Bruno Després Numerical Methods

Frontiers in Mathematics

Numerical Methods for Eulerian and Lagrangian

Conservation Laws

Bruno Després

Frontiers in Mathematics

Advisory Editorial Board

Leonid Bunimovich (Georgia Institute of Technology, Atlanta)

Benoît Perthame (Université Pierre et Marie Curie, Paris)

Laurent Saloff-Coste (Cornell University, Ithaca)

Igor Shparlinski (Macquarie University, New South Wales)

Cédric Villani (Institut Henri Poincaré, Paris)

William C. Chen (Nankai University, Tianjin, China)Y.

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Wolfgang Sprö ig (TU Bergakademie Freiberg)ß

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Bruno Després

Numerical Methods

Conservation Lawsfor Eulerian and Lagrangian

ISSN 1660-8046 ISSN 1660-8054 (electronic) Frontiers in Mathematics ISBN 978-3-319-50354-7 ISBN 978-3-319-50355-4 (eBook) DOI 10.1007/978-3-319-50355-4 Library of Congress Control Number: 2017946756

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Bruno Després

Laboratoire Jacques-Louis Lions

Université Pierre et Marie Curie

Paris, France

Mathematics Subject Classification (2010): 65M08, 65M12, 65Z05, 35L65, 35L67

http://www.birkhauser-science.com

Preface

(Entropy stability) =⇒ (Something nice)– Phil Roe

(HYP2006 conference, Lyon)

Systems of conservation laws are ubiquitous in compressible and incompressibleflows, and are encountered in numerous applications in hydrodynamics, aerody-namics, magnetohydrodynamics, plasma physics, combustion, traffic flow, and lin-ear and nonlinear elasticity. This list is far from exhaustive. The equations areintrinsically nonlinear and the mathematical framework is that of hyperbolic sys-tems of conservation laws, on which many excellent mathematical and numericaltextbooks exist already. The role of the entropy law is central.

In contrast to most of the material in the literature, which concentrates onEulerian formulations and Eulerian discrete schemes, i.e. on methods which aredeveloped and applied in a fixed reference frame, the present monograph focusesspecifically on Lagrangian formulations of conservation laws which have a physicalmotivation and originate from continuum mechanics, and on Lagrangian numericalmethods conceived in a moving frame attached to the flow. In other words, weare interested in relationships between the mathematical theory of conservationlaws and certain Lagrangian-based numerical methods used for compressible fluiddynamics (CFD). The Lagrangian approach has pros and cons which are mostly aconsequence of the kind of problems one wants to solve at the numerical level. Sobefore embarking on the study of Lagrangian methods per se and of the interplaybetween Lagrangian and Eulerian techniques, it is worth making a tentative listof the advantages of Lagrangian numerical methods so as to answer the question:why use the Lagrangian approach?

A first nice property of Lagrangian methods for compressible gas dynamics isthat the numerical transport is not solved like other partial differential equations,but rather is addressed by moving the mesh. It is thus possible to reduce thenumerical diffusion. This feature is of the highest value for convection-dominatedproblems, such as in application-oriented numerical simulations. In astrophysics,we refer the reader to the survey [185]. For the numerical modeling of direct-drive inertial confinement fusion (ICF), modern Lagrangian numerical methodshave enjoyed recent successes [28, 146]. In these fields and more generally in highenergy density physics (HEDP), standard Eulerian discretizations of the transportoperator may lead to unacceptable numerical errors, and the natural alternativeconsists of Lagrangian methods which exercise strong control over the numericaldiffusion associated with transport.

A second property is natural modeling of free boundaries and material in-terfaces. Indeed, there is no mass flux along a free boundary, so a Lagrangiannumerical method is natural since it easily respects such constraints. The situa-tion is similar for internal material interfaces which separate different fluids, since

v

vi Preface

the masses are constant on both sides of the material interface. This is also thecase in ICF flows and more generally in fluid–structure interaction. This feature ofLagrangian methods has been identified in the seminal work of von Neumann andRichtmyer [195], but in contrast to this fundamental reference, which concernsstaggered Lagrangian numerical discretization of non-viscous compressible fluiddynamics, cell-centered numerical methods are considered in this book.

A third property, which is closely related to the previous one, is that La-grangian formulations are a natural starting point for arbitrary Lagrange-Euler(ALE) numerical discretization techniques. The limiting case comprises the so-called Lagrange+remap techniques, which are actually Eulerian techniques.

The fourth property is of mathematical nature and is an extension of theapproach proposed in [68]. It asserts that the structure of Lagrangian models isvery particular. In fact, written in adapted variables, a Lagrangian flux is linear-quadratic with respect to a particular entropy variable, denoted by Ψ in thismonograph. One typically has in dimension d = 1,

ρDt U +

⎛⎝

MΨ

− 1

2(MΨ, Ψ)

⎞⎠ = 0 (1)

where M is a constant matrix. In the case of Lagrangian compressible gas dy-namics, one sees that U = (τ, u, e) (specific volume, velocity, total energy), Ψ =

(p, −u) (pressure, opposite of the velocity) and M =

(0 11 0

). Additional no-

tation includes the density ρ = τ −1 and the Lagrangian or material derivativeDt = ∂t + u∂x . Not surprisingly the justification of this structure is based uponcompatibility with invariance principles (translation invariance, for example) andwith the entropy law, which reads Dt S = 0 for smooth flows (and DtS ≥ 0 inthe general case). A nice consequence is that algebraic aspects of any method ofnumerical discretization for Lagrangian compressible gas dynamics immediatelyextend to more complex Lagrangian models. That is, not only are numerical La-grangian methods efficient for CFD in the cases mentioned above, but at the levelof principles they can be extended quite easily to more complex Lagrangian mod-els, such as ideal magnetohydrodynamics (MHD). It should be noted that thegeneral structure (1) also shows that these models are generically non-strictly hy-perbolic for size(U) ≥ 4. The consequence of this fact are manifold: one is thedefinition of an additional potential called the enthalpy of the system to analyzethe structure. A multidimensional generalization is proposed.

Another attractive property is that Lagrangian models (1) are endowed witha general theory of first-order entropy-increasing schemes. These schemes are a gen-eralization of the acoustic solver of Godunov [100]. This property gives systematiclinear and nonlinear stability of the numerical methods based on a rigorous dis-crete entropy inequality. Using a Lagrange+remap strategy, which offers the mostnatural way to decouple the physics from the advection, one gets back Eulerian

Preface vii

numerical methods which inherit the entropy inequality, and hence have similarstrong linear and nonlinear stability properties. The discretization by entropy-satisfying numerical methods also has the major advantage of solving the trickyquestion of multidimensional Lagrangian numerical methods, which was more orless an open problem for Lagrangian CFD before 2000. This point was first ob-served in Mazeran’s PhD thesis [69], where some questions addressed in Loubere’sPhD thesis [138] were answered, and has since been generalized in manydirections.Comparing with the standard numerical theory for systems of conservation laws, amore surprising consequence is that the correct generalization of multidimensionalLagrangian Riemann solvers is corner-based and not edge-based.

So, at the end of this tour of the assets of the Lagrangian approach in thecontext of cell-centered numerical methods, our reformulation of Roe’s aphorismquoted at the beginning of this preface could be:

(Lagrangian structure) =⇒ (multidimensional discrete entropy stability).

However, the above pros of the Lagrangian approach are strongly counterbal-anced by the fact that a moving mesh can easily become pathological. Even if thenonlinear stability afforded by discrete entropy inequalities provides a way to con-trol such pathological cases, there exist situations where the physics brings stronglimitations to Lagrangian methods. The most evident situation is a vorticity-dominated flow. In this case a Lagrangian method alone will never be able tocompute the solution and it is vital to address this issue. The answer is wellknown at the level of numerical methods: one equips the Lagrangian solver witha convenient remeshing strategy (a standard approach is ALE); in practice sucha method can give good results and in some cases excellent results. So one hasto mitigate this conclusion about the conflict with vorticity-dominated flows. Inpractice Lagrangian methods provide a solid starting point even for the design ofEulerian or ALE-based methods.

Another puzzling property of Lagrangian systems of PDEs is that they aregenerically weakly hyperbolic because the underlying physical problem containsshear velocities. The standard hyperbolic theory is not enough to explain all thefeatures of Lagrangian systems of conservation laws in dimensions higher than one.Therefore one can consider the mathematical foundations of Lagrangian PDEs forcontinuum mechanics as still needing to be reinforced. The very brief discussionof these issues in this monograph is not enough.

The organization of this text aims to follow a natural approach, from themathematical foundations to the numerical methods. Even though it is naturalfrom a theoretical perspective, this way of presenting Lagrangian numerical meth-ods is rarely followed, mostly because a more direct path focusing on applicationsand using mechanical analogies is possible. However, I believe that one can go

viii Preface

further and deeper starting from solid foundations, hence the plan of this mono-graph. More specifically, the two first chapters present a selection of well-knownmathematical features of conservation laws, intended to serve as preparation forthe next three chapters, which are dedicated to the analysis and discretization ofLagrangian systems. Illustrations are given to demonstrate the efficiency of thenumerical methods. Exercises at the end of each chapter introduce more material,and comments are given with additional references. The presentation of the newcorner-based Lagrangian finite volume techniques in the final chapter is based onjoint research with Emmanuel Labourasse and Stephane Delpino. This last chaptercan also be read first, since it is somewhat independent.

None of the results and methods presented here could have been obtainedwithout the constant support of the Commissariat a l’Energie Atomique over theyears, nor without hours of passionate discussion with numerous colleagues andfriends on both sides of the Atlantic. A preliminary and elementary version hasbeen published in French [70]. The author is particularly indebted, for variousreasons, to Constant Mazeran, Herve Jourdren, Remi Sentis, and Bruno Scheurer.Discussions with Pierre-Henri Maire were always valuable. Warm thanks are dueto Emmanuel Labourasse, Stephane Delpino, Stephane Jaouen, Guillaume Moreland Gautier Dakin for their kind help in correcting numerous errors. Responsibilityfor the remaining ones is mine.

Sorbonne Universites,UPMC University Paris 06,UMR 7598, Laboratoire Jacques-Louis Lions,F-75005, Paris, France,

and Institut Universitaire de France,

the 15th of September 2016.

Contents

Preface v

List of Figures xiii

List of Tables xvii

1 Models 11.1 Balance law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

1.1.1 Traffic flow . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.1.2 Shallow water . . . . . . . . . . . . . . . . . . . . . . . . . . 41.1.3 Compressible gas dynamics . . . . . . . . . . . . . . . . . . 81.1.4 Canonical form of a system of conservation laws . . . . . . 11

1.2 Lagrangian coordinates . . . . . . . . . . . . . . . . . . . . . . . . 111.2.1 General change of coordinates in balance laws . . . . . . . . 121.2.2 Lagrangian gas dynamics in dimension d = 1 . . . . . . . . 151.2.3 Lagrangian gas dynamics in dimension d = 2 . . . . . . . . 171.2.4 Hui’s formulation . . . . . . . . . . . . . . . . . . . . . . . . 191.2.5 Lagrangian gas dynamics in dimension d = 3 . . . . . . . . 19

1.3 Frame invariance . . . . . . . . . . . . . . . . . . . . . . . . . . . . 201.3.1 Naive method . . . . . . . . . . . . . . . . . . . . . . . . . . 211.3.2 A general method . . . . . . . . . . . . . . . . . . . . . . . 23

1.4 Linear stability and hyperbolicity . . . . . . . . . . . . . . . . . . . 241.4.1 Classification in dimension d = 1 . . . . . . . . . . . . . . . 251.4.2 A useful property . . . . . . . . . . . . . . . . . . . . . . . . 281.4.3 Generalization to dimension d ≥ 2 . . . . . . . . . . . . . . 291.4.4 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30

1.5 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 381.6 Bibliographic notes . . . . . . . . . . . . . . . . . . . . . . . . . . . 40

2 Scalar conservation laws 412.1 Strong solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 422.2 Weak solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46

ix

x Contents

2.3 Entropy weak solutions . . . . . . . . . . . . . . . . . . . . . . . . 502.3.1 Entropic discontinuities . . . . . . . . . . . . . . . . . . . . 542.3.2 Shocks and contact discontinuities . . . . . . . . . . . . . . 562.3.3 Rarefaction fans . . . . . . . . . . . . . . . . . . . . . . . . 582.3.4 The entropic solution of the Riemann problem . . . . . . . 59

2.4 Peculiarities of Lagrangian traffic flow . . . . . . . . . . . . . . . . 612.4.1 Application and physical interpretation . . . . . . . . . . . 63

2.5 Numerical computation of entropy weak solutions . . . . . . . . . . 672.5.1 Notion of a conservative finite volume scheme . . . . . . . . 672.5.2 Finite volume scheme . . . . . . . . . . . . . . . . . . . . . 702.5.3 Construction of the flux using the method

of characteristics . . . . . . . . . . . . . . . . . . . . . . . . 712.5.4 Definition of a generic flux . . . . . . . . . . . . . . . . . . . 762.5.5 Convergence . . . . . . . . . . . . . . . . . . . . . . . . . . 792.5.6 Scheme optimization . . . . . . . . . . . . . . . . . . . . . . 82

2.6 More schemes for the traffic flow equation . . . . . . . . . . . . . . 832.6.1 Numerical illustrations . . . . . . . . . . . . . . . . . . . . . 85

2.7 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 892.8 Bibliographic notes . . . . . . . . . . . . . . . . . . . . . . . . . . . 91

3 Systems and Lagrangian systems 933.1 Generalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 94

3.1.1 The Godunov theorem . . . . . . . . . . . . . . . . . . . . . 973.1.2 Entropy weak solutions . . . . . . . . . . . . . . . . . . . . 101

3.2 Lagrangian systems in dimension d = 1 . . . . . . . . . . . . . . . . 1033.2.1 Systems with a zero entropy flux . . . . . . . . . . . . . . . 1043.2.2 A more general Lagrangian structure . . . . . . . . . . . . . 111

3.3 Examples of Lagrangian systems . . . . . . . . . . . . . . . . . . . 1153.3.1 Ideal MHD . . . . . . . . . . . . . . . . . . . . . . . . . . . 1153.3.2 Compressible elasticity . . . . . . . . . . . . . . . . . . . . . 1193.3.3 Landau model for superfluid helium . . . . . . . . . . . . . 1233.3.4 A multiphase model . . . . . . . . . . . . . . . . . . . . . . 126

3.4 Self-similar solutions and the solution of theRiemann problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1303.4.1 Rarefaction fans . . . . . . . . . . . . . . . . . . . . . . . . 1313.4.2 Entropy discontinuities . . . . . . . . . . . . . . . . . . . . 1333.4.3 Lax theorem in the space U . . . . . . . . . . . . . . . . . . 1403.4.4 A Lagrangian Lax theorem in the space W . . . . . . . . . 144

3.5 Multidimensional Lagrangian systems . . . . . . . . . . . . . . . . 1473.6 More on compressible gas dynamics . . . . . . . . . . . . . . . . . . 152

3.6.1 Rarefaction fans . . . . . . . . . . . . . . . . . . . . . . . . 1533.6.2 Discontinuities . . . . . . . . . . . . . . . . . . . . . . . . . 1543.6.3 The Riemann problem for gas dynamics . . . . . . . . . . . 158

3.7 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 158

Contents xi

3.8 Bibliographic notes . . . . . . . . . . . . . . . . . . . . . . . . . . . 163

4 Numerical discretization 1654.1 Compressible gas dynamics . . . . . . . . . . . . . . . . . . . . . . 165

4.1.1 Principle of a Lagrange+remap scheme in one dimension . . 1674.1.2 Principle of an entropy Lagrangian solver . . . . . . . . . . 1684.1.3 Entropy Lagrangian solver based on matrix splitting . . . . 1694.1.4 An optimal splitting for fluid dynamics . . . . . . . . . . . 1744.1.5 Moving grid . . . . . . . . . . . . . . . . . . . . . . . . . . . 1794.1.6 Remapping . . . . . . . . . . . . . . . . . . . . . . . . . . . 1804.1.7 Eulerian formulation of a Lagrange+remap scheme . . . . . 1814.1.8 Boundary conditions . . . . . . . . . . . . . . . . . . . . . . 1844.1.9 A simple numerical result . . . . . . . . . . . . . . . . . . . 1854.1.10 Pure Lagrange and ALE methods in one dimension . . . . . 185

4.2 Linearized Riemann solvers and matrix splittings . . . . . . . . . . 1934.2.1 Solution of the Lagrangian linearized Riemann problem . . 1974.2.2 One-state solvers . . . . . . . . . . . . . . . . . . . . . . . . 1984.2.3 Two-state solvers . . . . . . . . . . . . . . . . . . . . . . . . 1994.2.4 Optimality of the two-state solver . . . . . . . . . . . . . . 207

4.3 Extension to multidimensional Lagrangian systems . . . . . . . . . 2114.3.1 A generic discrete entropy inequality . . . . . . . . . . . . . 2114.3.2 Cylindrical and spherical gas dynamics . . . . . . . . . . . . 2154.3.3 Lagrange+remap MHD in dimension d > 1 . . . . . . . . . 216

4.4 Lagrangian gas dynamics in dimension d = 2 . . . . . . . . . . . . 2234.4.1 Elementary considerations on moving meshes . . . . . . . . 2234.4.2 Some notation . . . . . . . . . . . . . . . . . . . . . . . . . 2254.4.3 Compatibility with Piola identities . . . . . . . . . . . . . . 2264.4.4 Compatibility with Hui’s formulation . . . . . . . . . . . . . 2284.4.5 First attempt and geometrical obstruction . . . . . . . . . . 2284.4.6 Solving the geometrical obstruction:

GLACE and EUCCLHYD . . . . . . . . . . . . . . . . . . . 2314.4.7 Comparison with a scheme on a staggered mesh . . . . . . 2414.4.8 Well-balanced hydrostatic cell-centered

Lagrangian schemes . . . . . . . . . . . . . . . . . . . . . . 2454.4.9 Mesh considerations and numerical examples . . . . . . . . 252

4.5 Calculation of Lagrangian multi-material problems . . . . . . . . . 2554.6 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2594.7 Bibliographic notes . . . . . . . . . . . . . . . . . . . . . . . . . . . 260

5 Starting from the mesh 2635.1 Axiomatization of mesh features . . . . . . . . . . . . . . . . . . . 264

5.1.1 Planar geometries . . . . . . . . . . . . . . . . . . . . . . . 2655.1.2 The reference cell method . . . . . . . . . . . . . . . . . . . 2685.1.3 Nodal control volumes . . . . . . . . . . . . . . . . . . . . . 274

xii Contents

5.1.4 Axisymmetric geometry . . . . . . . . . . . . . . . . . . . . 2765.2 Cell-centered Lagrangian schemes . . . . . . . . . . . . . . . . . . . 278

5.2.1 Construction of the scheme . . . . . . . . . . . . . . . . . . 2795.2.2 Time discretization and extensions . . . . . . . . . . . . . . 283

5.3 Stability of the mesh for simplexes . . . . . . . . . . . . . . . . . . 2875.4 Weak consistency of the gradient and divergence operators . . . . . 290

5.4.1 Additional inequalities . . . . . . . . . . . . . . . . . . . . . 2915.4.2 Gradient . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2925.4.3 Divergence . . . . . . . . . . . . . . . . . . . . . . . . . . . 296

5.5 Weak consistency of Lagrangian schemes . . . . . . . . . . . . . . . 2975.5.1 Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2985.5.2 The density equation . . . . . . . . . . . . . . . . . . . . . . 2995.5.3 The momentum equation . . . . . . . . . . . . . . . . . . . 3015.5.4 The energy equation . . . . . . . . . . . . . . . . . . . . . . 3025.5.5 The entropy inequality . . . . . . . . . . . . . . . . . . . . . 302

5.6 Stabilization with subzonal entropies . . . . . . . . . . . . . . . . . 3025.6.1 Lagrangian properties of volume fractions . . . . . . . . . . 3055.6.2 Building a scheme with subzonal entropies . . . . . . . . . . 3085.6.3 Consistency of subzonal entropies . . . . . . . . . . . . . . . 3115.6.4 Numerical illustration . . . . . . . . . . . . . . . . . . . . . 312

5.7 Constraints and quadratic formulation of fluxes . . . . . . . . . . . 3155.7.1 Quadratic functionals . . . . . . . . . . . . . . . . . . . . . 3155.7.2 Application to contact problems . . . . . . . . . . . . . . . 3185.7.3 Non-conformal meshes, hanging nodes

and internal constraints . . . . . . . . . . . . . . . . . . . . 3235.8 Bibliographic notes . . . . . . . . . . . . . . . . . . . . . . . . . . . 327

Bibliography 331

Subject Index 347

List of Figures

1.1 Time variation of N (t, x0, x1) . . . . . . . . . . . . . . . . . . . . . 21.2 LWR law for traffic flow . . . . . . . . . . . . . . . . . . . . . . . . 41.3 Column of water . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.4 Forces on the boundaries of the water column . . . . . . . . . . . . 71.5 Elementary quantity of gas in a moving interval. . . . . . . . . . . 91.6 Translation of the referential. . . . . . . . . . . . . . . . . . . . . . 211.7 Small perturbations for the traffic flow . . . . . . . . . . . . . . . . 31

2.1 Characteristic lines . . . . . . . . . . . . . . . . . . . . . . . . . . 452.2 Characteristic lines which do not cross . . . . . . . . . . . . . . . . 462.3 Test function with compact support . . . . . . . . . . . . . . . . . 472.4 Decomposition of the domain . . . . . . . . . . . . . . . . . . . . . 492.5 Discontinuous solutions in the (x, t) plane . . . . . . . . . . . . . . 502.6 Illustration of the Oleinik condition. . . . . . . . . . . . . . . . . . 562.7 Reversible discontinuity . . . . . . . . . . . . . . . . . . . . . . . . 572.8 Rarefaction fan . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 582.9 Oleinik solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 602.10 Trafic jam . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 642.11 Olmos-Munos model for traffic flow . . . . . . . . . . . . . . . . . . 652.12 Trafic jam in Bogota . . . . . . . . . . . . . . . . . . . . . . . . . . 652.13 Buckley-Leverett flux . . . . . . . . . . . . . . . . . . . . . . . . . . 672.14 Solution of the Buckley-Leverett Riemann problem . . . . . . . . . 68

2.15 Conservative and non-conservative schemes, t < Tshock . . . . . . . 692.16 Conservative and non-conservative schemes, t > Tshock . . . . . . . 692.17 Upwinded flux . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 712.18 A case of ambiguous upwinding . . . . . . . . . . . . . . . . . . . . 732.19 Lagrangian scheme and particle discretization . . . . . . . . . . . . 852.20 Traffic jam: entry and exit . . . . . . . . . . . . . . . . . . . . . . . 862.21 Non-entropic shock for LWR . . . . . . . . . . . . . . . . . . . . . 872.22 Lagrangian result for LWR . . . . . . . . . . . . . . . . . . . . . . 882.23 Eulerian result for Olmos-Munos model . . . . . . . . . . . . . . . 882.24 Numerical solution of the Buckley-Leverett equation . . . . . . . . 892.25 Non-entropic solutions for the Buckley-Leverett equation . . . . . . 90

xiii

xiv List of Figures

3.1 Irreversible and reversible processes . . . . . . . . . . . . . . . . . . 1043.2 Rarefaction fans and discontinuities . . . . . . . . . . . . . . . . . 1303.3 The minimum of a Kulikovski function . . . . . . . . . . . . . . . . 1393.4 Waves at a left state . . . . . . . . . . . . . . . . . . . . . . . . . . 1423.5 Waves at a right state . . . . . . . . . . . . . . . . . . . . . . . . . 1423.6 Structure of the solution of the Riemann problem for n = 3 . . . . 1433.7 Structure of the solution of the Riemann problem . . . . . . . . . . 1443.8 Lagrangian Riemann problem in the (m, t) plane . . . . . . . . . . 1463.9 Lagrangian Riemann problem in the (x, t) plane . . . . . . . . . . . 1463.10 Cylindrical invariance . . . . . . . . . . . . . . . . . . . . . . . . . 1493.11 Spherical invariance . . . . . . . . . . . . . . . . . . . . . . . . . . 1493.12 Hugoniot curve for compressible gaz . . . . . . . . . . . . . . . . . 1573.13 Reference solution of the Sod tube test problem . . . . . . . . . . . 159

4.1 Cartesian mesh . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1664.2 Principle of a Lagrange+remap . . . . . . . . . . . . . . . . . . . . 1684.3 Wall boundary condition . . . . . . . . . . . . . . . . . . . . . . . . 1844.4 Sod test problem computed with a Lagrange+remap scheme . . . . 1864.5 Harten test problem computed with a Lagrange+remap scheme . . 1874.6 Lagrangian grid velocity . . . . . . . . . . . . . . . . . . . . . . . . 1924.7 Pure Lagrangian simulation of Sod tube test problem . . . . . . . 1934.8 ALE simulation of Sod tube test problem . . . . . . . . . . . . . . 1944.9 A 2D Sod test problem: density and velocity . . . . . . . . . . . . . 1954.10 A 2D Sod test problem: density and entropy . . . . . . . . . . . . . 1964.11 Structure of the Lagrangian Riemann problem . . . . . . . . . . . 1974.12 One state versus two states solver . . . . . . . . . . . . . . . . . . . 2024.13 Cylindrical Sod tube test problem . . . . . . . . . . . . . . . . . . 2174.14 Spherical Sod tube test problem . . . . . . . . . . . . . . . . . . . 2184.15 MHD on a Cartesian mesh . . . . . . . . . . . . . . . . . . . . . . . 2184.16 AMR and MHD . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2224.17 The swept region . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2244.18 Edge-based flux versus corner-based flux . . . . . . . . . . . . . . . 2254.19 Notations for corner-based flux . . . . . . . . . . . . . . . . . . . . 2264.20 Displacement of a triangular cell . . . . . . . . . . . . . . . . . . . 2284.21 Structure of a tentative nodal flux . . . . . . . . . . . . . . . . . . 2294.22 Principle of a corner-based Riemann problem . . . . . . . . . . . . 2314.23 Delocalization of nodal pressures . . . . . . . . . . . . . . . . . . . 2324.24 Imposed external pressure at the boundary . . . . . . . . . . . . . 2354.25 Sliding boundary condition . . . . . . . . . . . . . . . . . . . . . . 2364.26 Coupling of two different boundary conditions . . . . . . . . . . . . 2374.27 2D Lagrangian numerical Sod tube test problem . . . . . . . . . . 2374.28 More delocalization . . . . . . . . . . . . . . . . . . . . . . . . . . . 2404.29 Saltzmann piston problem . . . . . . . . . . . . . . . . . . . . . . . 2534.30 Two evolutions of a cell . . . . . . . . . . . . . . . . . . . . . . . . 253

List of Figures xv

4.31 Unstable numerical modes for the Noh test problem . . . . . . . . 2544.32 Fall in the air of a drop of a water . . . . . . . . . . . . . . . . . . 2554.33 Free Lagrange technique . . . . . . . . . . . . . . . . . . . . . . . . 2564.34 Multi-material Sod shock tube test problem . . . . . . . . . . . . . 2574.35 Water-air shock tube test problem . . . . . . . . . . . . . . . . . . 2584.36 Sod problem with two γ’s in 2D . . . . . . . . . . . . . . . . . . . . 261

5.1 Corner vectors in 2D . . . . . . . . . . . . . . . . . . . . . . . . . . 2675.2 Barycentric functions . . . . . . . . . . . . . . . . . . . . . . . . . . 2695.3 Hexahedron with warped faces . . . . . . . . . . . . . . . . . . . . 2715.4 A generic convex polygon . . . . . . . . . . . . . . . . . . . . . . . 2735.5 Nodal control volumes . . . . . . . . . . . . . . . . . . . . . . . . . 2755.6 Completion of the corner vectors on a Cartesian mesh . . . . . . . 2825.7 Example of corner vectors degeneracy . . . . . . . . . . . . . . . . 2825.8 Sedov problem in dimension d = 3 . . . . . . . . . . . . . . . . . . 2845.9 Kidder problem in dimension d = 3 . . . . . . . . . . . . . . . . . . 2855.10 Displacement of the mesh in space-time . . . . . . . . . . . . . . . 3005.11 Sub-cell decomposition of a quadrangle . . . . . . . . . . . . . . . . 3035.12 Pathological evolution of a quadrangular cell . . . . . . . . . . . . 3035.13 Volume fractions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3055.14 Stabilization of a 2D Sod test problem . . . . . . . . . . . . . . . . 3135.15 Stabilization of a 2D Sod test problem: zoom . . . . . . . . . . . . 3145.16 Fluid-wall impact . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3185.17 Numerical results for plane wall . . . . . . . . . . . . . . . . . . . . 3205.18 Numerical impact of a fluid on a convex obstacle . . . . . . . . . . 3215.19 Numerical loss of total energy for fluid-wall impact . . . . . . . . . 3225.20 Numerical impact of a fluid on a concave obstacle . . . . . . . . . . 3225.21 Non-uniqueness of the minimum of J . . . . . . . . . . . . . . . . . 3235.22 Non-conformal mesh . . . . . . . . . . . . . . . . . . . . . . . . . . 324

List of Tables

1.1 Typical values of the constant γ. . . . . . . . . . . . . . . . . . . . 81.2 Linear systems. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

3.1 Sod tube test problem: numerical values at contact discontinuity. . 158

4.1 Experimental CFL condition . . . . . . . . . . . . . . . . . . . . . 2104.2 Multi-material Sod tube test problem: numerical values at contact

discontinuity. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2564.3 Water-air shock tube test problem: numerical values at contact dis-

continuity. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 257

5.1 Relative L1 -error and convergence order for a regular mesh. . . . . 3145.2 Relative L1 -error for a random displacement of 30%. . . . . . . . . 3145.3 Relative L1 -error for a random displacement of 50%. . . . . . . . . 315

xvii

Chapter 1

Models

The simplicity of models, compared with reality,lies in the fact that only the relevant propertiesof reality are represented. I’m no model lady. Amodel’s just an imitation of the real thing.

– Mae West

This chapter presents the basic concepts that will be used in the rest of the mono-graph. It starts with balance laws and systems of conservation laws. The examplesand equations are nonlinear. Since they come from continuum mechanics, theyhave to satisfy certain invariance principles, such as Galilean invariance. Then weprove that the structure of conservation laws is preserved through a change ofcoordinates, using the Piola identities. This principle is used to rewrite conserva-tion laws in Eulerian coordinates as conservation laws in Lagrangian coordinates.Eulerian coordinates are traditionally defined as the usual space coordinates, thatis, the coordinates of an exterior observer; in contrast, Lagrangian coordinates arethose attached to the flow. The notion of a linearly well-posed system of a hyper-bolic system and of a weakly hyperbolic system of conservation laws are introducedand used to discuss some differences between Eulerian and Lagrangian conserva-tion laws. A fundamental result is that Lagrangian compressible gas dynamics isweakly hyperbolic in dimensions two and higher.

1.1 Balance law

For simplicity we consider firstly the one-dimensional case of d = 1. A quantity ofinterest is denoted by ρ(t, x) ∈ R . It is a function of the time variable t ∈ R+ andthe space variable x ∈ R . The integral of ρ between two given points x0 ∈ R andx1 ∈ R is

N (t, x0, x1) =

∫ x1

x0

ρ(t, x)dx, x0 < x1.

The rate of change with respect to time is given by

© Springer International Publishing AG 2017

B. Després, Numerical Methods for Eulerian and Lagrangian Conservation Laws,

Frontiers in Mathematics, DOI 10.1007/978-3-319-50355-4_1

1

2 Chapter 1. Models

xx0 x1

f (t, x0) f (t, x1)

N (t, x0, x1)

Figure 1.1: Time variation of N (t, x0, x1).

d

dtN (t, x0, x1) =

∫ x1

x0

∂tρ(t, x)dx. (1.1)

Introducing a time dependence so that x0(t) and x1(t) can move, a more generalformula for the variation of N (x0(t), x1(t), t) is

d

dtN (t, x0(t), x1(t)) =

∫ x1(t)

x0(t)

∂tρ(t, x) dx+x′1(t)ρ(t, x1(t))−x′

0(t)ρ(t, x0 (t)). (1.2)

We make the hypothesis that the gain and loss can only be through the endpointsof the interval [x0, x1]. Moreover we consider fixed endpoints, i.e.x′

0(t) = x′1(t) = 0.

One can write an additional balance equation during the time interval ∆t > 0,

N (t + ∆t, x0, x1) = N (t, x0, x1) − f(t, x1)∆t + f(t, x0)∆t + o(∆t)

where f(t, x0) and f(t, x1) represent the gains or losses (depending on the sign)at the endpoints. Passing formally to the limit ∆t → 0+, one obtains

d

dtN (t, x0, x1)+ f(t, x1)− f(t, x0) =

d

dtN (t, x0, x1)+

∫ x1

x0

∂x f(t, x)dx = 0. (1.3)

Next, combine (1.1) and (1.3) to obtain∫ x1

x0

∂tρ(t, x)dx +

∫ x1

x0

∂xf(t, x) dx = 0.

Since this integral identity holds for all x0 < x1, one obtains the equivalent differ-ential identity

∂t ρ(t, x) + ∂x f(t, x) = 0. (1.4)

This relation is called a conservation law. Even if the formula is symmetric withrespect to the time and space variables, one generally treats the variables as notplaying the same roles. As a consequence, the quantity ρ under the differentialtime operator will be called the unknown. The other quantity f is the flux.

The previous method is very general and can be extended immediately to anyspatial dimension. For example, one has the following conservation law in threedimensions, (x, y, z) ∈ R 3 :

∂tρ(t, x, y, z) + ∂xf(t, x, y, z) + ∂yg(t, x, y, z) + ∂z h(t, x, y, z) = 0.

1.1. Balance law 3

In the aboveequation the number of equations is one, while the number of unknownquantities ρ, f, g, h is four. So it remains to specify the fluxes f , g and h as functionsof the unknown ρ to obtain a closed equation.

1.1.1 Traffic flow

Let us consider traffic flow for which the main unknown is the density ρ(t, x) ofvehicles along an infinite road x ∈ R . The number of vehicles between x0 and x1

is by definition

N (t, x0, x1) =

∫ x1

x0

ρ(t, x)dx, x0 < x1.

One observes that N (t, x0 , x1) is in general a real number, as is usual in suchmodels. Let us denote the velocity of the vehicles by u(t, x). What was referred toas the gain or loss is f = ρu. One obtains the conservation law

∂tρ + ∂xρu = 0. (1.5)

It is standard to make a modeling hypothesis: we assume that a reasonable driveradapts the speed of the vehicle according to the density of vehicles around his ownvehicle. More precisely, the denser the surrounding traffic, the slower the speed. Incontrast, one tends to drive fast if the local traffic density is low. The consequenceis that the velocity u is determined as a function of the density ρ. One thus obtainsthe equation ∂tρ+∂x f(ρ) = 0 with a flux f(ρ) = ρu(ρ). The so-called LWR (whichstands for Lighthill-Whitham-Richards [133]) model corresponds to

ρ → u(ρ) ≡ umax

(1 − ρ

ρmax

),

where the constants umax and ρmax can be determined by basic considerations.Typically the maximal velocity on highways is umax = 130 km/h in Europe, andthe maximal density can be estimated in terms of the mean length l of vehicles:ρmax = 1/l. The LWR law is illustrated in figure 1.2. The conservation law fortraffic flow takes the form

∂tρ + ∂xf(ρ) = 0, f(ρ) = ρu(ρ) = umax

(ρ − ρ2

ρmax

). (1.6)

The non-dimensional version of the equation is obtained for umax = 1 and ρmax = 1as

∂tρ + ∂x (ρ − ρ2) = 0.

Let us define a new unknown v = 12

−ρ, which satisfies the equation ∂tv+∂xv2 = 0.One more transformation of the time variable, t → 2t, yields the Burgers equation

∂tv + ∂xv2

2= 0. (1.7)

4 Chapter 1. Models

ρmaxumax

2

ρmax

2

f (ρ)

ρ

ρmax

u(ρ)

ρmax

ρ

Figure 1.2: The LWR law ρ → f(ρ) = ρu(ρ) for traffic flow.

The Burgers equation is nonlinear. This means that if v1 and v2 are two solutionsof the Burgers equation, then the function v3 = v1 + v2 is a priori not a solutionof the Burgers equation. That is, the superposition principle does not hold fornonlinear equations.

Nevertheless, the Burgers equation is scale invariant. Let v be a solution ofthe Burgers equation and let λ ∈ R . Then the function w = λv is a solution of

∂s w + ∂xw2

2= 0 after rescaling of the time variable by s = λt.

1.1.2 Shallow water

We show in this subsection how to derive the shallow water equations, also knownas the Saint Venant equations, starting from minimal assumptions.

Consider the two-dimensional cross-section of a lake or river represented infigure 1.3. The two components of the velocity of the fluid are written as u =(u, v); the first component is the horizontal velocity and the second is the verticalvelocity. Water is a priori considered an incompressible fluid, that is, the densityis constant: ρ = ρc > 0. The incompressibility condition on the velocity field reads∂x u + ∂yu = 0. Let us denote by h(t, x) the height of a water column. The balancelaw technique yields

N (t, x0, x1) = ρc

∫ x1

x0

h(t, x)dx.

The time variation of N (t, x0 , x1) due to the boundary fluxes is given by (1.3),

d

dtN (t, x0, x1) + f(t, x1) − f(t, x0) = 0.

1.1. Balance law 5

y

h(t, x)

x0 x1 x

Figure 1.3: A column of water between x0 and x1.

The left and right fluxes are naturally

f(t, x0) = ρc

∫ h(t,x0 )

0

u(t, x0 , y)dy and f(t, x1) = ρc

∫ h(t,x1 )

0

u(t, x1 , y)dy.

After dividing through by ρc,

d

dt

∫ x1

x0

h(t, x)dx +

∫ h(t,x1 )

0u(t, x1 , y)dy −

∫ h(t,x0 )

0u(t, x0 , y)dy = 0.

It is convenient to define the mean horizontal velocity of the column of water as

u(t, x) =

∫ h(t,x)0

u(t, x, y)dy

h(t, x),

so that one can write

d

dt

∫ x1

x0

h(t, x)dx + h(t, x1)u(t, x1) − h(t, x0)u(t, x0) = 0.

Since this holds for all x0 < x1, it yields a first conservation law

∂th + ∂x(hu) = 0. (1.8)

This conservation law is very similar to the traffic flow equation. The differenceis that there is no reason for the mean horizontal velocity to be a function of theheight h. This means that we must derive at least one more equation in order toobtain a closed system.

It is known that this additional equation exists. It is also a conservation law,where the main unknown is u. To construct it, one can use the formula (1.2) for

6 Chapter 1. Models

the time variation of a mass between moving boundaries in the form

N (t, x0(t), x1(t)) = ρc

∫ x1(t)

x0(t)h(t, x)dx,

where the left boundary is defined by x0(0) = X0 and x′0(t) = u(t, x0(t). Similarly,

the right boundary is defined by x1(0) = X1 and x′1(t) = u(t, x1(t)). Therefore

d

dtN (t,x0(t), x1(t))

= ρc

(∫ x1

x0

∂th(t, x) dx + x′1(t)h(t, x1(t)) − x′

0(t)h(t, x0(t))

)

= ρc

(∫ x1

x0

∂th(t, x) dx + u(t, x1(t))h(t, x1(t)) − u(t, x0(t))h(t, x0(t))

)

= ρc

∫ x1

x0

(∂t h(t, x) + ∂x (h(t, x)u(t, x))) dx = 0,

which shows that the mass is constant in the moving interval. It evokes the clas-sical analogy where the water column is like an individual particle with constantmass N (t, x0(t), x1(t)), to which Newton’s law applies. The horizontal inertial mo-mentum of the water column is

I(t, x0(t), x1 (t)) = ρc

∫ x1 (t)

x0 (t)hu dx.

The sum of forces on the boundaries is the right-hand side of Newton’s law

d

dtI(t, x0(t), x1(t)) = F (t, x1(t)) + F (t, x0(t)). (1.9)

The force can be interpreted as the integral over the vertical line of the hydrostaticpressure. Taking account of the sign, this yields

F (t, x0) =

∫ h(t,x0 )

0

p(t, x0 , y)dy and F (t, x1) = −∫ h(t,x1 )

0

p(t, x1 , y)dy.

The hydrostatic pressure at altitude y is proportional to the height of water abovey, so

p(t, x, y) = ρc

∫ h(t,x)

yg dy = ρcg(h(t, x) − y)

where g is the local gravitational constant. So the forces are

F (t, x0) =g ρc

2h2(t, x0), F (t, x1) = − g ρc

2h2(t, x1).

1.1. Balance law 7

x1x0

F1F0

p = 0

The total pressure integratedover the vertical line is

p = g

2ρch2

Figure 1.4: Sketch of the left force F0 and the right force F1 on the boundaries ofthe water column.

Plugging these into (1.9), we obtain

d

dt

∫ x1 (t)

x0 (t)

hu dx +1

ρc

∫ x1 (t)

x0 (t)

∂xF dx =d

dt

∫ x1(t)

x0 (t)

hu dx +g

2

∫ x1 (t)

x0(t)

∂x h2 dx = 0.

The formula (1.2) and the definitions of the velocities x′0(t) and x′

1(t) show that

d

dt

∫ x1 (t)

x0(t)hu dx =

∫ x1 (t)

x0 (t)∂t(hu)dx + (hu2)(t, x1(t)) − (hu2)(t, x0(t))

=

∫ x1 (t)

x0 (t)

[∂t (hu) + ∂x(hu2)

]dx.

So ∫ x1 (t)

x0 (t)

∂t(hu)dx +

∫ x1 (t)

x0(t)

∂x

(hu2 +

g

2h2)

dx = 0.

Since x0(t) and x1(t) are arbitrary, this yields a second conservation law

∂t(hu) + ∂x

(hu2 +

g

2h2

)= 0. (1.10)

Finally, we obtain the shallow water system defined by equations (1.8) and (1.10):

⎧⎨⎩

∂th + ∂x(hu) = 0,

∂t(hu) + ∂x

(hu2 +

g

2h2)

= 0, g > 0.(1.11)

This is a closed system of two equations with two unknowns.

8 Chapter 1. Models

1.1.3 Compressible gas dynamics

One can construct the system of compressible non-viscous gas dynamics usingthe same method as for the traffic flow and shallow water equations. However,an additional hypothesis of a thermodynamical nature is needed: it relates themacroscopic coefficient of the pressure law to its microscopic features. We willgive at the end of this chapter an indirect validation of this property. It must besaid that thermodynamics is understood hereafter in an extremely crude way. Thereader interested in the physical foundations of thermodynamics can find moredetails in classical physics textbooks such as [17, 33] or in many mathematicaltextbooks at the intersection of mathematics and continuum mechanics; see [57,188].

The thermodynamic assumption is that the pressure of a gas can be writtenas a function of two independent parameters, which are the density ρ and thetemperature T ; that is,

p = p(ρ, T ).

In this notation, the value of the pressure is confounded with the function itself.This will be useful for further manipulation. We also assume that the temperaturecan be computed in terms of the density and another thermodynamical quantityreferred to as the internal energy ε. Denote the velocity of the gas by u. Thetotal energy per unit mass is the sum of the internal (or potential) energy and thekinetic energy e = ε+ 1

2|u|2. A standard pressure law is that of a perfect polytropic

gas:

p = (γ − 1)ρε, ε = CvT, Cv > 0, γ > 1. (1.12)

Nature of the gas O2, N2 Air H2 He, Kr, Xe Ar CO2 SF6

γ 1.4 1.4 1.405 1.66 1.67 1.3 1.09

Table 1.1: Typical values of the constant γ.

Many other pressure laws exist. We mention just a few of them. For example,the stiffened gas pressure law is

p = (γ − 1)ρε − γΠ. (1.13)

Water is not a gas of course, but it can be modeled quite accurately by takingγ = 5.5 and Π = 4921.15 bars. Another example is the van der Waals pressure law

p =aε

τ − b− c

τ 2, a, b, c > 0, τ =

1

ρ, (1.14)

where the variable τ denotes the specific volume. The van der Waals law is usedfor phase transitions.

1.1. Balance law 9

Whatever the pressure law, the equations can be constructed with the bal-ance law method. For simplicity we consider the one-dimensional configurationdepicted in figure 1.5. An elementary (infinitesimal) quantity of gas is containedin the interval [x0(t), x1(t)]. The points can move, that is, x′(t, X) = u(t, x(t, X)),x(0, X) = X.

x1(t + ∆t)x0(t + ∆t)

t

t + ∆t

x0(t) x1(t)

Figure 1.5: Elementary quantity of gas in a moving interval.

At any time, the mass in the interval is

N (x0(t), x1(t), t) =

∫ x1(t)

x0(t)

ρ(t, x)dx.

The total impulse is

I(x0(t), x1(t), t) =

∫ x1 (t)

x0 (t)ρ(t, x)u(t, x) dx.

The mechanical forces on the edges of the moving interval are f = p on x0(t)and f = −p on x1(t). Using again the method described above, we obtain twoequations

∂tρ + ∂x(ρu) = 0,

∂t(ρu) + ∂x

(ρu2 + p

)= 0.

However, this system is not closed. In order to derive an additional equation, weconsider the total energy in the moving interval,

E(x0(t), x1(t), t) =

∫ x1(t)

x0(t)ρ(t, x)e(t, x) dx,

where the total energy per unit mass is the sum of the internal energy and thekinetic energy e = ε + 1

2u2. Energy considerations can be used to determine the

work exerted by the forces. Consider a small time interval ∆t. By definition thework w is the product of the force and the length of the interval on which it acts,i.e. w = ±pu∆t. One obtains

E(x0(t + ∆t), x1(t + ∆t), t + ∆t) = E(x0(t), x1(t), t) − ∆tp(t, x1(t))u(t, x1(t))

+ ∆tp(t, x0(t))u(t, x0(t)) + o(∆t).

10 Chapter 1. Models

Passing formally to the limit ∆t → 0, one gets

d

dtE(x0(t), x1(t), t) + p(t, x1(t))u(t, x1 (t)) − p(t, x0(t))u(t, x0(t)) = 0.

Combining this with formula (1.2) gives

∫ x1 (t)

x0 (t)

∂t(ρe) dx +

∫ x1(t)

x0(t)

∂x (ρue + pu) = 0.

Since the above is true for any pair (x0(t), x1(t)), it yields a new conservation law

∂t(ρe) + ∂x(ρue + pu) = 0.

One finally obtains the system of compressible non-viscous gas dynamics, alsoknown as the system of Euler equations:

⎧⎪⎨⎪⎩

∂tρ + ∂x (ρu) = 0,

∂t(ρu) + ∂x

(ρu2 + p

)= 0,

∂t(ρe) + ∂x(ρue + pu) = 0.

(1.15)

This system is closed since the pressure p can be calculated as a function of ρ andε = e − 1

2u2.

In two dimensions, one obtains by tensorization

⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩

∂tρ + ∂x(ρu) + ∂y(ρv) = 0,

∂t(ρu) + ∂x

(ρu2 + p

)+ ∂y (ρuv) = 0,

∂t(ρv) + ∂x (ρuv) + ∂y

(ρv2 + p

)= 0,

∂t(ρe) + ∂x(ρue + pu) + ∂y(ρve + pv) = 0.

(1.16)

The difference is mostly in the velocity field, which is a vector u = (u, v) withtwo components. The pressure p is a function of the density ρ and internal energyε = e − 1

2(u2 + v2). In three dimensions, one readily obtains

⎧⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎩

∂tρ + ∂x(ρu) + ∂y(ρv) + ∂z (ρw) = 0,

∂t(ρu) + ∂x

(ρu2 + p

)+ ∂y (ρuv) + ∂z (ρuw) = 0,

∂t(ρv) + ∂x (ρuv) + ∂y

(ρv2 + p

)+ ∂z (ρvw) = 0,

∂t(ρw) + ∂x (ρuw) + ∂y (ρvw) + ∂z

(ρw2 + p

)= 0,

∂t(ρe) + ∂x (ρeu + pu) + ∂y(ρev + pv) + ∂y(ρew + pw) = 0,

(1.17)

where the vectorial velocity is u = (u, v, w) and the pressure p is a function of thedensity ρ and internal energy ε = e − 1

2(u2 + v2 + w2).

1.2. Lagrangian coordinates 11

1.1.4 Canonical form of a system of conservation laws

Let us write the spatial coordinates as x = (x1, . . . , xd) ∈ Rd. All the previoussystems of conservation laws can be written in the canonical form

∂tU(x, t) + ∇ · f(U(x, t)) = 0

where U : Rd ×R → Rn is the unknown. The flux f : Rd → Rn×d is matrix-valued.Its divergence is

∇ · f(U) =

⎛⎜⎜⎝

∑dj =1 ∂xj f1j(U)

...∑dj =1 ∂xj fnj(U)

⎞⎟⎟⎠ ∈ R

n.

For example, in d = 2 and n = 4 the system (1.16) corresponds to

U =

⎛⎜⎜⎜⎝

ρ

ρu

ρv

ρe

⎞⎟⎟⎟⎠ ∈ R

4 and f(U) =

⎛⎜⎜⎜⎝

ρu ρv

ρu2 + p ρuv

ρuv ρv2 + p

ρue + pu ρve + pv

⎞⎟⎟⎟⎠ ∈ R

4×2.

1.2 Lagrangian coordinates

The traffic flow model, the shallow water equations and the Euler system arewritten in so-called Eulerian coordinates. Eulerian coordinates correspond to thecoordinates of a fixed observer.

In contrast, Lagrangian coordinates are, in some sense, attached to the localflow velocity. Local means that the change of coordinates is different from one pointto another because the velocity takes different values in different parts of the fluid.Therefore Lagrangian coordinates can be identified with the Eulerian coordinatesat another time, which is usually and arbitrarily taken to be the initial time, i.e.tini = 0. In dimension d = 1 the coordinates take the form

x(t = 0, X) = X and ∂tx(t, X) = u(x(t, X), t).

Our goal is to use this transformation to rewrite the original Eulerian equationsin Lagrangian coordinates. It will give us some insights into the structure of theequations.

However, the algebra associated with the Euler-to-Lagrange transformationis not completely evident, as stressed in [94, 93, 175]. We will use the so-calledPiola identities to detail this transformation. The following presentation is close tothat in [57], where the geometrical intuition or perspective is emphasized. Otherapproaches are possible, such as the ones in [192, 182] or [174]. The Piola identitiesare also called the geometric conservation laws.


1.2.1 General change of coordinates in balance laws

Consider a transformation

x → ϕ(x) = X = (X1 , . . . , Xd) ∈ Rd (1.18)

from Rd to Rd. This transformation is smooth with continuous derivatives, typi-cally C1. The Jacobian matrix of the transformation is

∇ϕ =

(∂Xi

∂xj

)

1≤i,j≤d

=

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

∂X1

∂x1

∂X1

∂x2. . .

∂X1

∂xd∂X2

∂x1

∂X2

∂x2. . .

∂X2

∂xd

......

.. ....

∂Xd

∂x1

∂Xd

∂x2. . .

∂Xd

∂xd

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

.

Assuming this matrix is non-singular, det(∇ϕ) = 0, the transformation is locallyinvertible. The inverse transformation is

X → ψ(X) = x = (x1 , . . . , xd) ∈ Rd ,

so thatψ(ϕ(x)) = x and ϕ(ψ(X)) = X.

Differentiation shows that the Jacobian matrices are inverse to one another:

∇ψ(ϕ(x))∇ϕ(x) = Id and ∇ϕ(ψ(X))∇ψ(X) = Id .

This relation is often written in the more compact form ∇ψ = (∇ϕ)−1

. To gofurther we need the comatrix.

Definition 1.2.1 (Comatrix). The comatrix com (M) ∈ Rd×d of a matrix M ∈ Rd×d

is the matrix of cofactors. The coefficient in position (i, j) in com(M) is equal to(−1)i+j times the determinant of the (d − 1) × (d − 1) submatrix obtained afterelimination of column j and line i.

Therefore M tcom (M) = det(M)Id for all M ∈ Rd×d. If M is non-singular,

then com(M) = det(M)× (M t )−1. Transposition of the comatrix yields the adju-gate of M : adj (M) = com (M)

t. The main result about changes of coordinates in

systems of balance laws can be now formulated using the comatrix of the inversetransformation com(∇ψ) and the determinant of the Jacobian of the transforma-tion,

J = det(∇ψ).

Theorem 1.2.2. The system of balance laws with a source S : Rd → Rn,

[∇ · f(U)] (x) = S(x), x ∈ Rd, (1.19)


is equivalent to the system of balance laws in the new frame X = ψ−1(x) ∈ Rd,

[∇ · (f(U(ψ)) com(∇ψ)) ] (X) = [JS(ψ)] (X), X ∈ Rd. (1.20)

Moreover, one has the Piola identity

[∇ · com(∇ψ)] (X) = 0, X ∈ Rd. (1.21)

Remark 1.2.3. The divergence is taken with respect to x in (1.19) and with respectto X in (1.20). Note that the dimension of the product of matrices is correct, thatis, f(U(ψ)) com(∇ψ) ∈ Rn×d.

Proof. Let Ω ⊂ Rd be any smooth open subset of Rd. Its boundary is denoted by∂Ω. The outward exterior normal vector on the boundary is denoted by n ∈ R

d.The boundary measure is dσ. One has the Stokes formula

∫

x∈Ω

∇ · f(U(x)) dx =

∫

x∈∂Ω

f(U)(x)n dσ.

After integration over Ω the system of balance laws (1.19) is formally equivalentto ∫

x∈∂Ω

f(U(x))n dσ =

∫

x∈Ω

S(x) dx ∀Ω ⊂ Rd ,

which is now written for all smooth open subsets Ω ⊂ Rd. Upon performing thechange of coordinates x = ψ(X) and using the Nanson formula (1.22) for pointsx ∈ ∂Ω and X = ϕ(x) ∈ ∂ϕ(Ω), one gets

∫

X∈∂ϕ(Ω)

f(U(ψ(X))) com(∇2ψ)(X)nX dσX =

∫

X∈ϕ(Ω)

S(ψ(X))J(X) dX.

A change of notation θ = ϕ(Ω) ⊂ Rd shows that∫

X∈∂θ

f(U(ψ(X)))com(∇ψ)(X)nX dσX =

∫

X∈θ

S(ψ(X))J(X) dX, θ ⊂ Rd.

Rewriting the left-hand side with the Stokes formula gives∫

X∈θ

∇ ·[f(U(ψ(X))) com(∇2ψ)(X)

]dX =

∫

X∈θ

S(ψ(X))J(X)dX, θ ⊂ Rd.

Since this identity holds for any smooth θ, it yields the first part of the claim,(1.20). The Piola identity (1.21) is obtained by taking f = Id ∈ Rd×d and avanishing source S = 0 in (1.20). The proof is thus complete.

One can see that the proof of the theorem relies heavily on Nanson’s formula(1.22), which is of a purely geometrical nature since it expresses the effect ondifferential elements of a general transformation of the space coordinates. This isa well-established formula [63] in continuum mechanics.


Proposition 1.2.4 (Nanson’s formula). The fol lowing identity holds between differ-ential elements:

n dσ = com(∇ψ)nX dσX . (1.22)

Proof. The proof relies on two well-known formulas. The first formula is a multi-dimensional generalization of (1.2). Let V (t) ⊂ Rn be a moving smooth boundeddomain in dimension d. Assume that the points on the boundary x ∈ ∂V movewith velocity u(x). One has

d

dt

∫

x∈V (t)

h(x, t)dx =

∫

x∈V (t)

∂th(x, t)dx +

∫

x∈∂V

h(x, t)(u(x), n(x)) dσ. (1.23)

The second formula is the formula for change of coordinates in integrals

|V | =

∫

x∈V

dx =

∫

X∈ψ(V )

det(∇ψ)dX. (1.24)

Hence the rate of change of the volume of a moving domain computed in thereference frame x with formula (1.23) (taking h = 1) is

d

dt|V (t)| =

∫

x∈∂V (t)

(u(x), n(x))dσ.

It can also be computed in the reference frame X as

d

dt|V (t)| =

d

dt

∫

X∈ϕ(V (t))

det(∇ψ)dX =

∫

X∈∂ϕ(V (t))

det(∇ψ)(uX (X), nX )dσX .

In this expression uX (X) is the velocity (measured in frame X) of points X ∈∂ϕ(V (t)). The chain rule yields

uX(ϕ(x)) = ∇ϕ(x)u(x), x ∈ ∂V (t).

So∫

x∈∂V (t)

(u(x), n(x))dσ =

∫

X∈∂ϕ(V (t))

det(∇ψ)(X)(

u(x),(∇ϕ(X)t nX(X)

))dσX

The velocity field is arbitrary. Therefore

n(x) dσ = det(∇ψ)(X)(∇ψ(X)t nX(X)

)dσX

where x ∈ ∂V (t) and X = ϕ(x) ∈ ∂ϕ(V (t)). By definition of the comatrix,

det(∇ψ)(X)∇ϕ(X)t = det(∇ψ)(X)∇ψ(X)−t = com(∇ψ).

This completes the proof.


1.2.2 Lagrangian gas dynamics in dimension d = 1

Let us rewrite the system (1.15) as a time-space divergence

∇tx · (U, f(U )) = 0 ∈ Rn .

Since U = (ρ, ρu, ρe)t and f(U) = (ρu, ρu2 + p, ρue + pu)t, one gets

∇tx ·

⎛⎜⎝

ρ ρu

ρu ρu2 + p

ρe ρeu + pu

⎞⎟⎠ = 0.

Consider the Euler-to-Lagrange change of space-time coordinates from (t, x) to(t′ , X), where

t′ = t and∂x(t′ , X)

∂t′ = u(t′, x(t′, X)) with x(0, X) = X.

This transformation is regular if u is a smooth enough function. We apply thetransformation (1.20) to the space-time free-divergence equation. To do so onemust determine the comatrix of the transformation. We have

∇(t′,X)(t, x) =

(1 0

u J

), J =

∂x

∂X.

So com(∇(t′,X)(t, x)

)=

(J −u

0 1

). Therefore equation (1.20) becomes

∇t′,X ·[

(U, f(U))

(J −u

0 1

)]= 0,

that is, ⎧⎪⎨⎪⎩

∂t′ (ρJ) = 0,

∂t′ (ρuJ) + ∂X p = 0,

∂t′ (ρeJ) + ∂X (pu) = 0.

This system is not closed since the determinant J of the Jacobian matrix appears.It shows the necessity of the Piola transform, which is used to close the system.Using (1.21), the Piola identity is just one equation, ∂t′ J − ∂X u = 0, which isalso obtained by differentiation of ∂t′ x(t′, X) = u with respect to X. We cannow replace t′ with the usual notation t for time. This yields a system of fourconservation laws ⎧

⎪⎪⎪⎨⎪⎪⎪⎩

∂t(ρJ) = 0,

∂t(ρuJ) + ∂X p = 0,

∂t(ρeJ) + ∂X(pu) = 0,

∂tJ − ∂X u = 0.

(1.25)

It is convenient to define the mass variable.


Theorem 1.2.5 (Lagrangian formulation with the mass variable in 1D). Considerthe Euler system (1.25) for compressible gas dynamics. Assume that the densityis positive, ρ > 0. Define the specific volume τ = 1

ρand the mass variable dm =

ρ(0, X)dX. Then the Euler system can be written in Lagrangian coordinates inthe form ⎧

⎪⎨⎪⎩

∂t τ − ∂mu = 0,

∂t u + ∂mp = 0,

∂t e + ∂m(pu) = 0.

(1.26)

Proof. The mass variable is just one more change of coordinates,

m(X) =

∫ X

0ρ(0, y) dy,

which can also be recast as dm = ρ0 dX. The first equation in (1.25) yields(ρJ)(t, X) = J(0, X)ρ(0, X). By definition, J(0, X) = ∂X

∂X= 1. So (ρJ)(t, X) =

ρ(0, X), which does not depend on the time t. One obtains for example that

∂t J − ∂Xu = 0 =⇒ ρ0∂tτ − ∂X u = 0 =⇒ ∂t τ − 1

ρ0∂X u = 0 =⇒ ∂tτ − ∂mu = 0.

The algebra is similar for the two last equations. One has

∂t (ρuJ)+∂X p = 0 =⇒ ρ0∂t u+∂X p = 0 =⇒ ∂tu+1

ρ0∂X p = 0 =⇒ ∂tu+∂mp = 0

and

∂t(ρeJ) + ∂X (pu) = 0 =⇒ ρ0∂t e + ∂X (pu) = 0

=⇒ ∂te +1

ρ0∂X (pu) = 0

=⇒ ∂te + ∂m(pu) = 0.

The proof is therefore complete.

Proposition 1.2.6. The differential operators ∂t and ∂m in the Lagrangian system(1.26) are Galilean invariant.

Remark 1.2.7. It is easier to understand the property using a notation standardin mechanical sciences. We let ∂a|b denote partial differentiation with respect tothe variable a with a frozen variable b. This enables us to make a clear distinctionbetween ∂t|x where the frozen variable is x and ∂t|m where the frozen variable isX or m. So the proposition means more precisely that ∂t|m and ∂m|t are Galileaninvariant.


Proof. One needs to refer to the chain formula (1.40) for the Galilean transforma-tion (1.39). One has

⎧⎨⎩

∂t|m = ∂t|x + u∂x|t = ∂t′ |x′ + (u + v)∂x′ |t′ = ∂t′|x′ + u′∂x′ |t′ = ∂t′|m′ ,

∂m|t =1

ρ∂x|t =

1

ρ∂x′|t′ = ∂m′|t′ .

So ∂t = ∂t′ and ∂m = ∂m′ .

Other notation for the time derivative with respect to the mass variableincludes

∂t|m = ∂t + u∂x =d

dt= Dt.


Start from (1.16). Define

A = ∂X x, B = ∂X y, L = ∂Y x, M = ∂Y y. (1.27)

The Jacobian matrix of the space-time transformation is

∂(t, x, y)

∂(t, X, Y )=

⎛⎜⎝

1 0 0

u A L

v B M

⎞⎟⎠ ,

so

com

(∂(t, x, y)

∂(t, X, Y )

)=

⎛⎜⎝

J −uM + vL uB − vA

0 M −B

0 −L A

⎞⎟⎠ , J = AM − BL.

The matrix–matrix product in (1.20) is

⎛⎜⎜⎜⎝

ρ ρu ρv

ρu ρu2 + p ρuv

ρv ρuv ρv2 + p

ρe ρue + pu ρve + pv

⎞⎟⎟⎟⎠

⎛⎜⎝

J −uM + vL uB − vA

0 M −B

0 −L A

⎞⎟⎠

=

⎛⎜⎜⎜⎝

ρJ 0 0

ρuJ pM −pB

ρvJ −pL pA

ρeJ puM − pvL −puB + pvA

⎞⎟⎟⎟⎠ .


One obtains⎧⎪⎪⎪⎨⎪⎪⎪⎩

∂t(ρJ) = 0,

∂t(ρJu) + ∂X (pM) + ∂Y (−pB) = 0,

∂t(ρJv) + ∂X (−pL) + ∂Y (pA) = 0,

∂t(ρJe) + ∂X (puM − pvL) + ∂Y (pvA − puB) = 0.

(1.28)

The Piola identities are⎧⎪⎨⎪⎩

∂t J − ∂X (uM − vL) − ∂Y (vA − uB) = 0,

∂X M − ∂Y B = 0,

− ∂X L + ∂Y A = 0.

(1.29)

The two last compatibility relations are evident, but this is not the case for thefirst one. We add the definition of the Lagrange-to-Euler transformation

∂tx(t, X, Y ) = u, x(0, X, Y ) = X,

∂ty(t, X, Y ) = v, y(0, X, Y ) = Y.(1.30)

Next, we ask ourselves if a basic notion of mass variable is possible in dimensiond = 2 (and higher). That is, we look for two other variables, denoted by α and β,such that

∂X = ρ0∂α and ∂Y = ρ0∂β . (1.31)

If such a transformation exists, it would be possible to simplify the equationsin system (1.28). For example, the second equation could be written as ∂tu +∂α (pM) + ∂β (−pB) = 0, and so on. But the answer to this question is negative inthe general case.

Proposition 1.2.8. The only differentiable solutions to (1.31) are the trivial ones.

Proof. The chain rule yields

⎧⎪⎨⎪⎩

∂X =∂α

∂X∂α +

∂β

∂X∂β,

∂Y =∂α

∂Y∂α +

∂β

∂Y∂β ,

=⇒

⎧⎪⎨⎪⎩

∂α

∂X∂α +

∂β

∂X∂β = ρ0∂α,

∂α

∂Y∂α +

∂β

∂Y∂β = ρ0∂β .

Since it must be true for all (α, β) at least locally in a subset of R 2 , one gets

∂α

∂X= ρ0 ,

∂α

∂Y=

∂β

∂X= 0 and

∂β

∂Y= ρ0 .

So

∂Y ρ0 = ∂Y

(∂α

∂X

)= ∂X

(∂α

∂Y

)= 0.

Similarly, we can show that ∂X ρ0 = 0. Therefore ρ0(X, Y ) = ρ is a constant. Thisis a trivial solution.


In very restricted situations such as in the purely 1D case, mass variablesolutions may nevertheless exist. Consider for example

∂X = ρ0∂α and ∂Y = 0.

The condition ∂Y = 0 indicates that only one-dimensional solutions of (1.28)–(1.30) are considered, typically ∂Y ρ0 = 0. In this case, one can of course use theone-dimensional mass variable defined above.

1.2.4 Hui’s formulation

The Hui’s formulation [112, 111, 108, 106, 113] of Lagrangian gas dynamics isslightly different from (1.28)–(1.30) and has been studied in [138] and [72]. Its ob-jective is to provide a closed system of conservation laws convenient for numericaldiscretization. Indeed, one can notice that A, B , L and M constitute the gradient(1.27) of the transformation. But by combining (1.27) and (1.30) one gets with-out difficulty the evolution equations ∂t A = ∂X u, ∂tB = ∂X v, ∂tL = ∂Y u and∂t M = ∂Y v. This yields Hui’s formulation

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

∂t (ρJ) = 0, J = AM − BL,

∂t (ρJu) + ∂X(pM) + ∂Y (−pB) = 0,

∂t (ρJv) + ∂X (−pL) + ∂Y (pA) = 0,

∂t (ρJe) + ∂X (puM − pvL) + ∂Y (pvA − puB) = 0,

∂t A = ∂Xu,

∂t B = ∂X v,

∂t L = ∂Y u,

∂t M = ∂Y v.

(1.32)

This system is closed. It will be used to discuss the weak hyperbolicity of La-grangian gas dynamics.


The system of Lagrangian gas dynamics in dimension d = 3 exhibits greater com-plexity, essentially because the comatrix becomes quadratic with respect to gradi-ent of the deformation. This makes interpretation of the equations less straightfor-ward and is probably the reason why it is tricky to use them directly for the designof numerical methods. The reader interested primarily in a practical perspective onmultidimensional Lagrangian methods is advised to compare the equations belowwith the methods developed in the final chapter.

The deformation gradient is

J=

⎛⎜⎝

∂X x ∂Y x ∂Zx

∂X y ∂Y y ∂Zy

∂X z ∂Y z ∂Zz

⎞⎟⎠ =

⎛⎜⎝

A L P

B M Q

C N R

⎞⎟⎠ .


Define J = det(J) and

com(J) =

⎛⎜⎝

MR − NQ −BR + CQ BN − CM

−LR + NP AR − CP −AN + CL

LQ − MP −AQ + BP AM − BM

⎞⎟⎠ .

In natural compact notation, the gradient of the space-time transformation is

∂(t, x, y, z)

∂(t, X, Y, Z)=

(1 0

u J

), u =

⎛⎜⎝

u

v

w

⎞⎟⎠ . (1.33)

So

com

(∂(t, x, y, z)

∂(t, X, Y , Z)

)=

(J −utcom(J)

0 com(J)

). (1.34)

The calculation of (1.20) reduces to

⎛⎜⎝

ρ ρut

ρu ρu ⊗ u + pId

ρe ρeut + put

⎞⎟⎠(

J −utcom(J)

0 com(J)

)=

⎛⎜⎝

ρJ 0

ρuJ p com(J)

ρeJ putcom(J)

⎞⎟⎠ .

(1.35)Therefore the Lagrangian system of gas dynamics in dimension d = 3 is made upof five physical conservation laws,

∇t,X ·

⎛⎜⎝

ρJ 0

ρuJ p com(J)

ρeJ putcom(J)

⎞⎟⎠ = 0, (1.36)

together with four scalar Piola identities

∇t,X.

(J −utcom(J)

0 com(J)

)= 0 (1.37)

and the definition of the Lagrange-to-Euler transformation

∂tx(t, X) = u, x(0, X) = X. (1.38)

1.3 Frame invariance

Frame invariance principles consist of a set of transformations that, when appliedto a given system of equations, leave invariant the structure of the equations. Moreprecisely, frame invariance is closely related to translational invariance (Galilean

1.3. Frame invariance 21

invariance) and invariance with respect to rotation of the axis. However, an addi-tional transformation of the unknowns is often needed to express frame invariance.This is why the property is not a simple instance of the previous method of changeof coordinates. We use two methods below: one is called the naive since it relies onelementary manipulations in dimension one; the other relies on the more powerfulgeneral method of change of coordinates. The naive method is nevertheless quitenatural.

1.3.1 Naive method

The following vague statement of Galilean invariance will be sufficient for what isneeded in this study. It will be made more rigorous at the end of this section.

Definition 1.3.1 (Vague Galilean invariance principle in dimension d = 1). A modelin one dimension satisfies the principle of Galilean invariance if the equations arethe same after a combination of a transformation of the space-time structure

t′ = t, x′ = x + vt, v ∈ R , (1.39)

and an additional change of variable dictated by the underlying physics.

x′

x

−v

tt′

Figure 1.6: Translation of the referential.

As is visible in figure 1.6, the translation of the referential (t′ , x′) with respectto the referential (t, x) is −v. Partial derivatives are calculated according to thechain rule:

∂t = ∂tt′ ∂t′ + ∂tx

′ ∂x′ = ∂t′ + v∂x′ ,

∂x = ∂xt′ ∂t′ + ∂x x′ ∂x′ = ∂x′

(1.40)

Proposition 1.3.2. The traffic flow model (1.6), the shal low water model (1.11)and the system of Euler equations (1.15) satisfy the Galilean invariance principle.


Proof. This a consequence of the chain rule formulas (1.40). We begin by provingthe principle for the traffic flow model which reads

∂t′ ρ + v∂x′ ρ + ∂x′ (ρu(ρ)) = 0.

Define u′(ρ) = u(ρ) + v; then we get

∂t′ ρ + ∂x′ (ρu′(ρ)) = 0.

This change of velocity is compatible with the additivity of velocity, which is a fun-damental characteristic of Galilean invariance. Therefore the traffic flow equationis Galilean invariant.

Let us now consider the shallow water system (1.11). The first equation ischanged into

∂t′ h + ∂x′ (hu′) = 0, u′ = u + v,

which has nevertheless the same form as in the original system. The second equa-tion of the system can be rewritten using (1.40) as

∂t′ (hu) + v∂x′ (hu) + ∂x′ (hu2 + p(h)) = 0, p(h) =g

2h2.

We subtract v (∂t′ h + ∂x′ (hu′)) = 0 from both sides. Then

∂t′ (hu′) + v∂x′ (hu) + ∂x′ (hu2 + p(h)) + v∂x′ (hu) = 0,

which turns into ∂t′ (hu′) + ∂x′ (hu′2 + p(h)) = 0. Since the second equation isglobally the same after the change of coordinates, the shallow water system isGalilean invariant.

Turning to the Euler system for non-viscous incompressible gas, it is clearlyan extension of the shallow water system. Therefore the first two equations canbe rewritten as

∂t′ ρ + ∂x′ (ρu′) = 0,

∂t′ (ρu′) + ∂x′ (ρ(u′)2 + p) = 0.(1.41)

So it is sufficient to show that the energy equation is Galilean invariant to obtainGalilean invariance of the Euler system as a whole. Start from

∂t′ (ρe) + v∂x′ (ρe) + ∂x′ (ρue + pu) = 0.

Set e′ = ε + 12u′2 = ε + 1

2u2 + uv + 1

2v2 = e + uv + 1

2v2. Combine this with (1.41)

and get

∂t(ρe′ ) + v∂x′ (ρe) + ∂′x(ρue + pu) + v∂x′ (ρ(u′ )2 + p) +

1

2v2∂x′ (ρu′) = 0.

After rearrangement this yields ∂t(ρe′ ) + ∂x′ (ρu′e′ + pu′) = 0. The proof is thuscomplete.

1.3. Frame invariance 23

1.3.2 A general method

The frame invariance principle can also be analyzed using the general method ofchange of coordinates. We consider a general change of coordinates in dimensiond = 3 which is the composition of a rotation of the axis (rotation matrix R ∈R

3×3, with unit determinant RtR = I3) and a translation with constant velocity(v ∈ R 3), that is,

x = RX + vt. (1.42)

Going back to section 1.2.5 and the general formulas in dimension d = 3, one seesthat J= R, so

J = det(J) = 1,

which means that the transformation is incompressible. Moreover, the comatrix is

com(J) = det(J)(Jt )−1 = R.

These two simple formulas explain the central role of frame-invariant principles.Much more can be found in [192].

Proposition 1.3.3 (Frame invariance of the Euler system). The Euler system ofcompressible gas dynamics (1.16) in dimension d = 3 is invariant with respect toGalilean transformations (1.42) for al l vectors v and al l rotation matrices R.

Proof. The formulas (1.33) to (1.35) are

∂(t, x, y, z)

∂(t, X, Y, Z)=

(1 0

v R

),

com

(∂(t, x, y, z)

∂(t, X, Y, Z)

)=

(1 −vtR

0 R

)

and

⎛⎜⎝

ρ ρut

ρu ρu ⊗ u + pId

ρe ρeut + put

⎞⎟⎠

(1 −vtR

0 R

)

=

⎛⎜⎝

ρ ρ(u − v)tR

ρu (ρu ⊗ u + pId)R − ρu ⊗ vtR

ρe (ρeut + put) R − ρevt R

⎞⎟⎠ .

(1.43)

Define a new velocity variable w = Rt(u − v). The density equation becomes∂t ρ+ ∇X · (ρw) = 0. The second equation can be combined with the first one and


multiplied by Rt on the left. It yields

∂t(ρw) = Rt∂t ρu − Rtv∂t ρ

= −Rt∇X ·((ρu ⊗ u + pId) R − ρu ⊗ vt R

)− Rtv∇X ·

(ρ(u − v)t

)R

= −∇X ·(ρRt(u − v) ⊗ (Rt(u − v)) + pId

)

= −∇X · (ρw ⊗ w + pId) .

One obtains the equation ∂t(ρw) + ∇X · (ρw ⊗ w + pId) = 0. The last equation

needs a decomposition of the total energy e = ε + 12

|u|2 so that the naturaldefinition of the total energy in the new frame is

e′ = ε +1

2|w|2 = ε +

1

2|u − v|2 = e − v · u +

1

2|v|2

.

Now take the third line of (1.43) minus the scalar product of the second line with

v, plus the first line multiplied by 12 |v|2 . This yields ∂t(ρe′ ) + ∇X · m = 0 with

m =[(

ρeut + put)

R − ρevt R]

− vt[(ρu ⊗ u + pId) R − ρu ⊗ vtR

]+

1

2|v|2

[ρ(u − v)t R

]= ρe′wt + pwt .

In summary, the equations can be rewritten as⎧⎪⎨⎪⎩

∂tρ + ∇X · (ρw) = 0,

∂t(ρw) + ∇X · (ρw ⊗ w + pId) = 0,

∂t(ρe′ ) + ∇X ·(ρe′ wt + pwt

)= 0,

which indeed shows that the equations are frame invariant.

1.4 Linear stability and hyperbolicity

Linear stability is a fundamental notion in dynamical systems. The idea is toadd a small perturbation to small constant initial data. After linearization of theequations, study of the time evolution of the small perturbation yields informationabout the linear stability (or instability) of the system.

Consider a system of conservation laws

∂t U + ∂x f(U) = 0, U, f(U ) ∈ Rn. (1.44)

Assume that the flux is differentiable and define the Jacobian matrix of the fluxby

A(U0) = ∇Uf(U)(U0) ∈ Rn×n, U0 ∈ R

n . (1.45)

Let Uε be a certain solution obtained from perturbing around the state U0 ,

Uε(t, x) = U0 + εV (t, x) + o(ε). (1.46)

1.4. Linear stability and hyperbolicity 25

Expansion of all terms of the equation ∂tUε + ∂x f(Uε ) = 0 in ascending powers ofε yields

(∂tU0 + ∂x f(U0)) + ε(∂tV + ∂x(A(U0)V )) + o(ε)

= ε(∂t V + ∂x(A(U0)V )) + o(ε) = 0.

Neglecting high-order terms, the perturbation V is a solution of the linear equation

∂tV (t, x) + A∂x V (t, x) = 0, V (t, x) ∈ Rn , A = A(U0) = ∇Uf(U0) ∈ R

n×n.(1.47)

The linear stability is based on studying bounded solutions of this equation. Thestandard notion of hyperbolicity is as follows.

Definition 1.4.1 (Hyperbolic nonlinear system of conservation laws). A nonlinearsystem of conservation laws (1.44) is said to be strongly hyperbolic in the domainΩ ⊂ Rn if and only if the companion linear system (1.47) is strongly linearly stablefor all U0 ∈ Ω.

It remains of course to define what is strong linear stability; this is done inthe next section. An important result in the context of this monograph will be thatLagrangian gas dynamics is not strongly hyperbolic, but only weakly hyperbolicin the general case.

1.4.1 Classification in dimension d = 1

The most general method for establishing the stability property of the linear sys-tem is based on Fourier-Laplace modes:

V (t, x) = ei(kx−ωt) W, W ∈ Rn , k ∈ R .

Plugging such a representation into (1.47), one obtains that W is the solution ofa matrix eigenproblem −iωei(kx−ωt) W + ikA(U0)ei(kx−ωt) W = 0, that is,

A(U0 )W = λW, W ∈ Rn , λ =

ω

k∈ C . (1.48)

It is clear that the location of the eigenvalues λ ∈ C in the complex plane matters.Indeed, let us assume that there exists an eigenvalue λ ∈ R , so λ has a non-zeroimaginary part: this has an immediate implication for the behavior of V (t, x) =eik(x−λt)W with respect to the time variable. If Im(λ) > 0, then, by increasingk ≫ 1, Re(−ikλ) = kRe(−iλ) > 0 can be made arbitrarily large. If Im(λ) < 0,Re(−ikλ) > 0 can also be made arbitrarily large by decreasing k ≪ −1. Moreover,the conjugate λ of a non-real eigenvalue λ ∈ R of a real matrix is also an eigenvalue.This yields additional exponentially growing solutions. In all these cases, one getsexponential growth in time with an arbitrarily large exponential growth factor:this behavior characterizes ill-posed problems.


Definition 1.4.2 (Strong linear instability). The linear system (1.47) is said to bestrongly unstable at U0 if and only if there exist non-real eigenvalues λ ∈ R to theeigenproblem (1.48).

Definition 1.4.3 (Linear stability). The linear system (1.47) is said to be stableat U0 if and only if all eigenvalues λ of the eigenproblem (1.48) are real. Theeigenvalues are interpreted as velocities of the Fourier modes.

The interpretation stems from the expression V (t, x) = eik(x−λt)W, whichshows that λ has the dimension of a velocity. Next we detail the structure of theeigenvectors.

Definition 1.4.4 (Linear stability, strong or weak). Assume that the linear system(1.47) is stable. It is said to be strongly stable if the eigenvectors span Rn. If theeigenvectors do not span Rn , the linear system is weakly stable.

These notions are easily justified with the Jordan representation of a squarematrix A ∈ Rn×n,

A = P (D + T )P −1 (1.49)

where P = P t is a complex unitary change-of-basis matrix (it is an orthogonalchange of basis when it is real-valued), D is a diagonal matrix whose diagonalcoefficients are the eigenvalues of A with multiplicity, and T is an upper-triangularmatrix with vanishing diagonal such that DT = TD. Since T is upper triangularwith vanishing diagonal, it is also a nilpotent matrix whose order is the smallest1 ≤ r ≤ n such that T r = 0. A Fourier representation of a solution V (t, x) =eikx Z(t) of (1.47) yields the equation

Z′(t) + ikAZ(t) = 0.

The general solution is easily written with the matrix exponential formula [104]as

Z(t) = e−iktAZ0 = e−iktP(D+T )P−1

Z0 = P e−ikt(D+T ) P −1Z0.

Since DT = TD, one gets

e−ikt(D+T ) = e−iktDe−ikT = e−iktDr−1∑

p=0

(−iktT )p

p!.

Therefore one obtains the general representation formula

V (t, x) =r−1∑

p=0

(eikxP e−iktDT pP −1Z0

) (−ik)p

p!tp . (1.50)

Since r is the nilpotent order of T , there exists a vector Z0 ∈ Rn such thatT pP −1Z0 = 0, p = r − 1. Since the matrix D = diag(dj )1≤j≤n is real diagonal,the matrix e−iktD is also diagonal with coefficients

(e−iktD

)j j

= e−iktdj, dj ∈ R .


Upon inspection of (1.50), the quadratic norm (in Cn) of V (t, x) is of order tr−1

for k = 0 :‖V (t, x)‖ = O(tr−1). (1.51)

Moreover, r− 1 is the maximal integer with such growth. So we can now interpretDefinition 1.4.4 as follows:

• If r = 1, meaning that T = 0 and A can be made diagonal with a change-of-basis matrix, then the norm of Fourier solutions is uniformly bounded intime. This is the notion of strong linear stability.

• Assume 1 < r ≤ n. There exist Fourier solutions with polynomial growth intime. Moreover, the growth order in time, tr−1, corresponds to a multipli-cation by kr−1 . Since such a multiplication in Fourier space corresponds tor − 1 differentiations with respect to x, we observe that weak linear stabilityalso implies loss of r − 1 derivatives. More precisely, the asymptotic behaviorfor large t is

‖V (t, x)‖ ≈ Ct(r−1)∥∥∂r−1

x V (0, x)∥∥ , C > 0. (1.52)

Definition 1.4.5 (Order of weak linear stability). The order of weak linear stabilityis the integer q = r − 1 ∈ N , where r is the nilpotent order of T in the Jordanrepresentation (1.49) of the matrix A = A(U0).

∂t

(uv

)+ ∂x

(0 1

−1 0

)(uv

)= 0 unstable

∂t

(uv

)+ ∂x

(0 11 0

)(uv

)= 0 strongly stable, hyperbolic

∂t

(uv

)+ ∂x

(0 01 0

)(uv

)= 0 weakly stable, of order 1

∂t

(uv

)+ ∂x

(0 01 0

)(uv

)= 0 well-prepared data satisfy ∂xu = 0

Table 1.2: Linear systems.

Definition 1.4.6 (Well-prepared data). By well-prepared data for a weakly stablelinear system (1.47) we mean any Fourier mode W (x) = eikxZ0 with TP −1Z0 = 0.


Of course the definition makes sense only if T = 0; more precisely, if T = 0 alldata are well prepared. For well-prepared data, the general representation formula(1.50) can be rewritten as

V (t, x) = eikx P e−iktDP −1Z0 for V (0, x) = W (x), (1.53)

which turns into the uniform-in-time estimate

‖V (t, x)‖ = O(1), TP −1V (0, x) = 0. (1.54)

Moreover, we get from (1.53) that

TP −1V (t, x) = TP −1(eikxP e−iktDP −1Z0

)

= Te−iktD P −1Z0 = e−iktD TP −1Z0 = 0,

which means that if V (0, x) is well-prepared data at the initial time, it remainswell prepared at any time.

The technicalities associated with the discussion of the eigenvector structureof a given matrix A might be avoided by using the strict hyperbolicity criterion,which goes back to [125].

Definition 1.4.7. A matrix A = A(U0 ) is said to be strictly hyperbolic if and onlyif all its eigenvalues are real and distinct.

Indeed, n distinct real eigenvalues yield n linearly independent real eigenvec-tors, that is, strong linear stability.

Remark 1.4.8. However, we will show in section 4.1 that the presence of multipleeigenvalues is the general situation for many systems of conservation laws thatcome from continuum mechanics. This rules out using the strict hyperbolicity forgeneral purposes.

1.4.2 A useful property

The previous concepts provide a framework for the linear stability analysis of ageneral system of conservation laws. But it remains to compute the eigenstructureof the Jacobian matrix of the flux. In practice this can be quite tricky. A usefulproperty is that the eigenstructure does not change if one performs a change ofunknowns. So, for a given system, this suggests looking for astute changes ofvariables.

Let us start with the system of conservation laws

∂t U + ∂x f(U) = 0. (1.55)

We introduce a change of unknown of the form W = φ(U) ∈ Rn where φ : Rn →Rn is a differentiable transformation of the phase space. The transformation is


non-singular, i.e. det(∇φ) = 0. Firstly we rewrite the system in a quasi-linearform

∂tU + A(U)∂x U = 0, A = ∇f(U).

Secondly we perform the change of unknowns and get another quasi-linear form

∂t W + Q−1A(U )Q ∂x W = 0, A = ∇f(U) ∈ Rn×n , Q = ∇φ(U) ∈ Rn×n.(1.56)

Proposition 1.4.9. The linear stability/instability (strong or weak) properties ofthe initial system (1.55) are the same as the linear instability/stability (strong orweak) properties of the equation (1.56) written with the unknown W .

Proof. The matrices A(U) and Q−1A(U)Q are similar, so they have exactly thesame eigenstructure: the eigenvalues are equal, counting multiplicities, and theeigenvectors coincide. Most of all, the number of missing eigenvectors is the same,together with the order of weak stability.

In practice, one can try to determine a convenient variable W inspired by theunderlying physics so that the structure of Q−1A(U )Q is simpler. This methodis helpful for avoiding the brute force calculation of the characteristic polyno-mial of A, which yields the eigenvalues. It is also quite useful for calculations ofeigenvectors.

1.4.3 Generalization to dimension d ≥ 2

Consider∂t U + ∂xf(U) + ∂yg(U) = 0 (1.57)

where both fluxes U → f(U) and U → g(U) are differentiable. Since the problemis additive with respect to the directions, the corresponding linearized system ata given state U0 reads

∂tV + A∂x V + B∂yV = 0, A = ∇f(U0) and B = ∇g(U0). (1.58)

Rotate the coordinates by an arbitrary angle θ ∈ R :

x′ = cos θx + sin θy,

y′ = − sin θx + cos θy.⇐⇒

x = cos θx′ − sin θy′ ,

y = sin θx′ + cos θy′ .

Studying the solutions V which are independent with respect to the directiony′ is equivalent to adding the constraint that ∂y′ = 0, i.e. using the chain rule∂y′

∂x∂x + ∂y′

∂y∂y = − sin θ∂x + cos θ∂y = 0. One also has that

∂x =∂x′

∂x∂x′ +

∂y′

∂x∂y′ = cos θ∂x′ and ∂y =

∂x′

∂y∂x′ +

∂y′

∂y∂y′ = sin θ∂x′ .


In this case the equation (1.58) is written in the one-dimensional form

∂tV + (cos θA + sin θB) ∂x′ V = 0.

DefineA(θ) = cos θA + sin θB, θ ∈ R .

We say that the problem (1.57) is hyperbolic at U0 if and only if the matrix A(θ)is hyperbolic for all real θ. The generalization to any dimension is as follows.

Definition 1.4.10. Consider the nonlinear system of conservation laws

∂t U +d∑

i=0

∂xifi(U) = 0

where the fluxes are differentiable. This problem is hyperbolic (at U0) if and onlyif the matrix

A(α) =d∑

i=0

αi∇fi(U0) ∈ Rn×n

is hyperbolic for all α = (α1, . . . , αd) ∈ Rd.

All previous definitions of strong and weak stability generalize as well, to-gether with the notion of ill-posedness.

1.4.4 Examples

We consider the different equations and systems that have been constructed earlierin this chapter, and show that they correspond to linearly well-posed systems. Wealso determine the eigenvalues of the Jacobian matrix for each system. Theseeigenvalues correspond to the velocity of small perturbations.

Traffic flow

Consider firstly the equation (1.6) for traffic flow. Linearize all quantities about agiven density ρ0; that is, ρε(t, x) = ρ0 + εμ(t, x) + o(ε). The linearized equationfor the perturbation μ is

∂tμ + a∂x μ = 0, a = umax

(1 − 2

ρ0

ρmax

).

The explicit solution is μ(t, x) = μ(x − at). Hence we identify the velocity of theperturbation: it is λ = a.

Let ρc be the critical density defined by ρc = ρmax

2: we note that if ρ0 < ρc

then the velocity is positive, a > 0. On the contrary, if ρ0 > ρc then the velocityis negative ,a < 0. This reveals a major difference between the velocity of vehicles,which is always non-negative, and the velocity of small linear perturbations, whichcan take either sign. This is illustrated in figure 1.7.


ρc

a < 0

ρ0

ρ′

0

x

ρ

a > 0

Figure 1.7: Velocity of small perturbations for the traffic flow model: ρ′0 < ρc < ρ0.

Shallow water

Lemma 1.4.11. The flux of the shal low water system (1.11) is differentiable pro-vided h = 0. If h > 0, the model is strictly hyperbolic. If h < 0, the model islinearly unstable.

Remark 1.4.12. A negative height h < 0 does not make sense physically. So it isreassuring that instability is associated with such non-physical data.

Proof. Set a = h and b = hu. Then the flux of the shallow water system (1.11)

reads f(a, b) =

(b

b2

a +g2a2

). One has

A =

(0 1

− b2

a2+ ga

2b

a

), tr(A) =

2b

a= 2u, det(A) =

b2

a2− ga = u2 − gh.

The eigenvalue equation is λ2−tr(A)λ+det(A) = 0. Therefore the two eigenvaluesare

λ =2u ±

√(2u)2 − 4(u2 − gh)

2= u ± c, c =

√gh,

where c is identified as the velocity of small perturbation in the local fluid frame.

• If h > 0 the eigenvalues are different which yields strict hyperbolicity.

• If h < 0 then c ∈ iR∗ . In this case the eigenvalues are pure imaginary andcomplex conjugate, so the linear system is unstable.

Moreover, the flux is not differentiable for h = 0. Even if it does not ex-actly correspond to the previous definitions, one can nevertheless study A0 =


limh→0+ A, which admits the eigenvalue u with multiplicity two. But A0 = uId.Therefore A0 is not diagonalizable. So A0 is weakly hyperbolic.

A numerical application is as follows. One considers the mean height in oceansto be approximately 4000 m. This can be used to calculate the velocity of tsunamis,with c =

√gh ≈

√9.81 × 4000 ≈ 200 m s−1 = 720 km h−1. This is a reasonable

value.

Eulerian compressible gas dynamics in dimension d = 1

Lemma 1.4.13. Consider the Euler system (1.15) with the polytropic pressure law(1.12) and a positive density ρ > 0.

Then: (a) the flux is differentiable; (b) if ε > 0, the model is strictly hyperbolicwith eigenvalues

λ1 = u − c, λ2 = u, λ3 = u + c, c =√

γ(γ − 1)ε, (1.59)

where c is the speed of sound. The model is linearly unstable for ε < 0.

Proof. Set a = ρ, b = ρu and c = ρe. The flux is

f(a, b, c) =

⎛⎜⎜⎜⎜⎜⎝

b

b2

a+ (γ − 1)

(c − b2

2a

)=

3 − γ

2

b2

a+ (γ − 1)c

bc

a+ (γ − 1)

(bc

a− b3

2a2

)= γ

bc

a− γ − 1

2

b3

a2

⎞⎟⎟⎟⎟⎟⎠

.

The Jacobian matrix of the flux is

A =

⎛⎜⎜⎜⎜⎝

0 1 0

− 3 − γ

2

b2

a2(3 − γ)

b

a(γ − 1)

−γbc

a2+ (γ − 1)

b3

a3γ

c

a− 3γ − 3

2

b2

a2γ

b

a

⎞⎟⎟⎟⎟⎠

.

The three invariants of A are

tr(A) = 3b

a= 3u, ∆2(A) =

γ2 − γ + 6

2

b2

a2− γ(γ − 1)

c

a= 3u2 − γ(γ − 1)εu

and the determinant

det(A) =γ2 − γ + 2

2

b3

a3− γ(γ − 1)

bc

a2= u3 − γ(γ − 1)ε.

The eigenvalue equation det(A − λI3) = 0 reads

λ3 − 3uλ2 + (3u2 − γ(γ − 1)ε)λ − u3 + γ(γ − 1)ε = 0.


An obvious solution is λ = u. Factorization yields

(λ − u)(λ2 − 2uλ + u2 − γ(γ − 1)ε) = 0.

Therefore the eigenvalues of A are real and as given in (1.59). If ε > 0 the systemis strictly hyperbolic. If ε < 0 it is linearly unstable. The proof is complete.

The quantity c =√

γ(γ − 1)ε =√

γpρ

is the speed of sound. Under standard

conditions one can measure ρ, p and c. Therefore one has access to the value ofγ, which is actually related to the microscopic structure of the gas. A numericalapplication is as follows. The density of air is ρ = 1.28 × 103 g m−3. The pressureat the surface of the earth is p = 1 atm = 1.013 bar = 1.013 × 108 g m−1s−2. Since

γair = 1.4, one finds that c =√

γairpρ ≈ 332.88 ms−1, which is in good agreement

with the experimental values.

It is instructive to follow what Newton and Poisson did in their time byneglecting the influence of temperature. This gives rise to what is called the Boyle

pressure law, p = μρ where μ > 0 is a constant. We would find c√

μ =√

pρ

≈281.31 m s−1. This value does not correspond to measurements. This refutation ofthe Boyle pressure law provides an indirect justification of a polytropic pressurelaw p = (γ − 1)ρε for air, with γair = 1.4.

Eulerian compressible gas dynamics in dimension d > 1

We first consider the Euler system (1.16) in dimension d = 3.

Proposition 1.4.14. Consider the system (1.17) with a positive density and positive

pressure described by a polytropic equation of state. Define c =√

γpρ

.

Then the system is hyperbolic. The five eigenvalues of the Jacobian matrixin the direction α ∈ R 3 , |α| = 1, are

uα − c, uα, uα, uα, uα + c with uα = u · α.

Proof. We study the stability using the method of section 1.4.3 and with a prelim-inary rotation of the axis as allowed by the frame invariance principle of Propo-sition 1.3.3. It is therefore sufficient to study the solutions of the Euler systemwhich are invariant with respect to y and z, taking into account that the horizon-tal velocity stands for uα = u · α. We use a new variable

W = (p, u, v, w, ε)t

and the notation ddt

= ∂t + u∂x for the material derivative. One gets, after some


algebra,⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

d

dtρ + ρ∂xu = 0,

ρd

dtu + ∂xp = 0,

ρd

dtv = 0,

ρd

dtw = 0,

ρd

dtε + p∂x u = 0,

which turns into ⎧⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎩

∂tp + u∂xp + ρc2∂xu = 0,

∂tu + u∂xu +1

ρ∂xp = 0,

∂tv + u∂x v = 0,

∂tw + u∂xw = 0,

∂tε + u∂xε + p∂x u = 0.

(1.60)

Therefore the quasi-linear system for the new variable W is ∂tW + B∂xW = 0with

B = uI +

⎛⎜⎜⎜⎜⎜⎜⎝

0 ρc2 0 0 01

ρ0 0 0 0

0 0 0 0 0

0 0 0 0 0

0 p 0 0 0

⎞⎟⎟⎟⎟⎟⎟⎠

.

The eigenvectors of B are

s1 =

⎛⎜⎜⎜⎝

ρc−1000

⎞⎟⎟⎟⎠ , s2 =

⎛⎜⎜⎜⎝

00100

⎞⎟⎟⎟⎠ , s3 =

⎛⎜⎜⎜⎝

00010

⎞⎟⎟⎟⎠ , s4 =

⎛⎜⎜⎜⎝

00001

⎞⎟⎟⎟⎠ , s5 =

⎛⎜⎜⎜⎝

ρc1000

⎞⎟⎟⎟⎠ ,

with eigenvalues u − c, u with multiplicity 3 and u + c. Since the eigenvectors arelinearly independent for ρc = 0, this problem admits five real eigenvalues and fivereal eigenvectors. This ends the proof.

Nevertheless, the system is not strictly hyperbolic since the eigenvalue uhas multiplicity 3. Another method for calculation of the eigenstructure of thelinearized matrix will be developed in section 4.1.


Lagrangian compressible gas dynamics in dimension d = 1

Proposition 1.4.15. Consider a pressure law written as (τ, ε) → p(τ, ε). Assumepτ − ppε > 0, τ > 0 and define c = 1

ρ

√pτ − ppε > 0. Then the Lagrangian gas

dynamics system (1.26) in the mass variable dimension one is hyperbolic. Theeigenvalues of the Jacobian matrix are −ρc, 0 and ρc.

Proof. Consider the system with a pressure law p = p(ρ, ε):

⎧⎪⎪⎨⎪⎪⎩

∂t τ − ∂mu = 0,

∂t u + ∂mp = 0,

∂t e + ∂m(pu) = 0, e = ε +1

2u2.

The Jacobian matrix of the flux calculated with respect to the variables (τ, u, e)is

A =

⎛⎝

0 −1 0

pτ −upε pε

upτ p − u2pε upε

⎞⎠ .

The characteristic polynomial is det(A − λI) = −λ3 − (pτ − ppε)λ. Set c =1ρ

√pτ − ppε. One can check that a polytropic pressure law yields c =

√γpτ =√

γpρ

. The roots of the characteristic polynomial are λ1 = −ρc , λ2 = 0 and

λ3 = ρc. Since the eigenvalues are real and different, the system is strictly hyper-bolic. The proof is complete.

Lagrangian compressible gas dynamics in dimension d = 2

It appears that the hyperbolicity of Eulerian gas dynamics does not imply thehyperbolicity of Lagrangian gas dynamics. The reason lies in the additional un-knowns of the Lagrangian system, which are associated with a loss of eigenvectorsin dimension d > 1. However, the analysis depends on how one takes into accountthe Piola identities. To simplify the discussion, consider Hui’s formulation (1.32)with invariance in the direction Y,

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

∂t (ρJ) = 0,

∂t (ρJu) + ∂X (pM) = 0,

∂t (ρJv) + ∂X(−pL) = 0,

∂t (ρJe) + ∂X (puM − pvL) = 0,

∂t A − ∂X u = 0,

∂t B − ∂X v = 0,

∂t L = 0,

∂t M = 0,


and add the identities ∂X L = ∂X M = 0. So L and M are now constant coefficientsof the system ⎧

⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎩

∂t (ρJ) = 0,

∂t (ρJu) + M∂X p = 0,

∂t (ρJv) − L∂X p = 0,

∂t (ρJe) + ∂X (puM − pvL) = 0,

∂t A − ∂X u = 0,

∂t B − ∂X v = 0.

(1.61)

We notice that the third Piola identity ∂tJ = ∂X (uM − vL) = M∂X − L∂X v canbe obtained as a consequence of (1.61), so there is no need to incorporate it in thesystem.

Proposition 1.4.16. The Lagrangian gas dynamics system (1.61) is weakly hyper-bolic with wave velocities −ρc, 0, ρc. The multiplicity of the eigenvalue 0 is 3. Theorder of weak hyperbolicity is 1.

Proof. To avoid unnecessary cumbersome calculations, we use the variable W =(ρJ, p, u, v, A, B) and write ρJ = ρ0. One has on the one hand

0 = ∂t(ρJ) = J∂tρ + ρ∂tJ = J∂tρ + ρ (M∂X u − L∂X v) ,

and on the other hand ∂t p = pε (∂te − u∂tu − v∂t v)+pρ∂tρ. Set μ2 =(ppε − ρ2pρ

).

One finds ⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

∂t(ρJ) = 0,

∂tp +M

ρ0μ2∂X u − L

ρ0μ2∂X v = 0,

∂tu +1

ρ0M∂X p = 0,

∂tv − 1

ρ0L∂X p = 0,

∂tA − ∂Xu = 0,

∂tB − ∂X v = 0.

The Jacobian matrix B ∈ R 6×6 of the flux with respect to W is

B =1

ρ0

⎛⎜⎜⎜⎜⎜⎜⎜⎝

0 0 0 0 0 0

0 0 μ2M −μ2L 0 00 M 0 0 0 0

0 −L 0 0 0 0

0 0 −ρ0 0 0 0

0 0 0 −ρ0 0 0

⎞⎟⎟⎟⎟⎟⎟⎟⎠

. (1.62)

Denote a generic eigenvector by r = (r1, . . . , r6). To organize the discussion, webegin with the assumption that the corresponding eigenvalue is non-zero, i.e. λ =


0. In this case Br = λr yields⎧⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎩

0 = ρ0λr1 ,

μ2(Mr3 − Lr4) = ρ0λr2 ,

Mr2 = ρ0λr3 ,

−Lr2 = ρ0λr4 ,

−r3 = λr5,

−r4 = λr6.

Elimination of r3 and r4 in the second expression yields μ2(M2 +L2)r2 = (ρ0λ)2r2,and all other quantities can be determined in terms of r2. The solutions are

λ = ± μ√

M2 + L2

ρ0= 0, (1.63)

with a corresponding unique eigenvector. Assuming now the eigenvalue vanishes(i.e. λ = 0), one gets

⎧⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎩

μ2(Mr3 − Lr4) = 0,

Mr2 = 0,

−Lr2 = 0,

−ρ0r3 = 0,

−ρ0r4 = 0.

⇐⇒

r3 = 0,r4 = 0,r2 = 0.

So the codimension of the zero eigenspace is 3 and its dimension is 6 − 3 =3, meaning that there exist three, not four, linearly independent correspondingeigenvectors.

One can identify the space of missing eigenvectors by studying the equationB2s = 0. One has

B2 =1

ρ20

⎛⎜⎜⎜⎜⎜⎜⎜⎝

0 0 0 0 0 0

0 μ2(M2 + L2) 0 0 0 0

0 0 μ2M2 −μ2LM 0 0

0 0 −μ2LM μ2L2 0 0

0 −ρ0M 0 0 0 0

0 ρ0L 0 0 0 0

⎞⎟⎟⎟⎟⎟⎟⎟⎠

. (1.64)

The equation B2s = 0 yields⎧⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎩

μ2(M2 + L2)s2 = 0,

μ2M(Ms3 − Ls4) = 0,

−μ2L(Ms3 − Ls4) = 0,

ρ0(Ms2 + ps8) = 0,

ρ0(Ls2 + ps7) = 0,

⇐⇒

⎧⎪⎨⎪⎩

μ2(Ms3 − Ls4) = 0,

Ms2 = 0,

Ls2 = 0,


that is, μ2(Ms3 − Ls4) = 0 and s2 = 0. So the dimension of the space B2s = 0is equal to 4. This shows that the order of weak hyperbolicity is 1. There exists avector in Ker(B2) which is not in Ker B . Additionally, the space of well-prepareddata is given by

s ∈ Ker(T ) ⇐⇒ s = Span(0, 0, L, M, 0, 0) .

The proof is thus complete.

Remark 1.4.17 (Interpretation of weak hyperbolicity). The geometrical variablesare the cause of the loss of strong hyperbolicity; see also [106]. It is possible to showthat shear velocities, which trigger shear between layers of flows, are archetypesof the phenomenon.

The key point is the deformation gradient, which is very sensitive to shearvelocities. In Lagrangian formulations, the deformation gradient is an unknown.In Eulerian formulations, the deformation gradient is not an unknown. This is thereason why Lagrangian gas dynamics produces only a weakly hyperbolic system,while Eulerian gas dynamics is strongly hyperbolic.

Indeed, consider the nonlinear equations. The results are the same for the lin-earized equations. One constructs a specific initial data for the Lagrangian system(1.32): the density and pressure are constants, ρ0 = ρc, p0 = pc; the vertical veloc-ity vanishes identically, v0 = 0, and the horizontal velocity is a function of the ver-tical variable, which is written with an abuse of notation as u0 = u0(Y ); the othervariables are initialized as A0 = M0 = 1, B0 = L0 = 0 and J0 = A0M0−B0L0 = 1.

Since the pressure is a constant, it is immediately evident that such an initialcondition generates shears between layers of flows. The physical solution is ρ = ρc

and p = pc; the velocity is constant on horizontal lines; the Euler-to-Lagrangetransformation is

x = X + tu0(Y ) and y = Y.

Therefore A = ∂X x = 1, B = ∂X y = 0, M = ∂Y y = 1 and, most importantly,

L = ∂Y x = tu′0(Y ).

The norm of L increases linearly in time and exhibits loss of one derivative with re-spect to the initial data. This behavior is typical of a weakly hyperbolic system; seeequation (1.52) and Definition 1.4.6. In this case a well-prepared initial conditioncorresponds to u′

0 = 0, which means that the horizontal velocity is independent ofthe vertical coordinate.

1.5 Exercises

Exercise 1.5.1. Consider a conservation law ∂tu + ∂x f(u) = 0 with f ′′(u) = 0 forall u. Find a function ϕ(u) such that v = ϕ(u) is solution of the Burgers equation

∂tv + ∂x

(v2

2

)= 0. (1.65)

1.5. Exercises 39

Exercise 1.5.2. Consider the shallow water system with g = 0. The system is calledthe pressureless gas dynamics system

∂th + ∂x(hu) = 0,

∂t(hu) + ∂x

(hu2

)= 0.

Show that this system is weakly hyperbolic and u is formally a solution of theBurgers equation.

Exercise 1.5.3. This exercise looks at a generalization of the linearization techniquearound a non-constant solution of ∂t u + ∂x f(u) = 0. Assume that uε(t, x) =u(t, x) + εv(t, x) + o(ε) is solution of ∂tu + ∂xf(u) = ∂tuε + ∂x f(uε) = 0 for all ε.Show that the pair (u, v) is formally a solution of the system

∂t

(uv

)+ ∂x

(f(u)

a(u)v

)= 0, a(u) = f ′(u). (1.66)

Show that this system is weakly hyperbolic for a′ (u)v = 0.

Exercise 1.5.4. Consider

∂tu + ε(a∂x u + b∂xv) = 0,

∂tv + ∂x u + ε(c∂x u + d∂xv) = 0,

where (a, b, c, d) ∈ R 4 with b < 0. Show that if ε > 0 is small, then the system isstrictly hyperbolic. Show that if ε < 0, then ill-posedness holds.

Exercise 1.5.5. Start from the Euler system in one dimension. Consider a non-constant-in-time translation of the reference frame:

t′ = t and ∂t x′(t, X) = v(t) with x(0, X) = X.

Show ⎧⎪⎨⎪⎩

∂t′ ρ + ∂x′ (ρu′) = 0,

∂t′ (ρu′) + ∂x′ (ρ(u′ )2 + p) = ρg(t),

∂t′ (ρe) + v∂x′ (ρe) + ∂x′ (ρue + pu) = ρg(t)u′ ,

where the acceleration is g(t) = ddt

v(t).

Exercise 1.5.6. This exercise yields another and more systematic proof of theGalilean invariance principle in dimension d = 1. We consider the change of refer-ential (1.39) with a given velocity v ∈ R . Show that the comatrix (1.39) is

com(M) =

(1 0v 1

).

Let us write

A =

(ρ ρu

ρu ρu2 + pρe ρue + pu

)


so that the Euler system becomes ∇t,xA = 0. Show that ∇t′,x′ (A × com(M)) = 0.Define

T =

⎛⎜⎜⎝

1 0 0

v 1 0

v2

2v 1

⎞⎟⎟⎠ .

Show that the system ∇t′,x′ (T × A × com(M)) = 0 is equivalent to the systemwritten in the reference frame (t′ , x′). Compare this with the result of Proposi-tion 1.3.2.

1.6 Bibliographic notes

The LWR model is derived from the seminal contributions of Lighthill [133],Whitham [203] and Richards. The Burgers equation comes from [31]. The La-grangian formulation of Eulerian equations is sometimes considered obscure [94,93, 175]. However, it is known that a systematic correspondence is possible: thereader can refer to [57] for a modern mathematical treatment. See also [200] in onedimension. The systematic derivation of such equations (in view of numerical meth-ods) can be found in the papers of Hui and collaborators [112, 111, 108, 106, 113]as well as in [208], and it is also studied in [138] and [72]. Galilean invariance is astandard property of conservation laws and gas dynamics [81]. Stability and hyper-bolicity are standard notions [94, 175, 57]. However, weak hyperbolicity, necessaryfor the discussion of Lagrangian equations, is rarely developed; see [115] and [79]to complement the modest introduction to the topic presented in this chapter.

Chapter 2

Scalar conservation laws

Nothing is lost, nothing is created, everything istransformed.

– Antoine Lavoisier(Elementary Treatise of Chemistry, 1789)

Once a model is obtained, one tries to determine its solutions by either theoret-ical or numerical means. In this context, it should be noticed that the nonlinearstructure of the equations induces the existence of discontinuous solutions.

In this chapter we analyze the structure of solutions to the scalar equation

∂t u + ∂x f(u) = 0, t > 0, x ∈ R , (2.1)

with initial data u(0, x) = u0(x) for all x ∈ R . Many excellent textbooks presentthe theory for such equations, some of which are listed at the end of the chapter.In what follows, the presentation is focused on a selection of elementary propertiesthat will be extended to the Lagrangian systems studied in forthcoming chapters.The plan of the chapter is as follows.

• Firstly we will construct the solution of (2.1) using the method of character-istics, with the restriction that the initial data u0 is smooth and the time ofexistence of a smooth solution is bounded above: t ∈ [0, T ) with T > 0. Thiswill show that the nonlinearity generates discontinuous solutions, typicallyat time t = T < ∞. The notion of a weak formulation is then naturally in-troduced to give meaning to the solution at any time t ∈ R . Entropies will bediscussed. The main result will be that entropy weak solutions are admissiblesolutions.

• Secondly we define a natural numerical scheme that will be proved to be sta-ble in the maximal norm and entropy consistent under a Courant-Friedrichs-Lewy (CFL) constraint

c∆t

∆x≤ 1.




41

42 Chapter 2. Scalar conservation laws

This guarantees that the numerical scheme will not capture the wrong non-entropy weak solutions. We illustrate the method with numerical simulationswhere incorrect solutions are interpreted as violations of the entropy criterion.The convergence of the discrete solution to an entropy weak solution will beproved, thus establishing the existence of entropy weak solutions.

• Finally, the results of a traffic flow simulation in Lagrangian coordinates willbe presented, highlighting the connection between the Lagrangian formula-tion and the modeling of particles.

2.1 Strong solutions

Consider the Cauchy problem

∂t u + ∂x f(u) = 0,

u(0, x) = u0(x),(2.2)

and assume that the initial data is differentiable, u0 ∈ C1(R ). Assuming thatthe solution u is also differentiable, one can write the equation in the quasi-linearnon-divergent form

∂tu + a(u)∂x u = 0 where by definition a(u) = f ′(u).

Let us consider the change of coordinates

⎧⎨⎩

∂x(t′, X)

∂t′ = a (u(t′ , x(t′, X))) , x(0, X) = X,

t = t′,(2.3)

with the chain rule

∂t′ = ∂t′ t ∂t + ∂t′ x ∂x = ∂t + a∂x ,

∂X = ∂X t ∂t + ∂X x ∂x = J∂x.(2.4)

By definition J = ∂X|t′ x. So ∂tu + a(u)∂x u = 0 is equivalent to

∂t′ u(t′, x(t′, X)) = 0.

This shows that u in constant along the characteristic curves

u(t′, x(t′, X)) = u(0, x(t′ , 0)) = u0(X)

and implies that∂x(t′, X)

∂t′ = a(u0(X)).

2.1. Strong solutions 43

The characteristic curves are actually straight lines of the form

x(t, X) = X + ta (u0(X)) , a(u) = f ′(u). (2.5)

The construction of the solution by the method of characteristics consistsof first solving the equation (2.3) and then propagating the solution along thecharacteristics. Considering that t and x = x(t, X) are given, the characteristicequation is a nonlinear equation for X:

x(t, X) = X + ta0(X), a0(X) = a(u0(X)),

Proposition 2.1.1. Assume that the function a0 ∈ C0(R ) ∩ L∞(R ) is piecewisedifferentiable, with everywhere a derivative on the right and on the left:

C ≤ a′0(X−), a′

0(X+) ≤ L ∀X ∈ R , 0 < L. (2.6)

Define the time T∗ by

T∗ =1

−Cif C < 0 and T∗ = ∞ if C ≥ 0. (2.7)

Then there exists a unique solution (t, X) → x(t, X) of the characteristic equationfor al l times 0 ≤ t < T∗.

Proof. The proof is an easy consequence of the fixed point theorem for a contrac-tive function. Rewrite the equation as

X = gα(X) with gα(X) =1

1 + α(x + αX − ta(X)) and α ≥ 0.

The derivatives of gα are

g′α(X±) =

1

1 + α(α − ta′(X± )) ∈

[α − tL

1 + α,

α − tC

1 + α

].

The contraction requirement ‖g′α‖L∞(R) < 1 needed in the fixed point theorem can

be expressed as −1 < α−tL1+α

and α−tC1+α

< 1. It yields the conditions −1−α < α−tLand α − tC ≤ 1 + α.

The second inequality is true if and only if −tC < 1. If C ≥ 0 this holds forall time t. If C < 0 it yields the condition t < T∗ = 1

−Cof the claim. The first

inequality can always be made true by taking a parameter α > 0 sufficiently large.The proof is complete.

Proposition 2.1.2 (Strong solutions). Assume that the initial data x → u0(x) iscontinuous with a piecewise continuous derivative that satisfies (2.6).

Then the continuous (with piecewise continuous derivative) function (t, x) →u(x, t) = u0(X(t, x)) is a solution of the Cauchy problem (2.2) for 0 ≤ t < T∗


Proof. One has that

∂t u(t, x) + ∂x f(u(t, x)) = ∂tu(t, x) + a(u(t, x))∂x u(t, x) = ∂t′|X u0(X) = 0.

This equality holds except at points where u is not C1.

We present some elementary examples of the use of the method of character-

istics for the Burgers equation ∂tu + ∂xu2

2= 0. Consider the initial condition

⎧⎪⎨⎪⎩

u0(x) = 1, x < −1,

u0(x) = −x, − 1 < x < 0,

u0(x) = 0, x > 0.

(2.8)

The characteristic lines are depicted in figure 2.1. The solution of the characteristicequation (2.5) is ⎧

⎪⎨⎪⎩

x(t, X) = X + t, X < −1,

x(t, X) = X − tX, − 1 < X < 0,

x(t, X) = X, X > 0.

Since the strong solution is defined by u(t, x(t, X)) = u0(X), one obtains

0 ≤ t < T = 1

⎧⎪⎪⎨⎪⎪⎩

u(t, x) = 1, x − t < −1,

u(t, x) = − x

1 − t, − 1 < x − t, x < 0,

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

(2.9)

Observe that the construction is valid only for 0 ≤ t < T∗ with T∗ = 1. Notethat the numerical value of T∗ is exactly that predicted by the definition (2.7). Attime t = T∗ the characteristic lines cross, and the solution of (2.5) becomes multi-valued. This shows that the condition t < T∗ in Proposition 2.1.2 is necessary andsufficient.

Another example is with the initial condition⎧⎪⎨⎪⎩

u0(x) = 0, x < −1,

u0(x) = 1 + x, − 1 < x < 0,

u0(x) = 1, x > 0,

for which the characteristic lines are⎧⎪⎨⎪⎩

x(t, X) = X, X < −1,

x(t, X) = X + t(1 + X), − 1 < X < 0,

x(t, X) = X + t, X > 0

(2.10)

depicted in figure 2.2. This defines the solution u where

for 0 ≤ t < T∗ = ∞

⎧⎪⎪⎨⎪⎪⎩

u(t, x) = 0, x < −1,

u(t, x) = 1 +x − t

1 + t, − 1 < x < t,

u(t, x) = 1, x − t > 0.

2.1. Strong solutions 45

t = 0

x−1

1

u0(x)

t =1

2

x−1

1

u(t, x)

t = T = 1

x−1

1

u(t, x)

In the (x, t) plane

−1 x

T = 1

t

Figure 2.1: Characteristic lines which cross at t = T∗ for the Cauchy data (2.8).


1

x−1

u0(x)

−1 x

T = 1

t

Figure 2.2: In the situation of (2.10) the characteristic lines never cross.

2.2 Weak solutions

The construction of a strong solution with the method of characteristics fails for acertain time t = T∗ . Even worse, the threshold time can vanish as well, i.e. T∗ = 0.In this case the method of characteristics does not construct anything, which showsthat the framework of strong solutions is too restrictive. It is therefore necessaryto extend the notion of solution. This is done by introducing the concept of weaksolutions.

The question is how to give a meaning to discontinuous solutions of theCauchy problem

∂tu + ∂xf(u) = 0, t > 0, x ∈ R ,

u(0, x) = u0(x), x ∈ R .(2.11)

To this end, assume that u is a strong solution. Consider an additional function(t, x) → ϕ(t, x) which is C1 and has compact support in space-time: typicallyϕ(t, x) ≡ 0 if t > T or |x| > A. The function ϕ is a priori non-zero for −A < x < A.The set of such functions will be denoted by

C10 =

ϕ ∈ C1(R+ ×R ) : ϕ has compact support

, R

+ = [0, ∞).

2.2. Weak solutions 47

t

x

T

A−A

Figure 2.3: Test function ϕ ∈ C10 with compact support in [0, T ) × (−A, A) .

Multiply the equation (2.11) by ϕ ∈ C10 and integrate in space and time:

∫

R

∫

0<t

(∂t u + ∂xf(u))ϕ dt dx =

∫

−A<x<A

∫ T

0<t

(∂t u + ∂xf(u))ϕ dt dx = 0.

Integrate by parts to get

−∫

−A<x<A

∫ T

0<t

(u∂tϕ + f(u)∂x ϕ) dx dt −∫

−A<x<A

u0(x)ϕ(0, x) dx = 0.

Notice that the derivatives of u and f(u) do not show up anymore. This means thatthe above integral expression enables us to manipulate discontinuous functions. Itis the weak formulation that we were looking for.

Definition 2.2.1 (Weak solutions). Let u ∈ L∞(R × R+). It is a weak solution ofthe Cauchy problem (2.11) if and only if

∫

R

∫

0<t(u∂tϕ + f(u)∂x ϕ) dx dt +

∫

R

u0(x)ϕ(0, x)dx = 0 (2.12)

holds for all ϕ ∈ C10 .

The equation (2.12) is also called a weak formulation.

Proposition 2.2.2. Strong solutions are also weak solutions. A weak solution whichis continuous and piecewise differentiable is also a strong solution.

Proof. That strong solutions are weak solutions is evident. To show the secondclaim, integrate the weak formulation (2.12) by parts for ϕ ∈ C1

0 . This yields∫ ∫

(∂tu + ∂xf(u)) ϕ(t, x)dt dx +

∫

R

(u0(x) − u(0, x)) ϕ(0, x)dx = 0. (2.13)


Since the space C10 is large enough, one can choose particular test functions. Let us

first consider a test function which vanishes identically at t = 0, that is, ϕ(0, x) = 0for all x ∈ R . Then

∫ ∫(∂tu + ∂xf(u)) ϕ(t, x)dt dx = 0

for all such test functions, which yields ∂tu+∂xf(u) = 0 for all x ∈ R and all t > 0.Therefore one can simplify (2.13) to obtain

∫R

(u0(x) − u(0, x)) ϕ(0, x) dx = 0 forall admissible test functions ϕ. In particular one can now take test functions whichdo not vanish identically at t = 0. This shows that u0(x)−u(0, x) = 0 for all x ∈ R .The proof is thus complete.

Theorem 2.2.3 (Characterization of discontinuous weak solutions). Assume thatu ∈ L∞(R × R+) and there exists a smooth curve Γ : t → x(t) such that u ispiecewise differentiable on both sides of the curve. Then u is a weak solution of(2.12) if and only if

(a) u is a strong solution separately on both sides of Γ;

(b) the fol lowing jump relation is satisfied for al l x(t) ∈ Γ:

−x′(t) [u] + [f(u)] = 0. (2.14)

Remark 2.2.4. The convention is that [g] = g+ − g− is the jump of g across thecurve.

Proof. It is sufficient to check (2.14). Let ϕ ∈ C10 with a support that contains a

piece of the curve Γ = t → x(t). Split the space into two parts Ω− = (t, x) :x < x(t) and Ω+ = (t, x) : x > x(t).

For simplicity we assume that ϕ(0, x) ≡ 0 for all x ∈ R since it plays no role.So (2.12) becomes

∫ ∫

Ω−

(u∂tϕ + f(u)∂x ϕ) dx dt +

∫ ∫

Ω+

(u∂tϕ + f(u)∂x ϕ)dx dt = 0.

Applying the Stokes formula to both sides, one gets

−∫ ∫

Ω−

(∂tu + ∂x f(u))ϕ +

∫

Γ

((f(u), u)− , n−)ϕdσ

−∫ ∫

Ω+

(∂t u + ∂xf(u))ϕ +

∫

Γ

((f(u), u)+ , n+)ϕdσ = 0.

The function u is a strong solution locally in Ω− and Ω+, so

∫ ∫

Ω−

(∂t u + ∂xf(u))ϕ =

∫ ∫

Ω+

(∂t u + ∂xf(u))ϕ = 0.

2.2. Weak solutions 49

x

t

Ω−

n+

Ω+

n−

t

Figure 2.4: The outward normals from Ω± are denoted by n± ∈ R 2.

Therefore∫

Γ

((f(u), u)− , n−)ϕdσ +

∫

Γ

((f(u), u)+ , n+)ϕdσ = 0.

Since ϕ is arbitrary, one gets the necessary and sufficient condition

((f(u), u)− , n−) + ((f(u), u)+ , n+) = 0 on Γ.

The tangent and normal vectors are

t =1

a

(x′(t)

1

), n+ =

1

a

(1

−x′(t)

)= −n−, a =

√1 + x′(t)2.

This yields −x′(t)[u] + [f(u)] = 0.

It is convenient to consider a discontinuity that moves at a constant velocityσ:

u(t, x) = uL for x < σt, u(t, x) = uR for x > σt.

Definition 2.2.5 (Rankine-Hugoniot relation). The triplet (σ, uL, uR) satisfies theRankine-Hugoniot relation if and only if

−σ(uR − uL) + (f(uR ) − f(uL)) = 0. (2.15)

The Rankine-Hugoniot relation is a rephrasing of (2.14) for σ = x′(t). Onealso notes that −σ[u] + [f(u)] = 0. Triplets which satisfy the Rankine-Hugoniotrelations therefore define weak solutions.


x = σt + x0

t

uL

uR

x

Figure 2.5: Discontinuous solutions in the (x, t) plane.

For example, discontinuous solutions of the Burgers equation satisfy

σ =

[u2

2

]

[u]=

uR + uL

2. (2.16)

This is a compatibility relation between the right and left states and the velocityof the discontinuity.

2.3 Entropy weak solutions

The use of weak solutions may give rise to a paradox in certain cases. Indeed,the space of weak solutions is by definition much larger than the space of strongsolutions. Therefore it is possible that the weak formulation admits additionalsolutions, even in the case where only a single solution is constructed by themethod of characteristics. To gain better insight into this paradox, consider thefollowing situation.

Proposition 2.3.1. The Burgers equation with initial data

u0(x) = 0, x < 0 ; u0(x) = 1, x > 0

admits two different weak solutions. One is a discontinuity with velocity σ = 12

.The other is

0 ≤ t < T = ∞

⎧⎪⎪⎨⎪⎪⎩

u(t, x) = 0, x < 0,

u(t, x) =x

t, 0 < x < t,

u(t, x) = 1, x − t > 0.

(2.17)

Proof. The first function satisfies the criterion (2.16), so it is indeed a weak solu-tion. The second function is bounded, 0 ≤ u(x, t) ≤ 1 for all x ∈ R and t ∈ R+.

2.3. Entropy weak solutions 51

To show that it is a weak solution we study the sum (2.12) of the two integrals inthe definition of the weak solution. Taking account of the three different zones in(2.17), for ϕ ∈ C1

0 the solution reads

I(ϕ) =

∫ ∫

0<x<t

(x

t∂tϕ(x, t) +

x2

2t2∂xϕ(x, t)

)dx dt

+

∫ ∫

0<t<x

(∂tϕ(x, t) +

1

2∂xϕ(x, t)

)dx dt +

∫

0<xϕ(x, 0)dx.

We need to show it is exactly zero for all admissible ϕ.

Consider the vector field G(x, t) =(

x2

2t2 ϕ(x, t), xtϕ(x, t)

)defined in the zone

0 < x < t. One has

∇xt · G = ∂x

(x2

2t2ϕ

)+ ∂t

(x

tϕ)

=x

t2ϕ +

x2

2t2∂xϕ − x

t2ϕ +

x

t∂tϕ =

x2

2t2∂x ϕ +

x

t∂tϕ.

So by the Stokes formula the first term in I(ϕ) is

I1(ϕ) ≡∫ ∫

0<x<t

(x2

2t2∂x ϕ(x, t) +

x

t∂tϕ(x, t)

)dx dt

=

∫ ∫

0<x<t

∇xt · G dx dt =

∫

∂(0<x<t)

G · n dσ,

where the exterior normal is n = 1√2(1, −1) and the measure is dσ =

√2 dx. That

is,

I1(ϕ) =

∫ ∞

0

(x2

2t2∂xϕ(x, t) − x

t∂tϕ(x, t)

)dx = − 1

2

∫ ∞

0

ϕ(x, x)dx.

A simple application of the Stokes formula shows that the second part of I(ϕ) is

I2(ϕ) ≡∫ ∫

0<t<x

(∂tϕ(x, t) +

1

2∂xϕ(x, t)

)dx dt

=1

2

∫ ∞

0

ϕ(x, x)dx −∫

0<x

ϕ(x, 0)dx.

So I(ϕ) = 0 for all ϕ ∈ C10 , which shows that this function is also a weak solution,

completing the proof.

Since it is not possible to distinguish between these two solutions solely onthe basis of weak formulations, a natural question is whether one can find anadditional criterion that would select only one solution of the two. Such a criterionexists and is formulated in terms of entropies. It is derived by considering that


admissible functions u should be the limit of viscous functions uε with a dissipativeregularizing operator in the right-hand side:

∂t uε + ∂x f(uε) = ε∂xx uε, t > 0,

uε(0, x) = u0,ε(x).(2.18)

The dissipative regularizing operator is ε∂xx with a vanishing parameter ε → 0+,and we assume that uε is naturally a very smooth function (which can be justified[183, 94, 175, 57]).

Let us make an additional hypothesis which can also be justified: we assumethat uε is a regular differentiable uniformly bounded function which tends touε ∈ L1

loc in the sense that

limε→0+

‖u − uε‖L1loc

([0,T [×R) = 0, ‖uε‖L∞([0,T [×R) ≤ C for all ε (2.19)

and thatlim

ε→0+‖u0 − u0,ε‖L1

loc(R) = 0. (2.20)

The functions uε are called viscous solutions. To identify all the limit equalitiesand inequalities satisfied by u as the limit of uε, we need entropies.

Definition 2.3.2 (Entropy and entropy flux). A function η : R → R which is twicedifferentiable and convex is called an entropy. The corresponding entropy fluxfunction ξ : R → R is defined up to a constant by

η′(u)f ′ (u) = ξ′(u), ξ(u) =

∫η′ (v)f ′ (v)dv.

Consider for example the Burgers equation with flux f(u) = u2

2 . Any function

ηp(u) = u2p

2p+ α u2

2with p ∈ N , p ≥ 2, and α > 0 is an entropy. The entropy flux

is ξp(u) = u2p+1

2p+1+ α u3

3since

η′p(u)f ′(u) = (u2p−1 + u)u = u2p + u2 = ξ′

p(u).

Let us define the space C10,+ = C1

0 ∩ ϕ ≥ 0.

Theorem 2.3.3. Let u be the limit of uε in the sense of (2.18)–(2.20). Then thelimit function u satisfies the fol lowing two properties:

(a) It is a weak solution (2.12).

(b) For al l entropy-entropy flux pair (η, ξ) and al l ϕ ∈ C10,+, one has the weak

entropy inequality

−∫

R

∫

0<t(η(u)∂tϕ + ξ(u)∂x ϕ) dx dt −

∫

R

η(u0(x))ϕ(0, x)dx ≤ 0. (2.21)


Proof. From (2.18) one has ∂t uε + f ′(uε)∂x uε = ε∂xx uε. Multiplying through byη′ (uε) gives

∂tη(uε ) + η′(uε)f ′ (uε)∂x uε = εη′ (uε)∂xx uε.

The entropy-entropy flux relation yields

∂tη(uε) + ∂xξ(uε) = ε∂xx η(uε) − εη′′ (uε)(∂x uε)2 ≤ ε∂xx η(uε ).

Multiply by ϕ ∈ C10,+ to get

∫ ∫(∂tη(uε ) + ∂x ξ(uε)) ϕdx dt ≤ ε

∫ ∫(∂xx (η(uε ))) ϕdx dt.

Integrating by parts leads to

−∫

R

∫

0<t

(η(uε )∂tϕ + ξ(uε)∂x ϕ) dx dt −∫

R

η(u0,ε)ϕ(0, ·) dx

≤ ε

∫

R

∫

0<t

η(uε )∂xx ϕdx dt.

It remains to show the convergence of the integrals with respect to ε. Notice thetest function ϕ is constant. One has

∣∣∣∣∫

R

∫

0<t

η(uε )∂t ϕdx dt −∫

R

∫

0<t

η(u)∂t ϕdx dt

∣∣∣∣

≤(

max |∂tϕ| max|v|≤C

|η′(v)|)

‖uε − u‖L1(supp(ϕ) ,

where the support is supp(ϕ) = (t, x) ∈ R+ × R : ϕ(t, x) > 0. Therefore

∫

R

∫

0<t

η(uε )∂tϕdx dt →∫

R

∫

0<t

η(u)∂t ϕdx dt.

The other two integrals converge for the same reasons. Finally, the right-hand sidetends to zero since it depends linearly on ε. The proof is complete.

Definition 2.3.4 (Entropy weak solutions). A weak solution u ∈ L∞(R×R+) whichsatisfies the weak entropy inequalities (2.21) for all entropy-entropyflux pairs (η, ξ)is called an entropy weak solution.

Theorem 2.3.5 (Characterization of discontinuous entropy weak solutions). Letu ∈ L∞(R × R+) and assume there exists a smooth curve Γ : t → x(t) such thatu is piecewise differentiable on both sides of the curve. Then u is an entropy weaksolution (2.12) if and only if

(a) u is a strong solution local ly on both sides of Γ;


(b) one has the jump relation

−x′(t) [u] + [f(u)] = 0, t > 0; (2.22)

(c) the jump relation holds

−x′(t) [η(u)] + [ξ(u)] ≤ 0, t > 0 (2.23)

for al l entropy-entropy flux pairs

Proof. Theorem 2.3.3 already shows (a) and (b). The last assertion (c) is provedby an easy extension of the proof of (b), where one notices that the strong relation∂t η(u) + ∂xξ(u) = 0 holds separately on both sides of Γ.

2.3.1 Entropic discontinuities

Consider a triplet (σ, uL, uR) ∈ R 3 that is a solution of the Rankine-Hugoniotrelation, with σ the velocity of the discontinuity, uL the left state and uR the rightstate. Since an entropy weak solution satisfies the jump inequality (2.23) for allentropies, one obtains a system consisting of one equality plus an infinite numberof inequalities:

−σ(uR − uL) + f(uR ) − f(uL) = 0,

−σ (η(uR ) − η(uL)) + ξ(uR ) − ξ(uL) ≤ 0 ∀(η, ξ) with η′′ ≥ 0 and ξ′ = η′f ′.

(2.24)Our goal hereafter is to characterize all solutions of this system.

Proposition 2.3.6 (Convex flux). Assume that the flux is twice differentiable andstrictly convex, f ′′ (u) > 0. Let (σ, uL, uR ) be an entropy discontinuity satisfying(2.24). Then al l entropy inequalities are equivalent to the single inequality

uL ≥ uR. (2.25)

Proof. We consider the case uL = uR.Define the two functions g(u) = f(u)− f(uR)− σ(u − uR) and H(u) = ξ(u)−

ση(u). The Rankine-Hugoniot relation can be rewritten as g(uR) = g(uL) = 0.The entropy inequality can be rewritten as H(uR) − H(uL) ≤ 0, that is,

∫ uR

uL

H ′(v)dv =

∫ uR

uL

(ξ′(v) − ση′ (v))dv =

∫ uR

uL

η′ (v)g′ (v)dv ≤ 0, (2.26)

which can be rewritten as∫ uR

uL(η(v) − αv − β)′g′(v)dv ≤ 0 where α and β are

arbitrary. Take α and β such that η(uR ) − αuR − β = η(uL) − αuL − β = 0, i.e.

β =uLη(uR ) − uRη(uL)

uL − uRand α =

η(uR ) − η(uR )

uL − uR.


An integration by parts yields

−∫ uR

uL

(η(v) − αv − β)g′′(v)dv =

∫ uR

uL

(−η(v) + αv + β)f ′′(v)dv ≤ 0.

Notice that the weight is f ′′(v) > 0.The function v → −η(v) + αv + β is strictly concave and vanishes at the

end-points uR and uL: it is therefore positive for intermediate values, that is,min(uR, uL) < v < max(uR , uL). This shows that the term under the integral ispositive. Therefore uL > uR. The proof is complete.

For a strictly convex flux, only a single entropy inequality is sufficient tocharacterize entropy discontinuities.

Proposition 2.3.7 (Oleinik condition for a general differentiable flux). Assume thatthe flux is differentiable. Let (σ, uL , uR) be an entropy discontinuity (2.24). Twocases occur:

• uR < uL . Then the graph of the flux f is below the chord v → f(uR )+ σ(v − uR)for uR < v < uL.

• uL < uR. Then the graph of the flux f is above the chord v → f(uR)+ σ(v − uR)for uL < v < uR.

Proof. Integrating by parts the inequality (2.26) yields −∫ uR

uLη′′(v)g(v) dv ≤ 0 for

all entropies. So

−∫ uR

uL

ϕ(v) (f(u) − f(uR) − σ(u − uR)) dv ≤ 0 ∀ϕ = η′′ > 0.

Assume uL > uR, so f(u) ≤ f(uR) + σ(u − uR), meaning that the graph of f isbelow the chord. If uL < uR we have the opposite situation and f is above thechord. The proof is complete.

A third criterion exists for the characterization of entropic shocks. It is calledthe Lax criterion and reads

f ′(uL) ≥ σ ≥ f ′(uR). (2.27)

This inequality compares the sound speed before and after the shock with theshock velocity.

Proposition 2.3.8 (Comparison of the different entropy criteria). The Oleinik con-dition implies the Lax condition. If the flux function is twice differentiable andstrictly convex, the Oleinik condition and the Lax condition are equivalent to thecondition uL ≥ uR.

Proof. From both graphs in figure 2.6, it is clear that f ′(uL) ≥ σ ≥ f ′(uR ). Sothe Oleinik condition implies the Lax condition.

Assume now that the flux f is a convex function. Then the chord is abovethe curve, which corresponds to uL ≥ uR. This is an equivalence, so the proof isfinished.


uL > uR

Chord with slope σ

uLuuR

f (u)

uL < uR

Chord with slope σ

uuLuR

f (u)

Figure 2.6: Illustration of the Oleinik condition.

2.3.2 Shocks and contact discontinuities

To motivate the discussion of shocks versus contact discontinuities, let us considerthe situation depicted in figure 2.7, where a solution is made up of two successivediscontinuities. We say that this solution is reversible because the value of thefunction after the two shocks is the initial value before the two shocks: uR →uL → uR.

The right discontinuity satisfies, for all entropy-entropy flux pairs (η, ξ),

−σ(uR − uL) + f(uR ) − f(uL) = 0,

−σ[η(uR ) − η(uL)] + ξ(uR) − ξ(uL) ≤ 0.

The left discontinuity satisfies, for all entropy-entropy flux pairs (η, ξ),

−σ′(uL − uR) + f(uL) − f(uR ) = 0,

−σ′[η(uL) − η(uR )] + ξ(uL) − ξ(uR) ≤ 0.


uR

x

σσ′

uL

uR

Figure 2.7: Reversible discontinuity.

Assuming the discontinuity is not trivial, uR = uL, the condition for reversibilityis σ = σ′ , and

−σ[η(uL) − η(uR )] + ξ(uL) − ξ(uR) = 0 (2.28)

for all entropy-entropy flux pairs (η, ξ).

Definition 2.3.9. Let (σ, uL, uR) be an entropic discontinuity.

• It is called an entropic shock if and only if there exists one entropy-entropyflux pair (η, ξ) such that

−σ[η(uR ) − η(uL)] + ξ(uR) − ξ(uL) < 0.

• It is said to be a contact discontinuity if and only if

−σ[η(uL) − η(uR )] + ξ(uL) − ξ(uR ) = 0

for all entropy-entropy flux pairs (η, ξ).

Proposition 2.3.10. Let (σ, uL , uR) be a contact discontinuity. Then the flux isaffine between uL and uR and σ is the slope of f.

Proof. A contact discontinuity satisfies (2.28). Moreover, the triplets (σ, uL, uR )and (σ, uR, uL) define entropic discontinuities. The Oleinik condition of Proposi-tion 2.3.7 shows that the graph of f is both below and above the chord, and so isequal to the chord in the interval min(uR, uL) ≤ u ≤ max(uR, uL).

Let us compare the different discontinuous solutions for the Burgers equation

∂t u + ∂xu2

2 = 0 and the linear transport equation ∂t u + a∂xu = 0. For the Burgersequation, the only admissible discontinuities are entropic shocks. In contrast, thelinear transport equation only admits contact discontinuities.


2.3.3 Rarefaction fans

Rarefaction fans are particular smooth solutions which behave mathematicallylike rarefaction fans in compressible gas dynamics, hence the name. The functiondefined in (2.17) is actually a rarefaction fan. A general definition is as follows.

Definition 2.3.11 (Rarefaction fans for conservation laws). A rarefaction fan is astrong solution which is constant on lines x−x0

t= a ∈ (α, β).

t

x

Figure 2.8: A rarefaction fan.

Proposition 2.3.12 (Equation of a rarefaction fan). The equation of the rarefactionfan can be written as

f ′ (u) =x − x0

t, f ′(uL) = α, f ′(uR ) = β, (2.29)

where uL and uR are the values of u at the edge of the fan. Provided the functionf ′ is invertible, the solution is

u(x, t) = (f ′)−1

(x − x0

t

).

Proof. The method of characteristics (2.3) shows that ∂x∂t = a = f ′(u), from which

the equation (2.29) of the fan is deduced.

Assume that f ′′ (u) > 0, which means that f ′ is invertible on its domain ofdefinition. Let α, β ∈ f ′(R ) with α < β. One can solve the equation α ≤ x−x0

t≤ β

to obtain the solution u(t, x) = (f ′)−1(

x−x0

t

).


Conversely consider a solution of (2.29). Then

∂t u + ∂xf(u) =1

f ′′(u)(f ′′ (u)∂t u + f ′ (u)f ′′ (u)∂x u)

=1

f ′′(u)

[∂tf

′ (u) + ∂x(f ′ (u)2

2

]

=1

f ′′(u)

[∂t

(x − x0

t

)+ ∂x

((x − x0)2

2t2

)]

=1

f ′′(u)

[− x − x0

t2+

x − x0

t2

]= 0.

So u is indeed a strong solution where it is defined. The proof is complete.

As a simple application, consider the flux f = u4

4. The equation of rarefaction

fans reads u3 = x−x0

t , from which we deduce the general form of such solutions:

u(t, x) =

(x − x0

t

) 13

.

2.3.4 The entropic solution of the Riemann problem

The solution of the Riemann problem is a building block for conservation laws.The Riemann problem is a Cauchy problem with a special initial condition

u0(x) = uL for x < 0, u0(x) = uR for x > 0.

The entropy weak solution of the Riemann problem is an ensemble of entropicdiscontinuities and rarefaction fans.

Proposition 2.3.13 (Solution of the Riemann problem for a convex flux). Assumef ′′ (u) > 0. The fol lowing function is an entropy weak solution of the Riemannproblem:

If uL < uR, a rarefaction fan. If x/t ≤ f ′(uL) then u(t, x) = uL . If x/t ≥ f ′(uR )then u(t, x) = uR. If f ′(uL) ≤ x/t ≤ f ′(uR) then f ′(u(t, x)) = x/t.

If uL > uR, an entropic shock. The shock velocity is σ = f(uR)−f(uL)uR−uL

.

Proof. Use (2.29) in the first case, or (2.24)–(2.25) in the second case.

The general case is covered by the Oleinik solution and is illustrated in fig-ure 2.9.

Proposition 2.3.14 (Oleinik solution of the Riemann problem, general case). As-sume that the flux is differentiable. The fol lowing function is an entropy weaksolution of the Riemann problem. It is described as a curve in the plane (u, f):

If uL > uR, the curve is the convex envelope of the graph of f for v ∈ [uR, uL ].


(u, f) plane

M

f (u)

uL u uR

R

QP

rarefactionrarefactionshock

(x, t) plane

rarefaction

x

t

rarefaction

shock

Figure 2.9: Oleinik solution for uL < uR, in the (u, f(u)) plane on the top andin the (x, t) plane below. The first branch MP is a rarefaction fan. The secondbranch P Q is a shock. The third branch is also a rarefaction fan.

2.4. Peculiarities of Lagrangian traffic flow 61

If uL < uR, the curve is the concave envelope of the graph of f for v ∈ [uL, uR].

Proof. We refer to [94] for a proof.

2.4 Peculiarities of Lagrangian traffic flow

The traffic flow equation is a paradigm for the discussion of shocks and contactdiscontinuities and for the comparison of Eulerian and Lagrangian formulations.To motivate the discussion, we start from the Eulerian equation

∂tρ + ∂xf(ρ) = 0, f(ρ) = ρu(ρ) (2.30)

and add the assumption that u′(ρ) ≤ 0, which expresses the fact that the velocityof a driver is adapted to the density: the denser the flow the slower the velocity.Using the Euler-to-Lagrange method allowing us to rewrite the one-dimensionalsystem of gas dynamics as the purely Lagrangian system (1.26), one obtains theLagrangian form of the traffic flow equation:

∂tτ − ∂mv(τ ) = 0, v(τ ) = u(ρ), ρ =1

τ. (2.31)

The mass variable is dm = ρ0(x)dx.• A first remark is, of course, that a zero initial density is a problem for the La-grangian formulation, which becomes nonsense. However, one mayinstead considera modified initial density ρε

0 = max(ρ0(x), ε) for a small value of the parameterε > 0. One expects from basic physical considerations some continuity propertiesof the solution with respect to the initial data (actually this is what we observeon the highway).• A second remark is about the speed of sound.

Proposition 2.4.1. Assuming u′ ≤ 0, the speed of sound a(τ ) of the Lagrangiansystem (2.31) has a constant sign:

a(τ ) = −ρ2u′(ρ) ≥ 0.

Proof. Indeed, a(τ ) = −v′(τ ) =d

dτu

(1

τ

)= − 1

τ 2u′(

1

τ

)= −ρ2u′(ρ).

This property has the important consequence that it allows us to use numer-ical methods with simpler structure. It will be generalized to systems in the nextchapter.• Our third remark concerns a comparison of the Rankine-Hugoniot relation ofthe Eulerian system,

−σ (ρR − ρL) + (f(ρR ) − f(ρL)) = 0, (2.32)

with the Rankine-Hugoniot relation of the Lagrangian system,

−D(τR − τL) − (v(τR ) − v(τL)) = 0. (2.33)


Proposition 2.4.2. The Eulerian and Lagrangian Rankine-Hugoniot relations areequivalent. Moreover, τR,L = ρ−1

R,L and one has the fol lowing correspondence be-tween the shock velocities D and σ:

D = ρR(σ − uR) = ρL(σ − uL). (2.34)

Moreover, the Eulerian and Lagrangian entropy criteria are equivalent.

Proof. The first part is easy. Indeed, the equation (2.32) with f = ρu can berewritten as ρR(σ − uR) = ρL(σ − uL). The interpretation is that the mass fluxacross the shock, whose dimension is the product of a density times a velocity, isconstant. Set D = ρR (σ − uR) = ρL(σ − uL). One has

−D (τR − τL) = −ρR(σ − uR)τR + ρL(σ − uL)τL = (uR − σ)− (uL − σ) = uR − uL.

The second part is proved as follows. First we consider the entropy criterion forthe Eulerian equation,

−σ (η (ρR) − η (ρL)) + (ξ (ρR) − ξ (ρL)) ≤ 0 (2.35)

for all entropies with η′′ > 0 and entropy fluxes such that ξ′ = f ′η′. Secondly wedefine, for 0 < τ < ∞,

η(τ ) = τη

(1

τ

)and ξ(τ ) = ξ

(1

τ

)− u

(1

τ

)η

(1

τ

)

and show that (η, ξ) is an entropy-entropy flux pair for the Lagrangian system.The concavity criterion is checked as follows:

d2

dτ 2η(τ ) =

d2

dτ 2

(τη

(1

τ

))=

d

dτ

(η

(1

τ

)− 1

τη′(

1

τ

))

= − 1

τ 2η′(

1

τ

)+

1

τ 2η′(

1

τ

)+

1

τ 3η′′

(1

τ

)=

1

τ 3η′′

(1

τ

)≥ 0,

so the function η is indeed convex. The relation ξ′(ρ) = (u(ρ) + ρu′(ρ)) η′(ρ) canbe used to verify the identity

d

dτξ(τ ) =

d

dτ

[ξ

(1

τ

)− u

(1

τ

)η

(1

τ

)]

= − 1

τ 2

[ξ′(

1

τ

)− u′

(1

τ

)η

(1

τ

)− u

(1

τ

)η′(

1

τ

)]

= − 1

τ 2

[1

τu′(

1

τ

)η′(

1

τ

)− u′

(1

τ

)η

(1

τ

)]

= − 1

τ 2u′(

1

τ

)[1

τη′(

1

τ

)− η

(1

τ

)].


But by definition η(τ ) = τη(

1τ

), which yields η′ (τ ) = η

(1τ

)− 1

τ η′ ( 1τ

). Therefore

d

dτξ(τ ) =

1

τ 2u′(

1

τ

)η′ (τ ) =

d

dτv (τ )

d

dτη(τ ),

with v the Lagrangian flux (2.31). This shows that ξ is indeed the Lagrangianentropy flux.

We can now transform (2.35) with the help of these identities. One rewrites(2.35) as

−(σ − uR)η (ρR) − (σ − uL)η (ρL) + ξ (τR) − ξ (τL) ≤ 0

and as

−(σ − uR)ρR η (τR) − (σ − uL)ρL η (τL) + ξ (τR) − ξ (τL) ≤ 0.

Using the formula (2.34) for the Lagrangian shock velocity get

−D (η (τR) − η (τL)) + ξ (τR) − ξ (τL) ≤ 0. (2.36)

In summary, all Eulerian entropy relations (2.35) are equivalent to all Lagrangianentropy relations (2.36). This establishes the last part of the proposition.

2.4.1 Application and physical interpretation

The following examples illustrate how the previous tools can be used to constructand interpret the solution of the Riemann problem for some cases of interest.

LWR model

The non-dimensional equation of the LWR model is

∂tρ + ∂xf(ρ) = 0, f(ρ) = (ρ − ρ2).

The flux function f(ρ) is strictly concave. Consider the initial data (0 ≤ a < b ≤ 1)

ρ0(x) =

⎧⎪⎨⎪⎩

a, x < 0,

b, 0 < x < 1,

a, 1 < x.

(2.37)

That is, the density of vehicles is initially discontinuous.The mathematical solution is composed of a shock at x = 0 and a rarefaction

fan at x = 1. The shock velocity is σ = 1 − (a+ b). This is locally compatible withthe entropy criterion since a < b. At x = 1 it is not possible to have a shock sinceb > a.

The crucial point is that the features of this mathematical solution are com-patible with observations. In real life discontinuities are observed on the road,with vehicles entering the traffic jam at the rear and exiting the traffic jam witha continuous acceleration. Once again this agrees with the usual observations.


a ba

t

x

x = σt

x = st + 1

Figure 2.10: Traffic jam structure for the LWR equation.

The Olmos-Munos model

The LWR model has been modified in manydirections. In [159], Olmos and Munosconsider traffic flow in the city of Bogota. We will refer to their model as the OMmodel. The main point is that Olmos and Munos used real data and extracted asimple flux function to describe the data. As they explain in their paper, driversin Bogota seem to have an aggressive way of driving, and this is reflected in theflux function. We consider hereafter an even simpler flux.

This function ρ → f(ρ) is continuous and assumed to be piecewise affine.Using non-dimensional notation, the flux is made up of three branches:

⎧⎪⎨⎪⎩

f(ρ) = αρ, 0 ≤ ρ ≤ ρ1,

f(ρ) = αρ1 + β(ρ − ρ1), ρ1 ≤ x ≤ ρ2,

f(ρ) = αρ1 + β(ρ2 − ρ1) − γ(ρ − ρ2), ρ2 ≤ x ≤ 1.

(2.38)

The parameters are the threshold densities 0 < ρ1 < ρ2 < 1 and three slopes 0 <α, β, γ with the constraint β < α. Moreover, f(1) = αρ1+β(ρ2−ρ1)−γ(1−ρ2) = 0.

This function is concave and the derivative is bounded. One can easily con-struct the solution of the Riemann problem with the Oleinik method. In contrastto the LWR model, there is no rarefaction fan for the OM model, since f ′′ = 0 ona full interval. Instead, rarefaction fans are replaced by contact discontinuities.

To illustrate, we consider the initial data

ρ0(x) =

⎧⎪⎨⎪⎩

a < ρ1, x < 0,

b > ρ2 , 0 < x < 1,

a, 1 < x.

The solution of the Cauchy problem still consists of a shock at the rear, with shock


ρ1

f (ρ)

ρρmaxρ2

u(ρ)

ρ1

ρρmaxρ2

Figure 2.11: Flux and velocity for the OM model. One can see the aggressive wayof driving at low densities.

velocity

σ =f(ρ1) − f(ρ2)

ρ1 − ρ2.

The shock velocity is non-negative for the chosen data. But, as is visible in fig-ure 2.12, the rarefaction fan in the LWR model is replaced by three contact dis-continuities with velocities −γ, β and α.

a ba

t

xx0

x = σt x = βt + x0

x = αt + x0

x = −γt + x0

Figure 2.12: Traffic jam in Bogota city.

Buckley-Leverett equation

This one-dimensional model arises in petroleum engineering, where it is commonto push underwater oil by injecting water into a well. Let u ∈ [0, 1] be the watersaturation in a mixture of water and oil. Consider the following initial data at


time t = 0:

u0(x) =

1 = uL, x < 0,

0 = uR, 0 < x.(2.39)

It models a situation where there is only water for x < 0 and only oil for x > 0.We assume that a second collecting well is at xwell > 0. A natural problem isto determine the composition of water and oil at the collecting well, bearing inmind that the higher the oil composition the better. A common model for suchsituations is the Buckley-Leverett model

∂tu + ∂xf(u) = 0, f(u) =u2

u2 + A(1 − u)2,

where A > 0 is a parameter. Observe that the flux is neither convex nor concave:

f ′(u) =2Au(1 − u)

(u2 + A(1 − u)2)2≥ 0, f ′(0) = f ′(1) = 0, f ′

(1

2

)> 0.

Let 0 < u∗ < 1 be such that

f ′(u∗) =f(u∗) − f(uR)

u∗ − uR=

f(u∗)

u∗ .

The equation with one physical solution can be recast as

(u∗)2 + A(u∗ − 1)2 = 2A(1 − u∗) ⇐⇒ u∗ =

√A

1 + A∈ (0, 1) the meaningful root.

The flux is depicted in figure 2.13.The entropy weak solution of the Riemann problem (2.39) consists of an

entropic shock from uR = 0 to u∗ and a rarefaction fan from u∗ to uL = 1. Thesolution (t, x) → u(t, x) is displayed in figure 2.14. The dynamics of the solutionis such that the collecting well suddenly sees a mixture of water and oil (withsaturation u∗), and then a continuum where the water saturation never reachesthe value 1. Therefore by this method one can collect all of the oil, but in infinitetime: this is bad news for engineers.

Lagrangian traffic flow

Consider the equation ∂tτ − ∂mu = 0 where u = 1 − ρ = 1 − 1τ

, with initial data(2.37) rewritten as

τ0(x) =

⎧⎪⎨⎪⎩

a−1, m < 0,

b−1, 0 < m < b,

a−1, b < m.

(2.40)

The flux is convex since d2

dτ2 (−u) = 2τ3 > 0. Admissible shocks satisfy τL > τR.

This corresponds to ρL < ρR, which reflects the fact that the density of vehicles

2.5. Numerical computation of entropy weak solutions 67

f (u)

uu∗ u = 1

Figure 2.13: Flux of the Buckley-Leverett equation.

inside traffic jams is greater than the incoming density of vehicles. Therefore theshock is at m = 0. A rarefaction is generated at m = b. This is the same solutionas for the Eulerian traffic flow equation.

2.5 Numerical computation of entropy weak solutions

The need to calculate correct entropy weak solutions gives rise to important con-straints on numerical methods. In this monograph we consider only first-orderfinite volume schemes, since they are the simplest schemes for which a compre-hensive construction is possible.

We will concentrate mostly on construction issues. Two cases are considered.The first is for the situation where the sign of the sound velocity is constant, whichultimately corresponds to Lagrangian traffic flow. In this situation the structureof the numerical scheme is the simplest. The second case corresponds to a changeof sign of the sound velocity and allows one to discretize the Eulerian traffic flowequation.

Numerical examples illustrate the theoretical properties. In particular we willshow why a violation of the correct CFL condition maygenerate incorrect solutionswhich cannot converge to entropy weak solutions.

2.5.1 Notion of a conservative finite volume scheme

It is known that finite volume (FV) schemes are among the most efficient onesfor the calculation of discontinuous weak solutions [129, 130, 131, 94]. To justifythis claim, we start from three versions of a basic finite difference scheme and


u∗

t

u = 0u = 1

x = σt

xCollecting well

Figure 2.14: Solution of the Buckley-Leverett Riemann problem. The shock isattached to the rarefaction fan.

show that if the exact solution is discontinuous, only one version of the schemeis correct. The correct version is the one which can be reinterpreted as an FVscheme. This highlights the notion of a conservative scheme.

The example is based on the Burgers equation ∂tu + ∂xu

2

2= 0 with initial

data ⎧⎪⎨⎪⎩

u(0, x) = u0(x) = 1 for x < 0.5,

u(0, x) = u0(x) = 1 + (0.5 − x) for 0.5 < x < 1.5,

u(0, x) = u0(x) = 0 for 1.5 < x.

For small times t < 1, the entropy weak solution contains a ramp between x = 1+tand x = 2. At t = 1 an entropy shock is formed. After that time, the shockpropagates with velocity σ = 1

2.

Let us now discretize the equation with a general numerical method in finitedifference form

un+1j − un

j

∆t+ an

j

unj − un

j −1

∆x= 0, (2.41)

but with three different evaluations of the velocity:⎧⎪⎪⎪⎨⎪⎪⎪⎩

Choice 1 anj =

unj + un

j −1

2,

Choice 2 anj = un

j −1,

Choice 3 anj = un

j .

(2.42)

In the classical finite difference sense and for smooth reference solutions, the dis-crete velocity is clearly consistent with the exact velocity. This is left to the readerto verify. Let us now present in figure 2.15 the numerical results at time t = 0.5and in figure 2.16 the numerical results at time t = 1.5.


0 1 1.5 2

0.2

0.4

0.6

0.8

0

1

0.5 x

Choice 1, 2 and 3

u

Figure 2.15: The numerical solution at t = 0.5 for the three numerical velocities.The discrete solutions are so close to one another that they are indistinguishable.

0 0 0.5 1 1.5 2

0.8

0.6

0.4

0.2

1

u

Choice 3

Choice 2

Choice 1

x

Figure 2.16: The numerical solution at t = 1.5 for the three numerical velocities.At this later time the numerical solutions are very different. Only choice 1 yields areasonable prediction of the theoretical shock position: xshock = 1.5+(1.5−1)/2 =1.75.

The reason why only choice 1 yields a correct numerical approximation of theshock is related to the notion of conservation. Indeed, the numerical scheme withthe first velocity in (2.42) can be recast as

Choice 1 :un+1

j − unj

∆t+

(unj )2

2− (un

j−1 )2

2

∆x= 0.

The scheme is said to be conservative since formal conservation of the total massholds, i.e.

Choice 1 :∑

j

un+1j =

∑

j

unj . (2.43)

It is clear from inspection of figure 2.16 that conservation of the total mass isstrongly correlated with accurate numerical prediction of the shock velocity. Inthat sense, preservation of the total mass seems to be a necessary condition for


accurate shock calculations. A scheme which has a property like (2.43) is called aconservative scheme.

2.5.2 Finite volume scheme

The previous example shows the importance of having a procedure which guaran-tees by construction the preservation of the total mass, (2.43). To achieve this end,we consider a space-time predefined grid with mesh size ∆x > 0 and time step∆t > 0. The so-called volumes of the FV scheme are the intervals

(xj −1

2, xj + 1

2

)

where xj +12

= (j + 12)∆x for all j ∈ Z. The scheme can be expressed in the general

form

un+1j − un

j

∆t+

fnj +1

2

− fnj −1

2

∆x= 0 ∀j ∈ Z, ∀n ∈ N , (2.44)

with the mean value initial data

u0j =

1

∆x

∫ xj−

12

xj−

12

u0(x)dx ∀j ∈ Z. (2.45)

The numerical solution in cell j and at time step n is unj . This scheme is explicit,

that is, one can determine explicitly the new value in the cell as a function of theprevious one and of the fluxes fn

j ± 12

un+1j = un

j +∆t

∆xfn

j − 12

− ∆t

∆xfn

j + 12

, j ∈ Z.

Therefore the construction of the scheme ultimately relies on the determinationof the numerical flux fn

j +12

for j ∈ Z.

Proposition 2.5.1. Whatever the fluxes, the scheme (2.44) is formal ly conservative,that is, ∑

j ∈Zun+1

j =∑

j ∈Zun

j .

Proof. Indeed,

∑

j ∈Zun+1

j −∑

j ∈Zun

j =∑

j ∈Z

(un+1

j − unj

)=

∆t

∆x

∑

j ∈Z

(fn

j −12

− fnj +1

2

)= 0,

since it is a telescopic sum. This is only formal in the sense that summabilityis needed to justify the calculations. A convenient summability assumption is∑

j ∈Z∣∣un

j

∣∣ +∑

j ∈Z

∣∣∣fnj +1

2

∣∣∣ < ∞.


2.5.3 Construction of the flux using the method of characteristics

The method of characteristics allows us to construct fluxes and FV schemes whichare consistent, stable and optimal under some simple conditions on the time stepand accuracy. Nevertheless, we will see that an important restriction arises, whichis that the sign of the sound velocity must be constant. This constraint will berelaxed later.

First case: the advection equation

t

xun

j un+1j

un+ 1

2

j

t

x

unj

un+1

ju

n+1

2

j

Figure 2.17: The flux is upwinded following the characteristic lines.

The first example is the advection equation ∂tu + a∂x u = 0, for which the exactflux is f(u) = au with a ∈ R . The characteristic lines are plotted in figure 2.17. Inboth cases the numerical flux is the value of the exact flux in the vertical segment,taking into account that this value is equal to the value at the starting point ofthe characteristic line. With this construction, the numerical flux is defined as

⎧⎪⎪⎨⎪⎪⎩

if a > 0, fnj +1

2

= f(unj ) = aun

j ,

if a < 0, fnj +1

2

= f(unj +1) = aun

j +1 ,

if a = 0, fnj +1

2

≡ 0.


The analysis of the resulting scheme is detailed below in Proposition 2.5.2 andCorollary 2.5.4.

Second case: the Eulerian LWR traffic flow model

We wish to use the same method for the Eulerian traffic flow equation

∂tρ + ∂x (ρ − ρ2) = 0.

The sound speed is now a(ρ) = ddρ

(ρ− ρ2) = 1 − 2ρ. This leads to the critical value

ρcr = 12

such that ⎧⎪⎨⎪⎩

if ρ < ρcr, a(ρ) > 0,

if ρ = ρcr, a(ρ) = 0,

if ρcr < ρ, a(ρ) < 0.

Assume that the initial density is lower than the critical value, ρ0j ≤ ρcr for all j .

In this case we can use the method of characteristics to construct the flux, whichis upwinded to the left. The result is

fnj +1

2

= f(ρn

j

). (2.46)

In contrast, if the initial data is everywhere higher than the critical value, i.e.ρ0

j ≥ ρcr for all j , then application of the method of characteristics yields a fluxupwinded to the right,

fnj +1

2

= f(ρn

j +1

). (2.47)

Notice that, unfortunately, the method of characteristics is not able to predict ina unique and non-ambiguous way a flux for the two cases illustrated in figure 2.18,since a(ρj )a(ρj +1) < 0. This case must be dealt with by other methods such asthe ones presented in section 2.5.4.

Third case: the Lagrangian traffic flow model

Consider the LWR Lagrangian traffic flow equation ∂t τ + ∂mf(τ ) = 0 with fluxf(τ ) = −u(τ ) = τ −1 − 1. Since f ′(τ ) = −τ −2 ≤ 0 is non-positive for all valuesof τ , one can readily apply the method of characteristics for the determination ofthe flux,

fnj +1

2

= f(τ nj +1). (2.48)

This yields the scheme

τ n+1j − τ n

j

∆t+

fnj +1

2

− fnj − 1

2

∆m= 0, j ∈ Z. (2.49)


ρnj

x

t

ρn+1

j

t

xρn

j ρn+1

j

Figure 2.18: The slopes of the characteristic lines have different signs, making theuse of the method of characteristics for the determination of the flux ambiguous.

Properties

The properties of the flux based on the method of characteristics are establishedhereafter for the general scheme (2.44) with the cellwise initialization (2.45). Wedefine

m = infj

(u0

j

)and M = sup

j

(u0

j

). (2.50)

Set

c = maxm≤u≤M

|f ′(u)| . (2.51)

We make the fundamental assumption that

∀u ∈ [m, M ], f ′(u) ≥ 0 (2.52)

and take a numerical flux

fnj +1

2= f

(un

j

)∀j, n. (2.53)


Proposition 2.5.2. Consider the scheme (2.44)–(2.45). Assume (2.52) and take theflux (2.53). Assume that the time is restricted by the CFL constraint c ∆t

∆x≤ 1.

Then the numerical solution satisfies the maximum principle

m ≤ un+1j ≤ M, j ∈ Z, n ∈ N , (2.54)

and is consistent with the entropy condition in the sense that

η(un+1j ) − η(un

j )

∆t+

ξ(unj ) − ξ(un

j −1)

∆x≤ 0 (2.55)

for al l entropy-entropy flux pairs.

Remark 2.5.3. One has similar results if f ′ ≤ 0 and the flux is upwinded to theright. See Corollary 2.5.4.

Proof. Define ν = ∆t∆x

. One has

un+1j = un

j − ν(f(un

j

)− f

(un

j −1

))= un

j − νanj

(un

j − unj −1

)(2.56)

with

anj =

f(un

j

)− f

(un

j −1

)

unj − un

j −1

=

∫ 1

0

f ′ [unj −1 + t(un

j − unj −1)

]dt.

At the first time step (n = 0) the initial data is such that a0j ≥ 0 for all j . Moreover,

the CFL constraint guarantees that νanj ≤ 1 remains true for n = 0. Therefore

un+1j =

(1 − νan

j

)un

j + νanj un

j −1

is a convex combination and the maximum principle (2.54) holds at n = 1. It istherefore true by iteration for all n.

Consider now two real numbers a, b ∈ [m, M ] and define the function

H(z) = η [a − ν (f (a) − f (z))] − η(a) + ν [ξ(a) − ξ(z)] ,

so that proving the entropy inequality is equivalent to proving H(b) ≤ 0. Weobserve that H(a) = 0. One has the series of first-order Taylor expansions

H(b) = (b − a)H ′(z)= (b − a) [νf ′(z)η′ [a − ν (f (a) − f (z))] − νξ′(z)]= (b − a) [νf ′(z)η′ [a − ν (f (a) − f (z))] − νf ′(z)η′ (z)]= (b − a)νf ′ (z) [η′ [a − ν (f (a) − f (z))] − η′(z)]= (b − a)νf ′ (z)η′′ (w) [a − ν (f (a) − f (z)) − z]

= (b − a)νf ′ (z)η′′ (w)

[1 − ν

f (a) − f (z)

z − a

](a − z).

By the CFL condition one has the inequality 1−νf(a)−f(z)

z−a ≥ 0. Therefore H(b) =R(b − a)(z − a) with R ≥ 0 and z between a and b. This shows that H(b) ≤ 0under CFL, completing the proof.


Corollary 2.5.4. Consider the scheme (2.44)–(2.45). Assume f ′(u) ≤ 0 for u ∈[m, M ] and take the flux

fnj +1

2= f

(un

j +1

)∀j, n. (2.57)

Assume that the time is restricted by the CFL constraint c ∆t∆x

≤ 1.Then the numerical solution satisfies the maximum principle (2.54) and is

consistent with the entropy condition for al l entropy-entropy flux pairs,


j )

∆t+

ξ(unj +1) − ξ(un

j )

∆x≤ 0. (2.58)

Proof. Consider (fj)j ∈Z and define g = (gj)j ∈Z with gj = f−j. This reverses thedirection of the upwinding in Proposition 2.5.2, and so yields the results.

Strong control of the oscillations is also guaranteed. This is expressed with thebounded variation (BV) criterion, for which we refer the reader to [131, 94, 175].

Proposition 2.5.5. Consider the hypotheses of Proposition 2.5.2 or Corol lary 2.5.4.Then the numerical solution satisfies the inequalities

∑

j ∈Z

∣∣un+1j +1 − un+1

j

∣∣ ≤∑

j ∈Z

∣∣unj +1 − un

j

∣∣ , n ∈ N , (2.59)

and

∑

0≤n∆t≤T

⎛⎝∆t

∑

j ∈Z

∣∣un+1j +1 − un+1

j

∣∣ + ∆x∑

j ∈Z

∣∣un+1j − un

j

∣∣⎞⎠ (2.60)

≤ T

(1 + c

∆t

∆x

)∑

j ∈Z

∣∣u0j +1 − u0

j

∣∣ .

Proof. The proof is based on the Harten formalism [109]. The recurrence relation(2.56) is rewritten in Harten’s notation as

un+1j = un

j + Cnj

(un

j −1 − unj

), Cn

j =∆t

∆xan

j .

From the maximum principle, one has the uniform bound 0 ≤ Cnj ≤ νc. So

un+1j − un+1

j −1 =(1 − Cn

j

) (un

j − unj −1

)+ Cn

j −1

(un

j −1 − unj −2

).

The control conferred by the maximum principle and the CFL condition yield0 ≤ Cn

j ≤ νc ≤ 1 for all j ∈ Z and all n ∈ N . So

∣∣un+1j − un+1

j −1

∣∣ ≤(1 − Cn

j

) ∣∣unj − un

j −1

∣∣ + Cnj −1

∣∣unj −1 − un

j −2

∣∣ .


After summation over j ∈ Z, one obtains the first inequality (2.59),

∑

j

∣∣un+1j − un+1

j −1

∣∣ ≤∑

j

(1 − Cn

j

) ∣∣unj − un

j −1

∣∣ +∑

j

Cnj −1

∣∣unj −1 − un

j −2

∣∣

=∑

j

∣∣unj − un

j −1

∣∣ .

The left-hand side in the second inequality is the sum of two terms. The first isimmediately controlled by the first inequality, but multiplied by n∆t ≤ T . Thesecond is obtained, after recasting (2.56), as

un+1j − un

j = −Cnj

(un

j − unj −1

),

which allows us to obtain the bound

∑

j ∈Z

∣∣un+1j − un

j

∣∣ ≤ νc∑

j

∣∣unj − un

j −1

∣∣ .

This ends the proof.

2.5.4 Definition of a generic flux

In order to discretize general cases where the sign of the sound speed is arbitrary,we consider

fnj +1

2

=1

2

(f(un

j +1) + f(unj ))

+c

2(un

j − unj +1), j ∈ Z, n ∈ N . (2.61)

This flux is the half-sum of the fluxes on both sides plus a term c2

(unj − un

j +1),which is called a viscous term. This formula is referred to in the literature as theRusanov flux [190].

Proposition 2.5.6. Consider the scheme (2.44)–(2.45). Assume (2.52) and take thegeneric flux (2.61). Assume the parameter c is sufficiently large such that

maxm≤x≤M

|f ′(x)| ≤ c. (2.62)

Assume the CFL constraint c ∆t∆x

≤ 1.

Then the discrete solution satisfies the maximum principle

m ≤ un+1j ≤ M, j ∈ Z, n ∈ N .

Remark 2.5.7. Notice that c does not vanish in (2.62), except in the trivial casewhere f is constant.


Proof. Note that ν = ∆t∆x . One has

un+1j = un

j +νc

2(un

j +1 − 2unj + un

j −1) − ν

2

[f(un

j +1) − f(unj −1)

]

= unj +

νc

2(un

j +1 − 2unj + un

j −1) − ν

2

[an

j (unj +1 − un

j −1)]

.

By construction one has

anj =

f(unj +1) − f(un

j −1)

unj +1 − un

j −1

= f ′ (znj ), min(un

j +1, unj −1) ≤ zn

j ≤ max(unj +1, un

j −1).

Soun+1

j = [1 − νc] unj +

ν

2

[c − an

j

]un

j +1 +ν

2

[c + an

j

]un

j −1. (2.63)

Assume that the coefficients in front of unj , un

j +1 and unj −1 are all non-negative.

Since the sum is equal to 1, un+1j is a convex combination of un

j , unj +1 and un

j −1.The CFL constraint at n = 0 ensures the non-negativity in the first step. Byiteration this property propagates from one time to the next. This proves themaximum principle and ends the proof.

Proposition 2.5.8. Make the assumptions of Proposition 2.5.6. Define the numer-ical entropy flux

ξnj + 1

2

=ξ(un

j +1) + ξ(unj )

2+

c

2(η(un

j ) − η(unj +1)). (2.64)

Then the numerical solution satisfies the discrete entropy inequality


j )

∆t+

ξnj +1

2

− ξnj − 1

2

∆x≤ 0, j ∈ Z, n ∈ N .

Proof. The proof is split into three steps.

Step 1. Rewrite the scheme as

un+1j =

1

2

(un

j + νc(−unj + un

j −1) − ν[f(un

j ) − f(unj −1)

])

+1

2

(un

j + νc(unj +1 − un

j ) − ν[(f(un

j +1) − f(unj )])

.

The convexity of η yields

η(un+1j ) ≤ 1

2η(un

j + νc(−unj + un

j −1) − ν[f(un

j ) − f(unj −1)

])

+1

2η(un

j + νc(unj +1 − un

j ) − ν[f(un

j +1) − f(unj )])

.

So one can write


j ) + ν(

ξnj + 1

2

− ξnj −1

2

)≤ 1

2ϕ(un

j +1) +1

2φ(un

j −1)


where

ϕ(w) = η(un

j + νc(w − unj ) − ν

[f(w) − f(un

j )])

− η(unj ) − νc(η(w) − η(un

j )) + ν[ξ(w) − ξ(un

j )]

and

φ(z) = η(

unj +

νc

2(−un

j + z) − ν

2

[f(un

j ) − f(z)])

− η(unj ) − νc

2(−η(un

j ) + η(z)) +ν

2

[ξ(un

j ) − ξ(z)]

.

So establishing the entropy inequality can be done by proving separately thatϕ(w) ≤ 0 and φ(z) ≤ 0. These inequalities are similar to H(z) ≤ 0 in theproof of Proposition 2.5.2.

Step 2. One has ϕ(unj ) = 0 and

ϕ′(w) = ν (c − f ′(w)) ×(η′ (un


[f(w) − f(un

j )])

− η′(w))

.

The first term is non-negative. The second term is the product of a non-negative term and

(un


[f(w) − f(un

j )])

−w. Using the CFLcondition, this term is itself the product of a non-negative term and un

j − w.Therefore one has

ϕ′(w) = k(w)(unj − w), k(w) ≥ 0 ∀w.

This shows that ϕ(w) ≤ 0.

Step 3. One has φ(unj ) = 0. Differentiation yields

φ′(z) = ν(c + f ′(z)) ×(η′ (un

j + νc(−unj + z) − ν

[f(un

j ) − f(z)])

− η′(z))

.

The first term is non-negative and the second term is the product of unj − z

and a non-negative term under CFL. So one has φ(z) ≤ 0.

The proof is complete.

Proposition 2.5.9. Make the assumptions of Proposition 2.5.6 and impose a morerestrictive CFL condition c ∆t

∆x≤ 1

2. Then the numerical solution satisfies the

inequalities ∑

j ∈Z

∣∣un+1j +1 − un+1

j

∣∣ ≤∑

j ∈Z

∣∣unj +1 − un

j

∣∣ , n ∈ N (2.65)

and∑

0≤n∆t≤T

⎛⎝∆t

∑

j ∈Z

∣∣un+1j +1 − un+1

j

∣∣ + ∆x∑

j ∈Z

∣∣un+1j − un

j

∣∣⎞⎠ (2.66)

≤ T

(1 + 2c

∆t

∆x

)∑

j ∈Z

∣∣u0j +1 − u0

j

∣∣ .


Proof. The recurrence relation (2.63) is rewritten in Harten’s notation as

un+1j = un

j + Cnj

(un

j −1 − unj

)+ Dn

j

(un

j +1 − unj

)

with Cnj = ν

2(c + an

j ) and Dnj = ν

2(c − an

j ). So

un+1j − un+1

j −1 =(1 − Cn

j − Dnj −1

)(un

j − unj −1

)

+ Cnj −1

(un

j −1 − unj −2

)+ Dn

j

(un

j +1 − unj

).

Under the restricted CFL condition of the proposition, one has

0 ≤ min(Cn

j −1, Dnj , 1 − Cn

j − Dnj −1

).

So∑

j

∣∣un+1j − un+1

j −1

∣∣ ≤∑

j

(1 − Cn

j − Dnj −1

) ∣∣unj − un

j −1

∣∣

+∑

j

Cnj −1

∣∣unj −1 − un

j −2

∣∣ +∑

j

Dnj

∣∣unj +1 − un

j

∣∣

=∑

j

∣∣unj − un

j −1

∣∣ .

The second inequality comes from the fact that (2.63) yields∣∣un+1

j − unj

∣∣ ≤ ν

2(c − an

j )∣∣un

j +1 − unj

∣∣+ν

2(c + an

j )∣∣un

j − unj −1

∣∣

≤ c∣∣un

j +1 − unj

∣∣ + c∣∣un

j − unj −1

∣∣ .

This proves the assertions.

2.5.5 Convergence

The question addressed in this section is the convergence of the discrete solutionto a certain limit, as the mesh parameters ∆x and ∆t tend to zero. We splitthe answer to this question into two parts. A first part is the identification ofthe limit. That is, if we assume that the discrete solution admits a limit, can weidentify the equations by the limit? The answer is not trivial, in particular becausenonlinear equations may have multiple solutions. The second part concerns whythe discrete solution admits a limit. For scalar conservation laws, the natural BVbounds of Propositions 2.5.5 and 2.5.9 guarantee the existence of a limit (at leastfor subsequences); see [131, 183, 94, 129, 57, 175, 131]. This the reason why weconcentrate on the properties of the limit. It will be shown in the final chapterthat this strategy is convenient on general grids in higher dimensions.

Define a function u∆x,∆t : R+ × R → R by

u∆x,∆t (t, x) = unj (2.67)

where (j − 12)∆x < x < (j + 1

2)∆x and n∆t ≤ t < (n + 1)∆t. Some of these strict

inequalities can be replaced by non-strict inequalities.


Theorem 2.5.10 (Lax-Wendroff). Consider u∆x,∆t provided by the scheme (2.44)–(2.61) with initial data u0 ∈ L∞(R ). Assume the CFL condition of Proposition2.5.6. Assume there exists u ∈ L1

loc(R+ × R ) such that

lim(∆x,∆t)→0

‖u∆x,∆t − u‖L1loc

(R+×R) = 0

andlim

(∆x,∆t)→0‖u∆x,∆t(0, ·) − u0‖L1

loc( R) = 0.

Assume the parameter c is such that

c∆x → 0 as ∆x → 0. (2.68)

Then u ∈ L∞ (R+ × R ) is an entropy weak solution.

Remark 2.5.11. The condition (2.68) is necessary as shown in Exercise 2.7.4. Takec = maxm≤u≤M |f ′(u)| with −∞ < m = inf u0 and M =

∑u0 < ∞; in this case

the condition is trivially satisfied.

Proof. The proof is performed in two steps.

Step 1. The CFL condition guarantees a uniform bound for unj . So

‖u∆x,∆t‖L∞(R+×R) ≤ ‖u0‖L∞(R).

Take a smooth function ϕ ∈ C20 with compact support and note that ϕn

j =ϕ(n∆t, j∆x). Hence

∑

j,n

(un+1

j − unj

∆t+

fnj +1

2

− fnj − 1

2

∆x

)ϕn

j ∆x∆t = 0,

which comes from a discrete integration of the numerical scheme. Let us nowperform discrete integration by parts of all terms. One has

−∑

j,n

unj

ϕnj − ϕn−1

j

∆t∆x∆t −

∑

j

u0jϕ0

j∆x −∑

j,n

fnj +1

2

ϕnj +1 − ϕn

j

∆x∆x∆t = 0.

The first sum becomes

∑

j,n

unj

ϕnj − ϕn−1

j

∆t∆x∆t =

∫ ∫u∆x,∆t(t, x)w(t, x) dt dx,

where w∆x,∆t(t, x) =ϕn

j −ϕn−1

j

∆t if and only if (j − 12)∆x < x < (j + 1

2 )∆x andn∆t < t < (n + 1)∆t. One has the convergence w∆x,∆t → ∂tϕ uniformly inthe support of ϕ. Therefore

∫ ∫u∆x,∆t(t, x)w(t, x)dt dx →

∫ ∫u(t, x)∂t ϕ(t, x)dt dx.


The second sum is treated similarly, giving∑

j u0jϕ0

j∆x →∫R

u0(x)ϕ(0, x) dx.For the third sum one has

∑

j,n

fnj +1

2

ϕnj +1 − ϕn

j

∆x∆x∆t =

∑

j,n

f(unj )

ϕnj +1 − ϕn

j −1

2∆x∆x∆t

+ c∑

j,n

unj

−ϕnj +1 + 2ϕn

j − ϕnj −1

2∆x∆x∆t.

Define f∆x,∆t(t, x) = f(unj ) for (j − 1

2)∆x < x < (j + 1

2)∆x and n∆t < t <

(n + 1)∆t. Since

|f∆x,∆t (t, x) − f(u(t, x))| ≤ max |f ′| × |u∆x,∆t (t, x) − u(t, x)| ,

the general assumptions imply that f∆x,∆t tends to f(u) in L1loc(R+ ×R ). So

∑

j,n

f(unj )

ϕnj +1 − ϕn

j −1

2∆x∆x∆t →

∫ ∫f(u(t, x))∂x ϕ(t, x)dx dt.

The additional term is

c∑

j,n

unj

−ϕnj +1 + 2ϕn

j − ϕnj −1

2∆x∆x∆t

=c∆x

2

∑

j,n

unj

−ϕnj +1 + 2ϕn

j − ϕnj −1

∆x2∆x∆t.

Since ϕ is twice differentiable with compact support, terms like

−ϕnj +1 + 2ϕn

j − ϕnj −1

∆x2

are uniformly bounded. By hypothesis the coefficient c∆x tends to 0, so theadditional term as a whole tends to 0. This shows that

∑

j,n

fnj +1

2

ϕnj +1 − ϕn

j

∆x∆x∆t →

∫ ∫f(u(t, x))∂x ϕ(t, x)dx dt.

Finally, u satisfies

−∫ ∫

u(t, x)∂t ϕ(t, x)dt dx −∫

R

u0(x)ϕ(0, x) dx

−∫ ∫

f(u(t, x))∂x ϕ(t, x) dx dt = 0.

The above equation holds for ϕ ∈ C20 , and hence, by density, for ϕ ∈ C1

0 .Therefore we have proved that u is a weak solution.


Step 2. It remains to shows that u is an entropy weak solution. Start from

∑

j,n

(η(un+1

j ) − η(unj )

∆t+

ξnj +1

2

− ξnj − 1

2

∆x

)ϕn

j ∆x∆x ≤ 0

with ϕ ∈ C20,+ which is non-negative, ϕ ≥ 0. One can use the same techniques

as of the first step to get

−∫ ∫

η(u(t, x))∂t ϕ(t, x)dt dx −∫

R

η(u0(x))ϕ(0, x) dx

−∫ ∫

ξ(u(t, x))∂x ϕ(t, x)dx dt = 0.

Therefore u is indeed an entropy weak solution.

2.5.6 Scheme optimization

The scheme (2.44) with the flux (2.61) can easily be optimized. To this end weconsider

fnj +1

2

=1

2

(f(un

j +1) + f(unj ))

+cn

j +12

2(un

j − unj +1), (2.69)

where the parameter to optimize is cnj +1

2

.

Optimization of the stability constraint

The general theory shows that the CFL constraint takes the form c ∆t∆x

≤ 1. Theoptimal value of c is defined as the one that maximizes the time step. Define

mn = minj

(unj ) and Mn = max

j(un

j ).

Setcn

j +12

≡ cn = maxmn≤x≤M n

|f ′(x)| . (2.70)

It is easy to show that this formula is compatible with Proposition 2.5.6. So theoptimal value (2.70) is such that

· · · ≤ mn−1 ≤ mn ≤ mn+1 ≤ · · ·

and· · · ≤ Mn+1 ≤ Mn ≤ Mn−1 ≤ · · · .

Therefore

0 ≤ · · · ≤ cn+1 ≤ cn ≤ cn−1 ≤ · · · .

Notice that the technical condition cn∆x → 0 of Theorem 2.5.10 holds.

2.6. More schemes for the traffic flow equation 83

Optimization with respect to the accuracy

It is an extremely delicate task to determine a priori the accuracy of a numericalmethod for nonlinear equations. This is why we focus hereafter on a simple ap-proach based on the consistency of flux. In practice this heuristic approach can beused to study and compare many methods from the vast literature on the topic.

The consistency error of the flux is snj + 1

2

, where

sj + 12

= f(uj + 12) − 1

2(f(uj) + f(uj +1)) −

cj + 12

2(uj − uj +1)

withuj = u(j∆x), uj +1

2= u((j + 1

2)∆x), uj +1 = u((j + 1)∆x).

Assume that u is smooth and that the flux is also a smooth function. A Taylorexpansion with respect to ∆x yields

f(uj + 12) − 1

2(f(uj ) + f(uj +1)) = O(∆x2) and uj − uj +1 = O(∆x).

Sosj +1

2= O(∆x2) + cj + 1

2O(∆x).

The usual interpretation is that the half-sum of the fluxes on both sides is centeredand of second order, whereas the viscous additional part is only of first order.Therefore, for accuracy considerations, the smaller cj + 1

2is the better.

An optimal choice

Comparing the two criteria for optimality, it appears that the smallest cj + 12

thatrespects the stability constraint is the optimal solution. Noting that

mnj +1

2

= minj

(unj , un

j +1) and Mnj + 1

2

= maxj

(unj , un

j +1),

the optimal choice reads

cnj +1

2

= maxmn

j+ 12

≤x≤M n

j+12

|f ′(x)| . (2.71)

It is left to the reader to generalize Proposition 2.5.6, Proposition 2.5.8 and The-orem 2.5.10. This new flux yields a scheme which satisfies the maximum principleand is entropic under the CFL condition.

2.6 More schemes for the traffic flow equation

Two numerical schemes have already been defined for the traffic flow equation.The first scheme (2.48)–(2.49) is Lagrangian and uses a flux which is upwind to


the right. The second scheme is Eulerian,

ρn+1j − ρn

j

∆x+

fnj +1

2

− fnj −1

2

∆x= 0, j ∈ Z,

and uses the general flux formula (2.61) adapted for situations where the sign ofthe slopes of characteristic lines is arbitrary.

Discrete equivalence

Consider the system ∂t (ρJ) = 0,

∂t J − ∂X u = 0,(2.72)

which is equivalent to

∂tτ − ∂mu = 0, dm = ρdx = ρ0dX. (2.73)

An interesting question for the development of numerical methods is to determinewhether different numerical discretizations for (2.72) and for (2.73) can be provedto be equivalent.

We start the discussion from the Lagrangian scheme

τ n+1j − τ n

j

∆t−

unj + 1

2

− unj −1

2

∆mj= 0 (2.74)

with the mass defined by

∆mj = ρ0j∆Xj , ∆Xj = Xj +1

2− Xj − 1

2= ∆x0

j.

A natural definition of the mesh displacement is

xn+1j +1

2

= xnj + 1

2

+ ∆tunj+1

2

, x0j + 1

2

= Xj + 12. (2.75)

The discrete Jacobian of the transformation is

Jnj =

xnj +1

2

− xnj − 1

2

x0j +1

2

− x0j − 1

2

.

Proposition 2.6.1. The Lagrangian scheme in mass variables in the form (2.74)–(2.75) is equivalent to the discretization of the system (2.72),

⎧⎪⎪⎪⎨⎪⎪⎪⎩

ρn+1j Jn+1

j − ρnj Jn

j

∆t= 0,

Jn+1j − Jn

j

∆t−

unj + 1

2

− unj −1

2

∆Xj= 0.

(2.76)


Proof. The mesh displacement (2.75) can be rewritten as

xn+1j +1

2

− xn+1j − 1

2

= xnj + 1

2

− xnj −1

2

+ ∆t(

unj +1

2

− unj − 1

2

).

Division by x0j +1

2

− x0j −1

2

= ∆Xj yields the second equation of (2.76).

Consider (2.74) at the first time step n = 0. Rewrite it as

ρ0j∆x0

jτ 1j − (∆x0

j + ∆t(u0j +1

2

− u0j − 1

2

)) = 0,

that is,ρ0

j

ρ1j

∆x0j − ∆x1

j = 0. This yields the first equation of (2.76). The proof is

completed by induction on n.

Interpretation as a particle discretization

The analogy between Lagrangian discretization and particular methods is straight-forward if one accepts the natural idea that a vehicle or group of vehicles behaveslike a particle or group of particles in kinetic equations. This is illustrated infigure 2.19.

unj+

1

2

= unj

group jvelocity un

j

group j + 1velocity un

j+1

Figure 2.19: Interpretation of the Lagrangian scheme as a particle discretization.

2.6.1 Numerical illustrations

We illustrate our results by applying them to some basic problems.

Eulerian LWR traffic flow

Consider the Eulerian LWR traffic flow equation ∂t ρ + ∂x(ρ − ρ2) = 0 with theinitial data ⎧

⎪⎨⎪⎩

ρ0 = 0.4 for x < 0.3,

ρ0 = 1 for 0.3 < x < 0.7,

ρ0 = 0 for 0.7 < x.

(2.77)

The numerical solution is computed with the scheme (2.44), (2.45) and (2.61). Wetake three meshes of 100, 200 and 400 cells on the interval [0, 1]. The constant ofthe flux is c = 2. The CFL condition is c∆t = 1

10∆x.


The results are displayed in figure 2.20 at the final time t = 0.2. According tothe theory, the solution is made up of a rarefaction for the rightmost part, whichcharacterizes exit from the traffic jam, and a shock for the leftmost part, typicalof entry into a traffic jam. We observe that the numerical solution satisfies theentropy condition.

0 0.4 0.6 0.8 1 0

1

0.8

0.6

0.4

0.2

0.2

100 cells

x

ρ

shock

fan

50 cells

200 cells

Figure 2.20: Entry into and exit from a traffic jam: numerical convergence for 100,200 and 400 cells.

However, it must be noted that wrong solutions show up easily if the discreteentropy inequalities are not satisfied. To illustrate this phenomenon, consider theinitial data

u0 = 1 for 0.4 < x < 0.6, u0 = 0 elsewhere.

Moreover, instead of computing correctly A = max0≤u≤1 |f ′(u)|, which takes thevalue A = 1, we replace A with the lazy prediction

B = maxuj =uj+1

∣∣∣∣f(uj +1) − f(uj )

uj +1 − uj

∣∣∣∣ .

Even worse, we prioritize implementation considerations and use

Bε = max|uj−uj+1|≥ε>0

∣∣∣∣f(uj +1) − f(uj )

uj +1 − uj

∣∣∣∣ ,

where ε > 0 is a purely artificial number, so as to avoid division by zero. Let ustake for example ε = 10−6. If one uses this value, it is easy to check that thenumerical evaluation gives the numerical value Bε = 0.

Let us now assume that one uses the scheme with a CFL constant equal to 1and the numerical parameter in the flux set as c = Bε = 0. With these parametersone obtains the result figure 2.21, where the numerical solution is stable in L∞.

Notice that the right discontinuity is not admissible from the entropy per-spective since 0 = uR < uL = 1. Since the numerical solution is independent of


the mesh parameters, the Lax-Wendroff theorem (Theorem 2.5.10) can be invokedto prove the convergence to a weak solution as ∆x → 0. This weak limit solutionis not an entropy weak solution.

0 0.2 0.4 0.6 0.8 1

1

0.8

0.6

0.4

0.2

0

non-entropic

x

ρ

shock

shockentropic

Figure 2.21: Non-entropic shock for the LWR model after 50 iterations.

Lagrangian LWR traffic flow

A Lagrangian simulation on a mobile mesh is presented in figure 2.22. The initialdata is a slight perturbation of the data (2.77):

⎧⎪⎨⎪⎩

ρ0 = 0.4 for x < 0.3,

ρ0 = 1 for 0.3 < x < 0.7,

ρ0 = 0.001 for 0.7 < x.

The modification is made to allow computation of the specific volume τ = 1ρ

forsmall but non-zero density. This modification has no impact on the time step forthis calculation.

Interpretation of the result is easy considering the visible mesh displacement.Indeed, the cells are stretched in the rarefaction fan and compressed at the shock.This feature can be compared with the Eulerian result in figure 2.20, where no suchmesh modification is possible of course. The numerical solution is quite similar evenif the numerical methods are extremely different. Note, however, the importantdilatation of the mesh near the foot of the fan.

An Eulerian result for the OM traffic flow model

We consider the OM model with initial data⎧⎪⎨⎪⎩

ρ0 = 0.4 for x < 0.3,

ρ0 = 0.9 for 0.3 < x < 0.7,

ρ0 = 0.1 for 0.7 < x.

(2.78)


0 0.4 0.6 0.8 1 1.2 0

1

0.8

0.6

0.4

0.2

0.2 x

ρ

shock

200 cells

rarefaction fan

50 cells

mesh distortion

Figure 2.22: Lagrangian result for the LWR model with 50, 100 an 200 cells. Thisshould be compared with the Eulerian result in figure 2.20.

The flux is continuous but not C1. The parameters are ρ1 = 0.5, ρ2 = 0.6, α = 1and β = 0.5: γ is recomputed as a function of α and β . A typical Eulerian resultis the one shown in figure 2.23.

Comparison with figure 2.20 shows that entry into the traffic jam is qualita-tively the same. However, exit from the traffic jam is very different. We observethat the exit is more sudden, with two intermediate stages at ρ ≈ 0.6 = ρ1 andρ ≈ 0.5 = ρ2 . This type of discontinuity is a contact discontinuity.

0 1 0.6 0.4 0.2 0

1

0.8

0.6

0.4

0.2

0.8

contact discontinuities

x

ρ

α

β

−γ

shock

Figure 2.23: Olmos-Munos model

2.7. Exercises 89

The Buckley-Leverett model

Consider the Buckley-Leverett equation with A = 1. The initial data representspure water on the left and pure oil on the right:

u0(x) = 1 for x < 0.5, u0(x) = 0 for 0.5 < x.

An Eulerian simulation at time t = 0.2 is shown is figure 2.24, from which one can

see a shock attached to a rarefaction fan. The critical value u∗ =√

AA+1

≈ 0.7 is

visible on the graph.

0 0 0.2 0.4 0.6 0.8 1

0.8

0.6

0.4

0.2

1

u

u∗

rarefaction fan

x

shock

Figure 2.24: Buckley-Leverett equation. Propagation of a mixture of oil and water.

Once again an approximate evaluation of max |f ′| may yield incorrect results.To illustrate this phenomenon, we perform additional calculations with parametervalues c = 2 (corresponding to the theory), then c = 1, c = 0.25 and finallyc = 0.1, the last three of which do not correspond to the theory. One sees that thenumerical solution for c = 0.1 is completely wrong as the maximum principle isnot satisfied. Moreover, the wrong solutions are non-entropic.

2.7 Exercises

Exercise 2.7.1. Consider the conservation law ∂tu + ∂xf(u) = 0. Use formula (2.3)to rewrite it as

∂t′ (Ju) + ∂X(f(u) − a(u)u) = 0,

∂t′ J − ∂X a = 0.

Simplify the system and show that ∂t′ u = 0.

Exercise 2.7.2. Generalize Exercise 2.7.1 to the system (1.66).


0 0.6 0.8 1 0.2 0

1.2

1

0.8

0.6

0.4

0.2

0.4

c = 2

incorrect

x

u

correct solution

solutions

c = 0.1

c = 0.25

c = 1

Figure 2.25: Buckley-Leverett equation: non-entropic solutions (c < 2) comparedwith entropic solutions (c = 2).

Exercise 2.7.3. Consider a solution u ∈ C∞ of the Burgers equation

∂tu + ∂x

(u2

2

)= 0.

The initial data is u(0, x) = u0(x) for x ∈ R . Show that

∂(n)t u = (−1)n∂(n)

x

(un+1

n + 1

).

Show the following formal identity due to Wiener (1938) [204]:

u(t, x) =

∞∑

n=0

(−t)n

(n + 1)!∂(n)

x un+10 (x).

Study the convergence of the series for u0(x) = −x (creation of a shock) andu0(x) = x (rarefaction fan).

Exercise 2.7.4. Consider the formula cnj +1

2

= ∆x∆t

for the Eulerian flux (2.61). Show

that it gives the Lax-Friedrichs scheme

un+1j =

unj +1 + un

j −1

2− ∆t

2∆x

(f(un

j +1

)− f

(un

j −1

)).

Verify that the Lax-Friedrichs scheme does not satisfy the technical condition ofTheorem 2.5.10 in cases where the time step tends to zero faster than the meshsize ∆x. For example, one can consider

∆t = O(∆x2).

2.8. Bibliographic notes 91

Rewrite the scheme as

un+1j − un

j

∆t+

f(un

j +1

)− f

(un

j −1

)

2∆x=

(∆x2

2∆t

)un

j +1 − 2unj + un

j −1

∆x2,

and explain the default case of consistency of the Lax-Friedrichs scheme in theregime ∆t = O(∆x2).

Exercise 2.7.5. The Godunov scheme consists of solving exactly the Riemannproblem at the interface between two cells in the time interval ∆t, and then takingthe mean value over cells.

Show that the Godunov scheme for the Burgers equation can be written asa general explicit scheme with the flux fn

j +12

= f(unj + 1

2

), where the state unj + 1

2

is

given by the following procedure:

If unj > un

j +1: Set σ =un

j +unj+1

2. If σ > 0 then un

j +12

= unj . If σ < 0 then un

j +12

=

unj +1.

If unj < un

j +1: If unj ≤ 0 ≤ un

j +1 then unj +1

2

= 0. If 0 ≤ unj ≤ un

j +1 then unj +1

2

= unj .

If unj ≤ un

j +1 ≤ 0 then unj +1

2

= unj +1.

Exercise 2.7.6. Show that the Godunov scheme is entropic with entropy fluxξn

j + 12

= ξ(un

j + 12

).

Exercise 2.7.7. Consider the traffic flow equation with a flux f that is affine withrespect to ρ. It can be one branch of the OM model

f(ρ) = a + bρ, a, b ∈ R .

Let ρ be a discontinuous solution, with a discontinuity velocity equal to σ ∈ R .Compute σ as a function of a and b. Write the Lagrangian equation. Determinethe Lagrangian discontinuity velocity D in terms of a and b.

Exercise 2.7.8. Analyze the numerical solution of the Olmos-Munos model withthe Oleinik entropy condition.


The concepts presented in this chapter are now classical; see [131, 183, 94, 129] andmany other similar references. A slight difference in our presentation lies in thedistinction made between a shock and a contact discontinuity, which is the same aswill be done for systems. Moreover, this enables correct interpretation of the OMmodel [159] and is in accordance with the Oleinik principle. The numerical sectionplaced more emphasis on the design of all-purpose numerical methods with provenentropy properties. In this direction the reader can refer to [160, 187]. The so-calledLax or Lax-Wendroff theorem [126] is a generic result that can be extended to thesystems discussed in forthcoming chapters. Much more on the numerical analysisof conservative schemes can be found in [84].

Chapter 3

Systems and Lagrangian systems

The science must be dogmatic, i.e. it must proveits conclusions strictly a priori from the secureprinciples.

– Immanuel Kant

(Critique of Pure Reason, 1781)

This chapter is devoted to a tentative classification of Lagrangian systems of con-servation laws. Such a definition is of course far from unique. For instance, onemight state that a Lagrangian system comes from continuum mechanics and iswritten in Lagrangian variables. This possible definition, which in some sense cor-responds to the material presented in chapter 1 is not the one employed in thepresent chapter. Instead we will rely on the entropy of the system, since it is anatural notion in continuum mechanics and is also central in the mathematicaltheory of systems of conservation laws. What characterizes entropies for mostsystems written in Lagrangian coordinates is the associated entropy flux, whichis actually zero. This is fundamentally related to the fact that an infinitesimalvolume of matter can be considered as a “closed” system in the theory of closedthermodynamical systems: “closed” means that the system has constant mass and,most importantly, in view of the material developed in this chapter, possesses con-stant entropy if the solution is smooth. For most systems coming from continuummechanics, the mathematical entropy is the mechanical energy, which is oppositeto thermodynamical entropy. The definition of Lagrangian system that we use hereis precisely a system of conservation laws with a zero entropy flux.

This chapter is organized as follows. In the first part we present some basicnotions such as entropy, the Godunov theorem and the entropy variable. Thesecond part is specifically devoted to systems of conservation laws with a zeroentropy flux. We will show that, in conjunction with simple Galilean invarianceprinciples, this framework yields a canonical representation of such systems inone dimension. We then present examples to illustrate the theory. The third partdiscusses a possible extension to multidimensional Lagrangian systems.




93

94 Chapter 3. Systems and Lagrangian systems

3.1 Generalities

The notion of convexity that we use for the analysis of entropies for systems ofconservation laws is actually pointwise strict convexity. Less stringent definitionsare possible [57, 175, 93].

Definition 3.1.1 (Strict convexity). Let η be a function defined on an open subsetof Rn in R , with continuous second derivatives. The function η is said to be strictlyconvex at U ∈ Rn if and only if

∇2η(U) > 0 (3.1)

in the sense of symmetric matrices. By continuity, ∇2η ≥ α > 0 in a neighborhoodof U .

The condition (3.1) means that (X, ∇2η(U)X) > 0 for all non-zero vectorsX ∈ Rn. It is equivalent to saying that all eigenvalues of the symmetric matrix arepositive. For a function η of class C2, this notion is equivalent to local α-convexity,that is,

η(θa + (1 − θ)b) − θη(a) − (1 − θ)η(b) ≤ − α

2θ(1 − θ)|a − b|2 (3.2)

with a, b in a neighborhood of U and α > 0 which depends on the neighborhood.Consider now smooth solutions of the shallow water equations and let us

show the existence of an entropy for this system.

Proposition 3.1.2. Let (h, u) be a solution of the shal low water system (1.11) ofclass C1 in space-time. Then the fol lowing additional conservation law is satisfied:

∂t

(gh2 + hu2

)+ ∂x

((2gh2 + hu2

)u)

= 0. (3.3)

Moreover, for g, h > 0, the function η(h, hu) = gh2 + hu2 is strictly convex withrespect to h and hu.

Proof. To obtain this relation we rewrite (1.11) as⎧⎨⎩

(∂t h + u∂xh) + h∂xu = 0,

h (∂tu + u∂xu) + ∂x

(g

2h2)

= 0

with aid of the identity ∂thu+∂x hu2 = h (∂tu + u∂x u). Multiply the first equationby gh and the second by 2u to obtain

gh(∂t h + u∂xh) + gh2∂xu = 0,

h(∂tu

2 + u∂x u2)

+ u∂x

(gh2

)= 0.

Use h(∂tu

2 + u∂xu2)

= ∂t(hu2)+ ∂x (hu3) and gh(∂t h + u∂xh) = ∂tgh2 + ∂xguh2,and then add; the sum yields (3.3). Notice that the fact that the solution (h, hu)is C1 with h > 0 is crucial.

3.1. Generalities 95

Define η(a, b) = ga2 + b2

a such that η(h, hu) = gh2 + hu2. One has

∇2η = 2

⎛⎜⎝

g +b2

a3− b

a2

− b

a2

1

a

⎞⎟⎠ .

This is a symmetric matrix with positive trace and positive determinant D = 2ga

>0. So ∇2η > 0, which completes the proof of the strict convexity.

One says that η is an entropy for the shallow water equations. The Eulersystem of compressible gas dynamics satisfies a similar property.

Proposition 3.1.3. Smooth solutions of the Euler equations⎧⎪⎨⎪⎩

∂t ρ + ∂x(ρu) = 0,

∂t (ρu) + ∂x

(ρu2 + p

)= 0,

∂t (ρe) + ∂x(ρue + pu) = 0

(3.4)

with a perfect gas pressure law satisfy an additional conservation law

∂t(ρS) + ∂x(ρuS) = 0, where S = log ετ γ−1 with ε = e − 1

2u2. (3.5)

Proof. The proof is given for ρ > 0. Rewrite the Eulerian system as⎧⎪⎨⎪⎩

ρDt τ − ∂xu = 0,

ρDt u + ∂x p = 0,

ρDt e + ∂xpu = 0,

with Dt = ∂t+u∂x the material derivative. Let (τ, u, e) be a C1 space-time solutionwith a perfect gas pressure law p = (γ − 1)ρε = (γ − 1) ε

τ. One can easily verify

the differential identity

TdS = dε + p dτ, T = ε, S = log(ετ γ−1). (3.6)

Notice that S is, physically speaking, a thermodynamical entropy per unit massfor a polytropic gas. The differential identity is clearly related to the fundamentalprinciple of thermodynamics. The function T is the temperature. Assume T > 0which is physically relevant. One also has

TdS = de − u du + p dτ. (3.7)

This yields for smooth functions TDtS = Dte − uDtu + pDtτ = 0, which turnsinto

DtS = 0. (3.8)

This expression implies ∂t (ρS) + ∂x (ρuS) = ρDt S + S (∂t ρ + ∂x(ρu)) = 0. Theproof is thus complete.


A consequence of the following proposition is that the function

(ρ, ρu, ρe) → η(ρ, ρu, ρe) ≡ −ρS(ρ, ρu, ρe) (3.9)

is strictly convex for ρ > 0 and ε > 0.

Proposition 3.1.4. Let the function Z := (τ, ε) → S(Z) be of class C2, i.e. ∇2ZS <

0, with ∂εS = 1T

> 0. Then the functions η1, given by

W = (τ, u, e) → η1(W ) = −S

(τ, e − 1

2u2

), (3.10)

and η2, given by

U = (ρ, ρu, ρe) → η2(U) = −ρS

(τ, e − 1

2u2

), (3.11)

are strictly convex.

Remark 3.1.5. Since the function −S defined in (3.6) is evidently strictly convex,this implies that the function (3.9) is a strictly convex entropy for the system ofcompressible gas dynamics.

Proof. The verification of this property can be quite cumbersome. In our case werely on the characterization (3.2).

Step 1: comparison of η1 and η2. The α-convexity of η2 reads

η2(θU1 +(1−θ)U2) ≤ θη2(U1)+(1−θ)η2(U2)−αθ(1−θ)|U1−U2|2 , θ ∈ [0, 1],(3.12)

with α > 0, U1 =

(ρ1

ρ1u1

ρ1e1

)and U2 =

(ρ2

ρ2u2

ρ2e2

). Observe the equivalence

⎧⎪⎨⎪⎩

ρ = θρ1 + (1 − θ)ρ2 ,

ρu = θ(ρ1u1) + (1 − θ)(ρ2u2),

ρe = θ(ρ1e1) + (1 − θ)(ρ2 e2)

⇐⇒

⎧⎪⎨⎪⎩

τ = μτ1 + (1 − μ)τ2 ,

u = μu1 + (1 − μ)u2,

e = μe1 + (1 − μ)e2

(3.13)

with τ = ρ−1 and μ =θρ1

θρ1+(1−θ)ρ2∈ [0, 1]. Divide (3.12) by ρ = θρ1+(1−θ)ρ2

to get

η1(μW1 + (1 − μ)W2) ≤ μη1(W1) + (1 − μ)η1(W2) − α

ρθ(1 − θ)|U1 − U2 |2,

for θ ∈ [0, 1]. However, αθ(1 − θ)|U1 − U2|2 ≥ βμ(1 − μ)|W1 − W2 |2 ≥γθ(1 − θ)|U1 − U2 |2, where β > 0 and γ > 0 are conveniently chosen. Thetransformation U → W is a diffeomorphism. It shows that

c1|U1 − U2 |2 ≥ |W1 − W2 |2 ≥ c2|U1 − U2|2 , (3.14)

where c1 > 0 and c2 > 0 are conveniently chosen. Therefore the equivalencebetween (3.10) and (3.11) is established.


Step 2: comparison of S and η1 . Set Z =

(τε

), Z1 =

(τ1

ε1

)and Z2 =

(τ2

ε2

),

where (3.13) is used to obtain the expressions ε = e − 12u2, ε1 = e1 − 1

2u2

1

and ε2 = e2 − 12u2

2 . One has

ε = μ

(ε1 +

1

2u2

1

)+ (1 − μ)

(ε2 +

1

2u2

2

)− 1

2(μu1 + (1 − μ)u2)

2

= με1 + (1 − μ)ε2 +μ(1 − μ)

2(u1 − u2)2.

So

η1(μW1 + (1 − μ)W2) = −S

(τ, με1 + (1 − μ)ε2 +

μ(1 − μ)

2(u1 − u2)2

).

By hypothesis S is strictly increasing with respect to its second variable.Therefore, for c3 > 0,

η1(μW1 + (1 − μ)W2) ≤ −S (τ, με1 + (1 − μ)ε2 ) −[c3

μ(1 − μ)

2(u1 − u2)2

].

Moreover,

S (τ, με1 + (1 − μ)ε2) ≥ μS(τ1 , ε1) + (1 − μ)S(τ2 , ε2)

+ c4μ(1 − μ)(

|τ1 − τ2|2 + |ε1 − ε2 |2)

, c4 > 0.

This yields, for c5 > 0,

η1(μW1 + (1 − μ)W2) ≤ μη1(W1) + (1 − μ)η1(W2)

− c5μ(1 − μ)(

|τ1 − τ2|2 + |ε1 − ε2 |2 + |u1 − u2|2)

.

Going back to (3.14), we see that this establishes the strict convexity of η1.

This finishes the proof.

3.1.1 The Godunov theorem

As suggested by the previous two examples, it is possible that for systems like

∂t U + ∂x f(U) = 0, U, f(U ) ∈ Rn, (3.15)

a certain nonlinear function of the unknown satisfies an additional conservationlaw for all C1 solutions. This additional equation is

∂tη(U) + ∂xξ(U) = 0, η(U), ξ(U ) ∈ R . (3.16)

The Godunov theorem states that the mere existence of this additional conserva-tion law is sufficient to prove the hyperbolicity of the system (3.15). The objective


of this section is to give a proof of the Godunov theorem based on the entropyvariable, which is also introduced for further use.

We will need some notation and assumptions. The functions f, η, ξ are as-sumed to be of class C2. Vectors are written in columns such as

U =

⎛⎜⎜⎝

U1

U2

...Un

⎞⎟⎟⎠ , f(U) =

⎛⎜⎜⎝

f1(U)f2(U)

...fn(U)

⎞⎟⎟⎠ .

The gradient of a vector with respect to the variable U is written as ∇U =(∂U1 , . . . , ∂Un ), that is, a horizontal vector. With this notation, one has

∇η(U) = (∂1η, . . . , ∂nη), ∂i = ∂Ui (1 ≤ i ≤ n).

The transpose of a vertical vector is a horizontal vector, and vice versa. With thisconvention, the gradient or Jacobian matrix of the flux is a square matrix

∇f(U) =

⎛⎝

∂1f1 . . . ∂nf1

.... ..

...∂1fn . . . ∂nfn

⎞⎠ ∈ R

n×n.

Definition 3.1.6 (Entropy of a system). The pair (η(U), ξ(U)) is an entropy-entropyflux pair for the system of conservation laws (3.16) if and only if the functionssatisfy the following two properties:

(a) The function η is strictly convex.

(b) The compatibility relation between first derivatives holds:

∇η(U)∇f(U) = ∇ξ(U). (3.17)

Note that it is possible to relax the hypothesis of strict convexity. The fullexpansion of the second condition b) reads

n∑

1≤j

∂jη(U )∂ifj(U) = ∂iξ(U), 1 ≤ i ≤ n.

Proposition 3.1.7. Assuming the compatibility relation (3.17) between gradients, al lC1 solutions of the system of conservation laws (3.15) satisfy the entropy relation(3.16).

Proof. Solutions of class C1 of (3.16) are also solutions of the quasi-linear system

∂tU + ∇f(U)∂x U = 0.


Take the scalar product with ∇Uη(U ) and get

∂tη(U) + ∇η(U)∇f(U)∂x U = 0.

This expression makes sense since ∇η(U) is a horizontal vector of length n, ∇f(U)is a square matrix of size n×n and ∂x U is a vertical vector of length n: the result isscalar. Using the compatibility relation (3.17), one gets ∂t η(U) + ∇ξ(U)∂x U = 0.The proof is complete.

One might wonder how to find the entropy-entropy flux pair for a givensystem of conservation laws. Most of the time, the physical intuition will be anefficient guide in the context of continuum mechanics.

Definition 3.1.8 (Entropy variable). Let (η(U), ξ (U)) be an entropy-entropy fluxpair for the system of conservation laws (3.16). The column vector

V = (∇Uη(U))t ∈ R

n

will be called the entropy variable.

Proposition 3.1.9. The transformation U → V is a diffeomorphism. If the entropyis α-convex, the transformation is global ly invertible.

Proof. It is sufficient to note that ∇UV = ∇2η(U) = ∇UV t > 0 is a non-singularmatrix to prove that the transformation is locally invertible. The global propertiesof the transformation are shown as follows.

Injectivity: The transformation is injective from U−1(K) in all convex sets K ⊂Rn . Indeed, integration over straight lines shows that

V (U1) − V (U2) =

∫ 1

t=0[∇UV (U1 + t(U2 − U1))] U2 − U1)dt.

So

(U1 − U2 , V (U1) − V (U2)) =

∫ 1

t=0(U1 − U2, [∇UV (U1 + t(U2 − U1))] U2 − U1)dt.

Therefore V (U1) − V (U2) = 0 yields

(0 = U1 − U2 , [∇UV (U1 + t(U2 − U1))] U2 − U1) ≥ c‖U1 − U2‖2, c > 0,

that is, U1 = U2 .

Surjectivity: By hypothesis the entropy is α-convex in Rn. Let λ ∈ Rn . So thefunction U → ηλ(U) ≡ η(U) − (λ, U) is α-convex over Rn . It admits a uniqueminimum which is a solution of ∇Uηλ(U) = V (U) − λ = 0.

Proposition 3.1.10. Consider the polar transform of the entropy, V → η∗(V ) =(U(V ), V ) − η(U(V )), and the polar transform of the entropy flux, V → ξ∗(V ) =(f(U (V )), V ) − ξ(U(V )).

Then (∇η∗)t = U and (∇ξ∗)t = f(U) for al l V .


One has to take care that the gradients are taken with respect to the entropyvariable, that is, ∇η∗ = ∇V η∗ and ∇ξ∗ = ∇V ξ∗.

Proof. By construction, ∇UV is the Hessian matrix of the entropy η, which isstrictly convex. It is a symmetric matrix, ∇V U = (∇V U)t. Furthermore, one hasthe chain rule formula ∇V (Ψ (U (V ))) = ∇UΨ(U(V ))∇V U where Ψ : Rn → Rn

is a given smooth vector field. One also has, for all vector fields A, B : Rn → Rn,∇V (A, B) = At∇V B + B t∇V A. This yields

∇V η∗ = U t∇V V + V t∇V U − ∇Uη(U)∇V U = U t

and

∇V ξ∗ = f(U)t∇V V + V t∇V f(U(V )) − ∇Uξ(U)∇V U

= f(U)t + V t∇V f(U) − ∇Uη(U)∇U f(U)∇V U = f(U)t .

The proof is finished.

Moreover, the Hessian matrices are inverse to one another:

∇2V η∗(V ) × ∇2

Uη(U(V )) = ∇V U t × ∇UV t = (∇UV × ∇V U)t

= It = I.

Godunov made the extremely important remark that the sole existence of anentropy-entropy flux pair guarantees linear stability, that is, stability around con-stant states, as stated in Definition 1.4.1. This is of paramount importance, sincemany systems coming from physics are endowed with such an entropy-entropy fluxpair. This principle establishes a connection between the existence of an entropy-entropy flux pair which comes from the underlying physical context and the math-ematical well-posedness of the linearized system.

Theorem 3.1.11 (Godunov theorem). Assume that the system of conservation laws(3.16) is endowed with an entropy-entropy flux pair. Then it is hyperbolic.

Proof. The proof is particularly simple with the entropy variable. Indeed, the chainrule yields ∇f(U) = ∇V f(U) ∇UV . So one can rewrite the Jacobian matrix ofthe flux as the product of two matrices in the form ∇Uf(U) = BC−1 where

B = ∇2V ξ∗(V ) = B t and C = ∇2

V η∗(V ) = ∇V U = Ct > 0. (3.18)

The eigenvectors of the Jacobian matrix ∇Uf(U)r = λr give the eigenproblem

BC−1r = λr ⇐⇒ Bs = λCs, s = C−1r.

Since the matrix B is symmetric and the matrix C is symmetric positive, this eigen-problem admits a complete set of real eigenvectors and real eigenvalues. Thereforethe linearized problem is well posed, which is the criterion (Definition 1.4.5) ofhyperbolicity.


3.1.2 Entropy weak solutions

Consider the system of conservation laws

∂t U + ∂x f(U) = 0, U, f(U ) ∈ Rn, (3.19)

endowed with an entropy-entropy flux pair. Assume that the initial data is

U(0, x) = U0(x), x ∈ R . (3.20)

Consider the space-time domain Ω defined by

Ω = (−A, A) × [0, T ), T, A > 0.

If U ∈ C1(Ω) and satisfies the initial condition (3.20), then it is called a strongsolution of the system (3.19) in Ω.

It is possible to relax the smoothness requirement; for example, U ∈ C1(Ω)and piecewise C1 is enough.

Definition 3.1.12 (Weak entropic solution). The bounded function U ∈ L∞(Ω) isa weak solution of (3.19) with initial data (3.20) if and only if

∫

R

∫

0<t

((U, ∂t ϕ) + (f(U ), ∂x ϕ)) dx dt +

∫

R

(U0 (x), ϕ(0, x)) dx = 0 (3.21)

for all functions ϕ ∈ (C10(Ω))n .

Of course, as is the case for scalar conservation laws, a strong solution is alsoa weak solution, and a smooth weak solution is also a strong solution. Next, weanalyze limits of viscous strong solutions and show that they correspond to theentropy weak solutions. However, a difficulty arises, which is due to the fact thatthere may exist many types of viscous solutions for systems. Indeed, a viscoussolution depends on a viscous tensor or matrix, and, as is usual for systems, thestructure of the viscous matrix may influence the limit (if it exists). Fortunatelyfor the simplicity of the theory, some properties of the limit are independent ofthe choice of viscous tensor.

This is why we start from a general definition of a viscous solution, i.e. witha generic viscous tensor proportional to the identity matrix. We will study morephysical viscous tensors at the end of the section. Consider Uε , a sequence ofsmooth C2(Ω) strong solutions of the system with evanescent viscosity

∂tUε + ∂xf(Uε ) = ε∂xx Uε, ε → 0+. (3.22)

Rewrite this as ∂tUε + (∇Uf(Uε)) ∂xUε = ε∂xx Uε , ε → 0+. One is interested inthe calculation of the variation of entropy, ∂tη(Uε ). Take the scalar product with∇Uη(Uε ) to get

∂tη(Uε ) + ∂x ξ(Uε) = ε((∇Uη(Uε )) , ∂xx Uε),


that is,

∂tη(Uε) + ∂xξ(Uε ) = ε∂xx η(Uε ) − ε(∂xUε ,

(∇2

Uη(Uε ))

∂xUε

)

and∂tη(Uε ) + ∂x ξ(Uε) ≤ ε∂xx η(Uε ).

Notice that the compatibility relation (3.17) is used to obtain ∂x ξ(Uε), and theconvexity of the entropy is used to derive the inequality

(∂x Uε, (∇2

Uη)(Uε )∂x Uε

)≥

0. It remains to multiplythrough by a smooth non-negative function with compactsupport ϕ ∈ C1

0,+ and integrate by parts as in (2.21). Standard assumptions (2.19)–(2.20) are used to formalize the fact that Uε tends to the limit U .

Definition 3.1.13 (Entropy weak solutions). Let U be a weak solution as in (3.21).One says it is an entropy weak solution if and only if

−∫

R

∫

0<t

(η(U)∂t ϕ + ξ(U)∂x ϕ)dx dt −∫

R

η(U0)(x)ϕ(0, x)dx ≤ 0 ∀ϕ ∈ C10,+.

(3.23)

As mentioned earlier, the abstract viscosity tensor (3.22) is mathematicallyconvenient but has no physical basis. That is why it is necessary to complementthe previous analysis of viscous solutions by at least one example which showsthe generality of Definition 3.1.6. The example is based on the compressible Eulersystem with viscosity and thermic dissipation:

⎧⎪⎨⎪⎩

∂tρν,κ + ∂x (ρν,κuν,κ) = 0,

∂t(ρν,κuν,κ) + ∂x

(ρν,κu2

ν,κ + pν,κ

)= ν∂xxuν,κ,

∂t(ρν,κeν,κ) + ∂x(ρν,κuν,κeν,κ + pν,κuν,κ) = ν∂x(uν,κ∂x uν,κ) + κ∂xxTν,κ .(3.24)

The term ν∂xxuν,κ is the viscosity scaled by a viscous parameter ν > 0. Note thatthe viscosity has a counterpart in the right-hand side of the energy equation. Thethermic dissipation ∂xx Tν,κ is scaled by κ > 0 and shows up in just one equation.Both terms are evanescent for ν, κ → 0+. Rewrite the system in terms of thematerial derivative Dt = ∂t + uν∂x as

⎧⎪⎨⎪⎩

ρνDtτν,κ − ∂xuν,κ = 0, τν,κ = ρ−1ν,κ,

ρν,κDtuν,κ + ∂xpν,κ = ν∂xxuν,κ,

ρν,κDteν,κ + ∂x(pν,κuν,κ) = ν∂x(uν,κ∂xuν,κ) + κ∂xx Tν,κ.

Let us assume for simplicity a perfect gas pressure law for which the entropy is(3.6). This yields

ρν,κTν,κDtSν,κ = ρν,κpν,κDtτν,κ − ρν,κuν,κDt uν,κ + ρν,κDteν,κ

= pν,κ∂xuν,κ − uν,κ (ν∂xx uν,κ − ∂xpν,κ)

+ ν∂x(uν,κ∂xuν,κ) − ∂x (pν,κuν,κ)

= ν (∂xuν,κ)2

+ κ∂xxTν,κ.

3.2. Lagrangian systems in dimension d = 1 103

One obtains, after dividing by the temperature and rearranging the right-handside,

ρν,κDt Sν,κ =

(ν

Tν,κ(∂xuν,κ)

2+ κ

∂xT 2ν,κ

T 2ν,κ

)+ κ∂xx ln Tν,κ, (3.25)

which can be rewritten in Eulerian form as

∂t (ρν,κSν,κ) + ∂x (ρν,κuν,κSν,κ) ≥ κ∂xx ln Tν,κ .

Noting that κ∂xx ln Tν,κ tends to zero in the sense of distributions, after passing tothe limit (ν, κ) → 0 one obtains ∂t (ρS)+∂x (ρuS) ≥ 0 in the sense of distributions.This is exactly (3.23), upon taking η = −ρS and ξ = −ρuS. We have thereforeestablished for this particular example that the weak limit with physically baseddissipation is also an entropic weak limit in the sense of the definition.

3.2 Lagrangian systems in dimension d = 1

In this section we come to the heart of the matter, that is, the structure of La-grangian systems of conservation laws.

Some features are common to both Eulerian and Lagrangian systems of con-servation laws in continuum mechanics. For example, both satisfy Galilean invari-ance principles; both obey the general theory of hyperbolic systems. In particular,discontinuous solutions such as shocks or contact discontinuities must be includedin the theory. There is nevertheless an important property which is different inEulerian or Lagrangian coordinates, and it is related to the entropy law. Indeed,many of the systems of conservation laws in continuum mechanics are endowedwith a density equation and an entropy equation (for smooth solutions) which arevery similar:

∂tρ + ∂x(ρu) = 0 and ∂t(ρS) + ∂x(ρuS) = 0.

In Lagrangian coordinates, i.e. with a mass variable in one dimension, a systemsuch as

∂tU + ∂mf(U ) = 0 (3.26)

is often endowed with a physical entropy relation

∂tS = 0 with a zero entropy flux.

That is, the mathematical entropy is η(U) = −S and the entropy flux is ξ(U) = 0.Most of the questions addressed in this work are related to this fact. This propertyis, when considering examples, so general that it can even be stated as a definition.

Hypothesis 3.2.1 (Lagrangian systems and zero entropy flux). One property ofLagrangian systems of conservation laws is the zero entropy flux: ξ(U) ≡ 0. Goingback to Definition 3.1.6, it is characterized by

∇η(U)∇f(U) = 0 ∀U. (3.27)


This relation has a very strong physical meaning if one considers that thephysical entropy measures some degree of irreversibility in the physical processsustained by the matter. Irreversible processes correspond to a strict increase of thelocal entropy, whereas the entropy is constant for reversible processes. In the caseof compressible gas dynamics, one can compare (3.5), which expresses reversibilityfor smooth solutions, to (3.25), which expresses the possibility of an irreversibleprocess.

t

x

t1

t2

t3

t4

Figure 3.1: Irreversible process: S1 < S2 < S3 < S4 . Reversible process has S1 =S2 = S3 = S4.

3.2.1 Systems with a zero entropy flux

We examine some consequences of the hypothesis of a zero entropy flux.

Proposition 3.2.2. Assume that the entropy flux is zero (3.27). Then the flux f(U)can be expressed as an homogeneous function of degree zero with respect to theentropy variable V .

Proof. The definition of a homogeneous function g of degree p is that g(λx) =λpg(x) for all λ ∈ R . If g is moreover differentiable, one has the following equivalentcharacterization in terms of the Euler relation,

∇xg(x) x = pg(x). (3.28)

To prove the claim, note that V = ∇η(U)t = −∇S(U)t. Using the chain rule∇Uf(U(V )) = ∇V f(U(V )) ∇UV , the relation (3.27) is equivalent to

∇Uη(U) ∇V f(U(V )) ∇UV = 0,

or∇Uη(U) ∇V f(U(V )) = 0


after multiplying through by the inverse of the non-singular matrix ∇UV . Trans-position yields

[∇V f(U(V ))]tV = 0.

But f(U(V )) = ∇V ξ∗(V ) where ξ∗ = (V, f(U)) − ξ(U) = (V, f(U )) is the polartransform of the entropy flux, so the matrix ∇V (f(U(V )) is symmetric. One thusobtains the Euler relation

[∇V f(U (V ))] V = 0.

Therefore the flux f is homogeneous of degree p = 0 with respect to V . The proofis complete.

One can make the additional assumption that the last component of theentropy vector V is non-zero, that is, Vn = 0. It will be shown later that this is ageneric assumption. In this case one can express the flux as f(U) = g(Ψ), whereg : Rn−1 → Rn is some function of the variable

Ψ =

(V1

Vn,

V2

Vn, . . . ,

Vn−1

Vn

)t

∈ Rn−1. (3.29)

To be even more specific, let us consider the system of Lagrangian gas dynamics⎧⎪⎨⎪⎩

∂t τ − ∂mu = 0,

∂t u + ∂mp = 0,

∂t e + ∂m(pu) = 0,

(3.30)

for which the differential relation (3.7) shows that V = − 1T

(p, −u, 1)t. The negativesign is present because η = −S. Upon substituting this into (3.29) one gets Ψ =(p, −u)t . The flux is a function of two independent variables, which are the pressurep and the opposite of the velocity −u. Keep in mind that the pressure law is notnecessarily a perfect gas. Any pressure law or equation of state is admissible,provided it is compatible with the zero entropy flux property.

Remark 3.2.3. Inspection of (3.30) reveals that the structure of the flux is ex-tremely simple, in the sense that it is only a linear-quadratic function of Ψ =(p, −u)t .

In order to show that the linear-quadratic structure of the flux is a generalproperty, one can rely on certain Galilean invariance properties. To do so, weassume that the main unknown U can be decomposed into three parts as

U =

⎛⎜⎝

v ∈ Rn−d−1

u ∈ Rd

e ∈ R

⎞⎟⎠ ∈ Rn, (3.31)

where the vector v groups density-like variables, u groups velocity-like variableand e is the total energy. The internal energy is

ε = e − 1

2|u|2.


Notice that the necessity of having a vectorial velocity variable is easily understoodby considering the one-dimensional system

⎧⎪⎪⎪⎨⎪⎪⎪⎩

∂t τ − ∂mu = 0,

∂t u + ∂mp = 0,

∂t v = 0,

∂t e + ∂m(pu) = 0,

(3.32)

for which u = (u, v).

Hypothesis 3.2.4 (Lagrangian Galilean invariance). Let U be any smooth solutionof the system of conservation laws (3.26) having the structure (3.31). Galileaninvariance means that for all u0 ∈ Rd,

Uu0 =

(v, u + u0, ε +

1

2|u + u0|2

)t

is also a smooth solution of (3.26).

Hypothesis 3.2.5 (Lagrangian reversibility). Let U be a smooth solution of thesystem of conservation laws (3.26) having the structure (3.31). Reversibility meansthat U− = (v, −u, e)t is also a solution of

−∂tU− + ∂mf(−U−) = 0.

Hypothesis 3.2.6 (Entropy as a function of v and ε). The physical entropy Scan be written as a function of the variables v and ε, and so the mathematicalentropy can be written as η(v, u, e) = −S

(v, e − 1

2u2

). Moreover, we assume that

Vn < 0, which generalizes the case of compressible gas dynamics, since in this caseVn = − 1

T .

Definition 3.2.7 (Precise definition of Ψ). Assume Hypothesis 3.2.6 and define

w(v, ε) =∇vS(v, ε)

∇εS(v, ε). (3.33)

This yields the following representation of the reduced entropy variable (3.29):

Ψ = (w(v, ε), −u)t ∈ Rn−1.

Theorem 3.2.8 (Representation formula of the flux). Consider the system of con-servation laws (3.26) and assume Hypotheses 3.2.1, 3.2.4, 3.2.5 and 3.2.6.

Then the flux of the system is linear-quadratic in Ψ. More precisely, thereexists a symmetric matrix M = M t ∈ Rn−1×n−1 such that

f(U) =

⎛⎝

MΨ

− 1

2(Ψ, MΨ)

⎞⎠ . (3.34)


Moreover, there exists a matrix N ∈ Rn−1−d×d such that

M =

(0 N

N t 0

). (3.35)

Remark 3.2.9. The internal structure (3.35) in the representation formula will berelaxed in certain cases where Hypotheses 3.2.1, 3.2.4, 3.2.5 and 3.2.6 need somemodifications.

Remark 3.2.10. The Lagrangian system of gas dynamics (3.30) corresponds to

M =

(0 11 0

)and Ψ =

(p

−u

),

so N = 1. Therefore the theorem means that many Lagrangian systems comingfrom continuum mechanics have the same structure.

Proof of the representation theorem 3.2.8. The proof proceeds in four steps.

Step 1. Using Hypothesis 3.2.6, we begin by writing f(U) =

(h(Ψ) ∈ Rn−1

g(Ψ) ∈ R

)

where Ψ = (w(v, ε), −u)t ∈ Rn−1. So the first n − 1 equations of the systemof conservation laws are

∂t

(vu

)+ [∇Ψh(w, −u)] ∂m

(w

−u

)= 0.

Step 2. Hypothesis 3.2.4 allows us to write

∂t

(v

u + u0

)+ [∇Ψh(w, −u − u0)] ∂m

(w

−u − u0

)= 0 ∀u0 ∈ R

d.

That is,

∂t

(vu

)+ [∇Ψh(w, −u − u0)] ∂m

(w

−u

)= 0 ∀u0 ∈ R

d.

So the matrix [∇Ψh(w, −u − u0)] is independent of u0:

∇u|w [∇Ψh(w, −u)] = ∇Ψ

[∇u|wh(w, −u)

]= 0.

Therefore the matrix ∇u|wh(w, −u) is independent of Ψ: one can write[∇u|wh(w, −u)

]= B for a given matrix B ∈ Rn−1×d. Integration yields

h(w, −u) = −Bu + l(w) where the function w → l(w) is unknown at thisstage of the analysis. The technique of separation of variables yields

h(w, −u) =

(−B1u + l1(w) ∈ Rn−1−d

−B2u + l2(w) ∈ Rd

).


Step 3. Hypothesis 3.2.5 implies

±∂t

(vu

)+ ∂m

(∓B1u + l1(w)∓B2u + l2(w)

)= 0.

Therefore l1 ≡ 0 and B2 ≡ 0. More precisely, ∂ml1 = 0 so one can eliminatel1. A similar relation holds for B2. So one can write

f(U) =

( −B1ul2(w)g(Ψ)

).

Step 4. The polar transform of the entropy flux is

ξ∗(V ) = (V, f(U )) = Vn [−(w, B1u) − (u, l2(w)) + g(w, −u)] .

One has that ∇V ξ∗ = f(U). Let us differentiate with respect to

(V1, . . . , Vn−1−d) = Vnw

and then with respect to (Vn−d, . . . , Vn−1) = −Vnu. This gives

−B1u − ∇wl2(w)tu + ∇wg(w, −u) = −B1u,

B t1w + l2(w) + ∇−ug(w, −u) = l2(w).

So ∇wg(w, −u) = (∇wl2(w))t u,

∇−ug(w, −u) = −B t1w.

The Maxwell relation for cross-derivatives yields

∇−u

(∇wl2(w)t u

)=

(∇w

(−B t

1w))t

,

i.e. ∇wl2(w) = B t1, whose solution is l2(w) = B t

1w. Furthermore, g(w, −u) =(u, B t

1w). One can characterize this result for the flux using the notation

N = B1:

f(U) =

(MΨ

− 1

2(Ψ, MΨ)

)with M =

(0 N

N t 0

). (3.36)


Next we study a consequence of the structure (3.34)–(3.35) on the eigen-structure of the Jacobian matrix. Consider the Lagrangian system with the flux(3.36): ⎧

⎪⎨⎪⎩

∂tv − N∂mu = 0,

∂tu + N t∂mw = 0,

∂te + ∂m (w, Nu) = 0.

(3.37)


In view of the computation of some properties of the wave velocities, one can thinkof using Proposition 1.4.9 with the variable W = (w, u, S)t . Since ∂t S = 0, oneobtains the quasi-linear system

⎧⎪⎨⎪⎩

∂tw −[∇w|Sv

]−1N∂mu = 0,

∂tu + N t∂mw = 0,

∂tS = 0.

(3.38)

So the wave velocities are the eigenvectors of the matrix

B =

⎛⎝

0 −[∇w|Sv

]−1N 0

N t 0 00 0 0

⎞⎠ ∈ Rn×n .

Let us write a generic eigenvector as (a, b, c) ∈ Rn−1−d × Rd × R , so

⎧⎪⎨⎪⎩

−[∇w|Sv

]−1Nb = μa,

N ta = μb,

0 = μc.

(3.39)

Then one obtains a first trivial eigenvector (a, b, c) = (0, 0, 1) for the eigenvalueμ = 0. Other null eigenvectors are described in the following proposition.

Proposition 3.2.11 (Zero spectrum and non-strict hyperbolicity). If n − 1 − d =d then the system (3.37) is not strictly hyperbolic and the nul l eigenvalue hasmultiplicity p ≥ 1 + max(n − 1 − 2d, 2d + 1 − n) ≥ 2.

Proof. The size of N is n− 1 − d × d. So the condition assumed in the propositionimplies that N is in general a rectangular matrix. The rank of N t is rk(N ) ≤min(n − 1 − d, d); therefore N and N t admit a non-trivial eigenvector for thenull eigenvalue. This means that there exists a non-trivial eigenvector of B forthe null eigenvalue. Together with the trivial eigenvector (0, 0, 1), it generates aneigenspace for the null eigenvector of dimension at least 2.

Assume for instance that the size of the velocity variable is d = 1. Thiscorresponds to a system which can be considered truly one-dimensional. Then thesystem cannot be strictly hyperbolic for n ≥ 4. This is already the case for theexample (3.32), for which n = 4 and d = 1.

Proposition 3.2.12. One has the formula

−[∇w|Sv

]−1= ∇2

v|Sε =(

∇2v|Sε

)t> 0. (3.40)

Proof. The result is proved using a locally well-posed change of variables and aLegendre transform.


Step 1. The change of variables involves changing (v, ε) into (v, S). Since ∂ε|vS =Vn = 0 by hypothesis, the implicit function theorem shows that one candefine ε as a function of v and S, at least locally in a certain open set I ⊂ R .So

S(v, ε(v, S )) = S, S ∈ I.

Partial differentiation with respect to v while holding S constant yields

∇v|εS(v, ε(v, S)) +(∇ε|vS(v, ε(v, S ))

)∇

v|Sε(v, S) = 0.

A second differentiation yields

∇2v|εS(v, ε(v, S)) + ∇ε|v

(∇v|εS(v, ε(v, S))

)∇

v|Sε(v, S)

+ ∇v|ε(∇ε|vS(v, ε(v, S))

)∇

v|Sε(v, S)

+(

∇2ε|vS(v, ε(v, S ))

)∇v|Sε(v, S) ⊗ ∇v|Sε(v, S)

+(∇ε|vS(v, ε(v, S ))

)∇2

v|Sε(v, S) = 0.

Rewrite the above as

(−Vn)∇2v|Sε(v, S ) = −

[In−1−d, ∇

v|Sε(v, S)]

∇2v,εS

[In−1−d, ∇

v|Sε(v, S )]t

.

Since −Vn = 1T

> 0 and S is strictly concave by hypothesis, the internalenergy ε is a convex function of v, S being constant. Therefore one has the

formula ∇2v|Sε(v, S) =

(∇2

v|Sε(v, S))t

> 0, which is the last part of (3.40).

Step 2. By the definition of w one has that dS = ∇εS [(w, dv) + dε] , so ∇v|Sε =−w. Therefore [

∇w|Sv]−1

= ∇w|Sw = −∇2v|Sε.


Theorem 3.2.13. The non-zero eigenvalues of the Jacobian matrix of the La-grangian system (3.38) are the plus and minus square roots of the eigenvaluesof the symmetric non-negative matrix A = At ≥ 0,

A = N t(

∇2v|Sε

)N ∈ R

d. (3.41)

Proof. The eigenvalue equation (3.39) for non-zero μ can be rewritten as⎧⎨⎩

(∇2

v|Sε)

Nb = μa,

N ta = μb.

Elimination of a yields N t(

∇2v|Sε

)Nb = μ2b, which is the claim for non-zero μ.

For non-zero μ, this relation is necessary and sufficient.


Remark 3.2.14. Theorem 3.2.13 has an important implication for the practicalcalculation of the wave speeds. Most problems in classical continuum mechanicsare posed in the physical space Rd with 1 ≤ d ≤ 3. So the matrix A is a symmetricmatrix of size at most 3. This means that it is always possible to rely on exactformulas to determine the roots of the characteristic polynomial, for example usingthe Cardan formulas.

3.2.2 A more general Lagrangian structure

In certain physical situations, it is necessary to introduce a modification of thegeneral structure (3.34) and (3.35). Indeed, a careful examination often showsthat some of the assumptions of Theorem 3.2.8 may not hold, which calls forsubtle modifications of the structure of the matrix M . The velocity is involved.For the Landau superfluid model, the difference of velocity does not satisfy As-sumption 3.2.4; the same for the Godunov-type multiphase model, not to mentionthe Lorentz-invariant Euler system (3.139) which is of course not compatible withGalilean invariance. In all these cases we observe that (3.34) is still true even if(3.35) may not hold. This is the reason to focus on systems like

∂tU + ∂m

(MΨ

− 1

2(Ψ, MΨ)

)= 0. (3.42)

The only assumption on M ∈ Rn−1 is that it is a symmetric matrix, M = M t. Thevector Ψ ∈ Rn−1 is deduced from the entropy via the entropy variable V = ∇US:

Ψi =Vi

Vn, 1 ≤ i = n − 1.

Definition 3.2.15 (Enthalpy of the Lagrangian system). Define another entropyvariable

W =

(Ψ ∈ Rn−1

S

)∈ Rn

together with the enthalpy H : Rn → R of the system,

H(W ) =(V, U)

Vn. (3.43)

The matrix D ∈ Rn−1×n−1 with

D = −∇2Ψ|SH = Dt > 0 (3.44)

is called the metric of the enthalpy. That D is a positive matrix is a consequenceof the identity (3.45).

For the Lagrangian gas dynamics system (3.30) one has W = (p, −u, S)t andH = T

(pT

τ − uT

u + 1T

e)

= ε + pτ − 12

u2 = h − 12u2.


Proposition 3.2.16. One has the identities

(∇WV )t ∇V U∇WV = (∇WV )

t ∇WU =

⎛⎝

−VnD 0

0

(∂V

∂S|Ψ ,∂U

∂S|Ψ

)⎞⎠ (3.45)

and

(∇WV )t

(M 0

−ΨtM 0

)=

(VnM 0

0 0

). (3.46)

Proof. Firstly, (∇WV )t ∇V U∇WV = (∇WV )

t ∇WU is just the chain rule. Sec-ondly, we adopt a block decomposition

(∇WV )t ∇WU =

( (∇Ψ|S V

)t ∇Ψ|S U(∇Ψ|SV

)t ∇S |ΨU(∇S |ΨV

)t ∇Ψ|S U(∇S |ΨV

)t ∇S |ΨU

)

and determine what these blocks are.

Step 1. Consider(∇Ψ|SV

)t ∇Ψ|S U ∈ Rn−1×n−1. Using (3.50)–(3.44) one has

∇Ψ|S U =

(−D

∇Ψ|SUn

)∈ R

n×n−1

and

∇Ψ|S V = Vn

(I0

)+ V ∇Ψ|SVn ∈ R

n×n−1,

in which we used

∂Vi

∂Ψj=

∂ (ΨiVn)

∂Ψj= Vnδij + Ψi

∂Vn

∂Ψj. (3.47)

So (∇Ψ|S V

)t ∇Ψ|S U = −VnD +(∇Ψ|SVn

)tV t∇Ψ|S U.

Since dS = (V, dU) one has

V t∇Ψ|SU = ∇Ψ|S S = 0. (3.48)

It shows that(∇Ψ|SV

)t ∇Ψ|SU = −VnD.

Step 2. Consider the lower diagonal block ∇S |ΨV t∇Ψ|SU . One has

∂Vi

∂S=

∂ (ΨiVn)

∂S= Ψi

∂Vn

∂S=

Vi

Vn

∂Vn

∂S. (3.49)

This shows that ∇S |ΨV =(

1Vn

∂Vn

∂S |Ψ

)V . Use (3.48) to get

∇S|ΨV t∇Ψ|SU =

(1

Vn

∂Vn

∂S|Ψ

)V t∇Ψ|SU = 0.


Step 3. By symmetry, the off-diagonal part is also null. This yields (3.45) sincethe bottom-right coefficient is trivial.

Step 4. It remains to show (3.46). Using the previous results one gets

∇WV =( ∇Ψ|SV, ∇S |ΨV

)

=

(Vn

(I0

)+ V ∇Ψ|S Vn,

(1

Vn

∂Vn

∂S|Ψ

)V

)

= Vn

(I0

)+ V K, where K =

(∇Ψ|SVn ,

1

Vn

∂Vn

∂S|Ψ

).

So

(∇WV )t

(M 0

−ΨtM 0

)=

(VnM 0

0 0

)+ KtV t

(M 0

−ΨtM 0

).

But

V t

(M 0

−ΨtM 0

)= Vn

(Ψt, 1

) ( M 0−ΨtM 0

)= 0.


Next we provide a simple characterization of the gradients using the variableU , which consists of the n − 1 unknowns of the main vectorial variable U :

U =

(U ∈ Rn−1

Un = e ∈ R

)∈ R

n .

Proposition 3.2.17. The enthalpy is the opposite of the Legendre transform of thetotal energy. One has the identities

∇Ψ|S H = U t, ∇U|Se = −Ψt and ∇2

U|Se = D−1 > 0. (3.50)

Proof. The definition (3.43) of the enthalpy implies that H =(

Ψ, U)

+ e and

dH =(

dΨ, U)

+(

Ψ, dU)

+ de =(

dΨ, U)

+1

Vn(V, dU) =

(dΨ, U

)+

1

Vnde.

So ∇Ψ|SH = U t. Similarly, dS = (V, dU) = Vn

[(Ψ, dU

)+ de

]yields ∇

U|Se =

−Ψt. One has ∇2Ψ|S H = −D < 0. The last property can be verified as follows:

∇2

U|S e = −∇U|SΨ = −∇Ψ|SU−1 = −(−D)−1 = D−1 > 0.



Theorem 3.2.18. The eigenvalues of the Jacobian matrix of the flux of the La-grangian system

∂tU + ∂m

(MΨ

− 1

2(Ψ, MΨ)

)= 0 (3.51)

consist of the nul l eigenvalue and the opposite of the eigenvalues of M in theenthalpy metric.

Proof. This is shown using the variable W . From (3.51) one gets

∇WU∂t W +

(M 0

−ΨtM 0

)∂mW = 0.

Multiply on the left by the matrix ∇WV t and use identities (3.45)–(3.46) to obtain

(−VnD 0

0 α < 0

)∂tW +

(VnM 0

0 0

)∂mW = 0,

that is,

∂tW +

(−D−1M 0

0 0

)∂mW = 0.

This quasi-linear form shows that the eigenvalues are 0 and the eigenvalues of−D−1M , as claimed.

One can show that the eigenvalues λj of −D−1M have the same sign as theeigenvalues μj of M . It is obvious from the min-max principle that

−λj = mindimP=i

[maxz∈P

((z, Mz)

(z, Dz)

)]and − μj = min

dimP=i

[maxz∈P

((z, Mz)

(z, z)

)].

For the system with structure (3.35) one has

Mx = λx ⇐⇒(

0 NN t 0

)(x1

x2

)= λ

(x1

x2

)⇐⇒

Nx2 = λx1,

N tx1 = λx2.

Non-zero eigenvalues correspond to the eigenproblem

N tNx2 = μx2, μ = λ2.

Similarly, the eigenvalue Mrj(0) = −λj(U)D(U)rj (U) can be evaluated with theRayleigh quotient

λj(U) = − (rj(U), M rj(U))

(rj(U), D(U )rj (U)). (3.52)

One can differentiate in a given direction z ∈ Rn. Since M is constant this yields(using the notation ′ = d

dz)

Mr′j(U) = −λ′

j(U)D(U)rj (U) − λj(U)D′ (U)rj (U) − λj(U)D(U )r′j (U).

3.3. Examples of Lagrangian systems 115

Taking the scalar product with the eigenvector rj(U) yields the formula

λ′j(U) = λj(U)

(rj (U), D′ (U)rj (U))

(rj (U), D(U)rj (U)). (3.53)

3.3 Examples of Lagrangian systems

We present four examples for which Ψ, H and D will be explicitly given. Thefirst example is the magnetohydrodynamics (MHD) system in one dimension. Thesecond models compressible elasticity. The third is the superfluid model of Landau.The fourth example is a multiphase model with two velocities, a la Godunov.

3.3.1 Ideal MHD

We start with the three-dimensional Eulerian MHD model written in conservativeform: ⎧

⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎩

∂tρ + ∇ · ρu = 0,

∂ρu + ∇ · (ρu ⊗ u) + ∇P − ∇ · B ⊗ B

μ= 0,

∂tB + ∇ ∧ (u ∧ B) = 0,

∂ρe + ∇ ·(

ρue + P u − 1

μB(B, u)

)= 0.

(3.54)

The magnetic field is B ∈ R 3 and u ∧ B denotes the vector product of u and B.The operator ∇∧ is the curl operator in dimension d = 3. By definition μ = 4πand

P = p +1

2μ|B|2.

The total energy is the sum of the internal thermal energy, the kinetic energy andthe magnetic energy:

ρe = ρεth +1

2ρ|u|2 +

1

2μ|B|2 . (3.55)

Correct solutions must satisfy the divergence-free constraint on the magnetic field,∇ · B = 0. Furthermore, the magnetic equation implies that

∂t∇ · B = −∇ · (∇ ∧ (u ∧ B)) = 0,

so if the magnetic field is divergence-free at t = 0, it remains divergence-free. Thenatural physical entropy relation for smooth solutions is

∂tρS + ∇ · (ρuS) = 0. (3.56)

Make the hypothesis that the flow is invariant in directions y and z:

∂y = ∂z = 0.


Let us decompose the magnetic field as B = (Bx, By , Bz) with

∂t

(Bx

By

Bz

)+

(∂x

00

)∧(

uyBz − uxBy

uz Bx − uxBz

uxBy − uyBx

)= 0,

not forgetting the divergence-free constraint

∂xBx + ∂yBy + ∂zBz = ∂xBx = 0.

So Bx is constant in time and space, ∂xBx = ∂t Bx = 0. One can simplify thesystem to ⎧

⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

∂tρ + ∂x ρux = 0,

∂ρux + ∂x(ρu2x ) + ∂xP = 0,

∂ρuy + ∂x

(ρuxuy − BxBy

μ

)= 0,

∂ρuz + ∂x

(ρuxuz − BxBz

μ

)= 0,

∂tBy + ∂x (uxBy − uyBx) = 0,

∂tBz − ∂x(ux Bz − uz Bx) = 0,

∂ρe + ∂x

(ρue + P ux − Bx(Byuy + Bzuz)

μ

)= 0,

(3.57)

with

P = P − 1

μB2

x = p +1

2μ(−B2

x + B2y + B2

z).

It is striking that the components of the magnetic field play the same role in thetotal pressure P. The associated Lagrangian formulation is

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

∂tτ − ∂mux = 0,

∂tτBy − ∂muyBx = 0,

∂tτBz − ∂muz Bx = 0,

∂tux + ∂mP = 0,

∂tuy − ∂mByBx

μ= 0,

∂tuz − ∂mBzBx

μ= 0,

∂te + ∂m

(P ux − Bx(Byuy + Bzuz)

μ

)= 0.

(3.58)

The Lagrangian entropy law for smooth solutions reads ∂tS = 0; therefore onecan use the general method to identify the entropic variable V and the reducedentropic variable Ψ.


Starting from the fundamental law of thermodynamics, TdS = dε+ p dτ, andthe definition of e, one gets

TdS = de − u du + P dτ − By

μdBy − Bz

μdBz.

One obtains immediately that

U =

⎛⎜⎜⎜⎜⎜⎜⎝

ττBy

τBz

ux

uy

uz

e

⎞⎟⎟⎟⎟⎟⎟⎠

, V =1

T

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

P

− By

μ

− Bz

μ−ux

−uy

−uz

1

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

, Ψ =

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

P

− By

μ

− Bz

μ−ux

−uy

−uz

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

.

Rewrite the Lagrangian system (3.58) as

∂t U + ∂m

(MΨ

− 1

2(Ψ, MΨ)

)= 0,

where the matrix M is

M = M t =

⎛⎜⎜⎜⎜⎝

0 0 0 1 0 00 0 0 0 Bx 00 0 0 0 0 Bx

1 0 0 0 0 00 Bx 0 0 0 00 0 Bx 0 0 0

⎞⎟⎟⎟⎟⎠

.

This matrix is indeed constant because the horizontal component of the magneticfield Bx is constant. The tangential components By and Bz are unknowns. Thedecomposition (3.35) holds with

N =

(1 0 00 Bx 00 0 Bx

).

The enthalpy of the system is

H = e − τB2

y + B2z

μ−(u2

x + u2y + u2

z

)+ P τ = ε + pτ − 1

2

(u2

x + u2y + u2

z

).

The internal energy, defined as the total energy minus the kinetic energy, is

ε = e − 1

2

(u2

x + u2y + u2

z

)

= εth +τ

2μ

(B2

x + B2y + B2

z

)= εth +

B2x

2μτ +

(τ By)2

2μτ+

(τBz)2

2μτ.


The Hessian matrix of internal energy reads

F = ∇2(τ,τBy ,τBz )|S ε = −∇(τ,τBy,τBz )|S

(P, − By

μ, − Bz

μ

)

=

⎛⎜⎜⎜⎜⎜⎜⎝

ρ2c2 +B2

y + B2z

μτ− By

μτ− Bz

μτ

− By

μτ

1

μτ0

− Bz

μτ0

1

μτ

⎞⎟⎟⎟⎟⎟⎟⎠

and the speed of sound c is given by ρ2c2 = − ∂p∂τ|S . Therefore the matrix A (see

(3.41)) is

A =

⎛⎜⎜⎜⎜⎜⎜⎝

ρ2c2 +B2

y + B2z

μτ− BxBy

μτ− BxBz

μτ

− BxBy

μτ

B2x

μτ0

− BxBz

μτ0

Bx

μτ

⎞⎟⎟⎟⎟⎟⎟⎠

.

A trivial eigenvalue of A is ϕa =B2

x

µτfor the eigenvector sa =

(0

Bz

−By

). Denote

by ϕs and ϕf the other two eigenvalues. One has the identities

ϕs + ϕf = tr(G) − ϕa = ρ2c2 +B2

x + B2y + B2

z

μτ

and ϕsϕf = det(B)/ϕa = ρ2c2 B2x

µτ. So the eigenvalue equation becomes

ϕ2 −(

ρ2c2 +B2

x + B2y + B2

z

μτ

)ϕ + ρ2c2 B2

x

μτ= 0.

Set a =√

c2 +B2

x+B2y +B2

z

µρ. Therefore

ϕs =ρ2

2

(a2 −

√a4 − 4c2

B2x

μρ

)

and

ϕf =ρ2

2

(a2 +

√a4 − 4c2

B2x

μρ

).


These define the velocities of slow waves, Alfven waves and fast waves:⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

c2s =

ϕs

ρ2=

1

2

(a2 −

√a4 − 4c2

B2x

μρ

),

c2a =

ϕa

ρ2=

B2x

μρ,

c2f =

ϕf

ρ2=

1

2

(a2 +

√a4 − 4c2

B2x

μρ

),

with the inequalities c2s ≤ c2

a ≤ c2f . The slow eigenvector ss associated with ϕs and

the fast eigenvector sf associated with ϕf are

ss =

⎛⎜⎜⎜⎜⎝

ϕs − ϕa

− BxBy

μτ

− BxBz

μτ

⎞⎟⎟⎟⎟⎠

and sf =

⎛⎜⎜⎜⎜⎝

ϕf − ϕa

− BxBy

μτ

− BxBz

μτ

⎞⎟⎟⎟⎟⎠

. (3.59)

The wave velocities of the Eulerian system (3.57) are

−cf + u ≤ −ca + u ≤ −cs + u ≤ u ≤ cs + u ≤ ca + u ≤ cs.

3.3.2 Compressible elasticity

Tensor notation such as ρ, u, σ, E is used hereafter. Following [121], note that

many models exist for nonlinear elasticity, such as those in [92, 95, 107, 96, 176]and many others. A reference hypoelastic model is the Wilkins model [207]. Itcouples the nonlinear Euler system for gas dynamics with the linear elasticitysystem for transverse waves. This transverse system adds new unknowns s, whichare the components of the deviatoric stress tensor

s = σ + pI , tr(s) = 0, s = st.

The model in three dimensions is⎧⎪⎪⎨⎪⎪⎩

∂t ρ + ∇ · (ρu) = 0,

∂t ρu + ∇ · (ρu ⊗ u + pI − s) = 0,

∂t ρe + ∇ · (ρup + pu − s u) = 0,

Dt s − 2μD = J,

(3.60)

with the material derivative Dt = ∂t + u · ∇. The right-hand side J models theobjective derivative correction (or Jaumann derivative) for obtaining a model in-variant by rotation:

J = −(

3∑

p=1

sipϕpj + spjϕpi

)

1≤i,j≤3


with

ϕ =1

2

(∂ui

∂xj− ∂uj

∂xi

)

1≤i,j ≤3

.

Here D is such that

Dij

=1

2(∂iuj + ∂jui) − 1

3∇ · u,

so tr(D) = 0. The original model is non-conservative, which makes the mathemat-ical analysis of shocks problematic.

Assume that α = ρμ is constant and recast Dts − 2μD = J as

ρDts − αD = ρJ ⇐⇒ ∂tρs + ∇ · (ρsu − αE) = ρJ. (3.61)

The tensor E is defined by ∇ · E = D. So the model for nonlinear elasticity that

we consider is ⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩

∂tρ + ∇ · (ρu) = 0,

∂tρu + ∇ · (ρu ⊗ u + pI − s) = 0,

∂tρe + ∇ · (ρup + pu − s u) = 0,

∂tρs + ∇ · (ρsu − 2αE) = ρJ .

(3.62)

The energy is the sum of the internal energy, the kinetic energy and the elasticenergy,

e = ε +1

2u2 +

1

4αE2.

Consider planar flows for which ∂y = ∂z = 0,

D =

⎛⎜⎜⎜⎜⎜⎝

2

3∂x1 u1

1

2∂x1 u2

1

2∂x1 u3

1

2∂x1 u2 − 1

3∂x1u1 0

1

2∂x1 u3 0 − 1

3∂x1 u1

⎞⎟⎟⎟⎟⎟⎠

and

ϕ =

(0 −∂x1 u2 −∂x1u3

∂x1 u2 0 0∂x1 u3 0 0

).

The equations for u2, u3, σ12 and σ13 are

⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩

ρDt u2 − ∂x1σ12 = 0,

ρDt u3 − ∂x1σ13 = 0,

ρDt σ12 − α∂x1 u2 = ρJ12

,

ρDt σ13 − α∂x1 u3 = ρJ13

.


Additionally, assume that v = w = 0, σ12 = s12 = 0, and σ23 = s13 = 0 att = 0. Therefore ϕ = 0 and J = 0 at t = 0. So v = w = 0, σ12 = s12 = 0 and

σ23 = s13 = 0 all vanish. One gets the conservative model

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

∂tρ + ∂x(ρu) = 0,

∂tρu + ∂x (ρu2 + p − s1) = 0, p = p(τ, ε), τ =1

ρ,

∂tρs1 + ∂x

(ρus1 − 4

3αu

)= 0,

∂tρs2 + ∂x

(ρus2 +

2

3αu

)= 0,

∂tρs3 + ∂x

(ρus3 +

2

3αu

)= 0,

∂tρσ23 + ∂x (ρuσ23) = 0,

∂tρe + ∂x(ρue + pu − s1u) = 0,

(3.63)

where the pressure is determined as a function of the specific volume and theinternal energy. The total energy is the sum of the internal energy, the kineticenergy and the elastic energy:

e = ε +1

2u2 +

1

4α

(s2

1 + s22 + s2

3 + 2σ223

).

The Lagrangian model associated with (3.63) is

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

∂t τ − ∂mu = 0,

∂t u + ∂m(p − s1) = 0, p = p(τ, ε), τ =1

ρ,

∂t s1 + ∂m

(− 4

3αu

)= 0,

∂t s2 + ∂m

(2

3αu

)= 0,

∂t s3 + ∂m

(2

3αu

)= 0,

∂t σ23 = 0,

∂t e + ∂x(pu − s1u) = 0.

(3.64)

In this system, s1 , s2 and s3 are not independent since the traceless conditionreads

tr(s) = s1 + s2 + s3 = 0.

So one can eliminate s3 and get, after permutation (just for convenience) of the


lines,⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

∂tτ − ∂mu = 0,

∂ts1 + ∂m

(− 4

3αu

)= 0,

∂ts2 + ∂m

(2

3αu

)= 0,

∂tσ23 = 0,

∂tu + ∂m(p − s1) = 0, p = p(τ, ε), τ =1

ρ,

∂te + ∂x(pu − s1u) = 0, e = ε +1

2u2 +

1

4α

(s2

1 + s22 + (s1 + s2)2 + 2σ2

23

).

(3.65)Since compressible elasticity is compatible with the basic principles of thermody-namics, one can compute the adjoint vector Ψ. One has

TdS = dε + p dτ = d

(e − 1

2u2 − 1

4α

(s2

1 + s22 + (s1 + s2)2 + 2σ2

23

))+ p dτ

= de − u du − 1

2α(2s1 + s2)ds1 − 1

2α(s1 + 2s2)ds2 − 1

ασ23 dσ23 + p dτ.

So one can write

U =

⎛⎜⎜⎜⎜⎝

τs1

s2

σ23

ue

⎞⎟⎟⎟⎟⎠

, V =1

T

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

p

− 1

2α(2s1 + s2)

− 1

2α(s1 + 2s2)

− 1

ασ23

−u1

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

and Ψ =

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

p

− 1

2α(2s1 + s2)

− 1

2α(s1 + 2s2)

− 1

ασ23

−u

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

.

Set

M = M t =

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

0 0 0 0 1

0 0 0 04

3α

0 0 0 0 − 2

3α

0 0 0 0 0

14

3α − 2

3α 0 0

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

so that (3.65) can be rewritten in the canonical form

∂t U + ∂m

(MΨ

− 1

2(Ψ, MΨ)

)= 0.


The decomposition (3.35) holds with

N =

⎛⎜⎜⎜⎜⎜⎝

14

3α

− 2

3α

0

⎞⎟⎟⎟⎟⎟⎠

.

The Hessian matrix of the total internal energy is

F = ∇2∂(τ,s1,s2 ,σ23 |Sε =

⎛⎜⎜⎜⎜⎜⎜⎜⎝

ρ2c2 0 0 0

01

α

1

2α0

01

2α

1

α0

0 0 01

α

⎞⎟⎟⎟⎟⎟⎟⎟⎠

.

The matrix A in (3.41) is actually a scalar,

A = ρ2c2 +4

3α.

So the wave speeds of the Lagrangian model (3.65) for compressible elasticity are

0 (with multiplicity 4) and ±√

ρ2c2 + 43α.

3.3.3 Landau model for superfluid helium

The following example, which can be treated as an exercise, is based on the Landaumodel [128, 127] for superfluidity.

The Landau model is Galilean invariant, but it does not have a zero entropyflux in the initial natural Lagrangian formulation. Instead, another formulationallows us to enforce the zero entropy flux property. Among the properties thatdistinguish this model from classical fluid dynamics, some are fundamental andare described in the following hypotheses.

Hypothesis 3.3.1. There exists a critical temperature Tc > 0 (the value for super-fluid helium Tc = 2.17 K = −270.83C) such that the following properties hold.

• If T > Tc the fluid is in a normal state and it satisfies the classical Eulerequations of a compressible non-viscous fluid.

• If T < Tc the fluid is no longer a classical fluid. It is composed of two parts:one is classical or normal, and the other one is superfluid. Following Landau,the normal part is indicated with a subscript “n” and the superfluid one witha subscript “s”.


Hypothesis 3.3.2. The superfluid part exhibits a surprising behavior. Typicallythe superfluid velocity is curl-free ∇ ∧ us = 0 in dimension d = 2 or d = 3.By induction it satisfies an equation which preserves such a property. There islittle choice except to postulate an equation like ∂tus + ∇ϕ = 0 with ϕ a certainpotential to be determined. In one dimension, the equation is ∂tus + ∂xϕ = 0.

Hypothesis 3.3.3. The fraction cn of the normal part (cs = 1 − cn is the fractionof the superfluid part) is a function of three variables, which are two independentthermodynamical variables and the velocity difference un − us.

The system proposed by Landau [127] satisfies these three requirements.The reasoning can be summarized as follows. One first chooses two independentthermodynamical variables which are the density ρ and the total entropy S. Thethird variable is un − us. One looks for a representation of the form

⎧⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎩

μ = μ(ρ, S, un − us),

e0 = e0(ρ, S, un − us),

T = T (ρ, S, un − us),

p = p(ρ, S, un − us),

cn = cn(ρ, S, un − us),

(3.66)

where μ is a potential, T is the temperature, p is the pressure and cn is the massfraction of the normal part. The mass fraction of the superfluid part is cs = 1 − cn.The internal energy will be denoted by e0 . The quantities in (3.66) are connectedby some thermodynamical relations such as

μ = e0 + τp − TS − cn(un − us)2, equation (130.12) in [127], (3.67)

and

dρe0 = μdρ + Td(ρS) + (un − us)d(ρcn (un − us)), equation (130.9) in [127].(3.68)

Note that the first relation is the classical definition of the chemical potential ingeneral thermodynamics textbooks. The second relation is a generalization to thesuperfluid of the fundamental principle of thermodynamics. By compatibility withthe assumptions, one has that if T > Tc then cn = 1 and us = un . Combining(3.67)–(3.68), one gets

TdS = de0 + p dτ − (un − us)d(cn(un − us)). (3.69)

One also has the relations

⎧⎨⎩

u = cnun + csus,

e = e0 +u2

s

2+ cn(un − us)us = e0 +

1

2u2 − c2

n

2(un − us)

2 .(3.70)


One obtains the one-dimensional Landau model⎧⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎩

∂t (ρ) + ∂x (ρu) = 0,

∂t (ρu) + ∂x (ρcnu2n + ρcsu2

s + p) = 0,

∂t (us) + ∂x

(u2

s

2+ μ

)= 0,

∂t (ρe) + ∂x

((μ +

u2s

2

)ρu + TρSun + ρcnu2

n(un − us)

)= 0.

(3.71)

It can be proved that this is a closed system, which means that the flux can beevaluated as a function of (ρ, ρu, us, ρe). But since this property is not criticalto the remaining analysis and will distract us from the main point, we leave itsverification to the interested reader. See also [70]. It is explained in [127] that thesystem (3.71) possesses an additional conservation law (for smooth solutions) ofthe form

∂t ρS + ∂x(ρunS) = 0. (3.72)

That is, the entropy is convected at the normal velocity and not at the globalvelocity. As a consequence, one needs a special Lagrangian formulation.

Let us define the Euler-to-Lagrange change of reference frame by

∂x(t, X)

∂t= un , x(0, X) = X. (3.73)

We obtain the following new Lagrangian formulation of (3.71):

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

∂tJ − ∂X un = 0,

∂t(ρJ) + ∂X (ρ(u − un)) = 0,

∂t(ρJu) + ∂X (ρcnu2n + ρcsu2

s + p − ρuun) = 0,

∂t(Jus) + ∂X

(u2

s

2+ μ − usun

)= 0,

∂t(ρJe) + ∂X

((μ +

u2s

2

)ρu + TρSun + ρcnu2

n(un − us) − ρune

)= 0.

(3.74)The first equation is the Piola identity and the other five equations come from(3.71). Using (3.69) one gets

TdS = de − (u + cs(un − us))du +cs(un − us)

τd(τ us) +

(p − cs(un − us)

τus

)dτ

or, after multiplication by ρJ,

Td(ρJS) = d(ρJe) − und(ρJu) +cs(un − u)

τd(Jus)

+

(p − cs(un − us)

τus

)dJ − μ′ d(ρJ),

(3.75)


where μ′ = e − TS − unu + pτ . The unknown of the system (3.74) is

U = (J, ρJ, ρJu, Jus, ρJe)t .

The entropy variable is

V = ∇U(ρJS) =1

T

((p − cs(u − un)

τus

), −μ′, −un ,

un − u

τ, 1

)t

.

It defines

Ψ =

⎛⎜⎜⎜⎜⎜⎝

p − cs(un − us)

τ−μ′

−un

un − u

τ

⎞⎟⎟⎟⎟⎟⎠

and M =

⎛⎜⎝

0 0 1 00 0 0 −11 0 0 10 −1 0 0

⎞⎟⎠ . (3.76)

One can then rewrite (3.74) as

∂t U + ∂X

(MΨ

− 1

2(Ψ, MΨ)

)= 0. (3.77)

Nevertheless, Assumption 3.2.6 is not true for this model. Smooth solutions of(3.77) satisfy

T∂t(ρJS) = (V, ∂tU) = 0. (3.78)

This yields a proof of the Eulerian entropy law (3.72) after going back to Eulerianvariables. One can derive other Eulerian conservation laws in a similar fashion.Starting from (3.78) written as T∂t h(ρJS) = 0 for any smooth function h, we get

∂t

(h(ρJS)

J

)+ ∂x

(un

h(ρJS)

J

)= 0. (3.79)

Taking h(z) = z recovers the original Landau entropy law (3.72).

3.3.4 A multiphase model

We consider the multiphase model

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

∂t(ρ) + ∂x (ρu) = 0,

∂t(ρc2) + ∂x (ρc2u2) = 0,

∂t(ρu) + ∂x(ρu2 + P ) = 0,

∂t(w) + ∂x

(uw + μ1 − μ2 − (1 − 2c2)

2w2

)= 0,

∂t(ρe) + ∂x

(ρue + P u + ρw(1 − c2)c2(μ1 − μ2 − (1 − 2c2)

2w2)

)= 0.

(3.80)


The total density is ρ, and the mass fractions of the different species are c1 = 1−c2

and c2. The two velocities, one per species, are denoted by u1 and u2. The totalpressure is P , while μ1 and μ2 denote some thermodynamical potentials. Finally,w = u1 − u2 is the velocity difference and e is the total energy. Such a systemis intended to be representative of multiphase-multispecies fluid dynamics andplasmas. The theoretical motivation is clearly to obtain a system in conservativeform that is compatible with the general principles of thermodynamics [67]. Thismodel belongs to a more general family which first appeared in the seminal worksof Godunov and Romensky [96] and Godunov [98].

The system (3.80) is endowed with closure relations⎧⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎩

1

ρ= c1τ1 + c2τ2,

u = c1u1 + c2u2,

ε = c1ε1 + c2ε2 ,

e = ε +1

2u2 +

c2(1 − c2)

2w2 .

(3.81)

For example, the first of these equations expresses the additivity of volumes in anon-miscible mix. An additional assumption is that each fluid has its own ther-modynamical consistency. We add, quite arbitrarily, the condition that the globalfluid is also consistent with the basic principles of thermodynamics: the total en-tropy is greater than the sum of partial entropies,

S ≥ c1S1 + c2S2.

To achieve this, it is sufficient to add a mixing entropy of the form

S = c1S1 + c2S2 + Smix(c1, c2),

where Smix is a non-negative and concave function. A first option is to consider aBoltzmann-like function Smix ≈ −c1 log c1 − c2 log c2. To avoid technical problemsnear c1 ≈ 0 and c2 ≈ 0, take a smooth function such as

Smix = kc1c2 = kc1(1 − c1) = kc2(1 − c2), k ≥ 0. (3.82)

Additional compatible relations are⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

p1(τ1, ε1) = p2(τ2 , ε2),

T1(τ1, ε1) = T2(τ2 , ε2) = T,

μ1 = −T∂

∂c1(c1S1 + c2S2 + Smix) = −TS1 + ε1 + pτ1 − kTc2 ,

μ2 = −T∂

∂c2(c1S1 + c2S2 + Smix) = −TS2 + ε2 + pτ2 − kTc1 ,

P = p1 + c2(1 − c2)ρ(u1 − u2)2 .

(3.83)


The associated Lagrangian formulation is

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

∂tτ − ∂mu = 0,

∂tc2 − ∂m(ρw(1 − c2)c2) = 0,

∂tu + ∂mP = 0,

∂t(τw) + ∂m

(μ1 − μ2 − (1 − 2c2)

2w2

)= 0,

∂te + ∂m

(P u + ρw(1 − c2)c2(μ1 − μ2 − (1 − 2c2)

2w2)

)= 0.

(3.84)

It is possible to model other phenomena, including phase transitions and dragforces as in

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

∂tτ − ∂mu = 0,

∂tc2 − ∂m(ρw(1 − c2)c2) = −A

(c2 − 1

2

),

∂tu + ∂mP = 0,

∂t(τw) + ∂m

(μ1 − μ2 − (1 − 2c2)

2w2

)= −Bw,

∂te + ∂m

(P u + ρw(1 − c2)c2

(μ1 − μ2 − (1 − 2c2)

2w2

))= 0.

(3.85)

The coefficient A > 0 is used to get an accurate scaling of the phase transition.Similarly, the term −Bw (with B > 0) models the drag force between the twofluids. It is easy to check that the system has the Lagrangian structure (3.42).Indeed,

Td(c1S1 + c2S2 + Smix(c1 , c2)

)

=

(TS1 +

∂Smix

∂c1

)dc1 +

(TS2 +

∂Smix

∂c2

)dc2

+ c1 (dε1 + p dτ1) + c2 (dε2 + p dτ2)

= (μ1 − μ2)dc2 + dε + p dτ

=

(μ1 − μ2 − c1 − c2

2(u1 − u2)2

)dc2 + de − u du

+ p dτ − c1c2(u1 − u2)d(u1 − u2)

= de − u du +

(μ1 − μ2 − c1 − c2

2(u1 − u2)2

)dc2

+

(p +

c2c2(u1 − u2)2

τ

)dτ − c1c2(u1 − u2)

τd(τ (u1 − u2)).


Therefore, for the system (3.84) one has

U =

⎛⎜⎜⎜⎝

τc2

uτwe

⎞⎟⎟⎟⎠ , V =

1

T

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

p +c2c2(u1 − u2)2

τ

μ1 − μ2 − c1 − c2

2(u1 − u2)2

−u

− c1c2(u1 − u2)

τ1

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

and

Ψ =

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎝

p +c2c2(u1 − u2)2

τ

μ1 − μ2 − c1 − c2

2(u1 − u2)2

−u

− c1c2(u1 − u2)

τ

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎠

, M =

⎛⎜⎝

0 0 1 00 0 0 11 0 0 00 1 0 0

⎞⎟⎠ .

Elementary calculations show that

D−1 = ∇2(τ,c2,u,τ(u1 −u2))|S ε = ∇(τ,c2 ,u,τ(u1−u2))|SΨ

=

⎛⎜⎜⎜⎜⎝

ρ2c2 + 3ρ2c1c2(u1 − u2)2 a 0 b

a∂(μ2 − μ1)

c2− (u1 − u2)2 0 c

0 0 1 0b c 0 c1c2ρ2

⎞⎟⎟⎟⎟⎠

with

a = −(1 − 2c2)ρ(u1 − u2)2 +∂p

∂c2,

∂p

∂c2=

∂(μ1 − μ2)

∂τ,

and

b = −2c1c2ρ(u1 − u2)2, c = (c1 − c2)ρ(u1 − u2).

The matrix D is symmetric but its positivity is less evident. This is a commonfeature of many systems of conservation laws for multiphase flows, where loss ofhyperbolicity is very often encountered. Finally, note that the variable τ (u1 − u2)changes sign upon inversion of the time arrow, but is left invariant by Galileantranslation. Therefore it is neither strictly a density-like variable nor strictly avelocity variable, as defined in Assumptions 3.2.4 and 3.2.5.


3.4 Self-similar solutions and the solution of theRiemann problem

We consider hereafter self-similar solutions of systems of conservation laws (La-grangian or not) in the form

U(t, x) = U(y), y =x

t. (3.86)

Functions of this kind are the building blocks to the solution of the Riemannproblem of Lax’s theorem. The classical Lax’s theorem has strict hyperbolicity asan important hypothesis. Since Lagrangian systems of large size are genericallynon-strictly hyperbolic, as explained in Proposition 3.2.11, the reformulation ofLax’s theorem for Lagrangian systems is an interesting question from a theoreticalperspective. The answer will be to construct the solution in the space W = (Ψ, S)t .

Our construction of the solution of the Riemann problem follows Lax’s sem-inal idea, but we do it with the variable V . We will construct rarefaction curvesand discontinuity curves. After that, an easy change of variable yields the form ofthe curves in the variable U : it gives the standard Lax theorem for strictly hyper-bolic systems. The use of the variable V in the discussion of rarefaction fans anddiscontinuities brings some simplifications relative to the standard method, sinceit naturally allows us to use symmetric matrices in some parts of the analysis.In the last part of this section we turn to an extension of the Lax theorem forLagrangian systems using the variable W , which is well adapted to the kind ofnon-strictly hyperbolic systems studied in this monograph.

rarefaction fan

x

t

UG

UD

y = b

y = at

UG

UD

x

x = σt

Figure 3.2: Illustration of the structure of rarefaction fans and discontinuities.

131


A rarefaction fan is a smooth solution U ∈ C1[a, b] of the equation

∂tU( x

t

)+ ∂xf

(U(x

t

))= 0, a ≤ x

t≤ b,

where the edges of the fan are a < b. Simplifications yield the self-similar equation

−yd

dyU(y) +

d

dyf(U(y)) = 0, a ≤ y ≤ b. (3.87)

The value on the left edge is U(a) = UL and the value on the right edge isU(b) = UR . The problem that we seek to solve is, given a value of UL ∈ Rn, todetermine all UR with a rarefaction fan that connects UL and UR .

To solve this problem, we will first use the variable V . Let V(y) = V (U(y)).Equation (3.87) is rewritten as

[B(V(y)) − yC(V(y))] V′(y) = 0 (3.88)

where the matrices are symmetric,

B(V ) = ∇2V ξ∗(V ) = B(V )t and C(V ) = ∇2

V η∗(V ) = C(V )t > 0.

Definition 3.4.1 (Generalized eigenvalue problem for rarefaction fans). The gener-alized eigenvalue problem at V is

B(V )r = λC(V )r, λ ∈ R , 0 = r ∈ Rn, (3.89)

where the eigenvector is r = 0 and the eigenvector is λ. The solutions B(V )rj(V ) =λj(V )C(V )rj(V ) of the generalized eigenproblem are treated on a hierarchicalbasis,

λ1(V ) ≤ λ2(V ) ≤ · · · ≤ λn(V )

and normalized as(ri(V ), C(V )rj(V )) = δij .

Equation (3.88) means that V′(y) is collinear to a certain eigenvector, i.e.

V′(y) = α(y) × ri(V(y)), λj(V(y)) = y. (3.90)

This is a first-order differential equation which has a natural solution in the frame-work of the Cauchy-Lipschitz theorem. We decide to fix

V′(a) = VG, (3.91)

where, once again, the line x = at is on the left of the rarefaction fan.But an important technical difficulty arises in the analysis of (3.90). In fact,

it will be necessary to differentiate a certain number of times the eigenvectorsand eigenvalues. This is a fundamental difficulty for matrices. To overcome thisdifficulty we will assume the following.

3.4. Self-similar solutions and the solution of the Riemann problem


Hypothesis 3.4.2. We make the assumption that the eigenvalues and eigenvectorsof the generalized eigenproblem (3.89) are infinitely continuously differentiable.

If the problem is strictly hyperbolic, this assumption is easy to prove. How-ever, to be more general we will not assume strict hyperbolicity.

Differentiating λj(V(y)) = y, one gets (∇V λj(V(y)), V′ (y)) = 1. Eliminationof V ′(y) using the second part of (3.90) yields

α(y) × (∇V λj(V(y)), rj (V(y))) = 1.

A necessary condition for a non-trivial solution being α(y) = 0, one gets thecondition

(∇V λj(V ), rj(V )) = 0. (3.92)

This is the reason for the following definition.

Definition 3.4.3. One says that the jth field is truly nonlinear if and only if thenon-degeneracy condition (3.92) holds for all V .

Since in this case α(y) = 0, either α > 0 or α < 0. Define s =∫ y

aα(z)dz.

Under general conditions, the Cauchy-Lipshitz theorem states that there exists anintegral curve for the jth vector field,

V ′j (s) = rj(Vj(s)), −ε < s < ε, Vj(0) = VL. (3.93)

It defines V(y) = Vj(s) = Vj

(∫ ya

α(z)dz), which is a solution of (3.88).

Proposition 3.4.4. Let VG ∈ Rn. Then the integral curve (3.93) admits the localTaylor expansion

Vj(s) = VL + srj (VL) +s2

2∇V rj(VL)rj(VL) + O(s3), (3.94)

and the eigenvalue admits the Taylor expansion

λj(Vj(s)) = λj(VL) + s(∇V λj(VL), rj (VL)) + O(s2). (3.95)

Proof. Equation (3.93) implies V ′j (0) = rj(VL), the first term of the expansion

(3.94). Differentiate a second time to get

V ′′j (s) =

d

dsrj(Vj (s)) = ∇V rj(V (s)) V ′

j (s) = ∇V rj(V (s)) rj(V (s)).

Then V ′′j (0) = ∇V rj(VL) rj(VL). This shows (3.93). One also has the chain rule

λ′j(Vj (s)) = (∇V λj(Vj(s)), V ′

j (s)). So[λ′

j(Vj(s))]

(0) = (∇V λj(VL), rj(VL)) ats = 0.

Since one aims to construct a self-similar solution which begins at V (a) = VL,one must keep the branch (3.93) such that y = λj > a (see figure 3.2).

133

Proposition 3.4.5. Rarefaction fans (3.90)–(3.91) have the form V(y) = Vj(s).

If (∇V λj(VL), rj(VL)) > 0, one must keep the half-branch 0 ≤ s < ε and rejectthe other part.

If (∇V λj(VL), rj (VL)) < 0, one must keep the half-branch −ε < s ≤ 0 and rejectthe other part.

3.4.2 Entropy discontinuities

A discontinuous solution has the form

U(y) = UL for y < σ, U(y) = UR for y > σ, (3.96)

where σ is the velocity of the discontinuity. The problem we are interested in is,given UL , to determine all UR with an entropy discontinuity that connects UL

and UR. Just by copying the proofs of Theorems 2.2.3 and 2.3.5, one obtains thecelebrated Rankine-Hugoniot relations.

Definition 3.4.6 (Rankine-Hugoniot relations for systems). A triplet (σ, UL , UR) ∈R ×Rn × Rn is a solution of the Rankine-Hugoniot relations if and only if

−σ(UR − UL) + f(UR ) − f(UL) = 0. (3.97)

It is an entropy solution if, moreover,

−σ(η(UR ) − η(UL)) + ξ(UR) − ξ(UL) ≤ 0. (3.98)

The velocity of the discontinuity is σ.

Next we introduce a difference between shocks and contact discontinuities.As for conservation laws, it is based on the entropy inequality.

Definition 3.4.7. A solution of the Rankine-Hugoniot relations is a shock if andonly if the entropy inequality is strict:

−σ(η(UR ) − η(UL)) + ξ(UR) − ξ(UL) < 0.

Definition 3.4.8. A solution of the Rankine-Hugoniot relations is a contact discon-tinuity if and only if the entropy inequality is an equality:

−σ(η(UR ) − η(UL)) + ξ(UR) − ξ(UL) = 0.

Analysis of the Rankine-Hugoniot relations (3.97)

Considering that UL is given, one can try to construct some curves in Rn suchthat all states on these curves are admissible right states for (3.97).



To this end, note that ∇2V ξ∗(V ) = ∇V f(U(V )) and ∇2

V η∗(V ) = ∇V U(V ).So the Rankine-Hugoniot relations can be rewritten as

BL(V )(V − VL) = σCL(V )(V − VL) (3.99)

with

BL(V ) =

∫ 1

0

∇2V ξ∗(VL + t(V − VL))dt = BL(V )t

and

CL(V ) =

∫ 1

0∇2

V η∗(VL + t(V − VL))dt = CG(V )t > 0.

Equation (3.99) can be interpreted as a certain generalized eigenproblem. Moreprecisely, V −LG = βjrL

j is collinear to a certain eigenvector of a second generalizedeigenproblem.

Definition 3.4.9 (Generalized eigenvalue problem for discontinuities). For a givenVL , the generalized eigenvalue problem at V is

BL(V )r = λCL(V )r, λ ∈ R , 0 = r ∈ Rn , (3.100)

where the eigenvector is r and the eigenvalue is λ. The solutions are treated on ahierarchical basis,

λL1 (V ) ≤ λL

2 (V ) ≤ · · · ≤ λLn (V ),

and normalized, (rL

i (V ), CL(V )rLj (V )

)= δij .

The form of this new eigenproblem is very similar to the one in (3.89). Moreprecisely, the matrices in (3.89) and (3.100) satisfy the relations

⎧⎪⎨⎪⎩

BL(VL) = B(VL), ∇V BL(VL) =1

2∇V B(VL),

CL(VL) = C(VL), ∇V CL(VL) =1

2∇V C(VL).

The factor 12

comes from the fact that if a function satisfies the Taylor expansion

f(x) = a0 + a1x + O(x2), then∫ 1

0f(sx) ds = a0 + 1

2a1x + O(x2). Thus

BL(V ) = B

(V +

1

2(V − VL)

)+ O(V − VL)2 (3.101)

and

CL(V ) = C

(V +

1

2(V − VL)

)+ O(V − VL)2 . (3.102)

But as before we will need to differentiate the generalized eigenproblem. In orderto avoid technical difficulties, we make the following assumption.

135

Hypothesis 3.4.10. We assume that the eigenvalues and eigenvectors of the gener-alized eigenproblem (3.100) are infinitely continuously differentiable. Considering(3.101)–(3.102), this implies that

rLi (V ) = ri

(VL +

1

2(V − VL)

)+ O(V − VL)2 (3.103)

and σLi (V ) = λi

(VL + 1

2(V − VL))

+ O(V − VL)2 .

Again, if strict hyperbolicity holds, then it can be shown that σL1 (V ) <

σL2 (V ) < · · · < σL

n (V ), which ultimately rules out the possibility of eigenvalues ofmultiplicity 2 or higher and simplifies considerably the analysis of the differentia-bility of the eigenvectors and eigenvalues.

With the above notation and assumptions, the condition V − VL = βjrLj is

equivalent to

(V − VL , CL(V )rLi (V )) = 0 ∀i = j, 1 ≤ i ≤ n. (3.104)

Moreover, σ = λLj (V ). For a given i = j , scalar equation in (3.104) characterizes

a hyper-surface. So the locus defined in (3.104) is the intersection of n − 1 hyper-surfaces, which is a priori a curve in Rn . This can be made more precise with thechange of variables

V ∈ Rn → W = (Wi)1≤i≤n, Wi = (V − VL , CL(V )rL

i (V )),

which is invertible at V = VL. The invertibility is ultimately a consequence of theimplicit function theorem and of the invertibility of the Jacobian matrix of thetransformation

∇V W (VL) =

⎛⎜⎜⎜⎝

CL(VL)rL1 (VL)

CL(VL)rL2 (VL)

...CL(VL)rL

n (VL)

⎞⎟⎟⎟⎠ .

This matrix is clearly invertible since its lines form a system of n linearly inde-pendent vectors which are moreover orthonormalized for the matrix CL(VL)−1.Therefore (3.104) is locally (in a neighborhood of VL) the reciprocal image of aline. So it is a curve.

Proposition 3.4.11. The curve defined by (3.104) is such that

Vj(s) = VL + srj(VL) +s2

2∇V rj(VL)rj (VL) + O(s3) (3.105)

where s is an abscissa. The shock velocity is

σLj (s) = λj(VL) +

s

2(∇V λj(VL), rj (VL)) + O(s2). (3.106)



Remark 3.4.12. The interpretation is that Rankine-Hugoniot curves are tangentat second order to rarefaction fan curves. The shock velocity σL

j (s) varies moreslowly than the wave velocity (3.95).

Proof. This classical result is obtained by a careful analysis of the properties ofthe curves around VL. It is proved in several steps.

Step 1. Start from σ = λGj (V ) and (3.104) and let s → 0. This yields σG

j (0) =λj(VG), Vj(0) = VG and V ′

j (0) = rj(VG).

Step 2. Differentiate the Rankine-Hugoniot relation

−σGj (s) [U(Vj (s)) − UL ] + f(U(V G

j (s))) − f(U(VG)) = 0

with respect to s. To simplify the notation, omit the superscripts L andsubscripts j . One gets

−σ′(s) (U(V (s)) − UL) − σ(s)C(s)V ′(s) + B(s)V ′(s) = 0

where B(s) = ∇V f(U(V (s))) and C(s) = ∇V U(V (s)). Differentiate a secondtime and evaluate the result at s = 0:

−2σ′(0)C(0)V ′(0) − σ(0) [C′ (0)V ′(0) + C(0)V ′′(0)]

+ B′(0)V (0) + B(0)V ′′(0) = 0.(3.107)

Step 3. Look at relation (3.89) for V = V Gj (s). This relation reads

B(s)rj(s) − λj(s)C(s)rj (s) = 0

where rj(s) = rj(V Lj (s)). Note that B(s) and C(s) are defined in (3.89).

Differentiate with respect to s and evaluate at s = 0:

B ′(0)rj (0) + B(0)r′j(0) − λ′

j(0)C(0)rj(0)

− λj(0)C ′ (0)rj (0) − λj(0)C(0)r′j (0) = 0.

(3.108)

Step 4. Note that B(0) = B(0) and C(0) = C(0). Subtract (3.108) from (3.107)to get

B(0)(V ′′(0) − r′

j(0))

− λj(0)C(0)(

V ′′(0) − r′j (0)

)

− (2σ′(0) − λ′j(0))C(0)rj (0) = 0.

Take the scalar product with rj(0). The symmetry of B(0) and C(0) yields2σ′(0) − λ′

j(0) = 0, which proves (3.106). So the relation is simplified to

B(0)(V ′′(0) − r′

j(0))

− λj(0)C(0)(V ′′(0) − r′

j(0))

= 0.

137

The classical way to analyze this relation is based on strict hyperbolicity. Inthis case V ′′(0) − r′

j(0) is necessarily proportional to rj(0), that is,

V ′′ (0) = r′j(0) + βrj(0). (3.109)

Since in our case we do not make the assumption of strict hyperbolicity, alittle more is needed. One has

(V (s) − VG , CG(s)rG

i (s))

= 0. Differentiatetwice and evaluate at s = 0 to get

(V ′′ (0), CG(0)rG

i (0))

+ 2(V ′(0), C ′

G(0)rGi (0)

)+ 2

(V ′(0), CG(0)(rG

i )′ (0))

= 0.

One also has (rj(s), C(s)ri(s)) = 0. Differentiate one time and evaluate ats = 0

(r′

j(0), C(0)ri (0))

+ (rj(0), C ′ (s)ri (0)) + (rj(0), C(0)r′i (0)) = 0.

One already knows from the relations (3.102)–3.103 that C ′G(0) = 1

2C ′(0)

and (rGi )′(0) = 1

2r′

i(0). So

(V ′′(0) − r′

j(0), CG(0)rGi (0)

)= 0 ∀i = j.

So one obtains (3.109) even if the system is not hyperbolic, just by making useof the weaker assumption that all eigenvectors and eigenvalues are infinitelydifferentiable.

Step 5. Since (3.109) holds, one has Vj(s) = VG + srj (VG)+ s2

2∇V rj(VG)rj(VG)+

O(s3). Let us redefine the abscissa along the curve by taking s = s + s2

2β .

Therefore

Vj(s) = VG + srj(VG) +s2

2∇V rj (VG)rj (VG) + O(s3).

The expansion of the discontinuity velocity is the same.

Analysis of the entropy inequality (3.98)

Entropy discontinuities satisfy the vectorial Rankine-Hugoniot relations and thescalar entropy inequality (3.98). To analyze the entropy inequality, define the func-tions

η∗L(V ) = (V, U) − η(U(V )) − (V, UL) + η(UL )

and

ξ∗L(V ) = (V, f(U )) − ξ(U(V )) − (V, f(UL)) + ξ(UL),

which are some polar transforms. The function V → η∗L(V ) is strictly convex at

least in a neighborhood of VL since ∇2V η∗(VL) = ∇V U(VL) > 0.



Proposition 3.4.13 (Reformulation of Rankine-Hugoniot relations). The Rankine-Hugoniot relation (3.97) expresses the fact that the iso-surface ξ∗

L(V ) = ξ∗G(VL) is

tangent at VL to the iso-surface η∗G(V ) = η∗

G(VL). The entropy inequality (3.98)becomes −ση∗

G(VR ) + ξ∗G(VR ) ≥ 0.

Proof. By construction,

∇V η∗L(V )t = U − UL and ∇V ξ∗

L(V )t = f(U) − f(UL ). (3.110)

So the shock velocity is a Lagrange multiplier which expresses the fact that thetwo gradients are proportional. Since the gradients are parallel, the iso-surfacesare tangent. Concerning the entropy inequality, a simple calculation shows that

−ση∗L(VR ) + ξ∗

L(VR) = − σ ((VR , UR) − η(UR ) − (VR , UL) + η(UL))

+ ((VR , f(UR )) − ξ(UR ) − (VR, f(UL)) + ξ(UL))

= (VR, −σ(UR − UL) + f(UR ) − f(UL))

+ σ(η(UR ) − η(UL)) − ξ(UR) + ξ(UL),

that is,

−ση∗L(VR) + ξ∗

L(VR ) = σ(η(UR ) − η(UL)) − ξ(UR ) + ξ(UL) ≥ 0. (3.111)


Definition 3.4.14 (Kulikovski generating function). Assume a given left state VL

and a given discontinuity velocity σ. The Kulikovski generating function [121] is

KL,σ(V ) = −ση∗L(V ) + ξ∗

L(V ).

Proposition 3.4.15. The stationary points of the Kulikovski generating functionare solutions of the Rankine-Hugoniot relations (3.97).

Proof. Indeed, ∇KL,σ(V ) = −σ(U − UL) + f(U) − f(UL).

Next we analyze the branch of the curve (3.104) which satisfies (3.105)–(3.106) with the help of the Kulikovski generating function,

ϕGj (s) = KL,σL

j(s) (Vj(s)).


d

dsϕG

j (s) = −η∗G(Vj (s))

d

dsσG

j (s). (3.112)

139

Proof. Starting from (3.110) one has

d

dsϕL

j (s) = − d

dsσL

j (s)η∗L (Vj(s))

+

(−σL

j (s)(U(V Lj (s)) − UR ) + f(U(V L

j (s))) − f(UR ),d

dsV L

j (s)

)

=

(− d

dsσL

j (s)

)× η∗

L(Vj(s)),

(3.113)

because the state U(V Lj (s)) is a solution of the Rankine-Hugoniot relation for a

discontinuity velocity equal to σLj (s).

s

Figure 3.3: The function s → η∗G(Vj(s)) ≥ 0 reaches a local minimum at s = 0.

Proposition 3.4.17. Smal l shocks for truly nonlinear fields are described by (3.105)and (3.106) solutions of (3.97)–(3.98), with the restriction that only half of thebranch is admissible.

If (∇V λj(VG), rj (VG)) > 0, one must keep the half-branch −ε < s < 0 and rejectthe other part.

If (∇V λj(VG), rj(VG)) < 0, one must keep the half-branch 0 < s < ε and rejectthe other part.

The good news is that shocks and rarefaction fans are complementary. Theentropy condition for shocks reads

∫ s

0

d

dsσG

j (t) × η∗G(Vj(t)) dt ≤ 0,

that is, upon using (3.108),

ϕGj (s) =

∫ s

0

[1

2(∇V λj(VG), rj(VG)) + O(s)

]× η∗(Vj(t)) dt ≤ 0 (3.114)

with η∗G(Vj(t)) > 0 for t = 0.



Proposition 3.4.18. The entropy jump is of third order for smal l s, that is,

−σ(η(UR ) − η(UL)) + ξ(UR) − ξ(UL) = O(s3)

for σ = σj(s) and UR = U(Vj (s)).

Proof. Expand (3.114).

Finally we turn to the analysis of contact discontinuities.

Definition 3.4.19 (Contact discontinuity). If (∇V λj(V ), rj (V )) = 0 for all V , onesays that the jth field is linearly degenerate.

Proposition 3.4.20. Smal l contact discontinuities for linear degenerate fields aredescribed by the curves (3.94) and (3.95) or (3.105) and (3.106) solutions of(3.97)–(3.98) with −ε < s < ε.

Proof. The proof is in three stages. We start from (3.94) and (3.95) for a linearlydegenerate field.

Step 1. Since λ′j(V (s)) = (∇V λj(V (s)), rj (V (s))) = 0, the wave velocity is con-

stant along the curve.

Step 2. One checks that all states on the curve satisfy the Rankine-Hugoniot re-lation with a discontinuity velocity σ = λj(VG). One obtains

d

ds(−λj(VG)(U (V (s)) − UL) + f(U(V (s))) − f(UL ))

= −λj(VG)C(V (s))V ′(s) + B(V (s))V ′(s)

= −λj(VG)C(V (s))rj (V (s)) + B(V (s))rj (V (s)) = 0.

Step 3. Finally, relation (3.112) shows that the entropy inequality is an equality.


3.4.3 Lax theorem in the space U

The curves constructed in the space V are easily transported to the space U bythe mapping V → U . Since the mapping is regular, the form of the local Taylorexpansion of all these curves is essentially the same. In what follows we firstdescribe rarefaction fans, shocks and contact discontinuities, and then formulatethe Lax theorem for the solution of the Riemann problem.

Rarefaction fans

Consider the rarefaction fan (Proposition 3.4.4)

Vj(s) = VL + srj (VL) +s2

2∇V rj(VL)rj(VL) + O(s3).

141

Write Uj (s) = U(Vj (s)). Notice that rj(UL ) = ∇V U(VL)rj (VL) is an eigenvectorof the eigenproblem

∇Uf(UL )rj(UL) = λj(0)rj(UL ),

and compare (3.18) and (3.89). This yields the rarefaction fan in the variable U :

Uj (s) = UL + srj(UL) +s2

2Aj + O(s3)

where Aj ∈ R can be calculated if desired. Similarly, the wavecurve of the theorem,

λj(s) = λj(0) + s(∇V λj(VL), rj (VL)) + O(s2),

can be expressed in the form

λj(s) = λj(0) + s(∇Uλj(V (UL)), rj(VL)) + O(s2),

which comes from

(∇V λj(VL), rj (VL)) = (∇Uλj(V (UL)), rj (VL)) .

Therefore the criterion that expresses the nonlinearity, or otherwise, of the jthfield admits the same form in the space U as in the space V . For the constructionof the solution of the Riemann problem, only the branch λj(s) > λj(0) needs tobe considered.

Shocks

Let us now consider the polar transform of the shock issuing from UL . It reads

U shockj (s) = UL + srj(UL) +

s2

2Aj + O(s3),

which is tangent at second order to the rarefaction fan. The entropy condition hasthe restriction that only one half of the curve must be retained. The branch that isretained is the one complementary to the half-branch of the rarefaction fan. Thisis illustrated in figure 3.4. The shock velocity is expanded as

σj(s) = λj(0) +s

2(∇Uλj(V (UL)), rj(VL)) + O(s2).

Contact discontinuities

Let us assume that(∇Uλj(V (UL)), rj(UL)) = 0

for all U in a neighborhood of UL, which means that the jth field is locally linearlydegenerate. The curve of the contact discontinuity can be expressed as

U contactj (s) = UL + srj(UL) +

s2

2Aj + O(s3) (3.115)



UL

half-branch of the shock curve

half-branch of the rarefaction fan curve

Figure 3.4: Rarefaction fan and shock at UL .

and one keeps both branches,

σj(s) = λj(s) = λj(0).

UL

contact curve

contact curve

Figure 3.5: Curve of the contact discontinuity from UL .

Solution of the Riemann problem

The previous results in the construction of the curves of rarefaction fans, shocksand contact discontinuities in phase space U are used to determine the solution ofthe Riemann problem ⎧

⎪⎨⎪⎩

∂tU + ∂x f(U) = 0,

U(0, x) = UL , x < 0,

U(0, x) = UR , 0 < x.

(3.116)

143

This is the celebrated Lax theorem [125], which nevertheless needs an assumptionof strict hyperbolicity. This will be relaxed later.

Theorem 3.4.21. Consider a system of conservation laws strictly hyperbolic at UL .Assume that al l fields are either truly nonlinear or linearly degenerate.

Then there exists a neighborhood NL ⊂ Rn of UL with the fol lowing prop-erties. For al l UR ∈ VG there exists a solution of the Riemann problem: it is aself-similar entropy weak solution, composed of n + 1 constant states separated byrarefaction fans, shocks or contact discontinuities.

U3

t

U4 = UDU1 = UG

U2

Figure 3.6: Structure of the general solution of the Riemann problem for n = 3.

Proof. The strict hyperbolicity and the continuity of the spectrum of a matrixallows us to obtain a strict ordering of the eigenvalues:

λ1(U) < λ2(U) < · · · < λn(U) ∀U ∈ NL.

Determine first a curve in phase space U1(s1 ; UL) with U1(0) = UL . Then deter-mine a second curve U2(s2 ; U1(s1)) with U2(0) = U1(s1), i.e. the starting point ofthe second curve is the end-point of the first curve. By iteration we construct thenext curves. At the end of this construction one has s = (s1 , s2 , . . . , sn) ∈ Rn suchthat

U(s) = Un(sn ; Un−1(sn−1 ; Un−2(· · · ; U1(s1 ; UL)))). (3.117)

This is a mapping between s ∈ Rn and U(s) ∈ Rn . For small s, an expansion tofirst order of U(s) yields

U(s) = UL + s1r1(UL ) + · · · + snrn(UL) + O(s2 ).

The Jacobian of the transformation at s = 0 is

∇sU(0) =

⎛⎝

r1(UL)...

rn(UL)

⎞⎠ .



Since the vectors ri(UL) are orthonormal for the matrix C(UL), the Jacobian isinvertible at s = 0 ∈ Rn and, by continuity for s in a small neighborhood of 0.So the transformation ∇sU(s) is invertible for 0 ≤ |s| < ǫ. Therefore one deducesthat for all UR ∈ NL there exists a unique solution s ∈ Rn of U(s) = UR . Goingback to the definition of U(s), this defines the n + 1 constant states

UL , U1(s1 ; UL), U2(s2 ; U1(s1 ; UL)), . . . ,

UR = Un(sn ; Un−1(sn−1 ; Un−2(· · · ; U1(s1 ; UR )))).

These n + 1 constant states are connected from one to the next by rarefactionfans, shocks or contact discontinuities. Define the corresponding self-similar func-tion (t, x) → U(x

t ). By construction it is an entropy weak solution. The proof iscomplete.

UL

U2

U3 = URU1

Figure 3.7: Structure of the solution of the Riemann problem for n = 3.

3.4.4 A Lagrangian Lax theorem in the space W

The previous Lax theorem is based on the hypothesis that the system is strictlyhyperbolic. In view of figure 3.7 this assumption is clearly an important partof the proof since it allows a natural ordering of the eigenvalues so that thereis no ambiguity in the construction of U(s) in (3.117). But if one relaxes thestrict hyperbolicity assumption, as we need to do, the construction falls apart.Unfortunately non-hyperbolicity is the generic situation for a Lagrangian systemof large size. If n − 1 − d = d, the multiplicity of the eigenvalue zero is greaterthan 2; see Proposition 3.2.11. To overcome this difficulty we rely on the followingevident result.

Proposition 3.4.22. The nul l eigenvalue of the Jacobian matrix of the quasi-linearLagrangian system issued from (3.42) has multiplicity n− rank(M). Moreover, theassociated eigenvectors are constant in the phase space W .

Proof. Let us recall that in the phase space associated with the variable W =(Ψ, S)t , the Lagrangian system admits the quasi-linear form

∇Ψ|SU∂tΨ + M∂mΨ = 0,

∂tS = 0.

145

So the null eigenvalue has multiplicity n−1−rank(M) plus 1, that is n−rank(M).The eigenvectors are of the form (r ∈ Rn−1, 0) with Mr = 0 and (0n−1, 1). Theyare indeed constant.

This property greatly simplifies the structure of contact discontinuities forthe null eigenvalue, but written in the space W . Instead of the Taylor expansion(3.115), one has

W contact,0j (s) = WG + srW

j (3.118)

with rWj ∈ Rn one of the constant eigenvectors described in the above proposition.

Note that if one starts from (3.118) and rewrites the equation in the space U ,one recovers (3.115). But (3.118) removes all ambiguities attached to the nulleigenvalue with multiplicity 2 or higher. Indeed, one can use the same constructionsince the ordering with which one constructs the solution for the null eigenvaluedoes not change the result. The result of the theorem is illustrated in figure 3.8.

Theorem 3.4.23. Consider a Lagrangian system of conservation laws (3.42). As-sume that it is hyperbolic at UL . Assume that non-zero eigenvalues are simple andthe associated fields are either truly nonlinear or linearly degenerate.

Then there exists a neighborhood NL ⊂ Rn of UL with the fol lowing prop-erties. For al l UR ∈ VG there exists a solution of the Riemann problem: it is aself-similar entropy weak solution, composed of constant states separated by rar-efaction fans, shocks or contact discontinuities.

Proof. With the correspondence λj(U) = μj(W ), the ordering of the eigenvaluesis

μ1(W ) < · · · < μr (W ) < μr+1(W ) = 0 = · · ·= μn−s(W ) < μn−s+1(W ) < · · · < μn(W ),

where r ∈ N is the number of negative eigenvalues, s ∈ N is the number of positiveeigenvalues and n−r−s is the multiplicity of the null eigenvalue. The constructionbecomes

W (s) = Wn(sn ; Wn−1(sn−1 ; Wn−2(· · · ; W1(s1 ; WL)))).

The main point is that internal states satisfy

Wn−s(sn−s; Wn−s−1(sn−s−1; Wn−s−2(· · · ; Wr+1(sr+1 ; Wr )))) = Wr +n−s∑

j =r

sjrWj .

Since the eigenvectors rWj are constant for r ≤ j ≤ n− s, the result is independent

of the ordering with which one constructs Wn−s from Wr. Therefore there is noambiguity in this construction. The rest of the proof is the same as in the globallystrictly hyperbolic case.

One can express the result in the Eulerian configuration. This is illustratedin figure 3.9. It is sufficient for our purposes to note that the velocity u is constant



null eigenvalue with multiplicity pt

m

U1 = UL

U2U3

U4 = UR

Figure 3.8: Solution of the Lagrangian Riemann problem in the (m, t) plane for|UL − UR| < ǫ.

across the null contact discontinuity. This is due to the equation ∂tτ − ∂mu = 0.It is therefore sufficient to transport the discontinuity in the Eulerian plane withthe velocity of the contact.

U3 = UD

x

U2

U1

t

UG

Figure 3.9: Solution of the Riemann problem in the (x, t) plane for |UL − UR| < ǫ.

Shocks

It is easy to derive an algebraic equation for Lagrangian shocks.

Proposition 3.4.24. Consider a shock for the Lagrangian system (3.51) with a non-zero shock velocity σ = 0. Then one has the algebraic Rankine-Hugoniot relation

((ΨL + ΨR

2, 1

), UR − UL

)= 0. (3.119)

The entropy inequality is SL > SR for σ > 0 and SL < SR for σ < 0.

3.5. Multidimensional Lagrangian systems 147

Proof. The general Rankine-Hugoniot relation is

−σ(UR − UL) +

⎛⎝

MΨD − MΨG

− 1

2(ΨD, MΨD) +

1

2(ΨG , MΨG)

⎞⎠ = 0.

Take the scalar product with the vector Z =(

ΨD +ΨG

2, 1). Thanks to the symmetry

of the matrix M , the difference of the fluxes vanishes. The entropy law is evident.

For one-dimensional Lagrangian gas dynamics, this relation takes the form

pG + pD

2(τR − τL) +

−uL − uR

2(uR − uL) + (eD − eG) = 0,

that is,

εD − εG +pG + pD

2(τR − τL) = 0.

This is the celebrated Rankine-Hugoniot relation. It was originally published byRankine [162] and independently by Hugoniot [105]. It is by analogy with thisequation that −σ[U ] + [f(U)] = 0 is called a Rankine-Hugoniot relation, for anysystem of conservation laws.

3.5 Multidimensional Lagrangian systems

A natural question is to extend to dimensions d > 1 some of the results for one-dimensional Lagrangian systems. In particular one might think of developing asystem of multidimensional axioms which extends what was done in section 3.2.It is clear that if such a convenient system of axioms exists, rotational invarianceshould play a role. But, in view of the derivation of Lagrangian numerical methods,it is not necessary to develop such a multidimensional theory with the same degreeof exactitude as was done in one dimension. Instead we rely on the followingformulation, which is enough to express the entropy principle for a large varietyof multidimensional systems.

Definition 3.5.1 (Multidimensional Lagrangian systems). Consider a system writ-ten in dimension d ≥ 1 as

∂t U +∑

1≤i≤d

∂Xi

(MiΨ

− 1

2(Ψ, MiΨ)

)=

(M0Ψ

0

), (3.120)

where U ∈ Rn is the unknown, and Ψ ∈ Rn−1 is a reduced entropy variablecomputed with the help of an entropy function S : Rn → R as follows: V = ∇USand Ψi = Vi

Vnfor 1 ≤ i = n − 1 (assuming Vn = 0). Assume that the matrices


Mi ∈ Rn−1 are symmetric Mi = M t

i for 1 ≤ i ≤ d. Assume that the matrices,which are not necessarily constant, satisfy the compatibility relation

∑

1≤i≤d

∂Xi Mi = M0 + M t0 . (3.121)

We call this a multidimensional Lagrangian system.

Notice that the function η(U ) = −S(U) is not a mathematical entropy in thestrict sense. Indeed, we never state that the matrices Mi are functions of U . Thismeans that the system (3.120) is not necessarily closed. However, if one considersa one-dimensional system with M0 = 0 then M1 is constant in space; if one addsthe condition that M1 is constant in time as well, one recovers the structure ofLagrangian systems in one dimension.

The main advantage of the structure (3.120) is its compatibility with theentropy law in the following sense.

Theorem 3.5.2. Assume that M0 , M1 , M2, . . . are smooth matrices. Then smoothsolutions of (3.120) satisfy ∂tS = 0.

Proof. Since the matrices are differentiable by hypothesis, smooth solutions canbe written as

∂tU +∑

1≤i≤d

(Mi (∂XiΨ) + (∂XiMi) Ψ

− 1

2(Ψ, ∂XiMiΨ) − (Ψ, Mi∂Xi Ψ)

)=

(M0Ψ

0

).

One has that ∂t S = Vn

((Ψ1

), U

), so

∂tS +Vn

2

∑

i

(Ψ, ∂Xi MiΨ) = Vn (Ψ, M0Ψ) =Vn

2

(Ψ,

[M0 + M t

0

]Ψ)

.

Therefore ∂tS = − Vn

2(Ψ, [M0 + M t

0 − ∑i∂Xi Mi] Ψ) = 0 using the compatibility

relation (3.121). The proof is complete.

We illustrate below the general structure with three examples.

Lagrangian gas dynamics in cylindrical and spherical coordinates

Lagrangian gas dynamics in cylindrical or spherical coordinates is an interestingcase where M0 = 0. To illustrate this, assume invariance of the flow with respectto rotation around a given axis (cylindrical invariance; see figure 3.10) or arounda given point (spherical invariance; see figure 3.11). For cylindrical invariance,invariance along the axis is also assumed.


z

r

Figure 3.10: Cylindrical invariance.

r

Figure 3.11: Spherical invariance.

It can be shown that compressible gas dynamics can be represented alongthe radius r ≥ 0 as

⎧⎪⎨⎪⎩

∂t(rd ρ) + ∂r(rd ρu) = 0,

∂t(rd ρu) + ∂r

(rdρu2 + rdp

)= drd−1p,

∂t(rd ρe) + ∂r (rdρue + rdpu) = 0.

(3.122)

One takes d = 1 for cylindrical invariance and d = 2 for spherical invariance. Thedimension of the initial problem is d+ 1. The velocity u is the radial velocity. Theplane or slab geometry is covered by the case d = 0. The material derivative is

d

dt= ∂t + u∂r .


So one rewrites (3.122) as

⎧⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎩

rdρd

dtτ − ∂r (rdu) = 0,

rdρd

dtu + ∂r

(rdp

)= drd−1p,

rdρd

dte + ∂r (rdpu) = 0.

(3.123)

Define the Euler-to-Lagrange change of coordinates

∂t|R r(t, R) = u(t, r(t, R)), r(0, R) = R,

where the radius at time t = 0 is R. The mass coordinate is defined by

dm = rdρdr = Rdρ0dR.

With the above notation, the system (3.122) or (3.123) can be rewritten as

⎧⎪⎪⎪⎨⎪⎪⎪⎩

∂tτ − ∂m(rdu) = 0,

∂t u + ∂m

(rdp

)=

d

ρrp,

∂te + ∂r (rdpu) = 0.

(3.124)

Define the matrices

M0 =d

ρr

(0 01 0

)and M1 = rd

(0 11 0

).

Proposition 3.5.3. The Lagrangian system (3.124) in cylindrical or spherical ge-ometry is of the form (3.120).

Proof. Indeed, Ψ = (p, −u)t and the compatibility relation (3.121) holds sincedm = ρrd dr and ∂

∂mM1 = M0 + M t

0 .

Quasi-Lagrangian ideal MHD

Let us discuss the system (3.54) of ideal MHD written in quasi-Lagrangian form,where quasi-Lagrangian means that a slight modification is made to the generalform (3.120). Let us consider ideal MHD in dimension d = 3 rewritten as

⎧⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎩

∂tρ + ∇ · ρu = 0,

∂tρu + ∇ · (ρu ⊗ u) + ∇P − ∇ · B ⊗ B

μ= 0,

∂tB + ∇ · (u ⊗ B − B ⊗ u) = 0,

∂ρe + ∇ ·(

ρue + P u − 1

μB(B, u)

)= 0.

(3.125)


Quite artificially we decide to freeze certain occurrences of the magnetic field:⎧⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎩

∂tρ + ∇ · ρu = 0,

∂tB + ∇ · (u ⊗ B − C ⊗ u) = 0,

∂tρu + ∇ · (ρu ⊗ u) + ∇P − ∇ · C ⊗ B

μ= 0,

∂ρe + ∇.

(ρue + P u − 1

μC(B, u)

)= 0,

(3.126)

where C could be any given vector field provided the divergence-free conditionholds,

∇ · C = 0. (3.127)

Clearly (3.126)–(3.127) is equivalent to (3.126) if and only if C = B. The corre-sponding formulation with the material derivative reads

⎧⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎩

ρDt τ − ∇ · u = 0,

ρDt τ B − ∇ · (C ⊗ u) = 0,

ρDt u + ∇P − ∇ · C ⊗ B

μ= 0,

ρDt e + ∇ ·(

P u − 1

μC(B, u)

)= 0.

(3.128)

Define, for any direction w = x, y or z,

Mw =

(0 Nw

N tw 0

), Nx =

⎛⎜⎝

1 0 0Cw 0 00 Cw 00 0 Cw

⎞⎟⎠ . (3.129)

One can take the entropy of a perfect gas. From TdS = dε + p dτ one obtains

TdS = de − u · du + P dτ − B

μdB.

So

Ψ =

(P , − Bx

μ, − By

μ, − Bz

μ, −ux, −uy , −uz

)t

.

Proposition 3.5.4. The system (3.128) can be written in the form

ρDt U + ∂x

(Mx Ψ

− 12(Ψ, Mx Ψ)

)+ ∂y

(MyΨ

− 12(Ψ, MyΨ)

)+ ∂z

(Mz Ψ

− 12(Ψ, Mz Ψ)

)= 0.

(3.130)

Proof. Obvious: notice that the matrices Mx, My and Mz are symmetric andsatisfy the compatibility relation

∂xMx + ∂z Mz + ∂z Mz = 0. (3.131)



Lagrangian gas dynamics in dimension d = 2

This example is inspired by [69, 72]. Let us start from (1.28). To simplify thenotation, we write

ρ0(X, Y ) = ρJ.

We note that the first of the Piola identities takes the form

ρ0 ∂t τ − ∂X (uM − vL) − ∂Y (vA − uB) = 0.

So one can write a unique system for the above equality, the momentum equationand the total energy equation:

⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩

ρ0 ∂tτ − ∂X (uM − vL) − ∂Y (vA − uB) = 0,

ρ0 ∂tu + ∂X(pM) + ∂Y (−pB) = 0,

ρ0 ∂tv + ∂X (−pL) + ∂Y (pA) = 0,

ρ0 ∂te + ∂X (puM − pvL) + ∂Y (pvA − puB) = 0.

(3.132)

It is an example of the general structure with

U =

⎛⎜⎝

τuve

⎞⎟⎠ , V =

1

T

⎛⎜⎝

p−u−v1

⎞⎟⎠ and Ψ =

(p

−u−v

).

Define

MX =

(0 M −L

M 0 0−L 0 0

)and MY =

(0 −B A

−B 0 0A 0 0

).

The matrices are symmetric and satisfy the compatibility relation

∂X MX + ∂Y MY = 0.

3.6 More on compressible gas dynamics

To end this chapter we review for completeness some well-known properties of thesolution of the Riemann problem for compressible gas dynamics,

⎧⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎩

∂tρ + ∂x (ρu) = 0,

∂t(ρu) + ∂x(ρu2 + p) = 0,

∂t(ρv) + ∂x (ρuv) = 0,

∂t(ρw) + ∂x (ρuw) = 0,

∂t(ρe) + ∂x(ρue + pu) = 0,

3.6. More on compressible gas dynamics 153

with the perfect gas pressure law p = (γ − 1)ρε where γ > 1. For smooth solutionsthis system admits the following reformulation close to (1.60):

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎩

∂tp + u∂xp + ρc2∂xu = 0,

∂tu + u∂xu +1

ρ∂xp = 0,

∂tv + u∂x v = 0,

∂tw + u∂xw = 0,

∂tS + u∂xS = 0.


Letting y = xt, one gets the equations for rarefaction fans,

⎧⎪⎪⎪⎨⎪⎪⎪⎩

− yp′ + up′ + ρc2u′ = 0,

− yu′ + uu′ +1

ρp′ = 0,

− yv′ = −yw′ = −yS′ = 0,

from which one deduces(u − y)2p′ = c2p′.

So y = u ± c. For y = u + c one gets

−cp′ + ρc2u′ = 0.

The other equations are

(u − y)S′ = (u − y)v′ = (u − y)w′ = 0.

Proposition 3.6.1. Assume a perfect gas pressure law. Then, along rarefaction fancurves, ⎧

⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎩

c = cL ± γ − 1

2(u − uL),

p = pL

(1 ± γ − 1

2

u − uL

cL

) 2γ

γ−1

,

ρ = ρL

(1 ± γ − 1

2

u − uL

cL

) 2γ−1

.

(3.133)

The other velocity components v and w and the entropy S are constant alongrarefaction fan curves.

Proof. Since u − y = −c < 0, we have S′ = 0, which implies that S is a constant.Therefore the density ρ and the sound velocity c can be expressed as functions ofa single variable p along the curve. One finds that

− 1

ρcp′ + u′ = 0.


This relation can be integrated exactly. For a perfect gas pressure law p = (γ −1)ρε, c2 = γ p

ρ, one has that p

ργ = pL

ργ

L

since the entropy is constant. So

ρc =√

γ

√ρL

p1

2γ

L

p1+ 1

γ

2 andp′

ρc=

2

γ − 1c′.

This integrates exactly to give − 2γ−1

c′+u′ = 0. The rest of the proof is evident.

The case − 1ρc

p′ + u′ = 0 corresponds to the plus sign in (3.133). The case1

ρcp′ + u′ = 0 corresponds to the minus sign.

Consider the case − 1ρcp′ + u′ = 0. If p increases so does u. On the other

hand, ρ = ρL

(p

pL

) 1γ

. So the density has the same monotonicity as p and u. From

c2 = γ pρ

= γp

1−1γ p

1γ

L

ρL, the sound velocity c also has the same monotonicity as p and

u. So u + c has the same monotonicity as p. Since y = u + c increases starting fromthe point on the left, the current pressure is greater than the pressure on the left.In a perfect gas the temperature T is proportional to the internal energy, which isitself proportional to c2. So the temperature decreases in the rarefaction fan. Onerecovers the physical phenomenon of the cooling of a gas in a rarefaction fan.

3.6.2 Discontinuities

The Rankine-Hugoniot equations read⎧⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎩

− σ[ρ] + [ρu] = 0,

− σ[ρu] + [ρu2 + p] = 0,

− σ[ρv] + [ρuv] = 0,

− σ[ρw] + [ρuw] = 0,

− σ[ρe] + [ρue + pu] = 0,

(3.134)

not forgetting the entropy inequality

−σ[ρS] + [ρuS] ≥ 0.

Contact discontinuities

The vectorial eigenspace for contact discontinuities associated with the null La-grangian eigenvalue has dimension 3. In differential form it can be written as(p, u, v, w, S)′ = (0, 0, α, β, γ).

Proposition 3.6.2. The pressure and normal velocity are continuous across a con-tact discontinuity. Both v and w components of the velocity and the entropy Shave an arbitrary jump.

Proof. Take σ = uL = uR in (3.134).


Shocks

Recall that by our definition of shocks, they satisfy the Rankine-Hugoniot relation(3.134) and the strict entropy inequality

−σ[ρS] + [ρuS] > 0.

This implies (σ − uL)ρLSL > (σ − uR)ρR SR, which de facto eliminates contactssince σ = uL = uR is no longer possible.

Proposition 3.6.3. The Rankine-Hugoniot relation for shocks implies

εR − εL +pR + pL

2(τR − τL) = 0. (3.135)

Moreover, the entropy condition is equivalent to: (a) SL > SR if ρL(σ − uL) =ρR (σ − uR) > 0; (b) SL < SR if ρL(σ − uL) = ρR(σ − uR) < 0. The transversevelocity components are constant across the shock.

Proof. Assume ρL(σ − uL) = ρR(σ − uR) > 0, which means that the fluid velocitymeasured in the shock frame is positive, uR − σ < 0 and uL − σ < 0. So the shocktransforms the right state R into a left state L. This is confirmed by the fact thatthe entropy inequality simplifies to SL > SR . One has the same interpretation buton the other side for ρL(σ − uL) = ρR(σ − uR) < 0.

It remains to show the thermodynamical Rankine-Hugoniot relation (3.135).A possibility is to note that it is exactly (3.119) for Lagrangian systems, as alreadyremarked. But it is instructive to prove (3.135) directly from algebraic manipula-tions of the system (3.134).

To do this, notice that the first equation of (3.134) can be rewritten as

j = ρL(σ − uL) = ρR(σ − uR).

It represents a constant flow of mass across the shock and yields

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎩

− j(τR − τL) − (uR − uL) = 0,

− j(uR − uL) + (pR − pL) = 0,

− j(vR − vL) = 0,

− j(wR − wL) = 0,

− j(eR − eL) + (pR uR − pLuL) = 0,

− j(SR − SL) > 0.

Since j = 0 one has vR − vL = wR − wL = 0. So eR − eL = εR − εL + 12(u2

R − u2L).

Eliminate the kinetic energy to get

−ju2

R − u2L

2+ (pR − pL)

uR + uL

2= 0,


and subtract from the total energy equation to obtain

−j(εR − εL) + (pRuR − pLuL) − (pR − pL)uR + uL

2= 0.

One has the identity (pR uR − pLuL) − (pR − pL)uR+uL

2= (uR − uL)pR+pL

2. One

can eliminate the difference of velocities uR − uL using the first equation. Thisyields

−j(εR − εL) − jpR + pL

2(τR − τL) = 0.

Dividing through by j = 0 yields the claim.

The analysis of the Rankine-Hugoniot relation for a perfect gas pressure lawis fundamental for the analysis of shocks in aeronautical applications.

Proposition 3.6.4. Assume a perfect gas pressure law. Then the pressure, normalvelocity, entropy and sound speed are strictly greater after a shock. The maximumcompression factor is γ+1

γ−1.

Proof. Assume j > 0 and start from

εR − εL +γ − 1

2

(εR

τR+

εL

τL

)(τR − τL) = 0.

Step 1. One has

εL = εR ×1 − (γ − 1) τR−τL

τR

1 + (γ − 1) τR−τL

τL

.

Define the inverse of the compression factor, z =ρR

ρL= τL

τR. One has

εL = εR × 2 + (γ − 1)(1 − z)

2 − (γ − 1)(z−1 − 1)= εR × (γ + 1) − (γ − 1)z

(γ + 1) − (γ − 1)z−1.

For a perfect gas pressure law pL = pRz−1εL , so

pL = pR(γ + 1) − (γ − 1)z

(γ + 1)z − (γ − 1).

Since the pressure is positive, γ−1γ+1

≤ z ≤ γ+1γ−1

. This shows that the compres-

sion factor is strictly bounded above by γ+1γ−1

.

Step 2. Let us now study the Lagrangian shock velocity j2 = − pR−pL

τR−τL, j > 0, with

respect to z. One has

j2 = − pR

τR

[1 − pL/pR

1 − τL/τR

]

= − pR

τR

⎡⎣1 − (γ+1)−(γ−1)z

(γ+1)z−(γ−1)

1 − z

⎤⎦ = γ

pR

τR

2

(γ + 1)z − (γ − 1).


+

− isentrope curve

p

τ

state before shock

state after shock

Rayleigh line

Hugoniot curve

Figure 3.12: Hugoniot curve in the (τ, p) plane. The state before the shock ismarked −, and the state after the shock marked +. The isentrope curve is tangentto the Hugoniot curve.

One easily determines the direction of variation of j with respect to z. Sincej > 0, d

dzj > 0 for z < 1. On the other hand, it is known that the variation of

j is related to the entropy production. To obtain that, we differentiate alongthe Hugoniot curve:

2jj′ =−(pR − pL)τ ′

L + (τR − τL)p′L

(τR − τL)2.

Differentiate the Rankine-Hugoniot relation to get

ε′L =

p′L

2(τR − τL) − pR + p

2τ ′

L .

The fundamental principle of thermodynamics yields

TLS′L = ε′

L + pLτ ′L =

p′L

2(τR − τL) − pR + pL

2τ ′

L + pLτ ′L

=p′

L

2(τR − τL) +

−pR + pL

2τ ′

L = (τR − τL)2jj ′.

Since SL > SR , this implies j ′ > 0 and z < 1. Therefore j increases in thesame direction as the entropy. For the same reasons, the pressure and normalvelocity increase across the shock. Notice that this is fully compatible withthe physical observation that a gas heats up across shocks.

We now discuss the Mach number in relation to the notion of supersonic andsubsonic states. Let us begin with a direct consequence of the previous analysis.

Proposition 3.6.5. Assume a positive Lagrangian shock velocity. One has the in-equalities c+ > σ − u+ and σ − u− > c−, where − refers to the state before theshock and + to the state after the shock.


Remark 3.6.6. One can analyze these inequalities by observing that the shockvelocity σ − u+ > 0 is subsonic after the shock since

σ−u+

c+< 1. On the other side

the shock velocity σ − u− is supersonic sinceσ−u−

c−> 1.

Proof. Assume for example that j > 0. Examination of the locus of possible statesafter a shock shows that the slope of the Rayleigh line is greater than the slope ofthe Hugoniot curve before the shock. Since the Hugoniot curve is tangent to theisentrope curve, one gets

j2 > − ∂p

∂τ |S−= (ρ−c−)2 ⇐⇒ j > ρ−c−.

By a similar reasoning, one obtains ρ+c+ < j . Moreover, j = ρ+(σ − u+) =ρ−(σ − u−), hence the claim.

This is more commonly expressed in terms of the Mach number, which is by

definition M =|u−σ|

c . One can rewrite Remark 3.6.6 as M− =|σ−u−|

c−> 1 and

M+ =|σ−u+|

c+< 1.

3.6.3 The Riemann problem for gas dynamics

Instead of detailing the construction of the Riemann problem for gas dynamics, werefer to the classical textbooks [93, 125, 127, 175] and give the numerical solutionof a simple Riemann problem referred to as the Sod problem.

This example was designed by Sod [184] for the purpose of benchmarkingnumerical methods. Take γ = 1.4 with the Riemann initial data

ρL = pL = 1, uL = vL = 0 and ρR = 0, 125, pR = 0, 1, uR = vR = 0.

The solution at time t = 0.14 is provided to five digits in figure 3.13. The exactvalues to five decimal places are provided in table 3.1. Refer to [189] for moredetails.

p∗ u∗ ρ∗L ρ∗

R v∗R = v∗

L

0.30313 0.92745 0.42632 0.26557 0

Table 3.1: Sod tube test problem: numerical values at the contact discontinuity.

3.7 Exercises

Exercise 3.7.1. Consider the system⎧⎨⎩

∂t u + ∂xu2

2= 0,

∂t v + a∂xv = 0, a ∈ R .

3.7. Exercises 159

0

0.2

0.4

0.6

0.8

1

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

’sod.ro’

0

0.2

0.4

0.6

0.8

1

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

’sod.p’

ρ p

0

0.2

0.4

0.6

0.8

1

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

’sod.u’

0

0.2

0.4

0.6

0.8

1

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

’sod.s’

u S

Figure 3.13: Reference solution of the Sod tube test problem at time t = 0.14. Therarefaction fan is visible in the left part of the profile for density, pressure andvelocity, but not for entropy. The contact discontinuity is visible on the densityand entropy profiles, but not for pressure and normal velocity. The shock is visibleon the right for all variables.

Show that it has one truly nonlinear field and one linearly degenerate field. Doesthe Lax theorem hold for the state UL = (a, a)?


∂tu + ∂xup

p= 0, p ≥ 2.

Show that the field is truly nonlinear if and only if p = 2.

Exercise 3.7.3. Consider ∂tu + ∂xv = 0,

∂tv + ∂x f(u, v) = 0.

Determine explicitly the wave velocities.


Exercise 3.7.4. The Awe-Rascle second-order system [10] for traffic flow is

∂tρ + ∂x ρu = 0,

ρ(∂t + u∂x)(u + αρ) = 0,

where α is a parameter. Write the conservative form of the system and showhyperbolicity for all α ∈ R . Find a parametrization such that one recovers theLWR model. Solve the Riemann problem for a general α ∈ R .

Exercise 3.7.5. Solve the Riemann problem for the shallow water system. Showthat the entropy condition can be interpreted as a hydraulic jump.

Exercise 3.7.6. Show that air can be compressed by a factor of at most 6 by aplanar shock.

Exercise 3.7.7. Consider a homogeneous column of air which expands in vacuum.Show that the velocity at the boundary between air and vacuum is 5c0 where c0

is the speed of sound in air at t = 0.

Exercise 3.7.8. Consider the system [88]

∂ta + ∂x

[(a2 − 1)b

]= 0,

∂tb + ∂x

[(b2 − 1)a

]= 0.

Study the hyperbolicity domain.

Exercise 3.7.9. Consider the system

∂tρ + ∂xρu = 0,

∂t(ρu) + ∂x(ρu2 + p) = 0,

with pressure law p = −ρ−1. Show that the Lagrangian form is

∂tτ − ∂mu = 0,

∂tu − ∂mτ = 0.

Solve analytically the Lagrangian form, then the Eulerian form.


∂tu + ∂x

((u2 + v2)u

)= 0,

∂tv + ∂x

((u2 + v2)v

)= 0.

Take (u, v) = (0, 0). Show that the states connected by shocks are either on thecircle u2 + v2 = u2 + v2 or on the line uv = uv. Assume (u, v) = (0, 0), and showthat all states in R 2 can be connected by a shock.

3.7. Exercises 161

Exercise 3.7.11. Consider the Keyfitz-Kranzer system [118]⎧⎪⎨⎪⎩

∂tu + ∂x

(u2 − v

)= 0,

∂tv + ∂x

(u3

3− u

)= 0.

Show that the two fields are truly nonlinear. Show that there is no strictly convexentropy. Show that there is no solution of the Riemann problem in the class ofbounded functions if the left and right initial data are sufficiently far from eachother. Compare with the Lax theorem.

Exercise 3.7.12 (Born-Oppenheimer Ti − Te model). Consider the system for ion–electron interaction ⎧

⎪⎪⎪⎨⎪⎪⎪⎩

∂tρ + ∂x (ρu) = 0,

∂t(ρu) + ∂x

(ρu2 + p

)= 0,

∂t(ρSe ) + ∂x (ρuSe) = 0,

∂t(ρe) + ∂x(ρue + pu) = 0,

(3.136)

where e = εi + εe + 12u2, p = pi + pe, pi = (γi − 1)ρεi with γi = 7/5 and pe =

(γe −1)ρεe with γe = 5/3. Such equations are often used in plasma physics to modela two-temperature plasma, where the electronic temperature Te may be differentfrom the ionic temperature Ti. One easily shows that η = −ρ(Si + Se) is a globalmathematical entropy. Show that smooth solutions satisfy ∂t(ρSi )+ ∂x(ρuSi) = 0.Show that the limit of viscous solutions satisfies

∂t(ρSi ) + ∂x(ρuSi ) ≥ 0. (3.137)

This reveals a difference between the effects of shocks on ions, S+i > S−

i , andon electrons S+

e = S−e . In the case of the electrons this is related to the famous

Born-Oppenheimer adiabatic hypothesis for electrons.Write the Rankine-Hugoniot relations for (3.136). Write the Lagrangian form.

What are the values of n, d and p (refer to (3.2.11))?

Exercise 3.7.13. Assume that a reasonable viscous approximation of compressibleturbulence is given by

⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩

∂tρ + ∂x(ρu) = 0,

ρ(∂t + u∂x)u + ∂x (p + pk) = (ν + νk)∂xx u,

ρ(∂t + u∂x)ε + p∂xu = ν (∂xu)2

,

ρ(∂t + u∂x)k + pk∂xu = νk (∂xu)2

.

Here pk = (γk − 1)ρk is the turbulent pressure and k is the density of turbulentenergy. One takes γk = 5

3and e = ε + k + 1

2u2. The classical viscosity is ν and the

turbulent viscosity is νk. We parametrize

νk

Tk= λ

ν

T, λ ∈ R

+,


where T and Tk are the temperatures:

TdS = dε + p dτ, TkdSk = dk + pkdτ.

Construct the conservative equation for the total energy ρe. Show that the limit(if it exists) of viscous solutions satisfies

⎧⎪⎪⎪⎨⎪⎪⎪⎩

∂tρ + ∂x (ρu) = 0,

∂t(ρu) + ∂x

(ρu2 + p + pk

)= 0,

∂t(ρ(Sk − λS)) + ∂x (ρu(Sk − λS)) = 0,

∂t(ρe) + ∂x (ρue + (p + pk)u) = 0,

with the entropy inequality (in the weak sense)

∂t(ρS) + ∂x(ρuS) ≥ 0.

Determine Ψ, n and d for the Lagrangian formulation.

Exercise 3.7.14. Detail (3.119) for one-dimensional ideal MHD.

Exercise 3.7.15. Compressible gas dynamics with Lorentz invariance reads⎧⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎩

∂tρ + ∂x(ρu) = 0,

∂t

(γ0

(1 +

h

c2

)ρu

)+ ∂x

(γ0

(1 +

h

c2

)ρu2 + p

)= 0,

∂t

(γ0

(1 +

h

c2

)ρ − p

c2

)+ ∂x

(γ0

(1 +

h

c2

)ρu

)= 0,

(3.138)

where c is the speed of light. Since this model is supposed to be compatible withthe principles of special relativity, one observes a clear distinction between thedensity of mass ρ seen by a fixed observer and the density of mass ρ0 evaluatedin the comoving frame. They are related by ρ = γ0ρ0. Set τ = ρ−1 and τ0 = ρ−1

0

such that τ = 1γ 0

τ0. The relativistic coefficient is

γ0 =1√

1 − u2

c2

≥ 1.

Let the thermodynamical enthalpy h in the moving frame be h = ε + pτ0. With aperfect gas pressure law p = (Γ − 1)ρ0ε, this yields h = Γε. We assume that thissystem is invariant with respect to Lorentzian transformations.

Define the modified velocity u = γ0

(1 + h

c2

)u and the modified total energy

e = c2γ0

(1 + ε

c2 + u2

c2

pτ0

c2

)− c2. Show that (3.138) is equivalent to

⎧⎪⎨⎪⎩

∂tρ + ∂x(ρu) = 0,

∂t (ρu) + ∂x (ρuu + p) = 0,

∂t (ρe) + ∂x (ρue + pu) = 0.

(3.139)


Set λ = uc , assume ε

c2 = O(λ2) and show that

u = u + O(λ2) with e = ε +1

2u2 + O(λ2). (3.140)

Start from TdS = dε + p dτ0 for a classical thermodynamical entropy. Show that

T

γ0dS = de − u du + p dτ. (3.141)

One can show first the identities de − u du + p dτ = d (e − uu + pτ ) + u du − τ dpand dγ0 = u

c2 γ30du. Show that smooth solutions satisfy

∂t(ρS) + ∂x(ρuS) = 0. (3.142)

Write (3.139) in Lagrangian form. Show that Ψ and M are the same as for classicalGalilean-invariant compressible fluids.

Exercise 3.7.16. Show that the Lorentz invariant system (3.138) is covariant, thatis, invariant under a change of reference frame

⎧⎪⎨⎪⎩

t = γ

(t′ +

β

cx′)

,

x = γ (x′ + βct′) ,

combined with a physically based change of unknown. Show that the velocity inthe new reference frame is

u′ =u − v

1 − uvc2

.

Compare with Exercise 1.5.6.


One can find in [127], the connection between thermodynamical entropy and math-ematical entropy of a system of conservation laws. The general theory of systemsof conservation laws is established in the seminal monograph of Lax [125]; see also[175, 94, 57, 82]. An important development in the symmetrizable formulation ofsystems of conservation laws is due to Godunov [99, 96], with extension to ro-tational invariance in [97]. Conditions for the differentiability of the eigenvaluesand eigenvectors of a matrix are given in [117]; these were used to characterizesome eigenproperties of a matrix which is not strictly hyperbolic. The Kulikovsipotential function is introduced in [121]. The Lagrangian multidimensional formu-lation comes from [68, 72]. We have not discussed the Liu admissibility conditionfor shocks; see [175] for a comprehensive introduction. Discussion of the Riemannproblem for non-convex equations of states appears in [151]. Shock stability canbe found in [134]. The multilayer shallow water system from [88] is related to asimilar model in [76].

Chapter 4

Numerical discretization

There is no doubt that computer experimentationwil l become a way of life in most parts of mathe-matical research.

– Peter Lax

(Mathematical Perspectives, 2008)

This chapter is dedicated to the development and analysis of families of cell-centered finite volume numerical schemes for the discretization of systems of con-servation laws which admit a Lagrangian formulation. The presentation does notmake clear a distinction between pure Lagrangian and Lagrange+remap schemes.The chapter is divided into three parts. The first section concerns the discretizationof compressible gas dynamics system, since it is fundamental to many applications.In the second section we develop a general numerical theory for Lagrangian systemsin one dimension. The third section is dedicated to multidimensional extensions. Inthese three sections, the emphasis is on the use of the general Lagrangian structurein relation to the analysis of entropy-satisfying numerical methods.

4.1 Compressible gas dynamics

Compressible gas dynamics systems are of central importance due to their ubiquityin applications. Our goal is to analyze some numerical methods which apply tothe discretization of

⎧⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎩

∂tρ + ∂x(ρu) + ∂y(ρv) + ∂z (ρw) = 0,

∂t(ρu) + ∂x(ρu2 + p) + ∂y(ρuv) + ∂z(ρuw) = 0,

∂t(ρv) + ∂x(ρuv) + ∂y(ρv2 + p) + ∂z (ρvw) = 0,

∂t(ρw) + ∂x(ρuw) + ∂y(ρvw) + ∂z(ρw2 + p) = 0,

∂t(ρe) + ∂x(ρue + pu) + ∂y(ρve + pv) + ∂z (ρwe + pw) = 0.

(4.1)




165

166 Chapter 4. Numerical discretization

Vertical direction

Horizontal direction

Figure 4.1: Cartesian mesh in two dimensions.

The total energy is the sum of the internal energy and the kinetic energy, e =ε + 1

2(u2 + v2 + w2). The pressure is assumed to be a function of the density and

internal energy, p = p(ρ, ε). One fundamental assumption is that there exists athermodynamical entropy S with the following two properties:

(a) It is strictly concave with respect to ε and τ = 1ρ.

(b) One has the fundamental principle of thermodynamics and there exists afunction T , called temperature, such that

TdS = dε + p dτ, T > 0. (4.2)

The system (4.1) is invariant with respect to rotation of the frame. In fact,this principle is widely used for the development of numerical methods. For exam-ple, consider the discretization of

⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩

∂tρ + ∂x (ρu) + ∂y(ρv) = 0,

∂t(ρu) + ∂x (ρu2 + p) + ∂y(ρuv) = 0,

∂t(ρv) + ∂x (ρuv) + ∂y(ρv2 + p) = 0,

∂t(ρe) + ∂x(ρue + pu) + ∂y(ρve + pv) = 0

(4.3)

on the Cartesian mesh of figure 4.1. All cells of the mesh can be referenced withtwo scalar indices j, k ∈ Z. We will consider a directional splitting technique. That

4.1. Compressible gas dynamics 167

is, we first discretize the system

⎧⎪⎪⎪⎨⎪⎪⎪⎩

∂tρ + ∂x (ρu) = 0,

∂t(ρu) + ∂x(ρu2 + p) = 0,

∂t(ρv) + ∂x (ρuv) = 0,

∂t(ρe) + ∂x(ρue + pu) = 0

(4.4)

in the horizontal direction, i.e. on every horizontal line of the Cartesian mesh.Then we solve the system

⎧⎪⎪⎪⎨⎪⎪⎪⎩

∂tρ + ∂y(ρv) = 0,

∂t(ρu) + ∂y(ρuv) = 0,

∂t(ρv) + ∂y(ρv2 + p) = 0,

∂t(ρe) + ∂y(ρve + pv) = 0

(4.5)

on all vertical lines of the Cartesian mesh. This splitting is done at every time stepand is decomposed into two intermediate steps. The extension to three dimensionsis obvious. This method is widely used for the discretization of gas dynamics andsimilar systems. It has the great advantage that the multidimensional featuresof the problem do not show up and only one-dimensional problems must be dis-cretized.

We will develop next a systematic numerical analysis of the so-called La-grange+remap strategy. This strategy consists of first discretizing the internalLagrangian structure associated with the Eulerian system (4.5) and then remap-ping the mesh to obtain an Eulerian scheme. Most of our efforts are focused onthe entropy properties of the numerical scheme; that is, the design principle is theentropy principle, not the theory of linearized Riemann solvers. Note that the sec-ond section will be devoted to showing that among all solvers with good entropyproperties, the optimal ones are nevertheless those which correspond to the use ofa certain linearized Riemann solver.

4.1.1 Principle of a Lagrange+remap scheme in one dimension

The basic principle starts from⎧⎪⎨⎪⎩

∂t τ − ∂mu = 0,

∂t u + ∂mp = 0,

∂t e + ∂m(pu) = 0,

(4.6)

which is a Lagrangian form of (4.4), with the mass variable defined by dm =ρ0dX = ρdx. One adds

∂tx = u, ρJ = ρ(0, X), (4.7)

which represents the displacement of the frame. Let us consider an initial mesh


1

∆x

∆t

6542 31

∆x

Mesh at time step tn = n∆t

654321

2 36

54

Figure 4.2: Principle of a Lagrange+remap scheme in dimension d = 1. Noticethe different scenarios during the first Lagrange stage: cell number 1 translatesto the left; cell number 2 expands; cells 3 and 4 translate to the right; cell 6 iscompressed. A time step restriction is needed to guarantee that the cells do notcross. At the end of this Lagrangian stage, one gets the cells j ′ for j = 1, . . . , 6.The second stage consists in remapping the Lagrangian mesh at the end of thefirst stage onto the initial Eulerian mesh.

with mesh size ∆x > 0. In a first stage one solves the Lagrangian system (4.6). Ina second stage one uses the discrete equivalent of (4.7). This finally yields a consis-tent discretization of the Eulerian system. The principle of the mesh displacementis depicted in figure 4.2.

4.1.2 Principle of an entropy Lagrangian solver

Consider a finite volume Lagrangian scheme written as⎧⎪⎪⎨⎪⎪⎩

mjτ ′j(t) − u∗

j +12

+ u∗j −1

2

= 0,

mju′j(t) + p∗

j +12

− p∗j −1

2

= 0,

mje′j (t) + p∗

j + 12

u∗j +1

2

− p∗j −1

2

u∗j − 1

2

= 0,

(4.8)

and assume that the fluxes are chosen as⎧⎪⎪⎨⎪⎪⎩

p∗j +1

2=

1

2(pj + pj +1) +

(ρc)j +12

2(uj − uj +1) ,

u∗j +1

2

=1

2(uj + uj +1) +

1

2 (ρc)j +12

(pj − pj +1) .(4.9)

These formulas will be justified later; however, it is already clear that they arean extension of similar formulas for the scalar case (2.61). The system is in semi-discrete form, that is, continuous in time. An important property is the following.


Proposition 4.1.1. The scheme (4.8) satisfies the identity

mjTjS′j(t) = −

(p∗

j + 12

− pj

) (u∗

j + 12

− uj

)+(

p∗j −1

2

− pj

) (u∗

j −12

− uj

). (4.10)

With the fluxes (4.10), it satisfies the entropy inequality S′j (t) ≥ 0.

Proof. The proof is only a matter of basic algebra using the fundamental principleof thermodynamics (4.2). Indeed, one has

mjTjS′j = mj

(ε′

j + pjτ′j

)= mj

(e′

j − uju′j + pjτ ′

j

)

= −(

p∗j +1

2

u∗j + 1

2

− p∗j −1

2

u∗j −1

2

)+ uj

(p∗

j +12

− p∗j − 1

2

)+ pj

(u∗

j + 12

− u∗j −1

2

)

=[−p∗

j +12

u∗j +1

2

+ ujp∗j +1

2

+ pju∗j +1

2

− pjuj

]

+[p∗

j − 12

u∗j −1

2

− ujp∗j −1

2

− pju∗j −1

2

+ pjuj

]

=[−(

p∗j +1

2

− pj

) (u∗

j +12

− uj

)]+[(

p∗j − 1

2

− pj

)(u∗

j − 12

− uj

)].

This proves the first part of the proposition.The second part easily follows from the remark that the fluxes can be rewrit-

ten as the solution of the specific linear system⎧⎪⎨⎪⎩

p∗j +1

2

− pj + (ρc)j +12

(u∗

j +12

− uj

)= 0,

p∗j +1

2

− pj +1 − (ρc)j +1

2

(u∗

j +12

− uj +1

)= 0.

(4.11)

Upon substituting this into (4.10) one obtains, after elimination of the pressures,

mjTjS′j (t) = (ρc)

j +12

(u∗

j +12

− uj

)2+ (ρc)

j −12

(u∗

j −12

− uj

)2≥ 0.

The velocities can be eliminated as well to arrive at the same conclusion.

4.1.3 Entropy Lagrangian solver based on matrix splitting

We are interested in matrix splitting because the flux (4.9) can be rewritten asa certain splitting which ultimately helps us to better understand the discreteentropy property. This will be generalized in section 4.2. To introduce this notionwe recall that the Lagrangian system (4.6) can be written as

∂tU + ∂m

⎛⎝

MΨ

− 1

2(Ψ, MΨ)

⎞⎠ = 0, (4.12)

where U = (τ, u, e)t , Ψ = (p, −u)t, M =

(0 1

1 0

)∈ R 2×2 and ∇US = 1

T(Ψ, 1).


Definition 4.1.2 (Matrix splitting). The pair of symmetric matrices M+ ∈ Rp×p

and M− ∈ Rp×p is said to be a splitting of a given symmetric matrix M = M t ∈Rp×p if

M+ =(M+

)t ≥ 0, M− =(M−)t ≤ 0 and M = M+ + M−.

One can also write |M | = M+ − M−. Notice that for a given symmetricmatrix M = M t, the decomposition M+ = 1

2M + 1

εI and M− = 1

2M − 1

εI is

always an admissible matrix splitting since the sign condition on M± is satisfiedfor ε > 0 small enough.

Proposition 4.1.3. The flux (4.9) can be rewritten as

MΨj +12

= M−j +1

2

Ψj + M+j + 1

2

Ψj +1

with(

M−j + 1

2

, M+j +1

2

)a splitting of M .

Proof. Take

M±j + 1

2

=1

2

⎛⎝ ± 1

(ρc)j + 12

1

1 ±(ρc)j + 12

⎞⎠ ,

which are indeed symmetric matrices. The matrix M+j +1

2

is non-negative with

eigenvalues 0 and 12

(1

(ρc)j+ 12

+ (ρc)j + 12

). The matrix M−

j + 12

is non-positive with

eigenvalues 0 and − 12

(1

(ρc)j+ 1

2

+ (ρc)j + 12

). The sum is equal to M . Moreover, one

can check directly that

Ψj +12

= M−1M−j + 1

2

Ψj + M−1M+j + 1

2

Ψj +1

gives back the flux (4.9). The proof is thus complete.

One can now give a more abstract proof of the entropy property for a stillsemi-discrete scheme

mjU ′j(t) + f∗

j +12

− f∗j +1

2

= 0 (4.13)

with a flux based on the matrix splitting

f∗j + 1

2

=

⎛⎝

M+j + 1

2

Ψj +1 + M−j +1

2

Ψj

− 1

2

(Ψj +1, M+

j + 12

Ψj +1

)− 1

2

(Ψj, M−

j + 12

Ψj

)⎞⎠ . (4.14)

Proposition 4.1.4. The Lagrangian scheme (4.13) with the flux based on the matrixsplitting (4.14) satisfies the entropy inequality

TjS′j ≥ 0.


Proof. Indeed, using the notation T∇US = (ψ, 1)t and U = (U , e)t , one obtains

mjTjS′j = mjΨj · U ′

j + mje′j

= − Ψj ·(

M+j +1

2

Ψj +1 + M−j + 1

2

Ψj

)

+

(1

2(Ψj +1, M+

j + 12

Ψj +1) +1

2(Ψj , M−

j + 12

Ψj)

)

+ Ψj ·(

M+j − 1

2

Ψj + M−j − 1

2

Ψj −1

)

−(

1

2(Ψj , M+

j − 12

Ψj) +1

2(Ψj −1, M−

j − 12

Ψj −1)

).

(4.15)

One can add to the first line the quantity (Ψj , MΨj), which can be convenientlydecomposed using the decomposition property of a matrix splitting,

M = M−j − 1

2

+ M+j − 1

2

= M−j + 1

2

+ M+j + 1

2

.

At the same time one subtracts the same quantity from the second line. One thusobtains, for the first line in the right-hand side of (4.15),

first line + (Ψj , MΨj) = − Ψj ·(

M+j + 1

2

Ψj +1 + M−j + 1

2

Ψj

)

+

(1

2(Ψj +1, M+

j +12

Ψj +1) +1

2(Ψj , M−

j + 12

Ψj)

)

+

(1

2(Ψj , M+

j +12

Ψj) +1

2(Ψj , M−

j +12

Ψj)

)

=1

2

(Ψj +1 − Ψj, M+

j +12

(Ψj +1 − Ψj))

≥ 0.

For the second line, a similar manipulation yields

second line − (Ψj, MΨj) = − 1

2

(Ψj −1 − Ψj, M−

j + 12

(Ψj −1 − Ψj))

≥ 0.

Therefore

mjTjS′j = first line + second line

= [first line + (Ψj, MΨj)] + [second line − (Ψj, MΨj)] ≥ 0.


The previous proposition yields the entropy inequality for the continuous-in-time scheme. At the same time we notice that only the entropy law for the firstderivatives has been used to prove the inequality. It is striking to observe that theother condition on the second derivatives, namely that S is strictly concave, can


be used to gain control on the CFL condition of the fully discrete scheme. To seethis, consider the fully discrete scheme

mj

ULj − Un

j

∆t+ fn

j +12

− fnj − 1

2= 0 (4.16)

for the discretization of the general Lagrangian system (4.12). The solution at theend of the Lagrangian time step is denoted by a superscript L (for Lagrange). Weuse the fully explicit flux

fnj +1

2

=

⎛⎜⎝

M+j +1

2

Ψnj +1 + M−

j + 12

Ψnj

− 1

2

(Ψn

j +1, Mj +1

2

+

Ψnj +1

)− 1

2

(Ψn

j , Mj +1

2

−Ψn

j

)⎞⎟⎠ (4.17)

which comes directly from (4.14).

Theorem 4.1.5 (Influence of the CFL condition on the discrete entropy inequality).Consider the explicit scheme (4.16) with the Lagrangian flux (4.17). Assume thatthe entropy is strictly concave with continuous second-order derivatives. Then foral l j ∈ Z there exists a positive constant cn

j > 0 such that if the CFL condition

cnj

∆t∆x

≤ 1 holds, the entropy is non-decreasing:

SLj ≥ Sn

j .

Proof. Define gj(α) = S(Un

j + α(ULj − Un

j ))

such that

gj(0) = Snj and gj(1) = SL

j .

Taylor expansion to second-order gives

gj(1) = gj(0) + g′j(1) − 1

2g′′

j (θ),

for some θ ∈ (0, 1). The chain rule yields

g′j(1) =

(∇US(UL

j ), ULj − Un

j

)

and

g′′j (θ) =

(UL

j − Unj , ∇2

US(Uθj )(UL

j − Unj ))

, Uθj = Un

j + θ(ULj − Un

j ).

Since the function U → S(U) is strictly concave, one obtains − 12g′′

j (θ) ≥ 0. On


the other hand one has

g′j(1) = − ∆t

mj

(∇US(UL

j ), fnj +1

2

− fnj −1

2

)= − ∆t

T Lj mj

((ΨL

j

1

), fn

j +12

− fnj −1

2

)

= − ∆t

T Lj mj

[(ΨL

j , M+j +1

2

Ψnj+1 + M−

j + 12

Ψnj

)

− 1

2

(Ψn

j +1, M+j + 1

2

Ψnj +1

)− 1

2

(Ψn

j , M−j +1

2

Ψnj

)−(

ΨLj , M+

j − 12

Ψnj + M−

j −12

Ψnj−1

)

+1

2

(Ψn

j , M+j − 1

2

Ψnj

)+

1

2

(Ψn

j −1, M−j − 1

2

Ψnj −1

)]

= − ∆t

T Lj mj

[(ΨL

j , M+j +1

2

Ψnj+1 + M−

j + 12

Ψnj

)

− 1

2

(Ψn

j +1, M+j + 1

2

Ψnj +1

)− 1

2

(Ψn

j , M−j +1

2

Ψnj

)

+(

ΨLj ,(

M+j + 1

2

+ M−j +1

2

)ΨL

j

)−(

ΨLj ,(

M+j − 1

2

+ M−j − 1

2

)ΨL

j

)

−(

ΨLj , M+

j − 12

Ψnj + M−

j − 12

Ψnj −1

)

+1

2

(Ψn

j , M+j − 1

2

Ψnj

)+

1

2

(Ψn

j −1, M−j − 1

2

Ψnj−1

)]

= − ∆t

2T Lj mj

[−(

Ψnj +1 − ΨL

j , M+j + 1

2

(Ψn

j +1 − ΨLj

))

−(

Ψnj − ΨL

j , M−j + 1

2

(Ψn

j − ΨLj

))

+(

Ψnj − ΨL

j , M+j − 1

2

(Ψn

j − ΨLj

))+(

Ψnj −1 − ΨL

j , M−j − 1

2

(Ψn

j −1 − ΨLj

))]

≥ − ∆t

2T Lj mj

[−(

Ψnj − ΨL

j , M−j + 1

2

(Ψn

j − ΨLj

))+(

Ψnj − ΨL

j , M+j − 1

2

(Ψn

j − ΨLj

))]

≥ − ∆t

2T Lj mj

(Ψn

j − ΨLj , |Mj |

(Ψn

j − ΨLj

)),

where we have defined|Mj | = M+

j − 12

− M−j + 1

2

≥ 0. (4.18)

So one gets the inequality

SLj ≥ Sn

j +1

T Lj

(Aj − ∆t

mjBj

)(4.19)

where Aj = − T L

j

2 g′′j (θ) ≥ 0 and Bj = 1

2(Ψnj − ΨL

j , |Mj |(Ψnj − ΨL

j )) ≥ 0. Twonoticeable features of this inequality are: (a) it is a purely local inequality whereall terms can be evaluated in cell j ; (b) Aj is a quadratic form function of UL

j −Unj ,


and Bj is another quadratic form function of ΨLj − Ψn

j . Since the function U → Ψ

is smooth, there exists a constant c > 0 such that∣∣ΨL

j − Ψnj

∣∣ ≤ c∣∣UL

j − Unj

∣∣. It

is then sufficient to take ∆tmj

small enough so that Aj − ∆tmj

Bj ≥ 0 as a quadratic

form. The proof is complete.

4.1.4 An optimal splitting for fluid dynamics

For practical computations, it is of course necessary to choose one particular split-ting, which leads to the question of the optimality of splittings. Our viewpointis that an optimal criterion is minimization of the constant cn

j that appears inTheorem 4.1.5. Indeed, if the constant is minimized, one can take a larger timestep, which is of great value in practical computations. Once this general principleis assumed, it remains to make it constructive. To this end we develop a lineariza-tion procedure for the quadratic forms Aj and Bj in inequality (4.19). It yieldsthe more explicit approximation cn

j ≈ E with E given by the Rayleigh quotient(4.21) (note that the indices j and n are omitted). It is then easy to determinethe coefficients which minimize E in Proposition 4.1.7.

Since the quadratic forms depend on many coefficients (and ultimately also on∆t through the intermediary of UL

j ), one linearizes to obtain the approximations

A = − T

2

(∆U, ∇2

US∆U)

+ O(∆t3), with ∆U = ULj − Un

j = O(∆t)

and

∆t

∆mB =

∆t

2ρj∆x(∆Ψ, |M |∆Ψ)+ O

(∆t4

∆x

), with ∆Ψ = ΨL

j − Ψnj = O(∆t).

Keeping in mind that a CFL constant will be chosen at the end in the form ∆t∆x ≤ c,

both approximations are of essentially the same order, O(∆t3). So the conditionA− ∆t

∆mB ≥ 0 which comes from (4.19) is simplified, upon retaining the dominant

approximation order, to

(max

∆U ∈Rn

((∆Ψ, |M |∆Ψ)

−Tρ (∆U, ∇2US∆U)

))∆t

∆x≤ 1. (4.20)

The maximum is over all ∆U . The difference ∆Ψ = (δp, −δu, −δv) in the numer-ator is actually dependent on ∆U as

∆Ψ = (∇UΨ)∆U + O(∆t2).

It is possible to invert this linearization procedure and express ∆U ∈ Rn as asecond–order approximation in terms of ∆Ψ ∈ Rn−1 and an auxiliary quantity,which is ∆S. That is, one linearizes ∆U as a function of ∆W .


Proposition 4.1.6. With the assumptions of Proposition 3.1.4 one has the relation

(∆U, ∇2US∆U) = − 1

T(∆u2 + ∆v2) − 1

Tρ2c2∆p2 − 1

T

∂T

∂S|p ∆S2 + O(∆t3)

where ∂T∂S |p > 0.

Proof. We have

C = (∆U, ∇2U S∆U) = (∆U, ∇U V ∆U) + O(∆t3) = (∆U, ∆V ) + O(∆t3)

= ∆τ ∆p

T− ∆u∆

u

T− ∆v∆

v

T+ ∆e∆

1

T+ O(∆t3)

= − 1

T(∆u2 + ∆v2) + ∆τ ∆

p

T+ ∆ε∆

1

T+ O(∆t3)

= − 1

T(∆u2 + ∆v2) + D + O(∆t3)

using e = ε + 12(u2 + v2), where

D = ∆τ ∆p

T+ ∆ε∆

1

T= (p∆τ + ∆ε)∆

1

T+

1

T∆τ ∆p + O(∆t3)

= T∆S∆1

T+

1

T∆τ ∆p + O(∆t3)

= T∆S∆1

T+

1

T

(∂τ

∂p|S ∆p +∂τ

∂S|p ∆S

)∆p + O(∆t3)

= − 1

Tρ2c2∆p2 + ∆S

(T∆

1

T+

1

T

∂τ

∂S|p ∆p

)+ O(∆t3).

One also has

∆1

T= − 1

T 2∆T + O(∆t2) = − 1

T 2(

∂T

∂S|p ∆S +∂T

∂p|S∆p) + O(∆t2).

This yields

D = − 1

Tρ2c2∆p2 +

1

T∆S

(− ∂T

∂S|p∆S − ∂T

∂p|S ∆p +∂τ

∂S|p ∆p

)+ O(∆t3).

The fundamental law of thermodynamics can be rewritten as

TdS + τ dp = d(ε + pτ ).

The Maxwell relation for cross-derivatives yields

∂T

∂p|S =∂τ

∂S|p .


This leads to a simplification

D = − 1

Tρ2c2∆p2 − 1

T

∂T

∂S|p ∆S2 + O(∆t3).

Therefore one obtains

C = − 1

T(∆u2 + ∆v2) − 1

Tρ2c2∆p2 − 1

T

∂T

∂S|p ∆S2 + O(∆t3).

This is a negative quadratic form provided that ∂T∂S|p > 0, which is indeed the case

thanks to

TdS =

(p

1

)· d

(τ

ε

)=

(p

1

)·[∇( p

T , 1T )(τ, ε)

]d

⎛⎝

p

T1

T

⎞⎠

=

(p

1

)·[∇( p

T , 1T )(τ, ε)

] ((p

1

)d

1

T+

(1T

0

)dp

).

Hence

T 3 ∂S

∂T |p = −(

p

1

)·[∇( p

T, 1

T)(τ, ε)

] ( p

1

).

This is a positive quantity since[∇( p

T , 1T )(τ, ε)

]is negative by hypothesis. So

∂T∂S |p > 0, which ends the proof.

The previous algebra shows that the constant in front of ∆t∆x

in the generalizedCFL condition (4.20) can be evaluated with the formula

E(|M |) = max∆W∈R4

⎛⎝ (∆Ψ, |M |∆Ψ)

ρ[(∆u2 + ∆v2) + 1

ρ2c2 ∆p2 + ∂T∂S |p∆S2

]

⎞⎠ . (4.21)

So the linearized CFL condition takes the form E ∆t∆x

≤ 1. Consider the simpleexample of

M+ =1

2M +

1

εI and M− =

1

2M − 1

εI

with ε ≤ 2 to ensure the non-negativity condition M± ≥ 0. One has

E = max∆W∈R4

⎛⎝

2ε(∆p2 + ∆u2 + ∆v2)

ρ[(∆u2 + ∆v2) + 1

ρ2c2 ∆p2 + ∂T∂S |p∆S2

]⎞⎠ = max

(2

ρε,

2ρc2

ε

).


It is then clear that the choice of the splitting has an effect on the simplified CFLconstant, and moreover that one cannot take ε too small. The optimal value ofε = 2 yields

E = max

(1

ρ, ρc2

)= c max

(1

ρc, ρc

).

Since max(a−1, a) ≥ 1 for a > 0, this shows that E ≥ c.

Proposition 4.1.7. One has the inequality E(|M |) ≥ c for al l admissible splittings.The minimal value is attained for

M+ =

⎛⎜⎜⎜⎜⎝

1

2ρc

1

20

1

2

ρc

20

0 0 0

⎞⎟⎟⎟⎟⎠

, M− =

⎛⎜⎜⎜⎝

− 1

2ρc

1

20

1

2− ρc

20

0 0 0

⎞⎟⎟⎟⎠ .

Proof. For the optimal splitting one has

Eopt = max∆W

(1

ρc∆p2 + ρc∆u2

ρ(∆u2 + ∆v2) + 1ρc2 ∆p2 + ρ ∂T

∂S |p∆S2

)= c.

Let us now consider another splitting M = M+ + M−. Denote the constant ofthis splitting by

E = max∆W

⎛⎝

(∆Ψ, |M |∆Ψ

)

ρ(∆u2 + ∆v2) + 1ρc2 ∆p2 + ρ ∂T

∂S |p∆S2

⎞⎠ .

Note that by definition |M| ≥ M . So

E ≥ max∆W

((∆Ψ, M∆Ψ)

ρ(∆u2 + ∆v2) + 1ρc2 ∆p2 + ρ ∂T

∂S |p∆S2

)

= max∆W

(−2∆p∆u

ρ(∆u2 + ∆v2) + 1ρc2 ∆p2 + ρ ∂T

∂S |p∆S2

)

= max∆u,∆p

(−2∆p∆u

ρ∆u2 + 1ρc2 ∆p2

)= c,

using the elementary relation

ρc∆u2 +1

ρc∆p2 ≥ −2∆u∆p.

So E ≥ Eopt. The proof is finished.


The flux associated with this optimal splitting is

M+Ψj +1 + M−Ψj

=

⎛⎜⎜⎜⎜⎝

1

2ρc

1

20

1

2

ρc

20

0 0 0

⎞⎟⎟⎟⎟⎠

⎛⎜⎝

pj +1

−uj +1

−vj+1

⎞⎟⎠ +

⎛⎜⎜⎜⎜⎝

− 1

2ρc

1

20

1

2− ρc

20

0 0 0

⎞⎟⎟⎟⎟⎠

⎛⎜⎝

pj

−uj

−vj

⎞⎟⎠

=

⎛⎜⎜⎜⎜⎝

− 1

2(uj + uj +1) − 1

2ρc(pj − pj +1)

1

2(pj + pj +1) +

ρc

2(uj − uj +1)

0

⎞⎟⎟⎟⎟⎠

.

The flux of the energy equation is

− 1

2

(Ψj +1, M+Ψj +1

)− 1

2

(Ψj, M−Ψj

)

= − 1

4ρcp2

j +1 +1

2pj +1uj +1 − ρc

4u2

j +1 +1

4ρcp2

j +1

2pjuj +

ρc

4u2

j

=

(1

2(pj + pj +1) +

ρc

2(uj − uj +1)

)(1

2(uj + uj +1) +

1

2ρc(pj − pj +1)

).

It is a consistent approximation of pu. The constant ρc in the flux could be chosenglobally, but in practice it seems highly desirable to adopt a local value; we useρc = (ρc)n

j + 12

at the interface between cell j and cell j + 1. The following value is

convenient for most calculations;

(ρc)nj + 1

2

=1

2

[(ρc)n

j + (ρc)nj +1

].

Additional considerations will be presented in section 4.2. One obtains the flux

fnj +1

2

=

⎛⎜⎜⎜⎜⎝

−u∗j +1

2

p∗j + 1

2

0

(pu)∗j + 1

2

= p∗j + 1

2

u∗j +1

2

⎞⎟⎟⎟⎟⎠

where ⎧⎪⎪⎪⎨⎪⎪⎪⎩

u∗j +1

2

=1

2(un

j + unj +1) +

1

2(ρc)nj + 1

2

(pnj − pn

j +1),

p∗j +1

2

=1

2(pn

j + pnj +1) +

(ρc)nj + 1

2

2(un

j − unj +1).

(4.22)


It is actually identical to (4.9). One obtains

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

ρnj ∆x

∆t(τ L

j − τ nj ) − u∗

j + 12

+ u∗j −1

2

= 0,

ρnj ∆x

∆t(uL

j − unj ) + p∗

j +12

− p∗j − 1

2

= 0,

vLj − vn

j = 0,

ρnj ∆x

∆t(eL

j − enj ) + p∗

j +12

u∗j + 1

2

− p∗j −1

2

u∗j − 1

2

= 0.

(4.23)

A practical CFL condition is(maxj cn

j

)∆t∆x

≤ CFL where the coefficient CFL < 1is a safety factor.

4.1.5 Moving grid

Since the Lagrangian phase is a discretization of (4.6), it is natural to require theLagrangian scheme to be compatible with a discrete version of (4.7). This is indeedthe case as shown below.

Define xnj +1

2

= (j + 12)∆x to be the edge between cell j and cell j + 1 at the

beginning of the time step. Since a velocity is provided by the Lagrangian flux(4.22), it is natural to define

xLj +1

2

= xnj +1

2

+ ∆tu∗j+ 1

2

. (4.24)

Proposition 4.1.8. The first equation of the Lagrangian scheme (4.23) and the griddisplacement (4.24) are compatible in the sense that the local mass is constant,

ρLj

(xL

j +12

− xLj − 1

2

)= ρn

j ∆x.

Proof. By substitution this relation is equivalent to

1

∆x

(∆x + ∆t(u∗

j +12

− u∗j −1

2

))

=ρn

j

ρLj

,

i.e. 1ρL

j

− 1ρn

j

= 1ρn

j ∆x(∆t(u∗

j + 12

− u∗j −1

2

)) which is the first equation of (4.23).

Define the length

∆xLj = xL

j +1

2

− xLj − 1

2

, j ∈ Z.

Since the mass is constant, ∆xLj ρL

j = ∆xρnj = ∆mj , the Lagrangian scheme (4.23)

is conservative in the sense that the following four formally conservative relations


hold: ⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

∑

j ∈Z∆mjτ L

j =∑

j ∈Z∆mjτ

nj ,

∑

j ∈Z∆mjuL

j =∑

j ∈Z∆mju

nj ,

∑

j ∈Z∆mjv

Lj =

∑

j ∈Z∆mjv

nj ,

∑

j ∈Z∆mje

Lj =

∑

j ∈Z∆mje

nj .

4.1.6 Remapping

The principles of remapping are simple: they essentially consist of writing thediscrete equations that correspond to the second stage of the scheme depicted infigure 4.2. The main constraints stem from natural conservativity and stabilityrequirements. The remapping can be achieved in a straightforward manner asshown below. Note, however, that it can be formulated as a splitting of operators;see section 4.1.10.

Consider figure 4.2 again. The projection corresponds to the computation ofmean values inside the cells of the fixed mesh. The only information to take careof is the sign of the velocities u∗

j −12

and u∗j + 1

2

, which determine the evolution of

cell j . A summary of the main cases is as follows:

j = 1 : ∆xρn+1j = (∆x + ∆tu∗

j +12

)ρLj − ∆tu∗

j+ 12

ρLj +1,

j = 2 : ∆xρn+1j = ∆xρL

j ,

j = 3, 4, 5 : ∆xρn+1j = (∆x − ∆tu∗

j −12

)ρLj + ∆tu∗

j −12

ρLj −1,

j = 6 : ∆xρn+1j = (∆x − ∆tu∗

j −12

+ ∆tu∗j +1

2

)ρLj − ∆tu∗

j+12

ρLj +1

+∆tu∗j−1

2

ρLj −1.

(4.25)

Replacing the densities ρ by their analogues ρu, ρv and ρe, one obtains the otherrelations. More precisely, ρn+1

j and ρLj are replaced by ρn+1

j un+1j and ρL

j uLj , and

so forth. For example, in cell j = 1 one gets

∆xρn+1j un+1

j = (∆x − ∆tu∗j +1

2

)ρLj uL

j + ∆tu∗j+1

2

ρLj +1uL

j +1.

Notice that (4.25) can be generically rewritten as

∆xρn+1j = (∆x − ∆tu∗

j−12

+ ∆tu∗j +1

2

)ρLj − ∆tu∗

j +12

ρL

j +12

+ ∆tu∗j −1

2

ρL

j − 12

(4.26)

where ρLj + 1

2

(resp. ρLj +1

2

uLj + 1

2

or ρLj + 1

2

eLj +1

2

) is the density (resp. momentum or total

energy) which is upwinded according to the sign of the edge velocity u∗j +1

2

. That


is, ⎧⎪⎪⎨⎪⎪⎩

if u∗j +1

2

> 0, ρL

j +12

= ρLj ,

if u∗j +1

2

< 0, ρLj +1

2

= ρLj +1,

if u∗j +1

2

= 0, arbitrary since the product vanishes.

Notice that (4.26) is equivalent to

∆xρn+1j = ∆xL

j ρLj − ∆tu∗

j+1

2

ρLj + 1

2

+ ∆tu∗j− 1

2

ρLj −1

2

. (4.27)

Inspection of the design principle in figure 4.2 shows that a criterion is needed toguarantee that xj +1

2does not cross more than one cell. This condition reads

(max

j

∣∣∣u∗j +1

2

∣∣∣)

∆t

∆x≤ 1. (4.28)

A slightly more stringent constraint can be used as an extra guarantee that noLagrangian cell will have a zero or negative length. Considering cell number 6 inthe figure, one gets the stricter condition

(max

j|u∗

j+ 12|)

∆t

∆x≤ 1

2,

where the maximal time step is divided by a factor of 2.

Proposition 4.1.9. The remapping is formal ly conservative up to boundaries:

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

∑

j ∈Z∆xρn+1

j =∑

j ∈Z∆xL

j ρLj =

∑

j ∈Z∆xρn

j ,

∑

j ∈Z∆xρn+1

j un+1j =

∑

j ∈Z∆xL

j ρLj uL

j =∑

j ∈Z∆xρn

j unj ,

∑

j ∈Z∆xρn+1

j vn+1j =

∑

j ∈Z∆xL

j ρLj vL

j =∑

j ∈Z∆xρn

j vnj ,

∑

j ∈Z∆xρn+1

j en+1j =

∑

j ∈Z∆xL

j ρLj eL

j =∑

j ∈Z∆xρn

j enj .

(4.29)

Proof. Geometrically evident in figure 4.2.

4.1.7 Eulerian formulation of a Lagrange+remap scheme

A Lagrange+remap scheme is a particular discretization of the Eulerian formula-tion of compressible gas dynamics. To make this statement clearer, one can rewrite


the two-stage Lagrange+remap scheme as a one-step Eulerian scheme in the form

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

ρn+1j − ρn

j

∆t+

u∗j +1

2

ρLj + 1

2

− u∗j −1

2

ρLj − 1

2

∆x= 0,

ρn+1j un+1

j − ρnj un

j

∆t+

u∗j +1

2

ρL

j + 12

uL

j +12

− u∗j − 1

2

ρL

j −12

uL

j − 12

∆x

+p∗

j+12

− p∗j − 1

2

∆x= 0,

ρn+1j vn+1

j − ρnj vn

j

∆t+

u∗j +1

2

ρLj +1

2

vLj +1

2

− u∗j − 1

2

ρLj −1

2

vLj− 1

2

∆x= 0,

ρn+1j en+1

j − ρnj en

j

∆t+

u∗j +1

2

ρLj + 1

2

eLj + 1

2

− u∗j −1

2

ρLj − 1

2

eLj − 1

2

∆x

+p∗

j+ 12

u∗j +1

2

− p∗j −1

2

u∗j − 1

2

∆x= 0.

(4.30)

The first equation comes from (4.27) after elimination of ∆xLj ρL

j = ∆xρnj . The

remaining three can be obtained from the generalization of (4.27) to the variablesρu, ρv and ρe. For example, the equation for u is obtained from

∆xρn+1j un+1

j = ∆xLj ρL

j uLj − ∆tu∗

j +12

ρLj +1

2

uLj + 1

2

+ ∆tu∗j−1

2

ρLj −1

2

uLj − 1

2

.

The second equation of the Lagrangian scheme (4.23) reads

1

∆t

(∆xL

j ρLj uL

j − ρnj ∆xun

j

)+ p∗

j + 12

− p∗j −1

2

= 0.

Simple manipulations give the second equation of (4.30). Similar algebra yieldsthe last two equations.

Note that the compact Eulerian formulation (4.30) is clearly conservativeup to boundary conditions. This yields another proof of the Eulerian part of theconservative relations (4.29). An important property in terms of stability is thefollowing.

Theorem 4.1.10. Assume the Lagrangian CFL constraint of Proposition 4.1.5 andthe remapping CFL constraint (4.28). Then the Lagrange+remap scheme (4.30)satisfies the entropy inequality

ρn+1j Sn+1

j − ρnj Sn

j

∆t+

u∗j +1

2

ρLj +1

2

SLj + 1

2

− u∗j − 1

2

ρLj −1

2

SLj − 1

2

∆x≥ 0 (4.31)

which is a discrete counterpart of ∂tρS + ∂x ρuS ≥ 0.


Proof. Rewrite (4.26) as⎧⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎩

ρn+1j = αρL

j + βρLj +1

2

+ γρLj − 1

2

,

ρn+1j un+1

j = αρLj uL

j + βρLj +1

2uL

j +12

+ γρLj − 1

2uL

j −12,

ρn+1j vn+1

j = αρLj vL

j + βρLj +1

2

vLj +1

2

+ γρLj − 1

2

vLj −1

2

,

ρn+1j en+1

j = αρLj eL

j + βρL

j +12

eL

j + 12

+ γρL

j − 12

eL

j − 12

,

making sure that β ≥ 0 and γ ≥ 0. Indeed, if the corresponding coefficients in(4.26) are negative, it is always possible to incorporate them into α, which is afactor of the central terms. Thanks to the CFL condition,α ≥ 0. So the coefficients(α, β, γ) define a convex combination since α, β, γ ≥ 0 and α + β + γ = 1 (notethat γ here has nothing to do with the constant of a perfect gas). So Un+1

j at

time step n + 1 is a convex combination of ULj −1, UL

j and ULj +1 at the end of

the Lagrangian time step L. Since the function ρS is concave with respect to itsarguments, it yields ρn+1

j Sn+1j ≥ αρL

j SLj + βρL

j +12

SL

j + 12

+ γρL

j −12

SL

j −12

. Using the

Lagrangian entropy inequality (4.23) SLj ≥ Sn

j , one gets

ρn+1j Sn+1

j ≥ αρLj Sn

j + βρLj +1

2

Snj + 1

2

+ γρLj −1

2

Snj − 1

2

. (4.32)

Multiplying by ∆x gives

∆xρn+1j Sn+1

j ≥ ∆xρnj Sn

j − ∆tu∗j +1

2

ρLj +1

2

SLj + 1

2

+ ∆tu∗j −1

2

ρLj − 1

2

SLj − 1

2

as claimed.

Proposition 4.1.11. Under the assumptions of Theorem 4.1.10, one has anotherdiscrete entropy inequality,

Sn+1j ≥ min

(Sn

j −1, Snj , Sn

j +1

). (4.33)

Proof. This is an easy consequence of (4.32) and the identity

ρn+1j = αρL

j + βρLj +1

2

+ γρLj − 1

2

.

This inequality yields some sort of nonlinear stability. For example, considera perfect gas for which S = log(ετ γ−1), γ > 1. Then the inequality implies

εnj

(ρnj )γ−1

≥ C0 = minj

(ε0

j

(ρ0j)γ−1

)> 0.

This means that the ratio of certain quantities which must remain non-negative forphysical correctness indeed remains positive. This is almost an a priori estimatefor the non-negativity of ρ and ε separately.

Remark 4.1.12. A remarkable feature of the Lagrange+remap scheme is the ex-istence of discrete entropy inequalities which have been proved for any generalpressure law, provided they admit a strictly concave entropy.


4.1.8 Boundary conditions

More general boundary conditions will be discussed in section 4.4.6. Here we con-centrate on the discretization of a normal velocity condition, also called a wallcondition. In dimension d = 1 it takes the form

u = 0 at the boundary of the computational domain. (4.34)

Since this condition is simple, numerous discretization approaches are possiblewhich all lead to the same result. Consider the example in figure 4.3, where thelast cell on the right of the domain of computation is indexed by J and an artificialunknown state is introduced behind the wall, with index J + 1. Once the artificialstate (often called a ghost cell) is determined, one just uses the standard schemefor the determination of the numerical flux at the interface J + 1

2between cell J

and J + 1, i.e. at the wall.

exterior ”cell” J + 1J − 1 J

uJ+1

2

= 0

right boundary J + 1

2

Figure 4.3: Wall boundary condition at J + 12.

Of course, now the question is how to determine a reasonable value for thisartificial state. In theory it requires four values, for ρJ+1 , uJ+1, vJ+1 and eJ+1 .However, in practice not four but just one numerical value is needed to obtain aconsistent discretization of all fluxes.

The solution is based on the fact that the Lagrangian velocity at the boundaryis

u∗J+ 1

2

=1

2

(un

J +1

ρcpn

J

)− 1

2

(−un

J+1 +1

ρcpn

J+1

), with a priori ρc = (ρc)n

j .

So the wall boundary condition u∗J+ 1

2

= 0 yields

−unJ+1 +

1

ρcpn

J+1 = unJ +

1

ρcpn

J ⇐⇒ pnJ+1 − ρcun

J+1 = pnJ + ρcun

J .


Then we observe that the generic Lagrangian flux can be written as

p∗J+ 1

2

=1

2(pn

J + ρcunJ ) +

1

2

(pn

J+1 − ρcunJ+1

)

whose solution is

p∗J+ 1

2

= pnJ − ρcun

J.

Finally, the Lagrangian flux of the energy equation is the product u∗J+ 1

2

p∗J+ 1

2

,

it is null. With this method, the wall boundary condition is discretized for allequations.

Remark 4.1.13. One obtains the same result with an anti-symmetrization of thevelocity variable and a symmetrization of the pressure, that is,

unJ+1 = −un

J and pnJ+1 = pn

J.

But, as noted above, only the linear combination pnJ+1 − ρcun

J+1 matters.

The remapping stage poses no difficulty for the wall boundarycondition, sincethe displacement of the mesh (see figure 4.2) is zero at the wall. The associatedfluxes vanish in (4.30).

Remark 4.1.14. The discrete entropy law (4.31) still holds for the rightmost cellJ in the domain of computation. The second discrete entropy inequality (4.33) ismodified to Sn+1

J = min(Sn

J−1, SnJ

).

4.1.9 A simple numerical result

We display in figure 4.4 numerical results for the Sod shock tube test problempresented section 3.6.3. Apart from the smearing effect of numerical methods, thenumerical solution is a good approximation of the true solution. With an initialcondition uL = uR = a = 1 this is called the Harten test problem: in figure 4.5one sees a similar solution, but translated to the right at a uniform velocity a = 1.Note that the Harten problem was defined to illustrate the entropy defect of someschemes [189]. One observes that the Lagrange+remap scheme is not polluted byany such non-entropic defect. This is a direct consequence of the discrete entropyinequalities (4.31) and (4.33).

4.1.10 Pure Lagrange and ALE methods in one dimension

ALE methods consider a grid velocity, denoted by v in the following, which canbe different from the fluid velocity. In dimension d = 1 the main interest in ALEmethods is pedagogical. In dimension d > 1, ALE is needed to regularize La-grangian grids which can deform to unacceptable proportions. This is discussedat the end of the chapter.


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’sod.ro’

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’sod.p’

ρ p

0

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’sod.u’

0

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0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

’sod.s’

u S

Figure 4.4: Sod test problem computed with a Lagrange+remap scheme and 200cells. Final time is t = 0.14.

We use the notation of section 1.2.2. Start from the Euler system in dimensiond = 1: ⎧

⎪⎨⎪⎩

∂tρ + ∂x (ρu) = 0,

∂t(ρu) + ∂x

(ρu2 + p

)= 0,

∂t(ρe) + ∂x(ρue + pu) = 0.

Consider the change of coordinates

t′ = t,∂x(t′ , X)

∂t′ = v(t′, x(t′ , X))

where (t, x) → v(t, x) is an arbitrary velocity called the grid velocity. Writing theequations in the set of coordinates (t′, X) amounts to computing the Jacobianmatrix

∇(t′,X)(t, x) =

(1 0

v J

), J =

∂x

∂X.

The comatrix (Definition 1.2.1) is com(∇(t′,X) (t, x)

)=

(J −v

0 1

). One obtains


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1

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

’sod.ro’

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1

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

’sod.p’

ρ p

0.9

1

1.1

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1.4

1.5

1.6

1.7

1.8

1.9

2

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

’sod.u’

0

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0.5

0.6

0.7

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

’sod.s’

u S

Figure 4.5: Harten test problem computed with a Lagrange+remap scheme and200 cells. Notice the Galilean invariance of the solution, up to some extra numericalsmearing.

from (1.20) the equations

⎧⎪⎨⎪⎩

∂t′ (ρJ) + ∂X (ρ(u − v)) = 0,

∂t′ (ρuJ) + ∂X (ρu(u − v) + p) = 0,

∂t′ (ρeJ) + ∂X (ρe(u − v) + pu) = 0.

The Piola identity (1.21) reads ∂t′ J − ∂X v = 0. To simplify, we use from now onthe same notation for the time variable, that is, t′ = t. One gets the closed system

⎧⎪⎪⎪⎨⎪⎪⎪⎩

∂t′ (ρJ) + ∂X (ρ(u − v)) = 0,

∂t′ (ρuJ) + ∂X (ρu(u − v) + p) = 0,

∂t′ (ρeJ) + ∂X (ρe(u − v) + pu) = 0,

∂t J − ∂X v = 0.

Since we wish to present ALE methods in terms of splitting of operators, we prefer


to write the system as

∂t

⎛⎜⎜⎜⎝

ρJ

J

ρuJ

ρeJ

⎞⎟⎟⎟⎠ + ∂X

⎛⎜⎜⎜⎝

0

−u

p

pu

⎞⎟⎟⎟⎠ + ∂X

⎛⎜⎜⎜⎝

ρ(u − v)

u − v

ρu(u − v)

ρe(u − v)

⎞⎟⎟⎟⎠ = 0. (4.35)

This formulation highlights the fact that the total flux is the sum of a Lagrangiancontribution and another contribution of convective nature. The initialization ofJ is naturally J(t = 0) ≡ 1. It is now clear that three main cases occur. A fourthexotic possibility may also arise.

Eulerian formulation: v = 0. Taking v = 0 is of course equivalent to the originalsystem (4.35). The second equation is eliminated.

Lagrangian formulation: v = u. The first equation in (4.35) becomes an ordinarydifferential equation and is therefore eliminated after exact integration, ρJ =ρ0. One can define naturally the mass variable dm = ρ0dx.

ALE formulation: v = 0 and v = u. This third case is neither pure Eulerian norpure Lagrangian. This is the arbitrary Lagrange-Euler formulation.

Energy-Lagrange system: v =(

1 + pρe

)u. This is the fourth and more exotic case.

It corresponds to the definition of the grid velocity v =(

1 +p

ρe

)u, which

makes the total flux of the energy equation zero. For a perfect gas in usualconditions,

1 ≤ p

ρe≤ γ − 1

so this definition makes sense. Using u − v = − p

ρeu one obtains the system

⎧⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎩

∂t(ρJ) − ∂X

(p

ρeu

)= 0,

∂t(ρuJ) + ∂X

(p − p

eu2

)= 0,

∂tJ − ∂X

((1 +

p

ρe

)u

)= 0.

A convenient mass variable is dm = ρ0e0dX. One obtains⎧⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎩

∂t1

e− ∂m

(p

ρeu

)= 0,

∂tu

e+ ∂m

(p − p

eu2)

= 0,

∂t1

ρe− ∂m

((1 +

p

ρe

)u

)= 0.


The use of this energy-Lagrange system for practical computations is an openproblem and will not be considered in this monograph.

Numerical discretization

The idea pursued below is to discretize the system (4.35) with a splitting strategy.One first discretizes

∂t

⎛⎜⎜⎜⎝

ρJ

J

ρuJ

ρeJ

⎞⎟⎟⎟⎠ + ∂X

⎛⎜⎜⎜⎝

0

−u

p

pu

⎞⎟⎟⎟⎠ = 0 (4.36)

during a time step ∆t. Secondly one discretizes

∂t

⎛⎜⎜⎜⎝

ρJ

J

ρuJ

ρeJ

⎞⎟⎟⎟⎠ + ∂X

⎛⎜⎜⎜⎝

ρ(u − v)

u − v

ρu(u − v)

ρe(u − v)

⎞⎟⎟⎟⎠ = 0 (4.37)

during the same time step ∆t.

Discretization of the first part (4.36)

Consider an infinite (j ∈ Z) initial grid with intermediate points Xj + 12,

∆Xj = Xj +12

− Xj − 12.

Start from (4.22) and (4.23) and simply adapt the notation: the Lagrangian fluxesare ⎧

⎪⎨⎪⎩

u∗j +1

2

=1

2(un

j + unj +1) +

1

2ρc(pn

j − pnj +1),

p∗j +1

2

=1

2(pn

j + pnj +1) +

ρc

2(un

j − unj +1).

(4.38)

The scheme is

⎧⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎩

ρn+ 1

2

j Jn+ 1

2

j − ρnj Jn

j = 0,

∆Xj

(J

n+ 12

j − Jnj

)− ∆t

(u∗

j +12

− u∗j − 1

2

)= 0,

∆Xj

(ρ

n+ 12

j Jn+ 1

2

j un+ 1

2

j − ρnj Jn

j unj

)+ ∆t

(p∗

j +12

− p∗j − 1

2

)= 0,

∆Xj

(ρ

n+ 12

j Jn+ 1

2

j en+ 1

2

j − ρnj Jn

j enj

)+ ∆t

(p∗

j + 12

u∗j +1

2

− p∗j −1

2

u∗j − 1

2

)= 0.

(4.39)


Discretization of the second part (4.37)

Definewn

j +12

= u∗j + 1

2

− vnj +1

2

,

with the understanding that u∗j +1

2

is an explicit value at time step n and vnj +1

2

is

the velocity of the grid point. The discretization of (4.37) with a correct definitionof the flux according to the sign of the differential velocity w = u − v yields thescheme⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

∆Xj

(ρn+1

j Jn+1j − ρ

n+ 12

j Jn+ 1

2

j

)+ ∆t

(ρ

n+ 12

j + 12

wnj +1

2

− ρn+ 1

2

j −12

wnj −1

2

)= 0,

∆Xj

(Jn+1

j − Jn+ 1

2

j

)+ ∆t

(wn

j +12

− wnj −1

2

)= 0,

∆Xj

(ρn+1

j Jn+1j un+1

j − ρn+ 1

2

j Jn+ 1

2

j un+ 1

2

j

)

+∆t(

ρn+ 1

2

j + 12

un+1

2

j +12

wnj +1

2

− ρn+ 1

2

j −12

un+ 1

2

j −12

wnj −1

2

)= 0,

∆Xj

(ρn+1

j Jn+1j en+1

j − ρn+ 1

2

j Jn+ 1

2

j en+ 1

2

j

)

+∆t(

ρn+ 1

2

j + 12

en+ 1

2

j + 12

wnj +1

2

− ρn+ 1

2

j − 12

en+ 1

2

j − 12

wnj −1

2

)= 0.

(4.40)

The convention is that fn+ 1

2

j +12

= fn+ 1

2

j for wnj +1

2

≥ 0 and fn+ 1

2

j +12

= fn+ 1

2

j +1 for wnj +1

2

<

0.

Reformulation on a moving grid

The grid displacement is naturally defined by

xn+ 1

2

j +1

2

= xnj + 1

2

+ ∆tu∗j+1

2

(4.41)

for the first stage and by

xn+1j +1

2

= xn+ 1

2

j +12

+ ∆t(

vnj +1

2

− u∗j +1

2

)(4.42)

for the second stage. One gets the total displacement

xn+1j + 1

2

= xnj +1

2

+ ∆tvnj +1

2

. (4.43)

Note the initial condition x0j +1

2

= Xj + 12. Define

∆xnj = xn

j +12

− xnj − 1

2

.

One checks that, by construction, the variation in time of ∆XjJnj is equal to the

variation in time of ∆xnj . So

Jnj =

∆xnj

∆Xj. (4.44)


Define the mass in the cell by ∆Mnj = ∆xn

j ρnj . It is therefore easy to rewrite the

first stage (4.39) in the form

⎧⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎩

∆Mnj

∆t

(τ

n+ 12

j − τ nj

)− u∗

j +12

+ u∗j −1

2

= 0,

∆Mnj

∆t

(u

n+ 12

j − unj

)+ p∗

j +12

− p∗j −1

2

= 0,

∆Mnj

∆t

(e

n+ 12

j − enj

)+ p∗

j +12

u∗j + 1

2

− p∗j −1

2

u∗j −1

2

= 0,

which coincides with the Lagrangian scheme (4.23). So the CFL condition of thisstage can be written as

maxj

(cn

j

∆xnj

)∆t ≤ CFL < 1. (4.45)

A similar entropy inequality holds under the CFL condition:

Sn+ 1

2

j ≥ Snj . (4.46)

The second stage (4.40) is analyzed as follows. Define ∆xn+1

2

j as the length of theLagrangian cell at the end of the Lagrangian time step. The first equation of (4.40)becomes

∆xn+1j ρn+1

j − ∆xn+ 1

2

j ρn+ 1

2

j + ∆t(

ρn+ 1

2

j + 12

wnj + 1

2

− ρn+ 1

2

j −12

wnj −1

2

)= 0.

Assume for simplicity that wnj +1

2

≥ 0 and wnj − 1

2

≥ 0. The compatibility between

ρn+ 1

2

j + 12

and ρn+ 1

2

j − 12

implies that

ρn+ 1

2

j + 12

= ρn+ 1

2

j and ρn+ 1

2

j − 12

= ρn+ 1

2

j −1 .

Therefore

∆xn+1j ρn+1

j =(

∆xn+ 1

2

j − ∆twnj+ 1

2

)ρ

n+ 12

j + ∆twnj −1

2ρ

n+ 12

j −1 . (4.47)

For the second equation one obtains similarly

∆xn+1j =

(∆x

n+12

j − ∆twnj +1

2

)+ ∆twn

j −12

=⇒ ∆xn+1

2

j − ∆twnj +1

2

= ∆xn+1j − ∆twn

j−12

.

Substitution into (4.47) yields

∆xn+1j ρn+1

j =(

∆xn+1j − ∆twn

j −12

)ρ

n+ 12

j + ∆twnj −1

2

ρn+ 1

2

j −1 . (4.48)


From comparison with (4.25) it is clear that, in the example for the cells indexed by3, 4 and 5, u∗

j +12

≥ 0 has been replaced by wnj +1

2

≥ 0. Besides this only difference,

the general situation is unchanged. The other cases for wnj +1

2

≤ 0 can be analyzed

with the same method.

Proposition 4.1.15. The scheme (4.40) is equivalent to a geometric projection ontothe mesh of figure 4.6.

Proof. The different cases can be represented in figure 4.6.

n + 1

2

n

n + 1

u∗

w = v − u∗

Figure 4.6: The grid velocity is u∗j +1

2

in the Lagrangian first stage. It is equal to

wnj + 1

2

= vnj +1

2

− u∗j +1

2

in the second remapping stage.

A natural stability constraint for the second stage is(

maxj

|w∗j+ 1

2|)

∆t

∆x≤ 1

2. (4.49)

This prevents any crossings. The following result is a generalization of Theo-rem 4.1.10 to the ALE configuration.

Theorem 4.1.16. Assume the two CFL conditions (4.45) and (4.49). Then the ALEscheme (4.39)–(4.40) satisfies the discrete entropy inequality

∆xn+1j ρn+1

j Sn+1j − ∆xn

jρnj Sn

j + ∆t(

w∗j +1

2

ρn+ 1

2

j + 12

Sn+ 1

2

j +12

− w∗j −1

2

ρn+ 1

2

j − 12

Sn+ 1

2

j − 12

)≥ 0.

(4.50)

Numerical illustrations

The results of the Lagrange+remap scheme for the Sod tube test problem pre-sented in figure 4.4 are complemented here by results for the same test problem,but in pure Lagrangian mode in figure 4.7 and in ALE mode in figure 4.8. Inpure Lagrangian mode, there is no numerical smearing at the contact discontinu-ity, and the numerical defaults at the interface are preserved by the scheme. Thisis especially visible for the density profile, which exhibits a small spike referredto as the wall-heating phenomenon. The spike is more evident on the entropyprofile. Velocity and pressure profiles do not have noticeable oscillations. The re-sults of an ALE calculation are given in figure 4.8 with an arbitrary grid velocity

4.2. Linearized Riemann solvers and matrix splittings 193

vj +12

= 0.5 sin(4πx)sin(2πt). These results are comparable to those of the pure La-grange computation and of the Eulerian computation in Lagrange+remap mode.

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’sod.S’

u S

Figure 4.7: Sod tube test problem: pure Lagrangian simulation with 200 cells andfinal time t = 0.14. Notice the compression of the mesh behind the shock and thenumerical discrepancy (called wall-heating) at the contact discontinuity.

We also present the result of a Sod tube test problem in d = 2 dimensionscomputed with the Lagrange+remap strategy combined with a directional splittingtechnique. That is, the two-dimensional problem is solved with a series of one-dimensional numerical methods. Note that the entropy inequality still holds inhigher dimensions.

The initial data is a Sod test problem from both sides of the interface definedby

√x2 + y2 = 0.5. Considerable numerical smearing is visible on the density

profile at the contact discontinuity.

4.2 Linearized Riemann solvers and matrix splittings

The notion of a Riemann solver was not addressed in the previous discussion.We will now pay much more attention to the general structure of Lagrangian


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0

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0.5

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0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

’sod.S’

u S

Figure 4.8: Same set-up as in figure 4.7, except that the calculation is done in ALEmode with grid velocity vj +1

2= 0.5 sin(4πx)sin(2πt).

Riemann solvers, in view of the definition of more efficient fluxes. When appliedto compressible gas dynamics, this discussion will also highlights the differencesbetween the Riemann solver

⎧⎪⎨⎪⎩

p∗ =pL + pR

2+

ρ∗c∗

2(uL − uR),

u∗ =uL + uR

2+

1

2ρ∗c∗ (pL − pR)(4.51)

where ρ∗c∗ is a local approximation of the acoustic impedance, typically

ρ∗c∗ =1

2(ρLcL + ρRcR ) , (4.52)

and another linearized Riemann solver given by⎧⎪⎪⎨⎪⎪⎩

p∗∗ =ρRcRpL + ρLcLpR

ρLcL + ρRcR+

ρLcLρR cR

ρLcL + ρRcR(uL − uR),

u∗∗ =ρLcLuL + ρRcRuR

ρLcL + ρRcR+

1

ρLcL + ρRcR(pL − pR).

(4.53)


Densite 0.8 0.6 0.4 0.2

0 0.1

0.2 0.3

0.4 0.5

0.6 0.7

0.8 0.9

1 0 0.1

0.2 0.3

0.4 0.5

0.6 0.7

0.8 0.9

1

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9

1

Vitesse 0.8 0.6 0.4 0.2

0 0.1

0.2 0.3

0.4 0.5

0.6 0.7

0.8 0.9

1 0 0.1

0.2 0.3

0.4 0.5

0.6 0.7

0.8 0.9

1

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9

1

Figure 4.9: Density ρ and modulus of the velocity√

u2 + v2 at the final timeT = 0.2. It is instructive to compare these results with the Lagrangian calculationin figure 4.27.

This solver is the original acoustic solver of Godunov. One could argue that thedifference between (4.51) and (4.53) is formally small. This is true except in twosituations, which validate a priori the use of (4.53).

• One situation is where a strong gradient exists in the computational domain.It can be due to a shock or, even simpler, is could be a manifestation of


Pression 0.8 0.6 0.4 0.2

0 0.1

0.2 0.3

0.4 0.5

0.6 0.7

0.8 0.9

1 0 0.1

0.2 0.3

0.4 0.5

0.6 0.7

0.8 0.9

1

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9

1

Entropie 0.6 0.4 0.2

0 0.1

0.2 0.3

0.4 0.5

0.6 0.7

0.8 0.9

1 0 0.1

0.2 0.3

0.4 0.5

0.6 0.7

0.8 0.9

1

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

Figure 4.10: Pressure p and entropy S at the final time T = 0.2.

the discontinuity of some initial data for the computation of a Riemannproblem. In this case we will see that (4.53) is more robust in terms of theCFL condition than (4.51). This is a generic result for all Lagrangian systems.

• The second situation concerns multi-material computations, where the pres-sure law may be different on the two sides. In this case one might expect thateven when the pressure and density are continuous, pL = pR and ρL = ρR,one nevertheless has cL = cR . Consequently the average value (4.52) does


not make sense on physical grounds, and so the formula (4.53) is appealingin this case.

4.2.1 Solution of the Lagrangian linearized Riemann problem

We construct an approximation solution, described in figure 4.11, of the system

∂tU + ∂m

⎛⎝

MΨ

− 1

2(Ψ, MΨ)

⎞⎠ = 0 (4.54)

with initial data U(x, 0) = U0(x), where

U0(x) = UL for m < 0, U0(x) = UR for m > 0. (4.55)

Intermediate flux (MΨ)∗

UL UR

t

m

Figure 4.11: Structure of the exact solution of the Lagrangian Riemann problem.The linearized solution focuses mainly on the determination of an approximatevalue of the intermediate flux (Mψ)∗ .

We base our approximation on the exact solution of a convenient linearizationof the system (4.54) where the variable U has been eliminated in the function sothat W = (Ψ, S). Indeed, the general formula (3.50) yields that smooth solutionsof (4.54) can be rewritten as smooth solutions of

⎧⎨⎩

− D∂tΨ + M∂mΨ = 0, D =(

∇2

U|S e)

= Dt > 0,

∂t S = 0.(4.56)

The matrix D of this linearization procedure is a priori function of the mass (orspace) variable, that is,

D(m) = DL for m < 0, D(m) = DR for m > 0. (4.57)


Here DL and DR are the degrees of freedom of this approach. We will distinguishtwo cases: the first corresponds to a constant D and is called a one-state solver;the second is for the more general situation (4.57).

4.2.2 One-state solvers

We begin with the simpler situation where the matrix, denoted by D∗ = (D∗)t > 0,is constant in space. For example, one may take

D∗ =1

2(DL + DR ) = (D∗)t > 0.

The linearized system is−D∗∂t Ψ + M∂mΨ = 0. (4.58)

It is a linear system, and both matrices are constant in space. Such a system iscalled a Friedrichs system. It is easy to construct the exact solution.

Denote by (−λ∗i , r∗

i ) the eigenpairs of M with respect to the matrix D∗ :

Mr∗i = −λ∗

i D∗r∗i , (r∗

i , D∗rj) = δij .

Take the scalar product of (4.58) with the eigenvector λ∗i r∗

i corresponding to anon-zero eigenvalue λ∗

i = 0. One has

−λ∗i (r∗

i , D∗∂tΨ) + λ∗i (r∗

i , M∂mΨ) = 0,

or∂t (r∗

i , MΨ) + λ∗i ∂m (r∗

i , MΨ) = 0.

This is a transport equation with velocity λ∗i . The solution is

⎧⎪⎪⎨⎪⎪⎩

if λ∗i > 0, (r∗

i , MΨ) moves to the right,

if λ∗i < 0, (r∗

i , MΨ) moves to the left,

if λ∗i = 0, (r∗

i , MΨ) = 0 by a direct computation.

This yields the value of (MΨ)∗ we are looking for.

Definition 4.2.1 (One-state solvers). A one-state solver is the unique solution ofthe linear system

⎧⎪⎨⎪⎩

(r∗i , (MΨ)∗) = (r∗

i , MΨL) for λ∗i > 0,

(r∗i , (MΨ)∗) = (r∗

i , MΨR) for λ∗i < 0,

(r∗i , (MΨ)∗) = 0 for λ∗

i = 0.

(4.59)

Since the eigenvectors are orthonormal, i.e. (r∗i , D∗ rj) = δij, this solution can be

expressed as

(MΨ)∗ =∑

λ∗i >0

(r∗i , MΨL)D∗r∗

i +∑

λ∗i <0

(r∗i , MΨR)D∗r∗

i . (4.60)


As an example of this method, consider the system of compressible gas dy-namics where

D∗ =

⎛⎝

1

(ρ∗c∗)20

0 1

⎞⎠ and M =

(0 1

1 0

).

The normalized eigenvalues of Mr = −λD∗r satisfy

det

⎛⎝

λ

(ρ∗c∗)21

1 λ

⎞⎠ =

λ2

(ρ∗c∗)2− 1 = 0.

So λ+ = ρ∗c∗ and λ− = ρ∗c∗ as expected. The eigenvectors are r+ = (ρ∗c∗ , −1)t

and r− = (ρ∗c∗, 1)t. Define, in the natural notation,

(−u∗

p∗

)= (MΨ)∗ .

The system (4.59) can then be rewritten as

ρ∗c∗(−u∗) − p∗ = ρ∗c∗(−uL) − pL,

ρ∗c∗(−u∗) + p∗ = ρ∗c∗(−uR) + pR,(4.61)

and its solution is precisely (4.51) or (4.22).

4.2.3 Two-state solvers

We describe in this section a more general solution based on the system

⎧⎪⎨⎪⎩

−D∗∗(m)∂t Ψ + M∂mΨ = 0,

D∗∗(m) = DL for m < 0,

D∗∗(m) = DR for m > 0.

This equation has discontinuous coefficients a priori, which may present funda-mental technical difficulties that we do not wish to treat in full detail. Instead weconsider a more direct generalization of the formulas (4.59) in the form (4.62).

Denote by (s∗∗i ) the eigenfamily consisting of the left eigenvectors with pos-

itive eigenvalues,

s∗∗i = rL

i : MrLi = −λL

i DLrLi , λL

i > 0,

the right eigenvectors with negative eigenvalues,

s∗∗i = rR

i : MrRi = −λR

i DR rRi , λR

i < 0,


and the null eigenvectors,

s∗∗i = r∗

i = rLi = rR

i : Mr∗i = 0, λ∗

i = λRi = λG

i = 0.

The number of elements in the family (s∗∗i ) is equal to the size of M , that is, equal

to n − 1. The reason is that the number of positive, negative and null eigenvaluesof the eigenproblem Mr = −λDr is independent of the matrix D = Dt > 0.

Definition 4.2.2 (Two-state solvers). A two-state solver is the unique solution ofthe linear system

⎧⎪⎨⎪⎩

(rLi , (MΨ)∗∗) = (rL

i , MΨL) for λLi > 0,

(rRi , (MΨ)∗∗) = (rR

i , MΨR) for λRi < 0,

(r∗i , (MΨ)∗∗) = 0 for λ∗

i = 0.

(4.62)

Remark 4.2.3. Even though it uses different notation, the system (4.62) expressesthat the linearized Riemann invariants are constant, and that they are integratedalong the characteristics on both sides. Let us write the Riemann invariants indifferential form as lt dΨ = 0. One obtains the formal system

ltLΨ∗∗ = ltLΨL for λL > 0,

ltR Ψ∗∗ = ltRΨR for λR < 0,

which can be directly identified with (4.62).

The eigenvectors are upwinded in two-state solvers and are centered (or av-eraged) in one-state solvers: this is the only difference. Since the linear system(4.62) has n − 1 unknowns and n − 1 equations, the solution exists and is uniqueif and only if the family (s∗

i ) is linearly independent. This is actually the case, sothe definition of two-state solvers makes sense.

Proposition 4.2.4. The family (s∗∗i ) is linearly independent.

Proof. Consider a vanishing linear combination

∑

i

αis∗∗i =

∑

λLi

>0

αirLi +

∑

λRi

<0

αirRi +

∑

M r∗i =0

αir∗i = 0. (4.63)

To prove linear independence, it is necessary and sufficient to prove that αi = 0for all i. Define the vector

z =∑

λLi

>0

αirLi +

∑

M r∗i =0

αir∗i = −

∑

λRi

<0

αirRi

and determine the scalar product (z, Mz) using the two different expressions forz.


• Firstly, the eigenvectors (rLi , r∗

i ) are all eigenvectors of the left eigenproblemMrL

i = −λLi DL rL

i . With a normalization hypothesis

(rLi , DLrL

i ) = δij ,

this yields

(z, Mz) =

⎛⎝ ∑

λLi >0

αirLi +

∑

M r∗i=0

αir∗i , M

⎛⎝ ∑

λLi >0

αirLi +

∑

M r∗i

=0

αir∗i

⎞⎠⎞⎠

=

⎛⎝

∑

λLi

>0

αirLi +

∑

M r∗i=0

αir∗i ,

∑

λLi

>0

αiλLi DLrL

i

⎞⎠ =

∑

λLi

>0

λLi |αi|2 ≥ 0.

• Secondly, one has

(z, Mz) =

⎛⎝

∑

λRi <0

αirRi , M

⎛⎝

∑

λRi <0

αirRi

⎞⎠⎞⎠ = −

∑

λRi <0

λDi |αi|2 ≤ 0.

So (z, Mz) = 0 =∑

λLi >0 λL

i |αi|2 = −∑

λRi <0 λR

i |αi|2 . This shows that αi =

0 for all indices i such that λLi > 0 or λR

i < 0. One then observes that (4.63)simplifies to 0 =

∑i αir

∗i , which finally implies αi = 0 for all i.

Remark 4.2.5. The application of these ideas to the system of compressible gasdynamics modifies the linear system (4.61), which becomes

ρLcL(−u∗∗) − p∗∗ = ρLcL(−uL) − pL,

ρRcR(−u∗∗) + p∗∗ = ρR cR(−uR) + pR.(4.64)

The solution of this system is provided by the formulas (4.53). Of course it is thesame as (4.51) if the coefficients are the same on the right and on the left. Thisnew flux is referred to in the literature as the acoustic flux.

The next step is to rewrite the solution of a two-state solver with a matrixsplitting. Define on the left side

M+L =

∑

λLi

>0

λLi

(DLrL

i

)⊗(DGrL

i

), M−

L =∑

λLi

<0

λLi

(DLrL

i

)⊗(DLrL

i

)(4.65)

and on the right side

M+R =

∑

λRi

>0

λRi

(DRrR

i

)⊗(DRrR

i

), M−

R =∑

λRi

<0

λRi

(DRrR

i

)⊗(DRrR

i

), (4.66)

where by convention the eigenvectors of M for the null eigenvalue are expressedas Mr∗

i = 0.


(p∗∗, u∗∗)two states

p

u

(pR, uR)(pL, uL)

(p⋆, u⋆)one state

Figure 4.12: Graphical interpretation of the one-state solver (4.61) and two-statesolver (4.64). In both cases the flux is at the intersection point of two lines. Thedifference lies in the slopes of the lines. For a one-state solver the slope of the twolines is the same in absolute value. For a two-state solver the slopes are a prioridifferent (in absolute value). In the figure ρLcL > ρRcR.

Proposition 4.2.6. Consider the solution (MΨ)∗∗ of the two states solver (4.62).

There exist two vectors ΨL, ΨR ∈ Rn−1 such that it can be written in the matrixsplitting form as

(MΨ)∗∗ = M+L ΨL + M−

L ΨL = M+R ΨR + M−

R ΨR. (4.67)

Proof. Set b = M+L ΨL =

∑λL

i >0 λLi

(DGrG

i , ΨL

)(DLrL

i

)where the operator is

defined in (4.65). By the definitions of (MΨ)∗∗ and b, and using the orthonormalityof the eigenvectors, one has that for λj > 0,

(rL

j , (MΨ)∗∗ − M+L ΨL

)=

(rL

j , (MΨ)∗∗)−(rL

j , M+L ΨL

)

=(rL

j , MΨL

)−

∑

λLi >0

λLi

(DLrL

i , ΨL

) (DLrL

i , rLj

)

=(rL

j , MΨL

)− λL

j

(DLrL

j , ΨL

)

=(rL

j , MΨL

)−(MrL

j , ΨL

)= 0.

Similarly, for the null eigenvalues one has(r∗

j , (MΨ)∗∗ − M+L ΨL

)=(r∗

j , (MΨ)∗∗) −(r∗

j , M+L ΨL

)= 0 − 0 = 0.

Therefore the vector (MΨ)∗∗ − M+L ΨL belongs to the orthogonal complement of

the space spanned by the rLj for λL

j > 0 and the r∗j for null eigenvalues, that is,

(MΨ)∗∗ − M+L ΨL =

∑

λLj <0

αjDLrLj , αj ∈ R .


One can rewrite

∑

λLj <0

αjDLrLj = −

∑

λLj <0

αj

λLj

MrLj = M

⎛⎜⎝−

∑

λLj <0

αj

λLj

rLj

⎞⎟⎠ = M−

L ΨL

withΨL = −

∑

λLj

<0

αj

λLj

rLj .

This shows the first part of the claim, namely that (MΨ)∗∗ = M+L ΨL + M−

L ΨL.The second part is proved with the same arguments.

The next proposition is obvious if M is non-singular.

Proposition 4.2.7. There exists a solution Ψ∗∗ ∈ Rn−1 of the equation

MΨ∗∗ = (MΨ)∗∗ .

This vector is defined up to an element of the kernel of M .

Proof. A necessary condition is that the right-hand side satisfies the compatibilityrelation (MΨ)∗∗ ∈ Ker (M t)⊥ = Ker(M)⊥ . This is the case since by definition(r∗

i , (MΨ)∗∗) = 0 for all vectors r∗i such that Mr∗

i = 0. This yields the existence ofΨ∗∗, which is clearly defined up to the addition of a vector in the kernel of M .

Proposition 4.2.8. With the notation and results of Proposition 4.2.6, one has

(Ψ∗∗, MΨ∗∗) =(M+

L ΨL, ΨL

)+(

M−L ΨL, ΨL

)=

(M+

R ΨR, ΨR

)+(M−

R ΨR, ΨR

).

(4.68)

Proof. The proof relies on the construction of two representatives of Ψ∗∗. The firstis based on a decomposition along the eigenvectors rL

i of the relation

MΨ∗∗ = M+L

ΨL + M−L

ΨL,

where M+L =

∑λL

i >0λL

i

(DLrL

i

)⊗

(DLrL

i

), M−

L =∑

λLi <0

λLi

(DLrL

i

)⊗

(DLrL

i

)

and M = M+L + M−

L . This yields the formulas

M+L ΨL =

∑

λLi >0

λLi αL

i

(DLrL

i

), αL

i =(DLrL

i , ΨL

)for λL

i > 0,

M−L ΨL =

∑

λLi

<0

λLi αL

i

(DLrL

i

), αL

i =(

DLrLi , ΨL

)for λL

i < 0.

DefineΨ∗∗

L = −∑

λLi =0

αLi rL

i


with the identity

MΨ∗∗L = −

∑

λLi

=0

αLi MrL

i =∑

λLi

=0

λLi αL

i DLrLi = M+

L ΨL + M−L ΨL.

Since the eigenvectors satisfy orthogonality relations, namely (DLrLi , rL

j ) = δij

and (MrLi , rL

j ) = −λiδij , one has

(Ψ∗∗L , MΨ∗∗

L ) = −∑

λLi =0

λLi

(αL

i

)2,

(ΨL, M+

L ΨL

)=

∑

λLi

>0

λLi

(αL

i

)2,

and (ΨL, M−

L ΨL

)=

∑

λLi <0

λLi

(αL

i

)2.

Thus (Ψ∗∗L , MΨ∗∗

L ) =(ΨL, M+

L ΨL

)+(

ΨL, M−L ΨL

). Since Ψ∗∗ and Ψ∗∗

L differ only

by an element in the kernel of M , this yields the first part of (4.68),

(Ψ∗∗, MΨ∗∗) = (Ψ∗∗L , MΨ∗∗

L ) =(ΨL, M+

L ΨL

)+(

ΨL, M−L ΨL

).

The second part is proved with a similar decomposition based on the right eigen-vectors.

Remark 4.2.9 (Illustration and verification of formula (4.68)). It is useful to illus-trate the non-intuitive identity (4.68) with a simple and more intuitive example.Take formula (4.64), the solution of which is (p∗∗, u∗∗) given by (4.53). In this casethe matrices are

M±L =

⎛⎜⎝

± 1

2(ρc)L

1

21

2± (ρc)L

2

⎞⎟⎠ and M±

R =

⎛⎜⎝

± 1

2(ρc)R

1

21

2± (ρc)R

2

⎞⎟⎠ .

The equation of the auxiliary left state MΨ∗∗ = M+L ΨL+M−

L ΨL can be expressedas (

−u∗∗

p∗∗

)=

1

2

⎛⎝

1

(ρc)LpL − uL

pL − (ρc)LuL

⎞⎠ +

1

2

⎛⎝ − 1

(ρc)LpL − uL

pL + (ρc)L uL

⎞⎠ ,

that is,⎧⎪⎨⎪⎩

u∗∗ = − 1

2(ρc)L(pL − (ρc)LuL) +

1

2(ρc)L(pL + (ρc)L uL) ,

p∗∗ =1

2(pL − (ρc)LuL) +

1

2(pL + (ρc)L uL) .


This relation implicitly defines the auxiliary state (pL , uL), whose exact value isnevertheless not needed. The formula (4.68) can now be checked directly. Indeed,(Ψ∗∗ , MΨ∗∗) = −2u∗∗p∗∗. By substitution one obtains the additive representation

−2u∗∗p∗∗ =1

2(ρc)L(pL − (ρc)LuL)

2 − 1

2(ρc)L(pL + (ρc)L uL)

2.

On the other hand, one can check directly that

(ΨL, M+

L ΨL

)=

1

2(ρc)L(pL − (ρc)LuL)

2

and (ΨL, M−

L ΨL

)= − 1

2(ρc)L(pL + (ρc)L uL)

2.

This yields the formula

(Ψ∗∗, MΨ∗∗) =(ΨL, M+

L ΨL

)+(

ΨL, M−L ΨL

).

The verification is similar for the right states ΨR and ΨR.

The representation formulas simplify the proof of the following result, whichis a generalization of Theorem 4.1.5. We consider the scheme

mj

ULj − Uk

j

∆t+ fk

j +12

− fkj −1

2= 0 (4.69)

with the flux

fkj +1

2

=

⎛⎝

(MΨ)∗∗j + 1

2

− 1

2(Ψ, MΨ)

∗∗j +1

2

⎞⎠ . (4.70)

In this formula it should be understood that (MΨ)∗∗j + 1

2

is defined by (4.62) or

equivalently by (4.68) with L = j and R = j + 1, and that the energy flux is thendefined by (4.68).

Theorem 4.2.10. Consider the generic Lagrangian numerical scheme with a two-state flux at interface j + 1

2given by (4.62) or (4.68). There exists a positive

constant ckj such that if the CFL condition

ckj

∆t

∆mkj

≤ 1

holds, then the scheme satisfies the entropy inequality

S(Uk+1

j

)≥ S

(Uk

j

).


Proof. The proof is essentially the same as that of Theorem 4.1.5, but with oneimportant difference: the states Ψk

j +1 and Ψkj −1 in Theorem 4.1.5 are replaced by

the auxiliary states Ψkj +1 = ΨL on the left and Ψk

j −1 = ΨR on the right (be carefulwith the notation, the auxiliary states are determined on the same interface).

Define gj(α) = S(Uk

j + α(Uk+1j − Uk

j ))

with gj(0) = Skj and gj(1) = Sk+1

j .As in the proof of Theorem 4.1.5, one has the following formula for the secondderivative:

g′′j (θ) =

(UL

j − Ukj , ∇2

US(Uθj )(UL

j − Ukj ))

, Uθj = Uk

j + θ(ULj − Uk

j ).

The first derivative at the end-point is

g′j(1) = − ∆t

∆mkj

(∇US(Uk+1

j ), fkj + 1

2

− fkj −1

2

)

= − ∆t

T k+1j ∆mk

j

((Ψk+1

j

1

), fk

j +12

− fkj −1

2

)

= − ∆t

T k+1j ∆mk

j

[(Ψk+1

j , ML,+j +1

2

Ψkj+1 + M

L,−j + 1

2

Ψkj

)− 1

2

(Ψk

j +1, ML,+j + 1

2

Ψkj +1

)

− 1

2

(Ψk

j , ML,−j + 1

2

Ψkj

)−(

Ψk+1j , MR,+

j − 12

Ψkj + MR,−

j − 12

Ψkj −1

)

+1

2

(Ψk

j , MR,+j − 1

2

Ψkj

)+

1

2

(Ψk

j −1, MR,−j − 1

2

Ψkj−1

)]

= − ∆t

T k+1j ρk

j ∆x

[(Ψk+1

j , ML,+

j + 12

Ψkj +1 + M

L,−j + 1

2

Ψkj

)− 1

2

(Ψk

j +1, ML,+

j + 12

Ψkj +1

)

− 1

2

(Ψk

j , ML,−j + 1

2

Ψkj

)+(

Ψk+1j ,

(M

L,+j + 1

2

+ ML,−j + 1

2

)Ψk+1

j

)

−(

Ψk+1j ,

(M

R,+j − 1

2

+ MR,−j − 1

2

)Ψk+1

j

)−(

Ψk+1j , M

R,+j − 1

2

Ψkj + M

R,−j − 1

2

Ψkj −1

)

+1

2

(Ψk

j , MR,+

j − 12

Ψkj

)+

1

2

(Ψk

j −1, MR,−j − 1

2

Ψkj−1

)]

= − ∆t

2T k+1j ρk

j ∆x

[−(

Ψkj +1 − Ψk+1

j , ML,+

j + 12

(Ψk

j +1 − Ψk+1j

))

−(

Ψkj − Ψk+1

j , ML,−j + 1

2

(Ψk

j − Ψk+1j

))+(

Ψkj − Ψk+1

j , MR,+j − 1

2

(Ψk

j − Ψk+1j

))

+(

Ψkj −1 − Ψk+1

j , MR,−j − 1

2

(Ψk

j −1 − Ψk+1j

))]

≥ − ∆t

2T k+1j ρk

j ∆x

×[−(Ψk

j − Ψk+1j , M

L,−j +1

2

(Ψkj − Ψk+1

j )) + (Ψkj − Ψk+1

j , MR,+

j − 12

(Ψkj − Ψk+1

j )]

≥ − ∆t

2T k+1j ρk

j ∆x(Ψk

j − Ψk+1j , |M |j(Ψk

j − Ψk+1j ))


where in the last line

|M |j = MR,+j − 1

2

− ML,−j + 1

2

. (4.71)

So one can write

Sk+1j ≥ Sk

j +1

T k+1j

(Aj − ∆t

∆mkj

Bj

)(4.72)

with Aj = − 12T k+1

j g′′(θ) ≥ 0 and Bj = − 12(Ψk

j − Ψk+1j , |M |j(Ψk

j − Ψk+1j )) ≤ 0.

This local formula generalizes (4.19). Thus one obtains the same conclusion andthe proof is finished.

4.2.4 Optimality of the two-state solver

Our goal in this section is to establish an optimality result for the two-state solver.We generalize the hypotheses and methods of section 4.1.4. That is we linearizethe quadratic forms Aj and Bj that appear in inequality (4.72). The linearizationis with respect to the time step ∆t and all coefficients are explicit at the beginningof the time step; for example, ρj in Aj is for ρk

j .

One obtains

Aj =Tj

2

(∆U, ∇2

US(Ukj )∆U

)+ O(∆t3), ∆U = UL

j − Ukj ∈ R

n,

and

Bj =1

2(∆Ψ, |M |j∆Ψ), ∆Ψ = ΨL

j − Ψkj ∈ R

n−1,

where |M |j = MR,+j − 1

2

− ML,−j + 1

2

. The first-order linear relation between ∆U and

∆Ψ can be written as ∆Ψ = ∇UΨj∆U with an error of order O(∆t2). With thisnotation the CFL condition Aj − ∆t

∆mkj

Bj admits a linearized form

(max

∆U ∈Rn, ∆U =0

(∆Ψ, |M |j∆Ψ)

Tj (∆U, ∇2USj ∆U)

)∆t

∆mj≤ 1.

Another form is based on the elimination of ∆U . Indeed, using the chain rule∆U = (∇WU)

j∆W one gets

(∆U, ∇2

USj∆U)

=(

∆W,[(∇WU)

tj (∇UV )j (∇WU)j

]∆W

)+ O(∆t3)

=(

∆W,[(∇WV )

tj (∇W U)j

]∆W

)+ O(∆t3).

The last matrix [. . . ] can be expressed using (3.45). So

(∆U, ∇2

USj ∆U)

=1

Tj(∆Ψ, Dj∆Ψ)+ βj∆S2 + O(∆t3), βj > 0.


We therefore obtain a second linearized formulation for the CFL condition,(

max∆W=(∆Ψ,∆S)∈Rn , ∆W =0

(∆Ψ, |M |j∆Ψ)

(∆Ψ, Dj∆Ψ) + γj∆S2

)∆t

∆mj≤ 1, γj > 0.

It is clear that the maximum occurs for ∆S = 0. We obtain a third linearizedformulation for the CFL condition,

(max

∆Ψ∈Rn−1 , ∆Ψ=0

(∆Ψ, |M |j∆Ψ)

(∆Ψ, Dj∆Ψ)

)∆t

∆mj≤ 1. (4.73)

At this stage there is nothing to optimize since all parameters of the two-state solver have been fixed. But it is actually possible to relax a very importantparameter, namely the metric D used to diagonalize M . This can be seen informulas (4.65) and (4.65), where the splitting is defined in terms of three matrices,which are, with the notation needed to distinguish between the left interface at j− 1

2and the right interface at j + 1

2, the pair (DL , DR) = (Dj −1, Dj) at interface j − 1

2and the pair (DL , DR ) = (Dj , Dj +1) at interface j + 1

2. To obtain a more general

method and to be able to compare the one-state and two-state solvers, we relaxthis construction and use instead four matrices, which are (DL , DR ) = (C, D−) atinterface j − 1

2and (DL , DR ) = (D+ , E) at interface j + 1

2. The obvious constraints

are that these four matrices should belong to R (n−1)×(n−1) and be symmetric andpositive.

With this generalization, all elements of the general Lagrangian scheme arewell defined and the scheme is entropic under the CFL condition as in the previoustheorem. The difference lies in the CFL condition, which takes the linearized formthat is an extension of (4.74):

(max

∆Ψ∈Rn−1 , ∆Ψ=0

(∆Ψ, M∆Ψ)

(∆Ψ, Dj∆Ψ)

)∆t

∆mj≤ 1 (4.74)

where M = M+ − M− . One has

M− ≤ 0 ≤ M+ . (4.75)

By construction there exist two other symmetric matrices M− ≤ 0 and M+ ≥ 0such that

M = M+ + M− = M+ + M− .

Therefore one also hasM− ≤ M ≤ M+ . (4.76)

The exact form of M+ can be deduced from (4.65) with D+ instead of DL (thisis on interface j + 1

2), and the exact form of M− can be derived from (4.66) withD− instead of DR (this is on interface j − 1

2). Since M is a function of D− and

D+ , we define

Jj (D− , D+) = max∆Ψ∈Rn−1 , ∆Ψ=0

(∆Ψ, M∆Ψ)

(∆Ψ, Dj∆Ψ). (4.77)


Definition 4.2.11 (Minimization problem related to (4.77)). Find two positive sym-metric matrices D− and D+ such that Jj(D− , D+) attains its minimum.

The answer to this minimization problem can formulated in terms of the spec-tral radius of the Jacobian matrix. For a Lagrangian system (see Theorem 3.2.13)the spectral radius is defined by (notice that we keep the minussign for consistencywith previous notation)

Λj = max |λj | over all λj such that Mrj = −λjDj rj with r ∈ Rn,∗ .

Theorem 4.2.12 (Optimality of the two-state solver). One has

Λj ≤ Jj(D− , D+) for al l admissible (D− , D+) (4.78)

andΛj = Jj(Dj , Dj ), Dj =

(∇2

U|Se)

j. (4.79)

Proof. The proof is in two steps.

Step 1. By (4.75)–(4.76), one has M ≤ M+ ≤ M+ − M− = M. Therefore

max∆Ψ∈Rn−1 , ∆Ψ=0

(∆Ψ, M∆Ψ)

(∆Ψ, Dj∆Ψ)≤ Jj(D− , D+),

which shows that the maximal eigenvalue 12ρj

D−1j M is less than Jj (D− , D+).

Similarly, −M ≤ −M− ≤ M+ − M− = M, so

max∆Ψ∈Rn−1 , ∆Ψ=0

(∆Ψ, −M∆Ψ)

(∆Ψ, Dj∆Ψ)≤ Jj (D− , D+),

showing that the maximal eigenvalue − 12ρj

D−1j M is less than Jj (D− , D+).

This proves the claim (4.78).

Step 2. Take D− = D+ = Dj . Then formulas (4.65)–(4.66) show that

M =∑

λi>0

λi (Djri) ⊗ (Dj ri) −∑

λi<0

λi (Djri) ⊗ (Djri)

with ri normalized eigenvector solutions of Mri = −λiDjri. Any vector∆Ψ ∈ Rn−1 can be decomposed into

∆Ψ =∑

λi>0

αiri +∑

λi<0

αiri +∑

λi=0

αiri.

Then(∆Ψ, M∆Ψ) =

∑

λi>0

λi(αi)2 −

∑

λi<0

λi(αi)2


and(∆Ψ, Dj∆Ψ) = 2

∑

i

(αi)2.

Therefore the maximum over all (αi) of the ratio of these two quantities isequal to

Jj(Dj , Dj ) = maxj

|λi| = Λj.


Corollary 4.2.13. The two-state solver minimizes the criterion Jj for al l j .

Proof. Indeed, the two-state solver uses at every interface with index j + 12

thematrices Dj on the left in cell j and Dj +1 on the right in cell j + 1. Therefore thematrices are optimal in the sense of the previous theorem in all cells. So the CFLcondition is optimal in all cells, i.e.

Λj∆t

∆mj≤ 1 ∀j.

The claimed result is a consequence of this.

a flux CFL flux CFL

1 (4.51) 0.5 (4.53) 0.5

10−1 (4.51) 0.5 (4.53) 0.5

10−2 (4.51) 0.002 (4.53) 0.5

10−3 (4.51) 0.002 (4.53) 0.5

10−4 (4.51) 10−5 (4.53) 0.5

Table 4.1: Experimental CFL conditions. The two-state solver (4.53) is clearlymuch more robust than the one-state solver (4.51) with respect to importantdifferences between the states.

The next example (with results recorded in table 4.1) is a numerical illustra-tion of the above theorem and corollary. Let us calculate the numerical solutionof the Riemann problem with ρL = pL = 1, uL = vL = 0 and ρR = 0.125 × a,pR = 0.1×a, uR = vR = 0, for a perfect gas law γ = 1.4. The parameter 0 < a ≤ 1modulates the ratio of mass and the ratio of density in the initial data. If we takea = 1 we recover the Sod tube test problem. A series of calculations is performedfor decreasing values of a, and we note the experimental maximal value of theCFL condition at which the calculation blows up, in practice because the inter-nal energy becomes negative. For the one-state solver one observes that the CFLcondition is very small for small a. In contrast, the CFL condition for the two-state solver is approximately uniform with respect to a, even for very small a. Weinterpret this behavior as a consequence of Corollary 4.2.13.

4.3. Extension to multidimensional Lagrangian systems 211

4.3 Extension to multidimensional Lagrangian systems

Our objective in this section is to extend the previous methods and results to themultidimensional problems discussed in section 3.5. We begin with an abstractgeneric discrete entropy inequalityand explain how the methods of matrix splittingcan be used in this context. The technique will be applied first to the discretizationof Lagrangian gas dynamics in cylindrical and spherical coordinates and to thequasi-Lagrangian MHD system, and then in the next section to the discretizationof Lagrangian gas dynamics on a moving two-dimensional grid. This method forLagrangian gas dynamics, developed in Mazeran’s PhD thesis and published [69]in 2003, was the first to use corner-based fluxes for cell-centered Lagrangian gasdynamics, and most modern treatments of cell-centered Lagrangian gas dynamicsare now based on this idea. A more direct presentation of the topic will be givenin the next chapter.

4.3.1 A generic discrete entropy inequality

Let us start from multidimensional Lagrangian systems written in the form (3.120).Suppose that the vector of unknowns U is discretized in cells indexed by j ∈ N

with area sj > 0. At time step k∆t and in cell j , the discrete unknown is denotedby Uk

j . We consider an almost explicit discretization of the form

sj

Uk+1j − Uk

j

∆t+∑

r

lj rfkj r = sjRk+1

j , (4.80)

where the fluxes fkj r are explicit:

fkj r =

⎛⎝

N +j rΨ

kj r + N −

j rΨkj

− 1

2

(Ψk

j r, N +j rΨ

kj r

)− 1

2

(Ψk

j , N −j rΨk

j

)

⎞⎠ . (4.81)

The fluxes are fully abstract objects at this level of presentation. They dependon the explicit entropy variable Ψk

j , on an auxiliary entropy variable Ψkj r (this is

the two-state method), and on two matrices N ±j r ∈ R (n−1)×(n−1) whose properties

are described in the hypothesis below. Note that a real number lj r ≥ 0, with lreferring to a length, is used to obtain a more natural interpretation in terms ofstandard finite volume methods. The right-hand side is a discretization of the term(

M0Ψ

0

)in (3.120). It reads

Rk+1j =

(M0Ψk+1

j

0

). (4.82)


The structure (4.80)–(4.82) should be viewed as a theoretical guide which is devel-oped to better understand the influence of the complex Lagrangian multidimen-sional structure on the discrete entropy inequality.

Notice that the right-hand side is implicit. This is unfortunate since it intro-duces complexity into the method, but it is nevertheless absolutely necessary forobtaining the generic entropy inequality. Since this term is non-differential, somenatural methods can be used to obtain explicit evaluations in practical situations.This will be detailed for cylindrical and spherical gas dynamics. The example ofbi-dimensional Lagrangian gas dynamics will show that the matrices may dependon the time variable (in this case another super-index k would be needed to de-scribe correctly the time dependence), which is another theoretical constraint thatwill be relaxed.

We now add an assumption which is both the discrete counterpart of (3.121)and an extension of the two-state method.

Hypothesis 4.3.1. We assume that there exists symmetric Nj r = N tj r satisfying

the compatibility relation

∑

r

lj rNj r = sj

(M0 + M t

0

). (4.83)

Moreover, the matrices N ±j r form a matrix splitting for Nj r, namely

Nj r = N −j r + N +

j r. (4.84)

Theorem 4.3.2. The scheme (4.80)–(4.84) satisfies a discrete local entropy inequal-ity S(Uk+1

j ) ≥ S(Ukj ) under the CFL condition.

Proof. The structure of the proof is very similar to that of Theorem 4.2.10. There-fore we concentrate on the essential part, which is the use of Hypothesis 4.3.1.Define

gj(α) = S(Uk

j + α(Uk+1

j − Ukj

)).

The second derivative of gj is evaluated using the concavity of the entropy. Thedifference is the first derivative g′(1) where the implicit right-hand side of (4.80)

necessarily shows up. One has g′j(1) =

∆t

T k+1j sj

(W, Z) with W =

(Ψk+1

j

1

)and

Z = sjRk+1j −

∑

r

lj r

⎛⎝

N +j rΨj r + N −

j rΨj

− 1

2

(Ψj r, N +

j rΨj r

)− 1

2

(Ψj , N −

j rΨj

)

⎞⎠ .

One has the identity

(W, sjRk+1

j

)= sj

(Ψk+1

j , M0Ψk+1j

)=

sj

2

(Ψk+1

j ,(M0 + M t

0

)Ψk+1

j

).


Eliminate the symmetric matrix M0 + M t0 by using relation (4.83). Hence

(W, sjRk+1

j

)=

1

2

(W,

( ∑r

Nj rΨk+1j

0

))

=1

2

∑

r

lj r

(Ψk+1

j , N +j rΨk+1

j

)+

1

2

∑

r

lj r

(Ψk+1

j , N −j rΨk+1

j

)

thanks to (4.84). So one can decompose the scalar product (W, Z) as

(W, Z) =1

2

∑

r

lj r

[(Ψk+1

j , N +j rΨk+1

j

)− 2

(Ψk+1

j , N +j rΨj r

)+(

Ψjr, N +j rΨj r

)]

+1

2

∑

r

lj r

[(Ψk+1

j , N −j rΨk+1

j

)− 2

(Ψk+1

j , N +j rΨ

kj

)+(Ψk

j , N +j rΨk

j

)]

=1

2

∑

r

lj r

(Ψk+1

j − Ψj r, N +j r

(Ψk+1

j − Ψj r

))

+1

2

∑

r

lj r

(Ψk+1

j − Ψkj , N −

j r

(Ψk+1

j − Ψkj

)).

Therefore

g′(1) ≥ ∆t

2T k+1j sj

∑

r

lj r

(Ψk+1

j − Ψkj , N −

j r

(Ψk+1

j − Ψkj

)).

One obtains

S(Uk+1

j

)≥ S

(Uk

j

)+

1

T k+1j

(Aj − ∆tBj)

with Uθj = Uk

j + θ(Uk+1

j − Ukj

)(0 ≤ θ ≤ 1),

Aj =−T k+1

j

2

(Uk+1

j − Ukj , ∇2

US(Uθj )(U k+1

j − Ukj ))

, (4.85)

and

∆tBj ≥ ∆t

2sj

∑

r

lj r

(Ψk+1

j − Ψkj , N −

j r

(Ψk+1

j − Ψkj

)). (4.86)

The final part of the proof is the same as for Theorem 4.2.10.

We observe that, as before, the entropy inequality is purely local and thatno conservation property is needed. It is nevertheless natural to ask for someconservation properties in view of the differential formulation (3.120). For example,one can assume that the auxiliary states satisfy the identity

∑

j

fkj r = 0 ∀r, k. (4.87)


If this assumption holds, it yields the abstract balance relation∑

j

sjUk+1j =

∑

j

sjUk+1j + ∆t

∑

j

sjRk+1j , (4.88)

which is the discrete counterpart of the integral balance relation deduced from(3.120),

d

dt

∫U(x, t) dx =

∫(M0Ψ(x, t)) dx.

Finally, we state a simple stability corollary in L2 which is fully justified forthe conservative case, written as

∑

j

sjUkj =

∑

j

sjU0j ∀k. (4.89)

In view of the discrete balance law (4.88) this equation probably imposes thecondition M0 = 0, but this is of no real importance for the analysis below. Definethe mean value

U∗ =

∑j sjU0

j∑j sj

, (4.90)

which makes sense for a domain with finite measure, 0 <∑

j sj < ∞.

Proposition 4.3.3. Assume the discrete entropy inequality (4.89). Then one hasthe global inequality∑

j

sj

[S(Uk

j

)− S (U∗) −

(∇US(U ∗), Uk

j − U∗)] ≥∑

j

sj

[S(U0

j

)− S (U∗)

].

(4.91)

The proof is evident.To exploit this inequality we note that

akj = S

(Uk

j

)− S (U∗) −

(∇US(U∗), Uk

j − U∗) .

Notice that the (strict) concavity of S yields that 0 ≥ akj and that 0 = ak

j if and

only if Ukj = U∗. Indeed, one has the classical convexity inequality for η = −S:

η(b) − η(a) + (∇η(a), b − a) =1

2

(b − a, ∇2η(c)(b − a)

)≥ α

2|b − a|2

.

Proposition 4.3.4 (L2 stability). Assume that the Hessian of the mathematicalentropy is uniformly bounded from below, ∇2η = −∇2

US ≥ αI with α > 0. Thenone has L2 stability around U∗:

α

2

∑

j

sj

∣∣Ukj − U∗∣∣2 ≤

∑

j

sj

[−S

(U0

j

)+ S (U∗)

]. (4.92)

The proof is evident.


4.3.2 Cylindrical and spherical gas dynamics

We consider in this section discretization of the cylindrical-spherical Lagrangiangas dynamics system (3.124).

At the beginning of the time step k∆t, the cells Ωj have a certain position

Ωj =(

rkj − 1

2

, rkj +1

2

). The Lagrangian mass in the cell is

∆mj = ρkj

∫ rk

j+ 12

rk

j− 12

rd dr = ρkj

(rk

j + 12

)d+1−(

rkj − 1

2

)d+1

d + 1.

We discretize the matrix M1 = rd

(0 1

1 0

)in (3.124) as

N kj + 1

2

=(

rkj +1

2

)d(

0 1

1 0

).

The compatibility relation (4.83) reads

N kj +1

2

− N kj − 1

2

∆mj=

(d

ρr

)k

j

(0 1

1 0

),

where by definition

(d

ρr

)k

j

=

(rk

j +12

)d−(

rkj − 1

2

)d

∆mj.

One obtains a possible Lagrangian scheme⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

∆mj

∆t(τ k+1

j − τ kj ) −

(rk

j +12

)du∗

j +12

+(

rkj −1

2

)du∗

j −12

= 0,

∆mj

∆t(uk+1

j − ukj ) +

(rk

j + 12

)dp∗

j +12

−(

rkj − 1

2

)dp∗

j − 12

=

((rk

j + 12

)d

−(

rkj − 1

2

)d)

pk+1j ,

∆mj

∆t(ek+1

j − ekj ) +

(rk

j + 12

)d

p∗j +1

2

u∗j +1

2

−(

rkj −1

2

)d

p∗j − 1

2

u∗j −1

2

= 0.

(4.93)

The fluxes u∗j +1

2

and p∗j + 1

2

can be taken as in (4.22). Unfortunately for the sim-

plicity of the method, the scheme is implicit and so requires a specific procedureto be used in practice. This procedure can be a Newton method or an exact so-lution if a perfect gas pressure law is used: in the latter case the algebra reducesto calculation of the roots of a second-order polynomial. We detail hereafter thisinstructive situation. The pressure law is

p = (γ − 1)ρε = (γ − 1)ρe − γ − 1

2ρu2.


Since ρ = 1τ and e are explicitly given by the scheme at the end of the time step as

functions of the other quantities, the pressure pk1j at the end of the time step is a

function of(uk+1

j

)2. Therefore uk+1

j is the solution of a second-order polynomial,of which only the physically relevant root is to be considered.

Another possibility is to sacrifice the entropy inequality by choosing anotherdiscretization of the right-hand side in the form

∆mj

∆t(uk+1

j − ukj) +

(rk

j +12

)dp∗

j + 12

−(

rkj − 1

2

)dp∗

j −12

=

((rk

j +12

)d−(

rkj − 1

2

)d)

pkj .

(4.94)

Notice that one can also replace the explicit pressure on the right-hand side bythe half-sum of the fluxes,

((rk

j +12

)d

−(

rkj − 1

2

)d) p∗

j +1

2

+ p∗j − 1

2

2. (4.95)

One obtains

∆mj

∆t(uk+1

j − ukj ) +

1

2

((rk

j + 12

)d

+(

rkj −1

2

)d)

×(

p∗j +1

2

− p∗j −1

2

)= 0. (4.96)

All of the schemes (4.93), (4.94) and (4.96) preserve rest states. More precisely, if atiteration k the pressure is constant in space and the velocity vanishes everywhere,then these features are preserved at iteration k + 1.

A numerical solution computed from (4.96) with the initial data of the Sodtube test problem, but in cylindrical geometry, is presented in figure 4.13. Thecomputation is done in a Lagrange+remap configuration where the Lagrange stepin plane geometry is easily adapted to the cylindrical geometry. One observes thedecomposition of the initial discontinuity into local solutions which are similar toa rarefaction fan, a contact discontinuity and a shock. However, the solution is notself-similar and the constant states are replaced by ramps. The results in sphericalgeometry are presented in figure 4.14. They are similar, except that the slope ofthe ramps is even more pronounced.

4.3.3 Lagrange+remap MHD in dimension d > 1

The second example is based on the model (3.130) and is from [77], where anentropy-consistent discretization of ideal MHD in dimension d = 2 is proposed.Numerical results are shown in figure 4.16. The general principles are true in alldimensions.

A important idea is that the magnetic field is discretized twice. A first rep-resentative of the magnetic field B is centered inside each cell. A second represen-tative is on the edges of the cells, where it is the normal value of the magnetic


0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

’sod.ro’

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

’sod.p’

x → ρ(x) x → p(x)

0

0.2

0.4

0.6

0.8

1

1.2

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

’sod.u’

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

’sod.s’

x → u(x) x → S(x)

Figure 4.13: Cylindrical Sod tube test problem, computed in a Lagrange+remapmode with (4.96), final time T = 0.14, 1000 cells.

field C that is discretized. The scheme is decomposed into three steps, which arethe Lagrangian phase, the remap phase and the Eulerian discretization of C. Thisstructure is depicted on a Cartesian mesh in figure 4.15. The circles represent cell-centered discrete unknowns such as ρ, u, B and e. The normal components (C, n)are discretized on the edges.

Lagrange step

Consider the matrix Mx given in (3.128)–(3.129). Even if it is possible to rely onthe solution of the linearized Riemann problem to determine an accurate splitting,the method described here is more direct. The matrix Mx is split into two parts,


0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

’sod.ro’

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

’sod.p’

x → ρ(x) x → p(x)

0

0.2

0.4

0.6

0.8

1

1.2

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

’sod.u’

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

’sod.s’

x → u(x) x → S(x)

Figure 4.14: Spherical Sod tube test problem, computed in a Lagrange+remapmode with (4.96), final time T = 0.14, 1000 cells.

i, j

i, j +1

2

i +1

2, j

i, j −

1

2

i −

1

2, j

i, j − 1

i − 1, j

i, j + 1

i + 1, j

Figure 4.15: Cartesian mesh in dimension d = 2. The magneticfield B is discretizedtwice: both components of B are discretized in the middle of the cells; the normalcomponents are discretized a second time at the edges.


a symmetric non-negative part

M+x =

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

1

2α10 0 0

1

20 0

0C2

x

2α20 0

Cx

20 0

0 0C2

x

2α20 0

Cx

20

0 0 0C2

x

2α20 0

Cx

21

2

Cx

20 0

α1 + α2

20 0

0 0Cx

20 0

α2

20

0 0 0Cx

20 0

α2

2

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

,

and a symmetric non-positive part

M−x =

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

− 1

2α10 0 0

1

20 0

0 − C2x

2α20 0

Cx

20 0

0 0 − C2x

2α20 0

Cx

20

0 0 0 − C2x

2α20 0

Cx

21

2

Cx

20 0 − α1 + α2

20 0

0 0Cx

20 0 − α2

20

0 0 0Cx

20 0 − α2

2

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

.

Notice the introduction of two parameters α1 and α2. By analogy with the one-dimensional case, two choices seem natural. The first possibility is to set α1 = α2 =ρcf with cf the fast MHD velocity. The second possibility is α1 = ρcf , α2 = ρca

with ca the Alfven velocity. In practice these velocities are evaluated on the edgesby taking a local mean. After lengthy calculations, the flux in the x direction isfound to be

fi+1/2,j,k =

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

−u∗x i+1/2,j,k

−Cx i+1/2,j,kui+1/2,j,k⎛⎜⎝

P ∗i+1/2,j,k

0

0

⎞⎟⎠ − Cx i+1/2,j,kBi+1/2,j,k

μ

P ∗i+1/2,j,ku∗

x i+1/2,j,k −Cx i+1/2,j,kui+1/2,j,k.Bi+1/2,j,k

μ

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

, (4.97)


where⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

u∗x i+1/2,j,k =

1

2(ux i+1,j,k + ux i,j,k) − 1

2α1 i+1/2,j,k(Pi+1,j,k − Pi,j,k) ,

ui+1/2,j,k =1

2(ui+1,j,k + ui,j,k) +

Cx i+1/2,j,k

2α2 i+1/2,j,kμ(Bi+1,j,k − Bi,j,k),

Bi+1/2,j,k =1

2(Bi+1,j,k + Bi,j,k) +

μα2 i+1/2,j,k

2Cx i+1/2,j,k(ui+1,j,k − ui,j,k),

P ∗i+1/2,j,k =

1

2(Pi+1,j,k + Pi,j,k) − α1 i+1/2,j,k

2(ux i+1,j,k − ux i,j,k).

It can be checked that this flux does not preserve the one-dimensional solution forwhich Bx is constant. Indeed, the fluxes have the form

⎧⎪⎪⎨⎪⎪⎩

u∗x i+1/2 =

1

2(ux i+1 + ux i) − 1

2α1 i+1/2(Pi+1 − Pi) ,

ux i+1/2 =1

2(ux i+1 + ux i).

So the flux of the equation for τ multiplied by Bx/μ is formally different from theflux of the equation for τBx/μ. In general this has the consequence that Bx willno longer be uniform at the next time step, even if it was constant at the initialtime.

A modification has been proposed in [77] to respect these special solutionsfor which Bx is constant. The idea is to add to M+

x (and, by symmetry, subtractfrom M−

x ) a new matrix Lx defined by

Lx =

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

γ

2β

γ

20 . . . 0

γ

2

βγ

20 . . . 0

0 0 0 . . . 0...

......

. .....

0 0 0 . . . 0

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

,

where β and γ are some coefficients used to optimize the scheme. With these newterms, u∗

x and ux become⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

u∗x i+1/2,j,k =

1

2(ux i+1,j,k + ux i,j,k) +

γi+1/2,j,k

2μ(Bx i+1,j,k − Bx i,j,k)

−(

1

2α1 i+1/2,j,k+

γi+1/2,j,k

2βi+1/2,j,k

)(Pi+1,j,k − Pi,j,k) ,

ux i+1/2,j,k =1

2(ux i+1,j,k + ux i,j,k) −

γi+1/2,j,k

2Cx i+1/2,j,k(Pi+1,j,k − Pi,j,k)

+

(Cx i+1/2,j,k

2μα2 i+1/2,j,k+

βi+1/2,j,kγi+1/2,j,k

2μCx i+1/2,j,k

)(Bx i+1,j,k − Bx i,j,k).


The condition which guarantees u∗x = ux in one-dimensional configurations reads

1

α1+

γ

β=

γ

Cx,

that is,

γ

(1

Cx− 1

β

)=

1

α1. (4.98)

It implies that γ, β and Cx have the same sign and that |β | ≥ |Cx|. A simplesolution is

β = 2Cx and γ =2Cx

α1. (4.99)

We continue the presentation with this choice of parameters, and we also omit theindices j and k. The flux in the x direction is

fi+1/2 =

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

−u∗x i+1/2

−Cx i+1/2ui+1/2⎛⎜⎝

P ∗i+1/2

0

0

⎞⎟⎠ − Cx i+1/2Bi+1/2

μ

P ∗i+1/2

u∗x i+1/2

− Cx i+1/2ui+1/2Bi+1/2

μ+Di+1/2

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

, (4.100)

with⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

u∗x i+1/2 =

1

2(ux i+1 + ux i) −

(1

2α1 i+1/2+

1

2α1 i+1/2

)(Pi+1 − Pi)

+Cx i+1/2

α1 i+1/2µ(Bx i+1 − Bx i),

ux i+1/2 =1

2(ux i+1 + ux i)

+Cx i+1/2

2μ

(1

α2 i+1/2+

4

α1 i+1/2

)(Bx i+1 − Bx i)

− 1

α1 i+1/2(Pi+1 − Pi),

Di+1/2 =1

4

(2α2 i+1/2

α1 i+1/2+ 1

)(ux i+1 − ux i)

×(

2Cx i+1/2

μ(Bx i+1 − Bx i) − (Pi+1 − Pi)

).

The quantities in boldface represent terms that have been added to the originalflux (4.97).


Figure 4.16: The initial data is a high-pressure bubble inside a uniform constantnon-zero magnetic field. The final result was computed with the AMR technology(where AMR stands for adaptive mesh refinement). The AMR technology allowsfor dynamic adaptation of the mesh in order to track the discontinuities with a finermesh. This calculation was performed at the Commissariat a l’Energie Atomique.

Note that the abstract compatibility condition (4.83) can be written in thesimple form

Cki+ 1

2 ,j− Ck

i− 12 ,j

+ Cki,j +1

2

− Cki,j − 1

2

= 0.

It also expresses an important physical relation which is that the magnetic fieldis divergence-free, ∇ · B = 0. It appears that the directional splitting method isnot compatible with such a relation. This is why is seems preferable not to usedirectional splitting, but rather a more direct implementation where all Lagrangianfluxes in all directions are used in one stage.

Remap stage

Once the Lagrangian stage is performed, the remap can be done in all directionsseparately.

Discretization of C

A new issue is the time integration of C. Described below is an Eulerian method,which has the natural ability to preserve the discrete divergence-free condition on

4.4. Lagrangian gas dynamics in dimension d = 2 223

the Cartesian grid of figure 4.15. The update of the normal fluxes is discretized inthe x direction as

Ck+1i+ 1

2 ,j− Ck

i+ 12 ,j

∆t+

qi+ 12 ,j + 1

2− qi+ 1

2 ,j −12

∆x= 0

and in the y direction as

Ck+1i,j +1

2

− Cki,j +1

2

∆t+

qi+ 12

,j + 12

− qi− 12

,j +12

∆x= 0.

Here q is a local approximation of Cxuy − Cyux.

4.4 Lagrangian gas dynamics in dimension d = 2

The following presentation of the topic is based on [69, 72]. It shows how thematrix splitting technique was used to design the first ever [69, 72] cell-centeredLagrangian scheme with corner-based fluxes, and more precisely how it overcomesthe geometrical obstruction of standard finite volume Lagrangian solvers by intro-ducing a natural corner-based extension of the flux. The price to pay is a non-trivialmodification of the discrete structure of the finite volume method. This schemewill be referred to as GLACE (acronym for Godunov LAgrange Conservation intotal Energy). Two other Lagrangian schemes will be discussed within the sameformalism. The first one [3], called EUCCLHYD, is a popular extension of thecorner-based technique which is closer in essence to the notion of an edge-basedRiemann solver; it also introduces the mechanical notation which substantiallysimplifies the presentation of such methods. The second scheme [34, 38] is devel-oped mainly for pedagogical purposes; it is a staggered scheme. Note that thistopic will be treated with a much more general and powerful method in chapter 5.

The idea is presented below in dimension d = 2. One starts from (4.80)–(4.81) and considers the constraints introduced by a moving grid. Among theseconstraints, an important one is that the method must be expressible both in theinitial X = (X, Y ) Lagrangian frame and in the current x = (x, y) Eulerian frame.This means that a discrete counterpart of the Piola identities must be part of themethod.

4.4.1 Elementary considerations on moving meshes

Let us begin with some elementary considerations about moving meshes.Consider the mesh of figure 4.17. Denote by u1 = (u1 , v1) and u2 = (u2, v2)

the velocities of A and B . Denote by n the exterior normal vector on the edge AB .The area of the swept region during a time step ∆t is A = l∆t

∣∣(u1+u2

2, n

)∣∣. Onealso has A = l∆t |(u∗, n)| , where the velocity u∗ in the middle of the segment is


O

B

A A

B

Figure 4.17: The segment AB moves to A′B ′. The swept region is marked withdashed lines.

defined by linear interpolation as

u∗ =u1 + u2

2. (4.101)

The two formulas for the evaluation of the area A of the swept region are the sameif and only if

(u∗, n) =1

2(u1, n) +

1

2(u2, n) . (4.102)

This observation is central to the discussion of the geometrical features of a bi-dimensional moving mesh, and it explains why standard finite volume methodswith edge-based fluxes are not satisfactory for that purpose. Indeed, let us imaginea cell-centered Lagrangian discrete scheme with edge-based fluxes. In this case(u∗, n) is given by the numerical fluxes. The main issue is to define the meshdisplacement. If one wishes to define the velocity of nodes with equation (4.102),then one obtains a linear system for the unknowns ur, r = 1, . . . , which are thenode velocities. The number of equations is of the same order as the numberof edges. This a global linear system that couples all the node velocities, andin general it does not correspond to a square matrix. Approximate solutions ofthis difficult geometrical problem using least squares methods have been proposedwithin the Caveat suite [7, 191].

In contrast, suppose that the fluxes of the cell-centered scheme are corner-based; then the definition of the velocity at the middle of the edges is explicit. Thisis the reason why we will abandon the paradigm of edge-based multidimensionalfluxes, in favor of corner-based multidimensional fluxes. This strong modificationof edge-based is necessary to solve the compatibility issue on a moving grid. Asummary of the discussion is illustrated in figure 4.18.


Figure 4.18: Geometrical difference between a standard edge-based flux (on theleft) and a corner-based flux (on the right).

4.4.2 Some notation

Consider a mesh where the individual cells have generic index j (or k). The nodesof the mesh have generic index r (or s). Denote by ej k = ekj ≥ 0 the length ofthe interface between cell j and cell k. It may vanish if the cells have no edge incommon. If they have an edge in common, define nj k = −nkj to be the exteriornormal vector on the edge, from cell j to cell k. The closed contour relation holds:

∑

k

ej knj k = 0 ∀j.

Define the nodal normal and nodal length according to the design principle shownin figure 4.18. Let r be the index of a generic node on the boundary of cell j ,and denote by r+ (resp. r−) the node in the counterclockwise (resp. clockwise)direction, as illustrated in figure 4.19.

Definition 4.4.1 (Nodal length and normal). The nodal length and nodal normalare defined by the local relation

lj rnj r =1

2ej knj k +

1

2ej pnj p =

1

2

(yr+ − yr−

xr− − xr+

). (4.103)

One has by definition, for all cells j ,

∑

r

lj rnj r = 0. (4.104)

Proposition 4.4.2 (Area of a cell). The area of a given cel l is

sj =1

2

∑

r

(xryr+ − xr+ yr) . (4.105)


node r+

corner normal vector

node r

cell p

cell k

cell j

node r−

Figure 4.19: Notation: nodes r− , r and r+ are oriented counterclockwise.

The convention is that the sum is listed counterclockwise around the current cel lj . Let xr = (xr, yr) be a given node; then xr+ is the next node (counterclockwise)and xr− is the previous one (clockwise).

Proof. One can consider the center of the mesh O = (0, 0) to be inside the cell. Sothe cell can be decomposed into a sum of triangles with corners O, xr = (xr, yr )and xr+ = (xr+ , yr+ ). Since 1

2

∑r (xryr+ − xr+ yr) is the area of the triangle

OMr Mr+ , the result is proved by summation over all triangles.

Suppose that the velocity of the nodes is a continuous and differentiablefunction of the time variable. It defines the node velocity ur(t) = (ur(t), vr (t)). Sothe area is also a function of the time variable, t → sj (t).

Proposition 4.4.3. One has the relation

s′j(t) =

∑

r

lj r (nj r, ur) =∑

k

lj k (nj k, uk) . (4.106)

Proof. Differentiate the area formula (4.105) and rearrange to get

s′j(t) =

∑

r

(ur

yr+ − yr−

2− vr

xr+ xr−

2

).

From inspection of figure 4.19 one can see that

nj r =

(yr+ − yr−

2, − xr+ − xr−

2

).

4.4.3 Compatibility with Piola identities

The previous relations can be used to prove a simplified version of the Piolaidentities. Recall that τj = 1

ρjis the specific volume.


Proposition 4.4.4. Let ρj(t) > 0 be the density in cel l j as a function of the timevariable. There is equivalence between local conservation of the Lagrangian mass,

(ρjsj)′(t) = 0,

and the discrete conservation law

Mjτ ′j(t) −

∑

r

lj r (nj r, ur) = 0. (4.107)

Proof. Let Mj(t) = ρj(t)sj (t) be the total mass in the cell. Then

s′j (t) =

d

dt(Mj(t)τj (t)) .

So (4.106) shows the equivalence.

We now discuss the discrete-in-time case. Consider a discrete evolution

xk+1r = xk

r + ∆tukr ⇐⇒ xk+1

r = xkr + ∆tuk

r and yk+1r = yk

r + ∆tvkr. (4.108)

Proposition 4.4.5. Assume ρk+1j > 0 and ρk

j > 0. One has equivalence betweenpreservation of the Lagrangian mass,

sk+1j ρk+1

j = skj ρk

j , (4.109)

and the discrete conservation law

Mj

τ k+1j − τ k

j

∆t−∑

r

lk+ 1

2

j r

(n

k+ 12

j r , ukr

)= 0, (4.110)

where the geometrical term is centered in time, i.e.

lk+ 1

2

j r nk+ 1

2

j r =1

2

(lkj rn

kj r + lk+1

j r nk+1j r

).

Proof. Define the linear interpolation

xr(t) = xk

r + tukr ,

yr(t) = ykr + tvk

r .

Integrate relation (4.107) over one time step ∆t. By definition (4.103), the geo-metrical quantity lj r(t)nj r(t) is a linear function of the time variable. Therefore

∫ ∆t

0

lj r(t)nj r(t) =1

2

(lkj rnk

j r + lk+1j r nk+1

j r

).



After

‘

Before

x, yX, Y

Figure 4.20: Displacement of a triangular cell.

4.4.4 Compatibility with Hui’s formulation

The ingredients of Hui’s formulation are easily recovered on the triangular meshof figure 4.20.

The Eulerian coordinate x = (x, y) at all time steps k∆t is, in cell j , an affinefunction of the initial Lagrangian variable X = (X, Y ), that is,

xkj (X, Y ) = ak

j + bkj X + ck

jY, ykj (X, Y ) = dk

j + ekj X + fk

j Y.

Since the deformation is necessarily continuous across edges, it is a P 1 function infinite element method terminology. It is enough to know the final position of nodesto determine all coefficients (a, b, c, d, e, f) in a cell. One determines immediatelythe deformation gradient

Akj = ∂Xx = bk

j , Bkj = ∂X y = ek

j , Lkj = ∂Y x = ck

j , Mkj = ∂Y y = fk

j .

One also has continuity of the normal fluxes of the vector fields (L, −A) and(M, −B) on edges.

4.4.5 First attempt and geometrical obstruction

For pedagogical purposes only, we describe a first attempt to define a cell-centeredLagrangian discretization of compressible fluid dynamics. It relies on two ingre-dients: the first is the splitting matrix technique, which guarantees a conveniententropy inequality; the second is the use of corner-based objects and fluxes as ex-plained above. However, it will be clear at the end of the construction that anotheringredient is needed to close the construction, and that without this ingredient,all the material described in this section is still not enough. Even though thisapproach can be considered naive, it illustrates all the difficulties.


Let us try to construct a scheme of the form⎧⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎩

Mjτ k+1j = Mjτ k

j + ∆t∑

r

lk+ 1

2

j r

(n

k+ 12

j r , u∗r

),

Mjuk+1j = Mjuk

j − ∆t∑

r

lk+ 1

2

j r nk+ 1

2

j r p∗r ,

Mjek+1j = Mjek

j − ∆t∑

r

lk+ 1

2

j r

(n

k+ 12

j r , u∗r

)p∗

r .

(4.111)

This form is an application to gas dynamics of the general structure (4.80). Theconstruction relies on determination of the nodal flux, which depends on the geo-

metrical quantities lk+ 1

2

j r nk+ 1

2

j r and the physical quantities

Ψ∗r =

(p∗

r

−u∗r

)=

⎛⎜⎝

p∗r

−u∗r

−v∗r

⎞⎟⎠ ,

as sketched in figure 4.21.

j

nm

q

(ur , vr, pr)r

Figure 4.21: A tentative nodal flux (ur, vr, pr). Cells with indices j , n, q and mhave in common the node with index r.

The splitting matrix method is used to determine entropy-producing schemes.The first task is nevertheless to identify the matrices.

Proposition 4.4.6. The scheme (4.111) can be written in the abstract form (4.80)–(4.84) where the matrices are

Nj r =

⎛⎜⎜⎝

0 cos θk+ 1

2

j r sin θk+ 1

2

j r

cos θk+ 1

2

j r 0 0

sin θk+ 1

2

j r 0 0

⎞⎟⎟⎠ , n

k+ 12

j r =(

cosθk+ 1

2

j r , sin θk+ 1

2

j r

),


the flux is

fkj r =

⎛⎝

Nj rΨ∗r

− 1

2(Ψ∗

r, Nj rΨ∗r)

⎞⎠ ,

and the right-hand side vanishes Rk+1j = 0 and M0 = 0.

Proof. One can check that this form is compatible with (4.110).

The next step of the construction is to decide how to split the symmetricmatrix Nj r and obtain a closure relation like

Nj rΨ∗r = N +

j rΨj r + N −j rΨj (4.112)

where Ψj r are some auxiliary states. The splitting of the matrix Nj r is not thedifficult part. One can take, for example,

N +j r =

1

2

⎛⎜⎜⎜⎜⎝

1αj

cos θk+ 1

2

j r sin θk+ 1

2

j r

cos θk+ 1

2

j r αj

(cos θ

k+ 12

j r

)2

αj cos θk+ 1

2

j r sin θk+ 1

2

j r

sin θk+ 1

2

j r αj cos θk+ 1

2

j r sin θk+ 1

2

j r αj

(sin θ

k+ 12

j r

)2

⎞⎟⎟⎟⎟⎠

and

N −j r =

1

2

⎛⎜⎜⎜⎜⎝

− 1αj

cos θk+ 1

2

j r sin θk+ 1

2

j r

cosθk+ 1

2

j r −αj

(cos θ

k+ 12

j r

)2

−αj cos θk+ 1

2

j r sin θk+ 1

2

j r

sin θk+ 1

2

j r −αj cos θk+ 1

2

j r sin θk+ 1

2

j r −αj

(sin θ

k+ 12

j r

)2

⎞⎟⎟⎟⎟⎠

,

with the acoustic impedanceαj = ρjcj > 0. (4.113)

These two matrices are symmetric, the first being non-negative and the secondnon-positive.

Proposition 4.4.7. The vectorial relation (4.112) is equivalent to just one (and onlyone) scalar relation

p∗r + αj

(n

k+ 12

jr , u∗r

)= pk

j + αj

(n

k+ 12

j r , ukj

). (4.114)

Proof. The proof is in two steps.

• Take the scalar product of (4.112) with the eigenvector of the matrix N +j r,

z =(

−αj, cos θk+ 1

2

j r , sin θk+ 1

2

j r

)t

.

This yields (4.114).


• Conversely, assume that (4.114) holds. Then (4.112) can be considered as a

linear equation with unknown Ψj r. So there exists one solution (or more) ifthe right-hand side is in the range of the matrix N +

j r. Since the matrix issymmetric, it is sufficient to check that the right-hand side is orthogonal toall vectors in the kernel of N +

j r, which is the case.


The identity (4.114) can be interpreted as a linear acoustic Riemann solverin the corner direction, as illustrated in figure 4.22. Unfortunately one has thefollowing obstruction result which ruins the construction.

fictitious cell

cell j

xr

njr

boundary betweencell j and fictitious cell

Figure 4.22: Principle of a Riemann problem in the corner direction nj r betweenthe current cell with index j and a fictitious cell.

Proposition 4.4.8 (Geometrical obstruction). The linear system given by (4.114)for al l cel ls j around the node r is overdetermined in dimension d = 2 on a genericmesh.

Proof. Indeed, (4.114) is a linear relation with unknowns(ur , pr). This means threescalar unknowns since the velocity is ur ∈ R 2 . For a generic mesh the number ofcells j = 1, 2, . . . , J impinging on a node is J > 3. This is why the system isoverdetermined.

A regular triangular mesh has 6 cells per node; a regular quadrangular meshhas 4 cells per node. An exception is a regular hexagonal mesh, which has 3 cellsper node; such a mesh meets the requirement of the proof, namely J = 3. However,it is not generic and cannot be considered as a general solution.

4.4.6 Solving the geometrical obstruction:GLACE and EUCCLHYD

Since the linear system (4.114) is overdetermined in general, the natural option isto modify the method so as to generate additional unknowns for the linear system.All the methods recently published follow this approach.


It is not possible to change the velocities, so let us add more pressures at thenodes. We modify the flux to the form

Ψ∗j r =

(p∗

j r

−u∗r

)=

⎛⎜⎝

p∗j r

−u∗r

−v∗r

⎞⎟⎠ .

The unique node pressure p∗r is replaced by delocalized pressures p∗

j r, where theindices refer to cell j and node r; see figure 4.23.

ur = (ur , vr)

pmr

pqr

pnr

pjr

Figure 4.23: The nodal pressures are delocalized on cells with indices j , n, q and m:one pressure per neighboring cell. The central black circle represents the vectorialvelocity. The smaller black circles represent the delocalized pressures at the node.

One modifies the system (4.114) to the form

p∗j r + αj

(n

k+ 12

j r , u∗r

)= pk

j + αj

(n

k+ 12

j r , ukj

).

Since the delocalized pressures are different, it seems natural to require

∑

i

lk+ 1

2

j r nk+ 1

2

j r p∗j r = 0 ∈ R

2. (4.115)

This relation expresses that the local sum of forces vanishes at a given node, andis related to momentum conservation.

One obtains the following linear system at node r where the unknowns arethe

(Ψ∗

j r

)j:

⎧⎪⎪⎨⎪⎪⎩

p∗j r + αj

(n

k+ 12

j r , u∗r

)= pk

j + αj

(n

k+ 12

j r , ukj

), j = 1, . . . , J,

∑

i

lk+ 1

2

j r nk+ 1

2

j r p∗j r = 0.

(4.116)

Theorem 4.4.9. Assume αj > 0 for al l j , and that the mesh is local ly non-singular.Then the linear system (4.116) admits a unique solution.


Proof. A non-singular mesh is such that lj r > 0 for all cells j adjacent to the noder. Moreover, the number of cells is greater than or equal to 3 in 2D, so the cornervectors nj r span R 2. With this in mind, let us prove the claim.

Eliminate p∗j r in the second equation of (4.116) by substitution from the first

equation. One obtains the linear system

Ar u∗r =

∑

j

(lk+ 1

2

j r αjnk+

12

j r ⊗ nk+12

j r ukj

)+∑

j

lk+ 1

2

j r nk+ 1

2

j r pkj , (4.117)

where Ar =∑

j

(lk+ 1

2

j r αjnk+12

j r ⊗ nk+12

j r

)= At

r ∈ R 2. Let us temporarily assume

that Ar is singular. So there exists u ∈ R 2 such that

Aru = 0 =⇒ (u, Ar u) = 0 =⇒∑

j

lk+ 1

2

j r αj

(n

k+ 12

j r , u)2

= 0

=⇒(

nk+ 1

2

j r , u

)= 0 for all j =⇒ u = 0,

since the mesh is non-singular. Therefore Ar is non-singular and ur is uniquelydetermined by (4.117). Finally, the nodal pressures pj r are uniquely determinedby the first equation of (4.116). The proof is complete.

Theorem 4.3.2 can be invoked to show that the scheme (4.111)–(4.116) isentropy consistent under the CFL condition. Unfortunately it is only a theoretical

scheme since it is implicit because of the terms lk+ 1

2

j r . Nevertheless, it is possibleto define an explicit and useful scheme at the price of a slight weakening of thetheoretical entropyproperties. The main idea is to abandon the strict compatibilitybetween (4.109) and (4.110) by using explicit geometric quantities lkj rnk

j r instead

of lk+ 1

2

j r nk+ 1

2

j r .

GLACE scheme

One obtains an explicit nodal solver [72, 150] which may be written as

⎧⎪⎨⎪⎩

p∗j r + αj

(nk

j r, u∗r

)= pk

j + αj

(nk

j r, ukj

), j = 1, . . . , J,

∑

j

lkj rnkj rp

∗j r = 0. (4.118)

One takes a priori αj = ρkj ck

j > 0. This solver is coupled to (4.108) for the meshdisplacement, to (4.119) for the mass conservation, which allows the recalculationof the new density once the new area sk+1

j is calculated,

ρk+1j =

skj

sk+1j

ρkj ⇐⇒ Mj = sk

j ρkj = sk+1

j ρk+1j , (4.119)


and to the velocity and total energy equations for the calculation of uk+1j and

ek+1j , ⎧

⎪⎪⎨⎪⎪⎩

Mjuk+1j = Mj uk

j − ∆t∑

r

lkj rnkj rp

∗r,

Mj ek+1j = Mj ek

j − ∆t∑

r

lkj r

(nk

j r, u∗r

)p∗

r .(4.120)

This scheme has been called GLACE, for Godunov-LAgrange-Conservative-in-Energy [42].

Proposition 4.4.10. The GLACE scheme (4.118)–(4.120) is conservative in localmass and formal ly conservative in total momentum and total energy.

Proof. The local mass preservation evidently comes from (4.119). The total mo-mentum conservation can be checked as follows:

∑

j

Mj uk+1j −

∑

j

Mjukj = ∆t

∑

j

(∑

r

lkj rnkj rp

∗r

)= ∆t

∑

r

(∑

r

lkj rnkj rp

∗r

)= 0,

thanks to the second equation of (4.118). This verification did not take intoaccountthe possible exterior boundaries. The algebra is very similar for the conservationof total energy:

∑

j

Mjek+1j −

∑

j

Mj ekj = −∆t

∑

j

∑

r

lkj r

(nk

j r, u∗r

)p∗

r

= −∆t∑

r

(∑

r

lkj rnkj rp

∗r, u∗

r

)= 0.


Boundary conditions for GLACE

Boundary conditions have been invoked implicitly in Proposition 4.4.10. We detailhere some basic boundary conditions often encountered in Lagrangian calculations.We adopt a practical viewpoint and do not analyze the resulting scheme, neitherin terms of conservation nor in relation to stability or entropy inequalities.

Imposed velocity. Consider a node with index r on the boundary of the La-grangian computation domain. The imposed velocity boundary condition is

u∗r = ugiven ∈ R

2.

In this case it is sufficient to evaluate pj r accordingly to

p∗j r = −αj

(nk

j r, u∗r

)+ pk

j + αj

(nk

j r, ukj

). (4.121)


xr

pext

pext

pext

pext

Figure 4.24: Imposed external pressure at the boundary of the three triangularcells.

Imposed external pressure. Consider the situation described in figure 4.24, wherethe linear system (4.118) is modified on the boundary.

We write

∑

j

lkj rnkj rp∗

j r +

⎛⎝−

∑

j

lkj rnkj r

⎞⎠pext = 0. (4.122)

This has a natural mechanical interpretation in terms of the action-reactionlaw around the node r, where the external pressure pext is applied to theexterior of a fictitious cell with fictitious boundary corresponding to

lkext,rnkext,r = −

∑

j

lkj rnkj r.

Elimination of the pressures in (4.122) yields a new linear system like (4.117):this system is non-singular provided the corner normals are locally a basisof R 2 . Once the velocity determined, the pressures p∗

j r are easily computedwith (4.121).

Vanishing normal velocity. Assume that a vanishing normal velocity is imposedon the boundary and that the normal is the exterior discrete vector nk

j r. Thegeneric situation is sliding along a flat wall.

Let t = n⊥ be the tangent vector on the boundary. The sliding boundarycondition can be written as

ukr = u∗

r tk.

Consider the mechanical analogy of a ring sliding on a bar. The interactionsbetween the ring and the bar are determined by the projection of Newton’slaw in the direction t of the bar. We write

⎛⎝∑

j

lkj rnkj rp

∗j r −

∑

j

lkj rnkj rpext, tk

r

⎞⎠ = 0. (4.123)


n

(u, n) = 0

Figure 4.25: Vanishing normal velocity, which is a sliding condition.

Here pext is the exterior pressure, which in the case of the bar is identical

to the reaction force of the bar on the ring. Since(∑

j lkj rnkj r, tk

r

)= 0 by

definition, the unknown pressure vanishes in (4.123). Elimination of the p∗j r

by expressing them as functions of u∗r yields the linear scalar equation with

structure ⎛⎝∑

j

lkj rαj

(nk

j r, tkr

)2

⎞⎠ u∗

r = cr ∈ R . (4.124)

So the value of u∗r can be obtained and hence the p∗

j r values.

Other boundary conditions can be derived or constructed from these basicconditions. One of the most interesting situations, from the viewpoint of the con-struction of a numerical method, is when two different boundary conditions meetat a corner. This is illustrated in figure 4.26, where a horizontal sliding boundaryline with tangent vector t is connected to another boundary line to which an exter-nal pressure pext is applied on the right. It corresponds for example to an externalpressure applied to a piston in the combustion chamber of an engine. The issue isthe node at the corner. A simple solution exists which involves using (4.123) where

pext is the given external pressure. A priori(∑

j lkj rnkj r, tk

r

)= 0, so this term gives

a non-zero contribution. The sliding condition is again ukr = u∗

rtk. This is enoughto calculate the numerical value of u∗ and then the numerical value of the p∗

j r.Notice that in all cases, the pressures p∗

j r are calculated with the expression(4.121).

Numerical analysis of the CFL constraint

The linearized analysis of the entropy inequality, which is also a stability criterion,is based on evaluation of the quadratic forms Aj in (4.85) and Bj in (4.86). It hasbeen proved in dimension d = 1 that the optimal value of the free parameter αj


sliding line

corner

pext

n

Figure 4.26: A vanishing horizontal velocity boundary condition meets at the cor-ner node an external pressure boundary condition.

Figure 4.27: Example of a numerical simulation performed with the GLACEscheme with a triangular mesh. The initial data is a 2D Sod tube test problem.The final time is T = 0.2. The shock, contact discontinuity and 2D rarefaction fanare visible on the mesh structure.


is the local speed of sound. This is the reason why, by analogy, we set

αj = ρjcj > 0 in formula (4.113).

However, it must be pointed out that there is no theoretical reason so far whythis value should be optimal in dimension d = 2. In the following we analyze thisquestion. The main result will be that multidimensional factors show up.

Start from the quadratic forms Aj in (4.85) and Bj in (4.86). Denote by∆p ∈ R and ∆u = (∆u, ∆v) ∈ R 2 the variations of pressure and velocity. Usingthe linearization procedure with respect to ∆t from section 4.2.4, one obtains

Aj + O(∆t3) =1

2ρj

(∆p2

ρ2c2+ |∆u|2

)

and

Bj + O(∆t3) =1

4sj

∑

r

lj r

(αj r

(1

αj r∆p − (nj r, ∆u)

)2)

.

Let us remind the reader that analysis of the time step control is based on

∆tBj ≤ Aj ∀(∆p, ∆u) ∈ R3. (4.125)

To simplify a little the analysis of Aj and Bj , consider

αj r = αj = λjρjcj

where λ > 0 is an arbitrary non-dimensional parameter. In dimension d = 1, onetypically takes λj = 1 at the end of the analysis. A more convenient variable is∆p = (ρc)j ∆q so that

Aj + O(∆t3) =1

2ρj

(∆q2 + |∆u|2

).

The identity∑

r lj rnjr = 0 yields the simplification

Bj + O(∆t3) =1

4sj(ρc)j

∑

r

lj r

(1

λ∆q2 + λ (nj r, ∆u)

2

).

The ratio of these quantities is

Bj + O(∆t3)

Aj + O(∆t3)=

1

4sj(ρc)j

∑

r

lj r

(1

λ∆q2 + λ (nj r, ∆u)

2

)

1

2ρj

(∆q2 + |∆u|2

)

=cj

2sj

∑

r

lj r

(1

λ∆q2 + λ (nj r, ∆u)

2

)

∆q2 + |∆u|2.


Define Λj > 0 to be the largest eigenvalue of the quadratic form

Cj =∑

r

lj r

(1

λ∆q2 + λ (nj r, ∆u)

2

). (4.126)

A convenient linearized time step criterion then reads

cΛj

sj∆t ≤ 1. (4.127)

Notice that Λj encodes the geometry of the cell. To formulate the next result, wewill need

aj =∑

r

lj r cos θ2j r, bj =

∑

r

lj r cos θj r sin θj r,

cj =∑

r

lj r sin θ2j r and Λ′

j =a + c +

√(a − c)2 + 4b2

2.

Proposition 4.4.11. The optimal value of the geometrical parameter is

Λj =1

2

(Λ′

j

∑

r

lj r

)1

2

.

Proof. Upon inspection of (4.126), the optimal value Λ is the maximum of 1λ

∑r lj r

and Λ′j , which is the maximal eigenvalue of the quadratic form

∆u → λ∑

r

lj r

((nj r, ∆u)

2)

=

(∆u,

[∑

r

lj rnj r ⊗ nj r

]∆u

).

One has Λ′j =

aj+cj +√

(aj−cj)2 +4b2j

2. So Λj = 1

2max

(1λ j

∑r lj r, λjΛ′

j

). The opti-

mal value of λj minimizes Λj , that is,

λj =

(∑r lj r

Λ′j

)12

.

A substitution completes the proof.

It is convenient to express the CFL condition with μj = 1λj

.

Proposition 4.4.12. The CFL condition (4.125) admits the linear approximation

μj

(cj

∑r lj r

2sj∆t

)≤ 1 (4.128)

where the μj in front of the standard CFL terms is given by μj =

(Λ′

j∑r

ljr

) 12

.


Proof. Obvious.

In one dimension, one has Λ′ = 2,∑

r lj r and 12sj

∑r lj r = 1

∆x: in this case

μj = 1. That is why μj can be viewed as a multidimensional geometrical correctionto the usual CFL constraint.

Note that we have the bounds 1√2

≤ μj ≤ 1. Indeed,

Λ′j ≥ aj + cj

2=

1

2

∑

r

lj r,

so 1√2

≤ μj . On the other side, the Cauchy-Schwarz inequality yields b2j ≤ ajcj ,

so (aj − cj)2 + 4b2j ≤ (aj + cj)2 , which gives Λ′ ≤ ∑

r lj r. So μj ≤ 1.

EUCCLHYD scheme

Soon after the publication of GLACE, it was observed by Maire that anothersolution exists to overcome the geometrical obstruction. It is now called EUC-CLHYD, for which the first published work is [3], where it is referred to by adifferent name. The fluxes are based on a different choice for the corner pressures.The initial idea was to get closer to the usual (“usual” in the realm of hyperbolictheory for conservation laws) interpretation in terms of acoustic linearized Rie-mann solvers through edges, while still maintaining corner-based solvers. This isillustrated in figure 4.28.

ur

p+

jrp−

jr

j

Figure 4.28: The corner-based pressures are delocalized twice. They are denotedby p−

j r and p+j r for the top corner.

At cell j and node r, one constructs two delocalized pressures p−j r and p+

j r

(orientation is counterclockwise). The discrete form of an acoustic Riemann solverbetween the current cell with index j and auxiliary states in directions nk

j r± yields


the two first equations:

⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩

p∗j r+ + αj

(nk

j r+, u∗r

)= pk

j + αj

(nk

j r+, ukj

), j = 1, . . . , J,

p∗−j r− + αj

(nk

j r−, u∗r

)= pk

j + αj

(nk

j r− , ukj

)j = 1, . . . , J,

∑

j

(lkj r+nk

j r+p∗j r+ + lkj r− nk

j r−p∗j r−

)= 0.

(4.129)The last equation is a direct copy of (4.118). Compared to GLACE the other equa-tions are the same. That is, one uses equation (4.108) for the mesh displacementand equation (4.119) for mass conservation after recalculation of the new area ofthe cell,

ρk+1j =

skj

sk+1j

ρkj ⇐⇒ Mj = sk

j ρkj = sk+1

j ρk+1j , (4.130)

and then one updates the new velocity uk+1j and new energy ek+1

j by

⎧⎪⎪⎨⎪⎪⎩

Mj uk+1j = Mjuk

j − ∆t∑

r

(lkj r+nk

j r+p∗j r+ + lkj r− nk

j r−p∗j r−

),

Mjek+1j = Mjek

j − ∆t∑

r

(lkj r+

(nk

j r+, u∗r

)p∗

j r+ + lkj r−

(nk

j r−, u∗r

)p∗

j r−

).

(4.131)The value of αj used in most calculations is the standard one, αj = ρk

jckj , i.e.

without the geometrical factor λj .One might postulate that the difference between GLACE and EUCCLHYD

is comparable to using different quadrature formulas in the theory of the finite ele-ment method. In practice, different behaviors havebeen observed from the GLACEscheme (4.118)–(4.120) and the EUCCLHYD scheme (4.129)–(4.131) for quadran-gular meshes. The GLACE scheme seems to be less viscous than the EUCCLHYDscheme; see [72, 150, 120]. The EUCCLHYD scheme is much less sensitive to hour-glass modes; this was explained for the first time in [163]. On triangular meshes noreal differences have been observed. This very preliminary understanding of multi-dimensional Lagrangian cell-centered schemes will be substantially augmented inchapter 5.

4.4.7 Comparison with a scheme on a staggered mesh

The schemes discussed previously are cell-centered. It is interesting to compare thestructure of these numerical methods with that of a more traditional staggeredscheme for which the velocity is discretized at nodes. Historically the foundationsof staggered Lagrangian schemes were laid out by von Neumann and Richtmyer[195] in one dimension and further extended to higher dimensions by Wilkins[205, 206]. A recent overview of the topic is given in [170]. The hydro-compatibleversion was recently proposed in [34, 38] and completed in [37, 35, 139]. Much


more can be found in the review [149]. The hydro-compatible variant discussedbelow uses a convenient integration-in-time procedure which makes the schemeconservative in total energy.

We start from the semi-discrete identity (4.106) for the time variation of thearea of a bi-dimensional cell,

s′j(t) =

∑

r

lj r (nj r, ur) .

The Lagrangian mass Mj =sj(t)τj(t)

being constant, the time evolution of the specific

volume is given by

Mjτ ′j(t) −

∑

r

lj r (nj r, ur) = 0. (4.132)

The total energy equation is not discretized directly in most staggered schemes.Instead one starts from the entropy equation discretized as

ε′j(t) + (pj + qj) τ ′

j(t) = 0. (4.133)

Here the term qj ≥ 0 is an artificial viscosity. The introduction of this term isrelated to the identity

TjS′j(t) = ε′

j (t) + pjτ ′j(t) = −qjτ ′

j(t).

In numerical shocks the evolution is mostly compressive, that is, −τ ′j(t) > 0. Since

the entropy production needs to be positive in shocks, −qjτ ′j(t) > 0, one sees that

qj > 0 is necessary for modeling shocks. The nearly optimal value of artificialviscosity was identified in the early work of von Neumann and Richtmyer [195]:

q ≈ ∆x(

C1 |∇u| + C2 |∇u|2)

. (4.134)

Refer to [170] for multidimensional extensions.Let us now concentrate on the momentum equation. Since the method is

staggered, the velocity variable is stored at the nodes. A discrete Newton’s lawyields

Mru′r(t) = −

∑

j

Drj (pj + qj) , (4.135)

where the geometrical vectors Drj , still to be defined, are a priori functions of themesh. Notice that the pressure gradient is approximated as

∇p ≈∑

j

Drj (pj + qj) .

Such a scheme is said to be conservative in total momentum if∑

r

Mr u′r(t) = 0


up to the boundaries. An important feature of the hydro-compatible staggeredscheme is the definition of the total energy, which is necessarily a blend of cellquantities and nodal quantities. One uses

Mjej = Mjεj +∑

r

Mj r1

2|ur|2 ,

where the zonal masses Mj r are still to be defined. A priori one has to complywith

∑r Mj r = Mj in order to get a natural definition of the total energy in the

cell. Such a scheme is said to be conservative in total energy if

∑

j

Mje′j (t) = 0

up to the boundaries.

Proposition 4.4.13. Assume that Drj = −lj rnj r and Mr =∑

j Mj r with Mj r

constant in time. Then the hydro-compatible scheme is formal ly conservative intotal momentum and total energy.

Proof. One has

Mj e′j(t) = Mjε′

j (t) +∑

r

M ′j r(t)

1

2|ur|2 +

∑

r

Mj rMj r (ur, u′r(t))

= − (pj + qj)∑

r

lj r (nj r, ur ) +∑

r

M ′j r(t)

1

2|ur|2

+∑

r

Mj r

Mr

⎛⎝ur, −

∑

j ′

Drj ′ (pj ′ + qj ′)

⎞⎠ ,

where the relations (4.132)–(4.135) have been used. So

Mj e′j(t) =

∑

r

M ′j r(t)

1

2|ur|2 −

∑

r

(ur, lj rnj r

(pj + qj

)+

Mj r

Mr

∑

j ′

Drj ′

(pj ′ + qj ′

)).

The first term vanishes for Lagrangian zonal masses such that M ′j r(t) = 0. So the

conservation identity∑

j Mje′j (t) = 0 holds if

0 =∑

j

⎛⎝lj rnj r (pj + qj) +

Mj r

Mr

∑

j ′

Drj ′ (pj ′ + qj ′)

⎞⎠ .

Rewrite this as⎛⎝∑

j

Mj r

Mr

⎞⎠∑

j

Drj (pj + qj) =∑

j

(lj rnj r (pj + qj))


using the hypothesis, that is, Mr =∑

j Mj r and Drj = −lj rnj r. In this case theabove identity is always true whatever the pressure pj and artificial viscosity qj

are. So the conservation of total energy is proved. Rewrite the momentum equationas Mr u′

r(t) =∑

j lj rnj r (pj + qj). Then

∑

r

Mru′r(t) =

∑

j

((pj + qj)

∑

r

lj rnj r

)= 0

due to (4.104). The proof is complete.

Remark 4.4.14. The Lagrangian zonal masses Mj r are in practice defined at thebeginning of the calculation. Typically Mj r = ρj(0)s0

j r with ρj(0) the initial den-

sity and s0j r a control area such that

∑r s0

j r = s0j .

The next step is to discretize in time. The hydro-compatible variant man-dates absolute respect of the conservation of total energy since it is necessary forobtaining correct numerical shocks. The result is a one-step scheme in the sensethat all unknowns are computed at time step (k + 1)∆t from the unknowns attime step k∆t. Let us now describe the workflow of the method. One starts from

skj , τ k

j , εkj and uk

r

known at time step k∆t either in cell j or at node r. The first step is to move thenodes, xk+1

r = xkr + tuk

r. The second step is to recalculate the area of the cell sk+1j

and the new density ρk+1j =

Mj

sk+1

j

. The third step is to update the nodal velocities

with a natural explicit Euler version of (4.135):

Mruk+1

r − ukr

∆t=

∑

j

f kj r with f k

j r = −Dkrj

(pk

j + qkj

). (4.136)

The vector f kj r can be physically interpreted as the force that cell j exerts on node

r. It remains to update the internal energy so that a global conservation law holdsfor the total energy. This is the main point of the method. One way to do this is

εk+1j − εk

j

∆t= −

∑

r

(f kj r,

ukr + uk+1

r

2

). (4.137)

Proposition 4.4.15. The relations (4.136)–(4.137) yield a scheme which is formal lyconservative in total energy.

Proof. Start from the total energy in the cell Mjekj = Mjεk

j +∑

r Mj r12

∣∣ukr

∣∣2 . One


has

Mj

ek+1j − ek

j

∆t= Mj

εk+1j − εk

j

∆t+∑

r

Mj r

(uk+1

r + ukr

2,

uk+1r − uk

r

∆t

)

= −∑

r

(f kj r,

ukr + uk+1

r

2

)+∑

r

Mj r

Mr

⎛⎝uk+1

r + ukr

2,∑

j ′

f kj ′r

⎞⎠

=∑

r

⎛⎝−f k

j r +Mj r

Mr

∑

j ′

f kj ′r,

ukr + uk+1

r

2

⎞⎠ .

Since∑

j

⎛⎝−f k

jr +Mj r

Mr

∑

j ′

f kj ′r

⎞⎠ = 0,

one has∑

j

(Mj

ek+1

j −ekj

∆t

)= 0.

In other words, the work done by the force is the scalar product of theexplicit term f k

j r with the half-sum of the velocities. This is the key to obtaining acorrect increment of the energy. Let us mention finally that an important degreeof freedom in such methods is the artificial viscosity.

4.4.8 Well-balanced hydrostatic cell-centered Lagrangian schemes

Gravity source terms are essential in scientific computing for geophysical appli-cations. The use of such source terms in the definition of the fluxes is an activearea of research. See [27, 101] and references therein for much more material onwell-balanced techniques for problems in geophysics and various other fields. Ourobjective is to show that an elegant strategy originally designed by Cargo andLeroux [41] in dimension d = 1 can be adapted to the Lagrangian setting indimensions d = 1 and d = 2.

The one-dimensional case

The Euler system for compressible fluids with gravity in dimension d = 1 reads⎧⎪⎪⎨⎪⎪⎩

∂tρ + ∂x (ρu) = 0,

∂t(ρu) + ∂x

(ρu2 + p

)= −ρg,

∂t(ρe) + ∂x (ρue + pu) = −ρgu.

(4.138)

Here g > 0 is the constant gravitational acceleration, oriented by convention inthe direction of decreasing x. Our attention will be focused on numerical methodswhich respect stationary states.


Definition 4.4.16. Stationary states of the Euler system with gravity (4.138) aredefined by

u = 0 and ∂xp = −ρg.

Let us use Cargo and Leroux’s [41] idea, which consists in defining a potentialq such that

∂x q = −ρg and ∂tq = ρgu.

These relations are compatible with ∂t ρ + ∂x(ρu) = 0. It is easy to verify theequation

∂t(ρq) + ∂x (ρuq) = ρ (∂t + u∂x) q = 0.

This means that the non-homogeneous Euler system can be written as a conser-vative system of equations

⎧⎪⎪⎪⎨⎪⎪⎪⎩

∂tρ + ∂x(ρu) = 0,

∂t(ρq) + ∂x (ρuq) = 0,

∂t(ρu) + ∂x

(ρu2 + Φ

)= 0, Φ = p − q,

∂t(F ) + ∂x (ρuF + Φu) = 0, F = ρe + q.

Physically, the potential q is the gravitational energy and F is the total energy,which is the sum of the thermic energy ρε, the kinetic energy 1

2ρu2 and the grav-

itational energy. The Lagrangian formulation is

⎧⎪⎪⎪⎨⎪⎪⎪⎩

∂tτ − ∂mu = 0,

∂tΦ = 0,

∂tu + ∂mΦ = 0,

∂tf + ∂m(Φu) = 0, f = e + τ q.

(4.139)

Notice that it is compatible with the entropy law ∂tS = 0 for smooth solutions:this is natural since gravitation has no interaction with the thermodynamics forsuch systems. Therefore the assumptions of Theorem 3.2.8 are satisfied: one canuse the whole machinery developed in this chapter for the definition of Lagrangiannumerical discretization.

From the fundamental principle of thermodynamics one obtains

TdS = dε+p dτ = de−u du+p dτ = d(f −τ q)−u du+p dτ = df −u du−τ dq+Φ dτ.

This yields

−V = (∇US)t

=1

T

⎛⎜⎜⎜⎝

Φ

−τ

−u

1

⎞⎟⎟⎟⎠ , Ψ =

⎛⎜⎝

Φ

−τ

−u

⎞⎟⎠ and M =

⎛⎜⎝

0 0 1

0 0 0

1 0 0

⎞⎟⎠ .


The two-state Lagrangian scheme reads⎧⎨⎩

(Φ∗

j +12

− Φj

)+ ρjcj

(u∗

j +12

− uj

)= 0,

(Φ∗

j +12

− Φj +1

)− ρj +1cj +1

(u∗

j +12

− uj +1

)= 0.

(4.140)

Denoting as usual by Mj the Lagrangian mass in the cell, one obtains the La-grangian scheme

⎧⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎩

Mj

(τ k+1

j − τ kj

)−(

u∗j + 1

2

− u∗j −1

2

)= 0,

Mj

(qk+1

j − qkj

)= 0,

Mj

(uk+1

j − ukj

)+(

Φ∗j +1

2

− Φ∗j − 1

2

)= 0,

Mj

(fk+1

j − fkj

)+(

Φ∗j +1

2

u∗j +1

2

− Φ∗j − 1

2

u∗j −1

2

)= 0.

(4.141)

Proposition 4.4.17. The scheme (4.141) respects the family of stationary statesuj = 0 and Φj = Φ ∈ R for al l j .

Proof. The key point is that stationary states of the discrete system are easyto analyze since the system (4.139) is in conservative or divergent form. Uponinspection of the fluxes in (4.141), these discrete stationary states satisfy u∗

j − 1

2

and Φ∗j + 1

2

= Φ ∈ R for all j . One obtains, by substitution into (4.140),

(Φ − Φj) − ρjcjuj = 0,

(Φ − Φj +1) + ρj +1cj +1uj +1 = 0,for all j.

That is, (Φ − Φj) − ρjcjuj = 0,

(Φ − Φj) + ρjcjuj = 0,for all j.

This yields the claim.

It is of great interest to reformulate the scheme (4.141) with standard vari-ables. The velocity is

u∗j −1

2

=ρjcjuj + ρj +1cj +1uj +1

ρjcj + ρj +1cj +1+

ρjcjρj +1cj +1

ρjcj + ρj +1cj +1(Φj − Φj +1) .

The viscous term is Φj − Φj +1 = pj − pj +1 + qj +1 − qj . Assume that the schemeis initialized with

qj +1 − qj = −g∆xjρj + ∆xj +1ρj +1

2,

which is consistent with the stationary equation ∂xq = −ρg. So

qj +1 − qj = − 1

2gMj − 1

2gMj +1


and

Φj − Φj +1 =

(pj − 1

2gMj

)−(

pj +1 +1

2gMj +1

).

Plug in the velocity

u∗j −1

2

=ρjcjuj + ρj +1cj +1uj +1

ρjcj + ρj +1cj +1

+ρjcjρj +1cj +1

ρjcj + ρj +1cj +1

((pj − 1

2gMj

)−(

pj +1 +1

2gMj +1

)).

Observe that a modification of the pressures, where pj becomes pj − 12gMj and

pj +1 becomes pj +1+ 12gMj +1, is enough to get the value of the velocity flux u∗

j −1

2

.

Definition 4.4.18 (Hydrostatic two-state solver). The hydrostatic two-state solveris an extension of the two-state formula in the form

⎧⎪⎨⎪⎩

(p∗

j +1

2

− pj

)+ ρjcj

(u∗

j + 1

2

− uj

)= − 1

2gMj ,

(p∗

j +12

− pj +1

)− ρj +1cj +1

(u∗

j +12

− uj +1

)=

1

2gMj +1 .

(4.142)

The hydrostatic solver is by construction equivalent to the formula withpotential q and unknown f . One can write the momentum equation as

Mj

(uk+1

j − ukj

)+(

Φ∗j +1

2− Φ∗

j − 12

)= 0 ⇐⇒

Mj

(uk+1

j − ukj

)+(

p∗j +1

2

− p∗j − 1

2

)=

(p∗

j +12

− p∗j −1

2

)−(

Φ∗j +1

2

− Φ∗j − 1

2

).

(4.143)

By comparison of the first lines in (4.140) and (4.142) one gets

[(Φ∗

j + 12

− Φj

)+ ρjcj

(u∗

j +12

− uj

)]−[(

p∗j +1

2

− pj

)+ ρjcj

(u∗

j + 12

− uj

)]=

1

2gMj ,

that is,

p∗j +1

2

− Φ∗j +1

2

= − 1

2gMj + pj − Φj . (4.144)

Similarly, comparison of the second lines in (4.140) and (4.142) yields

[(Φ∗

j +12

− Φj +1

)− ρj +1cj +1

(u∗

j +12

− uj +1

)]

−[(

p∗j + 1

2

− pj +1

)− ρj +1cj +1

(u∗

j +12

− uj +1

)]= − 1

2gMj +1 ,

that is,

Φ∗j +1

2

− p∗j +1

2

= − 1

2gMj +1 + Φj +1 − pj +1.


One gets, upon shifting the index,

Φ∗j −1

2

− p∗j − 1

2

= − 1

2gMj + Φj − pj . (4.145)

Substitute (4.144) and (4.145) into (4.143) to get

Mj

(uk+1

j − ukj

)+(

p∗j + 1

2

− p∗j −1

2

)= −gMj . (4.146)

Proposition 4.4.19. The scheme

⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩

Mj

(τ k+1

j − τ kj

)−(

u∗j +1

2

− u∗j −1

2

)= 0,

Mj

(uk+1

j − ukj

)+(

p∗j + 1

2

− p∗j −1

2

)= −gMj ,

Mj

(ek+1

j − ekj

)+(

p∗j + 1

2

u∗j +1

2

− p∗j −1

2

u∗j − 1

2

)= −guk

jMj

(4.147)

with the fluxes (4.142) is wel l-balanced.

Proof. Evident. Note that the energy equation can be modified while still main-taining the well-balanced hydrostatic property.

The 2D case

The Euler system for a compressible fluid with gravity reads, in dimension d = 2,⎧⎪⎪⎨⎪⎪⎩

∂t ρ + ∇ · (ρu) = 0,

∂t (ρu) + ∇ · (ρu ⊗ u) + ∇p = ρg, u ∈ R 2,

∂t (ρe) + ∇ · (ρeu + pu) = ρ(g, u),

(4.148)

where the vectorial gravitational acceleration is a priori aligned in the verticaldirection:

g = (0, −g)t ∈ R 2, g > 0.

Our objective is to generalize the one-dimensional well-balanced scheme (4.120)of the form

⎧⎪⎪⎨⎪⎪⎩

Mj uk+1j = Mjuk

j − ∆t∑

r

lkj rnkj rp

∗j r + Mj ∆tg,

Mjek+1j = Mjek

j − ∆t∑

r

lkj r

(nk

j r, u∗r

)p∗

j r + Mj∆t(g, uk

j

).

(4.149)

As noted earlier, the numerical evaluation of the work done by gravity(g, uk

j

)can

be modified to (g,

ukj + uk+1

j

2

).


Definition 4.4.20. The scheme (4.149) is said to be partially well-balanced if itpreserves non-trial states,

uk+1j = uk

j = 0 and ek+1j = ek

j . (4.150)

Remark 4.4.21. A fully well-balanced Lagrangian scheme would preserve in addi-tion u∗

r = 0. The partially well-balanced property is simpler to analyze.

The analysis showing that a standard solver like (4.118) is not well-balancedis as follows. Concerning the velocity equation, a well-balanced scheme satisfies

0 = −∆t∑

r

lkj rnkj rp

∗j r + Mj ∆tg. (4.151)

Using a standard corner-based solver like (4.118) yields p∗j r + αj

(nk

j r, u∗r

)= pk

j +

αj

(nk

j r, ukj

). One obtains three relations which express the compatibility with the

hydrostatic equilibrium (4.150):

⎧⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎩

p∗j r = pk

j (comes from (4.118)),

−∆t∑

r

lkj rnkj rp

∗j r + Mj∆tg = 0 (comes from (4.151)),

∑

i

lkj rnkj rp

∗j r = 0 (comes from (4.118)).

(4.152)

Using the closed-contour formula∑

r lkj rnkj r = 0, these three relations boil down

to two independent relations

⎧⎪⎨⎪⎩

Mj∆tg = 0,∑

j

lkj rnkj rp

kj = 0. (4.153)

Proposition 4.4.22. The standard corner-based solver (4.118) is not partial ly wel l-balanced.

Proof. Indeed, only trivial equilibria with zero gravity are solutions to (4.153).The proof is finished.

The problem is clearly with the first equation in (4.153). To propose a solu-tion, we directly generalize the one-dimensional hydrostatic solver (4.142) to thecorner-based Lagrangian structure. We denote by xk

r =(xk

r , ykr

)the node r at

time step k and by

xkj =

(xk

j , ykj

)

the center of mass of cell j at the same time step.


Definition 4.4.23. The hydrostatic Lagrangian well-balanced solver is

⎧⎪⎨⎪⎩

p∗j r + αj

(nk

j r, u∗r

)= pk

j + αj

(nk

j r, ukj

)+ ρk

j

(g, xk

r − xkj

)∀j,

∑

i

lkj rnkj rp∗

j r = 0, (4.154)

with αj = ρkj ck

j .

The new term in the well-balanced solver is ρkj

(g, xk

r − xkj

). It is a direct

generalization of ± 12gMj in dimension d = 1. Notice that the linear system (4.154)

is non-singular, so u∗r and the p∗

j r are correctly defined.

Proposition 4.4.24. The new corner-based solver (4.154) is partial ly wel l-balanced.

Proof. Indeed, the stationary relations (4.150) are equivalent to

⎧⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎩

p∗j r = pk

j + ρkj

(g, xk

r − xkj

),

− ∆t∑

r

lkj rnkj rp

∗j r + Mj ∆tg = 0,

∑

j

lkj rnkj rp

∗j r = 0.

After eliminating the p∗j r, one gets the two compatibility relations

⎧⎪⎪⎨⎪⎪⎩

−∑

r

lkj rnkj r

(g, xk

r − xkj

)+ sk

j g = 0,

∑

j

lkj rnkj r

(pk

j + ρkj

(g, xk

r − xkj

))= 0.

(4.155)

The first compatibility relation can be rewritten as

skj g =

∑

r

lkj rnkj r

(g, xk

r

).

It holds for all g thanks to the identity in Proposition 4.4.25; see also (5.17) in amore general context. The proof is complete.

The second compatibility relation corresponds to full well-balancedness. It iscurrently still an open problem to determine general solutions on arbitrary gridsfor arbitrary pressure laws. Nevertheless, numerical results have already shownthe advantage of partial well-balancedness.

Proposition 4.4.25. One has the geometrical identity∑

r lj rnj r ⊗ xr = sj I.


Proof. A more general proof of this fundamental relation is given in the nextchapter. One has

sj =

∫

Ωj

(∂x x)dx dy =

∫

∂Ωj

(n, (x, 0)) dσ

=∑

k

lj k cos θj kxk+ + xk−

2=∑

r

lj r cos θj rxr

where xk+ and xk− are the first coordinates of the endpoints of the segment jk.By symmetry sj =

∑r lj r sin θj ryr. One also has

0 =

∫

Ωj

(∂xy) dx dy =∑

r

lj r cosθj ryr

and, by symmetry, 0 =∑

r lj r sin θj rxr. The proof is complete.

4.4.9 Mesh considerations and numerical examples

The choice of mesh at the initialization of a Lagrangian computation has an im-portant influence on the quality of the computation, in particular on the dynamicquality of the mesh. For example, consider a challenging test problem called theSatzmann piston problem calculated with a quadrangular mesh; see figure 4.29.The bottom panel shows that the velocity of the numerical shock is correct andthe mesh after the shock is still reasonable. For this problem one might think thatit is only by chance that the mesh is correct. The discussion below is kept to aminimum; more will be said on this important topic in chapter 5.

Triangles versus quadrangles

The type of cells may have a fundamental effect on the mesh stability of thecomputation. Indeed, triangular meshes are endowed with an intrinsic stabilityproperty, which is relatively easy to understand from the pictures in figure 4.30.

On the left of the figure, a spurious behavior of a quadragular cell is depicted:two nodes cross each other while maintaining a positive total area of the quadran-gle. Such behavior is impossible for the triangular cell on the right. In this case,two nodes which come together induce a very strong compression of the cell. Sothe density would increase by virtue of a constant Lagrangian mass. This wouldresult in an increase of pressure, which in turn would generate a strong force onthe cells in the neighborhood. At the end of this scenario the triangular cell willexpand. A rigorous proof of this scenario is given in the next chapter.

Unstable mesh modes

Unstable mesh modes are particularly dangerous for Lagrangian computations.We consider here two types of unstable modes. The first is well documented in


Figure 4.29: Saltzmann piston problem calculated on a moving quadrangularmesh;initial mesh is on top. The piston velocity is up = 1.

Figure 4.30: Two evolutions of a cell.

the literature. These so-called hourglass modes are visible on quadrangular meshessuch as the one shown in figure 4.31.

For the Noh test problem, the matrix Ar for the solution of the corner-based


Figure 4.31: Noh test problem. The initial data has ρ = 1 and p = 10−6, witha velocity oriented toward the center with modulus 1. For a perfect pressure lawwith γ = 7

5, the exact solution is a plateau of density ρp = 16. The result on the

left is full of hourglass unstable modes. The result on the right is stabilized by anincrease of the speed of sound in the two-state solver.

solver is

Ar =∑

j

(lk+ 1

2

j r αjnk+12

j r ⊗ nk+12

j r

)≈ 0

since αj is proportional to ρjcj ≈ 0. A possible remedy is to increase the speedof sound to a value closer to the value of the exact solution. Another remedymight be to use the EUCCLHYD solver, which is more dissipative and so morerobust. This problem will be addressed in more detail in chapter 5. Another typeof unstable mesh mode occurs when the usual Lagrangian corner-based solver isused for a hydrostatic initial condition, but without employing the hydrostaticsolver discussed in section 4.4.8.

Shears and large mesh deformations

Finally, let us consider the simulations of figure 4.32, where a heavy fluid (water)is in a light gas (air) with a gravity field. Lagrangian schemes such as GLACEor EUCCLHYD ultimately behave in the same manner for such problems. Evenwith a triangular mesh and the hydrostatic partially well-balanced solver, largedeformations of the mesh make the computation globally incorrect after a certaintime. These large deformations are related to shear layers at the boundary of thedrop. Notice, however, that the two materials are still perfectly separated, whichis one of the main properties of a purely Lagrangian computation.

4.5. Calculation of Lagrangian multi-material problems 255

Figure 4.32: A drop of water falling in air [20]. The gravity is oriented toward thebottom. The presence of shears at the boundary of the drop makes the computationglobally unacceptable after a certain time, even with a triangular mesh.

The consequence is nevertheless that a purely Lagrangian computation isnever able to compute the flow of figure 4.32 without additional techniques. Thisis where ALE is needed to regularize the mesh in the air. A difficulty is to pre-serves the internal boundary between water and air. Another technique that is welladapted to this problem is free-Lagrange, where the connectivity of the mesh canchanged at certain time steps. This is typically what is done in simulations suchas that reported in figure 4.33. A vast literature exists on ALE and free-Lagrangemethods.

4.5 Calculation of Lagrangian multi-material problems

A multi-material Riemann problem involves different pressure laws. The theory ofsuch problems is essentially equivalent to the theory with only one pressure law,provided one uses a Lagrangian formulation. Indeed, as the Lagrangian formulationis written in the fluid frame, it is naturally suited to this problem. Additionalreferences are provided at the end of the chapter.

The two-state solver (4.53) used in the following calculations turns out tobe quite efficient: actually the one-state solver also gives good results for theseproblems. We consider two problems.

Sod problem with two γ values: Consider the Riemann initial data

γL = 2, ρL = pL = 2, uL = vL = 0


Figure 4.33: Same problem as in figure 4.32, but computed with a free-Lagrangetechnique on a Delaunay-Voronoı mesh (from [20]).

and

γR = 1.4, ρR = 0.125, pR = 0.1, uR = vR = 0.

This problem has two different pressure laws which are still perfect gases. But thecoefficients are different: γ = γL = 2 on the left, for x < 0.5, and γ = γL = 1.4on the right, where 0.5 < x. The solution is defined and studied in detail in [116]for the purpose of numerical benchmarking. In our calculations, a huge amount ofwall-heating is visible on the entropy profile: the integral of the spurious part canbe reduced by decreasing the mesh size. But the amplitude of the defect remainsapproximately constant and independent of the mesh size. The exact solution (tosix decimal places) on both sides of the contact discontinuity is provided in table4.2.


R v∗R = v∗

L

0.430332 1.127571 0.463860 0.325380 0

Table 4.2: Multi-material Sod tube test problem: numerical values at the contactdiscontinuity.

Water-air shock tube: This is also a classical test problem. It involves a stiffenedgas pressure for water (the parameters are γL = 4.4 and ΠL = 6 × 108) and aperfect gas pressure law for air (γR = 1.4). The other Riemann initial values are

ρL = 1000, pL = 109, uL = vL = 0

4.5. Calculation of Lagrangian multi-material problems 257

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

’wi.rho’

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

’wi.p’

x → ρ(x) x → p(x)

0

0.2

0.4

0.6

0.8

1

1.2

1.4

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

’wi.u’

0.6

0.7

0.8

0.9

1

1.1

1.2

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

’wi.S’

x → u(x) x → S(x)

Figure 4.34: Sod shock tube test problem with two γ values, computed in Lagrangemode using (4.53), with final time T = 0.2 and 100 cells.

andρR = 50, pR = 106, uR = vR = 0.

The results computed with 250 cells using the linearized Riemann solver with twostates are displayed in figure 4.35.


R v∗R = v∗

L

1.59868 × 107 4.81391 × 102 8.04979 × 102 2.20407 × 102 0

Table 4.3: Water-air shock tube test problem: numerical values at the contactdiscontinuity.

Sod problem with two γ values in 2D: we consider dimension d = 2. Consider theRiemann initial data

γL = 3, ρL = pL = 1, uL = vL = 0,√

x2 + y2 < 0.5,


0

100

200

300

400

500

600

700

800

900

1000

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

’wi.rho’ using 1:2

0

1e+08

2e+08

3e+08

4e+08

5e+08

6e+08

7e+08

8e+08

9e+08

1e+09

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

’wi.p’ using 1:2

x → ρ(x) x → p(x)

0

50

100

150

200

250

300

350

400

450

500

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

’wi.u’ using 1:2

-10

-8

-6

-4

-2

0

2

4

6

8

10

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

’wi.S’ using 1:2

x → u(x) x → S(x)

Figure 4.35: Water-air shock tube test problem, computed in Lagrange mode using(4.53), with final time T = 2.2 × 10−4 and 250 cells.

and

γR = 1.4, ρR = 0.125, pR = 0.1, uR = vR = 0, 0.5 <√

x2 + y2.

The numerical solution is computed with the GLACE scheme and the mesh is plot-ted at time t = 0.2 in figure 4.36. For such problems, multidimensional Lagrangiancalculation allows exact tracking of the interface between the two fluids (referredto as the two materials). In the second calculation, the Lagrangian scheme is post-processed at every time step with a free-Lagrange-Delaunay remeshing algorithmto get a more regular mesh.

The issue of solvers for multi-material flows on Eulerian grids is addressed inmany works. In our context, a general solution approach is to couple Lagrangiandiscretization with some interface tracking or interface reconstruction techniquein the remaping stage; for convenience of the reader we provide some references(which are far from exhaustive) on the topic [116, 75, 122, 144, 202, 168, 5, 52,169, 158, 181, 180, 199, 186, 85].

4.6. Exercises 259

4.6 Exercises

Exercise 4.6.1. Define a Lagrange+remap scheme for the shallow water system.

Exercise 4.6.2. Define an ALE scheme for the LWR traffic flow equation.

Exercise 4.6.3. Define a two-state solver for the linear wave equation in dimensiond = 1.

Exercise 4.6.4. Rewrite (1.25) in the Lagrangian variable X. Verify that the un-known is U t = (ρJ, J, ρJu, ρJe). Start from the entropy η = −ρJS and show thatthe generic Lagrangian structure has matrix

M =

⎛⎜⎝

0 0 0

0 0 1

0 1 0

⎞⎟⎠

and a vector Ψ to be determined. Construct entropy schemes starting from (4.59)or (4.62).

Exercise 4.6.5. Consider the Lagrangian system with constant gravity⎧⎪⎨⎪⎩

∂tτ − ∂mu = 0,

∂tu + ∂mp = g,

∂te + ∂mpu = gu.

Rewrite it as ⎧⎪⎪⎪⎨⎪⎪⎪⎩

∂t τ − ∂mu = 0,

∂tx = u,

∂tu + ∂mp = g,

∂t(e − q) + ∂mpu = 0,

where the potential q is to be determined. Show that this new system can be putin the form (3.120). Design an entropic solver by comparison with (4.80)–(4.82).

Exercise 4.6.6. Consider the scheme (4.118)–(4.120) for a flow initially at constantvelocity uk

j = U and constant pressure pkj = P . Show that the infinite mesh is

translated with velocity U and the pressure remains constant.

Exercise 4.6.7. Consider the GLACE scheme, defined either as in this chapter orwith the more general notation of the next chapter. Assume that the dual mesh(defined at nodes) is made up of simplexes. Show that pj r = pr for all j , whichmeans one single corner pressure.

Exercise 4.6.8. Consider a multidimensional Lagrangian system of size n and di-mension d, with M0 = 0. Recast it as

∂t u(t, x1, . . . , xd) +d∑

j =1

∂xj fj(u(t, x1 , . . . , xd)) = 0, (4.156)

with the property that∑d

j =1 ∂xj fj(v) = 0 for all v ∈ Rn.



Research on staggered Lagrangian hydrodynamics is still very active; we refer to[137] and references therein.

When applied to compressible gas dynamics, the two-state acoustic solveris equivalent to the linearization of the Godunov Lagrangian solver [100]. Othertextbooks on numerical methods for hyperbolic equations arising from physicsare [2, 190, 82, 189, 93, 94]. The review article [19] on numerical methods forcompressible fluids is highly recommended. An important reference on LagrangianRoe solvers [166] for gas dynamics is [155]: the one-state Lagrangian can actuallybe constructed as a Roe solver. A new direction of research concerns low-MachLagrange+remap schemes [45].

Numerical MHD is discussed in [21, 60, 77, 87, 161, 58, 59]. Other referencescan be found in [121]. The mathematical structure of hyperbolic equations withdifferential constraints, also called involute constraints, is discussed in [23].

The entropy inequality for a one-state solver is from [68]; see also [8]. A num-ber of different techniques give discrete entropy inequalities. One popular method,using relaxation techniques, can be found in the work of Coquel [55, 43, 16, 56].See also Gallice [39, 86, 40]. We refer to Bouchut [26, 27] for discussion of sometwo-state solvers in the context of Suliciu’s systems and well-balanced schemes. Astate-of-the-art review on well-balanced methods is the recent synthesis by Gosse[102]. Use of the Lagrangian structure for relaxation of fluid models can be foundin [54], and for the definition of linearized models with associated discretizationin [114]. More material on well-balanced solvers includes the seminal paper [103],where the objective is to incorporate the sources in the fluxes; see also [101] and[27]. Numerical methods, and specifically staggered schemes for gas dynamics, areintroduced in the first textbook ever on such topics, [165]. Some techniques forthe definition of satisfactory pseudo-viscosities for Lagrangian staggered schemesare discussed in different contexts in [157, 35, 83]. A different family of schemesis proposed in [196]. See also [174, 173, 171, 172] for Lagrangian finite elementschemes. The first cell-centered methods for Lagrangian gas dynamics can be foundin [100, 191], but without a consistent treatment of the GCL (geometric conser-vation law), another name for the Piola identity. These methods are finite volumetechniques with standard edge-based fluxes. Some information on high-order ex-tensions of cell-centered methods can be found in [145, 42]. A interesting pathto high-order Lagrangian discretization with discontinuous Galerkin techniques isdescribed in [64, 65, 193, 194, 164].

It should be mentioned that corner/vertex-based solvers are rarely studiedin the literature, with the notable exception of some schemes due to Roe; forfluctuation schemes see [4] and also [6]. An important difference from the materialpresented in this monograph is that fluctuation schemes are designed to optimizeapproximation properties. For Lagrangian schemes they are introduced to respectthe discrete GCL and the Piola identities. An introduction to the mathematicaltheory of numerical methods on hyperbolic manifolds can be found in [130, 18, 9].


Figure 4.36: Sod problem with two γ values in 2D, computed in Lagrange modewith GLACE (top) and with GLACE equipped with a free-Lagrange-Delaunayremeshing technique (bottom). Final time is T = 0.2, with 2 × 80 × 80 cells.

0

0.2

0.4

0.6

0.8

1

0 0.2 0.4 0.6 0.8 1

’mesh20.Sod’

0

0.2

0.4

0.6

0.8

1

0 0.2 0.4 0.6 0.8 1

’mesh20.Sod2’

Chapter 5

Starting from the mesh

Dans l’instant suivant, le paral lelepipede chan-gera a la fois de place et de forme, mais la massedm demeurera la meme.In the next instant, the cuboid changes both lo-cation and shape, but the mass dm remains thesame.

– Lagrange

(Memoire sur la theorie du mouvement des fluides, 1781)

In this chapter we take a completely different viewpoint on Lagrangian compress-ible fluid dynamics and Lagrangian numerical methods. In the previous chaptersthe starting point was the equations. The difficulty was firstly to establish somesort of equivalence between the Eulerian and Lagrangian equations and secondlyto discretize the mathematical structure of the Lagrangian equations in a way thatnumerically preserves the entropy inequalities. The emphasis was heavily on theentropy of the system.

We reconsider the seminal idea of Lagrange [123, 124], which turns out tobe surprisingly close to what is needed for practical computations. In our modernterms Lagrange’s idea consists of, in the first stage, deriving ordinary differentialequations associated with the displacement of a single element of fluid and, inthe second stage, considering the partial differential equations. Here we adopt thisapproach: we start from the ordinary differential equations for a single volumedescribed with nodes, faces, edges, etc., since this is what is needed for meshes inscientific computing. These ordinary differential equations are used to discretizethe mass, momentum and energy balances in a given volume together with thepreservation of an entropy inequality. Only afterwards will we explain how torecover a numerical scheme from these objects. One interesting aspect of the ap-proach is that it allows a simpleclassification of some modern Lagrangian methods.A general weak convergence result will justify the whole construction.




263

264 Chapter 5. Starting from the mesh

5.1 Axiomatization of mesh features

Since we wish to clarify some fundamental structures of moving meshes in scientificcomputing, we adopt a somewhat axiomatic description. A mesh consists of a setof points, also called vertices or nodes, in a given domain Ω ⊂ Rd where d = 1, 2, 3is the space dimension; our framework will also cover any dimension d > 3. It maybe convenient for pedagogical purposes to assume that Ω is a torus, i.e. withoutboundaries.

A node, also called a point or vertex, with index r ∈ N at time step n ∈ N

is denoted by xnr ∈ Ω. The collection of these points is (xn

1 , . . . , xnr , . . . ) ∈ (Rd)Nv ,

with Nv being the total number of vertices. For the sake of simplicity, we shall alsoassume that the points are differentiable functions with continuous derivatives andmostly forget about the time step index.

We writeur(t) = x′

r(t) ∈ C0(R

d)

, 1 ≤ r ≤ Nv .

The domain is covered by volumes Ωj ⊂ Ω for 1 ≤ j ≤ Nc, with Nc being thenumber of cells. The measure of a volume is Vj = |Ωj |, so that one can write

|Ω| =∑

j

Vj. (5.1)

This notation implicitly assumes that |Ω| < ∞, which is the case if Ω is a boundeddomain without boundaries, such as a torus in dimension d. In practical computa-tions, Ω is always bounded. The connection between points and volume is describedby a countable number of continuously differentiable functions ψj : (Rd)P → R+

such that Vj = ψj (x). A necessary assumption is that these volumes are measur-able, which gives meaning to Vj ≥ 0. For simplicity the notation ψj will not beused any more, and only Vj will be used. That is, we identify the volume Vj withthe function. We write

Vj = Vj(x1 , . . . , xr , . . . ).

Definition 5.1.1 (Corner vectors). Under the previous assumptions, the cornervectors are defined as

Cj r = ∇xrVj ∈ Rd ∀j, r. (5.2)

A basic property is of course that the time variation of a volume attached tomoving points is given by the chain rule:

d

dtVj =

∑

r

(Cj r, ur ) ∀j. (5.3)

Proposition 5.1.2. Assume that the volumes from a partition of Ω so that (5.1)holds. Assume that the total volume |Ω| is constant in time. Then

∑

r

Cj r = 0 ∀j. (5.4)

5.1. Axiomatization of mesh features 265

Proof. Indeed, one has

0 =d

dt|Ω| =

∑

j

∑

r

(Cj r, ur) =∑

r

⎛⎝∑

j

Cj r, ur

⎞⎠ .

Since the node velocities are independent vectors, one picks a particular index rand decides that us = 0 for s = r. This yields

⎛⎝∑

j

Cj r, ur

⎞⎠ = 0 ∀ur ∈ R

d,

which shows the claim.

5.1.1 Planar geometries

The following properties need additional assumptions on the mesh and correspondto planar geometry in dimension 1 ≤ d ≤ 3. The results in this section do not holdfor axisymmetric geometries which are examined later.

Proposition 5.1.3. Assume translation invariance of Vj . Then

∑

r

Cj r = 0 ∀j. (5.5)

Proof. Translation invariance yields ddt

Vj = 0 if Vj is moved by translation. Denoteby a ∈ Rd the vector of translation. Translation invariance corresponds also to

Vj(· + a) = Vj(·). So by the chain rule (5.3), 0 =∑

j(Cj r, a) =

(∑j

Cj r, a)

.

Since this holds for all a, it yields the claim.

The volumeadmits a scaling natural parameter due to the homogeneityof thespace. This homogeneity property can be written formally as Vj(λx) = λdVj(x)for λ > 0. It holds for generic meshes in dimensions d = 2 and d = 3.

Proposition 5.1.4. Assume the homogeneity property in dimension d. Then

1

d

∑

r

(Cj r, xr ) = Vj ∀j. (5.6)

Proof. This is the Euler relation for homogeneous functions.

It is possible to combine (5.5) and (5.6) to obtain

1

d

∑

r

(Cj r, xr − xref) = Vj

where xref is an arbitrary point. This shows that (5.6) is also translation invariant.


Let us now construct Cj r for meshes composed of simplexes. Consider firstthe one-dimensional situation where the mesh is defined by points on the line. . . , xr, xr+1 , . . . and the cell Ωj has volume

Vj = xj +1 − xj .

Differentiation yields∂Vj

∂xj +1= 1 and

∂Vj

∂xj= −1.

The points xr are often written with half-indices xr = xj + 12

for r = j, but wedo not use this notation for the sake of compatibility with the multidimensionalsituation. One obtains the correspondence

dimension d = 1:

⎧⎪⎨⎪⎩

Cj r = 1, r = j + 1,

Cj r = −1, r = j,

Cj r = 0, r = j + 1 and r = j.

So, in dimension d = 1, Cj r is just another notation for the exterior normal vectorof a segment.

Proposition 5.1.5. In dimension d = 2, one has

Cj r =1

2

(−yr−1 + yr+1

xr−1 − xr+1

)= lj rnj r (5.7)

where the nodal length lj r and nodal normal vector nj r are defined in (4.103).Moreover, Cj r = 0 if xr is not a vertex of the cel l.

Proof. Consider the decomposition in (5.1). By convention the vertices are listedcounterclockwise as xr−1 , xr , xr+1 , . . . with coordinates xr = (xr, yr). The quan-tity 1

2(xryr+1 − yrxr+1) is the oriented area of the triangle with vertices xr, xr+1

and 0 = (0, 0). The sum of these oriented areas is the total area, namely

Vj =∑

r

1

2(xryr+1 − yrxr+1) .

In fact, Vj = sj in the notation of the previous chapter. Partial differentiationgives

∂Vj

∂xr=

1

2(yr+1 − yr−1) and

∂Vj

∂yr=

1

2(−xr+1 + xr−1) .


Proposition 5.1.6. In dimension d = 3, each of the four corner vectors Cj r of atetrahedron is opposite to the normal vector of the opposite face, with norm equalto one third of the area of the opposite face.


jCell

xr

Cell

Cell

xr+1

xr−1

C⊥

jr

p

k

Cjr

Figure 5.1: A mesh in dimension d = 2. The vector C⊥j r which is orthogonal to

Cj r joins the midpoints of the two edges.

Proof. Consider a generic tetrahedron with vertices r = 1, 2, 3 and 4. One has

Vj =1

6

∣∣∣∣∣∣∣

x2 − x1 x3 − x1 x4 − x1

y2 − y1 y3 − y1 y4 − y1

z2 − z1 z3 − z1 z4 − z1

∣∣∣∣∣∣∣=

1

6

∣∣∣∣∣∣∣∣∣

x1 x2 x3 x4

y1 y2 y3 y4

z1 z2 z3 z4

1 1 1 1

∣∣∣∣∣∣∣∣∣.

Differentiating with respect to x1 yields

Cj 1 =1

6

⎛⎜⎝

∣∣∣∣∣∣∣

y2 y3 y4

z2 z3 z4

1 1 1

∣∣∣∣∣∣∣, −

∣∣∣∣∣∣∣

x2 x3 x4

z2 z3 z4

1 1 1

∣∣∣∣∣∣∣,

∣∣∣∣∣∣∣

x2 x3 x4

y2 y3 y4

1 1 1

∣∣∣∣∣∣∣

⎞⎟⎠

t

.

On the other hand, the exterior vector on the opposite face can be defined sd

S234 =1

2

⎛⎜⎝

x3 − x2

y3 − y2

z3 − z2

⎞⎟⎠ ∧

⎛⎜⎝

x4 − x2

y4 − y2

z4 − z2

⎞⎟⎠ .

By definition the norm of S234 is the area of the opposite face. Rearrangement ofthe formula for Cj 1 shows that

Cj 1 = − 1

3S234, (5.8)

which is the claim. The proof for the other vectors Cj r, r = 2, 3, 4, is identical upto a permutation of the indices.


5.1.2 The reference cell method

Let us come to a general definition of Vj , based on the reference element methodwhich is widely used in finite element methods [48]. One assumes a finite numbernc < ∞ of reference cells. In practice the reference cells belong to a finite catalogof basic cells, such as the unit hexahedron (0, 1)3 in three dimensions, a pyramid,a tetrahedron, etc. For example, the description of the cells needed for the tri-dimensional calculation reported in figure 5.8 requires two reference cells, whichare an hexahedron and a pyramid; in this case nc = 2. One general assumption isthat all reference cells are the convex hull of a finite number of vertices which aretheir corners. These corners are denoted by Xr with r = 1, . . . , nv(q), where nv(q)is the number of vertices of the reference cell for 1 ≤ q ≤ nc. To each referencecell we attach barycentric coordinates.

Definition 5.1.7 (Barycentric coordinates over the reference cell Ωq). For q =

1, . . . , nc, consider the reference cell Ωq. Barycentric coordinates λqr(X) : Ωq → R+

over the reference cell are piecewise differentiable functions such that ∀X ∈ Ωq,

nv(q)∑

r=1

λqr(X) = 1 and 0 ≤ λq

r(X) ≤ 1. (5.9)

The definition of barycentric functions for simplexes is known; see for ex-ample [48]. The point is that the definition of the barycentric functions may benon-unique if the reference cell is not a simplex but a union of simplexes: considerfor example the situation depicted in figure 5.2, where the pentagon is split into tri-angles in two different ways. On each of these triangles the barycentric functionsare easily defined as P 1 linear functions. Thus two different sets of barycentricfunctions are defined over the whole pentagon. This means that barycentric func-tions may not be intrinsic objects, even if the measure of the pentagon is uniquelydefined. The decomposition of the pentagon into simplexes generates barycentricfunctions which are piecewise affine.

Once a set of barycentric coordinates is chosen, it defines a natural mappingfrom a given reference cell Ωq (1 ≤ q ≤ nc) to the cell Ωj : we note the natural

mapping by ψqj : Ωq → Ωj ,

x = ψqj(X) ≡

nv(q)∑

r=1

λqr(X)xrj (r) ∈ Ωj , X ∈ Ωq. (5.10)

The notation is local, that is, one assumes that the Eulerian nodes xr are countedfrom 1 to nv(q). By construction,

xrq

j(r) = ψ

qj (Xr),

meaning that the nodes X1≤r≤nv(q) are mapped to the nodes(

xrq

j(r)

)1≤r≤nv(q)

via the mapping r → rqj(r). For simplicity one can always assume that the ordering


Figure 5.2: Non-uniqueness of the definition of the barycentric functions in dimen-sion d = 2. The pentagon is viewed as the union of 5 triangles on the left and asthe union of 3 triangles on the right. On each of the subtriangles the barycentricfunctions are the standard linear P 1 functions.

of nodes is such that rqj (r) = r. We additionally assume for convenience that the

mapping is non-singular:

det (∇Xx) > 0 ∀X ∈ Ωq. (5.11)

Remark 5.1.8. The property (5.11), needed for the correctness of the transfor-mation, is essential. It may be violated in Lagrangian computations with cellswhich are not simplexes. In this case the computation usually stops. A generalremedy is introduced in section 5.6 with a decomposition into simplexes for whicha dynamical procedure ensures that (5.11) is respected.

Definition 5.1.9. The reference element method yields the formula

Vj =

∫

Ωq

det (∇Xx) dX =

∫

Ωq

det

⎛⎝

nv(q)∑

r=1

xrj (q) ⊗ ∇λqr

⎞⎠ dX. (5.12)

This general formula defines explicitly the volume Vj as a function of thevertices x = (x1 , . . . ), so it can be used as a firm basis for the study of theproperties of the corner vectors Cj r. In what follows we use the notation

∑

s<t

=∑

s

∑

t>s

.

Proposition 5.1.10. In dimension d = 3, the reference cel l method yields the rep-resentation formula as a double sum

Cj r =∑

s<t

(xr

q

j (s) ∧ xrq

j (t)

∫

Ωq

det (∇λr, ∇λs, ∇λt) dX

). (5.13)


This is equivalent to

Cj r =∑

s<t

((xr

q

j(s) − xr

q

j(r)

)

∧(

xrq

j (t) − xrq

j (r)

) ∫

∂Ωq

det (nX, ∇λs, ∇λt)λr dσX

).

(5.14)

Proof. We omit the subscript j and superscript q for simplicity. We write thecomponents of a vertex as xr = (xr, yr, zr ). Omitting the indices j and q, theformula (5.12) can be rewritten as

V =

∫

Ωq

D dX with D = det

(∑

r

xr∇λr,∑

s

ys∇λs,∑

t

zt∇λt

),

where the matrix is made up of three column vectors. Differentiation of the deter-minant yields

Cr =∂

∂xrD =

∑

s

∑

t

yszt det (∇λr, ∇λs, ∇λt) ,

which can be rewritten as

Cr =∑

s<t

(yszt − zs yt) det (∇λr, ∇λs, ∇λt) .

The result is obtained after rearrangement, completing the proof.

Let us now consider a hexahedron with warped faces as depicted in figure 5.3.Such a situation is very often encountered in Lagrangian calculations. We decideto use the isoparametric representation of the hexahedron. This is based on theformula

x(α, β, γ) =

8∑

r=1

xr λr(α, β, γ)

where 0 ≤ α, β, γ ≤ 1. The functions λr are the eight barycentric functions of theunit reference cube: λ1 = (1 − α)(1 − β)(1 − γ), λ2 = α(1 − β)(1 − γ) and so on.

Proposition 5.1.11. The corner vectors Cj r of the isoparametric cel l are decom-posed as face integrals, and the value of the face integrals is equal to 1

12for positive

orientations r, s, t.

Remark 5.1.12. Let f be a face of the cell. This property can be recast as thedecomposition

Cj r =∑

f

Cfj r (5.15)


Figure 5.3: The faces of the hexahedron are warped. Such a cell is common inmultidimensional Lagrangian calculations with moving meshes.

where, for s < t,

Cfj r =

∑

xr ,xs,xt∈f

((xs − xr ) ∧ (xt − xr )

∫

∂Ωq

det(nX, ∇λs, ∇λt) λr dσX

).

(5.16)Similar decompositions are often obtained directly from the geometry, as in [34,89].

Proof. By symmetry it is sufficient to consider Cj 1 where X1 corresponds to α =β = γ = 0. The boundary of the unit cube is made up of six different faces denotedgenerically by f . Consider a face such that X1 ∈ f . Then λ1 vanishes identicallyonthe face. The integral on face f vanishes and so

∫f⊂∂Ωq

det(nX, ∇λs, ∇λt)λ1dX =

0. So it is sufficient to compute the integrals on the three faces which have thenode as a common vertex. Next, consider one of these remaining three faces, forexample the face delimited by the vertices with generic indices 1, 2, 3 and 4 withpositive orientation. The calculation of the integral can be split into three differentcases: r = 2 and s = 3; r = 2 and s = 4; r = 3 and s = 4. The outward normal isn = (0, 0, −1). The first case r = 2 and s = 3 corresponds to

A = −∫

0≤α,β≤1

λ1 (∂α λ2∂β λ3 − ∂β λ2∂αλ3)dα dβ.

Recall that λ1 = (1 − α)(1 − β), λ2 = α(1 − β), λ3 = αβ and λ4 = (1 − α)β . Basiccalculations yield A = 1

12 . All coefficients are computed by rotation of the indices.The proof is finished.

A formula which plays a fundamental role in the weak consistency analysisis the following.


Proposition 5.1.13. A consequence of the reference cel l formula is the identity

∑

r

Cj r ⊗ xr = Vj Id ∀j. (5.17)

Remark 5.1.14. One recovers the homogeneity property (5.6) by taking the traceof this identity.

Proof. Here linearity is equivalent to homogeneity of degree one in all directions.The list of vertices can be viewed as a rectangular matrix x ∈ RNv×d organizedin columns or in rows:

x = (x1 | · · · | xNv ) =

⎛⎜⎜⎝

a1

...

ad

⎞⎟⎟⎠ with ai ∈ RNv, 1 ≤ i ≤ d.

The vector ai is the collection of the ith components of all vertices. This repre-sentation can be immediately deduced from the reference cell formula (5.6).

• Due to the reference cell formula, the volume Vj is homogeneous of degree 1with respect to ai. So one has the Euler relation

Vj(x) = (∇aiVj(x), ai )RNv =

∑

r

Cj r|i xr|i

where |i indicates the ith component of a vector. This shows (5.6) for thediagonal part.

• Let us now prove the claim for the non-diagonal coefficients. Consider anotherset of vertices given by

yt =(

at2 at

2 at3 . . . at

d

),

that is, the first component of each vertex is replaced by the second compo-nent. This is the only difference between x and y. Of course Vj(y) = 0, sinceit is a determinant with equal rows. By linearity (also a consequence of theproperties of a determinant), ∇a1Vj(x) = ∇a1Vj(y). The homogeneity yields(∇a1Vj(y), a2)

RNv Vj(y) = 0. Therefore (∇aiVj(x), a2)

RNv = 0, which can be

expanded as ∑

r

Cj r|1 xr |2 = 0.

The proof is the same for all pairs of distinct indices.

The identity (5.17) is actually the generalization of a similar basic formula(5.18), which can be obtained as a consequence of the Stokes formula in dimensiond = 2. Let us consider the polygon depicted in figure 5.4. Denote the generic exte-


Ωj

xr+1/2

xr−1/2

xr−1

xr

xr+1

Figure 5.4: A generic convex polygon. In the local notation: corners are denotedby xr with integer r and midpoints of edges are denoted by xr+ 1

2.

rior normal vector on an edge by nr+ 12

=(

n1r+ 1

2

, n2r+ 1

2

)and the generic length of

the edge by lr+ 12. The midpoint of the edges is xr+ 1

2=

(x1

r+ 12

, x2r+ 1

2

)and the nodes

are xr for an integer value of r. One has the equality lr+ 12nr+ 1

2= 1

2 (xr+1 − xr)⊥

for all r. Let us write Cr+ 12

= lr+ 12nr+ 1

2. A consequence of Proposition 4.4.25 is

the following.

Proposition 5.1.15. The fol lowing identity holds for a polygon in dimension d = 2:

∑

r

Cr+ 12

⊗ xr+ 12

= VjI2. (5.18)

Proof. Indeed,

Vj =

∫

Ω

∂x1 x1dx1dx2 =

∫

∂Ω

n1x1dσ =∑

r

lr+ 12n1

r+ 12

x1r+ 1

2

.

Similarly one has that

Vj =

∫

Ω∂x2 x2dx1dx2 =

∫

∂Ωn2x2dσ =

∑

r

lr+ 12n2

r+ 12

x2r+ 1

2

,

0 =

∫

Ω

∂x1 x2dx1dx2 =

∫

∂Ω

n1x2dσ =∑

r

lr+ 12n1

r+ 12

x2r+ 1

2

and

0 =

∫

Ω∂x2x1dx1dx2 =

∫

∂Ωn2x1dσ =

∑

r

lr+ 12n2

r+ 12

x1r+ 1

2

.

The identity (5.18) is composed of these four relations.


Remark 5.1.16. For a polygon, it is even possible to derive (5.17) from (5.18).Indeed, by definition xr+ 1

2= 1

2(xr+1 + xr ), so it is possible to rearrange (5.18) to

obtain ∑

r

Cr ⊗ xr = Vj I2, Cr =1

2

(Cr+ 1

2+ Cr− 1

2

). (5.19)

Note, however, that (5.17) has been derived in a more general context.

Proposition 5.1.17. One has the inequality

‖Cj r‖ ≤ αq diam(Ωj)d−1, (5.20)

where the constant αq depends only on the reference cel l Ωq attached to cel l Ωj .

Proof. The proof in dimension d = 3 is a direct consequence of (5.14). The prooffor dimension d = 2 is left to the reader.

Proposition 5.1.18. Let s be the index of one vertex xs of the cel l Ωj. One has theinequality

Vj

diam(Ωj)‖v‖ ≤

∑

r=s

|(Cj r, v)| ∀v ∈ Rd . (5.21)

Proof. Rewrite (5.17) as∑

r Cj r ⊗ (xr − y) = Vj Id where y is an arbitrary point.Then Vjv =

∑r (Cj r, v) (xr − y) and Vj‖v‖ ≤ (

∑r |(Cj r, v)|) · maxj ‖xr − y‖.

Now take y = xs, so that ‖xr − xs‖ ≤ diam(Ωj ). The proof is finished.

5.1.3 Nodal control volumes

We introduce basic definitions and concepts about control volumes around nodes.These will be used mainly to formalize, in Proposition 5.3.1, an important positiv-ity property of some corner matrices. Since the identity (5.17) seems fundamental,we decide, somewhat arbitrarily, to adopt a similar abstract definition for thecontrol volumes.

Definition 5.1.19 (Abstract control volumes). Suppose that there exist points yj

attached to cells for all Ωj and Vr > 0 such that

∑

j

Cj r ⊗ (xr − yj ) = Vr Id ∀r. (5.22)

Then we say that there exists an abstract control volume at xr.

Remark 5.1.20. The node xr can be eliminated from this formula by using (5.4).Moreover, the definition of the points yj can be made more general by consideringsuch points being attached to cells and nodes. Denote these points by yj r. In thiscase it is sufficient for our purposes to establish a formula like

∑

j

Cj r ⊗ yj r = −Vr Id ∀r.


This will not be investigated further in the following.Note also that we do not require the control volumes to constitute a partition

of the domain.

This definition can be made more explicit for simple meshes, which is thecase that we examine now. Consider a mesh made up of simplexes, such as theone depicted in figure 5.5 in dimension d = 2.

yj+1 xr

yj

Figure 5.5: A mesh composed of triangles around node xr . The centers of mass ofthe triangles are denoted by . . . , yj −1, yj , yj +1 , . . . . The control volume is delim-ited by the dashed line.

Proposition 5.1.21. Consider a mesh made up of simplexes with positive volumeand denote by yj the center of mass of the simplex Ωj . Then the condition (5.22)is satisfied with Vr > 0.

Proof. This is trivial in dimension d = 1. Consider dimension d = 2. The principleof the construction depicted in figure 5.4 should be compared with the one infigure 5.4. The point zj which is the midpoint of the edge opposite to xr is suchthat

2

3(zj − xr ) = yj − xr

because yj is the center of mass. The exterior normal vector on the edge is parallelto Cj r. This can be shown using (5.7). More precisely, the exterior normal vectormultiplied by length of the edge is equal to −2Cj r. Then one can use the identity(5.18) for the polygon depicted in figure 5.5. So one has

∑

j

(−2Cj r) ⊗ (zj − xr ) =∑

j ∈C(r)

Vj ,

that is,∑

j

Cj r ⊗ (yj − xr ) = − 3

4

∑

j ∈C(r)

Vj .


Here C(r) stands for the cells in the neighborhood of r, i.e. j ∈ C(r) if and only ifCj r = 0. The extension to dimension d = 3 for a mesh made up of tetrahedrons isessentially the same, using formula (5.8) which shows that Cj r is parallel to theexterior normal vector on the opposite face.

5.1.4 Axisymmetric geometry

Axisymmetric geometry is a notorious source of many technical and fundamen-tal difficulties in Lagrangian computation, even for staggered discretizations [14,197, 198]. The goal here is to show how the geometry changes the definition andproperties of the corner vectors.

One can start in dimension d = 2 from the formula for the volume in theform

Vj =

∫

Ωq

det (∇Xx) XdXdY =

∫

Ωq

det

⎛⎝

nv(q)∑

r=1

∇Xλqr ⊗ xr

⎞⎠ Y dXdY, (5.23)

where X = (X, Y ), Ωq is the unique reference cell that corresponds to Ωj and theEuler-to-Lagrange correspondence is described as before by

x =

nv(q)∑

r=1

λqr(X)xr ∈ Ωj , X ∈ Ωq. (5.24)

The only difference from the planar case is the weight X in the measure XdXdY .This weight is the distance to the axis X = 0. The usual notation is R = X andZ = Y, which puts emphasis on the geometrical meaning. One also restricts thedomain to X > 0 for compatibility reasons. The volume 2πVj actually representsthe tri-dimensional measure of the annulus obtained by rotation in 3D of a 2Ddomain Ωj .

Since it is sufficient in practice to use polyhedrons, we consider below thesimpler situation where a polyhedron is decomposed into triangles.

Proposition 5.1.22. Consider a polyhedron. The axisymmetric corner vectors are

Caxij r =

1

6

((xr+1 + 2xr)yr+1 + (−xr+1 + xr−1)yr − (2xr + xr−1)yr−1

−x2r+1 − xr+1xr + xrxr−1 + x2

r−1

).

(5.25)They satisfy (5.3) and (5.4) but not (5.5), (5.6) and (5.17).

Proof. Denote the three vertices of a given triangle T by x1 =(x1

1, x21

), x2 =(

x12, x2

2

)and x3 =

(x1

3 , x23

)One has the formula

∫

T

x dx dy =1

3

(x1

1 + x12 + x1

3

)|T |. (5.26)


Consider Proposition 5.1.5. By summation over triangles of a polyhedron for which(5.26) holds, one has

Vj =∑

r

xr + xr+1

3× xryr+1 − yrxr+1

2.

Differentiation with respect to xr and yr yields (5.25).The properties (5.3) and (5.4) hold since they are satisfied independently of

the measure. Since Vj is homogeneous of degree 2 with respect to the first spacecoordinate xr, one has the Euler formula

∑

r

∂Vj

∂xrxr = 2Vj .

Since Vj is homogeneous of degree 1 with respect to the second space coordinateyr, one has ∑

r

∂Vj

∂yryr = Vj .

It is then easy to check that ∑

r

∂Vj

∂yrxr = 0.

The last term is

∑

r

∂Vj

∂xryr =

∑

r

(−xr+1 + xr)(−yr+1 + yr)yr .

Therefore the matrix∑

r Caxij r ⊗ xr is not diagonal, so (5.17) does not hold.

The new version of (5.6) is

∑

r

(Caxi

j r , xr

)= 3Vj ,

which is the consequence of the tri-dimensional nature of Vj .Finally, a direct computation yields

∑

r

Caxij r =

(sj

0

)(5.27)

where

sj =∑

r

xryr+1 − yrxr+1

2

is the area. The invariance of Vj with respect to translation along the y coordinateis evidenced by the null second component in the right-hand side of (5.27). Thiscompletes the proof.


5.2 Cell-centered Lagrangian schemes

Most of the newly developed cell-centered Lagrangian schemes can be thought ofas a unique family of methods based on preliminary determination of the cornervectors Cj r. To encompass an even larger class of methods, we will consider vectorsCj r that admit an additional decomposition.

Definition 5.2.1 (Decomposition of corner vectors). A decomposition of corner

vectors is any family of vectors(

Cfj r

)j,r,f

depending on an additional abstract

index f ≥ 1 such that ∑

f

Cfj r = Cj r ∀j, r. (5.28)

The letter “f” has been chosen because it is related to the faces in the caseof the EUCCLHYD scheme in dimension d = 3.

• The GLACE scheme corresponds formally to f = 1, which indicates that nodecomposition of the corner vectors is performed. This can be written in theabstract form

C1j r = Cj r,

where by convention Cfj r = 0 for f = 1.

• The EUCCLHYD scheme in dimension d = 2 involves only two corner vectorsin the decomposition, so non-zero vectors in the decomposition correspondto f = 1, 2. That is, an edge is identified with a face. Refer to figure 4.28 andnote that

C1j r = l+j rn+

j r and C2j r = l−j rn−

j r.

By construction the sum (5.28) is satisfied.

• An extension of EUCCLHYD to dimension d = 3 is described in [148]. It isrelated to the formula (5.13), which is valid in dimension d = 3. It correspondsto

Cfj r = s

fj rn

fj , f is the index of a true face.

See the decomposition (5.15)–(5.16).

• The compatibility relation (5.28) is satisfied by more methods than just theGLACE and EUCCLHYD schemes.

The difference between most of the methods presented below lies mainly inthe definition or choice of certain corner dissipation matrices which are constructedas functions of the Cf

j r. At the end of the analysis, such differences are reminiscentof different quadrature techniques used in finite element theory.

5.2. Cell-centered Lagrangian schemes 279

5.2.1 Construction of the scheme

The basic scheme presented below discretizes the system of compressible fluiddynamics in integral form:

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

d

dt

∫

V (t)

1 dV =

∫

V (t)

∇ · u dV (volume conservation),

d

dt

∫

V (t)ρdV = 0 (mass conservation),

d

dt

∫

V (t)

ρu dV +

∫

V (t)

∇p dV = 0 (momentum conservation),

d

dt

∫

V (t)ρedV +

∫

V (t)∇ · (pu)dV = 0 (total energy conservation),

(5.29)

where V (t) is a moving Lagrangian volume. Here Lagrangian refers to the massconservation equation. The physical variables are density ρ, velocity u, total energye and pressure p.

All the methods described below start from the first equation, which is dis-cretized as described earlier by

d

dtVj =

∑

r

(Cj r, ur) , (5.30)

where the vectors ur are the velocities of the vertices of the mesh. With thisabstract notation, the discretization of the momentum equation considered is

Mjd

dtuj = −

∑

r

∑

f

Cfj rp

fj r. (5.31)

The energy equation is

Mjd

dtej = −

∑

r

∑

f

(C

fj r, ur

)p

fj r. (5.32)

Proposition 5.2.2. Assume the formula (5.5) related to translation invariance. Thesemi-discrete scheme consisting of (5.30), (5.31) and (5.32), where the cornervectors satisfy the additivity relation (5.28), satisfies the local entropy identity

MjTjd

dtSj =

∑

r

∑

f

(C

fj r, ur − uj

)(pj − p

fj r

). (5.33)

Proof. Using the fundamental principle of thermodynamics and

d

dtVj = Mj

d

dtτj ,


one has

MjTjd

dtSj = Mj

(d

dtej −

(uj ,

d

dtuj

)+ pj

d

dtτj

)

= −∑

r

∑

f

(Cf

j r, ur

)pf

j r +

⎛⎝uj ,

∑

r

∑

f

Cfj rpf

j r

⎞⎠ + pj

∑

r

(Cj r, ur)

= −∑

r

∑

f

(C

fj r, ur

)p

fj r +

⎛⎝uj ,

∑

r

∑

f

Cfj rp

fj r

⎞⎠

+ pj

∑

r

(Cj r, ur) − pj

(∑

r

Cj r, uj

).

Using the additivity relation (5.28), we then get

Mj Tjd

dtSj = −

∑

r

∑

f

(Cf

j r, ur

)pf

j r +

⎛⎝uj ,

∑

r

∑

f

Cfj rpf

j r

⎞⎠

+ pj

∑

r

∑

f

(C

fj r, ur

)− pj

⎛⎝∑

r

∑

f

Cfj r, uj

⎞⎠ ,

from which the claim follows after factorization.

Non-negative entropy production is an attractive property for numerical sta-bility and correctness with respect to the underlying physics. A sufficient conditionis (

Cfj r, ur − uj

) (pj − p

fj r

)≥ 0 ∀j, r, e.

Many schemes in the literature rely on formula such as

pj − pfj r = αf

j r

(nf

j r, ur − uj

), αf

j r ≥ 0, (5.34)

where the normal vector nfj r is just a scaled version of C

fj r, that is,

nfj r =

Cfj r∣∣∣Cfj r

∣∣∣.

This definition makes sense only for∣∣∣Cf

j r

∣∣∣ > 0. To ensure the validity of (5.34) for

all f , it is sufficient to adopt the convention that

αfj r = 0 for C

fj r = 0.


The relation (5.34) is a scalar linear relation between pj − pfj r and ur − uj , and

can be seen a linear acoustic Riemann solver in the direction of the vector Cfj r.

A sufficient condition for the conservation of the momentum is∑

j

∑

f

Cfj rp

fj r = 0. (5.35)

Proposition 5.2.3. Consider a given vertex xr and assume the control volume iden-tity (5.22) as in Proposition 5.1.21. Then the vertex velocity ur and vertex-based

pressures pfj r (for al l r and al l f ) which solve (5.34) and (5.35) exist and are

unique. The velocity is the solution of the wel l-posed linear system

Arur = br (5.36)

where the matrix is symmetric and positive,

Ar =∑

j

∑

f

αfj r

∣∣∣Cfj r

∣∣∣nfj r ⊗ n

fj r = At

r > 0,

and the right-hand side is

br =∑

r

∑

f

Cfj r

(pj + αf

j r

(nf

j r, uj

)).

Moreover, the entropy is non-decreasing,

MjTjd

dtSj ≥ 0. (5.37)

Proof. In view of (5.34), the entropy inequality (5.37) holds provided one canprove that the linear system is well-posed.

Plug (5.34) into (5.35) to obtain

∑

j

∑

f

Cfj r

(pj + α

fj r

(n

fj r, uj − ur

))= 0

and then rearrange. This yields the linear system where the matrix Ar is symmetricand non-negative. It remains to show that Ar > 0 is positive to guarantee that ur

is well-defined. To this end it is sufficient to show that if Ar a = 0 then a = 0. ButAr v = 0 implies that (v, Ar v) = 0, so

αfj r|Cf

j r|(

nfj r, v

)2

= 0 ∀j, e.

Since αfj r > 0 by hypothesis, it means that |Cf

j r|(

nfj r, v

)= 0, that is,

(C

fj r, v

)=

0. Assume now that the directions

Cfj r

j,f

span Rd . In this case Ar is positive,

which proves the claim.


It remains to show that Spanj,f

(C

fj r

)= Rd. Consider the two situations

depicted in figure 5.6. For different reasons the corner vectors span R 2 in bothcases. Unfortunately the story is not finished, as one can see from figure 5.7 whereboth GLACE and EUCCLHYD become degenerate on this special degeneratepentagonal mesh.

cell j

Cjr

C2jr

cell j

C1jr

Figure 5.6: Cartesian mesh. On the left are the corner vectors for the GLACEscheme: the completion of the four corner vectors spans R 2 (two cells/vectors isthe minimum). On the right are the corner vectors of EUCCLHYD: the completionis achieved with just one cell.

Cjr

cell j

cell j ′

Cj ′r

C2j ′r

cell j

cell j ′

C1jrC2

jr

C1j ′r

Figure 5.7: Pathological mesh made up of two pentagons. One can observe the

degeneracy of Spanj f

(Cf

j r

)for both GLACE and EUCCLHYD.

It appears that the exceptional theoretical situation shown in figure 5.7 isruled out by our hypothesis based on Proposition 5.1.21, which is used to charac-terize non-degenerate situations. Indeed, the additivity identity (5.28) turns into(Cj r, v) = 0 for all j . In view of the control volume identity (5.22), one gets

Vrv =∑

j

(xr − yj) (Cj r, v) = 0 =⇒ v = 0


since Vr = 0. So the corner vectors span Rd and the proof is finished.

Proposition 5.2.4. The scheme (5.30)–(5.32) is formal ly conservative in total mo-mentum and total impulse and is local ly conservative in mass.

Proof. Since Mj = Vjτj is constant in cells by construction, the scheme is of courseconservative in local mass. The conservation of total momentum follows from

d

dt

∑

j

Mjuj = −∑

j

∑

r

∑

f

Cfj rpf

j r = −∑

r

⎛⎝∑

j

∑

f

Cfj rpf

j r

⎞⎠ = 0,

due to the relation (5.35) which is satisfied for every r (up to boundaries). Theconservation of total energy is very similar:

d

dt

∑

j

Mjej = −∑

j

∑

r

∑

f

(C

fj r, ur

)p

fj r = −

∑

r

⎛⎝∑

j

∑

f

Cfj rp

fj r, ur

⎞⎠ = 0,

also as a consequence of (5.35).

Remark 5.2.5. The proposition focuses on global conservation of momentum andenergy. A local conservation property (that is, a local finite volume formulation)can also be stated.

5.2.2 Time discretization and extensions

We first describe simple time discretization of the continuous-in-time generalscheme. Later we review some of the extensions for Lagrangian fluid dynamicsof the structure (5.30)–(5.32) published so far, which fall into two categories.

Explicit time discretization

Even though the notation here is restricted to the semi-discrete (continuous-in-time) scheme, the discretization in time is easy with an Euler procedure. Thestructure of the fully discrete explicit counterpart of (5.30)–(5.32) is a four-stepalgorithm.

First step: compute the nodal velocities ur with the linear vertex-based system(5.36) and update the nodal pressures p

fj r with (5.34).

Second step: update the cell-centered velocity and total energy by⎧⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎩

Mj

uk+1j − uk

j

∆t= −

∑

r

∑

f

Cfj rp

fj r,

Mj

ek+1j − ek

j

∆t= −

∑

r

∑

f

(C

fj r, uf

r

)p

fj r.

(5.38)


Third step: move the mesh by

xk+1r = xk

r + ∆tukr (5.39)

and update the volumes V k+1j .

Fourth step: update the density by

ρk+1j =

V kj ρk

j

V k+1j

. (5.40)

Control of the time step is ensured by a CFL condition. All cell-centered algorithmspublished so far respect this structure. Note that absolute preservation of themass is preferred. Second-order extension with a predictor–corrector technique isnatural.

A typical example computed with the GLACE scheme in dimension d = 3 isthe Sedov problem in figure 5.8. Another example is the solution of a Kidder prob-lem in dimension d = 3 shown in figure 5.9: Lagrangian methods are well adaptedto the calculation of such problems with free boundaries typically encountered innumerical ICF (inertial confinement fusion). The data can be found in [42]. Muchmore physically oriented test problems are listed in the bibliography at the end ofthe monograph.

Figure 5.8: Sedov problem in dimensiond = 3 computed with the explicit GLACEscheme. This calculation was performed at the Commissariat a l’Energie Atomique.


Figure 5.9: Kidder problem in dimensiond = 3 computed with the explicit GLACEscheme, for two different times. This calculation was performed at the Commis-sariat a l’Energie Atomique.

Implicit time discretization

An interesting problem is to develop an implicit version of this algorithm in orderto relax the CFL constraint. It is possible to propose such an implicit algorithmbased on the ideas of Coquel; see [44]. The starting point is a linearization ofthe discrete geometrical conservation law (5.30) with an isentropic assumptionddt

Vj ≈ − Mj

(ρjcj)2ddt

pj . One typically gets an implicit linear system where the cell-

centered quantities depend on the nodal quantities,

⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩

− Mj

(ρkj ck

j)2

pj − pkj

∆t=

∑

r

(Cj r, ur ) ,

Mj

uj − ukj

∆t= −

∑

r

∑

f

Cfj rp

fj r,

(5.41)

coupled with another implicit linear system where the nodal quantities depend onthe cell-centered quantities,

⎧⎪⎪⎨⎪⎪⎩

Arur =∑

r

∑

f

Cfj r

(pj + αf

j r

(nf

j r, uj

)),

pfj r = pj + α

fj r

(n

fj r, uj − ur

).

(5.42)


In certain cases, the global linear system can be proved to be well-posed. Itssolution is an implicit prediction of the quantities at the end of time step k + 1.One obtains the following algorithm.

First step: Compute the nodal quantities by solving the global linear system(5.41)–(5.42).

Next steps: Update the cell-centered quantities and geometry with (5.38)–(5.40).

The properties of this Coquel algorithm are an open problem in the context ofLagrangian discretization. It is possible in principle to rely on [44] to show someentropy inequalities; this is nevertheless an open problem.

Modification of the right-hand side

The first class of methods modifies the right-hand sides of the corner problem,which become

⎧⎪⎪⎨⎪⎪⎩

pfj r − p

fj r = α

fj r

(n

fj r, ur − u

fj r

), α

fj r ≥ 0,

∑

j

∑

f

Cfj rp

fj r = 0.

(5.43)

That is, the idea is to replace (pj , uj) in (5.34) with the pair(

pfj r, uf

j r

). The

solution is correctly defined. The first step is to solve

Ar ur = br, br =∑

r

∑

f

Cfj r

(p

fj r + α

fj r

(n

fj r, u

fj r

)).

By construction, the conservation properties of Proposition 5.2.4 are preserved,but, of course, the unconditional non-negative entropy production is a priori lost.This idea is pursued in particular in [135] for the development of a formally second-order extension.

Reinterpretation as a force

The idea is to make a connection with the modeling of the material strength,where p

fj rn

fj r is interpreted as a force (a force is a vector) which acts at the corner

xr or, more precisely, on the element Cfj r (a normal vector multiplied by an area

in dimension d = 2). With this in mind, the formulas (5.34)–(5.35) are replacedby ⎧

⎪⎨⎪⎩

αfj r (ur − uj) = pjnf

j − f fj r,∑

j

∑

r

f fj r = 0. (5.44)

5.3. Stability of the mesh for simplexes 287

Note that the solution of (5.44) is explicit:

ur =

∑j

∑f

(pjn

fj + α

fj ruj

)

∑j

∑f α

fj r

.

The scheme (5.30)–(5.32) is replaced by

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

d

dtVj =

∑

r

(Cj r, ur) ,

Mjd

dtuj = −

∑

r

∑

f

ffj r,

Mjd

dtej = −

∑

r

∑

f

(f fj r, ur

).

This idea is systematically pursued in [152] and related works. It is easy to checkthat Propositions 5.2.2 and 5.2.4 still hold.

5.3 Stability of the mesh for simplexes

We show that the dynamical system associated with the semi-discrete schemeon simplexes is unconditionally well-posed for all time. This means both thatthe physics is correct (for example, the density remains positive) and that themesh is correct, which is extremely important from the point of view of practicalLagrangian calculations.

The essence of this result was first proposed in [66] in dimension d = 1 andthen fully developed in dimension d = 2 in [72]. The following generalization coversall dimensions. Some assumptions are needed for the rigor of the reasoning.

• Consider a domain which is a torus T in dimension d. The torus can beidentified with the unit cube [0, 1]d ⊂ Rd with periodic boundary conditions.It also corresponds to a problem posed in the entire space, but with a pe-riodicity condition f(x + i) = f(x) for all x ∈ Rd and all i ∈ Zd , i.e. noboundaries.

• The mesh at time t = 0 is made up of simplexes with positive volume. Theglobal mesh covers the torus with the periodicity condition.

• Consider a perfect gas pressure law p = (γ − 1)ρε with γ > 1.

• The initial data is in bounded, (ρ0, u0, e0) ∈ L∞ (T )d+2 .

• The initial data is positive in the sense that ρ(x) ≥ ρ− > 0 and ε(x) ≥ ε− > 0for all x ∈ T .


• The initialization is cellwise for U = (ρ, u, e) ∈ Rd+2, that is,

Uj(0) =1

Vj

∫

Ωj

U0(x) dx ∀j.

• Use the generic Lagrangian scheme described in (5.30)–(5.32) and move themesh is with

x′r = ur ∀r, t ≥ 0.

The coefficients αfj r are computed with the acoustic prescription

αfj r = ρjcj ∀j, t ≥ 0,

with ρc =√

γpρ .

• The vectors Cfj r are C1 functions of the nodes (this is straightforward for

GLACE and EUCCLHYD in all dimensions).

Proposition 5.3.1 (Well-posedness of the Lagrangian schemes on simplexes). Underthe above assumptions, the solution of the Lagrangian scheme exists and is uniquefor al l times t ∈ [0, ∞).

Remark 5.3.2 (on the simplex structure). The fact that the mesh is made up ofsimplexes is used explicitly to invoke Propositions 5.1.21 and 5.2.3, which guaran-tee the invertibilityof the matrix Ar . Notice also that it guarantees the correctnessof the transformation given by (5.11). It is also sufficient to consider figure 5.7 tounderstand that such a hypothesis is mandatory for the correctness of the proof.In summary, the simplex structure of the mesh is a necessary and sufficient con-dition for the proof. Our interest in this hypothesis is reinforced by the fact thatpractitioners know very well that a non-simplex mesh can always blow up.

Remark 5.3.3. The proof consists on the one hand in a direct application ofthe Cauchy-Lipschitz theorem to the ordinary differential system defining thecontinuous-in-time scheme on a time interval t ∈ [0, T ), and on the other handin the use of the whole structure and the main properties we have emphasized,namely mass conservation, entropy condition, conservativity, and interpretationon a moving mesh, to prove that T = ∞.

Proof. Let us first show that the system admits a unique solution in the contextof the Cauchy-Lipschitz theorem on a maximal interval [0, T ).

The periodicity of both the initial data and the initial mesh allows us toconsider only Nv nodes and Nc cells. The scheme is equivalent to an initial value

5.3. Stability of the mesh for simplexes 289

problem with dNv + (d + 2)Nc unknowns. It can be written as

⎧⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎩

x′r(t) = ur(t), 1 ≤ r ≤ Nv ,

U′j(t) = − 1

Mj

⎛⎜⎜⎜⎝

∑r

∑f

(C

fj r, ur

)

−∑r

∑f

Cfj rp

fj r

−∑r

∑f

(C

fj r, ur

)p

fj r

⎞⎟⎟⎟⎠ , 1 ≤ j ≤ Nc.

(5.45)

To simplify the presentation we introduce the global unknown

z =(

(xr )1≤r≤Nv

, (Uj)1≤r≤Nc

)∈ RdNv+(2+d)Nc .

Define the phase space E as

E =

z ∈ RdNv+(2+d)Nc : τj > 0, εj > 0, Vj > 0 ∀j

where εj = ej − 12(u2

j + v2j) and Vj is the volume of the jth cell, the connectivity

of the mesh being fixed. The space E is open since it is the inverse image of the

open (R+∗)3Nc

under the smooth polynomial mapping

E −→ (R+∗)3Nc

z −→(

(τj)1≤r≤Nc

), (εj)1≤r≤Nc

), (Vj)1≤r≤Nc

),

With this notation, the system (5.45) may be rewritten as z′ = f(z).We claim that f is a C1 function of z in E. This is a consequence of the

fact that Cfj r is by assumption a C1 function of the vertices and that the fluxes

ur and pfj r are also C1 functions of z. The main points are that: the coefficients

of the matrices (Ar )1≤r≤Nn are smooth by assumption; and Ar > 0 is invertiblefor a mesh made up of simplexes with positive volume, thanks to Propositions5.1.21 and 5.2.3. By the Cauchy-Lipschitz theorem we deduce that there exists amaximal open interval [0, T ), 0 < T ≤ +∞, on which the initial value problemhas a unique solution t → z(t) ∈ C1(E).

Now we show that T = +∞ using a classical argument from the theory ofdynamical systems. A classical result in dynamical systems states that if T < ∞,then the solution z(t) escapes all compact sets in E. We will prove that this isnot the case here by means of a posteriori estimates for t < T based on globalconservation laws and use of the perfect gas pressure law.

Since the scheme is conservative one has that∑

j Mjτj(t) =∑

j Vj(t) = |T | ,which is the measure of the torus at initial time. This yields the upper bound

τj ≤ |T |Mj

. (5.46)


Define S− = minj ε0j τ 0

jγ−1

. The entropy criterion for a perfect gas law yields

S− ≤ εj(t)τγ−1j (t) =⇒

(Mj

|T |

)γ−1

S− ≤ εj(t). (5.47)

The conservation of total energy∑

j Mj

(εj(t) + 1

2|uj(t)|2

)= E, together with

the positivity of all internal energies (5.47), yields the bound

εj ≤ E

Mj≤ E

minj Mj. (5.48)

Therefore

0 <

(Mj

|T |

)γ−1

S− ≤ εj ≤ E

minj Mj,

which shows that the internal energy εj lives in a compact subinterval of (0, ∞).Another use of the entropy condition yields

0 <

(S− minj Mj

E

) 1γ−1

≤ τj(t). (5.49)

The two inequalities (5.46) and (5.49) imply that the specific volume τj lives in acompact subinterval of (0, ∞).

Finally, one easily shows that

0 < minj

(Mj)

(S− minj Mj

E

) 1γ−1

≤ Mj τj = Vj ≤ |Ω|,

which implies that the volume also lives in a compact subinterval of (0, ∞). Theseinequalities define a compact set in E from which the solution z(t) cannot escapefor t < T . Therefore T = ∞ and the proof is complete.

5.4 Weak consistency of the gradient and divergence

operators

To address the consistency or convergence of the method, we adopt the Lax strat-egy, which is to prove that if the numerical scheme converges to some limit, thenthe limit is a weak solution. It is more a proof of weak consistency. This strategywas introduced and used in Theorem 2.5.10 for scalar conservation laws.

In the Lagrangian context, this method can be split into two steps. The firststep is to show the weak consistency of the gradient (the term ∇p in the momentumequation) and the divergence operator (the term ∇· (pu) in the energy equation).This can be donewith a fixed mesh. It explains the fundamental role of the relation(5.17) in the convergence analysis. The second step, which takes place on movingmeshes, is discussed in the next section.

5.4. Weak consistency of the gradient and divergence operators 291

5.4.1 Additional inequalities

We need some basic continuity and stability estimates, which are consequences ofthe nodal solver definition. Recall that C(r) is the set of cells neighboring the vertexxr . The first inequality (5.50) shows that the nodal velocity ur is a mean value ofthe local velocity plus a contribution from the local difference of the pressures.

Proposition 5.4.1. Assume that 0 < α1 ≤ ρjcj ≤ α2 for al l j . Then there exists aconstant β1 > 0 such that the fol lowing bound holds for al l j ∈ C(r)

|(Cj r, ur − uj)| ≤ β1

(max

j ′∈C(r)‖uj ′ − uj‖ + max

j ′∈C(r)|pj ′ − pj |

)max

j ′∈C(r)‖Cj ′r‖ .

(5.50)

Proof. One has the identity

Aruj =∑

j ′

∑

r

ρj ′cj ′

⎛⎝uj ,

Cfj ′r∣∣∣Cfj ′r

∣∣∣

⎞⎠C

fj ′r.

One also has∑

j ′

∑f C

fj ′r pj = 0 since

∑j ′

∑f C

fj ′r =

∑j ′ Cj ′r = 0. So one can

rewrite the nodal solver (5.36) as Ary =∑

j ′

∑f C

fj ′ry

fj ′ where

y = ur − uj and yfj ′ = (pj ′ − pj) + ρj ′cj ′

⎛⎝uj ′ − uj ,

Cfj ′r∣∣∣Cfj ′r

∣∣∣

⎞⎠ .

Taking the scalar product with y, one gets (y, Ar y) =∑

j ′

∑f y

fj′

(C

fj ′r, y

). Un-

der the hypothesis, this yields the inequality

α1

∑

j ′

∑

f

(C

fj ′r, y

)2

∣∣∣Cfj ′r

∣∣∣≤∑

j ′

∑

f

∣∣∣yfj ′

∣∣∣∣∣∣(

Cfj ′r, y

)∣∣∣.

By the Cauchy-Schwarz inequality,

∑

j ′

∑

f

∣∣∣yfj ′

∣∣∣∣∣∣(

Cfj ′r, y

)∣∣∣ ≤

√√√√√√∑

j ′

∑

f

(C

fj ′r, y

)2

∣∣∣Cfj ′r

∣∣∣

√∑

j ′

∑

f

(yfj ′)2

∣∣∣Cfj ′r

∣∣∣.

Together with the previous inequality this gives

α1

√√√√√√∑

j ′

∑

f

(Cf

j ′r, y)2

∣∣∣Cfj ′r

∣∣∣≤√∑

j ′

∑

f

(yfj ′)2

∣∣∣Cfj ′r

∣∣∣.


Another use of the Cauchy-Schwarz inequality gives

∑

j ′

∑

f

∣∣∣(

Cfj ′r, y

)∣∣∣ ≤

√√√√√√∑

j ′

∑

f

(C

fj ′r , y

)2

∣∣∣Cfj ′r

∣∣∣

√∑

j ′

∑

f

∣∣∣Cfj ′r

∣∣∣.

Therefore one can write

∑

j ′

∑

f

∣∣∣(

Cfj ′r, y

)∣∣∣ ≤ 1

α1

√∑

j ′

∑

f

(yfj ′)2

∣∣∣Cfj ′r

∣∣∣√∑

j ′

∑

f

∣∣∣Cfj ′r

∣∣∣.

Since ∣∣∣yfj ′

∣∣∣ ≤ β0

(max

j ′∈V (r)‖uj ′ − uj‖ + max

j ′∈V (r)|pj ′ − pj |

)

for some auxiliary constant β0, the claim is proved after redefinition of the con-stant.

Another technical inequality is as follows. Note that it can be made sharperby keeping only a local difference of velocities.

Proposition 5.4.2. Under the assumptions of the previous proposition, there existsβ2 > 0 such that for al l j ∈ C(r),

∣∣∣Cfj rp

fj r

∣∣∣ ≤ β2

(max

j ′∈C(r)|uj ′| + max

j ′∈C(r)|pj ′|

)max

j ′∈C(r)

∣∣∣Cfj ′r

∣∣∣ . (5.51)

Proof. The definition of pfj r yields

Cfj rp

fj r = C

fj rpj + ρjcj

Cfj r

|Cfj r|

(ur − uj, C

fj r

).

Apply (5.50). The constant β2 is a function of the mesh. This completes theproof.

As a generalization of (5.20), the dimension of |Cfj r| is hd−1.

5.4.2 Gradient

Denote the indicator function of the cell Ωj by x → 1IΩj(x) and define thecell-centered gradient function x → B(x) almost everywhere by

B(x) =∑

j

⎡⎣∑

f

∑

r

Cfj rpf

j r

⎤⎦ 1IΩj(x)

Vj.


The cell-based pressure function is defined almost everywhere by

p(x) =∑

j

pj1IΩj(x).

A local neighborhood of cell j is denoted by N (j); this definition can be adapted ondemand. The important information is that all cells k ∈ N (j) are at a finite givenlogical distance of cell j . Some discrete bounds on the bounded variation (BV)of discrete functions are needed, as in (5.52). Let h be a characteristic length ofthe mesh. The BV assumptions below scale like hd−1 since they are homogeneouswith respect to the L1 norm of the gradients.

Theorem 5.4.3. Assume that the speed of sound is within the bounds 0 < α1 ≤ρjcj < α2 everywhere. Assume that the mesh is regular in the sense that α3hd ≤ Vj ,that diam(Ωj) ≤ α4h for al l cel ls and that a generalization of (5.20) holds in the

form∣∣∣Cf

j r

∣∣∣ ≤ α5h for al l j , r and f. Assume that (pj) and (uj) are bounded in

L∞ and are bounded in BV in the sense that

∑

j

∑

k∈N(j )

hd−1 |pj − pk| ≤ α6 and∑

j

∑

k∈N(j )

hd−1 |uj − uk| ≤ α7. (5.52)

The nodal pressure (pfj r) is computed with the nodal solver. Assume that al l con-

stants (α1, . . . , α6) are uniform with respect to h.Then B − ∇p = O(h) in the weak sense.

Remark 5.4.4. To be more precise, the final result is that for all ϕ ∈ C10(Ω)d there

exists a constant C(α1, . . . , α7, ϕ) such that

∣∣∣∣∫

(Bϕ+ p∇ · ϕ)

∣∣∣∣ ≤ C(α1, . . . , α7, ϕ)h.

Remark 5.4.5. A similar result is proved in [93] for a classical edge-based finitevolume scheme, with the restriction that the mesh needs to be composed of trian-gles in dimension d = 2. The proof below is more general because it is based uponthe fundamental identity (5.17).

Remark 5.4.6. The BV bounds (5.52) are similar in essence to (2.59) and (2.65);see also the space-time versions (2.60) and (2.66). The difference is of course thatthese bounds are a priori estimates for scalar conservation laws, but are onlyassumptions for multidimensional systems of conservation laws.

Proof. Let x → ϕ(x) be a smooth test function with compact support in thecomputational domain Ω. We need to prove that for all admissible test functionsϕ, one has ∫

Bϕdx +

∫p∇ϕdx = O(h).


By definition,

∫Bϕdx =

∑

j

⎛⎝∑

f

∑

r

Cfj rpf

j r

⎞⎠ 1

Vj

∫

Ωj

ϕdx.

Note that ∣∣∣∣∣1

Vj

∫

Ωj

ϕdx − ϕ(xj)

∣∣∣∣∣ ≤ Ch2

where xj is the center of mass of the cell and C is a constant that depends on α3

and α4. Since ϕ has compact support, the number of cells to take into account isbounded by

meas(supp(ϕ))

minj Vj≤ meas(supp(ϕ))

α3hd= O(h−d).

The constant depends on ϕ.The method consists in a series of discrete integration by parts and Taylor

expansions. We begin with the formula

∫Bϕdx =

∑

j

⎛⎝∑

f

∑

r

Cfj rp

fj r

⎞⎠ϕ(xj )

+∑

j

⎛⎝∑

f

∑

r

Cfj rp

fj r

⎞⎠

(1

Vj

∫

Ωj

ϕdx − ϕ(xj )

).

We want to show that the second contribution is O(h) under convenient hypothe-ses. Since one has the estimate (5.51) and

∣∣∣∣∣1

Vj

∫

Ωj

ϕdx − ϕ(xj)

∣∣∣∣∣ ≤ Ch2,

one can bound all terms in L∞ and multiply by the number of cells (in the supportof ϕ), which is bounded by O(h−d). One obtains

∫Bϕdx =

∑

j

⎛⎝∑

f

∑

r

Cfj rp

fj r

⎞⎠ϕ(xj ) + CO(hd−1)O(h2)O(h−d)

=∑

j

⎛⎝∑

f

∑

r

Cfj rp

fj r

⎞⎠ϕ(xj ) + O(h)

=∑

j

⎛⎝∑

f

∑

r

Cfj rp

fj r

⎞⎠ (ϕ(xj ) − ϕ(xr )) + O(h)


since ∑

j

∑

f

Cfj rpf

j r = 0.

One has the local Taylor expansion

ϕ(xr ) = ϕ(xj) + (∇ϕ(xj), xr − xj) + O(h2).

Using the same method as before to estimate the remainder, we get

∫Bϕdx =

∑

j

⎛⎝∑

f

∑

r

Cfj rp

fj r

⎞⎠ (∇ϕ(xj ), xj − xr ) + O(h)

=∑

j

⎛⎝∑

f

∑

r

Cfj rp

fj

⎞⎠ (∇ϕ(xj), xj − xr )

+∑

j

⎛⎝∑

f

∑

r

Cfj r

(p

fj r − pj

)⎞⎠ (∇ϕ(xj), xj − xr ) + O(h).

The first term can be recast as

∑

j

⎛⎝∑

f

∑

r

Cfj rpj

⎞⎠ (

∇ϕ(xj), xj − xr

)=∑

j

pj

∑

f

∑

r

Cfj r

(∇ϕ(xj ), xj − xr

)

= −∑

j

pj

⎛⎝∑

f

∑

r

Cfj r ⊗ xr

⎞⎠∇ϕ(xj).

Here one uses the fundamental relation (5.17), which shows that

∑

f

∑

r

Cfj r ⊗ xr =

∑

r

⎛⎝∑

f

Cfj r

⎞⎠ ⊗ xr =

∑

r

Cj r ⊗ xr = Vj Id .

So the first term is equal to

−∑

j

pj∇ϕ(xj)Vj = −∫

p∇ϕdx + O(h)

using our hypotheses and notation. The functionp is the constant-per-cell function,equal to pj in each cell. It remains to prove that

E =∑

j

(∑

r

Cj r (pj r − pj)

)(∇ϕ(xj), xj − xr ) = O(h). (5.53)


Denoting by V (ϕ) the (logical) support of ϕ, we observe that

∣∣∣∣∣∣

∑

j

(∑

r

Cj r (pj r − pj)

)(∇ϕ(xj), xj − xr )

∣∣∣∣∣∣≤ Ch

∑

j ∈supp(ϕ)

∑

r

|Cj r (pj r − pj)| .

From the first equation of the nodal solver, one gets the bound

|E| ≤ Ch‖ρc‖L∞

∑

j ∈V (ϕ)

∑

r

|(Cj r, ur − uj)| .

Using the estimates (5.50), the right-hand side is estimated in terms of the BVbounds, to which the hypothesis (5.52) applies, and in terms of the mesh, to which

the generalization∣∣∣Cf

j r

∣∣∣ ≤ α5h of (5.20) applies. One gets |E| ≤ Ch. The proof is

thus complete.

Remark 5.4.7. The cornerstone of the proof is the fundamental relation (5.17). Itsrole is evidenced by considering a simpler concept, which is mimetic consistency(see section 5.6.3 and take Qi

j r = 0 in (5.82)).

5.4.3 Divergence

A similar analysis holds for the divergence operator. Consider the cell-centereddivergence operator x → C(x) applied to the product of the pressure and thevelocity, which is defined almost everywhere by

C(x) =∑

j

⎡⎣∑

f

∑

r

(Cf

j r, ur

)pf

j r

⎤⎦ 1IΩj(x)

Vj.

Define the cell-based product of the pressure and the velocity almost everywhereby

pu =∑

j

pjuj1IΩj(x).

Theorem 5.4.8. Assume the hypotheses of Theorem 5.4.3. Then C−∇·(pu) = O(h)in the weak sense.

Proof. This is a generalization of the proof of Theorem 5.4.3. We only need tocheck that the node velocity ur is compatible with the different expansions andthe discrete integration by parts that we used in the proof of Theorem 5.4.3. A

5.5. Weak consistency of Lagrangian schemes 297

sketch of the proof is as follows:

∫Cϕ dx =

∑

j

⎛⎝∑

f

∑

r

(Cf

j r, ur

)pf

j r

⎞⎠ 1

Vj

∫

Ωj

ϕdx

=∑

j

⎛⎝∑

f

∑

r

(C

fj r, ur

)p

fj r

⎞⎠ ϕ(xj) + O(h)

=∑

j

⎛⎝∑

f

∑

r

(C

fj r, ur

)p

fj r

⎞⎠ (ϕ(xj ) − ϕ(xr )) + O(h)

=∑

j

⎛⎝∑

f

∑

r

(Cf

j r, ur

)pf

j r

⎞⎠ (∇ϕ(xj), xj − xr ) + O(h)

=∑

j

⎛⎝∑

f

∑

r

(C

fj r, uj

)pj

⎞⎠ (∇ϕ(xj ), xj − xr ) + O(h)

= −∑

j

(uj · ∇ϕ(xj) pjVj + O(h) = −∫

(pu, ∇ϕ)dx + O(h),

thanks to the fundamental relation (5.17).

5.5 Weak consistency of Lagrangian schemes

We show weak consistency of the Lagrangian schemes. This means that, underconvenient regularity and convergence assumptions in the spirit of the Lax theo-rem, the limit (if it exists) of a Lagrangian cell-centered scheme is a weak solutionof the Euler equations, that is,

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

∫

t∈R

∫

x∈Rd

(ρ∂tϕ + ρu · ∇ϕ)dt dx = 0 ∀ϕ ∈ D0,

∫

t∈R

∫

x∈Rd

(ρu · ∂tϕ + ρu ⊗ u + pI · ∇ϕ)dt dx = 0 ∀ϕ ∈ Dd0,

∫

t∈R

∫

x∈Rd

(ρe∂t ϕ + ρeu · ∇ϕ)dt dx = 0 ∀ϕ ∈ D0,

∫

t∈R

∫

x∈Rd

(ρS∂tϕ + ρSu · ∇ϕ)dt dx ≤ 0 ∀ϕ ∈ D+0 ,

(5.54)

where D0 is the set of smooth test functions with compact support, and D+0 ⊂ D0

is the set of smooth non-negative test functions with compact support.The use of the Eulerian formulation greatly simplifies the proof. It is perhaps

possible to obtain a similar result for the weak formulation of the Lagrangian


equations, i.e. (1.28), (1.29) and (1.32) or (1.36)–(1.38). But this would requireadditional assumptions to guarantee the convergence of the gradient of deforma-tion, something which is not needed with the Eulerian formulation (5.54).

5.5.1 Notation

The notation here is based on the reference cell method. It defines cell Ωj(t) byinterpolation:

x ∈ Ωj(t) ⇐⇒ x =

nv(q)∑

s=1

λqr(X) xrj(s) (t) ∈ Ωj, X ∈ Ωq. (5.55)

The vertices of Ωj are(xrj (s) (t)

)where 1 ≤ s ≤ nv(q) is the number of vertices

of the reference cell and of the current cell. The mapping rj takes care of thevertex numbering from the reference cell to the current cell. The collection ofinterpolated cells defines the interpolated mesh. One needs a stability hypothesisfor the interpolated mesh, which is expressed in the form of a stability condition

det

⎛⎝

nv(q)∑

s=1

xrj(s) (t) ⊗ ∇Xλqr(X)

⎞⎠ > 0 ∀X ∈ Ωq, ∀j, ∀t. (5.56)

As a consequence the Lagrange-Euler transformation X ↔ x is invertible locally inits cell. We also assume that the transformation is globally correct in the sense thatthe mesh is at every time well-defined. We define two different velocity functionsin Ωj(t). The first is defined by interpolation from the node velocities,

uj(t, x) =

nv(q)∑

s=1

λqr(X)urj (s) (t), X ∈ Ωq, (5.57)

and is a priori a polynomial (depending on the barycentric functions) in the cell.The second velocity function is uj(t) and is constant in the cell. One obtains thefollowing functions for t ≥ 0 and x ∈ T :

for x ∈ Ωj(t),

⎧⎪⎪⎪⎨⎪⎪⎪⎩

ρh(t, x) = ρj(t),

uh(t, x) = uj(t),

uh(t, x) = uj(t, x),

eh(t, x) = ej(t).

(5.58)

Note that ρh, uh and eh are constant in each cell and defined almost everywhere.On the other hand, the function uh is a priori a continuous function when thebarycentric functions can be defined continuously from any one cell to its neigh-bors.


5.5.2 The density equation

Most of the technicalities are contained in the analysis of the weak continuity ofthe density equation. This is moreover relevant for most Lagrangian schemes.


∂tρh + ∇ · (ρhuh) = A1

in the weak sense, where A1 is an interpolation error given by

A1 =∑

j

[ρj

(∇ · uj − 1

Vj(t)

∫

Ωj(t)

∇ · uj dx

)]1x∈Ωj(t). (5.59)

Proof. Denote by Mj = Vj(t)ρj (t) the constant mass of the cell. Consider a time-space point (t, x) such that x ∈ Ωj(t). Then

∂tρh = ∂tρj(t) = − Mj

V 2j (t)

V ′j (t) = − Mj

V 2j (t)

∫

Vj(t)

∇ · uj dx = − ρj

Vj

∫

Vj(t)

∇ · uj dx

due to the well-known formula that gives the variation of the volume as a functionof the divergence of the velocity. On the other hand one has

∇ · (ρhuh) = ρj∇ · uj,

which proves the claim for x ∈ Ωj(t). To finish the proof it is enough to showweak continuity of (ρj , ρhuh) at the boundaries of the cell Ωj(t), i.e. at all pointswhere the mesh has low regularity. Following the weak continuity criterion, adiscontinuous bounded vectorial function is in H(div) if the normal jump is zero.

We distinguish the exterior normals nx, as in figure 5.10. Consider the spatialboundary of Ωj(t), the lateral face in figure 5.10. The weak continuity conditionis (nx, ([ρh ], [ρhuh])) = 0. By construction the velocity uh is continuous on thelateral faces. The spatial weak continuity criterion reads

[ρh] (nx, (1, uh)) = 0.

The point on the boundary moves with velocity uh , which means that the vector(1, uh) is tangent to the lateral face. Therefore, by construction, (nx, (1, uh)) = 0.Thus the weak continuity condition is satisfied, which proves the claim.

Proposition 5.5.2. If the cel l is a line segment in dimension d = 1, a triangle indimension d = 2 or a tetrahedron in dimension d = 3, then the interpolation errorvanishes:

A1 ≡ 0.

Proof. In these cases the velocity is linear in each cell. Therefore the mean velocity

gradient is equal to the gradient, \∇ · uh = ∇ · uh, and so A1 = 0.


Ωj(t)

nt

nx

x

t

tk+1

tkΩj(tk)

Ωj(tk+1)

Figure 5.10: Displacement of the mesh within the time interval tk < t < tk+1 .

Proposition 5.5.3. Let x → ϕ be a smooth test function with compact support. Onehas the bound

∣∣∣∣∫

A1ϕdx

∣∣∣∣ ≤ C ‖ρ‖L∞

∥∥∥∇ · uh − \∇ · uh

∥∥∥L1

h,

where \∇ · uh is a constant-per-cel l function, equal in each cel l to the mean valueof ∇ · uh.

Proof. One has

∫A1ϕdx =

∫ ⎛⎝∑

j

[ρj

(∇ · uj − 1

Vj(t)

∫

Ωj(t)∇ · uj dx

)]1x

⎞⎠ ϕdx

=∑

j

∫

Ωj(t)

([ρj

(∇ · uj − 1

Vj(t)

∫

Ωj(t)∇ · uj dx

)])(ϕ − ϕj)dx

=∑

j

∫

Ωj(t)ρj

(∇ · uh − \∇ · uh

)(ϕ − ϕj)dx,

where ϕj =1

Vj(t)

∫Ωj(t)

ϕdx is the mean value in space of the test function. So

∣∣∣∣∫

A1ϕdx

∣∣∣∣ ≤ ‖ρh‖L∞

∥∥∥∇uh − ∇u

∥∥∥L1

‖ϕ − ϕ‖L∞

where ϕ is the local mean value of the test function. Since the size of the mesh isbounded by h, ‖ϕ − ϕ‖L∞ ≤ Ch. The density is bounded in L∞ by hypothesis.So the proof is complete.


It is enough to assume that the velocity field is BV as in (5.52) (which turns

into ‖∇ · uh‖L1 +∥∥∥\∇ · uh

∥∥∥L1

≤ C) to get that

A1 = O(h) in the weak sense.

In summary, the interpolation error A1 vanishes or is small under fairly generalassumptions.

5.5.3 The momentum equation


∂t (ρhuh) + ∇ · (ρhuh ⊗ uh) = B1 + B2 (5.60)

in the weak sense, where the interpolation error in space is

B1 =∑

j

[ρjuj

(∇ · uj − 1

Vj(t)

∫

Ωj(t)∇ · ujdx

)]1x∈Ωj(t)

and the approximation of −∇p is

B2 = −∑

j

⎡⎣∑

f

∑

r

Cfj rp

fj r

⎤⎦ 1IΩj

Vj.

Proof. The proof is similar to that of Proposition 5.5.1. In each cell one can write

∂t (ρjuj) + ∇ · (ρjuj ⊗ uj) = uj∂tρj + ρj∂tuj + ρjuj∇ · uj

= ρjuj

(∇ · uj − 1

Vj(t)

∫

Ωj(t)∇ · uj dx

)+

Mj

Vj(t)∂t uj

= ρjuj

(∇ · uj − 1

Vj(t)

∫

Ωj(t)∇ · uj dx

)− 1

Vj(t)

∑

f

∑

r

Cfj rp

fj r.

The first term corresponds to B1. The second corresponds to B2. As in the proofof Proposition 5.5.1, the weak continuity condition

(nx, ([ρhuh], [ρhuh ⊗ uh])) = (nx, (1, uh)) [ρhuh] = 0 × [ρhuh] = 0

holds on the faces of the moving cells.

Under fairly general assumptions, B1 tends to zero as the size of the mesh htends to zero. If the cell is a line segment, triangle or tetrahedron, then B1 ≡ 0.

The term B2 is just the negative of the integral in time of the gradientapproximation B defined in Theorem 5.4.3. So under reasonable assumptions suchas those discussed earlier, one has B2 → −∇p in the weak sense.


5.5.4 The energy equation

Generalizing without difficulties the techniques used in Propositions 5.5.1 and5.5.4, one gets the following result.


∂t (ρheh) + ∇ · (ρhuheh) = C1 + C2

in the weak sense, where the interpolation error in space is

C1 =∑

j

[ρjej

(∇ · uj − 1

Vj(t)

∫

Ωj(t)∇ · ujdx

)]1IΩj

and the discrete approximation of −∇ · (pu) is

C2 = −∑

j

⎡⎣∑

f

∑

r

(C

fj r, ur

)p

fj r

⎤⎦ 1IΩj

Vj.

The term C1 is small in the weak sense and vanishes on simplexes. The termC2 converges weakly to −∇(pu) under appropriate hypotheses such as those inTheorem 5.4.8.

5.5.5 The entropy inequality

The analysis of the weak entropy inequality presents no additional difficultiesprovided the scheme is entropy increasing, which is the case for the Lagrangianschemes described above, as shown by (5.37).

5.6 Stabilization with subzonal entropies

The stabilization of meshes which move dynamically under a strong coupling withthe physics poses immense difficulties. In the case of Lagrangian systems, this canbe traced back to the lack of strong hyperbolicity, which indicates that the entropydoes not provide convenientcontrol of the mesh. However, analyses of meshes madeup of simplexes show that, in this case, the entropy inequality is able to controlthe mesh. The method presented below tries to exploit this property by using sub-cell decompositions. Sub-cell decompositions are considered in the seminal workof Burton, Caramana and Shashkov [36], where subzonal pressures are describedin the context of staggered discretizations. However, the analysis of the subzonalstructure presented in this section is different since it uses what we call a subzonalentropy. It will become obvious that the basic ingredient is close to what is calleda mixing entropy for perfect gas equations of state in the thermodynamical theoryof mixing [17, 33].

5.6. Stabilization with subzonal entropies 303

To be more specific, consider the situation where a pathological quadrangleis split into four triangles. For example, the quadrangle Q in figure 5.11 is splitinto four triangles: Q = T1 ∪ T2 ∪ T3 ∪ T4. It is clear that the global pathological

AB

C

D

0

T1

T4T2

T3

Figure 5.11: Sub-cell decomposition of a quadrangle into four triangles.

evolution of the cell as depicted in figure 5.12 will be detected by triangle T4 sinceits volume fraction goes to zero.

B

C

A

D0

Figure 5.12: Pathological evolution of the quadrangular cell in figure 5.11. Thevolume fraction of sub-cell T4 goes to zero. The mesh is about to tangle.

It appears that a connection between subzonal entropies and the entropyof mixing discussed for the multiphase model (see section 3.3.4) is possible. Letus decide arbitrarily that the total mass in the quadrangle of figure 5.11 is splitinto four equal parts so that the mass of each triangle is mi = 1

4M. Specific

internal energy is an intensive variable, so εni = εn . The entropy in Ti is Si =

Cv log(

εiρ1−γi

). The density in triangle Ti is

ρi =mi

|Ti|=

1

4

M

|Ti|=

1

4

ρ

f i,

where f i =|Ti|V

is the volume fraction. So the entropy in Ti can be expressed as


Si = Cv log(ερ1−γ) + (γ − 1) Cv log(4f i). The total entropy in the global cell isassumed to be the average sum of the partial entropies, that is,

S =S1 + S2 + S3 + S4

4= S + (γ − 1)Cv

∑i log(4f i)

4. (5.61)

From this very basic example, one can see that sub-cell modeling is equivalent toreplacing the entropy S in the cell by a new entropy

S = S + ϕ(f), f = (f1, . . . , f4). (5.62)

Analyzing equation (5.61), we find that the entropy tends to −∞ when f i

tends to zero. This is convenient for gaining some geometrical control of the cell,since one volume fraction tending to zero is the signature of a nearly pathologi-cal mesh. From now on we will consider subzonal entropies independently of theentropy law S, which may be arbitrary.

Definition 5.6.1. A function of the volume fractions (f1 , f2, . . . ) → ϕ(f1 , f2, . . . ) ∈R is a subzonal entropy if it satisfies the following two conditions:

(i) It is a concave smooth function.

(ii) Whenever the volume fraction f i is such that f i → 0+, one has ϕ(f) → −∞.

In order to stabilize a given Lagrangian computation, the general principle isto introduce a subzonal entropy into the numerical method, which can be eitherstaggered [34] or cell-centered as in our case [42]. The next proposition explainsa fundamental advantage of subzonal entropies. Consider a Lagrangian schemewhich has the property that the entropy is non-decreasing, d

dtSj ≥ 0.

Proposition 5.6.2. Assume that we are able to modify this scheme in order tointroduce a subzonal entropy in such a way that

d

dtSj ≥ 0. (5.63)

Then the mesh cannot become pathological.

Proof. The proof is by contradiction. Starting from a valid mesh (i.e. a mesh forwhich all sub-triangles have positive measure) and a very small ∆t, a pathologicalmesh is such that one volume fraction tends to zero, vanishes and then becomesnegative. In this case ϕn+1 ≈ −∞, which contradicts the assumption (5.63). Thisends the proof.

The natural question that arises now is how to incorporate this new fictitiousentropy into a scheme. We present the ideas using the notation of the GLACEscheme, which is simpler. The immediate generalization to EUCCLHYD is left tothe reader. A preliminary task is to establish the properties of the volume fractionsduring the Lagrangian displacement of a cell.


5.6.1 Lagrangian properties of volume fractions

We adopt here the point of view developed in [73], which is based on the generaliza-tion of the corner vectors to volume fractions. We begin by considering figure 5.13,where a quadrangle is decomposed into four internal triangles, but the method ismore general.

x4

T1

T4

T2

x2

y

T3

x1

x3

Figure 5.13: Internal structure of quadrangle decomposed into four triangles.

The intuitive definition of the volume fractions is

f ij =

V ij

Vj

where Vj is the volume of the cell and V ij is a sub-volume, meaning it is the volume

of one of the sub-triangles. The first goal is to determine the generalization of thecorner vectors for the volume fraction of sub-volume number i in cell number j .

Definition 5.6.3. With a convenient rescaling, the gradient of f ij with respect to

xr isQi

j r = Vj∇xrf ij . (5.64)

Remark 5.6.4. By definition, the dimension of Qij r is the same as the dimension

of Cj r.

The chain rule yields

Vjd

dtf i

j =∑

r

(Qij r, ur). (5.65)

To be more explicit, let us consider again the example in figure 5.13 of a quadrangledecomposed into four triangles. We decide arbitrarily that the interior point y inthe current cell is simply the average of the corners of the cell:

y =1

4

4∑

r=1

xr. (5.66)


Using a local numbering of the vertices, the triangles are T1 = (y, x1 , x2), T2 =(y, x2 , x3), T3 = (y, x3 , x4) and T4 = (y, x4 , x1). The sub-volume of triangle Ti isV i = |Ti|. Using Definition 5.6.3 and the expression for the volume fraction, onecan expand the derivative of the product to get

Qij r = −

V ij

VjCj r + ∇xrV

ij = −f i

jCj r + ∇xrVi

j . (5.67)

Let us denote the differentiation with a δ (the letter d is already the space dimen-sion) and consider the first volume fraction, with i = 1. Then

δV 1 = δ|T1| = (∇y|T1|, δy) + (∇x1 |T1|, δx1) + (∇x2|T1|, δx2) .

The three vectors ∇y|T1|, ∇x1 |T1| and ∇x2 |T1| are actually the corner vectorsbut for the sub-mesh. One gets formulas [42] where each vector is parallel to thenormal of the opposite edge of the sub-triangle, that is,

∇x1 |T1| =1

2(y − x2)⊥, ∇x2 |T1| =

1

2(x1 − y)⊥ and ∇y|T1| =

1

2(x2 − x1)⊥ ,

and similarly for the three other triangles. Since δy = 14

(δx1 + δx2 + δx3 + δx4) ,we have

δ|T1| =

((∇x1 |T1| +

1

4∇y|T1|

), δx1

)+

((∇x2|T1| +

1

4∇y|T1|

), δx2

)

+1

4(∇y|T1 |, δx3) +

1

4(∇y|T1|, δx4) .

(5.68)

Plugging this into (5.67) we find that

Qi=1r=1 = −f1C1 + ∇x1 |T1| +

1

4∇y|T1|,

Qi=1r=2 = −f1C2 + ∇x2 |T1| +

1

4∇y|T1|,

Qi=1r=3 = −f1C3 +

1

4∇y|T1 |

and

Qi=1r=4 = −f1C4 +

1

4∇y|T1|.

All other vectors are obtained by permutation of the indices.

Remark 5.6.5. In 1D, and also for triangles in 2D and tetrahedrons in 3D, the Qij r

are zero. Indeed, in these cases, f ij = 1

d+1is a constant and V i

j =Vj

d+1. Then by

using equation (5.67) we immediately obtain

Qij r = − 1

d+ 1Cj r +

∇xrVj

d + 1= 0,

with d ≥ 1 being the dimension of the space.


Some properties of the Qij r are given in the next proposition. We will show

that the formula (5.71) has important consequences for the consistency analysisof the stabilized algorithm.

Proposition 5.6.6. Invariance under translation implies that

∑

r

Qij r = 0 ∀j, i. (5.69)

Another property is ∑

i

Qij r = 0 ∀j, r. (5.70)

Moreover, ∑

r

Qij r ⊗ xr = 0 ∀j, i. (5.71)

Proof. The verifications are as follows.

• If the velocity of all vertices is the same, i.e. ur = u, then the volume frac-tion is unchanged by the motion. Therefore 0 =

∑r(Qi

j r, u) for all u. Thisproves (5.69).

• The formula (5.70) comes from

∑

i

Qij r = Vj

∑

i

∇xrf ij = Vj∇xr

(∑

i

f ij

)= 0,

since∑

i f ij = 1 wherever the vertices are.

• The next formula needs more work and its proof is based on the example offigures 5.11–5.13. One has

∑

r

Qij r ⊗ xr =

∑

r

(−f i

jCj r + ∇xrVi

j

)⊗ xr

= −f ij

∑

r

Cj r ⊗ xr +∑

r

∇xrVi

j ⊗ xr.

One also has that∑

r Cj r ⊗ xr = VjI. So

∑

r

Qij r ⊗ xr = −f i

jVjI +∑

r

∇xrV ij ⊗ xr .

To study the remaining term, we consider triangle T1, as depicted in fig-ure 5.13. In this case the vectors ∇xrV

ij are given by the chain rule formula


(5.68); therefore

∑

r

∇xrVi

j ⊗ δxr

=

(∇x1|T1 | +

1

4∇y|T1 |

)⊗ δx1 +

(∇x2|T1| +

1

4∇y|T1|

)⊗ δx2

+1

4∇y|T1| ⊗ δx3 +

1

4∇y|T1| ⊗ δx4

= ∇x1|T1| ⊗ δx1 + ∇x2|T1| ⊗ δx2 + ∇y|T1| ⊗ δ

(x1 + x2 + x3 + x4

4

)

= ∇x1|T1| ⊗ δx1 + ∇x2|T1| ⊗ δx2 + ∇y|T1| ⊗ δy.

(5.72)

Thus we have recast the equation in terms of the vertices of the triangu-lar sub-cell T1 . Since by construction the partial derivatives of |T1 | satisfya relation similar to (5.17) but with Vj replaced by |T1|, this means that∑

r ∇xrVi

j ⊗ xr = |T1|I2 = V ij I2. Therefore

∑

r

Qij r ⊗ xr = −f i

jVjI2 + V ij I2 = 0,

which is (5.71).

5.6.2 Building a scheme with subzonal entropies

The Lagrangian properties of volume fractions are used below to develop thealgebra associated with the new entropy S defined in (5.62). The second principleof thermodynamics for S reads

TdS = TdS + Tdϕ = TdS +∑

i

(T

∂ϕ

∂f i

)df i.

The discrete version of this principle is, for cell j ,

MjTjd

dtSj = Mj Tj

d

dtSj +

∑

i

(ρjTj

∂ϕ

∂f ij

)Vj

d

dtf i

j.

Definition 5.6.7. The stabilizing terms are defined as functions of

qij = ρjTj

∂ϕ

∂f ij

. (5.73)

Remark 5.6.8. Since ϕ has the dimension of an entropy, the coefficient qij has the

dimension of a pressure. This term can be considered as a new type of pseudo-viscosity; see (4.134).


The new terms are introduced into the extended scheme⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

Mjτ ′j(t) =

∑

r

(Cj r, ur ) ,

Vjd

dtf i

j(t) =∑

r

(Qij r, ur) ∀i,

Mju′j(t) = −

∑

r

Cj rpj r −∑

i

∑

r

Qij rq

ij r,

Mje′j (t) = −

∑

r

(Cj r, ur )pj r −∑

i

∑

r

(Qi

j r, ur

)qi

j r.

(5.74)

The second equation is the continuity equation for the volume fractions f ij, so it

is similar in essence to the first equation, which is the usual continuity equation.The corresponding corner vectors Qi

j r are defined in (5.64).

Proposition 5.6.9. The entropy production of (5.74) is

MjTjd

dtS j =

∑

r

(pj − pj r) (ur − uj , Cj r) +∑

i

∑

r

(qij − qi

j r)(Qij r, (ur − uj)).

(5.75)

Proof. One has

[MjTjd

dtSj = Mj e′

j − (uj , Mju′j(t)) + pjMjτ ′

j(t) +∑

i

qijVj

d

dtf i

j

= pj

∑

r

(Cj r, ur) +∑

i

qij

∑

r

(Qij r, ur)

+

(uj ,

∑

r

Cj rpj r +∑

i

∑

r

Qij rq

ij r

)

−∑

r

(Cj r, ur) pj r −∑

i

∑

r

(Qi

j r, ur

)qi

j r.

One can subtract∑

r

(Cj r, uj)pj +∑

i

∑

r

(Qi

j r, uj

)qi

j = 0.

One gets, after rearrangement,

MjTjd

dtSj =

∑

r

|Cj r| (pj − pj r) (ur − uj , nj r)

+∑

r

∑

i

∣∣Qij r

∣∣ (qij − qi

j r

)(ur − uj , mj r) ,

(5.76)

where mij r =

Qij r∣∣Qij r

∣∣ and nj r =Cj r

|Cj r|. The proof is finished.


So, to get the non-negativity of the entropy production, it is natural to pos-tulate a new proportional relation between the new terms arising in the products.We assume that

pj r − pj + ρjcj (ur − uj , nj r) = 0,

qij r − qi

j + ρjaij (ur − uj , mj r) = 0 ∀i,

(5.77)

where aij ≥ 0 is to be defined (this term has the dimension of a velocity, like cj).

Proposition 5.6.10. The semi-discrete scheme (5.74) with the nodal solver (5.77)is compatible with the second principle of thermodynamics, that is, d

dtS j ≥ 0.

Proof. The identity (5.76) can be rewritten as

MjTjd

dtSj = ρjcj

∑

r

|Cj r| (ur − uj, nj r)2

+ ρj

∑

i

aij

∑

r

∣∣Qij r

∣∣ (ur − uj, nj r)2 ≥ 0.

(5.78)


The right-hand side of the momentum equation in (5.74) may be interpretedas the sum of certain forces which are defined at the vertices. The total momentumconservation requires also that the total sum of the forces equal zero. This is thecase if the sum of forces vanishes at all vertices independently, that is,

∑

j

Cj rpj r +∑

i

∑

j

Qij rq

ij r = 0. (5.79)

The linear system formed by (5.77) and (5.79) is obviously non-singular. Indeed,its solution is obtained by elimination of the pj r and the qi

j r in (5.79). One getsformally the same linear system

Arur = br,

but the matrix is now

Ar =∑

j

ρjcjCj r ⊗ Cj r

|Cj r|+∑

i

∑

j

ρjaij

Qij r ⊗ Qi

j r∣∣Qij r

∣∣ , (5.80)

and the right-hand side is

br =∑

j

(Cj rpj + ρjcj

Cj r ⊗ Cj r

|Cj r| uj

)+∑

i

∑

j

(Qi

j rqij + ρjai

j

Qij r ⊗ Qi

j r∣∣Qij r

∣∣ uj

).

(5.81)Once the velocity is computed, it is immediate to calculate pj r and qi

j r with (5.77).By Remark 5.6.5, the stabilization has no effect in 1D or on simplex meshes.


Proposition 5.6.11. By construction, the GLACE scheme with subzonal entropy islocal ly conservative in total mass, total momentum and total energy.

Proof. By construction the scheme is conservative for the mass of each cell, so thetotal mass is also preserved. Concerning the total momentum,

∑

j

Mjd

dtuj = −

∑

j

∑

r

Cj rpj r −∑

j

∑

i

∑

r

Qij rq

ij r

= −∑

r

⎛⎝∑

j

Cj rpj r +∑

j

∑

i

Qij rq

ij r

⎞⎠ = 0,

due to the nodal equation (5.79). Therefore the total momentum is indeed pre-served. Similar algebra shows that

∑

j

Mjd

dtej = −

∑

r

⎛⎝⎛⎝∑

j

Cj rpj r +∑

j

∑

i

Qij rq

ij r

⎞⎠ , ur

⎞⎠ = 0.

Therefore the total energy is preserved.

5.6.3 Consistency of subzonal entropies

It appears that subzonal entropies do not change the consistency of the scheme:this is a fundamental property. It can be explained within a simplified mimeticframework [177], or more rigorously in the weak sense.

Mimetic consistency

The idea is the following. One has to assume that the fluxes are linear quantitieswith respect to the space variable. Then one computes the local discrete gradientor the local discrete divergence. If the result is exact, we will say that the schemeis consistent in the mimetic sense. To be more specific we propose to use thefollowing definition, where the focus is on the gradient operator.

Assume that pj r is linear with respect to xr , that is, pj r = b+(c, xr ). We alsoassume that all the qi

j r are linear with respect to xr , that is, qij r = bi

j + (cij , xr). If

1

Vj

(∑

r

Cj rpj r +∑

i

∑

r

Qij rqi

j r

)= c, (5.82)

we will say that the gradient operator is consistent in the mimetic sense. Thisproperty is also referred to as linearity preserving.

Concerning the divergence operator, a similar property can be checked asa consequence of the fact that the divergence of a vector field is equal to thetrace of its gradient. So consistency in the mimetic sense of the gradient impliesconsistency in the mimetic sense of the divergence.


Proposition 5.6.12 (Mimetic consistency of subzonal entropies). The gradient op-erator is consistent in the mimetic sense.

Proof. One has

∑

r

Cj rpj r =∑

r

Cj rb +∑

r

Cj r(c, xr ) =

(∑

r

Cj r

)b +

(∑

r

Cj r ⊗ xr

)c.

One obtains 1Vj

∑r

Cj rpj r = 0 + 1Vj

VjIdc = c. On the other hand, for all i,

∑

r

Qij rq

ij r =

∑

r

Qij rb

ij +

∑

r

Qij r(c

ij , xr ) =

(∑

r

Qij r

)bi

j +

(∑

r

Qij r ⊗ xr

)ci

j .

Using (5.69)–(5.71) one obtains directly that∑

r Qij rqi

j r = 0. Therefore the gra-dient operator is consistent in the mimetic sense.

Weak consistency

The analysis of the weak consistency of the gradient and divergence operator iseasily extended. Apart from technical details, the fundamental point is the use ofthe consistency relation

∑r Qi

j r ⊗ xr = 0 for the gradient of the volume fractions.In some sense the weak consistency analysis is dual to the mimetic consistencyanalysis.

Consider

B(x) =∑

j

[∑

i

∑

r

Qij rq

qj r

]1IΩj(x)

Vj

and

C(x) =∑

j

[∑

i

∑

r

(Qi

j r, ur

)qi

j r

]1IΩj(x)

Vj.

Under the assumptions of Theorem 5.4.3, B + B − ∇p and C + C − ∇ · (pu) tendto zero in the weak sense.

The proof relies on (5.79), which is used to rearrange the weak formulation,and (5.69)–(5.71) to annihilate the contributions from B and C, so that the finalresult is obtained by using the methods of Theorems 5.4.3 and 5.4.8.

5.6.4 Numerical illustration

This numerical example is dedicated to the well-known Sod problem. It involvesa Riemann shock tube with a mild discontinuity. The solution consists of an ex-pending rarefaction at the left, a contact discontinuity, and a shock moving to theright. We use a computational domain of size 1 × 0.1 in the x and y directions.The contact discontinuity is initially at x = 0.5. The domain is filled with an ideal


gas with γ = 1.4. In the left part of the domain, p = 1 and ρ = 1 initially, while inthe right part p = 10−1 and ρ = 0.125. We perform a convergence study on thiscase. For our coarsest simulation, we use Nx = 20 and Ny = 6 cells respectively inthe x and y directions. We refine the mesh by a factor of 2 in each direction untila resolution of Nx = 640 and Ny = 96 is attained. The L1 error norm e is simplycomputed by

e =

∑j |ρj − ρanalytic(xj)| Vj∑

j Mj, (5.83)

where ρanalytic(xj ) accounts for the analytical value of the density at the center ofthe cell j . Since ρ is discontinuous at the contact discontinuity and at the shock,this formula is only order-1 accurate, but this is sufficient because we expect aconvergence rate lower than 1.

Figure 5.14: Mesh structures of the Sod shock tube problem with an initial per-turbation of the mesh. In the top panel the mesh tangles without stabilization.On the bottom, with stabilization, the mesh remains valid.

We first run the simulation on a classical perfectly regular planar mesh, i.e.with 0% perturbation. Nevertheless we use a subzonal entropy. Since the mesh iskept regular and planar during the computation, f i

j = 14

for all cells and at eachtime step. The effect of stabilization is a very slight increase in the magnitudeof Ar (5.80) and br (5.81). We observe that the results are as good as withoutstabilization. At least the numerical results converge to the correct solution ash → 0+ whatever μ is, as shown in the last row of table 5.1. Now we impose aperturbation on the initial mesh, which involves randomly moving the initiallyregularly spaced positions of the nodes. The maximal random displacement ischosen to be 30% and 50% of the regular spacing of the nodes, giving an almostsingular initial mesh for the latter case. Several values of the coefficient μ from0 (original GLACE scheme) to 1 (strong stabilization) are examined. Results aregiven in tables 5.1 to 5.3. Failed simulations correspond to tangled meshes.

General comments about these tables are the following: all calculations usingthe stabilization succeed to give correct results, while the original scheme doesnot succeed in simulating the problem for perturbed fine meshes; the order ofconvergence is between 0.6 to 0.82, whatever the level of stabilization, which is


Figure 5.15: Zoom on mesh structures of the Sod shock tube problem with aperturbed initial mesh. On the top the mesh tangles without stabilization. On thebottom with stabilization, the mesh remains valid.

μ 20 40 80 160 320 640 Order

1 1.37e–1 8.52e–2 5.61e–2 3.63e–2 2.29e–2 1.41e–2 0.66

10−2 4.94e–2 3.78e–2 2.34e–2 1.43e–2 8.63e–3 5.07e–3 0.66

10−3 4.72e–2 3.95e–2 2.647e–2 1.54e–2 9.29e–3 5.48e–3 0.62

10−5 5.03e–2 4.26e–2 3.023e–2 1.82e–2 1.09e–2 6.31e–3 0.6

0 5.52e–2 4.32e–2 3.09e–2 1.864e–2 1.11e–2 6.61e–3 0.61

Table 5.1: Relative L1 -error and convergence order for a regular mesh.

μ 20 40 80 160 320 640 Order

1 1.36e–1 8.28e–2 5.49e–2 3.52e–2 2.22e–2 1.36e–2 0.66

10−2 4.95e–2 3.61e–2 2.14e–2 1.29e–2 7.60e–3 4.50e–3 0.66

10−3 4.79e–2 3.79e–2 2.22e–2 1.31e–2 7.63e–3 4.51e–3 0.68

10−5 5.13e–2 4.30e–2 2.57e–2 1.41e–2 7.94e–3 4.51e–3 0.7

0 5.41e–2 4.36e–2 2.615e–2 1.89e–2 failed failed

Table 5.2: Relative L1-error for a random displacement of 30%.

comparable to the order of convergence of the non-stabilized scheme (0.61); errors

5.7. Constraints and quadratic formulation of fluxes 315

μ 20 40 80 160 320 640 Order

1 1.33e–1 7.96e–2 5.33e–2 3.34e–2 2.12e–2 1.30e–2 0.66

10−2 4.98e–2 3.36e–2 1.95e–2 1.15e–2 6.73e–3 3.98e–3 0.73

10−3 4.84e–2 3.34e–2 1.96e–2 1.11e–2 6.35e–3 3.7e–3 0.74

10−5 5.33e–2 3.82e–2 3.14e–2 1.03e–2 5.66e–3 3.81e–3 0.82

0 5.63e–2 3.99e–2 2.319e–2 3.163 failed failed

Table 5.3: Relative L1-error for a random displacement of 50%.

are significantly higher for μ = 1, but for lower values of μ errors are of the sameorder of magnitude as for the non-stabilized scheme.

5.7 Constraints and quadratic formulation of fluxes

It has recently been remarked [49, 50] that the corner linear systems can be re-cast as the minimization of particular quadratic functionals. The design of fluxesthrough convenient optimization-based formulations is addressed in the more gen-eral work [22]. In our context, this idea is very useful for the introduction ofconstraints (typically encountered in contact problems or for the treatment of in-ternal constraints in a Lagrangian code) into the definition of the fluxes and hasnice mathematical properties. For simplicity the idea is presented for (5.36), butit can be generalized easily.

5.7.1 Quadratic functionals

The initial observation is that the solution ur of (5.36) also realizes the minimumof the problem

ur = arg minv∈Rd

Jr(v), Jr(v) =1

2(v, Ar v) − (br, v) ,

since Ar = Atr > 0 is a symmetric positive matrix. Indeed,

∇vJr(v) = Ar v − br.

Since Jr is a quadratic form, the minimum of Jr is reached at the Lagrange-Eulercondition ∇vJr(ur) = 0.

It is appealing to have a more global definition. Set

u = (ur)r ∈(R

d)Nv


where Nv is the total number of vertices in the domain of calculation, and definethe global quadratic form

J(u) =∑

r

Jr(ur).

Let us consider a set of constraints

K ⊂(R

d)Nv

and the problem

u = arg minv∈K

J(v). (5.84)

This means that we aim to construct u which respects a certain number of con-straints that can be modeled with K . Examples for contact problems will be con-sidered below.

A general question is to determine sets K such that the resulting scheme(5.30)–(5.32) with the flux defined by (5.84) is endowed with good properties. Alist of such properties is given below. However, some examples will show that (ii),(iii) and (iv) are not always true.

(i) K is closed and 0 ∈ K .

(ii) K is convex.

(iii) K is preserved by translations. Write wa = (a, a, . . . , a); then v + wa ∈ K

for all v ∈ K and wa .

(iv) K is a cone. That is, λv ∈ K for all v ∈ K and all λ ≥ 0.

Assuming (i), there exists a minimizer umin ∈ K such that

J(umin ) ≤ J(v) ∀v ∈ K .

Assuming that all corner matrices Ar are positive, the functional J is strictlyconvex. Then property (ii) yields the uniqueness of the minimizer, so that one canwrite

umin = arg minv∈K

J(v). (5.85)

The following propositions characterize the formal conservation properties ofthe resulting scheme in terms of properties of the set K .

Proposition 5.7.1. If K has the property (iii), the minimization of the objectivefunction J within K formal ly preserves the total momentum.

Remark 5.7.2. In contrast to the previous situation, the conservation of totalmomentum is a global property in this case. It can be made local only if theconstraints are local.


Proof. The proof will use the GLACE notation for the sake of simplicity. On theone hand, the total momentum variation over a time step is

∑

j

Mjd

dtuj(t) = −

∑

r

⎛⎝∑

j

Cj rpj r

⎞⎠ . (5.86)

On the other hand, the minimum of the objective function J satisfies the Eulerinequality [1]

(∇J(umin), v − umin) ≥ 0 ∀v ∈ K . (5.87)

Take v = umin + wa; then

(∇J(umin), wa ) ≥ 0 ∀wa ∈ Rd.

Since this is true for both wa and w−a = −wa, one gets the annihilation of thegradient in the direction wa :

(∇J(umin), wa ) = 0 ∀wa ∈ Rd. (5.88)

The crux is the expression of the derivative of J with respect to ur:

∇urJ(u) = Arur − br

=∑

j

ρjcjCj r(Cj r, ur)

|Cj r|−∑

j

Cj r pj −∑

j

ρjcjCj r(Cj r, uj)

|Cj r|

=∑

j

Cj r

[−pj + ρjcj

(ur − uj ,

Cj r

|Cj r|

)]

= −∑

j

Cj rpj r.

Therefore (∇J(u), v) =∑

r

(−∑

jCj rpj r, vr

)for all v = (vr)r. The combination

of (5.86) and (5.88) yields⎛⎝∑

j

Mjd

dtuj(t), a

⎞⎠ = −

⎛⎝∑

r

⎛⎝∑

j

Cj rpj r

⎞⎠ , a

⎞⎠ = 0 ∀a ∈ R

d .

That is,∑j

Mjddt

uj(t) = 0, which ends the proof.

Proposition 5.7.3. If K is a cone, then the total energy is formal ly preserved.

Proof. Total energy variation over a time step can be recast as

∑

j

Mjd

dtej (t) = −

∑

r

⎛⎝∑

j

Cj rpj r, ur

⎞⎠ = (∇J(umin), umin) . (5.89)


The minimum is characterized by the Euler inequality

(∇J(umin), v − umin) ≥ 0 ∀v ∈ K .

Since K is a cone, take v = 0 × umin = 0 and v = 2umin. So (∇J(umin), umin) = 0,which ends the proof.

Remark 5.7.4 (Stability if the cone property does not hold). Examples show thatthe cone property is not always satisfied. In this case it is possible to rely on asimpler property, which is that O ∈ K. One can show nevertheless that

(∇J(umin), umin) ≤ 0,

which means that the total energy is non-increasing. The loss of total energy canbe extremely small as illustrated in figure 5.19.

5.7.2 Application to contact problems

y

xf (x, y) = 0

Figure 5.16: Schematics of the fluid-wall impact.

Consider the situation depicted in figure 5.16, where the equation of the wallsurface has the general form f(x) = 0 (with x = (x, y)), so that the fluid movesin the subset f(x) ≤ 0. For simplicity of presentation, the fluid domain in thesquare is meshed with triangular cells. The number of nodes will be denoted byNv . Again, we define the global velocity vector u = (u1 , u2, . . . ), which gathersthe velocity vectors of all nodes in the mesh. At the initial time, the fluid movesalong the x-direction with a constant velocity equal to 1. Its density is 1, and the


ambient pressure is 0. Since we work with a stiffened gas pressure law, the pressurein the matter is in equilibrium with the ambient pressure.

Some important differences exist between the continuouscase described in theabove theory and the discrete-in-time situation needed for implementation. Thereare essentially two possible methods. The first involves computing a time of impactof the fluid on the wall for all nodes, and decomposing the time step predicted bythe CFL condition into smaller time steps to match the time of impact. As thismethod may cost a lot in multidimensional situations, we will not investigate itfurther. The second method, described below, consists of incorporating the contactinto the definition of a convenient constraint set K n which varies from one timestep to the next, meaning that K n = K n+1 is possible.

The wall generates a constraint on the positions of nodes, which must remainin the subset Ω defined by

Ω = x ∈ R2 : f(x) ≤ 0. (5.90)

Obviously, the constraint for a specific node will be active only if the node impactsthe wall.

Plane wall

Consider a plane wall orthogonal to the x-direction. That is, f(x) = x so that thewall is the line x = 0. This leads in this particular case to the following discreteformulation of the constraint:

xn+1r ≤ 0 (1 ≤ r ≤ Nv),

⇐⇒ xnr + ∆t un

r ≤ 0 (1 ≤ r ≤ Nv),

⇐⇒ unr ≤ −xn

r

∆t(1 ≤ r ≤ Nv).

The set of admissible velocities K n is

K n =

u ∈ R

2Nv : ur ≤ −xnr

∆t, 1 ≤ r ≤ Nv

. (5.91)

This can also be written as Fr(ur) ≤ 0 with

Fr(ur ) = ur +xn

r

∆t, 1 ≤ r ≤ Nv . (5.92)

In this case, K n is closed and convex. But when the fluid is attached to the wall(xN = 0), property (iii) is lost since translations inducing a motion in the positivex direction are not allowed. The x-component of the momentum, equation (5.86),is therefore not preserved. In contrast, translations inducing a motion in the y-component are admissible, so the y-component of the momentum is preserved.Finally, we can show that the norm of the momentum decreases as long as impactoccurs.


Figure 5.17: Fluid impacting on a plane wall. Results (a), (b) and (c) correspondto three different time of computation: t = 0, t = 0.59 and t = 0.67.

As long as the fluid is attached to the wall, K n is a cone, and the totalenergy is preserved. Over the time step of impact, the 2D plane case can beenseen as P independent 1D impact problems, where P is the number of nodesimpacting the wall. As a consequence, the total energy will decrease as soon asa node impacts the wall. In the plane case, the P nodes impact the wall at thesame time, so that the total energy will decrease just over the time step of impact.Even if this is satisfactory in view of stability issues, it does not hold for the exactsolution. However, it can be seen in figure 5.19 that the loss of total energy isextremely small. It has been checked with numerical experiments that total energyloss converges to zero as the space length h tends to zero: this can be explainedby the fact that the single time step in which the loss of energy occurs tends tozero itself due to CFL constraints. Results of the computations are presented infigure 5.17 at three different times. As expected, once the impact has occurred,the x-component ur of the velocity vector of each node impacting the wall mustcancel out, while the y-component vr remains non-zero. The fluid then slides onthe wall.

Convex obstacle

We consider here a convex wall whose surface equation is given by the relation

f(x) ≡ y2 − x = 0, x = (x, y). (5.93)

The obstacle is convex for the domain f(x) ≥ 0. The form of the set K n is

K n =

u ∈ R2N : Fr(ur) ≥ 0, 1 ≤ r ≤ N

(5.94)


with

Fr(ur) = (ynr + ∆tvr)

2 − xnr + ∆tur ∀r. (5.95)

In contrast to the planar case, nodes impact the wall at different times,inducing for each impact a decrease in total energy. Between two consecutiveimpacts, the total energy is not preserved since we cannot find in K n any elementof the form (1 + μ) Umin with μ > 0; recalling the proof of Proposition 5.7.3, theconsequence is that total energy decreases between two consecutive impacts, asshown in figure 5.19. The results of the computation are shown in figure 5.18.

Figure 5.18: Fluid impacting on a convex obstacle boundary of the domain x+y2 ≥0. Results correspond to three different times of computation: (a) t = 0, (b)t = 0.59 and (c) t = 0.67.

Concave obstacle

Let us now consider a concave domain whose boundary is given by the relation

f(x, y) = x + y2 ≤ 0. (5.96)

The set K n is

K n =

u ∈ R2N : Fr(ur) ≤ 0 ∀r ∈ [1 : N ]

(5.97)

where

Fr(ur ) = xnr + ∆tur + (yn

r + ∆tvr)2

, 1 ≤ r ≤ N. (5.98)

In this case the domain is closed and non-empty, but it is not convex and thepossible non-uniqueness of the minimum of J is related to the non-uniqueness ofthe intersection point of the line issuing from xn

r , as illustrated in figure 5.21.


Figure 5.19: Numerical loss of total energy for 2D impact of a fluid on a wall.

Figure 5.20: Fluid impacting on a concave obstacle which is the boundary ofy2 − x ≥ 0. Results correspond to three different times of computation: (a) t = 0,(b) t = 0.59 and (c) t = 0.67.

The behavior of the fluid is nevertheless as expected, which can be seen infigure 5.20. While impacting the wall, the fluid slides over it. In this test case, theradius of curvature is sufficiently small to ensure that the minimum found in theminimization procedure is the one that corresponds to correct physical behaviorof the fluid. However, if the radius of curvature is much smaller, the minimummay be the wrong one and could lead a priori to the crash of the computation.We illustrate this situation in figure 5.21.

Consider a single point impacting a concave wall with a steep radius of cur-vature. The constraint expresses that the point must not enter the wall. In this


B

uA

Figure 5.21: Non-uniqueness of the minimum of J .

case, the position of the point X may be found within the set (−∞, A]∪[B, +∞),giving two admissible values for the velocity. The uniqueness of the minimum isthen not guaranteed. By reasonable control of the time step one is able to capturethe correct solution.

5.7.3 Non-conformal meshes, hanging nodesand internal constraints

In Lagrangian calculations an exceptional point is usually understood as a pointwhere the mesh is non-conformal. This point is different from the other points be-cause it is attached to its neighbors, i.e. it is written as a function of the two pointsat each end of the segment on which it is located. This is equivalent to the notionof non-conformal meshes, but with the additional difficulty that the exceptionalityor non-conformity must be preserved dynamically during mesh displacement.

Following the approach adopted in [51], we first show that it is sufficient toredefine the corner vectors to deal with such configurations. In a second stage weshow that a hanging node is an example of an internal constraint of the mesh forwhich the quadratic reformulation has a nice structure and can be used to obtainthe same solution.

Linear representation of the exceptional points

Consider the example depicted in figure 5.22, where the point M (resp. N) is an

exceptional point. One can use the formula M =1

2H +

1

2I (resp. N =

1

2G +

1

2H)

to define it.A more general axiomatization of what an exceptional point is as follows:

the point M is said to be an exceptional point if there is a smooth functionφ : Rd × Rd → Rd such that M = φ(H, I) and for which translation invarianceholds, i.e.

φ (x + a, y + a) = φ(x, y) + a ∀a ∈ Rd . (5.99)


H N

j1j2

j3j4

j5j6

A B C

I

DJEKF

GM

Figure 5.22: Non-conformal mesh with two exceptional points N and M.

A natural consistency requirement is

φ(x, x) = x for all x ∈ Rd . (5.100)

A point which is not exceptional is said to be free.Let us define the two constant matrices P1 = ∇xφ(0, 0) ∈ Rd×d and P2 =

∇yφ(0, 0) ∈ Rd×d. Taking the derivative of

φ (a, a) = φ(0, 0) + a

with respect to a, one gets

P1 + P2 = Id ∈ Rd×d. (5.101)

Proposition 5.7.5. All solutions of (5.99) can be expressed as

φ(x, y) = P1x + P2y + ψ(x − y), (5.102)

with ψ(0) = 0 and ∇ψ(0) = 0.

Remark 5.7.6. As a consequence of ψ(0) = 0 and ∇ψ(0) = 0, one has thatψ(x − y) = O(|x − y|2). Therefore, for nearby points, this term is a second-ordercorrection to the principal linear contribution P1x + P2y.

Proof. We define φ(x, y) = φ(x, y) − P1x − P2y. By construction, one has

φ(x + a, y + a) = φ(x, y) ∀a ∈ R .


Then, taking a = −y, we obtain φ(x, y) = φ(x − y, 0). We can define ψ(z) =

φ(z, 0), so (5.102) holds.Taking x = y in (5.102), one obtains φ(x, x) = P1x + P2x + ψ(0), that is, using(5.100) and (5.101) we have

ψ(0) = φ(x, x) − (P1 + P2)x = x − x = 0.

Taking y = 0 in (5.102), we get ψ(x) = φ(x, 0) − P1x. Thus, the gradient at theorigin is ∇ψ(0) = ∇xφ(0, 0) − P1 = 0.

We disregard for simplicity the nonlinear second-order correction ψ.

Definition 5.7.7 (Linear representation of exceptional points). A linear represen-tation formula for exceptional points is

φ(x, y) = P1x + P2y, (5.103)

with constraint (5.101).

Direct definition of the corner vectors

Let us consider the example depicted in figure 5.22. The volume Vj3 of the cellindexed by j3, for instance, is computed using points I, D, J and M, but M =φ(H, I) is an exceptional point. Therefore Vj3 is actually computed as a functionof the free points I, D, J and H, i.e.

Vj3 = V (I, D, J, φ(H, I)). (5.104)

We can directly extend the definition of corner vectors to such non-conformalmeshes. Some notation for the general case is as follows. We consider that the setof all points 1 ≤ r ≤ Nv is decomposed into two subsets: 1 ≤ r ≤ Nfree meansthat xr is a free point, so without constraint; and Nfree + 1 ≤ r ≤ Nv includes allother points, which are exceptional.

Definition 5.7.8 (Corner vectors with respect to free points). The corner vectorsare calculated by partial differentiation of the volume, but with respect to freepoints only, that is,

Cj r = ∇xrVj , 1 ≤ r ≤ Nfree . (5.105)

Remark 5.7.9. A striking feature of the definition is that it allows long-distanceinteraction through the mesh. For the example in figure 5.22, the corner vectorCj3 ,B is non-zero.

The chain rule provides an easy way to compute Cj r in terms of the partialderivatives with respect to all points. We illustrate this principle on the exam-ple depicted in figure 5.22. Let us write the generic corner vectors obtained bydifferentiation with respect to any type of point as

Dj r = ∇xrVj , xr is free or exceptional.


Therefore

dVj3

dt= (Dj3,I , uI) + (Dj3,D , uD) + (Dj3,J, uJ) + (Dj3,M , uM) . (5.106)

Point M is exceptional, i.e. M = P1H+P2I, which turns into uM = P1uH +P2uI.Using the definition of function φ, we obtain

dVj3

dt=(Dj3,I + P t

2Dj3,M , uI

)+ (Dj3,D , uD) + (Dj3,J, uJ) +

(P t

1Cj3,M , uH

).

(5.107)By comparison, we obtain the expressions

Cj3,I = Dj3 ,I + P t2Dj3,M , Cj3 ,D = Dj3,D ,

Cj3,J = Dj3 ,J, Cj3 ,H = P t1Dj3,M .

(5.108)

All other known situations can be treated with the same approach.

Proposition 5.7.10. One has the identities

Nfree∑

r=1

Cj r = 0 for 1 ≤ j ≤ Nc,Nc∑

j =1

Cj r = 0 for 1 ≤ r ≤ Nfree . (5.109)

Proof. This is a consequence of (5.4) and (5.5).

Expression as internal constraints

Another possibility is to think of the existence of hanging nodes as correspondingto internal constraints. In figure 5.22 one has that

M = P1H + P2I,

where one can take P1 = P2 = 12

for simplicity. Since the matrices are treatedas constant in time, one can differentiate. This yields a similar formula for thevelocities

uM = P1uH + P2uI. (5.110)

For the same reason one has that uN = P1uG + P2uH. A generalization of (5.110)reads

ur =

Nfree∑

s=1

Prsus, Nfree + 1 ≤ r ≤ Nv, (5.111)

where the matrices satisfy

Nfree∑

s=1

Prs = Id, Nfree + 1 ≤ r ≤ Nv . (5.112)


This defines the constraint set

K =

u = (u1, . . . , uNfree, uNfree+1 , . . . , uNv ) ∈ Rd×Nv, with (5.111)–(5.112)

.(5.113)

By definition this set of constraints satisfies all points (i), (ii), (iii) and (iv) insection 5.7.1. Therefore one can compute the nodal velocities by solving the min-imization problem (5.84). The result is a numerical method which is conservativein total momentum and total energy. It is an exercise to check that this solutionis the same as the one obtained by the direct method based on (5.105).

The use of a similar mechanism for sliding inside a Lagrangian mesh is ad-dressed in [50]. Even though it is much more technical due to the complicatedstructure of K , the general approach is the same.


The literature on cell-centered and related numerical methods for Lagrangian CFDis growing quite rapidly. Therefore the list of references mentioned below is nec-essarily incomplete.

The control of vorticity errors is a difficult problem in Lagrangian computa-tions (just think of a mesh on which is imposed a vortical flow – at some pointALE is mandatory); this is addressed in the seminal paper [83]. A recent exten-sion of the GLACE scheme with explicit preservation of the angular momentum isproposed in [74]. Such ideas or techniques may be used for the reduction of meshimprinting [143]. MUSCL-type second-order extensions can be found in most pa-pers in the reference list. High-order issues are addressed specifically in [145].Interesting comments on this topic are in [136]. It should be mentioned that acompletely different and more ambitious approach to high-order Lagrangian CFDis developed in the work of Rieben et al.; see [78]. The strong coupling of La-grangian discretization with remeshing is addressed in [142, 62]. Important ideasabout the use of large time steps are introduced in [53]: these ideas still need to befully developed for Lagrangian-based CFD. Additional details about a continuousvariational formulation of sliding inside a mesh are given in [61].

The modeling of compressible elasticity (a fundamental reference for modelsis [121]) in the context of Lagrangian computations takes advantage of the renewalof numerical Lagrangian studies. One can refer to [119, 120] for contributionsrelating to hyperelastic modeling, and to [167, 147] for cell-centered numerics inthe context of hypoelastic modeling. See also [89, 90, 91]. Hypoelastic models havethe form described in section 3.3.2. The central question of the hyperbolicity ofhyperelastic models is solved in [156]. The two main discretization ideas in [120]are the following (in the context of GLACE).


• Firstly, discretize the deformation gradient F kj ∈ R

3×3 as

F k+1j =

(I3 +

∆t

V kj

∑

r

ur ⊗ Ckj r

)Fk

j ,

which comes from a multiplicative representation (see [120]) of the solution of∂tF = ∇Xu. This is needed for hyperelastic models, but not for hypoelasticmodels. The analysis of mimetic or weak consistency of this equation shouldbe possible (just compare the above equation and the fundamental identity(5.116) below).

• Second, discretize the impulse equation as

Mj

uk+1j − uk

j

∆t=

∑

r

\σkj rC

kj r

where σkj r is a corner-based stress tensor. Using the notation of section 5.2.2,

it is clear that the force is f kj r = − \σk

j rCkj r. In this context, the main point is

the definition of the linear corner problems which give the values of the forcesin terms of the cell-centered variables. As for pure gas dynamics, there is nounique solution depending on the corner decomposition. A general formula[120] which is representative of difference approaches (e.g. [120, 167, 147])writes the force (omitting the superscript k) as

\σj rCj r = fj r = −σjCj r + |Cj r| Qj r (uj − ur) , (5.114)

where the viscosity matrix is a symmetric tensor

Qj r = Qtj r =

∑

r

∑

e

βej rC

ej r ⊗ Ce

j r,

with βej r > 0 a coefficient that gives the correct scaling of the result. Note

that the discrete “force” exerted by the cell on the corner is −σjCj r. Theequation (5.114) is closed with

∑

j

fj r = 0. (5.115)

It is clear that the nodal velocity ur can be obtained from the solution of thewell-posed linear system (5.114)–(5.115). In this approach the two remainingdegrees of freedom are the model that gives the value of the stress tensorσj in cells and the choice of the viscosity matrices, which is ultimately likethe difference between GLACE and EUCCLHYD. The analysis in terms ofentropy inequalities of such algorithm is easy to perform, and is rigorous forhyperelastic models.


• As remarked in [71], it is possible to rewrite Lagrangian ideal MHD as amodified hyperelastic model, which opens up the possibility of using numeri-cal methods for compressible elasticity to perform ideal MHD computations.This is a fully open problem.

A hybrid cell-centered/staggered variant of the main scheme is developed inthe work of Loubere et al. [140, 141]: the idea is to perform the gradient recon-struction inside the cells. See also [153, 154, 32, 201] for an interesting variant withfast development. Comparison with some modern staggered schemes, as in [137],is recommended. Symmetry-preserving techniques are discussed in [47]; see also[46]. The robustness of solvers on unstructured grids is discussed in [178, 179]. Avariant of all these methods is presented in [15].

The consistency criterion

∑

r

Cj r ⊗ xr = Vj Id ∀j (5.116)

is fundamental. Indeed, the same criterion comes up in the numerical analysis offinite volume techniques for the diffusion equation; see the fundamental paper [80]where a unified framework is developed based on similar formulas. More generally,the use of corner vectors for discretization of the diffusion equations is addressedin [30] within the context of the GLACE corner vectors (in 2D, with a proof ofconvergence) and in [29] for EUCCLHYD-like corner vectors.

An original direction of research is that of Balsara [11, 12, 13], who proposed adesign of new corner-based MHD linearized Riemann solvers. The aim is to designmultidimensional numerical MHD solvers that preserve divergence-free involutiveconstraints. Let us also mention the work of Dumbser et al. [25, 24] (and refer-ences therein) on the coupling of ALE and Lagrangian CFD with discontinuousGalerkin and ADER techniques. Lagrangian discontinuous Galerkin methods arealso investigated in [193, 132]. The question of Galilean-invariant discretizationin numerical astrophysics is thoroughly addressed in [185]. Remeshing/remappingof the velocity with an original VIP technique is conducted with the APITALImethod in [110]. There are connections with the method of [74] for the control ofangular momentum, which can be interpreted as a partial DG technique.

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Subject Index

Acoustic solver of Godunov, 195ALE

formulation, 188methods, 185

Artificial viscosity, 242Awe-Rascle second-order system, 160

Born-OppenheimerTi−Te model, 161Buckley-Leverett model, 89Burgers equation, 4, 68

Cartesian mesh, 166Characteristic curves, 43Comatrix, 12Compressible

elasticity, 119gas dynamics, 8

Conservative finite volumescheme, 67Constraints, 315Contact

discontinuities, 56discontinuities, 141problems, 318

CoordinatesEulerian, 11Lagrangian, 11

Directional splitting, 166Discontinuities

contact, 56entropic, 54

Energy-Lagrange system, 188Enthalpy of the Lagrangian system,

111Entropic discontinuities, 54

Entropy, 52flux, 52inequality, 182of a system, 98subzonal, 304variable, 99weak solutions, 53, 101

EUCCLHYD, 223, 278Eulerian

coordinates, 11formulation, 188

FormulationALE, 188Eulerian, 188Hui’s, 19, 228Lagrangian, 188

Frame invariance, 20

Galilean invariant operator, 16GLACE, 223, 278Godunov

acoustic solver, 195scheme, 91theorem, 97

Hanging nodes, 323Harten formalism, 75Hui’s formulation, 19, 228Hyperbolicity, 24

Ideal MHD, 115quasi-Lagrangian, 150

Invarianceframe, 20Galilean, 21



Frontiers in Mathematics, DOI 10.1007/978-3-319-50355-4

347

348 Subject Index

Lorenz, 162

Keyfitz-Kranzer system, 161Kulikovski generating function, 138

Lagrange+remap scheme, 167Lagrangian

coordinates, 11formulation, 188Galilean invariance, 106system, 103traffic flow, 61

Lawpressure, 8stiffened gas pressure, 8van der Waals, 8

Linear stability, 24, 26strong or weak, 26

Linearized Riemann solvers, 193Lorentz invariance, 162LWR model, 3

Mach number, 158Magnetohydrodynamics (MHD) sys-

tem, 115Mass variable, 15Matrix splitting, 169, 170, 193Maximum principle, 74Model

Born-Oppenheimer Ti − Te, 161Buckley-Leverett, 89LWR, 3multiphase, 126Olmos-Munos, 64

Moving grid, 179Multidimensional Lagrangian system,

147Multiphase model, 126

Nanson’s formula, 14Nodal control volumes, 274

Oleinik solution, 59Olmos-Munos model, 64One-state solvers, 198

Optimal splitting, 174

Particle discretization, 85Perfect polytropic gas, 8Piola

identities, 11, 226identity, 13

Pressure law, 8

Quasi-Lagrangian ideal MHD, 150

Rankine-Hugoniotrelation, 49relations, 133

Rarefaction fans, 58, 140Rayleigh line, 157Remapping, 180Riemann problem, 130Riemann solver, 193Rusanov flux, 76

SchemeConservative finite volume, 67Lagrange+remap, 167

Shallow water, 4Shocks, 56Simplexes, 287Solutions

entropy weak, 53, 101strong, 42weak, 46

Solverslinearized Riemann, 193one-state, 198two-state, 200

Stabilization of meshes, 302Stiffened gas pressure law, 8Strong

linear instability, 26solutions, 42

Subzonal entropy, 304Superfluid helium, 123System

Awe-Rascle second-order, 160

Subject Index 349

energy-Lagrange, 188Keyfitz-Kranzer, 161Lagrangian, 103magnetohydrodynamics(MHD),

115multidimensional Lagrangian, 147

Traffic flow, 3Lagrangian, 61

Translation invariance, 265Two-state solvers, 200

Unstable mesh modes, 252

van der Waals law, 8Volume fractions, 305

Weakconsistency, 290solutions, 46

Well-balanced techniques, 245Well-prepared data, 27

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