 󵄩󵄩 ) α2 󵄩 +

k−1 8τ 1 1 1 󵄩 󵄩2 󵄩2 2 󵄩 2 [( c92 + c32 c10 )󵄩󵄩󵄩Δh e1 󵄩󵄩󵄩 + ∑ ( c92 + c32 c10 + )󵄩󵄩󵄩Δh el 󵄩󵄩󵄩 ] α 2 2 2 l=2

+

k−2 4τ 󵄩󵄩 0 󵄩󵄩2 󵄩󵄩 1 󵄩󵄩2 󵄩󵄩 k−1 󵄩󵄩2 󵄩󵄩 k 󵄩󵄩2 󵄩 󵄩2 [󵄩󵄩e 󵄩󵄩 + 󵄩󵄩e 󵄩󵄩 + 󵄩󵄩e 󵄩󵄩 + 󵄩󵄩e 󵄩󵄩 + 2 ∑ 󵄩󵄩󵄩el 󵄩󵄩󵄩 ] α l=2

1 󵄩 󵄩2 󵄩 󵄩2 4 2 󵄩 󵄩2 2 󵄩 󵄩2 α 󵄩 󵄩2 + 2󵄩󵄩󵄩Δh e1 󵄩󵄩󵄩 + ( 󵄩󵄩󵄩qk 󵄩󵄩󵄩 + 󵄩󵄩󵄩qk−1 󵄩󵄩󵄩 + 󵄩󵄩󵄩Δh e1 󵄩󵄩󵄩 + 󵄩󵄩󵄩q2 󵄩󵄩󵄩 ) α α α 2 2α +

4τ k−1󵄩󵄩 l 󵄩󵄩2 4τ k 󵄩󵄩 l 󵄩󵄩2 ∑ 󵄩Δ q 󵄩 + ∑󵄩p 󵄩 , α l=2 󵄩 t 󵄩 α l=1󵄩 󵄩

1 ⩽ k ⩽ m.

Noticing (10.83), (10.58)–(10.60), (10.78) and (10.80), there are two constants c12 and c13 such that 󵄩󵄩 k+1 󵄩󵄩2 󵄩󵄩 k 󵄩󵄩2 󵄩󵄩Δh e 󵄩󵄩 + 󵄩󵄩Δh e 󵄩󵄩 k

2 󵄩 󵄩2 󵄩 󵄩2 ⩽ c12 τ ∑(󵄩󵄩󵄩Δh el 󵄩󵄩󵄩 + 󵄩󵄩󵄩Δh el−1 󵄩󵄩󵄩 ) + c13 (τ 2 + h12 + h22 ) , l=2

1 ⩽ k ⩽ m.

The application of the Gronwall inequality (Theorem 1.2(c)) produces 󵄩󵄩 k+1 󵄩󵄩2 󵄩󵄩 k 󵄩󵄩2 c kτ 2 2 2 2 󵄩󵄩Δh e 󵄩󵄩 + 󵄩󵄩Δh e 󵄩󵄩 ⩽ e 12 ⋅ c13 (τ + h1 + h2 ) ,

1 ⩽ k ⩽ m,

which implies 1 󵄩󵄩 k 󵄩󵄩 2 2 2 c T 󵄩󵄩Δh e 󵄩󵄩 ⩽ e 2 12 √c13 (τ + h1 + h2 ),

1 ⩽ k ⩽ m + 1.

(10.87)

Using Lemma 10.2, (10.83) and (10.87), one gets 1 1 1 󵄩󵄩 m+1 󵄩󵄩 󵄩 m+1 󵄩 󵄩 m+1 󵄩 󵄩󵄩e 󵄩󵄩∞ ⩽ 󵄩󵄩󵄩Δh e 󵄩󵄩󵄩 + √3[1 + ( + )]󵄩󵄩󵄩e 󵄩󵄩󵄩 2 L1 L2

1 1 1 1 ⩽ {e 2 c12 T √c13 + √3[1 + ( + )]c11 }(τ 2 + h12 + h22 ) 2 L1 L2

≡ c8 (τ 2 + h12 + h22 ),

which says that (10.73) is true for k = m + 1. By induction, (10.73) is true for all k (0 ⩽ k ⩽ n).

10.4 Three-level linearized compact difference scheme In this section, another higher-order three-level linearized difference scheme for solving (10.16)–(10.19) will be proposed.

10.4 Three-level linearized compact difference scheme

For any u ∈ 𝒱h , introduce the average operators: { { { { 𝒜1 uij = { { { { { { { { { 𝒜2 uij = { { { { {

5 u + 61 u1j , 6 0j 1 (u + 10uij + 12 i−1,j 1 + 65 um1 ,j , u 6 m1 −1,j 5 u + 61 ui1 , 6 i0 1 (u 12 i,j−1 1 u 6 i,m2 −1

i = 0, ui+1,j ),

+ 10uij + ui,j+1 ), +

5 u , 6 i,m2

1 ⩽ i ⩽ m1 − 1, i = m1 , j = 0, 1 ⩽ j ⩽ m2 − 1, j = m2 .

Denote 𝒜h = 𝒜1 𝒜2 ,

Λh = 𝒜2 δx2 + 𝒜1 δy2 .

Some useful lemmas will be prepared in advance. Lemma 10.5 ([21]). For any u ∈ 𝒱h , it holds that 5 ‖u‖2 ⩽ ‖𝒜1 u‖2 ⩽ ‖u‖2 , 12 5 ‖u‖2 ⩽ ‖𝒜2 u‖2 ⩽ ‖u‖2 , 12 25 ‖u‖2 ⩽ ‖𝒜h u‖2 ⩽ ‖u‖2 . 144 Lemma 10.6. For any u ∈ 𝒱h , it holds that 3 ‖Δh u‖2 ⩽ (3 + √2)‖Λh u‖2 . 7 Proof. By (10.13), one has 󵄩󵄩 2 󵄩󵄩2 󵄩󵄩 2 󵄩󵄩2 󵄩󵄩 󵄩2 󵄩󵄩δx u󵄩󵄩 + 󵄩󵄩δy u󵄩󵄩 ⩽ 󵄩󵄩Δh u󵄩󵄩󵄩 . Noticing Δh uij = Λh uij −

1 1 2 2 2 h δ δ u − h2 δ 2 δ 2 u , 12 2 y x ij 12 1 x y ij

one has 1 2 󵄩󵄩 2 2 󵄩󵄩 1 2 󵄩󵄩 2 2 󵄩󵄩 h 󵄩δ δ u󵄩 + h 󵄩δ δ u󵄩 12 2 󵄩 y x 󵄩 12 1 󵄩 x y 󵄩 1 󵄩 󵄩 󵄩 󵄩 ⩽ ‖Λh u‖ + (󵄩󵄩󵄩δx2 u󵄩󵄩󵄩 + 󵄩󵄩󵄩δy2 u󵄩󵄩󵄩) 3 1 󵄩 󵄩2 󵄩 󵄩2 ⩽ ‖Λh u‖ + √2(󵄩󵄩󵄩δx2 u󵄩󵄩󵄩 + 󵄩󵄩󵄩δy2 u󵄩󵄩󵄩 ) 3

‖Δh u‖ ⩽ ‖Λh u‖ +

� 337

338 � 10 Difference methods for the Cahn–Hilliard equation

⩽ ‖Λh u‖ +

√2 3

‖Δh u‖,

which implies ‖Δh u‖ ⩽

1 1−

√2 3

‖Λh u‖ =

3(3 + √2) ‖Λh u‖. 7

10.4.1 Derivation of the difference scheme Differentiating (10.16) with respect to x three times and using (10.20) with (10.22), we have vxxxxx |x=0 = 0,

vxxxxx |x=L1 = 0.

(10.88)

Differentiating (10.16) with respect to y three times and using (10.21) with (10.23), we have vyyyyy |y=0 = 0,

vyyyyy |y=L2 = 0.

(10.89)

Differentiating (10.17) with respect to x three times and using (10.18), (10.20), (10.22), we have uxxxxx |x=0 = 0,

uxxxxx |x=L1 = 0.

(10.90)

Differentiating (10.17) with respect to y three times and using (10.18), (10.21), (10.23), we have uyyyyy |y=0 = 0,

uyyyyy |y=L2 = 0.

(10.91)

Considering equations (10.16)–(10.17) at the point (xi , yj , t 1 ) gives 2

ut (xi , yj , t 1 ) = vxx (xi , yj , t 1 ) + vyy (xi , yj , t 1 ), 0 ⩽ i ⩽ m1 , 0 ⩽ j ⩽ m2 , { 2 2 2 { { v(xi , yj , t 1 ) = ϕ(u(xi , yj , t 1 )) − α[uxx (xi , yj , t 1 ) + uyy (xi , yj , t 1 )], { { 2 2 2 2 { { 0 ⩽ i ⩽ m1 , 0 ⩽ j ⩽ m2 . Performing the operator 𝒜1 𝒜2 on both the right- and left-hand sides of the above two equalities and using Lemma 2.2, (10.18), (10.20)–(10.23) with (10.88)–(10.91), we get 1

1

1

2 2 0 𝒜1 𝒜2 δt Uij2 = 𝒜2 δx Vij2 + 𝒜1 δy Vij2 + p̂ ij , 0 ⩽ i ⩽ m1 , 0 ⩽ j ⩽ m2 , { { { { 1 1 1 2 2 2 2 2 { ̂0 ̂ 𝒜 𝒜 V = 𝒜 𝒜 ϕ( u ) − α( 𝒜 δ U + 𝒜 δ U { 1 2 1 2 ij 2 1 x ij y ij ) + qij , { ij { { 0 ⩽ i ⩽ m1 , 0 ⩽ j ⩽ m2 ,

(10.92) (10.93)

10.4 Three-level linearized compact difference scheme

� 339

where τ û ij = u(xi , yj , 0) + ut (xi , yj , 0), 2

0 ⩽ i ⩽ m1 , 0 ⩽ j ⩽ m2 ,

and there is a constant c14 such that 󵄨󵄨 ̂ 0 󵄨󵄨 2 4 4 󵄨󵄨pij 󵄨󵄨 ⩽ c14 (τ + h1 + h2 ), { 󵄨 0󵄨 󵄨󵄨q̂ 󵄨󵄨 ⩽ c14 (τ 2 + h4 + h4 ), 1 2 󵄨 ij 󵄨

0 ⩽ i ⩽ m1 , 0 ⩽ j ⩽ m2 ,

(10.94)

0 ⩽ i ⩽ m1 , 0 ⩽ j ⩽ m2 .

(10.95)

Considering equations (10.16)–(10.17) at the point (xi , yj , tk ) yields ut (xi , yj , tk ) = vxx (xi , yj , tk ) + vyy (xi , yj , tk ), { { { { { 0 ⩽ i ⩽ m1 , 0 ⩽ j ⩽ m2 , 1 ⩽ k ⩽ n − 1, { { v(xi , yj , tk ) = ϕ(u(xi , yj , tk )) − α[uxx (xi , yj , tk ) + uyy (xi , yj , tk )], { { { { 0 ⩽ i ⩽ m1 , 0 ⩽ j ⩽ m2 , 1 ⩽ k ⩽ n − 1. Performing the operator 𝒜1 𝒜2 on both the right- and left-hand sides of the above two

equalities and using Lemma 2.2, (10.18), (10.20)–(10.23) with (10.88)–(10.91) again, we get k 2 k 2 k k 𝒜1 𝒜2 Δt Uij = 𝒜2 δx Vij + 𝒜1 δy Vij + p̂ ij , { { { { { { 0 ⩽ i ⩽ m1 , 0 ⩽ j ⩽ m2 , 1 ⩽ k ⩽ n − 1, { k k 2 k̄ 2 k̄ k { { 𝒜1 𝒜2 Vij = 𝒜1 𝒜2 ϕ(Uij ) − α(𝒜2 δx Uij + 𝒜1 δy Uij ) + q̂ ij , { { { { 0 ⩽ i ⩽ m1 , 0 ⩽ j ⩽ m2 , 1 ⩽ k ⩽ n − 1,

(10.96) (10.97)

where there is a constant c15 such that { { { { { { { { {

󵄨󵄨 ̂ k 󵄨󵄨 2 4 4 󵄨󵄨pij 󵄨󵄨 ⩽ c15 (τ + h1 + h2 ), 󵄨󵄨 ̂ k 󵄨󵄨 2 4 4 󵄨󵄨qij 󵄨󵄨 ⩽ c15 (τ + h1 + h2 ), 󵄨󵄨 ̂ k 󵄨󵄨 2 4 4 󵄨󵄨Δt qij 󵄨󵄨 ⩽ c15 (τ + h1 + h2 ),

0 ⩽ i ⩽ m1 , 0 ⩽ j ⩽ m2 , 1 ⩽ k ⩽ n − 1,

(10.98)

0 ⩽ i ⩽ m1 , 0 ⩽ j ⩽ m2 , 1 ⩽ k ⩽ n − 1,

(10.99)

0 ⩽ i ⩽ m1 , 0 ⩽ j ⩽ m2 , 2 ⩽ k ⩽ n − 2.

(10.100)

Omitting small terms in (10.92)–(10.93), (10.96)–(10.97) and noticing the initial value

condition

Uij0 = φ(xi , yj ),

0 ⩽ i ⩽ m1 , 0 ⩽ j ⩽ m2 ,

(10.101)

340 � 10 Difference methods for the Cahn–Hilliard equation we can deduce a difference scheme for solving (10.16)–(10.19), given by 1

{ { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { {

2

1

1

2

𝒜1 𝒜2 δt uij2 = 𝒜2 δx vij2 + 𝒜1 δy vij2 , 1 2

2

0 ⩽ i ⩽ m1 , 0 ⩽ j ⩽ m2 , 1 2

2

𝒜1 𝒜2 vij = 𝒜1 𝒜2 ϕ(û ij ) − α(𝒜2 δx uij + 𝒜1 δy uij ),

0 ⩽ i ⩽ m1 , 0 ⩽ j ⩽ m2 ,

k 𝒜1 𝒜2 Δt uij

=

k 𝒜1 𝒜2 vij

k 𝒜1 𝒜2 ϕ(uij )

2 k 𝒜2 δx vij

+

2 k 𝒜1 δy vij ,

0 ⩽ i ⩽ m1 , 0 ⩽ j ⩽ m2 , 1 ⩽ k ⩽ n − 1,

uij0

=

−

̄ α(𝒜2 δx2 uijk

+

2 k̄ 𝒜1 δy uij ),

0 ⩽ i ⩽ m1 , 0 ⩽ j ⩽ m2 , 1 ⩽ k ⩽ n − 1, = φ(xi , yj ),

(10.102)

1 2

0 ⩽ i ⩽ m1 , 0 ⩽ j ⩽ m2 .

(10.103) (10.104) (10.105) (10.106)

It follows from (10.102)–(10.105) that { { { { { { { { { { { { { { { { { { {

1

1

𝒜h δt uij2 = Λh vij2 ,

0 ⩽ i ⩽ m1 , 0 ⩽ j ⩽ m2 ,

(10.107)

𝒜h vij = 𝒜h ϕ(û ij ) − αΛh uij ,

0 ⩽ i ⩽ m1 , 0 ⩽ j ⩽ m2 ,

(10.108)

𝒜h Δt uij = Λh vij ,

0 ⩽ i ⩽ m1 , 0 ⩽ j ⩽ m2 , 1 ⩽ k ⩽ n − 1, (10.109)

1 2

1 2

k

k

k

k

𝒜h vij = 𝒜h ϕ(uij ) −

̄ αΛh uijk ,

0 ⩽ i ⩽ m1 , 0 ⩽ j ⩽ m2 , 1 ⩽ k ⩽ n − 1.

(10.110)

Performing the operator 𝒜h on both the right- and left-hand sides of (10.107) and substituting (10.108) into the resultant equality yield 2

1

1

𝒜h δt uij2 = Λh (𝒜h ϕ(û ij ) − αΛh uij2 ),

0 ⩽ i ⩽ m1 , 0 ⩽ j ⩽ m2 .

Performing the operator 𝒜h on both the right- and left-hand sides of (10.109) and substituting (10.110) into the resultant equality yield 2

k

k

k̄

𝒜h Δt uij = Λh (𝒜h ϕ(uij ) − αΛh uij ),

0 ⩽ i ⩽ m1 , 0 ⩽ j ⩽ m2 , 1 ⩽ k ⩽ n − 1.

Hence, a difference scheme for solving (10.1)–(10.3) can be derived in the form of 1

1

2 { 𝒜h δt uij2 = Λh (𝒜h ϕ(û ij ) − αΛh uij2 ), { { { { 2 k k k̄ 𝒜h Δt uij = Λh (𝒜h ϕ(uij ) − αΛh uij ), { { { { { 0 { uij = φ(xi , yj ),

0 ⩽ i ⩽ m1 , 0 ⩽ j ⩽ m2 ,

(10.111)

0 ⩽ i ⩽ m1 , 0 ⩽ j ⩽ m2 , 1 ⩽ k ⩽ n − 1,

(10.112)

0 ⩽ i ⩽ m1 , 0 ⩽ j ⩽ m2 .

(10.113)

10.4.2 Existence and uniqueness of the difference solution Theorem 10.9. The difference scheme (10.111)–(10.113) is uniquely solvable.

10.4 Three-level linearized compact difference scheme

� 341

Proof. Define uk = {uijk | 0 ⩽ i ⩽ m1 , 0 ⩽ j ⩽ m2 }.

The value of u0 is uniquely determined by (10.113). From (10.111), the linear system in u1 can be obtained. Consider its homogeneous one: 2

1 τ

1 2

1

1

𝒜h ( uij ) = Λh (−αΛh ( uij )),

0 ⩽ i ⩽ m1 , 0 ⩽ j ⩽ m2 .

Taking the inner product on both the right- and left-hand sides of the above equality with u1 produces 1 1 2 1 1 (𝒜 u , u ) + α(Λ2h u1 , u1 ) = 0, τ h 2 i. e., α 1 (𝒜 u1 , 𝒜h u1 ) + (Λh u1 , Λh u1 ) = 0, τ h 2 or 1 󵄩󵄩 α󵄩 1 󵄩2 1 󵄩2 󵄩𝒜 u 󵄩󵄩 + 󵄩󵄩󵄩Λh u 󵄩󵄩󵄩 = 0. τ󵄩 h 󵄩 2 In view of Lemma 10.5, we have 󵄩󵄩 1 󵄩󵄩 󵄩󵄩u 󵄩󵄩 = 0, which shows that (10.111) is uniquely solvable with respect to u1 . Now assume that the values of uk−1 and uk have been uniquely determined. Then the linear system in uk+1 can be obtained from (10.112). Consider its homogeneous one: 2

𝒜h (

1 1 k+1 u ) = Λh (−αΛh ( uijk+1 )), 2τ ij 2

0 ⩽ i ⩽ m1 , 0 ⩽ j ⩽ m2 .

Taking the inner product on both the right- and left-hand sides with uk+1 gives 1 󵄩󵄩 α󵄩 k+1 󵄩2 k+1 󵄩2 󵄩󵄩𝒜h u 󵄩󵄩󵄩 + 󵄩󵄩󵄩Λh u 󵄩󵄩󵄩 = 0. 2τ 2 It is easy from Lemma 10.5 to get 󵄩󵄩 k+1 󵄩󵄩 󵄩󵄩u 󵄩󵄩 = 0. Consequently, the value of uk+1 is uniquely determined by (10.112). By induction, the conclusion of this theorem is true.

342 � 10 Difference methods for the Cahn–Hilliard equation 10.4.3 Convergence of the difference solution Theorem 10.10. Suppose {Uijk | 0 ⩽ i ⩽ m1 , 0 ⩽ j ⩽ m2 , 0 ⩽ k ⩽ n} and {uijk | 0 ⩽ i ⩽ m1 , 0 ⩽ j ⩽ m2 , 0 ⩽ k ⩽ n} are solutions of the problem (10.1)–(10.3) and the difference scheme (10.111)–(10.113), respectively. Denote eijk = Uijk − uijk ,

0 ⩽ i ⩽ m1 , 0 ⩽ j ⩽ m2 , 0 ⩽ k ⩽ n.

Then there is a constant c16 such that when τ 2 + h14 + h24 ⩽ 󵄩󵄩 k 󵄩󵄩 2 4 4 󵄩󵄩e 󵄩󵄩∞ ⩽ c16 (τ + h1 + h2 ),

1 , c16

we have

0 ⩽ k ⩽ n.

(10.114)

Proof. Denote fijk = Vijk − vkij ,

0 ⩽ i ⩽ m1 , 0 ⩽ j ⩽ m2 , 0 ⩽ k ⩽ n.

Subtracting (10.102)–(10.106) from (10.92)–(10.93), (10.96)–(10.97), (10.101), respectively, the system of error equations can be obtained as { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { {

1

1

0

𝒜h δt eij2 = Λh fij2 + p̂ ij , 1 2

1 2

0

𝒜h fij = −αΛh eij + q̂ ij , k

k

k

𝒜h Δt eij = Λh fij + p̂ ij , k

k

0 ⩽ i ⩽ m1 , 0 ⩽ j ⩽ m2 ,

(10.115)

0 ⩽ i ⩽ m1 , 0 ⩽ j ⩽ m2 ,

(10.116)

0 ⩽ i ⩽ m1 , 0 ⩽ j ⩽ m2 , 1 ⩽ k ⩽ n − 1, k

𝒜h fij = 𝒜h (ϕ(Uij ) − ϕ(uij )) −

eij0

̄ αΛh eijk

+ q̂ ijk ,

0 ⩽ i ⩽ m1 , 0 ⩽ j ⩽ m2 , 1 ⩽ k ⩽ n − 1, = 0,

(10.117) (10.118)

0 ⩽ i ⩽ m1 , 0 ⩽ j ⩽ m2 .

(10.119)

From (10.119), it is obvious that ‖e0 ‖∞ = 0, so that (10.114) is true for k = 0. (I) Estimation of ‖𝒜h e1 ‖ and ‖Λh e1 ‖. (a) Taking the inner product on both the right- and left-hand sides of (10.115) with 1 𝒜h e 2 yields 1

1

1

1

1

(𝒜h δt e 2 , 𝒜h e 2 ) = (𝒜h e 2 , Λh f 2 ) + (𝒜h e 2 , p̂ 0 ). Taking the inner product on both the right- and left-hand sides of (10.116) with α1 𝒜h f produces 1 1 2 1 1 1 1 󵄩󵄩 󵄩 0 󵄩󵄩𝒜h f 2 󵄩󵄩󵄩 = −(𝒜h f 2 , Λh e 2 ) + (𝒜h f 2 , q̂ ). α α

Adding the above two equalities together gives

1 2

10.4 Three-level linearized compact difference scheme 1 2 1󵄩 1 󵄩󵄩 󵄩2 󵄩2 󵄩 󵄩 (󵄩󵄩𝒜h e1 󵄩󵄩󵄩 − 󵄩󵄩󵄩𝒜h e0 󵄩󵄩󵄩 ) + 󵄩󵄩󵄩𝒜h f 2 󵄩󵄩󵄩 2τ α 1 1 1 = (𝒜h e 2 , p̂ 0 ) + (𝒜h f 2 , q̂ 0 ) α 1 1 1 = (𝒜h e1 , p̂ 0 ) + (𝒜h f 2 , q̂ 0 ) 2 α 1 2 1 󵄩 1 1󵄩 󵄩2 τ 󵄩 󵄩2 󵄩 ⩽ ( 󵄩󵄩󵄩𝒜h e1 󵄩󵄩󵄩 + 󵄩󵄩󵄩p̂ 0 󵄩󵄩󵄩 ) + (󵄩󵄩󵄩𝒜h f 2 󵄩󵄩󵄩 + 2 2τ 2 α

� 343

1 󵄩󵄩 0 󵄩󵄩2 󵄩q̂ 󵄩 ); 4󵄩 󵄩

namely, τ 󵄩 0 󵄩2 󵄩󵄩 1 󵄩2 2 󵄩 0 󵄩2 󵄩󵄩𝒜h e 󵄩󵄩󵄩 ⩽ τ 󵄩󵄩󵄩p̂ 󵄩󵄩󵄩 + 󵄩󵄩󵄩q̂ 󵄩󵄩󵄩 . α The substitution of (10.94)–(10.95) will lead to τ 2 󵄩󵄩 2 4 4 2 1 󵄩2 2 󵄩󵄩𝒜h e 󵄩󵄩󵄩 ⩽ (τ + )c14 L1 L2 (τ + h1 + h2 ) . α

(10.120)

(b) Taking the inner product on both the right- and left-hand sides of (10.115) with 1

𝒜h δt e 2 yields

1 1 1 2 1 󵄩󵄩 󵄩 0 󵄩󵄩𝒜h δt e 2 󵄩󵄩󵄩 = (Λh f 2 , 𝒜h δt e 2 ) + (p̂ , 𝒜h δt e 2 ). 1

Taking the inner product on both the right- and left-hand sides of (10.116) with Λh δt e 2 produces 1

1

1

1

1

(𝒜h f 2 , Λh δt e 2 ) = −α(Λh e 2 , Λh δt e 2 ) + (Λh δt e 2 , q̂ 0 ). Adding the above two equalities together gives 1 2 1 1 1 1 󵄩󵄩 󵄩 0 0 󵄩󵄩𝒜h δt e 2 󵄩󵄩󵄩 + α(Λh e 2 , Λh δt e 2 ) = (p̂ , 𝒜h δt e 2 ) + (Λh δt e 2 , q̂ );

namely, 1 1 󵄩󵄩 α 󵄩 1 󵄩2 1 0 1 1 0 1 󵄩2 󵄩󵄩𝒜h e 󵄩󵄩󵄩 + 󵄩󵄩󵄩Λh e 󵄩󵄩󵄩 = (p̂ , 𝒜h e ) + (Λh e , q̂ ) 2 2τ τ τ τ 1󵄩 1 󵄩 󵄩2 󵄩2 1 󵄩 󵄩2 α 󵄩 󵄩2 ⩽ 2 󵄩󵄩󵄩𝒜h e1 󵄩󵄩󵄩 + 󵄩󵄩󵄩p̂ 0 󵄩󵄩󵄩 + 󵄩󵄩󵄩Λh e1 󵄩󵄩󵄩 + 󵄩󵄩󵄩q̂ 0 󵄩󵄩󵄩 . 4 4τ ατ τ Thus, 1 󵄩 󵄩2 τ 󵄩 󵄩2 4 󵄩 󵄩2 󵄩󵄩 1 󵄩󵄩2 4τ 1 󵄩󵄩 ̂ 0 󵄩󵄩2 ( 󵄩󵄩p 󵄩󵄩 + 󵄩󵄩󵄩q̂ 0 󵄩󵄩󵄩 ) = 󵄩󵄩󵄩p̂ 0 󵄩󵄩󵄩 + 2 󵄩󵄩󵄩q̂ 0 󵄩󵄩󵄩 . 󵄩󵄩Λh e 󵄩󵄩 ⩽ α 4 ατ α α The substitution of (10.94)–(10.95) will lead to

344 � 10 Difference methods for the Cahn–Hilliard equation τ 4 2 󵄩󵄩 1 󵄩󵄩2 2 4 4 2 󵄩󵄩Λh e 󵄩󵄩 ⩽ ( + 2 )c14 L1 L2 (τ + h1 + h2 ) . α α

(10.121)

(II) Assume that (10.114) is true for k from 0 to m (1 ⩽ m ⩽ n−1). When τ 2 +h14 +h24 ⩽ 󵄩󵄩 k 󵄩󵄩 2 4 4 󵄩󵄩e 󵄩󵄩∞ ⩽ c16 (τ + h1 + h2 ) ⩽ 1,

1 , c16

1 ⩽ k ⩽ m,

from which, one can get 󵄩󵄩 k 󵄩󵄩 󵄩 k 󵄩 k󵄩 󵄩 k󵄩 k k 󵄩 󵄩󵄩u 󵄩󵄩∞ = 󵄩󵄩󵄩U − (U − u )󵄩󵄩󵄩∞ ⩽ 󵄩󵄩󵄩U 󵄩󵄩󵄩∞ + 󵄩󵄩󵄩e 󵄩󵄩󵄩∞ ⩽ c0 + 1, 1 ⩽ k ⩽ m, (10.122) 󵄨󵄨 󵄨 k󵄨 k k 󵄨 (10.123) 󵄨󵄨ϕ(Uij ) − ϕ(uij )󵄨󵄨󵄨 ⩽ c9 󵄨󵄨󵄨eij 󵄨󵄨󵄨, 0 ⩽ i ⩽ m1 , 0 ⩽ j ⩽ m2 , 1 ⩽ k ⩽ m, and with the help of Lemma 10.3, we have 󵄨󵄨 󵄨 k󵄨 󵄨 k+1 󵄨 󵄨 k−1 󵄨 k k 󵄨 󵄨󵄨Δt [ϕ(Uij ) − ϕ(uij )]󵄨󵄨󵄨 ⩽ c9 󵄨󵄨󵄨Δt eij 󵄨󵄨󵄨 + c3 c10 (󵄨󵄨󵄨eij 󵄨󵄨󵄨 + 󵄨󵄨󵄨eij 󵄨󵄨󵄨), 0 ⩽ i ⩽ m1 , 0 ⩽ j ⩽ m2 , 1 ⩽ k ⩽ m − 1.

(10.124)

Now it is sufficient to verify that (10.114) is also true for k = m + 1. (a) Taking the inner product on both the right- and left-hand sides of (10.117) with ̄ 𝒜h ek yields (𝒜h ek , 𝒜h Δt ek ) = (𝒜h ek , Λh f k ) + (𝒜h ek , p̂ k ), ̄

̄

̄

1 ⩽ k ⩽ m.

Taking the inner product on both the right- and left-hand sides of (10.118) with α1 𝒜h f k produces 1 󵄩󵄩 1 1 k 󵄩2 k k k k k̄ k k 󵄩󵄩𝒜h f 󵄩󵄩󵄩 = (𝒜h (ϕ(U ) − ϕ(u )), 𝒜h f ) − (𝒜h f , Λh e ) + (𝒜h f , q̂ ), α α α

1 ⩽ k ⩽ m.

Adding the above two equalities together gives 1 󵄩󵄩 1󵄩 󵄩2 󵄩 󵄩2 󵄩2 (󵄩𝒜 ek+1 󵄩󵄩󵄩 − 󵄩󵄩󵄩𝒜h ek−1 󵄩󵄩󵄩 ) + 󵄩󵄩󵄩𝒜h f k 󵄩󵄩󵄩 4τ 󵄩 h α ̄ 1 1 = (𝒜h (ϕ(U k ) − ϕ(uk )), 𝒜h f k ) + (𝒜h ek , p̂ k ) + (𝒜h f k , q̂ k ) α α 1 󵄩 1 󵄩 󵄩2 󵄩2 ⩽ 󵄩󵄩󵄩𝒜h (ϕ(U k ) − ϕ(uk ))󵄩󵄩󵄩 + 󵄩󵄩󵄩𝒜h f k 󵄩󵄩󵄩 2α 2α 1 󵄩󵄩 1 󵄩󵄩 1 󵄩 󵄩2 󵄩2 1 󵄩󵄩 ̂ k 󵄩󵄩2 󵄩2 k̄ 󵄩 + 󵄩󵄩𝒜h e 󵄩󵄩 + 󵄩󵄩p 󵄩󵄩 + 󵄩󵄩𝒜h f k 󵄩󵄩󵄩 + 󵄩󵄩󵄩q̂ k 󵄩󵄩󵄩 , 1 ⩽ k ⩽ m. 2 2 2α 2α

(10.125)

By (10.123) and Lemma 10.5, we have 144 2 󵄩󵄩 󵄩󵄩 󵄩 󵄩2 k k 󵄩2 k k 󵄩2 2 󵄩 k 󵄩2 c 󵄩𝒜 ek 󵄩󵄩 , 󵄩󵄩𝒜h (ϕ(U ) − ϕ(u ))󵄩󵄩󵄩 ⩽ 󵄩󵄩󵄩(ϕ(U ) − ϕ(u ))󵄩󵄩󵄩 ⩽ c9 󵄩󵄩󵄩e 󵄩󵄩󵄩 ⩽ 25 9 󵄩 h 󵄩 1 ⩽ k ⩽ m. (10.126)

10.4 Three-level linearized compact difference scheme

� 345

Substituting (10.126) into (10.125) yields 1 󵄩󵄩 󵄩2 󵄩2 󵄩 (󵄩󵄩𝒜h ek+1 󵄩󵄩󵄩 − 󵄩󵄩󵄩𝒜h ek−1 󵄩󵄩󵄩 ) 4τ 1 144 2 󵄩󵄩 󵄩2 󵄩2 󵄩 󵄩2 1 󵄩 ⩽ ⋅ c 󵄩𝒜 ek 󵄩󵄩 + (󵄩󵄩󵄩𝒜h ek+1 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝒜h ek−1 󵄩󵄩󵄩 ) + 2α 25 9 󵄩 h 󵄩 4

1 󵄩󵄩 k 󵄩󵄩2 1 󵄩 k 󵄩2 󵄩p̂ 󵄩 + 󵄩󵄩󵄩q̂ 󵄩󵄩󵄩 , 2󵄩 󵄩 2α

1 ⩽ k ⩽ m.

Applying the Gronwall inequality (Theorem 1.2(a)) and noticing (10.120), (10.98), (10.99), there is a constant c17 such that 󵄩󵄩 l󵄩 2 4 4 󵄩󵄩𝒜h e 󵄩󵄩󵄩 ⩽ c17 (τ + h1 + h2 ),

1 ⩽ l ⩽ m + 1.

Using Lemma 10.5, we have 󵄩󵄩 l 󵄩󵄩 12 2 4 4 󵄩󵄩e 󵄩󵄩 ⩽ c17 (τ + h1 + h2 ), 5

1 ⩽ l ⩽ m + 1.

(10.127)

(b) Taking the inner product on both the right- and left-hand sides of (10.117) with

𝒜h Δt ek yields

󵄩󵄩 k 󵄩2 k k k k 󵄩󵄩𝒜h Δt e 󵄩󵄩󵄩 = (𝒜h Δt e , Λh f ) + (𝒜h Δt e , p̂ ),

1 ⩽ k ⩽ m.

Taking the inner product on both the right- and left-hand sides of (10.118) with Λh Δt ek produces (Λh Δt ek , 𝒜h f k ) = (𝒜h (ϕ(U k ) − ϕ(uk )), Λh Δt ek ) − α(Λh ek , Λh Δt ek ) ̄

+ (Λh Δt ek , q̂ k ),

1 ⩽ k ⩽ m.

Adding the above two equalities together gives α 󵄩 k+1 󵄩2 󵄩 k−1 󵄩2 󵄩󵄩 k 󵄩2 󵄩󵄩𝒜h Δt e 󵄩󵄩󵄩 + (󵄩󵄩󵄩Λh e 󵄩󵄩󵄩 − 󵄩󵄩󵄩Λh e 󵄩󵄩󵄩 ) 4τ = (𝒜h (ϕ(U k ) − ϕ(uk )), Λh Δt ek ) + (𝒜h Δt ek , p̂ k ) + (Λh Δt ek , q̂ k ), It follows from

that

1󵄩 󵄨󵄨 k 󵄩2 k k 󵄨 󵄨󵄨(𝒜h Δt e , p̂ )󵄨󵄨󵄨 ⩽ 󵄩󵄩󵄩𝒜h Δt e 󵄩󵄩󵄩 + 2

1 ⩽ k ⩽ m.

1 󵄩󵄩 k 󵄩󵄩2 󵄩p̂ 󵄩 2󵄩 󵄩

1 󵄩󵄩 α 󵄩 k+1 󵄩2 󵄩 k−1 󵄩2 k 󵄩2 󵄩𝒜 Δ e 󵄩󵄩 + (󵄩󵄩󵄩Λh e 󵄩󵄩󵄩 − 󵄩󵄩󵄩Λh e 󵄩󵄩󵄩 ) 2󵄩 h t 󵄩 4τ 1 󵄩 󵄩2 ⩽ (𝒜h (ϕ(U k ) − ϕ(uk )), Λh Δt ek ) + 󵄩󵄩󵄩p̂ k 󵄩󵄩󵄩 + (Λh Δt ek , q̂ k ), 2

1 ⩽ k ⩽ m.

Replacing the superscript k by l in the above result and then summing over l from 1 to k lead to 1 k 󵄩󵄩 󵄩2 α 󵄩 󵄩2 󵄩 󵄩2 󵄩 󵄩2 󵄩 󵄩2 ∑󵄩𝒜 Δ el 󵄩󵄩 + (󵄩󵄩󵄩Λh ek+1 󵄩󵄩󵄩 + 󵄩󵄩󵄩Λh ek 󵄩󵄩󵄩 − 󵄩󵄩󵄩Λh e1 󵄩󵄩󵄩 − 󵄩󵄩󵄩Λh e0 󵄩󵄩󵄩 ) 2 l=1󵄩 h t 󵄩 4τ

346 � 10 Difference methods for the Cahn–Hilliard equation k

⩽ ∑(𝒜h (ϕ(U l ) − ϕ(ul )), Λh Δt el ) + l=1

1 ⩽ k ⩽ m.

1 k 󵄩󵄩 l 󵄩󵄩2 k ∑󵄩p̂ 󵄩 + ∑(Λ Δ el , q̂ l ), 2 l=1󵄩 󵄩 l=1 h t

(10.128)

It follows from the first equality in Lemma 10.4 that k

∑(𝒜h (ϕ(U l ) − ϕ(ul )), Λh Δt el ) l=1

=

1 [(𝒜h (ϕ(U k ) − ϕ(uk )), Λh ek+1 ) + (𝒜h (ϕ(U k−1 ) − ϕ(uk−1 )), Λh ek ) 2τ − (𝒜h (ϕ(U 0 ) − ϕ(u0 )), Λh e1 ) − (𝒜h (ϕ(U 1 ) − ϕ(u1 )), Λh e0 )] k−1

− ∑ (Δt 𝒜h (ϕ(U l ) − ϕ(ul )), Λh el ), l=1

1 ⩽ k ⩽ m.

Applying the Cauchy–Schwarz inequality, Lemma 10.5, (10.123)–(10.124) and (10.119), we get k

∑(𝒜h (ϕ(U l ) − ϕ(ul )), Λh Δt el ) l=1

⩽

1 󵄩󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 (󵄩𝒜 (ϕ(U k ) − ϕ(uk ))󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩Λh ek+1 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝒜h (ϕ(U k−1 ) − ϕ(uk−1 ))󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩Λh ek 󵄩󵄩󵄩) 2τ 󵄩 h k−1

󵄩 󵄩 󵄩 󵄩 + ∑ 󵄩󵄩󵄩Δt 𝒜h (ϕ(U l ) − ϕ(ul ))󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩Λh el 󵄩󵄩󵄩 l=1

1 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 ⩽ (󵄩󵄩󵄩ϕ(U k ) − ϕ(uk )󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩Λh ek+1 󵄩󵄩󵄩 + 󵄩󵄩󵄩ϕ(U k−1 ) − ϕ(uk−1 )󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩Λh ek 󵄩󵄩󵄩) 2τ k−1

󵄩 󵄩 󵄩 󵄩 + ∑ 󵄩󵄩󵄩Δt (ϕ(U l ) − ϕ(ul ))󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩Λh el 󵄩󵄩󵄩 l=1

1 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 ⩽ (c9 󵄩󵄩󵄩ek 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩Λh ek+1 󵄩󵄩󵄩 + c9 󵄩󵄩󵄩ek−1 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩Λh ek 󵄩󵄩󵄩) 2τ k−1

󵄩 󵄩 󵄩 󵄩 󵄩 󵄩󵄩 󵄩 + ∑ (c9 󵄩󵄩󵄩Δt el 󵄩󵄩󵄩 + c3 c10 󵄩󵄩󵄩el+1 󵄩󵄩󵄩 + c3 c10 󵄩󵄩󵄩el−1 󵄩󵄩󵄩)󵄩󵄩󵄩Λh el 󵄩󵄩󵄩, l=1

1 ⩽ k ⩽ m.

(10.129)

As for the last term of (10.128) on the right-hand side, taking advantage of the second equality in Lemma 10.4, we obtain k

k

l=1

l=1

∑(Λh Δt el , q̂ l ) = ∑(q̂ l , Δt Λh el ) =

k−1 1 [(q̂ k , Λh ek+1 ) + (q̂ k−1 , Λh ek ) − (q̂ 2 , Λh e1 ) − (q̂ 1 , Λh e0 )] − ∑ (Δt q̂ l , Λh el ) 2τ l=2

⩽

1 󵄩󵄩 k 󵄩󵄩 󵄩󵄩 k+1 󵄩󵄩 󵄩󵄩 k−1 󵄩󵄩 󵄩󵄩 k 󵄩󵄩 󵄩󵄩 2 󵄩󵄩 󵄩󵄩 1 󵄩󵄩 k−1󵄩󵄩 l 󵄩󵄩 󵄩󵄩 l 󵄩󵄩 (󵄩q̂ 󵄩 ⋅ 󵄩Λ e 󵄩󵄩 + 󵄩󵄩q̂ 󵄩󵄩 ⋅ 󵄩󵄩Λh e 󵄩󵄩 + 󵄩󵄩q̂ 󵄩󵄩 ⋅ 󵄩󵄩Λh e 󵄩󵄩) + ∑ 󵄩󵄩Δt q̂ 󵄩󵄩 ⋅ 󵄩󵄩Λh e 󵄩󵄩. 2τ 󵄩 󵄩 󵄩 h l=2

(10.130)

10.4 Three-level linearized compact difference scheme

Inserting (10.129) and (10.130) into (10.128) and noticing (10.119) yield 1 k 󵄩󵄩 󵄩2 󵄩2 󵄩 󵄩2 α 󵄩 󵄩2 󵄩 ∑󵄩𝒜 Δ el 󵄩󵄩 + (󵄩󵄩󵄩Λh ek+1 󵄩󵄩󵄩 + 󵄩󵄩󵄩Λh ek 󵄩󵄩󵄩 − 󵄩󵄩󵄩Λh e1 󵄩󵄩󵄩 ) 2 l=1󵄩 h t 󵄩 4τ ⩽

1 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 (c 󵄩󵄩ek 󵄩󵄩 ⋅ 󵄩󵄩Λ ek+1 󵄩󵄩󵄩 + c9 󵄩󵄩󵄩ek−1 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩Λh ek 󵄩󵄩󵄩) 2τ 9 󵄩 󵄩 󵄩 h k−1

󵄩 󵄩 󵄩󵄩 󵄩 󵄩 󵄩 󵄩 + ∑ (c9 󵄩󵄩󵄩Δt el 󵄩󵄩󵄩 + c3 c10 󵄩󵄩󵄩el+1 󵄩󵄩󵄩 + c3 c10 󵄩󵄩󵄩el−1 󵄩󵄩󵄩)󵄩󵄩󵄩Λh el 󵄩󵄩󵄩 l=1

1 k 󵄩󵄩 l 󵄩󵄩2 1 󵄩󵄩 k 󵄩󵄩 󵄩󵄩 k+1 󵄩󵄩 󵄩󵄩 k−1 󵄩󵄩 󵄩󵄩 k 󵄩󵄩 󵄩󵄩 2 󵄩󵄩 󵄩󵄩 1 󵄩󵄩 ∑󵄩p̂ 󵄩 + (󵄩󵄩q̂ 󵄩󵄩 ⋅ 󵄩󵄩Λh e 󵄩󵄩 + 󵄩󵄩q̂ 󵄩󵄩 ⋅ 󵄩󵄩Λh e 󵄩󵄩 + 󵄩󵄩q̂ 󵄩󵄩 ⋅ 󵄩󵄩Λh e 󵄩󵄩) 2 l=1󵄩 󵄩 2τ

+

k−1

󵄩 󵄩 󵄩 󵄩 + ∑ 󵄩󵄩󵄩Δt q̂ l 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩Λh el 󵄩󵄩󵄩 l=2

1 α󵄩 α󵄩 󵄩2 2 󵄩 󵄩2 󵄩2 2 󵄩 󵄩2 ⩽ [( 󵄩󵄩󵄩Λh ek+1 󵄩󵄩󵄩 + c92 󵄩󵄩󵄩ek 󵄩󵄩󵄩 ) + ( 󵄩󵄩󵄩Λh ek 󵄩󵄩󵄩 + c92 󵄩󵄩󵄩ek−1 󵄩󵄩󵄩 )] 2τ 8 α 8 α k−1 1 25 󵄩󵄩 l 󵄩󵄩2 1 144 2 󵄩󵄩 l 󵄩󵄩2 + ∑ [( × c 󵄩Λ e 󵄩 ) 󵄩󵄩Δt e 󵄩󵄩 + × 2 144 2 25 9 󵄩 h 󵄩 l=1

1 󵄩 󵄩2 1 2 󵄩󵄩 l 󵄩󵄩2 1 󵄩 l−1 󵄩2 1 2 2 󵄩 l 󵄩2 + ( 󵄩󵄩󵄩el+1 󵄩󵄩󵄩 + c32 c10 󵄩󵄩Λh e 󵄩󵄩 ) + ( 󵄩󵄩󵄩e 󵄩󵄩󵄩 + c3 c10 󵄩󵄩󵄩Λh e 󵄩󵄩󵄩 )] 2 2 2 2

+

1 k 󵄩󵄩 l 󵄩󵄩2 1 α󵄩 󵄩2 2 󵄩 󵄩2 [( 󵄩󵄩󵄩Λh ek+1 󵄩󵄩󵄩 + 󵄩󵄩󵄩q̂ k 󵄩󵄩󵄩 ) ∑󵄩p̂ 󵄩 + 2 l=1󵄩 󵄩 2τ 8 α

α󵄩 α󵄩 1 󵄩 󵄩2 󵄩2 2 󵄩 󵄩2 󵄩2 + ( 󵄩󵄩󵄩Λh ek 󵄩󵄩󵄩 + 󵄩󵄩󵄩q̂ k−1 󵄩󵄩󵄩 ) + ( 󵄩󵄩󵄩Λh e1 󵄩󵄩󵄩 + 󵄩󵄩󵄩q̂ 2 󵄩󵄩󵄩 )] 8 α 2 2α +

1 k−1 󵄩󵄩 l 󵄩󵄩2 󵄩󵄩 l 󵄩󵄩2 ∑ (󵄩Δ q̂ 󵄩 + 󵄩Λ e 󵄩 ). 2 l=2 󵄩 t 󵄩 󵄩 h 󵄩

Moreover, we also have from the above result that 󵄩󵄩 k+1 󵄩󵄩2 󵄩󵄩 k 󵄩󵄩2 󵄩󵄩Λh e 󵄩󵄩 + 󵄩󵄩Λh e 󵄩󵄩

2 2 󵄩2 󵄩 󵄩2 4 2c 󵄩 󵄩2 2c 󵄩 ⩽ 2󵄩󵄩󵄩Λh e1 󵄩󵄩󵄩 + [ 9 󵄩󵄩󵄩ek 󵄩󵄩󵄩 + 9 󵄩󵄩󵄩ek−1 󵄩󵄩󵄩 α α α 1 󵄩 󵄩2 2 󵄩 󵄩2 2 󵄩 󵄩2 α 󵄩 󵄩2 + 󵄩󵄩󵄩q̂ k 󵄩󵄩󵄩 + 󵄩󵄩󵄩q̂ k−1 󵄩󵄩󵄩 + 󵄩󵄩󵄩Λh e1 󵄩󵄩󵄩 + 󵄩󵄩󵄩q̂ 2 󵄩󵄩󵄩 ] α α 2 2α

+

8τ k−1 1 󵄩󵄩 l+1 󵄩󵄩2 ∑ [ 󵄩e 󵄩 + α l=1 2 󵄩 󵄩

+

8τ 1 k−1󵄩󵄩 l 󵄩󵄩2 8τ 1 k 󵄩󵄩 l 󵄩󵄩2 k−1󵄩󵄩 l 󵄩󵄩2 ⋅ ∑ 󵄩Λ e 󵄩 + ⋅ (∑󵄩p̂ 󵄩 + ∑ 󵄩Δ q̂ 󵄩 ), α 2 l=2 󵄩 h 󵄩 α 2 l=1󵄩 󵄩 l=2 󵄩 t 󵄩

1 󵄩󵄩 l−1 󵄩󵄩2 72 2 2 2 󵄩 l 󵄩2 󵄩󵄩e 󵄩󵄩 + ( c9 + c3 c10 )󵄩󵄩󵄩Λh e 󵄩󵄩󵄩 ] 2 25 1 ⩽ k ⩽ m.

Noting (10.127), (10.98)–(10.100) and (10.121), there is a constant c18 such that

� 347

348 � 10 Difference methods for the Cahn–Hilliard equation 󵄩󵄩 k+1 󵄩󵄩2 󵄩󵄩 k 󵄩󵄩2 󵄩󵄩Λh e 󵄩󵄩 + 󵄩󵄩Λh e 󵄩󵄩 ⩽

k−1 4 144 2 2 󵄩 󵄩2 2 ( c9 + 2c32 c10 + 1)τ ∑ 󵄩󵄩󵄩Λh el 󵄩󵄩󵄩 + c18 (τ 2 + h14 + h24 ) α 25 l=1

⩽

k−1 2 144 2 2 󵄩2 󵄩 󵄩2 󵄩 2 ( c9 + 2c32 c10 + 1)τ ∑ (󵄩󵄩󵄩Λh el+1 󵄩󵄩󵄩 + 󵄩󵄩󵄩Λh el 󵄩󵄩󵄩 ) + c18 (τ 2 + h14 + h24 ) , α 25 l=0

1 ⩽ k ⩽ m.

Again making use of the Gronwall inequality (Theorem 1.2(c)), there is a constant c19 such that 󵄩󵄩 k 󵄩󵄩 2 4 4 󵄩󵄩Λh e 󵄩󵄩 ⩽ c19 (τ + h1 + h2 ),

1 ⩽ k ⩽ m + 1.

(10.131)

Combining Lemma 10.2, Lemma 10.5, Lemma 10.6 with (10.127), (10.131), there is a constant c16 such that 󵄩󵄩 k 󵄩󵄩 2 4 4 󵄩󵄩e 󵄩󵄩∞ ⩽ c16 (τ + h1 + h2 ),

1 ⩽ k ⩽ m + 1.

Noting that the constants c16 , c17 , c18 and c19 are all independent of τ, h1 and h2 , so that (10.114) is true for k = m + 1. By induction, the result of this theorem can be concluded.

10.5 Numerical experiments In this section, we will take the three-level linearized compact difference scheme (10.111)–(10.113) as an example to test the accuracy. Example 10.1. Let L1 = 1, L2 = 1, T = 0.1, α = 0.1, β = √2, γ = 1, φ(x, y) = cos(πx) cos(πy) in (10.1)–(10.3). Take h1 = h2 = h. We compute the numerical solution of the problem by the difference scheme (10.111)–(10.113). Denote the difference solution by {uijk (h, τ) | 0 ⩽ i ⩽ m1 , 0 ⩽ j ⩽ m2 , 0 ⩽ k ⩽ n}. For fixed sufficiently small τ, denote E(h, τ) =

max

0⩽i⩽m1 ,0⩽j⩽m2

󵄨󵄨 n 󵄨 n 󵄨󵄨uij (2h, τ) − u2i,2j (h, τ)󵄨󵄨󵄨,

Orderh = log2

E(2h, τ) . E(h, τ)

Orderτ = log2

E(h, 2τ) . E(h, τ)

For fixed sufficiently small h, denote E(h, τ) =

max

0⩽i⩽m1 ,0⩽j⩽m2

󵄨󵄨 2n 󵄨 n 󵄨󵄨uij (h, 2τ) − ui,j (h, τ)󵄨󵄨󵄨,

Some numerical results are presented in Tables 10.1 and 10.2. We can see that the difference scheme (10.111)–(10.113) is convergent with the order 4 in space and approximately order 2 in time.

10.6 Summary and extension

� 349

Table 10.1: Some numerical results of (10.111)–(10.113) when τ = T /1000. h

E(h, τ)

Orderh

�/� �/� �/�� �/�� �/�� �/���

2.168e-3 1.422e-4 8.855e-6 5.517e-7 3.281e-8

3.9300 4.0054 4.0046 4.0717

Table 10.2: Some numerical results of (10.111)–(10.113) when h1 = h2 = 1/128. τ

E(h, τ)

Orderτ

�/�� �/�� �/��� �/��� �/��� �/����

4.979e-5 4.663e-6 8.104e-7 1.518e-7 2.849e-8

2.8550 3.4163 2.5247 2.4163 2.4136

10.6 Summary and extension This chapter discussed the difference methods for solving the initial and boundary value problem of the Cahn–Hilliard equation. At the beginning, the solution of (10.1)–(10.3) is proved to satisfy the conservation of energy. Then a two-level nonlinear difference scheme, a spatial second-order three-level linearized difference scheme and a spatial fourth-order three-level linearized compact difference scheme are developed successively. For each proposed difference scheme, the existence and uniqueness of solutions are analyzed and the convergence of solutions in the maximum norm is proved. The content of this chapter is developed from [21] and [31]. The numerical experiment section in the chapter originates from [21]. Indeed, another two-level nonlinear difference scheme for solving (10.1)–(10.3) can be constructed: ψ(uijk+1 ) − ψ(uijk ) k+ 21 k+ 1 { { { δ u = Δ ( − αΔh uij 2 ), t ij h { k+1 k { uij − uij { { { 0 ⩽ i ⩽ m1 , 0 ⩽ j ⩽ m2 , 1 ⩽ k ⩽ n − 1, { { { { 0 { uij = φ(xj , yj ), 0 ⩽ i ⩽ m1 , 0 ⩽ j ⩽ m2 . It is easy to know

(10.132) (10.133)

350 � 10 Difference methods for the Cahn–Hilliard equation ψ(uijk+1 ) − ψ(uijk ) uijk+1 − uijk

= γ[

(uijk+1 )2 + (uijk )2 2

− β2 ]

uijk+1 + uijk 2

by noticing ψ(u) = γ4 (u2 −β2 )2 . It can be proved that the solution of the difference scheme (10.132)–(10.133) also satisfies the conservation of energy, and it is also solvable and convergent. Finally, it is worth mentioning that the method to prove Theorem 10.6 is also feasible to prove the convergence of the difference scheme (1.113)–(1.115) in the maximum norm.

11 Difference methods for the epitaxial growth model 11.1 Introduction Consider the finite difference method to solve the following 2D problem for the epitaxial growth model: {

ut + δΔ2 u − ∇ ⋅ (|∇u|2 ∇u) + Δu = 0, u(x, y, 0) = φ(x, y),

(x, y) ∈ ℛ2 , 0 < t ⩽ T,

(11.1)

(x, y) ∈ ℛ ,

(11.2)

2

where δ is a positive constant, ∇ and Δ are the gradient operator and the Laplace operator, respectively, u(x, y, t) is periodic with respect to (x, y) in ℛ2 with the period spatial domain Ω = [0, L1 ] × [0, L2 ]. Theorem 11.1. Let u(x, y, t) be the solution of the problem (11.1)–(11.2). Then it holds that 󵄩󵄩 󵄩2 󵄩 󵄩2 1 󵄩󵄩u(⋅, ⋅, t)󵄩󵄩󵄩 ⩽ 󵄩󵄩󵄩u(⋅, ⋅, 0)󵄩󵄩󵄩 + L1 L2 t, 2

0 < t ⩽ T.

Proof. Taking the inner product on both the right- and left-hand sides of (11.1) with respect to u on Ω, we have (ut , u) + δ(Δ2 u, u) − (∇ ⋅ (|∇u|2 ∇u), u) + (Δu, u) = 0. Applying the integration by parts and the periodic boundary condition gives 1 d 󵄩󵄩 󵄩2 󵄩 󵄩2 ⋅ (󵄩u(⋅, ⋅, t)󵄩󵄩󵄩 ) + δ󵄩󵄩󵄩Δu(⋅, ⋅, t)󵄩󵄩󵄩 + (|∇u|2 ∇u, ∇u) − (∇u, ∇u) = 0, 2 dt 󵄩 i. e., 1 d 󵄩󵄩 󵄩2 󵄩 󵄩2 󵄩 󵄩4 󵄩 󵄩2 ⋅ (󵄩u(⋅, ⋅, t)󵄩󵄩󵄩 ) + δ󵄩󵄩󵄩Δu(⋅, ⋅, t)󵄩󵄩󵄩 + 󵄩󵄩󵄩∇u(⋅, ⋅, t)󵄩󵄩󵄩4 − 󵄩󵄩󵄩∇u(⋅, ⋅, t)󵄩󵄩󵄩 = 0, 2 dt 󵄩 which can be rewritten as 2

1 1 d 󵄩󵄩 󵄩2 󵄩 󵄩2 󵄨 󵄨2 1 ⋅ (󵄩u(⋅, ⋅, t)󵄩󵄩󵄩 ) + δ󵄩󵄩󵄩Δu(⋅, ⋅, t)󵄩󵄩󵄩 + ∬(󵄨󵄨󵄨∇u(⋅, ⋅, t)󵄨󵄨󵄨 − ) dxdy = ∬ 1 dxdy. 2 dt 󵄩 2 4 Ω

Ω

Hence, 1 d 󵄩󵄩 1 󵄩2 ⋅ (󵄩u(⋅, ⋅, t)󵄩󵄩󵄩 ) ⩽ L1 L2 , 2 dt 󵄩 4

0 < t ⩽ T,

which implies 󵄩󵄩 󵄩2 󵄩 󵄩2 1 󵄩󵄩u(⋅, ⋅, t)󵄩󵄩󵄩 ⩽ 󵄩󵄩󵄩u(⋅, ⋅, 0)󵄩󵄩󵄩 + L1 L2 t, 2 https://doi.org/10.1515/9783110796018-011

0 < t ⩽ T.

352 � 11 Difference methods for the epitaxial growth model Theorem 11.2. Let u(x, y, t) be the solution of the problem (11.1)–(11.2). Denote t

E(t) =

δ 󵄩󵄩 2 󵄩2 󵄩 󵄨2 󵄩2 1 󵄨 󵄩Δu(⋅, ⋅, t)󵄩󵄩󵄩 + ∬(󵄨󵄨󵄨∇u(x, y, t)󵄨󵄨󵄨 − 1) dxdy + ∫󵄩󵄩󵄩us (⋅, ⋅, s)󵄩󵄩󵄩 ds. 2󵄩 4 0

Ω

Then it holds that E(t) = E(0),

0 < t ⩽ T.

(11.3)

Proof. Taking the inner product on both the right- and left-hand sides of (11.1) with ut on Ω, we have (ut , ut ) + δ(Δ2 u, ut ) − (∇ ⋅ (|∇u|2 ∇u), ut ) + (Δu, ut ) = 0. Applying the integration by parts and the periodic boundary condition arrives at ‖ut ‖2 +

δ d 1 d 1 d 󵄨 󵄨4 󵄨 󵄨2 ⋅ (‖Δu‖2 ) + ⋅ ∬󵄨󵄨∇u(x, y, t)󵄨󵄨󵄨 dxdy − ⋅ ∬󵄨󵄨∇u(x, y, t)󵄨󵄨󵄨 dxdy = 0. 2 dt 4 dt 󵄨 2 dt 󵄨 Ω

Ω

Hence, t

d δ 󵄩󵄩 2 󵄩 󵄩2 󵄩2 1 󵄨 󵄨2 [ 󵄩Δu(⋅, ⋅, t)󵄩󵄩󵄩 + ∬(󵄨󵄨󵄨∇u(x, y, t)󵄨󵄨󵄨 − 1) dxdy + ∫󵄩󵄩󵄩us (⋅, ⋅, s)󵄩󵄩󵄩 ds] = 0, dt 2 󵄩 4

0 < t ⩽ T,

0

Ω

i. e., dE(t) = 0, dt

0 < t ⩽ T,

which implies the truth of (11.3). By Theorem 11.1, Theorem 11.2 and the embedding theorem, there is a constant c0 such that 󵄩󵄩 󵄩 󵄩󵄩u(⋅, ⋅, t)󵄩󵄩󵄩∞ ⩽ c0 ,

0 ⩽ t ⩽ T.

11.2 Notation and basic lemmas Take three positive integers m1 , m2 and n. Denote h1 =

L1 , m1

h2 =

L2 , m2

τ=

T , n

xi = ih1 ,

Ωh1 ,h2 = {(xi , yj ) | 0 ⩽ i ⩽ m1 , 0 ⩽ j ⩽ m2 },

yj = jh2 ,

tk = kτ,

Ωτ = {tk | 0 ⩽ k ⩽ n}.

11.2 Notation and basic lemmas

Let 𝒲h = {u | u = {uij }, ui+m1 ,j = uij , ui,j+m2 = uij }.

For any u ∈ 𝒲h , introduce the following notation: 1 (u − uij ), h1 i+1,j

1 (u − uij ), h2 i,j+1 1 1 Δx uij = (u − ui−1,j ), Δy uij = (u − ui,j−1 ), 2h1 i+1,j 2h2 i,j+1 1 1 δx2 uij = (δx ui+ 1 ,j − δx ui− 1 ,j ), δy2 uij = (δy ui,j+ 1 − δy ui,j− 1 ), 2 2 2 2 h1 h2 T 2 2 2 ̄ ∇h uij = (Δx uij , Δy uij ) , Δh uij = δ uij + δ uij , Δh uij = (Δ + Δ2 )uij .

δx ui+ 1 ,j = 2

δy ui,j+ 1 = 2

x

y

x

y

Obviously, 1 Δx uij = (δx ui− 1 ,j + δx ui+ 1 ,j ), 2 2 2

1 Δy uij = (δy ui,j− 1 + δy ui,j+ 1 ). 2 2 2

For any mesh functions u ∈ 𝒲h and v ∈ 𝒲h , define the inner product m1 m2

(u, v) = h1 h2 ∑ ∑ uij vij i=1 j=1

and the Sobolev norms (or seminorms) by m1 m2

‖δx u‖ = √h1 h2 ∑ ∑(δx ui− 1 ,j )2 , i=1 j=1

2

m1 m2

‖δy u‖ = √h1 h2 ∑ ∑(δy ui,j− 1 )2 , i=1 j=1

2

|u|1 = √‖δx u‖2 + ‖δy u‖2 ,

‖u‖ = √(u, u),

m1 m2

‖∇h u‖ = √h1 h2 ∑ ∑ |∇h uij |2 , i=1 j=1

4

m1 m2

‖∇h u‖4 = √h1 h2 ∑ ∑ |∇h uij |4 , i=1 j=1

m1 m2

‖Δh u‖ = √h1 h2 ∑ ∑ |Δh uij |2 . i=1 j=1

Denote 0

1

n

𝒮τ = {w | w = (w , w , . . . , w ) is the mesh function defined on Ωτ }.

For any w ∈ 𝒮τ , introduce the following notation: 1 1 wk+ 2 = (wk + wk+1 ), 2 1 k ∇τ w = (wk − wk−1 ), τ

1 k+1 (w − wk ), τ 1 ∇2τ wk = (3wk − 4wk−1 + wk−2 ). 2τ 1

δt wk+ 2 =

� 353

354 � 11 Difference methods for the epitaxial growth model Lemma 11.1. For any u ∈ 𝒲h , it holds that −(Δh u, u) = |u|21 ,

(11.4)

2

‖∇h u‖ ⩽ ‖Δh u‖ ⋅ ‖u‖.

(11.5)

Proof. Using the integration by parts and the periodic boundary conditions, one has −(Δh u, u)

m1 m2

= −h1 h2 ∑ ∑(δx2 uij + δy2 uij )uij i=1 j=1

m2

m1

m1

m2

= h2 ∑[−h1 ∑(δx2 uij )uij ] + h1 ∑[−h2 ∑(δy2 uij )uij ] j=1

i=1

m2

m1

j=1

i=1

i=1

j=1

m1

m2

i=1

j=1

= h2 ∑[h1 ∑(δx ui− 1 ,j )2 ] + h1 ∑[h2 ∑(δy ui,j− 1 )2 ] 2

2

= |u|21 . Consequently, |u|21 = −(Δh u, u) ⩽ ‖u‖ ⋅ ‖Δh u‖. Noticing ‖∇h u‖ ⩽ |u|1 , we know (11.5) holds. Lemma 11.2. For any u ∈ 𝒲h and arbitrary ε > 0, it holds that 1 1 1 1 ‖u‖∞ ⩽ ε‖Δh u‖ + √3[ + ( + )]‖u‖. ε 2 L1 L2

(11.6)

Proof. Suppose ‖u‖∞ = |ui0 ,j0 |. By Lemma 1.1(e), we have m

m

1 1 1 1 2 ui20 ,j0 ⩽ εh1 ∑(δx ui− 1 ,j )2 + ( + )h1 ∑ ui,j 0 0 2 ε L 1 i=1 i=1

m

m

m

1 2 2 1 1 ⩽ εh1 ∑{εh2 ∑(δy δx ui− 1 ,j− 1 )2 + ( + )h2 ∑(δx ui− 1 ,j )2 } 2 2 2 ε L2 j=1 i=1 j=1

m

m

m

1 2 2 1 1 1 1 + ( + )h1 ∑{εh2 ∑(δy ui,j− 1 )2 + ( + )h2 ∑ uij2 } 2 ε L1 ε L2 i=1 j=1 j=1

⩽ ε2 ‖δy δx u‖2 + [ε(

1 1 1 1 1 1 + ) + 1]|u|21 + ( + )( + )‖u‖2 . L1 L2 ε L1 ε L2

(11.7)

� 355

11.2 Notation and basic lemmas

Noticing ‖Δh u‖2

m1 m2

2

= h1 h2 ∑ ∑(δx2 uij + δy2 uij ) i=1 j=1 m1 m2

m1 m2

m1 m2

i=1 j=1

i=1 j=1

i=1 j=1

m1 m2

m1 m2

2

2

= h1 h2 ∑ ∑(δx2 uij ) + 2h1 h2 ∑ ∑(δx2 uij )(δy2 uij ) + h1 h2 ∑ ∑(δy2 uij ) 2

m1 m2

2

= h1 h2 ∑ ∑(δx2 uij ) + 2h1 h2 ∑ ∑(δx δy ui− 1 ,j− 1 )2 + h1 h2 ∑ ∑(δy2 uij ) i=1 j=1

2

i=1 j=1

2

i=1 j=1

󵄩 󵄩2 󵄩 󵄩2 = 󵄩󵄩󵄩δx2 u󵄩󵄩󵄩 + 2‖δx δy u‖2 + 󵄩󵄩󵄩δy2 u󵄩󵄩󵄩 , it follows: 1 ‖δx δy u‖2 ⩽ ‖Δh u‖2 . 2

(11.8)

With the help of Lemma 11.1, for any ε1 > 0, it holds that |u|21 = −(Δh u, u) ⩽ ε1 ‖Δh u‖2 +

1 ‖u‖2 . 4ε1

(11.9)

Inserting (11.8) and (11.9) into (11.7) produces 1 1 1 1 1 1 1 1 ui20 ,j0 ⩽ ε2 ‖Δh u‖2 + [ε( + ) + 1](ε1 ‖Δh u‖2 + ‖u‖2 ) + ( + )( + )‖u‖2 . 2 L1 L2 4ε1 ε L1 ε L2 Taking ε1 such that [ε( L1 + 1

1 ) L2

+ 1]ε1 = 21 ε2 , and noticing 2

4 1 1 ⩽( + ) , L1 L2 L1 L2 it follows: ui20 ,j0 ⩽ ε2 ‖Δh u‖2 + { = ε2 ‖Δh u‖2 + [

[ε( L1 + 1

1 ) L2 2 2ε

+ 1]2

1 1 1 1 + ( + )( + )}‖u‖2 ε L1 ε L2 2

1 2 1 1 1 1 1 3 + ( + )+ ( + ) + ]‖u‖2 2 ε L L 2 L L L 2ε 1 2 1 2 1 L2 2

1 1 1 1 ⩽ ε2 ‖Δh u‖2 + 3[ + ( + )] ‖u‖2 . ε 2 L1 L2 Taking the square root on both the right- and left-hand sides arrives at 1 1 1 1 ‖u‖∞ ⩽ ε‖Δh u‖ + √3[ + ( + )]‖u‖. ε 2 L1 L2

356 � 11 Difference methods for the epitaxial growth model Remark 11.1. Comparing Lemma 11.2 with Lemma 10.2, one can find that almost the same conclusion has been drawn except the difference of the function spaces 𝒱h and 𝒲h .

11.3 Two-level nonlinear backward Euler difference scheme 11.3.1 Derivation of the difference scheme Let (

v ) = |∇u|2 ∇u. w

Then (11.1) is equivalent to ut + δΔ2 u − vx − wy + Δu = 0, { { { v = |∇u|2 ux , { { { 2 { w = |∇u| uy ,

(x, y) ∈ ℛ2 , 0 < t ⩽ T,

(11.10)

(x, y) ∈ ℛ2 , 0 < t ⩽ T,

(11.11)

(x, y) ∈ ℛ , 0 < t ⩽ T.

(11.12)

2

Define the mesh functions on 𝒲h : Uijk = u(xi , yj , tk ),

Vijk = v(xi , yj , tk ),

Wijk = w(xi , yj , tk ).

Consider equations (11.10)–(11.12) at the point (xi , yj , tk ). It follows from the Taylor expansion that ∇τ Uijk + δΔ2h Uijk − Δx Vijk − Δy Wijk + Δ̄ h Uijk−1 = Pijk , { { { { { { { 1 ⩽ i ⩽ m1 , 1 ⩽ j ⩽ m2 , 1 ⩽ k ⩽ n, 󵄨 󵄨2 { { Vijk = 󵄨󵄨󵄨∇h Uijk 󵄨󵄨󵄨 Δx Uijk + Qijk , 1 ⩽ i ⩽ m1 , 1 ⩽ j ⩽ m2 , 1 ⩽ k ⩽ n, { { { { { 󵄨󵄨 k k 󵄨󵄨2 k k { Wij = 󵄨󵄨∇h Uij 󵄨󵄨 Δy Uij + Rij , 1 ⩽ i ⩽ m1 , 1 ⩽ j ⩽ m2 , 1 ⩽ k ⩽ n,

(11.13) (11.14) (11.15)

where there is a positive constant c1 such that { { { { { { { { {

󵄨󵄨 k 󵄨󵄨 2 2 󵄨󵄨Pij 󵄨󵄨 ⩽ c1 (τ + h1 + h2 ), 󵄨󵄨 k 󵄨󵄨 2 2 󵄨󵄨Qij 󵄨󵄨 ⩽ c1 (h1 + h2 ), 󵄨󵄨 k 󵄨󵄨 2 2 󵄨󵄨Rij 󵄨󵄨 ⩽ c1 (h1 + h2 ),

1 ⩽ i ⩽ m1 , 1 ⩽ j ⩽ m2 , 1 ⩽ k ⩽ n,

(11.16)

1 ⩽ i ⩽ m1 , 1 ⩽ j ⩽ m2 , 1 ⩽ k ⩽ n,

(11.17)

1 ⩽ i ⩽ m1 , 1 ⩽ j ⩽ m2 , 1 ⩽ k ⩽ n.

(11.18)

1 ⩽ i ⩽ m1 , 1 ⩽ j ⩽ m2 ,

(11.19)

Noticing the initial value condition Uij0 = φ(xi , yj ),

and neglecting the small terms Pijk , Qijk , Rkij in (11.13)–(11.15), a difference scheme to solve (11.10)–(11.12) and (11.2) is derived as follows:

11.3 Two-level nonlinear backward Euler difference scheme

� 357

For 0 ⩽ k ⩽ n, find uk , vk , wk ∈ 𝒲h such that { { { { { { { { { { { { { { { { { { { { {

∇τ uijk + δΔ2h uijk − Δx vkij − Δy wijk + Δ̄ h uijk−1 = 0,

1 ⩽ i ⩽ m1 , 1 ⩽ j ⩽ m2 , 1 ⩽ k ⩽ n, 󵄨 󵄨2 = 󵄨󵄨󵄨∇h uijk 󵄨󵄨󵄨 Δx uijk , 1 ⩽ i ⩽ m1 , 1 ⩽ j ⩽ m2 , 1 ⩽ k ⩽ n, 󵄨 󵄨2 wijk = 󵄨󵄨󵄨∇h uijk 󵄨󵄨󵄨 Δy uijk , 1 ⩽ i ⩽ m1 , 1 ⩽ j ⩽ m2 , 1 ⩽ k ⩽ n,

vkij

uij0

= φ(xi , yj ),

1 ⩽ i ⩽ m1 , 1 ⩽ j ⩽ m2 .

(11.20) (11.21) (11.22) (11.23)

Substituting (11.21)–(11.22) into (11.20), a difference scheme to solve (11.1)–(11.2) is obtained as follows: For 0 ⩽ k ⩽ n, find uk ∈ 𝒲h such that 󵄨 󵄨2 ∇τ uijk + δΔ2h uijk − ∇h ⋅ (󵄨󵄨󵄨∇h uijk 󵄨󵄨󵄨 ∇h uijk ) + Δ̄ h uijk−1 = 0, { { { 1 ⩽ i ⩽ m1 , 1 ⩽ j ⩽ m2 , 1 ⩽ k ⩽ n, { { { 0 { uij = φ(xi , yj ), 1 ⩽ i ⩽ m1 , 1 ⩽ j ⩽ m2 .

(11.24) (11.25)

11.3.2 Boundedness of the difference solution Theorem 11.3. Suppose {uijk | 1 ⩽ i ⩽ m1 , 1 ⩽ j ⩽ m2 , 0 ⩽ k ⩽ n} ∈ 𝒲h is the solution of the difference scheme (11.24)–(11.25). Then it holds that 1 󵄩󵄩 k 󵄩󵄩2 󵄩 󵄩 0 󵄩2 󵄩 k 󵄩4 0 󵄩4 󵄩󵄩u 󵄩󵄩 + τ 󵄩󵄩󵄩∇h u 󵄩󵄩󵄩4 ⩽ 󵄩󵄩󵄩u 󵄩󵄩󵄩 + τ 󵄩󵄩󵄩∇h u 󵄩󵄩󵄩4 + L1 L2 tk , 2 E k ⩽ E k−1 , 1 ⩽ k ⩽ n,

1 ⩽ k ⩽ n,

where Ek =

m m

1 2 δ 󵄩󵄩 1 2 󵄨 k 󵄩2 k 󵄨2 󵄩󵄩Δh u 󵄩󵄩󵄩 + h1 h2 ∑ ∑(󵄨󵄨󵄨∇h uij 󵄨󵄨󵄨 − 1) . 2 4 i=1 j=1

Proof. (I) Taking the inner product on both the right- and left-hand sides of (11.24) with uk gives 󵄨 󵄨2 (∇τ uk , uk ) + δ(Δ2h uk , uk ) − (∇h ⋅ (󵄨󵄨󵄨∇h uk 󵄨󵄨󵄨 ∇h uk ), uk ) + (Δ̄ h uk−1 , uk ) = 0. Noticing τ󵄩 1 󵄩󵄩 k 󵄩󵄩2 󵄩󵄩 k−1 󵄩󵄩2 󵄩2 (󵄩u 󵄩 − 󵄩u 󵄩󵄩 ) + 󵄩󵄩󵄩∇τ uk 󵄩󵄩󵄩 , 2τ 󵄩 󵄩 󵄩 2 󵄩 󵄩2 (Δ2h uk , uk ) = 󵄩󵄩󵄩Δh uk 󵄩󵄩󵄩 , 󵄨 󵄨2 󵄨 󵄨2 󵄩 󵄩4 −(∇h ⋅ (󵄨󵄨󵄨∇h uk 󵄨󵄨󵄨 ∇h uk ), uk ) = (󵄨󵄨󵄨∇h uk 󵄨󵄨󵄨 ∇h uk , ∇h uk ) = 󵄩󵄩󵄩∇h uk 󵄩󵄩󵄩4 , (∇τ uk , uk ) =

358 � 11 Difference methods for the epitaxial growth model m1 m2

󵄨 󵄨 󵄨 󵄨 −(Δ̄ h uk−1 , uk ) = (∇h uk−1 , ∇h uk ) ⩽ h1 h2 ∑ ∑󵄨󵄨󵄨∇h uijk−1 󵄨󵄨󵄨 ⋅ 󵄨󵄨󵄨∇h uijk 󵄨󵄨󵄨 i=1 j=1

m1 m2

󵄨2 1 󵄨 󵄨2 󵄨 ⩽ h1 h2 ∑ ∑(󵄨󵄨󵄨∇h uijk−1 󵄨󵄨󵄨 󵄨󵄨󵄨∇h uijk 󵄨󵄨󵄨 + ) 4 i=1 j=1 1 1 󵄩 󵄩4 󵄩4 󵄩 ⩽ (󵄩󵄩󵄩∇h uk−1 󵄩󵄩󵄩4 + 󵄩󵄩󵄩∇h uk 󵄩󵄩󵄩4 ) + L1 L2 , 2 4

we have 1 󵄩󵄩 k 󵄩󵄩2 󵄩󵄩 k−1 󵄩󵄩2 τ󵄩 󵄩2 󵄩 󵄩2 󵄩 󵄩4 (󵄩u 󵄩 − 󵄩u 󵄩󵄩 ) + 󵄩󵄩󵄩∇τ uk 󵄩󵄩󵄩 + δ󵄩󵄩󵄩Δh uk 󵄩󵄩󵄩 + 󵄩󵄩󵄩∇h uk 󵄩󵄩󵄩4 2τ 󵄩 󵄩 󵄩 2 1 󵄩 1 󵄩4 󵄩 󵄩4 ⩽ (󵄩󵄩󵄩∇h uk−1 󵄩󵄩󵄩4 + 󵄩󵄩󵄩∇h uk 󵄩󵄩󵄩4 ) + L1 L2 . 2 4 Hence, 1 󵄩󵄩 k 󵄩󵄩2 󵄩 󵄩 k−1 󵄩2 󵄩 k 󵄩4 k−1 󵄩4 󵄩󵄩u 󵄩󵄩 + τ 󵄩󵄩󵄩∇h u 󵄩󵄩󵄩4 ⩽ 󵄩󵄩󵄩u 󵄩󵄩󵄩 + τ 󵄩󵄩󵄩∇h u 󵄩󵄩󵄩4 + L1 L2 τ, 2

1 ⩽ k ⩽ n.

By induction, it follows: 1 󵄩󵄩 k 󵄩󵄩2 󵄩 󵄩 0 󵄩2 󵄩 k 󵄩4 0 󵄩4 󵄩󵄩u 󵄩󵄩 + τ 󵄩󵄩󵄩∇h u 󵄩󵄩󵄩4 ⩽ 󵄩󵄩󵄩u 󵄩󵄩󵄩 + τ 󵄩󵄩󵄩∇h u 󵄩󵄩󵄩4 + L1 L2 kτ, 2

1 ⩽ k ⩽ n.

(II) Taking the inner product on both the right- and left-hand sides of (11.24) with ∇τ uk produces 󵄩󵄩 󵄨 k 󵄩2 2 k k k 󵄨2 k k k−1 k 󵄩󵄩∇τ u 󵄩󵄩󵄩 + δ(Δh u , ∇τ u ) − (∇h ⋅ (󵄨󵄨󵄨∇h u 󵄨󵄨󵄨 ∇h u ), ∇τ u ) + (Δ̄ h u , ∇τ u ) = 0. Noticing 1 󵄩 󵄩2 (Δ2h uk , ∇τ uk ) = (Δh uk , ∇τ (Δh uk )) = ∇τ (󵄩󵄩󵄩Δh uk 󵄩󵄩󵄩 ) + 2 󵄨 󵄨2 󵄨 󵄨2 −(∇h ⋅ (󵄨󵄨󵄨∇h uk 󵄨󵄨󵄨 ∇h uk ), ∇τ uk ) = (󵄨󵄨󵄨∇h uk 󵄨󵄨󵄨 ∇h uk , ∇τ ∇h uk )

τ 󵄩󵄩 k 󵄩2 󵄩󵄩∇τ (Δh u )󵄩󵄩󵄩 , 2

m1 m2

󵄨 󵄨2 = h1 h2 ∑ ∑󵄨󵄨󵄨∇h uijk 󵄨󵄨󵄨 (∇h uijk ) ⋅ ∇τ (∇h uijk ) i=1 j=1 m1 m2

󵄨 󵄨2 󵄨 󵄨2 1 = h2 h2 ∑ ∑󵄨󵄨󵄨∇h uijk 󵄨󵄨󵄨 ( ∇τ (󵄨󵄨󵄨∇h uijk 󵄨󵄨󵄨 ) + 2 i=1 j=1 m m

1 2 1 󵄨 󵄨2 󵄨 󵄨2 = h1 h2 ∑ ∑󵄨󵄨󵄨∇h uijk 󵄨󵄨󵄨 (∇τ (󵄨󵄨󵄨∇h uijk 󵄨󵄨󵄨 )) 2 i=1 j=1

m m

1 2 τ 󵄨 󵄨2 󵄨 󵄨2 + h1 h2 ∑ ∑󵄨󵄨󵄨∇h uijk 󵄨󵄨󵄨 ⋅ 󵄨󵄨󵄨∇τ (∇h uijk )󵄨󵄨󵄨 2 i=1 j=1

τ 󵄨󵄨 k 󵄨2 󵄨∇ (∇ u )󵄨󵄨 ) 2 󵄨 τ h ij 󵄨

11.3 Two-level nonlinear backward Euler difference scheme

� 359

m m

1 2 τ 1 1 󵄨 󵄨4 󵄨 󵄨2 2 = h1 h2 ∑ ∑[ ∇τ (󵄨󵄨󵄨∇h uijk 󵄨󵄨󵄨 ) + (∇τ (󵄨󵄨󵄨∇h uijk 󵄨󵄨󵄨 )) ] 2 2 2 i=1 j=1

m m

1 2 τ 󵄨2 󵄨 󵄨2 󵄨 + h1 h2 ∑ ∑󵄨󵄨󵄨∇h uijk 󵄨󵄨󵄨 ⋅ 󵄨󵄨󵄨∇h ∇τ uijk 󵄨󵄨󵄨 , 2 i=1 j=1

1 󵄩 󵄩2 −(Δ̄ h uk−1 , ∇τ uk ) = (∇h uk−1 , ∇h ∇τ uk ) = ∇τ (󵄩󵄩󵄩∇h uk 󵄩󵄩󵄩 ) − 2

τ 󵄩󵄩 k 󵄩2 󵄩∇ ∇ u 󵄩󵄩 , 2󵄩 h τ 󵄩

we have δ 1 1 󵄩 󵄩2 󵄩 󵄩4 󵄩 󵄩2 ∇ (󵄩󵄩Δ uk 󵄩󵄩 ) + ∇τ (󵄩󵄩󵄩∇h uk 󵄩󵄩󵄩4 ) − ∇τ (󵄩󵄩󵄩∇h uk 󵄩󵄩󵄩 ) ⩽ 0, 2 τ 󵄩 h 󵄩 4 2

1 ⩽ k ⩽ n,

i. e., ∇τ E k ⩽ 0,

1 ⩽ k ⩽ n.

Combining Theorem 11.3 with Lemma 11.2 leads to the following result. Theorem 11.4. Suppose {uijk | 1 ⩽ i ⩽ m1 , 1 ⩽ j ⩽ m2 , 0 ⩽ k ⩽ n} is the solution of the difference scheme (11.24)–(11.25). Then there is a constant c2 such that 󵄩󵄩 k 󵄩󵄩 󵄩󵄩u 󵄩󵄩∞ ⩽ c2 ,

0 ⩽ k ⩽ n.

11.3.3 Existence and uniqueness of the difference solution Theorem 11.5. If τ < 8δ, the solution of the difference scheme (11.24)–(11.25) exists. Proof. Assume that the value of uk−1 at the (k − 1)-th time level is determined. Then a nonlinear system in uk can be obtained from (11.24) as 1 k 󵄨 󵄨2 (u − uijk−1 ) + δΔ2h uijk − ∇h ⋅ (󵄨󵄨󵄨∇h uijk 󵄨󵄨󵄨 ∇h uijk ) + Δ̄ h uijk−1 = 0, τ ij 1 ⩽ i ⩽ m1 , 1 ⩽ j ⩽ m2 .

(11.26)

Define the operator Π : 𝒲 h → 𝒲 h by 1 (w − uijk−1 ) + δΔ2h wij − ∇h ⋅ (|∇h wij |2 ∇h wij ) + Δ̄ h uijk−1 , τ ij 1 ⩽ i ⩽ m1 , 1 ⩽ j ⩽ m2 .

Π(w)ij =

Taking the inner product of Π(w) with w yields 1 (‖w‖2 − (uk−1 , w)) + δ(Δ2h w, w) − (∇h ⋅ (|∇h w|2 ∇h w), w) + (Δ̄ h uk−1 , w) τ 1 = (‖w‖2 − (uk−1 , w)) + δ‖Δh w‖2 + (|∇h w|2 ∇h w, ∇h w) − (∇h uk−1 , ∇h w). τ

(Π(w), w) =

360 � 11 Difference methods for the epitaxial growth model Using Lemma 11.1, we have 1 1 󵄩󵄩 k−1 󵄩2 󵄩∇ u 󵄩󵄩󵄩 ⩽ ‖w‖ ⋅ ‖Δh w‖ + 2󵄩 h 2 1 1󵄩 󵄩2 ⩽ δ‖Δh w‖2 + ‖w‖2 + 󵄩󵄩󵄩∇h uk−1 󵄩󵄩󵄩 . 16δ 2

1 (∇h uk−1 , ∇h w) ⩽ ‖∇h w‖2 + 2

1 󵄩󵄩 k−1 󵄩2 󵄩∇ u 󵄩󵄩󵄩 2󵄩 h

Consequently, 1 1 1󵄩 󵄩 󵄩 󵄩2 (‖w‖2 − 󵄩󵄩󵄩uk−1 󵄩󵄩󵄩 ⋅ ‖w‖) − ‖w‖2 − 󵄩󵄩󵄩∇h uk−1 󵄩󵄩󵄩 τ 16δ 2 1 1 1󵄩 1 1󵄩 󵄩2 󵄩2 ⩾ ( ‖w‖2 − 󵄩󵄩󵄩uk−1 󵄩󵄩󵄩 ) − ‖w‖2 − 󵄩󵄩󵄩∇h uk−1 󵄩󵄩󵄩 τ 2 2 16δ 2 1 τ 2 󵄩 󵄩 󵄩 󵄩2 = (‖w‖2 − 󵄩󵄩󵄩uk−1 󵄩󵄩󵄩 − ‖w‖2 − τ 󵄩󵄩󵄩∇h uk−1 󵄩󵄩󵄩 ) 2τ 8δ τ 1 󵄩 󵄩2 󵄩 󵄩2 = [(1 − )‖w‖2 − (󵄩󵄩󵄩uk−1 󵄩󵄩󵄩 + τ 󵄩󵄩󵄩∇h uk−1 󵄩󵄩󵄩 )]. 2τ 8δ

(Π(w), w) ⩾

If

τ 8δ

< 1 and ‖w‖2 =

1 τ 1− 8δ

(‖uk−1 ‖2 + τ‖∇h uk−1 ‖2 ), (Π(w), w) ⩾ 0 is followed. By the

Browder theorem (Theorem 2.4), there is a w∗ ∈ 𝒲h satisfying ‖w∗ ‖2 ⩽

τ‖∇h uk−1 ‖2 ) such that

1 τ 1− 8δ

(‖uk−1 ‖2 +

Π(w∗ ) = 0, which reveals the existence of the solution to (11.26). Next, the uniqueness of the solution will be concerned. To this end, a useful lemma is prepared here. Lemma 11.3. For any u ∈ 𝒲h and v ∈ 𝒲h , denote εij = uij − vij . Then we have |∇h uij |2 − |∇h vij |2 = ∇h uij ⋅ ∇h εij + ∇h vij ⋅ ∇h εij ,

|∇h uij |2 ∇h uij − |∇h vij |2 ∇h vij = |∇h vij |2 ∇h εij + (∇h uij ⋅ ∇h εij + ∇h vij ⋅ ∇h εij )∇h uij . Proof. (I) |∇h uij |2 − |∇h vij |2

= ∇h uij ⋅ ∇h uij − ∇h vij ⋅ ∇h vij

= ∇h uij ⋅ (∇h uij − ∇h vij ) + (∇h uij − ∇h vij ) ⋅ ∇h vij

= ∇h uij ⋅ ∇h εij + ∇h vij ⋅ ∇h εij ; (II) |∇h uij |2 ∇h uij − |∇h vij |2 ∇h vij

11.3 Two-level nonlinear backward Euler difference scheme

� 361

= |∇h vij |2 (∇h uij − ∇h vij ) + (|∇h uij |2 − |∇h vij |2 )∇h uij

= |∇h vij |2 ∇h εij + (∇h uij ⋅ ∇h εij + ∇h vij ⋅ ∇h εij )∇h uij .

Theorem 11.6. The solution of the difference scheme (11.24)–(11.25) is unique. Proof. Suppose that (11.26) has another solution vk ∈ 𝒲h , i. e., there is a vk ∈ 𝒲h such that 1 k 󵄨 󵄨2 (v − uijk−1 ) + δΔ2h vkij − ∇h ⋅ (󵄨󵄨󵄨∇h vkij 󵄨󵄨󵄨 ∇h vkij ) + Δ̄ h uijk−1 = 0, τ ij 1 ⩽ i ⩽ m1 , 1 ⩽ j ⩽ m2 .

(11.27)

Let ρij = uijk − vkij . Subtracting (11.27) from (11.26) produces 1 󵄨 󵄨2 󵄨 󵄨2 ρ + δΔ2h ρij − ∇h ⋅ (󵄨󵄨󵄨∇h uijk 󵄨󵄨󵄨 ∇h uijk − 󵄨󵄨󵄨∇h vkij 󵄨󵄨󵄨 ∇h vkij ) = 0, τ ij 1 ⩽ i ⩽ m1 , 1 ⩽ j ⩽ m2 .

(11.28)

Taking the inner product on both the right- and left-hand sides of (11.28) with ρ gives 1 󵄨 󵄨2 󵄨 󵄨2 ‖ρ‖2 + δ‖Δh ρ‖2 + (󵄨󵄨󵄨∇h uk 󵄨󵄨󵄨 ∇h uk − 󵄨󵄨󵄨∇h vk 󵄨󵄨󵄨 ∇h vk , ∇h ρ) = 0. τ It follows from Lemma 11.3 that m m

1 2 1 󵄨 󵄨2 ‖ρ‖2 + δ‖Δh ρ‖2 + h1 h2 ∑ ∑(󵄨󵄨󵄨∇h vkij 󵄨󵄨󵄨 ∇h ρij + (∇h uijk ⋅ ∇h ρij + ∇h vkij ⋅ ∇h ρij )∇h uijk ) ⋅ ∇h ρij = 0; τ i=1 j=1

namely, m m

1 2 1 2 󵄨 󵄨2 󵄨 󵄨2 ‖ρ‖2 + δ‖Δh ρ‖2 + h1 h2 ∑ ∑[󵄨󵄨󵄨∇h vkij 󵄨󵄨󵄨 ⋅ 󵄨󵄨󵄨∇h ρij 󵄨󵄨󵄨 + (∇h uijk ⋅ ∇h ρij ) ] τ i=1 j=1

m1 m2

= −h1 h2 ∑ ∑(∇h vkij ⋅ ∇h ρij )(∇h uijk ⋅ ∇h ρij ) i=1 j=1 m m

1 2 1 2 2 ⩽ h1 h2 ∑ ∑[(∇h vkij ⋅ ∇h ρij ) + (∇h uijk ⋅ ∇h ρij ) ], 2 i=1 j=1

so that ‖ρ‖ = 0 follows.

362 � 11 Difference methods for the epitaxial growth model 11.3.4 Convergence of the difference solution Theorem 11.7. Suppose {Uijk | 1 ⩽ i ⩽ m1 , 1 ⩽ j ⩽ m2 , 0 ⩽ k ⩽ n} and {uijk | 1 ⩽ i ⩽ m1 , 1 ⩽ j ⩽ m2 , 0 ⩽ k ⩽ n} are solutions of the problem (11.1)–(11.2) and the difference scheme (11.24)–(11.25), respectively. Define the error function: ũ ijk = Uijk − uijk ,

1 ⩽ i ⩽ m1 , 1 ⩽ j ⩽ m2 , 0 ⩽ k ⩽ n.

Then there is a positive constant c3 such that 󵄩󵄩 ̃ k 󵄩󵄩 2 2 󵄩󵄩u 󵄩󵄩 ⩽ c3 (τ + h1 + h2 ),

0 ⩽ k ⩽ n.

(11.29)

Proof. Let vk ∈ 𝒲h and wk ∈ 𝒲h be defined by (11.21)–(11.22). Then the difference scheme (11.24)–(11.25) can be written as (11.20)–(11.23). Denote ṽkij = Vijk − vkij ,

w̃ ijk = Wijk − wijk ,

1 ⩽ i ⩽ m1 , 1 ⩽ j ⩽ m2 , 0 ⩽ k ⩽ n.

Subtracting (11.20)–(11.23) from (11.13)–(11.15), (11.19), respectively, we obtain { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { {

∇τ ũ ijk + δΔ2h ũ ijk − Δx ṽkij − Δy w̃ ijk + Δ̄ h ũ ijk−1 = Pijk , 1 ⩽ i ⩽ m1 , 1 ⩽ j ⩽ m2 , 1 ⩽ k ⩽ n, 󵄨 󵄨2 󵄨 󵄨2 = 󵄨󵄨󵄨∇h Uijk 󵄨󵄨󵄨 Δx Uijk − 󵄨󵄨󵄨∇h uijk 󵄨󵄨󵄨 Δx uijk + Qijk , 1 ⩽ i ⩽ m1 , 1 ⩽ j ⩽ m2 , 1 ⩽ k ⩽ n, 󵄨 󵄨2 󵄨 󵄨2 k w̃ ij = 󵄨󵄨󵄨∇h Uijk 󵄨󵄨󵄨 Δy Uijk − 󵄨󵄨󵄨∇h uijk 󵄨󵄨󵄨 Δy uijk + Rkij , 1 ⩽ i ⩽ m1 , 1 ⩽ j ⩽ m2 , 1 ⩽ k ⩽ n, ṽkij

ũ ij0

= 0,

1 ⩽ i ⩽ m1 , 1 ⩽ j ⩽ m2 .

(11.30) (11.31) (11.32) (11.33)

From (11.33), the truth of (11.29) for k = 0 is apparent. Taking the inner product on both the right- and left-hand sides of (11.30) with ũ k gives (∇τ ũ k , ũ k ) + δ(Δ2h ũ k , ũ k ) − (Δx ṽk , ũ k ) − (Δy w̃ k , ũ k ) + (Δ̄ h ũ k−1 , ũ k ) = (Pk , ũ k ), 1 ⩽ k ⩽ n.

(11.34)

It is easy to know that 1 󵄩󵄩 k 󵄩󵄩2 󵄩󵄩 k−1 󵄩󵄩2 (󵄩ũ 󵄩 − 󵄩ũ 󵄩󵄩 ) + 2τ 󵄩 󵄩 󵄩 󵄩 󵄩2 (Δ2h ũ k , ũ k ) = 󵄩󵄩󵄩Δh ũ k 󵄩󵄩󵄩 . (∇τ ũ k , ũ k ) =

It follows from (11.31)–(11.32) that

τ 󵄩󵄩 k 󵄩2 󵄩∇ ũ 󵄩󵄩 , 2󵄩 τ 󵄩

(11.35) (11.36)

11.3 Two-level nonlinear backward Euler difference scheme

� 363

−(Δx ṽk , ũ k ) − (Δy w̃ k , ũ k )

= (ṽk , Δx ũ k ) + (w̃ k , Δy ũ k ) 󵄨 󵄨2 󵄨 󵄨2 = (󵄨󵄨󵄨∇h U k 󵄨󵄨󵄨 Δx U k − 󵄨󵄨󵄨∇h uk 󵄨󵄨󵄨 Δx uk + Qk , Δx ũ k ) 󵄨 󵄨2 󵄨 󵄨2 + (󵄨󵄨󵄨∇h U k 󵄨󵄨󵄨 Δy U k − 󵄨󵄨󵄨∇h uk 󵄨󵄨󵄨 Δy uk + Rk , Δy ũ k ) 󵄨2 󵄨 󵄨2 󵄨 󵄨 󵄨2 = (󵄨󵄨󵄨∇h uk 󵄨󵄨󵄨 Δx ũ k + (󵄨󵄨󵄨∇h U k 󵄨󵄨󵄨 − 󵄨󵄨󵄨∇h uk 󵄨󵄨󵄨 )Δx U k + Qk , Δx ũ k ) 󵄨2 󵄨 󵄨 󵄨 󵄨 󵄨2 + (󵄨󵄨󵄨∇h uk 󵄨󵄨󵄨 Δy ũ k + (󵄨󵄨󵄨∇h U k 󵄨󵄨󵄨 − 󵄨󵄨󵄨∇h uk 󵄨󵄨󵄨 )Δy U k + Rk , Δy ũ k ) m1 m2

󵄨 󵄨2 󵄨 󵄨2 = h1 h2 ∑ ∑󵄨󵄨󵄨∇h uijk 󵄨󵄨󵄨 ⋅ 󵄨󵄨󵄨∇h ũ ijk 󵄨󵄨󵄨 i=1 j=1

m1 m2

󵄨 󵄨2 󵄨 󵄨2 + h1 h2 ∑ ∑(󵄨󵄨󵄨∇h Uijk 󵄨󵄨󵄨 − 󵄨󵄨󵄨∇h uijk 󵄨󵄨󵄨 )∇h Uijk ⋅ ∇h ũ ijk i=1 j=1

+ (Q , Δx ũ k ) + (Rk , Δy ũ k ). k

(11.37)

In addition, (Δ̄ h ũ k−1 , ũ k ) = −(∇h ũ k−1 , ∇h ũ k ).

(11.38)

Plugging (11.35)–(11.38) into (11.34) leads to m m

1 2 1 󵄩󵄩 k 󵄩󵄩2 󵄩󵄩 k−1 󵄩󵄩2 󵄩 󵄩2 󵄨 󵄨2 󵄨 󵄨2 (󵄩󵄩ũ 󵄩󵄩 − 󵄩󵄩ũ 󵄩󵄩 ) + δ󵄩󵄩󵄩Δh ũ k 󵄩󵄩󵄩 + h1 h2 ∑ ∑󵄨󵄨󵄨∇h uijk 󵄨󵄨󵄨 ⋅ 󵄨󵄨󵄨∇h ũ ijk 󵄨󵄨󵄨 2τ i=1 j=1

m1 m2

󵄨 󵄨2 󵄨 󵄨2 ⩽ −h1 h2 ∑ ∑(󵄨󵄨󵄨∇h Uijk 󵄨󵄨󵄨 − 󵄨󵄨󵄨∇h uijk 󵄨󵄨󵄨 )∇h Uijk ⋅ ∇h ũ ijk + (∇h ũ k−1 , ∇h ũ k ) i=1 j=1

k

− (Q , Δx ũ k ) − (Rk , Δy ũ k ) + (Pk , ũ k ),

1 ⩽ k ⩽ n.

(11.39)

Denote c4 =

max

̄ (x,y)∈Ω,0⩽t⩽T

{ux2 (x, y, t)} +

max

̄ (x,y)∈Ω,0⩽t⩽T

{uy2 (x, y, t)}.

(11.40)

Then we have 󵄨󵄨 k 󵄨2 󵄨󵄨∇h Uij 󵄨󵄨󵄨 ⩽ c4 . It follows from Lemma 11.3 that m1 m2

󵄨 󵄨2 󵄨 󵄨2 −h1 h2 ∑ ∑(󵄨󵄨󵄨∇h Uijk 󵄨󵄨󵄨 − 󵄨󵄨󵄨∇h uijk 󵄨󵄨󵄨 )∇h Uijk ⋅ ∇h ũ ijk i=1 j=1 m1 m2

= −h1 h2 ∑ ∑(∇h Uijk ⋅ ∇h ũ ijk + ∇h uijk ⋅ ∇h ũ ijk )∇h Uijk ⋅ ∇h ũ ijk i=1 j=1

(11.41)

364 � 11 Difference methods for the epitaxial growth model m1 m2

󵄨 󵄨 󵄨 󵄨󵄨 󵄨 󵄨 󵄨 󵄨 󵄨 󵄨 󵄨 ⩽ h1 h2 ∑ ∑(󵄨󵄨󵄨∇h Uijk 󵄨󵄨󵄨 ⋅ 󵄨󵄨󵄨∇h ũ ijk 󵄨󵄨󵄨 + 󵄨󵄨󵄨∇h uijk 󵄨󵄨󵄨 ⋅ 󵄨󵄨󵄨∇h ũ ijk 󵄨󵄨󵄨)󵄨󵄨󵄨∇h Uijk 󵄨󵄨󵄨 ⋅ 󵄨󵄨󵄨∇h ũ ijk 󵄨󵄨󵄨 i=1 j=1 m1 m2

󵄨2 󵄨 󵄨2 󵄨 = h1 h2 ∑ ∑󵄨󵄨󵄨∇h Uijk 󵄨󵄨󵄨 ⋅ 󵄨󵄨󵄨∇h ũ ijk 󵄨󵄨󵄨 i=1 j=1

m1 m1

󵄨 󵄨 󵄨 󵄨 󵄨 󵄨 󵄨 󵄨 + h1 h2 ∑ ∑(󵄨󵄨󵄨∇h uijk 󵄨󵄨󵄨 ⋅ 󵄨󵄨󵄨∇h ũ ijk 󵄨󵄨󵄨)(󵄨󵄨󵄨∇h Uijk 󵄨󵄨󵄨 ⋅ 󵄨󵄨󵄨∇h ũ ijk 󵄨󵄨󵄨) i=1 j=1

m1 m2

󵄨 󵄨2 󵄨 󵄨2 ⩽ h1 h2 ∑ ∑󵄨󵄨󵄨∇h Uijk 󵄨󵄨󵄨 ⋅ 󵄨󵄨󵄨∇h ũ ijk 󵄨󵄨󵄨 i=1 j=1

m m

1 2 󵄨 󵄨2 󵄨 󵄨2 1 󵄨 󵄨2 󵄨 󵄨2 + h1 h2 ∑ ∑(󵄨󵄨󵄨∇h uijk 󵄨󵄨󵄨 ⋅ 󵄨󵄨󵄨∇h ũ ijk 󵄨󵄨󵄨 + 󵄨󵄨󵄨∇h Uijk 󵄨󵄨󵄨 ⋅ 󵄨󵄨󵄨∇h ũ ijk 󵄨󵄨󵄨 ) 4 i=1 j=1

m m

m m

1 2 1 2 󵄨 󵄨2 󵄨 󵄨 󵄨2 󵄨2 󵄨 󵄨2 5 = h1 h2 ∑ ∑󵄨󵄨󵄨∇h uijk 󵄨󵄨󵄨 ⋅ 󵄨󵄨󵄨∇h ũ ijk 󵄨󵄨󵄨 + h1 h2 ∑ ∑󵄨󵄨󵄨∇h Uijk 󵄨󵄨󵄨 ⋅ 󵄨󵄨󵄨∇h ũ ijk 󵄨󵄨󵄨 4 i=1 j=1 i=1 j=1

m m

1 2 󵄩2 󵄨 󵄨2 󵄨 󵄨2 5 󵄩 ⩽ h1 h2 ∑ ∑󵄨󵄨󵄨∇h uijk 󵄨󵄨󵄨 ⋅ 󵄨󵄨󵄨∇h ũ ijk 󵄨󵄨󵄨 + c4 󵄩󵄩󵄩∇h ũ k 󵄩󵄩󵄩 . 4 i=1 j=1

In addition, 1󵄩 󵄩2 1 󵄩 󵄩2 (∇h ũ k−1 , ∇h ũ k ) ⩽ 󵄩󵄩󵄩∇h ũ k−1 󵄩󵄩󵄩 + 󵄩󵄩󵄩∇h ũ k 󵄩󵄩󵄩 , 2 2 󵄩 󵄩2 1 󵄩 󵄩2 󵄩 󵄩2 −(Qk , Δx ũ k ) − (Rk , Δy ũ k ) ⩽ 󵄩󵄩󵄩Qk 󵄩󵄩󵄩 + 󵄩󵄩󵄩Δx ũ k 󵄩󵄩󵄩 + 󵄩󵄩󵄩Rk 󵄩󵄩󵄩 + 4 󵄩󵄩 k 󵄩󵄩2 󵄩󵄩 k 󵄩󵄩2 1 󵄩󵄩 ̃ k 󵄩󵄩2 = 󵄩󵄩Q 󵄩󵄩 + 󵄩󵄩R 󵄩󵄩 + 󵄩󵄩∇h u 󵄩󵄩 , 4 1 2 2 󵄩 󵄩 󵄩 󵄩 k k k k (P , ũ ) ⩽ 󵄩󵄩󵄩P 󵄩󵄩󵄩 + 󵄩󵄩󵄩ũ 󵄩󵄩󵄩 . 4

1 󵄩󵄩 k 󵄩󵄩2 󵄩Δ ũ 󵄩 4󵄩 y 󵄩

Substituting the four inequalities above into (11.39) and using Lemma 11.1 arrive at 1 󵄩󵄩 k 󵄩󵄩2 󵄩󵄩 k−1 󵄩󵄩2 󵄩 󵄩2 (󵄩ũ 󵄩 − 󵄩ũ 󵄩󵄩 ) + δ󵄩󵄩󵄩Δh ũ k 󵄩󵄩󵄩 2τ 󵄩 󵄩 󵄩 3 5 󵄩 󵄩2 1 󵄩 󵄩2 1 󵄩 󵄩2 ⩽ ( + c4 )󵄩󵄩󵄩∇h ũ k 󵄩󵄩󵄩 + 󵄩󵄩󵄩∇h ũ k−1 󵄩󵄩󵄩 + 󵄩󵄩󵄩ũ k 󵄩󵄩󵄩 4 4 2 4 󵄩󵄩 k 󵄩󵄩2 󵄩󵄩 k 󵄩󵄩2 󵄩󵄩 k 󵄩󵄩2 + 󵄩󵄩P 󵄩󵄩 + 󵄩󵄩Q 󵄩󵄩 + 󵄩󵄩R 󵄩󵄩 3 5 󵄩 󵄩 󵄩 󵄩 1󵄩 󵄩 󵄩 󵄩 ⩽ ( + c4 )󵄩󵄩󵄩Δh ũ k 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩ũ k 󵄩󵄩󵄩 + 󵄩󵄩󵄩Δh ũ k−1 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩ũ k−1 󵄩󵄩󵄩 + 4 4 2 󵄩 󵄩2 󵄩 󵄩2 󵄩 󵄩2 + 󵄩󵄩󵄩Pk 󵄩󵄩󵄩 + 󵄩󵄩󵄩Qk 󵄩󵄩󵄩 + 󵄩󵄩󵄩Rk 󵄩󵄩󵄩

⩽

2

1 󵄩󵄩 k 󵄩󵄩2 󵄩ũ 󵄩 4󵄩 󵄩

1 3 5 1 1 󵄩󵄩 k−1 󵄩󵄩2 δ 󵄩󵄩 󵄩 k 󵄩2 δ 󵄩 k 󵄩2 k−1 󵄩2 ⋅ 󵄩ũ 󵄩󵄩 󵄩Δ ũ 󵄩󵄩 + ( + c4 ) 󵄩󵄩󵄩ũ 󵄩󵄩󵄩 + 󵄩󵄩󵄩Δh ũ 󵄩󵄩󵄩 + 2󵄩 h 󵄩 2δ 4 4 2 2δ 4 󵄩 1 󵄩 󵄩2 󵄩 󵄩2 󵄩 󵄩2 󵄩 󵄩2 + 󵄩󵄩󵄩ũ k 󵄩󵄩󵄩 + 󵄩󵄩󵄩Pk 󵄩󵄩󵄩 + 󵄩󵄩󵄩Qk 󵄩󵄩󵄩 + 󵄩󵄩󵄩Rk 󵄩󵄩󵄩 , 1 ⩽ k ⩽ n. 4

11.4 Two-level linearized backward Euler difference scheme

� 365

It follows by noticing (11.16)–(11.18) that 1 󵄩󵄩 k 󵄩󵄩2 󵄩2 󵄩 󵄩2 󵄩 󵄩2 󵄩 [(󵄩ũ 󵄩 + τδ󵄩󵄩󵄩Δh ũ k 󵄩󵄩󵄩 ) − (󵄩󵄩󵄩ũ k−1 󵄩󵄩󵄩 + τδ󵄩󵄩󵄩Δh ũ k−1 󵄩󵄩󵄩 )] 2τ 󵄩 󵄩 2

1 1 󵄩 1 3 5 2 󵄩2 󵄩 󵄩2 ⩽ [ + ( + c4 ) ]󵄩󵄩󵄩ũ k 󵄩󵄩󵄩 + 󵄩󵄩󵄩ũ k−1 󵄩󵄩󵄩 + 3L1 L2 c12 (τ + h12 + h22 ) , 4 2δ 4 4 8δ

1 ⩽ k ⩽ n.

Denote 󵄩 󵄩2 󵄩 󵄩2 F k = 󵄩󵄩󵄩ũ k 󵄩󵄩󵄩 + τδ󵄩󵄩󵄩Δh ũ k 󵄩󵄩󵄩 . Then it follows: 2

1 1 3 5 1 k 2 (F − F k−1 ) ⩽ [ + ( + c4 ) ](F k + F k−1 ) + 3L1 L2 c12 (τ + h12 + h22 ) , 2τ 4 2δ 4 4 1 ⩽ k ⩽ n, i. e., 2

1 1 3 5 {1 − [ + ( + c4 ) ]τ}F k 2 δ 4 4 2

1 1 3 5 2 ⩽ {1 + [ + ( + c4 ) ]τ}F k−1 + 6L1 L2 c12 τ(τ + h12 + h22 ) , 2 δ 4 4

1 ⩽ k ⩽ n.

When [ 21 + δ1 ( 43 + 45 c4 )2 ]τ ⩽ 31 , we have 2

1 1 3 5 2 F k ⩽ {1 + 3[ + ( + c4 ) ]τ}F k−1 + 9L1 L2 c12 τ(τ + h12 + h22 ) , 2 δ 4 4

1 ⩽ k ⩽ n.

Noticing F 0 = 0, with the help of the Gronwall inequality (Theorem 1.2(a)), one gets 󵄩󵄩 ̃ k 󵄩󵄩2 k 3[ 1 + 1 ( 3 + 5 c )2 ]T ⋅ 󵄩󵄩u 󵄩󵄩 ⩽ F ⩽ e 2 δ 4 4 4

1 2

3L1 L2 c12 +

1 3 ( δ 4

+

5 c )2 4 4

2

(τ + h12 + h22 ) ,

1 ⩽ k ⩽ n.

Taking the square root on both the right- and left-hand sides will yield the desired result.

11.4 Two-level linearized backward Euler difference scheme This section will focus on a two-level linearized backward Euler difference scheme for solving (11.1)–(11.2).

366 � 11 Difference methods for the epitaxial growth model 11.4.1 Derivation of the difference scheme Considering equations (11.10)–(11.12) at the point (xi , yj , tk ) and using the Taylor expansion, we have ∇τ Uijk + δΔ2h Uijk − Δx Vijk − Δy Wijk + Δ̄ h Uijk−1 = P̂ ijk , { { { { { { { 1 ⩽ i ⩽ m1 , 1 ⩽ j ⩽ m2 , 1 ⩽ k ⩽ n, 󵄨 󵄨2 { { Vijk = 󵄨󵄨󵄨∇h Uijk−1 󵄨󵄨󵄨 Δx Uijk + Q̂ ijk , 1 ⩽ i ⩽ m1 , 1 ⩽ j ⩽ m2 , 1 ⩽ k ⩽ n, { { { { { 󵄨󵄨 k k−1 󵄨󵄨2 k ̂k { Wij = 󵄨󵄨∇h Uij 󵄨󵄨 Δy Uij + Rij , 1 ⩽ i ⩽ m1 , 1 ⩽ j ⩽ m2 , 1 ⩽ k ⩽ n,

(11.42) (11.43) (11.44)

where there is a positive constant c5 such that { { { { { { { { {

󵄨󵄨 ̂ k 󵄨󵄨 2 2 󵄨󵄨Pij 󵄨󵄨 ⩽ c5 (τ + h1 + h2 ), 󵄨󵄨 ̂ k 󵄨󵄨 2 2 󵄨󵄨Qij 󵄨󵄨 ⩽ c5 (τ + h1 + h2 ), 󵄨󵄨 ̂ k 󵄨󵄨 2 2 󵄨󵄨Rij 󵄨󵄨 ⩽ c5 (τ + h1 + h2 ),

1 ⩽ i ⩽ m1 , 1 ⩽ j ⩽ m2 , 1 ⩽ k ⩽ n,

(11.45)

1 ⩽ i ⩽ m1 , 1 ⩽ j ⩽ m2 , 1 ⩽ k ⩽ n,

(11.46)

1 ⩽ i ⩽ m1 , 1 ⩽ j ⩽ m2 , 1 ⩽ k ⩽ n.

(11.47)

1 ⩽ i ⩽ m1 , 1 ⩽ j ⩽ m2 ,

(11.48)

Noticing the initial value condition Uij0 = φ(xi , yj ),

and omitting the small terms in (11.42)–(11.44), a difference scheme for solving (11.10)– (11.12) and (11.2) is built as follows: For 0 ⩽ k ⩽ n, find uk , vk , wk ∈ 𝒲h such that { { { { { { { { { { { { { { { { { { { { {

∇τ uijk + δΔ2h uijk − Δx vkij − Δy wijk + Δ̄ h uijk−1 = 0,

1 ⩽ i ⩽ m1 , 1 ⩽ j ⩽ m2 , 1 ⩽ k ⩽ n, 󵄨 󵄨2 vkij = 󵄨󵄨󵄨∇h uijk−1 󵄨󵄨󵄨 Δx uijk , 1 ⩽ i ⩽ m1 , 1 ⩽ j ⩽ m2 , 1 ⩽ k ⩽ n, 󵄨 󵄨2 wijk = 󵄨󵄨󵄨∇h uijk−1 󵄨󵄨󵄨 Δy uijk , 1 ⩽ i ⩽ m1 , 1 ⩽ j ⩽ m2 , 1 ⩽ k ⩽ n, uij0

= φ(xi , yj ),

1 ⩽ i ⩽ m1 , 1 ⩽ j ⩽ m2 .

(11.49) (11.50) (11.51) (11.52)

Substituting (11.50)–(11.51) into (11.49), a difference scheme for solving (11.1)–(11.2) is obtained as follows: For 0 ⩽ k ⩽ n, find uk ∈ 𝒲h such that 󵄨 󵄨2 ∇τ uijk + δΔ2h uijk − ∇h ⋅ (󵄨󵄨󵄨∇h uijk−1 󵄨󵄨󵄨 ∇h uijk ) + Δ̄ h uijk−1 = 0, { { { 1 ⩽ i ⩽ m1 , 1 ⩽ j ⩽ m2 , 1 ⩽ k ⩽ n, { { { 0 { uij = φ(xi , yj ), 1 ⩽ i ⩽ m1 , 1 ⩽ j ⩽ m2 .

(11.53) (11.54)

11.4 Two-level linearized backward Euler difference scheme

� 367

11.4.2 Boundedness of the difference solution Theorem 11.8. Suppose {uijk | 1 ⩽ i ⩽ m1 , 1 ⩽ j ⩽ m2 , 0 ⩽ k ⩽ n} is the solution of the difference scheme (11.53)–(11.54). Then it holds that 󵄩󵄩 k 󵄩󵄩2 󵄩󵄩 0 󵄩󵄩2 1 󵄩󵄩u 󵄩󵄩 ⩽ 󵄩󵄩u 󵄩󵄩 + L1 L2 tk , 2

0 ⩽ k ⩽ n.

Proof. Taking the inner product on both the right- and left-hand sides of (11.53) with uk gives 󵄨 󵄨2 (∇τ uk , uk ) + δ(Δ2h uk , uk ) − (∇h ⋅ (󵄨󵄨󵄨∇h uk−1 󵄨󵄨󵄨 ∇h uk ), uk ) + (Δ̄ h uk−1 , uk ) = 0. Noticing 1 󵄩󵄩 k 󵄩󵄩2 󵄩󵄩 k−1 󵄩󵄩2 (󵄩u 󵄩 − 󵄩u 󵄩󵄩 ) + 2τ 󵄩 󵄩 󵄩 󵄩 󵄩2 (Δ2h uk , uk ) = 󵄩󵄩󵄩Δh uk 󵄩󵄩󵄩 , (∇τ uk , uk ) =

τ 󵄩󵄩 k 󵄩2 󵄩∇ u 󵄩󵄩 , 2󵄩 τ 󵄩

m1 m2

󵄨 󵄨2 󵄨 󵄨2 󵄨 󵄨2 −(∇h ⋅ (󵄨󵄨󵄨∇h uk−1 󵄨󵄨󵄨 ∇h uk ), uk ) = h1 h2 ∑ ∑󵄨󵄨󵄨∇h uijk−1 󵄨󵄨󵄨 ⋅ 󵄨󵄨󵄨∇h uijk 󵄨󵄨󵄨 , i=1 j=1 m1 m2

−(Δ̄ h uk−1 , uk ) = (∇h uk−1 , ∇h uk ) = h1 h2 ∑ ∑ ∇h uijk−1 ⋅ ∇h uijk i=1 j=1 m1 m2

󵄨 󵄨 󵄨 󵄨 ⩽ h1 h2 ∑ ∑󵄨󵄨󵄨∇h uijk−1 󵄨󵄨󵄨 ⋅ 󵄨󵄨󵄨∇h uijk 󵄨󵄨󵄨 i=1 j=1 m m

1 2 󵄨 󵄨2 󵄨 󵄨2 1 ⩽ h1 h2 ∑ ∑(󵄨󵄨󵄨∇h uijk−1 󵄨󵄨󵄨 ⋅ 󵄨󵄨󵄨∇h uijk 󵄨󵄨󵄨 + ), 4 i=1 j=1

we have 1 󵄩󵄩 k 󵄩󵄩2 󵄩󵄩 k−1 󵄩󵄩2 1 (󵄩u 󵄩 − 󵄩u 󵄩󵄩 ) ⩽ L1 L2 , 2τ 󵄩 󵄩 󵄩 4

1 ⩽ k ⩽ n.

It is easy to know 󵄩󵄩 k 󵄩󵄩2 󵄩󵄩 0 󵄩󵄩2 1 󵄩 0 󵄩2 1 󵄩󵄩u 󵄩󵄩 ⩽ 󵄩󵄩u 󵄩󵄩 + L1 L2 kτ = 󵄩󵄩󵄩u 󵄩󵄩󵄩 + L1 L2 tk , 2 2

0 ⩽ k ⩽ n.

This completes the proof.

11.4.3 Existence of the difference solution Theorem 11.9. The difference scheme (11.53)–(11.54) is uniquely solvable.

368 � 11 Difference methods for the epitaxial growth model Proof. The value of u0 is determined by (11.54). Assume that the value of uk−1 at the (k − 1)-th time level has been determined. Then a linear system in uk can be obtained from (11.53). Consider its homogeneous one: 1 k 󵄨 󵄨2 u + δΔ2h uijk − ∇h ⋅ (󵄨󵄨󵄨∇h uijk−1 󵄨󵄨󵄨 ∇h uijk ) = 0, τ ij

1 ⩽ i ⩽ m1 , 1 ⩽ j ⩽ m2 .

(11.55)

Taking the inner product on both the right- and left-hand sides of the equality above with uk gives 1 󵄩󵄩 k 󵄩󵄩2 󵄩 󵄨 k 󵄩2 k−1 󵄨2 k k 󵄩u 󵄩 + δ󵄩󵄩󵄩Δh u 󵄩󵄩󵄩 + (󵄨󵄨󵄨∇h u 󵄨󵄨󵄨 ∇h u , ∇h u ) = 0. τ󵄩 󵄩 Thus, ‖uk ‖ = 0, which implies that (11.55) has only the trivial solution. By induction, the desired result is true.

11.4.4 Convergence of the difference solution Theorem 11.10. Suppose {Uijk | 1 ⩽ i ⩽ m1 , 1 ⩽ j ⩽ m2 , 0 ⩽ k ⩽ n} and {uijk | 1 ⩽ i ⩽ m1 , 1 ⩽ j ⩽ m2 , 0 ⩽ k ⩽ n} are solutions of the problem (11.1)–(11.2) and the difference scheme (11.53)–(11.54), respectively. Define the error function: ũ ijk = Uijk − uijk ,

1 ⩽ i ⩽ m1 , 1 ⩽ j ⩽ m2 , 0 ⩽ k ⩽ n.

Then there is a positive constant c6 such that 󵄩󵄩 ̃ k 󵄩󵄩 2 2 󵄩󵄩u 󵄩󵄩 ⩽ c6 (τ + h1 + h2 ),

0 ⩽ k ⩽ n.

(11.56)

Proof. Let {vk , wk } be defined by (11.50)–(11.51). Then the difference scheme (11.53)– (11.54) can be written as (11.49)–(11.52). Subtracting (11.49)–(11.52) from (11.42)–(11.44), (11.48), respectively, the system of error equations is produced as { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { {

∇τ ũ ijk + δΔ2h ũ ijk − Δx ṽkij − Δy w̃ ijk + Δ̄ h ũ ijk−1 = P̂ ijk , 1 ⩽ i ⩽ m1 , 1 ⩽ j ⩽ m2 , 1 ⩽ k ⩽ n, 󵄨 󵄨2 󵄨 󵄨2 = 󵄨󵄨󵄨∇h Uijk−1 󵄨󵄨󵄨 Δx Uijk − 󵄨󵄨󵄨∇h uijk−1 󵄨󵄨󵄨 Δx uijk + Q̂ ijk , 1 ⩽ i ⩽ m1 , 1 ⩽ j ⩽ m2 , 1 ⩽ k ⩽ n, 󵄨 󵄨2 󵄨 󵄨2 k w̃ ij = 󵄨󵄨󵄨∇h Uijk−1 󵄨󵄨󵄨 Δy Uijk − 󵄨󵄨󵄨∇h uijk−1 󵄨󵄨󵄨 Δy uijk + R̂ kij , 1 ⩽ j ⩽ m1 , 1 ⩽ j ⩽ m2 , 1 ⩽ k ⩽ n,

ṽkij

ũ ij0

= 0,

1 ⩽ i ⩽ m1 , 1 ⩽ j ⩽ m2 .

In view of (11.60), the truth of (11.56) is obvious for k = 0.

(11.57) (11.58) (11.59) (11.60)

11.4 Two-level linearized backward Euler difference scheme

� 369

Taking the inner product on both the right- and left-hand sides of (11.57) with ũ k gives (∇τ ũ k , ũ k ) + δ(Δ2h ũ k , ũ k ) − (Δx ṽk , ũ k ) − (Δy w̃ k , ũ k ) + (Δ̄ h ũ k−1 , ũ k )

= (P̂ k , ũ k ),

1 ⩽ k ⩽ n.

(11.61)

It follows from (11.58) and (11.59) that − (Δx ṽk , ũ k ) − (Δy w̃ k , ũ k )

= (ṽk , Δx ũ k ) + (w̃ k , Δy ũ k ) 󵄨 󵄨2 󵄨 󵄨2 = (󵄨󵄨󵄨∇h U k−1 󵄨󵄨󵄨 Δx U k − 󵄨󵄨󵄨∇h uk−1 󵄨󵄨󵄨 Δx uk + Q̂ k , Δx ũ k ) 󵄨 󵄨2 󵄨 󵄨2 + (󵄨󵄨󵄨∇h U k−1 󵄨󵄨󵄨 Δy U k − 󵄨󵄨󵄨∇h uk−1 󵄨󵄨󵄨 Δy uk + R̂ k , Δy ũ k ) 󵄨 󵄨2 󵄨 󵄨2 󵄨 󵄨2 = (󵄨󵄨󵄨∇h uk−1 󵄨󵄨󵄨 (Δx U k − Δx uk ) + (󵄨󵄨󵄨∇h U k−1 󵄨󵄨󵄨 − 󵄨󵄨󵄨∇h uk−1 󵄨󵄨󵄨 )Δx U k , Δx ũ k ) 󵄨 󵄨2 󵄨 󵄨2 󵄨 󵄨2 + (󵄨󵄨󵄨∇h uk−1 󵄨󵄨󵄨 (Δy U k − Δy uk ) + (󵄨󵄨󵄨∇h U k−1 󵄨󵄨󵄨 − 󵄨󵄨󵄨∇h uk−1 󵄨󵄨󵄨 )Δy U k , Δy ũ k ) + (Q̂ k , Δx ũ k ) + (R̂ k , Δy ũ k ) m1 m2

󵄨 󵄨2 󵄨 󵄨2 = h1 h2 ∑ ∑󵄨󵄨󵄨∇h uijk−1 󵄨󵄨󵄨 󵄨󵄨󵄨∇h ũ ijk 󵄨󵄨󵄨 i=1 j=1

m1 m2

󵄨 󵄨2 󵄨 󵄨2 + h1 h2 ∑ ∑(󵄨󵄨󵄨∇h Uijk−1 󵄨󵄨󵄨 − 󵄨󵄨󵄨∇h uijk−1 󵄨󵄨󵄨 )∇h Uijk ⋅ ∇h ũ ijk i=1 j=1

+ (Q̂ k , Δx ũ k ) + (R̂ k , Δy ũ k ), which yields after the substitution of the above equality into (11.61) that 1 󵄩󵄩 k 󵄩󵄩2 󵄩󵄩 k−1 󵄩󵄩2 (󵄩ũ 󵄩 − 󵄩ũ 󵄩󵄩 ) + 2τ 󵄩 󵄩 󵄩

τ 󵄩󵄩 k 󵄩2 󵄩∇ ũ 󵄩󵄩 2󵄩 τ 󵄩

m1 m2

󵄩 󵄩2 󵄨 󵄨2 󵄨 󵄨2 + δ󵄩󵄩󵄩Δh ũ k 󵄩󵄩󵄩 + h1 h2 ∑ ∑󵄨󵄨󵄨∇h uijk−1 󵄨󵄨󵄨 󵄨󵄨󵄨∇h ũ ijk 󵄨󵄨󵄨 i=1 j=1

m1 m2

󵄨 󵄨2 󵄨 󵄨2 = −h2 h2 ∑ ∑(󵄨󵄨󵄨∇h Uijk−1 󵄨󵄨󵄨 − 󵄨󵄨󵄨∇h uijk−1 󵄨󵄨󵄨 )∇h Uijk ⋅ ∇h ũ ijk + (∇h ũ k−1 , ∇h ũ k ) i=1 j=1

− (Q̂ k , Δx ũ k ) − (R̂ k , Δy ũ k ) + (P̂ k , ũ k ),

1 ⩽ k ⩽ n.

By Lemma 11.3 and (11.40), we have m1 m2

󵄨 󵄨2 󵄨 󵄨2 −h1 h2 ∑ ∑(󵄨󵄨󵄨∇h Uijk−1 󵄨󵄨󵄨 − 󵄨󵄨󵄨∇h uijk−1 󵄨󵄨󵄨 )∇h Uijk ⋅ ∇h ũ ijk i=1 j=1 m1 m2

= −h1 h2 ∑ ∑(∇h Uijk−1 ⋅ ∇h ũ ijk−1 + ∇h uijk−1 ⋅ ∇h ũ ijk−1 )∇h Uijk ⋅ ∇h ũ ijk i=1 j=1

(11.62)

370 � 11 Difference methods for the epitaxial growth model m1 m2

󵄨 󵄨 󵄨 󵄨󵄨 󵄨 󵄨 󵄨 󵄨 󵄨 󵄨 󵄨 ⩽ h1 h2 ∑ ∑(󵄨󵄨󵄨∇h Uijk−1 󵄨󵄨󵄨 ⋅ 󵄨󵄨󵄨∇h ũ ijk−1 󵄨󵄨󵄨 + 󵄨󵄨󵄨∇h uijk−1 󵄨󵄨󵄨 ⋅ 󵄨󵄨󵄨∇h ũ ijk−1 󵄨󵄨󵄨)󵄨󵄨󵄨∇h Uijk 󵄨󵄨󵄨 ⋅ 󵄨󵄨󵄨∇h ũ ijk 󵄨󵄨󵄨 i=1 j=1

m1 m2

󵄨 󵄨 󵄨 󵄨 ⩽ c4 h1 h2 ∑ ∑󵄨󵄨󵄨∇h ũ ijk−1 󵄨󵄨󵄨 ⋅ 󵄨󵄨󵄨∇h ũ ijk 󵄨󵄨󵄨 i=1 j=1

m1 m2

󵄨 󵄨 󵄨 󵄨 󵄨 󵄨 󵄨 󵄨 + h1 h2 ∑ ∑(󵄨󵄨󵄨∇h uijk−1 󵄨󵄨󵄨 ⋅ 󵄨󵄨󵄨∇h ũ ijk 󵄨󵄨󵄨)(󵄨󵄨󵄨∇h ũ ijk−1 󵄨󵄨󵄨 ⋅ 󵄨󵄨󵄨∇h Uijk 󵄨󵄨󵄨) i=1 j=1

m1 m2

󵄨 󵄨 󵄨 󵄨 ⩽ c4 h1 h2 ∑ ∑󵄨󵄨󵄨∇h ũ ijk−1 󵄨󵄨󵄨 ⋅ 󵄨󵄨󵄨∇h ũ ijk 󵄨󵄨󵄨 i=1 j=1 m m

1 2 󵄨 󵄨2 󵄨 󵄨2 1 󵄨 󵄨2 󵄨 󵄨2 + h1 h2 ∑ ∑(󵄨󵄨󵄨∇h uijk−1 󵄨󵄨󵄨 ⋅ 󵄨󵄨󵄨∇h ũ ijk 󵄨󵄨󵄨 + 󵄨󵄨󵄨∇h ũ ijk−1 󵄨󵄨󵄨 󵄨󵄨󵄨∇h Uijk 󵄨󵄨󵄨 ) 4 i=1 j=1

m m

1 2 c 󵄩 󵄩2 󵄩2 󵄩 󵄨 󵄩2 󵄨2 󵄨 󵄨2 c 󵄩 ⩽ h1 h2 ∑ ∑󵄨󵄨󵄨∇h uijk−1 󵄨󵄨󵄨 󵄨󵄨󵄨∇h ũ ijk 󵄨󵄨󵄨 + 4 (󵄩󵄩󵄩∇h ũ k−1 󵄩󵄩󵄩 + 󵄩󵄩󵄩∇h ũ k 󵄩󵄩󵄩 ) + 4 󵄩󵄩󵄩∇h ũ k−1 󵄩󵄩󵄩 . 2 4 i=1 j=1

Inserting the above inequality into (11.62) and using Lemma 11.1 arrive at 1 󵄩󵄩 k 󵄩󵄩2 󵄩󵄩 k−1 󵄩󵄩2 󵄩 󵄩2 (󵄩ũ 󵄩 − 󵄩ũ 󵄩󵄩 ) + δ󵄩󵄩󵄩Δh ũ k 󵄩󵄩󵄩 2τ 󵄩 󵄩 󵄩 3c 󵄩 󵄩2 c 󵄩 󵄩2 ⩽ 4 󵄩󵄩󵄩∇h ũ k−1 󵄩󵄩󵄩 + 4 󵄩󵄩󵄩∇h ũ k 󵄩󵄩󵄩 + (∇h ũ k−1 , ∇h ũ k ) 4 2 − (Q̂ k , Δx ũ k ) − (R̂ k , Δy ũ k ) + (P̂ k , ũ k )

3c4 󵄩󵄩 c 󵄩 1󵄩 1󵄩 k−1 󵄩2 k 󵄩2 k−1 󵄩2 k 󵄩2 󵄩∇ ũ 󵄩󵄩󵄩 + 4 󵄩󵄩󵄩∇h ũ 󵄩󵄩󵄩 + 󵄩󵄩󵄩∇h ũ 󵄩󵄩󵄩 + 󵄩󵄩󵄩∇h ũ 󵄩󵄩󵄩 4 󵄩 h 2 2 2 1󵄩 󵄩2 1 󵄩 󵄩2 1 󵄩 󵄩2 1 󵄩 󵄩2 1 󵄩 󵄩2 + 󵄩󵄩󵄩Δx ũ k 󵄩󵄩󵄩 + 󵄩󵄩󵄩Q̂ k 󵄩󵄩󵄩 + 󵄩󵄩󵄩Δy ũ k 󵄩󵄩󵄩 + 󵄩󵄩󵄩R̂ k 󵄩󵄩󵄩 + 󵄩󵄩󵄩ũ k 󵄩󵄩󵄩 + 2 2 2 2 2 c 3c4 1 󵄩󵄩 2 2 󵄩 󵄩 󵄩 + )󵄩󵄩∇h ũ k−1 󵄩󵄩󵄩 + ( 4 + 1)󵄩󵄩󵄩∇h ũ k 󵄩󵄩󵄩 =( 4 2 2 1 󵄩󵄩 k 󵄩󵄩2 1 󵄩󵄩 ̂ k 󵄩󵄩2 󵄩󵄩 ̂ k 󵄩󵄩2 󵄩󵄩 ̂ k 󵄩󵄩2 + 󵄩󵄩ũ 󵄩󵄩 + (󵄩󵄩P 󵄩󵄩 + 󵄩󵄩Q 󵄩󵄩 + 󵄩󵄩R 󵄩󵄩 ) 2 2 3c4 1 󵄩󵄩 c 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 + )󵄩󵄩Δh ũ k−1 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩ũ k−1 󵄩󵄩󵄩 + ( 4 + 1)󵄩󵄩󵄩Δh ũ k 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩ũ k 󵄩󵄩󵄩 ⩽( 4 2 2 1 󵄩 󵄩2 1 󵄩 󵄩2 󵄩 󵄩2 󵄩 󵄩2 + 󵄩󵄩󵄩ũ k 󵄩󵄩󵄩 + (󵄩󵄩󵄩P̂ k 󵄩󵄩󵄩 + 󵄩󵄩󵄩Q̂ k 󵄩󵄩󵄩 + 󵄩󵄩󵄩R̂ k 󵄩󵄩󵄩 ) 2 2 ⩽

2

⩽

1 󵄩󵄩 ̂ k 󵄩󵄩2 󵄩P 󵄩 2󵄩 󵄩

2

1 󵄩 k−1 󵄩2 δ 󵄩 1 3c 1 c δ 󵄩󵄩 󵄩 k 󵄩2 k−1 󵄩2 k 󵄩2 󵄩Δ ũ 󵄩󵄩󵄩 + ( 4 + ) 󵄩󵄩󵄩ũ 󵄩󵄩󵄩 + 󵄩󵄩󵄩Δh ũ 󵄩󵄩󵄩 + ( 4 + 1) 󵄩󵄩󵄩ũ 󵄩󵄩󵄩 2󵄩 h 2δ 4 2 2 2δ 2 1 󵄩 󵄩2 1 󵄩 󵄩2 󵄩 󵄩2 󵄩 󵄩2 + 󵄩󵄩󵄩ũ k 󵄩󵄩󵄩 + (󵄩󵄩󵄩P̂ k 󵄩󵄩󵄩 + 󵄩󵄩󵄩Q̂ k 󵄩󵄩󵄩 + 󵄩󵄩󵄩R̂ k 󵄩󵄩󵄩 ), 1 ⩽ k ⩽ n. 2 2

Denote 󵄩 󵄩2 󵄩 󵄩2 F k = 󵄩󵄩󵄩ũ k 󵄩󵄩󵄩 + τδ󵄩󵄩󵄩Δh ũ k 󵄩󵄩󵄩 .

11.5 Three-level linearized backward Euler difference scheme

� 371

Noticing (11.45)–(11.47), it follows: 1 k (F − F k−1 ) 2τ ⩽ max{

2

2

1 3c4 1 1 c 1 ( + ) , ( 4 + 1) + }(F k + F k−1 ) 2δ 4 2 2δ 2 2

3 2 + c52 L1 L2 (τ + h12 + h22 ) , 2

1 ⩽ k ⩽ n.

Let 2

2

1 1 c 1 3c c7 = max{ ( 4 + ) , ( 4 + 1) + 1}. δ 4 2 δ 2 Then 2

(1 − c7 τ)F k ⩽ (1 + c7 τ)F k−1 + 3c52 L1 L2 τ(τ + h12 + h22 ) ,

1 ⩽ k ⩽ n.

When c7 τ ⩽ 31 , it follows: 9 2 F k ⩽ (1 + 3c7 τ)F k−1 + c52 L1 L2 τ(τ + h12 + h22 ) , 2

1 ⩽ k ⩽ n.

The application of the Gronwall inequality (Theorem 1.2(a)) leads to F k ⩽ e3c7 kτ ⋅

3c52 L1 L2 2 (τ + h12 + h22 ) , 2c7

1 ⩽ k ⩽ n.

Thus, 2 2 󵄩󵄩 ̃ k 󵄩󵄩2 3c T 3c L1 L2 (τ + h12 + h22 ) , 󵄩󵄩u 󵄩󵄩 ⩽ e 7 ⋅ 5 2c7

1 ⩽ k ⩽ n.

The desired result is attained by taking the square root on both the right- and left-hand sides of the inequality above.

11.5 Three-level linearized backward Euler difference scheme 11.5.1 Derivation of the difference scheme From (11.1)–(11.2), it follows: 󵄨 󵄨2 ut (x, y, 0) = −δΔ2 φ(x, y) + ∇ ⋅ (󵄨󵄨󵄨∇φ(x, y)󵄨󵄨󵄨 ∇φ(x, y)) − Δφ(x, y) ≡ ψ(x, y). Denote φij = φ(xi , yj ),

ψij = ψ(xi , yj ).

372 � 11 Difference methods for the epitaxial growth model Noticing 1 1 [u (x, y, t1 ) + ut (x, y, t0 )] = [u(x, y, t1 ) − u(x, y, t0 )] + O(τ 2 ), 2 t τ one has ut (x, y, t1 ) =

2 [u(x, y, t1 ) − u(x, y, t0 )] − ut (x, y, t0 ) + O(τ 2 ). τ

Together with u(x, y, t1 ) = u(x, y, t0 ) + τut (x, y, t0 ) + O(τ 2 ), it follows by considering equations (11.10)–(11.12) at the point (xi , yj , t1 ) that 2∇τ Uij1 − ψij + δΔ2h Uij1 − Δx Vij1 − Δy Wij1 + Δ̄ h (Uij0 + τψij ) = P̆ ij1 , { { { { { { { 1 ⩽ i ⩽ m1 , 1 ⩽ j ⩽ m2 , 󵄨 󵄨2 { { Vij1 = 󵄨󵄨󵄨∇h (Uij0 + τψij )󵄨󵄨󵄨 Δx Uij1 + Q̆ ij1 , 1 ⩽ i ⩽ m1 , 1 ⩽ j ⩽ m2 , { { { { { 󵄨󵄨 󵄨󵄨2 1 0 1 ̆1 { Wij = 󵄨󵄨∇h (Uij + τψij )󵄨󵄨 Δy Uij + Rij , 1 ⩽ i ⩽ m1 , 1 ⩽ j ⩽ m2 ,

(11.63) (11.64) (11.65)

where there is positive constant c8 such that { { { { { { { { {

󵄨󵄨 ̆ 1 󵄨󵄨 2 2 2 󵄨󵄨Pij 󵄨󵄨 ⩽ c8 (τ + h1 + h2 ), 󵄨󵄨󵄨Q̆ 1 󵄨󵄨󵄨 ⩽ c8 (τ 2 + h2 + h2 ), 1 2 󵄨 ij 󵄨 󵄨󵄨 ̆ 1 󵄨󵄨 2 2 2 R ⩽ c (τ + h + h 8 1 2 ), 󵄨󵄨 ij 󵄨󵄨

1 ⩽ i ⩽ m1 , 1 ⩽ j ⩽ m2 ,

(11.66)

1 ⩽ i ⩽ m1 , 1 ⩽ j ⩽ m2 ,

(11.67)

1 ⩽ i ⩽ m1 , 1 ⩽ j ⩽ m2 .

(11.68)

Considering equations (11.10)–(11.12) at the point (xi , yj , tk ) (2 ⩽ k ⩽ n), and it follows from the Taylor expansion that ∇2τ Uijk + δΔ2h Uijk − Δx Vijk − Δy Wijk + Δ̄ h (2Uijk−1 − Uijk−2 ) = P̆ ijk , { { { { { { 1 ⩽ i ⩽ m1 , 1 ⩽ j ⩽ m2 , 2 ⩽ k ⩽ n, { { { { 2 k { { Vij = 󵄨󵄨󵄨󵄨∇h (2Uijk−1 − Uijk−2 )󵄨󵄨󵄨󵄨 Δx Uijk + Q̆ ijk , { { { 1 ⩽ i ⩽ m1 , 1 ⩽ j ⩽ m2 , 2 ⩽ k ⩽ n, { { { { 󵄨 󵄨2 { { W k = 󵄨󵄨∇h (2Uijk−1 − Uijk−2 )󵄨󵄨󵄨 Δy Uijk + R̆ kij , { { { ij 󵄨 { 1 ⩽ i ⩽ m1 , 1 ⩽ j ⩽ m2 , 2 ⩽ k ⩽ n,

(11.69) (11.70) (11.71)

where there is a positive constant c9 such that { { { { { { { { {

󵄨󵄨 ̆ k 󵄨󵄨 2 2 2 󵄨󵄨Pij 󵄨󵄨 ⩽ c9 (τ + h1 + h2 ), 󵄨󵄨 ̆ k 󵄨󵄨 2 2 2 󵄨󵄨Qij 󵄨󵄨 ⩽ c9 (τ + h1 + h2 ), 󵄨󵄨 ̆ k 󵄨󵄨 2 2 2 󵄨󵄨Rij 󵄨󵄨 ⩽ c9 (τ + h1 + h2 ),

1 ⩽ i ⩽ m1 , 1 ⩽ j ⩽ m2 , 2 ⩽ k ⩽ n,

(11.72)

1 ⩽ i ⩽ m1 , 1 ⩽ j ⩽ m2 , 2 ⩽ k ⩽ n,

(11.73)

1 ⩽ i ⩽ m1 , 1 ⩽ j ⩽ m2 , 2 ⩽ k ⩽ n.

(11.74)

11.5 Three-level linearized backward Euler difference scheme

� 373

Noticing the initial value condition Uij0 = φij ,

1 ⩽ i ⩽ m1 , 1 ⩽ j ⩽ m2 ,

(11.75)

and neglecting the small terms in (11.63)–(11.65) and (11.69)–(11.71), the following difference scheme for solving (11.10)–(11.12) and (11.2) is derived: For 0 ⩽ k ⩽ n, find uk , vk , wk ∈ 𝒲h such that { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { {

2∇τ uij1 − ψij + δΔ2h uij1 − Δx v1ij − Δy wij1 + Δ̄ h (uij0 + τψij ) = 0,

1 ⩽ i ⩽ m1 , 1 ⩽ j ⩽ m2 , 󵄨 󵄨2 = 󵄨󵄨󵄨∇h (uij0 + τψij )󵄨󵄨󵄨 Δx uij1 , 1 ⩽ i ⩽ m1 , 1 ⩽ j ⩽ m2 , 󵄨 󵄨2 wij1 = 󵄨󵄨󵄨∇h (uij0 + τψij )󵄨󵄨󵄨 Δy uij1 , 1 ⩽ i ⩽ m1 , 1 ⩽ j ⩽ m2 , ∇2τ uk + δΔ2 uk − Δx vk − Δy wk + Δ̄ h (2uk−1 − uk−2 ) = 0,

v1ij

ij

h ij

ij

ij

1 ⩽ i ⩽ m1 , 1 ⩽ j ⩽ m2 , 2 ⩽ k ⩽ n, 󵄨 󵄨2 k vij = 󵄨󵄨󵄨∇h (2uijk−1 − uijk−2 )󵄨󵄨󵄨 Δx uijk , 1 ⩽ i ⩽ m1 , 1 ⩽ j ⩽ m2 , 2 ⩽ k ⩽ n, 󵄨 󵄨2 k wij = 󵄨󵄨󵄨∇h (2uijk−1 − uijk−2 )󵄨󵄨󵄨 Δy uijk , uij0

ij

ij

(11.77) (11.78) (11.79) (11.80)

1 ⩽ i ⩽ m1 , 1 ⩽ j ⩽ m2 , 2 ⩽ k ⩽ n, = φij ,

(11.76)

(11.81)

1 ⩽ i ⩽ m1 , 1 ⩽ j ⩽ m2 .

(11.82)

Substituting (11.77)–(11.78) and (11.80)–(11.81) into (11.76) and (11.79), respectively, a difference scheme to solve (11.1)–(11.2) is established as follows: Find uk ∈ 𝒲h such that 󵄨 󵄨2 2∇τ uij1 − ψij + δΔ2h uij1 − ∇h ⋅ (󵄨󵄨󵄨∇h (uij0 + τψij )󵄨󵄨󵄨 ∇h uij1 ) + Δ̄ h (uij0 + τψij ) = 0, { { { { { { 1 ⩽ i ⩽ m1 , 1 ⩽ j ⩽ m2 , (11.83) { { { 󵄨󵄨 k 2 k k−1 k−2 󵄨󵄨2 k k−1 k−2 ∇2τ uij + δΔh uij − ∇h ⋅ (󵄨󵄨∇h (2uij − uij )󵄨󵄨 ∇h uij ) + Δ̄ h (2uij − uij ) = 0, { { { { { { 1 ⩽ i ⩽ m1 , 1 ⩽ j ⩽ m2 , 2 ⩽ k ⩽ n, (11.84) { { { 0 (11.85) { uij = φij , 1 ⩽ i ⩽ m1 , 1 ⩽ j ⩽ m2 . 11.5.2 Boundedness of the difference solution Theorem 11.11. Suppose {uijk | 1 ⩽ i ⩽ m1 , 1 ⩽ j ⩽ m2 , 0 ⩽ k ⩽ n} is the solution of the difference scheme (11.83)–(11.85). Denote 󵄩 󵄩2 󵄩 󵄩2 E k = 󵄩󵄩󵄩uk 󵄩󵄩󵄩 + 󵄩󵄩󵄩2uk − uk−1 󵄩󵄩󵄩 , Then it holds that

1 ⩽ k ⩽ n.

374 � 11 Difference methods for the epitaxial growth model 2 󵄩󵄩 τ 󵄩󵄩󵄩 󵄩 󵄩2 5 󵄩 E k ⩽ 6󵄩󵄩󵄩u0 + ψ󵄩󵄩󵄩 + 5󵄩󵄩󵄩u0 󵄩󵄩󵄩 + L1 L2 tk , 󵄩 󵄩󵄩 2 󵄩 4 2 󵄩 󵄩 󵄩 󵄩 1 󵄩󵄩 1 󵄩󵄩2 󵄩󵄩 0 τ 󵄩󵄩 󵄩󵄩u 󵄩󵄩 ⩽ 󵄩󵄩u + ψ󵄩󵄩 + L1 L2 t1 . 󵄩󵄩 2 󵄩󵄩 4

2 ⩽ k ⩽ n,

(11.86) (11.87)

Proof. (I) Taking the inner product on both the right- and left-hand sides of (11.83) with u1 gives 󵄨 󵄨2 2(∇τ u1 , u1 ) − (ψ, u1 ) + δ(Δ2h u1 , u1 ) − (∇h ⋅ (󵄨󵄨󵄨∇h (u0 + τψ)󵄨󵄨󵄨 ∇h u1 ), u1 ) + (Δ̄ h (u0 + τψ), u1 ) = 0. Noticing 2 󵄩 󵄩2 2 2(∇τ u1 , u1 ) − (ψ, u1 ) = 󵄩󵄩󵄩u1 󵄩󵄩󵄩 − ( u0 + ψ, u1 ), τ τ 󵄩 󵄩2 (Δ2h u1 , u1 ) = 󵄩󵄩󵄩Δh u1 󵄩󵄩󵄩 , 󵄨 󵄨2 󵄨 󵄨2 −(∇h ⋅ (󵄨󵄨󵄨∇h (u0 + τψ)󵄨󵄨󵄨 ∇h u1 ), u1 ) = (󵄨󵄨󵄨∇h (u0 + τψ)󵄨󵄨󵄨 ∇h u1 , ∇h u1 ) m1 m2

󵄨 󵄨2 󵄨 󵄨2 = h1 h2 ∑ ∑󵄨󵄨󵄨∇h (uij0 + τψij )󵄨󵄨󵄨 ⋅ 󵄨󵄨󵄨∇h uij1 󵄨󵄨󵄨 , i=1 j=1

−(Δ̄ h (u0 + τψ), u1 ) = (∇h (u0 + τψ), ∇h u1 ) m1 m2

󵄨 󵄨 󵄨 󵄨 ⩽ h1 h2 ∑ ∑󵄨󵄨󵄨∇h (uij0 + τψij )󵄨󵄨󵄨 ⋅ 󵄨󵄨󵄨∇h uij1 󵄨󵄨󵄨 i=1 j=1 m m

1 2 󵄨 󵄨2 󵄨 󵄨2 1 ⩽ h1 h2 ∑ ∑(󵄨󵄨󵄨∇h (uij0 + τψij )󵄨󵄨󵄨 ⋅ 󵄨󵄨󵄨∇h uij1 󵄨󵄨󵄨 + ) 4 i=1 j=1

m m

1 2 󵄨 󵄨2 󵄨 󵄨2 1 = h1 h2 ∑ ∑󵄨󵄨󵄨∇h (uij0 + τψij )󵄨󵄨󵄨 󵄨󵄨󵄨∇h uij1 󵄨󵄨󵄨 + L1 L2 , 4 i=1 j=1

we have 2 󵄩󵄩 1 󵄩󵄩2 󵄩 1 󵄩2 󵄩u 󵄩 + δ󵄩󵄩󵄩Δh u 󵄩󵄩󵄩 τ󵄩 󵄩 1 2 ⩽ ( u0 + ψ, u1 ) + L1 L2 τ 4 2 0 τ 1 1 = (u + ψ, u ) + L1 L2 τ 2 4 2 1 󵄩 󵄩2 1 󵄩󵄩󵄩 τ 󵄩󵄩󵄩 1 ⩽ 󵄩󵄩󵄩u1 󵄩󵄩󵄩 + 󵄩󵄩󵄩u0 + ψ󵄩󵄩󵄩 + L1 L2 . 󵄩 󵄩 τ τ󵄩 2 󵄩 4 Hence, 󵄩 󵄩2 1 󵄩󵄩 1 󵄩󵄩2 󵄩󵄩󵄩 0 τ 󵄩󵄩󵄩 󵄩󵄩u 󵄩󵄩 ⩽ 󵄩󵄩u + ψ󵄩󵄩 + L1 L2 t1 . 󵄩󵄩 2 󵄩󵄩 4

11.5 Three-level linearized backward Euler difference scheme

� 375

(II) Taking the inner product on both the right- and left-hand sides of (11.84) with uk produces 󵄨2 󵄨 (∇2τ uk , uk ) + δ(Δ2h uk , uk ) − (∇h ⋅ (󵄨󵄨󵄨∇h (2uk−1 − uk−2 )󵄨󵄨󵄨 ∇h uk ), uk ) + (Δ̄ h (2uk−1 − uk−2 ), uk ) = 0,

2 ⩽ k ⩽ n.

Noticing (∇2τ uk , uk ) 1 󵄩󵄩 k 󵄩󵄩2 󵄩󵄩 k−1 󵄩󵄩2 󵄩󵄩 k 󵄩2 = (󵄩󵄩u 󵄩󵄩 − 󵄩󵄩u 󵄩󵄩 + 󵄩󵄩2u − uk−1 󵄩󵄩󵄩 4τ 󵄩 󵄩2 󵄩 󵄩2 − 󵄩󵄩󵄩2uk−1 − uk−2 󵄩󵄩󵄩 + 󵄩󵄩󵄩uk − 2uk−1 + uk−2 󵄩󵄩󵄩 ) , 󵄩 󵄩2 (Δ2h uk , uk ) = 󵄩󵄩󵄩Δh uk 󵄩󵄩󵄩 , 󵄨 󵄨2 − (∇h ⋅ (󵄨󵄨󵄨∇h (2uk−1 − uk−2 )󵄨󵄨󵄨 ∇h uk ), uk ) 󵄨 󵄨2 = (󵄨󵄨󵄨∇h (2uk−1 − uk−2 )󵄨󵄨󵄨 ∇h uk , ∇h uk ) m1 m2

󵄨 󵄨2 󵄨 󵄨2 = h1 h2 ∑ ∑󵄨󵄨󵄨∇h (2uijk−1 − uijk−2 )󵄨󵄨󵄨 󵄨󵄨󵄨∇h uijk 󵄨󵄨󵄨 , i=1 j=1

(−Δ̄ h (2uk−1 − uk−2 ), uk )

= (∇h (2uk−1 − uk−2 ), ∇h uk ) m1 m2

= h1 h2 ∑ ∑ ∇h (2uijk−1 − uijk−2 ) ⋅ ∇h uijk i=1 j=1 m m

1 2 󵄨 󵄨2 󵄨 󵄨2 1 ⩽ h1 h2 ∑ ∑(󵄨󵄨󵄨∇h (2uijk−1 − uijk−2 )󵄨󵄨󵄨 ⋅ 󵄨󵄨󵄨∇h uijk 󵄨󵄨󵄨 + ) 4 i=1 j=1

m m

1 2 󵄨 󵄨2 󵄨 󵄨2 1 = h1 h2 ∑ ∑󵄨󵄨󵄨∇h (2uijk−1 − uijk−2 )󵄨󵄨󵄨 ⋅ 󵄨󵄨󵄨∇h uijk 󵄨󵄨󵄨 + L1 L2 , 4 i=1 j=1

we have 1 k 󵄩 󵄩2 1 (E − E k−1 ) + δ󵄩󵄩󵄩Δh uk 󵄩󵄩󵄩 ⩽ L1 L2 , 4τ 4

2 ⩽ k ⩽ n,

so that E k ⩽ E k−1 + τL1 L2 ⩽ ⋅ ⋅ ⋅ ⩽ E 1 + (k − 1)τL1 L2 , In view of 󵄩 󵄩2 󵄩 󵄩2 E 1 = 󵄩󵄩󵄩u1 󵄩󵄩󵄩 + 󵄩󵄩󵄩2u1 − u0 󵄩󵄩󵄩 󵄩 󵄩2 󵄩 󵄩2 = 5󵄩󵄩󵄩u1 󵄩󵄩󵄩 − 4(u0 , u1 ) + 󵄩󵄩󵄩u0 󵄩󵄩󵄩

1 ⩽ k ⩽ n.

376 � 11 Difference methods for the epitaxial growth model 󵄩 󵄩2 󵄩 󵄩2 ⩽ 6󵄩󵄩󵄩u1 󵄩󵄩󵄩 + 5󵄩󵄩󵄩u0 󵄩󵄩󵄩 2 󵄩󵄩 τ 󵄩󵄩󵄩 1 󵄩 󵄩 󵄩2 ⩽ 6(󵄩󵄩󵄩u0 + ψ󵄩󵄩󵄩 + L1 L2 t1 ) + 5󵄩󵄩󵄩u0 󵄩󵄩󵄩 , 󵄩󵄩 2 󵄩󵄩 4 it follows: 󵄩󵄩 󵄩 E k ⩽ 6󵄩󵄩󵄩u0 + 󵄩󵄩 󵄩󵄩 󵄩 ⩽ 6󵄩󵄩󵄩u0 + 󵄩󵄩

2 3 τ 󵄩󵄩󵄩󵄩 󵄩 󵄩2 ψ󵄩󵄩 + 5󵄩󵄩󵄩u0 󵄩󵄩󵄩 + [ τ + (k − 1)τ]L1 L2 2 󵄩󵄩 2 2 󵄩 τ 󵄩󵄩󵄩 󵄩 󵄩2 5 ψ󵄩 + 5󵄩󵄩󵄩u0 󵄩󵄩󵄩 + kτL1 L2 , 2 ⩽ k ⩽ n. 2 󵄩󵄩󵄩 4

Remark 11.2. The combination of (11.86) with (11.87) yields 󵄩2 󵄩󵄩 󵄩 0 󵄩2 5 󵄩󵄩 k 󵄩󵄩2 󵄩 0 τ 󵄩󵄩 󵄩󵄩u 󵄩󵄩 ⩽ 6󵄩󵄩󵄩u + ψ󵄩󵄩󵄩 + 5󵄩󵄩󵄩u 󵄩󵄩󵄩 + L1 L2 tk , 󵄩󵄩 2 󵄩󵄩 4

1 ⩽ k ⩽ n.

Remark 11.3. 󵄩󵄩2 󵄩󵄩 2 󵄩 󵄩2 󵄩 󵄩2 E k = (√2 − 1) (󵄩󵄩󵄩uk 󵄩󵄩󵄩 + 󵄩󵄩󵄩uk−1 󵄩󵄩󵄩 ) + 2󵄩󵄩󵄩√√2 + 1uk − √√2 − 1uk−1 󵄩󵄩󵄩 . 󵄩 󵄩 11.5.3 Existence of the difference solution Theorem 11.12. The difference scheme (11.83)–(11.85) is uniquely solvable. Proof. The value of u0 is determined by (11.85). (I) Equation (11.83) is a linear system in u1 . Consider its homogeneous one: 2 1 󵄨 󵄨2 u + δΔ2h uij1 − ∇h ⋅ (󵄨󵄨󵄨∇h (uij0 + τψij )󵄨󵄨󵄨 ∇h uij1 ) = 0, τ ij

1 ⩽ i ⩽ m1 , 1 ⩽ j ⩽ m2 .

Taking the inner product on both right- and left-hand sides of the equality with u1 gives 2 󵄩󵄩 1 󵄩󵄩2 󵄩 󵄨 󵄨2 1 󵄩2 0 1 1 󵄩u 󵄩 + δ󵄩󵄩󵄩Δh u 󵄩󵄩󵄩 + (󵄨󵄨󵄨∇h (u + τψ)󵄨󵄨󵄨 ∇h u , ∇h u ) = 0. τ󵄩 󵄩 Thus, ‖u1 ‖ = 0, which means that the value of u1 is uniquely determined by (11.83). (II) Assume that the values of uk−2 and uk−1 have been known. From (11.84), a linear system in uk is obtained. Consider its homogenous one: 3 k u + δΔ2h uijk − ∇h ⋅ (|∇h (2uijk−1 − uijk−2 )|2 ∇h uijk ) = 0, 2τ ij

1 ⩽ i ⩽ m1 , 1 ⩽ j ⩽ m2 .

Taking the inner product on both the right- and left-hand sides of the equality with uk produces

11.5 Three-level linearized backward Euler difference scheme

� 377

3 󵄩󵄩 k 󵄩󵄩2 󵄨 󵄩 k k k−1 k−2 󵄨2 k 󵄩2 󵄩󵄩u 󵄩󵄩 + δ󵄩󵄩󵄩Δh u 󵄩󵄩󵄩 + (󵄨󵄨󵄨∇h (2u − u )󵄨󵄨󵄨 ∇h u , ∇h u ) = 0, 2τ so that ‖uk ‖ = 0. Hence, the value of uk is uniquely determined by (11.84). By induction, the conclusion is true. 11.5.4 Convergence of the difference solution Theorem 11.13. Suppose {Uijk | 1 ⩽ i ⩽ m1 , 1 ⩽ j ⩽ m2 , 0 ⩽ k ⩽ n} and {uijk | 1 ⩽ i ⩽ m1 , 1 ⩽ j ⩽ m2 , 0 ⩽ k ⩽ n} are solutions of the problem (11.1)–(11.2) and the difference scheme (11.83)–(11.85), respectively. Define the error function: ũ ijk = Uijk − uijk ,

1 ⩽ i ⩽ m1 , 1 ⩽ j ⩽ m2 , 0 ⩽ k ⩽ n.

Then there is a constant c11 such that 󵄩󵄩 ̃ k 󵄩󵄩 2 2 2 󵄩󵄩u 󵄩󵄩 ⩽ c11 (τ + h1 + h2 ),

0 ⩽ k ⩽ n.

Proof. Let {vk , wk } be defined by (11.77)–(11.78) and (11.80)–(11.81). Then the difference scheme (11.83)–(11.85) can be reformulated as (11.76)–(11.82). Define ṽkij = Vijk − vkij ,

w̃ ijk = Wijk − wijk .

Subtracting (11.76)–(11.82) from (11.63)–(11.65), (11.69)–(11.71), (11.75), respectively, the system of error equations is obtained as { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { {

2∇τ ũ ij1 + δΔ2h ũ ij1 − Δx ṽ1ij − Δy w̃ ij1 = P̆ ij1 , 󵄨2

1 ⩽ i ⩽ m1 , 1 ⩽ j ⩽ m2 ,

󵄨 ṽ1ij = 󵄨󵄨󵄨∇h (Uij0 + τψij )󵄨󵄨󵄨 Δx ũ ij1 + Q̆ ij1 , 1 ⩽ i ⩽ m1 , 1 ⩽ j ⩽ m2 , 󵄨 󵄨2 w̃ ij1 = 󵄨󵄨󵄨∇h (Uij0 + τψij )󵄨󵄨󵄨 Δy ũ ij1 + R̆ 1ij , 1 ⩽ i ⩽ m1 , 1 ⩽ j ⩽ m2 , ∇2τ ũ ijk + δΔ2h ũ ijk − Δx ṽkij − Δy w̃ ijk + Δ̄ h (2ũ ijk−1 − ũ ijk−2 ) = P̆ ijk ,

1 ⩽ i ⩽ m1 , 1 ⩽ j ⩽ m2 , 2 ⩽ k ⩽ n, 󵄨 󵄨2 󵄨 󵄨2 = 󵄨󵄨󵄨∇h (2Uijk−1 − Uijk−2 )󵄨󵄨󵄨 Δx Uijk − 󵄨󵄨󵄨∇h (2uijk−1 − uijk−2 )󵄨󵄨󵄨 Δx uijk + Q̆ ijk , 1 ⩽ i ⩽ m1 , 1 ⩽ j ⩽ m2 , 2 ⩽ k ⩽ n, 󵄨 󵄨2 󵄨 󵄨2 w̃ ijk = 󵄨󵄨󵄨∇h (2Uijk−1 − Uijk−2 )󵄨󵄨󵄨 Δy Uijk − 󵄨󵄨󵄨∇h (2uijk−1 − uijk−2 )󵄨󵄨󵄨 Δy uijk + R̆ kij ,

ṽkij

1 ⩽ i ⩽ m1 , 1 ⩽ j ⩽ m2 , 2 ⩽ k ⩽ n,

ũ ij0 = 0,

1 ⩽ i ⩽ m1 , 1 ⩽ j ⩽ m2 .

(11.88) (11.89) (11.90) (11.91) (11.92) (11.93) (11.94)

From (11.94), it is easy to see that 󵄩󵄩 ̃ 0 󵄩󵄩 󵄩󵄩u 󵄩󵄩 = 0.

(11.95)

378 � 11 Difference methods for the epitaxial growth model (I) Taking the inner product on both the right- and left-hand sides of (11.88) with ũ 1 gives 2(∇τ ũ 1 , ũ 1 ) + δ(Δ2h ũ 1 , ũ 1 ) − (Δx ṽ1 , ũ 1 ) − (Δy w̃ 1 , ũ 1 ) = (P̆ 1 , ũ 1 ). Noticing −(Δx ṽ1 , ũ 1 ) − (Δy w̃ 1 , ũ 1 )

= (ṽ1 , Δx ũ 1 ) + (w̃ 1 , Δy ũ 1 ) 󵄨 󵄨2 󵄨 󵄨2 = (󵄨󵄨󵄨∇h (U 0 + τψ)󵄨󵄨󵄨 Δx ũ 1 + Q̆ 1 , Δx ũ 1 ) + (󵄨󵄨󵄨∇h (U 0 + τψ)󵄨󵄨󵄨 Δy ũ 1 + R̆ 1 , Δy ũ 1 ) 󵄨 󵄨2 = (󵄨󵄨󵄨∇h (U 0 + τψ)󵄨󵄨󵄨 ∇h ũ 1 , ∇h ũ 1 ) + (Q̆ 1 , Δx ũ 1 ) + (R̆ 1 , Δy ũ 1 ) and (11.66)–(11.68), we have 2 󵄩󵄩 1 󵄩󵄩2 󵄩 1 󵄩2 󵄩ũ 󵄩 + δ󵄩󵄩󵄩Δh ũ 󵄩󵄩󵄩 τ󵄩 󵄩 ⩽ −(Q̆ 1 , Δx ũ 1 ) − (R̆ 1 , Δy ũ 1 ) + (P̆ 1 , ũ 1 )

1󵄩 󵄩2 1 󵄩 󵄩2 1 󵄩 󵄩2 󵄩 󵄩2 󵄩 󵄩2 ⩽ 󵄩󵄩󵄩∇h ũ 1 󵄩󵄩󵄩 + 󵄩󵄩󵄩ũ 1 󵄩󵄩󵄩 + (󵄩󵄩󵄩Q̆ 1 󵄩󵄩󵄩 + 󵄩󵄩󵄩R̆ 1 󵄩󵄩󵄩 + 󵄩󵄩󵄩P̆ 1 󵄩󵄩󵄩 ) 2 2 2 1 󵄩󵄩 1 󵄩󵄩2 3 2 1 󵄩󵄩 1 󵄩󵄩 󵄩󵄩 2 1󵄩 󵄩 ⩽ 󵄩󵄩ũ 󵄩󵄩 ⋅ 󵄩󵄩Δh ũ 󵄩󵄩 + 󵄩󵄩ũ 󵄩󵄩 + c8 L1 L2 (τ 2 + h12 + h22 ) 2 2 2 1 󵄩 2 󵄩2 1 1 󵄩󵄩 ̃ 1 󵄩󵄩2 1 󵄩󵄩 ̃ 1 󵄩󵄩2 3 2 ⋅ 󵄩󵄩u 󵄩󵄩 + 󵄩󵄩u 󵄩󵄩 + c8 L1 L2 (τ 2 + h12 + h22 ) . ⩽ δ󵄩󵄩󵄩Δh ũ 1 󵄩󵄩󵄩 + 2 2δ 4 2 2

(11.96)

Hence, 1 1 2 󵄩 󵄩2 3 [1 − ( + 1)τ]󵄩󵄩󵄩ũ 1 󵄩󵄩󵄩 ⩽ c82 L1 L2 τ(τ 2 + h12 + h22 ) , 4 4δ 4 which is followed by 󵄩󵄩 ̃ 1 󵄩󵄩2 3 2 2 2 2 2 󵄩󵄩u 󵄩󵄩 ⩽ c8 L1 L2 τ(τ + h1 + h2 ) 2

(11.97)

1 if ( 4δ + 1)τ ⩽ 2. In view of (11.96) again, it follows:

δ 󵄩󵄩 1 󵄩 1 󵄩2 3 2 1 1 󵄩2 2 2 2 2 󵄩󵄩Δh ũ 󵄩󵄩󵄩 ⩽ ( + )󵄩󵄩󵄩ũ 󵄩󵄩󵄩 + c8 L1 L2 (τ + h1 + h2 ) 2 8δ 2 2 3 1 3 2 2 2 ⩽ ( + 1)c8 τL1 L2 (τ 2 + h12 + h22 ) + c82 L1 L2 (τ 2 + h12 + h22 ) , 4 4δ 2 so that 󵄩󵄩 ̃ 1 󵄩󵄩2 1 3 1 2 2 2 2 2 󵄩󵄩Δh u 󵄩󵄩 ⩽ [ ( + 1)τ + 3]c8 L1 L2 (τ + h1 + h2 ) . δ 2 4δ

(11.98)

11.5 Three-level linearized backward Euler difference scheme

� 379

(II) Taking the inner product on both the right- and left-hand sides of (11.91) with ũ k produces (∇2τ ũ k , ũ k ) + δ(Δ2h ũ k , ũ k ) − (Δx ṽk , ũ k ) − (Δy w̃ k , ũ k ) + (Δ̄ h (2ũ k−1 − ũ k−2 ), ũ k ) = (P̆ k , ũ k ).

(11.99)

Each term in the above equality will be analyzed as follows: 1 󵄩󵄩 k 󵄩󵄩2 󵄩󵄩 k 󵄩2 󵄩 󵄩2 󵄩 󵄩2 [(󵄩ũ 󵄩 + 󵄩2ũ − ũ k−1 󵄩󵄩󵄩 ) − (󵄩󵄩󵄩ũ k−1 󵄩󵄩󵄩 + 󵄩󵄩󵄩2ũ k−1 − ũ k−2 󵄩󵄩󵄩 ) 4τ 󵄩 󵄩 󵄩 󵄩 󵄩2 + 󵄩󵄩󵄩ũ k − 2ũ k−1 + ũ k−2 󵄩󵄩󵄩 ];

(∇2τ ũ k , ũ k ) =

󵄩 󵄩2 (Δ2h ũ k , ũ k ) = 󵄩󵄩󵄩Δh ũ k 󵄩󵄩󵄩 ; − (Δx ṽk , ũ k ) − (Δy w̃ k , ũ k )

= (ṽk , Δx ũ k ) + (w̃ k , Δy ũ k )

󵄨 󵄨2 󵄨 󵄨2 = (󵄨󵄨󵄨∇h (2U k−1 − U k−2 )󵄨󵄨󵄨 Δx U k − 󵄨󵄨󵄨∇h (2uk−1 − uk−2 )󵄨󵄨󵄨 Δx uk + Q̆ k , Δx ũ k ) 󵄨 󵄨2 󵄨 󵄨2 + (󵄨󵄨󵄨∇h (2U k−1 − U k−2 )󵄨󵄨󵄨 Δy U k − 󵄨󵄨󵄨∇h (2uk−1 − uk−2 )󵄨󵄨󵄨 Δy uk + R̆ k , Δy ũ k ) 󵄨 󵄨2 = (󵄨󵄨󵄨∇h (2uk−1 − uk−2 )󵄨󵄨󵄨 (Δx U k − Δx uk ) 󵄨 󵄨2 󵄨 󵄨2 + (󵄨󵄨󵄨∇h (2U k−1 − U k−2 )󵄨󵄨󵄨 − 󵄨󵄨󵄨∇h (2uk−1 − uk−2 )󵄨󵄨󵄨 )Δx U k , Δx ũ k ) 󵄨 󵄨2 + (󵄨󵄨󵄨∇h (2uk−1 − uk−2 )󵄨󵄨󵄨 (Δy U k − Δy uk ) 󵄨 󵄨2 󵄨 󵄨2 + (󵄨󵄨󵄨∇h (2U k−1 − U k−2 )󵄨󵄨󵄨 − 󵄨󵄨󵄨∇h (2uk−1 − uk−2 )󵄨󵄨󵄨 )Δy U k , Δy ũ k )

+ (Q̆ k , Δx ũ k ) + (R̆ k , Δy ũ k ) 󵄨 󵄨2 = (󵄨󵄨󵄨∇h (2uk−1 − uk−2 )󵄨󵄨󵄨 ∇h ũ k , ∇h ũ k ) 󵄨 󵄨2 󵄨 󵄨2 + ((󵄨󵄨󵄨∇h (2U k−1 − U k−2 )󵄨󵄨󵄨 − 󵄨󵄨󵄨∇h (2uk−1 − uk−2 )󵄨󵄨󵄨 )∇h U k , ∇h ũ k ) + (Q̆ k , Δx ũ k ) + (R̆ k , Δy ũ k ); − (Δ̄ h (2ũ k−1 − ũ k−2 ), ũ k ) = (∇h (2ũ k−1 − ũ k−2 ), ∇h ũ k ). Inserting the four equalities above into (11.99) leads to 1 󵄩󵄩 k 󵄩󵄩2 󵄩󵄩 k 󵄩2 󵄩 󵄩 󵄩 󵄩2 󵄩 󵄩2 [(󵄩ũ 󵄩 + 󵄩2ũ − ũ k−1 󵄩󵄩󵄩 ) − (󵄩󵄩󵄩ũ k−1 󵄩󵄩󵄩 + 󵄩󵄩󵄩2ũ k−1 − ũ k−2 󵄩󵄩󵄩 )] + δ󵄩󵄩󵄩Δh ũ k 󵄩󵄩󵄩 4τ 󵄩 󵄩 󵄩 󵄨 󵄨2 + (󵄨󵄨󵄨∇h (2uk−1 − uk−2 )󵄨󵄨󵄨 ∇h ũ k , ∇h ũ k ) 󵄨 󵄨2 󵄨 󵄨2 ⩽ −((󵄨󵄨󵄨∇h (2U k−1 − U k−2 )󵄨󵄨󵄨 − 󵄨󵄨󵄨∇h (2uk−1 − uk−2 )󵄨󵄨󵄨 )∇h U k , ∇h ũ k ) + (∇h (2ũ k−1 − ũ k−2 ), ∇h ũ k ) + (P̆ k , ũ k ) − (Q̆ k , Δx ũ k ) − (R̆ k , Δy ũ k ). (11.100)

380 � 11 Difference methods for the epitaxial growth model In view of (11.41), it follows: 󵄨2 󵄨2 󵄨 󵄨 −((󵄨󵄨󵄨∇h (2U k−1 − U k−2 )󵄨󵄨󵄨 − 󵄨󵄨󵄨∇h (2uk−1 − uk−2 )󵄨󵄨󵄨 )∇h U k , ∇h ũ k ) = −((∇h (2U k−1 − U k−2 ) ⋅ ∇h (2ũ k−1 − ũ k−2 ) + ∇h (2uk−1 − uk−2 ) ⋅ ∇h (2ũ k−1 − ũ k−2 ))∇h U k , ∇h ũ k ) m1 m2

󵄨 󵄨 󵄨 󵄨 󵄨 󵄨 󵄨 󵄨 ⩽ h1 h2 ∑ ∑󵄨󵄨󵄨∇h (2Uijk−1 − Uijk−2 )󵄨󵄨󵄨 ⋅ 󵄨󵄨󵄨∇h (2ũ ijk−1 − ũ ijk−2 )󵄨󵄨󵄨 ⋅ 󵄨󵄨󵄨∇h Uijk 󵄨󵄨󵄨 ⋅ 󵄨󵄨󵄨∇h ũ ijk 󵄨󵄨󵄨 i=1 j=1

m1 m2

󵄨 󵄨 󵄨 󵄨 󵄨 󵄨 󵄨 󵄨 + h1 h2 ∑ ∑󵄨󵄨󵄨∇h (2uijk−1 − uijk−2 )󵄨󵄨󵄨 ⋅ 󵄨󵄨󵄨∇h (2ũ ijk−1 − ũ ijk−2 )󵄨󵄨󵄨 ⋅ 󵄨󵄨󵄨∇h Uijk 󵄨󵄨󵄨 ⋅ 󵄨󵄨󵄨∇h ũ ijk 󵄨󵄨󵄨 i=1 j=1

m1 m2

󵄨 󵄨 󵄨 󵄨 ⩽ 3c4 h1 h2 ∑ ∑󵄨󵄨󵄨∇h (2ũ ijk−1 − ũ ijk−2 )󵄨󵄨󵄨 ⋅ 󵄨󵄨󵄨∇h ũ ijk 󵄨󵄨󵄨 i=1 j=1

m1 m2

󵄨 󵄨2 󵄨 󵄨2 + h1 h2 ∑ ∑(󵄨󵄨󵄨∇h (2uijk−1 − uijk−2 )󵄨󵄨󵄨 󵄨󵄨󵄨∇h ũ ijk 󵄨󵄨󵄨 i=1 j=1

1󵄨 󵄨2 󵄨 󵄨2 + 󵄨󵄨󵄨∇h (2ũ ijk−1 − ũ ijk−2 )󵄨󵄨󵄨 󵄨󵄨󵄨∇h Uijk 󵄨󵄨󵄨 ) 4 󵄨 󵄨2 ⩽ (󵄨󵄨󵄨∇h (2uk−1 − uk−2 )󵄨󵄨󵄨 ∇h ũ k , ∇h ũ k ) 󵄩 󵄩 󵄩 󵄩 1 󵄩 󵄩2 + 3c4 󵄩󵄩󵄩∇h (2ũ k−1 − ũ k−2 )󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩∇h ũ k 󵄩󵄩󵄩 + c4 󵄩󵄩󵄩∇h (2ũ k−1 − ũ k−2 )󵄩󵄩󵄩 . 4 Substituting the inequality above into (11.100) arrives at 1 󵄩󵄩 k 󵄩󵄩2 󵄩󵄩 k 󵄩2 󵄩 󵄩2 󵄩 󵄩2 󵄩 󵄩2 [(󵄩ũ 󵄩 + 󵄩2ũ − ũ k−1 󵄩󵄩󵄩 ) − (󵄩󵄩󵄩ũ k−1 󵄩󵄩󵄩 + 󵄩󵄩󵄩2ũ k−1 − ũ k−2 󵄩󵄩󵄩 )] + δ󵄩󵄩󵄩Δh ũ k 󵄩󵄩󵄩 4τ 󵄩 󵄩 󵄩 󵄩2 󵄩 󵄩 󵄩 󵄩 1 󵄩 ⩽ 3c4 󵄩󵄩󵄩∇h (2ũ k−1 − ũ k−2 )󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩∇h ũ k 󵄩󵄩󵄩 + c4 󵄩󵄩󵄩∇h (2ũ k−1 − ũ k−2 )󵄩󵄩󵄩 4 1 󵄩 󵄩 󵄩 󵄩 󵄩2 󵄩 󵄩2 󵄩 1 󵄩 󵄩2 󵄩 󵄩2 + 󵄩󵄩󵄩∇h (2ũ k−1 − ũ k−2 )󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩∇h ũ k 󵄩󵄩󵄩 + (󵄩󵄩󵄩ũ k 󵄩󵄩󵄩 + 󵄩󵄩󵄩P̆ k 󵄩󵄩󵄩 ) + (󵄩󵄩󵄩Δx ũ k 󵄩󵄩󵄩 + 󵄩󵄩󵄩Q̆ k 󵄩󵄩󵄩 ) 2 2 1 󵄩 󵄩2 󵄩 󵄩2 + (󵄩󵄩󵄩Δy ũ k 󵄩󵄩󵄩 + 󵄩󵄩󵄩R̆ k 󵄩󵄩󵄩 ) 2 1 󵄩 1 󵄩 󵄩2 󵄩 󵄩2 󵄩2 ⩽ (1 + 3c4 ) ⋅ (󵄩󵄩󵄩∇h (2ũ k−1 − ũ k−2 )󵄩󵄩󵄩 + 󵄩󵄩󵄩∇h ũ k 󵄩󵄩󵄩 ) + c4 󵄩󵄩󵄩∇h (2ũ k−1 − ũ k−2 )󵄩󵄩󵄩 2 4 1󵄩 󵄩2 1 󵄩 󵄩2 1 󵄩 󵄩2 󵄩 󵄩2 󵄩 󵄩2 + 󵄩󵄩󵄩∇h ũ k 󵄩󵄩󵄩 + 󵄩󵄩󵄩ũ k 󵄩󵄩󵄩 + (󵄩󵄩󵄩P̆ k 󵄩󵄩󵄩 + 󵄩󵄩󵄩Q̆ k 󵄩󵄩󵄩 + 󵄩󵄩󵄩R̆ k 󵄩󵄩󵄩 ) 2 2 2 3 1 7c4 󵄩󵄩 󵄩 󵄩2 1 󵄩 󵄩2 󵄩2 )󵄩󵄩∇h (2ũ k−1 − ũ k−2 )󵄩󵄩󵄩 + (1 + c4 )󵄩󵄩󵄩∇h ũ k 󵄩󵄩󵄩 + 󵄩󵄩󵄩ũ k 󵄩󵄩󵄩 =( + 2 4 2 2 1 󵄩󵄩 ̆ k 󵄩󵄩2 󵄩󵄩 ̆ k 󵄩󵄩2 󵄩󵄩 ̆ k 󵄩󵄩2 + (󵄩󵄩P 󵄩󵄩 + 󵄩󵄩Q 󵄩󵄩 + 󵄩󵄩R 󵄩󵄩 ) 2 3 1 7c4 󵄩󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 )󵄩Δ (2ũ k−1 − ũ k−2 )󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩2ũ k−1 − ũ k−2 󵄩󵄩󵄩 + (1 + c4 )󵄩󵄩󵄩Δh ũ k 󵄩󵄩󵄩 ⋅ 󵄩󵄩󵄩ũ k 󵄩󵄩󵄩 ⩽( + 2 4 󵄩 h 2 1 󵄩 󵄩2 1 󵄩 󵄩2 󵄩 󵄩2 󵄩 󵄩2 + 󵄩󵄩󵄩ũ k 󵄩󵄩󵄩 + (󵄩󵄩󵄩P̆ k 󵄩󵄩󵄩 + 󵄩󵄩󵄩Q̆ k 󵄩󵄩󵄩 + 󵄩󵄩󵄩R̆ k 󵄩󵄩󵄩 ) 2 2

11.5 Three-level linearized backward Euler difference scheme

� 381

2

󵄩2 󵄩2 1 1 7c 󵄩 󵄩2 δ 󵄩 󵄩 ⩽ ε󵄩󵄩󵄩Δh (2ũ k−1 − ũ k−2 )󵄩󵄩󵄩 + ( + 4 ) 󵄩󵄩󵄩2ũ k−1 − ũ k−2 󵄩󵄩󵄩 + 󵄩󵄩󵄩Δh ũ k 󵄩󵄩󵄩 4ε 2 4 3 2

1 3 3 󵄩 󵄩2 1 󵄩 󵄩2 󵄩 󵄩2 󵄩 󵄩2 + [ + (1 + c4 ) ]󵄩󵄩󵄩ũ k 󵄩󵄩󵄩 + (󵄩󵄩󵄩P̆ k 󵄩󵄩󵄩 + 󵄩󵄩󵄩Q̆ k 󵄩󵄩󵄩 + 󵄩󵄩󵄩R̆ k 󵄩󵄩󵄩 ). 2 4δ 2 2 Noticing 󵄩 k−2 󵄩2 󵄩 k−1 󵄩2 󵄩󵄩 ̃ k−1 k−1 k−2 k−2 󵄩2 󵄩󵄩2u − ũ 󵄩󵄩󵄩 = 4󵄩󵄩󵄩ũ 󵄩󵄩󵄩 − 4(ũ , ũ ) + 󵄩󵄩󵄩ũ 󵄩󵄩󵄩 󵄩 󵄩2 󵄩 󵄩2 󵄩 󵄩2 󵄩 󵄩2 ⩽ 4󵄩󵄩󵄩ũ k−1 󵄩󵄩󵄩 + (󵄩󵄩󵄩ũ k−1 󵄩󵄩󵄩 + 4󵄩󵄩󵄩ũ k−2 󵄩󵄩󵄩 ) + 󵄩󵄩󵄩ũ k−2 󵄩󵄩󵄩 󵄩 󵄩2 󵄩 󵄩2 = 5(󵄩󵄩󵄩ũ k−1 󵄩󵄩󵄩 + 󵄩󵄩󵄩ũ k−2 󵄩󵄩󵄩 ), 󵄩󵄩 󵄩 k−1 k−2 󵄩2 k−1 󵄩2 󵄩 k−2 󵄩2 󵄩󵄩Δh (2ũ − ũ )󵄩󵄩󵄩 ⩽ 5(󵄩󵄩󵄩Δh ũ 󵄩󵄩󵄩 + 󵄩󵄩󵄩Δh ũ 󵄩󵄩󵄩 ), one has 1 󵄩󵄩 k 󵄩󵄩2 󵄩󵄩 k 2 󵄩 󵄩2 󵄩2 󵄩 󵄩2 󵄩 󵄩2 [(󵄩󵄩ũ 󵄩󵄩 + 󵄩󵄩2ũ − ũ k−1 󵄩󵄩󵄩 ) − (󵄩󵄩󵄩ũ k−1 󵄩󵄩󵄩 + 󵄩󵄩󵄩2ũ k−1 − ũ k−2 󵄩󵄩󵄩 )] + δ󵄩󵄩󵄩Δh ũ k 󵄩󵄩󵄩 4τ 3 2

5 1 7c 󵄩 󵄩2 󵄩 󵄩 󵄩2 󵄩2 󵄩 󵄩2 ⩽ 5ε(󵄩󵄩󵄩Δh ũ k−1 󵄩󵄩󵄩 + 󵄩󵄩󵄩Δh ũ k−2 󵄩󵄩󵄩 ) + ( + 4 ) (󵄩󵄩󵄩ũ k−1 󵄩󵄩󵄩 + 󵄩󵄩󵄩ũ k−2 󵄩󵄩󵄩 ) 4ε 2 4 2

3 3 1 󵄩 󵄩2 1 󵄩 󵄩2 󵄩 󵄩2 󵄩 󵄩2 + [ + (1 + c4 ) ]󵄩󵄩󵄩ũ k 󵄩󵄩󵄩 + (󵄩󵄩󵄩P̆ k 󵄩󵄩󵄩 + 󵄩󵄩󵄩Q̆ k 󵄩󵄩󵄩 + 󵄩󵄩󵄩R̆ k 󵄩󵄩󵄩 ). 2 4δ 2 2 Taking ε =

δ 15

and noticing (11.72)–(11.74), we have 1 󵄩󵄩 k 󵄩󵄩2 󵄩󵄩 k 󵄩2 󵄩 󵄩2 󵄩 󵄩2 [(󵄩ũ 󵄩 + 󵄩2ũ − ũ k−1 󵄩󵄩󵄩 ) − (󵄩󵄩󵄩ũ k−1 󵄩󵄩󵄩 + 󵄩󵄩󵄩2ũ k−1 − ũ k−2 󵄩󵄩󵄩 )] 4τ 󵄩 󵄩 󵄩 δ 󵄩 󵄩2 󵄩 󵄩2 󵄩 󵄩2 + (2󵄩󵄩󵄩Δh ũ k 󵄩󵄩󵄩 − 󵄩󵄩󵄩Δh ũ k−1 󵄩󵄩󵄩 − 󵄩󵄩󵄩Δh ũ k−2 󵄩󵄩󵄩 ) 3 2

2

1 3 3 󵄩 󵄩2 75 1 7c 󵄩 󵄩2 󵄩 󵄩2 ⩽ [ + (1 + c4 ) ]󵄩󵄩󵄩ũ k 󵄩󵄩󵄩 + ( + 4 ) (󵄩󵄩󵄩ũ k−1 󵄩󵄩󵄩 + 󵄩󵄩󵄩ũ k−2 󵄩󵄩󵄩 ) 2 4δ 2 4δ 2 4 3 2 + c92 L1 L2 (τ 2 + h12 + h22 ) , 2 ⩽ k ⩽ n. 2 Replacing k by l in the inequality above and summing over l from 2 to k arrive at 1 󵄩󵄩 k 󵄩󵄩2 󵄩󵄩 k 󵄩2 󵄩 󵄩2 󵄩 󵄩2 [(󵄩ũ 󵄩 + 󵄩2ũ − ũ k−1 󵄩󵄩󵄩 ) − (󵄩󵄩󵄩ũ 1 󵄩󵄩󵄩 + 󵄩󵄩󵄩2ũ 1 − ũ 0 󵄩󵄩󵄩 )] 4τ 󵄩 󵄩 󵄩 +

k k k δ 󵄩 󵄩2 󵄩 󵄩2 󵄩 󵄩2 (2 ∑󵄩󵄩󵄩Δh ũ l 󵄩󵄩󵄩 − ∑󵄩󵄩󵄩Δh ũ l−1 󵄩󵄩󵄩 − ∑󵄩󵄩󵄩Δh ũ l−2 󵄩󵄩󵄩 ) 3 l=2 l=2 l=2 2

2

k k k 1 3 3 󵄩 󵄩2 75 1 7c 󵄩 󵄩2 󵄩 󵄩2 ⩽ [ + (1 + c4 ) ] ∑󵄩󵄩󵄩ũ l 󵄩󵄩󵄩 + ( + 4 ) (∑󵄩󵄩󵄩ũ l−1 󵄩󵄩󵄩 + ∑󵄩󵄩󵄩ũ l−2 󵄩󵄩󵄩 ) 2 4δ 2 4δ 2 4 l=2 l=2 l=2

3 2 + (k − 1)c92 L1 L2 (τ 2 + h12 + h22 ) , 2

2 ⩽ k ⩽ n,

382 � 11 Difference methods for the epitaxial growth model which can be rearranged as 1 󵄩󵄩 k 󵄩󵄩2 󵄩󵄩 k 󵄩2 󵄩 󵄩2 󵄩 󵄩2 [(󵄩󵄩ũ 󵄩󵄩 + 󵄩󵄩2ũ − ũ k−1 󵄩󵄩󵄩 ) − (󵄩󵄩󵄩ũ 1 󵄩󵄩󵄩 + 󵄩󵄩󵄩2ũ 1 − ũ 0 󵄩󵄩󵄩 )] 4τ δ 󵄩 󵄩2 󵄩2 󵄩 󵄩 󵄩2 󵄩2 󵄩 + (2󵄩󵄩󵄩Δh ũ k 󵄩󵄩󵄩 + 󵄩󵄩󵄩Δh ũ k−1 󵄩󵄩󵄩 − 2󵄩󵄩󵄩Δh ũ 1 󵄩󵄩󵄩 − 󵄩󵄩󵄩Δh ũ 0 󵄩󵄩󵄩 ) 3 2

2

k−2 k k−1 1 3 3 󵄩 󵄩2 󵄩 󵄩2 75 1 7c 󵄩 󵄩2 ⩽ [ + (1 + c4 ) ] ∑󵄩󵄩󵄩ũ l 󵄩󵄩󵄩 + ( + 4 ) ( ∑ 󵄩󵄩󵄩ũ l 󵄩󵄩󵄩 + ∑ 󵄩󵄩󵄩ũ l 󵄩󵄩󵄩 ) 2 4δ 2 4δ 2 4 l=0 l=2 l=1

3 2 + (k − 1)c92 L1 L2 (τ 2 + h12 + h22 ) , 2

2 ⩽ k ⩽ n.

Noticing (11.94), (11.97) and (11.98), there is a constant c10 such that k−1

󵄩󵄩 ̃ k 󵄩󵄩2 󵄩 l 󵄩2 2 2 2 2 󵄩󵄩u 󵄩󵄩 ⩽ c10 τ ∑ 󵄩󵄩󵄩ũ 󵄩󵄩󵄩 + c10 (τ + h1 + h2 ) , l=2

2 ⩽ k ⩽ n.

By the Gronwall inequality (Theorem 1.2(c)), it follows: 󵄩󵄩 ̃ k 󵄩󵄩2 c T 2 2 2 2 󵄩󵄩u 󵄩󵄩 ⩽ e 10 ⋅ c10 (τ + h1 + h2 ) ,

2 ⩽ k ⩽ n.

(11.101)

Combining (11.95), (11.97) with (11.101), the proof is completed.

11.6 Numerical experiments In this section, we compute two numerical examples using the second-order linearized difference scheme (11.83)–(11.85) to demonstrate the numerical accuracy and convergence. Example 11.1. In the problem (11.1)–(11.2), take Ω = [0, 2π] × [0, 2π], φ(x, y) = sin(x + y), and replace 0 on the right-hand side of (11.1) by [(4δ − 3)e−t + 12e−3t cos2 (x + y)] sin(x + y) with δ = 0.1. The analytic solution is u(x, y, t) = e−t sin(x + y). In the simulation, the linearized system at each time level is solved by the Gauss– Seidel iteration method. In our computation, the tolerance of the Gauss–Seidel iteration method is taken as 1e−8. To test the numerical accuracy and convergence, we take m1 = m2 = m, h1 = h2 = h and define the numerical error and order by m1 m2

2

E(h, τ) = √∑ ∑ h2 (uijn (h, τ) − u(xi , yj , tn )) , i=1 j=1

Order1 = log2

E(2h, 2τ) . E(h, τ)

11.6 Numerical experiments

� 383

Table 11.1 shows the errors and the convergence orders in time and space at time T = 1.0. From the table, we can see that the scheme (11.83)–(11.85) is second-order convergent both in time and space. Table 11.1: Errors and convergence rates in L2 -norm. m

n

E(h, τ)

Order�

10 20 40

10 20 40

�.����e−� �.����e−� �.����e−�

2.336 2.047

Example 11.2 ([20]). Consider the the problem (11.1)–(11.2) with the initial value condition φ(x, y) = 0.1(sin 3x sin 2y + sin 5x sin 5y) and the problem is subject to the periodic boundary condition on the domain [0, 2π] × [0, 2π] and δ = 0.1. This problem has been analyzed in [20] using the perturbation method to study the nonlinear instability. At each time level, the linearized system is solved by an algebraic multigrid method. Take m1 = m2 = m, or equivalently, h1 = h2 = h. In our computation, the tolerance for the algebraic multigrid method is taken as 10−8 . To test the numerical accuracy and convergence, define the numerical order Order2 = log2

F(2h, 2τ) , F(h, τ)

where m m

2

1 2 2n ( h , τ )) . F(h, τ) = √∑ ∑ h2 (uijn (h, τ) − u2i,2j 2 2 i=1 j=1

Table 11.2 shows the convergence order in time and space at time T = 0.01. From the table, we can see that the scheme (11.83)–(11.85) is the second-order convergent both in time and space. Table 11.2: Convergence rates in L2 -norm. m

n

Order�

20 40 80 160

20 40 80 160

2.1951 2.0507

384 � 11 Difference methods for the epitaxial growth model In Figure 11.1, we show the contour plots computed by the scheme with h = 2π/100 and τ = 0.001 at some selected time levels. Compared with the numerical solutions presented in [20], the good agreement can be observed. Figure 11.2 and Figure 11.3 show the development of energy and roughness. The energy has a rapid decay during the early stage. And then it follows the rough-smooth-rough pattern due to nonlinear interaction studied in [20]. Finally, it reaches the steady state after t = 15. Also, the similar phenomenon can be observed from the development of roughness of growth.

Figure 11.1: The contour plots of the numerical solutions at t = 2.5, 5.5, 8, 30.

Figure 11.2: The evolution of energy.

11.7 Summary and extension

� 385

Figure 11.3: The evolution of roughness.

11.7 Summary and extension This chapter discussed the numerical methods for solving the periodic boundary value problem of the epitaxial growth model (11.1)–(11.2). Three difference schemes have been developed: a two-level nonlinear backward Euler difference scheme, a two-level linearized backward Euler difference scheme and a three-level linearized backward Euler difference scheme. The boundedness, existence, uniqueness and convergence of the solution of each difference scheme have been analyzed. The content of the two-level linearized backward Euler difference scheme and the three-level linearized backward Euler difference scheme originated from [23] and [10], respectively. The numerical example section in this chapter originates from [10]. Indeed, a four-level linearized Crank–Nicolson difference scheme for solving (11.1)–(11.2) can be proposed [23]: Find uk ∈ 𝒲h such that 1 1 k+ 21 󵄨󵄨 2 k+ 2 ̂ k 󵄨󵄨󵄨2 ∇ uk+ 2 ) + Δ̄ ϕ̂ k = 0, { δ u + δΔ u − ∇ ⋅ ( ∇ ϕ 󵄨 { t h h h h h ij 󵄨 󵄨 ij ij ij { { { { { 1 ⩽ i ⩽ m , 1 ⩽ j ⩽ m , k = 0, 1, 1 2 { { { 1 ̂ k+ 21 k+ 1 󵄨 󵄨2 2 k+ 2 δt uij + δΔh uij − ∇h ⋅ (󵄨󵄨󵄨∇h û ijk 󵄨󵄨󵄨 ∇h uij 2 ) + Δ̄ h uijk = 0, { { { { { { 1 ⩽ i ⩽ m1 , 1 ⩽ j ⩽ m2 , 2 ⩽ k ⩽ n − 1, { { { { 0 { uij = φ(xi , yj ), 1 ⩽ i ⩽ m1 , 1 ⩽ j ⩽ m2 ,

where 1 ϕ̂ kij = u(xi , yj , 0) + (k + )τut (xi , yj , 0), 2

k = 0, 1,

(11.102) (11.103) (11.104)

386 � 11 Difference methods for the epitaxial growth model k− 21

û ijk = 2uij

k− 32

− uij

,

2 ⩽ k ⩽ n − 1.

The existence, boundedness and convergence of the solution of the difference scheme (11.102)–(11.104) were proved in [23]. Consider the gradient flow in the form of ∇ϕ ) − δΔ2 ϕ, 1 + |∇ϕ|2

(11.105)

|∇ϕ|2 ∇ϕ) − Δϕ − δΔ2 ϕ. 1 + |∇ϕ|2

(11.106)

ϕt = −∇ ⋅ ( which can further be written as ϕt = ∇ ⋅ (

When |∇ϕ|2 ≪ 1, the equation (11.106) can be replaced by ϕt = ∇ ⋅ (|∇ϕ|2 ∇ϕ) − Δϕ − δΔ2 ϕ, which is called the equation for the epitaxial growth model with slope selection, while (11.105) is called the equation for the epitaxial growth model without slope selection [37]. For the numerical methods for solving (11.105), please refer to [24] and [37].

12 Difference methods for the phase field crystal model 12.1 Introduction In this chapter, we consider the difference methods to solve the following phase field crystal model: ϕt = ∇ ⋅ (M(ϕ)∇μ), { { { μ = Δ2 ϕ + 2Δϕ + ϕ3 + (1 − γ)ϕ, { { { { ϕ(x, y, 0) = ψ(x, y),

(x, y) ∈ ℛ2 , 0 < t ⩽ T,

(12.1)

(x, y) ∈ ℛ , 0 < t ⩽ T,

(12.2)

(x, y) ∈ ℛ ,

(12.3)

2

2

where γ is a constant, 0 < γ < 1, ∇ and Δ are the gradient and Laplace operators, respectively, μ is the chemical potential, M(ϕ) > 0 is the mobility, ϕ is periodic with respect to (x, y) in ℛ2 with the period spatial domain Ω = [0, L1 ] × [0, L2 ]. At the beginning, the following energy identity is stated. Theorem 12.1. Let {ϕ(x, y, t), μ(x, y, t)} be the solution of the problem (12.1)–(12.3). Denote 1−γ 2 1 2 󵄨 󵄨2 1 E(ϕ(⋅, ⋅, t)) = ∬[ (Δϕ(x, y, t)) − 󵄨󵄨󵄨∇ϕ(x, y, t)󵄨󵄨󵄨 + ϕ4 (x, y, t) + ϕ (x, y, t)]dxdy. 2 4 2 Ω

Then it holds that d 󵄨 󵄨2 E(ϕ(⋅, ⋅, t)) + ∬ M(ϕ(x, y, t))󵄨󵄨󵄨∇μ(x, y, t)󵄨󵄨󵄨 dxdy = 0, dt

0 < t ⩽ T.

Ω

Proof. Taking the inner product on both the right- and left-hand sides of (12.1) with μ and integrating the result with respect to (x, y) on Ω give ∬ ϕt (x, y, t)μ(x, y, t)dxdy = ∬[∇ ⋅ (M(ϕ(x, y, t))∇μ(x, y, t))]μ(x, y, t)dxdy. Ω

Ω

Substituting (12.2) into the above equality leads to ∬[Δ2 ϕ(x, y, t) + 2Δϕ(x, y, t) + ϕ3 (x, y, t) + (1 − γ)ϕ(x, y, t)]ϕt (x, y, t)dxdy Ω

= ∬ [∇ ⋅ (M(ϕ(x, y, t))∇μ(x, y, t))]μ(x, y, t)dxdy. Ω

Applying the integration by parts and the periodic boundary conditions, one has 1−γ 2 1 d 2 󵄨 󵄨2 1 ϕ (x, y, t)]dxdy ∬[ (Δϕ(x, y, t)) − 󵄨󵄨󵄨∇ϕ(x, y, t)󵄨󵄨󵄨 + ϕ4 (x, y, t) + dt 2 4 2 Ω

https://doi.org/10.1515/9783110796018-012

388 � 12 Difference methods for the phase field crystal model 󵄨2 󵄨 + ∬ M(ϕ(x, y, t))󵄨󵄨󵄨∇μ(x, y, t)󵄨󵄨󵄨 dxdy = 0; Ω

namely, d 󵄨2 󵄨 E(ϕ(⋅, ⋅, t)) + ∬ M(ϕ(x, y, t))󵄨󵄨󵄨∇μ(x, y, t)󵄨󵄨󵄨 dxdy = 0, dt

0 < t ⩽ T.

Ω

It follows easily from Theorem 12.1 that E(ϕ(⋅, ⋅, t)) ⩽ E(ϕ(⋅, ⋅, 0)),

0 < t ⩽ T.

For simplicity, the case of M(ϕ) ≡ 1 will be considered hereinafter.

12.2 Notation and basic lemmas Take three positive integers m1 , m2 and n. Denote h1 = L1 /m1 ,

xi = ih1 ,

h2 = L2 /m2 ,

0 ⩽ i ⩽ m1 ;

τ = T/n;

yj = jh2 ,

0 ⩽ j ⩽ m2 ;

1 tk = kτ, 0 ⩽ k ⩽ n; tk+ 1 = (tk + tk+1 ), 0 ⩽ k ⩽ n − 1, 2 2 Ωh1 ,h2 = {(xi , yj ) | 0 ⩽ i ⩽ m1 , 0 ⩽ j ⩽ m2 }, Ωτ = {tk | 0 ⩽ k ⩽ n}, 𝒲h = {u | u = {uij }, ui+m1 ,j = uij , ui,j+m2 = uij }.

For any mesh functions u, v ∈ 𝒲h , denote δx ui+ 1 ,j = 2

δx2 uij =

1 (u − uij ), h1 i+1,j

1 (u − uij ), h2 i,j+1 1 δy2 uij = (δy ui,j+ 1 − δy ui,j− 1 ), 2 2 h2

δy ui,j+ 1 = 2

1 (δ u 1 − δx ui− 1 ,j ), 2 h1 x i+ 2 ,j

Δh uij = δx2 uij + δy2 uij , m1 m2

(u, v) = h1 h2 ∑ ∑ uij vij , i=1 j=1

‖u‖ = √(u, u),

m1 m2

⟨δx u, δx v⟩ = h1 h2 ∑ ∑(δx ui− 1 ,j )(δx vi− 1 ,j ), i=1 j=1

2

2

m1 m2

⟨δy u, δy v⟩ = h1 h2 ∑ ∑(δy ui,j− 1 )(δy vi,j− 1 ), j=1 j=1

|u|1 = √‖δx u‖2 + ‖δy u‖2 ,

2

2

‖δx u‖ = √⟨δx u, δx u⟩, ‖δy u‖ = √⟨δy u, δy u⟩,

12.2 Notation and basic lemmas m1 m2

(Δh u, Δh v) = h1 h2 ∑ ∑(Δh uij )(Δh vij ),

‖Δh u‖ = √(Δh u, Δh u),

i=1 j=1

m1 m2

� 389

2

2

|Δh u|1 = √h1 h2 ∑ ∑{[δx (Δh u)i− 1 ,j ] + [δy (Δh u)i,j− 1 ] }, 2

i=1 j=1

‖u‖∞ =

max

1⩽i⩽m1 ,1⩽j⩽m2

|uij |,

2

m1 m2

4

‖u‖4 = √h1 h2 ∑ ∑(uij )4 . i=1 j=1

Lemma 12.1 ([47]). For any u ∈ 𝒲h and ε > 0, it holds that ‖Δh u‖2 ⩽

1 2ε ‖u‖2 . |Δ u|2 + 3 h 1 3ε2

Proof. It is noted that u ∈ 𝒲h implies Δh u ∈ 𝒲h . By the integration by parts and the periodic boundary conditions, one has m1 m2

‖Δh u‖2 = h1 h2 ∑ ∑(Δh uij )(Δh uij ) i=1 j=1 m1 m2

= h1 h2 ∑ ∑(δx2 uij + δy2 uij )(Δh uij ) i=1 j=1

m1 m2

= −h1 h2 ∑ ∑[(δx ui− 1 ,j )δx (Δh ui− 1 ,j ) + (δy ui,j− 1 )δy (Δh ui,j− 1 )] i=1 j=1

2

2

2

1 ε ⩽ |Δh u|21 + |u|21 . 2 2ε

2

(12.4)

It follows from Lemma 11.1 that ε 1 |u|21 = −(Δh u, u) ⩽ ‖Δh u‖2 + ‖u‖2 . 2 2ε Substituting (12.5) into (12.4) gives ε 1 ε 1 ‖Δh u‖2 ⩽ |Δh u|21 + ( ‖Δh u‖2 + ‖u‖2 ), 2 2ε 2 2ε i. e., ‖Δh u‖2 ⩽

2ε 1 |Δ u|2 + ‖u‖2 . 3 h 1 3ε2

(12.5)

390 � 12 Difference methods for the phase field crystal model

12.3 Two-level nonlinear difference scheme Define the mesh functions Φn and U n on 𝒲h : Φnij = ϕ(xi , yj , tn ),

Uijn = μ(xi , yj , tn ).

Denote c0 =

max

(x,y)∈Ω,0⩽t⩽T

󵄨 󵄨󵄨 󵄨󵄨ϕ(x, y, t)󵄨󵄨󵄨.

12.3.1 Derivation of the difference scheme Considering equations (12.1) and (12.2) at the point (xi , yj , tk+ 1 ), we have 2

ϕt (xi , yj , tk+ 1 ) = Δμ(xi , yj , tk+ 1 ), 2 2 { { { 2 μ(xi , yj , tk+ 1 ) = Δ ϕ(xi , yj , tk+ 1 ) + 2Δϕ(xi , yj , tk+ 1 ) + ϕ3 (xi , yj , tk+ 1 ) { 2 2 2 2 { { +(1 − γ)ϕ(x , y , t 1 ). i j k+ { 2 By the Taylor expansion, it follows: 1

1

1

k+ k+ k+ { δt Φij 2 = Δh Uij 2 + Pij 2 , 1 ⩽ i ⩽ m1 , 1 ⩽ j ⩽ m2 , 0 ⩽ k ⩽ n − 1, (12.6) { { { { { k 2 k+1 2 1 k+ 21 k+ 21 (Φij ) + (Φij ) k+ 21 k+ 21 k+ 21 2 k+ 2 { { = Δ Φ + 2Δ Φ + (Φ ) + (1 − γ)Φ + Q , U { h h ij ij ij ij ij ij { 2 { { (12.7) { 1 ⩽ i ⩽ m1 , 1 ⩽ j ⩽ m2 , 0 ⩽ k ⩽ n − 1,

and there is a constant c1 such that { { { { {

󵄨󵄨 k+ 21 󵄨󵄨 2 2 2 󵄨󵄨Pij 󵄨󵄨 ⩽ c1 (τ + h1 + h2 ), 1 󵄨󵄨󵄨Qk+ 2 󵄨󵄨󵄨 ⩽ c1 (τ 2 + h2 + h2 ), 1 2 󵄨 ij 󵄨

1 ⩽ i ⩽ m1 , 1 ⩽ j ⩽ m2 , 0 ⩽ k ⩽ n − 1,

(12.8)

1 ⩽ i ⩽ m1 , 1 ⩽ j ⩽ m2 , 0 ⩽ k ⩽ n − 1.

(12.9)

Noticing the initial value condition Φ0ij = ψ(xi , yj ),

1 ⩽ i ⩽ m1 , 1 ⩽ j ⩽ m2 ,

(12.10)

a difference scheme for solving (12.1)–(12.3) can be derived: For 0 ⩽ k ⩽ n, find ϕk , μk ∈ 𝒲h such that k+ 1

k+ 1

{ δt ϕij 2 = Δh μij 2 , 1 ⩽ i ⩽ m1 , 1 ⩽ j ⩽ m2 , 0 ⩽ k ⩽ n − 1, { { { { { k 2 k+1 2 { { { k+ 21 k+ 1 k+ 1 k+ 1 (ϕij ) + (ϕij ) k+ 1 μij = Δ2h ϕij 2 + 2Δh ϕij 2 + ϕij 2 + (1 − γ)ϕij 2 , { 2 { { { { 1 ⩽ i ⩽ m1 , 1 ⩽ j ⩽ m2 , 0 ⩽ k ⩽ n − 1, { { { { 0 { ϕij = ψ(xi , yj ), 1 ⩽ i ⩽ m1 , 1 ⩽ j ⩽ m2 .

(12.11)

(12.12) (12.13)

12.3 Two-level nonlinear difference scheme

� 391

Inserting (12.12) into (12.11) will produce the following difference scheme: Find ϕk ∈ 𝒲h such that k 2 k+1 2 1 k+ 21 k+ 21 k+ 21 (ϕij ) + (ϕij ) k+ 1 { 2 k+ 2 { δt ϕij = Δh [Δh ϕij + 2Δh ϕij + ϕij + (1 − γ)ϕij 2 ], { { { 2 { 1 ⩽ i ⩽ m , 1 ⩽ j ⩽ m , 0 ⩽ k ⩽ n − 1, (12.14) { 1 2 { { { 0 (12.15) { ϕij = ψ(xi , yj ), 1 ⩽ i ⩽ m1 , 1 ⩽ j ⩽ m2 .

12.3.2 Boundedness of the difference solution Theorem 12.2. Suppose {ϕkij | 1 ⩽ i ⩽ m1 , 1 ⩽ j ⩽ m2 , 0 ⩽ k ⩽ n} is the solution of the difference scheme (12.14)–(12.15). Denote 1󵄩 󵄩2 󵄨 󵄨2 F k = 󵄩󵄩󵄩Δh ϕk 󵄩󵄩󵄩 − 󵄨󵄨󵄨ϕk 󵄨󵄨󵄨1 + 2

1 󵄩󵄩 k 󵄩󵄩4 1 − γ 󵄩󵄩 k 󵄩󵄩2 󵄩ϕ 󵄩 + 󵄩ϕ 󵄩 , 4 󵄩 󵄩4 2 󵄩 󵄩

0 ⩽ k ⩽ n.

Then it holds that F k ⩽ F 0,

1 ⩽ k ⩽ n.

Proof. Note that the solution {ϕkij | 1 ⩽ i ⩽ m1 , 1 ⩽ j ⩽ m2 , 0 ⩽ k ⩽ n} of the difference scheme (12.14)–(12.15) satisfies (12.11)–(12.13). Taking the inner product on both the right1 1 and left-hand sides of (12.11) with μk+ 2 and noticing the periodicity of μk+ 2 , we have 󵄨 󵄨2 (δt ϕk+ 2 , μk+ 2 ) = (Δh μk+ 2 , μk+ 2 ) = −󵄨󵄨󵄨μk+ 2 󵄨󵄨󵄨1 ⩽ 0, 1

1

1

1

1

the substitution of (12.12) into which gives 1

1

1

1

(δt ϕk+ 2 , Δ2h ϕk+ 2 + 2Δh ϕk+ 2 + ϕk+ 2 1

1 (ϕk )2 + (ϕk+1 )2 + (1 − γ)ϕk+ 2 ) ⩽ 0. 2

1

Noticing the periodicity of ϕk+ 2 and Δh ϕk+ 2 , it follows from the summation by parts that 1 󵄩󵄩 1 󵄨 󵄩2 󵄩 󵄨2 󵄨 󵄨2 󵄩2 (󵄩󵄩Δh ϕk+1 󵄩󵄩󵄩 − 󵄩󵄩󵄩Δh ϕk 󵄩󵄩󵄩 ) − (󵄨󵄨󵄨ϕk+1 󵄨󵄨󵄨1 − 󵄨󵄨󵄨ϕk 󵄨󵄨󵄨1 ) 2τ τ 1 󵄩 1 󵄩󵄩 k+1 󵄩󵄩4 󵄩󵄩 k 󵄩󵄩4 󵄩2 󵄩 󵄩2 + (󵄩󵄩ϕ 󵄩󵄩4 − 󵄩󵄩ϕ 󵄩󵄩4 ) + (1 − γ) (󵄩󵄩󵄩ϕk+1 󵄩󵄩󵄩 − 󵄩󵄩󵄩ϕk 󵄩󵄩󵄩 ) ⩽ 0, 4τ 2τ Obviously, 1 k+1 (F − F k ) ⩽ 0, τ

0 ⩽ k ⩽ n − 1.

Hence, F k ⩽ F 0,

1 ⩽ k ⩽ n.

0 ⩽ k ⩽ n.

392 � 12 Difference methods for the phase field crystal model Theorem 12.3. Suppose {ϕkij | 1 ⩽ i ⩽ m1 , 1 ⩽ j ⩽ m2 , 0 ⩽ k ⩽ n} is the solution of the difference scheme (12.14)–(12.15). Then there is a constant c2 such that ‖ϕk ‖∞ ⩽ c2 ,

0 ⩽ k ⩽ n.

Proof. In view of (ε2 − 1)2 ⩾ 0, one has 1 k 4 1 k 2 1 (ϕ ) ⩾ (ϕij ) − , 4 ij 2 4

1 ⩽ i ⩽ m1 , 1 ⩽ j ⩽ m2 .

Multiplying h1 h2 on both the right- and left-hand sides of the inequality above and summing over i, j give 1 k 4 1 k 2 L1 L2 ‖ϕ ‖4 ⩾ ‖ϕ ‖ − . 4 2 4 For any ε > 0, by the Cauchy–Schwarz inequality and (Δh ϕk , ϕk ) = −|ϕk |21 , it follows: ε 1 |ϕk |21 ⩽ ‖Δh ϕk ‖2 + ‖ϕk ‖2 . 2 2ε Denote a = 1 − γ > 0. It holds that 1 1 a F k = ‖Δh ϕk ‖2 − |ϕk |21 + ‖ϕk ‖44 + ‖ϕk ‖2 2 4 2 LL ε 1 1 a 1 ⩾ ‖Δh ϕk ‖2 − ( ‖Δh ϕk ‖2 + ‖ϕk ‖2 ) + ( ‖ϕk ‖2 − 1 2 ) + ‖ϕk ‖2 2 2 2ε 2 4 2 LL 1−ε 1 1 = ‖Δh ϕk ‖2 + (1 + a − )‖ϕk ‖2 − 1 2 . 2 2 ε 4 Take ε such that

1−ε 2

> 0 and 21 (1 + a − ε1 ) > 0, which is equivalent to 1−ε 1 1 = (1 + a − ), 2 2 ε

which has a unique positive root ε∗ = Fk ⩾

√a2 +4−a 2

1 ∈ ( 1+a , 1). Thus,

LL 1 − ε∗ (‖Δh ϕk ‖2 + ‖ϕk ‖2 ) − 1 2 . 2 4

It follows from Theorem 12.2 that LL 1 − ε∗ (‖Δh ϕk ‖2 + ‖ϕk ‖2 ) ⩽ F 0 + 1 2 , 2 4

0 ⩽ k ⩽ n.

1 1+a

< ε < 1. Let

12.3 Two-level nonlinear difference scheme

� 393

Then with the help of Lemma 11.2, there is a constant c2 such that ‖ϕk ‖∞ ⩽ c2 ,

0 ⩽ k ⩽ n.

12.3.3 Existence and uniqueness of the difference solution Theorem 12.4. When (2 +

25 4 c )τ 8 2

< 1, the difference scheme (12.14)–(12.15) has a solution.

Proof. The value of ϕ0 is given by (12.15). Now assume that the value of ϕk at the k-th time level has been known, then the nonlinear system in ϕk+1 is obtained from (12.14). k+ 21

In view of δt ϕij

k+ 21

= τ2 (ϕij k+ 21

obtained by ϕk+1 ij = 2ϕij

k+ 21

− ϕkij ), one can solve ϕij k+ 21

− ϕkij . Denote wij = ϕij

at first, then the value of ϕk+1 is

. It follows from (12.14) that

(2wij − ϕkij )2 + (ϕkij )2 2 (wij − ϕkij ) − Δh [Δ2h wij + 2Δh wij + wij + (1 − γ)wij ] = 0, τ 2 1 ⩽ i ⩽ m1 , 1 ⩽ j ⩽ m2 . (12.16) (I) Let 0 ⩽ s ⩽ 5c22 ,

s,

ρ(s) = {

5c22 , s > 5c22 .

Theorem 12.3 reveals that 󵄩󵄩 k 󵄩󵄩 󵄩󵄩ϕ 󵄩󵄩∞ ⩽ c2 ,

‖w‖∞ ⩽ c2 , which implies (2wij − ϕkij )2 + (ϕkij )2 2

⩽ 5c22 ,

1 ⩽ i ⩽ m1 , 1 ⩽ j ⩽ m2 .

Consequently, ρ(

(2wij − ϕkij )2 + (ϕkij )2 2

)=

(2wij − ϕkij )2 + (ϕkij )2 2

,

1 ⩽ i ⩽ m1 , 1 ⩽ j ⩽ m2 ,

which means that the system of nonlinear equations (12.16) is equivalent to the system of nonlinear equations: (2wij − ϕkij )2 + (ϕkij )2 2 (wij − ϕkij ) − Δh [Δ2h wij + 2Δh wij + wij ⋅ ρ( ) + (1 − γ)wij ] = 0, τ 2 1 ⩽ i ⩽ m1 , 1 ⩽ j ⩽ m2 . (12.17)

394 � 12 Difference methods for the phase field crystal model Thus, in order to prove the existence of the system of nonlinear equations (12.16), it suffices to prove the existence of the system of nonlinear equations (12.17). (II) Define the operator Π : 𝒲h → 𝒲h by (2wij − ϕkij )2 + (ϕkij )2 2 Π(w)ij = (wij − ϕkij ) − Δh [Δ2h wij + 2Δh wij + wij ⋅ ρ( ) + (1 − γ)wij ]. τ 2 Then (Π(w), w) 2 = [‖w‖2 − (w, ϕk )] τ − (Δh (Δ2h w + 2Δh w + w ⋅ ρ( =

2 [‖w‖2 − (w, ϕk )] τ − (Δ2h w + 2Δh w + w ⋅ ρ(

(2w − ϕk )2 + (ϕk )2 ) + (1 − γ)w), w) 2

(2w − ϕk )2 + (ϕk )2 ) + (1 − γ)w, Δh w). 2

(12.18)

Notice (w ⋅ ρ(

(2w − ϕk )2 + (ϕk )2 ), Δh w) ⩽ 5c22 ‖w‖ ⋅ ‖Δh w‖ 2

(12.19)

and − (Δ2h w, Δh w) = |Δh w|21 . Inserting (12.19)–(12.20) into (12.18) and using Lemma 12.1 lead to 2 (Π(w), w) ⩾ [‖w‖2 − (w, ϕk )] + |Δh w|21 − 2‖Δh w‖2 τ − 5c22 ‖w‖ ⋅ ‖Δh w‖ + (1 − γ)|w|21

2 3 1 ⩾ (‖w‖2 − ‖w‖ ⋅ ‖ϕk ‖) + (‖Δh w‖2 − 2 ‖w‖2 ) τ 2ε 3ε − 2‖Δh w‖2 − 5c22 ‖w‖ ⋅ ‖Δh w‖.

Taking ε = 21 , we have 2 4 (Π(w), w) ⩾ (‖w‖2 − ‖w‖ ⋅ ‖ϕk ‖) + 3(‖Δh w‖2 − ‖w‖2 ) τ 3 25 4 2 2 2 − 2‖Δh w‖ − (‖Δh w‖ + c2 ‖w‖ ) 4 2 25 󵄩 󵄩 = (‖w‖2 − ‖w‖ ⋅ 󵄩󵄩󵄩ϕk 󵄩󵄩󵄩) − (4 + c24 )‖w‖2 τ 4

(12.20)

12.3 Two-level nonlinear difference scheme

� 395

25 2 󵄩 󵄩 = {[1 − (2 + c24 )τ]‖w‖ − 󵄩󵄩󵄩ϕk 󵄩󵄩󵄩}‖w‖, τ 8 which is followed by (Π(w), w) ⩾ 0 when τ