104758472-gurtin-intro-to-continuum-mechanics.pdf

An Introduction to Continuum Mechanics

This is Volume 158 inMATHEMATICS IN SCIENCE AND ENGINEERINGA Series of Monographs and TextbooksEdited by RICHARD BELLMAN, University of Southern California

The complete listing of books in this series is available from the Publisherupon request.

An Introduction toContinuum Mechanics

MORTON E. GURTIN

Department of MathematicsCarnegie-Mellon University

Pittsburgh. Pennsylvania

1981

ACADEMIC PRESS

A Subsidiary of Harcourt Brace Jovanovich. Publishers

New York London Toronto Sydney San Francisco

For Amy and Bill

eOPYRIGHT © 1981, BY ACADEMIC PRESS, INC.ALL RIGHTS RESERVED.NO PART OF THIS PUBLICATION MAY BE REPRODUCED ORTRANSMITTED IN ANY FORM OR BY ANY MEANS, ELECTRONICOR MECHANICAL, INCLUDING PHOTOCOPY, RECORDING, OR ANYINFORMATION STORAGE AND RETRIEVAL SYSTEM, WITHOUTPERMISSION IN WRITING FROM THE PUBLISHER.

ACADEMIC PRESS, INC.III Fifth Avenue, New York, New York 10003

United Kingdom Edition published byACADEMIC PRESS, INC. (LONDON) LTD.24/28 Oval Road, London NWI 7DX

library of Congress Cataloging in PUblication Data

Gurtin, Morton E.An introduction to continuum mechanics.

II. Series.I. Title.BO-2335AACR2

(Mathematics in science and engineering)Bibliography: p.Includes index.1. Continuum mechanics.

QA80B.2.GB6 531ISBN 0-12-309750-9

AMS (MOS) 1970 Subject Classification: 73899

PRINTED IN THE UNITED STATES OF AMERICA

81 82 83 84 9 8 7 6 S 4 3 2 1

Contents

PrefaceAcknowledgments

Chapter I Tensor Algebra

1. Points. Vectors. Tensors2. Spectral Theorem. Cayley-Hamilton Theorem. Polar

Decomposition TheoremSelected References

Chapter IT Tensor Analysis

3. Differentiation4. Gradient. Divergence. Curl5. The Divergence Theorem. Stokes' Theorem

Selected References

Chapter ill Kinematics

6. Bodies. Deformations. Strain7. Small Deformations

v

ixxi

1117

19293740

4154

vi CONTENTS

8. Motions9. Types of Motions. Spin. Rate of Stretching

10. Transport Theorems. Volume. Isochoric Motions11. Spin. Circulation. Vorticity

Selected References

Chapter IV Mass. Momentum

12. Conservation of Mass13. Linear and Angular Momentum. Center of Mass

Selected Reference

Chapter V Force

14. Force. Stress. Balance of Momentum15. Consequences of Momentum Balance

Selected References

Chapter VI Constitutive Assumptions. Inviscid Fluids

16. Constitutive Assumptions17. Ideal Fluids18. Steady, Plane, Irrotational Flow of an Ideal Fluid19. Elastic Fluids

Selected References

Chapter VII Change in Observer. Invariance of Material Response

20. Change in Observer21. Invariance under a Change in Observer

Selected References

Chapter VIII Newtonian Fluids. The Navier-Stokes Equations

22. Newtonian Fluids23. Some Simple Solutions for Plane Steady Flow24. Uniqueness and Stability

Selected References

5867778085

879295

97108113

115119122130137

139143145

147156159164

CONTENTS

ChapterIX Finite Elasticity

25. Elastic Bodies26. Simple Shear of a Homogeneous and Isotropic Elastic Body27. The Piola-Kirchhoff Stress28. Hyperelastic Bodies29. The Elasticity Tensor

Selected References

ChapterX Linear Elasticity

30. Derivation of the Linear Theory31. Some Simple Solutions32. Linear Elastostatics33. Bending and Torsion34. Linear Elastodynarnics35. Progressive Waves

Selected References

Appendix

36. The Exponential Function37. Isotropic Functions38. General Scheme of Notation

References

Hints for Selected Exercises

Index

Vll

165175178184194198

199201205214220223226

227229239

243

247

261

This page intentionally left blank

Preface

This book presents an introduction to the classical theories of continuum me-chanics; in particular, to the theories of ideal, compressible, and viscous fluids,and to the linear and nonlinear theories of elasticity. These theories are impor-tant, not only because they are applicable to a majority of the problems in contin-uum mechanics arising in practice, but because they form a solid base uponwhich one can readily construct more complex theories of material behavior.Further, although attention is limited to the classical theories, the treatment ismodem with a major emphasis on foundations and structure.

I have used direct-as opposed to component-notation throughout. Whileengineers and physicists might at first fmd this a bit difficult, I believe that theadditionaleffort required is more than compensated for by the resulting gain inclarity and insight. For those not familiar with direct notation, and to make thebook reasonably self-contained, I have included two lengthy chapters on tensoralgebra and analysis.

The book is designed to form a one- or two-semester course in continuummechanics at the first-year graduate level and is based on courses I have taughtoverthe past fifteen years to mathematicians, engineers, and physicists at BrownUniversity and at Carnegie-Mellon University.

With the exception of a list of general references at the end of each chapter, Ihaveomitted almost all reference to the literature. Those interested in questionsof priority or history are referred to the encyclopedia articles of Truesdell andToupin [1], Truesdell and Noll [1], Serrin, [1], and Gurtin [1].

ix

Acknowledgments

lowe so much to so many people. To begin, this book would never have beenwritten without the inspiration and influence of Eli Sternberg, Walter Noll, andClifford Truesdell, who, through discussions, courses, and writings, both pub-lished and unpublished, have endowed me with the background necessary for aproject of this type.

I want to thank D. Carlson, L. Martins, I. Murdoch, P. Podio-Guidugli, E.Sternberg, and W. Williams for their detailed criticism of the manuscript, and fortheir many valuable discussions concerning the organization and presentation ofthe material. I would also like to thank J. Anderson, J. Marsden, L. Murphy, D.Owen, D. Reynolds, R. Sampaio, K. Spear, S. Spector, and R. Temam forvaluable comments.

The proof (p. 23) that the square-root function (on tensors) is smooth comesfrom unpublished lecture notes of Noll; the proof of the smoothness lemma (p.65) is due to Martins; Section 33, concerning bending and torsion of linear elas-tic bodies, is based on unpublished lecture notes of Sternberg; the first paragraphof Section 21 , concerning invariance under a change in observer, is a paraphras-ing of a remark by Truesdell [2].

Because this book has been written over a period of fifteen years, and becauseof a "fading memory", I have probably neglected to mention many othersources from which I have profited; if so, I apologize.

Finally, I am extremely grateful to Nancy Colmer for her careful and accuratetyping (and retyping!) of the manuscript.

xi

CHAPTER

I

Tensor Algebra

1. POINTS. VECTORS. TENSORS

The space under consideration will always be a three-dimensionaleuclidean point space $. The term point will be reserved for elements of $, theterm vector for elements of the associated vector space "Y. The difference

v=y-x

of two points is a vector (Fig. 1); the sum

y=x+v

ofa point x and a vector v is a point. The sum of two points is not a meaningfulconcept.

Figure 1 x

y

The inner product of two vectors u and vwill be designated by u . v, and wedefine

u2 = u· u.

2 I. TENSOR ALGEBRA

We use the symbol ~ for the reals, ~+ for the strictly positive reals.

Representation Theorem for Linear Forms. 1 Let l/J: "1/ ~ ~ be linear. Thenthere exists a unique vector a such that

l/J(v) = a' vfor every vector v.

A cartesian coordinate frame consists of an orthonormal basis {e;} ={el , e2, e3} together with a point 0 called the origin. We assume once and forall that a single, fixed cartesian coordinate frame is given. The (cartesian)components of a vector U are given by

Ui = u'ei ,

so that

U·V=LUiVi.i

Similarly, the coordinates of a point x are

Xi = (x - o)·ei •

The span sp{u, v, ... , w} ofa set {u, v, ... , w} ofvectors is the subspace of"1/consisting of all linear combinations of these vectors:

sp{u, v, ... , w} = {oeu + {3v + ... + yWIIX, {3, ... , y E ~}.

(We will also use this notation for vector spaces other than "1/.)Given a vector v, we write

{v}l- = {ulu- v = O}

for the subspace of "1/ consisting of all vectors perpendicular to v.We use the term tensor as a synonym for "linear transformation from "1-

into "1/." Thus a tensor S is a linear map that assigns to each vector u a vector

v = Su.

The set of all tensors forms a vector space if addition and scalar multiplicationare defined pointwise; that is, S + T and IXS (IX E ~) are the tensors defined by

(S + T)v = Sv + Tv,

(IXS)V = IX(SV).

The zero element in this space is the zero tensor 0 which maps every vector vinto the zero vector:

Ov = O.

1 Cf., e.g., Halmos [I, §67].


Another important tensor is the identity I defined by

Iv = v

for every vector v.The product ST of two tensors is the tensor

ST = SoT;

that is,

(ST)v = S(Tv)

for all v. We use the standard notation

3

S2 = SS, etc.

Generally, ST #- TS. If ST = TS, we say that Sand T commute.We write ST for the transpose of S; ST is the unique tensor with the

property

for all vectors u and v.It then follows that

(S + T)T = ST + TT,

(ST)T = TTST,

(ST)T = S.

A tensor S is symmetric if

skew if

(1)

Every tensor S can be expressed uniquely as the sum of a symmetric tensor Eand a skew tensor W:

S = E + W;

in fact,

E =!(S + ST),

W = t(S - ST).

We call E the symmetric part of S, W the skew part of S.

4 I. TENSOR ALGEBRA

The tensorproduct a (8) b of two vectors a and b is the tensor that assignsto each vector v the vector (b· v)a:

(a (8) b)v = (b· v)a.

Then

(a (8) b)T = (b (8) a),

(a (8) b)(c (8) d) = (b· c)a (8) d,

{O,

(e i (8) ei)(ej (8) e) =e, (8) e.,

L e, (8) e, = I.i

ii'j

i = i.(2)

Let e be a unit vector. Then e (8) e applied to a vector v gives

(v· e)e,

which is the projection of v in the direction of e, while I - e (8)e applied to vgives

v - (v· e)e,

which is the projection of v onto the plane perpendicular to e (Fig. 2).

Figure 2

The components Sij of a tensor S are defined by

Sij = ei . Sej.

With this definition v = Su is equivalent to

V; = L SijUj•j

Further,

S = L Sije; (8) eji.j

and

(3)


We write [S] for the matrix

5

It then follows that

CST] = [SF,

CST] = [S] [T],

and

[10 0]

[1]=010.001

The trace is the linear operation that assigns to each tensor S a scalar tr Sand satisfies

tr(u ® v) = u· v

for all vectors u and v. By (3) and the linearity of tr,

tr S = tr(~ Sije; ® ej) = ~ Sij tree; ® e)l,) I,)

= I Sije;· ej = I s..i,j i

Thus the trace is well defined:

trS = IS;;.i

This operation has the following properties:

tr ST = tr S,(4)

tr(ST) = tr(TS).

The space of all tensors has a natural inner product

which in components has the form

S·T = IS;J'ij'i,j

6

Then

I. TENSOR ALGEBRA

(5)

loS = tr S,

R 0 (ST) = (STR) 0 T = (RT T)0 S,

U 0 Sv = So (u ® v),

(a ® b) 0 (u ® v) = (a 0 u)(b 0 v).

More important is the following

Proposition

(a) IfS is symmetric,

SoT = SoTT = SO H(T + TT)}.

(b) IfW is skew,

WoT = -WoTT = WO {~T - TT)}.

(c) IfS is symmetric and W skew,

SoW=O.

(d) 1fT 0 S = Ofor every tensor S, then T = O.(e) 1fT 0 S = Ofor every symmetric S, then T is skew.(f) 1fT 0 W = 0 for every skew W, then T is symmetric.

We define the determinant of a tensor S to be the determinant of the matrix[S]:

det S = det[S].

This definition is independent of our choice of basis {e.}.A tensor S is invertible if there exists a tensor S- 1, called the inverse of S,

such that

SS-l = S-lS = I.

It follows that S is invertible if and only if det S =1= O.The identities

det(ST) = (det S)(det T),

det ST = det S,

det(S-l) = (det S)- \

(ST)-l = T- 1S-t,

(S-l)T = (ST)-l

(6)


will be useful. For convenience, we use the abbreviation

A tensor Q is orthogonal if it preserves inner products:

QU'Qv = u-v

7

for all vectors u and v. A necessary and sufficient condition that Q be ortho-gonal is that

or equivalently,

An orthogonal tensor with positive determinant is called a rotation. (Rota-tions are sometimes called proper orthogonal tensors.) Every orthogonaltensor is either a rotation or the product of a rotation with - I. If R =I I is arotation, then the set of all vectors v such that

Rv = v

forms a one-dimensional subspace of 1/ called the axis of R.A tensor S is positive definite provided

v'Sv > 0

for all vectors v =I O.Throughout this book we will use the following notation:

Lin = the set of all tensors;Lin + = the set of all tensors S with det S > 0;Sym = the set of all symmetric tensors;Skw = the set of all skew tensors;

Psym = the set of all symmetric, positive definite tensors;Orth = the set of all orthogonal tensors;

Orth + = the set of all rotations.

The sets Lin +, Orth, and Orth + are groups under multiplication; in fact,Orth + is a subgroup of both Orth and Lin ". Orth is the orthogonal group;Orth + is the rotation group (proper orthogonal group).

On any three-dimensional vector space there are exactly two crossproducts, and one is the negative of the other. We assume that one such crossproduct, written

u x v

8 I. TENSOR ALGEBRA

for all u and v, has been singled out. Intuitively, u x v will represent the right-handed cross product of u and v; thus if

then the basis {e;} is right handed and the components of u x v relative to{e;} are

Further,

u x v = -v x u,

u x u = 0,

u . (v x w) = w' (u x v) = v· (w xu).

When u, v, and ware linearly independent, the magnitude of the scalar

u· (v x w)

represents the volume of the parallelepiped 9 determined by u, v,w. Further, I

det S = Su' (Sv x Sw)u-fv x w)

and hence

Id I_ vol(S(9»

et S - vol(9) ,

which gives a geometrical interpretation of the determinant (Fig. 3). HereS(9) is the image of 9 under S, and vol designates the volume.

There is a one-to-one correspondence between vectors and skew tensors:given any skew tensor W there exists a unique vector w such that

Wv = w x v

for every v, and conversely; indeed,

(7)

[W] = [ ~-(3

corresponds to

WI = oc,

1 Cf., e.g., Nickerson, Spencer, and Steenrod [I, §S.2).

1. POINTS. VECTORS. TENSORS 9

Figure 3

We call w the axial vector corresponding to W. It follows from (7) that (forW =1= 0) the null space of W, that is the set of all v such that

Wv= 0,

is equal to the one-dimensional subspace spanned by w. This subspace iscalledaxis ofW.

We will frequently use the facts that "Y and Lin are normed vector spacesand that the standard operations of tensor analysis are continuous. Inparticular, on "Y and Lin, the sum, inner product, and scalar product arecontinuous, as are the tensor product on "Y and the product, trace, transpose,and determinant on suitable subsets of Lin.

EXERCISES

1. Choose a E "Y and let 1jJ: "Y ...... IR be defined by ljJ(v) = a' v. Show thata = Li ljJ(ei)ei'

2. Prove the representation theorem for linear forms (page I).

3. Show that the sum S + T and product ST are tensors.

4. Establish the existence and uniqueness of the transpose ST of S.

5. Show that the tensor product a ® b is a tensor.

6. Prove that

(a) Sea ® b) = (Sa) ® b,(b) (a ® b)S = a ® (STb).(c) Li (Sei) ® e, = S.

7. Establish (1), (2), (4), and (5).

10 I. TENSOR ALGEBRA

8. Show that

(a) v = Su is equivalent to Vi = Lj SijUj,(b) (ST)ij = Sji'(c) (ST)ij = Lk s., 1kj ,

(d) (a ® b)ij = a.b],(e) S . T = Li,j Sij Iij .

9. Prove that the operation S· T is indeed an inner product; that is, showthat

(a) S· T = T . S,(b) S . T is linear in T for S fixed,(c) S·S ~ 0,(d) S· S = 0 only when S = O.

10. Establish the proposition on p. 6.

11. Show that the trace of a tensor equals the trace of its symmetric part, sothat, in particular, the trace of a skew tensor is zero.

12. Prove that Q is orthogonal if and only if QTQ = I.13. Show that Q is orthogonal if and only if H = Q - I satisfies

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

14. Let ({J: iI x iI x iI -+ IR be trilinear and skew symmetric; that is, ({J islinear in each argument and

({J(u, v, w) = - ({J(v, U, w) = - ({J(u, w, v) = - ({J(w, v, u)

for all u, v, WE iI. Let SELin. Show that

({J(Se1, e2, e3) + ({J(et> Se2' e3) + ({J(el, e2, Se3) = (tr S)({J(e1, e2' e3)·

15. Let Q be an orthogonal tensor, and let e be a vector with

Qe = e.

(a) Show that

QTe = e.

(b) Let W be the axial vector corresponding to the skew part of Q.Show that w is parallel to e.

16. Show that if w is the axial vector of W E Skw, then

1[w] = j2,WI.

17. Let DE Psym, Q E Orth. Show that QDQT E Psym.

2. SPECTRAL, CAYLEY-HAMILTON, AND OTHER THEOREMS 11

2. SPECTRAL THEOREM. CAYLEY-HAMILTON THEOREM.POLAR DECOMPOSITION THEOREM

A scalar W is an eigenvalue of a tensor S if there exists a unit vector e suchthat

Se = we,

in which case e is an eigenvector. The characteristic space for S correspondingto W is the subspace of 1/ consisting of all vectors v that satisfy the equation

Sv = wv.

If this space has dimension n, then w is said to have multiplicity n. Thespectrum ofS is the list (Wt, W2, ..•), where Wt :5; W2 :5; •.. are the eigenvaluesofS with each eigenvalue repeated a number of times equal to its multiplicity.

Proposition

(a) The eigenvalues ofa positive definite tensor are strictly positive.(b) The characteristic spaces ofasymmetric tensor are mutually orthoqonal.

Proof Let W be an eigenvalue of a positive definite tensor S, and lete be a corresponding eigenvector. Then, since Se = we and [e] = 1, W =e-Se > O.

To prove (b) let wand A. be distinct eigenvalues of a symmetric tensor S,and let

Su = wu, Sv = A.V,

so that u belongs to the characteristic space of w, v to the characteristic spaceof A. Then

wu • v = v· Su = u . Sv = AU' v,

and, since ca =f; A, U • v = O. 0

The next result! is one of the central theorems oflinear algebra.

Spectral Theorem. Let S be symmetric. Then there is an orthonormal basis for1/ consisting entirely of eigenvectors of S. Moreover, for any such basise., e2' e3 the corresponding eigenvalues Wt, W2' W3' when ordered, form theentire spectrum ofS and

S = LWjej (8)ej.j

(1)

I Cf., e.g., Bowen and Wang [I, §27]; Halmos [1, §79]; Stewart [I, §37]. The first two refer-ences state the theorem in terms of perpendicular projections onto characteristic spaces (cf.Exercise 4).

12 I. TENSOR ALGEBRA

Conversely, ifS has the form (1) with {e.} orthonormal, then WI' w z , W 3 areeigenvalues ofS with e., e2, e3 corresponding eigenvectors. Further:

(a) S has exactly three distinct eigenvalues ifand only if the characteristicspaces ofS are three mutually perpendicular lines through o.

(b) S has exactly two distinct eigenvalues if and only if S admits therepresentation

(2)

In this case WI and W2 are the two distinct eigenvalues and the correspondingcharacteristic spaces are sp{e} and {e}', respectively. Conversely, if sp{e}and {elL (lei = 1) are the characteristic spaces for S, then S must have theform (2).

(c) S has exactly one eigenvalue ifand only if

S = wI. (3)

In this case W is the eigenvalue and "1/ the corresponding characteristic space.Conversely, if"l/ is a characteristic space for S, then S has the form (3).

The relation (1) is called a spectral decomposition of S. Note that, by (1),the matrix ofS relative to a basis {eJ of eigenvectors is diagonal:

Further, in view of (a)-(c), the characteristic spaces of a symmetric tensor Sare three mutually perpendicular lines through 0; or a line 1through 0 and theplane through 0 perpendicular to I; or S has only one characteristic space, "1/itself.Thus if Olt~ (01: = 1, ... , n :::;; 3) denote the characteristic spaces ofS, thenany vector v can be written in the form

v = L v~,~

(4)

Commutation Theorem. Suppose that two tensors Sand T commute. Then Tleaves each characteristic space ofS invariant; that is, ifv belongs to a character-istic space ofS, then Tv belongs to the same characteristic space. Conversely, ifT leaves each characteristic space ofa symmetric tensor S invariant, then Sand Tcommute.

Proof Let Sand T commute. Suppose that Sv = wv. Then

S(Tv) = T(Sv) = w(Tv),

so that Tv belongs to the same characteristic space as v.


To prove the converse assertion choose a vector v and decompose v as in(4). 1fT leaves each characteristic space 0/./,% ofS invariant, then Tv, E IJIJ,% and

S(Tv,%) = w,%(Tv,%) = T(w,%v,%) = T(Sv,%),

where w'% is the eigenvalue corresponding to 0/./,%' We therefore conclude, withthe aid of (4), that

STy = L STv,% = LTSv,% = TSv.'% '%

Thus, since v was chosen arbitrarily, ST = TS. 0

There is only one subspace of "Y that every rotation leaves invariant;namely "Y itself. We therefore have the following corollary of the spectraltheorem.

Corollary. A symmetric tensor S commutes with every rotation if and only ifS = wI.

Crucial to our development of continuum mechanics is the polar de-composition theorem; our proof of this theorem is based on the

Square-Root Theorem. Let C be symmetric and positive definite. Then there isa unique positive definite, symmetric tensor U such that

U 2 = C.

We write .JC for U.

Proof (Existence) Let

C = LWiei ®eii

be a spectral decomposition of C E Psym and define U E Psym by

U=L~ei®ei'i

(Since co, > 0, this definition makes sense.) Then U 2 = C is a direct con-sequence of (1.2h.

(Uniqueness') Suppose

U 2 = V2 = C

with U, V E Psym. Let e be an eigenvector ofC with W > 0 the correspondingeigenvalue. Then, letting A = JOj,

o= (U2- wI)e = (U + AI)(U - AI)e.

I Stephenson (1).

14

Let

Then

I. TENSOR ALGEBRA

v = (U - ..1.I)e.

Uv = -..1.V

and v must vanish, for otherwise -A. would be an eigenvalue of U, an im-possibility since U is positive definite and A. > O. Hence

Ue = ..1.e.

Similarly,

Ve = ..1.e,

and Ue = Ve for every eigenvector e of C. Since we can form a basis ofeigenvectors of C (cf. the spectral theorem), U and V must coincide. 0

Polar Decomposition Theorem. Let F E Lin +. Then there exist positivedefinite, symmetric tensors D, V and a rotation R such that

F = RU = YR.

Moreover, each of these decompositions is unique; infact,

(5)

U = JFTF, V = JFFT• (6)

We call the representation F = RU (resp. F = VR) the right (resp. left)polar decomposition ofF.

Proof. Our first step will be to show that FTF and FFTbelong to Psym.Both tensors are clearly symmetric. Moreover,

v • FTFv = (Fv) • (Fv) ~ 0,

and this inner product can equal zero only ifFv = 0, or equivalently, since F isinvertible, only if v = o. Thus FTF E Psym. Similarly, FFTE Psym.

(Uniqueness) Let F = RU be a right polar decomposition of F. Thensince R is a rotation,

FTF = URTRU = U 2•

But by the square-root theorem there can be at most one U E Psym whosesquare is FTF. Thus (6)1 holds and U is unique; since R = FU-l, R is alsounique. On the other hand, if F = VR is a left polar decomposition, then

FFT = V2,

and V is determined uniquely by (6)z, R by the relation R = V- 1F.


(Existence) Define U E Psym by (6)1 and let

R = FU- 1 .

To verify F = RU is a right polar decomposition we have only to show thatR EOrth + . Since det F > 0 and det U > 0 (the latter because all eigenvaluesofU are strictly positive), det R > o. Thus we have only to show that R EOrth.But this follows from the calculation

RTR = U- 1FTFU- t = U- 1U2U- 1 = I.

Finally, define

v = RURT.

Then V E Psym (Exercise 1.17) and

VR = RURTR = RU = F. D

Given a tensor S, the determinant of S - wI admits the representation

det(S - wI) = _w3 + 11(S)W2 - 12(S)W + 13(S) , (7)

for every wEIR, where

We call

11(S) = tr S,

12(S) = t[(tr S)2 - tr(S2)],

liS) = det s.(8)

(9)

(10)

the list of principal invariants' of S. When S is symmetric J s is completelycharacterized by the spectrum (w 1, W2' w3 ) ofS. Indeed, a simple calculationshows that

11(S) = W 1 + W2 + W3'

12(S) = W tW2 + W 2W3 + W tW3,

liS) = W tW2W3·

Moreover, the above characterization is a one-to-one correspondence. To seethis note first that WEIR is an eigenvalue ofa tensor S if and only if W satisfiesthe characteristic equation

det(S - wI) = 0,

[ liS) are called invariants of S because of the way they transform under the orthogonalgroup: lk(QSQT) = liS) for all QE Orth [cf. (37.3)].

16

or equivalently,

I. TENSOR ALGEBRA

(11)

Further, when S is symmetric the multiplicity of an eigenvalue w is equal to itsmultiplicity as a root of (11).1 Thus we have the following

Proposition. Let Sand T be symmetric and suppose that

f s = fT'

Then Sand T have the same spectrum.

More important is the

Cayley-Hamilton Theorem." Every tensor S satisfies its own characteristicequation:

(12)

EXERCISES

1. Determine the spectrum, the characteristic spaces, and a spectral de-composition for each of the following tensors:

A = IXI + pm @ m,

B = m@n+n@m.

Here IX and pare scalars, while m and n are orthogonal unit vectors.

2. Granted the validity of the portion of the spectral theorem ending with(1), prove the remainder of the theorem.

3. Let D E Sym, Q E Orth. Show that the spectrum ofD equals the spectrumof QDQT. Show further that if e is an eigenvector of D, then Qe is aneigenvector of QDQT corresponding to the same eigenvalue.

4. A tensor P is a perpendicular projection if P is symmetric and p 2 = P.

(a) Let n be a unit vector. Show that each of the following tensors is aperpendicular projection:

I, 0, n@n, 1- n@n. (13)

(b) Show that, conversely, if P is a perpendicular projection, then Padmits one of the representations (13).

1 Cf., e.g., Bowen and Wang [1, Theorem 27.4); Halmos [I, §78, Theorem 6).2 cr. e.g., Bowen and Wang [1, Theorem 26.1); Halmos[1, §58).

SELECTED REFERENCES 17

5. Show that a (not necessarily symmetric) tensor S commutes with everyskew tensor W if and only if

S = wI. (14)

6. Let F = RU and F = VRdenote the right and left polar decompositionsof FE Lin+.

(a) Show that U and V have the same spectrum (WI' W2' w3) .

(b) Show that F and R admit the representations

F = L Wi f i e e.,i

where e, and f i are, respectively, the eigenvectors of U and V cor-responding to Wi'

7. Let R be the rotation corresponding to the polar decomposition ofF E Lin +. Show that R is the closest rotation to F in the sense that,

IF - RI < IF - QI (15)

for all rotations Q =I' R.

The result (15) suggests that an alternate proof" of the polar decompositioncan be based on the following variational problem: Find a rotation R thatminimizes IF - Q lover all rotations Q.

SELECTED REFERENCES

Bowen and Wang [1].Brinkman and Klotz [1].Chadwick [1, Chapter 1].Halmos [1].Martins and Podio-Guidugli [1].Nickerson, Spencer, and Steenrod [1, Chapters 1-5].Stewart [1].

1 Such a proof is given by Martins and Podio-Guidugli [1].

CHAPTER

II

Tensor Analysis

3. DIFFERENTIATION

In this section we introduce a notion of differentiation sufficientlygeneralto include scalar, point, vector, or tensor functions whose arguments arescalars, points, vectors, or tensors. To accomplish this we use the fact that IR,"Y, and Lin are normed vector spaces, and where necessary phrase our def-initions in terms of such spaces.

Let fJII and "If'" denote normed vector spaces, and let f be defined in aneighborhood of zero in fJII and have values in "If'". We say that f(u) approacheszerofaster thanu, and write

or, more simply,

f(u) = o(u) as u -+ 0,

if

f(u) = o(u),

lim Ilf(u)II = o.• ->0 [u]."'0

19

20 II. TENSOR ANALYSIS

(1)

[Here 11£(0)11 is the norm of £(0) on 11', while 11011 is the norm of 0 on 0lI.]Similarly,

£(0) = g(o) + 0(0)

signifies that

£(0) - g(o) = 0(0).

Note that this latter definition also has meaning when f and g have values inC, since in this instance f - g has values in the vector space "Y.

As an example consider the function cp(t) = r, Then

cp(t) = o(t)

if and only if IX > 1.Let g be a function whose values are scalars, vectors, tensors, or points,

and whose domain is an open interval f0 of IJl The derivative g(t) of gat t, if itexists, is defined by

g(t) = dd g(t) = lim! [g(t + IX) - g(t)].t ~-->O IX

Ifthe values of g are points, then g(t + IX) - g(t) is a point difference and hencea vector. Thus the derivative of a point function is a vector. Similarly, thederivative of a vector function is a vector, and the derivative of a tensor func-tion is a tensor. We say that g is smooth if g(t) exists at each t E f0, and if thefunction gis continuous on f0.

Let g be differentiable at t. Then (1) implies that

lim! [g(t + IX) - g(t) - IXg(t)] = 0,~-o IX

or equivalently,

Clearly,

is linear in IX; thus

g(t + IX) = g(t) + IXg(t) + p(IX).

g(t + IX) - g(t)

(2)

is equal to a term linear in IX plus a term that approaches zero faster than IX. Indealing with functions whose domains lie in spaces of dimension greater thanone the most useful definition of a derivative is based on this result; we define

3. DIFFERENTIATION 21

the derivative to be the linear map which approximates get + a) - get) forsmall a.

Thus let Ol/ and "fII be finite-dimensional normed vector spaces, let q; bean open subset of Ol/, and let

g: q; ----+ "fII.

We say that g is differentiable at x E q; if the difference

g(x + u) - g(x)

is equal to a linear function of u plus a term that approaches zero faster than u.More precisely, g is differentiable at x if there exists a linear transformation

Dg(x): Ol/ ----+ if"

such that

g(x + u) = g(x) + Dg(x) [u] + o(u)

as u ----+ O. If Dg(x) exists, it is unique; in fact, for each u,

Dg(x) [u] = lim! [g(x + au) - g(x)] = dd g(x + auHx=o-cx-+O a acxelRl

(3)

(4)

We call Dg(x) the derivative of g at x, Since any two norms on a finite-dimensional vector space are equivalent, Dg(x) is independent ofthe choice ofnorms on iJIt and "fII. If g is differentiable at each x E q;, then Dg denotes themap x f-+ Dg(x) whose domain is q; and whose codomain is the space oflinear transformations from iJIt into if". This space is finite dimensional andcan be normed in a natural manner; thus it makes sense to talk about thecontinuity and differentiability of Dg. In particular, we say that g is of class C 1

,

or smooth, if g is differentiable at each point of q; and Dg is continuous. Con-tinuing in this manner, we say g is of class C2 if g is of class C1 and Dg issmooth, and so forth.

Ofcourse, the three-dimensional euclidean space is is not a normed vectorspace. However, when the domain q; of g is contained in is the above defini-tion remains valid provided we replace iJIt in (3) by the vector space 1-associated with is. Similarly, when g has values in is we need only replace if'

in (3) by "1"'.Note that by (2), when the domain q; of g is contained in ~,

for every a E ~.

Dg(t) [a] = ag(t) (5)


Smooth-Inverse Theorem. 1 Let ~ be an open subset of a finite-dimensionalnormed vector space 0/1. Further, let g: ~ -+ 0/1 be a one-to-one function ofclassC" (n ~ 1) and assume that the linear transformation Dg(x): 0/1 -+ 0/1 isinvertible at each x E~. Then g-l is ofclass C".

Often, the easiest method of computing derivatives is to appeal directlyto the definition. We demonstrate this procedure with some examples.Consider first the function tp: "Y" -+ ~ defined by

qJ(V) = v· v.

Then

qJ(V + u) = v· v + 2v' u + u· u = qJ(v) + 2v' U + o(u),

so that

DqJ(v) [u] = 2v' u.

Similarly, for G: Lin -+ Lin defined by

G(A) = A 2

we have

G(A + V) = A2 + AV + VA + V 2 = G(A) + AV + VA + o(V),

so that

DG(A) [V] = AV + VA.

Next, let G: Lin -+ Lin be defined by

G(A) = A 3•

Then

G(A + V) = (A + V)3 = A 3 + A2V + AVA + VA 2 + o(V)= G(A) + A2V + AVA + VA 2 + o(V),

and hence

As our last example, let L: 0/1 -+ "If'" be linear. Then

L(x + u) = L(x) + L(u),

(6)

1 This result is a direct consequence of the inverse function theorem (cf. Dieudonne [I,Theorem 10.2.5]).

3. DIFFERENTIATION

so that, trivially,

DL(x) = L.

The next two results involve far less trivial computations.

23

(7)

Theorem (Derivative of the determinant). Let cp be defined on the set ofallinvertible tensors A by

cp(A) = det A.

Then cp is smooth. Infact,

Dcp(A) [V] = (det A) tr(VA -1)

for every tensor V.

(8)

Proof By (2.7) with co = -1,

det(I + 8) = I + tr 8 + 0(8)

as 8 --+ O. Thus, for A invertible and V E Lin,

det(A + V) = det[(I + VA -1)A] = (det A) det(I + VA -1)

= (det A) [1 + tr(VA -1) + o(V)]

= det A + (det A) tr(VA -1) + o(V)

as V --+ O. Therefore, since the map

V f-+ (det A) tr(VA -1)

is linear, (8) must be valid. The proof that Dcp is continuous follows from thecontinuity of the determinant, trace, and inverse operations. 0

Theorem (Derivative of the square root). Thefunction H: Psym --+ Psymdefined by

H(C) =.jCis smooth.

Proof By the square-root theorem (page 13), the function H is one-to-one with inverse G: Psym --+ Psym defined by

G(A) = A2•

Clearly, G is smooth with derivative DG(A): Sym --+ Sym given by (6). Thus,in view of the smooth-inverse theorem, to complete the proof it suffices toshow that DG(A) is invertible at each A E Psym, or equivalently, that

DG(A) [V] = 0 implies V = O.

Assume the former holds, so that, by (6),

AU + UA = O.

Let Abe an eigenvalue of A with e a corresponding eigenvector. Then

AUe + UAe = AUe + AUe = 0

and

A(Ue) = - A(Ue).

If Ue # 0, then - A is an eigenvalue of A. But A is positive definite, so that Aand - A cannot both be eigenvalues. Thus Ue = 0, and this must hold forevery eigenvector e of A. By the spectral theorem, there is a basis for 1/ ofeigenvectors of A. Thus U = O. D

It will frequently be necessary to compute the derivative of a productn(f, g) of two functions f and g. In tensor analysis there are many differentproducts available; for example, the product of a scalar cp and a vector v

n(cp, v) = cpv,

the inner product of two vectors

n(u, v) = u • v,

the tensor product of two vectors

n(u, v) = u ® v,

the action of a tensor S on a vector v

n(S, v) = Sv,

and so forth. The above operations have one property in common, bilinearity.Therefore, in order to establish a product rule valid in all cases of interest, weconsider the general product operation

n: ff x <§ -+ "IIf,

which assigns to each fo E ff and go E <§ the product n(fo, go) E "IIf. Here ff,<§, and 'IY are finite-dimensional normed spaces and n is a bilinear map.Within this general framework the product h = n(f, g) of two functions

f:fI)-+ff,

is the function

defined by

hex) = n(f(x), g(x))


for all x E ~. Assume now that the common domain ~ of f and g is an opensubset of a finite-dimensional normed space dlt (or of our euclidean pointspace). We then have the

Product Rule. Let f and g be differentiable at x E~. Then their producth = 1t(f, g) is differentiable at x and

Dh(x) [u] = 1t(f(x), Dg(x) [u]) + 1t(Df(x) [u], g(x» (9)

for all u E dlt.

The proof of this result will be given at the end of the section.Note that when f is constant (9) reduces to

Dh(x) [u] = 1t(f,Dg(x) [u]),

with a similar result when g is constant. We can use this fact to interpret (9).Thus let Xo be a given point of~, let foand go denote the constant functions on~ with values f(xo) and g(xo), and let

Then (9) becomes

(10)

that is, the derivative of1t(f,g) at Xo is the derivative of the product holding fconstant at its value f(xo) plus the derivative ofthe product holding g constant atits value g(xo).

When the common domain off and g is an open subset of R, then (5), (9)(with x replaced by t), and the bilinearity of 1t imply that

h(t) = 1t(f(t), g(t» + 1t(f(t), g(t».

Thus we have the following

Proposition. Let tp, v, W, S, andT be smoothfunctions on an open subset of~

with qJ scalar valued, v and W vector valued, and Sand T tensor valued. Then

(qJV)" = qJv + (pv,

(V'W)" = v·w + v·w,(TS)" = TS + ts,

(T • S)" = T . S + t .S,

(Su)' = So + Su,


(11)

Figure I

Another theorem which we will use frequently is the chain rule. In orderto state this result in sufficient generality, let 0/1, f§, and fF denote finite-dimensional normed linear (or euclidean point) spaces, let rc and ~ denoteopen subsets of f§ and VII, respectively, and let

g: ~ -. f§, f: rc -. fF,

with the range of g contained in rc (Fig. 1).

Chain Rule.1 Let g be differentiable at x E~, and let f be differentiable aty = g(x). Then the composition

h = fog

is differentiable at x and

Dh(x) = Df(y) 0 Dg(x).

The relation (11), in less abbreviated form, reads

Dh(x) [uJ = Df(g(x» [Dg(x) [u]]

for every u E 0/1. In the case 0/1 = ~, g, and hence h, is a function of a realvariable; writing t in place of x,

Dh(t) [aJ = ah(t), Dg(t) [aJ = ag(t)

(12)

for every a E ~. Thus, since h(t) = f(g(t», we have the important relation

:t f(g(t» = Df(g(t» [g(t)J.

1 See, e.g., Bartle [I, Theorem 40.2].

Proposition. Let S be a smooth tensor-valued function on an open subset g ofIR. Then

and ifS(t) is invertible at each t E g,

(det S)' = (det S) tr(SS -1).

Proof Let L: Lin -+ Lin be defined by

L(A) = AT.

Then L is linear, so that (7) and (12) imply

(STy = [L(S)J = L(S) = (S)T,

(13)

(14)

which is (13). The result (14) is a direct consequence of (8) and (12). 0

We close this section with the

Proofofthe Product Rule. Let "0 denote the maximum of 111t(e, k)II'overall unit vectors e and k. This maximum certainly exists, since the set of allunit vectors is compact. (Recall that the underlying spaces :F and t'§ arefinite dimensional.) Assume a #- 0, b #- 0, and let e and k denote the unitvectors

Since 1t is bilinear,

e = a/llall, k = b/llbll.

and

1t(a, b) = Ilallllbll1t(e, k)

111t(a, b)11 ~ I<:ollalillbll·

This inequality is also satisfied when a = °or b = 0, as is clear from thebilinearity of 1t. A similar argument yields the existence of constants"1and 1<:2such that

IIDf(x) [u] II ~ 1<:111ull,

for all u E 1JIi.To establish (9) we note first that

IIDg(x) [u] II ~ 1<:211ull

f(x + u) = f(x) + Df(x) [u] + o(u),

g(x + u) = g(x) + Dg(x) [u] + o(u),

28

and

II. TENSOR ANALYSIS

Iln(Df(x) [0], Dg(x) [0])11 ~ KOK1K2110112,

Iln(Df(x) [0], 0(0»11 ~ KoKlllollo(o),

I/n(o(o), Dg(x) [0])11 s KoK21101/0(0).

Thus, since all ofthe above terms are 0(0), we conclude from the bilinearity ofn that

h(x + 0) = n(f(x + 0), g(x + u)

= n(f(x), g(x» + n(f(x), Dg(x) [0])

+ n(Df(x) [0], g(x» + 0(0),

which yields the desired result, because the first term on the right side is h(x),while the second and third terms are linear in o. 0

EXERCISES

1. Compute DG(A) for each of the following functions G: Lin --+ Lin.

(a) G(A) = (tr A)A,(b) G(A) = ABA (B a given tensor),(c) G(A) = ATA,(d) G(A) = (0· Ao)A (0 a given vector).

2. Let G be defined on the set of all invertible tensors by G(A) = A- I.

Assuming that G is differentiable, show that

DG(A)[H] = -A-1HA- 1 •

3. Let cp be defined on the set of all invertible tensors by cp(A) = det(A 2).Compute Dsp (A),

4. Let cp(v) = ev 2 for all vE 0/1. Compute Dcp(v).

5. Let G: Lin --+ Lin be defined by

G(A) = K(A)AT,

where K: Lin --+ Lin is differentiable. Show that if G(A) is symmetric foreach A and ifK(I) = 0, then DK(I) has symmetric values (i.e., DK(I) [H]= DK(I) [HF for every HELin).

6. Let Q: ~ --+ Orth be differentiable. Show that Q(t)Q(t)T is skew at eachte ~.

7. Let G: Lin -+ Lin be differentiable and satisfy

QG(A)QT = G(QA)

4. GRADIENT. DIVERGENCE. CURL 29

for all A E Lin and Q E Orth. Show that

G(A)WT + WG(A) = DG(A) [WA]

for all A E Lin and WE Skw.

8. Let S: Lin -+ Lin be smooth.

(a) Define G: Lin -+ Lin by

G(A) = S(Ao + IXA),

where IX E IR and Ao E Lin. Show that

DG(A) [U] = IX DS(Ao + IXA) [U]

for every U E Lin.(b) Define qJ: Lin -+ IR by

cp(A) = r- S(Ao + IXA) d«.

Compute Dip, (It is permissible to differentiate under the integral.)(c) Assume that DS(A) is symmetric at each A in the sense that

B •DS(A) [C] = C • DS(A) [B]

for all tensors Band C. Show that

Dcp(A) [U] = S(Ao + A) . U

for every U E Lin.

9. Compute the derivatives of the principal invariants 1\,12,13: Lin -+ R

4. GRADIENT. DIVERGENCE. CURL

We now consider functions defined over an open set &I in euclidean space.A function on &I is called a scalar, vector, tensor, or point fieldaccording as itsvalues are scalars, vectors, tensors, or points.

Let cp be a smooth scalar field on &I. Then for each x E &I, Dcp(x) is a linearmapping of j/O into IR, and by the representation theorem for linear formsthere exists a vector a(x) such that Dcp(x) [0] is the inner product of a(x) withu, We write Vcp(x) for the vector a(x), so that

Dcp(x) [0] = Vcp(x)·0,

and we call Vcp(x) the gradient of cp at x. In this case the expansion (3.4) hasthe form

cp(x + D) = cp(x) + Vcp(x)· 0 + 0(0).


Similarly, ifv is a smooth vector or point field on fll, then Dv(x) is a lineartransformation from 1/ into 1/ and hence a tensor. In this case we use thestandard notation Vv(x) for Dv(x) and write

Vv(x) u = Dv(x) [u].

The tensor Vv(x) is the gradient of v at x.Given a smooth vector field v on fll, the scalar field

div v = tr Vv

is called the divergence ofv. We can use this operator to define the divergence,div S, of a smooth tensor field S. Indeed, div S is the unique vector field withthe following property:

(div S)· a = div(STa) (1)

for every vector a. The reason for defining div S in this manner will becomeclear when we establish the divergence theorem for tensor fields.

Proposition. Let tp, v, w,and S be smooth fields with qJ scalar valued, v and W

vector valued, and S tensor valued. Then

V(qJv) = qJ Vv + v ® VqJ,

div(qJv) = qJ div v + v· VqJ,

V(v· w) = (VW)T V+ (VV)T w,

div(v ® w) = v div w + (Vv)w,

div(STv) = S· Vv + v· div S,

div(qJS) = qJ div S + S VqJ.

Proof Let h = qJV. Then by the general product rule (3.9),

Vh(x) u = qJ(x) Vv(x) u + (VqJ(x)• u)v(x)

= [qJ(x) Vv(x) + v(x) ® VqJ(x)]u,

which implies (2)1' Taking the trace of (2)1' we arrive at (2h.Next, let rjJ = v· w. Then by (3.9),

VrjJ(x) . u = v(x) . Vw(x) u + w(x)· Vv(x) u

= [Vw(x) T v(x) + Vv(x)T w(x)] • u,

(2)

and we have (2h.To prove (2)4 note that, by (1) and (2)2'

a . div(v ® w) = div[(w ® v)a] = div[(v· a)w] = (v - a) div w + w· V(v· a).


But by (2h,

Thus

a· div(v (8) w) = (v· a) div w + w· (VvTa) = [v div w + (Vv)w]· a,

which implies (2)4'Before proving (2)s, we note that

V(Av) = A w

for any (constant) tensor A and any smooth vector field v. Indeed,

v(x + u) = v(x) + Vv(x) U + o(u)

and hence

31

(3)

Av(x + u) = Av(x) + A Vv(x) u + o(u),

which implies (3). [This result is also a direct consequence of (3.7) and thechain rule (3.11).] Taking the trace of (3), we arrive at the identity

div(Av) = AT. v-. (4)

We are now in a position to prove (2)s. Clearly,

div(STv) = tr V(STV),

and by the product rule in the form (3.10), V(STV) (xo) is the sum of two terms:the gradient holding S constant at the value So = S(xo) plus the gradientholding v constant at Vo = v(xo). Therefore

div(STv)(xo) = tr[V(S~ v)(x o) + V(STVO)(XO)]= div(S~v)(xo) + div(STvo}(xo)'

By (4) with A = SJ,

div(SJv) = So • Vv.

On the other hand, since Vo is constant, (1) implies

div(STvo) = Vo •div S.

Thus

div(STv)(xo) = So· Vv(xo) + Vo • div S(xo),

which implies (2)5'

The proof of (2)6 is similar. Using the above argument it is clear thatdiv(q>S)(xo) is the divergence holding q> constant at q>o = q>(xo) plus thedivergence holding S constant at So = S(xo):

div(q>S)(xo) = div(q>oS)(x o) + div(q>So)(xo).

Clearly,

div(q>o S) = q>o div S.

Also, for any vector a,

a· div(q>So) = div(q>SJa),

and by (2h with v = SJa this becomes

Vq> • SJa = a . So Vq>;

hence

div(q>So) = So Vq>.

Therefore

div(q>S)(xo) = q>o div S(xo) + So Vq>(xo),

which implies (2)6' 0

Another important identity, for a class C 2 vector field x, is

div(Vv T) = V(div v). (5)

We postpone the proof of (5) until later.The curl of v, denoted curl v, is the unique vector field with the property

(Vv - VvT)a = (curl v) x a (6)

for every vector a. Thus curl v(x) is the axial vector corresponding to the skewtensor Vv(x) - VV(X)T.

Let be a scalar or vector field of class C2• Then the laplacian Ll<l> of is

defined by

Ll<l> = div V.

If

Ll<l> = 0,

then is harmonic.

Proposition. Let v be a class C2 vector field with

div v = 0, curl v = O.

Then v is harmonic.

Proof Since curl v = 0, (6) implies

Vv - VvT = o.Therefore, by (5),

0= div(Vv - VvT) = .1v - V div v = .1v. D

Let <D be a smooth scalar, vector, or tensor field. Then

D<D(x) [ea = lim ~ {<D(x + lXei) - <D(x)}.11-+0 IX

33

Ifx has components (Xl' X2' X3), then x + lXel, say, has components (Xl + IX,

X 2, X3)' Thus the limit on the right is simply the partial derivative of <D withrespect to Xi:

D<D(x) [ea = O;(x).ox,

This fact can be used to establish the following component representations:

Ocp(Vcp)i = ox.',

d. '\' OViIV V = L.-;-'

i uXi

OV·(VV)ij = ox'.'

J

(div S)i = L OSij,j OX j

(7)

Further, curl v has components (IX, f3, y) with

OV3 oV2IX----- oX2 ox 3'

and if W is the skew part of Vv, then

The verification of (5) furnishes an excellent example of the use of theabove identities. The components of VvTare

T OVj(VV)ij = ;--;

ox,

34

hence

II. TENSOR ANALYSIS

[div(VvTn = 4: a~. ;:~ = a~. 4: ;:i. = a~. (div v) = [V(div -n.J J 1 1 J J I

which implies (5).We now list some obvious consequences ofthe chain rule. First let

h = fa g

with f and g smooth point fields. Then (3.11) takes the form

Vh(x) = Vf(y) Vg(x),

where y = g(x). On the other hand, for a smooth scalar field (fJ, a smoothvector field v, and a smooth point-valued function g of a real variable, (3.12)reads

d .dt (fJ(g(t)) = V(fJ(g(t)) • g(t),

ddt v(g(t)) = Vv(g(t)) g(t).

A curve c in f!Il is a smooth map

c: [0, 1] --+ f!Il

with cnever zero; c is closed if c(O) = c(l); the length of c is the number

length(c) = f Ic(O") Ido

(see Fig. 2). Let v be a continuous vector field on f!Il. Then the integral of varound c is defined by

1v(x) • dx = fV(C(O")) • c(O") da.

Figure 2


Similarly, for a continuous tensor field S on fYI,

iS(x) dx = fS(c«(J»c«(J) do,

Note that if cp is a smooth scalar field on fYI,

i v cp(X) . dx = i1

Vcp(c«(J»' c«(J) do = i1

!!- cp(c«(J» docoo d(J

= cp(c(l» - cp(c(O»;

thus

35

(8)

whenever C is closed.Given a subset fYI of rff, we write afYI for its boundary and ~ for its interior.

We say that fYI is connected if any two points in fYI can be connected by a curvein fYI; simply connected if any closed curve in fYI can be continuously deformedto a point without leaving fYI (i.e., if given any closed curve c in fYI there exists asmooth function f: [0, 1] x [0, 1] --+ fYI and a point y E fYI such that, for all(J E [0, 1], f«(J, 0) = c«(J), f'(o, 1) = y, and f(O, (J) = f(l, (J». An open region is aconnected, open set in rff; the closure of an open region is called a closedregion.

Let fYI be a closed region. A field et> is smooth on fYI if et> is smooth on Ii, andifet> and Vet> have continuous extensions to all of fYI; in this case we also write et>and Vet> for the extended functions. An analogous interpretation applies to thestatement" et> is of class eN on fYI"; note that (for fYI open or closed) this willbe true if and only if the components of et> have continuous partial derivativesof all orders 5. N on fYI.

A vector field of the form v = Vcp satisfies curl v = 0 (Exercise 5a). Ournext theorem, which we state without proof, gives the converse of this result.

Potential Theorem.' Let v be a smooth point field on an open or closed,simply connected region fYI, and assume that

curl v = O.

Then there is a class e2 scalar field cp on fYI such that

v = Vcp.

We close this section by proving that vector fields with constant gradientsare affine.

1 Cf., e.g., Fleming [I, Corollary 2, p. 279].

Proposition. Let f be a smooth point- or vector-valuedfield on an open or closedregion !71, and assume that F = Vf is constant on !71. Then

f(x) = f(y) + F(x - y) (9)

for all x, y E !71.

Proof Choose x, y E !71. Since !71 is connected there is a curve c in !71 fromy to x. Thus

Ii d Iif(x) - f(y) = 0 da f(c(a)) do = 0 Vf(c(a))c(a) da

= F f c(a) do = F(x - y). D

By (9) any vector field fwith constant gradient F can be written in the form

f(x) = a + F(x - xo)

with a E 1/ and Xo E $. Moreover, the point Xo can be arbitrarily chosen: (Ofcourse, a depends on the choice of xo.)

EXERCISES

1. Let 0(, <p, D, V, w, and S be smooth fields with 0( and <p scalar valued; D, v,and W vector valued; and S tensor valued. Establish identities, similar to(2), for

(a) V(O(<p),(b) V[(D· v)w],(c) div(<pSv),(d) ~(v • w) (with v and w of class C2

) .

2. Establish the existence and uniqueness of the divergence, div S, of asmooth tensor field S.

3. Establish the component representations (7).

4. Use (7) to deduce (2)4 and (2)5.

5. Let <p and v be class C 2• Show that

(a) curl V<p = 0,(b) div curl v = o.

In Exercises 6-8,

r(x) = X - o.

5. THE DIVERGENCE THEOREM. STOKES' THEOREM

6. (a) Show that Vr = I.(b) Let e = r/lrl. Compute (Ve)e.

7. Let a E"f-, SELin, and define qJ: Iff ~ ~ by

qJ = a . (r x Sr).

Compute VqJ.

8. Let u be the vector field on Iff - {o} defined by

u = r/lrl 3•

Show that u is harmonic. Find a scalar field whose gradient is u.

9. Let u be a class C2 vector field. Show that

(a) div{(Vu)u} = Vu' VuT + u· (V div u),(b) Vu' VuT = div{(Vu)u - (div u)u} + (div U)2.

10. Let u and v be smooth. Show that

div(u x v) = v· curl u - u· curl v.

5. THE DIVERGENCE THEOREM. STOKES' THEOREM

37

We use the term regular region in the sense of Kellogg [1].1 Roughlyspeaking, a regular region is a closed region [Jt with piecewise smoothboundary aiYt. It is important to note that .JfI may be bounded or unbounded.In the former case we write

vol([Jt)

for the volume of [Jt.

Divergence Theorem. Let [Jt be a bounded regular region, and let qJ: fYt ~ ~,

v: [Jt ~ 1/, and S: rJt ~ Lin be smooth fields. Then

r qJn dA = rVqJ dV,JiJ9I J9I

r v - n dA = rdiv v dV,JiJ9I J9I

r Sn dA = rdiv S dV,JiJ9I J9I

where n is the outward unit normal field on arJt.

1 See also Gurtin [1, §5].

Proof The results concerning cp and v are classical! and will not beverified here. To establish the last relation, let a be a vector. Then

a' f SndA = fa. Sn dA = f (STa) · n dAMt iJ9t iJ9t

= f div(STa) dV = f (div S)· a dV = a' f div S dV,9t 9t 9t

which implies the desired result, since a is arbitrary. 0

Our reason for defining div S the way we did should be clear from theforegoing proof.

We will often find it important to deduce local field equations from globalformulations of balance laws. This procedure is greatly facilitated by the

Localization Theorem. Let be a continuous scalar or vector field on an openset fJIl in S. Then given any Xo E fJIl,

(xo) = lim I~O ) i dV,a~O vo a no

(I)

where Oa (8 > 0) is the closed ball of radius 8 centered at Xo' Therefore, if

for every closed ball 0 c fJIl, then

 = o.Proof Let

t, = I(xo) - ~ i dV I,va no

where Va = vol(Oa)' Then

t, ::;; ~ i 1(xo) - (x)j dVx ::;; sup I(xo) - (x) IVa no xeno

which tends to zero as 8 -+ 0, since is continuous. 0

1 See, e.g., Kellogg [1, Chapter 4].

5, THE DIVERGENCE THEOREM. STOKES' THEOREM 39

Note that the localization theorem and the divergence theorem togetheryield the following interesting interpretation of the divergence:

div v(x) = lim l~n) r v . 0 dA,~-o vo ~ Jilfi6

div S(x) = lim l~n) r So dA.~-o vo ~ Jilfi6

We will make use of Stokes' theorem, but only in the following veryspecial form.

Stokes' Theorem.' Let v be a smooth vector field on an open set [j£ in Iff.Further, let n be a disc in [j£ (Fig. 3), let 0 be a unit normal to n, and let thebounding circle c(O"), 0 ~ 0" ~ 1, be oriented so that

[c(O) x c(O")] • 0 > 0, 0<0"<1.

Then

The integral

In(curl v) . 0 dA = i v . dx.

i V'dX

represents the circulation of varound c; it sums the tangential component of varound the curve c and can be used to give a physical interpretation of curl v.Indeed, if we let Xo denote the center and b the radius of n, and write n = n~,

c = c~, then by Stokes' theorem and an argument similar to that used toderive (1),

• 1 ( ) - I' JC6 V • dxn cur v X o - im A(A) ,~-o ilo"<5

where A(n~) is the area ofn~. Considering different choices for 0, we concludethat curl v lies perpendicular to the plane in which the circulation per unit

n

Figure 3 C

I See, e.g., Kellogg [I, Chapter 4, §4].


area is greatest, and that the magnitude of curl v is the circulation per unitarea in that plane.

EXERCISES

1. Let [lA be a bounded regular region, and let v, w: [lA -4 "r and S: [lA -4 Linbe smooth. Show that:

(a) 10'" v ® 0 dA = I", v- dV,(b) Ioo'll (So) ® v dA = Io'll [(div S) ® v + SVvT

] dV,(c) 10'" v· So dA = I", (v -div S + S· Vv) dV,(d) 10'" v(w' 0) dA = Io'll [v div w + (Vv)w] dV.

2. Let v be a smooth vector field on an open region [lA. Show that

r v'odA = 0J09

for every regular region {!/J c [lA if and only if div v = o.

SELECTED REFERENCES

Bartle [1, Chapter 7].Chadwick [1, Chapter 1].Dieudonne [1, Chapters 8, 10].Gurtin [1, §§5, 6].Kellogg [1, Chapter 4].Nickerson, Spencer, and Steenrod [1, Chapters 6-8, 10].Truesdell and Toupin [1, Appendix by J. L. Ericksen].

CHAPTER

III

Kinematics

6. BODIES. DEFORMAnONS. STRAIN

Bodies have one distinct physical property: they occupy regions ofeuclidean space G. Although a given body will occupy different regions atdifferent times, and no one of these regions can be intrinsically associatedwith the body, we will find it convenient to choose one such region, ffl say,as reference, and to identify points of the body with their positions in ffl.Formally, then, a body ffl is a (possibly unbounded) regular region in G. Wewill sometimes refer to ffl as the reference configuration. Points p E ffl arecalled material points; bounded regular subregions of P'A are called parts.

Continuum mechanics is, for the most part, a study of deforming bodies.Mathematically, a body is deformed via a mapping f that carries eachmaterial point p into a point

x = f(p).

The requirement that the body not penetrate itself is expressed by theassumption that f be one-to-one. As we shall see later, det Vf represents,locally, the volume after deformation per unit original volume; it is thereforereasonable to assume that det Vf # O. Further, a deformation withdet Vf < 0 cannot be reached by a continuous process starting in the referenceconfiguration; that is, by a continuous one-parameter family f.,. (0 ~ (J ~ 1)

41

42 III. KINEMATICS

of deformations with fo the identity, f1 = f, and det Vfa never zero. Indeed,since det Vfa is strictly positive at (T = 0, it must be strictly positive for all (T.

We therefore require that

det Vf > o. (1)

The above discussion should motivate the following definition. By adef ormation of fJI we mean a smooth, one-to-one mapping f which maps fJIonto a closed region in C, and which satisfies (1). The vector

u(p) = f(p) - p (2)

represents the displacement ofp (Fig. 1).When u is a constant, f is a translation;in this case

f(p) = p + u.

The tensor

F(p) = Vf(p) (3)

is called the deformation gradient and by (I) belongs to Lin ". A deformationwith F constant is homogeneous. In view of (4.9), every homogeneous defor-mation admits the representation

f(p) = f(q) + F(p - q) (4)

for all p, q E fJI, and conversely, a point field f on fJI that satisfies (4) withF E Lin + is a homogeneous deformation.

For any given value of q the right side of (4) is well defined for all p E Iff.Thus any homogeneous deformation of fJI can be extended to form a homo-geneous deformation of Iff.We therefore consider homogeneous deformationsas defined on all of Iff.

For future use we note the following properties of homogeneous deforma-tions:

(i) Given a point q and a tensor F E Lin +, there is exactly one homo-geneous deformation f with Vf = F and q fixed [i.e., f(q) = q].

Figure 1

6. BODIES. DEFORMATIONS. STRAIN 43

(ii) If f and g are homogeneous deformations, then so also is fog and

V(f 0 g) = (Vf)(Vg).

Moreover, if f and g have q fixed, then so does fog.

Proposition. Let f be a homogeneous deformation. Then given any point q wecan decompose f as follows:

f=d1og=god2 ,

where g is a homogeneous deformation with qfixed, while d, and d2 are trans-lations. Further, each of these decompositions is unique.

Proof (Uniqueness) Assume that the first decomposition f = d, 0 gholds. Then Vf = (Vd1)(Vg) and Vd1 = I (because d1 is a translation), sothat Vf = Vg. Therefore, by property (i) above, g is uniquely determined.Moreover, since d1 = fog-I, this implies that d1 is uniquely determined.That f = g 0 d2 is also unique has an analogous proof.

(Existence) By hypothesis,

f(p) = f(q) + F(p - q).

Since g must fix q and have Vg = Vf (= F) (cf. the previous paragraph),

g(p) = q + F(p - q).

Define

To complete the proof we must show that d1 and d2 are translations. Let

00 = f(q) - q.

Then, since

we have

d1(p) = f(q) + F(q + F-1(p - q) - q) = p + 00'

d2(p) = q + F-1(f(q) + F(p - q) - q) = p + F-1oO ' 0

The last proposition allows us to concentrate on homogeneous deforma-tions with a point fixed. An important example of this type of deformationis a rotation about q:

f(p) = q + R(p - q)

with R a rotation.

44 III. KINEMATICS

- - - -- - - -- - - -- - • - -- - q - -- - - -- - - -Figure 2

A second example is a stretch from q, for which

f(p) = q + U(p - q)

with U symmetric and positive definite. If, in particular,

U = I + (A. - l)e ® e

with A. > 0 and IeI = 1, then f is an extension of amount A. in the direction e.Here the matrix of U relative to a coordinate frame with e = e1 has thesimple form

[

A. 0 0][U] = 0 1 0,001

and the corresponding displacement, shown in Fig. 2, has components(u 1, 0, 0) with

Properties (i) and (ii) of homogeneous deformations, when used inconjunction with the polar decomposition theorem, yield the following

Proposition. Let f be a homogeneous deformation with q.fixed. Then f admitsthe decompositions

r = g 0 SI = S2 0 g,

where g is a rotation about q, while SI and S2 are stretches from q. Further, eachof these decompositions is unique. In fact, if F = RU = VR is the polar de-composition ofF = Vf, then

Vg = R,

Thus any homogeneous deformation (with a fixed point) can be de-composed into a stretch followed by a rotation, or into a rotation followedby a stretch. The next theorem yields a further decomposition of either ofthese stretches into a succession of three mutually orthogonal extensions.


Proposition. Every stretch f from q can be decomposed into a succession ofthree extensions from q in mutually orthogonal directions. The amounts anddirections of the extensions are eigenvalues and eigenvectors of V = Vf, andthe extensions may be performed in any order.

Proof Since V E Psym, we conclude from the spectral theorem that

V = L Aiei (8) e,i

with {ei } an orthonormal basis of eigenvectors and Ai > 0 the eigenvalueassociated with e, (Ai> 0 since V is positive definite). In view of the identities(1.2h,4'

where

Vi = I + (Ai - l)ei (8) e..

Let f i (i = 1, 2, 3) be the extension from q of amount Ai in the direction e., Byproperty (ii) of homogeneous deformations, f 1 0 f 2 0 f 3 is a homogeneousdeformation with q fixed and deformation gradient V 1V 2 V 3 = U. But falso has q fixed and Vf = U. Thus by property (i) of homogeneous deforma-tions,

f = f 1 0 f 2 0 f 3 .

As a matter of fact, it is clear that

f = fa( l ) 0 fa (2 ) 0 fa( 3 )

for any permutation (J of {I, 2, 3}. D

In view of the last proposition every stretch can be decomposed into asuccession of extensions, the amounts of the extensions being the eigenvalues..1.1' ..1.2, ..1.3 of U. For this reason we refer to the Ai as principal stretches. Notethat, since the stretch tensors V and V have the same spectrum, the stretchs, (ofthe proposition on p. 44) has the same principal stretches as the stretchS2. Note also that by (2.10) the principal invariants of V take the form

'l(V) = ..1.1 + ..1.2 + ..1. 3,

liV) = ..1. 1..1.2 + ..1.2..1.3 + ..1. 3..1.1,

'3(V) = ..1. 1..1.2..1.3•

We now turn to a study of general deformations of f!l. To avoid repeatedhypotheses we assume for the remainder of the section that f is a deformationof f!l. Since f is one-to-one its inverse f- 1

: f(f!l) -. f!l exists. Moreover, by (1),

46 III. KINEMATICS

Vf(p) is invertible at each point p of fJI, and we conclude from the smooth-inverse theorem (page 22) that f - 1 is a smooth map. Two other importantproperties of fare

°f(fJI) = f(fJlt,

f(ofJI) = of(fJI),(5)

where f(fJI)o denotes the interior of f(fJI). We leave the proof of (5) as anexercise.

The concept of strain is most easily introduced by expanding the deforma-tion f about an arbitrary point q E fJI:

f(p) = f(q) + F(q)(p - q) + o(p - q),

where F is the deformation gradient (3). Thus in a neighborhood ofa point qand to within an error of o(p - q) a deformation behaves like a homogeneousdeformation. This motivates the following terminology: let

F = RU = VR

be the pointwise polar decomposition of F; then R is the rotation tensor, Uthe right stretch tensor, and V the left stretch tensor for the deformation f.R(p) measures the local rigid rotation of points near p, while U(p) and V(p)measure local stretching from p. Since U and V involve the square roots ofFTF and FFT, their computation isoften difficult.For this reason weintroducethe right and left Cauchy-Green strain tensors, C and B, defined by

in components,

C_ ~ ofm ofm

ij - L...--'m 0Pi OPj

Note that

thus, since R is a rotation,

B = V2 = FFT;

B = RCRT;

(6)

(7)

(cf. (2.10) and Exercise 2.3).Recall that the angle () between two nonzero vectors u and v is defined by

u·vcos () =--

[u/lvl(0 ~ () ~ zr).


Cf(p + ae)

f(p)9,

f(.") f(p + ad)

(8)

Figure 3

Proposition. Let d and e be unit vectors and let p E £i. Then as ex --+ 0,

If(p + exe) - f(p) I --+ IU( ) Ilexl p e ,

and the angle between f(p + «d) - f(p) and f(p + exe) - f(p) tends to the anglebetween U(p)d and U(p)e (Fig. 3).

Proof For convenience, we write F for F(p) and U for U(p). Then givenany vector u,

f(p + zu) = f(p) + exFu + o(ex)

as ex --+ O. Let

Then

and

d", = f(p + exd) - f(p),

d, = exFd + o(ex),

e", = f(p + exe) - f(p).

e", = zFe + o(ex),

d", • e", --+ Fd • Fe = RUd . RUe = Ud • Ue,ex 2

since the rotation tensor R is orthogonal. Taking d = e leads us to (8).Next, let ()'" designate the angle between d, and e",. Then

cos () = d",' e", = d", • e", • ~ •~ --+ Ud· Ue'" Id",lIe,,1 ex2 Id",1 [e, I IUdIIUel'

which is the cosine of the angle between Ud and Ue. (Note that since U isinvertible, Ud, Ue ¥= 0, and since f is one-to-one, d, e, ¥= 0 for ex ¥= O. Hencethe last computation makes sense.) Finally, since the cosine has a continuousinverse on [0, n], ()'" tends to the angle between Ud and Ue. 0

48 III. KINEMATICS

Figure 4

The above proposition shows that the stretch tensor U measures localdistance and angle changes under the deformation. In particular, since Iex Iis the distance between p and q = p + exe, (8) asserts that to within an errorthat vanishes as ex approaches zero, IU(p)el is the distance between p and qafter the deformation per unit original distance. The next result shows thatU also determines the deformed length of curves in Pi. In this regard, notethat given any curve c in Pi, f 0 c is the deformed curve f(c(o)), 0 :::; a :::; 1(Fig. 4).

Proposition. Given any curve c in i!4,

length(f 0 c) = f IU(c(u))c(u)I da.

Proof. By definition,

length(f 0 c) = f Id~ f(c(u))\ da.

But by the chain rule,

:u f(c(u)) = F(c(u))c(u) = R(c(u))U(c(u))c(u),

where R(p) is the rotation tensor. Thus, since R is orthogonal,

Id~ f(C(U))! = IU(c(u))c(u)!. 0

(9)

A deformation that preserves distance is said to be rigid. More precisely,f is rigidif

If(p) - f(q)1 = Ip - ql (10)

6. BODIES. DEFORMA nONS. STRAIN 49

(11)

for all p, q E fJl. This condition imposes severe restrictions; indeed, as thenext theorem shows, a deformation f is rigid if and only if: (i) f is homo-geneous and (ii) Vf is a rotation.

Theorem (Characterization of rigid deformations). The following areequivalent:

(a) f is a rigid deformation.(b) f admits a representation of the form

f(p) = f(q) + R(p - q)

for all p, q E fJl with R a rotation.

(c) F(p) is a rotation for each p E fJl.(d) U(p) = Ifor each p E fJl.(e) For any curve c in fJl, length(c) = length(f 0 c).

Proof We will show that (a) ~ (b) =(c) =(d) =(e) =(b).

(a) ~ (b). Let f be a rigid deformation. If we use (4.2h and (4.3) todifferentiate

[f(p) - f(q)J . [f(p) - f(q)J = (p - q) . (p - q),

first with respect to q and then with respect to p, we find that

Vf(q)T[f(p) - f(q)J = p - q,

Vf(q)TVf(p) = I.

Taking q = pin (11h we see that Vf(p) is orthogonal at each p; hence (11himplies that

Vf(p) = Vf(q)

for all p and q, so that Vf is constant. Finally, since det Vf > 0, Vfis a rotation.Thus f is a homogeneous deformation with R = Vf a rotation. Conversely,assume that (b) holds. Then, since R is orthogonal,

[f(p) - f(q)J . [f(p) - f(q)J = R(p - q) . R(p - q) = (p - q) . (p - q)

and (a) follows.(b) = (c) = (d). If (b) holds, then Vf = R, so that (c) is satisfied. Assume

next that F(p) is a rotation. Then C(p) = F(plF(p) = I. But U(p)2 = C(p),and by the square-root theorem (page 13) the tensor U(p) E Psym whichsatisfies this equation is unique; hence U(p) = I.

(d) = (e). This is an immediate consequence of (9).(e) = (b). This is the most delicate portion of the proof. Assume that (e)

holds. It clearly suffices to show that

F is a constant rotation. (12)

50 Ill. KINEMATICS

Thus choose PoE rj and let 0 be an open ball centered at Po and sufficientlysmall that f(O) is contained in an open ball r in f(91) (cf. the discussion at theend of the proof). Let p, q E 0 (p :F q), and let c be the straight line from qto p:

Then, trivially,

c(o) = q + e(p - q), O::;u::;1.

Ip - ql = length(c),

and since the end points of f 0 care f(q) and f(p),

length(f 0 c) ~ If(p) - f(q)I;hence (e) implies that

If(p) - f(q) I ::; Ip - q]. (13)

Next, since f(q) and rep) lie in the open ball r c £(91), the straight line bfrom Ceq) to f(p) lies in f(91). Consider the curve c in 91 that maps into b:

c(u) = £-l(b(u», O::;u::;1.

Then the argument used to derive (13) now yields the opposite inequality

If(p) - f(q)1 ~ Ip - ql·

Thus (10) holds for all p, q E 0, and f restricted to 0 is a rigid deformation.The argument given previously [in the proof of the assertion (a) => (b)]therefore tells us that (12) holds on o.

We have shown that (12) holds in some neighborhood of each point ofBi. Thus, in particular, the derivative of F exists and is zero on rj; since 91 isconnected, this means that F is constant on 91. Thus (12) holds on 91. D

We remark that U in (d) can be replaced by C, V, or B without impairingthe validity of the theorem.

We now construct the open ball 0 used in the above proof. By (5)1' £(91)contains an open ball r centered at f(po). Moreover, since f is continuous,f- l(n is an open neighborhood of Po and hence contains an open ball 0centered at Po. Trivially, C(O) c r, so that 0 has the requisite properties.

It follows from (b) of the last theorem that every rigid deformation is atranslation followed by a rotation or a rotation followed by a translation,and conversely; thus (as a consequence of the first two propositions of thissection) every homogeneous deformation can be expressed as a rigid de-formation followed by a stretch, or as a stretch followed by a rigid deforma-tion.

6. BODIES. DEFORMAnONS. STRAIN 51

(14)

It is often important to convert integrals over f(glI) to integrals over glI.The next proposition, which we state without proof, gives the correspondingtransformation law.1

Proposition. Let f be a deformation of f!4, and let qJ be a continuous scalarfield on f(f!4). Then given any part glI of f!4,

f qJ(x) dVx = f q>(f(p)) det F(p) dVp ,

J'(i¥) i¥

f qJ(x)m(x) dAx = f q>(f(p))G(p)n(p) dA p ,

iJf(i¥) iJi¥

where

G = (det F)F-r,

while m and n, respectively, are the outward unit normal fields on of(glI) and ogll.

Given a part glI,

vol(f(glI)) = f dVJ'(i¥)

represents the volume of glI after it is deformed under f. In view of (14)1>

vol(f(glI)) = Ldet F dV,

and therefore, by the localization theorem (5.1),

(15)

(16)

where (lei is the closed ball of radius b and center at p. Thus det F gives thevolume after deformation per unit original volume.

We say that f is isochoric (volume preserving) if given any part glI,

vol(f(glI)) = vol(glI).

An immediate consequence of this definition is the following

Proposition. A deformation is isochoric if and only if

det F = 1. (17)

1 For (14)1' cf., e.g., Bartle [I, Theorem 45.9]; for (14h, cf., e.g., Truesdell and Toupin[I, Eq. (20.8)].

52 III. KINEMATICS

EXERCISES

1. Establish properties (i) and (ii) of homogeneous deformations.

2. A homogeneous deformation of the form

Xl = PI + YP2'

is called a pure shear. For this deformation compute:

(a) the matrices of F, C, and B;(b) the list § c of principal invariants of C (or B);(c) the principal stretches.

3. Compute C, B, and § c for an extension of amount A in the direction e.

4. Show that a deformation is isochoric if and only if det C = 1.

5. Show that

C = I + Vu + VuT + VuT Vu.

6. Show that a deformation is rigid if and only if § c = (3, 3, 1).

7. Show that the principal invariants of C are given by

11(C) = Ai + A~ + AL

lic) = AiA~ + A~A~ + A~Ai,

13(C) = AiA~A~

with Ai the principal stretches.

8. A deformation of the form

Xl = fl(PI, P2),

X2 = f2(PI, P2),

is called a plane strain. Show that for such a deformation the principalstretch A3 (in the P3 direction) is unity. Show further that the deformationis isochoric if and only if the other two principal stretches, A" and Ap,

satisfy

1A" = T.

fJ


9. Let f l and f2 be deformations of fJl with the same right Cauchy-Greenstrain tensors. Show that there exists a rigid deformation g such that

f2 = go fl'

10. Establish the following analogs of (14h:

f v(x) . m(x) dA" = f v(f(p))' G(p)n(p) dA p ,

of(&') o&'

f T(x)m(x) dA" = f T(f(p))G(p)n(p) dA p , (18)of(&') o.~

f (x - 0) x T(x)m(x) dA" = f (f(p) - 0) x T(f(p))G(p)n(p) dA p •

of(&') o&'

Here v and T are continuous fields on f(fJl) with v vector valued and Ttensor valued.

11. Consider the hypothesis and notation of the proposition on page 47.The number

Id", x e",1IlXd x lXel

represents the ratio of the area dA" at x = f(p) spanned by d, =f(p + IXd) - f(p) and e", = f(p + lXe) - f(p) to the area dAp spanned bylXe and IXd (Fig. 5). Define

dA" = lim Id", x e",l.dA p ",-+0 IlXd x lXel

Use the identity

(Sa) x (Sb) = (det S)S-T(a x b)

f •

Figure 5

54 III. KINEMATICS

to show that

dA; G(m(x) dA = p)o(p),p

where m(x) and o(p) are the unit normals

I. d", x e",

m(x) = tm ld I'",-0 '" x e",

dxeo(p) = [d x e]"

[Cf. (14h,J

12. Establish (5).

13. Let f!l be the closed half-space

f!l = {piO :;;; PI < co}

and consider the mapping f on f!l defined by

1Xl = - --,

PI + 1

X3 = P3'

(a) Verify that f is one-to-one and det Vf > O.(b) Compute f(f!l) and use this result to demonstrate that f is not a

deformation.(c) Show that (5)2 is not satisfied.

7. SMALL DEFORMATIONS

We now study the behavior of the various kinematical fields when thedisplacement gradient Vu is small. Since

f(p) = p + u(p),

it follows that

F = I + Vu; (1)

hence the Cauchy-Green strain tensors C and B, defined by (6.6), obey therelations

C = I + Vu + VuT + VuT Vu,

B = I + Vu + VuT + VuVuT•

(2)

7. SMALL DEFORMAnONS

When the deformation is rigid, C = B = I and

Vu + VuT + VuVuT = O.

Moreover, in this case Vu is constant, because F is.The tensor field

is called the infinitesimal strain; clearly

C = I + 2E + VuT Vu,

B = I + 2E + VuVuT.

55

(3)

(4)

(5)

Proposition. Let f. (0 <s < eo) be a one-parameter family of deformationswith

IVu.1 = e.

Then

2E. = C. - I + o(e) = B. - I + o(e)

as s ~ O. Further, if each f. is rigid, then

VUE = - Vui' + o(e).

(6)

(7)

Proof The result (6) is a trivial consequence of (2), while (7) followsfrom (3). 0

This proposition asserts that to within an error of order o(e) the tensors2E., C. - I, and B. - I coincide. It asserts, in addition, that to within thesame error the displacement gradient corresponding to a rigid deformationis skew.

The above discussion should motivate the following definition: Aninfinitesimal rigid displacement of f!4 is a vector field u on f!4 with Vuconstantand skew; or equivalently, a vector field u that admits the representation

u(p) = u(q) + W(p - q) (8)

for all p, q E f!4, where W is skew (cf. the proposition on page 36). Of course,using the relation between skew tensors and vectors, we can also write u inthe form

u(p) = u(q) + co x (p - q)

with cothe axial vector corresponding to W.

56 III. KINEMATICS

(9)

Theorem (Characterization of infinitesimal rigid displacements). Let u bea smooth vector field on f!4. Then the following are equivalent:

(a) u is an infinitesimal rigid displacement.(b) u has the projection property: for all p, q E f!4,

(p - q) . [u(p) - u(q)] = o.(c) Vu(p) is skew at each p E f!4.(d) The infinitesimal strain E(p) = 0 at each p E f!4.

Proof (a) => (b). Let u be rigid. Then (8) implies

(p - q) • [u(p) - u(q)J = (p - q) . W(p - q) = 0,

since W is skew.(b) => (a). If we differentiate the expression in (b) with respect to p, we

arrive at

u(p) - u(q) + VU(p)T(p - q) = 0,

and this result, when differentiated with respect to q, yields

- Vu(q) - Vu(p? = O.

Taking p = q tells us that Vu(p) is skew; hence (9) implies that

Vu(p) = Vu(q)

for all p, q E 14, and Vu is constant. Thus (a) holds.(a) ¢> (c). Trivially, (a) implies (c). To prove the converse assertion

assume that H(p) = Vu(p) is skew at each p E 14. We must show that H isconstant. Let n be an open ball in f!4. Choose p, q E n and let

c(a) = q + a(p - q), 0 ~ a ~ 1,

so that c describes the straight line from q to p. Then

u(p) - u(q) = i Vu(x) dx = fH(c(a»c(a) do = fH(c(a»(p - q) do,

so that

(p - q) • [u(p) - u(q)] = f(p - q) . H(c(a»(p - q) do = 0,

since H is skew. Thus u has the projection property on n, and the argumentgiven previously [in the proof of the assertion (b) => (a)] tells us that H isconstant on n.But n is an arbitrary open ball in f!4; thus H is constant on f!4.

(c) ¢> (d). This is a trivial consequence of (4). 0

7. SMALL DEFORMAnONS

EXERCISES

1. Under the hypotheses of the proposition containing (6) show that

E. = U. - I + 0(8) = V. - I + 0(8),

det F. - 1 = div U. + 0(8).

57

Give a physical interpretation of det F. - 1 in terms of the volumechange in the deformation fe :

2. Let u and v be smooth vector fields on rJI and suppose that u and vcorrespond to the same infinitesimal strain. Show that u - v is an in-finitesimal rigid displacement.

For the remaining exercises u is a smooth vector field on rJI and E is thecorresponding infinitesimal strain. Also, in 3 and 5, rJI is bounded.

3. Define the mean strain E by

vol(rJI)E = LE dV.

Show that

- 1 rvol(rJI)E = "2 J

oal(u ® n + n ® u) dA,

so that E depends only on the boundary values of u.

4. Let

w = t(Vu - VuT) .

Show that

IEI2 + IWl 2 = IVuI2,

IEl 2- IWI 2 = Vu' VuT

•

5. (Korn's inequality) Let u be of class C2 and suppose that

Show that

u=O on i}rJI.

58 III. KINEMATICS

6. Consider the deformation defined in cylindrical coordinates

Xl = r cos e,X 2 = r sin e,

PI = R cos e,P2 = R sin e,P3 = Z,

by

r = R, e= e + rLZ, z = z.A deformation of this type is called a pure torsion; it describes a situationin which a cylinder with generators parallel to the Z-axis is twisteduniformly along its length with cross sections remaining parallel andplane. The constant rL represents the angle of twist per unit length. Showthat the corresponding displacement is given by

uI(p) = PI(COS /3 - 1) - P2 sin /3, /3 = rLP3'

uip) = P2(COS /3 - 1) + PI sin /3,U3(P) = O.

Show further that both Vu and u approach zero as a -+ 0, and that, infact,

as rL -+ O.

UI(P) = -rLP2P3 + o(rL),

uip) = rLPIP3 + o(rL)

8. MOTIONS

(10)

Let fJd be a body. A motion of fJd is a class C3 function

x:fJd x ~-+g

with x(', t), for each fixed t, a deformation of fJd (Fig. 6). Thus a motion is asmooth one-parameter family ofdeformations, the time t being the parameter.We refer to '

x = x(p, t)

as the place occupied by the material point p at time t, and write

fJdt = x(fJd, t)

1 We carefully distinguish between the motion x and its values x, and between the referencemap p and material points p.

8. MOTIONS

• x = X(P. t)

Figure 6

59

for the region of space occupied by the body at t. It is often more convenientto work with places and times rather than with material points and times,and for this reason we introduce the trajectory

At each t, x(', t) is a one-to-one mapping of flI onto flit; hence it has aninverse

p(', t): flit ~ flI

such that

Given (x, t) E !Y,

x(p(x, r), t) = x, p(x(p, r), t) = p.

p = p(x, t)

is the material point that occupies the place x at time t. The map

p: !Y ~ flI

so defined is called the reference mapof the motion.We call

. ax(p, t) = at x(p, t)

the velocity and

x(p, t) = :t22 x(p, t)

60 III. KINEMATICS

the acceleration. Using the reference map p we can describe the velocityx(p, r) as a function vex, t) of the place x and time t. Specifically,

v: ,'Y -+ 'f~

is defined by

vex, t) = x(p(x, t), t)

and is called the spatial description of the velocity. The vector vex, t) is thevelocity of the material point which at time t occupies the place x.

More generally, any field associated with the motion can be expressed asa function of the material point and time with domain fJ4 x IR, or as a functionof the place and time with domain ,'Y. We therefore introduce the followingterminology: a material field is a function with domain fJ4 x IR; a spatial fieldis a function with domain :Y. The field i is material, the field v is spatial. Itis a simple matter to transform a material field into a spatial field, and viceversa. We define the spatial description <lid ofa material field (p, t) f---+ <II(p, t) by

<II.(x, t) = <II(p(x, t), t),

and the material description Q", of a spatial field (x, t) f---+ Q(x, t) by

Q",(p, t) = Q(x(p, t), t).

Clearly,

(Q"')d = Q.

Smoothness Lemma. The reference map p is of class c: Thus a materialfield is ofclass en (n ~ 3) ifand only if its spatial description is ofclass en,

The proof of this lemma will be given at the end of the section.Given a material field <II we write

odl(p, t) = ot <II(p, t)

for the derivative with respect to time t holding the material point p fixed,and

V<II(p, t) = Vp <II(p, t)

for the gradient with respect to p holding t fixed. <1> is called the materialtime derivative of <II, V<II the material gradient of <II. In particular, the materialfield

F = Vx

8. MOTIONS 61

(1)

is the deformation gradient in the motion x. Since the mapping p ~ x(p, t) isa deformation of 14,

det F > O.

Similarly, given a spatial field 0 we write

aO'(x, t) = at O(x, t)

for the derivative with respect to t holding the place x fixed, and

grad O(x, t) = V"O(x, t)

for the gradient with respect to x holding t fixed. 0' is called the spatial timederivative of 0, grad 0 the spatial gradient of O.

We define the spatial divergence and the spatial curl, div and curl, to bethe divergence and curl operations for spatial fields, so that grad is the under-lying gradient. Similarly, Div and Curl designate the material divergenceand the materialcurlcomputed using the material gradient V.

The notation introduced above is summarized in Table 1.It is also convenient to define the material time derivative nof a spatial

field O. Roughly speaking, n represents the time derivative of 0 holding thematerial point fixed. Thus to compute n we transform 0 to the materialdescription, take the material time derivative, and then transform back tothe spatial description:

that is,Q = «0",)").;

aQ(x, t) = at O(x(p, t), t)lp~p(",tl'

(2)

The next proposition shows that the material time derivative commuteswith both the material and spatial transformations.

Table 1

DomainArguments

Gradient with respect tofirst argument

Derivative with respect tosecond argument (time)

DivergenceCurl

Material field l1>

~x~

Material point pand time t

<bDivl1>Curll1>

Spatial field n

.rPlace x and

time t

grad n

n'divncurl n

(3)

62 III. KINEMATICS

Proposition. Let et> be a smooth material field, Q a smooth spatial field. Then

(<1»,,= (et>,,)' == <1>",

(0)", = (Q",)' == 0 ....

Proof Ifwe take the material description of (2), we arrive at (3)2' Also,by definition,

(et>,,). = «et>,,)...)')., = (<1>).,. D

Note that, by (3h with Q = v,

(v)... = (v",), = ii,

so that vis the spatial description of the acceleration.The relation between the material and spatial time derivatives is brought

out by the following

.Proposition. Let cp and u be smooth spatial fields with cp scalar valued andu vector valued. Then

ciJ = cp' + v . grad cp,

U = u' + (grad u)v.

Thus, in particular,

v= Vi + (grad v)v.

Proof By the chain rule,

ciJ(x, t) = :t cp(x(p, t), t)lp=p(".t)

= [grad cp(x, t)] • x(p(x, r), t) + cp'(x, t)

= v{x, t) • grad cp(x, t) + cp'(x; t),

au(x, t) = at u(x(p, t), t) Ip=p(". t)

= [grad u(x, t)]x(p(x, t), t) + u'(x, t)

= [grad u(x, t)]v(x, t) + u'(x, t). D

(4)

(5)

A simple application of (4) is expressed in the next result, which gives thematerial time derivative of the position vector r: tff -+ "f/ defined by

r(x) = x - o.

8. MOTIONS 63

Proposition. Consider the position vector as a spatial field by definingr(x, t) = r(x)for every (x, t) E .'Y. Then

t = v. (6)

Proof Since r' = 0 and grad r = I, (4)2 with U = r yields (6). [Thisresult can also be arrived at directly by noting that r... (p, z) = x(p, t) - 0.] 0

Proposition. Let u be a smooth spatial vector field. Then

V(u...) = (grad u)",F,

where F is the deformation gradient.

Proof By definition,

u...(p, t) = u(x(p, t), t);

that is,

(7)

(8)

u",C t) = u(', r) a X(', t).

Thus the chain rule (3.11) tells us that V(u...) is the gradient grad u ofu timesthe gradient F = Vx of x. 0

The spatial field

L=gradv

is called the velocity gradient.

Proposition

F= L... F,

F = (grad v)...F.

Proof Since x is by definition C3,

. 0F(p, t) = ot Vx(p, t) = Vit(p, t) = Vv...(p, t),

and (8)1 follows from (7) with u = v. Similarly,

F(p, t) = Vx(p, t) = Vv",(p, t),

and taking u = vin (7) we are led to (8h. 0

Given a material point p, the function s: IR --+ S defined by

s(t) = x(p, t)

is called the path line of p. Clearly, s is a solution of the differential equation

S(t) = v(s(t), t),

64 Ill. KINEMATICS

and conversely every maximal solution of this equation is a path line. (Asolution is maximal provided it is not a portion of another solution.) On theother hand, if we freeze the time at t = r and look at solution curves of thevector field v(', r), we get the streamlines of the motion at time r, Thus eachstreamline is a maximal solution s of the differential equation

S(A) = V(S(A), r).

An example of a motion x (of 8) is furnished by the mapping defined incartesian components by

Xl = PIel >,X 2 = P2 el,

The deformation gradient F is given by

[

12

[F(p, I)] ~ e~

while the velocity i has components

o.

Thus, since the reference map p is given by

the spatial description of the velocity has components

vI(x, t) = 2x l t,

V2(X, t) = X2,

and the velocity gradient L has the matrix

[

2t

[L(x, t)J = ~

8. MOTIONS

The streamlines of the motion at time t are solutions of the system

Sl(A) = 2tS 1(A),

S2(A) = S2(A),

S3(A) = 0,

so that

65

Sl(A) = y 1e2 t\

S2(A) = Y2e',

S3(A) = Y3

is the streamline passing through (Yl, Y2, Y3) at A = O.We close this section by giving the

Proofofthe Smoothness Lemma. It suffices to show that p is of class C 3,

for then the remaining assertion in the lemma follows trivially.Consider the mapping

defined by

'I'(p, t) = (x(p, t), t).

It follows from the properties of x that 'I' is class C3 and one-to-one; in fact,

'I'-l(X, t) = (p(x, r), t).

Thus to complete the proof it suffices to show that the derivative

D'I'(p, t): 11 x IR --+ 11 x IR

is invertible at each (p, t), for then the smooth-inverse theorem (page 22) tellsus that 'I' - 1 is as smooth as '1', and hence that p is of class C3

•

Since

x(p + h, t + r) = x(p, t) + F(p, t)h + x(p, t)r + 0(8)

as 8 = (h2 + r 2) 1!2 --+ 0, it follows that

'I'(p + h, t + r) = 'I'(p, t) + (F(p, t)h + x(p, t)r, r) + 0(8).

Thus

D'I'(p, t) [h, r] = (F(p, t)h + x(p, t)r, r)

for all h E 11 and r E IR.

(9)

66 III. KINEMATICS

To show that D'I'(p, t) is invertible, it suffices to show that

D'I'(p, t) [h, r] = 0 (10)

implies

h = 0, r = o.Thus assume that (10) holds. Then by (9),

r = 0, F(p, t)h = 0,

and, since F(p, t) is invertible, h = 0. 0

EXERCISES

1. A motion is a simple shear if the velocity field has the form

v(x, t) = v1(x2)el

in some cartesian frame. Show that for a simple shear

div v = 0, (grad v)v = 0, v= v',

In the next two exercises D and W, respectively, are the symmetric and skewparts of grad v.

2. Prove that

C = 2FTD",F.

3. Let v be a class C2 velocity field. Show that

div v= (div v): + IDI2- /WI 2

•

4. Consider the motion of $ defined by

X3 = P3'

in some cartesian frame. Compute the spatial velocity field v and deter-mine the streamlines.

5. Consider the motion x defined by

x(p, t) = Po + U(t)[p - Po],

where3

U(t) = L Q(j(t)ej ® e,i= 1

9. TYPES OF MOTIONS. SPIN. RATE OF STRETCHING 67

with !Xi> 0 smooth. (Here {ei } is an orthonormal basis.) Compute p,v, and L, and determine the streamlines.

6. Define the spatial gradient and spatial time derivative of a material fieldand show that

X' = 0, grad x = I.

7. Consider a surface [I' in fJI of the form

[I' = {p E .@lqJ(p) = O},

where .@ is an open subset of fJI and qJ is a smooth scalar field on .@ withVqJ never zero on [1'. Let x be a motion of fJI. Then, at time t, [I' occupiesthe surface

[1', = {x E .@,Il/J(x, t) = O},

where .@, = x(.@, t) and

l/J(x, t) = qJ(p(x, t».

Show that:

(a) VqJ(p) is normal to [I' at p E [1';

(b) grad l/J(x, t) is normal to [1', at x E [1',;

(c) VqJ = FT (grad l/J)... , and hence grad l/J(x, t) never vanishes on Y't;(d) IVqJ1

2 = (grad l/J)... • B(grad l/J)_, where B = FFT is the leftCauchy-Green strain tensor;

(e) l/J' = - v • grad l/J.

9. TYPES OF MOTIONS. SPIN. RATE OF STRETCHING

A motion x is steady if

for all time t and

v' = 0

everywhere on the trajectory ff. Note that !!I = fJlo x IR, because the bodyoccupies the same region fJlo for all time. Also, since the velocity field v isindependent of time, we may consider v as a function xr->v(x) on fJlo. Thusin a steady motion the particles that cross a given place x all cross x with thesame velocity v(x). Of course, for a given material point p the velocity willgenerally change with time, since x(p, t) = v(x(p, t».

Consider now a (not necessarily steady) motion x and choose a point pwith x(p, T)E ofJI, at some time T. Then (6.5)2 with f = x(', T) implies that

68 III. KINEMATICS

(1)

P E iJ!!J and a second application of (6.5h, this time with f = x(', r), tells usthat x(p, t) E iJ!!Jt for all t. Thus a material point once on the boundary lies onthe boundary for all time. If the boundary is independent of time (iJ!!Jt = iJ!!Jofor all r), as is the case in a steady motion, then x(p, t), as a function of t,describes a curve on iJ!!Jo, and x(p, t) is tangent to iJ!!Jo. Thus we have thefollowing

Proposition. In a steady motion the velocity field is tangent to the boundary;i.e., v(x) is tangent to iJ!!Jo at each x E iJ!!Jo.

In a steady motion path lines and streamlines satisfy the same autonomousdifferential equation

s(t) = v(s(t».

Thus, as a consequence of the uniqueness theorem for ordinary differentialequations, we have the following

Proposition. In a steady motion every path line is a streamline and everystreamline is a path line.

Let be a smooth field on the trajectory of a steady motion. Then issteady if

<1>' = 0

[in which case we consider as a function x ....... (x) on !!JoJ.

Proposition. Let qJ be a smooth, steady scalar field on the trajectory of asteady motion. Then the following are equivalent:

(a) qJ is constant on streamlines; that is, given any streamline s,

ddt qJ(s(t» = 0

for all t.

(b) q, = O.(c) v· grad qJ = O.

Proof Note first that, by (8.4)1 and (1),

q, = v • grad qJ;

thus (b) and (c) are equivalent. Next, for any streamline s,

:t qJ(s(t» = s(t) • grad qJ(s(t» = v(s(t» • grad qJ(s(t», (2)

so that (c) implies (a). On the other hand, since every point of !!Jo has astreamline passing through it, (a) and (2) imply (c). D


x(q, t)

69

Figure 7 x(P, t)

The number

bet) = Ix(p, t) - x(q, t) I

x(a, t)

(3)

represents the distance at time t between the material points p and q. Similarly,the angle OCt) at time t subtended by the material points a, p, q is the anglebetween the vectors x(a, t) - x(p, t) and x(q, t) - x(p, t). (See Fig. 7.)

A motion x is rigid if

aat Ix(p, t) - x(q, t)! = 0 (4)

for all materials points p and q and each time t. Thus a motion is rigid if thedistance between any two material points remains constant in time.

Theorem (Characterization of rigid motions). Let x be a motion, and let vbe the corresponding velocity field. Then the following are equivalent:

(a) x is rigid.(b) At each time t, v(', t) has the form ofan infinitesimal rigid displacement

of fJI,; that is, v(', t) admits the representation

Vex, t) = v(y, t) + W(t)(x - y) (5)

for all x, y E fJI"

where Wet) is a skew tensor.(c) The velocity gradient L(x, t) is skew at each (x, t) E .'!I.

Proof. If we use (3) to differentiate b(t)2, we find that

bet) J(t) = [x(p, t) - x(q, t)] • [x(p, t) - x(q, t)],

or equivalently, letting x and y denote the places occupied by p and q at time t,

bet) J(t) = (x - y) • [vex, t) - v(y, t)J. (6)

By (6), (4), and the fact that bet) 'I- 0 for p 'I- q, x is rigid if and only if v(-, t)has the projection property at each time t. The equivalence of (a), (b), and(c) is therefore a direct consequence of the theorem characterizing infinitesi-mal rigid displacements (page 56). D

70 III. KINEMATICS

Let ro(t) be the axial vector corresponding to W(t); then (5) becomes

v(x, t) = v(y, t) + ro(t) x (x - y),

which is the classical formula for the velocity field of a rigid motion. Thevector function ro is called the angular velocity. Note that

curl v = 2ro,

which gives a physical interpretation of curl v, at least for rigid motions.For convenience, we suppress the argument t and write

v(x) = v(y) + ro x (x - y).

Assume ro ¥- O. Then for fixed y the velocity field

XHro x (x - y)

vanishes for x on the line

{y + O(rolO( E ~}

and represents a rigid rotation about this line. Thus given any fixed y, v isthe sum of a uniform velocity field with constant value

v(y)

and a rigid rotation about the line through y spanned by roo For this reasonwe calli = sp{ro} the spin axis. For future use, we note that 1= l(t) can alsobe specified as the set of all vectors e such that

We=O.

As we have seen, a rigid motion is characterized (at each time) by a velocitygradient which is both constant and skew. We now study the case in whichthe gradient is still constant, but is symmetric rather than skew. Thus con-sider a velocity field of the form

v(x) = D(x - y)

with D a symmetric tensor. By the spectral theorem D is the sum of threetensors of the form

O(e ® e, lei = 1,

with corresponding e's mutually orthogonal. It therefore suffices to limit ourdiscussion to the velocity field

v(x) = O(e ® e)(x - y).

Relative to a coordinate frame with e = e., v has components

(7)

(Vl' 0, 0),

- - - -- - - -- - - -- - • - -y -- - -- - - -- - - -Figure 8

and is described in Fig. 8. What we have shown is that every velocity fieldwith gradient symmetric and constant is (modulo a constant field) the sum ofthree velocity fields of the form (7) with" axes" e mutually perpendicular.

Now consider a general velocity field v. Since L = grad v, it follows that

v(x) = v(y) + L(y)(x - y) + o(x - y)

as x --+ y, where y is a given point, and where we have suppressed the argu-ment t. Let D and W, respectively, denote the symmetric and skew parts ofL:

D = !<L + U) = !<grad v + grad vT) ,

W = !<L - U) = !<grad v - grad vT).

Then

L=D+W

and

v(x) = v(y) + W(y)(x - y) + D(y)(x - y) + o(x - y).

Thus in a neighborhood ofa given point y and to within an error ofo(x - y) ageneral velocity field is the sum ofa rigid velocity field

X 1--+ v(y) + W(y)(x - y)

and a velocity field of the form

x 1--+ D(y)(x - y).

For this reason we call W(y, t) and D(y, r), respectively, the spin and thestretching, and we use the term spinaxis at (y, t) for the subspace I of ~ con-sisting of all vectors e for which

W(y, t)e = O.

[Of course, I has dimension one when W(y, t) ¥- O.J The term stretching isfurther motivated by the following

72 III. KINEMATICS

.11,

Figure 9

oProposition. Let x E iJlt with t afixed time, let e be a unit vector, and let c5,,.(r)(for IX sufficiently small) denote the distance at time t between the materialpoints that occupy the places x and x + lXe at time t (Fig. 9). Then

1. J~(t) )im s: ( ) = e . D(x, t e.

~-o u~ t(8)

Further, ifd is a unit vector perpendicular to e, and ifO~(t) is the angle at timer subtended by the material points that occupy the places x + ze, x, x + IXd attime t, then

lim e~(t) = -2d' D(x, t)e.~-o

Proof Since c5aCt) is the distance between x and x + lXe, which is IX, (6)with y = x + lXe implies

J~(t) e • [v(x + lXe, t) - vex, t)]

c5~(t) IX

But

lim! [v(x + lXe, t) - vex, t)] = L(x, t)ee-e OIX

and

e : L(x, t)e = e • D(x, t)e

[since W(x, t) is skew]. Thus (8) holds.

(9)


Next, let p, q, and a denote the material points that occupy the places x,y = x + ze, and z = x + ad, respectively, at time t. Further, let

u(r) = x(q, r) - x(p, r),

w(.) = x(a, r) - x(p, r),

so that

Thus

u(t) = «e,

w(t) = ad,

li(t) = v(y, t) - v(x, t),

w(t) = v(z, t) - v(x, t).

Further,

1- (u 0 w)o(t) = eo [v(z, t) - v(x, t)J + d 0 [v(y, t) - v(x, t)J.a

u oW

cos Oa. = [ullw!'

and, as u and ware orthogonal at time t,

(u 0 w)o(t)(cos Oa.)o(t) = Iu(t) IIw(t) I

On the other hand, since sin Oit) = 1,

(cos Oa.)°(t) = - Oa.(t),

and the above relations imply that

-aOa.(t) = eo [v(x + ad, t) - v(x, t)J + do [v(x + ae, t) - v(x, t)J.

Ifwe divide by a and let a --+ 0, we conclude, with the aid of(9) and its counter-part for d, that

lim Oit) = -e 0 L(x, t)d - do L(x, t)e

= -d 0 [L(x, t) + L(x, t?Je

= -2d 0 D(x, t)e. 0

Using the spin W we can establish the following important relations forthe acceleration v.Proposition

v= V' + t gradiv') + 2Wv,

v= v' + t grad(v2) + (curl v) x v.

(10)

74 III. KINEMATICS

Proof Since

2Wv = (grad v - grad vT)v = (grad v)v - t gradtv"), (11)

(10)1 follows from (8.5). The result (10)2 follows from (10)1 and the fact thatcurl v is twice the axial vector corresponding to W. 0

A motion is plane if the velocity field has the form

v(x, t) = V1(X 1, X2' t)e l + V2(XI, X 2, t)e2

in some cartesian frame.

Proposition. In a plane motion

WD + DW = (div v)W.

Proof Clearly, D and W have matrices of the form

[D] = [~ ~ ~], [W] = [~y ~ ~]o 0 0 0 0 0

(relative to the above frame), and a trivial computation shows that

[0 Y(IX + fJ) 0]

[WD + DW] = -Y(IXO+ fJ) ~ ~ = (IX + P)[W]·

But

div v = tr L = tr D = IX + P,and the proof is complete. 0

EXERCISES

(12)

1. It is often convenient to label material points by their positions at a giventime r. Suppose that a material point p occupies the place y at r and X

at an arbitrary time t (Fig. 10):

y = x(p, r), X = x(p, t).

Roughly speaking, we want x as a function of y. Thus, since

p = p(y, r),

we have

x = x(p(y, r], t).


-X,(', I)

Figure 10

We call the function

defined by

75

xlY, t) = x(p(y, r), t)

the motion relative to time r ; xly, t) is the place occupied at time t by thematerial point that occupies y at time r, Let

Fly, t) = Vyx.(y, t),

where Vy is the gradient with respect to y holding t fixed. Also let

Ft=RtUt

denote the right polar decomposition of F" and define

C, = (Ut)2,

(a) Show thata

v(x, t) = at xlY, t) (13)

provided x = xt(y, t).(b) Use the relation xC t) = x.(-, t) 0 x(', r) to show that

Fly, t)F(p, r) = F(p, r),

where y = x(p, r), and then appeal to the uniqueness of the polardecomposition to prove that

Fly, r) = Uly, r) = Rly, r) = I.

76 III. KINEMATICS

(c) Show that

(d) Show that

C(p, t) = F(p, ,yciy, t)F(p, T). (14)

(17)

aL(y, T) = at FlY, t)lt=t

= [~ Diy, t) + ~ RlY, t)] , (15)ot ot t=t

a aD(y, T) = at Diy, t)lt=t' W(y, T) = at Riy, Olt=t· (16)

(e) Show that

an + 2

:l n+2 Fiy, t)lt=t = grad aInley, T),ut

where a(nl is the spatial description of the material time derivativeof x of order n + 2.

2. Let x be a motion and suppose that for some fixed T,

xt(y, t) = q(t) + Q(t)(y - z),

with z and q(t) points and Q(t) E Orth+. Show that x is rigid.

3. Let x be a rigid motion. Show that x, has the form (17).

Exercises 2 and 3 assert that, given a motion x and a time T, xl, t) is a rigiddeformation at each t if and only if x is a rigid motion.

4. Let x be a COO motion. The tensors

anAnCy, T) = atn Ct(y, t)lt=t (n = 1,2, ...) (18)

are called the Rivlin-Ericksen tensors.

(a) Show that At = 2D.(b) Show, by differentiating (14) with respect to t, that

(19)

where c(nl is the nth material time derivative of C, and where wehave omitted the subscript {) from C and F.

(c) Verify that

10. TRANSPORT THEOREMS. VOLUME. ISOCHORIC MOTIONS 77

5. Show that the acceleration field of a rigid motion has the form

v(x, t) = V(y, t) + m(t) x (x - y) + ro(t) x [CO(t) x (x - y)]

with co the angular velocity.

10. TRANSPORT THEOREMS. VOLUME.ISOCHORIC MOTIONS

Let x be a motion of !!l. Given a part f!j>, we write

for the region of space occupied by f!j> at time t. Thus

represents the volume of f!j> at time t. Using the deformation gradient F = Vxwe can also express vol(f!j>t) as an integral over f!j> itself [cf. (6.15)J:

Thus

:t vol(f!j>t) = fa. (det F)· dY.

Next, (3.14) and (8.8)1 imply that

(det F)" = (det F) tr(FF- 1) = (det F) tr L",.

But

tr L = tr grad v = div v,

so that

(det F)· = (det F)(div v)",

and (1) becomes

dd vol(f!j>t) = f (div v)",det F dV = f div v dY.t ~ ~,

(1)

(2)

78 III. KINEMATICS

Thus we have the following

Theorem (Transport of volume). For any part f!IJ and time t,

:t vol(f!IJ,) = f (det F)' dV = f div v dV = f v· n dA. (3)~ ~, Jo~,

Thus (det F)' and div v represent rates of change of volume per unitvolume: (det F)' is measured per unit volume in the reference configuration;div v is measured per unit volume in the current configuration.

We say that x is isochoric if

ddt vol(f!IJ,) = 0 (4)

for every part f!IJ and time t. As a direct consequence of (3) and (4) we have thefollowing

Theorem (Characterization of isochoric motions). The following areequivalent:

(a) . x is isochoric.(b) (det F)' = O.(c) div v = O.(d) For every part f!IJ and time t,

f v· ndA = O.Jo~,

Warning: For a motion to be isochoric the volume of each part must beconstant throughout the motion; it is not necessary that the volume of a partduring the motion be equal to its volume in the reference configuration.

Note that rigid motions are isochoric, a fact which follows from (c) of theabove theorem and (c) of the theorem on page 69.

The computations leading to (3) are easily generalized; the result is

Reynolds' Transport Theorem. Let be a smooth spatial field, and assumethat is either scalar valued or vector valued. Thenfor any part f!IJ and time t,

: f dV = f (d> + div v) dV,t ~, ~t

: f dV = f <1>' dV + f <l>v • n dA.t ~t e, JO~t

(5)

10. TRANSPORT THEOREMS. VOLUME. ISOCHORIC MOTIONS 79

Proof In view of (2),

dd i <D dV = dd i <D",det F dV = i (<D...det F)'dVt WI, t WI WI

= r (ci> + <D div v)", det F dV = i (ci> + <D div v) dV,JWI WI,

which is (5)1' To derive (5h assume that <D is scalar valued. Then (8.4)1 yieldsthe identity

ci> + <D div v = <D' + div(<Dv),

and (5h follows from (5)1 and the divergence theorem. The proof for <Dvector valued is exactly the same. 0

Note that

i <D'dV = i : <D(x, t) dv" = [: i <D(x, r) dv"J .WI, WI, t r WI, t=t

Thus (5h asserts that the rate at which the integral of <D over f!lJt is changingis equal to the rate computed as if f!lJ t were fixed in its current position plusthe rate at which <D is carried out of this region across its boundary.

EXERCISES

1. Let Pbe a smooth spatial scalar field with p= 0 and define so that in an isochoric motionwith v = 0 on of!Jt ,

i WvdV = O.91,

3. Derive (5)2 for <D vector valued.

80 III. KINEMATICS

11. SPIN. CIRCULATION. VORTICITY

As we have seen, a rigid motion is characterized by a skew velocitygradient; that is, the velocity field is determined (up to a spatially uniformvector field) by its spin

W = -!(grad v - grad vT).

More generally, given any motion the spin W(x, t) describes the local rigidrotation of material points currently near x. As our next theorem shows, theevolution of W with time is govemed by the field

J = -!(grad v- grad vT) ,

which represents the skew part of the acceleration gradient. The statementand proof of this theorem are greatly facilitated if we introduce the notation

(1)

(2)

for any spatial tensor field G, where F is the deformation gradient.

Theorem (Transport of spin). The spin W satisfies the differential equations

(WF) · = J F ,

W+DW+WD=J,

where D is the stretching.

Proof Recall (8.8):

L = FF- 1, F = (grad t)F,

where for convenience we have omitted the subscript m from L and grad v.Since 2W = L - LT,

and hence

2WF = FTP - PTF = FTgrad vF - FTgrad vTF,

which implies (2)1' Next, by (1) and (2)1'

FTWF + FTWF + FTWF = FTJF.

Thus

W + F-TFTW + WFF- 1 = J,

and, since L = FF- 1 and LT = F-TFT, this equation implies

W + LTW + WL = J.

Thus as

II. SPIN. CIRCULATION. VORTICITY 81

it follows that

L = D + W,

W+ (D - W)W + W(D + W) = J,

which implies (2h. 0

We now introduce two important (and somewhat related) definitions. Amotion is irrotational if

W=o,

or equivalently, if

curl v = o.A spatial vector field g is the gradient of a potential if there exists a spatialscalar field tl. such that

g(x, t) = grad z(x, t)

for all (x, t) on the trajectory of the motion.For a large class of fluids, in particular, inviscid fluids under a conservative

body force, the acceleration vis the gradient of a potential. When this is thecase the potential, tl. say, is C2 in x, because vis C1

, and thus grad grad tl. is sym-metric; therefore

J = t[grad grad tl. - (grad grad tl.)T] = 0,

and we have the following consequence of (2)1.

Lagrange-Cauchy Theorem. A motion with acceleration the gradient of apotential is irrotational if it is irrotational at one time.

Proof: Since J = 0, we conclude from (2)1 that

(WF) " = 0, (3)

so that WF(p, t) is independent of t. But at some time r, W(x, r) = 0 for allx in f!4" and hence WF(p, r) = 0 for all P in f!4. Thus WF(p, t) = 0 for all pand t, and, since F is invertible, (1) implies that W = O. Hence the motion isirrotational. 0

As is clear from (9.12) and (2h, for plane, isochoric motions a resultstronger than (3) holds.

Proposition. For a plane, isochoric motion with acceleration the gradient ofa potential,

W=o.

82 III. KINEMATICS

Figure 11

Let x be a motion of !!A. By a material curve we mean a curve c in !!A.Choose an arbitrary material point c(O') on c. At time t, c(O') will occupy theplace

x(c(O'), t).

Thus the material points of the curve form a curve

ct(O') = x(c(O'), t), (4)

in !!At (Fig. 11). When c (and hence e.) is closed,

f. v(x, t) • dxCt

gives the circulation around C at time t; this integral sums the tangentialcomponent of the velocity around the curve Ct.

Theorem (Transport of circulation). Let C be a closed material curve.Then

:t f. v(x, t) • dx = f. v(x, t) • dx.Ct C:t

Proof By definition,

f. v(x, t) • dx = i l

v(ct(O'), t) . : ct(O') da;_ 0 uO'

(5)

(6)

our proof will involve differentiating the right side under the integral. Wetherefore begin by investigating the partial derivatives (%t)v(ct(O'), t) and(02/ot oO')ct(O'). Note that ct(O') has derivatives with respect to 0'and t whichare jointly continuous in (0', t). In particular, by (4),

:t ct(O') = i(c(O'), t) = v(ct(O'), t).

11. SPIN. CIRCULATION. VORTICITY 83

Clearly, this relation has a derivative with respect to a which is jointly con-tinuous in (a, t). Thus we can switch the order of differentiation; that is,

a2

at aa ct(a)

exists and by (6)

a2 a2 aat aa ct(a) = aa at ct(a) = aa v(ct(a), t).

Note also that (6) implies

aat v(ct(a), t) = x(c(a), t) = v(ct(a), t).

These identities can be used to transform the left side of (5) as follows:

d f. d fl act(a)dt c, v(x, t) . dx = dt 0 v(cla), t) . -----a;;- do

= f. v(x, t) • dx + fl v(ct(a), t) • aa v(ct(a), t) doc, 0 a

= f. v(x, t) • dx + !{v2(ct(l), t) - v2(c

t(O), t)}.c,

But c (and hence ct) is closed; thus ct(1) = ct(O) and the last term vanishes. D

We say that the motion preserves circulation if

: f. v(x, t) • dx = 0t c,

for every closed material curve c and all time t. When v= grad IX the right sideof (5) vanishes, as c, is closed [cf. (4.8)], and we have

Kelvin's Theorem. Assume that the acceleration is the gradient ofa potential.Then the motion preserves circulation.

A curve h in !!It is a vortex line at time t if the tangent to h at each pointx on h lies on the spin axis of the motion at (x, t). Since the spin axis at (x, t)is the set of all vectors e such that

W(x, t)e = 0,

h is a vortex line if and only if

W(h(a), t) d:~) = 0

for 0 s a ~ 1.

84 III. KINEMATICS

(7)

Theorem (Transport of vorticity). Assume that the acceleration is thegradient of a potential. Then vortex lines are transported with the motion;that is,for any material curve c, if e, is a vortex line at some time t = r, then c,is a vortex linefor all time t.

Proof Let C be a material curve. Then by (4),

ddu ct(u) = F(c(a), t)k(a),

k(a) = d~~).

If c, is a vortex line, then

d0= W(c.(a), r) du ct(a) = W(c.(a), -r)F(c(u), -r)k(a),

so that trivially

WF(c(u), -r)k(a) = O. (8)

Moreover, by (3), WF is constant in time, so that (8) is valid for any t, There-fore, multiplying by F(c(a), t)-T and using (1),

W(ct(u), t)F(c(u), t)k(u) = 0,

and this relation, with (7), implies that c, is a vortex line for all t. 0

We close this section by listing an important property of irrotational,isochoric motions; this result follows from (c) of the theorem on page 78 andthe proposition on page 32.

Theorem. The velocity field ofan irrotational, isochoric motion is harmonic:

~v = O.

EXERCISES

We use the notation

w = curl v,

1. Show that in a plane motion

u = (det Ft.

(uW)" = uJ,

so that when it is the gradient of a potential,

(uW)" = O.

SELECTED REFERENCES

2. Establish the identity

(ow): = uLw + u curl v.Note that when vis the gradient of a potential this reduces to

(ow): = uLw.

85

(9)

3. Assume that vis the gradient of a potential. Show, as a consequence of(9), that

w(x, t) = [det Fly, t)] - 1Fly, t)w(y, t),

where x is the place occupied at time t by the material point whichoccupies y at time r (cf. Exercise 9.1).

4. Let u be a smooth spatial vector field and c a material curve. Show that

~ f. u· dx = f. (Ii + LTu) . dx.dt c, c,

5. Let v be the gradient of a potential q>. Show that

v= grad ( q>' + v;). (10)

so that the acceleration is also the gradient of a potential.

SELECTED REFERENCES

Chadwick [1, Chapter 2].Eringen [1, Chapter 2].Germain [1, Chapters 1, 5].Serrin [1, §§1l, 17,21-23,25-29].Truesdell [1, Chapter 2].Truesdell and Noll [1, §§21-25].Truesdell and Toupin [1, §§13-149].

CHAPTER

IVMass. Momentum

12. CONSERVATION OF MASS

One of the most important properties of bodies is that they possess mass.We here consider bodies whose mass is distributed continuously. No matterhow severely such a body is deformed, its mass is the integral of a density field;that is, given any deformation f there is a density field Pr on f( fJi) such that themass m(&l) of any part &l is given by

m(&l) = f. Pr dV.r(&')

Since the mass of a part cannot be altered by deforming the part, m(&l) isindependent of the deformation f.

We now make the above ideas precise. A mass distribution for fJi is afamily of smooth density fields

Pr: f(fJi) -+ IR+,

one for each deformation f, such that

f. Pr dV = f. PI dV == m(&l)r(&') 1(&')

87

(1)

88 IV. MASS. MOMENTUM

for any part f!/' and all deformations f and g. The number Pr(x) represents thedensity at the place x E f(~) in the deformation f, and equation (I) expressesconservation ofmass.

We denote by Po the density field Pr when f(p) = p for all p E~. ThusPo(p) gives the density at p when the body is in the reference position. Notethat, by the localization theorem,

. m(nel)

Po(p) = lim l(n )'el-O VO s

where n el is the ball of radius /j centered at p. As our next result shows, thereference density Po determines the density in all deformations.

Proposition. Let f be a deformation of~ and let F = Vf. Then

Pr(x) det F(p) = Po(p)

provided x = f(p).

Proof By (1) and the definition of Po,

r Pr(x) dY., = f Po(p) dVp 'Jr(9'l 9'

(2)

On the other hand, if we change the variable of integration on the left sidefrom x to p, we arrive at

Lpr(f(p» det F(p) dv.,.

Thus

L[p,(f(p» det F(p) - Po(p)] dVp = 0

for every part f!/', and (2) follows from the localization theorem. 0

Given a motion x of ~ we will always write p(x, t) for the density at theplace x E ~t in the deformation x(., t). Thus

p::r -+ IR+

is defined by

p(x, r) = PX(.,t)(x);

we will refer to P as the density in the motion x. In view of (1),

m(f!/') = f p(x, r) dV" == f P dV,9', 9',

12. CONSERVATION OF MASS 89

and we have the following

Theorem. For every part f!J' and time t,

dd f p dV = O. (3)t ~t

If F = Vx is the deformation gradient in the motion, then (2) takes theform

p(x, t) det F(p, t) = Po(p) (4)

(5)p' + div(pv) = O.

provided x = x(p, t). Thus p is the spatial description of po/det F, and weconclude from the smoothness lemma that p is smooth on !Y.

Theorem (Local conservation of mass)

p + p div v = 0,

Proof By (4),

p(det F): + p(det F). = 0,

which with (10.2) yields (5k Next, by (8.4)1'

P= p' + v· grad p,

and (5)1' when combined with this relation, implies (5h. D

Since a motion is isochoric if and only if div v = 0, (5) has the following

Corollary. A motion is isochoric if and only ifp = O.

Equation (3)expresses conservation ofmass for a region f!jJt that moves withthe body. It is often more convenient, however, to work with afixed region &/called a control volume. Of course, material will generally flow into and out offit, a fact which will become clear in the derivation of the resulting conserva-tion law.

By a control volume at time t we mean a bounded regular region &/ with

&/ c !fl t

for all r in some neighborhood of t, Thus for J sufficiently small we have thesituation shown in Fig. 1. Let n denote the outward unit normal to 8&/. Bythe divergence theorem,

rdiv(pv) dV = r pv : n dA.J9I J091

90

Further,

IV. MASS. MOMENTUM

Figure 1

Lp'(X, t) dVx = L :t p(x, t) dv., = :t Lp(X, t) dVx '

where the last step uses the fact that .o/l is independent of t. Conservation ofmass (5h, when combined with the relations above, yields the following

Theorem (Conservation of mass for a control volume). Let ~ be a controlvolume at time t. Then

dd f p(x, t) dVx = - f p(x, t)v(x, t)· n{'x) dA x •t ~ o~

(6)

Since n is the outward unit normal to a~, pv • n represents the mass flow,per unit area, out of ~ across its boundary. Thus (6) asserts that the rate ofincrease ofmass in ~ is equal to the mass flow into ~ across its boundary.

Lemma. Let <II be a continuous spatial field. Then given any part 9,

f <Il(x, t)p(x, t) dv., = f <II",(p, t)po(p) dv...9, 9

(7)

Proof We simply change the variable of integration from x to p in theleft side of (7); the result is

f <II(x, t)p(x, t) dVx = f <II",(p, t)p",(p, t) det F(p, t) dVp ,9, 9

and, in view of (4), this relation implies (7). D

If we differentiate (7) with respect to t we arrive at the integral of <Dmpoover 9, or equivalently [by (7) with <II replaced by ] the integral of p over9,. Thus we have the following important

12. CONSERVATION OF MASS

Theorem. Let be a smooth spatial field. Then given any part f!J',

: f p dV = f <bp dV.t &', &',

In other words, to differentiate

f p dV&',

91

(8)

with respect to time we simply differentiate under the integral sign treating the"mass measure" p dV as a constant.

Note that if we take == 1 in (8) we recover (3).

EXERCISES

1. Derive (5)1 and (6) using (3) and Reynold's transport theorem (l0.5).

2. Show that a deformation f is isochoric if and only if

Pr(x) = Po(p)

whenever X = f(p).

3. Let be a smooth spatial field. Show that for any part f!J',

f (x, t)p(x, t) dVx = f (x.(y, t), t)p(y, T)dYy,&', &"

where x., defined in Exercise 9.1, is the motion relative to time T.

4. Prove Kelvin's theorem: Of all motions ofa body which, at a given time t,

(a) correspond to a given steady density p,(b) have;?lt assigned,(c) have v· n prescribed on iJ;?Io

one with velocity the gradient of a potential yields the least kinetic energyat time t. More precisely, let f!It be a bounded regular region of space, letp > 0 be a smooth scalar field on f!It, let A. be a scalar field on iJf!It, let d bethe set of all smooth vector fields v on f!It that satisfy

div(pv) = 0 in f!It, v •n = A. on iJf!It,

and define the kinetic energy % on d by

%{v} = L~ pdV.

Show that if v E d has the form

v = grad cp,


then

X{v} ~ X{g}

for all g E d, with equality holding only when v = g.

13. LINEAR AND ANGULAR MOMENTUM.CENTER OF MASS

Let x be a motion of fll. Given a part 9 of fll, the linear momentum 1(9, t)and the angular momentum a(9, t) (about the origin 0) of 9 at time taredefined by

1(9, t) = f vp dV,@,

a(9, t) = f r x vp dV,@,

where r: Iff ~ r is the position vector

r(x) = x - o.

Proposition. For every part 9 and time t,

i(9, t) = f vp dV,@,

a(9, t) = f r x vp dV.@,

Proof The identity (12.8) yields (3)1 and

a(9, t) = f (r x v)"pdV.@,

But by (8.6),

(1)

(2)

(3)

(r x v)" = r x v+ v x v,

and, since v x v = 0, (3}z follows. 0

Assume now that fJ8 is bounded, so that m(fll) is finite. Then the centerof mass e(r) at time t is the point of space defined by

e(r) - 0 = m(~) L,rp dV. (4)

13. LINEAR AND ANGULAR MOMENTUM. CENTER OF MASS 93

(It is not difficult to show that this definition is independent of the choice oforigin 0.) If we differentiate (4) with respect to t and use (12.8) and (8.6), wefind that

ci(t) = m(~) LtVP dV,

so that ci represents the average velocity of the body. This result and (1)1imply that

l(gjI, t) = m(gjI)ci(t). (5)

(6)

Thus the linear momentum of a body gjI is the same as that of a particle ofmass m(gjI) attached to the center ofmass ofgjI.

EXERCISES

In these exercises :14 is bounded.

1. Show that

IX(t) - z = _1_ f (x - z)p(x, t) dV,.,m(gjI) ga,

for any point z, so that IX(t) is independent of the choice of origin.

2. Other types of momenta of interest are the angular momentum a.(t)relative to a moving point z(t) and the spin angular momentum aspin(t):

Here

asPin(t) = f r" x V"P dV.gat .

rz(x, t) = x - z(t)

(7)

(8)

is the position vector from z(t), r,,(x, t) is the position vector from IX(t), and

V" = i" = V - ci

is the velocity relative to IX. For convenience, let I(t) = l(gjI, t) anda(t) = a(gjI, t). Show that

az = aspin + (IX - z) x I,

a = aspin + (IX - 0) x i.


The term «x - z) x I is usually referred to as the orbital angular momen-tum about z; it represents the angular momentum /JI would have if all ofits mass were concentrated at the center of mass.

3. Consider a rigid motion. By (9.17),

Xo(y, t) = q(t) + Q(t)(y - z) (9)

for all y E /JIo and all t. (Here Xo is x, at r = 0; i.e., Xo is the motion takingthe configuration at time r = 0 as reference.)

(a) Show that(X(t) = q(t) + Q(t)[(X(O) - z]

and hence [noting that xo(y, t) is well defined for all y E C]

!X(t) = xo((X(O), t).

(10)

What is the meaning of this result?

(b) Show that the angular velocity co(t) is the axial vector ofQ(t)Q(t)Tand that

v.. = co x r...

(c) A vector function k on ~ rotates with the body if

k(t) = Q(t)k(O)

(11)

(12)

for all t. [Note that Q(O) = I, since xo(y, 0) = y for all y.] Show thatk rotates with the body if and only if

k = co x k. (13)(d) Use the identity

f x (d x f) = (f21 - f ® f)d

to show thatas p in = Jco,

where

J(t) = f (r;1 - r.. ® r..)p dVBI,

is the inertia tensor of /JI t relative to the center of mass.(e) Show that

J(t) = Q(t)J(O)Q(t)T,

(14)

(15)

and use this fact to prove that the matrix [J(t)] of J(t)-relative toany orthonormal basis {e;(t)} that rotates with the body-isindependent of t.

SELECTED REFERENCE 95

(f) Construct an orthonormal basis {ei(t)} that rotates with the bodyand has

[

J 1 0 0 ][J] = 0 J 2 O.

o 0 J 3

In this case {ei(t)} is called a principal basis and the correspondingnumbers J; are called moments of inertia. Let w;(t) denote the com-ponents of ro(t) with respect to {e;(t)}. Show that the components ofaspin(t) relative to this basis are given by

(aspin)! = J lw! + (J 3 - J 2 )W 2 W 3 ,

(a.Pin) 2 = J 2 W2 + (J 1 - J 3)WIW3,

(a.Pinh = J 3 W3 + (J 2 - J l )W I W 2 '

4. Define the kinetic energy S(t) and the relative kinetic energy S ,,(t) by

S ,,(t) = t r v; p dV.JDI,

(a) Prove Konig's theorem:

S = SOl + tm(81)ci2.

(b) Consider a rigid motion. Use the identity

(d x £)2 = d· (£21 - f (8) £)d

to prove that

5. Show that

J = (tr M)I - M,

where

M(t) = r f" (8) f"P dVJDI,

is called the Euler tensor.

SELECTED REFERENCE

Truesdell and Toupin [1, §§155-171].

CHAPTER

vForce

14. FORCE. STRESS. BALANCE OF MOMENTUM

During a motion mechanical interactions between parts of a body orbetween a body and its environment are described by forces. Here we shall beconcerned with three types offorces: (i) contact forces between separate partsof a body; (ii) contact forces exerted on the boundary of a body by its en-vironment; (iii) body forces exerted on the interior points of a body by theenvironment.

One of the most important and far reaching axioms in continuummechanics is Cauchy's hypothesis concerning the form of the contact forces.Cauchy assumed 1 the existence of a surface force density s(n, x, t) defined foreach unit vector n and every (x, t) in the trajectory ff of the motion (Fig. 1).This field has the following property: Let!/' be an oriented surface in !!4t withpositive unit normal n at x. Then s(n, x, t) is the force, per unit area, exertedacross!/' upon the material on the negative side of!/' by the material on thepositive side. Cauchy's hypothesis is quite strong; indeed, if ~ is an orientedsurface tangent to !/' at x and having the same positive unit normal there, thenthe force per unit area at x is the same on ~ as on !/' (Fig. 2).

1 Noll [3) has shown that Cauchy's hypothesis actually follows from balance of linearmomentum under very general assumptions concerning the form of the surface force s. (Cf.also Truesdell [I, p. 136); Gurtin and Williams [1).)

97

98

Figure I

+

V. FORCE

Figure 2

s(n, x, I)

To determine the contact force between two separate parts rY' and!!} attime t (Fig. 3) one simply integrates s over the surface of contact

!/t = rY't n !!}t;

thus

f s(o", X, t) dA" == f s(o) dASf', Sf',

gives the force exerted on rY' by!!}at time t. Here 0" is the outward unit normalto orY't at x. For points on the boundary of !!It,s(o, X, t)-with n the outwardunit normal to o!!lt at x-gives the surface force, per unit area, applied to thebody by the environment. This force is usually referred to as the surfacetraction. In any case, given a part rY',

i s(o) dAiJ~,

represents! the total contact force exerted on rY' at time t (Fig. 4).The environment can also exert forces on interior points of 14, a classical

example being the force field due to gravity. Such forces are determined by avector field b on .r; b(x, t) gives the force, per unit volume, exerted by theenvironment on x. Thus for any part rY' the integral

f b(x, t) dV" == i b dVJ~, ~,

gives that part of the environmental force on rY' not due to contact.

1 Here it is tacit that, given any part [7; and time t, 5(n., x, r) is an integrable function ofxon of!l'" This assumption of integrability actually follows from the momentum balance laws (I)and our hypotheses (i) and (ii) concerning force systems, Indeed, (i) trivially implies integrabilityon surfaces of polyhedra, and this is all that is needed to establish the existence of the stress T;(9) and (11) then yield integrability on of!l', for any part [7;,

14. FORCE. STRESS. 8ALANCE OF MOMENTUM 99

Figure 3 Figure 4

The above discussion should motivate the following definitions. Let AI bethe set of all unit vectors. By a system of forces for 9B during a motion (withtrajectory ff) we mean a pair (s, b) of functions

s: AI x ff ~ "f/, b: ff ~ "//,

(1)

with(i) s(n, x, t), for each n E .K and t, a smooth function of x on 9B1 ;

(ii) b(x, t), for each t, a continuous function of x on 9Bt •

We call s the surface force and b the body force; and we define the force f(&>, t)

and moment m(&>, t) (about 0) on a part &> at time t by

f(&>, t) = i s(n) dA + f b dV,ofJ', fJ',

m(&>, t) = i r x s(n) dA + f r x b dV.o!¥., fJ',

Here n is the outward unit normal to iJ&>t and r is the position vector (13.2).The basic axioms connecting motion and force are the momentum

balance laws. These laws assert that for every part &>and time t,

f(&>, t) = l(&>, t),

m(&>, t) = a(&>, t);

they express, respectively, balance of linear momentum and balance ofangular momentum [cf. (13.1)J.

Tacit in the statement of axiom (1) is the existence of an observer (frame ofreference) relative to which the motion and forces are measured. The existenceof such an observer is nontrivial, since the momentum balance laws aregenerally not invariant under changes in observer (cf. Exercise 21.3). Ob-servers relative to which (1) hold are called inertial; in practice the fixed starsare often used to define the class of inertial observers.

100 v. FORCE

An obvious consequence of (1)1 and (13.5) is that

f(PA, t) = m(PA)iX(t),

provided PA is bounded. Thus the totalforce on afinite body is equal to the massof the body times the acceleration of its mass center.

By virtue of (13.3) the laws of momentum balance can be written asfollows:

r sen) dA + f b dV = f vp dV,Jo&'t fP, e,

r r x sen) dA + f r x b dV = f r x vp dV.J0&', fP, fP,

If we introduce the total body force

b* = b - pv,

which includes the inertial body force - pv, and define

f*(&" t) = r sen) dA + f b* dV,JofP, fP,

m*(&', t) = r r x sen) dA + f r x b* dV,JofP, fP,

then (1) takes the simple form

(2)

(3)

(4)

f*(&" t) = 0, (5)

Our next result gives a far less trivial characterization of the momentumbalance laws. Recall that an infinltesimal rigid displacement (ofS) is a mappingw: S -+ 1/ of the form

w(x) = wo + W(x - 0) (6)

with Wo a vector and Wa skew tensor [cf. (7.8)].

Theorem of Virtual Work. Let (s, b) be a system offorces for PA during amotion. Then a necessary and sufficient condition that the momentum balancelaws be satisfied is that given any part &' and time t,

r s(n)· w dA + f b*· w dV = 0JOfPt fP,

for every infinitesimal rigid displacement w.

Proof Note first that by (13.2) we can write (6) in the form

w = Wo + ro x r,

(7)

(8)

14. FORCE. STRESS. BALANCE OF MOMENTUM 101

with co the axial vector corresponding to W. Let cp(wo, co) denote the left sideof (7) with w given by (8). Then cp(wo, co) involves integrals of the fields

s • w = s· Wo + s . (co x r),

where we have written s for s(n), and since

k·(co x r) = co·(r x k),

it follows that

s· w = Wo •S + m- (r x s),

h, . w = Wo • b; + co· (r x b*).

Thus by (4),

cp(wo, co) = Wo • f*(&" t) + co· m*(&', t),

and cp(Wo, co) = 0 for all vectors Wo and co if and only if (5) hold. 0

The next theorem is one ofthe central results ofcontinuum mechanics. Itsmain assertion is that s(n) is linear in n.

Cauchy's Theorem (Existence of stress). Let (s, b) be a system offorcesforf1l during a motion. Then a necessary and sufficient condition that the momentumbalance laws be satisfied is that there exist a spatial tensor field T (called theCauchy stress) such that

(a) for each unit vector n,

s(n) = Tn; (9)

(10)

(b) T is symmetric;(c) T satisfies the equation of motion

div T + b = pt.

Proof In stating condition (c) we do not assume explicitly that

T(x, t) is smooth in x, (11)

so that, a priori, it is not clear whether or not (c) makes sense. We thereforebegin our proof by showing that (11) is a direct consequence of (9). Indeed, by(9), for an orthonormal basis {eJ,

L s(ei) e e, = L (Tei) ® ei;i i

thus, in view of Exercise 1.6c,

T(x, t) = L s(e;, x, r) ® e.,

and (11) follows from (12) and property (i) of force systems.

(12)

102 v. FORCE

We are now in a position to establish the necessity and sufficiency of(a)-(c).

N ecessity. Assume that (1) are satisfied. For convenience, we fix the timet and suppress it as an argument in most of what follows. The proof willproceed in a number of steps.

Assertion I. Given any x E [fBI> any orthonormal basis {e.], and anyunit vector k with

if follows that

(i = 1,2,3), (13)

s(k, x) = - L (k· ei)s(-ei' x).i

(14)

Proof First let x belong to the interior of [fBr' Let b > 0 and consider thetetrahedron '§s with the following properties: the faces of '§~ are 9'~, 9'1s»

9' 2~, and 9' 3~' where k and - e, are the outward unit normals to (J'§s on 9'sand 9'i~' respectively; the vertex opposite to 9'~ is x; the distance from x to 9'~

is b (Fig. 5). Clearly, '§~ is contained in the interior of [fBI for all sufficientlysmall s, say 0 s 00 .

Next, b*(x, t), defined by (3), is continuous in x, since b(x, r), p(x, t), andv(x, t) have this property; hence b*(·, t) is bounded on '§~o (for t fixed). Thus ifwe apply (5) 1 to the part f!jJ which occupies the region '§sat time t, and use (4)1,

we conclude that

IL,,,s(n) dA I :s; "vol(,§~)for all 0 :s; 00 , where" is independent of o.

~----~~.e2

Figure 5

(15)


Let A(15) denote the area of [1'6' Since A(15) is a constant times 152, while

vol(~6) is a constant times 15 3, we may conclude from (15) that

A~15) Lf~(O) dA -+ 0

as 15 -+ O. But

f s(o) dA = f s(k) dA + L: f s( -ei) dA,JiJ~6 9'6 i 9"6

and, since s(o, x) is continuous in x for each fixed 0,

A~15) f9'6S(k) dA -+ s(k, x),

A~15) L'6S( -e;) dA -+ (k· ei)s( -ei> x),

where we have used the fact that the area of [l'i6 is A(15)k • ei' Combining theabove relations we conclude that (14) holds as long as x lies in the interior offIlt • But x 1-+ s(o, x) is continuous on f!4t • Thus, by continuity, (14) must holdeverywhere on f!4t •

Assertion 2 (Newton's law of action and reaction). For each x E f!4 t themap 01-+ s(o, x) is continuous on .AI and satisfies

s(o, x) = -s( -0, x), (16)

Proof Since the right side of(14) is a continuous function ofk, s(k, x) iscontinuous on the set of all unit vectors k consistent with (13). Thus, since thisis true for every choice of orthonormal basis {e.},s(k, x) must be a continuousfunction ofk everywhere on JV: Therefore, ifin (14) we let k -+ e 1, we arrive ats(e1, x) = -s( -e1, x), and again, since the basis rei} is arbitrary, (16) musthold.

Assertion 3. Given any x E f!4t and any orthonormal basis {eJ,

s(k, x) = L: (k· ei)s(ei, x)i

for all unit vectors k.

(17)

Proof Choose an orthonormal basis {e.} and a unit vector k that doesnot lie in a coordinate plane (that is, in a plane spanned by two e.), Then thereis no i such that k· e, = 0, and we can define a new orthonormal basis rei} by

ei = [sgn(k· ei)]ei'

104 V. FORCE

Then k 0 ei > 0 for i = 1,2,3, and Assertion 1 together with (16) applied tothe basis {eJ yields

s(k, x) = - I (k 0 i\)s( -ei, x) =. I (k 0 ei)s(ei' x) = I (k 0 e;)s(ei , x).iii

Thus (17) holds as long as k does not lie in a coordinate plane. But the mapOl--+s(o, x) is continuous on A: Thus (17) holds for all unit vectors k, andAssertion 3 is proved.

Choose an orthonormal basis {ei} and let T(x, t) be defined by (12). ThenT(x, t) is smooth in x, and, in view of (17), (9) holds. By (9), balance of linearmomentum (2)1 takes the form

i To dA + f b dV = f vp dV,a~ ~ ~

or equivalently, using the divergence theorem,

f (divT + b - pv)dV = O.~,

By the localization theorem, this relation can hold for every part & and time tonly if the equation of motion (10) is satisfied.

To complete the proofofnecessity we have only to establish the symmetryofT. Let w be any smooth vector field on !!Jr' We then conclude, with the aid ofthe divergence theorem and (4.2)5' that

i ToowdA = i (TTw)oodA = f div(TTw)dViJ~t iJ~t ~t

= f (w 0 div T + To grad w) dV.~t

Thus, by (3), (9), and (10),

i s(o) 0 w dA + f b* 0 w dV = f To grad w dV. (18)iJ~t ~t e,

In particular, if we take w equal to the infinitesimal rigid displacement (6), thengrad w = Wand (7) implies

f ToW dV = 0~t

for every part &, so that

ToW = O.

Since this result must hold for every skew tensor W, (f) of the proposition onpage 6 yields the symmetry of T.

14. FORCE. STRESS. BALANCE OF MOMENTUM

Tn

105

Figure 6

Sufficiency. Assume that there exists a symmetric spatial tensor field Tconsistent with (9) and (10). Clearly, (18) holds in the present circumstances.Since T is symmetric, and since the gradient of an infinitesimal rigid displace-ment is skew, when w is such a field the right side of(18) vanishes. We thereforeconclude from the theorem of virtual work that the momentum balance lawshold. 0

Actually, one can show that (a) and (c) are equivalent to balance oflinearmomentum (1)1; and that granted (1)1' the symmetry of T is equivalent tobalance of angular momentum (1)2.

Let T = T(x, t) be the stress at a particular place and time. If

Tn = ern, [n] = 1,

then a is a principal stress and n is a principal direction, so that principalstresses and principal directions are eigenvalues and eigenvectors ofT. SinceT is symmetric, there exist three mutually perpendicular principal directionsand three corresponding principal stresses.

Consider an arbitrary oriented plane surface with positive unit normal nat x (Fig. 6). Then the surface force Tn can be decomposed into the sum of anormal force

(0 • Tn)n = (n ® o)Tn

and a shearing force

(I - n ® n)Tn,

and it follows that n is a principal direction if and only if the correspondingshearing force vanishes.

A fluid at rest is incapable of exerting shearing forces. In this instance Tn isparallel to n for each unit vector n, and every such vector is an eigenvector ofT.In view of the discussion given in the paragraph following the spectraltheorem, T has only one characteristic space, 1/ itself, and (c) of the spectraltheorem implies that

T =-nI

106 v. FORCE

Pressure Pure tension

Figure 7

10-tI tI tI tI t------

Pure shear

with n a scalar. n is called the pressure of the fluid. Note that in this case theforce (per unit area) on any surface in the fluid is -no.

Two other important states of stress are:

(a) Pure tension (or compression) with tensile stress a in the direction e,where lei = 1:

T = rr(e ® e).

(b) Pure shear with shear stress, relative to the direction pair (k, 0),where k and 0 are orthogonal unit vectors:

T = -r(k ® 0 + 0 ® k),

The surface force fields corresponding to the above examples (with T con-stant) are shown in Fig. 7.

EXERCISES

In Exercises 1, 4, and 8, fll is bounded.

1. The moment mit) about a moving point z(t) is defined by

mz(t) = r r, x s(o) dA + f r, x b dV,JiJ9I, 91,

where rz is the position vector from z [cf. (13.8)].

(a) Let f(t) = f(fll, t). Show that, for y: ~ -+ $,

m, = Illy + (y - z) x f.

(b) Let l(t) = l(fll, t). Show that [cf. (13.7)]

2. Prove that the momentum balance laws (1) hold for every choice of origin(constant in time) if they hold for one such choice.


3. Show that the Cauchy stress is uniquely determined by the force system.

4. Consider a rigid motion with angular velocity roo Let {ej(t)} denote aprincipal basis of inertia and let J, denote the corresponding moments ofinertia (relative to e) (cf. Exercise 13.3f). Further, let wlt) and mj(t)denote the components of ro(t) and m..(t) relative to {ei(t)}. DeriveEuler's equations

m1 = J 1Wl + (J 3 - JZ)WZ W3'

mz = Jzwz + (J 1 - J 3)WIW3,

m3 = J 3 W3 + (Jz - JdW I W Z '

These relations supplemented by

f = m(91)li

constitute the basic equations of rigid body mechanics. When f and m..are known they provide a system of nonlinear ordinary differentialequations for ro and ~.

5. Let T be the stress at a particular place and time. Suppose that thecorresponding surface force on a given plane !/ is perpendicular to g,while the surface force on any plane perpendicular to g vanishes. Showthat T is a pure tension.

6. Let T be a pure shear. Compute the corresponding principal stresses anddirections.

7. Prove directly that s(o, x) = -s( -0, x) by applying (2)1 to the part [JJ

that occupies the region fYtoat time t. Here fYto is the rectangular regionwhich is centered at x, which has dimensions (j x (j X (jz, and which hasn normal to the (j x (j faces (Fig. 8).

Figure 8 Ii

8. Prove Da Silva's theorem: At a given time the total moment on a bodycan be made to vanish by subjecting the surface and body forces to asuitable rotation; that is, there exists an orthogonal tensor Q such that

i r x Qs(o) dA + f r x Qb dV = O.O~t ~t

108 V. FORCE

Show in addition that

r (QTr) x s(n) dA + i (QTr) x b dV = O.JoflB• flB,

What is the meaning of this relation?

9. Suppose that at time t the surface traction vanishes on the boundary ofJBt •

Show that at any x E afJBt the stress vector on any plane perpendicular toafJB t is tangent to the boundary.

15. CONSEQUENCES OF MOMENTUM BALANCE

By Cauchy's theorem (and Exercise 14.3), to each force system consistentwith the momentum balance laws there corresponds exactly one symmetrictensor field T consistent with (14.9) and (14.10). Conversely, the force system(s, b) is completely determined by the stress T and the motion x. Indeed, thesurface force s is given by (14.9), while the body force b is easily computedusing the equation of motion (14.10) [with p computed using (12.4)]:

s(n) = Tn, b = pv - div T.

This observation motivates the following definition. By a dynamical processwe mean a pair (x, T) with .

(a) x a motion,(b) T a symmetric tensor field on the trajectory ff of x, and(c) T(x, t) a smooth function of x on fJBt •

Further, ifv and p are the velocity field and density corresponding to x, thenthe list (v, p, T) is called a flow. In view of the above discussion, to each forcesystem consistent with the momentum balance laws there corresponds exactlyone dynamical process (or equivalently, exactly one flow), and conversely.

For the remainder ofthis section (v, p, T) is a flow and (s, b) is the associatedforce system.

Theorem (Balance of momentum for a control volume). Let 9i be a controlvolume at time t. Then at that time

r s(n) dA + rb dV = dd rvp dV + r (pv)v' n dA,Jout Jut t Jut Jout

(1)

r r x s(n) dA + rr x b dV = dd rr x vp dV + f r x (pv)v' n dA.Jout Jut t Jut Jout

15. CONSEQUENCES OF MOMENTUM BALANCE 109

Figure 9

Proof We will prove only (1)\. First of all, by (4.2)4' (8.5), and (12.5h,

pv = pv' + p(grad v)v = (pv)' - p'v + (grad v)(pv)= (pv)' + v div(pv) + (grad v)(pv) = (pV)' + div(v ® pv).

Thus, since

Ldiv T dV = 191Tn dA = 191s(n) dA, i~~ ..

L(pv)' dV = :t Lpv dV, !::f; -,

f div(v ® pv) dV = r (v ® pv)n dA = r (pv)v· n dA,91 JiJ91 JiJ91

if we integrate the equation of motion (14.10) over f7l we arrive at (1k D

Equation (1)\ asserts that the totalforce on the control volume f7l is equal tothe rate at which the linear momentum ofthe material in f7l is increasing plus therate ofoutflow ofmomentum across of7l. Equation (1h has an analogous inter-pretation.

A flow (v, p, T) is steady if fJl1 = fJlo for all t and

Vi = 0, o' = 0, T' = O.

In this case wecall fJI0 the flowregion. Note that our definitions are consistent:in a steady flow the underlying motion is a steady motion (cf. Section 9).

As an example of the last theorem consider the steady flow of a fluidthrough a curved pipe (Fig. 9), and let f7l be the control volume bounded bythe pipe walls and the cross sections marking the ends ofthe pipe. Assume thatthe stress is a pressure and that the velocity, density, and pressure at theentrance and exit have the constant values

and

respectively, with e\ perpendicular to the entrance cross section and ezperpendicular to the exit cross section. Since the flow is steady,

~ f vpdV = O.dt 91

110 v. FORCE

Further, v- n must vanish at the pipe walls; therefore

i (pv)v-ndA = P2v~A2e2 - PlviA1e l,iJEJt

where AI and A 2 are the areas of the entrance and exit cross sections, re-spectively. Let f denote the total force exerted by the fluid on the pipe walls.Then the total force on the control volume 91 is

and (1)1 yields

f = (1t 1 + Plvi)A1e1 - (1t2 + P2v~)A2e2.

A similar analysis, based on conservation of mass (12.6), tells us that

P1v1A1 = P2V2A2 == em,

and hence

f = (1tlAI + emv1)e1 - (1t2A2 + mv2)e2.

Thus we have a simple expression for the force on the pipe walls in terms ofconditions at the entrance and exit. Although the assumptions underlyingthis example are restrictive, they are, in fact, good approximations for a largeclass of physical situations.

Theorem of PowerExpended. For every part (!J and time t,

i sen)- v dA + f b -v dV = f T -D dV + : f ~ P dV, (2)o&lt &It &It t &It

where

D = !(grad v + grad vT)

is the stretching.

Proof Since T is symmetric, (a) of the proposition on page 6 implies

T -grad v = T - D.

Also, by (12.8),

so that (14.3) yields

f b. -v dV = f b -v dV - ~ f v2

P dV.&It &It dt &It 2

15. CONSEQUENCES OF MOMENTUM BALANCE 111

Equation (14.18) (with w = v) and the above relations yield the desiredidentity. D

The terms

f ~ p dV and fT. D dV~. ~.

are called, respectively, the kineticenergy and the stress power of fJ/at time t.The theorem of power expended asserts that the power expended on fJ/ by thesurface and body forces is equal to the stress power plus the rate of change ofkinetic energy.

A flow is potential if the velocity is the gradient of a potential:

v = grad (fJ.

Note that (fJ is a spatial field of class C2 (because v is C2) ; hence

curl v = curl grad tp = 0

and potential flows are irrotational. Conversely, by the potential theorem(page 35), if a flow is irrotational, and if {fit is simply connected at some (andhence every) time t, then the flow is potential.

For a body force b, the field b/p represents the body force per unit mass.We say that the body force is conservative with potential [3 if

b/p = -grad [3. (3)

(5)

(6)

(4)

If the flow is also steady the equation of motion (14.10) implies that b' = 0;in this case we will require that

[3' = O.

Bernoulli's Theorem. Consider afiow (v, p, T) with stress a pressure -nI andbody force conservative with potential [3.

(a) Ifthefiow is potential,

grad ((fJ' + ~2 + [3) + ~ grad n = O.

(b) Ifthefiow is steady,

v . grad(~ + [3) + ~ v . grad n = O.

(c) Ifthefiow is steady and irrotational,

grad(~ + [3) + ~ grad n = O.

112

Proof By hypothesis,

so that

V. FORCE

T = -nI,

div T = -grad n;

thus the equation of motion (14.10) takes the form

- grad n + b = pv,

or equivalently, by (3),

(7)

(8)v= - .!. grad n - grad p.p

On the other hand, by (9.10)1' for a potential (and consequently irrotational)flow

for a steady flow

v = v' + ! gradtv") = grad ( q/ + ~2); (9)

v •v= v • grad(~) (10)

(where we have used the fact that v· Wv = 0, since W is skew); for a steady,irrotationa1 flow

v = grad(~} (11)

The relations (9)-(11), when combined with (8), yield the desired results(4)-(6). 0

EXERCISES

1. Consider a statical situation in which a (bounded) body occupies theregion fJlo for all time. Let b: fJlo -+ 1/ and T: fJlo -+ Sym with T smoothsatisfy

divT + b = O.

Define the mean stress T through

vol(fJloyr = r T dV.J£to


(a) (Signorini's theorem) Show that T is completely determined by thesurface traction Tn and the body force b as follows:

vol(86'o)T = i (Tn ® r) dA + r b ® r dV.o~o J~o

(b) Assume that b = 0 and that iJ86'o consists of two closed surfaces Yoand Y 1 with Y 1 enclosing Yo (Fig. 10).Assume further that Yoand Y 1 are acted on by uniform pressures no and n1' so that

sen) = -non

sen) = -n1n

on Yo,

on Y 1 ,

Figure 10

with no and n1 constants. Show that T is a pressure of amount

where 00 and 01 are, respectively, the volumes enclosed by Yo andY 1•

2. Prove (l)z.

3. Derive the following counterpart of (2) for a control volume 9l:

i s(n)· v dA + rb· v dVo~ J~

I d I v2 I pv

2

= T . D dV + -d - P dV + - (v . n) dA.~ t ~ 2 o~ 2

SELECTED REFERENCES

Gurtin and Martins [1].Noll [3, 4].Truesdell [1, Chapter 3].Truesdell and Toupin [1, §§195-238].

CHAPTER

VI

Constitutive Assumptions.Inviscid Fluids

16. CONSTITUTIVE ASSUMPTIONS

The axioms laid down for force-namely the laws ofmomentum balance-are common to most bodies in nature. These laws, however, are insufficient tofully characterize the behavior of bodies because they do not distinguishbetween different types of materials. Physical experience has shown that twobodies of the same size and shape subjected to the same motion will generallynot have the same resulting force distribution. For example, two thin wiresof the same length and diameter, one of steel and one of copper, will requiredifferent forces to produce the same elongation. We therefore introduceadditional hypotheses, called constitutive assumptions, which serve to dis-tinguish different types of material behavior.

Here we will consider three types of constitutive assumptions.

(i) Constraints on the possible deformations the body may undergo. Thesimplest constraint is that only rigid motions be possible, a constraint whichforms the basis of rigid body mechanics. Another example is the assumptionof incompressibility, under which only isochoric deformations are permissible.This assumption is realistic for liquids such as water under normal flow

115

116 VI. CONSTITUTIVE ASSUMPTIONS. INVISCID FLUIDS

conditions and even describes the behavior of gases like air provided thevelocities are not too large.

(ii) Assumptions on the form of the stress tensor. The most widely usedassumption of this type is that the stress be a pressure, an assumption valid formost fluids when viscous effects are negligible.

(iii) Constitutive equations relating the stress to the motion. Here aclassical example is the equation of state of a gas giving the pressure as afunction of the density.

We now make these ideas precise.A material body is a body f!J together with a mass distribution and a

family C(J of dynamical processes. C(J is called the constitutive class of the body;it consists of those dynamical processes consistent with the constitutiveassumptions of the body.

A dynamical process (x, T) is isochoric if

x(', t) is an isochoric deformation (1)

at each t; a material body is incompressible if each (x, T) E C(J is isochoric. By(1), given any part f!JJ,

vol(f!JJt) = vol(f!JJ)

for all t, which implies (10.4). Thus every motion of an incompressible body isisochoric. The restriction (1), however, implies more: not only does eachmotion preserve volume, but the volume of any part throughout the motionmust be the same as its volume in the reference configuration. Further, inview of (6.17) and (c) of the theorem characterizing isochoric motions (page78), every flow (v, p, T) of an incompressible body must have

det F = 1,(2)

div v = o.

Also, by (2)1 and (12.4), p(x, t) = Po(p) for x = x(p, t). Thus, in particular,when Po is constant,

in this case we refer to Po as the density of the body.A dynamical process (x, T) is Eulerian if the stress T is a pressure; that is, if

T = -1tI

with n a scalar field on the trajectory of x.

16. CONSTITUTIVE ASSUMPTIONS 117

Figure 1

An ideal fluid is a material body consistent with the following two con-stitutive assumptions:

(a) The constitutive class is the set of all isochoric, Eulerian dynamicalprocesses.

(b) The density Po is constant.

Thus an ideal fluid is an incompressible material body for which the stress is apressure in every flow.

Note that for an ideal fluid the pressure is not determined uniquely by themotion: an infinite number of pressure fields correspond to each motion.This degree of flexibility in the pressure is common to all incompressiblematerials. That such a property is physically reasonable can be inferred fromthe following example. Consider a ball composed of an ideal fluid under atime-independent uniform pressure 11: (Fig. 0, and assume, for the moment,that all body forces, including gravity, are absent. Then the ball should remainin the same configuration for all time. Moreover, since the ball is incom-pressible, an increase or decrease in pressure should not result in a deforma-tion. Thus the same "motion" should correspond to all uniform pressurefields. This argument is easily extended to nonuniform pressure fields andarbitrary motions, but for consistency the requirement of null body forcesmust be dropped to insure satisfaction of the equation of motion.

We next introduce a material body, called an elastic fluid, in whichcompressibility effects are not ignored, and for which the pressure is com-pletely specified by the motion. Here the constitutive class is defined by asmooth response function

giving the pressure when the density is known:

11: = ft(p). (3)

That is, the constitutive class is the set of all Eulerian dynamical processes(x, - nI) which obey the constitutive equation (3).

We will consistently write

(v, p, 11:) in place of (v, p, - nI)


for the flow of an ideal fluid or of an elastic fluid. (Of course, P = Po in theformer case.)

An inviscid fluid is often characterized by the requirement that it beincapable of exerting shearing forces; in this sense both ideal and elasticfluids are inviscid.

EXERCISE

1. Consider a material body f:!4 with constitutive class C(j. A simple constraintfor f:!4 is a function

y: Lin" -+ ~

such that each dynamical process (x, T) E C(j satisfies

y(F) = O. (4)

For such a material one generally lays down the following constraintaxiom: the stress is determined by the motion only to within a stress Nthat does no work in any motion consistent with the constraint. [The rateat which a stress N does work is given by the stress power, per unitvolume,

N'D

(cf. page 111), where D is the stretching.]We now make this idea precise. Let PlJ be the set of all possible stretch-

ing tensors; that is, PlJ is the set of all tensors D with the following prop-erty: for some C2 function F: ~ -+ Lin + consistent with (4), D is thesymmetric part of

L = F(t)F(t)-l

at some fixed time t. Let

fJt = PlJ.1;

that is,

fJt = {NESymIN'D = 0 forall DEPlJ}.

Then the constraint axiom can be stated as follows: If (x, T) belongs to C(j,

then so also does every dynamical process of the form (x, T + N) with

N(x, t) E fJt

for all (x, t) in the trajectory of x. We call fJt the reaction space.

17. IDEAL FLUIDS 119

An incompressible material can be defined by the simple constraint

y(F) = det F - 1. (5)

(1)

Show that this constraint satisfies the constraint axiom if and only if thecorresponding reaction space is the set of all tensors of the form - nI,n E IR; i.e., if and only if the stress is determined by the motion at most towithin an arbitrary pressure field. Show further that the constitutiveassumptions of an ideal fluid are consistent with the constraint axiom.

17. IDEAL FLUIDS

Consider an ideal fluid with density Po. The basic equations are the equa-tion of motion (15.7) and the constraint equation (16.2)z:

-grad n + b = Pov,

div v = O.

When the body force is conservative with potential p, (15.8) implies

v= -grad(~ + p),and the acceleration is the gradient of a potential. We therefore have thefollowing corollary of the results established in Section 11.

Theorem (Properties of ideal fluid motion). The flow of an ideal fluidunder a conservative body force has the following properties:

(a) If the flow is irrotational at one time, it is irrotational at all times.(b) The flow preserves circulation.(c) Vortex lines are transported with the fluid.

As we shall see, the equations of motion of a fluid are greatly simplifiedwhen the flow is assumed to be irrotational. The above theorem implies thatthe motion of an ideal fluid under a conservative body force is irrotational ifthe flow starts from a state of rest. This furnishes the usual justification forthe assumption ofirrotationality. For a plane flow in an infinite region a muchstronger assertion can be made. Indeed, if the flow is steady and in a uniformstate at infinity [i.e., if grad vex) ~ 0 as x ~ 00], and if each streamline beginsor ends at infinity, then, since W is constant on streamlines (cf. the propositionon page 81), the flow must necessarily be irrotational.


(3)

(2)

Bernoulli's Theorem for IdealFluids, Let (v, Po, n) be a fiow ofan idea/fluidunder a conservative body force with potential 13.

(a) If the flow is potential (v = grad cp),

grad( cp' + ~ + ~ + fJ) = o.

(b) If the flow is steady,

( v2

+ ~ + 13)' = 0,2 Po

so that (v212) + (nlPo) + fJ is constant on streamlines.(c) If the flow is steady and irrotational,

v2 n- + - + 13 = const2 Po

everywhere.

Proof. Conclusion (a) follows from (15.4) (with P = Po). To establish (b)and (c) let

v2 nI'[ = - + - + 13·

2 Po

For a steady flow

1]' = 0,

and by (15.5),

v' grad I'[ = 0;

hence

~ = 1]' + v ' grad I] = 0,

which is (2). Finally, for a steady, irrotational flow (15.6) yields

grad I] = 0,

which, with 1]' = 0, implies (3). D

By Bernoulli's theorem, for a steady, irrotational flow under a conservativebody force equations (1) reduce to

div v = 0, curl v = 0,

v2 n-2 + - + 13 = const.

Po

(4)

17. IDEAL FLUIDS 121

(6)

These equations are also sufficient to characterize this type of flow. Indeed,suppose that (v, n) is a steady solution of (4) on a region of space flIo. Then,since W = 0, (9.10)1 implies

v= v' + grad(~) = grad(~) = -grad(~ + p),and the basic equations (1) (with b/po = -grad p) are satisfied.

In a steady motion the velocity is tangent to the boundary (cf. the prop-osition on page 68); thus (4) should be supplemented by the boundarycondition

v'n=O on ofllo. (5)

When the flow is nonsteady we are left with the system (1) to solve. Thebasic nonlinearity of these equations is obscured by the material time deriva-tive in (1)1. Indeed, in terms of spatial operators these equations take the form

v' + (grad v)v = - grad n + b,

div v = 0,

where, for convenience, we have written n for n/po and b for b/po. Equations(6) are usually referred to as Euler's equations.

EXERCISES

1. Show that in the flow of an ideal fluid the stress power is zero (cf. page111 ).

2. Consider the flow of a bounded ideal fluid under a conservative bodyforce, and suppose that for each t and each x E ofllt> v(x, t) is tangent toofllr • Show that

d f 2-d v dV = 0,t 111,

so that the kinetic energy is constant.

3. Consider a homogeneous motion of the form

x(p, t) = Po + F(t) [p - Po]'

Show that if x represents a motion of an ideal fluid under conservativebody forces, then F(t)F(t) -1 must be symmetric at each t.

4. Consider a steady, irrotational flow of an ideal fluid over an obstacle fJl

(Fig. 2); that is, ~ is a bounded regular region whose interior lies outsidethe flow region flIo and whose boundary is a subset of ofllo. Assume that

122 VI. CONSTITUTIVE ASSUMPTIONS. IN VISCID FLUIDS

Figure 2

the body force is zero. Show that the total force exerted on fJi by thefluid is equal to

18. STEADY, PLANE, IRROTATIONAL FLOWOF AN IDEAL FLUID

For a steady, plane flow the velocity field has the form

vex) = Vt(Xt, x2)e t + V2(Xt, x2)e2, (1)

and the flow region can be identified with a region fJi in ~2. For convenience,we identify v and x with vectors in ~2 and write, in place of (1),

vex) = (Vt(x), vix»,Then the basic equations (17.4) reduce to

oV t + oV2 _ 0oXt OX2 - ,

(2)

1t!(vf + vD + - = constPo

with 1t also a field on fJi. Here, for convenience, we have assumed that thebody force is zero.

For the moment let

{J = -V2'

18. STEADY, PLANE, IRROTATIONAL FLOW OF IDEAL FLUID 123

Then the first two relations in (2) are the Cauchy-Riemann equations:

orx OpOX2 = - OX1;

these are necessary and sufficient conditions that

g(z) = rx(X1' X2) + ip(X1' x 2)

be an analytic function of the complex variable

on r7l (considered as a region in the complex plane C). Thus we have thefollowing

Theorem. Consider a steady, plane, irrotational flow of an ideal fluid. Letg: ~ --+ C be defined by

(3)

Then g is an analytic function. Conversely, any analyticfunction g generates, inthe sense of(3), a solution of(2)1.2'

The function g is called the complex velocity.Let c = (C1' C2) be a curve in~. Then c can also be considered as a curve

in C. Given any complex function h on ~ we write

i h(z) dz = f h(e(u»c(u) do ;

in particular, we define

r(c) = ig(Z) dz.

To interpret r(c) note first that

k(u) = (C2(U), -c 1(u»

(4)

(5)

is normal to c at c(u). Therefore when c is the boundary curve of ~ the re-quirement that v· n = 0 on o~ [cf. (17.5)] becomes

(6)

Thus in this instance

ig(Z)dZ = f(V 1 - iV2)(C1 + iC2)du = f(V 1C1 + v2c2)du,


-ll1I

Figure 3

(8)

and

r(c) = i V(X) • dx

is the circulation around c.Next,

Ikl = Jci + c~,

so that Ik(O") Ido is the element of arc length along c. The vector

f(c) = - fn(c(O"»k(O") da (7)

therefore represents the integral along c of the surface force - no (0 = k/ Ik I)with respect to arc length. In fact, for a simple closed curve oriented counter-clockwise as shown in Fig. 3, f(c) gives the total force (per unit length in the X3-

direction) on the material inside c.

Blasius-Kutta-Joukowski Theorem. Consider a steady, plane, irrotationalflow ofan ideal fluid in a region qj whose boundary is a simple closed curve c.Let g be the complex velocity ofthe flow. Then

ipo f 2fl(C) - ific) = 2 "g(z) dz.

If in addition qj is the region exterior to c and the velocity is uniform at infinityin the sense that

g(z) -+ a

with a real, then

fl(c) = 0,

Proof Since c is closed,

as z-+oo (9)

(10)

fkdO" = 0,

18. STEADY, PLANE, IRROTATJONAL FLOW OF IDEAL FLUID 125

and we conclude from (7) and the Bernoulli equation (2h that

f(e) = P2° f v2k da.

Thus

fl(e) - if2(e) = Po I1

V2(C2 + ic1) do = ipo I\2(C1 - ic2) do,2 0 2 0

On the other hand,

192 dz = f (Vl - iV2)2(l\ + ic2) da,

and a simple calculation using (6) yields

(vi + V~)(Cl - ic2) = (Vl - iV2)2(C1 + ic2)·

Thus (8) holds.Assume now that fJl is exterior to e and that (9) holds. Assume further

that the origin 0 lies inside e. Then, since g is analytic and (9) holds, theLaurent expansion of g about the origin has the form

IX 1 1X2g(z) = U + - + - + ...Z Z2

for any z E fJl. Hence (4) and Cauchy's theorem of residues imply

F(c) = 2ni1X1.

Next, since g is analytic, we can compute g2 by termwise multiplication. Thus

( )2 2 2UIX1 lXi + 2UIX2

gz =u +--+ +"',z Z2

and a second application of Cauchy's theorem tells us that

192 dz = 4niuIX1 = 2ur(e).

Therefore, in view of (8), (10) holds. 0

If fJl is simply connected, then g can be written as the derivative of ananalytic function w:

dwg = dz


(This result extends to multiply connected regions provided one is willing toadmit multivalued functions.) The function w: ~ -+ C is called the complexpotential; the real and imaginary parts of w, denoted qJ and t/J, respectively,generate the velocity v through

Let s = (SI' S2) be a streamline so that

s(t) = v(s(t».

Then

(11)

and t/J is constant on streamlines. For this reason t/J is called the streamfunction.

Let c be a curve. Then c is essentially a segment of a streamline s if there is asmooth one-to-one mapping r of [0, IJ onto a closed interval of IR such that

c(a) = s(-r(a», O~a~1. (13)

Proposition. Consider a flow with a complex potential. Let c be a curve in theflow region and assume that v is nowhere zero on c. Then the following areequivalent:

(a) vl(c(a»cia) = vic(a»cl(a)for 0 ~ a ~ 1.(b) v(c(a» is parallel to c(a)for 0 ~ a ~ 1.(c) c is essentially a segment ofa streamline.(d) t/J is constant on c; that is,for 0 ~ a ~ 1,

dda t/J(c(a» = O.

Proof We will show that (d) - (a) - (b) - (c).

(d) - (a). This follows from the identity [cf. (12)J

dda

t/J(c(a» = -vic(a»c l(a) + vl(c(a»cia).

(a) - (b). Here we use the fact that (a) and (b) are each equivalent to theassertion that k(a)· v(c(a» = 0 for 0 ~ a ~ 1, where k, defined by (5), isnormal to the curve c.


(c) => (b). If c is essentially a segment of a streamline s, then (11), (13),and the chain rule yield

c(o) = s(r(u»t(u) = v(s("C(u»)t(u) = i(u)v(c(u», (14)

and (b) is satisfied.(b) => (c). Assume that (b) holds. Then, since v(c(u» and c(u) never

vanish,

c(u) = tX(u)v(c(u», o s o s 1,

with tX a continuous function on [0, 1] which never vanishes. Let

"C(u) = J:tX().) d)',

let s be the streamline which passes through c(O) at time t = 0, so that ssatisfies (11) and

s(O) = c(O);

and let d be the curve in [R2 defined by

d(u) = s("C(u», 0 :s; a :s; 1.

Then

d(O) = c(O)

(15)

and, using the argument (14),

ci(u) = tX(u)v(d(u», O:S;u:S;1.

Thus c and d satisfy the same differential equation and the same initialcondition. We therefore conclude from the uniqueness theorem for ordinarydifferential equations that c = d, so that, by (15),c is essentially a segment of astreamline. 0

For problems involving flows about a stationary region one usuallyassumes, as the boundary condition, that the boundary of the region be astreamline. In viewofthe last proposition, this insures that the velocity fieldbetangent to the boundary, or equivalently, that the boundary be impenetrableto the fluid (cf. the proposition on page 68).

As an example consider the region exterior to the unit circle and assumethat the velocity is uniform and in the xrdirection at infinity, so that (9) issatisfied. Consider the flow generated by the complex potential

Figure 4

In terms of cylindrical coordinates z = re" the stream function tf; has theform

tf;(r, e) = a(r - ~) sin e,

so that tf; has the constant value zero on the unit circle. Further, the complexvelocity

satisfies (9). The streamlines are shown in Fig. 4. For this flow

F(unit circle) = 0,

and by the Blasius-Kutta-Joukowski theorem there is no net force on theboundary.

Another flow with the desired properties is generated by

w(z) = a(z + ~ + iy log z).

This potential has a stream function

tf;(r, e) = a[(r - ~) sin e + y log rJ'which again vanishes on the circle r = 1. Further, w generates a complexvelocity

(1 iY)g(z) = a 1 - Z2 + ~ ,

which satisfies (9).A point at which g(z) = 0 is called a stagnation point. Points z on the unit

circle are given by the equation z = e". A necessary and sufficient conditionthat there exist a stagnation point at z = ei9 is that

. II Ysm o = - 2'

and hence that - 2 ~ Y ~ 2. Assume Y ~ O. Then: (a) for 0 ~ Y < 2 there aretwo stagnation points on the cylinder; (b) for Y = 2 there is one stagnation


(a)

Figure 5

point on the cylinder; (c) for y > 2 there are no stagnation points on thecylinder. These three cases are shown in Fig. 5 with the stagnation pointsdenoted by S. For these flows

r(unit circle) = - 2n«y

provided the unit circle has a counter-clockwise orientation; thus there is nodrag (force in the xI-direction), but there is a lift (force in the x2-direction)equal to

2npo«2y.

The problem of flow about an airfoil is far more complicated than thesimple flowpresented in Fig. 5,but the results are qualitatively the same. Therethe circulation is adjusted as shown in Fig. 6 to insure two stagnation pointswith the aft stagnation point coincident with the (sharp) trailing edge (Kutta-Joukowski condition).

Figure 6


EXERCISES

1. Show that the complex potential

w(z) = _Ce-inpzP+l

represents the flow past a wedge (Fig. 7) of angle 21X, where C and {3({3 > -1) are real constants and

{3nIX = 1 + {3'

Determine the stagnation points (if any).

Figure 7

2. Sketch the streamlines of the complex potential

w(z) = Cz 2,

where C is a real constant, and show that the absolute value of thevelocity is proportional to the distance from the origin.

3. Consider a steady, plane, irrotational flow of an ideal fluid. Let c be acurve in the flow region with

Let no be the pressure at a point at which the velocity is Vo' Show that theforce (8) on c is given by

( 2) . f.. Po Vo . IPO 2fl(C) - if2(C) = - no + -2- [(YA - YB) + I(XA - x B)] + 2 cg dz.

19. ELASTIC FLUIDS

The basic equations for the flow of an elastic fluid are the equation ofmotion

-grad n + b = pt, (1)

19. ELASTIC FLUIDS

conservation of mass

p' + div(pv) = 0,

and the constitutive equation

n = ft(p).

131

(2)

(3)

(4)

We assume that ft has a strictly positive derivative, and we define functions" > 0 and e on ~ + by

2( ) _ dft(p)tc p - ----;[P'

e(p) = r ,,2(0 e:p. (

where p* is an arbitrarily chosen value of the density. The function ,,(p) iscalled the sound speed; the reason for this terminology will become apparentwhen we discuss the acoustic equations.

In a flow e(p) may be interpreted as a spatial field. We may therefore con-clude from the chain rule that

,,2(p) 1 dft(p) 1grad e(p) = --grad p = - -d- grad p = - grad n. (5)

p p p P

Thus when the body force is conservative with potential 13, (1) takes the form

v= - grad(e(p) + 13),and the acceleration is the gradient of a potential. We therefore have thefollowing corollary of the results established in Section 11.

Theorem (Properties of elastic fluid motion). The flow of an elastic fluidunder a conservative body force has the following properties:

(a) If the flow is irrotational at one time, it is irrotational at all times.(b) The flow preserves circulation.(c) Vortex lines are transported with the fluid.

Thus, in particular, the flow of an elastic fluid under a conservative bodyforce is irrotational provided the flow starts from rest.

Bernoulli's Theorem for Elastic Fluids. Let (v, p, n) be a flow of an elasticfluid under a conservative body force with potential 13. Let e be defined by (4).

(a) If the flow is potential,

grad(llJ' + ~ + e(p) + 13) = o.


(b) If the flow is steady,

(~ + e(p) + p)" = 0,

so that (v2j2) + e(p) + pis constant on streamlines.(c) If the flow is steady and irrotational,

v2

2 + e(p) + p = const

everywhere.

Proof Conclusion (a) follows from (15.4) and (5). Let

v2

'1 = 2 + e(p) + p.

For a steady flow

(6)

(7)

(8)

'1' = 0, v • grad '1 = 0,

where we have used (15.5) and (5); hence

~ = '1' + v • grad '1 = 0,

which is (6). Finally, for a steady, irrotational flow, (5) and (15.6) yieldgrad '1 = 0, which with '1' = °implies (7). 0

If we define

then (1)-(3) take the form

v' + (grad v)v + rt.(p) grad p = bjp,

p' + div(pv) = 0.

Equations (8) constitute a nonlinear system for p and v and are generallyquite difficult to solve. Further, for many problems of interest the solution willnot be smooth because of the appearance of shock waves (surfaces acrosswhich the velocity suffers a jump discontinuity), and the notion of a weaksolution must be introduced. A careful discussion ofthese matters, however, isbeyond the scope of this book.

We now consider flows which are close to a given rest state with constantdensity Po. We therefore assume that Ip - PoI,Igrad p I, lvi, and Igrad vi aresmall.' Let [)be an upper bound for these fields. Then, since

rt.(p) grad p = rt.(Po) grad p + [rt.(p) - rt.(Po)] grad p,

I Since our interest lies only in deriving the asymptotic form of the field equations in thelimit as (j ..... 0, it is not necessary to work with dimensionless quantities.

19. ELASTIC FLUIDS

and since rx. is continuous, we have, formally,

rx.(p) grad p = rx.(Po) grad p + 0(<»

as <> ~ O. Similarly,

133

div(pv) = Po div v + 0(<»,

(grad v)v = 0(15).

Thus if we assume b = 0 and neglect terms of 0(<» in (8), we arrive at thelinear system

v' + rx.(Po) grad p = 0,

p' + Po div v = O.

These equations are usually called the acoustic equations; they constitutean approximate system of field equations appropriate to small departuresfrom a state of rest. If we take the divergence of the first equation and thespatial time derivative of the second, and then eliminate the velocity, weobtain the classical wave equation

p" = K2(po) !:ip,

where A = div grad is the spatial laplacian. Thus within the framework of thelinearized acoustic equations disturbances about a rest state with density Popropagate with speed K(Po). This should motivate our use of the term "soundspeed" for K(p).

We now return to the general nonlinear system (8). The ratio

vm=--

K(p)

of fluid speed

v = Ivj

to sound speed is called the Mach number, and a flow is subsonic, sonic, orsupersonic at (x, t) according as m(x, t) is < 1, = 1, or > 1.

Consider a steady flow with b = O. Since p' = 0,

p = v· grad p,

and (8)1 implies that

K2(p) K

2(p) .V • V= - -- v •grad p = - -- p.

p p

Thus

v • (pv)' = v· (pv + pv) = p(v • v)(1 - m 2) .


Further, if we differentiate v 2 = v· v, we find that

vv = v· v.Similarly,

v(pvr = v· (pv)".

The last three identities yield the following

Proposition. In a steady flow of an elastic fluid under vanishing body forces

(9)

whenever v #- O.

Equation (9) demonstrates one of the chief qualitative differences betweensubsonic and supersonic flow. Consider an arbitrary streamline 0 (Fig. 8) anda point x on a. Then the quantity p(x)v(x) represents the mass flow, per unitarea, across the plane through x perpendicular to a. For subsonic flow anincrease in v along 0 is associated with an increase in mass flow; for supersonicflow an increase in v corresponds to a decrease in mass flow. Thus if v in-creases along 0 from subsonic to supersonic values, then the mass flowincreases in the subsonic range to a maximum at m = 1, and subsequentlydecreases along the remainder of 0, which is supersonic. This explains, atleast qualitatively, the difference between subsonic and supersonic nozzles.Consider the nozzles shown in Fig. 9. Assume that the density and velocityare constant over each cross section, so that the total mass flow across eachsuch section is pvA, with A the corresponding area. (Although this assumptioncan be satisfied exactly only when A is constant, it is reasonable when Avaries slowly with length.) By conservation of mass (cf. the anlysis on page110),pvA is independent of position along the nozzle, and hence pv is inverselyproportional to A. In the subsonic nozzle the flow is always in the subsonicrange, and the decreasing area from entrance to exit results in an increasingmass flow and a concomitant increase in velocity. A nozzle which increases thevelocity from subsonic to supersonic values cannot have this design; indeed,in the supersonic region a decreasing area would result in a decreasing velocity.A nozzle with the desired properties is the de Laval nozzle. There the velocityis subsonic at the entrance and increases with decreasing area to a value of

Subsonic, 1''' increasesm= I

Supersonic. p" decreases

Figure 8. Streamline with v increasing.

19. ELASTIC FLUIDS

~l Y",<"c""U" U '('M ""C''',"

1 II I

Entrance Exit• II II I

~>"""" """""""","'"Subsonic nozzle

~I I

Entrance Exiti I

;~r' '1de Laval nozzle

Figure 9 (subsonic-supersonic)

135

(10)

'in = 1 at the throat (point of minimum area); the subsequent increasing areathen accelerates the fluid into the supersonic range.

By Bernoulli's theorem, when the flow is steady and irrotational, andwhen the body force is conservative,

div(pv) = 0, curl v = 0,

v2

2 + e(p) + p= C = const.

Ofcourse, as for an ideal fluid, (10) should be supplemented by the boundarycondition

v•n = 0 on 0fJI0 • (11)

Equations (10) also suffice to characterize this type of flow. Indeed, (10)1implies (2), and if 1t is defined by (3), then (10h and (9.10)1 imply

v = grad(~) = -grad(e(p) + P),

which, by (5), implies (1).For a steady potential flow the basic equations reduce to a single nonlinear

second-order equation for the velocity potential cpo To derive this equationnote first that, by (10)1'

v· grad p + p div v = 0, (12)

and if we take the inner product of (8)1 with v and use (12) to eliminate theterm involving grad p, we arrive at

v • (grad v)v = K 2(p) div V.


Here, for convenience, we have assumed that b = O. Thus, if

v = grad cp,

then cp satisfies the partial differential equation

grad cp' (grad/ cp) grad cp = ,,2(p) Acp,

where grad? = grad grad, or equivalently, in components,

L ocp ocp 02cp = ,,2(p) L 02~.i.] OXi oXj OXi oXj i OXi

Further, by (11) and (13), the corresponding boundary condition is

ocp = 0 :lf1]Jon U"""o,00where

(13)

(14)

(15)

Ocp- = 0° grad cp00

is the normal derivative on o[JIo'

The sound speed K(p) in (14) can also be expressed as a function of grad cpby solving the Bernoulli equation

e(p) = C _ Igrad cpl22

[cf. (10h with P= 0] for p as a function of grad cp as follows. Since K > 0, (4himplies that dejdp > 0, so that e(p) is an invertible function of p ; we maytherefore write

p = e- 1 ( C _Igra~ C(12)

and use this relation to express K(p) as a function

A( d) (-l(C Igrad CPI2))K gra tp = K s - 2

of grad cpo Of Course C and hence R will generally vary from flow to flow.

EXERCISES

1. An ideal gas is an elastic fluid defined by a constitutive equation of theform

(16)


where x > 0 and y > 1 are strictly positive constants. [Note that an idealgas is not an ideal fluid. Actually, (16) represents the isentropic behavior ofan ideal gas with constant specific heats.]

(a) Show jhat the sound speed K = K(p) is given by

K2 = yn/p.

(b) Show that the relation (15) is given by

y - 1j(2(grad qJ) = -2- (v 2

- [grad qJ1 2)

with v a constant.

2. Consider the ideal gas (16) in equilibrium (v = 0) under the gravitationalbody force

b = -pge3'

where g is the gravitational constant. Assume that

n(x) = no = constant at X3 = O.

Determine the pressure distribution as a function of height x 3 •

3. Show that for an elastic fluid the stress power of a part &> at time t is

f T'DdV = -f ttbp di/~, ~,

(cr. page 111), where

1v=-

P

is the specific volume (i.e., the volume per unit mass).

SELECTED REFERENCES

Courant and Friedrichs [1].Hughes and Marsden [1, §§9-11, 13, 14].Lamb [1, Chapters 1-10].Meyer [1, §6].Serrin [1, §§15-18, 35,45,46].

CHAPTER

VII

Change in Observer. Invarianceof Material Response

20. CHANGE IN OBSERVER

Two observers viewing a moving body will generally record differentmotions for the body; in fact, the two recorded motions willdiffer by the rigidmotion which represents the movement of one observer relative to the other.We now make this idea precise.

Let x and x* be motions of fJI. Then x and x* are related by a change inobserver if

x*(p, t) = q(t) + Q(t) [x(p, r) - 0] (1)

for every material point p and time t, where q(t) is a point of space and Q(t) isa rotation. That is, writing

f(x, t) = q(t) + Q(t)(x - 0), (2)

then at each time t the deformation x*(., t) is simply the deformation x(', t)followed by the rigid deformation f~·, t):

x*(', t) = f(·, t) 0 x(', t).

139

(3)

140 VII. OBSERVER CHANGE. INVARIANCE OF RESPONSE

oX(o,'>j ,,*(,/)

f(o, I)

•

Figure I

For convenience, we write !!Ai for the region of space occupied by !!A at time tin the motion x*:

!!Ai = x*(!!A, t);

then the relation between x and x* is illustrated by Fig. 1.We now determine the manner.in which the various kinematical quantities

transform under a change in observer. Letting

F = Vx, F* = \lx*

and differentiating (1) with respect to p, we arrive at

F*(p, t) = Q(t)F(p, r), (4)

which is the transformation law for the deformation gradient. Note that, sincedet Q = 1, (4) implies

det F* = det F.

Next, let

(5)

F = RU = VR, F* = R*U* = V*R*

be the polar decompositions ofF and F*. Then (4) implies

F* = R*U* = QRU,

and, since QR is a rotation, we conclude from the uniqueness of the polardecomposition that

Also,

R* = QR, U* = u.

Figure 2

20. CHANGE IN OBSERVER

~

:/~I91, • fJI~

f( -, I)

141

and therefore, since Y = RURT,

y* = QYQT.

From these relations it follows that the Cauchy-Green strain tensors

C = U 2, B = v-, C* = U*2, B* = y*2

transform according to

C* = C,

Intuitively it is clear that if x E !!AI and x* E !!A~ are related through

x* = f(x, t), (6)

(7)

then x and x* must correspond to the same material point. To verify that this isindeed the case, note that, by (3),

x(-, t) = f(·, t) - 1 0 x*(-, t),

which is easily inverted to give

pC t) = p*C t) 0 f(·, t),

where p and p* are the reference maps corresponding to the motions x and x*,respectively (Fig. 2). Thus

p(x, t) = p*(f(x, t), t),

or equivalently, by (6),

p(x, t) = p*(x*, t),

so that x and x* correspond to the same material point.As a consequence of (1),

x*(p, t) = q(t) + Q(t)x(p, t) + Q(t) [x(p, t) - o].

If we take p = p*(x*, t) in the left side and p = p(x, t) in the right side of thisrelation-a substitution justified by (7) provided x* and x are related by(6)-we conclude that

v*(x*, t) = cj'<t) + Q(t)v(x, t) + Q(t)(x - 0), (8)


where

are the spatial descriptions of the velocity in the two motions. Equation (8) isthe transformation law for the velocity.

Now let

L = grad v, L* = grad v*.

Ifwe take x* = f(x, t) in (8) and differentiate with respect to x using the chainrule and the fact that grad f = Q, we arrive at

L*(x*, t)Q(t) = Q(t)L(x, t) + Q(t),

so that the velocity gradient transforms according to

L*(x*, t) = Q(t)L(x, t)Q(t)T + Q(t)Q(t)T.

Since QQT = I, it follows that

QQT + QQT = 0,

or equivalently that

(9)

QQT = _(QQT)T,

and the tensor QQT is skew. Thus the symmetric part of (9) is the followingtransformation law for the stretching:

where

D*(x*, t) = Q(t)D(x, t)Q(t?, (10)

EXERCISES

1. Show that the spin W transforms according to

W*(x*, t) = Q(t)W(x, t)Q(t? + Q(t)Q(t)T.

2. A spatial tensor field A is indifferent if, during any change in observer, Atransforms according to

A*(x*, t) = Q(t)A(x, t)Q(t)T.

Assume that A is smooth and indifferent.

(a) Show that Ais indifferent if and only if A(x, t) = P(x, t)1 for all xand t.

21. INVARIANCE UNDER A CHANGE IN OBSERVER

(b) Show that the tensor fields

A = A- WA + AW,

A= A + LTA + AL

143

o 6.

are indifferent. In the case of the stress T [cf. (21.1)], T and Tarecalled, respectively, the corotational and convected stress rates.

(c) Show thato •

W=W,6. •

W=W+DW+WD. (11)

3. Use (9.14) to show that

C:(y*, t) = Q(T)C.(y, t)Q(r?,

and use this result to show that the Rivlin-Ericksen tensors (9.18) areindifferent.

21. INVARIANCE UNDER A CHANGE IN OBSERVER

One of the main axioms of mechanics is the requirement that materialresponse be independent ofthe observer. This axiom is never stated explicitly inelementary books on mechanics, probably because it seems so obvious;nonetheless it is instructive to look at a simple example from that subject. Athin elastic string weighted at the end is spun with constant angular velocityand suffers an elongation of amount D. The axiom above asserts that the forcein the string which produces this extension is the same as the force required toproduce the elongation {) when the string is held in one place. Indeed, the twomotions of the string differ only by a change in observer; that is, an observerspinning with the string sees the string undergoing the same motion as a "stillobserver" sees when he extends the string by hand.

We now give a precise statement of this axiom within our general frame-work. To do this we must first deduce the manner in which we would expectthe stress to transform under a change in observer. Thus let (x, T) and (x*, T*)be dynamical processes with x and x* related through (20.1). IfT and T* arealso related through this change in observer, then the corresponding surfaceforce fields should transform as shown in Fig. 3. Thus if

we would expect that

n* = Qn, Q = Q(t)

s*(n*) = Qs(n).


Figure 3

But

and hence

s(n) = Tn, s*(n*) = T*n*,

T*n* = QTQTn*.

Since this relation should hold for every unit vector n*, we would expect thefollowing transformation law for the stress tensor:

T*(x*, t) = Q(t)T(x, t)Q(t)T. (1)

Of course, in (1) it is understood that x* = f(x, t) with f given by (20.2).This discussion should motivate the following definition: (x, T) and

(x*, T*) are related by a change in observer if there exist C3 functions

q:IR-.S, Q: IR -. Orth '

such that

(a) the transformation law (20.1) holds for all p E fJI and t E IR;(b) the transformation law (1) holds for all (x, t) in the trajectory of x.

We say that the response of a material body is independentof the observerprovided its constitutive class ce has the following property: if a process(x, T) belongs to ce, then so does every dynamical process related to (x, T) by achange in observer. The content of this definition is, of course, the axiomdiscussed previously.

Theorem. The response ofideal fluids and elastic fluids are independent oftheobserver.

The proof of this result makes use of the following lemma, which is adirect consequence of (16.2) and (20.5).


Lemma. A dynamical process related to an isochoric dynamical process by achange in observer is itself isochoric.

Proofofthe Theorem. Let ~ be the constitutive class of an ideal fluid, let(x, T) E~, and let (x*, T*) be related to (x, T) by a change in observer. To showthat (x*, T*)E~ we must show that (x*, T*) is isochoric and Eulerian. Theformer assertion follows from the lemma; to verify the latter note that, sinceT = - nI, (l) yields

T* = QTQT = _nQQT = -ni.

Therefore ideal fluids are independent of the observer. The proof that elasticfluids also have this property proceeds in the same manner and is left as anexercise. D

EXERCISES

1. Prove that the response of an elastic fluid is independent of the observer.

In the next two exercises we consider the change in observer defined by (20.1).

2. Show that

div.; T*(x*, t) = Q(t) div, T(x, t).

3. Show that the acceleration transforms according to

v*(x*, t) = Q(t)v(x, t) + q(t) + 2Q(t)v(x, t) + Q(t)(x - 0).

Show further that if the body force transforms according to

b*(x*, t) = Q(t)b(x, t),

then

div T* + b* + k = p*v*,

where

k(x*, t) = p(x, t)[q(t) + 2Q(t)v(x, t) + Q(t)(x - 0)].

Because of the additional term k, the equation of motion is not invariantunder all changes in observer. The observers for whom this equationtransforms "properly" are exactly those with Q and q constant. Suchobservers are called Galilean and have the property of being accelera-tionless with respect to the underlying inertial observer.

SELECTED REFERENCES

Noll [1, 21-Truesdell and Noll [1, §§17-19J.

CHAPTER

VIII

Newtonian Fluids.The Navier-Stokes Equations

22. NEWTONIAN FLUIDS

Friction in fluids generally manifests itself through shearing forces whichretard the relative motion of fluid particles. The fluids discussed thus farnever exhibit shearing stress, and for this reason are incapable of describingfrictional forces of this type. A measure of the relative motion of fluid particlesis furnished by the velocity gradient

L = grad v,

a fact which motivates our considering constitutive equations of the form

T = -nI + CELl (1)

Materials defined by relations of this type with C linear are called Newtonianfluids; they furnish the simplest-and most useful-model of viscous fluidbehavior. Here we will use the term Newtonian fluid to mean incompressibleNewtonian fluid; thus, since tr L = div v, we limit our discussion to fields Lwith

tr L = O.

147

148 VIII. NEWTONIAN FLUIDS. NAVIER-STOKES EQUATIONS

Since C is linear, C[O] = 0; thus

T = -ni

when L = 0, and a Newtonian fluid at rest behaves like an ideal fluid. Also,as for an ideal fluid, the pressure tt is arbitrary (i.e., not determined by themotion). This flexibility in n leads to a certain ambiguity regarding theresponse function C; given any linear, scalar-valued function [3(L) of L, wecan rewrite (1) in the form

T = - {n + [3(L)}I + Cp[L],

where

Cp[L] = C[L] + [3(L)I.

Since n is arbitrary, n can absorb the term [3(L). Thus our constitutiveequation (1) is essentially unaltered by replacing the term C[L] by Cp[L].To remove this ambiguity, we normalize C by demanding that

tr C[L] = O. (2)

(This is equivalent to replacing C in (1) by Cp with [3(L) = -~ tr C[L].) ThenC has domain

Lin, = {L E Linltr L = O}

and codomain

Sym, = {T E Sym Itr T = OJ.

In view of the restriction (2), the pressure n is uniquely determined bythe stress. In fact, if we take the trace of (1) and use (2), we conclude that

n = -1trT.

We define the extra stress To by

To = T + nI = T - 1(trT)I,

so that To is the traceless part of the stress tensor. Our constitutive equation(1) then takes the simple form

To = C[L]. (3)

The above discussion should motivate our next definition.A Newtonian fluid is an incompressible material body consistent with the

following constitutive assumptions:

(a) There exists a linear response function

C: Lin, --+ Sym.,

22. NEWTONIAN FLUIDS 149

such that the constitutive class is the set of all isochoric dynamical processes(x, T) which obey the constitutive equation (3).

(b) The density Po is constant.

The dimensions of the spaces Lin., and Sym., are 8 and 5, respectively, sothat the matrix of C relative to a coordinate frame has 40 entries. Thus atfirst sight it would appear that it .akes 40 independent constants to specifya Newtonian fluid. The next theorem shows that, in actuality, the responseis determined by a single constant.

Theorem. A necessary and sufficient condition that the response of a N ew-tonian fluid be independent of the observer is that its response function C havethe form

(4)

for every L E Lin., , where

D = l(L + LT) .

T he scalar constant Jl is called the viscosity of the fluid.

Proof (Sufficiency) Assume that (4) holds. Let (x, T) belong to theconstitutive class ~ of the fluid. Then (x, T) is isochoric and

To = 2JlD.

Let (x*, T*) be related to (x, T) by a change in observer. Then (x*, T*) isisochoric (cf. the lemma on page 145). Moreover, by (20.10) and (21.1),

and (1.4h yields

tr T* = tr(QTQT) = tr(TQTQ) = tr T.

Therefore

T6 = T* - ~tr T*)I = QTQT - !{tr T)QQT = QToQT

= Q(2JlD)QT = 2JlD*.

Thus (x*, T*) E ~ and the response is independent of the observer.

The proof of necessity is facilitated by the following

(5)

Lemma. Let Lo E Ling. Then there exists a motion x with velocity gradient

L= L o

and with x(', t) isochoric at each time t.

(6)


Proof Take

F(t) = eLot

so that F is the unique solution of

F(O) = I. (7)

By (36.2),det F = e(lrLo)t = 1.

Thusx(p, t) = q + F(t)[p - q]

defines a motion with deformation gradient F and with x(·, t) isochoric ateach t. Further, (6) follows from (7)1 and the identity (8.8)1.

We now return to the proof of the theorem. To establish the necessity of(4) we assume that

the response is independent of the observer. (8)

Let LoE Lin, be arbitrary, let x be the motion constructed in the lemma, andlet T = To be the constant field defined by (3). Then, clearly, (x, T) E CC. Let(x*,T*) be related to (x, T) by a change in observer. Then by (8), (x*, T*) E CCand

T~ = C[L*].

But

T~ = QToQT, L* = QLQT + QQT

[cf. (20.9), (21.1), and (5)]; hence (9) yields

QToQT = C[QLQT + QQT],

and we conclude from (3) and (6) that

QC[LO]QT = C[QLoQT + QQT].

(9)

(10)

Clearly, this relation holds for every Lo E Lin, (the domain of C) and everyC3 function Q: ~ --+ Orth +. Fix Lo and take

Q(t) = e- Wo t,

where

Wo = 1(Lo - L~).

Then Q(t) is a rotation, since W0 is skew, and

Q(O) = I, Q(O) = -Wo


(cf. the discussion in Section 36). Using this function Q in (10) at t = 0 yields

where

C[Lo] = C[Lo - W 0] = C[Do], (11)

Do = t(Lo + L~).

Thus C is completely determined by its restriction to Symo.Next, let Q be a constant function with value in Orth+. Then (10) with

Lo = Do (Do E Syrnj) implies that

(12)

(14)

Since this relation must hold for every Do E Sym., and every Q E Orth +, therestriction of C to Sym., is isotropic; we therefore conclude from the repre-sentation (37.26) that

for all Do E Sym.,; D

Note that the results (11) and (12) are valid also for nonlinear functionsC; we appealed to the linearity of C only in the last step of the argument.

By (4) the constitutive equation (1) takes the form

T = -7tI + 2.uD. (13)

This equation must be supplemented by the equation of motion

Po[v' + (grad v)v] = div T + b

and the incompressibility condition

div v = O.

In view of (4.5),

2 div D = div(grad v + grad vT) = dV + grad div v = dV,

where d = div grad is the spatial laplacian. Thus the above equations reduceto

Po[v' + (grad v)v] = .u dv - grad n + b,

div v = O.

These relations are the Navier-Stokes equations; given .u, Po, and b theyconstitute a nonlinear system of partial differential equations for the velocityv and the pressure n. If we define the kinematical viscosity v by

v = .u/Po

and write

no = n/po, bo = b/po,

(15)


we can rewrite the Navier-Stokes equations in the form

v' + (grad v)v = v ~v - grad 1to + bo,

div v = O.

If the flow is steady, and if we neglect the nonlinear term (grad v)v, then(15) reduce to

v ~v = grad 1to - bo,

div v = O.

Solutions ofthis equation are called Stokes flows and are presumed to describeslow or creeping flows of Newtonian fluids.

We now return to the general Navier-Stokes equations (15). One of thebasic differences between flows described by these equations and flows ofideal or elastic fluids is the way circulation and spin are transported.

Theorem (Transport of spin and circulation). Consider the flow of aNewtonian fluid under a conservative body force. Then

W+ DW + WD = v~W,

and given any closed material curve c,

:t i v(x, t) .dx = v i M(x, t) • dx.Ct Ct

If in addition the flow is plane,

w = v~W.

(16)

(17)

(18)

Proof Since the left side of (15)1 represents V, and since bo = - grad /3,

v= v ~v - grad(1to + /3).If we substitute this equation into (11.5), we recover (17), since c is closed.Further,

grad v= v ~ grad v - grad grad(1to + /3),because ~ grad = grad ~. (Here and in what follows we need the additionalassumption that, as functions of position, v is of class C3

, while 1to and /3 areof class C2

. ) If we take the skew part of this equation and use the fact thatgrad grad(1to + /3) is symmetric, we arrive at

!<grad v- grad vT) = v ~W.

This equation and (11.2h imply (16). Finally, since div v = 0, (16) and (9.12)yield (18). 0


Both ideal and elastic fluids preserve circulation. In contrast, (17) tells usthat circulation in a Newtonianfluid generally varies with time.

Equation (18) asserts that in plane flow the spin satisfies a diffusionequation. Further, if in (16) we neglect terms of second order in grad v, weagain arrive at (18), but here for flows which are not necessarily plane. Thismotivates the standard assertion that spin diffuses in a viscous fluid. Finally,note that by (20.11h we can rewrite (16) as

6.

W = V~W.

Suppose that the flow takes place in a region fJI. For the inviscid fluidsconsidered previously the boundary condition was that v . 0 = 0 on afJI,which is simply the requirement that the boundary be impenetrable to thefluid. For a Newtonian fluid we add the restriction that the fluid adhere,without slipping, to the boundary. For a stationary boundary this meansthat v = 0 on afJI. If the boundary moves, then at each point on the boundarythe fluid velocity must coincide with the velocity of the boundary.

The theorem of power expended has an interesting consequence forNewtonian fluids. Indeed, by (13),

T· D = -nl' D + 2J.lIDj2 = 2J.lIDI2,

since I . D = tr D = div v = 0, and (15.2) yields the following

Theorem (Balance of energy for a viscous fluid). Consider the flow of aNewtonianfluid. Then

r s(o)· v dA + f b· v dV = :t f v; Po dV + 2J.l f IDI2 dV (19)Jo@ , @, @, e,

for every part 9.

The term

represents the rate at which the fluid in 9 dissipates energy. The energyequation (19) asserts that the total power expended on 9 must equal the rateofchange of kinetic energy plus the rate ofenergy dissipation. Note that for afinite body, if v vanishes on afllt for all time, and if b == 0, then (assumingJ.l > 0)


so that the kinetic energy decreases with time. This result represents a typeof stability inherent in viscous fluids; in Section 24 we will establish a strongerform of stability: we will show that under these conditions the kinetic energyactually decreases exponentially with time.

It is instructive to write the Navier-Stokes equations in dimensionlessform. Consider a solution of (15) with

b = O.

Let I and v be numbers with I a typical length (such as the diameter of a body)and v a typical velocity (such as a mainstream velocity). Further, identifypoints x with their position vectors x - 0 from a given origin 0, and definethe dimensionless position vector

the dimensionless time

tvl =1'

the dimensionless velocity

1v(x, l) = - v(x, t),

v

and the dimensionless pressure

Then

Igrad v = - grad v,

v

IY'=2 V' ,

v

Igrad ito = 2 .grad no

v

[where grad v(x, l) = V;;y(x, l), etc.], so that (15) reduces to

1v' + (grad v)V = Re L\v - grad ito,

div v = 0,

where

IvRe=-

v

is a dimensionless quantity called the Reynolds number of the flow.

(20)


Equations (20) show that a solution of the Navier-Stokes equations witha given Reynolds number can be used to generate solutions which have differ-ent length and velocity scales, but the same Reynolds number. This fact allowsone to model a given flow situation in the laboratory by adjusting the lengthand velocity scales and the viscosity to give an experimentally tractableproblem with the same Reynolds number.

Some typical values for the kinematical viscosity are! :

water: v = 1.004 X 10- 2 cm 2jsec,

air: v = 15.05 X 10- 2 cm 2jsec.

EXERCISES

1. A Reiner-Rivlin jluid 2 is defined by the constitutive assumptions of theNewtonian fluid with the assumption of linearity removed. Use (37.15)to show that the response of a Reiner-Rivlin fluid is independent of theobserver if and only if the constitutive equation has the form

T = -nI + (l(o(.J"o)D + (l(l(.J"o)D2

with (l(o(.J"o) and (l(l(.J"O) scalar functions of the list.J"o = (0, liD), 13(D ))of principal invariants of D.

2. Consider a Newtonian fluid in a fixed, bounded region §I of space andassume that

v=o on G.ry(

for all time.

(a) Show that the rate at which the fluid dissipates energy, i.e.,

can be written in the alternative forms

indicating that all of the e,nergy dissipation is due to spin.

1 At 20°C and atmospheric pressure. Cf., e.g., Bird, Stewart, and Lightfoot [I, p. 8].2 There are better models of non-Newtonian fluids (cf. Truesdell and Noll [I, Chapter E]

and Coleman, Markovitz, and Noll [I]).


(b) The surface force exerted on of7t by the fluid is given by the simpleexpression

s = no - 2JlWo = no + JlO x curl v,

where 0 is the outward unit normal to of7t. Establish this relationfor a plane portion of the boundary.

23. SOME SIMPLE SOLUTIONS FOR PLANESTEADY FLOW

For steady flow in the absence ofbody forces the Navier-Stokes equations(22.15) reduce to

po(grad v)v = Jl .1v - grad n,

div v = o.Consider the plane velocity field

v(x) = Vl(X l, X 2)el'

Then (1)2 implies that

(I)

(2)

so that Vl = V1(X2) and v is a simple shear (see Fig. I). Further, the matrix ofgrad v is

and

(3)

div v = 0,

Therefore (I) reduces to

02V 1 onJl-=-

ox~ OXl'

(grad v)v = o.

(4)

23. SOME SIMPLE SOLUTIONS FOR PLANE STEADY FLOW 157

X 2

Figure 1

We now consider two specific problems consistent with the above flow.

Problem 1 (Flow between two plates). Consider the flow between twoinfinite flat plates, one at x 2 = 0 and one at Xz = h, with the bottom plateheld stationary and the top plate moving in the x 1 direction with velocity v

(Fig. 2). Then the boundary conditions are

v1(0) = 0, v1(h) = v. (5)

We assume in addition that there is no pressure drop in the xcdirection, sothat

11: == const.Then (4)1 yields

Vi = (X + 1hz ,

and this relation satisfies the boundary conditions (5) if and only if (X = 0 andP= v/h. Thus the solution of our problem is

Vi = ox-fh. (6)

In view of (22.13), (3), and (6), the stress T is the constant field

[T] = -11:[~ ~ ~] + fl;' [~ ~ ~];o 0 1 0 0 0

v

Figure 2

that is,

1l'UTI 2 = T2 1 = h'

Thus the force per unit area exerted by the fluid on the top plate has a normalcomponent tt and a tangential (shearing) component - 1l'U/h. This factfurnishes a method of determining the viscosity of the fluid.

Problem 2 (Flow between two fixed plates under a pressure gradient).Again we consider the above configuration, but now we hold the two platesfixed in space (Fig. 3). Then the relevant boundary conditions are

(7)

If the pressure were constant, then the solution of the first problem wouldtell us that the velocity is identically zero. We therefore allow a pressure drop;in particular, since v = V(X2), (4)1 implies that

~= -<5oX l

with b (the pressure drop per unit length) constant. For this case (4)1 has thesolution

bx~VI = - - + <X + f3X2'

211

which satisfies (7) if and only if <X = 0 and f3 = <5h/21l. Thus

bX2VI = - (h - x 2 )

211

and the velocity distribution is parabolic. Further, we conclude from (22.13)that the nonzero components of the stress are

Tll = T2 2 = T3 3 = -n,

TI 2 = T2 1 = <5(~ - X 2}

Thus the shearing stress is a maximum at the walls and vanishes at the center.

Figure 3

24. UNIQUENESS AND STABILITY 159

The volume discharge Q per unit time through a cross section one unitdeep is obtained by integrating V1(X l) from Xl = 0 to Xl = h; the result is

Q = h3 b/12/l.

Since the discharge Q is easily measured, as is the pressure drop b, thisformula yields a convenient method of determining the viscosity.

EXERCISES

1. Consider the flow of a Reiner-Rivlin fluid (cf. Exercise 1 of Section 22)between two flat plates. Show that the linear velocity profile (2), (6)remains a solution of the underlying equations with no constant. Showfurther that, in contrast to the linear theory, the normal stresses are nolonger equal.

2. Consider the plane, steady flow of a Newtonian fluid of depth h down aflat surface inclined at an angle IX to the horizontal (Fig. 4). Thus the fluidflows under the influence of the gravitational force

b = pog(e 1 sin IX - el cos IX),

and the boundary conditions are

v = 0 at Xl = 0, Tel = - Cel at Xl = h,

where C is the (constant) atmospheric pressure. Assume that the flowhas the form (2). Determine the velocity and stress fields in the fluid.

Figure 4

24. UNIQUENESS AND STABILITY

In this section we establish theorems of uniqueness and stability for theNavier-Stokes equations. Aside from their intrinsic interest, the proofs ofthese theorems utilize techniques which are important in their own right.

(1)


The classical viscous flow problem for the Navier-Stokes equations canbe posed as follows:

Given: a bounded regular region ~, a kinematic viscosity v > 0, a bodyforce field bon fJl x [0, 00), an initial velocity distribution Vo on ~, and aboundary velocity distribution von iJfJl x [0, 00).

Find: a class CZ velocity field v and a smooth pressure field tt on~ x [0, 00) that satisfy the Navier-Stokes equations

Vi + (grad v)v = v M - grad tt + b,

div v = 0,

the initial condition

v(x,O) = vo(x)

for every x E fJl, and the boundary condition

v = v on iJfJl x [0, 00).

A pair (v, rr) with these properties will be called a solution. (Here, for con-venience, we have written nand b for what we earlier called no and bo.)

Uniqueness Theorem. Let (VI' n l ) and (vz, nz) be solutions of the (same)viscous flow problem. Then

with IY. spatially constant:

grad IY. = O.

The proof of this theorem is based on the following

Lemma. Let w be a smooth vector field on fJl, let {J be a smooth scalar fieldon fJ1/, and assume that

div w = °and that at each point ofiJfJl either w = 0 or {J = 0. Then

Jw· grad {J dV = 0.."f

Proof The proof follows from the identity

div({Jw) = {J div w + w . grad {J,

the divergence theorem, and the hypotheses. 0

24. UNIQUENESS AND STABILITY

Proof of Theorem. Let

161

Then, since (VI' 1[d and (vz, 1[z) are solutions,

u(X,O) = 0, u = 0 on o;]lt x [0, (0), div u = 0. (2)

Further, subtracting (1)1 with V = Vz, 1[ = 1[z from (1)1 with V = VI' 1[ = 1[1'

we obtain the relation

u' + (grad VI)V I - (grad vz)vz = v ~u - grad a,

and, since

we have

u' + (grad u)vI + (grad vz)u = v ~u - grad a. (3)

U we take the inner product of (3) with u, and use the identities

u· ~u = div[(grad U)TU] - Igrad ulZ,

u· (grad u)vI = VI • (grad U)TU = VI • grad(~z). (4)

u . (grad vz)u = u • Du,

where

D = !(grad Vz + grad V1),

we conclude that

!(uz), + VI • grad(~) + u· Du = v div[(grad u)Tu] - vl grad ulz

- u· grad fl..

U we integrate this relation over ;]It, use the lemma twice (first with w = VI'

f3 = UZ/2, then with w = u, f3 = a), and appeal to the divergence theorem, we

conclude, with the aid of (2), that

where

1 d r2dt [u]? + J

9fu· DudV s 0, (5)

Since div v = 0, tr D = 0; thus the lowest proper value of the symmetrictensor D(x, t) is sO. Let -y(x, t) denote this proper value, so that y ~ °and

Now choose r > °and let

A. = 2 sup y(x, t).XE91

0" r s t

(Since V2 is smooth, D is continuous and °s A. < 00.) Thus

A. 2u·Du> --u- 2

and (5) becomes

on 9i x [0, r]

Therefore

and hence

on [0, r].

on [0, r],

But by (2)1'

thus IluI1 2(r) = °and

(6)

u(x, r) = 0

for every x E 9i. Since r was arbitrarily chosen, u == 0 and v1 = ~2' Finally,(3) implies that grad (X = O. 0

StabilityTheorem. Consider the viscous flow problem with vanishing bound-ary data and conservative body forces:

v= 0 on 89i x [0, (0),

b = -grad p.

Let (v, n) be the solution (if it exists) of this problem. Then there exists aconstant A. > °such that

(7)

24. UNIQUENESS AND STABILITY 163

(8)

Proof In view of (6), (v, n) satisfies the system

v' + (grad v)v = v dv - grad IX,

div v = 0,

v = 0 on i}fJl x [0, (0),

v(x, 0) = vo(x),

where IX = n + {3. If we take the inner product of (8)1 with v and use theidentities (4) 1, 2, we arrive at the relation

t(v2)' + v· gradt« + tv2

) = v div{(grad V)T V} - v Igrad v1 2.

Ifwe integrate this relation over fJl,we conclude, with the aid of the divergencetheorem, (8h, 3, and the lemma, that

ddt IIvl1

2 + 2vllgrad vl1 2= 0,

where

[grad vl1 2 = LIgrad vl2 dV.

By the Poincare inequality (Exercise 1) there exists a constant ..1.0 > 0 suchthat

Thus

where A. = VA.o, and hence

which integrates to give

or equivalently,

Ilvll(t) s Ilvll(O)e-l ' .

In view of (8)4' this inequality clearly implies (7). 0

Remark. It should be emphasized that the preceding two theoremsrequire that the solution be smooth for all time and therefore may be mis-leading, since there is no global regularity theorem for the Navier-Stokesequations. 1

1 cr. Ladyzhenskaya [I, §6]; Temam [I, Chapter III].

164 VIII. NEWTONIAN FLUIDS. NAVIER~STOKES EQUATIONS

EXERCISES

l. Supply the details of the following proof of the Poincare inequality. Lettp be a smooth scalar field on fJf with cp = 0 on a~. Since fJf is bounded,cp can be extended to a box ~ (with sides parallel to coordinate planes)by defining cp = 0 on ~ - fJf. Suppose that for x E ~,x\ varies betweeno: and fJ. Define cp\ = acpjax\. Then

cp2(X) = 2 flcp(e,X2,X3)cp\(e,X2,X3)de,

and hence

r cp2 dV = 2 J ('cp(e, x2, x 3)cp\(e, x2, x 3) de dVxJ!R ~ J",

~ 2 (L fcp2(e, x 2, x 3) de dVx ) \/2

X (L f CPT(e, x 2, X3) de dVx) 1/2

s 2(fJ - rx)(L cp2 dV yl2(L [grad cp 12 dV yl2

which implies the Poincare inequality

L cp2 dV ~ 4(fJ - rx)2 L1grad cpl2 dV.

Show that this implies an analogous result for a smooth vector field von fJf with v = 0 on afJf.

2. Establish (4).

SELECTED REFERENCES

Bird, Stewart, and Lightfoot [l, Part I].Hughes and Marsden [I, §§ 16, 17].Ladyzhenskaya [I].Lamb [I, Chapter I].Noll [I].Serrin [l, Chapter G].Temam [I].

CHAPTER

IX

Finite Elasticity

25. ELASTIC BODIES

In classical mechanics the force on an elastic spring depends only on thechange in length of the spring; this force is independent of the past history ofthe length as well as the rate at which the length is changing with time. For abody the deformation gradient F measures local distance changes, and ittherefore seems natural to define an elastic body as one whose constitutiveequation gives the stress T(x, t) at x = x(p, r) when the deformation gradientF(p, t) is known:

T(x, t) = t(F(p, r), p). (1)

Formally, then, an elastic body is a material body whose constitutive class C(j

is defined by a smooth response function

t: Lin" x fJ4 --+ Sym

as follows: C(j is the set of all dynamical processes (x, T) consistent with (1).Note that for an elastic body the function t is completely determined by

the response of the material to time-independent homogeneous motions ofthe form

x(p, r) = q + F(p - Po);

that is, to homogeneous deformations.

165

(2)

166 IX. FINITE ELASTICITY

When convenient, and when there is no danger of confusion, we will write

T(F) in place of T(F, p).

Proposition. A necessary and sufficient condition that the response of anelastic body be independent of the observer is that the response function tsatisfy

QT(F)QT = T(QF) (3)

for every F E Lin + and every Q E Orth + •

Proof Choose FE Lin + arbitrarily, let x be the motion (2), and let Tbe defined by (1), so that (x, T) E ~, the constitutive class of the body. Assumethat the response is independent of the observer. Then every (x*, T*) relatedto (x, T) by a change in observer must also belong to ~. By (20.4) and (21.1),this can happen only if, given any rotation Q,

QTQT = T(QF).

Since T = T(F), this clearly implies (3). Conversely, if (3) is satisfied, then,in view of (20.4) and (21.1), the response is independent of the observer. D

We assume henceforth that the response is independent of the observer, sothat (3) holds.

The importance of the strain tensors U and C is brought out by the nextresult, which gives alternative forms for the constitutive equation (1).

Corollary (Reduced constitutive equations). The response function T iscompletely determined by its restriction to Psym; infact,

T(F) = RT(U)RT (4)

for every F E Lin +, where R E Orth+ is the rotation tensor and U E Psym theright stretch tensor corresponding to F; that is, F = RU is the right polar

* - -decomposition ofF. Further, there exist smooth response functions T, T, and Tfrom Psym into Sym such that

T(F) = Ft(U)FT,

T(F) = RT(C)R T,

T(F) = FT(C)FT,

(5)

with C = U 2 = FTF the right Cauchy-Green strain tensor corresponding to F.

Proof To derive (4) we simply choose Q in (3) equal to RT. Next, sinceF = RU, (4) can be written in the form

T(F) = FU- 1T(U)U- 1FT.

Thus if we define

25. ELASTIC BODIES 167

(7)

(6)T(U) = U- 1T(U)U- 1,

we are led to (5h. On the other hand, if we let

T(C) = T(C 1/2 ) ,

T(C) = T(C1/2 ) ,

for C E Psym, then (4) and (5)1 imply (5h,3' Finally, by (6),.T is smooth,while (7) and the smoothness ofthe square root (page 23) yield the smoothnessofT and T. 0

The converse is also true: each of the constitutive equations in (4) and (5)is independent of the observer. We leave the proof of this assertion as anexercise.

* - -It is important to note that the response functions T, T, and T depend alsoon the material point p under consideration.

Suppose that we rotate a specimen of material and then perform anexperiment upon it. If the outcome is the same as ifthe specimen had not beenrotated, then the rotation is called a symmetry transformation. To fix ideasconsider the two-dimensional example shown in Fig. 1; there an elastic ringis connected by identical mutually perpendicular elastic springs which meetat the center ofthe ring. The forces required to produce any given deformationare the same for any prerotation of the ring by a multiple of 90°, so that suchrotations are symmetry transformations. Moreover, they are the only sym-metry transformations, since any other prerotation can be detected by somesubsequent deformation. We now apply these ideas to the general elasticbody 114.

Choose a point p of 114 and let fr denote the homogeneous deformationfrom p with deformation gradient F:

f,(q) = p + F(q - p)

for every q E 114. Consider two experiments:

(1) Deform 114 with the homogeneous deformation fr .(2) Rotate 114 with the rotation fQ (Q E Orth +) about p and then deform

the rotated body with the homogeneous deformation fF (Fig. 2). In this casethe total deformation is fr ° fQ and the deformation gradient is FQ.

Figure 1

The stress at p is

Figure 2

and

T(F, p) in Experiment 1

T(FQ, p) in Experiment 2.

If for Q fixed these two stress tensors coincide for each F, then Q is called asymmetry transformation; Q has the property that the response of thematerial at p is the same before and after the rotation fQ • Thus, more suc-cinctly, a symmetry transformation at p is a tensor Q E Orth + such that

T(F, p) = T(FQ, p) (8)

for every F E Lin ", We write <§p for the set of all symmetry transformationsat p and call <§p the symmetry group at p. The next proposition shows thatthis terminology is consistent.

Proposition. <§p is a subgroup of Orth + .

Proof. Let us agree to write

<§ in place of <§p

whenever there is no danger of confusion. [Recall our previous agreement towrite T(F) for T(F, p).] Clearly, <§ is a subset of Orth + and I E <§. To showthat <§ is a subgroup of Orth + it therefore suffices to show that <§ is closedunder both inversion and multiplication:

Q E <§ => Q-l E <§,

Q, H E <§ => QH E <§.

Choose Q E <§ and FE Lin + arbitrarily. Then (8) with F replaced by FQ- 1

yields


so that Q-1 E '§. Next, choose F E Lin + and Q, H E '§. Then

T(F) = T(FQ) = T«FQ)H) = T(FQH)

so that QH E '§. D

Our next result shows that (3) (which expresses invariance under observerchanges) and (8) together imply the invariance ofT under '§ (cf. Section 37).

"'" - * -Proposition. The response functions T, T, T, and T are invariant under '§.

Thus, in particular,

QT(F)QT = T(QFQT),

QT(C)QT = T(QCQT),

for every Q E '§, F E Lin +, and C E Psym.

Proof Choose Q E '§. Then QT E '§ and (8) implies

(9)

We therefore conclude, with the aid of (3), that (9)1 holds.To prove (9h choose C E Psym arbitrarily and let U be the unique square

root ofC:

By (6) and (7h,

Thus, since

(QUQT)2 = QU2QT = QCQT,

(QUQT)-1 = QU- 1QT,

we conclude from (9)1 that

QT(C)QT = QU- 1QTQT(U)QTQU- 1QT

= (QUQT)-1T(QUQT)(QUQT)-1 = T«QUQT)2) = T(QCQT).

We leave as an exercise the proof of the assertions concerning t and T. D

We say that the material at p is isotropic if

'§p = Orth"

(so that every rotation is a symmetry transformation), anisotropic if

'§p =I- Orth +.


This definition, the last proposition, and the proposition on page 230 implythe following

Proposition. Assume that the material at p is isotropic. Then each of theresponse functions t, t, 1', and f (at p) is an isotropicfunction.

The stress

T R = 1'(1)

is called the residual stress at p; T R is the stress at p when the body is unde-formed. Since C = U = R = I when F = I, (5) implies that

* - -TR = T(I) = T(I) = T(I). (10)

(11)

Proposition. If the material at p is isotropic, then TR is a pressure.

Proof Let Q E Orth. Then, since l' is isotropic,

QTRQT = QT(I)QT = T(QIQT) = 1'(1) = T R ,

or, equivalently, QTR = T R Q, so that TR commutes with every orthogonaltensor. We therefore conclude from the corollary on page 13 that T R = -niwith 1t a scalar. D

By (6.7) the right and left stretch tensors U and V and the right and leftCauchy-Green strain tensors C and B are related by

V = RURT, B = RCRT

,

where R E Orth + is the corresponding rotation tensor. Therefore when thematerial is isotropic, (4), (5)z, and the proposition containing (9) yield

T(F) = RT(U)RT = T(RURT) = T(V),

T(F) = RT(C)RT = T(RCRT) = T(B),

and thus the constitutive relation T = T(F) can be written in the alternativeforms

T = T(V),

T = feB),

with l' and Tisotropic functions. In view of(37.21), these comments have thefollowing consequence:

Theorem (Constitutive equation for an isotropic material). Assume thatthe material at p is isotropic. Then the constitutive relation can be written inthe form

T = Po(.fB)1 + PI(.FB)B + Pi.FB)B- I, (12)

where B = FFT is the left Cauchy-Green strain tensor and Po, PI' and pz arescalar functions ofthe list .FB ofprincipal invariants ofB.

(13)

Further, as is clear from (11) and (37.15), the constitutive equation (12)for an isotropic material point can also be expressed in the forms

T = lXo(JB)I + IXt(fB)B + 1X2(fB)B2,

T = Ko(fv)I + Kt(fv)V + K2(fv)V-t,

T = oo(fv)I + 0t(fv)V + 02(fv)V2,

where V = B t/2 is the left stretch tensor.

We now return to the general theory and list the complete system of fieldequations; this consists of the constitutive equation

T = J?lr(<:))?T,

C = )?T)?,

the equation of motion

and balance of mass

divT + b = pt,

p det F = Po,

(14)

(15)

where Po is the density in the reference configuration. When the body isisotropic (13) may be replaced by

T = Po(fB)I + Pt(fB)B + P2(fB)B-t,

B = )?)?T.

We say that the body is homogeneous provided both Po(p) and T(J?, p) areindependent ofthe material point p. In this case each of the response functions,as well as the symmetry group I'§p, is independent of p.

Consider the homogeneous deformation (2). For a homogeneous bodythe corresponding stress T = t(J?) is constant, because F is constant.Therefore T satisfies the equation of equilibrium

div T = 0

and (x, T) is a solution of (13)-(15) with b = O. Thus a homogeneous bodycan be deformed homogeneously without body force.

EXERCISES

1. Show that each of the constitutive equations in (5) is independent of theobserver.

2. Show that the response functions t and f are invariant under I'§.


3. Show that an elastic fluid is an isotropic elastic body.

4. Define the extended symmetry group .Yep to be the set of all tensorsA E Lin + such that

T(F, p) = T(FA, p)

for all F E Lin +.

(a) Show that .Yep is a subgroup of Lin +.(b) Give a physical interpretation of .Yep.(c) Let Unim denote the proper unimodular group; that is,

Unim = {A E Lin" [det A = I}.

(16)

Show that if.Yepis not contained in Unim, then there exist sequences{Fn} and {G n} with Fn, Gn E Lin+ such that

(i) det F, --. 00 but T(Fn , p) is the same for all n;(ii) det Gn --. 0 but T(Gn , p) is the same for all n.

Why is this physically unreasonable?

For the remainder of this section we assume that

.Yep c Unim.

(d) Show that an elastic fluid has

.Yep = Unim

(17)

(18)

for every p E fJl.

(e) Show that, conversely, an elastic body whose response functionobeys (18) is an elastic fluid. [For bodies exhibiting more generaltypes of behavior, such as viscoelasticity, the condition (18) furnishesa useful definition of a fluid.]

5. It is often convenient to use a deformed configuration as reference (seeFig. 3). With this in mind, let g be a deformation of fJl, let fbe a deforma-tion of g(fJl), and let

G = Vg(p), F = Vf(q), q = g(p).

Then (under fog) the material point p experiences the stress

T = T(V(f 0 g)(p), p) = T(FG, p).

Let Ta be defined by

Ta(F, p) = T(FG, p)

25. ELASTIC BODIES

Figure 3

173

for every FE Lin + ; f l(-, p) is the response function for p when the de-formed configuration (under the deformation g) is used as reference.

(a) Show that G E Unim belongs to Jrpifandonlyif,foreveryF ELin +,

fl(F, p) = f(F, p)

for every deformation g of ~ with Vg(p) = G.(b) Define Jrig) to be the extended symmetry group taking g as

reference; that is, Jrig) is the set of all HELin+ such that

fl(F, p) = fl(FH, p)

for all F E Lin +. Show that

Jrp(g) = GJrpG- I ,

(l9)

(20)

where G = Vg(p). (Here GJrpG- I is the set of all tensors of theform GHG- I

, H E Jrp.)(c) Show that

Jrp(g) c Unim.

Show further that if

Jrp(g) = Unim (21)

for some deformation g, then (21) holds for every deformation g.


(d) A material is a solid if there exists a configuration from which anychange in shape (i.e., any nonrigid deformation) is detectable bysome subsequent experiment. More precisely, the material at p issolid if there exists a deformation g such that

Jt'peg) c Orth +,

in which case g is undistorting at p. Assume that p is solid and thatthe identity map on f!J is undistorting at p (so that Jt'p c Orth "),Let g be any other deformation which is undistorting at p. Show that

Jt'peg) = RJt'pR - 1,

where R is the rotation in the polar decomposition of G = Vg(p).Show further that if, in addition, the material at p is isotropic, then

Jt'peg) = Jt'p(= Orth +),

and G is a similarity transformation:

G = A.Q

with A. > 0 and Q E Orth ".

6. An incompressible elastic body is an incompressible material body definedby a constitutive equation of the form

T = -nI + t(F),

or, more precisely, for x = x(p, t),

T(x, t) = -1t(p, t)I + t(F(p, t), p)

with

(22)

t: Unim x f!J -+ Sym

smooth. (As in the case of an incompressible fluid the pressure 1t is notuniquely determined by the motion.) For such a body:

(a) Determine necessary and sufficient conditions that the response beindependent of the observer.

(b) Define the notions of material symmetry and isotropy.(c) Show that in the case of isotropy the constitutive equation can be

written in the form

(23)

with fa = (ll(D), 12(D), 1). [The quantities 1tappearing in (22) and(23) are not necessarily the same.]

26. SIMPLE SHEAR 175

Two important examples of (23) are the Mooney-Rivlin material forwhich

T = -nI + PoB + PIB- I

with Po and PI constants, and the neo-Hookean material for which

T = -nI + PoB

with Po constant.

26. SIMPLE SHEAR OF A HOMOGENEOUS ANDISOTROPIC ELASTIC BODY

Let f1l be a homogeneous, isotropic body in the shape of a cube. Considerthe deformation x = x(p) defined (in cartesian components) by

Xl = PI + ypz,

where

Y = tan 8

is the shearing strain (Fig. 4). Since this deformation is homogeneous, thecorresponding stress T is constant. Thus (x, T) represents a solution of thebasic equations with body force b = o. (Cf. the discussion given at the endof the last section.) The matrix corresponding to the deformation gradientF is given by

[

I Y 0][F]=OIO

001

x •e2 e2 x(~)

0 e, 0 e.

Figure 4

and the matrix for the left Cauchy-Green strain tensor B = FFT is

[B]~[l~y' IHA simple calculation shows that

[Br' ~ [~Y 1 ~YY' Hdet{B - wI) = _w3 + {3 + y2)W2 - {3 + y2)W + 1,

and hence the list of principal invariants of B is

§B = (3 + yl,3 + y2, 1).

We therefore conclude from (25.12) that

T = Po{y2)I + Pl{y2)B + pz<y2)B - l,

Hence Tl 3 = T23 = 0 and

where

(l)

(2)

(3)

is the generalized shear modulus. The result (1) asserts that the shear stressTl2 is an odd function of the shear strain y.

In linear elasticity theory (cf. page 202) the normal stresses Tl l , T2 2 , andT3 3 in simple shear are zero. Here

Tll = y2Pl + r

T22 = y2P2+ r

26. SIMPLE SHEAR 177

Thus for the normal stresses to vanish both /31 and /32 would have to be zero,and this, in turn, would imply that J1 = 0. Thus if

for y #- 0, (4)

which is a reasonable assumption, then it is impossible to produce a simpleshear by applying shear stresses alone. Further (2)-(4) imply that, for y #- 0,

which is the Poynting effect. In fact, (1)-(3) yield the important result

This relation is independent of the material properties of the body; it issatisfied by every isotropic elastic body in simple shear.

EXERCISES

1. Show that the components of the unit normals on the slanted faces of thedeformed cube are given by

n l = ±(I + y2)-1/2

n2 = +: y(I + y2)-1/2,

n3 = 0,

and use this fact to show that

T _ yT12 and TI 2

22 1 + y2 1 + y2

represent, respectively, normal and tangential components of the surfaceforce Tn on these faces.

2. Consider the uniform extension defined by

x 3 = WP3'

Show that the corresponding stress T is a pure tension in the directione.; i.e.,


if and only if

(J = Po + Pl;.2 + P2;.-2,

0= Po + PlQi + P2 W -2,

where the scalar functions Po, Pl' and P2 are each functions of;.2 and w2(through the invariants J B) .

27. THE PIOLA-KIRCHHOFF STRESS

The Cauchy stress T measures the contact force per unit area in thedeformed configuration. In many problems of interest-especially thoseinvolving solids-it is not convenient to work with T, since the deformedconfiguration is not known in advance. For this reason we introduce a stresstensor which gives the force measured per unit area in the reference con-figuration.

Let (x, T) be a dynamical process. Then given a part 9, we can use (6.18hto write the total surface force on 9 at time t (see Fig. 5) as an integral over09; the result is

r Tm dA = r (det F)T",F-Tn dA,Jot?', Jot?'

where m and n, respectively, are the outward unit normal fields on 09, and09, while T", is the material description of T. Thus if we let

then

We call the field

r TmdA = r SndA.Jot?', Jot?'

(1)

(2)

(3)

S: PA x ~ -. Lin

defined by (1) the Piola-Kirchhoff stress.' By (2), Sn is the surface forcemeasured per unit area in the reference configuration.

Similarly, if b is the body force corresponding to (x, T), then

r b dV = rb",(det F) dV = rbo dV,Jt?'t Jt?' Jt?'

1 In the literature S is often referred to as thefirst Piola-Kirchhoffstress.

where

27. THE PIOLA-KIRCHHOFF STRESS

~-+.D(P)

Figure S

bo = (det F)b....

179

(4)

We call bo the reference body force; bo gives the body force measured per unitvolume in the reference configuration.

Proposition. The Piola-Kirchhoffstress S satisfies the balance equations

f SndA + f bo dV = f xPodV,J8~ J~ ~

(5)

f (x - 0) X SndA + f (x - 0) X bodV = f (x - 0) X xpo dVJ8~ J~ ~

for every part 9.

Proof. By (12.7),

f tp dV = f xPodV.~, ~

This relation, (2), and (3), when combined with balance of linear momentum(14.2)1' imply (5)1. Similarly, (6.18h, (12.7), (14.2h, and an analogous argu-mentyields (5)2. 0

Proposition. S satisfies the field equations

Div S + bo = Po X,

SFT = FST•

Proof. By (5)1 and the divergence theorem,

L(Div S + bo - Po x) dV = 0,

(6)


and, since this relation must be satisfied for every part &>, (6)1 must hold.Further, (1) implies

T", = (det F)-ISFT,

and (6)z follows from the symmetry of T. 0

It is important to note that, by (6)z, S generally is not symmetric.Next, by the symmetry ofT, (a) of the proposition on page 6, (8.8)1' and

(1.5)z,

and therefore

f T . D dV = f (det F)(T",F- T) • F dV = f s- F dV.

~. .Thus the stress power of &> at time t is given by

LS· F dV.

Also, (1), (4), (6.18)1' and (6.14)1 imply

f Tm. v dA = r (Tv)· m dA = f Sn· idA,lB., lB., lB.

f b· v dV = f bo • i dV,., .while (12.7) yields

(7)

We therefore have the following alternative version of (15.2).

Theorem of Power Expended. Given any part &>,

r Sn· i dA + f bo • it dV = f S • F dV + dd r i2

2

Po dV. (8)lB. • • t .I.The results established thus far are consequences of balance of momentum

and are independent of the particular constitution of the body.Assume now that the body is elastic, so that by (1) and (25.1),S is given by

a constitutive equation of the form

S = S(F) (9)

with

27. THE PIOLA-l(IRCHHOFF STRESS 181

S(F) = (det F)T(F)F- T. (10)

(As before, we do not mention the explicit dependence of S on the materialpoint p.) Choose Q E Orth ". Then

det(QF) = det F, (QF)-T = QF- T,

so that (25.3) and (10) imply

S(QF) = det(QF)T(QF)(QF)-T = (det F)QT(F)QTQF-T = QS(F).

Thus

S(QF) = QS(F) (11)

for every F E Lin + and Q E Orth". This relation expresses the requirementthat the response be independent of the observer; it can be used to deduceconstitutive equations for S analogous to (25.5). It is easier, however, toproceed directly from (25.5h. Indeed, (25.5h, (9), and (10) imply that

and if we define

then, since

(12) reduces to

S = (det F)FT(C),

seC) = Jdet C fCc),

S = FS(C).

(12)

(13)

(14)

(15)

We leave it as an exercise to show that this constitutive equation is indepen-dent of the observer.

Note that, by (13) and the smoothness off (cf. the corollary on page 166),S is a smooth mapping ofPsym into Sym.

In view of (o), and (14), we can rewrite the basic system offield equations(25.13)-(25.15) in the form

S = FS(C),

C = FTF, F = Vx

Div S + bo = PoX.

The relation (6h follows automatically from the fact that S has symmetricvalues. Also, since the density enters (15) only through its reference value Po,which is assumed known a priori, balance of mass (25.15) need not be includedin the list of field equations.


All of the fields in (15) have fJI x IR as their domain, and the operatorDiv is with respect to the material point p in fJI. In contrast, some of the fieldsin (25.13)-(25.15) are defined on the trajectory ff, and, more importantly,the operator div is with respect to the place x in the current configuration.For this reason the formulation (15) is more convenient than (25.13)-(25.15)in problems for which the trajectory is not known in advance.

The boundary-value problems of finite elasticity are obtained by adjoiningto (15) suitable initial and boundary conditions. As initial conditions oneusually specifies the initial motion and velocity:

x(p, 0) = xo(p), x(p, 0) = vo(p) (16)

with Xo and Vo prescribed functions on fJI. To specify the boundary conditionsone considers complementary regular1 subsets [/1 and [/2 of ofJI (so that

owhere [/IX is the relative interior of [/IX) and then prescribes the motion on[/1' the surface traction on [/2:

x = x on [/1 x [0, (0), on [/2 x [0, (0) (17)

with x and sprescribed vector fields on [/1 X [0, (0) and [/2 X [0, (0),respectively.

Another form of boundary condition arises when one specifies the surfacetraction Tm on the deformed surface X([/2' t). A simple example of this typeof condition arises when one considers the effects of a uniform pressure no.Here

T(x, t)m(x) = -nom(x)

for all x E X([/2' t) and all t ~ 0. We can also write this condition as arestriction on the Piela-Kirchhoff stress S; the result is (cf. Exercise 6)

on [/2 x [0, (0). (18)

Clearly, this is not a special case of (17) because F(p, t) is not known inadvance. We can, of course, generalize (17) to include this case by allowing sto be a function s(F, p, t) ofF, p, and t, rather than a function ofp and t only.[Actually, one can show that the combination (det F)F-Tn depends only onthe tangential gradient of x on [/2']

1 Cf. Kellogg [I]; Gurtin [I, p. 14]. Roughly speaking, each' 5/'. is a relatively closed,piecewise smooth surface with boundary a piecewise smooth closed curve.

27. THE PIOLA-KIRCHHOFF STRESS 183

In the statical theory all of the fields are independent of time and theunderlying boundary-value problem consists in finding a deformation f thatsatisfies the field equations

S = FS(C),

C = FTF, F = Vf,

Div S + bo = 0,

and the boundary conditions

(19)

on 9'1' So = s on 9'2 (20)

with f and sprescribed functions on 9'1 and 9'2' respectively.When tractions are prescribed over the entire boundary, that is, when

(20) has the form

(5)1 implies that

So = s on i}[fI, (21)

r sdA + i bo dV = 0.JiJl1I 111

This relation involves only the data and furnishes a necessary condition forthe existence of a solution. On the other hand, (5)z yields

r (f - 0) x sdA + i (f - 0) x bo dV = 0,JiJl1I 111

and, because of the presence of the deformation f, is not a restriction on thedata, but rather a compatibility condition which will automatically besatisfied by any solution of the boundary-value problem (19), (21). Thisdifference between force and moment balance is illustrated in Fig. 6. As long

----~

Figure 6

-


as the forces acting on f!4 obey balance of forces, we expect a solution, eventhough moments are not balanced in the reference configuration; the bodywill simply deform to insure balance of moments in the deformed configura-tion.

EXERCISES

1. Give the details of the proof of (5h.

2. Show that

S(FQ) = S(F)Q (22)

(23)

for every Q in the symmetry group f§ and every F E Lin +. Show furtherthat Sand S are invariant under f§.

3. Use (11) to show that

S(Q)· Q = 0

for any smooth function (of time) Q with values in Orth +.

4. Let f!4 be bounded. Consider a dynamical process and define

qJ = ~ Lu2Po dV,

where u(p, t) = x(p, t) - P is the displacement. Show that

iP = f iJ2 podV - f S· Vu dV + r u· So dA.~ ~ Jo~

5. Show that the constitutive equation (14) is independent of the observer.

6. Equation (6.18h holds for any sufficiently regular subsurface Y' of of!):

i T(x)m(x) dA; = f T(f(p»G(p)o(p) dA p •f(9') 9'

Use this fact to establish (18).

28. HVPERELASTIC BODIES

Thus far the only general restriction we have placed on constitutiveequations is the requirement that material response be independent of theobserver. Thermodynamics also serves to constrain constitutive behavior,and a standard thermodynamic axiom (consistent with a purely mechanicaltheory) is the requirement of nonnegative work in closed processes. We nowstudy the implications of this axiom for elastic bodies.

28. HYPERELASTIC BODIES 185

Let (x, T) be a dynamical process, and let b be the corresponding bodyforce. Then given any part f!I', the work on f!I' during a time interval [to, t 1]

is given by

f' {r Tn' v dA + f b· v dV} dt,to J{j~t ~t

or equivalently, in view of (27.7), by

I' {fa~ Sn • i dA + Lbo' i dV} dt (1)

with S the Piola-Kirchhoff stress and bo the reference body force. Let usagree to call (x, T) closedduring [to, t 1] provided

x(p, to) = x(p, t1), (2)

(3)

(4)

for all p E 911. Ifwe integrate (27.8) between to and t 1 and use (2h, we see thatfor processes of this type (1) reduces to

ft l f S . F dV dt.

to ~

We consider now an elastic body and restrict our attention to dynamicalprocesses belonging to the constitutive class <c ofthe body. For convenience,we will use the term process as a synonym for "dynamical process in <C." Wesay that the work is nonnegative in closed processes if given any part f!I' andany time interval [to, t 1] ,

ft ' f S· F dV dt ~ 0to ~

in any process which is closed during [to, tIl

Proposition. The work is nonnegative in closed processes if and only if givenany p E 911 and any time interval [to, t1],

ft '

S(p, t) • F(p, t) dt ~ 0to

in any process which is closed during [to, tIl

Proof Let A(p) denote the left side of (4). Clearly, A(p) ~ 0 for all pimplies (3). Conversely, assume that (3) holds for every part f!I'. Then, byinterchanging the integrations in (3), we may conclude that

LA dV ~ 0

for every part f!I'. In view of the localization theorem (5.1), this implies thatA(p) ~ 0 for all p. 0


Our next step will be to describe the class of elastic materials for which thework is nonnegative in closed processes. Anticipating the result, we introducethe following definition: An elastic body is hyperelastic if the Piela-Kirchhoffstress S(F, p) is the derivative of a scalar function a(F, p); that is,

S(F, p) = Du(F, p), (5)

where the derivative is with respect to F holding p fixed. The scalar function

u: Lin+ x fJi --+ ~

is called the strain-energy density. (Note that S determines uonly up to anarbitrary function of palone.)

For (5) to make sense we must interpret the derivative Du as a tensor. Bydefinition Du(F, p) is a linear mapping of Lin into ~,and hence Du(F, p) [A]can be written as the inner product! of a tensor K with A. We here identifyDu(F, p) with K and write

Du(F, p) . A in place of Du(F, p) [A].

With this interpretation Du(F, p) is a tensor and (5) has meaning. Thecomponents of this tensor are

oDu(F, P)ij = of.. u(F, p).

'}

Theorem. An elastic body is hyperelastic ifand only ifthe work is nonnegativein closed processes.

Proof. Assume that fJi is hyperelastic. Consider a closed process during[to, t1]. Then

d . ~ .dt u(F(p, t), p) = Du(F(p, t), p) • F(p, t) = ;:,(F(p, r), p) • F(p, t), (6)

and, since

[cf. (2)1]'

fit ft. dS(p, t) • F(p, t) dt = d u(F(p, r), p) dt

to to t

= u(F(p, t 1 ) , p) - u(F(p, to), p) = O.

Thus the work is nonnegative (in fact, zero) in closed processes.Conversely, assume that the work is nonnegative in closed processes.

(7)

1 The representation theorem for linear forms (page 1) is valid on any finite-dimensionalinner-product space.

28. HYPERELASTIC BODIES

Assertion 1. Let F: IR _ Lin + be a C2 function with

187

(8)

Then

fl l

S(F)· F dt = 010

(9)

(Here and in what follows we suppress the dependence of S on p.)

Proof Let!F = !F{to, t1) denote the set of all C2 functions F: IR - Lin +

which satisfy (8). As our first step we observe that given any F E !F,

for if x is the motion

fl l

S(F) . F dt ;::: 0,10

(10)

(11)

x(p, t) = Po + F(t)(p - Po),

then the corresponding process is closed during [to, t1] and (4) implies (10).Choose F E !F and let F*: IR - Lin + be defined by

F*(t) = F{to + t 1 - t),

so that F* represents the reversal (in time) of F. Then, by (8),

F*{to) = F(t 1) = F{to) = F*(t 1) ,

and, since

. d .F*(t) = dt F(to + t1 - t) = -F(to + t 1 - t), (12)

it follows that

F*{to) = -F(t 1 ) = -F(to) = F*{t1)' (13)

By (11) and (13), F* E!F; thus we may use (10) with F replaced by F* and(12) as follows:

Os fIIS(F*) • 11'* dt = - fIIS(F(to + t1 - t» • F{to + t1 - t) dtto to

= - fIIS(F(r» • F(r) dt:10

This inequality is clearly compatible with (10) only if (9) holds.Our next step will be to show that we can drop condition (8h without

affecting the validity of (9). We do this by applying (9) to a family of functions

It I < s

which satisfy both of (8), but whose limit satisfies only (8)1. This family isconstructed in

Assertion 2. Let F be a piecewise smooth,' closed curve in Lin + ; thatis, let F: [0, 1] --+ Lin + be piecewise smooth and satisfy

F(O) = F(I) == A.

Then there exists a one-parameter family Fs (0 < fJ < fJ o) of COO functionsF ~: IR --+ Lin + such that:

(a) F~ = A on (-00, -fJ] u [1 + fJ, (0);(b) IF ~ I and IF~ Iare bounded on IR with bounds independent of fJ;(c) as fJ --+ 0, F~ --+ F everywhere on [0, 1], while F~ --+ F at points of

continuity of F on (0, 1).

Proof We use the Friedrichs-Sobolev method ofmollifiers to constructthe family F~. For each fJ > 0 let p~: IR --+ IR be a Coo function with .thefollowing properties:

(i) p~ ~ 0;(ii) p~(t) = 0 whenever It I ~ fJ;

(iii) f~oo pit) dt = l.

An example of such a family of functions is furnished by

{ ,, ~ e x p ( - fJ2 1 2)'pit) = - t

0,

with ,,~ chosen to insure satisfaction of (iii).Using P~ we define a Coo family F~ as

F~(t) = f~oo p~(t - -r)F(r) dt (14)

for all t E IR, where we have extended the domain of F to IR by defining

F = A on (- 00,0) u (1, (0).

By (ii) and (iii), for t ~ - fJ or t ~ 1 + fJ,

Fit) = 5t H

p~(t - -r)A dt = A,t-~

and (a) follows.

1 F is continuous on [0, I], while Fexists and is continuous at all but a finite number ofpointswhere it suffers, at most, jump discontinuities.

Next, (14) and an integration by parts using (ii) yield

Fit) = f:ooPtl(t - T)F(T)dT = f:oo [- :TPtl(t - T)]F(T)dT

= 5:00 pit - T)F(T) dt: (15)

In view of (i) and (iii), we conclude from (14) and (15) that

lFit)1 s sup IFI foo pit - T) dt = sup IFI,[0,1] - 00 [0,1]

IFit) I s sup IFI,(O,l]

and these bounds yield (b).Finally, by (14), (ii), and (iii),

Ftl(t) - F(t) = f:oo Ptl(t - T)F(T) dt - F(t) f:oo Ptl(t - T) dt

= J~:tl pit - r)[F(r) - F(t)] dt;

hence (ii) and (iii) imply that

IFtl(t) - F(t) I ~ sup IF(T) - F(t)I,t€(t-tl,t+tl)

Similarly, using (15),

IFtl(t) - F(t) I ~ sup IF(T) - F(t)l,t€(t-tl,t+tl)

and the last two inequalities imply (c). Moreover, since F is uniformlycontinuous on IR (F is continuous on IR and constant outside a compactinterval), F tl ~ F uniformly on IR; thus, since det F > 0, there must exist a150 > °such that det Ftl > 0 on IR for 0 < 15 < bo. For this range of 15, F,can be considered as a mapping of IR into Lin +, This completes the proof ofAssertion 2.

Our last step is

Assertion 3. The body is hyperelastic.

Proof Let F be an arbitrary piecewise smooth, closed curve in Lin+ ,

and let Ftl (0 < 15 < 150 ) be the family established in Assertion 2. By (a) ofAssertion 2,

as long as C> < 1. Let C>I = min(1, C>o). Then for 0 < C> < C>I, each F" satisfiesthe hypotheses of Assertion 1 (with to = -1 and t l = 2); hence

rI ({J,,(t) dt = 0, (16)

where

(17)

(18)

By (a) of Assertion 2 the integral from - 1to - C> and from 1 + C> to 2 vanishes;we can therefore rewrite (16) as

II fO fl+"({J" dt + ({J" dt + ({J" dt = o.o -" I

By (b) and (c) of Assertion 2 in conjunction with Lebesgue's dominatedconvergence theorem I (recall that F has at most a finite number of dis-continuities),

f ({J" dt -. f S(F) • F dt.

On the other hand, (b) of Assertion 2 and (17) imply that the remainingintegrals in (18) converge to zero as C> -. O. Thus

fS(F) • F dt = O.

What we have shown is that the integral of Sover any piecewise smooth,closed curve in Lin + vanishes. Since Lin + is an open, connected subset of(the vector space) Lin, a standard theorem- in vector analysis tells us that Sis the derivative of a smooth scalar function b on Lin +. Thus (5) holds andthe proof is complete. 0

Hyperelastic materials have several interesting properties; an example isthe following direct consequence of (7).

Proposition. For a hyperelastic material the work is zero in closed processes.

If we write

IT(p, t) = b(F(p, r), t)

for the values of b in a process, then (6) implies

0- = S· F,

I Cf., e.g., Natanson [I, p. 161].2 cr., e.g., Nickerson, Spencer, and Steenrod [I, Theorem 8.4].


and the theorem of power expended (27.8) has the following importantcorollary.

Theorem (Balance of energy for hyperelastic materials). Each dynamicalprocess for a hyperelastic body satisfies the energy equation

L~ SO • xdA + Lbo· xdV = :t L(a + Po ~) dV

for each part &>. Here S is the Piola-Kirchhoffstress, Po is the reference density,and bo is the reference body force (27.4).

The term

Ladv

represents the strain energy of &>. The energy equation asserts that the powerexpended on &> must equal the rate at which the total energy of&> is changing.

As a direct consequence of the above theorem we have the followingimportant

Corollary (Conservation of energy). Assume that the body is finite andhyperelastic. Consider a dynamical process for the body corresponding to bodyforce b = 0, and suppose that

So· x= 0 on afJI

for all time. Then the total energy is constant:

L(a + Po ~) dV = const.

EXERCISES

1. Consider a hyperelastic body with strain-energy density fJ. (For con-venience, we suppress dependence on the material point p.)

(a) Use (27.23) and (5) to show that

fJ(Q) = fJ(I) (19)

for every Q E Orth+ . (Here you may use the connectivity of Orth +

to insure the existence of a smooth curve R: [0, 1] -+ Orth + withR(O) = I and R(l) = Q.)

(b) Use (27.11) and (5) to show that

fJ(QF) = fJ(F)

for every Q E Orth + and F E Lin +.


(c) Use (27.22) and (5) to show that

&(FQ) = &(F)

for every Q in the symmetry group and every .FE Lin + .

(d) Show that

&(F) = &(U)

and that there exists a function iJ such that

&(F) = iJ(C).

Here U is the right stretch tensor, C the right Cauchy-Greenstrain tensor.

(e) Show that

S = 2DiJ,

where DiJ(C) E Sym with components

o~ .. iJ(C)1)

has an interpretation analogous to D&(F).(f) Assume that the material at p is isotropic. Show that

iJ(C) = iJ(B)

with B the left Cauchy-Green strain tensor, and that

[cf. (37.4)J. Let

Show that the Piela-Kirchhoff and Cauchy stress tensors are givenby

S =2{iJ1F + iJ2 [ (tr B)I - BJF + (det B)iJ3F-T

} ,

T = 2(det F)-1{(det B)iJ3 1 + [iJ1 + (tr B)iJ2JB - iJ2B2 },

where we have omitted the arguments of iJk'

2. Extend the results of this section to incompressible elastic bodies (cf.Exercise 25.6).

3. Consider a Newtonian fluid. Show that the work is nonnegative in closedprocesses if and only if the viscosity u is nonnegative.


4. Consider the statical (time-independent) behavior of a homogeneous hyperelastic body without body forces. Show that

Div[a(F)I - FTS] = 0

and hence that

Ia9[6(F)n - FTSn] d A = 0

for every part 9 of 9. (This result is of importance in fracture mechanics.)

(Principle of stationary potential energy) For a hyperelastic body the (statical) mixed problem discussed in Section 27 can be stated as follows: find a C2 deformation f such that

5.

S = S(F) = Db(F), F = Vf,

Div S + b, = 0

on 9 and

f = P on Y,, S n = B on Y,. (20)

Let us agree to call a C2 function f : 9 4 8 with det Vf > 0 kinematically admissible i f f satisfies the boundary condition (20),. Assume that 9 is bounded. Define the potential energy @ on the set of kinematically admissible functions by

@{f} = Ia8(Vf) dV - IBb, - u dV - I - u dA, L where u(p) = f(p) - p, We say that the variation of @ is zero at f, and write

6@{f} = 0,

if

d da - @{f + ag} l a = , = 0

for every g with f + ag in the domain of @ for all sufficiently small a (that is, for every C 2 function g : 9d -+ Y with g = 0 on Yl). Show that

b@{f) = 0

provided f is a solution of the mixed problem.


29. THE ELASTICITY TENSOR

The behavior of the constitutive equation

S = S(F)

[cr. (27.9)] near F = I is governed by the linear transformation

C: Lin --+ Lin

defined by

C = DS(I). (1)

(As before we have suppressed mention of the dependence on the materialpoint p.) C is called the elasticity tensor for the material point p; TOughlyspeaking, C is the derivative of the Piela-Kirchhoff stress with respect toF at F = I. The importance of this tensor will become apparent in the nextsection, where we deduce the linearized theory appropriate to small deforma-tions from the reference configuration.

For convenience, we assume throughout this section that the residual stressvanishes:

S(I) = t(I) = O. (2)

Our first result shows that, because of (2), C could also have been definedas the derivative of r, the response function for the Cauchy stress.

Proposition

Proof By (27.10),

C = Dt(I).

S(F)FT = qJ(F)t(F),

(3)

where qJ(F) =det F. If we differentiate this relation with respect to F, usingthe product rule, we find that

S(F)HT + DS(F)[H]FT = qJ(F) Dt(F) [H] + DqJ(F) [H]t(F)

for every HELin. Evaluating this expression at F = I, we conclude, with theaid of (2), that

DS(I) [H] = Dt(I) [H],

which implies (3). 0

29. THE ELASTICITY TENSOR

Proposition (Properties of the elasticity tensor)

(a) C[H] E Symfor every HELin;(b) C[W] = Ofor every W E Skw.

Proof. Assertion (a) follows from the relation

dC[H] = drx 1'(1 + rxHH.=o

195

and the fact that l' has symmetric values. To prove (b) choose W E Skw andtake

Q(t) = eW t,

so that Q(t) E Orth+ (cf. Section 36). Then (25.3) with F = I and (2) imply

t(Q(t» = 0,

and this relation, when differentiated with respect to t, yields

Dt(Q(t» toon = O.

Since Q(O) = I and Q(O) = W, if we evaluate this expression at t = 0 anduse (3), we are led to (b). 0

Let

denote the symmetric part of HELin. Then H = E + W, with W skew, and(b) and the linearity of C imply that

C[H] = C[E];

hence C is completely determined by its restriction to Sym.Our next step is to establish the invariance properties of C.

(4)

Proposition. C is invariant under the symmetry group '§ for the material at p.

Proof. In view of (25.9)1' the last theorem in Section 37, and (3), forHELin and Q Ee,

QC[H]QT = QDt(l) [H]QT = Dt(QIQT) [QHQT]= Dt(l) [QHQ~] = C[QHQT],

and C is invariant under '§. 0


Since C has values in Sym [cf. (a) above], this proposition and (37.22)have the following obvious but important consequence.

Theorem. Assume that the material at p is isotropic. Then there exist scalarsJl and A such that

C[E] = 2JlE + A(tr E)I (5)

for every symmetric tensor E. The scalars Jl = Jl(p) and A = A(p) are called theLame moduli at p.

We say that C is symmetric if

H'C[G] = G' C[H]

for all tensors Hand G.Since C[W] = 0 for all skew W, C can never be positive definite in the

usual sense. There are, however, important situations in which C restrictedto Sym has this property. Thus let us agree to call C positive definite if

E' C[E] > 0

for all symmetric tensors E #- O.

Proposition. Assume that the material at p is isotropic. Then C is symmetric.Moreover, C is positive definite if and only if the Lame moduli obey the in-equalities:

u > 0, 2Jl + 3A > O. (6)

Proof Let Hand G be tensors with symmetric parts Us and Gs ' respec-tively. Then, by (a) and (b) of the proposition on page 195, and since I· H, =tr H, (5) implies

H • C[G] = H s ' C[Gs] = 2JlHs ' Gs + A(tr H)(tr G) = G . C[H],

so that C is symmetric. Next, choose a symmetric tensor H and let.

so that

a=!trH,

H = H, + aI,

H o = H - aI

tr H o = O.

Then, since I • H, = 0,

H • C[H] = 2Jl(1H o12 + 3ct2

) + 9A.a2 = 2Jll H, 12 + 3a2(2ti + 3A).

Trivially, (6) implies that C is positive definite. Conversely, if C is positivedefinite, then by choosing H = aI we conclude that 2Jl + 3A > 0, and bychoosing H with tr H = 0 we see that Jl > O. 0

29. THE ELASTICITY TENSOR 197

Proposition. Assume that the material at p is hyperelastic. Then C is sym-metric.

Proof Since

[ef. (28.5)], the proposition follows from the symmetry of the second deriva-tive. To see this note that

oof3 ~(I + txH + f3G) = SCI + txH + f3G) . G,

and hence

02

Otx of3 ~(I + txH + f3G) !a=P=o = DS(I)[H]· G = C[H] . G;

by switching the order of differentiation we see that this expression equals

02

of3 ctx ~(I + txH + f3G ) la=P=o = C[G]' H,

and the symmetry of C follows. D

EXERCISES

1. Show that, as a consequence of (27.11),

DS(F) [WF] = WS(F) (7)

for all F E Lin + and W E Skw.

2. Relate C = DS(I) and L = Dt(I) for the case in which (2) is not satisfied.

3. C is strongly elliptic if

A· C[A] > 0

whenever A has the form A = a ® e, a =1= 0, C =1= O.

(a) Show that if C is positive definite, then C is strongly elliptic.(b) Show that for an isotropic material C is strongly elliptic if and only

if

2Jl. + A. > O.


SELECTED ~ERENCES

Coleman and Noll [1].Green and Zerna [1].Sternberg and Knowles [1].Truesdell and Noll [1, Chapter OJ.Wang and Truesdell [1].

CHAPTER

xLinear Elasticity

30. DERIVATION OF THE LINEAR THEORY

We now deduce the linearized theory appropriate to situations in whichthe displacement gradient Vu is small. The crucial step is the linearization ofthe general constitutive equation

(1)

for the Piela-Kirchhoff stress [cf. (27.9)]. In order to discuss the behavior ofthis equation as

0= Vu

tends to zero, we consider S(F) as a function of H using the relation

F = I + 0[cf. (7.1)].

Theorem (Asymptotic form of the constitutive relation). Assume that theresidual stress vanishes. Then

S(F) = C[E] + 0(0)

as H -+ 0, where C is the elasticity tensor (29.1) and

E = 1(H + HT)

is the infinitesimal strain (7.4).

199

(2)

(3)

200 x. LINEAR ELASTICITY

Proof Since the residual stress vanishes, we may conclude from (29.1),(29.2), and (29.4) that

S(F) = S(I + H) = S(I) + DS(I) [HJ + o(H)

= C[HJ + o(H)

= C[EJ + o(H). D

Using (2) we can write the asymptotic form of the constitutive equation(1) as

S = C[EJ + o(Vu). (4)

Therefore, if the residual stress in the reference configuration vanishes, then towithin terms ofo(Vu) as Vu -. 0 the stress S is a linearfunction ofthe infinitesimalstrain E. Also, since C has symmetric values [cf. (a) of the proposition on page195J, to within the same error S is symmetric.

The linear theory of elasticity is based on the stress-strain law (4) with theterms of order o(Vu) neglected, the strain-displacement relation (3), and theequation of motion (27.6)1:

S = C[EJ,

E = t(Vu + VuT) ,

Div S + bo = Poii.

Note that these equations are expressed in terms of the displacement

u(p, t) = x(p, t) - p,

(5)

rather than the motion X.

It is important to emphasize that the formal derivation of the linearizedconstitutive equation (5)1 was based on the following two assumptions:

(a) The residual stress in the reference configuration vanishes.(b) The displacement gradient is small.

Note that, by (5)1 and the theorem on page 56, E = S = 0 in an infini-tesimal rigid displacement. This is an important property of the linearizedtheory.

Given C, Po, and bo, (5) is a linear system of partial differential equationsfor the fields u, E, and S. By (29.5), when the body is isotropic (5)1 may bereplaced by

S = 2J-lE + A.(tr E)I.

Moreover, when the body is homogeneous, Po, p; and A. are constants.

(6)

31. SOME SIMPLE SOLUTIONS

Assume now that f!l is homogeneous and isotropic. Then, since

Div(Vu + VUT) = L\u + V Div u

and

201

tr E = Div u,

the equations (5)2.3 and (6) are easily combined to give the displacementequation ofmotion

Ji. L\u + (A + Ji.)V Div u + bo = Poii.

In the statical theory ii = 0 and we have the displacement equation ofequilibrium

Ji. L\u + (A + Ji.)V Div u + bo = O.

EXERCISES

(8)

1. Let u be a C4 solution of (8) with bo = O. Show that Div u and Curl u areharmonic functions. Show further that u is biharmonic; that is,

L\L\u = O.

2. Let u be a C4 solution of (7) with bo = O. Show that Div u and Curl usatisfy wave equations.

3. (Boussinesq-Papkovitch-Neuber solution) Let qJ and g be harmonicfields on f!l (with qJ scalar valued and g vector valued) and define

u = g - exV(r . g + qJ),

Ji.+Aex = 2(2Ji. + A)'

where r(p) = p - o. Show that u is a solution of (8) with bo = O.

4. Consider the case in which the residual stress S(I) =F O. Show that

S(F) = 8(1) + W8(1) + C[E] + o(H).


Assume now that the body is homogeneous and isotropic. Then anystatical (i.e., time-independent) displacement field u with E constant gener-ates a solution of the field equations (30.5h. 3 and (30.6) with S constant and

202

LI

x. LINEAR ELASTICITY

Figure I

bo = O. We now discuss some particular solutions of this type, using cartesiancoordinates where convenient.

(a) Pure shear. Let

u(p) = YP2 el

(Fig. I), so that the matrices of E and S are

1[0 Y 0][E] = 2 Y 0 0,

000

with

't = p.y.

~ ~],o 0

Thus p. determines the response of the body in shear, at least within the lineartheory, and for this reason is called the shearmodulus. Note that, in contrast tothe general nonlinear theory (cf. Section 26), the normal stresses are all zero;the only stress present is the shear stress 'to

(b) Uniform compression or expansion. Here

u(p) = B(p - 0).

Then

E = BI,

S = -nl,

n = -3KB,

where

is the modulus of compression.For our third solution it is simpler to work with the stress-strain law (30.6)

inverted to give E as a function of S. This inversion is easily accomplishedupon noting that

tr S = (2p. + 3A.) tr E,


and hence

E = 21

[S - A. A. (tr S)IJ.J1. 2J1. + 3

(c) Pure tension. Here we want the stress to have the form

[

(1 0 0][S] = 0 0 O.

000

The corresponding strain tensor is then given by

o 0]I 0o I

with

203

(1)

and

1G = E (1,

E = J1.(2J1. + 3..1.)J1. + A. '

1= -VG,

Note that the strain E corresponds to a displacement field of the form

u(p) = GP1e1 + lPzez + lp3e3.

The modulus E is obtained by dividing the tensile stress (1 by the longitudinalstrain G produced by it. It is known as Young's modulus. The modulus v is theratio of the lateral contraction to the longitudinal strain of a bar under puretension. It is known as Poisson's ratio.

If we write Eo and So for the traceless parts of E and S, that is,

Eo = E - i{tr E)I, So = S - :ktr S)I, (2)

then the isotropic constitutive relation (30.6) is equivalent to the followingpair of relations:

So = 2J1.Eo,

tr S = 3K tr E.

Another important form for this stress-strain relation is the one taken by theinverted relation (1) when Young's modulus and Poisson's ratio are used:

1E = E [(1 + v)S - v(tr S)I]. (3)

204 X. LINEAR ELASTICITY

Since an elastic solid should increase its length when pulled, shoulddecrease its volume when acted on by a pure pressure, and should respond to apositive shearing strain by a positive shearing stress, we would expect that

E > 0, K> 0, Il > 0.

Also, a pure tensile stress should produce a contraction in the directionperpendicular to it; thus

v> O.

Even though these inequalities are physically well motivated, we will notassume that they hold, for in many circumstances other (somewhat weaker)assumptions are more natural. In particular, we will usually assume that C ispositive definite. A simple computation, based on (29.6), shows that thisrestriction is equivalent to either of the following two sets of inequalities:

(i) Il > 0, K > 0;(ii) E > 0, - 1 < v < 1-Some typical values for Young's modulus E and Poisson's ratio v are '

carbon steel: E = 2.1 X lOll N/m2,

copper: E = lOll N/m2,

glass: E = 0.55 X lOll N/m2,

EXERCISES

v = 0.29,v = 0.33,v = 0.25.

1. Assume that C is symmetric and define

~(E) = tE· C[E]

for every E E Sym.

(a) Show that the stress-strain law S = C[E] can be written as

S = m(E).

(b) Show that, for an isotropic material,

A.~(E) = IlIEI2 + "2 (tr E)2,

K~(E) = IIIEo1

2 + "2 (tr E)2,

~IEI2 s ~(E) ~ PIEI2,

where ~ is the smaller and Pthe larger of the numbers Il and 3K/2.

I Cf., e.g., Sokolnikoff [I, p. 70].

32. LINEAR ELASTOSTATICS 205

2. Show that C is positive definite if and only if either of the two sets ofinequalities (i) or (ii) hold.

32. LINEAR ELASTOSTATICS

The system offield equations for the statical behavior of an elastic body-within the framework ofthe linear theory-consists ofthe strain-displacementrelation

E = 1<Vu + VoT) ,

the stress-strain relation

S = C[E],and the equation ofequilibrium

Div S + b = 0

(1)

(2)

(3)

(4)

(where we have written b for bo). The elasticity tensor C, which is a linearmapping of tensors into symmetric tensors, will generally depend on positionp in 84; writing Cp to emphasize this dependence, we assume henceforth thatCp is a smooth function of p on 84.

A list [0, E, S] of fields which are smooth on 84 and satisfy (1)-(3) for agiven body force b will be called an elastic state corresponding to b. Note thatby (1), (2), and the properties of C, the fields E and S are symmetric.

We assume throughout this section that 84 is bounded.

Theorem of Work and Energy. Let [u, E, SJ be an elastic state correspondingto the body force b. Then

r se- 0 dA + f b' u dV = 2OU{E},JiJ~ IJI

where

OU{E} = ~ LE' C[E] dV (5)

is the strain energy.

The proof of this theorem is an immediate consequence of the following

Lemma. Let S be a smooth symmetric tensor field on 84, let ii be a smoothvector field on 84, and let

div S + b = 0,

206

Then

X. LINEAR ELASTICITY

r SO' ii dA + f b· ii dV = f S •EdV.Jo~ ~ ~

(6)

Proof By the symmetry ofS, the divergence theorem, (4.2)5' and (a) ofthe proposition on page 6,

r So' ii dA = r (Sii)' 0 dA = f Div(Sii) dV = f (ii • Div S + S . Vii) dV,Jo~ Jo~ ~ ~

S· Vii = S· {!<Vii + ViiT)} = S· E.These relations clearly imply the desired result (6). 0

We can interpret the left side of(4) as the work done by the external forces;(4) asserts that this work is equal to twice the strain energy. Note that when Cis positive definite, dIJ{E} ~ 0, and the work is nonnegative.

The next theorem is one of the major results of elastostatics; in essence, itexpresses the fact that the underlying system offield equations is self-adjointwhen C is symmetric.

Betti's Reciprocal Theorem. Assume that C is symmetric. Let [u, E, S] and[ii, E, S]be elastic states corresponding to body force fields band b, respectively.Then

r So' ii dA + f b . ii dV = r So' u dA + f b' u dVJo~ ~ Jo~ ~

= LS'EdV = LS'EdV. (7)

Proof Since C is symmetric, we conclude from the stress-strain relationthat

S'E = C[E]'E = C[E]'E = S'E,

and (7) follows from (6) and the analogous relation with the roles of the twostates reversed. 0

Betti's theorem asserts that given two elastic states, the work done by theexternal forces ofthe first over the displacement ofthe second equals the workdone by the second over the displacement of the first.

Let!/' 1 and!/'2 denote complementary regular subsets ofthe boundary offfl, so that


owith f/ Il the relative interior of f/ Il' Then the mixed problem of elastostaticscan be stated as follows:

Given: fJI, f/ l' f/2' an elasticity tensor C on fJI, a body force field b on fJI,surface displacements Uon f/ 1> surface tractions son f/ 2 •

Find: An elastic state [u, E, S] that corresponds to b and satisfies theboundary conditions

(8)

An elastic state with these properties will be called a solution.

Uniqueness Theorem. Assume that C is positive definite. Let [u l , E l , SI]and [u2, E2, S2] be solutions of the (same) mixed problem. Then

where w is an infinitesimal rigid displacement of fJI.

Proof Let

Then [w, E, S] is an elastic state that corresponds to null body forces andsatisfies the boundary conditions

w=o on f/l'

Thus

Sn·w = 0

and we conclude from (4) that

on iJfJI,

LE • C[E] dV = O.

Since C is positive definite, this relation can hold only if E = 0; this in turnimplies that S = 0 and that w is an infinitesimal rigid displacement of fJI (cf.the theorem on page 56). 0

When f/ 1 = iJfJI (f/2 = 0) the boundary condition (8) takes the form

u=Uand the mixed problem is referred to as the displacement problem. On theother hand, when f/ 2 = iJfJI (f/ 1 = 0), so that

Sn = s on iJfJI,

we have the traction problem. This last definition, (3), and Cauchy's theorem(page 101) have the following immediate consequence:

Proposition. A necessary condition that the traction problem have a solution isthat

i sdA + f b dV = 0,(JlJI Bit

i r x sdA + f r x b dV = 0,oBit Bit

(9)

where r(p) = p - 0 is the position vector.

Equations (9) insure equilibrium of the external forces applied to fJ4. It isinteresting to compare these conditions with the analogous restrictions ofthegeneral nonlinear theory discussed at the end of Section 27. There the appliedforces need not obey balance of moments in the reference configuration[cf. the discussion following (27.21)], as in the nonlinear theory the body isallowed to undergo the possibly large deformation needed to insure momentbalance in the deformed configuration.

We now show that the solution of the mixed problem, if it exists, can becharacterized as the minimum value of a certain functional.

By a kinematically admissible state we mean a list" = [0, E, S] with u, E,and S smooth fields on fJ4 that satisfy the field equations

E = t<Vu + VoT) ,

and the boundary condition

S = C[E],

u=o on [/'1.

Let ell be the functional defined on the set ofkinematically admissible states by

ell{,,} = Olt{E} - f b'udV - f s·odA.Bit .'72

(10)

Principle of Minimum Potential Energy. Assume that C is symmetric andpositive definite. Let" = [0, E, S] be a solution of the mixed problem. Then

ell{o} s ell{a}

for every kinematically admissible state a = [ii, E,S],and equality holds only ifii = 0 + w with w an infinitesimal rigid displacement of fJ4.

Proof Let

w=ii-o, E=E-E.

Then, since <1 is a solution and <3 is kinematically admissible,

E = -t(Vw + VWT),

W = 0 on !/'t.Further, since C is symmetric and S = C[E],

E . C[E] = E' C[E] + E' C[E] + E • C[E] + E . C[E]= E' C[E] + E' C[E] + 2S •E;

hence

209

(11)

o/!{E} - o/!{E} = o/!{E} + Ls' E dV.

Because <1 is a solution, we conclude from (11h and (6) with ii and Ereplacedby wand E that

rS· E dV = r So' w dA + rb· w dV = f s· w dA + rb· w dV.JBI JOBI JBI f/, JBI

In view of (10) and the last two relations,

cI>{<3} - cI>{<1} = o/!{E}.

Thus, since C is positive definite,

cI>{<1} S; cI>{<3}

and

cI>{<1} = cI>{<3} only when E = 0;

that is, only when w = ii - u is an infinitesimal rigid displacement. D

In words, the principle of minimum potential energy asserts that thedifference between the strain energy and the work done by the body force andprescribed surface traction assumes a smaller value for the solution of themixed problem than for any other kinematically admissible state.

A standard technique for obtaining approximate solutions is based on theprinciple of minimum potential energy. We now demonstrate this method,but for convenience limit our attention to the traction problem. Consider anapproximate solution of the form

N

u(p) = L: IXng,.(p),n=l

(12)

where gl' gz, ... , gN are given vector fields on fJI, and where the scalar con-stants 1X1, IXz, .•. , IXN are chosen to make cI>(1X1' IXz, ... , IXN) a minimum. Here

Gn = -!(Vg" + VgJ),

<D(tx l, tx2, .•. , txN) is the function obtained by substituting (12) into the rightside of (10):

where

K mn = LGm • C[Gn] dV,

qJn = f b· g" dV + r s· g" dA.fJI JofJI

Let K be the matrix with entries K mn • If C is symmetric and positive definite,K will be symmetric and positive semidefinite; thus <D(txl' tx2 , ••• , txN)will be aminimum at a "vector" (txl' tx2 , ••• , txN)ifand only ifthe vectoris a solution ofthe equation

KO[ = <p,

where 0[ and <p are the column vectors with entries txn and qJn' respectively. Theproblem is therefore reduced to solving a matrix equation; the correspondingsolution represents the displacement field of the form (12) which is "nearest inenergy" to the true solution. The matrix K, which characterizes the responseof the system, is usually called the stiffness matrix.

The crucial part of this approximation technique lies in the choice of thefunctions gl' g2' ... , gN (one specific family of choices generates what iscommonly called the finite-element method); in particular, an importantconsideration is convergence as N -+ 00. This matter, however, is beyond thescope of this book.

EXERCISES

1. Show that if the traction problem has a solution [u, E, S], then [u + w,E, S] is a solution for every infinitesimal rigid displacement w. Does theanalogous result hold for the mixed problem in general?

2. Show that for a homogeneous, isotropic elastic body the "infinitesimalvolume change"

r u- D dA = rdiv u dVJO fJI JfJI

in an elastic state is given by

where r(p) = p - o.

For the remaining exercises C is symmetric.

211

3. Assume that C is positive definite. Then C restricted to symmetrictensors is invertible. Let K denote the corresponding inverse, so that thestress-strain relation S = C[E] can be inverted to give

E = K[S].

(a) Show that K is symmetric.(b) Show that the strain energy can be written in the form

O/tK{S} = ~ Ls. K[S] dV.

(c) (Principle of minimum complementary energy) A statically ad-missible stress field S is a smooth symmetric tensor field thatsatisfies the equilibrium equation

Div S + b = 0

and the boundary condition

on [/2'

Let 'II be the functional defined on the set of statically admissiblestress fields by

'P{S} = CfIK {S } - f So· fj dA.9',

Show that if [u, E, S] is a solution of the mixed problem, then

'P{S} ~ 'P{S}

for every statically admissible stress field S.(d) Show that if o = [u, E, S] is a solution of the mixed problem, then

{{j} + 'P{S} = O.

4. (Hu-Washizu principle) An admissible state is a list a = [u, E, S] ofsmooth fields on 84 with E and S symmetric. Let A be the functionaldefined on the space of admissible states by

A{a} = IJlt{E} - fS·EdV - f (DivS + b)·udV + f So·(kdA~ ~ Y,

+ f (So - s) . u dA.Y2

Show that

(jA{a} = 0,

provided a is a solution of the mixed problem (cf. Exercise 28.5).

5. (Hellinger-Prange-Reissner principle) Assume that C restricted tosymmetric tensors is invertible with inverse K. Let d be the set of alladmissible states that satisfy the strain-displacement relation (1) anddefine e on d by

era} = IJltI({S} - f S·EdV + f b·udV + f So·(u - u)dA~ ~ Y,

+ f s·u dA.Y2

Show that

(je{a} = °provided a is a solution of the mixed problem.

6. Show that for a kinematically admissible,

A{a} = <D{a}.

7. Consider a one-parameter family of (homogeneous and isotropic)elastic states [u,; En S.J with Poisson's ratio v as parameter. Thesestates all correspond to the same shear modulus and to vanishing bodyforces.

(a) Assuming that [U., E., SvJ tends to a limit [u, E, S] as v -+ 1, showthat

S = -nI + 2flE, tr E = 0,

where t: = -t tr S. Give an argument to support the claim thatv = 1corresponds to incompressibility. (Note that, as in the case ofideal and Newtonian fluids, the "pressure" 1t is not uniquelydetermined by the deformation; that is, E does not uniquelydetermine n.)


(b) Assuming that the limit state [u, E, S] is approached in a mannerwhich is sufficiently regular to justify interchanges of limits andderivatives, show that

{t ~u - Vn = 0,

Div u = O.

(Cf. the equations on page 152 describing Stokes flows.)

8. An elastic state [u, E, S] is a state of plane strain if u has the form

in some cartesian frame. Consider such a state, and assume that thebody is isotropic and that b = O.

(a) Show that E and S are functions of (PI' P2), and that

while

E"p = 1(u",p + up,,,),

S"p = 2jJ.E"p + A.c5"p(E11 + E2 2 ) ,

2

L S"p,p=O,P=l

E13 = E2 3 = E3 3 = 0,

S33 = A.(E11 + E22) = V(Sl1 + S22)'

(13)

Here (X and p have the range of the integers (1, 2), c5"p = 1 when(X = pand zero otherwise, and

ou"u",p = opp' etc.

(b) Show that E (when C2) satisfies the compatibility equation

2E12, 12 = E 11 , 2 2 + E22,l1' (14)

For the remaining exercises fJt is an open simply connected region in 1R2•

9. Let E"p (= Ep,,) be a C2 solution of (14) on fJt. Show that there existdisplacements u" such that (13)1 hold.

10. Let E"p and S"p be C2 functions on fJt consistent with (13h, 3' Show that(14) is satisfied if and only if

(15)


[Thus for a simply connected region plane elastic states are completelycharacterized by (13h and (15). Indeed, when these equations aresatisfied, Ea.(J can be defined by (13h (assuming invertibility) and, since(14) holds, there exist Ua. that satisfy (13)1']

11. Let qJ be a scalar field of class C4 on fYt. Define

(16)

Show that Sa.(J satisfies (13h. Show further that (15) holds if and only if qJ

is biharmonic:

The field qJ is called an Airy stress function.

12. Let Sa.(J (= S(Ja.) be a C2 solution of (13)3 and (15) on fYt. Show that thereexists a biharmonic function qJ on fYt such that (16) holds.

33. BENDING AND TORSION

A. BENDING OF A BAR

Consider a (homogeneous, isotropic) cylindrical bar with generatorsparallel to the P3-axis (Fig. 2). Let the end faces [/0 and [/, be located at P3 = 0and P3 = I, respectively, with the origin at the centroid of [/0 and with the PI-and Pl-axes principal axes of inertia:

f PI dA = f P2 dA = f PIP2 dA = O.!/'0 !/'0 !/'0

(1)

We assume that the bar is loaded only on the end faces, and by opposingcouples about the P2-axis. More precisely, we assume that

(a) body forces are zero;(b) the lateral surface 2 is traction-free; that is,

So = 0

or equivalently, since n3 = 0 on 2,

on 2,

on 2; (2)

33. BENDING AND TORISON

P,

Figure 2

P3

215

(c) the total force on go vanishes and the total moment on go isequipollent to a moment of magnitude m about the negative P2-axis; that is,

f SodA = 0,9'0

f r x So dA = - me2'9'0

or equivalently, since 0 = -e3 and r(p) = p - 0,

f S13 dA = S S23 dA = f S33 dA = 0,9'0 9'0 9'0

f P2S33 dA = f (Pl S23 - P2 S13) dA = 0, (3)9'0 9'0

f PlS33 dA = -m.9'0

We do not specify the loading on g" since balance of forces and momentsrequire that (Exercise 1)

(d) the total force on g, vanish and the total moment on gl equal me2;in fact, (3) hold with go replaced by gl.

There are many sets of boundary conditions consistent with (c) and (d).We take the simplest possible and assume that

SI3(P) = S23(P) = 0, S33(P) = "0 + "IPl + "2P2 at P3 = 0, l,

with "0' "1' and "2 constants. Then (1) and (3) imply that

"1 = -mil,where

I = f pi dA9'0


is the moment of inertia of [/0 about the P2-axis. Our boundary conditionstherefore consist of (2) and

at P3 = 0, l. (4)

A stress field compatible with (2) and (4) is obtained by taking

Sl1(P) = S22(P) = S12(P) = S13(P) = S23(P) = 0,

mplS33(P) = - -/-

for all P in the body. This field clearly satisfies

Div S = o.Thus to show that S is actually a solution of our problem we need onlyconstruct the corresponding displacement field.

Using the stress-strain law in the form (31.3), we see that

E 12(p) = E 13(P) = E 23(p) = 0,

and the strain-displacement relation (32.1) is easily integrated to give(Exercise 2)

(5)

mu3(p) = - E/ P1P3 + W3(P)

with w(p) an arbitrary infinitesimal rigid displacement.To compare our results with classical beam theory we "fix" [/0 at P1 =

P2 = °by requiring that

u(o) = Vu(o) = O.

Then w(p) == 0 and the displacement of the centroidal axis (i.e., the P3-axis)takes the form

mp~Ul(O, 0, P3) = 2EI'

U2(0, 0, P3) = U3(0, 0, P3) = 0.

33. BENDING AND TORISON 217

[Actually, to insure that w(p) == 0 it suffices to assume that u(o) = 0 andVu(o) = VU(O)T.]

In this solution the maximum stress occurs at points of If for which IpIIattains its largest value, e say, and this maximal stress has magnitude

mcfl,

Moreover, the maximum deflection ofthe centroidal axis occurs at P3 = I andis given by

ml2

2EI

B. TORSION OF A CIRCULAR CYLINDER

Consider a circular cylinder of length I and radius a (Fig. 3). As before, theaxis of the cylinder coincides with the P3-axis, while f/0 and f/, correspond toP3 = 0 and P3 = I. We assume that the end face f/ 0 is held fixed, while the endface f/, is rigidly rotated about the P3-axis through an angle {3. Thus, assumingthat the lateral surface If is traction-free, our boundary conditions consist of(2) and the requirement that

n(p) = 0 at P3 = 0,

uI(p) = -rxIP2' uip) = rxlpl' U3(P) = 0 at P3 = I,

with rx = {3/1 the angle of twist per unit length. (The boundary condition atP3 = I represents an infinitesimal rigid rotation of f/, about the P3-axis [cf.(7.10) with P3 = I]).

It seems reasonable to expect that under this type of loading the cylinderwill twist uniformly along its length; thus, in view of (7.10), we consider adisplacement field of the form

ul(p) = -rxP2P3,

U2(P) = rxPIP3,

U3(P) = o.

P2

Figure 3 PI

-t......._p,

(6)


(7)

The corresponding stress field, computed using the strain-displacementrelation (32.1) and the stress-strain relation (30.6), is given by

S11(P) = S2ip) = S33(P) = S12(P) = 0,

S23(P) = /JlXPl, S13(P) = -/JlXP2,

and clearly satisfies

Div S = O.

Further, S satisfies the boundary condition (2); indeed (2)1.2 are satisfiedtrivially, and, since o(p) on !fl is proportional to (PI, P2' 0), it follows that

on !fl, (8)

which is (2h. Thus (6) and (7) comprise the solution of our problem.Next, by (1) and (7),

f S13 dA = f S23 dA = f S33 dA = 0,~o ~o ~o

so that

f SodA = 0,~o

and the total force on each end face vanishes. In addition,

(9)

(10)

where]0 = na4/2 is the polar moment ofinertia of the cross section. Therefore

f r x So dA = - me3~o

with

m = /JlX]o'

Further, by balance of moments,

f r x So dA = me3'~I

Thus the cylinder is twisted by equal and opposite couples about the P3-axis.

33. BENDING AND TORSION

EXERCISES

219

1. Establish (d).

2. Derive (5).3. Show that w(p) == 0 in (5) provided u(o) = 0 and Vu(o) = VU(O)T.

In the next two exercises we consider the torsion of an arbitrary cylindricalbar (cf. Fig. 2).

4. Show that (7) is compatible with (8) only if the cross section is circular.

5. Consider the displacement field

u1(p) = -r.xP2P3,

U2(P) = r.xPIP3,

u3(p) = r.xCP(Pl' P2);

cp is called the warpingfunction.

(a) Compute the corresponding stress fieldS and show that Div S = 0 isequivalent to

Acp = O.

Here

etc.

(b) Show that (2) is equivalent to

for all p E !fl, where

(11)

is the normal derivative of cp on !fl.(c) Show that (9) and (10) hold, where

m = Kr.x,

K=f.l f (Pi+P~+PICP'2-P2CP'1)dA;J.9'o

K is called the torsional rigidity of the cross section.


34. LINEAR ELASTODYNAMICS

(1)

In this section we will establish some of the basic results of linear elasto-dynamics. Here the basic equations, as derived previously, are

E = t(V'u + V'uT) ,

S = C[E],

Div S + b = pii

(where we have written p for Po and b for bo).We assume that f!4 is bounded and that p is continuous.Let [u, E, S] be a list offields on f!4 x [0, 00)with u ofclass C 2and E and S

smooth, and suppose that (1) holds with b a given body force field on f!4 x[0, 00).Then [u, E, S] is called an elastic process corresponding to b. Since E istime-dependent, the strain energy

Ol/{E} = t LE' C[E] dV

depends on time. In fact, when C is symmetric,

t(E' C[E]r = teE' C[E] + E • C[E]) = E· C[E] = S . E.

Thus

(Ol/{E})' = LS'EdV, (2)

so that the rate of change of strain energy is equal to the stress power.

Theorem of Power and Energy. Assume that C is symmetric. Let [u, E, S] bean elastic process corresponding to the body force b. Then

where

f Sn·aidA + f b'udV = (Ol/{E} + f{ti})·,oBI BI

(3)

is the kinetic energy.

Proof In view of the lemma on page 205 with ii replaced by 0 and b byb - po,

f Sn- 0 dA + f b·odV = f S . E dV + f pii . 0 dV.oBI BI BI BI

34. LINEAR ELASTODYNAMICS

This relation, (2), and the identity

(f{o}Y = Lpn'o dV

221

imply (3). D

This theorem asserts that the power expended by the external forcesequalsthe rate at which the total energy is changing.

Corollary. Assume that C is symmetric. Let [u, E, SJ be an elastic processcorresponding to body force b = 0, and suppose that

Sn'o=O

Then the total energy is constant:

on iJf!4. (4)

OlI{E} + f{o} = const. (5)

Another interesting corollary may be deduced as follows. Consider anelastic process which starts from an unstrained rest state. Then

E(·,O) = 0, 0(·,0) = 0,

and the total energy is initially zero. Assume that C is positive definite andp > 0, so that the total energy is always nonnegative. Then, ifwe integrate (3)with respect to time from 0 to r, we arrive at

Itr Sn'o dA dt + It rb·o dV dt ~ O.o JiJBI 0 JBI

(6)

The left side of (6) represents the work done by the external forces in the timeinterval [0, 't]; (6) asserts that for an elastic process starting from an un-strained rest state this work is always nonnegative.

The mixed problem of elastodynamics can be stated as follows:

Given: f!4, complementary regular subsets.Z, and !72 of iJf!4, an elasticitytensor C on f!4, a density field p on f!4, a body force field b on f!4 x [0, co),surface displacements uon !7I X [0, co),surface tractions § on !72 X [0, co),an initial displacement field Uo on f!4, an initial velocity field Vo on f!4.

Find: An elastic process [u, E, SJ that corresponds to b, satisfies theinitial conditions

u(p,O) = uo(p), o(p,O) = vo(p)

for every p E f!4, and satisfies the boundary conditions

u = u on !7I X [0, co), Sn = § on !72 X [0, oo).

An elastic process with these properties will be called a solution.


(7)

Uniqueness Theorem. The mixed problem has at most one solution provided Cis symmetric and positive definite and p > O.

Proof Let [u, E, S] denote the difference between two solutions. Then[u, E, S] is an elastic process corresponding to b = 0 and satisfies

u(-, 0) = u(', 0) = 0,

u = 0 on II'1 X [0, (0), Sn = 0 on II'2 X [0, (0).

Thus (4) holds and we conclude from (5) that

O//{E} + Jr'{u} = 0 (8)

for all time. Here we have used the fact that, by (7)1,2' the total energy isinitially zero. But since C is positive definite and p > 0, both the strain energyand kinetic energy are nonnegative; hence (8) implies that Jr' {u} = 0, andthis in turn yields u= 0 on ~ x [0, (0). This fact and (7)1 imply that

u = 0 on ~ x [0, (0),

and the proof is complete. 0

EXERCISES

Throughout these exercises C is symmetric and ~ bounded.

1. Consider an elastic process and define

e = !E· C[E] + !pu2,

q = -SU.

Show thate= - Div q + b· U.

2. (Brun's theorem) Consider an elastic process with null initial data, i.e.,with

u(p, 0) = 0,

for all p E~. Show that

u(p, 0) = 0 (9)

O//{E} - Jr'{u} = cI>,

where

1 itcI>(t) = - [q>(t + 1', t - 1') - q>(t - 1', t + 1')] di,2 0

q>(rx, /3) = r s(p, rx) • u(p, /3) dA p + rb(p, c) • u(p, /3) dVp ,

Jo~ J~with s = Sn.

(10)

35. PROGRESSIVE WAVES 223

(11)

3. Use (8) and (10) to establish uniqueness for the mixed problem when C issymmetric (but not necessarily positive definite) and p > O.

4. Show that under the null initial data (9) the equation of motion

Div S + b = pii

can be written in the form

). * (Div S + b) = pu,

where ).(t) = t and *denotes convolution; i.e.,if A and 'P are functions on14 x [0, (0),

(A * 'P)(p, t) = EA(P, t - r)'P(p, r) dt:

5. (Graffi's reciprocal theorem) Let us agree to use the notation (11) whenA and 'P are vector fields, but here the product in the integrand is the innerproduct A(p, t - r)· 'P(p, r), Show that if [u, E, S]_and [ii, E, S] areelastic processes corresponding to body forces band b, respectively, andto null initial data, then

r s * ii dA + f b * ii dV = r s*u dA + f b* u dV, (12)Jo~ ~ Jo~ ~

where s = Sn and s = Sn.

35. PROGRESSIVE WAVES

Sinusoidal progressive waves form an important class of solutions to theequations of linear e1astodynamics. We will study these waves under theassumptions of homogeneity and isotropy, and in the absence of body forces.The underlying field equation is then the displacement equation of motion

Jl .6u + (A. + Jl)V Div u = pii.

A vector field u of the form

(1)

u(p, t) = a sin(r . m - ct), r = p - 0 (2)

with Im I = 1is called a sinusoidal progressive wavewith amplitude a, directionm, and velocity c. We say that u is longitudinal if a and m are parallel, trans-verse if a and m are perpendicular.

We now determine conditions which are necessary and sufficient that (2)solve (1). By the chain rule,

Vu = (a ® m) cos tp,

224

where

and hence

Thus the wave is

Next,

x. LINEAR ELASTICITY

cp(p, t) = r . m - ct,

Div u = (a . m) cos cp,

Curl u = (m x a) cos tp,

longitudinal ¢;> Curl u = 0,

transverse ¢;> Div u = o.

L1u = -a sin cp,

V Div u = -(a' m)m sin cp,

ii = - c2a sin cp,

and u satisfies (1) if and only if

p,a + (A. + p,)(a . m)m = pc2a.

We call the tensor

1A(m) = - [p,I + (A. + p,)m ® m]

p

the acoustic tensor; with this definition (3) takes the simple form

A(m)a = c2a.

(3)

(4)

Thus a necessary and sufficient condition that u satisfy (1) is that c2 be aneigenvalue and a a corresponding eigenvector of the acoustic tensor A(m).

A simple computation shows that

A(m) = (A.: 2P,)m ® m + ~ (I - m ® m).

But this is simply the spectral decomposition of A(m), and we conclude from(b) of the spectral theorem (page 11) that

and p,jp

are the eigenvalues of A(m),while theline spanned by m and the plane (through0) perpendicular to m are the corresponding characteristic spaces. Thus wehave the following

35. PROGRESSIVE WAVES 225

Theorem. A sinusoidal progressive wave with velocity c will be a solution of(1)if and only if either

(a) c2 = (2J1 + )..)/p and the wave is longitudinal, or(b) c2 = J1/P and the wave is transverse.

This theorem asserts that for a homogeneous and isotropic material onlytwo types of sinusoidal progressive waves are possible: longitudinal andtransverse. The two corresponding speeds

J(2J1 + )..)jp and

are called, respectively, the longitudinal and transverse sound speeds of thematerial. (Note that these speeds are real when the elasticity tensor is positivedefinite.) For an anisotropic body the situation is far more complicated. Apropagation condition of the form (4) still holds (Exercise 1), but the waveswill generally be neither longitudinal nor transverse, and they will generallypropagate with different speeds in different directions.

EXERCISES

1. For an anisotropic but homogeneous body the displacement equation ofmotion has the form

Div C[Vu] = pii (5)

with C and p independent of position. (Here we have assumed zero bodyforces.)

(a) Show that the sinusoidal progressive wave (2) solves (5) if and onlyif the propagation condition (4) holds, where now A(m) is the tensordefined by

1A(m)k = - C[k ® m]m

p

for every vector k.(b) Show that A(m) is positive definite for each direction m ifand only if

C is strongly elliptic (cf. Exercise 29.3).(c) Show that

QA(m)QT = A(Qm)

for any symmetry transformation Q.

226

Fichera [1].Gurtin [1].Love [1].Sokolnikoff [1].Villaggio [1].

X. LINEAR ELASTICITY

SELECTED REFERENCES

(1)

(2)

Appendix

36. THE EXPONENTIAL FUNCTION

The exponential of a tensor can be defined either in terms of its seriesrepresentation 1 or in terms of a solution to an ordinary differential equation.For our purposes the latter is more convenient. Thus let A be a tensor andconsider the initial-value problem

X(t) = AX(t), t > 0,

X(O) = I,

for a tensor function X(t), 0 ~ t < 00. The existence theorem for lineardifferential equations tells us that this problem has exactly one solutionX: [0, 00) ~ Lin, which we write in the form

X(t) = e'".

Proposition. For each t ~ 0, eAt belongs to Lin + and

det(eAt) = e1trA)t.

Proof Let X(t) = e", Since X is continuous and

det X(O) = 1,

1 See, e.g., Hirsch and Smale (I, Chapter 5, §3].

227

(3)

228 APPENDIX

det X(t) > 0 in some nonempty interval [0, r). Let r be as large as possible;that is, let

r = sup{Aldet X(t) > 0 for 0 ~ t ~ A}.

To show that X(t) E Lin + for all t ~ 0 we must show that r = 00. Suppose, tothe contrary, that r is finite. Then, since X is continuous,

det X(r) = O. (4)

We will show that this leads to a contradiction. Since X(t) is invertible for allt E [0, r), (1)1 implies

A = X(t)X(t)-1

on (0, r), and we conclude from (3.14) that

(det X)" = (tr A) det X

on (0, r). This equation with the initial condition (3) has the unique solution

det X(t) = e(trAlt (5)

for 0 ~ t < r. In view of the continuity of X, this result clearly implies thatdet X(r) > 0, which contradicts (4). Thus r = 00 and (5) implies (2). 0

Proposition. Let W be a skew tensor. Then eW t is a rotation for each t ~ O.

Proof Let X(t) = eW t for t ~ O. Then, by definition,

Let

X=WX, X(O) = I.

Then, since W is skew,

Z = XXT + XXT = WXX T + XXTWT = WXXT - XXTW,

and Z satisfies

Z=WZ-ZW, Z(O) = I.

This initial-value problem has the unique solution Z(t) = I for all t ~ O. Thus

X(t)X(t)T = I

and X(t) is orthogonal. But from the preceding proposition X(t) E Lin +. ThusX(t) is a rotation. 0

37. ISOTROPIC FUNCTIONS

EXERCISE

1. Let A E Lin and Vo E "Y. Show that the function

vet) = eA1vo

satisfies the initial-value problem

229

vet) = Av(t),

v(O) = Vo'

t > 0,

SELECTED REFERENCE

Hirsch and Smale [1].


Let ~ c Orth. A set d c Lin is invariant under ~ if QAQTEd wheneverAEdandQE~.

Proposition. Thefollowing sets are invariant under Orth:

Lin, Lin +, Orth, Orth +, Sym, Skw, Psym.

Proof We will give the proof only for Lin +. Choose A E Lin andQ E Orth. Then

det(QAQT) = (det A)(det Q)2 = det A,

since [det QI = 1. Thus A E Lin" implies QAQT E Lin+. 0

Let d c Lin. A scalar function

qJ: d --+ IR

is invariant under ~ if d is invariant under ~ and

qJ(A) = qJ(QAQT)

for every A E d and Q E ~. Similarly, a tensor function

G: d --+ Lin

is invariant under ~ if d is invariant under ~ and

(1)

(2)

for every A E d and Q E ~. An isotropic function is a function invariant underOrth.

230 APPENDIX

(3)

Proposition. Let <1l be a scalar or tensor function with domain in Lin. Then <1lis isotropic if <1l is invariant under Orth +.

Proof The proof follows from the identity

(-Q)A( _Q)T = QAQT

and the fact that for Q E Orth either Q or - Q belongs to Orth+. 0

Theorem(a) det and tr, considered as functions on Lin, are isotropic.(b) The list (2.9) of principal invariants is isotropic; that is,

J A = JQAQT

for every A E Lin and Q E Orth,

Proof That det is isotropic follows from (1). Next, for Q E Orth,

tr(QAQT) = tr(AQTQ) = tr A,

so that tr is isotropic. By (2.8) and (2.9), since A f-+ (tr A)2 is isotropic, toestablish (b) we have only to show that A f-+ tr(A2) is isotropic. ChooseA E Lin and Q E Orth. Then

(QAQT)2 = QAQTQAQT = QA2QT,

and hence

For convenience, we write

for the set of all possible lists J A as A ranges through the set d. Of course,J(d) c lR 3

•

Next weestablish several important representation theorems for functionswith domain .91 in Sym. We assume for the remainder of this section that .91is invariant under Orth.

Representation Theorem for Isotropic Scalar Functions. A function

(.91 c Sym)

is isotropic if and only if there exists a funcionii: J(d) -+ lR such that

(4)

for every A E d.

37. ISOTROPIC FUNCTIONS 231

Proof Assume that qJ is isotropic. To establish the representation (4) itsuffices to show that

whenever

qJ(A) = qJ(B) (5)

.FA = .Fa. (6)

Thus let A, BE Sym and assume that (6) holds. By (6) and the proposition onpage 16,A and B have the same spectrum. Thus by the spectral theorem thereexist orthonormal bases {ei} and {fi} such that

A = L (JJiei <8> e;,i

B = L (JJifi <8> f i ·i

Let Q be the orthogonal tensor carrying the basis {fil into the basis {e.}:

Qfi = ei ·

Then, since

it follows that

QBQT = A.

But since qJ is isotropic, qJ(QBQT) = qJ(B); thus (5) holds and qJ admits therepresentation (4).

The converse assertion, that (4) defines an isotropic function, is a trivialconsequence of (3). D

Transfer Theorem. Let

G: d -+ Lin (d c Sym)

be isotropic. Then every eigenvector of A E d is an eigenvector of G(A).

Proof Let e be an eigenvector of A E .91, and let Q E Orth be the reflectionacross the plane perpendicular to e:

Qe = -e, Qf= f if f· e = O. (7)

Then by the spectral theorem, Q leaves invariant the characteristic spaces ofA; hence we may conclude from the commutation theorem that

QAQT = A.

Thus, since G is isotropic,

232 APPENDIX

(8)

so that Q commutes with G(A). Therefore

QG(A)e = G(A)Qe = - G(A)e,

and Q transforms G(A)e into its negative. But by (7) this can happen only ifG(A)e is parallel to e:

G(A)e = we.

Thus e is an eigenvector of G(A). D

The following lemma, whose statement is based on the spectral theorem,will be extremely useful in what follows.

Wang's Lemma. Let A E Sym.

(a) Consider the spectral decomposition

A = LWiei®ei,i

and assume that the eigenvalues Wi are distinct. Then the set {I, A, A2} is

linearly independent and

lei = 1.

(b) Assume that A has exactly two distinct eigenvalues, so that

A = w1e®e + w2(I - e®e),

Then {I, A} is linearly independent and

sp{I, A} = sp{e ® e, I - e ® e}.

(9)

(10)

(11)

Proof We prove only (a). To establish the linear independence of{I, A, A2} we must show that

exA2 + f3A + yI = 0 (12)

implies

ex = f3 = y = O. (13)

Assume (12) holds. Acting with (12) on the eigenvector e, leads to the relation

(exwf + f3Wi + y)ei = 0,

so that

exwf + f3Wi + Y = 0,

and the Wi are roots of a quadratic equation; since the co, are distinct, this ispossible only if (13) hold. Thus {I, A, A2} is linearly independent.


Next, the subspace

.Ye = sple, ® e l , e2 ® e2 , e3 ® e3 }

of Lin has dimension 3. On the other hand, since

A2 = L ro;e i ® ei ,i

233

(14)

as is clear from (1.2h and (8), we may conclude from (1.2)4 and (8) thatI, A, A2 E .Ye. But 3 linearly independent vectors in a vector space ofdimension3 must necessarily span the space. Thus.Ye = sp{I, A, A2}and the proofof(a)is complete. We leave the remainder of the proof as an exercise. 0

We are now in a position to establish several important representationtheorems for isotropic tensor functions.

First Representation Theorem for Isotropic Tensor Functions. A function

G: .91 ---. Sym (.91 c Sym)

is isotropic if and only if there exist scalar functions ({Jo, ({Jl' ({J2: j(d) ---.!R suchthat

(15)

for every A E d.

Proof Assume first that G admits the representation (15). ChooseA E .91 and Q E Orth. Then by (3),

G(QAQT) = ({Jo(..¢"QAQT)I + ({Jl(..¢"QAQT)QAQT + ({J2(..¢"QAQT)QAQTQAQT

= ({JO(..¢"A)QQT + ({Jl(..¢"A)QAQT + ({J2(..¢"A)QA2QT = QG(A)QT,

so that G is isotropic.To prove the converse assertion assume that G is isotropic. Choose

AEd.

Case 1. A has exactly three distinct eigenvalues. Let (8) be the spectraldecomposition of A. By the transfer theorem

G(A) = L Piei ® e.,i

and we conclude from (9) that there exist scalars IXo(A), IXl (A), and IXiA) suchthat

(16)

234 APPENDIX

Case 2. A has exactly two distinct eigenvalues. Then A admits therepresentation (10), and the characteristic spaces for A are sp{e}and {e}.L. Bythe transfer theorem each of these two subspaces must be contained in acharacteristic space for G(A). This is possible only if either: (i) G(A) has ascharacteristic spaces sp{e} and {e}', or (ii) -r:is the only characteristic spacefor G(A). By (b) and (c) of the spectral theorem,

(17)

in either case [withfi, =F p2incase(i),p1 = p2incase(ii)].Inviewof(1l),(17)can be written in the form (16) with

(18)

Case 3. A has exactly one distinct eigenvalue. Then by (c) of the spectraltheorem in conjunction with the transfer theorem, f is the characteristicspace for A and G(A), so that G(A) = PI, and G(A) again admits the repre-sentation (16) with oco(A) = Pand oct(A) = oc2(A) = O.

Since these three cases are exhaustive, we have shown that G, whenisotropic, admits the representation (16). In view of the representationtheorem for isotropic scalar functions, to complete the proof we have onlyto show that OCo, oc1,and OC2 are isotropic:

(k = 0, 1,2) (19)

for every A E.rd and Q E Orth. Thus choose A E.rd, Q E Orth. Since G isisotropic,

or equivalently,

and, since

it follows that

[oco(A) - oco(QAQT)]I + [oc1(A) - OCt(QAQT)]A

+ [ociA) - oc2(QAQT)]A2 = O. (20)

We consider again the three cases studied previously.

Case 1. By (a) of Wang's lemma, {I, A, A2} is linearly independent, sothat (20) implies (19).


Case 2. By (3) and the proposition on page 16, A and QAQT have thesame spectrum. Thus A and QAQTeach have exactly two distinct eigenvaluesand (18) yields C\:iA) = C\:2(QAQT) = O. Moreover, by (b) of Wang's lemma,{I, A} is linearly independent, and (20) again yields (19).

Case 3. Here A = A.I, so that A = QAQT and (19) holds trivially. D

By the Cayley-Hamilton theorem (2.12),

A2 = 11(A)A - liA)I + 13(A)A- l,

provided, of course, A is invertible. Thus, by (2.9), the previous theorem hasthe following corollary:

Second Representation Theorem for Isotropic Tensor Functions. Let ,?ibea subset ofthe set ofall invertible, symmetric tensors. Then

G: d -+Sym

is isotropic ifand only if there exist scalar functions Po, Pi'P2: J(d) -+ IR suchthat

(21)

for every A E d.

For linear functions there is a far simpler result.

Representation Theorem for Isotropic Linear Tensor Functions. A linearfunction

G: Sym -+ Sym

is isotropic if and only if there exist scalars f:l and Asuch that

G(A) = 2f:lA + A.(tr A)I

for every A E Sym.

(22)

Proof Let % be the set of all unit vectors. For e E % the tensor e ® ehas spectrum {O, 0, I} and characteristic spaces sp{e} and {e}1. • Thus the sameargument used to arrive at (17) now leads to the conclusion that there existfunctions f:l, A: % -+ IR such that

G(e ® e) = 2f:l(e)e ® e + A(e)I (23)

236 APPENDIX

for every e E.iV. Choose e, f E.iV, and let Q be any orthogonal tensor such thatQe = f. (Clearly, at least one such Q exists.) Since

Q(e ® e)QT = f® f,

and since G is isotropic,

o = QG(e ® e)QT - G(f ® f) = 2[/l(e) - tl(f)]f ® f + [A(e) - A(f)]I.

But {I, f ® f} is linearly independent; thus

/l(e) = J.l(f), A(e) = A(f).

Therefore /l and Amust be scalar constants, and we conclude from (23) that

G(e ® e) = 2/le ® e + AI. (24)

Next, choose A E Sym arbitrarily. By the spectral theorem A admits therepresentation

A = L OJi e, ® e,i

with {e.} orthonormal; therefore, in view of (24) and the linearity of G,

G(A) = L OJiG(ei ® e.) = 2/lA + A.(OJ l + OJ2 + OJ 3)1. (25)i

Since

tr A = OJ1 + OJ2 + OJ3,

(25) implies the desired result (22). The converse assertion, that (22) deliversan isotropic function, is left as an exercise. D

Corollary. Let

Sym, = {Ae Symltr A = O},

and let

G: Sym., -+ Sym

be linear. Then G is isotropic if and only if there exists a scalar J1 such that

(26)

for every A E Symo'

Proof Clearly (26) defines an isotropic function. To prove the converseassertion assume that G is isotropic. For A E Sym, A - j{tr A)I belongs toSymo. Thus we can extend G from Sym., to Sym by defining

C(A) = G(A - j{tr A)I)

for every A E Sym. Since G is isotropic, for every A E Sym and Q E Orth,

QG(A)QT = G(QAQT - !<tr A)I)= G(QAQT - t tr(QAQT)I)= G(QAQT),

where we have used the fact that tr is isotropic. Thus G is isotropic, and weconclude from (22) and the fact that G and G coincide on Sym., that

G(A) = G(A) = 2JlA + A(tr A)I = 2JlA

for every A E Syrnj , 0

Our next step is to discuss the invariance properties of the derivative of atensor function. Thus let d be an open subset of a subspace 0/1 of Lin, andlet G: d -+ Lin be smooth.

Theorem (Invariance of the derivative). Let G be invariant under I'§. Then

(27)

for every A E d, U E 0/1, and Q E I'§.

Proof For (27) to make sense we must show that both d and 0/1 areinvariant under I'§. Since d is the domain of G and G is invariant, d is (bydefinition) invariant under I'§. To see that 0/1 has this invariance, chooseU E 0/1, A E d, and Q E I'§. Then for all sufficiently small a, (A + !XU) E d, sothat Q(A + !XU)QT E d and hence (QAQT + CXQUQT) E d c 0/1. Thus,since 0/1 is a subspace, QUQT E 0/1, and 0/1 is invariant under I'§.

We have only to show that (27) is satisfied. For A E d, U E 0/1, and Q E I'§,

G(Q(A + U)QT) = G(QAQT + QUQT)

= G(QAQT) + DG(QAQT)[QUQT] + o(U)

as U -+ O. But since G is invariant under I'§,

G(Q(A + U)QT) = QG(A + U)QT= QG(A)QT + QDG(A)[U]QT + o(U),= G(QAQT) + QDG(A)[U]QT + o(U)

as U -+ O. Thus by the uniqueness of the derivative,

which is (27). 0

238 APPENDIX

EXERCISES

1. Show that the sets Orth, Orth + , Sym, Skw, and Psym are invariant underOrth.

2. A scalar function qJ: "Y - ~ is isotropic if

qJ(v) = qJ(Qv)

for all v E "Y and Q E Orth. Show that qJ is isotropic if and only if thereexists a function tp: [0, (0) - ~ such that

qJ(v) = tp(lvl)

for all v E "Y.

3. A vector function q: "Y - "Yis isotropic if

Qq(v) = q(Qv)

for all v E "Y and Q E Orth. Show that q is isotropic if and only if thereexists a function qJ: [0, (0) - ~ such that

q(v) = qJ( IvJ)v

for all v E "Y.

4. Let G: Lin - Lin be defined by

G(A) = A"

Show that G is isotropic.

5. Show that the mapping

(n ~ 1 an integer).

from Lin + into Lin + is isotropic.

6. Show that the function G: Sym - Sym defined by (22) is isotropic.

7. Establish (b) of Wang's lemma.

SELECTED REFERENCES

Gurtin [2].Martins and Podio-Guidugli [2].Serrin [1, §59].Truesdell and Noll (1, §§10-13].

38. GENERAL SCHEME OF NOTATION 239

38. GENERAL SCHEME OF NOTATION

INDEX OF FREQUENTLly USED SYMBOLS

Place ofdefinitionSymbol Name or first occurrence

a angular momentum 92.i4 body 41fJI, region occupied by body at time t 58b body force 98bo reference body force 179B left Cauchy-Green strain tensor 46C right Cauchy-Green strain tensor 46c curve 34C elasticity tensor 194curl curl, spatial curl 32,61div divergence, spatial divergence 30,61Div material divergence 61D derivative 21D stretching 71det determinant 6E Young's modulus 203E infinitesimal strain 558 euclidean point space If deformation 42F deformation gradient 42g complex velocity 123C§ symmetry group 168H displacement gradient 1995 s list of principal invariants of a tensor S 15I identity tensor 3%{Ii} kinetic energy 220L velocity gradient 63

~ linear momentum 92Lin set of all tensors 7Lin" set of all tensors S with det S > 0 7m(&') mass of &' 87m subscript indicating material description, also mach

number 60,133n outward unit normal to boundary 37o(u) symbol for "small order u" 190 origin 2Orth set of all orthogonal tensors 7Orth " set of all rotations (proper orthogonal tensors) 7p material point 41p reference map 59&' part of :J4 41Psym set of all symmetric, positive definite tensors 7Q orthogonal tensor 7R rotation tensor 46

position vector 92

240 APPENDIX

Place ofdefinitionSymhol Name or first occurrence

IR reals IIR+ strictly-positive reals I

S Piola-Kirchhoff stress 178s stress vector 97Sym set of all symmetric tensors 7Skw set of all skew tensors 7,; subscript indicating spatial description, also

kinematically admissible state 60,208sp span 2tr trace 5t time 58T Cauchy stress 101or trajectory 59u displacement 42'f/{E} strain energy 205U right stretch tensor 46V left stretch tensor 46v spatial description of the velocity 60v speed 133r vector space associated with r! Ivol volume 37W spin 71w complex potential 126x place 58x motion 58

or. center of mass 92p potential of body force IIIcp potential of flow III{,; } potential energy 208liS) principal invariants of a tensor S 15K(p) sound speed 131Ak principal stretches 45). Lame modulus 196tt pressure 106Po reference density 88P density in motion 888(F) strain-energy density 186Jl. viscosity, also shear modulus 149,196v kinematic viscosity, also Poisson's ratio 151.203ro angular velocity 70V gradient, material gradient 29,60~ laplacian 32Q9 tensor product 4x cross product 7

ep material time derivative of cp 61cp' spatial time derivative of cp 61

Expression

~agp~gp U ,Fgp n .17gpc:FXEgp

j: gp .......'1"

x ....... j(x)

jog

{xIR(x) holds}

38. GENERAL SCHEME OF NOTATION

GENERAL NOTATION

Meaning

interior of gpboundary of gpclosure of gpunion of gp and .Fintersection of gp and .Fgp is a subset of .Fx is an element of the set gpj maps the set gp into the set .'1"; gp is the domain, :J7 the

codomainthe mapping that carries x into j(x); e.g.,x ....... Xl is the

mapping that carries every real number x into itssquare

composition of the mappingsj andg; that is,(f 0 g) (x)= j(g(x»

the set of all x such that R(x) holds; e.g.,{x10 ~ x ~ I}is the interval [0, 1]

241

References

R. G. Bartle[I] The Elements of Real Analysis, 2nd ed. New York: Wiley (1976).

R. B. Bird, W. E. Stewart, and E. N. Lightfoot[I] Transport Phenomena. New York: Wiley (1960).

R. M. Bowen and C. C. Wang[1] Introduction to Vectors and Tensors, Vol. 1. New York: Plenum (1976).

H. W. Brinkmann and E. A. Klotz[I] Linear Alqebra and Analytic Geometry. Reading: Addison-Wesley (1971).

P. Chadwick[1] Continuum Mechanics. New York: Wiley (1976).

B. D. Coleman, H. Markovitz, and W. Noll[I] Viscometric flows of non-Newtonian fluids. Springer Tracts in Natural Philosophy 5.

New York: Springer-Verlag (1966).B. D. Coleman and W. Noll

[I] Material symmetry and thermostatic inequalities in finite elastic deformations. Arch.Rational Mech. Anal. 15,87-111 (1964).

R. Courant and K. O. Friedrichs[1] Supersonic Flow and Shock Waves. New York: Wiley (Interscience) (1948).

J. Dieudonne[1] Foundations of Modern Analysis. New York: Academic Press (1960).

A. C. Eringen[I] Nonlinear Theory of Continuous Media. New York: McGraw-Hili (1962).

G. Fichera[1] Existence theorems in elasticity. Handbuch der Physik 692' Berlin: Springer-Verlag

(1972).

243

244 REFERENCES

W. Fleming[I] Functions of Several Variables. Reading: Addison-Wesley (1965).

P. Germain[I] Cours de Mecanique des Milieux Continus, Vol. 1. Paris: Masson (1973).

A. E. Green and W. Zerna[I] Theoretical Continuum Mechanics. London and New York: Oxford Univ. Press (1954).

M. E. Gurtin[I] The linear theory ofelasticity. Handbuch der Physik 6a2 . Berlin: Springer-Verlag (1972).[2] A short proof of the representation theorem for isotropic, linear stress-strain relations.

J. Elasticity 4, 243-245 (1974).M. E. Gurtin and L. C. Martins

[I] Cauchy's theorem in classical physics. Arch. Rational Mech. Anal. 60,305-324 (1976).M. E. Gurtin and W. O. Williams

[I] An axiomatic foundation for continuum thermodynamics. Arch. Rational Mech. Anal.26, 83-117 (1967).

P. R. Halmos[I] Finite-Dimensional Vector Spaces, 2nd ed. New York: Van Nostrand-Reinhold (1958).

M. W. Hirsch and S. Smale[I] Differential Equations, Dynamical Systems and Linear Algebra. New York: Academic

Press (1974).T. J. R. Hughes and J. E. Marsden

[I] A Short Course in Fluid Mechanics. Boston: Publish or Perish (1976).

O. D. Kellogg[I] Foundations of Potential Theory. Berlin and New York: Springer-Verlag (1929). Re-

printed New York: Dover (1953).O. A. Ladyzhenskaya

[I] The Mathematical Theory of Viscous Incompressible Flow. New York: Gordon &Breach (1963).

H. Lamb[I] Hydrodynamics, 6th ed. London and New York: Cambridge Univ. Press (1932). Re-

printed New York: Dover (1945).A. E. H. Love

[I] A Treatise on the Mathematical Theory of Elasticity, 4th ed. London and New York:Cambridge Univ, Press (1927). Reprinted New York: Dover (1944).

L. C. Martins and P. Podio-Guidugli[I] A variational approach to the polar decomposition theorem. Rend. Accad. Naz. Lincei

(Classe Sci. Fis., Mat. Natur.) 66(6), 487-493 (1979).[2] A new proof of the representation theorem for isotropic, linear constitutive relations.

J. Elasticity 8, 319-322 (1978).R. E. Meyer

[I] Introduction to Mathematical Fluid Dynamics. New York: Wiley (lnterscience) (1971).I. P. Natanson

[I] Theory of Functions ofa Real Variable. New York: Ungar (1955).W. Noll

[I] On the continuity of the solid and fluid states. J. Rational Mech. Anal. 4,3-81 (1955).[2] A mathematical theory of the mechanical behavior of continuous media. Arch. Rational

Mech. Anal. 2, 197-226 (1958). Reprinted in W. Noll, The Foundations of Mechanicsand Thermodynamics. Berlin: Springer-Verlag (1974).

[3] The foundations of classical mechanics in the light of recent advances in continuummechanics. The Axiomatic Method, with Special Reference to Geometry and Physics.

REFERENCES 245

Amsterdam: North-Holland Pub!. (1959). Reprinted in W. Noll, The Foundations ofMechanics and Thermodynamics. Berlin: Springer-Verlag (1974).

[4] La rnecanique classique, basee sur un axiome d'objectivite. La Methode Axiomatiquedans les Mecaniques Classiques el Nouvelles. Paris: Gauthier-Villars (1963). Reprintedin W. Noll, The Foundations ofMechanics and Thermodynamics. Berlin: Springer-Verlag(1974).

H. K. Nickerson, D. C. Spencer, and N. E. Steenrod[I] Advanced Calculus. New York: V<'11 Nostrand-Reinhold (1959).

J. Serrin[I] Mathematical principles of classical fluid mechanics. Handbuch der Physik 8

"Berlin:

Springer-Verlag (1959).I. S. Sokolnikoff

[I] Mathematical Theory ofElasticity, 2nd ed. New York: McGraw-Hill (1956).R. Stephenson

[I] On the uniqueness of the square-root ofa symmetric positive-definite tensor. J. Elasticity10,213-214 (1980).

E. Sternberg and J. K. Knowles[I] On the existence of an elastic potential for a simple material without memory. Arch.

Rational Mech. Anal. 70, 19-30 (1979).F. M. Stewart

[I] Introduction 10 Linear Algebra. New York: Van Nostrand-Reinhold (1963).R. Temam

[I] Naoier-Stokes Equations, Theory and Numerical Analysis. Amsterdam: North-HollandPub!. (1977).

C. Truesdell[I] A First Course in Rational Continuum Mechanics, Vo!' I. New York: Academic Press

(1977).[2] Rational mechanics of deformation and flow (Bingham Medal Address). Proc, 4th lntl.

Conqr, Rheol., 1963, 2, 3-50 (1965).C. Truesdell and W. Noll

[I] The non-linear field theories of mechanics. Handbuch der Physik 33 , Berlin: Springer-Verlag (1965).

C. Truesdell and R. A. Toupin[I] The classical field theories. Handbuch der Physik 3,. Berlin: Springer-Verlag (1960).

P. Villaggio[I] Qualitatiie Methods in Elasticity. Leyden: Noordhoff(l977).

C. C. Wang and C. Truesdell[I] Introduction 10 Rational Elasticity. Leyden: Noordhoff(1973).

Hints for Selected Exercises

SECTION 1

1. Consider o/ev) with v = L Viejo

2. Take 8 = L o/(e;)ei'

3. Show that F = S + T is linear:

F(u + v) = Fu + Fv,

F(ocu) = ocFu

for all vectors U, v and scalars oc.

4. Fix v and consider Su . v as a function of u. Show that this function islinear and hence can be written as the inner product of a vector a,(which depends on v) and u:

Show that a, is a linear function of v and hence can be written asa, = Av with A a tensor. Define ST = A.

6. Apply each side of the identity to an arbitrary vector v.

247

248 HINTS FOR SELECTED EXERCISES

7. To verify (1)1 prove that

u . (S + T)v = (ST + TT)U . v,

which establishes ST + TT as the transpose of S + T. To verify (2)zapply each side of the identity to an arbitrary vector v.

14. Use the identity

to show that

SECTION 2

2. For part (a) use (b) of the first proposition. To derive (2) and (3) use(1) and (1.2)3'

3. Show that QDQT E Sym and f D = fQDQT.

4b. Prove that each eigenvalue of P is either °or 1 and then (with the aidof the spectral theorem) consider separately the following fourpossibilities for the spectrum of P: (0,0,0), (1, 1, 1), (0, 1, 1), (0, 0, 1).

5. (only if) Let S commute with every WE Skw. Choose a vector wandlet W be the skew tensor whose axial vector is w. Show that

W(Sw) = SWw = 0,

so that Sw belongs to the axis of W and is hence parallel to w. Thus

Sw = aw.

(Of course, a may depend on the choice of w.) For S E Sym use thespectral theorem and the fact that w is arbitrary to establish (14). ForS E Skw,

°= w . Sw = aw2=> a = °

and (14) holds trivially. We therefore have the desired result forS E Sym and S E Skw. Complete the proof by showing that if Scommutes with every WE Skw, then so also do the symmetric andskew parts of S.

7. Choose Q E Orth +, Q #- R. Use the identity

IF - Q 12 = IF 1

2- 2F • Q + 3

to show that

SECTION 5 249

1F - Q 12

- IF - R 12 = 2U . (I - Qo),

where F = RU is the right polar decomposition of F and

Qo = RTQ =F I.

Next, establish the identity

2U • (I - Qo) = U· (Qo - I)(Qo - I)T = tr{(Qo - I)TU(Qo - I)},

and show that U •(I - Qo) > 0 by showing that (Qo - I?U(Qo - I) E

Psym.

SECTION 3

2. Differentiate AA- 1 = I.

3. Use the chain rule.

6. Differentiate QQT = I.

7. Take Q(t) = eW t•

8c. Show that

Dcp(A) [U] = U· f {S(Ao + IXA) + IX DS(Ao + IXA) [A]} da,

and show further that the integrand is equal to

ddlX [IXS(Ao + IXA)].

SECTION 4

2. Show that div(STa) is linear in a and use the representation theoremfor linear forms.

SECTION 5

la. Use the fact that, given any vector a,

(1 v ® n dA)a = 1(v ® n)a dA = 1(v ® a)n dA.o~ o~ o~


SECTION 6

6. Use (2.7) and (2.10) to show that .fc = (3, 3, 1) if and only if thespectrum of Cis (1, 1, 1).

9. Let

and

be right polar decompositions of F 1 = Vf1 and F 2 = Vf2 , and notethat U 1(p) = U 2(p) for all P E .sf. Define g = f2 0 fi 1.Choose q E f1(.sf)arbitrarily and let P = fi 1(q). Show that

Vg(q) = R2(p)R1(p?

and appeal to the characterization theorem for rigid deformations.

10. To establish (18)1 let v = ~>iei' write

r v . m dA = L: ei' r Vim dA,Jol(go) i Jol(go)

and use (14.2). For (18)z let a be a vector, write

a' f Tm dA = r (TTa)· m dA,Jol(go) Jol(go)

and use (18)1' For (18h use the identity

a . (r x Tm) = (Tm) . (a x r) = m . TT(a x r)

to write

[r(x) = x - 0]

a' r r x Tm dA = r m . TT(a x r) dAJol(go) JiJf(go)

and appeal to (18)1 again.

12. To verify (5)1 use the continuity of f and f- 1 together with the factthat for a continuous function the inverse image of an open set is open.To prove (5)2 note that, since fJI is closed,

f(fJI) = f(~) u f(J.sf),

and since f(fJI) is closed,

f(fJI) = f(fJlt u Jf(.sf).

Use these identities with (5)1 to verify (5)z.

13b. Show that f(fJI) is not closed.

SECTION 8

SECTION 7

I. Consider C and U as functions of H = Vu:

251

C = t(H), U = O(H).

Use the chain rule and the identity t = 0 2 to conclude that

Dt(O) [S] = 2 DO(O) [S]

for all tensors S, and then conclude from (5)1 that

O(H) = I + E + o(H).

5. Use Exercise 4 in conjunction with Exercise 4.9b.

6. Use the trigonometric identities

cos(E> + {3) = cos E> cos {3 - sin E> sin {3,

sin(E> + {3) = sin E> cos {3 + cos E> sin {3,

together with the estimates

cos {3 = 1 + 0({3),

sin {3 = {3 + 0({3).

[It is a simple matter to verify that Vu -+ 0 as a -+ 0 without explicitlycomputing the gradient. Indeed, one simply notes that u, as a functionof a and p, is smooth for (a, p) E ~ X 8; one then obtains the abovelimit by first evaluating u at a = 0 and then taking its gradient.]

SECTION 8

3. Compute div v using (5) and Exercise 4.9a.

7a. Choose PEg and let c(o), 0::;; 0" ::;; 1, be a smooth curve in g withc(O) = p. Use the relation

ddO" cp(c(O"» = 0

to show that Vcp(p) is normal to c(O).


SECTION 9

Id, Show that the two terms in the right side of (15) are symmetric andskew, respectively, and use the uniqueness of the expansion L =D + W of L into symmetric and skew parts.

3. Show that

IxiY, t) - xiz, t)1 = Iy - z],

so that xl, t) is a rigid deformation.

4c. By (19),

(n + I) T (!!1 • T •C = F An + 1F = (C) = (F An F) .

SECTION 10

2. Use (4.2)4 and (9.11).

SECTION 11

1. Use (9.12), (10.2), and (2h.

2. Take the curl of (9.lOh and use the identity

curl(w x v) = (grad w)v - (grad v)w + (div v)w - (div w)v

and (10.2).

3. Fix Yand r, and let

g(t) = v(xt(y, t), t)w(xiY, t), r).

It suffices to show that

g(t) = Fly, t)g('t).

By (9), g satisfies a certain ordinary differential equation. Show that

f(t) = Fly, t)g('t)

satisfies the same differential equation and f('t) = g('t). Appeal to theuniqueness theorem for ordinary differential equations.

5. Use (9.10).

3. Show that

SECTION 13

SECTION 12

253

p(y, r) = p(x, t) det Fly, t)

and then proceed as in the proof of (7).

4. Let f = g - v and use the divergence theorem to prove that :f({g} =f{v} + f{f}.

SECTION 13

3a. Use (9) and (12.7) to convert (4) to an integral over PAo.

3b. Use (10) to eliminate q(t) from (9) and derive

xo(y, t) = a(t) + Q(t)[y - a(O)].

By (9.13),

v(x o(y, t), t) = xo(y, t) = !i(t) + Q(t)[y - a(O)].

Choose x E PAt arbitrarily, and let y be such that x = xo(y, t), i.e.,

y - a(O) = Q(t)T[X - a(t)].

Then

v(x, t) = !i(t) + Q(t)Q(t?[x - a(t)].

Compare the gradients (with respect to x) of the above equation and(9.7).

3c. (If) Define g(t) = Q(t)k(O) and show that g and k satisfy the samedifferential equation and g(O) = k(O). Appeal to the uniquenesstheorem for ordinary differential equations.

3e. To establish (15) transform (14) to an integral over PAo using the samesteps as in (3a). Then use (12) (with k = e) and (15) to show that

ei(t) • J(t)eit) = ej(O) • J(O)ej(O).

3f. Choose {ei(O)} to be an orthonormal basis of eigenvectors. Since

a.pin(t) = L Jijwit)ej(t),i.]

where Jij are the components of J(t) relative to {ej(t)}, we may con-clude, with the aid of (13), that

a.pin = L [JijWjei + JijWj(co x ej)].i.j

254

4a. Show that

HINTS FOR SELECTED EXERCISES

i v,,-pdV = 0iJIJ,

and then compute .Y( with v = v"- + lX.

4b. Use (11).

SECTION 14

8. Prove that it suffices to find a tensor Q such that

G(Q) = r r ® Qs(n) dA + i r ® Qb dVJo@ , iJIJ,

is symmetric. Show that G(Q) = G(I)QT, take the left polar de-composition of G(I), and choose QT to make G(I)QT symmetric. Hereyou will need the following extended version of the polar decom-position theorem 1: given F E Lin there exist unique positive semi-definite, symmetric tensors D, V and a (not necessarily unique)orthogonal tensor R such that

F = RD = VR.

9. We must show that

n· T(x, t)k = 0,

whenever 0 is normal to-and k tangent to-ofJlt at x. Use the sym-metry of T and the fact that ofJlt is traction-free.

SECTION 15

la. Use Exercise 5.lb.

lb. Use Exercise 5.la.

2. Use components.

I cr., e.g., Halmos [I, §83].

1. Show that

SECTION 22

SECTION 16

q) = {D E Syml tr D = O}.

255

Given any tensor N E Sym, write

N = -1tI + No,

where 1t = - -! tr N, so that tr No = O. Show that N . D = 0 for everyDE q) if and only if No = O.

SECTION 17

2. Use the divergence theorem to express the term involving b in (15.2)as an integral over iJtJ4t •

3. Compute grad v.4. Use Bernoulli's theorem.

SECTION 20

2a. Use Exercise 2.5.

SECTION 21

2. Use (20.1), (6.18)2' and (1) to show that

f T*n* dA = Q(t) f Tn dA.iJ&t iJ&,

Apply the divergence theorem to both sides and then use (6.14)1 toconvert the integral over fY'~ to an integral over fY'.

SECTION 22

2a. Use Exercises 4.9b and 7.4.

2b. Use coordinates with, say, e3 = n together with the relation div v = 0to show that (grad V)Tn = O.


SECTION 25

4c. For HE Yep with det H # 1 take F, = H" and Gn = H- n (or viceversa).

4e. Choose FE Lin + arbitrarily and take H = (det F)1/3F- 1 to reduceT to a function of the density. Show next that T is isotropic and usethis isotropy to reduce the stress to a pressure (cf. the corollary onpage 13).

5b. Choose H E Yep. By (16) and (19),

Tg(FG- 1) = T(F) = T(FH) = Tg(FHG- 1)

for every F E Lin +, so that

fg(F) = Tg(FGHG- 1)

for every F E Lin ". Thus H E Yep implies GHG- 1E Yep(g) andGYepG - 1 C Yep(g). Similarly, Yep(g) c GYepG -I.

SC. Use (17) and (20).

5d. Let G = RU be a right polar decomposition of G. Then

Yep(g) = RUYepU-1R- 1•

Choose Q E Yep. Then

Q = RUQU- 1R- 1

belongs to Yep(g). Thus, since Q, QE Orth + (why?),

(QR)U = (RQ)(QTUQ) (0()

represent polar decompositions ofthe same tensor; hence Q = RQRT.

Thus Q E Yep implies RQRTE Yep(g), and RYepRT c Yep(g). Similarly,Yep(g) c RYepRT. Equation (0() also implies that U = QTUQ. Whenp is isotropic this result and the corollary on page 13 tell us thatU = AI, so that G = AR.

6a. Without loss in generality add the constraint

tr T(F) = 0

for all F E Unim (cf. the remarks on page 148).

6c. Definef(F) = T«det F) - 1/3F) ({3)

for all FE Lin +. Then f(F) = T(F) for all FE Unim, so that fprovides an extension of T to Lin +. Show that f(F) must reduce toan isotropic function of B = FFT and use (37.21).

SECTION 29

SECTION 27

6. Transform

257

f (T + 1toI)m dA.(9',t)

to an integral over!/' ( c !/'2) and use the fact that!/' is arbitrary.

SECTION 28

1a. Integrate u(R)o along a path from R = I to R = Q.

lb. Substitute (5) into (27.11) and use the chain rule to verify that

DF[u(QF) - u(F)] = O.

Integrate this from F = I [using (19)].

Ie, Use the chain rule to differentiate

u(F) = u(FTF)

with respect to F and appeal to (5) and (27.14).

2. Use the incompressibility of the body to show that

(nI)· D = 0,

so that the term involving the pressure 11: does not enter the expressionfor the work. Then use the extension (p) to reduce the problem to theone studied in this section.

3. Show that the work done on a part {!IJ during [to, t 1] is given by

It

1

fT. D dV dtto iP,

provided the process is closed during this time interval.

4. Use components.

SECTION 29

1. Extend the proof of (b) of the theorem on page 195.


SECTION 30

4. Use (29.7) with F = I.

SECTION 32

2. Use Exercise 15.1.

9. The following theorem is needed: if w, are class C" (N ~ 1) fields on[]It with

then there exists a class C" + 1 field t/J on []It such that

The compatibility equation in the form

(E ll , 2 - E 12 , l ) ' 2 = (E 12 , 2 - E 22 , l ) ' l

establishes the existence of a field t/J such that

E ll , 2 = (E 12 + t/J),l,

and hence fields Ua such that

Ell = U 1,l'

E2 2 = U2,2'

E 12 + t/J = U 1,2'

E 12 - t/J = U2,l'

11. Equation (13h, written in the form

implies the existence of fields t/J and () such that

and there exists a qJ such that

SECTION 34

SECTION 33

259

4. Assume that the boundary of the cross section is parametrized byPI(a) and pz«(1) with (1 the arc lerigth. Then the vector with components

is normal to this boundary at each (1. Use this fact, (7), and (8) to showthat

5c. Show that

and use the divergence theorem on g>0 (as a region in [RZ) and (11) toverify that

f S13dA = O.Yo

Similarly,

f SZ3dA = O.Yo

SECTION 34

2. Show that

L{{It(t - T)' C[E(t + T)J

- E(t + T) . C[E(t - T)]} dt dV = 2'¥t{E}(t),

L{P{Ii(t - T)' ii(t + T) - o(t + T) • o(t - r)} dt dV = -2.Jt'"{u}(t)

(where we have suppressed the argument p), and that the left sides sumtoct>(t). To derive these identities write the integrands on the left sidesas derivatives with respect to r.


5. Show that the convolution of the left side of (12) with A(t) = t is equalto

A* LS * EdV + LpU * ii dV,

where the convolution of two tensor fields is defined using the innerproduct of tensors in the integrand in (11). Show further, using thecommutativity of the convolution operation and the symmetry of C,that

s*E=s*E.In view of the above remarks we have the identity obtained by takingthe convolution of (12) with I, and ifwe differentiate this relation twicewith respect to time, we arrive at (12).

SECTION 37

2. Choose U and v with IU I = Iv I, let Q be the orthogonal tensor carryingu into v, and show that qJ(u) = qJ(v).

3. Choose v and let Q be any rotation about v. Show that for this choiceof Q, Qq(v) = q(v), and use this fact to prove that q(v) is parallel to v.Thus q(v) = qJ(v)v. Show that qJ is isotropic and use Exercise 1.

4. Use induction.

A

Index

C

Acceleration, 60Acoustic equations, 133Acoustic tensor, 224Airy stress function, 214Angle of twist, 58Angular velocity, 70Anisotropic material point, 169Axial vector, 9

B

Balance of energyfor hypere1astic material, 191for linear elastic material, 220for viscous fluid, 153

Bending of bar, 214-217Bernoulli's theorem, 111

for elastic fluid, 131for ideal fluid, 120

Betti's reciprocal theorem, 206Blasius-Kutta-Joukowski theorem, 124Body, 41Body force, 99Boundary, 35Boussinesq-Papkovich-Neuber solution,

201

Cartesian coordinate frame, 2Cauchy stress, see Stress, CauchyCauchy's theorem, 101Cayley-Hamilton theorem, 16Center of mass, 92Chain rule, 26Characteristic space, 11Circulation, 82Closed region, 350',21Commutation of tensors, 3Commutation theorem, 12Compatibility equation, 213Complex potential, 126Complex velocity, 123Connected set, 35Conservation of mass, 88

for control volume, 90local, 89

Conservative body force, 111Constitutive assumption, 115-119Control volume, 89Cross product, 7Curl, 32

material, 61spatial, 61

261

262 INDEX

Curve, 34c1osed,34length of, 34

D

Da Silva's theorem, 107Deformation, 42Deformation gradient, 42Density, 88Derivative, 20-21

invariance of, 237Determinant, 6

derivative of, 23Differentiable, 21Displacement, 42Displacement equation of equilibrium, 201Displacement equation of motion, 201Displacement problem of elastostatics, 207Divergence

material, 61spatial, 61of tensor field, 30of vector field, 30

Divergence theorem, 37Dynamical process, 108

E

Eigenvalue, 11Eigenvector, 11Elastic body, 165

incompressible, 174Elastic fluid, 117, 130-137Elasticity tensor, 194-197Elastic process, 220Elastic state, 205Equation of motion, 101Euclidean point space, 1Eulerian process, 116Euler's equations for ideal fluids, 121Euler tensor, 95Exponential function, 227-229Extended symmetry group, 172Extension, 44Extra stress, 148

F

Finite elasticity, 165-198Flow, 108Flow region, 109

G

Gradientmaterial, 60of point field, 30of scalar field, 29spatial, 61of vector field, 30

Graffi's reciprocal theorem, 223

H

Harmonic field, 32Hellinger-Prange-Reissner principle, 212Homogeneous body, 171Homogeneous deformation, 42Hu-Washizu principle, 212Hyperelastic body, 186

I

Ideal fluid, 117, 119-130Incompressible body, 116Inertia tensor, 94Infinitesimal rigid displacement, 55

characterization of, 56Infinitesimal strain, 55Inner product

of tensors, 5of vectors, 1

Integral of vector field around curve, 34Interior of set, 35Invariance of function, 229Invariance of set, 229Invariance under change in observer, 143-

145for elastic body, 166for Newtonian fluid, 149

Inverse of tensor, 6Irrotational motion, 81Isochoric deformation, 51Isochoric dynamical process, 116Isochoric motion, 78Isotropic function

definition, 229representation theorem for linear tensor

functions, 235, 236representation theorem for scalar func-

tions,230representation theorem for tensor func-

tions, 233, 235Isotropic material point, 169

K

Kelvin's theorem, 83Kinematically admissible state, 208Kinematical viscosity, 151Kinetic energy, lIl, 220Konig's theorem, 95Kom's inequality, 57

INDEX

N

Navier-Stokes equations, 151Newtonian fluid, 148Normal force, 105

o

263

L

Lagrange-Cauchy theorem, 81Lame moduli, 196Laplacian, 32Left Cauchy-Green strain tensor, 46Left stretch tensor, 46Lin, 7Lint, 7Linear elasticity, 199-226Linear form, representation theorem, 1Localization theorem, 38Longitudinal sound speed, 225

M

Mach number, 133Mass, 87Mass distribution, 87Material body, 116Material curve, 82Material description, 60Material field, 60Material point, 41Material time derivative, 60, 61Mean strain, 57Minimum complementary energy, principle

of,211Minimum potential energy, principle of,

193,208Mixed problem of elastodynamics,

221Mixed problem of elastostatics, 207Modulus of compression, 202Momentum

angular, 92balance for control volume, 108balance of, 99linear, 92spin angular, 93

Motion, 58Motion relative to time T, 75Multiplicity of eigenvalue, 11

Observer change, 139Open region, 35Origin, 2Orth,7Orth", 7Orthogonal tensor, 7

p

Part of body, 41Path line, 63Piola-Kirchhoff stress, see Stress, Piola-

KirchhoffPlace, 58Plane motion, 74poincare inequality ,I 163, 164Point, 1Poisson's ratio, 203Polar decomposition of tensor, 14Polar decomposition theorem, 14Position vector, 62Positive definite tensor, 7Potential flow, 111Potential theorem, 35Power expended, theorem of, 110, 180Pressure, 106Principal invariants, 15

list of, 15Principal stress, see Stress, principalPrincipal stretch, 45Progressive wave, 223-226Psym, 7Pure torsion, 58

R

Reduced constitutive equation, 166Reference body force, 179Reference configuration, 41Reference density, 88Regular region, 37Reiner-Rivlin fluid, 155Reynolds number, 154Reynolds' transport theorem, 78

264 INDEX

Right Cauchy-Green strain tensor, 46Right stretch tensor, 46Rigid deformation, 48

characterization of, 49Rigid motion, 69Rivlin-Ericksen tensors, 76Rotation, 7

about point, 43tensor, 46

s

Shearing force, 105Shear modulus, 202Signorini's theorem, 113Simple shear

of elastic body, 175-178of Newtonian fluid, 156-159

Simply connected set, 35Skew tensor, 3Skw, 7Small deformation, 54-58Smooth, 20Smooth-inverse theorem, 22Smoothness lemma, 60Sonic flow, 133Sound speed, 131Span, 2Spatial description, 60Spatial description of velocity, 60Spatial field, 60Spatial time derivative, 61Spectral decomposition, 12Spectral theorem, 11Spectrum, 11Spin, 71

axis, 71transport of, 80

Square root of tensor, 13derivative of, 23

Square-root theorem, 13Stability theorem for Navier-Stokes equa-

tions, 162Stagnation point, 128Stationary potential energy, principle of, 193Steady flow, 109Steady motion, 67Stiffness matrix, 210Stokes flow, 152Stokes' theorem, 39Strain energy, 205

Strain-energy density, 186Stream function, 126Streamline, 64Stress

Cauchy, 101Piola-Kirchhoff, 178principal, 105shear, 106tensile, 106

Stress power, IIIStretch, 44Stretching, 71Subsonic flow, 133Supersonic flow, 133Surface force, 99Surface traction, 98Sym, 7Symmetric tensor, 3Symmetry group, 168Symmetry transformation, 168

T

Tensor, 2skew part of, 3symmetric part of, 3

Tensor product, 4Time, 58Torsion of circular cylinder, 217-219Torsional rigidity, 219Total body force, 100Trace, 5Traction problem of elastostatics, 208Trajectory of motion, 59Transfer theorem, 231Transformation law

for Cauchy-Green strain tensors, 141for Cauchy stress, 144for deformation gradient, 140for rotation tensor, 140for spin, 142for stretch tensors, 140, 141for stretching, 142for velocity gradient, 142

Transverse sound speed, 225

u

Unimodular group, 172Uniqueness

for linear elastodynamics, 222

for linear elastostatics, 207for Navier-Stokes equations, 160

v

Vector, 1Velocity gradient, 63Virtual work, theorem of, 100Viscosity, 149Viscous flow problem, 160Volume of set, 37

INDEX

Vortex line, 83

W

Wang's lemma, 232Wave equation, 133Work in closed processes, 185

Y

Young's modulus, 203

265

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