tam 551 solid mechanics i -...

TAM 551 Solid Mechanics I

R. B. Haber

Fall 2014

This document is based on a set of hand-written notesdeveloped by Prof. D. E. Carlson. While I have introduced

various modifications and additions, the well-organizedstructure is primarily due to Prof. Carlson. I’m also indebtedto him for many helpful discussions and comments. However, I

bear full responsibility for the quality and accuracy of thisdocument. Please bring any errors to my attention. —RBH

Contents

1 Mathematical Preliminaries 11.1 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 Indicial Notation . . . . . . . . . . . . . . . . . . . . . . . . . 21.3 The Summation Convention . . . . . . . . . . . . . . . . . . . 51.4 Kronecker’s Delta and the Alternating Symbol . . . . . . . . . 71.5 Determinants . . . . . . . . . . . . . . . . . . . . . . . . . . . 91.6 The Inverse of a 2-index Symbol . . . . . . . . . . . . . . . . . 141.7 Linear Algebraic Equations . . . . . . . . . . . . . . . . . . . 151.8 Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

1.8.1 Vectors in three-dimensional Euclidean space . . . . . . 171.8.2 Representation of vectors w.r.t. a Cartesian coordinate

frame . . . . . . . . . . . . . . . . . . . . . . . . . . . 191.8.3 Real vector spaces . . . . . . . . . . . . . . . . . . . . 23

1.9 Change of Coordinate Frame . . . . . . . . . . . . . . . . . . . 241.10 Inner Products and Norms . . . . . . . . . . . . . . . . . . . . 27

1.10.1 The inner product of Euclidean vectors . . . . . . . . . 271.10.2 Inner product and normed vector spaces . . . . . . . . 30

1.11 Second-Order Tensors . . . . . . . . . . . . . . . . . . . . . . . 331.11.1 Second–order tensors as linear operators . . . . . . . . 331.11.2 Cartesian component representation of second-order ten-

sors . . . . . . . . . . . . . . . . . . . . . . . . . . . . 371.11.3 Transformation rules for second-order tensors . . . . . 401.11.4 The transpose and symmetry of a second-order tensor . 421.11.5 The product of two second-order tensors . . . . . . . . 441.11.6 The determinant of a second-order tensor . . . . . . . . 461.11.7 The trace of a second-order tensor . . . . . . . . . . . . 471.11.8 The inner product of two second-order tensors . . . . . 48

iii

iv CONTENTS

1.11.9 The inverse of a second-order tensor . . . . . . . . . . . 501.11.10 Orthogonal second-order tensors. . . . . . . . . . . . . 53

1.12 Higher-Order Tensors . . . . . . . . . . . . . . . . . . . . . . . 561.12.1 Introduction to mth-order tensors . . . . . . . . . . . . 561.12.2 Components of mth-order tensors; transformation rules 611.12.3 The alternating tensor; restrictions on changes of frame 641.12.4 Another view of mth-order tensors; tensor contraction . 65

1.13 The Vector (Cross) Product . . . . . . . . . . . . . . . . . . . 681.14 The Scalar Triple Product . . . . . . . . . . . . . . . . . . . . 711.15 Special Cases of Second-Order Tensors . . . . . . . . . . . . . 74

1.15.1 Skew second-order tensors . . . . . . . . . . . . . . . . 741.15.2 Symmetric second-order tensors . . . . . . . . . . . . . 761.15.3 Positive-definite second-order tensors . . . . . . . . . . 86

1.16 Tensor Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . 91

2 Kinematics 972.1 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 972.2 Bodies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 982.3 Finite Deformations . . . . . . . . . . . . . . . . . . . . . . . . 100

2.3.1 Deformation, displacement, deformation gradient . . . 1002.3.2 Rigid deformations . . . . . . . . . . . . . . . . . . . . 1042.3.3 Geometric properties of deformations . . . . . . . . . . 109

2.4 Infinitesimal Deformation Theory . . . . . . . . . . . . . . . . 1232.4.1 Restriction to small displacement gradients . . . . . . . 1232.4.2 Analysis of infinitesimal strain . . . . . . . . . . . . . . 129

2.5 Motions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1432.6 The Transport and Localization Theorems . . . . . . . . . . . 149

3 Balance Laws and Stress Tensors 1553.1 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1553.2 Conservation of Mass . . . . . . . . . . . . . . . . . . . . . . . 155

3.2.1 Finite deformations . . . . . . . . . . . . . . . . . . . . 1553.2.2 Infinitesimal motions . . . . . . . . . . . . . . . . . . . 158

3.3 Force, Stress and Balance of Momentum . . . . . . . . . . . . 1603.4 The Cauchy Stress Tensor . . . . . . . . . . . . . . . . . . . . 1633.5 Cauchy’s Stress Equations of Motion . . . . . . . . . . . . . . 1703.6 Some Properties of the Cauchy Stress Tensor . . . . . . . . . . 1763.7 The Piola-Kirchhoff Stress Tensors . . . . . . . . . . . . . . . 178

CONTENTS v

3.8 Area Metrics and Nanson’s Relation . . . . . . . . . . . . . . . 185

4 Elastic Response 1874.1 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1874.2 The Elastic Response Function . . . . . . . . . . . . . . . . . 1874.3 Principle of Material Frame-Indifference . . . . . . . . . . . . 1924.4 Material Symmetry; Isotropy . . . . . . . . . . . . . . . . . . . 1984.5 Hyperelasticity . . . . . . . . . . . . . . . . . . . . . . . . . . 2044.6 Elastic Response to Infinitesimal Motions . . . . . . . . . . . . 211

5 Linearized Elasticity 2335.1 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2335.2 Summary of the Linearized Field Equations . . . . . . . . . . 2345.3 Linearized Traction Boundary Conditions . . . . . . . . . . . . 2375.4 Linearized Isotropic Stress-Strain Relation . . . . . . . . . . . 243

5.4.1 General Considerations . . . . . . . . . . . . . . . . . . 2435.4.2 Pure Pressure . . . . . . . . . . . . . . . . . . . . . . . 2455.4.3 Pure Shear . . . . . . . . . . . . . . . . . . . . . . . . 2465.4.4 Pure Tension . . . . . . . . . . . . . . . . . . . . . . . 2475.4.5 Relations Between the Elastic Moduli . . . . . . . . . . 2495.4.6 Principal Axes Coincide . . . . . . . . . . . . . . . . . 249

5.5 Stored Energy; Work-Energy Theorems . . . . . . . . . . . . . 250

6 Linearized Elastodynamics 2596.1 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2596.2 Field Equations; Elastic Processes . . . . . . . . . . . . . . . . 259

7 Linearized Elastostatics 2637.1 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2637.2 Field Equations; Elastic States . . . . . . . . . . . . . . . . . . 2637.3 Boundary-Value Problems . . . . . . . . . . . . . . . . . . . . 2667.4 Work-Energy Theorems. The Reciprocal Theorem . . . . . . . 2677.5 Uniqueness . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2727.6 Displacement Form of the Mixed Problem . . . . . . . . . . . 2747.7 Stress Form of the Traction Problem . . . . . . . . . . . . . . 2767.8 Minimum Principles . . . . . . . . . . . . . . . . . . . . . . . 281

7.8.1 Assumptions . . . . . . . . . . . . . . . . . . . . . . . . 2827.8.2 The Principle of Minimum Potential Energy . . . . . . 282

vi CONTENTS

7.8.3 The Principle of Minimum Complementary Energy . . 2857.8.4 Converses of the Minimum Principles . . . . . . . . . . 288

7.9 Variational Principles . . . . . . . . . . . . . . . . . . . . . . . 2947.10 The Principle of Virtual Work . . . . . . . . . . . . . . . . . . 300

7.10.1 The Principle of Virtual Work for Elastostatics . . . . 3007.10.2 The Principle of Virtual Work for Elastodynamics . . . 305

Chapter 1

Mathematical Preliminaries

1.1 References

1. R. M. Bowen and C.-C. Wang, Introduction to Vectors and Tensors,Vols. 1 and 2, Plenum, 1976.

2. P. Chadwick, Continuum Mechanics, Halsted, 1976.

3. R. Courant and F. John, Introduction to Calculus and Analysis, Vols.1 and 2, Interscience, 1965 and 1974.

4. C. L. Dym and I. H. Shames, Energy and Finite Element Methods inStructural Mechanics, McGraw-Hill, 1985.

5. M. E. Gurtin, An Introduction to Continuum Mechanics, Academic,1981.

6. M. E. Gurtin, E. Fried and L. Anand, The Mechanics and Thermody-namics of Continua, Cambridge University Press, 2010.

7. H. Jeffreys and B. S. Jeffreys, Methods of Mathematical Physics , 3rd

Ed., Cambridge, 1956.

8. A. D. Martin and V. J. Mizel, Introduction to Linear Algebra, McGraw-Hill, 1966.

9. W. Noll, Finite-Dimensional Spaces, Martinus Nijhoff, 1987.

10. W. Prager, Introduction to Mechanics of Continua, Ginn, 1961.

1

2 CHAPTER 1. MATHEMATICAL PRELIMINARIES

11. J. G. Simmonds, A Brief on Tensor Analysis, Springer, 1982.

12. A. J. M. Spencer, Continuum Mechanics, Longman, 1980.

1.2 Indicial Notation

Indicial notation is a compact and powerful method for representing orderedlists of real numbers by a single symbol decorated with one or more indices.Typically, the indices range over the integer set 1, 2, 3, corresponding to thethree spatial dimensions. Occasionally, other ranges will be defined explicitly.Thus, the symbol aj denotes the three numbers

a1 a2 a3.

The letter chosen for an index has no significance. That is, the symbols ai,aj and ak each represent the same list of three real numbers.

When an integer — rather than a letter — appears as an index, we takeit at “face value”. That is, a3 represents a single real number.

Multiple indices are used to represent ordered arrays of real numbers.Each distinct unspecified (i.e., letter) index varies independently: aij denotesthe array

a11 a12 a13

a21 a22 a23

a31 a32 a33

;

a2k denotes the three numbers

a21 a22 a23.

Note the relation between the two-index symbol aij and the matrix [aij].Similarly, a2k corresponds to the second row of a 3 x 3 matrix, and ak2

corresponds to the second column. An indexed symbol can carry an unlimitednumber of unspecified indices; e.g., aplmno is a valid symbol that denotes35 = 243 real numbers.

Addition of indexed symbols requires agreement between the unspecifiedindices in each term. For example, ai + aj, ai + bij and a2 + aj have nomeaning. If the (free, c.f. Section 1.3) unspecified indices are in agreement,

1.2. INDICIAL NOTATION 3

then addition is defined in the logical way, with each letter index varyingindependently: ai + bi denotes the three numbers

a1 + b1, a2 + b2, a3 + b3.

Agreement does not require that the order of the unspecified indices be thesame. For example, the expression aij + bji is a legitimate expression thatcorresponds to the sum of two matrices, [aij] + [bij]

t.Scalar multiplication is an operation between a real number and an in-

dexed variable. Suppose λ ∈ (“is in”, “is a member of” or “belongs to”) <(the set of real numbers). Then λai denotes the three numbers

λa1, λa2, λa3.

Similarly, a2ai denotes the three numbers

a2a1, a2a2, a2a3,

since a2 denotes a single real number.Outer multiplication is an operation that takes two arguments with dis-

tinct unspecified indices. For example, aibj denotes the nine numbers

a1b1 a1b2 a1b3

a2b1 a2b2 a2b3

a3b1 a3b2 a3b3

.

Note the similarity to the matrix product abt, where a and b are 3 x 1column vectors.

Indexed equations provide a compact notation for multiple equations.E.g., ai = bi denotes the three equations

a1 = b1, a2 = b2, a3 = b3.

The unspecified indices on each side of an equality must agree. So the equa-tions, aij = ck and ai + bj = cij, have no meaning, while am1l = blm − cml isa legitimate equation.

Subtraction is an operation defined in terms of addition and scalar mul-tiplication in the usual way,

akl − bkl = akl + (−1)bkl = akl + (−bkl),


in which −bkl := (−1)bkl is the negative of bkl. [Question: How many equa-tions does this represent?]

The arithmetic defined above follows a number of familiar rules.Commutative Rules:

ai + bi = bi + ai

aijbk = bkaij

Associative Rules:

ai + (bi + ci) = (ai + bi) + ci = ai + bi + ci

ai(bjkcl) = (aibjk)cl = aibjkcl

Distributive Rule:

aij(bk + ck) = aijbk + aijck

Substitution Rules: Given ai = bi, then

λai = λbi

aicj = bicj.

Definition 1 The symbol aij...n...m...k is symmetric w.r.t. (“with respect to”)the indices m and n iff (“if and only if”)

aij...n...m...k = aij...m...n...k;

it is antisymmetric (or skewsymmetric or simply skew) w.r.t. m and n iff

aij...n...m...k = −aij...m...n...k.

Theorem 1 If the 2-index symbol aij is skew w.r.t. its indices, then a11 =a22 = a33 = 0.

Proof.

aji = −aij (aij is skew)

⇒ (“implies”) a11 = −a11 (choose i = j = 1)

⇒ 2a11 = 0 (add a11 to both sides)

⇒ a11 = 0 (multiply both sides by1

2)

Similarly, a22 = a33 = 0. (indicates end of the proof)

1.3. THE SUMMATION CONVENTION 5

Theorem 2 An arbitrary 2-index symbol aij has a unique decomposition asthe sum of a symmetric symbol a(ij) and a skew symbol a[ij]. That is, aij =a(ij) + a[ij], where

a(ij) ≡1

2(aij + aji)

is the symmetric part of aij, and

a[ij] ≡1

2(aij − aji)

is the skew part of aij.

Exercise 1 Prove that the decomposition given in Theorem 2 of an arbitrary2-index symbol aij into symmetric and skew parts exists and is unique. Thatis, first show that a(ij) is symmetric, that a[ij] is skew and that aij = a(ij)+a[ij]

(this demonstrates existence by example). Then show that if aij = bij +cij, where bij is symmetric and cij is skew, then it must be true that bij= 1

2(aij + aji) and cij = 1

2(aij − aji) .

1.3 The Summation Convention

Summation convention: An unspecified index that appears exactly two timesin a monomial term implies summation over the range of that index – unlesscancelled by writing “no sum” or by underlining one of the repeated indices.Thus,

aii =3∑i=1

aii = a11 + a22 + a33,

aijbj =3∑j=1

aijbj = ai1b1 + ai2b2 + ai3b3.

Note that the letter used for a repeated index has no significance: aii =akk = amm. Therefore, repeated indices (in the same monomial term) aretermed dummy indices, while unspecified indices that are not repeated arecalled free indices. Note that there are no dummy indices in the expressionaij + bij,while there are no free indices in aijbij. A monomial with m freeindices denotes a total of 3m real numbers.

Contraction is the process of setting equal two (initially distinct) unspec-ified indices in a monomial and then carrying out the indicated summation.


Each contraction operation reduces by a factor of 32 (or, in general, the rangeof the indices squared) the number of real numbers denoted by the monomial.For example, the contraction over i and k of aibjck (27 real numbers) is

aibjci =3∑i=1

aibjci (3 real numbers, for j = 1, 3).

Contraction on both sides of an equation preserves the equality. Suppose weare given aij = bij. Then

contraction of aij over i and j = aii (definition of contraction)

= a11 + a22 + a33 (summation convention)

= b11 + b22 + b33 (given: aij = bij)

= bii

= contraction of bij over i and j

That is, aij = bij ⇒ aii = bii.Inner multiplication of two indexed symbols is a two-step operation that

involves first taking the outer product of the two symbols and then contract-ing the result w.r.t. one unspecified index from each symbol. The result ofthis operation is called an inner product. The inner product of aijk and brsw.r.t. k and s is written as, aijkbrk.

Example 1 Suppose u and v are vectors with Cartesian components ui andvi. Then the inner product (or “dot product”) of the two vectors is equal tothe inner product of their Cartesian components: u · v = uivi.

Example 2 Matrix multiplication can also be defined in terms of inner mul-tiplication. The product of two square matrices [aij] and [bij] is the matrix[cij] with elements

cij = aikbkj.

The product of the square matrix [aij] and the column matrix [xi] is the columnmatrix [bi] with elements

bi = aijxj.

[It is best to always specify either the column matrix [xi] =

x1

x2

x3

or the

row matrix [xi] =[x1 x2 x3

]to avoid ambiguity.]

1.4. KRONECKER’S DELTA AND THE ALTERNATING SYMBOL 7

Theorem 3 If aij...m...n...k is symmetric w.r.t. m and n and if bpq...m...n...r isskew w.r.t. m and n, then

aij...m...n...kbpq...m...n...r = 0.

[Note: The important issue here involves the relations between the dummyindices m and n. The free indices, whether present or not, have no bearingon the result.]

Theorem 4 For arbitrary 2-index symbols aij and bij, aijbji = a(ij)b(ji) +a[ij]b[ji].

Exercise 2 Prove Theorem 3 for the case of 2-index symbols. That is, giventhat aij is symmetric and that bij is skew, show that aijbij = 0.

Exercise 3 Prove Theorem 4.

1.4 Kronecker’s Delta and the Alternating Sym-

bol

Definition 2 Kronecker’s delta is the 2-index symbol δij defined by

δij =

1 if i = j (no sum)0 if i 6= j

.

Kronecker’s delta corresponds to the identity element for matrix multiplica-tion. The properties of Kronecker’s delta include:

1. δij = δji (symmetry);

2. δii = 3;

3. δii = 1 (no sum);

4. δijap...j...q = ap...i...q; ap...j...qδjk = ap...k...q (contraction with Kronecker’sdelta has the effect of renaming an unspecified index)

5. δijaj = ai;

6. δijajk = aik, aijδjk = aik;


1

2 3

even odd

Figure 1.1: Even and odd permutations of 123.

7. δijaij = aii;

8. δijδij = 3.

The proofs are either self-evident or useful exercises if this is new to you.

The contraction of a monomial ap...i...j...q w.r.t. the indices i and j can beobtained by forming an inner product with δij:

δijap...i...j...q = ap...i...i...q;

ap...i...j...qδij = ap...j...j...q.

Definition 3 The even permutations of the integers 123 are 123, 231 and312 . The odd permutations of the integers 123 are 321, 213 and 132. Anylist that repeats an integer is not a permutation.

An even number of pair-wise interchanges to the list 123 always producesan even permutation. An odd number of interchanges produces on odd per-mutation. These two observations provide a simple means to determine thesense of a list of three non-repeating indices (see also Figure 1.1).

Definition 4 The alternating symbol is the 3-index symbol εijk defined by

εijk =

1 if ijk is an even permutation of 123−1 if ijk is an odd permutation of 1230 if ijk is not a permutation of 123

1.5. DETERMINANTS 9

Theorem 5 The alternating symbol is skew w.r.t. any two of its indices.That is, each pair-wise interchange of indices changes the sign of the alter-nating symbol; i.e.,

εjik = εkji = εikj = −εijk.

Proof. Case I (ijk contains a repeated index (no sum)): Then ijk is nota permutation and neither are jik, kji and ikj. Then by definition, εjik =εkji = εikj = −εijk = 0.

Case II (ijk contains no repeated indices): Then ijk is a permutation,and so are jik, kji and ikj – each of which is obtained through one additionalpair-wise interchange relative to ijk. Thus if ijk is odd, then jik, kji andikj must each be even, and vice versa. Therefore, the permutations haveopposite sense, and εjik = εkji = εikj = −εijk.

Exercise 4 Evaluate ci = εijkajk when

1. ajk = δjk,

2. ajk = djbk,

3. [ajk] =

1 4 72 5 83 6 9

.1.5 Determinants

Recall that the determinant of a 3× 3 matrix [aij] is given by

det [aij] =

∣∣∣∣∣∣a11 a12 a13

a21 a22 a23

a31 a32 a33

∣∣∣∣∣∣= a11a22a33 + a12a23a31 + a13a21a32 − a31a22a13

−a32a23a11 − a33a21a12.

This familiar result can be written in compact, indicial form using the alter-nating symbol.

Theorem 6det [aij] = εijka1ia2ja3k = εijkai1aj2ak3.


Proof. The first indicial expression is the row expansion of the determinantand the second is the column expansion. It follows that if we use ai, bi andci to form the rows (columns) of a matrix, then∣∣∣∣∣∣

a1 a2 a3

b1 b2 b3

c1 c2 c3

∣∣∣∣∣∣ = εijkaibjck and

∣∣∣∣∣∣a1 b1 c1

a2 b2 c2

a3 b3 c3

∣∣∣∣∣∣ = εijkaibjck.

This result foreshadows a theorem below concerning the determinants of amatrix and its transpose. We use the indicial expressions and the propertiesof the alternating symbol to derive properties of the determinant.

Theorem 7 Properties of the determinant include:

1. If all the elements of a row (column) of [aij] are zero, then det [aij] = 0.

Proof. If one of the rows is zero (a1i = 0 or a2i = 0 or a3i = 0), thenεijka1ia2ja3k = 0 by substitution. If one of the columns is zero (ai1 = 0or ai2 = 0 or ai3 = 0) then εijkai1aj2ak3 = 0 by substitution.

2. If two rows (columns) of [aij] are equal, then det [aij] = 0.

Proof. Suppose the first two rows are equal (a1i = a2i). Thus, det [aij] =εijka1ia2ja3k = εijka1ia1ja3k = 0 by Theorem 3, because εijk is skeww.r.t. i and j, while a1ia1ja3k is symmetric. Similar logic holds if thesecond and third rows (a2i = a3i) or first and third rows (a1i = a3i) areidentical. The same logic applies to repeated columns if we work withthe column expansion, det [aij] = εijkai1aj2ak3.

3. Suppose that [bij] is formed by interchanging two rows (columns) of[aij]. Then det [bij] = − det [aij] .

Proof. For example, suppose that the first two rows are interchanged(b1i = a2i, b2i = a1i, b3i = a3i). Then

det [bij] = εijkb1ib2jb3k (row expansion)

= εijka2ia1ja3k (substitution)

= εijka1ja2ia3k (commutative rule)

= εjika1ia2ja3k (rename dummy indices)

= −εijka1ia2ja3k (εijk is skew)

= − det [aij] .

1.5. DETERMINANTS 11

4. Suppose [bij] is formed by scaling one row (column) of [aij] by λ andleaving the remaining rows unaltered. Then det [bij] = λ det [aij] .

Proof. E.g., suppose the first row is scaled (b1i = λa1i, b2i = a2i, b3i =a3i). Then


= εijkλa1ia2ja3k (substitution)

= λ det [aij] (commutative rule, row expansion).

5. Suppose matrices [aij], [bij] and [cij] share the same elements, except forthe kth rows (columns) where we only assume that the kth row (column)of [aij] is equal to the sum of the kth rows (columns) of [bij] and [cij].Then det [aij] = det [bij] + det [cij].

Proof. Suppose that the first row of [aij] is the sum of the first rowsof [bij] and [cij] (i.e., k = 1). Then a1i = b1i + c1i; a2i = b2i = c2i; anda3i = b3i = c3i. Then

det [aij] = εijka1ia2ja3k (row expansion)

= εijk (b1i + c1i) a2ja3k (substitution)

= εijkb1ia2ja3k + εijkc1ia2ja3k (distributive rule)

= εijkb1ib2jb3k + εijkc1ic2jc3k (substitution)

= det [bij] + det [cij] (row expansions).

6. Suppose [bij] is formed by adding a multiple of one row (column) of[aij] to a different row (column). Then det [bij] = det [aij].

Proof. For example, suppose λ times the first row is added to thesecond (b1i = a1i, b2i = λa1i + a2i, b3i = a3i). Then


= εijka1i (λa1j + a2j) a3k (substitution)

= εijka1iλa1ja3k + εijka1ia2ja3k (distributive rule)

= λ (0) + εijka1ia2ja3k (part 2 of this theorem)

= det [aij] (row expansion).


7. Suppose[atij]

is the transpose of [aij]; i.e.; atij = aji. Then det[atij]

=det [aij] .

Proof.

det[atij]

= εijkat1ia

t2ja

t3k (row expansion)

= εijkai1aj2ak3 (substitution)

= det [aij] (column expansion).

Theorem 8 Given any 3× 3 matrix [aij],

εijkapiaqjark = det [amn] εpqr

andεijkaipajqakr = det [amn] εpqr.

Exercise 5 Prove Theorem 8. Note that εijkapiaqjark is the row expansionfor det [aij] when pqr = 123. Then consider other possible sequences (odd,even permutations and non-permutations) for pqr.

The following theorem is an important corollary of Theorem 8.

Theorem 9 If cij = aikbkj, then det [cij] = det [aij] det [bij] . That is, thedeterminant of the product of two matrices is the product of the determinantsof the factors.


Theorem 10 For an arbitrary 3× 3 matrix [aij] ,

εijkεpqr det [amn] =

∣∣∣∣∣∣aip aiq airajp ajq ajrakp akq akr

∣∣∣∣∣∣ .Exercise 7 Prove Theorem 10.

1.5. DETERMINANTS 13

Theorem 10, in turn, has the corollary

Theorem 11

1. εijkεpqr =

∣∣∣∣∣∣δip δiq δirδjp δjq δjrδkp δkq δkr

∣∣∣∣∣∣ ,2. εijkεiqr = δjqδkr − δjrδkq, (this result is worth remembering)

3. εijkεijr = 2δkr,

4. εijkεijk = 6.

Proof. Set amn = δmn in Theorem 10.

Now, combining Theorem 8 (2) and 11 (4), we obtain

Theorem 12 For an arbitrary 3× 3 matrix [aij]

det [amn] =1

6εijkεpqraipajqakr.

Proof.

εijkaipajqakr = det [amn] εpqr (Theorem 8 (2))

⇒ εijkεpqraipajqakr = det [amn] εpqrεpqr (multiply by εpqr)

⇒ εijkεpqraipajqakr = 6 det [amn] (Theorem 11 (4)).

The following expression for the derivative of a determinant follows di-rectly from the definition of the determinant.

Theorem 13 Assume that all the elements of [aij] are each differentiablew.r.t. a common real variable and use primes to denote derivatives w.r.t.that common variable. [In other words, assume aij = fij (x) , where fij is a

differentiable function w.r.t. x, and a′ij ≡dfijdx.] Then

(det [aij])′ ≡ d (det [aij])

dx=

∣∣∣∣∣∣a′11 a′12 a′13

a21 a22 a23

a31 a32 a33

∣∣∣∣∣∣+

∣∣∣∣∣∣a11 a12 a13

a′21 a′22 a′23

a31 a32 a33

∣∣∣∣∣∣+

∣∣∣∣∣∣a11 a12 a13

a21 a22 a23

a′31 a′32 a′33

∣∣∣∣∣∣=

∣∣∣∣∣∣a′11 a12 a13

a′21 a22 a23

a′31 a32 a33

∣∣∣∣∣∣+

∣∣∣∣∣∣a11 a′12 a13

a21 a′22 a23

a31 a′32 a33

∣∣∣∣∣∣+

∣∣∣∣∣∣a11 a12 a′13

a21 a22 a′23

a31 a32 a′33

∣∣∣∣∣∣


1.6 The Inverse of a 2-index Symbol

Definition 5 A 2-index symbol, bij, is an inverse of aij iff it satisfies

bikakj = aikbkj = δij.

The notation a−1ij denotes an inverse of aij.

Theorem 14 The inverse of a 2-index symbol, if it exists, is unique.

Proof. Suppose bij and cij are inverses of aij. Then

bikakj = δij (definition of inverse)

⇒ bikakjcjl = δijcjl = cil (property of Kronecker’s delta)

⇒ bikδkl = cil (definition of inverse)

⇒ bil = cil (property of Kronecker’s delta).

Theorem 15 If aij has an inverse, then det [akj] 6= 0.

Proof. Suppose aij has an inverse a−1ij . Then by definition,

a−1ik akj = δij (definition of inverse)

⇒ det[a−1ik akj

]= det [δij]

⇒ det[a−1ik

]det [akj] = 1 (Theorem 9)

⇒ det [akj] 6= 0.

So we have established that det [aij] 6= 0 is a necessary condition for theexistence of the inverse of aij and that the inverse must be unique if it exists.We would also like to show that det [aij] 6= 0 is a sufficient condition forexistence of the inverse. A “shortcut” proof of sufficiency makes use of anexplicit expression for the inverse that is valid, provided that det [aij] 6= 0(the derivation of this expression is not addressed here).

Exercise 8 If det [aij] 6= 0, then we can define bij := 12 det[ars]

εipqεjmnampanq.

Show by direct calculation that bij = a−1ij . That is, show that

bikakj = aikbkj = δij.

1.7. LINEAR ALGEBRAIC EQUATIONS 15

Thus we have

Theorem 16 Given a 2-index symbol, aij 3 (“such that”) det [aij] 6= 0, then∃ (“there exists”) a unique inverse of aij, denoted by a−1

ij and given by

a−1ij =

1

2 det [ars]εipqεjmnampanq.

1.7 Linear Algebraic Equations

Now we are ready to solve systems of 3 linear equations in 3 unknowns.

Theorem 17 Given a 3× 3 matrix, [aij] 3 det [aij] 6= 0, then the system oflinear algebraic equations

aijxj = bi

in the 3 unknowns xi has the unique solution

xi = a−1ij bj =

1

2 det [ars]εipqεjmnampanqbj.

Proof. We know that[a−1ij

]exists, since det [aij] 6= 0. Now suppose xi solves

the system. Then we have

aijxj = bi

⇒ a−1ki aijxj = a−1

ki bi

⇒ δkjxj = a−1ki bi

⇒ xk = a−1ki bi.

This establishes uniqueness, since if a solution exists it must have the formxk = a−1

ki bi. Next, we need to demonstrate existence — it will suffice to showthat xi = a−1

ij bj is a solution.

aijxj = aij(a−1jk bk

)(substitute the proposed solution)

=(aija

−1jk

)bk (associative rule)

= δikbk (defininition of inverse)

= bi

Therefore, xi = a−1ij bj is the unique solution.

Now consider the homogeneous case where bi = 0. The previous resultindicates that if det [aij] 6= 0, then the unique solution is the trivial solution,xi = 0. This suggests the following theorem.


Theorem 18 Given a 3 × 3 matrix [aij] , then det [aij] = 0 is a necessaryand sufficient condition for the existence of a nontrivial solution to the ho-mogeneous, linear algebraic system

aijxj = 0.

Proof. It is evident that det [aij] = 0 is a necessary condition for theexistence of a nontrivial solution, because if det [aij] 6= 0 then the trivialsolution xi = 0 is the unique solution according to the previous theorem.Next consider the system in expanded form:

a11x1 + a12x2 + a13x3 = 0

a21x1 + a22x2 + a23x3 = 0

a31x1 + a32x2 + a33x3 = 0.

We can apply Gaussian elimination (with pivoting as needed and makinguse of det [aij] = 0) to arrive at the following parametric expressions for thegeneral solution (including non-trivial solutions)

x1 = αx2 = βx3 = γ

orx1 = b1σ + b2τx2 = σx3 = τ

orx1 = c1νx2 = c2νx3 = ν

where it might have been necessary to reorder the equations so that x3 is notexplicitly zero. Here α, β, γ, σ, τ and ν are parameters that can be adjustedto obtain different solutions, while b1, b2, c1 and c2 are constants determinedin the course of Gaussian elimination. The first solution holds when aij = 0;the second holds when there is only one linearly independent equation in thesystem; and the third solution holds when there are two linearly independentequations. Whichever one of these general, nontrivial solutions is producedby Gaussian elimination can be plugged back into the original system todemonstrate that a solution exists. Thus, det [aij] = 0 is a necessary andsufficient condition, and the proof is complete.

1.8 Vectors

We begin with a purely geometrical approach to vectors in three-dimensionalEuclidean point space, and then introduce the standard component represen-tation relative to a rectangular Cartesian coordinate frame. Next we developan algebra for vectors, as an instance of a real (abstract) vector space andrelate it to the algebra of 1-index symbols introduced above.

1.8. VECTORS 17

P

QE

Figure 1.2: The vector from point P to point Q

1.8.1 Vectors in three-dimensional Euclidean space

Definition 6 Let P and Q be points in a three-dimensional Euclidean point

space E. The vector−→PQ is the directed line segment that originates at P and

terminates at Q (Fig. 1.2). We refer to P as the tail of−→PQ and to Q as the

head of−→PQ. We shall denote vectors by bold symbols.

Definition 7 The magnitude of−→PQ is the (non-negative) distance between

points P and Q. [A Euclidean point space possesses a measure that establishesthe distance between any two points.] The magnitude of a vector a is denoted|a|.

A vector, as described so far, has three intrinsic properties:

1. a direction (established by the relative positions of P and Q)

2. a magnitude

3. and an absolute position, (say, defined by the position of the tail P ).

We ignore the absolute position in developing our vector algebra. That is,we take all vectors with the same direction and magnitude to be equivalent(i.e., the same vector) and ignore the absolute position. However, keep inmind that there are situations in mechanics where absolute position does


make a difference (e.g., a force acting through a point). Vectors that aredifferentiated by absolute position are called bound vectors. For now, assumethat a vector is completely defined by its magnitude and direction. Thus,

Definition 8 Two vectors a and b are equal iff they have the same magni-tude and direction. Then we write a = b.

Definition 9 The zero vector is the vector with zero magnitude and unde-fined direction generated when P and Q are coincident. The zero vector isdenoted by a bold-face zero, 0.

Our basic description of vectors is complete. Next we develop an algebrafor vectors, continuing for now to avoid any reference to a coordinate frame.

Definition 10 Let λ be a real number. The scalar multiple of a vector a,denoted by λa, is a vector with magnitude |λ| |a| (read absolute value of λtimes the magnitude of a). The direction of λa is equal to the direction of aif λ > 0 and |a| 6= 0, opposite to the direction of a if λ < 0 and |a| 6= 0, andundefined if either λ = 0 or |a| = 0.

Theorem 19 Properties of the scalar multiple of a vector include:

1. λ = 0⇒ λa = 0 for all vectors a,

2. λ = 1⇒ λa = a for all vectors a,

3. a = 0⇒ λa = 0 for all real numbers λ.

Proof. The proofs follow directly from the definitions of scalar multiplicationof a vector, vector equality and the zero vector.

Definition 11 The sum of two vectors a and b , denoted a + b, is definedby the parallelogram rule.

Definition 12 Vector subtraction is defined in terms of scalar multiplicationand vector addition. That is, a− b ≡ a + (−1) b.

Theorem 20 The vector −a ≡ (−1) a is the negative of a in the sense thata +−a = a− a = 0.

1.8. VECTORS 19

aa

b

b

a + b

Figure 1.3: Parallelogram rule for vector addition


We obtain the following rules of vector algebra.

Theorem 21 Vector addition is

1. commutative: a + b = b + a, and

2. associative: (a + b) + c = a + (b + c) = a + b + c.

The zero vector is the additive identity element: a + 0 = a. Scalar mul-tiplication of vectors is

1. associative: (λµ) a =λ (µa) = λµa,

2. distributive w.r.t. vector addition: λ (a + b) = λa+λb, and

3. distributive w.r.t. scalar addition: (λ+ µ) a =λa + µa.

These can be proven by appealing to the geometric definitions of vectoraddition, scalar multiplication and the zero vector.

1.8.2 Representation of vectors w.r.t. a Cartesian co-ordinate frame

So far we have described vectors as geometric objects, independent of anycoordinate system. This is desirable because the principle of frame indiffer-ence requires that the intrinsic properties of vectors should not depend on


P

Q

a3

a2

a1

a3e3

X1

X3

X2

a2e2

a1e1

E a

Figure 1.4: Components of a vector a with respect to a Cartesian coordinateframe

the choice of an arbitrary coordinate frame. On the other hand, coordinateframes and the component vector representations they support are a tremen-dous aid in practical computations. Thus, we next consider the componentrepresentation of vectors and the algebra of vector components.

Let X denote a rectangular Cartesian coordinate frame in a three-dimensionalEuclidian space E . We denote the origin of the frame as O and the coordinateaxes as Xi. Further, Let ei denote the base vectors for X — the vectors of

unit magnitude in the Xi directions, and let a =−→PQ be a vector in E .

Definition 13 Let Pi and Qi be the coordinates of the perpendicular projec-tions of points P and Q on the Xi axes. The rectangular Cartesion compo-nents of the vector a w.r.t. X, denoted ai, are the real numbers ai = Qi−Pi.

1.8. VECTORS 21

Theorem 22 For any vector a and for any frame X,

a = a1e1 + a2e2 + a3e3 = aiei.

The vectors aiei are called the vector components of a w.r.t. X, and the realnumbers ai are called the scalar components.

Theorem 23 The magnitude of any vector a is given by

|a| =√aiai.

(3D Pythagorean theorem).

Next we consider the description of the direction of a w.r.t. X. Here itwill be convenient to make use of the fact that the absolute position of thedirected segment defining a vector is immaterial. That is, we can always

describe the vector in terms of a rigid translation of−→PQ (without rotation)

to the origin:

a =−→PQ =

−−→OQ∗,

where the coordinates of Q∗ are given by Q∗i = Qi − Pi. Note that−→PQ =

−−→OQ∗ ⇒ that

−→PQ and

−−→OQ∗ share the same magnitude and direction; i.e., they

are the same vector.

Definition 14 The ith direction angle of a vector a =−−→OQ∗ w.r.t. X is the

smallest angle θi between−−→OQ∗ and the Xi coordinate axis. The direction

cosines of the vector a w.r.t. X are

cos θi =ai|a|.

The direction angle (cosines) of the zero vector are undefined.

Definition 15 Two vectors have the same direction iff they have the samedirection angles (cosines). They have the opposite direction iff the directioncosines of one vector are the negatives of the direction cosines of the other.

Theorem 24 A vector a = 0 iff the components of a w.r.t. any frame areai = 0.


Proof. Let a =−→PQ. By the definition of the zero vector,

a = 0⇔ P and Q are coincident

⇔ Pi = Qi,

⇔ ai = Qi − Pi = 0.

Theorem 25 Two vectors are equal iff their components w.r.t. any givenframe are equal. That is,

a = b⇔ ai = bi.

Proof. Suppose a = b 6= 0, then by Theorem 8 , |a| = |b| and ai|a| = bi

|b| ⇔ai = bi. If a = b = 0, then the directions of a and b are undefined. However,we have ai = bi = 0 by Theorem 24. Conversely, suppose ai = bi. Then

|a| =√aiai =

√bibi = |b|

andai|a|

=bi|b|

,

if |a| = |b| 6= 0. In this case, a and b have the same magnitude and direction,and hence a = b. If ai = bi = 0, then |a| = |b| = 0 and the directions of aand b are undefined ⇒ a = b = 0.

Theorem 26 The components of a vector w.r.t. a given frame are unique.

Proof. Suppose a = aiei = aiei. Then,

aiei − aiei = a− a

⇔ (ai − ai) ei = 0 (distributive rule, Theorem 20)

⇔ ai − ai = 0 (Theorem 24)

⇔ ai = ai.

Next we show that addition of two vectors corresponds to addition of the1 -index symbols constructed from their respective scalar components and,similarly, scalar multiplication of a vector corresponds to scalar multiplicationof a 1-index symbol of components.

1.8. VECTORS 23

Theorem 27

(λa)i = λai

(a + b)i = ai + bi.

Exercise 10 Prove Theorem 27. Use definitions 10, 14 and 15 for the firstpart. A graphical proof based on definitions 11 and 13 is acceptable for thesecond part.

1.8.3 Real vector spaces

In view of the close correspondence between scalar multiplication and addi-tion defined over vectors and 1-index symbols, it is not surprising that theyshare the same algebraic properties (commutative, associative, distributive).Indeed, vectors in Euclidean three-space and n-index symbols are each in-stances of the more general abstract class of real vector (linear) spaces. Areal vector space consists of

1. the set of real numbers <,

2. the standard algebraic operations of addition and multiplication definedover < (each of these maps two real numbers into a real number),

3. a set G of elements called vectors (not necessarily vectors in Euclideanspace!), that contains a zero vector, denoted 0,

4. an operation called vector addition, denoted−→+, that maps two vectors

into a vector,

5. an operation called scalar multiplication, denoted ∗, that maps a realnumber and a vector into a vector,

such that the following relations hold:

1. a−→+b = b

−→+a ∀ (“for all”) a,b ∈ G. (vector addition is commutative)

2.(a−→+b)−→

+c = a−→+(b−→+c)∀ a,b, c ∈ G. (vector addition is associative)

3. a−→+0 = a ∀ a ∈ G. (property of the zero vector)


4. (λµ) ∗ a =λ ∗ (µ ∗ a) ∀ a ∈ G and ∀ λ, µ ∈ <. (scalar multiplication isassociative)

5. λ ∗(a−→+b)

= (λ ∗ a)−→+ (λ ∗ b) ∀ a,b ∈ G and ∀ λ ∈ <. (scalar multi-

plication is distributive w.r.t. vector addition)

6. (λ+ µ) ∗ a = (λ ∗ a)−→+ (µ ∗ a) ∀ a ∈ G and ∀ λ, µ ∈ <. (scalar multi-

plication is distributive w.r.t. addition of real numbers)

7. 0 ∗ a = 0; 1 ∗ a = a ∀ a ∈ G.

Remark 1 Here we use distinct symbols to denote addition of real numbersand vectors as well as multiplication of real numbers and scalar multiplicationof vectors. This is to emphasize that these are distinct operations. However,it is common practice to use “+” for both forms of addition and to omitthe operator for both forms of mutiplication. The specific operator can thenbe inferred from the arguments. In general, we follow this more compactconvention in these notes.

The following cancellation rules and identity can be useful when workingwith real linear (vector) spaces.

Theorem 28 Let a,b, c ∈ G and λ, µ ∈ <. Then

1. a−→+b = a

−→+c⇒ b = c.

2. λ ∗ a = λ ∗ b and λ 6= 0⇒ a = b.

3. λ ∗ a = µ ∗ a and a 6= 0⇒ λ = µ.

4. λ ∗ 0 = 0.

1.9 Change of Coordinate Frame

Although the intrinsic properties of a vector (magnitude and direction) existindependent of any coordinate frame, the components of a vector have nomeaning unless we specify a coordinate frame. We are free to select anycoordinate frame that suits our purposes. Thus, it is useful to develop asystematic approach to transforming the components of a vector specified

1.9. CHANGE OF COORDINATE FRAME 25

w.r.t. a given frame X to the components w.r.t. a second frame X′. Of

course, the vector’s magnitude and direction should remain unaltered underthe transformation.

Let X and X′

be distinct Cartesian coordinate frames for the 3-D Eu-clidean point space E . The base vectors for the two frames are denoted eiand e

′i, respectively. To simplify matters, and without loss of generality (in

view of the lack of significance of a vector’s absolute position), we assumethat X and X

′share a common origin. If λij are the cosines of the angles

between the coordinate axes Xj and X′i (or equivalently, the base vectors ej

and e′i), then the base vectors of one frame can be expanded in terms of the

base vectors of the other frame.

Theorem 29

e′

i = λijej

ei = λjie′

j.

Proof. See Definition 13 and Theorem 22.Let a be an arbitrary vector with components ai w.r.t. X and a

′i w.r.t.

X′. Referring once again to Theorem 22, we have

a = aiei = a′

ie′

i

⇔ aiλjie′

j = a′

je′

j

⇔ a′

j = λjiai (Theorem 26).

Similarly,

a = aiei = a′

ie′

i

⇔ ajej = a′

iλijej

⇔ aj = λija′

i (Theorem 26).

To summarize:

Theorem 30 The transformation rules for the rectangular Cartesian com-ponents of a vector a are

a′

i = λijaj, ai = λjia′

j

where a = aiei = a′ie′i and λij are the direction cosines between e

′i and ej.


The nine direction cosines λij have special properties that derive from theorthogonality of the base vectors for each frame.

Theorem 31 The direction cosines satisfy the orthogonality conditions

λjiλjk = λijλkj = δik.

Proof. Let a be an arbitrary vector. Then we have

ai = δikak (property of Kronecker’s delta),

and ai = λjia′

j = λjiλjkak (Theorem 30, both parts).

⇒ (δik − λjiλjk) ak = 0 ∀ ak⇔ δik − λjiλjk = 0

⇔ λjiλjk = δik.

Similarly, a′i = λijaj = λijλkja

′

k leads to λijλkj = δik.

Exercise 11 Show that the orthogonality conditions imply that

det [λij] = ±1.

Exercise 12 Use the transformation rules and the orthogonality conditionsto show that the magnitude of a vector is invariant under a change of Carte-sian coordinate frames. Refer to Theorem 23.

The orthogonality conditions can be rewritten as

λtijλjk = λijλtjk = δik.

Thus, the inverse of the matrix of direction cosines is its transpose.

Definition 16 An orthogonal matrix is a matrix whose transpose is its in-verse. If its determinant is +1, then it is called proper orthogonal; if itsdeterminant is −1, then it is called improper orthogonal.

Theorem 32 Let there be a set of three real numbers associated with ev-ery rectangular Cartesian coordinate frame in a three-dimensional Euclideanpoint space E. For example, consider the pairings X, ai and

X′, a′i

where

X and X′

are arbitrary frames. Then ∃ (“there exists”) a vector a given bya = aiei = a

′ie′i iff the indexed symbols ai and a

′i satisfy the transformation

rulesa′

i = λijaj, ai = λjia′

j

for all choices of X, ai and

X′, a′i

.

1.10. INNER PRODUCTS AND NORMS 27

Proof. Suppose ai and a′i are the components of a single vector a w.r.t. X

and X′. Then the components must be related by the transformation rules,

a′i = λijaj and ai = λjia

′j, according to Theorem 30. Conversely, let a = aiei

and a′= a

′ie′i be two possibly distinct vectors whose components are related

by the transformation rules. We must show that a′

= a to complete theproof. We have

a′

= a′

ie′

i

= (λijaj) e′

i

= aj

(λije

′

i

)(Theorem 29)

= ajej

= a.

Remark 2 Theorem 32 is sometimes used as a definition for vectors. Thatis, a vector is a mathematical object whose rectangular Cartesian componentstransform under a change of frame according to the rules of Theorem 30. Thisdefinition is equivalent to the geometric definition we have employed here.

Remark 3 There are cases where we encounter triplets of numbers that donot transform as vector components.

1.10 Inner Products and Norms

We first describe the inner product for Euclidean vectors, and then extend theconcept to define abstract inner product spaces and normed vector spaces.

1.10.1 The inner product of Euclidean vectors

The inner product1 is an operation that maps two vectors into a real number(or scalar). We begin with a geometric definition of the inner product andthen examine its component representation.

Definition 17 The inner product of two vectors a and b, denoted a · b, isthe real number

a · b = |a| |b| cos θ

where θ (0 ≤ θ ≤ π) is the angle between a and b.

1Some authors refer to this operation as the scalar product.


Remark 4 The result of the inner product operation is a scalar because itsnumeric representation is independent of the choice of a specific coordinateframe. Contrast this to the case of vectors, where the numeric representationof vector components (and higher-order tensor components, as we shall see)is dependent on the choice of coordinate frame.

Although the algebraic properties of the inner product of two vectors canbe developed from the above definition, we will first introduce the componentrepresentation of the inner product and use that form to derive the properties.

Theorem 33 Let ai and bi be the components of the vectors a and b w.r.t.an arbitrary rectangular Cartesian coordinate frame X. Then

a · b = aibi.

Proof. If either a = 0 or b = 0, then the proposition, |a| |b| cos θ = aibi,reduces directly to 0 = 0. Next we show that the value of the expression aibiis the same for all coordinate frames. Let ai, bi be the components in X, andlet a

′i, b

′i be the components in a second frame X

′, Then

aibi = λjia′

jλkib′

k (Theorem 30)

= (λjiλki) a′

jb′

k

= δjka′

jb′

k (orthogonality condition, thm. 31)

= a′

jb′

j = a′

ib′

i.

Thus we are free to choose any frame X without affecting the value of eitherthe r.h.s. (this supports the assertion that the inner product is a scalar)or the l.h.s. (which is frame independent by definition). So proving thetheorem for any specific choice of X is equivalent to proving it ∀ X. Hencefor the case a 6= 0 and b 6= 0, we choose X so that a lies along the X1

axis. Then a1 = |a| , a2 = a3 = 0 and b1 = |b| cos θ. This leads directly toa · b = aibi = |a| |b| cos θ (the values of b2 and b3 don’t matter).

Remark 5 Theorem 33 establishes the close correspondence between the in-ner product of two 1-index symbols and the inner product of two vectors.Question: Are these the same operation?

Theorem 34 Properties of the inner product:


1. b · a = a · b (commutativity),

2. a · (b + c) = a · b + a · c (distributivity w.r.t. vector addition),

3. (λa) · b = λ (a · b) (homogeneity).

4. a · a ≥ 0 with a · a = 0 iff a = 0. (positive definiteness)

Proof.

1. b · a = biai = aibi = a · b (commutativity of multiplication of realnumbers),

2. a · (b + c) = ai (bi + ci) = aibi + aici = a · b + a · c (distributivity ofmultiplication of real numbers w.r.t. addition),

3. (λa) · b = (λai) bi = λ (aibi) = λ (a · b) (associativity of multiplicationof real numbers).

4. a · a = aiai =∑

i=1,3 (ai)2 ≥ 0. ∴ a · a = 0⇔ (ai)

2 = 0⇔ ai = 0. (thesquare of a real number is positive definite)

Our development also leads to a relation between the magnitude of avector a and its inner product with itself.

Theorem 35 |a| =√

a · a.

Proof. |a| = √aiai =√

a · a

Theorem 36 Let X be a rectangular Cartesian coordinate frame (r.C.c.f.)with base vectors ei. Then the components of a vector a w.r.t. X are

ai = ei · a.

Proof. We have

ai = |a| cos θi (def. of direction angle)

=∣∣ei∣∣ |a| cos θi (where |ei| := |e1| , |e2| , |e3| = 1, 1, 1 )

= ei · a (def. of inner product)


Theorem 37 Let a and b be non-zero vectors separated by an angle θ. Then

a · b = 0

iff a and b are orthogonal (i.e., θ is π/2 ).


Theorem 38

ei · ej = δij.

Proof. Follows directly from |ei| = (1)i , Theorem 37 and the definition ofKronecker’s delta.

Remark 6 This result can be used to prove the orthogonality conditions ofTheorem 31.

Theorem 39 Let a and b be fixed vectors, independent of an arbitrary vectorc. Then

a · c = 0 ∀ c ∈ V ⇔ a = 0.

Also,

a · c = b · c ∀ c ∈ V ⇔ a = b.


1.10.2 Inner product and normed vector spaces

We can generalize the concept of an inner product to abstract vector spaces.

Definition 18 An inner product is an operation that maps two vectors (el-ements of a real vector space) into a real number and that conforms to thefour properties described in Theorem 34.

Definition 19 A real inner-product space2 is a real vector space with all theusual properties (see Section 1.8.2) equipped with an inner product.

2Sometimes an inner-product space is called a pre-Hilbert space or a unitary space.


Definition 20 The magnitude of a vector a in a real inner-product space isa non-negative real number, denoted |a|, given by

|a| =√

a · a.

We will encounter additional examples of inner-product spaces (not based onEuclidean vectors) when we discuss variational formulations. There are twoimportant inequality theorems associated with inner-product spaces.

Theorem 40 (Cauchy-Schwarz (or Bunyakovskii) inequality) Let Vbe a real inner-product space. Then

|a · b| ≤ |a| |b| ∀ a,b ∈ V .

Proof. Case I: Suppose a = 0 and/or b = 0. Then |a · b| = 0, and (|a| = 0and/or |b| = 0). Thus, |a · b| = |a| |b| = 0.

Case II: Suppose a 6= 0 and b 6= 0. Let α and β be arbitrary real numbers.Then αa+βb is a vector in V

⇒ (αa+βb) · (αa+βb) ≥ 0 (inner product is pos. def.)

⇔ α2a · a + αβa · b + βαb · a + β2b · b ≥ 0 (distributivity and homogenity)

⇔ α2 |a|2 + 2αβa · b + β2 |b|2 ≥ 0 (commutativity, def. of magnitude)

Since α and β are arbitrary, we can choose α = |b| and β = − (a · b) / |b| .

⇒ |b|2 |a|2 − 2 (a · b)2 + (a · b)2 ≥ 0 (substitute for α and β)

⇒ |a|2 |b|2 − (a · b)2 ≥ 0

⇒ (a · b)2 ≤ |a|2 |b|2

⇒ |a · b| ≤ |a| |b|(monotonicity of positive square root; i.e., x ≤ y ⇔

√x ≤ √y ∀x ≥ 0)

Theorem 41 (triangle inequality) Let V be a real inner-product space.Then

|a + b| ≤ |a|+ |b| ∀ a,b ∈ V .

Exercise 15 Prove the triangle inequality. Hint: Use the Cauchy-Schwarzinequality to show that |a + b|2 ≤ (|a|+ |b|)2 , which implies the desired resultby the monotonicity of the positive square root.


ab

a+b

Figure 1.5: The triangle inequality

Definition 21 Let V be a real vector space. A norm is an operator, de-noted ‖·‖ , that maps a vector a ∈ V into a non-negative real number withthe properties

1. ‖λa‖ = |λ| ‖a‖ ∀ scalars λ and ∀ vectors a ∈ V (homogeneity),

2. ‖a‖ ≥ 0 ∀ vectors a ∈ V with ‖a‖ = 0 iff a = 0, (positive definiteness)

3. ‖a + b‖ ≤ ‖a‖ + ‖b‖ ∀ vectors a,b ∈ V (satisfaction of the triangleinequality)

Definition 22 A normed vector space is a real vector space equipped with anorm.

When dealing with an inner-product space, it is very common to definethe norm in terms of the inner product:

‖a‖ =√

a · a.

Thus, inner-product spaces are often assumed to be normed spaces by default.However, it is possible to adopt a different definition of the norm on an inner-product space, and it is also possible to define a norm on a space that is notan inner-product space (not all normed spaces are inner-product spaces). Inthe case of our Euclidean vector space, the norm of a vector coincides withits magnitude. That is, we have the Euclidean norm: ‖a‖ = |a| .

1.11. SECOND-ORDER TENSORS 33

Exercise 16 Show that the common definition of the norm in terms of aninner product is homogeneous.

There is one remaining point to consider before proceeding on to second-order tensors. That is, the inner product allows us to interpret a (Euclidean)vector as a linear operator that maps a vector into a real number (scalar).In fact, this is the defining property of a first-order tensor, and vectorsare indeed first order tensors. To illuminate this point, let V be the set ofall vectors in some Euclidean point space E . Now consider a specific vectora ∈ V , where the overbar indicates that we hold a fixed. We can define afunction fa that maps a vector into a scalar by taking the inner product ofa and any vector b ∈ V . That is,

fa (b) ≡ a · b.A review of the properties of the inner product shows that fa is indeed alinear operator. In fact, the Riesz representation theorem states that everylinear function on V to < can be represented in this fashion (by varying ourchoice of the fixed vector a)! We use a similar approach in the next sectionto define second-order tensors as a special class of linear operators.

1.11 Second-Order Tensors

Second-order tensors are common mathematical objects in mechanical de-scriptions of nature. For example, mechanical stress and strain are repre-sented by second-order tensors. In this section we define second-order ten-sors as linear operators that map vectors into vectors (i.e., first-order tensorsinto first-order tensors). This definition is sufficient to develop all of theproperties of second-order tensors without reference to a coordinate frame.Later, we introduce the Cartesian component representation of second-ordertensors for purposes of calculation with indicial notation.

1.11.1 Second–order tensors as linear operators

Definition 23 Let V be the set of all vectors in some Euclidean point spaceE. A second-order tensor T is a linear operator (or function) on V into V.We write Tu to denote the value of the function T at u ∈ V. That T is alinear function implies

T (αu + βv) = αTu + βTv ∀α, β ∈ < and ∀ u,v ∈ V .


Remark 7 We use light-face letters to denote scalars (e.g., a), bold-facelower-case letters for vectors (e.g., a) and bold-face upper-case letters forsecond-order tensors (e.g., A).

Remark 8 We denote the set of all linear functions on V into V as Lin V .That is, Lin V is the set of all second-order tensors on V .

Theorem 42 All second-order tensors map the zero vector into itself; thatis

T0 = 0 ∀ T ∈ Lin V .

Proof. The zero vector may be represented as 0 = 0u + 0v, according tothe properties of the scalar multiple of a vector. Then

T0 = T (0u + 0v) = 0Tu + 0Tv = 0

Definition 24 Two second-order tensors S, T ∈ Lin V are equal iff theyare the same function:

S = T⇔ Su = Tu ∀ u ∈ V .

Definition 25 Let S, T ∈ Lin V and α ∈ < . Then the sum of S and T, thescalar multiple of T by α, the zero second-order tensor 0, and the identitysecond-order tensor I are all second-order tensors defined, respectively, as

1. (S + T) u = Su + Tu ∀ u ∈ V ,

2. (αT) u = α (Tu) ∀u ∈ V ,

3. 0u = 0 ∀ u ∈ V ,

4. Iu = u ∀ u ∈ V .

Definition 26 Subtraction of second-order tensors is defined through

S−T = S + (−T) .

where −T ≡ (−1) T is the negative of the tensor T.

Theorem 43 Second-order tensor addition has the basic properties


1. S + T = T + S (commutativity),

2. (S + T) + U = S + (T + U) (associativity);while scalar multiplication of second-order tensors has the properties

3. (αβ) T =α (βT) (associativity),

4. α (S + T) = αS + αT (distributivity w.r.t. tensor addition),

5. (α + β) T = αT + βT (distributivity w.r.t. scalar addition).

Exercise 17 Prove Theorem 43. Use the definition of tensor equality above.That is, show that two tensors are equal by showing that they represent thesame mapping of an arbitrary vector into a vector. Do not use componentsin your proof, since they have not yet been introduced and because these prop-erties hold independent of any coordinate frame.

Additional properties of second-order tensors are summarized as

Theorem 44 Let T ∈ Lin V and λ ∈ <. Then

1. 0T = 0,

2. λ0 = 0,

3. 1T = T,

4. 0 + T = T + 0 = T,

5. T + (−T) = (−T) + T = 0.


The following definition and theorem give an important result: that asecond-order tensor can be constructed from two vectors.

Definition 27 The tensor (or dyadic) product of u,v ∈ V, denoted u⊗ v,is defined by

(u⊗ v) w ≡ (v ·w) u ∀ w ∈ V .

Theorem 45 Let u,v ∈ V. The tensor product of u and v, u ⊗ v, is asecond-order tensor on V.


Proof. We must show that u⊗v is a function that maps vectors into vectorson V and that the mapping has the linear properties required by Definition 23.The r.h.s. of the expression in Definition 27 is a vector, so u⊗v does indeedmap an arbitrary vector w ∈ V into a vector. To demonstrate linearity,consider arbitrary a,b ∈ V and arbitrary α, β ∈ <. Then

(u⊗ v) (αa + βb) = [v · (αa + βb)] u (definition of tensor product),

= [α (v · a) + β (v · b)] u

(distrib. and homog. of inner product),

= α (v · a) u + β (v · b) u (distrib. of scalar multiple),

= α (u⊗ v) a + β (u⊗ v) b (def. of tensor product).

So the tensor product u ⊗ v is indeed a linear function on V to V and,therefore, a second-order tensor.

Theorem 46 The tensor product is homogeneous and distributive w.r.t.addition:

1. (αu)⊗ (βv) = αβ (u⊗ v) ∀ u,v ∈ V and ∀ α, β ∈ <,

2. u ⊗ (v + w) = u ⊗ v + u ⊗ w and (u + v) ⊗ w = u ⊗ w + v ⊗ w ∀u,v,w ∈ V .


The tensor product is not commutative. That is, u ⊗ v 6= v ⊗ u, since(u⊗ v) w = (v ·w) u while (v ⊗ u) w = (u ·w) v. This demonstrates inequal-ity because (v ·w) u 6= (u ·w) v if, for example, u and v have distinctdirections. We have now established the frame-independent properties ofsecond-order tensors. For convenience of calculation, we next introduce acomponent representation of second-order tensors w.r.t. a given Cartesiancoordinate frame.

Exercise 20 Show that the set<,+,× ,

Lin V ,−→+

, ∗

is a real vector

space, where + and × are addition and multiplication on <, and−→+ and

∗ denote addition and scalar multiplication of second-order tensors as perDefinition 25.


1.11.2 Cartesian component representation of second-order tensors

Let X be a rectangular Cartesian coordinate frame with base vectors ei ina Euclidean point space E , let V denote the set of all vectors in E and letT ∈ Lin V . The following definition and theorem establish the componentrepresentation of a second-order tensor in terms of a two-index symbol andtensor products of the base vectors.

Definition 28 The rectangular Cartesian components of the tensor T w.r.t.X are

Tij = ei ·Tej ≡ ei · (Tej) (i.e., compute Tej first).

Note that we have nine components for a second-order tensor in three di-mensions.

Theorem 47 An arbitrary second-order tensor T ∈ Lin V can be representedin component form w.r.t. any rectangular Cartesian coordinate frame X as

T = Tijei ⊗ ej.

Proof. We will show that T and Tijei ⊗ ej define the same mapping on anarbitrary vector u ∈ V . Thus,

Tu = (Tu)i ei = [ei · (Tu)] ei (vector component)

= ei · [T (ujej)] ei (substitution)

= [ei · (ujTej)] ei (linearity of T)

= [uj (ei ·Tej)] ei (homogeneity and distributivity of inner product)

= [(ej · u) (ei ·Tej)] ei (definition of vector comp.)

= [(ei ·Tej) (ej · u)] ei (commutativity of multiplication in <)

= (ei ·Tej) [(ej · u) ei] (distributivity of scalar multiplication in V)

= (ei ·Tej) [(ei ⊗ ej) u] (definition of tensor product)

= [(ei ·Tej) ei ⊗ ej] u (defs. of tensor addition and scalar mult.)

= (Tijei ⊗ ej) u (definition of tensor components).


Theorem 48 The components of the second-order zero tensor, the second-order identity tensor, the scalar multiple of a second-order tensor, the sumof two second-order tensors and the tensor product of two vectors are

1. 0ij = (0)ij ,

2. Iij = δij,

3. (λT)ij = λTij,

4. (S + T)ij = Sij + Tij,

5. (u⊗ v)ij = uivj,

respectively.


We close this section with two theorems that demonstrate that the com-ponents uniquely characterize a second-order tensor plus one additional theo-rem that provides a component expression for second-order tensor mappingson Lin V to Lin V .

Theorem 49 Two second-order tensors are equal iff their components w.r.t.a given frame are equal; i.e.,

S = T⇔ Sij = Tij.

Proof. Suppose S = T. Then

Sij = ei · Sej (definition of 2nd-order tensor components),

= ei ·Tej (definition of S = T),

= Tij (definition of 2nd-order tensor components).

Conversely, if Sij = Tij, then

S = Sijei ⊗ ej,

= Tijei ⊗ ej, (substitution),

= T


Theorem 50 The components of a second-order tensor are unique w.r.t. agiven frame.

Proof. Suppose the second-order tensor T has components Tij and Tij w.r.t.

the same frame X. Then T = Tijei ⊗ ej = Tijei ⊗ ej. Then, by Theorem 49,

Tij = Tij so the components are unique w.r.t. the given frame.(Tij − Tij

)ei ⊗ ej = Tijei ⊗ ej − Tijei ⊗ ej (distr. of scalar product),

= T−T (by above assumption),

= 0 (Theorem 44),

= (0)ij ei ⊗ ej (theorems 47 and 48),

⇔ Tij − Tij = (0)ij (theorem 49),

⇔ Tij = Tij.

Theorem 51 Let T ∈ Lin V with components Tij w.r.t. the r.C.c.f. X, andlet u,v ∈ V with components ui, vi w.r.t. X. Then

Tu = v⇔ Tijuj = vi.

Proof. Suppose Tu = v. Then

vi = ei · v (components of v),

= ei · (Tu) (supposition),

= ei · [(Tklek ⊗ el) u] (component representation of T),

= ei · [Tkl (ek ⊗ el) u] (defns. of tensor + and scalar mult.),

= ei · [Tkl (el · u) ek] (defn. of ⊗ ),

= ei · [Tkl (ulek)] (components of u),

= Tklul (ei · ek) (distr. and homog. of · ),

= Tklulδik (orthonormality of base vectors),

= Tilul = Tijuj (property of δik and labeling).

∴ Tu = v⇒ Tijuj = vi.


Conversely, suppose Tijuj = vi. Then

v = viei (components of v),

= Tijujei (supposition),

= [Tij (ej · u)] ei (components of u),

= Tij [(ej · u) ei] (distr. of scalar mult.),

= Tij [(ei ⊗ ej) u] (defn. of ⊗),

= [Tij (ei ⊗ ej)] u (defns. of tensor + and scalar mult.),

= Tu (component representation of T).

∴ Tijuj = vi ⇒ Tu = v.

1.11.3 Transformation rules for second-order tensors

As with vectors, the components of a second-order tensor have no meaningunless we specify a coordinate frame, and there is a systematic method forconverting the Cartesian components of a second-order tensor from one frameto another. Let X,ei and

X′, e′i

be arbitrary pairs of associated rectan-

gular Cartesian coordinate frames and base vectors for the 3-D Euclideanpoint space E . The following theorem, analogous to Theorem 30, describesthe transformation of components defined w.r.t. one frame to the other.

Theorem 52 The transformation rules for the rectangular Cartesian com-ponents of a second-order tensor T are

T′

ij = λikλjlTkl; Tij = λkiλljT′

kl,

where T = Tijei⊗ej = T′ije′i⊗e

′j and λij are the cosines of the angles between

e′i and ej.

Proof. We have

T = T′

ije′

i ⊗ e′

j = Tijei ⊗ ej,

= Tij

(λkie

′

k

)⊗(λlje

′

l

)(substitution),

= (λkiλljTij) e′

k ⊗ e′

l (distrib. & homog. props. of ⊗ ; props. of <),

= (λikλjlTkl) e′

i ⊗ e′

j (labeling).

∴ T′

ij = λikλjlTkl (Theorem 50).

Similarly, Tij = λkiλljT′

kl.


Exercise 22 Fill out the details for the second half of the proof of Theorem52. That is, show that Tij = λkiλljT

′

kl.

The following theorem continues the parallel structure with our resultsfor the transformation of vectors.

Theorem 53 Let there be a set of nine real numbers associated with ev-ery r.C.c.f. in a three-dimensional Euclidean point space E. For example,consider the sets X, ei, Tij and

X′, e′i, T

′ij

where X and X

′are arbitrary

frames with respective base vectors ei and e′i. Then ∃ (“there exists”) T ∈

Lin V given by T = Tijei ⊗ ej = T′ije′i ⊗ e

′j iff the two-index symbols Tij and

T′ij satisfy the transformation rules

T′

ij = λikλjlTkl; Tij = λkiλljT′

kl.

for all choices of X, ei, Tij and

X′, e′i, T

′ij

, where λij are the cosines of

the angles between e′i and ej.

Proof. Suppose Tij and T′ij are the components of a single second-order

tensor T w.r.t. X and X′. Then the components must be related by the

transformation rules, T′ij = λikλjlTkl and Tij = λkiλljT

′

kl, according to Theo-

rem 52. Conversely, let T = Tijei ⊗ ej and T′

= T′ije′i ⊗ e

′j be two possibly

distinct tensors whose components are related by the transformation rules.We must show that T

′= T to complete the proof. We have

T′

= T′

ij

(e′

i ⊗ e′

j

),

= (λikλjlTkl) e′

i ⊗ e′

j (substitution)

= Tkl

[(λike

′

i

)⊗(λjle

′

j

)](distrib. & homog. props. of ⊗ ; props. of <)

= Tkl (ek ⊗ el) (theorem 29)

= T (substitution).

Remark 9 Theorem 53 is sometimes used as a definition for second-ordertensors. That is, a second-order tensor is a mathematical object whose rect-angular Cartesian components transform under a change of frame accordingto the rules of Theorem 52. This definition is equivalent to the direct defi-nition of a second-order tensor as a linear operator on Lin V to Lin V givenin Definition 23.


1.11.4 The transpose and symmetry of a second-ordertensor

We begin with the direct, frame-independent definition of the transpose ofa tensor, and then consider the component representation of the transpose.This leads to the definition of symmetric and skew second-order tensors andto the decomposition of an arbitrary second-order tensor into symmetric andskew parts.

Definition 29 Let T ∈ Lin V . Then the transpose of T, denoted Tt, is amapping from V into V with the property

u ·Ttv = v ·Tu ∀ u,v ∈ V .

Theorem 54 Given T ∈ Lin V, then Tt ∈ Lin V (that is, Tt is a second-order tensor).

Proof. According to the definition, Tt is a function on V into V . It remainsto prove that it is linear.

w ·Tt (αu + βv) = (αu + βv) ·Tw ∀ u,v,w ∈ V ; ∀ α, β ∈ <(def. of transpose),

= α (u ·Tw) + β (v ·Tw) ∀ u,v,w ∈ V ; ∀ α, β ∈ <(distrib. and homog. of inner product),

= α(w ·Ttu

)+ β

(w ·Ttv

)∀ u,v,w ∈ V ; ∀ α, β ∈ <

(def. of transpose),

= w ·(αTtu

)+ w ·

(βTtv

)∀ u,v,w ∈ V ; ∀ α, β ∈ <

(homog. of inner product),

= w ·(αTtu + βTtv

)∀ u,v,w ∈ V ; ∀ α, β ∈ <

(distrib. of inner product).

Hence,

Tt (αu + βv) = αTtu + βTtv ∀ u,v ∈ V ; ∀ α, β ∈ <(consider arbitrary w).

∴ Tt ∈ Lin V (def. of second-order tensor)

The following theorem summarizes some of the properties of the transpose.


Theorem 55 Let α ∈ <; S,T ∈ Lin V and u,v ∈ V. Then

1. (Tt)t

= T;

2. (S + T)t = St+Tt;

3. (αT)t = αTt (homogeneity of transpose w.r.t. scalar mult.);

4. It = I; (where I is the second-order identity tensor);

5. 0t = 0; (where 0 is the second-order zero tensor);

6. (u⊗ v)t = v ⊗ u.


The following theorem describes the component form of the transposeoperator.

Theorem 56 Let Sij and Tij be the components of two second-order tensorsS and T w.r.t. an arbitrary r.C.c.f. X, ei. Then

S = Tt iff Sij = Tji.

Proof. Suppose S = Tt. Then

S = Tt ⇔ u · Sv = v ·Tu ∀ u,v ∈ V (definition of transpose),

⇔ uiSijvj = viTijuj ∀ ui,vi (substitution of theorems 33 and 51),

⇔ (uivj)Sij = (uivj)Tji ∀ ui,vi (labeling, props. of <),

⇔ Sij = Tji (consider permutations of ui,vi = (1, 0, 0) , (0, 1, 0) , (0, 0, 1) .∴ S = Tt ⇔ Sij = Tji.

Theorem 57 The components of the transpose w.r.t. two r.C.c.f.’s X, eiand

X′, e′i

transform as(

T t)′ij

= λikλjlTtkl = λikλjlTlk.

Proof. Tt is a second-order tensor by definition; so it must follow thestandard transformation rules. Substitution of Theorem 56 completes theproof.


We are now ready to introduce symmetric and skew second-order tensors.

Definition 30 A second-order tensor T is symmetric iff

Tt = T⇔ Tji = Tij;

and skew (or skewsymmetric or antisymmetric) iff

Tt = −T⇔ Tji = −Tij.

Theorem 58 A second-order tensor T admits the unique decomposition intothe sum of a symmetric and a skew second-order tensor,

T = symT + skewT,

where symT := 12

(T + Tt) and skewT := 12

(T−Tt

)are called the sym-

metric and skew parts of T.

Proof. This follows directly from theorems 2 and 49 if we work with com-ponent notation.

Exercise 24 Prove Theorem 58 using direct notation.

Sometimes we wish to refer to the sets of all symmetric (or skew) second-order tensors. Thus we define the sets

Sym ≡T ∈ Lin V : Tt = T

and Skew ≡

T ∈ Lin V : Tt = −T

.

Remark 10 The notation x ∈ A : P1 (x) , P2 (x) , . . . is to be read, “the setof all x in the set A with the properties P1 (x) , P2 (x) , . . . .” This notationallows us to specify subsets of a larger set by specifying one or more propertiesthat identify the elements of the subset.

1.11.5 The product of two second-order tensors

Definition 31 Let S,T ∈ Lin V. Their product, denoted ST, is definedby their composition

(ST) u = S (Tu) ∀ u ∈ V .


Theorem 59 Let S,T ∈ Lin V. Then ST ∈ Lin V (the product of twosecond-order tensors is a second-order tensor).


The following theorem summarizes some of the properties of the productof two second-order tensors.

Theorem 60 Let R,S,T ∈ Lin V and α, β ∈ <. Also, let I be the second-order identity tensor. Then

1. (RS) T = R (ST) =: RST (associativity),

2. T (R + S) = TR + TS; (R + S) T = RT + ST (distributivity w.r.t.tensor addition),

3. (αR) (βS) = αβ (RS) (homogeneity),

4. IT = TI = T (I is the identity element for multiplication of second-order tensors),

5. (RS)t = StRt (the transpose of the product equals the reverse–orderproduct of the transposes).


Theorem 61 Let S,T ∈ Lin V. The rectangular Cartesian components ofthe product ST w.r.t. a given r.C.c.f. X, ei are

(ST)ij = SikTkj,

where Sik and Tkj are the components of S and T. w.r.t. X.

Proof. According to Definition 28, the components of ST w.r.t. the frameX, ei are given by

(ST)ij = ei · (ST) ej

= ei · S (Tej) (definition of tensor product),

= ei · S (Tej)k ek (component form of vector Tej),

= ei · S [ek · (Tej)] ek (Theorem 36),

= ei · S (Tkjek) (Definition 28),

= ei · (TkjSek) (linearity of S),

= Tkjei · Sek (linearity of inner product),

= TkjSik = SikTkj (Definition 28, properties of <).


Remark 11 The product of two second-order tensors is not commutative.

1.11.6 The determinant of a second-order tensor

We define the determinant of a second-order tensor as a scalar (i.e., a frame-independent real number).

Definition 32 Given T ∈ Lin V with components Tij w.r.t. an arbitraryr.C.c.f. X, ei , the determinant of T, denoted det T, is given by

det T = det [Tij] .

Theorem 62 The determinant of a second-order tensor is a scalar. Thatis, the determinant is invariant under a change of coordinate frame.

Proof. Let T ∈ Lin V have components Tij and T′ij w.r.t. r.C.c.f.’s X, ei

and

X′, e′i

. Then by Definition 32,

det T = det[T′

ij

],

= εijkT′

1iT′

2jT′

3k (row expansion),

= εijk (λ1rλilTrl) (λ2mλjnTmn) (λ3pλkqTpq) (transformation rules),

= (εijkλilλjnλkq) (TrlTmnTpq) (λ1rλ2mλ3p) (properties of <),

= (det [λst] εlnq) (TrlTmnTpq) (λ1rλ2mλ3p) (Theorem 8),

= det [λst] (εlnqTrlTmnTpq) (λ1rλ2mλ3p) (properties of <),

= det [λst] (det [Tuv] εrmp) (λ1rλ2mλ3p) (Theorem 8),

= det [λst] det [Tuv] (εrmpλ1rλ2mλ3p) (properties of <),

= det [λst] det [Tuv] det [λpq] = det [λst]2 det [Tuv] (row expansion),

= det [Tuv] (exercise 11).

The properties of the determinant of a second-order tensor can be easilyascertained from the properties of the determinant of a matrix.

Theorem 63 Let S,T ∈ Lin V . Then

1. det (αT) = α3 det T;


2. det (Tt) = det T;

3. det (ST) = det S det T;

4. det I =1;

5. det 0 =0;

6. det (u⊗ v) = 0 ∀u,v ∈ V .


1.11.7 The trace of a second-order tensor

Definition 33 Let T ∈ Lin V. Then the trace of T, denoted tr T, is a realnumber with the properties

tr (αS + βT) = α tr S + βtr T ∀ α, β ∈ <;∀S,T ∈ LinV (linearity);

tr (u⊗ v) = u · v ∀ u,v ∈ V .

Theorem 64 tr (u⊗ v) = tr ( v ⊗ u) ∀ u,v ∈ V .

Proof. Let u,v ∈ V . Then

tr (u⊗ v) = u · v (Definition 33),

= v · u (commutativity of inner product),

= tr (v ⊗ u) (Definition 33).

Theorem 65 Given T ∈ Lin V with components Tij w.r.t. an arbitraryr.C.c.f. X, ei ,

tr T =Tii.

Proof.

tr T = tr (Tijei ⊗ ej) (comp. rep. of 2nd-order tensor),

= Tij tr (ei ⊗ ej) (Definition 33 – linearity),

= Tijei · ej (Definition 33),

= Tijδij (orthogonality of base vectors),

= Tii (property of Kronecker’s delta).


Theorem 66 The trace of a second-order tensor is invariant under a changeof coordinate frame. That is, the trace is a scalar.

Proof. Let T ∈ Lin V have components Tij and T′ij w.r.t. r.C.c.f.’s X, ei

and

X′, e′i

. Then

tr T = T′

ii (theorem 65),

= λikλilTkl (transformation rule),

= δklTkl (orthogonality conditions),

= Tkk = Tii (prop. of Kronecker’s delta, labeling).

Theorem 67 Properties of the trace operator include

1. tr Tt= tr T;

2. tr (ST) = tr (TS) ;

3. tr I = 3;

4. tr 0 = 0.

Exercise 28 Use Theorem 65 to prove Theorem 67.

1.11.8 The inner product of two second-order tensors

Here we extend the concept of the inner product operation to second-ordertensors. It will come as no surprise that this leads us to define an innerproduct space for second-order tensors.

Definition 34 The inner product of two second-order tensors S,T ∈ LinV, denoted S ·T, is a real number with the property

S ·T = tr(STt

).

Theorem 68 Properties of the inner product include

1. T · S = S ·T (commutativity);

2. R· (S + T) = R · S + R ·T (distributivity w.r.t. tensor addition);


3. (αS) ·T =α (S ·T) (homogeneity);

Exercise 29 Prove Theorem 68 without resorting to component representa-tion. Hint: You might find some of the properties of the trace and of theproduct of two second-order tensors helpful.

Theorem 69 Given S,T ∈ Lin V with components Sij, Tij w.r.t. an arbi-trary r.C.c.f. X, ei ,

S ·T = SijTij.


Theorem 70 T ·T ≥ 0 with T ·T = 0 iff T = 0 (positive definiteness).

Proof. T ·T = TijTij, which is the sum of the squares of the components.The sum must be greater than zero iff any of the components are non-zero,in which case T 6= 0. The sum equals zero iff all the components are zero, inwhich case T = 0.

Theorem 71 The inner product of two second-order tensors is a scalar.

Proof. Given S,T ∈ Lin V with components Sij, Tij w.r.t. frame X, eiand components S

′ij, T

′ij w.r.t. frame

X′, e′i

. Then

S ·T = S′

ijT′

ij (theorem 69),

= (λimλjnSmn) (λikλjlTkl) (Theorem 53),

= (λimλikSmn) (λjlλjnTkl) (properties of <),

= (δmkSmn) (δlnTkl) (orthogonality condition),

= SknTkn = SijTij (prop. of Kronecker’s delta, labeling).

We can construct an inner product space by equipping the real vectorspace Lin V with the inner product of Definition 34. Then the magnitude ofa second-order tensor T ∈Lin V , denoted |T| , is given by

|T| =√

T ·T,

and the standard inner-product norm is given by

‖T‖ = |T| =√

T ·T.

Exercise 31 Show that the real vector space Lin V equipped with the innerproduct defined in 34 is an inner product space.


1.11.9 The inverse of a second-order tensor

We will again use the direct approach to define the inverse before developingthe component representation as a theorem.

Definition 35 Let T ∈ Lin V with det T 6= 0. Then the inverse of T, de-noted T−1, is a mapping on V to V, with the property

T−1 (Tu) = T(T−1u

)= u ∀ u ∈ V.

Remark 12 The requirement that det T 6= 0 guarantees the existence of theinverse.

Theorem 72 Let T ∈ Lin V with det T 6= 0. Then

T−1T = TT−1 = I.

Proof. By the definition of the product of two second-order tensors, thedefining property is equivalent to,(

T−1T)

u =(TT−1

)u = u ∀ u ∈ V ,

which in turn implies the desired result by the definition of the identitytensor.

Theorem 73 Let u,v ∈ V and T ∈ Lin V with det T 6= 0. Then

Tu = v⇔ T−1v = u.

Proof. Suppose Tu = v. Then

T−1v = T−1 (Tu) (substitution),

=(T−1T

)u (def. of tensor multiplication),

= Iu (Theorem 72),

= u (definition of the identity tensor).

Thus, Tu = v⇒ T−1v = u. Now suppose T−1v = u. Then

Tu = T(T−1v

)(substitution),

=(TT−1

)v (def. of tensor multiplication),

= Iv (Theorem 72),

= v (definition of the identity tensor).

Thus, T−1v = u⇒ Tu = v. The two results together demonstrate thatTu = v⇔ T−1v = u.


Theorem 74 Let T ∈ Lin V with det T 6= 0. Then T−1 is a second-ordertensor.

Proof. By definition, the structure of the mapping T−1 qualifies it to bea second-order tensor. It remains to show that it is linear. To this end, leta,b ∈ V and α, β ∈ <. Also, define u ≡ T−1a and v ≡ T−1b. Note thatu,v ∈ V by the definition of T−1. Then

a = Tu and b = Tv (Theorem 73).

Hence,

T (αu + βv) = α Tu + βTv (linearity of T),

= αa + βb (substitution).

Therefore,

T−1 (αa + βb) = αu + βv (def. of the inverse),

= αT−1a + βT−1b (substitution).

We have thus demonstrated linearity, which proves that T−1 is a second-ordertensor.

The next theorem summarizes some additional properties of the inverseof a second-order tensor.

Theorem 75 Let S,T ∈ Lin V with det S 6= 0 and det T 6= 0. Then

1. det (T−1) = (det T)−1 6= 0;

2. (T−1)t

= (Tt)−1

=: T−t;

3. det (ST) 6= 0 and (ST)−1 = T−1S−1.

Proof.

1. We have

det(T−1

)det T = det

(T−1T

)(Theorem 63),

= det I (Theorem 72),

= 1 (Theorem 63).

Hence, det (T−1) det T = 1. Accordingly, det (T−1) = (det T)−1 6= 0.


The remainder of the proof is left to the following exercises.

Exercise 32 Prove Theorem 75.2.

Exercise 33 Prove Theorem 75.3.

Remark 13 The astute reader will notice a close correspondance betweenthe properties of the inverse of a second-order tensor and the properties ofthe inverse of the matrix containing the tensor’s components. This is, ofcourse, no accident.

Remark 14 Next we consider the component representation of the inverseof a second-order tensor.

Theorem 76 Given a r.C.c.f. X, ei , let T = Tijei⊗ ej such that det T 6=0. Then the rectangular Cartesian components of T−1 w.r.t. X are(

T−1)ij

=: T−1ij =

1

2 det TεipqεjmnTmpTnq. (cf. Theorem 16.)

Proof. The object T−1 = T−1ij ei⊗ej is clearly a second-order tensor (since it

represents a sum of scalar multiples of the nine dyadic products ei⊗ej, whichare themselves second-order tensors by Theorem 45). Thus, this descriptionof T−1 satisfies the linearity requirement of the definition of the inverse. Itremains for us to show that T−1 (Tu) = T (T−1u) = u ∀ u ∈ V . Given anarbitrary u ∈ V we have

T−1 (Tu) =(T−1ij ei ⊗ ej

)[Tkl (ulek)] (component forms of T−1 and Tu),

= T−1ij Tklul (ei ⊗ ej) ek

(linearity of tensors, homogeneity of tensor multiplication),

= T−1ij Tklul (ej · ek) ei (Definition 27),

= T−1ij Tklulδjkei (orthogonality),

= T−1ij Tjlulei (property of Kronecker’s delta),

= δilulei (Definition 5 and Theorem 16),

= uiei (property of Kronecker’s delta),

= u (component form of u).

Similarly, T (T−1u) = u ∀ u ∈ V .


To close this subsection, we define two subsets of LinV . First, the set ofall second-order tensors on V with strictly positive determinants,

LinV+ := T ∈LinV : det T > 0 ,

and the set of all invertible second-order tensors on V ,

InvV := T ∈LinV : det T 6= 0 .

1.11.10 Orthogonal second-order tensors.

Continuing the parallel development of second-order tensors and matrices,we now consider orthogonal second-order tensors (c.f. Definition 16).

Definition 36 A second-order tensor T is orthogonal iff

TtT = TTt= I.

Definition 37 The set of all orthogonal second-order tensors, denoted OrthV , is defined by

OrthV ≡ T ∈LinV : T is orthogonal .

Theorem 77 Let T ∈ Orth V. Then

det T =± 1.

Proof. By Definition 36 and T ∈ Orth V we have

TTt = I (Definition 36);

⇒ det(TTt

)= det I = 1 (Theorem 63.4);

⇒ det T det Tt = 1 (Theorem 63.3);

⇒ det T det T = (det T)2 = 1 (Theorem 63.2);

⇒ det T = ±1.

Definition 38 An orthogonal second-order tensor T is proper iff

det T = +1,

and improper iffdet T = −1.


Definition 39 The set of all proper orthogonal second-order tensors, de-noted Orth V+, is defined by

Orth V+ ≡ T ∈Orth V : det T = +1 .

Orthogonal tensors possess special geometrical properties that will playan important role in our development of kinematics. The following theoremstates that these properties are equivalent characterizations of an orthogonalsecond-order tensor.

Theorem 78 Let T ∈LinV . Then the following statements are equivalent.

1. T ∈OrthV ,

2. T preserves inner products ; i.e.,

Tu ·Tv = u · v ∀ u,v ∈ V ,

3. T preserves magnitudes ; i.e.,

|Tu| = |u| ∀ u ∈ V ,

4. T preserves distances ; i.e.,

|Tu−Tv| = |u− v| ∀ u,v ∈ V ,

Proof. It suffices to show that

1. 78.1⇔78.2,

2. 78.2⇒78.3,

3. 78.3⇒78.4 and

4. 78.4⇒78.2.

The circular implication structure of the last three statements implies thatthe statements 78.2, 78.3 and 78.4 are equivalent.


First we show that 78.1⇔78.2. Suppose T ∈OrthV , and select arbitraryu,v ∈ V . Then

Tu ·Tv = v·[Tt (Tu)

](Definition 29 of tensor transpose),

= v·[(

TtT)u]

(definition of tensor multiplication),

= v· [(I) u] (T ∈OrthV),

= v · u (definition of I),

= u· v (commutativity of inner product).

Next, suppose that T preserves inner products and again select arbitraryu,v ∈ V . Then

u · v = Tu ·Tv

= v·[Tt (Tu)

](Definition 29 of tensor transpose),

= v·[(

TtT)u]

(definition of tensor multiplication),

=[(

TtT)u]·v (commutativity of inner product).

⇔(TtT

)u = u ∀ u ∈ V (Theorem 39),

⇔ TtT = I (Definition 25.4 of identity tensor), (*)

⇔ TTt = I (see exercise below), (**)

⇔ T ∈OrthV . ( def. of OrthV , (*) and (**))

We have shown that 78.1⇒78.2 and that 78.2⇒78.1; hence, 78.1⇔78.2.Next we show that 78.4⇒78.2. Suppose that T preserves distances; that

is, |Tu−Tv| = |u− v| ∀ u,v ∈ V . This implies

|Tu| = |u| ∀ u ∈ V (set v = 0),

⇒ |Tu|2 = |u|2 ∀ u ∈ V (square both sides),

⇒ Tu ·Tu = u · u ∀ u ∈ V (*).

Also,

|Tu−Tv| = |u− v| ∀ u,v ∈ V ,

⇒ (Tu−Tv) · (Tu−Tv) = (u− v) · (u− v) ∀ u,v ∈ V ,

⇒ Tu ·Tu− 2Tu ·Tv + Tv ·Tv = u · u− 2u · v + v · v ∀ u,v ∈ V ,

⇒ Tu ·Tv = u · v ∀ u,v ∈ V (apply (*) twice, divide by − 2).

Thus we have shown the desired result: |Tu−Tv| = |u− v| ∀ u,v ∈ V ⇒Tu ·Tv = u · v ∀ u,v ∈ V .

The remainder of the proof is left to the following exercises.


Exercise 34 Show that TtT = I⇔ TTt = I.

Exercise 35 Show that Tu ·Tv = u · v ∀ u,v ∈ V ⇒ |Tu| = |u| ∀ u ∈ V .

Exercise 36 Show that |Tu| = |u| ∀ u ∈ V ⇒ |Tu−Tv| = |u− v| ∀ u,v ∈V .

1.12 Higher-Order Tensors

This section extends our results for scalars, vectors and second-order tensorsto tensors of arbitrary order m. Once again, we use the direct approachto introduce higher-order tensors as linear transformations, independent ofany coordinate frame. Then we define their components w.r.t. a r.C.c.f.and the generalization of the coordinate transformation rules. We introducethe alternating tensor, the tensor analogue of the alternating symbol, as anexample of a third-order tensor. After a discussion of the vector product, thesection closes with an expanded view of tensors as multiple linear functionsthat map tensors to tensors.

1.12.1 Introduction to mth-order tensors

Based on our definition of vectors, we introduce a recursive definition ofhigher-order tensors.

Definition 40 Given a set of Euclidean vectors V and an integer m ≥ 1, anmth-order tensor is a linear function that maps elements of V into tensorsof order m − 1. Tensors of order 1 are vectors, and tensors of order 0 are(formally) defined to be scalars.

We have already seen how this definition applies to the case m = 1, wherevectors can be viewed as linear functions that map vectors into scalars viathe inner product, as described by the Riesz representation theorem. Ourdirect definition 23 of second-order tensors also conforms to this extendeddefinition. We shall denote a typical mth -order tensor (m > 2) as

m

T, and its

value at u ∈ V asm

T u.

Definition 41 Two mth-order tensorsm

S andm

T are said to be equal iff

m

S u =m

T u ∀ u ∈ V.

1.12. HIGHER-ORDER TENSORS 57

Definition 42 The sum of two mth-order tensorsm

S andm

T is the mappingon V to tensors of order m− 1 defined by(m

S +m

T)

u =m

S u+m

T u ∀ u ∈ V.

Definition 43 Let λ ∈ <. Then the scalar multiple of an mth-order tensorm

T by λ is the mapping on V to tensors of order m− 1 defined by(λ

m

T)

u = λ(mT u

).

Definition 44 The zero mth-order tensor, denotedm

0, is the mapping on Vto tensors of order m− 1 defined by

m

0 u =m−1

0 ∀ u ∈ V.

Remark 15 The identity tensor exists only for the case m = 2. Why canthere be no identity tensor for m > 2, according to Definition 40? Weshall see that there can be higher-order identity tensors for certain values ofm > 2, if we adopt the expanded definition of an mth-order tensor advancedin subsection 1.12.4.

Remark 16 The requirement in Definition 40 that any mth-order tensorm

Tbe a linear function implies that

m

T (αu + βv) = αm

T u + βm

T v ∀ α, β ∈ < and ∀ u,v ∈ V.

The following theorem generalizes previous results for second-order ten-sors.

Theorem 79 Letm

S,m

T andm

U be mth-order tensors, and let α, β ∈ <. Thenthe algebraic properties of mth-order tensors include

1.m

S +m

T=m

T +m

S (commutativity of tensor addition),

2.(mS +

m

T)

+m

U=m

S +(mT +

m

U)

(associativity of tensor addition),

3. (αβ)m

T= α(β

m

T)

(associativity scalar multiple w.r.t. multiplication

of scalars),


4. α

(m

S +m

T

)= α

m

S +αm

T (distributivity of scalar multiple w.r.t. ten-

sor addition),

5. (α + β)m

T= αm

T +βm

T (distributivity of scalar multiple w.r.t. scalaraddition).


Further, we have

Theorem 80 For any mth-order tensorm

T and for any λ ∈ <,

1. 0m

T=m

0,

2. λm

0=m

0,

3. 1m

T=m

T,

4.m

0 +m

T=m

T +m

0=m

T,and

5.m

T +(−

m

T)

= −m

T +m

T=m

0, where −m

T:= (−1)m

T .


Theorem 81 The sum of two mth-order tensors, the scalar multiple of anmth-order tensor and the zero mth-order tensor are all mth-order tensors.

Proof. (1) Letm

S andm

T be mth -order tensors. Then the sum,m

S +m

T, hasthe right structure to be an mth-order tensor: a mapping on V into tensorsof order m − 1. We must also demonstrate linearity. For arbitrary u,v ∈ Vand for arbitrary α, β ∈ < we have(mS +

m

T)

(αu + βv) =m

S (αu + βv) +m

T (αu + βv) (defn. of sum),

= αm

S u + βm

S v + αm

T u + βm

T v (linearity),

=(αm

S u + αm

T u)

+(βm

S v + βm

T v)

(comm., assoc.),

= α(mS u+

m

T u)

+ β(mS v+

m

T v)

(distrib. of scalar mult. w.r.t. tensor addition),

= α(mS +

m

T)

u+β(mS +

m

T)

v (defn. of sum).


Thus the sum is a linear function and hence an mth-order tensor. The proofof the remaining parts of the theorem are left as an exercise.

Exercise 39 Use direct notation to prove that the scalar multiple of anmth-order tensor and the zero mth-order tensor are both mth -order tensors.

Before describing component representations of mth-order tensors, we in-troduce the mth-order version of the tensor product. The definition is in-ductive. Starting with vectors and the inner product of two vectors, themth-order tensor product for m ≥ 2 is defined in terms of the (m− 1)th-order tensor product.

Definition 45 Let u1,u2,. . . ,um ∈ V for m ≥ 2. Then the mth -order tensorproduct (or polyad), denoted u1 ⊗ u2⊗ . . . ⊗ um, is the function that mapselements of V to (m− 1)th-order tensor products according to

(u1 ⊗ u2 ⊗ . . .⊗ um) v = (um · v) u1 ⊗ u2 ⊗ . . .⊗ um−1.

For the case m = 2 we have

(u1 ⊗ u2) v = (u2 · v) u1,

which coincides with our previous definition of the dyadic product. Thisresult only involves the familiar inner product and scalar product for vectors.This second-order result allows us to interpret the defintion for m = 3:

(u1 ⊗ u2 ⊗ u3) v = (u3 · v) u1 ⊗ u2.

Thus, by induction we can interpret the definition of an mth-order tensorproduct for any integer value of m ≥ 2..

Theorem 82 The mth-order tensor product u1 ⊗ u2 ⊗ . . . ⊗ um is an mth-order tensor.

Proof. The function specified in Definition 45 has the right structure; ourobjective here is to prove linearity. We have previously shown that the dyadicproduct (i.e., tensor product with m = 2) is a second-order tensor and there-fore linear. Accordingly, this proof is obtained by induction if we can proveu1 ⊗ u2 ⊗ . . . ⊗ um is an mth-order tensor given that u1 ⊗ u2 ⊗ . . . ⊗ um−1

is an (m− 1)th-order tensor. Thus, suppose that u1 ⊗ u2 ⊗ . . . ⊗ um−1 is


an (m− 1)th-order tensor. Then for arbitrary v,w ∈ V and for arbitraryα, β ∈ <,

u1 ⊗ u2 ⊗ . . .⊗ um (αv+βw) = um · (αv+βw) (u1 ⊗ u2 ⊗ . . .⊗ um−1)

(definition),

= (αum · v+βum ·w) (u1 ⊗ u2 ⊗ . . .⊗ um−1)

(props. of inner prod.),

= αum · v (u1 ⊗ u2 ⊗ . . .⊗ um−1)

+βum ·w (u1 ⊗ u2 ⊗ . . .⊗ um−1)

(Theorem 79.5),

= α (u1 ⊗ u2 ⊗ . . .⊗ um) v+β (u1 ⊗ u2 ⊗ . . .⊗ um) w

(definition).

∴ u1⊗ u2⊗ . . .⊗ um−1 is an (m− 1)th-order tensor ⇒ u1⊗ u2⊗ . . .⊗ um isan mth-order tensor. This completes the proof by induction.

Theorem 83 All mth-order tensor products are homogeneous

(α1u1)⊗ (α2u2)⊗ . . .⊗ (αmum) = α1α2 . . . αm (u1 ⊗ u2 ⊗ . . .⊗ um) ,

and distributive w.r.t. vector addition

u1 ⊗ u2 ⊗ . . .⊗ um−1 ⊗ (v + w) = u1 ⊗ u2 ⊗ . . .⊗ um−1 ⊗ v

+u1 ⊗ u2 ⊗ . . .⊗ um−1 ⊗w;

(v + w)⊗ u1 ⊗ u2 ⊗ . . .⊗ um−1 = v ⊗ u1 ⊗ u2 ⊗ . . .⊗ um−1

+w ⊗ u1 ⊗ u2 ⊗ . . .⊗ um−1,

where α1, α2, . . . , αm ∈ < and u1,u2, . . . ,um,v,w ∈ V are arbitrary.

Remark 17 By the inductive definition of the mth-order tensor product, thedistributive property in Theorem 83 holds for any term in the tensor product.For example,

u1 ⊗ (v + w)⊗ u2 = u1 ⊗ v ⊗ u2 + u1 ⊗w ⊗ u2.



1.12.2 Components of mth-order tensors; transforma-tion rules

Here we extend the rectangular Cartesian component representation of second-order tensors to component representations of tensors of an order m ≥ 2.

Definition 46 Let X, ei be a r.C.c.f. in a Euclidean point space, and letm

Tbe an arbitrary mth-order tensor with m ≥ 2. Then the rectangular Cartesian

components ofm

T w.r.t. X, denotedm

T i1i2...im , are the m-index symbol definedby

m

T i1i2...im= ei1 ·[(((m

T eim

)eim−1

). . . ei2

)],

where i1, i2, . . . , im are distinct unspecified indices.

Theorem 84 Letm

T be an arbitrary mth -order tensor, m ≥ 2, with rectan-

gular Cartesian componentsm

T i1i2...im w.r.t. a given frame X, ei . Then the

rectangular Cartesian component representation ofm

T is given by the equality

m

T=m

T i1i2...im ei1 ⊗ ei2 ⊗ . . .⊗ eim .

Note thatm

T is, in general, a weighted sum of polyads and that this resultis consistent with Theorem 47 for m = 2. Let’s prove the theorem for m = 3.Let u ∈ V be arbitrary and recall that a third-order tensor maps a vector


into a second-order tensor. We have ∀u ∈ V ,

3

T ijk (ei ⊗ ej ⊗ ek) u = ei ·(

3

T ek

)ej

(ei ⊗ ej ⊗ ek) u

(defn. of 3rd-order tens. comps.),

= ei ·(

3

T ek

)ej

(ek · u) (ei ⊗ ej)

(defn. of 3rd-order tens. prod.),

= ei ·(

3

T ek

)ej

uk (ei ⊗ ej)

(defn. of vector comp.),

= ei ·[

3

T (ukek)

]ej

ei ⊗ ej

(linearity of3

T and inner prod.),

= ei ·[(

3

T u

)ej

]ei ⊗ ej (comp. rep. of vector),

=

(3

T u

)ij

ei ⊗ ej (defn. of 2nd-order tensor comps.),

=3

T u (comp. rep. of 2nd-order tensor).

∴3

T ijk (ei ⊗ ej ⊗ ek) =3

T (defn. of tensor equality).

Thus, things look good for m = 2, 3. The proof for m = 3 provides thepattern for a proof by induction.

Exercise 41 Complete the proof of Theorem 84 by induction. That is, as-sume the theorem holds for tensors of order m, and then show that it mustalso hold for tensors of order m+ 1.

The next three theorems parallel our development for second-order ten-sors and are stated here without proof. Of course, the diligent student willtake these on as exercises in her abundant spare time!

Theorem 85 Two mth-order tensors are equal iff their components w.r.t. agiven r.C.c.f. are equal.


Theorem 86 The components of an mth-order tensor w.r.t. a given r.C.c.f.are unique.

Theorem 87 The rectangular Cartesian components of the mth-order zerotensor, the scalar multiple of an mth-order tensor, the mth-order tensor sumand the tensor product of m vectors are

1.m

0i1i2...im= (0)i1i2...im (that is, all components are zero),

2.(α

m

T)i1i2...im

= αm

T i1i2...im ∀α ∈ <,

3.(mS +

m

T)i1i2...im

=m

Si1i2...im +m

T i1i2...im ,

4. (u1 ⊗ u2 ⊗ . . .⊗ um)i1i2...im = (u1)i1 (u2)i2 . . . (um)im .

In contrast to our development for second-order tensors, the latter theoremomits any mention of the identity tensor because, in general, there is nomth-order identity tensor.

Next we consider the transformation rules for the components of an mth

-order tensor.

Theorem 88 Let X,ei and

X′, e′i

be arbitrary pairs of associated rect-

angular Cartesian coordinate frames and base vectors for the 3-D Euclideanpoint space E . The transformation rules for the rectangular Cartesian com-

ponents of an mth -order tensorm

T are

m

T ′i1i2...im = λi1j1λi2j2 . . . λimjmm

T j1j2...jm ,m

T i1i2...im = λj1i1λj2i2 . . . λjmimm

T ′j1j2...jm ,

wherem

T =m

T i1i2...im ei1 ⊗ ei2 ⊗ . . .⊗ eim =m

T ′i1i2...im e′i1⊗ e

′i2⊗ . . .⊗ e

′im and

λij are the cosines of the angles between e′i and ej.

Proof. Essentially the same as Theorem 52.

Theorem 89 Let there be a set of 3m real numbers associated with everyr.C.c.f. in a three-dimensional Euclidean point space E. For example, con-

sider the sets

X, ei,m

T i1i2...im

and

X′, e′i,m

T ′i1i2...im

where X and X

′are


arbitrary frames with respective base vectors ei and e′i. Then ∃ an mth-order

tensor given bym

T=m

T i1i2...im ei1⊗ei2⊗. . .⊗eim =m

T ′i1i2...im e′i1⊗e

′i2⊗. . .⊗e

′im

iff the indexed symbolsm

T i1i2...im andm

T ′i1i2...im satisfy the transformation rules

of Theorem 88 for all choices of

X, ei,m

T i1i2...im

and

X′, e′i,m

T ′i1i2...im

,

where λij are the cosines of the angles between e′i and ej.

Proof. Essentially the same as Theorem 53.

1.12.3 The alternating tensor; restrictions on changesof frame

An important example of a higher-order tensor is the third-order alternatingtensor.

Definition 47 Let X, ei be a r.C.c.f. in a three-dimensional Euclidean

point space E. Then the alternating tensor, denoted3

E, is the third-ordertensor defined by

3

E= εijkei ⊗ ej ⊗ ek.

It turns out that the alternating tensor has a very simple transformationrule; especially if we restrict our choice of coordinate frames to those thatare related by proper orthogonal matrices (i.e., det [λij] = +1).

Theorem 90 Let X, ei and

X′, e′i

be rectangular Cartesian frames such

that λij = e′i · ej defines a proper orthogonal transformation. Then the com-

ponents of the alternating tensor are the same in both systems. That is,

ε′

ijk = εijk.

Proof.

ε′

ijk = λipλjqλkrεpqr (Theorem 88),

= det [λst] εijk (Theorem 8),

= +1εijk = εijk (Theorem 11, proper by assumption).

If we had not restricted the choice of coordinate systems to those relatedby proper orthogonal transformations, we would have the more general, but


less convenient, result ε′

ijk = ±εijk. To avoid this ambiguity and to ensurethat the components of the alternating tensor are frame invariant, henceforthwe restrict our attention to right-handed coordinate frames. This restrictionguarantees that all coordinate transformations are proper orthogonal.

Definition 48 An isotropic tensor is one whose components w.r.t. all co-ordinate frames are the same.

The second-order identity tensor and the alternating tensor are examples ofisotropic tensors.

1.12.4 Another view of mth-order tensors; tensor con-traction

In this subsection we develop an alternative view of mth-order tensors as lin-ear functions that map nth-order tensors (m ≥ n) to tensors of order m− n.This view is important in developing the constitutive relations in elastic-ity. For example, the fourth-order material stiffness tensor (m = 4) mapsthe second-order strain tensor (n = 2) into the second-order stress tensor(m−n = 2). We begin by extending the notion of the tensor product (previ-ously described for products of vectors only) and introduce a new operationcalled tensor contraction that is analogous to the contraction operation forindexed symbols. These two operations are then combined to define thedesired mapping.

Definition 49 Letm

U := u1⊗ u2⊗ . . .⊗ um andn

V := v1⊗ v2⊗ . . .⊗ vn betwo tensors defined by polyads of order m and n. Then the tensor product

ofm

U andn

V , denotedm

U ⊗n

V , is the mapping given by the (m+ n)th-orderpolyad

m

U ⊗n

V = u1 ⊗ u2 ⊗ . . .⊗ um ⊗ v1 ⊗ v2 ⊗ . . .⊗ vn.

Exercise 42 Show thatm

U ⊗n

V , as defined above, is an (m+ n)th-ordertensor.

Definition 50 Letm

U := u1⊗u2⊗ . . .⊗ur⊗ . . .⊗us⊗ . . .⊗um be a tensor

of order m. Then the contraction ofm

U w.r.t. the vectors ur and us, denoted


C(ur,us)

(mU)

is the tensor of order m− 2 given by

C(ur,us)

(mU)

= (ur · us) u1 ⊗ u2 ⊗ . . .⊗ ur−1 ⊗ ur+1 ⊗. . .⊗ us−1 ⊗ us+1 ⊗ . . .⊗ um.

The contraction operation can be applied sequentially (up to m/2 timesto an even-order tensor) to obtain polyads of progressively lower order.

For example, we introduce the notation C(ur−1,us−1)(ur,us)(ur+1,us+1)

(mU)

:=

C(ur+1,us+1)

C(ur,us)

[C(ur−1,us−1)

(mU)]

, etc. That is,

C(ur−1,us−1)(ur,us)(ur+1,us+1)

(mU)

= (ur−1 · us−1) (ur · us) (ur+1 · us+1) u1 ⊗ u2 ⊗. . .⊗ ur−2 ⊗ ur+2 ⊗ . . .⊗ us−2 ⊗ us+2 ⊗. . .⊗ um.

The composition of m contraction operations on a (2m)th -order tensor yieldsa scalar.

To illustrate the relation with the contraction of an indexed symbol, con-

sider the Cartesian component representation of an mth-order tensorm

T w.r.t.a given r.C.c.f. X, ei:

m

T =Ti1i2...ir...is...imei1 ⊗ ei2 ⊗ . . .⊗ eir ⊗ . . .⊗ eis ⊗ . . .⊗ eim .

Except in special degenerate cases,m

T requires a weighted sum of polyads.

Next, take the contraction ofm

T w.r.t. the base vectors eir and eis in eachpolyad in the sum. We obtain

C(eir ,eis )

(mT)

= Ti1i2...ir...is...im (eir · eis) ei1 ⊗ ei2 ⊗ . . .⊗ eir−1 ⊗ eir+1 ⊗. . .⊗ eis−1 ⊗ eis+1 ⊗ . . .⊗ eim

= Ti1i2...ir...is...imδirisei1 ⊗ ei2 ⊗ . . .⊗ eir−1 ⊗ eir+1 ⊗. . .⊗ eis−1 ⊗ eis+1 ⊗ . . .⊗ eim

= Ti1i2...k...k...imei1 ⊗ ei2 ⊗ . . .⊗ eir−1 ⊗ eir+1 ⊗. . .⊗ eis−1 ⊗ eis+1 ⊗ . . .⊗ eim .

That is, the scalar components of the contracted tensor are the same as thecontraction of the indexed symbol Ti1i2...ir...is...im over the indices ir and is.

We are ready to introduce the extended definition of an mth-order tensoras a linear mapping of an nth-order tensor.


Definition 51 An mth-order tensor is a linear mapping of nth-order tensorsinto tensors of order m − n, where m ≥ 1 and 0 ≤ n ≤ m. Given a r.C.c.f.

X, ei , letm

S =Si1i2...imei1⊗ei2⊗. . .⊗eim andn

T =Tj1j2...jnej1⊗ej2⊗. . .⊗ejn .

Then the value of the mappingm

S atn

T , denotedm

S( nT), is given by

m

S( nT)

= C(eim−n+1,ejn)

. . . C(eim−1

,ej2)

[C(eim ,ej1)

(mS ⊗

n

T)]

= Si1i2...im−njnjn−1...j1Tj1j2...jnei1 ⊗ ei2 ⊗ . . .⊗ eim−n .

Remark 18 According to this definition, a higher-order tensor of order msimultaneously defines m + 1 mappings — one for each admissable value ofn.

Remark 19 When m = n = 2 ⇐⇒ S,T ∈ Lin V, we use S(T) todistinguish the mapping of T by S into a scalar from the product, ST, thatgenerates a new second-order tensor, as defined in Definition 31.

Example 3 This definition coincides with our previous definition of an mth-order tensor (i.e., as a mapping of vectors into tensors of order m− 1) whenn

T is a vector ( n = 1). For example, letm

S= Si1i2...imei1 ⊗ ei2 ⊗ . . .⊗ eim and

supposen

T:= u =ukek. Then

m

S (u) = C(eim ,ek) [(Si1i2...imei1 ⊗ ei2 ⊗ . . .⊗ eim)⊗ (ukek)]

= Si1i2...imuk (eim · ek) ei1 ⊗ ei2 ⊗ . . .⊗ eim−1

= Si1i2...imukδimkei1 ⊗ ei2 ⊗ . . .⊗ eim−1

= Si1i2...im−1kukei1 ⊗ ei2 ⊗ . . .⊗ eim−1

=(mS u)i1i2...im−1

ei1 ⊗ ei2 ⊗ . . .⊗ eim−1 .

Exercise 43 Consider the case m = 2 and n = 1. Let S = Sijei ⊗ ej andu = ukek be component representations w.r.t. some r.C.c.f. X, ei . Findthe component representation of Su := S (u) w.r.t. X, ei .

Example 4 Here we consider the important case of a fourth-order tensor

mapping second-order tensors to second-order tensors. Let4

S =Sijklei⊗ ej ⊗


ek ⊗ el and T =Tmnem ⊗ en. Then

4

S (T) = C(ek,en)

C(el,em) [(Sijklei ⊗ ej ⊗ ek ⊗ el)⊗ (Tmnem ⊗ en)]

,

= SijklTmn (el · em) (ek · en) ei ⊗ ej,

= SijklTmnδlmδknei ⊗ ej,

= SijklTlkei ⊗ ej,

=

[4

S (T)

]ij

ei ⊗ ej.

The extended Definition 51 allows us to define higher-order identity ten-sors of order 2m that map tensors of order m into tensors of order m.

Definition 52 The identity tensor of order 2m, denoted2m

I , is the tensorthat maps tensors of order m into themselves. That is,

2m

I(mS)

=m

S ∀ mth-order tensorsm

S .

Remark 20 We cannot define identity tensors of odd order. Why?

Exercise 44 Let S and T be two second-order tensors. Show that an al-ternative definition of the inner product of two second-order tensors is givenby

S ·T = S(Tt).

Construct extended definitions of the transpose and inner product that holdfor tensors of order m. Refer to the definitions for tensors of order 2 as aguide. Confirm that your extended definition of the inner product satisfiesall the required properties.

1.13 The Vector (Cross) Product

In this section we employ the alternating tensor to develop the vector (orcross) product of two vectors. We use the third-order alternating tensor tomap one of the vectors into a second-order tensor, which in turn maps theremaining vector into the vector result. We begin with the direct definition.

1.13. THE VECTOR (CROSS) PRODUCT 69

Definition 53 The vector ( cross) product of two vectors u,v ∈ V , denotedu× v , is the vector

u× v =

(3

E v

)u.

Theorem 91 Let X, ei and

X′, e′i

be right-handed r.C.c.f.’s, and let

u,v ∈ V have components uj, vk w.r.t. X and u′j, v

′

k w.r.t. X′. Then the

component representations w.r.t. X and X′

of the vector product u × v aregiven by

u× v = εijkujvkei = εijku′

jv′

ke′

i.

Proof. We have

u× v =

(3

E v

)u (defn. 53),

= [(εijkei ⊗ ej ⊗ ek) v] u (defn. 47),

= (εijk (v · ek) ei ⊗ ej) u (defn. 45),

= (εijkvkei ⊗ ej) u (thm. 36),

= εijkvk (u · ej) ei (defn. 45),

= εijkvkujei (thm. 36).

∴ u× v = εijkujvkei. (rearrange).

Similarly, we have u× v = ε′

ijku′jv′

ke′i = εijku

′jv′

ke′i (thm. 90).

Theorem 92 Let u,v,w ∈ V , and let λ ∈ <. The properties of the vectorproduct include

1. v × u = −u× v (anticommutativity),

2. u× (v + w) = u× v + u×w (distributivity w.r.t. addition),

3. (λu)× v =λ (u× v) (homogeneity).

Exercise 45 Use Theorem 91 to prove Theorem 92.

Theorem 93 The vector product is not associative:

(u× v)×w 6= u× (v ×w) ,

indeed

(u× v)×w = (u ·w) v− (v ·w) u,

u× (v ×w) = (u ·w) v− (u · v) w.


u = u1e1

n = e3

v = v1e1 + v2e2

v2e 2

v1e1 θ

X1

X2

X3

Figure 1.6: Geometric interpretation of the vector product


The next theorem gives the familiar geometric interpretation of the vectorproduct.

Theorem 94 Let θ be the angle (0 ≤ θ ≤ π) between two vectors u and v,and let n be the unit vector ⊥ (“orthogonal”) to the plane of u and v 3 theordered triple u,v,n is right-handed. Then

u× v = |u| |v| sin θn.

Proof. If either u = 0 or v = 0, then the theorem holds trivially. So con-sider the case u 6= 0 and v 6= 0. Now select a right-handed r.C.c.f. X 3 u isaligned with the X1-axis and the X2-axis lies in the u-v plane with v in thefirst or second quadrant of the X1-X2 plane. Then

u1 = |u| , u2 = u3 = 0;

v1 = |v| cos θ, v2 = |v| sin θ, v3 = 0;

n = e3.

1.14. THE SCALAR TRIPLE PRODUCT 71

By Theorem 91,

u× v = εijkujvkei = |u| |v| sin θn.

Remark 21 Our identification of e3 with n in the proof requires that boththe frame X and the triple u,v,n are right-handed. Beyond this, our choiceof the frame X does not limit the generality of the proof, since the vector prod-uct u×v (see Definition 53) and the normal vector n are defined independentof any coordinate frame. Things would also work out if we had chosen to con-sider only left-handed frames and if n were defined to form a left-handed tripleu,v,n . The critical point is to avoid mixing left-handed and right-handedsystems (unless appropriate sign corrections are introduced systematically).Here, we shall continue to assume everything is right-handed.

Remark 22 Recall the common mnemonic for the cross product

u× v =

∣∣∣∣∣∣e1 e2 e3

u1 u2 u3

v1 v2 v3

∣∣∣∣∣∣ ;but note that this is not really a determinant. Why?

Theorem 95 Given two non-zero vectors u and v, then u×v = 0 iff u andv are parallel (i.e., the angle between them is θ = 0 or θ = π).


1.14 The Scalar Triple Product

Definition 54 The scalar triple product of three vectors u, v and w, is thereal number given by

u· (v ×w) .

Exercise 48 Show that

u· (v ×w) = εijkuivjwk,

and

u· (v ×w) =

∣∣∣∣∣∣u1 u2 u3

v1 v2 v3

w1 w2 w3

∣∣∣∣∣∣ .


u

v

w

P

Figure 1.7: Parallelepiped determined by the right-handed vector tripletu,v,w


u· (v ×w) = v· (w × u) = w· (u× v) .

The next theorem relates the scalar triple product to the volume of aparallelepiped constructed from its three constituent vectors.

Theorem 96 Let V be the volume of the parallelepiped P determined by theordered triple of vectors u,v,w . Then

V = u· (v ×w) (or V = −u· (v ×w) ),

if the ordered triple u,v,w is right-handed (or left-handed).

Proof. Let A be the area of the parallelogram determined by the vectors vand w, and let θ be the angle between them. Then

A = |v| |w| sin θ= |v ×w| (Theorem 94).

1.14. THE SCALAR TRIPLE PRODUCT 73

Now let φ be the angle between u and the vector v ×w (which is ⊥ to theplane of the face with area A ). Then

V = A cosφ |u|= cosφ |u| |v ×w|

=

u· (v ×w) if u,v,w is right-handed−u· (v ×w) if u,v,w is left-handed

(definition of inner product).

The next theorem states that if we transform a right-handed frame witha proper orthogonal transformation, then the resulting frame will also beright-handed. Thus, if we start with a right-handed frame then we only needto check that any new coordinate frame is related to the original by a properorthogonal transformation to ensure that the new frame is also right-handed.

Theorem 97 Let X, ei be a right-handed r.C.c.f. and let [λij] be a properorthogonal matrix. Then any frame

X′, e′i

obtained by the transformation

e′i = λijej,

is also a right-handed frame. If [λij] is improper orthogonal, then

X′, e′i

is

left-handed.

Proof. The orthogonal matrix [λij] corresponds to the components of anorthogonal second-order tensor. The transformed basis e′i must be an or-thonormal basis, since (by theorem 78) the transformation must maintainthe angles and magnitudes of the original basis ei. Thus, the volume of theparallelepiped defined by the vectors e′i is V ′ = 1. According to the transfor-mation,

e′1· (e′2 × e′3) =

∣∣∣∣∣∣λ11 λ12 λ13

λ21 λ22 λ23

λ31 λ32 λ33

∣∣∣∣∣∣= 1

if [λij] is proper orthogonal. Thus, V ′ = e′1· (e′2 × e′3) which implies thate′i is right-handed by Theorem 96. If [λij] is improper orthogonal, then−e′1· (e′2 × e′3) = V ′ which implies that e′i is left-handed.


1.15 Special Cases of Second-Order Tensors

This section reviews properties of certain special cases of second-order ten-sors. These properties play important roles in our subsequent developmentof solid mechanics.

1.15.1 Skew second-order tensors

Examination of the form of a skew second-order tensor reveals only threeindependent components. So it is only somewhat surprising that the mappingof a vector by a skew tensor can be represented by a mapping based on a singlevector (which also has three components). In this subsection, we develop theconcept of the axial vector of a skew tensor which defines the mapping viathe vector product.

Our first result shows that skew tensors map vectors into vectors orthog-onal to themselves.

Theorem 98 Let W ∈ Skew . Then

u ·Wu = 0 ∀ u ∈ V .

In other words,Wu ⊥ u ∀ u ∈ V .

Proof. Let u be an arbitrary vector in V . Then

u ·Wu = u ·Wtu (definition of transpose),

= u · (−W) u (definition of skew),

= −u ·Wu,

⇒ u ·Wu =0.

The orthogonal mapping of vectors by skew tensors is reminiscent of theaction of the vector product. Indeed, it turns out that the mapping of vectorsby any skew second-order tensor is replicated by a mapping defined in termsof an associated “axial” vector and the vector product.

Theorem 99 Given W ∈ Skew with components Wij w.r.t. a r.C.c.f. X, ∃a unique w ∈ V 3

Wu = w × u ∀ u ∈ V ,

1.15. SPECIAL CASES OF SECOND-ORDER TENSORS 75

and the components of w w.r.t. X are

wi = −1

2εijkWjk.

Conversely, given w ∈ V with components wi w.r.t. a r.C.c.f. X, ∃ a uniqueW ∈ Skew 3

w × u = Wu ∀ u ∈ V ,

and the components of W w.r.t. X are

Wij = −εijkwk.

The vector w is called the axial vector of the skew tensor W, and W is theaxial tensor of the vector w. We notate this as

w = ax W

W = ax w.

Proof. We first show that if an axial vector exists, it must have the statedcomponent relation with the skew tensor. That is, given W ∈ Skew, suppose∃ w ∈ V 3

Wu = w × u ∀ u ∈ V .

This is equivalent to the component expression

Wikuk = εijkwjuk ∀ ui ∈ <,⇔ Wik = εijkwj (ui is arbitrary).

Next, premultiply by εlik to obtain

εlikWik = εlikεijkwj ,

= εikl (−εikj)wj (props. of εijk),

= −2δljwj (thm. 11.3),

= −2wl, (props. of δij),

⇒ wi = −1

2εijkWjk.

So if w exists, it must have the correct component relation to the tensor W.Given the unambiguous form of this relation, it must also be unique. Next,


we demonstrate the existence of w by confirming that it has the desired axialproperty. That is,

w × u = εijkwjukei (thm. 91),

= εijk

(−1

2εjmnWmn

)ukei (subst.),

=1

2εjikεjmnWmnukei (props. of εijk),

=1

2(δimδkn − δinδkm)Wmnukei (thm. 11.2),

=1

2(Wik −Wki)ukei (props. of δij, props. of <),

= (Wikuk) ei (W ∈ Skew ),

= (Wu)i ei (thm. 51),

= Wu (thm. 22).

∴ w = ax W. Proof of the converse assertion is left to the following exercise.

Exercise 50 Complete the proof of Theorem 99.

1.15.2 Symmetric second-order tensors

This subsection covers properties of symmetric second-order tensors, S ∈Sym . In particular, we are concerned with the eigenproblem that gives riseto the characteristic values and vectors of the tensor.

Definition 55 A unit vector n (with components ni w.r.t. X) is a princi-pal direction (or eigenvector or characteristic vector) of T ∈ Lin V (notnecessarily symmetric, with components Tij w.r.t. X) iff ∃ τ ∈ < 3

Tn = τn (Tijnj = τni) ,

where τ is a principal value (or eigenvalue or characteristic value) of T, andthe pair τ ,n is called a principal pair (or eigenpair or characteristic pair)of T.

Theorem 100 Given a principal pair τ ,n of T ∈ Lin V , then

1. τ = n ·Tn,


2. τ ,−n is a principal pair of T.


Remark 23 There is a unique principal value for a given principal directionaccording to Theorem 100.1, but the converse is not true in general.

Things become much easier if we restrict our attention to symmetricsecond-order tensors. Accordingly, for the remainder of this subsection weconsider only tensors S ∈ Sym .

Theorem 101 Given the r.C.c.f. X, ei and S ∈ Sym, the base vector e1

is a principal direction of S (with components Sij w.r.t. X) and τ is theassociated principal value iff

[Sij] =

τ 0 00 S22 S23

0 S32 S33

.Similar results hold for e2 and e3.

Exercise 52 Prove the theorem. If e1 is a principal direction of S and τ isthe associated principal value, use definition 55, Theorem 51 and S ∈ Sym toshow that [Sij] has the form indicated. Then assume that [Sij] has the formindicated, and show that e1 is a principal direction with associated principalvalue τ .

Theorem 102 Let σ,n and σ, n be principal pairs of S ∈ Sym 3 σ 6= σ.Then n ⊥ n (i.e., n · n = 0).

Proof. We have n · Sn = n · Sn since S ∈ Sym. Thus

0 = n · Sn− n · Sn,

= n·σn− n·σn (defn. 55),

= σ (n · n)− σ (n · n) (homogeneity of inner prod.),

= (σ − σ) (n · n) (commutativity and distributivity),

⇒ n · n = 0 (σ 6= σ).


Next, we rearrange the equation in Definition 55 to obtain the standardform

(S− σI) n = 0 or (Sij − σδij)nj = (0)i .

For a given principal value σ, the component form is a system of three homo-geneous simultaneous equations for the components of the principal directionvector. By Theorem 18, ∃ a non-trivial solution consistent with |n| = 1, iff

det (S− σI) = det [Sij − σδij] = 0.

By linearity, any solution n′ can be scaled to obtain n so that |n| = 1. Thus,we have

Theorem 103 Let σ,n be a principal pair of S ∈ Sym . Then σ satisfies

det (S− σI) = det [Sij − σδij] = 0.

Conversely, suppose σ ∈ < satisfies the above equation. Then σ,n is aprincipal pair of S, where n is the solution to

(S− σI) n = 0 or (Sij − σδij)nj = (0)i ,

subject to |n| = 1.

The following key theorem describes the construction of a principal rect-angular Cartesian coordinate frame based on the eigenproblem associatedwith a symmetric second-order tensor.

Theorem 104 Every S ∈ Sym has at least one principal frame∗X, con-

sisting of a right-handed triplet of mutually orthogonal principal directions∗e1,

∗e2,

∗e3

associated with at most three distinct principal values. The prin-

cipal values are the roots of the characteristic polynomial

det (S− σI) = det [Sij − σδij] = −σ3 + I1σ2 − I2σ + I3,

with coefficients

I1 = tr S = Sii,

I2 =1

2

[(tr S)2 − tr

(S2)]

=1

2(SiiSjj − SijSji) ,

I3 = det S = det [Sij] ,


called the fundamental invariants of S, where S2 := SS. The zeros of thecharacteristic polynomial are three real numbers, denoted σi, which need not

be distinct. The components w.r.t.∗X of the principal directions

∗ek associated

with the principal values σk solve

(Sij − σkδij)(∗ek

)j

= (0)i (no sum on k),

subject to(∗ek

)i

(∗ek

)i

= 1. The principal axes∗ek are not necessarily unique.

There are three cases.

1. All three principal values are distinct. Then the corresponding principalaxes are unique except for sense.

2. Exactly two prinicpal values are distinct; say, σ1 6= σ2 = σ3. Then∗e1

is unique to within sense. Let∗X be a right-handed frame containing

∗e1 . Then every rotation of

∗X about

∗e1 is also a principal frame.

3. All three principal values are equal. Then every right-handed frame isa principal frame.

The proof of this lengthy theorem is lengthy; it is not repeated here forthe sake of brevity.

Theorem 105 Let ∗

X,∗ek

be a principal frame of S ∈ Sym . Then the

components of S w.r.t.∗X are given by

[S∗ij]

=

σ1 0 00 σ2 00 0 σ3

,where σk are the principal values associated with the principal axes

∗ek . Fur-

thermore, the spectral resolution of S is expressed as

S = σ1∗e1 ⊗

∗e1 + σ2

∗e2 ⊗

∗e2 + σ3

∗e3 ⊗

∗e3 .



Exercise 54 Show that the fundamental invariants of S ∈ Sym are expressedin terms of the principal values as

I1 = σ1 + σ2 + σ3,

I2 = σ1σ2 + σ2σ3 + σ3σ1,

I3 = σ1σ2σ3.

Exercise 55 Suppose that the components of S w.r.t. frame X are given by

[Sij] =

5 −3 0−3 4 20 2 3

.Find the principal values of S and all associated principal directions.

Exercise 56 Suppose that the components of S w.r.t. frame X are given by

[Sij] =

0 1 11 0 11 1 0

.Find the principal values of S and all associated principal directions.

Consider the fundamental invariants of S ∈ Sym: I1, I2, I3. As the nameimplies, these quantities are all scalar invariants that do not depend on thechoice of coordinate frame. In particular, I1 and I2 are defined in terms ofthe trace operator, which is invariant according to Theorem 66. I3 is definedin terms of the determinant operator which is also invariant according toTheorem 62. Further, as described in the next theorem, any scalar invariantof S ∈ Sym can be expressed as a function of the three fundamental invariants(this is the sense of “fundamental”).

Theorem 106 Let

f (Sij) := f (S11, S22, S33, S12, S13, S23)

be a scalar invariant of S ∈ Sym (that is f (Sij) = f(S′ij

), where Sij and

S′ij are components of S w.r.t. two frames X and X

′). Then ∃ a unique

real-valued function g of three real variables 3

f (Sij) = g (I1 (S) , I2 (S) , I3 (S))

where Ii (S) are the fundamental invariants of S. Conversely, any functionof this form is a scalar invariant of S.


Proof. Since f(Sij) is a scalar invariant we can choose any frame. Selecting

the principal frame∗X, we obtain

f(S∗ij) = f(S∗11, S∗22, S

∗33, S

∗12, S

∗13, S

∗23)

= f(σ1, σ2, σ3, 0, 0, 0) (Theorem 105)

=: h(σ1, σ2, σ3)

But the principal values are the roots of the characteristic polynomial deter-mined by the fundamental invariants. Thus, we can write

σi = qi(I1(S), I2(S), I3(S)),

and therefore

f(Sij) = h (q1(I1(S), I2(S), I3(S)), q2(−), q3(−))

=: g(I1(S), I2(S), I3(S)).

To show uniqueness, suppose there is a second function g that maps threereal numbers to a real number 3

f(Sij) = g(I1(S), I2(S), I3(S)) ∀ S ∈ Sym .

Then clearly,

g(I1(S), I2(S), I3(S)) = g(I1(S), I2(S), I3(S)) ∀ S ∈ Sym .

Thus, g and g are the same function (at least when the arguments are re-stricted to the scalar invariants of a symmetric second-order tensor). Theconverse argument follows from the above discussion of the invariance of thetrace and determinant operators.

The remainder of this subsection is concerned with the extremal prop-erties of the principal values of a symmetric tensor, a topic that is of greatimportance in the analysis of stress and strain. Indeed, this brings us back“full circle” to Mohr, beloved by all students of strength of materials.

Let S ∈ Sym have components Sij w.r.t. a given r.C.c.f. X, and let n(with components ni w.r.t. X) be a unit vector normal to a plane P . Wedenote the mapping of n by S into another vector as,

s(n) := Sn or si(n) = Sijnj.


The projections of s(n) to the direction n and the plane P are denoted N(n)nand t(n), respectively, where

N(n) := n · s(n) = n · Sn = Sijninj

is the component of s(n) in the n-direction, and

t(n) = s(n)−N(n)n.

By the orthogonality of n and P , we have

n · t(n) = 0.

Let T (n) = |t(n)| , so that by the Pythagorean theorem,

|s(n)|2 = [N(n)]2 + [T (n)]2 .

Now, leaving n arbitrary, select the principal Cartesian frame∗X, so that

[S∗ij]

=

σ1 0 00 σ2 00 0 σ3

,and

s(n) = S∗ij∗nj e∗i = σ1

∗n1∗e1 +σ2

∗n2∗e2 +σ3

∗n3∗e3,

⇒ N (n) = n · s(n) = σ1∗n

2

1 +σ2∗n

2

2 +σ3∗n

2

3 .

Collecting these results with |n| = 1, we have

[N(n)]2 + [T (n)]2 = |s(n)|2 = σ21

∗n

2

1 +σ22

∗n

2

2 +σ23

∗n

2

3,

N(n) = σ1∗n

2

1 +σ2∗n

2

2 +σ3∗n

2

3,

1 =∗n

2

1 +∗n

2

2 +∗n

2

3,

which is a system of three equations in the unknowns∗n

2

1,∗n

2

2 and∗n

2

3: σ21 σ2

2 σ23

σ1 σ2 σ3

1 1 1

∗n

2

1∗n

2

2∗n

2

3

=

N2 + T 2

N1

.


Since the∗n

2

i≥ (0)i , these equations impose bounds on N(n) and T (n), de-pending on the principal values which are assumed to be ordered:

σ1 ≥ σ2 ≥ σ3.

As before, there are three cases to investigate:

1. All principal values are distinct. Then∣∣∣∣∣∣σ2

1 σ22 σ2

3

σ1 σ2 σ3

1 1 1

∣∣∣∣∣∣ = − (σ1 − σ2) (σ2 − σ3) (σ3 − σ1) 6= 0,

and by Theorem 17 the solution is

0 ≤ ∗n

2

1=T 2 + (N − σ2) (N − σ3)

(σ1 − σ2) (σ1 − σ3),

0 ≤ ∗n

2

2=T 2 + (N − σ3) (N − σ1)

(σ2 − σ3) (σ2 − σ1),

0 ≤ ∗n

2

3=T 2 + (N − σ1) (N − σ2)

(σ3 − σ1) (σ3 − σ2).

Based on the assumed order of the principal values, these results ⇒

T 2 + (N − σ2) (N − σ3) ≥ 0,

T 2 + (N − σ3) (N − σ1) ≤ 0,

T 2 + (N − σ1) (N − σ2) ≥ 0.

Then we complete the squares and rewrite these inequalities as

T 2 +

[N − σ2 + σ3

2

]2

≥(σ2 − σ3

2

)2

,

T 2 +

[N − σ3 + σ1

2

]2

≤(σ3 − σ1

2

)2

,

T 2 +

[N − σ1 + σ2

2

]2

≥(σ1 − σ2

2

)2

.

These last results describe Mohr’s circle in 3 dimensions if we plot theadmissable values in the N–T plane. That is, we obtain three semi-circles in the first two quadrants (because T ≥ 0 by definition). Theadmissable region is the area between the circles.


N

T

AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA

AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA

σ1,0( )σ 2 ,0( )σ3 ,0( )σ2 + σ3( )

2, 0

σ3 + σ1( )

2, 0

σ1 + σ2( )

2,0

Figure 1.8: Mohr’s Circle plot of extremal values for N and T

2. Exactly two principal values are the same; say, σ1 ≥ σ2 = σ3. Now oursystem reduces to

N2 + T 2 = σ21

∗n

2

1 +σ22

( ∗n

2

2 +∗n

2

3

),

N = σ1∗n

2

1 +σ2

( ∗n

2

2 +∗n

2

3

),

1 =∗n

2

1 +( ∗n

2

2 +∗n

2

3

).

Then substituting for( ∗n

2

2 +∗n

2

3

)from the third equation into the first

two we obtain

N2 + T 2 =(σ2

1 − σ22

) ∗n

2

1 +σ22,

N = (σ1 − σ2)∗n

2

1 +σ2.

Then eliminating∗n

2

1 we find

T 2 +

[N −

(σ1 + σ2

2

)]2

=

(σ1 − σ2

2

)2

.

This describes a single semi-circle in the N–T plane, consistent withthe notion of collapsing the admissable region for Case 1 according toσ3 → σ2.


3. All three principal values are the same; say, σ1 = σ2 = σ3 = σ. Thenour equations reduce to

N = σ,

T = 0;

which plots as a single point in the N–T plane, consistent with thenotion of collapsing the semi-circular admissable region for Case 2 ac-cording to σ2 → σ1.

It should now be clear that Mohr’s circle is simply a graphical means ofdescribing the mapping of unit vectors by a symmetric second-order tensor.We also observe that the extreme values of N and T over all unit vectors nare given by

max|n|=1

N (n) = σ1 ⇒ T = 0,

min|n|=1

N (n) = σ3 ⇒ T = 0,

max|n|=1

T (n) =σ1 − σ3

2⇒ N =

σ1 + σ3

2,

min|n|=1

T (n) = 0⇒ N = σ1 or N = σ2 or N = σ3

To find the specific vectors n that generate these extrema, we substitute theextreme values of N and T into our system of equations and solve for the

components∗ni . Then we find

1. maxN occurs at∗n1= ±1,

∗n2=

∗n3= 0,

2. minN occurs at∗n1=

∗n2= 0,

∗n3= 1,

3. maxT occurs at∗n1= ±1/

√2,∗n2= 0,

∗n3= ±1/

√2.

Exercise 57 Let S ∈ Sym have components Sij w.r.t. an arbitrary r.C.c.f.X, and let σ1 ≥ σ2 ≥ σ3 be the principal values of S. Show that

maxX,i

Sii = σ1

minX,i

Sii = σ3.


1.15.3 Positive-definite second-order tensors

This section introduces positive-definite second-order tensors, a subset ofSym (the set of symmetric second-order tensors). We show that we can definea “square root” for this class of tensors. These notions lead to the PolarDecomposition Theorem, an important tool in kinematic and constitutivetheory.

Definition 56 S ∈ Sym is positive definite iff

u · Su ≥ 0 ∀ u ∈ V 3 u · Su = 0 iff u = 0.

The set of all positive definite tensors is denoted Psym .

Definition 56 is limited to symmetric tensors, because only the symmetricpart of a second-order tensor contributes to the quadratic form u · Su. Toillustrate, suppose T ∈ Lin V .

u ·Tu = u · (sym T + skew T) u (thm. 58),

= u · sym Tu + u · skew Tu (defn. sum, distrib.),

= u · sym Tu + u ·[

1

2

(T−Tt

)]u (subst.),

= u · sym Tu +1

2

(u ·Tu− u ·Ttu

)(defn. sum, distrib.,homog.),

= u · sym Tu +1

2(u ·Tu− u ·Tu) (defn. of transpose),

= u · sym Tu.

Thus, we have the following theorem.

Theorem 107 Given T ∈ Lin V , then

u ·Tu = u · sym Tu ∀ u ∈ V .

Theorem 108 Let S ∈ Sym . Then S ∈ Psym iff each of its principal valuesis strictly positive.

Proof. Suppose S ∈ Psym with principal pair σ,n . Then

n · Sn > 0 (defn. 56, |n| = 1⇒ n 6= 0),

⇒ σ > 0 (substitution, Theorem 100).


Conversely, suppose that the principal values σi of S are all strictly positive

(σi > 0), and that ∗

X, e∗i

is a principal frame for S. Then S admits the

spectral resolution

S = σ1e∗1 ⊗ e∗1 + σ2e

∗2 ⊗ e∗2 + σ3e

∗3 ⊗ e∗3.

Accordingly,

u · Su = u · (σ1e∗1 ⊗ e∗1 + σ2e

∗2 ⊗ e∗2 + σ3e

∗3 ⊗ e∗3) u (subst.),

= u · [σ1 (e∗1 ⊗ e∗1) u + σ2 (e∗2 ⊗ e∗2) u + σ3 (e∗3 ⊗ e∗3) u] (defn. tensor sum),

= u · [σ1 (e∗1 · u) e∗1 + σ2 (e∗2 · u) e∗2 + σ3 (e∗3 · u) e∗3] (defn. of ⊗ ),

=(∗u1 e∗1+

∗u2 e∗2+

∗u3 e∗3

)·(σ1∗u1 e∗1 + σ2

∗u2 e∗2 + σ3

∗u3 e∗3

)(vect. comps.),

= σ1∗u

2

1 +σ2∗u

2

2 +σ3∗u

2

3≥ 0.

Further, u·Su = 0⇒ ∗u

2

i= 0⇒ ∗ui= 0⇒ u = 0. Obviously, u = 0⇒ u·Su =

0. Thus σi > 0⇒ S ∈ Psym .

Theorem 109 Let S ∈ Psym with principal frame ∗

X, e∗i

such that S ad-

mits the spectral resolution

S = σ1e∗1 ⊗ e∗1 + σ2e

∗2 ⊗ e∗2 + σ3e

∗3 ⊗ e∗3.

Then ∃ a unique U ∈ Psym 3

U2:= UU = S,

where the tensor U admits the spectral resolution

U =√σ1e

∗1 ⊗ e∗1 +

√σ2e

∗2 ⊗ e∗2 +

√σ3e

∗3 ⊗ e∗3.

The tensor U is called the positive definite square root of S, and is denotedU =√

S.

Proof. We first prove uniqueness. Suppose that U is a positive definitesquare root of S, and let σ,n be a principal pair of S. Then

0 = (S− σI) n,

=(U2 − σI

)n,

=(U +

√σI) (

U−√σI)

n.


This implies (U−√σI) n = 0. Otherwise, we would have (U +

√σ I) v = 0

(where v := (U−√σI) n), which would imply that the negative number

−√σ is a principal value of U, a contradiction of Theorem 108. Now,

(U−√σI) n = 0 ⇒

√σ is a principal value of U, so that U must admit

the spectral resolution,

U =√σ1e

∗1 ⊗ e∗1 +

√σ2e

∗2 ⊗ e∗2 +

√σ3e

∗3 ⊗ e∗3.

The unambiguous form of this expression for U implies uniqueness. The restof the proof is left to an exercise.

Exercise 58 Complete the proof by showing that U =√σ1e

∗1⊗e∗1 +

√σ2e

∗2⊗

e∗2 +√σ3e

∗3 ⊗ e∗3 is indeed a positive definite square root of S.

Theorem 110 Psym ⊂ Lin V+ ⊂ Inv V. That is, U ∈ Psym ⇒ det U >0⇒ U ∈ Inv V.


Theorem 111 Let F ∈ Inv V, and define C := FtF. Then C ∈ Psym.

Proof. First we show that C ∈ Sym (or Ct= C).

Ct = Ft(Ft)t

(Thm. 60.5),

= FtF (Thm. 55.1),

= C.

Next, we show that C must be positive definite. We have

u ·Cu = u ·[(

FtF)

u]

(substitution),

= u ·[Ft (Fu)

](defn. of tensor mult.),

= (Fu) · (Fu) (defn. of transpose),

= |Fu|2 (defn. of magnitude),

≥ 0 with u ·Cu = 0 iff Fu = 0 (pos. def.).

Thus, (i) u ·Cu ≥ 0 ∀ u ∈ V ; (ii) u = 0 ⇒ u ·Cu = 0; and (iii) u ·Cu =0⇒ Fu = 0⇒ u = 0, since F ∈ Inv V . Therefore, C ∈ Psym.

We are now ready to state the main result of this subsection. The PolarDecomposition Theorem states that every invertible second-order tensor canbe expressed as a product of a symmetric positive definite tensor and anorthogonal tensor.


Theorem 112 (Polar Decomposition Theorem) Let F ∈ Inv V. Then∃ a unique pair of tensors U,V ∈Psym and a unique R ∈ Orth V 3

F = RU = VR.

Moreover, det R = + 1 or det R = −1, depending as det F > 0 or < 0.

Proof. We begin with uniqueness. Suppose ∃ U ∈ Psym, V ∈ Lin V andR ∈ Orth V 3 F = RU = VR. Then F ∈ Inv V ⇒ FtF ∈Psym by Theorem111, so

FtF =(UtRt

)(RU) (subst., Thm. 60.5),

=(URt

)(RU) ( U ∈ Psym ⊂ Sym ),

= U(RtR

)U (associativity of tensor mult.),

= U2 (R ∈ Orth V)

⇒ U =√

FtF is unique (Theorem 109).

Further, ∃ a unique U−1 by Theorem 110. Therefore, F = RU⇒ R = FU−1

uniquely determines R. R must have a unique inverse, since R ∈ Orth V .Then F = VR⇒ V =FR−1 uniquely defines V.

Thus, we have uniqueness of U, V and R; we still must prove existence.We have

FtF ∈ Psym

⇒ ∃ U :=√

FtF ∈ Psym (theorem 109),

⇒ U ∈ Inv V (Theorem 110),

⇒ ∃ R := FU−1

⇒ ∃ V := FRt


Next we show that these definitions ⇒ R ∈ Orth V . We have

RtR =(FU−1

)t (FU−1

)(subst.),

=[(

U−1)t

Ft] (

FU−1)

(Thm. 60.5),

=[(

Ut)−1

Ft] (

FU−1)

(Thm. 75.2),

=(U−1Ft

) (FU−1

)(U ∈ Sym ),

= U−1(FtF

)U−1 (assoc.),

= U−1 (UU) U−1 (see above),

=(U−1U

) (UU−1

)(assoc.),

= II = I (Thm. 72),

∴ R ∈ Orth V .

It easy to show that these definitions satisfy the required equalities. We have

RU =(FU−1

)U (substitution),

= F(U−1U

)(associativity),

= FI (def. of inverse),

= F;

and

VR =(FRt

)R (substitution),

= F(RtR

)(associativity),

= FI (R ∈ Orth V),

= F.

We can also demonstrate the required relations for det R, recalling that byTheorem 110, U ∈ Psym⇒ det U > 0.

det R = det(FU−1),

= det F det(U−1) (Thm. 63.3),

= det F (det U)−1 (Thm. 75.1),

∴ sign(det R) = sign(det F) ( det U > 0).

1.16. TENSOR FIELDS 91

We also need to show that V ∈ Psym. We first show V ∈ Sym.

Vt =(FRt

)t(subst.),

=(Rt)t

Ft (Thm. 60.5),

= RFt (Thm. 55.1),

= R (RU)t (subst.),

= R(UtRt

)(Thm. 60.5),

= R(URt

)(U ∈ Sym ),

= (RU) Rt (assoc.),

= FRt (subst.),

= V (subst.),

∴ V ∈ Sym

We show that V is positive definite to complete the proof.

u ·Vu = u·(FRt

)u (subst.),

= u·[(RU) Rt

]u (subst.),

= u·[(RU)

(Rtu

)](defn. of tensor mult.),

=(Rtu

)· (RU)t u (defn. of transpose),

=(Rtu

)·(Ut Rt

)u (Thm. 60.5),

=(Rtu

)·Ut

(Rtu

)(defn. of tensor mult.),

=(Rtu

)·U(Rtu

)(U ∈ Sym ),

≥ 0 with u ·Vu = 0 iff Rt u = 0 (U ∈ Psym ).

⇒ u ·Vu ≥ 0 with u ·Vu = 0 iff u = 0 (Rt ∈ Inv V)

∴ V ∈ Psym .

1.16 Tensor Fields

In our subsequent treatment of solid mechanics, we shall be concerned withfunctions that attach a tensor-valued quantity (i.e., a scalar, a vector or anmth-order tensor) to each point in a domain embedded in a Euclidean pointspace. These functions might refer to temperature, velocity, stress, strain,etc. This section introduces these tensor fields, their derivatives and somefamiliar differential operators.


Definition 57 Let D denote a subset of a three-dimensional Euclidean pointspace E (we write D ⊂ E), and let P be an arbitrary point in D. A functionf that maps D to a set of tensors of order m is called an mth-order tensorfield, where D is called the domain of f . The value of f at P ∈ D is denotedf(P).

Suppose the position vector x =−→OP (i.e, the vector from the origin of

some r.C.c.f. X to P) has the component representation, x = xiei. For conve-nience, many authors write f(P) = f(x) = f(x1, x2, x3), although these mustreally be distinct functions given the differences in their domains. We adoptthis somewhat imprecise notation for the sake of convenience and tradition.

Sometimes we wish to define a tensor field with respect to a rotatedcoordinate frame X

′that shares the same origin as X. According to Theorem

30 the components of a position vector x w.r.t. X′

are given by

x′

i = e′

i · x = (λijej) · x =λij (ej · x) = λijxj.

Supposem

T is an mth-order tensor field on domain D. Then it may be

convenient to describe the value ofm

T at P in terms of its 3m componentsw.r.t. X, Ti1i2...im , which are themselves scalar-valued fields on D. Thus, wemay write

m

T (P) = Ti1i2...im(P) ei1 ⊗ ei2 ⊗ . . .⊗ eim .

According to Theorem 88, the components of the fieldm

T (P) w.r.t. X are

Ti1i2...im(P) = λj1i1λj2i2 . . . λjmimT′j1j2...jm

(P).

We omit the label m above the scalar components, because the order of thetensor can be determined by counting the subscripts.

We next consider the derivatives and the degree of smoothness of a tensorfield defined on an open domain D (from here on, all domains are assumedto be open, except where explicitly indicated otherwise).

Definition 58 Let f be a real-valued scalar function defined on D. Then fis said to be in (continuity) class CM on D (we write f ∈ CM(D)) iff fand all its partial derivatives up to and including order M are continuouseverywhere on D. A tensor field is in class CM(D) iff all of its componentfields are in CM(D).


Remark 24 We shall see shortly that the continuity class of a tensor fielddoes not depend on the choice of frame X. That is, M is the same for a giventensor field for all choices of X.

Next we introduce the “comma notation”, a convenient method for no-tating the partial derivatives of the components of a tensor field.

Definition 59 (Comma Notation) Suppose an mth-order tensor fieldm

Ton D has rectangular Cartesian components

Tij...k ∈ CM(D) w.r.t. X and T′

ij...k ∈ CM(D) w.r.t. X′.

Then

Tij...k,p(P) = Tij...k,p(x) = Tij...k,p(x1, x2, x3) :=∂

∂xpTij...k(x1, x2, x3),

Tij...k,pq(P) = Tij...k,pq(x) = Tij...k,pq(x1, x2, x3) :=∂

∂xqTij...k,p(x1, x2, x3),

. . .

T′

ij...k,p(P) = T′

ij...k,p(x) = T′

ij...k,p(x′

1, x′

2, x′

3) :=∂

∂x′pT′

ij...k(x′

1, x′

2, x′

3),

. . . ;

where M is assumed to be large enough so that the indicated partial derivativesexist.

The next theorem states that the partial derivative of an mth-order tensorfield is itself a tensor field of order m+ 1.

Theorem 113 Suppose an mth-order tensor fieldm

T on D has rectangularCartesian components

Tij...k ∈ C1(D) w.r.t. X,

T′

ij...k ∈ C1(D) w.r.t. X′.

Define

Sij...kl(x1, x2, x3) = Tij...k,l(x1, x2, x3) w.r.t. X,

S′

ij...kl(x′

1, x′

2, x′

3) = T′

ij...k,l(x′

1, x′

2, x′

3) w.r.t. X′.

Then the indexed symbols Sij...kl and S′

ij...kl are the rectangular Cartesian

components w.r.t. X and X′

of a tensor field of order m+ 1 on D.


Proof. By the transformation rules for tensor and vector components wehave

T ′ij...k(x′1, x′2, x′3) = λimλjn . . . λkpTmn...p(x1, x2, x3)

Next, use the chain rule to differentiate this result w.r.t. x′

l. Noting that theλ’s are constant on D, and that xq = λuqx

′u, we obtain

T ′ij...k,l(x′1, x′2, x′3) = λimλjn . . . λkpTmn...p,q(x1, x2, x3)

∂

∂x′l

(λuqx

′

u

),

= λimλjn . . . λkpTmn...p,q(x1, x2, x3)λuq∂x′u

∂x′l

,

= λimλjn . . . λkpTmn...p,q(x1, x2, x3)λuqδul,

= λimλjn . . . λkpTmn...p,q(x1, x2, x3)λlq,

= λimλjn . . . λkpλlqTmn...p,q(x1, x2, x3).

Thus by substitution,

S ′ij...kl(P) = λimλjn . . . λkpλlqSmn...pq(P).

This result conforms to the standard transformation rule for tensors of orderm+ 1, so by Theorem 89 we have that Sij...kl and S ′ij...kl are the components

of a tensor field of order m+ 1 w.r.t. X and X′.

Remark 25 This result demonstrates that the existence of partial derivativesof the components of a tensor field is independent of the choice of coordinateframe.

The identification of the partial derivatives of the components of tensorswith the components of higher-order tensors justifies the tensorial represen-tation of various differential operators in the following definition.

Definition 60 Given a scalar field ϕ, a vector field v, and a second-ordertensor field T, each sufficiently smooth on D, then we define the

1. gradient of ϕ: gradϕ = ∇ϕ = ϕ,iei,

2. Laplacian of ϕ: 4ϕ = ∇2ϕ = ϕ,ii,


3. gradient of v: grad v = ∇v = vi,jei ⊗ ej,

4. divergence of v: div v = ∇ · v = tr(grad v) = vi,i,

5. curl of v: curl v = ∇× v = εijkvk,jei,

6. Laplacian of v: 4v = ∇2v = (4vi) ei = vi,jjei,

7. divergence of T: div T = ∇ ·T = Tij,jei.

The following theorems, which summarize standard relations involvingfirst and second-order differential operators, can be easily proven using indi-cial notation.

Theorem 114 Let the real scalar fields θ and ϕ, the vector fields u and vand the second-order tensor field T all be of class C1(D). Then

1. ∇(θϕ) = ϕ∇θ + θ∇ϕ,

2. ∇ · (θu) = ∇θ · u+θ∇ · u,

3. ∇× (θu) = ∇θ × u+θ∇× u,

4. ∇ · (u× v) = v·∇ × u− u·∇ × v,

5. ∇ · (θT) = T∇θ+θ∇ ·T,

6. ∇ · (Tv) = (∇ ·Tt) · v + T· (∇v)t .

Proof. (1):

[∇(θϕ)]i = (θϕ),i (defn. 60.1),

= θ,iϕ+ θϕ,i (deriv. of a product),

= ϕ (∇θ)i + θ (∇ϕ)i (defn. 60.1),

∴ ∇(θϕ) = ϕ∇θ + θ∇ϕ (thm. 27).

Exercise 60 Prove the remaining identities of Theorem 114.

Theorem 115 Let the real scalar fields θ and ϕ as well as the vector field uall be of class C2(D). Then

1. 4 (θϕ) = (4θ)ϕ+ θ4ϕ+ 2∇θ · ∇ϕ,


2. ∇ · ∇θ = 4θ,

3. ∇×∇θ = 0,

4. ∇ · ∇ × u = 0,

5. ∇× (∇× u) = ∇ (∇ · u)−4u.

Proof. (5): Let v = ∇× u. Then

[∇× (∇× u)]i = (∇× v)i ,

= εijkvk,j (defn. 60.5),

= εijk (εkmnun,m),j (defn. 60.5),

= εijkεkmnun,mj,

= εkijεkmnun,mj,

= (δimδjn − δinδjm)un,mj (thm. 11.2),

= uj,ij − ui,jj,= uj,ji − ui,jj (invert order of derivs.)

= (uj,j),i − ui,jj (defn. of 2nd deriv.)

= (∇ · u),i −4ui (defn. 60.(2,4)),

= [∇ (∇ · u)−4u]i (defn. 60.(3,6)),

∴ ∇× (∇× u) = ∇ (∇ · u)−4u (thm. 27).

Exercise 61 Prove the remaining identities of Theorem 115.

We close with the Divergence theorem which plays a key role in severalaspect of mechanics theory.

Theorem 116 (Divergence Theorem) Given an open domain B with asuitably regular boundary equipped with a unit outward normal vector, n,a vector field w ∈ C1(B,V) on B, and a second-order tensor field, T ∈C1(B,Lin V), ∫

B∇ ·w dV =

∫∂B

w · n dA∫B∇ ·T dV =

∫∂B

T(n) dA.

where w and T are interpreted in the sense of the trace operator in theboundary integrals.

Chapter 2

Kinematics

2.1 References

1. S. S. Antman, Nonlinear Problems of Elasticity, Springer, 1995.

2. R. J. Atkin and N. Fox, An Introduction to the Theory of Elasticity,Longman, 1980.

3. P. Chadwick, Continuum Mechanics, Halsted, 1976.

4. P. G. Ciarlet, Mathematical Elasticity, Volume I: Three-DimensionalElasticity, North-Holland, 1988.

5. Y. C. Fung, Foundations of Solid Mechanics, Prentice-Hall, 1965.

6. M. E. Gurtin, An Introduction to Continuum Mechanics, Academic,1981.

7. M. E. Gurtin, The Linear Theory of Elasticity, Handbuch der Physik,VIa/2, Springer, 1972.

8. M. E. Gurtin, E. Fried and L. Anand, The Mechanics and Thermody-namics of Continua, Cambridge University Press, 2010.

9. A. E. H. Love, The Mathematical Theory of Elasticity, 4th ed., Cam-bridge, 1927.

10. L. E. Malvern, Introduction to the Mechanics of a Continuous Medium,Prentice-Hall, 1969.

97

98 CHAPTER 2. KINEMATICS

11. J. T. Oden and L. F. Demkowicz, Applied Functional Analysis, CRCPress, 1996.

12. R. W. Ogden, Non-linear Elastic Deformations, Wiley, 1984.

13. I. S. Sokolnikoff, Mathematical Theory of Elasticity, 2nd ed., McGraw-Hill, 1956.

14. C. Truesdell, The Elements of Continuum Mechanics, Springer, 1966.

15. C. Truesdell, A First Course in Rational Continuum Mechanics, Vol.1, 2nd ed., Academic, 1992.

16. C. Truesdell and W. Noll, The Non-linear Field Theories of Mechanics,Handbuch der Physik, III/3, Springer, 1960.

17. C. Truesdell and R. A. Toupin, The Classical Field Theories, Handbuchder Physik, III/3, Springer, 1960.

18. C.-C. Wang and C. Truesdell, Introduction to Rational Elasticity, No-ordhoff, 1973.

2.2 Bodies

This chapter presents a mathematical description for the motions of de-formable bodies. Our first task is to describe the geometry of a body andto introduce the constraints that limit the class of admissable geometries.These limitations are designed to ensure that the remainder of our theory iswell-defined and well-behaved. Although this may seem to be a very basicissue, the problem of precisely describing the most general class of geometriesfor which our mechanics theory is valid remains an open research issue.1

We begin with the deformation of certain geometric entities from topologytheory. We use the concepts to describe the region of physical space occupiedby a body at each instant of time.

Definition 61 Let d(P,Q) denote the distance between points P and Q in athree-dimensional Euclidean point space, E. Then the ball of radius rcentered

1See, for example, W. Noll, The geometry of contact, separation, and reformation ofcontinuous bodies, Arch. Rational Mech. Anal. 122, (1993), 197-212.

2.2. BODIES 99

at P, denoted as B(P, r), is the set of all points Q ∈ E separated from P bya distance less than r. That is,

B(P, r) = Q ∈ E : d(P,Q) < r .

Definition 62 A set A ⊂ E is bounded iff ∃ a ball, B(P, r), with finiteradius 3 A ⊂ B(P, r).

Definition 63 Let A ⊂ E. A point P ∈ A is an interior point of A iff ∃ areal number ε > 0 3 B(P, ε) ⊂ A.

Definition 64 The interior of a set A ⊂ E , denoted Int A, is the union ofall interior points of A. That is,

IntA = P ∈ A : P is an interior point of A .

Definition 65 A set A ⊂ E is open iff IntA = A.

Definition 66 Let A ⊂ E . A point Q ∈ E (Q need not be in A) is anaccumulation point of A iff ∀ ε > 0, B(Q, ε) contains at least one point inA distinct from Q.

Roughly speaking, an accumulation point may be either on the interior ofA or on the boundary of A. In fact, we are now ready to describe these twoconcepts precisely.

Definition 67 Let A ⊂ E, and let A be the set of all accumulation pointsof A. The closure of A, denoted A , is

A = A ∪ A.

Note: A * A if A includes isolated points, so we cannot simply write A = A.

Definition 68 A set A ⊂ E is closed iff A = A.

Definition 69 Let A ⊂ E. The boundary of A , denoted ∂A, is the set ofall accumulation points of A that are not elements of IntA. That is,

∂A = A − IntA.


Definition 70 A set (open or closed) A ⊂ E is a region iff ∀ P,Q ∈ A ∃ aC1 curve connecting points P and Q that is wholly contained in A.

This definition contains two key points. First, all points in a region mustbe connected. Second, the C1 requirement of the connecting curves willfacilitate our subsequent construction of line integrals between any pair ofpoints in the region (e.g., as when integrating a strain field to obtain relativedisplacements).

We sidestep a careful definition of the non-trivial concept of regularity.2

Instead, it is sufficient for our purposes to say that a region is regular if itsboundary is geometrically “nice” enough to admit application of the diver-gence theorem. With that said, we are ready to describe the requirements ofa geometrically admissable body for our subsequent development.

Definition 71 A body is identified by the region B ⊂ E that it occupiesat some reference time, t = 0. For admissability, we require that B be the

closure of a bounded, open region0

B ⊂ E and that B be regular.

This construction – the closure of an open region – ensures that B containsno lower-dimensional appendages. Occasionally, we relax the prohibition onunbounded regions to consider infinite domains.

2.3 Finite Deformations

This section is concerned with finite deformations of a body. We begin withthe deformation mapping and its relation to the displacement and the de-formation gradient fields. Then we consider the special case of rigid-bodydeformations, before considering the geometric aspects of deformation andtheir relation to various strain measures. The restriction of the finite defor-mation theory to infinitesimal deformations is covered in the next section.

2.3.1 Deformation, displacement, deformation gradi-ent

Now we consider changes in shape, as when a system of forces acts on a body.This leads to the following definition of deformations.

2See O.D. Kellogg, Foundations of Potential Theory, Dover, 1953.

2.3. FINITE DEFORMATIONS 101

Definition 72 Let0

B be an open, bounded, regular region of a Euclidean

point space E. A deformation f is a mapping (function) of points in0

B ontoanother open region of E with the properties

1. f is one-to-one; i.e., f(x) = f(y)⇒ x = y ∀ x,y ∈0

B,

2. f ∈ C2(0

B), f−1 ∈ C2(f(0

B)),

3. det∇f(x) > 0 ∀ x ∈0

B.

The notation f(0

B) refers to the mapped region, which is called the image

of the set0

B under f .

Remark 26 That f is a mapping implies that a single point of0

B can only

map into a single point of f(0

B). Thus, the present theory does not admitfracture-like motions. The one-to-one property is a necessary and sufficientcondition for the existence of the inverse mapping f−1. Further, it implies that

distinct points on0

B cannot map into the same point on f(0

B). This preventsself-penetration and infinite strains (good). However, it does not rule outself-contact between distinct boundary points (also good), because the open

set0

B does not include its boundary. Some authors define f as a one-to-one mapping on a closed region B . This approach precludes self-contact(an undesirable limitation), and presents a number of other difficulties inthe context of differentiation and weak formulations of mechanics. To avoidthese problems, we require that f be defined on an open region.

We use position vectors defined w.r.t. a r.C.c.f. X, ei to identify points

in0

B. It is natural to use the same approach to identify points in the image

set f(0

B). Thus, we write

z = f(x) = fi(x)ei = fi(x1, x2, x3)ei,

or

zi = fi(x) = fi(x1, x2, x3).


Remark 27 The one-to-one property and the continuity requirement (prop-

erties 1 and 2) are sufficient to ensure that det∇f(x) 6= 0 ∀ x ∈0

B; a prerequi-site for applying certain results from advanced calculus. The further require-ment of a strictly positive determinant (property 3), ensures that the deforma-tion is orientation preserving (see section 2.3.3 below). That is, right-handedtriads of vectors map to right-handed triads under the deformation, so the

body cannot turn “inside out”. The definition of0

B, the one-to-one mappingassumption and the two additional properties are sufficient to ensure that the

image set f(0

B) is also an open, bounded, regular region.

Although we define the deformation as a mapping on the open region0

Bfor very good reasons, we are nonetheless interested in describing the motionof points on the boundary of the region, (as for example, when we wish todescribe displacement boundary conditions to limit the set of admissable de-formations). The notions of the extension, restriction and trace of a function

allow us to derive the boundary motion from f(0

B). We define these terms forthe particular case of real-valued functions with specified levels of smooth-ness. As usual, E denotes a Euclidean point space.

Definition 73 Given A ⊂ B ⊂ E, let g be a function on B. Then a functionon A is called the restriction of g to A, denoted g|A , iff g|A (x) = g(x) ∀x ∈ A.

Definition 74 Given A ⊂ B ⊂ E, let f be a function on A. A functionon B is called the extension of f to B, denoted ExtB f , iff (ExtB f)|A = f.Typically, ExtB f is required to have similar continuity properties on B as fhas on A.

Definition 75 Let0

B= IntB, where B ⊂ E is a closed, regular region. Fur-

ther, let f be a real-valued function on0

B with “well-behaved” first partialderivatives.3 The trace of f, denoted γf, is a real-valued function on ∂Bdefined by γf = (ExtB f)|∂B.4

3More precisely, we require that f be in the Hilbertian Sobolev space H1(0

B).4This definition provides only a rough interpretation of the trace operator; a precise

treatment requires a knowledge of Sobolev spaces. See S. C. Brenner and L. R. Scott, The


Remark 28 The requirement of well-behaved first partial derivatives sup-ports the unambiguous extension of f to the boundary ∂B. Inductively, the

trace operator “evaluates” a function f ∈ CM(0

B) and its partial derivatives

up to order M − 1 on ∂B. Specifically, for any deformation f ∈ C2(0

B), thetrace allows us to “evaluate” the components fi and the partial derivativesfi,j on ∂B. This is sufficient for a complete kinematic description of theclosed body B. These arguments are associated with the following ExtensionTheorem.

Theorem 117 (Extension Theorem) Let the vector field h ∈C0(B). Then

h = 0 on0

B ⇒ h = 0 on B.

Proof. Since B =0

B ∪∂B, we need to show that h = 0 on ∂B. To this end,let x∗ ∈ ∂B. Since h is continuous at x∗ we have, given any ε > 0 ∃ δ > 0 3

x ∈ B(x∗, δ) ∩ B ⇒ |h(x)− h(x∗)| < ε (ε− δ definition of continuity).

Thus, we have

x ∈ B(x∗, δ)∩0

B ⇒ |0− h(x∗)| = |h(x∗)| < ε (0

B⊂ B; h = 0 on0

B ),

⇒ |h(x∗)| = 0, (see below)

⇒ h = 0 on ∂B ( x∗ arbitrary on ∂B).

The penultimate step is justified by the following argument. Suppose|h(x∗)| 6= 0 ⇒ 0 < |h(x∗)| < ε. But the result must hold for any ε > 0;e.g., ε = |h(x∗)| /2. This contradicts |h(x∗)| < ε, so the supposition must befalse.

In practice, the extension theorem is often applied to the difference oftwo functions, and similar results are available for scalar-valued as well asmth-order-tensor-valued functions.

Next consider the displacement and the deformation gradient fields.

Mathematical Theory of Finite Element Methods, Springer-Verlag, 1994; J. L. Lions and E.Magenes, Non-homogeneous Boundary-Value Problems and Applications, Springer-Verlag,1972; and R. E. Showalter, Hilbert Space Methods for Partial Differential Equations, Pit-man Publishing Ltd., 1977 for a more complete exposition.


Definition 76 Let x ∈0

B, and let y be the mapped position vector of x undera deformation f . Then the corresponding displacement, denoted u(x), is

u(x) = y − x = f(x)− x,

or ui(x) = fi(x)− xi.

Remark 29 Clearly, the displacement field inherits the continuity properties

of the deformation field, i.e., u ∈ C2(0

B).

Definition 77 Let f be a deformation field on0

B, and let u be the correspond-ing displacement field. Then the deformation gradient and the displacementgradient fields are defined by

F(x) := ∇f(x) [Fij(x) = fi,j(x)] ,

H(x) := ∇u(x) [Hij(x) = ui,j(x)] .

Theorem 118 The deformation gradient and displacement gradient fieldsare related by

F = I + H.


2.3.2 Rigid deformations

Definition 78 A deformation f of0

B is rigid iff it preserves distance between

all pairs of points in0

B; i.e.,

|f(x1)− f(x2)| = |x1 − x2| ∀ x1,x2 ∈0

B .

Theorem 119 Let f be a rigid deformation of0

B; let O ∈0

B be the origin ofa r.C.c.f. X, ei; and define the relative deformation field w.r.t. O as

ξ(x) = f(x)− f(0), x ∈0

B,

where x is the position vector of an arbitrary point in0

B w.r.t. X. Then ∀x,y ∈

0

B,

|ξ(x)| = |x| ,ξ(x) · ξ(y) = x · y,


and ∃ Q ∈ Orth V 3ξ(x) = Qx.

Thus, the inner product and magnitudes of position vectors are preserved un-der any rigid relative deformation w.r.t. O, and a rigid relative deformationw.r.t. O can always be represented by an orthogonal second-order tensor.

Proof. First ∀ x ∈0

B,

|ξ(x)| = |f(x)− f(0)| (subst.),

= |x− 0| (definition of rigid def.),

= |x| . (*)

Then ∀ x,y ∈0

B,

2ξ(x) · ξ(y) = ξ(x) · ξ(x) + ξ(y) · ξ(y)− |ξ(x)− ξ(y)|2

(defn. of |·| , props. of inner prod.),

= |ξ(x)|2 + |ξ( y)|2 − |ξ(x)− ξ(y)|2 (defn. of |·| ),= |x|2 + |y|2 − |ξ(x)− ξ(y)|2 (*),

= |x|2 + |y|2 − |(f(x)− f(0))− (f(y)− f(0))|2 (subst.),

= |x|2 + |y|2 − |f(x)− f(y)|2 (comm. & assoc. of vect. +),

= |x|2 + |y|2 − |x− y|2 (definition of rigid def.),

= x · x + y · y− (x · x− 2x · y + y · y) (defn. of |·| ),= 2x · y (comm. & assoc. of vect. +).

∴ ξ(x) · ξ(y) = x · y ∀ x,y ∈0

B . (**)

To complete the proof, we first construct an orthonormal basis at O. Since

O ∈0

B, we can find non-zero position vectors εi (possibly having very small

magnitudes) along each of the Xi-axes to points in0

B. Then by (**) and theorthogonality of Xi,

ξ(εi) · ξ( εj) = εi · εj =

|εi|2 for i = j, no sum0 for i 6= j,

.

Thus, we can construct an orthonormal basis for the entire space V (not just

for0

B) as

ei =ξ(εi)

|ξ(εi)|,


so that the components of the relative deformation ξ(x) w.r.t. the frameX, ei are (according to Theorem 36),

ξi(x) = ei · ξ(x) (defn. vector component),

=ξ(εi)

|ξ(εi)|· ξ(x) (subst.),

=εi · x|ξ(εi)|

, (**)

∴ ξ(x) =

(εi · x|ξ(εi)|

)ei. (***)

By the linearity of the inner product, we then have

ξ(αx + βy) = αξ(x) + βξ(y),

so ξ is linear. Although we started with ξ as a function on0

B only, it iseasy to consider its extension to all of V as defined by (***). The extendedfunction is a linear mapping on V to V , so ξ is a second-order tensor. More

precisely, ξ is the restriction of a second-order tensor field to0

B. Since ξpreserves inner products, it must be an orthogonal tensor field according toTheorem 78. Thus, we have shown ∃ Q ∈ Orth V 3

ξ(x) = Qx ∀x ∈0

B .

Theorem 120 A deformation f of0

B is rigid iff for any given r.C.c.f. ∃ aunique pair c,Q 3

f(x) = c + Qx ∀ x ∈0

B ,

where c is a uniform vector and Q is a uniform proper orthogonal tensor.

Proof. This result follows directly from Theorem 119 if our origin lies on0

B.

We have ∀ x ∈0

B ,

f(x)− f(0) = ξ(x) = Qx

∴ f(x) = c + Qx, (§),


where c := f(0) is a uniform vector (i.e., independent of the argument x),and Q ∈ Orth V is also uniform. This result is easily extended to arbitraryr.C.c.f.’s with origins anywhere in V . Let O be any such origin associatedwith a frame X, and define d as the vector from O to O. Then the positionvectors x and x of a given point w.r.t. the two frames are related by

x =x+d.

Therefore, the mapped position vectors under the same physical deformationare related by f(x) = f(x) + d, so that

f(x) = f(x)− d

= c + Qx− d (§),= c + Q (x+d)−d (subst.),

= (c + Qd− d) + Qx (linearity, comm., assoc.),

= c + Qx (subst.),

where c := c + Qd− d is a uniform vector. Thus we obtain the same form,independent of our choice of origin.

We still must show that Q is proper and that c and Q are unique. Dif-ferentiation of f(x) = c + Qx leads to

Q = ∇f(x), x ∈0

B,

which uniquely determines Q for a given deformation f . Note further that Qis independent of the choice of origin. Our definiton, c := f(0), also uniquelydetermines c for a given deformation f (but only for a fixed origin!). The

requirement that det∇f(x) > 0 ∀ x ∈0

B (Definition 72) ensures that Q is aproper orthogonal tensor.

Now suppose that we are given a mapping f(x) = c+Qx, where Q ∈ OrthV+ and c ∈ V are uniform. To prove that this mapping is a rigid deformationwe need to show that:

1. f is one-to-one. Assume that f(x) = f(y) ∀x,y ∈0

B . Then

c + Qx = c + Qy (subst.),

⇒ Qx = Qy

⇒ x = y (premult. by Q−1 ).

∴ f(x) = f(y) ⇒ x = y.


That f is a mapping implies that x = y⇒ f(x) = f(y). Thus, f isone-to-one.

2. f preserves distances. See exercise 63.

3. f ∈C2(0

B), f−1∈C2(f(0

B)). From the supposition and ∀ x ∈0

B,

f(x) = c + Qx,

⇔ fi(x) = ci +Qijxj,

⇒ fi,j(x) = Qij,

⇒ fi,jk(x) = 0.

Thus, fi, fi,j and fi,jk are continuous functions on0

B, and ∴ f ∈C2(0

B).For the inverse mapping, y := f(x) = c + Qx⇒

x = Q−1 (y − c) ,

= Qt (y − c) , (Q ∈ OrthV+ ⊂ OrthV),

from which we obtain ∀ y ∈ f(0

B)

f−1(y) = Qt (y − c)

⇔ f−1i (y) = Qji (yj − cj)

⇒ f−1i,j (y) = Qji

⇒ f−1i,jk(y) = 0.

Thus, f−1i , f−1

i,j and f−1i,jk are continuous functions on f(

0

B), and ∴

f−1∈C2(f(0

B)).

4. det∇f(x) > 0 ∀ x ∈0

B . We have shown that ∀ x ∈0

B

f(x) = c + Qx

⇒ fi,j(x) = Qij

⇒ ∇f(x) = Q

⇒ det∇f(x) = det Q =1 > 0 (Q ∈ OrthV+).

∴ det∇f(x) > 0 ∀ x ∈0

B .


Thus, f is a rigid deformation.The last part of this proof contains the proof of the next theorem.

Theorem 121 Let f be a deformation 3 f(x) = c + Qx on0

B . Then

Q = ∇f(x) ∀ x ∈0

B .

Exercise 63 Given a mapping f(x) = c + Qx on0

B where Q ∈ Orth V+,

show that |f(x)− f(y)| = |x− y| ∀ x,y ∈0

B .

Remark 30 For simplicity of notation, we may sometimes denote a defor-mation as f(B), where the value of the deformation on the boundary ∂B isunderstood in terms of the trace operator.

2.3.3 Geometric properties of deformations

This subsection describes the motions, the changes in length and the changesin angle of curves embedded in a body subject to a deformation. In partic-ular, we will develop the properties of and relations between the rotationtensor, the left and right stretch tensors, the left and right Cauchy-Greendeformation tensors and the Green-St. Venant strain tensor associated witha deformation and its gradient.

Definition 79 A material point is a particle that is simultaneously identifiedby a point P in the closed region B at time t = 0 and its image f(P) under a(possibly time-dependent) deformation. Points Q ∈ f(B) and P ∈ B identifythe same material point iff

f−1(Q) = P.

Consider a smooth (continuity class C1) material curve C ⊂0

B, comprisedof a set of material points. We specify this set of points via the parametricrepresentation,

C = x : x = c(τ),τ∈ [0, 1] ,

where c ∈C1 ([0, 1]) . The undeformed length of C is

length C =

∫ 1

0

∣∣∣c′(τ)∣∣∣ dτ ,


f

Bf(B )

C

f(C)

Figure 2.1: Mapping of a material curve under a deformation

where c′(τ) := dc(τ)/dτ is a vector tangent to C at x = c(τ) in the direction

of the parameterization. The image of C under the deformation f is the curve

f(C) = y : y = f(c(τ)),τ∈ [0, 1]

and

length f(C) =

∫ 1

0

∣∣∣d′(τ)∣∣∣ dτ (*),

where d(τ) := f(c(τ) and d′(τ) := d[f(c(τ))]/ dτ is a vector tangent to theimage curve f(C) at x = f(c(τ)) in the direction of the parameterization.Now, by the Chain Rule, we have

d′(τ) = ∇f(c(τ))c′(τ) [d′i(τ) = fi,j(c(τ))c′j(τ)],

= F(c(τ))c′(τ) [d′i(τ) = Fij(c(τ))c′j(τ)] (Definition 77).

Next, define C ∈ LinV on0

B by

C(x) = Ft(x)F(x),


in which Ft(x) denotes [F(x)]t. 5 Then,

|d′(τ)|2 = [F(c(τ))c′(τ)] · [F(c(τ))c′(τ)] (thm. 35),

= c′(τ)· [F(c(τ))]t [F(c(τ))c′(τ)] (defn. of transpose),

= c′(τ)·

[F(c(τ))]t [F( c(τ))]

c′(τ) (defn. of tensor mult.),

= c′(τ) ·C(c(τ))c′(τ) (subst.),

∴ |d′(τ)| =

√c′(τ) ·C(c(τ))c′(τ) (§),

We have C ∈ Psym, since det F(x) 6= 0 (defns. 77 and 72). ∴ C(x) ∈ Psym(see proof of the Polar Decomposition Theorem, thm. 112) has a positivedefinite square root

U(x) =√

C(x),

by thm. 109. 6 Thus,

|d′(τ)|2 = c′(τ) ·U2( c(τ))c′(τ) (subst.),

= c′(τ) ·U(c(τ))[U(c(τ ))c′(τ)

](defn. of tensor mult.),

=[U(c(τ ))c′(τ)

]· [U( c(τ))]t c′(τ) (defn. of transpose),

=[U(c(τ ))c′(τ)

]·[U( c(τ))c′(τ)

](U(x) ∈ Sym ),

=∣∣U(c(τ ))c′(τ)

∣∣2 (thm. 35),

∴ |d′(τ)| =∣∣U(c(τ ))c′(τ)

∣∣ (§§),.

Then substituting (§) and (§§) into (*), we have proved the following theorem.

Theorem 122 Let C = x : x = c(τ),τ∈ [0, 1] be a smooth curve in0

B, let

f be a deformation of0

B, and define F =∇f . Then the tensor fields

C : = FtF,

U : =√

C

are Psym on0

B, and the length of the image curve f(C) is given by

length f(C) =

∫ 1

0

√c′(τ) ·C(c(τ))c′(τ) dτ

=

∫ 1

0

∣∣U(c(τ ))c′(τ)∣∣ dτ .

5Later, we identify C as the right Cauchy-Green deformation tensor.6Later, we identify U as the right stretch tensor.


f

B

f(B )

x C1

C2

s2

s1

θ(s1,s2)θ(t1,t2)

f(C2

)

f(C1)

t2

t1

f(x)^

Figure 2.2: Change in angle between two material curves due to a deforma-tion

This theorem indicates that the tensor fields C and U contain all theinformation needed to evaluate the change in length of an arbitrary curveunder a given deformation. Next, we demonstrate that they also determinethe angle change between two intersecting curves under the deformation.

Let Cα= x : ∃ τα ∈ [0, 1] 3 x = cα(τα), for α = 1, 2, be two smooth

curves in0

B that intersect at a point x identified by the parametric coordinatesτα ∈ [0, 1]. That is, the intersection point is given by

x = c1(τ 1) = c2(τ 2).

It is convenient to define F = ∇f(x), C = FtF and U =√

C. The vectorstangent to the curves Cα at x in the direction of parameterization are providedby c′α := c′α(τα) (no sum). Similarly, vectors tangent to the image curvesf(Cα) at f(x) are given by

d′α := d′α(τα) = F(cα(τα))c′α(τα) = Fc′α (no sum),

in which the derivative of dα(τα) := f(cα(τα)) (no sum) follows the derivationin the previous theorem. Thus, the unit tangents of the undeformed and


deformed curves are given by

sα =c′α∣∣c′α∣∣ (no sum),

tα =Fc′α∣∣∣Fc′α

∣∣∣ (no sum).

This leads to the following theorem concerning the angle change describedby a deformation field.

Theorem 123 Let s1 and s2 be unit tangents at x in the directions of pa-

rameterization to two smooth material curves in0

B which intersect at x. Lett1 and t2 be the corresponding unit tangents to the image curves under thedeformation f at f(x). Then

cos θ(t1, t2) =Us1∣∣∣Us1

∣∣∣ · Us2∣∣∣Us2

∣∣∣=

s1 · Cs2√s1 · Cs1

√s2 · Cs2

,

where θ(t1, t2) denotes the angle between t1 and t2.

Proof. From the development of Theorem 122 we have

∣∣∣Fc′α

∣∣∣ =

√c′α · Cc

′α

(no sum),

=∣∣∣Uc

′α

∣∣∣ .


Then

cos θ(t1, t2) = t1 · t2(t1 and t2 are unit vectors),

=Fc′1∣∣∣Uc′1

∣∣∣ · Fc′2∣∣∣Uc′2

∣∣∣ (substitution),

=c′1·Cc

′2∣∣∣Uc

′1

∣∣∣ ∣∣∣Uc′2

∣∣∣ (Exercise 64a),

=Uc′1∣∣∣Uc′1

∣∣∣ · Uc′2∣∣∣Uc′2

∣∣∣ (Exercise 64b),

=Us1∣∣∣Us1

∣∣∣ · Us2∣∣∣Us2

∣∣∣ (Exercise 64c),

=s1 · Cs2√

s1 · Cs1

√s2 · Cs2

(Exercise 64d).

Exercise 64 Fill in the intermediate steps to justify parts a, b, c and d inthe preceding proof.

It is natural to require that a deformation f have the property that theimage under f of any right-handed triad be a right-handed triad, and theimage of any left-handed triad be a left-handed triad. That is, a deformationshould preserve orientations.

The following lemma and theorem demonstrate that the property

det F(x) > 0 ∀ x ∈0

B ,

which is part of the definition of a deformation, is sufficient to ensure thatthis property holds.

Lemma 124 Let f be a mapping on an open region0

B⊂ E to E, let Ci be a

triplet of smooth material curves in0

B that intersect at x, let c′i be a triplet

of vectors tangent to the curves Ci at x ∈0

B 3 the parallelipiped constructed


from c′i has non-zero volume, and let d

′i be the triplet of vectors tangent to

the image curves f(Ci) at f(x) given by

d′

i = Fc′

i.

Thend′

1 ·(d′

2 × d′

3

)= det F

[c′

1 ·(c′

2 × c′

3

)].

Proof. We have

d′

1 ·(d′

2 × d′

3

)= Fc

′

1 ·(Fc′

2 × Fc′

3

)(subst.),

= εijk

(Fc′

1

)i

(Fc′

2

)j

(Fc′

3

)k

(exercise 48),

= εijkFil

(c′

1

)lFjm

(c′

2

)mFkn

(c′

3

)n

(Theorem 51),

= εijkFilFjmFkn

(c′

1

)l

(c′

2

)m

(c′

3

)n

(props. of real numbers),

= det[Fpq

]εlmn

(c′

1

)l

(c′

2

)m

(c′

3

)n

(Theorem 8),

= det Fεlmn

(c′

1

)l

(c′

2

)m

(c′

3

)n

(Theorem 32),

= det F[c′

1 ·(c′

2 × c′

3

)](exercise 48).

Theorem 125 A deformation f preserves orientations.

Proof. Let c′i and d

′i be triplets of tangent vectors to curves Ci and im-

age curves f(Ci), as described in Lemma 124. Let V (ai) be the volume ofthe parallelipiped constructed from the vector triplet ai. By Definition 72,

det F(x) > 0 ∀ x ∈0

B⇒ det F > 0. Suppose that c′i is right-handed. Then

noting that volume is a non-negative quantity, we have

0 < V (c′

i) (V (c′

i) 6= 0 by assumption),

= c′

1 ·(c′

2 × c′

3

)(Theorem 96, c

′

i right-handed),

=1

det Fd′

1 ·(d′

2 × d′

3

)(Lemma 124),

⇒ 0 < d′

1 ·(d′

2 × d′

3

)( det F > 0),

⇒ V (d′

i) = d′

1 ·(d′

2 × d′

3

)(Theorem 96, V ≥ 0),

⇒ d′

i is right-handed (Theorem 96).


f

B

f(B )

x

e

f(x)^

lL (x,e,l)^

f(L (x,e,l))

^

Figure 2.3: Extensional strain in the direction of the unit vector e

A similar line of reasoning shows that d′i must be left-handed if c

′i is left-

handed. Thus, the property det F(x) > 0 ∀ x ∈0

B ensures that deformationspreserve orientation.

The next theorems relate C and U to conventional strain measures ofelementary mechanics. We shall make use of the notation L(x, e,l) to denote

the straight line of length l in0

B that emanates from x ∈0

B in the directionof the unit vector e.

Theorem 126 Let f be a deformation of0

B . Then the corresponding exten-

sional strain at x ∈0

B in the direction e is

ε(x, e) := liml→0

length f(L(x, e,l))− ll

,

=√

e · Ce− 1,

and the extensional strain at x in the coordinate direction Xi is

ε(x, ei) =

√Cii − 1 (no sum).

Proof. We construct the line analytically as

L(x, e,l) = x : x = c(τ) = x + τ le, τ ∈ [0, 1] ,


so that c′(τ) = le. Then by Theorem 122,

length f(L(x, e,l)) = l

∫ 1

0

√e · [C(x + τ le)] e dτ ,

= l√

e · [C(x + τ le)] e for some τ ∈ [0, 1]

(Mean Value Theorem for integrals),

where the value of τ depends on l. Then letting C(l) denote C(x + τ le) andC denote C(x), we obtain

ε(x, e) = liml→0

√e · C(l)e− 1 (subst.),

=√

e · Ce− 1 (continuity at l = 0 of C(l)),

⇒ ε(x, ei) =

√Cii − 1 (Theorem 28).

Theorem 127 Let f be a deformation of0

B . Then the corresponding shear

strain at x ∈0

B between the directions s1 and s2, that is the angle decrease

γ(s1, s2) := θ(s1, s2)− θ(t1, t2),

in which tα := f(sα), α = 1, 2, is determined by

cos [θ(s1, s2)− γ(s1, s2)] =s1 · Cs2√

s1 · Cs1

√s2 · Cs2

;

the shear strain at x ∈0

B between the coordinate directions Xi and Xj (i 6= j)is determined by

sin γ(ei, ej) =Cij√

Cii

√Cjj

.


We can apply the Polar Decomposition Theorem to the deformation gra-dient to identify a number of useful kinematic tensors.


Definition 80 Let the deformation gradient F =∇f of the deformation f of0

B have the polar decompostion

F(x) = R(x)U(x) = V(x)R(x)

∀ x ∈0

B, where U(x),V(x) ∈Psym and R(x) ∈Orth V+. The following ter-minology is standard.

• R(x) — the rotation tensor at x;

• U(x) — the right stretch tensor at x;

• V(x) — the left stretch tensor at x;

• C(x) = Ft(x)F(x) — the right Cauchy-Green deformation tensor at x;

• B(x) = F(x)Ft(x) — the left Cauchy-Green deformation tensor at x.

Theorem 128 Let f be a deformation on0

B . Then

1. C = U2, B = V2,

2. V = RURt, U = RtVR;

3. B = RCRt, C = RtBR.


Some discussion of the geometric interpretations of R, U, and V is inorder at this point. The mapping by the deformation f of differential vectorsdx in the neighborhood of the undeformed position x can alternatively bedecomposed into successive transformations as

1. “Right” Path

(a) a “stretch” of undeformed vectors dx by the operator U (lengthand angle changes prior to rotation),

(b) a rigid rotation about x of stretched vectors Udx by the operatorR,


(c) a rigid translation of stretched and rotated vectors RUdx from xto f(x),

2. “Left” Path

(a) a rigid translation from x to f(x),

(b) a rigid rotation about f(x) of translated vectors dx by the operatorR,

(c) a “stretch” of translated and rotated vectors Rdx at f(x) by theoperator V (length and angle changes after rotation).

The magnitudes and directions of the differential vectors are unaffected bythe translation operations. If the undeformed vectors dx are aligned withthe principal directions of U, then the stretch defined by U consists solelyof length changes. Similarly, if the rotated vectors Rdx are aligned with theprincipal directions of V, then the stretch defined by V consists solely oflength changes.

The right and left paths (including inverse transformations as needed) de-fine a circular set of transformation relations. These can be used to constructalternative representations of individual transformations. For example, theaction of the left stretch tensor in step 2c can be represented as the compo-sition of steps 1c, 1b, 1a, (2a)−1 and (2b)−1. In other words,

Vdx = (x− f(x)) +RURtdx+ (f(x)− x)

= RURtdx

⇒ V = RURt

The next theorem provides a simple criterion for rigid-body deformationsbased on the right Cauchy-Green deformation tensor.

Theorem 129 A deformation f of0

B is rigid iff C(x) = I ∀ x ∈0

B .

Proof. Suppose that f of0

B is rigid. Then

f(x) = c + Qx ∀ x ∈0

B , Q ∈OrthV+.


We can differentiate at any x ∈0

B to obtain

F(x) = ∇f(x) = Q,

∴ C(x) = Ft(x)F(x) (subst.),

= QtQ (subst.)

= I (Q ∈OrthV+).

Conversely, suppose that C(x) = I ∀ x ∈0

B ⇒ everywhere on0

B we have

Cij = fm,ifm,j = δij (1)⇒0 = fm,ikfm,j + fm,ifm,jk (2) (differentiate),

= fm,ijfm,k + fm,ifm,kj (3) (swap j and k in (2)),

∴ 0 = fm,ikfm,j − fm,ijfm,k (4) (subtract (3) from (2), fm,jk = fm,kj).

Also,

0 = fm,kifm,j + fm,kfm,ji (5) (swap i and k in (2)),

∴ 0 = fm,jfm,ik (6) (add (4) and (5), fm,ji = fm,ij).

Equation (6) can be viewed as a system of three homogeneous, linear al-gebraic equations in the unknowns fm,ik for each fixed choice of i and k.However, only the trivial solutions

fm,ik = 0 (7)

are possible, since

det [fm,j] = det F 6= 0 (in fact, det F > 0).

Next, we use fm,ik = 0 on0

B with the fact that0

B is a connected region to

show that fm,i = constant .7 We have0

B is arc-wise connected by definitions

71 and 70. Thus, ∃ a smooth curve lying entirely in0

B that connects any point

x ∈0

B to a fixed point x ∈0

B . We can describe such a curve parametricallyas

z = g(τ), g ∈C1([0, 1] ,0

B ), g(0) = x, g(1) = x.

7This result cannot be obtained from fm,ik = 0 alone, if0

B is not a connected region.Why?


We use the connecting curves to define functions on [0, 1] to < in termsof the components of the deformation gradient. as

ϕmi(τ) = fm,i(g(τ)).

Then by the Chain Rule,

ϕ′

mi(τ) = fm,ik(g(τ))g′

k(τ)

∴ ϕ′

mi(τ) = 0 (8) (see (7)).

Next, we integrate along the curve to find an expression for the components

of the deformation gradient at an arbitrary x ∈0

B.

fm,i(x) = fm,i(g(1)) (subst.),

= ϕmi(1) (subst.),

= ϕmi(0) +

∫ 1

0

ϕ′

mi(τ)dτ (Fundamental Thm. of calculus),

= ϕmi(0) (see (8)),

= fm,i(g(0)) (subst.),

= fm,i(x) (subst.),

∴ fm,i(x) = constant =: Qmi. (9)

Noting the identity, xj,i = δji, we can write

(fm(x)−Qmjxj),i = fm,i(x)−Qmjδji,

= Qmi −Qmi ((9), prop. of Kronecker’s delta),

∴ (fm(x)−Qmjxj),i = 0.

Then, using a similar integration argument to the one that led from (7) to(9), we conclude that

fm(x)−Qmjxj = constant =: cm,

orfm(x) = cm +Qmjxj. (10)

Referring to (1) and (9), we have

QmiQmj = δij.


∴ [Qij] is orthogonal. Since f is a deformation, (9) ⇒

det [Qij] > 0;

so [Qij] is proper orthogonal.

Returning to direct notation, we have shown that ∀ x ∈0

B ,

f(x) = c + Qx, Q ∈OrthV+.

We have shown that the right Cauchy-Green deformation tensor C carriesa great deal of important kinematic information. However, this tensor maybe inconvenient for developing stress-strain relations, since it does not vanishwhen the motion is rigid (a rigid motion should induce no change in the stressstate). Accordingly, it is sometimes replaced by a tensor that does vanishunder rigid motion.

Definition 81 The Green–St. Venant strain tensor (or simply, the Greenstrain tensor) at x is

G(x) =1

2[C(x)− I] ,

or Gij(x) =1

2[Cij(x)− δij] .

Clearly, G ∈ Sym. This definition and Theorem 129 lead directly to

Theorem 130 A deformation is rigid iff

G(x) = 0 ∀x ∈0

B .

The Green strain tensor may also be represented in terms of the displace-ment gradient (see Definition 77).

Theorem 131

G =1

2

(H + Ht + HtH

).


Remark 31 The theorems of this section directly pertain only to the open

region0

B . As we will be working with boundary-value problems, it is naturalto ask whether these results can be extended to the closed body B, specifically,

to the boundary ∂B. Since by definition, fm ∈ C2(0

B ), and because only first-order partial derivatives are involved, the trace operator readily extends thesekinematic results to the boundary.

2.4. INFINITESIMAL DEFORMATION THEORY 123

2.4 Infinitesimal Deformation Theory

Linearized elasticity requires an assumption of “infinitesimal deformations”to justify a strain measure with a linear dependence on the deformation (ordisplacement) field. As we shall see, this common designation is somethingof a misnomer. It is actually the displacement gradient H = ∇u, rather thanthe deformation f , that must be limited to support the approximation of alinear strain measure. The next subsection develops the concept of smalldisplacement gradients and the associated infinitesimal strain measure. Thesecond subsection is devoted to an analysis of the infinitesimal strain measure.

2.4.1 Restriction to small displacement gradients

We begin with an explanation of what “small” means in the context of dis-placement gradients in the infinitesimal deformation theory.

Definition 82 Given a displacement gradient field H derived from a defor-

mation f on0

B, a norm of the displacement gradient field,8 denoted ε, is

ε = ‖H‖ := sup

x∈0B

i,j∈1,2,3

|Hij(x)| .

Remark 32 Here “sup” denotes the supremum operator, which generatesthe least upper bound of the argument over the indicated range. It would

be nicer to simply take the max of |Hij(x)| , but since0

B is open, |Hij(x)|need not have a maximum on

0

B .9 Thus, we use the sup operator instead.

This presents no particular problem here, since f ∈C2(0

B) ⇒ H ∈C1(0

B) ⇒the trace γH is unambiguously defined on ∂B. Thus, ExtBHij(x) is clearly

8More than one norm can be defined. We use one here that is convenient for ourpurposes, but that is by no means the standard.

9A continuous (i.e., bounded) function on an open, bounded domain might not attaina maximum value on that domain. E.g., consider the function f(x) = x, on ]0, 1[. Does itattain a maximum value anywhere on the open domain?


defined via the trace operator, and we can write10

sup

x∈0B

i,j∈1,2,3

|Hij(x)| = maxx∈B

i,j=1,2,3

∣∣∣ExtBHij(x)

∣∣∣ .Remark 33 This norm might appear to be problematic, since it is definedin terms of the components of H, which in turn depend on the choice ofcoordinate frame. Although this norm is not frame-invariant, this dependenceis benign for our purposes, since we will only be concerned with order-of-magnitude arguments. Given two r.C.c.f.s, X and X

′, we have

H′

ij = λikλjlHkl (tensor transformation rule),⇒∣∣∣H ′ij∣∣∣ = |λi1λj1H11 + λi1λj2H12 + . . .+ λi3λj3H33| (expansion, 9 terms),

≤ |λi1| |λj1| |H11|+ |λi1| |λj2| |H12|+ . . .+ |λi3| |λj3| |H33| (props. of <),

≤ 9ε (defn. 82, λij are direction cosines),

∴ ε′

:= sup

x∈0B

i,j∈1,2,3

∣∣∣H ′ij(x)∣∣∣ ≤ 9ε.

This result is strong enough to conclude that

ε << 1⇒ ε′<< 1,

which is sufficient for our ultimate purposes.

The following definition supports order-of-magnitude comparisons.

Definition 83 Let ϕ, θ be real-valued functions of a displacement gradientfield H derived from a deformation f . Then ϕ is of order εn (or big oh ofεn), denoted

ϕ = O (εn) ,

iff ∃ K ∈ <, independent of H, 3

|ϕ(H)| ≤ Kεn ∀ admissible H.

10A continuous (i.e., bounded) function on a closed, bounded domain has a maximumvalue somewhere on that domain.


Also,ϕ = θ +O (εn)

meansϕ− θ = O (εn) .

For vector and tensor-valued fields,

v = O (εn) means vi = O (εn) ∀ i,T = O (εn) means Tij = O (εn) ∀ ordered pairs (i, j) .

We next define the infinitesimal strain tensor of elementary mechanics.

Definition 84 Given a displacement field u on0

B, the infinitesimal strain

tensor at x ∈0

B is

E(x) := sym H(x) =1

2

[H(x) + Ht(x)

]=

1

2

[∇u(x) + (∇u(x))t

],

or Eij(x) =1

2(Hij(x) +Hji(x)) =

1

2(ui,j(x) + uj,i(x)) .

Theorem 132 Given a displacement field u on0

B,

1. E =O (ε) ,

2. F = I +O (ε) ,

3. C = I + 2E +O (ε2) ,

4. G = E +O (ε2) .

Proof. Parts 1, 2 and 4 are reserved for exercises. For part 3, we have bydefinitions 81 and 84 with Theorem 131,

C = I + 2G = I + 2E + HtH.

Thus, after rearranging in component form, we have

|Cij − δij − 2Eij| = |HkiHkj|= |H1iH1j +H2iH2j +H3iH3j|≤ |H1i| |H1j|+ |H2i| |H2j|+ |H3i| |H3j|≤ 3ε2.


Thus, by Definition 83 with K = 3,

Cij − δij − 2Eij = O(ε2),

or

C = I + 2E +O(ε2).

Exercise 68 Prove parts 1, 2 and 4 of Theorem 132.

Part 4 of the theorem states that the infinitesimal strain tensor and theGreen–Saint Venant strain tensor agree to within O (ε2) . This point is fun-damental to the infinitesimal deformation theory. In particular, we have thatE is a good approximation to G iff ε << 1, since the error is O (ε2) .

Recall that Theorem 130 states that the Green–Saint Venant strain tensorG vanishes iff a deformation f is rigid. This implies that the infinitesimalstrain tensor E vanishes to within O (ε2) — but not identically! — undera rigid deformation. Conversely, if E vanishes identically (E = 0), then Gvanishes to within O (ε2) , and the deformation is nearly rigid if ε << 1. Thislatter observation motivates the notion of an infinitesimal rigid deformation.

Definition 85 A deformation (or displacement field) on0

B is infinitesimalrigid iff

E = 0 on0

B .

Theorem 133 A displacement field u on0

B is infinitesimal rigid iff it hasthe form

u(x) = u + Wx,

where u ∈ V and W ∈ Skew are both uniform.

Proof. Suppose u is infinitesimal rigid on0

B . Then E = 0 on0

B by Definition85, and

ui,jk = ui,kj (u ∈ C2(0

B)),

= −uk,ij (E = 0⇒ ui,j = −uj,i).


Also,

ui,jk = −uj,ik (E = 0),

= −uj,ki (u ∈ C2(0

B)),

= uk,ji (E = 0),

= uk,ij (u ∈ C2(0

B)).

Thus we have ui,jk = uk,ij and ui,jk = −uk,ij, which ⇒

ui,jk = 0 on0

B .

Following the logic of the proof of Theorem 129, this leads to

ui(x) = ui + Wijxj on0

B,

where ui and Wij are uniform. Now, E = 0⇒

0 = ui,j + uj,i,

= Wij + Wji,

∴ Wij = −Wji ⇒ W ∈ Skew .

Thus, we have shown that u is infinitesimal rigid on0

B ⇒ u(x) = u + Wxwhere u ∈ V and W ∈ Skew are uniform. We still need to prove the converse,but this is left to the following exercise.

Exercise 69 Suppose that u is a displacement field on0

B of the form

u(x) = u + Wx,

where u ∈ V and W ∈ Skew are both uniform. Show that u is infinitesimal

rigid on0

B according to Definition 85.

According to Theorem 99, we can obtain an alternative representationof an infinitesimal rigid deformation by replacing the skew tensor W with aform based on its axial vector.


Theorem 134 A displacement field u on0

B is infinitesimal rigid iff it hasthe form

u(x) = u + w × x,

where u, w ∈ V are uniform.

Proof. This result follows directly from theorems 133 and 99.It turns out that any displacement gradient field has a unique decompo-

sition in terms of an infinitesimal strain tensor and an infinitesimal rotationtensor, as demonstrated by the following definitions and theorem.

Definition 86 Given a displacement field u on0

B (not necessarily infinites-

imal rigid), the infinitesimal rotation tensor at x ∈0

B is

W(x) = skew H(x) =1

2

[H(x)−Ht(x)

]=

1

2

[∇u(x)− (∇u(x))t

],

or Wij(x) =1

2(Hij(x)−Hji(x)) =

1

2(ui,j(x)− uj,i(x)) ,

and the infinitesimal rotation vector at x is

w(x) = ax W(x),

or wi(x) = −1

2εijkWjk(x).

Theorem 135 Let u be a displacement field on0

B . Then

1. H = E + W, and

2. w =12

curl u,

everywhere on0

B .


Exercise 71 Determine the infinitesimal rotation tensor of the infinitesimalrigid displacement field

u(x) = u + w × x.


2.4.2 Analysis of infinitesimal strain

This subsection presents a study of the relation between displacement fields

and infinitesimal strain fields. Given a displacement field u ∈C2(0

B), thedefinition

E := sym∇u on0

B⇔ Eij =1

2(ui,j + uj,i) on

0

B,

provides a unique calculation of the corresponding strain field. Here we are

concerned with the inverse problem, given a strain field E on0

B, find a cor-

responding displacement field u on0

B that satisfies the system of linear, in-homogeneous, first-order partial differential equations in the definition. Thisis a more interesting problem that requires some care. Below, we consideruniqueness (in a limited sense), existence, etc.

We do not expect the infinitesimal strain E alone to uniquely determinethe displacement field, since rigid-body motions are not reflected in E. How-ever, the following theorem provides a limited, but useful, notion of unique-ness for the problem.

Theorem 136 Let u1,u2∈C1(0

B) be displacement fields that correspond to

an infinitesimal strain field E on0

B; i.e.,

sym∇u1= sym∇u2= E on0

B .

Then u1 and u2 agree to within an infinitesimal rigid displacement field; i.e.,u1 − u2 is an infinitesimal rigid displacement field.

Proof. We have

E (u1 − u2) = sym∇ (u1 − u2) (subst.),

= sym∇u1− sym∇u2 (linearity),

= E− E (subst.),

= 0.

∴ u1 − u2 is an infinitesimal rigid displacement field by Definition 85.Next, we turn our attention to the compatibility conditions. These arise

as required conditions for the existence of a displacement field that corre-sponds to a given infinitesimal strain field. Note that more stringent continu-ity conditions are required in the following development than those requiredby our definition of a deformation.


Theorem 137 Let E ∈ C2(0

B, Sym). A necessary condition that ∃ u ∈C3(

0

B,V) 3sym∇u = E on

0

Bis that the components of E satisfy the compatibility conditions

εipmεjqnEpq,mn = (0)ij on0

B .

Proof. Given E ∈ C2(0

B, Sym), suppose ∃ u ∈ C3(0

B,V) 3

Epq =1

2(up,q + uq,p) .

Then partial differentiation on0

B and premultiplication by εipmεjqn ⇒

εipmεjqnEpq,mn =1

2εipmεjqnup,qmn +

1

2εipmεjqnuq,pmn,

= (0)ij (Theorem 3).

where we have made use of the fact that εjqn and εipm are skew w.r.t. q, n andp,m respectively. Symmetry of up,qmn w.r.t. q, n and uq,pmn w.r.t. p,m follow

from u ∈ C3(0

B,V) and an interchange in the order of partial differentiation.There are alternative equivalent forms for the compatibility conditions.

Theorem 138 Let E ∈ C2(0

B, Sym). Then the following systems of equa-tions are equivalent.

1. εipmεjqnEpq,mn = (0)ij ;

2. Eij,kl + Ekl,ij − Eil,jk − Ejk,il = (0)ijkl ;

3. Eij,kk + Ekk,ij − Eik,jk − Ejk,ik = (0)ij .

Proof. We can apply the logic that if (1)⇒ (2)⇒ (3)⇒ (1), then (1)⇔ (2)⇔ (3). To show that (1) ⇒ (2), we relabel i, j to r, s in (1) and premultiplyby εrikεsjl to obtain

(0)ijkl = εrikεsjlεrpmεsqnEpq,mn,

= (δipδkm − δimδkp) (δjqδln − δjnδlq)Epq,mn, (Theorem 11(2)),

= (δipδkm − δimδkp) (Epj,ml − Epl,mj) (props. of Kronecker’s delta),

= Eij,kl + Ekl,ij − Eil,kj − Ekj,il (props. of Kronecker’s delta),

= Eij,kl + Ekl,ij − Eil,jk − Ejk,il, (E ∈ C2(0

B, Sym)).


We simply contract over k and l in (2) to show that (2)⇒ (3). To show that(3) ⇒ (1), recall that by Theorem 11(1),

εipmεjqn =

∣∣∣∣∣∣δij δiq δinδpj δpq δpnδmj δmq δmn

∣∣∣∣∣∣ .Thus,

εipmεjqnEpq,mn = (δijδpqδmn − δijδmqδpn − δiqδpjδmn+δiqδmjδpn + δinδpjδmq − δinδmjδpq)Epq,mn

= δijEpp,mm − δijEpq,qp − Eji,mm + Epi,jp + Ejq,qi − Epp,ji(props. of Kronecker’s delta),

= δijA− Aij,

where A := Epp,mm − Epq,qp and Aij := Eji,mm − Epi,jp − Ejq,qi + Epp,ji.Contraction over i and j in (3) and interchanges of the order of partialdifferentiation leads to A = 0. Applying symmetry of Eij and interchangesin the order of partial differentiation, we find that Aij is equal to the l.h.s.of (3), and thus Aij = 0. Therefore, we have εipmεjqnEpq,mn = (0)ij .

We would like to show that the compatibility conditions are also suffi-cient for the existence of a solution u to the infinitesimal strain-displacementrelation,

sym∇u = E on0

B .To this end, we introduce the Cesaro Line Integral Representation (CLI rep-resentation) for the displacement field u.11 Given a sufficiently smooth in-finitesimal strain field E, any displacement field that satisfies the infinitesimalstrain-displacement relations must admit the CLI representation. In otherwords, the CLI representation is a necessary condition for the existence ofa displacement solution. Furthermore, if the given strain field satisfies thecompatibility conditions on a simply-connected region, the CLI representa-tion defines a corresponding displacement solution to the infinitesimal strain-displacement relation. In other words, the compatibility conditions are suffi-cient to ensure the existence of a displacement solution (under the specifiedconditions). Since the caveats differ for the necessary and sufficient asser-tions of the existence of a displacement solution, we present them in separatetheorems to maintain generality.

11See Sokolnikoff, 1956.


˜ x

x

ξ ( τ )

B

0

Figure 2.4: Parametric description of curve Cx,x in the Cesaro Line Integral

Theorem 139 (Cesaro Line Integral Representation - I) Given E ∈C2(

0

B, Sym), let ξ(τ), τ ∈ [0, 1] be a parametric representation of any piece-

wise smooth curve Cx,x in an open, regular, bounded region0

B from a fixedpoint x to a variable position x 3 x = ξ(0) and x = ξ(1).12 Then a necessary

condition that ∃ u ∈ C2(0

B,V) 3

sym∇u = E on0

B

12 In other words, Cx,x =

x ∈

0

B: ∃ τ ∈ [0, 1] 3 x = ξ(τ), x = ξ(0),x = ξ(1)

.


is that u13 admit the Cesaro line integral representation

ui(x) =

∫ 1

0

Uik(ξ(τ),x)ξ′

k(τ)dτ + ui(x) +Wij(x) (xj − xj) ,

where

Wij =1

2(ui,j − uj,i) ,

Uik(ξ,x) = Eik(ξ) +(xj − ξj

)[Eik,j(ξ)− Ejk,i(ξ)] .

Proof. Suppose that u ∈ C2(0

B,V) is a solution to sym∇u = E on0

B .We know that a smooth curve of the form described in the theorem can be

constructed from any fixed point x ∈0

B and any other point x ∈0

B because0

Bis a region. It must also be possible to construct a piecewise smooth curve ofthe same parametric form. Then, by the Fundamental Theorem of Calculus∫ 1

0

d

dτ[ui(ξ(τ))] dτ = ui(ξ(1))− ui(ξ(0)) = ui(x)− ui(x).

Applying the Chain Rule we obtain,

d

dτ[ui(ξ(τ))] = ui,j(ξ(τ))ξ

′

j(τ)

= [Eij(ξ(τ)) +Wij(ξ(τ))] ξ′

j(τ) (Theorem 135(1)).

∴ by substitution,

ui(x) = ui(x) +

∫ 1

0

[Eij(ξ(τ)) +Wij(ξ(τ))] ξ′

j(τ)dτ .

Next, integrate the last term by parts to obtain14∫ 1

0

Wij(ξ(τ))ξ′

j(τ)dτ =Wij(ξ(τ))

[ξj(τ)− xj

]∣∣10−∫ 1

0

[ξj(τ)− xj

]Wij,k(ξ(τ))ξ

′

k(τ)dτ

= Wij(x) (xj − xj)−∫ 1

0

[ξj(τ)− xj

]Wij,k(ξ(τ))ξ

′

k(τ)dτ .

13It would appear that E ∈ C2(0

B,Sym) and sym∇u = E on0

B ⇒ u ∈ C3(0

B,V).

While this is reasonable, the proof of this theorem only requires u ∈ C2(0

B,V), so the lessstringent condition is assumed. Furthermore, displacement fields inherit the continuity ofthe associated deformation, and this level of continuity is consistent with the minimumlevel of continuity required by the definition of a deformation.

14Since ddτ

Wij(ξ(τ))

[ξj(τ)− xj

]= Wij,k(ξ(τ))ξ

′

k(τ)[ξj(τ)− xj

]+Wij(ξ(τ))ξ

′

j(τ)


Also,

Wij,k =1

2(ui,j − uj,i),k

=1

2(ui,jk − uj,ik) +

1

2(uk,ji − uk,ji) (add 0),

=1

2(ui,kj + uk,ij)−

1

2(uj,ki + uk,ji) (u ∈ C2(

0

B,V)),

= Eik,j − Ejk,i (E = sym∇u).

Assembling the above results to obtain an expression for ui(x) completes theproof.

Remark 34 The term ui(x)+Wij(x) (xj − xj) in the preceding theorem rep-resents an infinitesimal rigid displacement field. Since the infinitesimal strainfield E only determines the displacement field to within an an infinitesimalrigid displacement, we drop these terms in the subsequent development forsimplicity. Thus we shall write

ui(x) =

∫ 1

0

Uik(ξ(τ),x)ξ′

k(τ)dτ .

So far we have shown that a displacement solution to the infinitesimalstrain-displacement relation for a given infinitesimal strain field E, neces-sarily admits the CLI representation. Only C2 smoothness of E was assumed,the compatibility conditions did not enter the proof. The following theoremasserts that (subject to certain additional hypotheses) the compatibility con-ditions on E are sufficient for the existence of a displacement field definedby the CLI representation that satisfies the infinitesimal strain-displacementrelation.

Theorem 140 (Cesaro Line Integral Representation - II) Let E ∈ CN(0

B, Sym) satisfy the compatibility conditions, where N ≥ 2 and

0

B is simply-

connected. Then ∃ a vector field u ∈ CN+1(0

B,V)15 3

sym∇u = E on0

B .15Now u satisfies the continuity condition implied by its relation to the infinitesimal

strain field.


Moreover, such a field is uniquely defined to within an infinitesimal rigid fieldby the Cesaro line integral

ui(x) =

∫ 1

0

Uik(ξ(τ),x)ξ′

k(τ)dτ

of Theorem 139 in which ξ(τ) is now only required to be piecewise smooth.

Proof. First, we must show that the CLI defines a proper function on0

B; i.e.,that ui(x) are well-defined functions of x alone — specifically, that the valueof the integral does not depend on the choice of the curve ξ(τ) connectingthe fixed point x to x. For fixed choices of i and x, the CLI is a conventionalline integral on the curve ξ(τ), which need only be piecewise smooth for theintegral to be well-defined for a particular curve ξ(τ). Thus, we have thatthe displacement components ui(x) are well-defined as functions of x aloneif we can show that the CLI from x to x is path independent. Sufficientconditions for path independence are

1.0

B is a simply-connected region. [A region is simply-connected iff anyclosed, regular curve in the region can be continuously contracted toa point in the region such that the curve remains entirely within theregion throughout the contraction.]

2. The curl of the vector field Uikek (for i = 1, 2, 3) vanishes; i.e.,

∂Uik∂ξl− ∂Uil∂ξk

= 0 ∀k, l.

The first condition is assumed, and the second follows from the definition


of Uik in Theorem 139:

Uik(ξ,x) = Eik(ξ) +(xj − ξj

)[Eik,j(ξ)− Ejk,i(ξ)]⇒

∂Uik∂ξl

(ξ,x) = Eik,l(ξ)− δjl [Eik,j(ξ)− Ejk,i(ξ)]

+(xj − ξj

)[Eik,jl(ξ)− Ejk,il(ξ)] ,

= Elk,i(ξ) +(xj − ξj

)[Eik,jl(ξ)− Ejk,il(ξ)]

(prop. of Kronecker’s delta),

∴∂Uik∂ξl

(ξ,x)− ∂Uil∂ξk

(ξ,x) = Elk,i(ξ) +(xj − ξj

)[Eik,jl(ξ)− Ejk,il(ξ)]

−Ekl,i(ξ)−(xj − ξj

)[Eil,jk(ξ)− Ejl,ik(ξ)] ,

=(xj − ξj

)[Eik,jl(ξ) + Ejl,ik(ξ)

−Ekj,il(ξ)− Eil,jk(ξ)]

(E ∈ Sym ),

=(xj − ξj

)[Eik,jl(ξ) + Ejl,ik(ξ)

−Ekj,il(ξ)− Eil,kj(ξ)]

(E ∈ C2(0

B)),

= 0 (compatibility, thm. 138(2))

Thus, we have shown that the CLI describes well-defined functions ui(x) on0

B for the displacement components. Next, we show that this displacementfield satisfies the infinitesimal strain-displacement relations.

We construct a path ξ(τ) from x to x+hej in two parts. The first part isan arbitrary smooth curve η(τ) from x to x; the second part is a line segmentfrom x to x + hej 3 ξ(1) = x + hej. Thus,

ξ(τ) =

η(τ) for τ ∈ [0, 1/2] , 3 η(0) = x, η(1/2) = xx + 2 (τ − 1/2)hej for τ ∈ [1/2, 1]

.

We can always select h small enough 3 the curve lies entirely within the open

region0

B, so that ξ(τ) is well-defined. The overall curve is clearly piecewisesmooth, so that ξ

′

k(τ) is well defined within each of the two curve segments:

ξ′

k(τ) =

η′

k(τ) for τ ∈ [0, 1/2)2hδjk for τ ∈ (1/2, 1]

.


˜ x

x

B

0

x +he j

ξ ( τ) = η( τ), τ ∈ [0,1 2]

ξ ( τ) = x + 2(τ −1 2)he j ,

τ ∈ [1 2,1]

Figure 2.5: Piecewise Cesaro line integral in two parts

Also,

Uik(ξ,x + hej) = Eik(ξ) + (xl + hδjl − ξl) [Eik,l(ξ)− Elk,i(ξ)]

= Eik(ξ) + (xl − ξl) [Eik,l(ξ)− Elk,i(ξ)]

+h [Eik,j(ξ)− Ejk,i(ξ)] .

Substituting these results, we evaluate the CLI to obtain an expression forui(x + hej).

ui(x + hej) =

∫ 1/2

0

Eik( η(τ)) + (xl − ηl(τ)) [Eik,l( η(τ))− Elk,i(η(τ))] η′k(τ)dτ

+

∫ 1/2

0

h [Eik,j(η(τ))− Ejk,i( η(τ))] η′

k(τ)dτ

+

∫ 1

1/2

Eik(x + 2 (τ − 1/2)h ej) +

[xl −

(xl + 2 (τ − 1/2)hδjl

)][Eik,l(x + 2 (τ − 1/2)hej)− Elk,i(x + 2 (τ − 1/2)hej)]

+ h[Eik,j(x + 2 (τ − 1/2)hej)− Ejk,i(x + 2 (τ − 1/2)hej)

]2hδjkdτ .


The first integral is ui(x), as is demonstrated in exercise 72 below. Aftersome other simplifications we obtain

ui(x + hej) = ui(x) + h

∫ 1/2

0

[Eik,j(η(τ))− Ejk,i(η(τ))] η′

k(τ)dτ

+2h

∫ 1

1/2

Eij(x + 2 (τ − 1/2)hej)

−2 (τ − 1/2)h[Eij,j(x + 2 (τ − 1/2)hej)

−Ejj,i(x + 2 (τ − 1/2)hej)]

+[hEij,j(x + 2 (τ − 1/2)h ej)

− Ejj,i(x + 2 (τ − 1/2)hej)]

dτ ,

so that

1

h[ui(x + hej)− ui(x)] =

∫ 1/2

0

[Eik,j(η(τ))− Ejk,i( η(τ))] η′

k(τ)dτ

+2

∫ 1

1/2

Eij(x + 2 (τ − 1/2)hej)dτ

−4h

∫ 1

1/2

τ[Eij,j (x + 2 (τ − 1/2)hej)

− Ejj,i(x + 2 (τ − 1/2)h ej)]

dτ

+4h

∫ 1

1/2

[Eij,j (x + 2 (τ − 1/2)hej)

− Ejj,i(x + 2 (τ − 1/2)h ej)]

dτ .


Recalling Eij ∈ C2(0

B), we have by the Mean Value Theorem for integrals∫ 1

1/2

Eij(x + 2 (τ − 1/2)hej)dτ =1

2Eij(x + 2 (τ 1 − 1/2)hej),∫ 1

1/2

τ[Eij,j(x + 2 (τ − 1/2)hej)− Ejj,i(x + 2 (τ − 1/2)hej)

]dτ

=1

2τ 2

[Eij,j(x + 2 (τ 2 − 1/2)hej)− Ejj,i(x + 2 (τ 2 − 1/2)hej)

],∫ 1

1/2

[Eij,j(x + 2 (τ − 1/2)hej)− Ejj,i(x + 2 (τ − 1/2)hej)

]dτ

=1

2


],

where τ 1, τ 2, τ 3 ∈ [1/2, 1] . Thus,

ui,j(x) = limh→0

1

h[ui(x + hej)− ui(x)]

= lim

h→0

∫ 1/2

0


k(τ)dτ

+Eij(x + 2 (τ 1 − 1/2)hej)

−2hτ 2


]+2h


]= Eij(x) +

∫ 1/2

0


k(τ) dτ .

This line integral is path independent, c.f. exercise 73 below, so the followingresults are independent of the choice of connecting curve between x and x.Substituting we obtain

1

2(ui,j(x) + uj,i(x)) =

1

2Eij(x) + Eji(x) +

∫ 1/2

0

[Eik,j(η(τ)) + Ejk,i(η(τ))

−Ejk,i(η(τ))− Eik,j(η(τ))] η′

k(τ)dτ

=1

2Eij(x) + Eji(x)

= Eij(x) (E ∈ Sym ).


Therefore, the vector field defined by the CLI representation does indeedsatisfy the infinitesimal strain-displacement relations.

Next, we consider higher-order partial derivatives of the displacementfield. An analysis similar to the one above (c.f. exercise 74) applied to

ui,j(x) = Eij(x) +

∫ 1/2

0

[Eik,j( η(τ))− Ejk,i(η(τ))] η′

k(τ)dτ

leads toui,jk = Eij,k + Eik,j − Ejk,i.

This last result trivially leads to

ui,jkl = Eij,kl + Eik,jl − Ejk,il,

since at a minimum, E ∈ C2(0

B). We may be able to take additional partialderivatives, depending on the value of N in our continuity assumption for E.

The present results suggest that E ∈ CN(0

B)⇒ u ∈ CN+1(0

B). In particu-lar, the continuity assumptions on E ⇒ continuity of the partial derivatives

of ui of order 2 through N + 1 on0

B . However, further analysis is required toverify that the line integral representations ensure continuity of ui and ui,j

on0

B .Consider, for example, the continuity of ui,j on

0

B . From above, we have

ui,j(x) = Eij(x) + Γij(x) on0

B,

where Γij is the path-independent integral

Γij(x) =

∫ 1/2

0


k(τ) dτ ,

and η(τ) is a piecewise smooth curve from x to x. Next, consider the value of

ui,j at x ∈0

B, a neighbor of x. We construct a piecewise smooth curve η(τ)from x to x in two parts. The first is an arbitrary smooth curve ζ1(τ) from xto x; the second is an arbitrary smooth curve ζ2(τ) from x to x. Specifically,

η(τ) =

ζ1(τ) for τ ∈ [0, 1/4] 3 η(0) = x, η(1/4) = xζ2(τ) for τ ∈ [1/4, 1/2] 3 η(1/4) = x, η(1/2) = x

.


Of course, ζ1 and ζ2 must lie entirely in0

B .We know that Eij(x) is continuous

on0

B by assumption, so to prove that ui,j(x) is continuous on0

B, it remains

to show that Γij(x) is continuous on0

B. Thus, for fixed i and j,

Γij(x)− Γij(x) =

∫ 1/2

1/4

ak(τ) (ζ2)′

k (τ)dτ ,

where ak(τ) := Eik,j(ζ2(τ))− Ejk,i( ζ2(τ)). Therefore,

|Γij(x)− Γij(x)| ≤∫ 1/2

1/4

∣∣∣ak(τ) (ζ2)′

k (τ)∣∣∣ dτ (

∣∣∣∣∫ f

∣∣∣∣ ≤ ∫ |f | ),=

∫ 1/2

1/4

∣∣∣a(τ) · ζ ′2(τ)∣∣∣ dτ (component rep. of · ),

≤∫ 1/2

1/4

|a(τ)|∣∣∣ζ ′2(τ)

∣∣∣ dτ (Cauchy-Schwarz),

≤(

maxτ∈[1/4,1/2]

|a(τ)|)∫ 1/2

1/4

∣∣∣ζ ′2(τ)∣∣∣ dτ

(defns. of max and

∫),

=

(max

τ∈[1/4,1/2]|a(τ)|

)l(x,x),

≤

sup

z∈0B

√[Eik,j(z)− Ejk,i(z)] [Eik,j(z)− Ejk,i(z)]

l(x,x)

(ζ2 ⊂0

B, use sup since0

B is open, no sum on i and j),

in which l(x,x) := the length of the path ζ2(τ) from x to x. Since theterm in parentheses is a constant independent of x, we can make this resultas small as desired by taking x sufficiently close to x, and selecting a linesegment for ζ2 (recall Γij is path independent). Thus, Γij is continuous on0

B⇒ ui,j is continuous on0

B . A similar analysis leads to the conclusion that

the components ui themselves are also continuous on0

B . Combining all of

these results we have shown that u ∈ CN+1(0

B).


Finally we note that since the displacement field satisfies E = sym∇u on0

B, it must be unique to within an infinitesimal rigid deformation by Theorem136.


ui(x) =

∫ 1/2

0

Eik(η(τ)) + (xl − ηl(τ)) [Eik,l(η(τ))− Elk,i(η(τ))] η′k(τ)dτ .

Exercise 73 Show that the line integral∫ 1/2

0


k(τ)dτ

is path independent.


ui,j(x) = Eij(x) +

∫ 1/2

0


k(τ)dτ

⇒ ui,jk = Eij,k + Eik,j − Ejk,i.

Remark 35 The previous theorem applies only to the open region0

B . Sincewe are interested in solving boundary value problems, it is natural to ask,“Dowell-defined continuous extensions of u and its partial derivatives to the

closed region B exist?” Given that u ∈ CN+1(0

B), the trace operator sup-ports continuous extensions to B of the components of u and their partialderivatives up to order N . Thus, under our present assumption, N ≥ 2, weare assured that u ∈ C2(B).

Exercise 75 Given the infinitesimal strain field

[Eij(x)] =

0 − b2x3

b2x1

− b2x3 0 0

b2x1 0 0

, b = constant,

use the Cesaro line integral to find the corresponding displacement field. Notethat the strain components are all linear functions of position, so the specifiedstrain field automatically satisfies the compatibility conditions.

2.5. MOTIONS 143

Remark 36 Before closing this section, it is worthwhile to comment on thephysical significance of the compatibility conditions. We have seen that thecompatibility conditions are sufficient to ensure that the displacement fieldderived from a strain field via the Cesaro Line Integral Representation is a

single-valued function on0

B (c.f. proof of Theorem 140). Thus, the compat-ibility conditions ensure that a given infinitesimal strain field gives rise to acoherent deformed geometry.

2.5 Motions

To some people, the trip is as important as what you find when you get there.So rather than consider a deformation as an isolated, instantaneous event,we now consider the kinematics of how a body moves continuously from anundeformed configuration to a deformed state. Here we return to the generalsetting of finite deformation theory.

Definition 87 A motion of a body is a family of deformations ordered by asingle real parameter called time, denoted t. We introduce a reference time t0associated with the undeformed state of the body.16 Then a motion is denotedby

f(·, t) , t ∈ [t0,∞),

wherey = f(x, t)

is the position vector at time t of the material point identified by the positionvector x in the undeformed state at time t0. A motion inherits all the requiredproperties of a deformation, except that the numbered properties in Definition72 are superceded by the requirements

1. f(x, t0) = x;

2. f ∈ C2(0

B × [t0,∞),V).17

16Typically, t0 = 0, but this is not mandatory.

17The notation0

B × [t0,∞) denotes the Cartesian set product

0

B ×[t0,∞) =

(x, t) : x ∈

0

B, t ∈ [t0,∞)

.


The first requirement means that the body occupies the region0

B at timet0. The second provides the necessary mathematical structure for developingthe concepts of velocity and acceleration. As we shall see, these requirements

are also sufficient to ensure that det∇f > 0 ∀ (x, t) ∈0

B × [t0,∞) (c.f.requirement 2 in Definition 72).

Definition 88 Given a motion f(·, t) ,of a body0

B, the configuration ofthe body at time t, denoted Bt, is the region of E occupied by the body at timet. We write

Bt = f(0

B, t),

where f(0

B, t) is the image of0

B under the deformation f(·, t).

Definition 89 The trajectory = of a body under a motion f(·, t) is the setof ordered pairs

= = (y, t) : y ∈ Bt, t ∈ [t0,∞) .

Thus, a trajectory is a space-time region in a four-dimensional Euclideanspace.

Definition 90 Given a motion f(·, t) ,of a body0

B, the velocity and accel-eration at time t of the material point identified by the position vector x attime t0 are

v(x, t) =∂f

∂t(x, t)

and

a(x, t) =∂v

∂t(x, t) =

∂2f

∂t2(x, t).

These fields are called the referential velocity and acceleration fields.

Remark 37 We use ∂/∂t to denote partial differentiation with respect tothe time variable, where any dependence of the position vector on time isignored. E.g.,

∂f

∂t(x, t) = lim

h→0

f(x, t+ h)− f(x, t)

h.

A motion can therefore be viewed as a function defined on a four-dimensional Cartesianspace, and we can write

f = f(x1, x2, x3, t).

This representation allows us to apply the usual definition of C2 continuity.

2.5. MOTIONS 145

We use the comma notation for partial differentiation w.r.t. the componentsof the position vector. E.g.,

fi,3(x, t) = limh→0

fi(x1, x2, x3 + h, t)− fi(x1, x2, x3, t)

h.

The most common choice in solid mechanics is to use time and the un-deformed coordinates x at time t0 as the independent variables. However,sometimes in solid mechanics (and typically in fluid mechanics) we choosetime and the coordinates y in the deformed state at the current time t as theindependent variables. It is always possible to transform from one approachto the other, since the deformation f(·, t) at any time t is invertible. Thus,if y = f(x, t) is the position at time t of the material point that occupiesposition x at time t0, then

x = f−1(y, t)

is the undeformed coordinate at t0 of the material point that occupies positiony at time t. Accordingly,

Definition 91 Given a motion f(·, t) , the velocity and acceleration at timet of the material point that has position vector y at time t are

v(y, t) = v(f−1(y, t), t),

anda(y, t) = a(f−1(y, t), t).

These fields are called the spatial velocity and acceleration fields.

Remark 38 It should be evident that this definition is equivalent to

v(f(x, t), t) = v(x, t),

anda(f(x, t), t) = a(x, t).

Definition 92 The pair (x, t) are called referential or material variables,while the pair (y, t) are called spatial variables.18

18Material variables are often called Lagrangian variables, and spatial variables are oftencalled Eulerian variables. However, this attribution to Euler and Lagrange is historicallyinaccurate.


It is important to note that

a 6=∂v

∂t(y, t),

because the difference quotient,

∂v

∂t(y, t) = lim

h→0

1

h[v(y, t+ h)− v(y, t)] ,

compares the velocities of the two distinct material points that occupy po-sition y at the two distinct times t and t + h. To obtain the accelerationof a particle, we must examine the change in velocity of the same materialpoint over a specified time interval. This generates convective terms in thespatial representation of the acceleration field as developed in the next twotheorems.

Theorem 141 Let f(·, t) be a motion. Then the corresponding spatialvelocity field v ∈ C1 (=) .

Proof. Since by definition f(·, t) ∈ C2(0

B ×[t0,∞),V), we have that v ∈C1(

0

B × [t0,∞),V). The rest of the proof is an application of the InverseFunction Theorem.19

The C1 continuity of v supports the following theorem.

Theorem 142 The spatial acceleration field is given in terms of the spatialvelocity field by

a =∂v

∂t+ (∇v) v on =.

Proof. We write v(x, t) = v(f(x, t), t) where x = f−1(y, t) to track thechange in velocity of the material point that occupies position y at time t.Then

ai(y, t) =∂vi(f

−1(y, t), t)

∂t(substitution),

= vi,j(f(f−1(y, t), t), t)∂fj∂t

(f−1(y, t), t) +∂vi∂t

(f(f−1(y, t), t), t)

(Chain Rule, v ∈ C1 (=) ),

= vi,j(y, t)vj(f−1(y, t), t) +

∂vi∂t

(y, t) (f(f−1(y, t), t) = y, def. of v),

= vi,j(y, t)vj(y, t) +∂vi∂t

(y, t) (definition of v ).

19See p. 60 of Gurtin’s An Introduction to Continuum Mechanics

2.5. MOTIONS 147

This is the indicial form of the theorem, which is stated in direct notation.

Next we consider the Jacobian determinant of a motion, a quantity relatedto changes in a differential volume.

Definition 93 Let f(·, t) be a motion. The Jacobian determinant, denotedJ(x, t), is a real-valued function defined by

J(x, t) := det F(x, t) on0

B ×[t0,∞).

Theorem 143 Let f(·, t) be a motion. Then

J(x, t) > 0 on0

B ×[t0,∞).

Proof. Our definition of a motion requires that for any fixed value of t,

the mapping f(·, t) is a deformation on0

B . Therefore, f(·, t) is invertible ⇒its Jacobian determinant J(x, t) 6= 0 on

0

B ×[t0,∞). Since J(x, t) must be acontinuous function of both position and time, the requirement J(x, t) 6= 0

on0

B ×[t0,∞) ⇒ either J(x, t) > 0 or J(x, t) < 0 everywhere on0

B ×[t0,∞).At t = t0 we have

f(x, t0) = x⇒ fi,j(x, t0) = δij.

Thus,

J(x, t0) = det F(x, t0)

= det [fi,j(x, t0)]

= det [δij]

= 1.

∴ J(x, t) > 0 everywhere on0

B ×[t0,∞).The next theorem relates the Jacobian determinant to the divergence of

the spatial velocity field.


∂J

∂t(x, t) = J(x, t) div v(f(x,t), t) on

0

B ×[t0,∞).


Proof. We apply the Chain Rule to the identity

vl(x,t) =vl(f(x,t),t)

to obtainvl,i(x,t) =vl,j(f(x,t),t)fj,i(x,t).

For each fixed choice of l, this provides a system of 3 linear algebraic equationsin the 3 unknowns vl,j(f(x,t),t) in which the coefficient matrix is [fj,i(x, t)]

t =[fi,j(x, t)]. Thus, the determinant of the coefficient matrix is

det [fi,j(x,t)] = det F(x, t) = J(x, t) 6= 0.

So, by Theorem 16, the unique solution is

vl,i =1

2Jεijkεpqrfj,qfk,rvl,p.

Now consider the case where l = 1 and i = 1. Then

v1,1 =1

2Jε1jkεpqrfj,qfk,rv1,p,

=1

2Jεpqrv1,p (f2,qf3,r − f3,qf2,r) ,

=1

2J(εpqrv1,pf2,qf3,r − εpqrv1,pf3,qf2,r) (distrib.),

=1

2J(εpqrv1,pf2,qf3,r + εprqv1,pf3,qf2,r) (skew w.r.t. q, r),

=1

2J(εpqrv1,pf2,qf3,r + εpqrv1,pf3,rf2,q) (relabel q, r),

=1

Jεpqrv1,pf2,qf3,r,

=1

Jεpqr

∂f1,p

∂tf2,qf3,r (subst.).

Then recalling the row expansion of a 3× 3 determinant we have

Jv1,1 =

∣∣∣∣∣∣∂f1,1∂t

∂f1,2∂t

∂f1,3∂t

f2,1 f2,2 f2,3

f3,1 f3,2 f3,3

∣∣∣∣∣∣ .Similarly,

Jv2,2 =

∣∣∣∣∣∣f1,1 f2,1 f3,1∂f2,1∂t

∂f2,2∂t

∂f2,3∂t

f3,1 f3,2 f3,3

∣∣∣∣∣∣

2.6. THE TRANSPORT AND LOCALIZATION THEOREMS 149

and

Jv3,3 =

∣∣∣∣∣∣f1,1 f2,1 f3,1

f2,1 f2,2 f2,3∂f3,1∂t

∂f3,2∂t

∂f3,3∂t

∣∣∣∣∣∣ .Summing these three relations we obtain

Jvi,i =∂

∂tdet [fi,j] ,

or∂J

∂t(x, t) = J (x, t) div v(f (x, t) , t) on

0

B × [t0,∞).

Exercise 76 Consider the motion defined on E× [t0,∞) by

y1 = f1 (x, t) = x1ekt,

y2 = f2 (x, t) = x2 + ct,

y3 = f3 (x, t) = x3,

where k and c are constants.

1. Confirm that J (x, t) > 0 and find x = f−1 (y, t) .

2. Find v (x, t) and a (x, t) .

3. Find v (y, t) .

4. Find a (y, t).

2.6 The Transport and Localization Theorems

The balance laws of continuum mechanics are presented in the next chapterfor arbitrary subdomains of the overall body under consideration. Here weintroduce a formal definition of a subbody and the Transport Theorem for thetemporal partial derivative of an integral quantity on the subbody. We alsoexamine the Localization Theorem which converts balance laws in integralform to pointwise relations.


t = 0 t = t

B

0

B t

P P t

x y = f x, t( )

f ⋅, t( )

O

Figure 2.6: Motion of a body and a subbody for the Transport Theorem

Definition 94 Let f(·, t) be a motion of a body0

B. P ⊂0

B denotes the

open, regular region occupied by a subbody of0

B at time t0. The region occu-pied by the same subbody at time t is denoted by

Pt = f(P , t).

We do allow for the possibility that P =0

B .

Theorem 145 (Transport Theorem) Let g ∈ C1(=,<) be a spatial scalarfield. Then

d

dt

∫Pt

g(y, t)dVy =

∫Pt

[∂g

∂t(y, t) + g,i(y, t)vi(y, t) + g(y, t)vi,i(y, t)

]dVy

=

∫Pt

∂g

∂t(y, t) + [gvi],i (y, t)

dVy

=

∫Pt

∂g

∂t(y, t)dVy +

∫∂Pt

g(y, t) [v(y, t) · n(y, t)] dAy,

where n(y, t) is the outward unit normal to ∂Pt at y.20

20This theorem is sometimes called the Reynolds transport theorem after its originator.


Proof. Define

I(t) =

∫Pt

g(y, t)dVy.

Since Pt is a variable domain dependent on time, we change the variable ofintegration from y to x to obtain a region of integration that is invariantw.r.t. time.21

I(t) =

∫Pt

g(y, t)dVy =

∫Pg(f(x, t), t) |det [fi,j(x, t)]| dVx

=

∫Pg(f(x, t), t)J(x, t)dVx,

where we have made use of Theorem 143. Next we use the Chain Rule todifferentiate under the integral to get

I(t) =

∫P

[g,i(f(x, t), t)

∂fi∂t

(x, t) +∂g

∂t(f(x, t), t)

]J(x, t)

+ g(f(x, t), t)∂J

∂t(x, t)

dVx

=

∫P

[∂g

∂t(f(x, t), t) + g,i(f(x, t), t)vi(x, t)

+ g(f(x, t), t)vi,i(f(x, t), t)] J(x, t)dVx

(defn. 91 and thm. 144).

Transforming back to Pt yields

I(t) =

∫Pt

[∂g

∂t(y, t) + g,i(y, t)vi(y, t) + g(y, t)vi,i(y, t)

]dVy,

=

∫Pt

∂g

∂t(y, t) + [gvi],i (y, t)

dVy.

The third alternative form is obtained via the Divergence Theorem22 whichis supported by our assumption that P (and therefore Pt) is regular.

Remark 39 We have presented the Transport Theorem for a scalar fieldg. There are variants of the Transport Theorem for vector and tensor fieldswhich can be derived by applying the scalar version to the vector or tensorcomponents.

21See p. 403 of Courant and John, vol. 2.22See p. 601 of Courant and John, vol. 2.


Remark 40 The surface integral in the third expression of Theorem 145requires that g and v be interpreted via the trace operator when ∂Pt ∩ ∂Bt isnot empty. By assumption, both functions possess sufficient smoothness tosupport the trace operator.

Exercise 77 Given a spatial vector field g ∈ C1(=,V), show that

d

dt

∫Pt

g(y, t)dVy =

∫Pt

∂g

∂t(y, t) +∇g(y, t)v(y, t) + [div v(y, t)] g(y, t)

dVy

=

∫Pt

∂g

∂t(y, t) + [div (g ⊗ v)] (y, t)

dVy

=

∫Pt

∂g

∂t(y, t)dVy +

∫∂Pt

[v(y, t) · n(y, t)] g(y, t)dAy.

The general balance laws of continuum mechanics are developed in termsof integral expressions over arbitrary subbodies. The following localizationtheorem allows us to convert these into pointwise relations.

Theorem 146 (Localization Theorem — referential description) Let

h ∈ C0(0

B,<). If ∫Ph(x)dVx = 0 ∀ P ⊂

0

B,

then

h = 0 on0

B .

Proof. Suppose h 6= 0 on0

B . Then ∃ some x ∈0

B 3 h(x) 6=0, say h(x) >0.Then continuity of h ⇒ ∃ a ball B centered at x 3

h > 0 everywhere on B ∩0

B (see exercise below).

Now choose P = B ∩0

B to get∫Ph(x)dVx > 0.

This contradicts the hypothesis. ∴ h = 0 on0

B .


Exercise 78 Let h ∈ C0(0

B,<) with h(x) >0 for some x ∈0

B . Prove that

∃ a ball B centered at x 3 h > 0 everywhere on B ∩0

B . Hint: Recall thedefinition of continuity of h at x: given any ε > 0 ∃ δ > 0 3 x ∈ B(x, δ)

∩0

B ⇒ |h(x)−h(x)| < ε. Use the definition of the absolute value operator toshow that

|h(x)−h(x)| < ε⇒ −ε < h(x)−h(x) < ε.

Then add h(x) to the inequalities and choose ε = h(x).

Theorem 147 (Localization Theorem — spatial description) Let k ∈C0(=,<). If ∫

Pt

k(y, t)dVy = 0 ∀ Pt ⊂ Bt, t ∈ [t0,∞)

thenk = 0 on =.

Exercise 79 Prove Theorem 147. Hint: Apply the Change of Variable The-orem to obtain a form where the referential version (Theorem 146) is appli-cable.

Exercise 80 Let f(·, t) be a motion of a body0

B. Show that

dVy(f(x, t), t) = J(x, t)dVx(x) on0

B,

where dVx(x) is the differential volume associated with a triad of differential

vectors dxi at x ∈0

B, and dVy(f(x, t), t) is the differential volume associatedwith the triad of differential vectors dyi(f(x, t), t) = F(x, t)dxi.

Chapter 3

Balance Laws and StressTensors

This section develops fundamental balance laws relating to conservation ofmass and balance of momentum. In addition, the chapter also introducesthe Cauchy Equations of motion and various stress tensors that arise in thestudy of solid mechanics.

3.1 References

Same as for Chapter 2.

3.2 Conservation of Mass

3.2.1 Finite deformations

Definition 95 Let f(·, t) be the motion of a body0

B . Then the mass den-

sity of0

B is a function ρ ∈ C1(=,<+) 3∫Pt

ρ(y, t)dVy

is the mass of part P at time t, where the mass of part P represents theamount of material in the part.1

1<+ = x ∈ < : x > 0 .

155

156 CHAPTER 3. BALANCE LAWS AND STRESS TENSORS

The following axiom is the Principle of Conservation of Mass.

Axiom 1 (Principle of Conservation of Mass – CM) The mass of eachpart of a body is conserved at all times; i.e.,

d

dt

∫Pt

ρ(y, t)dVy = 0 ∀ parts P ⊂0

B, t ∈ [t0,∞).

Theorem 148 Mass is conserved iff∫Pρ0(x)dVx =

∫Pt

ρ(y, t)dVy ∀ parts P ⊂0

B, t ∈ [t0,∞),

whereρ0 = ρ(·, t0)

is the mass density function of the reference configuration (i.e., of the bodyat time t0).


Theorem 149 (Referential Density Equation) Mass is conserved iff

ρ(f(x, t), t) =ρ0(x)

J(x, t)∀ x ∈

0

B, t ∈ [t0,∞).

Proof. First, assume that mass is conserved. Then by Theorem 148, for

each part P ⊂0

B and for all times t ∈ [t0,∞),∫Pρ0(x)dVx =

∫Pt

ρ(y, t)dVy

=

∫Pρ(f(x, t), t)J(x, t)dVx (change of integration variable).

∴ by the Localization Theorem (Theorem 146)ρ0(x) = ρ(f(x, t), t)J(x, t).Conversely, suppose that the referential density equation holds. Integrate

over an arbitrary part P to obtain∫Pρ0(x)dVx =

∫Pρ(f(x, t), t)J(x, t)dVx

=

∫Pt

ρ(y, t)dVy (change of integration variable).

Thus, mass is conserved by Theorem 148.

3.2. CONSERVATION OF MASS 157

Remark 41 In other words, the ”differential volume ratio” represented by

J(x, t) =dVydVx

=ρ0(x)

ρ(f(x, t), t)

is precisely the ratio needed to maintain a constant differential mass:

dm = ρ(y, t)dVy

= ρ(f(x, t), t)J(x, t)dVx (c.f. exercise 80),

= ρ0(x)dVx (subst.).

The Spatial Density Equation follows from an application of the Trans-port Theorem (Theorem 145) to the Principle of Conservation of Mass.

Theorem 150 (Spatial Density Equation) Mass is conserved iff

∂ρ

∂t(y, t) +∇ρ(y, t) · v(y, t) + ρ(y, t)∇ · v(y, t) = 0 on =,

or∂ρ

∂t(y, t) +∇ · (ρv) (y, t) = 0 on =.


Remark 42 The Spatial Density Equation is called the Continuity Equationby most authors.

We commonly encounter situations in mechanics where the distributionof some quantity is described in terms of its specific density (i.e., an amountper unit mass). For example, suppose that g ∈ C1(=,<) is a function thatdescribes the specific density of some scalar quantity G (i.e., g describes Gper unit mass). Then the spatial volume density of G (i.e., G per unit volume)is the product g(y, t)ρ(y, t). Thus, the following result is useful for trackingthe amount of G in a subbody.

Theorem 151 (Reduced Transport Theorem) Let g ∈ C1(=,<). Then

d

dt

∫Pt

g(y, t)ρ(y, t)dVy =

∫Pt

[∂g

∂t(y, t) + g,i(y, t)vi(y, t)

]ρ(y, t)dVy.

=

∫Pt

[∂g

∂t(y, t) +∇g(y, t) · v(y, t)

]ρ(y, t)dVy


From here on, we shall assume that mass is always conserved (i.e., thataxiom 1 is in force).


3.2.2 Infinitesimal motions

This subsection investigates the Referential Density Equation (Theorem 149)for motions with small displacement gradients. To begin, we introduce aformal definition for such a motion.

Definition 96 An infinitesimal motion is a motion f(·, t) for which ∃ε ∈ (0, 1) (i.e., 0 < ε < 1) 3

|Hij(x, t)| ≤ ε ∀ (x, t) ∈0

B ×[0,∞),∀ i, j ∈ 1, 2, 3 and ∀ frames X.

Remark 43 We must consider all frames X, because the components of Hare frame-dependent; c.f., Remark 33.

Here it is convenient to extend our definition of the displacement gradientnorm from a deformation to a motion. Thus,

ε = ‖H‖ := sup

x∈0B

t∈[t0,∞)

i,j∈1,2,3

|Hij(x, t)| .

By definition 96, it follows that 0 < ε < 1 no matter what frame we chooseto work in.

Remark 44 Of course, we are most interested in the case of infinitesimalmotions for which ε << 1, for then our infinitesimal kinematic variables aregood approximations to their finite deformation counterparts.

Theorem 152 For an infinitesimal motion,

J(x, t) = 1 + tr H(x, t) +O(ε2).

Proof. We have

J(x, t) = det [Fij(x, t)] (defns. 93 and 32),

= det [δij +Hij(x, t)] (Theorem 118),

= εijk (δ1i +H1i) (δ2j +H2j) (δ3k +H3k) (Theorem 6).

On expansion, this yields

J = det [δij] +Hii + ε1jkH2jH3k + εi2kH1iH3k + εij3H1iH2j + det [Hij] .

From here, the assertion follows from det [δij] = 1, theorems 65 and 6, anddefinition 82.

3.2. CONSERVATION OF MASS 159

Theorem 153 For an infinitesimal motion with∣∣tr H(x, t) +O(ε2)∣∣ < 1,

conservation of mass ⇒

ρ(y, t) = ρ(f(x, t), t) = ρ0(x)− ρ0(x) tr H(x, t) +O(ε2).

Conversely, if the above equation holds, then mass is conserved in the sensethat the Referential Density Equation is satisfied to within O(ε2), i.e.,

ρ(f(x, t), t) =ρ0(x)

J(x, t)+O(ε2).

Proof. Let z = tr H(x, t) +O(ε2), so that

J(x, t) = 1 + z.

by Theorem 152 and |z| < 1 by assumption. The binomial series expansionprovides,

(1 + z)−1 = 1− z + z2 − . . . ,if |z| < 1. Thus, since tr H(x, t) = O(ε), we can write

J−1(x, t) = 1− tr H(x, t) +O(ε2). (*)

Suppose that

ρ(f(x, t), t) = ρ0(x)− ρ0(x) tr H(x, t) +O(ε2).

Then

ρ(f(x, t), t) = ρ0(x)[1− tr H(x, t) +O(ε2)

],

= ρ0(x)[J−1(x, t) +O(ε2)

](*),

=ρ0(x)

J(x, t)+O(ε2),

and mass is conserved to within O(ε2) by Theorem 149. Now suppose thatmass is conserved. Then Theorem 149 requires that

ρ(f(x, t), t) =ρ0(x)

J(x, t)⇒

ρ(f(x, t), t) = ρ0(x)[1− tr H(x, t) +O(ε2)

](*),

= ρ0(x)−ρ0(x) tr H(x, t) +O(ε2).


3.3 Force, Stress and Balance of Momentum

Force is a primitive (i.e., undefined) concept in mechanics associated withthe action of one body on another that causes motion. We consider a body0

B undergoing a motion f(·, t) . At each instant t ∈ [t0,∞), each part P ∈0

Bis assumed to be subjected to two types of force:2

1. Surface Force. Force distributed over the boundary ∂Pt due to the

contact of part Pt with the rest of0

B and/or other bodies.3

2. External Body Force. Force distributed throughout the region Pt due

to the presence of bodies other than0

B (e.g., gravitational forces on abody due to the presence of the earth and other heavenly bodies.

These concepts are developed in the following axioms.

Axiom 2 (Cauchy’s Stress Principle) For every unit vector u, ∃ a vector-valued function tu on the trajectory = 3 the total surface force acting on partP at time t is given by ∫

∂Pt

tn(y,t)(y,t)dAy,

where n(y,t) is the outward unit normal to ∂Pt at y. The vector tn(y,t) iscalled the stress vector (or traction) at the location y on the surface throughy with unit normal n at y at time t. We assume that for each choice of uand t, tu(·, t) ∈ C1(Bt). We may think of tu(y, t) as a surface force per unitarea.

Axiom 3 ∃ a vector-valued function b on = 3 the total external force actingon part P at time t is given by∫

Pt

b(y,t)ρ(y,t)dVy.

2In this chapter we use0

B to refer to the undeformed configuration of the material

volume that occupies the region Bt at time t. Similarly, P ⊂0

B refers to a part of the

material volume0

B.

3We assume that any fields defined on ∂0

B can be evaluated via the trace operator asneeded, provided the field has sufficient smoothness.

3.3. FORCE, STRESS AND BALANCE OF MOMENTUM 161

O

y

n y, t( )t n y,t( )

y, t( )

P t

B t

Figure 3.1: Cauchy’s Stress Principle — stress vector

O

y

P t

B t

b y ,t( )ρ y,t( )

Figure 3.2: Body force density at point y at time t.


The vector b(y,t) is called the body force density at the point y at time t.We assume that for each t, b(·,t) ∈ C0(Bt). We may think of b(y,t) as theexternal body force per unit mass.

The following balance of momentum principles, known as Euler’s Laws,establish relations between the forces acting on a body and its motion. Thefollowing caveats apply.

1. The motion of the underlying frame of reference of the Euclidean pointspace with respect to the “fixed stars” must be at most a constant-velocity translation. For engineering purposes, it generally suffices tofix the frame of reference to a point on the earth.4

2. Velocities are small relative to the speed of light.

3. Bodies are not too small (subatomic).

Definition 97 The linear momentum of part P at time t is∫Pt

v(y,t)ρ(y,t)dVy.

Axiom 4 (Principle of Balance of Linear Momentum) ∀ P ⊂0

B and∀ t ∈ [t0,∞),∫

∂Pt

tn(y,t)(y,t)dAy +

∫Pt

b(y,t)ρ(y,t)dVy =d

dt

∫Pt

v(y,t)ρ(y,t)dVy.

That is, the total force acting on any part of a body at any time t is balancedby the instantaneous time rate of change of the part’s linear momentum.

Definition 98 The angular momentum of part P about0x at time t is∫

Pt

(y− 0

x)× v(y,t)ρ(y,t)dVy,

where0x is the position vector of an arbitrary fixed point of E .

4However, caution is sometimes required, because such a frame does not quite satisfythe caveat.

3.4. THE CAUCHY STRESS TENSOR 163

Axiom 5 (Principle of Balance of Angular Momentum) ∀ P ⊂0

B and∀ t ∈ [t0,∞),∫

∂Pt

(y− 0

x)× tn(y,t)(y,t)dAy +

∫Pt

(y− 0

x)× b(y,t)ρ(y,t)dVy

=d

dt

∫Pt

(y− 0

x)× v(y,t)ρ(y,t)dVy,

where0x is the position vector of an arbitrary fixed point of E . That is, the

total moment of the forces acting on any part of a body about an arbitraryfixed point at any time t is balanced by the instantaneous time rate of changeof the part’s angular momentum about the same point.

Exercise 84 Assume that the forces acting on a body and its velocity sat-isfy the Principle of Balance of Linear Momentum. Show that under thisassumption, if the angular momentum about any one fixed point is balanced,then it is balanced about all fixed points in E (not necessarily in the body).

[Hint: Begin by demonstrating that, for any two fixed points0x and

0z,∫

Pt

(y− 0

z)× v(y,t)ρ(y,t)dVy =

∫Pt

(y− 0

x)× v(y,t)ρ(y,t)dVy

+

∫Pt

(0x − 0

z)× v(y,t)ρ(y,t)dVy.


d

dt

∫Pt

v(y,t)ρ(y,t)dVy =

∫Pt

a(y,t)ρ(y,t)dVy.

Exercise 86 Show that, for any fixed point0x,

d

dt

∫Pt

(y− 0

x)× v(y,t)ρ(y,t)dVy =

∫Pt

(y− 0

x)× a(y,t)ρ(y,t)dVy.

3.4 The Cauchy Stress Tensor

We again consider a body0

B undergoing a motion f(·, t) in response to asystem of forces.


O

P t

δh

δ2 h

δh

l

m

u

z

Figure 3.3: ”Pillbox” for Cauchy’s Action/Reaction Lemma

Theorem 154 (Cauchy’s Action/Reaction Lemma) For any given unitvector u,

t−u(y,t) =− tu(y,t) ∀ (y,t) ∈ =.

Proof. Select arbitrary u and any z ∈ Bt. Choose P ⊂0

B3 Pt ⊂ Bt is therectangular prism with faces normal to the mutually orthogonal unit vectorsl, m and u given by

Pt =

y = z + αl + βm + γu : −δh

2< α, β <

δh

2,

−δ2h

2< γ <

δ2h

2

,

where h is a length and δ > 0 is dimensionless. The parameter δ can alwaysbe selected sufficiently small to ensure that Pt ⊂ Bt, because z is an interiorpoint of the open region Pt. The face of Pt with outward normal u is de-noted U+ and the opposite face is denoted U−. Similarly, L+ is the face withoutward normal l, etc.


Next, apply balance of linear momentum and exercise 85 to obtain∫L+

tl(y,t )dAy +

∫L−

t−l(y,t)dAy

+

∫M+

tm(y,t)dAy +

∫M−

t−m(y,t)dAy

+

∫U+

tu(y,t) dAy +

∫U−

t−u(y,t)dAy

+

∫Pt

[b(y,t)− a(y,t)] ρ(y,t)dVy = 0.

Apply the Mean Value Theorems for surface and volume integrals to the ith

component of this vector equation to get

δ2h2δ (tl(y1,t))i + δ (t−l(y2,t))i

+ δ (tm(y3,t))i + δ (t−m(y4,t))i+ (tu(y5,t))i + (t−u(y6,t))i+ δ2h [bi(y7,t)− ai(y7,t)] ρ(y7 ,t)

= 0,

where y1 ∈ L+, y2 ∈ L−, y3 ∈ M+, y4 ∈ M−, y5 ∈ U+, y6 ∈ U−, and y7 ∈Pt. Both sides of this equation are continuous at δ = 0. Thus, after dividingboth sides by δ2h2, we can take the limit as δ → 0 to obtain

(tu(z,t))i + (t−u(z,t))i = 0.

Therefore,tu(z,t) = −t−u(z,t) for z ∈ Bt.

Remark 45 By definition, tu has sufficient smoothness to extend this theo-rem to the boundary ∂Pt via the trace operator.

Remark 46 To understand the terminology “action/reaction”, consider theinteraction of two disjoint deformed subbodies, Pt and P ′t , 3 their boundarieshave a single point of contact, y =∂Pt∩∂P

′t , and u = n is the outward normal

to Pt at y.

Remark 47 Cauchy’s Action/Reaction Lemma is the continuum analog ofNewton’s Third Law in particle mechanics. However, the continuum versionarises as a theorem, whereas the particle version is introduced as an axiom.


OX1

X2

X3

AAAAAAAAAAAAAAAAAAAAAAAAAAAAAA

T11

T21

T31

Figure 3.4: Components of the traction vector acting on the 2 - 3 Cartesiancoordinate plane

We are ready to proceed with the development of the Cauchy stress ten-sor. We introduce a notation for the components of the traction vector onrectangular Cartesian coordinate planes.

Definition 99Tij(y,t) := ei · tej

(y,t), (y,t) ∈ =.Thus, Tij(y,t) is the ith component of the traction vector at position y attime t acting on a surface through y with unit normal ej at y at time t.The components Tii (no sum) are called normal stresses, and the terms Tij(i 6= j) are called shear stresses. According to our continuity assumptionsfor each traction t,

Tij(·, t) ∈ C1(Bt).

Remark 48 CAUTION!!! This definition is the transpose of that used bymany (perhaps most) authors. It is used here because it makes things a bitsimpler. This difference is easy to miss, because it turns out that Tij is usuallysymmetric.

Theorem 155 (Cauchy’s Theorem on the Existence of the Stress Tensor)∃ a unique second-order tensor field T on the trajectory =, 3 ∀ unit vectorsu,

tu(y,t) = T(y,t)u ∀ (y,t) ∈ =.


T(y,t) is called the Cauchy stress tensor at time t and position y ∈ Bt, andits Cartesian component representation is

T(y,t) = Tijei ⊗ ej.

For each t ∈ [t0,∞), T(·,t) ∈ C1(Bt).5

Proof. We first show that, for any unit vector u,

(tu(y,t))i = Tij(y,t)uj on =. (*)

The easy case is when u is parallel to one of the coordinate axes, say u = e1.Then

(te1(y,t))i = ei · te1(y,t) (calculation of components),

= Ti1(y,t) (defn. 99),

= Tij(y,t)δj1 (prop. of Kronecker’s delta),

= Tij(y,t) (e1)j (subst.).

Cauchy’s Action/Reaction Lemma readily extends this result to cases whereu is the negative of one of the base vectors.

Now we consider the general case where u may have any orientation.To make things easier, we assume that u points into the first octant of ther.C.c.f. X so that ui > 0. For any fixed t ∈ [t0,∞), choose an arbitraryz ∈ Bt and construct the part P 3 Pt is a tetrahedron of altitude h withthree faces Si that are respectively orthogonal to ei and which intersect atz. The remaining face S is orthogonal to u at a distance h from z. Since z isan interior point, h can always be selected small enough to ensure that theentire tetrahedron is contained in Pt. Let A denote the area of S. Then thearea of Si is Aui, and the volume of Pt is 1

3Ah.

Apply the Balance of Linear Momentum to part Pand use exercise 85 toobtain∫S

tu(y,t)dAy +3∑j=1

∫Sj

t−ej(y,t)dAy +

∫Pt

[b(y,t)− a(y,t)] ρ(y,t)dVy = 0.

5If we had used the standard definition of the Cauchy stress components, then thetheorem would read,

tu(y,t) = Tt(y,t)u ∀ (y,t) ∈ =.


O

u

zh

P tS

Figure 3.5: Tetrahedron of altitude h located at z ∈ Bt

Then by the Action/Reaction Lemma and definition 99, the ithcomponent ofthis vector equations is∫S

(tu(y,t))i dAy −3∑j=1

∫SjTij (y,t)dAy +

∫Pt

[bi(y,t)−ai(y,t)] ρ(y,t)dVy = 0.

The Mean Value Theorems for surface and volume integrals yield

(tu(y0,t))iA− Tij(yj,t)Auj + [bi(y4,t)−ai(y4,t)] ρ(y4,t)Ah

3= 0,

where y0 ∈ S, yj ∈ Sj, and y4 ∈ Pt.6 Divide by A and take the limitas h → 0 (c.f. the justification of a similar step in the proof of Cauchy’sAction/Reaction Lemma) to get

(tu(z,t))i = Tij(z,t)uj for z ∈ Bt.

Again, this result can be extended to the boundary ∂Bt via the trace operator.If we exercise care with signs, we can extend this part of the proof to vectorsuthat point into any of the other octants.

Next we need to show that the Tij in (*) are components of a second-ordertensor. We have from (*) in each frame X,

(tu)i = Tijuj

6The summation in the term Tij(yj ,t)Auj is to be interpreted according to TijAuj .


for arbitrary unit vectors u, where tu is known to be a vector. Thus, by theinverse transformation rule,

(tu)i = Tij

(λlju

′

l

)⇒

λkiλljTiju′

l = λki (tu)i (mult. by λki, props. of <),

= (tu)′

k (trasformation rule),

= T′

kl u′

l ((*) in X′),

where T′

kl := e′

k · te′l

consistent with definition 99. Considering that u is

an arbitrary unit vector, this last result implies

T′

kl = λkiλljTij

— the transformation rule for the components of a second-order tensor. Thus,by Theorem 53, the Tij are indeed tensor components. Reverting to directnotation with

T := Tijei ⊗ ej = T′

ije′

i ⊗ e′

j = . . . ,

(*) becomestu = Tu.

The required smoothness of the stress tensor T = Tijei ⊗ ej follows fromthe definition of Tij and the assumed smoothness of the traction vector tu.

We have only to establish the uniqueness of T to complete the proof.Suppose ∃ another second-order tensor field T on = 3 for any unit vector u,

tu(y,t) = T(y,t)u.

We also havetu(y,t) = T(y,t)u,

and ∴T(y,t)u = T(y,t)u

for any unit vector u. This result leads to the conclusion (see the followingexercise) that T = T on =.

Exercise 87 Let T,T be second-order tensor fields on =. Show that T(y,t)u =T(y,t)u ∀ unit vectors u implies T = T on =.


Exercise 88 An observant student would point out that there is a gap in ourproof. We have considered the cases of u parallel to any coordinate axis andu pointing into any octant of the frame X. However, we have not coveredthe case where u lies in the plane of any pair of coordinate axes 3 u is notparallel to either axis. In this case, Cauchy’s tetrahedron construction fails,so we must treat this as a third case. We can construct a triangular “plate”of uniform thickness and height h at z ∈ Bt and follow a similar argument toCauchy’s tetrahedron development above. Use this approach to complete theproof that (tu(z,t))i = Tij(z,t)uj ∀ u at z ∈ Bt. Be sure to maintain a fullythree-dimensional setting throughout.

So far, we have developed the Cauchy stress tensor as a function of thespatial variables y on a deformed region Bt. This proves to be an awkwardapproach in solid mechanics, because Bt is generally unknown a priori, andfurthermore, this region varies with time. Thus, it is often convenient todevelop the Cauchy stress as a function of the referential variables.

Definition 100 Given a motion f(·,t) , the referential description of theCauchy stress field is

T(x,t) = T(f(x,t),t), (x,t) ∈0

B × [t0,∞).

That is, T(x,t) is the Cauchy stress tensor at time t at the deformed locationy = f(x,t) occupied by the material point that occupies position x at time t0.

Remark 49 This referential stress description will suffice for the develop-ment of elastic constitutive theory in Chapter 4. However, in our later devel-opment of the linearized theory and in the formulation of boundary/initial-value problems it will be necessary to work with the more elaborate Piola-Kirchhoff stress tensors intorduced in Section 3.7 below.

3.5 Cauchy’s Stress Equations of Motion

Section 3.3 introduced Euler’s Principles of Balance of Linear and AngularMomentum. These consist of integral expressions that determine the rela-tions between the motion of a body and the system of forces acting on it.This section develops Cauchy’s equations of motion — the correspondinglocal forms of these relations.

3.5. CAUCHY’S STRESS EQUATIONS OF MOTION 171

Theorem 156 (Cauchy’s First Stress Equation of Motion) If the lin-ear momentum of a body is balanced, then

∇ ·T(y, t) + ρ(y, t)b(y, t) = ρ(y, t)a(y, t) on =.

Conversely, if ∃ a tensor field T ∈ C1(=,LinV) 3 the stress vector field isgiven by

tu = Tu on =,and if

∇ ·T(y, t) + ρ(y, t)b(y, t) = ρ(y, t)a(y, t) on =,then the linear momentum of the body is balanced.

Proof. Suppose that linear momentum is balanced. Then by exercise 85,∫∂Pt

tn(y,t)(y,t)dAy+

∫Pt

ρ(y,t) [b(y,t)− a(y,t)] dVy = 0 ∀ parts P ⊂0

B . (*)

We can use Theorem 155 to rewrite the surface integral as∫∂Pt

tn(y,t)(y,t)dAy =

∫∂Pt

T(y,t)n(y,t)dAy ⇔∫∂Pt

(tn(y,t)(y,t)

)idAy =

∫∂Pt

Tij(y,t)nj(y,t)dAy

=

∫Pt

Tij,j(y,t)dVy (Divergence Thm. 116)

=

∫Pt

(∇ ·T)i (y,t)dVy.

Substituting this result we obtain∫Pt

[∇ ·T(y,t)+ρ(y,t)b(y,t)−ρ(y,t)a(y,t)] dVy = 0 ∀ parts P ⊂0

B, (**)

and by the Localization Theorem,

∇ ·T(y,t) + ρ (y,t)b(y,t)− ρ(y,t)a(y,t) = 0 on =. (***)

Conversely, suppose that (***) holds. Integrate over Pt corresponding to

an arbitrary part P ⊂0

B to get (**). Then tu = Tu on = and the DivergenceTheorem 116 yield (*), which implies that linear momentum is balanced viaexercise 85.


Theorem 157 (Cauchy’s Second Stress Equation of Motion) If the lin-ear momentum is balanced, then the angular momentum is also balanced iff

Tt(y,t) =T(y,t) on =;

i.e., the Cauchy stress tensor is symmetric on =.

Proof. We assume throughout that the linear momentum is balanced. Inaddition, we first suppose that the angular momentum is balanced. Thenby exercise 84, the angular momentum about the origin must be balanced.Futher, by exercise 86, we have∫

∂Pt

y×tn(y,t)(y,t)dAy +

∫Pt

y× [b(y,t)− a(y,t)] ρ(y,t)dVy = 0.

(3.1)where the surface integral written in terms of the Cauchy stress tensor is∫

∂Pt

y×tn(y,t)(y,t)dAy =

∫∂Pt

y× [T(y,t)n(y,t)] dAy ⇒∫∂Pt

(y×t n(y,t)(y,t)

)idAy =

∫∂Pt

εijkyj [T(y,t)n(y,t)]k dAy

=

∫∂Pt

εijkyj [Tkl(y,t)nl(y,t)] dAy

=

∫Pt

εijk [yjTkl(y,t)] ,l dVy (Divergnc. Thm. 116),

=

∫Pt

εijk [δjlTkl (y,t) + yjTkl,l(y,t)] dVy (calculus),

=

∫Pt

εijk [Tkj(y,t ) + yjTkl,l(y,t)] dVy.

Then substituting into the component form of (3.1), we obtain∫Pt

εijkyj [Tkl,l(y,t)+ρ(y,t)bk(y,t)−ρ(y,t)ak(y,t)] dVy+

∫Pt

εijkTkj(y,t)dVy = 0.

(3.2)The first integral vanishes by Cauchy’s First Stress Equation of Motion,because the linear momentum is balanced. This yields∫

Pt

εijkTkj(y,t)dVy = 0.


Since this must hold ∀ parts P ⊂0

B, the Localization Theorem ⇒

εijkTkj(y,t) = 0 on =. (3.3)

This result is equivalent to (see exercise below)

Tij(y,t) = Tji(y,t) on =; (3.4)

so T is symmetric.Conversely, assume that (3.4) holds. Then (3.3) holds, and with Cauchy’s

First Equation of Motion we get (3.2). Then (3.1) follows from the DivergenceTheorem and Cauchy’s Theorem on the Existence of the Stress Tensor. Thus,we have that the angular momentum is balanced.

Exercise 89 Show that εijkTkj(y,t) = 0 ⇔ Tij(y,t) = Tji(y,t).

Definition 101 The kinetic energy of a part P ⊂0

B at time t ∈ [t0,∞) is

1

2

∫Pt

|v(y,t)|2 ρ(y,t)dVy.

Definition 102 The stress power of a part P ⊂0

B at time t ∈ [t0,∞) is∫Pt

T(y,t) · L(y,t)dVy,

where L(y,t) := ∇v(y,t) is the spatial velocity gradient.

Definition 103 The power of the forces acting on a part P ⊂0

B at timet ∈ [t0,∞) is∫

∂Pt

tn(y,t)(y,t) · v(y,t)dAy +

∫Pt

b(y,t) · v(y,t)ρ(y,t)dVy.

Definition 104 The rate of deformation tensor is the symmetric part of thespatial velocity gradient. That is,

D(y,t) := sym L(y,t)


Theorem 158 (Theorem of Expended Power) If the linear momentum

is balanced, then ∀ parts P ⊂0

B and ∀ times t ∈ [t0,∞),

d

dt

∫Pt

1

2|v(y,t)|2 ρ(y,t)dVy =

∫∂Pt

tn(y,t)(y,t) · v(y,t)dAy

+

∫Pt

b(y,t) · v(y,t)ρ(y,t)dVy

−∫Pt

T(y,t) · L(y,t)dVy.

In other words, the time rate of change of the kinetic energy of any part ofa body with balanced linear momentum is equal to the power of the forcesacting on the part minus the part’s stress power. If in addition the angularmomentum is balanced, then

d

dt

∫Pt

1


∫∂Pt


+

∫Pt


−∫Pt

T(y,t) ·D(y,t)dVy.

That is, the stress power of the part simplifies to∫Pt

T(y,t) ·D(y,t)dVy.

Proof. By the Reduced Transport Theorem (Theorem 151),

d

dt

∫Pt

1


∫Pt

v(y,t) · a(y,t)ρ(y,t)dVy (see exercise below),

=

∫Pt

v· (∇ ·T + ρb) dVy (1st Cauchy Eqn. Of Motion).

We also have,∫Pt

v· (∇ ·T) dVy =

∫Pt

vi (∇ ·T)i dVy

=

∫Pt

viTij,jdVy (defn. of ∇ ·T),

=

∫Pt

[(viTij),j − vi,jTij

]dVy (product rule).


By the Divergence Theorem 116,∫Pt

(viTij),j dVy =

∫∂Pt

viTijnjdAy

=

∫∂Pt

vi (tn)i dAy Cauchy’s thm.),

=

∫∂Pt

v · tndAy

=

∫∂Pt

tn · vdAy.

Further,

vi,jTij = (∇v)ij Tij

= LijTij

= L ·T (defn. of · in Lin V),

= T · L (commutativity).

Combining these results we obtain

d

dt

∫Pt

1


∫∂Pt


+

∫Pt


−∫Pt


Now suppose that the angular momentum is also balanced. Then the stresstensor is symmetric by Cauchy’s Second Stress Equation of Motion (theorem157). Therefore,

T · L = TijLij (defn. of · in LinV),

= TijL(ij) (theorems 2 and 3),

= TijDij (subst.),

= T ·D (defn. of · in Lin V),

which yields the desired simplification.


Exercise 90 Show that ∀ parts P ⊂0

B and ∀ times t ∈ [t0,∞),

d

dt

∫Pt

1


∫Pt

v(y,t) · a(y,t)ρ(y,t)dVy.

Remark 50 From here on, we assume that both the linear and angular mo-mentum are balanced.

3.6 Some Properties of the Cauchy Stress Ten-

sor

The Cauchy stress tensor T is symmetric, according to Cauchy’s SecondStress Equation of Motion (Theorem 157) and our assumption that the an-gular momentum is balanced. Thus, a number of properties of symmetrictensors can immediately be ascribed to the Cauchy stress tensor.

Theorem 159 For each time t and for each point y ∈ Bt, ∃ at least one

frame∗X, called the principle axes of stress, 3

[ ∗T ij(y, t)

]=

τ 1(y, t) 0 00 τ 2(y, t) 00 0 τ 3(y, t)

.The principal values of T, denoted τ i, are called the principal stresses.

Definition 105 Decompose the stress vector tn(y, t) 3

tn(y, t) = Nn(y, t)n + sn(y, t)

wheren · sn(y, t) = 0.

Then Nn(y, t) is the (scalar) normal stress, and

Sn(y, t) := |sn(y, t)|

is the (scalar) shear stress at time t exerted on the surface at y having outwardunit normal n.

3.6. SOME PROPERTIES OF THE CAUCHY STRESS TENSOR 177

y

n

Nnn

t n

sn

Figure 3.6: Decomposition of the stress vector

Theorem 160 The scalar normal stress is given by the quadratic form

Nn(y, t) = n ·T(y, t)n.


Theorem 161 Suppose the principal stresses are ordered 3

τ 1 ≥ τ 2 ≥ τ 3.

Then

maxn∈V|n|=1

Nn(y, t) = τ 1(y, t),

minn∈V|n|=1

Nn(y, t) = τ 3(y, t),

maxn∈V|n|=1

Sn(y, t) =1

2[τ 1(y, t)− τ 3(y, t)] .



3.7 The Piola-Kirchhoff Stress Tensors

The Cauchy stress tensor is a natural choice for formulating problems in thespatial description. However, the spatial configuration of a body is generallyunknown a priori in solid mechanics, so we generally prefer to work in thereferential description. An obvious (but as it turns out, not fully successful)response is to rewrite the Cauchy stress as a function of the referential vari-ables. Thus, suppose f(·, t) is the motion of a body. Then we define the“referential” Cauchy stress tensor field as

T(x, t) := T(f(x, t), t) ∀ (x, t) ∈0

B ×[t0,∞);

⇔ T(y, t) = T(f−1(y, t), t) ∀ (y, t) ∈ Bt × [t0,∞).

Theorem 162

(∇ ·T(f(x, t), t))i = f−1k,j (f(x, t), t)Tij,k(x, t).

Proof. By the Chain Rule we have

(∇ ·T(y, t))i = Tij,j(y, t) = Tij,k(f−1(y, t), t)f−1

k,j (y, t);

so that(∇ ·T(f(x, t), t))i = f−1

k,j (f(x, t), t)Tij,k(x, t).

Now, definingb(x, t) := b(f(x, t), t),

and recallinga(f(x, t), t) = a(x, t)

and the Referential Density Equation (Theorem 149), we can rewrite Cauchy’sFirst and Second Stress Equations of Motion in terms of T as

J(x, t)f−1k,j (f(x, t), t)Tij,k(x, t) + ρ0(x)bi(x, t) = ρ0(x)ai(x, t) on

0

B ×[t0,∞);

Tt(x, t) = T(x, t) on0

B ×[t0,∞).

This result is good as far as it goes, but it is inconvenient because the firstterm does not have the form of a divergence. Ideally, we would like to rewriteCauchy’s two equations of motion in terms of the divergence of a symmetric

3.7. THE PIOLA-KIRCHHOFF STRESS TENSORS 179

referential stress tensor. It turns out that we can only accomplish one of theseobjectives at a time, and these alternatives will lead us to the first and secondPiola-Kirchhoff stress tensors. We next introduce a useful identity and thefirst Piola-Kirchhoff stress tensor in preparation for developing alternativeforms of the stress equations of motion.

Theorem 163 (Piola’s Identity) For any motion f(·, t) ,

∂

∂xj

[J(x, t)f−1

j,k (f(x, t), t)]

= 0 on0

B ×[t0,∞),

or in direct notation

∇ ·(JF−t

)= 0 on

0

B ×[t0,∞).

Proof. Recall y = f(x, t)⇔ x = f−1(y, t). ∴

yi = fi(x, t) = fi(f−1(y, t), t);

and partial differentiation w.r.t. yk yields

∂yi∂yk

= fi,j(f−1(y, t), t)f−1

j,k (y, t) = δik. (3.5)

For each fixed value of k, and at each (y, t) ∈ =, we have a system of threelinear simultaneous equations in the three unknowns f−1

j,k (y, t), where thecoefficient matrix is fi,j(f

−1(y, t), t). This system has the unique solution,

f−1i,k =

1

2Jεimnεkpqfp,mfq,n

since

det [fi,j] = det F = J 6= 0.

Thus,

J(x, t)f−1j,k (f(x, t), t) =

1

2εjmnεkpqfp,m(x, t)fq,n(x, t).

Then differentiating w.r.t. xj, we find the desired result (see the followingexercise).


To get at the direct form of the theorem, note that (3.5) ⇒[f−1i,j

]= [fi,j]

−1

= [Fij]−1

=[(

F−1)ij

].

∴, from our indicial result,

(0)k =∂

∂xj

(Jf−1

j,k

)=

∂

∂xj

(J(F−1

)jk

)=

∂

∂xj

(J(F−t)kj

)=

∂

∂xj

((JF−t

)kj

)=

(∇ ·(JF−t

))k

⇒ ∇ ·(JF−t

)= 0.


∂

∂xj

[1

2εjmnεkpqfp,m(x, t)fq,n(x, t)

]= 0.

Definition 106 The first Piola-Kirchhoff stress tensor is

S(x, t) := J(x, t)T(x, t)F−t(x, t),

orSij(x, t) := J(x, t)f−1

j,k (f(x, t), t)Tik(x, t).

The next theorem restates Cauchy’s stress equations of motion in termsof the first Piola-Kirchhoff stress tensor.

Theorem 164 Cauchy’s stress equations of motion are equivalent to

∇ · S+ρ0b =ρ0aFSt = SFt

on

0

B ×[t0,∞).


Proof. We have previously shown that Cauchy’s first equations of motionis equivalent to

J(x, t)f−1k,j (f(x, t), t)Tij.k(x, t) + ρ0(x)bi(x, t) = ρ0(x)ai(x, t) on

0

B ×[t0,∞).

Thus, to prove the first equivalence, we only need to show that

(∇ · S)i (x, t) = Sij,j(x, t) = J(x, t)f−1k,j (f(x, t), t)Tij,k(x, t).

Accordingly, we have

Sij,j(x, t) =∂

∂xj

[J(x, t)f−1

j,k (f(x, t), t)]Tik(x, t)

+J(x, t)f−1j,k (f(x, t), t)Tik,j(x, t)

= J(x, t)f−1j,k (f(x, t), t)Tik,j(x, t) (thm. 163),

= J(x, t)f−1k,j (f(x, t), t)Tij,k(x, t) (relabel).

Note that definition 106 for S is equivalent to

T(x, t) =1

J(x, t)S( x, t)Ft(x, t).

We have previously shown that Cauchy’s second stress equation of motion is

equivalent to Tt(x, t) = T(x, t) on0

B ×[t0,∞)

⇒ 1

J(x, t)F(x, t)St(x, t) =

1

J(x, t)S(x, t)Ft(x, t)

(subst., transpose of product)

⇒ F(x, t)St(x, t) = S(x, t)Ft(x, t)

The next theorem motivates a physical interpretation of the first Piola-Kirchhoff stress tensor.

Theorem 165 Let n(y, t), y ∈ ∂Pt be the unit outward normal at y to∂Pt, and let m(x), x ∈ ∂P be the unit outward normal at x to ∂P (i.e., anundeformed normal vector). Then the total surface force and the total bodyforce acting on part P at time t are∫

∂Pt

tn(y,t)( y, t)dAy =

∫∂P

S(x, t)m(x)dAx


and ∫Pt

b(y, t)ρ(y, t)dVy =

∫P

b(x, t)ρ0(x)dVx,

respectively.

Proof. The total surface force acting on part P at time t is∫∂Pt

tn(y,t)( y, t)dAy =

∫∂Pt

Tij(y, t)nj(y, t)dAy (Cauchy’s Theorem 155),

=

∫Pt

Tij,j(y, t)dVy (Divergence Theorem),

=

∫PTij,j(f(x, t), t)J(x, t)dVx (change of variable),

=

∫PTij,k(x, t)f

−1k,j (f(x, t), t)J(x, t)dVx (thm. 162),

=

∫PSij,j(x, t)dVx (c.f. proof of thm. 164),

=

∫∂PSij(x, t)mj(y, t)dAx (Divergence Theorem).

The total body force acting on part P at time t is∫Pt

b(y, t)ρ(y, t)dVy =

∫P

b(f(x, t), t)ρ(f(x, t), t)J(x, t)dVx (change of variable),

=

∫P

b(x, t)ρ(f(x, t), t)J(x, t)dVx (defn. of b),

=

∫P

b(x, t)ρ0(x)dVx (Ref. Density Eqn., thm. 149).

In words, we have shown that S maps the unit outward normal to the un-deformed surface ∂P at x (i.e., m) to the surface traction acting at y = f(x, t)on the deformed surface ∂Pt at time t, but measured per unit undeformedarea. Sometimes S is called the engineering or nominal stress, because it isreferenced to the undeformed area (the Cauchy stress T is sometimes calledthe true stress). We have also shown that b(x, t) is the body force acting onthe deformed part Pt at time t at y = f(x, t), but measured per unit mass ofthe undeformed body P at x.

We do not obtain an elegant form for the second Cauchy stress equationof motion, an unfortunate consequence of the asymmetry of S.


Theorem 166 The total moment about x due to the surface force acting onpart P at time t and the total moment about x due to the body force actingon part P at time t are∫

∂Pt

(y − x)× tn(y,t)(y, t)dAy =

∫∂P

(f(x, t)− x)× S(x, t)m(x)dAx

and ∫Pt

(y − x)× b(y, t)ρ(y, t)dVy =

∫P

(f(x, t)− x)× b(x, t)ρ0(x)dVx,

respectively.

Exercise 94 Prove thm. 166.

To summarize our results so far, the spatial Cauchy stress tensor T has theadvantages that it appears through its divergence in the first stress equationof motion and that the second stress equation of motion reduces to symmetryof the stress tensor. However, it has the distinct disadvantage that it is basedon the unknown spatial configuration. The referential Cauchy stress tensorT overcomes the latter drawback, but it no longer enters the first stressequation of motion through its divergence. The first Piola-Kirchhoff (PK-I)stress tensor S is defined on the known undeformed geometry and appears viaits divergence in the first stress equation of motion. Its main disadvantageis its asymmetry and the complicated form of the second stress equation ofmotion. Thus, sometimes it is advantageous to work with the second Piola-Kirchhoff (PK-II) stress tensor P (introduced below). As we shall see, thesecond stress equation of motion reduces to symmetry of P, and the firststress equation of motion involves the divergence of a simple mapping of P(albeit, not simply the divergence of P).

Definition 107 The second Piola-Kirchhoff stress tensor is

P(x, t) := F−1(x, t)S(x, t)

= J(x, t)F−1(x, t)T(x, t)F−t(x, t),

or

Pij(x, t) := J(x, t)f−1i,l (f(x, t), t)f−1

j,k (f(x, t), t)Tlk(x, t).


ConsideringP(x, t)m(x) = F−1(x, t)S(x, t)m(x),

we see that P maps the unit outward normal to the undeformed surface ∂Pat x (i.e., m) into a force vector that is related to the surface traction actingat y = f(x, t) on the deformed surface ∂Pt at time t per unit undeformedarea in the same way that an undeformed material vector dX is related to itsdeformed image dx at time t (i.e., dX(x, t) = F−1(x, t)dx).

Theorem 167 Cauchy’s stress equations of motion are equivalent to

∇ · (FP) +ρ0b =ρ0aPt = P

on

0

B ×[t0,∞).


We close the section with two theorems that provide referential expres-sions for expended power and kinetic energy.

Theorem 168 (Theorem of Expended Power, Referential Description)

If the linear momentum is balanced, then ∀ parts P ⊂0

B and ∀ times t ∈[t0,∞),

d

dt

∫P

1

2|v(x,t)|2 ρ0(x) dVx =

∫∂P

[S(x, t)m(x)] · v(x, t) dAx

+

∫P

b(x, t) · v(x, t)ρ0(x) dVx

−∫P

S(x, t) · ∇v(x, t) dVx.

Exercise 96 Prove Theorem 168. Base your proof on the statement of localbalance of linear momentum in the referential description,

∇ · S + ρ0b = ρ0a on0

B ×[t0,∞).

Calculated

dt

∫P

1

2|v|2 ρ0 dVx,

referring to the proof of Theorem 158 as a guide.

3.8. AREA METRICS AND NANSON’S RELATION 185

Theorem 169 For each part P ⊂0

B and for each time t ∈ [t0,∞),∫P

1

2|v(x,t)|2 ρ0(x)dVx =

∫Pt

1

2|v(y,t)|2 ρ(y,t)dVy.


3.8 Area Metrics and Nanson’s Relation

In many applications (for example, those involving use of the divergencetheorem), it is necessary to form the product of the normal vector and adifferential surface area. In particular, we might wish to know the relationbetween this product defined on the undeformed configuration of a bodyand the analogous property on the deformed geometry. Here we develop thenecessary relations.

Let0n (x) be the unit normal to ∂

0

B at x ∈ ∂0

B, and let f(x, t) bethe trace of a given motion on the boundary at x and time t. Consider

differential vectors d0r and d

0s in the tangent plane of ∂

0

B at x that define

a differential area d0a,

d0a = |d 0

r ×d0s|.

Further, let y = f(x, t), let n be the unit normal to the deformed surfaceat y, and consider the “deformed differential vectors” dr and ds at (y, t) =(f(x, t), t) defined by

dr = F(x, t)d0r

ds = F(x, t)d0s

which define a deformed differential area da = |dr× ds|. Thus,

d0r = F−1 (y,t) dr

d0s = F−1 (y,t) ds.

Recall also,

J−1εlmn = εijkf−1i,l f

−1j,mf

−1k,n;

J =ρ0

ρ.


Now we write

0n d

0a = d

0r ×d

0s

= F−1dr× F−1ds

or in indicial form,

⇒ 0ni d

0a = εijkf

−1j,m (dr)m f

−1k,n (ds)n

⇒ f−1i,l

0ni d

0a = εijkf

−1i,l f

−1j,mf

−1k,n (dr)m (ds)n

= J−1εlmn (dr)m (ds)n

⇒ F−t0n d

0a =

ρ

ρ0

dr×ds

=ρ

ρ0

nda.

Thus, we have

n da =ρ0

ρF−t

0n d

0a

which is known as Nanson’s relation. Sometimes the normal vectors anddifferential areas are combined to form a scaled normal vector, such that

d0a:=

0n d

0a and da := n da. Then we have

da =ρ0

ρF−td

0a .

Chapter 4

Elastic Response

The discussion in this chapter is predicated on the assumptions

• that the body in question is undergoing a motion f(·, t) under theaction of a system of forces,

• that mass is conserved,

• that linear and angular momentum are balanced.

4.1 References


4.2 The Elastic Response Function

So far we have considered general principles of continuum mechanics — inparticular, the kinematic relations and the balance laws. This chapter devel-ops the constitutive laws which establish the relations between the kinematicfields and the force-related fields that are peculiar to elasticity. These rela-tions are built upon a few essential notions of elastic response:

1. The stress in a body depends only on the present (time t) configurationof the body and is independent of the motion by which that presentconfiguration is attained.

187

188 CHAPTER 4. ELASTIC RESPONSE

2. The present configuration is described by a deformation of a definedreference configuration.

3. Since stress represents a contact force, the stress at a material point de-pends only on the present configuration in the immediate neighborhoodof that point. Since the “local” information is carried in a deforma-tion’s gradient, the stress should be a function of the local deformationgradient.

These ideas lead to the following definition of elastic response.

Definition 108 A body0

B is (Cauchy) elastic if ∃ a function

G : Lin V+×0

B→ Sym

3 the referential Cauchy stress field corresponding to the motion f(·, t) isgiven by

T(x, t) = G(F(x, t),x), (x, t) ∈0

B ×[t0,∞).

and the corresponding spatial Cauchy stress field has the smoothness property

T(y, t) := T(f−1(y, t), t) ∈ C1(Bt).

The function G is called the elastic response function, and the above equationis called the elastic constitutive equation.

Remark 51 Given a motion f(·, t), the elastic constitutive equation pro-duces the referential stress field T. The spatial Cauchy stress field is computedon the trajectory = as

T(y, t) = T(f−1(y, t), t).

Now by Cauchy’s Theorem on the Existence of the Stress Tensor (Theorem155), for each t ∈ [t0,∞), the spatial stress field has the smoothness property

T(·, t) ∈ C1(Bt).

Therefore, the elastic response function is required to generate a stress fieldwith this level of continuity, which supports the calculation of the divergence∇ ·T in Cauchy’s first stress equation of motion.

4.2. THE ELASTIC RESPONSE FUNCTION 189

Remark 52 Lin V+×0

B is the proper domain for the elastic response func-

tion, since the deformation gradient F(x, t) ∈ Lin V+ and x ∈0

B. Thecodomain is Sym, in compliance with Cauchy’s second stress equation of mo-tion which requires the Cauchy stress to be a symmetric tensor. The explicitdependence on the position x is retained in our definition of the domainto allow for heterogeneous (or inhomogeneous) bodies for which the stressresponse function varies from point to point in the body. There is no such de-pendence for homogeneous bodies, and then we may write G : Lin V+ → Symand T(x, t) = G(F(x, t)).

Definition 109 An elastic body is homogeneous w.r.t. stress response if ∃a reference configuration

0

B 3 the elastic response function defined relative to

deformations on0

B (let’s denote it G0(·,x)) is the same at all material pointsx; i.e.,

G0(F,x) = G0(F, x) ∀ x, x ∈0

B and ∀ F ∈LinV+.

We say that the body is homogeneous w.r.t. stress response relative to refer-

ence configuration0

B .

Remark 53 In general, the choice of reference configuration is arbitrary.Note that the elastic response function of even a homogeneous elastic bodywill generally not be uniform w.r.t. position for all possible choices of thereference configuration. For example, the elastic response function will benon-uniform for a reference configuration derived from a deformation of astress-free configuration if the deformation has a non-uniform deformationgradient. The arbitrariness in the reference configuration is reduced somewhatif we require the reference configuration to be a stress-free state. However, ...

Remark 54 It is commonly (but by no means universally) assumed that thereference configuration represents a natural state of the body 3 the stressfield vanishes (i.e., there is no initial stress). This assumption of courserequires that

G(I,x) = 0 ∀ x ∈0

B .

Although this assumption is not always appropriate, we shall nonethelessadopt it for reasons of simplicity from here on.


x (2)

x (1)

y

O

f (1)⋅,t( )

f (2)⋅,t( )

g

B(1)

B(2 )

B t

Figure 4.1: Alternative reference configurations for elastic response

We have seen that the property of homogeneity is manifest only in cer-tain special choices of the reference configuration, and that there are alsospecial configurations called natural states. Thus, the character of the elas-tic response function is very much dependent on the choice of the referenceconfiguration. This begs the question, “If ∃ an elastic response function for abody w.r.t. one reference configuration, does ∃ an elastic response functionw.r.t. all configurations of the body?”

Theorem 170 If a body is elastic w.r.t. one reference configuration, thenit is elastic w.r.t. all reference configurations. Specifically, let B(1) and B(2)

denote two reference configurations, such that g is the deformation from B(1)

to B(2). I.e.,

B(2) = g(B(1)

).

Let the motion of the body be described by

y = f (1)(x(1), t) = f (2)(x(2), t) = f (2)(g(x(1)), t),

4.2. THE ELASTIC RESPONSE FUNCTION 191

where x(1) ∈ B(1), x(2) ∈ B(2) and t ∈ [t0,∞). Then if

T(1)(x(1), t) = G(1)(F(1)(x(1), t),x(1))

where F(1)(x(1), t) := ∇f (1)(x(1), t), then

T(2)(x(2), t) = G(2)(F(2)(x(2), t),x(2))

where F(2)(x(2), t) := ∇f (2)(x(2), t) and the response function for the secondconfiguration is given in terms of the first response function by

G(2)(F(2)(x(2), t),x(2)) = G(1)(F(2)(x(2), t)∇g(g−1(x(2))),g−1(x(2))).

Proof. First we develop a useful identity relating the deformation gradientsF(1) and F(2). We apply the chain rule to the equality

y = f (1)(x(1), t) = f (2)(g(x(1)), t)

to obtainf

(1)i,j (x(1), t) = f

(2)i,k (g(x(1)), t)gk,j(x

(1)),

which is equivalent to

F(1)(x(1), t) = F(2)(g(x(1)), t)∇g(x(1)) (*)

in direct notation.Now suppose the body is elastic w.r.t. the first reference configuration.

Then by Definition 108, ∃ a function G(1) : Lin V+ × B(1) → Sym 3

T(1)(x(1), t) = G(1)(F(1)(x(1), t),x(1))

= G(1)(F(2)(g(x(1)), t)∇g(x(1)),x(1)) (subst. (*)),

= G(1)(F(2)(x(2), t)∇g(g−1(x(2))),g−1(x(2))) (subst.),

= : G(2)(F(2)(x(2), t),x(2)) (definition).

For the l.h.s. we have by Definition 155,

T(1)(x(1), t) = T(f (1)(x(1), t), t)

= T(y, t) (subst.),

= T(f (2)(x(2), t), t) (subst.),

= T(2)(x(2), t) (defn. 155).


Combining these results we have

T(2)(x(2), t) = G(2)(F(2)(x(2), t),x(2))

with G(2)(F(2)(x(2), t),x(2)) := G(1)(F(2)(x(2), t)∇g(g−1(x(2))),g−1(x(2))). Thatis, the body is elastic w.r.t. the second configuration and the second elasticresponse function has the expected relation to the first.

The next theorem is a consequence of the preceding one.

Theorem 171 If an elastic body is homogeneous w.r.t. stress response rel-

ative to reference configuration0

B, then the body is also homogeneous w.r.t.

stress response relative to any reference configuration obtained from0

B by ahomogeneous deformation.1


4.3 Principle of Material Frame-Indifference

This section explores the notion that material response is invariant under(indifferent to) superposed rigid motions and shifts in the origin of the timescale. Only invariance under superposed rigid motion is relavent in the con-text of elasticity theory which does not include memory effects. We beginwith the notion of equivalent motions.

Definition 110 Two motions of a body, f(·, t) and∗

f (·, t), are equiv-

alent w.r.t. material response if they differ by a rigid deformation for eacht ∈ [t0,∞); i.e., ∃ functions c : [t0,∞)→ V and Q : [t0,∞)→ Orth V+ 3

∗f (x, t) = c(t) + Q(t)f(x, t) ∀ (x, t) ∈

0

B × [t0,∞).

Next we compare the stress vectors of an elastic body associated with a

pair of equivalent motions, f(·, t) and∗

f (·, t)

. For the motion f(·, t) ,consider a plane through point y ∈ Bt with unit normal n and stress vector

tn(y, t). The corresponding items for the equivalent motion∗

f (·, t)

are

denoted∗y,∗n and

∗t∗n

(∗y, t).

4.3. PRINCIPLE OF MATERIAL FRAME-INDIFFERENCE 193

y

O

*y

*n

nt*

*n (y

*,t )

t n (y,t )

Figure 4.2: Surface tractions from equivalent motions

y

O

*y

*n

n

y y *

Figure 4.3: Relation between normals for equivalent motion


Let y = y + n be the position vector to the terminal point of the unit

vector n located at y. Similarly,∗y=

∗y +

∗n is the position vector to the

terminal point of∗n located at

∗y . Then we appeal to Definition 110 to find

the mapping of n into∗n . We have

∗n =

∗y −

∗y

= (c(t) + Q(t)y)− (c(t) + Q(t)y) (subst.),

= Q(t) (y − y)

= Q(t)n (subst.).

Material frame-indifference corresponds to the assumption that∗t∗n

and tn are

related in the same way as∗n and n. Thus, we assume

∗t∗n

( ∗y, t)

= Q (t) tn (y, t) .

This result establishes a relation between the Cauchy stress tensors associatedwith the two motions. We write

∗T( ∗y, t) ∗

n =∗t∗n

( ∗y, t)

(Cauchy’s thm.),

= Q(t)tn (y, t) (assumption of frame indifference),

= Q(t) (T (y, t) n) (Cauchy’s thm.),

= Q(t)[T (y, t)

(Qt(t)

∗n)]

(∗n= Qn, Q ∈ OrthV+),

=(Q (t) T (y, t) Qt (t)

) ∗n (defn. of tensor mult.).

In view of the arbitrary selection of n (and therefore∗n) we see that material

frame-indifference implies that∗T( ∗y, t)

= Q (t) T (y, t) Qt (t) .

Thus, we shall adopt the Principle of Material Frame-Indifference.

Axiom 6 (Principle of Material Frame-Indifference) 2 The Cauchy stress

tensors T and∗T corresponding to equivalent motions f(·, t) and

∗f (·, t)

,

where∗f (x, t) = c(t) + Q(t)f(x, t), are related through

∗T(∗y, t) = Q(t)T(y, t)Qt(t) ∀ (x, t) ∈

0

B × [t0,∞).

1A homogeneous deformation is a deformation with a uniform deformation gradient.2Sometimes this principle is referred to as Objectivity or Observer Independence.


This assumption leads to certain restrictions on the form of the elastic re-sponse function, as indicated by the following theorem.

Theorem 172 The elastic constitutive equation is consistent with the Prin-ciple of Material Frame-Indifference iff the elastic response function G sat-isfies

G(QF,x) = QG(F,x)Qt ∀ Q ∈ OrthV+, F ∈ LinV+, x ∈0

B .

Proof. First suppose that the elastic constitutive equation satisfies the Prin-

ciple. The referential stress field due to the motion∗

f (·, t)

is given by

∗T (x, t) = G(

∗F (x, t),x).

Recalling,∗fi (x, t) = ci(t) +Qij(t)fj(x, t), we find that

∗Fik (x, t) =

∗f i,k (x, t) = Qij(t)fj,k(x, t)

⇔∗F (x, t) = Q(t)F(x, t)

∴∗T (x, t) = G(Q(t)F(x, t),x).

Combining this result with

∗T (x, t) =

∗T (

∗y, t)

= Q(t)T(y, t)Qt(t)

= Q(t)T(x, t)Qt(t)

= Q(t)G(F(x,t),x)Qt(t),

we obtainG(Q(t)F(x, t),x) = Q(t)G(F(x, t),x)Qt(t).

Conversely, suppose that G(QF,x) = QG(F,x)Qt holds. For any (x, t) ∈0

B ×[t0,∞),

∗T (x, t) = G(

∗F (x, t),x) (constitutive equation),

= G(Q(t)F(x, t),x) (subst.),

= Q(t)G(F(x, t),x)Qt(t) (subst.),

= Q(t)T(x,t)Qt(t) (constitutive equation).


Rewriting this result in the spatial description, we find

∗T(∗y, t) = Q(t)T(y,t)Qt(t) ,

which means the stresses generated by the elastic constitutive equation satisfythe Principle of Material Frame-Indifference.

The condition G(QF,x) = QG(F,x)Qt required for frame-indifferenceimposes certain restrictions on the form of the elastic response function. Thenext theorem describes admissable forms for the dependence of G on thedeformation gradient F that are consistent with frame-indifference. As weshall see, the the polar decomposition, F = RU, leads to certain reducedforms that automatically satisfy frame-indifference.

Theorem 173 If the elastic constituitive equation

T(x, t) = G(F(x, t),x) (4.1)

is consistent with the Principle of Material Frame-Indifference, then it canbe written in any of the following reduced forms:

T(x, t) = R(x, t)G(U(x, t),x)Rt(x, t); (4.2)

T(x, t) = F(x, t)G(U(x, t),x)Ft(x, t); (4.3)

T(x, t) = F(x, t)G(C(x, t),x)Ft(x, t); (4.4)

where G : Psym×0

B→ Sym and G : Psym×0

B→ Sym .

Conversely, an elastic response function written in any of these reducedforms is consistent with the Principle of Material Frame-Indifference ∀ choicesof G, G and G.

Proof. For simplicity and without loss of generality, select any fixed (x, t) ∈0

B×[t0,∞) and suppress the dependence on (x, t) in the notation. Now sup-pose the elastic response function is consistent with the Principle. Then byTheorem 172, we have,

QG(F)Qt = G(QF) ∀ F ∈ LinV+and ∀ Q ∈ OrthV+.


This implies (premultiply by Qt, postmultiply by Q)

G(F) = QtG(QF)Q (QQt = QtQ = I),

= QtG(Q(RU))Q (polar decomp.),

⇒ T = RG(U)Rt (invoke (4.1); choose Q = Rt; proves (4.2)),

= FU−1G(U)U−1Ft (R = FU−1, U−1 ∈ Psym ),

= FG(U)Ft (G(U) := U−1G(U)U−1; proves (4.3)),

= FG(√

C)Ft (thm. 128),

= FG(C)Ft (G(C) := G(√

C); proves (4.4)).

Further, we must have G, G : Psym×0

B→ Sym to ensure that G(F) yieldsa symmetric stress tensor.3

To illustrate the proofs of the converse assertions, suppose that elasticconstitutive equation has the first reduced form

T(x, t) = R(x, t)G(U(x, t),x)Rt(x, t).

For a motion∗

f (·, t)

equivalent to f(·, t),

∗F = QF, (see proof of thm. 172),

= Q (RU) (polar decomp. of F),

= (QR) U.

Now, Q,R ∈ OrthV+ ⇒ QR ∈ OrthV+ (see exercise below). Therefore, bythe uniqueness of the polar decomposition,

∗F=

∗R∗U where

∗R= QR and

∗U= U.

Thus, T(x, t) = R(x, t)G(U(x, t),x)Rt(x, t)⇒ that the stress corresponding

to the motion∗

f (·, t)

is

∗T =

∗R G(

∗U)

∗Rt

= (QR) G(U) (QR)t (subst.),

= Q(RG(U)Rt

)Qt ( (AB)t = BtAt, assoc.),

= QTQt

(subst., 1st reduced form).

3Of course, Psym is consistent with the arguments U and C.


∴The stresses satisfy the Principle of Material Frame-Indifference. Similar,proofs hold for the two other reduced forms.

Exercise 99 Given Q,R ∈ OrthV+, show that QR ∈ OrthV+.

Exercise 100 Show that an elastic constitutive relation of the form,

T(x, t) = F(x, t)G(C(x, t),x)Ft(x, t),

where G : Psym×0

B→ Sym, satisfies the Principle of Material Frame-Indifference.

Remark 55 The third reduced form is generally preferred for computations,because the kinematical variables F and C are easy to compute for a givenmotion. For theoretical development, the non-reduced form is usually theeasiest to work with.

Remark 56 From here on, we shall assume that our elastic constitutiveequation satisfies the Principle of Material Frame-Indifference.

4.4 Material Symmetry; Isotropy

In continuum mechanics, a material point of an unconstrained body4 hasonly two material properties: mass density in a given configuration and itsstress response w.r.t. the given configuration. In this section, we considertransformations of a body that leave these properties unaltered. This leadsto definitions of concepts such as material isotropy, orthotropy, etc. that aremore rigorous than, for example, “material isotropy means that the materialproperties are the same in all directions.”

Definition 111 A transformation of an object that leaves some property ofthe object invariant is called a symmetry transformation.

For example, the topography of the earth’s surface is invariant undera 360o rotation of the earth about its polar axis. Thus, a 360o rotationis a symmetry transformation for the surface topography. In the presentcontext, transformations which take one configuration of a body into anotherconstitute the relevant class of transformations. Thus, the material symmetryof a material point of a body is characterized by the class of deformationmappings which leave the mass density and stress response at the materialpoint unaltered.

4An unconstrained body has no internal constraints; e.g., rigidity, incompressibility,etc.

4.4. MATERIAL SYMMETRY; ISOTROPY 199

x (2)

x (1)

y

O

f (1)⋅,t( )

f (2)⋅,t( )

g

B(1)

B(2 )

B t

Figure 4.4: Alternative reference frames for development of material symme-try


We use the same framework developed for Theorem 170 to develop theconcept of material symmetry. That is, let B(1) and B(2) denote two referenceconfigurations, such that g is the mapping from B(1) to B(2); i.e., B(2) =g(B(1)

). Consider the material point identified by x(1) ∈ B(1). The position

of the material point in configuration Bt at time t is given by

y = f (1)(x(1), t) = f (2)(x(2), t) = f (2)(g(x(1)), t),

where x(2) ∈ B(2) and t ∈ [t0,∞).According to the Referential Density Equation, the mass density in the

second configuration is

ρ(2)(x(2)) = ρ(2)(g(x(1))) =ρ(1)(x(1))

det∇g(x(1)).

Therefore, it is necessary and sufficient that

∇g(x(1)) ∈ Unim V+ :=H ∈ Lin V+ : det H = 1

(4.5)

in order that the change in reference configuration leave the mass densityunaltered (ρ(2)(x(2)) = ρ(1)(x(1))).5 Thus, any mapping g 3 ∇g ∈ Unim V+

is a symmetry transformation of B(1) w.r.t. mass density.Next, consider the stress response. According to Theorem 170, the elastic

response functions in the two reference configurations are related by

G(2)(F(2)(x(2), t),x(2)) = G(1)(F(2)(x(2), t)∇g(x(1)),x(1)).

where we have made use of g−1(x(2)) = x(1). In order that the stress re-sponse of the material point be the same when defined w.r.t. the referenceconfigurations B(1) and B(2), it is necesarry and sufficient that

G(2)(·,x(2)) = G(1)( ·,x(1)),

which, in view of the previous relation is equivalent to

G(1)(F∇g(x(1)),x(1)) = G(1)(F,x(1)) ∀ F ∈ Lin V+. (4.6)

Note that the two necessary and sufficient conditions (equations 4.5 and4.6) for a material symmetry transformation involve only the gradient of the

5The elements of Unim V+ are called proper unimodular tensors.


deformation g that links two reference configurations (because only ∇g de-termines the changes in the material properties). Accordingly, the followingdefinition of the set of material symmetry transformations at some locationx is defined in terms of the gradient of the linking deformation.

Definition 112 (Noll, 1958) Given an elastic body and a reference config-

uration that corresponds to the region0

B, the material symmetry group at thematerial point identified by x in the reference configuration is the set

Msgx

=H ∈ Unim V+ : G(FH,x) = G(F,x) ∀ F ∈ Lin V+

.

Again, it should be emphasized that the material symmetry group is char-acterized by tensors H that correspond to the gradients at x of deformations— not the deformations themselves. This is because the mass density and theelastic response function in the second reference configuration depend onlyon the gradient of the connecting deformation. Also, note that H ∈ Msgx isnot a tensor field, but rather the value of a tensor field at x.

The following theorem presents a property of all orthogonal elements ofMsgx that derives from the Principle of Material Frame-Indifference.

Theorem 174 Let Q ∈ Orth V+. Then Q ∈ Msgx iff

G(QFQt,x) = QG(F,x)Qt ∀ F ∈ Lin V+.

Proof. Suppose that Q ∈ Msgx . Then by definition 112,

G(F,x) = G(FQ,x) ∀ F ∈ Lin V+.

It can be shown that QFQt ∈ Lin V+ ∀ F ∈ Lin V+ (see exercise below), sowe are free to replace any F ∈ Lin V+ with QFQt in the preceding equation.Hence,

G(QFQt,x) = G((QFQt

)Q,x)

= G(QF,x) (assoc., Q ∈ OrthV+),

= QG(F,x)Qt (Material Frame Indiff. via thm. 172).

Conversely, suppose that

G(QFQt,x) = QG(F,x)Qt ∀ F ∈ Lin V+.


Then by Theorem 172,

G(QFQt,x) = G(QF,x) ∀ F ∈ Lin V+.

The proof is completed by a special choice of F, similar to the one above (seeexercise below).

Exercise 101 Let Q ∈ Orth V+ and F ∈ Lin V+. Show that QFQt ∈ LinV+.

Exercise 102 Suppose Q ∈ Orth V+ and

G(QFQt,x) = G(QF,x) ∀ F ∈ Lin V+.

Show that Q ∈ Msgx .

In the context of an application, we may need to know the conditionsthat material symmetry imposes on the form of the response function in thereduced constitutive equation (4.4).

Theorem 175 The material symmetry group at the material point identi-fied by x in the reference configuration is described in terms of the reducedconstitutive equation (4.4)

T(x, t) = G(F(x, t),x) = F(x, t)G(C(x, t),x)Ft(x, t)

by

Msgx

=H ∈ Unim V+ : HG(HtCH,x)Ht = G(C,x) ∀ C ∈ Psym

.

Exercise 103 Prove Theorem 175. (Recall that equality of sets A and B⇔ A ⊂ B and B ⊂ A.)

The reduced analog of Theorem 174 is

Theorem 176 Let Q ∈ Orth V+. Then Q ∈ Msgx iff

G(QCQt,x) = QG(C,x)Qt ∀ C ∈ Psym .



We now have a general framework for describing various forms of materialsymmetry. Each particular form of material symmetry at a material pointx is defined by requiring that an appropriate class of transformations bea subset of Msgx . For example, material isotropy means that the materialproperties are invariant under all rotations of the reference configuration.Since the set Orth V+ contains the gradients of all of the deformations thatdefine these rotations, we have

Definition 113 An elastic body is isotropic at the material point identifiedby x in the reference configuration iff

OrthV+ ⊂ Msgx

.

Exercise 105 Prove that

OrthV+ ⊂ Unim V+,

which is, of course, a requirement for the definition of isotropy to make sense.

We can also use the reduced form of the constitutive equation to get atthe concept of material isotropy.

Theorem 177 An elastic body is isotropic at the material point identifiedby x in the reference configuration iff

G(QCQt,x) = QG(C,x)Qt ∀ Q ∈ OrthV+ and C ∈ Psym .

Proof. Combine Definition 113 and Theorem 176.

Remark 57 The condition in Theorem 177 is the common mathematicaldefinition of an isotropic function on Sym to Sym .

Remark 58 We can define other familiar forms of material symmetry, suchas transverse isotropy, orthotropy, and various forms of crystal classes, byreplacing OrthV+ in Definition 113 with an appropriate subgroup of OrthV+.For example, orthotropy is defined using the subgroup consisting of the gra-dients of rotations of 0o and multiples of 180o about any principal axis.


4.5 Hyperelasticity

This section introduces a special case of elasticity called hyperelasticity. Thedescription derives from a special case of the Principle of Balance of Energy,an axiom of continuum thermomechanics.

Axiom 7 (Principle of Balance of Energy — no thermal effects) Suppose

that thermal effects are negligable. Then ∀ parts P ⊂0

B and ∀ times t ∈[t0,∞), the time rate of change of the total energy of part P equals the powerof the loads acting on it; i.e.,

d

dt

∫Pt

[1

2|v(y,t)|2 + ε(y,t)

]ρ(y,t)dVy =

∫∂Pt

tn(y,t)(y,t ) · v(y,t)dAy

+

∫Pt

b(y,t) · v(y,t)ρ(y,t)dVy,

where ε(y,t) ∈ C1(=) is the internal energy per unit mass at point y at timet.

From here on, we assume that thermal effects are negligable. Then Bal-ance of Linear Momentum leads to the local statement,

Theorem 178 (Local Balance of Mechanical Energy)[∂ε

∂t(y, t) +∇ε(y, t) · v(y, t)

]ρ(y, t) = T(y,t) · L(y,t) on =.

Proof. By the Theorem of Power Expended (158) and our assumption thatlinear momentum is balanced, we have

d

dt

∫Pt

1


∫∂Pt


+

∫Pt


−∫Pt

T(y,t) · L(y,t)dVy..

Combining this result with Axiom 7, we obtain

d

dt

∫Pt

ε(y,t)ρ(y,t)dVy =

∫Pt


4.5. HYPERELASTICITY 205

The Reduced Transport Theorem 151 leads to

d

dt

∫Pt

ε(y,t)ρ(y,t)dVy =

∫Pt

[∂ε

∂t(y, t) +∇ε(y, t) · v(y, t)

]ρ(y, t)dVy,

and therefore, by the Localization Theorem 147, the local energy balance isdescribed by[

∂ε

∂t(y, t) +∇ε(y, t) · v(y, t)

]ρ(y, t) = T(y,t) · L(y,t).

Next we introduce a referential internal energy density function

ε(x,t) := ε(f(x,t),t) = ε(y,t),

so that by the Chain Rule we find that the material time rate of change ofthe internal energy density is

∂ε

∂t(x, t) = ∇ε(f(x,t), t) · ∂f

∂t(x, t) +

∂ε

∂t(f(x,t), t)

= ∇ε(f(x,t), t) · v(x, t) +∂ε

∂t(f(x,t), t)

= ∇ε(y, t) · v(y, t) +∂ε

∂t(y, t).

Definition 114 An elastic body0

B is hyperelastic (or Green elastic) iff The-orem 178 holds and ∃ a function

ε : Lin V+×0

B→ <

3 the internal energy density corresponding to a motion f(·, t) is given bythe elastic constitutive equation for the internal energy

ε(x, t) = ε(F(x, t),x) on =.

In view of the scalar character of the internal energy density function, therelevant version of the Principle of Material Frame-Indifference is


Axiom 8 (Scalar Principle of Material Frame-Indifference) Let ε and∗ε be the internal energy densities (in the spatial description) corresponding to

a pair of equivalent motions f(·,t) and∗

f (·,t). Then the corresponding

referential functions for the internal energy are related by

∗ε (x, t) = ε(x, t).

Theorem 179 If the elastic constitutive equation for the internal energy

ε(x, t) = ε(F(x, t),x)

satisfies the Principle of Material Frame-Indifference, then it can be writtenin either of the following reduced forms:

ε(x, t) = ε(U(x, t),x); (4.7)

ε(x, t) = ε(C(x, t),x). (4.8)

Conversely, each of these reduced constitutive equations automatically satis-fies the scalar Principle of Material Frame-Indifference ∀ ε or ε.

Exercise 106 Prove the preceding theorem.

Next we use (4.8) to compute the material time derivative of the internal

energy density, ∂ε/∂t(x, t). Note that the domain of ε is Psym×0

B, so the 9components of C are not independent. Thus, we can define 6 independentparameters (suppressing the dependence on (x, t))

C1 = C11, C2 = C22, C3 = C33,

C4 =C12 + C21

2, C5 =

C23 + C32

2, C6 =

C13 + C31

2,

and write

ε(C) = εS(C1, C2, C3, C4, C5, C6)

= : εL(C11, C22, C33, C12, C21, C23, C32, C13, C31).

By the Chain Rule we have

∂εL∂Cij

=6∑

Γ=1

∂εS∂CΓ

∂CΓ

∂Cij,


so, in particular,∂εL∂C12

=∂εS∂C4

∂C4

∂C12

=1

2

∂εS∂C4

.

and∂εL∂C21

=1

2

∂εS∂C4

.

In general, we have∂εL∂Cpq

=∂εL∂Cqp

.

With this background, we introduce the notations

∂ε

∂C:=

∂εL∂Cij

ei ⊗ ej ∈ Sym

and

C =∂Cij∂t

ei ⊗ ej.

Putting all this together with the Chain Rule (and restoring the dependenceon position and time), we have

∂ε

∂t(x, t) =

∂ε

∂C(C(x, t),x) · C(x, t).

The next theorem provides a general kinematic result.


F(x, t) = L(f(x, t), t) F(x, t)

andC(x, t) = 2Ft(x, t)D(f(x, t), t)F(x, t).

Proof. First, we have

Fij =∂

∂tFij

=∂

∂t(fi,j) (subst.),

=

(∂fi∂t

),j

(f ∈ C2),

= vi,j (subst.).


Now by definition,vi(x, t) = vi(f(x, t), t)

⇒vi,j(x, t) = vi,k(f(x, t), t)fk,j(x, t) (chain rule).

Combining these results yields

Fij(x, t) = vi,k(f(x, t), t)fk,j(x, t)

orF(x, t) = L(f(x, t), t)F(x, t).

Next, recall C := FtF to obtain

C =(F)t

F + FtF (see exercise below),

= (LF)t F + Ft (LF) (see above),

=(FtLt

)F + Ft (LF)

= Ft(Lt + L

)F (assoc., distrib.),

= Ft (2 sym L) F (subst.).

= 2FtDF (subst.).

Exercise 107 Show that C =(F)t

F + FtF.

Definition 115 Given a body0

B and a prescription of referential mass den-sity ρ0, the set of three functions

1. f(·, t) a motion,

2. T the Cauchy stress field, and

3. ε the specific internal energy field

constitute a mechanical process iff all of our smoothness hypotheses are sat-isfied, mass is conserved and linear momentum is balanced.6 A mechanical

process is local if the domains are restricted to0

B × (t1, t2), t0 < t1 < t2 <∞;it is admissable if it is consistent with the relevant constitutive assumptions.

6Normally, we would also require that angular momentum is balanced, but this turnsout to be a theorem in the case of hyperelasticity.


Theorem 181 Every admissable, local mechanical process corresponding tothe constitutive equations

T(x, t) = T(F(x, t),x)

andε(x, t) = ε(C(x, t),x)

satisfies the local balance of mechanical energy (LBME)

ρ(y, t)ε(y, t) = T(y, t) · L(y, t)

iff

T(F(x, t),x) = 2ρ(f(x, t), t)F(x, t)∂ε

∂C(C(x, t),x)Ft(x, t).

[Here, we do not assume symmetry of T.]

Proof. Suppose ε(x, t) = ε(C(x, t),x). Then by our previous calculation,

∂ε

∂t(x, t) =

∂ε

∂C(C(x, t),x) · C(x, t),

=∂ε

∂C· 2FtDF (thm. 180),

= 2

(F∂ε

∂CFt

)·D (see exercise below).

Now, since

F∂ε

∂CFt =

(F∂ε

∂CFt

)t(∂ε

∂C∈ Sym ),

our previous result implies that

ε(y, t) =∂ε

∂t(x, t)=2

(F∂ε

∂CFt

)·D (subst.)

= 2

(F∂ε

∂CFt

)· sym L (subst.)

= 2

(F∂ε

∂CFt

)· L. (symmetry of term in parens.)

Now suppose that in addition

T(F(x, t),x) = 2ρ(f(x, t), t)F(x, t)∂ε



Then clearly,

ρε = T · L

∀ admissable, local mechanical processes.Conversely, suppose that LBME is satisfied ∀ admissable, local mechani-

cal processes. Then our previous calculations yield[T(F(x, t),x)− 2ρ(f(x, t), t)F(x, t)

∂ε

∂C(C(x, t),x)Ft(x, t)

]·L(f(x, t), t) = 0.

Here we need the following lemma (see exercise below).

Lemma 182 Given arbitrary∗F ∈ Lin V+ and

∗L ∈ LinV, ∃ an admissable,

local mechanical process 3 for any given∗x ∈

0

B and any given∗t ∈ (0,∞)

F( ∗x,∗t)

=∗F and L

(f( ∗x,∗t),∗t)

=∗L .

By this lemma we may substitute the arbitrary uniform fields∗F and

∗L

and arbitrary choices∗x and

∗t into our previous result to obtainT

( ∗F,∗x)− 2

ρ0

( ∗x)

det∗F

∗F∂ε

∂C

( ∗C,∗x) ∗

Ft

· ∗L = 0 ∀∗L∈ LinV ,

∗F∈ LinV+,

∗x∈

0

B,

where we have applied the Referential Density Equation and∗C :=

∗Ft ∗

F .

Since∗L ∈ LinV is arbitrary, it follows that

T(∗F,∗x) = 2

ρ0(∗x)

det∗F

∗F∂ε

∂C(∗C,∗x)

∗Ft

on LinV+×0

B .

Then given a motion f(·, t) , choose∗F = F(x, t) and apply the Referential

Density Equation once more to obtain

T(F(x, t),x) = 2ρ(f(x, t), t)F(x, t)∂ε


4.6. ELASTIC RESPONSE TO INFINITESIMAL MOTIONS 211


∂ε

∂C· 2FtDF =2

(F∂ε

∂CFt

)·D.

Exercise 109 Prove Lemma 182.

To summarize, the preceding theorem demonstrates that the constitutiveequation for stress in a hyperelastic body is (suppressing dependence onposition)

T = T(F) = 2ρ0

det FF∂ε

∂C(C)Ft. (4.9)

We have also shown that

F∂ε

∂C(C)Ft ∈ Sym .

Therefore, satisfaction of balance of angular momentum is a theorem for ahyperelastic body, rather than an axiom.

Exercise 110 Show that the hyperelastic constitutive equation, (4.9), meetsthe Principle of Material Frame-Indifference.

Exercise 111 Show that (4.9) reduces to

S = 2ρ0F∂ε

∂C(C) and P = 2ρ0

∂ε

∂C(C).

when rewritten in terms of the Piola-Kirchhoff stress tensors.

4.6 Elastic Response to Infinitesimal Motions

This section develops the asymptotic approximation of the first Piola-Kirchhoffstress S in an elastic body when the displacement gradient H is small. Wecombine the reduced elastic constitutive equation (4.4),7

T = FG(C)Ft,

7In the hyperelastic case, it follows that

G(C) = 2ρ0√

detC

∂ε

∂C(C).


with Definition 106,

S := JTF−t

(in which dependence on (x, t) has been suppressed), to obtain

S =JFG(C). (4.10)

Exercise 112 Recall that G(C) ∈ Sym . Prove that (4.10) yields values forS that automatically satisfy

FSt = SFt;

i.e., balance of angular momentum is automatically satisfied.

Next we estimate the r.h.s. of (4.10) for an infinitesimal motion. That is,under the assumption that, independent of our choice of coordinate frame,the norm of the displacement gradient field has the property 0 < ε < 1,where ε is the norm of the associated displacement gradient field describedin subsection 3.2.2:

ε = ‖H‖ := sup

x∈0B

t∈[0,∞)

i,j=1,2,3

|Hij(x, t)|

(c.f. definitions 96 and 82).89 The following theorems develop estimates forthe individual terms before arriving at an overall estimate for (4.10).

Theorem 183 Let f(·, t) be an infinitesimal motion where ε is the normof the associated displacement gradient field described in subsection 3.2.2.Then

J(x, t) = 1 +O(ε).

Proof. We have by Theorem 152,

J(x, t) = 1 + tr H(x, t) +O(ε2)

= 1 +O(ε)

8Of course, we are most interested in the case ε << 1, but our results shall holdwhenever ε ∈ (0, 1).

9Do not confuse the parameter ε here with the internal energy density introduced insection 4.5.


since

|tr H| = |H11 +H22 +H33|≤ |H11|+ |H22|+ |H33| ≤ 3ε

and ∣∣O(ε2)∣∣ ≤ Kε2 = ε (Kε) ≤ Kε.

Recall that (Theorem 132), for an infinitesimal motion,

F(x, t) = I +O(ε).

Next we consider the term G(C). This term involves the mappings Gij :Psym → <, which we assume to be continuously differentiable at I. Theprecise meaning of this continuity and our following analysis require the useof the “little oh” notation, so we pause here to introduce it.

Definition 116 Let Ψ : A ⊂ Lin V → < have the property Ψ(0) = 0, andlet ε := ‖A‖, A ∈A.10 Then we say that Ψ is little oh of ε,11 denoted

Ψ(A) = o(ε) as ε→ 0,

iff

limε→0ε 6=0

|Ψ(A)|ε

= 0.

If Θ,Φ : A→ < 3 Θ(0) = Φ(0), then

Θ(A) = Φ(A) + o(ε) as ε→ 0

means

Θ(A)− Φ(A) = o(ε) as ε→ 0.

Vector and tensor-valued functions of A are o(ε) as ε → 0 iff each of theircomponents are.12

10Here we assume that an appropriate norm has been defined over the set A.11Alternatively, we say that Ψ(A) approaches 0 faster than ε approaches 0.12Recall that for A ∈ Lin V, |A| =

√A ·A =

√AijAij .


Remark 59 In the interest of convenience, the notation o(ε) as ε → 0 (orsimply o(ε)) is sometimes used to represent a function that has the little ohproperty. This slight abuse of notation facilitates manipulations of expres-sions such as

Θ(A) = Φ(A) + o(A).

Remark 60 Similar definitions hold for cases where the argument of thefunction Ψ is a scalar, vector or higher-order tensor.

Theorem 184

O(ε2) = o(ε) as ε→ 0.

Proof. First, O(ε2) = 0 for ε = 0. By the definition of big oh and for ε > 0,

0 ≤ limε→0ε6=0

|O(ε2)|ε

≤ limε→0ε 6=0

Kε2

ε

= limε→0

Kε

= 0.

∴

limε→0ε6=0

|O(ε2)|ε

= 0,

so O(ε2) = o(ε) as ε→ 0.

Theorem 185 Given 0 < k <∞,

δijo(ε) = (o(ε))ij ;

ko(ε) = o(ε);

o(ε) + o(ε) = o(ε).


We shall also have use of the following results.


Theorem 186

δikO(ε) = (O(ε))ik ; (4.11)

O(ε) +O(ε) = O(ε); (4.12)

O(ε)O(ε) = O(ε2); (4.13)

O(ε) +O(ε2) = O(ε). (4.14)


Theorem 187O(ε)o(ε) = o(ε) as ε→ 0.

Proof.

|O(ε)o(ε)|ε

=|O(ε)| |o(ε)|

ε( |ab| = |a| |b| ),

≤ Kε |o(ε)|ε

(definition of O(ε)),

= K |o(ε)|→ 0 as ε→ 0 (definition of o(ε)).

Therefore, by the definition of o(ε), O(ε)o(ε) = o(ε) as ε→ 0.

Next we use the little oh concept to define a continuously differentiablefunction of a second-order tensor.

Definition 117 Let Ψ : A ⊂ Lin V → <; A,B ∈A; and ε := ‖A−B‖.Then Ψ is continuously differentiable at B ∈A iff

Ψ(A) = Ψ(B) +∂Ψ

∂Aij(B)(A−B)ij + o(ε) as ε→ 0.

Vector and tensor-valued functions are continuously differentiable at B iffeach of their components are.

We next proceed with our analysis of the term G(C). We assume that

for each x ∈0

B, the scalar functions

Gij (·,x) : Psym→ <


are continuously differentiable at I (∈ Psym). Since the domain of G isPsym, we here encounter the same issues in taking the partial derivatives ofGij(C) w.r.t. the mutually dependent components Cij that we encounteredin differentiating the internal energy density ε(C) in Section 4.5. Applyingthe same machinery, we define 6 independent parameters (suppressing thedependence on (x, t)):

C1 = C11, C2 = C22, C3 = C33,

C4 =C12 + C21

2, C5 =

C23 + C32

2, C6 =

C13 + C31

2,

and write

Gij(C) = GSij(C1, C2, C3, C4, C5, C6)

= : GLij(C11, C22, C33, C12, C21, C23, C32, C13, C31).

By the Chain Rule we have

∂GLij

∂Cpq=

6∑Γ=1

∂GSij

∂CΓ

∂CΓ

∂Cpq,

so, in particular,∂GL

ij

∂C12

=∂GS

ij

∂C4

∂C4

∂C12

=1

2

∂GSij

∂C4

.

and∂GL

ij

∂C21

=1

2

∂GSij

∂C4

.

In general, we have∂GL

ij

∂Cpq=∂GL

ij

∂Cqp.

With this background, we introduce the notation

∂Gij

∂C:=

∂GLij

∂Cpqep ⊗ eq ∈ Sym,

and introduce the following estimate for G(C).


Theorem 188 Let f(·, t) be an infinitesimal motion of an elastic bodywith zero initial stress, where ε is the norm of the displacement gradient fielddescribed in Definition 82. Then an estimate of the body’s elastic responsefunction is provided by

Gij(C,x) = GLij(C,x) = 2

∂GLij

∂Cpq(I,x)Epq(C) + (o(ε))ij as ε→ 0.

Proof. For convenience, we suppress the dependence on position x in thenotation. The fact that the body has zero initial stress implies

GLij(I) = (0)ij .

Let ε := |C− I|.13 Then our assumption that the elastic response functionis continuously differentiable at I implies that

GLij(C) = GL

ij(I+ (C− I)) (4.15)

= GLij(I) +

∂GLij

∂Cpq(I) (C− I)pq + (o (ε))ij as ε→ 0 (4.16)

=∂GL

ij

∂Cpq(I) (C− I)pq + (o (ε))ij as ε→ 0(no initl. stress).(4.17)

Here we need a relation between o(ε) and o (ε). Recalling C = FtF andF = I + H,

C− I = H + Ht+HtH,

so that

|C− I|2 =∣∣H + Ht+HtH

∣∣2=

(H + Ht+HtH

)·(H + Ht+HtH

)= H ·H + H ·Ht + H·

(HtH

)+ Ht ·H + Ht ·Ht + Ht·

(HtH

)+(HtH

)·H+

(HtH

)·Ht +

(HtH

)·(HtH

)= 2H ·H+2H ·Ht + 2H·

(HtH

)+ 2Ht·

(HtH

)+(HtH

)·(HtH

)≤

∣∣2H ·H+2H ·Ht + 2H·(HtH

)+ 2Ht·

(HtH

)+(HtH

)·(HtH

)∣∣≤ 2 |H ·H|+2

∣∣H ·Ht∣∣+ 2

∣∣H· (HtH)∣∣+ 2

∣∣Ht·(HtH

)∣∣+∣∣(HtH

)·(HtH

)∣∣ ,13 The magnitude of a second-order tensor is computed via the inner product. In the

case of H, the magnitude is distinct from the special norm, ε.


where the last step follows from multiple applications of the triangle inequal-ity. Now consider estimates for representative terms on the r.h.s. of the lastinequality.∣∣H ·Ht

∣∣ = |HijHji|= |H11H11 +H12H21 + . . .+H33H33|≤ |H11H11|+ |H12H21|+ . . .+ |H33H33|≤ |H11| |H11|+ |H12| |H21|+ . . .+ |H33| |H33|≤ 9ε2 (infinitesimal motion).

Similarly, we find that |H ·H| ≤ 9ε2, |H· (HtH)| = |Ht· (HtH)| ≤ 27ε3 and|(HtH) · (HtH)| ≤ 81ε4. Hence,

|C− I|2 ≤ 4(9ε2)

+ 4(27ε3

)+ 81ε4

< 225ε2 (ε < 1⇒ ε2 > ε3 > ε4).

∴ for any infinitesimal motion,

ε := |C− I| < 15ε,

and hence

ε→ 0 as ε→ 0.

Then for ε 6= 0 and using o (ε) to represent some function with the o (ε)as ε→ 0 property, we can write

|o (ε)|ε

=|o (ε)|ε

ε

ε

< 15|o (ε)|ε

.

Then, referring to Definition 116, we find

limε→0ε 6=0

|o (ε)|ε

= 0 ⇔ limε→0ε6=0

|o (ε)|ε

= 0

⇔ limε→0ε6=0

|o (ε)|ε

= 0


since ε → 0 as ε → 0. Referring to the little oh definition once more, thislast result implies that o (ε) as ε→ 0⇔ o (ε) as ε→ 0, or

o (ε) = o (ε) .

Substituting this result into (4.17), we obtain

GLij(C) =

∂GLij

∂Cpq(I) (C− I)pq + (o (ε))ij as ε→ 0.

Now

C− I = H + Ht+HtH

= 2E + HtH (defn. 84),

= 2E +O(ε2) (see exercise below),

and ∴∂GL

ij

∂Cpq(I) ( C− I)pq = 2

∂GLij

∂Cpq(I)Epq +

∂GLij

∂Cpq(I)Apq,

where Apq = (O(ε2))pq . Let

∆ij = maxp,q

∣∣∣∣∣∂GLij

∂Cpq(I)

∣∣∣∣∣ .Then ∣∣∣∣∣∂GL

ij

∂Cpq(I)Apq

∣∣∣∣∣ =


∂C11

(I)A11 + . . .+∂GL

ij

∂C33

(I)A33

∣∣∣∣∣≤


∂C11

(I)

∣∣∣∣∣ |A11|+ . . .+


∂C33

(I)

∣∣∣∣∣ |A33|

≤ ∆ijK11ε2 + . . .+ ∆ijK33ε

2 (Apq =(O(ε2)

)pq

)

= ∆ij (K11 + . . .+K33) ε2;

so ∣∣∣∣∣∂GLij

∂Cpq(I)Apq

∣∣∣∣∣ =(O(ε2)

)ij,


and

GLij(C) = 2

∂GLij

∂Cpq(I)Epq +

(O(ε2)

)ij

+ (o (ε))ij as ε→ 0,

= 2∂GL

ij

∂Cpq(I)Epq + (o (ε))ij + (o (ε))ij as ε→ 0, (thm. 184),

= 2∂GL

ij

∂Cpq(I)Epq + (o (ε))ij as ε→ 0, (thm. 185).

Exercise 115 Show that for an infinitesimal deformation, HtH =O(ε2).

We are now ready to assemble our estimate for the elastic constitutiverelation for the first Piola-Kirchhoff stress tensor for an infinitesimal motion.

Theorem 189 Let f(·, t) be an infinitesimal motion of an elastic bodywith zero initial stress, where ε is the norm of the displacement gradient fielddescribed in Definition 82. Then the constitutive equation of the body reducesto

Sij(x, t) = Cijkl(x)Ekl(x, t) + (o(ε))ij as ε→ 0,

where

Cijkl(x) = 2∂GL

ij

∂Ckl(I,x),

[This 4-index symbol should not be confused with the components of thesecond-order right Cauchy-Green deformation tensor C.]

Proof. Returning to (4.10), we have

Sij = JFikGkj(C).


Then, appealing to Theorems 183, 132 and 188, we obtain

Sij = [1 +O(ε)] [δik + (O(ε))ik]

[2∂GL

kj

∂Cpq(I)Epq + (o (ε))kj

]as ε→ 0,

= [δik + (O(ε))ik + δikO(ε) +O(ε) (O(ε))ik]

×

[2∂GL

kj


]as ε→ 0,

= [δik + (O(ε))ik]

[2∂GL

kj


]as ε→ 0

(thm. 186, all parts),

= 2∂GL

ij

∂Cpq(I)Epq + 2

∂GLkj

∂Cpq(I)Epq (O(ε))ik + δik (o (ε))kj

+ (O(ε))ik (o (ε))kj as ε→ 0

= 2∂GL

ij

∂Cpq(I)Epq + 2

∂GLkj

∂Cpq(I)Epq (O(ε))ik + (o (ε))ij as ε→ 0

(thms. 185 and 187).

Now, in view of the facts that Eij = H(ij) and that |Hij| is bounded by (ε)ijfor an infinitesimal motion (trivially, ⇒

∣∣H(ij)

∣∣ is bounded by (ε)ij), we have

|Epq (O(ε))ik| = |Epq| |(O(ε))ik| ( |ab| = |a| |b| ),=

∣∣H(pq)

∣∣ |(O(ε))ik| (subst.),

≤ (ε)pq |(O(ε))ik| (infinitesimal motion),

≤ (ε)pqKikε (def. of O(ε)).

∴

|Epq (O(ε))ik|ε

≤ Kik (ε)pq → (0)pqik as ε→ 0;

so

Epq (O(ε))ik = (o (ε))pqik .


Then in view of our assumption that GLij(C) is continuously differentiable at

I, we have

2∂GL

kj

∂Cpq(I)Epq (O(ε))ik = 2

∂GLkj

∂Cpq(I) (o (ε))pqik as ε→ 0,

= (o (ε))ij as ε→ 0. (2∂GL

kj

∂Cpq(I) <∞, thm. 185).

Substituting this result into our previous result for Sij, we have

Sij = 2∂GL

ij

∂Cpq(I)Epq + (o (ε))ij + (o (ε))ij as ε→ 0

= 2∂GL

ij

∂Cpq(I)Epq + (o (ε))ij as ε→ 0 (thm. 185)

= CijklEkl + (o (ε))ij as ε→ 0 (subst. and relabel).

Remark 61 This looks a lot like Hooke’s Law (to within o(ε)), but only ifwe overlook the difference between the Cauchy and PK-I stress tensors.

Take careful note of:

Theorem 190 The indexed symbol Cijkl(x) of Theorem 189 are the compo-nents of a fourth-order tensor called the elasticity tensor at x.

Proof. Consider a second coordinate frame X′. Then we have

C′

ijkl(x) = 2∂(GL)′ij

∂C′kl

(I,x).

Applying the transformation rules for second-order tensors, we obtain

(GL)′ij

(C′

rs,x) = λimλjnGLmn

(Crs

(C′),x).


in which Crs(C′)

= λprλqsC′pq. Next, we use the Chain Rule to find

∂(GL)′ij

∂C′kl

(C′

rs,x) = λimλjn∂GL

mn

∂Cab

(Crs

(C′),x) ∂Cab∂C

′kl

= λimλjn∂GL

mn

∂Cab

(Crs

(C′),x) ∂

∂C′kl

(λpaλqbC

′

pq

)= λimλjn

∂GLmn

∂Cab

(Crs

(C′),x)λpaλqbδpkδql

= λimλjnλkaλlb∂GL

mn

∂Cab

(Crs

(C′),x)

= λimλjnλkaλlb∂GL

mn

∂Cab

(Crs

(C′),x).

Evaluate at C = I and multiply by 2 to get

C′

ijkl(x) = λimλjnλkaλlbCmnab(x).

Thus, the terms Cmnab are the components of a fourth-order tensor.

Remark 62 We have shown that ∃ a fourth-order tensor given by

4

C (x) = Cijkl(x)ei ⊗ ej ⊗ ek ⊗ el,

3 by Theorem 189 and our definition of fourth-order tensors as mappings ofsecond-order tensors into second-order tensors,

S(x, t) =4

C (x) (E(x, t)) + o(ε)

orSij(x, t) = Cijkl(x)Ekl(x, t) + (o(ε))ij .

Apparently, we have a mapping4

C (x) : Sym → Lin V. However, due to thesymmetry properties of the elasticity tensor developed immediately below, we

actually have a mapping4

C (x) : Sym→ Sym.

Theorem 191 The elasticity tensor has the minor symmetries

Cjikl = Cijkl;

Cijlk = Cijkl.


Proof. By definition (c.f. Theorem 189),

Cijkl(x) = 2∂GL

ij

∂Ckl(I,x).

Then, Cjikl = Cijkl follows from the fact that values of GL are in Sym; i.e.,GLij = GL

ji. The second result follows from our earlier finding that

∂GLij

∂Clk(C) =

∂GLij

∂Ckl(C).

The following theorem states that the elasticity tensor exhibits anotherform of symmetry in the special case of hyperelasticity.

Theorem 192 For a hyperelastic body with zero initial stress, the elasticitytensor has the major symmetry

Cklij(x) = Cijkl(x).

Proof. Recall the component form of the hyperelastic response function

GLij(C) = 2ρ0

(1√

det C

)L∂εL

∂Cij(C),

where the superscript “L” indicates that the function is adjusted to account

for the symmetry of C. It can be shown that the term(

1/√

det C)L

is

continuously differentiable w.r.t. the components of C; thus, the assumptionthat GL

ij ∈ C1 ⇒ εL ∈ C2. With this knowledge we differentiate to obtain

∂GLij

∂Ckl(C) = 2ρ0

∂

∂Ckl

(1√

det C

)L∂εL

∂Cij(C) + 2ρ0

(1√

det C

)L∂2εL

∂Ckl∂Cij(C).


Next, evaluate this equation at C = I to obtain

∂GLij

∂Ckl(I) = 2ρ0

∂

∂Ckl

(1√

det C

)L∣∣∣∣∣C=I

∂εL

∂Cij(I) + 2ρ0

∂2εL

∂Ckl∂Cij(I),

= 2ρ0

∂

∂Ckl

(1√

det C

)L∣∣∣∣∣C=I

1

2ρ0

GLij(I) + 2ρ0

∂2εL

∂Ckl∂Cij(I) (subst.),

= 2ρ0

∂

∂Ckl

(1√

det C

)L∣∣∣∣∣C=I

× 0 + 2ρ0

∂2εL

∂Ckl∂Cij(I), (zero initial stress),

= 2ρ0

∂2εL

∂Ckl∂Cij(I).

Thus, by Theorem 189,

Cijkl = 4ρ0

∂2εL

∂Cij∂Ckl(I),

= 4ρ0

∂2εL

∂Ckl∂Cij(I), (εL ∈ C2),

= Cklij (thm. 189).

Exercise 116 Assume that Theorem 191 holds. Show that the major sym-metry of the elasticity tensor is equivalent to

A·4

C (B) = B·4

C (A) ∀ A,B ∈ Sym .

Theorem 193 If Q ∈ Orth V+ and Q ∈ Msgx, then

QipQjqQkrQlsCpqrs(x) = Cijkl(x).

Proof. Suppose Q ∈ Orth V+ and Q ∈ Msgx . Then by Theorem 176

G(QCQt,x) = QG(C,x)Qt ∀ C ∈Psym,

or in component form

GLij(QCQt,x) =QimQjnG

Lmn(C,x) ∀ C ∈Psym .


Differentiation w.r.t. Ckl yields

QimQjn∂GL

mn

∂Ckl(C,x) =

∂GLij

∂Cpq(QCQt,x)

∂(QCQt)pq∂Ckl

,

=∂GL

ij

∂Cpq(QCQt,x)

∂

∂Ckl(QprQqsCrs) ,

=∂GL

ij

∂Cpq(QCQt,x)QprQqsδrkδsl,

=∂GL

ij

∂Cpq(QCQt,x)QpkQql,

and evaluation at C = I⇒

QimQjn∂GL

mn

∂Ckl(I,x) =

∂GLij

∂Cpq(I,x)QpkQql ⇒

QimQjnCmnkl = CijpqQpkQql ⇒QimQjnQrkQslCmnkl = CijpqQpkQrkQqlQsl (multiply by QrkQsl),

= Cijpqδprδqs

= Cijrs.

Then relabel to obtain

QipQjqQkrQlsCpqrs(x) = Cijkl(x).

Combining this result with Definition 113, we directly obtain

Theorem 194 If an elastic body is isotropic at x, then

QipQjqQkrQlsCpqrs(x) = Cijkl(x) ∀ Q ∈ OrthV+.

Exercise 117 Show that the isotropy condition of the preceding theorem isequivalent to

4

C (QEQt) = Q4

C (E)Qt ∀ Q ∈ OrthV+ and ∀ E ∈ Sym .


Remark 63 Since Q ∈ OrthV+, the terms Qij can be viewed as the directioncosines corresponding to the transformation of a given frame X to a newframe X

′. Then by the transformation rules for a fourth-order tensor

QipQjqQkrQlsCpqrs(x) = C′

ijkl(x);

which implies that if the body is isotropic, then

Cijkl(x) = C′

ijkl(x) ∀ X′.

That is, the components of the fourth-order elasticity tensor4

C are the samein all frames.

Remark 64 It is clear from the above development, that the elasticity tensorof an isotropic body can be viewed as a linear function on Sym to Sym, and,furthermore, that linear function is isotropic.

A major simplification of the form of the elastic constitutive relation ispossible in the case of isotropic elastic bodies.

Theorem 195 (Representation Theorem of Isotropic Elasticity) Theelasticity tensor at x is isotropic iff it has the form

Cijkl(x) = λ(x)δijδkl + µ(x) (δikδjl + δilδjk) ,

where the scalars λ(x) and µ(x) are called the Lame moduli.14

Proof. First, suppose that the elasticity tensor is isotropic; i.e.,

Cijkl = QipQjqQkrQlsCpqrs,

where we have dropped the dependence on position for convenience. Weproceed by investigating certain special choices for Q, using the interpretationthat Q corresponds to a rotation of the coordinate frame.15

14µ(x) is also called the shear modulus.15There are more elegant ways to carry out this proof, but they require a bit more math-

ematical machinery. See K. A. Pericak-Spector and S. J. Spector, ”On the representationtheorem for linear, isotropic tensor functions. J. Elasticity 39 (1995) 181-185.


1. Rotation of 180o about e3.

We have

Qij = e′

i · ej =

−1 0 00 −1 00 0 1

⇒C1123 = Q1pQ1qQ2rQ3sCpqrs

= (−1) (−1) (−1) (1)C1123

= −C1123 ⇒C1123 = 0.

Similarly, we have a zero component whenever index 3 appears once orthree times:

0 = C1131 = C2223 = C2231 = C3323 = C3331

= C2311 = C2322 = C2333 = C2312 = C3111

= C3122 = C3133 = C3112 = C1223 = C1231.

In addition, we have zero for any of the terms obtained from theseresults and the minor symmetries Cijkl = Cjikl = Cijlk.


We have

Qij = e′

i · ej =

1 0 00 −1 00 0 −1


= (1) (−1) (−1) (−1)C1233

= −C1233 ⇒C1233 = 0.

Similarly, whenever index 1 appears once or three times:

0 = C1112 = C2212 = C3312 = C2331 = C3123

= C1211 = C1222

(plus minor symmetries and previous results).



We have

Qij = e′

i · ej =

1 0 00 0 10 −1 0


= (1) (1) (1) (1)C1133 ⇒C1122 = C1133.

Similarly,

C2211 = C3311, C2233 = C3322, C1212 = C1313, C2222 = C3333.

(plus minor symmetries).


We have

Qij = e′

i · ej =

0 1 0−1 0 00 0 1


= (1) (1) (−1) (−1)C2211 ⇒C1122 = C2211.

Similarly,

C1111 = C2222, C1133 = C2233, C3311 = C3322, C2323 = C3131.

(plus minor symmetries).



We have

Qij = e′

i · ej =

1/√

2 1/√

2 0

−1/√

2 1/√

2 00 0 1


=1√2Q2qQ1rQ2s (C1qrs + C2qrs)

=1

2Q1rQ2s (−C11rs + C12rs − C21rs + C22rs)

=1

2√

2Q2s (−C111s − C112s + C121s + C122s − C211s

−C212s + C221s + C222s)

=1

4(C1111 − C1112 + C1121 − C1122 − C1211

+C1212 − C1221 + C1222 + C2111 − C2112

+C2121 − C2122 − C2211 + C2212 − C2221 + C2222)

⇒C1212 =

1

2(C1111 − C1122) (by previous results).

Collecting all these results together, and setting λ = C1122 and µ = C1212,we obtain

Cijkl = λδijδkl + µ (δikδjl + δilδjk) .

Conversely, suppose that the elasticity tensor has the form

Cijkl = λδijδkl + µ (δikδjl + δilδjk) .

Then

QipQjqQkrQlsCpqrs = QipQjqQkrQls [λδpqδrs + µ (δprδqs + δpsδqr)]

= λδijδkl + µ (δikδjl + δilδjk) (see exercise below),

= Cijkl;

i.e., the elasticity tensor is isotropic.

Exercise 118 Show that ∀ Q ∈ Orth V+,

QipQjqQkrQls [λδpqδrs + µ (δprδqs + δpsδqr)] = λδijδkl + µ (δikδjl + δilδjk) .



Cijkl = λδijδkl + µ (δikδjl + δilδjk)⇔4

C (E) = 2µE + λ (tr E) I ∀ E ∈ Sym .

The next theorem states that the first Piola-Kirchhoff stress tensor issymmetric (within o(ε)) for infinitesimal motions of an elastic body withzero initial stress.

Theorem 196 For infinitesimal motions of an elastic body with zero initialstress,

St= S + o(ε) as ε→ 0 on0

B ×[t0,∞).

Proof. Recall the estimate of Theorem 189 for infinitesimal motions of anelastic body with zero initial stress,

Sij = CijklEkl + (o(ε))ij as ε→ 0,

or

S =4

C (E) + o(ε) as ε→ 0.

The theorem is a direct result of the minor symmetry,

Cijkl = Cjikl.

The final theorem of the chapter states that for infinitesimal motionsof an elastic body with zero initial stress, the first Piola-Kirchhoff stresstensor and the Cauchy stress tensor agree to within o(ε) when evaluated atcorresponding arguments.

Theorem 197 For infinitesimal motions of an elastic body with zero initialstress,

T(f(x, t), t) = S(x, t) + o(ε) as ε→ 0 on0

B ×[t0,∞).

Exercise 120 Combine

S =4

C (E) + o(ε) as ε→ 0

with

T =1

JSFt

to prove Theorem 197.

Chapter 5

Linearized Elasticity

In the preceding chapter, we explored asymptotic approximations to the non-linear equations of elasticity under the assumption of infinitesimal deforma-tions. In particular, we considered the case of small displacement gradientsunder the assumption that ε(u) < 1. Here we consider the case where ε(u)is much less than one. In this case, we can drop terms that are o(ε), yieldingthe equations of linearized elasticity.

5.1 References

1. P. G. Ciarlet, Mathematical Elasticity, Volume I: Three-DimensionalElasticity, North-Holland, 1988.

2. M. E. Gurtin, The Linear Theory of Elasticity, Handbuch der Physik,VIa/2, Springer, 1972.

3. A. E. H. Love, The Mathematical Theory of Elasticity, 4th ed., Cam-bridge, 1927.

4. I. S. Sokolnikoff, Mathematical Theory of Elasticity, 2nd. ed., McGraw-Hill, 1956.

233

234 CHAPTER 5. LINEARIZED ELASTICITY

5.2 Summary of the Linearized Field Equa-

tions

Typically, the data for problems in solid mechanics include a specificationof the undeformed configuration that a body occupies at time t0, while thedeformed configurations are unknown and must be determined as part of thesolution process. Thus, it is natural to choose the undeformed configurationas the reference configuration and to work with referential field descriptions— in particular, the first Piola-Kirchhoff stress tensor of Theorem 106.

We have seen that the asymptotic approximation to the reduced form ofthe general elastic constitutive equation for the PK-I stress,

S = JFG(C),

is in the absence of initial stress

S =4

C (E) + o(ε) as ε→ 0 (Theorem 189).

In the present linearized theory, we discard the o(ε) term and write thegeneralized Hooke’s Law,

S =4

C (E) .

The tensor4

C always displays the minor symmetry

Cjikl = Cijkl (Theorem 191),

so the first Piola-Kirchhoff stress is always symmetric in the linearized theory :

St = S.

There is no approximation in the definition of the infinitesimal straintensor as

E =1

2

[∇u + (∇u)t

]= sym∇u.

However, the geometric interpretations of the infinitesimal strain tensor de-pend on our exact geometric interpretations of the right Cauchy-Green de-formation tensor C (c.f. Section 2.3.3), and the relation between E and Cdepends on the approximation

E =1

2(C− I) + o(ε) as ε→ 0.

5.2. SUMMARY OF THE LINEARIZED FIELD EQUATIONS 235

The velocity and acceleration are given by

v =∂f

∂tand a =

∂v

∂t,

or in terms of the displacements

v =∂u

∂tand a =

∂2u

∂t2.

Since the displacements u are a referential field, we have

v = u and a = u,

where a superposed dot indicates the material time rate of change. Thus,the First Stress Equation of Motion can be written (without approximation)as (c.f. Theorem 164)

∇ · S + ρ0b = ρ0u.

However, we must exercise caution if we wish to substitute the asymptoticconstitutive equation,

S =4

C (E) + o(ε) as ε→ 0,

into this equation. We need to show that the term ∇ · o(ε) = o(ε) as ε→ 0to justify the linearized form of the First Stress Equation of Motion (i.e.,the local expression of the Balance of Linear Momentum in the referentialdescription):

∇·4

C (E) + ρ0b = ρ0u. (5.1)

Prof. D. E. Carlson points out that this step requires an additional restric-tion on the class of admissable motions beyond those that we have introducedfor infinitesimal motions. Specifically, controls on the derivatives of H (orthe second derivatives of u) are required to justify this step. The class ofmotions that obey these controls are called superinfinitesimal motions. Inthe interest of brevity, we shall not develop this topic in detail here. Theinterested student is directed to Prof. D. E. Carlson for a complete treat-ment. For our present purposes, it will suffice to assume that our motionsare superinfinitesimal whenever we invoke the linearized form of (5.1).

Now consider the Second Stress Equation of Motion

FSt= SFt.


In the full theory, this condition is satisfied automatically by the constitutiverelation S = JFG(C) with G(C) ∈ Sym . We must confirm that the sructure

of our stress estimate, S =4

C (E) + o(ε) with4

C (E) ∈ Sym, maintains thisrelation to within o(ε). Thus, consider

FSt − SFt = (I + H)

(4

C (E) + o(ε)

)t−(

4

C (E) + o(ε)

)(I + H)t (subst.)

= (I + H)4

C (E)−4

C (E) (I + H)t + (I + H) o(ε)− o(ε) (I + H)t

(expansion; minor symmetry, thm. 191),

= H4

C (E)−4

C (E) Ht + Ho(ε)−o(ε)Ht (cancellation)

= O(ε)O(ε) +O(ε)O(ε) + o(ε) (O(ε) +O(ε)) (H,Ht,E =O(ε))

= O(ε2) +O(ε2) + o(ε)O(ε) (Thm. 186)

= o(ε) + o(ε) + o(ε) (Thms. 184 and 187)

= o(ε) as ε→ 0 (Thm. 185).

Thus we have shown that

S =4

C (E) + o(ε) plus minor symmetry of4

C⇒ FSt = SFt + o(ε).

That is, our stress estimate automatically satisfies the second stress equationof motion to within the accuracy of the linearized theory (but not exactly!).

The Principle of Conservation of Mass is incorporated in the referentialform of the First Stress Equation of Motion through the referential densityequation. It does not warrant further consideration, because only the refer-ential density ρ0 appears in our referential linearized equations.

That leaves the Principle of Material Frame-Indifference which was in-corporated in our starting constitutive assumption. Due to our linearization,

we cannot expect S =4

C (E) + o(ε) to meet the Principle of Material Frame-Indifference exactly.

To summarize, the field equations of the classical linearized theory of

elasticity are on0

B ×[t0,∞)

1. the (Infinitesimal) Strain-Displacement Relation

E =1

2

[∇u + (∇u)t

],

5.3. LINEARIZED TRACTION BOUNDARY CONDITIONS 237

2. the Stress-Strain Relation

S =4

C (E)

(where Cijkl = Cjikl = Cijlk and generally4

C=4

C (x).), and

3. the Stress Equation of Motion

∇ · S + ρ0b = ρ0u.

These equations are predicated on the assumptions that the motion is in-finitesimal (i.e., ∇u is small) and that the initial stress vanishes. If, inaddition, the motion is superinfinitesimal and the elastic response functionis class C2, then we are assured that the Displacement Equation of Motion

∇·4

C (sym∇u) + ρ0b = ρ0u

(based on the substitution of the first two equations into the third) is thelocal expression of the Balance of Linear Momentum to within the accuracyof the linearized theory.

5.3 Linearized Traction Boundary Conditions

Cauchy’s Theorem on the Existence of the Stress Tensor (Theorem 155)states that (in the spatial description) the traction applied to a regular pointy of the deformed boundary ∂Bt at time t is

tn(y, t) = T(y, t)n,

where n is the outward unit normal to ∂Bt at y, and the value of T(y, t)must be interpreted in the sense of the trace operator. Our objective hereis to develop a referential description of this relation in terms of the firstPiola-Kirchhoff stress tensor S(x, t) and the unit outward normal m to the

undeformed boundary ∂0

B at x = f−1(y, t). Then we will consider the asymp-totic approximation and the linearization of the resulting relationship.

The first step is to establish the geometric relationship between the nor-mal vectors m and n. We proceed by describing a patch of the undeformed

boundary ∂0

B as the zero set of some differentiable potential function θ(x).


That is, if x is a regular point of ∂0

B, then it must belong to a surface patchS which admits the description

S = x ∈ D : θ(x) = 0

and then

m(x) =∇θ(x)

|∇θ(x)|.

The image of the surface patch S under the motion f(·, t) is

f(S, t) =y ∈ f(D, t) : φ(y, t) := θ(f−1(y, t)) = 0

,

so that

n(y, t) =∇φ(y, t)

|∇φ(y, t)|.

The following theorem gives the desired relation between the normal vectorsm and n.

Theorem 198 Under the motion f(·, t) , the outward unit normals m and

n to ∂0

B and ∂Bt, respectively, are related by

n(y, t) =F−t(x, t)m(x)

|F−t(x, t)m(x)|,

where y = f(x, t).

Proof. Using the preceding set-up, we apply the Chain Rule to obtain

φ,i(y, t) = θ,j(f−1(y, t))f−1

j,i (y, t)

= θ,j(f−1(y, t))

(F−1(x, t)

)ji

;

or in direct notation

∇φ(y, t) = F−t(x, t)∇θ(x)

= |∇θ(x)|F−t(x, t)m(x) (substitution),


where y = f(x, t). Hence,

n(y, t) =∇φ(y, t)

|∇φ(y, t)|

=|∇θ(x)|F−t(x, t)m(x)

||∇θ(x)|F−t(x, t)m(x)|(subst.),

=|∇θ(x)|F−t(x, t)m(x)

|∇θ(x)| |F−t(x, t)m(x)|

=F−t(x, t)m(x)

|F−t(x, t)m(x)|

The next step is to find the asymptotic approximation to F−tm for smalldisplacement gradients Based on F = I + H,

F−t =(I + Ht

)−1

= J(K(H)) := L(H),

whereJ(K) := K−1 and K(H) = I + Ht.

Assume for the moment that J and K are class C1 functions. Then L is alsoclass C1 by the Chain Rule. Thus,

Lij(H) = Lij(0) +∂Lij∂Hkl

(0)Hkl + o(ε) as ε→ 0

= δij +∂Lij∂Hkl

(0)Hkl + o(ε) as ε→ 0 (defn. of L(H)).

Also by the Chain Rule,

∂Lij∂Hkl

(0) =∂Jij∂Kmn

(K(0))∂Kmn

∂Hkl

(0)

=∂Jij∂Kmn

(I)∂Kmn

∂Hkl

(0) (defn. of K(H)).

Since, by definition,Kmn(H) = δmn +Hnm;

we have∂Kmn

∂Hkl

(H) = δnkδml,


which demonstrates the fact that K ∈ C∞(Lin V ,LinV).We must also examine the differentiability of J(K) = K−1. Recalling that

Jij(K) =1

2 det KεipqεjmnKmpKnq

anddet K = εijkK1iK2jK3k,

it is clear that the partial derivative

∂Jij∂Kmn

(K)

exists. However, rather than carry out the calculations suggested by theabove formulas, it is more convenient to differentiate the product

KijJjk = δik

implied by the definition of J. Differentiation w.r.t. Kmn via the productrule yields

(0)imnk = δimδjnJjk(K) +Kij∂Jjk∂Kmn

(K)

= δimJnk(K) +Kij∂Jjk∂Kmn

(K).

Multiplication by (K−1)li yields

(0)lmnk =(K−1

)liδimJnk(K) +

(K−1

)liKij

∂Jjk∂Kmn

(K)

=(K−1

)lm

(K−1

)nk

+ δlj∂Jjk∂Kmn

(K)

=(K−1

)lm

(K−1

)nk

+∂Jlk∂Kmn

(K),

or∂Jlk∂Kmn

(K) = −(K−1

)lm

(K−1

)nk.

This is a significant result. We have a simple formula for the first par-tial derivatives of J and, furthermore, we find that J(K) = K−1 is a classC∞(Inv,LinV) function.1

1Strictly speaking, we also need to establish that Inv is an open subset of Lin V. It is,so let’s proceed.


Now assembling our results, we have

∂Lij∂Hkl

(0) = − (δimδjn) (δnkδml) = −δilδjk,

and∂Lij∂Hkl

(0)Hkl = −Hji = −(Ht)ij.

Thus, (F−t)ij

= Lij(H) = δij −(Ht)ij

+ o(ε) as ε→ 0,

orF−t = I−Ht + o(ε) as ε→ 0.

∴F−tm = m−Htm + o(ε) as ε→ 0.

We also need an asymptotic expansion for∣∣F−tm∣∣ =√

(F−tm) · (F−tm).

By substitution,(F−tm

)·(F−tm

)=

[m−Htm + o(ε)

]·[m−Htm + o(ε)

]= m ·m−2m ·Htm +

(Htm

)·(Htm

)+ o(ε)

= 1− 2m ·Hm+o(ε) + o(ε)

= 1− 2m ·Hm+o(ε) as ε→ 0.

Then, ∣∣F−tm∣∣−1= [1− 2m ·Hm+o(ε)]−1/2

= 1 + m ·Hm+o(ε) as ε→ 0

by the binomial series expansion.Finally, we have

n = [1 + m ·Hm+o(ε)][m−Htm+o(ε)

]= m−Htm+ (m ·Hm) m− (m ·Hm) Htm+o(ε)

= m−Htm+ (m ·Hm) m−o(ε)+o(ε)= m−Htm+ (m ·Hm) m+o(ε) as ε→ 0.

This proves the following theorem which is a purely kinematic result (i.e., itis independent of the material properties.


Theorem 199 Under an infinitesimal motion f(·, t) , the outward unit

normals m and n to ∂0

B and ∂Bt, respectively, are related by

n(y, t) = m(x)−Ht (x, t) m(x)+ [m(x) ·H (x, t) m(x)] m(x)+o(ε) as ε→ 0,

where y = f(x, t).

Applying this result to the case of elasticity, we get our desired asymptoticapproximation of the traction vector acting on the deformed surface in termsof the first Piola-Kirchoff stress tensor and the outward unit normal to theundeformed surface.

Theorem 200 For an elastic body with zero initial stress undergoing aninfinitesimal motion f(·, t) ,

T(y, t)n(y, t) = S(x, t)m(x)+o(ε) as ε→ 0,

where m and n are the outward unit normals to ∂0

B and ∂Bt, respectively,and x = f−1(y, t).

Proof. By theorems 197 and 199,

Tn = [S + o(ε)][m−Htm+ (m ·Hm) m + o(ε)

]= Sm + S

[−Htm+ (m ·Hm) m + o(ε)

]+ o(ε)

= Sm +

[4

C (E) + o(ε)

] [−Htm+ (m ·Hm) m + o(ε)

]+ o(ε)

(thm. 189),

= Sm + o(ε) as ε→ 0.

The significance of this theorem is that to within the accuracy of the lin-earized theory, the traction vector tn(y, t) = T(y, t)n(y, t) acting at positiony ∈ ∂Bt at time t is equal to S(x, t)m(x) — the desired referential expressionin terms of the PK-I stress tensor.

5.4. LINEARIZED ISOTROPIC STRESS-STRAIN RELATION 243

5.4 Linearized Isotropic Stress-Strain Rela-

tion

5.4.1 General Considerations

A body is isotropic at material point x if the elasticity tensor and, conse-quently, the linearized stress-strain relation at x are

Cijkl(x) = λ(x)δijδkl + µ(x) (δikδjl + δilδjk)

andS(x, t) = 2µ(x)E(x, t) + λ(x) [tr E(x, t)] I.

An isotropic elastic body is one that is isotropic at all of its material points.If an isotropic body is homogeneous w.r.t. stress reponse, then the Lameparameters λ and µ are uniform.

First we consider the volume change due to an infinitesimal motion.

Theorem 201 The volume change of a part P ⊂0

B under an infinitesimalmotion f(·, t) is given by∫

Pt

dVy −∫P

dVx =

∫P

tr E(x, t)dVx + o(ε) as ε→ 0.

Proof. Since tr E = tr H and by Theorem 152, the Jacobian determinant

J(x, t) = 1 + tr E(x, t) + o(ε) as ε→ 0

for an infinitesimal motion. Also, by the Change of Variable Theorem forVolume Integrals ∫

Pt

dVy =

∫PJ(x, t)dVx.

∴ ∫Pt

dVy −∫P

dVx =

∫P

[1 + tr E(x, t) + o(ε)] dVx −∫P

dVx (subst.),

=

∫P

[tr E(x, t) + o(ε)] dVx

=

∫P

tr E(x, t)dVx + o(ε) as ε→ 0.


Thus, to within the accuracy of the linearized theory, the local change involume is given by the dilatation:

ϑ := tr E = tr H = Ekk = uk,k = ∇ · u.

For an isotropic material, the dilatation is proportional to the trace of thefirst Piola-Kirchhoff stress tensor.

Theorem 202 Let Θ := tr S =Skk. Then for an isotropic material,

Θ = (3λ+ 2µ) tr E,

and therefore, the dilatation is given by

ϑ = tr E =Θ

(3λ+ 2µ).


Thus, we may invert the linearized isotropic stress-strain relation to ob-tain

E =1

2µ[S− λ (tr E) I]

=1

2µ

[S− λ

(3λ+ 2µ)(tr S) I

],

provided thatµ 6= 0 and 3λ+ 2µ 6= 0.

There is strong experimental evidence for a one-to-one correspondence be-tween stress and strain (at least for some materials — we call these elasticmaterials). Therefore, we shall assume that the inversion is possible andaccordingly require that the Lame parameters satisfy the above inequalities.To sum up,

Theorem 203 For an isotropic material,

E =1

2µS− λ

2µ (3λ+ 2µ)(tr S) I,

where µ 6= 0 and 3λ+ 2µ 6= 0.

We consider certain special stress states to get a physical interpretationof the Lame parameters as well as alternative choices for the isotropic elasticmoduli.


5.4.2 Pure Pressure

We assume that the PK-I stress has the form

S = −pI.

Then by Theorem 203,

E =−p

(3λ+ 2µ)I.

By Theorem 132, the right Cauchy-Green deformation tensor is

C(x, t) = I + 2E(x, t) + o(ε)

=

[1− 2p(x, t)

3λ(x) + 2µ(x)

]I + o(ε) as ε→ 0.

Thus, to within o(ε), there is no shear strain between the coordinate direc-tions (Theorem 127); and the extensional strain in each coordinate directionis (Theorem 126) √

1− 2p(x, t)

3λ(x) + 2µ(x)− 1 .

For p > 0, 3λ+2µ > 0 and 2p < 3λ+2µ, we have the same negative extensionin all three coordinate directions. Thus, positive pressure causes a materialcube to shrink to a smaller cube.

To physically interpret the stress state, consider the corresponding Cauchystress. We have

T(y, t) = S(x, t) + o(ε) = −p(x, t)I + o(ε) as ε→ 0.

So, to within the accuracy of the theory, the Cauchy stress is also a purepressure.

From E = −p(3λ+2µ)

I, we find

p = −1

3(3λ+ 2µ) tr E =− 1

3(3λ+ 2µ)ϑ;

so for pure pressure loading, the pressure is proportional to the dilatation.

Definition 118 The bulk modulus or modulus of compression at the ma-terial point x is defined by

k(x) =1

3[3λ(x) + 2µ(x)] .


We requirek > 0⇔ 3λ+ 2µ > 0,

to ensure that positive pressures decrease the volume (i.e., tr E < 0, c.f.Theorem 202).

5.4.3 Pure Shear

Here we assume the PK-I stress has the form

[Sij(x, t)] =

0 τ(x, t) 0τ(x, t) 0 0

0 0 0

= [Tij(y, t)] + o(ε) as ε→ 0.

So, to within o(ε), the Cauchy stress is also pure shear.Theorem 203⇒

[Eij(x, t)] =1

2µ

0 τ(x, t) 0τ(x, t) 0 0

0 0 0

.Thus, in pure shear, the shear strains between the e1 and e2 directions areproportional to the Cauchy shear stresses (e.g., T12 = 2µE12). Accordingly,µ is called the shear modulus (often denoted G in the literature).

By Theorem 132,

[Cij(x, t)] =

1 τ(x,t)µ(x)

0τ(x,t)µ(x)

1 0

0 0 1

+ o(ε) as ε→ 0.

Thus, to within o(ε), there is no extensional strain in the coordinate direc-tions (Theorem 126), no shear strain between the e1 and e3 directions andbetween the e2 and e3 directions, and the shear strain between the e1 ande2 dirctions is given by (Theorem 127)

sin γ(e1, e2) =τ(x, t)

µ.

Thus, we get a positive strain for τ > 0 and µ > 0. This maps an undeformedcube into a skewed prism. To understand the Cauchy stress state, it is easierto envision a cube in the deformed state (corresponding to a “negatively


skewed” prism in the undeformed state). Cauchy’s Theorem on the Existenceof the Stress Tensor then determines the positive sense of the shear tractionson the deformed cube. We require

µ > 0

to ensure that the shear tractions and the shear strain have correspondingdirections.

Next we consider an alternative representation of the linearized isotropicstress-strain relation involving Young’s modulus and the Poisson ratio.

Theorem 204 For an isotropic material with 3λ + 2µ > 0 and µ > 0, wehave

λ+ µ > 0

and

E =1

β[(1 + ν) S− ν (tr S) I] ,

where

β :=µ (3λ+ 2µ)

λ+ µ> 0

is called Young’s modulus, and

ν :=λ

2 (λ+ µ)

is called Poisson’s ratio.2

Exercise 122 Prove that 3λ+ 2µ > 0 and µ > 0⇒ λ+ µ > 0.

Exercise 123 Complete the proof of Theorem 204.

5.4.4 Pure Tension

To interpret β and ν, we consider the pure tension stress state in the e1

direction

[Sij(x, t)] =

σ(x, t) 0 00 0 00 0 0

= [Tij(y, t)] + o(ε) as ε→ 0.

2Gurtin uses β to denote Young’s modulus. It is typically denoted by E. SometimesPoisson’s ratio is denoted by σ.


Thus, to within o(ε) the Cauchy stress state is also a pure tension in the e1

direction.

Theorem 204⇒

[Eij(x, t)] =1

β(x)

σ(x, t) 0 00 −νσ(x, t) 00 0 −νσ(x, t)

.Thus, to within the accuracy of our theory,

T11 = β(x)E11

in pure tension. In other words, the normal stress in the direction of puretension is proportional to the normal strain in the same direction, wherethe Young’s modulus is the constant of proportionality. This is, of course,Hooke’s Law.

By Theorem 132,

[Cij(x, t)] =

1 + 2σ(x,t)β(x)

0 0

0 1− 2ν σ(x,t)β(x)

0

0 0 1− 2ν σ(x,t)β(x)

+ o(ε) as ε→ 0.

Thus, to within o(ε), there is no shear between the coordinate directions(Theorem 127), and the extensional strain in the e1 direction is positivefor σ > 0, while there is a lateral contraction in the other two coordinatedirections, provided that

ν > 0,

which we assume to be the case. Then ν = λ/2(λ+ µ)⇒

λ > 0.

The lateral contraction is often expressed as

E22 = E33 = −νE11;

it is proportional and opposite to the extensional strain in the tension direc-tion — the Poisson’s Ratio Effect.


5.4.5 Relations Between the Elastic Moduli

We have introduced five elastic moduli —- λ, µ, β, ν and k — to describe thelinearized isotropic elastic stress-strain relation. Any two of these are suffi-cient to uniquely characterize the stress-strain law. The following theoremdescribes how each parameter is defined in terms of any two of the others.

Theorem 205 Any one of the elastic moduli λ, µ, β, ν and k can be ex-pressed in terms of any two of the others in accordance with the followingchart.

λ µ β ν k

λ, µ - - µ(3λ+2µ)λ+µ

λ2(λ+µ)

3λ+2µ3

λ, β -

(β−3λ)4

+√(β−3λ)2+8λβ

4

-

−(β+λ)4λ

+√(β+λ)2+8λ2

4λ

(3λ+β)6

+√(3λ+β)2−4λβ

6

λ, ν - λ(1−2ν)2ν

λ(1+ν)(1−2ν)ν

- λ(1+ν)3ν

λ, k - 3(k−λ)2

9k(k−λ)2

λ3k−λ -

µ, β (2µ−β)µβ−3µ

- - β−2µ2µ

µβ3(3µ−β)

µ, ν 2µν(1−2ν)

- 2µ (1 + ν) - 2µ(1+ν)3(1−2ν)

µ, k 3k−2µ3

- 9kµ3k+µ

12

[3k−2µ3k+µ

]-

β, ν νβ(1+ν)(1−2ν)

β2(1+ν)

- - β3(1−2ν)

β, k 3k(3k−β)9k−β

3βk9k−β - 1

2

[3k−β

3k

]-

ν, k 3kν1+ν

3k(1−2ν)2(1+ν)

3k (1− 2ν) - -

Exercise 124 Prive Theorem 205. (Just prove row 8 (β, ν) to keep thishumane.)

5.4.6 Principal Axes Coincide

For a general elastic material, the principal axes of stress and strain can bedistinct. However, we get the following convenient result for isotropic elasticmaterials.

Theorem 206 For an isotropic elastic material, the principal axes of stressand strain coincide.


Proof. We must show that Sij = (0)ij (i 6= j)⇔ Eij = (0)ij (i 6= j). From

S = µE + λ (tr E) I,

we have

Eij = (0)ij (i 6= j)⇒ Sij = (0)ij (i 6= j).

From

E =1

β[(1 + ν) S−ν (tr S) I] ,

we have

Sij = (0)ij (i 6= j)⇒ Eij = (0)ij (i 6= j).

5.5 Stored Energy; Work-Energy Theorems

We first encountered the internal energy density function in our discussionof hyperealsticity in the general large-deformation theory of elasticity. Inthat case, we found that the stress is obtained by differentiating the internalenergy density w.r.t. strain and that the hyperelastic elasticity tensor hasthe major symmetry. In this section, we find similar results for the storedenergy function in the linearized theory and, further, that the two functionsare identified with each other in the linearized theory.

Theorem 207 An elastic body defined by S =4

C (E) possesses a stored en-ergy function

σ(·,x) ∈ C2(Sym,<)

at x 3S(x, t) =

∂σ

∂E(E(x, t),x)

iff

Cijkl(x) = Cklij(x).

In this case,

σ(E,x) =1

2E·

4

C (E) =1

2Cijkl(x)EijEkl.

5.5. STORED ENERGY; WORK-ENERGY THEOREMS 251

Proof. Suppose there is a stored energy function σ(·,x) ∈ C2(Sym,<) atx 3 S(x, t) = ∂σ

∂E(E(x, t),x). We must account for the symmetry of E when

we consider the partial derivatives of σ w.r.t. the components of E (c.f.Sections 4.5 and 4.6 ). Thus, we write in the same fashion

∂σ

∂E:=

∂σL

∂Eijei ⊗ ej.

Now,

∂2σL

∂Eij∂Ekl=

∂

∂Eij

(∂σL

∂Ekl

)(definition),

=∂

∂Eij(Skl) =

∂

∂Eij(CklpqEpq) (substitution),

= Cklpqδpiδqj = Cklij.

Similarly,

∂2σL

∂Ekl∂Eij= Cijkl.

Since σ(·,x) ∈ C2(Sym,<)

∂2σL

∂Eij∂Ekl=

∂2σL

∂Ekl∂Eij

⇒

Cklij = Cijkl at x.

Coneversely, suppose Cijkl(x) = Cklij(x). Define

σ : Sym×0

B→ <

by

σ(E,x) =1

2Cijkl(x)EijEkl.


Note that this implies σ(·,x) ∈ C∞(Sym,<) ⊂ C2(Sym,<). Differentiationleads to

∂σL

∂Epq=

1

2CijklδipδjqEkl +

1

2CijklEijδkpδlq

=1

2CpqklEkl +

1

2CijpqEij

=1

2CpqklEkl +

1

2CpqijEij (assumed major symmetry),

=1

2CpqklEkl +

1

2CpqklEkl = CpqklEkl (relabel and sum),

= Spq (substitution).

Thus, σ(E,x) = 12Cijkl(x)EijEkl is a stored energy function at x.

To interpret S = ∂σ/∂E, recall the referential description of the Theoremof Expended Power (Theorem 168)

d

dt

∫P

1

2|u|2 ρ0dVx =

∫∂P

[Sm] ·udAx +

∫P

b · uρ0dVx −∫P

S·∇udVx,

wherein we have substituted u for v. This general result of continuum me-chanics is based solely on the relation

∇ · S + ρ0b =ρ0a,

which holds exactly in the linearized theory of elasticity.Consider the stress-power term S·∇u. In the linearized theory, S ∈ Sym;

and ∴S·∇u = S· sym∇u = S · E.

The next exercise links this result with the time rate of change of the storedenergy function.

Exercise 125 Let4

C have the major symmetry. Show that

S · E =σ.

Definition 119 The kinetic energy of part P at time t is

K(P , t) =

∫P

1

2|u(x, t)|2 ρ0(x)dVx,

i.e., the referential description of the kinetic energy.


Definition 120 The strain energy of part P at time t (in the linearizedtheory only) is

U(P , t) =1

2

∫P

S(x, t) · E(x, t)dVx

=1

2

∫P

E(x, t)·4

C (x) [E(x, t)] dVx.

Theorem 208 Let the elasticity tensor4

C (x) have the major symmetry ∀x ∈

0

B . Then for each part P ⊂0

B:

1.

U(P , t) =

∫Pσ(E(x, t),x)dVx;

2.

U(P , t) =

∫P

S(x, t) · E(x, t)dVx;

3.

·

(K(P , t) + U(P , t)) =

∫∂P

[S(x, t)m(x, t)] · u(x, t)dAx

+

∫P

b(x, t) · u(x, t)ρ0(x)dVx.

Proof.

1. Definition 120 and Theorem 207.

2. Theorem 1 and exercise 125.

3. Theorem 168, definition 119, and 2.

Comparison of theorems 1 and 3 with The Principle of Balance of Energy(axiom 7) makes it clear that

σ(E(x, t),x) = ρ0(x)ε(x, t),


the anticipated identification of the stored energy function with the internalenergy density of hyperelasticity.

Next, we consider the relation between the work done by the externalforces acting on an elastic body during a motion and the energy stored in

the body. Consider the motion of an elastic body0

B with a stored energyfunction σ under the action of a system of external forces. The body startsat rest in a unstressed reference configuration at time t0, and returns to restat time t∗ in some configuration that might be distinct from the referenceconfiguration. By Theorem 3 and the Fundamental Theorem of Calculus,∫ t∗

t0

[∫∂

0B

(Sm) · udAx +

∫0B

b · uρ0dVx

]dt =

(K(

0

B, t) + U(0

B, t))∣∣∣∣t=t∗

t=t0

= U(0

B, t∗) (see exercise below),

=

∫0Bσ(E(x, t∗),x)dVx (Theorem 1).

The l.h.s. represents the total work done on the body by the externalforces during the motion from t0 to t∗. We expect this work to be positive;and, consequently, we also expect the strain energy at time t∗,

U(0

B, t∗) =1

2

∫0B

E(x, t∗)·4

C (x) [E(x, t∗)] dVx

to be positive. This observation is closely associated with the positive-definiteproperty of the elasticity tensor.

Definition 121 The elasticity tensor4

C (x) at x (with or without the majorsymmetry) is positive semi-definite iff

E·4

C (x) [E] ≥ 0 ∀ E ∈ Sym;

it is positive definite iff it is positive semi-definite and

E·4

C (x) [E] = 0 iff E = 0.

The next theorem treats the special case of a body that returns to restin the unstressed reference configuration.


Theorem 209 Suppose an elastic body0

B with a stored energy function σstarts at rest in an unstressed reference configuration at time t0, undergoes amotion under the action of a system of external forces, and returns to restin the reference configuration at time t∗. Then the work of the external forcesvanishes for the closed cycle; i.e.,∫ t∗

t0

[∫∂

0B

(Sm) · udAx +

∫0B

b · uρ0dVx

]dt = 0.

Thus, there is no energy dissipation in an elastic body with a stored energyfunction.

Proof. Since the body occupies the reference configuration0

B at time t∗, we

have E(·, t∗) = 0 on0

B . Then by Theorem 207,

σ(E(x,t∗),x) =1

2E (x,t∗)·

4

C (x)(E(x,t∗)) = 0.

Therefore, by our previous discussion,∫ t∗

0

[∫∂

0B

(Sm) · udAx +

∫0B

b · uρ0dVx

]dt =

∫0Bσ(E(x, t∗),x)dVx

= 0.

Finally, consider the special case of isotropy.

Theorem 210 If an elastic body is isotropic at x, then it has a stored energyfunction at x. Moreover,

σ(E,x) =1

2

[λ(x) (tr E)2 + 2µ(x)E · E

]=

β(x)

2 [1 + ν(x)]

[ν(x)

1− 2ν(x)(tr E)2 + E · E

].


The following decomposition of the strain tensor facilitates the analysisof the stored energy of isotropic materials.


Definition 122 The infinitesimal strain tensor may be decomposed into itsdeviatoric and spherical parts as

E =0

E +1

3(tr E) I,

where the deviatoric part is given by

0

E:= E−1

3(tr E) I,

and 13

(tr E) I is the spherical part. Note that the trace of the deviatoric partvanishes identically.

Theorem 211 Let an elastic body be isotropic at x. Then its elasticity tensoris positive definite iff either of the following conditions hold.

1.

µ(x) > 0, 3λ(x) + 2µ(x) > 0,

2.

β(x) > 0, − 1 < ν(x) <1

2.

Proof. For an isotropic material,

E·4

C (E) = 2σ

= λ (tr E)2 + 2µE · E

by theorems 207 and 210. Applying the decomposition of definition 122, wehave

E·4

C (E) = 2µ0

E ·0

E +1

3(3λ+ 2µ) (tr E)2 (see exercise below). (5.2)

Now suppose that4

C is positive definite. Then the choice

[Eij] =

1 0 00 −1 00 0 0

⇒ tr E = 0,0

E = E,0

E ·0

E= 2;


whence by (5.2) and the positive-definite assumption,

E·4

C (E) = 4µ > 0⇒ µ > 0.

Similarly, the choice

E = I⇒ tr E = 3,0

E= 0;

and by 5.2,

E·4

C (E) = 3 (3λ+ 2µ) > 0⇒ 3λ+ 2µ > 0.

Conversely, suppose that µ > 0 and 3λ+ 2µ > 0. Then by equation 5.2

E·4

C (E) = 2µ0

E ·0

E +1

3(3λ+ 2µ) (tr E)2 ≥ 0

since0

E ·0

E≥ 0 by the positive definiteness of the tensor inner product. Forthe same reason,

E·4

C (E) = 0⇒0

Eij = 0 and tr E = 0⇒E = 0.

Thus,4

C is positive definite.That 1⇔2 follows from Theorem 205 (see exercise below).

Exercise 127 Show that, for an isotropic material,

λ (tr E)2 + 2µE · E =2µ0

E ·0

E +1

3(3λ+ 2µ) (tr E)2 .

Exercise 128 Use Theorem 205 to prove that 1⇔2.

Chapter 6

Linearized Elastodynamics

This chapter formulates the boundary/initial-value problem of the linearizedtheory of elasticity in the dynamic (i.e., time-dependent) case. A more ex-tended treatment is reserved for the case of elastostatics in the followingchapter.

6.1 References

Same as for Chapter 5. For a typical treatment of the topic, see

• J. D. Achenbach, Wave Propagation in Elastic Solids, North-Holland,1975.

Gurtin’s treatise on linearized elasticity provides a more complete accountof the general theory from a viewpoint similar to the one developed in thesenotes.

6.2 Field Equations; Elastic Processes

As developed in Section 5.2 and in Theorem 165, the fundamental field equa-tions that govern the motion of an elastic body in the linearized theory (smalldisplacement gradients, zero initial stress, as well as small second displace-ment gradients — if we wish to guarantee that the Displacement Equation of

259

260 CHAPTER 6. LINEARIZED ELASTODYNAMICS

Motion represents local balance of linear momentum to within the accuracyof the theory) are1

• the Strain-Displacement Relation

E =1

2

[∇u + (∇u)t

],

• the Stress-Strain Relation

S =4

C (E) ,

• the Stress Equation of Motion

∇ · S + ρ0b = ρ0u,

• and the Surface Force-Stress Relation

s = Sm,

which applies on the boundary of the region. Therefore, S must beinterpreted in the sense of the trace operator.

We assume that the body is given. This means that the open, bounded

regular region0

B is specified, as well as the referential mass density field

ρ0 ∈ C0(0

B,<+)

and the elasticity tensor

4

C with Cijkl = Cjikl = Cijlk ∈ C1(0

B,<).

The requirement that the underlying fields are sufficiently smooth to sup-port our formulation lead to the next two definitions.

1Beware, the development here deviates from the standard approach in the continuummechanics community, in that we introduce the surface force s as an independent field.Accordingly, we also include the relation s = Sm as a fourth field equation. In the standardapproach, s is treated as a dependent variable defined by s := Sm, so only three equationsare included in the set of fundamental field equations. See, for example, Gurtin’s treatise.

6.2. FIELD EQUATIONS; ELASTIC PROCESSES 261

Definition 123 An admissible process on0

B is an ordered 4-tuple u,E,S, swith

1. u ∈ C2(0

B ×(t0,∞),V),

u, u, u, sym∇u, sym∇u ∈ C0(0

B ×[t0,∞));

2. E ∈ C0(0

B ×[t0,∞), Sym);

3. S ∈ C1,0(0

B ×(t0,∞), Sym), 2

S,∇ · S ∈ C0(0

B ×[t0,∞));

4. s ∈ C0(∂0

Br ×[t0,∞)), where ∂0

Br denotes the set of regular points of

∂0

B and s is called the surface force field.3

This definition does not assume any relation between the fields in the4-tuple.

Definition 124 An elastic process on0

B corresponding to the body force field

b is an admissable process on0

B that meets the fundamental field equations

(for b ) on0

B ×[t0,∞). The ordered pair

b, s

is called the external force

system for the elastic process.

Theorem 212 Let b be the body force field for an elastic process on0

B cor-responding to b. Then

b ∈ C0(0

B ×[t0,∞)).

Proof. The continuity of b follows from the Stress Equation of Motion andthe smoothness of the other fields therein.

2The first and second superscripts in the notation C1,0 denote the level of continuityfor spatial and temporal derivatives respectively.

3It should be emphasized that the surface force field is an unknown solution field; itis distinct from the prescribed surface traction field used to define the stress boundaryconditions below.

262 CHAPTER 6. LINEARIZED ELASTODYNAMICS

Theorem 213 The displacement field of an elastic process on a homoge-neous and isotropic elastic body (with Lame constants λ and µ) correspondingto the body force field b meets the Cauchy-Navier Displacement Equation ofMotion

µ∇ · ∇u+ (λ+ µ)∇∇ · u+ρ0b = ρ0u on0

B ×[t0,∞),

or in component form,

µui,jj + (λ+ µ)uj,ji + ρ0bi = ρ0ui.


Remark 65 We would still need to specify initial conditions and boundaryconditions to complete the statement of the elastodynamic initial/boundary-value problem.

Chapter 7

Linearized Elastostatics

This chapter formulates the boundary-value problem of the linearized theoryof elasticity in the static (i.e., time-independent) case and establishes theuniqueness and various other properties of the solutions. The developmentparallels the formulation of the elastodynamic case in the previous chap-ter. The lengthier treatment of elastostatics reflects its maturity relative toelastodynamics.

7.1 References


7.2 Field Equations; Elastic States

As developed in Section 5.2 and in Theorem 165, the fundamental field equa-tions that govern the equilibrium of an elastic body in the linearized theory(small displacement gradients, zero initial stress, as well as small second dis-placement gradients — if we wish to guarantee that the Displacement Equa-tion of Equilibrium represents local balance of linear momentum to withinthe accuracy of the theory) are1

1Again, the reader is cautioned that the inclusion of the fourth fundamental field equa-tion and the surface force as an independent field variable is non-standard. However, thisapproach provides a more general framework that admits certain variational formulationsthat are not addressed in the standard approach.

263

264 CHAPTER 7. LINEARIZED ELASTOSTATICS

• the Strain-Displacement Relation

E =1

2

[∇u + (∇u)t

],

• the Stress-Strain Relation

S =4

C (E) ,

• the Stress Equation of Equilibrium

∇ · S + ρ0b = 0,

• and the Surface Force-Stress Relation

s = Sm,

which applies on the boundary of the region. Therefore, S must beinterpreted in the sense of the trace operator.

We assume that the body is given. This means that the open, bounded

regular region0

B is specified, as well as the referential mass density field

ρ0 ∈ C0(0

B,<+)

and the elasticity tensor

4

C with Cijkl = Cjikl = Cijlk ∈ C1(0

B,<).

Unless noted otherwise, we assume that4

C is positive definite on0

B.The requirement that the underlying fields are sufficiently smooth to sup-

port our formulation leads to the next two definitions.

Definition 125 An admissible state on0

B is an ordered 4-tuple u,E,S, swith

1. u ∈ C2(0

B,V)⇒ u, sym∇u ∈ C0(0

B);

2. E ∈ C0(0

B, Sym);

7.2. FIELD EQUATIONS; ELASTIC STATES 265

3. S ∈ C1(0

B, Sym)⇒ S,∇ · S ∈ C0(0

B);

4. s ∈ C0(∂0

Br) where ∂0

Br denotes the set of regular points of ∂0

B .

This definition does not assume any relation between the four fields in the 4-

tuple. Two admissible states on0

B, u,E,S, s and u′,E′,S′, s′ , are equaliff u = u′, E = E′, S = S′ and s = s′.

Definition 126 An elastic state on0

B corresponding to the body force field

b is an admissable state on0

B that meets the fundamental field equations (for

b ) on0

B . The ordered pair

b, s

is called the external force system for

the elastic state.

The following two theorems are direct and obvious analogs of their elas-todynamic counterparts in the previous chapter, so their proofs are omittedhere.

Theorem 214 Let b be the body force field for an elastic state on0

B corre-sponding to b. Then

b ∈ C0(0

B).

Theorem 215 The displacement field of an elastic state on a homogeneousand isotropic elastic body (with Lame constants λ and µ) corresponding tothe body force field b meets the Cauchy-Navier Displacement Equation ofEquilibrium

µ∇ · ∇u+ (λ+ µ)∇∇ · u+ρ0b = 0 on0

B,or in component form,

µui,jj + (λ+ µ)uj,ji + ρ0bi = (0)i .

Proof. The continuity of b follows from the Stress Equation of Equilibriumand the smoothness of the fields therein.

Remark 66 In the general, anisotropic homogeneous setting we have

∇·4

C (sym∇u) + ρ0b = 0 on0

B

for the Displacement Equation of Equilibrium.


The following definitions and theorem are associated with the linearity ofthe fundamental field equations for linearized elastostatics.

Definition 127 The sum of two admissible states on0

B, u,E,S, s andu′,E′,S′, s

′, is defined by

u,E,S, s+ u′,E′,S′, s′ =u + u′,E + E′,S + S

′, s + s′

.

The scalar multiple of the admissible state u,E,S, s on0

B by a constantc ∈ < is defined by

c u,E,S, s = cu,cE,c S,cs .

Theorem 216 (Principle of Superposition) Addition and scalar multi-

plication of admissible states on0

B result in admissible states. Moreover, if

u,E,S, s andu′,E′,S′, s

′are elastic states on

0

B corresponding to the

body force fields b and b′, respectively, and if c and c′ are real constants,then

c u,E,S, s+ c′ u′,E′,S′, s′

is an elastic state on0

B corresponding to the body force field cb + c′b′.


7.3 Boundary-Value Problems

We assume for the remainder of this chapter, that in addition to the specifica-

tion of the body (the region0

B, ρ0 and4

C), the following data are prescribed:

1. the body force field, b ∈ C0(0

B);

2. the surface displacement field, u ∈ C0(Su);

3. the surface traction field, s ∈ C0(Ssr).

The sets Su and Ss are complementary subsurfaces of ∂0

B, and Ssr is thesubset of regular points of Ss.

7.4. WORK-ENERGY THEOREMS. THE RECIPROCAL THEOREM267

Definition 128 Given the above data, the mixed problem of linearized elas-

tostatics is to find an elastic state u,E,S, s on0

B corresponding to b thatsatisfies the displacement boundary condition

u = u on Su

and the traction boundary condition

s = s on Ssr.

We call such an elastic state a solution of the mixed problem. If Ss = ∅,so that ∂

0

B= Su, we have the displacement problem; and, if Su = ∅ so that

∂0

B= Ss, we have the traction problem.

Remark 67 It follows from Theorem 200 that, to within the accuracy of thelinearized theory,

S(x)m(x) = T(y)n(y).

That is, S(x)m(x) may be interpreted as the surface force at the point y = f(x)of the deformed body.

Remark 68 Since our problem is defined on the open region0

B, the displace-ments u in the first boundary condition of theorem 128 must be interpretedin the sense of the trace operator. This is not necessary in the case of thesurface force field s which is defined as a boundary field. However, note thatthe trace of S is required to interpret the fourth field equation for s.

7.4 Work-Energy Theorems. The Reciprocal

Theorem

The energetic developments of section 5.5 need separate treatment in thestatic case. The following general result (deformation not necessarily small,body not necessarily elastic) is the statical analog of the Theorem of powerExpended. Note that the static case is degenerate in the sense that nokinematical fields appear in the Stress Equation of Equilibrium.


Theorem 217 (Work-Energy Theorem) Let u and S be admissible dis-placement and stress fields, respectively, in the sense of Definition 125, exceptthat here the values of S are not required to be symmetric tensors. If S satis-fies the Stress Equation of Equilibrium for the body force field b, then ∀ parts

P ⊂0

B ∫∂P

(Sm) · udAx +

∫P

b · uρ0dVx =

∫P

S · ∇udVx.

If, in addition, s is an admissible surface force field on ∂0

B that satisfies theSurface Force-Stress Relation, then∫

∂0B

s · udAx +

∫0B

b · uρ0dVx =

∫0B

S · ∇udVx.


The l.h.s. of the equation in Theorem 217 can be interpreted as the workdone by the forces external to part P , so it is natural to identify the term∫

PS · ∇udVx

with the work of the internal forces and to call it the stress work. Indeed,Theorem 217 is often identified as a variant of the Principle of Virtual Work.

If we require that S ∈ Sym (as in linearized elasticity), then the stresswork can be expressed as ∫

PS · sym∇udVx,

which for linearized elasticity becomes∫P

S · EdVx =

∫P

E·4

C (E)dVx.

By Definition 120, this is twice the strain energy. At this point, it is con-venient to introduce a different notation for the strain energy that is moreconvenient for the elastostatic case.

Definition 129 The strain energy of the elastic body0

B corresponding to thestrain field E is

U (E) :=1

2

∫0B

E(x)·4

C (x)(E(x))dVx.


Then by the above development we have

Theorem 218 (Elastic Work-Energy Theorem) 2 Let u,E,S, s be an

elastic state on0

B corresponding to the body force field b. Then∫∂

0B

s · udAx +

∫0B

b · uρ0dVx = 2U (E) ;

i.e., the work is twice the strain energy.

As a second corollary of Theorem 217, we have

Theorem 219 (Betti’s Reciprocal Theorem, 1872) Let u,E,S, s and

u′,E′,S′, s′ be elastic states on0

B corresponding to the body force fields band b′, respectively, and let the elasticity tensor have the major symmetry

Cijkl = Cklij on0

B .

Then ∫∂

0B

s′ · udAx +

∫0B

b′ · uρ0dVx =

∫0B

S′ · EdVx

=

∫0B

S · E′dVx

=

∫∂

0B

s · u′dAx +

∫0B

b · u′ρ0dVx.

Proof. The first and third equalities follow from Theorem 217. The secondequality is a consequence of the major symmetry of the elasticity tensor (seeExercise 116).

In words, Betti’s Reciprocal Theorem asserts that, given two elastic states,the work done by the external force system of the second system over thedisplacement of the first equals the work done by the external force systemof the first system over the displacement of the second.

Several interesting results concerning volume averages can be obtainedfrom Betti’s Reciprocal Theorem via special choices of the elastic states. Forexample,

2Attributed to Lame, 1852


Theorem 220 Let u,E,S, s be an elastic state on0

B corresponding to thebody force field b. Then the (infinitesimal) volume change (c.f. Theorem 201and the discussion of the dilatation below it) for an isotropic homogeneousbody is ∫

0BϑdVx =

1

3k

(∫∂

0B

s · xdAx +

∫0B

b · xρ0dVx

).

Proof. By Theorem 201,∫0BϑdVx =

∫0B

tr EdVx

=

∫0B

I · EdVx.

If we choose S′ = I, we have∫0BϑdVx =

∫0B

S′ · EdVx.

For an isotropic material, S′ = I⇒

E′ =1

3λ+ 2µI (thm. 203),

=1

3kI (defn. 118).

A corresponding displacement field is given by

u′(x) =1

3kx,

provided that k is uniform. The stress field S′ = I satisfies the Stress Equa-tion of Equilibrium if b′ = 0. The corresponding surface force field is

s′ = S′m = Im = m.

Thus, u′,E′,S

′, s′

=

1

3kx,

1

3kI, I,m

is an elastic state on

0

B corresponding to the body force field b′ = 0.


Then the Reciprocal Theorem, in the form∫0B

S′ · E dVx =

∫∂

0B

s · u′dAx +

∫0B

b · u′ρ0dVx,

yields ∫0BϑdVx =

∫∂

0B

s ·(

1

3kx

)dAx +

∫0B

b ·(

1

3kx

)ρ0dVx

=1

3k

(∫∂

0B

s · xdAx +

∫0B

b · xρ0dVx

).

This theorem provides a means to calculate the volume change from aknowledge of the external force system without having to solve for the dis-placement field. (Note, however, that this knowledge is only available a prioriin the case of the traction problem.) Consequently, it provides a means todetermine the bulk modulus k.

Exercise 132 Recall that a strain field only determines the displacementfield to within an infinitesimal rigid deformation. By Theorem 136, the mostgeneral displacement field corresponding to the strain field

E′(x) =1

3kI

is given by

u′(x) =1

3kx+

0u +

0w ×x,

where0u and

0w are constant vectors. Investigate the use of this more general

displacement field in the proof of Theorem 220.

Exercise 133 Apply Betti’s theorem in its most common form,∫∂

0B

s′ · udAx +

∫0B

b′ · uρ0dVx =

∫∂

0B

s · u′dAx +

∫0B

b · u′ρ0dVx,

to prove Theorem 220.


7.5 Uniqueness

The following uniqueness theorem for linearized elastostatics is an importantcorollary of the Elastic Work-Energy Theorem (Theorem 218). Note thatneither homogeneity nor isotropy are assumed.

Theorem 221 (Uniqueness Theorem — Kirchhoff, 1859) Let the elas-

ticity tensor be positive definite throughout0

B . Then

1. the true mixed problem (i.e., Su 6= ∅ and Ss 6= ∅) has at most onesolution;

2. the displacement problem has at most one solution;

3. any two solutions to the traction problem are equal to within an in-finitesimal rigid displacement field.

Proof. Suppose that u′,E′,S′, s′ andu′′,E′′,S′′, s′′

are two solutions tothe mixed problem. Then by the Principle of Superposition (Theorem 216),

their difference is an elastic state u,E,S, s on0

B corresponding to the bodyforce field

b = 0 on0

B,

and which satisfies

u = 0 on Su

and

s = 0 on Ssr.

Next we apply the Elastic Work-Energy Theorem (Theorem 218) to thisdifference state:

2U (E) =

∫∂

0B

s · udAx +

∫0B

b · uρ0dVx

=

∫Su

s · udAx +

∫Ss

s · udAx +

∫0B

b · uρ0dVx

=

∫Su

s · 0dAx +

∫Ss

0 · udAx +

∫0B

0 · uρ0dVx

= 0;

7.5. UNIQUENESS 273

that is, ∫0B

E(x)·4

C (x)(E(x))dVx = 0.

This implies

E = 0 on0

B,

since4

C is positive definite. Then by definition 85 and Theorem 136,

u(x) =0u +

0w × x, x ∈

0

B,

where0u and

0w are constants. Note that the analysis to this point covers the

displacement problem, the traction problem and the mixed problem.

If Su 6= ∅ (true for both the displacement problem and the mixed prob-lem), the condition u = 0 on Su ⇒

0u= 0,

0w= 0 (see exercise below),

so that u = 0 on0

B . Finally, both u = 0 and u(x) =0u +

0w × x⇒

E = 0 (E = sym∇u)

⇒S = 0 (S =

4

C (E))

⇒s = 0 (s = Sm),

so that the elastic state for the difference is

u,E,S, s =

0,0,0,0 mixed and disp. probs.

0u +

0w × x,0,0,0

traction prob.

Exercise 134 Show that u = 0 on Su ⇒0u = 0 and

0w = 0, according to the

set-up for Theorem 221.


7.6 Displacement Form of the Mixed Prob-

lem

This section presents a single-field version of the mixed problem obtained byapplying three of the fundamental field equations to eliminate three of thefour field variables that describe an admissible state.

Theorem 222 Let u be an admissible displacement field in the sense ofDefinition 125. Then u corresponds to a solution of the mixed boundary-value problem iff

1. Displacement Equation of Equilibrium

∇·4

C (sym∇u)+ρ0b = 0 on0

B[(Cijkluk,l),j +ρ0bi = (0)i],

2. Displacement Boundary Condition

u = u on Su

[ui = ui on Su],

and

3. Traction Boundary Condition

4

C (sym∇u)m = s on Ssr

[Cijkluk,lmj = si].

Furthermore, when the body is homogeneous and isotropic, these equationsreduce to

1. Cauchy-Navier Displacement Equation of Equilibrium

µ∇ · ∇u+ (λ+ µ)∇∇ · u+ρ0b = 0 on0

B

2. Displacement Boundary Condition

u = u on Su,

7.6. DISPLACEMENT FORM OF THE MIXED PROBLEM 275

3. Traction Boundary Condition

2µ(sym∇u)m + λ (∇ · u) m = s on Ssr.

Proof. Suppose that u corresponds to a solution of the mixed problem.Then, by Definition 128,

E = sym∇u on0

B, (7.1)

S =4

C (E) on0

B, (7.2)

s = Sm on ∂0

B, (7.3)

∇ · S+ρ0b = 0 on0

B, (7.4)

u = u on Su, (7.5)

s = s on Ssr. (7.6)

Substitution of (7.1) into (7.2) yields

S =4

C (sym∇u) on0

B,

and subsitution of this result into (7.3) yields

s =4

C (sym∇u)m on ∂0

B .

Then substitution of these two results into equations 7.4 - 7.6 completes thefirst part of the proof. The component form of the equations employs theminor symmetry Cijkl = Cijlk to write

Cijklu(k,l) = Cijkluk,l.

Conversely, suppose that u is an admissible displacement field that meetsthe stated equations, including the displacement boundary conditions on Su.

Define E on0

B throughE = sym∇u.

Then E is an admissible strain field. Define S on0

B through

S =4

C (E) =4

C (sym∇u).


Then S ∈ C1(0

B, Sym) is an admissible stress field. Define s on ∂0

B through

s = Sm := S|∂

0B

m =4

C (sym∇u)

∣∣∣∣∂

0B

m,

where we have made use of the C1 property of S to invoke the trace operator.Then s is an admissible surface force field that meets the traction boundarycondition

s = s on Ssr.

In terms of S, the Displacement Equation of Equilibrium becomes

∇ · S + ρ0b = 0 on0

B .

Thus, u,E,S, s is a solution of the mixed problem.The reduction to the homogeneous and isotropic case is a straightforward

calculation.

Remark 69 We obtain alternative forms of the mixed problem by employingvarious subsets of the field equations to eliminate corresponding subsets of thefield variables. We thus obtain alternative forms with one to four indepen-dent fields and a corresponding number of “free” field equations. For example,the following section develops an alternative single-field problem based on thestress. The various alternative forms are significant in the context of vari-ational formulations wherein the remaining free field equations are weaklyenforced. These variational problems are the basis of important approximatesolution methods, such as the finite element method. For example, the stiff-ness method is obtained by weak enforcement of the Displacement Equation ofEquilibrium in the displacement form of the mixed problem described above.

7.7 Stress Form of the Traction Problem

This section presents a single-field version of the traction problem (not thegeneral mixed problem!), obtained by applying three of the fundamentalfield equations to eliminate three of the four field variables that describe anadmissible state.

7.7. STRESS FORM OF THE TRACTION PROBLEM 277

We follow common practice and assume that the linearized stress-strain

relation S =4

C (E) is invertible. Thus, we have the existence of a function

E : Sym→ Sym

3S =

4

C (E)⇔ E = E(S).

Exercise 135 Show that the linearity of4

C⇒ that E is linear; i.e.,

E(c1S1+c2S2) = E(c1S1) + E(c2S2) ∀ c1, c2 ∈ < and S1,S2 ∈ Sym .

The linearity of E implies that ∃ a unique fourth-order tensor4

K (x)called the compliance tensor at the material point x 3

E =4

K (S) or Eij = KijklSkl.

Since E ∈ Sym,4

K necessarily has the minor symmetry

Kjikl = Kijkl.

Since S ∈ Sym,Eij = KijklSkl = Kij(kl)Skl;

so there is no loss of generality in assuming the additional minor symmetry

Kijlk = Kijkl.

Exercise 136 Suppose that the elasticity tensor4

C has the major symmetry

Cijkl = Cklij.

Does the compliance tensor4

K also exhibit the major symmetry? Prove yourresult.

Theorem 223 Assume that the elasticity tensor is invertible everywhere on0

B, so that

Eij = KijklSkl on0

B,

and that Kijkl ∈ C2(0

B). Also, let the region0

B be simply-connected. Then

S ∈ C2(0

B, Sym) corresponds to a solution of the traction problem with u ∈C3(

0

B) iff


1. Stress Equations of Equilibrium

Sij,j + ρ0bi = (0)i on0

B,

2. Stress Equations of Compatibility

εipmεjqn(KpqklSkl),mn = (0)ij on0

B,

and

3. Stress Traction Boundary Conditions

Sijmj = si on ∂0

Br .

Furthermore, when the body is homogeneous and isotropic and the StressEquations of Equilibrium hold, the Stress Equations of Compatibility areequivalent to the Beltrami-Michell Compatibility Equations,3

Sij,kk +1

1 + υSkk,ij +

(ρ0bi

),j

+(ρ0bj

),i

+υ

1− υ

(ρ0bk

),kδij = (0)ij on

0

B .

Proof. Suppose that S ∈ C2(0

B, Sym) corresponds to a solution of the

traction problem with u ∈ C3(0

B). Then, by Definition 126,

Eij =1

2(ui,j + uj,i) on

0

B, (7.7)

Sij = CijklEkl on0

B, (7.8)

Sij,j + ρ0bi = (0)i on0

B, (7.9)

si = Sijmj on ∂0

Br, (7.10)

si = si on ∂0

Br . (7.11)

3Beltrami in 1892 for b = 0, Donati in 1894 and Michell in 1900 for general b. Peopledo not always get the credit they deserve!

7.7. STRESS FORM OF THE TRACTION PROBLEM 279

Combining equations 7.10 and 7.11 we obatain

Sijmj = si on ∂0

Br

where Sij must be interpreted in the sense of the trace operator. We invert(7.8) to get

Eij = KijklSkl on0

B, (7.12)

where Eij ∈ C2(0

B), in view of our smoothness assumptions for Kijkl and Skl.Then by (7.7) and Theorem 137, Eij meets the Compatibility Conditions


B,

which with (7.12)⇒ the Stress Equations of Compatibility on0

B . Of course,(7.9) is one of the desired results.

Note that we did not need the restriction to simply-connected regions forthe first part of the proof.

Conversely, suppose that S ∈ C2(0

B, Sym) meets the stated equations.

Define E on0

B throughEij = KijklSkl.

Then E ∈ C2(0

B, Sym), and by the Stress Equations of Compatibility,


B .

Since0

B is assumed to be simply-connected, it follows from Theorem 140

(Cesaro Line Integral II) that ∃ u ∈ C3(0

B) 3

sym∇u = E on0

B .

Use the trace operator to define s = Sm on ∂0

Br . Then by the Stress TractionBoundary Conditions,

si = si on ∂0

Br .

Consequently, u,E,S, s is a solution of the traction problem with u ∈C3(

0

B).


Suppose0

B is homogeneous and isotropic and the Stress Equations ofEquilibrium as well as the Stress Equations of Compatibility hold. By ourprevious results, we know that Eij must meet the various forms of the com-patibility conditions. Also, by Theorem 204

Eij =1

β[(1 + ν)Sij − νSkkδij]

with β and ν uniform. Substitute this into the alternative form of the com-patibility conditions (Theorem 138)

Eij,kk + Ekk,ij − Eik,jk − Ejk,ik = 0,

to obtain (see exercise below)

Sij,kk +1

1 + νSkk,ij =

ν

1 + νSkk,llδij + Sik,jk + Sjk,ik. (7.13)

These equations could be called the “Stress Equations of Compatibility” inthe homogeneous, isotropic case. However, they are usually simplified usingthe Stress Equations of Equilibrium. From (7.9) and the supposition that

Sij ∈ C2(0

B),

Sik,jk = −(ρ0bi

),j. (7.14)

From (7.13), we get (see exercise below)

Sii,kk =1 + ν

1− νSik,ik,

which with (7.14) ⇒Sii,kk = −1 + ν

1− ν

(ρ0bi

),i. (7.15)

Substitution of equations 7.14 and 7.15 into (7.13) yields the Beltrami-MichellCompatibility Equations.

Conversely, suppose that0

B is homogeneous and isotropic and that theStress Equations of Equilibrium and the Beltrami-Michell Compatibility Equa-tions hold. From the Beltrami-Michell Compatibility Equations we get (seeexercise below) (

ρ0bk

),k

= −1− ν1 + ν

Sll,kk. (7.16)

7.8. MINIMUM PRINCIPLES 281

From the Stress Equations of Equilibrium (7.9), we obtain (7.14), as before.Substituting equations 7.16 and 7.14 into the Beltrami-Michell CompatibilityEquations yields (7.13). It can then be shown (see yet another exercise below)that (7.13) ⇒ the Stress Equations of Compatibility. (in the homogeneousand isotropic case).

Exercise 137 Combine Eij = 1β

[(1 + ν)Sij − νSkkδij] and Eij,kk + Ekk,ij −Eik,jk − Ejk,ik = 0 to obtain (7.13).

Exercise 138 Use (7.13) to obtain

Sii,kk =1 + ν

1− νSik,ik.

Exercise 139 Use the Beltrami-Michell Compatibility Equations to obtain(7.16).

Exercise 140 Show that, in the homogeneous and isotropic case, (7.13) ⇒the Stress Equations of Compatibility.

Exercise 141 Write the Beltrami-Michell Compatibility Equations in directnotation.

Remark 70 Compatibility equations play a major role in the development ofthe stress form of the traction problem. Note, however, that the compatibilityconditions are not an issue in the displacement form of the mixed problem.

7.8 Minimum Principles

This section explores the Principle of Minimum Potential Energy and itscounterpart, the Principle of Minimum Complementary Energy. These prin-ciples provide an important energetic interpretation for solutions to the boundary-value problems of linearized elasticity. In addition, they are the basis of avariety of approximate solution methods.


7.8.1 Assumptions

Throughout this section, we assume that ∀ x ∈0

B, the elasticity tensor4

C ispositive definite and that it has the major symmetry. We shall also modifyour notation for the strain energy with a subscript “C”, to emphasize itsdependence on the elasticity tensor and to distinguish it from the stressenergy that will be introduced later in this section. Thus, we write

UC (E) =1

2

∫0B

E(x)·4

C (x)(E(x))dVx

for the strain energy corresponding to a strain field E. We view UC (·) as afunctional that maps strain fields into scalar strain energies.

For purposes of developing the Principle of Minimum Complementary

Energy, we assume that ∀ x ∈0

B, the elasticity tensor4

C is invertible, so that

S =4

C (E)⇔ E =4

K (S),

where4

K (x) is the compliance tensor at x. Recall that the compliance tensorpossesses the same symmetries as the elasticity tensor. Futhermore,

Theorem 224 If the elasticity tensor at x ∈0

B is positive definite, then thecompliance tensor at x is also positive definite.


Thus,4

K is positive definite by our previous assumption for4

C.

7.8.2 The Principle of Minimum Potential Energy

Definition 130 A kinematically admissible displacement field is an admis-

sible displacement field (in the sense of Definition 125; i.e., u ∈ C2(0

B)) that

meets the additional smoothness hypothesis ∇ · (4

C (sym∇u)) ∈ C0(0

B) andthe Displacement Boundary Condition

u = u on Su.

We denote the set of all kinematically admissible displacement fields as Ju.


Definition 131 The functional Φ(·), defined on Ju by

Φ(u) := UC (sym∇u)−∫

0Bρ0b · udVx −

∫Ss

s · udAx, u ∈ Ju

is called the potential energy functional. The value Φ(u), u ∈ Ju, is calledthe potential energy of the kinematically admissible displacement field u.

Theorem 225 (Principle of Minimum Potential Energy) If u corre-sponds to a solution of the mixed boundary-value problem, then

Φ (u) ≤ Φ (u) ∀ u ∈ Ju;

moreover,Φ (u) = Φ (u)

only if u = u to within an infinitesimal rigid displacement field.

Proof. Let u correspond to the solution u,E,S, s of the mixed prob-lem. Then u is a kinematically admissible displacement field (i.e., u ∈ Ju;see exercise below). Let u be arbitrary in Ju, and define the admissibledisplacement field

u′ := u− u⇔ u = u + u′

⇒ u′ = 0 on Su.

Then we have

UC (sym∇u) = UC (sym∇u+ sym∇u′)

= UC (E + E′)

where E′ := sym∇u′. Now, by the assumed major symmetry of4

C (seeexercise below),

UC (E + E′) = UC (E) + UC (E′) +

∫0B

E′·4

C (E)dVx.

Therefore, since u,E,S, s is a solution,

UC (sym∇u)− UC (sym∇u) = UC (sym∇u′) +

∫0B

S · ∇u′dVx, (7.17)


where we have used S ∈ Sym to write the last term. By Theorem 217 andrecalling u′ = 0 on Su, the last term can be expressed as∫

0B

S ·∇u′dVx =

∫∂

0B

s ·u′dAx+

∫0B

b ·u′ρ0dVx =

∫Ss

s ·u′ dAx+

∫0B

b ·u′ρ0dVx.

(7.18)Now consider the potential energy difference,

Φ (u)− Φ (u) = UC (sym∇u)− UC (sym∇u)

−∫

0Bρ0b· (u− u) dVx −

∫Ss

s· (u− u) dAx

= UC (sym∇u)− UC (sym∇u)

−∫

0Bρ0b · u

′dVx −

∫Ss

s · u′dAx (u′ := u− u),

= UC (sym∇u′) +

∫0B

S · ∇u′dVx −

∫0B

S · ∇u′dVx

(eqns. 7.17 and 7.18),

= UC (sym∇u′) .

Now, by the positive definiteness of4

C, UC (sym∇u′) ≥ 0, with UC (sym∇u′) =0 only if sym∇u′ = 0⇔ u′ is infinitesimal rigid (by Definition 85). That is,

Φ (u) ≤ Φ (u)

with Φ (u) = Φ (u) only if u− u is an infinitesimal rigid displacement field.

Exercise 143 Show that u corresponds to the solution u,E,S, s of themixed problem ⇒ u is a kinematically admissible displacement field.

Exercise 144 Show that UC

(E + E

′)

= UC (E)+UC (E′)+∫

0B

E′·4

C (E)dVx.

Remark 71 In words, the Principle of Minimum Potential Energy assertsthat solutions to the mixed problem are minimizers over the set of kinemati-cally admissible displacement fields (i.e., Ju) of the potential energy (or thedifference between the strain energy (internal energy) and the work of theprescribed forces).


Remark 72 The converse of the Principle of Minimum Potential Energy ismore important in applications (as well as in existence theory) than the directstatement of the principle. The converse statement asserts that a kinemat-ically admissible displacement field that minimizes the potential energy is asolution to the mixed problem. The proof of the converse principle is moreinvolved, and is presented below.

Remark 73 Uniqueness of the solution to the mixed problem (to within aninfinitesimal displacement field) can be obtained as a corollary of the Principleof Minimum Potential Energy. Suppose that u and u are both displacementfields corresponding to solutions to the mixed problem (⇒ u, u ∈ Ju). By thePrinciple of Minimum Potential Energy, we have

Φ (u) ≤ Φ (u) and Φ (u) ≤ Φ (u) ,

⇒Φ (u) = Φ (u) .

Thus in general, by the last part of the Principle, u and u agree to within aninfinitesimal rigid displacement field. In the mixed and displacement prob-lems, this gives full equality as a result of the Displacement Boundary Con-dition. In all cases, we have

E = E, S = S and s = s.

7.8.3 The Principle of Minimum Complementary En-ergy

We now consider a minimum principle formulated in terms of the stress. Tobegin, we manipulate the strain energy functional to obtain an equivalentexpression in terms of the stress.

Definition 132 A statically admissible stress field is an admissible stress

field in the sense of Definition 125 (i.e., S ∈ C1(0

B, Sym)⇒ S,∇·S ∈ C0(0

B))that meets the Stress Equation of Equilibrium

∇ · S + ρ0b = 0 on0

Band the Stress Traction Boundary Condition (via Sm = s)

Sm = s on Ss.

We denote the set of all statically admissible stress fields as JS.


Definition 133 The stress energy corresponding to the statically admissible

stress field S of the elastic body0

B is

UK (S) :=1

2

∫0B

S(x)·4

K (x)(S(x))dVx.

Theorem 226 If E =4

K (S) (or equivalently, S =4

C (E)), then

UK (S) = UC (E) ;

i.e., the stress energy is equal to the strain energy.

Proof. Substitution of the inverse and direct forms of the Stress-StrainRelation.

Exercise 145 Show that for a body comprised of a linear elastic isotropicmaterial,

UC (E) =1

2

∫0B

[λ (I · E)2 + 2µE · E

]dVx,

and

UK (S) =1

2

∫0B

1

β

[(1 + ν) S · S− ν (I · S)2] dVx.

Definition 134 The functional Ψ (·) , defined on JS by

Ψ (S) := UK (S)−∫Su

(Sm) ·u dAx, S ∈ JS

is called the complementary energy functional. The value Ψ (S) , S ∈ JS, iscalled the complementary energy of the statically admissible stress field S.

Theorem 227 (Principle of Minimum Complementary Energy) If Scorresponds to a solution of the mixed boundary-value problem, then

Ψ(S) ≤ Ψ(S) ∀ S ∈ JS;

moreover,Ψ(S) = Ψ(S)

only if S = S.


Proof. Let S correspond to the solution u,E,S, s of the mixed problem.Then S is a statically admissible stress field (i.e., S ∈ JS). Let S be arbitraryin JS, and define the admissible stress field

S′ := S− S⇔ S = S + S′

corresponding to

b′ = 0 on0

B

and

s′ = 0 on Ss

Then we have

UK(S) = UK (S + S′)

= UK(S) + UK(S′) +

∫0B

S′·4

K (S)dVx,

by Exercise 144. Therefore, since u,E,S, s is a solution,

UK(S)−UK(S) = UK(S′)+

∫0B

S′·4

K (S)dVx = UK(S′)+

∫0B

S′ ·EdVx. (7.19)

By Theorem 217, the last term can also be expressed as∫0B

S′ ·E dVx =

∫∂

0B

(S′m)·udAx+

∫0Bρ0b

′ ·udVx =

∫Su

(S′m) ·udAx (7.20)

since b′ = 0 on0

B, S′m = s′ = 0 on Ss and u = u on Su. Now consider thecomplementary energy difference,

Ψ(S)−Ψ (S) = UK(S)− UK (S)−∫Su

[(S− S

)m]· udAx

= UK(S)− UK (S)−∫Su

(S′m) · udAx (S′ := S− S),

= UK(S′) +

∫0B

S′ · EdVx −∫

0B

S′ · EdVx (eqns. 7.19 and 7.20),

= UK(S′).


Now, by the positive definiteness of4

K, UK(S′) ≥ 0, with UK(S′) = 0 only if

S′ = 0 ⇔ S = S. That is, we have that

Ψ (S) ≤ Ψ(S)

with Ψ(S) = Ψ(S) only if S = S.

Remark 74 In words, the Principle of Minimum Complementary Energyasserts that stress solutions to the mixed problem are minimizers over the setof statically admissible stress fields (i.e., JS) of the complementary energy(or the difference between the stress energy (internal energy) and the workdone by the surface forces over the prescribed displacements).

With certain additional assumptions, the “converse” of the Principle ofMinimum Complementary Energy holds. This converse principle is developedlater in this section.

Exercise 146 Develop a uniqueness theorem for the mixed problem as acorollary of the Principle of Minimum Complementary Energy.

7.8.4 Converses of the Minimum Principles

We require several additional mathematical tools, associated with the Funda-mental Lemma of the Calculus of Variations in preparation for our expositionof the converse principles.

Definition 135 Φh(x) denotes the class of infinitely smooth real-valued func-tions on E with positive values on the open ball of radius h centered at x (i.e.,the ball B(x, h) according to Definition 61) but that vanish outside of the ball;i.e.,

Φh(x) := ϕ ∈ C∞(E ,<) : ϕ > 0 on B(x, h), ϕ = 0 on E−B(x, h) .

To demonstrate that Φh(x) is not empty, consider the scalar field definedon E by

ϕ(x) =

e

[−1

h2−(x−x)·(x−x)

]for x ∈B(x, h)

0 for x /∈B(x, h).


Definition 136m

D denotes the set of infinitely smooth functions on0

B whosetrace as well as the traces of all partial derivatives of the function to arbitrary

order vanish on ∂0

B; i.e.,

m

D=

m

T∈ C∞(0

B) :m

T|∂

0B

=m

0;

(n

∇m

T

)|∂

0B

=n+m

0 , n = 1, 2, . . .

in which

n

∇ denotes the gradient operator applied n times. For the case of

vector-valued functions (m = 1), we simply write D :=1

D . Similarly,m

Du

denotes the set of infinitely smooth functions on0

B with traces that vanish onSu; i.e.,

m

Du=

m

T∈ C∞(0

B) :m

T|Su =m

0

.

Clearly,m

D ⊂m

Du. Also, for the vector case, Du :=1

Du . The set of scalar

functions0

D is often called the set of test functions on0

B .

We are now ready to introduce two variants of the Fundamental Lemmaof the Calculus of Variations that are suitable for our purposes.

Lemma 228 (Fundamental Lemma of the Calculus of Variations) Let

w ∈ C0(0

B,V) satisfy ∫0B

w · v dVx = 0 ∀ v ∈ D.

Then

w = 0 on0

B .

Proof. We offer a proof by contradiction. Suppose that w 6= 0 at some

position x ∈0

B . Then at least one component of w(x) is nonzero, say wk(x) 6=0 for k ∈ 1, 2, 3 . For definiteness, let us assume wk(x) > 0. Then by theassumed continuity of w, ∃ h > 0 3 wk > 0 on B(x, h). Define

v = ϕek,

where ϕ ∈ Φh(x)⇒ v ∈ C0(0

B,V). Furthermore, for any choice of x ∈0

B , we

can always select h small enough so that B(x, h) ⊂0

B — without violating


wk > 0 on B(x, h). Then, for h sufficiently small, v vanishes on ∂0

B ⇒ v ∈ D.Moreover, ∫

0B

w · vdVx =

∫0BϕwkdVx =

∫B(x,h)

ϕwkdVx > 0

since ϕ and wk are both positive everywhere on B(x, h). This result contra-

dicts the main hypothesis of the theorem. ∴ w = 0 on0

B .

Lemma 229 (Boundary Form of the Fundamental Lemma on Ss) Letw :Ss → V be piecewise continuous on Ss, continuous at regular points of Ss,and satisfy ∫

Ss

w · v dAx = 0 ∀ v ∈ Du.

Then

w = 0 on Ss.

Proof. Again, we offer a proof by contradiction. Suppose that w 6= 0 atsome regular interior point x ∈ Ss. Then at least one component of w(x)is nonzero, say wk(x) 6= 0 for k ∈ 1, 2, 3 . For definiteness, let us assumewk(x) > 0. Then by the assumed continuity of w at regular points of Ss (i.e.,

on Ssr), ∃ h > 0 3 B(x, h)∩ ∂0

B ⊂ Ssr and wk > 0 on B(x, h)∩ ∂0

B . Define

v = ϕek,

where ϕ ∈ Φh(x) ⇒ v ∈ C0(0

B,V). Furthermore, for any choice of x ∈ Ss,we can always select h small enough so that B(x, h) ∩ Su = ∅ — withoutviolating wk > 0 on B(x, h). Thus, for h sufficiently small, v vanishes on Su

⇒ v ∈ Du. Moreover,∫Ss

w · vdAx =

∫Ss

ϕwkdAx =

∫B(x,h)∩Ss

ϕwkdAx > 0

since ϕ and wk are both positive everywhere on B(x, h) ∩ Ss. This resultcontradicts the main hypothesis of the theorem. ∴ w = 0 on Ss.

Now we are ready to tackle


Theorem 230 (Converse of the Principle of Min. Pot. Energy) Letu be a kinematically admissible displacement field with the property that

Φ (u) ≤ Φ (u)

∀ kinematically admissible displacement fields u. Then u corresponds to asolution of the mixed problem.

Proof. Define E = sym∇u, S =4

C (E) and s = Sm. Let u′ ∈ Du. Thenu := u + u′ is a kinematically admissible field. By definition of the potentialenergy,

Φ (u)− Φ (u) = UC (sym∇u)− UC (sym∇u)

−∫

0Bρ0b· (u− u) dVx −

∫Ss

s· (u− u) dAx

= UC (sym∇u)− UC (sym∇u)

−∫

0Bρ0b · u

′dVx −

∫Ss

s · u′dAx.

By Exercise 144,

UC (sym∇u)− UC (sym∇u) = UC (sym∇u′) +

∫0B

(sym∇u′) ·4

C (sym∇u)dVx

= UC (sym∇u′) +

∫0B

(sym∇u′) · SdVx

= UC (sym∇u′) +

∫0BSklu

′k,ldVx (index form, S ∈ Sym),

= UC (sym∇u′) +

∫∂

0BSklu

′kml dAx

−∫

0BSkl,lu

′kdVx (Divergence Theorem 116),

= UC (sym∇u′) +

∫Ss

Sm · u′dAx −∫

0B

(∇ · S) · u′dVx

(direct notation, u′ ∈ Du ⇒ u′ = 0 on Su),

= UC (sym∇u′) +

∫Ss

s · u′dAx

−∫

0B

(∇ · S) · u′dVx.


Thus, we are left with

Φ (u)− Φ (u) = UC (sym∇u′)−∫

0B

(∇ · S+ρ0b

)·u′dVx

−∫Ss

(s− s) ·u′dAx ≥ 0

since Φ (u) ≤ Φ (u) .Since u′ ∈ Du ⇒ αu′ ∈ Du ∀ α ∈ <, the above inequality also holds when

αu′ is substituted for u′; i.e.,

α2a− αb ≥ 0 ∀ α ∈ <.

where

a := UC (sym∇u′)

b :=

∫0B

(∇ · S+ρ0b

)·u′dVx +

∫Ss

(s− s) ·u′dAx.

We have a > 0 by the positive-definite property of the elasticity tensor. Of

course the inequality holds in the trivial case, u′ = 0 on0

B . For the more

interesting case, u′ 6= 0 at least somewhere on0

B, we shall next show thatb = 0. Suppose that b > 0. Then we have

0 ≤(αa

b− 1)α ∀ α ∈ <.

For α = b/2a, we get 0 ≤ −b/4a — a contradiction⇒ b ≤ 0. The suppositionb < 0 also leads to a contradiction (see exercise below). Therefore, b = 0;hence∫

0B

(∇ · S+ρ0b

)·u′dVx +

∫Ss

(s− s) ·u′ dAx = 0 ∀ u′ ∈ Du; (7.21)

⇒ ∫0B

(∇ · S+ρ0b

)·u′ dVx +

∫Ss

(s− s) ·u′ dAx = 0 ∀ u′ ∈ D,

since D ⊂ Du. Further, since u′ ∈ D ⇒ u′∣∣Ss

= 0, we have∫0B

(∇ · S+ρ0b

)·u′dVx = 0 ∀ u′ ∈ D.


Then by the Fundamental Lemma of the Calculus of Variations, Lemma 228,

∇ · S+ρ0b = 0 on0

B .

That is, the stress field defined in terms of u through4

C (sym∇u) satisfiesthe Stress Equation of Equilibrium for the body force field b. In view of thisresult, (7.21) reduces to∫

Ss

(s− s) ·u′dAx = 0 ∀ u′ ∈ Du,

and by the Boundary Form of the Fundamental Lemma on Ss 229 we obtain,

s− s = 0 on Ss.

That is, the traction boundary condition is satisfied. Hence u,E,S, s is asolution to the mixed problem.

Remark 75 In words, the Converse of the Principle of Minimum PotentialEnergy (CPMPE) asserts that the minimizer of the potential energy over theclass of kinematically admissible displacement fields is the displacement fieldcorresponding to the solution of the mixed problem. The CPMPE providesthe foundation for a number of important approximate solution methods, in-cluding the Rayleigh-Ritz method and the displacement (or stiffness) finiteelement method. The approximations are obtained by minimizing the po-tential energy over a finite-dimensional solution space, rather than the infi-nite dimensional space Du. Notably, less stringent smoothness conditions areimposed on the solution space — in particular, the displacements must besmooth enough to support the integration of the strain energy. The CPMPEalso provides a vehicle for establishing existence of the solution to the mixedproblem.

Theorem 231 (Converse of the Principle of Min. Compl. Energy)

Let the elasticity tensor be invertible everywhere in0

B 3 the compliance ten-

sor4

K∈ C2(0

B). Let0

B be simply-connected and convex w.r.t. Su. If S ∈ JS isa statically admissible stress field with the property that

Ψ (S) ≤ Ψ(S)∀ S ∈ JS,

then S corresponds to a solution to the mixed problem.


Proof. See §36 of Gurtin’s Handbuch article. An interesting part of the

proof is Donati’s Theorem which asserts that if E ∈ C2(0

B, Sym) and∫0B

S · EdVx = 0

∀ S ∈ C∞(0

B, Sym) with vanishing trace on ∂0

B and that meet ∇ · S = 0,then E satisfies the compatibility conditions.

7.9 Variational Principles

In this section, we assume that the elasticity tensor has the major symmetry

everywhere in0

B . However, in contrast to the previous section describingminimization principles, we do not require that the elasticity tensor be pos-itive definite for the stationary principles described in this section.

Before approaching the stationary principles, we develop the mathemat-ical notion of the variation of a functional.

Definition 137 Let K be a subspace of a linear space L, and let z ∈ L bea fixed vector. Then the linear variety (or flat or affine linear subspace) Jformed by a translation of K by z is the subset of L given by

J = v ∈L : v = z + w, w ∈K .

Note that, in general, a linear variety is not closed under vector addition orscalar multiplication, so it is not a subspace. The exception is the degeneratecase where z ∈K; then J = K is a subspace of L.

Definition 138 Let K be a subspace of a linear space L, let z ∈ L and let Jbe the linear variety formed by a translation of K by z. Let Ω be a functionaldefined on J , and let w ∈K. Then if the limit

δwΩ (z) : = limλ→0

1

λ[Ω (z + λw)− Ω (z)]

=

[d

dλΩ (z + λw)

]∣∣∣∣λ=0

exists, it is called the first variation (or directional derivative or Gateauxderivative) of Ω at z w.r.t. w.

7.9. VARIATIONAL PRINCIPLES 295

Definition 139 If δwΩ (z) exists and = 0 ∀ w ∈K, we say that the firstvariation of Ω is zero (or Ω is stationary) at z and write

δΩ (z) = 0 over J .

We shall use the notion of a linear variety to describe the set of kinemat-ically admissible displacement fields and to construct admissible variationsof the same.

Definition 140 Lu is the linear space of admissible displacement fields,consistent with Definition 125.

Definition 141 The space of kinematically admissible displacement varia-tions, denoted Ku ⊂ Lu, is given by

Ku =

u ∈Lu : ∇ · (

4

C (sym∇u)) ∈ C0(0

B); u|Su= 0

.

That is, Ku contains admissible displacement fields that meet the additionalsmoothness requirement and whose trace vanishes on Su.

Definition 142 Let u ∈Lu be a kinematically admissible displacement field.Then the linear variety of kinematically admissible displacement fields isdenoted Ju and is given by

Ju= v ∈Lu : v = u + w, w ∈Ku .

Furthermore, we have that the potential energy functional Φ (·) is a mappingon Ju to <, and the first variation of Φ at u ∈ Ju w.r.t. w ∈Ku is

δwΦ (u) := limλ→0

1

λ[Φ (u + λw)− Φ (u)] .

Theorem 232 (Principle of Stationary Potential Energy) Let u ∈ Ju

(i.e., a kinematically admissible displacement field) correspond to a solutionof the mixed problem. Then

δΦ (u) = 0 over Ju ⇔ δwΦ (u) = 0 ∀ w ∈Ku.


Proof. Let u,E,S, s be the solution to the mixed problem correspondingto u. Recall,

Φ(u) = UC (sym∇u)−∫

0Bρ0b · udVx −

∫Ss

s · udAx.

Since u + λw ∈Ju ∀ w ∈Ku and λ ∈ <, we can write

Φ(u + λw) = UC (sym∇ (u + λw))−∫

0Bρ0b· (u + λw) dVx −

∫Ss

s· (u + λw) dAx

= UC (sym∇u) + UC (λ sym∇w) +

∫0B

(λ sym∇w) ·4

C (sym∇u)dVx

−∫

0Bρ0b· (u + λw) dVx −

∫Ss

s· (u + λw) dAx (Exercise 144),

= UC (sym∇u) + λ2UC (sym∇w) + λ

∫0B

S· (sym∇w) dVx

−∫

0Bρ0b · udVx − λ

∫0Bρ0b ·wdVx −

∫Ss

s · udAx − λ∫Ss

s ·wdAx.

Thus,

δwΦ (u) = limλ→0

(λUC (sym∇w) +

∫0B

S· (sym∇w) dVx

−∫

0Bρ0b ·wdVx −

∫Ss

s ·wdAx

)=

∫0B

S· (sym∇w) dVx −∫

0Bρ0b ·wdVx −

∫Ss

s ·wdAx

=

∫0B

[S· (sym∇w)− ρ0b ·w

]dVx −

∫∂

0B

s ·wdAx (w = 0 on Su)

= 0 ∀ w ∈Ku (Work–Energy Thm. 217).

Next, we assume that the elasticity tensor is invertible on0

B and considerthe stationary analog of the Principle of Minimum Complementary Energy.To this end, we need to develop descriptions of the linear variety of staticallyadmissible stress fields and the subspace of admissible stress variations.

Definition 143 LS is the linear space of admissible stress fields, consistentwith Definition 125.


Definition 144 The space of statically admissible stress variations, denotedKS ⊂ LS, is given by

KS =

W ∈ LS : ∇ ·W = 0 on

0

B, W|Ssm = 0 on Ss

.

That is, KS contains admissible stress fields with zero divergence and whichgenerate vanishing surface forces on Ss via the trace operator.

Definition 145 Let S ∈LS be a statically admissible stress field. Then thelinear variety of statically admissible stress fields is denoted JS and is givenby

JS= V ∈LS : V = S + W, W ∈KS .

Theorem 233 (Principle of Stationary Complementary Energy) LetS ∈ JS (i.e., a statically admissible stress field) correspond to a solution ofthe mixed problem. Then

δΨ (S) = 0 over JS ⇔ δWΨ (S) = 0 ∀ W ∈ KS.

Proof. Let u,E,S, s be the solution to the mixed problem correspondingto S. Then S + λW is a statically admissible stress field; i.e., S + λW ∈ JS,∀ λ ∈ < and W ∈ KS. Thus,

Ψ (S + λW) = UK (S + λW)−∫Su

[(S + λW) m] · udAx

is well-defined ∀W ∈ KS. By Exercise 144,

UK (S + λW) = UK (S) + UK (λW) +

∫0BλW·

4

K (S)dVx

= UK (S) + UK (λW) +

∫0BλW· sym∇udVx.

∴

Ψ (S + λW) = UK (S) + λ2UK (W) + λ

∫0B

W· sym∇udVx

−∫Su

(Sm) · udAx − λ∫Su

(Wm) · udAx,


whence,

δWΨ (S) = limλ→0

[λUK (W) +

∫0B

W· sym∇udVx −∫Su

(Wm) · u dAx

]=

∫0B

W· sym∇u dVx −∫Su

(Wm) · u dAx

=

∫0B

W· sym∇u dVx −∫∂

0B

(Wm) · u dAx (W|Ss = 0; u|Su=u),

= 0 ∀W ∈ KS (Work–Energy Thm. 217 with b = 0).

The converses of the preceding stationary principles are most useful forsolving boundary-value problems in elasticity, so we consider these next.

Theorem 234 (Converse of the Principle of Stationry. Pot. Energy)Let u ∈ Ju 3

δΦ (u) = 0 over Ju ⇔ δwΦ (u) = 0 ∀ w ∈ Ku.

Then u corresponds to a solution of the mixed problem.

Proof. The displacement field u ∈ Ju satisfies the Displacement Boundary

Condition by the definition of Ju. Define E = sym∇u on0

B, S =4

C (E) on0

B and s = Sm on Ss. By hypothesis and by calculations in the proof of thePrinciple of Stationary Potential Energy, we have

δwΦ (u) =

∫0B

S· (sym∇w) dVx −∫

0Bρ0b ·wdVx −

∫∂

0B

s ·wdAx

=

∫0B

S·∇wdVx −∫

0Bρ0b ·wdVx −

∫Ss

s ·wdAx

(S ∈ Sym; w = 0 on Su),

= 0 ∀ w ∈ Ku


Now for w ∈ Ku,∫0B

S·∇wdVx =

∫0BSklwk,ldVx (indicial notation),

=

∫∂

0BSklwkmldAx −

∫0BSkl,lwkdVx (divergence thm. 116),

=

∫∂

0B

Sm ·wdAx −∫

0B

(∇ · S) ·wdVx (direct notation),

=

∫∂

0B

s ·wdAx −∫

0B

(∇ · S) ·wdVx (subst.),

=

∫Ss

s ·wdAx −∫

0B

(∇ · S) ·wdVx (w = 0 on Su).

Hence,

δwΦ (u) = −∫

0B

(∇ · S+ρ0b

)·wdVx +

∫Ss

(s− s) ·wdAx = 0 ∀ w ∈ Ku.

(7.22)

Since D ⊂ Du ⊂ Ku (functions whose trace vanishes on all of ∂0

B must also

have a vanishing trace on Su ⊂ ∂0

B; C∞ functions satisfy the less stringentcontinuity requirements of Ku ⊂ Lu), we may write

δwΦ (u) = −∫

0B

(∇ · S+ρ0b

)·wdVx = 0 ∀ w ∈ D

since w ∈ D ⇒ w|Ss= 0. Then by Lemma 228,

∇ · S+ρ0b = 0 on0

B;

so the Stress Equation of Equilibrium is satisfied and (7.22) reduces to

δwΦ (u) =

∫Ss

(s− s) ·wdAx = 0 ∀ w ∈ Ku,

⇒δwΦ (u) =

∫Ss

(s− s) ·wdAx = 0 ∀ w ∈ Du

since Du ⊂ Ku. Hence, by Lemma 229, we have

s− s = 0⇔ s = s on Ss;

i.e., the Traction Boundary Conditions are satisfied on Ss. Thus, u,E,S, sis a solution to the mixed problem.


Remark 76 In words, the Converse of the Principle of Stationary PotentialEnergy asserts that, if u is a kinematically admissible displacement field,and the first variation of the potential energy vanishes at u for all admissibledisplacement variations, then u solves the mixed problem.

Theorem 235 (Converse of Principle of Statnry. Compl. Energy) Let

the elasticity tensor be invertible on0

B 34

K ∈ C2(0

B); let0

B be simply con-nected and convex w.r.t. Su; and let S ∈ JS 3

δΨ (S) = 0 over JS ⇔ δWΨ (S) = 0 ∀ W ∈ KS.

Then S corresponds to a solution of the mixed problem.

The proof is involved and we shall not present it here. See Gurtin’s proofof the Converse of the Principle of Minimium Complementary Energy in hisHandbuch article, and his 1963 paper on variational principles in viscoelas-ticity for guidelines on how to construct a proof (Gurtin does not offer aproof of this theorem in his Handbuch article).

Remark 77 Typically, the smoothness requirements in the converse theo-rems are relaxed when they are used to construct approximate solution tech-niques. This leads to significant practical advantages when one is faced withconstructing the approximate solution spaces.

7.10 The Principle of Virtual Work

7.10.1 The Principle of Virtual Work for Elastostatics

In general, the Principle of Virtual Work expresses a balance of forces. How-ever, the forces that are balanced in the principle vary from place to place inthe literature. Accordingly, we here examine three different versions of thePrinciple that express

1. global balance of forces on a part,

2. local balance of forces,

3. local balance of forces for an elastic body.

7.10. THE PRINCIPLE OF VIRTUAL WORK 301

Theorem 236 (Principle of Virtual Work — Global Force Balance)Let s be continuous on ∂P , and let ρ0 and b be continuous on P . Then theforces on part P are balanced, i.e.∫

∂Ps dAx +

∫Pρ0b dVx = 0,

iff the virtual work of the external forces acting on P vanishes for everytranslation of P, i.e.∫

∂Ps ·w dAx +

∫Pρ0b ·w dVx = 0 ∀ w ∈ V ,

where w is a uniform displacement field that describes the translation.

Proof. Suppose that the forces acting on P are balanced; i.e.,∫∂P

s dAx +

∫Pρ0b dVx = 0,

⇒ [∫∂P

s dAx +

∫Pρ0b dVx

]·w = 0 ∀ w ∈ V ,

⇒ ∫∂P

s ·w dAx +

∫Pρ0b ·w dVx = 0 ∀ w ∈ V .

Conversely, suppose that the virtual work vanishes for every translationof P . Then ∫

∂Ps ·w dAx +

∫Pρ0b ·w dVx = 0 ∀ w ∈ V ,

⇒ [∫∂P

s dAx +

∫Pρ0b dVx

]·w = 0 ∀ w ∈ V ,

⇒ ∫∂P

s dAx +

∫Pρ0b dVx = 0, (Theorem 39).


Theorem 237 (Principle of Virtual Work — Local Force Balance)

Let the Surface Force-Stress Relation, s = Sm, hold on ∂0

B . Then the StressEquation of Equilibrium,

∇ · S + ρ0b = 0,

holds on0

B and the Traction Boundary Condition,

s = s,

holds on Ss iff the virtual work of the prescribed external forces on0

B equals

the virtual work of the stresses on0

B ∀ virtual displacement fields w ∈ Ku;i.e., ∫

Ss

s ·w dAx +

∫0Bρ0b ·w dVx =

∫0B

S·∇w dVx ∀ w ∈Ku.

Proof. Suppose that both the Stress Equation of Equilibrium and the Trac-tion Boundary Condition hold. Let w be arbitrary in Ku. Then

∫Ss

s ·w dAx +

∫0B

b ·wρ0 dVx =

∫Ss

s ·w dAx +

∫0B

b ·wρ0 dVx

(Traction Boundary Condition),

=

∫∂

0B

s ·w dAx +

∫0B

b ·wρ0 dVx ( w|Su= 0),

=

∫0B

S · ∇wdVx (Work–Energy Thm. 217).

So,

∫Ss

s ·w dAx +

∫0B

b ·wρ0 dVx =

∫0B

S · ∇w dVx ∀ w ∈Ku. (7.23)


Conversely, suppose that (7.23) holds. Integrate the r.h.s. by parts:∫0B

S · ∇wdVx =

∫0BSijwi,jdVx(index notation),

=

∫0B

[(Sijwi),j − Sij,jwi

]dVx (product rule),

=

∫∂

0BSijwimjdAx −

∫0BSij,jwidVx (divergence thm. 116),

=

∫Ss

SijmjwidAx −∫

0BSij,jwidVx (w|Su = 0),

=

∫Ss

(Sm) ·wdAx −∫

0B

(∇ · S) ·wdVx (direct notation),

=

∫Ss

s ·wdAx −∫

0B

(∇ · S) ·wdVx (Cauchy relation).

Thus, (7.23) ⇒∫Ss

(s− s) ·wdAx +

∫0B

(∇ · S + ρ0b

)·wdVx = 0 ∀ w ∈Ku. (7.24)

Since D ⊂ Ku, (7.24) reduces to∫0B

(∇ · S + ρ0b

)·wdVx = 0 ∀ w ∈D,

since w|Ss= 0 ∀ w ∈D. This result implies

∇ · S + ρ0b = 0 on0

B

by the Fundamental Lemma 228. Applying this result to (7.24) and recallingthat Du ⊂ Ku, we have∫

Ss

(s− s) ·wdAx = 0 ∀ w ∈Du,

which impliess = s on Ss

by the Boundary Form of the Fundamental Lemma 229.


Remark 78 Theorems 236 and 237 are general results in continuum me-chanics that are independent of the constitutive model and are not restrictedto “small-deformation” theory. We return to this point later in the next sub-section to develop their counterparts in the spatial description in a dynamicalsetting.

Exercise 147 In lieu of the above analysis of the r.h.s. of (7.23) using in-dicial notation, use the direct-notation identity,

∇ · (Av) = (∇ ·At) · v + tr(A∇v),

to prove∫0B

S · ∇w dVx =

∫Ss

s ·w dAx −∫

0B

(∇ · S) ·w dVx ∀ w ∈ Ku

Next, we consider the Principle of Virtual Work in the setting of linearizedelastostatics.

Theorem 238 (Virtual Work for Linearized Elastostatics) A kinemat-ically admissible displacement field u ∈ Ju corresponds to a solution of themixed problem iff∫

Ss

s ·w dAx +

∫0B

b ·wρ0 dVx =

∫0B

4

C (sym∇u) · ∇w dVx ∀ w ∈Ku.

Proof. Suppose that u corresponds to a solution of the mixed problemu,E,S, s . Then u ∈ Ju and

E = sym∇u on0

B,

S =4

C (E) =4

C (sym∇u) on0

B,

∇ · S + ρ0b = ∇·4

C ( sym∇u)+ρ0b = 0 on0

B,

s = Sm on ∂0

B,s = s on Ss.

Then Theorem 237 ⇒∫Ss

s ·wdAx +

∫0B

b ·wρ0dVx =

∫0B

4

C (sym∇u) · ∇wdVx ∀ w ∈Ku. (7.25)


Conversely, assume that (7.25) holds with u ∈ Ju. Define

E = sym∇u on0

B,

S =4

C (sym∇u) on0

B,

and

s = Sm on ∂0

B .Then Theorem 237 ⇒

∇·4

C ( sym∇u)+ρ0b = 0 on0

B,

ands = s on Ss.

Thus, u corresponds to a solution u,E,S, s of the mixed problem.

Remark 79 Theorem 238 is of great significance. It serves as the founda-tion for approximate solution methods (e.g., the finite element method); itis the variational form used to establish existence, uniqueness and continu-ous dependence on the data (i.e., boundedness) for solutions to the linearizedproblem of elastostatics.

7.10.2 The Principle of Virtual Work for Elastody-namics

Now we turn our attention to spatial formulations of the Principles of VirtualWork for Global and Local Force Balance in a dynamical setting. The basicidea is to treat the acceleration terms in the momentum balance laws in thesame way as a body force field. Thus, in view of Exercises 85 and 86, wecan write Euler’s Principles of Balance of Linear and Angular Momentum(Axioms 4 and 5) as∫

∂Pt

tn(y,t) (y, t) dAy +

∫Pt

[b (y, t)− a (y, t)] ρ (y, t) dVy = 0,

and∫∂Pt

y × tn(y,t) (y, t) dAy +

∫Pt

y× [b (y, t)− a (y, t)] ρ (y, t) dVy = 0.

Thus, the acceleration enters both equations in the same way as the bodyforce. Following d’Alembert, we are led to


Definition 146 The total body force field b is defined on the trajectory =by

b (y, t) = b (y, t)− a (y, t) ,

where −a is called the inertial body force (per unit mass).

Thus by direct substitution in the above equations, we obtain

Theorem 239 Euler’s Principles of Balance of Linear and Angular Momen-tum adopt the statical forms (in terms of the total body force field b)∫

∂Pt

tn(y,t) (y, t) dAy +

∫Pt

b(y, t)ρ(y, t) dVy = 0

and ∫∂Pt

y × tn(y,t)(y, t) dAy +

∫Pt

y × b(y, t)ρ(y, t) dVy = 0

for each part P ∈0

B and each time t ∈ [t0,∞).

We are ready for the dynamical versions of the Virtual Work Principles.

Theorem 240 (Principle of Virtual Work for Global Balance of Linear andAngular Momentum, Piola 1833) The linear and angular momentum of a

part P ∈0

B are balanced at time t ∈ [t0,∞) iff∫∂Pt

tn(y,t)(y, t) ·w(y) dAy +

∫Pt

b(y, t) ·w(y)ρ(y, t) dVy = 0

∀ infinitesimal-rigid displacement fields w on Bt.

Proof. Recall that an infinitesimal rigid displacement field has the form

w (y) = u + w × y,

where u and w are uniform vector fields. Consequently, the virtual workof the external forces acting on part P at time t over the infinitesimal-rigiddisplacement field w is∫∂Pt

tn ·w dAy +

∫Pt

b ·w ρ dVy = u ·(∫

∂Pt

tn dAy +

∫Pt

bρ dVy

)+ w ·

(∫∂Pt

y × tn dAy +

∫Pt

y × b ρ dVy

),

(7.26)


where we have made use of the vector identity u · (v×w) = v · (w× u) (cf.Exercise 49).

Suppose that the linear and angular momentum are both balanced. Then,from Theorem 239 and (7.26),∫

∂Pt

tn ·wdAy +

∫Pt

b ·wρdVy = 0 (7.27)

∀ infinitesimal rigid displacement fields w on Bt.Conversely, suppose that (7.27) holds ∀ infinitesimal rigid displacement

fields w on Bt. Then, by 7.26,

u·(∫

∂Pt

tn dAy +

∫Pt

b ρ dVy

)+w·

(∫∂Pt

y × tn dAy +

∫Pt

y × b ρ dVy

)= 0

(7.28)∀ u, w ∈V . Mathematically, (7.28) has the form

a · b + c · d = 0 ∀ a, c ∈ V ;

this implies b = d = 0. [To see this, first take c = 0 and apply Theorem 39to get b = 0⇒ c · d = 0 ∀ c ∈ V ; then apply Theorem 39 once more to getd = 0.] Thus, we have that∫

∂Pt

tndAy +

∫Pt

bρdVy = 0

and ∫∂Pt

y × tndAy +

∫Pt

y × bρdVy = 0;

i.e., both the linear and the angular momentum are balanced.For the spatial, dynamical counterpart of Theorem 237 we have

Theorem 241 (Principle of Virtual Work for Local Balance of Linear Mo-mentum) Let t = Tn on ∂Bt ∀ t ∈ [t0,∞). Then Cauchy’s First StressEquation of Motion holds on the trajectory =

∇ ·T+ρb = 0 on =,

and the Traction Boundary Condition holds on (∂Bt)s for each time t ∈[t0,∞)

t = t on (∂Bt)sr ,


where (∂Bt)s is a regular surface of ∂Bt and (∂Bt)sr is the subset of regularpoints of (∂Bt)s , iff for each t ∈ [t0,∞)∫

(∂Bt)st ·w dAy +

∫Bt

b ·wρ dVy =

∫Bt

T·∇w dVy

for every w ∈ C1(Bt,V) 3 w|∂Bt−(∂Bt)s= 0.

Proof. Suppose that∇ ·T+ρb = 0 on =

andt = t on (∂Bt)sr for each time t ∈ [t0,∞).

Then it follows from Theorem 217 that, at each time t ∈ [t0,∞) and for anyadmissible displacement field w on Bt,∫∂Bt

t ·w dAy +

∫Bt

b ·wρ dVy =

∫(∂Bt)s

t ·w dAy +

∫Bt

b ·wρ dVy

( w|∂Bt−(∂Bt)s= 0),

=

∫(∂Bt)s

t ·w dAy +

∫Bt

b ·wρ dVy (subst.),

=

∫Bt

T · ∇w dVy.

That is, ∫(∂Bt)s

t ·w dAy +

∫Bt

b ·wρ dVy =

∫Bt

T · ∇w dVy (7.29)

for every w ∈ C1(Bt,V) 3 w|∂Bt−(∂Bt)s= 0.

Conversely, suppose that (7.29) holds at each time t ∈ [t0,∞). Integratingthe r.h.s. by parts (c.f. the proof of Theorem 237), we get, for each timet ∈ [t0,∞), ∫

(∂Bt)s

(t− t

)·w dAy +

∫Bt

(∇ ·T + ρb

)·w dVy = 0

∀ w ∈ C1(Bt,V) 3 w|∂Bt−(∂Bt)s= 0. Then we use the Fundamental Lemmas,

228 and 229, to conclude

∇ ·T + ρb = 0 on =t = t on (∂Bt)sr for each time t ∈ [t0,∞).

tam 551 solid mechanics i -...

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