Vanishing Viscosity Method: Solutions to Nonlinear Systems 9783110494273, 9783110495287

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Boling Guo, Dongfen Bian, Fangfang Li, Xiaoyu Xi Vanishing Viscosity Method

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Boling Guo, Dongfen Bian, Fangfang Li, Xiaoyu Xi

Vanishing Viscosity Method Solutions to Nonlinear Systems

Mathematics Subject Classification 2010 Primary: 35K59, 35L02, 76W05; Secondary: 65M12, 35D30 Authors Prof. Boling Guo Laboratory of Computational Physics Institute of Applied Physics and Computational Mathematics 6 Huayuan Road Haidian District 100088 Beijing People’s Republic of China Dr. Dongfen Bian Beijing Institute of Technology School of Mathematics and Statistics Beijing Key Laboratory on MCAACI 5 Zhongguancun South Street Haidian District 100081 Beijing People’s Republic of China [email protected]

Dr. Fangfang Li Laboratory of Computational Physics Institute of Applied Physics and Computational Mathematics 6 Huayuan Road Haidian District 100088 Beijing People’s Republic of China Dr. Xiaoyu Xi Laboratory of Computational Physics Institute of Applied Physics and Computational Mathematics 6 Huayuan Road Haidian District 100088 Beijing People’s Republic of China

ISBN 978-3-11-049528-7 e-ISBN (PDF) 978-3-11-049427-3 e-ISBN (EPUB) 978-3-11-049257-6 Set-ISBN 978-3-11-049428-0 Library of Congress Cataloging-in-Publication Data A CIP catalog record for this book has been applied for at the Library of Congress. 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.

8

© 2017 Walter de Gruyter GmbH, Berlin/Boston Cover image: Chong Guo Typesetting: Integra Software Services Pvt. Ltd. Printing and binding: CPI books GmbH, Leck  Printed on acid-free paper Printed in Germany www.degruyter.com

Preface Nonlinear evolution equations include quasilinear hyperbolic equation and equations, nonlinear parabolic equation and equations containing integrable system with soliton solution and those which has chaotic solution and so on. With important physical background, these nonlinear partial differential equations reveal essential characteristics of nonlinear phenomena and are of great importance to nonlinear science. From the qualitative research of partial differential equations, the research methods of these equations have common characteristics. For instance, the existence, uniqueness and smoothness of global solution, asymptotic property as t → ∞ and blow-up in finite time of solution can be studied. However, these equations also display unique characteristics. Some mathematicians generalize these equations into three categories: firstly, dissipative system, which is represented by Burgers equation; secondly, hyperbolic conservation system, which is represented by quasilinear hyperbolic equation ut + uux = 0; thirdly, nonlinear dispersion system (integrable system), which is represented by the KdV equation. The research methods of these three kinds of equations are also different. The purpose of writing this book is to introduce and compare the characteristics and research methods of these three kinds of equations. Apart from expounding the basic concepts and methods, the book presents distinctive work, the latest developments and achievements on correlational research direction, based on research results of the author and his collaborators in the last decade. Due to length limitation, instead of extending various methods to find solution for these equations, we focus on the application of vanishing viscosity method in these nonlinear evolution equations. As literally suggested, the so-called vanishing viscosity method is parabolic regularization method, which requires us to find the global solution u% of the approximate equation with viscosity, and then use uniformly bounded estimation of u% and modulus of some of its derivatives on the small parameters %; let % → 0 be the solution we need. This book also presents specific physical background of these nonlinear equations with viscosity. When applying vanishing viscosity method for the numerical solution obtained by discreting nonlinear equation, we consider the viscous problems of corresponding finite difference method with artificial viscosity and the viscous effects caused by the discretization of difference schemes itself. As we all know, famous mathematicians J. von Neumann and R. D. Richtmyer successfully introduced an artificial viscosity term (dissipative term) in the numerical calculation of one-dimensional hydrodynamic equations with Lagrangian form in 1949; and such method has been widely used in the differential calculation of generalized solutions of quasilinear hyperbolic equations, establishing the so-called “artificial viscosity method”. In addition, Godunov, Lax-Friedrichs and other conservative difference schemes also account for an important position in the calculation of hyperbolic equations, although these difference schemes imply viscosity rather than adding artificial viscosity. We investigate

VI

Preface

all kinds of viscous problems of the difference scheme thoroughly, and analyze and compare them qualitatively and numerically, as most of the results have not yet been published. We hope the publishing of this book will help the readers to sort out issues that are not clarified in previous books and literatures, enabling the readers to carry out basic research by referring to the relevant references when they are interested in specific aspect of the problem. Due to the limitation of the book’s length, we have not been able to discuss the limitations of vanishing viscosity method itself. In addition, we value the input of our readers who are quite welcomed to contact us for the improvement of the work. Finally, our gratitude also goes to professor Hsiao Ling and professor Shui Hongshou, for their consistent interest and enthusiasm in writing this book, and for putting forward valuable suggestions. I would like to express our gratitude to professor Yunlin Zhou, who guided us and helped us in the difficult years. Boling Guo 18 August 2016, Beijing

Contents 1 1.1 1.1.1 1.1.2 1.1.3 1.2 1.3 1.4 1.5

Sobolev Space and Preliminaries 1 Basic Notation and Function Spaces 1 Basic Notation 1 Function Spaces 1 Some Basic Inequalities 3 Weak Derivatives and Its Properties, Wpm (K) and H j,p (K) Spaces Sobolev Embedding Theorem and Interpolation Formula 6 Compactness Theory 24 Fixed Point Principle 37

2 2.1

39 The Vanishing Viscosity Method of Some Nonlinear Evolution System Periodic Boundary and Cauchy Problem for High-Order Generalized KdV System in Dimension One 39 Some KdV System with High-Order Derivative Term 62 High-Order Multivariable KdV Systems and Hirota Coupled KdV Systems 72 Initial Boundary Value Problem for Ferrimagnetic Equations 88 Existence and Uniqueness of Smooth Solution to Ferrimagnetic Equations and Other Problems for High-Dimensional Ferrimagnetic 100 Equations with Nonlinear Boundary Conditions Periodic Boundary Problem and Initial Value Problem for the Coupling KdV–Schrödinger Equations 117 Initial Value Problem for the Nonlinear Singular Integral and Differential Equations in Deep Water 132 Initial Value Problem for the Nonlinear Schrödinger Equations 159 Initial Value Problem and Boundary Value Problem for the Nonlinear Schrödinger Equation with Derivative 165 Initial Value Problem for Boussinesq Equations 187 Initial Value Problem for Langmuir Turbulence Equations 210

2.2 2.3 2.4 2.5

2.6 2.7 2.8 2.9 2.10 2.11 3 3.1 3.2 3.3 3.4 3.5 3.6 3.7

4

233 The Vanishing Viscosity Method of Quasilinear Hyperbolic System Generalized Solutions to the First-Order Quasilinear Hyperbolic Equation in One Dimension 233 Existence and Uniqueness of the General Solution to First-Order Multivariable Quasilinear Equations 246 Convergence of Solutions to the Multidimensional Linear Parabolic System with Small Parameter 264 On Gradient Quasilinear Parabolic Equations and Viscous Isentropic Gas Hydrodynamic Equations 274 On Some Results of Some Quasilinear Parabolic Equations 308 On Traveling Wave Solutions of Some Quasilinear Parabolic Equations with Small Parameter 318 On General Solutions of Some Diagonal Quasilinear Hyperbolic Equations 337

VIII

Contents

The Compensated Compactness Method 349 The Existence of Generalized Solutions for the First-Order Quasilinear Hyperbolic System 373 Convergence of Solutions to Some Nonlinear Dispersive Equations 389

3.8 3.9 3.10 4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 4.9

4.10 4.11 5

415 Physical Viscosity and Viscosity of Difference Scheme Ideal Fluid, Viscous Fluid and Radiation Hydrodynamic Equations 415 The Artificial Viscosity of Difference Scheme 424 The Fundamental Difference Between Linear and Nonlinear Viscosity Qualitatively 427 von Neumann Artificial Viscosity and Godunov Scheme Implicit Viscosity 431 Several Difference Schemes with Mixed Viscosity 442 Artificial Viscosity Problem for Hydrodynamic Equations with Radiation Conductivity Transfer Term 450 Qualitative Analysis of Singular Points of Some Artificial Viscosity Equation 454 Numerical Calculation Results and Analysis 460 Local Comparison of Different Viscosity Method to Initial Discontinuity Problem for One-Dimensional Radiation Hydrodynamic Equations with 470 Different Medium Implicit Viscosity of PIC Method 477 Two-Dimensional “Artificial Viscosity” Problem 480

5.3

Convergence of Lax–Friedrichs Scheme, Godunov Scheme and Glimm Scheme 485 Convergence of Lax–Friedrichs Difference Scheme 485 The Convergence of Hyperbolic Equations in Lax–Friedrichs Scheme 501 Convergence of Glimm Scheme 513

6 6.1 6.2 6.3 6.4 6.5 6.5.1 6.5.2 6.5.3 6.5.4 6.5.5 6.6

531 Electric–Magnetohydrodynamic Equations Introduction 531 Definition of the Finite Energy Weak Solution 534 The Faedo–Galerkin Approximation 536 The Vanishing Viscosity Limit 540 Passing to the Limit in the Artificial Pressure Term 541 Passing to the Limit 541 The Effective Viscous Flux 544 The Amplitude of Oscillations 546 The Renormalized Solutions 547 Strong Convergence of the Density 549 Large-Time Behavior of Weak Solutions 553

5.1 5.2

References

557

1 Sobolev Space and Preliminaries In this chapter, we collect a number of preparatory results. We introduce some function spaces, interpolation theory, compactness principle, the fixed point principle and so on. We shall not attempt, however, to give an exhaustive treatment of the subject, since this is beyond the scope of the book. Therefore, the reader who wants more details is referred to the specialized literature quoted throughout. As a rule, we give proofs where they are elementary or relevant to the development of the subject, or also when the result is new or does not seem to be widely known. See also Refs [52, 81, 116].

1.1 Basic Notation and Function Spaces 1.1.1 Basic Notation For convenience, we introduce some notations through this book. Rn denotes the ndimensional spaces, where x = (x1 , x2 , . . . , xn ) ∈ Rn denotes the element of Rn . K ⊂ Rn is bounded domain, with the boundary ∂K. If ! is n-tuple of nonnegative integers !i , we set |!| =

n 

!i

i=1

and ∂ |!| u

D! u(x) =

! ∂x1 1 ⋯∂xn!n

.

1.1.2 Function Spaces The space Lp (K): Let K be a domain in Rn and let p be a positive real number. We denote by Lp = Lp (K) the class of all measurable functions u defined on K for which  p K |u(x)| d, < ∞, the normal Lp norm is defined as u(x)Lp

⎧ ⎨( |u(x)|p dx) p1 , 1 ≤ p < ∞; K = ⎩sup |u(x)|, p = ∞.

(1.1.1)

x∈K

The space Lp (K) (1 ≤ p ≤ ∞) with the norm (1.1.1) is a Banach space. It is not a norm if 0 < p < 1. If p = 2, L2 (K) is a Hilbert space with respect to the inner product  (u, v) = u(x)v(x)dx. K

DOI 10.1515/9783110494273-001

2

1 Sobolev Space and Preliminaries

The Sobolev space Wpm (K): the class of all locally summable functions u : U → R such that for each multiindex ! with |!| ≤ k, D! u exists in the weak sense and belongs to Lp (K). The Sobolev space Wpm (K) is a Banach space, with its norm ⎛ u(x)Wpm (K)

=⎝

⎞1

p



p Dj uLp (K) ⎠

|j|≤m

.

If p = 2, W2m (K) = H m (K) is a Hilbert space. D(K)(D(K)) consists of all those functions in C∞ (K)(C∞ (K)) that have compact support in K(K). m W0,p (K) denotes the closure of D(K) in the space Wpm (K). If p = 2, then m (K) = H0m (K). W0,2

Lp (a, b; X) is a Lp function space from (a, b) to Banach space X, with its norm  u(x)Lp (a,b;X)(K) =

b

a

p1 p u(t)X dt

.

Lp (a, b; X) is also a Banach space. The function spaces Lp (a, b; Wpm (K)), Wpm (0, T; Lr (K)) and so on can be similarly defined. If p = ∞, L∞ (a, b; X) consists of all functions that are measurable and essentially bounded. It is a Banach space with respect to the norm u(x)L∞ (a,b;X)(K) = sup esst∈(a,b) u(t)X . Similarly, if –∞ < a < b < +∞, C([a, b]; X) is a space consisting of all continuous functions from [a, b] to Banach space X. Ck ([a, b]; X), k ∈ N is a space consisting of all partial derivatives D! u of orders |!| ≤ k are continuous from [a, b] to Banach space X. They are Banach spaces with norms given by u(x)C([a,b];X) = sup u(t)X , t∈(a,b)

u(x)Ck ([a,b];X) =

k  j=0



dj u C([a,b];X) . dtj

Cl (K) is a space consisting of all partial derivatives D! u of orders |!| ≤ l in K. If 0 < + ≤ 1, we define Cn,+ (K) to the subspace of Cm (K) consisting of those functions u for which, for 0 ≤ |!| ≤ l, D! u satisfies in K a Hölder condition of exponent +. The Cn,+ (K) norm is defined as

1.1 Basic Notation and Function Spaces



u(x)Cn,+ (K) = u(x)Cn (K) +

3

H+ (D! u),

|!|=n

where H+ (>) = max x,y∈K

|>(x) – >(y)| . |x – y|+

1.1.3 Some Basic Inequalities (1) Cauchy’s inequality   aij .i .j aij 'i 'j

|aij .i 'j |
0, there exists some constant C' = C(', X0 , X, X1 ) such that vX ≤ 'vX0 + C' vX1

∀ v ∈ X0 .

(1.4.15)

Proof. We argue by contradiction. Suppose that eq. (1.4.15) is not true, that is, there exists some ' > 0 such that for each c ∈ R, vX ≥ vX0 + cvX1 , for some v ∈ X0 . Then, we have a sequence by taking c = m vm X ≥ vm X0 + mvm X1 . For the normalized sequence wm =

vm , vm X0

1.4 Compactness Theory

29

one has wm X ≥ ' + mwm X1

∀ m.

(1.4.16)

{wm } is bounded in X since wm X0 = 1. This together with eq. (1.4.16) implies wm X1 → 0,

m → ∞.

(1.4.17)

By eq. (1.4.14), we derive that wm is compact in X. Thus, we can extract a subsequence {w, } from {wm } such that it converges strongly in X. Moreover, this limit is zero by eq. (1.4.17). However, this contradicts with eq. (1.4.16) w, X ≥ ' > 0

∀ ,. ∎

Therefore, Lemma 1.4.13 follows. Assume that X0 , X and X1 are Banach spaces satisfying X0 ⊂ X ⊂ X1 .

(1.4.18)

and the above embedding is continuous, and Xi is reflexive,

i = 0, 1,

(1.4.19)

mapping X0 → X is precompact.

(1.4.20)

Let T > 0 be a fixed finite number, and !0 , !1 > 1 be finite numbers. Now, we consider the space Y =Y(0, T; !0 , !1 ; X0 , X1 ),   dv Y = v ∈ L!0 (0, T; X0 ), U ′ = ∈ L!1 (0, T; X1 ) , dt

(1.4.21) (1.4.22)

equipped with the norm vY := vL!0 (0,T;X0 ) + vL!1 (0,T;X1 ) .

(1.4.23)

Then, Y is a Banach space and it is obvious that the mapping Y ⊂ L!0 (0, T; X) is continuous. Indeed, we will show that this mapping is compact. Theorem 1.4.14. Under assumptions (1.4.18)–(1.4.21), the mapping Y ⊂ L!0 (0, T; X) is compact.

30

1 Sobolev Space and Preliminaries

Proof. (1) For any bounded sequence {um } in Y, we need to show that there exists a subsequence {u, } such that it converges strongly in L!0 (0, T; X). Since Xi (i = 0, 1) is reflexive and 1 ≤ !i < +∞, L!i (0, T; Xi )(i = 0, 1) is also reflexive. Thus, Y is reflexive. Hence, there exists some u ∈ Y and a subsequence {u, } such that u, ⇀ u,

in Y,

, → ∞.

u, ⇀ u

in L!0 (0, T; X0 ),

u′,

in L!1 (0, T; X0 ).

(1.4.24)

This means

⇀u

′

(1.4.25)

Thus, it suffices to show that v, := u, – u → 0 in L! (0, T; X0 ).

(1.4.26)

(2) Equation (1.4.26) can be reduced to prove v, → 0 in L!0 (0, T; X0 ).

(1.4.27)

In fact, using Lemma 1.4.13, we have v, L!0 (0,T;X) ≤ 'v, L!0 (0,T;X0 ) + c' v, L!0 (0,T;X1 ) . This together with {v, } bounded in Y implies that v, L!0 (0,T;X) ≤ c' + c' v, L!0 (0,T;X) .

(1.4.28)

Taking , → ∞ and by eq. (1.4.27), we obtain lim v, L!0 (0,T;X) ≤ c'.

,→∞

(1.4.29)

This limit is zero since ' > 0 is arbitrary. So eq. (1.4.26) follows. (3) Now, we turn to show eq. (1.4.27). It is easy to check that the mapping Y ⊂ C([0, T]; X1 )

(1.4.30)

is continuous. This implies v, X1 ≤ c

∀ t ∈ [0, T],

∀ m.

(1.4.31)

By Lebesgue’s theorem, we only need to prove that v, (t) → 0 in X1 , as , → ∞, a.e. t ∈ [0, T].

(1.4.32)

31

1.4 Compactness Theory

We only verify eq. (1.4.32) with t = 0, since the proof of other t is similar. Noting that  v, (0) =v, (t) – =

1 s

 0

t

v,′ (4) d4

0

 s

s

v, (t) dt –

0

t

0

 v,′ (t) d4 dt ,

we estimate v, (t) = a, + b, ,

(1.4.33)

with 1 a, = s



s

0

1 b, = – s

v, (t) dt,

 0

s

(s – t)v,′ (t) dt.

(1.4.34)

For fixed % > 0, we take s > 0 such that  b, X1 ≤

s

0

% v,′ (t)X1 dt ≤ . 2

Hence, for this fixed s, a, converges weakly to zero in X0 as , → ∞. Thus, a, converges strongly in X1 and for , sufficiently large, % a, X1 ≤ . 2 And then eq. (1.4.32) follows. Therefore, we complete the proof of Theorem 1.4.14.

∎

Next, we consider the above compact theory for the space with fractional derivative. Suppose that X0 , X and X1 are Hilbert spaces satisfying X0 ⊂ X ⊂ X1 ,

(1.4.35)

and the above embedding is continuous, and the mapping X0 → X

(1.4.36)

is compact. Let v : R → X1 , and vˆ denotes its Fourier transform  vˆ (4) =

+∞

e–2i0i4 v(t) dt.

(1.4.37)

–∞

The #-order derivative in t for v is the inverse Fourier transform of (20i4)ˆv, in other words,

32

1 Sobolev Space and Preliminaries

'# v(4) = (20i4)ˆv. D i

(1.4.38)

For # > 0(0 < # ≤ 1 in general), we define the space # $ # H# (R; X0 , X1 ) := v ∈ L2 (R; X0 ), Dt v ∈ L2 (R; X1 ) ,

(1.4.39)

equipped with the norm (2 ( # vH# (R;X0 ,X1 ) := v2L2 (R;X0 ) + (|4|# vˆ (L

$1

2 (R;X1 )

2.

Then, H# (R; X0 , X1 ) is a Hilbert space. # For any subset K ⊂ R, we denote the subset HK ⊂ H# which consists of the # functions u in H with compact support contained in K. Theorem 1.4.15. Let X0 , X, X1 be Hilbert spaces satisfying eqs (1.4.35) and (1.4.36). Then, # for any bounded set K and # > 0, the mapping HK (R; X0 , X1 ) → L2 (R; X) is compact. #

Proof. (1) Fix K and #. Let the sequence {um } ⊂ HK (R; X0 , X1 ) be bounded. We are reduced to show that there exists a subsequence converging strongly in L2 (R, X). Since H# (R; X0 , X1 ) is the Hilbert space, for the sequence {um }, there exists a # subsequence {u, } converging weakly to some u ∈ HK . Letting v, = u, – u, #

then the sequence {v, } is bounded in HK (R; X0 , X1 ) and converges weakly to zero, that is v, → 0 weakly in #ˆ

|4| v, → 0

weakly in

L2 (R; X0 ),

(1.4.40)

L2 (R; X1 ).

(1.4.41)

Then, Theorem 1.4.15 follows, provided that we can show that u, converges strongly to u, that is v, → 0 in L2 (R; X).

(1.4.42)

(2) Equation (1.4.42) can be further reduced to prove v, → 0 in L2 (R; X1 ).

(1.4.43)

In fact, we have by Lemma 1.4.13 v, L2 (R;X) ≤ 'v, L2 (R;X0 ) + C' v, L2 (R;X1 ) .

(1.4.44)

33

1.4 Compactness Theory

Combining this with the fact that v, is bounded in L2 (R; X0 ), we obtain v, L2 (R;X) ≤ C' + C' v, L2 (R;X1 ) .

(1.4.45)

Letting , → ∞ and by eq. (1.4.43), one has lim v, L2 (R;X) ≤ C'.

,→∞

And so eq. (1.4.42) follows since ' in Lemma 1.4.13 can be arbitrary small. (3) Now, we turn to show eq. (1.4.43). It follows from Parseval identity  +∞  +∞ I, = v, (t)2X1 dt = ˆv, (4)2X1 d4, –∞

(1.4.46)

–∞

where vˆ , denotes the Fourier transform of v, . Then, eq. (1.4.43) is equivalent to I, → 0,

as

, → ∞.

(1.4.47)

To prove this, we rewrite I,   ! " d4 1 + |4|2# ˆv, (4)2X1 I, = ˆv, (4)2X1 d4 + (1 + |4|2# ) |4|≤M |4|>M  c ≤ + ˆv, (4)2X1 d4, 1 + M 2# |4|≤M where we use the fact that v, is bounded in H# in the above. For fixed % > 0, we take M such that % c ≤ . 1 + M 2# 2 Hence,  I, ≤

% ˆv, (4)2X1 d4 + . 2 |4|≤M

It is easy to see that eq. (1.4.47) follows, provided that we can prove that  |4|≤M

ˆv, (4)2X1 d4 → 0,

as

, → ∞,

(1.4.48)

for any fixed M. Denote 7 to be the characteristic function on K. Then, v, 7 = v, and  vˆ , (4) =

+∞

–∞

e–2i0t4 7(t)v, (t) dt.

Hence, ˆv, (4)X1 ≤v, L2 (R;X1 ) e–2i0t4 7L2 (R) ≤ const.

(1.4.49)

34

1 Sobolev Space and Preliminaries

On the other hand, for each 3 ∈ X0 and fixed 4, we know that the term !

vˆ , (4), 3

"∗ X0

 :=

+∞ !

–∞

v, (t), e–2i0t4 7(t)3

" X0

dt

tends to zero by eq. (1.4.40). Thus, the sequence vˆ , (4) converges weakly to zero in X0 and converges strongly in X and X1 . This together with eq. (1.4.49) and Lebesgue theorem yields eq. (1.4.48). ∎ Theorem 1.4.16. Under assumptions (1.4.25) Y(0, T; 4, 1; X0 , X1 ) → L2 (0, T; X) is compact.

and

(1.4.26),

the

mapping

Proof. Let the sequence {um } be bounded in Y. We denote the function u˜ m defined in R, and u˜ m =

 um , 0,

t ∈ [0, T], t ∈/ [0, T].

From Theorem 1.4.15, we only need to show that u˜ m is bounded in H# (R; X0 , X1 ) for some # > 0. By Lemma 1.4.13, each um is continuous from [0, T] to X1 up to a set with zero measure. More precisely, the mapping Y → C([0, T]; X1 ) is continuous. Since u˜ m has two discontinuous points 0 and T, the weak derivative of u˜ m is du˜ m = g˜ m + um (0)$0 – um (T)$T , dt

(1.4.50)

where $0 and $T denote the delta function in 0 and T, respectively, and gm = u′m = the derivative of um in [0, T].

(1.4.51)

Applying Fourier transformation in eq. (1.4.50) gives 2i04uˆ m (4) = gˆ m (4) + um (0) – um (T)e–2i04T ,

(1.4.52)

where gˆ m and uˆ m denotes the Fourier transformation of gm and um , respectively. Since gm is bounded in L1 (0, T; X1 ), g˜ m is bounded in L1 (R; X1 ) and gˆ m is bounded in L∞ (R; X1 ): ˆgm (4)X1 ≤ const,

∀ m,

∀ 4 ∈ R.

Noting that the mapping Y → C([0, T]; X1 ) is continuous, we have um (0)X1 ≤ const,

um (T)X1 ≤ const.

(1.4.53)

35

1.4 Compactness Theory

Combining this with eq. (1.4.52), we obtain |4|2 uˆ m (4)2X1 ≤ c

∀ m,

∀ 4 ∈ R.

(1.4.54)

For fixed # < 21 , we get |4|2# ≤ c0 (#)

1 + 42 1 + |4|2(1–#)

∀ 4 ∈ R.

Hence, 

+∞ –∞

 +∞ 1 + 42 |4|2# uˆ m (4)2X1 d4 ≤c0 (#) uˆ m (4)2X1 d4 2(1–#) –∞ 1 + |4|  +∞  +∞ d4 ≤c1 + c (#) uˆ m (4)2X1 d4. 0 2(1–#) –∞ 1 + |4| –∞

The integral 

+∞

–∞

d4 1 + |4|2(1–#)

converges by # < 21 . On the other hand, we have by Parseval’s identity 

+∞ –∞

 uˆ m (4)2X1 d4 =

0

T

um (t)2X1 dt.

This integral is bounded. And so, 

+∞

–∞

|4|2# uˆ m (4)2X1 d4 ≤ c2 (#).

(1.4.55)

Thus, the sequence u˜ m is bounded in H# (R; X0 , X1 ), and its support is contained in [0, T]. Therefore, we complete the proof of Theorem 1.4.16. ∎ Remark 1.4.17. Suppose that X1 is a Hilbert space, and X0 , X1 are Banach spaces. Then, by the same argument as above, we derive that the mapping from Y(0, T; !0 , 1; X0 , X1 ) to L!0 (0, T; X) for the bounded number ! > 1 is compact. One can utilize the theory of real variable function to show the following theorem. Theorem 1.4.18. Suppose that the function f (x, u) is measurable in the set {x ∈ K, u ∈ (–∞, ∞)}, and is continuous for almost everywhere x ∈ K. If the sequence {uk (x)} ⊂ L1 (K) converges almost everywhere to u(x) ∈ L1 (K), and

36

1 Sobolev Space and Preliminaries

f (x, uk (x))Lq (K) ≤ C,

q > 1,

(1.4.56)

then the function f (x, uk (x)) converges strong in Lq∗ (K)(q∗ < q) and converges weakly in Lq (K). Assume in addition, f (x, uk (x))Lq (K′ ) ≤ ,(mes K′ ),

(1.4.57)

with ,(4) being a continuous function in 4 > 0, ,(0) = 0, K′ being the measurable subset of K. Then, f (x, uk (x)) converges strongly to f (x, u(x)) in Lq (K). As we know, in some cases, for some nonlinear equation or system (such as quasilinear hyperbolic system), we can only obtain a weak prior estimate, such as the uniformly prior estimate of L∞ -norm or Lq -norm. In this case, Helly’s election principle fails. One needs to study the more weak precompact theory. Up to now, there has been the following compensate precompact principle. Theorem 1.4.19. Suppose that K ⊂ Rm , K ⊂ R are open bounded sets, and f (x) : Rm → R is continuous. Then, we have (i) If us : K → Rm satisfies us (x) ∈ K, a.e., then there exists a subsequence {us }∞ s=1 and a family of probability measure {vx }x∈K so that supp vx ∈ K¯

and f (us ) converges weak∗ to f¯ in L∞ (K),

with f¯ (x) =

 Rm

vx (+)f (+) d+.

(1.4.58)

(ii) On the contrary, if there exists vx satisfying the above property, then there exists m a sequence {us }∞ s=1 (us : K → R ), us (x) ∈ K, a.e. such that for any continuous function f : K → R f (us ) converges weak∗ to f¯ in L∞ (K), with f¯ (x) defined in eq. (1.4.58). ∗

→ u. Then, Corollary 1.4.20. Assume that us converges weak∗ to u in L∞ , written as us  us converges strongly to u in Lp (p < ∞) if and only if vx = $u(x) . Regarding the compensate precompact theory and its applications, we will discuss them explicitly in Chapter 3.

1.5 Fixed Point Principle

37

1.5 Fixed Point Principle In this section, we list some common fixed point principles. These fixed point principles play important roles in the study of partial differential equations. Since the proof of these theory can be found in a lot of books, we just list these principles without proof here. Theorem 1.5.1 (Contraction mapping principle). Suppose that (X, d) is a complete metric space, and the function f : X → X satisfies ! " d f (x), f (y) ≤ kd(x, y), k < 1

∀x, y ∈ X.

Then there exists a unique fixed point x0 ∈ X such that f (x0 ) = x0 . Theorem 1.5.2. Assume that the mapping T : M → M is “Boyd–Wong contraction”, that is, there exists a function >(t) such that (i) (ii) (iii)

> < t ∀ t ∈ R+ , ! " d(Tx, Ty) < > d(x, y) ∀ x, y ∈ M, > is right upper semicontinuous.

Then, the mapping T has a fixed point in M. Theorem 1.5.3. Let (X, d) be a complete metric space, and the mapping f : X → X be continuous. Suppose that there exists a lower semicontinuous > : X → R+ such that for any x ∈ X, there holds d(x, f (x)) < >(x) – >(f (x)). Then, f has a fixed point in X. Theorem 1.5.4 (Brouwer’s fixed point theorem). Let Bn = (x, x ∈ Rn , x ≤ 1), f : Bn → Bn is continuous. Then, f has a fixed point in Bn . Theorem 1.5.5 (H. Poincare’s fixed point theorem). Suppose that the mapping f : En → En is continuous, and for some r > 0 and any + > 0, we have f (u) + +u ≠ 0, ∀u u = r. Then, there exists one point u0 with u0  = r such that f (u0 ) = 0. Theorem 1.5.6 (Schauder fixed point theorem). Assume that the bounded set C is convex in X, and the mapping f : C → C is compact. Then, there exists x0 ∈ C such that f (x0 ) = x0 .

38

1 Sobolev Space and Preliminaries

Theorem 1.5.7. Let the set C be compact convex in C, and the mapping f : C → C is closed. Then, there exists x0 ∈ C such that f (x0 ) = x0 . Theorem 1.5.8. Suppose that C is a reflexive Banach space, and the mapping T : X → X ∗ is closed with   Tx, x ≥ cx2

∀ x ∈ X,

for some c > 0. Then, Tx = x. Theorem 1.5.9. Let X be a reflexive Banach space. Assume that the mapping B : X ×X → C is linear in the first variable and antilinear in the second variable, and satisfies (i) (ii)

|B(x, y)| ≤ c1 x ⋅ y ∀x, y ∈ X, |B(x, y)| ≥ c2 x2 ∀x, y ∈ X.

Then, for each x ∈ X, there exists a unique point x∗ ∈ X ∗ such that X ∗ (y) = B(y, x). Theorem 1.5.10 (Leray–Schauder theorem). Let E be a Banach Space. Suppose that the nonlinear operator equation x – A(x, +) = 0, x ∈ E, + ∈ I = [0, 1] satisfies (i) (ii) (iii)

For + ∈ I, the operator A : E×I → E is completely continuous. And for the bounded set M ⊂ E, the operator A is continuous in +. For + ∈ I, all solutions x to x – A(x, +) = 0 are uniformly bounded in E. When + = 0, A(x, 0) = x0 ∈ E, ∀ x ∈ E.

Then, for each + ∈ I, there is at least one solution to x – A(x, +) = 0.

2 The Vanishing Viscosity Method of Some Nonlinear Evolution System In this chapter, we mainly present the application of vanishing viscosity method in some nonlinear evolution equations, including Korteweg–de Vries (KdV) system, nonlinear Schrödinger equation, ferromagnetic chain equations (i.e., generalized Landau–Lifshitz equation), ZaxapoB equation, the nonlinear singular integral– differential equation of deep water, Boussinesq equation and some coupling system between them. The so-called vanishing viscosity method states that through adding dissipative term with suitable small parameter in the original nonlinear evolution equation, one can obtain the corresponding parabolic equation. From the known result of the parabolic equations, we can easily get the smooth solution for such parabolic equation with small parameter. Then, using the uniform estimate of the solution to this equation with respect to small parameter and letting such small parameter tends to zero, we obtain the solution to original equation in some sense. In mathematics, this method is also called parabolic regularization method. In general, the result given by this method is very nice. In physics, the regularized equation has practical significance. The small parameter term represents dissipative effect. The disappearing process of this term reflects the actual approximation process under some physic assumption. Certainly, the vanishing viscosity method has its limitation. For example, for some nonlinear evolution equations, it is hard to obtain the uniform estimate in small parameter. Sometimes, one can get some uniform estimate. However, such estimate is not enough to ensure the convergence of the solution. In addition, for some irregular coupling system, the other methods (such as Galerkin method) are much more convenient. One can refer to Refs [34–41, 48, 49, 105–114] for this chapter.

2.1 Periodic Boundary and Cauchy Problem for High-Order Generalized KdV System in Dimension One In the study of soliton, the representative equation – KdV equation ut + !uux + "uxxx = 0

(2.1.1)

plays an important role. This equation has been researched by many mathematics and physics. The complex KdV equation and high-order KdV system are also proposed from physics. We consider the following high-order generalized KdV system: ut + (grad >(u))x + ux2p+1 = A(x, t)u + g(x, t), (2.1.2) " ! where u(t, x) = u1 (x, t), . . . , uJ (x, t) is a J-dimensional vector function, >(u) = ! " >(u1 , . . . , uJ ) is a scalar function of u, g(x, t) = g1 (x, t), . . . , gJ (x, t) is a J-dimensional DOI 10.1515/9783110494273-002

40

2 The Vanishing Viscosity Method of Some Nonlinear Evolution System

vector function of (x, t), A(x, t) = (aij (x, t)) is J × J-matrix function and p ≥ 1 is an integer. For system (2), we consider the periodic boundary problem in the region {(x, t)| – ∞ < x < +∞, 0 ≤ t ≤ T}:  u(x, 0) = u0 (x),

–∞ < x < ∞,

u(x + 2D, t) = u(x, t),

–∞ < x < ∞,

0 ≤ t ≤ T,

(2.1.3)

where u0 = (u01 (x), . . . , u0J (x)) is a J-dimensional vector function with period 2D and D > 0 is a constant. We also consider the Cauchy problem u(x, 0) = u0 (x),

–∞ < x < +∞,

(2.1.4)

with u0 (x) being a J-dimensional vector function defined in (–∞, +∞). For p = 1, we discuss the similar problem for the system like ! " ut + grad>(u) x + uxxx = f (u).

(2.1.5)

We will utilize the vanishing viscosity method to study the well-posedness of all the above problems. First, we study the global generalized and classical solution for the periodic boundary problem of the system with small parameter term ! " ut + %uxxx + grad>(u) x + uxxx = f (u),

(2.1.6)

! " ut + (–1)p+1 %ux2p+ + grad>(u) x + ux2p+1 = A(x, t)u + g(x, t),

(2.1.7)

and

with % > 0 and p ≥ 1. As % → 0, the limit of solutions u% (x, t) corresponds to the solution u(x, t) of the periodic boundary problem for systems (2.1.2) and (2.1.5), respectively. As the period D → ∞, we can get the global solution of the Cauchy problem for systems (2.1.2) and (2.1.5). Now, we turn to discuss the small parameter system (2.1.7) with periodic boundary (2.1.3). Lemma 2.1.1. Assume that the linear partial differential equation vt + %(–1)p+1 vx2p+2 + vx2p+1 = f (x, t),

(2.1.8)

with periodic boundary condition 

v(x + 2D, t) = v(x, t),

v(x, 0) = v0 (x),

(2.1.9)

2.1 Periodic Boundary and Cauchy Problem for High-Order Generalized KdV System

41

satisfies v0 

(p+1)

W2

(–D,D)

+ f L2 ((–D,D)×(0,T)) < ∞,

(2.1.10)

where f (x, t) and v0 (x) are both functions with period 2D in variable x, p ≥ 1 and D > 0. Then, there exists a unique solution v to eqs (2.1.8) and (2.1.9) in the region [–D, D]×[0, T]. (2p+2) Moreover, the solution v belongs to L2 (0, T; W2 (–D, D)) ∩ W2(1) (0, T; L2 (–D, D)) and satisfies v

(p+1)

L∞ (0,T;W2

!

≤ c v0 

(–D,D))

+ vt L2 ((–D,D)×(0,T)) + v (2p+2) L2 (0,T;W2 (–D,D)) " + f L2 ((–D,D)×(0,T)) ,

(2.1.11)

(p+1) W2 (–D,D)

where the constant c depends on % and p. Proof. The existence of solution to eqs (2.1.8) and (2.1.9) can be shown by separation variable method. Estimation (2.1.11) follows easily by the integral estimate. ∎ Corollary 2.1.2. Assume that for k ≥ 0, h ≥ 0, we have Dkx Dht f (x, t) ∈ L2 ((–D, D) × (0, T)) ((p+1)(2h+1)+k) and v0 (x) ∈ W2 (–D, D). Then, the solution v to eqs (2.1.8) and (2.1.9) satisfies (2p+1) (p+2) h k (–D, D)) ∩ W2(1) (0, T; L2 (–D, D)) ∩ L2 (0, T; W2 (–D, D)) Dx Dt v(x, t) ∈ L∞ (0, T; W2 and the similar estimate as eq. (2.1.11) also holds. Lemma 2.1.3. Suppose the nonlinear partial differential system (2.1.7) with periodic boundary condition (2.1.3) satisfies the following conditions: (i) u0 (x) is a function with period 2D and u0 

(p+1)

W2

(–D,D)

< ∞ (p ≥ 1).

(2.1.12)

(ii) For any 1 ≤ k ≤ p, |∂xk A(x, t)| + |∂xk g(x, t)| < ∞ in [–D, D] × [0, T]. And all ∂xk A(x, t), ∂xk g(x, t) are functions with period 2D. (iii) The function >(u) is 3-times continuously differentiable in vector variable u = (u1 , . . . , uJ ). Besides, it satisfies (a) |>(u)| ≤ c|u|l ,   (b) grad >(u) ≤ c|u|l–1 ,  2  " !   (c)  ∂∂u>(u)  ≤ c |u|l–2 + 1 (i, j = 1, . . . , J), where l = 4p + 2 – $, $ > 0, c is a i ∂uj constant. Then, there is a unique global generalized solution u(x, t) = (u1 (x, t), . . . , uJ (x, t) to sys(p+1) tems (2.1.7) and (2.1.3) satisfying uj ∈ L∞ (0, T; W2 (–D, D)) ∩ W2(1) (0, T; L2 (–D, D)) ∩ (2p+2) L2 (0, T; W2 (–D, D)).

42

2 The Vanishing Viscosity Method of Some Nonlinear Evolution System

Proof. Let B := L∞ (0, T; W2(1) (–D, D)) and # $ B∗ := u = (u1 , . . . , uJ )|uj ∈ B, j = 1, 2, . . . , J . Similarly, (p+1)

Z := L∞ (0, T; W2

(2p+2)

(–D, D)) ∩ W2(1) (0, T; L2 (–D, D)) ∩ L2 (0, T; W2

(–D, D))

and # $ Z ∗ := u = (u1 , . . . , uJ )|uj ∈ Z, j = 1, 2, . . . , J . In the functional vector space B∗ , we define the operator T+ : B∗ → B∗ (0 ≤ + ≤ 1) as follows: for each v = (v1 , . . . , vJ ) ∈ B∗ , assume that the components uj (j = 1, . . . , J) of u = T+ v(0 ≤ + ≤ 1) are the solutions to the problem ujt + (–1)p+1 %ujx2p+2 + ujx2p+1 = +hj (x, t, v1 , . . . , vJ ),

(1.8)j

j = 1, 2, . . . , J,

uj (x, 0) = u0j (x),

(1.9)j

with

hj (x, t, v1 , . . . , vJ ) =

J 

aij (x, t)vi + gj (x, t) –

i=1

 ∂>(v , . . . , v )  1 J . x ∂uj

Since >(u) is 2-times continuously differentiable and v ∈ B∗ , we derive that v is bounded. Thus, the term +hj on the right-hand side of eq. (1.8)j is uniformly bounded in L2 ((–D, D) × (0, T)). It follows from Lemma 2.1.1 that there exists a unique solution uj (x, t) ∈ Z to eqs (1.8)j and (1.9)j satisfying ! uj Z ≤ c u0j 

(p+1)

W2

(–D,D)

" ++hj L2 ((–D,D)×(0,T)) , j = 1, . . . , J.

(1.10)j

And so uj ∈ Z ⊂ B(j = 1, . . . , J), that is u ∈ Z ∗ ⊂ B∗ . From the above argument, we know that u ∈ Z ∗ for v ∈ B∗ . This implies that (p+1) u(x, t1 ) – u(x, t2 ) ∈ W2 (–D, D) for any t1 , t2 ∈ [0, T]. By interpolation, one has ! sup |u(x, t1 ) – u(x, t2 )| ≤ cu(⋅, t1 ) – u(⋅, t2 )1–! L2 (–D,D) u(⋅, t1 ) – u(⋅, t2 ) 1–" u(⋅, t1 ) 2 (–D,D)

sup |ux (x, t1 ) – ux (x, t2 )| ≤ cu(⋅, t1 ) – u(⋅, t2 )L

–D≤x≤D

(p+1)

W2

–D≤x≤D

– u(⋅, t2 )

(–D,D)

" (p+1)

W2

(–D,D)

, ,

43

2.1 Periodic Boundary and Cauchy Problem for High-Order Generalized KdV System

with ! =

1 , 2(p+1)

3 , 2(p+1)

"=

p ≥ 1. Observing that 

u(⋅, t1 ) – u(⋅, t2 )2L2 (–D,D) = 

D   t2

= –D

 

t1

D

|u(x, t1 ) – u(x, t2 )|2 dx

–D

 2  ut (x, 4) d4 dx ≤ |t1 – t2 |

D  t2 –D

|ut (x, 4)|2 d4 dx

t1

≤ |t1 – t2 | ⋅ ut 2L2 ((–D,D)×(0,T)) and u(⋅, t1 ) – u(⋅, t2 )

(p+1)

W2

(–D,D)

≤ 2 sup u(⋅, t) 0≤t≤T

(p+1)

W2

(–D,D)

= 2u

(p+1)

L∞ (0,T;W2

(–D,D))

,

we estimate 2p+1

2p+1

1

u 2p+2 sup |u(x, t1 ) – u(x, t2 )| ≤ c|t1 – t2 | 4p+4 ut L2p+2 ((–D,D)×(0,T)) 2

–D≤x≤D

2p–1

2p–1 2

(–D,D))

,

3

u 2p+2 sup |ux (x, t1 ) – ux (x, t2 )| ≤ c|t1 – t2 | 4p+4 ut L2p+2 ((–D,D)×(0,T))

–D≤x≤D

(p+1)

L∞ (0,T;W2

(p+1)

L∞ (0,T;W2

(–D,D))

.

2p+1 2p–1 This shows that the functions u(x, t) and ux (x, t) are 4p+4 -order and 4p+4 -order Hölder continuous in the region [–D, D] × [0, T], respectively. Combining this with eq. (1.10)j , we deduce that the operator T+ : B∗ → B∗ is uniformly complete continuous in +. For +, +¯ ∈ [0, 1], v ∈ B∗ , we have u = T+ v, u¯ = Tv¯ v. Then the components wj of the difference vector w = u – u¯ solve

¯ j (x, t, v) wjt + (–1)p+1 %wjx2p+2 + wjx2p+1 = (+ – +)h and conditions wj (x + 2D, t) = wj (x, t), wj (x, 0) = 0,

j = 1, 2, . . . , J.

It follows from eq. (1.10)j ¯ ⋅ hj L ((–D,D)×(0,T)) . wj Z = uj – u¯ j Z ≤ c|+ – +| 2 For any bounded subset of B∗ , the operator T+ : B∗ → B∗ is uniformly continuous in +(0 ≤ + ≤ 1). When + = 0, T0 (B∗ ) is a fixed element of B∗ . Now, we consider the solution u(x, t) to the nonlinear system with parameter 0≤+≤1 ! " ut + (–1)p+1 %ux2p+2 + ux2p+1 + + grad >(u) x = +f (x, t, u),

(1.7)+

44

2 The Vanishing Viscosity Method of Some Nonlinear Evolution System

with periodic boundary condition (2.1.3) which is uniformly priori estimate in B∗ with respect to 0 ≤ +1, and f (x, t, u) = A(x, t)u + g(x, t). Multiplying eq. (1.7)+ by u, and integrating in x from –D to D, we get 

d dt



D

–D



D

(u, u) dx + 2% –D

(uxp+1 , uxp+1 ) dx = 2+

D

(u, f ) dx, –D

where we use the fact that 



D!

! " " u, grad>(u) x dx =

–D

D !

–D

"  u, grad>(u) – >(u) x dx = 0.

Since f = A(x, t)u + g(x, t), we obtain  t 0

D

 (u, f ) dx ≤

–D

C+

1 2



t 0

1 u(⋅, t)2L2 (–D,D) dt + g2L2 ((–D,D)×(0,1)) , 2

with   C = max max aij (x, t). i,j

(x,t)

This together with 0 ≤ + ≤ 1 yields that  u(⋅, t)2L2 (–D,D) ≤ B

t 0

u(⋅, t)2L2 (–D,D) dt + B′ ,

where B = 2C + 1, B′ = u0 2L (–D,D) + g2L ((–D,D)×(0,1)) . Thus, we have by Gronwall’s 2 2 inequality sup u(⋅, t)2L2 (–D,D) ≤ B′ eBT .

0≤t≤T

Next, we turn to estimate the term of the derivative uxp . Multiplying eq. (1.7)+ by uxp , and integrating in x from –D to D, one has d dt





D

(uxp , uxp ) –D

 p+1

+ 2(–1)

+

D

dx + 2% –D

D! –D

(ux2p+1 , ux2p+1 ) dx "

grad>(u), ux2p+1 dx = 2+



D

p

(Dx f , uxp ) dx, –D

(1.11)x

2.1 Periodic Boundary and Cauchy Problem for High-Order Generalized KdV System

∂ ∂x .

with Dx =



A simple computation gives

D!

(–1)p+1

–D D!

 =(–1)p

" grad>(u), ux2p+1 dx

" grad>(u), ut + %(–1)p+1 ux2p+2 + +(grad>(u))x – +f dx

–D



d =(–1)p dt

45

D

 >(u) dx + %

–D

D! –D

" (grad>(u))x , ux2p+1 dx + (–1)p+1 +



D!

" grad>(u), f dx

–D

and    2

D! –D

(grad>(u))x , ux2p+1

"

   D  dx ≤ (ux2p+1 , ux2p+1 ) dx+ –D

D!

" (grad>(u))x , (grad>(u))x dx,

–D

and 

D!

" (grad>(u))x , (grad>(u))x dx =



–D

D

–D

 ≤C

J 

 >ij ujx >jk ukx dx

i,j,k=1 D!

" ′ |u|2l + 1 |ux |2 dx

–D

(2l′ ! " ≤ C u(⋅, t)(L∞ (–D,D) + 1



D

|ux |2 dx,

–D

with l′ = l – 2 = 4p – $, $ > 0. Notice that 4p+1 1 ( u(⋅, t)(L∞ (–D,D) ≤ Cu(⋅, )L4p+2 u(⋅, t) 4p+2 (2p+1) (–D,D) 2

W2

2p

(–D,D)

,

1

u(⋅, t) 2p+1 ux (⋅, )L2 (–D,D) ≤ Cu(⋅, )L2p+1 (2p+1) (–D,D) 2

W2

(–D,D)

,

we derive that 

D!

" (grad>(u))x , (grad>(u))x dx

–D 4p+1 $p–2$ 4p+2

≤ Cu(⋅, )L

2 (–D,D)

u(⋅, t)

≤ ux2p+1 (⋅, t)2L2 (–D,D)

$ 2– 2p+1 (2p+1) W2 (–D,D)

4p

2

+ C′ u(⋅, )L2p+1 u(⋅, t) 2p+1 (2p+1) (–D,D) 2

W2

+ C(p, $),

where C(p, $) depends on p, $ and sup u(⋅, )L2 (–D,D) . Hence 0≤t≤T

   2

"  (grad>(u))x , ux2p+1 dx ≤ 2ux2p+1 (⋅, t)2L2 (–D,D) + C(p, $).

D! –D

(–D,D)

46

2 The Vanishing Viscosity Method of Some Nonlinear Evolution System

Equation (1.11)x can be rewritten as  D  d >(u) dx + 2%(1 – +)ux2p+1 (⋅, t)2L2 (–D,D) uxp (⋅, t)2L2 (–D,D) + 2+(–1)p dt –D  D  D ! " ! p " 2 p grad>(u), f dx + 2+ Dx f , uxp dx, ≤ 2%+C(p, $) + 2+ (–1) –D

–D

with 0 ≤ + ≤ 1. Integrating the above inequality in t gives  uxp (⋅, t)

2 L2 (–D,D)

≤ C + 2+2 (–1)p

+ 2+(–1)

>(u(x, t)) dx

–D

 t 0

D

p

D!

" grad>(u), f dx dt + 2+

–D

 t 0

D!

" p Dx f , uxp dx dt,

(1.11)′

–D

with  C = 2%+C(p, $)T + u0xp (⋅, t)2L2 (–D,D) + 2+(–1)p

D

>(u0 (x)) dx.

–D

The last term on the right-hand side of eq. (1.11)′ can be estimated as  t    0

–D

0

–D

 t   = 

t

≤C 0

 "  p Dx f , uxp dx dt

D!

p  D  k=0

   p k  Dx Auxp–k + gxp , uxp dx dt k

1 uxp (⋅, t)2L2 (–D,D) dt + gxp 2L2 ((–D,D)×(0,T)) + C′ , 2

  where C, C′ depends on max max max Dkx aij (x, t) and sup u(⋅, t)L2 (–D,D) . 1≤i,j≤J 0≤k≤p (x,t)

0≤t≤T

It follows from assumption (iii) that 

D

2+  t  2 2+  0

! " > u(x, t) dx ≤ 2+C



–D

1 |u|l dx ≤ uxp (⋅, t)2L2 (–D,D) + C(p, $), 2 –D

  t  grad>(u), f dx dt ≤ uxp (⋅, t)2L2 (–D,D) dt + C(p, $).

D! –D

D

"

0

Thus, eq. (1.11)′ can be rewritten as  uxp (⋅, t)2L2 (–D,D) ≤ C

0

t

uxp (⋅, t)2L2 (–D,D) dt + C′ ,

  where C, C′ depends on max max max Dkx aij (x, t), gxp L2 ((–D,D)×(0,T)) and is inde1≤i,j≤J 0≤k≤p (x,t)

pendent of %, +. Hence, we obtain

2.1 Periodic Boundary and Cauchy Problem for High-Order Generalized KdV System

47

sup uxp (⋅, t)L2 (–D,D) ≤ C,

0≤t≤T

where p ≥ 1, and the constant C is independent of %, D. From the above argument, we know that all solutions to eq. (1.7)+ with periodic boundary condition (2.1.3) are uniformly bounded in B∗ with respect to 0 ≤ + ≤ 1. Thus, by Leray–Schauder’s fixed point theorem, there exists solution u+ (x, t) to eqs (1.7)+ and (2.1.3). This means that there is a solution to eqs (1.7)+ and (2.1.3) when + = 1. And this solution u(x, t) belongs to Z ∗ by Lemma 2.1.1. Assume that there are two solutions u(x, t) and v(x, t) to eqs (2.1.7) and (2.1.3). Let w(x, t) := u(x, t) – v(x, t). Then w(x, t) solves ⎧ p+1 ⎪ ⎪ ⎨wt + %(–1) wx2p+1 + wx2p+1 + Rwx + Sw = Aw, w(x + 2D, t) = w(x, t), ⎪ ⎪ ⎩w(x, 0) = 0,

(1.12)′

where R=

 ∂ 2 >(u)  ∂ui ∂uj

ij

,

S=

J  k=1

 vkx 0

1

∂ 3 >(4u + (1 – 4)v)  d4 . ij ∂ui ∂uj ∂uk

Hence, ! " ! " Rwx + Sw = grad>(u) x – grad>(v) x . Multiplying eq. (1.12)′ by w, and then integrating in x from –D to D, integration by parts implies that d w(⋅, t)2L2 (–D,D) + 2%wxp+1 (⋅, t)2L2 (–D,D) = dt



D!

" w, (2A + Rx – 2S)w dx,

–D

where the matrix R is symmetric. Thus, d w(⋅, t)2L2 (–D,D) ≤ Cw(⋅, t)2L2 (–D,D) , dt where C = 2A + Rx – 2SL∞ (–D,D) . This implies w(x, t) ≡ 0, that is u(x, t) = v(x, t). And so uniqueness follows. We conclude the proof of Lemma 2.1.3. ∎ Corollary 2.1.4. Under the assumptions of Lemma 2.1.3, and for k ≥ 0, h ≥ 0 ((p+1)(2h+1)–k) (i′ ) u0 (x) ∈ W2 (–D, D); (ii′ )

max

0≤i≤k+p,0≤j≤h

j

Dix Dt A(x, t)L∞ ((–D,D)×(0,T)) < ∞ and

48

2 The Vanishing Viscosity Method of Some Nonlinear Evolution System

j

max

0≤i≤k+p,0≤j≤h

Dix Dt g(x, t)L2 ((–D,D)×(0,T)) < ∞;

(iii′ ) >(u) is k + h + 3-times continuously differentiable function. Then, the solution u(x, t) to system (2.1.7) with periodic boundary condition (2.1.3) has the smooth property: Dkx Dht u(x, t) ∈ Z ∗ . In order to use the limit process of the solution u% (x, t) to eqs (2.1.7) and (2.1.3) as % → 0 to study the solution to generalized high-order KdV system (2.1.2) with periodic boundary condition (2.1.3), we need a series of uniform estimate in % for the sequence of vector function {u% (x, t)}. From the proof of Lemma 2.1.3, we know that all solution u+% (x, t) to eqs (1.7)+ and (p) (2.1.3) is uniformly bounded in L∞ (0, T; W2 (–D, D)) with respect to 0 ≤ + ≤ 1, and also to % ≥ 0 (even to D > 0). Thus, these estimates are the priori estimate for the possible solution to eqs (2.1.7) and (2.1.3). Lemma 2.1.5. Under the assumptions of Lemma 2.1.3, the solution u% (x, t) to eq. (2.1.7) with periodic boundary condition (2.1.3) satisfies sup u% (⋅, t)

0≤t≤T

(p)

W2 (–D,D)

≤K

(p ≥ 1),

(2.1.13)

where K is independent of % and D. Moreover, sup |u% (x, t)| ≤ K.

0≤t≤T –D≤x≤D

Lemma 2.1.6. Suppose that the assumptions of Lemma 2.1.3 hold for p = 1, and ( j ( j (ii′ ) max (∂x A(x, t)(L∞ ((–D,D)×(0,T)) < ∞ and max ∂x g(x, t)L2 ((–D,D)×(0,T)) < ∞; 0≤j≤2

0≤j≤2

(iii′ ) >(u) is 4-times continuously differentiable.

Then, the solution u% (x, t) to eqs (2.1.7) and (2.1.3) (p = 1) satisfies sup u%xx (x, t)L2 (–D,D) ≤ K,

0≤t≤T

where K is independent of % and D. Proof. A simple computation shows that (uxx , uxx )t = – 2%(uxx , uxxxx )x + 2%(uxxx , uxxxx )x – 2%(uxxxx , uxxxx )x – 2(uxx , uxxxx )x + (uxxx , uxxx )x  + 2(uxx , D2x f ) – 2 >ijkl uixx ujx ukx ulx –6

 i,j,k

i,j,k,l

>ijk uixx ujxx ukx – 2

 i,j

>ij uixx ujxxx

(2.1.14)

2.1 Periodic Boundary and Cauchy Problem for High-Order Generalized KdV System

49

and   " ! (grad>(u))x , ux t = – % >ijk >ijk uix ujx ukxxxx – 2% >ij uix ujxxxx i,j,k

–



>ijk uix ujx ukxxx – 2



i,j,k

–3



i,j

>ijk >kl uix ujx ulx – 2

i,j,k,l

+



>ijk uix ujx fk + 2

ijk



i,j

>ij uix ujxxxx



>ij >jk ujx ukxx

i,j,k

>ij uix Dx fj ,

i,j

and 

>ij uixx ujxx

i,j

 

>ij uix ujxxx

i,j

>ijk uixx ujx ukx

 x

= =

x

=



>ij uixx ujxxx ,

i,j





>ijk uix ujx ukxxx +

i,j,k



i,j,k

>ijk uixx ujxx ukx + 2

i,j,k

 x





>ij uixx ujxxx +

i,j

>ijkl uixx ujx ukx ulx + 2

i,j,k,l





>ij uix ujxxxx ,

i,j

>ijk uixx ujxx ukk +

i,j,k



>ijk uixxx ujx ukx ,

i,j,k

 where denotes to be the summation from 1 to J for every index, and >ij , >ijk denotes to be the partial derivative of >(u) for ui , uj , uk , . . . . To cancel the terms     i,j,k >ijk uixx ujxx ukx , i,j >ij uixx ujxxx , i,j,k >ijk uix ujx ukxxx , i,j >ij uix ujxxxx , multiplying the above equalities by 3, –5, 8, –10, 5 and summing together yield " ! 3(uxx , uxx )t – 5 (grad>(u))x , ux t + 6%(uxxxx , uxxxx )  + 6%(uxx , uxxxxx ) – 6%(uxxx , uxxxx ) + 6(uxx , uxxxx ) – 3(uxxx , uxxx )     >ij uixx ujxx – 10 >ij uix ujxxx + 5 >ijk uix ujx ukxx x +8 i,j

= 6(uxx , D2x f ) + 5%



– 10



i,j

>ij uix Dx fj – 5

i,j

>ijk uix ujx ukxxxx + 10%

i,j,k

+ 15







i,j,k

>ijk uix ujx fk

i,j,k

>ij uix ujxxxxx –

i,j

>ijk >kl uix ujx ulx + 10

i,j,k,l





>ijkl uixx ujx ukx ulx

i,j,k,l

>ij >jk ujx ukxx .

(2.1.15)

i,j,k

Integrating the right-hand side of eq. (2.1.15) in x, we easily get the estimations of the first three terms and last two terms as follows:   D   (uxx , D2x f ) dx ≤ C1 uxx (⋅, t)2L2 (–D,D) + C2 ,  –D

50

2 The Vanishing Viscosity Method of Some Nonlinear Evolution System

        

D

–D i,j

D

–D i,j,k



D

–D i,j,k,l

  

  >ij uix Dx fj dx ≤ C1 uxx (⋅, t)2L2 (–D,D) + C2 ,

  >ijk uix ujx fk dx ≤ C1 uxx (⋅, t)2L2 (–D,D) + C2 ,

  >ijk >kl uix ujx ulx dx ≤ C1 uxx (⋅, t)2L2 (–D,D) + C2 ,

D

–D i,j,k

  >ij >jk ujx ukxx dx ≤ C1 uxx (⋅, t)2L2 (–D,D) + C2 ,

where C1 , C2 depends on max ∂xk A(x, t)L∞ (–D,D) , 0≤k≤2

max ∂xk g(x, t)L2 (–D,D) and

0≤k≤2

sup u(⋅, t)L2 (–D,D) . And we have the estimates for the middle three terms as

0≤t≤T

follows:   



D

–D i,j,k,l

       >ijkl uixx ujx ukx ulx dx ≤ sup >ijkl ujx ukx  ⋅  i,j,k,l –D≤x≤D

D

–D

  ulx uixx dx

≤ Cux (⋅, t)2L∞ (–D,D) ux (⋅, t)L2 (–D,D) uxx (⋅, t)L2 (–D,D) ≤ C3 uxx (⋅, t)2L2 (–D,D) + C4 , where C3 , C4 depend on sup ux (⋅, t)L2 (–D,D) , and we use the fact that 0≤t≤T

1

1

ux (⋅, t)L∞ (–D,D) ≤ Cux (⋅, t)L2

2 (–D,D)

1

ux (⋅, t) 2 (1)

W2 (–D,D)

≤ C5 uxx (⋅, t)L2

2 (–D,D)

+ C6 .

On the other hand, we have   

 2  >ijk uix ujx ukxxxx dx ≤ uxxxx (⋅, t)2L2 (–D,D) + C7 , 5 –D i,j,k    D    D       > u u dx >ijk uix ujx + >ik uixx ukxxxx dx  =  ij ix jxxxxx D

–D i,j

–D

i,j,k

i,k

2 ≤ uxxxx (⋅, t)2L2 (–D,D) + C8 uxx (⋅, t)2L2 (–D,D) + C9 , 5 where C7 , C8 , C9 depend on sup u(⋅, t)W (1) (–D,D) or sup |u(x, t)|. 2

0≤t≤T

(x,t)

Thus, the integral formula of eq. (2.1.15) can be rewritten as d 3uxx (⋅, t)2L2 (–D,D) – 5 dt



D –D i,j

 >ij uix ujx dx ≤ C10 uxx (⋅, t)2L2 (–D,D) + C11 .

51

2.1 Periodic Boundary and Cauchy Problem for High-Order Generalized KdV System

Integrating in t implies that  3uxx (⋅, t)2L2 (–D,D) ≤ 5

D –D i,j

 t t  >ij uix ujx dx + C10 uxx (⋅, t)2L2 (–D,D) dt 0

0

+ C11 T + 3u′′0 2L2 (–D,D) . This together with   

D –D i,j

  >ij uix ujx dx ≤ C12 ux (⋅, t)2L2 (–D,D) ≤ C13

yields  uxx (⋅, t)2L2 (–D,D) ≤ C14

0

t

uxx (⋅, t)2L2 (–D,D) dt + C15 . ∎

Combining this with Gronwall’s inequality, we obtain eq. (2.1.14).

Corollary 2.1.7. Under the assumptions of Lemmas 2.1.3 and 2.1.6, and p ≥ 1, the solution to eqs (2.1.7) and (2.1.3) satisfies u%x L∞ ((–D,D)×(0,T)) ≤ K,

(2.1.16)

where K is independent of % and D. By the similar argument as above, we can show that sup0≤t≤T u%xxxx (⋅, t)L2 (–D,D) is uniformly bounded in % when p = 1, and sup0≤t≤T u%x2p+2 (⋅, t)L2 (–D,D) is uniformly bounded in % when p ≥ 2. Lemma 2.1.8. Under the assumptions of Lemmas 2.1.3 and 2.1.6, and u0 (x) (2p+2) W2 (–D, D), and

∈

! j " j max ∂x ∂tk A(x, t)L∞ ((–D,D)×(0,T)) + ∂x ∂tk g(x, t)L2 ((–D,D)×(0,T)) < ∞,

0≤j+k≤2

then the solution to eqs (2.1.7) and (2.1.3) satisfies sup ut (⋅, t)W (1) (–D,D) ≤ K,

0≤t≤T

(2.1.17)

2

where K is independent of % and D. Proof. Derivating system (2.1.7) in t, we derive that v := ut solves ! " vt + %(–1)p+1 vx2p+2 + vx2p+1 + grad>(u) tx = Dt f ,

(2.1.18)

52

2 The Vanishing Viscosity Method of Some Nonlinear Evolution System

with  ! "   grad>(u) tx i = >ijk vj ukx + >ij vjx j,k

(i = 1, . . . , J).

(2.1.19)

j

Multiplying eq. (2.1.18) by vx2h (h ≥ 0), and then integrating in x, we get d v h (⋅, t)2L2 (–D,D) + 2%vxh+p+1 (⋅, t)2L2 (–D,D) + 2 dt x  D =2 (vxh , Dhx Dt f ) dx.



D!

–D

" ! " vxh , grad>(u) txh+1 dx (2.1.20)

–D

The integrand in the last term can be written as  ! " 1 ! " ! 1" >ij vixh vjxh + h + >ijk vixh vjxh ukx vxh , grad>(u) txh+1 = x 2 2 i,j

+

i,j,k

h   i,j !=2

+



 h+1 >ijk vixh vj ukxh+1 (D! >ij )vixh vjxh+1–! + ! i,j,k

h    h i,j,k "=1

"

(D" >ijk )vixh vj ukxh+1–" .

(2.1.21)

If h = 0, then there only appear the first two terms in the right-hand side. And if h = 1, there is no third term. Now, we derive the integral estimate of the above terms. For h ≥ 0, one has  D     D      >ijk vixh vjxh ukx dx ≤ sup |>ijk ukx | ⋅  vixh vjxh dx  –D i,j,k

–D

i,j,k (x,t)

≤ C1 ux L∞ (–D,D) vxh (⋅, t)2L2 (–D,D) ; and for h ≥ 2,    ≤

h  D  –D i,j !=2

  h+1  (D! >ij )vixh vjxh+1–! dx !

 h   h+1 i,j !=2

!

  sup |D! >ij | ⋅  (x,t)

D

–D

  vixh vjxh+1–! dx

≤ C2 vxh (⋅, t)2L2 (–D,D) + C3 v(⋅, t)2L2 (–D,D) + C4 ; and for h ≥ 1,   

h  D 

–D i,j,k "=1

  h  (D" >ijk )vixh vj ukxh+1–" dx ≤ C5 vxh (⋅, t)2L2 (–D,D) + C6 v(⋅, t)2L2 (–D,D) + C7 , "

53

2.1 Periodic Boundary and Cauchy Problem for High-Order Generalized KdV System

"

where Cj (1 ≤ j ≤ 7) depends on sup(x,t) |D!x >ij |, sup(x,t) |Dx >ijk |(! = 2, . . . , h; " = 1, . . . , h), and ux L∞ , uxh L∞ (h ≥ 1), and >(u) is h + 3-times continuously differentiable. For the remaining terms, when h = 1,   

D –D i,j,k

     >ijk vixh vj ukxx dx ≤ sup |>ijk vj | ⋅ 

D

–D

i,j,k (x,t)

  vixh ukx2 dx

≤ C8 vx (⋅, t)2L2 (–D,D) + C9 v(⋅, t)2L2 (–D,D) uxx (⋅, t)4L

2 (–D,D)

+ C10 v(⋅, t)2L2 (–D,D) uxx (⋅, t)2L2 (–D,D) 1

where v(⋅, t)L∞ (–D,D) ≤ Cv(⋅, t)L2

2

  

D

–D i,j,k

1

v(⋅, t) 2 (1) (–D,D)

W2 (–D,D)

     >ijk vixh vj ukxh+1 dx ≤ |>ijk vj | i,j,k

D

–D

. When h ≥ 2,

  vixh ukxh+1 dx

! " ≤ C11 vL∞ (–D,D) vxh (⋅, t)2L2 (–D,D) + uxh+1 (⋅, t)2L2 (–D,D) ,

where the above constants are all independent of % and D. When h = 0, eq. (2.1.20) can be written as d v(⋅, t)2L2 (–D,D) ≤ Cv(⋅, t)2L2 (–D,D) + C′ , dt where C, C′ depend on ux L∞ ((–D,D)×(0,T)) . When h = 1, eq. (2.1.20) can be rewritten as d ˜ vx (⋅, t)2L2 (–D,D) ≤ Cvx (⋅, t)2L2 (–D,D) + C, dt where C depends on ux L∞ , and C˜ depends on v(⋅, t)L2 (–D,D) and uxx (⋅, t)L2 (–D,D) , and both of them are independent of % and D. Therefore, we conclude Lemma 2.1.8 by Gronwall’s inequality. ∎ Now, we use the above integral estimate to prove that as % → 0, the solution u% (x, t) to system (2.1.7) with periodic boundary condition (2.1.3) converges to the global generalized and classical solution to KdV system (2.1.2) with periodic boundary condition (2.1.3). Theorem 2.1.9. Assume that system (2.1.2) with periodic boundary condition (2.1.3) satisfies: (i) u0 (x) is a J-dimensional vector function with period 2D, and u0 (x) ∈ (3p+2) W2 (–D, D), p ≥ 1, D > 0. ! " (ii) The element of A(x, t) = aij (x, t) satisfies (2p+2)

aij (x, t) ∈ L∞ (0, T; W2

¯ (p)

(–D, D)) ∩ W2(1) (0, T; W2 (–D, D)).

(2.1.22)

54

2 The Vanishing Viscosity Method of Some Nonlinear Evolution System

The component of g(x, t) satisfies (2p+2)

gj (x, t) ∈ L2 (0, T; W2

¯ (p)

(–D, D)) ∩ W2(1) (0, T; W2 (–D, D)),

(2.1.23)

with p¯ = max{2, p}, i, j = 1, . . . , J. And all these functions are periodic in 2D. The function >(u) is p + 4-times continuously differentiable in u = (u1 , . . . , uJ ), and

(iii)

|>(u)| ≤ A|u|l ,   grad>(u) ≤ C|u|l–1 ,  ∂ 2 >(u)  " !    ≤ C |u|l–2 + 1 ,  ∂ui ∂uj where l = 4p + 2 – $, $ > 0 and C is a constant. Then there exists a unique global generalized solution u(x, t) = (u1 (x, t), . . . , uJ (x, t)) to system (2.1.2) with periodic boundary condition (2.1.3). Moreover, (2p+2)

u ∈ L∞ (0, T; W2

¯ (p)

(1) (–D, D)) ∩ W∞ (0, T; W2 (–D, D)),

p¯ = max(2, p).

(2.1.24)

Therefore, ∂xk u(x, t)(0 ≤ k ≤ 2p + 1) are Hölder continuous in the region [–D, D] × [0, T]. Proof. Under the assumptions of this theorem, the solution u% (x, t) to systems (2.1.7) ¯ (2p+2) (1) (0, T; W (p) and (2.1.3) are uniformly bounded in L∞ (0, T; W2 (–D, D)) ∩ W∞ 2 (–D, D)) with respect to % and D. Thus, there exists a constant 0 < ! < 1 which is independent of % and D, such that ∂xk u% (x, t) (0 ≤ k ≤ 2p+1) are Hölder continuous of !-order. Moreover, these Hölder coefficients are uniformly bounded in % and D. In fact, by interpolation, we get when p¯ ≤ k ≤ 2p + 1   u k (x, t1 ) – u k (x, t2 ) ≤ u k (⋅, t1 ) – u k (⋅, t2 )L (–D,D) x x x x ∞ 1–" u(⋅, t1 ) 2 (–D,D)

≤ Cuxk (⋅, t1 ) – uxp (⋅, t2 )L ≤ C|t1 – t2 |

1–"

sup 0≤t≤T

1–" uxp (⋅, t)L (–D,D) 2

– u(⋅, t2 )

"

sup u(⋅, t)

0≤t≤T

(2p+2)

W2

(–D,D)

" (2p+2)

W2

(–D,D)

;

when 0 ≤ h ≤ p¯ – 1,   "¯ u h (x, t1 ) – u h (x, t2 ) ≤ C|t1 – t2 | sup u(⋅, t)1–"¯ u(⋅, t) x x L (–D,D) 0≤t≤T

1–"=

2

¯ (p)

W2 (–D,D)

2p – k + 32 h + 21 (0, ≤ h ≤ p¯ – 1), (p¯ ≤ k ≤ 2p + 1), "¯ = p¯ p+2

for any t1 , t2 ∈ [0, T] and x ∈ [–D, D]. Hence, ∂xk u% (x, t) (0 ≤ k ≤ p¯ – 1) are Lipschitz continuous in the region [–D, D]×[0, T], while u%xk (p¯ ≤ k ≤ 2p+1) are Hölder continuous of 2p–k+ 32 p+2

-order. Then, taking ! =

1 2p+4 , the

functions ∂xk u% (x, t) (0 ≤ k ≤ 2p + 1) are Hölder

55

2.1 Periodic Boundary and Cauchy Problem for High-Order Generalized KdV System

continuous of !-order. And it is easy to see that their Hölder coefficients are uniformly bounded with respect to % and D. There exists a subsequence {u%i (x, t)} such that it converges uniformly to some vector function u(x, t) = (u1 (x, t), . . . , uJ (x, t)) in [–D, D] × [0, T] as %i tends to zero. Moreover, the derivative vector functional sequences {u%i xk (x, t)}, 0 ≤≤ 2p + 1 con¯ and verge uniformly to {uxk (x, t)}, respectively. Besides, {u%i txh (x, t)}(0 ≤ h ≤ p) {u%j x2p+2 (x, t)} converge weakly to {utxh (x, t)} and {ux2p+2 (x, t)}, respectively. Hence, u ∈ ¯ (p)

(2p+2)

(1) (0, T; W (–D, D)). For any test vector function 8(x, t) = L∞ (0, T; W2 (–D, D)) ∩ W∞ 2 (81 (x, t), . . . , 8J (x, t)) with period 2D in variable x, as %i → 0, the limitation of the integral equality

 t 0

D 

–D

! "   u%i t + %i (–1)p u%i x2p+2 + u%i x2p+1 + grad>(u%i ) x – Au%i – g , 8 dx dt = 0

should be  t 0

D 

–D

! "   ut + ux2p+1 + grad>(u) x – Au – g , 8 dx dt = 0,

(2.1.25)

with 0 ≤ t ≤ T. The vector function u(x, t) is the global generalized solution to KdV system (2.1.2) with periodic boundary condition (2.1.3). Suppose that there are two global generalized solution to systems (2.1.2) and (2.1.3), denoted by u(x, t) and v(x, t). Let w(x, t) = u(x, t) – v(x, t). Then it solves  t 0

D 

–D

! " ! "   wt + wx2p+1 + grad>(u) x – grad>(v) x – Aw , 8 dx dt = 0,

for any test vector function 8 with period 2D in variable x. Taking 8 = w, we obtain w = 0 by the same argument as in the proof of uniqueness in Lemma 2.1.3, that is u = v. Therefore, the solution to systems (2.1.2) and (2.1.3) is unique. ∎ Theorem 2.1.10. Suppose that we have the assumptions as in Theorem 2.1.9. Then, as % → 0, the global solution u% (x, t) to systems (2.1.7) and (2.1.3) converges to the global (2p+1) generalized solution u(x, t) to systems (2.1.2) and (2.1.3) in L∞ (0, T; W∞ (–D, D)). Moreover, it follows 1

! 1– k+ 2 " u%xk (x, t) – uxk (x, t)L∞ ((–D,D)×(0,T)) = O (% 2p+2

(2.1.26)

for any 0 ≤ k ≤ 2p + 1. Proof. For any test vector function 8 with period 2D in variable x, the vector function z(x, t) := u% (x, t) – u(x, t) satisfies the integral equality

56

2 The Vanishing Viscosity Method of Some Nonlinear Evolution System

 t 0

D 

  zt + %(–1)p+1 u%x2p+2 + zx2p+1 + R% zx + S% z – Az , 8 dx dt = 0,

–D

where R% = (>ij (u% )) and 

S% =

 u%kx

1

0

k

! "  >ijk 4u% + (1 – 4)u d4 , z(x, 0) = 0.

Taking 8 = z, we have z(⋅, t)2L2 (–D,D)

 t = 0

D ! –D





"

2A + R% – 2S% z, z dx dt + 2(–1)

p

D!

–D

" %u%x2p+2 , z dx dt

or  z(⋅, t)2L2 (–D,D) ≤ C

t

0

z(⋅, t)2L2 (–D,D) dt + %2 u%x2p+2 2L2 ((–D,D)×(0,T)) ,

where C depends on the norm of 2A + R% – 2S% . Hence, sup z(⋅, t)2L2 (–D,D) ≤ %2 u%x2p+2 2L2 ((–D,D)×(0,T)) eCT ;

0≤t≤T

in other words, zL∞ (0,T;L2 (–D,D)) = O(%). This together with interpolation formula 1–" " z(⋅, t) (2p+2) , 2 (–D,D) W (–D,D)

zxk (⋅, t)L∞ (–D,D) ≤ Cz(⋅, t)L

"=

2

k + 21 , 0 ≤ k ≤ 2p + 1 2p + 2

implies that ! 1– k+ 21 " zxk (⋅, t)L∞ ((–D,D)×(0,T)) = O % 2p+2 . ∎

So eq. (2.1.26) follows.

We say that u(x, t) is a global weak solution to systems (2.1.2) and (2.1.3), if it satisfies  t 0

D 

–D

 t

+ 0

! "   ut + grad>(u) x – Au – g , 8 dx dt

D 

–D

p+1

%uxp+1 + (–1)





uxp , 8xp+1 dx dt = 0

for any test vector function 8 with period 2D in variable x, and 0 ≤ t ≤ T.

(2.1.27)

57

2.1 Periodic Boundary and Cauchy Problem for High-Order Generalized KdV System

Theorem 2.1.11. Assume that system (2.1.2) with periodic boundary condition (2.1.3) satisfies: (i) u0 (x) is a J-dimensional vector function with period 2D, and u0 (x) ∈ (p+1) W2 (–D, D), p ≥ 1, D > 0. ! " (ii) The element of A(x, t) = aij (x, t) satisfies (p+1)

aij (x, t) ∈ L∞ (0, T; W2

(1) (1) (–D, D)) ∩ W∞ (0, T; W∞ (–D, D)).

The component of g(x, t) satisfies (p+1)

gj (x, t) ∈ L2 (0, T; W2

(–D, D)) ∩ W2(1) (0, T; W2(1) (–D, D)).

(iii) The function >(u) is 4-times continuously differentiable in u = (u1 , . . . , uJ ), and   |>(u)| ≤ C|u|l , grad>(u) ≤ C|u|l–1 ,  ∂ 2 >(u)  " !    ≤ C |u|l–2 + 1 ,  ∂ui ∂uj where l = 4p + 2 – $, $ > 0 and C is a constant. Then there exists a unique global weak solution u(x, t) = (u1 (x, t), . . . , uJ (x, t)) to system (2.1.2) with periodic boundary condition (2.1.3). Moreover, (p+1)

u ∈ L∞ (0, T; W2

(1) (–D, D)) ∩ W2(1) (0, T; W∞ (–D, D)).

(2.1.28)

Therefore, ∂xk u(x, t)(0 ≤ k ≤ p) are Hölder continuous in the region [–D, D] × [0, T]. Theorem 2.1.12. Under the assumptions of Theorem 2.1.11, as % tends to zero, the global solution u% (x, t) to systems (2.1.7) and (2.1.3) converges to the global weak solution u(x, t) (p) to systems (2.1.2) and (2.1.3) in L∞ (0, T; W∞ (–D, D)). For the global classical solution, we have the following theorems. Theorem 2.1.13. Suppose that the assumptions of Theorem 2.1.9 hold, and the elements ∂t2 aij (x, t) are bounded, and ∂t2 gi (⋅, t)L2 ((–D,D)×(0,T)) (i, j = 1, 2 . . . , J) are bounded, and (4p+4)

(–D, D). Then there is a unique global classical solution to systems (2.1.2) u0 (x) ∈ W2 and (2.1.3) satisfying (2p+2)

u ∈ L∞ (0, T; W2

¯ (p)

(1) (2) (–D, D)) ∩ W∞ (0, T; W2 (–D, D)) ∩ W∞ (0, T; L2 (–D, D)). (2.1.29)

Therefore, u(x, t) and all derivatives of u appearing in eq. (2.1.2) are Hölder continuous.

58

2 The Vanishing Viscosity Method of Some Nonlinear Evolution System

Theorem 2.1.14. Under the assumptions of Theorem 2.1.13, as % tends to zero, the global classical solution u% (x, t) to systems (2.1.7) and (2.1.3) converges to the global classical ¯ (1) (0, T; W (p–1) solution u(x, t) to systems (2.1.2) and (2.1.3) in W∞ ∞ (–D, D)). Now, we consider the Cauchy problem for KdV system (2.1.2) with initial data (2.1.4), i.e. 

ut + (grad>(u))x + ux2p+1 = A(x, t)u + g(x, t),

u(x, 0) = u0 (x),

(x, t) ∈ (–∞, ∞) × [0, T],

x ∈ (–∞, ∞),

with u0 (x) is a known J-dimensional function in (–∞, ∞). First, we use the solution sequence u% (x, t; Ds ) to system (2.1.7) with eq. (2.1.3) to approximate the solution u% (x, t) to eq. (2.1.7) with eq. (2.1.4) by taking Ds → ∞. Noting that the priori estimates for system (2.1.7) with eq. (2.1.3) are independent of Ds , we can obtain the following theorem. Theorem 2.1.15. Assume that system (2.1.7) with initial data (2.1.4) satisfies: (p+1) (i) u0 (x) ∈ W2 (–∞, ∞), p ≥ 1. ! " (ii) The element of A(x, t) = aij (x, t) satisfies ( ( max (∂xk aij (x, t)(L∞ ((–∞,∞)×(0,T)) < +∞.

0≤k≤p

The component of g(x, t) satisfies ( ( max (∂xk gj (x, t)(L

0≤k≤p

(iii)

2 ((–∞,∞)×(0,T))

< +∞.

The function >(u) is 3-times continuously differentiable in u = (u1 , . . . , uJ ), and   |>(u)| ≤ C|u|l , grad>(u) ≤ C|u|l–1 ,  ∂ 2 >(u)  " !    ≤ C |u|l–2 + 1 , i, j = 1, 2, . . . , J,  ∂ui ∂uj

where l = 4p + 2 – $, $ > 0 and C is a constant. Then there exists a unique global generalized solution u% (x, t) to system (2.1.7) with eq. (2.1.4) satisfying (p+1)

u ∈ L∞ (0, T; W2

(2p+2)

(–∞, ∞)) ∩ W2(1) (0, T; L2 (–∞, ∞)) ∩ L2 (0, T; W2

(–∞, ∞)).

Similar to Corollary 2.1.4, by improving the regularity of >(u), u0 (x), g(x, t) and A(x, t) in eqs (2.1.7) and (2.1.4), we can obtain the corresponding global classical solution. Since the above priori estimates are independent of % and D, the solution u% (x, t) to eqs (2.1.7) and (2.1.4) in the region (–∞, ∞)×[0, T] and its derivative have the estimates

2.1 Periodic Boundary and Cauchy Problem for High-Order Generalized KdV System

59

independent of %. Thus, we can show the existence and uniqueness of global weak, generalized and classical solution to systems (2.1.2) and (2.1.4) under some different conditions. Moreover, as % tends to zero, the limit of global solution to system (2.1.7) with eq. (2.1.4) corresponds to system (2.1.2) with eq. (2.1.4). Theorem 2.1.16. Assume that system (2.1.2) with initial data (2.1.4) satisfies: (3p+2) (i) u0 (x) ∈ W2 (–D, D), p ≥ 1. ! " (ii) The element of A(x, t) = aij (x, t) satisfies (2p+2)

aij (x, t) ∈ L∞ (0, T; W2

¯ (1) (p) (–∞, ∞) ∩ W∞ (0, T; W∞ (–∞, ∞)).

The component of g(x, t) satisfies (2p+2)

gj (x, t) ∈ L2 (0, T; W2

(iii)

¯ (p)

(–∞, ∞)) ∩ W2(1) (0, T; W2 (–∞, ∞)),

with p¯ = max{2, p}, i, j = 1, 2 . . . , J. The function >(u) is p + 4-times continuously differentiable, and   |>(u)| ≤ C|u|l , grad>(u) ≤ C|u|l–1 ,  ∂ 2 >(u)  " !    ≤ C |u|l–2 + 1 ,  ∂ui ∂uj

where l = 4p + 2 – $, $ > 0, i, j = 1, 2, . . . , J and C is a constant. Then there exists a unique global generalized solution u(x, t) = (u1 (x, t), . . . , uJ (x, t)) to system (2.1.2) with initial data (2.1.3). Moreover, (2p+2)

u ∈ L∞ (0, T; W2

¯ (p)

(1) (–∞, ∞)) ∩ W∞ (0, T; W2 (–D, D)).

Theorem 2.1.17. Under the assumptions of Theorem 2.1.15, as % tends to zero, the global classical solution u% (x, t) to systems (2.1.7) and (2.1.4) converges to the global generalized (2p+1) solution u(x, t) to systems (2.1.2) and (2.1.4) in L∞ (0, T; W∞ (–∞, ∞)). Similarly, one can show the similar results as Theorems 2.1.11–2.1.14 concerning the existence and uniqueness of global weak and classical solution to system (2.1.2) with initial data (2.1.4). Moreover, we can obtain the similar theorems that as % tends to zero, the global solution u% (x, t) to systems (2.1.7) and (2.1.4) converges to the global weak and classical solution u(x, t) to systems (2.1.2) and (2.1.4) in some norm, respectively. For p = 1, we consider the KdV equation of the form ! " ut + uxxx + grad>(u) x = f (u).

(2.1.5)

By the same argument as above, we use the solution to the corresponding system with small parameter

60

2 The Vanishing Viscosity Method of Some Nonlinear Evolution System

! " ut + uxxx + %uxxxx + grad>(u) x = f (u)

(2.1.6)

to approximate the solution to eq. (2.1.5). In order to apply the fixed point theorem to prove the existence of solution to eq. (2.1.6) with periodic boundary condition (2.1.3), and the convergence as % → 0, we need to establish the priori estimates of the solution to the system ! " ut +uxxx +%uxxxx ++ grad>(u) x = +f (u),

(2.1.6)+

with periodic boundary condition (2.1.3), which are independent of 0 ≤ + ≤ 1, % > 0 and sometimes also of D > 0. Assume that f (u) has semi-bound Jacobi derivative matrix ∂f∂u(u) , that is, there exists a constant b > 0 such that for any J-dimensional vector . and J-dimensional vector value u, there holds 

.,

∂f (u)  . ≤ b(. , . ). ∂u

(2.1.30)

Multiplying eq. (2.1.6)+ by u, and then integrating in D, we get d dt





D



D

(u, u) dx + 2%

D

(uxx , uxx ) dx = 2+

–D

–D

(u, f (u)) dx, –D

with 

D

 (u, f (u)) dx ≤



D

1

u,

–D

–D

!

≤ b+

" 1 2



0 D

 ∂f (4u)  d4 u + f (0) dx ∂u

(u, u) dx + D|f (0)|2 .

–D

Hence, ! " u(⋅, t)2L2 (–D,D) ≤ e(2b+1)T u0 2L2 (–D,D) + 2D|f (0)|2 , where 0 ≤ + ≤ 1 and u0 (x) is a known function. If f (u) is a homogenous vector function, that is, f (0), then u(⋅, t)2L2 (–D,D) ≤ e(2b+1)T u0 2L2 (–D,D) . Derivating eq. (2.1.6)+ in variable x, and then multiplying by ux , and then integrating in x, we obtain  D  D  ! " d D grad>(u), uxxx dx (ux , ux ) dx + 2% (uxxx , uxxx ) dx + 2+ dt –D –D –D  D ∂f (u)  =2+ ux , ux dx, ∂u –D

61

2.1 Periodic Boundary and Cauchy Problem for High-Order Generalized KdV System

For the third term on the left-hand side, we have 

D!

" grad>(u), uxxx dx

–D D

 =

–D

=–

d dt

 ! " grad>(u), –ut – %uxxxx – + grad>(u) x + +f (u) dx



D

 >(u) dx + %

–D

D!

" (grad>(u))x , uxxx dx + +

–D



D!

" grad>(u), f (u) dx.

–D

Suppose that the inequality ! " grad>(u), f (u) ≤ C|u|l ,

(2.1.31)

then we can estimate by the same argument as in Lemma 2.1.3 ux (⋅, x)L2 (–D,D) ≤ K, where K depends on b, C and u(⋅, t)L2 (–D,D) . Thus, K is independent of + and %. Especially, K is also independent of D when f (0) = 0. Therefore, the solution to system (2.1.6) with periodic boundary condition (2.1.3) is unique. And we can obtain the uniform estimates of high-order derivative of solution to systems (2.1.6) and (2.1.3) with respect to %. Then, taking % → 0, we derive the existence of global solution to system (2.1.5) with periodic boundary condition (2.1.3). Theorem 2.1.18. Assume that system (2.1.5) with periodic boundary condition (2.1.3) satisfies (i) u0 (x) is a J-dimensional vector function with period 2D, and u0 (x) ∈ W2(5) (–D, D). (ii) f (u) is 4-times continuously differentiable and has semi-bound Jacobi derivative matrix ∂f∂u(u) . (iii) The function >(u) is 5-times continuously differentiable, and   |>(u)| ≤ C|u|l , grad>(u) ≤ C|u|l–1 ,  ∂ 2 >(u)   "  !    ≤ C |u|l–2 + 1 , (grad>(u), f (u)) ≤ C|u|l ,  ∂ui ∂uj where l = 6 – $, $ > 0, i, j = 1, 2, . . . , J and C is a constant. Then there exists a unique global generalized solution u(x, t) = (u1 (x, t), . . . , uJ (x, t)) to system (2.1.5) with periodic boundary condition (2.1.3). Moreover, (1) (0, T; W2(2) (–D, D)). u ∈ L∞ (0, T; W2(4) (–D, D)) ∩ W∞

Therefore, u, ux , uxx , uxxx are Hölder continuous in the region [–D, D] × [0, T].

62

2 The Vanishing Viscosity Method of Some Nonlinear Evolution System

Theorem 2.1.19. Under the assumptions of Theorem 2.1.18, as % tends to zero, the global weak solution u% (x, t) to systems (2.1.6) and (2.1.3) converges to the global generalized (3) (–D, D)). solution u(x, t) to systems (2.1.5) and (2.1.3) in L∞ (0, T; W∞ Taking D → ∞, one can obtain the similar theorems for system (2.1.5) with initial data (2.1.4). Theorem 2.1.20. Suppose that assumptions (ii) and (iii) in Theorem 2.1.18 holds true, and (i′ ) u0 (x) ∈ W2(5) (–∞, ∞), (ii′ ) f (0) is zero vector. Then there exists a unique global generalized solution u(x, t) = (u1 (x, t), . . . , uJ (x, t)) to system (2.1.5) with initial data (2.1.4). Moreover, (1) (0, T; W2(2) (–∞, ∞)). u ∈ L∞ (0, T; W2(4) (–∞, ∞)) ∩ W∞

Theorem 2.1.21. Under the assumptions of Theorem 2.1.20, as % tends to zero, the global generalized solution u% (x, t) to systems (2.1.6) and (2.1.4) converges to the global weak (3) (–∞, ∞)). solution u(x, t) to systems (2.1.5) and (2.1.4) in L∞ (0, T; W∞ Similarly, one can show the similar theorems concerning the existence and uniqueness of global weak and classical solution to system (2.1.5), with periodic boundary condition (2.1.3) or initial data (2.1.4). Moreover, we can obtain the similar theorems that as % tends to zero, the global solution u% (x, t) to system (2.1.6) with periodic boundary condition (2.1.3) or initial data (2.1.4) converges to the global weak and classical solution u(x, t) to system (2.1.5) with periodic boundary condition (2.1.3) or initial data (2.1.4) in some norm, respectively.

2.2 Some KdV System with High-Order Derivative Term In the physics problems, there are many other KdV systems with high-order derivative terms, such as ut + 120u2 ux + 20(uxxx + ux uxx ) + uxxxxx = 0, ut + 180u2 ux + 30(uxxx + ux uxx ) + uxxxxx = 0, " ! ut = uxx ± 3u2 ux + 32 u5 , ut –

15 2 8 u ux

+ 45 ux uxx + 85 uuxxx –

1 16 uxxxxx

= 0.

Here, we study the more general KdV system which includes the above derivative terms as follows: ut + ux2p+1 + G(u, ux , . . . , ux2s )x = A(x, t)u + g(x, t),

(2.2.1)

2.2 Some KdV System with High-Order Derivative Term

63

where the nonlinear term G(u, ux , . . . , ux2s )x has the form G(u, ux , . . . , ux2s )x =

s  ! " (–1)k gradk F(u, ux , . . . , uxs ) xk ,

(2.2.2)

k=0

and u = (u1 , . . . , uN ) is an N-dimensional value unknown function, A(x, t) is an N × N matrix, g(x, t) is an N-dimensional value function, F(p0 , p1 , . . . , ps ) is a function of p0 , p1 , . . . , ps ∈ RN , p ∈ Z≥1 and “gradk ”(k = 0, 1, . . . , s) denotes the gradient operator of pk = (pk1 , pk2 , . . . , pkN )(k = 0, 1, . . . , s). In this section, we will study system (2.2.1) with periodic boundary condition  u(x, t) = u(x + 2D, t),

(2.2.3)

u(x, 0) = >(x) and with initial data u(x, 0) = >(x).

(2.2.4)

We will utilize the vanishing viscosity method to prove the existence and uniqueness of the global solution to eqs (2.2.1), (2.2.3), and (2.2.1), (2.2.4). That is, by introducing the corresponding nonlinear parabolic system with small parameter ut + (–1)p+1 %ux2p+2 + ux2p+1 + G(u, ux , . . . , ux2s )x = A(x, t)u + g(x, t),

(2.2.5)

we show that as % → 0, the solution to eqs (2.2.5) and (2.2.3) approximates to the solution to eqs (2.2.1) and (2.2.3). Lemma 2.2.1. Let F(p0 , p1 , . . . , ps ) ∈ Cs+2 and satisfy

(F)

 Fp

k1 i1 ,pk2 i2

r ⎛ ⎞l  (2kj +1) s j=0  1   |pj | 2j+1 ⎠ , ,...,pkr ir (p0 , . . . , ps ) ≤ K1 ⎝

j=1

with l = 4p + 2 – $, $ > 0, k1 , k2 , . . . , kr = 0, 1, . . . , s; i1 , i2 , . . . , is = 1, 2, . . . , N, 0 ≤ r ≤ s + 2, pk (pk1 , pk2 , . . . , pkN ), k = 0, 1, . . . , s, K1 being a constant. Then, for any ' > 0, we have ( ( (G(v, vx , . . . , v 2s )x ( x L

2 (–D,D)

≤ 'vx2p+1 L2 (–D,D) + K2 vL2 (–D,D)

(2.2.6)

and F(v, vx , . . . , vxs )L2 (–D,D) ≤ 'vxp 2L2 (–D,D) + K3 vL2 (–D,D) ,

(2.2.7)

64

2 The Vanishing Viscosity Method of Some Nonlinear Evolution System

where 4p > 2s2 + 5s, v(x) ∈ H 2p+1 (–D, D) is a vector function with period 2D, and the constants K2 and K3 depend on ' and vL2 (–D,D) . Proof. The term in G(v, vx , . . . , vx2s )x !

"

gradk F(u, ux , . . . , uxs )

(k = 0, 1, . . . , s)

xk+1

can be written as Fp k

1 i1

,pk2 i2 ,...,pkr ir pki (v, vx , . . . , vxs )

× (vi1 xk1 +1 vi2 xk2 +1 . . . vir xkr +1 )xk–r+1

or P = F pk

1 i1

,pk2 i2 ,...,pkr ir pki (v, vx , . . . , vxs )v

′

v

′

k +k +1 k +k +1 i1 x 1 1 i2 x 2 2

. . . vi xkr +kr′ +1 , r

where k1 , k2 , . . . , kr = 0, 1, . . . , s; i, i1 , . . . , ir = 1, . . . , N; 1 ≤ r ≤ k + 1; k1′ + k2′ + . . . + kr′ = k – r + 1, v = (v1 , . . . , vN ). Thus, PL2 (–D,D) ≤ Fpk

,pk2 i2 ,...,pkr ir pki (v, vx , . . . , vxs )L∞ (–D,D)

1 i1

× v

k +k′ +1 L∞ (–D,D) . . . v k +k′ +1 L∞ (–D,D) vi xkr +kr′ +1 L2 (–D,D) . i1 x 1 1 ir–1 x r–1 r–1 r

By the assumption (F), we have Fpk

1 i1

,pk2 i2 ,...,pkr ir pki (v, vx , . . . , vxs )L∞ (–D,D)

r s (  1 l– j=1 (2kj +1)–(2k+1) ( ( ( 2j+1 ≤ K1 ( |vxj | ( j=1

≤ C1

s  

2j+1  C¯ j vL (–D,D) vH4p+2 2p+1 (–D,D) 2 2j+1 1– 4p+2

L∞ (–D,D)

l–

r (2k +1)–(2k+1) j=1 j 2j+1

j=1 r (2k +1)–(2k+1)  s j=1 j 1 – 1 )(l–r (2k +1)–(2k+1))   ( 2j+1 j 4p+1 j=1 4p+2 ≤ C2 vH 2p+1 (–D,D) vL (–D,D) . 2 l–

j=1

On the other hand, we get by interpolation for m = 1, 2, . . . , r – 1: vi

k +k′ +1 L∞ (–D,D) mx m m

2k +2k′ +3 1– m4p+2m

≤ C3 vL

2 (–D,D)

′ +3 2km +2km

4p+2 vH 2p+1 (–D,D)

and k +2k′ +3 kr +kr′ +3 1– r 2p+1r 2p+1 v . H 2p+1 (–D,D) 2 (–D,D)

vi xkr +kr′ +1 L2 (–D,D) ≤ C3 vL r

Plugging these inequalities into the right-hand side of inequality about PL2 (–D,D) gives

65

2.2 Some KdV System with High-Order Derivative Term

s $ –1  r+ 4p+2

1– $

4p+2 PL2 (–D,D) ≤ C5 vH 2p+1 vL (–D,D)

2 (–D,D)

r

(l–

vL

j=1 (2kj +1)–(2k+1)/(2j+1))

2 (–D,D)

.

j=1

Hence, $ $ ! " 1– 4p+2 4p+2 Gx L2 (–D,D) ≤ vH 2p+1 v C vL2 (–D,D) , L (–D,D) 6 (–D,D) 2

where C6 is a continuous function of vL2 (–D,D) . And then eq. (2.2.6) follows. Next, it follows from assumption (F) that ⎛ ⎞ s  l   F(v, vx , . . . , vxs ) ≤ C7 ⎝ |vxj | 2j+1 ⎠ . j=0

Thus, F(v, vx , . . . , vxs )L2 (–D,D) s  D  l ≤ C7 |vxj | 2j+1 dx –D

j=0

$ 2– 2p

⎧ s ⎨

$ 2p

≤ C8 vH p (–D,D) vL

2 (–D,D)

≤ 'v2H p (–D,D) + C9 (')

⎩

⎧ s ⎨ ⎩

l –2 2j+2

vL

j=0

j=0

2 (–D,D) ⎭

l –2 2j+2

vL

⎫ ⎬

⎫ 4' ⎬$

2 (–D,D) ⎭

v2L2 (–D,D) ,

where ' > 0, and the constant C9 (') depends on '. Therefore, we conclude Lemma 2.2.1. ∎ Theorem 2.2.2. Assume that (1) F(p0 , p1 , . . . , ps ) ∈ Cs+2 and satisfies the assumption (F) as in Lemma 2.2.1. (2) >(x) ∈ H p+1 (–D, D) and has the period 2D. (p)

(3) A(x, t) ∈ L∞ (0, T; W∞ (–D, D), and (A. , . ) ≤ b|. |2 , . ∈ RN , b = const., g(x, t) ∈ L2 (0, T; H p (–D, D))

and

g(x, t) = g(x + 2D, t).

Then there exists a unique global generalized solution u(x, t) to the nonlinear parabolic system (2.2.5) with periodic boundary condition (2.2.3). Moreover, the solution u(x, t) ∈ (2p+2,1) W2 (QT ) ∩ L∞ (0, T; H p+1 (–D, D)) with QT = (–D, D) × (0, T). (2p+2,1)

Proof. Let B = L∞ (0, T; L2 (–D, D)) ∩ L2 (0, T; H 2p+1 (–D, D)) and Z = W2 (QT ) ∩ L∞ (0, T; H p+1 (–D, D)). For any N-dimensional vector function v = (v1 , . . . , vN ) ∈ B, we

66

2 The Vanishing Viscosity Method of Some Nonlinear Evolution System

define the vector function u = (u1 , . . . , uN ) = T+ v ∈ Z to be the generalized solution to system ut + %(–1)p+1 ux2p+2 + ux2p+1 = +Av + +g – +G(v, vx , . . . , vx2s )x , with periodic boundary condition  u(x, t) = u(x + 2D, t), u(x, 0) = +>(x),

(2.2.5)′

(2.2.3)+ ,

with 0 ≤ + ≤ 1. By Lemma 2.2.1, we know that the right-hand side of (2.2.5)′ is square integral in QT . Hence, there is a unique weak solution u(x, t) ∈ Z to systems (2.2.5)′ and (2.2.3)+ . Thus, T+ : B → Z ⊂ B,

0 ≤ + ≤ 1.

Since the embedding Z ↪ B is completely continuous, the operator T+ is completely continuous for 0 ≤ + ≤ 1. It is easy to see that for any bounded subset M ⊂ B, the operator T+ is uniformly continuous with respect to 0 ≤ + ≤ 1, and T0 B = 0. In order to obtain the weak solution to system (2.2.5) with periodic boundary condition (2.2.3), we need to verify the uniform bound in 0 ≤ + ≤ 1 for all solutions to the nonlinear parabolic system ut + %(–1)p+1 ux2p+2 + ux2p+1 + +G(u, ux , . . . , ux2s )x = +Au + +g,

(2.2.5)+

with periodic boundary condition (2.2.3)+ . Multiplying eq. (2.2.5)+ by u, and integrating this equality in Qt (0 ≤ t ≤ T), we estimate u(⋅, t)2L2 (–D,D) – +2 >2L2 (–D,D) + 2%uxp+1 2L2 (Qt )  t ≤ +(2b + 1) u(⋅, 4)2L2 (–D,D) d4 + +g2L2 (Qt ) , 0

where we use the following facts that   D (u, Gx ) dx = – –D

D

Fx dx = 0,

–D

and (. , A. ) ≤ b|. |2

∀ ∈ RN .

Using Gronwall’s inequality, we obtain sup u(⋅, t)L2 (–D,D) ≤ K2 ,

0≤t≤T

where the constant K2 is independent of % > 0 and D > 0.

2.2 Some KdV System with High-Order Derivative Term

67

Multiplying eq. (2.2.5)+ by ux2p , and integrating this equality in Qt (0 ≤ t ≤ T), we deduce that uxp (⋅, t)2L2 (–D,D) – +2 >(p) 2L2 (–D,D) + 2%ux2p+1 2L2 (Qt ) + (–1)p + =+(–1)p

 t 0

 t 0

D

–D

(ux2p , Gx ) dx dt

D

–D

(ux2p , Au + g) dx dt.

(2.2.8)

Observing that  t

 t

D

0

–D D

0

–D

(ux2p , Gx ) dx dt = –

 t 

D

–D

(ux2p+1 , G) dx dt

(ut + %(–1)p+1 ux2p+2 + ux2p+1 + +Gx – +Au – +g, G) dx dt

= =

0

D



F(u, ux , . . . , uxs ) dx –

–D p+1

+ %(–1)

 t 0

D

F(+>, . . . , +>(s) ) dx

–D D

–D

(ux2p+1 , Gx ) dx dt – +

 t

D

(Au + g, G) dx dt, 0

–D

we obtain by Lemma 2.2.1  t  D  1 %   % (ux2p+1 , Gx ) dx dt ≤ %ux2p+1 2L2 (Qt ) + Gx 2L2 (Qt ) 2 2 0 –D 1 ≤ %ux2p+1 2L2 (Qt ) + %C10 u2L2 (Qt ) , 2 s  t D     t  D ! "     (Au + g)xk gradk F dx dt (Au + g, G) dx dt ≤    0

–D

k=0



≤ C11 0

0

t

–D

uxp (⋅, 4)2L2 (–D,D) d4 + C12 ,

and   

 1  F(u, . . . , uxs ) dx ≤ uxp (⋅, t)2L2 (–D,D) + C13 , 2 –D  t   t  D   (ux2p , Au + g) dx dt ≤ C14 uxp (⋅, 4)2L2 (–D,D) d4 + C15 ,  0

–D

D

0

where the above constants C are independent of % > 0, + > 0 and D > 0. This means that the vector function u(x, t) is uniform bounded in B with respect to 0 ≤ + ≤ 1. Therefore, by Leray–Schauder’s fixed point theorem, we derive that in space Z ⊂ B, there exists at least one solution to systems (2.2.5) and (2.2.3). The uniqueness of the weak solution can be shown by the general method. Thus, we complete the proof of Theorem 2.2.2. ∎

68

2 The Vanishing Viscosity Method of Some Nonlinear Evolution System

As a consequence of Theorem 2.2.2, we have the following lemma. Lemma 2.2.3. Under the assumptions of Lemma 2.2.1, the weak solution u(x, t) ∈ (2p+2,1) W2 (QT ) ∩ L∞ (0, T; H p+1 (–D, D)) to systems (2.2.5) and (2.2.3) satisfies sup u(⋅, t)H p+1 (–D,D) +

0≤t≤T

√ %u(x, t)

(2p+2,1)

W2

(QT )

≤ K3 ,

(2.2.9)

with the constant K3 being independent of D > 0 and % > 0. To obtain the weak solution to systems (2.2.1) and (2.2.3), we need to show the uniform estimate in % for the weak solution to systems (2.2.5) and (2.2.3). Lemma 2.2.4. Under the assumptions of Theorem 2.2.2, the weak solution u(x, t) to systems (2.2.5) and (2.2.3) has the uniform estimate as follows: sup uxp (⋅, t)L2 (–D,D) ≤ K4 ,

(2.2.10)

0≤t≤T

where the constant K4 is independent of D > 0 and % > 0. Lemma 2.2.5. Under the assumptions of Theorem 2.2.2 and let g(x, t) ∈ L∞ (QT ). Then the weak solution u(x, t) to systems (2.2.5) and (2.2.3) has the uniform estimate as follows: sup ut (⋅, t)H –(p+2) (–D,D) ≤ K5 ,

(2.2.11)

0≤t≤T

where the constant K5 is independent of D > 0 and % > 0, and depends on T > 0. p+2

Proof. Taking v(x) ∈ H0 (–D, D), we have by simple computation 

D

v(x)ut (t, x) dx –D d

 =%

 v(p+2) (x)uxp (x, t) dx + (–1)p

–D

+

s   k=0

D

v(p+1) (x)uxp (x, t) dx

–D D

v –D

(k+1)

(x)gradk F(u(x, t), . . . , uxs (x, t) dx +



D

! " v(x) A(x, t)u(x, t) + g(x, t) dx.

–D

It is easy to show that every term on the right-hand side is bounded for any 0 ≤ t ≤ T. And so eq. (2.2.11) follows by duality. ∎

69

2.2 Some KdV System with High-Order Derivative Term

As a consequence of interpolation for Hilbert space with negative order and Lemma 2.2.4, we easily obtain the following lemma. Lemma 2.2.6. Under the assumption of Lemma 2.2.5, we have for k = 0, 1, . . . , p – 1 p–k

uxk (⋅, t + Bt) – uxk (⋅, t)L2 (–D,D) ≤ K6 Bt 2p+2 ,

(2.2.12)

where the constant K6 is independent of Bt > 0, % > 0 and D > 0. Definition 2.2.7. We say that N-dimensional vector function u(x, t) ∈ L2 (0, T; H p (–D, D)) is weak solution to generalized KdV system (2.2.1) with periodic boundary condition (p+2,1) (2.2.3), if for any test function 8(x, t) ∈ W2 (QT ) with period 2D and 8(x, T) = 0, there holds 

T



0

D

–D

8t u + (–1)p 8xp+1 uxp + 

+ 8(Au + g) dx dt +

s  k=0  D

8xk+1 gradk F(u, ux , . . . , uxs ) (2.2.13) 8(x, 0)>(x) dx = 0.

–D

We have already known that the generalized solution set {u% (x, t)} (for any % > 0) to systems (2.2.5) and (2.2.3) is uniformly bounded in space L∞ (0, T; H p (–D, D)) ∩ (0, p )

(1) (0, T; H –(p+2) (–D, D)) or L (0, T; H p (–D, D)) ∩ C 2p+2 (0, T; L (–D, D)). Hence, we W∞ ∞ 2 can extract a subsequence {u%i (x, t)} from {u% (x, t)} such that there exists some Ndimensional vector function u(x, t), as i → ∞, %i → 0, and the subsequence {u%i xp (x, t)} weakly converges to uxp (x, t) in L2 (–D, D), and the subsequence {u%i xk (x, t)}(k = 0, 1, . . . , p – 1) uniformly converges to uxk (k = 0, 1, . . . , p – 1). And so, u(x, t) ∈ (0, p )

L∞ (0, T; H p (–D, D)) ∩ C 2p+2 (0, T; L2 (–D, D)). For the weak solution u% (x, t) to systems (2.2.5) and (2.2.3), we have the integral equality  0

T



D –D

8t u% + %8xp+2 u%xp + (–1)p 8xp+1 u%xp +   + 8(Au% + g) dx dt +

s 

8xk+1 gradk F(u% , u%x , . . . , u%xs )

k=0 D

8(x, 0)>(x) dx = 0,

–D (p+2,1)

(2.2.13)%

where the test function 8(x, t) ∈ W2 (QT ) and 8(x, T) = 0. Since the subsequence {u%i xk (x, t)}(k = 0, 1, . . . , p – 1) uniformly converges to uxk in QT and p > s, the subsequence {gradj F(u%i (x, t), u%i x (x, t), . . . , u%i xs (x, t))} uniformly converges to gradj F(u(x, t), ux (x, t), . . . , uxs (x, t)) in QT . Thus, as %i → 0, the limit of eq. (2.2.13)%

70

2 The Vanishing Viscosity Method of Some Nonlinear Evolution System

is the integral equality (2.2.13). This shows that u(x, t) is the weak solution to systems (2.2.1) and (2.2.3). Theorem 2.2.8. Under the assumptions of Lemma 2.2.5, there exists a generalized weak solution u(x, t) to the generalized KdV system (2.2.1) with periodic boundary condition (2.2.3). Moreover, u(x, t) ∈ L∞ (0, T; H p (–D, D)) ∩ C

p (0, 2p+2 )

(0, T; L2 (–D, D)).

From Lemma 2.2.4, we know that the priori estimate of the solution to systems (2.2.5) and (2.2.3) is independent of D. Hence, as D → ∞, we can obtain the global weak solution of system (2.2.5) with initial data (2.2.4). Theorem 2.2.9. Suppose that the assumptions of Theorem 2.2.2 hold with ∞ instead of D. Then there exists a unique global generalized solution u(x, t) to the nonlinear parabolic system (2.2.5) with initial data (2.2.4). Moreover, u(x, t) ∈ L∞ (0, T; H p+1 (R)) ∩ (2p+2,1) ∗ W2 (QT ) with Q∗T = {|x| < ∞, 0 ≤ t ≤ T}. Under the assumptions of Theorem 2.2.9, Lemmas 2.2.4–2.2.6 become as follows: Lemma 2.2.10. Suppose that the assumptions of Theorem 2.2.9 hold and g(x, t) ∈ L∞ (Q∗T ). Then, we have sup u(⋅, t)H p (R) + sup ut (⋅, t)H –(p+2) (R) ≤ K7 ,

0≤t≤T

(2.2.14)

0≤t≤T

( ( (u k (⋅, t + Bt) – u k (⋅, t)( x x L

p–k

2 (R)

≤ K8 Bt 2p+2 , l = 0, 1, . . . , p – 1,

where the constants K7 , K8 are independent of %. Definition 2.2.11. We say that N-dimensional vector function u(x, t) ∈ L2 (0, T; H p (R)) is weak solution to generalized KdV system (2.2.1) with initial data (2.2.4), if for any test (p+2,1) ∗ function 8(x, t) ∈ W2 (QT ), there holds 

T 0



∞

–∞

8t u + (–1)p 8xp+1 uxp + 

+ 8(Au + g) dx dt +

s 

8xk+1 gradk F(u, ux , . . . , uxs )

k=0  ∞

(2.2.15) 8(x, 0)>(x) dx = 0.

–∞

Theorem 2.2.12. Under the assumptions of Lemma 2.2.10, there exists at least a weak solution u(x, t) ∈ L∞ (0, T; H p (R)) ∩ C (2.2.1) with initial data (2.2.4).

p (0, 2p+2 )

(0, T; L2 (R)) to the generalized KdV system

71

2.2 Some KdV System with High-Order Derivative Term

Suppose that the matrix A(x, t) is negative, that is, there exists negative constant –b such that (. , A. ) ≤ –b|. |2 ∀ . ∈ RN ,

b > 0.

Then, the above generalized solution and weak solution have the asymptotic property u(⋅, t)L2 (–D,D) , u(⋅, t)L2 (R) → 0,

as

t → ∞.

Theorem 2.2.13. Assume that the N × N matrix A(x, t) is negative, that is, (. , A. ) ≤ –b|. |2

∀ . ∈ RN ,

b > 0,

(2.2.16)

and g(x, t)L2 (Q∞ ) < ∞. Suppose the assumptions of Theorem 2.2.2 or Theorem 2.2.8 or Theorem 2.2.9 or Theorem 2.2.12 hold for T = 0. Then, the generalized solution to systems (2.2.5) and (2.2.3) (or systems (2.2.1) and (2.2.3)) and the weak solution to systems (2.2.5) and (2.2.4) (or systems (2.2.1) and (2.2.3)) have the asymptotic property lim u(⋅, t)L2 (–D,D) = 0

t→∞

or

lim u(⋅, t)L2 (–R) = 0.

t→∞

(2.2.17)

Next, we consider the KdV system with high-order derivative term of the form ut + ux2p+1 + G(u, ux , . . . , ux2s )x = f (x, t, u),

(2.2.18)

with periodic boundary condition or initial data. Assume that f (x, t, u) satisfies 

D

–D

" ! (u, f (x, t, u)) dx ≥ C0 u(⋅, t)2L2 (–D,D) 8 u(⋅, t)L2 (–D,D) ,

(2.2.19)

where D is bounded or infinity, C0 < 0, and 

∞

C1

dz < ∞. z8(z)

(2.2.20)

Multiplying eq. (2.2.18) by u, and then integrating in x ∈ (–D, D), we get d u(⋅, t)2L2 (–D,D) = 2 dt



D

(u, f (x, t, u)) dx. –D

Hence, d ¯ ≥ C0 w(t)8( ¯ ¯ w(t) w(t)), dt ¯ with w(t) = u(⋅, t)L2 (–D,D) . This implies that w(t) → ∞ as t → t0 with t0 being some finite positive time.

72

2 The Vanishing Viscosity Method of Some Nonlinear Evolution System

Theorem 2.2.14. Suppose that f (x, t, u) satisfies eqs (2.2.19) and (2.2.20). Then the solution u(x, t) to system (2.2.18) with periodic boundary condition (2.2.3) (D < ∞) or initial data (2.2.4) (D = ∞) has the asymptotic behavior lim u(⋅, t)L2 (–D,D) = ∞ (D ≤ ∞).

t→t0

(2.2.21)

2.3 High-Order Multivariable KdV Systems and Hirota Coupled KdV Systems We discuss the high-order multivariable KdV system as follows: ut + $$2p u + $grad>(u) = A(x, t)u + g(x, t),

(2.3.1)

where u(x, t) = (u1 (x, t), . . . , uN (x, t)) is an N-dimensional value unknown function of x ∈ Rn t ∈ R+ , >(u) ≡ >(u1 , . . . , uN ) is a scale function of u ∈ RN , A(x, t) is an  N × N matrix, g(x, t) is an N-dimensional value function, $k = ni=1 Dki , Di = ∂x∂ (i = i 1, 2, . . . , n), $ = $1 , “grad” denotes the gradient operator of u. At the same time, we also consider the system with small dissipative term ut + (–1)p+1 %$2p+2 u + $$2p u + $grad>(u) = A(x, t)u + g(x, t).

(2.3.2)

Obviously, this system is parabolic. Let K ⊂ Rn be an n-dimensional cube with length 2l, that is ¯ ={x = (x1 , . . . , xn )| |xi | ≤ l, i = 1, 2, . . . , n}, K ¯ T ={x ∈ K, ¯ 0 ≤ t ≤ T}, T ≤ ∞. Q We will discuss systems (2.3.1) and (2.3.2) and (2.3.1) and (2.3.2) in the cylindrical domain QT (n + 1 dimension) with periodic boundary condition u(x, t) = u(x + 2lei , t), u(x, 0) = u0 (x),

(2.3.3)

where x + 2lei = (x1 , . . . , xi–1 , xi + 2l, xi+1 , . . . , xn ), i = 1, 2, . . . , n, and u0 (x) is an N¯ ⊂ Rn . We also study the above systems with dimensional vector periodic function in K initial data u(x, 0) = u0 (x),

x ∈ Rn ,

(2.3.4)

where u0 (x) is an N-dimensional vector function of x ∈ Rn . In order to apply the fixed point principle to study the generalized solution to systems (2.3.1) and (2.3.2) with periodic boundary condition (2.3.3), we first consider the following linear parabolic equation

73

2.3 High-Order Multivariable KdV Systems and Hirota Coupled KdV Systems

vt + %(–1)p–1 $2p+2 v + $$2p v = f (x, t)

(2.3.5)

in the domain QT = K × [0, T], with periodic boundary condition v(x, t) = v(x + 2lei , t), v(x, 0) = v0 (x), i = 1, 2, . . . , n, (p+1)

where f (x, t) ∈ L2 (QT ), v0 (x) ∈ W2

(2.3.6)

(K), p ∈ Z≥1 .

Lemma 2.3.1. For any function u(x) ∈ H |!| (K) satisfying periodic condition, we have D! uL2 (K) ≤

n 

|!|

!i

Di uL|!|(K) ,

(2.3.7)

2

i=1 !

!

where ! = (!1 , . . . , !n ), D! = D1 1 D2 2 . . . D!nn , |!| = nonnegative integer.

n

i=1 !i ,

!i (i = 1, 2, . . . , n) is

Corollary 2.3.2. Under the assumption of Lemma 2.3.1, we have D! uL2 (K) ≤

n  !i |!| D uL2 (K) , |!| i

(2.3.8)

i=1

with ! = (!1 , . . . , !n ). Lemma 2.3.3. For the generalized solution to linear parabolic equation (2.3.5) with periodic boundary condition (2.3.6), there holds sup v(⋅, t)H p+2,1 (K) + v

0≤t≤T

(2p+2,1)

W2

(QT )

! " ≤ K1 v0 H p+1 (K) + f L2 (QT ) ,

(2.3.9)

where the constant K1 is independent of l > 0 and depends on T < ∞ and % > 0. Based on the priori estimate in Lemma 2.3.3, one can use the parameter continuation (2p+2,1) method to show the existence of the generalized solution v(x, t) ∈ W2 (QT ) to linear parabolic equation (2.3.5) with periodic boundary condition (2.3.6). The uniqueness of generalized solution follows by eq. (2.3.9), since the equation is linear. Thus, we obtain the existence and uniqueness of the generalized solution to eqs (2.3.5) and (2.3.6). Lemma 2.3.4. Let % > 0, f (x, t) ∈ L2 (QT ) and v0 (x) ∈ H p+1 (K). Then, there is a unique generalized solution v(x, t) to linear parabolic equation (2.3.5) with periodic boundary (2p+2,1) condition (2.3.6). Moreover, v(x, t) ∈ Z := W2 (QT ) ∩ L∞ (0, T; H p+1 (K)). Theorem 2.3.5. Suppose system (2.3.2) (% > 0) and vector periodic function u0 (x) satisfy the following conditions:

74

2 The Vanishing Viscosity Method of Some Nonlinear Evolution System

(1) u0 (x) = u0 (x + 2lei ) and u0 (x) ∈ H p+1 (K). (p)

(2) The elements aij (x, t)(i, j = 1, 2, . . . , N) of A(x, t) belong to space L∞ (0, T; W∞ (K). The components gi (x, t)(i = 1, 2, . . . , N) of g(x, t) belong to space L2 (0, T; H p (K)). All these functions are periodic function of x = (x1 , . . . , xn ), and the period is 2l. (3) The scale function >(u) is 2-times continuously differentiable, and (i) |>(u)| ≤ K|u|s+2 , (ii) |grad>(u)| ≤ K|u|s+1 ,  ∂ 2 >(u)    (iii)   ≤ K(|u|s + 1) (i, j = 1, 2, . . . , N), ∂ui ∂uj

(2.3.10)

where s = 4p n – $ > 0, $ > 0 and K is a constant. Then there exists at least one global generalized solution u(x, t) to system (2.3.2) with periodic boundary condition (2.3.3). Moreover, (2p+2,1)

u(x, t) ∈ Z = W2

(QT ) ∩ L∞ (0, T; H p+1 (K)), p ≥ 1.

Proof. Taking base space B to be L∞ (0, T; L2 (K) ∩ L2 (0, T; H 2p+1 (K)). Let B∗ = {u = (u1 , . . . , uN )| ui ∈ B, i = 1, 2, . . . , N}, Z ∗ = {u = (u1 , . . . , uN )| ui ∈ Z, i = 1, 2, . . . , N}. For 0 ≤ + ≤ 1, we define the mapping T+ : B∗ → B∗ as follows: For each v = (v1 , . . . , vN ) ∈ B∗ , let the component uk (k = 1, 2, . . . , N) of u = T+ v be the solution to the following parabolic system with periodic boundary condition: ⎧ p+1 ⎪ ⎪ ⎨ukt + (–1) %$2p+2 uk + $$2p uk = +hk (x, t, v1 , . . . , vN ) (2.3.11) uk (x, t) = uk (x + 2lei , t) ⎪ ⎪ ⎩u (x, 0) = +u (x), k

0,k

with hk (x, t, v1 , . . . , vN ) =

N 

aki (x, t)vi + gk (x, t) –

i=1

N  i=1

Di

 ∂>(v , . . . , v )  1 N . ∂uk

(2.3.12)

First, we verify the right-hand side of eq. (2.3.11). Equation (2.3.12) belongs to L2 (QT )space: ( ( ($grad>(v)(2 L

2 (QT )

=

N  n   i,j=1 m,k=1 0



≤ C5 0

T



K

T

 K

∂ 2 >(v) ∂ 2 >(v) Di vl Dj vm dx dt ∂vl ∂vk ∂vm ∂vk

! 2s " |v| + 1 |Dv|2 dx dt,

75

2.3 High-Order Multivariable KdV Systems and Hirota Coupled KdV Systems

where C5 is a constant and Dv denotes one derivative of v in x. Hence, 

T

K

0

with

1 .

+

1 '





T

|v|2s |Dv|2 dx dt ≤ 0

Dv(⋅, t)2L

2. (K)

v(⋅, t)2s L2s' (K) dt,

= 1. On the other hand, we have by interpolation n(. –1) 2p 2p+1 – . (4p+2)

Dv(⋅, t)L2. (K) ≤ C6 v(⋅, t)L

2 (K)

n(. –1) 1 2p+1 + . (4p+2)

v(⋅, t)H 2p+1 (K)

n(s'–1)

1– s'(4p+2)

v(⋅, t)L2s' (K) ≤ C7 v(⋅, t)L

2 (K)

,

n(s'–1) s'(4p+2)

v(⋅, t)H 2p+1 (K) ,

with . , ' satisfying 0


n . 4p

(2.3.14)

Equation (2.3.13) implies that s' < Since s =

4p n

n . n – (4p + 2)

(2.3.15)

– $ > 0, when $n –1 n n 4p + 2 –1  1– , H 1 (R1 ) + f L2 (Q∗ ) , T

0≤t≤T

T

where the constant C2 depends on the norms bL∞ (Q∗ ) , cL∞ (Q∗ ) and % > 0, but not T T on +(0 ≤ + ≤ 1). Thus, under the above conditions, the solutions to problem (2.7.11) and (2.7.7) satisfy sup u(⋅, t)H 1 (R1 ) + uxx L2 (Q∗ ) + ut L2 (Q∗ ) T

0≤t≤T

#

$

≤ C3 >H 1 (R1 ) + f L2 (Q∗ ) . T

T

(2.7.12)

136

2 The Vanishing Viscosity Method of Some Nonlinear Evolution System

In other words, # $ uW (2,1) (Q∗ ) ≤ C3 >H 1 (R1 ) + f L2 (Q∗ ) , 2

T

T

where the constant C3 depends on the norms bL∞ (Q∗ ) , cL∞ (Q∗ ) and % > 0, but not T T on +(0 ≤ + ≤ 1). Using this estimate, we can show that the set E is closed in [0, 1]. Let +˜ ∈ [0, 1] ˜ be the limit point of E, that is, there is a sequence {+k } ⊂ E such that lim +k = +. k→∞

Denote uk (x, t) to be the unique generalized solution of L+k uk = f (x, t), k = 0, 1, . . .. Thus, we can choose a subsequence {+k } and {uk (x, t)} such that as k → ∞, +k → +˜ ˜ t) ∈ W2(2,1) (Q∗T ). Moreover, {uk (x, t)} converges uniand {uk (x, t)} converges to u(x, ˜ t) in any compact subset of Q∗T , and {ukx (x, t)} converges strongly to formly to u(x, ˜ t) in Lp (0, T; L2 (R))(2 ≤ p < ∞). But {ukxx (x, t)} and {ukt (x, t)} converge weakly u(x, ˜ t) ∈ to u˜ xx (x, t) and u˜ t (x, t) in L2 (Q∗T ), respectively. Therefore, the limit function u(x, (2,1) ∗ W2 (QT ) is the unique global generalized solution to problem (2.7.11) and (2.7.7). ˜ This means that the set E is closed in [0, 1]. L+ u = f (x, t), + = +. Next, we prove that the set E is open in [0, 1]. In fact, we define the map A+ : W2(2,1) (Q∗T ) → W2(2,1) (Q∗T ) as follows: for each v ∈ W2(2,1) (Q∗T ) ⊂ W2(2,0) (Q∗T ), denote u = A+ v to be the unique global generalized solution to the linear parabolic equation L+0 u ≡ ut – %uxx + +0 Huxx + b(x, t)ux + c(x, t)u = f (x, t) + (+0 – +)Hvxx , with +0 ∈ E. For v1 , v2 ∈ W2(2,1) (Q∗T ), we have u1 = A+ v1 , u2 = A+ v2 ∈ W2(2,1) (Q∗T ), and u1 – u2 solves the linear equation L+0 (u1 – u2 ) ≡ (u1 – u2 )t – %(u1 – u2 )xx + +0 H(u1 – u2 )xx + b(u1 – u2 )x + c(u1 – u2 ) = (+0 – +)H(v1 – v2 )xx , with initial data u1 (x, 0) – u2 (x, 0) = 0. We have by eq. (2.7.12) u1 – u2 W (2,1) (Q∗ ) ≤ C3 |+0 – +| ⋅ v1 – v2 W (2,0) (Q∗ ) 2

T

2

T

≤ C3 |+0 – +| ⋅ v1 – v2 W (2,1) (Q∗ ) . 2

T

For |+0 – +| sufficiently small, the map A+ : W2(2,1) (Q∗T ) → W2(2,1) (Q∗T ) is contractive. Hence, there is a unique solution u(x, t) ∈ W2(2,1) (Q∗T ) such that u = A+ u. This implies + ∈ E. Therefore, the set E is open in [0, 1].

137

2.7 Initial Value Problem for the Nonlinear Singular Integral and Differential Equations

In sum, E ≡ [0, 1]. Especially for + = 1, problem (2.7.11) and (2.7.7) has a unique global generalized solution. That is Theorem 2.7.4. Let b(x, t), c(x, t) ∈ L∞ (Q∗T ), f (x, t) ∈ L2 (Q∗T ) and >(x) ∈ H 1 (R1 ). Then there is a unique global generalized solution u(x, t) ∈ W2(2,1) (Q∗T ) to the initial value problem (2.7.11) and (2.7.7). Corollary 2.7.5. Under the assumptions of Theorem 2.7.4, the global generalized solution u(x, t) to the initial value problem (2.7.11) and (2.7.7) satisfies # $ uW (2,1) (Q∗ ) ≤ K1 >H 1 (R1 ) + f L2 (Q∗ ) , 2

(2.7.13)

T

T

where the constant K1 depends on the norms bL∞ (Q∗ ) , cL∞ (Q∗ ) and % > 0. T

(k,[ k ])

T

(k,[ k ])

Corollary 2.7.6. Let b(x, t), c(x, t) ∈ W∞ 2 (Q∗T ), f (x, t) ∈ W2 2 (Q∗T ) and >(x) ∈ H k+1 (R1 ), k ≥ 1. Then there is a unique global generalized solution u(x, t) ∈ (k+2,[ k2 ]+1)

W2

(Q∗T ) to the initial value problem (2.7.11) and (2.7.7).

Now, we turn to show the existence of the global generalized solution to the initial value problem (2.7.8) and (2.7.7). Theorem 2.7.7. Let b(x, t), c(x, t) ∈ L∞ (Q∗T ), f (x, t) ∈ L2 (Q∗T ) and >(x) ∈ H 1 (R1 ). Then there is a unique global generalized solution u(x, t) ∈ W2(2,1) (Q∗T ) to the initial value problem (2.7.8) and (2.7.7). It solves eq. (2.7.8) in the sense of generalized solution, and it satisfies eq. (2.7.7) in the classical sense. Furthermore, there holds sup u(⋅, t)H 1 (R1 ) + uxx L2 (Q∗ ) + ut L2 (Q∗ ) T

0≤t≤T

#

T

$

(2.7.14)

≤ K2 >H 1 (R1 ) + f L2 (Q∗ ) , T

where the constant K2 depends on the norms bL∞ (Q∗ ) , cL∞ (Q∗ ) and % > 0, T > 0. T

T

Proof. We argue by fixed point principle. Define the mapping T+ : B → B = L∞ (Q∗T )(0 ≤ + ≤ 1) as follows: for each v(x, t) ∈ B, u = T+ v is the global generalized solution to the linear parabolic equation ut – %uxx + Huxx + 2vux + bux + cu = +f ,

(2.7.15)

u(x, 0) = +>(x).

(2.7.16)

with initial data

138

2 The Vanishing Viscosity Method of Some Nonlinear Evolution System

When v(x, t) ∈ B, u(x, t) ∈ W2(2,1) (Q∗T ). Since the operator W2(2,1) (Q∗T ) ↪ B is precompact, the mapping T+ : B → B defined by u = T+ v is completely continuous for any v ∈ B and 0 ≤ + ≤ 1. When + = 0, T0 (B) = 0. To verify the existence of the global generalized solution to the initial value problem (2.7.8) and (2.7.7), we only need to show that in the base function space B, the fixed point of the mapping T+ : B → B is uniformly bounded with respect to 0 ≤ + ≤ 1. To do this, we have to set up the priori estimates for the solutions u+ (x, t) to the nonlinear parabolic equation ut – %uxx + Huxx + 2uux + bux + cu = +f ,

(2.7.17)

with initial data condition (2.7.16). Making inner product for u(x, t) and eq. (2.7.17) gives 

∞

! " u ut – %uxx + Huxx + 2uux + bux + cu – +f dx = 0.

–∞

Like the same estimate before, we obtain by the above equality # $ d u(⋅, t)2L (R1 ) + ux (⋅, t)2L (R1 ) ≤ C4 u(⋅, t)2L (R1 ) + f (⋅, t)2L (R1 ) . 2 2 2 2 dt Hence, # $ sup u(⋅, t)L2 (R1 ) + ux L2 (Q∗ ) ≤ C5 >L2 (R1 ) + f L2 (Q∗ ) ,

0≤t≤T

T

T

where the constants C4 and C5 depend on the norms bL∞ (Q∗ ) , cL∞ (Q∗ ) and % > 0, T T but not on 0 ≤ + ≤ 1. Making inner product for uxx and eq. (2.7.17) shows 

∞

! " uxx ut – %uxx + Huxx + 2uux + bux + cu – +f dx = 0.

–∞

Similarly, # $ sup ux (⋅, t)L2 (R1 ) + uxx L2 (Q∗ ) ≤ C6 >H 1 (R1 ) + f L2 (Q∗ ) ,

0≤t≤T

T

T

where C6 is independent of 0 ≤ + ≤ 1. This shows that the solutions to the initial value problem (2.7.17) and (2.7.7) is uniformly bounded in the base space B with respect to 0 ≤ + ≤ 1. From Leray– Schauder’s fixed point principle, we know that there is a global generalized solution u(x, t) ∈ W2(2,1) (Q∗T ) to the initial value problem (2.7.17) and (2.7.7). Therefore, we conclude Theorem 2.7.7. ∎

139

2.7 Initial Value Problem for the Nonlinear Singular Integral and Differential Equations

Theorem 2.7.8. The global generalized solution u(x, t) ∈ L∞ (0, T; H 2 (R1 )) to the initial value problem (2.7.17) and (2.7.7) is unique. Proof. Let u(x, t), v(x, t) ∈ L∞ (0, T; H 2 (R1 )) be two global generalized solutions to the initial value problem (2.7.17) and (2.7.7). Then the difference function w(x, t) = u(x, t) – v(x, t) solves the linear equation wt – %wxx + Hwxx + (b + u + v)wx + (c + ux + vx )w = 0, with initial data condition w(x, 0) = 0, Since the coefficients of wx and w are bounded, we easily get w(x, t) ≡ 0 by energy estimate. And so we obtain the uniqueness. ∎ In order to obtain the existence of the global generalized solution to the initial value problem (2.7.6) and (2.7.7), we need to establish the uniform priori estimates with respect to % for the smooth solution to the initial value problem (2.7.8) and (2.7.7). To do this, we set up the following lemma. Lemma 2.7.9. Let b(x, t), bx (x, t), c(x, t) ∈ L∞ (Q∗T ), f (x, t) ∈ L2 (Q∗T ) and >(x) ∈ L2 (R1 ). Then for the global generalized solution u% (x, t) ∈ W2(2,1) (Q∗T ) to the initial value problem (2.7.8) and (2.7.7), there holds # $ sup u% (⋅, t)L2 (R1 ) ≤ K3 >L2 (R1 ) + f L2 (Q∗ ) , (2.7.18) T

0≤t≤T

where the constant K3 is independent of % and depends on bx L∞ (Q∗ ) and cL∞ (Q∗ ) . T

T

Proof. Making inner product for eq. (2.7.8) and u gives  ∞ ! " u ut + 2uux + Huxx – %uxx + bux + cu – f dx = 0. –∞

Note that



∞

1 buux dx = – 2 –∞



∞

bx u2 dx,

–∞

we get d u(⋅, t)2L (R1 ) + 2%ux (⋅, t)2L (R1 ) 2 2 dt # $ ∗ ∗ ≤ bx L∞ (Q ) + 2cL∞ (Q ) + 1 u(⋅, t)2L T

Hence, Lemma 2.7.9 follows.

T

1 2 (R )

+ f (⋅, t)2L

1 . 2 (R )

∎

140

2 The Vanishing Viscosity Method of Some Nonlinear Evolution System

Next, we turn to estimate the term u%x (⋅, t)L2 (R1 ) . By direct computation, and using the properties of Hilbert operator, we obtain a series of inequalities: d dt

 ∞ u2x dx = – 2 uxx ut dx –∞  ∞–∞   =2 uxx 2uux + Huxx – %uxx + bux + cu – f dx.



∞

–∞

Noting that 

∞

uxx Huxx dx = 0, –∞ ∞

 1 ∞ bux uxx dx = – bx u2x dx, 2 –∞ –∞  ∞  ∞ cuuxx dx = – (cu2x + cx uux ) dx, –∞ –∞  ∞  ∞ uxx f dx = – ux fx dx, –∞

–∞

we obtain  ∞  ∞  d ∞ 2 ux dx = 4 uux uxx dx – 2%uxx (⋅, t)2L (R1 ) – (bx 2c)u2x dx 2 dt –∞ –∞ –∞  ∞  ∞ –2 cx uux dx + 2 ux fx dx. –∞

(2.7.19)

–∞

On the other hand, a direct computation gives d dt



∞ –∞

d dt  =2



u2 Hux dx =

∞!

" 2uut Hux + u2 Huxt dx

–∞ ∞

–∞



=–2

 uHux + H(uux ) ut dx

∞

 ! " uHux + H(uux ) ⋅ 2uux + Huxx – %uxx + bux + cu – f dx.

–∞

It follows from property (3) of Hilbert transform that 

∞

uux H(uux ) dx = 0. –∞

By Lemma 2.7.1, one has 



∞

∞

H(uux )Huxx dx = –∞ ∞

uux uxx dx, –∞ ∞

  1 uHux Huxx dx = u (Hux )2 x dx 2 –∞ –∞   1 ∞ 1 ∞ 2 =– ux (Hux ) dx = ux H(ux Hux ) dx. 2 –∞ 2 –∞

141

2.7 Initial Value Problem for the Nonlinear Singular Integral and Differential Equations

From property (2) of Hilbert transform, we get H(fHf ) = H(HfH 2 f ) + (Hf )2 + fH 2 f . This together with property (1) of Hilbert transform yields 1 1 H(fHf ) = (Hf )2 – f 2 . 2 2 Hence,  ∞ ux H(ux Hux ) dx = –∞

1 2



∞

ux (Hux )2 dx –

–∞ ∞

=– 

d dt

∞

1 2



∞

 –∞ ∞

uHux Huxx dx + –∞  ∞

u2 Hux dx = – 4

–∞

uux uxx dx, –∞



8 3

u2 ux Hux dx –

 –∞ ∞ –2

u3x dx

!

∞

uux uxx dx – 2 –∞



bux H(uux ) dx + 2% –∞ ∞



∞

buux Hux dx –∞

∞!

" uHux + H(uux ) uxx dx

–∞

"

uHux + H(uux ) (cu – f ) dx.

–2

(2.7.20)

–∞

Again, we have  ∞  ∞  ! " d ∞ but Hux + buHuxt dx buHux dx = bt uHux dx + dt –∞ –∞  ∞  ∞–∞ ! "  = bHux + H(bu)x 2uux + Huxx – %uxx + bux + cu – f dx. bt uHux dx – –∞

–∞

Noting that 



∞

∞

uux H(bu)x dx = – –∞ ∞

(bx u + bux )H(uux ) dx, –∞



1 ∞ bHux Huxx dx = – bx (Hux )2 dx, 2 –∞ –∞  ∞  ∞ H(bu)x Huxx dx = (bu)x uxx dx –∞ –∞  ∞ ! " 3 bxx uux + bx u2x dx, =– 2 –∞ we deduce that d dt





∞



∞

buHux dx = – 2 –∞

–∞



∞



∞

buux Hux dx + 2

∞

bux H(uux ) dx + –∞

1 +2 bx uH(uux ) dx + 2 –∞



bt uHux dx –∞

∞

–∞

 bx (Hux )2 dx +

(2.7.21) ∞

bxx uux dx –∞

142

2 The Vanishing Viscosity Method of Some Nonlinear Evolution System

+

3 2 



–

∞

–∞ ∞!

 bx u2x dx + %

∞!

" bHux + H(bu)x uxx dx

–∞

" bHux + H(bu)x (bux + cu – f ) dx.

–∞

Similarly, 

d dt

d dt

∞

 u4 dx = 12

–∞



∞

 u2 ux Hux dx + 4%

–∞  ∞

–4  ∞ bu3 dx = 6

∞

u3 ux dx

–∞ 3

u (bux + cu – f ) dx,  ∞  ∞ buux Hux dx + bt u3 dx – 6 bu3 ux dx –∞ –∞ –∞  ∞  ∞ 2 +3 bx u Hux dx + 3% bu2 uxx dx –∞ –∞  ∞ –3 bu2 (bux + cu – f ) dx.

–∞

(2.7.22)

–∞ ∞

(2.7.23)

–∞

We remark that in the above equalities, we used the fact that the global solution u% (x, t) and its first derivative tend to zero as |x| → ∞, and the basic properties of Hilbert transformation. In order to eliminate the four terms 



∞

∞

uux uxx dx, –∞

 u2 ux Hux dx,

–∞



∞

∞

buux Hux dx, –∞

bux H(uux ) dx, –∞

and making the linear combination of the five equalities (2.7.19)–(2.7.23), we obtain 

∞

 u4 + 2bu3 + 3buHux + 3u2 Hux + 2u2x dx –∞  ∞ # $ = – 4%uxx (⋅, t)2L (R1 ) + % 6(uHux + H(uux )) + 3(bHux + H(bu)x + 4u3 + 6bu2 ) uxx dx 2 –∞  ∞   ∞   5 3 2 + u3 bux dx bx – 4c ux + bx (Hux )2 – 3b2 ux Hux dx – 16 2 2 –∞ –∞  ∞   2 2 2 (bx – c)(u Hux + uH(uux )) – b u ux dx +6  –∞ ∞  +3 (–bc + bt )uHux + (bxx – 4cx )uux – bux H(bx u) – cuH(bux ) dx –∞  ∞ ∞   3bfHux + 3fH(bux ) + 4ux fx dx +6 f (uHux + H(uux )) dx + –∞ –∞  ∞  ∞  ∞ ∞ 4 –4 cu dx + (2bt – 6bc)u3 dx + 4 fu3 dx + 6 bfu2 dx –∞ –∞ –∞ –∞  ∞  ∞ –3 cuH(bx u) dx + 3 fH(bx u) dx. (2.7.24) d dt

–∞

–∞

2.7 Initial Value Problem for the Nonlinear Singular Integral and Differential Equations

143

Now, we turn to estimate the contribution of each term in eq. (2.7.24). Writing J(x, t) = 6(uHux + H(uux )) + 3(bHux + H(bu)x + 4u3 + 6bu2 ), then, we have   

 1 1  Juxx dx ≤ uxx (⋅, t)2L (R1 ) + J(⋅, t)2L (R1 ) . 2 2 2 2 –∞ ∞

(2.7.25)

A simple computation shows  J(⋅, t)2L

1 2 (R )



∞ –∞

∞#

u2 (Hux )2 + (H(uux ))2 + b2 (Hux )2 $ + (H(bu)x )2 + u6 + b2 u4 dx,

≤ C7

(2.7.26)

–∞

u2 (Hux )2 dx ≤ u(⋅, t)2L

2

1 Hux (⋅, t)L (R1 ) ∞ (R ) 2

= u(⋅, t)2L

2

1 ux (⋅, t)L (R1 ) ∞ (R ) 2



≤ 'uxx (⋅, t)2L (R1 ) + C8 (')u(⋅, t)10 , L2 (R1 ) 2  ∞ ∞ (H(uux ))2 dx = u2 u2x dx ≤ 'uxx (⋅, t)2L (R1 ) + C8 (')u(⋅, t)10 , L (R1 )

–∞ ∞ –∞

2

–∞

2

b2 (Hux )2 dx ≤ b2L∞ (Q∗ ) ux (⋅, t)2L

1 2 (R )

T

≤ 'uxx (⋅, t)2L

1 2 (R )

+ C9 (')b4L∞ (Q∗ ) u(⋅, t)2L

1 2 (R )

T

and 

∞

" ! (H(bu)x )2 dx ≤ 'uxx (⋅, t)2L (R1 ) + C10 (') bx 2L∞ (Q∗ ) + b4L∞ (Q∗ ) u(⋅, t)2L (R1 ) , 2 2 T T –∞  ∞ u6 ≤ 'uxx (⋅, t)2L (R1 ) + C11 (')u(⋅, t)10 , L2 (R1 ) 2 –∞  ∞ 8 14 b2 u4 dx ≤ 'uxx (⋅, t)2L (R1 ) + C12 (')bL3∞ (Q∗ ) u(⋅, t)L3 (R1 ) . 2

–∞

T

2

Plugging these inequalities into eq. (2.7.26), we obtain J(⋅, t)2L

1 2 (R )

≤ 6C7 'uxx (⋅, t)2L

1 2 (R )

+ C13 (')u(⋅, t)2L

1 , 2 (R )

where C13 (') depends on the norms bW (1,0) (Q∗ ) and sup0≤t≤T u(⋅, t)L2 (R1 ) . Taking ' ∞

T

sufficiently small such that 6C7 ' = 1, we derive that

144

2 The Vanishing Viscosity Method of Some Nonlinear Evolution System

J(⋅, t)2L (R1 ) ≤ uxx (⋅, t)2L (R1 ) + C14 u(⋅, t)2L (R1 ) , 2 2 2   ∞ 1   Juxx dx ≤ uxx (⋅, t)2L (R1 ) + C14 u(⋅, t)2L (R1 ) ,  2 2 2 –∞ where

the

constant

C14 (')

depends

on

the

norms

(2.7.27)

bW (1,0) (Q∗ ) ∞

and

T

sup0≤t≤T u(⋅, t)L2 (R1 ) . By interpolation formula, we can estimate the other terms in eq. (2.7.24) as follows:   

   5 3  bx – 4c u2x + bx (Hux )2 – 3b2 ux Hux dx 2 2 –∞ ! " ≤ 4bx L∞ (Q∗ ) + 4cL∞ (Q∗ ) + 3b2L∞ (Q∗ ) ux (⋅, t)2L (R1 ) , T T 2 T   ∞   bu3 ux dx ≤ C15 bL∞ (Q∗ ) u(⋅, t)2L (R1 ) ux (⋅, t)2L (R1 ) ,  T 2 2 –∞   ∞    (bx – c)(u2 Hux + uH(uux )) – b2 u2 ux dx  –∞ ! "! " ≤ C16 bx L∞ (Q∗ ) + cL∞ (Q∗ ) + b2L∞ (Q∗ ) ux (⋅, t)2L (R1 ) + u(⋅, t)6L (R1 ) , T T 2 2 T  ∞    (–bc + bt )uHux + (bxx – 4cx )uux – bux H(bx u) – cuH(bux ) dx  ∞ 

–∞

! 1 1 ≤ bL∞ (Q∗ ) cL∞ (Q∗ ) + bL∞ (Q∗ ) bx L∞ (Q∗ ) + bxx L∞ (Q∗ ) T T T T T 2 2 "! " 1 2 2 + 2cx L∞ (Q∗ ) + bt L∞ (Q∗ ) ux (⋅, t)L (R1 ) + u(⋅, t)L (R1 ) , T T 2 2 2  ∞    4 2 2 6 cu dx ≤ cL∞ (Q∗ ) ux (⋅, t)L (R1 ) + C17 u(⋅, t)L (R1 ) ,  2 2 T –∞  ∞  ! "!   (2bt – 6bc)u3 dx ≤ C18 bt L∞ (Q∗ ) + bL∞ (Q∗ ) cL∞ (Q∗ ) ux (⋅, t)2L  T

–∞

T

T

1 + 2 (R )

10 " u(⋅, t)L3 (R1 ) , 2  ∞    cuH(bx u) dx ≤ cL∞ (Q∗ ) bx L∞ (Q∗ ) u(⋅, t)2L (R1 ) ,  T T 2

–∞

where the constants Cij are independent of % > 0. For the terms involving f (x, t) and fx (x, t), we have the following estimates:   

  3bfHux + 3fH(bux ) + 4ux fx dx

∞ –∞

9  ≤ b2L∞ (Q∗ ) + 2 ux (⋅, t)2L (R1 ) + 2f (⋅, t)2H 1 (R1 ) , 2 T 2   ∞ 1   fu3 dx ≤ C19 u(⋅, t)4L (R1 ) ux (⋅, t)2L (R1 ) + f (⋅, t)2L (R1 ) ,  2 2 2 2 –∞  ∞  1 ! " 1   bfu2 dx ≤ b2L∞ (Q∗ ) ux (⋅, t)2L (R1 ) + u(⋅, t)6L (R1 ) + f (⋅, t)2L (R1 ) ,  2 2 2 T 2 2 –∞  ∞  1 1   fH(bx u) dx ≤ bx 2L∞ (Q∗ ) u(⋅, t)2L (R1 ) + f (⋅, t)2L (R1 ) ,  2 2 T 2 2 –∞

145

2.7 Initial Value Problem for the Nonlinear Singular Integral and Differential Equations

furthermore, we have   ∞   fuHux dx ≤ u(⋅, t)L2 (R1 ) f (⋅, t)L∞ (R1 ) Hux (⋅, t)L2 (R1 )  –∞

1 1 ≤ ux (⋅, t)2L (R1 ) + f (⋅, t)2L (R1 ) u(⋅, t)2L (R1 ) ∞ 2 2 2 2 1 ≤ ux (⋅, t)2L (R1 ) + C20 f (⋅, t)2H 1 (R1 ) , 2 2 and   

∞

–∞

  fH(uux ) dx ≤ Hf (⋅, t)L∞ (R1 ) u(⋅, t)L2 (R1 ) ux (⋅, t)L2 (R1 )

1 ≤ ux (⋅, t)2L (R1 ) + C20 u(⋅, t)2L (R1 ) Hf (⋅, t)L2 (R1 ) Hfx (⋅, t)L2 (R1 ) 2 2 2 1 2 2 ≤ ux (⋅, t)L (R1 ) + C21 f (⋅, t)H 1 (R1 ) , 2 2 where the constant C21 depends on the norm sup0≤t≤T u(⋅, t)L2 (R1 ) . Plugging these inequalities into the right-hand side of eq. (2.7.24), we deduce that d dt



∞

 u4 + 2bu3 + 3buHux + 3u2 Hux + 2u2x dx + 3%uxx (⋅, t)2L

1 2 (R )

–∞

≤ C22 u(⋅, t)H 1 (R1 ) + C23 f (⋅, t)2H 1 (R1 ) ,

(2.7.28)

where the constants C22 and C23 depend on the norms bW (2,1) (Q∗ ) , cW (1,0) (Q∗ ) and ∞

sup0≤t≤T u(⋅, t)L2 (R1 ) , but do not depend on % > 0. Integrating eq. (2.7.28) in t ∈ [0, t] implies 

∞

–∞



≤ C22  +

T

∞

 u4 + 2bu3 + 3buHux + 3u2 Hux + 2u2x dx t

ux (⋅, 4)2L

0 ∞!

1 2 (R )

–∞

# d4 + C24 >2L

1 2 (R )

T

(2.7.29)

+ f W (1,0) (Q∗ ) 2

$

T

" >4 + 2b(x, 0)>3 + 3b(x, 0)>H>x + 3>2 H>x + 2>2x dx,

where the constant C24 is independent of % > 0. The last integral term can be controlled by   

"  >4 + 2b(x, 0)>3 + 3b(x, 0)>H>x + 3>2 H>x + 2>2x dx

∞!

–∞

! " ≤ C25 b(⋅, 0)L∞ (R1 ) + 1 >x 2L

1 2 (R )

! + C25 >4L

1 2 (R )

4 " + >L3 (R1 ) + 1 >2L 2

≤ C26 >2H 1 (R1 ) , where the constant C26 depends on the norms b(⋅, 0)L∞ (R1 ) and >L2 (R1 ) .

1 2 (R )

146

2 The Vanishing Viscosity Method of Some Nonlinear Evolution System

Similarly, the left-hand side of eq. (2.7.29) has the lower bound ux (⋅, t)2L

1 2 (R )

– C27 u(⋅, t)2L

1 , 2 (R )

where C27 is dependent on the norms bL∞ (Q∗ ) and sup u(⋅, t)L2 (R1 ) , but independT

0≤t≤T

ent of % > 0. Thus, eq. (2.7.29) becomes  ux (⋅, t)2L (R1 ) 2

≤ C22

t

0

ux (⋅, 4)2L

1 2 (R )

# d4 + C28 >2H 1 (R1 ) + f 2

$

W2(1,0) (Q∗T )

,

(2.7.30)

where the constants C22 and C28 depend on the norms bW (2,1) (Q∗ ) , cW (1,0) (Q∗ ) and ∞

sup0≤t≤T u(⋅, t)L2 (R1 ) , but do not depend on % > 0. It follows from eq. (2.7.30) that ux (⋅, t)2L

1 2 (R )

# ≤ C29 >2H 1 (R1 ) + f 2

$

W2(1,0) (Q∗T )

T

∞

T

∀0 ≤ t ≤ T,

where C29 is independent of % > 0. Therefore, we have the following lemma. (2,1) (Q∗ ), c(x, t) ∈ W (1,0) (Q∗ ), f (x, t) ∈ W (1,0) (Q∗ ) and >(x) ∈ Lemma 2.7.10. Let b(x, t) ∈ W∞ ∞ 2 T T T 1 1 H (R ). Then for the global generalized solution u% (x, t) ∈ W2(2,1) (Q∗T ) to the initial value problem (2.7.8) and (2.7.7), there holds

# $ sup u%x (⋅, x)L2 (R1 ) ≤ K4 >H 1 (R1 ) + f W (1,0) (Q∗ ) , 2

0≤t≤T

(2.7.31)

T

where the constant K4 is independent of % > 0. Lemma 2.7.11. Under the assumptions of Lemma 2.7.10, for the global generalized solution u% (x, t) ∈ W2(2,1) (Q∗T ) to the initial value problem (2.7.8) and (2.7.7), we have # $ u% L∞ (Q∗ ) ≤ K5 >H 1 (R1 ) + f W (1,0) (Q∗ ) , T

2

(2.7.32)

T

where the constant K5 is independent of % > 0. To estimate the term u%xx (⋅, t)L2 (R1 ) , we have the similar integral equality. It is easy to see that (we ignore the subscript % in the following) d dt



∞ –∞

 u2xx dx =2

∞

uxx uxxt dx –∞



∞

=–2 –∞

! " uxxxx 2uux + Huxx – %uxx + bux + cu – f dx.

2.7 Initial Value Problem for the Nonlinear Singular Integral and Differential Equations

147

Observing that 

∞

uux uxxxx dx = 

–∞ ∞



5 2

–∞

uxxxx Huxx dx =0, –∞   ∞ bux uxxxx dx = –∞

∞

ux u2xx dx,

∞

bxx ux uxx dx +

–∞

3 2



∞

–∞

bx u2xx dx,

we obtain  ∞  ∞  d ∞ 2 uxx dx = – 10 ux u2xx dx – 2%uxxx (⋅, t)2L (R1 ) – 2 bxx ux uxx dx 2 dt –∞ –∞ –∞  ∞  ∞ –3 bx u2xx dx – 2 uxx (cu – f )xx dx. (2.7.33) –∞

–∞

Next, we have d dt





∞

uux Huxx dx =  –∞ ∞



–∞ ∞



= =

∞!

" ut ux Huxx + uuxt Huxx + uux Huxxt dx

–∞

 uHuxxx + H(uux )xx ut dx ! " uHuxxx + H(uux )xx 2uux + Huxx – %uxx + bux + cu – f dx.

–∞

We easily check that 

∞

 u2 ux Huxxx dx = –

–∞ ∞

∞

(u2 ux )x Huxx dx,

–∞

uux H(uux )xx dx = 0, –∞ ∞ uHuxx Huxxx dx = – –∞ ∞

1 2 



∞

H(uux )xx Huxx dx = –∞

ux (Huxx )2 dx,

–∞ ∞

(uux )xx uxx –∞

5 dx = 2



∞ –∞

ux u2xx dx.

Hence,  5 ∞ uux Huxx ux (Huxx ) dx + ux u2xx dx 2 –∞ –∞ –∞  ∞  ∞     uHuxxx + H(uux )xx uxx dx – u(bux + cu – f ) x Huxx dx –% –∞ –∞  ∞ ∞ 2 –2 (u ux )x Huxx dx – H(uux )x (bux + cu – f )x dx.

d dt



∞

–∞

1 dx = – 2



∞

–∞

2

(2.7.34)

148

2 The Vanishing Viscosity Method of Some Nonlinear Evolution System

Next, we compute the equality d dt



∞

–∞



= –2  =2

 u2x Hux dx =

∞!

–∞

∞

" 2ux uxt Hux + u2x Huxt dx

 (ux Hux )x – H(ux uxx ) ut dx

–∞ ∞

! " (ux Hux )x – H(ux uxx ) 2uux + Huxx – %uxx + bux + cu – f dx.

–∞

It is easy to see that 



∞

∞

H(ux uxx )Huxx dx = –∞ ∞

 –∞ ∞

ux u2xx dx,

(ux Hux )x Huxx dx = uxx Hux Huxx dx + –∞ –∞  ∞  ∞ =– ux H(uxx Huxx ) dx + ux (Huxx )2 dx –∞ –∞   1 ∞ 1 ∞ = ux (Huxx )2 dx + ux u2xx dx, 2 –∞ 2 –∞ 1 1 H(uxx Huxx ) = (Huxx )2 – u2xx . 2 2



∞

ux (Huxx )2 dx

–∞

Hence, d dt



∞

–∞



– 2%  +4

 u2x Hux dx = ∞

∞

 ux (Huxx )2 dx –

–∞



∞

–∞



ux u2xx dx ∞

(ux Hux )x – H(ux uxx ) uxx dx + 4 uux (ux Hux )x dx –∞ –∞  ∞ ∞   (ux Hux )x – H(ux uxx ) (bux + cu – f ) dx. ux uxx H(uux ) dx + 2 –∞

(2.7.35)

–∞

From equality (2.7.33) to (2.7.35), we can eliminate the following two terms: 

∞ –∞

 ux u2xx dx and

∞

ux (Huxx )2 dx.

–∞

Thus, 

∞!

" 2u2xx + 5u2x Hux + 10uux Huxx dx + 4%uxxx (⋅, t)2L (R1 ) 2 –∞  ∞   uHuxxx + H(uux )xx + (ux Hux )x – H(ux uxx ) uxx dx = –10%  ∞ –∞   (6bx + 4c)u2xx + 10(bu + 2u2 )uxx Huxx + 10buxx H(uuxx ) dx – –∞  ∞   (2u + b)ux Hux – bH(u2x ) – 12ux H(uux ) uxx dx + 10

d dt

–∞

(2.7.36)

2.7 Initial Value Problem for the Nonlinear Singular Integral and Differential Equations

149

 ∞  ∞ – 20 uu2x Huxx dx – 10 bux H(ux uxx ) dx –∞  ∞–∞  ∞   10cuHux – 4(2bxx + cx ) uxx dx – 10 + (bx – c)uux Huxx dx –∞ –∞  ∞  ∞ – 10 (bx + c)ux H(uuxx ) dx – 10 cuH(ux uxx ) dx –∞  –∞ ∞ ∞  H(ux uxx ) – uxx Hux f dx – 4 + 10 (cxx u – fxx )uxx dx –∞ –∞  ∞ ∞ – 10 (cx u2 – fx u)Huxx dx – 10 (cx u + fx )H(uuxx ) dx –∞ –∞  ∞ ∞ – 10 (bx + c)ux H(u2x ) dx – 10 (cx u – fx )H(u2x ) dx. –∞

–∞

We denote Jk (k = 1, 2, . . . , 15) to be the kth integral on the right-hand side of eq. (2.7.36). For J1 , we have 

∞

J1 = 20%

(uxx uxxx Hu – uuxx Huxxx ) dx. –∞

Hence, # |J1 | ≤ 20% HuL∞ (Q∗ ) uxx (⋅, t)L2 (R1 ) uxxx (⋅, t)L2 (R1 ) T

$ + uL∞ (Q∗ ) uxx (⋅, t)L2 (R1 ) Huxxx (⋅, t)L2 (R1 ) T " ! ≤ 4%uxxx (⋅, t)2L (R1 ) + 50% u2L∞ (Q∗ ) + Hu2L∞ (Q∗ ) uxx (⋅, t)2L 2

T

1 . 2 (R )

T

On the other hand, Hu2L∞ (Q∗ ) = sup Hu(⋅, t)2L T

1 ∞ (R )

0≤t≤T

≤ C30 sup Hu(⋅, t)L2 (R1 ) Hux (⋅, t)L2 (R1 ) 0≤t≤T

≤ C30 sup u(⋅, t)L2 (R1 ) sup ux (⋅, t)L2 (R1 ) 0≤t≤T

0≤t≤T

is uniformly bounded with respect to %. For J2 , we have |J2 | ≤ 6bx + 4cL∞ (Q∗ ) uxx (⋅, t)2L

1 2 (R )

T

2

+ 10bu + 2u L∞ (Q∗ ) uxx (⋅, t)L2 (R1 ) H(uxx (⋅, t))L2 (R1 ) T

10bL∞ (Q∗ ) uxx (⋅, t)L2 (R1 ) H(u(⋅, t)uxx (⋅, t))L2 (R1 ) T

+ " ! ≤ 6bx L∞ (Q∗ ) + 4cL∞ (Q∗ ) + 20bL∞ (Q∗ ) uL∞ (Q∗ ) + 20u2L∞ (Q∗ ) uxx (⋅, t)L2 (R1 ) . T

T

T

T

T

150

2 The Vanishing Viscosity Method of Some Nonlinear Evolution System

For the remaining terms of Ji , we have ! "! " |J3 | + |J4 | + |J5 | ≤ C31 uL∞ (Q∗ ) + bL∞ (Q∗ ) uxx (⋅, t)2L (R1 ) + ux (⋅, t)6L (R1 ) , T T 2 2 " ! |J6 | + |J7 | + |J8 | + |J9 | ≤ 10bx L∞ (Q∗ ) + 20cL∞ (Q∗ ) uL∞ (Q∗ ) T T T ! " + 4bxx L∞ (Q∗ ) + 2cx L∞ (Q∗ ) uxx (⋅, t)2L (R1 ) + ux (⋅, t)2L (R1 ) , T T 2 2 ! |J11 | + |J12 | + |J13 | ≤ 12uxx (⋅, t)2L (R1 ) + 2cxx 2L∞ (Q∗ ) 2 T " 2 2 2 + 10cx L∞ (Q∗ ) uL∞ (Q∗ ) u(⋅, t)L (R1 ) T

+

2

T

10u2L∞ (Q∗ ) fx (⋅, t)2L (R1 ) 2 T

+ fxx (⋅, t)2L

1 , 2 (R )

10 ! "! " |J14 | ≤ C32 bx L∞ (Q∗ ) + cL∞ (Q∗ ) uxx (⋅, t)2L (R1 ) + ux (⋅, t)L3 (R1 ) , T T 2 2 ! " |J15 | ≤ C33 uxx (⋅, t)2L (R1 ) + ux (⋅, t)6L (R1 ) + cx 2L∞ (Q∗ ) u(⋅, t)2L (R1 ) + fx (⋅, t)2L 2

2

1 . 2 (R )

2

T

Finally, we turn to estimate the term J10 by   |J10 | = 10

"  ux uxx Hf + fuxx Hux dx

∞!

–∞

≤ 10Hf (⋅, t)L∞ (R1 ) ux (⋅, t)L2 (R1 ) uxx (⋅, t)L2 (R1 ) + 10f (⋅, t)L∞ (R1 ) Hux (⋅, t)L2 (R1 ) uxx (⋅, t)L2 (R1 ) ≤ 10uxx (⋅, t)2L

1 2 (R )

+ C34 f (⋅, t)2H 1 (R1 ) ,

where the constant C34 is dependent on sup0≤t≤T u(⋅, t)L2 (R1 ) . Plugging these inequalities into eq. (2.7.36) implies d dt



∞!

" 2u2xx + 5u2x Hux + 10uux Huxx dx

–∞

≤ C35 uxx (⋅, t)2L

1 2 (R )

+

C38 f (⋅, t)2H 1 (R1 )

+ C36 ux (⋅, t)2L

1 2 (R )

+

+ C37 u(⋅, t)2L

1 2 (R )

fxx (⋅, t)2L (R1 ) , 2

where the above constants Cij depend on the norms bW (2,0) (Q∗ ) , cW (2,0) (Q∗ ) and ∞

∞

T

T

sup0≤t≤T u(⋅, t)H 1 (R1 ) , but do not depend on % > 0. Integrating the above inequality in time t ∈ [0, t], and using Lemmas 2.7.9 and 2.7.10, we obtain 

∞! –∞

" 2u2xx + 5u2x Hux + 10uux Huxx dx



≤ C35  +

t

uxx (⋅, 4)2L

0 ∞!

–∞

1 2 (R )

# d4 + C39 >2H 1 (R1 ) + f 2

" 2>2xx + 5>2x H>x + 10»x H>xx dx,

W2(2,0) (Q∗T )

$

2.7 Initial Value Problem for the Nonlinear Singular Integral and Differential Equations

151

where the constants C35 and C39 are independent of % > 0. Similarly, the last integral term in the right hand side of the above inequality is C40 >2H 2 (R1 ) . And the left-hand side of the above inequality has the lower bound   2 2 2 uxx (⋅, t)L (R1 ) – C41 >H 1 (R1 ) + f  (2,0) ∗ , 2

(QT )

W2

where C40 and C41 depend on the norm >2H 1 (R1 ) and the norm of the coefficients in eq. (2.7.8). Hence,    t uxx (⋅, t)2L (R1 ) ≤ C35 uxx (⋅, 4)2L (R1 ) d4 + C42 >2H 2 (R1 ) + f 2 (2,0) ∗ , 2

2

0

W2

(QT )

where C42 is independent of % > 0. Therefore, we obtain the following lemma. (2,1) (Q∗ ), c(x, t) ∈ W (2,0) (Q∗ ), f (x, t) ∈ W (2,0) (Q∗ ) and Lemma 2.7.12. Let b(x, t) ∈ W∞ ∞ 2 T T T 2 1 >(x) ∈ H (R ). Then for the global generalized solution u(x, t) ∈ W2(2,1) (Q∗T ) to the initial value problem (2.7.8) and (2.7.7), there holds





sup u%xx (⋅, t)L2 (R1 ) ≤ K6 >H 2 (R1 ) + f W (2,0) (Q∗ ) 2

0≤t≤T

(2.7.37)

T

where the constant K6 depends on the norms bW (2,1) (Q∗ ) , cW (2,0) (Q∗ ) and >H 1 (R1 ) , ∞

but do not depend on % > 0.

∞

T

T

As a consequence of Lemma 2.7.12, eq. (2.7.8) and interpolation, we have the following lemmas. Lemma 2.7.13. Under the conditions of Lemma 2.7.12, for the global generalized solution u(x, t) to the initial value problem (2.7.8) and (2.7.7), there holds   sup u%t (⋅, t)L2 (R1 ) ≤ K7 >H 2 (R1 ) + f W (2,0) (Q∗ ) 2

0≤t≤T

(2.7.38)

T

where the constant K7 depends on the norms bW (2,1) (Q∗ ) , cW (2,0) (Q∗ ) and >H 1 (R1 ) , ∞

but do not depend on % > 0.

∞

T

T

Lemma 2.7.14. Under the assumptions of Lemma 2.7.12, for the global generalized solution u(x, t) to the initial value problem (2.7.8) and (2.7.7), we have   ∗ (2.7.39) u%x (⋅, t)L∞ (Q ) ≤ K8 >H 2 (R1 ) + f W (2,0) (Q∗ ) T

where the constant K8 is independent of % > 0.

2

T

152

2 The Vanishing Viscosity Method of Some Nonlinear Evolution System

Lemma 2.7.15. Under the assumptions of Lemma 2.7.12, for the global generalized solution u(x, t) to the initial value problem (2.7.8) and (2.7.7), we have |u% (¯x, t) – u% (x, t)| ≤ K9 |¯x – x|,

(2.7.40)

|u% (x, t¯) – u% (x, t)| ≤ K10 |t¯ –

3 t| 4 ,

(2.7.41)

|u%x (¯x, t) – u%x (x, t)| ≤ K11 |¯x –

1 x| 2 ,

(2.7.42)

|u%x (x, t¯) – u%x (x, t)| ≤ K12 |t¯ –

1 t| 4 ,

(2.7.43)

where x¯ , x ∈ R, t¯, t ∈ [0, T], and the constants Kij are independent of % > 0. Proof. It is easy to see that eq. (2.7.40) is a direct consequence of eq. (2.7.39). By interpolation 3

1

|u(x, t¯) – u(x, t)| ≤ Cu(⋅, t¯) – u(⋅, t)L4 (R1 ) uxx (⋅, t¯) – uxx (⋅, t)L4 (R1 ) , 2

2

and 

u(⋅, t¯) – u(⋅, t)2L

1 2 (R )

∞

=

|u(x, t¯) – u(x, t)|2 dx

–∞  ∞   t¯

= –∞

 

t

2  ut (x, 4) d4 dx

≤ |t¯ – t|2 sup ut (⋅, t)2L

1 , 2 (R )

0≤t≤T

uxx (⋅, t¯) – uxx (⋅, t)L2 (R1 ) ≤ 2 sup uxx (⋅, t)L2 (R1 ) , 0≤t≤T

we obtain eq. (2.7.41) by Lemmas 2.7.12–2.7.13. Equation (2.7.42) can be derived by   |ux (¯x, t) – ux (x, t)| ≤ 

x¯

x

 1  uxx (. , t) d.  ≤ |¯x – x| 2 uxx (⋅, t)L2 (R1 ) .

Finally, by the same argument, we obtain eq. (2.7.43) as follows: 1

3

|ux (x, t¯) – ux (x, t)| ≤ C2 u(⋅, t¯) – u(⋅, t)L4 (R1 ) uxx (⋅, t¯) – uxx (⋅, t)L4 (R1 ) 2

≤ 2C2 |t¯ –

1 t| 4

2

1 4

sup ut (⋅, t)L

0≤t≤T

2

3

(R1 )

sup uxx (⋅, t)L4 (R1 ) .

0≤t≤T

2

∎

And so Lemma 2.7.15 follows.

From the uniform estimates (2.7.40)–(2.7.43), we know that the function sequences (1, 3 )

(1,1)

{u% (x, t)} and {u%x (x, t)} are uniformly Hölder continuous in Cx,t 4 (Q∗T ) and Cx,t2 4 (Q∗T ), respectively. Thus, the sets {u% (x, t)} and {u%x (x, t)} are uniformly bounded and

2.7 Initial Value Problem for the Nonlinear Singular Integral and Differential Equations

153

equicontinuous in domain Q∗T . And so there exists a sequence {%i }, as i → ∞, % → 0, (1,0) ∗ and there exists a function u(x, t) ∈ Cx,t (QT ) such that the sequences {u%i (x, , t)} and {u%i x (x, , t)} converge uniformly to u(x, t) and ux (x, t) in any compact subset of Q∗T . Hence, u(x, t) satisfies the initial data condition (2.7.7) in the classical sense. Obviously, (1, 3 )

(1,1)

u(x, t) ∈ Cx,t 4 (Q∗T ), ux (x, t) ∈ Cx,t2 4 (Q∗T ). On the other hand, the sequence {u%i (x, , t)} is (1) (0, T; L (R1 )), and the also uniformly bounded in the space Z := L∞ (0, T; H 2 (R1 )) ∩ W∞ 2 sequences {u%i xx (x, , t)} and {u%i t (x, , t)} are uniformly bounded in L∞ (0, T; L2 (R1 ) with respect to %i . As a consequence, we know that there exists a subsequence of {%i } (we still denote by {%i }) such that the subsequences {u%i xx (x, , t)} and {u%i t (x, , t)} converge weakly to uxx and ut in L2 (Q∗T ), respectively, and also in Lp (0, T; L2 (R1 ))(2 ≤ p < ∞). Since uxx (x, t) and ut (x, t) are uniformly bounded in Lp (0, T; L2 (R1 )) with respect to 2 ≤ p < ∞, we obtain uxx (x, t), ut (x, t) ∈ L∞ (0, T; L2 (R1 )). Therefore, the limit function u(x, t) ∈ Z. Now, we prove that the limit function u(x, t) solves the singular integral– differential equation (2.7.6) in the sense of distributions. Let >(x, t) be the text function in L2 (Q∗T ). The global generalized solution u% (x, t) of the initial value problem (2.7.8) and (2.7.7) satisfies the integral identity    > u%t + 2u% u%x + Hu%xx – %u%xx + bu%x + cu% – f dx dt = 0 ∀ % > 0. Q∗T

It follows from Lemma 2.7.12 that      % >u%xx dx ≤ %>L2 (Q∗ ) u%xx L2 (Q∗ ) ≤ C3 % → 0. T T Q∗T

Using the identity  Q∗T

" ! > Hu%xx – Huxx dx dt = –

 Q∗T

" ! H> u%xx – uxx dx dt,

we deduce that the sequence {Hu%i xx } converges weakly to Huxx (x, t) in L2 (Q∗T ). Letting %i → 0 in the integral identity of u% , we obtain    > ut + 2uux + Huxx + bux + cu – f dx dt = 0. (2.7.44) Q∗T

Thus, u(x, t) ∈ Z, and it solves eq. (2.7.6) in the sense of distributions. This shows the proof of the existence of the global generalized solution the initial value problem (2.7.6) and (2.7.7). Thus, we have the following theorem. (2,1) (Q∗ ), c(x, t) Theorem 2.7.16 (Existence theorem). Let b(x, t) ∈ W∞ ∈ T (2,0) ∗ ∗ (2,0) 2 1 W∞ (QT ), f (x, t) ∈ W2 (QT ) and > ∈ H (R ). Then there exists at least one

154

2 The Vanishing Viscosity Method of Some Nonlinear Evolution System

global generalized solution u(x, t) ∈ Z to the initial value problem for the nonlinear singular integral–differential equation with Hilbert operator (2.7.6) and (2.7.7). The solution u(x, t) satisfies eq. (2.7.6) in the sense of distribution, and the initial data condition (2.7.7) in the classical sense. For the uniqueness of the global generalized solution, we have the following theorem. (1,0) (Q∗ ), c(x, t) ∈ L (Q∗ ). Then Theorem 2.7.17 (Uniqueness theorem). Let b(x, t) ∈ W∞ ∞ T T the global generalized solution u(x, t) ∈ Z of the initial value problem (2.7.6) and (2.7.7) is unique.

Proof. Let u(x, t), v(x, t) ∈ Z be two global generalized solutions of the initial value problem (2.7.6) and (2.7.7). Writing w(x, t) = u(x, t) – v(x, t), it solves the homogenous linear equation wt = Hwxx + (u + v + b)wx + (ux + vx + c)w = 0, with zero initial data w(x, 0) = 0. Hence, d w(⋅, t)2L (R1 ) + 2 dt



∞

(3ux + 3vx + 2c – bx )w2 dx = 0.

–∞

Since the above integral coefficient is bounded, we get w(⋅, t)L2 (R1 ) = 0 ∀ t ∈ [0, T]. This shows the uniqueness of the global generalized solution. ∎ Now, we turn to discuss the convergence and the rate of convergence for the generalized solution to the nonlinear equation (2.7.8) involving small parameter % > 0 with initial data (2.7.7). We have known that the global generalized solution {u% (x, t)} of the initial value problem (2.7.8) and (2.7.7) is uniformly bounded in Z = L∞ (0, T; H 2 (R1 )) ∩ (1) (0, T; L (R1 )). Then there exists some sequence % → 0 such that {u (x, t)} conW∞ 2 %i i verges weak ∗ to the generalized solution u(x, t) ∈ Z of the initial value problem (2.7.6) and (2.7.7) in space Z, and {u%i (x, t)} and {u%i x (x, t)} converge uniformly to u(x, t) and ux (x, t) in any compact subset of Q∗T , respectively. {u%i xx (x, t)} and {u%i t (x, t)} converge weakly to uxx (x, t) and ut (x, t) in Lp (0, T; L2 (R1 ))(2 ≤ p < ∞). Since the generalized solution of the initial value problem (2.7.8) and (2.7.7) is unique, the above convergence is not only in terms of a defined sequence, but also for any sequence %i → 0. Therefore, we can replace %i → 0 by % → 0.

2.7 Initial Value Problem for the Nonlinear Singular Integral and Differential Equations

155

Theorem 2.7.18. Suppose that the assumptions of Theorem 2.7.16 hold true. As % → 0, the global generalized solution u% (x, t) ∈ W2(2,1) (Q∗T ) of the initial value problem (2.7.8) and (2.7.7) converges to the unique global generalized solution u(x, t) ∈ Z of initial value problem (2.7.6) and (2.7.7). Moreover, {u% (x, t)} and {u%x (x, t)} converge uniformly to u(x, t) and ux (x, t) in any compact subset of Q∗T , and {u%xx (x, t)} and {u%t (x, t)} converge weakly to uxx (x, t) and ut (x, t) in Lp (0, T; L2 (R1 ))(2 ≤ p < ∞), respectively. Theorem 2.7.19. Under the conditions of Theorem 2.7.16, the approximate order which the global generalized solution u% (x, t) ∈ W2(2,1) (Q∗T ) of the initial value problem (2.7.8) and (2.7.7) converges to the unique global generalized solution u(x, t) ∈ Z of initial value problem (2.7.6) and (2.7.7) is sup u% (⋅, t) – u(⋅, t)L2 (R1 ) ≤ K13 %,

(2.7.45)

0≤t≤T

3

u% – uL∞ (Q∗ ) ≤ K14 % 4 ,

(2.7.46)

1 ux (⋅, t)L2 (R1 ) ≤ K15 % 2

(2.7.47)

T

sup u%x (⋅, t) –

0≤t≤T

and 1

u%x – ux L∞ (Q∗ ) ≤ K16 % 4 , T

where the above constants K are independent of % > 0. Proof. Writing w(x, t) = u% (x, t) – u(x, t), it solves the linear equation wt + Hwxx + (u% + u + b)wx + (u%x + ux + c)w = %u%xx , with initial data w(x, 0) = 0. Multiplying the above equation by w, and then integrating in x ∈ R1 , we get d w(⋅, t)2L (R1 ) ≤ C1 w(⋅, t)2L (R1 ) + %2 u%xx (⋅, t)2L (R1 ) . 2 2 2 dt This implies sup w(⋅, t)L2 (R1 ) ≤ C%.

0≤t≤T

(2.7.48)

156

2 The Vanishing Viscosity Method of Some Nonlinear Evolution System

And so eq. (2.7.45) follows. Estimates (2.7.46), (2.7.47) and (2.7.48) follow from the following interpolation: 3

1

w(⋅, t)L∞ (R1 ) ≤ C3 w(⋅, t)L4 (R1 ) wxx (⋅, t)L4 (R1 ) ,

2 2 1 1 2 2 wx (⋅, t)L2 (R1 ) ≤ C4 w(⋅, t)L (R1 ) wxx (⋅, t)L (R1 ) , 2 2 1 3 wx (⋅, t)L∞ (R1 ) ≤ C5 w(⋅, t)L4 (R1 ) wxx (⋅, t)L4 (R1 ) . 2 2

∎ Since T is arbitrary given constant in the above argument, we can extend the above results to the unbounded domain R+ = [0, ∞). We have the following theorem. Theorem 2.7.20. Suppose that the assumptions of Theorem 2.7.16 hold true for any T > 0. Then there is a unique global generalized solution u(x, t) ∈ L∞,loc (R+ , H 2 (R1 )) ∩ (1) W∞,loc (R+ ; L2 (R1 )) to the initial value problem (2.7.6) and (2.7.7). If the coefficient b(x, t), c(x, t), the free term f (x, t) and initial function >(x) have more regularity, then the regularity of the generalized solution to the initial value problem (2.7.6) and (2.7.7) can be correspondingly improved. If we assume that b(x, t) ∈ (M,1) (Q∗ ), c(x, t) ∈ W (M,0) (Q∗ ), f (x, t) ∈ W (M,0) (Q∗ ) and > ∈ H M (R1 ), M ≥ 2, then, W∞ ∞ 2 T T T by common method, we can derive that the global smooth solution of the initial value problem (2.7.6) and (2.7.7) belong to L∞ (0, T; H M (R1 )) from the uniform boundedness of the global solution {u% (x, t)} of the initial value problem (2.7.8) and (2.7.7) in L∞ (0, T; H M (R1 )). In fact, we have by eq. (2.7.8)  ∞  d ∞ 2 uxM dx = (–1)M 2 ux2M ut dx dt –∞ –∞  ∞ ! " = (–1)M+1 2 ux2M 2uux + Huxx – %uxx + bux + cu – f dx –∞ ∞ M+1 = (–1) 4 uux ux2M dx – 2%ux2M+1 (⋅, t)2L (R1 ) 2 –∞  ∞  ∞ + (–1)M+1 2 bux ux2M dx – 2 uxM (cu – f )uxM dx. –∞

By integration by parts, we get   ∞ uux ux2M dx = (–1)M –∞

= (–1)M 

M–1     M k=2

∞ –∞

k

∞

–∞

∞

–∞

(uux )xM uxM dx

! " uxk uxM+1–k uxM dx + (–1)M M + 21

–∞

 M

bux ux2M dx = (–1)

∞

–∞

(bux )xM uxM dx



∞

–∞

ux u2xM dx,

2.7 Initial Value Problem for the Nonlinear Singular Integral and Differential Equations

M    M

= (–1)M

k

k=2

∞

–∞

! " bxk uxM+1–k uxM dx + (–1)M M – 21



∞

–∞

157

bx u2xM dx.

Hence,   ∞   uux ux2M dx ≤ C1 uxM (⋅, t)2L (R1 ) + C2 uxx (⋅, t)2L (R1 ) ,  2 2 –∞  ∞    bux ux2M dx ≤ C3 uxM (⋅, t)2L (R1 ) + C4 ux (⋅, t)2L (R1 ) ,  2 2 –∞

where the constants C3 and C4 depend on bW (M,0) (Q∗ ) . Similarly, ∞

  

∞

–∞

  

T

  (cu)xM uxM dx ≤ C5 uxM (⋅, t)2L

1 2 (R )

+ C6 u(⋅, t)2L

1 , 2 (R )

 1 1  uxM fxM dx ≤ uxM (⋅, t)2L (R1 ) + fxM (⋅, t)2L (R1 ) , 2 2 2 2 –∞ ∞

where the constants C5 and C6 depend on cW (M,0) (Q∗ ) . Thus, ∞

T

# $ d uxM (⋅, t)2L (R1 ) ≤ C7 uxM (⋅, t)2L (R1 ) + C8 >2H 2 (R1 ) + f 2H M (R1 ) , 2 2 dt where the constants C7 and C8 depend on bW (M,0) (Q∗ ) and cW (M,0) (Q∗ ) , but not on ∞

T

∞

T

% > 0. This shows that sup0≤t≤T uxM (⋅, t)L2 (R1 ) is uniformly bounded with respect to %. Therefore, we obtain the regular theorem for the solution to the initial value problem of Benjamin–Ono equation (2.7.4) with initial data (2.7.7). Theorem 2.7.21. Let > ∈ H M (R1 ), M ≥ 2. Then there is a unique global smooth solution u(x, t) to the initial value problem of Benjamin–Ono equation (2.7.4) and (2.7.7). Moreover, [M 2 ]

u(x, t) ∈

-

(k) W∞,loc (R+ ; H M–2k (R1 )), uxr ts (x, t) ∈ L∞,loc (R+ ; L2 (R1 )), 0 ≤ 2s + r ≤ M.

k=0

For the initial value problem of the more general nonlinear singular integral– differential equation ut – auxx + !Huxx + "Hux + #Hu + >(u)x + bux + cu = f ,

(2.7.49)

u(x, 0) = u0 (x),

(2.7.50)

with initial data

158

2 The Vanishing Viscosity Method of Some Nonlinear Evolution System

we can obtain the global smooth solution by the priori estimates and Leray–Schauder’s fixed point principle by the same argument as above. Here, a > 0, ! is a constant, "(x, t), #(x, t), b(x, t), c(x, t) and f (x, t) are known functions, and H is Hilbert operator. Then, we have the following theorems. Theorem 2.7.22. Suppose that (1) b(x, t), c(x, t), "(x, t), #(x, t) ∈ L∞ (Q∗T ); (2) f (x, t) ∈ L2 (Q∗T ), u0 (x) ∈ H 1 (R1 ); (3) >(u) ∈ C2 and |>′ (u)| ≤ c(1 + u2 ), c = const > 0. Then there is a unique global generalized solution u(x, t) ∈ W2(2,1) (Q∗T ) to the initial value problem of eqs (2.7.49) and (2.7.50) satisfying " ! sup u(⋅, t)2H 1 (R1 ) + uxx 2L2 (Q∗ ) + ut 2L2 (Q∗ ) ≤ C u0 2H 1 (R1 ) + f 2L2 (Q∗ ) . T

0≤t≤T

T

T

(2.7.51)

Theorem 2.7.23. Suppose that (k,[ k2 ])

(1) b(x, t), c(x, t), "(x, t), #(x, t) ∈ W∞ (k,[ k2 ])

(2) f (x, t) ∈ W2

(R2+ );

(R2+ ), u0 (x) ∈ H k+2 (R1 );

(3) >(u) ∈ Ck+2 and |>′ (u)| ≤ c(1 + u2 ). Then there is a unique global classical solution u(x, t) to the initial value problem of eqs (2.7.49) and (2.7.50) satisfying u(x, t) ∈

-

! " s (0, ∞); H r (R) . Hloc

r+2s≤k+2

Finally, we have the asymptotic behavior for the solution to the initial value problem of the following equation: ut – auxx + !Huxx + >(u)x = 0, u(x, 0) = u0 (x),

(2.7.52) (2.7.53)

as follows. Theorem 2.7.24. Let u0 (x) ∈ H k+2 (R1 ), >(u) ∈ Ck+2 (R1 ). Then the solution u(x, t) of the initial value problem (2.7.52) and (2.7.53) enjoys the following asymptotic behavior: lim uxl (⋅, t)L∞ (R1 ) = 0,

t→∞

l = 0, 1, . . . , k + 2.

(2.7.54)

2.8 Initial Value Problem for the Nonlinear Schrödinger Equations

159

2.8 Initial Value Problem for the Nonlinear Schrödinger Equations We consider the nonlinear Klein–Gordon equation of nonrelativistic approximation h2 h2 mc2 8tt – B8 + 8 + +|8|p–1 8 = 0, x ∈ Rd , t ∈ R, 2 2mc 2m 2

(2.8.1)

where h is Plank’s constant, m is the mass of particle and c is the speed of light, + ∈ R. Let 8 = 6eimc

2 t/h

.

Inserting it into eq. (2.8.1) gives h2 h2 6tt + ih6t – B6 + +|6|p–1 6 = 0. 2 2mc 2m

(2.8.1′ )

Letting c → ∞, one can obtain the standard nonlinear Schrödinger equation: ih6t –

h2 B6 + +|6|p–1 6 = 0. 2m

(2.8.2)

In this section, our goal is to show that the solution of eq. (2.8.1′ ) converges to the solution of eq. (2.8.2) in some sense. For convenience, let h = 1, m = 21 , % = c1 . Then we can get the following singular perturbation problem: %2 6tt + i6t – B6 + +|6|p–1 6 = 0, x ∈ Rd , t ∈ R,

(2.8.3)

d

(2.8.4)

d

(2.8.5)

6(x, 0) = 60% (x), x ∈ R , 6t (x, 0) = 61% (x), x ∈ R . Writing t

u(x, t) = 6(%x, %2 t)ei 2 with 6 being the solution of eq. (2.8.3), we have utt – Bu +

1 u + +%2 |u|p–1 u = 0. 4

On the contrary, if u solves eq. (2.8.6), then  x t  it – , 2 e 2%2 % %

6(x, t) = lim u %→0

satisfies the nonlinear Schrödinger equation i6t – B6 + +|6|p–1 6 = 0. We will show the following theorem.

(2.8.6)

160

2 The Vanishing Viscosity Method of Some Nonlinear Evolution System

Theorem 2.8.1. Let d = 2, 2 ≤ p ≤ 3, + > 0, and 60% ∈ H 2 (R2 ), 61% ∈ H 1 (R2 ), 60 ∈ H 2 (R2 ) ( ( lim 60% – 60 H 2 (R2 ) = 0 and lim (i61% – B60% + +|60% |p–1 60% (L

%→0

2 2 (R )

%→0

= 0.

Let >% (x, t) be the solution to the initial value problem of eqs (2.8.3)–(2.8.5), and 6(x, t) be the solution of initial value problem 

i6t – B6 + +|6|p–1 6 = 0, x ∈ R2 , t ∈ R, 6(x, 0) = 60 (x), x ∈ R2 .

(2.8.7)

Then, we have lim 6% (⋅, t) – 6(⋅, t)L2 (R2 ) = 0.

%→0

Proof. For convenience, we take + = 1. From the assumptions of Theorem 2.8.1 and known results, we know that there is a unique solution 6% (x, t) to problem (2.8.3)– (2.8.5) satisfying  d k dt

6% ∈ L∞ (0, T; H 2–2k (R2 )), k = 0, 1.

And there is a unique solution 6(x, t) to problem (2.8.7):  d k dt

6 ∈ L∞ (0, T; H 2–2k (R2 )), k = 0, 1.

¯ t , and then integrating in x ∈ R2 , we obtain by taking the Multiplying eq. (2.8.3) by 6 real part %2

d dt

 R2

|6%t |2 dx +

d dt

 R2

|∇6% |2 dx +

2 d p + 1 dt

 R2

|6% |p+1 dx = 0.

This implies %6%t L∞ (0,T;L2 (R2 )) ≤ C,

(2.8.8)

∇6% L∞ (0,T;L2 (R2 )) ≤ C,

(2.8.9)

6% L∞ (0,T;Lp+1 (R2 )) ≤ C,

(2.8.10)

where the constant is independent of % ∈ (0, %0 )(%0 > 0). ¯ t and 6, respectively, and minus each other, Multiplying eqs (2.8.3) and (2.8.3) by 6 and then integrating in x, we obtain d % dt



2

R2

! " ¯ % 6%t – 6% 6 ¯ %t dx + i d |6% |2 dx = 0. 6 dt

161

2.8 Initial Value Problem for the Nonlinear Schrödinger Equations

Hence 6% L∞ (0,T;L2 (R2 )) ≤ C.

(2.8.11)

For convenience, we ignore the subscript % of 6% temporarily. Multiplying eq. (2.8.3) by ¯ t , and integrating in x ∈ R2 , and integrating by part, we estimate by taking real part B6 %2



d dt

R2





d dt

|∇6t |2 dx +

R2

|B6|2 dx =

 R2

 ¯ t B(|6|p–1 6) + 6t B(|6|p–1 6) ¯ dx. 6

The right-hand side can be controlled by 



 ¯ t B(|6|p–1 6) + 6t B(|6|p–1 6) ¯ dx 6   |6|p–1 |B6| ⋅ |6t | dx + 2p(p – 1) ≤ 2p R2

R2

p–1

≤ C6L∞

p–1 ≤ C6L∞

 

R2

|B6|2 dx 2

R2

|B6| dx

1  2

1 

R2

2

R2

R2

|6t |2 dx |6t |2 dx

|6|p–2 |∇6|2 |6t | dx

1

2

p–2



+ C6L∞

R2

1

2

|∇6|4 dx

1  2

R2

|6t |2 dx

1

2

,

where we used the Sobolev inequality 1

∇6L4 (R2 ) ≤ CB6L2

1

2 2 (R )

6L2

2 . ∞ (R )

Thus, d % dt 2



d |∇6t | dx + 2 dt R



2

2

R2

|B6| dx ≤

p–1 6L∞



2

R2

|B6| dx

1  2

R2

|6t |2 dx

1

2

.

(2.8.12)

Making derivation of eq. (2.8.3) in t gives %2 6ttt + i6tt – B6t + (|6|p–1 6)t = 0.

(2.8.13)

¯ tt , and integrating in x ∈ R, and integrating by part, we Multiplying eq. (2.8.3) by 6 obtain by taking the real part    d 3–p d d |6tt |2 dx + |∇6t |2 dx + |6|p–1 |6t |2 dx dt R2 dt R2 2 dt R2  p–1 d ¯ t + 66 ¯ t |2 dx |6|p–3 |66 + 4 dt R2  |6|p–2 |6t |3 dx, ≤C %2

R2

(2.8.14)

162

2 The Vanishing Viscosity Method of Some Nonlinear Evolution System

where we used the following identity: ¯ tt + (|6|p–1 6) ¯ t 6tt (|6|p–1 6)t 6 p – 1 p–3 2 ¯ ¯ tt ¯ tt + |6|p–1 6t 6 = |6| (6 6t + |6|2 6t )6 2 p – 1 p–3 ¯t +6 ¯ t 6tt ¯ 2 6t )6tt + |6|p–1 6 + |6| (|6|2 6 2  p – 1 p–3 d  ¯ ¯ t 2 + 3 – p |6|2 d |6t |2 = |6| 66t + 66 4 dt 2 dt ! " ¯ t + 6|6 ¯ t |2 6t . – (p – 1)|6|p–3 6|6t |2 6 By Sobolev inequality, eq. (2.8.8), we control the right-hand side of eq. (2.8.14) by p–2 C6L∞ p–2

≤ C6L∞

 |6t |2 dx

R2



R2

 C p–2 6L∞ %2 %

≤

|∇6t |2 dx

1  2

R2

 R2

|∇6t |2 dx

|6t |2 dx

1  2

R2

1

2

|6t |2 dx

1

2

.

Thus,    3–p 2 d d d |6tt |2 dx + %2 |∇6t |2 dx + |6|p–1 |6t |2 dx % dt R2 dt R2 2 dt R2  p–1 2 d ¯ t + 66 ¯ t |2 dx |6|p–3 |66 + % 4 dt R2   1  1 2 2 p–2 |∇6t |2 dx |6t |2 dx . ≤ C6L∞ %2 %4

R2

(2.8.15)

R2

¯ t , and integrating in x ∈ R2 , and integrating by part, we Multiplying eq. (2.8.13) by 6 get by taking the imaginary part 2%2

d Im dt

 R2

¯ t dx + 6tt 6

d dt



p–1

R2

|6t |2 dx ≤ C6L∞

 R2

|6t |2 dx.

(2.8.16)

Combining eqs (2.8.12), (2.8.15) and (2.8.16), we deduce that 

 ¯ t + 9%2 |∇6t |2 + |6t |2 + |B6|2 dx 8%4 |6tt |2 + 2%2 Im6tt 6 R2    d ¯ t + 66 ¯ t |2 dx 4(3 – p)|6|p–1 |6t |2 + 2(p – 1)|6|p–3 |66 + %2 2 dt R 1  1   ! 2 2 p–1 " 2 ≤ C 1 + 6L∞ |B6| dx |6t |2 dx d dt



  + %2

R2

2

R2

|∇6t | dx

R2

1  2

2

R2

|6t | dx



1

2

+ R2

 |6t |2 dx .

(2.8.17)

2.8 Initial Value Problem for the Nonlinear Schrödinger Equations

163

On the other hand, it follows from the assumptions of Theorem 2.8.1 and eq. (2.8.3) that ( ( %2 6%tt L2 ≤ 61% L2 + B60% L2 + (|60% |p–1 60% (L ≤ C, 2

(2.8.18)

where we used the fact that ( ( (|60% |p–1 60% (

p

L2

p

= 60% L2p ≤ 60% H 1 ≤ C.

Integrating eq. (2.8.17) in t ∈ (0, t), we obtain by eq. (2.8.8)   4  ¯ t + 9%2 |∇6t |2 + |6t |2 + |B6|2 dx 8% |6tt |2 + 2%2 Im6tt 6 R2    ¯ t + 66 ¯ t |2 dx 4(3 – p)|6|p–1 |6t |2 + 2(p – 1)|6|p–3 |66 + %2 R2 t!



 p–1 " 1 + 6(s)L∞ B6(s)L2 6t (s)L2 0  2 + % ∇6t (s)L2 6t (s)L2 + 6t (s)2L2 ds.

≤C + C

(2.8.19)

Denote E(t) by E(t) := 6t (s)2L2 + %2 ∇6t (t)2L2 + B6(t)2L2 + %4 ∇6tt (t)2L2 . Since   ¯ t  ≤ 2%4 |6tt |2 + 1 |6t |2 , 2%2 Im6tt 6 2 the left-hand side of eq. (2.8.19) has the lower bound   4  1 1 6% |6tt |2 + |6t |2 + 9%2 |∇6t |2 + |B6|2 dx ≥ E(t). 2 2 R2 This together with eq. (2.8.19) yields that  t ! p–1 " 1 + 6(s)L∞ E(s) ds. E(t) ≤ C + C

(2.8.20)

0

By Brezis–Gallouet’s inequality ! "1 uL∞ ≤ C 1 + log(1 + BuL2 ) 2 , uH 1 ≤ C we get  E(t) ≤ C + C

t

!

1

" p–2

1 + log(1 + E(s) 2 )

0

2

E(s) ds.

(2.8.21)

164

2 The Vanishing Viscosity Method of Some Nonlinear Evolution System

When 1 ≤ p ≤ 3, we have sup E(t) ≤ C.

0≤t≤T

And so %2 6%tt L∞ (0,T;L2 (R2 )) ≤ C,

(2.8.22)

6%t L∞ (0,T;L2 (R2 )) ≤ C,

(2.8.23)

%B6%t L∞ (0,T;L2 (R2 )) ≤ C,

(2.8.24)

B6% L∞ (0,T;L2 (R2 )) ≤ C.

(2.8.25)

Using eqs (2.8.11) and (2.8.25), we obtain 6% L∞ (0,T;L∞ (R2 )) ≤ C.

(2.8.26)

From eqs (2.8.3) and (2.8.7), we know that %2 6%tt + i(6% – 6)t – B(6% – 6) + |6% |p–1 ⋅ 6% – |6|p–1 6 = 0. Multiplying the above equation by 6% – 6, and integrating in x ∈ R2 , and integrating by part, we get by taking the imaginary part d 6% (t)–6(t)2L2 +2%2 Im dt



! " 2 ¯ % –6 ¯ d 6% dx = –2Im 6 dt2 R2



! R2

"! " ¯ % –6 ¯ |6% |p–1 ⋅6% –|6|p–1 6 dx. 6

Hence,  ! " d d ¯% –6 ¯ 6%t dx 6 6% (t) – 6(t)2L2 + 2%2 Im dt dt R2   ! " ! "! " ¯ %t – 6 ¯% –6 ¯ t 6%t dx – 2Im ¯ |6% |p–1 ⋅ 6% – |6|p–1 6 dx 6 6 = 2%2 Im !

R2

"

R2

p–1 p–1 " ≤ 2% 6%t L2 + 6t L2 6%t L2 + 2 6% L∞ + 6L∞ 6% – 62L2 2

≤ C%2 + C6% (t) – 6(t)2L2 ,

!

0 ≤ t ≤ T.

Integrating the above inequality in t, and using eqs (2.8.11) and (2.8.23), we estimate  6% (t) – 6(t)2L2 ≤ 60% – 60 2L2 + C%2 + C

0

t

6% (s) – 6(s)2L2 ds.

Thus, 6% (t) – 6(t)2L2 ≤ C60% – 60 2L2 + C%2 ,

0 ≤ t ≤ T,

2.9 Initial Value Problem and Boundary Value Problem for the Nonlinear Schrödinger

which tends uniformly to zero as % → 0 with respect to t ∈ [0, T]. Therefore, we complete the proof of Theorem 2.8.1.

165

∎

Corollary 2.8.2. The above result also holds true for the initial value problem and homogenous boundary problem of eqs (2.8.3)–(2.8.5) and (2.8.7).

2.9 Initial Value Problem and Boundary Value Problem for the Nonlinear Schrödinger Equation with Derivative The nonlinear Schrödinger equation with derivative ut = iuxx + (|u|2 u)x + g(x, t),

(x, t) ∈ R2

(2.9.1)

has been proposed in many problems of plasma physics. And there is a lot of work on its infinite conservation law and inverse scattering method. In this section, we study the initial value problem and boundary value problem for eqs (2.9.1), (2.9.2) or (2.9.2)′ , (2.9.3): ∂ku (x, t) → 0, as |x| → ∞, k = 0, 1, . . . , ∂xk ∂ku ∂ku (x + 1, t) = (x, t), ∀ (x, t) ∈ R2 , k = 0, 1, 2, . . . , ∂xk ∂xk u(x, 0) =u0 (x), x ∈ R.

(2.9.2) (2.9.2′ ) (2.9.3)

We argue by the vanishing viscosity method. That is, we utilize the existence and uniqueness of the weak, strong and smooth solution to the parabolic regularity equation u%t = (i + %)u%xx + (|u% |2 u% )x + g(x, t), % > 0

(2.9.4)

to approximate eq. (2.9.1). Since the conservation function is not regular in eq. (2.9.1) with g(x, t) ≡ 0, we need the sufficient smallness of initial data and force to get the global solution. Lemma 2.9.1. Let (m, j) ∈ Z2 : 0 ≤ j ≤ m, 0 ≤ a < 1, 1 ≤ p < ∞, and 1  1 1 = a – m + j + (1 – a) 2 . p 2

(2.9.5)

Then for each v ∈ H m , there exists a constant C1 = C1 (p, m, j) independent of v such that # $a d Dj vLp ≤ C1 Dm vL2 + vL2 v1–a . L2 , D = dx

(2.9.6)

166

2 The Vanishing Viscosity Method of Some Nonlinear Evolution System

! Proof. Since H s ⊂ Lp s =

1 2

–

1 p

" ≥ 0 , we have

Dj vLp ≤ CDj vH s ≤ CvH am .

(2.9.7)

This together with the interpolation [H m , H 0 ]1–a = H am ,

H 0 = L2 ,

(2.9.8)

implies that Dj vLp ≤ CvH am ≤ CvaH m v1–a L2 .

(2.9.9)

And so eq. (2.9.6) follows, since the norm vH m is equivalent to the norm Dm vL2 + vL2 . ∎ Lemma 2.9.2. For any v ∈ H 1 , there exists a constant C2 independent of v such that 1 ! "1 vL∞ (R) ≤ C2 DvL2 + vL2 2 vL22 .

(2.9.10)

Lemma 2.9.3. Let s, r ∈ R : s > 1, r > 21 , u ∈ H max(s,r+1) , v ∈ H max(s,r) . There holds " ! uvH s ≤ C(s, r) uH s vH r + uH r vH s + uH 1+r vH s–1 ,

(2.9.11)

where the constant C(s, r) is independent of u and v. Proof. Since the proof for the periodic function is the discrete form of the function in R, we only need to prove the case in R. For u, v ∈ >(R), we have  uvH s =  =

∞

  (1 + |. |2 )s 

–∞ ∞ ∞ –∞

 

–∞

∞

–∞

2  1 2  ˆ – ')ˆv(') d' d. u(.

2  1 s 2  ˆ – ')ˆv(') d' d. . (1 + |. |2 ) 2 u(.

Noting that     (1 + |. |2 ) 2s – (1 + |'|2 ) 2s  = 

 2 " s ∂ !   1 + ' + ((. – ') 2 d( 0 ∂(  s–1 1 s ≤ C(s) (1 + |'|2 ) 2 (1 + |. – '|2 ) 2 + (1 + |. – '|2 ) 2 , 1

2.9 Initial Value Problem and Boundary Value Problem for the Nonlinear Schrödinger

we get by the Hausdorff–Young inequality  ∞  ∞  ˆ )| d. uvH s ≤ C(s) uH s |ˆv(. )| d. + vH s |u(. –∞ –∞  ∞  1 ˆ )| d. . + vH s–1 (1 + |. |2 ) 2 |u(.

167

(2.9.12)

–∞

Using the Cauchy–Schwartz inequality, we obtain  ∞  ∞ r r 1 ˆ ˆ )|(1 + |. |2 )– 2 d. ≤ C(r)uH r , r > . |u(. )| d. = (1 + |. |2 ) 2 |u(. 2 –∞ –∞

(2.9.13) ∎

And so, we obtain the desired estimate (2.9.11). Corollary 2.9.4. Let s > 32 . Then for u, v ∈ H s , there holds uvH s ≤ C(s)uH s vH s .

Thus, H s is a Banach algebra. ∂2 2 Writing A := – ∂x 2 , its domain D(A) = H . The operator A is nonnegative selfs conjugate in L2 . We define A by As u = F –1 (|. |2s F u), u ∈ H 2s (R), s ≥ 0, As u =

∞ 

|k|2s uk e2i0kx , u ∈ H 2s (T),

(2.9.14) (2.9.15)

k=–∞

where F denotes Fourier transformation, H 2s (T) denotes the complex s-order Sobolev space with periodic 1: u(x) =

∞ 

uk e20ikx ,

k=–∞

with 

∞ 

,1

2

2 s

(1 + |k| ) |uk |

2

< +∞.

k=–∞

In view of definition, we have the following lemmas. Lemma 2.9.5. (1) As1 +s2 u = As1 As2 u, u ∈ H 2(s1 +s2 ) (s1 , s2 ≥ 0). (2) As Du = DAs u, u ∈ H 2(s+1) . "1 ! (3) The norm [u]s = As/2 u2L2 + u2L2 2 is equivalent to the norm uH s . Lemma 2.9.6. Let r, s ∈ R : s > 21 , r > 21 . Then for u, v ∈ D(As )(= H 2s ) ∩ H r+1 , we have " ! As (uv) – uAs vL2 ≤ C(s, r) uH 2s vH r + uH r+1 vH 2s–1 .

(2.9.16)

168

2 The Vanishing Viscosity Method of Some Nonlinear Evolution System

Lemma 2.9.7. Suppose that the functions 3(t) and 6(t) are nonnegative measured, t ∈ [0, T](T > 0), and 6(t)3(t) is integrable in [0, T], and a is a nonnegative constant. If  t 6(s)3(s) ds, 0 ≤ t ≤ T, 32 (t) ≤ a2 + 2 0

then we have

 3(t) ≤ a +

t

6(s) ds,

0 ≤ t ≤ T.

0 2

∂ If we let A = – ∂x 2 , then the operator iA is skew-self-conjugate in L2 (R)(L2 (T)). It is the generation of strong continuous U-group with simple parameter. Writing

J0 u =

∂ (|u|2 u), ∂x

u ∈ H1,

then J0 is a nonlinear operator from H 1 to L2 . We can rewrite eqs (2.9.1) and (2.9.3) as ut = – iAu + J0 u + g, u(0) = u0 .

(2.9.17) (2.9.18)

Since the nonlinear term J0 u has large singularity, one cannot directly apply the Segal theory of nonlinear semi-group to eqs (2.9.17) and (2.9.18). This mainly comes from the fact that the operator J0 (u) is not continuous from H s to H s (s ≥ 1). First, we consider the regular parabolic equation of the form ut = – A% u + Ju + g, u(0) = u0 ,

(2.9.19) (2.9.20)

where A% u = (i + %)Au + u, % > 0, Ju = J0 u – u.

(2.9.21) (2.9.22)

Obviously, D(A% ) = D(A). It is easy to see that A% is an m-sector operator with ver! " tex zero and half-size 02 – tan–1 % in view of Kato’s definition. Hence, A% forms holomorphic semi-group. Similarly, we can define A!% by   (2.9.23) A!% = F –1 P {(i + %). 2 + 1}! F u, with D(A!% ) = H 2! (R) = D(A! ) or A!% u =

∞ 

  uk P {(i + %)k2 + 1}! e2i0kx ,

k=–∞

with D(A!% ) = H 2! (T) = D(A! ), P(z! ) is the principal value of z! .

(2.9.24)

2.9 Initial Value Problem and Boundary Value Problem for the Nonlinear Schrödinger

Suppose that J satisfies: there exist !, ", 0 ≤ ! < 1, ! ≤ " ≤ 1 such that " ! " " " Ju – JvL2 ≤ h A% uL2 + A% vL2 A!% u – A!% vL2 ∀ u, v ∈ D(A% ),

169

(2.9.25)

where h(s) is a positive and nondecreasing known function, and s ∈ [0, ∞). Then from the known theory for the semilinear parabolic equations, we can obtain the local existence theorem for the initial value problem (2.9.19) and (2.9.20). Theorem 2.9.8. Let 21 ≤ ! < 1, T > 0, g(⋅, t)L2 ∈ Ct1–! ([0, T]) and u0 ∈ D(A!% ). Then for % > 0, there is a unique local solution u(x, t) to the initial value problem (2.9.19) and (2.9.20) satisfying u ∈ C(0, T0 ; D(A!% )) ∩ C1 (0, T0 ; L2 ) ∩ C(0, T0 ; D(A% )),

(2.9.26)

where T0 (0 < T0 ≤ T) is some constant depending on A!% u0 L2 . Proof. It suffices to verify that J satisfies condition (2.9.25). By Lemma 2.9.2 and uH 1 ≤ CA!% uL2 ,

1 !≥ , 2

we derive that ( ( ( ( ( ( Ju – JvL2 ≤ (u2 (u¯ x – v¯ x )(L + (v¯ x (u + v)(u – v)(L + 2(|u|2 (ux – vx )(L 2 2 2 ( ( 2 2 ( ( + 2 (|u| – |v| )vx L + u – vL2 2 " ! ≤ h uH 1 + vH 1 u – vH 1 " ! " " ≤ h A% uL2 + A% vL2 A!% u – A!% vL2 , which implies eq. (2.9.25). ∎ Remark 2.9.9. T0 = T0 (A!% u0 L2 ) is a decreasing function of A!% u0 L2 . Lemma 2.9.10. Let w(t) ∈ D(A!% ), A!% w(t) ∈ C([0, T]), ! > 0. Set  v(t) =

t

e–(t–4)A% w(z) d4.

(2.9.27)

0 !+"

Then for any " ∈ (0, 1), we have v(t) ∈ D(A% ), t ∈ [0, T], and !+"

A% v(t) = !+"



t 0

Moreover, A% v(t) ∈ C1–! ([0, T]).

"

A% e–(t–4)A% A!% w(4) d4, 0 ≤ t ≤ T.

(2.9.28)

170

2 The Vanishing Viscosity Method of Some Nonlinear Evolution System

Theorem 2.9.11. Let g ∈ C∞ (0, T; D(A∞ )), u0 ∈ D(A∞ ). Then for the solution to the initial value problem (2.9.19) and (2.9.20), there holds u(x, t) ∈ C∞ (0, T0 ; D(A∞ )), with D(A∞ ) =

2 !

(2.9.29)

D(A! ).

Proof. It follows from eqs (2.9.19) and (2.9.20) that 



t

u(t) = e–A% t u0 +

e–(t–4)A% Ju(4) d4 +

0

t

e–(t–4)A% g(4) d4.

(2.9.30)

0

By the assumptions on u0 and g, we know that the first term and the third term on the right-hand side of eq. (2.9.30) belong to C∞ (0, T; D(A∞ )). For u ∈ C(0, T0 ; D(A!% )), if we take

3 4

!– 21

< ! < 1, then we get Ju(t) ∈ D(A%

claim that

!– 1 A% 2 Ju(t)

) by Lemma 2.9.3 and Corollary 2.9.4. We

∈ C([0, T0 ]). In fact,

( !– 21 ! "( (A% Ju(t) – Ju(t′ ) ( L2 ( ! 2 ( ( !– 1 ( ! ≤ (A% (|u| u)(t) – A% (|u|2 u)(t′ )(L + (A% 2 (u(t) – u(t′ ))(L 2 2 ( ( " ! ≤ C(!) u(t)2H 2! + u(t′ )2H 2! u(t) – u(t′ )H 2! + (A!% (u(t) – u(t′ ))(L 2 ( ( ≤ C′ (!) max A!% u(t)2L2 (A!% (u(t) – u(t′ ))(L , 2

0≤t≤T0

(2.9.31)

where A!% u(t)L2 is equivalent to uH 2! . From Lemma 2.9.10 and eq. (2.9.30), we obtain !+"– 1

!+"–1

u ∈ C(0, T0 ; D(A% 2 )) for any " ∈ (0, 1). Hence, Ju ∈ D(A% Using Lemma 2.9.10 again, we get for any "′ ∈ (0, 1), !+"+"′ –1

u ∈ C(0, T0 ; D(A%

)).

!+"–1

), A%

Ju(t) ∈ C([0, T0 ]).

(2.9.32)

By induction, one has u ∈ C(0, T0 ; D(A∞ % )).

(2.9.33)

This together with eq. (2.9.19) implies ut ∈ C(0, T0 ; D(A∞ % )).

(2.9.34)

The further regularity in t can be shown by eqs (2.9.33) and (2.9.34). Thus, we conclude Theorem 2.9.11. ∎ Remark 2.9.12. D(A∞ ) = H ∞ =

2

sH

s.

2.9 Initial Value Problem and Boundary Value Problem for the Nonlinear Schrödinger

171

Remark 2.9.13. Theorems 2.9.8 and 2.9.11 still hold true for the more general nonlinear term Ju satisfying eq. (2.9.25) and ( !– 21 ( " ! (A% (Ju – Jv)( ≤ h A! uL + A! vL A! (u – v)L , % % % 2 2 2 L

(2.9.35)

2

where ! > 43 , and h(s) is a positive and nondecreasing function. Theorem 2.9.14. Let ! ∈ R : ! > 43 , T > 0, u0 ∈ D(A! ), g ∈ L1 (0, T; D(A! )). Then there exists a positive constant T1 ∈ (0, T) such that there is a unique local solution u(x, t) to the initial value problem (2.9.17) and (2.9.18) in the interval [0, T1 ], and u ∈ L∞ (0, T1 ; D(A! )) ∩ C(0, T1 ; D(A!–1 )), !–1

ut ∈ L1 (0, T1 ; D(A where T1 depends on [u0 ]# and T1 is uniform for

T 0

(2.9.36)

)),

[g(t)]# dt, # is any fixed constant in

(2.9.37) !3

"

2 , 2!

. Moreover,

 T   3 [g(t)]# dt ≤ b, < # < 2! {u0 , g} ∈ {u, g} : [u]# ≤ a, 2 0 with a, b being fixed positive constants. Proof. Since the proof of uniqueness is easy, we omit it. Due to the dense of D(A∞ ) in D(A! ) = H 2! , we can choose the sequence {u0% } ⊂ D(A∞ ), {g% } ⊂ C(0, T; D(A∞ )) such that lim u0% – uD(A! ) = 0,

%→0

lim g% – gL1 (0,T;D(A! )) = 0.

%→0

(2.9.38) (2.9.39)

Let u% (t) be the solution to the initial value problem: u%t = – A% u% + Ju% + g% , u% (0) = u0% .

(2.9.40) (2.9.41)

Then by Theorem 2.9.11, we get u% ∈ C∞ (0, T; D(A∞ )). Making inner product for eq. (2.9.40) and u% in L2 , and adding its complex conjugate, we obtain d u% 2L2 + 2%A1/2 u% 2L2 = 2Re(g% , u% ) ≤ 2g% L2 u% L2 . dt

(2.9.42)

172

2 The Vanishing Viscosity Method of Some Nonlinear Evolution System

This together with Lemma 2.9.7 yields that  u% L2 ≤ u0% L2 + 

t

2% 0

T 0

g% (4)L2 d4, 0 ≤ t ≤ T0 ,

  %A1/2 u% (4)2L2 d4 ≤ u0% 2L2 + 2 u0% 2L2 +

T

0



T

⋅ 0

(2.9.43)



g% (4)L2 d4

g% (4)L2 d4.

(2.9.44)

√ Hence, {u% } and { %A1/2 u% } are bounded in L∞ (0, T; L2 ) and L2 (0, T; L2 ), respectively, that is u% L∞ (0,T;L2 ) ≤ C1 , √ %A1/2 u% L2 (0,T;L2 ) ≤ C2 ,

(2.9.45) (2.9.46)

where the constants Ck (k = 1, 2) and C appearing in the following are all independent of % ∈ [0, %0 ]. Applying As on both sides of eq. (2.9.40), and then making inner product with s A u% , we get by taking real part d s 2 A u% L2 + 2%As+1/2 u% 2L2 = 2Re(As Ju% , As u% ) + 2Re(As g% , As u% ). dt

(2.9.47)

We can rewrite the first term on the right-hand side as " ! " ! 4Re As (|u% |2 Du% ) – |u% |2 As Du% , As u% + 4Re |u% |2 As Du% , As u% " ! " ! + 2Re As (u2% Du¯ % ) – u2% As Du% , As u¯ % + 2Re u2% Du¯ % , As u% .

(2.9.48)

Letting s > 21 , and using Lemmas 2.9.2–2.9.6, we can control the first term and the third term in eq. (2.9.48) by   C u2% H 2s Du% H r + u2% H r+1 Du% H 2s–1 As u% L2  ! " ≤ C 2u% H 2s u% H r + u% H 2s–1 u% H r+1 u% H r+1 + u% H r+1 u% H 2s ⋅ As u% L2 ≤ Cu% 2H r+1 [u% ]22s . (2.9.49) For any r > 21 , since A can exchange with D, integrating by part gives " ! 4Re |u% |2 As Du% , As u% = –2(As u% D|u% |2 , As u% ).

(2.9.50)

Thus, the second term of eq. (2.9.48) can be controlled by 1 Cu% L∞ Du% L∞ As u% 2L2 ≤ Cu% 2H r+1 [u% ]22s , r > , 2

(2.9.51)

2.9 Initial Value Problem and Boundary Value Problem for the Nonlinear Schrödinger

173

where we used the Sobolev embedding: H r ⊂ L∞ (r > 21 ). Similarly, the fourth term of eq. (2.9.48) can be controlled by Cu% 2H r+1 [u% ]22s . Inserting eqs (2.9.48)–(2.9.51) into eq. (2.9.47), we obtain d s 2 A u% L2 + 2%As+1/2 u% 2L2 ≤ Cu% 2H r+1 [u% ]22s + 2As g% L2 As u% L2 . dt

(2.9.52)

This together with eq. (2.9.42) yields that d [u% ]22s + 2%[u% ]22s+1 ≤ Cu% 2H r+1 [u% ]22s + 2[g% ]2s [u% ]2s , dt

(2.9.53)

with s, r > 21 . We can choose s, r : 2s = r + 1 = # < 2! and d [u% ]2# + 2%[u% ]2#+1 ≤ C[u% ]4# + 2[g% ]# [u% ]# , dt

(2.9.54)

d [u% ]# ≤ C[u% ]3# + [g% ]# . dt

(2.9.55)

which implies

Integrating eq. (2.9.55) in t ∈ [0, t](0 ≤ t ≤ T0 ) shows  [u% (t)]# ≤ [u0% ]# +

0



t

[g% (4)]# d4 + C

0

t

[u% (4)]3# d4.

(2.9.56)

On the other hand, from Opial’s inequality, we know that the solution of the integral equation  y(t) = ' + C 0

' = [u0% ]# +

t

y3 (4) d4,

 0

(2.9.57)

T0

[g% (4)]# d4

(2.9.58)

only exists on 0 ≤ t ≤ T∗ = 21 C'2 . Thus, {u% } keep bounded in L∞ (0, T1 ; H # ) as % → 0. We can take T1 < min(T0 , T∗ ) such that u% L∞ (0,T1 ;H # ) ≤ C3 .

(2.9.59)

Integrating eq. (2.9.54) in t gives √

%u% L2 (0,T1 ;H #+1 ) ≤ C4 .

(2.9.60)

174

2 The Vanishing Viscosity Method of Some Nonlinear Evolution System

Taking s = !, r + 1 = # in eq. (2.9.52). we get by eq. (2.9.59) d [u% ]22! + 2%[u% ]22!+1 ≤ C′ [u% ]22! + [g% ]22! , 0 ≤ t ≤ T, dt

(2.9.61)

where the constant C′ depends on C3 , but not on %. Using Gronwall’s inequality, we obtain u% L∞ (0,T;D(A! )) ≤ C5 .

(2.9.62)

A% u% L∞ (0,T1 ;D(A!–1 )) ≤ C6 ,

(2.9.63)

This implies

Ju% 

L∞ (0,T1 ;D(A

!– 21

))

≤ C7 .

(2.9.64)

These together with eq. (2.9.19) yield u%t L1 (0,T1 ;D(A!–1 )) ≤ C8 .

(2.9.65)

Now, we turn to prove the convergence of the approximate solution {u% (t)}. Writing w(t) = u% (t) – u$ (t), we have by eq. (2.9.40) wt = –A% w + Ju% – Ju$ + g% – g$ + (% – $)Au$ .

(2.9.66)

Making inner production for eq. (2.9.66) and w, and adding its complex conjugate, we obtain d w2L2 +2%A1/2 w2L2 = 2Re(Ju% – Ju$ , w) dt + 2Re(g% – g$ , w) + 2(% – $)Re(A1/2 u$ , A1/2 w).

(2.9.67)

Integrating by parts shows ! " ! " ¯ 2Re(Ju% – Ju$ , w) = wD(|u% |2 + |u$ |2 ), w + Re wD(u % u$ ), w . Hence,   ! 2Re(Ju% – Ju$ , w) ≤ 2u% L Du% L + 2u$ L Du$ L + Du% L u$ L ∞ ∞ ∞ ∞ ∞ ∞ " + u% L∞ Du$ L∞ w2L2 " ! ≤ C u% 2H 2! + u$ 2H 2! w2L2 . By Sobolev embedding: H s ⊂ L∞ (s > 1/2), we obtain for 0 ≤ t ≤ T1   2Re(Ju% – Ju$ , w) ≤ C′′ w2 , L2

175

2.9 Initial Value Problem and Boundary Value Problem for the Nonlinear Schrödinger

where C′′ only depends on C5 . Plugging into eq. (2.9.67) implies d w2L2 ≤ C′′ w2L2 + 2g% – g$ L2 wL2 + 2C′′′ (% + $), dt

(2.9.68)

′′

where C′′′ only depends on C5 . Multiplying eq. (2.9.68) by e–C t , and then integrating in t ∈ [0, t)(0 < t < T1 ), we have w2L2 ≤ eC

′′ T !

" ′′ w(0)2L2 + 2C′′′ (% + $)T1 /C′′ + 2eC T



t

0

g% (4) – g$ (4)L2 w(4)L2 d4.

Using Lemma 2.9.7, we deduce that w(t)2L2

  ≤ C u0% – u0$ L2 + % + $ +

 g% (4) – g$ (4)L2 d4 , 0 ≤ t ≤ T1 .

T 0

(2.9.69)

This shows that {u% } is a Cauchy sequence in C(0, T1 ; L2 ). Hence, there exists a function u(x, t) such that lim u% – uC(0,T1 ;L2 ) = 0.

(2.9.70)

%→0

Furthermore, there exists a subsequence {u% }(still denoted by u% }) such that u% → u, almost everywhere in [0, T1 ] × R.

(2.9.71)

From eqs (2.9.62) to (2.9.64), we know that u% converges weakly ∗ to u in L∞ (0, T1 ; D(A! )),

(2.9.72)

!–1

)),

(2.9.73)

!– 21

)).

(2.9.74)

A% u% converges weakly ∗ to iAu + u in L∞ (0, T1 ; D(A Ju% converges weakly ∗ to Ju in L∞ (0, T1 ; D(A

Making inner product for eq. (2.9.40) and v ∈ D(A∗% ) = D(A) = H 2 , and then integrating in t ∈ (0, t), we obtain  (u% (t), v) = (u0% , v) –

t 0

!

u% (4), A∗% v

"





t

d4 + 0

(Ju% , v) d4 +

t

!

" g% (4), v d4.

0

Letting % → 0 gives 

t

!

∗

"



u(4), (–iA )v d4 +

(u(t), v) = (u0 , v) – 0



t

t

(J0 u, v) d4 + 0

!

" g(4), v d4.

0

Thus,  t 3 4 ! " ¯ u(t), v = u0 + – iAu(4) + J0 u(4) + g(4) d4, v 0

∀ v ∈ D(A),

176

2 The Vanishing Viscosity Method of Some Nonlinear Evolution System

where ,  denotes the dual inner product between D(A)′ = H –2 and D(A) = H 2 , A¯ is the unique continuous extension of A from L2 to D(A)′ . We can view A¯ as the continuous 1 operator from D(A! ) to D(A!–1 ), and D(A! ) ⊂ D(A!– 2 ) ⊂ D(A!–1 ). Hence,  t ! " ¯ – iAu(4) + J0 u(4) + g(4) d4, t ∈ (0, T1 ). u(t) = u0 + 0

This implies that u(t) is strongly abstract continuous in D(A!–1 ), and strongly differential in D(A!–1 ) for almost everywhere t ∈ [0, T1 ], and du ¯ + J0 u + g ∈ L1 (0, T1 ; D(A!–1 )). = –iAu dt ∎

Therefore, we complete the proof of Theorem 2.9.14. Remark 2.9.15. u(t) is weakly continuous in [0, T1 ], and is well defined in D(A! ).

Remark 2.9.16. The conclusion of Theorem 2.9.14 can be extended to ∂ ¯ ∂u J0 u = ∂x (|u|p u), p > 1 or more general form: J0 (u) = F(u, u) ∂x with F(x, t) being smooth function. Remark 2.9.17. Assume g ∈ L2 (0, T; D(A! )). Then the result of Theorem 2.9.14 is true. Meanwhile, we replace ut ∈ L1 (0, T1 ; D(A!–1 )) by ut ∈ L2 (0, T1 ; D(A!–1 )). In fact, by eq. (2.9.19) and estimates (2.9.62)–(2.9.64), we have u%t L2 (0,T1 ;D(A!–1 )) ≤ C. This together with precompact principle yields lim u% – uL2 (0,T1 ;D(A!–' )) = 0, 0 ≤ ' < 1, ' ≤ !,

%→0

which implies lim u% – uL2 (Q) = 0,

%→0

for any compact subset Q ⊂ R × [0, T1 ]. Moreover, u%t converges weakly to ut in L2 (0, T1 ; D(A!–' )). Corollary 2.9.18. Suppose that the initial data is sufficiently small in some sense. Then there is global solution to the initial value problem (2.9.17) and (2.9.18), or (2.9.2), (2.9.2)′ and (2.9.3). Theorem 2.9.19. Let u0 ∈ D(A1/2 ), g ∈ L∞ (0, T; L2 ) ∩ L2 (0, T; D(A1/2 )), and ∃ k > 0 s.t. [u0 ]1/2 ≤ k, gL1 (0,T;L2 ) ≤ k.

(2.9.75)

2.9 Initial Value Problem and Boundary Value Problem for the Nonlinear Schrödinger

177

Then there is a unique global weak solution u to the initial value problem (2.9.17) and (2.9.18). It is weakly continuous in [0, T], and u ∈ L∞ (0, T; D(A1/2 )), 1/2 ′

ut ∈ L∞ (0, T; D(A ) ).

(2.9.76) (2.9.77)

Remark 2.9.20. It is obvious that the constant k is determined by [u0 ]1/2 and gL1 (0,T;L2 ) . Remark 2.9.21. The weak solution u to the initial value problem (2.9.17) and (2.9.18) satisfies eq. (2.9.17) in the sense that ¯ + J0 u + g, 0 < t < T, ut = –iAu

(2.9.78)

where A¯ is the unique linear extension of A from D(A1/2 ) = H 1 to D(A1/2 )′ = H –1 . Before proving Theorem 2.9.19, we state the conservation function of the solution to eq. (2.9.1) with g ≡ 0. Lemma 2.9.22. Let u(x, t) be the smooth solution to eq. (2.9.1), and g ≡ 0. Then we have the two quantities independent of t as follows: I1 (u(t)) = u(t)2L2 ,

(2.9.79)

" 1 3 ! I2 (u(t)) = Du(t)2L2 + u(t)6L6 – Im |u(t)Du(t), u(t) . 2 2

(2.9.80)

Proof. A direct computation gives d d I1 (u(t)) = I2 (u(t)) = 0. dt dt ∎ The proof of Theorem 2.9.19. We need to find the sequence {u0% } ⊂ D(A∞ ), {g% } ⊂ C∞ (0, T; D(A∞ )) such that lim u0% – u0 D(A1/2 ) = 0,

(2.9.81)

lim g% – gL∞ (0,T;L2 )∩L2 (0,T;D(A1/2 )) = 0.

(2.9.82)

%→0 %→0

Let u% (t) be the solution to eqs (2.9.40) and (2.9.41). Then u% ∈ C∞ (0, T0 ; D(A∞ )). By eqs (2.9.43) and (2.9.44) and L2 (0, T; L2 ) ⊂ L1 (0, T; L2 ), we get u% L∞ (0,T0 ;L2 ) ≤ C1 , √ %A1/2 u% L2 (0,T0 ;L2 ) ≤ C2 , where constants Ck (k = 1, 2, 3, . . .) and C are independent of T0 , % ∈ (0, %).

(2.9.83) (2.9.84)

178

2 The Vanishing Viscosity Method of Some Nonlinear Evolution System

Derivative I2 (u% (t)) in t, and integrating by part, we obtain by eq. (2.9.19) " ! " ! d I2 (u% (t)) + 2%Au% 2L2 = 3%Re |u% |4 u% , Au% – 6%Im |u% |2 Du% , Au% dt " ! " ! + 2Re(Du% , Dg% ) + 3Re |u% |4 u% , g% – 6Im |u% |2 Du% , g% . (2.9.85) Using Lemma 2.9.1, eq. (2.9.83), the first term on the right-hand side of eq. (2.9.85) can be controlled by " ! 3%u% 5L10 Au% L2 ≤ 3%C15 (10, 2, 0) Au% L2 + u% L2 u% 4L2 Au% L2 ≤ 3%C14 C15 (10, 2, 0)Au% 2L2 + %,Au% 2L2 +

C , ,

(2.9.86)

where 0 ≤ t ≤ T0 , , is arbitral positive number. By Hölder’s inequality, Lemma 2.9.1 and eq. (2.9.83), the second term on the right-hand side of eq. (2.9.85) can be controlled by " ! 6%u% 2L∞ Du% L2 Au% L2 ≤ 6%C12 (8, 2, 0)C1 (4, 2, 1) Au% L2 + u% L2 u% 2L2 Au% L2 ≤ 6%C12 C12 (8, 2, 0)C1 (4, 2, 1)Au% 2L2 + %,Au% 2L2 +

C . ,

(2.9.87)

Similarly, the fourth and fifth term terms on the right-hand side of eq. (2.9.85) can be controlled by 10 ! 2 "5 "1 ! 3u% 5L g% L6 ≤ 3C26 (6, 1, 0) Du% L2 + u% L2 3 u% L32 Dg% L2 + g% L2 3 g% L32 6

≤ CDu% 2L2 + C[g% ]21 + C, 6g% L∞ u% 2L4 Du% L2 ≤ ≤ CDu% 2L2

+

C[g% ]21

+ C,

(2.9.88) 1 ! 6C12 (4, 1, 0)g% H 1 g% L22 Du% L2 1 2

+ u% L2

0 ≤ t ≤ T0 ,

"1

1 2 u  2 Du  % L2 % L2

(2.9.89)

where we used the estimate: g% L∞ (0,T;L2 ) ≤ C. Combining the above estimates with eq. (2.9.85), we deduce that # $ d I2 (u% (t)) + % 2 – 3C14 C15 (10, 2, 0) – 6C12 C12 (8, 2, 0)C1 (4, 2, 1) – 2, Au% 2L2 dt C ≤ CDu% 2L2 + C[g% ]21 + C + , (2.9.90) , where the constant C1 depends on u0 L2 and gL1 (0,T;L2 ) . It is easy to see that for # $ {u0 , g} ⊂ (u0 , g) : u0 L2 ≤ k, gL1 (0,T;L2 ) ≤ k , there exist k > 0 and sufficiently small C1 such that 2 – 3C14 C15 (10, 2, 0) – 6C12 C12 (8, 2, 0)C1 (4, 2, 1) ≥ $ > 0,

(2.9.91)

2.9 Initial Value Problem and Boundary Value Problem for the Nonlinear Schrödinger

179

with $ being sufficiently small positive number. Since , is an arbitrary positive number, we can take , = 4$ . Then, we have by eq. (2.9.90) $% d I2 (u% (t)) + Au% 2L2 ≤ CDu% 2L2 + C[g% ]21 + C. dt 2

(2.9.92)

Integrating in t ∈ [0, t] gives  " $% t 1 3 ! 2 6 + u% L6 + Im |u% | Du% (t), u% (t) + Au% (4)2L2 d4 2 2 2 0  t  t ≤C Du% (4)2L2 d4 + C [g% (4)]21 d4 + CT + I2 (u0% ), 0 ≤ t ≤ T0 ≤ T. Du% (t)2L2

0

(2.9.93)

0

Observing that the third term on the left-hand side of eq. (2.9.93) can be controlled by 3 3 3 u% (t)2L6 ⋅ Du% (t)L2 ≤ u% (t)6L6 + Du% (t)2L2 , 2 4 4 we get  Du% (t)2L2

+ 2$% 0

 ≤ u% (t)6L6 + C

t 0

t

Au% (4)2L2 d4 

Du% (4)2L2 d4 + C

0

t

(2.9.94) [g% (4)]21 d4 + CT, 0 ≤ t ≤ T0 ≤ T.

On the other hand, by Lemma 2.9.1, the first term on the right-hand side of eq. (2.9.94) can be estimated by "2 ! u% (t)6L6 ≤ C16 (6, 1, 0) Du% L2 + u% L2 u% 4L2 . Hence, " ! 1 – 2C16 (6, 1, 0)C14 Du% 2L2  t  t 6 6 2 ≤ 2C1 C1 (6, 1, 0) + C Du% (4)L2 d4 + C [g% (4)]21 d4 + CT  ≤C 0

0

t

Du% (4)2L2

0

d4 + C.

(2.9.95)

Taking k sufficiently small such that 1 – 2C16 (6, 1, 0)C14 ≥ $ > 0,

(2.9.96)

then we have by eq. (2.9.5) $Du% (t)2L2

$% + 2

 0

t

 Au% (4)2L2

d4 ≤ C 0

t

Du% (4)2L2 d4 + C.

(2.9.97)

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2 The Vanishing Viscosity Method of Some Nonlinear Evolution System

Thus, Du% L∞ (0,T0 ;L2 ) ≤ C3 , √ %Au% L2 (0,T0 ;L2 ) ≤ C4 .

(2.9.98) (2.9.99)

Since the constants Cj (j = 1, 2, 3, 4) are independent of T0 , it follows from eqs (2.9.83), (2.9.84), (2.9.98) and (2.9.99) that the solution u% (t) to initial value problem (2.9.19) and (2.9.20) can be extended to any T. Hence, estimates (2.9.83), (2.9.84), (2.9.98) and √ (2.9.99) also hold true replacing T0 by T, that is, {u% } and { %Du% } keep bounded in L∞ (0, T; D(A1/2 )) and L2 (0, T; D(A1/2 )), respectively, as % → 0. Therefore, there is a function u(x, t) and a subsequence {u% (x, t)} such that as % → 0, u% converges weakly ∗ to u in L∞ (0, T; D(A1/2 )),

(2.9.100)

u%t converges weakly to ut in L2 (0, T; L2 ).

(2.9.101)

By precompact lemma, we have lim u% – uL2 (Q) = 0,

%→0

(2.9.102)

for any compact subset Q ⊂ R × [0, T], and u% converges to u almost everywhere in R × [0, T].

(2.9.103)

It follows from eqs (2.9.83) and (2.9.98) that {Ju% } is bounded in L∞ (0, T; L2 ). Thus, we have by eq. (2.9.103) Ju% converges weakly ∗ to Ju in L∞ (0, T; L2 ).

(2.9.104)

This implies that u(x, t) is the weak solution to problem (2.9.17) and (2.9.18). Therefore, we conclude Theorem 2.9.19. Furthermore, using the conservation law of eq. (2.9.1), we can obtain the existence of the global strong and smooth solution to eqs (2.9.17) and (2.9.18). Lemma 2.9.23. Let v(x, t) be the smooth solution to eq. (2.9.1) with g ≡ 0. Then we have the three quantities independent of t: I1 (v(t)) = v(t)2L2 ,

(2.9.105)

1 3 I2 (v(t)) = Dv(t)2L2 + v(t)6L6 – Im(|v(t)|2 Dv(t), v(t)), (2.9.106) 2 2 ( " ! 7 25 ( (|v(t)|2 Dv(t)(2 + 5Re |v(t)|2 v(t)2 , (Dv(t))2 I3 (v(t)) = D2 v(t)2L2 + v(t)10 L10 + L 2 8 2 " 35 ! 2 2 (2.9.107) + 5Im |v(t)| D v(t), Dv(t) – Im(|v(t)|6 v(t), Dv(t)). 8

181

2.9 Initial Value Problem and Boundary Value Problem for the Nonlinear Schrödinger

Proof. It is easy to get eq. (2.9.105). Equations (2.9.106) and (2.9.107) will be shown lately. ∎ Theorem 2.9.24 (Existence of global strong solution). Let u0 ∈ D(A), g ∈ L∞ (0, T; L2 )∩ L2 (0, T; D(A)). Then there is k > 0 for [u0 ]1 ≤ k, gL1 (0,T;L2 ) ≤ k, and the initial value problem (2.9.17) and (2.9.18) has the unique global solution u(t), u(x, t) ∈ L∞ (0, T; D(A)) ∩ C(0, T; L2 ), ut (x, t) ∈ L∞ (0, T; L2 ).

Proof. Let u(t) be the smooth solution to the initial value problem (2.9.17) and (2.9.18), 0 ≤ t ≤ T0 . Multiplying eq. (2.9.17) by u, and integrating in x, we obtain by taking real part 1 d u(t)2L2 = Re(g(t), u(t)) ≤ g(t)L2 u(t)L2 . 2 dt This implies that  u(t)2L2 ≤ u0 2L2 +

0

T

g(t)L2 dt,

0 < t < T0 .

(2.9.108)

Hence, uL∞ (0,T0 ;L2 ) ≤ C1 .

(2.9.109)

Derivative I2 (u(t)) in t, and integrating by part, we get by eq. (2.9.17) d I2 (u(t)) = 2Re(Du, Dg) + 3Re(|u|4 u, g) – 6Im(|u|2 Du, g) dt ≤ CDu2L2 + C[g]21 + C, 0 ≤ t ≤ T0 .

(2.9.110)

Integrating in t ∈ [0, t] gives 1 3 Du2L2 + u6L6 – Im(|u|2 Du, u) 6 2  T  T 2 Du(s)L2 ds + C [g(s)]21 ds + CT + initial data, ≤C 0

0

for any 0 ≤ t ≤ T0 < T. By Hölder’s inequality and Young’s inequality, we have 3  3 3 3    Im(|u|2 Du, u) ≤ u3L6 DuL2 ≤ u6L6 + Du2L2 . 2 2 4 4 Since 2 "1 ! uL6 ≤ K DuL2 + uL2 3 uL32 ,

(2.9.111)

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2 The Vanishing Viscosity Method of Some Nonlinear Evolution System

we obtain  (1 – C14 k6 )Du(t)2L2 ≤ C

0

T

Du(s)2L2 ds + C.

(2.9.112)

Taking k such that 1 – C14 k6 ≥ $ > 0, then for u0 L2 ≤ k, gL1 (0,T;L2 ) ≤ k, we estimate by Gronwall’s inequality and eq. (2.9.112) DuL∞ (0,T0 ;L2 ) ≤ C2 .

(2.9.113)

Set 35 8 35 5 25 2 ¯ ¯ 2 u3 – |u|4 D2 u – (Du) |u| u + |u|2 u(Du) 8 2 2 2 35 2 2¯ 2 2 2 2¯ + 5|u| D u ⋅ u – 15|u| |Du| u – 5iD uDu ⋅ u – i|u|6 Du 2 2 ¯ 2 u ⋅ u – 10iuDuD ¯ + 5i|Du|2 Du – 5iDuD u + 5i|u|2 D3 u,

Q(u) = D4 u +

where u¯ denotes the complex conjugate of u. Making inner product for eq. (2.9.17) and Q(u), we obtain by taking real part d I3 (u(t)) = Re(g(t), Q(u(t))). dt

(2.9.114)

By eqs (2.9.109) and (2.9.113) and H 1 ⊂ L∞ , we have d I3 (u(t)) ≤ D2 u(t)2L2 + CD2 g(t)2L2 . dt

(2.9.115)

Integrating in t ∈ [0, t] gives ( " ! 7 25 ( (|u|2 Du(2 + 5Re |u|2 u2 , (Du)2 u10 L10 + L 2 8 2 " 35 ! 2 2 + 5Im |u| D u, Du – Im(|u|6 u, Du) 8  t D2 u(s)2L2 ds + Cg2L2 (0,T;D(A)) + C. ≤C D2 u2L2 +

(2.9.116)

0

Using eqs (2.9.109) and (2.9.113), one has  1  " ! " 35 !   5Re |u|2 u2 , (Du)2 + 5Im |u|2 D2 u, Du – Im(|u|6 u, Du) ≤ D2 u(t)2L2 + C. 8 2

2.9 Initial Value Problem and Boundary Value Problem for the Nonlinear Schrödinger

183

Hence,  2

D

u(t)2L2

≤C 0

t

D2 u(s)2L2 ds + Cg2L2 (0,T;D(A)) + C.

This yields D2 uL∞ (0,T0 ;L2 ) ≤ C3 ,

(2.9.117)

where C3 only depends on u0 and g. Therefore, by this priori estimate, we deduce the global strong solution of the initial value problem (2.9.17) and (2.9.18). ∎ Theorem 2.9.25 (Existence of global smooth solution). Let u0 ∈ D(A! ), g ∈ L∞ (0, T; L2 ) ∩ L2 (0, T; D(A)) ∩ L1 (0, T; D(A! )), ! ≥ 1. Then there is k > 0 for [u0 ]1 ≤ k, gL1 (0,T;L2 ) ≤ k, and the initial value problem (2.9.17) and (2.9.18) has the unique global solution u(t), u(x, t) ∈ L∞ (0, T; D(A! )) ∩ C(0, T; D(A!–1 )), ut (x, t) ∈ L∞ (0, T; D(A!–1 )) ∩ L∞ (0, T; L2 ). Proof. Let u be the smooth solution to problem (2.9.17) and (2.9.18), t ∈ [0, T0 ]. Applying As in eq. (2.9.17), and then making inner product with As u, we obtain by taking real part d s 2 A uL2 = Re(As Ju, As u) + Re(As g, As u). dt

(2.9.118)

By the same argument as in Theorem 2.9.14, we have uL∞ (0,T0 ;D(A! )) ≤ C4 . Since C4 is independent of T0 , we can extend the solution u(t) to T. We complete the proof of theorem. ∎ Corollary 2.9.26. Let u0 ∈ D(A∞ ), g ∈ C∞ (0, T; D(A∞ )), and [u0 ]1 ≤ k, gL1 (0,T;L2 ) ≤ k. Then the solution u(x, t) to problem (2.9.17) and (2.9.18) satisfies u ∈ C∞ (0, T; D(A∞ )). Under the assumptions of Theorem 2.9.25, we consider the initial value problem (2.9.17) and (2.9.18) with g ≡ 0. We define the map St : u(0) = u0 ↦ u(t), which maps from D(A! ) to itself (! ≥ 1).

184

2 The Vanishing Viscosity Method of Some Nonlinear Evolution System

Theorem 2.9.27 (Continuous dependence of solution). St is local Hölder continuous of 1 + + + 2 -order, that is, there is a continuous function L : R × R → R such that 1 " ! 2 [St u0 – St v0 ]2! ≤ L [u0 ]2!+ 1 , [v0 ]2!+ 1 [u0 – v0 ]2! . 2

2

1

Remark 2.9.28. If we view St as the operator from D(A!+ 2 ) to D(A! )(! ≥ 1), then St is local Lipschitz continuous, i.e. " ! [St u0 – St v0 ]2! ≤ L [u0 ]2!+1 , [v0 ]2!+1 [u0 – v0 ]2! . Remark 2.9.29. Let gi ∈ L∞ (0, T; L2 ) ∩ L2 (0, T; D(A! ))(i = 1, 2), u, v correspond to the solution of eqs (2.9.17) and (2.9.18) with g = g1 and g = g2 , respectively. Then 1 " ! 2 + Cg1 – g2 L2 (0,T;D(A! )) , [u(t) – v(t)]2! ≤ L [u0 ]2!+ 1 , [v0 ]2!+ 1 [u0 – v0 ]2! 2 2 " ! [u(t) – v(t)]2! ≤ L [u0 ]2!+1 , [v0 ]2!+1 [u0 – v0 ]2! + Cg1 – g2 L2 (0,T;D(A! )) .

Finally, we return to prove the conservation law of eq. (2.9.17) with g ≡ 0, that is, eqs (2.9.106) and (2.9.107). Let & be a complex parameter. We consider the total differential equations dv1 = F1 dx + G1 dt,

(2.9.119)

dv2 = F2 dx + G2 dt,

(2.9.120)

where F1 = –i& 2 v1 + q(x, t)& v2 ,

F2 = i& 2 v2 + r(x, t)& v1 ,

G1 = A(x, t, & )v1 + B(x, t, & )v2 , G2 = C(x, t, & )v1 – A(x, t, & )v2 . We can rewrite as v1x = – i& 2 v1 + q(x, t)& v2 , 2

(2.9.121)

v2x = i& v2 + r(x, t)& v1 ,

(2.9.122)

v1t = A(x, t, & )v1 + B(x, t, & )v2 ,

(2.9.123)

v2t = C(x, t, & )v1 – A(x, t, & )v2 .

(2.9.124)

By the integrable conditions for eqs (2.9.119) and (2.9.120), we have ⎧ ⎪ ⎪ ⎨Ax = & (qC – rB),

Bx + 2i& 2 B = & qt – 2& qA, ⎪ ⎪ ⎩C – 2i& 2 C = & r + 2& rA. x t

(2.9.125)

2.9 Initial Value Problem and Boundary Value Problem for the Nonlinear Schrödinger

185

As a special case, we take ⎧ 4 2 ⎪ ⎪ ⎨A = –2i& – iqr& ,

B = 2q& 3 + q2 r& + iqx & , ⎪ ⎪ ⎩C = 2r& 3 + qr2 & – ir & , x

(2.9.126)

which is one of the solutions to eq. (2.9.125). We remark that if we set r = q¯ , then we can get the nonlinear Schrödinger equation with derivative from eqs (2.9.125) and (2.9.126) qt = iqxx + (q2 q¯ )x .

(2.9.127)

First, we derive the circulation formula for the infinite conservation function. We have by eq. (2.9.125) v2 =

1 (v1x + i& 2 v1 ). &q

(2.9.128)

Plugging into eq. (2.9.122) shows 1

q

q

 v1x

x

– i& 2

qx v1 + & 4 v1 = & 2 qrv1 . q

(2.9.129)

Suppose that the solution of eq. (2.9.129) has the form   2 v1 (x, t, & ) = exp – i& x +

x

 E(s, t; & ) ds .

(2.9.130)

On the other hand, inserting eq. (2.9.128) into eq. (2.9.123) gives v x & v1  1 +i . &q q

v1t = Av1 + B

This together with eq. (2.9.130) implies that 

x

 B  Et (s, t; & ) dsv1 = A + E v1 . &q

Hence,  ∂ ∂  B E(x, t; & ) = A + E(x, t; & ) . ∂t ∂x &q This means that the quantity



E(x, t; & ) dx is independent of time t.

(2.9.131)

186

2 The Vanishing Viscosity Method of Some Nonlinear Evolution System

It follows from eqs (2.9.129) and (2.9.130) that 1   i 1  2 E(x, t; & ) . + q E(x, t; & ) – qr = E(x, t; & ) x 2 2i& 2 q Writing E –

(2.9.132)

!i"

qr = z, & 2 = . , we get

2

 z(x, t; . ) i   1  i 2 z(x, t; . ) + qr + q + r . 2i. 2 q 2 x

z(x, t; . ) =

(2.9.133)

Since eq. (2.9.133) is Riccati type, it enjoys the asymptotic expansion z=



z(n) (x, t; . )(zi. )–n

(n = 0, 1, 2, . . .),

n

where z(i) (x, t; . ) is determined by the following circulation formula: z(i+1) =

i–1 

z(k) z(i–k) + iqrz(i) + q

k=1

∂  1 (i)  z , (i = 1, 2, 3, . . .), ∂x q

z(0) = qr,

(2.9.134) (2.9.135)

1 2 2 i (2.9.136) q r + qrx , 4 2 1 i i (2.9.137) z(2) = – qqx r2 – q2 rrx – q3 r3 + qrxx , 4 4 2 5 5 3 3 1 z(3) = q4 r4 – q2 rx2 – q3 rrxx – qqx rrx – qqxx r2 16 4 2 2 4 3 i – 2iq3 r2 rx – iq2 qx r3 + qrxxx , (2.9.138) 4 2 29 9 11 z(4) = 4q4 r3 rx + q3 qx r4 – q2 rx rxx – 2q2 rrxxx – qqx rx2 16 2 4 1 47 5 5 15 3 2 25 2 – 3qqx rrxx – qqxxx r + iq r – iq r rxx – iq3 rrx2 4 16 4 4 3 2 3 i 2 2 2 3 – 8iq qx r rx – iq qxx r – iqqx r + rxxxx q. (2.9.139) 4 2  Noting that each z(i) (x, t) dx is conserved, we obtain eqs (2.9.106) and (2.9.107) from eqs (2.9.137) and (2.9.139), respectively. For example, setting r = q¯ in eq. (2.9.137), we have z(1) = –

 z(2) (q) dx =

i 2

 q¯qxx dx –

i 4



 |q|6 dx –

|q|2 q¯qx dx –

1 4

 |q|2 qx q¯ dx. (2.9.140)

Observing that  0=

∂ |q|4 dx = 2 ∂x



 |q| qx q¯ dx + 2 2

|q|2 q¯qx dx,

(2.9.141)

2.10 Initial Value Problem for Boussinesq Equations

187

and integrating by part, we can rewrite eq. (2.9.140) as       i i 3 z(2) (q) dx = – |qx |2 dx – |q|6 dx + |q|2 qx q¯ dx – |q|2 q¯ x q dx . 2 4 8 And so we get eq. (2.9.106). Similarly, we can obtain eq. (2.9.107) from eqs (2.9.139) and (2.9.141).

2.10 Initial Value Problem for Boussinesq Equations We consider the existence of weak solutions to the following equations:  1t + !ux + "(u1)x = 0, ut + #1x + $uux – -vxxt = 0, where ", $, - → 0, we choose ! = # = 1, " = $ = 3-, which describe the propagation of the long surface wave in the pipe with constant depth. This model was introduced by Boussinesq, u(x, t) denotes the velocity, 9(x, t) = 1 + $1(x, t) denotes the altitude from the bottom to the free surface of the flow. Boussinesq equations can be considered as the perturbation of the one-dimensional wave, where the dispersion and the nonlinear effect are of the same order. For simplicity, we set ! = " = # = $ = - = 1. We consider the following equations:  1t + ux + (u1)x = 0, (2.10.1) ut + 1x + uux – uxxt = 0, In order to construct the existence theorem, we consider the parabolic regularization of the above equations  1t + ux + (u1)x = :1xx , : > 0. (2.10.2) ut + 1x + uux – uxxt = 0, We first prove that eq. (3.10.7) admits a global smooth solution for fixed : > 0 and derive the a priori estimates in the entropy inequality independent of : for u in H 1 space and 1 in Orlicz space with respect to the function 1 ln 1 to deduce the existence of the weak solutions to the initial value problem (3.10.6). We first prove the existence of the local smooth solutions to the initial value problem (3.10.7) by the contraction mapping principle. j We denote C0k = {g : g ∈ Ck , lim|x|→∞ ∂x g = 0, 0 ≤ j ≤ k}, Cck = {g : g ∈ Ck , with compact support}, ||g|| = sup |g(x)|. x

188

2 The Vanishing Viscosity Method of Some Nonlinear Evolution System

Theorem 2.10.1. Assume 1¯ , u¯ are fixed constants, for any initial value 10 (x), u0 (x) such that ¯ ∈ C02 × Cc2 , ( 10 – 1¯ , u0 – u) then there exists a constant t0 > 0, which depends on ||10 ||, ||u0 ||, 1¯ , u¯ and :, such that there exists a unique solution (1, u) to the initial value problem (3.10.7) with the initial values 10 (x), u0 (x), 1 – 1¯ , u – u¯ ∈ C02 , 0 ≤ t ≤ t(0) . ¯ then by eq. (3.10.7), we have Proof. We set 1˜ = 1 – 1¯ , u˜ = u – u, 

˜ 1 + 1¯ )]x = :˜1xx , 1˜ t + u˜ x + [(u¯ + u)(˜ ¯ u˜ x – u˜ xxt = 0 u˜ t + 1˜ x + (u˜ + u)

(2.10.3)

and 1˜ 0 (x) = 10 (x) – 1¯ , ¯ u˜ 0 (x) = u0 (x) – u.

(2.10.4)

We note that 1˜ 0 , u˜ 0 ∈ Cc2 . We can prove that eqs (3.10.8) admit a unique solution ˜ 1˜ , u˜ ∈ C02 . Our main idea is that: we apply the linear segment of the first (˜1, u), equation, and write the integral expression of the solution as  1˜ (x, t) =  t

∞

+ 0

G(x – z, t)˜10 (z)dz

–∞ ∞

¯ 1˜ + 1¯ )]Gx (x – z, t – s)dzds, [u˜ + (u˜ + u)(

–∞

where x2 1 G(x, t) = √ e– 4:t . 40:t

Equation (3.10.13) is derived since the boundary integrals of ˜ z) + u(s, z)] = 1¯ u, ¯ lim [˜1(s, z) + 1(s, z)][u(s,

|z|→∞

˜ z) = 0, lim G(x – z, t – s) = 0 lim u(s,

|z|→∞

|z|→∞

vanish. For the second equation of eq. (3.10.8), we rewrite it as   ˜ 2 (u¯ + u) (I – ∂xx )u˜ t = –∂x 1˜ + , 2

(2.10.5)

2.10 Initial Value Problem for Boussinesq Equations

189

which has the formal solution u˜ t (x, t) = –

1 2

  ˜ 2 (u¯ + u) e–|x–z| ∂z 1˜ + dz, 2 –∞



∞

after integrating by parts, we have ¯ t) = u˜ 0 (x) + u(x,

  ˜ 2 (u¯ + u) K(x – z) 1˜ + dzds, 2 –∞

 t 0

∞

(2.10.6)

where 1 K(x) = sgn(x) exp(–|x|). 2 For any t0 > 0, we set Kt0 = {(1, u) ∈ C(0, t0 ; C02 × C02 ); ||1 – G(t) ∗ 10 || +||u(t) – u0 || ≤ ||10 || + ||u0 ||, 0 ≤ t ≤ t0 }. Therefore, if (1, u) ∈ Kt0 , then ||1(t)|| + ||u(t)|| ≤ 2(||˜10 || + ||u˜ 0 ||), 0 ≤ t ≤ t0 . Now we define the mapping A = (A1 , A2 ) from C(0, t0 ; C02 × C02 ) into itself, where  ˜ A1 (˜1(t), u(t)) = G(t) ∗ 1˜ 0 +

t

¯ ∗ Gx (t – s)ds, [u˜ + (˜1 + 1¯ )(u˜ + u)] 5 6  t ˜ 2 (u¯ + u) ˜ A2 (˜1(t), u(t)) = u˜ 0 + K ∗ 1˜ + ds. 2 0 0

(2.10.7)

By some direct computations, there exists t0 , which depends only on ||10 ||, ||u0 ||, 1¯ , u¯ and :, such that A maps the closed set Kt0 into itself, and A is the contraction map on Kt0 , thus by the contraction mapping principle, the initial value problem for eq. ˜ (3.10.8) admits a unique solution. We deduce the uniqueness of the solution (˜1, u) from the integral expression (3.10.13) and (2.10.6) and the Gronwall inequality. We deduce the existence and uniqueness of the solutions to the initial value problem to eq. (3.10.7) from ˜ 1 = 1¯ + 1˜ , u = u¯ + u. This completes the proof. In order to prove the global solution, we need the following auxiliary result.

∎

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2 The Vanishing Viscosity Method of Some Nonlinear Evolution System

¯ ∈ C(0, t0 ; C02 × C02 ), (1, u) is a solution to eqs (3.10.7) Lemma 2.10.2. Assume (1 – 1¯ , u – u) with the initial value (10 , u0 ). If 1 + 10 (x) > 0, then 1 + 1(x, t) ≥ 0

∀x ∈ R, t ≥ 0.

(2.10.8)

Proof. We set 9 = 1 + 1, then the first equation of eq. (3.10.7) is 9t + (9u)x = :9xx , then by the maximum principle, when 9(0, x) = 1 + 10 (x) ≥ 0, we have 9(x, t) ≥ 0 ∀x ∈ R, t > 0. ∎ In order to construct the global existence of the solutions of eqs (3.10.7) for fixed : > 0, and through the limit as : → 0, we construct the existence of the global weak solutions to eqs (3.10.6), we need to construct the a priori estimates independent of :. For this purpose, for the hyperbolic equations ⎧ ⎪ ⎨ 1t + (u + u1)x = 0, u2 ⎪ ⎩ ut + (1 + )x = 0, 2

(2.10.9)

we construct a convex and positive entropy, and deduce an entropy inequality, which imply estimates for u in H 1 and for the function 1 log 1 in the Orlicz space of 1. As we know, for the quasi-linear hyperbolic equations ut + f (u)x = 0,

(2.10.10)

where u = u(x, t) ∈ Rn , f : Rn → Rn is smooth, we say that eq. (2.10.10) is hyperbolic, if ∇f has an absolute set of n real character values and character vectors. We say that two functions ', q are entropy and entropy flow, respectively, if the smooth solutions of eq. (2.10.10) satisfy the additional conservation law: '(u)t + q(u)x = 0.

(2.10.11)

We note that the compatible condition of eqs (2.10.10) and (2.10.11) is ∇' ⋅ ∇f = ∇q.

(2.10.12)

In order to construct estimates for u in H 1 and for the function 1 log 1 in the Orlicz space of 1, we introduce the signs

2.10 Initial Value Problem for Boussinesq Equations

¯ = 1 + 1¯ , 9 = 1 + 1, 9

191

(2.10.13) ′

¯ = 3(9) ¯ + 3 (9)(9 ¯ ¯ 3(9) = 9 log 9, 3L (9, 9) – 9),

(2.10.14)

¯ = 3(9) – 3L (9, 9) ¯ = 9(log 9 – log 9) ¯ +9 ¯ – 9. 30 (9, 9)

(2.10.15)

¯ that yields the We note that 3(9) is convex, which subtracts the linear part 3L (9, 9) ¯ as 9 → 9; ¯ it has the order 9 – 9 ¯ and increases as positive convex function 30 (9, 9) 9 log 9 at ∞. Theorem 2.10.3. Assume 1¯ , u¯ are given constants, and assume that 10 (x), u0 (x) are two ¯ ∈ C02 × C02 , with the initial data (10 , u0 ), there functions, such that (10 (x) – 1¯ , u0 (x) – u) ¯ such that the initial value exists a constant c0 , which depends only on 10 , 1¯ , u0 and u, problem (3.10.7) admits a smooth solution u, 9 satisfying 

∞ –∞

¯ 2 (u – u) dx + 2



∞

–∞

u2x dx + 2



∞

30 (x)dx ≤ c0 ,

(2.10.16)

–∞

¯ where 30 (x) = 30 (9, 9). Proof. Taking the transformation 9 = 1 + 1, eq. (2.10.12) becomes ⎧ ⎪ ⎨ 9t + (9u)x = :9xx , u2 ⎪ ⎩ ut + (9 + )x = uxxt , 2

(2.10.17)

9(0, x) = 90 (x) = 1 + 10 (x), u(0, x) = u0 (x).

(2.10.18)

with the initial data

The corresponding hyperbolic equations (2.10.9) become ⎧ ⎪ ⎨ 9t + (9u)x = 0, u2 ⎪ ⎩ ut + (9 + )x = 0. 2

(2.10.19)

The procedures are as follows: assume the mapping f : R2 → R2 ,   u2 . f (9, u) = 9u, 9 + 2 Assume that '(9, u), q(9, u) are the entropy and entropy flow, for the above given f , the compatible condition (2.10.12) holds. For all solutions of eq. (2.10.17), which satisfy the additional equation '(9, u)t + q(9, u)x = :'9 9xx + 'u uxxt .

(2.10.20)

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2 The Vanishing Viscosity Method of Some Nonlinear Evolution System

Now we choose the entropy with the form '(9, u) =

u2 + !(9), 2

(2.10.21)

where !(9) is a function to be determined later, for the entropy of this form, eq. (2.10.20) becomes '(9, u)t + q(9, u)x = :'9 9xx + 'u uxxt = :!′ (9)9xx + uuxxt 1 = –:!′′ (9)92x + :!(9)xx – (u2x )t + (uux )x . 2

(2.10.22)

If we deduce the entropy of the form similar as eq. (2.10.21), which is convex and positive, then by eq. (2.10.22), we can deduce the a priori estimate. We choose !(9) = 3(9) = 9 log 9. A simple computation yields the compatible condition (2.10.12), which holds for this !(9). We have found the convex entropy (2.10.22) which is not positive. For this pur˜ which deduces from ' after subtracting pose, we construct positive convex entropy ', ¯ here ¯ u), its linear part at (9, ¯ = lim (1 + 10 (x)) = 1 + 1¯ , u¯ = lim u0 (x). 9 |x|→∞

|x|→∞

Therefore, ¯ – ∇'(9, ¯ ¯ ˜ ¯ u) ¯ u)[(9, ¯ u)] '(9, u) = '(9, u) – '(9, u) – (9, 1 ¯ 2 + 9[log9 – log 9] ¯ +9 ¯ –9 = (u – u) 2 1 ¯ 2 + 30 (9, 9). ¯ = (u – u) 2

(2.10.23)

˜ In order to let '(9, u) be the entropy of eq. (2.10.17), we need to find the entropy flow ˜q(9, u). We set ¯ – ∇'(9, ¯ (9, u) – f (9, ¯ ¯ u) ¯ u)[f ¯ u)]. q˜ (9, u) = q(9, u) – q(9, A simple computation yields ∇˜' ⋅ ∇f = ∇˜q. For simplification, we denote '˜ = ', q˜ = q. ', q satisfy eq. (2.10.22), which can be regarded as the starting point of the a priori estimates. Integrating eq. (2.10.22) over (x, t) yields

2.10 Initial Value Problem for Boussinesq Equations

193

R = {(x, s) : –N1 ≤ x ≤ N2 , 0 ≤ s ≤ t, N1 , N2 > 0, }. Using the divergence theorem,  (qx + 't )dxdt = (q, ') ⋅ ndl R ∂R  2 6  5 ux + :3(9)xx + (uuxt )x dxdt, =: – 3′′ 92x – 2 t R



where n denotes the exterior normal. We thus have 

N2

 ['(x, t) – '(x, 0)]dx +

–N1

t

[q(s, N2 ) – q(s, –N1 )]ds 0

 3

= –: R

′′

  92x dxdt

– 

R

u2x 2





t

dxdt + :

[3(9(s, N2 ))x 0

t

t

–3(9(s, –N1 ))x ]ds +

[uuxt (s, N2 ) – uuxt (s, N1 )]ds. 0

If for every t, as N1 , N2 → ∞, by the Lebesgue-dominated convergence theorem and ¯ – q(9, ¯ = 0, ¯ u) ¯ u) lim [q(s, N2 ) – q(s, –N1 )] = q(9,

N1 ,N2 →∞

¯ x – 3(9) ¯ x = 0, lim [3(9(s, N2 )) – 3(9(s, –N1 ))]x = 3(9)

N1 ,N2 →∞

we have 

∞



∞

u2x dx –∞ –∞ 2  ∞ 2  t ∞  ∞ u0x '(x, 0)dx + 30′′ (9)u2x dxdt. dx – : = –∞ –∞ 2 0 –∞ '(x, t)dx +

Since 30′′ (9) ≥ 0, we have 



 ∞ ¯ 2 u2x (u – u) 'dx + dx = dx 2 –∞ –∞ 2 –∞  ∞ 2  ∞ ux ¯ 30 (9, 9)dx ≤ c0 , + dx + –∞ 2 –∞ ∞

∞

(2.10.24)

where 

∞

c0 = –∞

 '(90 , u0 )dx +

∞ –∞

u20x dx. 2

(2.10.25)

¯ ¯ u), Since ' is the convex function, which is the remaining part of its linear part at (9, therefore

194

2 The Vanishing Viscosity Method of Some Nonlinear Evolution System

¯ 2 ). ¯ 2 + |u0 – u| '(90 , u0 ) ≤ c(|90 – 9| ¯ = (10 – 1¯ , u0 – u) ¯ ∈ Cc2 × Cc2 , thus since ¯ u0 – u) By virtue of the assumptions (90 – 9, c0 given by eq. (2.10.25) is bounded, estimate (2.10.24) implies estimate (2.10.16). This completes the proof. ∎ Corollary 2.10.4. Assume 10 , 1¯ , u0 , u¯ satisfy the conditions of Theorem 2.10.3, if (1 – ¯ ∈ C(0, t; C02 × C02 ), then the solution (1, u) to problem (3.10.7) with the initial data 1¯ , u – u) (10 , u0 ) satisfies sup |u(x, t)| = ||u(t)||L∞ ≤ c0 , x  |1|dx ≤ cA ,

(2.10.26) (2.10.27)

A

¯ where c0 denotes any positive constant, which depends on 10 , u0 , 1¯ , u,  ¯ A), cA = cA (10 , u0 , 1¯ , u,

dx < ∞. A

Proof. Estimate (2.10.26) is derived from eq. (2.10.16) and the Sobolev inequality 1/2 sup |u(x, t)| ≤ ||u(t)||1/2 L2 ||ux (t)||L2 . x

Inequality (2.10.27) is deduced as the middle corollary of the a priori estimate (2.10.16). ∎ Using the integral expression of 1(x, t) and the a priori estimate for ||u(t)||L∞ in eq. (2.10.26) and the Gronwall inequality, we can deduce the a priori estimate for ||1(t)||L∞ =supx |1(x, t)|, which depends on :. Theorem 2.10.5. Assume 10 , 1¯ , u0 , u¯ satisfy the conditions of Theorem 2.10.3, then the solution (1, u) to problem (3.10.7) with the initial data (10 , u0 ) satisfies  7  t sup |1(x, t)| ≤ c0 ||10 ||L∞ exp c0 . : x

(2.10.28)

Proof. We regard the first equation of eqs (3.10.7) as the nonhomogeneous heatconducting equation, then 1(x, t) is of the form 

∞

1(x, t) =

–∞

10 (z)G(x – z, t)dz +

 t

(u(s, z) 0

+ 1(s, z)u(s, z))Gx (x – z, t – s)dzds, where G(x, t) is the heat kernel

∞ –∞

(2.10.29)

2.10 Initial Value Problem for Boussinesq Equations

G(x, t) = √



1 4:0t

exp

–

195

 x2 . 4:t

By eq. (2.10.29), we have 7

t ||1(x)||L∞ ≤ ||10 ||L∞ + c0 :  t  + c0 ||1(s)||∞ 0

≤ ||10 ||L∞

∞

|Gx (z – x, t – s)|dzds 8 7  t 1 t + c0 c ||1(s)||∞ ds + c0 , :(t – s) : 0 –∞

(2.10.30)

where we have used the fact 

∞

8

1 |Gx (z – x, t – s)|dz ≤ c :(t – s) –∞ c ≤  . :(t – s)



∞

|G(z – x, t – s)|dz

–∞

By the Gronwall inequality and eq. (2.10.30), we deduce that  7  t . ||1(t)||L∞ ≤ c0 ||10 ||L∞ exp c0 : ∎ Now we extend the local solution of the initial value problem (3.10.7) to the global solution. Theorem 2.10.6. Assume 10 , 1¯ , u0 , u¯ satisfy the conditions of Theorem 2.10.3, then there exists a unique global solution (1, u) to problem (3.10.7) with the initial data ¯ ∈ C([0, ∞); C02 × C02 ) and (10 , u0 ), such that (1 – 1¯ , u – u) 1 + 1(x, t) ≥ 0 ∀t > 0, x ∈ R. Proof. The global existence and uniqueness of the solution to the initial value problem (3.10.7) come from the local existence,uniqueness and the estimates of Corollary 2.10.15 and Theorem 2.10.5. The estimate for the lower bound comes from Lemma 2.10.2. ∎ In order to deduce a weak solution to eqs (3.10.6) with the initial data (10 , u0 ), we prove that for the solution sequence {1: , u: } to eqs (3.10.7) with the initial data (10 , u0 ), there exists a subsequence {1: , u: }, such that 1: converges weakly to 1, for every t, u: converges strongly to u in L2 , and the limit equation (1, u) is a weak solution to eqs (3.10.6) with the initial data (10 , u0 ). In order to derive the weak convergence of 1: ,

196

2 The Vanishing Viscosity Method of Some Nonlinear Evolution System

we apply some a priori estimates for 1: independent of : and the weak continuously compact Dunford theorem on L1 ; in order to deduce the convergence of u: on L2 , we use the fact that H 1 ↪ L2 compactly and some continuous conditions with respect to t, we first prove a convergent theorem. Theorem 2.10.7. Assume 10 , 1¯ , u0 , u¯ satisfy the conditions of Theorem 2.10.3, then the solution (1: , u: ) to problem (3.10.7) with the initial data 1: (0, x) = 10 (x),

u: (0, x) = u0 (x),

¯ ∈ C([0, ∞); C02 × C02 ), then there exists a subsequence 1: that converges (1: – 1¯ , u: – u) weakly on the compact set in the L1 sense, and there exists a subsequence u: such that for every t it converges strongly on L2 , that is, there exist 1, u such that (1) for any > ∈ L∞ (R+ × R) with compact support 

∞ ∞

(1: (x, t) – 1(x, t))>(x, t)dxdt = 0;

(2.10.31)

|u: (x, t) – u(x, t)|2 dx = 0 ∀t ∈ [0, T].

(2.10.32)

lim

:→0 0

–∞

(2) for every T > 0,  lim

∞

:→0 –∞

Proof. We first prove that on any fixed compact set S, for any > ∈ L∞ (R+ × R) and supp > ⊂ S, eq. (2.10.31) holds true, namely, there exists a subsequence 1: , such that 1: |S is convergent weakly. For this purpose, we apply the following convergent results: Lemma 2.10.8 (Dunford-Schwartz). If the set K is weakly sequentially compact, then  lim f (s),(ds) = 0 ,(E)→0 E

holds uniformly for f ∈ K. Otherwise, if ,(S) < ∞, for a bounded set K to become sequentially compact, this condition is sufficient. We set , = dxdt, G = {A : A ⊂ S, A, is measurable}, K = {1: |S }. By Dunford lemma, it’s sufficient to prove  lim f (t, x)dxdt = 0 ∀f ∈ K. ,(E)→0

E

We note that, if for some function >(1), we get that  E >(1)d, < ∞, then  1d, = 0.

lim

,(E)→0

E

>(1) 1

→ ∞(as1 → ∞) and

2.10 Initial Value Problem for Boussinesq Equations

197

For our case, >(1) = 1 log 1, we set E = Ek ∪ Ekc , E ⊂ S, here, Ek = {(x, t) ∈ E|1: ≤ k}, then    1: |S dxdt = 1: dxdt + 1: dxdt. (2.10.33) E

We note that

E∩Ekc

E∩Ek

 E∩Ekc

1: dxdt ≤

1 log k

 E∩Ekc

1: log 1: dxdt

By inequality (2.10.16), for sufficiently large k, we have  1: log 1: dxdt ≤ c0 |Pt E|, E∩Ekc

(2.10.34)

(2.10.35)

where Pt E = projection of E in t direction, |Pt E| = Lebesgue measure of Pt E. Therefore, by virtue of eqs (2.10.33), (2.10.34) and (2.10.35), for sufficiently large k, we have  c0 1: |S dxdt ≤ k,(Ek ∪ E) + |Pt E|. log k E Since |Pt E| is bounded, we deduce that  lim 1: |S dxdt = 0 ,(E)→0

E

holds uniformly for any 1: |S ∈ K. We deduce the boundedness of K on L1 from Corollary 2.10.15, eq. (2.10.27). Therefore, for any fixed compact set S, we choose a subsequence 1: that is weakly convergent in the L1 (S) sense. In order to get the weak convergence for any compact set in the L1 sense, we use the following diagonal procedure and set Sn = [0, n] × [–n, n]. Assume that 1n: is the corresponding L1 (Sn ) weakly convergent subsequence, we k choose these subsequences {1n:k } ⊃ {1n+1 :k }. Therefore, the diagonal sequence {1:k } weakly converges to some function 1(x, t) for every n in the L1 sense. In other words, assume > ∈ L∞ (R+ × R), supp > = S, choose n large enough, S ⊂ Sn , we still denote 1k:k = 1: , then for > ∈ L∞ (Sn ) = L1 (Sn ),  (1: – 1)>dxdt = 0. lim :→0

0

Therefore, we have constructed the subsequence {1: } satisfying eq. (2.10.31).

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2 The Vanishing Viscosity Method of Some Nonlinear Evolution System

Now we prove the strong convergence of the subsequence of {u: }. By the a priori ¯ ∈ H 1 , since the imbedding of H 1 ↪ L2 is estimate (2.10.16), we see that (u: (⋅, t) – u) compact; therefore, for any t, there exists a subsequence u: (t), and a function u(x, t) such that  ∞ lim |u:(t) (x, t) – u(x, t)|2 dx = 0, :(t)→0 –∞

and u– u¯ ∈ L2 , by the diagonal procedure, we can deduce the subsequence u: satisfying  lim

∞

:→0 –∞

|u: (x, t) – u(x, t)|2 dx = 0∀ finite t ∈ [0, T].

Since the weak convergence for every t, combined with the compactness, yields the strong convergence, namely, we have u: (x, t) → u(x, t) in L2 ∀ t ∈ [0, T]. Therefore, it’s sufficient to prove for any > ∈ Cc∞ (R),  lim

:→0 R

(u: (x, t) – u(x, t))>dx = 0,

which is derived from the fact that  lim

:→0 a

b

(u: (x, t) – u(x, t))dx = 0,

(2.10.36)

where a, b are any finite constants, in order to deduce eq. (2.10.36), it’s sufficient to prove some regularity with respect to t for u: (x, t), we will prove there exists a constant ¯ a, b) such that for all t1 , t2 ≥ 0, c = c(10 , 1¯ , u0 , u,  b     (u: (x, t2 ) – u: (x, t1 ))dx ≤ c|t2 – t1 |. 

(2.10.37)

a

Inequality (2.10.37) combined with the strong convergence for any finite t ∈ [0, T] and the uniform boundedness for u: in the sense of L∞ module yields eq. (2.10.36); we thus deduce the proof of weak convergence. In fact, assume eq. (2.10.37) holds, we have  b   b         ¯ (u: (x, t) – u:¯ (x, t))dx ≤  (u: (x, t) – u: (x, t))dx  a a   b   b        ¯ ¯ ¯ (u: (x, t) – u:¯ (x, t))dx +  (u:¯ (x, t) – u:¯ (x, t))dx + a

a

1  2 ≤ c|t – t¯| + |b – a| |u: (x, t¯) – u:¯ (x, t¯)|2 dx ,

2.10 Initial Value Problem for Boussinesq Equations

199

where t¯ is a real number that approximates t sufficiently. The convergence of u: (x, t¯) in L2 implies that u: (x, t¯) is a Cauchy sequence; therefore, u: (x, t) is convergent weakly for any t. In order to complete the proof, it’s sufficient to testify eq. (2.10.37). As we know, u: satisfies the following integral equation:  t ∞ ! 1 " (2.10.38) K(x – z) 1: + u2: dzds, u: (x, t) = u0 (x) + 2 0 –∞ where 1 K(x) = sgn(x)exp(–|x|). 2 For u: (x, t2 ) – u: (x, t1 ), integrating over [a, b] with respect to x, we have  b (u: (x, t2 ) – u: (x, t1 ))dx a



b  t2



t1 b  t2



= 

a

= a

t1

∞

! 1 " K(x – z) 1: + u2: dzdsdx 2 –∞ ∞

  1 K(y) 1: (s, y + x) + u2: (s, y + x) dydsdx, 2 –∞

(2.10.39)

where we take the transformation z = y + x, and the above integral is bounded by  t2  ∞  b c0 |b – a| ⋅ |t2 – t1 | + |K(y)| |1: (s, y + x)|dxdyds, (2.10.40) t1

we can prove the estimate 

b

a

–∞

a

|1: (s, y + x)|dx ≤ c0 (1 + |b – a|).

(2.10.41)

In fact, we take the transformation r = y + x, and the left-hand side of eq. (2.10.41) is divided into two parts:  y+b  b |1: (s, y + x)|dx = |1: (s, r)|dr a y+a   |1: (s, r)|dr + |1: (s, r)|dr, (2.10.42) = I∩PEr0

I∩(PEr0 )c

where Er0 = {(s, r) : 1(s, r) ≤ r0 }, PEr0 = projection of Er0 in R, I = [y + a, y + b].

200

2 The Vanishing Viscosity Method of Some Nonlinear Evolution System

By virtue of the a priori estimate of eq. (2.10.16), we choose r0 large enough such that  1: log 1: dx ≤ c0 . (PEr0 )c

Therefore, the right-hand side of eq. (2.10.42) is bounded by  1 1: log 1: dx ≤ c0 (1 + |b – a|). |b – a|r0 + log r0 I∩(PEr0 )c By eqs (2.10.40) and (2.10.41), we deduce inequality (2.10.37) and complete the proof. ∎ Theorem 2.10.9. Assume 10 , 1¯ , u0 , u¯ satisfy the conditions of Theorem 2.10.3, u, 1 are the derived limit functions of Theorem 2.10.7, then (1, u) is a weak solution to the Boussinesq equations with the initial data (10 , u0 ). Proof. It’s sufficient to prove that for any > ∈ Cc∞ (R+ × R),  ∞ ∞  ∞ ∞ 1>t dxdt + (u + 1u) >x dxdt = 0, 0 –∞ 0 –∞   ∞ ∞   ∞ ∞ u2 u>t dxdt + >x dxdt, 1+ 2 0 –∞ 0 –∞  ∞ ∞ = u>xxt dxdt. 0

(2.10.43)

–∞

Assume that 1: , u: are the convergent subsequences derived in Theorem 2.10.7, as we know, for each :, we have  ∞ ∞  ∞ ∞ 1: >t dxdt + (u: + 1: u: ) >x dxdt 0 –∞ 0 –∞  ∞  ∞ 1: >xx dxdt =: 0 –∞   ∞ ∞  ∞ ∞ u2 u: >t dxdt + 1: + : >x dxdt, 2 0 –∞ 0 –∞  ∞ ∞ = u: >xxt dxdt. (2.10.44) 0

We note that

∞ ∞

 0

–∞

–∞

 1: >xx dxdt =

S

1: >xx dxdt, S = supp >,

by Corollary 2.10.15, eq. (2.10.27), we deduce that 

∞ ∞

lim :

:→0

0

–∞

1: >xx dxdt = 0,

2.10 Initial Value Problem for Boussinesq Equations

201

which combined with eq. (2.10.44) implies that in order to prove that (1, u) satisfies eq. (2.10.43), it’s sufficient to construct the identity: 6 5 ∞  ∞  ∞ ∞ lim (1: – 1)>t dxdt + (u: + 1: >: – u – 1u)>x dxdt = 0, (2.10.45) :→0

5

0

–∞ ∞ ∞

0



–∞ ∞ ∞ 2 u:

(u: – u)>t dxdt + :→0 0 –∞ 0 6  ∞ ∞ + (u: – u)>xxt dxdt = 0, lim

0

–∞

+ 1: –

2

 u2 – 1 >x dxdt 2 (2.10.46)

–∞

By virtue of the weak convergence of 1: and the strong convergence of u: , in order to derive eq. (2.10.45), it’s sufficient to prove  ∞ ∞ lim (1: u: – 1u)>x dxdt = 0. (2.10.47) :→0 0

–∞

We note that 

∞ ∞ 0

(1: u: – 1u)>x dxdt  1: (u: – u)>x dxdt + u: (1: – 1)>x dxdt.

–∞



= S

(2.10.48)

S

By virtue of the strong convergence on L2 , there exists a subsequence (not relabeled) of u: which converges pointwisely, and the corresponding 1: satisfies the first equation of eqs (3.10.7); so we prove that eqs (2.10.45) and (2.10.46) hold true for this subsequence. Since u: (x, t) → u(x, t) a.e. for x, and u: ∈ L∞ , therefore, for fixed " > 0, there exists :0 > 0, such that when : ≤ :0 , we have |u: (x, t) → u(x, t)| ≤ ", t ∈ [0, T], a.e. convergent for x, u ∈ L∞ . Therefore, we have 

 lim

:→0

S

1: (u: – u)>x dxdt ≤ "

S

|1: |>x dxdt ≤ "CS ,

where ¯ u0 , S). CS = CS (¯1, 10 , u, Since u ∈ L∞ , u>x ∈ L∞ with compact support, by Theorem 2.10.7, we have  ∞ ∞ lim u(1: – 1)>x dxdt = 0. :→0 0

–∞

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2 The Vanishing Viscosity Method of Some Nonlinear Evolution System

Then the right-hand side of eq. (2.10.48) converges to zero as : → 0. In order to deduce eq. (2.10.46), it’s sufficient to prove that  ∞ ∞ lim (u2: – u2 )6dxdt = 0, :→0 0

–∞

which is derived from the dominated convergence theorem and  ∞ (u2: (x, t) – u2 (x, t))dx = 0, lim :→0 –∞  ∞ ∞ |u2: – u2 ||>|dxdt ≤ c0 . 0

–∞

In order to prove the theorem, we must prove that the solution achieves the given initial data. In detail, lim u(x, t) = u0 (x), holds a.e. with respect to t, t→0  ∞ lim (1(x, t) – 10 (x))>(x)dx = 0 ∀> ∈ Cc∞ (R). t→0 –∞

(2.10.49) (2.10.50)

In order to prove eq. (2.10.49), we notice that |u(x, t) – u0 (x)| ≤ |u(x, t) – u: (x, t)| + |u: (x, t) – u0 (x)|. It’s sufficient to prove that limt→0 u: (x, t) = u0 (x) holds uniformly with respect to :. From the expression of u: (x, t) in eq. (2.10.38), we see that  t  ∞    u2 K(x – z)(1: + : )dzds |u: (x, t) – u0 (x)| ≤  2 0 –∞   t   t ∞ 2   u: K(x – z)1: dzds dzds +  ≤ k 2 0

–∞

0

 t + 0

 t 0

+

1: dzds

(PErk0 )c

≤ c0 t + r0 1 logr0

 t 0

PEr0

PErk0

(PErk0 )c

|K(x – z)|dzds 1: log 1: dzds

≤ c0 t + c. In order to deduce eq. (2.10.50), we need the following two auxiliary lemmas. Lemma 2.10.10. Assume S is a compact set, N(S) is the neighborhood with finite measure of S, if tn is an arbitrary sequence, which satisfies limn→∞ tn = 0, then there exists a subsequence (not relabeled), such that

203

2.10 Initial Value Problem for Boussinesq Equations



b

lim

tn →0 a

(1: (x, tn ) – 10 (x))dx = 0,

(2.10.51)

where the above equality holds true for any a, b in N(S) with respect to :. Proof. We subtract 10 from the integral expression (2.10.29) and integrate with respect to x, we get 

b b

(1: (x, tn ) a ∞ b

a



– 10 (x))dx

G(x – z, tn )(10 (z) – 10 (x))dzdx

=

–∞ a  tn  ∞  b

Gx (x – z, tn – s)

+ –∞

0

a

× [1: (z, s) × u: (z, s) + u: (z, s)]dxdsdx. We consider the left-hand side of the above equality as the functions with respect to a, b, and choose its absolute value, then integrating with respect to a, b, here a ∈ N(S), b ∈ N(S); then we get the following inequality: 

 N(S)

 

 a

N(S)





 

≤ N(S)



b



∞ b

–∞

N(S)



 

+ N(S)

 (1: (x, tn ) – 10 (x))dxdbda

N(S)



a tn



 G(x – z, tn )(10 (z) – 10 (x))dzdxdbda ∞ b

–∞

0

 + u: (s, z))dzdsdxdbda.

a

Gx (x – z, tn – s)(1: (s, z)u: (s, t) (2.10.52)

Our main motivation is to prove that the right-hand side of eq. (2.10.52) converges to zero uniformly as tn → 0, from which we deduce that  b     (1: (x, tn ) – 10 (x))dx lim tn →0  a

= 0,

L1 (N(S)×N(S))

holds uniformly with respect to :, then there exists a subsequence tn such that  lim

tn →0 a

b

(1: (x, tn ) – 10 (x))dx = 0,

holds uniformly with respect to : a.e. on N(S)×N(S); therefore, the proof of this lemma is completed. For this purpose, namely, the right-hand side of eq. (2.10.52) converges to zero, we analyze every term of the integral. Obviously, the first term converges to zero as tn → 0, the second term

204

2 The Vanishing Viscosity Method of Some Nonlinear Evolution System





N(S)



b  tn

a

N(S)



0

2

≤ c0 |N(S)| |b – a|  ≤ c0 tn .

∞

–∞  tn

|Gx (z – x, tn – s)u: (z, s)|dzdsdxdbda 

∞

|Gx (z – x, tn – s)|dzds

–∞

0

Thus it’s sufficient to prove that 

 N(S)

N(S)

   

b  tn

a



∞

–∞

0

  Gx (z – x, tn – s)1: (z, s)u: (z, s)dzdsdxdbda

converges to zero as tn → 0. By Fubini’s theorem and eq. (2.10.16), we deduce that 

 N(S)

  b ∞    1 (z, s)u (z, s)[G(b – z, t – s) – G(a – z, t – s)]dzds : n n N(S)  a –∞ :  dbda,

(2.10.53)

which is bounded by 5





tn



∞

c0 

N(S)



N(S)

0 tn





–∞ ∞

+ N(S)

N(S)

–∞

0

|1: (z, s)|G(b – z, tn – s)dzdsdbda

6 |1: (z, s)|G(a – z, tn – s)dzdsdbda .

(2.10.54)

We take the transformation y = b – z in the first integral of the above formulation and integrate with respect to b, then we take the transformation 9 = a – z in the second integral and integrate with respect to a, and then the last two integrals are bounded by 5



tn





∞

c0

G(y, tn – s) 

N(S)



0 tn



–∞ ∞



N(S)

G(9, tn – s)

+ N(S)

0

–∞

N(S)

|1: (y + b, s)|dbdsdyda 6 |1: (9 + a, s)|dad9dsdb .

(2.10.55)

By eq. (2.10.27), we have  N(S)

 |1: (y + b, s)|db +

N(S)

|1: (9 + a, s)|da ≤ CN(S) ,

¯ N(S)). Then integral (2.10.55) is bounded by tn CN(S) ; where CN(S) = CN(S) (10 , u0 , 1¯ , u, thus, as tn → 0, the right-hand side of eq. (2.10.52) converges to zero, and the proof of this lemma is completed. ∎

2.10 Initial Value Problem for Boussinesq Equations

205

Lemma 2.10.11. Assume > ∈ Cc∞ (R), supp > ⊂ S, and t > 0, tn ∈ [0, T], then there exists a subsequence 1: of the solution to problem (3.10.7), such that 

∞

lim

:→0 –∞

(1: (x, tn ) – 1(x, tn ))>dx = 0.

(2.10.56)

Proof. By Dunford lemma and the a priori estimates (2.10.16) and (2.10.27), then for every t, there exists a subsequence 1: , such that 

∞

lim

:→0 –∞

(1: (x, t) – 1(x, t))>dx = 0,

then by the diagonal procedure, we deduce eq. (2.10.56).

∎

Now we apply Lemmas 2.10.10 and 2.10.11 to prove that the solution 1(x, t) achieves the initial data. More precisely, for any > ∈ Cc∞ (R) and a > 0, then there exists a positive constant N, such that for every n ≥ N, we have    

∞

–∞

  (1(x, tn ) – 10 (x))>dx ≤ a,

(2.10.57)

where tn is the subsequence in Lemma 2.10.10. supp > = S. We set that sm (x) is the step function sequence, which converges to >. supp sm (x) is in the neighborhood N(S) of S, sm (x) =



"r 7r ,

(2.10.58)

r=0

where 7r is the character function on the interval [ar , br ], without loss of generality, we assume that ar , br such that eq. (2.10.51) does not hold, and the measure is not zero. Now we estimate the left-hand side of eq. (2.10.57),  ∞     (1(x, tn ) – 10 (x))>(x)dx     –∞      (1(x, tn ) – 1: (x, tn ))>dx +  1: (x, tn )(> – sm )dx ≤  N(S) N(S)         + (1: (x, tn ) – 10 )sm dx +  10 (sm – >)dx N(S)

N(S)

= I + II + III + IV, where the sequence 1: is given by Lemma 2.10.11. By Lemma 2.10.11, we choose : such that I ≤ a/4.

206

2 The Vanishing Viscosity Method of Some Nonlinear Evolution System

 We notice that since N(S) |1: |dx ≤ c, eq. (2.10.27) and the assumption  N(S) |10 (x)|dx ≤ c0 , then we choose m sufficiently large such that II + IV ≤ a/2. For the boundedness of the integral III, we notice that  lim (1: (x, tn ) – 10 (x))sm (x)dx tn →0 N(S)

=





"r lim

br

tn →0 ar

(1: (x, tn ) – 10 (x))dx = 0

holds uniformly with respect to :. Here we have used the results in Lemma 2.10.10; thus, there exists N such that for every n > N, III ≤ a/4, and thus we complete the proof of the theorem.

∎

Assume that 30 (1 + 1) = 30 (1 + 1, 1 + 1¯ ) is a convex positive function defined by eq. (2.10.15), we set   ∞  D30 = 1 : 30 (1 + 1)dx < ∞ . –∞

Then we will prove that there exists a solution to problem (3.10.6) with the initial data D30 × H 1 in the functional space L∞ (0, ∞; D30 × H 1 ). ¯ 10 – 1¯ have compact Theorem 2.10.12. Assume 1¯ , u¯ are given constants, and u0 – u, 1 supports, 10 ∈ D30 , u0 – u¯ ∈ H , then there exists a weak solution to eqs (3.10.6) with the initial data (10 , u0 ). Proof. We first smooth the initial data, and then take the limit. Assume that > ∈ Cc∞ (R) such that  ∞ >(x)dx = 1, > ≥ 0, supp > ⊂ {x : |x| ≤ 1}. –∞

We define >, (x) = ,–1 >(,–1 x), ,

,

10 = 10 ∗ >, , u0 = u0 ∗ >, .

(2.10.59)

We choose , = :. Assume that (1: , u: ) is a solution to eqs (3.10.7) with the initial data (1:0 , u:0 ), we will prove that there exists a subsequence (1: , u: ) which converges to the weak solution of the Boussinesq equations (3.10.6) with the initial data 10 , u0 . In fact, we notice that if we can prove the following inequality

2.10 Initial Value Problem for Boussinesq Equations



∞



∞

u2:x dx –∞ –∞ 2  ∞ 2  ∞ u0x '(90 , u0 )dx + dx, ≤ –∞ –∞ 2 '(9: , u: )dx +

207

(2.10.60)

where ' is entropy (2.10.23), ¯ 2 (u – u) ¯ + 30 (9, 9), 2 9 = 1 + 1, 90 = 1 + 10 , 90, = 90 ∗ >, , 9: = 1 + 1: .

'(9, u) =

We repeat the proof of Theorems 2.10.7 and 2.10.9, such that there exists a subsequence (1: , u: ), which converges weakly to the solution (1, u) of eq. (3.10.6). In order to prove that (1, u) achieves the initial data (10 , u0 ), it’s sufficient to prove that for any 8 ∈ Cc∞ (R), we have   ∞       (1(x, t) – 1(x, 0))8dx ≤  (1(x, t) – 1: (x, t))8dx –∞  ∞    –∞     ∞ :  : + (1: (x, t) – 10 )8dx +  (10 – 10 )8dx.

   

∞

–∞

(2.10.61)

–∞

Similarly, we have    

  ∞     (u(x, t) – u(x, 0))8dx ≤  (u(x, t) – u: (x, t))8dx –∞  –∞   ∞    ∞ :   :    (u: (x, t) – u0 (x))8dx +  (u0 (x) – u0 )8dx. + ∞

–∞

(2.10.62)

–∞

Obviously, for given ! > 0, there exist :0 , ", such that : ≤ :0 , |t| ≤ ", the three integrals on the right-hand side of eqs (2.10.61) and (2.10.62) are smaller than !/3, thus 

∞

lim

(1(x, t) – 10 )8dx = 0,

lim

(u(x, t) – u0 )8dx = 0.

t→0 –∞  ∞ t→0 –∞

In order to complete the proof, we need to construct the entropy inequality (2.10.60). By the a priori estimates derived by eq. (2.10.16), we have 

∞



∞

u2:x dx –∞ –∞ 2  ∞ : 2  ∞ (u0 )x '(9:0 , u:0 )dx + dx. ≤ 2 –∞ –∞ '(9: , u: )dx +

(2.10.63)

208

2 The Vanishing Viscosity Method of Some Nonlinear Evolution System

By Jensen’s inequality, we get that  ∞  : : '(90 , u0 )dx +

(u:0 )2x dx 2 –∞ –∞  ∞ 2  ∞ (u0 )x '(90 , u0 )dx + dx. ≤ 2 –∞ –∞ ∞

Thus we complete the proof of the theorem. In fact, we set   z–x d:(x) = 8: dz. :

(2.10.64)

(2.10.65)

¯ is a positive convex function, which is defined by We set that 30 (9) = 30 (9, 9) eq. (2.10.15), and #1 , #2 are convex functions, ¯ 2 (u – u) u2 , #2 = , 2 2 ¯ 2 (u – u) '(9, u) = + 30 (9). 2 #1 (u) =

Therefore, we have '(9:0 , u:0 ) +

   ∞ u0 d: + #2 u0x d: –∞ –∞   ∞ + 30 90 d: .

(u:0 )2x = #1 2



∞

–∞

By Jensen’s inequality of convex functions, we have     ∞  ∞  ∞ u0 d:(x) + #2 u0x d:(x) + 30 90 d:(x) #1 –∞ –∞ –∞  ∞  ∞  ∞ ≤ #1 (u0 )d:(x) + #2 (u0 )d:(x) + 30 (90 )d:(x). –∞

–∞

–∞

Thus,  ∞ : 2  ∞ ∞ (u0 )x '(9:0 , u:0 )dx + #1 (u0 )d:(x)dx dx ≤ 2 –∞ –∞ –∞ –∞  ∞ ∞  ∞ ∞ + #2 (u0x )d:(x)dx + 30 (90 )d:(x)dx. 

∞

–∞

–∞

–∞

–∞

By Fubini’s theorem, changing the order of integration of the right-hand side of the inequality yields that 

∞ –∞



≤

∞

–∞



(u:0 )2x dx 2 –∞    ¯ 21 x–z (u0 (z) – u) > dxdz 2 : |x–z|≤: :

'(9:0 , u:0 )dx +

∞

2.10 Initial Value Problem for Boussinesq Equations

209



 x–z dxdz : –∞ |x–z|≤:     ∞ 1 x–z 30 (9d(z)) > dxdz + : |x–z|≤: : –∞  ∞ 2  ∞ u0x '(90 , u0 )dx + = dx. –∞ –∞ 2 

+

∞

u20z (z) 1 2 :



>

∎

Thus we complete the proof of the theorem.

Theorem 2.10.13. Under the assumptions of Theorem 2.10.12, assume that (1, u) is a ¯ ∈ L∞ (0, ∞; D30 × H 1 ). weak solution to eq. (3.10.6), then (1, u – u) Proof. Assume that (1, u) is a solution derived in Theorem 2.10.12. Since H 1 is a reflexive space, whose bounded module sequence has a subsequence that converges to an element in H 1 , therefore, for every t, there exists a subsequence u:(t) , such that ¯ = u – u, ¯ weakly in H 1 . lim (u:(t) – u)

:(t)→0

Since H 1 module is lower semi-continuous, we have 

∞

–∞

 ¯ 2 + u2x )dx ≤ lim ((u – u)

∞

:(t)→0 –∞

¯ 2 + u2:(t)x )dx ≤ c0 , ((u:(t) – u)

the last term of which comes from estimate (2.10.16). Thus u – u¯ ∈ L∞ (0, ∞; H 1 ). In order to prove that 1 ∈ D30 in the solution (1, u), we need to apply with the following lemma. Lemma 2.10.14. Assume that E is a convex set in the local convex space, then the weak ¯ closure E¯ 9 of E is equal to its strong closure E. Now we apply with this lemma, we set X = L1 (Kn ), Kn = (–n, n),    E = 1 : 30 (1 + 1)dx ≤ c0 , where 30 (1 + 1) = 30 (1 + 1, 1 + 1¯ ), which is defined by eq. (2.10.15). Since 30 is convex, then E is convex. By the a priori estimate (2.10.16), for any : > 0, we have 

∞

–∞

30 (1 + 1: )dx ≤ c0 ,

where (1: , u: ) is the solution to eqs (3.10.7), thus 1: |Kn ∈ E.

210

2 The Vanishing Viscosity Method of Some Nonlinear Evolution System

For fixed t, we set that 1:(t) converges weakly to 1 in L1 , then by Lemma 2.10.14, there exists a sequence -k ∈ E, which converges strongly to 1|Kn in L1 (Kn ). Therefore, lim -k (x) = 1|Kn (x, t), a.e. on Kn with respect to x.

k→∞

Then lim 30 (-k + 1) = 30 (1|Kn (x, t) + 1), a.e. on Kn with respect to x.

k→∞

Since 30 is nonnegative, by Fatou’s lemma, 

n

 30 (1(x, t) + 1)dx =

30 (1 + 1)dx  30 (-k (x) + 1)dx ≤ c0 . ≤ lim

–n

Kn

k→∞ Kn

Therefore, 

n

30 (1 + 1)dx ≤ c0 .

–n

We deduce 1 ∈ D30 as n → ∞.

∎

Since D30 is contained in L1,loc (R), we have Corollary 2.10.15. Under the assumptions of Theorem 2.10.9, the solution (1, u) to eq. ¯ ∈ L1,loc (R) × H 1 (R). (3.10.6) satisfies that (1, u – u)

2.11 Initial Value Problem for Langmuir Turbulence Equations We consider the initial value problem for Langmuir turbulence Zakharov equations with a small parameter : ≥ 0:

(S: )

⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩

i

∂E + △E = nE, x ∈ Rd , d = 1, 2, 3, ∂t :2 ntt – △n = △(|E|2 ), t ∈ R+ ,

E(x, 0) = E0 (x), n(x, 0) = n0 (x), nt (x, 0) = n1 (x).

We will investigate the asymptotic behavior of the solution (E: , n: ) to the problem S: as : → 0. In fact, (E: , n: ) converges to (E, –|E|2 ), where E is the solution to the initial value problem for the nonlinear Schro¨ dinger equations

2.11 Initial Value Problem for Langmuir Turbulence Equations

211

⎧ ⎨ i ∂E + △E + |E|2 E = 0, x ∈ Rd , t ∈ R+ , ∂t (NLS) ⎩ E(x, 0) = E0 (x), d = 1, 2, 3. Moreover, we can prove that in the three-dimensional case, the smooth solution to the problem (S: ) is defined on [0, Tmax (:)), where Tmax (:) → ∞, : → 0. Now we first consider the weak convergence of the following S: equations: ⎧ : iEt + △E: = n: E: , x ∈ Rd , d = 1, 2, 3, ⎪ ⎪ ⎪ ⎪ ⎨ :2 n: – △n: = △|E: |2 , t ∈ R+ , tt ⎪ E: (x, 0) = E0 (x) ∈ H m (Rd ), n(x, 0) = n0 (x) ∈ H m–1 (Rd ), ⎪ ⎪ ⎪ ⎩ nt (x, 0) = n1 (x) ∈ H m–2 (Rd )

(2.11.1) (2.11.2) (2.11.3)

as : → 0, where m is an arbitrary nonzero integer. Assume that there exists some function u0 (x), such that n1 (x) = △u0 (x), ∇u0 (x) ∈ (L2 (Rd ))d .

(2.11.4)

We introduce the potential function V : (x, t) satisfying ∂n: + divV : = 0, ∂t ∂V : :2 + ∇n: + ∇|E: |2 = 0, V : (x, 0) = –∇u0 (x). ∂t

(2.11.5) (2.11.6)

Similarly as the method to investigate the Zakharov equation in Section 9, we deduce the following invariant variables: 1 1 N : = ||E: ||2L2 , H : = ||∇E: ||2L2 + ||:V : ||2L2 + ||n: ||2L2 , 2 2  + n: |E: |2 dx.

(2.11.7)

Rd

Applying with the conservation law, we can easily construct the existence theorem of the global weak solution to the S: equations, we have Theorem 2.11.1. Assume E0 (x) ∈ H 1 (Rd ), n0 (x) ∈ L2 (Rd ), n1 (x) ∈ H –1 (Rd ), and n1 (x) satisfies eq. (2.11.4). When d = 2, ||E0 ||L2 ≤ ||8||L2 , where 8 is a positive solution to the equation △ u – u + u3 = 0.

(2.11.8)

When d = 3, ||E0 ||H 1 is small enough, then the equations S: admit the global weak solution

212

2 The Vanishing Viscosity Method of Some Nonlinear Evolution System

E: ∈ L∞ (R+ , H 1 (Rd )), n: ∈ L∞ (R+ , L2 (Rd )). In order to construct the weak convergence of equations (S: ), we need to introduce the following known lemma. Lemma 2.11.2. Assume that B0 , B, B1 are three reflexive Banach spaces, and the imbedding B0 → B is compact. We set   ∂v W = v ∈ Lp0 (0, T; B0 ), ∈ Lp1 (0, T; B1 ) , ∂t T < ∞, 1 < p0 , p1 < ∞. W is a Banach space, with the module ||v||W = ||v||Lp0 (0,T;B0 ) + ||vt ||Lp1 (0,T;B1 ) , then the imbedding of W ↪ Lp0 (0, T; B) is compact. Lemma 2.11.3. Assume K is an open set in Rn , g, g: ∈ Lp (Rn ), 1 < p < ∞, such that g: → g a.e., ||g: ||Lp (K) ≤ C. Then g: → g weakly in Lp (K). Theorem 2.11.4. Under the assumptions of Theorem 2.11.1, (E: , n: ) is an arbitrary solution of (S: ), then as : → 0 (E: , n: ) → (E, –|E|2 ), weakly in L∞ (R+ , H 1 (Rd )) × L∞ (R+ , L2 (Rd )), where E(x, t) is the unique solution to initial value problem for the nonlinear Schrödinger equation (NLS) i

∂E + △E + |E|2 E = 0, E(x, 0) = E0 (x) ∈ H 1 (Rd ). ∂t

Proof. (i) Under the assumptions of Theorem 2.11.1, the invariant relationship (2.11.7) implies that ||E: ||L∞ (R+ ,H 1 (Rd )) , ||n: ||L∞ (R+ ,L2 (Rd )) and ||:V : ||L∞ (R+ ,L2 (Rd )) are bounded uniformly with respect to :; therefore, there exists a subsequence of (E: , n: , :V : ) which converges weakly to the limit (E, n, 9). In detail, E: → E weakly ∗ in L∞ (R+ , H 1 (Rd )), :

+

d

n → n weakly ∗ in L∞ (R , L2 (R )), :

+

d

:V → 9 weakly ∗ in L∞ (R , L2 (R )).

(2.11.9) (2.11.10) (2.11.11)

2.11 Initial Value Problem for Langmuir Turbulence Equations

213

We notice that the following mappings are continuous (d = {1, 2, 3}): H 1 (Rd ) → L4 (Rd ), H 1 (Rd ) × L2 (Rd ) → H –1 (Rd ) u → u, (u, v) → u, v.

(2.11.12)

We deduce from eqs (2.11.9) and (2.11.10) that ||△ E: ||L∞ (R+ ,H –1 (Rd )) , |||E: |2 ||L∞ (0,T;L2 (Rd )) and ||n: E: ||L∞ (R+ ,H –1 (Rd )) are bounded uniformly with respect to :, thus we have n: E: → z weakly in L∞ (R+ , H –1 (Rd )), : 2

d

(2.11.13)

|E | → f weakly in L∞ (0, T; L2 (R ))

(2.11.14)

△E: → △E weakly ∗ in L∞ (R+ , H –1 (Rd )),

(2.11.15)

and

which combined with eqs (2.11.9) and (2.11.13), and eq. (2.11.1) yields that ∂E ∂E: → weakly ∗ in L∞ (R+ , H –1 (Rd )), ∂t ∂t

(2.11.16)

From the above results, if we can construct n + |E|2 = 0, 2

Z + E|E| = 0,

(2.11.17) (2.11.18)

then the proof of the theorem is completed. We will prove eqs (2.11.17) and (2.11.18) successively. (ii) Proof of eq. (2.11.17). Assume K is an arbitrary bounded subset in Rd , we first note that the embedding of H 1 (K) ↪ L4 (K) is compact,

(2.11.19)

and for any Banach space X, the embedding of L∞ (R+ , X) ↪ L2 (0, T; X)is continuous,

(2.11.20)

thus, by virtue of eqs (2.11.9), (2.11.16) and Lemma 2.11.2, where B0 = H 1 (K), B = L4 (K), B1 = H –1 (K), the fact that the subsequence E: |K converges strongly to E|K in L2 (0, T; L4 (K)) implies that :→0

E: → E, strongly in L2 (0, T; L4loc (Rd )), : :→0

E → E, a.e. (x, t) ∈ (Rd × (0, T));

(2.11.21) (2.11.22)

214

2 The Vanishing Viscosity Method of Some Nonlinear Evolution System

by virtue of eqs (2.11.14), (2.11.20) and (2.11.22) and Lemma 2.11.3, we see that f = |E|2 . Since ||:V : ||L∞ (R+ ,L2 (Rd )) is bounded with respect to :. By eq. (2.11.6), we have :→0

grad(n: + |E: |2 ) → 0, in D ′ (R+ , L2 (Rd )). Therefore, we get that grad(n + |E|2 ) = 0. Since (n + |E|2 ) ∈ L∞ (R+ , L2 (Rd )), then n + |E|2 = 0.

(2.11.23)

(iii) Proof of eq. (2.11.18). We can prove that n: E: → –E|E|2 weakly in L2 (0, T; H –1 (Rd )), which combined with eqs (2.11.13) and (2.11.20) yields that z = –E|E|2 . In fact, let 6 be some test function in L2 (0, T; H 1 (Rd )), which vanishes outside a compact set K ⊂ Rd , then we have  T (n: E: + E|E|2 )6dxdt 0



Rd T

= =

n: (E: – E)6dxdt +

0 K I:1 (6) + I:2 (6).



T

0

 K

(n: + |E|2 )E6dxdt

Since |I:1 (6)| ≤ ||n: ||L∞ (R+ ,L2 (Rd )) ||6||L2 (0,T;L4 ) × ||E: – E||L2 (0,T;L4 ) → 0,

it is derived from the fact that K is bounded and eqs (2.11.10) and (2.11.21). Since E6 ∈ L1 (0, T; L2 ), ||E6||L1 (0,T;L2 ) ≤ ||E||L2 (0,T;L4 ) ||6||L2 (0,T;L4 ) < ∞. Therefore, by using the fact that n: → –|E|2 weakly * in L∞ (0, T; L2 ), we deduce that :→0

I:2 (6) → 0, then for any test function 6, 

T 0

 Rd

(n: E: + E|E|2 )6dxdt → 0, : → 0.

On the other hand, ||n: E: + E|E|2 ||L2 (0,T;H –1 (Rd )) is bounded uniformly with respect to :; by the Ascoli theorem, we deduce that n: E: → –E|E|2 weakly in L2 (0, T; H –1 (Rd )), and thus we complete the proof of the theorem. ∎ Obviously, if the initial data is more smooth, then we will get higher regularity. But if the initial data does not satisfy the compatible condition, namely, |E0 (x)|2 + n0 (x) ≠ 0, then n: does not converge to –E2 , when the initial boundary layer occurs, which will " ! be revised by the function n˜ : (x, t) = n˜ x, :t , where n˜ is the solution to the initial value problem for the wave function

215

2.11 Initial Value Problem for Langmuir Turbulence Equations

∂ 2 n˜ – △˜n = 0, x ∈ Rd , t ∈ R+ , ∂t2 ∂ n˜ ˜ 0) = |E0 |2 + n0 = n˜ 0 (x), (x, 0) = 0. n(x, ∂t

(2.11.24)

Remark. For the two-dimensional and three-dimensional cases for problem (2.11.24), the following L∞ – L1 estimates ˜ t)||L∞ ≤ ||D! n(⋅,

c (1 + t)

d–1 2

||n˜ 0 ||W 3+|!|

(2.11.25)

1

hold. Now we start to prove the strong convergence of F : = E: – E, u: = n: + |E|2 – n˜ : , where (E: , n: ) is the solution of equation S: , E is the solution to the equation NLS, n˜ is the solution of eq. (2.11.24), then (F : , u: ) satisfies the following equations and initial data: ⎧ : iF + △F : = (n: + n˜ : )(E + F : ) – |E|2 F : , x ∈ Rd , t ∈ R+ , (2.11.26) ⎪ ⎪ ⎪ t ⎨ 2 : : : 2 : 2 2 ¯ (2.11.27) (T: ) : utt – △u = △(|F | + 2ReEF ) + : |E|tt , ⎪ : 2 ⎪ ⎪ ⎩ F : (x, 0) = u: (x, 0) = 0, ∂u (x, 0) – n1 = ∂|E| (x, 0). (2.11.28) ∂t ∂t ¯ then ∂ |E|2 (x, t) = div g. If n1 (x) satisfies We notice that, if we set g(x, t) = 2Im E∇E, ∂t condition (2.11.4), then by introducing the potential function V : (x, t) we testify that ∂V : + ∇u: + ∇(|F : |2 + 2ReEF¯ : ) = 0, ∂t ∂u: + div V : = div g, V : (x, 0) = –∇u0 . ∂t

:2

(2.11.29) (2.11.30)

As we know, in the one-dimensional and two-dimensional cases, if E0 ∈ H m , n0 ∈ H m–1 , n1 ∈ H m–2 (m ≥ 3), and ||E0 ||L2 ≤ ||8||L2 (two-dimensional case), then there exists a unique global solution (E: , n: ) to equation (S: ), E: ∈ Lloc,∞ (R+ , H m ), n: ∈ Lloc,∞ (R+ , H m–1 ), and there exists a unique global solution E ∈ C0 (R+ , H m ) to equation NLS. In order to prove that F : → 0, u: → 0 (strong convergence), we need higher regularity for the initial data. Theorem 2.11.5 (One-dimensional case). If the initial data satisfies the following conditions: E0 ∈ H m+2 (R), n0 ∈ H m+1 (R), n1 ∈ H m (R), m ≥ 3, n1 satisfies eq. (2.11.4), i.e., n1 = △u0 , ∇u0 ∈ L2 (R),

(H1)

216

2 The Vanishing Viscosity Method of Some Nonlinear Evolution System

then there exists a function M(t) ∈ Lloc,∞ (R+ ), such that √ ||E: (⋅, t) – E(⋅, t)||H m (R) ≤ M(t)(: + ||n0 + |E0 |2 ||H m :), ! t" √ ||n: (⋅, t) + |E|2 (⋅, t) – n˜ ⋅, ||H m–1 (R) ≤ M(t)(: + ||n0 + |E0 |2 ||H m :). : Theorem 2.11.6 (Two-dimensional case). If the initial data satisfies the following conditions: E0 ∈ H m+2 (R2 ), n0 ∈ W1m+3 (R2 ), n1 ∈ H m (R2 ), m ≥ 3, ||E0 ||L2 < ||>||L2 , n1 satisfies eq. (2.11.4),

(H2)

then there exists a local bounded function M(t), such that √ ||E: (⋅, t) – E(⋅, t)||H m (R2 ) ≤ M(t)(: + ||n0 + |E0 |2 ||W m+3 :), 1

! t" √ ||n (⋅, t) + |E| (⋅, t) – n˜ ⋅, ||H m–1 (R2 ) ≤ M(t)(: + ||n0 + |E0 |2 ||W m+3 :). 1 : :

2

Theorem 2.11.7 (Three-dimensional case). If the initial data satisfies the following conditions: E0 ∈ H m+3 (R3 ), n0 ∈ W1m+3 (R3 ), n1 ∈ H m (R3 ), m ≥ 3, ||E0 ||H 1 is sufficiently small, n1 satisfies eq. (2.11.4),

(H3)

then (1) the life time of the solution (E: , n: ) is T ∗ (:) → ∞, : → 0. (2) (E: , n: ) converges to (E, –|E|2 ) in the following sense: there exists a function M ∈ Lloc,∞ (R+ ), such that for any finite time interval [0, T], there exists :(T) > 0, such that ∀t ∈ [0, T], ∀: > 0, : < :(T) ⇒ ||E: (⋅, t) – E(⋅, t)||H m (R3 ) ≤ M(t)(: + ||n0 + |E0 |2 ||W m+3 :log:), 1

! t" ||n (⋅, t) + |E| (⋅, t) – n˜ ⋅, ||H m–1 (R3 ) ≤ M(t)(: + ||n0 + |E0 |2 ||W m+3 :log:), 1 : :

2

where E is the global solution to NLS. We will first construct estimates in three dimensions, then in the other dimensional cases. Now we give the proofs of the above three theorems. Step 1. We first prove that ||F : ||H 1 ≤ M(t)|: + a: (t)|, ||u: ||L2 ≤ M(t)|: + a: (t)|,

2.11 Initial Value Problem for Langmuir Turbulence Equations

217

where a: (t) =



t 0

||n˜ : E||H 1 d4.

We note that by assumptions H 1 , H 2 , H 3 and the invariant variable (2.11.7), we have ||E: (t)||H 1 , ||n: ||L2 ≤ C, for each : and t, ||E(t)||H m+2 ≤ M(t) and ||n˜ : (⋅, t)||H m ≤ ||n0 + |E0 |2 ||H m ≤ ||n0 ||H m + ||E0 ||2H m (m ≥ 3). From which we deduce that ||F : ||H 1 ≤ M(t), ||u: ||L2 ≤ M(t).

(2.11.31)

We set 62 (t) = ||F : ||2L2 , 82 (t) = ||∇F : ||2L2 + 21 (||u: ||2L2 + ||:V : ||2L2 ). (1) Multiplying eq. (2.11.26) by F¯ : , integrating and choosing the imaginary part, we get that  d : 2 (u: + n˜ : )EF¯ : dx ||F ||L2 = 2Im dt Rd ≤ C(||E||L∞ ||u: ||L2 + ||En˜ : ||L2 )||F : ||L2 , namely, d62 ≤ M(t)68 + 6||En˜ : ||L2 . dt

(2.11.32)

(2) Multiplying eq. (2.11.26) by F¯ t: , integrating over Rd and choosing the real part, we deduce that  d u: (|F : |2t + 2ReEF¯ t: )dx ||∇F : ||2L2 = d dt R   2 ¯: 2 n˜ : (E + F : )F¯ t: dx = 0. |E| |F |t dx + 2Re – Rd

Rd

Multiplying eq. (2.11.29) by (V : – g), using eq. (2.11.30), we have    1 d 1 ||:V : ||2L2 + ||u: ||2L2 – :2 V : gdx dt 2 2   : : 2 : ¯ + ut (|F | + 2ReEF )dx + :2 V : gt dx = 0.

218

2 The Vanishing Viscosity Method of Some Nonlinear Evolution System

From above estimates, we deduce that       d 2 2 : 2 : 2 : : 2 : : ¯ 8 – : V gdx – |E| |F | dx + u |F | dx + 2Re EF u dx dt   = –:2 V : gt dx – |E|2t |F : |2 dx   (2.11.33) + 2Re (Et F¯ : )u: dx – 2Re n˜ : (E + F : )F¯ t: dx. For the right-hand side of eq. (2.11.33), we deduce that  2

:  

2 2 |V : gt |dx ≤ :8||gt ||L∞ ≤ C:8||E||W∞ 3 ≤ M(t)(: + 8 ),

|E|2t |F : |2 dx ≤ 62 |||E|2t ||L∞ ≤ M(t)62 , |Et F¯ : u: |dx ≤ ||Et ||L∞ ||F : ||L2 ||u: ||L2 ≤ M(t)(62 + 82 ).

On the other hand, we deduce from eq. (2.11.26) that 

n˜ : (E + F : )F¯ t: dx| ≤ ||∇F : ||L2 ||∇(n˜ : E)||L2

|2Re

2 : ˜ : E||L2 . + ||F : ||2H 1 ||n˜ : ||W∞ 1 + ||E||L ||F ||L2 ||n ∞

Therefore, the right-hand side of eq. (2.11.33) is less than M(t)[:2 + 62 + 82 +



62 + 82 ||n: E||H 1 ],

and for the left-hand side of eq. (2.11.33), we have the following estimates: 

 82 , :2 V : gdx ≤ :8||g||L∞ ≤ M(t):2 + 8     |E|2 |F : |2 dx ≤ ||E||2 62 ≤ M(t)62 , L∞  2    EF¯ : u: dx ≤ ||E||L 68 ≤ M(t)62 + 8 , ∞ 8  |u: ||F : |2 dx ≤ ||u: ||L2 ||F : ||2L4 ≤ ||u: ||L2 (61–d/4 8d/4 )2 .  

From eq. (2.11.31), we see that ||u: (t)||L2 ≤ M(t). Therefore,  



 82 u: |F : |2 dx ≤ M(t)62 + . 8

219

2.11 Initial Value Problem for Langmuir Turbulence Equations

By virtue of the initial data F : (x, 0) = u: (x, 0) = 0, V : (x, 0) = –∇u0 , we deduce from eq. (2.11.33) that 382 82 (t) ≤ M(t)[:2 + 62 ] + + M(t) 8  " + ||n˜ : E||H 1 62 + 62 d4.



t

! 2 (6 + 62 )

0

(2.11.34)

(3) We set 72 = 62 + 82 , we deduce from eqs (2.11.32) and (2.11.34) that  72 ≤ M(t) :2 +



t 0

 (72 + 7||n˜ 2 E||H 1 )d4 .

(2.11.35)

We denote 7(t) by supt∈[0,t] 7(t) and relabel it still by 7(t), then eq. (2.11.35) becomes  ! 72 (t) ≤ M(t) :2 +

 0

t

||n: E||H 1 dt

"2

 +

t

 72 dt .

(2.11.36)

0

From which we deduce results of step 1. + Remark. As : → 0, a: (t) → 0 in L∞ loc (R ). In particular, 7(t) is a local bounded function. Step 2. We will give the following estimates: 2 7m (t) = ||Dm F : ||2L2 +

||Dm–1 u: ||2L2 + ||:Dm–2 Ft: ||2L2 2

,

where Dm = △p , m = 2p; Dm = ∇△p , m = 2p + 1. (1) Multiplying eq. (2.11.26) by △m–1 F¯ t: , and choosing the real part, we get that  d ||Dm F : ||2L2 = –2Re Dm–1 [u: (E + F : )]Dm–1 F¯ t: dx dt  + 2Re Dm [n˜ : E + (n˜ : – |E|2 )F : ]Dm–2 F¯ t: dx. Multiplying eq. (2.11.27) by △m–2 u:t , we get that   m–1 : 2  m–2 : 2 d ||D u ||L2 + ||:D Ft ||L2 + (Dm–1 u: ) ⋅ Dm–1 (|F : |2 + 2ReEF¯ : )dx dt 2  m–1 : = 2Re (D u ) ⋅ Dm–1 [(E + F : )F¯ t: ]dx   m–1 : m–1 : 2 ¯ (Dm–2 |E|2tt )Dm–2 u:t dx. + 2Re (D u ) ⋅ D (Et F )dx + :

220

2 The Vanishing Viscosity Method of Some Nonlinear Evolution System

Then 6 5  d 2 m–1 : m–1 : 2 : ¯ 7 + (D u ) ⋅ D (|F | + 2ReEF )dx dt m  = 2Re (Dm–1 u: ) ⋅ Dm–1 (Et F¯ : )dx  + 2Re (Dm–1 u: ⋅ Dm–1 [(E + F : )F¯ t: ]  – Dm–1 [u: (E + F : )]Dm–1 F¯ t: )dx + :2 (Dm–2 |E|2tt )Dm–2 u:t dx  + 2Re Dm [n˜ : E + (n˜ : – |E|2 )F : ]Dm–2 F¯ t: dx,

(2.11.37)

which yields the following lemma: Lemma 2.11.8. If 7m (t) is identified with sup[0,t] 7m (4), then the following Gronwall inequality: 7m (t)2 ≤ M(t)[:2 + a:m (t)2 + where ! = d/(2(m – 1)), a:m (t) =

t 0

 0

t

2 2 ! 1–! (7m + 7m 7m 7 )d4],

(2.11.38)

||n˜ : E||H m d4.

From this lemma, we deduce the following corollary: Corollary 2.11.9. If tM(t) < – log[:2 + a:m (t)2 ], then 7m (t)2 ≤

M(t)[:2 + a:m (t)2 ]  . (e–tM(t) – :2 + a:m (t)2 )2/!

(2.11.39)

We will give proofs of above lemmas and corollary. √ Step 3. (One-dimensional case). (1) First, we will prove that a:m ≤ M(t) :, which  2 t : t ˜ m ≤ t 0 ||n˜ : E||2H m d4, thus a:m (t)2 is holds for a: (t). We get that a:m (t)2 = 0 ||n E||H dt less than summation of the following terms:   2  Di n˜ x, 4  |Dj E(x, 4)|2 dxd4, 0 ≤ i + j ≤ m.  :  –∞

 t t 0

˜ t/:) = As we know, n(x,

∞

1 2



! " ! " n˜ 0 x + :t + n˜ 0 x – :t .

2.11 Initial Value Problem for Langmuir Turbulence Equations

221

We will prove that, I : ≤ M(t):, where 2   t ∞   i D n˜ 0 x + 4  |Dj E(x, 4)|2 dxd4 I: =  :  0 –∞  t/:  ∞ |Di n˜ 0 (x + 4)|2 |Dj E(x, 4)|2 dxd4. =: 0

–∞

From equation NLS, we deduce that d j ¯ j △E + 2ImDj (|E|2 E)(Dj E) ¯ = Fj (x, s), |D E(x, s)|2 = 2ImDj ED dt then  |Dj E(x, :4)|2 ≤ |Dj E0 (x)|2 +

t

|Fj (x, s)|ds, (:4 ≤ t).

0

Assume that f (x, t) denotes the right-hand side of the inequality, and H(1) implies that ||E(t)||H m+2 ≤ M(t), thus f (⋅, t) ∈ L1 , ||f (⋅, t)||L1 ≤ M(t). Similarly, we deduce that |Di n˜ 0 (⋅, t)|2 ∈ L1 , i ≤ m; therefore,  ∞ ∞ : |Di n˜ 0 (x + 4)|2 |f (x, t)|dxd4 ≤ :M(t)||n˜ 0 ||2H m . I ≤: 0

–∞

Then √ a:m (t) ≤ M(t) :. √ : (2) Now  we fix T > 0, and consider that am (t) ≤ M(t) :, we deduce from eq. (2.11.39) that :2 + :||n˜ 0 ||H m → 0, 7m (t) → 0 uniformly with respect to t ∈ [0, T]. We complete the proof of Theorem 2.11.5. Step 4. (Two-dimensional case). We can deduce the estimate for a:m (t) from prosperities of the wave equations (2.11.24) and (2.11.25), in fact, m ||n˜ : E(4)||H m ≤ ||E(4)||H m ||n˜ : ||W∞ √ : M(4)||n˜ 0 ||W m+3 . ≤ √ 1 :+4

Then a:m (t) ≤

√



t

: 0

√ M(4) d4 ≤ M(t) :. √ :+4

We deduce results similar as that of the one-dimensional case.

222

2 The Vanishing Viscosity Method of Some Nonlinear Evolution System

Step 5. (Three-dimensional case). (1) Applying estimate (2.11.24) to d = 3, we have ||n˜ : E||H m ≤

: M(t)||n˜ 0 ||W m+3 . 1 :+t

Then a:m (t) ≤ :M(t)



t 0

d4 ≤ –:log:M(t). :+4

(2) In the three-dimensional case, the global existence of the solution (E: , n: ) has not been proved, but can be proved if e–tM(t) – M(t)(: – :log:) < 0, i.e. M(t)etM(t) ≤

1 , : – :log:

namely t < f (:), where lim:→0 f (:) = +∞. The a priori estimate (2.11.39) implies the boundedness of ||F : (t)||H m and ||u: (t)||H m–1 , since the assumption (H3) (||E0 ||H 1 is sufficiently small) implies that the global existence of E, E: = F : + E, n: = u: – |E|2 + n˜ : . It’s seen that as : → 0, the lifetime T ∗ (:) of the smooth solution (E: , n: ) of equations S: goes to ∞. (3) Using the above proofs, the second part of Theorem 2.11.6 is derived similarly as the one-dimensional case by eq. (2.11.39) (we fix T and set : → 0). Remark. If the compatible condition is satisfied, that is, n0 +|E0 |2 = 0, then n˜ : (x, t) and a:m (t) vanish. The convergent order for the three-dimensional case is M(t):. Step 6. Now we start to prove Lemma 2.11.8 and Corollary 2.11.9. For this purpose, we need the following lemma. Lemma 2.11.10. Assume 1 ≤ q, r ≤ +∞, 0 ≤ j < k, then k d ||Dj u||Lp ≤ C||Dk u||aLr ||u||1–a Lq , u ∈ Wr (R ), 1 p

where j/k ≤ a < 1,

=

j d

+

1 q

!1

+a

r

–

k d

–

1 q

"

.

Lemma 2.11.11. Assume f , g are some functions, such that the modules that arise in the sequel are bounded, then (i) ||Dk (fg)||Lp ≤ Ck (||f ||Lr ||Dk g||Lr′ + ||Dk f ||Ls ||g||Ls′ ), (ii) ||Dk (fg) – fDk g||Lp ≤ Ck (||Df ||Lr ||Dk–1 g||Lr′ + ||Dk f ||Ls ||g||Ls′ ), (iii) ||Dk (fg) – fDk g – gDk f ||Lp ≤ Ck (||Df ||Lr ||Dk–1 g||Lr′ + ||Dk–1 f ||Ls ||Dg||Ls′ ), where k ≥ 2,

1 p

=

1 r

+

1 r′

=

1 s

+

1 s′ ,

1 ≤ p ≤ +∞.

This lemma is the general case of that derived by Klainerman.

223

2.11 Initial Value Problem for Langmuir Turbulence Equations

Proof. (1) ||Dk (fg)||Lp ≤ Ck

k  i=0

i ks′

+

k–i kr′ ,

||Di f ||Lpi ||Dk–i g||Lqi , where

1 p

1 pi

=

+

1 1 qi , pi

=

i ks

+

k–i 1 kr , qi

=

then by Lemma 2.11.10, when d = 1, m = k, we have k i/k ||Di f ||Lpi ≤ C||f ||1–i/k Lr ||D f ||Ls , i/k ||Dk–i g||Lqi ≤ C||Dk g||1–i/k L ′ ||g||L ′ . r

s

Therefore, ||Dk (fg)||Lp ≤ Ck

k  (||f ||Lr ||Dk g||Lr′ )1–i/k (||Dk f ||Ls ||g||Ls′ )i/k i=0

≤ Ck (||f ||Lr ||Dk g||Lr′ + ||Dk f ||Ls ||g||Ls′ ). (2) ||Dk (fg) – fDk g||Lp ≤ Ck

k–1  i=0

||Di Df ||Lpi ||Dk–i–1 g||Lqi , where

1 i i k–i–1 k–i–1 1 = = + . + , pi (k – 1)s (k – 1)r qi (k – 1)s′ (k – 1)r′ By Lemma 2.11.10, when m = k – 1, we have ||Dk f ||i/(k–1) , ||Di Df ||Lpi ≤ C||Df ||1–i/(k–1) Lr Ls ||Dk–i–1 g||Lqi ≤ C||Dk–1 g||1–i/(k–1) ||g||i/(k–1) . L′ L′ r

s

Then we deduce (2). (3) ||Dk (fg) – fDk g – gDk f ||Lp ≤ Ck

k–2 

||Di (Df )||Lpi ||Dk–2–i (Dg)||Lqi ,

i=0

where 1 i i k–i–2 k–i–2 1 = = + . + , pi (k – 2)s (k – 2)r qi (k – 2)s′ (k – 2)r′ By Lemma 2.11.10, when m = k – 2, we have ||Dk–1 f ||i/(k–2) , ||Di (Df )||Lpi ≤ C||Df ||1–i/(k–2) Lr Ls ||Dk–i–2 (Dg)||Lqi ≤ C||Dk–1 g||1–i/(k–2) ||Dg||i/(k–2) , L′ L′ r

s

similarly as the proof of (1), we complete the proof of this lemma.

∎

224

2 The Vanishing Viscosity Method of Some Nonlinear Evolution System

Lemma 2.11.12. Assume F : , Ft: and u: satisfy the following estimates: ! , != ||u: ||L∞ ≤ 71–! 7m

d , 2(m – 1)

(2.11.40)

! , != ||F : ||L∞ ≤ 7 + || △F : ||L2 ≤ 7 + 71–! 7m

1 , m–1

(2.11.41)

! , ||DF : ||L∞ ≤ 71–! 7m

||Ft: ||L6

≤

b M(t)[71–b 7m

(2.11.42) ˜:

+ 7 + ||n E||H m ], 1 d b= + , m – 1 3(m – 1) ||Dm–2 Ft: ||L2 ≤ M(t)[7m + 7 + ||n˜ : E||H m ], m–2 d c ||Dm–1 F : ||L3 ≤ 71–c 7m , c= + . m – 1 6(m – 1)

(2.11.43)

(2.11.44) (2.11.45)

Proof. Since ||E||H m+2 ≤ M(t), and from 72 ≤ M(t)[:2 + a:m (t)], 72 = ||F : ||2H 1 +

||u: ||2L2 + ||:V : ||2L2 2

,

we deduce that 7(t) ≤ M(t). We can deduce estimates (2.11.40), (2.11.41), (2.11.42) and (2.11.45) directly from Lemma 2.11.10. (1) Proof of eq. (2.11.43). From eq. (2.11.26), we have ||Ft: ||L6 ≤ || △F : ||L6 + ||n˜ : – |E|2 ||L∞ ||F : ||L6 + ||n˜ : E||L6 + (||E||L∞ + ||F : ||L∞ )||u: ||L6 b ! ≤ 71–b 7m + M(t)7 + ||n˜ : E||H m + (M(t) + 7 + 71–! 7m )||u: ||L6 . m–1 u: ||e ≤ 71–e 7e , e = d/3(m – 1), and e + a = On the other hand, ||u: ||L6 ≤ ||u: ||1–e m L2 ||D L2 1 m–1

+

d 3(m–1)

= b. We have e b e/b b 71–e 7m = 71–e/b (71–b 7m ) ≤ 7 + 71–b 7m ,

therefore, b + 7 + ||n˜ : E||H m ). ||Ft: ||L6 ≤ M(t)(71–b 7m

(2) Proof of eq. (2.11.44). : ||Dm–2 Ft: ||L2 ≤ ||Dm–2 △ F : ||L2 + |||E|2 ||W∞ m–2 ||F ||H m–2

: : m–2 : + ||n˜ : E||H m–2 + ||E||W∞ u ||L2 m–2 ||u ||H m–2 + ||F ||L∞ ||D

+ ||u: ||L2 ||Dm F : ||L∞

≤ M(t)(7m + 7 + ||n˜ : E||H m–2 ) + ||F : ||L∞ ||Dm–2 u: ||L2 + ||u: ||L2 ||Dm–2 F : ||L∞ ,

2.11 Initial Value Problem for Langmuir Turbulence Equations

225

the last term of which is less than ! + 7(7 + 7m ) ≤ M(t)(7 + 7m ), f = (7 + 71–! 7m )71–f 7m f

m–2 . m+1 ∎

Then we deduce eq. (2.11.44). Lemma 2.11.13.      2Re Dm (n˜ : E + (n˜ : – |E|2 )F : )Dm–2 F¯ : dx t   ≤ (||Dm F : ||L2 + ||Dm–2 [u: (E + F : )]||L2 )(||Dm [n˜ : E + (n˜ : – |E|2 )F : ]||L2 ) ≤ M(t)(7 + 7m )(||n˜ : E||H m + 7m ). Proof. We set z = n˜ : E + (n˜ : – |E|2 )F : , from eq. (2.11.26), we have  2Re Dm–2 Ft: Dm z¯ dx  = 2Im (Dm F : – Dm–2 u: (E + F : ))Dm z¯ dx  –2Im Dm–2 zDm z¯ dx  = 2Im (Dm F : – Dm–2 u: (E + F : ))Dm z¯ dx Then we deduce the proof of this lemma by eq. (2.11.44).

∎

Now we start to prove Lemma 2.11.8. First, we notice the right-hand side of eq. (2.11.37):      2Re (Dm–1 u: ) ⋅ Dm–1 (Et F¯ : )dx ≤ 7m ||Et || m–1 ||F : || m–1 H H   2 ≤ M(t)(72 + 7m ), (m ≥ 3),    2Re Dm (n˜ : E + (n˜ : – |E|2 )F : )Dm–2 F¯ : dx t  2 ≤ M(t)(72 + 7m + ||n˜ : E||H m 7 + ||n˜ : E||H m 7m ).

(see Lemma 2.11.13),     2  m–2 2 m–2 :  2 : |E| D u dx D tt t   ≤ :|||E|tt ||H m–2 7m 2 ≤ M(t)(:2 + 7m ).

Since E0 ∈ H m+2 , and  2Re

[(Dm–1 u: )Dm–1 (E + F : )F¯ t:

–(Dm–1 F¯ t: )Dm–1 (u: (E + F : ))]dx

226

2 The Vanishing Viscosity Method of Some Nonlinear Evolution System

 = 2Re

{Dm–1 u: [Dm–1 ((E + F : )F¯ t: ) – (E + F : )Dm–1 F¯ t: ]

–Dm–1 F¯ t: [Dm–1 (u: (E + F : )) – Dm–1 u: (E + F : )]}dx, the absolute value of the first term of the right-hand side of which is less than 7m [||D(E + F : )||L∞ ||Dm–2 Ft: ||L2 + ||Ft: ||L6 (||Dm–1 E||L3 + ||Dm–1 F : ||L3 )] (see Lemma 2.11.11); hence, by Lemma 2.11.10, which is less than 2 ! 2 M(t)[72 + 7m + ||n˜ : E||H m 7m + 71–! 7m 7m ].

We deduce after integrating the second term by parts that 

Dm–2 F¯ t: [Dm [u: (E + F : )] – Dm u: (E + F : )]dx  –2Re (Dm–2 F¯ t: )Dm–1 u: D(E + F : )dx,

2Re

the absolute value of which is less than ||Dm–2 F¯ t: ||L2 (2||D(E + F : )||L∞ ||Dm–1 u: ||L2 + ||Dm (E + F : )||L2 ||u: ||L∞ ) (see Lemma 2.11.11); hence, by Lemma 2.11.10, which is less than 2 ! 2 M(t)[72 + 7m + ||n˜ : E||H m 7m + 71–! 7m 7m ].

Therefore, the right-hand side of eq. (2.11.37) is less than 2 ! 2 M(t)[72 + 7m + ||n˜ : E||H m 7m + 71–! 7m 7m + ||n˜ : E||H m 7 + :2 ].

(2.11.46)

Now we estimate the left-hand side of eq. (2.11.37). 2 (0) = :2 ||Dm–2 (n + g(0))|| ≤ K:2 . (1) t = 0, 7m 1 L2 (2) On the other hand, we have 

|Dm–1 u: Dm–1 (|F : |2 + 2ReEF¯ : )|dx

≤ 7m (||F : ||L∞ ||Dm–1 F : ||L2 + ||F : ||H m–1 ||E||H m–1 ). m : v 1–v 7v , v = (m – 2)/(m – 1), and ||F : || Since ||Dm–1 F : ||L2 ≤ ||DF : ||1–v L∞ ≤ m L2 ||D F ||L2 ≤ 7 1–u u : m : ||F ||L ||D F ||L2 , u = d/2(3m – d), then the above inequality is less than 6

2 (1+u+v)/2 2 (1–u–v)/2 2 (1+v)/2 2 (1–v)/2 M(t)(7m ) (7 ) + M(t)(7m ) (7 ) .

2.11 Initial Value Problem for Langmuir Turbulence Equations

227

Since (1 + u + v)/2 < 1, m > 1, then   2    (Dm–1 u: ) ⋅ Dm–1 (|F : |2 + 2ReEF¯ : )dx ≤ 7m + M(t)72 .   4

(2.11.47)

Equations (2.11.37), (2.11.46) and (2.11.47) yield that 2 (t) ≤ M(t)[:2 + 72 + a:m (t)2 + 7m



t

0

2 ! 2 (7m + ||n˜ : E||H m 7m + 71–! 7m 7m )d4].

If we regard that 7m (t) equals sup[0,t] 7m (4), then we complete the proof of Lemma 2.11.8, namely 2 7m (t) ≤ M(t)[:2 + a:m (t)2 +



t

0

2 ! 2 (7m + 71–! 7m 7m )d4].

(2.11.48)

Finally, we will prove Corollary 2.11.9. We assume t ∈ [0, T], T > 0, then eq. (2.11.48) becomes 2 7m (t)

2

≤ M(T)[: +

We set 6(t) = :2 + a:m (T)2 +

t

2 0 (7m

a:m (T)2

 + 0

t

2 ! 2 (7m + 71–! 7m 7m (T))d4].

! 72 (T))d4, then + 71–! 7m m !

6′ (t) ≤ M(t)6(t) + M(T)71–! (T)6(t)1+ 2 , if we set 8(t) = 6–!/2 (t), then 8 satisfies 8′ ≥ M(T), 8 + 71–! (T)

(2.11.49)

from which we deduce Corollary 2.11.9. Since equations S: and T : are complicated, we will consider their “good main part”. We successfully change the nonlinear term and consider the compatible and noncompatible initial value cases, respectively. A. Compatible initial value cases. ˜ t) = n˜ : (x, t) = 0, a:m (t) = 0. Then from previNow we have n0 + |E0 |2 = 0, thus n(x, ous results, the convergence order of ||E: – E||H m as well as that of ||n: + |E|2 ||H m–1 is :. Theorem 2.11.14. Assume n0 + |E0 |2 = 0, and assumptions (H1), (H2), (H3) hold, the the following problem

228

2 The Vanishing Viscosity Method of Some Nonlinear Evolution System

⎧ : : : 2 : d + ⎪ ⎨ iGt + △G = v E – |E| G , x ∈ R , t ∈ R , ¯ (P: ) :2 vtt: – △v: = 2Re △ (EG¯ : ) + :2 divgt , (divg = 2Im E∇E), ⎪ ⎩ : G (x, 0) = v: (x, 0) = 0, vt: (x, 0) = n1 + divg(0)

(2.11.50) (2.11.51) (2.11.52)

admits a unique solution (G: , v: ) ∈ H m+1 × H m , with the following estimates: for any T > 0, there exists :(T) such that ∀: < :(T), ∀t ∈ [0, T], ||E: (t) – E(t) – G: (t)||H m (Rd ) ≤ M(T):2 , ||n: (t) + |E|2 (t) – v: (t)||H m–1 (Rd ) ≤ M(T):2 . Proof. Step 1. Since E0 ∈ H m+2 , n0 ∈ H m+1 , we prove that ∀t ≥ 0, ||G: (t)||H m+1 ≤ M(t):, ||v: (t)||H m (Rd ) ≤ M(t):. For this purpose, we introduce the potential function V : (x, t), which satisfies :2 Vt: + ∇v: + ∇(2ReEG¯ : ) = 0,

(2.11.53)

vt:

(2.11.54)

:

:

+ divV = divg, V (x, 0) = –∇u0 .

(1) We set 621 = ||G: ||2L2 , 622 = ||∇G: ||2L2 + 21 ||V : ||2L2 + 21 ||:v: ||2L2 , we deduce from eq. (2.11.50) that d62 ≤ M(t)(621 + 622 ), dt   d ||∇G: ||2L2 + 2Re EG¯ : v: dx – |E|2 |G: |2t dx = 0. dt

(2.11.55) (2.11.56)

By eqs (2.11.53) and (2.11.54), we have      d 1 : 2 1 ||v ||L2 + ||:V : ||2L2 – :2 V : gdx + 2Re (EG¯ : )vt: dx = –:2 V : gt dx. dt 2 2

(2.11.57)

Using eqs (2.11.55), (2.11.56) and (2.11.57), we have      d 621 + 622 + 2Re (EG¯ : )V : dx – |E|2 |G: |2 dx – :2 V : gdx dt ≤ M(t)(621 + 622 ) + ||Et ||2L∞ ||G: ||2L2 + 2||Et ||L∞ ||G: ||L2 ||v: ||L2 + ||gt ||L2 :||:v: ||L2 ≤ M(t)(621 + 622 + :2 ). We consider the left-hand side of eq. (2.11.58).

(2.11.58)

2.11 Initial Value Problem for Langmuir Turbulence Equations

229

 For t = 0, :2 ||V : (0)||2L2 – : V : (0)g(0)dx ≤ K:2 ,              2Re (EG¯ : )V : dx +  |E|2 |G: |2 dx + : :V : gdx       ≤ M(t)(621 + 62 61 + :62 ) ≤ M(t)(621 + :2 ) +

622 . 2

We deduce from eq. (2.11.58) that  2

t

2

6(t) ≤ M(t)[: + 0

62 (4)d4], 62 = 621 + 622 ,

and 6(t) ≤ M(t):.

(2.11.59)

(2) Then we estimate 62m (t) = ||Dm–1 G: ||2L2 + 21 ||Dm v: ||2L2 + 21 ||:Dm–1 vt: ||L2 , we deduce from eq. (2.11.50) that  d m+1 : 2 ||D G ||L2 + 2Re Dm (Ev: )Dm G¯ :t dx, dt  (2.11.60) – 2Re Dm+1 (|E|2 G: )Dm–1 G¯ :t dx = 0. By eq. (2.11.51), we have  m–1 : 2 m : 2 d ||:D vt ||L2 + ||D v ||L2 + 2Re Dm (EG¯ : )Dm vt: dx, dt 2  + : Dm–1 (divgt ):Dm–1 vt: dx = 0.

(2.11.61)

From eqs (2.11.60) and (2.11.61), we have  " d! 2 6m + 2Re Dm (EG¯ : )Dm v: dx dt ≤ ||E||H m+2 6m (||Dm–1 G:t ||L2 + :) ≤ M(t)(62m + :2 ). We consider the left-hand side of eq. (2.11.62). For t = 0, :2 ||n1 + divg(0)||H m–1 ≤ :2 ,    2Re Dm (EG¯ : )Dm v: dx ≤ M(t)6m ||DG: ||1–! ||Dm+1 G: ||! , L2 L2 ≤ M(t):2 +

62m m–1 , != . 2 m

(2.11.62)

230

2 The Vanishing Viscosity Method of Some Nonlinear Evolution System

Thus we deduce from eq. (2.11.62) that    t 62m (4)d4 , 62m (t) ≤ M(t) :2 + 0

which combined with eq. (2.11.59) yields that ||G: ||H m+1 + ||v: ||H m + ||:vt: ||H m–1 ≤ M(t):.

(2.11.63)

Step 2. We set H : = F : – G: , 9: = u: – v: , equations (T : ) combined with equations (P: ) yield that (H : , 9: ) satisfies equations iHt: + △H : = 9: (E + G: + H : ) + v: (G: + H : ) – |E|2 H : , :2 9: – △9: = △(|H : |2 + 2Re(E + G: )H¯ : + |G: |2 ), tt

:

:

H (x, 0) = 9 (x, 0) =

9:t (x, 0)

= 0.

(2.11.64) (2.11.65) (2.11.66)

We introduce the potential function V : , which satisfies :Vt: + ∇9: + ∇(|H : |2 + 2Re(E + G: )H¯ : + |G: |2 ) = 0,

(2.11.67)

9:t

(2.11.68)

:

:

+ divV = 0, V (x, 0) = 0.

We set 821 = ||H : ||2L2 , 822 = ||∇H : ||2L2 + 21 ||9: ||2L2 + 21 ||:V : ||2L2 , 8 = 821 + 822 , then from eqs (2.11.64), (2.11.67) and (2.11.68), we deduce that d 2 8 ≤ M(t)(81 82 + :2 82 ), (||v: (G: + H : )||L2 = ||v: F : ||L2 ≤ :2 ) dt 1  ! : :2 " " d! 2 9 (|H | + |G: |2 ) + 2Re(E + G: )H¯ : dx 82 + dt ≤ |||G: |2t ||L∞ ||9: ||L2 + ||∇f ||L2 ||∇H : ||L2 +||f ||L2 ||9: ||L2 ||G: + H : + E||L∞ + ||G:t + Ht: ||L∞ ||H : ||L2 ||9: ||L2 , where f = 9: (G: + H : ) – |E|2 H : , (||f ||H 1 ≤ M(t)(822 + :2 )). Then ! d82 ≤ M(t) :4 + dt



t

" 82 (4)d4 .

0

We have 82 (t) ≤ M(t):4 .

(2.11.69)

(2) We set 82m (t) = ||Dm H : ||2L2 + 21 ||Dm–1 9: ||2L2 + 21 ||:Dm–2 9:t ||2L2 , As same as (2) in the first step, from eqs (2.11.64) and (2.11.65), we deduce the following inequalities:

231

2.11 Initial Value Problem for Langmuir Turbulence Equations

d 2 (8 + dt m



%

Dm–1 (|H % |2 + 2Re(E + G% )H + |G% |2 )Dm–1 9% dx)

≤ Dm f L2 Dm–2 Ht% L2 + (Dm–1 |G% |2t ||L2 + Et + G%t H m–1 H % H m–1 ) ⋅ Dm–1 w% L2  % + |2Re Dm–1 w% (Dm–1 (E + G% + H % )H t )

(2.11.70)

%

– (Dm–1 H t )Dm–1 [(E + G% + H % )w% ]dx|. By Lemma 2.11.11 and eq. (2.11.63), we have  82m (t) ≤ M(t)(%4 +

t 0

82m (4)d4).

(2.11.71) ∎

Thus Theorem 2.11.14 can be deduced from eqs (2.11.69) and (2.11.71). B. n0 + |E0 |2 ≠ 0.

It’s seen that the convergence order of the initial layer 9 n(x, %t ), F % H m and u% H m+1 is O(% + a%m (t)). On the other hand, as we know, if k ≥ 3, we have a%k (t) ≤ M(t)f (%), f (%) =

√

%, d = 1 or 2,

(2.11.72)

f (%) = –% log %, d = 3.

n% E will appear in the main part, we therefore need more As n0 + |E0 |2 ≠ 0, the term 9 regularity for initial data. Theorem 2.11.15. Assume the conditions (H1), (H2), (H3) are satisfied, n0 +|E0 |2 ≠ 0 and E0 ∈ H m+3 , n0 ∈ H m+2 (one-dimensional case), E0 ∈ H

m+4

, n0 ∈

W1m+4 (two-dimensional

(H4)

case, three-dimensional case).(H5)

then there exists a unique solution (G% , v% ) ∈ H m+1 × H m+1 to the following equation: ⎧ % % % n% – |E|2 )G% + 9 n% E, ⎪ ⎨ iGt + BG = v E + (9 % (R ) %2 vtt% – Bv% = 2ReB(EG¯ % ), ⎪ ⎩ % G (x, 0) = v% (x, 0) = vt% (x, 0) = 0

(2.11.73) (2.11.74) (2.11.75)

which satisfies following estimates: for any T > 0, there exists %(T) > 0, such that ∀% < %(T), ∀t ∈ [0, T], E% (t) – E(t) – G% H m ≤ M(T)%, t n% (⋅, t) + |E|2 (t) – 9 n(⋅, ) – v% H m–1 ≤ M(t)%. %

232

2 The Vanishing Viscosity Method of Some Nonlinear Evolution System

Proof. The procedures of proof are similar as before; first, we prove G% H m+1 ≤ M(t)a%m+1 (t), v% H m ≤ M(t)a%m+1 (t).

(2.11.76)

Then H % = E% – E – G% ,w% = n% + |E|2 – 9 n% – v% satisfy the following equations: ⎧ % % % % % % % % % ⎪ n% – |E|2 )H % , ⎨ iHt + BH = w (E + G + H ) + v H + v G + (9 % % 2 % 2 2 % 2 % % ¯ (R ) % wtt – Bw = B[|H | + |G | + 2Re(E + G )H ] + %2 divgt , ⎪ ⎩ H % (x, 0) = w% (x, 0) = 0, w% (x, 0) = n (x) + divg(x, 0). 1 t Using eq. (2.11.76) and Lemma (2.11.11) and setting 7m (t) = H % 2H m–1 + w% 2H m–1 + wt% 2H m–2 , we get  2 (t) 7m

≤

2 M(t)[7m (0)

2

t

4

+ % + f (%) + 0

2 7m (4)d4],

(2.11.77)

where t ∈ [0, T],% < %(T). Whereas the convergence order of 7m (t) is O(%) because of the initial data wt% (x, 0) ≠ 0 and the term %2 divgt . ∎

3 The Vanishing Viscosity Method of Quasilinear Hyperbolic System For details, refer to Refs [12–15, 29, 50, 57–61, 64, 65, 71–73, 76, 79, 88, 91, 93, 94, 101, 105–114].

3.1 Generalized Solutions to the First-Order Quasilinear Hyperbolic Equation in One Dimension In this section, the asymptotic property of the quasilinear parabolic equation ∂u ∂>(u) ∂ 2u + = : 2 , x ∈ R1 , t > 0 ∂t ∂x ∂x

(3.1.1)

u|t=0 = u0 (x), x ∈ R1 ,

(3.1.2)

with the initial condition

where : > 0 is a small parameter, will be considered as : → 0. Here >(u) is a sufficiently smooth function with respect to the variable u, and u0 (x) is a bounded measurable function. For the existence of the smooth solutions to the initial value problem (5.1.1) and (5.1.2), we require the following important results. Theorem 3.1.1. If the initial value problem ∂u ∂ 2u ∂u = A(x, t, u) + B(x, t, u) + F(x, t, u), 2 ∂x ∂t ∂x u|t=0 = u0 (x), x ∈ R1

(3.1.3) (3.1.4)

satisfy the following initial conditions: (1) For all u,A(x, t, u) ≥ a > 0, Fu′ ≥ c, |F(x, t, 0)| ≤ M1 , in S = {–∞ < x < ∞, 0 ≤ t ≤ T}. ′

3,! (2) u0 (x) ∈ Cloc (R1 ); u0 (x), u0 (x) are bounded for –∞ < x < ∞.

(3) A(x, t, u), B(x, t, u), F(x, t, 0), Fu′ (x, t, u) together with their first derivatives with respect to u and second derivatives with respect to x, which are uniformly bounded in  Q1 =

 – ∞ < x < ∞, 0 ≤ t ≤ T, |u| ≤ max

M1 e!T max |u0 (x)|, –∞ 0. If ′ ′′ ′′′ ′′′ ′′′ ′′′ ′′′ Axt , Aut , Axtu , Atuu , Auxx , Auux , Auuu and similar derivatives for B, F, Fu with respect to x, DOI 10.1515/9783110494273-003

234

3 The Vanishing Viscosity Method of Quasilinear Hyperbolic System

t, u satisfy the Lipschitz condition in the finite closed domain Q∗1 ⊂ Q1 , then there exists a unique smooth solution to the initial value problem (5.1.3) and (5.1.4); besides the 3 ∂2u derivatives in eq. (5.1.3), derivatives ∂∂xu3 , ∂t∂x are also continuous. According to Theorem 3.1.9, for the initial value problem (5.1.1) and (5.1.2), we have ′

Lemma 3.1.2. If >(u) ∈ C5 , u0 (x) ∈ C4 , x ∈ R1 and |u0 (x)| < k, |u0 (x)| < k, then there exists a unique smooth solution u: (x, t), and

∂ 3 u: ∂ 2 u: ∂u: ∂ 2 u: , , , ∂x2 ∂x3 ∂t∂x ∂t

are all continuous, and

|u: (x, t)| < M,

(3.1.5)

where M is independent of :. Lemma 3.1.3 (Maximum Principle to Linear Parabolic Equations). If u(x,t) satisfy the following conditions: (1) u, ux , ut , uxx are continuous on t √ ≥ 0; (2) for all x ∈ R1 , t ≥ 0, u(x, t) ≥ A(t) x2 + 1, where A(t) is continuous; 2 ∂u (3) L: u = : ∂∂xu2 – ∂u ∂t + a(x, t) ∂x ≤ 0, where a(x, t) is a bounded function; so if u(x, 0) ≥ 0, we have u(x, t) ≥ 0. Proof. The proof is by contradiction. Assume that there exists a point (x0 , t0 ) such 0) that ux0 ,t0 < 0 for t0 > 0. We set auxiliary function W(x, t) = (M + t) ch!(x–x with l M = max0≤t≤t0 |A(t)|, !, l are constants which will be determined later. Let W(x, t) satisfy the following conditions: (i) W(x, t) ≥ 0, it’s obvious; (ii) let ! be small enough and L: W = :Wxx – Wt + !Wx < 0, in fact, 0) L: W = ch!(x–x [:!2 (M + t) – 1 + a!(M + t)th!(x – x0 )], we choose such small l ! that L: W < 0; (iii) let l be sufficiently large so that W(x0 , t0 ) + u(x0 , t0 ) =

M + t0 + u(x0 , t0 ) < 0, l

and ch!l  > (l + |x0 |)2 + 1. l Now we consider the function v(x, t) = W(x, t) + u(x, t) on a rectangle R = {0 ≤ t ≤ t0 , |x – x0 | ≤ l}. When t = 0, v(x, 0) = W(x, 0) + u(x, 0) ≥ 0. When |x – x0 | = l,

3.1 Generalized Solutions to the First-Order Quasilinear Hyperbolic Equation

v(x, t) = W(x, t) + u(x, t) ≥ M ≥M

235

√ ch!l + A(t) x2 + 1 l

 ch!l – M (|x0 |2 + l)2 + 1 ≥ 0. l

While v(x0 , t0 ) < 0 for t0 > 0, assuming that v(x, t) has a negative minimum value at the point (x′ , t′ ) in R or on t = t0 , we have ∂v ∂v |(x′ ,t′ ) = 0, |(x′ ,t′ ) ≤ 0, ∂x ∂t ∂ 2v |(x′ ,t′ ) ≥ 0; ∂x2 consequently, L: v|(x′ ,t′ ) ≥ 0, a contradiction to the fact that L: v = L: W + L: u < 0. This completes the proof of the lemma. ∎ Lemma 3.1.4. If the assumptions in Lemma 3.1.2 are satisfied, and >′′ (u) ≥ , > 0, we have ∂u: (x, t) ≤ K1 ∀x ∈ R1 , t ≥ 0, ∂x

(3.1.6)

where the constant K1 is independent of :. Proof. Set v: =

∂u: ∂x ,

then by eq. (5.1.1), we have

∂ 2 v: ∂v: ∂v: = + >′ (u: ) + >′′ (u: )v:2 , ∂x2 ∂t ∂x ∂ 2 v: ∂v: ∂v: + >′ (u: ) = >′′ (u: )v:2 ≥ 0, L: v: ≡ : 2 – ∂x ∂t ∂x v: (x, 0) = u′0 (x). :

Since u0 (x) is bounded, we have u′0 (x) ≤ K1 . We deduce eq. (5.1.6) according to Lemma 3.1.3. ∎ Lemma 3.1.5. Under the assumptions in Lemma 3.1.4, ∀ x ∈ R1 , we have    ∂u:   :  ∂x  ≤ K2 ,

(3.1.7)

where the constant K2 is independent of :. Proof. Using variable transformation x = :x′ , t = :t′ , together with eq. (5.1.1), we get ∂u: ∂>(u: ) ∂ 2 u: + = ′ . ∂t′ ∂x′ ∂x 2

236

3 The Vanishing Viscosity Method of Quasilinear Hyperbolic System

When t = 0,        ∂u:   = : ∂u0  ≤ :K1 ≤ K ∗ ,  1  ∂x   ∂x′  t=0 so we have      ∂u:   ∂u:  1   = :   ∂x′   ∂x  ≤ K2 , x ∈ R , t > 0. ∎ Theorem 3.1.6. Under the assumptions in Lemma 3.1.4, the solution u: to problem (5.1.1) and (5.1.2) converges in L1 to some bounded measurable function u(x,t) on any beeline t = t0 > 0 or any finite domain [a,b] when : → 0, that is, 

b a

|u: (x, t0 ) – u(x, t0 )|dx → 0, : → 0.

(3.1.8)

Moreover, for any bounded domain G on t ≥ 0,  lim

:→0

Proof. We first prove lim:→0

G

 X2

|u: (x, t) – u(x, t)|dxdt = 0.

u: (x, t)dx exists; from eq. (5.1.1), we get

X1

  ∂u: ∂ ∂u: : – >(u: ) – = 0. ∂x ∂x ∂t On any smooth contour A, we set  5 A

:

6 ∂u: – >(u: ) dt + u: dx = 0, ∂x

 V: (x, t) =

(x,t) 5

:

(0,0)

6 ∂u: – >(u: ) dt + u: dx. ∂x

Obviously, we have ∂V: ∂u: ∂V: = u: , =: – >(u: ). ∂x ∂t ∂x We consider the problem    X ∂V: ∂V: ∂ 2 V: u0 (x)dx. =: 2 –> , V: (x, 0) = ∂t ∂x ∂x 0

(3.1.9)

3.1 Generalized Solutions to the First-Order Quasilinear Hyperbolic Equation

237

Assume :1 < :2 and define 9(x, t, :1 , :2 ) = V:1 (x, t) – V:2 (x, t). Then ⎧ 2 ⎪ ⎨ : ∂ 9 = ∂9 + >′ ∂9 + (: – : ) ∂u:2 , 1 2 1 u ∂x2 ∂t ∂x ∂x ⎪ ⎩ 9(x, 0) = 0, where >′u is a bounded function. We set V(x, t, :1 , :2 ) = 9(x, t, :1 , :2 ) + (:2 – :1 )K1 t. Therefore, :1 Since

∂u: ∂x

∂u:2 ∂ 2 V ∂V ∂V = + >′u + (:2 – :1 ) – (:2 – :1 )K1 . ∂x2 ∂t ∂x ∂x

≤ K1 (Lemma 3.1.4), so L: V = :1

∂ 2 V ∂V ∂V – – >′u ≤ 0, ∂x2 ∂t ∂x

and V|t=0 = 0. Since V(x, t, :1 , :2 ) increases less slowly than linear function, according to Lemma 3.1.3, we get V(x, t, :1 , :2 ) ≥ 0, that is, V:1 – :1 K1 t – (V:2 – :2 K1 t) ≥ 0,

:1 < :2 .

Therefore, V: – :kt is a monotone function for fixed (t, x), and we deduce from Lemma 3.1.5 that    

 6  ∂u: – >(u: ) dt + u: dx – :K1 t ≤ K ∗ . : ∂x

(x,t) 5

(0,0)

Therefore, lim V: (x, t) – :kt exists for any x, t.

:→0

And  lim[V: (x2 , t0 ) – V: (x1 , t0 )] = lim

:→0

exists, because of

∂u: ∂x

x2

:→0 x 1

≤ K1 , |u: | ≤ M, then we have

u: (x, t0 )dx

238

3 The Vanishing Viscosity Method of Quasilinear Hyperbolic System

 a

    b     ∂u: dx = K1 – ∂u: – K1 dx  ∂x    ∂x a   b  b ∂u: K1 dx ≤ dx + K1 – ∂x a a

b

≤ 2K1 (b – a) – u: |ba ≤ 2K1 (b – a) + 2M ≤ C.

(3.1.10)

So we use Helly’s alternative theorem to deduce that {u: (x, t)} is sequentially compact pointwise on t = t0 . Therefore, the subsequence of {u: (x, t)} is also sequentially compact in L1 . Because of the weak convergence of {u: (x, t)}, eq. (5.1.8) holds for all :. Setting  E:n (t) =

b a

|u:n (x, t) – u(x, t)|dx,

it’s easy to see that E:n (t) is measurable, and |E:n (t)| ≤ 2M(b – a), E:n (t) → 0(n → ∞). Using Lebesgue convergence theorem, eq. (3.1.9) holds.

∎

Lemma 3.1.7. Under the assumptions in Lemma 3.1.4, for x ∈ R1 , t ∈ [0, T], T is an arbitrary positive constant, then ∂u: E ≤ , ∂x t

(3.1.11)

where the constant E is independent of :, but depends on the upper bound M of u: and T. Proof. Define .h (x) ∈ C3 , .h (x) = .h (–x), satisfying 0 ≤ .h (x) ≤ 1, .h (x) = 1 for |x| ≤ h; .h (x) = 0 for |x| ≥ h + 1, |.h′ (x)|2 ≤ a.h (x), |.h′ | + |.h′′ | + |.h′′′ | ≤ b, h < |x| < k + 1, .h (x) = .h–1 (x – 1), where a, b are positive constants. Obviously, .h (x) exists; setting  Vh = .h u =

in view of eq. (5.1.1), we have

u,

for |x| ≤ h,

0,

for |x| ≥ h + 1,

(3.1.12)

3.1 Generalized Solutions to the First-Order Quasilinear Hyperbolic Equation

239

∂Vh ∂Vh ∂ 2 Vh – : 2 + Bh + Ah Vh = 0, ∂t ∂x ∂x where Ah = –2: Bh = 2:

(.h′ )2 .h2

+:

.h′′ >′ (u).h′ – , .h .h

.h′ + >′ . .h

Define 9h = tVh =

 tu, 0,

for |x| ≤ h, for |x| ≥ h + 1,

then ∂9h ∂9h ∂ 2 9h – : 2 + Bh + Ah 9h – Vh = 0, ∂t ∂x ∂x 9h (x, 0) = 0. Now estimate

∂9h ∂x ,

in view of

∂9hx ∂ 2 9hx ∂9hx + Bh –: + Ahx 9h + Bhx 9hx + Ah 9hx – Vhx = 0, ∂t ∂x2 ∂x x ∈ (–h – 1, h + 1), 0 < t < T, 9hx (x, 0) = 0, –h – 1 ≤ x ≤ h + 1, 9hx (±(h + 1), t) = 0, then 9hx satisfies either (1) 9hx don’t have positive value, that is, 9hx ≤ 0, or else (2) 9hx has positive value in the interior. Assume that there exists a positive maximum at (x0 , t0 ), where –h – 1 < x0 < h + 1, 0 < t0 ≤ T, then   ∂9hx  ∂9hx  = 0, ≥ 0, ∂x (x0 ,t0 ) ∂t (x0 ,t0 )  ∂ 2 9hx  ≤ 0, ∂x2 (x0 ,t0 ) and we have at (x0 , t0 ) Ahx 9h + Bhx 9hx + Ah 9hx – Vhx ≤ 0,

240

3 The Vanishing Viscosity Method of Quasilinear Hyperbolic System

that is :h 9hx + E9h ≤ 0, 92hx + D where 6 5  ′ ′ .h′ .h t.h 1 ′′ ′′ ′ – > (u) u + A – (u). u , 2: + > h h >′′ (u) .h .h t   t. 2 .′ E9h = ′′ h Ahx + >′′ (u) h2 Vhx 9h . > (u) .h

:h = D

In view of the construction of .h (x) and the assumptions, we see that :h | + |E9h | < C, |D where the constant C is independent of : and h, but only depends on M, T. We deduce from the second-order equation satisfied by 9hx that :h  2 : :  D :h /4 – E9h ≤ |Dh | + |Dh | + |E9h | < E. 9hx (x0 , t0 ) ≤ – + D 2 2 2 that is 9hx (x, t) ≤ 9hx (x0 , t0 ) ≤ E. Thus, we have ∂u: E ≤ . ∂x t This completes the proof.

∎

Lemma 3.1.8. Under the assumptions in Lemma 3.1.2, assume that [–a,a] is any segment on the beeline t = t0 and ! is an arbitrary positive real constant, then there must exist positive constants T and $ that are independent of : such that    

x2 x1

  (u1: (x0 , t0 ) – u2: (x, t0 ))dx ≤ ! ∀ x1 , x2 ∈ [a, b],

(3.1.13)

provided  x    1 2   < $, |x| ≤ N. (u (x, 0) – u (x, 0))dx : :   0

(3.1.14)

3.1 Generalized Solutions to the First-Order Quasilinear Hyperbolic Equation

241

Proof. We set (x,t) 5

 v:i (x, t)

= 0

6 ∂ui: i – >(u: ) dt + ui: dx, i = 1, 2, : ∂x

then

 x ∂v:i ∂v:i ∂ 2 v:i i ui: (x, t)dx. +> : , v: (x, 0) = : 2 = ∂x ∂t ∂x 0 Setting 9: = v:1 – v:2 , then ∂ 2 9: ∂9: ∂9: L: 9 = : 2 – – >′u = 0, ∂x ∂t ∂x  x (u1: (x, 0) – u2: (x, 0))dx, 9: (x, 0) = 0

where >′h is a bounded function, and setting function 9(x, t) = rch x ⋅ e"t > 0, r, " will be fixed later. Choose " sufficiently large so that L: 9 = 9(: – " – >h th x) < 0, and choose # sufficiently small such that # ch ae"t0 ≤ !2 , so #, " can be fixed. Since  9: (x, 0) =

0

x

(u1: (x, 0) – u2: (x, 0))dx,

|9: (x, 0)| ≤ 2M|x|, 9(x, 0) = rch x, we can find N such that |9: (x, 0)| ≥ |9: (x, 0)|, provided |x| ≥ N; choose $ = #, when |x| ≤ N, in addition to eq. (4.1.12), then we have |9: (x, 0)| < $ = # ≤ rch x = 9(x, 0), L: (9 ± 9: ) = L: 9 ± L: 9: < 0, t – 0, 9(x, 0) ± 9: (x, 0) ≥ 0, ∀ x ∈ R1 and the increase of the index of |9: | at ∞ is slower than the linear function. Hence, according to the maximum principle (Lemma 3.1.3), we see that 9(x, t) ± 9: (x, t) ≥ 0 ∀ x, t ≥ 0.

242

3 The Vanishing Viscosity Method of Quasilinear Hyperbolic System

Therefore, if |9: (x, t)| ≤ 9(x, t) = # ≤ rch xe"t , |9: (x, t0 )| ≤ rch xe"t0 ≤ #ch ae"t0 ≤

! , x ∈ [–a, a], 2

then    

x2

(u1: (x, t0 )

x1

–

 

u2: (x, t0 ))dx

= |9: (x1 , t0 ) – 9: (x2 , t0 )| ≤ !, ∎

and so completes the proof.

Theorem 3.1.9. Let >(u) ∈ C2 , >uu ≥ , > 0, u0 (x) be a bounded measurable function, then there exists a general solution to the initial value problem ut +

∂>(u) = 0, x ∈ R1 , t > 0, ∂x u|t=0 = u0 (x), x ∈ R1 ,

(3.1.15) (3.1.16)

that is, there exists a bounded measurable function u(x,t) such that (1) For any finite function f (x, t) ∈ C1 , then 

T 0



 [u(x, t)ft (x, t) + >(u)fx ]dxdt + R

f (x, 0)u0 (x)dx = 0,

(3.1.17)

R

(2) u(x2 , t) – u(x1 , t) ≤ k, t > 0. x2 – x1 Proof. Setting sequence un0 (x) to be sufficiently smooth and evenly bounded, it’s possible to define the average function of u0 (x) that un0 (x) → u0 (x) on any finite domain, then we have |un0 (x)| ≤ M, |u(n) 0x | ≤ kn . So consider the following initial value problem: ∂>(u) , ∂x n u(x, 0) = u0 (x), :uxx = ut +

(3.1.18) (3.1.19)

we deduce from Theorem 3.1.9 that there exist smooth solutions un: (x, t) to problem (3.1.18) and (3.1.19), and in view of Lemmas 3.1.4 and 3.1.7, we have

3.1 Generalized Solutions to the First-Order Quasilinear Hyperbolic Equation

|un (x, t)| ≤ M,

∂un: ∂un: E ≤ kn , ≤ , ∂x ∂x t

243

(3.1.20)

where the constant E is independent of M, n. So un: (x, t) is a function of bounded variation with respect to : uniformly on t = t0 . According to the Helly theorem, un: (x, t) is sequentially compact pointwise, and thus sequentially compact in L1 , so we deduce from Lemma 3.1.8 that  x2  x2  x2 un: (x, 0)dx = un0 (x)dx → u0 (x)dx, n → ∞; x1

x1

x1

then    

x2

x1

(um : (x, t0 )

–

 

m, n > 9 N,

un: (x, t0 ))dx,

this means that there exists a limit function u: (x, t) to un: (x, t) in L1 , that is, for any finite domain on t ≥ 0, we have un: (x, t) → u: (x, t), in L1 , from eqs (3.1.18) to (3.1.20), we have 

T



 (:un: fxx + un: ft + >(un: )fx )dxdt +

0 R un: (x1 , t)

R

f (x, 0)un0 (x)dx = 0,

– un: (x2 , t) E ≤ , t > 0. x1 – x2 t

(3.1.21) (3.1.22)

Let n → ∞, we have 

T 0



 R

(:u: fxx + u: ft + >(u: )fx )dxdt +

u: (x1 , t) – u: (x2 , t) E ≤ , t > 0. x1 – x2 t

f (x, 0)u0 (x)dx = 0,

(3.1.23)

R

(3.1.24)

According to |u: (x, t)| ≤ M, and E ∂u: |t=t0 ≤ , ∂x t0 where M, E are independent of :, then in view of Helly’s theorem, u: (x, t) converges weakly pointwise (and thus in L1 ) to the bounded measurable function u(x, t) on t = t0 . According to Lemma 3.1.7 and the above proof, we have

244

3 The Vanishing Viscosity Method of Quasilinear Hyperbolic System



x2

x1

 |u:1 (x, t0 ) – u:2 (x, t0 )|dx ≤

x2

|u:1 (x, t0 ) – un:1 (x, t0 )|dx

x1



x2

+ x1  x2

+ x1

|un:1 (x, t0 ) – un:2 (x, t0 )|dx, |un:2 (x, t0 ) – u:2 (x, t0 )|dx → 0,

so u: (x, t) converges in L1 on the finite domain [a, b] on t = t0 > 0, and thus converges weakly to bounded measurable function u(x, t). Since      (>(u: ) – >(u))fx dx ≤ ||>u ||L∞ ||f x ||L∞ ⋅ ||u: – u|| 1 → 0, L   R

then performing the limit to eqs (3.1.23) and (3.1.24), we see that the bounded measurable function u(x, t) is the general solution to the initial value problem (3.1.15) and (3.1.16). ∎ Theorem 3.1.10. Let the general solution u(x, t) to the initial value problem (3.1.15) and (3.1.16) satisfy the following entropy conditions: u – u– >(u+ ) + u+ – u– u– < u < u+ , u – u– >(u+ ) + >(u) ≤ + u – u– u+ < u < u– , >(u) ≥

u+ – u >(u– ), u+ – u– u+ – u >(u– ), u+ – u– (3.1.25)

where u+ (x, t) = u(x + 0, t), u– (x, t) = u(x – 0, t). Proof. Multiplying eq. (5.1.1) by h(u: ) ∈ C1 (R), we have ∂I(u: ) ∂6(u: ) + = :I(u: )xx – :h′ (u: )u2:x , ∂t ∂x

(3.1.26)

where  I(u: ) = 6(u: ) =



h(u: )du: , h(u: )d>(u: ) =



h(u: )>′ (u: )du: ,

:h(u: )u:xx = :(u:x h(u: ))x – :h′ (u: )u2:x , = :I(u: )xx – :h′ (u: )u2:x . Multiplying eq. (3.1.26) by f (x, t) with finite support, and integrating on t ≥ 0, we have

3.1 Generalized Solutions to the First-Order Quasilinear Hyperbolic Equation



245



t>0

[I(u: )ft + 6(u: )fx ]dxdt = –:

 t>0

+: t>0

I(u: )fxx dxdt, h′ (u: )u2:x fdxdt.

(3.1.27)

We pass to limits as m → ∞, since u: (x, t) → u(x, t), and |u: (x, t)| ≤ M, it’s easy to test that for each finite interval at t ≡ const, I(u: (x, t)) → I(u(x, t)). Similarly, 6(u: (x, t)) → 6(u(x, t)). Therefore, the right-hand side of eq. (3.1.27) is uniformly bounded as : → 0, and then converges to the responding integration of u(x, t). It’s easily seen that the first part converges to zero and the second part is always positive when h′ (u)f ≥ 0; therefore, the limit function u(x, t) satisfies the following inequality:  [I(u)ft + 6(u)fx ]dxdt ≥ 0,

(3.1.28)

t>0

that is, the general solution to the initial value problem (3.1.15) and (3.1.16) satisfies inequality (3.1.28) when h′ (u)f ≥ 0. Now choose domains on t ≥ 0: D$ = {t0 – $ ≤ t ≤ t0 + $, x(t) – : ≤ x < x(t) + :}, D$ = D+$ (x ≥ x(t)) + D–$ (x ≤ x(t)), A$ : f (x, t) = 0. From eq. (3.1.28), 

5 I(u)

D$

6   6   5 ∂f ∂f ∂f ∂f + + 6(u) dxdt = dxdt I(u) + 6(u) ∂t ∂x ∂t ∂x D+$ D– $ 5 6   ∂I ∂6 f [I(u)dx – 6(u)dt] – f + dxdt = ∂t ∂x A+$ D+$ 5 6   ∂I ∂6 f [I(u)dx – 6(u)dt] – f + dxdt + ∂t ∂x A– D– $ $  t0 +$ dx = f (x(t), t)[(I(u+ ) – I(u– )) dt t0 –$ –(6(u+ ) – 6(u– ))]dt ≥ 0

(3.1.29)

exists for ∀t0 > 0, $ > 0, h′ (u) ≥ 0, f (x, t) ≥ 0; so from eq. (3.1.28), we have (I(u+ ) – I(u– ))

dx – (6(u+ ) – 6(u– )) ≥ 0. dt

(3.1.30)

246

3 The Vanishing Viscosity Method of Quasilinear Hyperbolic System

Similarly for eq. (3.1.15), we deduce the Hugoniot condition dx >(u+ ) – >(u– ) . = dt u+ – u– Choose h(u) as additional function: h(u) =

 0,

u < v,

1,

u > v.

Therefore,  I(u) =  6(u) =

0,

u ≤ v,

u – v,

u ≥ v,

0,

u ≤ v,

>(u) – >(v),

u ≥ v.

If u– < v < u+ , then I(u+ ) = u+ – v, 6(u+ ) = >(u+ ) – >(v), I(u– ) = 0, 6(u– ) = 0. Then utilizing eq. (3.1.30), we deduce >(v) ≥

u+ – v v – u– + >(u ) + >(u– ). u+ – u– u+ – u–

If u+ < v < u– , then I(u– ) = u– – v, 6(u– ) = >(u– ) – >(v), I(u+ ) = 0, 6(u+ ) = 0. Therefore, >(v) ≤

u+ – v v – u– + >(u ) + >(u– ), u+ – u– u+ – u–

and so completes the proof.

∎

3.2 Existence and Uniqueness of the General Solution to First-Order Multivariable Quasilinear Equations We consider the following initial value problem of first-order multivariable quasilinear equations:

3.2 Existence and Uniqueness of the General Solution to First-Order Multivariable

ut +

n  d >i (x, t, u) + 8(x, t, u) = 0, dxi

247

(3.2.1)

i=1

u|t=0 = u0 (x), x ∈ Rn ,

(3.2.2)

where d >i (x, t, u) = >xi + >u uxi , x ∈ Rn . dxi We use vanishing viscosity method to prove the existence of general solutions to the initial value problem, and discuss the stability and uniqueness, that is, we use the smooth solutions to the following parabolic equations with small parameter: ut +

n  d >i (x, t, u) + 8(x, t, u) = : △ u, dxi

(3.2.3)

i=1

u|t=0 = u0 (x)

(3.2.4)

to approach the general solution to the initial value problem (5.2.2) and (5.2.3) as : → 0. So we must give the definition of general solutions to the first-order multivariable quasilinear equations, and the necessary a priori estimates independent of : for the smooth solutions to problem (3.2.3) and (3.2.4). Definition 3.2.1. We call a bounded measurable function u(x,t) the general solution to problem (5.2.2) and (5.2.3) on Rn × [0, T] if (1) For each constant k and f (x, t) ≥ 0 with smooth support, the following inequality holds:   |u(x, t) – k|f t + sign(u(x, t) – k)[>i (x, t, u(x, t)) – >i (x, t, k)]fxi FT

 –sign(u(x, t) – k)[>ixi (x, t, k) + 8(x, t, u(x, t))]f dxdt ≥ 0,

(3.2.5)

(2) There exists a measurable set : with Lebesgue measure zero on [0, T], for t ∈ [0, T] :, u(x,t) is defined a.e. in Rn , we have for each ball Kr = {|x| ≤ r} ⊂ Rn that  |u(x, t) – u0 (x)|dx = 0. (3.2.6) lim t→0 Kr t∈[0,T]/:

Obviously, when k = ± sup |u(x, t)|, for any smooth function f (x, t), we deduce from inequality (3.2.5) that the general solution to problem (5.2.2) and (5.2.3) satisfies the following common integral inequality:  0

T

 Rn

5 uft +

n  i=1

6 >i (x, t, u)fxi – 8(x, t, u)f dxdt = 0.

(3.2.7)

248

3 The Vanishing Viscosity Method of Quasilinear Hyperbolic System

Moreover, similar entropy conditions can be deduced from inequality (3.2.5); in fact, when the general solution is piecewise smooth function, integrating inequality (3.2.5) around the discontinuity surface and choosing any function f (x, t) with support, then for each constant k and along the discontinuity surface, we have |u+ – k|cos(v, t) + sign(u+ – k)|>i (x, t, u+ ) – >i (x, t, k)|cos(v, xi ) ≤ |u– – k|cos(v, t) + sign(u– – k)[>i (x, t, u– ) – >i (x, t, k)]cos(v.xi ), where v is the normal vector at (x, t) on the discontinuity surface, u+ (respectively u– ) is the limit of the general solution at (x, t) with respect to the plus side (respectively minus side) of the discontinuity surface. It’s easily seen that for n = 1, inequality (3.2.8) is the entropy condition of the one-dimensional general solution. In order to prove the existence of the general solution to problem (5.2.2) and (5.2.3), we make the following assumptions. (1) >i (x, t, u) is twice continuously differentiable. (2) >iu (x, t, u) is uniformly bounded in DM = FT × [–M, M]. (3) J(x, t, u) = >ixi (x, t, u) + 8(x, t, u) is twice continuously differentiable and is uniformly bounded in DM , and sup |J(x, t, 0)| ≤ C0 = const, sup | – Ju (x, t, u)| ≤ C1 = const.

(3.2.8) (3.2.9)

(x,t)∈FT

(4) For each bounded measurable function u0 (x) in Rn , u0 (x) ≤ M0 , and assume that >i (x, t, u) is independent of x, and >iut (x, t, u), 8u (x, t, u), 8xj (x, t, u), 8t (x, t, u) are uniformly bounded in DM . In order to get the existence of the smooth solutions to the quasilinear equation (3.2.3) with the initial value condition (3.2.4), we have to suppose that u0 (x) is bounded until to the third derivative. Factually, we have Theorem 3.2.2. Under the conditions assumed in (1), (2), (3), (4), u0 (x) and its derivatives until the third order are bounded, then there exists a unique classical solution u: (x, t) to problem (3.2.3) and (3.2.4), and u: (x, t) and its derivatives in eq. (3.2.3) are all bounded and are uniformly H o¨ lder continuous. Then we start to give the uniform estimates for the classical solution to problem (3.2.3), (3.2.4), and rewrite eq. (3.2.3): ut +

n 

>iu uxi + J(x, t, u) = : △ u.

(3.2.10)

i=1

˜ Since J(x, t, u) = J(x, t, 0) + Ju (x, t, u)u, consider conditions (3.2.8) and (3.2.9), then by the maximum principle, we have

3.2 Existence and Uniqueness of the General Solution to First-Order Multivariable

|u: (x, t)| ≤ const = (M0 + C0 t)eC1 T = M.

249

(3.2.11)

Now we construct the L1 -continuous norm estimate for u: (x, t), choose some vector z ∈ Rn and set 9(x, t) ≡ u: (x + z, t) – u: (x, t), obviously, 9(x, t) satisfies 9t + (ai 9)xi + c9 + ei zi = : △ 9,

(3.2.12)

where 

1

ai (x, t) =

>iu (t, !u: (x + z, t) + (1 – !)u: (x, t))d!,

0



1

c(x, e) =

Ju (!(x + z) + (1 – !)x, t, !u: (x + z, t)

0



:

1

+(1 – !)u (x, t))d! ≡

Ju (⋅, ⋅, ⋅)d!

(3.2.13)

0



1

ei (x, t) = 0

Jxi (⋅)d!,

n  (|ai | + |ei |) + |c| ≤ const; i=1

obviously, all functions ai , c, ei satisfy the Lip condition on any compact set of FT , multiplying eq. (6.3.5) by g(x, t) with support (gt , gxi , gxi xj are all continuous functions) on F4 ⊂ FT , and integrating with respect to x, we deduce by integrating by parts that   9g|t=4 dx – L(g)9dxdt Rn



= Rn

F4



9g|t=0 dx –

i Ft

zi ei gdxdt,

(3.2.14)

where L(g) ≡ gt + ai gxi – cg + : △ g,

(3.2.15)

Lemma 3.2.3. Assume that q(x,t) is continuous in F4 , satisfying the inequality L(g) ≥ 0. In addition, assume that |q(x, t)| ≤ q0 , q(x, 4) ≡ 0, |x| ≥ r, where q0 , r are some constants, then q(x, t) ≤ q0 exp[:–1 (H(4 – t) + r – |x|) + 4 sup |c| + (t – 4) inf c] F4

F4

≡ Q: (x, t), (x, t) ∈ K = {(x, t)||x| ≥ r + H(4 – t), 0 ≤ t ≤ 4},

(3.2.16)

250

3 The Vanishing Viscosity Method of Quasilinear Hyperbolic System

where H = 1 + sup

 n

F4

1

2

a2i

.

i=1

Proof. It’s easy to verify that L(Q: ) ≤ 0, in K, Q: ||x|=r+H(4–t) ≥ q0 , Q: |t=4 ≥ 0; therefore, by the maximum principle, we see that q(x, t) ≤ Q: (x, t) a.e. in K, and thus finish the proof. ∎ Now we fix some constant r > 1, define qh (x, t) as the solution of the following initial value problem 

(x, t) ∈ F4 ,

L(qh ) = 0, qh (4, x) =

"h (x),

where "(x) = sign 9(4, x), |x| ≤ r – h, "(x) ≡ 0, |x| > r – h, "h (x) is the average function of "(x). Obviously, by the maximum principle, we have |qh (x, t)| ≤ const = q0 . In eq. (3.2.14), we set  g(x, t) = qh (x, t)'m (|x|), 'm (3) = 1 –

3

$(3 – m)d3,

(3.2.17)

–∞

where $(3) ∈ C∞ (R), $(3) ≥ 0, $(3) ≡ 0, |3| ≥ 1, and

 $(3)d3 = 1,

(2.17’)

R

m is some natural number, so we have 

 5 xi xi 9qh 'm |t=4 dx = – ai $(|x| – m)9 + 2:9xi $(|x| – m). |x| |x| Rn F4 6   zi ei qh 'm dxdt + 9qh 'm |t=0 dx, +:9 △ 'm qh dxdt – F4

Rn

by Lemma 5.2.2, for : ∈ (0, 1), we have  |qh (x, t)| ≤ const exp

–

 |x| . 2

(3.2.18)

3.2 Existence and Uniqueness of the General Solution to First-Order Multivariable

251

¯ > 1, we have For R ≥ r¯ = r + (1 + N)T  Rn \KR

|qh (x, 0)|dx ≤ const Rn–1 exp[:–1 (¯r – R)],

where N¯ = sup

 n

F4

1 >2iu (x, t, u)

2

.

i=1

Let m → ∞ in eq. (3.2.18), and let h → 0, then we have  Kr

|9(x, 4)|dx ≤ const{|z| + 9R (|z|) + Rn–1 exp[:–1 (¯r – R)]} = +:R (|z|).

If we assume at the initial time that  |u0 (x + △x) – u0 (x)|dx ≤ 9r (| △ x|) ∀r > 0, Jr (u0 (x), △x) =

(3.2.19)

(3.2.20)

Kr

where 9r (3) is the modulus of continuity function; when 3 ≥ 0, it’s continuous nondecreasing function, and 9r (0) = 0, by eqs (3.2.19) and (3.2.20), we deduce for 0 ≤ t ≤ T that Jr (u: , △x) ≤ min +:k (| △ x|) = 9xr (| △ x|), R≥¯r

(3.2.21)

where the function 9xr (3) is independent of :. In order to estimate the modulus of continuity with respect to t, we construct the following lemma: Lemma 3.2.4. Assume that (i) u(x,t) is measurable in the cylindricality {(x, t)} = Kr+1 × [0, T], (0 < 21 < r), |u(x, t)| ≤ M, (ii) when 0 ≤ t ≤ T, | △ x| ≤ 1, Jr (u(x, t), △x) ≤ 9xr (| △ x|), (iii) for each second-order smooth function g(x) with support, satisfying the inequality      g(x)[u(x, t + △t) – u(x, t)]dx  Kr 6 5 n  ≤ cr △ t max |g|+|gx |+ |gxi xj | , ∀t, t + △t ∈ [0, T], △t > 0, (3.2.22) x∈Kr

i,j=1

252

3 The Vanishing Viscosity Method of Quasilinear Hyperbolic System

then we have  Ir (u(x, t), △t) =

|u(x, t + △t) – u(x, t)|dx 5 6 △t x min h + 9r (h) + 2 , 0 0, n h Rn h  n  $(xi ) ≥ 0, +(x) = 1. +(x) ≡ (sign v(x))h =

i=1

Rn

In fact ||v(x)| – v(x) sign v( y)| = ||v(x)| – |v( y)| – [v(x) – v( y)]sign v( y)|, ≤ 2|v(x) – v( y)|, thus

(3.2.24)

3.2 Existence and Uniqueness of the General Solution to First-Order Multivariable

253

 ||v(x)| – v(x)(sign v(x))h |dx   x–y h–n +( ≤ )||u(x)| – u(x)sign u( y)|dydx n h Kr R   +(z) |u(x) – u(x – hz)|dxdz ≤ 29r (h), ≤2 Kr

Rn

Kr

utilizing eq. (3.2.24) to 9(x) in k2r–h , we have 5 Ir (u(x, t), △t) ≤ const h +

9xr (h)

6 △t + 2 , h ∈ (0, 1), h ∎

and thus finish the proof.

By using Lemma 5.2.3, we deduce the estimate of the modulus of continuity 9r (x) with respect to space variables, and thus the L1 estimate of modulus of continuity of the solution to eq. (3.2.3) with respect to t. As a matter of fact, multiplying eq. (3.2.3) by the function g(x) with support, we deduce inequality (3.2.22), where the constant cr = sign rn , 0 < : < 1, then 5 6 △t Ir (u: (x, t), △t) ≤ 9tr (△t) = sign min h + 9xr (h) + 2 , 0≤h≤1 h

(3.2.25)

By estimates (3.2.21) and (3.2.25), it’s easy for us to construct the existence theorem of the general solutions to problem (5.2.2) and (5.2.3). Theorem 3.2.5. Under the conditions assumed in (1), (2), (3), (4), if >i (x, t, u) is independent of x, assume that >ixi xj (x, t, u), >iuxj (x, t, u), >ixj t (x, t, u), >iut (x, t, u) and Ju (x, t, u), Jxj (x, t, u), Jt (x, t, u) are bounded in DM , then there exists a general solution to problem (5.2.2) and (5.2.3). Proof. Under the condition that the initial value u0 (x) is sufficiently smooth, especially satisfying the Lipschitz condition in L1 (KR ) ( for each R > 0), and >i is independent of x, we have proven that there exist functions 9xr (3), 9tr (3) independent of :, for 0 ≤ t ≤ T: Jr (u: , △x) + Ir (u: , △t) ≤ 9xr (| △ x|) + 9tr (| △ t|).

(3.2.26)

254

3 The Vanishing Viscosity Method of Quasilinear Hyperbolic System

Since >i depends on x, we deduce from the function u0 (x) satisfying Lipschitz condition in L1 (KR ) with smooth initial data that  |u0x |dx ≤

√ nc(R, + 1), c = sign, , = sign > 0.

(3.2.27)

Kr

Setting vk (x, t) = u:xk (x, t), vk (x, t) satisfies the following parabolic system: vtk +

d [>iu (x, t, u: )vk ] + >iuxk vi + >ixk xi + Ju vk + Jxk dxi

= : △ vk , k = 1, 2, . . . , n.

(3.2.28)

Multiplying the kth equation of eq. (3.2.28) by a sufficiently smooth function g k (x, t) with support, integrating in Fr , summing from k = 1 to n and integrating by parts, we have    k k k v g |t=4 dx – Lk (g)v dxdt = vk g k |t=0 dx Rn Fr Rn  (>ixi xk + Jxk )g k dxdt, (3.2.29) – Fr

where Lk (g) = gtk + ai gxki – (>kxi u + $ik Ju )g i + : △ g k ai = >iu (x, t, u: ), g = (g 1 , . . . , g n ), k = 1, 2, . . . , n. Similarly to the previous proof, we have the following estimate: Jr (u: , △x) ≤ const | △ x| = 9xr (| △ x|), 1

Ir (u: , △x) ≤ const | △ t| 3 = 9tr (| △ t|),

(3.2.30)

for the general bounded measurable function u0 (x), we can use its mean value function to approximate, for the solution vh: (x, t) to eq. (3.2.3) with the initial condition u:h (x, 0) = uh0 (x) (0 ≤ h ≤ 1), since  KR

|uh0 (x + △x) – uh0 (x)|dx ≤ const h–1 Rn | △ x|,

from the above discussion, we deduce

3.2 Existence and Uniqueness of the General Solution to First-Order Multivariable

Jr (u:h , △x) ≤

255

1 const const | △ x|, Ir (u:h , △t) ≤ | △ t| 3 , h h

If we estimate the function 9(x, t) = u:h (x, t) – u: (x, t):  Kr

|u:h (x, t) – u: (x, t)|dx ≤ const



|x|

e– 2 |uh0 (x) – u0 (x)|dx Rn 5  6 R ≤ const 9k (h) + Rn–1 exp – ∀ k ≥ 1, 2

then we have 9R (h) + Rn–1 exp

0i (x, t, u(x, t)) – >i (x, t, k)]fxi 

–sign(u(x, t) – k)[>ixi (x, t, k) + 8(x, t, u(x, t))]f dxdt ≥ 0.

(3.2.31)

In fact, let I(u) ∈ C2 and be the concave function, –∞ < u < ∞. Multiplying eq. (3.2.3) by I′ (u)f (x, t) (f (x, t) is a function supported on FT , f (x, t) ∈ C2 , f (x, t) ≥ 0), and integrating on FT , utilizing integrating by parts and the fact that I′′ (u)uxi uxi f ≥ 0, we deduce that 

 FT

 I(u)ft + k

5

u

+ k

u

I′ (u)>iu (x, t, u)dufxi – I′ (u)>ixi f

6  I′ (u)>iuxi du – I′ (u)8 f + :I(u) △ f dxdt ≥ 0,

where k is some constant, approximating |u – k| with second-order continuous functions yielding

256

3 The Vanishing Viscosity Method of Quasilinear Hyperbolic System

 FT

{|u – k|( ft + : △ f ) + sign(u – k)[>i (x, t, u) – >i (x, t, k)]fxi

–sign(u – k)[>ixi (x, t, k) + 8(x, t, u)]f }dxdt ≥ 0,

(3.2.32)

as : → 0, we get inequality (3.2.31). Since the function f (x, t) ≥ 0, f (x, t) ∈ C1 can be uniformly approximated by the supported function f (x, t) ∈ C2 , f (x, t) ≥ 0, the first demand of the general solutions holds; obviously, for 1 ∈ [0, T]\:, the subsequence {u:m (x, t)} in Rn converges to u(x, t), where : denotes the zero measurable set in [0, T]. Meanwhile, for any r > 0, 1 ∈ [0, T], we have  |u:m (1, x) – u0 (x)|dx ≤ 9tr (1), (3.2.33) Kr

as :m → 0 and 1 → 0, we deduce  |u(x, t) – u0 (x)|dx = 0,

lim

t→0 t ∈ [0, T]\:

(3.2.34)

Kr

that is, the limit function u(x, t) satisfies the second demand of the general solutions. This completes the proof. ∎ Theorem 3.2.6. Assume that u(x, t), v(x, t) is a general solution to problem (5.2.2) and (5.2.3) with the initial conditions u0 (x), v0 (x), and |u(x, t)| ≤ M, v(x, t) ≤ M hold almost everywhere on (x, t) ∈ KR × [0, T]; assume that >i (x, t, u), 8(x, t, u) are continuous and differentiable on the domain {(x, t) ∈ FT , –∞ < u < ∞}, >ixj (x, t, u), >it (x, t, u) satisfy Lip condition on any compactly supported set with respect to u; and assume that # = max[–8u (x, t, u)], for almost every t ∈ [0, T0 ], we have on the domain {(x, t) ∈ K, |u| ≤ M} that   rt |u(x, t) – v(x, t)|dx ≤ e |u0 (x) – v0 (x)|dx, (3.2.35) st

s0

where K denotes the character cone: {(x, t)||x| ≤ K – Nt , 0 ≤ t ≤ T0 = min(T, KN –1 )}, ; n iu (x, t, u) , (x, t) ∈ KR × [0, T] |u| ≤ M

i=1

s4 denotes the section of the plane t = 4 on the cone K, 4 ∈ [0, T0 ]. Proof. Let the smooth function g(x, t, y, 4) ≥ 0 be the finite function supported on FT × FT . Set k = v(y, 4), f = g(x, t, y, 4) in inequality (3.2.31), where (y, 4) is a fixed point and the function v(y, 4) is defined a.e. on FT ; integrating on FT (i.e., with respect to (y, 4)), we have

3.2 Existence and Uniqueness of the General Solution to First-Order Multivariable



257



FT

FT

{|u(x, t) – v(y, 4)|g4 + sign(u(x, t) – v(y, 4))

⋅[>i (x, t, u(x, t)) – >i (x, t, v(y, 4))]gxi –sign(u(x, t) – v(y, 4))[>ixi (x, t, v(y, 4)) +8(x, t, u(x, t))]g}dxdtdyd4 ≥ 0.

(3.2.36)

Similarly, for the function v(y, 4), set k = u(x, t), f = g(x, t, y, 4) in inequality (3.2.31), and integrating on FT (i.e. with respect to (x, t)), we have 



FT

FT

{|v(y, 4) – u(x, t)|g4 + sign(v(y, 4) – u(x, t))

⋅[>i (y, 4, v(y, 4)) – >i (y, 4, u(x, t))]gyi –sign(v(y, 4) – u(x, t))[>iyi (y, 4, u(x, t)) +8(y, 4, v(y, 4))]g}dyd4dxdt ≥ 0,

(3.2.37)

summing eqs (3.2.36) and (3.2.37), and utilizing identical transformation in the integral expression, then for any smooth function g(x, t, y, 4) ≥ 0 supported on FT × FT , we have the following inequality: 

 FT

FT

{|u(x, t) – v(y, 4)|(gt + g4 ) + sign(u(x, t) – v(y, 4))

⋅[>i (x, t, u(x, t)) – >i (y, 4, v(y, 4))](gxi + gyi ) +sign(u(x, t) – v(y, 4))([>i (y, 4, v(y, 4)) – >i (x, t, v(y, 4))]gxi –>ixi (x, t, v(y, 4))g + [>i (y, 4, u(x, t)) – >i (x, t, u(x, t))]gxi +>iyi (y, 4, u(x, t))g + sign(u(x, t) – v(y, t)) ⋅[8(y, 4, v(y, 4)) – 8(x, t, u(x, t))]g}dxdtdyd4   {I1 + I2 + I3 + I4 }dxdtdyd4. = FT

(3.2.38)

FT

We first consider the simpler case, that is, >i (x, t, u) in eq. (5.2.2) depends only on u, that is, >i (x, t, u) = >i (u), where I3 = I4 = 0, and we have the following inequality: 

 FT

FT

{|u(x, t) – v(y, 4)|(gt + g4 ) + sign(u(x, t) – v(y, 4))

⋅[>i (u(x, t)) – >i (v(y, 4))](gxi + gyi )}dxdtdyd4 ≥ 0.

(3.2.39)

Assume that f (x, t) is an arbitrary test function, which equals zero except in the cylinder {(x, t)} = [1, T – 21] × Kr–21 , 21 ≤ min(T, r).

258

3 The Vanishing Viscosity Method of Quasilinear Hyperbolic System

Set  g(x, t, y, 4) = f n  i=1

$h

   x+y t+4 t–4 , $h , 2 2 2

x – y  . i i ≡ f (⋅ ⋅ ⋅)+h (..), h ≤ 1, 2

(3.2.40)

.. x–y t–4 t+4 where (⋅ ⋅ ⋅) ≡ ( x+y 2 , 2 ), (.) = ( 2 , 2 ). According to eq. (2.17’), define $h (3) as follows: for any h > 0, set $h (3) = h–1 $(h–1 3), $h (3) ≥ 0, $h (3) = 0, |3| ≥ h,  ∞ $h (3)d3 = 1. |$h (3)| ≤ h–1 const, –∞

Therefore, gt + g4 = ft (⋅ ⋅ ⋅)+h , gxi + gyi = fxi (⋅ ⋅ ⋅)+h . As h → 0, by eq. (3.2.39), we have  {|u(x, t) – v(x, t)|f t (x, t) + sign(u(x, t) – v(x, t)) FT

⋅[>i (u(x, t)) – >i (v(x, t))]fi (x, t)}dxdt ≥ 0.

(3.2.41)

In fact, every part of the right-hand side of eq. (3.2.39) can be represented by the following form: . ph (x, t, y, 4) ≡ F(x, t, y, 4, u(x, t), v(y, t))+h (..),

(3.2.42)

where F satisfies Lip condition with respect to all variables, ph equals zero except on the domain  t+4 |t – 4| {(x, t, y, 4)} = 1 ≤ ≤ T – 21, ≤ h, 2 2  |xi – yi | |x + y| ≤ r – 21, ≤h , 2 2     ph dxdtdyd4 = [F(x, t, y, 4), u(x, t), v(y, 4)) FT

FT

FT

FT

. –F(x, t, x, t, u(x, t), v(x, t))]+h (..)dxdtdyd4   . + F(x, t, x, t, u(x, t), v(x, t))+h (..)dxdtdyd4 FT

≡ J1 (h) + J2 .

FT

3.2 Existence and Uniqueness of the General Solution to First-Order Multivariable

259

Noting the obvious estimate . |+h (..)| ≤ const h–(n+1) , and the properties of the function F, we have 5 |J1 (h)| ≤ C h +

1



 |t–4| 2 ≤h |xi –yi | ≤h 2

hn+1

1 ≤ t+4 2 ≤T–1 x+y | 2 |≤r–1

6 |v(x, t) – v(y, 4)|dxdtdyd4 .

Here the constant C is independent of h; first, we have J1 (h) → 0, in fact, set vh =

1



 |t–4| 2 ≤h |x–y| 2 ≤h

hn+1

1 ≤ t+4 2 ≤T–1 x+y | 2 |≤r–1

|v(x, t) – v(y, 4)|dxdtdyd4.

Setting t+4 t–4 x+y x–y = !, = ", = ', = ., 2 2 2 2 then we have  vh = 2n+1 Gh (!, ') =

1 hn+1

1≤!≤T–1 |'| ≤ r – 1

G! (!, ')d'd!,



|"| ≤ h |. | ≤ h

|v(! + ", ' + . ) – v(! – ", ' – . )|d. d".

Since v(x, t) is bounded measurable in the cylinder Q = [0, T] × Kr , for almost all points in the cylinder [1, T – 1] × Kr–1 , (!, ') is the Lebesgue point of the function v(!, '), then we have |v(! + ", . + ') – v(! – ", ' – . )| ≤ |v(! + ", . + ') – v(!, ')| +|v(!, ') – v(! – ", ' – . )|. Therefore, as h → 0, Gh (!, ') → 0 and |Gh (!, ')| ≤ C(n) sup |v|, then performing the limit leads to limh→0 vh = 0, so as h → 0, J1 (h) → 0, for J2 , set t = !, t–4 2 = ", x = x–y ', 2 = . , and obviously, we have 

h –h

Therefore

Rn

+h (", . )d. d" = 1.

260

3 The Vanishing Viscosity Method of Quasilinear Hyperbolic System



 J2 = 2n+1 

FT

h h

–h

F(!, ', !, ', u(!, '), v(!, '))

 +h (", . )d. d" d'd!

 R

= 2n+1 FT

F(x, t, x, t, u(x, t), v(x, t))dxdt.

Moreover, since 



lim

h→0

FT

FT



= 2n+1 FT

ph dxdtdyd4

F(x, t, x, t, u(x, t), v(x, t))dxdt,

by eq. (3.2.39), inequality (3.2.41) holds as h → 0. Now let H denote some characteristic cone, :u , :v are zero sets of general solutions u, v on [0, T] of u, v, respectively, and let :, denote the set of Lebesgue points on [0, T] of a function  ,(t) = |u(x, t) – v(x, t)|dx, (3.2.43) st

which is not bounded measurable, set :0 = :u ∪ :v ∪ :, , obviously, mes :0 = 0, define  3 $h (3)d3, !′h (3) = $h (3) ≥ 0. !h (3) = –∞

Choose two constants 1, 4 ∈ [0, T0 ]\:0 , 1 < 4, in eq. (3.2.41), and set f = [!h (t – 1) – !h (t – 4)]7(x, t), h < min(1, T0 – 4), where 7 = 7: (x, t) ≡ 1 – !: (|x| + Nt – R + :), : > 0. Therefore, outside the characteristic cone H, 7(x, t) = 0, for (x, t) ∈ H, we have 3 = 7t + N|7x | ≥ 7t +

>i (u) – >i (v) 7xi . u–v

Therefore, by eq. (3.2.41), we have  [$h (t – 1) – $h (t – 4)]7: (x, t)|u(x, t) – v(x, t)|dxdt ≥ 0, FT0

when : → 0 in eq. (3.2.44), we have

(3.2.44)

3.2 Existence and Uniqueness of the General Solution to First-Order Multivariable



T0



 |u(x, t) – v(x, t)|dx dt ≥ 0,

 [$h (t – 1) – $h (t – 4)]

0

261

st

since 1, 4 are Lebesgue points of the function ,(t), as h → 0, we have  |u(x, 4) – v(x, 4)|dx

,(4) = 

s4

|u(x, 1) – v(x, 1)|dx = ,(1).

≤

(3.2.45)

s1

In fact, at the point t = 1, since |$h (3)| ≤

const , h

when h ≤ min(1, T0 – 1), we have    

T0

0

    $h (t – 1),(t)dt – ,(1) =  

≤ const h–1

0

1+h

1–h

T0

  $h (t – 1)[,(t) – ,(1)]dt

|,(t) – ,(1)|dt.

Since |u(1, x) – v(1, x)| ≤ |u(1, x) – u0 (x)| +|v(1, x) – v0 (x)| + |u0 (x) – v0 (x)|, as 1 → 0 in eq. (3.2.45), we get estimate (3.2.35). Now we consider the general case, by eq. (3.2.38), and we deduce the following inequality that is similar to eq. (3.2.41):  FT

{|u(x, t) – v(x, t)|f t + sign(u(x, t) – v(x, t))

⋅[>i (x, t, u(x, t)) – >i (x, t, v(x, t))]fxi – sign(u(x, t) – v(x, t)) ⋅[8(x, t, u(x, t)) – 8(x, t, v(x, t))]f }dxdt ≥ 0.

(3.2.46)

In fact, in eq. (3.2.38), I1 , I2 , I4 have similar integral as the form of function (3.2.42), utilizing stated method above; we can verify as h → 0 that 

 FT

FT

[I1 + I2 + I4 ]dxdtdyd4 → 2n+2 ,

multiplying by the right-hand side of eq. (3.2.46). Therefore, in order to prove eq. (3.2.46), it’s sufficient to prove that I3 → 0 as h → 0, so we consider

262

3 The Vanishing Viscosity Method of Quasilinear Hyperbolic System

 Ih =

 FT

FT

f (⋅ ⋅ ⋅)sign(u(x, t) – v(y, 4))

{[>i (y, 4, v(y, 4)) – >i (x, t, v(y, 4))](+h )xi –>ixi (x, t, v(y, 4))+h + [>i (y, 4, u(x, t)) –>i (x, t, u(x, t))](+h )yi +>iyi (y, 4, u(x, t))+h }dxdtdyd4.

(3.2.47)

Since the first-order derivative of >i (x, t, u) is uniformly continuous in any compact sets, we have (here +h is denoted by + for simplification, and $ij denotes Kronecker sign): [>i (y, 4, v(y, 4)) – >i (x, t, v(y, 4))]+xi – >ixi (x, t, v(y, 4))+ = >i4 (y, 4, v(y, 4))(4 – t)+xi + >iyj (y, 4, v(y, 4)) ⋅[(yj – xj )+xi – $ij +] + :i +xi + :0 + ≡ >i4 (y, 4, v(y, 4))((4 – t)+)xi +>iyj (y, 4, v(y, 4))((yj – xj )+)xi + :i +xi + :0 +. Similarly, since +yj = –+xi , we get [>i (y, 4, u(x, t)) – >i (x, t, u(x, t))]+yi + >iyi (y, 4, u(x, t))+ = >i4 (y, 4, u(x, t))(4 – t)+yi + >iyj (y, 4, u(x, t)) ⋅[(yj – xj )+yj – $ij +] + "i +yi ≡ >i4 (y, 4, v(y, 4))((4 – t)+)xi –>iyj (y, 4, u(x, t))((yi – xi )+)xi + "i +yi , where |:0 | +

n  (|:i | + |"i |) ≤ d:(d), i=1

d = |t – 4| + |x – y|. As d → 0, :(d) → 0, since + = +h = 0, |t – 4| ≥ 2h or |xi – yi | ≥ 2h, |+xi | + |+yi | ≤ const h–(n+2) , |f (⋅ ⋅ ⋅) – f (y, 4)| ≤ const(|t – 4| + |x – y|); therefore,  Ih =

FT

 FT

f (y, 4)sign(u(x, t) – v(y, 4)){[>i4 (y, 4, v(y, 4))

–>i4 (y, 4, u(x, t))]((4 – t)+)xi

3.2 Existence and Uniqueness of the General Solution to First-Order Multivariable

263

[>iyj (y, 4, v(y, 4)) – >iyj (y, 4, u(x, t))] ((yi – xi )+)xi }dxdtdyd4 + "(h),

(3.2.48)

where "(h) → 0 as h → 0. Denote by Bh the integrand expression of eq. (3.2.48), obviously Bh = Fi (y, 4, u(x, t), v(y, 4))((t – 4)+h f (y, 4))xi +Gij (y, 4, u(x, t), v(y, 4))((yj – xj )+h f (y, 4))xi , where Fi , Gij satisfy the Lip condition with respect to u, since   {Fi (y, 4, u(y, 4), v(y, 4))((4 – t)+h f (y, 4))xi FT

FT

+Gij (y, 4, u(y, 4), v(y, 4))((yj – xj )+h f (y, 4))xi }dxdtdyd4 = 0, by eq. (3.2.48), we have (   ( ( ( ( |Ih – "(h)| = ( B dxdtdyd4 h ( ( FT FT   f (y, 4)[+h + (|t – 4| + |x – y|) ≤ const FT

FT

⋅|(+h )x ||u(y, t) – u(y, 4)|dxdtdyd4   const ≤ n+1 |u(x, t) – u(y, 4)|dxdtdyd4 → 0 |t–4| | t+4 2 ≤h h 2 |≤T–1 x+y |x –y | i

i ≤h

2

| 2 |≤r–1

(similarly as the proof of J1 (h) → 0 in eq. (3.2.42)). Therefore, as h → 0, Ih → 0, then eq. (3.2.46) holds. Now choose two constants 1, 4 ∈ [0, T]\:0 , 0 < 1 < 4 < T0 , similarly as the proof of the simple case, we have  {[$h (t – 1) – $h (t – 4)]7: (x, t)|u(x, t) – v(x, t)| FT

+#7: (x, t)|u(x, t) – v(x, t)|}dxdt ≥ 0,   |u(x, 4) – v(x, 4)|dx ≤ |u(x, 1) – v(x, 1)|dx ,(t) = s4 4

s1

 +#

1

|u(x, t) – v(x, t)|dxdt,

st

in the set [0, T]\:0 , as 1 → 0, for 4 ∈ [0, T]\:0 , we have  4 ,(4) ≤ ,(0) + # ,(t)dt; 0

therefore, eq. (3.2.35) holds, and this completes the proof of Theorem 3.2.6.

∎

264

3 The Vanishing Viscosity Method of Quasilinear Hyperbolic System

Let K be the characteristic cone with the radius R, |u| ≤ M, set R–1 NM (R) → 0, R → ∞,

(3.2.49)

where NM (R) = max

|u|≤M [0,t]×KR

; n 

2iu (x, t, u)

,

i=1

from which we deduce obviously that each point (x, t) ∈ FT can be contained in its characteristic cone. From Theorem 3.2.6, we have Theorem 3.2.7. The general solution to problem (5.2.2) and (5.2.3) in FT is unique. Similarly as the derivation of estimate (3.2.35), for the general solutions u(x, t) and v(x, t) to problem (5.2.2) and (5.2.3), we have   I(u(x, t) – v(x, t))dx ≤ ert I(u0 (x) – v0 (x))dx, (3.2.50) st

s0

where I(3) ≡ 3 + |3|; therefore, we deduce the monotonic dependence on the initial data of the general solution to problem (5.2.2) and (5.2.3), we have Theorem 3.2.8. Assume that u(x, t), v(x, t) is the general solution to problem (5.2.2) and (5.2.3) with the initial conditions u0 (x), v0 (x), and u0 (x) ≤ v0 (x) holds a.e. in Rn , then in FT , we have u(x, t) ≤ v(x, t).

(3.2.51)

3.3 Convergence of Solutions to the Multidimensional Linear Parabolic System with Small Parameter We consider the linear parabolic equations with small parameter : > 0: Li: u ≡

∂uj ∂ui + aij uj = fi – : △ ui + akij ∂t ∂xk

(3.3.1)

(i, j = 1, 2, . . . , N, k = 1, 2, ⋅ ⋅ ⋅, n), satisfying the following initial conditions: ui |t=0 = >i (x), i = 1, 2, ⋅ ⋅ ⋅, N, x ∈ Rn .

(3.3.2)

Here we assume that akij (x, t) = akji (x, t); >i (x), fi (x, t) are all finite supported functions. Theorem 3.3.1. For the solutions u:i of problem (5.3.7) and (5.3.10), we have the following inequality:

3.3 Convergence of Solutions to the Multidimensional Linear Parabolic System

 t N  (u:i (x, t))2 dx + :



Rn i=1

5

N  (u:i (x, 0))2 dx

Rn i=1

 t

N 

+

Rn i=1

0

 (u:ixk (x, t))2 dxdt

Rn i,k

0

≤ c(t)

265

6 fi2 (x, t)dxdt , t ∈ [0, T],

where c(t) depends on t and the maximum value of coefficients akij , aij , but not on :0 .

(3.3.3) ∂akij ∂xk

in Rn × [0, T],

Proof. We first assume that the solution u:i ||x|=R = 0. Multiplying eq. (5.3.7) by ui , and integrating with respect to (x, t), then summing from 1 to N, we have  t

N 

|x|≤R i=1

0

1 2

ui Li: udxdt =



|x|≤R i=1

u2i dx|t=t t=0

5  1 ∂ u2ixk + (ak ui uj ) : 2 ∂xk ij |x|≤R

 t 0

N 

i,k

6 1 ui uj + aij uj ui dxdt 2 ∂xk  t fi ui dxdt. = ∂akij

–

0

(3.3.4)

|x|≤R

Since

 I1 =

Rn

∂ (ak ui uj )dx = 0, ∂xk ij

by eq. (5.3.16), we have 

1 2

N 

|x|≤R i=1

u2i dx|t=t t=0 +

≤c

|x|≤R

0

 N

 t 0

 t

|x|≤R

u2j

:

+

i=1



u2ixk dxdt

i,k N 

fi2

 dxdt,

i=1

then we get 

N  |x|≤R i=1

5 ≤ c(t)

u2i (x, t)dx + N 

|x|≤R i=1

 t 0

|x|≤R

u2i (x, 0)dx +

:



u2xk (x, t)dxdt

i,k

 t 0

N 

|x|≤R i=1

6 fi2 (x, t)dxdt .

(3.3.5)

266

3 The Vanishing Viscosity Method of Quasilinear Hyperbolic System

The solution u:i (x, t) to problem (5.3.7) and (5.3.10) can be taken as the limit of u:iR as R → ∞, (5.3.14) holds for any R < ∞; when R → ∞, we deduce inequality (5.3.15). This completes the proof. ∎ :

By eq. (5.3.15), there exists a subsequence {:k } such that as :k → 0 ui k converges in L2 (Rn × [0, T]) to some function ui (x, t), which can be proved as the general solution to the linear parabolic system with initial values ∂vi ∂vi + aij vj = fi , + akij (x, t) ∂t xk = >i (x), i, j = 1, ⋅ ⋅ ⋅, N, k = 1, ⋅ ⋅ ⋅, n.

Mi vi =

(3.3.6)

vi |t=0

(3.3.7)

In fact, multiplying eq. (5.3.7) by the continuous differentiable supported function Ii (x, t) with Ii (x, T) = 0, and summing from i = 1 to N, then integrating with respect to x ∈ Rn , t ∈ [0, T], through integrating by parts, we have  0

T

 ∂akij I ∂I ∂ui ∂Ii + uj – aij uj Ii dxdt –: ui ∂t ∂xk ∂xk ∂xk Rn  >i (x)Ii (x, 0)dx +







Rn T

=– Rn

0

fi (x, t)Ii (x, t)dxdt,

when performing the limit : = :k → 0, the second part of the right-hand side of the equality vanishes; therefore, for the limit function u(x, t), we have  T  ui Mi∗ (I)dxdt + >i I(x, 0)dx Rn

0



T



=– 0

Rn

Rn

fi (x, t)Ii (x, t)dxdt,

(3.3.8)

where Mi∗ (I) = fi are conjugate equations of eq. (3.3.6), which means that ui (x, t) is the general solution to problem (5.3.7) and (5.3.10) in L2 (Rn × [0, T]). utilizing the solvability of the initial value problem (3.3.6) and (3.3.7) and the corresponding conjugate problem M ∗ (I) = Ji , Ii |t=T = 0.

(3.3.9)

By solvability in the classical sense, we can deduce the uniqueness of the general solution to problem (3.3.6) and (3.3.7). In fact, by eq. (3.3.8), assume two general solutions ui (x, t), vi (x, t) with  T (ui – vi )Mi∗ (I)dxdt = 0, 0

Rn

3.3 Convergence of Solutions to the Multidimensional Linear Parabolic System

267

and choose Ji (x, t) as any smooth supported function; by eq. (3.3.9), we have 

T

 Rn

0

(ui – vi )Ji dxdt = 0.

Therefore, ui (x, t) = vi (x, t), and we deduce the following theorem: Theorem 3.3.2. Under the assumptions of Theorem 5.3.7, as : → 0, the solution u:i (x, t) of the initial value problem (5.3.7) and (5.3.10) converges weakly to the unique general solution of the initial value problem (3.3.6) and (3.3.7) in L2 (Rn × [0, T]). For the following second-order linear parabolic equation with the small parameter :>0 L: (u) = :

n 

 ∂ 2u ∂u + ai (x, t) ∂xi ∂xj ∂xi n

aij (x, t)

j,i=1

i=1

+b(x, t)u – ut = f (x, t),

(3.3.10)

with Cauchy condition u(x, :, 0) = y(x).

(3.3.11)

We can consider the behavior of solutions of problem (3.3.10) and (3.3.11) by the fundamental solution method as : → 0. Under quite weak conditions, by this method, if we can get satisfactory results about solutions, which can give the exact estimates of the convergence order, and the convergence is pointwise, then we introduce some main results as follows. When : → 0, problem (3.3.10) and (3.3.11) become L0 (v) = f (x, t), t > 0,

(3.3.12)

v(x, 0) = g(x),

(3.3.13)

where L0 (v) =

n  i=1

ai (x, t)

∂v ∂v + b(x, t)v – = f (x, t). ∂xi ∂t

(3.3.14)

Theorem 3.3.3. Assume that the following conditions are satisfied: (1) aij (x, t) = aji (x, t), and there exist l∗1 , l∗2 > 0 such that l∗1 +T + ≤

n 

aij +i +j ≤ l∗2 +T +

i,j=1

holds for (x, t) ∈ Qt′′ = Rn × [t′ , t′′ ] and any real n-dimensional vector +, where t′ , t′′ are two fixed finite numbers.

268

3 The Vanishing Viscosity Method of Quasilinear Hyperbolic System

(2) aij (x, t), ai (x, t), b(x, t) all satisfy H o¨ lder conditions with respect to x, with H o¨ lder index # and the common H o¨ lder constant N. (3) if

∂ ∂xj ai (x, t) exists in QT

and satisfies H o¨ lder condition with respect to x, with H o¨ lder

index #, then for any (x, t), (. , ') ∈ Qt′′ and t > ' and any : > 0, there exists a fundamental solution A(x, t; . , ', :) to L: (u) = 0,

(3.3.15)

A(x, t; . , ', :) = G(x, t, . , ', :)  t  d4 A(x, t, s, 4, :)8(s, 4, . , ', :)ds, +

(3.3.16)

with

'

where A(x, t, . , ', :) has the form  A(x, t, . , ', :) = h(t, . , ', :) exp

–(

 1 T )z F(t, . , ')z , 4:

z(x, t, . , ') = x – I(. , ', t), – 1   t 2 A(t, . , ', :) = 2–n Fn :n detC(t, . , ') exp b(v, . , ')dv, '

F(t, . , ') =

IT. (t, . , ')C–1 (t, . , ')I. (t, . , ').

I(t, . , ') is the unique solution of L0 (u) character equation dx = –a(x, t), dt

x = (x1 , ⋅ ⋅ ⋅, xn )T ,

a = (a1 , ⋅ ⋅ ⋅, an )T ,

(3.3.17)

satisfying the initial value I(', . , ') = . , 6. (t, . , ') =

∂ 6i (t, . , '), I. (t, . , ') = 6–1 . (t, . , '), ∂.j

C(t, . , ') is some definite symmetric matrix  C(t, . , ') =

t '

I. (t, . , ')A(v, . , ')IT. (v, . , ')dv,

A(v, . , ') = (aij (v, . , ')), J(x, t, . , ', :) is the following integration equation: J(x, t, . , ', :) = L: [G(x, t, . , ', :)]  t  d4 L: [G(x, t, s, 4, :)]J(s, 4, . , ', :)ds. + '

(3.3.18)

3.3 Convergence of Solutions to the Multidimensional Linear Parabolic System

269

Proof. We consider the equation along the character line L0 (u): D: (u) ≡ :aij (t, . , ')u′ jk + {aj (t, . , ') + (xk – 6k )ajk (t, . , ')}u′ j +b(t, . , ')u – u′ t = 0,

(3.3.19)

where ∂ 2u ∂u ∂u , u′ j = , u′ t = , ∂xi ∂xj ∂xj ∂t aik (t, . , ') = aki (6(t, . , '), t), u′ ij =

6k = 6k (t, . , '), (. , ') ∈ QT . Taking the Fourier transformation of the solution u(x, t, . , ', :) to eq. (3.3.19), we have  v(s, t, . , ', :) =

T

e–ix s u(x, t, . , ', :)dx,

then v(s, t, . , ', t) satisfies 5 6   d6 sT Bv′ s + v′ t = v – isT + B6 – :sT As + b – trB , dt

(3.3.20)

where 

 ∂v ∂v , ⋅ ⋅ ⋅, , A(t, . , ') = (aij (t, . , ')), ∂s ∂sn   ∂ai B(t, . , ') = (t, . , ') , ∂xj v′ s =

where trB denotes the trace of B. Consider the solution of eq. (3.3.20) with the initial condition T

v|t=' = :–i. ! , ! = (!1 , ⋅ ⋅ ⋅, !n )T ,

(3.3.21)

then we get the following formulation:  v = exp

 – isT 6 – :sT 6. c6T. s +

t

 (b – trB)dv .

(3.3.22)

'

Therefore, we deduce the inverse Fourier transformation of v 5 t 6 [b(v, . , ') – trB(v, . , ')]dv G(x, t, . , ', :) = (20)–n exp '  ⋅ exp[isT (x – 6) – :sT 6. c6T. s]ds.

(3.3.23)

270

3 The Vanishing Viscosity Method of Quasilinear Hyperbolic System

Modifying G(x, t, . , ', :) to the form in Theorem 5.3.15 and testifying D: (G) = 0, t > ', : > 0, then making some estimates for the differentiability of G(x, t, . , '), we have  t  L: (T) = L: (G) + d4 {:aij (x, t)G′ jk '

+aj G′ j + bG – Gt }8ds – 8  t = L: (G) – 8 + L: (G)8ds = 0 '

and

 lim

t→'+

A(x, t, . , ', :)g(. )d. = g(x).

Therefore, A(x, t, . , ', :) is the fundamental solution of L: (u).

(3.3.24) ∎

Theorem 3.3.4. Assume that g(x) is continuous in Rn , f (x, t) is continuous in QT and is H o¨ lder continuous with respect to x for t ∈ [t′ , t′′ ] uniformly, under the assumptions of Theorem 5.3.7, and assume that there exist constants m1 > 0, m2 ≥ 0 such that |g(x)| ≤ m1 em2 |x|2 , |f (x, t)| ≤ m1 em2 |x|2 ,

x ∈ Rn , (x, t) ∈ QT ,

then for any : > 0 and t ∈ [t′ , t′′ ], we deduce that  u(x, t, y, :) = A(x, t, . , ', :)g(. )d.  t  d' A(x, t, . , ', :)f (. , ')d. –

(3.3.25)

'

is the solution of the initial value problem L: (u) = f (x, t), t > y, u|t=y = g(x), : > 0

(3.3.26)

in Qt′′ , where 0 ≤ t – y ≤ t˜(,, y) = min(t′′ – y, (ˆr – ,)/32:m2 ), l 0 < , < rˆ = exp{–2p(t′′ – t)}, 8 1

l = d1 (d2 l∗2 ), p = n 2 N, and in Qt′′ (where 0 < t – y ≤ tˆ(,, y)), we have lim u(x, t, y, :) = v∗ (x, t, y),

:→0+

(3.3.27)

3.3 Convergence of Solutions to the Multidimensional Linear Parabolic System

271

where v∗ is the weak solution to the initial value problem L0 (v) = f , t > y, v|t=y = g,

(3.3.28)

and estimate u(x, t, . , :) which is a solution of the initial value problem (3.3.26) 2

|u(x, t, y, :)| ≤ k1 ek2 |x| , (x, t) ∈ Qt′′ ,

(3.3.29)

where k1 , k2 > 0 are fixed constants.  Proof. We can prove u∗ (x, t, y, :) = A(x, t, . , ', :)g(. )d. , L: (u∗ ) = 0 and u∗ → g(x) as t → y+ , that is, u∗ satisfies eq. (3.3.26) (f = 0) and the following estimate: lim u∗ (x, t, y, :) = g(6(y, x, t)) exp



:→0

t

b(v, x, t)dv, y

(x, t) ∈ QT , 0 < t – y < tˆ(,, y).

(3.3.30)

On the other hand, set 



t

¯ t, y, :) = – u(x,

d'

A(x, t, . , ', :)f (. , ')d. ,

y

then we have ¯ t, y, :)] = f (x, y). L: [u(x, Therefore, u(x, t, y, :) = u∗ + u¯ is the solution of L: (u) = f (x, y), and 



t

lim

:→0 y

A(x, t, . , ', :)f (. , ')d.

d'

 t



=

lim

'

:→0+

 t = '

 A(x, t, . , ', :)f (. , ')d. d'

f (6(', x, t), ') exp

 '

t

 b(v, x, t)dv d'.

(3.3.31)

By eq. (3.3.30), we have lim u(x, t, y, :) = v∗ (x, t, y).

:→0+

Here v∗ is the weak solution to the problem L0 (v) = f , V|t=y = g.

∎

272

3 The Vanishing Viscosity Method of Quasilinear Hyperbolic System

In order to improve the results of Theorem 5.3.16, we can replace the H o¨ lder continuous of f with respect to x with Dini continuous to deduce the following theorem. Theorem 3.3.5. Assume that g(x) is continuous uniformly in Rn , f (x, t) is continuous uniformly in QT and is Dini continuous uniformly with respect to x ∈ Rn in Qt′′ , under the assumptions of Theorem 5.3.7, and there exist N1 , N2 > 0 such that |g| ≤ N1 , x ∈ Rn , |f | ≤ N2 , (x, t) ∈ Qt′′ , then Theorem 5.3.16 holds. Moreover, there exists a constant k > 0 which is independent of f , g such that :(t – y) > 0 is sufficiently small and 0 < $ < 1, we have #

$

|u(x, t, y, :) – v∗ (x, t, y)| ≤ k[N1 [:(t – y)] 2 + 9(g, [:(t – y)] 2 ) #

$

+(t – y){N2 [:(t – y)] 2 + 9( f , [:(t – y)] 2 )}], (x, t) ∈ Qt′′ ,

(3.3.32)

where t > y, : > 0, 9( f , #) denotes the continuous modulus of f with respect to x. Now we consider the following initial value problem of the first-order linear parabolic equations with small viscosity term: ∂u = Ku + :Bu + g, ∂t u|t=0 = f ,

(3.3.33) (3.3.34)

where K = K(x, t, Dx ) is the first-order differentiable operator in the compact manifold M, whose main part K1 (x, t, . ) has different pure imaginary character values, and B = B(x, t, Dx ) is the second-order elliptic differentiable operator. Definition 3.3.6. We call that K + :B is dissipative uniformly, if there exists a sequence of self-conjugate operators R(:, x, t, Dx ) such that ||Ru||H s ≥ cs ||u||H s – c′s ||u||H s–1

(3.3.35)

Re(R(K + :B)u, u) ≤ –c:||u||2H 1 + c′ ||u||2L2 ,

(3.3.36)

and

where the constants cs , c′s , c, c′ are independent of :, t. Obviously, if K1 (x, t, . ) is antisymmetric, B(x, t, . ) ≤ – c|. |2 is negative self-conjugate, then when R = I, the above conditions are satisfied. Theorem 3.3.7. Assume that K + :B is dissipative uniformly, f ∈ H s (M), g ∈ Cj ([0, T], H s (M)), then there exists a unique solution u: (x, t) to initial value problem

3.3 Convergence of Solutions to the Multidimensional Linear Parabolic System

273

(3.3.33) and (3.3.34), and u: (x, t)(0 ≤ : ≤ 1) is bounded in Cl ([0, T], H s–2l (M)) (0 ≤ l ≤ j), that is sup ||Dlt u: ||H s–2l (M) ≤ cs,j,l [||f ||H s (M) + ||g||Cj ([0,T],H s (M)) ],

(3.3.37)

0≤t≤T

where the constant cs,j,l is independent of :. Proof. Since eq. (3.3.33) is a parabolic evolution equation when : > 0 and is symmetric hyperbolic equation when : = 0, whose existence and uniqueness have been proved, using the standard method, we can construct inequality (3.3.37); first, set s = 0, then we have d (Ru: , u: ) = (Ru′: , u: ) + (Ru: , u′: ) + (R′ u: , u: ) dt = (R(Ku: + :Bu: + g), u: ) + (Ru: , Ku: + :Bu: + g) + (R′ u: , u: ) ≤ 2Re(R(K + :B)u: , u: ) + c||u: ||2L2 + c||g||2L2 ≤ –c:||u: ||2H 1 + c||u: ||2L2 + c||g||2L2 . We can add a first-order operator to R such that R is positive definite, so we choose c′s = 0 in eq. (3.3.35), which has no effect on eq. (3.3.36), then we have d (Ru: , u: ) + c′ :||u: ||2H 1 (M) ≤ c(Ru: , u: ) + c||g||2L2 , dt

(3.3.38)

if we omit the term :||u: ||2H 1 (M) , by the Gronwall inequality 5 ||u: ||2L2

≤ ct ||f

 ||2L2

+ 0

t

||g(s)||2L2 ds

6 .

(3.3.39)

: For l = 0, s = 0, we get eq. (3.3.37), by using the equation ∂u ∂t = Ku: +:Bu+g repeatedly; for l ≤ j, s = 0, we construct inequality (3.3.37), in order to deduce eq. (3.3.27) with the general s; we set v: = D: u: multiplying eqs (3.3.33) and (3.3.34) by D–s , and noting that D–s KDs and D–s BDs have the same main part as K snd B, we can also get the necessary results by energy estimates. ∎

Theorem 3.3.8. Under the assumptions of Theorem 3.3.6, we have u: → u0 in Cl ([0, T], H s–2l (M)), : → 0, 0 ≤ l ≤ j. Proof. Assume that S: denotes the solution operator of problem (3.3.33) and (3.3.34), and ( f , g) → u: , (: → 0). Let j ≥ 1, since the mapping Cj–1 ([0, T], H s–1–2j (M)) is compact, for any sequence u:j (:j → 0), there exists a subsequence that is strongly convergent in

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3 The Vanishing Viscosity Method of Quasilinear Hyperbolic System

Cj–1 ([0, T], H s–1–2j (M)) with the limit v. Obviously, v is the solution of problem (3.3.33) and (3.3.34) (: = 0), because of the uniqueness, we have v ≡ u0 . Therefore, u: → u0 in Cj–1 ([0, T], H s–1–2j (M)), which is weaker than the conclusion of this theorem. However, if g ∈ C∞ ([0, T] × M), f ∈ C∞ (M), then we have u: → u0 in C∞ ([0, T] × M) as : → 0. Since E = {( f , g) : f ∈ C∞ , g ∈ C∞ ([0, T] × M)} is dense, with which to approximate Ej,s = {( f , g) : f ∈ H 2 , g ∈ Cl ([0, T], H s )}, and this completes the proof. ∎

3.4 On Gradient Quasilinear Parabolic Equations and Viscous Isentropic Gas Hydrodynamic Equations We first consider the following mixed boundary value problem for gradient parabolic equations: ∂ grad>(u) = 0, t > 0, ∂x u|t=0 = u0 (x), x1 < x < x2 ,

(3.4.2)

u|x=x1 = u|x=x2 = c = const,

(3.4.3)

ut – ,uxx +

(3.4.1)

where u = u(u1 , ⋅ ⋅ ⋅, un ) is an unknown n-dimensional vector function, >(u) = >(u1 , ⋅ ⋅ ⋅, un ) is a known function, , = const > 0, u0 (x) is a sufficiently smooth vector function and u0 (x1 ) = u0 (x2 ) = c; assume that >(u) is sufficiently smooth, and satisfies the following conditions: M2 (1 + max|u|≤M |>′′ (u)|2 ) where

|>′′ (u)|

→ ∞, M → ∞,

denotes the operator modulus of the matrix

>′′ (u)

(3.4.4)  =

∂ 2 >(u) ∂ui ∂uj

 in Rn .

Under the above assumptions, we have Theorem 3.4.1. There exists a solution to problem (3.4.1)–(3.4.3), as t → ∞, the vector value function u(x, t) → c for x ∈ [x1 , x2 ] uniformly. Proof. By taking the transformation u = u′ + c, the form of eq. (3.4.1) also holds. Setting c = 0, multiplying eq. (3.4.1) by u and integrating on RT1 = {x1 ≤ x ≤ x2 , 0 ≤ t ≤ T1 }, T1 > 0, by the equality  ∂ ∂ [(u, grad >(u)) – >(u)] u, grad >(u) = ∂x ∂x



and the boundary conditions (3.4.3) (c = 0), through integrating by parts, we have 1 2



x2 x1



T1

u2 (x, T1 )dx + , 0



x2 x1

 2  x2  ∂u    dxdt = 1 u2 (x)dx.  ∂x  2 x1 0

(3.4.5)

3.4 On Gradient Quasilinear Parabolic Equations and Viscous Isentropic Gas

275

Differentiating eq. (3.4.1) with respect to x, we have ∂ 2 ux ∂2 ∂ux –, + 2 >′ (u) = 0, ∂t ∂x ∂x

(3.4.6)

′ where ux = ∂u ∂x , > (u) = grad >(u). Multiplying eq. (3.4.6) by ux , integrating on RT1 and integrating by parts, we have

 2 2 ∂ u   +,  ∂x2  dxdt x1 0 x1  T 1  x2  ′ 2   1 x2 ′ 2 ∂> ∂ u = (u0 ) dx + , dxdt 2 x1 ∂x ∂x2 0 x1   T1  2 ∂ u ∂>′ ∂u x=x2 dt. + , , 2 – ∂x ∂x ∂x 



1 2

x2



u2x (x, T1 )dx

T1



x2

0

(3.4.7)

x=x1

By eqs (3.4.1) and (3.4.3), we deduce that the last term of the right-hand side of eq. (3.4.7) equals zero and 

     ∂>′ ∂ 2 u 1 ∂ 2 u 2 1 ∂>′ 2 + , , ≤ , ∂x ∂x2 2 ∂x2 2, ∂x 2  2  ′ 2   ∂u  ∂u ∂> ≤ >′′  ≤ |>′′ (u)| , ∂x ∂x ∂x

so we have 

∂>′ ∂ 2 u , ∂x ∂x2



   2 1 ∂ 2 u 2 1 ′′ 2 ∂u + (u)| . |> ≤ , 2 ∂x2 2, ∂x

(3.4.8)

Assume that T1 ≤ T, T > 0 are arbitrary constants, and M = maxt≤T |u(x, t)|, by eqs (3.4.7), (3.4.8) and (3.4.5), we have 

 2 2 ∂ u   +,  ∂x2  dxdt x1 0 x1  x2  x2 1 (u′0 )2 dx + 2 max |>′′ (u)|2 u20 dx. ≤ 2, |u|≤M x1 x1 x2



u2x (x, T1 )dx

T1



x2

(3.4.9)

By the boundary condition u(x1 , t) = 0, we have   x ∂ ∂u (u, u)dx = 2 , u dx ∂x x1 ∂x x1 1   x2   2  x2 2 ∂u dx u2 dx . ≤2 ∂x x1 x1 

|u|2 = (u, u) =

x

(3.4.10)

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3 The Vanishing Viscosity Method of Quasilinear Hyperbolic System

By eqs (3.4.5), (3.4.9) and (3.4.10), we deduce (x1 ≤ x ≤ x2 , 0 ≤ T1 ≤ T) |u(x, T1 )|2 ≤

1 c0 max |>′′ (u)| + c0 , , |u|≤M

(3.4.11)

where the constant c0 depends only on the initial value u0 (x) of problem (3.4.1)–(3.4.3), but not on the constant T0 ; by eq. (3.4.11), we get inequalities M2 =

1 c0 max |>′′ (u)| + c0 ,0 |u|≤M

and  M

2

′′

–1

max |> (u)| + ,

|u|≤M

≤ ,–1 c0 .

(3.4.12)

We deduce from eqs (3.4.12) and (3.4.4) that M ≤ M0 < ∞, where M0 depends only on the initial data u0 (x) and ,, but not on T. Therefore, in R{x1 ≤ x ≤ x2 , t ≥ 0}, we have |u(x, t)| ≤ M0 .

(3.4.13)

Utilizing the following existence theorem of the general parabolic equations, we deduce the first half of the theorem. Theorem 3.4.2. For the initial value problem of the following equations ∂ 2 ui ∂ui  ∂ui = bij (x, t, u1 , ⋅ ⋅ ⋅, uN ) + 2 ∂x ∂t ∂x N

j=1

+

N 

cij (x, t, u1 , ⋅ ⋅ ⋅, uN )uj + fi (x, t),

(3.4.14)

j=i (j)

ui (x, 0) = u(0) i (x), ui (xj , t) = ui (t) i = 1, ⋅ ⋅ ⋅, N, j = 1, 2,

(3.4.15)

then on the rectangle RT {x1 ≤ x ≤ x2 }, 0 ≤ t ≤ T, there exists a unique solution, which is continuous on RT . Derivatives that arise in equations on RT are bounded and are continuous in RT , if the following conditions are satisfied: (1) For (x, t) ∈ RT and any u1 , ⋅ ⋅ ⋅, uN , |bij (x, t, u1 , ⋅ ⋅ ⋅, uN )| < B, N 

cij (x, t, u1 , ⋅ ⋅ ⋅, uN )+i +j ≥ c

i,j=1

where B, c are bounded constants.

N  i,j=1

+2i ,

(3.4.16)

3.4 On Gradient Quasilinear Parabolic Equations and Viscous Isentropic Gas

277

(2) For (x, t) ∈ RT and N  i=1

⎧ ⎪ ⎨

⎫ ⎪ N ⎬ 2  NF (j) u2i ≤ M = max max (ui )2 e2cT , e2cT , ⎪ ⎪ C ⎩ RT ⎭

(3.4.17)

j=0,1,2 i=1

where C > B2 N 2 /2 + 2|c| + 1, F = max |f i (x, t)|, the coefficients bij (x, t, u1 , ⋅ ⋅ ⋅, uN ) and their first-order and second-order derivatives with respect to x, uk are continuous, the derivatives ∂ 2 bij ∂ 2 bij ∂ 3 bij ∂ 3 bij , , , , ∂x∂t ∂uk ∂t ∂x2 ∂t ∂x∂t∂uk ∂ 3 bij ∂ 3 bij ∂ 3 bij ∂ 3 bij , , , 2 ∂x∂uk ∂ul ∂t∂uk ∂ul ∂x ∂uk ∂un ∂uk ∂ul

(3.4.18)

are continuous in RT , cij (x, t, u1 , ⋅ ⋅ ⋅, uN ) and fi (x, t) have the same smoothness as bij . (3) The function u(0) i (x) has the second-order derivative, which is bounded on x1 ≤ x ≤ (2) x2 and is continuous on x1 < x < x2 , the functions u(1) i (t), ui (t) have first-order continuous derivative: (j)

u(0) i (xj ) = ui (0), i = 1, ⋅ ⋅ ⋅, N, j = 1, 2. Theorem 3.4.2 can be proved by the Rothe method. By Theorem 3.4.2 and estimate 3.4.13, we can deduce the existence of the classical solutions to problem (3.4.1)– (3.4.3), and now we consider the asymptotic property of the solutions to problem (3.4.1)–(3.4.3) as t → ∞, we first prove 

x2

z(t) = x1

u2x (x, t)dx → 0, t → ∞.

In fact, since T1 in eq. (3.4.5) is arbitrary, we have  ∞ z(t)dt < ∞.

(3.4.19)

0

Moreover,    ∂u ∂u x2 ∂u ∂ 2 u , dx = 2 , ∂x ∂x∂t ∂x ∂t x1 x1    x2  2  x2  2 ∂ u ∂u ∂ u ∂u –2 , , dx = –2 dx. ∂x2 ∂t ∂x2 ∂t x1 x1

dz =2 dt



x2



By eq. (3.4.1), we have dz = –2, dt



x2 x1



∂ 2u ∂x2

2



x2

dx + 2 x1



 ∂ 2 u ′′ ∂u , > (u) dx. ∂x2 ∂x

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3 The Vanishing Viscosity Method of Quasilinear Hyperbolic System

By eq. (3.4.8), we have 

     dz dt ≤ 3,  dt 

∞  x2

∞ 0

+

0



x1

∂ 2u ∂x2

1 max |>′′ (u)|2 , |u|≤M0

2 dxdt 

∞  x2



x1

0

∂u ∂x

2 dxdt.

(3.4.20)

By eqs (3.4.9) and (3.4.13) and since T1 in eq. (3.4.9) is arbitrary, we have  0

∞  x2



x1

∂ 2u ∂x2

2 dxdt < ∞.

By eq. (3.4.19), the second term of the right-hand side of eq. (3.4.20) is finite, so we have 

    dz dt < ∞.  dt 

∞

0

We deduce from the convergence of the integration 0 as t → ∞, by eq. (3.4.5), T1 = t, so we have 

x2

 u2 (x, t)dt
0, and we deduce from eq. (3.4.21) and z(t) → 0 (t → ∞) that u(x, t) → 0, t → ∞ ∀x ∈ [x1 , x2 ].

(3.4.22) ∎

This completes the proof of Theorem 3.4.1. For the initial value problem of eq. (3.4.1) u|t=0 = u0 (x), x ∈ R1 ,

(3.4.23)

where u0 (x) is a sufficiently smooth vector function; similarly, we deduce the following theorem: Theorem 3.4.3. Assume that condition (3.4.4) holds, and the functions u0 (x), u′0 (x), ′′′′ 1 u′′0 (x), u′′′ 0 (x), u0 (x) are bounded in R , and 

 2

|u0 (x) – c| dx and R

R

|u′0 (x)|2 dx

3.4 On Gradient Quasilinear Parabolic Equations and Viscous Isentropic Gas

279

are convergent, then there exists a classical solution u(x, t) to the initial value problem (3.4.1) and (3.4.23), and the modulus of u(x, t), ux , ut , uxx , uxt , utt , uxxx , uxxt , uxxxx is bounded on t ≥ 0. Theorem 3.4.4. Under the assumptions of Theorem 3.4.3, and 

∞

–∞

|u′′0 (x)|2 dx < ∞,

(3.4.24)

then for the solution of the initial value problem (3.4.1) and (3.4.23), we have u(x, t) → c, t → ∞ ∀ x ∈ R1 .

(3.4.25)

We can extend the above conclusion to more general gradient quasilinear parabolic system: ∂ ∂ ∂2 grad L(u) – , 2 grad L(u) + grad >(u) = 0, ∂t ∂x ∂x

(3.4.26)

where the functions L(u), >(u) are sufficiently smooth, satisfying min (L′′ (u). , . ) = #(M) > 0, . ∈ Rn ,

(3.4.27)

M 2 #(M)(max |>′′ (u)| + 1)–1 → ∞, M → ∞,

(3.4.28)

M#(M) → ∞, M → ∞.

(3.4.29)

|. |=1 |u|≤M

|u|≤M

Now we consider the following quasilinear semi-parabolic systems: :uxx = ut + uux + vx ,

(3.4.30)

0 = vt + vux + uvx ,

(3.4.31)

u|t=0 = u0 (x), v|t=0 = v0 (x), x ∈ [x1 , x2 ],

(3.4.32)

u(x1 , t) = u(x2 , t) = 0.

(3.4.33)

with the initial conditions

We first prove the existence of the solution of problem (3.4.30), (3.4.31), (3.4.32) and (3.4.33), for which we make the transformation  9=

x x1

vdx, z = :u – 9.

(3.4.34)

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3 The Vanishing Viscosity Method of Quasilinear Hyperbolic System

Then eqs (3.4.30) and (3.4.31) become :

∂ 2 z ∂z 9 + z ∂z ∂9 9 + z ∂9 = + , 0= + , ∂x2 ∂t : ∂x ∂t : ∂x

(3.4.35)

with the corresponding boundary conditions  x v0 (x)dx = 90 (x), 9|t=0 = x1

z|t=0 = :u0 (x) – 90 (x) = z0 (x), 9|x=x1 = –z|x=x1 = 0,  x2 vdx = const. 9|x=x2 = –z|x=x2 =

(3.4.36)

x1

Obviously, if there exists a solution to problem (3.4.35) and (3.4.36), then the solution to problem (3.4.30)–(3.4.33) is u=

9+z , v = 9x . :

(3.4.37)

Lemma 3.4.5. If the following conditions hold: (1) 90 (x), z0 (x) ∈ C6 , (2) 90 (x! ) + z0 (x! ) = 0, ! = 1, 2; z0′′ (x! ) + 9′′0 (x! ) = 0, (3) T0 < 5(9˜:ln2 , x +˜zx ) where

  |9| + |z| ′ ˜ , 9x = max max 90 (x), max x=x1 ,x2 t=0 :   |9| + |z| z˜ x = max max z0′ (x), max , x=x1 ,x2 t=0 :

then there exists a unique local classical solution to problem (3.4.35) and (3.4.36), where the derivatives that arise in eq. (3.4.35) are continuous on RT0 {x1 ≤ x ≤ x2 , 0 ≤ t ≤ T0 }. Proof. By successive approximation method, assume that z0 (x, t) = z0 (x), 90 (x, t) = 9(x), zn (x, t), 9n (x, t) satisfy :

∂zn 9n–1 + zn–1 ∂zn ∂ 2 zn = + , ∂x2 ∂t : ∂x ∂9n 9n–1 + zn–1 ∂9n 0= + ∂t : ∂x

(3.4.38)

and condition (3.4.36), n = 1, 2, ⋅ ⋅ ⋅, we first prove the existence and smoothness of the solution to problem (3.4.38) and (3.4.36). Since z0 (x, t), 90 (x, t) ∈ C4 , in RT0 , we assume that zn–1 , 9n–1 have the following continuous derivatives:

3.4 On Gradient Quasilinear Parabolic Equations and Viscous Isentropic Gas

281

(zn–1 )x , (zn–1 )t , (zn–1 )xx , (zn–1 )tt , (zn–1 )xxx , (zn–1 )xxxx , (9n–1 )x , (9n–1 )t , . . . .

(3.4.39)

Now we start to prove that 9n , tn have the same smoothness as the initial functions. The first-order equation 0=

∂9n 9n–1 + zn–1 ∂9n + , ∂t : ∂x

(3.4.40)

which is equivalent to the ordinary differential equation dx 9n–1 + zn–1 d9n = , = 0, dt : dt

(3.4.41)

the integration of eq. (3.4.41) x0 = g(x, t)(x0 = x(0)) has the second-order derivative with respect to t and the fourth-order derivative with respect to x. Meanwhile, 9n (x, t) = 90 (x0 ) = 90 (g(x, t)) has the same smoothness. Now we consider the smoothness of the solution to :

∂ 2 zn ∂zn 9n–1 + zn–1 ∂zn = + . ∂x2 ∂t : ∂x

Set zxx = zt + b(x, t)zx .

(3.4.42)

Assume that bx , bxt , bxxt exist in RT , and consider the initial value problem z|t=0 = z0 (x), z|x=x1 = 0, z|x=x2 = c. If the initial function z0 (x) ∈ C6 , z(x, t) satisfies z(x1 , 0) = 0, z(x2 , 0) = c, z0′′ (x! , 0) = b(x! , 0)z0′ (x! , 0), z0′′′ (x! , 0) = b(x! , 0)z′′′ (x! , 0) +2bx (x! , 0)z0′′ (x! , 0) + bxx (x! , 0)z0′ (x! , 0),

(3.4.43)

then the solution of eqs (3.4.42) and (3.4.43) is bounded on RT0 ; if b(x, t) has the same form of continuous derivatives as that of eq. (3.4.39), then the corresponding solution has the same derivative with the same smoothness. Therefore, we have proved the existence and the same smoothness of the solution to eqs (3.3.38) and (3.4.36). Now we start to prove 9n , zn and the corresponding derivatives are even bounded. It’s obvious to verify that 9n (x, t), zn (x, t) are even bounded, in view of the characteristic line of 9n (x, t), Then |9n | ≤ max |90 (x)|. Since zn (x, t)

282

3 The Vanishing Viscosity Method of Quasilinear Hyperbolic System

satisfies the parabolic equation (3.4.38), the maximum can’t be achieved except on the boundary. ∂9n n Now consider the even boundedness of znx = ∂z ∂x , 9nx = ∂x , 9nx and znx are bounded on t = 0, by the barrier function estimation on the boundary, we can deduce the boundedness of znx on x = x1 , x = x2 ; we therefore have |znx (x, t)| ≤

1 max |9n + zn |, : x=x1 ,x2

now we estimate znx , 9nx in RT0 , and set ¯ n. zn = ect z¯ n , 9n = ect 9

(3.4.44)

¯ nx , z¯ nx satisfy the equation Then 9 :

∂ z¯ nx 9n–1 + zn–1 ∂ z¯ nx ∂ 2 z¯ nx = + ∂x2 ∂t : ∂x 9n–1x + zn–1x z¯ nx + c¯znx , + : ¯ nx 9n–1 + zn–1 ∂ 9 ¯ nx 9n–1x + zn–1x ∂9 ¯ nx . ¯ nx + c9 0= + + 9 ∂t : ∂x :

(3.4.45)

Set zx = max |znx |, 9x = max |9nx |. x=x1 x2 t=0

If we choose T0 =

:ln2 , 5(9x +zx )

x=x1 x2 t=0

then for t ≤ T0

max |zn–1x | ≤ 2zx , max |9n–1x | ≤ 29x . ¯ nx are positive, then Let c = 5: (9x + zx ), then in eq. (3.4.45), the coefficients of z¯ nx , 9 ¯ nx | can only we deduce: |¯znx | can only achieve the maximum on x = x1 , x2 or t = 0, |9 achieve the maximum on t = 0. Therefore, we have |9nx | < ecT0 9x , |znx | < ecT0 zx , since eCT0 < 2, we get |9nx | < 29x , |znx | < 2zx ,

(3.4.46)

which hold for n – 1, and thus hold for all n. For estimates of the other derivatives in eq. (3.4.39) with respect to 9, z, it’s sufficient to differentiate the original equation. For example, to estimate 9nxx , znxx , we need to differentiate the original equation two times, that is,

3.4 On Gradient Quasilinear Parabolic Equations and Viscous Isentropic Gas

:

283

∂ z¯ nxx 9n–1 + zn–1 ∂ z¯ nxx ∂ 2 z¯ nxx 9n–1x + zn–1x z¯ nxx = + +2 ∂x2 ∂t : ∂x : 9n–1xx + zn–1xx z¯ nx + c¯znxx , + : ¯ nxx 9n–1 + zn–1 ∂ 9 ¯ nxx ∂9 9n–1x + zn–1x ¯ nxx 0= + +2 9 ∂t : ∂x : 9n–1xx + zn–1xx ¯ nxx . ¯ nx + c9 + 9 :

When c = 5: (9x + zx ) and t ≤ T0 , by the maximum theory, we have  2     d z0  max |9n–1xx + zn–1xx | |znxx | ≤ max 2 max  2 , , dx c:  2     d 90  max |9n–1xx + zn–1xx | , |9nxx | ≤ max 2 max  dx2  c: or  2   2    d 90   d z0   ,  |9nxx | + |znxx | ≤ max 2 max  2  + 2 max  dx dx2   max |9n–1xx + zn–1xx | . Then we deduce when t ≤ T0 that  2   2   d z0   d 90    + 2 max  2 , |znxx | + |9nxx | ≤ 2 max  dx2  dx the even boundedness of 9nt , znt can be deduced from eq. (3.4.38). The proof of the even boundedness of 9nxt , znxt , 9nxxx , znxxx is similar to the above, and the even boundedness of the other relative derivatives can be derived by differentiating eq. (3.4.38). Using estimates of the even boundedness of all derivatives with respect to n in eq. (3.4.39) and sequential compactness theory, we deduce easily the existence of the local classical solution to problem (3.4.35) and (3.4.36); now we start to prove the uniqueness of the solution to problem (3.4.35) and (3.4.36). Suppose that there exist two solutions: z1 , 91 , z2 , 92 , set ¯ = (91 – 92 )e–!t . z¯ = (z1 – z2 )e–!t , 9 Then :

¯ + z¯ ∂z2 ∂ 2 z¯ ∂ z¯ 91 + z1 ∂ z¯ 9 = + + + !¯z, 2 ∂x ∂t : ∂x : ∂x ¯ 9 ¯ 91 + z1 ∂ 9 ¯ + z¯ ∂92 ∂9 ¯ + + + !9 0= ∂t : ∂x : ∂x

(3.4.47)

284

3 The Vanishing Viscosity Method of Quasilinear Hyperbolic System

satisfy the zero initial condition and the boundary condition. By the second term of ¯ = 0(¯z); therefore, we deduce from the first term that z¯ can’t eq. (3.4.47), we deduce 9 achieve the maximum value in RT0 when ! is sufficiently large, hence z¯ = 0, and ¯ ≡ 0. 9 ∎ Lemma 3.4.6. Assume that 9x , zx in problem (3.4.35) and (3.4.36) satisfy |zx | < M, |9x | < M,

(3.4.48)

and all the assumptions in Lemma 3.4.5 are satisfied with v0 (x) > 0. Then there exists a global classical solution to problem (3.4.35) and (3.4.36) on any RT . Proof. The solution of problem (3.4.35) and (3.4.36) can be taken as the limit of the solution of problem (3.4.38) and (3.4.36) when n → ∞, and 9nx → 9x , znx → zx hold true evenly on the domain where 9, z exist, if |zx | < M, |9x | < M(t ≤ T1 ), then we can find n0 such that when n ≥ n0 , we have |znx | < 2M, |9nx | < 2M. Take the transformation ˜ n ec(t–T1 ) , zn = z˜ n ec(t–T1 ) , 9n = 9 Consider the classical solution of problem (3.4.38) and (3.4.36) on RT2 , where T2 = T1 + :ln2/(8M). Assume that |zn–1x | < 4M, |9n–1x | < 4M on RT2 , we use inductive method to prove |znx | < 4M, |9nx | < 4M. Obviously, this holds true for n = 0; assume c = 8M/: and estimate 9nx , znx on RT2 – RT1 , we have ˜ nx |ec(T2 –T1 ) , |9nx | ≤ max |9 t=T1

|znx | ≤ max |˜znx |ec(T2 –T1 ) . t=T1 x=x1 ,x2

Note that ˜ nx | = max |9nx | < 2M, max |9 t=T1

t=T1

max |˜znx | = max |znx | < 2M,

t=T1 x=x1 ,x2

t=T1

and because of ec(T2 –T1 ) < 2, we therefore have on RT2 that |9nx | < 4M, |znx | < 4M,

(3.4.49)

3.4 On Gradient Quasilinear Parabolic Equations and Viscous Isentropic Gas

285

similarly, we can prove derivatives of the other forms are all bounded, and hence we can deduce the existence of the classical solution to problem (3.4.35) and (3.4.36) on RT2 ; by this step, we can extend the solution globally on any RT . When 9′0 (x) ≥ 0, i.e., v0 (x) ≥ 0, we construct estimate (3.4.48). Consider the equation satisfied by 9x 0=

∂9x 9 + z ∂9x zx + 9x + + 9x . ∂t : ∂x :

(3.4.50)

Since 9x |t=0 ≥ 0, thus we have 9x (x, t) ≥ 0, t > 0. In addition, 9x |t=0 is bounded, assume that 9x achieve the maximum value on t > 0, then on this point, ∂9x ∂9x ≥ 0, = 0. ∂t ∂x We therefore have (9x + zx )9x ≤ 0, that is 9x + zx ≤ 0, |9x | ≤ max |zx |. Now we estimate |zx |, and take the transformation 

z¯

z = >(¯z) = k

s2

e– 2 ds.

0

Whence 8 k=

2 z˜ , z˜ ≥ max |z|, T1

we have  ′′ ′  3 ∂ 2 z˜ x 9 + z ∂ z˜ x ∂ z˜ x > ∂ z¯ L˜zx ≡ : 2 – – = –: ∂x : ∂x ∂t >′ ∂x >′′ ∂ 2 z¯ ∂ z¯ 9x + zx z¯ x . –2: ′ 2 + > ∂x ∂x : On the maximum point of z¯ x , we have z¯ x L¯zx =

:¯zx4

9x + zx 2 z¯ x , + :



>′′ >′

′

= –1, z¯ xx = 0.

Then |9x | ≤ max |zx | ≤ max |>′ (¯z)||¯zx |,

286

3 The Vanishing Viscosity Method of Quasilinear Hyperbolic System

that is |9x | ≤ k max |¯zx |, 0 ≥ z¯ x L¯zx ≥ :¯zx4 –

2k 3 z¯ . : x

Therefore, |¯zx | ≤

2k . :2

The boundedness of zx is deduced from zx = >′ (¯z)¯zx ; we finally have |9x | ≤

2k2 , :2

from which we get estimate (3.4.48).

(3.4.51) ∎

We deduce from Lemma 3.4.6 and estimate (3.4.48) that Theorem 3.4.7. Under the assumptions of Lemma 3.4.5 and v0 (x) > 0, there exists a global classical solution to problem (3.4.35) and (3.4.36) on any RT , that is, the solution is continuous on RT , and derivatives in eq. (3.4.35) are also continuous. For the initial value problem of eq. (3.4.30) u(x, 0) = u0 (x), v(x, 0) = v0 (x), x ∈ R.

(3.4.52)

Similarly, we have Theorem 3.4.8. Assume that the following conditions are satisfied: (1) The initial functions u0 (x), v0 (x) ∈ C4 , and the corresponding derivatives are bounded.  (2) v0 (x) ≥ 0, R v0 (x)dx < ∞, then there exists a unique classical solution to problem (3.4.30), (3.4.31) and (3.4.52) on RT {x ∈ R, 0 ≤ t ≤ T}. Proof. Set u0m (x), v0m (x) on Rm {–m ≤ x ≤ m, 0 ≤ t ≤ T} such that they are sufficiently smooth on Rm–1 , u0m (x) = u0 (x), v0m (x) = v0 (x), and are zero outside Rm , dk u0 (x) dk u0m (x) → dxk dxk dk v0 (x) dk v0m (x) → , –m < x < m, k = 0, 1, 2, 3, 4. dxk dxk In addition, their derivatives up to fourth order are bounded.

3.4 On Gradient Quasilinear Parabolic Equations and Viscous Isentropic Gas

287

Now we consider eqs (3.4.30), (3.4.31) and u|t=0 = u0m (x), v|t=0 = v0m (x), u|x=–m = u|x=m = 0.

(3.4.53)

Assume that the solution of problem (3.4.30), (3.4.31) and (3.4.52) is of the form u(m) (x, t), v(m) (x, t). Setting  9(m) (x)(x, t) =

x

v(m) (x, t)dx,

–m

z(m) (x, t) = :u(m) (x, t) – v(m) (x, t), which satisfy the equations of the form (3.4.35), we can prove that zn(m) , 9(m) n and their relative derivatives are bounded uniformly with respect to m, n, as m, n → ∞, and we deduce the solution to the initial value problems (3.4.30), (3.4.31) and (3.4.52). Now we start to prove the uniqueness, and suppose that there exist two solutions to problems (3.4.30), (3.4.31) and (3.4.52), the corresponding equations are z1 , 91 , z2 , 92 . Setting ¯ = e–!t (91 – 92 ), z¯ = e–!t (z1 – z2 ), 9 which satisfy :

¯ + z¯ ∂z2 ∂ 2 z¯ ∂ z¯ 91 + z1 ∂ z¯ 9 = + + + !¯z, 2 ∂x ∂t : ∂x : ∂x ¯ 9 ¯ 91 + z1 ∂ 9 ¯ + z¯ ∂92 ∂9 ¯ + + + 29 0= ∂t : ∂x : ∂x

and zero initial condition, then ¯ ≤ max |¯z|. max |9| Set v = x2 + kt, where k is a constant that will be fixed later, then when k is sufficiently large, we have Lv ≡ :vxx – vt –

¯ + z¯ 91 + z1 9 vx – z2x – !v < 0. : :

Consider F1 = '

v(x, t) v(x, t) + z¯ , F2 = –' + z¯ , ' > 0 v(x0 , t0 ) v(x0 , t0 )

on RM {–M ≤ x ≤ M, 0 ≤ t ≤ T}, choose M sufficiently large, such that on the boundary of RM ,

288

3 The Vanishing Viscosity Method of Quasilinear Hyperbolic System

F1 ≥ 0, F2 ≤ 0

and

LF1 < 0, LF2 > 0.

We deduce from LF1 < 0 that at the minimum point, we have    ∂z2  ¯ + max |¯z| 1 max |9| .  F1 ≥ – max  ! : ∂x  Then    ∂z2  2 max |¯z| .  F1 ≥ – max  ! : ∂x  Therefore, on (x0 , t0 ) ∈ Rm , we have    ∂z2  2 , max |¯z| max  !: ∂x 

z¯ (x0 , t0 ) ≥ –' – and hence z¯ (x0 , t0 ) ≤ ' +

   ∂z2  2 .  max |¯z| max  !: ∂x 

So when ! is sufficiently large, we have max |¯z| ≤ ' + q max |¯z|, 0 < q < 1, ' max |¯z| ≤ . 1–q Since ' is an arbitrary number, thus we have ¯ = 0. z¯ = 0, 9 This completes the proof of the theorem.

∎

Now we consider the following viscous gas hydrodynamic equations: ut – (k(v)ux )x + >(v)x = 0,

(3.4.54)

v t – ux = 0

(3.4.55)

endowed with the initial conditions u|t=0 = u0 (x), v|t=0 = v0 (x).

(3.4.56)

We will prove the existence of the global smooth solution to problem (3.4.54)–(3.4.56) and the asymptotic behavior as t → ∞. We make the following assumptions:

3.4 On Gradient Quasilinear Parabolic Equations and Viscous Isentropic Gas

289

(1) Functions k(v), >(v) are smooth enough, and k(v) > 0, v > 0; >′ (v) < 0, v > 0,  c  k(v) 8(v, c)dv = ∞,

(3.4.57)

0

where 

v

8(v, c) =

(>(c) – >(s))ds,

(3.4.58)

c

or 

+∞

c



c

(3.4.59)

 k(v) 8(v, c)dv ≥ c0 , c0 is some definite constant,

0



 k(v) 8(v, c)dv = ∞,

+∞

 k(v) 8(v, c)dv ≥ c0 .

(3.4.60) (3.4.61)

c

(2) The initial functions u0 (x), v0 (x) are sufficiently smooth, and their all derivatives are bounded as x → ±∞, u0 (x) → c1 , v0 (x) → c(x → ±∞), where c1 , c are some real numbers:  ∞  ∞ 2 (u0 – c1 ) dx < ∞, (u′0 )2 dx < ∞, –∞ –∞  ∞  ∞ (v0′ )2 dx < ∞, 8(v0 , c)dx < ∞. (3.4.62) –∞

–∞

Moreover, v0 (x) ≥ h = const > 0.

(3.4.63)

In order to prove the asymptotic behavior of the solution as x → ∞, we assume  ∞  ∞ 2 (u′′′ ) dx < ∞, (v0′′ )2 dx < ∞. (3.4.64) 0 –∞

–∞

Obviously, when k(v) ≡ 0, systems (3.4.54) and (3.4.55) are one-dimensional isentropic const gas hydrodynamic equations, when k(v) = const v , 8(v) = v# , # = const ≥ 1, k(v), 8(v) satisfy the above condition (1). Assume RT = {–∞ < x < ∞, 0 < t < T}. We set |f |T0 = sup |f (x, t)|, (x,t)∈RT

|f |T! = |f |T0 +

sup (x,t),(x′ ,t′ )∈RT

|f (x, t) – f (x′ , t′ )| , |x′ – x|! + |t′ – t|!/2

290

3 The Vanishing Viscosity Method of Quasilinear Hyperbolic System

 T  ∂f  |f = |f +   , ∂x !  2 T  T ∂ f   ∂f  |f |T2+! = |f |T0 +  2  +   ∂x ! ∂t ! |T1+!

|T0

For fixed T > 0, by Nash estimate we can prove that there exists a local classical solution to problem (3.4.54), (3.4.55) and (3.4.56) on R4 (0 < 4 < T) such that v ≥ v1 = const > 0, |u|41+! , |u|42+! , |v|41+! (0 < ! < 1) are bounded; thus, in order to deduce the global classical solution to problem (3.4.54)–(3.4.56), it’s sufficient to estimate |u|41+! , |u|42+! , |v|41+! < c0 (T), 0 < ! < !0 (T),

(3.4.65)

v > h > 0,

(3.4.66)

where c0 (T), !0 (T) denote constants that depend only on T and the initial data. In fact, we give new initial condition at t = 4, then the solution can be extended to R4+$ , since in eqs (3.4.65) and (3.4.66), $ is independent of 4. Thus the solution can be extended to RT with arbitrary T, that is, we deduce the global smooth solution on t ≥ 0. Therefore, we will construct estimates (3.4.65) and (3.4.66), for simplification, we set c1 = 0, otherwise, which can be deduced by transformation u′ = u – c1 , v′ = v. Multiplying eq. (3.4.54) by u, eq. (3.4.55) by 8(c) – >(v) and adding the resultant equations together, by virtue of the following identities: 6 5 ∂u ∂v ∂ 1 2 + [>(c) – >(v)] = u + 8(v, c) , ∂t ∂t ∂t 2 ∂> ∂u ∂ u – [>(c) – >(v)] = [u(>(v) – >(c))], ∂x ∂x ∂x      2 ∂ ∂u ∂ ∂u ∂u . u k = ku –k ∂x ∂x ∂x ∂x ∂x

u

Integrating over the rectangle Rt1 = {x1 ≤ x ≤ x2 , 0 ≤ t < t1 } t1 ≤ 4 with respect to x, t, we have   2  t1  x2 ∂u 1 2 k(v) dxdt u + 8(v, c) dx|t=t1 + 2 ∂x x1 0 x1   t1  x2  1 2 x=x u[kux + >(c) – >(v)]|x=x21 dt. u0 + 8(v0 , c) dx + = 2 x1 0 

x2



(3.4.67)

Since >′ (v) < 0, we have 

v

8(v, c) ≡

[>(c) – >(s)]ds ≥ 0.

c

Since the right-hand side of eq. (3.4.67) is uniformly bounded with respect to x1 , x2 , and the left-hand side in nonnegative, so we have

3.4 On Gradient Quasilinear Parabolic Equations and Viscous Isentropic Gas

 1 2 u + 8(v, c) dx < ∞, 2



∞



–∞ t1  ∞

k(v)u2x dxdt < ∞.

–∞

0

291

∞ 2 By virtue of the boundedness of –∞ u2 (x, t)dx < ∞ and ∂u ∂x , we deduce that, for any t, 0 ≤ t ≤ 4, u(x, t) → 0 as x → ±∞. In fact, if there exist t and : > 0, u2 (xi , t) ≥ 2: > 0, here xi → ∞, then for each i, there exists a domain Ii which contains xi such that u2 (Ii , t) ≥ : > 0. Since u2 ∈ L1 (R), the diameter Ii → 0, i → ∞, which destroys the boundedness of (u2 )x . Since ut is bounded on R4 , thus u(x, t) → 0 uniformly for t as |x| → ∞. In fact, we choose countable set [tn ] which is dense on [0, 4]. If |ut | < M, we choose tn such that M|t – tn | < :2 , then let k be sufficiently large such that |u(x, tn )| < 42 , |x| > k, then when 0 ≤ t ≤ 4, we have |u(x, t)| ≤ |u(x, t) – u(x, tn )| + |u(x, tn )| < :, |x| > k. From eq. (3.4.67), as x1 → –∞, x2 → +∞, we have 

  t1  ∞ 1 2 k(v)(ux )2 dxdt u + 8(v, c) dx|t=t1 + 2 ∞ –∞ 0   ∞ 1 2 u + 8(v0 , c) dx ≡ C. 2 0 ∞ ∞

(3.4.68)

Now we construct the energy inequality for vx , and set 

v

k(s)ds,

k(v) = 0

then eq. (3.4.54) becomes ut – kxt + >′ (v)vx = 0.

(3.4.69)

Multiplying eq. (3.4.69) by kx , integrating over the rectangle R = {x1 ≤ x ≤ x2 , 0 ≤ t ≤ t1 } and integrating by parts with respect to x, t, by eq. (3.4.55), we have 1 2

 



x2

k2 vx2 dx|t=t1 –

x1 x2

x1

+ 0

0

k(v)vx udx|t=t1 +

= 

t1

t1

x (kut )|x21 dt



x2

+ x1



x2

>′ (v)k(v)vx2 dxdt

x1 t1  x2

 0

x1

x

ku2x dxdt – (uk)|x21 |t=t1

u′0 k(v0 )dx

1 + 2



x2

x1

k2 (v0 )(v0′ )2 dx.

By virtue of eq. (3.4.68) and the following inequality 

x2

x1

ukvx dx ≤

1 4



x2 x1

 k2 vx2 dx +

x2

x1

u2 dx.

(3.4.70)

292

3 The Vanishing Viscosity Method of Quasilinear Hyperbolic System

By virtue of eq. (3.4.70), we have 1 4



x2

x1

 k2 vx2 dx|t=t1 –

t1



0

x2 x1

>′ (v)k(v)vx2 dxdt

 t1 x x (kut )|x21 dt ≤ –[u(x, t1 )k(v(x, t1 ))]|x21 + 0   x2 1 x2 2 ′ u0 (x)k(v0 (x))dx + k (v0 )(v0′ (x))2 dx + 2 x1 x1  x2 [u20 (x) + 28(v0 (x), c)]dx + 

x1 t1

+ 0

x

[kuux + u[>(c) – >(v)]]|x21 dt.

(3.4.71)

Let x1 → –∞, x2 → +∞ in eq. (3.4.65), note that u(x, t) → 0 as |x| → ∞, v > const > 0 and 

t1

kut dt = 0

t uk|01



t1

ukvt dt → 0, |x| → ∞,

– 0

then when 0 ≤ t1 ≤ 4 < T, we have 1 4



∞

–∞

 h2 vx2 dx|t=t1

t1



∞

– –∞

0

>′ (v)k(v)vx2 dxdt ≤ c0 .

(3.4.72)

Now we define auxiliary function 

v

f (v) =

 k(s) 8(s, c)ds.

c

By eq. (3.4.68), we have

∞

–∞ 8(v, c)dx

< ∞, and

∂8 = (>(c) – >(v))vx ∂x is bounded; thus 8(v, c) → 0 as |x| → ∞, that is, v → c, and thus f (v) → 0 as |x| → ∞. Since 

x

f (v) = –∞

∂f dx = ∂x



x

 vx k(v) 8(v, c)dx.

–∞

In addition, by eqs (3.4.68) and (3.4.72), we have  |f (v)| ≤

∞ ∞

1  k2 vx2 dx

2

∞

∞

1 8(v, c)dx

2

≤ c0 .

(3.4.73)

3.4 On Gradient Quasilinear Parabolic Equations and Viscous Isentropic Gas

293

By virtue of eqs (3.4.58) and (3.4.59) or (3.4.60) and (3.4.61), from eq. (3.4.73), there exist constants h and H such that 0 < h < v < H < ∞,

(3.4.74)

where h, H depend only on the initial data, but not on T. From eq. (3.4.74) we deduce that 0 < k1 < k(v) < k2 , 0 < k3 < –>′ (v)k,

(3.4.75)

where k1 , k2 , k3 depend only on the initial data, but not on T. We continue to estimate ux . Therefore, we will deduce energy inequality of difference quotient △u △x , where △u = u(x + △x, t) – u(x, t). We deduce from eq. (3.4.54) that     ∂ △u ∂ △(kux ) △>x – =– . ∂t △x ∂x △x △x Multiplying this inequality by △u △x , and integrating over the rectangle R = {x1 ≤ x ≤ x2 , 0 ≤ t ≤ t1 ≤ 4}, using the identity 6 5 △ux △k △(kux ) △ux △ux ⋅ = k((vx + △x, t)) + ux (x, t) △x △x △x △x △x and the inequality : 1 ab ≤ a2 + b2 , 2 2: then we have 1 2

 △ux 2 dxdt △x x1 0 x1    t1  x2  1 1 △u0 2 △v 2 dx + dxdt ≤ max |>′ (v)|2 2 x1 △x 2: h≤v≤H △x 0 x1   t1  x2  △v 2 2 ′ 2 1 ux dxdt + max |k (v)| 2: 0 x1 △x h≤v≤H  6  t1 5 △u △kux △> x2 + dt. – △x △x △x x1 0 



△u △x  x2 

x2

2

If we set △x → 0, we have



dx + (k1 – :)

t1



x2



(3.4.76)

294

3 The Vanishing Viscosity Method of Quasilinear Hyperbolic System



1 2



x2

x1

u2x dx + (k1 – :)

t1



0

x2

(uxx )2 dxdt

x1

 t1  x2 1 (vx )2 dxdt max |>′ (v)|2 2: h≤v≤H x1 0 x1  t1  x2 ′ 2 1 v2 u2 dxdt + max |k (v)| 2: 0 x1 x x h≤v≤H  t1 x (ux (kux )x – >x )|x21 dt. + 1 2

≤



x2

(u′0 )2 dx +

(3.4.77)

0

Since ux , uxx and vx are bounded on R and eqs (3.4.68), (3.4.72), (3.4.74) and (3.4.75), we see that the right-hand side of eq. (3.4.77) are bounded uniformly with respect to x1 , x2 . Therefore, 

∞

–∞

 u2x (x, t1 )dx < ∞,



t1

∞

–∞

0

u2xx dxdt < ∞.

Because of the convergence of the first above, and the boundedness of (u2x )x , we deduce that ux → 0 as |x| → ∞. Therefore, in eq. (3.4.77), we set x1 → –∞, x2 → +∞, for 0 < : < k1 , then 1 2



∞

–∞

1 2

 u2x (x, t1 )dx + (k1 – :)



∞

t1



∞

u2xx dxdt

–∞

0

1 max |>′ (v)|2 2: h≤v≤H –∞  t1  ∞ 1 + max |k′ (v)|2 v2 u2 dxdt. 2: 0 –∞ x x h≤v≤H

≤

(u′0 (x))2 dx +



t1



∞

–∞

0

vx2 dxdt (3.4.78)

Since eqs (3.4.72) and (3.4.75) hold and the first two terms of the right-hand side of eq. (3.4.78) are bounded by a constant, which depends only on the initial data, when x → –∞, ux → 0. Then  u2x = 2

x

 ux uxx dx ≤ 2

–∞

1 ≤ 2 :



∞

–∞ ∞

–∞

1  2

u2x dx

 u2x dx + :2

∞

–∞ ∞

–∞

1 u2xx dx

2

u2xx dx.

Therefore, by eqs (3.4.72), (3.4.68) and (3.4.75), we have 1 :

 0

t1



∞

–∞

u2x vx2 dxdt ≤

  1 ∞ 2 v dx dt :2 –∞ x –∞ 0   ∞   t1   ∞ u2xx dx vx2 dx dt +: 1 :



t1

0



∞



u2x dx

–∞

1 ≤ 2 c0 + :c0 :



t1 0



–∞

∞

–∞

u2xx dxdt.

(3.4.79)

295

3.4 On Gradient Quasilinear Parabolic Equations and Viscous Isentropic Gas

From eqs (3.4.78) and (3.4.79), when 0 ≤ t1 ≤ 4, and : is sufficiently small, we have 

∞

–∞

 u2x (x, t1 )dx +

t1



∞

u2xx dxdt < c0 .

–∞

0

(3.4.80)

Since  2

x

u (x, t) = 2



∞

uux dx ≤ 2

1 

∞

2

2

u dx

–∞

–∞

–∞

1 u2x dx

2

,

then from eqs (3.4.68) and (3.4.80), we have |u(x, t)| < c0 .

(3.4.81)

Now we estimate |vx |, multiplying eq. (3.4.55) by k(v), then Kt = kux , where K =

v c

(3.4.82)

k(s)ds, by eq. (3.4.82), we have 

t

K(v) = K(v0 ) +

kux dt1 , 0

Kx = kvx = k(v0 )v0′ +



t

0

(k(ux ))x dt1 .

By virtue of eq. (3.4.54), we have kvx = k(v0 )v0′ + u – u0 +



t

>′ (v)vx dt1 .

(3.4.83)

0

By eqs (3.4.83), (3.4.75), (3.4.81) and (3.4.74), we have   t  |vx |dt1 . |vx | ≤ 1 +

(3.4.84)

0

Then |vx | ≤ C0 (T).

(3.4.85)

We regard eq. (3.4.54) as linear parabolic equation of u(x, t), by eqs (3.4.74), (3.4.75), (3.4.85), and coefficients are bounded; therefore, by Schauder-type estimate, we have |u|41+! ≤ C0 (T), 0 < ! < !0 (T).

(3.4.86)

Furthermore, we estimate |vxx |, through the difference quotient transformation with respect to x, and eq. (3.4.83) becomes

296

3 The Vanishing Viscosity Method of Quasilinear Hyperbolic System

△vx △k △(v0′ k(v0 )) △u0 = –vx (x1 , t) + – △x △x △x △x  t  t ′ △v △> △u x >′ (v(x2 , t)) vx (x1 , t) + + dt + dt. △x △x △x 0 0

k(v(x2 , t))

(3.4.87)

Then by eqs (3.4.74), (3.4.75), (3.4.85) and (3.4.86), we have       t  △vx   △vx   < C0 (T) 1 +     △x dt .  △x  0 Then    △vx     △x  < C0 (T), 0 ≤ t < 4 ≤ T.

(3.4.88)

Then we deduce that vxx exists a.e. on R4 , thus |vxx | < c0 (T).

(3.4.89)

Next we will estimate |u|42+! . Then by employing the average function v# that denotes the average of v on the nuclear radius #, when t ≤ 0, we define v(x, t) = v0 (x) + u′0 (x)t. Assume that u# (x, t) satisfies ⎧ #  # ⎪ ⎨ ∂u – ∂ k(v# ) ∂u = – ∂ >(v# ), ∂t ∂x ∂x ∂x (3.4.90) ⎪ ⎩ u# | = u (x). t=0 0 From eqs (3.4.74), (3.4.75), (3.4.85) and (3.4.89), we obtain that when # is sufficiently small 1 1 h < v# < 2H, k1 < k(v# ) < 2k2 , 2 2 # |v# | < c0 (T), |vxx | < c0 (T).

(3.4.91) (3.4.92)

Now we consider eq. (3.4.90) as the linear equation of u# , by maximum principle and estimates (3.4.91) and (3.4.92), we have |u# | ≤ C0 (T).

(3.4.93)

From eqs (3.4.91), (3.4.92), (3.4.93) and Schauder-type estimates, we have |u# |T1+! ≤ C0 (T), 0 < ! < !0 (T).

(3.4.94)

Now we differentiate eq. (3.4.90) with respect to x, then #

#

#

∂ 2 ux ∂ 2 >(v# ) ∂ux # ∂ux # # # . – k(v# ) 2 – 2k′ (v# )vx = [k′′ (v# )(vx )2 + k′ (v# )vxx ]ux – ∂t ∂x ∂x ∂x2

(3.4.95)

3.4 On Gradient Quasilinear Parabolic Equations and Viscous Isentropic Gas

297

#

We consider eq. (3.4.95) as the linear equation of ux , using Schauder-type estimates, eqs (3.4.91)–(3.4.93) imply |u# |42+! ≤ c0 (T), 0 < ! < !0 (T).

(3.4.96)

We set △u = u# – u. If eq. (3.4.90) subtracts eq. (3.4.54), we have ∂ △u ∂ △u #2 △ u – k′ (v)vx – k(v) = f# (x, t), 2 ∂t ∂x ∂x

(3.4.97)

where ∂ 2 u# # # + (k′ (v# )vx – k′ (v)vx )ux ∂x2 # +>′ (v)vx – >′ (v# )vx .

f# (x, t) = (k(v# ) – k(v))

We will prove f# (x, t) → 0 as # → 0, which holds for x, t ∈ R4 uniformly. In fact, we see that v → c as |x| → ∞, since vt = vx is bounded. Therefore, this convergence holds uniformly with respect to t, 0 ≤ t ≤ 4. Hence, v# → v(# → 0) uniformly on (x, t) ∈ R4 . Equations (3.4.72) and (3.4.82) imply 

∞

–∞

vx2 dx < +∞, 0 ≤ t ≤ 4,

which combined with the boundedness of (vx2 )x , we obtain vx → 0, (|x| → ∞) (0 ≤ t ≤ 4). With the boundedness of vxt = uxx , we deduce that this convergence holds uniformly for t, 0 ≤ t ≤ 4. Hence, vxr → vx (r → 0) holds uniformly for (x, t) ∈ R4 . # # From v# → v uniformly, and the boundedness of ux , uxx uniformly with respect to #, we deduce that f# (x, t) → 0(# → 0) uniformly for (x, t) ∈ R4 . Now we consider eq. (3.4.97) as the linear equation of △u. Since ||f r ||L∞ (R4 ) → 0(r → 0), △u(x, 0) = 0, by maximum principle, we obtain that △u → 0(ur → u) uniformly on R4 , which combined with eq. (3.4.96), we have |u|42+! < C0 (T), 0 < ! < !0 (T). Since uxx is bounded, by vxt = uxx , from eqs (3.4.74), (3.4.85) and (3.4.89), we obtain |v|41+! ≤ c0 (T), we therefore have Theorem 3.4.9. If the following conditions are satisfied:

298

3 The Vanishing Viscosity Method of Quasilinear Hyperbolic System

(1) Functions k(v), >(v) are sufficiently smooth, and k(v) > 0, v > 0, >′ (v) < 0, v > 0,  ∞  c   k(v) 8(v, c)dv = ∞, k(v) 8(v, c)dv = ∞, 0

c

where 

v

8(v, c) =

(>(c) – >(s))ds ≥ 0.

c

(2) The initial functions u0 (x), v0 (x) are sufficiently smooth, their all derivatives pass to zero as |x| → ∞. v0 (x) ≥ h = const > 0, v (x) → c, u0 (x) → 0, |x| → ∞, 0 ∞ (u20 (x) + u′0 (x))2 + (v0′ (x))2 + 8(v0 , c)]dx < ∞. –∞

Then there exists a global classical solution u(x, t), v(x, t)(t ≥ 0) to problem (3.4.54), (3.4.55) and (3.4.56). Now we consider the behavior of the solution to problem (3.4.54)–(3.4.56) as t → ∞. We first deduce energy inequality of the difference quotient △u u(x, t) – u(x, t – △t) = , △t ≡ const > 0, △t △t assume t ≥ △t, eq. (3.4.54) implies 6 5 ∂ △u ∂ △> ∂ ∂ △u △k – k(v(x, t)) + ux (x, t – △t) + = 0. ∂t △t ∂x ∂x △t △t ∂x △t

(3.4.98)

Multiplying eq. (3.4.98) by △u △t and integrating over the domain {–∞ < x < ∞, △t ≤ t ≤ t1 }, where t1 is an arbitrary constant, t1 > △t; hence 1 2

   t1  ∞  ∂ △u 2 △u 2 dx|t=t1 + k dxdt △t ∂x △t –∞ △t –∞   t1  ∞   ∂ △u △k 1 ∞ △u 2 dx|t=△t – = ux (x, t – △t) dxdt 2 –∞ △t ∂x △t △t –∞ △t    t1  ∞ △> △u dxdt. + △t x △t –∞ △t



∞

By eqs (3.4.99), (3.4.74) and (3.4.75), we have

(3.4.99)

3.4 On Gradient Quasilinear Parabolic Equations and Viscous Isentropic Gas

299

   t1  ∞  ∂ △u 2 △u 2 dx|t=t1 + 2(k1 – :) dxdt △t △t –∞ ∂x △t    ∞  t1  ∞  1 △u 2 △v 2 2 dx|t=△t + max [k′ (v)]2 ux (x, t – △t)dxdt ≤ △t : h≤v≤H –∞ △t –∞ △t   t1  ∞  1 △v 2 dxdt. (3.4.100) + max [>′ (v)]2 : h≤v≤H △t –∞ △t

 

Now we estimate the first term of the right-hand side of eq. (3.4.100), we set u(x, t) = u0 (x) + t[k(v0 )u′0 – >(v0 )]′ , t ≤ 0, then when 0 ≤ t ≤ △t, it holds that ut (x, t – △t) – (k(v0 )u′0 ) + (>(v0 ))′ = 0. Multiplying by eq. (3.4.54) by △u △t , integrating over R△t {–∞ < x < ∞, 0 ≤ t ≤ △t}, after integrating by parts, one obtains that     △t  ∞ u – u0 2 △u 2 dx|t=△t + k(v) dxdt △t △t –∞ 0 –∞ x   △t  ∞   △t – t 1 ∞ △u 2 dx|t=0 – 2 k(v) = [k(v0 )u′0 – >(v0 )]′′ 2 –∞ △t △t 0 –∞      △t  ∞ u – u0 △t – t 2 ⋅ dxdt – k(v) [(k(v0 )u′0 )′ – >′ (v0 )]2 dxdt △t △t 0 –∞ x  5 6  △t  ∞  u – u0 >(v) – >(v0 ) ′ k(v) – k(v0 ) + – u0 dxdt △t △t △t 0 –∞ x  △t  ∞ △t – t [k(v0 )u′0 – >(v0 )]′′ + △t 0 –∞ 5 6 >(v) – >(v0 ) ′ k(v) – k(v0 ) ⋅ – u0 dxdt. (3.4.101) △t △t 1 2



∞

It follows from eqs (3.4.101), (3.4.74), (3.4.75) and relevant assumptions about u0 , v0 , >(v), k(v) that 

   △t  ∞  △u 2 u – u0 2 dx|t=△t + (k1 – :) dxdt △t △t –∞ 0 –∞ x 6  5  △t  ∞  v – v0 2 ≤ c0 1 + dxdt , 0 < : < k1 . △t 0 –∞ ∞

(3.4.102)

300

3 The Vanishing Viscosity Method of Quasilinear Hyperbolic System

Noting eq. (3.4.55), when t ≤ △t, we have    2  1 v – v0 2 1 ∂ = v(x, +t)dx △t △t 0 ∂x  1 ≤ u2x (x, +t)d+. 0

So eqs (3.4.102) and (3.4.80) imply that   ∞ △u 2 dx|t=△t ≤ C0 . △t –∞

(3.4.103)

Now we estimate the second and the third terms of the right-hand side of eq. (3.4.100), then   t1  ∞  △v 2 2 ux (x, t – △t)dxdt △t –∞ △t  1   t1  ∞ 1   t1  ∞   2 2 △v 4 4 k dxdt vx (x, t – △t)dxdt , ≤ △t △t –∞ △t –∞ 3

1

since ||vx ||L4 ≤ c||vx ||L42 ||vxx ||L42 , by eq. (3.4.80), we have 

t1



△t

≤ c0

∞

u4x (x, t – △t)dxdt

–∞  t1



△t

∞

–∞

u2xx (x, t – △t)dxdt < c′0 .

Furthermore, eq. (3.4.55) yields △v = △t



1

ux (x, t + (+ – 1) △ t)d+.

0

Therefore,  △v 4 dxdt △t –∞ △t  1  t1  ∞ ≤ u4x (x, t + (+ – 1) △ t)dxdtd+ < c0 .



t1



∞

0

△t

–∞

It follows from eq. (3.4.80) that  △v 2 dxdt △t –∞ △t  1  t1  ∞ ≤ u2x (x, t + (+ – 1) △ t)dxdtd+ < c0 .



t1



∞

0

△t

–∞

(3.4.104)

3.4 On Gradient Quasilinear Parabolic Equations and Viscous Isentropic Gas

301

By eq. (3.4.100), we get ∞



–∞

△u △t

2 dx < c0 .

Letting △t → 0, one obtains that 

∞

–∞

2

∂u ∂t

dx < c0 .

(3.4.105)

Assuming T as an arbitrary positive constant, we deduce from eq. (3.4.54) and estimates (3.4.105), (3.4.74), (3.4.75), (3.4.85) and (3.4.80) that 

∞ –∞

u2xx (x, t)dx < c0 (T), 0 ≤ t ≤ T.

(3.4.106)

∞ We set z(t) = –∞ u2x (x, t)dx. We can prove z(t) → 0 as t → ∞. In fact, it follows from eqs (3.4.105) and (3.4.106) that when 0 ≤ t ≤ △t,      △z   ∞ △u   = (u (x, t – △t) + u (x, t)) dx xx xx  △t   △t  –∞ 1   ∞  ∞ 2 2 2 uxx (x, t – △t)dx + 2 uxx (x, t)dx ≤ 2 –∞

  × 2

∞

–∞

which yields

dz dt

△u △t

–∞

1

2

2

dx

< c0 (T),

(3.4.107)

exists almost everywhere. For any t1 ≥ 0, t2 ≥ 0, we have 

t2

z(t2 ) = z(t1 ) + t1

dz dt. dt

(3.4.108)

Obviously, it’s easily seen that 

△u △t

2



1

≤ 0

u2t (x, t + (+ – 1) △ t)d+.

(3.4.109)

So when 0 < △t < $, we have 

T



0

∞

–∞

△u △t

2



T+△t

dxdt ≤ 0



∞ –∞

u2t (x, t)dxdt.

(3.4.110)

It follows from eqs (3.4.109), (3.4.107) and (3.4.80) that 

T 0

  1   T+△t  ∞ 2  △z  2 dt ≤ c1  u (x, t)dxdt . t  △t  0 –∞

(3.4.111)

302

3 The Vanishing Viscosity Method of Quasilinear Hyperbolic System

We deduce from eqs (3.4.54), (3.4.74), (3.4.75), (3.4.80) and (3.4.72) that 

T+△t 0





∞

u2t (x, t)dxdt

–∞



T+△t

< C1 1 + 0



∞ –∞

u2x vx2 dxdt

 .

(3.4.112)

Since  u2x ≤ 2

1 

∞

u2x dx

–∞

2

∞

–∞

1 u2xx dx

2

,

then eqs (3.4.72), (3.4.75), (3.4.68) and (3.4.80) yield that 

T+△t



0

∞

–∞

u2x vx2 dxdt 

∞

≤ 2 max

0≤t≤T+△t –∞

 ×

T+△t



0

∞

–∞

 vx2 dx

t+△t 0

u2xX dxdt

< c1 .

We deduce from eqs (3.4.110) to (3.4.113) that 

T 0

   △z     △t dt < c1 .

Letting △t → 0, we have 

T

0

   dz   dt < c1 .  dt 

Since T is arbitrary, we have 

    dz dt < c1 .  dt 

∞ 0

Equations (3.4.68) and (3.4.75) yield that 

∞

∞

–∞

1

2



z(t)dt < ∞,

0

which gives z(t) → 0 (t → ∞).

1 u2x dxdt

2

(3.4.113)

3.4 On Gradient Quasilinear Parabolic Equations and Viscous Isentropic Gas

303

By the inequality 

1

∞

u2 ≤ 2z 2 (t)

1 u2 dx

2

–∞

and eq. (3.4.68), we deduce that u(x, t) → 0 (t → ∞) ∀x ∈ (–∞, ∞). We set 

∞

y(t) = –∞

vx2 (x, t)dx.

It follows from eqs (3.4.73), (3.4.68) and (3.4.75) that  |f (v)| < c0 y(t).

(3.4.114)

Noting eqs (3.4.55), (3.4.72), (3.4.75) and (3.4.106), we have    ∞  1  ∞ 2  △y  2 2 < 2  v (x, t – △t)dx + 2 v (x, t)dx x x  △t  –∞ –∞  1  ∞ 1 2 × d+ u2xx (x, t + (+ – 1) △ t)dx < c0 (T). 0

(3.4.115)

–∞

we therefore deduce for each derivative y′ (t)(t ≥ 0), and t1 ≥ 0, t2 ≥ 0 that 

t2

y(t2 ) = y(t1 ) +

y′ (t)dt.

t1

By eqs (3.4.114), (3.4.72), (3.4.80) and (3.4.75), we have 

+∞

|y′ (t)|dt < ∞,

0

which together with eq. (3.4.116) yields y(t) → 0, t → ∞. We deduce from the formulation of f (u) that v(x, t) → c, t → ∞ ∀x ∈ R1 exists uniformly. Then we deduce the following theorem:

(3.4.116)

304

3 The Vanishing Viscosity Method of Quasilinear Hyperbolic System

Theorem 3.4.10. Under the assumptions of Theorem 3.4.9, and assume 

∞ –∞

2 (u′′′ 0 (x)) dx +



∞

–∞

(u′′0 (x))2 dx < ∞,

then there exists a solution u(x, t), v(x, t) to problem (3.4.54), (3.4.55) and (3.4.56) satisfying u(x, t) → 0, v(x, t) → c, t → ∞ ∀x ∈ R1 .  c –! –$ Remark: If >(v) = c 1 v , k(v) = c2 v , then when 2$ + ! ≥ 3, 0 k(v) 8(v, c)dv = +∞. If ∞ $ < 32 , then c k(v) 8(v, c)dv = ∞. If $ = 1, and ! ≥ 1, then this condition is satisfied for isentropic gas hydrodynamic equations. We consider the following artificial viscosity p equations: vt – ux = :vxx , ut + p(v)x = :uxx , : > 0,

(3.4.117)

with initial conditions (v(x, 0), u(x, 0)) = (v0 (x), u0 (x)),

(3.4.118)

lim (v0 (x), u0 (x)) = (h, 0),

|x|→∞

where p′ (v) < 0, p′′ (v) > 0, v > 0, v0 (x) ≥ $ > 0, h is a constant, assume 

∞

–∞

where p(v, $) =

v

$ (p($)

[u20 + p(v0 , $) + (u′0 )2 + (v0′ )2 ]dx < ∞,

(3.4.119)

– p(s))ds ≥ 0.

Theorem 3.4.11. Under the above conditions, there exists a global solution u(x, t), v(x, t) for t ≥ 0 to the initial value problem (3.4.117) and (3.4.118). Proof. It’s sufficient to prove that the solutions u, v of problem (3.4.13) and (3.4.14) are bounded. In fact, if we introduce the fundamental solution 1 G(x, t; . , 4) = √ 1 exp 1 2 0: 2 (t – 4) 2

 –

 (x – . )2 , t ≥ 4, 4:(t – 4)

then x–. ∂ exp G(x, t; . , 4) = √ 3 3 ∂. 4 0: 2 (t – 4) 2 A direct computation yields

 –

 (x – . )2 . 4:(t – 4)

3.4 On Gradient Quasilinear Parabolic Equations and Viscous Isentropic Gas



∞

  ∂ G(x, t; . , 4)d. = √ 1 . 0:(t – 4) –∞ ∂.

305

(3.4.120)

The solution of the problem (3.4.13) and (3.4.14) can be written as the equivalent integral equation  ∞  t ∞ ∂ u(x, t) = G(x, t; . , 4)u0 (. )d. + G(x, t; . , 4)f(u(. , 4))d. d4, (3.4.121) –∞ 0 –∞ ∂.

u p(v) . From the integral equations (3.4.121) and (3.4.120), ,u = where f(u) = v –u by the simple iteration method, we deduce that there exists a positive constant t0 > 0 such that for 0 ≤ t ≤ t0 , –∞ < x < ∞, there exists a unique local classical solution to problems (3.4.13) and (3.4.14). In order to obtain the solution in large scale to initial value problem (3.4.13) and (3.4.14), differentiating eq. (3.4.121) with respect to x, and integrating by parts, we get  t ∞  ∞ ∂u ∂u ∂ G(x, t; . , 0)u′0 (. )d. + = G(x, t; . , 4)f ′ (u(. , 4)) d. d4, ∂x ∂. –∞ 0 –∞ ∂.

If |u(. , c)| ≤ M0 , by eq. (3.4.120), set

   ∂u  U(t) = sup  , ∂x x

then we have 

1

U(t) ≤ c0 + c1 (:) 0

1 1

(t – 4) 2

U(4)d4,

where c0 is a constant independent of : and c1 (:) is a constant dependent on :, by Gronwall’s inequality, we deduce the estimate about U(t). Therefore, the local solution can be extended on the upper half plane 0 < t < ∞, –∞ < x < ∞. Therefore, in order to deduce the global classical solution to problem (3.4.117) and (3.4.118), the key is to deduce |u(x, t)| ≤ M, |v(x, t)| ≤ M, where the constant M depends only on the initial function, but not on t ∈ [0, T]. By invariant set theory or turning eq. (3.4.117) into diagonal form and then using maximum principle, we obtain the lower bound of v(x, t), that is v(x, t) ≥ $′ > 0, (x, t) ∈ R × R+ , where $′ ≤ $. Therefore, it’s sufficient to get estimates for the upper bound of v(x, t) and the upper and lower bounds of u(x, t). Assume that T > 0 is an arbitrary constant, 0 ≤ t ≤ T, multiplying the second equation of eq. (3.4.117) by u, and the first equation by [p(h) – p(v)], then summing the resultant equations, we have

306

3 The Vanishing Viscosity Method of Quasilinear Hyperbolic System

6 5  ∂ ∂ u2 v(x,t) (p(h) – p(s))ds + [u(p(v) – p(h))] ∂t 2 h ∂x = :[uuxx + (p(h) – p(v))vxx ], integrating over RT = {0 ≤ t ≤ T, –∞ < x < ∞}, we have 

∞5 2 u



6

v



(p(h) – p(s))ds |t=T dx + 2 –∞ h 6  v0  ∞5 u0 (p(h) – p(s))ds dx + 2 –∞ h  [uuxx + (p(h) – p(v))vxx ]dxdt. +:



T

∞

u(p(v) – p(h))

+

0

dt –∞

RT

When |x| → ∞, u → 0, the second term of the left-hand side vanishes, and integrating the last term of the right-hand side by parts, it follows from the above equation that 

∞5 2 u

2 

–∞

 +

v(x,t)

6  (p(h) – p(s))ds |t=T dx – :

h

+: RT

u2x dxdt

RT

p′ (v)vx2 dxdt

˜ = k,

(3.4.122)

where k˜ depends only on the initial function, then differentiating each equation of system (3.4.117) with respect to x, multiplying the second equation by ux , multiplying the first equation by vx and integrating over RT with respect to x, t, we have  (u2x + vx2 )|t=T dx + (ux pxx – vx uxx )dxdt –∞ RT  = k1 + : (ux uxxx + vx vxxx )dxdt, 

1 2

∞

(3.4.123)

RT

where k1 depends only on the initial data. Integrating the following equation by parts: 



T

(ux pxx – :ux uxxx )dxdt = RT

 –

 (uxx px –

RT

:u2xx )dxdt

RT

T

=

 –

0

0

(ux px – :ux uxx )|∞ –∞ dt

ux (–ut )|∞ –∞ dt

 (uxx px – :u2xx )dxdt =

RT

(uxx px – :u2xx )dxdt.

Similarly, we have 

 (–:vx vxxx – vx uxx )dxdt = RT

RT

2 (:vxx + :vxx ux )dxdt.

307

3.4 On Gradient Quasilinear Parabolic Equations and Viscous Isentropic Gas

So by eq. (3.4.123), we have 1 2

 2 (u2x + vx2 )|t=T dx + : (u2xx + vxx )dxdt –∞ RT  = k1 + (uxx px – vxx ux )dxdt



∞



RT



≤ k1 + RT

 p2 : 2 1 : 2 + u2x dxdt. uxx + x + vxx 2 2: 2 2:

Then  2 (u2x + vx2 )2 |t=T dx + : (u2xx + vxx )dxdt –∞ RT  1 ≤ 2k1 + (p2 + u2x )dxdt. : RT x



∞

(3.4.124)

Since p′ < 0, p′′ > 0, p2x = p′ (v)2 vx2 ≤ (p′ ($))2 vx2 , it follows from eq. (3.4.121) that 1 :



+

RT

1 :

(p2x

+

u2x )dxdt

 RT

u2x dxdt ≤

p′ ($) ≤– :



(–p′ (v)vx2 )dxdt

RT

–p′ ($) k˜ k˜ ⋅ + 2. : : :

(3.4.125)

Therefore, by eq. (3.4.124), we have 

∞

–∞

 (u2x + vx2 )|t=T dx + :

≤ 2k1 +

RT

2 (u2xx + vxx )dxdt

k˜ (1 – p′ ($)), :2

(3.4.126)

Now we define auxiliary function f 

v



1

y

2

(p(h) – p(s))ds

f (v) = h

dy.

h

Note that f (h) = 0, then     x   ∂f   x ′ f (v)vx dx dx =  |f (v)| =  –∞ ∂x –∞ 1  ∞ 1  ∞ 2 2 ′ 2 2 f (v) dx vx dx ≤ –∞



∞ v

= –∞

h

–∞

 1  2 [p(h) – p(s)]ds dx ⋅

∞

–∞

1 vx2 dx

2

.

308

3 The Vanishing Viscosity Method of Quasilinear Hyperbolic System

By eqs (3.4.121) and (3.4.125), we have  1 1 k˜ 2 ′ ˜ 2 |f (v)| ≤ k 2k1 + (1 – p ($)) 2 , :

(3.4.127)

Set A = {(x, t) ∈ RT , v > 3h}, then on A, we have 

v

|f (v)| =

5

61

y

2

(p(h) – p(s))ds 

>

h v

5

h

61

y

2

(p(h) – p(s))ds 2h

dy dy

2h

61 2 (p(h) – p(2h))ds dy > 2h 2h  v  1 = p(h) – p(2h) (y – 2h) 2 dy 

v

5

y

2h

3 2 p(h) – p(2h)(v – 2h) 2 ≥ c1 v3/2 , = 3

where c1 depends only on k, so by eq. (3.4.127), when v > 3h, we have  1 1 k˜ 2 c1 v3/2 ≤ k˜ 2 2k1 + (1 – p′ ($)) 2 . : Then we derive v ≤ M0 (:), and the constant M0 depends not on T. Since T is arbitrary, we therefore have $ ≤ v(x, t) ≤ M0 , x ∈ R, t > 0.

(3.4.128)

Now we estimate u, by eqs (3.4.121) and (3.4.123), and we have  u2 = 2

x –∞

≤

˜ 21 2(2k)

 uux dx ≤ 2

∞

–∞

1  2 u dx ⋅ 2

1 2 k˜ 2k1 + 2 (1 – p′ ($)) . :



∞

–∞

1 u2x dx

2

Since we give the estimate for u(x, t), v(x, t), we complete the proof of the theorem. ∎

3.5 On Some Results of Some Quasilinear Parabolic Equations In this section, we mainly state some interesting results about some quasilinear parabolic equations.

3.5 On Some Results of Some Quasilinear Parabolic Equations

309

1. For equations ∂uj ∂ui  ∂ 2 ui = bij (x, t, u1 , ⋅ ⋅ ⋅, uN ) + 2 ∂x ∂t ∂x N

j=1

+

N 

cij (x, t, u1 , ⋅ ⋅ ⋅, uN )uj + fi (x, t),

(3.5.1)

j=1

with initial conditions 1 ui (x, 0) = u(0) i (x), i = 1, 2, ⋅ ⋅ ⋅, N, x ∈ R

(3.5.2)

and boundary conditions (j)

ui (xj , t) = ui (t), i = 1, 2, ⋅ ⋅ ⋅, N, j = 1, 2,

(3.5.3)

we have the following theorem. Theorem 3.5.1. Assume that the following conditions are satisfied: (1) (x, t) ∈ ST = {–∞ < x < +∞, 0 ≤ t ≤ T}, |f i (x, t)| < F, (2) (x, t) ∈ ST , for each uk , we have |bij (x, t, u)| < B, N 

cij (x, t, u)+i +j ≥ c

i,j=1

N 

+2i ,

i=1

where B, F, c are constants. (3) ∀(x, t) ∈ ST , we have N  i=1



u2i

 N 2  (0) 2 2cT NF 2cT ≤ M0 = max max (ui ) e , e , C i=1

2 2

where C > N 2B + 2|c| + 1. The coefficients bij , cij , fi and their first and second derivatives with respect to x, uk are continuous and bounded, derivatives ∂ 2 bij ∂ 2 bij ∂ 3 bij ∂ 3 bij , , , , ∂x∂t ∂uk ∂t ∂uk ∂ul ∂t ∂x2 ∂t ∂ 3 bij ∂ 3 bij ∂ 3 bij ∂ 3 bij , , , ∂x∂t∂uk ∂x2 ∂uk ∂uh ∂ul ∂um ∂x∂uk ∂ul are continuous on (x, t) ∈ ST , and the similar derivatives of cij , fi are also bounded in ST . (4) Function u(0) i (x) has the second-order continuous, bounded derivatives.

310

3 The Vanishing Viscosity Method of Quasilinear Hyperbolic System

Then there exists a unique classical global solution to problem (3.5.1) and (3.5.2), which is continuous on ST , and all derivatives that arise in eq. (3.5.1) are continuous in ST . Proof of Theorem 3.5.1. We can regard the solution to the initial value problem (3.5.1) and (3.5.2) as the limit of the sequence of solutions to the solvable boundary value problem (3.5.1) and (3.5.3). In order to prove existence of the limit, we need to use the solution to the boundary value problem (3.5.1), (3.5.3) and uniform estimates for its derivatives.

Theorem 3.5.2. Assume that the following conditions are satisfied: (1) On the whole space of {u1 , ⋅ ⋅ ⋅, uN } except for some closed set s (which may contain one point, its all coordinates are finite), it holds that smooth conditions of (3) in Theorem 3.5.1 and the inequality N 

cij +i +j ≥ c

i,j=1

N 

+2i .

i=1

(2) Except for s, function u(0) i (x) has second-order derivatives and is bounded on x ∈ (2) [x1 , x2 ] and continuous in x ∈ (x1 , x2 ), and functions u(1) i (t), ui (t) have first-order continuous derivatives. (j)

(3) The compatible condition u(0) i (xj ) = ui (0), i = 1, 2, ⋅ ⋅ ⋅, N, j = 1, 2. Moreover, all solutions to problem (3.5.1) and (3.5.3) satisfy 1(u, s) ≥ a > 0,

(3.5.4)

where 1(u, s) denotes the distance between solution sets u of problem (3.5.1) and (3.5.3) and the closed set s. Then there exists a global classical solution to the initial value problem (3.5.1) and (3.5.3) on RT (T is arbitrary). Theorem 3.5.3. For the equations ∂u2 ∂u ∂>(v) = – , ∂x2 ∂t ∂x ∂v2 ∂v ∂u – , : 2 = ∂x ∂t ∂x

:

(3.5.5) (3.5.6)

with initial boundary conditions u|t=0 = u0 (x), v|t=0 = v0 (x), x1 ≤ x ≤ x2 ,

(3.5.7)

3.5 On Some Results of Some Quasilinear Parabolic Equations



u|x=x1 = u1 (t), v|x=x1 = v1 (t), u|x=x2 = u2 (t), v|x=x2 = v2 (t).

311

(3.5.8)

If the following conditions are satisfied: and has fourth-order continuous (1) The function >(v) is defined on v ≥ v0∗ ≥ –∞ v  derivatives, >′ (v) > 0, >′′ (v) ≤ 0, F(v) = –∞ >′ (v)dv → –∞, (v → v2∗ ); (2) v0 (x), v1 (t), v2 (t) > v2∗ ; (3) u0 (x), v0 (x) ∈ C2 [x1 , x2 ], ui (t), vi (t)(i = 1, 2) ∈ C1 , and u0 (xj ) = uj (0), v0 (xj ) = vj (0), j = 1, 2, then there exists a unique global classical solution to problem (3.5.5), (3.5.6), (3.5.7) and (3.5.8). Proof. In order to prove the existence of the global classical solution to problem (3.5.5)–(3.5.8), it’s sufficient to prove that for solutions u, v to problem (3.5.5)–(3.5.8), we have v ≥ v0∗ > v2∗ . Therefore, transforming eqs (3.5.5) and (3.5.6) (: = 0) to the diagonal form ∂f +  ′ ∂f + ∂f –  ′ ∂f – + > (v) = 0, – > (v) = 0, ∂t ∂x ∂t ∂x

(3.5.9)

where f + , f – are Riemann invariants f + = –F(v) + u, f – = –F(v) – u.

(3.5.10)

For eqs (3.5.5) and (3.5.6) (: = 0), by virtue of the transformation (u, v) → ( f + , f – ), we have   ∂ 2 f + ∂f –  ′ ∂f + ∂ 2 f + ∂v 2 : 2 – , = > (v) +: 2 ∂x ∂t ∂u ∂v ∂x    ∂ 2 f – ∂f – ∂f – ∂ 2 f – ∂v 2 : 2 – . = – >′ (v) +: 2 ∂x ∂t ∂u ∂v ∂x Since >′′ (v) ≤ 0, we have ∂ 2f – ∂ 2f + ≥ 0, ≥ 0. ∂v2 ∂v2 Using maximum principle, we deduce from eqs (6.5.17) and (3.5.12) that

(3.5.11) (3.5.12)

312

3 The Vanishing Viscosity Method of Quasilinear Hyperbolic System

f + ≤ M + = max f + , f – ≤ M – = max f – , x = x1 , x2 t=0

F(v) = –

x = x1 , x2 t=0

f+ + f– M+ + M– ≥– . 2 2

(3.5.13) (3.5.14)

 Since F ′ (v) = >′ (v) > 0, we deduce from eq. (3.5.14) that v ≥ v3∗ , and that v3∗ > v2∗ from monotonicity of the function F(v) and F(v2∗ ) = –∞. Therefore, existence of solutions to problem (3.5.5)–(3.5.8) can be derived. ∎ Using the solution of the first boundary value problem, we can deduce the solution of the initial value problem (3.5.5) and (3.5.6). Theorem 3.5.4. Assume that the following conditions are satisfied: (1) The function >(v) is defined on v ≥ v0∗ ≥ –∞. >(v) ∈ C4 , >′ (v) > 0, >′′ (v) ≤ 0, F(v) → –∞(v → v0∗ ); (2) v0 (x) ≥ v1∗ > v0∗ ; (3) u0 (x), v0 (x) ∈ C2 , and their first and second order derivatives are bounded on –∞ < x < ∞, then under the following initial conditions u(x, 0) = u0 (x), v(x, 0) = v0 (x), –∞ < x < +∞,

(3.5.15)

there exists a global classical solution to problem (3.5.5) and (3.5.6) on t ≥ 0. Now we consider the following systems ∂ 2 u ∂u ∂>(u, v) = + , ∂x2 ∂t ∂x ∂ 2 v ∂v ∂8(u, v) + , : 2 = ∂x ∂t ∂x :

(3.5.16) (3.5.17)

with the initial boundary conditions u|t=0 = u0 (x), v|t=0 = v0 (x),

(3.5.18)

u|x=x1 = u|x=x2 = 0, v|x=x1 = v|x=x2 = 0.

(3.5.19)

We suppose that its first-order system (: = 0) is hyperbolic system. We consider the auxiliary equation >v Fuu – (>u – 8v )Fuv – 8u Fvv = 0,

(3.5.20)

then the type of eq. (3.5.20) is analogous to that of systems (3.5.16) and (3.5.17)(: = 0). In fact, the classification discriminant of eq. (3.5.20): △ = (>u – >(v))2 + 4>v 8u ,

3.5 On Some Results of Some Quasilinear Parabolic Equations

313

which is also the discriminant of the characteristic equation +2 – (>u + 8v )+ + (>u 8v – 8u >v ) = 0 for systems (3.5.16) and (3.5.17) (: = 0). Theorem 3.5.5. Assume that F(u, v) is the solution to equation (3.5.20), and for solutions u, v to problem (3.5.16)-(3.5.19), it holds that Fuu . 2 + 2Fuv .' + fvv '2 ≥ ,(u, v)(. 2 + '2 ), , > 0,

(3.5.21)

then we have the following integral inequality: 

x2

 F(u(x, 4), v(x, 4))dx ≤

x1

x2

F(u0 (x), v0 (x))dx.

(3.5.22)

x1

Proof. Multiplying eq. (3.5.16) by Fu , and eq. (3.5.17) by Fv , summing the resultant equations and integrating over RT = {x1 ≤ x ≤ x2 , 0 ≤ t ≤ T}, we have 5

 :

Fu RT

5



6  ∂ 2u ∂ 2v ∂F + F dxdt dxdt = v ∂x2 ∂x2 RT ∂t

Fu (>u ux + >v vx ) + Fv (8u ux + 8v vx )dxdt.

+

(3.5.23)

RT

Since F(u, v) is the solution of problem (3.5.20), the last term of the right-hand side of equality (3.5.23) can be written as ∂G(u,v) ∂x . Choose F(0, 0) = Fu (0, 0) = Fv (0, 0) = 0.

(3.5.24)

Then  RT

∂G dxdt = 0. ∂x

Therefore, integrating eq. (3.5.23) by parts, we get 5



 Fuu

–: RT  x2

∂u ∂x

2

 + 2Fuv



x1

then we deduce inequality (3.5.22).



 + Fvv

∂v ∂x

2 6 dxdt

x2

F|t=T dx –

=

∂u ∂v ∂x ∂x

F|t=0 dx,

(3.5.25)

x1

∎

314

3 The Vanishing Viscosity Method of Quasilinear Hyperbolic System

We set B(M) = max (|>u |, |>v |, |8u |, |8v |), x, t ∈ RT |u|, |v| ≤ M

B1 (M) = max (|D2 >|, |D2 8|), x, t ∈ RT |u|, |v| ≤ M

where D2 denotes the second-order derivatives with respect to all variables: f (|u|) = min(min F(u, v), min F(–u, v)), v

v

g(|v|) = min(min F(u, v), min F(u, –v)). u

u

Theorem 3.5.6. Assume that coefficients and boundary conditions for systems (3.5.16) and (3.5.17) are sufficiently smooth, the solution F(u, v) of eq. (3.5.20) satisfies eqs (3.5.21) and (3.5.24) and    M B(M) + MB1 (M) = o f , 2 (3.5.26)    M B(M) + MB1 (M) = o g , 2 then there exists a global classical solution to the initial boundary value problem (3.5.16)–(3.5.19) on t ≥ 0. Proof. As it’s noted above, in order to prove this theorem, it’s sufficient to construct the a priori estimates for solutions u(x, t), v(x, t) of problem (3.5.16)–(3.5.19) in C0 . Then we set M = max (|u|, |v|), x,t∈RT      ∂u   ∂v  M1 = max  ,   . ∂x ∂x x,t∈RT Using Schauder-type estimates of the parabolic equations, we have M1 ≤ k(MB(M) + M 2 B1 (M)).

(3.5.27)

Now we estimate M = max |u|. We deduce from estimate (3.5.22) that 

x2 x1

 f (|u|)|t=T dx ≤

x2

F(u0 (x), v0 (x))dx = A. x1

We define the set of line segment [x1 , x2 ] by E, where M ≤ |u(x, t)| ≤ M, x ∈ E. 2

(3.5.28)

3.5 On Some Results of Some Quasilinear Parabolic Equations

315

From eq. (3.5.28), we deduce A Mes E ≤ ! M " . f 2 Then ! "    ∂u  Mf M2   , M1 ≥ max   ≥ ∂x A

(3.5.29)

which combined with eq. (3.5.27) yields Mf

!M" 2

A

≤ M1 ≤ k(MB(M) + M 2 B1 (M)).

(3.5.30)

If M = max |v|, similarly, we have ! " Mg M2 ≤ M2 ≤ k(MB(M) + M 2 B1 (M)). A

(3.5.31)

We deduce the boundedness of M from estimates (3.5.30) and (3.5.31) and the condition (3.5.26). ∎ Theorem 3.5.7. Assume that the growth order of >(u), 8(u) is p and coefficients and the boundary function of eqs (3.5.16) and (3.5.17) satisfy conditions in Theorem 3.5.6, then there exists a global classical solution to problem (3.5.16)-(3.5.19). Proof. Equation (3.5.20) is of the form >′ (v)Fuu = 8′ (u)Fvv , we therefore have 

u

F(u, v) =



v

8(u)du +

0

>(v)dv.

0

Since eqs (3.5.16) and (3.5.17) are hyperbolic equations when : = 0, if >′ (v), 8′ (u) have the same sign, without loss of generality, then we set >′ (v) > 0, 8′ (u) > 0, and >(0) = 8(0) = 0. Therefore, condition (3.5.24) satisfies 

u

f (|u|) = min



0

 g(|v|) = min 0

0

8(u)du, v

 8(u)du ,

–u 0

 >(v)dv,

–v

 >(v)dv .

316

3 The Vanishing Viscosity Method of Quasilinear Hyperbolic System

Then the growth order of f (M), g(M) is M p+1 . The growth order of the function B(M) + MB1 (M) is M p–1 , where condition (3.5.26) holds true. Therefore, all conditions in Theorem 3.5.6 are satisfied, and thus the proof of Theorem 3.5.7 is complete. ∎ For the solutions of a kind of viscous hydrodynamic equations ∂1 ∂ 2u = : 2, ∂x ∂x ∂ 21 1t + 1ux + u1x = : 2 , : > 0, ∂x

ut + uux + #1#–2

(3.5.32) (3.5.33)

with periodic boundary conditions or first boundary conditions, we have the following estimates: Theorem 3.5.8. For # ≥ 3, the solutions u, 1 to problem (3.5.32) and (3.5.33) with periodic boundary conditions and first boundary conditions satisfy |u| ≤ M1 , 0 < 1 ≤ M2 ,

(3.5.34)

where M1 , M2 are independent of :. We consider the initial boundary value problem for multidimensional quasilinear parabolic equations in divergence form: n  i,j

∂u  ∂ ∂ 2u = (gradIi (u)), + ∂xi ∂xj ∂t ∂xi n

Aij

(3.5.35)

i=1

u|(x∈∂K)∩(t=0) = u0 (x, t),

(3.5.36)

kl kl where u = (u1 , ⋅ ⋅ ⋅, uN ), Aij = (Akl ij ) is N × N matrix Aij = Aji , n n   (Aij .i , .j ) ≥ a |.i |2 , a > 0, i,j=1

i=1

.i = (.i1 , ⋅ ⋅ ⋅, .iN ), DT = K × [0, T], K ∈ Rn .    ∂2I  2 i   Theorem 3.5.9. Assume I(M) = max i, j, k  ∂u ∂u . If I(M) = O(M p ), p < n+2 , and j k |u| ≤ M Aij , Ii (u), u0 (x, t) are sufficiently smooth, then there exists a global classical solution to problem (3.5.35) and (3.5.36). If u0 (x, t) ≡ 0, p
b1

QT

bT

b2 b – b1

max |u(x, 0)| = K

6 =M

and for each pk , functions aij (x, t, u), bi (x, t, u, pk ), bl (x, t, u, pk ), continuous, and satisfy the following inequalities: -(|u|)

n  i=1

.i2 ≤

n 

aij (x, t, u).i .j ≤ ,(|u|)

i,j=1

n 

∂aij (x,t,u) ∂aij , ∂ul ∂xk

.i2 , - > 0,

are

(3.5.42)

i=1

|bi (x, t, u, pk )| ≤ ,(|u|)(1 + p),

(3.5.43)

l

2

|b (x, t, u, pk )| ≤ ,(|u|)(: + P(p))(1 + p) , ∂aij (x, t, u) ∂aij | , | ≤ -(|u|), ∂xk ∂ul

(3.5.44) (3.5.45)

where P(p) → 0, p → ∞, : > 0 is sufficiently small, and determined only by M = maxQT |u|. ¯ T , |u| ≤ M, |p| ≤ M1 , the first-order derivatives of functions (3) For (x, t) ∈ Q ∂2a

aij (x, t, u), bi (x, t, u, pk ), bl (x, t, u, pk ) and ∂ul ∂uijm , ous. " ¯ (4) u0 (x) ∈ C2 (K) satisfies the compatible conditions

∂ 2 aij

∂ul ∂xk

,

∂ 2 aij

∂ul ∂t

,

∂ 2 aij ∂xk ∂t

are continu-

318

3 The Vanishing Viscosity Method of Quasilinear Hyperbolic System

>|∂K = 0, u0 |∂K = 0, –aij (x, t, 0)>lxi xj + bi (x, t, 0, >xk )>lxi + bl (x, t, 0, >xk )|∂K = 0. "

(5) ∂K ∈ C2 , then there exists a unique global classical solution u(x, t) to the initial ",

"

¯ T ). boundary value problem (3.5.37)–(3.5.39), and u(x, t) ∈ C2,12 (Q

3.6 On Traveling Wave Solutions of Some Quasilinear Parabolic Equations with Small Parameter We consider narrow quasilinear hyperbolic equations: ∂u ∂>(u) + = 0, ∂t ∂x

(3.6.1)

where u = (u1 , ⋅ ⋅ ⋅, un ), >(u) = (>1 (u), ⋅ ⋅ ⋅, >n (u)). Adding small parameter : viscous term yields   ∂u: ∂u: ∂>(u: ) ∂ + =: B(u: ) , ∂t ∂x ∂x ∂x

(3.6.2)

where B(u: ) is an n ×n matrix. We hope that after adding small parameter viscous term B(u: ) the parabolic equations (3.6.2) satisfy the following properties: (1) The initial value problem of eq. (3.6.2) is valid, that is, the solution of eq. (3.6.2) is stable in the sense of Hadamard, that is, when initial perturbation is small, for any finite moment, the perturbation of the solution is also small. (2) There exists a smooth solution to the initial boundary value problem of eq. (3.6.2). (3) For : → 0, the solution u: (x, t) of eq. (3.6.2) converges to the stable general solution u(x, t) of the corresponding hyperbolic equations (3.6.1). It’s easy to verify that B(u: ) satisfies the second requirement in view of the known existence theorem of classical solutions to the quasilinear parabolic equations. In general, it’s difficult to prove the third requirement, and there exist results only for some particular equations. Here, we mainly consider when eq. (3.6.2) will satisfy condition (1). Assume that eqs (3.6.1) and (3.6.2) are linear equations > = Au, where A, B are constant matrixes. Take the transformation  u: (x, t) = u then by virtue of eq. (3.6.2), u(x, t) satisfies

 x t , , : :

3.6 On Traveling Wave Solutions of Some Quasilinear Parabolic Equations

∂u ∂u ∂ 2u +A = B 2. ∂t ∂x ∂x

319

(3.6.3)

Suppose that the particular solution of eq. (3.6.3) is u = u0 ei(+t+-x) ,

(3.6.4)

where u0 is a constant vector, + is a constant and - is a real constant, adding eq. (3.6.4) to eq. (3.6.3), we have (i+ + i-A)u0 = Bi2 -2 u0 ,

(3.6.5)

 –i+ i B + A u0 = 2 u0 . -

(3.6.6)

that is 

Then –i+ is the character value of the matrix B + -i A, u0 is the corresponding right -2 character vector. Assume that "j is the character value of B, then for |-| → ∞, we have + = i"j -2 + 0(-). In view of the above condition (1), when u(x, 0) → 0, u(x, t) → 0, by eq. (3.6.4), it’s necessary that Re "j > 0 j = 1, 2, ⋅ ⋅ ⋅, n.

(3.6.7)

In fact, if Re "j0 < 0, then we can choose initial function sequence u- (x, 0) =

u0 i-x e , -

when - → ∞, u- (x, 0) → 0, by eq. (3.6.4), |u- (x, t)| → ∞(t → 0) - → ∞. Now we consider the behavior of + when - is small. Assume that !j is the character value of A. rj , lj are the corresponding right and left character vectors, respectively. (Assume that !j is a real value with adverse signs.) Then + = –!j - + i-2 (lj Brj ) + o(-3 ).

(3.6.8)

If there exists some j = j0 such that lj0 Brj0 < 0,

(3.6.9)

then by eq. (3.6.9), there exists a particular solution u: = e

-2 |lj Brj |t 0 0 :

v: ,

(3.6.10)

320

3 The Vanishing Viscosity Method of Quasilinear Hyperbolic System

where v: is a bounded function(: → 0). Therefore, we deduce from eq. (3.6.10) that u: → ∞ as : → 0. Thus, by virtue of the third requirement lj Brj ≥ 0 ∀j,

(3.6.11)

then we have the following theorem: Theorem 3.6.1. The initial value problem of parabolic equations (3.6.2) with small parameter is valid and their solutions converge to the stable general solution of hyperbolic equations (3.6.1) as : → 0 provided that Re "j > 0, j = 1, 2, ⋅ ⋅ ⋅, n,

(3.6.12)

lj Brj ≥ 0, j = 1, 2, ⋅ ⋅ ⋅, n,

(3.6.13)

and

where "j is the character value of matrix B(u: ); lj , rj are the left and right character vectors of the matrix (>iuj ), respectively. Now we make simple analysis to the traveling wave solution of eq. (3.6.2), and assume that there exists a particular solution to eq. (3.6.2):  u: = u:



x – Dt :

= u: (. ), . =

x – Dt , :

(3.6.14)

which combined with eq. (3.6.2) yields du: d>(u: ) du: du: d B(u: ) = –D = [A(u: ) – DE] , d. d. d. d. d.

(3.6.15)

where E is the identity matrix, and integrating eq. (3.6.15) yields B(u: )

du: = >(u: ) – Du: – C = I(u: ), d.

(3.6.16)

where C is an arbitrary constant vector. Now we need that the solution u: to eq. (3.6.16) satisfies the boundary condition: |. | → ∞ ⋅ u(. ) → constant or  u: (. ) → and

du: (. ) d.

u– , . → –∞, u+ , . → +∞,

(3.6.17)

→ 0 as |. | → ∞, then I(u+ ) = I(u– ) = 0. We therefore deduce H condition >(u– ) – >(u+ ) = D(u– – u+ ).

(3.6.18)

321

3.6 On Traveling Wave Solutions of Some Quasilinear Parabolic Equations

Now we ask that when there exists a unique solution to the boundary value problem of the ordinary differential equations (3.6.16) and (3.6.17). We can list the following two supplemental necessary conditions: (1) u– , u+ are two neighbor zero points.

(3.6.19)

In fact, otherwise we have I(u∗ ) = 0, where u∗ ∈ [u+ , u– ], then u = u∗ is the solution of equations (3.6.16), which obviously does not satisfy the boundary condition (3.6.17). (2) The matrix B–1 (u– )[A(u– ) – DE] is nonnegative, and B–1 (u+ )[A(u+ ) – DE] is nonpositive.

(3.6.20)

In fact, eq. (3.6.16) can be rewritten as du. = B–1 (u: )I(u: ). d.

(3.6.21)

We consider around the left critical point u: = u– that   du ∂I(u: ) (u: – u– ) = B–1 (u– ) |u=u– (u: – u– ) d. ∂u +O(|u: – u|–2 ) = B–1 (u– )[A(u– ) – DE](u: – u– ) +O(|u: – u|–2 ), which multiplied by (u: – u– ) yields d (u: – u– )2 /2 = (u: – u– )B–1 (u– )[A(u– ) – DE](u: – u– ) + O(|u: – u– |3 ). d. – 2

) does not decrease If B–1 (u– )[A(u– ) – DE] is nonnegative, then as & increases, (u: –u 2 – around u = u . Therefore, if there exists the integral line of eq. (3.6.16) through u– , then the matrix B–1 (u– )[A(u– ) – DE] can’t be negative; similarly, B–1 (u+ )[A(u+ ) – DE] can’t be positive. Then from this condition, we will deduce the stability condition >′ (u– ) > D > ′ > (u+ ) when eq. (3.6.16) degenerates into one equation. If there exists a solution u: (. ) to the boundary value problem (3.6.16) and (3.6.17), then the limit of u: (x, t)

 lim u: (x, t) =

:→0

u– , x – Dt < 0, u+ , x – Dt > 0,

is a discontinuous solution, which satisfies condition H on x = Dt; meanwhile, eq. (3.6.18) satisfies the conservation law. However, although there exists a solution to the

322

3 The Vanishing Viscosity Method of Quasilinear Hyperbolic System

problem (3.6.16) and (3.6.17), it can not guarantee that the solution converges to stable general solution, as : → 0. Then we will give some counter examples to illustrate that (1) may converge to different general solutions with respect to different viscous matrix, and (2) may converge to an unstable general solution even if some necessary conditions are satisfied.  Example 1. We consider eq. (3.6.2) and assume that >(u) = ⎛√ B=⎝

u21 u22 2 , 2

 ⎞ √ 2– + 2 ! 1–! ⎠. √1 1–! 2

 ,



1 2

(3.6.22)

For 0 < ! < 1 – 2√1 2 , it’s easy to verify that the matrix is positive definite and symmetric, the condition (3.6.13) is satisfied; it’s easy to testify that there exists a solution to eq. (3.6.2) satisfying 2u0 √ u1 + u01 1x u2 = – 2u1 , = e !: . 0 u1 – u1

(3.6.23)

Choose u01 , from eq. (3.6.23) and deduce that u1 = u1 (. ) = u1 an unstable general solution:  u1 (x) =

–u01 , x < 0,

!x" :

, as : → 0, converges to

(3.6.24)

u01 , x > 0,

and u2 = u2 (. ) is obviously stable. Example 2. In eq. (3.6.2), we choose >(u) = {–u2 , p(u1 )}, B =

a0 cb

,

(3.6.25)

where p′ (u1 ) < 0, p′′ (u1 ) > 0, a, b, c > 0. Assume that u+ = {1, 0}, D > 0, u–1 > 1, u–2 < 0, then when D is small enough, it’s easy to verify that with conditions (3.6.19) and (3.6.20) the integral curve of eqs (3.6.16) and (3.6.17) is a nodal point at u– and is a saddle point at u+ , as shown in Fig. 3.1. For : → 0, the solution u: (x, t) of (3.6.16) and (3.6.17) converges to unstable shock rarefaction wave (i.e., but density or entropy decreases, which occurs when p = p(v) Poisson adiabatic curve is abnormal upward convexly. This substance will generate this wave around the critical point of vapor phase and liquid phase). It’s easy to verify that condition (3.6.12) holds but eq. (3.6.13) does not hold true.

3.6 On Traveling Wave Solutions of Some Quasilinear Parabolic Equations

323

u2

O

u–1

u+

u1

u2–

u

Fig. 3.1

Example 3. In eq. (3.6.2), we choose  >(u) =

 1 v 5 2 , , –u , u = (u, v), B(u) = v 2 –8 –3

(3.6.26)

with initial conditions u(x, 0) = u0 (x/:), v(x, 0) = v0 (x/:),

(3.6.27)

⎧ 11 ⎪ ⎪ ⎨ u0 (–∞) = – , u0 (+∞) = 0, 30 5 6 ⎪ ⎪ ⎩ v0 (–∞) = , v0 (+∞) = . 6 5

(3.6.28)

where

It’s easy to verify that the solution to problem (3.6.26) and (3.6.27) is  u: (x, t) = u0

   x+t x+t , v: (x, t) = v0 , : :

which converge to unstable shock rarefaction wave as : → 0.

(3.6.29)

324

3 The Vanishing Viscosity Method of Quasilinear Hyperbolic System

Example 4. There exists different limit solution to the following quasilinear hyperbolic equations:   ⎧ 30 ∂u ∂u ⎪ ⎪ – cos u = 0, ⎨ ∂t ∂x 2 2 ⎪ ⎪ ⎩ ∂v – ∂v = 0, ∂t ∂x

(3.6.30)

with different viscous matrix B(u). We consider the fixed solution problem:     ∂u ∂u 30 ∂ ∂u ∂v – cos u = : b(u) – , ∂t ∂x 2 ∂x ∂x ∂x   ∂ ∂u ∂v ∂v ∂v2 – =: – + , ∂x ∂x ∂x ∂x ∂x 4x 4x , v|t=0 = 2th . u|t=0 = th : :

(3.6.31) (3.6.32) (3.6.33)

Then u: (x, t) → u(x, t) = sign x, v: (x, t) → v(x, t) = 2sign x,

(3.6.34)

as : → 0, on the other hand, the fixed solution problem:   ∂u ∂u 30 ∂ 2u – cos u = : 2 , ∂t ∂x 2 ∂x

(3.6.35)

∂v ∂v2 ∂ 2v – = : 2, ∂t ∂x ∂x 4x 4x , v|t=0 = 2th . u|t=0 = th : : ⎧ x ⎪ sign x,   > a, ⎪ ⎨ t   u: (x, t) → u(x, t) = x 2 x 2 ⎪ –1   < a, ⎪ , ⎩ sin 30 30 t t v: (x, t) → v(x, t) = 2sign x,

(3.6.36) (3.6.37)

where a satisfies the equation 5 a

2

2 30

2



2 2a + 1– sin–1 30 30

2 6 = 1.

Obviously, the limits of problem (3.6.35)–(3.6.37) and that of problem (3.6.32) and (3.6.33) are different.

3.6 On Traveling Wave Solutions of Some Quasilinear Parabolic Equations

325

Now we consider simpler case, in eq. (3.6.2), B(u: ) ≡ I, that is ∂u: ∂f (u: ) ∂ 2u + = : 2, ∂t ∂x ∂x

(3.6.38)

where there exist some different real character roots +1 (u) < +2 (u) < ⋅ ⋅ ⋅ < +n (u) to the hyperbolic conservation law ut + :

f (u) = 0. ∂x

(3.6.39)

We consider Riemann problem for eq. (3.6.39): u(x, 0) = ul , x < 0; u(x, 0) = ur , x > 0.

(3.6.40)

Assume that the entropy condition is satisfied, that is, shock wave velocity s with respect to k shock wave satisfies xk–1 (ul ) < s < +k (ul ), +k (ur ) < s < +k+1 (ur ),

(3.6.41)

where + denotes ith character value of grad f . P.D. Lax has proved that under condition (3.6.41) there exists a unique single parameter family of left state ul (:), which can be connected with right state via k weak shock wave, where the shock wave velocity is 1 s(:) = +k + :. 2

(3.6.42)

Then we will prove that as : → 0, the traveling wave solution u: of eq. (3.6.38) converges to the general solution u(x, t) of eq. (3.6.39), whose left and right states can be connected via k weak shock wave. We first consider the formal traveling wave solution of eq. (3.6.38), assume  u(3) = u

 x – st , :

(3.6.43)

which combined with eq. (3.6.38) yields :

du = f (u) – f (u4 ) – s(u – u4 ), d4

(3.6.44)

where 4 = :3. Assume that u(4, :) can be expanded with respect to : by power series expansion, and for simplicity, we denote +k by +, add s = 21 : + + to eq. (3.6.44), and differentiate eq. (3.6.44) with respect to :, then

326

3 The Vanishing Viscosity Method of Quasilinear Hyperbolic System

⎧ 1 du du˙ 1 ⎪ ⎪ ˙ +: = (A – +)u˙ – (u – u4 ) – :u, ⎪ ⎪ d4 d4 2 2 ⎪ ⎪ ⎨ ˙ 1 du d¨u ˙ u˙ + (A – +)¨u – u˙ – :u, ˙ 2 +: = (A′ u) ⎪ d4 d4 2 ⎪ ⎪ ... ⎪ ⎪ 1 ... ... 3 ⎪ ⎩ 3 d¨u + : du = ([A′′ u] ˙ u) ˙ u˙ + (A′ u¨ )u˙ + 2(A′ u)¨ ˙ u + (A – +)u – u¨ – :u, d4 d4 2 2

(3.6.45)

where A denotes grad f , “⋅” denotes the differentiation with respect to :, “′ ” denotes that grad. Equation (3.6.39) is a hyperbolic equation, which has n different real roots +i , whose left and right character vectors are li , ri , respectively, after standardizing them we have li ri = $ij , and setting : = 0 in eq. (3.6.45) yields (A – +)u˙ = 0, du˙ ˙ u, ˙ + u˙ – (A′ u) d4 d¨u 3 ... ˙ u) ˙ u˙ – (A′ u¨ )u˙ – 2(Au)¨ ˙ u. (A – +)u = 3 + u¨ – ([A′′ u] d4 2

(A – +)¨u = 2

For the matrix equation of the form (A – +)f = g admits the solution f =

 ! !≠k

1 (l! g)r! + crk . +! – +

The compatible condition holds lg = 0. Therefore, we deduce from eq. (3.6.46) that ˙ u(0) = c0 (4)rk . c0 is derived from the compatible condition of eq. (3.6.46). In fact, we have ˙ 2 ac2 dc0 c0 4Arc 0 + = = 0, d4 2 2 2 which admits the solution c0 (4) =

1 4

a – ke 2

,

(3.6.46)

3.6 On Traveling Wave Solutions of Some Quasilinear Parabolic Equations

˙ ≠ 0, k = [ac0 (0) – 1]/c0 (0). For 0 < c0 (0) < where a = lAr continuous with respect to 4, then ˙ 0) = u(4,

rk 4

a(1 + e 2 )

1 a,

.

327

the solution curve is

(3.6.47)

In order to derive the solution u¨ , by eq. (3.6.46), we have u¨ = –c20

 ! !≠k

1 ˙ ! + c1 rk , (l! Ar)r +! – +

where c1 is to be determined later. In order to confirm c1 , by the compatible condition of eq. (3.6.46), we have dc1 1 1 ˙ u] ˙ u˙ + (A′ u¨ )u˙ + 2(A′ u)¨ ˙ u}, + c1 = l{[(A′′ u) d4 2 3

(3.6.48)

  dc1 1 + c1 – ac0 = bc30 , d4 2

(3.6.49)

which yields

integrating eq. (3.6.49) implies 1 c1 (4) = b sech 4 4



4

2

0

1 cosh2 4c30 d4. 4

(3.6.50)

Continuing this procedure yields the Taylor expansion of u(4, :) with respect to :. From the existence theorem of solutions to eq. (3.6.44) to be proved later, we see that this expansion is convergent asymptotically. We deduce from what have been discussed above that ⎧ u(4, :) = ur + Ev(4, :), ⎪ ⎪ ⎪  ⎪ ⎪ ⎪ &! r! + O(:2 ), v(4, :) = c0 rk + Ec1 rk + Ec20 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ 1 ⎪ ⎪ c0 = ⎪ 4 , ⎪ ⎨ a(1 + e 2 ) ˙ = lB(r, r), a = 4Ar ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ –l! B(r, r) ⎪ ⎪ , &! = ⎪ ⎪ +! – + ⎪ ⎪  4 ⎪ ⎪ ⎪ 1 1 ⎪ ⎩ c1 = b sech2 4 cosh2 4c30 d4, 4 0 4 where 4 = :3.

(3.6.51)

328

3 The Vanishing Viscosity Method of Quasilinear Hyperbolic System

Now we prove existence of solutions of the equation :

du = f (u) – f (ur ) – s(u – ur ), d4

(3.6.52)

when : is small enough. Then u(4, :) = ur + :v(4, :).

(3.6.53)

Substituting s = + + 21 : into eq. (3.6.52) and expanding f (u) yield 1 1 (A – +) dv =– v+ B+ v + :R, d4 2 2 :

(3.6.54)

where A denotes grad f (ur ), and B(v, v) the second-order bilinear form, R(:, v) denotes remainder term of the expansion. Since A has different real character values +i ranked in order, which correspond to identity right character vectors ri , choose left character vectors li , such that li rj = $ij , then we choose v as v(4, :) =



(li v)ri =



&i ri .

(3.6.55)

Multiplying the left-hand side of eq. (3.6.54) by li and transposing the third term of the right-hand side yield that u&i +i – + 1 1 – &i = – &i + li B + :Ri = Si (v). d4 : 2 2

(3.6.56)

Introducing the sign a!"# = l! B(r" , r# ), and we divide &k into two functions, one of which depends only on 4 &k = &k (4, 0) + &k (4, :) – &k (4, 0) = &0 (4) + '(4, :),

(3.6.57)

where &0 (4) = &k (4, 0); '(4, :) = &k (4, :) – &k (4, 0). Applying the above signs, we write the expression 1 ! 2 [a & + 2a!k" &k &" + a!"# &# &" ] 2 kh k 1 1 = a!kh &02 + a!kh &0 ' + a!kh '2 + a!k" &0 &" 2 2 1 ! ! +ak" &" ' + a"# &# &" . 2

l! B(v, v) =

(3.6.58)

Summing indexes ", # except for k in eq. (3.6.58), Greek character: 1, 2, ⋅ ⋅ ⋅ k – 1, k + 1, ⋅ ⋅ ⋅, n, Latin alphabet: 1, ⋅ ⋅ ⋅, k – 1, k + 1, ⋅ ⋅ ⋅, n.

3.6 On Traveling Wave Solutions of Some Quasilinear Parabolic Equations

329

Substituting eqs (3.6.57) and (3.6.58) into the kth equation of eq. (3.6.56), we deduce two equations, one of which does not depend on : 1 d&0 1 + &0 = a'2 , d4 2 2  1 1 d' +' – a&0 = a'2 + akk" &k &" + ak"# &" &# + :Rk = sk (v), d4 2 2

(3.6.59) (3.6.60)

where a = lB(r, r) ≠ 0. Equation (3.6.59) has a nonconstant solution 1

&0 =

4

a(1 + e 2 )

.

(3.6.61)

Set (+! – +)/: = ,! , then for ! ≠ k, multiplying the !th equation of eq. (3.6.56) by e–,! 4 and integrating from ±∞ when ,! ≷ 0 to 4 yield &! = e,! 4



4

±∞

e–,! 4 s! (v)d4.

(3.6.62)

Similarly, multiplying eq. (3.6.60) by cosh2 44 , and integrating from 0 to 4 yield ' = sech

24

4



4 0

4 cosh sk (v)d4. 4

(3.6.63)

Equations (3.6.56) and (3.6.60) are equivalent to the integral equations (3.6.62) and (3.6.63), respectively. We write eqs (3.6.62) and (3.6.63) as operator forms v = H(v),

(3.6.64)

or &! = H! (v) = T! s! (v),

(3.6.65)

' = Hk (v) = Tk sk (v),

(3.6.66)

where s! , sk denote nonlinear functions on the right-hand side of eqs (3.6.56) and (3.6.60), respectively, while T! and Tk denote integral operators that arise in eqs (3.6.62) and (3.6.63), respectively. Now existence problem of solutions for eq. (3.6.52) is formulated as to find the solutions for the integral equation (3.6.64), (& , ') = (&1 , &2 , ⋅ ⋅ ⋅, &k–1 , ', &k+1 , ⋅ ⋅ ⋅, &n ). Assume x is a vector in space B, which consists of n-dimensional differentiable functions f (4), –∞ < 4 < ∞, and df df (+∞) = f (∞) = (–∞) = 0. d4 d4

330

3 The Vanishing Viscosity Method of Quasilinear Hyperbolic System

For every function f (4), we define ||f || = max4 |f |, denote x = (& , '), where & is an n – 1-dimensional vector, and define ||& || = max! ||&! ||, then   1 ||x||L = max ||& ||, ||'|| , x ∈ B, 2L

(3.6.67)

where L denotes a constant determined later. Obviously, the space defined by eq. (3.6.67) is Banach space. Our problem lies in to find a fixed point in the Banach space. Consider the integral operators T! and Tk defined in eqs (3.6.62) and (3.6.63), then : , |+! – +| ||Tk || = 2, ||T! || =

(3.6.68) (3.6.69)

where ||T|| is the maximum modulus, now we set ||v||L < $(i.e.||& || < $, ||'|| < 2L$). Substituting eq. (3.6.58) into eq. (3.6.56) yields 1 1 S! (v) = – &! + a!kh (&02 + 2&0 ' + '2 ) + a!k" &0 &" 2 2 1 +a!k" '&" + a!"# &# &" + :R! . 2

(3.6.70)

For ||v||L < $, ||S! (v)|| ≤ O($) +

a!kh + O(L$) + O(L2 $2 ) 2|a|2

+O($2 ) + O(L$2 ) + O($2 ) + O(:). In view of the above estimates, and using eq. (3.6.68), we have ||H! (v)|| ≤

6 5 ! akh : + O($) + O(:) . |+! – +| 2|a|2

(3.6.71)

By virtue of eqs (3.6.60) and (3.6.69), we have ||Hk (v)|| ≤ 2

5  |ak | " k" |a|

6 + O($) $ + O(:).

(3.6.72)

Now we choose  L = 1+

k " |ak" |

|a|

.

Choose : small enough, such that for each !, we have

(3.6.73)

3.6 On Traveling Wave Solutions of Some Quasilinear Parabolic Equations

:≤

2|+! – +||a|2 $ . a!kh

331

(3.6.74)

and : ≤ $2 ,

(3.6.75)

||H(v)||L < $, ||v||L < $.

(3.6.76)

then

Consider H satisfies Lip condition, from eq. (3.6.64), we have ||H(v1 ) – H(v2 )||L ≤ ||T||||s(v1 ) – s(v2 )||L . For ||v1 || < $, ||v2 ||L < $, we have ||H! (v1 ) – H! (v2 )|| ≤

 !  |akh |  : + O($) + O(:) ||'1 – '2 || |+! – +| |a|  !  |a  1 " k" |  + + O($) ||& 1 – & 2 || . 2 |a|

By virtue of the choice of :, satisfying eq. (3.6.74), from the above inequality, we get ||H! (v1 ) – H! (v2 )|| ≤ O($)||'1 – '2 || + O($)||& 1 – & 2 ||.

(3.6.77)

For the ith equation, we deduce from eq. (3.6.60) that ||Hk (v1 ) – Hk (v2 )|| ≤ [O(L$) + O(:)]||'1 – '2 || 5  |a! | 6 k" + + O($) + O(:) ||& 1 – & 2 || |a| or ||Hk (v1 ) – Hk (v2 )|| ≤ O($)||'1 – '2 || + 2L||& 1 – & 2 ||.

(3.6.78)

By eqs (3.6.77) and (3.6.78), we see that the operator H(v) satisfies Lipschitz condition. Thus when : is small enough, by fixed point theory, there exists a unique solution to the operator equation (3.6.64). We therefore have the following theorem. Theorem 3.6.2. Assume that ul and ur are two states that can be connected by weak k shock wave for hyperbolic equations (3.6.39) with conservation, then there exists a solution to the parabolic equations (3.6.38) with viscous term, and as : → 0, the solution converges to the general solution to the hyperbolic system (3.6.39).

332

3 The Vanishing Viscosity Method of Quasilinear Hyperbolic System

For one-dimensional isothermal gas hydrodynamic equations ut + p(v)x = 0,

(3.6.79)

vt – ux = 0,

(3.6.80)

we consider the traveling wave solution to its corresponding viscous equations Ut + F(U)x = :PUxx ,

(3.6.81)

p(v) u . We deduce from , F(U) = U(x, t) = U = U(. ), where U = –u v eq. (3.6.81) that the second-order constant coefficient matrix is 



x–st :

dU = BV(U), d. where B =

P–1

=

(3.6.82)



p(v) – sv + c1 !" , V(U) = , the boundary conditions for $# –sv – u + c2

eq. (3.6.82) are lim v(. ) = U0 ,

. →–∞

lim U(. ) = U1 .

. →+∞

(3.6.83)

For the boundary value problem (3.6.82) and (3.6.83) of the ordinary differential equation, we have the following results.

!" , :0 = p′ (v– ), :1 = p′ (v+ ), B = Theorem 3.6.3. Assume $# : = limv→∞ p′ (v). Here (u+ , v+ ) is the saddle point of equations (3.6.82) and (u– , v– ) is the nodal point of equations (3.6.82). Then (1) If " > 0 and

p′ (v) < 0,

p′′ (v) > 0,

– !s – ": – $ – #s < 0

(3.6.84)

then there exists a solution to problem (3.6.82) and (3.6.83), if " < 0 and – !s – ":0 – $ – #s < 0, –!s – 1:1 – $ – #s > 0,

(3.6.85)

then there exists no solutions to problem (3.6.82) and (3.6.83) in any case, that is, the solution for problem (3.6.81) does not converge to the general solution to problem (3.6.79) and (3.6.80). For the symmetric case, " = $, ! > 0, # > 0, then (2) If :0s+1 ≥ 21 , then for each positive-definite symmetric matrix B, eqs (3.6.82) and (3.6.83) admit a solution; if :1s+1 < 21 , then there exists a positive-definite symmetric matrix B such that there exists no solutions to eqs (3.6.82) and (3.6.83).

333

3.6 On Traveling Wave Solutions of Some Quasilinear Parabolic Equations

(3) Assume that B is symmetric positive definite, then there exist u– , u+ connected by a solution of the shock wave, such that the traveling wave solution for its corresponding parabolic equations does not converge to the given shock wave solution. Theorem 3.6.3 contains more general case than Theorem 3.6.2 (in which strong shock wave case is contained). Now we consider eq (3.6.39) that is not necessarily a genuine nonlinear case (in the sense of P.D. Lax), then we introduce the corresponding viscous equation: ut + f (u)x = :tuxx .

(3.6.86)

We set . = xt , then u(x, t) = u(. ) satisfies the following boundary value problem: ˙ ) – . u(. ˙ ), :¨u(. ) = f0 (u(. ))u(. –

+

u(–∞) = u , u(∞) = u ,

(3.6.87) (3.6.88)

where “⋅” denotes differentiation with respect to . . First, for fixed : > 0, we prove that there exists a solution u: (. ) to problem (3.6.86) and (3.6.87), whose total variation is bounded uniformly with respect to :. Then by Helly’s choice theorem, we choose a subsequence {u:n (. )} which is bounded and convergent, and converges to bounded ! " variation function u(. ) (when n → ∞, :n → 0), then u xt is the general solution to Riemann problem (3.6.39) and (3.6.40). In order to prove the existence of solutions to problem (3.6.87) and (3.6.88) for every fixed : > 0, we consider the following double-parameter boundary value problem: ˙ :¨u(. ) = ,f˙ (u(. )) – . u, –

(3.6.89) +

u(–L) = ,u , u(L) = ,u ,

(3.6.90)

where , is a parameter, , ∈ [0, 1], L ≥ 1. We have Lemma 3.6.4. Assume that for all possible solutions u(. ) to problem (3.6.89) and (3.6.90), sup |u(. )| < M,

(3.6.91)

–L≤. ≤L

where the constant M depends on u– , u+ , : and f , then there exists a solution u: (x, t) to problem (3.6.89) and (3.6.90). Proof. We define the operator u = Tv : C0 ([–L, L]; Rm ) → C0 ([–L, L]; Rm )   .  –& 2 1 . 1 e– 2: !& + f (v(& ))d& – 2 u(. ) = , u– + 9 : : –L –L   . & 42 –& 2 ⋅ 4f (v(4))e 2: d4d& , –L

0

(3.6.92)

334

3 The Vanishing Viscosity Method of Quasilinear Hyperbolic System

where 5 9=

L

–L

+

1 :2

6–1 5 exp(–& 2 /2:)d& 

L &

–L

u+ – u – –

4f (v(4))e

42 –& 2 2:

6 d4d& .

1 :



L

f (v(& ))d& –L

(3.6.93)

0

Obviously, the operator T is continuous, we deduce from eq. (3.6.92) that    42 –. 2 1 1 . ˙ ) = , 9exp(–& 2 /2:) + f (v(. )) – 2 u(. 4f (v(4))e 2: d4 . : : 0 From which we see that the operator T maps the bounded set in C0 ([–L, L], Rm ) to the equivalent continuous set in C0 ([–L, L], Rm ); thus T is compact. We see from eq. (3.6.91) that each fixed point (, ∈ [0, 1]) of T is contained in an open ball that centers in zero point with the radius M. Therefore, by Leray–Schauder’s fixed point theorem, there exists a fixed point u(. , L)(, = 1) such that it’s a solution to problem (3.6.87) and (3.6.88). ∎ We need to prove the existence of solutions to problem (3.6.87) and (3.6.88). Since u(. , L) solves the equation 5 6 .2 .2 d 1 ˙ , L)e 2: = ∇f (u(. ˙ , L))u(. ˙ , L)e 2: , u(. d. :

(3.6.94)

from which we have ˙ , L)| ≤ |u(0, ˙ L)|e |u(.

2!|. |–. 2 2:

,

–L ≤ . ≤ L,

(3.6.95)

where ! = sup{|∇f (v)| : v ∈ Rm , |v| < M}; on the other hand, from the simple integral of eq. (3.6.89), we deduce  ˙ L) u(0,

1 –1

.2

e– 2: d. = u(1, L) – u(–1, L) –

1 :



1

f (u(. , L))d. –1

 1 .2 1 e– 2: d. + f (u(0, L)) : –1  2    1 1 . & – .2 + 2 & f (u(& , L)) exp d& d. . : –1 0 2:

(3.6.96)

˙ L)|. We extenBy virtue of eq. (3.6.91), we deduce the uniform boundedness L of |u(0, ded the definition domain of u(⋅, L) to the whole real axis, u(. , L) = u– , . < – L, u(. , L) = u+ , . > L. From eq. (3.6.95) we see that the set {u(⋅, L); L ≥ 1} is sequential compact; thus, there exists a sequence of {Ln }(Ln → ∞, n → ∞) and a function u(. ) ∈

3.6 On Traveling Wave Solutions of Some Quasilinear Parabolic Equations

335

C0 ((–∞, ∞), Rm ) such that u(. , Ln ) → u(. ), N → ∞, ∀. ∈ (–∞, ∞). It’s easy to prove that u(. ) is a solution to problem (3.6.87) and (3.6.88). For Riemann problem (3.6.39) and (3.6.40), we have the following lemma. Lemma 3.6.5. For fixed : > 0, assume that u: (. ) is the solution to problem (3.6.87) and (3.6.88). Assume that the set {u: (⋅); 0 < : < 1} are uniformly bounded variation functions, ! " then there exists a bounded variation function u(. ), such that u xt is the general solution to problem (3.6.39) and (3.6.40). Proof. By Helly’s theorem, there exists a sequence {:n }(:n → 0, n → ∞) and a bounded ! " variation function u(. ) , such that u:n (. ) → u(. ), n → ∞, . ∈ (–∞, ∞); thus, u:n xt → !x" !x" u t , n → ∞. (x, t) ∈ {(–∞, ∞) × (0, ∞)}. Since u:n t is a solution of eq. (3.6.86) ! " (: = :n ), it’s easy to see that u xt is a weak solution to eq. (3.6.39), and it remains to ! " verify that u xt satisfies the initial condition (3.6.87). In fact, similarly to the estimates of eqs (3.6.95) and (3.6.96), we have |u˙ : (. )| ≤ |u˙ : (0)|e

2!|. |–. 2 2:

, –∞ < . < ∞,

(3.6.97)

where ! = sup{|∇f (u: (. ))| : –∞ < . < ∞, 0 < : < 1} and 

 1 1 e d. = u: (1) – u: (–1) – f (u: (. ))d. : –1 –1  1   .2 & 2 –. 2 1 1 1 . + f (u: (0)) e– 2: d. + 2 & f (u: (& ))e 2: d. d& . : : –1 0 –1

u˙ : (0)

1

.2

– 2:

(3.6.98)

Calculate the independence on : of every term of eq. (3.6.98), by eq. (3.6.97), we deduce that 5

|u˙ : (. )| ≤ K:– 2 |e

2!|. |–. 2 2:

, –∞ < . < ∞,

(3.6.99)

where constants k, ! are independent of :, u:n (. ) → u(. ), n → ∞, ∀. ∈ (–∞, ∞), we deduce from eq. (3.6.99) that u(. ) = u– , . < –2!, u(. ) = u+ , . > 2!. In particular, u

!x" t

solves eq. (3.6.87).

∎

Now we consider the case of two equations, assume the equations 

vt + g(v, 9)x = 0, 9t + h(v, 9)x = 0,

(3.6.100)

336

3 The Vanishing Viscosity Method of Quasilinear Hyperbolic System

where g, h are continuous differentiable functions, satisfying the hyperbolic condition g9 hv > 0. Assume g9 (v, 9) < 0, hv (v, 9) < 0, ∀(v, 9) ∈ R2 .

(3.6.101)

Equations (3.6.89) and (3.6.90) have the following form: 



:¨v = ,˙g (v, 9) – . v˙ , ˙ 9) – . 9; ¨ = ,h(v, ˙ :9

v(–L) = ,v– , v(L) = ,v+ , 9(–L) = ,9– , 9(L) = ,9+ ,

(3.6.102)

(3.6.103)

For each bounded subset U ⊂ R, the set {g(v, 9) : v ∈ R, 9 ∈ U} and the set {h(v, 9) : v ∈ U, 9 ∈ R} are bounded and 

inf |g(v, 9)| → ∞, |9| → ∞, inf |h(v, 9)| → ∞, |v| → ∞.

(3.6.104)

Under the above assumptions, we have the following lemma: Lemma 3.6.6. Assume that (u(. ), u(. )) is a solution to problem (3.6.102) and (3.6.103). Then (1) there exists a constant M, which depends on :, v– , v+ , 9– , 9+ , g, h, but not on L ≥ 1, , ∈ [0, 1], such that ⎧ ⎪ sup |v(. )| ≤ M, ⎪ ⎨ –L≤. ≤L (3.6.105) ⎪ ⎪ ⎩ sup |9(. )| ≤ M; –L≤. ≤L

(2) for , = 1, L = ∞, there exists a constant N, which depends on v– , v+ , 9– , 9+ , g, h, but not on :, : ∈ (0, 1), such that ⎧ ⎪ ⎪ ⎨ sup |v(. )| ≤ N, –∞ 0, there exists a convergent subsequence for the solutions sequence {rn: (x, t)}, {s:n (x, t)} of problem (3.7.16)–(3.7.20), such that L1

L1

rn: - (x, t) → r: (x, t), s:n- (x, t) → s: (x, t), n- → ∞, and there exists a subsequence of { ∂rn: ′ -

∂x

L1 weak

→

∂rn: ∂x

}, {

∂s:n∂x

}, such that

: ∂r: ∂sn-′ L1 weak ∂s: ′ → , , n- → ∞, ∂x ∂x ∂x

where {r: (x, t)}, {s: (x, t)} are monotone functions with respect to x. Proof. By Helly’s theorem and the expression of solutions of problem (3.7.16)–(3.7.20) rn: - (x, t) = (1 – x),1 (t) + x,2 (t)  t x – G: (x, t, . , 4)f1 (. , 4)d. d4 

0

0

1

+ 0

G: (x, t, . , 0)¯r0 (. )d. ,

s:n- (x, t) = (1 – x)-1 (t) + x-2 (t)  t x – G: (x, t, . , 4)f2 (. , 4)d. d4 

0

0

1

+ 0

G: (x, t, . , 0)¯s0 (. )d. ,

344

3 The Vanishing Viscosity Method of Quasilinear Hyperbolic System

where f1 (x, t) = +1 f2 (x, t) = +2

∂rn: -

∂x ∂s:n∂x

G: (x, t, . , 4) =

+ (1 – x),′1 (t) + x,′2 (t), + (1 – x)-1′ (t) + x-2′ (t),

∞ 2  –m2 02 :(t–4) e sin m0x sin m0. , 0 m=0

r¯0 (x) = r0 (x) – (1 – x),1 (0) – x,2 (0), s¯ 0 (x) = s0 (x) – (1 – x)-1 (0) – x-2 (0), as n- → ∞, we deduce the conclusion of this lemma.

∎

Theorem 3.7.13. Under the conditions of Lemma 3.7.10, 3.7.11 and 3.7.12, there exists a general solution to the quasilinear hyperbolic equations ∂r ∂r ∂s ∂s + +1 (r, s) = 0, + +2 (r, s) = 0, ∂t ∂x ∂t ∂x

(3.7.24)

with the first initial boundary conditions r|x=0 = ,1 (t), r|x=1 = ,2 (t), r|t=0 = r0 (x), s|x=0 = -1 (t), s|x=1 = -2 (t), s|t=0 = s0 (x),

(3.7.25)

that is, for the following integral identical equation: 

T 0

 15 0



T

+

6  1 ∂r ∂f1 f1 (x, 0)r0 (x)dx r(x, t) – +1 (r, s)f1 (x, t) dxdt + ∂t ∂x 0

f1 (0, t)+1 (,1 (t), -1 (t)),1 (t)dt

0



T

– 0



T 0

f1 (1, t)+1 (,2 (t), -2 (t)),2 (t)dt = 0,

 15 0



T

+

6  1 ∂s ∂f2 f2 (x, 0)s0 (x)dx s – +2 f2 dxdt + ∂t ∂x 0

f2 (0, t)+2 (,1 (t), -1 (t))-1 (t)dt

0

 –

T

f2 (1, t)+2 (,2 (t), -2 (t))-2 (t)dt = 0,

0

where fi (x, t) ∈ C2 , fi (x, T) = 0, i = 1, 2. Proof. We complete the proof similarly as Lemma 3.7.12 and the regular argument.

∎

345

3.7 On General Solutions of Some Diagonal Quasilinear Hyperbolic Equations

Theorem 3.7.14. If the following conditions are satisfied: (1) r0 (x), s0 (x) are monotone functions with respect to x, and 0 ≤ r0′ (x), s′0 (x) ≤ K; (2) ,i (t), -i (t) ∈ C2 , +i ∈ C2 , i = 1, 2; (3)

∂+i ∂r

≥ 0,

∂+i ∂s

≥ 0, i = 1, 2;

(4) the compatible conditions r0 (0) = ,1 (0), r0 (1) = ,2 (0), s0 (0) = -1 (0), s0 (1) = -2 (0) hold. (5) +1 (,1 (t), -1 (t)) ≥ l1 > 0, +1 (,2 (t), -2 (t)) ≤ l2 < 0, +2 (,1 (t), -1 (t)) ≥ m1 > 0, +2 (,2 (t), -2 (t)) ≤ m2 < 0, then there exists a general solution to the initial value problem (3.7.24) and (3.7.25). Now we consider the existence of general solutions for the first-order initial boundary value problem for quasilinear hyperbolic equations with the initial condition in the sense of the general bounded measurable functions. In the above proof, we give the essential assumption: the monotonicity of the initial functions, then we will choose another successive approximate sequence in order to omit this assumption. Now we construct a successive approximate sequence {rn: (x, t)}, {s:n (x, t)}, which satisfies equations ∂r: ∂ 2 r: ∂rn: + +1 (rn: , s:n–1 ) n = : 2n , ∂t ∂x ∂x : 2 s: ∂s ∂ ∂s:n : , s:n ) n = : 2n , + +2 (rn–1 ∂t ∂x ∂x

(3.7.26) (3.7.27)

with the initial boundary conditions rn: |t=0 = r0 (x), rn: |x=0 = ,1 (t), rn: |x=1 = ,2 (t),

(3.7.28)

s:n |t=0

(3.7.29)

= s0 (x),

s:n |x=0

= -1 (t),

s:n |x=1

= -2 (t),

and choose r0 (x, t) = r0 (x), s0 (x, t) = s0 (x). Theorem 3.7.15. If the following conditions are satisfied: (1) +i (r, s) ∈ C4 ; (2) r0 (x), s0 (x) ∈ C4 , ,i (t), -i (t) ∈ C2 , i = 1, 2; (3) the compatible conditions r0 (0) = ,1 (0), r0 (1) = ,2 (0), s0 (0) = -1 (0), s0 (1) = -2 (0), hold, then there exists a classical solution rn: (x, t), s:n (x, t) to the initial boundary value ∂3 ∂2 problem (3.7.26)–(3.7.29), which have continuous derivatives with the type of ∂x 3 , ∂x∂t .

346

3 The Vanishing Viscosity Method of Quasilinear Hyperbolic System

Lemma 3.7.16. We have the following estimates for the solution rn: (x, t), s:n (x, t) to the initial boundary value problem (3.7.26)–(3.7.29): |rn: (x, t)| ≤ K, |s:n (x, t)| ≤ K, (x, t) ∈ RT ,

(3.7.30)

where max{|r0 (x)|, |s0 (x)|, |,i (t)|, |-i (t)|} ≤ K. RT i=1,2

Proof. We set r¯n: = e–!t rn2 (! > 0) and deduce from the maximum theory that |rn: (x, t)| ≤ e!t max{|r0 (x)|, |,i (t)||}. RT i=1,2

Since ! > 0 is an arbitrary positive number, as ! → 0, we get |rn: (x, t)| ≤ max{|r0 (x)|, |,i (t)||} ≤ K. RT i=1,2

Similarly, we deduce that |s:n (x, t)| ≤ K. ∎ Lemma 3.7.17. If the following conditions are satisfied: ¯ s′ (x) ≤ K, ¯ |,′ (t)| ≤ K, ¯ |-′ (t)| ≤ K, ¯ i = 1, 2; (1) r0′ (x) ≤ K, 0 i i (2) +1 (,1 (t), -1 (t)) ≥ l1 > 0, +1 (,2 (t), -2 (t)) ≤ l2 < 0, +2 (,1 (t), -1 (t)) ≥ m1 > 0, +2 (,2 (t), -2 (t)) ≤ m2 < 0; (3) ∂+1 ∂+1 > 0, ≤ 0, ∂r ∂s ∂+2 ∂+2 > 0, ≤ 0, ∂s ∂r

∂+1 = ∂+1 | | | ≤ 1, ∂s ∂r ∂+2 = ∂+2 | | | | ≤ 1, ∂r ∂s

|

then we have the following estimates for the solution rn: (x, t), s:n (x, t) for problem (3.7.26)–(3.7.29) ∂rn: ∂s: ≤ E, n ≤ E, ∂x ∂x $ # ¯ K¯ , K¯ , K¯ , K¯ . where the constant E = max :K, l1 |l2 | m1 |m2 | Proof. Assume ∂rn: ∂x

≤ 0,

∂s:n ∂x

: ∂rn–1 ∂x

≤ E,

∂s:n–1 ∂x

≤ E, which hold true obviously when n = 1. If

≤ 0, the lemma holds. If

∂rn: ∂x

achieves positive maximum value at x = 0,

3.7 On General Solutions of Some Diagonal Quasilinear Hyperbolic Equations

347

then 

|,′1 (t)| l1

∂r:

Thus, ∂xn |x=0 ≤ then by virtue of

≤ E1 . Similarly, if 

we have

∂rn: ∂x |x=1

≤

  ∂rn:  ∂ 2 rn:  ∂rn: : : =: 2 ≤ 0. + +1 (rn , sn–1 ) ∂t ∂x x=0 ∂x x=0 ∂rn: ∂x

achieves positive maximum value at x = 1,

  ∂r:  ∂ 2 r:  ∂rn: + +1 (rn: , s:n–1 ) n  = : 2n  ≥ 0, ∂t ∂x x=1 ∂x x=1

|,′2 (t)| l2

≤ E. If

∂rn: ∂x

achieves positive maximum value in the interior of

RT , then differentiating eq. (3.7.26) with respect to x, we set

∂rn: ∂x

= u:n ,

∂s:n ∂x

= vn: , then



 u: ∂+1 : ∂+1 : ∂ 2 u: un + vn–1 u:n + +1 n = : 2n , ∂r ∂s ∂x ∂x  ∂u:n  ∂ 2 u:n ∂u:n ≤ 0. : 2 – – +1 ∂x ∂t ∂x (x′ ,t′ ) ∂u:n + ∂t

Therefore, we deduce that     ∂+1 : ∂+1 : ∂+1 : ∂+1 : un + vn–1 u:n ≤ 0, un + vn–1 ≤ 0. ∂r ∂s ∂r ∂s Thus, u:n

  ∂rn: ∂+1 = ∂+1 : : = ≤ vn–1 ≤ E. ≤– ⋅ vn–1 ∂x ∂s ∂r

Similarly, we can prove that

∂s:n ∂x

≤ E. We complete the proof.

∎

Theorem 3.7.18. Under the conditions of Lemmas 3.7.16 and 3.7.17, then we can choose a convergent subsequence (rn:-- (x, t), s:n-- (x, t)) from the solution sequence {rn: (x, t)}, {s:n (x, t)} of the initial boundary value problem (3.7.26)–(3.7.29) on t ∈ [0, T], as L1

L1

n- → ∞, :- → 0, rn:-- (x, t) → r(x, t), s:n-- (x, t) → s(x, t) and there exists a subsequence of

∂rn:-∂x

,

∂s:n-∂x

(not relabeled) such that ∂rn:-- L1 weak ∂r ∂s:n-- L1 weak ∂s → → , , ∂x ∂x ∂x ∂x

where the limit functions r(x, t), s(x, t) are bounded variation functions with respect to x and have general derivatives with respect to x. Theorem 3.7.19. If the following conditions are satisfied: (1) r0 (x), s0 (x) are bounded measurable functions;

348

3 The Vanishing Viscosity Method of Quasilinear Hyperbolic System

(2) +1 (,1 (t), -1 (t)) ≥ l1 > 0, +1 (,2 (t), -2 (t)) ≤ l2 < 0, +2 (,1 (t), -1 (t)) ≥ m1 > 0, +2 (,2 (t), -2 (t)) ≤ m2 < 0; (3) ∂+1 ∂+1 > 0, ≤ 0, ∂r ∂s ∂+2 ∂+2 > 0, ≤ 0, ∂s ∂r

∂+1 = ∂+1 | | | ≤ 1, ∂s ∂r ∂+2 = ∂+2 | | | | ≤ 1; ∂r ∂s

|

(4) +i (r, s) ∈ C2 , ,i (t), -i (t) ∈ C2 , i = 1, 2; (5) r0 (0) = ,1 (0), r0 (1) = ,2 (0), s0 (0) = -1 (0), s0 (1) = -2 (0), then there exists a general solution to the initial boundary value problem (3.7.24)– (3.7.25). As for the deduction of Theorem 3.7.19, we have the following two important conclusions. Theorem 3.7.20. For the initial boundary value problem   ∂r #+1 3 – # ∂r + r+ s = 0, ∂t 2 2 ∂x   #+1 3 – # ∂s ∂s + s+ r = 0, ∂t 2 2 ∂x with the initial boundary conditions r|t=0 = r0 (x), s|t=0 = s0 (x), 0 ≤ x ≤ 1, r|x=0 = ,1 (t), r|x=1 = ,2 (t), s|x=0 = -1 (t), s|x=1 = -2 (t), if the following conditions are satisfied: (1) r0 (x), s0 (x) are bounded measurable functions; (2)

#+1 2 ,1 (t)

+

3–# 2 -1 (t)

≥ l1 > 0,

#+1 2 ,2 (t)

+

3–# 2 -2 (t)

≤ l2 < 0,

#+1 2 -1 (t)

+

3–# 2 ,1 (t)

≥ m1 > 0,

#+1 2 -2 (t)

+

3–# 2 ,2 (t)

≤ m2 < 0;

3.8 The Compensated Compactness Method

349

(3) ,i (t), -i (t) ∈ C2 , i = 1, 2, r0 (0) = ,1 (0), r0 (1) = ,2 (0), s0 (0) = -1 (0), s0 (1) = -2 (0); (4) # ≥ 3, then the problem admits a general solution r(x, t), s(x, t). Theorem 3.7.21. Under the conditions of Theorem 3.7.20, for one-dimensional isothermal nonstationary hydrodynamic equations with Lagrangian form: ∂v ∂u – = 0, ∂t ∂x p′ (v) < 0, p′′ (v) > 0. ∂u ∂p(v) – = 0, ∂t ∂x

(3.7.31)

The initial boundary value problem for the corresponding Riemann invarianty equations ∂r ∂r ∂s ∂s – c(v) = 0, + c(v) =0 ∂t ∂x ∂t ∂x admits a general solution r(x, t), s(x, t), where  ∞ r(x, t) = u(x, t) – c(')d', v ∞ ∂p c(')d', c2 (v) = – . s(x, t) = u(x, t) + ∂v v Remark: Theorem 3.7.19 holds true also for the initial value problem (3.7.24).

3.8 The Compensated Compactness Method In Section 3.1, for the quasilinear hyperbolic equation: ∂u ∂f (u) + = 0, ∂t ∂x

(3.8.1)

We use the vanishing viscosity method, that is, we add the viscosity term with a small parameter to eq. (3.8.15) ∂u ∂f (u) ∂ 2u + = : 2, ∂t ∂x ∂x

(3.8.2)

whose solution is u: , which will converge to the general solution to eq. (3.8.15) as : → 0. For this purpose, we must give the uniform estimate for u: with respect to : : and the uniform estimate of ∂u ∂x ≤ E, by using the Helly alternative theorem, and

350

3 The Vanishing Viscosity Method of Quasilinear Hyperbolic System

we deduce that u: converges strongly to the limit function u, which is the general solution to eq. (3.8.15). It’s complicated to consider the quasilinear hyperbolic equations. Obviously, it’s easily seen that the smooth solution for the parabolic equations is given by the discussion in Section 3.4, but the maximum principle does not exist for general parabolic equations; thus it’s not easy to derive the uniform estimate with respect to : for u: and that for the first-order derivative of u: . Generally speaking, we M : √ can derive the estimate for || ∂u , in which case, it’s difficult to apply the ∂x ||L2 ≤ : compensated compactness method to prove the existence of the general solution to eq. (3.8.15) in the old frame of sequential compactness. For the other nonlinear evolution equations, we also frequently meet the following problem: when the sequences {un }, {vn } converge weakly to u, v, respectively, whether un vn converges weakly to uv or not. Generally speaking, if the functional sequence {un } converges weakly to u, the nonlinear functional sequence f (un ) converges weakly to l, but f (u) ≠ l, for example, weak

un → u, u2n → l ≠ u2 ; thus we introduce the following problem: in what circumweak

stances f (u) = l, as un → u, f (un ) → l. The compensated compactness method as the new compactness frame implies that f (lim un ) = lim f (un ) for the nonlinear evolution equations under the conditions of fewer a priori estimates for the solutions. Definition 3.8.1. We say the functional sequence {vn } converges weakly to the function v, if and only if  K

 vn 6dx →

K

v6dx ∀6 ∈ D(K),

where D(K) denotes all infinite differentiable functions with compact support in K, we say the functional sequence {vn } converges weakly ∗ to the function v, if and only if 

 K

vn 6dx →

K

v6dx ∀6 ∈ L1 (K).

Lemma 3.8.2. Assume D is a right angle polyhedron in Rn , f ∈ Lp (D), p ≥ 1, after extending every component of f periodically to the whole space Rn , we deduce that f- = f (-x)

Lp (D) ⇀

1 mesD



f (x)dx = f¯ , - → ∞, D

as p = ∞, we have

∗

f- = f (-x) ⇀

1 mesD



f (x)dx = f¯ , - → ∞. D

Proof. Without loss of generality, we consider D as the unit cube {x : 0 < xi < 1, i = 1, 2, ⋅ ⋅ ⋅, n}, obviously f- ∈ Lp (D). We choose an arbitrary right angle polyhedron D1 = {x : ai < x < bi , i = 1, 2, ⋅ ⋅ ⋅, n} ⊂ D and denote 7D1 by the characteristic function of D1 ,

3.8 The Compensated Compactness Method

351

that is 

1, x ∈ D1 ,

7D1 =

0, x ∉ D1 .

Since D1 is arbitrary, it’s sufficient to prove that 

 7D1 (x)f- (x)dx =

lim

-→∞ D

¯ 7D1 (x)fdx.

D

Since 

 D

7D1 (x)f- (x)dx =





b1

D1

a1

=

1 -n

bn

dx1 ⋅ ⋅ ⋅

f (-x)dx =

f (-x)dxn an



-b1

-a1



dy1 ⋅ ⋅ ⋅

(3.8.3)

-bn

f ( y)dyn ,

-an

(3.8.4)

where 1 -



-bi

-ai

1 = =





[-ai ]+1

[-ai ]+2

+ -ai

 +⋅⋅⋅+

[-ai ]+1

[-bi ] – [-ai ] – 1 -



1

+ 0

1 -



-bi



[-bi ] [-ai ]+1

+ -ai

1 -



-bi

, [-bi ]

[+] denotes the largest integer which is less than +, as - → ∞ in eq. (3.8.12), then the limit of the right-hand side of eq. (3.8.12) is  1  1  n  ¯ (bi – ai ) dy1 ⋅ ⋅ ⋅ f ( y)dyn = 7D1 fdx. 0

i=1

0

D

∎

This completes the proof. Theorem 3.8.3. Set  F(u, K) =

K

f (u(x))dx,

where K ⊂ Rn is an open bounded set, and f : Rm → R is continuous, and we have (1) F is lower semi-continuous for every weak ∗ convergent sequence in every bounded open set K, if and only if f is a convex function, in other words, for the weak ∗ ∗ convergent sequence {u(-) } ∈ L∞ (K)m , u(-)  → u, then when f is a convex function 

 lim

-→∞ K

f (u(-) (x))dx ≥

K

f (u(x))dx,

(3.8.5)

352

3 The Vanishing Viscosity Method of Quasilinear Hyperbolic System

whereas if f is continuous, for every bounded open set K, and every weak ∗ convergent sequence {u(-) }, eq. (3.8.8) holds, then f is a convex function. (2) F is continuous for every weak ∗ convergent sequence in every bounded open set ∗ → u in (L∞ (K))m , and K, if and only if f is a linear function, that is, for every u(-)  ∗ f (u(-) )  → l in L∞ (K), we have l = f (u). Proof. (1) Necessity. Assume that for every sequence {u- }(- = 1, 2, . . .), we have  lim inf u →¯u



-

K

f (u (x))dx ≥

K

¯ f (u(x))dx.

(3.8.6)

We must prove that for any v, 9 ∈ Rm , + ∈ [0, 1], we deduce that f (+v + (1 – +)9) ≤ +f (v) + (1 – +)f (9).

(3.8.7)

Assume D is a unit cube in Rn , D1 is an open subset of D, such that mes D1 = +. We define that 71 = 7D1 is the characteristic function of D1 , then we have  71 (x) =

1, x ∈ D1 , 0, x ∉ D1 .

We extend 71 periodically from D1 to Rn . Applying Lemma 3.8.2, we have in (L∞ (D))m that 71- (x)

∗ = 71 (-x) ⇀

 D

7D1 (x)dx = mes D1 = +, - → ∞.

Define u- (x) = 71 (-x)v + (1 – 71 (-x))9, ∞ m then by eq. (3.8.14) and the definition of 71 , we have in L∞ m (D) = (L (D)) that

∗

u- ⇀ +v + (1 – +)9. We have in L∞ that

∗

f (u- ) = 71- f (v) + (1 – 71- )f (9) ⇀ +f (v) + (1 – +)f (9);

(3.8.8)

3.8 The Compensated Compactness Method

353

thus by eq. (3.8.11), we have 

f (u- (x))dx = [+f (v) + (1 – +)f (9)]mes D ∗ D u- ⇀ u¯  ¯ f (u(x))dx = f (+v + (1 – +)9)mes D. ≥

lim inf

D

(2) Sufficiency. We set  L = lim inf

∗

K

u- ⇀ u¯

f (u- (x))dx.

We need to prove that when f is a convex function, we have that  L≥

K

¯ f (u(x))dx.

Without loss of generality, we choose the subsequence (not relabeled) of {u- } such that  L = lim

∗

K

u- ⇀ u¯

f (u- (x))dx.

(3.8.9)

We introduce Mazur’s lemma, for any fixed : > 0, and an arbitrary natural number -,  N there exists a convex combination of u- : Nk=- !k uk (!k ≥ 0, k=- !k = 1) converges ¯ that is strongly to u, N 

!k uk → u¯ a.e, (N → ∞),

k=-

such that 

 K

¯ f (u(x))dx ≤

K

f

 N

 !k uk (x) dx + :.

(3.8.10)

k=-

By the convexity of f  K

f

 N

  N  !k uk (x) dx ≤ !k f (uk )dx

k=-

k=-

K

and the definition of L, as - → ∞, we have by eq. (3.8.10) that  K

¯ f (u(x))dx ≤ L.

For f and –f , we apply (1) to deduce (2).

∎

354

3 The Vanishing Viscosity Method of Quasilinear Hyperbolic System

Definition 3.8.4. We say that the function f : Rm → R A-quasiconvex function, provided 

 f (, + & (x))dx ≥ D

f (,)dx = f (,)mes D,

(3.8.11)

D

where , ∈ Rm , the cube D ⊂ Rn , & ∈ L(D),    L(D) = & ∈ {L∞ (D)}m ; & (x)dx = 0, & ∈ Ker A , & ∈ Ker A :

 j,k

D

∂&j ai,j,k = 0. ∂xk

we give some restrictions for the derivations of the sequence {u- }. Assume f : Rm → R is continuous, K ⊂ Rm is a bounded open set, we give some assumptions on the sequence {u- } ⊂ {L∞ (u)}m , u- = (u-1 , ⋅ ⋅ ⋅, u-m ): ⎧ - ∗ m ⎪ ⎪ ⎪ u ⇀ u in {L∞ (K)} , ⎪ ⎪ ⎪ m  n ⎨  ∂u-j ai,j,k is bounded in {L2 (K)}q , i = 1, 2, ⋅ ⋅ ⋅, q, H : Au = ∂xk ⎪ ⎪ j=1 k=1 ⎪ ⎪ ⎪ ⎪ ⎩ - ∗ f (u ) ⇀ l in L∞ (K). Theorem 3.8.5. For any functional sequence {u- } satisfying the condition H, l ≥ f (u) holds, that is  lim inf -→∞

K

f (u- (x))dx ≥

 K

f (u(x))dx.

(3.8.12)

so f is a A-quasiconvex function. Proof. Suppose D is a unit cube, & ∈ L(D), & extend to the whole space according to each variable with periodic 1, for the definition of integer & - (x) = & (-x). Thus, ∗ 0 is in {L∞ (D)}m , &- ⇀

and & - ∈ KerA, namely we have  j,k

aijk

∂&j∂xk

= 0, i = 1, 2, . . . , q.

3.8 The Compensated Compactness Method

355

Because & has periodic 1, then 

 1 f (, + & (x))dx = n f (, + & ( y))dy D  -D f (, + & ( y))dy. = -

(3.8.13)

D

Set - → ∞ in eq. (3.8.13) and because of eq. (3.8.12), we have 



f (, + & - ( y))dy ≥ f (,)mesD.

f (, + & ( y))dy = lim inf -→∞

D

D

∎

The proof is completed.

By Theorems 3.8.3 and 3.8.4 we find the convexity of f ⇒ lower semicontinuity of f ⇒ A-quasiconvexity of f . Theorem 3.8.6. Suppose u- , u satisfy condition H, and u- – u ∈ KerA. If f is A-quasiconvex, then eq. (3.8.12) holds, that is  lim inf -→∞

K

f (u- (x))dx ≥

 K

f (u(x))dx ∀K ⊂ Rn .

Proof. If we use the sum set of square with the boundary of 1/k to approximate K, namely ⎧ I > ⎪ ⎪ ⎪ = Dki , H ⎪ k ⎪ ⎪ ⎨ i=1 mes(K – Hk ) → 0, k → ∞, ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ mesD = 1 , 1 ≤ i ≤ I. ki kn

(3.8.14)

For x ∈ Hk , let uk (x) =

1 mesDki

 Dk

u(& )d& , x ∈ Dki , 1 ≤ i ≤ I,

(3.8.15)

i

f (u- ) – f (u) = f (u + (u- – u)) – f (uk + (u- – u)) + f (uk + (u- – u)) – f (uk ) + f (uk ) – f (u).

(3.8.16)

356

3 The Vanishing Viscosity Method of Quasilinear Hyperbolic System

From eq. (3.8.15), for any given % > 0, k is sufficiently large, we get  % |f (u + (u- – u)) – f (uk + (u- – u))|dx ≤ , 2 Hk  % |f (u) – f (uk )|dx ≤ . 2 Hk From eqs (3.8.16) to (3.8.18) it follows that   f (u (x))dx – Hk

(3.8.18)

f (u(x))dx + %

Hk



(3.8.17)

(3.8.19)

[f (uk + (u- – u)) – f (uk )]dx.

≥ Hk

From hypotheses H and H0 , we obtain ∗ 0 in L∞ & - ≡ u- – u ⇀ m (Hk ),

and & - ∈ KerA. Setting '- (x) = & - (x) –



& - ( y)dy, x ∈ Hk ,

(3.8.20)

Hk

Thus, ∗ 0 in {L∞ (Hk )}m , - → ∞, '- ⇀ m '- (x)dx = 0, '- ∈ KerA,

(3.8.21) (3.8.22)

Hk

|'- (x) – & - (x)| → 0, - → 0. From eq. (3.8.19) it follows that   f (u- (x))dx – Hk

(3.8.23)

f (u(x))dx + %

Hk



[f (uk + '- (x)) – f (uk + '- (x))]dx

≥ Hk



+

f (u + '- (x))dx –

Hk



(3.8.24)

 f (uk )dx .

Hk

Because f is A-quasiconvex, the last curly braces item on the right of (3.8.24) is nonnegative in Dki , thus it is also nonnegative in Hk . So let - → ∞ in eq. (3.8.22) and note (3.8.23), we have   f (u (x))dx ≥ f (u(x))dx – %. (3.8.25) lim inf -→∞

Hk

Hk

3.8 The Compensated Compactness Method

357

Because of eq. (3.8.14) and the arbitrary of %, then eq. (3.8.12). The proof of theorem is completed. ∎ Define operator B(. ) : Rm → Rq (. ∈ Rn ) B(. )+ =

n m  

aijk +j .k .

j=1 k=1

Let N = {(+, . ) ∈ Rm × Rn : B(. )+ = 0} ⊂ Rm × Rn , D = {+ ∈ Rm , ∃. ∈ Rn – {0}, (+, . ) ∈ N} ⊂ Rm . Theorem 3.8.7. (1) For each sequence {u- } satisfying condition (H), if l ≥ f (u), for any H ∈ Rm and A ∈ D, f (H + tA) is convex function of t. (2) For each sequence {u- } satisfying condition (H) there are l = f (u), then f (H + tA) is a linear function of t, where H ∈ Rm , A ∈ D. Proof. (2) is the deduction of (1), we only need to prove (1). Suppose + ∈ (0, 1), for A ⊂ D, set F = H + A, G = H –

+ A; 1–+

thus, +F + (1 – +)G = H, F – G ∈ D. Because A ∈ D, there exists 0 ≠ . ∈ Rn such that m  n 

aijk Aj .k = 0 ∀i = 1, 2, . . . , q.

j=1 k=1

Let D be a unit cube in Rn , D1 ⊂ D, mesD1 = +, if 6 : R → R, 6 ∈ L∞ , and ⎧ x ∈ D1 , ⎨ 1, 6(x ⋅ . ) = –+ ⎩ , x ∈ D – D1 . 1–+ Let & (x) = A6(x⋅. ) , then & ∈ L(D), that is,

 & ∈ L∞ (D),

& (x)dx = 0, D

 j,k

aijk

∂&j = 0. ∂xk

358

3 The Vanishing Viscosity Method of Quasilinear Hyperbolic System

Applying Theorem 3.8.5, then  f (H + & (x))dx ≥ f (H)mesD = f (H). D

From the structure of & (x), we know the above inequality is   f (F)dx + f (G)dx ≥ f (+F + (1 – +)G). D1

D–D1

So we have +f (F) + (1 – +)f (G) ≥ f (+F + (1 – +)G). ∎

The proof of theorem is completed. Theorem 3.8.8. Suppose that M : Rm → Rm is a symmetric matrix, let f (a) = Ma, a,

where a ∈ Rm , ⋅, ⋅ denotes quantitative product in Rm . If sequence {u% } satisfies condition (H1 ): ⎧ ⎪ u% ⇀ u((L2 (K))m , % → 0), ⎪ ⎪ ⎪ ⎪ ⎨ f (u% ) ⇀ l(in the distribution sense, % → 0), (H1 )  ∂u%j ⎪ ⎪ –1 ⎪ Au% = a in Wloc,2 (K), i = 1, 2, ⋅ ⋅ ⋅, q. ⎪ ijk ⎪ ⎩ ∂xk jk

Set D=

⎧ ⎨

+ ∈ Rm : ∃. ∈ rn – {0}, such that

⎩

 jk

⎫ ⎬

aijk +j .k = 0 , ⎭

thus, (1) If f (+) ≥ 0, ∀+ ∈ D, we have l ≥ f (u). (2) If f (+) = 0, ∀+ ∈ D, we have l = f (u). Proof. We only need to apply the conclusions of (1) about f , –f , respectively, then we can deduce (2). Now we give the proof of (1). Set v% = u% – u, so for 6 ∈ C0∞ (K) we have w% = 6v% .

3.8 The Compensated Compactness Method

359

It is not difficult to verify ⎧ % w ⇀ u((L2 (K))m , % → 0), ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ∂wj% aijk → 0 (W2–1 (K), i = 1, 2, ⋅ ⋅ ⋅, q, % → 0), ∂x ⎪ k ⎪ ⎪ jk ⎪ ⎪ ⎩ % w is a certain compactly supported functions on K ⊂ Rn .

(3.8.26)

We only need to prove  lim inf %→0

Rn

Mw% ; w% dx ≥ 0.

(3.8.27)

Step 1: we apply Fourier transform on w% (x): ' w% (. ) =



w% (x)e–20i. ⋅x dx, Rn

By eq. (3.8.26) (because e–20i. ⋅x ∈ L2 (K)), we have 

' w% (. ) → 0 a.e,

(3.8.28)

|' w% (. )| ≤ 3. So 3 is a constant, thus ' w% → 0 (strongly converges in Lloc,2 (Rn ), % → 0).

(3.8.29)

By hypothesis (3.8.26), we have 1  aijk ' wj% (. )%k → 0({L2 (Rn )}q , % → 0). 1 + |. |

(3.8.30)

jk

Step 2: we extend f (w) = Mw; w from Rm to Cm and define 9 f (w) = Mw; w.

(3.8.31)

For + = +1 + i+2 ∈ D + iD, we have Re9 f (+) = Re(M+; +) = M+1 ; +1  + M+2 ; +2  ≥ 0. By Plancherel formula, and because f (w% (x)) is real, 

%



f (w (x))dx = Rn

Rn

9 f (' w% (. ))d. =

 Rn

Re9 f (' w% (. ))d. .

(3.8.32)

360

3 The Vanishing Viscosity Method of Quasilinear Hyperbolic System

So we only need to prove  lim inf %→0

Rn

Re9 f (' w% (. ))d. ≥ 0.

(3.8.33)

Step 3: we will prove that there exists a constant C! > 0 for ! > 0 such that Re9 f (+) ≥ –!|+|2 – C!

  q  2  aijk +j 'k  . i=1

(3.8.34)

jk

Otherwise, then we have !0 > 0, C! = -, +- ∈ Cm , |+-|=1 and '- ∈ Rn , |'- | = 1 such that  2   - 2 - - 9  Re f (+ ) < –!0 |+ | – aijk +j 'k  . (3.8.35)  i

jk

We select subsequences +- → +∞ , '- → '∞ (- → ∞) and divide eq. (3.8.35) by -, and let - → ∞, then 2     - -  aijk +j 'k  → 0, - → ∞. (3.8.36)  i

jk

So we have 

∞ aijk +∞ j 'k = 0;

(3.8.37)

jk

hence, +∞ ∈ D + iD. Because Re9 f (+) ≥ 0, + ∈ D + ID, then we have Re9 f (+∞ ) ≥ 0. But by eq. (3.8.35) we have Re9 f (+∞ ) ≤ –!0 < 0. This contradiction led to the establishment of eq. (3.8.34). Step 4: because of    Re9 f (' w% (. ))d. = Re9 f (' w% )d. + Re9 f (' w% )d. , |. |≤1

Rn

(3.8.38)

|. |>1

and ' w% → 0 (strongly converges in L2loc (Rn ), % → 0) in eq. (3.8.28), we have  |. |≤1

Re9 f (' w % )d. → 0, % → 0.

(3.8.39)

3.8 The Compensated Compactness Method

361

By the result of step 3 we take + = ' w% (. ), ' = . /|. |, then we have    .k 2 %  ' Re9 f (' w% (. )) ≥ –!|' w% (& )|2 – C! a (. ) . w ijk j  |. |  i

Integrating the equation, then   Re9 f (' w% (. )) ≥ – ! |. |>1

|. |>1

|' w% (. )|2 d.

 – C!

By eq. (3.8.30), we have 

Because ! is arbitrary and

|. |>1

   .k 2 %  ' a (. ) d. . w ijk j  |. |  i

Re9 f (' w% (. ))d. ≥ –!

|. |>1



w% (. )|2 d. |. |>1 |'  lim inf %→0

|. |>1

jk

(3.8.40)

jk

 |. |>1

|' w% (. )|2 d. .

is bounded, thus

Re9 f (' w% (. ))d. ≥ 0.

(3.8.41)

We obtain eq. (3.8.33) by combining eqs (3.8.29) and (3.8.41). The proof is completed.

∎

Corollary 3.8.9. Let u% (x1 , x2 , ⋅ ⋅ ⋅, xn ) = (v1% , ⋅ ⋅ ⋅, vn% , w1% , ⋅ ⋅ ⋅, wn% ) ⇀ (v, w)(L2 (K)), and if div v% =

n  ∂v% i

i=1

rot w% =



∂xi

is bounded in L2 (K),

% ∂wi% ∂wj – ∂xj ∂xi

 is bounded in L2 (K),

then there is v% , w%  → v, w(in the distribution sense).

(3.8.42)

Proof. U = (u, w) ∈ Rn × Rn (m = 2n), Au = (div v, rot w), then we have ⎧ n  ⎪ ⎪ ⎨ +i .i = 0, ⎪ ⎪ ⎩

i=1

,j .i – ,i .j = 0, i, j = 1, 2, ⋅ ⋅ ⋅, n.

(3.8.43)

362

3 The Vanishing Viscosity Method of Quasilinear Hyperbolic System

So 

V = {((+,), . ) ∈ R2n × Rn ; +. , ,. }, D = {(+, ,) ∈ R2n , +,} ⊂ R2n .

(3.8.44)

By Theorem 3.8.7, we know that the necessary condition that makes f to be weak continuity is that f (a + t+, b + t,) be a linear function of t, ∀a, b ∈ Rn , (+, ,) ∈ D. So we can take f (v, w) = v, w. Because f satisfies all conditions of Theorem 3.8.8, we obtain the conclusion of corollary. ∎ Corollary 3.8.10. Let u% (x1 , x2 ) = (v% (x1 , x2 ), w% (x1 , x2 )) ⇀ (v, w)({L2 (K)}2 , % → 0), and ∂v% ∂w% ∂x1 and ∂x2 is bounded in L2 (K), then there is v% w% ⇀ vw(in the distribution sense, % → 0).

(3.8.45)

Proof. At this time D = {(+, ,) ∈ R2 , + = 0 or , = 0}. If f (v, w) = vw, then it satisfies all hypothesis of Theorem 3.8.8, thus eq. (3.8.45) holds. ∎ Now if we consider the convergence problem of f (u% ) for weakly convergent sequence {u% }(% → 0), in order to introduce parameterized measure, then - is a probability measure on space Rm , + ⊂ Rn , f : Rm → R is continuous, which can be denoted by  -, f (+) =

f (+)-(d+). Rm

Now we suppose that the above probability measure depends on x, x ∈ K ⊂ Rm , which is denoted by -x , -x is measured depending on x which refers -x (E) to be a measurable function of x for every Borel set E ⊂ Rm . Theorem 3.8.11. If there are bounded set K ⊂ Rm , open set K ⊂ Rm . Suppose that f : Rm → R is continuous, then (1) Suppose that the sequence {us } : K → Rm , us (x) ∈ K a.e., then there exists a subsequence of {us } which is also denoted by {us } and a family of parameterized probability measures {-x }x∈K , such that supp -x ⊂ R,

(3.8.46)

f (us ) ⇀ f (L (K), s → ∞),

(3.8.47)

∗

∞

3.8 The Compensated Compactness Method

363

where  f = Rm

-x (+)f (+)d+.

(3.8.48)

(2) On the contrary, if vx is a family probability measure which depends on x, supp -x ⊂ K, then there exists sequence {us }(us : K → Rm , us (x) ∈ K, a.e.), such that for any continuous function f : Rm → R, f (us ) ⇀∗ f , s → ∞, where f satisfies eq. (3.8.48). Proof. (1) For 6(x, +) ∈ C0 (K × Rm ), the integral  K

6(x, us (x))dx

defines a linear bounded function on C0 (K × Rm ). Due to Riesz theorem, there exists the unique regular Borel meausre ,s such that  ,s , 6(x, +) =

K

6(x, us (x))dx.

(3.8.49)

Denote the space consisting of all the regular Borel measure on K × Rm by M(K × Rm ). The set {,s } is bounded in M(K × Rm ). Take a weak ∗ convergence subsequence, also denoted by {,s }, which has the limit , ∈ M(K × Rm ), namely ,s ⇀ ,, i.e. ,s , 6 → ,, 6 ∀6 ∈ C0 (K × Rm ). It is easy to verify the following properties of ,: (!) , ≥ 0, namely for any 6 ≥ 0, we have  ,, 6 = lim (,s , 6) = lim s→∞

s→∞ K

6(x, us (s))dx ≥ 0.

( ") supp(,) ⊂ K × K, namely for any 6 such that 6 = 0 on K × K, we have  ,, 6 = lim

s→∞ K

6(x, us (s))dx = 0.

( #) projK , = dx, namely if 6(x, +) = 8(x), we have 

 ,, 6 = lim

s→∞ K

8(x)dx =

K

8(x)dx.

(3.8.50)

364

3 The Vanishing Viscosity Method of Quasilinear Hyperbolic System

It follows from properties (!), ("), (#) that , is absolutely continuous on the Lebesgue measure. With the help of the Radon–Nikodym theorem, there exists a family of probability measure {-x }, supp -x ⊂ K, such that  ,= -x dx, (3.8.51) K

thus

  ,, 6 =

K

Rm

6(x, +)-x (+)d+dx.

(3.8.52)

For any f : Rm → R, f ∈ C0 (Rm ), let 8(x) ∈ C0 (K), then due to eqs (3.8.50) and (3.8.52),    ,s , 8f  = 8(x)f (,s (x))dx → ,, 8f  = 8(x)f (+)-x (+)dxd+. K

K

So

Rn

 -x (+)f (+)d+.

f (x) = Rm

(2) On the other hand, set

 ,=

K

-x dx.

We are going to prove that there exists a sequence {us }(us : K → K), such that for any continuous function f : K → R,  f (+)ux (+)d+ (L∞ (K)). f (us ) ⇀∗ f = K

Thus, we only need to prove that M = {m ∈ M(K × Rm ) : m corresponds to a measurable function u : K → K}, N = {, ∈ M(K × Rm ) : (!) , ≥ 0, (") supp , ⊂ K × K, (#) projK , = dx}. M = N in the weak sense, which will complete the proof. In the proof of the first part, we have actually proved that M ⊂ N. So it is only needed to prove N ⊂ M. Thus, first we prove that M is convex and then prove that N ⊂ conv M (the convexity closure of M). Step 1. Suppose m1 , ⋅ ⋅ ⋅, mp ∈ M, then there exists the measurable function u1 , ⋅ ⋅ ⋅,up : K → K such that  6(x, ui (x))dx ∀6 ∈ C0 (K × Rm ). mi , 6 = K

We only need to prove 7i , such that

P

i–1 +i mi

∈ M, ∀+i ≥ 0,



+i = 1. It is easy to find eigenfunction

3.8 The Compensated Compactness Method

x

7i% (x) = 7i

%

∗ +i (L∞ (K)), ⇀

i = 1, ⋅ ⋅ ⋅, p.

365

(3.8.53)

Let v% (x) =

P 

7i% (x)ui (x),

i–1

then for each 6 ∈ C0 (K × Rm ), 6(x, v% (x)) =

P 

7i% (x)6(x, uj (x)).

(3.8.54)

i–1

Let % → 0, it follows from eqs (3.8.53) and (3.8.54) that 

%

K

6(x, v (x))dx →

P   i–1

K

+i 6(x, ui (x))dx =

P 

+i mi , 6.

(3.8.55)

i–1

Let m% have the following form: m% , 6 =

 K

6(x, v% )dx,

 then eq. (3.8.55) indicates that Pi–1 +i mi ∈ M. Step 2. We prove convex closure convM of N ⊂ M. Due to the Hahn–Banach theorem, the convex closure of M is the intersection of all the closed half-space containing M, which can be characterized by 60 ∈ C0 (K × Rm ), Q0 ∈ R satisfying ,, 60 (x, +) + a0 ≥ 0

∀, ∈ M,

namely  K

60 (x, u(x))dx + a0 ≥ 0 ∀, : K → K.

Thus, -

{m, 60  + a0 ≥ 0, ∀m ∈ M}  -  = 60 (x, u(x))dx + a0 ≥ 0, for all the measurable function u : K → K ,

conv M =

K

(3.8.56) where 60 ∈ C0 (K × Rm ), a0 ∈ R. Set , ∈ N, we need to prove , ∈ conv M which is equivalent to prove that: if 60 ∈ C0 (K × Rm ), a0 ∈ R, then

366

3 The Vanishing Viscosity Method of Quasilinear Hyperbolic System

,, 60  + a0 ≥ 0.

(3.8.57)

Define 80 (x) = inf(60 (x, +)), +∈K

70 (x, +) = 60 (x, +) – 80 (x) ≥ 0, thus eq. (3.8.57) implies  K

80 (x)dx + a0 ≥ 0.

Therefore, it follows from , ∈ N and eq. (3.8.58) that ,, 60  + a0 = ,, 70  + ,, 80  + a0 ≥ ,, 80  + a0 =

(3.8.58)

 K

80 (x)dx + a0 ≥ 0.

This means eq. (3.8.57), namely N ⊂ conv M, which completes the proof.

∎

∗ u(L (K)), then u → u(strongly) in L (K)(p < ∞), if and Corollary 3.8.12. Suppose us ⇀ ∞ s p only if -x = $u(x) ($u(x) denotes the Dirac measure at u(x)).

Proof. (1) Suppose us → u (strongly) in Lp , then it follows from Theorem 3.8.11,  f (u(x)) = Rm

-x (+)f (+)d+

(3.8.59)

for any f , thus -x = $u(x) . (2) If -x = $u(x) , let f (+) = |+|2 , then due to Theorem 3.8.11, ∗ u2s ⇀ |u|2 (L∞ (K)) ∗ u(L (K)), we deduce that u converges strongly to u in {L (K)}m . Similarly, if and us ⇀ ∞ s 2 4 f = |+| , |+|8 , . . . in proper order, one can get the strong convergence for any p < ∞. This completes the proof. ∎

In the following we will apply some results about the theory of compensated compactness to the first-order quasilinear hyperbolic equations and prove the existence results of generalized solutions. Consider the first-order quasilinear hyperbolic equation ∂u ∂f (u) + = 0, ∂t ∂x

x ∈ R, t > 0,

(3.8.60)

where f : R → R is a given continuous function. Consider the Cauchy problem, namely it satisfies eq. (3.8.60) with the initial condition

3.8 The Compensated Compactness Method

u(x, 0) = u0 (x),

x ∈ R.

367

(3.8.61)

Generally speaking, there is no globally smooth solution of problem (3.8.60) and (3.8.61), even the initial function u0 (x) is sufficiently smooth. Thus, we consider the weak solution of eq. (3.8.60): u(x, t) is a bounded measurable function satisfying  ∞ ∞ (u6t + f (u)6x )dxdt = 0, (3.8.62) 0

–∞

where 6 ∈ C0∞ (R × (0, ∞)). Similarly, consider the weak solution of problem (3.8.60) and (3.8.61). u is the bounded measure function satisfying  ∞ ∞  ∞ (u6t + f (u)6x )dndt + u0 (x)6(x, 0)dx = 0. (3.8.63) 0

–∞

–∞

As the requirements of a physicist and mathematician, to ensure the uniqueness of the weak solution of eqs (3.8.60) and (3.8.61), we require the solution satisfying the entropy condition: u(x, t) satisfies the following in the generalized sense: '(u)t + q(u)x ≤ 0,

(3.8.64)

where '(u) is convex function named convex entropy. q(u) is called entropy flow and satisfies the compatible condition q′ (u) = f ′ (u)'′ (u).

(3.8.65)

Theorem 3.8.13. Suppose that K → R2 is a bounded open set, f : R → R is a smooth function. Moreover suppose ∗ u(L∞ (u), % → 0), (1) the sequence u% : u% ⇀ (2) –1 '(u% )t + q(u% )x is compact in W2,loc .

(3.8.66)

where '(u) is any convex function that satisfies eq. (3.8.65). Then we have (1) ∗ f (u), f (u% ) ⇀

L∞ (u), % → 0

(3.8.67)

(2) f ′ (u% ) → f ′ (u) is strongly convergent in Lp (K)

∀p < ∞.

(3.8.68)

Moreover, if it does not exists as an interval, f is a linear function on it, then u% → u

is strongly convergent in Lp (K)

∀p < ∞.

(3.8.69)

368

3 The Vanishing Viscosity Method of Quasilinear Hyperbolic System

Proof. Step 1: we first prove eq. (3.8.67), for any fixed convex ', consider sequence U % = (u% , f (u% ), '(u% ), q(u% )) ∈ {L∞ (K)}4 .

(3.8.70)

Because {u% } is bounded in L∞ , there exists a subsequence such that ∗ U% ⇀ U(u, v, w, z)

in

{L∞ (K)}4 .

(3.8.71)

We are going to prove v(x, t) = f (u(x, t)) a.e.

(3.8.72)

It follows from eq. (3.8.66), and let v% = f (u% ),

w% = '(u% ),

z% = q(u% ),

(3.8.73)

then ∂u% ∂v% –1 . + is compact in W2,loc ∂t ∂x

(3.8.74)

∂w% ∂z% –1 . + is compact in W2,loc ∂t ∂x

(3.8.75)

Set '(x) = x in eq. (3.8.74) and D = {(!, ", #, $) ∈ R4 , !$ – "# = 0},

(3.8.76)

then it follows from Theorem 3.8.8 that ∗ Q(U % ) = u% z% – v% w% ⇀ uz – vw

(L∞ (K)).

(3.8.77)

(L∞ (K)).

(3.8.78)

It follows from eq. (3.8.73) that ∗ u% q(u% ) – f (u% )'(u% ) ⇀ uz – vw

We rewrite eq. (3.8.78) as the parameterized measure form, due to Theorem 3.8.11, there exists family of probability measure -xt such that ⎧ a.e. u(x, t) = -xt ; + ⎪ ⎪ ⎪ ⎪ ⎨ v(x, t) = -xt ; f (+) a.e. ⎪ w(x, t) = -xt ; '(+) a.e. ⎪ ⎪ ⎪ ⎩ z(x, t) = -xt ; q(+) a.e. Due to eq. (3.8.78) and Theorem 3.8.8 that

(3.8.79)

3.8 The Compensated Compactness Method

-; +q(+) – f (+)'(+) = -, +-, q(+) – -, f (+)-, '(+),

369

(3.8.80)

where - = -xt . Combining eqs (3.8.79)–(3.8.80), we get -; (+ – ,)q(+) – ( f (+) – -)'(+) = 0.

(3.8.81)

The above equation is established for all convex function ' and q′ (+) = f ′ (+)'′ (+). Now we specifically choose ' as '(+) = |+ – u|.

(3.8.82)

We immediately obtain  q(+) =

f (u) – f (+), + ≤ u, f (+) – f (u), + ≥ u.

(3.8.83)

So we have (+ – u)q(+) – ( f (+) – v)'(+) = (v – f (u))|+ – u|. Substituting eqs (3.8.82)–(3.8.83) into eq. (3.8.81), we have (v – f (u))-; |+ – u| = 0.

(3.8.84)

Due to eq. (3.8.84) v = f (u). In fact (i) If -; |+ – u| ≠ 0, we have v = f (u). (ii) If -; |+ – u| = 0, we have - = $u ; then v = f (u). Step 2. In order to prove eqs (3.8.68)–(3.8.69), we prove that the support set of contains in interval that made f a linear function. Without loss of generality, u(x, t) = f (u(x, t)) = 0 on (x, t). Now eq. (3.8.81) becomes -; +q(+) – f (+)'(+) = 0

(3.8.85)

for all convex function '. By eqs (3.8.79)–(3.8.72) and u = f (u) = 0, we have 

-; + = 0, -; f (+) = 0.

(3.8.86)

If we take !, " then conv (supp -) = [!, "],

(3.8.87)

370

3 The Vanishing Viscosity Method of Quasilinear Hyperbolic System

where conv M denotes the convex closed set of M. It follows from eq. (3.8.86) that there must be ! ≤ 0 ≤ ". If ! = 0, then " = 0. Now - is the Dirac measure and the problem has been solved. Thus, we only consider the case ! < 0 < " in the following. Define g(+), h(+) bounded variation function. Let  g(+) =

+

!

 h(+) =

!

,-d,,

(3.8.88)

f (,)-d,.

(3.8.89)

+

Due to eq. (3.8.86), we can set g, h = 0 outside [!, "]. It follows from eq. (3.8.88) that g(+) < 0, + ∈ (!, "). It follows from eq. (3.8.85) that g′ ; q(+) – h′ ; '(+) = 0.

(3.8.90)

–g; q′  + h; '′  = 0.

(3.8.91)

h – gf ′ ; '′  = 0.

(3.8.92)

So

It follows from q′ = f ′ '′ that

The above is correct for any convex function '. Thus, for any increasing function '′ is correct. Since eq. (3.8.92) is linear with respect to '′ , so does the difference of two increasing functions. Thus it is correct for any smooth functions. Therefore, h – gf ′ = 0.

(3.8.93)

f (+)g′ – +h′ = 0.

(3.8.94)

From eqs (3.8.88)–(3.8.89),

Combine eqs (3.8.93)–(3.8.94), then ( f (+)g – +h)′ = 0.

(3.8.95)

Since g, h are zero outside [!, "], thus from eq. (3.8.95) f (+)g – +h = 0.

(3.8.96)

Due to eq. (3.8.93) and g(+) < 0, + ∈ (!, "). It follows from eq. (3.8.96) that f (+) – +f ′ (+) = 0,

+ ∈ (!, ").

(3.8.97)

371

3.8 The Compensated Compactness Method

So f (+) = c+

∀+ ∈ (!, ").

(3.8.98)

Thus f ′ is constant on (!, "). Therefore f ′ (u% ) converges strongly to f ′ (u). In the end, if there is no interval on which f (u) is a linear function, then from eqs (3.8.87) and (3.8.98) we have -xt = $u(x,t) . Then eq. (3.8.69) is correct from Corollary 3.8.12. This completes the proof. ∎ Theorem 3.8.14 (Existence Theorem). Let u0 ∈ W 1,∞ (R), then 

ut + f (u)x = 0,

x ∈ R, t > 0,

u(x, 0) = u0 (x)

(3.8.99)

has a weak solution u ∈ L∞ . Proof. Step 1. Consider parabolic regularized problem of eq. (3.8.99): 

u%t + f (u% )x – %u%xx = 0, u% (x, 0) = u0 (x).

(3.8.100)

For problem (3.8.100), using the above results, we know that problem (3.8.100) has a classical solution u% and for any open set K ⊂ R × R+ u% is uniformly bounded in L∞ (K), √ % %ux is uniformly bounded in L2 (K).

(3.8.101) (3.8.102)

Step 2. Choose a subsequence of eq. (3.8.101) such that u% ⇀ u (L∞ (K)).

(3.8.103)

Our goal is to prove that the limit function u(x, t) is a weak solution of eq. (3.8.99). So in order to prove Theorem 3.8.14, we need to prove –1 'u% t + q(u% )x is compact in W2,loc (K),

(3.8.104)

where ' is a convex function and q(u) satisfies eq. (3.8.65). First, we note that u% is a classical solution for problem (3.8.100), so we have √ '(u% )t + q(u% )x = %'(u% )xx – '′′ (u% )( %u%x )2 .

(3.8.105)

372

3 The Vanishing Viscosity Method of Quasilinear Hyperbolic System

–1 (K), we In order to verify that the right-hand side of eq. (3.8.105) is compact in W2,log will use the following Murat’s lemma.

Lemma 3.8.15 (Murat). Suppose that K is open set in Rn , there are generalized function sequences {g % } satisfying –1 (K), r > 2, (1) {g % } is bounded in Wr,loc –1 (K), q > 1, {g % } is a bounded set in M(K), (2) g % = g1% + g2% , {g1% } is a compact set on Wq,loc 2 where M(K) denotes the space composed by regular Borel measure. Then {g% } is –1 (K). compact set in W2,loc –1 (K), we only need to Therefore, in order to prove that eq. (3.8.105) is compact in W2,loc proof enough –1 ∪ the bounded set of M(K)) '(u% )t + q(u% )x ∈ (the compact set of W2,loc –1 ( the bounded set of W∞ )

(3.8.106)

–1 (K) is obvious, because u% is uniformly The bounded set of '(u% )t + q(u% )x ∈ W∞ ∞ bounded in L (K). Therefore, we only need to prove that %'(u% )xx in eq. (3.8.105) √ is a compact set in W2–1 (K), '′′ (u% )( %u%x )2 is a bounded solution for M(K). But √ '′′ (u% )( %u%x )2 is uniformly bounded in L1 (K) because of estimate (3.8.102). So we only need to prove

%'(u% )xx W –1 → 0, 2

In fact, %'(u% )xx W –1 = sup 2

6∈N

 K

% → 0.

 %'(u% )xx 6(x, t)dxdt ,

(3.8.107)

(3.8.108)

where N = {6 ∈ H01 (K), 6W 2 (K) ≤ 1}. Using partial integral formula and Schwarz 1 inequality in eq. (3.8.108), we have %'(u% )xx W –1 2 √ √ ≤ % %u%x L2 '′ (u% )L∞ sup (6x L2 ) → 0, % → 0. 6∈N

Thus eq. (3.8.106) holds. Using Theorem 3.8.13, we get the conclusion of the theorem. In fact, from eq. (3.8.100) we have    ∂6 ∂6 ∂ 26 + f (u% ) + %u% 2 dxdt u% ∂t ∂x ∂x R2+ (3.8.109)  ∞ + –∞

Take the limit % → 0.

u0 (x)>(x, 0)dx = 0 ∀> ∈ C0∞ (R2+ ).

373

3.9 The Existence of Generalized Solutions for the First-Order Quasilinear Hyperbolic

If f is not a linear function in any interval, u% strongly converges to u in Lp , ∀6 ∈ C0∞ (R2+ ), > ≥ 0. For any convex entropy and entropy flow, it is '(u), q(u) obtained by eq. (3.8.105):    ∂> ∂> ∂ 2> + q(u% ) + %'(u% ) 2 dxdt '(u% ) ∂t ∂x ∂x R2+  % 2  ∂u %'′′ (u% ) dxdt ≥ 0. = 2 ∂x R+ If % → 0, we have  R2+

 ∂> ∂> + q(u) dxdt ≥ 0, '(u) ∂t ∂x



(3.8.110)

namely the weak solution u satisfies the entropy condition.

∎

3.9 The Existence of Generalized Solutions for the First-Order Quasilinear Hyperbolic System Now we consider applying compensated compactness method to the following firstorder quasilinear hyperbolic equation: ∂u ∂f (u) + = 0, ∂t ∂x

(3.9.1)

where u and f (u) are all n-dimensional vectors, u = (u1 , ⋅ ⋅ ⋅, un )T , f = ( f1 , ⋅ ⋅ ⋅, fn )T . T is transpose and f (u) is sufficiently smooth. We denote Jacobi matrix with f ′ = grad f . If the eigenvalue of f ′ are all real numbers, eq. (3.9.1) is called hyperbolic, if the eigenvalue of f ′ are all real numbers and not equal, eq. (3.9.1) is called strictly hyperbolic. With +i (u), li (u), ri (u)(i = 1, 2, ⋅ ⋅ ⋅, n), respectively, denote ith eigenvalues, left eigenvector and right eigenvector, for a component i, at u ∈ Rn we have +′i (u) ⋅ ri (u) ≠ 0,

(3.9.2)

Equation (3.9.2) at u for the ith group eigenvalues is genuinely nonlinear. If for all i, j = 1, 2, ⋅ ⋅ ⋅, n and the considered set K, u ∈ K, (3.9.2) also holds, then eq. (3.9.1) is genuinely nonlinear. If the function W : Rn → R satisfies the equation W ′ (u) ⋅ ri (u) = 0,

(3.9.3)

W is called an ith group Riemann invariant for eq. (3.9.1). As with a single equation, we can define entropy and entropy flow for eq. (3.9.1). The functions U : Rn → R and F : Rn → R are called a pair of entropy and entropy flow, if any differentiable solution u of eq. (3.9.1) satisfy

374

3 The Vanishing Viscosity Method of Quasilinear Hyperbolic System

∂U(u) ∂F(u) + = 0; ∂t ∂x

(3.9.4)

and we obtain compatibility condition by comparing with eq. (3.9.1) U ′f ′ = F′.

(3.9.5)

Obviously, the necessary and sufficient condition that makes U and F are a pair of entropy and entropy flow is that they satisfy eq. (3.9.5). These are equations that contain two unknown functions and n equations. When n ≥ 3, eq. (3.9.5) is the overdetermined equation and generally does not have nontrivial solutions. When n = 1, take any differentiable function as U(u), we can ensure  F(u) =

U ′ (u)f ′ (u)du.

When n = 2, eq. (3.9.5) is hyperbolic, Lax construct the asymptotic solution of eq. (3.9.5) with the following form: U = ek6(u) F = ek6(u)

∞  n=0 ∞ 

Vn /kn ,

(3.9.6)

Hn /kn .

(3.9.7)

n=0

So k > 0. Substituting eqs (3.9.6) and (3.9.7) into eq. (3.9.5), the same power containing k is equal and we have V0 6′ f ′ = H0 6′ , ′ ′

Vj 6 f +

′ ′ Vj–1 f

(3.9.8) ′

= Hj 6 +

′ Hj–1 ,

j ≥ 1.

(3.9.9)

Equation (3.9.8) indicates that 6′ is left eigenvector of matrix f ′ , we might as well allow it to be the second family: 6′ f ′ = +2 6′ , thus H0 = +2 V0 .

(3.9.10)

6′ and right eigenvector corresponding to eigenvalue +1 are orthogonal: 6′ ⋅ r1 = 0, so 6 is a first family Riemann invariant. Multiplying eq. (3.9.9) with r1 by right, we have ′ ′ ′ Vj–1 f r1 = Hj–1 r1 .

(3.9.11)

3.9 The Existence of Generalized Solutions for the First-Order Quasilinear Hyperbolic

375

Let j = 1, we note f1′ r1 = +1 r1 , by eqs (3.9.10) and (3.9.11), we obtain (+2 – +1 )V0′ ⋅ r1 + (+′2 ⋅ r1 )V0 = 0. The equation is a first-order differential equation which V0 satisfies. It only need to select positive initial value along noncharacteristic curve, we can obtain its solution V0 > 0, and so we can get H0 . From the above we can find that phase function 6 of traveling wave solution is invariant along the first family characteristic curve, and V0 , H0 are only related to eigenvalue, eigenvector of f ′ = grad f , but they do not depend on the special selection of phase angle function 6. Due to eq. (3.9.9) ′ ′ ′ (+2 Vj – Hj )6′ = Hj–1 – Vj–1 f .

(3.9.12)

This can be summarized as finding out all Vj , Hj . Now we rewrite eqs (3.9.6) and (3.9.7) as U = ek6(u)

N  Vj j=0

F = ek6(u)

kj

N  Hj j=0

kj

+ UN (u),

(3.9.13)

+ FN (u).

(3.9.14)

So UN and FN satisfy equation UN′ f ′ = FN′ + ek6 RN /kN .

(3.9.15)

If we solve eq. (3.9.15) in bounded region G ⊂ R2 and make curve A so that A is outside G and not characteristic line of eq. (3.9.15), G is in decision region of A, and give initial value UN = FN = 0 on A. By eq. (3.9.15) we have UN , FN . It is not difficult to see UN , FN = O(ek6 /kN+1 ) (3.9.15)∗ as k → ∞. Now we consider the initial value problem for strictly hyperbolic equations containing two unknown variables ∂u ∂f (u) + = 0, ∂t ∂x u(x, 0) = u0 (x), where n = 2, u0 (x) ∈ W21,∞ (R), f is sufficiently smooth.

(3.9.16) (3.9.17)

376

3 The Vanishing Viscosity Method of Quasilinear Hyperbolic System

For this purpose, we consider the initial value problem for parabolic equation with a small parameter ∂u% ∂f (u% ) ∂ 2 u% + =% 2 , ∂t ∂x ∂x u% (x, 0) = u0 (x).

(3.9.18) (3.9.19)

Suppose the solution u% to the initial value problem (3.9.18) and (3.9.19) have consistency estimator on % sup u% (t)L∞ (R) ≤ M,

(3.9.20)

t∈[0,T]

where constant M does not depend on %. Suppose eq. (3.9.16) in the region {u ∈ R2 , |ui | ≤ M, i = 1, 2} have at least one pair of strictly convex entropy and entropy flow. Similar to the relevant proof of Lemma 8.8, we have Lemma 3.9.1. Suppose U, F are any pair of entropy and entropy flow,   ∂U(u% ) ∂F(u% ) + ∂t ∂x

(3.9.21)

–1 . is compact in Hloc

Now if w1 , w2 are a pair of Riemann invariant corresponding to the eigenvalue +1 , +2 . Suppose they are defined in {u ∈ R2 , |ui | ≤ M, i = 1, 2}, and mapping u ↦ w = (w1 , w2 ) is bidirectional single value, we denote the minimum rectangle containing support of -xt with G = {w; wj– ≤ wj ≤ wj+ , j = 1, 2}. Lemma 3.9.2. If w1– < w1+ , there are points on two closed vertical edges I + = {w; w1 = w1+ , w2– ≤ w ≤ w2+ }, I – = {{w; w1 = w1– , w2– ≤ w2 ≤ w2+ } of G to make grad +2 ⋅ r2 = 0. Proof. By eqs (3.9.13) and (3.9.14), let Uk = ek6(u) +



Vj /kj + U (1) (u),

j=0,1

Fk = ek6(u) +



Hj /kj + F (1) (u),

(3.9.22)

j=0,1

where 6(u) is the first family Riemann invariant. Take 6 = w1 , by eq. (3.9.15) there exists U (1) , F (1) = O(ekw1 /k2 ). Define probability measure ,±k on G as follows:

(3.9.23)

377

3.9 The Existence of Generalized Solutions for the First-Order Quasilinear Hyperbolic

,±k , h =

-xt , hU±k  ∀h(u, v) ∈ C0 (G). -xt , U±k 

(3.9.24)

Because {,±k } is bounded sequence, we can select subsequence such that lim ,±k , h = ,± , h ∀h ∈ C0 (G).

k→∞

(3.9.25)

Taking any pair of entropy and entropy flow U, F, similar to eq. (3.8.80) we have -xt , Uk F – Fk U -xt , Fk  = -xt , F – -xt , U . -xt , Uk  -xt , Uk 

(3.9.26)

In view of eqs (3.9.10) and (3.9.22), (3.9.23) we have    1 Fk = +2 + O vk . k

(3.9.27)

Let k → ∞ in eq. (3.9.26); we note eqs (3.9.25) and (3.9.27) and find ,± , F – +2 U = -xt , F – -xt , U,± , +2 . Let +±2 = ,± , +2 , so we have ,± , F – +2 U = -xt , F – +±2 U. Similar to eq. (3.9.26) we have ,± , Fk U–k – F–k Uk  -xt , Fk  -xt , F–k  = – . -xt , Uk -xt , U–k  -xt , Uk  -xt , U–k  Let k → +∞, we get ++2 = +–2 , so ,+ , F – +2 U = ,– , F – +2 U. Take F = Fk , U = Uk , and due to  Fk – +2 Uk = ekw1

  1 H1 – +2 V1 +O 2 , k k

(3.9.28)

378

3 The Vanishing Viscosity Method of Quasilinear Hyperbolic System

we have

   1 + ekw1 ,+ , H1 – +2 V1  + O k    1 – = ekw1 ,– , H1 – +2 V1  + O . k

(3.9.29)

Due to eq. (3.9.9) V1 w1′ f ′ + V0′ f ′ = H1 w1′ + H0′ . Multiplying the above equation by right with r2 , we have (V1 w1′ + V0′ )+2 r2 = (H1 w1′ + H0′ )r2 , that is, H1 – +2 V1 =

(+2 V0′ – H0′ )r2 . w1′ r2

By eq. (3.9.10) H1 – +2 V1 = –

+′2 V0 r2 , w1′ r2

because w1′ r1 = 0, we find w1′ r2 ≠ 0, and from known V0 > 0. Now we prove that +′2 r2 has zero point on I + , I – . Otherwise, we might as well know that +′2 r2 do not equal zero on I + . Choosing k sufficiently large, there must be constant C1 > 0 so that the absolute value of the left of eq. (3.9.29) is     1  + kw1+  + ≥ C1 ekw1 . (3.9.30) e , , H1 – +2 V1  + O k  Meanwhile, the absolute value of the right of eq. (3.9.29) is     1  – kw1–  – ≤ C1 ekw1 , e , , H1 – +2 V1  + O k  that is +

–

C1 ekw1 ≤ C2 ekw1 . Taking k sufficiently large, by w1+ > w1– , that is the contradiction, the proof is completed.

∎

Similar to Lemma 3.9.2, there are points satisfying +′1 r1 = 0 on two closed horizontal edges of G when w1– < w2+ . Meanwhile there is

3.9 The Existence of Generalized Solutions for the First-Order Quasilinear Hyperbolic

379

W2

P G

Q O

W1

Fig. 3.2

Corollary 3.9.3. If eq. (3.9.1) is genuinely nonlinear, -xt is a function of $. Proof. If eq. (3.9.1) is genuinely nonlinear, we get wi– = wi+ (i = 1, 2) by Lemma 3.9.2; thus -xt is a function of $. ∎ Corollary 3.9.4. If eq. (3.9.1) is genuinely nonlinear at the rest of the points except the points on strictly monotone curve w2 = 8(w1 ), thus -xt is a function of $. Proof. If not so, the curve must pass through four edges of the rectangle G, so w1– < w2+ , w2– < w2+ , and the curve passes through two diagonal vertexes of G. Set the vertexes as P, Q, as shown in Fig. 3.2. Now ,+ and ,– are all the functions of $, and supporting sets are P, Q, otherwise, so by eq. (3.9.30), we also can deduce contradiction. From eq. (3.9.28) we have F(P) – +2 (P)U(P) = F(Q) – +2 (Q)U(Q), as well as F(P) – +1 (P)U(P) = F(Q) – +1 (Q)U(Q), that is (+2 (P) – +1 (P))U(P) = (+2 (Q) – +1 (Q))U(Q). Note u = (u1 , u2 ), so U(u) = au1 + bu2 are all entropy for arbitrary constants a, b, and +2 – +1 ≠ 0; therefore, there can be only P = Q, and the proof is completed. ∎

380

3 The Vanishing Viscosity Method of Quasilinear Hyperbolic System

Now we consider the initial value problem for nonlinear equations: ∂u ∂3(v) – = 0, ∂t ∂x ∂v ∂u – = 0, x ∈ R1 , t > 0, ∂t ∂x u(x, 0) = u0 (x), v(x, 0) = v0 (x), x ∈ R1 ,

(3.9.31) (3.9.32) (3.9.33)

where 3′ > 0, v3′′ > 0, and the initial value problem for nonlinear parabolic equation with small parameter corresponding to it is ∂u ∂3(v) ∂ 2u – = % 2, ∂t ∂x ∂x ∂ 2v ∂v ∂u x ∈ R1 , t > 0, – = % 2, ∂t ∂x ∂x u(x, 0) = u0 (x), v(x, 0) = v0 (x), x ∈ R1 .

(3.9.34) (3.9.35) (3.9.36)

For initial problem (3.9.31)–(3.9.33), we have the following existence theorem: 1 (R), the solution sequences {u% }, {v% } to the initial Theorem 3.9.5. If u0 (x), v0 (x) ∈ W∞ value problem show subsequences that are weakly ∗ convergent in L∞ (R2+ ), and strongly convergent to u and v, 1 < p < +∞, in Lp,loc (R2+ ); u, v are general solutions to problem (3.9.31)–(3.9.33).

Proof. First, it is easy to know eq. (3.9.31) is a hyperbolic equation as 3′ (v) > 0, and its Riemann invariant is  w1 = u –

  6v 3′ (v)dv, w2 = u +

T



3′ (v)dv,

and it has convex entropy 1 ' = u2 + 2

 3(v)dv

(3.9.37)

and entropy flow q = –u3.

(3.9.38)

Utilizing estimation of invariant region method for eqs (3.9.34)–(3.9.36), we know that the set 

= {(w1 , w2 ) : |wj | ≤ w, j = 1, 2}

is invariant set of eqs (3.9.34) and (3.9.35). So the solution of problem (3.9.34)–(3.9.36) |wj | ≤ w just as t = 0, that is |wj (x, t)| ≤ w, t > 0, so we have the estimate

3.9 The Existence of Generalized Solutions for the First-Order Quasilinear Hyperbolic

u% L∞ + v% L∞ ≤ M,

381

(3.9.39)

where constant M does not depend on %. Because of similitude transformation and maximum estimate of derivative of parabolic group, for the solution of problem (3.9.34)–(3.9.36) we have ( %(  ( % ( ( ∂u ( ( ∂v ( ( ( ( +( % ( ≤ M. ∂x (L∞ ( ∂x (L∞

(3.9.40)

On the other hand, due to eqs (3.9.37) and (3.9.38) we have  2  ∂ 2v ∂ u ∂' ∂q + = % u 2 + 3(v) 2 . ∂t ∂x ∂x ∂x

(3.9.41)

Integrating eq. (3.9.41) in x and t on RT = {–z ≤ x ≤ z, 0 ≤ t ≤ T}, we have 

z

 '(x, t)dx –

–z

z



–z

=–%

 T   0

–z

 T

T

'(x, 0)dx + ∂u ∂x

2

{q(z, t) – q(–z, t)}dt 0



∂v + 3 (v) ∂x ′

2  dxdt

∂v ∂u (z, t) + 3(v(z, t)) (z, t) ∂x ∂x 0  ∂u ∂v – u(–z, t) (–z, t) – 3(v(–z, t)) (–z, t) dt. ∂x ∂x +%

u(z, t)

(3.9.42)

Because of estimates (3.9.39) and (3.9.41), by eq. (3.9.42) we obtain the estimate ( % (2 ( ∂u ( ( % ( ( ∂x (

( % (2  ( ∂v ( ( +( ≤ M. ( ∂x ( L2 (RT ) L2 (RT )

(3.9.43)

So by estimates (3.9.39) and (3.9.43) we can verify ∂'(u% , v% ) ∂q(u% , v% ) –1 (R2T ). + is compact in Hloc ∂t ∂x

(3.9.44)

Due to eq. (3.9.39) we can select subsequence u% , v% such that ∗ ∗ u(L∞ (K)), v% ⇀ v(L∞ (K)), u% ⇀

(3.9.45)

where K ⊂ R × R+ is any bounded open set. We can verify eqs (3.9.31) and (3.9.32) are genuinely nonlinear except the point on the straight line w1 = w2 , so we can obtain the result by the method in the proof of Theorem 3.8.13 and Corollary 3.9.2. ∎

382

3 The Vanishing Viscosity Method of Quasilinear Hyperbolic System

For the following the initial value problem for one-dimensional isentropic hydrodynamic equations in Euler coordinates: 1t + (1u)x = 0,

(3.9.46) #

2

(1u)t + (1u + p)x = 0, p = A1 ,

(3.9.47)

1|t=0 = 10 (x), u|t=0 = u0 (x),

(3.9.48)

We can rewrite eqs (3.9.46) and (3.9.47) into the following form: 1t + mx = 0,

(3.9.49)

2

mt + (m /1 + p)x = 0,

(3.9.50)

where m = 1u. Accordingly we consider the initial value problem for viscous equations adding small parameter for eqs (3.9.49) and (3.9.50): 1t + mx = %1xx ,

(3.9.51)

2

mt + (m /1 + p)x = %mxx ,

(3.9.52) 1

1|t=0 = 10 (x), m|t=0 = 10 (x)u0 (x), x ∈ R .

(3.9.53)

For the existence problem of global classical solution to the initial boundary value problems (3.9.51)–(3.9.53) for the quasilinear parabolic equations, if we can prove the solution to problem (3.9.51)–(3.9.53) 1% (x, t) ≥ $% (t) > 0 when 10 (x) ≥ $ > 0, the initial value problem (3.9.51)–(3.9.53) is regular initial value problem for the parabolic equations, we can obtain the existence of classical solution by general estimate method, that is, if |1% (⋅, t)|∞ ≤ m(t, |10 |∞ , |m0 |∞ ), |m% (⋅, t)|∞ ≤ m(t, |10 |∞ , |m0 |∞ ), we have |1% (⋅, t)|Ck ≤ pk (t, |10 |Ck , |m0 |Ck ), |m% (⋅, t)|Ck ≤ pk (t, |10 |Ck , |m0 |Ck ), |1% (⋅, t)|H k ≤ qk (t, |10 |H k , |m0 |H k ), |m% (⋅, t)|H k ≤ qk (t, |10 |H k , |m0 |H k ). Next we estimate |1% |L∞ , |u% |∞ . It is easy to know that the eqs (3.9.49) and (3.9.50) are hyperbolic, their Riemann invariant and eigenvalues are  m w = m + c/1d1 +2 = + c, 1  (3.9.54) m – c, z = m – c/1d1 +1 = 1 where c =



p′ (1). For the polytropic gas 1 = 1# /#, we have

383

3.9 The Existence of Generalized Solutions for the First-Order Quasilinear Hyperbolic

#+1 m 3–# + 1( /(, +2 = w+ z, 1 4 4 #+1 3–# m – 1( /(, +1 = z+ w, z= 1 2 4 w=

(3.9.55)

where ( = ( # – 1)/2. It is easy to verify w, z as a function of ( 1, m) is quasiconvex, so we have r1 ⋅ ∇w = 0, ∇2 w(r1 , r1 ) ≥ 0, r2 ⋅ ∇z = 0, –∇2 z(r2 , r2 ) ≥ 0, where r1 , r2 are the right eigenvector corresponding to eigenvalues +1 , +2 , respectively. By the invariant region theory we know w% (x, t) ≤ sup w0 (x), –z% (x, t) ≤ sup(–z0 (x)). x

x

So we obtain the estimate |u% (⋅, t)|∞ ≤ M, 0 ≤ 1% (x, t) ≤ M.

(3.9.56)

where constant M does not depend on %, just depends on |u0 (x)|∞ and |10 (x)|∞ . Next if 10 (x) ≥ $ > 0,

(3.9.57)

we can prove that the solution 1% (x, t) to problem (3.9.51)–(3.9.53) has the estimate 1% (x, t) ≥ $(t, %) > 0.

(3.9.58)

Next we suppose  10 ≥ $ > 0,

∞

(( 10 – 1)2 + (u0 – u)2 )dx < ∞,

(3.9.59)

–∞

where 1 = limx→±∞ 10 (x), u = limx→±∞ u0 (x). Multiplying eq. (3.9.51) by the derivative h′ (1) of convex function h(1), we have ht + (h′ 1u)x – h′′ 11x u = %hxx – %h′′ 12x . Integrating eq. (3.9.60) about x, t on G{–∞ ≤ x ≤ ∞, 0 ≤ 4 ≤ t}, we have 

∞

(h(t) – h(0))dx + %

–∞

 t



∞

0 –∞ t ∞

h′′ 11x udxdt.

= 0

h′′ 12x dxdt

–∞

(3.9.60)

384

3 The Vanishing Viscosity Method of Quasilinear Hyperbolic System

1

If we denote the upper bound of 1 by M 2 , taking advantage of Young’s inequality, we have % ′′ 2 2M ′′ 2 2 h 1x + h 1u, 2M 2  t ∞  ∞ % (h(t) – h(0))dx + h′′ 12x dxd4 2 –∞ 0 –∞   2M t ∞ ′′ 2 2 ≤ h 1 u dxd4. % 0 –∞ h′′ 11x u ≤

(3.9.61)

We select h(1) ∈ C2 which is strictly convex function and has the following properties: h(1) = h′ (1) = 0, h(1) = 1–! ,

0 < ! < 1.

So we hypothesize that



h ≤ c(1 – 1)2 , 1 is close to 1, 1

12 h′′ ≤ c1,

1/2 ≤ 1 ≤ M 2 ,

12 h′′ ≤ ch(1),

0 < 1 < 1/2.

(3.9.62)

Because of |u| ≤ M, utilizing the properties (3.9.62) of h, we have h′′ 12 u2 ≤ c(1u2 + h).

(3.9.63)

So from eq. (3.9.61) and the boundedness of 1u2 , we have 

  % t ∞ ′′ 2 h 1x dxdt 2 0 –∞  t ∞ ct ≤ h(1)dxdt. +c 2 0 –∞

∞

(h(t) – h(0))dx + –∞

(3.9.64)

Utilizing Gronwall’s inequality and because of eq. (3.9.64) we have the estimate 

 t

∞

∞

h(t)dt + –∞

0

–∞

h′′ 12x dxdt ≤ Cect/% ,

(3.9.65)

where constant C does not depend on initial value. In order to obtain the lower bound estimate of 1(x, t) according to the point state, we can use the following lemma:

3.9 The Existence of Generalized Solutions for the First-Order Quasilinear Hyperbolic

385

Lemma 3.9.6. If 6(t) is a nonnegative function and satisfies 1

6(t) – 6(s) ≥ C1 (t – s) 2 , t > s,  T 6–! dt ≤ C2 ,

(3.9.66) (3.9.67)

0

so 6(t) ≥ C3 , t ∈ (0, T),

(3.9.68)

where constant C3 depends on C1 , C2 , T and !. Let 6(t) = min 1(x, t), x

then the solution of eq. (3.9.51) can be written in integral expression:  ∞ % 1 = z(x – y, t – s)1(y, s)dy –∞

 t

∞

+ s

zx (x – y, t – 4)u1(y, 4)dyd4 = I1 + I2 ,

(3.9.69)

–∞

where z= 

1 1

(40(t – 4)) 2

∞

exp{–(x – y)2 /4(t – 4)}, 1

|zx (x – y, t – 4)|dy ≤ c(t – 4)– 2 .

(3.9.70)

–∞

Thus, we have 1

I1 ≥ min 1(y, s), |I2 | ≤ C2 (t – s) 2 . y

So we have 1

6(t) – 6(s) ≥ c(t – s) 2 . By estimate (3.9.65) we have  t 0

∞ –∞

(h2 + h2x )dxdt ≤ cect/% .

(3.9.71)

In fact, we can replace h with h2 in eq. (3.9.65), so we find the estimate of the first term on the left-hand side of eq. (3.9.71) and for the second term of the left-hand side of eq. (3.9.71), we can replace the function h with g, which satisfies

386

3 The Vanishing Viscosity Method of Quasilinear Hyperbolic System

h′2 ≤ g ′′ , so we have  t 0

∞

–∞

h2x dxdt =

 t 

∞

0

–∞ t ∞

0

–∞

≤

h′2 12x dxdt g ′′ 12x dxdt ≤ cect/% .

Therefore, estimate (3.9.71) holds. Due to Sobolev’s embedding theorem we have 

t

sup h(1(x, t))dt ≤ cect/% , x

0

or 

t

sup 1–! (x, t)dt ≤ cect/% ,

(3.9.72)

x

0

that is, 

t

6–! (t)dt ≤ cect/% ,

0

So by Lemma 3.9.6 we have 1(x, t) ≥ $(t, %) > 0,

(3.9.73)

where constant $(t, %) depends on initial value 10 , % and 0 ≤ t ≤ T. Because we have the upper and lower bounded estimates of m% (x, t), 1% (x, t), in particular, 1% (x, t) ≥ $ > 0. We use the usual method of estimation and can obtain the existence of classical solution to the initial value problem (3.9.51)–(3.9.53). Theorem 3.9.7. If (1 – 1, u0 – u) ∈ C2 ∩ H 2 , 10 ≥ $ > 0, there exists global classical solution 1% , m% = 1% u% to the initial value problem (3.9.51)–(3.9.53), (1% (⋅, t) – 1, u% (⋅, t) – u) ∈ C2 ∩ H 2 , 1% ≥ $(t, %) > 0. In order to obtain the initial value problem (3.9.46)–(3.9.48) for quasilinear hyperbolic equations or the generalized solution to initial value problem (3.9.49), (3.9.50) and (3.9.53), we use method of vanishing viscosity, that is, we can consider the limit function (1(x, t), m(x, t)) of solution sequence {1% }, {m% } to the initial value problem (3.9.51)– (3.9.53) as % → 0, which is generalized solution to problem (3.9.49), (3.9.50) and (3.9.53). First, for the solution of problem (3.9.51)–(3.9.53), from principle of invariant region we know that 1% , u% have the consistent estimation

3.9 The Existence of Generalized Solutions for the First-Order Quasilinear Hyperbolic

0 ≤ 1% ≤ M, |u% | ≤ M,

387

(3.9.74)

where constant M does not depend on %, thus we have ∗ ∗ 1% ⇀ 1, m% ⇀ m.

(3.9.75)

The problem becomes very complex in the case of containing vacuum 1 = 0, at this time the eigenvalues coincide and eigenvector coincide; therefore, hyperbolicity of equation vanishes. For example, in the case of polytropic gases, #+1 #+1 3–# 3–# z+ w, +2 = w+ z, 4 4 4 4 z = u – 1( /(, w = u + 1( /(, +1 = u – c =

( = #–1 2 , # ≥ 1. Because of 1 = 0, c = 0, +1 = +2 = u, w = z = u, the coefficient of progressive wave asymptotic series for Lax diverges when eigenvalue coincides as 1 = 0, so we should take the other different method of constructing entropy function. In fact, for equation ∂u ∂f (u) + = 0, ∂t ∂x

(3.9.76)

we have additional conservation law '(t)t + q(u)x = 0,

(3.9.77)

so (', q) must satisfy the consistency condition ∇'∇f = ∇q.

(3.9.78)

By eq. (3.9.78) and taking curl rot, we have rot(∇'∇f ) = 0.

(3.9.79)

Thus we find that ' satisfies Euler–Poisson–Darboux equation: 'zw –

+ ('w – 'z ) = 0, (w – z)

(3.9.80)

where + = – 21 (3 – #)/(# – 1). Equation (3.9.80) at (1, u) can be conversed into '11 = d2 'uu , d = c/1.

(3.9.81)

Set ' = h(1)eku , h = a(1)8(r), a(1) = 1(1–()/2 , r = k1( /(, k is a constant, and substituting eq. (3.9.81) we find that 8(r) satisfies Fuchsian equations

388

3 The Vanishing Viscosity Method of Quasilinear Hyperbolic System

8′′ – (1 + ,/r2 )8 = 0, where , = (1 – (2 )/4(2 . We use Frobenius method to solve eq. (3.9.82):  8= #n r n .

(3.9.82)

(3.9.83)

n≥j

Indicators equation has two different roots, the corresponding solutions are j+ = (( + 1)/2(, j– = (( – 1)/2(, 8+ = rj+ 8+( (r), 8– = rj– 8–( (r), where 6±( are even entire functions and have positive coefficient depending on (: 6±( =

∞ 

b±n (()r2n , b±0 (() = 1.

n=0

Thus Fuchsian structure has two kinds of waves near vacuum according to algebraic properties: weak homogeneous wave (corresponding to j+ ) disappears in the vacuum, that is, ' = a8+ eku = const 16+ eku ; strong homogeneous wave (corresponding to j+ ) does not vanish in the vacuum, ' = a8– eku = const 16– eku . Thus, we will study two subspaces divided by the difference structure for the space (', q) according to vacuum state v = {(1, u), 1 = 0}. We can prove that the mechanical energy 'm = 21 1u2 + 1E(1) is strictly convex entropy of 1, 1u = m as p′ (1) > 0; this is qm = u'm + up, and there exists |∇2 '| ≤ const∇2 'm for all weak homogeneous wave '. And it can be controlled by ∇2 'm , so there exists ∂'(u% , 1% ) ∂q(u% , 1% ) + is compact in H –1 . ∂t ∂x

(3.9.84)

In fact, for viscous equations ut + f (u)x = %uxx , u ∈ R2 .

(3.9.85)

Multiplying the above equality by ∇', ∇'(ut + f (u)x ) = !∇'uxx , 't + qx = %'xx – ∇2 'u2x .

(3.9.86)

3.10 Convergence of Solutions to Some Nonlinear Dispersive Equations

Integrating it and then  %

T 0

 2

∇

'u2x dxdt

=

389

 '(x, 0)dx –

'(x, T)dx.

(3.9.87)

Therefore, if |u% | ≤ M, ∇2 ' ≥ D > 0,

(3.9.88)

thus %

 t 0

u2x dxdt ≤ M,

(3.9.89)

where M does not depend on %, so we obtain eq. (3.9.84): -, '1 q2 – '2 q1  = -, '1 -, q2  – -, '2 -, q1 

(3.9.90)

holds for all weak homogeneous waves ('j , qj ), where - indicates measure. Using the integral expression of the partial differential equations satisfied by ', when # = 1 + 2 n , n = 2m + 1, m ≥ 2, Diperna proves: -xt is a function of $. Thus we use the similar method in Theorem 3.8.13, and we can prove the existence of generalized solution to eqs (3.9.49), (3.9.50) and (3.9.53). Theorem 3.9.8. If p = A1# , # = 1+ n2 , n = 2m+1, m ≥ 2, (1–1, u0 –u) ∈ C2 ∩H 2 , 10 ≥ $ > 0, where 1 = limx→±∞ 10 (x), u = limx→±∞ u0 (x), the solution sequence {1% , m% } to the initial value problem (3.9.51)–(3.9.53) is weak ∗ convergence to 1(x, t), m(x, t), respectively. ∗ 1(x, t), (L∞ (R2 )), m% (x, t) ⇀ ∗ m(x, t), (L∞ (R2 )), but the limit function That is 1% (x, t) ⇀ loc loc 1(x, t), m(x, t) are the generalized solution to problem (3.9.49), (3.9.50) and (3.9.53), and satisfy entropy condition, that is ∂'(1, u) ∂q(1, u) + ≥ 0, ∂t ∂x

(3.9.91)

in the sense of generalized function, where '(1, u) is convex function, q(1, u) satisfies the consistency condition. In China, xiakui Ding, guiqiang Chen, peizhu Luo proved the convergence of the Lax– Friedrichs scheme, and they improved the results to 1 ≤ # ≤ 5/3.

3.10 Convergence of Solutions to Some Nonlinear Dispersive Equations In the section, we will use compensated compactness method to consider convergence limit of solution u% (x, t) as % → 0, $ → 0 to Korteweg-de Vries (KdV)–Burgers equation:

390

3 The Vanishing Viscosity Method of Quasilinear Hyperbolic System

ut + f (u)x + $uxxx = %uxx ,

(3.10.1)

and Benjamin-Bona-Mahony (BBM)–Burgers equation: ut + f (u)x – $uxxt = %uxx ,

(3.10.2)

where f (u) is a sufficiently smooth and nonlinear function. We know that for the solution of KdV–Burgers equation ut + f (u)x + $uxxx = 0,

(3.10.3)

generally speaking, which does not tend to the shock solution for quasilinear hyperbolic equation as $ → 0, ut + f (u)x = 0.

(3.10.4)

Lax made a detailed study on the limit situations of the solution u$ (x, t) for KdV equations. But for the solution of Burgers equation as $ → 0 ut + uux = %uxx ,

(3.10.5)

E. Hopf has proved early that: u% (x, t) → u(x, t) as % → 0, u is a generalized solution of eq. (3.10.5). We demonstrate the following results by applying the generalized compensated compactness theory: (1) f (u) = u2 /2, then the traveling wave solution of eqs (3.10.1) and (3.10.2) is monotonous as $ ≤ k%2 (k is some positive constant); the traveling wave solution is choppy as $ > k%2 . (2) f (u) = u2 /2, $ ≤ k%2 , k depends on initial value, then the solution u%,$ (x, t) to initial value problem for eq. (3.10.1) weakly converges to the generalized solution u(x, t) to initial value problem for eq. (3.10.4). The situation is unknown when $ > k%2 . When f (u) = u2 /2, $ ≤ k%4 , then the solution u%,$ (x, t) to initial value problem for eq. (3.10.2) strongly converges to the generalized solution u(x, t) to initial value problem for eq. (3.10.4). (3) For the general f (u), we can find the similar result to (2) under some growth condition which f (u) satisfies. Next we will give the proof step by step. First, we must extend L∞ uniform boundedness of sequence u% refers to the compensated compactness method of Tartar to the Lp uniform boundedness, thereby we can promote the corresponding results. Theorem 3.10.1. Suppose K ⊂ Rm is bounded open set, u% : K → Rn is uniform boundedness sequence in Lp (K), p > 1, then there exists subsequence {u%k } and a family of

3.10 Convergence of Solutions to Some Nonlinear Dispersive Equations

391

probability measures {-x }x∈K on Rn such that if f ∈ C(Rn ) and satisfy f (u) = O(|u|p ), |u| → ∞, then in the distribution sense there are  f (u%k ) → -x , f (+) = f (+)d-x (+). Rn

Proof. The proof is divided into two steps, first we give the idea of the proof. (1) For every M > 0, we define a new sequence {v%M }, which is consistent with u% in the ball with radius M and is L∞ (K) uniformly bounded, which is a compact supported set function sequence. We apply Tartar’s theorem, namely Theorem 3.8.11, then for every M, there exists a family of probability measure {-xM }, and a subsequence {v%Mk } of {v%M } such that for all f ∈ C(Rn ), we have ∗ -xM , f (+), L∞ (K). f (v%Mk ) ⇀

(2) We identify a new family of probability measures {-x } as the limit of {-xM } when M → ∞, and establish the conclusion of the probability measure theorem of this family. % Step 1. If AM % (x) is a cutoff function of u (x), which is defined by

 AM % (x)

=

1,

|u% (x)| ≤ M,

0,

|u% (x)| > M.

Let % v%M (x) = AM % (x)u (x),

then v%M (x) ∈ BM = {+; |+| ≤ M}. It follows from Theorem 3.8.11 that there exists a family of probability measures {-x }x∈K for every M, its support set is in BM , the subsequence of v%M is still recorded as v%M such that for any f ∈ C(Rn ), we have ∗ f (v%M ) ⇀ -xM , f (+), L∞ (K).

Similar to eq. (3.8.49), we can define probability measures ,M % as  ,M x , 6(x, +) =

K

6(x, v%M (x))dx.

Without loss of generality, if K is bounded, we may choose a subsequence still reM = ,M (in the sense of weak ∗ convergence), corded as ,M % such that lim%→0 ,% that is M lim = ,M x , 6 = , , 6 ∀6 ∈ C(K) and has compact support.

%→0

392

3 The Vanishing Viscosity Method of Quasilinear Hyperbolic System

then ,M has the following property: (i) ,M ≥ 0, (ii) supp,M ⊂ K × BM , (iii) projK ,M = dx, i.e., If 6(x, +) = 8(x), then  ,, 6 = 8(x)dx. K

These properties can be easily obtained by the corresponding properties of ,M % . In fact, we have (i′ ) ,M % ≥ 0, M (ii′ ) supp,M % ⊂ graph V% , ′ M (iii ) projK , = dx. By (i′ ), (ii′ ) and (iii′ ), take limit % → 0, then we obtain (i), (ii) and (iii). By the properties (i), (ii) and (iii) of ,M and applying Radon–Nikodym’s theorem, then there exists a family of probability measures {-xM } such that  M -xM dx, a.e., , = K  ,M , 6(x, +) = -xM , 6(x, +)dx, K  (3.10.6) = lim 6(x, v%M (x))dx. %→0 K

Step 2. We define -x , 6 = lim -xM , 6(∀6 ∈ (K × Rn ) and is compact support set function.) (3.10.7) M→∞

limM→∞ -xM , 6 exists, this is due to eq. (3.10.6), so we have   M -xM , 68(x)dx = lim 8(x)6(x, v% 0 (x))dx %→0 K K  M = -x 0 , 68(x)dx, ∀8 ∈ C(K), 8 is compact support set function, M ≥ M0 , where M0 is a constant such that supp6 ⊂ K × BM0 . If f ∈ C(Rn ) satisfies the growth condition f (+) = o(|+|p ), |+| → ∞.

(3.10.8)

-x , f  = lim -x , f N , a.e.

(3.10.9)

Define N→∞

3.10 Convergence of Solutions to Some Nonlinear Dispersive Equations

393

where {f N } are continuous function sequences that approach to f and satisfy f N (+) = f (+), N

f (+) = 0,

|+| ≤ N, |+| ≥ N + 1,

N

|f (+)| ≤ |f (+)| ∀+. In order to establish the existence of limit limN→∞ -x , f N , we will prove that -x , f N  is a Cauchy sequence. In fact, -x , f N  – -x , f M  = [-x , f N  – -xM , f N ] +[-xM , f N  – -xM , f M ] + [-xM , f M  – -x , f M ].

(3.10.10)

So we only need to prove that -xM , f N  is a Cauchy sequence. Let 8 ∈ C(K) be a compact supported set function, if N ≤ M, then we have      [-M , f M  – -M , f N ]8(x)dx x x   K     =  lim[f M (v%M ) – f N (v%M )]8(x)dx K %→0  |f (v%M )| ⋅ |8(x)|dx, (3.10.11) ≤ 2 lim %→0 KN %

where KN% = {x : x ∈ K, |v%M (x)| ≥ N}. The boundedness of the right-hand side of inequality (3.10.11) can be deduced by the growth condition of f and Lp uniform boundedness of u% , in fact, we have   |f (v%M )| ⋅ |8(x)|dx ≤ %0 |8(x)| ⋅ |u% |p dx KN %

K

≤ %0 C0 |8|∞ , N ≫ 1.

(3.10.12)

Therefore, -xs , f N  is a Cauchy sequence, a.e. for x and all s > 0. In order to complete the proof of theorem, we must prove that limk→∞ f (u%k ) exists, and limk→∞ f (u%k ) = -x , f (+), for certain subsequence {u%k } it holds in the distribution sense. Without loss of generality, we may suppose the following: (1) K is bounded. (2) f (0) = 0. If {v%Nk } is a subsequence that satisfies ∗ -xN , f (+), (L∞ (K)). f (v%Nk ) ⇀

(3.10.13)

We record u%k = u% , v%Nk = v%N for some N ≫ 1, for the convenience of record, through direct calculation we have f (u% ) ⇀ -x , f (+).

(3.10.14)

394

3 The Vanishing Viscosity Method of Quasilinear Hyperbolic System

In fact, if 6 ∈ C(K) is compactly supported set function, then       [f (u% ) – -x , f (+)]6(x)dx ≤ |f (u% ) – f (v%M )| ⋅ |6(x)|dx   K K  + |f (v%M ) – -xM , f (+)| ⋅ |6(x)|dx K + |-xM , f (+) – -xM , f N (+)| ⋅ |6(x)|dx K + |-xM , f M (+) – -x , f N (+)| ⋅ |6(x)|dx K + |-x , f N (+) – -x , f N (+)| ⋅ |6(x)|dx.

(3.10.15)

K

Because four integrals on the right of inequalities (3.10.6) and (3.10.12), and (3.10.13) and (3.10.15) are all sufficiently small. In order to prove that the first integral on the right of inequality (3.10.15) tends to 0 as M → ∞, we take use of the growth condition of f (u) and Lp uniform boundedness of u% , that is  K

|f (u% – f (v%M ))| ⋅ |6(x)|dx ≤ %

 KM %

|u% |p ⋅ |6(x)|dx

≤ %0 C|6|∞ ∀6 ∈ C(K) has compact support, M ≫ 1. ∎

So this completes the proof.

Now we point out that the growth condition of the function f (u) in Theorem 3.10.1 is strict, which cannot be relaxed, namely we can construct function sequence {u% } which is uniformly bounded on Lp (p > 1), the growth condition of the function f (u) satisfies f (+) = O(|+|p ), f (+) ≠ o(|+|p ), then there does not exist a subsequence {u%k }, which exists as a family of probability measure -x such that f (u%k ) ⇀ -x , f (+). Example 1. If {w% } is a function sequence that approaches to $ function on L1 (K), that  is: (1) w% ∈ L1 , (ii) w% → $ is in the sense of generalized function, (iii) |w% |dx = 1. Set  1 u% = |w% | p , then |u% |p dx = 1. Set f (+) = +p , if there exists subsequence {u%k } and a family of probability measure -x such that f (u%k ) → -x , f (+), then we find $ = -x , f (+) is a measurable function, so Theorem 3.10.1 is strict. We already know for first-order quasilinear partial differential equations ut + f (u)x = 0.

(3.10.16)

We can define its entropy 6(u) and entropy flow 8(u) which satisfy equation ∂ ∂ 6(u) + 8(u) = 0 ∂t ∂x

(3.10.17)

395

3.10 Convergence of Solutions to Some Nonlinear Dispersive Equations

and consistency condition 8′ = 6′ f ′ .

(3.10.18)

Similar to the proof of Tartar, we can extend uniformly bounded results of {u% } in L∞ (K) to the results of u% in Lp (K) and then obtain the corresponding conclusions. Lemma 3.10.2. If K is bounded open set in R2 , sequence {u% } is uniformly bounded on Lp (K)(p > 1) and entropy condition ∂ ∂ –1 (K), 6(u% ) + 8(u% ) is compact in Hloc ∂t ∂x

(3.10.19)

where >, 8 satisfy the consistency condition (3.10.18), then we choose subsequence {u%k } that converges to the generalized solution of eq. (3.10.16). Lemma 3.10.3. If K is bounded open set in R2 , {w% }, {v% } are two vector function sequences in (L2 (K))N , if ∗ ∗ w% ⇀ w((L2 (K))N ), v% ⇀ v((L2 (K))N ),

and {div(w% )}, {rot(v% )} are compact in H –1 (K), so in the distribution sense there is w% ⋅ v% ⇀ wv. Lemma 3.10.4. If K is bounded open set in R2 , and f ∈ C1 satisfies the following growth condition: |f ′ (u)| ≤ c(1 + |u|p–1 ), p > 1.

(3.10.20)

If {u% } is a sequence of approximate solution for eq. (3.10.16), which is uniformly bounded on Lp (K), and for all entropy pair (6, 8) satisfies entropy condition (3.10.19), where 6 is convex function, and 8′ = 6′ f ′ , then there exists a subsequence {u%k } converging to the weak solution u for eq. (3.10.16), and in the generalized function sense we have f (u%k ) ⇀ f (u). Proof. We only need to replace the results of L∞ approximate solution of Tartar to the corresponding results of Lp . ∎ Next we will prove the results by the other method, especially entropy 6 must be the compactly supported set function, if we only need a convex function on a bounded interval, then we have

396

3 The Vanishing Viscosity Method of Quasilinear Hyperbolic System

Theorem 3.10.5. If K is bounded open set in R2 , and f ∈ C1 satisfy eq. (3.10.20), if {u% } is a sequence of approximate solution for eq. (3.10.16), which is uniformly bounded on Lp (K) and satisfies the entropy condition (3.10.19), where 6 ∈ C2 and has compact support set and is convex function in a bounded nonempty interval, then there exists a subsequence {u%k } such that in the distribution sense u%k ⇀ u, f (u%k ) ⇀ f (u), where u is the weak solution for eq. (3.10.16). Proof. The idea of proof is as follows: First, we define two function sequences {In } and {6n }, which have the compact support set and are convex functions on some bounded interval, In → 1 as n → ∞, {6n } tends to the absolute value of function. By the hypothesis we find that divIn (u% ), fn (u% ), rot8n (u% ), –6n (u% ) are compact in H –1 (K), where 8′n = 6′n f ′ , fn′ = In′ f ′ . By Lemma 3.10.3 we find inner product In (u% ), fn (u% ), 8n (u% ), –6n (u% ) converging in the generalized function sense. The convergence leads to equation containing In (u% ), fn (u% ), 6n (u% ) and 8n (u% ). And then let n → ∞ in all equations, the limit equation leads to the weak continuity of f . Set u = -xt , +, without loss of generality, and we may suppose u = f (u) = 0 at (x, t), if In (u) ∈ C2 such that ⎧  ⎪ u, |u| ≤ n, ⎪ ⎨ In (u) = 0, |u| ≥ 2n, ⎪ ⎪ ⎩ |In (u)| ≤ |u|, |In′ (u)| ≤ 2, then fn (u) =

n

′ ′ 0 In (s)f (s)ds



(3.10.21)

satisfies

fn (u) ∈ C1 ,

|f n (u) ≤ Cn ,

fn (u) ∈ u,

|u| ≤ n, |f n (u)| ≤ 2|f (u)|,

(3.10.22)

where Cn denotes a constant depending on n. Set 6n (u) ∈ C2 , and ⎧ ⎪ 6 (u) ≤ |u| + ,, |u| ≤ n, , > 0, ⎪ ⎨ n 6n (u) = 0, |u| ≥ 2n, ⎪ ⎪ ⎩ 6 (u) is convex function in [–n, n],

(3.10.23)

n

thus  8n (u) = 0

u

6′n (s)f ′ (s)ds, 8n ∈ C1 , |8n (u)| ≤ Cn .

Suppose {-xt } denotes a family of probability measure such that

(3.10.24)

3.10 Convergence of Solutions to Some Nonlinear Dispersive Equations

397

u%k → -xt , +. In order to simplify it, let u%k = u% , -xt = -, then we have un = -, In (+), f n = -, fn (+), 6n = -, 6n (+), 8n = -, 8n (+).

(3.10.25)

By eqs (3.10.21)–(3.10.24) and we note that sequences {In (u% )}, {f n (u% )}, {6n (u% )}, {8n (u% )} are bounded on L2 (K), and by Lemma 3.10.3, div{In (u% ), fn (u% )} and rot{8n (u% ), 6n (u% )} are compact in H –1 (K), therefore by Lemma 3.10.3, we have In (u% )8n (u% ) – 6n (u% )8n (u% ) → un 8n – f n 6n ,

(3.10.26)

where the convergence is in the generalized function sense. By Theorem 3.10.1, the left-hand side of eq. (3.10.26) tends to -, In (+)8n (+) – 6n (+)fn (+), % → 0. So -, In (+)8n (+) – 6n (+)fn (+) = -, In (+)-, 8n (+) –-, fn (+)-, 6n (+) = -, un 8n (+) – f n 6n (+), thus we have -, [In (+) – un ]8n (+) – [fn (+) – f n ]6n (+) = 0.

(3.10.27)

The above equation holds for all function sets In , fn , 6n , 8n which satisfy eqs (3.10.21) and (3.10.23) and fn′ = In′ f ′ , 8′n = 6′n f ′ . Equation (3.10.27) also holds for In , fn as defined before and the continuous function satisfying the following property: 6n (x) = |+|,

|+| ≤ n,

6n (x) = 0,

|+| > n,

8n (x) = f (+), 0 ≤ + ≤ n, 8n (+) = –f (+), –n ≤ + ≤ 0. For these functions, because of eq. (3.10.27) we have -, f (+)[un – +] – |+|[f n – f (+)] = 0, 0 ≤ + ≤ n, -, –f (+)[un – +] – |+|[f n – f (+)] = 0, –n ≤ + ≤ 0. So if |+| ≤ n, we have

398

3 The Vanishing Viscosity Method of Quasilinear Hyperbolic System

-, f (+)un – |+|f n  = 0.

(3.10.28)

If let n → ∞ in eq. (3.10.28), we can prove 0 = lim -, f (+)un – |+|f n  = –-, |+|f , n→∞

(3.10.29)

where f = -, f (+). If we make a translation such that u = 0, f (u) = f (0) at (x, t), then by eq. (3.10.29) we have f = 0 = f (0), theorem holds, or -, |+| = 0, so - = $0 , f (u% ) ⇀ $0 , f (+) = f (0) = 0. Therefore, in order to complete the proof of theorem, we only need to prove that eq. (3.10.29) holds. From the determinate property of {-xt } and the uniform boundedness of {u% } we have |In (+)| ≤ |+| ≤ c(1 + |+|p ),

(3.10.30) p

|f n (+)| ≤ 2|f (+)| ≤ c(1 + |+| ),  C(1 + |+|p )d- ≤ C.

(3.10.31) (3.10.32)

R

By Lebesgue-dominated convergence theorem and eqs (3.10.30)–(3.10.32), we have   In (+)d- = +d- = u, lim un = lim n→∞ n→∞ R R  fn (+)d- = f (+)d- = f . lim f n = lim n→∞

n→∞ R

R

So lim -, f (+)un – |+|f n  = lim [-, f (+)un  – -, |+|fn ]

n→∞

n→∞

= -, f (+)u – -, |+|f = -, |+|f . The last equality is obtained by u = 0. This completes the proof.

∎

Corollary 3.10.6. Suppose K ⊂ R2 is bounded open set, sequence {u% } and function f (u) are identified as Theorem 3.10.5, if for all function 6 ∈ C1 (it has compact set and is convex on the finite interval) we have

399

3.10 Convergence of Solutions to Some Nonlinear Dispersive Equations

∂ ∂ 6(u% ) + 8(u% ) ∈ { the compact set of H –1 (K)} ∪ { the bounded set of ,(K)}, ∂t ∂x where ,(K) denotes measure space, 8′ = 6′ f ′ , then the conclusion of Theorem 3.10.5 holds. Proof. We obtain the theorem by using the results of Theorem 3.10.5 and Murat lemma. ∎ Corollary 3.10.7. Under the hypothesis of Corollary 3.10.6, if f ′′ > 0, there exists u% → u strongly converging in Lq (K) for all 1 < q < p. Proof. From the proof of Theorem 3.10.5, we found f (u) = f or - = $u , because f ′′ > 0, we may prove that - = $u . In fact, f (+) spreads out near u and integrates with respect to -, we have 

Because



  f (+)d- = f (u) d- + f ′ (u) (u – +)d ′′ f (() + (u – +)2 d-. 2

d- = 1, f (u) = f (u) =



f (+)d-, u =





+d-, from the above equality we have

f ′′ (() (u – +)2 d- = 0. 2

Because of the convexity of f , there is - = $u , so by Theorem 3.10.1, we have f (u% ) → $u , f (+) = f (u).

(3.10.33)

The above equality holds for all f ∈ C2 , f (u) = o(|u|p ), |u| → ∞, f ′′ > 0. In order to complete the proof of Corollary 3.10.2, we have divided into three cases: Case 1. p > 2. Because of eq. (3.10.33) we have (u% )q ⇀ uq ∀q < p. In particular, (u% )2 ⇀ u2 , u% ⇀ u, so u% → u strongly converges in L2 (K). Because u% is bounded on Lp (K), then by general interpolation principle we have u% → u, strongly converges in Lq (K) ∀q < p.

400

3 The Vanishing Viscosity Method of Quasilinear Hyperbolic System

Case 2. 1 < p < 2. We will prove that 

|u% – u|q dx = 0, 1 < q < p.

lim

%→0 K

In order to this we divide K into K = Kr ∪ Kcr , where Kr = {x : |u% (x) – u(x)| ≤ r/(4|K|), % ≤ 1}, r ≪ 1, where |K| denotes the measure of K, select sufficiently small r, we only need to estimate  |u% – u|q dx. (3.10.34) Kcr

According to f (u) ∈ C2 , f ′′ (u) > 0, f (u) = o(|u|p ), |u| → ∞, and u%k ⇀ u, then by eq. (3.10.33) we have  lim

%→0 K

Qf (u% )dx = 0,

(3.10.35)

where Qf denotes square part of f close to u, that is Qf (u% ) = f (u% ) – f (u) – f ′ (u)(u% – u) =

f ′′ (() % (u – u)2 . 2

So if we take fn (u) ∈ C1 such that fn (u) = uq , |u| ≥ n1 , f ′′ (u) > 0, then 1 p(p – 1) (u% – u)2 ≤ Qfn (u% ) ∀n. 2 (|u% | + |u|)2–q Thus by eq. (3.10.35) we have  lim

%→0

(u% – u)2 dx = 0. (|u% | + |u|)2–q

In order to estimate eq. (3.10.34), set Kcr = KM ∪ KNM ∪ KN , where KM = {x; x ∈ Kcr , |u% (x)| + |u(x)| ≤ M}, KNM = {x; x ∈ Kcr , M ≤ |u% (x)| + |u(x)| ≤ N}, KN = {x; x ∈ Kcr , N ≤ |u% (x)| + |u(x)|}. By eq. (3.10.36) we have  KM

|u% – u|q dx < r/4, % ≪ 1.

(3.10.36)

3.10 Convergence of Solutions to Some Nonlinear Dispersive Equations

401

In order to obtain the estimate on KN , we note that {u% } is uniformly bounded on Lp (K), u ∈ Lp (K), and have   |u% |q dx ≤ lim N –(p–q) |u% |p dx = 0, lim N→∞ KN

N→∞

 lim

N→∞ KN

KN

uniformly holds with respect to ∀% |u|q dx = 0.

Because of Minkowski’s inequality  |u% – u|q dx ≤ r/4, n ≫ 1. KN

Finally, we estimate the integral in KNM (N is sufficiently large), and it follows from eq. (3.10.36). In fact we have    4|K|N q–2 r |u% – u|2–q ⋅ |u% – u|q ⋅ dx ≥ r 4 (|u% | + |u|)2–q KN M  2–q  r |u% – u|q dx ≥ 4|K|N KN M

for % ≪ 1. So we obtain

 K

|u% – u|q dx ≤ r, % ≪ 1.

This completes the proof.

∎

The flow function f is a convex function, and a family of entropy function {6n } is strictly convex function in interval [–n, n]; in this case, Theorem 3.10.5 holds, we have Corollary 3.10.8. Suppose K ⊂ R2 is bounded open set, {u% } is uniformly bounded on Lp (K) (p>2). If f ∈ C1 is convex function that satisfies the condition f ′ (u) = O(|u|p–s–2 ), |u| → ∞, s < p – 2, if there exists 6 ∈ C2 which is strictly convex and satisfies 6′ (u) = O(|u|s+1 ), |u| → ∞ and {6n } ∈ C1 , it is a entropy function and has compact support set, and satisfies 6n (u) = 6(u), |u| ≤ n, |6n (u)| ≤ |6(u)|, such that



   ∂ ∂ ∂ ∂ 6n (u% ) + 8n (u% ) , In (u% ) + fn (u% ) ∂t ∂x ∂t ∂x

is compact in H –1 (K), where In , fn and 8n as defined by Theorem 3.10.5, then there exists subsequence {u%k } and a family of probability measure -xt such that

402

3 The Vanishing Viscosity Method of Quasilinear Hyperbolic System

u%k ⇀ u = -xt , +, f (u%k ) ⇀ f (u). Furthermore, if f ′′ > 0 we have u% → u, strongly converges in Lq (K), q < p. Proof. First, we note that f and 6n are convex functions as |u| ≤ n, for |a|, |b| ≤ n we have Hn (a, b) = [8n (a) – 8n (b)][In (b) – In (a)] – [6n (a) – 6n (b)][fn (b) – fn (a)] ≤ 0.

(3.10.37)

Next we will give the idea and steps of proof: (1) Hn (u, u% ) → Hn (u) as u% → u, where Hn (u) = [8n (u)–8n ][un –In (u)]–[6n (u)–6n ][f n – fn (u)], un , f n , 6n , 8n are determined by eq. (3.10.25). (2) For given m > 0, there exists Nm such that  1 Hn (u, u% )dxdt ≤ ∀∧ ⊂ K m ∧

(3.10.38)

for all n ≥ Nm . In fact, let D = D1 ∪ ∧2 , where D1 = {(x, t); |u(x, t)| ≤ n, |u% (x, t)| ≤ n}, D2 = Dc1 , So by inequality (3.10.37) we have  ∧1

Hn (u, u% )dxdt ≤ 0.

Then using H¨older’s inequality, eq. (3.10.37) and the growth condition about f (u) and 6(u), the boundedness of u% and u ∈ Lp (K), we obtain  1 Hn (u, u% )dxdt ≤ ∀n ≥ Nm , m D2 (3) So eq. (3.10.38) holds. H(u) = lim Hn (u) = [6 – 6(u)][f – f (u)] ≤ 0, n→∞

where 6 =



6(+)d-, f =



f (+)d-.

(3.10.39)

403

3.10 Convergence of Solutions to Some Nonlinear Dispersive Equations

(4) H(u) = 0. It follows from the convexity of f (u) and 6(u), eq. (3.10.29) and Jensen’s inequality. (5) We can deduce f = f (u) or - = $u by the strict convexity of 6(u) and H(u) = 0. In fact, if f ≠ f (u), we find 6 = 6(u), that is 

 6(+)d- = 6

 +d- .

If we have - = $u by the strict convexity of 6 as the proof of Corollary 3.10.7, then we obtain the conclusion of the corollary. ∎ Now we take the above results related to compensated compactness to apply on KdV– Burgers equation and BBM–Burgers equation. Now we consider the following initial value for KdV–Burgers equation: ut + uux + $uxxxx = %uxx , u|t=0 =

u○ %$ (x),

(3.10.40)

1

x∈R

(3.10.41)

Theorem 3.10.9. If K = R × [0, T], T > 0, suppose that u%$ : K → R are smooth solution sequences to the initial value (3.10.40) and (3.10.41), if u%$ (x, t) and its derivatives all tend to 0 as |x| → ∞, the initial function u○ %$ has compact support set and satisfies u○ %$ H 2 (K) + u○ %$ L4 (K) ≤ C0 .

(3.10.42)

If $ = O(%2 ), there exists a subsequence {u%k } such that lim u%k (x, t) = u(x, t)

%k →0

(strongly converges in Lq (K), q < 4).

(3.10.43)

where limit function u(x, t) ∈ L4 (K) and is the generalized solution to the equation ut + uux = 0.

(3.10.44)

Proof. If the sufficient smoothness of solution u%$ (x, t) to the initial value (3.10.40) and (3.10.41) make all of the following formal operations strictly, then u%$ (x, t) ∈ L∞ ([0, T]; H 4 ). The following symbols said B,(K) = bounded set of measure space ,(K), A∈ A∈ A∈

Hc–1 (K) indicates that A is in compact set p Lc (K) : A is in the compact set of Lp (K), p Lb (K) : A is in the bounded set of Lp (K).

(3.10.45) –1

of H (K),

(3.10.46) (3.10.47) (3.10.48)

404

3 The Vanishing Viscosity Method of Quasilinear Hyperbolic System

By Corollary 3.10.7, in order to obtain the conclusion of theorem, we only need to prove: (i) {u%$ } ∈ L4b (K), (ii) {u%$ } satisfies the entropy condition,

∂ ∂ 6n (u%$ ) + 8n (u%$ ) ∈ Hc–1 (K) ∪ B,(K) , ∂t ∂x ∀(6n , 8n ) ∈ C2 and satisfies (3.10.23) and (3.10.24), where f (u) = u2 /2. In order to simplify the symbol, record u%$ = u% , and its subsequence is still recorded as {u% }. We note that {u% } satisfies eq. (3.10.40), and multiplying eq. (3.10.40) by 6′n (u% ), we have 6′n (u% )u%t + 8′n (u% )u2x = –$6′n (u% )u%xxx + %6′n (u% )u2xx . So the entropy condition (ii) can be written as –$6′n (u% )u%xxx + %6′n (u% )u2xx ∈ Hc–1 (K) ∪ B,(K) . Since $6′n (u% )u%xxx = $(6′n u%xx )x – $6′′n u%x u%xx , %6′n u%xx = %(6′n u%x )x – %6′′n (u%x )2, in order to obtain (ii), we can fully prove that {$6′n u%xx }, {%6′n u%x } ∈ L2c (K), {$6′′n u%x u%xx }, {%6′′n (u%x )2 } ∈ L2c (K). Therefore, in order to prove (ii), we only need to prove that {$u%xx }, {%u%x } ∈ L2c (K),

(3.10.49)

{$u%x u%xx },

(3.10.50)

{%(u%x )2 }

∈

L1b (K).

Because ∀n, 6n can be selected to make 6′n , 6′′n ∈ L∞ , in order to obtain (i), (3.10.49) and (3.10.50), the following two types of estimates are necessary: (I) L2 and L4 estimates of {u% }, (II) u% depend on % and L∞ estimate of $ and L2 estimate of {u%x }, {u%xx }. Equations (3.10.49) and (3.10.50) can be obtained by the latter type of estimate as $ = O(%2 ). We use energy estimate method and make the best of four conservation laws to obtain these estimates. In order to be convenient, record u% (x, t) = u(x, t), u% (x, 0) = u○ % = u0 . Multiplying eq. (3.10.40) by u, and integrating it with respect to x, we have

3.10 Convergence of Solutions to Some Nonlinear Dispersive Equations

d dt

405



 u2 dx = –% R

u2x dx.

R

Integrating above equation with respect to t: 



0

R





T

u2 dx + %

R

u2x dxdt =

R

u20 dx ≤ C0 ,

(3.10.51)

where constant C0 depends on u0 , in order to obtain the estimate of uL∞ , ux L2 , uxx L2 , we can multiply eq. (3.10.40) by 2$uxx + u2 , and integrate it with respect to x, we have d dt

  $u2x –

R

   u3 dx = –2$% u2xx dx + 2% uu2x dx. 3 R R

Integrating above equation with respect to t, we have   T u3 u2xx dxdt dx + 2$% 3 0 R  T uu2x dxdt. = C0 + 2%

 

$u2x –

R

0

R

By eq. (3.10.51) we have 

 $ R

u2x dx

T



+ 2$% 0

R

u2xx dxdt

    T 1 u2 dx + 2% u2x dxdt 3 R 0 R ≤ C0 (1 + |u|∞ ). ≤ C0 + |u|∞

(3.10.52)

In order to obtain the estimate of |u|∞ , using Schwarz’s inequality and eq. (3.10.52), we have  |u2 (x, t)| ≤ 2

x

1



|uux |dx ≤ 2$– 2

u2 dx

–∞

≤ C0 (1 +

R

1   1 2 2 $ u2x dx R

1 1 |u|∞ ) 2 $– 2 .

Therefore, 1

|u|∞ ≤ C0 $– 3 .

(3.10.53)

By eqs (3.10.52) and (3.10.53) we have 

 $ R

u2x dx

T



+ 2%$ 0

R

1

u2xx dxdt ≤ C0 $– 3 .

(3.10.54)

406

3 The Vanishing Viscosity Method of Quasilinear Hyperbolic System

The following estimates will depend on the ratio of % and $, especially $ = O(%2 ), we will obtain the consistent estimate of u% L4 , $u%x u%xx L1 , so we can multiply eq. (3.10.40) 2 by u3 + 3$u2x + 6$uuxx + 18 5 $ uxxxx , and integrate this with respect to x, by calculating directly then we have d dt

5

6    9 u4 dx – 3$ uu2x dx + $2 u2xx dx = –3% u2 u2x dx 5 R 4 R R R   15 2 2 uxxx dx + 6$% uux uxxx dx. – %$ 3 R R

(3.10.55)

We may utilize Schwarz’s inequality estimate on the last term of eq. (3.10.55), that is 

 uux uxxx dx ≤ 3!$2 %

6$% R

R

If we take 1 < ! < 65 , especially take ! = 

11 10 ,

u2xxx dx +

3% !

 R

u2 u2x dx.

now by eq. (3.10.55) we have



 9 2 u /4dx – 3$ + $ u2xx dx 5 R R R  t  t 3 3 u2x u2 dxdt – %$2 u2xxx dxdt. ≤ C0 – % 11 0 R 10 0 R 4

uu2x dx

Integrating above equation with respect to t, we have 

T



0

R

3 +% 11 + %$2

 0

3 10

 t

T

∞

0 –∞ T t

 0

0



T

≤ C0 T + 3$ 0

R



T

0

R

u2xx dxdt

u2x u2 dxdsdt

R





9 2 $ 5

u4 /4dxdt +

F(T) =

u2xxx dxdsdt

uu2x dxdt.

(3.10.56)

By eqs (3.10.51) and (3.10.53) we have 

T



3$ 0

R

 uu2x dxdt ≤ 3$|u|∞

T

0

 R

u2x dxdt

1

≤ C0 $(%–1 $– 3 ). 3

Therefore, if $ = O(% 2 ), namely the left-hand side F(T) of eq. (3.10.56) is bounded, that is F(I) ≤ C0 T.

(3.10.57)

3.10 Convergence of Solutions to Some Nonlinear Dispersive Equations

407

By this we have 

T

 u4 dxdt ≤ C0 T, T > 0.

0

R

By eq. (3.10.56) we still can obtain auxiliary estimate  t

%

0

R

u2x u2 dxdt ≤ C0 , t > 0.

(3.10.58)

By eq. (3.10.58) and L2 estimate of Ux% we obtain the uniform boundedness of $u%x u%xx L1 . In fact, multiplying eq. (3.10.40) by –%2 uxx and integrating this with respect to x, t, we have 



%2 2

R

u2x dx = c0 – %3 



0

T

+ %2

T

R



u2xx dxdt

uux uxx dxdt. 0

R

Applying Schwarz’s inequality to estimate the last integral term of the above equation: %2 2

  %3 T u2 dxdt 2 0 R xx   % T u2 u2x dxdt. + 2 0 R

 R

u2x dx ≤ C0 –

Utilizing eq. (3.10.58), we have  %2 /2 ⋅ R

u2x dx +

%3 2



T



0

∞

–∞

u2xx dxdt ≤ C0 .

(3.10.59)

By eqs (3.10.59) and (3.10.51), and Sobolev’s inequality, we have 

T

$ 0

 |ux uxx |dxdt R

  ≤$ %

T 0

 R

u2x dxdt

1   2 %3 0

T

1

 R

u2xx dxdt

2

%–2 ≤ c0 $%–2 .

Therefore, if $ = O(%2 ), then {$u%x u%xx } ∈ L1b (R × [0, T]). We use eqs (3.10.51) and (3.10.59) to prove %(u%x )2 ∈ L1b and {$u%xx }, {%u%x } ∈ L2c . In fact, by eq. (3.10.51) we have 

T

% 0

 R

u2x dxdt ≤ C0 ,

408

3 The Vanishing Viscosity Method of Quasilinear Hyperbolic System

1

So %(u%x )2 ∈ L1b (R×[0, T]), {% 2 u%x } ∈ L2b (R×[0, T]), thus {%u%x } ∈ L2c (R×[0, T]), by eq. (3.10.59) we have 

T

%3 0

 R

u2xx dxdt ≤ C0 .

If $ = O(%2 ), 

 $2 0

T

1

 R

u2xx dxdt

2

  1 ≤ k% 2 %3

T

0

1

 R

u2xx dxdt

2

1

≤ C0 % 2 ,

so {%u%xx } ∈ L2c (R × [0, T]). The proof of theorem is completed.

∎

For more extensive KdV–Burgers equation ut + f (u)x + $uxxx = %uxx ,

(3.10.60)

we have the following theorem: Theorem 3.10.10. If f ∈ C1 and satisfies growth condition f ′ (u) ≤ C(|u| + 1).

(3.10.61)

Suppose u% : R × [0, T] → R is smooth solution sequence for eq. (3.10.60), it is 0 as |x| → ∞ and satisfies initial condition u% (x, 0) = u%0 (x), then u%0 (x) satisfies the condition of Theorem 3.10.9, such as $ = O(%3 ), then there exists a subsequence {u%k } such that u%k ⇀ u, f (u%k ) ⇀ f (u), u(x, t) is a solution of equation ut + f (u)x = 0.

(3.10.62)

If we add f ′′ (u) > 0, then u%k → u strongly converges in Lp (K), 1 < p < 4. Proof. Similar to Theorem 3.10.9, we only need to prove that: (i) {u% } ∈ L4b (K), (ii) {%u%x }, {$u%xx } ∈ L2c (K), (iii) {$u%xx u%x }, {$(u%x )2 } ∈ L1b (K), where K = R × [0, R], T > 0. For the convenience of record, we can omit superscript %, that is u% = u, multiplying eq. (3.10.60) by u, integrating this with respect to x, t, we have  R

u2 dx + % 2



T 0

 R

u2x dxdt = C0 .

(3.10.63)

3.10 Convergence of Solutions to Some Nonlinear Dispersive Equations

409

So %u%x ∈ L%c (R × [0, T]), %(u%x )2 ∈ L%b (R × [0, T]). Next we estimate L2 norm of ux and L∞ norm of u, for this we multiply eq. (3.10.60) by $uxx + f (u) and integrate this on K, we have $ 2

 

 u2x dx –

R



u

R 0  T



0

u2xx dxdt

R

f ′ (u)u2x dxdt

+ 0

T

f (s)dsdx = C0 – %$

R

Utilizing Schwarz’s inequality and condition (3.10.61), there holds $ 2



 R

u2x dx

+ %$





T



0

R

u2xx dxdt 

3

|u| dx + %

≤ C0 1 +

T

0

R



 R

|u|u2x dxdt

≤ C0 (1 + |u|∞ ). From the final estimate and Sobolev’s inequality, we have  2

1   1 2 2 1 2 u dx ⋅ $ ux dx ⋅ $– 2 2

|u(x, t)| ≤ 2 R

≤ C0 (1 +

R

1 |u|∞ ) 2

– 21

⋅$

.

so 1

|u|∞ ≤ C0 $– 3 .

(3.10.64)

So this is an auxiliary estimate, and the following should be used. Multiplying eq. (3.10.52) by u3 – 2%2 uxx and integrating this on K:  R

  T  T u4 u2 u2x dxdt + %2 u2x dx + 2%3 u2xx dxdt dx + 3% 4 0 0 R R R  T  T u2 uxx ux dxdt + 2%2 uxx f ′ (u)ux dxdt. = C0 + 3$ 0

0

R

(3.10.65)

R

We can use Schwarz’s inequality, eq. (3.10.61) or (3.10.63) to estimate two integrals of the right-hand side of eq. (3.10.65), that is 

T



 2

3$

2

T



uxx f ′ (u)ux dxdt

u uxx ux dxdt + 2% 0

R

3 ≤ %3 4



T

0



3

0

 R

T

+ 3% $ 

R

u2xx dxdt



–3 2



+%

0

u2xx dxdt

R

T

+% 0

|u|2∞

 R

0

T

 R

u2 u2x dxdt

u2 u2x dxdt + C0 .

(3.10.66)

410

3 The Vanishing Viscosity Method of Quasilinear Hyperbolic System

Now let $ = C%3 , C ≪ 1, then by eq. (3.10.64) and as C is sufficiently small, we have 2

%–3 $2 |u|2∞ ≤ %–3 $2 C0 $– 3 ≤ %–3 C0 C4/3 ⋅ %4 ≤

3 %. 4

So the right-hand side of eq. (3.10.66) lies in 6 5    T 7 3 T u2xx dxdt + % u2 u2x dxdt . % 4 0 0 R R

C0 +

(3.10.67)

Hence, the boundedness of the left-hand side of eq. (3.10.65) can be determined by eqs (3.10.66) and (3.10.67), that is  u4 /4dx + R

+

5 4

T





R T

 0

1 3 % 4

 u2 u2x dxdt + %2

0

R

R

u2x dx

u2xx dxdt ≤ C0 .

By the above inequalities, we obtain the compactness that {u% } requires as the proof of Theorem 3.10.9. This completes the proof. ∎ For more general flow functions, we have the following corollary: Corollary 3.10.11. If f ∈ C1 and satisfies growth condition |f ′ (u)| ≤ C(1 + |u|2 ).

(3.10.68)

Suppose {u% } is smooth solution sequence for eq. (3.10.60), it vanishes as |x| → ∞ and has initial condition u% (x, 0) = u%0 (x), u%0 (x) satisfy the condition of Theorem 3.10.9 and u%0 ∈ L6b (K), if $ = o(%4 ), then there exists a subsequence u%k (x, t) such that u%k ⇀ u, f (u%k ) ⇀ f (u), where u is a solution of eq. (3.10.62), if f ′′ > 0, then u%k strongly converges to u in Lp (K) (1 < p < 6). Proof. As the proof of Theorem 3.10.10, we only need to give the following ideas and steps: (1) In order to obtain L2 estimate of ux , we multiply eq. (3.10.60) by $uxx – f (u) and integrate this on K, by the integration of the division and take advantage of condition (3.10.68), we have $ 2



 R

u2x dx +

%$T 0

 R

u2xx dxdt ≤ C0 (1 + |u|2∞ ).

Utilizing Sobolev’s inequality again, we have

411

3.10 Convergence of Solutions to Some Nonlinear Dispersive Equations

1



|u|2∞ ≤ $– 2

u2 dx R

1   1 2 2 1 ≤ $– 2 C0 (1 + |u|∞ ). $ u2x dx R

Therefore, 1

|u|∞ ≤ C0 $– 2 . (2) By utilizing |u|∞ estimate and multiplying eq. (3.10.60) by u5 – 3%2 uxx as Theorem 3.10.10, then we can obtain the compactness that {u% } requires. As a special case of the general KdV–Burgers equation, flow function f (u) is f (u) = –u2p+1 /(2p + 1), p ≥ 1, now we only need to suppose $ = O(%2 ).

∎

Corollary 3.10.12. If K = R × [0, T], T > 0, suppose u% : K → R is a solution of equation ut – u2p ux + $uxxx = %uxx ,

(3.10.69)

which vanishes as |x| → ∞ and has smooth initial condition u% (x, 0) = u%0 (x), and u%0 (x) has compact support set and satisfies u%0 H 2 + u%0 L2(p+1) ≤ C0 , such as $ = O(%2 ), then {u% } ∈ Lb

2(p+1)

(K) and there exists a subsequence {u%k } such that

u%k → u (strongly converges in Lq (K), q < 2(p + 1)), u ∈ L2(p+1) (K) is a solution of equation ut – u2p ux = 0.

(3.10.70)

Proof. Multiplying eq. (3.10.69) by u and integrating this with respect to x, we have 



T

2

u dx + %



0

R

u2x dxdt = C0 .

R

(3.10.71)

Because {%u%x } ∈ L2c (K), {%(u2x )} ∈ L1b (K), multiplying eq. (3.10.69) by u2p+1 – (2p + 1)$uxx and integrating this with respect to x, t, we have  R

u2(p+1) dx + 2(p + 1)$ 2(p + 1)



 R

u2x dx



T



+ 2(p + 1)$% 0

R

+ %(2p + 1) 0

u2xx dxdt = C0 .

T

 R

u2p u2x dxdt (3.10.72)

412

3 The Vanishing Viscosity Method of Quasilinear Hyperbolic System

Because of {u% } ∈ L2n (p + 1), such as $ = o(%2 ), then by eq. (3.10.72) we have 



T

$2 0

1

 R

2

u2xx dxdt

1



≤ C% 2 0

T

1

 R

2

u2xx %$dxdt

1

≤ % 2 C0 ,

Therefore {$u%xx } ∈ L2c (K), finally by estimates (3.10.71), (3.10.72) and Schwarz’s inequality, we have  $ 0

T







T

uxx ux dxdt ≤

%$

0

R

R

u2xx dxdt

1   2 %

T 0

1

 R

u2x dxdt

2

≤ C0 . So {$u%xx u%x } ∈ L1b (K), and the corollary holds.

∎

Now consider BBM–Burgers equation ut + uux – $uxxt = %uxx .

(3.10.73)

We need to prove that: if $ = O(%4 ), then we can obtain sequential compactness, for the general BBM–Burgers equation ut + f (u)x – $uxxt = %uxx .

(3.10.74)

We will do further research. Corollary 3.10.13. If K = R × [0, T], T > 0, suppose u% : K → R is a smooth solution sequence of eq. (3.10.73), it tends to 0 as |x| → ∞ and has initial condition u% (x, 0) = u%0 (x), and u%0 (x) is smooth and satisfies u%0 H 2 + u%0 L4 ≤ C0 . If $ = o(%4 ), then there exists sequence {u%k } such that u%k → u (strongly converges in Lp (K), p < 4), and u ∈ L4 (K) is a solution of eq. (3.10.44). Proof. If {u% } is sufficiently smooth, the following forms of operations are strict; for example, we may suppose {u% } ∈ L∞ ([0, T]; H 4 ), similar to the proof of Theorem 3.10.9, we only need to prove sufficiently: {u% } ∈ L4b (K),

(3.10.75)

{%u%xt }, {$u%x } ∈ L2c (K), {$u%x u%xt }, {$(u%x )2 } ∈ L1b (K),

(3.10.76) (3.10.77)

3.10 Convergence of Solutions to Some Nonlinear Dispersive Equations

413

we may omit superscript %, namely let u% = u, and still use energy method. Multiplying eq. (3.10.73) by u and integrating this with respect to x, t: 





u2 dx + $ R

R

u2x dx + 2%



T

0

R

u2x dxdt = C0 .

(3.10.78)

Using Sobolev’s inequality, we have 

 |u(x, t)|2 ≤ 2

u2 dx

|uux |dx ≤ 2 R

≤ C0 $

R

– 21

1   1 2 2 1 $ u2x dx ⋅ $– 2 R

.

So 1

$ 4 |u|∞ ≤ C0 .

(3.10.79)

If $ = o(%4 ), we obtain estimate of u4 L1 . In fact, multiplying eq. (3.10.73) by u3 + 2%ut and integrating this with respect to x, t: 



T

u4 /4dx + 3% R



0



 R

u2 u2x dxdt + 2%

T





R T

0

R

R

u2t dxdt

u2x dx 

T

ux u2 uxt dxdt – 2%

= C0 – 3$ 0



0

 u2xt dxdt + %2

+ 2%$

T

uux ut dxdt. 0

R

 (3.10.80)

R

Using Schwarz’s inequality, the right-hand side of eq. (3.10.80) lies in C0 +

 T   " T 3! 3 $|u|2∞ %–1 u2 u2x dxdt + $% u2xt dxdt 2 2 0 0 R R  T  T u2 u2x dxdt + % u2t dxdt. +% 0

0

R

(3.10.81)

R

If $ = C%4 and C ≪ 1, by eq. (3.10.79) we have 3 3 1 3 $|u|2∞ %–1 ≤ $ 2 %–1 C0 ≤ %. 2 2 2 So by eqs (3.10.80) and (3.10.81) we have 

 4

2

u /4dx + % R



T

+% 0

 R

R

u2x dx

% + 2

u2t dxdt

T





R T



%$ + 2

0

0

u2 u2x dxdt

R

u2xt dxdt ≤ C0 .

(3.10.82)

414

3 The Vanishing Viscosity Method of Quasilinear Hyperbolic System

Therefore, {u% } ∈ L4b (R×[0, T]). By eqs (3.10.82) and (3.10.78) we may obtain the estimate 

T

1



2

2

≤

($uxt ) dxdt 0



T

$ 0

3 C% 2



T



%$



T

ux uxt dxdt ≤ % $% 0

R

 R

1



0

R





u2xt dxdt

R

u2xt dxdt

1   2 %

T 0

2

3

≤ C0 % 2 ,

1

 R

u2x dxdt

2

≤ C0 %. So {$u%xt } ∈ L2c (R × [0, T]), {$u%xx u2x } ∈ L1b (R × [0, T]), from eq. (3.10.78) we find {%(u%x )2 } ∈ L1b (R × [0, T]), {%u%x } ∈ L2c (R × [0, T]). The proof of theorem is completed. ∎ We can extend Theorem 3.10.13 to further general BBM–Burgers equation (3.10.74). Theorem 3.10.14. If f ∈ C1 and satisfy growth condition |f ′ (u)| ≤ C(1 + |u|p–1 ), p ≥ 2.

(3.10.83)

Suppose u% : R × [0, T] → R is a smooth solution sequence for eq. (3.10.74), it vanishes as |x| → ∞ and take initial condition u%0 (x), u%0 (x) satisfy the condition of Theorem 3.10.13 and u%0 L2p ≤ C0 , let $ = O(%4 ), then there exists a subsequence {u%k } such that u%k ⇀ u, f (u%k ) ⇀ f (u), where u is solution for equation ut + f (u)x = 0; if we add condition f ′′ (u) > 0, then u%k strongly converges to u in Lq (K), q < 2p. Proof. We can complete the proof as the proof of Theorem 3.10.13; the difference is that we only need to use L2p (K) estimate instead of L4 (K) estimate. In order to this, we only need to multiply eq. (3.10.74) by u2p–1 + 2%ut and then make integral estimates. ∎

4 Physical Viscosity and Viscosity of Difference Scheme In this chapter, we introduce physical viscosity of unsteady fluid motion and viscosity and artificial viscosity of all kinds of main difference schemes in quasilinear hyperbolic equations; make a qualitative analysis on the viscosity; and compare the numerical calculation results. The contents of this chapter can be referred to Refs [1, 6, 7, 23, 24, 42–47, 53–55, 62, 63, 67, 68, 70, 75, 89, 115].

4.1 Ideal Fluid, Viscous Fluid and Radiation Hydrodynamic Equations We first establish the hydrodynamic equations of ideal fluid. Hydrodynamic study motion of liquids or gases, relates to particle velocity u(x1 , x2 , x3 , t) = (u1 , u2 , u3 ) except the thermodynamic quantities. All of the quantities are the function of time t and the coordinates x1 , x2 , x3 of point. Fluid motion satisfies hydrodynamic equations described by the quality, momentum and energy conservation laws of physics. (1)

Law of conservation of mass (Continuity equation)

Suppose K is a determined space region, A is its boundary, then the change of mass in this volume is   ∂ ∂1 1dx1 dx2 dx3 = dx1 dx2 dx3 ∂t K K ∂t in unit time (1 is density), and the change of mass should equal the mass from boundary surface flowing into the mass of volume in unit time. It can be denoted by the following surface integral:  – 1un d3, A

where un is the projection of velocity u along the outer normal of A and d3 is the surface element. By Gauss–Ostrogradsky formula   1un d3 = div(1u)dx1 dx2 dx3 , A

K

for any area, we find  5 K

DOI 10.1515/9783110494273-004

6 ∂1 + div(1u) dx1 dx2 dx3 = 0. ∂t

416

4 Physical Viscosity and Viscosity of Difference Scheme

This equation holds only when the integral expression is zero. So we have ∂1 + div( 1u) = 0 ∂t or write in the form of coordinates ∂1  ∂1uk = 0. + ∂t xk 3

(4.1.1)

k=1

(2)

Momentum conservation and Euler equation

We can write down Newton’s second law for the volume element dV of mass. The mass contained in this volume is 1dV and the acceleration of particle of volume is du dt , that is du ∂u  ∂u uk . = + dt ∂t ∂xk 3

(4.1.2)

k=1

The particle obtains acceleration because of the action of force. Under the condition without considering viscous, the only force acting on this volume is the pressure p. The pressure acting on a volume can be calculated by the following integral: 



– A

pnd3 = –

K

grad pdV,

We calculate the product between the mass contained in the volume and the acceleration obtained from above. This is equal to the force acting on this volume, so we have the equation 1

∂u ∂u = –grad p. + 1uk ∂t ∂xk

(4.1.3)

It is called Euler equation. We point out that the equation is of nondivergence type, namely it does not denote conservation laws. In order to obtain the equations of conservation form, we multiply eq. (4.1.1) by u, and then add it to eq. (4.1.3), so we obtain ⎧ 3  ⎪ ∂0ik ⎪ ⎨ ∂1ui + = 0 (i = 1, 2, 3), ∂t ∂xk k=1 ⎪ ⎪ ⎩ 0ik = p$ik + 1ui uk .

(4.1.4)

We can see that eq. (4.1.4) represents momentum conservation law. In fact, eq. (4.1.4) is equivalent to the integral equation

4.1 Ideal Fluid, Viscous Fluid and Radiation Hydrodynamic Equations

  ∂ 1ui dx1 x2 x3 + 1ni d3 ∂t A  K 1ui (u ⋅ n)d3 = 0 (i = 1, 2, 3), +

417

(4.1.5)

A

which has the following meaning:  K

1ui dx1 x2 dx3 ,

whose derivative indicates the change of component i of the momentum included in the determined volume K per unit time;  1ui (u ⋅ n)d3. A

It denotes component i of the momentum along with the flow flowing through the interface A and out of the flow region K at the same time;  pni d3 A

indicates the variation of the momentum caused by the acceleration of particle in the region K under the action of pressure. Tensor 0ik = 1ui uk + p$ik is called momentum flow tensor. (3)

Energy conservation law

The energy of the mass per unit volume is composed of internal energy 1e and kinetic 2 energy 1 u2 . After the unit time, the energy in the region K is changed into ∂ ∂t

 1u2 dx1 dx2 dx3 2 K    ∂ 1u2 ≡ 1e + dx1 dx2 dx3 . 2 K ∂t  

1e +

This variation is formed by the two portions of energy. The first part is the input of the energy caused by the transport of the energy flowing into the region, and can be expressed in the following integral:  1u2 un d3; 1e + 2

  – A

the second part is the increase of the energy caused by the work done by the pressure on the volume K, and can be written as  – pun d3. A

418

4 Physical Viscosity and Viscosity of Difference Scheme

So we obtain the equation: ∂ ∂t

 1u2 dx1 dx2 dx3 2 K    u2 + un d3 = 0. P + 1e + 2 A  

1e +

As continuity equation did, we can converse the surface integrals into volume integrals by Gauss–Ostrogradsky formula and take use of the arbitrary of the region K; for ideal fluid we obtain the following energy equation:    6 5  3 ∂ ∂ 1u2 u2 P 1e + + +e+ = 0. 1uk ∂t 2 ∂xk 1 2

(4.1.6)

k=1

So we finally obtained the ideal hydrodynamic equations: ⎧ 3 ⎪ ∂1  ∂1uk ⎪ ⎪ =0 + ⎪ ⎪ ⎪ ∂t ∂xk ⎪ k=1 ⎪ ⎪ ⎪ ⎪ 3 ⎨ ∂1u  ∂(p$ik + 1ui uk ) i = 0 (i = 1, 2, 3) + ⎪ ∂t ∂xk ⎪ k=1 ⎪ ⎪ ⎪ ⎪ 6 5  3 3 ⎪ ∂ !1e + 1 3 ui ui "   ⎪ ∂ ui ui P ⎪ i=1 2 ⎪ + +e+ ) = 0, 1uk ⎪ ⎩ ∂t ∂xk 1 2 k=1

(4.1.7)

i=1

together with the state equation e = e(v, S) and thermodynamic relation p(v, S) = –ev (v, S).

(4.1.8)

Construct the complete equations of five equations with five unknown functions, where specific volume is v = 11 . State function S is called entropy. If we study the smooth solution to eqs (4.1.7), from the equations we can deduce the following equation about entropy: ∂1S  ∂(1uk S) = 0. + ∂t ∂xk 3

(4.1.9)

k=1

We can replace eqs (4.1.7) by the equations that consist of eqs (4.1.1), (4.1.4) and (4.1.9) to continue research. When the flow has the constant temperature T, we can obtain the following isothermal motion equations:

4.1 Ideal Fluid, Viscous Fluid and Radiation Hydrodynamic Equations

419

⎧ 3  ⎪ ∂1uk ⎪ ⎪ ∂1 + = 0, ⎪ ⎪ ⎨ ∂t ∂xk k=1

3 ⎪ ⎪ ∂1ui  ∂1ui uk ∂p ⎪ ⎪ + = 0, ⎪ ⎩ ∂t + xk ∂xi

(i = 1, 2, 3),

(4.1.10)

k=1

where p = p( 1). Now consider the viscous fluid. At this point, we have to examine the internal friction in the derivation of equation. Obviously, the continuity equation ∂1  ∂1uk =0 + ∂t ∂xk 3

(4.1.11)

k=1

also holds for the ideal fluid, because leading-out of the equation is not related to the action of which force. For the ideal fluid, the momentum conservation equations have been obtained as follows: ∂1ui  ∂0ik = 0 (i = 1, 2, 3), + ∂t ∂xk 3

k=1

where momentum flow tensor is composed of momentum interaction ( 1ui uk ) of medium movement and pressure (p$ik ). Obviously, the general form of the equations is also maintained in the case of viscous fluid, only the friction term should also be added to the expression, namely 0ik = 1ui ub + p$ik – -ik , where -ik is tensor, which denotes the ith component of the viscous friction force that acts on the given point in unit area and is vertical to xk axis. We can prove that the general form of viscous tensor satisfying certain conditions is  -ik = '

 3 3  ∂ui ∂uk 2  ∂ue ∂ue + – $ik + 2. $ik , ∂xk ∂xi 3 e=1 ∂xe ∂xe e=1

(4.1.12)

where the coefficient of . and ', respectively, refers to as the first and second viscosity coefficients. Now we write the equations of energy conservation law for viscous fluid. For the ideal fluid, we know ∂ ∂t

  K

1e +

    1u2 1u2 dx1 dx2 dx3 + un d3 = 0. p + 1e + 2 2 A

420

4 Physical Viscosity and Viscosity of Difference Scheme

For the viscous fluid, applying the work done by the friction added to the second (surface) integral, namely adding the integral

–

  3 A i,k=1

ui -ki nk d3,

we obtain the equation  1u2 dx1 dx2 dx3 2 K  6   5 3  1u2 + ui -ki nk d3 = 0, un – p + 1e + 2 A

∂ ∂t

 

1e +

i,k=1

thereby we obtain the following energy conservation law in divergence type:      5  6 3 3 ∂ ∂ 1u2 u2 P ui -ki = 0. 1e + + +e+ – 1uk ∂t 2 ∂xk 1 2 k=1

i,k=1

So we finally obtain the viscous hydrodynamics equations as follows: 3 ∂1u ∂1  k = 0, + ∂t k=1 ∂xk 3 ∂0 ∂1ui  ik = 0 (i = 1, 2, 3), + ∂t k=1 ∂xk ∂ ∂t

 1e +

1u2 2

 +

3  k=1

∂ ∂xk

5

 1uk

P 1

+e+

u2 2

 –

3 

6 ui -ki = 0,

(4.1.12)∗

i,k=1

where ⎧ 0 = 1ui uk + p$ik – -ik , ⎪ ⎪ ⎨ ik   3 3  ∂ui ∂uk 2  ∂ue ∂ue ⎪ = ' + – $ $ik , + 2. ⎪ ik ik ⎩ ∂xk ∂xi 3 e=1 ∂xe ∂xe e=1

(4.1.13)

Now we consider hydrodynamic equations with radiation heat conduction. At this time, the basic form of the equations do not vary, and we should just increase the radiation pressure in the pressure term and increase the radiation energy in energy, and effects of radiation heat flux. Under the Rosseland approximate conditions of local radiation thermodynamic equilibrium and radiation heat conduction, we can obtain the following ideal radiation hydrodynamic equations:

4.1 Ideal Fluid, Viscous Fluid and Radiation Hydrodynamic Equations

⎧ 3 ⎪ ∂1  ∂1uk ⎪ ⎪ = 0, + ⎪ ⎪ ⎪ ∂t ∂xk ⎪ k=1 ⎪ ⎪ ⎪ ⎪ 3 ⎨ ∂1u  ∂0ik i = 0 (i = 1, 2, 3), + ⎪ ∂t ∂xk ⎪ k=1 ⎪ ⎪ ⎪ ⎪        3 ⎪ 2 ⎪ ∂ 1u2 lc ∂Up ⎪ ⎪ ∂ 1e + 1u + Up + + p + p u 1e + – = 0, ⎪ k ⎩ ∂t 2 ∂xk 2 3 ∂xk

421

(4.1.14)

k=1

where 0ik = 1ui uk + (p + pv )$ik , up = 3p- = 43T 4 /c,

(4.1.15)

c is sound velocity, 3 is Boltzmann constant and l is free distance of radiation. Next we consider the Lagrange coordinate system. We derive hydrodynamic equations, which are all derived from the Euler coordinate system. At this point the independent variable is taken as the time t and Descartes coordinates of space fixed point; of course, it can also be introduced to other coordinate systems, such as cylindrical or spherical coordinate system. Sometimes it is more convenient to use the other coordinate system, which is the state of the liquid or gas in a fixed (although it is in motion) particle. At this point, the position of the particle in the space, that is, its coordinates are determined by the function of time and the particle coordinates. Such a coordinate system is known as the Lagrange coordinate system. Consider the one-dimensional case, and suppose the Euler coordinates are t and x. We can naturally select time as one of the Lagrange coordinates; in order to avoid confusion, mark t′ , another Lagrange coordinates,  q=

x

1(x, t)dx.

(4.1.16)

0

Equation (4.1.16) denotes the mass of the fluid at the left of x, that is q and fluid particle has a relationship. Because the journey of the particle moved at the time dt′ is udt′ , for the mixed q, we have dx = udt′ . And by fixed time t, from eq. (4.1.16) we obtain dq = 1dx.

422

4 Physical Viscosity and Viscosity of Difference Scheme

Thus noting t′ = t, we get the differential relationship 1 dx = udt′ + dq, 1 dq = –1udt + 1dx.

(4.1.17) (4.1.18)

They give all derivatives from new coordinates to old coordinates, that is ⎧ ∂ ∂ ∂ ⎪ ⎪ ⎨ ∂t = ∂t′ – 1u ∂q , ∂ ∂ ⎪ ⎪ ⎩ =1 . ∂x ∂q

(4.1.19)

The continuity equation is lost when translating into the Lagrangian coordinate, because the continuity of the mass is contained in the essence of the coordinate q. The momentum and energy equations are transformed according to the formula. Then it leads to the integral condition of expression (4.1.17) to x, that is   ∂ ∂ 1 = ′ ∂q ∂t 1

(4.1.20)

be taken as an additional equation. In addition, if the problem needs to be aware of the coordinate x, we can get from eq. (4.1.17), namely ∂x 1 ∂x = u, = . ∂t′ ∂q 1

(4.1.21)

We can obtain it from any one of the above equations. These two equations are compatible because of eq. (4.1.20). Therefore, the ideal one-dimensional hydrodynamic equations in Lagrange coordinate are ⎧ ∂v ∂u ⎪ – = 0, ⎪ ⎪ ⎪ ∂t′ ∂q ⎪ ⎪ ⎪ ⎨ ∂u ∂p + = 0, (4.1.22) ∂t ∂q ⎪ ⎪ ⎪ 2 ⎪ ⎪ ⎪ ∂(e + u2 ) ∂(pu) ⎪ ⎩ + = 0. ∂t′ ∂q And general isothermal hydrodynamic equations are ⎧ ∂v ⎪ ⎪ ⎨ ∂t′ – ∂u ⎪ ⎪ ⎩ + ∂t′

∂u = 0, ∂q ∂p = 0. ∂q

(4.1.23)

4.1 Ideal Fluid, Viscous Fluid and Radiation Hydrodynamic Equations

423

We can also achieve the Lagrangian coordinate transformation from another view and are going to find a class of transformation closely related to the equation itself. It can translate any equations under conservation laws into the similar form. Continuity equation ∂1 ∂1u + =0 ∂t ∂x can be seen as dq = 1dx – 1udt,

(4.1.24)

which is the complete differential of a function q. q can be used as new coordinate. At this point it will automatically satisfy the continuity equation, but instead it is 1 dx = udt + dq, 1 which is the condition of complete differential. Applying this to the general equations ∂ui ∂fi (u) + = 0 (i = 1, 2, . . . , n). ∂t ∂x

(4.1.25)

We consider the equation of fixed index i, corresponding to eq. (4.1.25), and expression dq = ui dx – fi (u)dt

(4.1.26)

is complete differential. In order to get new coordinate q, eq. (4.1.26) can be given as a similar coordinate transformation with eq. (4.1.19) for the rest of the n – 1 equations: ⎧ ∂ ∂ ∂ ′ ⎪ ⎪ ⎨ ∂t = ∂t′ – fi (u) ∂q , ∂ ∂ ⎪ ⎪ ⎩ = ui . ∂x ∂q

(4.1.27)

And the ith equation is used instead of the right part of the equation: dx =

1 fi (u) dt + dq, ui ui

which is the condition of total differential, that is, we can use     ∂ 1 ∂ fi (u) – =0 ∂t ui ∂q ui to replace.

(4.1.28)

(4.1.29)

424

4 Physical Viscosity and Viscosity of Difference Scheme

4.2 The Artificial Viscosity of Difference Scheme We already know the initial value problem for first-order quasilinear hyperbolic equations and systems ∂u ∂f (u) + = 0. ∂t ∂x

(4.2.1)

Even if the initial function is sufficiently smooth, it may also appear as discontinuous solution. From the physical aspect, the shock wave phenomenon may occur for the ideal fluid motion. In 1922, physicist Becker demonstrated that any shock waves of all the original strength of the amount of change in the shock region are continuous when the fluid motion is allowed to have viscous effect, and it is proved that the solution of the parabolic equation and equations with a viscous term can be obtained even if the initial value is interrupted, and there are still likely to be a smooth solution. In order to study the properties of the discontinuous solutions to the quasilinear hyperbolic equations, we naturally think of using the corresponding smooth parabolic equations with a small parameter to approximate its solution; on this issue, we have discussed in great detail in Chapter 3. For the numerical calculation, in 1949 the famous mathematician J. von Neumann and R.D. Richtmyer studied the numerical calculation problem for hydrodynamic equations with Lagrange form in one dimensional ⎧ ∂u 1 ∂p ⎪ ⎪ =– , ⎪ ⎪ ∂t 1 0 ∂x ⎪ ⎪ ⎨ ∂v ∂: = –p , ⎪ ∂t ∂t ⎪ ⎪ ⎪ ⎪ 1 ∂u ∂v ⎪ ⎩ = , ∂t 10 ∂x

(4.2.2)

pv where u is fluid velocity, p is pressure, v is specific volume, : = #–1 is internal energy, # is adiabatic exponent and 10 is initial density. Their main idea is: for the above ideal hydrodynamic equations, they introduce an artificial viscosity term (dissipative term) so that the physical quantities are smooth after a few space step (three or four) of the shock wave transition zone, and then they construct a differential equation containing dissipative term, but the shock wave as smoothly and rapidly changing layer automatically displayed without special treatment. In particular, there are three requirements: (1) Add artificial viscosity term in the momentum and energy equations, thus some artificial dissipative mechanism was introduced in hydrodynamics motion, and the discontinuous solution of the shock wave becomes a sharp change in a relatively narrow transition region, but it is a continuous solution. (2) Because the artificial viscosity is added to the equation, it is required out of the transition zone of the shock wave, the effect of “artificial viscosity term” must tend to disappear, while the physical quantities on both sides of the transition zone approximately satisfy Hugoniot condition. (3) The range of the shock transition zone should be limited to a few space

4.2 The Artificial Viscosity of Difference Scheme

425

steps as the calculation of the transition zone is not expanded, and the velocity of the transition zone should be approachable to the shock velocity of the vacuum. For eqs (4.2.2), they put forward the concrete form of the artificial viscosity term according to these requirements, namely in momentum and energy equations transform pressure P into P = p + q, where ⎧  2 ∂u ∂u ⎪ ⎪ , < 0, ⎨ l2 p ∂x ∂x q= ⎪ ∂u ⎪ ⎩ 0, ≥ 0, ∂x

(4.2.3)

l is a constant and has dimension of length. It is generally written in the form of l = !Bx, Bx is step size, a is a dimensionless constant according to computing requirements and preselected, a ≈ 2. Consider the mass conservation equation in one-dimensional symmetric coordinate system, q can be also written as  q = l20 1

1 ∂v ⋅ v0 ∂t

2 ,

(4.2.4)

where v0 is a constant with a specific volume dimension which is used to transform from the Euler coordinates to the Lagrange coordinates; here l0 = aBr(r/R)- , r and R are the Lagrange coordinate and the Euler coordinate, respectively; - = 0, 1, 2 are, respectively, corresponding to the plane, cylindrical and spherical symmetry situation. Practice has proved that this artificial viscosity differential method proposed by von Neumann and Richtmyer for solving the unsteady compressible hydrodynamics appears multiple shock, a series of complex problems such as contact discontinuity are quite effective and very successful. This method has been developed to be a universal method for adding artificial viscosity to deal with discontinuous solution of the equation, just as the deepening of the research, the form of artificial viscosity term used in various methods is different, some of the methods are not obviously added as a term for a viscous term, but it implies that some kinds of terms actually play the role of viscous term. For example, the famous Lax–Friedrichs difference scheme, Godunov difference scheme and so on are all implicit in a viscous difference format. As we know, N–R viscosity is a nonlinear quadratic viscous; in many cases, it has calculated the images that are right, but in the reflection boundary (such as near the solid wall boundary, or shock wave light medium to heavy medium and then use discontinuous interaction), we calculate the results after adding N–R viscous term, and energy and density will appear less accurate images; in addition, if the viscous coefficient a in N–R viscosity select improperly, the oscillation of shock front cannot be eliminated. Therefore, in order to suppress the development of the wave oscillation,

426

4 Physical Viscosity and Viscosity of Difference Scheme

a lot of research is proposed to add linear artificial viscosity term, but linear artificial viscosity term shows shortcomings such as widening the width of the shock wave, and the shock wave and the shock wave intensity width are related and so on. Hence, Landschoff, B. Samarkii, Arsenin, B.F. Kurolatenko and so on have proposed a mixed viscosity successively, that is linear combination of linear and nonlinear viscosity, for example: 

q=

(# + 1) 2 ∂u h 4v ∂x

2

?; < @ @ (# + 1)  ∂u 2 2 #p  ∂u 2 A + + h2 . h2 4v ∂x v ∂x

   -p ∂u  As  #+1 h ) >> 4v ∂x v , it is a quadratic viscidity, that is q=

  # + 1 h2 ∂u 2 . 2 v ∂x

   -p ∂u  h ) 0, > 0, ⎨ ∂x ∂x b= ⎪ ∂u ⎪ ⎩ < 0. 0, ∂x

(4.3.2)

where

Looking for its traveling wave solution u = u(. ), . = x – Dt, substituting eq. (4.3.1), we have –D

d db du + f (u) = : . d. d. d.

Integrating the above equation with respect to . one times, we obtain   db :b + Du – f (u) = c, d.

(4.3.3)

(4.3.4)

428

4 Physical Viscosity and Viscosity of Difference Scheme

u

u = u+ u = u–

Fig. 4.1

where c is arbitrary constant, if u ≠ const, then we have 1 1 du = :– ! I(u) ! , d.

(4.3.5)

where I(u) = c – Du + f (u). In view of eq. (4.3.5), we get 1

1

d. = : ! I(u)– ! du. Suppose I(u+ ) = I(u– ) = 0.

(4.3.6)

u 1 It is easy to know u–+ I– ! (u)du converges as ! > 1, but diverges as 0 < ! ≤ 1. From this if we know the width of the shock wave is finite for the nonlinear viscosity, whose 1 u 1 length is B. = : ! u–+ I– ! (u)du, then obliterated discontinuities are infinite for the linear viscosity, namely the change of u(. ) occurs in the whole x-axis. It tends to u– and u+ only when . → ∞. Traveling wave solution for nonlinear viscous equations (4.3.1) are shown in Fig. 4.1. u– , u+ are adjacent singularities of function I(u), as shown in Fig. 4.2. This moment, u(. ) → u+ as . → +∞, u(. ) → u– as . → –∞, along discontinuity line x = Dt we have dx f (u+ ) – f (u– ) . =D= dt u+ – u–

(4.3.7)

We now consider the traveling wave solution f = f (s), s = x – Dt to the following onedimensional ideal hydrodynamic equations in Lagrange coordinate and with different artificial viscosity:

4.3 The Fundamental Difference Between Linear and Nonlinear Viscosity Qualitatively

429

Z

z = f(u)

u–

u+

z = C – Du

u

Fig. 4.2

⎧ ut + (p + q)x = 0, ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ v + ux = 0, ⎪ ⎪ t ⎪ 5 6 ⎪ 2 ⎪ ⎪ ⎨ :+ u + (p + q)u = 0, 2 t x ⎪ ⎪ ⎪ ⎪ p(+∞) = p1 = 0, p(–∞) = p2 , v(+∞) = v1 , v(–∞) = v2 , ⎪ ⎪ ⎪ ⎪ ⎪ u(+∞) = u1 = 0, u(–∞) = u2 , ⎪ ⎪ ⎪ ⎩ pv = (# – 1 ) :,

(4.3.8)

We can obtain the following by eq. (4.3.8) 1 qv = (# – 1)D2 (v1 – v)(v – v2 ), q(±∞) = 0. 2

(4.3.9)

We now take the form of q as qv = -|ux |1+, , 0 ≤ , ≤ 1, - is constant.

(4.3.10)

So we can find the following by eq. (4.3.9): v(s)q(s) = -D1 + ,



dv ds

1 + , .

(4.3.11)

Introducing a new function +(s) =

6 5 2 1 v – (v1 + v2 ) , Bv = v1 + v2 . Bv 2

(4.3.12)

430

4 Physical Viscosity and Viscosity of Difference Scheme

Then by eqs (4.3.9) and (4.3.11), we have 5 63 2 + (s) = a 1 – + (s) , ′

(4.3.13)

where 3=

5 63 1 #+1 , a = 0.5 (0.5DBv)1–, , 1+ , -

(4.3.14)

v = v1 , +(s) = 1; v = v2 , +(s) = –1.

(4.3.15)

and

If s1 = min{s; +(s) = 1}, s2 = min{s; +(s) = –1}, then let L = s2 – s1 be the width of shock wave. We obtain the following by integrating eq. (4.3.13): I = aL,

(4.3.16)

where 

1

I= –1

√ A(1 – 3) d+ = 0 , (1 – +2 )3 A(3/2 – 3)

(4.3.17)

L = 2-/(# + 1)0.5 0.

(4.3.18)

A(s) is Gamma function. When , = 1(3 = 0.5), we have

If let - = -0 h1+, , L = nh, by eqs (4.3.14) and (4.3.16) we obtain -0 = 0.5(# + 1)(0.5DBv)1 – , n1 + , I –1 – , .

(4.3.19)

As can be seen from eq. (4.3.19), when , = 1 we find -0 = 0.5(# + 1)

n2 . 02

(4.3.20)

4.4 von Neumann Artificial Viscosity and Godunov Scheme Implicit Viscosity

431

So -0 have nothing to do with the shock wave strength. In view of eq. (4.3.17), we know L is finite as 0 < , ≤ 1; L = ∞ as , = 0. Therefore, we can take effective shock width as , = 0: L: = s2 – s1 ,

(4.3.21)

where s1 = max{s; v(s) – v2 = 0.5Bv[1 + +(s)] = :Bv, : is a given number}, s2 = min{s; v(s) – v1 = –0.5Bv[1 – +(s)] = –:Bv, 0 < : < 1}.

(4.3.22)

So  aL: =

1–2: –1+2:

(1 – +2 )–1 d+, a =

0.25 (# + 1)DBv. -

Let L: = nh, -0 = -h, then we have D0 ≈ 0.25(# + 1)DB(n ln–1

(4.3.23)   1 . :

(4.3.24)

From the above we can see the fundamental difference between linear viscosity and nonlinear viscosity: (1) Shock width: As to add the linear viscosity, its width is definite; as to add the nonlinear viscosity, its width is finite as shown in Figs. 4.3 and 4.4, respectively. (2) For the quadratic nonlinear viscosity, shock width has nothing to do with the shock wave strength. For the linear viscosity, shock width has something to do with the shock wave strength.

4.4 von Neumann Artificial Viscosity and Godunov Scheme Implicit Viscosity The one-dimensional hydrodynamic equations in Lagrange coordinates are ⎧ ∂v ∂u ⎪ = , ⎪ ⎪ ⎪ ∂t ∂x ⎪ ⎨ ∂p ∂u =– , ⎪ ∂t ∂x ⎪ ⎪ ⎪ ⎪ ⎩ ∂: = – ∂pu , ∂t ∂x

(4.4.1)

432

4 Physical Viscosity and Viscosity of Difference Scheme

u

u–

u+

O

ξ

Fig. 4.3

u

u–

u+

O

ξ

Fig. 4.4

where v=

1 pv 1 , : = e + u2 , e = . 1 2 #–1

(4.4.2)

Suppose we introduce the artificial viscosity term pressure at the pressure P, ⎧ 2  2 ∂u l ∂u ⎪ ⎪ , < 0, ⎨ ∂x q = v ∂x ⎪ ∂u ⎪ ⎩ 0, ≥ 0, ∂x

(4.4.3)

4.4 von Neumann Artificial Viscosity and Godunov Scheme Implicit Viscosity

433

we have the hydrodynamic equations with artificial viscosity term ⎧ ∂v ∂u ⎪ ⎪ = , ⎪ ⎪ ∂t ∂x ⎪ ⎪ ⎨ ∂(p + q) ∂u =– , ⎪ ∂t ∂x ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ ∂: = – ∂(p + q)u . ∂t ∂x

(4.4.4)

Now consider the traveling wave solution to eqs (4.4.4), . = x – Dt. In view of eq. (4.4.4), we have ⎧ d ⎪ ⎪ (Dv + u) = 0, ⎪ ⎪ d. ⎪ ⎪ ⎪ ⎨ d (p + q – Du) = 0, ⎪ d. ⎪ ⎪ ⎪ ⎪ d ⎪ ⎪ ⎩ ((p + q)u – D:) = 0. d.

(4.4.5)

Suppose the boundary conditions of eq. (4.4.5) are as follows: . → +∞, v → v1 , . → –∞, v → v2 , v1 > v2 ; . → +∞, u → u1 , . → –∞, u → u2 ;

(4.4.6)

and q(±∞) = 0. From eqs (4.4.5) and (4.4.5), we have the solution 7 ⎧ √ v 1 + v2 v 1 – v 2 #+1 0 2l ⎪ ⎪ ⎪ , + sin . /l, |. | ≤ √ ⎪ ⎪ 2 2 2 r+1 ⎪ ⎪ ⎪ √ ⎨ 0 2l v(. ) = , v1 , .≥ √ ⎪ # + 12 ⎪ ⎪ ⎪ √ ⎪ ⎪ 0 2l ⎪ ⎪ ⎩ v1 , , . ≤ –√ # + 12 7 ⎧ √ 5 6 D #+1 0 2l ⎪ ⎪ ⎪ u , + – v ) 1– sin (v . /l , |. | ≤ √ 1 1 2 ⎪ ⎪ 2 2 r+1 ⎪ ⎪ ⎪ √ ⎨ 0 2l u(. ) = , u1 , .≥ √ ⎪ # + 12 ⎪ ⎪ ⎪ √ ⎪ ⎪ 0 2l ⎪ ⎪ ⎩ u1 , , . ≤ –√ # + 12

(4.4.7)

(4.4.8)

u(. ) is as shown in Fig. 4.5. First, we study the artificial viscosity implied in Godunov method, Godunov difference equations of eq. (4.4.1) are as follows:

434

4 Physical Viscosity and Viscosity of Difference Scheme

u u2

u1 ξ Fig. 4.5

⎧ Bt n ⎪ n+1 n n ⎪ ⎪ ⎪ Vm+ 21 = Vm+ 21 + Bx (Um+1 – Um ), ⎪ ⎪ ⎨ Bt n n unm+ 1 = unm+ 1 – ), (Pm+1 – Pm ⎪ Bx 2 2 ⎪ ⎪ ⎪ ⎪ Bt n ⎪ n n ⎩ :n+11 = :n 1 – Um ), (P U n – Pm m+ 2 m+ 2 Bx m+1 m+1

where

(4.4.9)

⎧ an 1 Pn 1 + an 1 Pn 1 + an 1 an 1 (un 1 – un 1 ) ⎪ m+ 2 m– 2 m– 2 m+ 2 m+ 2 m– 2 m– 2 m+ 2 ⎪ n ⎪ ⎪ , Pm = ⎪ n n ⎪ a 1 +a 1 ⎨ m+ m– 2

2

(4.4.10)

⎪ an 1 un 1 + an 1 un 1 + Pn 1 – Pn 1 ⎪ m– 2 m– 2 m+ 2 m+ 2 m– 2 m+ 2 ⎪ n ⎪ ⎪ , ⎪ Um = ⎩ an 1 + an 1 m+ 2

anm+ 1

2

m– 2

⎧ n n am+ 1 = Cm+ ⎪ 1 , linear sound velocity, ⎪ ⎪ 2 2 ⎨ ? @ : @ (# + 1)pm– 1 + (# – 1)pm+ 1 ⎪ n 2 2 A ⎪ a = , shock wave. ⎪ ⎩ m+ 21 2vm+ 1 2

an

m+ 21

: we only need to change the index m +

1 2

in expression of an

m+ 21

to m – 21 .

We can denote fjn = f (jBx, nBt). We make the Taylor expansion at the point ((m + 21 )Bx, nBt) of the differential equation (4.4.9) under the approximation of linear sound velocity, so we get the approximate equations ⎧ ∂v ∂(u + m) ⎪ ⎪ – = 0, ⎪ ⎪ ∂t ∂x ⎪ ⎪ ⎨ ∂u ∂(p + q) + = 0, ⎪ ∂t ∂x ⎪ ⎪ ⎪ ⎪ ⎪ ∂q ∂(p + q)(u + m) ⎩ + = 0, ∂t ∂x

(4.4.11)

4.4 von Neumann Artificial Viscosity and Godunov Scheme Implicit Viscosity

435

where ⎧ 1 ∂p ⎪ !(Bx), ⎨m = – c ∂x ⎪ ⎩ q = –c ∂u !(Bx), ∂x

(4.4.12)

!(h) is the infinitesimals of the same order of h. So we can regard eqs (4.4.11) as the viscosity equations of eqs (4.4.1). We consider the traveling wave solutions of eqs (4.4.11). Let . = x – Dt, by eqs (4.4.11) we have ⎧ d ⎪ ⎪ (u + m + Dv) = 0, ⎪ ⎪ d. ⎪ ⎪ ⎪ ⎨ d (p + q – Du) = 0, ⎪ d. ⎪ ⎪ ⎪ ⎪ d ⎪ ⎪ ⎩ (D: – (pu + qu + pm)) = 0. d.

(4.4.13)

Integrating eq. (4.4.13) we have ⎧ ⎪ u + m + Dv = c1 , ⎪ ⎨ Du – (p + q) = c2 , ⎪ ⎪ ⎩ D: – (pu + qu + pm) = c , 3

(4.4.14)

where ⎧ ⎪ c = Dv1 + u1 = Dv2 + u2 , ⎪ ⎨ 1 c2 = Du1 – p1 = Du2 – p2 , , ⎪ ⎪ ⎩ c = D: – p u = D: – p u . 3 1 1 1 2 2 2

(4.4.15)

From the first term of eqs (4.4.14), we have v=

1 (c1 – u – m), D

(4.4.16)

Substituting the third term of eq. (4.4.14), we get 5 6 p(c1 – u – m) u2 D + – p(u + m) – qu = c3 . (# – 1)D 2 We obtain the following by eq. (4.4.14):

(4.4.17)

436

4 Physical Viscosity and Viscosity of Difference Scheme

–

  # Du2 c1 – u c1 – u pm – q = c3 – + (p + q) u – #–1 #–1 2 #–1   2 Du c1 – u = c3 – + (Du – c2 ) u – 2 #–1 D(# + 1) = (u – u! )(u – u" ), 2(# – 1)

(4.4.18)

where u! , u" satisfy the following quadratic equation: u2 –

2(c1 + cD2 #) 2( c1Dc2 + cD3 (# – 1)) u+ = 0, #+1 #+1

(4.4.19)

5 6 2(c1 + cD2 #) 1 2(c1 + cD2 #) 2 ± 2(# + 1) 2 #+1   1 2 8 c1 c2 c3 – , + (# – 1)) #+1 D D

(4.4.20)

thereby u!," =

with c1 = Dv1 + u1 , c2 = Du2 – p2 , c3 = D

D1 v 1 Du2 – p1 u 1 + . #–1 2

(4.4.21)

Substituting into eq. (4.4.20), we obtain u!,"

  (c1 + cD2 #) p1 1 = ± Dv1 – # = u2,1 , #+1 #+1 D

(4.4.22)

so eq. (4.4.18) can be written as –

# c1 – u D(# + 1) pm – q= (u – u1 )(u – u2 ). #–1 #–1 2(# – 1)

(4.4.23)

Take advantage of eq. (4.4.14) and expression (4.4.12) of m, q, we have 5 D(# + 1) 1 dp = – (u – u1 )(u – u2 ) ′ c d. 2#p 6 (c1 – u)(Dp – p – c2 ) + = –>(u, p), #p du –c ′ = Dp – p – c2 = –8(u, p), d.

–

where . ′ =

. , !(Bx)

and eqs (4.4.24) and (4.4.25) can also be rewritten as

(4.4.24)

(4.4.25)

4.4 von Neumann Artificial Viscosity and Godunov Scheme Implicit Viscosity

du 8(u, p) v8(u, p) (c1 – u + >)8(u, p) = 2 = = . dp c >(u, p) #p>(u, p) D#p>(u, p)

437

(4.4.26)

We will discuss the several analytical integral situations of eqs (4.4.24), (4.4.25) or (4.4.26), for its general property of singular point, we will talk about it later. (I) m = 0, q ≠ 0. At the moment, because of eq. (4.4.23), we have –(c1 – u) D(# + 1) q= (u – u1 )(u – u2 ). #–1 2(# – 1)

(4.4.27)

In view of eq. (4.4.14), u + Dv = c1 and c1 = u1 + Dv1 , substituting them into eq. (4.4.27), we obtain qv =

#+1 2 D (v – v1 )(v2 – v), 2

(4.4.28)

then utilizing q = –c

dv du =D , '= ′ d. d'



1 ′ ′ d. , . = . /!, c

(4.4.29)

we have v1 dv # + 1 = D(1 – )(v2 – v). d' 2 v

(4.4.30)

Integrating eq. (4.4.30), we obtain  (#+1) (#+1) (v – v2 )v2 = e 2 D(v1 –v2 )' = e 2 D(v1 –v2 ) v (v1 – v) 1

d. ′ c

.

(4.4.31)

(1) When c ≈ const, (#+1) . (v – v2 )v2 = e 2 D(v1 –v2 ) !c . v 1 (v1 – v)

(2) When c ≈



#(p+q) v ,

substituting dv du q = –c ′ = cD ′ , d. d. 7 ′ 2 (c2 – D v)# c= , v

and p + q = Du – c2 = c1 D – D2 v – c2 into eq. (4.4.28), we have

(4.4.32)

438

4 Physical Viscosity and Viscosity of Difference Scheme

 dv #+1 #(c′2 – D2 v)v ′ = D(v – v1 )(v2 – v). d. 2

(4.4.33)

Integrating eq. (4.4.33), we obtain [D2 #v2 – #(c′2 – D2 v2 )]  sin–1 2 D2 # ⋅

[–2D2 #(v – v2 ) – (D2 #v2 – #(c′2 – D2 V2 ))]

[D2 #v2 – #(c′2 – D2 v2 )]2 + 4D2 #2 v2 (c′2 – D2 v2 ) 5 61/2   ′ 2 (c2 – D2 v2 )#2 vv2 (c′2 – D2 v) ′ 2 ln + (c2 – D v2 )#v2 v – v2  ′ 2 2 2#v2 (c2 – D v2 ) [D #v2 – #(c′2 – D2 v2 )]  + + #c′2 + v – v2 2 D2 # ⋅ sin–1 

[–2D2 #(v1 – v) – (D2 #v1 – #(c′2 – D2 V1 ))]

[D2 #v1 – #(c′2 – D2 v1 )]2 + 4D2 #2 v1 (c′2 – D2 v1 ) 5 61/2   ′ 2 (c2 – D2 v1 )#v1 (c′2 – D2 v)#v ′ 2 ln + (c2 – D v2 )#v2 v1 – v  ′ 2 2#v2 (c2 – D v1 ) #+1 + + #c′2 = D(v1 – v2 ). ′ . v1 – v 2

(3) When c ≈



#p) v ,

(4.4.34)

in view of eqs (4.4.14), we have p = (# – 1)[Du2 + c3 + c2 u]/(c1 – u).

In view of eq. (4.4.27), we have 7 du D(# + 1) Du2 = (u – u1 )(u – u2 )/ #(# + 1)D( + c3 + c2 u). ′ d. 2 2 Integrating the above equation, we obtain (c3 > 0) 6 Du√1 +c2 5√ 7 √ D 2 2D 2D u + c3 + c2 u + 2D(u – u1 ) + (Du1 + c2 ) 2 5√ 7 6– Du√2 +c2 √ D 2 2D ⋅ 2D u + c3 + c2 u + 2D(u2 – u) – (Du2 + c2 ) 2   5 √2D D u2 + c + c u c u + c u + D u2 3 2 3 2 1 2 2 1 ⋅ u – u1

(4.4.35)

4.4 von Neumann Artificial Viscosity and Godunov Scheme Implicit Viscosity

439

6–(c3+c2 u1 – D u2 )1/2 2 1 2(c3 + c2 u1 – D2 u21 ) + (Du1 + c2 ) u – u1 7 7 5√ D 2D D 2 ⋅ u + c3 + c2 u c3 u + c2 u2 + u22 u2 – u 2 2  6 c3 +c2 u2 – D2 u22 2(c3 + c2 u2 + D2 u22 ) + – (Du2 + c2 ) u2 – u

+

√

=e

(#+1) D . – √ (u2 –u1 ) !(Bx) 2 #(#–1)

(4.4.36)

(II) m ≠ 0, q = 0. In view of eq. (4.4.23), we have –# D(# + 1) pm = (u – u1 )(u – u2 ). #–1 2(# – 1)

(4.4.37)

dp Take advantage of p = Du – c2 and m = – c1 d. ′ , we find

du –(# + 1) = (u – u1 )(u – u2 ), d' 2#(Du – c2 ) where ' =



(4.4.38)

cd. ′ , integrating eq. (4.4.38), we have

(u –

c2 –u D 1 u u1 ) 1 –u2

⋅ (u2 –

c2 –u D 2 u u1 ) 1 –u2

=e

#+1

– 2#D '

.

(4.4.39)

If c ≈ const, we have #+1

(u – u1 )(c2 /D–u1 ) (u2 – u)(c2 /D–u2 ) = e 2#D

(u2 –u1 )c. ′ /!

.

(4.4.40)

We discuss the traveling wave solutions of the equations with implicit viscosity just in the approximation of linear sound velocity, for the Godunov differential scheme with expressions of shock conditions am+ 1 , we can similarly talk 2 about it. For the one-dimensional hydrodynamic equations in Euler coordinates ⎧ ∂1 ∂1u ⎪ ⎪ + = 0, ⎪ ⎪ ∂t ∂x ⎪ ⎪ ⎨ ∂1u ∂ + (1u2 + p) = 0, ∂t ∂x ⎪ ⎪    ⎪ ⎪ ∂  ⎪ u2 ∂ u2 ⎪ ⎩ 1 e+ + 1u e + p/1 + = 0, ∂t 2 ∂x 2 its Godunov differential scheme may be similar to write

(4.4.41)

440

4 Physical Viscosity and Viscosity of Difference Scheme

$ Bt # (1U)nm+1 – (1U)nm , Bx 2 2 $ Bt # (P + 1U 2 )nm+1 – (P + 1U 2 )nm , (1u)n+1 = (1u)nm+ 1 – m+ 21 Bx 2 5  5  6n+1 6n 2 u2 u = 1 e+ 1 e+ 2 2 m+ 21 m+ 21 5  5  6n 6n , Bt U2 U2 , – – 1U e + P/1 + 1U e + P/1 + Bx 2 2 m+1 m   n n n = p 1 , e Pm+ . 1 m+ 1 m+ 1

1n+1 = 1nm+ 1 – m+ 1

2

2

(4.4.42)

2

In view of the discontinuity of 1(x, t), eqs (4.4.42) can be approximated as 5 6 Bt n n n – 1nm– 1 Um 1m+ 1 Um+1 , Bx 2 2 2 2 5 Bt n n = (1u) – (1u)nm+ 1 Um+1 (1u)n+1 m+ 21 m+ 21 Bx 2 6 n n n –(1u)nm– 1 Um–1 + Pm+1 – Pm ,

= 1nm+ 1 – 1n+1 m+ 1

2

5 5  6n+1 6n 6n 5  u2 Bt u2 u2 = 1 e+ – Un 1 e+ 1e + 1 2 2 Bx 2 m+ 1 m+1 m+ 21 m+ 21 2 ,  n 2 u – 1e + 1 U n + Pm+1 Um+1 – Pm Un , 2 m– 1 m

(4.4.43)

2

where the expressions of discontinuous decomposition of Pm , Um are similar to eq. (4.4.10). We just need to change coefficient am± 1 correspondingly, for example, under 2  the approximation of linear sound velocity, we replace a 1 to 1c 1 , where c2 = ∂p  . m+ 2

m– 2

∂1 s

Similar to Godunov difference scheme, the viscous term is also implicit in the quality, momentum and energy equation in Rusanov difference scheme. For the convenience of calculation, we only list its one-dimensional difference scheme and corresponding approximate equations. The one-dimensional isothermal hydrodynamic equations in Lagrange coordinates are ⎧ ∂u ∂p(v) ⎪ ⎪ ⎨ + = 0, ∂t ∂x (4.4.44) ⎪ ∂v ∂u ⎪ ⎩ – = 0. ∂t ∂x The corresponding Rusanov difference scheme is

4.4 von Neumann Artificial Viscosity and Godunov Scheme Implicit Viscosity

441

Bt n n+1 n n Um = Um – ) (Pm+1 – Pm–1 2Bx 6 5 1 n n n n + !nm+ 1 (Um+1 – Um ) – !m– 1 (Um – Um–1 ) , 2 2 2 Bt n+1 n n n = Vm + – Um–1 ) Vm (Um+1 2Bx 6 5 1 n n n n + !nm+ 1 (Vm+1 – Vm ) – !m– 1 (Vm – Vm–1 ) , 2 2 2

(4.4.45)

where ⎧  1 ⎪ !nm+ 1 = 9nm+1 (u + c)nm+1 + 9nm (u + c)nm , ⎪ ⎪ ⎪ 2 2 ⎨  1 !nm– 1 = 9nm (u + c)nm + 9nm–1 (u + c)nm–1 , ⎪ ⎪ 2 2 ⎪ ⎪ ⎩ 9 is constant and satisfies the stability condition 9Bt(u + c) ≤ 1.

(4.4.46)

It is not difficult to find the differential equations approximated by differential equations (4.4.45), that is ⎧ ∂u ∂(p + q) ⎪ ⎪ + = 0, ⎨ ∂t ∂x ⎪ ∂v ∂(u + m) ⎪ ⎩ – = 0, ∂t ∂x

(4.4.47)

where ∂u B(h), ∂x ∂v m = 9(u + c) B(h), ∂x q = 9(u + c)

B(h) is the infinitesimals of same order of h. The one-dimensional hydrodynamic equations in Euler coordinate are ⎧ ∂1 ∂1u ⎪ + = 0, ⎪ ⎪ ⎪ ∂t ∂x ⎪ ⎨ ∂ ∂1u + (p + 1u2 ) = 0, ⎪ ∂t ∂x ⎪ ⎪ ⎪ ⎪ ⎩ ∂: + ∂ (: + p)u = 0. ∂t ∂x

(4.4.48)

It is easy to verify approximate equations of Rusanov difference scheme of eq. (4.4.48), that is

442

4 Physical Viscosity and Viscosity of Difference Scheme

5 6 ∂1 ∂1u ∂ ∂1 + = 9(u + c) B(h), ∂t ∂x ∂x ∂x 5 6 ∂ ∂ ∂1u ∂1u 2 + (p + 1u ) = 9(u + c) B(h), ∂t ∂x ∂x ∂x 5 6 ∂ ∂ ∂e ∂: + (: + p)u = 9(u + c) B(h). ∂t ∂x ∂x ∂x

(4.4.49)

4.5 Several Difference Schemes with Mixed Viscosity In this section, we introduce B.F. Kurolatenko’s several difference schemes with mixed viscosity, which are some basic improvements of S.K. Godunov scheme, and we only consider to calculate two kinds of waves: rarefaction wave and shock wave. (I) Consider one-dimensional hydrodynamic equations: ⎧ ∂v ∂u ⎪ – = 0, ⎪ ⎪ ⎪ ∂t ∂x ⎪ ⎨ ∂u ∂p (4.5.1) + = 0, p = f (v, e), ⎪ ∂t ∂x ⎪ ⎪ ⎪ ⎪ ⎩ ∂e + p ∂v = 0. ∂t ∂x Define two kinds of waves on the interval [xi , xi+1 ]: R(rarefaction wave) S(shock wave)

ui – ui+1 ≥ 0, xi – xi+1 ui – ui+1 < 0. xi – xi+1

We know that the Rankine–Hugoniot strong discontinuity conditions of eqs (4.5.1) are ⎧ 1 ⎪ v+ – v– = – (u+ – u– ), ⎪ ⎪ ⎪ 9 ⎪ ⎨ 1 (4.5.2) u – u– = (p+ – p– ), ⎪ + 9 ⎪ ⎪ ⎪ ⎪ ⎩ e – e = 1 (p + p )(v – v ), + – + – + – 2 where 9 is the velocity of shock wave. Set up the following difference scheme of eqs (4.5.1): 5 6 ⎧ Bt n+1 n n n ⎪ ⎪ u – u = – ) – (p ) (p + i+ 1 + i– 1 , ⎪ i i ⎪ Bx 2 2 ⎪ ⎪ ⎪ ⎪ ⎪ Bt ⎪ ⎪ vn+11 – vn 1 = (un – uni–1 ), ⎨ i– 2 i– 2 Bx i 5 6 1 ⎪ n+1 n n n n n+1 ⎪ ⎪ e – e = ) + (p ) (p + i– 1 + i– 1 (vei– 1 – vi– 1 ), ⎪ i– 21 i– 21 ⎪ 2 2 2 2 2 ⎪ ⎪   ⎪ ⎪ ⎪ ⎪ ⎩ en+11 = f vn+11 , en+11 , i– i– i– 2

2

2

(4.5.3)

4.5 Several Difference Schemes with Mixed Viscosity

where p+ select as follows:

n+ 21

n+1 As un+1 – un+1 i i–1 ≥ 0, namely it is R wave, take (p+ ) 1 = p n+1 As un+1 i – ui–1 < 0, namely it is S wave, take v– ±(un+1 – un+1 i i–1 ) in eqs (4.5.2), so we have

=

i– 2 i– 21 vn 1 , p– = i– 2

; pn 1 , e– = en 1 , u+ – u– = i– 2

⎧ 1 n ⎪ ⎪ (v+ )n+1 – vi– (un+1 – un+1 1 = ∓ ⎪ i i–1 ), i– 21 ⎪ 9 1 2 ⎪ i– ⎪ 2 ⎪ ⎪ 5 6 ⎪ ⎪ 1 ⎪ n+1 n+1 n+1 n ⎪ – u ) = ) – (p ) ± (u (p , ⎪ + + 1 1 i i–1 ⎨ i– 2 i– 2 9i– 1 2 5 65 6 ⎪ ⎪ 1 ⎪ n+1 n n+1 n n n+1 ⎪ ) – (e ) = ) + p – (v ) (p v (e ⎪ + i– 1 + i– 1 + i– 1 + i– 1 , ⎪ i– 21 i– 21 2 ⎪ 2 2 2 2 ⎪ ⎪   ⎪ ⎪ ⎪ ⎪ ⎩ (p+ )n+1 = f (v+ )n+1 , (e+ )n+1 . i– 1 i– 1 i– 1 2

2

443

i– 2

(4.5.4)

2

By eq. (4.5.4) we can solve (p+ )n+11 . If state equation is p = (# – 1)E/v, then we obtain i– 2

= pni– 1 + (p+ )n+1 i– 1 2

2

# + 1 Bu2 4 vn 1 i– 2

⎡⎛

⎞2

#pn

#+1 ⎠ + + ⎣⎝ 4 vn 1 vn Bu2

i– 21

⎤ 21 Bu

2⎦

(4.5.5) ,

i– 21

i– 2

where 2 (Bu)2 = (un+1 – un+1 i i–1 ) .

To write eq. (4.5.5) as continuous function form p+ = p + q,

(4.5.6)

where q=

 5  6 # + 1 2 ∂u 2 2 ∂u 2 + Bx ∂x 4v ∂x  2 1/2 #p ∂u + Bx2 . v ∂x #+1 2 Bx 4v



From the expression  of eq. (4.5.7), we can find  #+1 ∂u    As >> #p , Bx 4v

∂x

v

q=

# + 1 Bx2 2 v



∂u ∂x

2 ;

(4.5.7)

444

4 Physical Viscosity and Viscosity of Difference Scheme

   #p ∂u  As  #+1 4v Bx ∂x > – 2v ∂x Bx, m= 8 ⎪ 2vBx ∂p #p # + 1 ∂p ⎪ ⎪ – , as (u, p)]8(u, p) = , dp D#p>(u, p)

(4.8.9)

where ⎧ ⎪ ⎨ >(u, p) = D(# + 1) – u(u – u2 ) + (c1 – u)(Du – p) , 2#p #p ⎪ ⎩ 8(u, p) = p – Du.

(4.8.10)

So we have ⎧ dp p(6u2 – p(6 – u)) ⎪ ⎪ =9 , ⎪ ⎨ du (p(6 – u) + 3u2 )(p – 6u) u = 0, p = 0, ⎪ ⎪ ⎪ ⎩ u = 3, p = 18, 8 36 p dx =3 , du (p(6 – u) + 3u2 ) p – 6u

(4.8.11)

(4.8.12)

and the obtained solution is recorded as uq,m,A . (II) Artificial viscosity method for fluid mechanic equations with radiative radiation conductivity Take A = 30, c1 = 32, c2 = 600, cv = 90, R = 60, 4A′ a = 50, 11 = 1, T1 = 1, u1 = 30, l = 1.2. (i) von Neumann artificial viscosity method: ⎧ (90T + (32 – u2 )u – 600) dT ⎪ ⎪  = –0.72 , ⎪ ⎪ ⎪ uT 3 (32 – u)u – 60T ⎨ du 1 dx ⎪ = –1.2[(32 – u)u – 60T]– 2 , ⎪ ⎪ du ⎪ ⎪ ⎩ u = 30, T = 1; u = 10, T = 3.6667.

(ii)

(4.8.13)

The obtained numerical solutions are recorded as u#,N and T#,N . Godunov implicit viscosity method: ⎧ (90T + (32 – u2 )u – 600) dT ⎪ ⎪  = –0.15492 , ⎪ ⎪ ⎪ du T 2.5 (32 – u)u – 60T ⎪ ⎨ √ u 60T dx ⎪ = –0.03333 , ⎪ ⎪ du [(32 – u)u – 60T] ⎪ ⎪ ⎪ ⎩ u = 30, T = 1; u = 10, T = 3.6667.

The obtained numerical solutions are recorded as u#,A and T#,A .

(4.8.14)

4.8 Numerical Calculation Results and Analysis

463

The calculation results are shown in Tables 4.1–4.4: The above results of the calculation are shown in Figs 4.9–4.15. Next we analyze qualitatively the effects of wave width B. of traveling wave solutions for a variety of different artificial viscosity method of ideal hydrodynamics. These analyses show that the results are in agreement with the numerical results.

Table 4.1 Hydrodynamic equations without radiation heat conduction. .

uq,0,z

u0,m1 ,p

u0,m2 ,s

uq,0,N

uq,0,A

uq,m,A

–1 –0.75 –0.5 –0.25 0 0.1 0.2 0.25 0.5 0.75 1

2.39999 2.2199 2.01 1.77 1.5 1.38 1.26 1.2 0.9 0.615 0.345

2.795 2.605 2.307 1.92 1.5 1.34 1.180 1.076 0.651 0.240 0.155

2.34 2.16 1.965 1.74 1.5 1.395 1.290 1.245 0.975 0.735 0.495

2.8869 2.66055 2.3346 1.9365 1.5 1.3245 1.14345 1.0635 0.6654 0.33945 0.1131

2.5248

2.22 2.1 1.96 1.72 1.5

2.15 1.8786 1.5 1.3 1.26 1 3.32 0 0

1.26 0.98 0.7 0.4

Table 4.2 Derivative comparison. .

duq,0,z d.

du0,m1 ,p d.

du0,m2 ,s d.

duq,0,N d.

duq,0,A d.

0

–1.13

–1.695

–0.999

–1.77

–1.7

Table 4.3 With radiation conduction u# (. ). .

u#,A

u#,N

u#,c (without artificial viscosity)

–15 –14 –12 –10 –8 –2 0 1 2 2.9 3.8

30 29.9 29.4 28 26.9 23.7 22 19.7 15 11.3 10.2

30 30 29.3 28 26.9 23.9 22 17.7 12.7 10 10

29.7 29.4 28.1 27 25.9 23 22 10 10 10 10

464

4 Physical Viscosity and Viscosity of Difference Scheme

Table 4.4 With radiation conduction T# (. ). .

T#,A

T#,N

T#,c

–15 –13 –10 –7 –5 –3 0 1 2 3.2 3.7

1 1.06 1.76 2.36 2.72 3 3.38 3.50 3.60 3.65 3.67

1 1.14 1.86 2.46 2.80 3.08 3.48 3.56 3.64 3.67 3.67

1.1 1.58 2.26 2.78 3.06 3.33 3.67 3.67 3.67 3.67 3.67

Denote u(x) calculated by mixed viscosity method, where q=

γ+1 2 h 4v

∂u 2 + ∂x

2 γ + 1 2 ∂u 2 γp ∂u 2 + v h2 h ∂x ∂x 4v

Denote u(x) calculated by another mixed viscosity method, where ∂p – h ∂x q = 0,m = γp γ+1 ∂p h – 4 v ∂x ∂up Denote u(x) calculated by Godunov viscosity, where q = 0,m = – h c ∂x Denote u(x) calculated by von Neumann viscosity, where q =

l2 v

∂u 2 ∂x

u(ξ )

Denote u(x) calculated by Godunov viscosity, where m = 0, q = –c ∂u ∂x Denote exact solution Denote u(x) calculated by Godunov viscosity, where m = –c

∂p , q = –c ∂u ∂x ∂x

3

2

1

ξ 0 Fig. 4.9

–6

–4

–2

0

2

4.8 Numerical Calculation Results and Analysis

u(x) 3

2 u(x) is calculated by Godunov u(x) is calculated by PIC viscosity. Denote exact solution.

1 Denote approximate solution of u(x) ∂e ∂ calculated by PIC viscosity with ≠0 λ ∂x ∂x ∂e ∂ 0 ≠ Denote PIC viscosity with λ ∂x ∂x and Godunov viscosity (m = 0).

x –4

–3

–1

–2

0

1

T (u)

Fig. 4.10

T (u) figure of Godunov implied visous equation (7.2)

7

6 90 T + (32 – 21 u)u–600 = 0 5

(32 – u)u–60T = 0

4

3 Integral curves 2

1

0

4

8

12

20

16 u

Fig. 4.11

24

28

32

465

466

4 Physical Viscosity and Viscosity of Difference Scheme

p(u) figure of Godunov implied viscous equation (7.3) P = (6u)

20 18 16

P=

6u2 6–u

P = (u)

14 12 10 8 6

Integral curves

4 2 u 0

1

2

3

4

5

Fig. 4.12

u(ξ )

20

10

Denote exact solution. Denote velocity distribution calculated by von artificial viscosity method. Denote velocity distribution calculated by Godunov discontinuous decomposition method with radiation.

ξ 0 Fig. 4.13

–15

–10

–5

0

5

4.8 Numerical Calculation Results and Analysis

Denote exact solution. Denote temperature distribution calculated by von artificial viscosity method with radiation.

3

T(ξ )

Denote temperature distribution calculated by Godunov discontinuous decomposition method with radiation.

2

1

ξ

0

–15

–10

–5

0

u(ξ )

Fig. 4.14

6

5 Effects of different viscosity constants l on velocity u(ξ)

4 l = 1.2 velocity distribution

3

l = 2 velocity distribution l = 3 velocity distribution ξ = 0 velocity distribution with initial value u = 23.1, T = 2.595

2

ξ = 0 velocity distribution with initial value u = 23.1, T = 2.646

1

0 Fig. 4.15

5

10

15

ξ

5

467

468

4 Physical Viscosity and Viscosity of Difference Scheme

(i) von Neumann viscosity: √ 2l The width of stationary waves is B. = 2 √0#+12 , l = 1, 2, # = 3, B. = 2.6656. (i) Mixed viscosity ⎧   5  6   ⎪ (# + 1) 2 ∂u 2 2 #p 2 ∂u 2 1/2 # + 1 2 ∂u 2 ⎨ + + Bx , Bx Bx q= 4v ∂x 4v ∂x v ∂x ⎪ ⎩ m = 0. When # = 3, Bx = 1, u1 = p1 = 0, v1 = 1, we have   3–:  u(32 – u) 3+: √ 1 B. = du = – 3 sin–1 + 27 ln 2u(3 – u) 6 6 0 ; √ < √  √ 2 27 –:2 – 6: + 27 54 + – 6 – 27 ln 12 , : : : = 0.1, B. ≈ 4, : = 0.01, B. ≈ 6. ∂p (iii) Godunov viscosity (q = 0, m = – Bx c ∂x ) 5 6 ∞  3–: 7 √ 1 t+1  3 6–4 1 –1 B. = du = 2 3 ln + tg t  3+: 2 3u 3 – u 2 t – 1 0

3–:

: = 0.1, B. = 6.6 dy ) (iv) Godunov viscosity (m = 0, q = – d. 7 (#+1) 6 . 1 2(#–1) u = u2 1 – e , 2 : = 0.3, B. ≈ 4.5,

5

: = 0.1, B. ≈ 8.7. So we have Table 4.5. Table 4.5 Comparison of different viscosity method. B.

(: = 0.1)

Method von Neumann viscosity Mixed viscosity Godunov q = 0, m ≠ 0 Godunov q ≠ 0, m = 0

B. 2.6656 4 6.6 8.7

4.8 Numerical Calculation Results and Analysis

469

From the above results, we have the following several points: (I) The artificial viscosity method of hydrodynamics without radiation conduction term: (i) From the traveling wave solutions we can see that the difference in von Neumann viscosity method and all kinds of Godunov viscosity method is significant. The characteristics of the particle velocity calculated by Godunov viscosity method (except a strong wave method) are: wave area is very wide, and starts moving early with poor stamina, the rate of change is small. The von Neumann viscous method is opposite, the movement is too late, the change rate is large, and it tends to be more exact solution quickly. ∂p (ii) In various Godunov viscous scheme, one of the viscous schemes q = 0, m = – Bx c ∂x is the best. The essence of this scheme is obtained by strict shock relations; it is more close to the exact solution, wave area is relatively steep and narrow, but it can carry out the differential calculation. Its inconvenience is more troublesome when you want to carry out the iterative calculation. But this shortcoming can be overcomed, for example, we can simply take the front row value. That is   ∗(n+1) n n pi = max pi+ 1 , pi– 1 , 2

= unj+ 1 + 7 u∗(n+1) j 2

2

pn 1 , pn

i+ 21

i– 2

1 b 2 pi+ 1 [(#

+ 1)pn

2

i– 21

,

+ (# – 1)pn 1 ] i+ 2

(iii) Mixed viscosity method (II), namely q = 0, u∗ = u + 9, 9 = 

– ∂p ∂x Bx #p v

–

#+1 ∂p 2v ∂x Bx

have narrow wave area and close to von Neumann viscosity solution, exact solution; it seems to be able to use. (iv) By comparison with the exact solution, von Neumann viscosity solution which is more close to the exact solution can be seen; in addition to the short time formed by the decomposition of the initial discontinuity, von Neumann viscosity solution has higher accuracy. (II) Hydrodynamic artificial viscosity method with radiation conductivity term: (i) The difference between von Neumann artificial viscosity method and Godunov discontinuous decomposition method is not very significant, especially the impact on the speed difference is smaller, which may be due to the general widening of the wave area by the action of light radiation. Relatively speaking, von Neumann artificial viscosity method still has the characteristics of the front calculation without radiation conductivity term: wave area is very small, especially wave rear, more close to the exact solution.

470

4 Physical Viscosity and Viscosity of Difference Scheme

(ii) Calculation results show that, in von Neumann artificial viscosity method, as shown in Fig. 4.7, the influence of different viscosity coefficients l on the width of the wave area is not significant; of course, in principle, the bigger the l is, the greater the wave area should be pulled. (iii) The solution to the viscous ordinary differential equations with the radiation conductivity is very sensitive for the initial value, just like taking different initial value at . = 0: u1 = 23.1, T1 = 2.595; u2 = 23.1, T2 = 2.646, the results lead to a great difference: . 1.0 5.0 8.4 18.0

uI

uII

25.9 28.7 30.0 30.0

25.1 27.1 28.2 30.0

This may be due to the nonlinear terms T – 2.5 in eq. (4.4.2).

4.9 Local Comparison of Different Viscosity Method to Initial Discontinuity Problem for One-Dimensional Radiation Hydrodynamic Equations with Different Medium In the section, we use the radiation hydrodynamic equations (one-dimensional spherical Lagrange coordinate) for numerical calculation to investigate the difference caused by different artificial viscosities. To this end, we first give an exact solution to the initial discontinuity problem (without radiation conductivity). We investigate one-dimensional initial discontinuity problem and assume that the initial discontinuity image consists of shock wave, contact discontinuity and rarefaction wave as shown in Fig. 4.16, where u1 = u5 = 0, let

Shock

t 2

R

4 5

1 O

Fig. 4.16

3

x

4.9 Local Comparison of Different Viscosity Method to Initial Discontinuity Problem

P1 = 33 x 102 P5 = 124 x 102

ρ1 = 20 ρ5 = 10 γ1 = 1.4648

471

γ5 = 1.6 O

Z

Fig. 4.17

. = p2 /p1 , ' =

v1 , z = p5 /p1 . v2

(4.9.1)

By shock relation, we have 8 u2 =

(, – 1)p1 v1 #+1 (1 – . ), , = . ,. + 1 #–1

(4.9.2)

By rarefaction wave relation: 6 5  #–1 √ 2 #√ . 2# p5 v5 –1 , u3 = #–1 x

(4.9.3)

Utilizing contact discontinuity condition u 2 = u3 , to find out . , and then by shock relation, we have ⎧ ,. + 1 ⎪ ⎪ '= , ⎪ ⎪ ⎪ ,+. ⎪ ⎨ 7 , √ #p v (. – 1), D = 1 1 1+ ⎪ ⎪ , +1 ⎪ ⎪ ⎪ ⎪ 1 ⎩ v3 = v5 (x/. ) # . By rarefaction wave relation, we obtain ⎧   1x ,–1 √ ⎪ ⎪ c4 = #p5 v5 , + ⎪ ⎪ ⎪ ,t , ⎪ ⎪ ⎪  ⎨ 2 2 x (c4 – c5 ) = – c5 , u4 = #–1 #–1 t ⎪ ⎪ ⎪ ⎪   2 ⎪ ⎪ c4 #–1 ⎪ ⎪ . ⎩ 14 /15 = c5 Now consider the following specific model (as shown in Fig. 4.17): x > 0 : 15 = 10, #5 = 1.6, p5 = 124 × 102 , x > 0 : 11 = 20, #1 = 1.4648, p1 = 33 × 102 .

(4.9.4)

(4.9.5)

472

4 Physical Viscosity and Viscosity of Difference Scheme

By the above formula, we have ,1 = 5.302, ,5 = 3.999, x = p5 /p1 = 3.757, ⎧ 29.58 ⎪ ⎪ ⎨ u2 = √5.3023 + 1 (1 – . ), 5 6  ⎪ . 0.2583 ⎪ ⎩ u3 = 1.3647 –1 . 3.75

(4.9.6)

When u2 = u3 , we obtain ⎧ ⎪ . = 2.605, u2 = u3 = –12.33, ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ' = 1.873, D = 23.3016, ⎨ x v3 = 0.123, c4 = 0.25 + 34.5496, ⎪ t ⎪ ⎪ ⎪  3 ⎪ x  ⎪ c4 ⎪ ⎪ . – 45.46 , 14 = 10 ⎩ u4 = 0.75 t 45.66

(4.9.7)

Next we use the different model of radiation hydrodynamic equations (onedimensional spherical Lagrange coordinate) for numerical calculation and analyze the numerical results. Model (I) Material geometric structure: 0(A)20(B)30(C)40∗ Initial condition: 1 : 20, 30, 75. T : 5, 20, 5. u : 0, 0, 0. Model (II) Material geometric structure: 0(A)10(B)20(C)23 Initial condition: the same as model (I) Where ∗ denotes that it consists of three kinds of media A, B, C. 0(A)20 indicates the space occupied by matter A from # = 0 to # = 20, and the rest is similar to deduce. Now we list the numerical results in Tables 4.6–4.8. Table 4.6 Comparison of moving interface position. Model (I)

R = 20

t (time)

von viscosity (nonradiative)

von viscosity (radiation)

God (nonradiative)

Exact solution (nonradiative)

0.04 0.1 0.2 0.3

19.70 19.00 17.95 16.85

19.70 19.00 17.70 16.80

19.50 18.90 17.80 16.70

19.55 18.80 17.53

4.9 Local Comparison of Different Viscosity Method to Initial Discontinuity Problem

Model (II)

473

R = 10

t (time)

von viscosity (nonradiative)

von viscosity (radiation)

God (nonradiative)

God (radiation)

Exact solution (nonradiative)

0.04 0.1 0.2 0.3 0.4

9.70 9.1 7.95 7.15 6.6

9.70 9.0 17.70 7.2 6.8

9.55 8.80 7.85 7.1 6.9

9.55 8.9 7.9 7.3 7.1

9.55 8.80 7.53

Table 4.7 Comparison of interface velocity. Model (I) t 0 0.02 0.04 0.1 0.2 0.3

R = 20

uvon+radiation

uvon+nonradiative

uGod+nonradiative

0 –8.15 –11 –12.4 –11.3 –10.75

0 8.05 –10.55 –11.95 –11.10 –10.55

–12.1 –11.50 –11.1 –11.10 –10.95 –10.55

Model (II) t 0 0.02 0.04 0.1 0.2 0.3

R = 10

uGod+radiative

uvon+radiation

uvon+nonradiative

uGod+nonradiative

–12.10 –11.75 –11.45 –10.95 –8.75 –4.65

0 –8.1 –10.95 –12.05 –10.80 –5.9

0 –8.0 –10.50 –11.75 –10.00 –7.1

–12.10 –11.45 –11.00 –10.90 –9.6 –5.6

Table 4.8 Comparison of rarefaction wave region. uGod+nonradiative

uGod+radiative

uvon+radiation

uvon+nonradiative

uexact (isentropic)

uexact (isothermal)

–4.15

–3.31

–2.54

–3.36

–4.095

–3.79

These results are shown in Figs 4.18–4.23. By comparison (namely start from R = 20 and characteristic lines connect to the still region R – 20 = –45.46t, t ≤ 0.28) of effective local area the following can be seen:

474

4 Physical Viscosity and Viscosity of Difference Scheme

Artificial viscosity + radiation Godunov viscosity + radiation

t

Godunov viscosity without radiation Artifical viscosity without radiation

0.04

0.03

0.02

0.01 R (t) 0.00 10

20

Fig. 4.18

Artificial viscosity without radiation

t

Artificial viscosity + radiation 0.4

Discontinuous decomposition without radiation

0.3

0.2

R(t) figure (larger size)

0.1

R (t) 0.0 10 Fig. 4.19

20

4.9 Local Comparison of Different Viscosity Method to Initial Discontinuity Problem

–u (t) 12

10

8

6 Godunov viscosity + radiation Godunov viscosity without radiation

4

Artificial viscosity + radiation Artificial viscosity without radiation

2 t 0

0.0

0.1

0.2

0.3

Fig. 4.20

u (t) –12

–10

–8 Artificial viscosity + radiation

–6

Artificial viscosity without radiation Discontinuous decomposition without radiation

–4

–2 t

0

0.01

Fig. 4.21

0.1

0.2

475

476

4 Physical Viscosity and Viscosity of Difference Scheme

ρ(x) 40 Artificial viscosity + radiation Godunov viscosity without radiation Godunov viscosity + radiation

30

Artificial viscosity without radiation

20

10

x

0

10

Fig. 4.22

T (x) Artificial viscosity + radiation Godunov viscosity without radiation

20

Godunov viscosity + radiation Artificial viscosity without radiation

15

10

5

x 0 Fig. 4.23

10

4.10 Implicit Viscosity of PIC Method

(i)

(ii)

(iii)

477

Comparing von Neumann viscosity without radiation conduction and Godunov discontinuous decomposition method, discontinuous decomposition method is more close to the exact solution from interface position and interface movement speed, in particular, discontinuous decomposition method accords with the actual situation in the physical in a period of time from the initial formation of discontinuities t ≤ 0.04, but von Neumann artificial viscosity makes it far away from the exact solution due to continuous accelerated dramatically in the interface. Thus at the initial stage of a time, the faster speed of discontinuous decomposition motion is reasonable; of course, making an overview of the total impact on the interface, the difference is very little compared with the difference between the positions of the interface BR ≤ 0.5 cm. The impact of radiation on the interface movement was very little, so we can compare different viscosity methods without radiation in our specific questions, and the role of radiation just pushes the density higher, also raises some temperature, so it only has some effect on speed. As can be seen from the comparison of interface speed, spherically imploding term has a great influence on the speed. For model (I), it is close to the results of the two-dimensional plane. But for model (II), the role of spherically imploding makes the interface speed dropping a lot at R = 10 cm.

4.10 Implicit Viscosity of PIC Method We know that the PIC (particle in cell) method is a very effective method for calculating many kinds of compressible fluids. It uses Euler grids, tracks fluid with discrete particles and uses the motion of discrete particle to calculate transport capacity through the grid boundaries. We can write the equations approximated by PIC method of hydrodynamic equations in Euler coordinate as the following form: ⎧ ∂1 ∂1u ⎪ ⎪ + = 0, ⎪ ⎪ ∂t ∂x ⎪ ⎪ ⎪ ⎨ ∂1u ∂ + ( p + 1u2 ) = 0, ∂t ∂x ⎪ ⎪ ! ⎪ 2" 6 5   ⎪ ⎪ ∂1 e + u2 ∂ ∂e ∂ u2 ⎪ ⎪ + 1 e+ +p u= + , ⎩ ∂t ∂x 2 ∂x ∂x

(4.10.1)

where p=p–+

∂u 1 ∂u , + = 1uBx, –+ = q. ∂x 2 ∂x

(4.10.2)

Now consider the traveling wave solution of eq. (4.10.1). Let . = x – Dt, thus by eq. (4.10.1), we have

478

4 Physical Viscosity and Viscosity of Difference Scheme

⎧ d ⎪ ⎪ (1u – 1D) = 0, ⎪ ⎪ d. ⎪ ⎪ ⎪ ⎨ d (p + 1u2 – 1uD) = 0, d. ⎪ ⎪ ⎪ 6  5    ⎪ ⎪ u2 u2 d de d ⎪ ⎪ ⎩ 1 e+ +p u–D e+ = + . d. 2 2 d. d. Take Bx = 1, u0 = p0 = 0, # = 3, so by integrating eq. (4.10.3), we have ⎧ 1(D – u) = 10 D, ⎪ ⎪ ⎪ ⎨ p = 10 Du,   ⎪ ⎪ ⎪ + de = 1 D 1 u2 – e . ⎩ 0 d. 2

(4.10.3)

(4.10.4)

de First, if + d. = 0, by the third term of eq. (4.10.4), we deduce

1 (p – q) 1 2 u =e= p/1 = 2 #–1 (# – 1)1 (D – u)u 1 du = + u . (# – 1) 2(# – 1) d.

(4.10.5)

If we take D = 6, by eq. (4.10.5) we obtain the exact solution up (. ) = . – 1.5e4. .

(4.10.6)

Similarly, for the Godunov viscosity of one-dimensional equations (4.10.1) in Euler coordinate, m = 0, q = –1c ∂u ∂x , we can obtain the analytical solution uA (. ) = 3 – 1.5e

√2 . 3 .

(4.10.7)

de Now if + d. ≠ 0, then we can obtain equation

6(4 – u)

du d du + u = 8(6 – u)(u – 3). d. d. d.

(4.10.8)

Let u = 3 – 1.5e4(1+!). ,

(4.10.9)

where ! is an undetermined constant; substituting it into eq. (4.10.8), we can have ! ≈ 0.5 analogously, so we can obtain the approximate solution of eq. (4.10.8): up, modify ≈ 3 – 1.5e6. .

(4.10.10)

Now we make a qualitative discussion of eq. (4.10.8), and eq. (4.10.8) can be written as equations

4.10 Implicit Viscosity of PIC Method

⎧ du ⎪ ⎪ ⎨ d. = 9, ⎪ d9 8(6 – u)(u – 3) – 6(4 – u)9 – 92 ⎪ ⎩ = . d. u

479

(4.10.11)

Equation (4.10.11) can also be written as d9 8(6 – u)(u – 3) – 6(4 – u)9 – 92 = . du 9u

(4.10.12)

It is easy to see that (u, 9) = (6, 0) is also the singular point of eq. (4.10.12). Let u = u–3, so by eq. (4.10.12) we have d9 24u – 69 + o(92 + u2 ) = . du 39 + o(92 + u2 )

(4.10.13)

It is easy to know (u, 9) = (0, 0) is a saddle point, as shown in Fig. 4.15; another singular point (u, 9) = (6, 0) can be directly calculated by the equation du 9u = d9 8(6 – u)(u – 3) – 6(4 – u)9 – 92 for numerical calculation. Now we list the above calculation results in Table 4.9, we can also see Fig. 4.24.

W W = 4u

u

W = –4u Fig. 4.24

480

4 Physical Viscosity and Viscosity of Difference Scheme

Table 4.9 Comparison of different solutions. .

up

up,modify

up,A

–0.1 –0.5 –0.25 0 0.1 0.5 0.75

2.9725 2.7971 2.448 1.5 0.762 0 0

2.9963 2.92531 2.66535 1.5 0 0 0

2.4 2.04 1.5 0.65 0

4.11 Two-Dimensional “Artificial Viscosity” Problem Obviously, it is necessary to consider adding “artificial viscosity” during the calculation of two-dimensional hydrodynamics. It is natural that different types of one-dimensional “artificial viscosity” can be extended to two-dimensional problems, and now we adopt two-dimensional “artificial viscosity” as follows: (1) von Neumann viscosity (direct extension of one-dimensional situation), that is, adding the viscosity term in the pressure term  q=

C02 L2 1

(∇ ⋅ u)2

∇ ⋅ u ≤ 0,

0

∇ ⋅ u > 0,

(4.11.1)

where speed u = (u, v), L2 = Bx2 + By2 , C0 is a certain constant. We also can make ∂ 1 two-dimensional von Neumann viscosity in the form –1 ∂t 1 . Where L has the length dimension, above it is just a method of selecting; in fact, it is much more complicated in two dimensions, because on the two-dimensional grid, there is a lot of length dimensions, such as grid of four sides, two diagonals, the square root of the grid area. It can also be used to measure the size of the rotating body which is obtained by rotating the grid in cylindrical coordinates. Therefore, it is important to select L of the different problems. (2) Linear viscosity: namely we add a linear viscosity term to the pressure term, (i) q = C0 1CD,

(4.11.2)

   ∂u ∂v D = – min 0, Bx – min 0, By , ∂x ∂y

(4.11.3)

where 

where C is sound velocity and C0 is a certain determined constant.

4.11 Two-Dimensional “Artificial Viscosity” Problem

(ii)

481

Godunov, Rusanov linear viscosity If we take: qx = –1C

∂u , ∂x

qy = –1C

∂v . ∂y

(4.11.4)

The following is conservation equations in Euler coordinate system: ∂f ∂F x ∂F y + + + J = 0, ∂t ∂x ∂y

(4.11.5)

Using Rusanov scheme, where ⎞ ⎛ ⎛ ⎞ 1u 1 ⎜ p + 1u2 ⎟ ⎜ 1u ⎟ ⎟ ⎜ ⎜ ⎟ f = ⎜ ⎟, Fx = ⎜ ⎟, ⎝ 1uv ⎠ ⎝ 1v ⎠ (E + p)u E ⎛ ⎞ ⎛ ⎞ 1 1v ⎟ ⎜ 1uv ⎟ -v ⎜ ⎜ 1u ⎟ ⎜ ⎟ Fy = ⎜ ⎜ ⎟, ⎟, J = ⎝ p + 1v ⎠ y ⎝ 1v ⎠ E+p (E + p)V

(4.11.6)

where - = 0 corresponds to the plane coordinate system, - = 1 corresponds to the cylindrical coordinate system. Rusanov scheme of equations (4.11.5) are n+1 n fj,k = fj,k – BtJnj,k –

Bt x x – Fj–1,k )n (F 2Bx j+1,k

Bt y y – Fj,k–1 )n (F 2By j,k+1 Bt x + (I 1 – Ixj– 1 ,k ) Bx j+ 2 ,k 2 Bt y y + – I 1 ), (I j,k– 2 By j,k+ 21 –

(4.11.7)

where ⎧ ⎪ ⎪ x ⎪ ⎨I

j+ 21 ,k

⎪ ⎪ y ⎪ ⎩I

j,k+ 21

= !xj+ 1 ,k 2

=

!xj,k+ 1 2

n n (fj+1,k – fj,k )

Bx n n (fj,k+1 – fj,k ) By

, (4.11.8) .

Obviously, the equations that the difference equations (4.11.7) and (4.11.8) approxim∂ ∂f ∂ ∂f ! ∂x + ∂y ! ∂x , instead ate are constructed by eq. (4.11.5) with two-order diffusion term ∂x of 0 on the right side of eq. (4.11.5): !njk =

9 BxBy  (|u| + c)nj,k . 2 (Bx)2 + (By)2

(4.11.9)

482

4 Physical Viscosity and Viscosity of Difference Scheme

Here 9 is an appropriate selection of parameters and c is sound velocity. If we set  30n

= max j,k

(Bx)2 + (By)2 Bt(|u| + c)nj,k , BxBy

(4.11.10)

then according to the stability requirements of the difference scheme, it should satisfy 30n = 1, 30n ≤ 9 ≤

1 . 30n

(4.11.11)

(3) “Second-order” viscosity: Schulz thought that the uniform isentropic compression and nonuniform shock compression should be distinguished. He proposed the basic form of the artificial viscosity containing the two-order derivative factor:    ∂u  ∂ ∂u  , q! = –C02 1  ∂! ∂! ∂!     (4.11.12)  ∂u  2 ∂u  ∂ q" = –C0 1  , ∂" ∂" ∂"  where !, " are certain orthogonal directions; this q! (or q" ) is in a uniform compression   2 or expansion, and there is ∂u < 0, ∂ u2 = 0, while the original q! = –C2 1 ∂u  ∂u  cannot ∂!

0 ∂! ∂!

∂!

meet. (4) Consider the two-dimensional viscous of strain rate in the direction of the grid acceleration: ⎧  2    ds  ds ds ⎪ ⎪ + lL 1C , < 0, ⎨ l20 1 dt dt dt q9 = (4.11.13) ⎪ ds ⎪ ⎩ 0, ≥ 0, dt where ds dt is the strain rate in the direction of the grid acceleration, l0 = a0 L, lL = aL l, a0 ≈ 2, aL ≈ 1, c is sound velocity. Wilkim proposed to use the strain rate in the direction of the grid acceleration ∂v ds ∂u = cos2 ! + sin2 ! + dt ∂x ∂y



 ∂u ∂v + sin ! cos ! ∂y ∂x

(4.11.14)

instead of one-dimensional strain rate ∂u ∂x , where ! takes the angle between the direction of flow field acceleration and x-axis. In the original one-dimensional viscosity coefficient, the amount of the length dimension can be taken as (as shown in Fig. 4.25) L=

2A , d 1 + d2 + d3 + d4

(4.11.15)

where d1 , d2 , d3 , d4 are vertical distances of straight line that are caused by the flow direction of acceleration field from four corners to the center of the grid, respectively,

4.11 Two-Dimensional “Artificial Viscosity” Problem

483

y

d1

d2 d3

d4

x Fig. 4.25

and A is the area of grid. So we get the two-dimensional artificial viscosity term q9 with the form (4.11.13). It seems the two-dimensional artificial viscosity must pay attention to the following questions: (i) The characteristics of two-dimensional must be considered, that is direction of viscosity. From the point of practical problems, there should be directional. It is not very appropriate to use ∇ ⋅ u as a judge. (ii) It may be a feasible method to select the two-dimensional viscosity close to the physical viscosity. The momentum equation for the two-dimensional viscous fluid can be written as ∂(1u) + ∇(1uu) + ∇p = ∇(+∇ ⋅ u) + ,[∇∇u + ∇2 u] ∂t + (∇,)(∇ ⋅ u) + (∇u) ⋅ ∇,,

(4.11.16)

Selecting parameters + and , appropriately as its artificial viscosity, in fact, the selection of some previously described two-dimensional viscous term is a special case of it, such as q = +(∇ ⋅ u)2 , (iii)

  ∂u 2 q=+ , and so on. ∂n

The selection of viscosity should be made to ensure that the sticky entropy inequality holds; inequality that makes entropy not to reduce ensures the uniqueness of solutions to the equation in a one-dimensional quasilinear hyperbolic equation. A lot of selection of artificial viscosity scheme is to guarantee this condition in the case of one-dimensional hydrodynamic equations, for example, in Godunov discontinuous decomposition method, there is

484

4 Physical Viscosity and Viscosity of Difference Scheme

∂e ∂v ∂u ∂p +p = –q –m , ∂t ∂t ∂x ∂x where m=–

1 ∂p , C ∂x

q = –C

∂u . ∂x

Using thermodynamic relation TdS = de + pdv, there is  2   1 ∂p 2 ∂u dS + > 0. =C T dt ∂t C ∂x In von Neumann viscosity method, ∂e ∂v ∂v ∂u +p = –q = –q . ∂t ∂t ∂t ∂x Because ⎧ 2  2 ∂u e ∂u ⎪ ⎪ , < 0, ⎨ v ∂x ∂x q= ⎪ ∂u ⎪ ⎩ > 0, 0, ∂x

(4.11.17)

there is T

∂u ds = –q ≥ 0. dt ∂x

When selecting a two-dimensional artificial viscosity, it should also be considered to meet the condition that the entropy does not decrease when crossing the shock.

5 Convergence of Lax–Friedrichs Scheme, Godunov Scheme and Glimm Scheme 5.1 Convergence of Lax–Friedrichs Difference Scheme We study the first-order quasilinear hyperbolic equation as follows: ∂u ∂>(x, t, u) + + 8(x, t, u) = 0, ∂t ∂x

(5.1.1)

u|t=o = u0 (x), –∞ < x < +∞.

(5.1.2)

with initial value

The Lax–Friedrichs difference scheme of eq. (5.1.1) is uk+1 n –

ukn–1 +ukn+1 2

Bt >((n + 1)Bx, kBt, ukn+1 ) – >((n – 1)Bx, kBt, ukn–1 ) + 2Bx +8((n + 1)Bx, kBt, ukn+1 ) = 0,

(5.1.3)

k = 1, 2, . . . ; n = 0, 1, 2, . . . ; –1, –2, . . . ; where record j

ui = u(iBx, jBt). The solution of differential equation (5.1.3) is required to satisfy the initial conditions corresponding to eq. (5.1.2) u0n = u0 (nBx).

(5.1.4)

In order to prove that the solution of differential problems (5.1.3) and (5.1.4) converges to the generalized solution of the initial value problems (5.1.1) and (5.1.2), we make the following assumption: (i) For all –∞ < u < ∞ and (x, t) ∈ G = {–∞ < x < ∞, 0 ≤ t ≤ T ≤ ∞}, >x , >u , >xx , >xu , >uu , 8x , 8u are continuous functions. For bounded u and (x, t) ∈ G, above derivatives are bounded. (ii) For all –∞ < , < +∞, (x, t) ∈ G, >uu ≥ 0;

DOI 10.1515/9783110494273-005

(5.1.5)

486

5 Convergence of Three Different Schemes

For bounded u and 0 ≤ t ≤ #, we have >uu ≥ , > 0,

(5.1.6)

where 4 and , are some positive numbers. (iii) There are constant M and continuous differentiable function V(v) such that ′ max |>x + 8| ≤ V(v), V (v) ≥ 0 and (x,t)∈G,|u|≤v



M

m

d9 ≥ T, V(v) + !

(5.1.7)

where ! is a positive number. (iv) |u0 (nBx)| ≤ m.

(5.1.8)

Denote region {|u| ≤ M, (x, t) ∈ G} by K. A = max |>u |.

(5.1.9)

Lemma 5.1.1. Assume |u0n | ≤ m and ABt Bx < 1, if (i) and (iii) hold, then for the sufficiently small Bx, the solution ukn of difference problems (5.1.3) and (5.1.4) has the estimate |ukn | ≤ M, for all n and kBt ≤ T.

(5.1.10)

Proof. Difference equation (5.1.3) can be written as 6 Bt 1 + >u ((n + 1)Bx, kBt, (nk ) 2 2Bx 6 5 1 k k Bt + un+1 – >u ((n + 1)Bx, kBt, (n ) 2 2Bx 5

k uk+1 n = un–1

(5.1.11)

– >x ((n + 1)Bx, kBt, ukn+1 )Bt – 8((n + 1)Bx, kBt, ukn+1 )Bt – >xx (#nk , kBt, ukn+1 )BxBt, where (nk is a numerical value between ukn+1 and ukn–1 , and #nk is certain number between (n – 1)Bx and (n + 1)Bx. Let M k = max |ukn |, and M k ≤ M. Because of the condition ABt Bx ≤ 1, the coefficients k of un–1 and ukn+1 are all nonnegative on the right of eq. (5.1.11), and the summation is 1. So by eq. (5.1.11), we obtain M k+1 ≤ M k + BtV(M k ) + max |>xx |BtBx

5.1 Convergence of Lax–Friedrichs Difference Scheme

487

When Bx is sufficiently small, there is max |>xx |Bx ≤ !; thus M k+1 – M k ≤ V(M k ) + !. Bt

(5.1.12) ∎

Consider the solution Z(t) to initial value problem for ordinary differential equation dZ = V(Z) + ! dt

(5.1.13)

Z(0) = m,

(5.1.14)

2

since ddtZ2 = V ′ (Z)(V(Z) + !) ≥ 0, all solution of problem (5.1.13) and (5.1.14) are lower convex function, so by eq. (5.1.13) we obtain M k ≤ Z(kBt), kB ≤ T.

(5.1.15)

By equation 

Z(t)

m

dz =t V(z) + !

(5.1.16)

determine the function Z(t), due to the assumption (iii) we obtain Z(t) ≤ M, t ≤ T.

(5.1.17)

M k ≤ M, kBt ≤ T.

(5.1.18)

By these we deduce

This completes the proof. In the following discussion we assume ABt Bx < 1, and we denote the solution of difference problems (5.1.3) and (5.1.4) by ukn , and |u0 (nBx)| ≤ m. Lemma 5.1.2. There exists a constant E so that ukn – ukn–2 E ≤ , 2Bx kBt

(5.1.19)

for all n and kB ≤ T, where constant E does not depend on Bt and Bx. Proof. Let Znk =

ukn –ukn–2 2Bx .

By eq. (5.1.3) we can obtain Znk which satisfies the equation

488

5 Convergence of Three Different Schemes

k k + Zn+1 Zn–1 2>((n – 1)Bx, kBt, ukn–1 )Bt + 2 4Bx2 k >((n – 1)Bx, kBt, un–3 ) + >((n + 1)Bx, kBt, ukn+1 ) – Bt 4Bx2 8((n + 1)Bx, kBt, ukn+1 ) – 8((n – 1)Bt, kBt, ukn–1 ) – Bt. 2Bx

Znk+1 =

(5.1.20)

Take advantage of the Taylor expansion; by eq. (5.1.20) we have 5

Znk+1

Bt 1 + >u ((n – 1)Bx, kBt, ukn–1 ) 2 2Bx  –>xu ((n – 1)Bx, kBt, (n–k )Bt 5 Bt 1 k + Zn+1 – >u ((n – 1)Bx, kBt, ukn–1 ) 2 2Bx – >xu ((n – 1)Bx, kBt, (˜ nk )Bt  –8u ((n – 1)Bx, kBt, (ˆ nk )Bt =

k Zn–1

(5.1.21)

Bt 2 k – [>xx (#1 , kBt, un+1 ) + >xx (#2 , kBt, ukn–3 )

k 2 ) >uu ((n – 1)Bx, kBt, (1 ) – (Zn+1

+ 28x (#3 , kBt, ukn+1 )]

Bt 2

k 2 – (Zn+1 ) >uu ((n – 1)Bx, kBt, (2 )

Bt 2

We firstly estimate Znk in the case of kBt ≤ 4, here 4 > 0 is determined by the following; that is, >uu ≥ , > 0, as|u| ≤ M, 0 ≤ t ≤ r.

(5.1.22)

⎧ a = max |>xu | + max |8u |, ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ b = max |>xx | + max |8x |, a˜ = a + b, ⎪ ⎪   ⎪ ⎪ ˜ ⎪ , A(1 + aBt) ⎪ ⎩ c = min , . 2 4M

(5.1.23)

Record

k k , Zn+1 , 0}; when Bt is sufficiently small, the coefficients of the Assume 9 Znk = max{Zn–1 right of eq. (5.1.21) are all positive, so by eq. (5.1.21), we have

˜ + bBt – c(Z˜ nk )2 Bt. Znk+1 ≤ Z˜ nk (1 + aBt)

(5.1.24)

489

5.1 Convergence of Lax–Friedrichs Difference Scheme

Because |ukn | ≤ M, we have M Z˜ nk ≤ . Bx By condition

ABt Bx

(5.1.25)

aBt) < 1 and c < A (1+9 4M we deduce

˜ M ˜ k 1 + aBt , Zn < . Z˜ nk ≤ ABt 4Btc

(5.1.26)

˜ Consider polynomial H(y) = y(1 + aBt) + bBt – cBty2 . Obviously, ′

˜ – 2ycBt ≥ 0, as y < H (y) = 1 + aBt

˜ 1 + aBt . 2cBt

(5.1.27)

Let M k = max{max Znk , 0}; from the estimates (5.1.24) and (5.1.26) of Znk we deduce n

˜ + bBt – c(M k )2 Bt. Znk+1 ≤ M k (1 + aBt)

(5.1.28)

We know the right of eq. (5.1.28) is positive because of eq. (5.1.26). So ˜ M k+1 ≤ M k (1 + aBt) + bBt – c(M k )2 Bt.

(5.1.29)

Let V k = M k + 1, then by eq. (5.1.29) we have V k+1 ≤ V k (1 + 9 aBt + 2cBt) + bBt – 9 aBt – c(V k )2 Bt, ≤ V k (1 + 9 aBt) – cBt(V k )2 ,

(5.1.30)

where 9 c =9 a + 2c. When Bt is sufficiently small, there is 0 < 1 –9 cBt < 1; by eq. (5.1.30) we obtain V k+1 ≤ V k +

9 cBt V k – cBt(V k )2 . 1 –9 cBt

(5.1.31)

cBt)k V k ; multiplying eq. (5.1.31) by (1 –9 cBt)k+1 , we have Let W k = (1 –9 W k+1 ≤ W k – cBt(W k )2 (1 –9 cBt)–k–1 .

(5.1.32)

So when k ≥ 0, (1 –9 cBt)–k–1 ≥ 1, by eq. (5.1.32) we have W k+1 ≤ W k – cBt(W k )2 .

(5.1.33)

By inequality (5.1.33) it is easy to obtain the estimates of W k , so as to get the estimates of M k .

490

5 Convergence of Three Different Schemes

Consider the initial value problem for ordinary differential equation ⎧ ⎨ dW = –cW 2 , dt ⎩ W(0) = W 0 .

(5.1.34)

2

3 By ddtW 2 = 2cW we know all positive solutions of eq. (5.1.34) are lower convex function. The solution W(t) of problem (5.1.34) is

W(t) =

1 . ct + W10

(5.1.35)

It is easy to know W k ≤ W(kBt) ≤

1 , ckBt

(5.1.36)

then we have 1 , ckBt

(1 –9 cBt)k (M k + 1) ≤ E M ≤ , kBt

(5.1.37)

k

here the constant E does not depend on Bt and Bx. Therefore, lemma is obtained when kBt ≤ 4. When kBt > 4, then by eq. (5.1.21) we deduce Znk+1 ≤ 9 Znk (1 + aBt) + bBt, M k+1 ≤ M k (1 + aBt) + bBt.

(5.1.38)

So we have 4

M k ≤ B1 + B2 M [ Bt ] ,

(5.1.39)

where B1 and B2 do not depend on Bt and Bx. Thus the estimate (5.1.19) holds for all k ≤ T/Bt and Bx. ∎ Lemma 5.1.3 (Space Estimates). For arbitrary X > 0 and kBt ≥ ! > 0, there exists a constant C which depends on X and !, but has nothing to do with Bt and Bx, such that    uk – uk  < C, (5.1.40) n+2 n X |n|≤ Bx

where make summation about the integer n for all |n| ≤ X/Bx. Proof. Let vnk = ukn – c1 nBx, where c1 is sufficiently large so that E/! < c1 , then we have k vn+2 – vnk < 0, kBt ≥ !. In fact, by Lemma 5.1.2 we have

5.1 Convergence of Lax–Friedrichs Difference Scheme

491

k vn+2 – vnk = ukn+2 – ukn – 2c1 Bx,

E ⋅ 2Bx – c1 2Bx, kBt  E ≤ 2Bx – c1 < 0. ! ≤

Thus we obtain        uk – uk  ≤ vk – vk  + 2c1 Bx, n+2 n n+2 n |n|≤X/Bx

|n|≤X/Bx

|n|≤X/Bx

     vk – vk  + 2c1 Bx 2X , =– n+2 n Bx |n|≤X/Bx

≤ 2 max |vnk | + 4c1 X, |n|≤X/Bx

≤ 2M + 2c1 X + 4c1 X = 2M + 6c1 X. ∎

The proof of the Lemma 5.1.3 is completed.

Next we only consider the estimate of ukn when k – n is even number. This is due to the property of the difference equation (5.1.3) and the calculation of ukn when n – k is even number and has nothing to do with ukn when n – k is odd number. The set which makes n – k even number is denoted as S1 . Bt Lemma 5.1.4 (Time-Space Estimates). If Bx ≥ $ > 0, Bx, Bt ≤ 1, there exists a constant L > 0 which has nothing to do with Bx, Bt such that when k > p, k – p are even number, and pBt ≥ ! > 0, there is

   uk – upn Bx ≤ L(k – p)Bt. n

(5.1.41)

|n|≤X/Bx

When k – p is odd number, similar estimates also hold. p

Proof. We will denote ukn as the term of un , k – p is even number. In order to do this, we consider difference scheme (5.1.3) and (5.1.11), then k,k–1 k–1 k–1 ukn = ak,k–1 n,n–1 un–1 + an,n+1 un+1 + 'k–1 ,

(5.1.42)

k,k–1 where ak,k–1 n,n–1 + an,n+1 = 1, |'k–1 | ≤ 'Bt; ' is a constant which does not depend on Bt and n. Successively applying eq. (5.1.42), we have

ukn =

n+k–p  j=n–(k–p)

k,p p

an,j uj + 'p ,

(5.1.43)

492

5 Convergence of Three Different Schemes

where

⎧  k,p k,p an,j = 1, an,j ≥ 0, |'p | ≤ '(k – p)Bt, ⎪ ⎪ ⎪ ⎪ ⎪ j ⎪ ⎪ 6 5   ⎨ Bt k,p k,p+1 1 p an,j = an,j+1 + >u (j + 2)Bx, pBt; (j+1 , ⎪ 2 2Bx ⎪ ⎪ 6 5   ⎪ ⎪ ⎪ Bt k,p+1 1 p ⎪ ⎩ + an,j–1 – >u jBx, pBt; (j–1 . 2 2Bx

(5.1.44)

Taking advantage of eqs (5.1.43) and (5.1.44), we have    uk – upn 2Bx, n |n|≤X/Bx

≤

   

n+k–p 

|n|≤X/Bx j=n–(k–p)

  k,p p p  an,j uj – un  ⋅ 2Bx

+2X'(k – p)Bt,     n+k–p  k,p p p   ≤ an,j uj – un 2Bx  |n|≤X/Bx j=n–(k–p)

+2X'(k – p)Bt,     n+k–p   k,p p p   ≤ a – u u r r–2 2Bx n,j  |n|≤X/Bx j=n–(k–p)

+2X'(k – p)Bt. Because of

 j

(5.1.45)

k,p

an,j = 1, the right of inequality (5.1.45) is not more than

≤

   

n+k–p 



|n|≤X/Bx r=n–(k–p)

≤ 2Bt(k – p)



  p p ur – ur–2 2Bx + 2X'(k – p)Bt,    

n+k–p 

|r|≤X/Bx+k–p j=n–(k–p)



p ur

–

p ur–2

   + 2X'(k – p)Bt. 

(5.1.46)

By Lemma 5.1.3 and assuming Bt/Bx ≥ $ > 0, we obtain the right of inequality (5.1.46) is not more than 2(k – p)BtC + 2'(k – p)BtX ≤ L(k – p)Bt. ∎

This completes the proof.

Lemma 5.1.5 (Stability). Assume {ukn } and {vnk } are respectively the solutions of difference scheme (5.1.3) corresponding to the initial value {u0n }A; M{vn0 }. If sup |u0n | ≤ M, sup |vn0 | ≤ M, then if k > 0, we have n

n

5.1 Convergence of Lax–Friedrichs Difference Scheme

     uk – vk Bx ≤ (1 + BBt)k u0 – v0 Bx, n n n n |n|≤N

493

(5.1.47)

|n|≤N+k

where B = max |8u |, k ≤ T/Bt. K

Proof. Let wnk = ukn – vnk , so by difference scheme (5.1.3), we have 5   Bt k k–1 1 k–1 9 9n = 9n+1 – >u (n + 1)Bx, (k – 1)Bt, (n+1 2 2Bx   6 k–1 – >u (n + 1)Bx, (k – 1)Bt, 9 (n+1 Bt 6 5   1 Bt k–1 + 9k–1 + >u (n – 1)Bx, (k – 1)Bt, 9 , (n–1 n+1 2 2Bx

(5.1.48)

k

where 9 (nk , (n is a value between ukn and vnk . Because of ABt/Bx < 1, the coefficients of k–1 k–1 wn–1 and wn–1 in eq. (5.1.48) are all nonnegative. Thus,  |9kn | |n|≤N

5      Bt k–1  1 k–1 9  ≤ 9n+1  2 – >u (n + 1)Bx, (k – 1)Bt, (n+1 2Bx |n|≤N   6 k–1 – 8u (n + 1)Bx, (k – 1)Bt, (n+1 Bt 5  6    1  Bt k–1  + >u (n + 1)Bx, (k – 1)Bt, 9 + 9k–1 , ( n+1  n+1 2 2Bx  5   6    9k–1  1 – >u nBx, (k – 1)Bt, (k–1 Bt , ≤ n  n  |n|≤N+1

≤

   9k–1 (1 + BBt). n

|n|≤N+1

Thus, it is easy to find  |n|≤N

This completes the proof.

|9kn |Bx ≤

   90 (1 + BBt)k Bx. n

|n|≤N+k

∎

We are ready to prove the convergence of the solution to the difference problems (5.1.3) and (5.1.4). We know that difference approximate solutions are only defined at the grid points. We want it to be defined as a function on the entire half plane t ≥ 0. For this we can construct function family {UBt,Bx } from the difference solution {ukn }; the definition is as follows:

494

5 Convergence of Three Different Schemes

(n, k + 1)

(n + 1, k + 1)

Δt

(n + 1, k)

(n, k) Δx Fig. 5.1

UBt,Bx (x, t) = ukn , nBx ≤ x < (n + 1)Bx, kBt ≤ t < (k + 1)Bt.

(5.1.49)

That is the value of {UBt,Bx } in the rectangular nBx ≤ x < (n + 1)Bx; kBt ≤ t < (k + 1)Bt is equal to the value of the difference solution at a point (nBx, kBt) as shown in Fig. 5.1. We will prove that the function set {UBt,Bx } on a compact set in accordance with the L1 norm is compact. Lemma 5.1.6. There exists a subsequence {UBti ,Bxi } ⊂ {UBt,Bx } in the sense of L1 norm which converge to a measurable function u(x, t), namely for any X > 0, t > 0 and T > 0, we have    UBt ,Bx (x, t) – u(x, t)dx → 0, (5.1.50) i i 

|x|≤X T 0

|x|≤X

  UBt ,Bx (x, t) – u(x, t)dxdt → 0, i i

(5.1.51)

i → ∞, Bti → 0, Bxi → 0, Furthermore, sup |u(x, t)| ≤ M,

x∈R,t>0



x2 x1

|u(x, t) – v(x, t)| ≤

(5.1.52) 

x2 +At

|u0 (x) – v0 (x)|dx,

(5.1.53)

x1 –At

where u(x, t) and v(x, t), respectively, correspond to the limit function which the difference equation (5.1.3) with the different initial conditions u0n , vn0 tends to, A = max |>u |. |u|≤M

Proof. From Lemmas 5.1.1 and 5.1.3, we know that the function set {UBt,Bx } as a function of x is uniformly bounded, and it is a uniformly bounded variation on any bounded ′ interval of any t = const > 0. By Helley theorem, we can select a subsequence {UBt,Bx }

5.1 Convergence of Lax–Friedrichs Difference Scheme

495

such that it is pointwise convergence on any bounded interval of any t = const > 0. Ap′′ plying a diagonal subsequence argument, we may construct a subsequence {UBt,Bx }⊂ ′ {UBt,Bx } such that it converges at every point on t = const > 0 as Bt, Bx → 0. If {tm } is countable and dense in the interval (0, T), furthermore, applying a di′′ agonal subsequence argument, we can select a subsequence {UBti ,Bxi } ⊂ {UBt,Bx } such that it converges at every point on t = tm (m = 1, 2, . . .) as i → ∞ (that is, Bti , Bxi → 0). It converges at every point on every straight line t = tm (m = 1, 2, . . .). Let Ui = UBti ,Bxi . We will prove that the function sequence converges at every t ∈ [0, T], so its limit function u(x, t) defined on the interval 0 < t ≤ T. For this we firstly prove that for every t ∈ (0, T], we have 

X

Iij (t) =

  Ui (x, t) – Uj (x, t)dx → 0, i, j → ∞,

(5.1.54)

–X

so {Ui (⋅, t)} is a Cauchy sequence in L1 (|x| ≤ X). For t ∈ (0, T], we can find a subsequence {tms } ⊂ {tm } so that tms → t(s → ∞). Let 4s = tms , then we have 

X

Iij (t) ≤

  Ui (x, t) – Ui (x, 4s )dx

–X



X



–X X

+ +

  Ui (x, 4s ) – Uj (x, 4s )dx   Uj (x, 4s ) – Uj (x, t)dx,

–X

= I1 + I2 + I3 .

(5.1.55)

We know I2 → 0 as i, j → ∞ based on the selection of {tm } and the Lebesque bounded control theorem. Secondly, if we denote the maximum integer (3 ≥ 0) which is not greater than 3 by [3], so by the definition of Ui , we know Ui (x, t) = Ui (x, [t/Bti ]Bti ). Thus,        Ui x, [t/Bti ]Bti – Ui x, [4s /Bti ]Bti dx,   –X     (n+1)Bxi   Ui x, [t/Bti ]Bti ≤  

X

I1 =

|n|≤(X/Bxi )+1



nBxi

  – Ui x, [4s /Bti ]Bti dx,     [t/Bti ] [4s /Bti ]  un = – u n  Bx, |n|≤(X/Bxi )+1

5 6 5 6  t 4s  Bti . ≤ L – Bti Bti 

(5.1.56)

496

5 Convergence of Three Different Schemes

We have used Lemma 5.1.4 on the right side of the inequality (5.1.56). By the above inequality, we have I1 ≤ L|t – 4s |.

(5.1.57)

I3 ≤ L|t – 4s |.

(5.1.58)

I1 + I3 ≤ 2L|t – 4s |.

(5.1.59)

Similarly, we have

So

Now we can select 4s such that 4L|t – 4s | < % for the arbitrary % > 0. For this fixed s, we choose i, j so large that 2I2 < %. So for this i, j, by eqs (5.1.55) and (5.1.59), we have Iij (t) ≤ %.

(5.1.60)

This proves eq. (5.1.54). So we obtain that the subsequence {Ui (x, t)} has a limit function u(x, t) for each fixed t, 0 < t < T. Now we prove that Iij (t) uniformly tends to be 0 about t as i, j → ∞. In fact, if for a given t, we select a finite subset F ⊂ {tm }, which has the following properties: that is tm ∈ F so that 2L|t – tm | < %/2 as 0 ≤ t ≤ T. Then we can select sufficiently large i, j so that 2I2 < %. For all tm ∈ F , we have Iij < % for this i, j. This gives the consistency from Iij (t) → 0 to t. Using this uniform convergence, for any 4, 0 < 4 ≤ T, we have 

T

4

Iij (t)dt → 0, i, j → ∞.

(5.1.61)

Let 

T 0





X

  Ui (x, t) – Uj (x, t)dxdt,

–X 4 X

 Ui (x, t) – Uj (x, t)dxdt

= 0



–X T X

+ 4

  Ui (x, t) – Uj (x, t)dxdt.

(5.1.62)

–X

Select sufficiently small 4(0 < 4 ≤ T) such that 8MX4 < %. For the fixed 4, we select sufficiently large i, j so that the second integral on the right side of eq. (5.1.62) is less than %/2 (because of eq. (5.1.61), it is possible). So for this i, j we have 

T 0

Iij (t)dt < %.

(5.1.63)

5.1 Convergence of Lax–Friedrichs Difference Scheme

497

So the limit function u(x, t) can be measured, and eq. (5.1.51) holds. By local L1 convergence of the function sequence {Ui (x, t)} we can select a subsequence convergence almost everywhere. Because of |Ui (x, t)| ≤ M, the limit function u(x, t) also satisfies the same inequality. Therefore, eq. (5.1.52) holds. The inequality holds by Lemma 5.1.5. ∎ Next we prove that the limit function u(x, t) satisfies the entropy inequality. Lemma 5.1.7. The limit function u(x, t) constructed by Lemma 5.1.6 satisfies the entropy inequality u(x1 , t) – u(x2 , t) 2E < , x1 – x2 t

t > 0,

(5.1.64)

where constant E is defined by Lemma 5.1.2: –∞ < x1 , x2 < ∞. Proof. If x1 – x2 > 2Bxi , t > Bti , we only need to prove sufficiently 2E Ui (x1 , t) – Ui (x2 , t) < , x1 – x2 t – Bti

(5.1.65)

where Ui (x, t) is defined by the proof of Lemma 5.1.6, and constant E is defined by Lemma 5.1.2. It is due to the fact if eq. (5.1.65) holds, and let i → ∞, we have eq. (5.1.64). For x1 > x2 , we note that ! " Ui (xj , t) = Ui xj – 'j , [t/Bti ]Bti , j = 1, 2, where 0 ≤ 'j ≤ Bxi , and xj – 'j (j = 1, 2) belong to S1 , so  1 Ui (x1 , t) – Ui (x2 , t) = (uk – ukn–2 ), x1 – x2 (x1 – x2 ) n n where k = [t/Bti ], and we make the summation for all integers in the interval  x2 –'2 x1 –'1  Bx , Bx . By Lemma 5.1.2, we have Ui (x1 , t) – Ui (x2 , t) E(x1 – '1 – x2 + '2 ) ≤ , x1 – x2 [t/Bti ]Bti (x1 – x2 ) E(x1 – '1 – x2 + '2 ) , ≤ (t – Bti )(x1 – x2 ) E('2 – '1 ) E + , = t – Bti (t – Bti )(x1 – x2 ) 2E . ≤ t – Bti Because of '2 – '1 ≤ Bxi < x1 – x2 , this proves eq. (5.1.64).

498

5 Convergence of Three Different Schemes

Another kind of proof is based on the fact: We have proved that vnk = ukn – C1 nBx in the proof of Lemma 5.1.3 is a decreasing function of x for every fixed t. This fact is also set up for the limit function u(x, t) – C1 x. Thus, we can get eq. (5.1.64). ∎ Next we prove the limit function u(x, t) is the generalized solution of problems (5.1.1) and (5.1.2). Lemma 5.1.8. If Bxi → 0 as i → ∞, for f (x, t) ∈ C03 , there is  lim

∞

i→∞ –∞

[Ui (x, 0) – u0 (x)]f (x, 0)dx = 0.

(5.1.66)

So u(x, t) satisfies integral equality 6 ∂f ∂f – 8(x, t, u)f dxdt u + >(x, t, u) ∂t ∂x  ∞ u0 (x)f (x, 0)dx = 0, +

 5 G

(5.1.67)

–∞

∀f (x, t) ∈ C03 , namely u(x, t) is the generalized solution to initial value problem (5.1.1) and (5.1.2). Proof. The difference equation (5.1.3) on the set S1 can be rewritten as k uk – 2ukn + ukn–1 Bx2 uk+1 n – un ⋅ – n+1 Bt 2Bx2  Bt   > (n + 1)Bx, kBt, ukn+1 – > (n – 1)Bx, kBt, ukn–1 +  2Bx + 8 (n + 1)Bx, kBt, ukn+1 = 0,

(5.1.68)

where the set S1 denotes a grid point when k – n is even number, and the rest of the grid points records as S2 . Multiplying eq. (5.1.68) by fnk = f (nBx, kBt), we obtain k k fnk+1 – fnk fnk+1 uk+1 n – fn un – uk+1 n Bt Bt k k k 2 Bx 2fn+1 – fn+1 – fn–1 k ⋅ un + ⋅ 2Bt Bx2 f k uk – f k uk f k uk – f k uk + n+1 n n n–1 + n–1 n n n+1 2Bt 2Bt " ! k k + fn 8 (n + 1)Bx, kBt, un+1

5.1 Convergence of Lax–Friedrichs Difference Scheme

" k ! " ! k > (n + 1)Bx, kBt, ukn+1 – fn–1 > (n + 1)Bx, kBt, ukn–1 fn+1 + 2Bx k " ! f – fk – > (n + 1)Bx, kBt, ukn+1 n+1 n 2Bx k – fk " ! f n–1 – > (n + 1)Bx, kBt, ukn–1 n = 0. 2Bx

499

(5.1.69)

For the grid points set S2 , eq. (5.1.3) can be rewrite as uk+1 n –

ukn–1 +ukn+1 2

Bt " ! " ! > nBx, kBt, ukn+1 – > (n – 2)Bx, kBt, ukn–1 + 2Bx " ! + 8 (n + 1)Bx, kBt, ukn+1 = 0.

(5.1.70)

Now ukn at point (nBx, kBt) equals ukn–1 . Multiplying eq. (5.1.70) by fnk , we obtain k k fnk+1 – fnk fnk+1 uk+1 n – fn un – uk+1 n Bt Bt k k k – fn+1 – fn–1 Bx2 2fn+1 ⋅ ukn + ⋅ 2Bt Bx2 f k uk – f k uk f k uk – f k uk + n+1 n n n–1 + n–1 n n n+1 2Bt 2Bt " ! + fnk 8 nBx, kBt, ukn+1 " k ! " ! k fn+1 > nBx, kBt, ukn+1 – fn–1 > nBx, kBt, ukn–1 + 2Bx k " ! f – fk – 8 nBx, kBt, ukn+1 n+1 n 2Bx k " f k – fn–1 ! – > (n – 2)Bx, kBt, ukn–1 n = 0. 2Bx

(5.1.71)

We make summation of equality (5.1.69) about n, k when k + 1 – n(k > 0) is odd number, and make summation of equality (5.1.71) about n, k(k > 0) when k+1–n is even number. Because f (x, t) is a compactly supported set function, we can think that fnk equals 0 for all n as k > [T/Bt]. Multiplying BtBx again, we have –



u0n fn0 Bx + BtBx

n

5

uk+1 n

n,k>0

! k "6 k fn+1 – 2fnk + fn–1 – ⋅ 2Bt Bx2  " ! fnk 8 (n + 1)Bx, kBt, ukn+1 + Bx2

S1

ukn

fnk – fnk+1 Bt

500

5 Convergence of Three Different Schemes



+

" ! fnk 8 nBx, kBt, ukn+1

S2

 ! "fk – fk – > (n + 1)Bx, kBt, ukn+1 n+1 n 2Bx S1

 ! "fk – fk > (n – 1)Bx, kBt, ukn–1 n n–1 2Bx

–

S1

 ! "fk – fk – > nBx, kBt, ukn+1 n+1 n 2Bx S2

k  ! " f k – fn–1 – > (n – 2)Bx, kBt, ukn–1 n = 0. 2Bx

(5.1.72)

S2

By the definition of UBt,Bx (x, t) and the continuity of 0  n

u0n fn0 Bx =



u0 (2nBx)

n



∂f ∂x ,

we obtain

0 0 + f2n+1 f2n 2Bx 2

∞

= –∞

UBt,Bx (x, 0)f (x, 0)dx + O(Bx).

Because UBt,Bx (x, t) is a fragment constant, when f (x, t) ∈ C3 , f (x, T) = 0 and |x| is sufficiently large, f (x, t) = 0, then we have 

uk+1 n

n,k>0

 =

G

fnk+1 – fnk BtBx Bt

UBt,Bx

∂f dxdt ∂t

k  Bx2 f k – 2fnk + fn–1 BtBx, ukn ⋅ n+1 2Bt Bx2 n,k>0  Bx2 ∂ 2f = UBt,Bx 2 dxdt 2Bt ∂x G 5   fnk 8 (n + 1)Bx, kBt, ukn+1 + $2 BtBx

+ $1

+





S1

 6 fnk 8 nBx, kBt, ukn+1 ,

S2

= G

f 8(x, t, UBt,Bx )dxdt

+ $3 BtBx

5   k f – fk > (n + 1)Bx, kBt, ukn+1 ⋅ n+1 n 2Bx S1

501

5.2 The Convergence of Hyperbolic Equations in Lax–Friedrichs Scheme

 k   f – fk > (n – 1)Bx, kBt, ukn–1 ⋅ n n–1 2Bx S1    f k – fnk + > n – Bx, kBt, ukn+1 ⋅ n 2Bx S2 6  k   f – fk + > (n – 2)Bx, kBt, ukn–1 ⋅ n n–1 , 2Bx S2  ∂f >(x, t, UBt,Bx ) dxdt + $4 , = ∂x G +

where $1 , $2 , $3 , $4 → 0(Bt → 0, Bx → 0). For the function UBt,Bx equality (5.1.72) can be written as  – G

UBt,Bx

 ∂f ∂ 2f Bx2 UBt,Bx 2 dxdt dxdt – ∂t 2Bt ∂x G  + f 8(x, t, UBt,Bx )dxdt  G ∂f >(x, t, UBt,Bx ) dxdt – ∂x G  ∞ UBt,Bx (x, 0)f (x, 0)dx = $(Bt, Bx), –

(5.1.73)

–∞

where $(Bt, Bx) → 0 as Bt → 0, Bx → 0. And if we obtain integral equality (5.1.67).

Bx2 2Bt

→ 0, then by eqs (5.1.73) and (5.1.66) ∎

Remark 5.1.9. Because C03 is dense in C01 , then eq. (5.1.67) holds for f (x, t) ∈ C01 . So we obtain that u(x, t) is the generalized solution of problem (5.1.1) and (5.1.2).

5.2 The Convergence of Hyperbolic Equations in Lax–Friedrichs Scheme For the hyperbolic equation ut + f (u)x = 0.

(5.2.1)

Its Lax–Friedrichs difference scheme is un+1 j

  1 n + n n n – (uj+1 + uj–1 ) + f (uj+1 ) – f (uj–1 ) = 0, 2 2

(5.2.2)

Bt . We regard the Lax–Friedrichs scheme as the average value of the where + = Bx solution to the Riemann problem with initial value (unj–1 , unj+1 ).

502

5 Convergence of Three Different Schemes

un+1 j

1 = 2Bx



(j+1)Bx

(j–1)Bx

" ! 9 x, Bt, unj–1 , unj+1 dx,

(5.2.3)

where w(x, t, a, b) denotes the solution to the Riemann problem with the initial value (a, b) in the classical sense, namely seeking to the traveling wave solution which meets the equations and initial value.  a, x < 0, 9(x, 0) = b, x > 0. In fact, we can use the following function as an approximation of u0 (x): u(x, 0) = u0 (jBx), (j – 1)Bx < x < (j + 1)Bx, where u is a piecewise smooth solution of eq. (5.2.1) on the rectangular Kj = {(x, t),(j – 1)Bx ≤ x ≤ (j + 1)Bx, 0 < t < Bt}, so we have F {udx – f (u)dt} = 0, (5.2.4) ∂Kj

namely, we have u0j–1 Bx + u0j+1 Bx – f (u0j+1 )Bt  (j+1)Bx – u(x, Bt – 0)dx +

(j–1)Bx f (u0j–1 )Bt

= 0.

Comparing with eq. (5.2.2) (n = 0), we have  (j+1)Bx 1 u(x, Bt – 0)dx. u1j = 2Bx (j–1)Bx

(5.2.5)

(5.2.6)

So scheme (5.2.2) can be seen as the following two-step algorithm: (1) When t < nBt, –∞ < x < +∞, u(x, t) have been found. j + n is an even number. Let  (j+1)Bx 1 u(x, nBt) = u(x, nBt – 0)dx, 2Bx (j–1)Bx (j – 1)Bx ≤ x < (j + 1)Bx.

(5.2.7)

(2) Taking u(x, n △ t) as initial value, t ∈ [n △ t, (n + 1) △ t], we can find exact solutions to the Riemann problem for eq. (5.2.1). Namely by unj–1 , unj+1 , we can find w(x, △t, unj–1 , unj+1 ). Lemma 5.2.1. If (', q) is arbitrary a pair of convex entropy and entropy flow of eq. (5.2.1),

5.2 The Convergence of Hyperbolic Equations in Lax–Friedrichs Scheme

–1 (R2+ ), 't (un ) + q(un )x is compact in Hloc

503

(5.2.8)

where un is a solution of the Lax–Friedrichs scheme (5.2.2). Proof. If > is a compactly supported set function, we have   (>), (>t '(u) + >x q(u))dxdt = M(>) + L(>) + where

  ⎧ ⎪ M(>) ≡ >'(x, t)dx – >'(x, 0)dx, ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ n–1  ⎨  >(x, mBt)['m ]dx, L(>) ≡ ⎪ ⎪ m=1 ⎪ ⎪ ⎪  T ⎪  ⎪ ´ ⎪ ⎩ (>) = (>)dt;

(5.2.9)

(5.2.10)

0

# = (x(t), t) denotes the shock location; ['m ] denotes jump of 'm across the shock wave from the left leap, that is ['m ] = '{u(x, mBt – 0)} – '{u(x, mBt + 0)});   dx ´ {3['] – [q]}>(x(t), t), 3 = (>) = , dt all shocks

['] = '{u(x(t) – 0, t)} – '{u(x(t) + 0, t)}. For L(>) we have L(>) =

n–1  

=

n–1 

 >(jBx, mBt)

n–1   j

(j+1)Bx

['m ]dx

(j–1)Bx

m=1

j

+

>(x, mBt)['m ]dx,

(j–1)Bx

m=1

j

(j+1)Bx

m=1

(j+1)Bx

(5.2.11)

(>(x, mBt) – >(jBx, mBt))['m ]dx,

(j–1)Bx

≡ L1 (>) + L2 (>). For L1 (>) we have the estimate |L1 (>)| ≤ C maxx,t |>(x, t)|.

(5.2.12)

In fact, if we take > ≡ 1 in eq. (5.2.9), and assume |u| ≤ M0 , there is |M(1)| ≤ C, and by  convex entropy we have (1) ≥ 0.

504

5 Convergence of Three Different Schemes

L(1) =

n–1   j

m=1

(j+1)Bx

['(u(x, mBt – 0)) (j–1)Bx

– '(u(x, mBt + 0))]dx. By Taylor’s expansion, we have '(u(x, mBt – 0)) = '(u(x, mBt + 0)) + '′ (u(x, mBt + 0)) ⋅ (u(x, mBt – 0) – u(x, mBt + 0)) 1 + (u(x, mBt – 0) – u(x, mBt + 0))T 2 '′′ (9 u)(u(x, mBt – 0) – u(x, mBt + 0)).

(5.2.13)

Due to 

(j+1)Bx

[u(x, mBt – 0) – u(x, mBt + 0)]dx = 0,

(5.2.14)

(j–1)Bx

by eq. (5.2.13) we have 

(j+1)Bx

['(u(x, mBt – 0)) – '(u(x, mBt + 0))]dx

(j–1)Bx



(j+1)Bx

= (j–1)Bx

1 (u(x, mBt – 0) – u(x, mBt + 0))T 2

'′′ (9 u)(u(x, mBt – 0) – u(x, mBt + 0))dx. Thus, we get L(1) ≥ 0.

(5.2.15)

In view of eq. (5.2.1), we get M(1) + L(1) +



(1) = 0.

(5.2.16)

|u(x, mBt – 0) – u(x, mBt + 0)|2 dx ≤ c.

(5.2.17)

Thus, we can obtain the estimate n–1   j

m=1

(j+1)Bx (j–1)Bx

By eq. (5.2.17) we find estimate (5.2.12). Because of |L2 (>)| ≤

n–1  j

m=1

 >C!

(j+1)Bx

(j–1)Bx

    (Bx)! ['m ]dx.

(5.2.18)

505

5.2 The Convergence of Hyperbolic Equations in Lax–Friedrichs Scheme

Utilizing the inequality a2 + $b2 , $

ab ≤

(5.2.19)

set $ = Bx( (( > 0), by eq. (5.2.18) we have 6 (Bx)2! ( m 2 |L2 (>)| ≤ + (Bx) [' ] dx (Bx)( (j–1)Bx j m=1   (Bx)2!+1 ( + const(Bx) , ≤ >C! (Bx)2+( 1 ≤ ! < 1. 2 n–1 



>C!

(j+1)Bx

5

Because > has compact support set in eq. (5.2.20), the summation only has We select appropriate small ( such that |L2 (>)| ≤ C>C! (Bx)( .

(5.2.20)

1 Bx2

term.

(5.2.20)∗

By Embedding Theorem, for p > 2, from eq. (5.2.20)∗ we have |L2 (>)| ≤ C>1,p (Bx)( .

(5.2.21)

For a pair of convex entropy and entropy flow ', q, let H(u) = ('(u) – '(ul ))s(u) – (q(u) – q(ul )),

(5.2.22)

where ul = u(x(t) – 0, t), s = x′ (t), discontinuity line x = x(t) is a shock wave. Obviously H(ul ) == 0. By the Rankine–Hugoniont relation, we have (u – ul )s(u) = f (u) – f (ul ).

(5.2.23)

By compatibility condition of entropy and entropy flow of eq. (5.2.1) q′ (u) = '′ (u)f ′ (u) = '′ (u)(s(u) + (u – ul )s′ (u)), thus H ′ (u) = '′ (u)s(u) + ('(u) – '(ul ))s′ (u) – '′ (u)(s(u) + (u – ul )s′ (u)) = s′ (u)('(u) – '(ul ) – '′ (u)(u – ul )). By direct checking, we can prove H ′ (ul ) = H ′′ (ul ) = 0, H ′′ (ul ) = –s′ (ul ) ⋅ '′′ (ul ) and if f ′′ (u) > 0, by eq. (5.2.23) we have

506

5 Convergence of Three Different Schemes

s′ (u) =



1

f ′′ ((1 – t)ul + tu)dt > 0.

0

Because '(u) is a convex function, then H ′′′ (ul ) = –s′ (ul )'′′ (ul ) < 0. So as |u – ul | is small enough, we obtain H(u) ≥ C0 (ul – u)3 ,

(5.2.24)

where constant C0 > 0 only depends on the boundary of solution u(x, t). If '1 and q1 are any a pair of entropy (not necessarily strict convex) and entropy flow, let H1 (u) = ('1 (u) – '1 (ul ))s(u) – (q1 (u) – q1 (ul )). Similarly, there is H1 (ul ) = H1′ (ul ) = H1′′ (ul ) = 0, thus |H1 (u)| ≤ C1 (ul – u)3 .

(5.2.25)

So by eqs (5.2.24) and (5.2.25) we have |['1 ]s – [q1 ]| ≤ C([']s – [q]). 

(5.2.26)

(>) of estimate (5.2.10) can be estimated by eq. (5.2.26),       (>) ≤ C (1) max |>(x, t)|  x,t

(5.2.27)

≤ C max |>(x, t)|. x,t

So we obtain  

 ∂'(un ) ∂q(un ) + >dxdt ∂t ∂x

(5.2.28)

= –L1 (>) – L2 (>) – G(>). It is easy to see, by the left side of eq. (5.2.28) and |un | ≤ M, so ∂'(un ) ∂q(un ) + is bounded in 9–1 ∞ (K). ∂t ∂x  By eqs (5.2.12) and (5.2.21), L1 (>) + (>) is bounded in ,(K). By eq. (5.2.21) we find that L2 (>) is compact in wq–1 , where p1 + q1 = 1. By the relevant lemma of the third chapter, # n) ∂q(un ) $ –1 (R2 ). is compact in Hloc we know ∂'(u ∎ + ∂t + ∂x

5.2 The Convergence of Hyperbolic Equations in Lax–Friedrichs Scheme

507

Theorem 5.2.2. The solutions of the Lax–Friedrichs difference schemes for hyperbolic equation (5.2.1) are uniformly bounded, and there exists a subsequence {u(nk ) } ⊂ {u(n) } which is weakly ∗ convergent in L∞ (R2+ ) as k → ∞, and strongly converges to u in Lp,loc (R2+ ), 1 < p < ∞, so u is the solution of eq. (5.2.1) with the initial value. u|t=0 = u0 (x).

(5.2.29)

2

Proof. ∀> ∈ C0∞ (R+ ), we have   +∞ ∂> (nk ) ∂> (nk ) + u0 (x)>(x, 0)dx u f (u ) dxdt + ∂t ∂x –∞ K n–1  (j+1)Bx  ! >(x, mBt) u(nk ) (x, mBt – 0) =

 

(j–1)Bx

m=1

j

" – u(nk ) (x, mBt + 0) dx.

(5.2.30)

By Theorem 3.8.13 in Chapter 3, we obtain u(nk ) ⇀ u, (Lp (K), 1 < p < +∞), " ∗ ! f u(nk ) ⇀ f (u). (Lp (K), nk → ∞). And by the property of the Lax–Friedrichs scheme we find 

(j+1)Bx

!

" u(nk ) (x, mBt – 0) – u(nk ) (x, mBt + 0) dx = 0.

(j–1)Bx

So the right of eq. (5.2.30) is L(>) =

n–1   j

m=1

(j+1)Bx

!

"

>(x, mBt) – >(jBx, mBt)

(j–1)Bx

! (n ) " u k (x, mBt – 0) – u(nk ) (x, mBt + 0) dx. Similar to eq. (5.2.21), there is |L(>)| ≤ C>1,p (Bx)( , where ( > 0, and there is L(>) → 0 as Bx → 0. So we take the limit as nk → ∞ in eq. (5.2.30), namely there is  

  +∞ ∂> ∂> u0 (x)>(x, 0)dx = 0. u+ f (u) dxdt + ∂t ∂x –∞

Thus u(x, t) is a weak solution of eqs (5.2.1) and (5.2.29).

(5.2.31)

508

5 Convergence of Three Different Schemes

2

And take > ∈ C0∞ (R+ ), > ≥ 0, ' and q are a pair of strictly convex entropy and entropy flow. Similar to eq. (5.2.9), there is  ∂> ∂> '(u(nk ) ) + q(u(nk ) ) dxdt ∂t ∂x K  n–1  (j+1)Bx  >(x, mBt) '(u(nk ) (x, mBt – 0)) =  

m=1,j (j–1)Bx

    – '(u(nk ) (x, mBt – 0)) dx + [']>dx – [q]>dt

≡ L(>) + Because of



A

A



(>).

(5.2.32)

(>) ≥ 0, L(>) =



n–1 

>(jBx, mBt)

 '(u(nk ) (x, mBt – 0))

(j–1)Bx

m=1,j



– '(u +

(j+1)Bx

(nk )



(x, mBt + 0)) dx

(j+1)Bx



 >(x, mBt) – >(jBx, mBt)

(j–1)Bx

m,j

  '(u(nk ) (x, mBt – 0)) – '(u(nk ) (x, mBt + 0)) dx = L1 (>) + L2 (>). Because '(u) is a convex function, and by the Jensen inequality and the property of the Lax–Friedrichs scheme, there is 1 2Bx





(j+1)Bx

'(u(nk ) (x, mBt – 0))

(j–1)Bx

 – '(u(nk ) (x, mBt + 0)) dx,

 ≥

1 2Bx

– '(u



(j+1)Bx

 u(nk ) (x, mBt – 0)dx

(j–1)Bx

(nk )

(x, mBt + 0)) = 0.

But L2 (>) → 0 as Bx → 0; Thus, let nk → ∞, and by eq. (5.2.32) we have   K

 ∂> ∂> '(u) + q(u) dxdt ≥ 0, ∂t ∂x

namely u is admissible solution.

5.2 The Convergence of Hyperbolic Equations in Lax–Friedrichs Scheme

509

n+1 of the Godunov scheme at t = tn+1 can be seen as the Because the solution vj+1/2 mean value in interval xj < x < xj+1 of exact solution to eq. (5.2.1) which satisfies initial condition n u(x, tn ) = vj+ 1 , xj < x < xj+1 , j = 0, ±1, ±2, . . . 2

at 

t=t

n+1

Bx = t + Bt Bt < maxj |aj | n

   ∂f aj are the eigenvalues of the matrix ; ∂u

that is n+1 vj+ 1 = 2

1 Bx



xj+1

(5.2.32)∗

u(x, tn+1 )dx.

xj

Therefore, the convergence of the Godunov scheme is similar to the proof of the Lax– Friedrichs scheme. ∎ Next we prove that the limit solutions of the Lax–Friedrichs scheme and the Godunov scheme satisfy the entropy inequality. Theorem 5.2.3. Suppose that eq. (5.2.1) exists convex entropy '(u) and entropy flow q(u). If the solution un of the Lax–Friedrichs scheme (5.2.2) converges to the vector function Bt u(x, t) as Bt, Bx → 0, there exists a positive number +0 , as + = Bx ≤ +0 , and u(x, t) satisfies the entropy inequality ∂'(u) ∂q(u) + ≤ 0. ∂t ∂x

(5.2.33)

Proof. In fact, we just need to prove '(un+1 j )–

'(unj+1 )+'(unj–1 ) 2

Bt

+

q(unj+1 ) – q(unj–1 ) 2Bx

≤ 0,

(5.2.34)

multiplying the left side of eq. (5.2.34) by a nonnegative compactly supported function >nj , and then make the summation about j and n, and then let Bt, Bx → 0; there is eq. (5.2.33). Let un+1 = z, unj–1 = v, unj+1 = w, j so the scheme (5.2.2) can be written as 1 + Bt z = (v + w) + [f (v) – f (w)], + = . 2 2 Bx

(5.2.35)

510

5 Convergence of Three Different Schemes

But eq. (5.2.34) can be written as 1 + '(z) ≤ ('(v) + '(w)) + (q(v) – q(w)). 2 2

(5.2.36)

⎧ >(() = (v + (1 – ()w, 0 ≤ ( ≤ 1, ⎪ ⎪ ⎪ ⎪ ⎪ 1 + ⎪ ⎪ ⎨ & (() = [>(() + w] + [f (>(()) – f (w)], 2 2 ⎪ !(() = '(& (()), ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ "(() = 1 ['(>(()) + '(w)] + + [q(>(()) – q(w)]. 2 2

(5.2.37)

Constructing function

It is easy to know !(0) = "(0) = '(w), and !(1) and "(1), respectively, denote the left and right side of inequality (5.2.36). Thus we should prove "(1) – !(1) ≥ 0.

(5.2.38)

In view of 

1

["′ (() – !′ (()]d(, 6  15 d> + ′ d> d& 1 ′ ′ = ' (>) + q (>) – ' (& ) d(, 2 d( 2 d( d( 0 6  15 + ′ d& 1 ′ ′ ' (>)(v – w) + ' (>)(v – w) – ' (>) d(, = 2 2 d( 0

"(1) – !(1) =

0

(5.2.39)

consider d> 1 1 d> + ′ + d& = + f (>) = (v – w) + f ′ (>)(v – w), d( 2 d( 2 d( 2 2 and '′ f ′ = q′ , by eq. (5.2.39) we obtain  "(1) – !(1) = 0

1

1 ′ [' (>) – '′ (& )] 2

[I + +f ′ (>)](v – w)d(. Let 8(,, () = ,>(() + (1 – ,)w, 0 ≤ , ≤ 1, = ,(v + (1 – ,()w,

(5.2.40)

5.2 The Convergence of Hyperbolic Equations in Lax–Friedrichs Scheme

. (,, () =

511

1 [>(() + 8(,, ()] 2 + + [f (>(()) – f (8(,, ())]. 2

There holds 8(0, () = w, 8(1, () = >((), . (0, () = & ((), . (1, () = >((), so we obtain ′



′

' (>) – ' (& ) =

1 0

∂ ′ ' (. (,, ())d,. ∂,

Because of ∂. ∂ ′ ' (. (,, ()) = '′′ (. ) , ∂, ∂, ∂8 1 ∂8 + ′ ∂. = – f (8) ∂, 2 ∂, 2 ∂, ( = [I – +f ′ (8)](v – w), 2 then '′ (>) – '′ (& ) =



1 0

( (v – w)T [I – +f ′ (8)]T '′′ (. )d,. 2

From eq. (5.2.40) we get "(1) – !(1) =

 1 0

0

1

( (v – w)T [I – +f ′ (8)]T 4

(5.2.41)

'′′ (. )[I + +f ′ (8)][v – w]d(d,. The integrand function under the right integral sign in eq. (5.2.41) can be denoted by G. Because '(u) is convex function, the eigenvalue of '′′ (u) must be greater than 0. m and M are used to denote the minimum and maximum eigenvalues of '′′ , respectively. If f ′  = c, note v – w = y, then G = yT [I – +f ′ (8)]T [I + +f ′ (8)]y'′′ (. ), = yT '′′ (. )[I + +f ′ (8)]y – +yT f ′ (8)T '′′ (. )[I + +f ′ (8)]y, = yT '′′ (. )y + +yT '′′ (. )f ′ (>)y

512

5 Convergence of Three Different Schemes

– +yT f ′ (8)T '′′ (. )y – +2 yT f ′ (8)T '′′ (. )f ′ (>)y, ≥ m|y|2 – +cM|y|2 – +cM|y|2 – +2 c2 M|y|2 , = [m – (2c+ + c2 +2 )M]|y|2 . Therefore, we only need m – (2c+ + c2 +2 )M ≥ 0,

(5.2.42)

to obtain G ≥ 0. If we take +0 =

1 c

7

 m –1 , M

1+

eq. (5.2.24) holds when + ≤ +0 . And thus the proof of theorem is completed.

∎

For the Godunov scheme, by eq. (5.2.32)∗ we know n+1 vj+ 1 = 2

1 Bx



xj+1

u(x, tn+1 )dx,

(5.2.43)

xj

where u(x, tn+1 ) satisfies the initial conditions n u(x, tn ) = vj+ 1 , xj < x < xj+1 , j = 0, ±1, ±2, . . .

(5.2.44)

2

The exact solution of eq. (5.2.1) satisfies entropy inequality 

xj+1

! " ' u(x, tn+1 ) dx –

xj

 

xj+1

! " ' u(x, tn ) dx

xj tn+1

+ tn  tn+1

–

! " n n q u(0; vj+ 1 ; vj+ 3 ) dt 2

! " n n q u(0; vj– 1 ; vj+ 1 ) dx ≤ 0, 2

tn

(5.2.45)

2

2

where u(0; a, b) denotes the solution of eq. (5.2.1) with the discontinuous initial value  u(x, 0) =

a, x < 0 b, x > 0,

which is the solution of the Riemann problem. Thus, n n u(0; vj+ 1 ; vj+ 3 ) = u(xj+1 , t), 2

2

n n u(0; vj– 1 ; vj+ 1 ) = u(xj , t). 2

2

5.3 Convergence of Glimm Scheme

513

Because '(u) is a convex function, by the Jensen inequality we have   xj+1  1 n+1 ' u(x, t ) Bx xj    x j+1 ! " 1 ≥' u x, tn+1 dx Bx xj ! n+1 " = ' vj+ 1 . 2

By eq. (5.2.45) we find   n+1 n n n Bx '(vj+ 1 ) – '(vj+ 1 ) + Bt(qj+1 – qj ) ≤ 0. 2

2

Thus, we deduce that the solution of the Godunov difference scheme satisfies the entropy inequality.

5.3 Convergence of Glimm Scheme The Glimm scheme is a kind of difference scheme with randomly selected amount which is proposed by Glimm in order to make the convergence of its differential sequence. The main difference between it and the Godunov scheme is a random number which is introduced into the progress in seeking exact solution of the Riemann problem. Roughly speaking, this is as follows: If we divided the line t = nBt into many subintervals by the point kBx(k = 0, ±1, ±2, . . .). For every k, given a normal range uk on the interval (k –1)Bx ≤ x < (k +1)Bx. We solve the Riemann problem (5.2.1) in the region nBt < t ≤ (n+1)Bt with initial value,  u(x, 0) =

uk ,

x < kBx,

uk+1 , x > kBx

(5.3.1)

we can record the solution as vk (x, t). Define u(x, (n + 1)Bt) = vk ((, (n + 1)Bt), kBx ≤ x < (k + 2)Bx,

(5.3.2)

where ( is randomly selected point in interval kBx ≤ x < (k + 2)Bx. For actual calculation, the section of ( is very important. Generally, the values (1 , (2 , ⋯ , (i , ⋯ of the random variables in the calculation are as close as possible to the uniform distribution   in interval – 21 , 21 . We describe the Glimm scheme more detailedly and more strictly as follows: Given the bounded domain u1 of the state 9 v, if u3 ⊂ u2 ⊂ u1 is domain of 9 v so that ul , ur is in u3 , there exists a unique solution to the Riemann problem (ul , ur )), which has

514

5 Convergence of Three Different Schemes

intermediate state u1 , u2 , . . . , un–1 that are all in u2 . Select Bx and Bt that satisfy the stability condition   Bx sup |+j (u)| : u ∈ u3 , 1 ≤ j ≤ n < , (5.3.3) Bt where +j (u) is the eigenvalue of f ′ (u). Fixed Bx/Bt = c, so Bt = c–1 Bx. Set  Y = {(m, n) ∈ Z 2 : m + n is even number, n ≥ 0}, I = F(m,n)∈Y {[(m – 1)Bx, (m + 1)Bx] × {nBt}}.

(5.3.4)

Set each interval ∈ I with the adjacent probabilistic metric (1/2Bx Lebesque measure). And I has product measure dI. We choose a point am,n = mBx + (n Bx, where (n is randomly selected in [–1, 1], am,n that is mesh point. Now define the Glimm difference scheme: If u = u(x, t) is defined in points am–1,n–1 and am+1,n–1 . In order to define u(am,n ), we solve the Riemann problem as follows: vt + f (v)x = 0, (m – 1)Bx ≤ x ≤ (m + 1)Bx, (n – 1)Bt ≤ t < nBt, with initial value

 u(x, 0) =

u(am–1,n–1 ), (m – 1)Bx ≤ x < mBx, u(am+1,n–1 ), mBx < x ≤ (m + 1)Bx.

We define u(am,n ) = v(am,n ). Because u(am–1,n–1 ) and u(am+1,n–1 ) are all in u3 , then we can do that. The process can be repeated down, which is due to the fact that the obtained intermediate states are all in u3 . In order to prove the convergence of the Glimm scheme, we have to establish some estimates of basic wave interaction. We know the solution to the Riemann problem with initial value  ul , x < 0, u(x, 0) = (5.3.5) ur , x > 0 for eq. (5.2.1) is composed of many waves: shock wave, rarefaction wave, contact discontinuity, normality and so on; they are generally denoted by (ul = u0 , u1 , . . . , un = ur ). It can be recorded as uk = T%kk uk–1 . If %k < 0, then uk is connected to the right of uk–1 by k shocks; %k > 0, so uk is connected to the right of uk–1 by rarefaction wave. Sometimes it can be expressed by the (ul , ur ) which connects with the solution to the Riemann problem

5.3 Convergence of Glimm Scheme

(ul , ur ) = [(u0 , u1 , . . . , un )/(%1 , . . . , %n )].

515

(5.3.6)

|%k | is referred to as the intensity of k wave uk linking with uk–1 . Now if ul , um and ur are three states close to the given state u, let (ul , um ) = [(u′0 , u′1 , . . . , u′n )/(#1 , . . . , #n )], (um , ur ) = [(u′′0 , u′′1 , . . . , u′′n )/($1 , . . . , $n )],

(5.3.7)

be the solution to the Riemann problems (ul , um ) and (um , ur ), respectively. We say that j wave u′j and k wave u′′k are intersecting: If (i) j > k, or (ii) j = k, #k < 0 or $k < 0. In other words, the two waves are disjoint. (i) For the different family wave, the left side of the wave has a larger eigenvalue, or (ii) For the akin wave, at least one is shock wave, because the akin rarefaction waves are not interacting. Lemma 5.3.1. If ul , um and ur are three states close to u, and if eqs (5.3.6) and (5.3.7) hold, then there exists %i = #i + $i + O(|#||$|).

(5.3.8)

If there is a coordinate system {wj } nearby u so that Ri wj ≡ ri ⋅ ∇wj = 0, i ≠ j, there exists %i = #i + $i + O([|#| + |$|]3 ),

(5.3.9)

where rj is right eigenvector, and wj (j = 1, 2) is a Riemann invariant as n = 2, Rj = rj ⋅ ∇. Proof. We know that eqs (5.3.6) and (5.3.7) hold when |%| is sufficiently small. It is not difficult to obtain u˙ = rk , u¨ = r˙k . So there exist u r – um =



$j r j +

j



$j $i Rj ri (1 – $ij ) + O(|$|3 ),

j≤i

and u l – um =



#j ]rj +

j

 j≥i

  1 $i $j Rj ri 1 – $ij + O(|$|3 ). 2

(5.3.10)

The coefficients all take values at um . Then by eq. (5.3.10), there exists ur – u l =

 ($i + "i )ri + O([|#| + |$|]2 ). i

(5.3.11)

516

5 Convergence of Three Different Schemes

On the other hand ur – ul =



%i r i +

 j≤i

  1 $j $i Rj ri 1 – $ij + O(|$|3 ). 2

(5.3.12)

The coefficients all take values at ul . % can be considered as a function of # and $, and because of $ = # = 0, % = 0, we have % = O(|$| + |#|). So ri (ul ) = ri (um ) –



#j Rj ri (um ) + O(|$|2 ).

(5.3.13)

j

By eq. (4.3.12), we have u r – ul =



%i ri (um ) + O([|#| + |$|]2 ).

and compare with eq. (5.3.11), then %i = #i + $i + O([|#| + |$|]2 ).

(5.3.14)

Substituting eqs (5.3.13) and (5.3.14) into eq. (5.3.12), we have u r – ul =



–

%i ri (um ) +



 j≤i



1 $j $i Rj ri (um ) 1 – $ij 2

 (5.3.15)

%j #i Ri rj (um ) + O([|#| + |$|]3 ).

ij

Finally, compare eqs (5.3.15) and (5.3.10); we have 

(%i – #i – $i )ri =

i



#i $j [Ri rj – Ri ri ] + O([|#| + |$|]3 )

j 0. ′′ : there exists a unique solution, as shown in In the first case, take vk′ = vm = vk–1 Fig. 5.2. So we have %i = #i , i ≤ k, %i = $i , i > k. In the second case, there also exists a unique solution, as shown in Fig. 5.3. Therefore, we have %i = #i , i ≤ k – 1, %k = #k + $k , %i = #i , i > k. So eq. (5.3.16) holds as D = 0. Now consider general situation D ≠ 0. We use the inductive method to prove. Suppose the estimate (5.3.16) holds for all $ = ($1 , . . . , $2 , . . . , $p–1 , 0, . . . , 0). We prove it also holds for $ = ($1 , . . . , $p–1 , . . . , $p , 0, . . . , 0). Because of $ = (0, . . . , 0), D = 0; we have proved it holds. ∎

νkʹ–1

γk

δk+1

δk+1

γk νmʹ νkʺ

νm

ν ʺk–1

ν ʺk+1

νm

νl

ν ʺk+1 νr

Fig. 5.2

δk νk–1 γk – 1

Fig. 5.3

νkʹ = νm

νk–1

ν ʺk νm

εk

δk+1

ν ʺk+ 1

γk–1

δk+1

518

5 Convergence of Three Different Schemes

+

γ

δ

ε

Δ γ2

γn

δ2

γ1

δp – 1

δp

δ3

εn – 1

ε1

εn νpʺ– 1

νoʹ = νl

νnʹ = νm

νoʺ = νm

νnʺ = νr

ν1 νo = νl

νn = ν r

Fig. 5.4

Definition 5.3.3. B = ($1 , $2 , . . . , $p–1 , 0, . . . , 0), B0 = (0, . . . , 0, $p , 0, . . . , 0),

(5.3.18)

and $ = ($1 , $2 , . . . , $p–1 , $p , 0, . . . , 0) = B + B0 . % depends on vm , and denote it by % = %(#, $; vm ). We first consider the interaction between # and B. Define , = (,1 , . . . , ,n ) and - = (-1 , . . . , -n ) as follows:  %i (#, B; vm ), ,i = 0,  -i =

0, ,i = %i (#, B; vm ), %i (#, B; vm ), ,i = 0.

So we have ,i = 0, i > p; -i = 0, i < p; and , + - = %(#, B; vm ). The interaction between # and B generates % shock, namely # + B → , + -. We use v: m to denote intermediate state which connects , wave and wave. If #p or $p is a shock wave,9 vm = vp–1 , but9 vm = vp as #p ≥ 0, $p ≥ 0. ′′ plays a role We then consider the interaction between # and B0 . At this time vp–1 of vm . We have ′′ ). - + B0 → 0 = %(-, B0 ; vp–1

Make , and - interact. , + 0 → %(,, 0;9 vm ).

519

5.3 Convergence of Glimm Scheme

π Δ0 ν ṽm

νr

νpʺ– 1

ṽm

r

Fig. 5.5

By final solution of (vl , vr ) and the uniqueness of the Riemann problem, we have %(#, $; vm ) = %(,, 0;9 vm ).

(5.3.19)

Now by the hypothesis of inductive method, %i (#, B; vm ) = #i + Bi + D(#, B)O(1). Because D(#, B) ≤ D(#, $), we have %i (#, B; vm ) + $ip $p = #i + $i + D(#, $)O(1).

(5.3.20)

0i = -i + $ip $p + |-||$p |O(1)

(5.3.21)

By eq. (5.3.8) we have

as shown in Fig.5.5. In order to prove Lemma 5.3.2, we also need to prove the following lemma. Lemma 5.3.4. |-||$p | = D(#, $)O(1).

Proof. By definition, -i = 0, i < p. Therefore, |-||$p | = -p = %p (#, B; vm ); by eq. (5.3.20) we have



i≥p |-i ||$p |.i

= p, or -p = 0 or by

-p = #p + D(#, $)O(1). So |-p ||$p | = |#p ||$p | + D(#, $)O(1)|$p |, ≤ D(#, $)[1 + O(1)|$p |], ≤ D(#, $)O(1). Finally, if i > p, by eq. (5.3.20) we have -i = #i + D(#, $)O(1). Thus, |-i ||$p | = D(#, $)O(1). This completes the proof.

∎

520

5 Convergence of Three Different Schemes

By Lemma 5.3.3, we can improve estimates (5.3.21) to 0i = -i + $ip $p + D(#, $)O(1).

(5.3.22)

Because %(#, B; vm ) = , + -, eq. (4.3.20) implies #i + $i = ,i + -i + $ip $p + D(#, $)O(1).

(5.3.23)

Now define 9 0 = - + B0 , and then we have D(,,9 0 ) = 0. In fact, if i < p,9 0i = 0, i > p, ,i = 0, i = p, or ,p = 0 or -p = 0, and ,p , $p do not intersect. Because the lemma holds when D = 0, we have %(,,9 0 ;9 vm ) = , + 9 0. Therefore, %i (,,9 0 ;9 vm ) = ,i + -i + $ip $p . By eq. (5.3.23) we have %i (,,9 0 ;9 vm ) = #i + $i + D(#, $)O(1). Because % is C1 function, we have   %(,, 0;9 0 ;9 vm ) = |0 – 9 0|O(1). vm ) – %i (,,9 By eq. (5.3.22) we find |0 – 9 0 | = D(#, $)O(1). Thus, %i (,, 0;9 vm ) = #i + $i + D(#, $)O(1). We deduce eq. (5.3.16) by eq. (5.3.19). If eq. (5.3.9) holds, we have   %(,, 0;9 vm ) – %i (,,9 0 ;9 vm ) = |0 – 9 0|O(1), ≤ D(#, $)(|#| + |$|).

(5.3.24)

5.3 Convergence of Glimm Scheme

521

Instead of eq. (5.3.24) we find %i (,,9 0 ;9 vm ) = #i + $i + D(#, $)(|#|, |$|).

(5.3.25)

Similarly, we can obtain eq. (5.3.17). The proof of Lemma 5.3.2 is completed. Definition 5.3.5. Grille curve is an unbounded segmented linear curve located on the sides of diamond from W to N or from W to s as shown in Fig. 5.6. If I is any grille curve, then I splits the upper half plane t ≥ 0 into two parts: I + and I – . I – includes t = 0. I1 > I2 means that for any point of I1 , it is either on I2 or including in I2+ . We say I is close to J which means that I > J and all the points of I are in J except one point as shown in Fig. 5.7. In the end let O denote the only grille curve through the grid points t = 0 and t = Bt. Now consider the functional defined on the grille curve J, F(J) = F(u|J ), where u|J denotes all the shock waves and rarefaction waves intersecting with grille curve J and splitting by normalcy. Let ! is j wave crossing J and " is k wave crossing J. We say ! intersects " which means that (1) j ≠ k, and fast family waves (high target) lie on the left side of J. or (2) j = k, ! = ", and there is at least one shock wave. N

W

E

S Fig. 5.6

I

J Fig. 5.7

522

5 Convergence of Three Different Schemes

This definition consists of the previous one. Now for given u(,Bx , we define the functionals Q and L on the grille curve J as Q(J) = G{|!||"|; !, " cross and intersect J}, L(J) = G|!|; ! crosses J, where ! and " are the functions of u(,Bx . Theorem 5.3.6. Suppose that I and J are two grille curves and J > I. Let I be in the domain of definition of u(,Bx . If L(I) is small enough, then J satisfies the following in the domain of definition of u(,Bx : Q(J) ≤ Q(I),

(5.3.26)

L(J) + kQ(J) ≤ L(I) + kQ(I),

(5.3.27)

where k is a constant independent with J. If total variation of u0 TV(u0 ) is small, then u(,Bx can be defined on t ≥ 0. Proof. First, let J is close to I; thus, they have difference only on a diamond as shown in Fig. 5.8. G G Let I = I0 I ′ , J = I0 J ′ . We have L(I) = L(I0 ) + L(I ′ ) = L(I0 ) +



|#i | +

I′

L(J) = L(I0 ) + L(J ′ ) = L(I0 ) +





|$i |,

I′

(5.3.28)

|%i |.

I′

Due to Lemma 5.3.2, L(J) ≤ L(I) + k0 Q(I ′ ),

J´ Δ

I0

δ

γ

I´ Fig. 5.8

(5.3.29)

I0

5.3 Convergence of Glimm Scheme

523

where k0 is the constant O(1) in Lemma 5.3.1. On the other hand 

Q(I) = Q(I0 ) + Q(I ′ ) + Q(I0 , I ′ ), Q(J) = Q(I0 ) + Q(I0 , J ′ ),

(5.3.30)

where Q(I0 , I ′ ) denotes the summation of product for two kinds of intersecting waves crossing I0 and I ′ . Then from Lemma 5.3.2 we have 

Q(I0 , J ′ ) =

|%||!|

intersecting wave pairs ! cross I0

≤



(|#i | + |$i |)|!| + k0 Q(I ′ )L(I0 ),

≤ Q(I0 , I ′ ) + k0 Q(I ′ )L(I0 ), 1 ≤ Q(I0 , I ′ ) + Q(I ′ ), 2 where we have used the condition that L(I0 ) is sufficiently small such that k0 L(I0 ) < 1/2. Thus, Q(J) – Q(I) = Q(I0 , J) – Q(I0 , I ′ ) – Q(I ′ ), 1 1 ≤ Q(I ′ ) – Q(I ′ ) = – Q(I ′ ) ≤ 0. 2 2 This means that Q is decreasing. Further, due to eq. (5.3.29) and the above, we have 1 L(J) + kQ(J) ≤ L(I) + K0 Q(I ′ ) + kQ(I) – kQ(I ′ ), 2 ≤ L(I) + kQ(I). (k ≥ 2k0 ). So for J closing to I, eqs (5.3.26) and (5.3.27) hold. Furthermore, we have L(J) + kQ(J) ≤ L(0) + kQ(0), ≤ L(0) + kL(0)2 ,   1 ≤ 2L(0) L(0) < . k This indicates that if the total variation of initial data is small, then the variation of u(,$x on J is also small. This means that u(,$x can be defined on grille curve closing to J. Thus, if J is an arbitrary grille curve, J > I, we can let I be adjacent to J gradually. Q and L + kQ are non-increasing; thus u(,$x can be defined on J. If TV(u0 ) is small, then

524

5 Convergence of Three Different Schemes

L(J) + kQ(J) ≤ 2L(0). So sup{|+j (u(,Bx )|, 1 ≤ j ≤ n}
t + Bt. Define  9 u(y, t) =

u(,Bx (y, t), |y – x| ≤ |t – t′ |(Bx/Bt), x ∈ R, others x ∈ R.

0,

We use 9 u to denote the initial value of difference scheme at t. On the depending area ′ of (x, t ), 9 u = u(,$x ,9 u(x, t′ ) = u(,$x (x, t′ ). On this depending area, |u(,Bx (x, t) – u(,Bx (x, t′ )| = |9 u(x, t) – 9 u(x, t′ )| ≤ TV(9 u(⋅, t)) + TV(9 u(⋅, t′ )), ≤ const ⋅ TV(9 u(⋅, t)), = const ⋅ TV(u(,Bx (⋅, t) is on the X),   Bx Bx where X = y : |y – x| ≤ |t – t′ | . Since is fixed, Bt Bt 

+∞ –∞

|u(,Bx (x, t) – u(,Bx (x, t′ )|dx  ≤

+∞ 5  x+0(t′ –t)

–∞

≤ O(t′ – t)

x–0(t′ –t) +∞

6 |du(,Bx (⋅, t)| dx,



–∞

|du(,Bx (⋅, t)|dx,

= O(t′ – t)TV[u(,Bx (⋅, t)], = O(t′ – t)TV(u0 ). ∎

This completes the proof.

In the following we consider the convergence of the difference approximate solutions. Suppose that {u! (x, t)} satisfies the following: ⎧ ⎪ u (⋅, ⋅)L∞ ≤ M1 , ⎪ ⎨ ! TVu! (⋅, t) ≤ M2 , ⎪ ⎪ ⎩ u (⋅, t ) – u (⋅, t ) !

1

!

2

(I) (II) L2

≤ M3 |t1 – t2 |,

(III)

where M1 is independent of !, M2 is independent of t and !, and M3 is independent of t1 , t2 , !. Theorem 5.3.10. Suppose that {u! } satisfies conditions (I), (II) and (III), and then there exists a subsequence {un } that converges to u in L1,loc .

526

5 Convergence of Three Different Schemes

The proof of this theorem is obvious. The solution {u(,Bx : ( ∈ I, Bx > 0.} satisfies conditions (I), (II) and (III); thus, it is compact. Next we should prove that the limit function u(x, t) is indeed the solution of problems (5.2.1) and (5.2.19); thus, we should first establish the following corollary. Corollary 5.3.11. Suppose that {un } satisfies the conditions in Theorem 5.3.10, and f (u) is continuous, then there exists a subsequence {unj } ⊂ {un } such that f (unj ) → f (u) in L1,loc norm. Proof. Since un → u(L1,loc ), then there exists a subsequence {unj } ⊂ {un } such that unj → u. almost everywhere. So f (unj ) → f (u). {unj } is uniform bound, so does {f (unj )}. With the help of Lebesgue bounded convergence theorem, we deduce that f (unj ) → f (u) in L1,loc norm. For fixed ( ∈ I, record uBx = u(,Bx . If 8 ∈ C01 (t > 0) is the test function. Define functional L8 .  L8 (u, f (u)) =

∞  +∞

(u8t + f (u)8x )dxdt 

0 –∞ +∞

(5.3.31)

u0 (x)8(x, 0)dx.

+ –∞

Our purpose is to establish that u satisfies L8 (u, f (u)) = 0, ∀8 ∈ C01 . We know that uBx is a weak solution in every time interval lBt ≤ t ≤ (l + 1)Bt, so for every test function 8, we have 

(l+1)Bt

lBt



+∞

–∞ +∞



+ –∞ +∞

{8t uBx + 8x f (uBx )}dxdt

8(x, lBt)uBx (x, lBt + 0)dx

 –

–∞

8(x, (l + 1)Bt)uBx (x, (l + 1)Bt – 0)dx = 0.

By the summation of the above formula we have  L8 (uBx , f (uBx )) =

∞  +∞

0 –∞  +∞

8(x, 0)u0Bx (x)dx

+ +

{8t uBx + 8x f (uBx )}dxdt

–∞ ∞  +∞  l=1

–∞

(5.3.32)

8(x, lBt)[uBx ](x, lBt)dx = 0,

where [uBx ] = uBx (x, lBt + 0) – uBx (x, lBt – 0). By comparing eqs (5.3.32) and (5.3.31) we find that we need the following conclusion such that uBx satisfies L(u, f (u)) = 0. ∎

5.3 Convergence of Glimm Scheme

527

Corollary 5.3.12. If ( ∈ I, u = lim u(,Bxj is a solution for eqs (5.2.1) and (5.2.19), if it Bxj →0

satisfies the following two conditions, when Bxj → 0, u(,Bxj (⋅, 0) → u0 , 5 6 ∞  ∞      8(x, lBt) u(,Bxj (⋅, lBt) dx ≡ 8(x, lBt) u(,Bxj l dx → 0. l=1

R

l=1

(5.3.33) (5.3.34)

R

Eq. (5.3.33) is easy to verify; in order to prove eq. (5.3.34), for test function 8, ( ∈ I, Bx > 0, define  Jl = R

J=

  8(x, lBt) u(,Bx l dx ≡ Jl ((, Bx, 8),

∞ 

Jl ≡ J((, Bx, 8).

l=1

Lemma 5.3.13. If ( ∈ I, 8 ∈ C0 ∩ L∞ , suppose condition (II) holds, ! = ((, Bx). Then there exist constants M and M1 which are independent of 8, ( and Bx such that |Jl ((, Bx, 8)| ≤ MBx8∞ ,

(5.3.35)

|J((, Bx, 8)| ≤ M1 (branch set lengths of 8)8L∞ .

(5.3.36)

Proof. Eq. (5.3.36) is corollary of eq. (5.3.35). Because if the support set of 8 is in [a, b] × [0, T], Jl is nonzero only as l < T/BT. Therefore, nonzero term only appears in the summation symbol as (Bt)–1 =O((Bx)–1 )(the diameter of support set of 8). Now we are going to prove eq. (5.3.35).

|Jl ((, Bx, 8)| ≤

∞  

(m+1)Bx

|8(x, lBt)[u(,Bx ]l |dx,

m=–∞ (m–1)Bx   (m+1)Bx

≤ 8∞

(m–1)Bx

m

 u(,Bx ((m + (l )Bx, lBt – 0)

 – u(,Bx (x, lBt – 0)dx,  # Osc u(,Bx (⋅, lBt – 0) ≤ 8∞ m

 (m+1)Bx $ [(m – 1)Bx, (m + 1)Bx] ⋅ dx, (m–1)Bx   ≤ 8∞ TV u(,Bx (⋅, lBt – 0) m

528

5 Convergence of Three Different Schemes

 [(m – 1)Bx, (m + 1)Bx]

⋅ 2Bx,

= MBx8∞ . Next we will seek a zero measure set N in I which is independent of Bx such that ( ∈ I/N; there exists a sequence Bxi → 0 such that eqs (5.3.33) and (5.3.34) hold. ∎ Lemma 5.3.14. If 8 has a compact support set and 8 is a fragment constant in [(m – 1)Bx, (m + 1)Bx] × lBt, then if l1 ≠ l2 , there is Jl1 (⋅, Bx, 8) ⊥ Jl2 (⋅, Bx, 8), where the sense of “” (orthogonal) is that orthogonal under L2 inner product is in I space. Proof. We only need to prove that l1 < l2 , Jl1 is independent of (l2 and we have  Jl2 d(l2 = 0.

I

In fact, if eq. (5.3.37) holds, we have  

 Jl1 Jl2 d(l2 0l≠l2 d(l ,    = Jl1 Jl2 d(l2 0l≠l2 d(l ,

< Jl1 , Jl2 > =

= 0. Now we prove eq. (5.3.37). Let tl = lBt – 0, then we have  I

Jl2 ((, Bx, 8)d(l2

 =

∞  

(m+1)Bx

I m=–∞ (m–1)Bx

 8mt u(,Bx ((m + (l )Bx, tl )

 – u(,Bx (x, tl ) dxd(l . We exchange integral variable and then obtain ∞ 

  8mt

I

m=–∞



(m+1)Bx

(m–1)Bx (m+1)Bx

u(,Bx (x, tl )dxd(l



– (m–1)Bx

I

 u(,Bx ((m + (l )Bx, tl )d(l dx = 0.

(5.3.37)

5.3 Convergence of Glimm Scheme

529

Now we consider Bxk = 2–k , if 8 satisfies the hypothesis of Lemma 5.3.14 for some Bx, and then it is satisfied by any smaller Bx. ∎ Theorem 5.3.15. If the condition (II) is satisfied, where ! = ((, Bx), then there exist zero measure set N ⊂ I and sequence Bxi → 0 such that for ∀( ∈ I/N and 8 ∈ C01 (t > 0), there is J((, Bxi , 8) → 0,

i → ∞.

Remark: J((, Bxi , 8) → 0 is equivalent to eq. (5.3.34). Proof. If 8 satisfies Lemma 5.3.14, we have J(⋅, Bxi , 8)22 =



Jl (⋅, Bxi , 8)22 ,

l

≤



Jl (⋅, Bxi , 8)2∞ .

(5.3.38)

l

Because



I d(

= 1, by eq. (5.3.35)  l

Jl (⋅, Bxi , 8)2∞ ≤



M12 (Bxi )2 82∞ ,

l∈D

where D = {l : (R × lBti ) ∩ (the support set of 8) is nonempty }. So 

J(⋅, Bxi , 8)22 ≤ M12 Bti (support set lengths of 8)82∞ .

(5.3.39)

l

Therefore, fragment constant 8 has compact support set, J(⋅, Bxi , 8) → 0 in L2 norm as Bxi → 0. Secondly, for 8 ∈ L∞ ∩ C0 , by eq. (5.3.35), we have J(⋅, Bxi , 8)2 ≤ J(⋅, Bxi , 8)∞ ≤ const8∞ .

(5.3.40)

If {8- } is a fragment constant sequence and has compact support set, it is dense in the test function space in L∞ norm (for example, take Haar function). For every 8- there exist zero measure set N- ⊂ I and sequence Bxi → 0 such that J((, Bxi , 8- ) → 0,

( ∈ I/N- .

(5.3.41)

Set N = ∪- N- , ( ∈ I/N, then by selection of diagonal, we can select a subsequence which is still recorded as Bxi such that for all J((, Bxi , 8- ) → 0,

i → ∞.

530

5 Convergence of Three Different Schemes

Now if 9 8 is any test function, by eqs (5.3.40) and (5.3.41), for ( ∈ I/N, there is 8)| ≤ |J((, Bxi , 9 8 – 8- )| + |J((, Bxi , 8- )|, |J((, Bxi , 9 ≥ const9 8 – 8- ∞ + o(1), i → ∞. Firstly we select - such that 9 8 – 8- ∞ is sufficiently small, then select i sufficiently large such that the second term on the right side of the above inequality is very small, and the proof is completed. ∎ Theorem 5.3.16. If TV(u0 ) is sufficiently small, there exist zero measure set N ⊂ I and sequence Bxi → 0 such that if ( ∈ I/N; then u( = lim u(,Bxi is generalized solution to problems (5.2.1) and (5.2.19). Proof. First, we select subsequence of {Bxi } such that eq. (5.3.33) holds, and then select subsequence such that for any ( ∈ I/N, 8 ∈ C01 (t > 0), J((, Bxi , 8) → 0, i → ∞. And make {u(,Bxj } convergence. By Corollary 5.3.12, we obtain the proof of the theorem. ∎ For more details about the contents of this chapter refer to Refs [11–19, 31, 32, 60]

6 Electric–Magnetohydrodynamic Equations In this chapter, we are concerned with global existence and large-time behavior of solutions to the isentropic electric–magnetohydrodynamic equations in a bounded domain K ⊆ RN , N = 2, 3. We establish the existence and large-time behavior of global weak solutions through a three-level approximation, energy estimates on condition that the adiabatic constant satisfies # > 3/2.

6.1 Introduction In many astrophysical models, stars may be considered as compressible fluids (see Ref. [92]), and their dynamics is very often shaped and controlled by electric fields and magnetic fields. Under some physical assumptions (see Refs [22, 33]), such dynamics can be described by the following mathematical model: ⎧ ⎪ ∂t 1 + div(1u) = 0, ⎪ ⎪ ⎨ ∂ (1u) + div(1u ⊗ u) + ∇p – ,Bu – (, + +)∇div u = J × B, t ⎪ % 0 ∂t E – ∇ × B + J = 0, J = 3(E + u × B), ⎪ ⎪ ⎩ ∂t B + ∇ × E = 0, div B = 0.

(6.1.1)

In the above equations, 1, u, E, B and p denote the density, velocity field, electric field, magnetic field and pressure, respectively. The pressure p is a function of 1 and assumed to obey the polytropic #-law, that is, p(1) = a1# , where a > 0 is the entropy constant and # > 3/2 is the adiabatic exponent. We denote by + and , the two viscosity coefficients of the fluid, which are assumed to satisfy the physical restrictions: , > 0 and + + 2, N ≥ 0. In Maxwell’s equations, %0 is the dielectric constant, and 3 > 0 is the electric conductivity of the fluid. For simplicity, we treat 3 as a constant in this work. The fourth equation of eq. (6.1.1) can be reduced to E = -∇ × B – u × B. Indeed, since %0 is usually small, we can neglect the displacement current in the third equation of eq. (6.1.1) to get J = ∇ × B. Here, - := 1/3 is the magnetic diffusivity acting as a magnetic diffusion coefficient of the magnetic field. In this case, the third and fourth equations of eq. (6.1.1) can be reduced to ∂t B + ∇ × (-∇ × B) – ∇ × (u × B) = 0, div B = 0. DOI 10.1515/9783110494273-006

(6.1.2)

532

6 Electric–Magnetohydrodynamic Equations

Equations (6.1.1)1 , (6.1.1)2 and (6.1.2) are the standard magnetohydrodynamic (MHD) equations. Because of its physical importance, complexity, rich phenomena and mathematical challenges, many physicists and mathematicians have studied it; please refer to Refs [8, 9, 22, 25, 30, 68, 78, 80, 95, 97, 98, 103, 104] for details. For the one-dimensional problem, please see Refs [8, 9, 25, 68, 74, 83, 95]. However, many fundamental problems for the MHD equations are still open. Even for the onedimensional case, when all viscosity, heat conductivity and diffusivity coefficients are constant, the global existence of a classical solution to the full perfect MHD equations with large initial data remains unknown, although the corresponding problem for the Navier–Stokes equations was solved by Kazhikhov and Shelukhin [75] in 1977. For more about Navier–Stokes equations, please see Refs [21, 84–86, 99, 100]. Due to the presence of magnetic field and its interaction with the hydrodynamic motion in the MHD fluids of large oscillations, the MHD problem presents serious difficulties. Recently, Bian and Yuan have obtained local well-posedness in the critical Besov spaces [4] and supercritical Besov spaces [5] for the N-dimensional (N ≥ 2) compressible MHD equations. Bian and Guo obtained local well-posedness in the critical Besov spaces for the N-dimensional (N ≥ 2) full compressible MHD equations [2] and proved that any smooth radially symmetric solutions to the compressible MHD equations would blow up in finite time when the initial density and magnetic field have compact support [3]. Hu–Wang [69] obtained the global existence and large-time behavior of weak solutions to the three-dimensional MHD equations. Chen–Tan [10] and Pu– Guo [90] showed the global existence of smooth solutions to the three-dimensional compressible MHD equations with small initial data. Li et al. [80] showed the existence and uniqueness of local in time strong solution with large initial data for the three-dimensional compressible MHD isentropic flow with zero magnetic diffusivity. For the one-dimensional thermally radiative MHD equations, Zhang–Xie [102] proved the global existence of classical solutions. In this paper, we consider a general case (i.e., we consider the effect of the displacement current, i.e., %0 ≠ 0), which has only been studied by Guo–Zhang [33] in the previous work. More specially, Guo– Zhang [33] obtained the global existence of classical solutions for the one-dimensional thermally radiative electric–magnetohydrodynamic equations. We study in this paper the equations of the electric–magnetohydrodynamic isentropic flows (6.1.1) in a bounded domain K ⊂ RN , N = 2, 3, with the following boundary conditions: ⎧ ⎪ 1(x, 0) = 10 (x) ∈ L# (K), 10 (x) ≥ 0, ⎪ ⎪ ⎪ ⎪ 1 ⎪ ⎨ 1(x, 0)u(x, 0) = m0 (x) ∈ L (K), m0 = 0 if 10 = 0, B(x, 0) = B0 (x) ∈ L2 (K), div B0 = 0 in D ′ (K), ⎪ ⎪ ⎪ ⎪ E(x, 0) = E0 (x) ∈ L2 (K), ⎪ ⎪ ⎩ u| = 0, B| = 0, E| = 0. ∂K ∂K ∂K

|m0 |2 10

∈ L1 (K), (6.1.3)

Up to now, there is not any result about the two- and three-dimensional electric– magnetohydrodynamic equations. So we believe that our results obtained in this

6.1 Introduction

533

chapter make a big breakthrough and is of great importance to the research of fluid dynamics. Our main results are as follows. Theorem 6.1.1. Suppose that K ⊆ RN , N = 2, 3, is a bounded domain with a boundary of class C 2+* , * > 0, and # > 3/2. Then for any given T > 0, the initial–boundary value problem (6.1.1) and (6.1.3) has a finite energy weak solution (1, u, B, E) on K × (0, T). Theorem 6.1.2. Suppose that (1, u, B, E) is a finite energy weak solution to the initial– boundary value problem (6.1.1) and (6.1.3) obtained in Theorem 6.1.1. Then there exists a stationary state of density 1s which is a positive constant, a stationary state of velocity field us = 0, a stationary state of magnetic field Bs which is a positive constant and a stationary state of electric field Es = 0 such that, as t → ∞, ⎧ ⎪ 1(x, t) → 1s strongly in L# (K); ⎪ ⎪ ⎨ u(x, t) → us = 0 strongly in L2 (K); ⎪ B(x, t) → Bs strongly in H –1 (K); ⎪ ⎪ ⎩ E(x, t) → Es = 0 strongly in H –1 (K).

(6.1.4)

Remark 6.1.3. When B = 0 and E = 0, system (6.1.1) reduces to the compressible Navier–Stokes equations. For the compressible Navier–Stokes equations, Theorem 6.1.1 also holds. Moreover, in this case Theorem 6.1.1 covers that in Ref. [28]. When %0 = 0, system (6.1.1) reduces to the compressible MHD equations, and we obtain the same results as that in Ref. [69]. Remark 6.1.4. In order to overcome the difficulty arising from the coupling and interaction among the magnetic field, electric field and the fluid variables, we need energy estimates about magnetic field and electric field. Also, we apply the idea used in the incompressible Navier–Stokes equations to deal with the nonlinear terms. Notations. Throughout the chapter, C stands for a “harmless” constant and may be different from line to line. We denote that f p stands for f Lp , and the symbol F(v) ′ stands for a weak limit of {F(vn )}∞ n=1 . We say f ∈ C([0, T]; Xweak ), that is, for any 8 ∈ X , < f (t), 8 >X×X′ ∈ C([0, T]). This chapter is structured as follows. Section 2 is devoted to establishing the energy estimate, giving definition of the finite energy weak solution and two lemmas that will be used in the following sections. In Section 3, we establish the existence of solutions to the approximation problem following the method in Ref. [28]. In Section 4, we let % → 0 in the approximate solutions obtained in Section 3 to get Proposition 6.4.2. In Section 5, we let $ → 0 in eq. (6.4.1) and finish the proof of Theorem 1.1. Section 6 is devoted to prove Theorem 1.2.

534

6 Electric–Magnetohydrodynamic Equations

6.2 Definition of the Finite Energy Weak Solution In this section, we first derive the energy estimate, then give the definition of the finite energy weak solution. At last, we introduce two lemmas that will be used many times in the following sections. Multiplying eq. (6.1.1)2 by u, integrating over K and using the boundary conditions in eq. (6.1.3) give that     a # d 1 (6.2.1) 1|u|2 + 1 dx + ,∇u22 + (+ + ,)div u22 = (J × B) ⋅ udx. dt K 2 #–1 K Similarly, multiplying eq. (6.1.1)3 by E and eq. (6.1.1)4 by B, one can deduce   d 1 %0 |E|2 dx = (∇ × B – J) ⋅ Edx, dt K 2 K d dt

 K

1 2 |B| dx = – 2

(6.2.2)

 K

(∇ × E) ⋅ Bdx.

(6.2.3)

Adding eqs (6.2.1), (6.2.2) and (6.2.3), and using the fact that div (E × B) = (∇ × E) ⋅ B – (∇ × B) ⋅ E, ( J × B) ⋅ u – J ⋅ E = J ⋅ (B × u) – J ⋅ E = –-|J|2 , we get    a # 1 1 1 d 1|u|2 + 1 + %0 |E|2 + |B|2 dx + ,∇u22 + (+ + ,)div u22 + -J22 = 0, dt K 2 #–1 2 2 (6.2.4) which together with the assumptions on initial data and Lemma 3.3 in Ref. [87] gives the following a priori estimates: 1 ∈ L∞ ([0, T]; L# (K)), 1|u|2 ∈ L∞ ([0, T]; L1 (K)), u ∈ L2 ([0, T]; H01 (K)), B ∈ L∞ ([0, T]; L2 (K)), E ∈ L∞ ([0, T]; L2 (K)), J ∈ L2 ([0, T]; L2 (K)). ′

Multiplying the continuity equation (6.1.1)1 by b (1), we get the renormalized continuity equation ′

b(1)t + div(b(1)u) + (b (1)1 – b(1))div u = 0 for some function b ∈ C 1 (R+ ).

(6.2.5)

535

6.2 Definition of the Finite Energy Weak Solution

Now, following the strategy of Refs [26, 81, 82], we introduce the concept of finite energy weak solution (1, u, B, E) to the initial–boundary value problem (6.1.1) and (6.1.3). Definition 6.2.1. The solution (1, u, B, E) to the initial–boundary value problem (6.1.1) and (6.1.3) is called finite energy weak solution, if it holds in the following sense: (i) The density 1 is a nonnegative function, 1 ∈ C([0, T]; L1 (K)) ∩ L∞ ([0, T]; L# (K)), 1(x, 0) = 10 , 2# #+1

and the momentum 1u satisfies 1u ∈ C([0, T]; Lweak (K)). (ii) The velocity field u, the magnetic field B and the electric field E satisfy the following conditions: u ∈ L2 ([0, T]; H01 (K)), B ∈ L∞ ([0, T]; L2 (K)), E ∈ L∞ ([0, T]; L2 (K)), J ∈ L2 ([0, T]; L2 (K)), 1u ⊗ u is integrable on (0, T) × K, ′

and 1u(x, 0) = m0 , B(x, 0) = B0 , E(x, 0) = E0 , div B = 0 in D (K). ′

(iii) Equations (6.1.1) are satisfied in D (K × (0, T)) provided that 1, u, B and E are prolonged to be zero outside K. (iv) The continuity equation in eq. (6.1.1) is satisfied in the sense of renormalized solutions, that is, eq. (6.2.5) holds in D ′ (K × (0, T)) for any b ∈ C 1 (R+ ) satisfying ′

b (z) = 0 for all z ∈ R+ large enough, say, z ≥ z0 ,

(6.2.6)

where the constant z0 depends on the choice of function b. (v) The energy inequality  F(t) + 0

t

(,∇u22 + (+ + ,)div u22 + -J22 )dxds ≤ F(0)

holds for almost everywhere t ∈ [0, T], where  

 a # 1 1 2 1 2 2 F(t) = 1|u| + 1 + %0 |E| + |B| dx, #–1 2 2 K 2    a # 1 1 1 |m0 |2 + 10 + %0 |E0 |2 + |B0 |2 dx. F(0) = #–1 2 2 K 2 10 Finally, we introduce two lemmas that will be used many times in the following. Lemma 6.2.2. ([70, 81]) Let X be a reflexive Banach space, Y is a Banach space, and X ↪ Y, Y ′ is separable and dense in X ′ . Moreover, assume that fn satisfies

536

6 Electric–Magnetohydrodynamic Equations

fn ∈ L∞ (0, T; X), ∂t fn ∈ Lp (0, T; Y), for some 1 < p ≤ ∞, fn L∞ (0,T;X) ≤ C, ∂t fn Lp (0,T;X) ≤ C, for any n ≥ 1. Then fn is relatively compact in C([0, T], Xweak ). Lemma 6.2.3. ([70]) Let X and Y be Banach spaces such that X ↪ Y is compact. Let fn ∈ L∞ (0, T; X) and fn → f in C([0, T], Xweak ). Then fn → f in C([0, T], Y).

6.3 The Faedo–Galerkin Approximation The proof of Theorem 6.1.1 is based on the following approximation problem: ⎧ ∂t 1 + div (1u) = %B1, ⎪ ⎪ ⎪ ⎪ # " ⎪ ⎨ ∂t (1u) + div (1u ⊗ u) + a∇1 + $∇1 + %∇1 ⋅ ∇u – ,Bu –(, + +)∇div u = J × B, ⎪ ⎪ ⎪ %0 ∂t E – ∇ × B + J = %BE, J = 3(E + u × B), ⎪ ⎪ ⎩ ∂t B + ∇ × E = %BB, div B = 0,

(6.3.1)

complemented by the boundary conditions: ∇1 ⋅ n |∂K = 0, ∇E ⋅ n |∂K = 0, ∇B ⋅ n |∂K = 0,

(6.3.2)

u|∂K = 0, and modified initial data: ¯ 0 < 1 ≤ 10 (x) ≤ 1¯ , ∇10 ⋅ n| ¯ K = 0, 1(0) = 10 ∈ C 2+- (K), ¯ i = 1, 2, 3, (1u)(0) = m0 , m0 = [m10 , m20 , m30 ], mi0 ∈ C 2 (K), 2+- ¯ ¯ K = 0, E(0) = E0 ∈ C (K), ∇E0 ⋅ n|

(6.3.3)

¯ ∇B0 ⋅ n| ¯ K = 0. B(0) = B0 ∈ C 2+- (K), In this section, we establish the existence of solutions to eq. (6.3.1) following the method in Ref. [28] with extra effort to overcome the difficulty arising from the magnetic field and electric field. Let Xn = span{8j }nj=1 be the finite-dimensional space endowed with the L2 Hilbert space structure, where the functions 8j ∈ D(K), j = 1, 2, . . . , form a dense subset ¯ {8j } consists of the orthonormal base in L2 . The approximate velocity field in C02 (K). un ∈ C([0, T]; Xn ) satisfies the system of integral equations:

6.3 The Faedo–Galerkin Approximation





537

 t

 ,Bun – div(1un ⊗ un ) 1(t)un (t) ⋅ 8dx – m0,$ ⋅ 8dx = 0 K K K "  ! + ∇ (+ + ,)div un – a1# – $1" – %∇1 ⋅ ∇un + Jn × B ⋅ 8dxds,

(6.3.4)

with Jn = 3(E + un × B), for all t ∈ [0, T] and any function 8 ∈ Xn , where %, $ and " are fixed positive parameters. Next, we introduce a family of operators M[1] : Xn ↦ Xn∗ ,  < M[1]v, w >=

K

1vwdx ∀v, w ∈ Xn .

These operators are invertible provided 1 is strictly positive on K, and we have  –1 . M–1 [1]L(Xn∗ ,Xn ) ≤ inf 1(x) x∈K

Moreover, making use of the identity M–1 [11 ] – M–1 [12 ] = M–1 [12 ](M[12 ] – M[11 ])M–1 [11 ], one can see that the map 1 ↦ M–1 [1] mapping L1 (K) into L(Xn∗ , Xn ) is well defined and satisfies M–1 [11 ] – M–1 [12 ]L(Xn∗ ,Xn ) ≤ C(n, ')11 – 12 L1 (K)

(6.3.5)

for any 11 , 12 from the set N' = {1 ∈ L1 (K)| inf 1 ≥ ' > 0}. x∈K

Now, identity (6.3.4) can be rephrased as follows:  t   N [1(s), E(s), B(s), un (s)]ds , un (t) = M–1 [1(t)] (m0,$ )∗ + 0

where < (m0,$ )∗ , 8 >=

 K

m0,$ ⋅ 8dx

and   ,Bun – div(1un ⊗ un ) < N [1(s), E(s), B(s), un (s)], 8 >= K " !  + ∇ (+ + ,)div un – a1# – $1" – %∇1 ⋅ ∇un + Jn × B ⋅ 8dx,

(6.3.6)

538

6 Electric–Magnetohydrodynamic Equations

with Jn is defined as same as that in eq. (6.3.4). The density 1n = 1[un ] is determined uniquely as the solution of the Neumann initial–boundary value problem: ⎧ ∂t 1 + div(1un ) = %B1, ⎪ ⎪ ⎨ (6.3.7) ∇1 ⋅ n |∂K = 0, ⎪ ⎪ ⎩ 1(x, 0) = 10,$ (x); and similarly we can prove that (En , Bn ) = (E, B)[un ] is determined uniquely as the solution of the system ⎧ ⎪ %0 ∂t E – ∇ × B + J = %BE, J = 3(E + u × B), ⎪ ⎪ ⎨ ∂ B + ∇ × E = %BB, div B = 0, t (6.3.8) ⎪ ∇E ⋅ n |∂K = 0, ∇B ⋅ n |∂K = 0, ⎪ ⎪ ⎩ E(x, 0) = E0,$ (x), B(x, 0) = B0,$ (x). ¯ × C 3 (K) ¯ × C 3 (K) ¯ satisfies the homogeneous The initial data (10,$ , E0,$ , B0,$ ) ∈ C 3 (K) Neumann boundary conditions ∇10,$ ⋅ n |∂K = 0, ∇E0,$ ⋅ n |∂K = 0, ∇B0,$ ⋅ n |∂K = 0 and 1 – 2"

0 < $ ≤ 10,$ ≤ $

,

10,$ → 10 in L# (K), |{10,$ < 10 }| → 0 as $ → 0; furthermore, we let

 m0,$ =

m0 (x), if 10,$ ≥ 10 (x), 0, if 10,$ < 10 (x).

(6.3.9) (6.3.10)

(6.3.11)

Problem (6.3.6)–(6.3.8) can be solved locally in time by means of the Schauder fixed point technique (see section 7.2 in Ref. [26] and sections 2.1–2.3 in Ref. [28]). For more details, please see Lemma 2.2 in Ref. [28]. The following a priori estimates furnish the possibility of repeating the above fixed point argument to extend the local solution un to the whole time interval [0, T]. Then, by the solvability of eqs (6.3.7) and (6.3.8), we know the functions (1n , En , Bn ) on the whole time interval [0, T]. √ Lemma 6.3.1. Assume ( 10 u0 , u0 , E0 , B0 ) ∈ (L2 (K), L2 (K), L2 (K)), 10 ∈ L# (K) ∩ L" (K). Then 5   6 a # $ " 1 1 1 sup 1n |un |2 + 1n + 1n + %0 |En |2 + |Bn |2 dx #–1 "–1 2 2 0≤t≤T K 2  T (6.3.12) + ,∇un 22 + (+ + ,)div un 22 + -Jn 2 + %∇En 22 + %∇Bn 22 ds 

0

T



+ 0

K

#–2

"–2

(a#1n + $"1n )|∇1n |2 dxds ≤ C,

where the constant C is independent of n.

6.3 The Faedo–Galerkin Approximation

539

Lemma 6.3.2. Under the condition of Lemma 6.3.1, for the (1n (x, t), un (x, t), En (x, t), Bn (x, t)) of problem (6.3.6)–(6.3.8), it holds that

solution

(1n un )t L2 (0,T;H –2 (K)) + Ent L2 (0,T;H –1 (K))∩L∞ (0,T;H –2 (K))

(6.3.13)

+ Bnt L2 (0,T;H –1 (K))∩L∞ (0,T;H –2 (K)) ≤ C, where the constant C is also independent of n.

Remark 6.3.3. Lemmas 6.3.1 and 6.3.2 can be proved easily; hence, we omit the proofs. Now, passing to the limit as n → ∞ in the sequence of approximate solutions (1n , un , Bn , En ), we can derive an existence result for problem (6.3.1)–(6.3.2) as follows. Proposition 6.3.1. Suppose that K ⊆ RN , N = 2, 3, is a bounded domain of the class C 2+* , * > 0. Let % > 0, $ > 0 and " > max{4, #} be fixed. Assume the initial data 10 , q, E0 , B0 satisfy eq. (6.3.3). Then for any given T > 0, problem (6.3.1)–(6.3.3) admits at least one solution (1, u, B, E) in the following sense: (i) The density is a nonnegative function such that 1 ∈ Lr ([0, T]; W 2,r (K)), ∂t 1 ∈ Lr ([0, T] × K), for some r > 1, the velocity u belongs to the class L2 ([0, T]; H01 (K)), eq. (6.3.1)1 holds almost everywhere on K × (0, T), and the boundary condition as well as the initial data condition on 1 are satisfied in the sense of traces. Moreover, the total mass is conserved, especially, 

 K

1(x, t)dx =

K

10,$ (x)dx,

for all t ∈ [0, T]; and the following estimates hold: 

T

$

 K

0



T

% 0

1"+1 dxdt ≤ C(%),

 K

|∇1|2 dxdt ≤ C

with C independent of %. (ii) All quantities appearing in eq. (6.3.1)2 are locally integrable, and the equation is 2# #+1

satisfied in D ′ (K × (0, T)). Moreover, we have 1u ∈ C([0, T]; Lweak (K)), and 1u satisfies the initial condition. (iii) All terms in eqs (6.3.1)3 and (6.3.1)4 are locally integrable on K × (0, T). The magnetic field B and electric field E satisfy eqs (6.3.1)3 , (6.3.1)4 and the initial data in the sense of distribution. div B = 0 also holds in the sense of distribution. (iv) The energy inequality

540

6 Electric–Magnetohydrodynamic Equations

  K

 a # $ " 1 1 1 2 1u + 1 + 1 + %0 |E|2 + |B|2 dx 2 #–1 "–1 2 2  T ! " ,|∇u|2 + (+ + ,)|div u|2 + -|J|2 + %|∇E|2 + %|∇B|2 dxdt + K

0



T

+ 0   ≤

K



K

(a#1#–2 + $"1"–2 )|∇1|2 dxdt

 a # $ " 1 1 1 10 u20 + 10 + 10 + %0 |E0 |2 + |B0 |2 dx 2 #–1 "–1 2 2

holds almost everywhere t ∈ [0, T].

6.4 The Vanishing Viscosity Limit Following Chapter 7 in Ref. [26] or [28], letting % → 0 in the approximate solutions obtained in Section 3, we can get the following result. Proposition 6.4.2. Suppose that K ⊆ RN , N = 2, 3, is a bounded domain of the class 6# C 2+* , * > 0. Let $ > 0 and " > max{4, 2#–3 } be fixed. Then for given initial data 10 , m0 as in eqs (6.3.9)–(6.3.11) and B0 , E0 as in eq. (6.1.3), there exists a finite energy weak solution (1, u, B, E) of the problem: ⎧ ⎪ ∂t 1 + div (1u) = 0, ⎪ ⎪ ⎨ ∂ (1u) + div (1u ⊗ u) + a∇1# + $∇1" – ,Bu – (, + +)∇div u = J × B, t ⎪ %0 ∂t E – ∇ × B + J = 0, J = 3(E + u × B), ⎪ ⎪ ⎩ ∂t B + ∇ × E = 0, div B = 0,

(6.4.1)

satisfying the initial boundary conditions (6.3.9)–(6.3.11) and (6.1.3). Moreover, 1 ∈ L"+1 (K × (0, T)) and the first equation of eq. (6.4.1) holds in the sense of renormalized solutions. Furthermore, (1, u, B, E) satisfies the following uniform estimates: #

sup 1(t)L# (K) ≤ CF$ [10 , m0 , B0 , E0 ],

(6.4.2)

t∈[0,T]

$ sup 1(t) t∈[0,T]

" L" (K)

≤ CF$ [10 , m0 , B0 , E0 ],

(6.4.3)

 sup  1(t)u(t)2L2 (K) ≤ CF$ [10 , m0 , B0 , E0 ],

(6.4.4)

uL2 ([0,T];H 1 (K)) ≤ CF$ [10 , m0 , B0 , E0 ],

(6.4.5)

sup B(t)2L2 (K) ≤ CF$ [10 , m0 , B0 , E0 ],

(6.4.6)

sup E(t)2L2 (K) ≤ CF$ [10 , m0 , B0 , E0 ],

(6.4.7)

JL2 ([0,T];L2 (K)) ≤ CF$ [10 , m0 , B0 , E0 ],

(6.4.8)

t∈[0,T]

0

t∈[0,T]

t∈[0,T]

6.5 Passing to the Limit in the Artificial Pressure Term

541

where the constant C is independent of $ > 0 and   F$ [10 , m0 , B0 , E0 ] =

K

 a # $ " 1 1 1 |m0,$ |2 + 10,$ + 10,$ + %0 |E0 |2 + |B0 |2 dx. 2 10,$ #–1 "–1 2 2

6.5 Passing to the Limit in the Artificial Pressure Term Our ultimate goal is to let $ → 0 in eq. (6.4.1). By applying the technique which is similar to that in Refs [28, 69], we can derive that 

T



"

K

0

#

8($1$ + a1$ ) ln(1 + 1$ )dx dt ≤ C

(6.5.1)

and "

$1$ → 0 as $ → 0, in D ′ (K × (0, T)),

(6.5.2)

where the constant C is independent of $ > 0, and 1$ is the solution of system (6.4.1). 6.5.1 Passing to the Limit From Proposition 6.4.2 and eq. (6.5.1), we get 1$ ∈ L∞ ([0, T]; L# (K)) and ∂t 1$ ∈ L∞ ([0, T]; W

2#

–1, #+1

(K)).

Hence, by Lemma 6.2.2, it holds that #

1$ → 1 in C([0, T]; Lweak (K)), as $ → 0,

(6.5.3)

From Proposition 6.4.2, we can also obtain u$ → u weakly in L2 ([0, T]; H01 (K)),

(6.5.4)

B$ → B, E$ → E weakly in L∞ ([0, T]; L2 (K)),

(6.5.5)

J$ → J weakly in L2 ([0, T]; L2 (K)),

(6.5.6)

div B = 0 in D ′ (K × (0, T)).

(6.5.7)

From eq. (6.5.1) and Proposition 2.1 in Ref. [26], one can get, as $ → 0, #

1$ → 1# weakly in L1 ([0, T]; L1 (K)),

(6.5.8)

542

6 Electric–Magnetohydrodynamic Equations

subject to a subsequence. On the other hand, by virtue of the momentum balance in eq. (6.4.1) and estimates (6.4.2)–(6.4.8), we have 2#

1$ u$ ∈ L∞ ([0, T]; L #+1 ), ∂t (1$ u$ ) = –div(1$ u$ ⊗ u$ ) + ⋯ ∈ L∞ ([0, T]; W

6#

–1, 4#+3

) + ⋯.

Hence, by Lemma 6.2.2, we have 2# #+1

1$ u$ → 1u in C([0, T]; Lweak (K)), as $ → 0.

(6.5.9)

Similarly, from eqs (6.4.1)3 , and (6.4.1)4 , we can deduce that ∂t B$ ∈ L∞ ([0, T]; H –1 (K)), ∂t E$ ∈ L∞ ([0, T]; H –1 (K)), which together with eq. (6.5.5) and Lemma 6.2.2 gives that B$ → B, E$ → E in C([0, T]; L2weak (K)), as $ → 0.

(6.5.10)

Thus, the limits 1, 1u, B, E satisfy the initial conditions (6.1.3) in the sense of distribution. The above estimates together with J$ ∈ L2 ([0, T]; L2 (K)) and J$ = 3(E$ + u$ × B$ ) imply that E$ × B$ → E × B in D ′ (K × (0, T)). Moreover, from the above estimates, by the fact that u$ ∈ L2 ([0, T]; W01,2 (K)), we have u$ × B$ × B$ → u × B × B in D ′ (K × (0, T)). Hence, we obtain, as $ → 0, J$ × B$ → J × B in D ′ (K × (0, T). Since the exponent # > 3/2, we get 2# 6 > , #+1 5 which combining with eq. (6.5.9) and Lemma 6.2.3 implies that 1$ u$ → 1u in C([0, T]; H –1 (K)), as $ → 0.

(6.5.11)

6.5 Passing to the Limit in the Artificial Pressure Term

543

Hence, by eq. (6.5.4), one gets 1$ u$ ⊗ u$ → 1u ⊗ u in D ′ (K × (0, T)), as $ → 0.

(6.5.12)

Therefore, letting $ → 0 in eq. (6.4.1) and making use of eqs (6.5.1)–(6.5.12), we can prove that (1, u, B, E) satisfies ⎧ ⎪ ∂ 1 + div (1u) = 0, ⎪ ⎪ t ⎪ ⎪ ⎪ ⎨ ∂t (1u) + div (1u ⊗ u) + a∇1# – ,Bu – (, + +)∇div u = J × B, ⎪ ⎪ %0 ∂t E – ∇ × B + J = 0, J = 3(E + u × B), ⎪ ⎪ ⎪ ⎪ ⎩ ∂t B + ∇ × E = 0, div B = 0,

(6.5.13)

in D ′ (K × (0, T)). Hence, the only thing left to complete the proof of Theorem 6.1.1 is to show the strong convergence of 1$ in L1 or, equivalently, 1# = 1# . Since 1$ , u$ is a renormalized solution to the continuity equation of eq. (6.4.1) in ′ D (RN × (0, T)), we deduce that ′

Tk (1$ )t + div(Tk (1$ )u$ ) + (Tk (1$ )1$ – Tk (1$ ))div u$ = 0 in D ′ (RN × (0, T)),

(6.5.14)

where Tk is the cutoff functions defined as follows: Tk (z) = kT

z k

for z ∈ R, k = 1, 2, . . .

(6.5.15)

and T ∈ C ∞ (R) is concave and is chosen such that  z, if z ≤ 1, T(z) = 2, if z ≥ 3.

(6.5.16)

Since for any fixed k, Tk (1$ ) ∈ L∞ (K × [0, T]) and Tk (1$ ) → Tk (1) weak–∗ in L∞ (K × [0, T]); moreover, ∂t Tk (1$ ) = –div(Tk (1$ )u$ ) + ⋯ ∈ L2 ([0, T]; W

6p –1, p+6

) + ⋯.

We get as similarly as eq. (6.5.3) that p

Tk (1$ ) → Tk (1) in C([0, T]; Lweak (K)) for all 1 ≤ p < ∞,

(6.5.17)

Tk (1$ ) → Tk (1) in C([0, T]; H –1 (K)). Passing to the limit for $ → 0+ , we obtain ′

∂t Tk (1) + div(Tk (1)u) + (Tk (1)1 – Tk (1))div u = 0 in D ′ (RN × (0, T)),

(6.5.18)

544

6 Electric–Magnetohydrodynamic Equations

where ′

′

(Tk (1$ )1$ – Tk (1$ ))div u$ → (Tk (1)1 – Tk (1))div u weakly in L2 (K × (0, T)). 6.5.2 The Effective Viscous Flux In this section, we discuss the effective viscous flux p – (+ + 2,)div u. Similarly to Refs [26, 28, 82], one can deduce the following result. Lemma 6.5.1. Assume that 1$ , u$ are the sequence of approximation solutions obtained in Proposition 6.4.1. Then 

T

lim

$→0+



0 T

= 0

 8  8

#

K

K

6(a1$ – (+ + 2,)div u$ )Tk (1$ )dxdt 6(a1# – (+ + 2,)div u)Tk (1)dxdt,

for any 8 ∈ D(0, T) and 6 ∈ D(K). Proof. As in Refs [26, 28, 69], we also consider the operators Ai [v] = B–1 [∂xi v], i = 1, 2, 3,

where B–1 stands for the inverse of the Laplace operator on RN . To be more specific,  –i.  Ai can be expressed by their Fourier symbol Ai [⋅] = F –1 |. |2i F [⋅] , i = 1, 2, 3, with the following properties: Ai vW 1,s (K) ≤ C(s, K)vLs (RN ) , if 1 < s < ∞, 1 1 1 Ai vLq (K) ≤ C(q, s, K)vLs (RN ) , if 1 ≤ q < ∞, ≥ – , q s N Ai vL∞ (K) ≤ C(s, K)vLs (RN ) , if s > N, N ≥ 2. Multiplying the momentum equation of eq. (6.4.1) by the test functions >i (x, t) = 8(t)6(x)Ai [Tk (1$ )], with 8 ∈ D(0, T), 6 ∈ D(K), i = 1, 2, 3, yields that 

T

 8

0

#

K T

 =

"

6(a1$ + $1$ – (+ + 2,)div u$ )Tk (1$ )dxdt 

8

0



T

K



! # "" ∂i 6 (+ + ,)div u$ – a1$ – $1$ Ai [Tk (1$ )]dxdt 61$ ui$

– 0



+, 0

K

T K

!

′

"

∂t 8Ai [Tk (1$ ] + 8Ai [(Tk (1$ ) – Tk (1$ )1$ )div u$ ] dxdt

! " 8 ∂j 6∂j ui$ Ai [Tk (1$ )] – ui$ ∂j 6∂j Ai [Tk (1$ )] + ui$ ∂i 6Tk (1$ ) dxdt

(6.5.19)

545

6.5 Passing to the Limit in the Artificial Pressure Term



T



T



+ K

0

 –

K

0

! " j j 8ui$ Tk (1$ )Ri,j [61$ u$ ] – 61$ u$ Ri,j [Tk (1$ )] dxdt 

j 81$ ui$ u$ ∂j 6Ai [Tk (1$ )]dxdt

T



– K

0

86J$ × B$ ⋅ Ai [Tk (1$ )]dxdt,

where J$ = 3(E$ + u$ × B$ ), the operators Ri,j = ∂j Ai [v] = Rj,i . Analogously, we can repeat the above arguments for eq. (6.5.13) and choose >i (x, t) = 8(t)6(x)Ai [Tk (1)], i = 1, 2, 3, as the test functions. Simply computing gives that 

T

 8

0

K T

 =

6(a1# – (+ + 2,)div u)Tk (1)dxdt  8

0



K

T





T

" ! ∂i 6 (+ + ,)div u – a1# Ai [Tk (1)]dxdt

– K

0

+, 



K

0 T

i

i

i

K

(6.5.20)

"

8 ∂j 6∂j u Ai [Tk (1)] – u ∂j 6∂j Ai [Tk (1)] + u ∂i 6Tk (1) dxdt

! " 8ui Tk (1)Ri,j [61uj ] – 61uj Ri,j [Tk (1)] dxdt 



– 0

!

0 K T

+ 

! " ′ 61ui ∂t 8Ai [Tk (1)] + 8Ai [(Tk (1) – Tk (1)1)div u] dxdt

81ui uj ∂j 6Ai [Tk (1)]dxdt – 0

T

 K

86J × B ⋅ Ai [Tk (1)]dxdt,

with J = 3(E + u × B). As similarly as that in Refs [26, 28, 69], it can be shown that all the terms on the right of eq. (6.5.19) converge to their counterparts in eq. (6.5.20). In fact, by eqs (6.5.1)–(6.5.12) and the Sobolev embedding theorem, it is not difficult to see that  T ! " j j 8ui$ Tk (1$ )Ri,j [61$ u$ ] – 61$ u$ Ri,j [Tk (1$ )] dxdt 0 K (6.5.21)  T ! " 8ui Tk (1)Ri,j [61uj ] – 61uj Ri,j [Tk (1)] dxdt. → K

0

In fact, from eqs (6.5.9), (6.5.17) and Lemma 3.4 in Ref. [28], we know that j

j

Tk (1$ )Ri,j [61$ u$ ] – 61$ u$ Ri,j [Tk (1$ )] → Tk (1)Ri,j [61uj ] – 61uj Ri,j [Tk (1)], 2#p (2+p)#+p

weakly in C([0, T]; Lweak j

), i, j = 1, 2, 3. If

2#p (2+p)#+p

>

6 5

(that is, p >

6# 2#–3 ),

j

then we get

Tk (1$ )Ri,j [61$ u$ ] – 61$ u$ Ri,j [Tk (1$ )] → Tk (1)Ri,j [61uj ] – 61uj Ri,j [Tk (1)], strongly in C([0, T]; H –1 (K)). Hence, eq. (6.5.21) holds due to eq. (6.5.4). On the other hand, by the properties of Ai , we get Ai [Tk (1$ )] → Ai (Tk (1)) in C((0, T) × K),

546

6 Electric–Magnetohydrodynamic Equations

Ri,j [Tk (1$ )] → Ri,j (Tk (1)) weakly in Lp ([0, T] × K) for all 1 < p < ∞, and ′

′

Ai [(Tk (1$ ) – Tk (1$ )1$ )div u$ ] → Ai [(Tk (1) – Tk (1)1)div u]

weakly in L2 ([0, T]; H 1 (K)). "+(

By eq. (6.5.1), one gets $1$$ → 0 strongly in L " (K × [0, T]), with ( > 0 small enough such that 1($ ≤ ln(1 + 1$ ). Therefore, the proof of Lemma 6.5.1 is complete. ∎ 6.5.3 The Amplitude of Oscillations The main result of this section is taken from Refs [28, 69], which reads as follows. Lemma 6.5.2. There exists a constant C independent of k such that lim Tk (1$ ) – Tk (1)L#+1 ((0,T)×K) ≤ C.

$→0+

Proof. Due to the fact z ↦ z# is convex, Tk concave on [0, ∞), we get 1# ≥ 1# , Tk (1) ≥ Tk (1).

(6.5.22)

(Tk (z) – Tk (y))#+1 ≤ (z# – y# )(Tk (z) – Tk (y)) ∀z, y ≥ 0,

(6.5.23)

Note that

Indeed, because of the symmetry of z and y, we only need to prove eq. (6.5.23) holds ′ for z ≥ y ≥ 0. From |Tk (z)| ≤ 1, ∀z ≥ 0, and (z – y)# ≤ z# – y# , ∀z ≥ y ≥ 0, we have ′

(Tk (z) – Tk (y))# ≤ |Tk (. )||z – y|# ≤ |z – y|# ≤ z# – y# . By eqs (6.5.22) and (6.5.23), one can deduce that 

T



!

lim

$→0+

" # 1$ Tk (1$ ) – 1# Tk (1) dxdt

K

0



T



= lim

$→0+



0

T



+ 0



K

(1# Tk (1) – 1# Tk (1) – 1# Tk (1) + 1# Tk (1))dxdt

K T

= lim

$→0+



0

T



+ 0

K

#

(1$ Tk (1$ ) – 1# Tk (1) – 1# Tk (1) + 1# Tk (1))dxdt

K

#

(1$ – 1# )(Tk (1$ ) – Tk (1))dxdt

(1# – 1# )(Tk (1) – Tk (1))dxdt

(6.5.24)

6.5 Passing to the Limit in the Artificial Pressure Term

 ≥ lim

$→0+





T



K

0

≥ lim

$→0+

T

0

K

547

#

(1$ – 1# )(Tk (1$ ) – Tk (1))dxdt #

(1$ – 1# )(Tk (1$ ) – Tk (1))dxdt #+1

≥ lim Tk (1$ ) – Tk (1)L#+1 ((0,T)×K) . $→0+

On the other hand, by interpolation inequality and Young’s inequality, we can prove that  T ! " div u$ Tk (1$ ) – div uTk (1) dxdt lim $→0+

K

0



T



T



! " div u$ Tk (1$ ) – Tk (1) dxdt

= lim

$→0+

0



K

= lim

$→0+

K

0

(Tk (1$ ) – Tk (1) + Tk (1) – Tk (1))div u$ dxdt

≤ C sup div u$ L2 ((0,T)×K) lim (Tk (1$ ) – Tk (1)L2 ((0,T)×K)

(6.5.25)

$→0+

$

+ Tk (1) – Tk (1)L2 ((0,T)×K) ) ≤ C lim Tk (1$ ) – Tk (1)L2 ((0,T)×K) $→0+

≤C+

1 #+1 lim Tk (1$ ) – Tk (1)L#+1 ((0,T)×K) . 2 $→0+

By Lemma 6.5.1, one can deduce that 

T



lim

$→0+

0

K

1 = lim a $→0 –

! # " 1$ Tk (1$ ) – 1# Tk (1) dxdt



0

T



K

#

{[a1$ – (+ + 2,)div u$ + (+ + 2,)div u$ ]Tk (1$ )

(6.5.26)

[a1#

– (+ + 2,)div u + (+ + 2,)div u]Tk (1)}dxdt  T + + 2, = (div u$ Tk (1$ ) – div uTk (1))dxdt, lim a $→0 0 K which combining with eqs (6.5.24) and (6.5.25) implies the desired conclusion.

∎

6.5.4 The Renormalized Solutions Now, we apply Lemma 6.5.2 to prove the following result. Lemma 6.5.3. The limits 1, u satisfy the continuous equation of eq. (6.5.13)1 in the sense of renormalized solutions, that is,

548

6 Electric–Magnetohydrodynamic Equations

′

∂t b(1) + div (b(1)u) + (b (1)1 – b(1))div u = 0

(6.5.27) ′

holds in D ′ (RN × (0, T)), N = 2, 3, for any b ∈ C 1 (R) satisfying b (z) = 0 for all z ∈ R large enough, say, z ≥ z0 , where the constant z0 may depend on b. Proof. Regularizing eq. (6.5.18), one gets ′

∂t Sm [Tk (1)] + div (Sm [Tk (1)]u) + Sm [(Tk (1)1 – Tk (1))div u] = rm ,

(6.5.28)

where Sm [v] = vm ∗ v are the smoothing operators and rm → 0 in L2 ([0, T]; L2 (RN )), ′ for any fixed k. Multiplying eq. (6.5.28) by b (Sm [Tk (1)]) and letting m → ∞, we deduce ! ′ " ∂t b[Tk (1)] + div(b[Tk (1)]u) + b (Tk (1))Tk (1) – b(Tk (1)) div u ′

′

(6.5.29)

= –b (Tk (1))[(Tk (1)1 – Tk (1))div u] in D ′ ((0, T) × RN ). By virtue of Tk (1) – 1Lp (K×(0,T)) ≤ lim inf Tk (1$ ) – 1$ Lp (K×(0,T)) $→0+

(6.5.30)

and p

Tk (1$ ) – 1$ Lp (K×(0,T))  T 1  $ ≤ |kT – 1$ |p dxdt k 0 {1$ ≥k}  T 1  $ (|kT ≤ | + |1$ |)p dxdt k 0 {1$ ≥k}  p ≤3 |1$ |p dxdt  ≤ 3p

{(x,t)∈K×(0,T)||1$ |≥k}

(6.5.31)

#

1$ {(x,t)∈K×(0,T)||1$ |≥k} #

#–p dxdt

1$

≤ 3p kp–# 1$ L# (K×(0,T)) , where 1 ≤ p < #, as k → ∞, we have Tk (1) → 1 in Lp (K × (0, T)) for 1 ≤ p < #, as k → ∞. Thus, if we show ′

′

b (Tk (1))[(Tk (1)1 – Tk (1))div u] → 0 in L1 (K × (0, T)) as k → ∞, then eq. (6.5.29) implies eq. (6.5.27). To do it, let us denote Qk,M = {(x, t) ∈ K × (0, T) | Tk (1) ≤ M)}, for M large enough ′ such that b (z) = 0 if z ≥ M, then

549

6.5 Passing to the Limit in the Artificial Pressure Term



T



′

′

|b (Tk (1))[(Tk (1)1 – Tk (1))div u]|dxdt 0 K ′ ′ = |b (Tk (1))[(Tk (1)1 – Tk (1))div u]|dxdt Qk,M

′



′

≤ sup |b (z)| 0≤z≤M

Qk,M

|(Tk (1)1 – Tk (1))div u|dxdt

′

′

≤ sup |b (z)| lim inf (Tk (1$ )1$ – Tk (1$ ))div u$ L1 (Qk,M ) $→0+

0≤z≤M

′

′

≤ sup |b (z)| sup u$ L2 ([0,T];H 1 (K)) lim inf Tk (1$ )1$ – Tk (1$ )L2 (Qk,M ) $→0+

$

0≤z≤M

′

≤ C lim inf Tk (1$ )1$ – Tk (1$ )L2 (Qk,M ) $→0+

#–1 #+1   ′ ′ 2# 2# ≤ C lim inf Tk (1$ )1$ – Tk (1$ )L1 (K×(0,T)) Tk (1$ )1$ – Tk (1$ )L#+1 (Q ) k,M $→0+  #–1 #+1 ! # ≤ C lim inf (3k1–# sup 1$ L# ) 2# ⋅ 2 2# Tk (1$ ) – Tk (1)L#+1 (K×(0,T))

$→0+

$

+ Tk (1) – Tk (1)L#+1 (K×(0,T)) + Tk (1)L#+1 (Qk,M )

" #+1  2#

 (#–1)2 #+1  1 – ≤ C lim inf k 2# ⋅ (4C + 2M|Qk,M | #+1 ) 2# → 0, as k → ∞, $→0+

′

where we have used Lemma 6.5.2, interpolation inequality and Tk (z)z ≤ Tk (z). Thus, we finish the proof of Lemma 6.5.3. ∎

6.5.5 Strong Convergence of the Density We are going to finish the proof of Theorem 6.1.1. For that, we introduce a family of functions Lk ∈ C 1 (R+ ) ∩ [0, ∞):  Lk (z) =

z log(z), z z log(k) + z k

Tk (s) ds, s2

if 0 ≤ z < k, if z ≥ k.

(6.5.32)

For z large enough, Lk (z) is a linear function, that is, for z ≥ 3k, Lk (z) = "k z – 2z, with  3k T (s) "k = ln k + k ks2 ds + 32 . Denote bk (z) := Lk (z) – "k z, then Lk can be rewritten as Lk (z) = "k z + bk (z),

(6.5.33)

and bk satisfies the conditions in Lemma 6.5.3. By the fact that 1$ , u$ are renormalized solutions of eq. (6.4.1)1 , we deduce that ∂t Lk (1$ ) + div(Lk (1$ )u$ ) + Tk (1$ )div u$ = 0.

(6.5.34)

550

6 Electric–Magnetohydrodynamic Equations

By eq. (6.5.13)1 and Lemma 6.5.3, we have ∂t Lk (1) + div(Lk (1)u) + Tk (1)div u = 0 in D ′ ((0, T) × K).

(6.5.35)

Since Lk (z) is a linear function for z large enough, we get Lk (1$ ) ∈ L∞ (0, T; L# (K)) uniformly for $. Therefore, Lk (1$ ) → Lk (1) weak–∗ in L∞ (0, T; L# (K)), as $ → 0. On the other hand, ∂t Lk (1$ ) = –div(Lk (1$ )u) + ⋯ ∈ L2 (0, T; W Lemmas 6.2.2 and 6.2.3, it is easy to show that

6#

–1, #+6

) + ⋯. Using

#

Lk (1$ ) → Lk (1) in C([0, T]; Lweak (K)) ∩ C([0, T]; H –1 K), as $ → 0.

(6.5.36)

#

So, Lk (1$ ) and Lk (1) belong to C([0, T]; Lweak (K)). Letting eq. (6.5.34) minus eq. (6.5.35), and integrating about t, we deduce  K

(Lk (1$ ) –

Lk (1))6|t0 dx

 t

!

(Lk (1$ )u$ – Lk (1)u) ⋅ ∇6 " + (Tk (1)div u – Tk (1$ )div u$ )6 dxds,

=

0

K

(6.5.37)

for any 6 ∈ D(K). From eq. (6.5.36), it holds that [Lk (1$ ) – Lk (1)]|t=0 = Lk (10,$ ) – Lk (10 ) → 0, as $ → 0.

(6.5.38)

Letting $ → 0 in eq. (6.5.37), one gets 

 t K

(Lk (1) – Lk (1))6dx = 0

K

(Lk (1) – Lk (1))u ⋅ ∇6dxds  t

+ lim

$→0+

0

K

(6.5.39) (Tk (1)div u – Tk (1$ )div u$ )6dxds,

for any 6 ∈ D(K). Since the velocity u ∈ L2 ([0, T]; H01 (K)), one has |u| ∈ L2 ([0, T]; L2 (K)). dist[x, ∂K] Considering a family of functions 6m ∈ D(K) which approximates the characteristic function of K such that 0 ≤ 6m ≤ 1, 6m = 1 for all x such that dist[x, ∂K] ≥

1 , m

(6.5.40)

and |∇6m (x)| ≤ 2m for all x ∈ K. Then taking 6 = 6m as the test functions in eq. (6.5.39) and passing to the limit as m → ∞, one can get

6.5 Passing to the Limit in the Artificial Pressure Term

551

 K

Lk (1) – Lk (1)dx  t = 0

K

 t Tk (1)div udxds – lim

$→0+

 t = 0

K

0

K

Tk (1$ )div u$ dxds

 t 1 # [a1 – (+ + 2,)div u$ ] lim + + 2, $→0+ 0 K $  t # 1$ Tk (1$ )dxds lim

Tk (1)div udxds +

a + + 2, $→0+ 0 K  t  t 1 = Tk (1)div udxds + [a1# – (+ + 2,)div u] lim + + 2, $→0+ 0 K 0 K  t a # 1 Tk (1$ )dxds × Tk (1)dxds – lim + + 2, $→0+ 0 K $  t  t a # = [Tk (1) – Tk (1)]div udxds – 1 Tk (1$ )dxds lim + + 2, $→0+ 0 K $ 0 K  t a 1# Tk (1)dxds + lim + + 2, $→0+ 0 K  t ≤ [Tk (1) – Tk (1)]div udxds 0 K     + = [Tk (1) – Tk (1)]div udxds × Tk (1$ )dxds –

{1≥k}

(6.5.41)

{1≤k}

≤ Tk (1) – Tk (1)L2 ({1≥k}) div uL2 ({1≥k}) + Tk (1) – Tk (1)L2 ({1≤k}) div uL2 ({1≤k}) ≤ C(div uL2 ({1≥k}) + Tk (1) – Tk (1)L2 ({1≤k}) ), where we have used estimate (6.5.24). By eqs (6.5.30) and (6.5.31), Tk (1) ≥ Tk (1) and Tk (1) ≤ 1, we have Tk (1) – Tk (1)L1 ({1≤k}) = 1 – Tk (1)L1 ({1≤k}) ≤ 1 – Tk (1)L1 ([0,T]×K) ≤ 3Ck–(#–1) → 0, as k → ∞. From the interpolation inequality and Lemma 6.5.2, for any 1 ≤ p < # + 1, it holds that Tk (1) – Tk (1)Lp ({1≤k}) ≤ Tk (1) – Tk (1)!L1 ({1≤k}) Tk (1) – Tk (1)1–! L#+1 ({1≤k}) (6.5.42) ≤ Tk (1) –

Tk (1)!L1 ({1≤k})

→ 0, as k → ∞,

where ! > 0. Let k → ∞ in eq. (6.5.41), applying div u ∈ L2 ([0, T] × K) and eq. (6.5.42) for p = 2, one can prove that  K

(Lk (1) – Lk (1))(x, t)dx ≤ 0, t ∈ [0, T].

552

6 Electric–Magnetohydrodynamic Equations

On the other hand, because of the fact that Lk (1) is a linear function when 1 is large enough, Lk (1) increases slower than 1 ln 1. Therefore, it holds that  Lk (1) – 1 ln 1L1 ((0,T)×K) ≤

|Lk (1) – 1 ln 1|dxdt {1≥k}

 ≤

|Lk (1)| + |1 ln 1|dxdt

(6.5.43)

{1≥k}

 ≤C

|1 ln 1|dxdt → 0, as k → ∞. {1≥k}

From the definition of Lk (1$ ), it follows that Lk (1$ ) – 1$ ln 1$ L1 ((0,T)×K)  ≤ |Lk (1$ ) – 1$ ln 1$ |dxdt  ≤

{1$ ≥k}

|Lk (1$ )| + |1$ ln 1$ | #

1$

{1$ ≥k}



≤ C(')

#

1$ dxdt (6.5.44)

#

1$

dxdt (' > 0 small enough) #–1–' {1$ ≥k} 1 $  # ≤ C(')k–#+1+' 1$ dxdt {1$ ≥k}

≤ C(')k–#+1+' → 0, as k → ∞, uniformly in $. By the weak lower semicontinuity of the norm, we obtain Lk (1) – 1 ln 1L1 ([0,T]×K) ≤ lim inf Lk (1$ ) – 1$ ln 1$ L1 ([0,T]×K) → 0, $→0

uniformly in $,

(6.5.45)

which together with estimates (6.5.43) and (6.5.44) gives that  K

(1 ln 1 – 1 ln 1)(x, t)dx ≤ 0, for t ∈ [0, T] almost everywhere.

(6.5.46)

Because of the convexity of the function z → z ln z, one gets 1 ln 1 ≥ 1 ln 1, almost everywhere in K × [0, T], together with eq. (6.5.46), giving 1 ln 1(t) = 1 ln 1(t), for t ∈ [0, T] almost everywhere. This equality implies that 1$ → 1 strongly in L1 ((0, T) × K). The proof of Theorem 6.1.1 is completed. ◻

553

6.6 Large-Time Behavior of Weak Solutions

6.6 Large-Time Behavior of Weak Solutions In this section, our goal is to study the large-time behavior of the finite energy weak solutions, whose existence is ensured by Theorem 6.1.1. First, from Theorem 6.1.1, one can get 

∞

ess sup F(t) + t>0

! K

0

" ,|Du|2 + (+ + ,)|div u|2 + -|J|2 dxdt ≤ F(0).

(6.6.1)

Applying the method in Ref. [27], let us introduce a sequence for all m ∈ Z, t ∈ (0, 1), x ∈ K: ⎧ ⎪ 1m (x, t) := 1(x, t + m), ⎪ ⎪ ⎨ u (x, t) := u(x, t + m), m ⎪ B m (x, t) := B(x, t + m), ⎪ ⎪ ⎩ Em (x, t) := E(x, t + m). By eq. (6.6.1), we can obtain uniform estimates 1m ∈ L∞ ([0, 1]; L# (K)), Bm ∈ L∞ ([0, 1]; L2 (K)), 2# √ 1m um ∈ L∞ ([0, 1]; L2 (K)), 1m um ∈ L∞ ([0, 1]; L #+1 (K)), Em ∈ L∞ ([0, 1]; L2 (K)), which are independent of m. Moreover, we have  1 lim

m→∞ 0

  ∇um 2L2 (K) + -Jm 2L2 (K) dt = lim

m+1 

m→∞ m

 ∇u2L2 (K) + -J2L2 (K) ds

= 0. Hence, choosing a subsequence if necessary, we can assume that, as m → ∞, 1m (x, t) → 1s weakly in L# (K × (0, 1)); um (x, t) → us weakly in L2 ([0, 1]; H01 (K)); Bm (x, t) → Bs weakly in L2 (K × (0, 1)); Em (x, t) → Es weakly in L2 (K × (0, 1)); Jm (x, t) → Js weakly in L2 (K × (0, 1)). Furthermore, 

 K

1s dx ≤ lim inf m→∞

K

1m (t)dx ≤ C(E0 ).

(6.6.2)

554

6 Electric–Magnetohydrodynamic Equations

Therefore, from the Poincaré inequality and eq. (6.6.2), we know that 

1

lim

m→∞ 0

um 2L2 (K) dt = 0,

which combining with H 1 ↪ L2 is compact implies us = 0, almost everywhere in K × (0, 1).

(6.6.3)

In virtue of Sobolev’s inequality, Hölder’s inequality, and eqs (6.6.1) and (6.6.2), we get  lim

1!

1m |um |2 

m→∞ 0

3# L #+3



≤ 2 sup 1m L# lim

m→∞ 0

t

"

+ 1m |um |2

6# L #+6

1

dt

∇um 2L2 (K) dt ≤ 0,

which gives  1

2

1m |um | 

lim

m→∞ 0

3#

L #+3

+ 1m |um |

2 6#

dt = 0.

(6.6.4)

L #+6

Because 1, u are solutions to eq. (6.1.1) in the sense of renormalized solutions, we have 1t + div(1u) = 0 in D ′ (K × (0, T)).

(6.6.5)

Taking test functions >(x, t) = 8(t)6(x) in eq. (6.6.5), where 8(t) ∈ (0, 1), 6 ∈ D(K), integrating by parts, we get  1  0

K

 1  ′ 1m 6dx 8 (t)dt + 81m um ⋅ ∇6dxdt = 0 for all integer m. 0

K

Letting m → ∞ and using eq. (6.6.4), one can conclude   1  ′ 1s 6dx 8 (t)dt = 0, 0

K

which together with the arbitrariness of 8 yields that 1s must be independent of t. #+( Arguing as similarly as that in Section 5, we can get 1m is bounded in L1 (K × (0, 1)), independent of m > 0, for some ( > 0. Thus, we get #

1m → 1# weakly in L1 (K × (0, 1)).

(6.6.6)

Hence, passing to the limit in the second equation of eq. (6.1.1), and applying eqs (6.6.2) and (6.6.4), one can get ∇1# = 0 in D ′ (K).

(6.6.7)

555

6.6 Large-Time Behavior of Weak Solutions

Now, we prove that the convergence in eq. (6.6.6) is indeed strong. For that, as similarly ( as that in Ref. [27], we consider G(z) = z! , 0 < ! < min{ 2#1 , (+# }, so that b(z) = G(z# ) may #

#

be used in eq. (6.2.5). Introducing the vector functions [G(1m ), 0, 0, 0] and [1m , 0, 0, 0] of the time variable t and the spatial coordinates x, applying eqs (6.2.5), (6.6.2) and (6.6.4), one can get #

–1,q

div[G(1m ), 0, 0, 0] is precompact in Wloc 1 (K × (0, 1)),

(6.6.8)

for some q1 > 1 small enough. Similarly, making use of the second equation in eqs (6.1.1), (6.6.2) and (6.6.4), we obtain #

–1,q

curl[1m , 0, 0, 0] is precompact in Wloc 2 (K × (0, 1)),

(6.6.9)

for some q2 > 1. Moreover, we can assume #

G(1m ) → G(1# ) weakly in Lp2 (K × (0, 1)), # 1m

→ 1# weakly in L (K × (0, 1)), p1

(6.6.10) (6.6.11)

with p2 = !1 , p11 + p12 = 1r < 1. Applying the Lp version of the celebrated div-curl lemma, we drive from eqs (6.6.8) to (6.6.11) that G(1# )1# = G(1# )1# ,

(6.6.12) 1

together with that G = is strictly monotone, implying G(1# ) = G(1# ). Since L ! is uniformly convex, this implies strong convergence in eq. (6.6.6). Therefore, we have # 1m → 1s strongly in L# (K × (0, 1)), together with eqs (6.6.6) and (6.6.7), giving ∇1s = 0 in the sense of distributions, which implies that 1s is independent of the spatial variables. On the other hand, from the third equation and the fourth equation in eq. (6.1.1), we can get that ∂t Em and ∂t Bm belong to L2 (0, T; H –1 (K)). From the fact that L2 (K) ↪↪ H –1 (K), eqs (6.6.2) and (6.6.3), it holds that Em (x, t) → Es = 0, strongly in H –1 (K); Bm (x, t) → Bs , strongly in H –1 (K). Finally, denoting G(t) =

 !1 K

2 a # 2 1u + #–1 1

(6.6.13)

"

dx+ 21 %0 EH –1 (K) + 21 BH –1 (K) , by the energy in-

equality, we know that G(t) converges to a finite constant as t → ∞, G∞ := limt→∞ G(t). Using eqs (6.6.4) and (6.6.13), we have 

m+1 

lim

m→∞ m

K

1|u|2 dxdt = 0,

556

6 Electric–Magnetohydrodynamic Equations



m+1

BH –1 (K) dt = Bs H –1 ,

lim

m→∞ m

lim



m+1

m→∞ m

EH –1 (K) dt = 0.

Thus, G∞



m+1  



  a # 1 1 2 1 1u + 1 dx + %0 EH –1 (K) + BH –1 (K) dt 2 #–1 2 2

= limm→∞ m K  1 a # = 1s dx + Bs H –1 . 2 K #–1

Moreover, applying the first equation in eq. (6.1.1), we observe that 1(x, t) → 1s weakly in L# (9) as t → ∞. So we get 

1 a # 1s dx + Bs H –1 # – 1 2 K   1 a # ≤ lim inf 1 dx + BH –1 t→∞ 2 K #–1   1 a # ≤ lim sup 1 dx + BH –1 t→∞ 2 K #–1   1  a 1 1 ≤ limt→∞ 1u2 + 1# dx + %0 EH –1 (K) + BH –1 (K) #–1 2 2 K 2

G∞ =

= limt→∞ G(t) = G∞ , which implies  lim

t→∞ K

a # 1 dx = #–1

 K

a # 1s dx, #–1

and by the fact that the space L# is uniformly convex, eq. (6.1.4) follows. We complete the proof of Theorem 6.1.2. ∎

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