nhan phan-thien nam mai-duy understanding...

Understanding Viscoelasticity

Nhan Phan-ThienNam Mai-Duy

An Introduction to Rheology

Third Edition

Graduate Texts in Physics


Series editors

Kurt H. Becker, Polytechnic School of Engineering, Brooklyn, USAJean-Marc Di Meglio, Université Paris Diderot, Paris, FranceSadri Hassani, Illinois State University, Normal, USABill Munro, NTT Basic Research Laboratories, Atsugi, JapanRichard Needs, University of Cambridge, Cambridge, UKWilliam T. Rhodes, Florida Atlantic University, Boca Raton, USASusan Scott, Australian National University, Acton, AustraliaH. Eugene Stanley, Boston University, Boston, USAMartin Stutzmann, TU München, Garching, GermanyAndreas Wipf, Friedrich-Schiller-Universität Jena, Jena, Germany


Graduate Texts in Physics publishes core learning/teaching material for graduate- andadvanced-level undergraduate courses on topics of current and emerging fields withinphysics, both pure and applied. These textbooks serve students at the MS- orPhD-level and their instructors as comprehensive sources of principles, definitions,derivations, experiments and applications (as relevant) for their mastery and teaching,respectively. International in scope and relevance, the textbooks correspond to coursesyllabi sufficiently to serve as required reading. Their didactic style, comprehensive-ness and coverage of fundamental material also make them suitable as introductionsor references for scientists entering, or requiring timely knowledge of, a research field.

More information about this series at http://www.springer.com/series/8431

Nhan Phan-Thien • Nam Mai-Duy

UnderstandingViscoelasticityAn Introduction to Rheology

Third Edition

123

Nhan Phan-ThienDepartment of Mechanical EngineeringNational University of SingaporeSingaporeSingapore

Nam Mai-DuyFaculty of Health, Engineering and SciencesUniversity of Southern QueenslandToowoomba, QLDAustralia

ISSN 1868-4513 ISSN 1868-4521 (electronic)Graduate Texts in PhysicsISBN 978-3-319-61999-6 ISBN 978-3-319-62000-8 (eBook)DOI 10.1007/978-3-319-62000-8

Library of Congress Control Number: 2017945696

1st and 2nd edition: © Springer-Verlag Berlin Heidelberg 2002, 20133rd edition: © Springer International Publishing AG 2017This work is subject to copyright. All rights are reserved by the Publisher, whether the whole or partof the material is concerned, specifically the rights of translation, reprinting, reuse of illustrations,recitation, broadcasting, reproduction on microfilms or in any other physical way, and transmissionor information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilarmethodology now known or hereafter developed.The use of general descriptive names, registered names, trademarks, service marks, etc. in thispublication does not imply, even in the absence of a specific statement, that such names are exempt fromthe relevant protective laws and regulations and therefore free for general use.The publisher, the authors and the editors are safe to assume that the advice and information in thisbook are believed to be true and accurate at the date of publication. Neither the publisher nor theauthors or the editors give a warranty, express or implied, with respect to the material contained herein orfor any errors or omissions that may have been made. The publisher remains neutral with regard tojurisdictional claims in published maps and institutional affiliations.

Printed on acid-free paper

This Springer imprint is published by Springer NatureThe registered company is Springer International Publishing AGThe registered company address is: Gewerbestrasse 11, 6330 Cham, Switzerland

Preface to the Third Edition

In this third edition, we have updated on the Dissipative Particle Dynamics methodwith recent results and provided some MATLAB programs for students to try out influid/flow modelling. In addition, a solution manual is provided, to guide studentsin their attempts at the questions in the book. The book itself remains compact, butsufficient in content for a first-year graduate module.

We sincerely thank our wives, Kim-Thoa (N. Phan-Thien) and Oanh(N. Mai-Duy) for their encouragement and their unfailing support—this is theirachievement as well as ours!

Singapore Nhan Phan-ThienToowoomba, Australia Nam Mai-Duy

v

Preface to the Second Edition

In this second edition, typographical errors brought about by the conversion processto LATEX were corrected; my gratitude went to Brittany Bannish (University ofUtah) for painstakingly going through the first edition. My main aim in revising thisis to produce a still compact book, sufficient at the level of first-year graduate coursefor those who wish to understand viscoelasticity, and to embark in modellingviscoelastic multiphase fluids. To this end, I have decided to introduce a newchapter on Dissipative Particle Dynamics (DPD), which I believe is relevant inmodelling complex-structured fluids. All the basic ideas in DPD are reviewed, withsome sample problems to illustrate the methodology. My gratitude goes toA*STAR, the Agency for Science, Technology and Research, for fundingMultiphase Modelling Projects, and Prof. Khoo Boo Cheong, a colleague andabove all a friend, for his support, which made the writing of Chapter 9 possible.I wish to acknowledge Prof. Mai-Duy Nam and Dr. Pan Dingyi, for their contri-butions to the DPD research, and Prof. Yu Shaozheng, for his comments on therevised book. Lastly, my humble thanks to the continuing support and constantencouragement of my wife, Kim-Thoa—without her capable hands, normal dailytasks would be impossible, let alone revising this book! It has been good for me togo through this revision, and I sincerely hope that the readers find the book useful intheir research works.

Singapore Nhan Phan-ThienJune 2012

vii

Preface to the First Edition

This book presents an introduction to viscoelasticity, in particular, to the theories ofdilute polymer solutions and dilute suspensions of rigid particles in viscous andincompressible fluids. These theories are important, not just because they apply topractical problems of industrial interest, but because they form a solid theoreticalbase upon which mathematical techniques can be built, from which more complextheories can be constructed, to better mimic material behaviour. The emphasis is noton the voluminous current topical research, but on the necessary tools to understandviscoelasticity at a first-year graduate level.

Viscoelasticity, or Continuum Mechanics, or Rheology1 (certainly not to beconfused with Theology) is the science of deformation and flow. This definition wasdue to Bingham, who, together with Scott-Blair2 and Reiner,3 helped form TheSociety of Rheology in 1929. Rheology has a distinguished history involvinghigh-profile scientists. The idea that everything has a timescale and that if we areprepared to wait long enough, then everything will flow was known to the Greekphilosopher Heraclitus, and prior to him, to the Prophetess Deborah—TheMountains Flowed Before The Lord.4 Not surprisingly, the motto of the Society ofRheology is pamsaqei (everything flows), a saying attributed to Heraclitus.

From the rheological viewpoint, there is no clear distinction between solid andliquid; it is a matter between the relative timescale T of the experiment to thetimescale s of the material concerned. The timescale ratio, De ¼ s=T is called theDeborah number. If this ratio is negligibly small, then one has a viscous fluid (moreprecise definition later), and if it is large, a solid, and in-between, a viscoelasticliquid. The timescale of the fluid varies considerably, from 10−13 s for water, to a

1This word was coined by E.C. Bingham (1878–1946), Professor of Chemistry at LafayetteCollege, Pennsylvania. The Bingham fluid is named after him.2G.W. Scott Blair (1902–1987), Professor of Chemistry at the University of Reading. His maincontributions were in biorheology.3M. Reiner (1886–1976), Professor of Mathematics at the Technion University of Haifa, Israel. Heis remembered for contributing to the Reiner–Rivlin fluid.4The Book of Judges.

ix

few milliseconds for automotive oils, to minutes for polymer solutions and to hoursfor melts and soft solids.

Graduate students of Rheology naturally have the unenviable task of walking thebridge between solid mechanics and fluid mechanics, and at the same time trying tograsp the more significant and relevant concepts. They often find it hard (at least forme, during my graduate days) to piece together useful information from severalcomprehensive monographs and published articles on this subject. This set oflectures is an attempt to address this problem—it contains the necessary tools tounderstand viscoelasticity but does not insist on giving the latest piece of infor-mation on the topic.

The book starts with an introduction to the basic tools from tensor and dyadicanalysis. Some authors prefer Cartesian tensor notation, others, dyadic notation. Weuse both notations, and they will be summarised here. Chapter 2 is a review ofnon-Newtonian behaviour in flows; here, the elasticity of the liquid and its ability tosupport large tension in stretching can be responsible for variety of phenomena,sometimes counter-intuitive. Kinematics and the equations of balance are discussedin detail in Chapter 3, including the finite strain and Rivlin–Ericksen tensors. InChapter 4, some classical constitutive equations are reviewed, and the generalprinciples governing the constitutive modelling are outlined. In this Chapter, theorder fluid models are also discussed, leading to the well-known result that theNewtonian velocity field is admissible to a second-order fluid in plane flow.Chapter 5 describes some of the popular engineering inelastic and the linear elasticmodels. The inelastic models are very useful in shear-like flows whereviscosity/shear rate relation plays a dominant role. The linear viscoelastic model is alimit of the simple fluid at small strain—any model must reduce to this limit whenthe strain amplitude is small enough. In Chapter 6, we discuss a special class offlowsknown as viscometric flows in which both the kinematics and the stress are fullydetermined by the flow, irrespective of the constitutive equations. This class offlowsis equivalent to the simple shearing flow. Modelling techniques for polymer solu-tions are discussed next in Chapter 7. Here one has a set of stochastic differentialequations for the motion of the particles; the random excitations come from a whitenoise model of the collision between the solvent molecules and the particles. It is ourbelief that a relevant model should come from the microstructure; however, when themicrostructure is so complex that a detailed model is not tractable, elements ofcontinuum model should be brought in. Finally, an introduction to suspensionmechanics is given in Chapter 8. I have deliberately left out a number of topics:instability, processing flows, electro-rheological fluids, magnetised fluids and

x Preface to the First Edition

viscoelastic computational mechanics. It is hoped that the book forms a goodfoundation for those who wish to embark on the Rheology path.

This has been tested out in a one-semester course in Viscoelasticity at theNational University of Singapore. It is entirely continuous-assessment based, withthe assignments graded at different difficulty levels to be attempted—solvingproblems is an indispensable part of the education process. A good knowledge offluid mechanics is helpful, but it is more important to have a solid foundation inMathematics and Physics (Calculus, Linear Algebra, Partial Differential Equations),of a standard that every one gets in the first two years in an undergraduateEngineering curriculum.

I have greatly benefitted from numerous correspondence with my academicbrother, Prof. Raj Huilgol, and my mentor, Prof. Roger Tanner. Prof. Jeff Giacominread the first draft of this; his help is gratefully acknowledged.

Singapore Nhan Phan-ThienFebruary 2002

Preface to the First Edition xi

Contents

1 Tensor Notation. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.1 Cartesian Frame of Reference . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

1.1.1 Position Vector . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.2 Frame Rotation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

1.2.1 Orthogonal Matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.2.2 Rotation Matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

1.3 Tensors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.3.1 Zero-Order Tensors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.3.2 First-Order Tensor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.3.3 Outer Products. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.3.4 Second-Order Tensors . . . . . . . . . . . . . . . . . . . . . . . . . . . 61.3.5 Third-Order Tensors . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61.3.6 Transpose Operation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71.3.7 Decomposition. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71.3.8 Some Common Vectors. . . . . . . . . . . . . . . . . . . . . . . . . . 81.3.9 Gradient of a Scalar. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81.3.10 Some Common Tensors . . . . . . . . . . . . . . . . . . . . . . . . . 8

1.4 Tensor and Linear Vector Function . . . . . . . . . . . . . . . . . . . . . . . 101.4.1 Claim . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101.4.2 Dyadic Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

1.5 Tensor Operations. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111.5.1 Substitution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111.5.2 Contraction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111.5.3 Transpose . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121.5.4 Products of Two Second-Order Tensors . . . . . . . . . . . . . 12

1.6 Invariants . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131.6.1 Invariant of a Vector . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131.6.2 Invariants of a Tensor . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

xiii

1.7 Decompositions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141.7.1 Eigenvalue and Eigenvector . . . . . . . . . . . . . . . . . . . . . . 141.7.2 Square Root Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . 141.7.3 Polar Decomposition Theorem . . . . . . . . . . . . . . . . . . . . 151.7.4 Cayley–Hamilton Theorem . . . . . . . . . . . . . . . . . . . . . . . 15

1.8 Derivative Operations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161.8.1 Derivative of det(A) . . . . . . . . . . . . . . . . . . . . . . . . . . . . 171.8.2 Derivative of tr(A). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 171.8.3 Derivative of tr(A2) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

1.9 Gradient of a Field . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 181.9.1 Field . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 181.9.2 Cartesian Frame. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 181.9.3 Non-Cartesian Frames . . . . . . . . . . . . . . . . . . . . . . . . . . . 201.9.4 Spherical Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

1.10 Integral Theorems. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 231.10.1 Gauss Divergence Theorem. . . . . . . . . . . . . . . . . . . . . . . 231.10.2 Stokes Curl Theorem. . . . . . . . . . . . . . . . . . . . . . . . . . . . 241.10.3 Leibniz Formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

2 Rheological Properties. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 292.1 Viscosity. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

2.1.1 Shear-Rate Dependent Viscosity . . . . . . . . . . . . . . . . . . . 292.2 Normal Stress Differences. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

2.2.1 Weissenberg Rod-Climbing Effect . . . . . . . . . . . . . . . . . . 332.2.2 Die Swell . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 332.2.3 Flow down an Inclined Channel . . . . . . . . . . . . . . . . . . . 35

2.3 Transient Responses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 362.3.1 Small Strain Oscillatory Flow . . . . . . . . . . . . . . . . . . . . . 362.3.2 Stress Overshoot . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 372.3.3 Stress Relaxation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 372.3.4 Relaxation Modulus. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 382.3.5 Recoil . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38

2.4 Elongational Flows . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 392.4.1 Elongational Viscosity. . . . . . . . . . . . . . . . . . . . . . . . . . . 39

2.5 Viscoelastic Instabilities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

3 Kinematics and Equations of Balance . . . . . . . . . . . . . . . . . . . . . . . . . 433.1 Kinematics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43

3.1.1 Reference Configuration . . . . . . . . . . . . . . . . . . . . . . . . . 433.1.2 Velocity and Acceleration Fields . . . . . . . . . . . . . . . . . . . 443.1.3 Material Derivative . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45

xiv Contents

3.2 Deformation Gradient and Strain Tensors . . . . . . . . . . . . . . . . . . . 453.2.1 Deformation Gradient . . . . . . . . . . . . . . . . . . . . . . . . . . . 453.2.2 Cauchy–Green Strain Tensor . . . . . . . . . . . . . . . . . . . . . . 463.2.3 Relative Strain Tensors . . . . . . . . . . . . . . . . . . . . . . . . . . 483.2.4 Path Lines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 483.2.5 Oscillatory Shear Flow . . . . . . . . . . . . . . . . . . . . . . . . . . 49

3.3 Rivlin–Ericksen Tensors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 503.4 Small Strain . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 533.5 Equations of Balance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54

3.5.1 Reynolds Transport Theorem. . . . . . . . . . . . . . . . . . . . . . 543.5.2 Conservation of Mass . . . . . . . . . . . . . . . . . . . . . . . . . . . 563.5.3 Conservation of Momentum . . . . . . . . . . . . . . . . . . . . . . 573.5.4 Conservation of Energy . . . . . . . . . . . . . . . . . . . . . . . . . . 61

Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63

4 Constitutive Equation: General Principles. . . . . . . . . . . . . . . . . . . . . . 654.1 Some Well-Known Constitutive Equations . . . . . . . . . . . . . . . . . . 65

4.1.1 Perfect Gas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 654.1.2 Inviscid Fluid . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 664.1.3 Fourier’s Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 674.1.4 Hookean Solid. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 674.1.5 Newtonian Fluid . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 694.1.6 Non-Newtonian Fluid . . . . . . . . . . . . . . . . . . . . . . . . . . . 71

4.2 Weissenberg and Deborah Numbers . . . . . . . . . . . . . . . . . . . . . . . 714.2.1 Deborah Number . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 714.2.2 Weissenberg Number. . . . . . . . . . . . . . . . . . . . . . . . . . . . 71

4.3 Some Guidelines in Constitutive Modelling . . . . . . . . . . . . . . . . . 724.3.1 Oldroyd Approach. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 734.3.2 Principle of Material Objectivity . . . . . . . . . . . . . . . . . . . 734.3.3 Objectivity of the Stress . . . . . . . . . . . . . . . . . . . . . . . . . 744.3.4 Frame Indifference. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 744.3.5 Principle of Local Action . . . . . . . . . . . . . . . . . . . . . . . . 774.3.6 Principle of Determinism. . . . . . . . . . . . . . . . . . . . . . . . . 77

4.4 Integrity Bases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 774.4.1 Isotropic Scalar-Valued Functions . . . . . . . . . . . . . . . . . . 774.4.2 Isotropic Vector-Valued Functions . . . . . . . . . . . . . . . . . 784.4.3 Isotropic Tensor-Valued Functions . . . . . . . . . . . . . . . . . 79

4.5 Symmetry Restrictions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 804.5.1 Unimodular Matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 804.5.2 Symmetry Group . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 814.5.3 Isotropic Materials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81

4.6 Isotropic Elastic Materials. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82

Contents xv

4.7 The Simple Material . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 834.7.1 Simple Fluid . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 844.7.2 Incompressible Simple Fluid . . . . . . . . . . . . . . . . . . . . . . 844.7.3 Fading Memory. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85

4.8 Order Fluids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 854.8.1 Unsteady Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 864.8.2 Velocity Field in a Second-Order Fluid. . . . . . . . . . . . . . 87

4.9 Green–Rivlin Expansion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89

5 Inelastic Models and Linear Viscoelasticity . . . . . . . . . . . . . . . . . . . . . 955.1 Inelastic Fluids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95

5.1.1 Carreau Model. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 965.1.2 Power-Law Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97

5.2 Linear Viscoelasticity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 995.2.1 Simple Shear Flow . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1015.2.2 Step Strain. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1025.2.3 Relaxation Spectrum . . . . . . . . . . . . . . . . . . . . . . . . . . . . 104

5.3 Correspondence Principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1055.3.1 Quasi-Static Approximation. . . . . . . . . . . . . . . . . . . . . . . 1055.3.2 Circular Couette Flow . . . . . . . . . . . . . . . . . . . . . . . . . . . 106

5.4 Mechanical Analogs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 108Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 110

6 Steady Viscometric Flows . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1136.1 Kinematics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113

6.1.1 Steady Parallel Flow . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1156.1.2 Rectilinear Flow . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1156.1.3 Axial Fanned Flow . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1166.1.4 Helical Flow . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1166.1.5 Helicoidal Flow . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116

6.2 Stresses in Steady Viscometric Flows. . . . . . . . . . . . . . . . . . . . . . 1176.2.1 Controllable and Partially Controllable Flows . . . . . . . . . 118

Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 118

7 Polymer Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1237.1 Characteristics of a Polymer Chain. . . . . . . . . . . . . . . . . . . . . . . . 123

7.1.1 Random-Walk Model . . . . . . . . . . . . . . . . . . . . . . . . . . . 1237.2 Forces on a Chain. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1267.3 Fluctuation-Dissipation Theorem . . . . . . . . . . . . . . . . . . . . . . . . . 130

7.3.1 Langevin Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1307.3.2 Equi-Partition of Energy . . . . . . . . . . . . . . . . . . . . . . . . . 1317.3.3 Fluctuation-Dissipation Theorem . . . . . . . . . . . . . . . . . . . 1317.3.4 Diffusivity Stokes–Einstein Relation . . . . . . . . . . . . . . . . 1337.3.5 Fokker–Planck Equation . . . . . . . . . . . . . . . . . . . . . . . . . 1337.3.6 Smoothed-Out Brownian Force . . . . . . . . . . . . . . . . . . . . 134

xvi Contents

7.4 Stress Tensor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1357.4.1 Kramers Form . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 138

7.5 Elastic Dumbbell Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1387.5.1 Langevin Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1397.5.2 Average Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1407.5.3 Strong and Weak Flows . . . . . . . . . . . . . . . . . . . . . . . . . 1407.5.4 Upper-Convected Maxwell Model . . . . . . . . . . . . . . . . . . 1417.5.5 Oldroyd-B Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 142

7.6 Main Features of the Oldroyd-B Model . . . . . . . . . . . . . . . . . . . . 1437.6.1 Simple Flows . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1447.6.2 Multiple Relaxation Time UCM Model. . . . . . . . . . . . . . 146

Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 146

8 Suspensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1498.1 Bulk Suspension Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1508.2 Dilute Suspension of Spheroids . . . . . . . . . . . . . . . . . . . . . . . . . . 152Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 156

9 Dissipative Particle Dynamics (DPD) . . . . . . . . . . . . . . . . . . . . . . . . . . 1599.1 1-D Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1609.2 DPD Fluid . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 163

9.2.1 Langevin Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1639.2.2 Phase-Space Description: Fokker–Planck Equation . . . . . 1689.2.3 Distribution Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . 1699.2.4 Equation of Change. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1709.2.5 Conservation of Mass . . . . . . . . . . . . . . . . . . . . . . . . . . . 1719.2.6 Conservation of Linear Momentum . . . . . . . . . . . . . . . . . 1729.2.7 Energy Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 176

9.3 Some Approximate Results. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1799.3.1 High Damping Limit . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1799.3.2 Standard DPD Parameters . . . . . . . . . . . . . . . . . . . . . . . . 1859.3.3 Effective Size of a DPD Particle . . . . . . . . . . . . . . . . . . . 186

9.4 Modification of the Weighting Function . . . . . . . . . . . . . . . . . . . . 1889.5 Imposition of Physical Parameters . . . . . . . . . . . . . . . . . . . . . . . . 190

9.5.1 Time Scales . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1909.5.2 Imposition of Dimensionless Compressibility and

Time-Scale Ratio . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1919.5.3 Imposition of Viscosity and Dynamic Response . . . . . . . 193

9.6 Numerical Implementation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1969.6.1 Velocity Verlet Scheme. . . . . . . . . . . . . . . . . . . . . . . . . . 1969.6.2 Exponential Time Differencing Scheme. . . . . . . . . . . . . . 1979.6.3 Implementation of No-Slip Boundary Conditions . . . . . . 2019.6.4 Computation of Interparticle Forces. . . . . . . . . . . . . . . . . 2029.6.5 Calculation of Stress Tensor . . . . . . . . . . . . . . . . . . . . . . 2039.6.6 Complex-Structure Fluid . . . . . . . . . . . . . . . . . . . . . . . . . 203

Contents xvii

9.7 Flow Verifications and Some Typical Problems . . . . . . . . . . . . . . 2089.8 Matlab Program . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2199.9 Epilogue . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 223Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 224

Solutions to Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 229

References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 297

Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 301

xviii Contents

Chapter 1Tensor Notation

A Working Knowledge in Tensor Analysis

This chapter is not meant as a replacement for a course in tensor analysis, but it willprovide a sufficient working background to tensor notation and algebra.

1.1 Cartesian Frame of Reference

Physical quantities encountered are either scalars (e.g., time, temperature, pressure,volume, density), or vectors (e.g., displacement, velocity, acceleration, force, torque),or tensors (e.g., stress, displacement gradient, velocity gradient, alternating tensors– we deal mostly with second-order tensors). These quantities are distinguished bythe following generic notation:

s denotes a scalar (lightface italic)u denotes a vector (boldface)F denotes a tensor (boldface)

The distinction between vector and tensor is usually clear from the context. Whenthey are functions of points in a three-dimensional Euclidean spaceE, they are calledfields. The set of all vectors (or tensors) form a normed vector space U.

Distances and time are measured in the Cartesian frame of reference, or simplyframe of reference, F = {O; e1, e2, e3}, which consists of an origin O , a clock, andan orthonormal basis {e1, e2, e3}, see Fig. 1.1,

ei · e j = δi j , i, j = 1, 2, 3 (1.1)

where the Kronecker delta is defined as

© Springer International Publishing AG 2017N. Phan-Thien and N. Mai-Duy, Understanding Viscoelasticity,Graduate Texts in Physics, DOI 10.1007/978-3-319-62000-8_1

1

2 1 Tensor Notation

x3

x2

x1

e3e2

e1

u

(a) (b)

Fig. 1.1 a Cartesian frame of reference and b right hand rule

δi j ={1, i = j,0, i �= j.

(1.2)

We only deal with right-handed frames of reference (applying the right-hand rule,when the thumb is in direction 1, and the forefinger in direction 2, the middle fingerlies in direction 3, or an even permutation of this), where (e1 × e2) · e3 = 1.

The Cartesian components of a vector u are given by

ui = u · ei (1.3)

so that one may write

u =3∑

i=1

uiei = uiei . (1.4)

Here we have employed the summation convention, i.e., whenever there are repeatedsubscripts, a summation is implied over the range of the subscripts, from (1, 2, 3).For example,

Ai j B jk =3∑j=1

Ai j B jk . (1.5)

This short-hand notation is due to Einstein (Fig. 1.2), who argued that physical lawsmust not depend on coordinate systems, and therefore must be expressed in tensorialformat. This is the essence of the Principle of Frame Indifference, to be discussedlater.

The alternating tensor is defined as

εi jk =⎧⎨⎩

+1, if (i, j, k) is an even permutation of (1, 2, 3)−1, if (i, j, k) is an odd permutation of (1, 2, 3)0, otherwise.

(1.6)

1.1 Cartesian Frame of Reference 3

Fig. 1.2 Albert Einstein(1879–1955) got the NobelPrize in Physics in 1921 forhis explanation inphotoelectricity. He derivedthe effective viscosity of adilute suspension ofneutrally buoyant spheres,

η = ηs(1 + 52φ), ηs : the

solvent viscosity, φ: thesphere volume fraction

1.1.1 Position Vector

In the frame F = {O; e1, e2, e3}, the position vector is denoted by

x = xiei , (1.7)

where xi are the components of x.

1.2 Frame Rotation

Consider the two frames of references, F = {O; e1, e2, e3} and F ′ ={O; e′

1, e′2, e′

3}, as shown in Fig. 1.3, one obtained from the other by a rotation.

Hence,ei · e j = δi j , e′

i · e′j = δi j .

Define the cosine of the angle between(ei , e′

j)as

Ai j = e′i · e j .

Thus Ai j can be regarded as the components of e′i in F , or the components of e j in

F ′. We writee′p = Apiei Api Aqi = δpq .

4 1 Tensor Notation

Fig. 1.3 Two frames ofreference sharing a commonorigin

u

x1

x'1

x2

x'2

x3x'3

Oe1e'1

e2

e'2e'3

e3

Similarlyei = Apie′

p Api Apj = δi j .

1.2.1 Orthogonal Matrix

A matrix is said to be an orthogonal matrix if its inverse is also its transpose; fur-thermore, if its determinant is +1, then it is a proper orthogonal matrix. Thus [A] isa proper orthogonal matrix.

We now consider a vector u, expressed in either frame F or F ′,

u = uiei = u′je

′j .

Taking scalar product with either base vector,

u′i = e′

i · e j u j = Ai ju j ,

u j = e j · e′i u

′i = Ai ju

′i .

In matrix notation,

[A] =⎡⎣ A11 A12 A13

A21 A22 A23

A31 A32 A33

⎤⎦ , [u] =

⎡⎣ u1u2u3

⎤⎦ ,

[u′] =

⎡⎣ u′

1u′2

u′3

⎤⎦ ,

we have

[u′] = [A] . [u] , [u] = [A]T · [

u′] ,

u′i = Ai ju j , u j = Ai ju

′i . (1.8)

1.2 Frame Rotation 5

In particular, the position transforms according to this rule

x = x ′ie

′i = x je j x ′

i = Ai j x j or x j = Ai j x′i .

1.2.2 Rotation Matrix

The matrix A is called a rotation – in fact a proper rotation (detA = +1).

1.3 Tensors

1.3.1 Zero-Order Tensors

Scalars, which are invariant under a frame rotation, are said to be tensors of zero order.

1.3.2 First-Order Tensor

A set of three scalars referred to one frame of reference, written collectively asv = (v1, v2, v3), is called a tensor of first order, or a vector, if the three componentstransform according to (1.8) under a frame rotation.Clearly,

• If u and v are vectors, then u + v is also a vector.• If u is a vector, then αu is also a vector, where α is a real number.

The set of all vectors form a vector space U under addition and multiplication. Inthis space, the usual scalar product can be shown to be an inner product. With thenorm induced by this inner product, |u|2 = u · u, U is a normed vector space. Wealso refer to a vector u by its components, ui .

1.3.3 Outer Products

Consider now two tensors of first order, ui and vi . The product uiv j represents theouter product of u and v, and written as (the subscripts are assigned from left to rightby convention),

[uv] =⎡⎣ u1v1 u1v2 u1v3u2v1 u2v2 u2v3u3v1 u3v2 u3v3

⎤⎦ .

6 1 Tensor Notation

In a frame rotation, from F to F ′, the components of this change according to

u′iv

′j = Aim A jnumvn.

1.3.4 Second-Order Tensors

In general, a set of 9 scalars referred to one frame of reference, collectively writtenas W = [Wi j ], transformed to another set under a frame rotation according to

W ′i j = Aim A jnWmn, (1.9)

is said to be a second-order tensor, or a two-tensor, or simply a tensor (when theorder does not have to be explicit). In matrix notation, we write

[W′] = [A] [W] [A]T orW′ = AWAT or W ′

i j = AikWkl A jl .

In the direct notation, we denote a tensor by a bold face letter (without the squarebrackets). This direct notation is intimately connected to the concept of a linearoperator, e.g., Gurtin [34].

1.3.5 Third-Order Tensors

A set of 27 scalars referred to one frame of reference, collectively written as W =[Wi jk], transformed to another set under a frame rotation according to

W ′i jk = Ail A jm AknWlmn, (1.10)

is said to be a third-order tensor.Obviously, the definition can be extended to a set of 3n scalars, andW = [Wi1i2...in ]

(n indices) is said to be an n -order tensor if its components transform under a framerotation according to

W ′i1i2...in = Ai1 j1 Ai2 j2 · · · Ain jnW j1 j2... jn . (1.11)

We will deal mainly with vectors and tensors of second order. Usually, a higher-order (higher than 2) tensor is formed by taking outer products of tensors of lowerorders, for example the outer product of a two-tensor T and a vector n is a third-ordertensor T ⊗ n. One can verify that the transformation rule (1.11) is obeyed.

1.3 Tensors 7

1.3.6 Transpose Operation

The components of the transpose of a tensorW are obtained by swapping the indices:

[W]i j = Wi j , [W]Ti j = Wji .

A tensor S is symmetric if it is unaltered by the transpose operation,

S = ST , Si j = Sji .

It is anti-symmetric (or skew) if

S = −ST , Si j = −Sji .

An anti-symmetric tensor must have zero diagonal terms (when i = j).Clearly

• If U and V are two-tensors, then U + V is also a two-tensor.• If U is a two-tensor, then αU is also a two-tensor, where α is a real number. Theset of U form a vector space under addition and multiplication.

1.3.7 Decomposition

Any second-order tensor can be decomposed into symmetric and anti-symmetricparts:

W = 1

2

(W + WT

) + 1

2

(W − WT

),

Wi j = 1

2

(Wi j + Wji

) + 1

2

(Wi j − Wji

). (1.12)

Returning to (1.9), if we interchange i and j , we get

W ′j i = A jm AinWmn = A jn AimWnm .

The second equality arises because m and n are dummy indices, mere labels in thesummation. The left side of this expression is recognised as the components of thetranspose of W. The equation asserts that the components of the transpose of W arealso transformed according to (1.9). Thus, if W is a two-tensor, then its transposeis also a two-tensor, and the Cartesian decomposition (1.12) splits an arbitrary two-tensor into a symmetric and an anti-symmetric tensor (of second order).

We now go through some of the first and second-order tensors that will be encoun-tered in this course.

8 1 Tensor Notation

1.3.8 Some Common Vectors

Position, displacement, velocity, acceleration, linear and angular momentum, linearand angular impulse, force, torque, are vectors. This is because the position vectortransforms under a frame rotation according to (1.8). Any other quantity linearlyrelated to the position (including the derivative and integral operation) will also be avector.

1.3.9 Gradient of a Scalar

The gradient of a scalar is a vector. Let φ be a scalar, its gradient is written as

g = ∇φ, gi = ∂φ

∂xi.

Under a frame rotation, the new components of ∇φ are

∂φ

∂x ′i

= ∂φ

∂x j

∂x j

∂x ′i

= Ai j∂φ

∂x j,

which qualifies ∇φ as a vector.

1.3.10 Some Common Tensors

We have met a second-order tensor formed by the outer product of two vectors,written compactly as uv, with components (for vectors, the outer products is writtenwithout the symbol ⊗)

(uv)i j = uiv j .

In general, the outer product of n vectors is an n-order tensor.

Unit Tensor. The Kronecker delta is a second-order tensor. In fact it is invariant inany coordinate system, and therefore is an isotropic tensor of second-order. To showthat it is a second-order tensor, note that

δi j = Aik A jk = Aik A jlδkl,

which follows from the orthogonality of the transformation matrix. δi j are said tobe the components of the second-order unit tensor I. Finding isotropic tensors ofarbitrary orders is not a trivial task.

1.3 Tensors 9

Gradient of a Vector. The gradient of a vector is a two-tensor: if ui and u′i are the

components of u in F and F ′,

∂u′i

∂x ′j

= ∂xl∂x ′

j

∂

∂xl(Aikuk) = Aik A jl

∂uk∂xl

.

This qualifies the gradient of a vector as a two-tensor.

Velocity Gradient. If u is the velocity field, then ∇u is the gradient of the velocity.Be careful with the notation here. By our convention, the subscripts are assignedfrom left to right, so

(∇u)i j = ∇i u j = ∂u j

∂xi.

In most books on viscoelasticity including this, the term “velocity gradient” is takento mean the second-order tensor L = (∇u)T with components

Li j = ∂ui∂x j

. (1.13)

StrainRate andVorticity Tensors. The velocity gradient tensor can be decomposedinto a symmetric part D, called the strain rate tensor, and an anti-symmetric part W,called the vorticity tensor:

D = 1

2

(∇u + ∇uT), W = 1

2

(∇uT − ∇u). (1.14)

Stress Tensor and Quotient Rule. We are given that stress T = [Ti j ] at a point x isdefined by, (see Fig. 1.4),

t = Tn, ti = Ti j n j , (1.15)

where n is a normal unit vector on an infinitesimal surface ΔS at point x, and t isthe surface traction (force per unit area) representing the force the material on thepositive side of n is pulling on the material on the negative side of n. Under a framerotation, since both t (force) and n are vectors,

t′ = At, t = AT t′ n′ = An, n = ATn′,AT t′ = t = Tn = TATn′ t′ = ATATn′.

From the definition of the stress, t′ = T′n′, and therefore

T′ = ATAT .

So the stress is a second-order tensor.In fact, as long as t and n are vector, the 9 components Ti j defined in the manner

indicated by (1.15) form a second-order tensor. This is known as the quotient rule.

10 1 Tensor Notation

Fig. 1.4 Defining the stresstensor

x1

x2

x3

e1e2

e3 n

T.n

1.4 Tensor and Linear Vector Function

L is a linear vector function on U if it satisfies

• L (u1 + u2) = L (u1) + L (u2) ,

• L (αu) = αL (u) , ∀u,u1,u2 ∈ U , ∀α ∈ R

1.4.1 Claim

Let W be a two-tensor, and define a vector-valued function through

v = L (u) = Wu,

then L is a linear function. Conversely, for any linear function on U , there is a uniquetwo-tensor W such that

L (u) = Wu, ∀u ∈ U .

The first statement can be easily verified. For the converse part, given the linearfunction, let define Wi j through

L (ei ) = Wjie j .

Now, ∀u ∈ U ,v = L (u) = L (uiei ) = uiWjie jv j = Wjiui .

W is a second-order tensor because u and v are vectors. The uniqueness part of Wcan be demonstrated by assuming that there is another W′, then

(W − W′) u = 0, ∀u ∈ U ,

1.4 Tensor and Linear Vector Function 11

which implies that W′ = W.

In this connection, one can define a second-order tensor as a linear function, takingone vector into another. This is the direct approach, e.g., Gurtin [34], emphasisinglinear algebra.Weusewhatever notation is convenient for the purpose at hand. The setof all linear vector functions forms a vector space under addition and multiplication.The main result here is that

L (ei ) = Wei = Wjie j W ji = e j · (Wei ) .

1.4.2 Dyadic Notation

Thus, one may writeW = Wi jeie j . (1.16)

This is the basis for the dyadic notation, the eie j play the role of the basis “vectors”for the tensor W.

1.5 Tensor Operations

1.5.1 Substitution

The operation δi j u j = ui replaces the subscript j by i – the tensor δi j is thereforesometimes called the substitution tensor.

1.5.2 Contraction

Given a two-tensor Wi j , the operation

Wii =3∑

i=1

Wii = W11 + W22 + W33

is called a contraction. It produces a scalar. The invariance of this scalar under aframe rotation is seen by noting that

W ′i i = Aik AilWkl = δklWkl = Wkk .

This scalar is also called the trace of W, written as


trW = Wii . (1.17)

It is one of the invariants of W (i.e., unchanged in a frame rotation). If the trace of Wis zero, thenW is said to be traceless. In general, given an n-order tensor, contractingany two subscripts produces a tensor of (n − 2) order.

1.5.3 Transpose

Given a two-tensor W = [Wi j ], the transpose operation swaps the two indices

WT = (Wi jeie j

)T = Wi je jei ,[WT

]i j = Wji . (1.18)

1.5.4 Products of Two Second-Order Tensors

Given two second-order tensors, U and V,

U = Ui jeie j , V = Vi jeie j ,

one can form different products from them, and it is helpful to refer to the dyadicnotation here.

• The tensor product U ⊗ V is a 4th-order tensor, with component Ui j Vkl ,

U ⊗ V = Ui j Vkleie jekel . (1.19)

• The single dot product U.V is a 2nd-order tensor, sometimes written without thedot (the dot is the contraction operator),

U · V = UV = (Ui jeie j

) · (Vklekel)

= Ui jeiδ jkVklel = Ui j Vjleiel , (1.20)

with components UikVkl , just like multiplying two matrices Uik and Vkj . Thissingle dot product induces a contraction of a pair of subscripts ( j and k) inUi j Vkl ,and acts just like a vector dot product.

• The double dot (or scalar, or inner) product produces a scalar,

U : V = (Ui jeie j

) : (Vklekel) = (Ui jei

)δ jk · (Vklel)

= Ui j Vklδ jkδil = Ui j Vji . (1.21)

1.5 Tensor Operations 13

The dot operates on a pair of base vectors until we run out of dots. The end resultis a scalar (remember our summation convention). It can be shown that the scalarproduct is in fact an inner product.

• The norm of a two-tensor is defined from the inner product in the usual manner,

‖U‖2 = UT : U = Ui jUi j = tr(UTU

). (1.22)

The space of all linear vector functions therefore form a normed vector space.• One writes U2 = UU, U3 = U2U, etc.• A tensor U is invertible if there exists a tensor, U−1, called the inverse of U, suchthat

UU−1 = U−1U = I (1.23)

One can also define the vector cross product between two second-order tensors(and indeed any combination of dot and cross vector products). However, werefrain from listing all possible combinations here.

1.6 Invariants

1.6.1 Invariant of a Vector

When a quantity is unchanged with a frame rotation, it is said to be invariant. Froma vector, a scalar can be formed by taking the scalar product with itself, vivi = v2.

This is of course the magnitude of the vector and it is the only independent scalarinvariant for a vector.

1.6.2 Invariants of a Tensor

From a second-order tensor S, there are three independent scalar invariants that canbe formed, by taking the trace of S, S2 and S3,

I = trS = Sii , I I = trS2 = Si j S ji , I I I = trS3 = Si j S jk Ski .

However, it is customary to use the following invariants

I1 = I, I2 = 1

2

(I 2 − I I

), I3 = 1

6

(I 3 − 3I I I + 2I I I

) = det S.

It is also possible to form ten invariants between two tensors (Gurtin [34]).


1.7 Decompositions

We now quote some of the well-known results without proof, some are intuitivelyobvious, others not.

1.7.1 Eigenvalue and Eigenvector

A scalar ω is an eigenvalue of a two-tensor S if there exists a non-zero vector e,called the eigenvector, satisfying

Se = ωe. (1.24)

The characteristic space for S corresponding to the eigenvalue ω consists of allvectors in the eigenspace, {v : Sv = ωv}. If the dimension of this space is n, thenω is said to have geometric multiplicity of n. The spectrum of S is the ordered list{ω1,ω2, . . .} of all the eigenvalues of S.

A tensor S is said to be positive definite if it satisfies

S : vv > 0, ∀v �= 0. (1.25)

We record the following theorems:

• The eigenvalues of a positive definite tensor are strictly positive.• The characteristic spaces of a symmetric tensor are mutually orthogonal.• Spectral decomposition theorem: Let S be a symmetric two-tensor. Then there isa basis consisting entirely of eigenvectors of S. For such a basis, {ei , i = 1, 2, 3},the corresponding eigenvalues {ωi , i = 1, 2, 3} form the entire spectrum of S, andS can be represented by the spectral representation, where

S =3∑

i=1ωieiei , when S has three distinct eigenvalues,

S = ω1ee + ω2 (I − ee) , when S has two distinct eigenvalues,S = ωI, when S has only one eigenvalue.

(1.26)

1.7.2 Square Root Theorem

Let S be a symmetric positive definite tensor. Then there is a unique positive definitetensor U such that U2 = S. We write

U = S1/2.

The proof of this follows from the spectral representation of S.

1.7 Decompositions 15

1.7.3 Polar Decomposition Theorem

For any given tensor F, there exist positive definite tensors U and V, and a rotationtensor R, such that

F = RU = VR. (1.27)

Each of these representations is unique, and

U = (FTF

)1/2, V = (

FFT)1/2

. (1.28)

The first representation (RU) is called the right, and the second (VR) is called theleft polar decomposition.

1.7.4 Cayley–Hamilton Theorem

The most important theorem is the Cayley–Hamilton theorem: Every tensor S satis-fies its own characteristic equation

S3 − I1S2 + I2S − I3I = 0, (1.29)

where I1 = trS, I2 = 12

((trS)2 − trS2

), and I3 = det S are the three scalar invariants

for S, and I is the unit tensor in three dimensions.In two dimensions, this equation reads

S2 − I1S + I2I = 0, (1.30)

where I1 = trS, I2 = det S are the two scalar invariants for S, and I is the unit tensorin two dimensions.

Cayley–Hamilton theorem is used to reduce the number of independent tensorialgroups in tensor-valued functions. We record here one possible use of the Cayley–Hamilton theorem in two dimensions. The three-dimensional case is reserved as anexercise.

Suppose C is a given symmetric positive definite tensor in 2-D,

[C] =[C11 C12

C12 C22

],

and its square root U = C1/2 is desired. From the characteristic equation for U,

U = I−11 (U) [C + I3 (U) I] ,


so if we can express the invariants of U in terms of the invariant of C, we’re done.Now, if the eigenvalues of U are λ1 and λ2, then

I1 (U) = λ1 + λ2, I2 (U) = λ1λ2,

I1 (C) = λ21 + λ2

2, I2 (C) = λ21λ

22.

ThusI2 (U) = √

I2 (C),

I 21 (U) = I1 (C) + 2√I2 (C).

Therefore

U = C + √I2 (C)I√

I1 (C) + 2√I2 (C)

.

1.8 Derivative Operations

Suppose ϕ (u) is a scalar-valued function of a vector u. The derivative of ϕ(u) withrespect to u in the direction v is defined as the linear operator Dϕ (u) [v]:

ϕ (u + αv) = ϕ (u) + αDϕ (u) [v] + HOT ,

where HOT are terms of higher order, which vanish faster than α. Also, the squarebrackets enclosing v are used to emphasise the linearity of in v. An operationaldefinition for the derivative of ϕ(u) in the direction v is therefore,

Dϕ (u) [v] = d

dα[ϕ (u + αv)]α=0 . (1.31)

This definition can be extended verbatim to derivatives of a tensor-valued (of anyorder) function of a tensor (of any order). The argument v is a part of the definition.We illustrate this with a few examples.

Example 1 Consider the scalar-valued function of a vector, ϕ (u) = u2 = u · u. Itsderivative in the direction of v is

Dϕ (u) [v] = d

dαϕ (u + αv)α=0 = d

dα

[u2 + 2αu · v + α2v2

]α=0

= 2u · v.

Example 2 Consider the tensor-valued function of a tensor, G (A) = A2 = AA. Itsderivative in the direction of B is

1.8 Derivative Operations 17

DG (A) [B] = d

dα[G (A + αB)]α=0

= d

dα

[A2 + α (AB + BA) + O

(α2

)]α=0

= AB + BA.

1.8.1 Derivative of det(A)

Consider the scalar-valued function of a tensor, ϕ (A) = detA. Its derivative in thedirection of B can be calculated using

det(A + αB) = det αA(A−1B + α−1I

) = α3 detA det(A−1B + α−1I

)= α3 detA

(α−3 + α−2 I1

(A−1B

) + α−1 I2(A−1 · B) + I3

(A−1B

))= detA

(1 + αI1

(A−1B

) + O(α2

)).

Thus

Dϕ (A) [B] = d

dα[ϕ (A + αB)]α=0 = detAtr

(A−1B

).

1.8.2 Derivative of tr(A)

Consider the first invariant I (A) = trA. Its derivative in the direction of B is

DI (A) [B] = d

dα[I (A + αB)]α=0

= d

dα[trA + αtrB]α=0 = trB = I : B.

1.8.3 Derivative of tr(A2)

Consider the second invariant I I (A) = trA2. Its derivative in the direction of B is

DI I (A) [B] = d

dα[I I (A + αB)]α=0

= d

dα

[A : A + α (A : B + B : A) + O

(α2

)]α=0

= 2A : B.


1.9 Gradient of a Field

1.9.1 Field

A function of the position vector x is called a field. One has a scalar field, forexample the temperature field T (x), a vector field, for example the velocity fieldu(x), or a tensor field, for example the stress field S(x). Higher-order tensor fields arerarely encountered, as in themany-point correlation fields. Conservation equations incontinuummechanics involve derivatives (derivativeswith respect to position vectorsare called gradients) of different fields, and it is absolutely essential to know how tocalculate the gradients of fields in different coordinate systems. We also find it moreconvenient to employ the dyadic notation at this point.

1.9.2 Cartesian Frame

We consider first a scalar field, ϕ (x) . The Taylor expansion of this about point x is

ϕ (x + αr) = ϕ (x) + αr j∂

∂x jϕ (x) + O

(α2

).

Thus the gradient ofϕ (x) at point x, nowwritten as∇ϕ, defined in (1.31), is given by

∇ϕ [r] = r · ∂ϕ

∂x. (1.32)

This remains unchanged for a vector or a tensor field.

Gradient Operator. This leads us to define the gradient operator as

∇ = e j∂

∂x j= e1

∂

∂x1+ e2

∂

∂x2+ e3

∂

∂x3. (1.33)

This operator can be treated as a vector, operating on its arguments. By itself, it hasno meaning; it must operate on a scalar, a vector or a tensor.

Gradient of a Scalar. For example, the gradient of a scalar is

∇ϕ = e j∂ϕ

∂x j= e1

∂ϕ

∂x1+ e2

∂ϕ

∂x2+ e3

∂ϕ

∂x3. (1.34)

Gradient of a Vector. The gradient of a vector can be likewise calculated

∇u =(ei

∂

∂xi

) (u je j

) = eie j∂u j

∂xi. (1.35)

1.9 Gradient of a Field 19

In matrix notation,

[∇u] =

⎡⎢⎢⎢⎢⎢⎢⎣

∂u1∂x1

∂u2∂x1

∂u3∂x1

∂u1∂x2

∂u2∂x2

∂u3∂x2

∂u1∂x3

∂u2∂x3

∂u3∂x3

⎤⎥⎥⎥⎥⎥⎥⎦

.

The component (∇u)i j is ∂u j/∂xi ; some books define this differently.

Transpose of a Gradient. The transpose of a gradient of a vector is therefore

∇uT = eie j∂ui∂x j

. (1.36)

In matrix notation,

[∇u]T =

⎡⎢⎢⎢⎢⎢⎢⎣

∂u1∂x1

∂u1∂x2

∂u1∂x3

∂u2∂x1

∂u2∂x2

∂u2∂x3

∂u3∂x1

∂u3∂x2

∂u3∂x3

⎤⎥⎥⎥⎥⎥⎥⎦

.

Divergence of a Vector. The divergence of a vector is a scalar defined by

∇ · u =(ei

∂

∂xi

)· (u je j

) = ei · e j ∂u j

∂xi= δi j

∂u j

∂xi

∇ · u = ∂ui∂xi

= ∂u1∂x1

+ ∂u2∂x2

+ ∂u3∂x3

. (1.37)

The divergence of a vector is also an invariant, being the trace of a tensor.

Curl of a Vector. The curl of a vector is a vector defined by

∇ × u =(ei

∂

∂xi

)× (

u je j) = ei × e j

∂u j

∂xi= εki jek

∂u j

∂xi

= e1

(∂u3∂x2

− ∂u2∂x3

)+ e2

(∂u1∂x3

− ∂u3∂x1

)+ e3

(∂u2∂x1

− ∂u1∂x2

). (1.38)

The curl of a vector is sometimes denoted by rot.

Divergence of a Tensor. The divergence of a tensor is a vector field defined by

∇ · S =(ek

∂

∂xk

)· (Si jeie j

) = e j∂Si j∂xi

. (1.39)


Fig. 1.5 Cylindrical andspherical frame of references

x1

x2

x3

z

x1

x2

x3

r

r

1.9.3 Non-Cartesian Frames

All the above definitions for gradient and divergence of a tensor remain valid in anon-Cartesian frame, provided that the derivative operation is also applied to thebasis vectors as well. We illustrate this process in two important frames, cylindricaland spherical coordinate systems (Fig. 1.5); for other systems, consult Bird et al. [6].

Cylindrical Coordinates. In a cylindrical coordinate system (Fig. 1.5, left), pointsare located by giving them values to {r, θ, z}, which are related to{x = x1, y = x2, z = x3} by

x = r cos θ, y = r sin θ, z = zr = √

x2 + y2, θ = tan−1( yx

), z = z

The basis vectors in this frame are related to the Cartesian ones by

er = cos θex + sin θey, ex = cos θer − sin θeθ

eθ = − sin θex + cos θey, ey = sin θer + cos θeθ

Physical components. In this system, a vector u, or a tensor S, are represented by,respectively,

u = urer + uθeθ + uzez,

S = Srrerer + Srθereθ + Srzerez + Sθreθer+ Sθθeθeθ + Sθzeθez + Szrezer + Szθezeθ + Szzezez .

Gradient operator. The components expressed this way are called physical compo-nents. The gradient operator is converted from one system to another by the chainrule,


∇ = ex∂

∂x+ ey

∂

∂y+ ez

∂

∂z= (cos θer − sin θeθ)

(cos θ

∂

∂r− sin θ

r

∂

∂θ

)

+ (sin θer + cos θeθ)

(sin θ

∂

∂r+ cos θ

r

∂

∂θ

)+ ez

∂

∂z

= er∂

∂r+ eθ

1

r

∂

∂θ+ ez

∂

∂z. (1.40)

When carrying out derivative operations, remember that

∂

∂rer = 0,

∂

∂reθ = 0,

∂

∂rez = 0

∂

∂θer = eθ,

∂

∂θeθ = −er ,

∂

∂θez = 0

∂

∂zer = 0,

∂

∂zeθ = 0,

∂

∂zez = 0 (1.41)

Gradient of a vector. The gradient of any vector is

∇u =(er

∂

∂r+ eθ

1

r

∂

∂θ+ ez

∂

∂z

)(urer + uθeθ + uzez)

= erer∂ur∂r

+ ereθ∂uθ

∂r+ erez

∂uz

∂r+ eθer

1

r

∂ur∂θ

+ eθeθurr

+ eθeθ1

r

∂uθ

∂θ− eθer

uθ

r+ eθez

1

r

∂uz

∂θ+ ezer

∂ur∂z

+ ezeθ∂uθ

∂z

+ ezez∂uz

∂z

∇u = erer∂ur∂r

+ ereθ∂uθ

∂r+ erez

∂uz

∂r+ eθer

(1

r

∂ur∂θ

− uθ

r

)

+ eθeθ

(1

r

∂uθ

∂θ+ ur

r

)+ eθez

1

r

∂uz

∂θ+ ezer

∂ur∂z

+ ezeθ∂uθ

∂z

+ ezez∂uz

∂z. (1.42)

Divergence of a vector. The divergence of a vector is obtained by a contraction ofthe above equation:

∇ · u = ∂ur∂r

+ 1

r

∂uθ

∂θ+ ur

r+ ∂uz

∂z. (1.43)

1.9.4 Spherical Coordinates

In a spherical coordinate system (Fig. 1.5, right), points are located by giving themvalues to {r, θ,φ}, which are related to {x = x1, y = x2, z = x3} by


x = r sin θ cosφ, y = r sin θ sin φ, z = r cos θ,

r =√x2 + y2 + z2, θ = tan−1

(√x2 + y2

z

), φ = tan−1

( y

x

).

The basis vectors are related by

er = e1 sin θ cosφ + e2 sin θ sin φ + e3 cos θ,

eθ = e1 cos θ cosφ + e2 cos θ sin φ − e3 sin θ,

eφ = −e1 sin φ + e2 cosφ,

and

e1 = er sin θ cosφ + eθ cos θ cosφ − eφ sin φ,

e2 = er sin θ sin φ + eθ cos θ sin φ + eφ cosφ,

e3 = er cos θ − eθ sin θ.

Gradient operator. Using the chain rule, it can be shown that the gradient operatorin spherical coordinates is

∇ = er∂

∂r+ eθ

1

r

∂

∂θ+ eφ

1

r sin θ

∂

∂φ. (1.44)

We list below a few results of interest.Gradient of a scalar. The gradient of a scalar is given by

∇ϕ = er∂ϕ

∂r+ eθ

1

r

∂ϕ

∂θ+ eφ

1

r sin θ

∂ϕ

∂φ. (1.45)

Gradient of a vector. The gradient of a vector is given by

∇u = erer∂ur∂r

+ ereθ∂uθ

∂r+ ereφ

∂uφ

∂r+ eθer

(1

r

∂ur∂θ

− uθ

r

)

+ eθeθ

(1

r

∂uθ

∂θ+ ur

r

)+ eφer

(1

r sin θ

∂ur∂φ

− uφ

r

)

+ eθeφ1

r

∂uφ

∂θ+ eφeθ

(1

r sin θ

∂uθ

∂φ− uφ

rcot θ

)

+ eφeφ

(1

r sin θ

∂uφ

∂φ+ ur

r+ uθ

rcot θ

). (1.46)

Divergence of a vector. The divergence of a vector is given by


∇ · u = 1

r2∂

∂r

(r2ur

) + 1

r

∂

∂θ(uθ sin θ) + 1

r sin θ

∂uφ

∂φ. (1.47)

Divergence of a tensor. The divergence of a tensor is given by

∇ · S = er

[1

r2∂

∂r

(r2Srr

) + 1

r sin θ

∂

∂θ(Sθr sin θ) + 1

r sin θ

∂Sφr

∂φ(1.48)

− Sθθ + Sφφ

r

]+ eθ

[1

r3∂

∂r

(r3Srθ

) + 1

r sin θ

∂

∂θ(Sθθ sin θ)

+ 1

r sin θ

∂Sφθ

∂φ+ Sθr − Srθ − Sφφ cot θ

r

]+ eφ

[1

r3∂

∂r

(r3Srφ

)

+ 1

r sin θ

∂

∂θ

(Sθφ sin θ

) + 1

r sin θ

∂Sφφ

∂φ+ Sφr − Srφ + Sφθ cot θ

r

].

1.10 Integral Theorems

1.10.1 Gauss Divergence Theorem

Various volume integrals can be converted to surface integrals by the followingtheorems, due to Gauss (Fig. 1.6):

∫V

∇ϕdV =∫SϕndS, (1.49)

Fig. 1.6 Carl FriedrichGauss (1777–1855) was aProfessor of Mathematics atthe University of Göttingen.He made severalcontributions to NumberTheory, Geodesy, Statistics,Geometry, Physics. Hismotto was “few, but ripe”(Pauca, Sed Matura). He didnot publish several importantpapers because they did notsatisfy these requirements


Fig. 1.7 A region enclosedby a closed surface withoutward unit vector field

x3

x2

x1

n

∫V

∇ · udV =∫Sn · udS, (1.50)

∫V

∇ · SdV =∫Sn · SdS. (1.51)

The proofsmaybe found inKellogg [43]. In these,V is a bounded regular region,withbounding surface S and outward unit vector n (Fig. 1.7),ϕ, u, and S are differentiablescalar, vector, and tensor fields with continuous gradients. Indeed the indicial versionof (1.50) is valid even if ui are merely three scalar fields of the required smoothness(rather than three components of a vector field).

1.10.2 Stokes Curl Theorem

Various surfaces integrals can be converted into contour integrals using the followingtheorems: ∫

Sn · (∇ × u) dS =

∮Ct · udC, (1.52)

∫Sn · (∇ × S) dS =

∮Ct · SdC . (1.53)

In these, t is a tangential unit vector along the contourC . The direction of integrationis determined by the right-hand rule: thumb pointing in the direction of n, fingerscurling in the direction of C .

1.10 Integral Theorems 25

Fig. 1.8 Gottfried W.Leibniz (1646–1716) was aGerman philosopher andmathematician, whoindependently with Newton,laid the foundation forintegral and differentialcalculus in 1675

1.10.3 Leibniz Formula

Ifϕ is a field (a scalar, a vector, or a tensor) define on a region V (t), which is changingin time, with bounding surface S(t), also changing in time with velocity uS , then(Leibniz formula, Fig. 1.8)

d

dt

∫V

ϕdV =∫V

∂ϕ

∂tdV +

∫SϕuS · ndS. (1.54)

Problems

Problem 1.1 The components of vectors u, v, and w are given by ui , vi , wi . Verifythat

u · v = uivi ,

u × v = εi jkei u jvk,

(u × v) · w = εi jkuiv jwk,

(u × v) · w = u · (v × w) ,

(u × v) × w = (u · w) v − (v · w)u,

(u × v)2 = u2v2 − (u · v)2 ,

where u2 = |u|2 and v2 = |v|2 .


Problem 1.2 Let A be a 3 × 3 matrix with entries Ai j ,

[A] =⎡⎣ A11 A12 A13

A21 A22 A23

A31 A32 A33

⎤⎦ .

Verify that

det [A] = εi jk A1i A2 j A3k = εi jk Ai1A j2Ak3,

εlmn det [A] = εi jk Ail A jm Akn = εi jk Ali Amj Ank,

det [A] = 1

6εi jkεlmn Ail A jm Akn.

Problem 1.3 Verify that

εi jkεimn = δ jmδkn − δ jnδkm .

Given that two 3 × 3 matrices of components

[A] =⎡⎣ A11 A12 A13

A21 A22 A23

A31 A32 A33

⎤⎦ , [B] =

⎡⎣ B11 B12 B13

B21 B22 B23

B31 B32 B33

⎤⎦

verify that if [C] = [A] · [B] , then the components of C are Ci j = Aik Bkj . Thus if[D] = [A]T [B] , then Di j = Aki Bkj .

Problem 1.4 Show that, if[Ai j

]is a frame rotation matrix,

det[Ai j

] = (e′1 × e′

2) · e′

3 = 1.

[A]T [A] = [A] [A]T = [I] , [A]−1 = [A]T , det [A] = 1.

Problem 1.5 Verify that

εi jkuiv jwk = det

⎡⎣ u1 u2 u3

v1 v2 v3w1 w2 w3

⎤⎦ .

Consider a second-order tensor Wi j and a vector ui = εi jkW jk . Show that if W issymmetric, u is zero, and if W is anti-symmetric the components of u are twicethose of W in magnitude. This vector is said to be the axial vector of W.

Hence, show that the axial vector associated with the vorticity tensor of (1.14) is−∇ × u.

Problems 27

Problem 1.6 If D, S and W are second-order tensors, D symmetric and W anti-symmetric, show that

D : S = D : ST = D : 12

(S + ST

),

W : S = −W : ST = W : 12

(W − WT

),

D : W = 0.

Further, show that

if T : S = 0 ∀S then T = 0,if T : S = 0 ∀ symmetric S then T is anti-symmetric,if T : S = 0 ∀ anti-symmetric S then T is symmetric.

Problem 1.7 Show that Q is orthogonal if and only if H = Q − I satisfies

H + HT + HHT = 0, HHT = HTH.

Problem 1.8 Show that, if S is a second-order tensor, then I = tr S, I I = tr S2,I I I = det S are indeed invariants. In addition, show that

det (S − ωI) = −ω3 + I1ω2 − I2ω + I3.

If ω is an eigenvalue of S then det (S − ωI) = 0. This is said to be the characteristicequation for S.

Problem 1.9 Apply the result above to find the square root of the Cauchy-Greentensor in a two-dimensional shear deformation

[C] =[1 + γ2 γ

γ 1

].

Investigate the corresponding formula for the square root of a symmetric positivedefinite tensor S in three dimensions.

Problem 1.10 Write down all the components of the strain rate tensor and the vor-ticity tensor in a Cartesian frame.

Problem 1.11 Given that r = xiei is the position vector, a is a constant vector, andf (r) is a function of r = |r|, show that

∇ · r = 3, ∇ × r = 0, ∇ (a · r) = a, ∇ f = 1

r

d f

drr.

Problem 1.12 Show that the divergence of a second-order tensor S in cylindricalcoordinates is given by


∇ · S = er

(∂Srr∂r

+ Srr − Sθθ

r+ 1

r

∂Sθr

∂θ+ ∂Szr

∂z

)

+ eθ

(∂Srθ∂r

+ 2Srθr

+ 1

r

∂Sθθ

∂θ+ ∂Szθ

∂z+ Sθr − Srθ

r

)

+ ez

(∂Srz∂r

+ Srzr

+ 1

r

∂Sθz

∂θ+ ∂Szz

∂z

). (1.55)

Problem 1.13 Show that, in cylindrical coordinates, the Laplacian of a vector u isgiven by

∇2u = er

[∂

∂r

(1

r

∂

∂r(rur )

)+ 1

r2∂2ur∂θ2

+ ∂2ur∂z2

− 2

r2∂uθ

∂θ

]

+ eθ

[∂

∂r

(1

r

∂

∂r(ruθ)

)+ 1

r2∂2uθ

∂θ2+ ∂2uθ

∂z2+ 2

r2∂ur∂θ

]

+ ez

[1

r

∂

∂r

(r∂uz

∂r

)+ 1

r2∂2uz

∂θ2+ ∂2uz

∂z2

]. (1.56)

Problem 1.14 Show that, in cylindrical coordinates,

u · ∇u = er

[ur

∂ur∂r

+ uθ

r

∂ur∂θ

+ uz∂ur∂z

− uθuθ

r

]

+ eθ

[ur

∂uθ

∂r+ uθ

r

∂uθ

∂θ+ uz

∂uθ

∂z+ uθur

r

]

+ ez

[ur

∂uz

∂r+ uθ

r

∂uz

∂θ+ uz

∂uz

∂z

]. (1.57)

Problem 1.15 The stress tensor in a material satisfies ∇ · S = 0. Show that thevolume-average stress in a region V occupied by the material is

〈S〉 = 1

2V

∫S(xt + tx) dS, (1.58)

where t = n ·S is the surface traction. The quantity on the left side of (1.58) is calledthe stresslet (Batchelor [4]).

Problem 1.16 Calculate the following integrals on the surface of the unit sphere

〈nn〉 = 1

S

∫SnndS (1.59)

〈nnnn〉 = 1

S

∫SnnnndS. (1.60)

These are the averages of various moments of a uniformly distributed unit vector ona sphere surface.

Chapter 2Rheological Properties

Overall Material Properties and Flow Behaviour

Fluids with featureless microstructures are well described by the Newtonian consti-tutive equation, which states that the stress tensor is proportional to the shear rate ten-sor (these concepts will be made precise later). Fluids with complex microstructures,for example suspensions of particles or droplets (blood, paint, ink, asphalt, bitu-men, foodstuffs, etc.), polymer melts and solutions (molten plastics, fibre-reinforcedor particulate-filled plastics), exhibit a wide variety of behaviours. These are sum-marised here. For more information, consult Bird et al. [6] and Tanner [85].

2.1 Viscosity

2.1.1 Shear-Rate Dependent Viscosity

One of themost important fluid properties for engineering calculations is its viscosity.This quantity is defined as the ratio of the shear stress to the shear rate in a simpleshear flow.

Here, as shown in Fig. 2.1, the flow is generated by sliding one plate atop another,with the fluid in-between. The quantities of interest are the shear rate, γ = U/h (Uis the velocity of the top plate in the x-direction, the bottom plate is fixed, h is thedistance between the plates), and the shear stress, S = F/A (F is the shear forceon the top plate, A the fluid contact area). The shear stress is an odd function of theshear rate. In addition, with viscoelastic fluids, there may be a normal force on theplates. When a steady flow is established, the viscosity is defined as

η = S

γ. (2.1)


29

30 2 Rheological Properties

h

U

x

y

u = Uy/h

S = F/A

Normal force

Fig. 2.1 Shear flow generated by sliding one plate on top of another. Shear force as well as normalforce may be required to keep the plates at a fixed distance

Shear rate (s-1)10

-410

-310

-210

-110

010

110

210

310

4

(Pa.

s)

101

102

103

104

105

106

Fig. 2.2 Viscosity of a low density polyethylene melt at different temperature. Top curve,T = 388 K, and the bottom curve T = 513 K

For a Newtonian fluid, its viscosity is a constant (having units Pa.s) depending onlyon the temperature. For most fluids with long chain microstructure (polymer meltsand solutions), the viscosity is a decreasing function of the shear rate, sometimesreaching 10−3−10−4 of the zero-shear rate viscosity. This type of behaviour is calledshear thinning. Theopposite behaviour, shear thickening, is sometimes observedwithsome suspensions, primarily due to the formation of clusters.

A typical viscosity-shear rate curve is shown in Fig. 2.2 for a low-densitypolyethylene (LDPE) at different temperatures. It can be seen from this figure thatthis fluid viscosity is a strong function of its temperature: a 60 ◦C increase in the tem-perature induces a ten-fold decrease in its viscosity. In addition, it may be observedthat LDPE viscosity decreases by an order of magnitude (compared to its zero-shear-rate value) at a moderate value of shear rate. Note that a constant viscosity does notqualify a fluid to be Newtonian – the term Newtonian is much more restrictive in its

2.1 Viscosity 31

Fig. 2.3 Jean-Louis-MariePoiseuille (1797–1869) wasa French physician whoestablished experimentallythe pressure-drop/flow raterelationship of laminar flowin tubes

meaning, implying a whole class of constitutive behaviour in which the stress tensoris proportional to the strain rate tensor.

For some materials with a solid-like behaviour (for example, bread dough, bio-logical tissues), viscosity measurement makes no sense, since the shear stress justkeeps increasing with time until the sample breaks or flows out of the test cell, andwhat has been measured is not a material property, but an indication of the frictionbetween the sample and the test apparatus (which depends on the flow process).

With suspensions of particles with surface charges, one can get the viscosity tobehave in many different ways; even a discontinuity at a particular shear rate maybe induced.

Incidentally, the CGS units for viscosity is Poise (P), in honour of Poiseuille(Fig. 2.3), who provided flow rate/pressure drop experimental relationship for pipeflow in1846; (1Pa.s=10P).Viscosity ofwater is about 1 cP (1 centi-Poise = 10−2 P).

2.2 Normal Stress Differences

Normal stress differences refer to the differences between the unequal normal stressesin shear flow (for aNewtonianfluid in shear flow, the normal stresses are always equal,a consequence of its constitutive equation). With three normal stress components,we can form two independent quantities, the first and the second normal stressdifferences:

N1 = Sxx − Syy, N2 = Syy − Szz . (2.2)


Fig. 2.4 Viscometricfunctions of 6.8% ofpolyisobutylene in cetane at24 ◦C. Here σ is the shearstress

(s-1)

N

-N

1 10 102 103

10

102

103

104

1

1

2

(Pa.s)

N ,

-N ,

(P

a)1

2

These normal stress differences are even functions of the shear rate, and thereforeone defines the normal stress coefficients as

ν1 = N1

γ2, ν2 = N2

γ2. (2.3)

These normal stress coefficients are even functions of the shear rate. The normal stressdifferences and the shear viscosity are collectively called viscometric functions; theyare the material properties of the fluid in shear (viscometric) flow.

Figure2.4 shows some typical measurements of viscometric properties of a poly-isobutylene solution, the first (N1) and the second (N2) normal stress differences, theshear stress (σ) and the viscosity (η = σ/γ).

The second normal stress difference is not usually measured. In general, for poly-mer melts and solutions, it is negative, and about 10% of N1 in magnitude.

Suspensions have non-zero normal stress differences as well; however, our knowl-edge of them is still incomplete. Non-equal normal stresses are responsible forsome visually striking differences between Newtonian and non-Newtonian fluid. Wesummarise the key features here.1

1The filmRheological Behavior of Fluids, presented by Prof. HershelMarkovitz, should bewatchedat this point. It contains the main important non-Newtonian flow phenomena and can be found atthe site www.web.mit.edu/hml/ncfmf.html. This site is a depository of a large number of otherinteresting fluid mechanics films. The book by Boger and Walters [10] should also be consulted -it contains a large number of interesting photographs detailing non-Newtonian behaviours.

www.web.mit.edu/hml/ncfmf.html

2.2 Normal Stress Differences 33

Fig. 2.5 Weissenberg rodclimbing effect

2.2.1 Weissenberg Rod-Climbing Effect

When a rod rotates in viscoelastic fluid, the fluid climbs the rod. This phenomenonis called the Weissenberg2 rod-climbing effect (Fig. 2.5). Rod climbing is due to thefluid element being able to support a tension along a streamline (due to non-zeronormal stress differences), which forces the fluid up the rod. This effect can occurwithout the rod: if one rotates a disk at the bottom of the beaker, then the free surfacebulges up at the middle.

2.2.2 Die Swell

When a viscoelastic fluid exits from a capillary of diameter D, it tends to swellconsiderably more than a Newtonian fluid. For a viscous Newtonian fluid, the swellratio, DE/D, where DE is the extrudate diameter, is a function of the Reynoldsnumber and is at most 13%. For a polymer melt, the extrudate diameter could bea few times the capillary diameter. This phenomenon is called die swell (Fig. 2.6),and the dominant mechanism causing this is the first normal stress difference. Infact, Tanner [84] proposed the simple rule for capillary die swell, based on a simpleanalysis,

2Karl Weissenberg (1893–1976) contributed significantly to Rheology in the early years, and hasseveral phenomena named after him.


Fig. 2.6 Die swell – top aNewtonian fluid, bottom aviscoelastic fluid

Fig. 2.7 Delay die swell –increasing Reynolds numberfrom left to right

DE

D= 0.13 +

[1 + 1

2

(N1

2S

)2

w

]1/6

, (2.4)

where N1 and S are the first normal stress difference and the shear stress, bothevaluated at the wall (subscript w). Die swell is mainly due to the fluid elasticity(normal stress effects), but it can also occur with the shear thinning induced byviscous heating. Inertia (i.e., Reynolds number) tends to reduce the amount of swell,and to delay it, see Fig. 2.7.

2.2 Normal Stress Differences 35

Fig. 2.8 In a flow down aninclined channel, the freesurface will bulge up if N2 isnegative

Fig. 2.9 Reversedsecondary flow – leftNewtonian fluid, rightviscoelastic fluid

2.2.3 Flow down an Inclined Channel

The second normal stress difference, although small in magnitude compared to thefirst normal stress difference, is important in some cases. In the flow down an inclinedchannel, a Newtonian fluid is seen to have a nearly flat free surface, whereas a convexsurface is seen for a viscoelastic fluid with a negative second normal stress difference(Fig. 2.8).

Viscoelasticity is also responsible for the reversal of the secondary flow pattern;one such case is sketched in Fig. 2.9.


2.3 Transient Responses

Viscoelastic fluids have a relaxation time scale, and this can be quantified in severalways.

2.3.1 Small Strain Oscillatory Flow

In an oscillatory shear flow (Fig. 2.1), where the top plate oscillates sinusoidally withangular frequency ω, x = δ sinωt, the plate velocity is U = ωδ cosωt. The shearrate (γ) and the shear strain (γ) are given by, respectively,

γ = γ0 cosωt, γ = γ0 sinωt, (2.5)

γ0 = δω/h, γ0 = δ/h.

When the strain is small, the shear stress is also sinusoidal, but is not in phase witheither the strain or the strain rate,

S = S′ sinωt + S′′ cosωt.

The part that is in phase with the strain is used to define the storage modulus (G ′),or the storage viscosity (η′′)

G ′ = S′

γ0, η′′ = S′

γ0, G ′ = ωη′′, (2.6)

and the part that is in phase with the strain rate is used to define the loss modulus(G ′′), or the dynamic viscosity (η′),

G ′′ = S′′

γ0, η′ = S′′

γ0, G ′′ = ωη′. (2.7)

These are functions of the frequency, and they are collectively referred to as thedynamic properties of the fluid.

Figure2.10 show the storage and loss moduli of LDPE at different temperatures.The data have been collapsed into a master curve through the use of the time-temperature superposition principle, which involves scaling the frequency by anempirical shift factor aT . The dynamic properties contain time scale information onthe fluid expressed in the frequency domain. Large-amplitude oscillatory tests havealso been done, but their interpretation is less straightforward.

2.3 Transient Responses 37

Reduced angular frequency aT (s-1)

10-3

10-2

10-1

100

101

102

103

104

Stor

age

and

loss

mod

uli (

Pa)

101

102

103

104

105

106

G''

G'

Fig. 2.10 Dynamic properties of a low density polyethylene (LDPE) melt

2.3.2 Stress Overshoot

In a start-up of a shear flow, i.e., U = U0H (t) , or γ = γ0H (t) , where H(t) isthe Heaviside function, the shear stress increases with time, then overshoots beforeapproaching its steady value, sometimes with a few oscillations about the steadyvalue. This is seen with the normal stress differences as well. One can define theunsteady viscosity, in a similarmanner to the definition of the (steady-state) viscosity:

η+ (t, γ0) = S (t)

γ0. (2.8)

The amount of stress overshoot can be significant at high shear rates, and it may bethe main reason why certain biological fluid (e.g., synovial fluid) is a good lubricant.

2.3.3 Stress Relaxation

Corresponding to the start-up of shear flow is stress relaxation, where the fluidmotionthat has been undergoing a steady-state shear flow at a shear rate is suddenly stopped.The shear stress (and the normal stress differences) is monitored as it relaxes. Again,one can define the stress-relaxation viscosity

η− (t, γ0) = S (t)

γ0. (2.9)


2.3.4 Relaxation Modulus

There is another type of relaxation experiment, in which a large strain rate γ0 isapplied over a small interval Δt , so that the total strain is γ0 = γ0Δt, and the shearstress S (t, γ0) is monitored as it relaxes. This allows the relaxation modulus to bedefined:

G (t, γ0) = S (t, γ0)

γ0. (2.10)

At small enough strains, G (t, γ0) = G0(t) is independent of the strain, because ofthe linearity between the stress and the strain at low strains. The relaxation modulusof LDPE is shown in Fig. 2.11. The nearly parallelism of the curves (at differentstrains) suggests that the relaxation modulus can be factored in a function of strainand a function of time,

G(t, γ0) = h(γ0)G0(t), h(0) = 1. (2.11)

This is known as strain-time separability.

2.3.5 Recoil

If the loading is suddenly removed by cutting the liquid column, as seen in Fig. 2.12,the liquid retracts to some previous shape. The liquid is said to have memory (itremembers its original configuration). However, its memory is imperfect, as it can

Time (s)10-2 10-1 100 101 102 103

Rel

axat

ion

mod

ulus

(Pa)

100

101

102

103

104

105

G0(t)

Fig. 2.11 Relaxation modulus of low density polyethylene (LDPE) – increasing strain from top tobottom

2.3 Transient Responses 39

Fig. 2.12 The liquid recoils back into the beaker after Prof. A.S. Lodge cut the liquid column(University of Wisconsin, Madison)

only retract partially. In that sense it has fading memory. A Newtonian liquid has acatastrophic memory: the moment the loading is removed, the motion ceases imme-diately. An elastic solid has a perfect memory: upon removal of the loads, the solidparticles return to exactly the positions they occupied previously.

2.4 Elongational Flows

2.4.1 Elongational Viscosity

Elongational flows refer to flow where the velocity gradient is diagonal, i.e.,

u = ax, v = by, w = cz, (2.12)

where a + b + c = 0 for incompressibility (more about this later). These flows cor-respond to stretching or elongating a sample fluid specimen. When b = −a andc = 0 one has a planar elongational flow, and a uni-axial elongational flow whena = b = −c/2. This latter flow occurs in many processes; here a is termed the elon-gational rate. The elongational viscosity is defined as

ηE = Sxx − Syya

. (2.13)

Except at very low elongational rates, elongational viscosity does not usually reacha steady state (the sample elongates and fails). For a Newtonian fluid its elongationalviscosity is thrice its shear viscosity; but for a polymer solution, the elongationalviscosity can be orders of magnitude greater. The Trouton ratio is defined as the ratioof the elongation viscosity to the shear viscosity of the fluid


Trouton Ratio = ηE

η. (2.14)

A typical plot of the Trouton ratio for a polybutene solution is shown in Fig. 2.13.For a Newtonian fluid, the Trouton ratio is three, for a viscoelastic fluid, this ratiomay be very large.

The ability of a liquid filament to support a significant tensile stress is mainly whythe tubeless siphon experiment (Fig. 2.14) works.

Fig. 2.13 Transient Troutonratio for a high molecularweight polyisobutylenesolution – the extensionalrate is a = 2 s−1

Fig. 2.14 Tubeless siphon

2.5 Viscoelastic Instabilities 41

2.5 Viscoelastic Instabilities

Because of the non-linearity in the constitutive equations, viscoelastic flows are fullof instabilities. These instabilities may not depend on inertia; they are mainly drivenby the fluid normal stresses (elasticity), or by the nature of the boundary conditions.To name a few, we have instability in Taylor-Couette flow, in the torsional flowbetween two parallel disks, in the shear flow between cone-and-plate, in curvedpipe flow, in contraction flows, in the flows from extrusion dies, etc. The extrudatedistortion, commonly called melt fracture, is an example of instability due to theinterplay between viscoelasticity and the nature of the boundary conditions. Thereference [10] contains several photographs of this phenomenon.

Chapter 3Kinematics and Equations of Balance

A Quick Review of Continuum Mechanics

In this chapter, we review the kinematics and the equations of balance (conserva-tion equations), leaving the question of constitutive description to the next chapter.Bird et al. [6], Tanner [85], Huilgol and Phan-Thien [39], Morrison [59] provideadditional reading materials.

3.1 Kinematics

3.1.1 Reference Configuration

We deal with a continuous body B, which occupies a region consisting of pointsin E3. We refer to one particular configuration, BR , for example the configurationat time t = 0, as the reference configuration. The particle position in the referenceconfiguration is denoted by a capital letter X. This particle traces out a path in thethree-dimensional space E3 and its current position is denoted by the small letterx; x is a function of time. The particle is referred to by its position in the referenceconfiguration, X.

A motion is defined to be a twice-differentiable and invertible map (so that accel-eration field can be defined, and that every position xmust correspond to a particleX)

x = M (X, t) , xi = Mi(X j , t

), (3.1)

where t is the time. This is also called theLagrangian description of themotion (afterLagrange, Fig. 3.1). Note that X = M (X, 0), by definition. Collectively M (X, t) ,

X ∈ BR gives us the spatial description of the motion, called the current configura-tion. Since M is invertible,

X = M−1 (x, t) , Xi = M−1i

(x j , t

)(3.2)

gives us the reference in terms of the current configuration.


43

44 3 Kinematics and Equations of Balance

Fig. 3.1 Joseph-LouisLagrange (1736–1813) was aItalian/Frenchmathematician, who madeimportant contributions toanalytic mechanics, calculusof variations, and numbertheory. His book MécaniqueAnalytique was an approvedpublication by a committeewith members includingLaplace and Legendre in1788

3.1.2 Velocity and Acceleration Fields

The velocity and the acceleration fields are defined as

u = ∂

∂tM (X, t) , ui = ∂

∂tMi

(X j , t

), (3.3)

and

a = ∂2

∂t2M (X, t) , ai = ∂2

∂t2Mi

(X j , t

), (3.4)

respectively.It is customary to refer to velocity and acceleration fields as functions of the

current position, the so-called Eulerian description (after Euler, Fig. 3.2). This isaccomplished using (3.2).

Denoting u = u (x, t) as the Eulerian velocity field, we find that the Eulerianacceleration field is given by

a = ∂

∂tu (x, t)

∣∣∣∣X

= ∂u∂t

+ ∂u∂x

· ∂x∂t

∣∣∣∣X

= ∂u∂t

+ u · ∇u = ∂u∂t

+ L · u, (3.5)

where we have introduced the velocity gradient tensor (note the transpose operationin the definition)

L =(

∂u∂x

)T

= (∇u)T , Li j = ∂ui∂x j

. (3.6)

3.1 Kinematics 45

Fig. 3.2 Leonhard Euler(1707–1783) was one of thegreatest mathematicians ofthe 18th century. Heoccupied Daniel Bernouilli’schair of mathematics atSt. Petersburg. He perfectedthe integral calculus, workedin analytic geometry, theoryof lunar motion, introducedthe Euler’s identity and manyother mathematical symbolsthat we are familiar with, π,f (),

∑. He published more

than 856 articles and books

An Eulerian velocity is called steady if it does not depend on time, i.e., u = u (x) .

A steady Eulerian velocity field is thus not necessarily Lagrangian steady.

3.1.3 Material Derivative

The derivative on u, as implied in (3.5) is called the material derivative, or total timederivative, or simply time derivative (if there is no confusion),

d

dt(.) = ∂

∂t(.) + u · ∇(.). (3.7)

The symmetric part of the velocity gradient is called the strain rate tensor D, and itsanti-symmetric part is called the vorticity tensor,

L = D + W, D = 1

2

(L + LT

), W = 1

2

(L − LT

). (3.8)

3.2 Deformation Gradient and Strain Tensors

3.2.1 Deformation Gradient

The gradient of x with respect to X is called the deformation gradient,

F =(

∂x∂X

)T

, Fi j = ∂xi∂X j

. (3.9)


Note again our subscript convention. At time t = 0 the initial value of F is

F (0) = I, (3.10)

the identity tensor. The mass in a region V is

∫

Vρdx =

∫

V0

ρ |J | dX,

where J = det F and V0 is the region occupied by the reference configuration. Thus,we demand that det F > 0, so that the mapping is not degenerate. For an incompress-ible fluid, the kinematic constraint is of course

det F = 1. (3.11)

Because of the chain rule,

(∂X∂x

)T

= F−1, F−1i j = ∂Xi

∂x j. (3.12)

The connection between the deformation and the velocity gradients arises from theequality

∂

∂tFi j = ∂

∂t

(∂Mi

∂X j

)= ∂

∂X j

(∂Mi

∂t

)= ∂ui

∂X j.

Using the Eulerian description for the velocity,

∂ui∂X j

= ∂ui∂xk

∂xk∂X j

= Lik Fkj ,

one has,F = LF, F (0) = I, (3.13)

where the super dot denotes the (total) time derivative. This equation provides aninitial-value problem for F.

3.2.2 Cauchy–Green Strain Tensor

The concept of strain is introduced by comparing the length of a fluid element at thecurrent time to that in the reference configuration. We have from the definition of thedeformation gradient,

dx = FdX.

3.2 Deformation Gradient and Strain Tensors 47

Fig. 3.3 An element dX atX in the referenceconfiguration at time t = 0 ismapped to dx at x at time t

Fig. 3.4 Augustin Cauchy(1789–1857) was a prolificFrench mathematician. Hecontributed more than 16fundamental concepts andtheorems, and publishedmore than 800 papers. Hisname is one of the 72inscribed on the Eiffel tower

Here dX is a fluid element at point X, which at time t is mapped to dx at point x,see Fig. 3.3. Its current length is

dx2 = dx · dx = Fi jdX j FikdXk = dXFT · FdX = FTF : dXdX.

The tensorC = FTF (3.14)

is therefore ameasure of the strain the fluid experiences. It is called the right Cauchy–Green tensor (for a portray of Cauchy, see Fig. 3.4). The left Cauchy–Green tensoris defined as

B = FFT . (3.15)

The name refers to the right or left polar decompositions of F,

F = RU (right) = VR (left),


whereU andV are symmetric positive-definite tensors (right and left stretch tensors),and R is an orthogonal tensor. Thus

C = FTF = U2, B = FFT = V2.

The inverse of the Cauchy–Green tensor is also used; it is called the Finger straintensor.

3.2.3 Relative Strain Tensors

The reference configuration enjoys no particular mathematical status, although itmay have a physical significance (e.g., the stress-free state). Suppose the particle Xat time τ occupies the position ξ, then the relative deformation gradient is definedas

Ft (τ ) =(

∂ξ

∂x

)T

, (Ft (τ ))i j = ∂ξi

∂x j. (3.16)

Correspondingly, the right relative Cauchy–Green tensor is given by

Ct (τ ) = Ft (τ )T Ft (τ ) , (Ct (τ ))i j = ∂ξk

∂xi

∂ξk

∂x j. (3.17)

Similarly, the left Cauchy-Green tensor is given by

Bt (τ ) = Ft (τ )Ft (τ )T , (Bt (τ ))i j = ∂ξi∂xk

∂ξ j

∂xk. (3.18)

Because of the chain rule, the relative deformation gradient satisfies

Ft (τ ) =(

∂ξ

∂x

)T

=(

∂X∂x

· ∂ξ

∂X

)T

=(

∂ξ

∂X

)T

·(

∂X∂x

)T

= F (τ )F (t)−1 . (3.19)

3.2.4 Path Lines

To solve for the path lines of the particles, knowing the velocity field, we integratethe set of equations

dξ

dτ= u (ξ, τ ) , ξ (τ )|τ=t = x. (3.20)

When the flow is steady, the velocity field is independent of time, this system is anautonomous system.

3.2 Deformation Gradient and Strain Tensors 49

As an example, consider the case where the velocity is steady and homogeneous:

u (x) = Lx. (3.21)

The path lines are obtained by solving

x = Lx, x (0) = X. (3.22)

The solution to this isx = Φ (t)X, (3.23)

where Φ is called the fundamental matrix. It obeys

dΦ

dt= LΦ, Φ (0) = I. (3.24)

In fact, we find from (3.13) that F is indeed the fundamental matrix of system (3.22).The solution to (3.24) is

F (t) = exp (tL) , (3.25)

where the exponential function of a tensor is defined as [12]

exp (A) =∞∑

n=0

1

n!An = I + A + 1

2A2 + · · ·

Thus the path lines are all given by

ξ (τ ) = Φ (τ )X = Φ (τ )Φ (t)−1 x (t) ,

= e(τ−t)Lx. (3.26)

The problem of calculating the exponential of a constant tensor falls into three cate-gories, depending on the canonical form of the tensor, see Huilgol and Phan-Thien[39].

3.2.5 Oscillatory Shear Flow

We now consider the oscillatory shear flow, where the velocity depends explicitly ontime:

u = ωγa y cosωt, v = 0, w = 0. (3.27)


The equations to solve for the path lines are

x = ωγa y cosωt, y = 0, z = 0,x (0) = X, y (0) = Y, z (0) = Z .

Integrate these,

x (t) = X + γaY sinωt, y (t) = Y, z (t) = Z .

The path lines are then given by, note that ξ = (ξ, ψ, ζ),

ξ (τ ) = X + γaY sinωτ , ψ (τ ) = Y, ζ (τ ) = Z .

ξ (τ ) = x + γa y (sinωτ − sinωt) , ψ (τ ) = y, ζ (τ ) = z.(3.28)

Suppose the path lines have been determined, then the relative deformation gra-dient and the strain tensors may be calculated from (3.16) and (3.17). We illustratethis with the path lines for the simple shear flow (3.74) (Problem 3.2):

ξ= x + (τ − t)Lx,

which leads to the relative deformation gradient,

Ft (τ ) = (∇xξ)T = I + (τ − t)L,

from which the Cauchy–Green strain tensor can be calculated as

Ct (τ ) = Ft (τ )T Ft (τ ) = (I + (τ − t)L)T (I + (τ − t)L)

= I + (τ − t)(L + LT

) + (τ − t)2 LTL. (3.29)

3.3 Rivlin–Ericksen Tensors

Suppose the relative Cauchy–Green tensor has been determined. The n-th Rivlin–Ericksen tensor is defined as

An (t) = dn

dτ nCt (τ )

∣∣∣∣τ=t

, n = 1, 2, . . . (3.30)

Since Ct (τ )|τ=t = I, because Ft (t) = I, we may define

A0 = I, (3.31)

3.3 Rivlin–Ericksen Tensors 51

and extend the definition (3.30) to n = 0, 1, . . . In effect, the Rivlin–Ericksen tensorsare defined as the coefficients of the following Taylor series about t :

Ct (τ ) =∑

n=0

(τ − t)n

n! An (t). (3.32)

Rivlin–Ericksen tensors can be determined directly from the velocity field, withouthaving to find the strain tensor. This is shown below.

First, we note from (3.19),

d

dτFt (τ ) = d

dτ

[F (τ )F (t)−1

] = L (τ )F (τ )F (t)−1 = L (τ )Ft (τ ) .

Then,d

dτFt (τ )T = [L (τ )Ft (τ )]T = Ft (τ )T L (τ )T .

Therefore, from (3.17)

d

dτCt (τ ) = d

dτ

[Ft (τ )T Ft (τ )

]

= Ft (τ )T L (τ )T Ft (τ ) + Ft (τ )T L (τ )Ft (τ ) . (3.33)

When τ = t, the relative deformation gradient is the unit tensor, and we have

A1 (t) = d

dτCt (τ )

∣∣∣∣τ=t

= L (t) + L (t)T = 2D (t) , (3.34)

i.e., the first Rivlin–Ericksen tensor is twice the strain rate tensor. Higher-orderRivlin–Ericksen tensors can be obtained by taking derivatives of (3.33) repeatedly.However, it is more instructive to look at an alternative way of calculating the Rivlin–Ericksen tensors, which also reveals the nature of the tensors.

We start with the length square of a fluid element in the current time:

dx (t)2 = dx · dx = FTF : dXdX = C (t) : dXdX. (3.35)

Similarlydξ (τ )2 = C (τ ) : dXdX. (3.36)

Now, since

Ct (τ ) = Ft (τ )T Ft (τ ) = F (t)−T F (τ )T F (τ )F (t)−1

= F (t)−T C (τ )F (t)−1 ,


we havedn

dτ nCt (τ ) = F (t)−T dn

dτ nC (τ )F (t)−1 .

Thus

F (t)T[dn

dτ nCt (τ )

]F (t) = dn

dτ nC (τ ) .

Taking the scalar product of this with dXdX and τ = t, and recall (3.36) we obtain

dXTF (t)T[dn

dτ nCt (τ )

]F (t) dX =

[dn

dτ nC (τ )

]: dXdX

An : dxdx = dn

dτ ndξ (τ )2

∣∣∣∣τ=t

,

which relates the Rivlin–Ericksen tensors to the high-order stretching rate of a fluidelement. For example,

A1 : dxdx = d

dtdξ (t)2 . (3.37)

A recursive relation can be derived by noting that

An+1 : dxdx = d

dt

(dn

dtndξ (t)2

)

= d

dt(An : dxdx)

=(d

dtAn : dxdx + An : d

dt(dx) dx + An : dx d

dt(dx)

).

Butd

dt(dx) = d

dt(FdX) = LFdX = Ldx,

and therefore

An+1 : dxdx =(d

dtAn + AnL + LTAn

): dxdx,

which leads to the recursive formula due to Rivlin and Ericksen [78]

An+1 = d

dtAn + AnL + LTAn, A0 = I, n = 1, 2, . . . (3.38)

3.3 Rivlin–Ericksen Tensors 53

As an example, let’s calculate the Rivlin–Ericksen tensors for the simple shearflow (3.72). The first Rivlin–Ericksen tensor is twice the strain rate tensor:

[A1] =⎡

⎣0 γ 0γ 0 00 0 0

⎤

⎦ .

The second Rivlin–Ericksen tensor is obtained from the first using (3.38),

A2 = A1L + LTA1

=⎡

⎣0 γ 0γ 0 00 0 0

⎤

⎦

⎡

⎣0 γ 00 0 00 0 0

⎤

⎦ +⎡

⎣0 0 0γ 0 00 0 0

⎤

⎦

⎡

⎣0 γ 0γ 0 00 0 0

⎤

⎦ =⎡

⎣0 0 00 2γ2 00 0 0

⎤

⎦ .

All other higher-order Rivlin–Ericksen tensors are zero for this flow. In fact, in thisflow L2 = 0, and it can be verified from (3.29) that

Ct (τ ) = I + (τ − t)A1 + 1

2(τ − t)2 A2. (3.39)

3.4 Small Strain

When the strain is small, in the sense that the fluid particles remain close to theiroriginal positions in the reference configuration at all times, then the strain may becalculated by introducing the displacement function v:

v = x (X, t) − X. (3.40)

The deformation gradient is

F (t) = (∇Xx)T = I + E (t) , E (t) = (∇Xv)T . (3.41)

When the displacement gradient is small, terms of order ε2 = O(‖E‖2) and higher

can be neglected, one has

F (t)−1 = I − E (t) ,

C (t) = I + E (t) + E (t)T ,

Ft (τ ) = F (τ )F (t)−1 = I + E (τ ) − E (t) ,

Ct (τ ) = Ft (τ )T Ft (τ ) = I + E (τ ) + E (τ )T − E (t) − E (t)T . (3.42)


In terms of the infinitesimal strain tensor

ε (t) = 1

2

(E (t)+E (t) T

), (3.43)

we haveC (t) = I + 2ε (t) , Ct (τ ) = I + 2ε (τ ) − 2ε (t) . (3.44)

In the right polar decomposition of F, F = RU, U is the square root of C,

U = C1/2 = I + ε. (3.45)

ThusR = FU−1 = (I + E) (I − ε) = I + E − ε = I + ω,

where ω is the infinitesimal rotation tensor:

ω = 1

2

(E − ET

). (3.46)

3.5 Equations of Balance

The equations of balance are mathematical statements of the conservation of mass,linear and angular momentum, and energy.

3.5.1 Reynolds Transport Theorem

Theorem 1 Let Φ (x, t) be a field (scalar, vector or tensor) defined over a region Voccupied by the body B at time t . The Reynolds transport theorem states that (for aportray of Reynolds, see Fig. 3.5)

d

dt

∫

VΦdV =

∫

V

(dΦ

dt+ Φ∇ · u

)dV =

∫

V

(∂Φ

∂t+ ∇ · (Φu)

)dV (3.47)

where u is the velocity field, and d/dt is the material derivative (3.7). This is provedby expressing the volume integral in the reference configuration,

d

dt

∫

VΦdx = d

dt

∫

V0

Φ JdX, (3.48)

3.5 Equations of Balance 55

Fig. 3.5 Osborne Reynolds(1842–1912) introduced thelubrication theory andformulated the frameworkfor turbulence flow. He wasthe first Professor in the UKuniversity system to hold thetitle Professor ofEngineering at OwensCollege (now University ofManchester). The Reynoldsnumber and stresses arenamed after him

where J = det (∂x/∂X) = det F is the Jacobian of the transformation (we use dxinterchangeably with dV ), and V0 is the region occupied by the reference configu-ration.

Lemma We record the following Lemma

d J

dt= J∇ · u. (3.49)

This is proved by using a result obtained previously, Sect. 1.8.1, or by noting that

d J = det (F + dF) − det (F) = det[F

(I + F−1dF

)] − det F

= det F det(I + F−1dF

) − det F

= det F(1 + tr

(F−1dF

)) − det F

= J tr(F−1dF

).

Divide both sides by dt , and use F = LF, and note that

tr(F−1LF

) = F−1i j L jk Fki = δk j L jk = Lkk = ∇ · u

leading to the lemma (3.49). This lemma can be used directly in (3.48) to prove(3.47).

Another form for the Reynolds transport theorem which emphasises the flux ofΦ into the volume V bounded by the surface S is given below by recognizing that

http://dx.doi.org/10.1007/978-3-319-62000-8_1


∫

V

(dΦ

dt+ Φ∇ · u

)dV =

∫

V

(∂Φ

∂t+ u · ∇Φ + Φ∇ · u

)dV

=∫

V

(∂Φ

∂t+ ∇ · (Φu)

)dV

=∫

V

∂Φ

∂tdV +

∫

S(Φu) · ndS

Theorem 2

d

dt

∫

VΦdx =

∫

V

∂Φ

∂tdV +

∫

S(Φu) · ndS. (3.50)

This form allows a physical interpretation of the theorem: the first term on the rightrepresents the rate of creation of the quantity Φ, and the second term, the flux of Φ

into the volume V through its bounding surface.

3.5.2 Conservation of Mass

The mass in the volume V is conserved at all time, i.e.,

d

dt

∫

VρdV = 0,

where ρ (x, t) is the density field at time t . From Reynolds transport theorem (3.47),

∫

V

(∂ρ

∂t+ ∇ · (ρu)

)dV = 0.

Since the volume V is arbitrary, a necessary and sufficient condition for the conser-vation of mass is

∂ρ

∂t+ ∇ · (ρu) = dρ

dt+ ρ∇ · u = 0. (3.51)

For an incompressible material, the density is constant everywhere, and the conser-vation of mass demands that

∇ · u = ∂ui∂xi

= trL = trD = 1

2trA1 = 0. (3.52)

From a solid point of view, the conservation of mass requires that

ρdV = ρRdVR ρR = ρJ,


where J = det F is the Jacobian of the deformation. For an incompressible material,we have

J = det F = 1, detC = 1, (3.53)

at all time.

Theorem 3 As a corollary to (3.47) and (3.51) we have

d

dt

∫

VρΦdV =

∫

VρdΦ

dtdV . (3.54)

This is easily demonstrated by using (3.47) and (3.51) on the left hand side:

d

dt

∫

VρΦdV =

∫

V

(Φ

(dρ

dt+ ρ∇ · u

)+ ρ

dΦ

dt

)dV .

3.5.3 Conservation of Momentum

The forces acting on the body are surface forces t (tractions), and body forces b(those that act at a distance).

Body Force Density. An example of body force is gravitational. If b (x, t) is thebody force density defined on V, then the resulting force and moment (about a fixedpoint O) on V due to the body force field are given respectively by

∫

VρbdV,

∫

Vx × bdV .

Surface Force. Surface traction is a concept due to Cauchy. Consider a particle Xoccupying the position x at time t . Construct a surface St through this point withunit normal vector n (x, t) at point x, which separates the body into two regions:B+ is the region into which the unit normal n is directed and B− on the other side(see Fig. 3.6). t (x, t;n) is called the surface force density per unit (current) area ifthe force and moment (about O) exerted on B− by B+ are given respectively by

Fig. 3.6 The tractiont(x, t; n) is the force per unitarea exerted by B+ on B−


∫

St

t (x, t) dS,

∫

St

x × t (x, t) dS.

Balance of Linear Momentum. Newton’s second law (a postulate), as applied to avolume V occupied by the body, requires that

d

dt

∫

VρudV =

∫

StdS +

∫

VρbdV . (3.55)

Here S is the bounding surface of V . The first term on the right hand of (3.55)represents the surface force acting on V due the body outside V, and the second termis the net body force on V .

Corollary (3.54) can be used on the left side of the preceding equation, leading to

∫

Vρdudt

dV =∫

StdS +

∫

VρbdV . (3.56)

Balance of Angular Momentum. The balance of angular momentum (a postulate)can likewise be written down as, in the absence of body couple,

d

dt

∫

Vx × ρudV =

∫

Sx × tdS +

∫

Vx × ρbdV . (3.57)

Again, corollary (3.54) can be used on the left side of the preceding equation, leadingto ∫

Vx × ρ

dudt

dV =∫

Sx × tdS +

∫

Vx × ρbdV . (3.58)

Note that the term involving dx/dt = u does not contribute to the left hand sideof this because of the definition of the cross product. This should be contrasted tothe rigid body mechanics case where Newton’s second law is a postulate and thebalance of angular momentum is a consequence (a theorem). Here two postulatesare required (Newton’s 2nd law and the balance of angular momentum).

The term

a = dudt

= ∂u∂t

+ u · ∇u (3.59)

is recognised as the acceleration field. To convert the first term on the right into avolume integral, we need the concept of the stress tensorT(x;n), also due to Cauchy.

Cauchy Stress Tensor. The existence of the Cauchy stress tensor is guaranteed bythe following theorem.

• The traction vector (force per unit area) satisfies

t (x, t;−n) = −t (x, t;n) . (3.60)


Fig. 3.7 Existence of thestress tensor

• Further, there exists a second-order tensor field T (x, t) with the following prop-erties:

t (x, t;n) = T (x, t) · n, (3.61)

with components in the frame F = {e1, e2, e3} given by

Ti j (x, t) = t(x, t; e j

) · ei . (3.62)

The proof, due to Gurtin [34], lies in the construction of a one-parameter familyof tetrahedra, Fig. 3.7, with vertex at point x, height h (the parameter). The facenormal to n has area A; the face normal to −ei has area Ai . From the directionalcosine of n

Ai (h) = A(h)ni .

Furthermore the volume of the tetrahedron is Vh = 13hA(h).

Applying (3.56) to the tetrahedron (omitting the time argument t for brevity):

∫

Vh

ρa (y) dV (y) =∫

Vh

ρb (y) dV (y) +∫

A(h)

t (y;n) dS (y)

+3∑

j=1

∫

Ai (h)

t(y;−e j

)dS (y) .

From the continuity of all the field variables, and the mean-value theorem,

(ρa (x) + μ) Vh = (ρb (x) + α) Vh + (t (x;n) + β) A (h)

+3∑

j=1

[t(x;−e j

) + β j

]A (h) n j , (3.63)


where α (h) ,β (h) ,β j (h) ,μ (h) = o (1) , h → 0. Divide (3.63) by A (h) and leth → 0, we find

t (x;n) = −3∑

j=1

t(x;−e j

)n j . (3.64)

In the case where n = ei (fixed i)

t (x;−ei ) = −t (x; ei ) ,

which leads to (3.60) due the arbitrary configuration of the frame of reference. Thus

t (x;n) =3∑

j=1

t(x; e j

)n j . (3.65)

From (3.65), the components of t are given by

ti (x;n) = t (x;n) · ei =3∑

j=1

t(x; e j

) · ei n j

= Ti j (x) n j , (3.66)

where Ti j are defined as in (3.62). From the quotient rule (§1.3.10), Ti j are indeed thecomponents of a second-order tensor thus proving the existence of the stress tensor(3.61).

Conservation of LinearMomentum. Returning to the balance of linear momentum(3.56) and using the definition of the stress tensor (3.61), we have

∫

Vρdudt

dV =∫

ST · ndS +

∫

VρbdV ,

∫

Vρduidt

dV =∫

STi j n j dS +

∫

VρbidV .

The surface integral on the right hand of the preceding equation can be convertedinto a volume integral to obtain

∫

Vρdudt

dV =∫

V∇TT dV +

∫

VρbdV,

∫

Vρduidt

dV =∫

S

∂Ti j∂x j

dS +∫

VρbidV .


Since the integrand is continuouson an arbitraryV, the conservationof linearmomen-tum becomes

ρdudt

= ∇TT + ρb, ρduidt

= ∂Ti j∂x j

+ ρbi . (3.67)

Conservation of AngularMomentum. Returning to the balance of angular momen-tum (3.58), ∫

Vx × ρ

dudt

dV =∫

Sx × tdS +

∫

Vx × ρbdV

and examining the first term on the right,

∫

Sεi jk x j tkdS =

∫

Sεi jk x j TklnldS =

∫

V

∂

∂xl

(εi jk x j Tkl

)dV

=∫

V

(εi jkδ jl Tkl + εi jk x j

∂Tkl∂xl

)dV

=∫

V

(t Ai + εi jk x j

∂Tkl∂xl

)dV ,

where the “axial vector” is defined as t Ai = εi jkTk j .When these results are substitutedback into (3.58),

∫

Vx × ρ

dudt

dV =∫

V

(tA + x × ∇ · TT

)dV +

∫

Vx × ρbdV

∫

VtAdV =

∫

Vx ×

[ρdudt

− ∇ · TT − ρb]dV

From the conservation of linear momentum, and the continuity of the integrands inthe arbitrary volume V , it follows that

tA = 0 εi jkTk j = 0 T = TT or T is symmetric. (3.68)

A necessary and sufficient condition for the balance of angular momentum, in theabsence of body couples, is that the stress be symmetric. In deriving (3.68), both theconservation of mass and linear momentum are needed.

3.5.4 Conservation of Energy

We define the kinetic energy K , and internal energy E as

K =∫

V

1

2ρu2dV , E =

∫

VρεdV , (3.69)


where ε is the specific internal energy per unit mass. The rate of work done on thebody due to surface and body forces is given by

∫

St · udS +

∫

Vρb · udV .

If we define q to be the flux of energy out of S per unit area, and r the amount ofenergy created per unit mass, then the mathematical statement for the first law ofthermodynamics can be expressed as

d

dt

∫

V

(1

2ρu2 + ρε

)dV =

∫

S(t · u − q · n) dS

+∫

V(ρr + ρb · u) dV . (3.70)

The left of (3.70) can be expressed as, using Reynolds transport theorem (3.54),

d

dt

∫

V

(1

2ρu2 + ρε

)dV =

∫

Vρ (u · a + ε) dV .

The surface integral on the right of (3.70) is converted into volume integral as

∫

S(t · u − q · n) dS =

∫

V

(∂

∂x j

(ui Ti j

) − ∂qi∂xi

)dV

=∫

V

(u · ∇TT + T : L − ∇ · q)

dV .

Hence (3.70) becomes

∫

Vρ (u · a + ε) dV =

∫

V

(u · ∇TT + T : L − ∇ · q)

dV

+∫

V(ρr + ρb · u) dV .

Because of the conservation of linearmomentum, and the continuity of the integrandsin the arbitrary volume V , the conservation of energy is reduced to

ρε = T : D − ∇ · q + ρr. (3.71)

In deriving this, all the three balance equations for mass, linear momentum andangular momentum are required. The term T : D represents the rate of work done bythe stress, or the “stress power”. It is seen that

T : D = −PtrD + S : D


The rate of work done by the pressure for an incompressible fluid is zero, becausetrD = 0.

Problems

Problem 3.1 Using FF−1 = I for a deformation gradient F, show that

d

dtF−1 = −F−1L, F−1 (0) = I.

Problem 3.2 For a simple shear flow, where the velocity field takes the form

u = γy, v = 0, w = 0, (3.72)

show that the velocity gradient and its exponent are given by

[L] =⎡

⎣0 γ 00 0 00 0 0

⎤

⎦ , exp (L) = I + L. (3.73)

Show that the path lines are given by

ξ (τ ) = x + (τ − t)Lx. (3.74)

so that a fluid element dX can only be stretched linearly in time at most.

Problem 3.3 Repeat the same exercise for an elongational flow, where

u = ax, v = by, w = cz, a + b + c = 0. (3.75)

In this case, show that

[L] =⎡

⎣a 0 00 b 00 0 c

⎤

⎦ ,[eL

] =⎡

⎣ea 0 00 eb 00 0 ec

⎤

⎦ . (3.76)

Show that the path lines are given by

[ξ (τ )] =⎡

⎣ξψζ

⎤

⎦ =⎡

⎣ea(τ−t) 0 0

0 eb(τ−t) 00 0 ec(τ−t)

⎤

⎦

⎡

⎣xyz

⎤

⎦ . (3.77)

Conclude that exponential flow can stretch the fluid element exponentially fast.


Problem 3.4 Consider a super-imposed oscillatory shear flow:

u = γm y, v = 0, w = ωγa y cosωt.

Show that the path lines are

ξ (τ ) = x + γm (τ − t) y,

ψ (τ ) = y, (3.78)

ζ (τ ) = z + γ0y (sinωτ − sinωt) .

Problem 3.5 Calculate the Rivlin–Ericksen tensors for the elongational flow (3.75).

Problem 3.6 Calculate the Rivlin–Ericksen tensors for the unsteady flow (3.78).

Problem 3.7 Write down, in component forms the conservation of mass and linearmomentum equations, assuming the fluid is incompressible, in Cartesian, cylindricaland spherical coordinate systems.

Chapter 4Constitutive Equation: General Principles

Basic Principles and Some Classical ConstitutiveEquations

In isothermal flow where the conservation of energy is not relevant, there are fourscalar balance equations (one conservation of mass and three conservation of linearmomentum), and there are 10 scalar variables (3 velocity components, one pres-sure, and 6 independent stress components – thanks to the conservation of angularmomentum, the stress tensor is symmetric). Clearly we do not have a mathematicallywell-posed problem until 6 extra equations are specified. The constitutive equation,or the rheological equation of state, provides the linkage between the stresses andthe kinematics and provides the missing information. Modelling a complex fluid, orfinding a relevant constitutive equation for the fluid, is the central concern in rhe-ology. In this chapter, we review some of the well-known classical models, and thegeneral principles underlying constitutive modelling.

4.1 Some Well-Known Constitutive Equations

4.1.1 Perfect Gas

The most well known constitutive equation is the perfect gas law, due to Boyle(Fig. 4.1), where the state of the gas is fully specified by its volume V , its pressureP and its temperature T

PV = RT, (4.1)

where R is a universal gas constant.


65

66 4 Constitutive Equation: General Principles

Fig. 4.1 Robert Boyle(1627–1691) made severalimportant contributions toPhysics and Chemistry, thebest known is the PerfectGas Law. He employedRobert Hooke as his assistantin the investigation of thebehaviour of air. Hisexperiments led him tobelieve in vacuum, and rejectDescartes’ concept of ether

Fig. 4.2 Jean d’Alembert(1717–1783) was a Frenchmathematician. He madeseveral importantcontributions to mechanics.He is most famous for theD’Alembert principle

4.1.2 Inviscid Fluid

The perfect fluid concept of D’Alembert (Fig. 4.2) and Euler (Fig. 3.2) is anotherwell-known constitutive equation. In our notation, the stress is given by

T = −PI, Ti j = −Pδi j . (4.2)

Here, P is the pressure. The inviscid fluid model fails to account for the pressurelosses in pipe flow, and a better model is needed.

http://dx.doi.org/10.1007/978-3-319-62000-8_3

4.1 Some Well-Known Constitutive Equations 67

Fig. 4.3 Joseph Fourier(1768–1830) was a Frenchhistorian, administrator andmathematician. He wasfamous for his Fourier’sseries. His name is one of the72 inscribed on the Eiffeltower

4.1.3 Fourier’s Law

Students of thermodynamics would recognise Fourier’s law of heat conduction (Fig.4.3), linking the heat transfer rate q to the temperature gradient:

q = −k∇θ, qi = −k∂θ

∂xi, (4.3)

where k is the thermal conductivity and θ is the temperature field.

4.1.4 Hookean Solid

In 1678 Robert Hooke published his now famous law for material behaviour as asolution to an anagram that he published two years earlier, ut tensio sic vis, whichroughly translated as extension is proportional to the force (see Fig. 4.4). This idea hasgone through several revisions by several well-known scientists, including Young,Poisson, and Navier (Fig. 4.5), who thought that one needs only one elastic constant(the other constant, the Poisson’s ratio was thought to be 0.25). The concept of thestress tensor was introduced by Cauchy (Fig. 3.4), who also gave the correct versionof the constitutive equation for infinitesimal elasticity in 1827. In our notation, thestress tensor is given by

T = C · ε, Ti j = Ci jklεkl, (4.4)

http://dx.doi.org/10.1007/978-3-319-62000-8_3


Fig. 4.4 Robert Hooke (1635–1703) was perhaps the foremost experimental scientist in the 17thcentury. He was a noted architect, an inventor (Hooke universal joint, spring control in watches,reflecting telescope, etc.), a mathematician, a physicist, a chemist and contributed significantly toanatomy, astronomy, botany, chemistry – the term “cell” is due to him. No portrait survived him.He is most well-known for the law of elasticity that bears his name

where

ε = 1

2

(∇v + ∇vT), εi j = 1

2

(∂vi

∂x j+ ∂v j

∂xi

)(4.5)

is the infinitesimal strain tensor, v = x − X is the displacement vector, C is a 4thorder elasticity tensor, a tensorial material constant, which reflects all the anisotropyof the material. For an isotropic solid, the elastic constants are reduced to just two:


Fig. 4.5 C.L.M.H Navier(1785–1836), a Frenchengineer, developed aparticle model for an elasticsolid, which has a shearmodulus and a Poisson’sratio of 0.25 (see Love [54]).He obtained theNavier–Stokes equations bymolecular arguments. Hisname is one of the 72inscribed on the Eiffel tower

Ci jkl = λδi jδkl + μ(δikδ jl + δilδ jk

), (4.6)

where λ and μ are called Lamé moduli. Using this in (4.4) produces

T = λ (trε) I + 2με, Ti j = λεkkδi j + 2μεi j . (4.7)

An alternative form for this can be derived by first taking the traces of both sides of(4.7)

trT = (3λ + 2μ) trε,

and then substituting this back into (4.7) to obtain

ε = 1

2μT − λ

2μ (3λ + 2μ)(trT) I, εi j = 1

2μTi j − λ

2μ (3λ + 2μ)Tkkδi j . (4.8)

4.1.5 Newtonian Fluid

About nine years after Hooke published his paper on elasticity, Newton (Fig. 4.6)introduced the concept of “lack of slipperiness”, which is the important quantity thatwe now call viscosity. Then Navier, in 1827, derived the Navier–Stokes equation,

ρdudt

= −∇P + η∇2u + ρb. (4.9)

In this equation, the terms η∇2u represent the viscous forces, although Navier didnot attach much physical significance to η. Stokes (Fig. 4.7) gave the correct form


Fig. 4.6 Sir Isaac Newton(1643–1727) was adominating personality inScience. He (andconcurrently Leibnitz)invented differential andintegral calculus, and thegravitational theory. He wasappointed the LucasianProfessor at Cambridge atthe age of 26

Fig. 4.7 George GabrielleStokes (1819–1903) was aIrish mathematician. He wasappointed the LucasianProfessor at Cambridge atthe age of 30. He isremembered for Stokes flowand his contributions in theNavier–Stokes equations

to the constitutive equation that we call a Newtonian fluid:

T = −PI + Λ(trD) I + 2ηD, Ti j = −Pδi j + ΛDkkδi j + 2ηDi j . (4.10)

Here, η is the viscosity, Λ is the bulk viscosity, and D = (∇u+∇uT )/2 is the strainrate tensor. Stokes assumed that

Λ = −2

3η,

so that pure volumetric change does not affect the stress (tr T is independent of tr D).Furthermore, the terms ΛtrDδi j can be absorbed in the pressure term. This leads to


the familiar constitutive equation for Newtonian fluids:

T = −PI + 2ηD. (4.11)

4.1.6 Non-Newtonian Fluid

The term non-Newtonian fluid is an all-encompassing term denoting any fluid thatdoes not obey (4.11). To discuss constitutive relations for non-Newtonian fluids, weneed a convenient way to classify different flow regimes.

4.2 Weissenberg and Deborah Numbers

Most non-Newtonian fluids have a characteristic time scale λ. In a flow with acharacteristic shear rate γ and a characteristic frequency ω, or characteristic time T ,two dimensionless groups can be formed

Deborah number De = λω or λ/T,

Weissenberg number Wi = λγ(4.12)

4.2.1 Deborah Number

The Deborah number,1 the ratio between the fluid relaxation time and the flow char-acteristic time, represents the transient nature of the flow relative to the fluid timescale. If the observation time scale is large (small De number), the material respondslike a fluid, and if it is small (large De number), we have a solid-like response. Underthis viewpoint, there is no fundamental difference between solids and liquids; it isonly a matter of time scale of observation. In the limit, when De = 0 one has aNewtonian liquid, and when De = ∞, an elastic solid.

4.2.2 Weissenberg Number

The Weissenberg number is the ratio of elastic to viscous forces. It has been var-iously defined, but usually as given in (4.12). Thus one can have a flow with asmall Wi number and a large De number, and vice versa. We expect a significantnon-Newtonian behaviour in a large Wi number flow, and therefore the constitutive

1The terminology is due to M. Reiner.


Fig. 4.8 Pipkin diagramdelineates different flowregimes, and relevantconstitutive equations [72]

equation must contain the relevant non-Newtonian physics. A different definition ofthe Weissenberg number is explored in Problem 4.3.

Pipkin’s diagram (Pipkin and Tanner [72]) helps guide the choice of constitutiveequations. In Fig. 4.8, the vertical axis represents the Weissenberg number and thehorizontal axis the Deborah number. Newtonian response is represented by a singlepoint, at De = 0 = Wi. Elastic response is also represented by a single point, atDe = ∞. Nearly steady flows, at low De numbers, can be analysed by assumingviscometric motion (order fluids), and fast flows, at large De numbers, a rubber-likeresponse is expected - the relevant constitutive equation here is rubber-like elasticity.The region at low Wi numbers (at small strain amplitudes) can be handled by alinear viscoelastic model. The large domain in the middle of the diagram, markednon-linear viscoelasticity, is the constitutive modeller’s haven!

4.3 Some Guidelines in Constitutive Modelling

There are two alternatives for constitutive modelling: the continuum approach andthe microstructure approach. In the continuum approach, the material is assumed tobe a continuum, with no micro-inertial feature. The relevant variables are identified,and are related in a framework that ensured invariance under a change of frames.Different restrictions are then imposed to simplify the constitutive equation as far aspracticable. In the microstructure approach, a physical model of the microstructurerepresenting the material is postulated. Solving the deformation at that level usingwell-tested physical principles (Newton’s laws, conservation laws, etc.) allows theaverage stress and strain to be related producing a constitutive equation. In the con-tinuum approach one is usually left with a general constitutive equation, which mayhave some undetermined functions or functionals (loosely speaking, functionals arefunctions of functions). The details of these functions or functionalsmay be furnishedby relevant experiments. In the microstructure approach, the constitutive equationstend to be more specific and therefore more relevant to the material in question.

4.3 Some Guidelines in Constitutive Modelling 73

In the mid 1950s, there were some intense activities in setting up a rigoroustheoretical framework for continuum mechanics. Everything possible is set up in anaxiomatic format. This has been good in focusing on what is permissible. However,it has the unfortunate consequence that it leaves the students with the impressionthat all one needs is a set of relevant variables and some principles – and that wouldallows us to construct a general constitutive equation for any material.

It is generally believed that relevant constitutive equations should be based on a(simplified)model of themicrostructure.When the physics governing themicrostruc-ture interactions are complicated, one must not hesitate to introduce elements ofcontinuum modelling, but the continuum approach should not completely replacethe microstructure modelling.

4.3.1 Oldroyd Approach

The basic ideas behind constitutive modelling of finite deformation were well under-stood in the early 1950s, but these ideas have not been extended to all continuousmaterials undergoing large deformation. It wasOldroyd2 [64]who clearly enunciatedthat a constitutive equation must be based on

• the relative motion of the neighbourhood of a particle;• the history of the metric tensor (i.e., strain tensor) associated with the particle;• the convected coordinate system embedded in the material and deforming with it;• and the physical constants defining the symmetry of the material.

It was unfortunate that his work has been grossly overlooked, see Tanner andWalters [86] for an interesting historical account. The later influential work of Noll[61, 62] put these ideas in an axiomatic form that is elegant and appealing to thegeneration of graduate students at that time. We will go through the principles asdetailed by Noll, but always keep in mind the relevance of microstructure models.

4.3.2 Principle of Material Objectivity

Consider a change of frame

x′ = c (t) + Q (t) x, (4.13)

which consists of a spatial translation (by c) and a rotation (through an orthogonaltensor Q). A physical quantity is said to be objective, or frame-invariant, when it is

2James G. Oldroyd (1921–1982) was a Professor in Applied Mathematics at Universities of Walesand Liverpool. He made several important contributions to the constitutive equation formulation.The Oldroyd fluids (fluid A and fluid B) were named after him.


invariant under the transformation (4.13). Specifically, using the prime to denote thequantity in the new frame,

• a scalar φ is invariant when its value remains unchanged under a change of frame(4.13)

φ′ = φ, (4.14)

• a vector u is invariant when it transforms under a change of frame (4.13) accordingto

u′ = Q (t)u, (4.15)

• a tensorT is invariant when it transforms under a change of frame (4.13) accordingto

T′ = Q (t)TQ (t)T . (4.16)

Note that only Q (t) is involved in this transformation.

4.3.3 Objectivity of the Stress

The principle of material objectivity asserts that the stress tensor must be objectiveunder a change of frame (4.13).

This principle pre-supposes that the material has no inertial feature at themicroscale, that is, it is a continuous media. With micro-inertia, such as suspen-sions of microsized particles of a different density than the fluid, there will be acomponent of stress due to the particle inertia that is not objective. This is a “princi-ple” for inertialess microstructure only. Indeed, if a microstructural model violatesobjectivity, the reason can always be found in the physics of the microstructure –if the physics are sound, there may be a very good reason for the stress not beingobjective (Ryskin and Rallison [79]).

4.3.4 Frame Indifference

This principle enunciates that one does not obtain a new constitutive equation everytime there is a change in frame of reference: the constitutive operator is the same forall observers in relative motion. The objectivity and frame indifference principlesroughly correspond to Oldroyd’s third point.

There are kinematic quantities that are not objective. We give a few exampleshere.


Deformation Gradient Tensor. Consider the deformation gradient tensor,

F (t) =[∂x (t)

∂X

]T

, Fi j = ∂xi∂X j

.

Under the change of frame (4.13),

F ′i j = ∂x ′

i

∂X j= ∂x ′

i

∂xk

∂xk∂X j

= Qik Fkj , F′ = QF, (4.17)

i.e., F is not frame-invariant. The relative deformation gradient, Ft (τ ) = F (τ )

F−1 (t) , is not frame-invariant either,

F′t (τ ) = F′ (τ )F′ (t)−1 = Q (τ )F (τ ) [Q (t)F (t)]−1

= Q (τ )F (τ )F (t)−1 Q (t)T (4.18)

= Q (τ )Ft (τ )Q (t)T ,

since both Q (τ ) and Q (t) are involved in this transformation.Cauchy–Green Tensor. The left Cauchy–Green tensor is objective, but the rightCauchy–Green tensor is not:

B′ = F′F′T = QF (QF)T = QFFTQT = QBQT ,

C′ = F′TF′ = (QF)T QF = FQTQF = C.(4.19)

The left relative Cauchy–Green tensor is not objective (note the argument of Q):

B′t (τ ) = F′

t (τ )F′t (τ )T = Q (τ )Ft (τ )Q (t)T

(Q (τ )Ft (τ )Q (t)T

)T

= Q (τ )Ft (τ )Q (t)T Q (t)Ft (τ )T Q (τ )T (4.20)

= Q (τ )Bt (τ )Q (τ )T ,

but the right relative Cauchy–Green tensor is,

C′t (τ ) = F′

t (τ )T F′t (τ ) = (

Q (τ )Ft (τ )Q (t)T)T

Q (τ )Ft (τ )Q (t)T

= Q (t)Ft (τ )T Q (τ )T Q (τ )Ft (τ )Q (t) (4.21)

= Q (t)Ct (τ )Q (t)T .

Velocity Gradient. The velocity gradient is not objective. To see this, we note thatthe motion is transformed according to,

M′ (X, t) = c (t) + Q (t)M (X, t) , (4.22)


since themotion is given by x (t) = M (X, t) .Thus the velocity transforms accordingto

u′ (X, t) = c (t) + Q (t) u (X, t) + Q (t) x (X, t) . (4.23)

Expressing this in the Eulerian sense

u′ (x, t) = c (t) + Q (t)u (x, t) + Q (t) x. (4.24)

The velocity gradient thus transforms accordingly

L ′i j = ∂u′

i

∂x ′j

= ∂u′i

∂xk

∂xk∂x ′

j

=(Qil

∂ul∂xk

+ Qik

)QT

kj ,

that is,

L′ = QLQT + QQT . (4.25)

Since QQT = I, QQT + QQT = 0, and QQT = −QQT = − (QQT

)Tis anti-

symmetric, and thus the symmetric part of (4.25), or the strain rate tensor, is objectivewhile the anti-symmetric part of (4.25), or the vorticity tensor, is not:

D′ = QDQT ,

W′ = QWQT + QQT .(4.26)

Rivlin–Ericksen Tensors. All the Rivlin–Ericksen tensors are objective. We havejust seen that the first Rivlin–Ericksen tensor is objective,

A′1 = QA1Q

T . (4.27)

Let all the Rivlin–Ericksen tensors up to order 1 ≤ n be objective. The next Rivlin–Ericksen tensor of order n + 1 transforms according to

A′n+1 = d

dtA′

n + A′nL′ + L′ TA′

n

= d

dt

(QAnQ

T) + QAnQ

T(QLQT + QQT

)

+ (QLQT + QQT

)TQAnQ

T

= Q[d

dtAn + AnL + LTAn

]QT

+ QAnQT + QAnQT + QAnQ

T QQT + QQTQAnQT .

Since QQT = I,


Q = −QQTQ, QT = −QT QQT ,

and we find that

A′n+1 = Q

[d

dtAn + AnL + LTAn

]QT , (4.28)

i.e., the (n + 1)th-order Rivlin–Ericksen tensor is also objective. By the process ofinduction, all the Rivlin–Ericksen tensors are objective.

4.3.5 Principle of Local Action

The principle of local action embodies the idea that only particles near a point shouldbe involved in determining the stress at that point. This is consistent with the exclu-sion of long-range forces, which have already been included in body forces. This isOldroyd’s first point.

4.3.6 Principle of Determinism

This principle states the obvious, that the current stress state in the material is deter-mined by the past history of the motion. Future state of the motion has no say in thecurrent state of the stress; i.e., the material possesses no clairvoyance.

In addition to these principles, there may be restrictions imposed on the constitu-tive equation because of the symmetry of the material. These symmetry restrictionsare discussed separately. This is satisfied by Oldroyd’s second and third points.

4.4 Integrity Bases

4.4.1 Isotropic Scalar-Valued Functions

Consider a scalar-valued function of a vectoru, and suppose that this function satisfies

f (u) = f (Qu) , (4.29)

for every orthogonal tensor Q. Such a function is called isotropic. We are interestedin how f depends on u. The answer is simple because only the magnitude of u isinvariant under every orthogonal tensor. Thus f must be a function of u = |u| . Inthat sense, u = |u| is called an integrity basis for f (u) .


Likewise, a scalar-valued function of two vectors u and v satisfying

f (u, v) = f (Qu,Qv) , (4.30)

for every orthogonal Q, is called an isotropic function of its two arguments. Since

u · u, v · v, u · v (4.31)

are the only three invariants under rotation (their lengths and the angle between themare invariant, Weyl [90]),

f (u, v) = f (u · u, v · v,u · v) . (4.32)

The three scalar invariants form the integrity basis for f (u, v).Similarly, f is said to be an isotropic function of a tensor S if for every orthogo-

nal Q,f (S) = f

(QSQT

). (4.33)

Here, f must be a function of the three invariants of S,

trS, trS2, trS3, (4.34)

or, equivalently, of the three eigenvalues ofS. These three invariants form the integritybases for f (S).

The invariants that can be formed from two tensors A and B are

trA, trA2, trA3, trB, trB2, trB3,

trAB, trA2B, trAB2, trA2B2.(4.35)

Thus, an isotropic scalar-valued function of A and B must be a function of these 10invariants. These form the integrity basis for f (A,B).

We list another integrity basis for a scalar valued, isotropic function of a symmetricsecond order tensor S and two vectors u and v:

tr S, tr S2, tr S3, u · u, u · v, v · v, u · Su, u · S2u,

v · Sv, v · S2v, u · Sv, u · S2v. (4.36)

4.4.2 Isotropic Vector-Valued Functions

Suppose that w = g (v) is a vector-valued function of the vector v. It is calledisotropic if, for every orthogonal tensor Q,

4.4 Integrity Bases 79

Qg (v) = g (Qv) . (4.37)

Now, define a scalar-valued function of two vectors u and v through

f (u, v) = u · g (v) .

Thus, since g is isotropic,

f (Qu,Qv) = Qu · g (Qv) = Qu · Qg (v) = u · g (v) ,

since Qu · Qg = Qi ju j Qikgk = δ jku j gk = u · g. Consequently, f is an isotropicfunction of its two arguments, and thus it is a function of the invariants listed in (4.31).However, by its definition, f (u, v) is linear in its first argument, and therefore

f (u, v) = u · h (v · v) v.

It follows that

g (v) = h (v · v) v. (4.38)

4.4.3 Isotropic Tensor-Valued Functions

A symmetric tensor-valued function of a symmetric tensor B is isotropic if

QG (B)QT = G(QBQT

), (4.39)

for every orthogonal tensor Q.Now, define a scalar function of two symmetric tensors through

f (A,B) = tr [AG (B)] .

From its definition,

f(QAQT ,QBQT

) = tr[QAQTG

(QBQT

)]

= tr[QAQTQG (B)QT

]

= tr[QAG (B)QT

] = Qi j A jkGkl Qil

= tr [AG (B)] .

That is, f is isotropic in its two arguments. It is therefore a function of the teninvariants listed in (4.35). Since f (A,B) is linear in its first argument,

f (A,B) = tr[A

(g0I + g1B + g2B2)] ,


and consequently

G (B) = g0I + g1B + g2B2, (4.40)

where g0, g1, g2 are scalar-valued functions of the three invariants of B.The underlying principle is (Pipkin and Rivlin [71]): to find the form for isotropic

vector-valued, or a symmetric tensor-valued functions of a vector or a symmetrictensor, first form an artificial scalar product with a second vector or another symmet-ric tensor, which can be shown to be an isotropic scalar-valued function. Then findthe relevant integrity bases for this isotropic scalar-valued function. Finally, becauseof the linearity in its first arguments, non-linear terms in these arguments can bediscarded, arriving at the correct form for the original isotropic function. For func-tions which are isotropic, or transversely isotropic, or have crystal classes as theirsymmetric groups, see the review article by Spencer [82]. For functions which areinvariant under the full unimodular group, see Fahy and Smith [23].

4.5 Symmetry Restrictions

4.5.1 Unimodular Matrix

Let X and X ′ be two adjacent particles with positionsX andX+dX in the referenceconfiguration. Now consider a change in the local configuration so that X remainsat X, while X ′ goes to X + dX′. We assume that the gradient of this transformationis H, where

dX′ = HdX, (4.41)

andH is a proper unimodularmatrix, i.e., detH = 1 (the configuration change shouldnot lead to a change in volume).

Now if the motion is such that X goes to x, and X ′ goes to x + dx, then we havea new motion M ′ (with deformation gradient F′):

M ′ : X or X �→ x, X ′ or X + dX′ �→ x + dx,M : X or X �→ x, X or X + dX �→ x + dx.

(4.42)

The two mappings are different, because the two shapes about X are mapped intothe same shape about x (see Fig. 4.9). Now, since

dx = FdX = F′dX′ = F′HdX,

we haveF′H = F F′ = FH−1. (4.43)

4.5 Symmetry Restrictions 81

Fig. 4.9 A change in thelocal configuration leads tothe same shape afterdeformation

Xx

dxdX'

dX H

The strain measures for the motion M ′ can be calculated as

B′ = F′F′T = FH−1H−TFT , (4.44)

C′ = F′TF′ = H−TFTFH−1 = H−TCH−1, (4.45)

F′t (τ ) = F′ (τ )F′ (t)−1 = F (τ )H−1HF (t)−1 = Ft (τ ) , (4.46)

C′t (τ ) = F′

t (τ )T F′t (τ ) = Ct (τ ) . (4.47)

The relative strain measure Ct (τ ) is not sensitive to the unimodular changesabout X .

4.5.2 Symmetry Group

Suppose that we are interested in a certain constitutive property ℘, say the stresstensor, that depends on the kinematics. Moreover, suppose that some unimodularchanges of the local shape about X leave this quantity unchanged. Let

G℘ = {I,H1,H2, . . .} (4.48)

be the set of all the unimodular transformations that preserves ℘, then G℘ is agroup, called the ℘-symmetry group, i.e., the group of unimodular changes in theneighbourhood of X that leaves ℘ invariant.

4.5.3 Isotropic Materials

Different materials have different symmetry groups: there are isotropic groups, trans-versely isotropic groups, etc. We are mainly concerned with isotropic materials,where the symmetry group is the proper orthogonal group, H−1 = HT .


4.6 Isotropic Elastic Materials

An elasticmaterial is one inwhich the stress is a function of the deformation gradient:

T = f (F) , (4.49)

where f is a symmetric tensor-valued function of F. For an isotropic material, thesymmetry group is the full proper orthogonal group, that is, for every orthogonal H,we have

f (F) = f (FH) . (4.50)

Isotropic Constraint. Now F has the unique polar decomposition

F = VR,

where R is orthogonal. Thus

F′ = FH = VRH,

and because H is orthogonal, RH is orthogonal. In other words, F′ can be chosenin the set {V,VQ1,VQ2, . . .} , where Qi are orthogonal. Thus f is a function of Valone. Since B = FFT = V2, f is a function of the strain B:

T = f (F) = f (V) = f (B) . (4.51)

Objectivity Constraint. Objectivity imposes the following constraint on T:

T′ = QTQT , (4.52)

for every orthogonal Q. Since T′ = f(B′) = f

(QBQT

)(frame indifference of the

stress, objectivity of B) the requirement (4.52) becomes

Qf (B)QT = f(QBQT

). (4.53)

Thus f is an isotropic function of B (see (4.39)). The general form for f has beenfound in (4.40), thus

f (B) = α0I + α1B + α2B2, (4.54)

where the scalar coefficients are functions of the three invariants of B. One may usethe Cayley–Hamilton theorem and express B2 in terms of B and B−1.

Mooney and neo-Hookean Materials. Therefore, the general constitutive equationfor an isotropic elastic solid is given by

T = β0I + β1B − β2B−1, (4.55)

4.6 Isotropic Elastic Materials 83

where βi are functions of the three invariants of B. The term β0I can be absorbedin the hydrostatic pressure. The case when β1, β2 are constant is called theMooneymaterial [58] or sometimes the Mooney–Rivlin material, in deference to Rivlin. Inaddition, if β2 = 0 it is called the neo-Hookean material [87].

If the material behaves like a Mooney model, in a uniaxial deformation (Problem4.6), the plot of TZZ/(λ−λ2) against λ−1 should be a straight line, with slope β2 andintercept β1. The data from Rivlin and Saunder (see [87]) showed that the Mooneymodel is inadequate: in compression, their data indicated that β2 ≈ 0, whereas intension, their data showed that β2/β1 varies from 0.3 to 1. However, the Mooneymodel should be reasonable for most qualitative purposes.

4.7 The Simple Material

Noll [62] defined a simple material (solid or liquid) as one in which the currentstress is a functional (function of function) of the history of the deformation gradientF (τ ) , −∞ < τ ≤ t :

T (t) = G (F (τ )) , −∞ < τ ≤ t. (4.56)

Objectivity and frame indifference require,

Q (t)G (F (τ ))Q (t)T = G (Q (τ )F (τ )) , (4.57)

for all rotational histories Q (τ ) , −∞ < τ ≤ t.Recall now the polar decomposition for F, F (τ ) = R (τ )U (τ ) . Since Q is

arbitrary, we choose Q (τ ) = R (τ )T . Thus (4.57) requires

G (F (τ )) = R (t)G (U (τ ))R (t)T

= F (t)U (t)−1 G (U (τ ))U (t)−1 F (t)T .

Using the definition C = U2, we define a new functional through

F (C (τ )) = U (t)−1 G (U (τ ))U (t)−1 . (4.58)

Thus for a simple material,

T (t) = F (t)F (C (τ ))F (t)T , −∞ < τ ≤ t. (4.59)

In addition, from Problem 4.7 (4.104),

T (t) = G (F (τ )) = F (t)F (F (t)T Ct (τ )F (t)

)F (t)T , −∞ < τ ≤ t. (4.60)


This says that the current stress is a functional of the history of the right relativeCauchy–Green tensor, and the current value of the deformation gradient. That is, wemay define a new functional

G (F (τ )) = H (Ct (τ ) ,F (t)) . (4.61)

Replacing F(τ ) by F(τ )H, where H is unimodular, leaves Ct (τ ) unchanged, referto (4.47), and thus,

H (Ct (τ ) ,F (t)) = H (Ct (τ ) ,F (t)H) (4.62)

In addition, this functional must obey the objectivity restriction (4.57):

Q (t)H (Ct (τ ) ,F (t))Q (t)T = H (Q (t)Ct (τ )Q (t)T ,Q (t)F (t)

)(4.63)

noting the objectivity of the relative right Cauchy–Green strain tensor (4.21).

4.7.1 Simple Fluid

If the material is an isotropic fluid (note that fluid is isotropic in Noll’s definition),the stress is invariant under the full orthogonal group. Noll showed in this case, thecurrent stress is given by

T (t) = F (Ct (τ ) , ρ (t)) , −∞ < τ ≤ t. (4.64)

This functional must satisfy objectivity,

Q (t)F (Ct (τ ) , ρ (t))Q (t)T = F (Q (t)Ct (τ )Q (t)T , ρ (t)

), (4.65)

for all orthogonal tensors Q.

4.7.2 Incompressible Simple Fluid

Incompressibility has been introduced as a simplification of real material behaviour.This demands that

ρ (t) = ρR, det F (t) = 1, detC (t) = 1, ∇ · u = 0. (4.66)

Its adoption implies that the constitutive relation can only determine the stress up to anisotropic part (the hydrostatic pressure); this hydrostatic pressuremust be determinedby the equations of balance. We write

4.7 The Simple Material 85

T = −PI + S,

S (t) = F (Ct (τ )) , −∞ < τ ≤ t. (4.67)

S is called the extra stress. Of course the functional F must satisfy objectivity:

Q (t)F (Ct (τ ))Q (t)T = F (Q (t)Ct (τ )Q (t)T

). (4.68)

4.7.3 Fading Memory

The idea of fading memory embodies the notion that distant events in the past (largeτ ) should have less bearing on the current stress than events in recent past. This ideacan be implemented in the functional in various ways, through integral or differentialoperators. We close this chapter with two classes of constitutive relations obtainedby assuming that the fluid memory is instantaneous, and that the deformation is smallin some sense, i.e., the relative strain tensor hardly departs from the unit tensor.

4.8 Order Fluids

When the fluidmemory is catastrophic (i.e., instantaneous), we can assume that stressis an isotropic function of the Rivlin–Ericksen tensors An . Recall that the relativestrain tensor can be expressed as a Taylor series with coefficients An ,

S = f (A1,A2, . . . ,AN ) . (4.69)

The physical dimensions of An are T−n , where T is the time. A sequence of approx-imations to f , correct to order T n, n = 1, 2, . . . can be developed.To first order, the Newtonian fluid:

S = S(1) = η0A1. (4.70)

To second order, the second-order fluid model:

S = S(2) = S(1) + (ν1 + ν2)A21 − ν1

2A2. (4.71)

To third order, the third-order fluid model:

S = S(3) = S(2) + α0(trA2

1

)A1 + α1 (A1A2 + A2A1) + α3A3. (4.72)

Higher-order fluids can be developed in the same manner. These order fluids possessno memory, and using them to describe memory phenomena may lead to disaster.


4.8.1 Unsteady Motion

To see that the order fluids are unsuitable for discussing unsteadymotion,we considerthe flow of a second-order fluid (4.71) in a channel of width h,

u = u (y, t) , v = 0, w = 0. (4.73)

The first Rivlin–Ericksen tensor and its square are given by

[A1] =⎡

⎣0 ∂u/∂y 0

∂u/∂y 0 00 0 0

⎤

⎦ ,[A2

1

] =⎡

⎣(∂u/∂y)2 0 0

0 (∂u/∂y)2 00 0 0

⎤

⎦ .

The second Rivlin–Ericksen tensor is

[A2] = [A1

] + [A1L] + [LTA1

]

=⎡

⎣0 ∂2u/∂t∂y 0

∂2u/∂t∂y 0 00 0 0

⎤

⎦ +⎡

⎣0 0 00 2 (∂u/∂y)2 00 0 0

⎤

⎦ .

The stress components are

Sxy = η0∂u

∂y− ν1

2

∂2u

∂t∂y, Sxx = (ν1 + ν2)

(∂u

∂y

)2

,

Syy = ν2

(∂u

∂y

)2

, Szz = 0.

Now, the equations of balance read

ρ∂u

∂t= −∂P

∂x+ ∂Sxx

∂x+ ∂Sxy

∂y+ ∂Sxz

∂z,

0 = −∂P

∂y+ ∂Syx

∂x+ ∂Syy

∂y+ ∂Syz

∂z,

0 = −∂P

∂z+ ∂Szx

∂x+ ∂Szy

∂y+ ∂Szz

∂z.

With no pressure gradient, the equations of motion reduce to

ρ∂u

∂t= η0

∂2u

∂y2− ν1

2

∂3u

∂t∂y2, u (0, t) = 0, u (h, t) = 0. (4.74)

This is a linear third-order partial differential equation; we look for a solution byseparation of variables:

4.8 Order Fluids 87

u (y, t) =∑

n=1

φn (t)ψn (y),

where

ρφnψn = η0φnψ′′n − 1

2ν1φnψ

′′n .

We can try the Fourier series for ψn:

ψn = an sinnπy

h,

which leads to[ρ − 1

2ν1

n2π2

h2

]φn = −η0

n2π2

h2φn.

This has the solution

φn (t) = φ0n exp

(η0t

ν1/2 − ρh2/n2π2

). (4.75)

Clearly, for given material properties and geometry, we can always find n so thatthe exponent is positive (a positive exponent implies instability: the solution isunbounded in time); that is the solution (4.75) is unstable to any disturbance ofthe form stated. Only when ν1 = 0 (Newtonian fluid) is the solution (4.75) stable.Thus all unsteady flows are too fast for the second-order fluid to handle (and indeedfor all order fluids).

4.8.2 Velocity Field in a Second-Order Fluid

The Newtonian pressure field pN obeys

∇ pN = η0∇ · A1 − ρa, (4.76)

where a is the acceleration field. If the same Newtonian velocity field were to occurin the second-order fluid, then an additional pressure term pS will arise and this hasto satisfy

∇ pS = ∇ ·[(ν1 + ν2)A2

1 − 1

2ν1A2

]. (4.77)

There are a few special classes of flows for which the right side of (4.77) can beexpressed as a gradient of a scalar. We will consider two special cases.


Potential Flows. When the velocity field is a potential flow and, consequently u =∇φ, incompressibility demands that φ,i i = 0, where the comma denotes a partialderivative. Since

(A1)i j = 2φ,i j , (4.78)

it follows that

∇ · A1 = 0. (4.79)

Hence, the Newtonian pressure field is given by

pN = −ρ[φ,t + 1

2u · u]. (4.80)

The results from Problem 4.10, (4.111), show that in potential flows, the New-tonian and second-order fluid velocity fields are identical, with the second-orderpressure term given by

pS = 1

8(ν1 + 4ν2) tr A2

1. (4.81)

Plane Creeping Flows. The second case where the Newtonian velocity field is alsothe second-order fluid velocity field is the steady plane flow, where the velocity takesthe form u = u(x, y)i + v(x, y)j. The Cayley–Hamilton theorem and incompress-ibility together show that

A21 + (detA1)1 = 0. (4.82)

Thus,

detA1 = −1

2tr A2

1. (4.83)

Hence, we have the result

∇ · A21 = ∇

(1

2tr A2

1

). (4.84)

To show that the divergence ofA2 can be expressed as a gradient of a scalar, we needthe results of Giesekus [31] and Pipkin [70]

∇ · A2 = ∇(1

η0

dpNdt

+ 3

4tr A2

1

). (4.85)

Thus, a plane creeping flow in a Newtonian fluid is also a solution to the planecreeping flow problem in a second order fluid.

4.9 Green–Rivlin Expansion 89

4.9 Green–Rivlin Expansion

Green and Rivlin [32] proposed an expansion based on the integral of the strainhistory

G (s) = Ct (t − s) − I, 0 ≤ s < ∞, (4.86)

which is regarded to be small in some sense (i.e., the relative strainCt (t − s) is nearto identity, the undeformed state). The expansion is quite unwieldy, consisting ofmultiple integral terms. We record the first term here

S (t) =∫ ∞

0μ (s)G (s) ds, (4.87)

which is called the finite linear viscoelasticity integral model. Here, μ(s) is thememory kernel, a decreasing function of s (distant past is less important – fluid hasfading memory). Usually an exponential memory function is chosen

μ (s) = −G

λe−s/λ, (4.88)

where G is a modulus and λ is a relaxation time. A multiple relaxation (exponential)mode is sometimes used as well.

Problems

Problem 4.1 In a simple shear deformation of a linear elastic material (4.7), thedisplacement field takes the form

v1 = γy, v2 = 0, v3 = 0, (4.89)

where γ is the amount of shear. Find the elastic stress, in particular, the shear stress.This justifies calling μ the shear modulus.

Problem 4.2 In a uni-axial extension of a linear elastic material (4.7), the displace-ment field is given by

v1 = εx, v2 = −νεy, v3 = −νεz, (4.90)

where ε is the elongational strain, and ν is the amount of lateral contraction due tothe axial elongation, called Poisson’s ratio. If the lateral stresses are zero, show that


ν = λ

2 (λ + μ), μ = E

2 (1 + ν), (4.91)

where E is the Young’s modulus, i.e., Txx = Eε.

Problem 4.3 Show that a material element dX = dXP, where P is a unit vector, isstretched according to

dx2 = dX2C : PP,

where C is the right Cauchy-Green tensor. Thus when P is randomly distributed inspace, the average amount of stretch is

⟨dx2

⟩ = 1

3dX2trC,

and therefore 13 trC can be used as a definition of the Weissenberg number.

Problem 4.4 Let f be a vector-valued, isotropic polynomial of a symmetric tensorS and a vector v. Use the integrity basis in (4.36) to prove that

f(S, v) = [ f01 + f1S + f2S2]v,

where the scalar valued coefficients are polynomials in the six invariants involvingonly S and v in the list (4.36).

Problem 4.5 Consider a simple shear deformation of a rubber-like material (4.55),where

x = X + γY, y = Y, z = Z . (4.92)

Show that the Finger strain tensor B and its inverse are given by

[B] =⎡

⎣1 + γ2 γ 0

γ 1 00 0 1

⎤

⎦ ,[B−1] =

⎡

⎣1 −γ 0

−γ 1 + γ2 00 0 1

⎤

⎦ . (4.93)

Consequently, show that the stress tensor is given by

[T] = −P [I] +⎡

⎣β1

(1 + γ2

) − β2 (β1 + β2) γ 0(β1 + β2) γ β1 − β2

(1 + γ2

)0

0 0 β1 − β2

⎤

⎦ , (4.94)

where P is the hydrostatic pressure. Thus, show that the shear stress and the normalstress differences are

S = (β1 + β2) γ, N1 = (β1 + β2) γ2, N2 = −β2γ2. (4.95)

Problems 91

Deduce that the linear shear modulus of elasticity is

G = limγ→0

(β1 + β2) . (4.96)

The ratio

N1

S= γ (4.97)

is independent of the material properties. Such a relation is called universal.

Problem 4.6 In a uniaxial elongational deformation of a rubber-likematerial (4.55),where (in cylindrical coordinates)

R = λ1/2r, Θ = θ, Z = λ−1z, (4.98)

show that the inverse deformation gradient is

[F−1

] =⎡

⎣∂R∂r

1r

∂R∂θ

∂R∂z

0 Rr 0

∂Z∂r

1r

∂Z∂θ

∂Z∂z

⎤

⎦ =⎡

⎣λ1/2 0 00 λ1/2 00 0 λ−1

⎤

⎦ . (4.99)

Consequently, the Finger strain tensor B and its inverse B−1 are

[B−1

] = [F−TF−1

] =⎡

⎣λ 0 00 λ 00 0 λ−2

⎤

⎦ , [B] =⎡

⎣λ−1 0 00 λ−1 00 0 λ2

⎤

⎦ . (4.100)

Thus the total stress tensor for a rubber-like material (4.55) is

[T] = −P [I] +⎡

⎣β1λ

−1 − β2λ 0 00 β1λ

−1 − β2λ 00 0 β1λ

2 − β2λ−2

⎤

⎦ . (4.101)

Under the condition that the lateral tractions are zero, i.e., Trr = 0, the pressure canbe found, and thus show that the tensile stress is

Tzz = −P + β1λ2 − β2λ

−2 = β1λ2 − β2λ

−2 − β1λ−1 + β2λ

= (λ2 − λ−1

) (β1 + β2λ

−1). (4.102)

This tensile stress is the force per unit area in the deformed configuration. As r =λ1/2R, the corresponding force per unit area in the undeformed configuration is

TZZ = Tzzλ−1 = (

λ − λ−2) (

β1 + β2λ−1

). (4.103)


Problem 4.7 Show that

C (τ ) = F (t)T Ct (τ )F (t) . (4.104)

Problem 4.8 Consider a simple shear flow

u = γy, v = 0, w = 0. (4.105)

Show that the stress tensor in the second-order model is given by

[S] = η0

⎡

⎣0 γ 0γ 0 00 0 0

⎤

⎦ + (ν1 + ν2)

⎡

⎣γ2 0 00 γ2 00 0 0

⎤

⎦ − ν1

2

⎡

⎣0 0 00 2γ2 00 0 0

⎤

⎦ . (4.106)

Thus, the three viscometric functions are

S = η0γ, N1 = S11 − S22 = ν1γ2, N2 = S22 − S33 = ν2γ

2. (4.107)

Problem 4.9 Consider a second-order fluid in an elongational flow

u = εx, v = − ε

2y, w = − ε

2z. (4.108)

Show that the stress is given by

[S] = η0

⎡

⎣2ε 0 00 −ε 00 0 −ε

⎤

⎦ + (ν1 + ν2)

⎡

⎣4ε2 0 00 ε2 00 0 ε2

⎤

⎦ − ν1

2

⎡

⎣4ε2 0 00 ε2 00 0 ε2

⎤

⎦ . (4.109)

Consequently the elongational viscosity is given by

ηE = Sxx − Syyε

= 3η0 + 3(ν1

2+ ν2

)ε. (4.110)

Problem 4.10 Show that, for potential flows,

∇ · A21 = 1

2∇(tr A2

1), ∇ · A2 = 3

4∇(tr A2

1). (4.111)

Problem 4.11 For steady two-dimensional incompressible flows, a stream functionψ = ψ(x, y) can be defined such that the velocity components u and v can beexpressed as

u = ∂ψ

∂y, v = −∂ψ

∂x. (4.112)

Problems 93

Obtain the stresses in a second-order fluid in terms of ψ. Substitute the stresses intothe equations of motion and eliminating the pressure term through the use of theequality of the mixed partial derivatives, i.e., p,xy = p,yx , to obtain

η2ψ − ν1

2v · ∇(2ψ) = 0, (4.113)

where 2 is the two-dimensional biharmonic operator. Deduce that a Newtonianvelocity field is also a velocity field for the second-order fluid. The result is due toTanner [83].

Problem 4.12 In a simple shear flow,

u = γ (t) y, v = 0, w = 0,

show that the path lines ξ (τ ) = (ξ,ψ, ζ) are given by

ξ (τ ) = x + yγ (t, τ ) , ψ (τ ) = y, ζ (τ ) = z, (4.114)

where

γ (t, τ ) =∫ τ

tγ (s) ds. (4.115)

Find the relative strain tensor, the stress tensor for a finite viscoelastic integral fluid,and show that the shear stress and the normal stress differences are given by

S12 (t) =∫ ∞

0μ (s) γ (t, t − s) ds, (4.116)

N1 (t) =∫ ∞

0μ (s) [γ (t, t − s)]2 ds = −N2. (4.117)

Investigate the case where the shear rate is constant and sinusoidal in time for thememory function in (4.88).

Chapter 5Inelastic Models and Linear Viscoelasticity

Some Practical Engineering Models

We have seen some of the classical constitutive equations introduced in the lastthree centuries, and explored the general formulation of constitutive equations inthe last chapter. There, we mention that the general constitutive principles shouldbe taken as guidelines only; they should emerge from the physics of the fluids’microstructures. In engineering, the emphasis is to produce the analyses for thedesign process. Therefore we prefer simple models with the “right” physics to beincluded. “Right” physics here means to correctly account for the flow process tobe modelled: if the flow process does not call for certain behaviour, then it may beleft out in the constitutive modelling. In addition, engineers do not have any qualmin supplementing a constitutive equation with empirical data, as long as the correctphysical framework has already been incorporated in the constitutive equation. Weare aiming to strive for simplicity in a correct constitutive framework, with enoughempirical inputs as needed, to ensure a quantitative prediction to the physical flowphenomenon that we are modelling.

5.1 Inelastic Fluids

When the flow phenomena are dominated by viscosity effects, then it makes sense tomodel the viscosity function accurately. Inelastic, or generalised Newtonian, fluidsare those for which the extra stress tensor is proportional to the strain rate tensor, butthe “constant” of proportionality (the viscosity) is allowed to depend on the strainrate:

S = 2η (γ)D, (5.1)

where γ = √2trD2 is called the generalised strain rate (in a simple shear flow, this

quantity reduces to themagnitude of the shear rate). Inelasticmodel possesses neither


95

96 5 Inelastic Models and Linear Viscoelasticity

Fig. 5.1 Differentnon-Newtonian behaviours

Shear rate

Shear stress

dilatantwithout yield

Newtonian

pseudoplastic

plastic flowBingham flow

yield stress

memory nor elasticity, and therefore it is unsuitable for transient flows, or flows thatcall for elasticity effects. It is only useful in steady viscometric flows where anaccurate representation of the viscosity is paramount. Depending on the functionalform of η (γ) , one can get different non-Newtonian behaviours, see Fig. 5.1.

Bingham fluids are those that can support a yield stress. When the shear stressexceeds this yield value, the fluid flows like a Newtonian fluid, with a constantviscosity. Plastic fluids are yield-stress types of fluids, with flow viscosity decreasingwith shear rate (shear thinning). The term “pseudoplastic” means that the viscositydecreases with shear rates (shear thinning). The opposite of pseudoplastic is dilatant(shear thickening).

5.1.1 Carreau Model

Different forms for the viscosity have been proposed for pseudoplastic fluids; themost popular one is the Carreau model:

η (γ) = η∞ + η0 − η∞(1 + Γ 2γ2

)(1−n)/2. (5.2)

There are four parameters: η∞ is the infinite-shear-rate viscosity, η0 is the zero-shear-rate viscosity, n is called the power-law index (n − 1 is the slope of (η − η∞)/(η0 −η∞) versus γ in log-log plot), and Γ is a constant bearing the dimension of time(this constant has no relation to the relaxation time of the fluid). A graph of (5.2) isillustrated in Fig. 5.2.

The model is meant for shear-thinning fluids and therefore 0 < n ≤ 1. n = 1represents the Newtonian behaviour.

5.1 Inelastic Fluids 97

Shear rate (s-1)10-3 10-2 10-1 100 101 102 103 104 105

Visc

osity

(Pa.

s)

101

102

103 zero shear rate viscosity, 0 = 1000

slope (n - 1), n = 0.5

infinite shear-rate viscosity,

-1 = 0.1

Fig. 5.2 A typical plot of the Carreau viscosity

5.1.2 Power-Law Model

Included in the Carreau model is the power-law model:

η (γ) = k |γ|n−1 , (5.3)

where k is called the consistency and n the power-law index. When η∞ = 0 andat high shear rate, the Carreau model (5.2) reduces to the power-law model withpower-law index n and consistency k = η0Γ

n−1. The power-lawmodel breaks downin regions where the shear rate is zero - in these regions the stress is unbounded.

Simple Shear Flow. In a simple shearing flow, where the fluid is confined betweentwo plates and the top plate is moving with a velocity U , the velocity field takes theform

u = γy, v = 0, w = 0, γ = U/h.

The velocity gradient and the strain rate tensors are

[L] = [∇uT] =

⎡

⎣0 γ 00 0 00 0 0

⎤

⎦ , [D] = 1

2

(L + LT

) = 1

2

⎡

⎣0 γ 0γ 0 00 0 0

⎤

⎦ ,


and

[D2

] = 1

4

⎡

⎣0 γ 0γ 0 00 0 0

⎤

⎦

⎡

⎣0 γ 0γ 0 00 0 0

⎤

⎦ = 1

4

⎡

⎣γ2 0 00 γ2 00 0 0

⎤

⎦ .

The generalised shear rate is γ = √2trD2, and the stress tensor for the inelastic

model (5.1) is

[S] = η (γ)

⎡

⎣0 γ 0γ 0 00 0 0

⎤

⎦ . (5.4)

The only non-zero component of the stress is the shear stress,

Sxy = η (γ) γ. (5.5)

Elongational Flow. In an elongational flow, where

u = εx, v = − ε

2y, w = − ε

2z,

we have

[L] =⎡

⎣ε 0 00 −ε/2 00 0 −ε/2

⎤

⎦ = [D] ,[D2

] =⎡

⎣ε2 0 00 ε2/4 00 0 ε2/4

⎤

⎦ , γ = √3ε.

Thus the stress tensor for the inelastic model (5.1) is

[S] = 2η(√

3ε)

⎡

⎣ε 0 00 −ε/2 00 0 −ε/2

⎤

⎦ (5.6)

and the elongational viscosity is given by

ηE = N1

ε= 3η

(√3ε

). (5.7)

It has been known that the elongational viscosity and the shear viscosity have dis-similar shape. Therefore, in-elastic models may not be suitable in processes wherethere is a mixture of shear and elongational flow components.

5.2 Linear Viscoelasticity 99

5.2 Linear Viscoelasticity

The concept of linear viscoelasticity was originated with Maxwell (Fig. 5.3), whoproposed the following equation in 1867–68,

dσ

dt= E

dε

dt− σ

λ, (5.8)

where σ is the (one-dimensional) stress, ε is the (one-dimensional) strain, E is themodulus of elasticity and λ is a time constant. When the relaxation time is zero,keeping the product η = λE constant, the Newtonian model is recovered. And whenthe relaxation is infinitely large, a further integration yields the Hookean model.

Years later, Meyer (1874) introduced the equation

σ = Gγ + ηdγ

dt, (5.9)

combining the solid response (G is the elastic modulus, γ is the shear strain) andliquid response (η is the viscosity) in one equation. This equation is now known asthe Kelvin–Voigt body – it should be called Kelvin–Meyer–Voigt (a full account canbe found in Tanner and Walter [86]).

Boltzmann (Fig. 5.4) criticised the lack of generality in Maxwell’s and Meyer’swork and proposed that the stress at the current time depends not only on the currentstrain, but on the past strains as well. It was assumed that a strain at a distant pastcontributes less to the stress than a more recent strain. This is recognised as thefamiliar concept of fading memory. Furthermore, linear superposition was assumed:supposing that the strain between times t ′ and t ′ + dt ′, say dγ

(t ′), contributes

Fig. 5.3 J.C. Maxwell(1831–1879) published hisfirst scientific paper when hewas fourteen. He set upCavendish Laboratory atCambridge in 1874, and diedof cancer at an early age of48. He united electricity andmagnetism into the conceptof electromagnetism. He alsointroduced the concept ofstress relaxation in thekinetic theory of gases


Fig. 5.4 Ludwig Boltzmann(1844–1906) was mostfamous for his atomicviewpoint and his inventionof statistical mechanics. Heis said never to have failedany student taking his course

G(t − t ′

)dγ

(t ′)to the stress, then the total stress at time t is

σ (t) =∫ t

−∞G

(t − t ′

)dγ

(t ′) =

∫ t

−∞G

(t − t ′

)γ

(t ′)dt ′. (5.10)

Here, G(t) is a decreasing function of time, the relaxation modulus, and γ is theshear rate. A three-dimensional version of this is

S (t) = 2∫ t

−∞G

(t − t ′

)D

(t ′)dt ′, (5.11)

Si j (t) = 2∫ t

−∞G

(t − t ′

)Di j

(t ′)dt ′.

The Newtonian liquid is recovered with the delta memory function

G(t − t ′

) = η0δ(t − t ′

)

S (t) = 2∫ t

−∞η0δ

(t − t ′

)D

(t ′)dt ′ = 2η0D (t) . (5.12)

The most-often used relaxation modulus function is theMaxwell discrete relaxationspectrum:

G (t) =N∑

j=1

G je−t/λ j , (5.13)


which consists of a discrete spectrum of relaxation times{G j ,λ j

}. The linear vis-

coelastic constitutive model (5.11) is not objective; it is only valid at vanishinglysmall strains.

5.2.1 Simple Shear Flow

In a flow with constant strain rate D, the linear viscoelastic stress (5.11) is

S (t) = 2η (t)D, η (t) =∫ t

−∞G

(t − t ′

)dt ′. (5.14)

With the Maxwell relaxation modulus (5.13),

S (t) =N∑

j=1

2η j(1 − e−t/λ j

)D, η j = G jλ j . (5.15)

Now consider an oscillatory shear flow between two parallel plates at a distance hapart, where the bottom plate is stationary and the top plate is sinusoidally displacedby a small amount δ sinωt, δ � h. The top plate velocity is

U (t) = δω cosωt.

The shear rate and the shear strain are, respectively,

γ (t) = δ

hω cosωt = γ0 cosωt,

γ (t) = δ

hsinωt = γ0 sinωt, γ0 = δ

h� 1, γ0 = ωγ0. (5.16)

The only non-zero component of the stress is the shear stress,

S12 =∫ t

−∞G

(t − t ′

)γ0 cosωt ′dt ′,

=∫ ∞

0γ0G (s) cosω (t − s) ds,

=∫ ∞

0γ0G (s) [cosωt cosωs + sinωt sinωs] ds,

= G ′ (ω) γ0 sinωt + η′ (ω) γ0 cosωt,

where the coefficients in the strain, G ′ (ω) , the storage modulus, and in the strainrate, η′ (ω) , the dynamic viscosity, are material functions of the frequency; they are


defined by

G ′ (ω) =∫ ∞

0ωG (s) sinωsds, η′ (ω) =

∫ ∞

0G (s) cosωsds. (5.17)

The other two related quantities, the loss modulus G ′′, and the storage viscosity η′′are defined as

G ′′ (ω) = ωη′ (ω) , η′′ (ω) = G ′ (ω)

ω. (5.18)

Sometimes it is more convenient to work with complex numbers, and the complexmodulus G∗ and the complex viscosity η∗ are thus defined as

G∗ (ω) = G ′ (ω) + iG ′′ (ω) , η∗ (ω) = η′ (ω) − iη′′ (ω) . (5.19)

One can denote the shear rate as γ∗ = γ0eiωt , then

γ (t) = γ0Re(eiωt

), S12 (t) = Re

(η∗γ∗

0

). (5.20)

It should be remembered that the linear viscoelasticity model is valid only at smallstrains, that is, bothG ′ and η′ are independent of the strain amplitude γ0; this ought tobe tested before a frequency sweep is done. The strain amplitude for which linearityholds could be as large as 20% for polymer melts and solutions, and as small as 0.1%for biological materials, such as bread dough.

Figure5.5 is a plot of the storagemodulus and dynamic viscosity of a water-doughas functions of the frequency, where the strain amplitude was kept at 0.1%.

By performing the inverse Fourier transform of (5.17), the relaxation moduluscan be obtained from dynamic data as

G (s) = 2

π

∫ ∞

0η′ (ω) cosωsdω = 2

π

∫ ∞

0

G ′ (ω)

ωsinωsdω. (5.21)

Inverting the dynamic properties to obtain the relaxation modulus using (5.21) maybe an ill-conditioned problem (see Problem 5.3).

5.2.2 Step Strain

Consider the stress relaxation experiment after a step strain, Fig. 5.6. Suppose a shearstrain ofmagnitude γ is imposed at time t0 within a short periodΔ, as sketched in Fig.5.6. During this period the strain rate can be assumed constant, given by γ = γ/Δ.

The stress is (for t > t0 + Δ)


Frequency, (rd/s)

10-4 10-3 10-2 10-1 100 101 102 103 104

Dyn

amic

vis

cosi

ty

' (Pa

.s)

10-1

100

101

102

103

104

105

106

107

Stor

age

mod

ulus

, G' (

Pa)

102

103

104

105

G'

Fig. 5.5 Dynamic properties of flour-water dough. The strain amplitude was set at 0.1%

Fig. 5.6 Stress relaxationafter a step strain


S12 (t) =∫ t

−∞G (t − s) γ (s) ds =

∫ t0+Δ

t0

G (t − s)γ

Δds. (5.22)

Recall the mean-value theorem,

∫ t0+Δ

t0

f (s) ds = Δ f (t0 + ζ) , 0 ≤ ζ ≤ Δ.

When applied to (5.22), this yields

S12 (t) = γG (t − t0 − ζ) .

When Δ → 0, ζ → 0, and the stress relaxes as does the relaxation modulus (hencethe name). In particular, when t0 = 0,

S12 (t) = γG (t) . (5.23)

The relaxation modulus is a material function and can be measured routinely on arheometer.

5.2.3 Relaxation Spectrum

We have met the discrete Maxwell relaxation spectrum, (5.13). The continuous ver-sion of this is the relaxation spectrum H , defined by the relation

G (t) =∫ ∞

0H (λ) e−t/λ dλ

λ=

∫ ∞

−∞H (λ) e−t/λd ln λ. (5.24)

In this, the relaxation time λ is supposedly evenly distributed on a logarithmicscale. The quantity H(λ) is called the relaxation spectrum. Note that

G ′ (ω) =∫ ∞

0ω sinωs

∫ ∞

0H (λ) e−s/λ dλ

λds

=∫ ∞

0ωH (λ)

dλ

λ

∫ ∞

0e−s/λ sinωsds. (5.25)

The last integral can be evaluated, giving

G ′ (ω) =∫ ∞

−∞ω2λ2

1 + ω2λ2H (λ) d ln λ. (5.26)


Time (s)10-4 10-3 10-2 10-1 100 101 102 103 104 105

Log

rela

xatio

n sp

ectru

m, H

() (

Pa)

2

3

4

Time (s)10-3 10-2 10-1 100 101 102 103 104

Rel

axat

ion

mod

ulus

, G (P

a)

100

101

102

103

104

105

Fig. 5.7 Relaxation modulus and spectrum

Similarly,

η′ (ω) =∫ ∞

0

H (λ)

1 + ω2λ2dλ. (5.27)

These results should be compared to those obtained with the discrete Maxwell relax-ation modulus, Problem 5.4. Inversion of the dynamic data to find H according to(5.26) or (5.27) is an ill-conditioned problem (Problem 5.3). In Fig. 5.7 the relaxationmodulus, and its spectrum are shown for a dough-water flour, using a regularisationmethod of Weese [89].

5.3 Correspondence Principle

5.3.1 Quasi-Static Approximation

To solve boundary-value problems for a linear viscoelastic fluid, the quasi-staticapproximation is used. Here, one ignores the inertia terms ρu. This is possible ifthe characteristic frequency is not too high (De � 1). Let suppose the flow startsfrom time zero, before which the stress is zero. The linear viscoelastic stress and the


equations of motion are

T (t) = −PI + S = −PI +∫ t

0G

(t − t ′

) (∇u

(t ′) + ∇u

(t ′)T)

dt ′, (5.28)

∇ · T = −∇P + ∇ · S = 0, ∇ · u = 0, x ∈ V, (5.29)

subjected to a relevant boundary condition on the bounding surface S, for example,

u (x, t) = u0 (t) , x ∈ S. (5.30)

We can take the Laplace transform of all the above equations to arrive at

T (s) = −PI + G(∇u + ∇uT

), (5.31)

− ∇ P + G∇2u = 0, ∇ · u = 0, (5.32)

u (x, s) = u0 (s) , x ∈ S, (5.33)

where the overbar denotes a Laplace transform variable, i.e.,

φ (s) =∫ ∞

0e−stφ (t) dt . (5.34)

Equations (5.31)–(5.33) are identical to those of the corresponding Newtonian(Stokes) flow problem, except that the viscosity is now replaced by G. Thus thesolution in the Laplace transform domain matches the Stokes solution. This is theessence of the Correspondence Principle. The solution in the physical domain isthen obtained by inverting the Laplace transform.We give an example for the start-upof a circular Couette flow.

5.3.2 Circular Couette Flow

For a circular Couette flow, the velocity field in cylindrical coordinates (Fig. 5.8) is

u = {0, rΩ (r, t) , 0} , (5.35)

where the boundary conditions on the inner cylinder (angular velocityΩi , radius Ri )and the outer cylinder (stationary, radius Ro) are

Ω (Ri , t) = Ωi , Ω (Ro, t) = 0. (5.36)

5.3 Correspondence Principle 107

Fig. 5.8 Circular Couetteflow

The strain rate tensor is

[D] = 1

2

⎡

⎢⎢⎢⎣

0 r∂Ω

∂r0

r∂Ω

∂r0 0

0 0 0

⎤

⎥⎥⎥⎦

.

The only non-zero component of the linear viscoelastic stress is the shear component

Srθ =∫ t

0G

(t − t ′

)r∂Ω

∂r

(r, t ′

)dt ′.

The balance of linear momentum requires

∂

∂r

(r2Srθ

) = 0, Srθ = M (t)

2πr2,

where M(t) is a “constant” of integration. The torque on the inner cylinder is

Γ =∫ Ro

Ri

2πr2Srθdr = M (Ro − Ri ) . (5.37)

Taking the Laplace transform,

Srθ = Gr∂Ω

∂r

∂Ω

∂r= M

2πGr3.

Integrating

Ω = C − M

4πr2G,


where C is an integration constant. Applying the boundary conditions (5.36),

Ω = M

4πR2o G

(1 − R2

o

r2

), Ωi = M

4πR2o G

(1 − R2

o

R2i

). (5.38)

Note that

Ω

Ωi= 1 − R2

o/r2

1 − R2o/R

2i

, (5.39)

which is the Stokes solution (in Laplace transform domain). In the Newtonian case,G (t) = η0δ (t) , giving G = η0. Hence,

MN = 4πR2oη0Ωi

1 − R2o/R

2i

. (5.40)

For the linear viscoelastic case,

M = 4πR2o

1 − R2o/R

2i

GΩi ,

M (t) = 4πR2o

1 − R2o/R

2i

∫ t

0G

(t − t ′

)Ωi

(t ′)dt ′. (5.41)

5.4 Mechanical Analogs

In the older rheology literature, one finds mechanical analogs for linear viscoelasticbehaviours, springs for solidHookean behaviour and dashpots for viscousNewtonianbehaviour. We illustrate this with a few popular models in Fig. 5.9.

(a)

(b)

(c)

Fig. 5.9 Mechanical analogs of linear viscoelastic behaviours

5.4 Mechanical Analogs 109

Maxwell Element. Figure 5.9a shows theMaxwell model, where the spring elementrepresents a Hookean behaviour, and the dashpot element, a Newtonian viscousbehaviour. These elements are arranged in series with the understanding that thedisplacements (strains) are additive, and the forces (stresses) are equal across theelements. Then

x = y + (x − y)

= F

G+ F

η,

or

F + η

GF = ηx .

Thus, if F is identified with the stress Si j and x , the strain γi j , then one obtains theMaxwell model

Si j + η

GSi j = ηγi j , (5.42)

where λ = η/G may be identified as the relaxation time. The Maxwell model is afluid (it cannot support a shear stress without deforming).

Kelvin-Voigt-Meyer Element. Figure5.9b shows the Kelvin-Voigt-Meyer model,where the spring and the dashpot elements are arranged in parallel. Across theseelements, the displacements (strains) are equal, and the forces (stresses) are additive.Thus

Gx + ηx = F.

Identify x with the strain γi j , and F with the stress Si j , one obtains the Kelvin–Voigt–Meyer model:

Gγi j + ηγi j = Si j . (5.43)

The Kelvin–Voigt–Meyer material is a solid (it can support a shear stress indefinitelywithout deforming). Similar to the Maxwell model, λ = η/G is the relaxation timeof the model.

Four-ElementModel. In a similar manner, Fig. 5.9c shows a so-called four-elementmodel. The displacements are additive, the forces are the same across series elements;and across parallel elements, the displacements are the same, the forces are additive.Working in Laplace transform domain, with zero initial conditions, and then convertback to time domain, it may be shown that (Problem5.6)

F + a1 F + a2 F = b1 x + b2 x,


where

a1 = η1

G2

(1 + G2

G1+ η2

η1

), a2 = η1η2

G1G2, b1 = η1, b2 = η1η2

G2,

leading to the following stress-strain relation for the four-element model

Si j + a1 Si j + a2 Si j = b1γi j + b2γi j . (5.44)

Multimode Models. To each of the one-relaxation time models, we could define acorresponding multimode model. For example, the multimode Kelvin–Voigt–Meyermodel is written as

γi j =N∑

n=1

γ(n)i j , γ(n)

i j + λn γ(n)i j = Si j

Gn, (5.45)

where λn, n = 1, . . . , N , are the relaxation times.

Problems

Problem 5.1 Show that, with the relaxation modulus function (5.13), the relation(5.11) is equivalent to

S =N∑

j=1

S( j), (5.46)

S( j) + λ j S( j) = 2η jD, η j = G jλ j .

This relation is called the linear Maxwell equation. Equation (5.46) is equivalent to(5.9).

Problem 5.2 Show that the shear stress for (5.11) in an oscillatory flow, where theshear rate is γ = γ0 cos (ωt) , can be expressed as

S12 = ∣∣G∗∣∣ sin (ωt + φ) , tan φ = G ′′

G ′ , (5.47)

where tan φ is called the loss tangent.

Problem 5.3 Verify that for the spectrum

H (λ) = cos2 (nλ) ,

Problems 111

η′ (ω) =∫ ∞

0

H (ω)

1 + λ2ω2dω = π

4ω

(1 − e−2n/ω

).

At large n, the data η′ is smooth, but the spectrum is highly oscillatory. Concludethat the inverse problem of finding H (λ) , given the data η′ in the chosen form isill-conditioned – that is, a small variation in the data (in the exponentially small term)may lead to a large variation in the solution.

Problem 5.4 For the Maxwell discrete relaxation spectrum (5.13), show that

G (t) =N∑

j=1

G je−t/λ j , G ′ (ω) =

N∑

j=1

G jω2λ2

j

1 + ω2λ2j

, η′ (ω) =N∑

j=1

G jλ j

1 + ω2λ2j

. (5.48)

In particular, with one relaxation mode λ = λ1,

tan φ = 1

ωλ(5.49)

deduce that as ω = 0 → ∞, the response goes from fluid (φ = π/2) to solid behav-iour (φ = 0).

Problem 5.5 Suppose we have a Maxwell material with one relaxation time,

G (t) = η0

λe−t/λ.

and Ωi = constant. Show that the solution to the circular Couette flow problemconsidered in Sect. 5.3.2 is

M (t)

MN= 1 − e−t/λ. (5.50)

Problem 5.6 Working in Laplace transform domain, show that the mechanical ana-log of Fig. 5.9c leads to (5.44).

Chapter 6Steady Viscometric Flows

Shear Flows

There is a class of flows of the simple fluid, equivalent to the simple shearing flow,for which the kinematics and the stress can be completely determined. Ericksen [21]called them laminar shear flows, but the current term used to describe these flows isviscometric flows [16]. We review this class of flows here.

6.1 Kinematics

First, consider a simple shear flow with the kinematics

u = γy, v = 0, w = 0, (6.1)

where the shear rate γ is a constant. This flow has the velocity gradient tensor

[L] =⎡⎣0 γ 00 0 00 0 0

⎤⎦ ,

which obeys L2 = 0. For this flow, the only non-trivial Rivlin–Ericksen tensors areA1 and A2; the rest of the Rivlin–Ericksen tensors are nil,

[A1] =⎡⎣0 γ 0γ 0 00 0 0

⎤⎦ , [A2] =

⎡⎣0 0 00 2γ2 00 0 0

⎤⎦ .

Consequently, the relative right Cauchy–Green tensor is quadratic in the time lapse,recall (3.32),


113

http://dx.doi.org/10.1007/978-3-319-62000-8_3

114 6 Steady Viscometric Flows

Ct (t − s) = I − sA1 + s2

2A2, 0 ≤ s. (6.2)

Ericksen [21] referred to flows obeying (6.2) as laminar shear flows, Coleman [16]called them viscometric flows. Yin and Pipkin [92] embarked on a search for allsuch flows and now our knowledge of them is essentially complete. One can writefor (6.1),

u = γ (b · x) a, (6.3)

where a = e1 and b = e2. We now define viscometric flows as those where thevelocity field obeys (6.3), for three mutually orthogonal unit vectors a, b, and c,refer to Fig. 6.1.

The three directions a, b, and c are called the shear axes; a is the direction ofshear, b is the direction of shear rate, and c is the vorticity axis. The motion canbe visualised as the relative sliding motion of a stack of playing cards, each cardrepresents a slip surface b · x = constant.

(a) (b)

(c) (d)

(e)

Fig. 6.1 Viscometric flows

6.1 Kinematics 115

In Fig. 6.1, all such flows are sketched: (a) simple shearing flow, (b) steady parallelflow, (c) rectilinear flow, (d) circular flow, and (e) helical flow. Shear flow has alreadybeen considered, we now briefly look at the rest.

6.1.1 Steady Parallel Flow

In the steady parallel flow (Fig. 6.1b), the velocity field takes the form

u = w (x, y) k, (6.4)

where a = k is a unit vector in the z-direction, the slip surfaces are cylinders withconstant w (x, y) . The velocity gradient is

(∇u)T = ∂w

∂xki + ∂w

∂ykj = γab, (6.5)

with

γb = ∂w

∂xi + ∂w

∂yj, γ2 =

(∂w

∂x

)2

+(

∂w

∂y

)2

. (6.6)

The material derivative of γ is zero, i.e., the shear rate is constant along eachstreamline.

6.1.2 Rectilinear Flow

Another class of viscometric flows is the rectilinear flow in which the velocity fieldtakes the form (Fig. 6.1c)

u = u (z) i + v (z) j. (6.7)

These flows have parallel plane surfaces z = constant, like a pack of playing cards.The velocity gradient is

(∇u)T = ∂u

∂zik + ∂v

∂zjk = γab, (6.8)

where the direction of the shear rate is b = k, and

γa = ∂u

∂zi + ∂v

∂zj, γ2 =

(∂u

∂z

)2

+(

∂v

∂z

)2

. (6.9)

Again the shear rate remains constant along each streamline.


6.1.3 Axial Fanned Flow

In the axial fanned flow, the velocity field takes the form

u = cθk, 0 ≤ θ ≤ 2π, γ = c/r, (6.10)

where c is a constant and θ = tan−1 (y/x) .

6.1.4 Helical Flow

In the axial translation, rotation and screw motions of coaxial circular slip surfaces(Fig. 6.1e), the velocity field takes the form

u = rω (r) eθ + u (r) ez . (6.11)

The velocity gradient is

(∇u)T = r∂ω

∂reθer + ∂u

∂rezer = γab, (6.12)

where b = er , and

γa = r∂ω

∂reθ + ∂u

∂rez, γ2 = r2

(∂ω

∂r

)2

+(

∂u

∂r

)2

. (6.13)

The shear rate remains constant along each streamline. These flows include

1. Circular pipe flow, or Poiseuille flow, when the flow occurs in a circular pipe, orannular flow, when it occurs between two concentric cylinders. Here ω = 0.

2. Circular Couette flow, when the flow occurs between concentric cylinders, oneor both rotating, and u = 0.

3. Helical flow,when both rotational and translational components are present. Sincethe angular velocity isω, a particle covers 2π radians in 2π/ω seconds,while rising2πu/ω. This rise is constant on each cylinder surface.

6.1.5 Helicoidal Flow

If all the helical paths in the previous flow have the same rise per turn, then the slipsurfaces need not be helical but can be general helicoids. The velocity field takes theform

u = (reθ + cez)ω (r, z − cθ) , (6.14)

6.1 Kinematics 117

where c is a constant and ω is a function of r and z − cθ. All the helices have thesame rise per turn, i.e., 2π/c. The velocity gradient is

(∇u)T = [∇ (rωeθ + cωez)]T = (∇ω (reθ + cez) + ω (ereθ − eθer ))T

= (reθ + cez)∇ω + ω (eθer − ereθ)

= γab. (6.15)

6.2 Stresses in Steady Viscometric Flows

It is remarkable that the stresses in steady viscometric flows can be determinedcompletely for isotropic simple fluids. Take, for example, the simple shear flowwhere u = γyi. The general form of the stress tensor is

[S] =⎡⎣Sxx Sxy 0Sxy Syy 00 0 Szz

⎤⎦ .

The shear stress Sxy must be an odd function of the shear rate based on physicalgrounds alone.1 Thus we can write

Sxy = γη (γ) , (6.16)

where the viscosity, defined as η = Sxy/γ, must be an even function of the shearrate:

η (−γ) = η (γ) . (6.17)

Reversing the direction of shear will not change the normal stress components, andtherefore these will be even functions of the shear rate. The arbitrary pressure can beeliminated by taking the differences between these normal stresses. We thus definethe first and the second normal stress differences by

Sxx − Syy = N1 (γ) , Syy − Szz = N2 (γ) , (6.18)

respectively. These normal stress differences are even function of the shear rate, andthey vanish when the shear rate is zero. This is made explicit by writing

1The fluid has no way of knowing that the experimenter has suddenly changed his mind andre-defined x1 as −x1. It will continue merrily reporting the same shear rate and stress. To the exper-imenter, however, he will notice that the shear rate and the shear stress have the same magnitudes asbefore, but they have changed signs. He therefore concludes that the shear stress is an odd functionof the shear rate. The same story applies to normal stresses; they are even functions of the shearrate.


N1 (γ) = γ2ν1 (γ) , N2 (γ) = γ2ν2 (γ) , (6.19)

where ν1 and ν2 are called the first and second normal stress coefficients. They areeven functions of the shear rate. Collectively, η (γ) , N1 (γ) and N2 (γ) are calledviscometric functions. They are material properties for the fluid.

In a general steady viscometric flow, the above reasoning continues to hold, andwe write the stress tensor using the base vectors a, b, c as

T = −PI + γη (ab + ba) + (N1 + N2) (aa + bb) − N1bb.

Here, P is the hydrostatic pressure, I = aa + bb + cc is the unit tensor, and

A1 = γ (ab + ba) , A21 = γ2 (aa + bb) , (6.20)

A2 = A1L + LTA1 = γ2 (ab + ba) ab + γ2ba (ab + ba) = 2γ2bb,

Writing the total stress tensor asT = −PI+S, the extra stress tensor S can bewrittenas

S = ηA1 + (ν1 + ν2)A21 − ν1

2A2. (6.21)

This resembles the second-order fluid model (4.71), but with important differences.The second-order fluidmodel is a slow-flow approximation to the simple fluid, and allthe coefficients in themodel are constant, whereas (6.21) is a restriction on the simplefluid in steady viscometric flows, and therefore is valid only in steady viscometricflows for all simple fluids. All the coefficients in (6.21) are functions of the strainrate. More importantly, (6.21) is not a model of a fictitious fluid, but is a proventheorem for steady viscometric flows [17].

6.2.1 Controllable and Partially Controllable Flows

If the velocity field can be fully determined (with or without inertia), no matter whatform the viscometric functions may take, then the flow is said to be controllable.There are flows inwhich the kinematics are fully determined by the viscosity functionalone – the normal stress differences do not influence the velocity field. Such flowsare called partially controllable.

Problems

Problem 6.1 Show that the velocity gradient for (6.3) is

L = γab. (6.22)

http://dx.doi.org/10.1007/978-3-319-62000-8_4

Problems 119

Fig. 6.2 Shear flow betweeninclined planes

Consequently, show that all the flows represented by (6.3) are isochoric. Show thatthe shear rate is |γ| .Problem 6.2 Show that the shear rate for the helicoidal flow (6.14) is

γ2 = (r2 + c2

) ∇ω · ∇ω. (6.23)

Problem 6.3 Consider the shear flow between two tilted plates: the first plate is atrest and the second plate, tilted at an angle θ0 to the first plate, is moving with avelocity U in the k−direction, as shown in Fig. 6.2. Show that

u = Uθ

θ0ez . (6.24)

Show that the stress is given by

T = −PI + ηγ (ezeθ + eθez) + (N1 + N2) ezez + N2eθeθ, (6.25)

where the shear rate is γ = U/rθ0, and

P = P (r0) + I2 (γ) − I2 (γ0) , I2 (γ) =∫ γ

0γν2dγ. (6.26)

Suggest a way to measure N2 based on this.

Problem 6.4 The flow between two parallel, coaxial disks is called torsional flow.In this flow, the bottom disk is fixed, and the top disk rotates at an angular velocityof Ω . The distance between the disks is h. Neglecting the fluid inertia, show that

u = Ωrz

heθ, γ = Ω

r

h. (6.27)


Show that the torque required to turn the top disk is

M = 2π∫ R

0γη (γ) r2dr , (6.28)

where R is the radius of the disks. Show that the pressure is

P (r) =∫ γR

γ

γ (ν1 + ν2) dγ. (6.29)

From the axial stress, show that the normal force on the top disk is

F = πR2γ−2R

∫ γR

0γ (N1 − N2) dγ, (6.30)

where γR = ΩR/h is the shear rate at the rim r = R.

By normalizing the torque and the force as

m = M

2πR3, f = F

πR2, (6.31)

show that

η (γR) = m

γR

[3 + d lnm

d ln γR

], (6.32)

and

N1 (γR) − N2 (γR) = f

(2 + d ln f

d ln γR

). (6.33)

Relations (6.32), (6.33) are the basis for the operation of the parallel-disk viscometer.

Problem 6.5 In a pipe flow, of radius R and pressure drop/unit length ΔP/L , showthat the flow rate is

Q = 8π

(L

ΔP

)3 ∫ τw

0

τ 3

ηdτ , (6.34)

where τ is the shear stress, and τw is the shear stress at the wall. In terms of thereduced discharge rate,

q = Q

πR3, (6.35)

show that

dq

dτw

= 1

η (τw)− 3q

τw

, (6.36)

Problems 121

or

γw = q (τw)

[3 + d ln q

d ln τw

]. (6.37)

The relation (6.37) is due toRabinowitch [74] and is the basis for capillary viscometry.

Chapter 7Polymer Solutions

From Atoms to Flows

In the microstructure approach to the quest for a relevant constitutive equation forthe complex-structure fluid, a relevant model for the microstructure is postulated,and the consequences of the micromechanics are then explored at the macrostruc-tural level, with appropriate averages being taken to smear out the details of themicrostructure. The advantage of this is that the resulting constitutive equation isexpected to be relevant to the material concerned; and if a particular phenomenonis not well modelled, the microstructural model can be revisited and the relevantphysics put in place. This iterative model-building process is always to be preferredover the continuum approach. In this chapter, we will concentrate on the constitutivemodelling of dilute polymer solutions.

7.1 Characteristics of a Polymer Chain

Viscoelastic fluids are predominantly suspensions of long-chain polymer moleculesin a solvent. We will not be concerned with aspects of polymer chemistry here. It issufficient for us to know that a typical polymer, with a molecular weight of the order107 grams per mole, has about 105 repeating (monomer) units, and its monomermolecular weight is of the order 102 grams per mole.

7.1.1 Random-Walk Model

The simplest representation of a polymer molecule is a freely rotating chain with Nsegments, as illustrated in Fig. 7.1. Each segment has a constant bond length b, butis randomly oriented in space. The segments are not physical entities; they can crossover each other in space and can be freely rotating irrespective of the neighbouring


123

124 7 Polymer Solutions

Fig. 7.1 A random-walkmodel of a polymer chain

1

2

N

N+1

4

R

3

r4

R1

R2

R3

RN

segments. Its end-to-end vector is given by

R =N∑

j=1

R j . (7.1)

On the average, we expect that a randomly oriented vector will have zero mean, andits square is constant as a consequence of the constant bond length:

⟨R j

⟩ = 0,⟨R j · R j

⟩ = b2 (no sum). (7.2)

Here and elsewhere, the angular brackets denote the average with respect to theprobability density function of the variable concerned. Thus, if P (R, t) dR is theprobability of finding a segment of configuration between R and R + dR at time tthen the n-th moment of R is defined as

⟨RR . . .R

n times

⟩=

∫RR . . .R

n timesP (R, t) dR. (7.3)

Thus, a chain has zero end-to-end vector on average,

〈R〉 =N∑

j=1

⟨R j

⟩ = 0, (7.4)

and its mean square is

〈R · R〉 = ⟨R2⟩ =

N∑

j=1

N∑

k=1

⟨R j · Rk

⟩ =N∑

j=1

⟨R j · R j

⟩ +N∑

j=1

N∑

k �= j

⟨R j · Rk

⟩.

7.1 Characteristics of a Polymer Chain 125

Since R j is independent of Rk, k �= j,⟨R j · Rk

⟩ = 0, the last double sum on theright side is zero. From (7.2), we have

⟨R2

⟩ = Nb2. (7.5)

Non-freely-rotating chains, for example, chains where the bond angle between suc-cessive segments remains fixed, has been considered [26] – they all lead to

⟨R2

⟩ = kNb2, (7.6)

where k is a constant depending on the geometry.

Strong Flow. Thus, in the random walk model, a chain of extended contour length

Nb is expected to have a linear dimension of O(√

Nb). It is difficult to unravel a

polymer molecule, and flows that can do this are called strong.

Diffusion Equation. For a given chain of N segments, the end-to-end vector isa stochastic quantity1 and must be characterised by its probability density functionP (R; N ) .Now consider a chain of N segments, with an end-to-end vector ofR − b.The probability of N + 1 segments having an end-to-end vector R is precisely theprobability of the last segment (N + 1) having a bond vector b, conditional on thefirst N segments having an end-to-end vector of R − b:

P (R; N + 1) =∫

P (R − b; N ) Pb (b) db. (7.7)

This is the property ofMarkovian processes [52] – namely what happens at any giveninstant depends only on the instantaneous state of the system, not on its previoushistory; Pb (b) is called the transition probability, which depends on both currentand next states of the process. Here we assume that b is completely independent ofthe current state. The equation can be expanded in a Taylor series for |b| � |R| ,

P(R; N ) + ∂

∂NP (R; N ) + · · ·

=∫

Pb (b)

[P (R; N ) − b · ∇P (R; N ) + 1

2!bb : ∇∇P (R; N ) + · · ·]db.

The average with respect to b is taken, noting that its distribution is purely random,

∫Pb (b) db = 1,

∫bPb (b) db = 0,

∫bbPb (b) db = b2

3I.

1A stochastic process is a family of random variables X (t), where t is the time, X is a randomvariable, and X (t) is the value observed at time t . The totality of {X (t) , t ∈ R} is said to be arandom function or a stochastic process.


Thus, we obtain the following diffusion equation for the process R:

∂

∂NP (R; N ) = b2

6∇2P (R; N ) . (7.8)

The solution of this, subjected to the “initial condition”

limN→0

P (R; N ) = δ (R)

is the Gaussian distribution

P (R; N ) =(

3

2πNb2

)3/2

exp

(− 3R2

2Nb2

). (7.9)

This distribution is unrealistic in the sense that there is a finite probability for R >

Nb; a more exact treatment produces the Langevin distribution [26], which vanishesat R ≥ Nb as required.

7.2 Forces on a Chain

In simple models for dilute polymer solutions, such as the Rouse bead-spring modelshown in Fig. 7.2a, a polymer chain is discretised into several effective segments,called Kuhn segments,2 each of which has a point mass (bead) undergoing somemotion in a solvent (which is treated as a continuum). Each Kuhn segment may con-tain several monomer units. Each bead accelerates in response to the forces exertedon it by the solvent, the flow process, and the surrounding beads, and consequentlythe chain will adopt a configuration. The task here is to relate the microstructureinformation to a constitutive description of the fluid. When there are only two beads,the model is called the elastic dumbbell model [46], Fig. 7.2b, which has been mostpopular in elucidating the main features of the rheology of dilute polymer solutions.

The forces acting on the beads include:

Hydrodynamic forces: These arise from the average hydrodynamic resistance ofthe motion of the polymer through a viscous solvent. Since the relevant Reynoldsnumber based on the size of the polymer chain is negligibly small, the averagemotion of the chain is governed by Stokes equations, and Stokes resistance can beused to model this. The chain is usually treated as a number of discrete points ofresistance, each having a frictional coefficient. In the simplest model, a frictionalforce of

F(d)i = ζ (ui − ri ) , (7.10)

2W. Kuhn (1899–1963) was a Professor at the Technische Hochschule in Karlruhe, and later on, inBasel, Switzerland. He is most famous for the f-summation theorem in quantum mechanics.

7.2 Forces on a Chain 127

(a) Rouse Bead and Spring Model (b) Elastic Dumbbell Model

R1R i

r1

r2ri

ri+1

rN+1

r1

r2

R

Fig. 7.2 (a) Rouse model and (b) the elastic dumbbell model of a polymer chain

is assumed to be acting on the i-th bead, which has a velocity ri − ui relative tothe solvent, and ζ is a constant frictional coefficient, which is usually taken as6πηsa (Stokes drag on the bead), where ηs is the viscosity of the solvent, and arepresents the size of the bead. To obtain a more realistic model of the nature ofthe dependence of the frictional forces on the configuration and the deformationof the polymer chain, ζ can be allowed to depend on the length of the segmenti , or indeed it may be considered to be a second-order tensor, which reflectsthe physical idea that the resistance to the motion perpendicular to the chain ismuch higher than that along the chain. The simple frictional model (7.10) neglectshydrodynamic interaction with other beads, and is usually called the free-drainingassumption. Hydrodynamic interaction arises because of the solvent velocity thatappears in (7.10) contains disturbance terms due to the presence of other beads.

Tension in the chain: A chain in equilibrium will tend to curl up into a sphericalconfiguration, with the most probable state of zero end-to-end vector. However,if the chain ends are forcibly extended, then there is a tension or a spring force,arising in the chain, solely due to the fewer configurations available to the chain.To find the expression for the chain tension, we recall that the probability densityfunction is proportional to the number of configurations available to the chain(i.e., the entropy), and thus the Helmholtz free energy of the chain is [26]

Fr(r j

) = A (T ) − kT ln P(r j

), (7.11)

where A (T ) is a function of the temperature alone. The entropic spring forceacting on bead i is

F(s)i = −∂Fr

∂ri= kT

∂ ln P

∂ri. (7.12)


For the Gaussian chain (7.9), the tension required to extend the chain by a vectorR is

F = 3kT

Nb2R. (7.13)

This applies to individual beads of Fig. 7.2a. Thus if each segment consists of nKuhn segments, each of bond length b, then the force on bead i due to the chaintension is

F(s)i = 3kT

nb2(ri+1 − ri + ri−1 − ri ) = 3kT

nb2(Ri − Ri−1) . (7.14)

This implies that the beads are connected by linear springs of stiffness

H = 3kT

nb2. (7.15)

A distribution, which better accounts for the finite segment length, is the Langevindistribution,3 and this results in the so-called inverse Langevin spring law for thechain tension:

L

(bF

kT

)= coth

(bF

kT

)− bF

kT= r

nb, (7.16)

where F is the magnitude of the force, r the magnitude of the extension, and theLangevin function is defined as L(x) = coth x − x . A useful approximation ofthe Langevin spring law is the Warner spring [6]

Hi = 3kT

nb21

1 − (Ri/Li )2, (7.17)

where Li = nb is the maximum extended length of segment i . This stiffnessapproaches infinity as Ri → Li .

Brownian forces: Brownian forces are the cumulative effect of the bombardment ofthe chain by the solvent molecules.4 These forces have a small correlation timescale, typically the vibration period of a solvent molecule, of the order 10−13sfor water molecules. If we are interested in time scales considerably larger than

3P. Langevin (1872–1940) introduced the stochastic DE (7.20) in 1908, and showed that the particleobeys the same diffusion equation as described by Einstein (1905).4The random zig-zag motion of small particles (less than about 10µm) is named after R. Brown(1773–1858), an English botanist, who mistook this as a sign of life. He travelled with MatthewFlinders to Australia in 1801 on the ship Investigator as a naturalist. The correct explanation ofthe phenomenon was given by Perrin (Fig. 7.3). Brownian particles are those undergoing a randomwalk, or Brownian motion.

7.2 Forces on a Chain 129

Fig. 7.3 The Frenchphysicist Jean BaptistePerrin (1870–1942) gave thecorrect explanation to therandom motion of smallparticles as observed byBrown and confirmed thetheoretical calculations byEinstein. For this work hewas awarded the Nobel Prizefor Physics in 1926. He wasalso the founder of theCentre National de laRecherche Scientifique

this correlation time scale, then the Brownian forces acting on bead i , F(b)i , can

be considered as white noise having a zero mean and a delta autocorrelationfunction5:

⟨F(b)i (t)

⟩= 0,

⟨F(b)i (t + s)F(b)

j (t)⟩= 2δi jδ (s) f . (7.18)

This autocorrelation states that the strength of the Brownian forces is the measureof the integral correlation function over a time scale which is considerably greaterthan the correlation time scale of the Brownian forces:

2f =∫ ∞

−∞

⟨F(b)i (t + s)F(b)

i (t)⟩ds. (7.19)

The strength of the Brownian forces is not an arbitrary quantity determined by aconstitutive modelling process; it is in fact related to the mobility of the Brownianparticle - this is the essence of the fluctuation-dissipation theorem which wediscuss next.

5This approximation is called white noise, i.e., Gaussian noise of all possible frequencies uniformlydistributed. Sometimes it is called “rain-on-the-roof” approximation: two (or more) rain drops donot fall on the same spot on the roof.


7.3 Fluctuation-Dissipation Theorem

7.3.1 Langevin Equation

There are several fluctuation-dissipation theorems [47], relating the strength of thefluctuating quantity to the macroscopic “mobility” of the phenomenon concerned.The following development is patterned after Hinch [36].

All micro-mechanical models for a polymer chain in a dilute solution can bewritten as

mx + ζx + Kx = F(b) (t) , (7.20)

called the Langevin equation (for a portray of Langevin, see Fig. 7.4), where thesystem state is represented by the finite-dimensional vector x, such that its kineticenergy is 1

2m : xx, and its generalised linearmomentum ismx,m being a generalisedinertia tensor. The inertia tensorm is defined through the kinetic energy, and thereforethere is no loss of generality in considering only symmetricm. The system is acted onby a frictional force, which is linear in its state velocities, a restoring force (possiblynonlinear in x), and a Brownian force F(b) (t) . We assume that the frictional tensorcoefficient ζ is symmetric. This system can be conveniently started from rest at timet = 0.

Since theBrownian force has onlywell-defined statistical properties, the Langevinequation (7.20) must be understood as a stochastic differential equation [52]. Itcan only be “solved” by specifying the probability distribution W (u, x, t) of theprocess {u = x, x} defined so that W (u, x, t) dudx is the probability of finding the

Fig. 7.4 Paul Langevin(1872–1946) was a Professorof Physics at College deFrance. He studied underLord Kelvin and Pierre Curie

7.3 Fluctuation-Dissipation Theorem 131

process at the state between {u, x} and {u + du, x + dx} at time t . Prescribing theinitial conditions u(0) = u0, x(0) = x0 for (7.20) is equivalent to specifying a deltaprobability at time t = 0:

W (u, x, 0) = δ (u − u0) δ (x − x0) .

The distribution W (u, x, t) is the phase space description of the stochastic process{u, x}. The dependence of W on x or u can be eliminated by integrating out theunwanted independent variable. Then, we have either a velocity space, or a configu-ration space description, respectively.

7.3.2 Equi-Partition of Energy

The existence of the temperature T of the surrounding fluid demands that the distri-bution in the velocity space must satisfy the equi-partition energy principle:

limt→∞ 〈x (t) x (t)〉 = kTm−1, (7.21)

as demanded by the kinetic theory of gases, i.e., each mode of vibration is associatedwith a kinetic energyof 1

2kT .Wenowexplore the consequence of this on theLangevinsystem (7.20).

There are three time scales in this system:

1. τr the relaxation time scale of the chain in its lowest mode; this time scale is ofthe order

∣∣ζK−1∣∣ ;

2. τi the much shorter inertial relaxation time scale of the chain; this time scale isof the order

∣∣mζ−1∣∣ ;

3. τc the still shorter correlation time scale of the Brownian force – this is of thesame order as the relaxation time scale of a solvent molecule.

7.3.3 Fluctuation-Dissipation Theorem

In general we have τc � τi � τr , but the estimate of τi can vary considerably.To derive a fluctuation-dissipation theorem for the Langevin equation (7.20), it issufficient to consider only events on the time scale τi . In this time scale, m (x) andζ (x) can be replaced by their local values, i.e., regarded as constant, and the statevector can be re-defined to eliminate Kx so that (7.20) becomes

x + m−1ζx = m−1F(b) (t) , (7.22)


subjected to the initial rest state

x (0) = 0 = x (0) . (7.23)

Note that A · eAt = eAt · A,

d

dt

(eAtv

) = eAt (v + Av) .

Identify A = m−1ζ and v = x, the solution to (7.22) is therefore given by

x (t) =∫ t

0exp

{m−1ζ

(t ′ − t

)}m−1F(b)

(t ′)dt ′. (7.24)

This leads to the expectation

〈x (t) x (t)〉 =∫ t

0

∫ t

0exp

{m−1ζ

(t ′ − t

)}m−1

⟨F(b)

(t ′)F(b)

(t ′′

)⟩

· m−1 exp{ζm−1

(t ′′ − t

)}dt ′dt ′′.

If τc � τi , the white noise assumption for the Brownian force can be used, so that

⟨F(b)

(t ′)F(b)

(t ′′

)⟩ = 2δ(t ′ − t ′′

)f,

giving

〈x (t) x (t)〉 = 2∫ t

0exp

{m−1ζ

(t ′ − t

)}m−1fm−1 exp

{ζm−1

(t ′ − t

)}dt ′.

This can be integrated by parts to yield

〈x (t) x (t)〉 = −2 exp{−m−1ζτ

}ζ−1fm−1 exp

{−ζm−1t}

+ 2ζ−1fm−1 − ζ−1m 〈x (t) x (t)〉 ζm−1.

In the limit of t → ∞ (i.e., t τi but t � τr so that the equation of state remainslinear), the equi-partition of energy (7.21) holds, and we have

kTm−1 = 2ζ−1fm−1 − kTm−1,

or

f = kTζ. (7.25)


This is the fluctuation-dissipation theorem, relating the strength of the Brownianforce to the mobility of the Brownian system; any dependence on the configurationof f is inherited from that of ζ.

7.3.4 Diffusivity Stokes–Einstein Relation

The diffusivity of a Brownian particle is defined by

D = limt→∞

1

2

d

dt〈x (t) x (t)〉 . (7.26)

This is equivalent to

D = limt→∞

1

2

∫ t

0〈x (t) x (t − τ ) + x (t − τ ) x (t)〉 dτ . (7.27)

In Problem 7.1, it can be shown that

D = kTζ−1. (7.28)

7.3.5 Fokker–Planck Equation

Asmentioned earlier, the Langevin equation can only be considered solved when theprobability function of the process is specified. In the limit m → 0 it can be shownthat the configuration probability density function φ (x, t) satisfies

∂φ

∂t= lim

Δt→0

∂

∂x·[ 〈ΔxΔx〉

2Δt· ∂φ

∂x− 〈Δx〉

Δtφ

]. (7.29)

This is the Fokker–Planck,6 or Smoluchowski7 diffusion equation. A clear expo-sition of this can be found in Chandrasekhar [14], Fig. 7.5.

It can be shown that (Problem 7.2), in the limit of m → 0, the Fokker–Planckequation is

∂φ

∂t= ∂

∂x·[kTζ−1 ∂φ

∂x+ ζ−1 · Kxφ

]. (7.30)

6A.D. Fokker derived the diffusion equation for a Brownian particle in velocity space in 1914. Thegeneral case was considered by M. Planck (1858–1947) in 1917.7The general solution to the random walk problem in one dimension was obtained by M. vonSmoluchowski in 1906.


Fig. 7.5 SubrahmanyanChandrasekhar (1910–95)was an outstanding Indianastrophysicist. He worked onvarious aspects of stellardynamics and was awardedthe Nobel Prize in 1983

7.3.6 Smoothed-Out Brownian Force

Since the probability must satisfy

∫φ (x, t) dx = 1, (7.31)

an application of the Reynolds transport theorem yields

∂φ

∂t+ ∂

∂x· (xφ) = 0, (7.32)

assuming that we deal with an equivalent “deterministic” system x. By comparing(7.32)–(7.30), the velocity of this equivalent system must satisfy

x = −ζ−1 · Kx − kTζ−1 · ∂ ln φ

∂x.

That is,

ζx + Kx = −kT∂ ln φ

∂x. (7.33)

Comparing this to (7.20), it is as though the Brownian force has been replaced by

F(b) (t) = −kT∂ ln φ

∂x. (7.34)


Of course, this equation is notmathematicallymeaningful: the left side is a stochastic,and the right side is a deterministic quantity. It is so defined for the sole purpose ofgetting the correct diffusion equation (7.30). Written in the form (7.34), this force iscalled the smoothed-out Brownian force; it is a device employed inmost texts dealingwith kinetic theories of polymers, e.g., Bird et al. [6].

7.4 Stress Tensor

There are several ways to derive the expression for the stress tensor contributed bythe polymer chains in a dilute solution. One is the probabilistic approach (Bird etal. [6]), where the number of polymer chains straddling a surface and the net forceacting on that surface by the chains are calculated. The force per unit area can berelated to the stress tensor. Another approach is to calculate the free energy of thechain from its entropy, and the rate of work done can be related to the dissipation dueto the presence of the chains from which the expression for the stress tensor can bederived [48].We present a simple mechanistic approach here to derive the expressionfor the stress tensor.

Consider the bead-spring model for a polymer chain, as shown in Fig. 7.6. Thetension in the i-th Kuhn segment is denoted by fi . If the chain is Gaussian, then

fi = HiRi , (no sum) (7.35)

whereRi is the end-to-end vector, and Hi is the stiffness of segment i , given in (7.15).This bead-spring model is also called the Rouse model, and the chain is known asthe Rouse chain. Using the approach of Landau and Lifshitz [47] and Batchelor [4],the fluid is taken as an effective continuum made up of a homogeneous suspensionof Rouse chains. Its effective stress is simply the volume-averaged stress:

〈T〉 = 1

V

∫

VTdV = 1

V

∫

Vs

TdV + 1

V

∑

p

∫

Vp

TdV , (7.36)

Fig. 7.6 Connector force ina Rouse chain

R1 R i

ri

F1

F2

F

Fi

F3F2

F1 F3

N

FN

Fi

r1

r2

ri+1

rN+1


where T is the total stress, V is a representative volume containing several chains,and is made up of a solvent volume Vs and a polymer volume

∑Vp. In the solvent

volume, the stress is simply the solvent stress, and we have

1

V

∫

Vs

T(s)dV = 1

V

∫

VT(s)dV − 1

V

∑

p

∫

Vp

T(s)dV . (7.37)

With a Newtonian solvent, the first term on the right of (7.37) is simply a Newtonianstress,

1

V

∫

VT(s)dV = −p1I + 2ηsD, (7.38)

where p1 is the hydrostatic pressure, ηs the solvent viscosity, and D the strain ratetensor. The second term on the right of ( 7.37) is

1

V

∑

p

∫

Vp

T(s)dV = 1

V

∑

p

∫

Vp

[−pI + ηs(∇u + ∇uT

)]dV

= −p2I + ηs

V

∑

p

∫

Sp

(un + nu) dS. (7.39)

Since the chain is modelled as a series of discrete beads, where the interaction withthe solvent and other segments takes place, the surface of the chain p consists of thesurfaces of the beads. The connectors are entirely fictitious, they are allowed to crossone another; thus the model is sometimes called the phantom chain model.

On the bead surface, the velocity is regarded as uniform, and can be taken out ofthe integral. Thus ∫

Sp

undS = u∫

Sp

ndS = 0,

by an application of the divergence theorem. The contribution from (7.39) is thereforeonly an isotropic stress, which can be lumped into the hydrostatic pressure.

Next, if we consider the chain as a continuum as well, then from the force equilib-riumwemust have∇ · T = 0 in the chain, and thus Tik = ∂(Ti j xk)/∂x j .The volumeintegral can be converted into a surface integral, and the contribution to the effectivestress from the polymer chains is

1

V

∑

p

∫

Vp

TdV = 1

V

∑

p

∫

Sp

xT · ndS = ν

∫

Sp

xT · ndS, (7.40)

where Sp is the surface of a representative chain in V , T · n is the traction arising inthe chain due to the interaction with the flow, and ν is the number density of the chain

7.4 Stress Tensor 137

(number of chains per unit volume); the passage to the second equality is permissiblebecause of the homogeneity assumption which allows us to just consider one genericchain. We can now replace the integral in (7.40) by a sum of integrals over the beads:

∫

Sp

xT · ndS =∑

i

∫

bead ixT · ndS.

On bead i , x can be replaced by ri and taken outside the integral, and the remainingintegral of the traction on the surface of bead i is therefore the drag force, whichbead i exerts on the solvent, and is proportional to the velocity of the bead relativeto the solvent:

F(d)i = −ζ (ri − ui ) .

In the absence of inertia, this force is equal to the connector forces plus the Brownianforces acting on the beads:

∫

Sp

xT · ndS = −r1f1 + r2 (f1 − f2) + · · · + rN+1fN −N+1∑

i=1

riF(b)i (7.41)

=N∑

i=1

Ri fi −N+1∑

i=1

riF(b)i .

Next, the ensemble average with respect to the distribution function of Ri is taken.The contribution from the Brownian forces is only an isotropic stress, as can beshown either by using the expression for the smoothed-out Brownian forces, or byintegrating theLangevin equations directly. This is demonstrated using the expressionfor the smoothed-out Brownian force,

N+1∑

i=1

riF(b)i = −kT

N+1∑

i=1

∫ri

∂ ln φ

∂riφdr1 . . . drN+1

= −kTN+1∑

i=1

∫ri

∂φ

∂ridr1 . . . drN+1.

The integral can be evaluated by parts,

∫ri

∂φ

∂ridr1 . . . drN+1 =

∫∂

∂ri(riφ) dr1 . . . drN+1 − I

∫φdr1 . . . drN+1.

(7.42)

The first integral on the right of (7.42) is a volume integral over an unboundeddomain. It can be converted into surface integral at infinity. On this surface, φ → 0and therefore the resulting surface integral vanishes. The second integral on the right


of (7.42) is unity, because φ is the probability density function. Thus,

N+1∑

i=1

riF(b)i = kT (N + 1)I. (7.43)

7.4.1 Kramers Form

With all isotropic stresses absorbed in the pressure, the polymer-contributed stress(7.41) is

S(p) = ν

N∑

i=1

〈Ri fi 〉 = ν

N∑

i=1

〈HiRiRi 〉. (7.44)

This is called the Kramers form for the polymer-contributed stress. The total stresstensor in a dilute polymer solution is

〈T〉 = −pI + S(s) + S(p) = −pI + 2ηsD + ν

N∑

i=1

〈HiRiRi 〉. (7.45)

7.5 Elastic Dumbbell Model

The simplest model designed to capture the slowest, and in many ways, the mostimportant relaxation mode of a polymer chain, is the elastic dumbbell model firstproposed by Kuhn [46] (Fig. 7.2b). Here we care only about the end-to-end vector ofthe polymer chain, and all interactions between the solvent and the chain are localisedat two beads, located at the chain ends, r1 and r2. Each bead is associated with africtional factor ζ and a negligible mass m. We will assume a Gaussian chain, withthe constant spring stiffness

H = 3kT

Nb2,

where N is the number of effectiveKuhn segments in the dumbbell, each ofwhich hasan extended length b. Furthermore, the frictional coefficient ζ = 6πηsa is assumedto be constant, where ηs is the solvent viscosity and a represents the radius of thebeads. The model is also called the linear elastic dumbbell model to emphasise thelinear force law being used. Although the general equations have been developed inthe previous section, it is instructive to write down all the equations again, for thisparticular case.

7.5 Elastic Dumbbell Model 139

7.5.1 Langevin Equations

The equations of motion are,

mr1 = ζ (u1 − r1) + H (r2 − r1) + F(b)1 (t) ,

mr2 = ζ (u2 − r2) + H (r1 − r2) + F(b)2 (t) ,

(7.46)

where ui = u (ri ) is the fluid velocity evaluated at the location of the bead i , andF(b)i (t) is the Brownian force acting on bead i . The fluctuation-dissipation theorem

(7.25) can be used to relate the strength of the Brownian forces to the mobility of thebeads:

⟨F(b)

j (t)⟩= 0,

⟨F(b)i (t + s)F(b)

j (t)⟩= 2kT ζδ (s) δi j I. (7.47)

Let us now define the centre of gravity and the end-to-end vector of the dumbbellrespectively by

R(c) = 1

2(r2 + r1) , R = r2 − r., (7.48)

The solvent velocity can be expanded about the centre of gravity,

u1 = u(c) − 12R · ∇u(c) + 1

8RR : ∇∇u(c) + O(R3

),

u2 = u(c) + 12R · ∇u(c) + 1

8RR : ∇∇u(c) + O(R3

),

(7.49)

where the superscript c denotes an evaluation at the centre of gravity. From (7.46),

mR(c) = ζ(u(c) − R(c)

) + 18ζRR : ∇∇u(c) + F(b,c) (t) ,

mR = ζ(LR − R

) − 2HR + F(b) (t) ,(7.50)

where L = (∇u(c))T

is the velocity gradient evaluated at the centre of gravity of thedumbbell, and

F(b,c) = 1

2

(F(b)1 + F(b)

2

), F(b) = F(b)

2 − F(b)1 (7.51)

are the Brownian forces acting on the centre of gravity and the end-to-end vector.From (7.47),

⟨F(b,c) (t)

⟩ = 0,⟨F(b,c) (t + s)F(b,c) (t)

⟩ = kT ζδ (s) I,⟨F(b) (t)

⟩ = 0,⟨F(b) (t + s)F(b) (t)

⟩ = 4kT ζδ (s) I.(7.52)


With negligible mass, the Langevin equations (7.50) become

R(c) = u(c) + 18RR : ∇∇u(c) + ζ−1F(b,c) (t) ,

R = LR − 2ζ−1HR + ζ−1F(b) (t) .(7.53)

7.5.2 Average Motion

If the flow is homogeneous, i.e., L is constant, ∇∇u(c) = 0, and the dumbbell’scentre of gravity drifts just like a particle of fluid,

⟨R(c)

⟩ = ⟨u(c)

⟩ = L⟨R(c)

⟩ = u(⟨R(c)

⟩). (7.54)

A migration from the streamline of the centre of gravity will be induced by a non-homogeneous flow field. This migration is

⟨R(c) − u(c)

⟩ = 1

8〈RR〉 : ∇∇u(c). (7.55)

The average end-to-end vector evolves in time according to

⟨R

⟩ = L 〈R〉 − 2Hζ−1 〈R〉 , (7.56)

which consists of a flow-induced stretching (first term on the right) plus a restor-ing mechanism (second term on the right) due to the connector spring force. Theparameter

λ = ζ

4H= ζNb2

12kT(7.57)

is called the Rouse relaxation time.

7.5.3 Strong and Weak Flows

Equation (7.56) has been used as a basis for delineating between strong and weakflows: strong flows are those in which the flow-induced deformation overcomes therestoring force allowing the microstructure (as represented by 〈R〉) to grow expo-nentially in time. Otherwise the flow is weak.

Since we are more interested in the end-to-end vector, the process R(c) can nowbe discarded. The Fokker–Planck equation for the density distribution function forR reads


∂φ

∂t= ∂

∂R·[2kT

ζ

∂φ

∂R−

(LR − 2H

ζR

)φ

]. (7.58)

In many cases, there is no need to find the full probability distribution – all we wantis 〈RR〉 , or the equation governing the evolution of 〈RR〉 , since this quantity isrelated to the stress. This can be accomplished without solving for φ.

First, from (7.53)

d

dtRR = RR + RR (7.59)

= L · RR + RR · LT − 4Hζ−1RR + ζ−1(RF(b) + F(b)R

).

Secondly, R and F(b) have widely different time scales. Thus

⟨RF(b)

⟩ = ⟨R (Δt)F(b) (Δt)

⟩

=⟨([

LR − 2Hζ−1R]Δt + ζ−1

∫ Δt

0F(b) (t) dt

)F(b) (Δt)

⟩

= ζ−1∫ Δt

0

⟨F(b) (t)F(b) (Δt)

⟩dt = 4kT I

∫ Δt

0δ (t − Δt) dt

= 2kT I.

Thus, from (7.59),

d

dt〈RR〉 = L 〈RR〉 + 〈RR〉LT − 4H

ζ〈RR〉 + 4kT

ζI.

Re-arranging, and recalling (7.57),

〈RR〉 + λ

{d

dt〈RR〉 − L 〈RR〉 − 〈RR〉LT

}= 1

3Nb2I. (7.60)

Next using (7.45),

〈T〉 = −pI + S(s) + S(p) = −pI + 2ηsD + νH 〈RR〉 . (7.61)

7.5.4 Upper-Convected Maxwell Model

Since the polymer-contributed stress is νH 〈RR〉 , we can multiply (7.60) with νHto generate the equation for S(p):

S(p) + λ

{d

dtS(p) − LS(p) − S(p)LT

}= GI, (7.62)


where

G = 1

3Nb2νH = νkT . (7.63)

The derivative operator implicitly define in the braces (acting on S(p)) on the leftside of (7.62) is one of the many derivatives introduced by Oldroyd [64] to guaranteethe stress tensor objectivity. It is called the upper-convected derivative, δ/δt, and isdefined by

δAδt

= dAdt

− LA − ALT . (7.64)

We customarily re-define the polymer-contributed stress as

S(p) = GI + τ (p). (7.65)

Then, since

δ

δtI = − (

L + LT) = −2D,

we obtain

τ (p) + λδ

δtτ (p) = 2ηpD, (7.66)

where

ηp = Gλ = νζNb2

12= 1

2πνaNb2ηs (7.67)

is the polymer-contributed viscosity. The model (7.62), or (7.66), is called the UpperConvected Maxwell (UCM) model in honour of Maxwell, who introduced the linearversion in his kinetic theory of gases in 1867. Since S(p) is proportional to 〈RR〉,it is positive definite, whereas τ (p) is not. In some numerical applications, (7.62)may be preferred to (7.66), because the lack of positive definiteness in S(p) can beconveniently tested numerically; this lack can be used as an indication of impendingnumerical divergence.

7.5.5 Oldroyd-B Model

When the solvent and the polymer-contributed stresses are combined, c.f. (7.61),

S = S(s) + τ (p) = 2ηsD + τ (p), (7.68)


one has

(S − 2ηsD) + λδ

δt(S − 2ηsD) = 2ηpD

S + λ1δSδt

= 2η

(D + λ2

δDδt

), (7.69)

where λ1 = λ is the relaxation time, η = ηs + ηp is the total viscosity, λ2 = ληs/ηis the retardation time. The model (7.69) is called the Oldroyd fluid B, or Oldroyd-Bmodel.

The Oldroyd-B constitutive equation qualitatively describes many features of theso-called Boger fluids.8 In a steady state simple shear flow, this constitutive equationpredicts a constant viscosity, a first normal stress difference which is quadratic inthe shear rate, and a zero second normal stress difference. In an unsteady state shearflow, the stresses increase monotonically in time to their steady values, without stressovershoots which are sometimes observed with some dilute polymer solutions. In anelongational flow, the elongational viscosity becomes infinite at a finite elongationrate of 1/(2λ) – these will be explored in a series of problems.

7.6 Main Features of the Oldroyd-B Model

Recall the relative strain tensor

Ct (t − s) = Ft (s)T Ft (s) , Ct (s)

−1 = Ft (s)−1 Ft (s)

−T ,

Ft (s) = F (t − s)F (t)−1 , Ft (t − s)−1 = F (t)F (t − s)−1 ,

F (t) = L (t)F (t) , Ft (s)−1 = L (t)Ft (s) , Ft (s)

−T = Ft (s)L (t)T ,

Ct (s)−1 = L (t)Ct (s)

−1 + Ct (s)−1 L (t)T . (7.70)

From the results (7.70), it can be shown that theUCMmodel is solved by (Problem7.5)

S(p) (t) = G

λ

∫ t

−∞e(s−t)/λCt (s)

−1 ds = GI + τ (p). (7.71)

This integral version of the UCMmodel is called the Lodge rubber-like liquid model[53]. It was derived from a network of polymer strands, a model meant for concen-trated polymer solutions and melts. It is remarkable that two models for two distinctmicrostructures, a dilute suspension of dumbbells and a concentrated network ofpolymer strands, share a common constitutive framework.

8Dilute solutions of polymers in highly viscous solvents [9].


7.6.1 Simple Flows

In a simple shear flow, with a time-dependent shear rate γ (t) , the stress componentsof the UCM model (7.66) obey

τ(p)11 + λ

{τ

(p)11 − 2γτ

(p)12

}= 0,

τ(p)22 + λτ

(p)22 = 0,

τ(p)33 + λτ

(p)33 = 0,

τ(p)12 + λ

{τ

(p)12 − γτ

(p)22

}= ηpγ.

(7.72)

If the stresses start from zero initial states, then τ(p)22 = τ

(p)33 = 0 for all time, and the

only two non-trivial components are τ(p)11 and τ

(p)12 .

Start-Up Shear Flow. In a start-up of a shear flow, with a constant shear rate γ, thesolution to (7.72) is

τ(p)12 = ηpγ

(1 − e−t/λ

), (7.73)

τ(p)11 = 2ηpλγ2

(1 − e−t/λ

) − 2ηpt γ2e−t/λ. (7.74)

At steady state, τ (p)12 = ηpγ, τ

(p)11 = 2ηpλγ2 and thus the viscometric functions are

η = ηs + ηp, N1 = 2ηpλγ2, N2 = 0. (7.75)

Oscillatory Shear Flow. In an oscillatory flowwith shear rate γ = γ0�(eiωt

), where

� denotes the real part, (7.72) becomes

τ(p)11 + λτ

(p)11 = 2γ0�

(eiωt

)τ

(p)12 , τ

(p)12 + λτ

(p)12 = ηpγ0�

(eiωt

). (7.76)

We search for the steady solution

τ(p)12 = � (

S0eiωt

), τ

(p)11 = � (

N0e2iωt

). (7.77)

When these are substituted into (7.72), we find that

S0 = ηpγ0

1 + iλω, N0 = 2ηpλγ2

0

(1 + iλω) (1 + 2iλω). (7.78)

Consequently, the dynamic properties of the Oldroyd-fluid are

η∗ = ηs + ηp

1 + iλω, (7.79)

7.6 Main Features of the Oldroyd-B Model 145

η′ = ηs + ηp

1 + λ2ω2, η′′ = ηpλω

1 + λ2ω2, (7.80)

G ′ = Gλ2ω2

1 + λ2ω2, G ′′ = ηsω + Gλω

1 + λ2ω2. (7.81)

Elongational Flow. In the start-up of an uniaxial elongational flow, where the veloc-ity gradient is

[L] = diag (ε,−ε/2,−ε/2) ,

the stress components of the UCM model (7.66) obey (note that τ (p)22 = τ

(p)33 )

τ(p)11 + λ

(τ

(p)11 − 2ετ (p)

11

)= 2ηp ε, (7.82)

τ(p)22 + λ

(τ

(p)22 + ετ

(p)22

)= −ηp ε.

The solution is

τ(p)11 = 2ηp ε

1 − 2λε

(1 − e−(1−2λε)t/λ

), (7.83)

τ(p)22 = τ

(p)33 = − ηp ε

1 + λε

(1 − e−(1+λε)t/λ

).

Thus, if either λε ≥ 1/2 or λε ≤ −1, then at least one component of the stress growsunboundedly. This reflects the linear spring in the model that allows the end-to-endvector of the dumbbell to grow without bound in a strong flow. For −1 < λε < 1/2and at steady state, the elongational viscosity of the Oldroyd-B model can be shownto be, using (7.83),

ηE = S11 − S22ε

= 3ηs + 3ηp

(1 − 2λε) (1 + λε). (7.84)

The Trouton ratio ηE/η increases from the Newtonian value of 3, when ε = 0 andbecomes unbounded at λε approaches either −1 of 1/2.

The prediction of an infinite stress at a finite elongational rate is not physicallyrealistic. It is due to the linear dumbbell model being allowed to stretch infinitely.Constraining the dumbbell to amaximum allowable lengthwill fix this problem (e.g.,FENE dumbbell, Phan-Thien/Tanner model [6, 39, 85]).

The linear elastic dumbbell model is also inadequate in oscillatory flow: it predictsa shear stress proportional to the amplitude of the shear strain, irrespective of thelatter magnitude. This is unrealistic: in practice this proportionality is only foundwhen the shear strain is small (typically less than about 10% for polymer solutionsand melts).


7.6.2 Multiple Relaxation Time UCM Model

The frequency response of the dumbbell model is also inadequate, due to only onerelaxation time in the model. With multiple relaxation times, the Rouse model, Fig.7.1a, results in

τ (p) =N∑

j=1

τ ( j), (7.85)

τ ( j) + λ jδτ ( j)

δt= 2η jD,

where{λ j , η j

}is the discrete relaxation spectrum. The dynamic properties are now

much improved:

η′ = ηs +N∑

j=1

η j

1 + λ2jω

2, η′′ =

N∑

j=1

η jλ jω

1 + λ2jω

2, (7.86)

G ′ =N∑

j=1

G jλ2jω

2

1 + λ2jω

2, G ′′ = ηsω +

N∑

j=1

G jλ jω

1 + λ2jω

2.

In a steady shear flow, the model still predicts a constant viscosity, a quadratic firstnormal stress difference in the shear rate, and a zero second normal stress difference.The Boger fluids show little shear thinning over a large range of shear rates, but this isnodoubt due to the high solvent viscosity that completelymasks the contribution fromthe polymer viscosity; any amount of shear-thinning from the polymer contributionwould hardly show up on the total fluid viscosity. In general, dilute polymer solutionsusually show some degree of shear thinning. The fix is to adopt a more realistic forcelaw for the chain. One such model is the FENE (Finitely Extendable NonlinearElastic) model [6].

Problems

Problem 7.1 Use the solution (7.24) in (7.27) to show that

D = kTζ−1.

This is the Stokes–Einstein relation, relating the diffusivity to the mobility of aBrownian particle.

Problem 7.2 Starting from the Langevin equation in configuration space, in thelimitm → 0,

Problems 147

x = −ζ−1 · Kx + ζ−1F(b) (t) , (7.87)

show that

Δx (t) = −ζ−1 · KxΔt +∫ t+Δt

tζ−1F(b)

(t ′)dt ′. (7.88)

From this, show that

〈Δx〉 = −ζ−1 · Kx, 〈Δx (t)Δx (t)〉 = 2kTζ−1Δt (7.89)

and conclude that the Fokker–Planck equation is

∂φ

∂t= ∂


∂x+ ζ−1 · Kxφ

].

Problem 7.3 Investigate the migration problem in a plane Poiseuille flow.

Problem 7.4 Show that the solution to (7.56) is

〈R (t)〉 = e−t/2λeLtR0. (7.90)

Thus conclude that the flow is strong if

eigen (L) ≥ 1/2λ,

where eigen (L) is the maximum eigenvalue of L.

Problem 7.5 Using the result (7.70), show that the following solves the Maxwellequation (7.62):

S(p) (t) = G

λ

∫ t


−1 ds = GI + τ (p).

Chapter 8Suspensions

Particulates

Suspension is a termused to describe an effective fluidmade upof particles suspendedin a liquid; examples of such liquids abound in natural and man-made materials:blood, milk, paints, inks. The concept of a suspension is meaningful only when thereare two widely different length scales in the problem: l is a typical dimension of asuspended particle, L is a typical size of the apparatus, and l � L . When this isnot met, we simply have a collection of discrete individual particles suspended in aliquid. Most progress has been made with Newtonian suspensions, i.e., suspensionsof particles in a Newtonian liquid. The review paper byMetzner [57] contained mostof the relevant information on the subject.

If the particles are small enough (less than 10µm in size), then they will undergoBrownian motion, their micromechanics are described by a set of stochastic differ-ential equations, together with some relevant fluctuation-dissipation theorems, andthe full solution of the relevant equations can only be obtained by specifying theprobability distribution of the system.

The relative importance of Brownian motion is characterised by a Péclet number,such as Pe = O(ηs γl3/kT ), the ratio of viscous stress to stress induced by thermalexcitation, where γ is a typical strain rate, ηs is the solvent viscosity, and kT is theBoltzmann temperature. At low Péclet numbers, Brownian motion is strong, andthe particles’ orientation tends to be randomised, leading to a larger dissipation (i.e.,higher effective viscosity) thanwhen the Péclet number is large, theBrownianmotionis weak, and the particles tend to alignwith the flowmost of the time. Thus, we expectshear-thinning with the inclusion of Brownian motion (increasing shear rate leadsto an increase in the Péclet number). We will focus on non-Brownian flow regime,where the particles are large enough (typically of ˜10µm in size), but yet orders ofmagnitudes smaller than L .

With l � L , the microscale Reynolds number is small. Thus, the relevant equa-tions governing the micromechanics are the Stokes equations,

∇ · u = 0, −∇ p + η∇2u = 0. (8.1)


149

150 8 Suspensions

Stokes equations are linear and instantaneous in the driving boundary data. Conse-quently the microdynamics are also linear and instantaneous in the driving forces;only the present boundary data are important, not their past history. This does notimply that the overall response will have no memory, nor does it imply that themacroscaled Reynolds number is small. In most studies, the particle’s inertia isneglected, its inclusion may lead to a non-objective constitutive model, since thestress contributed by micro inertia may not be objective (Ryskin and Rallison [79]).

The linearity of the micromechanics implies that the particle-contributed stresswill be linear in the strain rate; in particular all the rheological properties (shearstress, first and second normal stress differences, and elongational stresses) will belinear in the strain rate. Several investigators have indeed found Newtonian behav-iour in shear for suspensions up to a large volume fraction. However, experimentswith some concentrated suspensions usually show shear-thinning behaviour, but theparticles in these experiments are in the µm range, where Brownian motion wouldbe important. Shear-thickening behaviour, and indeed, yield stress and discontinu-ous behaviour in the viscosity-shear-rate relation have been observed, e.g., Metzner[57]. This behaviour cannot be accommodated within the framework of hydrody-namic interaction alone; for a structure to be formed, we need forces and torques ofa non-hydrodynamic origin.

8.1 Bulk Suspension Properties

Consider now a volume V which is large enough to contain many particles butsmall enough so that macroscopic variables hardly change on the scale V 1/3, i.e.,l � V 1/3 � L . The effective stress tensor seen from a macroscopic level is simplythe volume-averaged stress [4, 47],

⟨σi j

⟩ = 1

V

∫

Vσi j dV = 1

V

∫

V f

σi j dV + 1

V

∫

Vp

σi j dV,

where V f is the volume occupied by the solvent, Vp is the volume of the particlesin V , and the angle brackets denote a volume-averaged quantity. If the solvent isNewtonian, we have

σi j (x) = −pδi j + 2ηs Di j ≡ σ( f )i j , x ∈ V f .

Thus

1

V

∫

V f

σi j dV = −〈p〉 δi j + 2ηs⟨Di j

⟩ − 1

V

∫

Vp

σ( f )i j dV .

Furthermore, from the equations of motion in the absence of inertia and the bodyforce,

8.1 Bulk Suspension Properties 151

∂

∂xk

(xiσk j

) = σi j ,

and we find that

1

V

∫

Vp

σi j dV = 1

V

∫

Vp

∂

∂xk

(xiσk j

)dV = 1

V

∫

Sp

xi t j dS,

where t j = σk j nk is the traction vector and Sp is the bounding surface of all theparticles. In addition,

1

V

∫

Vp

σ( f )i j dV = 1

V

∫

Vp

[−pδi j + ηs

(∂ui∂x j

+ ∂u j

∂xi

)]dV

= 1

V

∫

Sp

[ηs

(uin j + u jni

)]dS

to within an isotropic tensor which can be lumped into a generic hydrostatic pressureP , which is determined through the balance of momentum and the incompressibilityconstraint. The average stress is thus given by

⟨σi j

⟩ = −p′δi j + 2ηs⟨Di j

⟩

︸︷︷︸solvent

+ 1

V

∫

Sp

{xi t j − ηs

(uin j + u jni

)}dS

︸︷︷︸particles

, (8.2)

consisting of a solvent contribution, and a particle contribution; p′ is just a scalarpressure (the prime will be dropped from hereon). The particle contribution can bedecomposed into a symmetric part, and an antisymmetric part. The symmetric partis in fact the sum of the stresslets S(p)

i j defined by

Si j = 1

2

∫

Sp

{xi t j + x j ti − 2ηs

(uin j + u jni

)}dS =

∑

p

S(p)i j , (8.3)

and the antisymmetric part leads to the rotlet:

Ri j = 1

2

∫

Sp

(xi t j − x j ti

)dS = 1

2

∑

p

εi jkT(p)k , (8.4)

where T (p)k is the torque exerted on the particle p, and the summation is over all

particles in the volume V . The particle-contributed stress is therefore given by

σ(p)i j = 1

V

∑

p

(S(p)i j + 1

2εi jkT

(p)k

). (8.5)

152 8 Suspensions

The total rate of energy dissipation can be calculated by consider a large enoughvolume V to contain all the particles - the rate of energy dissipation in V is thus

Φ =∫

Vσi j Di j dV =

∫

V

∂

∂x j

(σi j ui

)dV =

∑

p

∫

Sp

σi j ui n j dS. (8.6)

The second equality comes from the balance of momentum, and the third one froman application of the divergence theorem, assuming that the condition at infinity isquiescent (the bounding surface of V consists of particles’ surfaces and surface atinfinity). For a system of rigid particles, the boundary condition on the surface of aparticle p is that

u = U(p) + Ω(p) × x, (8.7)

whereU(p) andΩ(p) are the translational and rotation velocities of the particle, whichcan be taken outside the integral in (8.6). The terms remaining can be identified withthe force F(p), and the torque T(p) imparted by the particle p to the fluid. Thus thetotal rate of energy dissipation is

Φ =∑

p

(U(p) · F(p) + Ω(p) · T(p)

). (8.8)

Note also that for a system of rigid particles, the integral

∫

Sp

(un + nu) dS = 0,

since∫

Sp

U(p)ndS = 0,∫

Sp

(Ω(p) × xn + nΩ(p) × x

)dS = 0,

by applications of the divergence theorem.

8.2 Dilute Suspension of Spheroids

Weconsider nowadilute suspension of force- and torque-freemonodispersed spheresin a general homogeneous deformation. The dilute assumption means the volumefraction

φ = ν4πa3

3� 1, (8.9)

8.2 Dilute Suspension of Spheroids 153

where ν is the number density of the spheres, of radius a each. In this case, in arepresentative volume V we expect to find only one sphere. Thus, the microscaleproblem consists of a single sphere in an effectively unbounded fluid; the superscriptp on the generic particle can be omitted, and the coordinate system can be conve-niently placed at the origin of the sphere. The boundary conditions for thismicroscaleproblem are

u = U + (D + W) · x, far from the particle, |x| → ∞, (8.10)

and

u = V + w · x, on the particle’s surface, |x| = a, (8.11)

where L = D+W is the far-field velocity gradient tensor;D is the strain rate tensor,W is the vorticity tensor, w is the skew-symmetric tensor such that wi j = −εi jkΩk ,with Ω being the angular velocity of the particle. The far-field boundary conditionmust be interpreted to be far away from the particle under consideration, but not farenough so that another sphere can be expected. The solution to this unbounded flowproblem is well known, [35]

u = U + L · x + a3

x3(w − W) · x +

(3a

4x+ a3

4x3

)(V − U) − a5

x5D · x (8.12)

+ 3(V − U) · xx2

(a

x− a3

x3

)x − 5D : xx

2x2

(a3

x3− a5

x5

)x, (8.13)

and

p = 3

2ηsa

(V − U) · xx3

− 5ηsa3D : xx

x5. (8.14)

The traction on the surface of the sphere is

t = σ · n|x=a = −3ηs2a

(V − U) − 3ηsa

(w − W) · x + 5ηs

aD · x. (8.15)

The force and the torque on the particle can be evaluated:

F =∫

Sσ · n dS = −6πηsa(V − U) (8.16)

and

T =∫

Sx × σ · ndS = −8πηsa

3(Ω − ω), (8.17)

where ωi = 12 εi jkW jk is the local vorticity vector. Thus, if the particle is force-free

and torque-free, then it will translate with U and spin with an angular velocity of ω.

154 8 Suspensions

Returning now to the particle-contributed stress, (8.5),

⟨σ

(p)i j

⟩= 1

V

∑

p

Si j = νSi j ,

where the stresslet is given in (8.3). From (8.12), and noting that

∫

Sx dS = 0,

∫

Sxx dS = 4πa4

31,

we find

Si j = 1

2

∫

S

(xi t j + x j ti

)dS = 5

ηs

a

∫

SDik xkx j dS = 5ηs

(4πa3

3

)Di j .

Recall that the volume fraction of particles is φ = 4πa3ν/3, the effective stress willnow become

〈σ〉 = −p1 + 2ηs

(1 + 5

2φ

)D. (8.18)

This is the celebrated Einstein’s result [20], who arrived at the conclusion from theequality of the dissipation at the microscale and the dissipation at the macroscale asdescribed by an effective Newtonian viscosity.

A similar theory has been worked out for a dilute suspension of spheroids by Lealand Hinch [51], when the spheroids may be under the influence of Brownian motion,using the solution for flow around a spheroid due to Jeffery [41]. Here, if p denotesa unit vector directed along the major axis of the spheroid, then Jeffery’s solutionstates that

p = W · p + R2 − 1

R2 + 1(D · p − D : ppp) , (8.19)

where R is the aspect ratio of the particle (major to minor diameter ratio). Theparticle-contributed stress may be shown to be

σ(p) = 2ηsφ{AD : 〈pppp〉 + B (D · 〈pp〉 + 〈pp〉 · D) (8.20)

+ CD + dRF 〈pp〉},

where the angular brackets denote the ensemble average with respect to the distrib-ution function of p; A, B, C and F are some shape factors, and dR is the rotationaldiffusivity. If the particles are large enough so that Brownian motion can be ignored,then the last term, as well as the angular brackets, can be omitted in (8.20). Theasymptotic values of the shape factors are given in Table8.1.

8.2 Dilute Suspension of Spheroids 155

Table 8.1 Asymptotic values of the shape factors

Asymptoticlimit

R → ∞ (rod-like) R = 1 + δ, δ � 1(near-sphere)

R → 0 (disk-like)

AR2

2(ln 2R − 1.5)

395

147δ2

10

3πR+ 208

9π2 − 2

B6 ln 2R − 11

R2

15

14δ − 395

588δ2 − 8

3πR+ 1 − 128

9π2

C 25

2

(1 − 2

7δ + 1

3δ2

)8

3πR

F3R2

ln 2R − 1/29δ − 12

πR

The rheological predictions of this constitutive equation have also been consideredby Hinch and Leal [37]. In essence, the viscosity is shear-thinning, the first normalstress difference is positive while the second normal stress difference is negative,but of a smaller magnitude. The precise values depend on the aspect ratio and thestrength of the Brownian motion. The predictions of the constitutive equation (8.20)are considered in Problems 8.2–8.4. In particular, the reduced elongational viscosityof the suspension can be shown to be

N1 − 3ηs γ

ηs γ= 2 (A + 2B + C) φ ≈ φR2

ln 2R − 1.5. (8.21)

Strictly speaking, the dilute assumptionmeans that the volume fraction is lowenough,so that a particle can rotate freely without any hindrance from its nearby neighbours.The distance Δ between any two particles must therefore satisfy l < Δ, so that avolume of l3 contains only one particle, where l is the length of the particle and d isits diameter. The volume fraction therefore satisfies

φ ∼ d2l

Δ3, φR2 < 1.

Thus, the reduced elongational viscosity is only O(1) in the dilute limit, not O(R2) assuggested by the formula (8.21). As the concentration increases, we get subsequentlyinto the semi-dilute regime, the isotropic concentrated solution, and the liquid crys-talline solution. The reader is referred to Doi and Edwards [18] for more details.Here, we simply note that the concentration region 1 < φR2 < R is called semi-concentrated. Finally, the suspension with φR > 1 is called concentrated, wherethe average distance between fibres is less than a fibre diameter, and therefore fibrescannot rotate independently except around their symmetry axes. Any motion of thefibre must necessarily involve a cooperative motion of surrounding fibres.

156 8 Suspensions

Problems

Problem 8.1 Use the instantaneous nature of the micromechanics to explain theshear reversal experiments of Gadala-Maria and Acrivos [30]. They found that ifshearing is stopped after a steady state has been reached in a Couette device, thetorque is reduced to zero instantaneously. If shearing is resumed in the same directionafter a period of rest, then the torque would attain its final value that corresponds tothe resumed shear rate almost instantaneously. However, if shearing is resumed inthe opposite direction, then the torque attains an intermediate value and graduallysettles down to a steady state. How would you classify the memory of the liquid,zero, fading or infinite?

Problem 8.2 Show that (8.19) is solved by

p = QQ

, (8.22)

where

Q = L · Q, L = L − 2

R2 + 1D. (8.23)

The effective velocity gradient tensor is L = L − ζD, where ζ = 2/(R2 + 1) is a‘non-affine’ parameter.

Problem 8.3 In the start-up of a simple shear flow, where the shear rate is γ, showthat

Q1 = Q10 cosωt +√2 − ζ

ζQ20 sinωt,

Q2 = Q20 cosωt −√

ζ

2 − ζQ10 sinωt,

and Q3 = Q30, where {Q10, Q20, Q30} are the initial components of Q, and thefrequency of the oscillation is

ω = 1

2γ√

ζ(2 − ζ) = γR

R2 + 1.

From these results, obtain the particle-contributed stress and the viscometric func-tions as

The reduced viscosity:

〈σ12〉 − ηs γ

ηs γφ= 2Ap21 p

22 + B

(p21 + p22

) + C, (8.24)

Problems 157

The reduced first normal stress difference:

N1

ηs γφ= 2Ap1 p2

(p21 − p22

), (8.25)

and the reduced second normal stress difference:

N2

ηs γφ= 2p1 p2

(Ap22 + B

). (8.26)

Thus, the particles tumble along with the flow, with a period of T = 2π(R2+1)/γR,spending most of their time aligned with the flow.

Problem 8.4 In the start-up of an elongational flow with a positive elongational rateγ, show that

Q1 = Q10 exp {(1 − ζ)γt} ,

Q2 = Q20 exp

{−1

2(1 − ζ)γt

},

Q3 = Q30 exp

{−1

2(1 − ζ)γt

},

so that the particle is quickly aligned with the flow in a time scale O(γ−1). At asteady state, show that the reduced elongational viscosity is given by

N1 − 3ηs γ

ηs γφ= 2A + 4B + 3C ≈ R2

ln 2R − 1.5. (8.27)

Chapter 9Dissipative Particle Dynamics (DPD)

A Particle-Based Method

Wehavediscussed constitutivemodelling techniques for some simplemicro-structuremodels in previous chapters, including dilute suspensions of dumbbells (as a modelfor polymer solutions) and spheroids (as a model for suspension of rigid particles).Clearly close form solutions for more complex models may be very limited and aredifficult to find. In these cases, a more suitable method is required that can handleboth the constitutive modelling and the flow problems. In this chapter, we discussa particle-based method called the Dissipative Particle Dynamics (DPD) method.The method, originally conceived as a mesoscale technique by Hoogerbrugge andKoelman [38], has its basis in statistical mechanics (Español andWarren [22], Marsh[56]). In the method, the fluid is modelled by a system of particles (called DPD parti-cles) in their Newton 2nd lawmotions. In the original technique, these DPD particlesare regarded as clusters of molecules, undergoing a soft pairwise repulsion, in addi-tion to other dissipative and random forces designed to conservemass andmomentumin the mean. The microstructure of the DPD fluid can be made as complicated aswe like: DPD particles can be connected to form strings to model polymer solutions(Kong et al. [44]), to form rigid particles to model suspensions (Boek et al. [8]),to form immiscible droplets to model multiphase fluids (Novik and Coveney [63]),to form vapour phase to model liquid/vapour interaction (Arienti et al. [1]). TheDPD system exists in continuous space, rather than on a lattice as the Lattice-GasAutomata (LGA) (e.g., Frisch et al. [28]), and hence it removes some of the isotropyand Galilean invariance problems facing LGA, but it still retains the computationalefficiency of the LGA. Another popular numerical method for simulating complex-structure fluids is the Brownian Dynamics Simulation (BDS) (e.g., Fan et al. [25]).In BDS, the bulk flow field kinematics are specified a-priori; then the effects of thefluidmicrostructure evolution on the flow field are taken into account by coupling theBDS to the kinematics in an iterative manner. Furthermore, BDS conserves particles(mass), but not momentum. In contrast, DPD method conserves both the number ofparticles and also the total momentum of the system; its transport equations are of thefamiliar form of mass and momentum conservation. Thus, both the flow kinematics,


159

160 9 Dissipative Particle Dynamics (DPD)

and the stress tensor (constitutive equation) can be found as a part of the solutionprocedure. In this point of view, DPD can be regarded as a particle-based solver forcontinuum problems – DPD particles are regarded as fictitious constructs to satisfyconservation laws.We are especially attracted to theDPDmethod because of the easeand flexibility of its modelling of a complex-structure fluid. The method conservesmass and momentum in the mean, and therefore is not only restricted to mesoscaleproblems – it is also applicable to problems of arbitrary scales and therefore it maybe regarded as a particle-based method for solving continuum flow problems. Wereview the technique here in details, together with some test problems of interest, tocomplete our adventure in the constitutive modelling of complex-structure fluids.

9.1 1-D Model

Although one-dimensional models do not make sense in fluid systems, they provideconsiderable pedagogic insights into the DPD method, and therefore we first con-sider a 1-D system under the action of a “conservative” force FC (r), a function ofthe particle’s position r , a dissipative force FD = γwD (r) v proportional to the parti-cle’s velocity v = r , with strength γwD (r), and a random force FR = σwR (r) θ(t),with strength σwR (r), and θ(t) is a Gaussian white noise, with zero mean and unitvariance:

dr

dt= v, m

dv

dt= Fc − γwDv + σwRθ (t) , r (0) = r0, v (0) = v0, (9.1)

〈θ (t)〉 = 0, 〈θ (t) θ (t + τ )〉 = δ (τ ) .

The zone of influence of the dissipative and random forces may be prescribed byspecifying the weighting functions wD(r) and wR(r) - they are defined to be dimen-sionless and are zero outside a certain cutoff radius rc, for r ≥ rc. The angularbrackets denote an ensemble average with respect to the distribution function of thequantity concerned. Of course, there is no need to separate the dissipative and ran-dom forces strength into a scalar and a configuration-dependent weighting functionin the manner indicated – this is done only to conform with existing DPD notation.

Langevin Equation Since the random force has only well-defined statistical proper-ties, the so-called Langevin equation (9.1) is understood as a stochastic differentialequation, its complete solution is specified by the joint probability distribution func-tion f (r, v, t) of the process {r, v}, defined so that f (r, v, t) drdv is the probabilityof finding the process at the states between {r, v} and {r + dr, v + dv} at time t .Specifying an initial state, as done in (9.1), is equivalent to prescribing an initialdelta probability function:

f (r, v, 0) = δ (r − r0) δ (v − v0) . (9.2)

9.1 1-D Model 161

A specification of f (r, v, t) leads to the phase-space description of the stochasticsystem (9.1). The velocity space v, or the configuration space r , may be integratedout of f (r, v, t), in which case we have a configuration-space, or a velocity-spacedescription of the stochastic system (9.1). The equation governing the probability issometimes known as the Fokker–Planck or Smoluchowski, or simply the diffusionequation of the process.

Fluctuation-Dissipation Theorem There are three time scales in the stochasticdifferential equation (9.1):

1. a fluctuation time scale τR of the random force, which is arbitrarily small;2. an inertial time scale τI = O

(mγ−1

) ;3. and a relaxation time scale τ = O

(γH−1

)where H is the stiffness of the system,

H = O (|∂r Fc|) .

In a typical physical system with small inertia, we have a natural separationbetween the time scales: τR � τI � τ . The random force fluctuations on the timescale τI inject kinetic energies into the system and raise its temperature, defined asits Boltzmann temperature kBT ,

1

2kBT = 1

2m

⟨v2(t)

⟩. (9.3)

In order to focus on events on the time scale τI we can regard the restoring force Fc

as constant, so that it can be absorbed in a re-definition of the system state system(by re-defining r and v appropriately) to obtain (in the new state space),

dv

dt= −m−1γwDv + m−1σwRθ (t) . (9.4)

A formal solution for this is

v (t) =∫ t

0em

−1γwD(t ′−t)m−1σwRθ(t ′)dt ′. (9.5)

With this result, the existence of the temperature assumption (9.3) demands that(Problem 9.1)

σ2w2Rγ−1w−1

D = 2kBT . (9.6)

Equation (9.6) is a consequence of the temperature assumption (9.3), sometimesknown as Equi-Partition Principle, one of the many fluctuation-dissipation theoremsin Statistical Fluid Mechanics ([47]).

It is customary in the DPD literature to choose,

wD (r) = [wR (r)

]2and γ = σ2

2kBT. (9.7)


This choice guarantees the thermodynamic temperature as defined by (9.3). Notethat the requirement (9.6), or (9.7), implies that the cutoff radius for the random andthe dissipative forces must be the same.

Phase-Space Description: Fokker–Planck Equation The phase-space descriptionof (9.1) is given by the Fokker–Planck equation, sometimes known as Liouvilleequation ([14]):

(∂

∂t+ v

∂

∂r

)W (r, v, t) = lim

Δt→0

∂

∂v

[ 〈ΔvΔv〉2Δt

∂

∂v− 〈Δv〉

Δt

]W (r, v, t) . (9.8)

Its derivation is based on the assumption of theMarkovian nature of the process (9.1),namely, what is going to happen at any given instant t depends on the current stateof the system at time t , not on what has already happened preceding time t ([14]).For the stochastic system (9.1),

Δv = m−1 (−γwDv + Fc)Δt +∫ Δt

0m−1σwRθ

(t ′)dt ′. (9.9)

Consequently,

〈Δv〉 = m−1 (−γwDv + Fc)Δt,

〈ΔvΔv〉 = O(Δt2

) +∫ Δt

0

∫ Δt

0m−2σ2w2

R

⟨θ(t ′)θ (t)

⟩dt ′dt

= O(Δt2

) +∫ Δt

0m−2σ2w2

Rdt′,

= O(Δt2

) + m−2σ2w2RΔt,

and the phase-space diffusion equation for the stochastic system (9.1) is

(∂

∂t+ v

∂

∂r

)W = ∂

∂v

[σ2w2

R

2m2

∂

∂v− m−1 (Fc − γwDv)

]W, (9.10)

subjected to some relevant initial condition, for example (9.2). From the phase spacedescription (9.10), one can derive balance, or conservation equations for the system,but it is not meaningful to do so for this simplistic 1D system.

Configuration Space: Fokker-Planck-Smoluchowski Equation Sometimes, itis more convenient to deal directly with the configuration-space distribution; thevelocity-space can be integrated out from (9.10) and the resulting equation is usu-ally called the Fokker–Planck equation in configuration space, or the Smoluchowskiequation. The Fokker–Planck equation in configuration space can be shown to be([14]):

∂

∂tW (r, t) = lim

Δt→0

∂

∂r

[( 〈ΔrΔr〉2Δt

∂

∂r− 〈Δr〉

Δt

)W (r, t)

]. (9.11)

9.1 1-D Model 163

In (9.11) limt→∞ 〈Δr/Δt〉 is called the drift velocity, and limt→∞ 〈ΔrΔr/2Δt〉 iscalled the diffusivity. In the limit of small inertia, the stochastic system (9.1) becomes

dr

dt= γ−1w−1

D Fc + γ−1w−1D σwRθ (t) . (9.12)

Problem 9.2 shows how to derive the following Fokker–Planck equation in configu-ration space:

∂

∂tW (r, t) = ∂

∂r

[(kBT

γwD

∂

∂r− Fc

γwD

)W (r, t)

]. (9.13)

9.2 DPD Fluid

9.2.1 Langevin Equations

Most of the essential ideas in DPD are contained in the 1-D theory discussed in 9.1.We continue our discussion of DPD in a 3D setting. We define a DPD fluid as theensemble of N particles, call DPD particles, each of massmi , i = 1, . . . , N , locatedat position ri , with velocity vi . Furthermore, we assume identical mass mi = m,without much loss of generality (Fig. 9.1). In the original viewpoint, a DPD particlemay be thought of as a cluster of fluid molecules; alternatively, it may be regarded asa fictitious construct to solve a flow problem of a complex-structure fluid – we willreturn to these different viewpoints later. The DPD particles interact with each otherin their Newton’s second law motions:

dridt

= vi , mdvidt

=∑

j

Fi j + Fe. (9.14)

Here, Fe is an external force on particle i (for example, gravity to simulate the effectof a pressure gradient), and Fi j is the pairwise additive interparticle force by particlej on particle i ; this force consists of three parts, a conservative force FC

i j , a dissipativeforce FD

i j , and a random force FRi j :

Fi j = FCi j + FD

i j + FRi j . (9.15)

In (9.14), the sum runs over all other particles except i (note, by definition Fi i = 0).Outside a certain cutoff radius rc, the interactions are zero. Here we may allow thecutoff radius to be different for different type of forces. There are some key ideas inthe DPD theory, which we will go through in details.

Key Idea 1: DPD is an MD-like Method The first key idea is that DPD looks verymuch like molecular dynamics simulation in the absence of dissipative and random


Fig. 9.1 A DPD fluid is made up of DPD particles of various connectivities

forces. In fact, the system (9.14), with FDi j = FR

i j = 0, is called the associate system.When the conservative forces are chosen appropriately, the associate system is thebasis for molecular dynamics (MD) simulation.

Key Idea 2: Conservative Force is Soft Repulsion The second key idea is that theconservative force is chosen to be a soft repulsion. This allows a much larger timestep to be employed (easily by a factor of 10), as compared to a much smaller timestep, when a standard MD molecular potential, such as Lennard-Jones potential, isused (Keaveny et al. [42]). Here the standard DPD soft conservative force, cutoffoutside a critical normalised radius rc = 1 is used:

FCi j = −∂ϕ

(ri j

)

∂ri=

{ai j

(1 − ri j

)ei j , ri j < 1,

0, ri j ≥ 1,(9.16)

where ai j is a coefficient of interaction, ri j is the distance between particles i and j ,and ei j is the unit vector directed from particles j to i :

ri j = ri − r j , ri j = ∣∣ri j∣∣ , ei j = ri j/ri j . (9.17)

The DPD particles can be of a complex type, for examples, interconnecting parti-cles to form (polymer) chains, inwhich case theremay be connector forces pertainingto the nature of the connectors, or forces arisen due to some rigidity constraints toform a rigid body. We may consider lumping these connector forces in the conser-vative forces FC

i j as a matter of convenience. The conservative forces (together withany connector forces) determine the rheology of the system.

Key 3: Dissipative Force is Centre-to-Centre The dissipative force slows downthe particle, extracts parts of the kinetic energy injected by the random force. It isdirected from centre-to-centre and is given by

FDi j = −γwD

(ri j

) (ei j · vi j

)ei j , vi j = vi − v j , (9.18)

9.2 DPD Fluid 165

Its strength is controlled by the parameter γ, and its zone of influence is governedby the weighting function wD

(ri j

).

Key Idea 4: Random Force is Centre-to-Centre The random force compensatesfor the loss of kinetic energy from dissipation; it is directed from centre-to-centre,its strength and domain of influenced are governed by the parameter σ, and theweighting function wR

(ri j

):

FRi j = σwR

(ri j

)θi j ri j , (9.19)

where θi j (t) = θ j i (t) is a Gaussian white noise with the properties

⟨θi j (t)

⟩ = 0,⟨θik (t) θ jl

(t ′)⟩ = (

δi jδkl + δilδ jk)δ(t − t ′

), i = k, j = l. (9.20)

Note that the random force is central, and it is anticipated that the strengths and theweighting functions of the random and the dissipative forces cannot be prescribedindependently: they act together in a precise manner to keep the Boltzmann tem-perature of the system constant. Furthermore, because of the central nature of therandom forces, ∑

i, j =i

FRi j = 0. (9.21)

Key Idea 5: Fluctuation-Dissipation Theorem Links the Dissipative and Ran-dom Forces Again, there are three time scales in the system: the fluctuation timescale of the random force, τR , which is arbitrarily small, the inertial time scale,τI = O

(m/γ−1w−1

D

), and the relaxation time scale, τ = O

(γwDH−1

), where

H = O(∣∣∂rFC

∣∣) is the stiffness of the system. Since the particles’ mass is typi-cally small for a physical system, we have the natural separation of the time scales,τR � τI � τ . The fluctuations of the random force inject kinetic energies into andheat up the system, which is then taken away and cooled down by the dissipativeforce. This balance allows a thermodynamic temperature to be defined. In order tosee this detailed balance, one needs only to consider the motion at time scale τI . Inthis time scale, the conservative and external forces can be regarded as constant, andcan be scaled out of the equations of motion (by re-define the state variables, in thesame manner as the 1D model discussed previously),

dvidt

+∑

j

m−1γwDei jei j · vi j =∑

j

m−1σwRθi jei j . (9.22)

Collecting the velocity vectors into a state variable equation (9.22) then results intoa linear system:


vα +∑

β

m−1Aaβ · vβ = m−1Bα, (9.23)

Aαβ = δαβ

∑

j

γwDeα jeα j − γwDeαβeαβ,

Bα =∑

j

σwRθα jeaj , α,β = 1, . . . , N .

Here, we reserve Greek indices to refer to state vectors and matrices; the Romanindices are used to enumerate the particles as usual – summations on them will beindicated explicitly to avoid confusion. It is to be noted that A is an [N , N ] matrix,with second-order tensors as elements. It is also a symmetric matrix of zero row andcolumn sums; it has zero as one eigenvalue, with corresponding unit eigen-vector.Likewise, B is an [N , 1] matrix, with three-dimensional vectors as elements. Notealso that, because (9.21), the centre of gravity of the system, rC = 1

N

∑i ri moves

deterministically.A formal solution of the linear system (9.23) can be derived, noting the commu-

nicative property of A and its exponent function exp (At) ,

vα =∑

β

∫ t

0exp

[m−1Aαβ

(t ′ − t

)] · m−1Bβdt′. (9.24)

Thus, the specific kinetic energy is

〈vαvν〉 = m−2∑

β,γ

∫ t

0

∫ t

0exp

[m−1Aαβ

(t ′ − t

)] · ⟨Bβ

(t ′)Bγ

(t ′′)⟩ ·

exp[m−1Aνγ

(t ′′ − t

)]dt ′dt ′′. (9.25)

But, from the property of the random force (9.20),

⟨Bβ

(t ′)Bγ

(t ′′)⟩ =

∑

j

∑

k

σβ jσγkeβ jeγk(δβγδ jk + δβkδγ j

)δ(t ′ − t ′′

),

where we have used the short-hand notation σi j to denote the scalar σwR(ri j

). This,

when used in (9.25), yields

〈vαvν〉 = m−2∑

β,γ

∫ t

0exp

[m−1Aαβ

(t ′ − t

)] · (9.26)

{

δβγ

∑

k

σβkσγkeβkeγk + σβγσγβeβγeγβ

}

· exp [m−1Aνγ

(t ′ − t

)]dt ′.

9.2 DPD Fluid 167

If the existence of a thermodynamic temperature is assumed, then the left hand issimply m−1kBT Iαν, where kBT is the Boltzmann temperature of the system, andIαν is the [N , N ] unit tensor (having diagonal elements as second-order unit tensors,and zero entries elsewhere). Thus

kBTAην =∑

β,γ

∫ t

0m−1Aηα exp

[m−1Aαβ

(t ′ − t

)] · (9.27)

{

δβγ

∑

k


}

· exp [m−1Aνγ

(t ′ − t

)]dt ′.

Similar to 1-D case, an integration by parts is done, recognising A exp (At) as thetime derivative of exp (At) and take the limit of large time (compared to the inertialtime scale),

kBTAην = limt→∞

∑

β,γ

exp[m−1Aηβ

(t ′ − t

)] ·{

δβγ

∑

k


}

· exp [m−1Aνγ

(t ′ − t

)]]t ′=t

t ′=0

−∑

β,γ

∫ t

0dt ′exp

[m−1Aηβ

(t ′ − t

)] ·{

δβγ

∑

k


}

· exp [m−1Aδγ

(t ′ − t

)]Aνδ.

The last integral on the right side of the preceding equation is recognised as theright side of (9.27), and the limit at large time (compared to the inertial time scale)can be taken to yield

2kBTAην = Iηβ ·{

δβγ

∑

k


}

· Iνγ,

or

2kBT

⎛

⎝δην

∑

j

γwDeη jeη j − γwDeηνeην

⎞

⎠ = δην

∑

k

σηkσηkeηkeηk

− σηνσηνeηνeην . (9.28)

Equation (9.28) is the consequence of the existence of the temperature, and may beregarded as the fluctuation-dissipation theorem for our DPD system. From the formof A, it is sufficient that


2kBTγwD (r) = σ2w2R (r) (9.29)

for the equality (9.28) to hold true – and this is the final result for the existence oftemperature. It is customary in DPD to take

wD (r) = w2R(r) and γ = σ2

2kBT(9.30)

to ensure a constant Boltzmann temperature kBT , or strictly speaking, a constantspecific kinetic energy of the system. The requirement (9.29), or (9.30) implies thatthe cutoff radii for the random and the dissipative forces must be the same.

As far as thermal energy is concerned, the random two-particle force, FRi j , rep-

resenting the results of thermal motion of all molecules contained in particles i andj , “heats up” the system. The dissipative force, FD

i j , reduces the relative velocity oftwo particles and removes kinetic energy from their mass centre to “cool down” thesystem. When the detailed balance, (9.29), is satisfied, the system temperature willapproach the given value. The dissipative and random forces act like a thermostat inthe conventional molecular dynamics (MD) system.

Clearly, the addition of the dissipative and random forces to a conservative systemmay appear to be artificial, but the versatility of DPD lies in its ability to satisfyconservation laws in the mean (to be shown later). It is possible to think of theDPD system as a coarse-grained model of a physical model. Therefore we couldconstruct models of complex-structure fluids by endowing the simple DPD particleswith features in a manner similar to modelling, for example, polymeric solution bya suspension of Rouse chains.

9.2.2 Phase-Space Description: Fokker–Planck Equation

The complete solution of the stochastic system (9.22) would require a specificationof the probability density distribution f (r1, v1, . . . , rN , vN , t) of the state spaceχ = {ri , vi , t} , i = 1, . . . , N ; f (χ, t) dχ is the probability of finding the process atthe state between χ and χ + dχ at time t . With the assumption of the Markoviannature of the process, Chandrasekhar [14] showed that this distribution functionobeys the following Fokker–Planck (or Liouville) equation,

∂ f

∂t+

∑

i

∂

∂ri· (vi f ) +

∑

i

∂

∂vi·(⟨

ΔviΔt

⟩f

)(9.31)

=∑

i, j

∂

∂vi·(⟨

ΔviΔv j

2Δt

⟩· ∂ f

∂v j

),

where the limit Δt → 0 is implied in the ensemble averages.

9.2 DPD Fluid 169

From the Langevin equation (9.14), the drift and the diffusion terms can be found(Problem 9.5), and the Fokker–Planck equation for the process is

∂ f

∂t+

∑

i

∂

∂ri· (vi f ) +

∑

i, j

∂

∂pi· (FC

i j f) = γ

∑

i, j

wDi j ei j

∂

∂pi· (ei j · vi j f

)

+ γkBT∑

i, j

wDi j ei j · ∂

∂pi·(ei j ·

(∂ f

∂pi− ∂ f

∂p j

)), (9.32)

where pi = mvi is the linear momentum of particle i.

9.2.3 Distribution Functions

Consider a function Q = Q (X) of the state X = {r1, v1, . . . , rN , vN }. Its ensembleaverage is defined to be

〈Q〉 =∫

. . .

∫Q f (r1, v1, . . . , rN , vN ) dr1dv1 . . . drNdvN (9.33)

=∫

Q (X) f (X, t) dX.

Sometimes, a full statistical description f (X, t) is not required, since the quanti-ties of interest only involve a subset of the state variable, the rest of the state variablecan be integrated out. In particular, we are interested in the one-particle, f1 (χ, t) =f1 (r, v, t) , χ = {r, v} , or the two-particles distribution function, f2

(χ,χ′, t

) =f2(r, v, r′, v′, t

), χ = {r, v} , χ′ = {

r′, v′} , defined as:

f1 (χ, t) =⟨∑

i

δ (χ − χi )

⟩

=⟨∑

i

δ (r − ri ) δ (v − vi )

⟩

, (9.34)

f2(χ,χ′, t

) =⟨∑

i, j =i

δ (χ − χi ) δ(χ′ − χ j

)⟩

(9.35)

=⟨∑

i, j =i

δ (r − ri ) δ (v − vi ) δ(r′ − r j

)δ(v′ − v j

)⟩

.

The one-particle distribution function, f1 (χ, t) , is the probability distribution func-tion of finding a particle at the state χ = {r, v} , at time t . The two-particles distri-bution function, f2

(χ,χ′, t

), is the probability distribution function of finding two

particles simultaneously at state χ = {r, v} , of position r and velocity v, and at state


χ′ = {r′, v′} , of position r′ and velocity v′ at time t . The velocities can be inte-

grated out of (9.34) and (9.35), in which case one has the configuration-dependentone-particle distribution function

f1 (r, t) =∫

f1 (r, v, t) dv, (9.36)

and the configuration-dependent two-particle, distribution function

f2(r, r′, t

) =∫

dv∫

dv′ f2(r, r′, v, v′, t

). (9.37)

9.2.4 Equation of Change

The equation of change for any dynamical process Q(X) can be derived by pre-multiplying Q with (9.32), and integrate the resulting equation, using Green’s theo-rem whenever necessary. First, we re-write (9.32) as

∂ f

∂t+ L f = 0, (9.38)

where the operator L is defined by

L f =∑

i

∂

∂ri· (vi f ) +

∑

i, j

∂

∂pi· ((FC

i j + FDi j

)f)

− γkBT∑

i, j

wDi j ei j · ∂

∂pi

(ei j ·

(∂

∂pi− ∂

∂p j

)f

).

Next, multiply Q with (9.38) and integrate over the state space to obtain

∂

∂t〈Q〉 = −

∫Q (X)L f (X, t) dX =

∫f (X, t)L′QdX = ⟨L′Q

⟩,

where the operator L′ is defined from applications of Green’s theorem (the surfaceintegrals resulted from these applications vanish, because the probability distributionfunction is assumed to vanish on these surfaces). In full,

∂

∂t〈Q〉 = ⟨L′Q

⟩ =∑

i

⟨vi · ∂

∂riQ

⟩+

∑

i, j

⟨FCi j · ∂

∂piQ

⟩(9.39)

− γ∑

i, j

⟨wD

i j vi j · ei jei j · ∂

∂piQ

⟩

9.2 DPD Fluid 171

+ γkBT∑

i, j

⟨wD

i j ei j · ∂

∂pi

(ei j ·

(∂Q

∂pi− ∂Q

∂p j

))⟩.

9.2.5 Conservation of Mass

If we take, as our dynamical variable

Q =∑

i

δ (r − ri ), (9.40)

then the element of this sum is zero, unless a DPD particle is at the position r. Thus,the average of this variable is the number density, which is precisely the one-particleconfiguration-dependent distribution function:

n (r, t) =⟨∑

i

δ (r − ri )

⟩

=∫

f1 (r, v, t) dv = f1 (r, t) . (9.41)

Now, if we takeQ =

∑

i

mδ (r − ri ), (9.42)

then its average is the fluid density:

ρ (r, t) =⟨∑

i

mδ (r − ri )

⟩

=∫

m f1 (r, v, t) dv = mn (r, t) . (9.43)

Now, noting that ∂ri f (r − ri ) = −∂r f (r − ri ) , we find, for our chosen dynamicalvariable (9.42),

⟨L′Q⟩ =

∑

i, j

⟨mv j · ∂

∂r jδ (r − ri )

⟩= − ∂

∂r·⟨∑

j

mv jδ(r − r j

)⟩

= − ∂

∂r· (ρu) .

The term

ρ (r, t)u (r, t) =⟨∑

j

mv jδ(r − r j

)⟩

=∫

mv f1 (r, v, t) dv (9.44)

can be identified as the linear momentum density. We thus obtain the usual equationfor the conservation of mass:


∂

∂tρ (r, t) + ∇ · (ρ (r, t)u (r, t)) = 0, ∇ = ∂/∂r. (9.45)

This is merely a statement of conservation of the mass probability (note that theprobability distribution is normalised to one, and (9.45) is simply aReynolds transporttheorem).

9.2.6 Conservation of Linear Momentum

Next, taking our dynamical variable Q the linear momentum (9.44),

Q =∑

j

mv jδ(r − r j

),

and note that

vi · ∂

∂riQ = vi · ∂

∂ri

∑

α

mvαδ (r − rα) = −∇ · (mviviδ (r − ri )) ,

∂

∂piQ = ∂

∂pi

∑

α

mvαδ (r − rα) =∑

α

Iαiδ (r − ri ),

and thus,

∂

∂t〈Q〉 = ⟨L′Q

⟩ = −∇ ·∑

i

〈mviviδ (r − ri )〉 +∑

i, j

⟨FCi jδ (r − ri )

⟩

+∑

i, j

⟨FDi j δ (r − ri )

⟩. (9.46)

There are three terms on the right side of (9.46), each leads to a stress tensor, afterrecasting them into suitable divergence forms (the first term is already in a divergenceform). Although the second and the third terms are similar in form, the conservativeforce in second term involves only the configuration whereas the dissipative force inthe third term involves both the configuration and the velocity; different treatmentswill be necessary.

Kinetic Pressure Tensor Although the first term on the right of (9.46) is already ina divergence form, it can be simplified further. We first define the peculiar velocityas

Vi = vi − u (9.47)

i.e., the velocity fluctuation of particle i with respect to themean field velocity definedby (9.44). The first term on the right of (9.46) is then re-written as

9.2 DPD Fluid 173

−∇ ·∑

i

〈mviviδ (r − ri )〉 = −∇ ·∑

i

〈mδ (r − ri ) (uu + ViVi )〉

= −∇ · (ρuu) − ∇ · PK , (9.48)

where the terms linear in the fluctuations are averaged to zero, because of theirdefinitions, and the kinetic pressure tensor is defined as

PK (r, t) =∑

i

〈mViViδ (r − ri )〉 =∫

m (v − u) (v − u) f1 (r, v, t) dv. (9.49)

Stress Tensors from Interaction Forces The second and third terms on the right of(9.46) can be written as

∑

i, j

⟨(FCi j + FD

i j

)δ (r − ri )

⟩ = 1

2

∑

i, j

⟨(FCi j + FD

i j

) (δ (r − ri ) − δ

(r − r j

))⟩,

(9.50)

where we have used the anti-symmetric nature of the interaction forces betweenparticles i and j . Irving and Kirkwood [40] showed how to recast the right side of(9.50) into a divergence form and we summarise their method here.

First, by expressingr − ri = r − r j − ri j ,

and expanding δ (r − ri ) in a Taylor’s series expansion about r − r j

δ (r − ri ) = δ(r − r j

) − ∇ ·[ri j

(1 − 1

2ri j · ∇ + · · ·

)δ(r − r j

)], (9.51)

it is seen that, from (9.50) and (9.51),

∑

i, j

⟨(FCi j + FD

i j

)δ (r − ri )

⟩ = −1

2∇ ·

∑

i, j

(FCi j + FD

i j

)

ri j

(1 − 1

2ri j · ∇ + · · ·

)δ(r − r j

)

= ∇ · S = ∇ · (SC + SD) , (9.52)

where the particle-interaction stress is given by

S (r, t) = −1

2

∑

i, j

⟨(FCi j + FD

i j

)ri j

(1 − 1

2ri j · ∇ + · · ·

)δ(r − r j

)⟩

= SC (r, t) + SD (r, t) . (9.53)


Here, the stress contributed from the conservative forces is defined as

SC (r, t) = −1

2

∑

i, j

⟨FCi jri j

(1 − 1

2ri j · ∇ + · · ·

)δ(r − r j

)⟩, (9.54)

and the stress from the dissipative forces is given by

SD (r, t) = −1

2

∑

i, j

⟨FDi j ri j

(1 − 1

2ri j · ∇ + · · ·

)δ(r − r j

)⟩. (9.55)

Thus, finally the conservation of linear momentum takes the usual familiar form

∂

∂t(ρu) + ∇ · (ρuu) = ∇ · T, (9.56)

where the total stress is

T (r, t) = −PK (r, t) + SC (r, t) + SD (r, t) . (9.57)

Key Idea 6: DPD Method Conserves Mass and Linear Momentum Herein liesthe power of DPD method: from our seemingly artificial construct (9.14) that bearsa great deal of resemblance to MD, it has been shown that its mean quantities sat-isfy both the conservation of mass and momentum, with the stress derived directlyfrom the microstructure through (9.49) and (9.53). Conversely, flow problems for acomplex-structure fluid may be solved by this particle-based method; the stress (i.e.,constitutive law) needs not be specified a-priori, but can be found by a post-processingapplication.

Of the total stress (9.57), the kinetic pressure has been recast into an averagewith respect to the one-particle distribution function (9.49) – likewise, the particle-interaction stress can be recast into an average with respect to the two-particle dis-tribution function, in the manner described by Irving and Kirkwood [40], and weoutline the development here below.

Stress from Conservative Forces We first re-write the conservative-force-contributed stress (9.54) as

SC (r, t) = −1

2

∑

i, j

⟨∫δ(R − ri j

)FCR

(1 − 1

2R · ∇ + · · ·

)δ(r − r j

)dR

⟩,

(9.58)

where any dependence on ri j of FC has been replaced by the dependence on R,because the pre-multiplied delta function δ

(R − ri j

)in the integration shifts to role

of ri j to R. Next the sum can be taken inside the integral:

9.2 DPD Fluid 175

SC (r, t) = −1

2

∫dRFCR

(1 − 1

2R · ∇ + · · ·

)∑

i, j

⟨δ(R − ri j

)δ(r − r j

)⟩.

Furthermore, the product δ(R − ri j

)δ(r − r j

)is recognised as equivalent to

δ (R + r − ri ) δ(r − r j

), because of the property of delta functions, resulting in

SC (r, t) = −1

2

∫dRFCR

{1 − 1

2R · ∇ + · · ·

}∑

i, j

⟨δ (R + r − ri ) δ

(r − r j

)⟩.

(9.59)

The sum in (9.59) is recognised as the two-point probability distribution function,

∑

i, j

⟨δ (R + r − ri ) δ

(r − r j

)⟩ = f2 (r + R, r, t) , (9.60)

and thus we obtain for the stress contributed from the conservative forces,

SC (r, t) = −1

2

∫dRFCR

{1 − 1

2R · ∇ + · · ·

}f2 (r + R, r, t) . (9.61)

Stress from Dissipative Forces The dissipative-force-contributed stress can betreated in the same manner,

SD (r, t) = −1

2

∑

i, j

⟨∫δ(R − ri j

)FD

(R, v − v′)R

{1 − 1

2R · ∇ + · · ·

}δ(r − r j

)dR

⟩,

where we have used the notation for convenience

FD(R, v − v′) = γwD (R) RR · (v − v′) , R = R/R, R = |R| . (9.62)

The extra complication here is due to the explicit appearance of the velocities. Asbefore, we note that

SD (r, t) = −1

2

∫dR

∫dv

∫dv′FD

(R, v − v′)R

{1 − 1

2R · ∇ + · · ·

}

.∑

i, j

⟨δ(R − ri j

)δ(r − r j

)δ (v − vi ) δ

(v′ − v j

)⟩

= −1

2

∫dR

∫dv

∫dv′FD

(R, v − v′)R

{1 − 1

2R · ∇ + · · ·

}

.∑

i, j

⟨δ (R + r − ri ) δ

(r − r j

)δ (v − vi ) δ

(v′ − v j

)⟩


The sum is recognised as the two-point probability distribution function; and thuswe obtain,

SD (r, t) = −1

2

∫dR

∫dv

∫dv′FD

(R, v − v′)R

{1 − 1

2R·∇ + · · ·

}

. f2(r + R, r, v, v′, t

). (9.63)

Irving and Kirkwood pointed out that all the terms inside the curly brackets in (9.61)and (9.63) beyond the first may be neglected, due to the fact that they are of higherorders (in O (R)).

Marsh’s Equivalent Results for the StressesMarsh [56] used a different techniquein deriving the stresses, which involves expressing the delta function (9.51) as anintegral. Problem9.7 explores this. From this, it can be shown that the kinetic pressuretensor, and the stresses from the interaction forces can be expressed as,

PK (r, t) =∑

i

〈mViViδ (r − ri )〉 =∫

m (v − u) (v − u) f1 (r, v, t) dv, (9.64)

SC (r, t) = −1

2

∫dv′

∫dv′′

∫FCRW2

(χ′,χ′′, t; r) dR, (9.65)

SD (r, t) = −1

2

∫dv′

∫dv′′

∫FD

(R, v′ − v′′)RW2

(χ′,χ′′, t; r) , (9.66)

where

W2(r′, v′, r′′, v′′, t; r) =

∫ 1

0f2(r + λR, r − (1 − λ)R, v′, v′′, t

)dλ. (9.67)

9.2.7 Energy Equation

Energy is not a conserved quantity, but one can write a statement of balance for thespecific energy e (r, t), defined as [56]

e (r, t) =⟨∑

i

eiδ (r − ri )

⟩

, ei = 1

2mivi ·vi + 1

2

∑

j

ϕ(ri j

), (9.68)

where ei is the energy per particle. First we define the kinetic energy

eK =⟨∑

i

1

2mivi ·viδ (r − ri )

⟩

= 〈K E〉 . (9.69)

9.2 DPD Fluid 177

Using Q = K E in the equation of change (9.39) yields

∂

∂teK = −∇.

⟨∑

i

1

2mvvi ·viδ (r − ri )

⟩

+⟨∑

i, j

FCi j ·viδ (r − ri )

⟩

(9.70)

−γ

⟨∑

i, j

wDi j vi j ·ei jei j ·viδ (r − ri )

⟩

+ γkBT

⟨∑

i, j

1

miwD

i j δ (r − ri )

⟩

.

The same can be done for the potential energy,

eP =⟨∑ 1

2ϕ(ri j

)δ (r − ri )

⟩

= 〈PE〉 . (9.71)

Again, by using Q = PE in the equation of change (9.39):

∂

∂teP =

⟨∑

i, j,k

1

2vk ·

[∂ϕ

(ri j

)

∂rkδ (r − ri ) + ϕ

(ri j

) ∂

∂rkδ (r − ri )

]⟩

(9.72)

=⟨∑

i, j

1

2

{−FCi j ·viδ (r − ri ) + FC

i j ·v jδ(r − r j

) − ∇·ϕ (ri j

)δ (r − ri )

}⟩

.

Thus,

∂

∂te = −∇·

⟨∑

i

vi eiδ (r − ri )

⟩

+⟨∑

i, j

1

2FCi j ·

(vi + v j

)δ (r − ri )

⟩

(9.73)

−γ

⟨∑

i, j

wDi j vi j ·ei jei j ·viδ (r − ri )

⟩

+ γkBT

⟨∑

i, j

1

miwD

i j δ (r − ri )

⟩

.

The first term on the right of (9.73) is already in the form of a divergence of a vector−∇ · qK , where

qK (r, t) =∑

i

vi eiδ (r − ri )

=∫

dv∫

dv′∫

dR(m

2v2 + 1

2ϕ (R)

)v f2

(χ,χ′, t

). (9.74)

The second term on the right of (9.73) can be converted into a divergence of a vectorby the use of (9.51):


⟨∑

i, j

1

2FCi j ·

(vi + v j

)δ (r − ri )

⟩

= −∇.∑

i, j

⟨1

4FCi j · (vi + v j

)ri j

(1 − 1

2ri j ·∇ + · · ·

)δ(r − r j

)⟩

= −∇·qC , (9.75)

where

qC =⟨∑

i, j

1

4FCi j ·

(vi + v j

)ri j

(1 − 1

2ri j ·∇ + · · ·

)δ(r − r j

)⟩

. (9.76)

This can be converted into average with respect to two-particle distribution functionas before (Problem 9.8).

The third term on the right of (9.73) can be re-cast into a source/sink term and aflux, using the fact that vi = vi j/2 + (

vi + v j)/2:

γ

⟨∑

i, j

wDi j vi j · ei jei j · viδ (r − ri )

⟩

= γ

2

⟨∑

i, j

wDi j

(vi j · ei j

)2δ (r − ri )

⟩

+ γ

2

⟨∑

i, j

wDi j vi j · ei jei j · (vi + v j

)δ (r − ri )

⟩

= ΛD + ∇·qD,

with

ΛD (r, t) = γ

2

⟨∑

i, j

wDi j

(vi j · ei j

)2δ (r − ri )

⟩

(9.77)

= −γ

2

∫dv

∫dv′

∫dRwD (R)

[R · (v − v′)

]2f2(χ,χ′, t

),

and

qD (r, t) = −γ

4

∑

i, j

⟨wD

i j vi j · ei jei j · ( vi + v j)ri j

(1 − 1

2ri j · ∇ + · · ·

)δ(r − r j

)⟩

= γ

4

∫dR

∫dv

∫dv′wD (R)

(v − v′) ·RR· (v + v′)R

.

{1 − 1

2R · ∇ + · · ·

}f2(r + R, r, v, v′, t

). (9.78)

9.2 DPD Fluid 179

Finally, the last term on the right of (9.73) is recognised as a source term due toinjection of kinetic energy via the random forces:

ΛR (r, t) = γkBT

⟨∑

i, j

1

miwD

i j δ (r − ri )

⟩

(9.79)

= γkBT

m

∫dv

∫dv′

∫dRwD (R) f2

(χ,χ′, t

).

The energy flow balance equation then takes the form

∂e

∂t= −∇· (qK + qC + qD) + ΛR − ΛD. (9.80)

The two source/sink terms will cancel each other when the temperature reaches thevalue dictated by the fluctuation-dissipation theorem, as can be recognised by theirforms:

ΛR − ΛD = γ

⟨∑

i, j

{kBT

mi− 1

2

(vi j · ei j

)2}

wDi j δ (r − ri )

⟩

. (9.81)

9.3 Some Approximate Results

9.3.1 High Damping Limit

Exclusion of Conservative ForcesMarsh [56] derived some important results in thelimit of high γ, where the conservatives forces are zero, FC

i j = 0.This limit is thoughtto be relevant to the general DPD case. In this limit, he showed that a compressibleNewtonian fluid is the solution to the resulting linearised Fokker-Planck-Boltzmannequations,

T (r, t) = −nkBT I + (ζK + ζD)∇ · uI (9.82)

+ (ηK + ηD)

(∇u + ∇uT − 2

3∇ · uI

),

with shear and bulk viscosities given by

η = ηK + ηD = 3mkBT

2γ[wD

]R

+ γn2[R2wD

]R

30

= ρD

2+ γn2

[R2wD

]R

30, (9.83)


ζ = ζK + ζD = mkBT

γ[wD

]R

+ γn2[R2wD

]R

18, (9.84)

where the diffusivity of a “tagged” DPD particle is

D = 3kBT

nγ[wD

]R

, (9.85)

and the square brackets define the following operation

[wD

]R =

∫dRwD (R) ,

[R2wD

]R =

∫dRR2wD (R) . (9.86)

This “compressible” Newtonian fluid is the analogue to an elastic solid with Lamémoduli η and ζ. We could define its “Poisson’s ratio” ν as,

ν = ζ − 23η

2(ζ + 1

3η) = X

1 + 4X, X = γ2n2

[wD

]R

[R2wD

]R

90mkBT. (9.87)

As this parameter X varies from zero to infinity, the fluid equivalent Poisson’s ratiois less than 0.25.

With the standard weighting function for DPD,

wD (r) ={

(1 − r/rc)2 , r < rc,

0, r ≥ rc,(9.88)

the integrations indicated in (9.86) can be performed, resulting in

[wD

]R = 2π

15r3c ,

[R2wD

]R = 4π

105r5c , (9.89)

which then leads to

η = 45mkT

4πγr3c+ 2πγn2r5c

1575, ζ = 15mkT


945, D = 45kBT

2πγnr3c. (9.90)

Inclusion of Conservative Forces In the limit of m/γ2 → 0 and no external force,(9.14) reduces to

0 =N∑

j=1, j =i

ai jwCei j −N∑

j=1, j =i

γwD(ei jei j · vi j

) +N∑

j=1, j =i

σwRθi jei j , (9.91)

where we note that N is the total number of DPD particles, and when necessary, weuse the superscript k on relevant variables to denote their values at the time level

9.3 Some Approximate Results 181

t = t k . This limit is not equivalent to a steady-state flow assumption - it only guar-antees that the inertial terms (as represented by Reynolds number) in the momentumequations are zero, but any other time-dependent behaviour inherited, for examplefrom the boundary conditions, has not been eliminated.

Taking the dissipative force to the left side, (9.91) becomes

N∑

j=1, j =i

γwDei jei j · (vi − v j) =

N∑

j=1, j =i

ai jwCei j +N∑

j=1, j =i

σwRθi jei j . (9.92)

The DPD system is obtained by letting i taking values from 1 to N in (9.92), in thematrix-vector notation,

Av = f, (9.93)

whereA is the systemmatrix, whose elements are second-order tensors, v the columnvector of unknown velocity vectors and f the column vector of right hand forcevectors:

A =

⎡

⎢⎢⎢⎢⎣

∑Nj=2 γwDe1 je1 j −γwDe12e12 · · · −γwDe1Ne1N−γwDe21e21

∑Nj=1, j =2 γwDe2 je2 j · · · −γwDe2Ne2N

......

. . ....

−γwDeN1eN1 −γwDeN2eN2 · · · ∑N−1j=1 γwDeN jeN j

⎤

⎥⎥⎥⎥⎦

, (9.94)

v =

⎛

⎜⎜⎜⎝

vk1vk2...

vkN

⎞

⎟⎟⎟⎠

, f =

⎛

⎜⎜⎜⎜⎝

∑Nj=2 a1 jwCe1 j + ∑N

j=2 σwRθ1 je1 j∑Nj=1, j =2 a2 jwCe2 j + ∑N

j=1, j =2 σwRθ2 je2 j...∑N−1

j=1 aN jwCeN j + ∑N−1j=1 σwRθN jeN j

⎞

⎟⎟⎟⎟⎠

. (9.95)

This system is singular since the row (and column) sum ofA is zero. The eigenvectorof the system, corresponding to an eigenvalue of 0, is {U, . . . ,U}T , where U is anarbitrary vector. Below are two schemes proposed to deal with this singularity.

Scheme 1

We first record a physical constraint for the DPD system. The centre of mass (formi = m),

Rc = 1

Nm

N∑

i=1

miri , (9.96)

satisfies

Nmd2Rc

dt2= 0 =

N∑

i=1

mid2ridt2

, (9.97)


due to pair-wise force interactions, and thus the centre of mass of the DPD systemmoves deterministically with constant velocity Uc. Without much loss to generality,one can take this constant as zero,

Uc = 1

N

N∑

r=1

vr = 0. (9.98)

By adding the term∑N

j=1, j =i γwDei jei j(∑N

r=1,r = j vr)to both sides of equation

(9.92), we obtain, in the matrix-vectorial notation,

N∑

j=1, j =i

γwDei jei j(vi − v j

) +N∑

j=1, j =i

γwDei jei j

⎛

⎝N∑

r=1,r = j

vr

⎞

⎠ =

N∑

j=1, j =i

ai jwCei j +N∑

j=1, j =i

γwDei jei j

⎛

⎝N∑

r=1,r = j

vr

⎞

⎠ +N∑

j=1, j =i


It can be rearranged as

N∑

j=1, j =i

γwDei jei j

⎛

⎝vi +N∑

r=1,r = j

vr

⎞

⎠ =N∑

j=1, j =i

ai jwCei j+

N∑

j=1, j =i

γwDei jei j

(N∑

r=1

vr

)

+N∑

j=1, j =i


Substitution of (9.98) into (9.100) yields

N∑

j=1, j =i

γwDei jei j

⎛

⎝vi +N∑

r=1,r = j

vr

⎞

⎠ =N∑

j=1, j =i

ai jwCei j +N∑

j=1, j =i

σwRθi jei j .

(9.101)The DPD system matrix corresponding to (9.101) is

A =

⎡

⎢⎢⎢⎢⎢⎢⎢⎣

2∑N

j=2 γwDe1 je1 j∑N

j=3 γwDe1 je1 j · · · ∑N−1j=2 γwDe1 je1 j

∑Nj=3 γwDe2 je2 j 2

∑Nj=1, j =2 γwDe2 je2 j · · · ∑N−1

j=1, j =2 γwDe2 je2 j

......

. . ....∑N−1

j=2 γwDeN jeN j∑N−1

j=1, j =2 γwDeN jeN j · · · 2∑N−1

j=1 γwDeN jeN j

⎤

⎥⎥⎥⎥⎥⎥⎥⎦

,

(9.102)


which is invertible, but fully populated and not diagonally dominant. The presentDPD system can be solved by a direct solver such as the one based on the LU fac-torisation with partial pivoting, with a high computational cost at large number ofparticles.

Scheme 2

The system matrix A in (9.93) can be decomposed into a diagonal block I andthe remainder K:

Iv + Kv = f, (9.103)

where

I =

⎡

⎢⎢⎢⎢⎣

∑Nj=2 γwDe1 je1 j 0 · · · 0

0∑N

j=1, j =2 γwDe2 je2 j · · · 0...

.... . .

...

0 0 · · · ∑N−1j=1 γwDeN jeN j

⎤

⎥⎥⎥⎥⎦

, (9.104)

and

K =

⎡

⎢⎢⎢⎣

0 −γwDe12e12 · · · −γwDe1Ne1N−γwDe21e21 0 · · · −γwDe2Ne2N

......

. . ....

−γwDeN1eN1 −γwDeN2eN2 · · · 0

⎤

⎥⎥⎥⎦

. (9.105)

Now, since I is diagonal block, its inverse is

I−1 =

⎡

⎢⎢⎢⎣

(∑Nj=2 γwDe1 je1 j

)−1 · · · 0...

. . ....

0 · · ·(∑N−1

j=1 γwDeN jeN j

)−1

⎤

⎥⎥⎥⎦

, (9.106)

and we can rewrite (9.103) asv + Hv = b, (9.107)

where H = I−1K and b = I−1f .Because of the zero row (and column) sum of the operator I + K, an eigenvector

forH isε = (U,U, ...,U)T /

√NU, U 2 = U · U, (9.108)

where U is any constant vector - this eigenvector corresponds to a normalised –1eigenvalue. Therefore solution of (9.107), if exist, is non-unique - any solution plusa multiple of the eigenvector (9.108) is also another solution.


Following the Wielandt’s deflation approach ([7, 68]), we can render a uniquesolution to (9.107), and at the same time, deflate the eigenvalue at –1 to zero, byconsidering the equivalent system,

v + Hv + εx [v, εx ] + εy[v, εy

] + εz[v, εz

] = b, (9.109)

where [·, ·] denotes the natural inner product of two vector elements in the underlyingspace, and

εx = ( Ux ,Ux , ...,Ux )T/√

NUx , Ux = (1, 0, 0)T , U 2x = Ux · Ux , (9.110)

εy = (Uy,Uy, ...,Uy)T/√

NUy, Uy = (0, 1, 0)T , U 2y = Uy · Uy, (9.111)

εz = (Uz,Uz, ...,Uz)T/√

NUz, Uz = (0, 0, 1)T , U 2z = Uz · Uz . (9.112)

This is seen by expanding the additional inner-product terms

εx [v, εx ] + εy[v, εy

] + εz[v, εz

] = 1

NU 2x

(Ux ,Ux , . . . ,Ux )T Ux ·

N∑

i=1

vi x+

1

NU 2y

(Uy,Uy, . . . ,Uy

)TUy ·

N∑

i=1

viy+

1

NU 2z

(Uz,Uz, . . . ,Uz)T Uz ·

N∑

i=1

vi z = 0,

(9.113)

and therefore they do not contribute to the governing equation; they represent theconstraint (9.98). However, by writing this in this fashion, we have (i) mapped theeigenvalue –1 to zero (ε is still the eigenvector, but with corresponding eigenvalueof zero); and (ii) rendered a unique solution to (9.103).

If the previous spectral radius (before deflation) is one, it is now less than one anditerative numerical implementation to (9.103) may work well; a simplest one is thePicard’s iteration:

vli = −Hivl−1 − εx 〈vl−1, εx 〉 − εy〈vl−1, εy〉 − εz〈vl−1, εz〉 + bi , (9.114)

where i = (1, 2, . . . , N ),

Hi =⎛

⎝N∑

j=1, j =i

γwDei jei j

⎞

⎠

−1

[−γwDei1ei1, . . . ,−γwDei(i−1)ei(i−1), 0, . . . ,−γwDei Nei N], (9.115)

and


bi =⎛

⎝N∑

j=1, j =i

γwDei jei j

⎞

⎠

−1 ⎛

⎝N∑

j=1, j =i

ai jwCei j +N∑

j=1, j =i

σwRθi jei j

⎞

⎠ . (9.116)

The iterative process stops when the difference of the solutions between two succes-sive iterations is less than a specified tolerance - a possible tolerance may be

norm(vl − vl−1) < 10−7norm(vl). (9.117)

It can be seen that Hi is a sparse matrix because (i) the construction of this matrixinvolves only particles within the interaction zone whose size is defined by the cutoffradius rc; and (ii) the value of rc is generally much smaller than a typical lineardimension of the flow domain. Furthermore, it is straightforward to implement theiterative scheme in parallel as (9.114) is applied to each particle, independent of therest of the particles.

For both Schemes 1 and 2, the Euler algorithm is employed to advance the positionof the particles

rk+1i = rki + vki Δt, (9.118)

where Δt = t k+1 − t k .

9.3.2 Standard DPD Parameters

It is a convenient point to mention the “standard” choice of DPD parameters. Thestandard weighting function (9.88) has been mentioned. Mass and cutoff radius arecommonly normalised to unity (m = 1 = rc), then the density depends on the initialarrangement of DPD particles – for an initial face centered cubic (FCC) arrangement,for example, the density is 4 (four DPD particles per unit cell, rc × rc × rc). Boltz-mann temperature is also normalised to unity, kBT .For a liquidwith a compressibilitylike that of water, Groot and Warren [33] recommended

ai j = 75kBT

nr4c, (9.119)

in 3D, and in 2D, the repulsive coefficients are given by

ai j = 57.23kBT

nr3c. (9.120)

They further recommended a random force strength ofσ = 3; consequently, from thefluctuation-dissipation theorem,γ = 4.5.Toohigh a value ofσmay render the systemunstable. We refer to the foregoing choice of DPD parameters as the “standard”choice.


9.3.3 Effective Size of a DPD Particle

A DPD particle is a point mass in its Newton’s 2nd law of motion, under the actionof inter-particle forces. However, it is endowed with a soft repulsive potential, andtherefore has an “effective” size, which is the exclusion zone of the particle. Onecould provide an estimate for this effective size of a DPD particle by various means.In what follows, we analyse roles of the DPD forces on the particle’s exclusion size.

Role of the Conservative Force The inclusion of the conservative force (repulsiveforce) into the DPD formulation is to provide an independent mean of controllingthe speed of sound (compressibility) to the number density and the temperature ofthe DPD system [56]. Keeping the dissipative and random forces unchanged, anincrease in the repulsion strength ai j will promote incompressibility of the DPDfluid [65]. However, at large values of ai j , the mean squared particle displacement⟨(r(t) − r(0))2

⟩is observed to be no longer linear in time (a characteristic of a Brown-

ian particle) with crystallisation occurring and the DPD system has a solid-like struc-ture. Here, we limit our attention to the case where the compressibility of the systemis matched to that of the water at room temperature [33].

If the weighting functionwC is fixed at its linear form, the size of solvent particlesinduced by the conservative forces will be controlled by means of the repulsionparameter ai j . A larger value of ai j results in a larger size of the particle and viceversa. From expressions (9.120) and (9.119), one can reduce the particle size byincreasing n, increasing rc or reducing kBT . These expressions also reveal that thelarger the DPD particle’s size (corresponding to larger ai j ), the coarser level (smallern) the DPD system will be (this comes from the scaling property of the DPD system[29]).

Role of the Dissipative and Random Forces Consider a generic “tagged” DPDparticle in a sea of other DPD particles undergoing a diffusion process. Its size aef fmay be estimated by the Stokes–Einstein relation

aef f = kBT

6πDη, (9.121)

where D is the diffusion coefficient of the tagged particle subject to Brownianmotionin an unbounded domain and η the shear viscosity of the surrounding fluid. This canbe equated to the diffusivity in (9.90) with FC

i j = 0 to yield

aef f = ργr3c135η

= 630 × 2πγ2n2r6c27

(1575 × 45ρkBT + 8π2γ2n3r8c

) . (9.122)

For our standard DPD parameters aef f ≈ 0.12. Note that expression (9.122) is estab-lished assuming the Stokes–Einstein relation, which is valid for a dispersion ofmeso-scopic particles in a continuous solvent. The DPD particles representing the solventphase are assumed not to be clustered, and are of a size considerably less than that


Fig. 9.2 DPD system, standard parameters: effects of the number density (left), cutoff radius(middle) and temperature (right) on the particle size (the dash line representing the zone sizecaused by the dissipative and random forces (9.122) and the solid line by the conservative force(9.119))

of the tagged particle that has no inertia. This latter condition is of course not sat-isfied here and the expression (9.122) can only be considered as best an estimate.However, the solvent particle size approaches zero as any one of the number densityn, the cutoff radius rc or the thermodynamic temperature kBT approaches infinity.

Although expression (9.122) is only an approximate result, but it serves as a meanto gauge the effect of different parameters. Figure9.2 (left to right) show respectivelythe effects of n, rc and kBT on the solvent particle size. Figure9.2 (left and middle)clearly indicate that the exclusion zones caused by the conservative force and bythe dissipative and random forces are both smaller with increasing either n or rc.However, in Fig. 9.2 (right), the solvent particle size is an increasing function ofkBT , for the conservative force, and a decreasing function of kBT , for the dissipativeand random forces. Special care is thus needed if one tries to control the solventparticle size via kBT . It is noted that (i) reducing kBT makes ai j smaller which canresult in the clustering of particles; and (ii) the particle size, defined in (9.122), isinversely proportional to r2c and n, and the value of ai j , defined in (9.119), is inverselyproportional to r4c and n. These observations imply that (i) controlling particles sizevia n and rc is clearly more effective than via kBT ; and (ii) increasing rc results in afaster decrease in the particle size than increasing n.

From this approximate analysis, we can see that there are many possible com-binations of n and rc that reduces aef f . One can employ rc = 1 with a large valueof n (finer coarse-graining level with a standard cutoff radius). Or one can employ,for example n = 3 with a larger value of rc > 1 (upper coarse graining level with alarger cutoff radius).

The determination of the exclusion size of a DPD particle can be accomplishedwith the concept of radial distribution function [76]

g(q) = 1

N/V

〈s〉4πq2Δq

, (9.123)


0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1−0.2

0

0.2

0.4

0.6

0.8

1

1.2

q

g(q)

Fig. 9.3 Radial distribution function for the DPD particles

where V is the volume of the domain of interest containing N particles, and 〈s〉 isthe average number of particles in a spherical shell of width Δq at a distance q froma particle in the fluid. Figure9.3 shows the variation of g(q), where Δq is chosen as0.05, for the standard DPD fluid. Let q be the value of q at which g(q) > 0.05. Theeffective radius of the solvent particles can be estimated as aef f = q/2 = 0.32/2 =0.16, a result close to Pan et al.’s [67].

9.4 Modification of the Weighting Function

The dissipative force directly affects the rate of momentum transfer of the systemand thus the dynamic properties of the system; a measure of this response is theSchmidt number, defined as the rate of the speed of momentum transfer to the speedof particle’s diffusion

Sc = η

ρD. (9.124)

Because the soft interaction between DPD particles, the speed of momentum transferis slow, of the same order as particle diffusion. Therefore the Schmidt number is aboutunity. For a real fluid of physical properties like those of water, the Schmidt numberis O(103), and therefore there is the need to improve on the dynamic behaviour ofthe DPD system.

A simpleway to do this has been proposed (Fan et al. [24]): the standardweightingfunction (9.88) for the dissipative force is to be replaced by

9.4 Modification of the Weighting Function 189

wD (r) = [wR (r)

]2 ={

(1 − r/rc)s , r < rc,

0, r ≥ rc.(9.125)

When s = 2, the weight function for the conventional DPD formulation is recovered.From (9.86), and using the new weighting function (9.125), it can be shown that

[wD

]R

= 4πr3c

[1

1 + s− 2

2 + s+ 1

3 + s

], (9.126)

[R2wD

]R = 4πr5c

[1

1 + s− 4

2 + s+ 6

3 + s− 4

4 + s+ 1

5 + s

],

leading to the following diffusivity and viscosity,

D = 3kBT

nγ[wD

]R

= 3kBT

4πnγr3c

[1

1 + s− 2

2 + s+ 1

3 + s

]−1

, (9.127)

η = 3mkBT

8πγr3c

[1

1 + s− 2

2 + s+ 1

3 + s

]−1

(9.128)

+2πr5c γn2

15

[1

1 + s− 4

2 + s+ 6

3 + s− 4

4 + s+ 1

5 + s

],

Consequently, Schmidt number is

Sc = 1

2+ 8π2γ2n2r8c

45mkBT

[1

1 + s− 4

2 + s+ 6

3 + s− 4

4 + s+ 1

5 + s

].

.

[1

1 + s− 2

2 + s+ 1

3 + s

](9.129)

The physical properties of two DPD systems, corresponding to the standardweighting function (s = 2), and a modified weighting function (s = 1

2 ) are tabu-lated in Table9.1. With the standard choice of DPD parameters, a simple change

Table 9.1 Properties of the two DPD systems

Formulation Conventional (s = 2) Modified (s = 1/2)

Diffusivity, D45kBT

2πγnr3c

315kBT

64πγnr3c

Viscosity, η45mkBT


1575

315mkBT

128πγr3c+ 512πγn2r

5c

51975

Bulk viscosity, ζ15mkBT


945

105mkBT

436πγr3c+ 512πγn2r

5c

31185

Schmidt number, Sc1

2+

(2πγnr4c

)2

70875mkBT

1

2+ 32 768

(πγnr4c

)2

16 372 125mkBT


from s = 2 to s = 12 increases the Sc number by a factor of 10, and thereby improve

on the dynamic response of the DPD fluid. Since Sc = O(γ2, r8c

), a modest increase

in either γ or rc results in a large increase in Sc. Increasing γ strengthens the randomforces, resulting in larger fluctuations of thermal energy injected into the system.Clearly, the most efficient way to increase Sc is to increase rc. However, the compu-tation demand of calculating the dissipative and random forces is O

(r3c), and can be

significantly increased with rc. We prefer to combine the modified weighting func-tion with a moderate increase in the cutoff radius for dissipative weighting function,so that a physical level of Sc can be reached with a moderate increase in the compu-tation cost. For s = 1

2 , rc = 2, and standard choice of the remaining DPD parameterswe find that Sc = 1.6 × 103.

9.5 Imposition of Physical Parameters

Equations (9.45) and (9.56) show that the motion of a DPD fluid is governed bythe Navier–Stokes equation, with its physical properties, such as viscosity, Schmidtnumber, isothermal compressibility, relaxation and inertia time scales, etc., in factits whole rheology, resulted from the choice of the DPD model parameters. In thissection, we will explore the response of a DPD fluid with respect to its parameterspace, where themodel input parameters can be chosen in advance so that (i) the ratiobetween the relaxation and inertia time scales is fixed; (ii) the isothermal compress-ibility of water at room temperature is enforced; and (iii) the viscosity and Schmidtnumber can be specified as inputs. These impositions are possible with some extradegrees of freedom in the weighting functions for the conservative and dissipativeforces.

9.5.1 Time Scales

Let us focus on a tagged, but otherwise arbitrary DPD particle in the system, and letα be the ratio of its relaxation time scale to inertia time scale

α = τ

τI= O

(γ2rcmai j

), (9.130)

where τ = O(γH−1), H = O (|∂r FC |) = O(ai jr−1

c

)and τI = O

(mγ−1

).

Substitution of (9.119) and (9.120) into (9.130) yield, respectively,

α = τ

τI∼ γ2r5c n

71.54mkBT, for 3D space, (9.131)

α = τ

τI∼ γ2r4c n

57.23mkBT, for 2D space, (9.132)

9.5 Imposition of Physical Parameters 191

which reveal the dependence of the dimensionless quantityα onm, rc, n, γ and kBT .For the case of small m (small τI , large α), the particles’ inertia can be neglected

leading to a fast response. Small time steps are required for a numerical simulation,but particles are well distributed because of a lowMach number. The low-mass DPDsystem will be further discussed in Sect. 9.6.2.

We now keep the mass and number density fixed (m = 1 and n = 4) and varythe cutoff radius (rc ≥ 1). With standard DPD input values (m = 1, kBT = 1, n =4,σ = 3 (γ = 4.5), rc = 1), where the DPD system is observed to well behave, theratio between the two time scales can be estimated as

α = τ

τI∼ 1.1322 = O(1), for 3D space, (9.133)

α = τ

τI∼ 1.4153 = O(1), for 2D space. (9.134)

Expressions (9.131) and (9.132) indicate a very strong influence of rc on the time-scale ratio. For example, with rc = 2.5, one has α = O(102). Also, as rc increasesfrom1 to2.5 (ai j , from (9.120), is reduced from14.30 to 0.91), numerical experimentsreveal that DPD particles tend to form local clusters. It appears that a large valueof α, due to large rc, may adversely affect the stability of the DPD system throughclustering. Our recommendation here is to make α constant (independently of theinput parameters) and having an appropriate value to avoid the clustering of particles.The optimal value of α can be determined numerically.

9.5.2 Imposition of Dimensionless Compressibilityand Time-Scale Ratio

Our goal here is to create a new form of the conservative force that can control boththe isothermal compressibility and time scale ratio. To do so, apart from ai j , there isa need for having another free parameter. A conservative force is proposed to be

Fi j,C = ai j

(1 − r

rc

)s

, s > 0, (9.135)

whose average gradient over 0 ≤ r ≤ rc is also ai j/rc. These two free parameters,ai j and s, can be designed to satisfy

α = τ

τI= γ2rc

mai j, (9.136)

κ−1 = 1

kBT

∂ p

∂n, (9.137)


where p and κ are the pressure and isothermal compressibility, respectively (α andκ are given constants, e.g., for water, κ = 1/15.98).

From the virial theorem, the pressure is computed as

p = nkBT + n2

2d

∫dRr Fi j,C(r)g(r), (9.138)

whered is theflowdimensionality and g(r) the radial distribution function - a possiblefunction may be g(r) = 1 corresponding to an infinite number of DPD particles.

Expression (9.138) results in

p = nkBT + 4πai j n2r4c(s + 1)(s + 2)(s + 3)(s + 4)

, (9.139)

∂ p

∂n= kBT + 8πai j nr4c

(s + 1)(s + 2)(s + 3)(s + 4), (9.140)

for 3D case, and

p = nkBT + πai j n2r3c(s + 1)(s + 2)(s + 3)

, (9.141)

∂ p

∂n= kBT + 2πai j nr3c

(s + 1)(s + 2)(s + 3), (9.142)

for 2Dcase. Equations (9.140) and (9.142) for the variable s canbe solved analyticallyand we are interested in only physical positive values of s.

Analytic solution to (9.136) and (9.137) can thus be found as

ai j = 1

α

γ2rcm

, (9.143)

s =√5 + 4

√C + 1 − 5

2, C = 8πai j nr4c

(κ−1 − 1)kBT, (9.144)

for 3D space, and

ai j = 1

α

γ2rcm

, (9.145)

s = 1

3B+ B − 2, B =

(C

2+

√C2

4− 1

27

)1/3

, C = 2πai j nr3c(κ−1 − 1)kBT

,

(9.146)

for 2D space. For (9.144) and (9.146) to have a physical value (i.e. s > 0), it requiresC > 24 and C > 6, respectively, which can be easily satisfied.


Alternatively, the condition of isothermal compressibility can be replaced withthe speed of sound

c2s = ∂ p

∂ρ, (9.147)

where ρ = mn is the density. The corresponding solutions ai j and s have the sameforms as (9.143)–(9.144) and (9.145)–(9.146), except that values of C in (9.144)and (9.146) are replaced with 8πai j nr4c /(mc2s − kBT ) and 2πai j nr3c /(mc2s − kBT ),respectively.

9.5.3 Imposition of Viscosity and Dynamic Response

Following [24], we employ

wD =(1 − r

rc

)s

, s > 0. (9.148)

The dissipative force now involves 2 free parameters, γ and s, and we utilise themto match the two thermodynamic properties: flow resistance and dynamic response.

As shown in (9.83), there are two contributions to the viscosity, the kinetic part ηK

(gaseous contribution) and the dissipative part ηD (liquid contribution). We are inter-ested in the case where the dissipative contribution is a dominant part, i.e. ηD � ηK

(liquid-like behaviour) under which the two following constraints are approximatelysatisfied

ηD = η, (9.149)ηD

ηK= 2Sc, (9.150)

or

γn2[R2wD]R2d(d + 2)

= η, (9.151)

2γη[wD]RdmkBT

= 2Sc, (9.152)

where η and Sc are the specified viscosity and Schmidt number, respectively. Thissystem can be solved analytically for the two variables γ and s. In 3D space, itssolution is


s = −9 + √1 + 4C

2, C = 6ScmkBTn2r2c

5η2 , (9.153)

γ = 5η(s + 1)(s + 2)(s + 3)(s + 4)(s + 5)

16πn2r5c. (9.154)

Since s > 0, it requires

η <

√3ScmkBTn2r2c

50for a given Sc, (9.155)

Sc >50η2

3mkBTn2r2cfor a given η. (9.156)

In 2D, the solution to the system takes the form

s = −7 + √1 + 4C

2, C = 3ScmkBTn2rc2

4η2 , (9.157)

γ = 4η(s + 1)(s + 2)(s + 3)(s + 4)

3πn2r4c. (9.158)

Since s > 0, it requires

η <

√ScmkBTn2r2c

16for a given Sc, (9.159)

Sc >16η2

mkBTn2r2cfor a given η. (9.160)

Verification The four parameters, namely isothermal compressibility, time-scaleratio, dynamic response and viscosity, can be prescribed as inputs to the DPD equa-tions. They define a particular fluid; but unlike conventional DPDs, a change in rcshould not affect the fluid characteristics. For a given shear rate γ, shear stresses inCouette flow can be computed as τxy = ηγ and we use these “exact values” to assessthe accuracy of the present DPD.

For a typical set of input parameters (η = 30, Sc = 400, rc = 2.5,m = 1, kBT =1, n = 4, γ = 0.2), we plot the standard deviations of n and τxy on the cross sectionand percentage errors of τxy against the time-scale ratio α (Fig. 9.4), which show thattheir minimum values occur within a range α = 10−1 − 101. For simplicity, α canbe taken as 1.

To examine the effects of enforcing a fixed time scale ratio, the obtained resultswith α = 1 are compared with those by the DPD using conventional conservativeforces (i.e. variable α). As shown in Table9.2, the former is more stable and accurate


Fig. 9.4 Couette flow: Standard deviations of the number density n and shear stress τxy on thecross section and percentage errors of τxy against time-scale ratio α = τ/τI

Table 9.2 Couette flow: mean and standard deviation of the number density n and shear stress τxyon the cross section

Mean (n) Std(n) Mean (τxy) Error (τxy) (%) Std (τxy)

rc = 2

Conventional FC 4.0000 0.2140 5.7873 3.5450 0.4293

Modified FC 4.0000 0.1043 6.0956 1.5933 0.4208

rc = 2.5

Conventional FC 4.0000 0.2672 5.7110 4.8167 0.4688

Modified FC 4.0000 0.1701 6.1176 1.9600 0.4022

rc = 3.0

Conventional FC 4.0000 0.2964 5.6590 5.6833 0.4119

Modified FC 4.0000 0.1396 6.0286 0.4767 0.3177

rc = 3.5

Conventional FC 4.0000 0.3958 5.6559 5.7350 0.6375

Modified FC 4.0000 0.1246 6.0648 1.0800 0.3372

rc = 4

Conventional FC 4.0000 0.3859 5.5519 7.4683 0.4635

Modified FC 4.0000 0.1464 6.0403 0.6717 0.4576

than the latter for all values of rc employed. In Table9.3, several viscosities corre-sponding to s between 0 and 2 (standard value) are imposed, and the case of usingconservative forces with a fixed α is seen to yield a better agreement.


Table 9.3 Couette flow: imposed and computed viscosities.

Conventional FC Modified FCImposed η Computed η Error (%) Computed η Error (%)

rc = 2.5

39 38.3277 1.72 38.8942 0.27

35 34.1332 2.47 35.0943 0.26

31 29.7771 3.94 31.4797 1.54

rc = 3.5

58 57.5698 0.74 58.0389 0.06

50 49.3714 1.25 49.8442 0.31

42 41.1593 2.00 42.0830 0.19

9.6 Numerical Implementation

A typical DPD simulation is done in a similar manner to that of an MD: an initialconfiguration of fluid and wall particles are generated separately by a pre-processingprogram and read in as input data. The total number of particles depends on the sizeand geometry of the flow domain and the densities of the fluid and wall materials.Typically, fluid particles are initially located at conveniently points, for example, atthe sites of a face-centred cubic (FCC) lattice. The initial velocities of the particlesare set randomly according to a given temperature but the wall particles are frozen.At the beginning of the simulation the particles are allowed tomovewithout applyingthe external force until a thermodynamic equilibrium is reached. Then the externalforce field is applied to fluid particles and the non-equilibrium simulation starts.There are many schemes to integrate the equations of motion for a DPD systemin time, including Verlet, leap-frog, predictor-corrector algorithms (Nikunena et al.[60]), but the standard one is the velocity Verlet scheme, in a modified form - it isreviewed here below.

9.6.1 Velocity Verlet Scheme

The standard velocity Verlet algorithm follows directly from a Taylor’s series expan-sion, omitting the subscript,

r (t + Δt) = r (t) + Δt r (t) + 1

2Δt2r (t) + 1

6Δt3

...r (t) + O

(Δt4

), (9.161)

where the acceleration in r is a known function, from the total force calculation.Similarly,

9.6 Numerical Implementation 197

r (t − Δt) = r (t) − Δt r (t) + 1

2Δt2r (t) − 1

6Δt3

...r (t) + O

(Δt4

). (9.162)

From these, the basic Verlet updating scheme is derived,

r (t + Δt) = 2r (t) − r (t − Δt) + Δt2r (t) + O(Δt4

). (9.163)

Since the evaluation of the acceleration (the forces) requires the current value of thevelocity field, a slightly modified version of (9.163) is required:

r (t + Δt) = r (t) + Δtv (t) + 1

2Δt2v (t) ,

v (t + Δt) = v (t) + λΔt v (t) ,

v (t + Δt) = v (r (t + Δt) , v (t + Δt)) ,

v (t + Δt) = v (t) + 1

2Δt [v (t) + v (t + Δt)] , (9.164)

where the acceleration v (t) (i.e. the total net force) is a known function, given theposition and velocity at instant t , v (t + Δt) is the velocity prediction at the instantt + Δt , λ is an empirically introduced parameter, which account for some additionaleffects of the stochastic interactions. If the total force is velocity independent, thestandard velocity-Verlet algorithm is recovered at λ = 1/2. Groot and Warren [33]found λ = 0.65 to be optimum. At this value, the time step can be increased toΔt ∼ 0.05 without a loss of temperature control.

9.6.2 Exponential Time Differencing Scheme

DPD fluid in its standard form is compressible in nature and its dynamic response israther slow - the Schmidt number (Sc) is about unity because of the soft interactionbetween particles in the DPD system. On the other hand, many practical applicationsinvolve incompressible flows that exhibit strongly viscous behaviour at low fluidinertia, (i.e. low Reynolds number (Re) flows). Incompressibility is a good approxi-mation in many practical flows at low Mach numbers (M < 0.3). One effective wayto induce an incompressible slow viscous flow in a DPD fluid and simultaneouslyenhance its dynamic response is to reduce the mass of the particles (m). In the limitofm → 0, the DPD system is singular with a corresponding loss of the highest orderderivative and a strongly overdamped system results, with Re → 0, M → 0 andSc → ∞.

At low mass,m → 0, the system (9.14) is stiff, and a different numerical integra-tion scheme designed for stiff systems, such as the Exponential Time Differencing(ETD) method, may be a better alternative. To illustrate different ETD schemes, it issufficient to consider the scalar system


dφ(t)

dt+ cφ(t) = A + Bθ (t) , (9.165)

where the parameter c is either large and positive or large and imaginary, with Arepresenting the deterministic part, and Bθ (t), the stochastic part of the forcingterms, with θ (t) a white noise of unit strength. In our DPD case, when the mass issignificantly small, c = O

(m−1

)is positively large, and the system (9.165) is stiff.

The solution in this case contains very different time scales occurring simultaneously(not counting the time scales inherent in θ (t)), and standard integration schemes,such as the velocity-Verlet, are not accurate enough [81].

We multiply (9.165) by the integrating factor ect , and then integrate the equationover a single time step from t to t + Δt to get

ec(t+Δt)φ (t + Δt) = ectφ (t) + ect∫ Δt

0ecτ (A (t + τ ) + B (t + τ ) θ (t + τ )) dτ ,

orφ (t + Δt) = e−cΔtφ (t) + ΔA + ΔB, (9.166)

where

ΔA (t;Δt) = e−cΔt∫ Δt

0ecτ A (t + τ ) dτ , (9.167)

and

ΔB (t;Δt) = e−cΔt∫ Δt

0ecτ B (t + τ ) θ (t + τ ) dτ . (9.168)

The basis of different ETD schemes lies in the approximations to the integrals (9.167)and (9.168).

First-Order Stochastic ETD Scheme In the simplest approximation, where A andB are treated as constants within this time step, one arrives at the first-order ETDscheme. In this case

ΔA = ΔA1 = e−cΔt∫ Δt

0ecτ A (t + τ ) dτ = A (t)

c

(1 − e−cΔt

), (9.169)

and

ΔB (t;Δt) = e−cΔt B (t)∫ Δt

0ecτ θ (t + τ ) dτ

= B (t)ΔW1, (9.170)

where ΔW1 is the Wiener process

ΔW1 (Δt) = e−cΔt∫ Δt

0ecτ θ (t + τ ) dτ , (9.171)


with zero mean,〈ΔW1 (Δt)〉 = 0, (9.172)

and auto-correlation function

〈ΔW1 (Δt)ΔW1 (Δt)〉 = e−2cΔt∫ Δt

0dt ′

∫ Δt

0dt ′′ect

′ ⟨θ(t + t ′

)θ(t + t ′′

)⟩ect

′′

= e−2cΔt∫ Δt

0e2ct

′dt ′

= 1

2c

(1 − e−2cΔt

). (9.173)

We call the resulting scheme (9.166), with (9.169) and (9.170), the first-orderstochastic exponential time differencing scheme. In summary, the scheme is definedas

φ (t + Δt) = e−cΔtφ (t) + ΔA1 (t;Δt) + B (t)ΔW1. (9.174)

Second-Order Stochastic EDT Scheme Next, we consider a higher-order approxi-mation scheme whereby A and B are approximated as linear functions in the interval0 < τ < Δt :

A (t + τ ) = A (t) + τ

Δt(A (t) − A (t − Δt)) = A (t) + Δa

Δtτ ,

B (t + τ ) = B (t) + τ

Δt(B (t) − B (t − Δt)) = B (t) + Δb

Δtτ ,

where Δa = A (t) − A (t − Δt) and Δb = B (t) − B (t − Δt) are the first-orderbackward differences for A and B. Then,

ΔA (t;Δt) = e−cΔt∫ Δt

0ecτ A (t + τ ) dτ

= A (t)

c

(1 − e−cΔt

) + Δa

Δte−cΔt

∫ Δt

0ecτ τdτ

= ΔA1 (t;Δt) + cΔt − 1 + e−cΔt

c2Δa

Δt= ΔA1 (t;Δt) + ΔA2 (t;Δt) , (9.175)

where ΔA1 (t;Δt) has been defined in (9.169) and

ΔA2 (t;Δt) = cΔt − 1 + e−cΔt

c2Δa

Δt. (9.176)


In addition ΔB (t;Δt) is given by

ΔB (t;Δt) = e−cΔt∫ Δt

0ecτ B (t + τ ) θ (t + τ ) dτ

= B (t)ΔW1 + Δb

Δte−cΔt

∫ Δt

0τecτ θ (t + τ ) dτ

= B (t)ΔW1 + Δb

ΔtΔW2, (9.177)

where the Wiener process ΔW1 has been defined in (9.171), and the Wiener processΔW2 is defined as

ΔW2 (Δt) = e−cΔt∫ Δt

0τecτ θ (t + τ ) dτ . (9.178)

It has zero mean〈ΔW2〉 = 0, (9.179)

and its mean square is given by

〈ΔW2ΔW2〉 = e−2cΔt∫ Δt

0dt ′

∫ Δt

0dt ′′t ′ect

′ ⟨θ(t + t ′

)θ(t + t ′′

)⟩t ′′ect

′′

= e−2cΔt∫ Δt

0dt ′t ′2e2ct

′

= 1

4c3e−2cΔt

[e2cΔt

(2c2Δt2 − 2cΔt + 1

) − 1]

= 1

4c3[2c2Δt2 − 2cΔt + 1 − e−2cΔt

]. (9.180)

These result in an integration scheme called the second-order stochastic exponen-tial time differencing, summarised as

φ (t + Δt) = e−cΔtφ (t) + ΔA1 (t;Δt) + cΔt − 1 + e−cΔt

c2Δa

Δt(t;Δt)

+b (t)ΔW1 (Δt) + Δb

ΔtΔW2 (Δt) . (9.181)

In our DPD simulations, with mass as low asm = 10−3, the first-order ETD schemeproved to be very accurate, and is to be preferred for its simplicity.


9.6.3 Implementation of No-Slip Boundary Conditions

Periodic boundary conditions are applied on thefluid boundaries of the computationaldomain. The solid wall is usually represented using frozen particles. Due to the softrepulsion betweenDPDparticles, it is possible for fluid particles to penetrate thewall.Near-wall particles may not be sufficiently slowed down and slip with respect to thewall may then occur. To prevent this, a higher wall density and larger repulsive forceshave been used to strengthen wall effects. These, however, result in large densitydistortions in the flow field near to the wall, similar to MD. Special treatments havebeen proposed to implement no-slip boundary condition in DPD simulation withoutusing frozen wall particles. Revenga et al. [77] used effective forces to represent theeffects of wall on fluid particles instead of using wall particles. For a planar wall, theeffective forces can be obtained analytically. But these forces are not guaranteed toprevent fluid particles from crossing the wall. When particles cross the wall, a wallreflection is used to reflect particles back to the fluid. Willemsen et al. [91] added anextra layer of particles outside of the simulation domain. The position and velocityof particles in this layer are determined by the particles inside the simulation domainnear the wall, such that the mean velocity of a pair of particles inside and outsidethe wall satisfies the given boundary conditions. The above mentioned methods maynot work when the geometry of wall is complicated. We still use frozen particlesto represent the wall. Near the wall we assume that there is a thin layer where theno-slip boundary condition holds. We enforce a random velocity distribution in thislayer with zero mean corresponding to the given temperature. Similar to Revenga etal.’s [77] reflection law, we further require that particles in this layer always leavethe wall. The velocity of particle i in the layer is

vi = vR + n(√

(n · vR)2 − n · vR)

, (9.182)

where vR is a random vector and n a unit vector normal to the wall and pointing tothe fluid. The thickness of this layer and the strength of the repulsion between walland fluid particles are chosen to minimise the velocity and density distortion. Thelayer thickness is chosen to be the minimum between 0.5% of the narrowest widthof channels in the flow domain and 0.5 of length unit. A thin layer is necessary toprevent the frozen wall to cool down the fluid. This method is more flexible whendealing with a complex geometry.

A simpler reflection law is used in Duong et al. [19] to ensure no-slip boundarycondition on a solid surface,

vnew = 2Vwall − vold , (9.183)

where Vwall is the wall velocity vector. With this reflection law, no-slip boundary isguaranteed; however there is still a density fluctuation near the wall. Duong et al.[19] recommended a two-layer structure as illustrated in Fig. 9.5.


Fig. 9.5 Boundary DPD particles

In this scheme, the distances from the wall layers to the boundary are proportionalto rc, say α1rc and α2rc, respectively, with 0 < α1 < α2 < 1. Those layers of frozenparticles also interact with free particles in the same way as the free particles interactbetween themselves. To prevent free particles from penetrating the wall, firstly wechose α1 sufficient small, and secondly when particles enter in the wall layers thena bounce back is performed as:

rnew = rold + 2drnw, (9.184)

where dr is the distance from the particle to the boundary and nw is the normalvector on the wall directing into the simulation domain. The purpose of secondlayer is not only to strengthen the wall effect but also to make repulsive forces thewall particles exerting on free particles almost uniform. This layer plays a key roleon maintaining a uniform density in the fluid near the wall. We typically use thisapproach in implementing no-slip boundary conditions in DPD.

9.6.4 Computation of Interparticle Forces

In simulation a large system the computational efficiency is an important considera-tion. The main effort in solving the equations of motion is to compute the interactionforces between the particles. Fortunately, there has been a considerable number ofworks done in MD, which are directly applicable to DPD simulations. In particular,the cell sub-division and linked-list approach as described in Rapaport [75] can beemployed to reduce the computation time in the force calculation. The linked listassociates a pointer with each data item and to provide a non-sequential path throughthe data. In the cell algorithm, the linked list associates particles with the cells inwhich they reside at any given instance. It needs a one dimension array to store thelist and makes the search of neighbouring particles more efficient. To further savecomputational time, a neighbour-list method can also be employed. The neighbour-list remains valid over a number of time steps, typically 10–20 time steps. Hence thecomputation time in force calculation can be reduced further, with periodic refreshingthe list.


9.6.5 Calculation of Stress Tensor

The flow domain is divided into grids and local data are collected in each bin. Theflow properties are calculated by averaging over all sampled data in each bin. Thestress tensor is calculated using the Irving–Kirkwood expression [40]

S = − 1

V

⎡

⎣∑

i

mViVi + 1

2

∑

i

∑

j =i

ri jFi j

⎤

⎦ = −n

(〈mVV〉 + 1

2〈rF〉

), (9.185)

where n is the number density of particles, and Vi is the peculiar velocity of particlei (this should be compared to (9.49) and (9.50)). The first term on the right side of(9.185) denotes the contribution to the stress from the momentum (kinetic) transferof DPD particles (c.f., (9.49)) and the second term from the interparticle forces. Ifparticle i is an element of a complex particle (for example, a bead in a chain), thetotal force in the second term on the right of (9.185) should include constraint forces.The constitutive pressure can be determined from the trace of the stress tensor:

p = −1

3trS. (9.186)

9.6.6 Complex-Structure Fluid

DPD particles are connected by various spring-force laws or constraints to simulatea complex-structure fluid. Various degrees of complexities in a complex-structurefluid may be incorporated in a DPD simulation. For example, a suspension may bemodelled by constraining some DPD particles as rigid bodies in a DPD fluid (as asolvent) (i.e. frozen particle model, spring model), or by using two different speciesof DPD particles with different exclusion sizes: one of smaller size may representsolvent particles, and the other of larger size may represent suspended particles (i.e.single particle model). A polymer solution may be modelled in the same manner: apolymer chain is represented by connecting some DPD particles by some connectorforce law. Several kinds of spring-bead chains have been used in polymer rheologyas coarse-grained models of macromolecules. For example, the Kramers’ bead-rodchain, the FENE (Finitely Extendable Non-linear Elastic) dumbbell [6], worm-likechain [50]. These two types for force laws and the spring model are briefly reviewedbelow.

FENE Chains The FENE chain is a simple coarse-grained model of a polymermolecule, which can capture most of the important nonlinear rheological propertiesof polymer solution. Its rheological properties arewell known [88]. In the frameworkof DPD, the beads in a FENE chain are replaced by DPD particles. In addition tothe three kinds of forces in (9.14), the connector (intermolecular) forces act on these


FENE particles and contribute to the right hand of (9.14). For the FENE chain, theforce on bead i due to bead j is

FSi j = − Hri j

1 − (ri j/rm

)2 , (9.187)

where H is the spring constant, ri j = ri − r j , ri j = ∣∣ri j∣∣, and rm is the maximum

length of one chain segment. The spring force increases unboundedly as ri j/rmapproaches unity. This model can capture the finite extensibility of a polymer chain,and yields a shear-rate dependent viscosity along with a finite elongational viscosity(as opposed to an infinite elongational viscosity when the force law is linear for theHookean dumbbell).

The time constant is important in characterizing molecular motion and can beformed from model parameters. Two constants with time dimension can be definedfor the FENE spring [88],

λH = ζ

4H, (9.188)

where ζ = 6πηaef f is the Stokes friction on a bead of size aef f , and

λQ = ζr2m12kBT

. (9.189)

The FENE parameter, b, is the ratio of these two time constants,

b = 3λQ

λH= Hr2m

kBT. (9.190)

Chain models (with more than two beads) have a spectrum of relaxation times [6].There is no close-form expression for the relaxation time spectrum for FENE chains.However, a modification of FENE chain, called the FENE-PM chain, has the samespectrum as the Rouse chain (Hookean spring chain) [6]. A time constant can bedefined for FENE-PM chain as [88]:

λ f ene = b

b + 3λH

N 2b − 1

3, (9.191)

where Nb is the number of beads in the chain. The contour length is an appropriateparameter to represent the molecular size. If Lc denotes the length of one segmentof a molecular chain, the contour length of the molecular chain is simply equal to(Nb − 1) Lc. We may use the equilibrium length of a segment as an estimation of Lc

or just use the maximal length of a segment, rm .


Fig. 9.6 A schematicdiagram for the springmodel: basic DPD particles(circles) connected toreference sites (squares) vialinear springs. The referencesites are supposed to move asa rigid body

Wormlike Chains It was found that mechanical properties of DNA molecules in anaqueous solution can be realistically modelled by the wormlike chain model [13].The spring force law of this model can be expressed as

FSi j = − kBT

4λe f fp

[(1 − ri j

l

)−2 + 4ri jl

− 1

]ri j , (9.192)

where l is the maximum length of one chain segment, ri j = ri j/ri j , and λe f fp is the

effective persistence length of the chain. If the total length of the chain is L and thenumber of bead in the chain is Nb, l = L/ (Nb − 1). Bustamante et al. [13] foundthe persistence length for their unlabelled DNA molecules, λp, is about 0.053 µm.Whenmodelling theDNAmoleculeswithwormlike spring chain, the beads increasesthe molecules’ flexibility, since they do not transmit bending moment. This increaseof the chain flexibility can effectively be compensated by slightly increasing thepersistence length [49]. In their simulation, the length of the DNA molecule is 67.2µm, Larson et al. [49] used λ

e f fp = 0.061 for a 40-bead DNA chain and λ

e f fp = 0.07

for a 80-bead DNA chain. Note that their chains are tethered, i.e., a bead in thetethered end is not required. Hence, the number of beads in their chains is equal tothe number of chain segments.

Suspended Particles by Spring Model In the spring model ([55, 69]), a colloidalparticle is modelled by using only a few basic DPD particles that are connected tothe reference sites (on that colloidal particle) by linear springs of very large stiffness(Fig. 9.6). For example, a hard discmay be simply constructed using 3 or 4 basic DPDparticles, whose associated reference sites are located at the vertices of an equilateraltriangle or square, while a spherical particle may be represented using 6 or 8 basicDPD particles with their reference sites at the vertices of either an octahedron or a


Fig. 9.7 A 2D circular disc can be modelled using 3 (equilateral triangle) and 4 (square) basicDPD particles (top), while a spherical particle can be represented using 6 (octahedron) and 8 (cube)basic DPD particles (bottom)

Fig. 9.8 2D circular disc, 4 basic DPD particles: contour plots of the superimposition of conserva-tive/repulsive shape functions (left) and the radial component of the total conservative force vector(right)

cube (Fig. 9.7). The rationale behind the spring model is that the shape of a colloidis actually defined by the repulsion force field generated by constituent particles ofthat colloid (Fig. 9.8). The reference sites, collectively modelling a rigid body, moveas a rigid body motion calculated through their Newton-Euler equations, using datafrom the previous time step,


Mkc

dVkc

dt= Fk(t − Δt), (9.193)

Ikdωk

dt= Tk(t − Δt), (9.194)

Fk(t − Δt) =p∑

i=1

N∑

j=1, j =i

[Fki j,C(t − Δt) + Fk

i j,D(t − Δt) + Fki j,R(t − Δt)

],

(9.195)

Tk(t − Δt) =p∑

i=1

(rki (t − Δt) − Rk

c (t − Δt))×

N∑

j=1, j =i

[Fki j,C(t − Δt) + Fk

i j,D(t − Δt) + Fki j,R(t − Δt)

], (9.196)

where Mkc , I

k , Rkc , r

ki , V

kc , ω

k and p are the mass, moment of inertia tensor, centre ofmass, position of particle i , centre-of-mass velocity, angular velocity, and the numberof constituent particles of the kth colloidal particle, respectively.

The velocities of their associated DPD particles are found by solving the DPDequations at the current time level

midvki (t)dt

=N∑

j=1, j =i

[Fki j,C(t) + Fk

i j,D(t) + Fki j,R(t)

] + Fki,S(t), i = (1, 2, . . . , p).

(9.197)

where Fki,S(t) = −H

[rki (t) − rki (t)

]is the spring force with H being the stiffness of

the spring and ri the position of the reference site i . Introducing springs into themodelallows the constituent particles to fully participate in the system dynamics, leadingto a better control of the system temperature. The spring stiffness is chosen largeenough to ensure a good representation of rigid particles, but not so large becauseof stability consideration (the value of 3000 is chosen after some experimentations).Note that in the stiff limit (H → ∞), the spring model exactly reduces to a frozenparticle model.

Compared to the single DPD particle models, the spring model works well withlarger time steps (only soft potentials are employed here), involves considerably lessparameters (the dispersion phase is also based on basic DPD particles and conse-quently, the solvent-colloidal interaction can be decomposed into a set of solvent-solvent interactions), and can be extended to the case of suspended particles ofnon-spherical or non-circular shapes more straightforwardly. It was reported in [66]that the single particle model employs a time step in the range of 0.0002–0.0005. Thetime steps used in the spring model are 0.005–0.01 for 2D and 0.001–0.005 for 3D(i.e. about one order of magnitude higher). Compared to the frozen particle models,the spring model employs only a few DPD particles to represent a suspended particleand can maintain the specified temperature throughout the flow domain.


One distinguishing feature of the spring model is that the solvent particles andthe constituent particles of the suspended particles use the same values of theDPD parameters including the Boltzmann temperature (contrasting to different setsof DPD parameters, solvent-solvent, solvent-colloidal, colloidal-colloidal, in one-single DPD particle models). Consequently, the resultant DPD system is based onidentical basic particles, and the particle volume fraction can be simply defined as

φ = N 0c

N 0c + Ns

, (9.198)

where N 0c and Ns are the numbers of the basic DPD particles used to represent the

solvent and colloidal phases, respectively. Numerical results show that this definitionof the volume fraction (9.198) (which is the number fraction) recovers Einstein’srelation at low volume fraction irrespective of values of the input DPD parameters,and therefore is appropriate for DPD suspensions.

9.7 Flow Verifications and Some Typical Problems

Couette Flow In a Couette flow, in which the velocity is

u = (u, 0, 0) , u = γy, γ = U/h, (9.199)

the stress for a simple fluid is

[T] =⎡

⎣T11 T12 0T12 T22 00 0 T33

⎤

⎦ , (9.200)

with the shear stress is an odd function of the shear rate γ, and the first (N1) and thesecond (N2) normal stress differences are even functions of the shear rate γ,

T12 = η (γ) γ, η (γ) = η (−γ) ,

N1 (γ) = T11 − T22 = N1 (−γ) ,

N2 (γ) = T22 − T33 = N2 (−γ) ,

(9.201)

where the viscosity is an even function of the shear rate.Our DPD flow domain consists of unit cells in the x , y and z directions, respec-

tively, Lx × Ly × Lz = 40 × 10 × 30. Initially, aFCCarrangement ofDPDparticleswas formed and was allowed to equilibrate before the simulation was started, withan initial density of 4 (4 particles per unit cell or volume); a total of 94400 DPDfluid particles and 12420 DPD wall particles were involved in the simulation. In oursimulation, the wall boundary is constructed using three layers of frozen particles.In addition, we assume that there is a thin layer near the wall, in which a random

9.7 Flow Verifications and Some Typical Problems 209

(a) (b)

(c) (d)

Fig. 9.9 Couette flow results a Density and temperature b Velocity u c Normal stress differencesd Viscosity as function of shear rates

velocity distribution with zero mean corresponding to a given temperature is gener-ated [19]. In order to prevent particles from penetrating the walls, we further requirethat the particles in this layer always leave the wall according to the reflection lawreported in [77]. The central flow region across z direction was divided into 300bins (for the averaging purpose), and the average was done in the bins over 104 timesteps. The choice of the standard DPD parameters, kBT = 1, γ = 4.5, n = 4, with amodified weighting function correspond to s = 1

2 leads to the temperature, density,velocity, normal stress differences and viscosity profiles as shown in Fig. 9.9, withthe expected fluctuations near the solid walls, using velocity-Verlet algorithm. Thenumerical results show a Newtonian behaviour with zero mean normal stress dif-ferences (except at the wall, where fluctuations occur), and constant viscosity withshear rate, in agreement with the expectation of a Newtonian behaviour of simpleDPD fluid.

To explore the effects of mass, we consider four values ofm, namely 1.0, 0.1, 0.01and 0.001. This results in a fluid of η = 2.4059 for m = 1, η = 2.2494 for m = 0.1,η = 2.2337 for m = 0.01 and η = 2.2322 for m = 0.001.


Table 9.4 Couette flow: comparison of the mean equilibrium temperature of the ETD and velocity-Verlet algorithms for the case of m = 0.1. The velocity-Verlet algorithm fails to converge at Δt >

0.009

kBT

Δt ETD Velocity-Verlet

0.01 1.003 –

0.009 1.002 –

0.007 1.002 0.9538

0.005 1.002 0.9386

0.003 1.002 0.9607

0.001 1.002 0.9869

0.0009 1.002 0.9884

Table 9.5 Couette flow: Comparison of themean equilibrium temperature of the ETD and velocity-Verlet algorithms for the case of m = 0.01. The velocity-Verlet algorithm fails to converge atΔt > 0.0009

ETD Velocity-Verlet

Δt kBT Error (%) kBT Error (%)

0.007 0.5680 43.19 – –

0.005 0.7633 23.66 – –

0.003 0.9356 6.44 – –

0.001 0.9835 1.64 – –

0.0009 0.9863 1.36 – –

0.0007 0.9916 0.83 0.9187 8.12

0.0005 0.9958 0.41 0.9322 6.77

0.0003 0.9987 0.12 0.9576 4.23

0.0001 0.9992 0.07 0.9840 1.60

Tables9.4, 9.5 and 9.6 show the behaviour of the temperature against the timestep for m = 0.1, m = 0.01 and m = 0.001, respectively, using the first-order ETDscheme (9.174). Results by the velocity-Verlet algorithm are also included. It can beseen that the first-order ETD algorithm works effectively for relatively-large timesteps. Furthermore, for a given small time step, the ETD algorithm is much moreaccurate than the velocity-Verlet algorithm. In the case of m = 0.1, the ETD algo-rithm produces the equilibrium temperature that is accurate up to 3 significant digits.In the case of m = 0.01 and m = 0.001, equipartition is consistently improved asthe time step reduces. Velocity-Verlet algorithm fails to converge except at smalltime steps, and the associated errors are much larger than those produced by theETD algorithm. Velocity, temperature and number density results obtained with thefirst-order ETD scheme are presented in Fig. 9.9. We obtain a linear velocity profilein the x direction, and uniform temperature and density using a time step Δt = 0.02


Table 9.6 Couette flow: comparison of the mean equilibrium temperature of the ETD and velocity-Verlet algorithms for the case of m = 0.001. The velocity-Verlet algorithm fails to converge atΔt > 0.00007

ETD Velocity-Verlet

Δt kBT Error (%) kBT Error (%)

0.0002 0.9501 4.98 – –

0.0001 0.9805 1.94 – –

0.00009 0.9837 1.62 – –

0.00008 0.9865 1.34 – –

0.00007 0.9891 1.08 – –

0.00006 0.9916 0.83 0.9127 8.72

0.00005 0.9934 0.65 0.9235 7.64

0.00004 0.9951 0.48 0.9365 6.34

0.00003 0.9958 0.41 0.9502 4.97

for m = 1.0, Δt = 0.005 for m = 0.1, Δt = 0.001 for m = 0.01 and Δt = 0.0002for m = 0.001.

Poiseuille Flow Poiseuille flow in a channel of Lx × Ly × Lz = 40 × 10 × 30 isalso assessed at four different values ofm, 1.0, 0.1, 0.01 and 0.001. Other parametersused are kept the same as before, rc = 1.0, n = 4, ai j = 18.75, σ = 3.0, s = 1/2and kBT = 1.0, resulting in a fluid of η = 2.4059 for m = 1, η = 2.2494 for m =0.1, η = 2.2337 for m = 0.01 and η = 2.2322 for m = 0.001. Pressure gradient issimulated by applying a body force F = (0.1, 0, 0)T to each particle. Analyticalparabolic velocity profile for a Newtonian fluid is

vx = nFx

2η

(Lz

2− z

)(Lz

2+ z

). (9.202)

Results for the velocity, temperature and number density are presented in Fig. 9.10.We obtain a parabolic velocity profile in the x direction, as expected. Percentageerrors (relative to the exact value obtained by (9.202) using the approximate viscosityvalues quoted) for the maximum value of vx at the centreline are 8.71% for m = 1and 10.57% for m = 0.001. Note also that the Sc number at m = 0.001 is estimatedto be of O

(6.8 × 103

),which is muchmore than a water-like liquid (Sc ∼ 2 × 103).

Convergent-Divergent Channel Flow Next we consider a more complex flow: asimple DPD fluid flows through a periodic channel with abrupt contraction and diffu-sion shown in Fig. 9.11. The DPD parameters used in the velocity-Verlet simulationare, rc = 1.5, n = 4, ai j = 18.75, σ = 3.0, s = 1/2, kBT = 1.0 and m = 1. Thetotal length of one period is 90 and the length of the contraction segment is 40.The width of the contraction segment is 20 and the contraction ratio is 9:4. When aparticle passes through this channel, it would be experienced acceleration and decel-eration, and the dynamic behaviour of the system becomes important. A system with


Fig. 9.10 Poiseuille flow: Profiles of velocity, temperature and number density for several valuesof the mass

Fig. 9.11 The geometry of aperiodic abrupt expansionand diffusion channel

-50 -40 -30 -20 -10 0 10 20 30 40 50X

-40

-30

-20

-10

0

10

20

30

40

Z


Fig. 9.12 Velocity profiles at two different stations as compared to Fluent’s predictions, Um is themean velocity

the total number of 87886 particles was simulated, in which there were 8688 wallparticles and 79198 fluid particles. To check the results of DPD simulation, we useda commercial software, Fluent, to simulate this flow. The same physical parametersfor the DPD fluid, flow geometry and periodic boundary conditions are used in theFluent simulation.

From the results in Fig. 9.12, it is clear that the DPD simulation captures the flowkinematics well, especially the dynamic behaviour of a particle as it moves throughan unsteady Lagrangian flow field.

Hookean Dumbbell Model in DPD Simulation A dilute polymer solution couldbe modelled by a dilute suspension of Hookean dumbbells, in which the connectorforce between particles i and j on a dumbbell is given by

FSi j = −Hri j , (9.203)

where H is a spring constant. In Sect. 7, it was shown that this microstructure leadsto the Oldroyd-B fluid in which the total stress tensor is

T = S(s) + S(p), L = (∇u)T

S(p) + λ

{d


}= GI, G = νkBT, (9.204)

where ν is the number density of dumbbells, and λ = ζ/4H is the relaxation time,with ζ the frictional factor of a dumbbell’s bead. ζ may be estimated as 6πηsae f f ,where aef f is the effective size of a dumbbell bead (DPD particle).

In a simple shear flow, the solution viscosity is a constant, its 1st normal stressdifference is quadratic in the shear rate, and its second normal stress difference iszero:


Fig. 9.13 DPD model for a dilute polymer solution

η = S12γ

= ηs + ηp, N1 = S(p)11 − S(p)

22 = 2ηpλγ2, N2 = S(p)22 − S(p)

33 = 0,

(9.205)

where the polymer-contributed viscosity is ηp = Gλ, which is proportional to thenumber density of the dumbbell. Note that λ may estimated from the normal stressto shear stress ratio,

λ = N1

2γS(p)12

. (9.206)

In the velocity-Verlet DPD simulation, a total of 94400 DPD fluid plus 12420wall DPD particles was used in a (40, 20, 30) cells domain, with the standard DPDparameters, rc = 1, n = 4, ai j = 18.75, σ = 3.0, s = 1/2, kBT = 1.0 andm = 1. Inthe simulation, a dumbbell is modelled by connecting twoDPDparticles with a linearspring of constant stiffness H . From the solvent and the total viscosity (Fig. 9.13a),the polymer-contributed viscosity can be calculated, and the results are plotted inFig. 9.13b. This quantity is seen proportional to the number density of the dumbbellsas expected. Furthermore, from the first normal stress difference, Fig. 9.13c, which


is quadratic in the shear rate, the relaxation time can be derived according to (9.206)and is plotted in Fig. 9.13c against the spring stiffness H . Plotted in the same figureis the predicted relaxation time from kinetic theory, ζ/(4H); a good agreement isclearly seen.

Particulate Suspensions In DPD,monodispersed suspensions are modelled throughtwo sets of particles: one for the solvent phase (basic DPD particles freely movable)and the other for the dispersion phase (e.g. constrained basic DPD particles for thespring model). The kinetic theory [56], confirmed by numerical results, shows thatthere is an exclusion zone associated with a basic DPD particle. The size of thisexclusion zone may be sensibly defined as the particle size. This particle size can beassessed by means of the radial distribution function. Note that the cutoff radius doesnot necessarily represent the size of theDPDparticle - a larger cutoff radiusmay resultin a smaller size of the particle. If the solvent particle size is significant (comparedto colloidal size), the maximal packing fraction of the colloidal particles will bereduced. Consequently, the relative viscosity is expected to diverge earlier as thevolume fraction increases approaching the maximal packing fraction, and the shearthinning behaviour becomes stronger. Controlling the solvent particle size to havea correct representation of the solvent phase (in the sense that the colloidal/solventsize ratio is very large) is a vital issue in the DPDmodelling of colloidal suspensions.

As analysed in Sect. 9.3.3 (i.e. roles of interaction forces), there are many possiblecombinations of n and rc that reduce aef f of a basic DPD particle. One can employrc = 1 with a larger value of n (fine coarse-graining level with a standard cutoffradius). Or one can employ, for example n = 3 with a larger value of rc > 1 (uppercoarse graining level with a larger cutoff radius).

In what follows, we investigate numerically the effects of the number den-sity, cutoff radius and thermodynamic temperature on the rheological properties ofmonodispersed suspensions. The problem domain is chosen as Lx × Ly = 20 × 20and the input parameters employed are σ = 3, s = 1/2, n = 3 − 9, rc = 1 − 2 andkBT = 0.25 − 1. To represent the suspended particles, we utilise the spring modelusing 4–6 basic particles per colloid with their spring stiffness H = 3000. Their rhe-ological properties are predicted by conducting the simulation in a simple shear flow.The relative viscosity (suspension/solvent) is calculated in an average sense from tensimulations - each simulation consists of 300,000 time steps. For “zero-shear-rate”viscosity, we compute it at a shear rate of 0.1 for the volume fraction φ ≤ 0.1 and0.01 for φ > 0.1.

Simulation of suspensions with large number densities

Awide range of the number density n are employed, while other DPD parameters aretaken as standard values (e.g., kBT = 1, σ = 3, rc = 1 and 4 basic DPD particles percolloid). Results concerning the zero-shear-rate relative viscosity against the volumefraction are shown in Figs. 9.14 and 9.15.

Figure9.14 reveals that the relative viscosity curves collapse onto a single curveat large values of the number density. It implies that the size effect of the particles


Fig. 9.14 Suspension:Effects of the numberdensity

Fig. 9.15 Suspension:Relative viscosities by DPD,SPH and empirical model

size ratio (colloidal/solvent) becomes negligible at fine coarse graining levels, i.e.at n ≥ 6. Theoretical estimate for the relative viscosity in the dilute regime [11] isincluded.

Figure9.15 shows a comparison of the present relative viscosities and those pre-dicted by the empirical model of Krieger and Dougherty [45], defined as

ηr =(1 − φ

φm

)−φm [η], (9.207)

where φm is the maximal packing fraction and [η] the intrinsic viscosity. In 2D, themaximal volume fraction is 0.91 for hexagonal close packing and the intrinsic vis-cosity is 2 for rigid cylinders. It can be seen that a fine coarse graining level n = 9


Fig. 9.16 Suspension:Effects of the cutoff radius inthe upper coarse-graininglevel n = 3

results in viscosities that are located along the empirical curve of φm = 0.91 overthe whole range of volume fraction employed. In contrast, results at a coarse level(n = 3) fail to follow the correlation; it is in close agreement with the correlationusing φm = 0.76 up to a semidilute regime, but under-predicts the correlation inthe concentrated regime. SPH results [5], where short range lubrication forces areincluded explicitly, are also shown in the figure. Thus it is seen that the DPDmethodat a fine coarse graining level follows the established correlation well.

Simulation of suspensions with large cutoff radii

Different values of the cutoff radius rc are employed with their effects being seenclearer for the case of using smaller number densities (e.g., an upper coarse grainingat n = 3). As the cutoff radius rc increases, the solvent particle size is significantlyreduced (Sect. 9.3.3). The relative viscosity - volume fraction relation for n = 3 isshown in Fig. 9.16, where a colloidal particle is model using 6 basic DPD particles.By increasing rc, a coarse level n = 3 is able to produce results following Krieger-Dougherty correlation.

Simulation of suspensions with low thermodynamic temperature

Unlike the number density n and cutoff radius rc, reducing the temperature kBTdoes not affect the solvent particle size significantly (Sect. 9.3.3). We now examinethe effects of kBT on the size of a colloidal particle modelled by 4 basic DPD par-ticles and the results obtained in a fine coarse graining level (n = 9) are shown inFig. 9.17. From these results, it is expected that changing kBT has little effect on thesize ratio of the colloidal to solvent particle, and this is confirmed in Fig. 9.18 for therelative viscosity versus volume fraction.

Remarks. In DPD, the fluid (solvent) and colloidal particles are replaced by a setof DPD particles and therefore their relative sizes (as measured by their exclusion


Fig. 9.17 Suspension: the colloidal particle size over a range of kBT

Fig. 9.18 Suspension: Effects of the thermodynamic temperature in the fine coarse graining leveln = 9

zones) can affect the maximal packing fraction of the colloidal particles. To bettermimic the physical system, the DPD system should be designed to have as small asolvent particle as possible in order to make the colloidal/solvent size ratio as largeas possible (e.g., a few orders of magnitude). The size of DPD particles is found tobe decided not only by the conservative force but also by the dissipative and randomforces. By keeping the compressibility of the system unchanged, it is shown thatthe solvent phase can be modelled correctly (in the sense just mentioned above)at both low (large number density) and high (low number density) coarse-graininglevels. In the former, one can simply employ standard values of the other input DPDparameters, while in the latter, a larger value of the cutoff radius is required. It isfound that the solvent particle size is a decreasing function of the cutoff radius andvarying the temperature is not an effective way of controlling the solvent particlesize. When the requirement of large colloidal/solvent particles size is met, the DPD


results for the reduced viscosity are basically identical for any values of the inputDPD parameters.

There are other complex flows that have been successfully simulated by DPD,or some variants of DPD method, including DNA, droplet suspension, gas/vapoursuspension, porous media flow, red blood cell modelling. We shall not review of therecent development in DPD in this compact book, but refer the reader to the reviewby Pivkin et al. [73].

9.8 Matlab Program

The following DPD code, written in MATLAB, is for the numerical simulation ofviscometric flows in two dimensions. Its main purpose is to illustrate how the DPDequations are solved in practice. For an easy reading, only plain MATLAB functions(i.e. no toolboxes/user-defined functions) are used. The program consists of 4 parts:input, initialisation, computation and output in a single file. Tasks for computingthe positions and velocities of the DPD particles and computing the velocities andstresses of the flow are highlighted. An efficient cell linked list approach is utilised tocompute the interaction forces between the particles - the most time consuming partin a simulation. One can choose Couette flow or Poiseuille flow through the inputvalues.

%% Dissipative Particle Dynamics% 2D viscometric flows and Periodic boundary conditions% Couette flow: specify the imposed velocity (e.g. Vx=1)% and set the body force zero (Fex=0)% Poiseuille flow: specify the body force (e.g. Fex=1)% and set the imposed velocity zero (Vx=0)% No flow: set Vx=0 and Fex=0close all; clear;

%%%%%%%%%%%%%%%%%%%% INPUT %%%%%%%%%%%%%%%%%%%%%%% Flow domain and Imposed conditions (Vx: velocity and Fex: body force)Lx=10; Ly=Lx;Vx=0; Fex=1;

% Input DPD parameters and Time steprCut=1; sigma=3; kBT=1;numDensityX=1; numDensityY=3; numDen=numDensityX*numDensityY;aij=57.23*kBT/numDen/rCutˆ3;deltaT=0.01;

% Step numbers for thermal equilibrium and for flow simulationstepEquil=500; stepSample=3000; stepLimit=stepEquil+stepSample;

%%%%%%%%%%%%%%%%%% INITILISATION %%%%%%%%%%%%%%%%% Number of DPD particles and Number of binsnFreeAtom=Lx*Ly*numDen;nxBin=4*Lx; nyBin=4*Ly; nBin=nxBin*nyBin;% Initial configuration


r=zeros(2,nFreeAtom);X=linspace(-Lx/2+1/2/numDensityX,Lx/2-1/2/numDensityX,numDensityX*Lx);Y=linspace(-Ly/2+1/2/numDensityY,Ly/2-1/2/numDensityY,numDensityY*Ly);[xm,ym]=meshgrid(X,Y);r(:,:)=[xm(:),ym(:)]’;% Initial velocitiesrv=zeros(2,nFreeAtom);velMag=sqrt(2*(1-1/nFreeAtom)*kBT);phim=2*pi*rand(numDensityX*Lx,numDensityY*Ly);vxm=cos(phim); vym=sin(phim);rv(:,:)=velMag*[vxm(:),vym(:)]’; vSum=sum(rv,2);rv(1,:)=rv(1,:)-vSum(1)/nFreeAtom;rv(2,:)=rv(2,:)-vSum(2)/nFreeAtom;figure; plot(r(1,:),r(2,:),’mo’); hold on;plot([-Lx/2,Lx/2,Lx/2,-Lx/2,-Lx/2],[-Ly/2,-Ly/2,Ly/2,Ly/2,-Ly/2],’b-’);quiver(r(1,:),r(2,:),rv(1,:),rv(2,:),’b’); hold off;title(’Initial configuration and Initial particles velocities’);% Initial accelerations and forcesra=zeros(2,nFreeAtom);rforce=zeros(3,nFreeAtom);% Friction strengthgamma=sigmaˆ2/(2*kBT);

areaBin=(Lx/nxBin)*(Ly/nyBin);velGrid=zeros(nBin,4); % n, uˆ2, ux, uystrsGrid=zeros(nBin,4); % Sxx, Syy, Sxy, pvoff=[0,1,1,0,-1,2;...

0,0,1,1,1,1];region=[Lx;Ly];cells=floor((1/rCut)*region); nCell=cells(1)*cells(2);

%%%%%%%%%%% COMPUTATION %%%%%%%%%%%%%%%%%%%%%%%%stepCount=0; timeNow=0;while stepCount < stepLimit

stepCount=stepCount+1; disp(stepCount);%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% Compute the particles’ positions and velocities%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%temp = deltaT*ra;% Particles’ positionsr = r + deltaT*(rv+0.5*temp);% Imposition of velocity conditions for Couette flowbdySlide=rem(2*Vx*(stepCount*deltaT)+Lx/2,Lx)-Lx/2;for i=1:nFreeAtom

if (abs(r(1,i))>Lx || abs(r(2,i))>Ly)error(’Particle moves too far, reduce time step’)

endif r(2,i) > 0.5*Ly

r(2,i)=r(2,i)-Ly;r(1,i)=r(1,i)-bdySlide;rv(1,i) = rv(1,i)-2*Vx;

elseif r(2,i) < -0.5*Lyr(2,i)=r(2,i)+Ly;r(1,i)=r(1,i)+bdySlide;rv(1,i) = rv(1,i)+2*Vx;

end

if r(1,i) > 0.5*Lxr(1,i)=r(1,i)-Lx;

elseif r(1,i) < -0.5*Lx

9.8 Matlab Program 221

r(1,i)=r(1,i)+Lx;end

endrvm = rv; rv = rv + 0.65*temp; rvm = rvm + 0.5*temp;% Interaction force calculationscellShiftX = fix(cells(1)*(1.-bdySlide/region(1))) - cells(1);ra=zeros(2,nFreeAtom); rforce=zeros(3,nFreeAtom);invWid = cells./region;cellList = -ones(nFreeAtom+nCell,1);rs = [r(1,:)+0.5*region(1); r(2,:)+0.5*region(2)];cc = ceil([rs(1,:)*invWid(1);rs(2,:)*invWid(2)]);c = (cc(2,:)-1)*cells(1)+cc(1,:) + nFreeAtom;for i=1:nFreeAtom

cellList(i) = cellList(c(i));cellList(c(i)) = i;

endfor m1y = 1:cells(2)

for m1x = 1:cells(1)m1v = [m1x;m1y];m1 = (m1v(2)-1)*cells(1)+m1v(1) + nFreeAtom;%--------------------------------------------------------------offsetHi=size(voff,2);if m1y < cells(2)

offsetHi=offsetHi-1;endfor offset=1:offsetHi

m2v = m1v+voff(:,offset);shift = zeros(2,1);velshift = zeros(2,1);if m1v(2)==cells(2) && voff(2,offset)==1

m2v(1)=m2v(1)+cellShiftX;shift(1)=bdySlide;if m2v(1) > cells(1)

m2v(1) = m2v(1)-cells(1);shift(1) = shift(1)+region(1);

elseif m2v(1) < 1m2v(1) = m2v(1)+cells(1);shift(1) = shift(1)-region(1);

endelse

if m2v(1) > cells(1)m2v(1)=1; shift(1)=region(1);

elseif m2v(1) < 1m2v(1)=cells(1); shift(1)=-region(1);

endendif m2v(2) > cells(2)

m2v(2)=1; shift(2)=region(2);velshift(1)=2*Vx;

elseif m2v(2) < 1m2v(2)=cells(2); shift(2)=-region(2);velshift(1)=-2*Vx;

end%----------------------------------------------------------m2 = (m2v(2)-1)*cells(1)+m2v(1) + nFreeAtom;j1=cellList(m1);while j1>=1

j2=cellList(m2);while j2>=1


if m1˜=m2 || j2<j1dr = r(:,j1)-(r(:,j2)+shift);rr=dr’*dr;if rr < rCutˆ2

rij = sqrt(rr); dr = dr/rij;rdvij=dot(dr,rv(:,j1)-(rv(:,j2)+velshift));weight = 1 - rij/rCut;weightc = weight;weightd = sqrt(weight);weightr = sqrt(weightd);fcVal = aij*weightc;fdVal = -gamma*weightd*rdvij;frVal=sigma*weightr*randn*(1/sqrt(deltaT));ftot = fcVal+fdVal+frVal;ra(:,j1) = ra(:,j1) + ftot*dr;ra(:,j2) = ra(:,j2) - ftot*dr;

fstl = (fcVal + fdVal)*rij;rforce(:,j1) = rforce(:,j1) + ...

fstl*[dr(1)*dr(1);dr(2)*dr(2);dr(1)*dr(2)];

rforce(:,j2) = rforce(:,j2) + ...fstl*[dr(1)*dr(1);dr(2)*dr(2);dr(1)*dr(2)];

endendj2=cellList(j2);

endj1=cellList(j1);

endend

endend% Imposition of body forces for Poiseuille flow

j=0;for i=1:nFreeAtom

if r(2,i)>0; j=j+1; endendfor i=1:nFreeAtom

if r(2,i)>0ra(1,i) = ra(1,i) + Fex*(0.5*nFreeAtom/j);

elsera(1,i) = ra(1,i) - Fex*(0.5*nFreeAtom/(nFreeAtom-j));

endend% Particles’ velocitiesrv = rvm + 0.5*deltaT*ra;%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% Compute the flow velocities and stresses%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%if stepCount > stepEquil

timeNow=timeNow+deltaT;for i=1:nFreeAtom

c=floor((r(2,i)+Ly/2)*nyBin/Ly)*nxBin + ...floor((r(1,i)+Lx/2)*nxBin/Lx) + 1;

velGrid(c,:)=velGrid(c,:)+[1,rv(:,i)’*rv(:,i),rv(:,i)’];strsGrid(c,1:3)=strsGrid(c,1:3)+ ...

[rv(1,i)*rv(1,i),rv(2,i)*rv(2,i),rv(1,i)*rv(2,i)]+0.5*rforce(:,i)’;endif (stepCount-stepEquil) == stepSample

9.8 Matlab Program 223

id = find(velGrid(:,1)>0);for j=2:4

velGrid(id,j)=velGrid(id,j)./velGrid(id,1);endvelGrid(:,2)=velGrid(:,2)-(velGrid(:,3).ˆ2+velGrid(:,4).ˆ2);

for j=1:3strsGrid(id,j)=strsGrid(id,j)./velGrid(id,1);

endstrsGrid(:,1) = strsGrid(:,1) - velGrid(:,3).ˆ2;strsGrid(:,2) = strsGrid(:,2) - velGrid(:,4).ˆ2;strsGrid(:,3) = strsGrid(:,3) - velGrid(:,3).*velGrid(:,4);

velGrid(:,1)=velGrid(:,1)/(stepSample*areaBin);strsGrid(:,1) = -strsGrid(:,1).*velGrid(:,1);strsGrid(:,2) = -strsGrid(:,2).*velGrid(:,1);strsGrid(:,3) = -strsGrid(:,3).*velGrid(:,1);strsGrid(:,4) = -(strsGrid(:,1) + strsGrid(:,2))/2;

endend

end%%%%%%%%%%%%%%%% OUTPUT %%%%%%%%%%%%%%%%%%% Plot the particles’ positions and velocities at the final timefigure; plot(r(1,:),r(2,:),’mo’); hold onplot([-Lx/2,Lx/2,Lx/2,-Lx/2,-Lx/2],[-Ly/2,-Ly/2,Ly/2,Ly/2,-Ly/2],’b-’);quiver(r(1,:),r(2,:),rv(1,:),rv(2,:)); hold off;title(’Configuration and Particles velocities at the final time’)axis off

% Plot the flow fields: velocity and stressY=-Ly/2+Ly/(2*nyBin):Ly/nyBin:Ly/2-Ly/(2*nyBin);R=zeros(nyBin,4);for k=1:nyBin

R(k,:)=sum(velGrid(nxBin*(k-1)+1:nxBin*k,:),1)/nxBin;endfigure;plot(Y,R(:,1),’b-’,Y,0.5*R(:,2),’m-.’,Y,R(:,3),’k--o’,Y,R(:,4),’g--’);legend(’n’,’T’,’u_x’,’u_y’); xlabel(’y’); title(’Flow field calculations’);

R=zeros(nyBin,4);for k=1:nyBin

R(k,:)=sum(strsGrid(nxBin*(k-1)+1:nxBin*k,:),1)/nxBin;endfigure; plot(Y,R(:,3),’b-o’,Y,R(:,1)-R(:,2),’r-s’);legend(’Shear stress’,’Normal stress difference’); xlabel(’y’);title(’Flow field calculations’);

9.9 Epilogue

This compact book outlines the main basic developments in viscoelasticity, fromcontinuum principles to microstructure modelling. We have not mentioned the rep-utation concept of Doi and Edwards [18], the modelling effort in fibre suspensions


[27], in biological materials [3], in electro-rheological fluids, as well as transitionphenomena.

The constant theme emphasised throughout is that relevant evolution equa-tions for the microstructure should be derived from well-established physics. This,together with relevant statistical mechanics linking the microstructure evolution toa macroscaled stress induced by the microstructure, should provide a useful consti-tutive equation for the fluid. Having a relevant constitutive relation is only half ofthe story; one needs to be able to make predictions with it, and that usually meansa numerical implementation, a vast open area that we only touch on briefly with theDissipative Particle Dynamics.

It is hoped that the readers find the book useful in their research works.

Problems

Problem 9.1 In Problems 9.1–9.4, we deal with the 1-D system (9.1). Re-definethe state variable to reduce (9.1) to (9.4) in the inertial time scale, and show that itsformal solution is

v (t) =∫ t

0em

−1γwD(t ′−t)m−1σwRθ(t ′)dt ′.

Then calculate the mean square velocity and show that

〈v (t) v (t)〉 = σ2

m2

∫ t

0

∫ t

0em

−1γwD(t ′−t)w2R

⟨θ(t ′)θ(t ′′)⟩em

−1γwD(t ′′−t)dt ′dt ′′

= σ2

m2

∫ t

0em

−1γwD(t ′−t)wRem−1γwD(t ′−t)wRdt

′,

= 1

2m−1σ2w2

Rγ−1w−1D . (9.208)

This leads directly to (9.6), assuming (9.3).

Problem 9.2 Show that the drift velocity and the diffusivity of the process (9.12)are given by

〈Δr〉Δt

= γ−1w−1D Fc,

〈ΔrΔr〉2Δt

= O (Δt) + kBTγ−1w−1D , (9.209)

leading to the Fokker–Planck equation (9.13).

Problems 225

Problem 9.3 Show that, from (9.4)

1

2m

d

dt

(d

dt

⟨r2⟩) − m

⟨v2⟩ + 1

2γwD

d

dt

⟨r2⟩ = σwR 〈θ (t) r〉 . (9.210)

Definee = d

⟨r2⟩/dt,

and show thate + m−1γwDe = 2kBTm

−1, e (0) = 0.

Show that this has the solution, for the assumed initial condition,

e = d

dt

⟨r2⟩ = 2kBTγ−1w−1

D

[1 − e−m−1γwDt

].

Consequently, if Δt � τI = O(m−1γwD

), show that

〈ΔrΔr〉2Δt

= kBTγ−1w−1D .

Problem 9.4 Define the velocity correlation as

R (τ ) = limt→∞ 〈v (t + τ ) v (t)〉 , (9.211)

where the limit refers to large time compared to the inertial time scale, but yet smallcompared to the relaxation time scale. From the solution (9.5), show that

R (τ ) = limt→∞

∫ t+τ

0dt ′

∫ t

0em

−1γwD(t ′−t−τ)(m−1σwR

)2

⟨θ(t ′)θ(t ′′)⟩em

−1γwD(t ′′−t)dt′′

(9.212)

= e−m−1γwDτ limt→∞ 〈v (t) v (t)〉 = e−m−1γwDτ R (0)

= kBTm−1e−m−1γwDτ . (9.213)

That is, the velocity correlation decays after an inertial time scale, after which thevelocity is independent to its previous state.

Next, the diffusivity can also be defined as

D = limt→∞

1

2〈v (t) r (t) + r (t) v (t)〉 .

Show that this leads to


D = limt→∞

∫ t

0〈v (t) v (t + τ )〉 dτ =

∫ ∞

0R (τ ) dτ

= kBTγ−1w−1D ,

consistent with previous results.

Problem 9.5 Show, with the aid of the Langevin equation (9.14), that the drift andthe diffusion of the process are given by

⟨ΔviΔt

⟩= −m−1

∑

j

(FCi j + FD

i j

) = − m−1∑

j

(FCi j − γwD

i j ei jei j · vi j), (9.214)

and

⟨ΔvαΔvβ

2Δt

⟩= γkBT

m2

[

δαβ

∑

k

wDαkeαkeβk − wD

αβeαβeαβ

]

. (9.215)

Thus show that the Fokker–Planck equation is given by (9.32).

Problem 9.6 Show that the equilibrium distribution of the associate system to (9.14)is

feq (χ, t) = 1

Zexp

⎡

⎣− 1

kBT

⎛

⎝∑

i

pi · pi2m

+ 1

2

∑

i, j

ϕ(ri j

)⎞

⎠

⎤

⎦ (9.216)

= 1

Zexp

[− HkBT

],

where Z is a normalizing constant, and

H=∑

i

pi · pi2m

+ 1

2

∑

i, j

ϕ(ri j

)(9.217)

is the Hamiltonian of the associate system to (9.14). Show that,

∑

i

vi · ∂ feq∂ri

= 1

kBT

∑

i, j

vi · FCi j feq ,

∑

i, j

FCi j · ∂ feq

∂pi= − 1

kBT

∑

i, j

FCi j · vi feq ,

γ∑

i, j

wDi j ei j

∂

∂pi· (ei j · vi j feq

) = γ∑

i, j

wDi j

(− 1

kBTei j · viei j · vi j + 1

m

)feq ,

Problems 227

γkBT∑

i, j

wDi j ei j · ∂

∂pi·(ei j ·

(∂

∂pi− ∂

∂p j

)feq

)

= γ∑

i, j

wDi j

(− 1

m+ 1

kBTei j · viei jvi j

),

and conclude that feq is also a stationary solution (i.e. solution that is independentof time) of the Fokker–Planck equation (9.32).

Problem 9.7 Show the equivalence between (9.61), (9.63), (9.65) and (9.66), byexpressing f2

(r + λR, r − (1 − λ)R, v′, v′′, t

)as a function of (r − εR), where

ε = 1 − λ, then integrating after taking a Taylor’s series in ε.

Problem 9.8 Show that

qC (r, t) = 1

4

∫dR

∫dv

∫dv′FC (R) · (v + v′)R

{1 − 1

2R · ∇ + · · ·

}

. f2(r + R, r, v, v′, t

). (9.218)

Problem 9.9 The stress contributed from the damping forces is, from (9.63),

SD (r, t) = −1

2

∫dR

∫dv

∫dv′γwD (R) RR · (v − v′)R {1 + O (R)}

. f2(r + R, r, v, v′, t

).

Show that, for an homogeneously shear flow, v − v′ = LR, where L is the velocitygradient, together with Groot and Warren’s approximation, f2 = n2 (1 + O (R)) ,

the stress contributed by the damping forces is

SD,αβ (r, t) = −γn2

2

⟨Rα Rβ Ri R j Li j

∫R2wD (R) 4πR2dR

⟩

= −2πγn2

15

(δαβδi j + δαiδβ j + δα jδβi

)Li j

∫R4wD (R) dR.

For the standard weighting function (9.88) adopted in DPD, show that

SD,αβ (r, t) = −2πγn2

15

(Lαβ + Lβα + Liiδαβ

) ∫ rc

0R4 (1 − R/rc)

2 dR

= 2πγn2r5c1575


), (9.219)

and consequently the damping-contributed viscosities are given by

ηD = 2πγn2r5c1575

, ζD = 5

3ηD = 2πγn2r5c

945. (9.220)


Problem 9.10 For the standard DPD weighting function (9.88), show that the vis-cosities and diffusivity are given as shown in (9.90).

Problem 9.11 For themodifiedDPDweighting function (9.125), derive the viscosi-ties and diffusivity of the DPD fluid.

Problem 9.12 Write a Matlab routine to compute the radial distribution function(RDF) for DPD particles and integrate it into the DPD main program in Sect. 9.8.Study the effects of n, rc and kBT on the particle’s exclusion zone for the case ofusing modified weighting function.

Problem 9.13 Write a Matlab routine to compute the mean square displacements(MSDs) of DPD particles and integrate it into the DPD main program in Sect. 9.8.Study the effects ofn, rc and kBT on the self-diffusion coefficient of theDPDparticlesfor a given noise amplitude σ = 3.

Solutions to Problems

Problems in Chap. 1

Problem 1.1

The components of vectors u, v, and w are given by ui , vi , wi . By expanding all thesummations, we have:

u · v = u1v1 + u2v2 + u3v3 = uivi . (S.221)

Similarly,

u × v = uiei × v je j= (u1v2 − u2v1) e3 + (u3v1 − u1v3) e2 + (u2v3 − u3v2) e1= εi jkei u jvk . (S.222)

And, by using (S.221) and (S.222),

(u × v) · w = εi jkei u jvk · wlel= εi jku jvkwi

= εi jkuiv jwk . (S.223)

From the cyclic properties of εi jk = ε jki , and use the last preceding result,

(u × v) · w = u · (v × w) . (S.224)

Next, consider the 4th order tensor εni jεnkm . Clearly, if i = j , or k = m, thistensor takes on a zero value. If i = k, the sum in n will return a zero value, unlessj = m, and εni j = εnkm = ±1, and the product εni jεnkm = +1. Now, if i = mand j = k the product εni jεnkm will return a zero value unless εni j = ±1, then

© Springer International Publishing AG 2017N. Phan-Thien and N. Mai-Duy, Understanding Viscoelasticity,Graduate Texts in Physics, DOI 10.1007/978-3-319-62000-8

229

http://dx.doi.org/10.1007/978-3-319-62000-8_1

http://dx.doi.org/10.1007/978-3-319-62000-8_1

230 Solutions to Problems

εnkm = εnji = ∓1, and the product εni jεnkm = −1. These are also the componentsof the tensor δikδ jm − δimδ jk . Hence

εni jεnkm = δikδ jm − δimδ jk . (S.225)

Using (S.225) in the following:

(u × v) × w = εi jkeiε jmnumvnwk = (δkmδin − δknδim) ei umvnwk

= ei ukviwk − ei uivkwk

(u × v) × w = (u · w) v − (v · w) u. (S.226)

This could also be derived by expanding components of both sides.Now, use the result (S.225) in the following

(u × v)2 = εi jku jvkεimnumvn = (δ jmδkn − δ jnδkm

)u jvkumvn

= u ju jvkvk − u jv jvkuk= u2v2 − (u · v)2 . (S.227)

Problem 1.2

Let A be a matrix with entries Ai j ,

[A] =⎡

⎣A11 A12 A13

A21 A22 A23

A31 A32 A33

⎤

⎦ .

By expanding

det [A] = A11A22A33 + A12A23A31 + A13A21A32 − A31A22A13

− A32A23A11 − A33A21A12

= A11 (A22A33 − A23A32) + A12 (A23A31 − A21A33)

+ A13 (A21A32 − A31A22)

= εi jk A1i A2 j A3k

= A11 (A22A33 − A23A32) + A21 (A32A13 − A12A33)

+ A31 (A12A23 − A22A13)

= εi jk Ai1A j2Ak3.

First, note that

ε123 det [A] = det [A] = ε123εi jk A1i A2 j A3k .

http://dx.doi.org/10.1007/978-3-319-62000-8_1

Solutions to Problems 231

If we swap two rows, or two columns, determinant of the resulting matrix will be thenegative of det[A]. For instance

εi jk A1i A2k A3 j = − det [A] = ε132 det [A] = ε132εi jk A1i A3 j A2k .

By inspection,

εlmn det [A] = εlmnεi jk A1i A2 j A3k = εi jk Ali Amj Ank .

Furthermore, since εlmnεlmn = 6,

εlmnεlmn det [A] = 6 det [A] = εlmnεi jk Ail A jm Akn,

det [A] = 1

6εi jkεlmn Ail A jm Akn. (S.228)

Problem 1.3

The following result has been shown in Problem 1.1:Given that two matrices of entries

[A] =⎡

⎣A11 A12 A13

A21 A22 A23

A31 A32 A33

⎤

⎦ , [B] =⎡

⎣B11 B12 B13

B21 B22 B23

B31 B32 B33

⎤

⎦ ,

Then, be expanding the summation: [C] = [A] · [B] , Ci j = Aik Bkj .

[D] = [A]T [B] , Di j = ATik Bkj = Aki Bkj .

Problem 1.4

Note that the components of e′i in frame F are Ai j , i.e., e′

i = Ai je j . Thus

(e′1 × e′

2).e′

3 = +1 = εi jk A1i A2 j A3k = det [A] .

Problem 1.5

By expanding, it can be seen that

εi jkuiv jwk = det

⎡

⎣u1 u2 u3v1 v2 v3w1 w2 w3

⎤

⎦ .

http://dx.doi.org/10.1007/978-3-319-62000-8_1

http://dx.doi.org/10.1007/978-3-319-62000-8_1

http://dx.doi.org/10.1007/978-3-319-62000-8_1

http://dx.doi.org/10.1007/978-3-319-62000-8_1


Consider a two-tensor Wi j and vector defined as ui = εi jkW jk . By expandingagain, it can be demostrated that, ifW is symmetric, Wi j = Wji , u is zero, and ifWis anti-symmetric, Wi j = −Wji ,

[W] =⎡

⎣0 W12 W13

−W12 0 W23

−W13 −W23 0

⎤

⎦ ,

⎡

⎣u1u2u3

⎤

⎦ =⎡

⎣ε1 jkW jk

ε2 jkW jk

ε3 jkW jk

⎤

⎦ =⎡

⎣2W23

2W31

2W12

⎤

⎦ .

the components of u are twice those of W. This vector is said to be the axial vectorof W. IfW represent the vorticity tensor,

[W] = 1

2

(∇uT − ∇u) = 1

2

⎡

⎢⎢⎢⎢⎢⎢⎣

0∂u1∂x2

− ∂u2∂x1

∂u1∂x3

− ∂u3∂x1

−∂u1∂x2

+ ∂u2∂x1

0∂u2∂x3

− ∂u3∂x2

−∂u1∂x3

+ ∂u3∂x1

−∂u2∂x3

+ ∂u3∂x2

0

⎤

⎥⎥⎥⎥⎥⎥⎦

Then the axial vector of this vorticity tensor is

[w] =

⎡

⎢⎢⎢⎢⎢⎢⎢⎣

∂u2∂x3

− ∂u3∂x2

∂u3∂x1

− ∂u1∂x3

∂u1∂x2

− ∂u2∂x1

⎤

⎥⎥⎥⎥⎥⎥⎥⎦

= −εi jk∇ j uk = [−∇ × u] .

Problem 1.6

If D, S, and W are two-tensors, D symmetric and W anti-symmetric,

D : S = Di j S ji = Dji S ji = Dji STi j = D : ST (symmetric D)

= 12Di j S ji + 1

2Dji STi j = 12Di j

(Sji + STji

)= D : 1

2

(S + ST

).

Furthermore,

W : S = Wi j S ji = −Wji S ji = −Wji STi j = −W : ST (anti-symmetric W)

= 12

(Wi j S ji − Wji STi j

)= 1

2Wi j

(Sji + STji

)= W : 1

2

(S − ST

),

D : W = W : 12

(D − DT

) = D : 12

(W + WT

) = 0.

http://dx.doi.org/10.1007/978-3-319-62000-8_1


In addition, if

T : S = 0 ∀S then T = 0. (S.229)

This is shown by choosing S to be unity at any particular entry ij and zero elsewhere.The corresponding ij component of T has to be zero. This implies T = 0.

If

T : S = 0 ∀ symmetric S then S:12

(T + TT

) = 0.

This leads to T + TT = 0, or T is anti-symmetric. In a similar manner,

if T : S = 0 ∀ anti-symmetric S then T is symmetric.

Problem 1.7

If Q is orthogonal,

Q−1 = QT

⇒ H + HT + HHT = Q + QT − 2I + QQT − Q − QT + I = 0

HHT = 2I − Q − QT = HTH

Conversely, ifH + HT + HHT = 0, HHT = HTH,

then

QQT − Q − QT + I = QTQ − QT − Q + I ⇒ QQT = QTQ.

Q + QT − 2I + QQT − Q − QT + I = 0 ⇒ QQT = I = QTQ,

and Q is orthogonal.

Problem 1.8

To show that I, II, III are invariants, we note that

I ′ = S′i i = Ai j Aik S jk = δ jk S jk = Sj j = I.

I I ′ = S′i j S

′j i = Aik A jm Skm A jn Air Snr = δkrδmnSkm Snr

= SkmSmk = I I.

http://dx.doi.org/10.1007/978-3-319-62000-8_1

http://dx.doi.org/10.1007/978-3-319-62000-8_1


I I I ′ = S′i j S

′jk S

′ki = AiαA jβSαβ A jγ AkδSγδAkεAiϕSεϕ

= δαϕδβγδδεSαβSγδSεϕ = SαβSβδSδα = I I I.

Next,

det [S − ωI] = det

⎡

⎣S11 − ω S12 S13S21 S22 − ω S23S31 S32 S33 − ω

⎤

⎦

= (S11 − ω)(S22S33 − ω (S22 + S33) + ω2

)+ S12S23S31+ S13S21S32 − S31S22S13 + ω (S31S13 + S32S23 + S21S12)

− S32S23S11 − S33S21S12= −ω3 + ω2 (S11 + S22 + S33) − ω(S11S22 + S11S33 + S22S33

− S31S13 − S32S23 − S21S12) + S11S22S33 + S12S23S31+ S13S21S32 − S31S22S13 − S32S23S11 − S33S21S12

= −ω3 + I1ω2 − I2ω + I3,

where

I1 = I = trS,

I2 = 12

(I 2 − I I

) = S11S22 + S11S33 + S22S33 − S12S21 − S13S31 − S23S32

I3 = 16

(I 3 − 3I.I I + 2I I I

) = det S.

If e is an eigenvector of S, with eigenvalue ω then

Se = ωe.

The condition for this to have non-trivial solutions is that

det [S − ωI] = 0 = −ω3 + I1ω2 − I2ω + I3. (S.230)

This is said to be the characteristic equation for S.According to the Cayley-Hamiltontheorem, S satisfies its own characteristic equation.

Problem 1.9

Consider the 2 × 2 matrix

[C] =[1 + γ2 γ

γ 1

].

http://dx.doi.org/10.1007/978-3-319-62000-8_1


Denote U = C1/2, U satisfies its own characteristic equation (in 2-D):

U2 − I1 (U)U + det [U] I = 0U = 1

I1 (U)(C + det (U) I) .

Expressed in eigenspace,

U =[

λ1 00 λ2

], C =

[λ21 00 λ2

2

],

I1 (U) = λ1 + λ2, I1 (C) = λ21 + λ2

2, I2 (U) = λ1λ2, I2 (C) = λ21λ

22,

I2 (U) = √I2 (C), I1 (U) =

√I1 (C) + 2

√I2 (C)

I1 (C) = 2 + γ2, I2 (C) = det [C] = 1 + γ2 − γ2 = 1.

Thus,

U = C1/2 = 1√4 + γ2

[2 + γ2 γ

γ 2

]. (S.231)

For the 3× 3 matric case:

U3 − I1 (U)U2 + I2 (U)U − I3 (U) I = 0 (S.232)

leading to

U4 + I2U2 − I3U = I1U3 = I 21U2 − I1 I2U + I1 I3I

U = 1I3−I1 I2

(C2 + (

I2 − I 21)C − I1 I3I

),

(S.233)

where I1, I2, I3 are the invariants ofU, and their dependence onU has been suppressedfor brevity. In terms of the eigenvalues:

I1 = λ1 + λ2 + λ3, I2 = λ1λ2 + λ1λ3 + λ2λ3, I3 = λ1λ2λ3.

These need to be expressed in terms of the invariants of C. First

I3 (U) = √I3 (C). (S.234)

Next, in terms of the eigenvalues of U,

I 21 (U) = (λ1 + λ2 + λ3)2 = I1 (C) + 2I2 (U) . (S.235)


Next, from (S.232),

λ41 + λ4

2 + λ43 = (I3 − I1 I2) (λ1 + λ2 + λ3) + (

I 21 − I2) (

λ21 + λ2

2 + λ23

)

+ 3I1 I3= (I3 − I1 I2) I1 + (

I 21 − I2) (

I 21 − 2I2)+ 3I1 I3

= 4I1 I3 − 4I 21 I2 + I 41 + 2I 22 . (S.236)

Left hand of (S.236) can be expressed as

I 21 (C) − 2I2 (C) = (λ21 + λ2

2 + λ23

)2 − 2(λ21λ

22 + λ2

1λ23 + λ2

2λ23

) = λ41 + λ4

2 + λ43

Thus,

I 21 (C) − 2I2 (C)

I 21 (C)= 1

I 21 (C)

[4I1 I3 − 4I 21 I2 + I 41 + 2I 22

]

With some further algebra,

2

[1 − I 21 (C) − 2I2 (C)

I 21 (C)

]+ 8

√I3 (C)

I 31 (C)

I1 (U)√I1 (C)

= 2

+ 1

I 21 (C)

[8I2 (U) I 21 (U) − 4I 22 (U) − 2I 41 (U)

]

= I 41 (U)

I 21 (C)− 2

I 21 (U)

I1 (C)+ 1

=[I 21 (U)

I1 (C)− 1

]2.

This is a quartic equation(x2 − 1

)2 = a + bx in x, with

x = I1 (U)√I1 (C)

, a = 2

[1 − I 21 (C) − 2I2 (C)

I 21 (C)

]= 2

[1 − trC2

(trC)2

],

b = 8

√I3 (C)

I 31 (C)= 8

√detC

(trC)3.

Once x is obtained, tr(U) is determined in term of tr(C), and from (S.235), thesecond invariant of U is determined in terms of C.


Problem 1.10

The components of the strain rate tensor, D = 12

(∇u + ∇uT), and the vorticity

tensor,W = 12

(∇uT − ∇u)are

[L] = [∇u]T =

⎡

⎢⎢⎢⎢⎢⎢⎣

∂u1∂x1

∂u1∂x2

∂u1∂x3

∂u2∂x1

∂u2∂x2

∂u2∂x3

∂u3∂x1

∂u3∂x2

∂u3∂x3

⎤

⎥⎥⎥⎥⎥⎥⎦

,

[D] = 1

2

⎡

⎢⎢⎢⎢⎢⎢⎣

2∂u1∂x1

∂u1∂x2

+ ∂u2∂x1

∂u1∂x3

+ ∂u3∂x1

∂u1∂x2

+ ∂u2∂x1

2∂u2∂x2

∂u2∂x3

+ ∂u3∂x2

∂u1∂x3

+ ∂u3∂x1

∂u2∂x3

+ ∂u3∂x2

2∂u3∂x3

⎤

⎥⎥⎥⎥⎥⎥⎦

,

[W] = 1

2

⎡

⎢⎢⎢⎢⎢⎢⎣

0∂u1∂x2

− ∂u2∂x1

∂u1∂x3

− ∂u3∂x1

−∂u1∂x2

+ ∂u2∂x1

0∂u2∂x3

− ∂u3∂x2

−∂u1∂x3

+ ∂u3∂x1

−∂u2∂x3

+ ∂u3∂x2

0

⎤

⎥⎥⎥⎥⎥⎥⎦

.

Problem 1.11

First, we express S in cylindrical coordinates as

S = Srrerer + Srθereθ + Srzerez + Sθreθer+Sθθeθeθ + Sθzeθez + Szrezer + Szθezeθ + Szzezez,

and taking the gradient operation,

∇ = er∂

∂r+ eθ

1

r

∂

∂θ+ ez

∂

∂z,

keeping in mind that

∂

∂rer = 0,

∂

∂reθ = 0,

∂

∂rez = 0

http://dx.doi.org/10.1007/978-3-319-62000-8_1

http://dx.doi.org/10.1007/978-3-319-62000-8_1


∂

∂θer = eθ,

∂

∂θeθ = −er ,

∂

∂θez = 0

∂

∂zer = 0,

∂

∂zeθ = 0,

∂

∂zez = 0

to result in

∇ · S = er

(∂Srr∂r

+ Srr − Sθθ

r+ 1

r

∂Sθr

∂θ+ ∂Szr

∂z

)

+ eθ

(∂Srθ∂r

+ 2Srθr

+ 1

r

∂Sθθ

∂θ+ ∂Szθ

∂z+ Sθr − Srθ

r

)

+ ez

(∂Srz∂r

+ Srzr

+ 1

r

∂Sθz

∂θ+ ∂Szz

∂z

). (S.237)

Problem 1.12

In cylindrical coordinates,

u = urer + uθeθ + uzez .

Thus

∇u =(er

∂

∂r+ eθ

1

r

∂

∂θ+ ez

∂

∂z

)(urer + uθeθ + uzez)

= ∂ur∂r

erer + ∂uθ

∂rereθ + ∂uz

∂rerez

+(1

r

∂ur∂θ

− uθ

r

)eθer +

(1

r

∂uθ

∂θ+ ur

r

)eθeθ + 1

r

∂uz

∂θeθez

+∂ur∂z

ezer + ∂uθ

∂zezeθ + ∂uz

∂zezez,

and

u · ∇u = er

[ur

∂ur∂r

+ uθ

r

∂ur∂θ

+ uz∂ur∂z

− uθuθ

r

]

+eθ

[ur

∂uθ

∂r+ uθ

r

∂uθ

∂θ+ uz

∂uθ

∂z+ uθur

r

]

+ez

[ur

∂uz

∂r+ uθ

r

∂uz

∂θ+ uz

∂uz

∂z

]. (S.238)

http://dx.doi.org/10.1007/978-3-319-62000-8_1


Additional problem: evaluate u · ∇S in cylindrical coordinates. Now with

S = Srrerer + Srθ (ereθ + eθer ) + Srz (erez + ezer )

+ Sθθeθeθ + Sθz (ezeθ + eθez) + Szzezez, (S.239)

∇S =(er

∂

∂r+ eθ

∂

r∂θ+ ez

∂

∂z

)S

= er

(erer

∂Srr∂r

+ (ereθ + eθer )∂Srθ∂r

+ (erez + ezer )∂Srz∂r

+ eθeθ∂Sθθ

∂r+ (ezeθ + eθez)

∂Sθz

∂r+ezez

∂Szz∂r

)

+ eθ

(erer

∂Srrr∂θ

+ Srrr

(eθer + ereθ) + (ereθ + eθer )∂Srθr∂θ

+ 2Srθr

(eθeθ − erer ) + (erez + ezer )∂Srzr∂θ

+ Srzr

(eθez + ezeθ)

+ eθeθ∂Sθθ

r∂θ− Sθθ

r(ereθ + eθer )

+ (ezeθ + eθez)∂Sθz

r∂θ− Sθz

r(ezer + erez) + ezez

∂Szzr∂θ

)

+ ez

(erer

∂Srr∂z

+ (ereθ + eθer )∂Srθ∂z

+ (erez + ezer )∂Srz∂z

+ eθeθ∂Sθθ

∂z+ (ezeθ + eθez)

∂Sθz

∂z+ ezez

∂Szz∂z

).

Thus

u · ∇S = ur

(erer

∂Srr∂r

+ (ereθ + eθer )∂Srθ∂r

+ (erez + ezer )∂Srz∂r

+eθeθ∂Sθθ

∂r+ (ezeθ + eθez)

∂Sθz

∂r+ezez

∂Szz∂r

)

+ uθ

(erer

∂Srrr∂θ

+ Srrr

(eθer + ereθ) + (ereθ + eθer )∂Srθr∂θ

+2Srθr

(eθeθ − erer ) + (erez + ezer )∂Srzr∂θ

+ Srzr

(eθez + ezeθ)

+eθeθ∂Sθθ

r∂θ− Sθθ

r(ereθ + eθer )

+ (ezeθ + eθez)∂Sθz

r∂θ− Sθz

r(ezer + erez) + ezez

∂Szzr∂θ

)


+ uz

(erer

∂Srr∂z

+ (ereθ + eθer )∂Srθ∂z

+ (erez + ezer )∂Srz∂z

+ eθeθ∂Sθθ

∂z+ (ezeθ + eθez)

∂Sθz

∂z+ ezez

∂Szz∂z

).

u · ∇S = erer

(ur

∂Srr∂r

+ uθ∂Srrr∂θ

+ uz∂Srr∂z

− 2uθSrθr

)

+ (ereθ + eθer )(ur

∂Srθ∂r

+ uθ∂Srθr∂θ

+ uz∂Srθ∂z

+uθSrrr

− uθSθθ

r

)

+ (erez + ezer )(ur

∂Srz∂r

+ uθ∂Srzr∂θ

+ uz∂Srz∂z

− uθSθz

r

)

+ eθeθ

(ur

∂Sθθ

∂r+ uθ

∂Sθθ

r∂θ+ uz

∂Sθθ

∂z+ 2uθ

Srθr

)

+ (ezeθ + eθez)(ur

∂Sθz

∂r+ uθ

∂Sθz

r∂θ+ uz

∂Sθz

∂z+ uθ

Srzr

)

+ ezez

(ur

∂Szz∂r

+ uθ∂Szzr∂θ

+ uz∂Szz∂z

).

Problem 1.13

The stress tensor in a material satisfies ∇ · S = 0. Thus

∇ j(xk Si j

) = Sik .

and

〈Sik〉 = 1V

∫V SikdV = 1

V

∫V ∇ j

(xk Si j

)dV = 1

V

∫S xk Si j n j d A

= 12V

∫S (xkti + xi tk) d A

after a symmetrisation.

Problem 1.14

We can regard

〈nn〉 = 1

S

∫

SnndS

http://dx.doi.org/10.1007/978-3-319-62000-8_1

http://dx.doi.org/10.1007/978-3-319-62000-8_1


and

〈nnnn〉 = 1

S

∫

SnnnndS

as averages of various moments of the normal unit vector, uniformly distributed inspace.

We further note that < nn > and < nnnn > are isotropic tensors (unchangedwith coordinate rotation) and thus their general forms are given by

〈nn〉 = αI = 1

S

∫

SnndS,

⟨nin jnknl

⟩ = β(δi jδkl + δikδ jl + δilδ jk

) = 1

S

∫

Snin jnknldS.

The scalars α and β are found by contracting indices and performing some simpleintegrations:

3α = 1

S

∫

SdS = 1 α = 1

3.

15β = 1

S

∫

SdS = 1 β = 1

15.

Another approach (here a is the radius of the sphere): for 〈nn〉:1

S

∫nin jdS = 1

aS

∫

Sxin jdS

= 1

aS

∫

V

∂

∂x j(xi ) dV

= 1

aSδi j

∫

VdV = V

aSδi j

= 1

3δi j

And for⟨nin jnknl

⟩:

1

S

∫nin jnknldS = 1

a3S

∫

Sxi x j xknldS

= 1

a3S

∫

V

∂

∂xl

(xi x j xk

)dV

= 1

a3S

∫

V

(δil x j xk + δ jl xi xk + δkl xi x j

)dV


Also, from

∂

∂xk

(xi x j xk

) = δik x j xk + δ jk xi xk + δkk xi x j = 5xi x j

Thus

1

a3S

∫

Vδil x j xkdV = 1

5a3S

∫

Vδil

∂

∂xm

(x j xk xm

)dV

= 1

5a3Sδil

∫

Sx j xk xmnmdS

= a3

5a3Sδil

∫

Sn jnkdS

= 1

15δilδ jk

Similarly

1

a3S

∫

Vδ jl xi xkdV = 1

15δ jlδik,

1

a3S

∫

Vδkl xi x j dV = 1

15δklδi j

and

1

S

∫nin jnknldS = 1

15

(δilδ jk + δ jlδik + δklδi j

).

Problems in Chap. 3

Problem 3.1

From FF−1 = I,

Fd

dtF−1 +

(d

dtF)F−1 = 0 F

d

dtF−1 = −

(d

dtF)F−1 = −LFF−1

yielding

d

dtF−1 = −F−1L, F (0) = I = F−1 (0) .

http://dx.doi.org/10.1007/978-3-319-62000-8_3

http://dx.doi.org/10.1007/978-3-319-62000-8_3


Problem 3.2

For a simple shear flow, where the velocity field takes the form

u = γy, v = 0, w = 0,

the velocity gradient is

[L] = [∇u]T =⎡

⎣0 γ 00 0 00 0 0

⎤

⎦ .

The deformation gradient obeys

dFdt

= LF, F (0) = I,

which has the solution for a constant L,

F (t) = exp (tL) = I +∑

n=1

1

n! (tL)n = I + tL for Ln = 0, n ≥ 2

The particle X is carried to x at time t according to

x (t) = F (t)X or X = F−1 (t) x

At time τ if the particle is at point ζ, then

ζ (τ ) = F (τ )X = F (τ )F−1 (t) x = exp ((τ − t)L) x

= x + (τ − t)Lx. (S.240)

This shows the linearly stretching nature of a fluid filament.

Problem 3.3

For an elongational flow, the velocity field is

u = ax, v = by, w = cz, a + b + c = 0 (for incompressibility). (S.241)

the velocity gradient is

[L] =⎡

⎣a 0 00 b 00 0 c

⎤

⎦ ,[etL

] =⎡

⎣eat 0 00 ebt 00 0 ect

⎤

⎦ .

http://dx.doi.org/10.1007/978-3-319-62000-8_3

http://dx.doi.org/10.1007/978-3-319-62000-8_3


The path lines are given by

[ξ(τ )] =⎡

⎣ξψζ

⎤

⎦ =⎡

⎣ea(τ−t) 0 0

0 eb(τ−t) 00 0 ec(τ−t)

⎤

⎦

⎡

⎣xyz

⎤

⎦ , (S.242)

and thus exponential flow can stretch the fluid element exponentially fast.

Problem 3.4

Consider a super-imposed oscillatory shear flow:

u = γm y, v = 0, w = ωγa y cosωt. (S.243)

The path lines are defined by

d

dt(x, y, z) = (γm y, 0,ωγa y cosωt) , x (0) = X, y (0) = Y, z (0) = Z

This has the solution

x (t) = X + t γmY, y (t) = Y, z (t) = Z + γaY sinωt

The path line, (ξ (τ ) ,ψ (τ ) , ζ (τ )) ,

(ξ (τ ) ,ψ (τ ) , ζ (τ )) = (X + τ γmY,Y, Z + γaY sinωτ )

= (x (t) + (τ − t) γm y, y, z (t) + γa y (cosωτ − cosωt)) .

Problem 3.5

We want to calculate Rivlin–Ericksen tensor for an elongation flow (S.241) wherethe velocity gradient is

[L] = diag (a, b, c) .

The first Rivlin–Ericksen tensor is

A1 = L + LT = 2L.

http://dx.doi.org/10.1007/978-3-319-62000-8_3

http://dx.doi.org/10.1007/978-3-319-62000-8_3


The subsequent two Rivlin–Ericksen tensors are, because of L being diagonal,

A2 = A1L + LTA1 = A21, A3 = A2L + LTA2 = A3

1.

By induction,

An = An1 = diag

((2a)n , (2b)n , (2c)n

).

Problem 3.6

For the velocity field of (S.243) takes the form

u = γm y, v = 0, w = ωγa y cosωt.

The velocity gradient tensor is

[L] =⎡

⎣0 γm 00 0 00 ωγa cosωt 0

⎤

⎦ .

The first Rivlin–Ericksen tensor is

[A1] = [L + LT

] =⎡

⎣0 γm 0γm 0 ωγa cosωt0 ωγa cosωt 0

⎤

⎦ .

The second Rivlin–Ericksen tensor is

[A2] = [ddtA1 + A1L + LTA1

] =⎡

⎣0 0 00 0 −ω2γa sinωt0 −ω2γa sinωt 0

⎤

⎦

+⎡


⎤

⎦

⎡


⎤

⎦

+⎡

⎣0 0 0γm 0 ωγa cosωt0 0 0

⎤

⎦

⎡


⎤

⎦

=⎡

⎣0 0 00 2

(γ2m + ω2γ2

a cos2 ωt

) −ω2γa sinωt0 −ω2γa sinωt 0

⎤

⎦ .

http://dx.doi.org/10.1007/978-3-319-62000-8_3


The third Rivlin–Ericksen tensor is

[A3] = [ ddtA2 + A2L + LTA2

] =⎡

⎣0 0 00 −2ω3γ2

a sin 2ωt −ω3γa cosωt0 −ω3γa cosωt 0

⎤

⎦

+⎡

⎢⎣

0 0 00 2

(γ2m + ω2γ2

a cos2 ωt


⎤

⎥⎦

⎡


⎤

⎦

+⎡


⎤

⎦

⎡

⎣0 0 00 2

(γ2m + ω2γ2

a cos2 ωt


⎤

⎦

=⎡

⎣0 0 00 −3ω3γ2


⎤

⎦ .

The 4th Rivlin–Ericksen tensor:

[A4] = [ ddtA3 + A3L + LTA3

] =⎡

⎣0 0 00 −6ω4γ2

a cos 2ωt ω4γa sinωt0 ω4γa sinωt 0

⎤

⎦

+⎡

⎣0 0 00 −3ω3γ2


⎤

⎦

⎡


⎤

⎦

+⎡


⎤

⎦

⎡

⎣0 0 00 −3ω3γ2


⎤

⎦

=

⎡

⎢⎢⎣

0 0 0

0 2ω4γ2a

(3 − 7 cos2 ωt

)ω4γa sinωt

0 ω4γa sinωt 0

⎤

⎥⎥⎦ .

The 5th Rivlin–Ericksen tensor:

[A5] = [ddtA4 + A4L + LTA4

] =⎡

⎣0 0 00 14ω5γ2

a sin 2ωt ω5γa cosωt0 ω5γa cosωt 0

⎤

⎦

+⎡

⎣0 0 00 2ω4γ2

a

(3 − 7 cos2 ωt

)ω4γa sinωt

0 ω4γa sinωt 0

⎤

⎦

⎡


⎤

⎦

+⎡


⎤

⎦

⎡

⎣0 0 00 2ω4γ2

a

(3 − 7 cos2 ωt

)ω4γa sinωt

0 ω4γa sinωt 0

⎤

⎦

=⎡

⎣0 0 00 15ω5γ2


⎤

⎦ .


The 6th Rivlin–Ericksen tensor is

[A6] =[d

dtA5 + A5L + LTA5

]=⎡

⎣0 0 00 30ω6γ2

a cos 2ωt −ω6γa sinωt0 −ω6γa sinωt 0

⎤

⎦

+⎡

⎣0 0 00 15ω5γ2


⎤

⎦

⎡


⎤

⎦

+⎡


⎤

⎦

⎡

⎣0 0 00 15ω5γ2


⎤

⎦

=⎡

⎣0 0 00 30ω6γ2

a cos 2ωt −ω6γa sinωt0 −ω6γa sinωt 0

⎤

⎦+⎡

⎣0 0 00 2ω6γ2

a cos2 ωt 0

0 0 0

⎤

⎦

=⎡

⎢⎣

0 0 0

0 −2ω6γ2a

(15 − 31 cos2 ωt

)ω6γa sinωt

0 ω6γa sinωt 0

⎤

⎥⎦ .

By induction, the general form for Rivlin–Ericksen tensors is (n ≥ 1)

[A2n+1

] =⎡

⎣0 0 00 (−1)n αω2n+1γ2

a sin 2ωt (−1)n ω2n+1γa cosωt0 (−1)n ω2n+1γa cosωt 0

⎤

⎦

[A2n+2

] =⎡

⎣0 0 00 2 (−1)n+1 ω2n+2γ2

a

(α − β cos2 ωt

)(−1)n+1 ω2n+2γa sinωt

0 (−1)n+1 ω2n+2γa sinωt 0

⎤

⎦

α = 3.5 . . . (2n + 1), β = 2α + 1.

Problem 3.7

Cylindrical Coordinates. Conservation of mass: ∂ρ∂t + ∇ · (ρu) = 0

∂ρ

∂t+ 1

r

∂

∂r(rρur ) + 1

r

∂

∂θ(ρuθ) + ∂

∂z(ρuz) = 0. (S.244)

Conservation of linear momentum ρa = ρb + (∇ · S)T where

a =(

∂

∂tu + u · ∇u

)

http://dx.doi.org/10.1007/978-3-319-62000-8_3


is the fluid acceleration, b is the body force and S is the (symmetric) stress tensor:

ar = ∂ur∂t

+ ur∂ur∂r

+ uθ

r

∂ur∂θ

− u2θr

+ uz∂ur∂z

,

aθ = ∂uθ

∂t+ ur

∂uθ

∂r+ uθur

r+ uθ

r

∂uθ

∂θ+ uz

∂uθ

∂z, (S.245)

az = ∂uz

∂t+ ur

∂uz

∂r+ uθ

r

∂uz

∂θ+ uz

∂uz

∂z,

ρar = ρbr + 1

r

∂

∂r(r Srr ) − Sθθ

r+ 1

r

∂

∂θSrθ + ∂

∂zSrz,

ρaθ = ρbθ + 1

r2∂

∂r

(r2Sθr

)+ 1

r

∂

∂θSθθ + ∂

∂zSθz, (S.246)

ρaz = ρbz + 1

r

∂

∂r(r Szr ) + 1

r

∂

∂θSzθ + ∂

∂zSzz .

Spherical Coordinates. Conservation of mass: ∂ρ∂t + ∇ · (ρu) = 0

∂ρ

∂t+ 1

r2∂

∂r

(r2ρur

)+ 1

r sin θ

∂

∂θ(ρuθ sin θ) + 1

r sin θ

∂

∂φ

(ρuφ

) = 0. (S.247)

Conservation of linear momentum ρa = ρb + (∇ · S)T where

a =(

∂

∂tu + u · ∇u

)

is the fluid acceleration, b is the body force and S is the (symmetric) stress tensor:

ar = ∂ur∂t

+ ur∂ur∂r

+ uθ

r

∂ur∂θ

− u2θr

+ uφ

(1

r sin θ

∂ur∂φ

− uφ

r

),

aθ = ∂uθ

∂t+ ur

∂uθ

∂r+ uθur

r+ uθ

r

∂uθ

∂θ

+uφ

(1

r sin θ

∂uθ

∂φ− uφ

rcot θ

), (S.248)

aφ = ∂uφ

∂t+ ur

∂uφ

∂r+ uθ

r

∂uφ

∂θ+ uφ

(1

r sin θ

∂uφ

∂φ+ ur

r+ uθ

rcot θ

),


ρar = ρbr + 1

r2∂

∂r

(r2Srr

)− Sθθ + Sφφ

r+ 1

r sin θ

∂

∂θ(Srθ sin θ)

+ 1

r sin θ

∂

∂φSrφ,

ρaθ = ρbθ + 1

r3∂

∂r

(r3Sθr

)+ 1

r sin θ

∂

∂θ(Sθθ sin θ) + 1

r sin θ

∂

∂φSθφ

− Sφφ

rcot θ, (S.249)

ρaφ = ρbφ + 1

r3∂

∂r

(r3Sφr

)+ 1

r sin θ

∂

∂θ

(Sφθ sin θ

)+ 1

r sin θ

∂

∂φSφφ

+ Sθφ

rcot θ.

Problems in Chap. 4

Problem 4.1

In a simple shear deformation of a linearly elastic body (4.7), in which the displace-ment field takes the form

v1 = γy, v2 = 0, v3 = 0, (S.250)

where γ is the amount of shear, the infinitesimal strain tensor is

[ε] = 1

2

⎡

⎣0 γ 0γ 0 00 0 0

⎤

⎦ , (S.251)

corresponding to the stress tensor

[T] = μ

⎡

⎣0 γ 0γ 0 00 0 0

⎤

⎦ . (S.252)

The only non-trivial stress component is the shear stress, T12 = μγ. Thus μ is calledthe shear modulus.

http://dx.doi.org/10.1007/978-3-319-62000-8_4

http://dx.doi.org/10.1007/978-3-319-62000-8_4

http://dx.doi.org/10.1007/978-3-319-62000-8_4


Problem 4.2

In a uni-axial extension of a linearly elastic material (4.7), in which the displacementfield is given by

v1 = εx, v2 = −νεy, v3 = −νεz, (S.253)

where ε is the elongational strain, and ν is the amount of lateral contraction due tothe axial elongation, called Poisson’s ratio. The infinitesimal strain tensor is

[ε] = diag [ε,−νε,−νε] , (S.254)

leading to the stress tensor

[T] = λ (1 − 2ν) εI + 2μ diag [ε,−νε,−νε] , (S.255)

or

Txx = [λ (1 − 2ν) + 2μ] ε,

Tyy = [λ (1 − 2ν) − 2μν] ε, (S.256)

Tzz = [λ (1 − 2ν) − 2μν] ε.

All the other components of T are zero. If the lateral stresses are zero, then

ν = λ

2 (λ + μ). (S.257)

In addition, if we set

λ (1 − 2ν) + 2μ = 2μ (1 + ν) = E, (S.258)

then

Txx = Eε. (S.259)

E is called the Young’s modulus of the material.

Problem 4.3

Consider a deformation characterised by

F = ∂x∂X

,

http://dx.doi.org/10.1007/978-3-319-62000-8_4

http://dx.doi.org/10.1007/978-3-319-62000-8_4

http://dx.doi.org/10.1007/978-3-319-62000-8_4


F takes an element dX into dx = FdX. Thus, if dX = dXP,where P is a unit vector,

dx2 = FikdXk Fild Xl = Fik Fil Pk PldX2

= Ckl Pk PldX2. (S.260)

If P is distributed randomly in space, then on the average, 〈Pk Pl〉 = δkl/3 and onehas

⟨dx2

⟩ = 1

3dX2tr C, (S.261)

and a measure of the amount of stretch (Weissenberg number) could be defined as

Wi = 1

3tr C. (S.262)

Problem 4.4

Let f be a vector-valued, isotropic polynomial of a symmetric tensor S and a vectorv. By definition, it satisfies, ∀ orthogonal Q,

Qf (S, v) = f(QSQT ,Qv

). (S.263)

Now, define a scalar function g = g (u,S, v) as

g (u,S, v) = u · f (S, v) . (S.264)

Then,

g(Qu,QSQT ,Qv

) = Qu · f (QSQT ,Qv), (S.265)

and from the properties of f, (S.263),

g(Qu,QSQT ,Qv

) = Qu · Qf (S, v)

= u · f (S, v) = g (u,S, v) (S.266)

Thus g (u,S, v) is an isotropic scalar function in all of its arguments. If f is a poly-nomial in its arguments then g is also a polynomial in its arguments - its integritybasis has been listed in (4.36):

tr S, tr S2, tr S3, u · u, u · v, v · v, u · Su, u · S2u,

v · Sv, v · S2v, u · Sv, u · S2v. (S.267)

http://dx.doi.org/10.1007/978-3-319-62000-8_4

http://dx.doi.org/10.1007/978-3-319-62000-8_4


But g is linear in u and cannot depend on any non-linear manner on u.Consequently,

g = u · f (tr S, tr S2, tr S3, v · v, v · Sv, v · S2v) (S.268)

= u · [ f01 + f1S + f2S2]v.

This choice is dictated by the fact the term multiplied with u is a vector. This leadsto the form

f(S, v) = [ f01 + f1S + f2S2]v, (S.269)

where the scalar valued coefficients are polynomials in the six invariants involvingonly S and v in the list (S.268).

Problem 4.5

Consider a simple shear deformation of a rubber-like material (4.55), where

x = X + γY, y = Y, z = Z . (S.270)

The deformation gradient is

[F] =⎡

⎣1 γ 00 1 00 0 1

⎤

⎦ ,

leading to the Finger tensor

[B] = [FFT

] =⎡

⎣1 γ 00 1 00 0 1

⎤

⎦

⎡

⎣1 0 0γ 1 00 0 1

⎤

⎦

=⎡

⎣1 + γ2 γ 0γ 1 00 0 1

⎤

⎦ (S.271)

[B−1] =

⎡

⎣1 −γ 0

−γ 1 + γ2 00 0 1

⎤

⎦ . (S.272)

Consequently, the stress tensor is given by

[T] = −P [I] +⎡

⎣β1(1 + γ2

)− β2 (β1 + β2) γ 0(β1 + β2) γ β1 − β2

(1 + γ2

)0

0 0 β1 − β2

⎤

⎦ , (S.273)

http://dx.doi.org/10.1007/978-3-319-62000-8_4

http://dx.doi.org/10.1007/978-3-319-62000-8_4


whereP is the hydrostatic pressure. The shear stress and the normal stress differencesare

S = (β1 + β2) γ, N1 = (β1 + β2) γ2, N2 = −β2γ2. (S.274)

Deduce that the linear shear modulus of elasticity is

G = limγ→0

(β1 + β2) . (S.275)

The ratio

N1

S= γ (S.276)

is independent of the material properties. Such a relation is called universal.

Problem 4.6

In a uniaxial elongational deformation of a rubber-like material (4.55), where(in cylindrical coordinates)

R = λ1/2r, Θ = θ, Z = λ−1z, (S.277)

show that the inverse deformation gradient is

[F−1] =

⎡

⎢⎢⎢⎢⎢⎣

∂R

∂r

1

r

∂R

∂θ

∂R

∂z

0R

r0

∂Z

∂r

1

r

∂Z

∂θ

∂Z

∂z

⎤

⎥⎥⎥⎥⎥⎦

=⎡

⎣λ1/2 0 00 λ1/2 00 0 λ−1

⎤

⎦ . (S.278)

Consequently, the strains are

[B−1

] = [F−TF−1

] =⎡

⎣λ 0 00 λ 00 0 λ−2

⎤

⎦ , [B] =⎡

⎣λ−1 0 00 λ−1 00 0 λ2

⎤

⎦ . (S.279)

Thus the total stress tensor for a rubber-like material (4.55) is

[T] = −P [I] +⎡

⎣β1λ

−1 − β2λ 0 00 β1λ

−1 − β2λ 00 0 β1λ

2 − β2λ−2

⎤

⎦ . (S.280)

http://dx.doi.org/10.1007/978-3-319-62000-8_4

http://dx.doi.org/10.1007/978-3-319-62000-8_4

http://dx.doi.org/10.1007/978-3-319-62000-8_4


Under the condition that the lateral tractions are zero, i.e., Trr = 0, the pressure canbe found

P = β1λ−1 − β2λ, (S.281)

and thus the tensile stress is

Tzz = −P + β1λ2 − β2λ

−2 = β1λ2 − β2λ

−2 − β1λ−1 + β2λ

= (λ2 − λ−1

) (β1 + β2λ

−1). (S.282)

This tensile stress is the force per unit area in the deformed configuration. As r =λ1/2R, the corresponding force per unit area in the undeformed configuration is

TZZ = Tzzλ−1 = (

λ − λ−2) (

β1 + β2λ−1). (S.283)

Problem 4.7

Note that

Ct (τ ) = Ft (τ )T Ft (τ ) , (S.284)

but (use subscript notation for clarity)

Ft (τ ) = ∂x (τ )

∂x (t)= ∂x (τ )

∂X∂X

∂x (t)= F (τ )F (t)−1

and therefore

Ct (τ ) = F (t)−T F (τ )T F (τ )F (t)−1 = F (t)−T C (τ )F (t)−1

leading to the desired result

C (τ ) = F (t)T Ct (τ )F (t) . (S.285)

Problem 4.8

Consider a simple shear flow

u = γy, v = 0, w = 0. (S.286)

http://dx.doi.org/10.1007/978-3-319-62000-8_4

http://dx.doi.org/10.1007/978-3-319-62000-8_4


The first and second Rivlin–Ericksen tensors are

[A1] =⎡

⎣0 γ 0γ 0 00 0 0

⎤

⎦ ,[A2

1

] =⎡

⎣γ2 0 00 γ2 00 0 0

⎤

⎦

[A2] = [A1L + LTA1

] =⎡

⎣0 0 00 2γ2 00 0 0

⎤

⎦ .

Thus, the stress tensor in the second-order model is given by

[S] = η0

⎡

⎣0 γ 0γ 0 00 0 0

⎤

⎦+ (ν1 + ν2)

⎡

⎣γ2 0 00 γ2 00 0 0

⎤

⎦− ν1

2

⎡

⎣0 0 00 2γ2 00 0 0

⎤

⎦ , (S.287)

and the three viscometric functions are

S = η0γ, N1 = S11 − S22 = ν1γ2, N2 = S22 − S33 = ν2γ

2. (S.288)

Problem 4.9

In an elongational flow

u = εx, v = − ε

2y, w = − ε

2z, (S.289)

the first and second Rivlin–Ericksen tensors are

[A1] =⎡

⎣2ε 0 00 −ε 00 0 −ε

⎤

⎦ ,[A2

1

] =⎡

⎣4ε2 0 00 ε2 00 0 ε2

⎤

⎦ ,

[A2] = [A1L + LTA1

] =⎡

⎣4ε2 0 00 ε2 00 0 ε2

⎤

⎦ .

http://dx.doi.org/10.1007/978-3-319-62000-8_4


Thus the stress is given by

[S] = η0

⎡

⎣2ε 0 00 −ε 00 0 −ε

⎤

⎦+ (ν1 + ν2)

⎡

⎣4ε2 0 00 ε2 00 0 ε2

⎤

⎦

−ν1

2

⎡

⎣4ε2 0 00 ε2 00 0 ε2

⎤

⎦ . (S.290)

Consequently the elongational viscosity is given by

ηE = Sxx − Syyε

= 3η0 + 3(ν1

2+ ν2

)ε. (S.291)

Problem 4.10

For potential flows, the velocity field is a gradient of a potential:

u = ∇φ, (S.292)

from which, incompressibility demands

∇ · u = ∇2φ = 0. (S.293)

Now, since

Li j = ∂ui∂x j

= φ,i j , A1i j = ∂ui∂x j

+ ∂u j

∂xi= 2φ,i j (S.294)

where the comma denotes a spatial derivative, it follows that

∇ · A1 = 0. (S.295)

Next,

(∇ · A21

)i= (

A1ik A1k j), j

= 4(φ,ikφ,k j

), j

= 4φ,k jφ,ik j

= 1

2

(4φ,k jφ,k j

),i

∇ · A21 = 1

2

(∇(tr A21)). (S.296)

http://dx.doi.org/10.1007/978-3-319-62000-8_4


Furthermore,

A2i j = uk A1i j,k + A1ik Lk j + Lki A1k j

= 2φ,kφ,i jk + 4φ,ikφ,k j

A2i j, j = 2φ,k jφ,i jk + 4φ,k jφ,ik j = 6φ,k jφ,i jk

= 3(φ,k jφ,k j

)i

= 3

4

(4φ,k jφ,k j

)i

∇ · A2 = 3

4∇(tr A2

1). (S.297)

In a second-order fluid model, the stress is given by

T = −PI + η0A1 + (ν1 + ν2)A21 − ν1

2A2,

and conservation of momentum yields

∇P = ∇ ·[η0A1 + (ν1 + ν2)A2

1 − ν1

2A2

]− ρa, (S.298)

where a is the acceleration field. If the same flow, of the same kinematics occurs ina Newtonian fluid of viscosity η0, then we must have

∇PN = η0∇ · A1 − ρa, (S.299)

where PN is the Newtonian pressure field. Thus, for this to occur, one must have an“extra” pressure field

∇PE = ∇ ·[(ν1 + ν2)A2

1 − ν1

2A2

]. (S.300)

From the results (S.296) and (S.297), this is possible, and thus in potential flows, thevelocity fields for a Newtonian and a second-order fluid are identical, with the extrapressure given by

∇PE = ∇ ·[(ν1 + ν2)A2

1 − ν1

2A2

]

=[1

2(ν1 + ν2) − 3ν1

8

]∇ (

trA21

),


or that

PE = 1

8(ν1 + 4ν2) trA2

1, (S.301)

to within a constant which can be absorbed in PN .

Problem 4.11

For steady two-dimensional incompressible flows, a stream function ψ = ψ(x, y)can be defined such that the velocity components u and v can be expressed as

u = ∂ψ

∂y, v = −∂ψ

∂x. (S.302)

If the fluid is incompressible Newtonian, then for a flow at zero Reynolds number,the Newtonian stresses are

[SN ] = η

[2ψ,xy ψ,yy − ψ,xx

ψ,yy − ψ,xx −2ψ,xy

]

and the balance of momentum requires

∂P

∂x= 2η

∂2u

∂x2+ η

(∂2u

∂y2+ ∂2v

∂x∂y

)

= η[2ψ,yxx + ψ,yyy − ψ,xxy

],

∂P

∂y= η

(∂2u

∂x∂y+ ∂2v

∂x2

)+ 2η

∂2v

∂y2

= η[ψ,yyx − ψ,xxx − 2ψ,xyy

].

For compatibility, the stream function must satisfy a bi-harmonic equation:

ψ,xxy y + ψ,yyyy = −ψ,xxxx − ψ,xxyy

∇2∇2ψ = Δ2ψ = 0. (S.303)

Now,

[L] =[

ψ,xy ψ,yy

−ψ,xx −ψxy

]

[A1] =[2ψ,xy ψ,yy − ψ,xx


], (S.304)

http://dx.doi.org/10.1007/978-3-319-62000-8_4


[A2

1

] = (4ψ2

,xy + ψ2,yy − 2ψ,xxψ,yy + ψ2

,xx

) [1 00 1

], (S.305)

[A2] = [u · ∇A1] + [A1L] + [LTA1

](S.306)

=(

ψy∂

∂x− ψx

∂

∂y

)[2ψ,xy ψ,yy − ψ,xx


]

+[2ψ2

,xy − ψ,xx(ψ,yy − ψ,xx

)2ψ,xyψ,yy − ψ,xy

(ψ,yy − ψ,xx

)

2ψ,xyψ,xx + ψ,xy(ψ,yy − ψ,xx

)2ψ2

,xy + ψ,yy(ψ,yy − ψ,xx

)]

+[2ψ2


)2ψ,xyψ,xx + ψ,xy

(ψ,yy − ψ,xx

)

2ψ,xyψ,yy − ψ,xy(ψ,yy − ψ,xx

)2ψ2


)]

,

from which the second-order fluid stresses (see 4.71) can be determined as

Sxx = 2ηψ,xy + (ν1 + ν2)(4ψ2


,xx

)(S.307)

−ν1

[(ψy

∂

∂x− ψx

∂

∂y

)ψ,xy + 2ψ2


)]

Sxy = η(ψ,yy − ψ,xx

)− ν1

2[(

ψy∂

∂x− ψx

∂

∂y

) (ψ,yy − ψ,xx

)

+2ψ,xy(ψ,yy + ψ,xx

)] (S.308)

Syy = −2ηψ,xy + (ν1 + ν2)(4ψ2


,xx

)

− ν1

[(ψx

∂

∂y− ψy

∂

∂x

)ψ,xy + 2ψ2


)]. (S.309)

The balance of momentum requires

∂P

∂x= ∂Sxx

∂x+ ∂Sxy

∂y,

∂P

∂y= ∂Sxy

∂x+ ∂Syy

∂y.

Compatibility requires

∂2

∂x∂y

(Sxx − Syy

)+(

∂2

∂y2− ∂2

∂x2

)Sxy = 0. (S.310)

http://dx.doi.org/10.1007/978-3-319-62000-8_4


The second-order fluid stresses can be substituted in the preceding results to yield

0 = ∂2

∂x∂y

[4ηψ,xy − ν1

(2

(ψy

∂

∂x− ψx

∂

∂y

)ψ,xy − (

ψ2,yy − ψ2

,xx

))]

+(

∂2

∂y2− ∂2

∂x2

)[η (ψ,yy − ψ,xx

)− ν1

2[(

ψy∂

∂x− ψx

∂

∂y

) (ψ,yy − ψ,xx

)

+2ψ,xy(ψ,yy + ψ,xx

)]].

This is simplified to

ηΔ2ψ − ν1

2u · ∇ (�2ψ

) = 0. (S.311)

This shows that a Newtonian velocity field (Δ2ψ = 0) is also a velocity field for thesecond-order fluid.

Problem 4.12

In a simple shear flow,

u = γ (t) y, v = 0, w = 0,

the path lines x satisfy

d

dt(x, y, z) = (γ (t) y, 0, 0) , x(0) = X.

That is,

x (t) = X + Y∫ t

0γ(t ′)dt ′, y = Y, z = Z .

Thus the path lines ¸ (τ ) = (ξ,ψ, ζ)

ξ (τ ) = X + Y∫ τ

0γ(t ′)dt ′

= x + y∫ τ

tγ(t ′)dt ′,

ψ = y,

ζ = z,

http://dx.doi.org/10.1007/978-3-319-62000-8_4


or

ξ (τ ) = x + yγ (t, τ ) , ψ (τ ) = y, ζ (τ ) = z, (S.312)

where

γ (t, τ ) =∫ τ

tγ (s) ds. (S.313)

The relative deformation gradient:

Ft (τ ) = (∇x¸)T =⎡

⎣1 γ (t, τ ) 00 1 00 0 1

⎤

⎦

Ct (τ ) = Ft (τ )T Ft (τ ) =⎡

⎣1 γ (t, τ ) 0γ (t, τ ) 1 + γ2 (t, τ ) 00 0 1

⎤

⎦ .

S (t) =∫ ∞

0μ (s) (Ct (t − s)−I) ds

=∫ ∞

0μ (s)

⎡

⎣0 γ (t, t − s) 0γ (t, t − s) γ2 (t, t − s) 00 0 0

⎤

⎦ .

That the shear stress and the normal stress differences are given by

S12 (t) =∫ ∞

0μ (s) γ (t, t − s) ds, (S.314)

N1 (t) = −∫ ∞

0μ (s) [γ (t, t − s)]2 ds = −N2. (S.315)

For a constant shear rate γ,

γ (t, τ ) = γ (τ − t) ,

and with the memory function μ (s) = −G

λe−s/λ,

S12 (t) =∫ ∞

0μ (s) γ (t, t − s) ds

= −∫ ∞

0μ (s)γsds

= G

λγ

∫ ∞

0e−s/λsds

= Gγ. (S.316)


N1 (t) = G

λγ2∫ ∞

0e−s/λs2ds

= 2G (λγ)2 . (S.317)

For a sinusoidal shear rate γ = γ0 cos (ωt)

γ (t, τ ) =∫ τ

tγ0 cosωtdt = γ0

ω(sinωτ − sinωt) .

S12 (t) =∫ ∞

0μ (s) γ (t, t − s) ds

= −Gγ0

λω

∫ ∞

0e−s/λ (sinω (t − s) − sinωt) ds

= Gγ0

λω

∫ ∞

0e−s/λ (sinωt − sinωt cosωs + cosωt sinωs) ds

= Gγ0

ω

[sinωt − sinωt

1 + λ2ω2+ ωλ cosωt

1 + λ2ω2

]

= Gγ0λλω sinωt + cosωt

1 + λ2ω2. (S.318)

N1 = Gγ20

λω2

∫ ∞

0e−s/λ (sinωt − sinωt cosωs + cosωt sinωs) 2ds

= Gγ20

λω2

∫ ∞

0e−s/λ[sin2 ωt − 2 sin2 ωt cosωs + sin 2ωt sinωs

+1

2sin 2ωt sin 2ωs + sin2 ωt cos2 ωs + cos2 ωt sin2 ωs]ds

= Gγ20

ω2

[sin2 ωt − 2 sin2 ωt

1 + λ2ω2+ ωλ sin 2ωt

1 + λ2ω2

+ωλ sin 2ωt

1 + 4λ2ω2+ sin2 ωt

(1 + 2λ2ω2

)

1 + 4λ2ω2+ 2λ2ω2 cos2 ωt

1 + 4λ2ω2

]

. (S.319)

Problems in Chap. 5

Problem 5.1

Assume the relaxation modulus function (5.13)

G (t) =N∑

j=1

G je−t/λ j ,

http://dx.doi.org/10.1007/978-3-319-62000-8_5

http://dx.doi.org/10.1007/978-3-319-62000-8_5

http://dx.doi.org/10.1007/978-3-319-62000-8_5


the relation (5.11) is

S (t) = 2∫ t

−∞

N∑

j=1

G je−(t−t ′)/λ jD

(t ′)dt ′

=N∑

j=1

S( j)

S( j) = 2∫ t

−∞G je

−(t−t ′)/λ jD(t ′)dt ′.

Each component S( j) satisfies, by direct differentiation

S( j) = 2G jD (t) − 2G j

λ j

∫ t

−∞e−(t−t ′)/λ jD

(t ′)dt ′,

or that

S( j) + λ j S( j) = 2η jD, η j = G jλ j .

This relation is called the linear Maxwell equation.

Problem 5.2

In an oscillatory flow where the shear rate and the shear strain are

γ = γ0 cosωt, γ = γ0 sinωt, γ0 = ωγ0 (S.320)

the only non-zero component of the stress is

S12 =∫ t

−∞G(t − t ′

)γ0 cosωt ′dt ′,

=∫ ∞

0γ0G (s) cosω (t − s) ds,

=∫ ∞

0γ0G (s) [cosωt cosωs + sinωt sinωs] ds,

= G ′ (ω) γ0 sinωt + η′ (ω) γ0 cosωt, (S.321)

where the coefficients in the strain is the storage modulus,

G ′ (ω) =∫ ∞

0ωG (s) sinωsds, (S.322)

http://dx.doi.org/10.1007/978-3-319-62000-8_5

http://dx.doi.org/10.1007/978-3-319-62000-8_5


and in the strain rate, the dynamic viscosity,

η′ (ω) =∫ ∞

0G (s) cosωsds.

The shear stress can be written as

S12 = G ′ (ω) γ0 sinωt + G ′′ (ω) γ0 cosωt, (S.323)

where the loss modulus G ′′ is defined as

G ′′ (ω) = ωη′ (ω) . (S.324)

Re-write the shear stress as

S12 = S012 sin (ωt + φ)

= S012 (sinωt cosφ + sin φ cosωt) .

Identify this with the previous result for S12 :

S012 cosφ = γ0G′ (ω) , S012 sin φ = γ0G

′′ (ω) ,

or that

S012 = γ0√G ′2 (ω) + G ′′2 (ω), tan φ = G ′′ (ω)

G ′ (ω), (S.325)

where tan φ is called the loss tangent. Sometimes it is more convenient to work withcomplex numbers, and the complex modulus G∗ and the complex viscosity η∗ arethus defined as

G∗ (ω) = G ′ (ω) + iG ′′ (ω) , η∗ (ω) = η′ (ω) − iη′′ (ω) . (S.326)

One can define the complex shear strain as

γ∗ = γ0e−iωt , (S.327)

then the shear stress is

S12 = Re(G∗γ∗) = γ0 Re

[(G ′ (ω) + iG ′′ (ω)

)(sinωt − i cosωt)

]

= γ0(G ′ (ω) sinωt + G ′′ (ω) cosωt

). (S.328)


Problem 5.3

Recall the definition of the spectrum H (λ) :

G ′ (ω) =∫ ∞

−∞ω2λ2

1 + ω2λ2H (λ) d ln λ, (S.329)

or, equivalently,

η′ (ω) =∫ ∞

0

H (λ)

1 + ω2λ2dλ. (S.330)

For the spectrum of the form

H (λ) = cos2 (nλ) ,

η′ (ω) =∫ ∞

0

H (ω)

1 + λ2ω2dω

= π

4ω

(1 − e−2n/ω

). (S.331)

At large n, the data η′ is smooth (e−2n/ω goes to zero), but the spectrum is highlyoscillatory. Thus, the inverse problemoffinding H (λ) ,given the dataη′ in the chosenform is ill-conditioned – that is, a small variation in the data (in the exponentiallysmall term) may lead to a large variation in the solution.

Problem 5.4

For the Maxwell discrete relaxation spectrum (5.13),

G (t) =N∑

j=1

G je−t/λ j

the storage modulus and the dynamic viscosity are given by (5.17):

G ′ (ω) =∫ ∞

0ωG (s) sinωsds

=N∑

j=1

∫ ∞

0ωG je

−s/λ j sinωsds

http://dx.doi.org/10.1007/978-3-319-62000-8_5

http://dx.doi.org/10.1007/978-3-319-62000-8_5

http://dx.doi.org/10.1007/978-3-319-62000-8_5

http://dx.doi.org/10.1007/978-3-319-62000-8_5


=N∑

j=1

ω2G jλ2j

1 + ω2λ2j

, (S.332)

η′ (ω) =∫ ∞

0G (s) cosωsds

=N∑

j=1

∫ ∞

0G je

−s/λ j cosωsds

=N∑

j=1

G jλ j

1 + ω2λ2j

. (S.333)

In particular, with one relaxation mode λ = λ1,

G ′ (ω) = ω2G1λ21

1 + ω2λ21

, η′ (ω) = G1λ1

1 + ω2λ21

, tan φ = ωη′ (ω)

G ′ (ω)= 1

ωλ1(S.334)

As as ωλ1 = 0, η′ (ω) = G1λ1, G ′ (ω) = 0 (Newtonian fluid-like response), andwhen ωλ1 → ∞, η′ → 0, G ′ → G1 corresponds to a solid-like response.

Problem 5.5

Suppose we have a Maxwell material with one relaxation time,

G (t) = η0

λe−t/λ.

and Ωi = constant. For the circular Couette flow problem, the torque on the innercylinder is Γ = M (Ro − Ri ) , where

M (t) = 4πR2o

1 − R2o/R

2i

∫ t

0G(t − t ′

)Ωi

(t ′)dt ′

= 4πR2oη0Ωi(

1 − R2o/R

2i

)λ

∫ t

0e−(t−t ′)/λdt ′

= 4πR2oη0Ωi(

1 − R2o/R

2i

)(1 − e−t/λ

).

With the Newtonian result (corresponds to λ → 0),

MN = 4πR2oη0Ωi(

1 − R2o/R

2i

) ,

http://dx.doi.org/10.1007/978-3-319-62000-8_5


we thus have

M (t)

MN= 1 − e−t/λ. (S.335)

Problem 5.6

Working in Laplace transform domain (s is the Laplace transform variable, and theoverbar denotes the Laplace transform function), and denote the displacement acrossthe spring G1 as x1, across the Kelvin-Voigt element (G2, η2) as x2 − x1, and acrossthe dashpot η1 as x − x2, the relationship between force and various displacementsso defined are

• across the spring G1:F

G1= x1,

• across the dashpot of η1:F

sη1= x − x2

• across the Kelvin–Voigt–Meyer element (G2, η2):

F

G2 + sη2= x2 − x1.

These may be summed up to yield

F

G1+ F

G2 + sη2+ F

sη1= x

[sη1 (G2 + sη2) + sη1G1 + G1 (G2 + sη2)] F = sη1G1 (G2 + sη2) x

leading to

[1 + a1s + a2s

2] F = (b1 + b2s) sx,

where

a1 = η1

G2

(1 + G2

G1+ η2

η1

), a2 = η1η2

G1G2, b1 = η1, b2 = η1η2

G2.

http://dx.doi.org/10.1007/978-3-319-62000-8_5


In time domain, this is equivalent to

F + a1 F + a2 F = b1 x + b2 x,

which corresponds to the following stress-strain relation for the four-element model

Si j + a1 Si j + a2 Si j = b1γi j + b2γi j . (S.336)

Problems in Chap. 6

Problem 6.1

The velocity field for (6.3) is

u = γ (b · x) a, (S.337)

where a, b, c are a set of orthonormal vectors. The velocity gradient is

Li j = ∂ui∂x j

= ∂

∂x j(γbkxkai )

= γb jai ,

leading to the stated result

L = γab. (S.338)

Since

tr L = 0,

all the flows represented by (6.3) are isochoric (volume-conserving). Now

4D2 = γ2 (ab + ba) · (ab + ba)

= γ2 (aa + bb) ,

the shear rate is√2tr D2 =

√γ2 = |γ| . (S.339)

http://dx.doi.org/10.1007/978-3-319-62000-8_6

http://dx.doi.org/10.1007/978-3-319-62000-8_6

http://dx.doi.org/10.1007/978-3-319-62000-8_6

http://dx.doi.org/10.1007/978-3-319-62000-8_6


Problem 6.2

For the helicoidal flow (6.14), the velocity field is

u = (reθ + cez)ω (r, z − cθ) , (S.340)

where c is a constant. The velocity gradient is

(∇u)T = [∇ (rωeθ + cωez)]T

= [∇ω (reθ + cez) + ω∇ (reθ + cez)]T

= [∇ω (reθ + cez) + ω (ereθ − eθer )]T

= (reθ + cez) ∇ω + ω (eθer − ereθ)

= γab = L (S.341)

We first note that

2D = (reθ + cez)∇ω + ∇ω (reθ + cez)

+ω (eθer − ereθ) + ω (−eθer + ereθ)

= (reθ + cez)∇ω + ∇ω (reθ + cez) . (S.342)

Thus

4D2 = (reθ + cez)∇ω · (reθ + cez)∇ω

+2∇ω (reθ + cez) · (reθ + cez) ∇ω

+∇ω (reθ + cez) · ∇ω (reθ + cez)

= (reθ + cez)(

∂ω

∂θ+ c

∂ω

∂z

)∇ω

+2(r2 + c2

)∇ω∇ω

+∇ω

(∂ω

∂θ+ c

∂ω

∂z

)(reθ + cez) .

Since ω = ω (r, z − cθ) is a function of r and z − cθ = Z ,

∂ω

∂θ+ c

∂ω

∂z= −c

∂ω

∂Z+ c

∂ω

∂Z= 0,

and the shear rate squared given by

γ2 = (r2 + c2

)∇ω · ∇ω. (S.343)

http://dx.doi.org/10.1007/978-3-319-62000-8_6

http://dx.doi.org/10.1007/978-3-319-62000-8_6


The vectors a and b are identified as

γa = LLT , γb = LTL, (S.344)

neither is constant.

Problem 6.3

This flow is one of the fan flows - the flow is uniaxial and by inspection of theboundary conditions we find that the velocity field is

u = Uθ

θ0ez . (S.345)

This velocity field may be written as

u = γ (b · x) a, (S.346)

where a = ez . To identify b we note

L = ∇uT =(eθ1

r

∂

∂θu)T

= ezeθU

rθ0= γab.

Clearly

γ = U

rθ0, b = eθ.

The stress is given by (see equation preceding 6.20)

T = −PI + γη (ab + ba) + (N1 + N2) (aa + bb) − N1bb,

or

T = −PI + ηγ (ezeθ + eθez) + (N1 + N2) ezez + N2eθeθ. (S.347)

In full, the non-trivial components for the stress are

Trr = −P, Tθθ = −P + N2, Tzz = −P + N1 + N2, Tzθ = Tθz = ηγ.

http://dx.doi.org/10.1007/978-3-319-62000-8_6

http://dx.doi.org/10.1007/978-3-319-62000-8_6


To find the pressure field, we note that the velocity field is uni-directional, with zeroinertia force u · ∇u = 0. Thus, the non-trivial component of the conservation ofmomentum equation is

0 = ∂Trr∂r

+ Trr − Tθθ

r+ 1

r

∂Tθr

∂θ+ ∂Tzr

∂z

∂P

∂r= −N2

r= − γ2ν2 (γ)

r

Noting that

γ = U

rθ0= r0γ0

r, γ0 = U

r0θ0

and dγ/γ = −dr/r, which can be used in integrating for the pressure:

P = P (r0) −∫ r

r0

γ2ν2 (γ)

rdr

= P (r0) +∫ γ

γ0

γν2 (γ) .

Alternatively,

P = P (r0) + I2 (γ) − I2 (γ0) , I2 (γ) =∫ γ

0γν2dγ. (S.348)

Suppose we have a pressure measurement P = P (γ) at different shear rates. Thenaccording to the preceding

dP

dγ= d I2

dγ= γν2 (γ) .

This may be used to deduce ν2, or N2 = γ2ν2 (γ) .

Problem 6.4

In the flow between two parallel, coaxial disks, or the torsional flow, the flow is asub-class of the helicoidal flow (6.14), and by the boundary conditions on the plates,the velocity may be shown that (another way would be to show that the kinematicsbelow satisfy conservation of mass and momentum)

u = Ωrz

heθ = γzeθ, γ = Ω

r

h. (S.349)

http://dx.doi.org/10.1007/978-3-319-62000-8_6

http://dx.doi.org/10.1007/978-3-319-62000-8_6


This velocity is already in the form (6.3)

u = γ (b · x) a, (S.350)

where

a = eθ, (S.351)

and

(b · x) = z, b = ez . (S.352)

The stress is

T = −PI + S = −PI + ηγ (ab + ba) + γ2 (ν1 + ν2) aa + γ2ν2bb, (S.353)

with the non-trivial components

Trr = −P, Tθθ = −P + N1 + N2, Tzz = −P + N2, Szθ = Sθz = ηγ. (S.354)

The conservation of momentum required, neglecting inertia terms,

∂P

∂r= ∂Srr

∂r+ Srr − Sθθ

r= −N1 + N2

r,

∂P

r∂θ= 0,

∂P

∂z= 0.

Consequently, P is a function of r only. It is given by

P (r) = C −∫ r

0

N1 (ζ) + N2 (ζ)

ζdζ (S.355)

where C is a constant of integration. This is found by the assumption that at the freesurface at r = R, the radial stress is zero:

Trr (R) = −P (R) = 0. (S.356)

This implies

C =∫ R

0

N1 (ζ) + N2 (ζ)

ζdζ,

http://dx.doi.org/10.1007/978-3-319-62000-8_6


or that

P (r) =∫ R

r

N1 (ζ) + N2 (ζ)

ζdζ

=∫ γR

γ(r)

N1 (γ) + N2 (γ)

γdγ

=∫ γR

γ

γ (ν1 + ν2) dγ, (S.357)

where γR = ΩR/h is the shear rate at the rim r = R.

The torque required to turn the top disk is γ = Ωr/h = γRr/R

M =∫ R

02πr2Szθdr

= 2π∫ R

0γη (γ) r2dr

= 2π∫ γR

0ηr3dγ

= 2πR3γ−3R

∫ γR

0γ3η (γ) dγ. (S.358)

From the axial stress and the result for the pressure,

Tzz = −P + N2

= N2 +∫ r

R

N1 (ζ) + N2 (ζ)

ζdζ, (S.359)

and therefore the normal force on the top disk is, by an integration by parts, keepingin mind that normal stresses are zero at the free surface r = R,

F = −2π∫ R

0Tzzrdr

= π

∫ R

0T ′zzr

2dr

= π

∫ R

0(−2r N2 + r (N1 + N2)) dr

= π

∫ R

0r (N1 − N2) dr (S.360)

Convert this into an integral with respect to the shear rate, the normal force is givenby


F = πR2γ−2R

∫ γR

0γ (N1 − N2) dγ. (S.361)

By normalizing the torque and the force as

m = M

2πR3, f = F

πR2, (S.362)

we have

m = γ−3R

∫ γR

0γ3η (γ) dγ.

Thus

dm

dγR= −3

m

γR+ η (γR)

or

η (γR) = m

γR

(3 + γR

m

dm

dγR

)

= m

γR

(3 + d lnm

d ln γR

), (S.363)

as required to show.In addition

f = γ−2R

∫ γR

0γ (N1 − N2) dγ.

And thus

d f

dγR= −2

f

γR+ N1 (γR) − N2 (γR) ,

or

N1 (γR) − N2 (γR) = f

(2 + d ln f

d ln γR

). (S.364)

Relations (S.363) and (S.364) are the basis for the operation of the parallel-diskviscometer.


Problem 6.5

Pipe flow is a special case for helical flow, where the velocity field is (uniaxial flow)

u = u (r) ez . (S.365)

This velocity is already in the viscometric form (6.3), and the non-trivial stresses arefunctions of r. The only non-trivial momentum equation (in the axial z direction) is

∂P

∂z= ∂Srz

∂r+ Srz

r= 1

r

d

dr(r Srz) , (S.366)

where

Srz = ηdu

dr.

The right side of (S.366) is a function of r alone, thus

P =(

∂Srz∂r

+ Srzr

)z + P0

= −ΔP

Lz + P0, (S.367)

where ΔP/L is the pressure drop per unit length. This leads to

1

r

d

dr(r Srz) = −ΔP

L

Srz = −ΔP

2Lr. (S.368)

noting the boundedness of the stress in r. Thus the shear rate is

du

dr= −ΔP

2Lrη−1. (S.369)

The flow rate is

Q = 2π∫ R

0ru (r) dr

= −π

∫ R

0r2

du

drdr

= πΔP

2L

∫ R

0r3η−1dr

http://dx.doi.org/10.1007/978-3-319-62000-8_6

http://dx.doi.org/10.1007/978-3-319-62000-8_6


Since r = −2Lτ/ΔP, where τ = Srz , a change in variable yields

Q = 8π

(L

ΔP

)3 ∫ τw

0τ 3η−1dτ (S.370)

where τw = −ΔPR/(2L) is the shear stress at the wall. In terms of the reduceddischarge rate,

q = Q

πR3= τ−3

w

∫ τw

0τ 3η−1dτ , (S.371)

and therefore

dq

dτw

= 1

η (τw)− 3q

τw

, (S.372)

or

η−1 (τw) = q

τw

(3 + d ln q

d ln τw

). (S.373)

Since η (τw) γw = τw, or η−1 (τw) = γw/τw

γw = q (τw)

[3 + d ln q

d ln τw

]. (S.374)

The relation (S.374) is due to Rabinowitch and is the basis for capillary viscometry.

Problems in Chap. 7

Problem 7.1

The solution to (7.22) is given by (7.24), reproduced here

x (t) =∫ t

0exp

{m−1ζ

(t ′ − t

)}m−1F(b)

(t ′)dt ′. (S.375)

Now, the diffusivity is defined

D = limt→∞

1

2

∫ t

0〈x (t) x (t − τ ) + x (t − τ ) x (t)〉 dτ . (S.376)

http://dx.doi.org/10.1007/978-3-319-62000-8_7

http://dx.doi.org/10.1007/978-3-319-62000-8_7

http://dx.doi.org/10.1007/978-3-319-62000-8_7

http://dx.doi.org/10.1007/978-3-319-62000-8_7


First, we form the correlation function

R (τ ) = 〈x (t + τ ) x (t)〉=∫ t

0

∫ t

0exp

{m−1ζ

(t ′ − t − τ

)}m−1

⟨F(b)

(t ′)F(b)

(t ′′)⟩

m−1 exp{m−1ζ

(t ′′ − t

)}dt

′dt ′′

=∫ t

0exp

{m−1ζ

(t ′ − t − τ

)}m−12fm−1 exp

{m−1ζ

(t ′−t

)}dt

′

= e−m−1ζτ 〈x (t) x (t)〉 . (S.377)

Thus

D = limt→∞

∫ t

0e−mζτ

⟨x(t ′)x(t ′)⟩dτ

= ζ−1mkTm−1

= kTζ−1. (S.378)

This is the Stokes–Einstein relation, relating the diffusivity to the mobility of aBrownian particle.

Problem 7.2

In the limitm → 0, the Langevin equation (7.20) becomes (note that all the materialmatrices are symmetric)

x = −ζ−1 · Kx + ζ−1F(b) (t) , (S.379)

which has the solution

Δx (t) = −ζ−1 · KxΔt +∫ t+Δt

tζ−1F(b)

(t ′)dt ′. (S.380)

From this,

〈Δx (t)〉 = −ζ−1 · KxΔt +∫ t+Δt

tζ−1 ⟨F(b)

(t ′)⟩dt ′

= −ζ−1 · Kx. (S.381)

http://dx.doi.org/10.1007/978-3-319-62000-8_7

http://dx.doi.org/10.1007/978-3-319-62000-8_7


In addition,

〈Δx (t)Δx (t)〉 = O(Deltat2

)

+∫ t+Δt

t

∫ t+Δt

tdt ′′ζ−1 ⟨F(b)

(t ′)F(b)

(t ′′)⟩

ζ−1dt ′

= 2∫ t+Δt

tdt ′ζ−1f ζ−1

= 2kTζ−1Δt. (S.382)

Thus the Fokker–Planck equation is

∂φ

∂t= ∂

∂x·[ 〈ΔxΔx〉

2Δt

∂φ

∂x+〈Δx〉

Δtφ

]

is simply

∂φ

∂t= ∂


∂x+ ζ−1 · Kxφ

]. (S.383)

Problem 7.3

In a non-homogeneous flow and using the dumbell model, the particle centre ofgravity will migrate from the streamline according to

⟨R(c) − u(c)

⟩ = 1

8〈RR〉 : ∇∇u(c). (S.384)

In the planar Poiseuille flow,

u(c) = u (y) i, u (y) = U(1 − y2/h2

)

⟨R(c) − u(c)

⟩ = 1

8

[⟨R21

⟩ ∂2

∂x2+ 2 〈R1R2〉 ∂2

∂x∂y+ 2 〈R1R3〉 ∂2

∂x∂z

+ ⟨R22

⟩ ∂2

∂y2+ 2 〈R2R3〉 ∂2

∂y∂z+ ⟨

R23

⟩ ∂2

∂z2

]u(c)

= 1

8

⟨R22

⟩ ∂2

∂y2u(c)

= − U

4h2⟨R22

⟩i. (S.385)

http://dx.doi.org/10.1007/978-3-319-62000-8_7


This must be solved together with the constitutive equation for R to obtain themigration velocity. In fact, puttingA = 〈RR〉 , and using the elastic dumbbell model,we have

A11 + λ(A11 − 2u′A12

) = 1

3,

A12 + λ(A12 − u′A22

) = 0,

A22 + λ A22 = 1

3.

Thus the migration is along a streamline, and of the amount

⟨R(c) − u(c)

⟩ = − U

12h2i. (S.386)

Of course, this is a simplistic model, to have cross-streamline migration, a bettermodel is required, see for example, Goh et al., J Chem Phys 81 (1985) 6259–6265and some of the references cited thereon.

Problem 7.4

The average end-to-end vector of a linear dumbbell evolves in time according to

⟨R⟩ = L 〈R〉 − 2Hζ−1 〈R〉 , (S.387)

This has the integrating factor et/2λeLt :d

dt

(et/2λe−Lt 〈R〉) = et/2λe−Lt

⟨R + 1

2λR − LR

⟩

= 0,

where λ = ζ/ (4H) is the relaxation time. This has the solution

〈R (t)〉 = e−t/2λeLtR0. (S.388)

Whether or not 〈R〉 decays to zero depends on the eigenvalues of L − I/2λ, if thisis positive, 〈R〉 is a run-away process, and if this is negative, then 〈R〉 will decay tozero. Thus a strong flow will result if

eigen (L) ≥ 1/2λ, (S.389)

where eigen (L) is the maximum eigenvalue of L. Otherwise we have a weak flow.

http://dx.doi.org/10.1007/978-3-319-62000-8_7


Problem 7.5

The upper-convected Maxwell model is written as

S(p) + λ

{d


}= GI, (S.390)

Now, consider the following integral model:

S(p) (t) = G

λ

∫ t


−1 ds, (S.391)

one has

S(p) (t) = G

λI + G

λ

∫ t

−∞− 1

λe(s−t)/λCt (s)

−1 ds

+G

λ

∫ t

−∞e(s−t)/λ

[L (t)Ct (s)

−1 + Ct (s)−1LT (t)

]ds,

or that

S(p) + λ

{d


}= GI.

We conclude that (S.391) indeed solves (S.390).

Problems in Chap. 8

Problem 8.1

In the shear reversal experiments of Gadala-Maria and Acrivos (1980), it was foundthat if shearing is stopped after a steady state has been reached in a Couette device,the torque is reduced to zero instantaneously. This is due to the insignificant inertiaof the suspended particles, and the micromechanics are governed by the Stokesequation (8.3), only the present boundary conditions matter. When the flow stops,all the forces, including the torque, go to zero instantaneously.

If shearing is resumed in the same direction after a period of rest, then the torquewould attain its final value that corresponds to the resumed shear rate almost instan-taneously. This is due to the equilibrated configuration of the suspended particles hasbeen achieved and frozen in place after the flow stops. If the flow starts in the samedirection, the particles are happy to stay in their previously preferred configuration,and the forces and torques instantaneously assume their previous values.

http://dx.doi.org/10.1007/978-3-319-62000-8_7

http://dx.doi.org/10.1007/978-3-319-62000-8_8

http://dx.doi.org/10.1007/978-3-319-62000-8_8

http://dx.doi.org/10.1007/978-3-319-62000-8_8


However, if shearing is resumed in the opposite direction, then the particles find adifferent preferred configuration corresponding to the reversed shear, and the forcesand torques go through a period of adjustment to their steady state values. Zero,fading or infinite memory is a convenient description - in this suspension case, it isirrelevant to think of fading memory - the whole rheology is what matters.

Problem 8.2

Jeffery’s solution for a unit vector p directed along the major axis of a spheroidalsupsended particle obeys

p = W · p + R2 − 1

R2 + 1(D · p − D : ppp) (S.392)

= L · p − 2

R2 + 1D · p − R2 − 1

R2 + 1D : ppp,

where R is the aspect ratio of the particle (major to minor diameter ratio), L is thevelocity gradient, D = (

L + LT)/2 is the strain rate, and W = (

L − LT)/2 is the

vorticity tensor. Denote the effective velocity gradient as

L = L − 2

R2 + 1D = L − ζD, (S.393)

then (S.392) can be simplified to

p = L · p − L : ppp. (S.394)

Now, consider the linear system

Q = L · Q. (S.395)

If we denote Q = Qp, where Q is the magnitude of Q, and p is a unit vector, then

Q = Qp + Qp = QL · p.

Since p · p = 0,

Q = QL : pp,

and thus

p = L · p − L : ppp.

http://dx.doi.org/10.1007/978-3-319-62000-8_8


We conclude that (S.395) solves (S.394) and therefore solves (S.392). The parameterζ = 2/(R2 + 1) is a ‘non-affine’ parameter, representing the straining inefficiencyof the flow.

Problem 8.3

In the start-up of a simple shear flow, the velocity gradient is

[L] =⎡

⎣0 γ 00 0 00 0 0

⎤

⎦ , [L] =⎡

⎢⎣0

(1 − ζ

2

)γ 0

− ζ2 γ 0 0

0 0 0

⎤

⎥⎦ ,

where the shear rate is γ, the process Q obeys

Q1 =(1 − ζ

2

)γQ2,

Q2 = −ζ

2γQ1, (S.396)

Q3 = 0.

This implies Q3 = Q30, a constant, and

Q1 + ζ

2

(1 − ζ

2

)γ2Q1 = 0,

Q2 + ζ

2

(1 − ζ

2

)γ2Q2 = 0.

The solutions are

Q1 = Q10 cosωt +√2 − ζ

ζQ20 sinωt,

Q2 = Q20 cosωt −√

ζ

2 − ζQ10 sinωt,

where {Q10, Q20, Q30} are the initial components of Q, and the frequency of theoscillation is

ω = 1

2γ√

ζ(2 − ζ) = γR

R2 + 1.

From these results, the result for p can be obtained.

http://dx.doi.org/10.1007/978-3-319-62000-8_8


The stress is

œ = 2ηsD + 2ηsφ{AD : pppp + B (D · pp + pp · D) + CD}, (S.397)

In full we have the reduced viscosity:

〈σ12〉 − ηs γ

ηs γφ= 2Ap21 p

22 + B

(p21 + p22

)+ C, (S.398)

the reduced first normal stress difference:

N1

ηs γφ= 2Ap1 p2

(p21 − p22

), (S.399)

and the reduced second normal stress difference:

N2

ηs γφ= 2p1 p2

(Ap22 + B

). (S.400)

Thus, the particles tumble along with the flow, with a period of T = 2π(R2+1)/γR,spending most of their time aligned with the flow.

Problem 8.4

In the start-up of an elongational flowwith a positive elongational rate γ, the velocitygradient is

[L] =⎡

⎣γ 0 00 −γ/2 00 0 −γ/2

⎤

⎦ , [L] = (1 − ζ)

⎡

⎣γ 0 00 −γ/2 00 0 −γ/2

⎤

⎦ ,

Q1 = (1 − ζ) γQ1,

Q2 = −1

2(1 − ζ) γQ2,

Q3 = −1

2(1 − ζ) γQ3,

which have the solutions

Q1 = Q10 exp {(1 − ζ)γt} ,

Q2 = Q20 exp

{−1

2(1 − ζ)γt

},

http://dx.doi.org/10.1007/978-3-319-62000-8_8


Q3 = Q30 exp

{−1

2(1 − ζ)γt

},

so that the particle is quickly aligned with the flow in a time scale O(γ−1),p → (1, 0, 0) . The stress is

œ = 2ηsD + 2ηsφ{AD : pppp + B (D · pp + pp · D) + CD}, (S.401)

In full we have:

σ11

ηs γ= 2 + 2φ

[A

(p21 − 1

2p22 − 1

2p23

)p21 + 2Bp21 + C

]

σ22

ηs γ= −1 + 2φ

[A

(p21 − 1

2p22 − 1

2p23

)p22 − 2Bp22 − 1

2C

]

σ33

ηs γ= −1 + 2φ

[A

(p21 − 1

2p22 − 1

2p23

)p23 − 2Bp23 − 1

2C

]

This yields the reduced elongational viscosity

N1 − 3ηs γ

ηs γφ= 2

[A

(p21 − 1

2p22 − 1

2p23

) (p21 − p22

)+ 2B(p21 + p22

)+ 3

2C

]

At a steady state, show that the reduced elongational viscosity is given by

N1 − 3ηs γ

ηs γφ= 2A + 4B + 3C ≈ 2A = R2

ln 2R − 1.5. (S.402)

This elongational viscosity could be several order of magnitudes greater than theshear viscosity, due to the term O

(R2).

Problems in Chap. 9

Problem 9.1

We start with the 1-D system

dr

dt= v, m

dv

dt= Fc − γwDv + σwRθ (t) , r (0) = r0, v (0) = v0, (S.403)

〈θ (t)〉 = 0, 〈θ (t) θ (t + τ )〉 = δ (τ ) .

In the inertial time scale, the displacement, the force Fc and the wighting functionsmay be regarded as constant. We may re-define the velocity as

http://dx.doi.org/10.1007/978-3-319-62000-8_9

http://dx.doi.org/10.1007/978-3-319-62000-8_9


v = u + Fc/γwD,

thus eliminating Fc in the governing equation altogether, thus only deal with a linearsystem

mdv

dt= −γwDv + σwRθ (t) . (S.404)

This system has the integrating constant em−1γwDt

d

dt

(em

−1γwDtv)

= em−1γwDt

(v + m−1γwDv

)

= em−1γwDtm−1σwRθ (t) ,

which can be integrated to yield

v (t) =∫ t

0em

−1γwD(t ′−t)m−1σwRθ(t ′)dt ′. (S.405)

Its mean square velocity is

〈v (t) v (t)〉 = σ2

m2

∫ t

0

∫ t

0em

−1γwD(t ′−t)w2R

⟨θ(t ′)θ(t ′′)⟩em

−1γwD(t ′′−t)dt ′dt ′′

= σ2

m2

∫ t

0em

−1γwD(t ′−t)wRem−1γwD(t ′−t)wRdt

′,

= 1

2m−1σ2w2

Rγ−1w−1D . (S.406)

Assuming the equi-partition principle

1

2m⟨v2(t)

⟩ = 1

2kBT, (S.407)

leads directly to

σ2w2Rγ−1w−1

D = 2kBT . (S.408)

Problem 9.2

In the case of small inertia, our main stochastic system becomes

dr

dt= γ−1w−1

D Fc + γ−1w−1D σwRθ (t) . (S.409)

http://dx.doi.org/10.1007/978-3-319-62000-8_9


This may be solved in one time step Δt

Δr = γ−1w−1D FcΔt +

∫ t+Δt

tγ−1w−1

D σwRθ(t ′)dt ′. (S.410)

From this, and the properties of white noise, the drift velocity is given by

〈Δr〉Δt

= γ−1w−1D Fc. (S.411)

In addition

〈ΔrΔr〉 = O(Δt2

)+ γ−2w−2D σ2w2

R

∫ t+Δt

tdt ′

∫ t+Δt

tdt ′′

⟨θ(t ′)θ(t ′′)⟩

= O(Δt2

)+ γ−2w−2D σ2w2

RΔt

= O(Δt2

)+ 2kBTγ−1w−1D Δt.

In the last step, the equipartition principle has been used. Thus

〈ΔrΔr〉2Δt

= O (Δt) + kBTγ−1w−1D , (S.412)

and the Fokker–Planck equation is

∂

∂tW (r, t) = lim

Δt→0

∂

∂r

[( 〈ΔrΔr〉2Δt

∂

∂r− 〈Δr〉

Δt

)W (r, t)

], (S.413)

or

∂

∂tW (r, t) = ∂

∂r

[(kBT

γwD

∂

∂r− Fc

γwD

)W (r, t)

]. (S.414)

Problem 9.3

In order to focus on events on the time scale τI we can regard the restoring force Fc

as constant, so that it can be absorbed in a re-definition of the system state system:

dv

dt= −m−1γwDv + m−1σwRθ (t) . (S.415)

Noting that

2vr = d

dt

(r2), 2

(vr + v2) = d2

dt2(r2),

http://dx.doi.org/10.1007/978-3-319-62000-8_9


we find, by multiplying (S.415) with r,

m(vr + v2

)− mv2 − γwDvr = σwRθ (t) r,

or,

1

2m

d

dt

(d

dt

⟨r2⟩)− m

⟨v2⟩+ 1

2γwD

d

dt

⟨r2⟩ = σwR 〈θ (t) r〉 . (S.416)

We now define

e = d⟨r2⟩/dt.

From the temperature definition, m⟨v2⟩ = kBT ; furthermore, 〈θ (t) r〉 = 0 due to

different time scales of θ (t) and r. Thus

e + m−1γwDe = 2kBTm−1, e (0) = 0.

This has the integration factor

d

dt

(e exp

(m−1γwDt

)) = 2kBTm−1 exp

(m−1γwDt

),

which has the solution, for the assumed initial condition,

e = d

dt

⟨r2⟩ = 2kBTγ−1w−1

D

[1 − e−m−1γwDt

]. (S.417)

Consequently, if Δt � τI = O(m−1γwD

), writing Δr = r (Δt) ,

〈ΔrΔr〉2Δt

= kBTγ−1w−1D . (S.418)

Problem 9.4

Define the velocity correlation as

R (τ ) = limt→∞ 〈v (t + τ ) v (t)〉 , (S.419)

where the limit refers to large time compared to the inertial time scale, but yet smallcompared to the relaxation time scale. A formal solution of (S.415) is

v (t) =∫ t

0em

−1γwD(t ′−t)m−1σwRθ(t ′)dt ′. (S.420)

http://dx.doi.org/10.1007/978-3-319-62000-8_9


From this solution, and for τ > 0

R (τ ) = limt→∞

∫ t+τ

0dt ′

∫ t

0em

−1γwD(t ′−t−τ)(m−1σwR

)2

⟨θ(t ′)θ(t ′′)⟩em

−1γwD(t ′′−t)dt′′

(S.421)

= e−m−1γwDτ limt→∞ 〈v (t) v (t)〉 = e−m−1γwDτ R (0)

= kBTm−1e−m−1γwDτ . (S.422)

That is, the velocity correlation decays after an inertial time scale, after which thevelocity is independent to its previous state.

Next, the diffusivity can also be defined as

D = limt→∞

1

2〈v (t) r (t) + r (t) v (t)〉 .

This is equivalent to

D = limt→∞

∫ t

0〈v (t) v (t + τ )〉 dτ =

∫ ∞

0R (τ ) dτ

= kBTγ−1w−1D ,

consistent with previous results.

Problem 9.5

Now, consider the Langevin system

dridt

= vi , mdvidt

=∑

j

Fi j . (S.423)

Here, Fi j is the pairwise additive interparticle force by particle j on particle i; thisforce consists of three parts, a conservative force FC

i j , a dissipative force FDi j , and a

random force FRi j :

Fi j = FCi j + FD

i j + FRi j . (S.424)

From (S.423), the drift velocity of the process is

⟨ΔviΔt

⟩= −m−1

∑

j

(FCi j + FD

i j

) = − m−1∑

j

(FCi j − γwD

i j ei jei j · vi j). (S.425)

http://dx.doi.org/10.1007/978-3-319-62000-8_9


Furthermore,

⟨ΔvαΔvβ

⟩ = O(Δt2

)

+∑

k,m

∫ Δt

dt ′∫ Δt

dt ′′σ2

m2wR

αkwRβm

(δαβδkm + δαmδβk

)δ(t ′ − t ′′

)eαkeβm,

or

⟨ΔvαΔvβ

⟩ = O(Δt2

)

+ Δtσ2

m2

[

δαβ

∑

k

(wR

αk

)2eαkeβk + (

wRαβ

)2eαβeβα

]

,

Using the fluctuation-dissipation theorem,

⟨ΔvαΔvβ

⟩ = O(Δt2

)+ 2kBTγ

m2Δt

[

δαβ

∑

k


αβeαβeαβ

]

,

⟨ΔvαΔvβ

2Δt

⟩= γkBT

m2

[

δαβ

∑

k


αβeαβeαβ

]

. (S.426)

The Fokker–Planck equation for the process is

∂ f

∂t+∑

i

∂

∂ri· (vi f ) +

∑

i

∂

∂vi·(⟨

ΔviΔt

⟩f

)(S.427)

=∑

i, j

∂

∂vi·(⟨

ΔviΔv j

2Δt

⟩· ∂ f

∂v j

),

where the limit Δt → 0 is implied. Using the results obtained above for the driftand the diffusivity, we finally obtain the Fokker–Planck equation for the process

∂ f

∂t+∑

i

∂

∂ri· (vi f ) +

∑

i, j

∂

∂pi· (FC

i j f) = γ

∑

i, j

wDi j ei j

∂

∂pi· (ei j · vi j f

)

+ γkBT∑

i, j

wDi j ei j · ∂

∂pi·(ei j ·

(∂ f

∂pi− ∂ f

∂p j

)), (S.428)

where pi = mvi is the linear momentum of particle i.


Problem 9.6

We note that the Hamiltonian of the associate system is

H=∑

i

pi · pi2m

+ 1

2

∑

i, j

ϕ(ri j), (S.429)

noting

∂H∂pα

= pα

m= vα,

∂H∂rαβ

= −FCαβ .

Thus, the equilibrium distribution of the associate system to is

feq (χ, t) = 1

Zexp

⎡

⎣− 1

kBT

⎛

⎝∑

i

pi · pi2m

+ 1

2

∑

i, j

ϕ(ri j)⎞

⎠

⎤

⎦ (S.430)

= 1

Zexp

[− HkBT

],

where Z is a normalizing constant.We further note that,

∑

i

vi · ∂ feq∂ri

= 1

kBT

∑

i, j

vi · FCi j feq ,

∑

i, j

FCi j · ∂ feq

∂pi= − 1

kBT

∑

i, j

FCi j · vi feq ,

γ∑

i, j

wDi j ei j

∂

∂pi· (ei j · vi j feq

) = γ∑

i, j

wDi j

(− 1

kBTei j · viei j · vi j + 1

m

)feq ,

γkBT∑

i, j

wDi j ei j · ∂

∂pi·(ei j ·

(∂

∂pi− ∂

∂p j

)feq

)

= γ∑

i, j

wDi j

(− 1

m+ 1

kBTei j · viei jvi j

),

and conclude that feq is also a stationary solution (i.e., solution that is independentof time) of the Fokker–Planck equation (S.428).

http://dx.doi.org/10.1007/978-3-319-62000-8_9


Problem 9.7

The stress contributed from the conservative forces,

SC (r, t) = −1

2

∫dRFCR

{1 − 1

2R · ∇ + · · ·

}f2 (r + R, r, t) . (S.431)

Marsh used a different technique in deriving the stresses, which involves expressingthe delta function as an integral. From this, the stresses from the conservative forcesis expressed as,

SC (r, t) = −1

2

∫dv′

∫dv′′

∫FCRW2

(χ′,χ′′, t; r) dR, (S.432)

where

W2(r′, v′, r′′, v′′, t; r) =

∫ 1

0f2(r + λR, r − (1 − λ)R, v′, v′′, t

)dλ. (S.433)

Expressing f2(r + λR, r − (1 − λ)R, v′, v′′, t

)as a function of (r − εR), where

ε = 1 − λ,f2 = f2

(r + R − εR, r − εR, v′, v′′, t

),

and taking a Taylor’s series in ε,

f2 = f2(r + R, r, v′, v′′, t

)− εR · ∇ f2(r + R, r, v′, v′′, t

)+ · · ·

Integrating

W2(r′, v′, r′′, v′′, t; r) =

∫ 1

0f2(r + λR, r − (1 − λ)R, v′, v′′, t

)dε

={1 − 1

2R · ∇ + · · ·

}f2(r + R, r, v′, v′′, t

).

This, substituted in (S.432) and the velocity spaces are integrated out, leads to (S.431).The remaining part of the question is demonstrated similarly.

Problem 9.8

The “heat flux” has been shown to be

qC =⟨∑

i, j

1

4FCi j ·(vi + v j

)ri j

(1 − 1

2ri j ·∇ + · · ·

)δ(r − r j

)⟩

. (S.434)

http://dx.doi.org/10.1007/978-3-319-62000-8_9

http://dx.doi.org/10.1007/978-3-319-62000-8_9


This can be demonstrated, in an almost verbatim manner to the treatment of thestress, to be

qC (r, t) = 1

4

∫dR

∫dv

∫dv′FC (R) · (v + v′)R

{1 − 1

2R · ∇ + · · ·

}

. f2(r + R, r, v, v′, t

). (S.435)

Problem 9.9

The stress contributed from the damping forces is,

SD (r, t) = −1

2

∫dR

∫dv

∫dv′γwD (R) RR · (v − v′)R {1 + O (R)}

. f2(r + R, r, v, v′, t

).

For a homogeneously shear flow, the velocity gradient is constant. By taking theparticles’ velocities as the fluid’s velocities at the particles’ locations: v− v′ = LR,

where L is the velocity gradient, together with Groot and Warren’s approximation,f2 = n2 (1 + O (R)) , the stress contributed by the damping forces is

SD,αβ (r, t) = −γn2

2

⟨Rα Rβ Ri R j Li j

∫R2wD (R) 4πR2dR

⟩

= −2πγn2

15


)Li j

∫R4wD (R) dR.

Here we have assume a completely isotropic distribution for the structure tensor:

⟨Rα Rβ Ri R j

⟩= 1

15


).

For the standard weighting function (9.88) adopted in DPD, one has

SD,αβ (r, t) = −2πγn2

15


) ∫ rc

0R4 (1 − R/rc)

2 dR

= 2πγn2r5c1575


), (S.436)

and consequently the damping-contributed viscosities are given by

ηD = 2πγn2r5c1575

, ζD = 5

3ηD = 2πγn2r5c

945. (S.437)

http://dx.doi.org/10.1007/978-3-319-62000-8_9

http://dx.doi.org/10.1007/978-3-319-62000-8_9


For the modified weighting function (9.125) with s = 1/2,

SD,αβ (r, t) = 512πγn2r5c51975


)(S.438)

and consequently

ηD = 512πγn2r5c51975

, ζD = 5

3ηD = 512πγn2r5c

31185. (S.439)

Problem 9.10

By using the kinetic theory or by considering the case of a homogeneously shearflow (as shown above), one can obtain an estimate for the viscosity and diffusivity

η = ηK + ηD = ρD

2+ γn2

[R2wD

]R

30,

D = 3kBT

nγ[wD

]R

,

where

[wD

]R =

∫ rc

04πR2wD (R) dR,

[R2wD

]R =

∫ rc

04πR4wD (R) dR .

With the standard weighting function,

[wD

]R

= 2π

15r3c ,

[R2wD

]R

= 4π

105r5c ,

which leads to

η = 45mkT


1575, D = 45kBT

2πγnr3c.

Problem 9.11

With the modified weighting function (s = 1/2),

[wD

]R = 64π

105r3c ,

[R2wD

]R = 1024π

3465r5c ,

http://dx.doi.org/10.1007/978-3-319-62000-8_9

http://dx.doi.org/10.1007/978-3-319-62000-8_9

http://dx.doi.org/10.1007/978-3-319-62000-8_9


which leads to

η = 315mkBT


51975, D = 315kBT

64πγnr3c.

Problem 9.12

%%%%%%%%%%%%%%%%%% INITILISATION %%%%%%%%%%%%%%%%% Array to store the number of times that the particles cross boundariesncc=zeros(2,nFreeAtom);% Array to store the mean square displacements against timeMSDs=zeros(stepSample,2);

%%%%%%%%%%% COMPUTATION %%%%%%%%%%%%%%%%%%%%%%%%for i=1:nFreeAtom

if r(1,i) > 0.5*Lxr(1,i)=r(1,i)-Lx; ncc(1,i) = ncc(1,i)+1;

elseif r(1,i) < -0.5*Lxr(1,i)=r(1,i)+Lx; ncc(1,i) = ncc(1,i)-1;

endif r(2,i) > 0.5*Ly

r(2,i)=r(2,i)-Ly; ncc(2,i) = ncc(2,i)+1;elseif r(2,i) < -0.5*Ly

r(2,i)=r(2,i)+Ly; ncc(2,i) = ncc(2,i)-1;end

end

if stepCount == stepEquil% Actual positions of particles at the thermal equilibriumr0=r+[Lx*ncc(1,:);Ly*ncc(2,:)];

end

if stepCount > stepEquiltimeNow=timeNow+deltaT;% Actual positions of particles at the current time trt = r+[Lx*ncc(1,:);Ly*ncc(2,:)];% Relative positionsDeltaR = rt-r0;% Mean Square DisplacementMSD = sum(sum(DeltaR.ˆ2,1))/nFreeAtom;MSDs(round(timeNow/deltaT),:) = [timeNow,MSD];

end

%%%%%%%%%%%%%%%% OUTPUT %%%%%%%%%%%%%%%%%%% Plot MSD and Diffusivity of DPD particlesfigure; plot(MSDs(:,1),MSDs(:,2),’b-’)xlabel(’Time’); ylabel(’MSD’)title(’Mean Square Displacements of DPD particles’);figure; plot(MSDs(:,1),(1/4)*(MSDs(:,2)./MSDs(:,1)),’b-’);xlabel(’Time’); ylabel(’D’);title(’Self-diffusion coefficient of DPD particles’);

http://dx.doi.org/10.1007/978-3-319-62000-8_9


Problem 9.13

%%%%%%%%%%%%%%%%%%%% INPUT %%%%%%%%%%%%%%%%%%%%%%% Number of annular elements and the domain sizenRDF=100; rRDF=1;

%%%%%%%%%%%%%%%%%% INITILISATION %%%%%%%%%%%%%%%%% Array to store RDF values against timeRDFs=zeros(stepSample,nRDF);

%%%%%%%%%%% COMPUTATION %%%%%%%%%%%%%%%%%%%%%%%%if stepCount > stepEquil

timeNow=timeNow+deltaT;

% Pick up particles in the core region for computing their RDFid = find(abs(r(1,:))<Lx/2-rRDF & abs(r(2,:))<Ly/2-rRDF);rs = r(:,id); ns=size(rs,2); RDF=zeros(ns,nRDF);for i=1:ns

DeltaR = [r(1,:)-rs(1,i);r(2,:)-rs(2,i)];DistR = sqrt(sum(DeltaR.ˆ2,1));for j=1:nRDF

Rmin=(j-1)*(rRDF/nRDF); Rmax=j*(rRDF/nRDF);id=find(DistR>Rmin & DistR<=Rmax); RDF(i,j)=length(id);

endendRDFs(round(timeNow/deltaT),:) = mean(RDF,1);

end

%%%%%%%%%%%%%%%% OUTPUT %%%%%%%%%%%%%%%%%%% Plot RDF of the DPD particlesR = mean(RDFs,1); dq=(rRDF/nRDF); q=dq*(1:nRDF);R(1)=R(1)/(numDen*pi*dqˆ2);R(2:end)=R(2:end)./(numDen*2*pi*(q(2:end)-dq)*dq);figure; plot(q,R,’b-o’)xlabel(’Distance’),ylabel(’RDF’)title(’Radial distribution function for DPD particles’)

http://dx.doi.org/10.1007/978-3-319-62000-8_9

References

1. M. Arienti, W. Pan, X. Li, G. Karniadakis, J. Chem. Phys. 134, 204114 (2011)2. H. Ashi, Numerical Methods for Stiff Systems. Ph.D thesis, University of Nottingham (Not-

tingham 1998)3. E.B. Bagley, Food Extrusion Science and Technology (Marcel Dekker, New York, 1992)4. G.K. Batchelor, J. Fluid Mech. 44, 545–570 (1970)5. X. Bian, M. Ellero, Comput. Phys. Commun. 85(1), 53–62 (2014)6. R.B. Bird, R.C. Armstrong, O. Hassager, Dynamics of Polymeric Liquids. Fluid Mechanics,

vol. 1, 2nd edn. (Wiley, New York, 1987)7. E. Bodewig, Matrix Calculus (North-Holland Publishing, Amsterdam, 1956)8. E. Boek, P. Coveney, H. Lekkerkerker, J. Phys. Cond. Matt. 8(47), 9509–9512 (1996)9. D.V. Boger, R. Binnington, Trans. Soc. Rheol. 21, 515–534 (1977)10. D.V. Boger, K. Walters, Rheological Phenomena in Focus (Elsevier, Amsterdam, 1993)11. J.F. Brady, Int. J. Multiphase Flow 10(1), 113–114 (1984)12. F. Brauer, J.A. Nohel, Qualitative Theory of Ordinary Differential Equations (Dover, New

York, 1989)13. C. Bustamante, J.F. Marko, E.D. Siggia, S. Smith, Science 265, 1500–1600 (1994)14. S. Chandrasekhar, Rev. Mod. Phys. 15, 1–89 (1943)15. S. Chen, N. Phan-Thien, B.C. Khoo, X.-J. Fan, Phys. Fluids 18, 103605 (2006)16. B.D. Coleman, Arch. Rat. Mech. Anal. 9, 273–300 (1962)17. W.O. Criminale Jr., J.L. Ericksen, G.L. Filbey Jr., Arch. Rat. Mech. Anal. 1, 410–417 (1957)18. M. Doi, S.F. Edwards, The Theory of Polymer Dynamics (Oxford University Press, Oxford,

1988)19. D. Duong-Hong, N. Phan-Thien, X.-J. Fan, Comput. Mech. 35, 24–29 (2004)20. A. Einstein, in Investigations on the Theory of the Brownian Movement, ed. by R. Fürth, R.,

Transl. A.D. Cowper (Dover, New York, 1956)21. J.L. Ericksen, in Viscoelasticity - Phenomenological Aspects, ed. by J.T. Bergen (Academic

Press, New York 1960)22. P. Español, P. Warren, Europhys. Lett. 30(4), 191–196 (1995)23. E. Fahy, G.F. Smith, J. Non-Newt, Fluid Mech. 7, 33–43 (1980)24. X.-J. Fan, N. Phan-Thien, S. Chen, X. Wu, T.Y. Ng, Phys. Fluids 18(6), 063102 (2006)25. X.-J. Fan, N. Phan-Thien, R. Zheng, J. Non-Newtonian Fluid Mech. 84(2–3), 257–274 (1999)26. P.J. Flory, Statistical Mechanics of Chain Molecules (Wiley Interscience, New York, 1969)27. F.P. Folgar, C.L. Tucker Jr., J. Reinf, Plast. Compos. 3, 98–119 (1984)28. U. Frisch, B. Hasslachcher, Y. Pomeau, Phys. Rev. Lett. 56, 1505–1508 (1986)


297

298 References

29. R.M. Füchslin, H. Fellermann, A. Eriksson, H.-J. Ziock, J. Chem. Phys. 130, 214102 (2009)30. F. Gadala-Maria, A. Acrivos, J. Rheol. 24, 799–814 (1980)31. H. Giesekus, Rheol. Acta 3, 59–81 (1963)32. A.E. Green, R.S. Rivlin, Arch. Rat. Mech. Anal. 1, 1–21 (1957)33. R.D. Groot, P.B. Warren, J. Chem. Phys. 107, 4423–4435 (1997)34. M.E. Gurtin, An Introduction to Continuum Mechanics (Academic Press, New York, 1981)35. J. Happel, H. Brenner, Low Reynolds Number Hydrodynamics (Noordhoff International Pub-

lishing, Leyden, 1973)36. E.J. Hinch, J. Fluid Mech. 72, 499–511 (1975)37. E.J. Hinch, L.G. Leal, J. Fluid Mech. 52, 683–712 (1972)38. P. Hoogerbrugge, J. Koelman, Europhys. Lett. 19(3), 155–160 (1992)39. R.R. Huilgol, N. Phan-Thien, Fluid Mechanics of Viscoelasticity: General Principles, Consti-

tutive Modelling, Analytical and Numerical Techniques (Elsevier, Amsterdam, 1997)40. J.H. Irving, J.G. Kirkwood, J. Chem. Phys. 18, 817–829 (1950)41. G.B. Jeffery, Proc. R. Soc. Lond. A 102, 161–179 (1922)42. E.E. Keaveny, I.V. Pivkin, M. Maxey, G.E. Karniadakis, J. Chem. Phys. 123, 104107 (2005)43. O.D. Kellogg, Foundations of Potential Theory (Dover, New York, 1954)44. Y. Kong, C. Manke, W. Madden, A. Schlijper, J. Chem. Phys. 107(2), 592–602 (1997)45. I.M. Krieger, T.J. Dougherty, Trans. Soc. Rheol. 3, 137–152 (1959)46. W. Kuhn, Kolloid-Zeit. 68, 2–15 (1934)47. L.D. Landau, E.M. Lifshitz, Fluid Mechanics, translated by J.B. Sykes, W.H. Reid (Pergamon

Press, New York 1959)48. R.G. Larson, Constitutive Equations for Polymer Melts and Solutions (Butterworth Publishers,

Boston, 1988)49. R.G. Larson, H. Hu, D.E. Smith, S. Chu, J. Rheol. 43(2), 267–304 (1999)50. R.G. Larson, T.T. Perkins, D.E. Smith, S. Chu, Phys. Rev. E 55(2), 1794–1797 (1997)51. L.G. Leal, E.J. Hinch, Rheol. Acta 12, 127–132 (1973)52. Y.K. Lin, Probabilistic Theory of Structural Dynamics (McGraw-Hill, New York, 1967)53. A.S. Lodge, Elastic Liquids (Academic Press, New York, 1964)54. A.E.H. Love, A Treatise in the Mathematical Theory of Elasticity, Note B (Dover, New York,

1944)55. N. Mai-Duy, N. Phan-Thien, B.C. Khoo, Comput. Phys. Commun. 189, 37–46 (2015)56. C. Marsh, Theoretical Aspects of Dissipative Particle Dynamics. Ph.D thesis, University of

Oxford (Oxford 1998)57. A.B. Metzner, J. Rheol. 29, 739–775 (1985)58. M. Mooney, J. Appl. Phys. 11, 582–592 (1940)59. F.A. Morrison, Understanding Rheology: Topic in Chemical Engineering (Oxford Unversity

Press, New York, 2001)60. P. Ninkunen, M. Karttunen, I. Vattulainen, Comput. Phys. Commun. 153, 407–423 (2003)61. W. Noll, J. Ratl. Mech. Anal. 4, 2–81 (1955)62. W. Noll, Ratl. Mech. Anal. 2, 197–226 (1958)63. L. Novik, P. Coveney, Int. J. Mod. Phys. C 8(4), 909–918 (1997)64. J.G. Oldroyd, Proc. Roy. Soc. Lond. A 200, 523–541 (1950)65. D. Pan, N. Phan-Thien, N. Mai-Duy, B.C. Khoo, J. Comput. Phys. 242, 196–210 (2013)66. W. Pan, B. Caswell, G.E. Karniadakis, Langmuir 26(1), 133–142 (2010)67. W. Pan, I.V. Pivkin, G.E. Karniadakis, EPL 84, 10012 (2008)68. N. Phan-Thien, S. Kim,Microstructures in ElasticMedia: Principles andComputationalMeth-

ods (Oxford University Press, New York, 1994)69. N. Phan-Thien, N. Mai-Duy, B.C. Khoo, J. Rheol. 58, 839–867 (2014)70. A.C. Pipkin, Lectures on Viscoelasticity Theory, 2nd edn. (Springer, Berlin, 1986)71. A.C. Pipkin, R.S. Rivlin, Arch. Rational Mech. Anal. 4, 129–144 (1959)72. A.C. Pipkin, R.I. Tanner, inMechanics Today, vol. 1, ed. by S. Nemat-Nasser (Pergamon, New

York, 1972), pp. 262–321

References 299

73. I.V. Pivkin, B. Caswell, G.E. Karniadakis, Reviews in Computational Chemistry 27, 85–110(2011)

74. B. Rabinowitch, Z. Physik, Chem. A 145, 1–26 (1929)75. D. Rapaport, The Art of Molecular Dynamics Simulation (Cambridge University Press, Cam-

bridge, 1995)76. L.E. Reichl, AModern Course in Statistical Physics (University of Texas Press, Austin, Texas,

1980)77. M. Revenga, I. Zuñiga, P. Español, Int. J. Mod. Phys. C 9(8), 1319–1328 (1998)78. R.S. Rivlin, J.L. Ericksen, J. Ration. Mech. Anal. 4, 323–425 (1955)79. G. Ryskin, J.M. Rallison, J. Fluid Mech. 99, 513–529 (1980)80. H. See, J. Phys. D Appl. Phys. 33, 1625–1633 (2000)81. L.F. Shampine, C.W. Gear, SIAM Rev. 21(1), 1–17 (1979)82. A.J.M. Spencer, in Continuum Physics, vol. 1, ed. by A.C. Eringen (Academic Press, New

York, 1971)83. R.I. Tanner, Phys. Fluids 9, 1246–1247 (1966)84. R.I. Tanner, J. Polym. Sci. A 2(8), 2067–2078 (1970)85. R.I. Tanner, Engineering Rheology, Revised edn. (Oxford University Press, Oxford, 1992)86. R.I. Tanner, K. Walters, Rheology: An Historical Perspective (Elsevier, Amsterdam, 1998)87. L.R.G. Treloar, The Physics of Rubber Elasticity, 3rd edn. (Clarendon Press, Oxford, 1975)88. L.E. Wedgewood, D.N. Ostrov, R.B. Bird, J. Non-Newtonian Fluid Mech. 40, 119–139 (1991)89. L. Weese, Comput. Phys. Commun. 77, 429–440 (1993)90. H. Weyl, Classical Groups (Princeton University Press, New Jersey, 1946)91. S. Willemsen, H. Hoefsloot, P. Iedema, Int. J. Mod. Phys. C 11(5), 881–890 (2000)92. W.-L. Yin, A.C. Pipkin, Arch. Rat. Mech. Anal. 37, 111–135 (1970)

Index

AAcceleration field, 44Acrivos, 156Alternating tensor, 2Arienti, 159Armstrong, 29Axial fanned flow, 116

BBatchelor, 28, 135Bingham fluid, 96Bird, 29, 135Body force, 57Boek, 159Boltzmann, 99Boyle, 65Brauer, 49Brenner, 153Brownian force, 128Bulk viscosity, 190

CCarreau model, 96Cauchy stress tensor, 58Cauchy–Green tensor

left, 47right, 47

Cayley–Hamilton theorem, 15Chandrasekhar, 133, 162Coleman, 114Configuration space, 131Conservation

of angular momentum, 61of energy, 61of linear momentum, 60

of mass, 56Conservative force, 164Controllable flow, 118Convergent-divergent channel flow, 211Correspondence principle, 105Couette flow, 208Coveney, 159Curl of a vector, 19Current configuration, 43Cylindrical coordinates, 20

DD’Alembert, 66Deborah number, 71Deformation gradient tensor, 45Derivative of

detA, 17tr A, 17tr A2, 17

Die swell, 33delay, 34

Diffusivity, 190Dilatant, 96Dissipative force, 164Divergence

cylindrical coordinates, 21of a tensor, 19of a vector, 19spherical coordinates, 22

Doi, 155Dumbbell model, 126, 138Duong, 201Dyadic notation, 11


301

302 Index

EEdwards, 155Eigenvalue, 14Eigenvector, 14Einstein, 2, 154Elongational viscosity, 39Energy equation, 176Entropic spring force, 127Equation of change, 170Equi-partition energy principle, 131Equi-partition principle, 161Ericksen, 113

recursive, 52Español, 159Euler, 44, 66Eulerian description, 44Exclusion size, 186Exponential time differencing scheme, 197

FFading memory, 39, 85Fahy, 80Fan, 159, 188, 201FENE dumbbell, 145FENE model, 203Field, 18Finger tensor, 48First normal stress coefficient, 117First normal stress difference, 31, 117Fluctuation-dissipation theorem, 130, 133,

161, 165Fokker–Planck equation, 133, 168Fokker-Planck-Smoluchowski equation,

162Fourier, 67Frame indifference, 74Frame of reference, 1Frame rotation, 3Free-draining assumption, 127Freely rotating chain, 125Frisch, 159

GGadala-Maria, 156Gauss, 23Gauss divergence theorem, 23Generalised Newtonian fluid, 95Generalised strain rate, 95Giesekus, 88Gradient

cylindrical coordinates, 20of a scalar, 8, 18

of a vector, 9, 18of velocity, 9operator, 18spherical coordinates, 22transpose of, 19

Green, 89Green–Rivlin expansion, 89Gurtin, 6, 59

HHappel, 153Hassager, 29Heat conduction, 67Helical flow, 116Helicoidal flow, 116High damping limit, 179Hinch, 130, 154, 155Hoogerbrugge, 159Hooke, 67Hookean solid, 67Huilgol, 43, 49

IIll-conditioned, 105Inelastic fluid, 95Integrity bases, 77Invariants, 13Inviscid fluid, 66Irving, 174Isotropic elastic materials, 82

JJeffery, 154

KKellogg, 24Kelvin–Voigt body, 99Kinetic pressure tensor, 172Kirkwood, 174Koelman, 159Kong, 159Kramers form, 138Kronecker delta, 1Kuhn, 126, 138Kuhn segment, 126

LLagrange, 43Laminar shear flows, 113Landau, 135, 161

Index 303

Langevin distribution, 126Langevin equation, 130, 160, 163Langevin spring law, 128Leal, 154, 155Leibniz, 25

formula, 25Lifshitz, 135, 161Linear vector function, 10Linear viscoelastic, 99Liouville equation, 162Lodge, 143

MMarkovian process, 125Markovitz, 32Marsh, 159, 176, 179Material derivative, 45Matrix

fundamental, 49orthogonal, 4rotation, 5unimodular, 80

Maxwell, 99relaxation spectrum, 100

Metzner, 149, 150Meyer, 99Mooney, 82

material, 82Motion, 43

NNavier, 67Neo-Hookean material, 83Newton, 69Newtonian fluid, 69Nikunena, 196Nohel, 49Noll, 73, 83No-slip boundary conditions, 201Novik, 159

OObjectivity, 73Oldroyd, 73, 142Oscillatory shear flow, 49

PPartially controllable flow, 118Path lines, 48Péclet, 149

Perfect fluid, 66Perfect gas law, 65Phan-Thien, 43, 49, 145, 159, 188, 201Phase space, 131Pipe flow, 120Pipkin, 72, 80, 88, 114Pipkin’s diagram, 72Plastic fluid, 96Poiseuille, 31Poiseuille flow, 211Polar decomposition theorem

left, 15right, 15

Polymer solutions, 123Power-law, 97Principle of determinism, 77Principle of local action, 77Pseudoplastic, 96

QQuotient rule, 9

RRabinowitch, 121Radial distribution function, 187Rallison, 150Random force, 165Random-walk model, 123Recoil, 39Rectilinear flow, 115Reference configuration, 43Relative strain tensors, 48Relaxation modulus, 38, 100, 104Relaxation spectrum, 104Revenga, 201Reynolds, 54

transport theorem, 54Rivlin, 80, 89

recursive, 52Rivlin–Ericksen tensor, 50

recursive, 52Rod-climbing, 33Rot of a vector, 19Rotlet, 151Rouse model, 126, 146Ryskin, 150

SSchmidt number, 188, 190Second normal stress coefficient, 117Second normal stress difference, 31, 117

304 Index

Second normal stress differences, 35Second-order fluid, 85

plane creeping flow, 88potential flow, 88

Shear thickening, 96Shear thinning, 96Simple fluid, 84Simple material, 83Small strain, 53Smith, 80Smoluchowski, 133Smoothed-out Brownian force, 134Spectral decomposition, 14Spencer, 80Spherical coordinates, 21Spring-dashpot models, 108Square root theorem, 14Standard DPD parameters, 185Steady parallel flow, 115Step strain, 102Stokes, 69Stokes curl theorem, 24Stokes–Einstein relation, 146Storage modulus, 36Strain rate tensor, 9, 45Strain-time separability, 38Stress relaxation, 102Stress tensor, 9Stresslet, 151Strong flow, 125, 140Summation convention, 2Surface force, 57Suspension

of spheroids, 152Symmetry group, 81

TTanner, 29, 33, 72, 73, 93, 99, 145Tensors, 5

anti-symmetric, 7contraction, 11decomposition, 7derivative, 16dot product of, 12invariants of, 13inverse of, 13

norm of, 13outer product of, 5product of, 12scalar product of, 13skew, 7substitution, 11symmetric, 7trace, 11transpose, 7, 12unit, 8

Third-order fluid, 85Torsional flow, 119Trouton ratio, 39

UUpper-convected Maxwell model, 141

VVelocity field, 44Velocity gradient tensor, 44Velocity Verlet scheme, 196Viscometric flows, 113Viscometric functions, 117Viscosity, 29, 117, 190

dynamic, 36shear-thinning, 30

Vorticity tensor, 9, 45

WWalter, 73, 99Warner spring law, 128Warren, 159Weak flow, 140Wedgewood, 203Weese, 105Weighting function, 188Weissenberg number, 71Weyl, 78Willemsen, 201Wormlike model, 205

YYin, 114

nhan phan-thien nam mai-duy understanding...

Documents