1 Introduction to Tensors

Home , Del, Symmetric tensor, Tensor field

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In elementary physics, we often come across two classes of quantities, namely scalars and vectors. Mass, density and temperature are examples of scalar quantities, while velocity and acceleration are examples of vector quantities. A scalar quantity’s value does not depend on the choice of the coordinate system. Similarly, although the components of a vector depend on a particular choice of coordinate system, the vector itself is invariant, and has an existence independent of the choice of coordinate system. In this chapter, we generalize the concept of a scalar and vector, to that of a tensor. In this general framework, scalars are considered as zeroth-order tensors and vectors as first-order tensors. Tensor quantities of order two and greater, similar to scalars and vectors, have an existence independent of the coordinate system. Their components, however, just as in the case of vectors, depend on the choice of coordinate system. We will see that the governing field equations of continuum mechanics can be written as tensorial equations. The advantage of writing the field equations in such ‘coordinate-free’ notation is that it is immediately obvious that these equations are valid no matter what the choice of coordinate system is. A particular coordinate system is invoked only while solving a particular problem, whence the appropriate form of the differential operators and the components of the tensors with respect to the chosen coordinate system are used. It must be borne in mind, however, that although using tensorial notation shows the ‘coordinate-free’ nature of the governing equations in a given frame of reference, it does not address the issue of how the equations transform under a change of frame of reference. This aspect will be discussed in greater detail later in this book. We now present a review of tensors. Throughout the text, scalars are denoted by light- face letters, vectors are denoted by boldface lower-case letters, while second and higher- order tensors are denoted by boldface capital letters. As a notational issue, summation over repeated indices is assumed, with the indices ranging from 1 to 3. Thus, for example, uivi represents u1v1 + u2v2 + u3v3, and Tijnj represents Ti1n1 + Ti2n2 + Ti3n3. The quantity on the right-hand side of a ‘:=’ symbol defines the quantity on its left-hand side. A function on ‘V V to V’ means that the function is defined in terms of two elements that belong to V, and× the result is also in V.

1.1 Vector Spaces

In what follows, we consider only real vector spaces. We denote the set of real numbers by .A vector space (or linear space) is a set, say V, equipped with an addition function on V< V to V (denoted by +), and a scalar multiplication function on V to V, which satisfy× the following conditions: < ×

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1. Commutativity: For all u, v V, ∈ u + v = v + u.

2. Associativity: For all u, v, w V, ∈ (u + v) + w = u + (v + w).

3. Existence of a zero element: There exists 0 V such that ∈ u + 0 = u.

4. Existence of negative elements: For each u V, there exists a negative element denoted u in V such that ∈ − u u = 0. − 5. Distributivity with respect to addition of vectors: For all α , and u, v V, ∈ < ∈ α(u + v) = αu + αv.

6. Distributivity with respect to scalar addition: For all α, β , and for all u V, ∈ < ∈ (α + β)u = αu + βu.

7. Associativity: For all α, β , and for all u V, ∈ < ∈ α(βu) = (αβ)u.

8. Identity in scalar multiplication: For all u V, ∈ 1u = u.

Since a vector space is a group with respect to addition (see Section 1.12) with 0 and u playing the roles of the neutral and reverse elements, respectively, all the results derived− for groups are applicable for vector spaces. In particular, the zero element of V, and the negative element u corresponding to a given u V are unique. Note also that αu + βv V for all α, β ,− and all u, v V. ∈ ∈ Perhaps, the∈ < most famous∈ example of a vector space is the n-dimensional coordinate space n, which is deﬁned by < n := (u , u ,..., un) : u . < { 1 2 i ∈ <} For n, addition and scalar multiplication are deﬁned by the relations <

u + v := (u1 + v1, u2 + v2,..., un + vn), αu := (αu1, αu2,..., αun).

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If p is a positive real number, then another example of a vector space is the space

Z 1 Lp[0, 1] := f : f p dx < ∞ , 0 | |

with addition and scalar multiplication deﬁned by

( f + g)(x) := f (x) + g(x), x [0, 1], ∈ (α f )(x) := α[ f (x)], x [0, 1]. ∈ To show that Lp[0, 1] is a vector space, we need to show that f + g Lp[0, 1] if f , g Lp[0, 1]. This follows from the inequality ∈ ∈

f + g p [2 max( f , g )]p 2p f p + g p . | | ≤ | | | | ≤ | | | | A subset u , u ,..., u of V is said to be linearly dependent if and only if there exist { 1 2 m} scalars α1, α2,..., αm, not all zero, such that

α u + α u + + α u = 0. 1 1 2 2 ··· m m Thus, a subset u , u ,..., u of V is linearly independent if and only if the equation { 1 2 m} α u + α u + + α u = 0, 1 1 2 2 ··· m m implies that α = α = = α = 0. A subset, say u , u ,..., u , 0 , which includes the 1 2 ··· m { 1 2 m } zero element is always linearly dependent even when the subset u1, u2,..., um is linearly independent, since the coefﬁcient of the zero element can be taken{ to be nonzero.} A subset e , e ,..., e of V is said to be a basis for V if { 1 2 n} 1. e , e ,..., e is linearly independent, and { 1 2 n}

2. Any element of V can be expressed as a linear combination of e1, e2,..., en , i.e., if u V, then { } ∈ u = u e + u e + + u e , 1 1 2 2 ··· n n

where the scalars u1, u2,..., un are known as the components of u with respect to the basis e , e ,..., e . { 1 2 n} If the bases have a ﬁnite number of elements, we have the following theorem:

Theorem 1.1.1. All bases for a given vector space contain the same number of elements.

Proof. Suppose that e , e ,..., e and e , e ,..., e are bases for a vector { 1 2 n} { 1∗ 2∗ m∗ } space. Every ei∗, since it is an element of the vector space, can be expressed in terms of the basis e , e ,..., e as β e , β . Thus, { 1 2 n} ij j ij ∈ <

αiei∗ = βijαiej = 0

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implies, by the linear independence of e , e ,..., e , that { 1 2 n}

βijαi = 0 i = 1, 2, . . . , m, j = 1, 2, . . . , n.

Let m > n. Then, the number of unknowns αi is more than the number of equations, so that it is possible to ﬁnd a nontrivial solution. Thus, there exist α1, α2, ..., αm, not all zero such that

αiei∗ = 0, i.e., e , e ,..., e is linearly dependent, contradicting the fact that it is a basis. { 1∗ 2∗ m∗ } Hence, m n. Next, suppose that m < n. Now reverse the roles of ei∗ and e in the≤ above argument to conclude that m n. Hence, m = n. { } { i} ≥

In view of the above result, a vector space V is said to be n-dimensional if it contains a basis with n elements. If no such finite integer n exists, then the vector space is said to be infinite-dimensional. For example, n is finite-dimensional with <

e1 = (1, 0, 0, . . . , 0), e2 = (0, 1, 0, . . . , 0), ...

en = (0, 0, 0, . . . , 1),

as the ‘canonical’ or natural basis. On the other hand, Lp[0, 1] is an infinite-dimensional vector space. Note that even in the finite-dimensional case, the basis need not be unique. When V is finite-dimensional, using the linear independence of e1, e2,..., en , it is easy to show that the components of any element u are unique. { } Let V be an n-dimensional vector space. A subset f , f ,..., f of V, where { 1 2 m} 1. m > n cannot be a basis for V, since, as can be shown by following the same methodology as used in the proof of Theorem 1.1.1, it is linearly dependent.

2. m < n cannot be a basis for V, since by Theorem 1.1.1, any basis has n elements. Although, the subset f 1, f 2,..., f m can be linearly independent in this case, an arbitrary element of V{cannot be expressed} in terms of the elements of this subset.

Thus, m = n is a necessary condition for the subset f 1, f 2,..., f m to be a basis. The following theorem is useful in ﬁnding if a given subset{ with n elements} is a basis for an n-dimensional vector space:

Theorem 1.1.2. Let f 1, f 2,..., f n be a subset of an n-dimensional vector space V, one of whose bases is given{ by e , e ,...,} e . Since f , f ,..., f is a subset of V, { 1 2 n} { 1 2 n} each f i can be expressed as a linear combination of the basis vectors e1, e2,..., en , i.e., f = β e , i, j = 1, 2, . . . , n, β . The following statements are{ equivalent: } i ij j ij ∈ < 1. f , f ,..., f is a basis. { 1 2 n}

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2. f , f ,..., f is linearly independent. { 1 2 n} 3. det [β ] = 0. ij 6 Proof. We prove that (i) = (ii) = (iii) = (i). ⇒ ⇒ ⇒ If f 1, f 2,..., f n is a basis, then it is linearly independent by deﬁnition. To{ prove that (ii)}= (iii), we show that “not (iii)” = “not (ii)”. Thus, let ⇒ ⇒ det [βij] = 0. From matrix algebra, it follows that there exist αi, i = 1, 2, . . . , n, not all zero, such that βijαi = 0. Thus, there exist αi, not all zero, such that

αi f i = αi βijej = 0,

which implies that the set f 1, f 2,..., f n is linearly dependent. To prove that (iii) = { (i), note that} α f = β α e = 0, implies, since ⇒ i i ij i j e1, e2,..., en is a basis (and hence a linearly independent set), that βijαi = 0, i{, j = 1, 2, . .} . , n, and since det [β ] = 0, it follows that all the α are zero, ij 6 i thus showing that f 1, f 2,..., f n is linearly independent. In addition, since det [β ] = 0, the matrix{ [β ] is invertible.} Let [γ ] denote the components of this ij 6 ij ij inverse, i.e., γijβjk = δik. Then

γij f j = γijβjkek = δikek = ei.

Since e , e ,..., e is a basis of V, every element u of V can be expressed as { 1 2 n} uiei, so that on using the above relation, we have

u = uiei = γijui f j.

Thus, the set f , f ,..., f satisﬁes the two conditions for qualifying as a basis. { 1 2 n}

A nonempty subset Vs of a vector space V is said to be a linear subspace if a linear combination of any two of its elements also lies in Vs, i.e., if (αu + βv) Vs for any arbitrary 2 3 ∈ u, v Vs, and α, β . For example, is a linear subspace of . By assuming the operations∈ of addition∈ and < scalar multiplication< to be the same as that< for the ‘parent space’ V, it can be shown by verifying the defining axioms that a linear subspace is itself a vector space. The set of all linear combinations of a subset u1, u2,..., um of V is called the linear span of u , u ,..., u , i.e., { } { 1 2 m} Lsp u , u ,..., u := u : u = α u + α u + + α u , α . { 1 2 m} { 1 1 2 2 ··· m m i ∈ <} From this definition, it immediately follows that Lsp u1, u2,..., um is a linear subspace of V. { } An inner product space (or Euclidean space) is a vector space V equipped with a function on V V to , denoted by (u, v), and called the inner product (also called scalar product or dot product)× < of u and v, that satisfies the following conditions: (u, v) = (v, u) u, v V, (1.1a) ∀ ∈ (αu + βv, w) = α(u, w) + β(v, w) α, β and u, v, w V, (1.1b) ∀ ∈ < ∈

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(u, u) 0 u V with (u, u) = 0 if and only if u = 0, (1.1c) ≥ ∀ ∈ The magnitude u is deﬁned by | | u := (u, u)1/2. | | From the above deﬁnition, and the properties of the inner product it is obvious that

αu = α u α and u V, | | | | | | ∀ ∈ < ∈ (1.2) u 0 with u = 0 if and only if u = 0. | | ≥ | | Both n and L2[0, 1] are inner product spaces. In the case of n, the inner product can be deﬁned< by <

(u, v) := u v + u v + + u v . 1 1 2 2 ··· n n The choice of inner product is not unique. If S is a symmetric, positive deﬁnite n n matrix, then another choice of inner product for n is × < n n (u, v) := ∑ ∑ Sijuivj. i=1 j=1

The canonical inner product for L2[0, 1] is deﬁned by

Z 1 ( f , g) := f (x)g(x) dx. 0 We now prove the following important inequality:

Theorem 1.1.3 (Cauchy–Schwartz Inequality). Let V be an inner product space. Then

(u, v)2 (u, u)(v, v) u, v V, (1.3) ≤ ∀ ∈ with equality if and only if u and v are linearly dependent.

Proof. If (u, v) = 0, then the above inequality is obvious. If it is not zero, then u and v are nonzero vectors, so that (u, u) and (v, v) are nonzero. By Eqn. (1.1c), we have

(u αv, u αv) 0 α . − − ≥ ∀ ∈ < Expanding this equation using the properties of the inner product, we get

(u, u) 2α(u, v) + α2(v, v) 0 α . (1.4) − ≥ ∀ ∈ < The left-hand side of the above inequality is a quadratic function in α, with a minimum at α = (u, v)/(v, v). Substituting this value of α into Eqn. (1.4), we

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get Eqn. (1.3). Alternatively, one can use a modiﬁed form of Eqn. (5.128a) that includes equality to directly obtain Eqn. (1.3). If u and/or v is the zero element, then by the remark following the deﬁni- tion of linear independence, u and v are linearly dependent, and we also have equality in Eqn. (1.3). If they are linearly dependent and nonzero, then there exists α such that u = αv, and equality in Eqn. (1.3) follows immediately. Conversely,∈ < if there is equality in Eqn. (1.3), and if u and/or v is the zero element, then they are linearly dependent. If there is equality, and both u and v are nonzero, then by letting α = (u, v)/(v, v), we see that

(u αv, u αv) = 0, − − which in turn implies that u = αv.

Applying the Cauchy–Schwartz inequality to n and L2[0, 1], we get <

n !2 n ! n ! 2 2 ∑ uivi ∑ ui ∑ vi , i=1 ≤ i=1 i=1 Z 1 2 Z 1 Z 1 f (x)g(x) dx f (x)2 dx g(x)2 dx . 0 ≤ 0 0

Using the Cauchy–Schwartz inequality, we now prove the following:

Theorem 1.1.4 (Triangle inequality). Let V be an inner product space. Then

A vector space V is a normed vector space, if we assign a nonnegative real number, u , called the norm of u, to each u V such that k k ∈ αu = α u α and u V, k k | | k k ∀ ∈ < ∀ ∈ u 0 u V with u = 0 if and only if u = 0, k k ≥ ∀ ∈ k k u + v u + v . k k ≤ k k k k

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One can show that Lp[0, 1], p 1 is a normed vector space. From Eqns. (1.2) and the triangle inequality, it is clear that≥ an inner product space is a normed vector space with the norm deﬁned by

u := u = (u, u)1/2. k k | | Since continuum mechanics is primarily the study of deformable bodies in three-dimensional space, we now specialize the results of this section to the case n = 3, and henceforth present results only for this case.

1.2 Vectors in 3 < 3 From now on, V denotes the three-dimensional Euclidean space . Let e1, e2, e3 be a ﬁxed set of orthonormal vectors that constitute the Cartesian basis.< We have{ }

e e = δ , i · j ij

where δij, known as the Kronecker delta, is deﬁned by ( 0 when i = j, δij := 6 (1.5) 1 when i = j.

The Kronecker delta is also known as the substitution operator, since, from the deﬁnition, we can see that xi = δijxj, τij = τikδkj, and so on. Note that δij = δji, and δii = δ11 + δ22 + δ33 = 3. Any vector u can be written as

u = u1e1 + u2e2 + u3e3, (1.6)

or, using the summation convention, as

u = uiei.

The inner product of two vectors is given by

(u, v) = u v := u v = u v + u v + u v . (1.7) · i i 1 1 2 2 3 3 Using Eqn. (1.6), the components of the vector u can be written as

u = u e . (1.8) i · i Substituting Eqn. (1.8) into Eqn. (1.6), we have

u = (u e )e . (1.9) · i i

We deﬁne the cross product of two base vectors ej and ek by

ej × ek := eijkei, (1.10)

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where eijk are the components of a third-order tensor E known as the alternate tensor (which we will discuss in greater detail in Section 1.7), and are given by

e123 = e231 = e312 = 1 e = e = e = 1 132 213 321 − eijk = 0 otherwise.

Taking the dot product of both sides of Eqn. (1.10) with em, we get

e (e × e ) = e δ = e . m · j k ijk im mjk Using the index i in place of m, we have

e = e (e × e ). (1.11) ijk i · j k The cross product of two vectors is assumed to be distributive, i.e.,

(αu) × (βv) = αβ(u × v) α, β and u, v V. ∀ ∈ < ∈ If w denotes the cross product of u and v, then by using this property and Eqn. (1.10), we have

w = u × v

= (ujej) × (vkek)

= eijkujvkei. (1.12)

It is clear from Eqn. (1.12) that

u × v = v × u. − Taking v = u, we get u × u = 0. The scalar triple product of three vectors u, v, w, denoted by [u, v, w], is deﬁned by

[u, v, w] := u (v × w). · In indicial notation, we have

[u, v, w] = eijkuivjwk. (1.13)

From Eqn. 1.13, it is clear that

[u, v, w] = [v, w, u] = [w, u, v] = [v, u, w] = [u, w, v] = [w, v, u] u, v, w V. − − − ∀ ∈ (1.14)

If any two elements in the scalar triple product are the same, then its value is zero, as can be seen by interchanging the identical elements, and using the above formula. From Eqn. (1.13), it is also clear that the scalar triple product is linear in each of its argument variables, so that, for example,

[αu + βv, x, y] = α [u, x, y] + β [v, x, y] u, v, x, y V. (1.15) ∀ ∈

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As can be easily veriﬁed, Eqn. (1.13) can be written in determinant form as   u1 u2 u3   [u, v, w] = det  v1 v2 v3  . (1.16) w1 w2 w3

Using Eqns. (1.11) and (1.16), the components of the alternate tensor can be written in determinant form as follows:     ei e1 ei e2 ei e3 δi1 δi2 δi3 · · · e = e , e , e = det e e e e e e  = det δ δ δ  . (1.17) ijk i j k  j · 1 j · 2 j · 3   j1 j2 j3  e e e e e e δ δ δ k · 1 k · 2 k · 3 k1 k2 k3 Thus, we have     δi1 δi2 δi3 δp1 δp2 δp3     eijkepqr = det δj1 δj2 δj3  det δq1 δq2 δq3 δk1 δk2 δk3 δr1 δr2 δr3     δi1 δi2 δi3 δp1 δq1 δr1     T = det δj1 δj2 δj3  det δp2 δq2 δr2 (since det T = det(T )) δk1 δk2 δk3 δp3 δq3 δr3     δ δ δ δ δ δ  i1 i2 i3 p1 q1 r1      = det δj1 δj2 δj3  δp2 δq2 δr2 (since (det R)(det S)=det(RS))    δk1 δk2 δk3 δp3 δq3 δr3    δimδmp δimδmq δimδmr   = det δjmδmp δjmδmq δjmδmr  δkmδmp δkmδmq δkmδmr   δip δiq δir   = det δjp δjq δjr  . (1.18) δkp δkq δkr

From Eqn. (1.18) and the relation δii = 3, we obtain the following identities (the ﬁrst of which is known as the e–δ identity):

e e = δ δ δ δ , (1.19a) ijk iqr jq kr − jr kq eijkeijm = 2δkm, (1.19b)

eijkeijk = 6. (1.19c)

Using Eqn. (1.12) and the e–δ identity, we get

(u × v) (u × v) = e e u v u v · ijk imn j k m n = (δ δ δ δ )u v u v jm kn − jn km j k m n

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= (u v u v u v u v ) m n m n − j m m j = (u u)(v v) (u v)2. (1.20) · · − · We now have the following result:

Theorem 1.2.1. For any two vectors u, v V, u × v = 0 if and only if u and v are linearly dependent. ∈

Proof. If u and v are linearly dependent, and u and/or v is zero, then it is obvious that u × v = 0. If they are linearly dependent and nonzero, then there exists a scalar α such that v = αu. Hence

u × v = αu × u = 0.

Conversely, if u × v = 0, then

0 = (u × v) (u × v) · = (u u)(v v) (u v)2, (by Eqn. (1.20)) · · − · which implies that

(u, v)2 = (u, u)(v, v).

But by Theorem 1.1.3, this implies that u and v are linearly dependent.

The vector triple products u × (v × w) and (u × v) × w, deﬁned as the cross product of u with v × w, and the cross product of u × v with w, respectively, are different in general, and are given by

u × (v × w) = (u w)v (u v)w, (1.21a) · − · (u × v) × w = (u w)v (v w)u. (1.21b) · − · The ﬁrst relation is proved by noting that

u × (v × w) = eijkuj(v × w)kei

= eijkekmnujvmwnei

= ekijekmnujvmwnei = (δ δ δ δ )u v w e im jn − in jm j m n i = (u w v u v w )e n n i − m m i i = (u w)v (u v)w. · − · The second relation is proved in an analogous manner. For scalar triple products, we have the following useful result:

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Theorem 1.2.2. The scalar triple product [u, v, w] is zero if and only if u, v and w are linearly dependent.

Proof. Since

u = u1e1 + u2e2 + u3e3, v = v1e1 + v2e2 + v3e3, w = w1e1 + w2e2 + w3e3,

by Theorem 1.1.2, u, v, w is linearly independent if and only if (see Eqn. (1.16)) { }   u1 u2 u3 [u, v, w] = det  v v v  = 0,  1 2 3  6 w1 w2 w3

which is equivalent to the statement of the theorem.

Another proof of the above theorem may be found in [43].

1.3 Second-Order Tensors

A second-order tensor is a linear transformation that maps vectors to vectors. We shall denote the set of second-order tensors by Lin. If T is a second-order tensor that maps a vector u to a vector v, then we write it as

v = Tu. (1.22)

T satisﬁes the property

T(ax + by) = aTx + bTy, x, y V and a, b . ∀ ∈ ∈ < By choosing a = 1, b = 1, and x = y, we get − T(0) = 0.

From the deﬁnition of a second-order tensor, it follows that the sum of two second-order tensors deﬁned by

(R + S)u := Ru + Su u V, ∀ ∈ and the scalar multiple of T by α , deﬁned by ∈ < (αT)u := α(Tu) u V, ∀ ∈ are both second-order tensors. The two second-order tensors R and S are said to be equal if

Ru = Su u V. (1.23) ∀ ∈

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The above condition is equivalent to the condition

(v, Ru) = (v, Su) u, v V. (1.24) ∀ ∈ To see this, note that if Eqn. (1.23) holds, then clearly Eqn. (1.24) holds. On the other hand, if Eqn. (1.24) holds, then using the bilinearity property of the inner product, we have

(v, (Ru Su)) = 0 u, v V. − ∀ ∈ Choosing v = Ru Su, and using Eqn. (1.1c), we get Eqn. (1.23). If we deﬁne the− element Z as

Zu = 0 u V, ∀ ∈ then we see that Z Lin, and is the zero element of Lin since ∈ (T + Z)u = Tu + Zu = Tu u V, ∀ ∈ which, by the deﬁnition of the equality of tensors, implies that

T + Z = T.

Henceforth, we simply use the symbol 0 to denote the zero elements of both Lin and V. With the above deﬁnitions of addition, scalar multiplication, and zero, Lin is a vector space, and hence, all the results that we have derived for vector spaces in Section 1.1 are valid for Lin. If we deﬁne the function I : V V by → Iu := u u V, (1.25) ∀ ∈ then it is clear that I Lin. I is called as the identity tensor. ∈ Choosing u = e1, e2 and e3 in Eqn. (1.22), we get three vectors that can be expressed as a linear combination of the base vectors ei as

Te1 = α1e1 + α2e2 + α3e3 Te2 = α4e1 + α5e2 + α6e3 (1.26) Te3 = α7e1 + α8e2 + α9e3,

where αi, i = 1 to 9, are scalar constants. Renaming the αi as Tij, i = 1, 3, j = 1, 3, we get

Tej = Tijei. (1.27)

The elements Tij are called the components of the tensor T with respect to the base vectors ej; as seen from Eqn. (1.27), Tij is the component of Tej in the ei direction. Taking the dot product of both sides of Eqn. (1.27) with ek for some particular k, we get

e Te = T δ = T , k · j ij ik kj or, replacing k by i,

T = e Te . (1.28) ij i · j

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By choosing v = ei and u = ej in Eqn. (1.24), it is clear that the components of two equal tensors are equal. From Eqn. (1.28), the components of the identity tensor in any orthonormal coordinate system ei are

I = e Ie = e e = δ . (1.29) ij i · j i · j ij Thus, the components of the identity tensor are scalars that are independent of the Carte- sian basis. Using Eqn. (1.27), we write Eqn. (1.22) in component form (where the components are with respect to a particular orthonormal basis e ) as { i}

viei = T(ujej) = ujTej = ujTijei,

which, by virtue of the uniqueness of the components of any element of a vector space, yields

vi = Tijuj. (1.30)

Thus, the components of the vector v are obtained by a matrix multiplication of the components of T, and the components of u. The transpose of T, denoted by TT, is deﬁned using the inner product as

(TTu, v) := (u, Tv) u, v V. (1.31) ∀ ∈ Once again, it follows from the deﬁnition that TT is a second-order tensor. The transpose has the following properties:

(TT)T = T, (αT)T = αTT, (R + S)T = RT + ST.

T If (Tij) represent the components of the tensor T, then the components of T are

(TT) = e TTe ij i · j = Te e i · j = Tji. (1.32)

The tensor T is said to be symmetric if

TT = T,

and skew-symmetric (or anti-symmetric) if

TT = T. − Any tensor T can be decomposed uniquely into a symmetric and an skew-symmetric part as (see Problem 5)

T = Ts + Tss, (1.33)

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Introduction to Tensors 15

where 1 T = (T + TT), s 2 1 T = (T TT). ss 2 − The product of two second-order tensors RS is the composition of the two operations R and S, with S operating first, and defined by the relation (RS)u := R(Su) u V. (1.34) ∀ ∈ Since RS is a linear transformation that maps vectors to vectors, we conclude that the product of two second-order tensors is also a second-order tensor. From the definition of the identity tensor given by (1.25) it follows that RI = IR = R. If T represents the product RS, then its components are given by T = e (RS)e ij i · j = e R(Se ) i · j = e R(S e ) i · kj k = e S Re i · kj k = S (e Re ) kj i · k = SkjRik

= RikSkj, (1.35) which is consistent with matrix multiplication. Also consistent with the results from matrix theory, we have (RS)T = ST RT, which follows from Eqns. (1.24), (1.31) and (1.34). 1.3.1 The tensor product We now introduce the concept of a tensor product, which is convenient for working with tensors of rank higher than two. We first define the dyadic or tensor product of two vectors a and b by (a ⊗ b)c := (b c)a c V. (1.36) · ∀ ∈ Note that the tensor product a ⊗ b cannot be defined except in terms of its operation on a vector c. We now prove that a ⊗ b defines a second-order tensor. The above rule obviously maps a vector into another vector. All that we need to do is to prove that it is a linear map. For arbitrary scalars c and d, and arbitrary vectors x and y, we have (a ⊗ b)(cx + dy) = [b (cx + dy)] a · = [cb x + db y] a · · = c(b x)a + d(b y)a · · = c[(a ⊗ b)x] + d [(a ⊗ b)y] , which proves that a ⊗ b is a linear function. Hence, a ⊗ b is a second-order tensor. Any second-order tensor T can be written as

T = Tijei ⊗ ej, (1.37)

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16 Continuum Mechanics

where the components of the tensor, Tij are given by Eqn. (1.28). To see this, we consider the action of T on an arbitrary vector u:

Tu = (Tu) e = [e (Tu)] e i i i · i = e T(u e ) e i · j j i = u e (Te ) e j i · j i = (u e ) e (Te ) e · j i · j i = e (Te ) (u e )e i · j · j i = e (Te ) (e ⊗ e )u i · j i j = e (Te ) e ⊗ e u. i · j i j Hence, we conclude that any second-order tensor admits the representation given by Eqn. (1.37), with the nine components Tij, i = 1, 2, 3, j = 1, 2, 3, given by Eqn. (1.28). The dyadic products ei ⊗ ej , i = 1, 2, 3, j = 1, 2, 3 constitute a basis of Lin since, as just mentioned, any element{ of Lin} can be expressed in terms of them, and since they are linearly independent (Tijei ⊗ ej = 0 implies that Tij = ei 0ej = 0). From Eqns. (1.29) and (1.37), it follows that ·

I = e¯i ⊗ e¯i, (1.38)

where e¯1,e ¯2,e ¯3 is any orthonormal coordinate frame. If T is represented as given by Eqn. (1.37),{ it follows} from Eqn. (1.32) that the transpose of T can be represented as

T T = Tjiei ⊗ ej. (1.39)

From Eqns. (1.37) and (1.39), we deduce that a tensor is symmetric (T = TT) if and only if Tij = Tji for all possible i and j. We now show how all the properties of a second-order tensor derived so far can be derived using the dyadic product. Using Eqn. (1.27), we see that the components of a dyad a ⊗ b are given by

(a ⊗ b) = e (a ⊗ b)e ij i · j = e (b e )a i · · j = aibj. (1.40)

The components of a vector v obtained by a second-order tensor T operating on a vector u are obtained by noting that

v e = T (e ⊗ e )u = T (u e )e = T u e , (1.41) i i ij i j ij · j i ij j i which in equivalent to Eqn. (1.30). Convention regarding complex vectors: In what follows, we will often encounter the dyadic product of two complex-valued vectors. We note here that we use the same definition for the dyadic product of complex-valued vectors as that for real-valued vectors, namely, the deﬁnition given by Eqn. (1.36). Using Eqn. (1.40), we get the components of a ⊗ b as aibj. However, it is possible to follow another convention in deﬁning the dyadic product, namely (a ⊗ b)c = (bˆ c)a (with the hat denoting complex conjugation), whereby ·

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Introduction to Tensors 17

one obtains the components as aibˆj. We will follow this rule of using the same definitions for the real and complex case even when we define the dyadic products A ⊗ B and A B of two complex-valued second-order tensors A and B. The advantage of our convention is that all relations that we derive for the real case are also valid for the complex case, since the definitions are the same. If we used the alternate convention, one would need to re- derive all the relations for the complex case, and this can be quite cumbersome, since we will be deriving dozens of relations (including rules for differentiation) involving tensor products. We emphasize that both conventions are correct, and, as long as one sticks to one convention consistently in all the derivations, the particular choice made is just a matter of convenience. 1.3.2 Principal invariants of a second-order tensor

Theorem 1.3.1. If (u, v, w) and (a, b, c) are two pairs of linearly independent vectors, and T is an arbitrary tensor, then

[Tu, v, w] + [u, Tv, w] + [u, v, Tw] [Ta, b, c] + [a, Tb, c] + [a, b, Tc] = , (1.42a) [u, v, w] [a, b, c] [Tu, Tv, w] + [u, Tv, Tw] + [Tu, v, Tw] [Ta, Tb, c] + [a, Tb, Tc] + [Ta, b, Tc] = , (1.42b) [u, v, w] [a, b, c] [Tu, Tv, Tw] [Ta, Tb, Tc] = . (1.42c) [u, v, w] [a, b, c]

Proof. Since u = uiei, v = vjej and w = wkek, the left-hand side of Eqn. (1.42a) can be written as uivjwk Tei, ej, ek + ei, Tej, ek + ei, ej, Tek LHS = [u, v, w]

eijkuivjwk [Te1, e2, e3] + [e1, Te2, e3] + [e1, e2, Te3] = { } [u, v, w]

= [Te1, e2, e3] + [e1, Te2, e3] + [e1, e2, Te3] .

Since the choice of vectors u, v and w is arbitrary, Eqn. (1.42a) follows. Equa- tions (1.42b) and (1.42c) are proved in a similar manner.

As a consequence of Theorem 1.3.1, corresponding to an arbitrary tensor T, there exist three scalars, I1, I2 and I3 such that

[Tu, v, w] + [u, Tv, w] + [u, v, Tw] = I1 [u, v, w] , (1.43a) [Tu, Tv, w] + [u, Tv, Tw] + [Tu, v, Tw] = I2 [u, v, w] , (1.43b) [Tu, Tv, Tw] = I3 [u, v, w] . (1.43c) The above equations are valid even when [u, v, w] = 0. To see this, consider Eqn. (1.43a). Following the proof of Theorem 1.3.1, we see that if [u, v, w] = 0, then [Tu, v, w] + [u, Tv, w] + [u, v, Tw] = [u, v, w] [Te , e , e ] + [e , Te , e ] + [e , e , Te ] = 0. { 1 2 3 1 2 3 1 2 3 } Equations (1.43b) and (1.43c) can be proved analogously.

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The scalars I1, I2 and I3 are called the principal invariants of T. The reason for calling them as the principal invariants is that any other scalar invariant of T can be expressed in terms of them. The ﬁrst and third invariants are referred to as the trace and determinant of T, respectively, and written as

I1 = tr T, I3 = det T.

Using the properties of the scalar triple product, it is clear that the trace is a linear operation, i.e.,

tr (αR + βS) = αtr R + βtr S α, β and R, S Lin. ∀ ∈ < ∈

To get the component form for tr T, choose u = e1, v = e2 and w = e3 in Eqn. (1.43a):

tr T = e Te + e Te + e Te = T + T + T = T . (1.44) 1 · 1 2 · 2 3 · 3 11 22 33 ii Similarly, we have

tr TT = e TTe + e TTe + e TTe 1 · 1 2 · 2 3 · 3 = e Te + e Te + e Te 1 · 1 2 · 2 3 · 3 = tr T. (1.45)

By letting T = a ⊗ b in Eqn. (1.44), we obtain

tr (a ⊗ b) = a b + a b + a b = a b = a b. (1.46) 1 1 2 2 3 3 i i · Using the linearity of the trace operator, and Eqn. (1.37), we get

tr T = tr T e ⊗ e = T tr (e ⊗ e ) = T e e = T , ij i j ij i j ij i · j ii which agrees with Eqn. (1.44). To prove

tr (RS) = tr (SR),

we choose u, v and w to be an orthonormal basis, so that

tr (RS) = (RSu, u) + (RSv, v) + (RSw, w) = (Su, RTu) + (Sv, RTv) + (Sw, RTw).

Substituting

RTu = (RTu, u)u + (RTu, v)v + (RTu, w)w = (u, Ru)u + (u, Rv)v + (u, Rw)w,

and similar expressions for RTv and RTw, we get

tr (RS) =(u, Ru)(u, Su) + (u, Rv)(v, Su) + (u, Rw)(w, Su)+ (v, Ru)(u, Sv) + (v, Rv)(v, Sv) + (v, Rw)(w, Sv)+

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Introduction to Tensors 19

(w, Ru)(u, Sw) + (w, Rv)(v, Sw) + (w, Rw)(w, Sw).

Interchanging R and S in the above equation gives an expression for tr (SR), which is the same as that for tr (RS), thus yielding the desired result. The proof using indicial notation is left as an exercise (Problem 6). Similar to the vector inner product given by Eqn. (1.7), we can deﬁne a tensor inner product of two second-order tensors R and S, denoted by R : S, by

T T T T (R, S) = R : S := tr (R S) = tr (RS ) = tr (SR ) = tr (S R) = RijSij. (1.47) The fact that R : S satisﬁes the inner product conditions in Eqn. (1.1) can be easily veriﬁed. Thus, Lin equipped with the above inner product is an inner product space, and enjoys all the properties derived in Section 1.1. In particular, if the magnitude associated with the inner product is given by

T = (T : T)1/2, | | then by the Cauchy–Schwartz inequality, we get

R : (ST) = (ST R) : T = (RTT) : S = (TRT) : ST, (1.48)

since

R : (ST) = tr (ST)T R = tr TT(ST R) = (ST R) : T = (RTS) : TT = tr ST(RTT) = (RTT) : S = (TRT) : ST.

From Eqn. (1.43c), it can be seen that

det I = 1, det(αT) = α3 det T.

It also follows from Eqn. (1.43c) that

det(RS) [u, v, w] = [RSu, RSv, RSw] = det R [Su, Sv, Sw] = (det R)(det S) [u, v, w] .

Choosing u, v and w to be linearly independent, the above equation leads us to the relation

det(RS) = (det R)(det S), (1.49)

from which we also conclude that det(RS) = det(SR). By choosing u = ej × ek = eijkei, v = ej and w = ek in Eqn. (1.43c), so that [u, v, w] = eijkeijk = 6 (by Eqn. (1.19c)), and using the fact that Tei = Tpiep, we get 1 1 det T = e e T T T = e e T T T , (1.50) 6 ijk pqr pi qj rk 6 ijk pqr ip jq kr

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20 Continuum Mechanics

where the second relation is obtained by interchanging the dummy indices. From Eqn. (1.50), it is immediately obvious that (see Section 1.3.4 for a proof without using indicial notation)

det TT = det T. (1.51)

By choosing u = ep, v = eq and w = er in Eqn. (1.43c), and using Eqn. (1.51), we also have

epqr(det T) = eijkTipTjqTkr = eijkTpiTqjTrk. (1.52)

By choosing (p, q, r) = (1, 2, 3) in the above equation, we get

det T = eijkTi1Tj2Tk3 = eijkT1iT2jT3k.

Using Eqns. (1.16), (1.49), (1.51), (1.340) and (1.341), we have

[u, v, w][p, q, r] = det [e ⊗ u + e ⊗ v + e ⊗ w] det [e ⊗ p + e ⊗ q + e ⊗ r] { 1 2 3 }{ 1 2 3 } = det [u ⊗ e + v ⊗ e + w ⊗ e ][e ⊗ p + e ⊗ q + e ⊗ r] { 1 2 3 1 2 3 } = det [u ⊗ p + v ⊗ q + w ⊗ r] . (1.53)

We now prove the following important theorem:

Theorem 1.3.2. Given a tensor T, there exists a nonzero vector n such that Tn = 0 if and only if det T = 0. Proof. If det T = 0, then by Eqn. (1.43c),

[Tu, Tv, Tw] = 0 u, v, w V, ∀ ∈ which, by Theorem 1.2.2, implies that Tu, Tv, Tw are linearly dependent. This means that there exist scalars α, β and γ, not all zero, such that

αTu + βTv + γTw = T(αu + βv + γw) = 0.

Thus, Tn = 0, where n = αu + βv + γw. Conversely, if there exists a nonzero vector n such that Tn = 0, choose v and w such that n,v and w are linearly independent, i.e., [n, v, w] = 0. Then, by Eqn. (1.43c), we have 6

det T [n, v, w] = [Tn, Tv, Tw] = [0, Tv, Tw] = 0,

which implies that det T = 0.

1.3.3 Inverse of a tensor In order to define the concept of a inverse of a tensor, it is convenient to first introduce the cofactor tensor, denoted by cof T, and defined by the relation

cof T(u × v) := Tu × Tv u, v V. (1.54) ∀ ∈

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Introduction to Tensors 21

cof T obviously maps a vector to a vector. To prove linearity, note that every nonzero vector w can be expressed as u × v for some nonzero u, v V (e.g., take v as a unit vector ∈ perpendicular to w, and u = v × w). We now show that if w1 and w2 are two arbitrary nonzero vectors, they can be expressed as u1 × v and u2 × v. If w1 and w2 are linearly dependent, then w2 = αw1, α , and then u2 = αu1. If w1 and w2 are linearly in- ∈ < 2 dependent, then let v = (w1 × w2)/ w1 × w2 . Using Eqn. (1.21a), we can show that 2 | 2| u1 = (w1 w2)w1 + w1 w2 and u2 = w2 w1 + (w1 w2)w2 yield u1 × v = w1 and u × v−= w· . Noting that| |T Lin, we now− | have| · 2 2 ∈

cof T(αw1 + βw2) = cof T[(αu1 + βu2) × v] = [T(αu1 + βu2)] × Tv = [αTu1 + βTu2] × Tv = α(Tu1 × Tv) + β(Tu2 × Tv) = αcof T(u1 × v) + βcof T(u2 × v) = αcof T w1 + βcof T w2,

thus proving that cof T Lin. Using Eqn. (1.54), we∈ now prove the following explicit formula for the cofactor:

(cof T)T = I I (tr T)T + T2. (1.55) 2 − Replacing u in Eqn. (1.43a) by Tu, we get h i tr T [Tu, v, w] = T2u, v, w + [Tu, Tv, w] + [Tu, v, Tw] ,

which on rearranging yields h i [Tu, Tv, w] + [Tu, v, Tw] = tr T [Tu, v, w] T2u, v, w . (1.56) − Using Eqn. (1.43b) and (1.54), we have

h T i I2 [u, v, w] = [Tu, Tv, w] + [Tu, v, Tw] + (cof T) u, v, w . (1.57)

Substituting Eqn. (1.56) into (1.57), using the fact that u, v and w are arbitrary, and invoking equality of tensors, we get Eqn. (1.55). It immediately follows from Eqn. (1.55) that cof T corresponding to a given T is unique. We also observe that

cof TT = (cof T)T, (1.58)

and that

(cof T)T T = T(cof T)T. (1.59)

The components of the cofactor tensor are given by

1 (cof T) = e e T T . (1.60) ij 2 imn jpq mp nq

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To prove this, let u = eq × ej and v = eq in Eqn. (1.54), and use Eqn. (1.21b) to get u × v = 2ej. On the right-hand side of Eqn. (1.54), use the relations eq × ej = epqjep, Tep = Tmpem and Teq = Tnqen. Taking the dot product with ei of both sides of the relation so obtained, we get the desired result. Equation (1.60) when written out explicitly reads   T T T T T T T T T T T T 22 33 − 23 32 23 31 − 21 33 21 32 − 22 31 [cof T] = T T T T T T T T T T T T  . (1.61)  32 13 − 33 12 33 11 − 31 13 31 12 − 32 11 T T T T T T T T T T T T 12 23 − 13 22 13 21 − 11 23 11 22 − 12 21 Equation (1.54) can be used to get a simple expression for I2, the second invariant of a tensor. We ﬁrst write Eqn. (1.43b) as

I2 [u, v, w] = [u, Tv, Tw] + [v, Tw, Tu] + [w, Tu, Tv] , and then use Eqn. (1.54) to write the above equation as

h T i h T i h T i I2 [u, v, w] = (cof T) u, v, w + (cof T) v, w, u + (cof T) w, u, v h i h i h i = (cof T)Tu, v, w + u, (cof T)Tv, w + u, v, (cof T)Tw = tr (cof T)T [u, v, w] , where the last step follows from Eqn. (1.43a). Using Eqn. (1.45), and choosing u, v and w to be linearly independent (so that [u, v, w] = 0), we get 6 I2 = tr (cof T). (1.62)

An alternative expression for the I2 can be found by directly taking the trace of both sides of Eqn. (1.55). Using Eqns. (1.45) and (1.62), and the fact that the trace is a linear operator, we get

I = 3I (tr T)2 + tr T2, 2 2 − which implies that 1 h i I = (tr T)2 tr T2 . (1.63) 2 2 −

The equivalence of the two formulae for I2 given by Eqns. (1.62) and (1.63) can also be shown using indicial notation (see Problem 10). Substituting Eqn. (1.54) into Eqn. (1.43c), we get (Tu, cof T(v × w)) = det T(Iu, v × w) u, v, w V, ∀ ∈ which implies that

((cof T)T Tu, v × w) = ((det T)Iu, v × w) u, v, w V. ∀ ∈ Any arbitrary vector z can be expressed as v × w for some v, w V (take w as a unit vector perpendicular to z, and v = w × z), so that from Eqn. (1.24), it∈ follows that (cof T)T T = (det T)I, which when combined with Eqn. (1.59) yields

T(cof T)T = (cof T)T T = (det T)I. (1.64)

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Similar to the result for determinants, the cofactor of the product of two tensors is the product of the cofactors of the tensors (see also Problem 9), i.e.,

cof (RS) = (cof R)(cof S).

This is because

[cof R cof S]T RS = (cof S)T(cof R)T RS = (det R)(cof S)T IS (by Eqn. (1.64)) = (det R)(det S)I (by Eqn. (1.64)) = det(RS)I, (by Eqn. (1.49))

and since the cofactor matrix is unique, the result follows. 1 The inverse of a second-order tensor T, denoted by T− , is deﬁned by

1 T− T = I, (1.65)

where I is the identity tensor. A characterization of an invertible tensor is the following:

Theorem 1.3.3. A tensor T is invertible if and only if det T = 0. The inverse, if it exists, is unique. 6

1 Proof. Assuming T− exists, from Eqns. (1.49) and (1.65), we have 1 (det T)(det T− ) = 1, and hence det T = 0. Conversely, if det T = 0, then from Eqn.6 (1.64), we see that at least one inverse exists, and is given by 6

1 1 T T− = (cof T) . (1.66) det T

1 1 1 1 Let T1− and T2− be two inverses that satisfy T1− T = T2− T = I, from which 1 1 1 it follows that (T1− T2− )T = 0. Choose T2− to be given by the expression in − 1 Eqn. (1.66) so that, by virtue of Eqn. (1.64), we also have TT2− = I. Multiplying 1 1 1 1 1 both sides of (T1− T2− )T = 0 by T2− , we get T1− = T2− , which establishes − 1 the uniqueness of T− .

From Eqns. (1.64) and (1.66), we have

1 1 T− T = TT− = I. (1.67)

Another characterization of an invertible tensor is the following:

Theorem 1.3.4. A tensor T is invertible if and only if the equation

Tu = v,

has a unique solution u V for each v V. ∈ ∈

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24 Continuum Mechanics

1 Proof. If T is invertible, then multiplying v = Tu by T− , we get

1 1 T− v = T− Tu = Iu = u

1 Since T− is unique, u is unique. To prove the converse, note that if u1 and u2 are two solutions of the equation Tu = v, then, by the assumed uniqueness, u1 = u2. Also, by assumption, for every v V, there exists a u such that Tu = v. Thus, the mapping T : V V is one-to-one∈ and onto, and hence, T 1 : V V exists. → − → Summarizing, if T is invertible, we have

1 Tu = v u = T− v, u, v V. ⇐⇒ ∈ 1 1 By the above property, T− clearly maps vectors to vectors. Hence, to prove that T− is a second-order tensor, we just need to prove linearity. Let a, b V be two arbitrary vectors, 1 1 1 ∈ and let u = T− a and v = T− b. Since I = T− T, we have 1 I(αu + βv) = T− T(αu + βv) 1 = T− [T(αu + βv)] 1 = T− [αTu + βTv] 1 = T− (αa + βb), which implies that

1 1 1 T− (αa + βb) = αT− a + βT− b a, b V and α, β . ∀ ∈ ∈ < The inverse of the product of two invertible tensors R and S is

1 1 1 (RS)− = S− R− , (1.68) since the inverse is unique, and

1 1 1 1 S− R− RS = S− IS = S− S = I. Similarly, if T is invertible, then TT is invertible since det TT = det T = 0. The inverse of the transpose is given by 6

T 1 1 T (T )− = (T− ) , (1.69) since

1 T T 1 T T (T− ) T = (TT− ) = I = I. Hence, without fear of ambiguity, we can write

T T 1 1 T T− := (T )− = (T− ) .

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Introduction to Tensors 25

1 From Eqn. (1.69), it follows that if T Sym, then T− Sym. ∈ 1 ∈ Although, no easy expression for (T1 + T2)− is known, the following Woodbury formula gives an expression for a particular kind of perturbation on an invertible tensor T, and which holds for any underlying space dimension n:

1 1 1 1 1 1 1 (T + UCV)− = T− T− U(C− + VT− U)− VT− . (1.70) − T U A1 B1 To prove this formula, consider the inversion of the matrix − 1 . If represents VC− C1 D1 the inverse of this matrix, then the condition " #" # " # A1 B1 T U I 0 − 1 = , C1 D1 VC− 0 I

leads to the equations

A1T + B1V = I, (1.71a) 1 A U + B C− = 0. (1.71b) − 1 1

From Eqn. (1.71b), we get B1 = A1UC, which on substituting into Eqn. (1.71a) yields

1 A1 = (T + UCV)− . (1.72)

On the other hand, from Eqn. (1.71a), we have

1 1 A = T− B VT− , (1.73) 1 − 1 which on substituting into Eqn. (1.71b) yields

1 1 1 1 B1 = T− U(C− + VT− U)− .

Substituting this expression into Eqn. (1.73) yields

1 1 1 1 1 1 A = T− T− U(C− + VT− U)− VT− . (1.74) 1 − Comparing Eqns. (1.72) and (1.74), we get Eqn. (1.70). If C = 1, then we get the following Sherman–Morrison formula:

1 1 1 1 T− (u ⊗ v)T− (T + u ⊗ v)− = T− . (1.75) − 1 + v T 1u · − 1 1 + v T− u should be nonzero for T + u ⊗ v to be invertible. · 1 1 Let S Sym be an invertible tensor, and letu ˜ = S− u andv ˜ = S− v. Note that because of the symmetry∈ of S, we have u v˜ = u˜ v. By applying Eqn. (1.75) twice, we get · · 1 1 1 (S + u ⊗ v + v ⊗ u)− = S− + [(u u˜ )v˜ ⊗ v˜ + (v v˜ )u˜ ⊗ u˜ (1 + u v˜ )(u˜ ⊗ v˜ + v˜ ⊗ u˜ )] , D · · − · where

D = (1 + u v˜ )2 (u u˜ )(v v˜ ). · − · ·

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D should be nonzero for S + u ⊗ v + v ⊗ u to be invertible. As a corollary, for S = I, we get

1 1 h 2 2 i (I + u ⊗ v + v ⊗ u)− = I + u v ⊗ v + v u ⊗ u (1 + u v)(u ⊗ v + v ⊗ u) , D | | | | − · where

D = (1 + u v)2 u 2 v 2 . · − | | | | Consider the tensor

H = I + u ⊗ v, (1.76)

and let the underlying space dimension be n. Let e1∗, e2∗,..., en∗ 1 be n 1 unit vectors that are perpendicular to v, and mutually perpendicular{ to each other.− } Then− from Eqn. (1.76), it is immediately evident that (1, e1∗), (1, e2∗),..., (1, en∗ 1), (1 + u v, u/ u ) are eigenvalues/eigenvectors of H. If u v = 0, then all the eigenvectors− are· linearly| | independent, while if u v = 0, then the ﬁrst· n6 1 eigenvectors are linearly independent, while the last can be expressed· in terms of them.− The characteristic equation is

n 1 (λ 1) − (λ 1 u v) = 0. − − − · On applying the Cayley–Hamilton theorem (see Theorem 1.3.5), we get

n 1 [H I] − [H (1 + u v)I] = 0. − − · The minimal polynomial is

[H I][H (1 + u v)I] = 0, − − · which can be immediately veriﬁed by using Eqn. (1.76). Multiplying the eigenvalues of H, we get

det H = 1 + u v. · Now consider the tensor that occurs in Eqn. (1.75), which is a generalization of the tensor H, namely,

M = T + u ⊗ v,

1 which can be written as T[I + (T− u) ⊗ v], assuming T to be invertible. Thus,

1 det M = det T[1 + (T− u) v]. · The result for arbitrary T is obtained by taking the limit in the above equation. We get

det M = det T + u [(cof T)v], · with cof T given by Eqn. (J.12). Consider a symmetric perturbation to I:

K = I + u ⊗ v + v ⊗ u.

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Introduction to Tensors 27

Let e1∗, e2∗,..., en∗ 2 be n 2 unit vectors that are perpendicular to u and v. Then (1, e1∗), { − } − (1, e2∗),..., (1, en∗ 2), (1 + u v + u v , u/ u + v/ v ), (1 + u v u v , u/ u v/ v ) are the eigenvalues/eigenvectors− · | of| | K|, so that| | | | · − | | | | | | − | |

det K = (1 + u v + u v )(1 + u v u v ) = (1 + u v)2 u 2 v 2 . (1.77) · | | | | · − | | | | · − | | | | By virtue of the Cauchy–Schwartz inequality,

det K 1 + 2u v. ≤ · Let S be a symmetric positive definite tensor, and let U = √S be its symmetric positive definite square-root. If Z is defined by

Z := S + u ⊗ v + v ⊗ u,

1 1 1 1 then it can be written as U[I + (U− u) ⊗ (U− v) + (U− v) ⊗ (U− u)]U, so that on using Eqn. (1.77), we get

2 1 1 1 det Z = det S[ 1 + (S− u) v (u S− u)(v S− v)]. · − · · Although our proof assumed S to be positive deﬁnite, actually, the above result holds for any invertible S Sym. ∈ 1.3.4 Eigenvalues and eigenvectors of tensors If T is an arbitrary tensor, a vector n is said to be an eigenvector of T if there exists λ such that

Tn = λn. (1.78)

Writing the above equation as (T λI)n = 0, we see from Theorem 1.3.2 that a nontrivial eigenvector n exists if and only if −

det(T λI) = 0. − This is known as the characteristic equation of T. Using Eqn. (1.43c), the characteristic equation can be written as

[Tu λu, Tv λv, Tw λw] = 0, − − − for linearly independent vectors u, v and w. Using Eqns. (1.14) and (1.15), we can write the characteristic equation as

λ3 I λ2 + I λ I = 0, (1.79) − 1 2 − 3

where I1, I2 and I3 are the principal invariants given by Eqns. (1.43). Since the principal invariants are real, Eqn. (1.79) has either one or three real roots. If one of the eigenvalues is complex, then it follows from Eqn. (1.78) that the corresponding eigenvector is also complex. By taking the complex conjugate of both sides of Eqn. (1.78), we see that the complex conjugate of the complex eigenvalue, and the corresponding complex eigenvector are

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also eigenvalues and eigenvectors, respectively. Thus, eigenvalues and eigenvectors, if complex, occur in complex conjugate pairs. If λ1, λ2, λ3 are the roots of the characteristic equation, then from Eqn. (1.79), it follows that

I1 = tr T = T11 + T22 + T33, = λ1 + λ2 + λ3, (1.80a)

h i 1 2 2 T11 T12 T22 T23 T11 T13 I2 = tr cof T = (tr T) tr (T ) = + + 2 − T21 T22 T32 T33 T31 T33

= λ1λ2 + λ2λ3 + λ1λ3, (1.80b) 1 h i I = det(T) = (tr T)3 3(tr T)(tr T2) + 2tr T3 = e T T T 3 6 − ijk i1 j2 k3 = λ1λ2λ3, (1.80c)

where . denotes the determinant. The set of eigenvalues λ1, λ2, λ3 is known as the spectrum| | of T. { } If λ is an eigenvalue, and n is the associated eigenvector of T, then λ2 is the eigenvalue of T2, and n is the associated eigenvector, since

T2n = T(Tn) = T(λn) = λTn = λ2n.

In general, λn is an eigenvalue of Tn with associated eigenvector n. The eigenvalues of TT and T are the same since their characteristic equations are the same. An extremely important result is the following:

Theorem 1.3.5 (Cayley–Hamilton Theorem). A tensor T satisﬁes an equation hav- ing the same form as its characteristic equation, i.e.,

T3 I T2 + I T I I = 0 T. (1.81) − 1 2 − 3 ∀ Proof. Multiplying Eqn. (1.55) by T, we get

(cof T)T T = I T I T2 + T3. 2 − 1 T Since by Eqn. (1.64), (cof T) T = (det T)I = I3 I, the result follows.

By taking the trace of both sides of Eqn. (1.81), and using Eqn. (1.63), we get 1 h i det T = (tr T)3 3(tr T)(tr T2) + 2tr T3 . (1.82) 6 − From the above expression and the properties of the trace operator, Eqn. (1.51) follows. We have

λ = 0, i = 1, 2, 3 = 0 tr (T) = tr (T2) = tr (T3) = 0. (1.83) i ⇐⇒ IT ⇐⇒ The proof is as follows. If all the invariants are zero, then from the characteristic equation given by Eqn. (1.79), it follows that all the eigenvalues are zero. If all the eigenvalues are zero, then from Eqns. (1.80a)–(1.80c), it follows that all the principal invariants are zero. IT

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Introduction to Tensors 29

If tr (T) = tr (T2) = tr (T3) = 0, then again from Eqns. (1.80a)–(1.80c) it follows that the principal invariants are zero. Conversely, if all the principal invariants are zero, then all j 3 j the eigenvalues are zero from which it follows that tr T = ∑i=1 λi, j = 1, 2, 3 are zero. Consider the second-order tensor u ⊗ v. By Eqn. (1.60) or by writing u ⊗ v as u ⊗ v + 0 ⊗ 0 + 0 ⊗ 0 and using Eqn. (1.333), it follows that

cof (u ⊗ v) = 0, (1.84)

so that the second invariant, which is the trace of the above tensor, is zero. Similarly, on using Eqn. (1.53), we get the third invariant as zero. The ﬁrst invariant is given by u v. Thus, from the characteristic equation, it follows that the eigenvalues of u ⊗ v are (0, 0,·u v). If u and v are perpendicular, u ⊗ v is an example of a nonzero tensor all of whose· eigenvalues are zero.

1.4 Skew-Symmetric Tensors

Let W Skw and let u, v V. Then ∈ ∈ (u, Wv) = (W Tu, v) = (Wu, v) = (v, Wu). (1.85) − − On setting v = u, we get

(u, Wu) = (u, Wu), − which implies that

(u, Wu) = 0. (1.86)

Thus, Wu is always orthogonal to u for any arbitrary vector u. By choosing u = ei and v = ej, we see from the above results that any skew-symmetric tensor W has only three independent components (in each coordinate frame), which suggests that it might be replaced by a vector. This observation leads us to the following result:

Theorem 1.4.1. Given any skew-symmetric tensor W, there exists a unique vector w, known as the axial vector or dual vector, corresponding to W such that Wu = w × u u V. (1.87) ∀ ∈ Conversely, given any vector w, there exists a unique skew-symmetric second-order tensor W such that Eqn. (1.87) holds. Proof. A skew-symmetric tensor W, like any other tensor, has at least one real eigenvalue, say λ. Let p be the associated eigenvector. Then W p = λp, and on taking the dot product of both sides with p, we get λ = p (W p). By virtue of Eqn. (1.86), λ = 0 (we show later in this section that this· is the only real eigenvalue). Let q, r be unit vectors that form an orthonormal basis with p. We have p = q × r, q = r × p, r = p × q. (1.88)

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Since W p = λp = 0, we have (q, W p) = (r, W p) = 0, and by Eqn. (1.85), we also have (p, Wq) = (p, Wr) = 0. Thus, using W = Wijei ⊗ ej referred to the basis p, q, r , we get { } W = γ(r ⊗ q q ⊗ r), − where γ = r (Wq) must be nonzero if W = 0. Let w = γp, and let u be an arbitrary vector.· Then 6

Wu w × u = γ (r ⊗ q)u (q ⊗ r)u p × [(u p)p + (u q)q + (u r)r] − { − − · · · } = γ [(u q)(r p × q) (u r)(q + p × r)] · − − · = 0,

where the last step follows from Eqn. (1.88). Thus, associated with every skew- symmetric tensor W, there is a vector w such that Eqn. (1.87) holds. Conversely, given w, let q and r be unit vectors such that w/ w , q and r form an orthonormal basis. Then | |

W = w (r ⊗ q q ⊗ r), (1.89) | | − is a skew-symmetric tensor with w as the axial vector of W, a fact that can be checked easily by verifying that Wu = w × u for any arbitrary vector u. Given W, uniqueness of the axial vector w can be shown by assuming the existence of two vectors w1 and w2. Now we have Wu = w1 × u and Wu = w2 × u for all vectors u. Hence,

(w w ) × u = 0 u V. 1 − 2 ∀ ∈

If w1 w2 is a nonzero vector, then we can choose u to be perpendicular to w −w , so that (w w ) × u is nonzero. But this leads to a contradiction 1 − 2 1 − 2 with the above property. Hence w1 = w2, and the uniqueness of the axial vector is established. Similarly, given w, uniqueness of the corresponding W can be established by assuming the existence of two tensors W1 and W2, such that W1u = w × u and W2u = w × u for all vectors u. Now we have

(W W )u = 0 u V, 1 − 2 ∀ ∈

which implies that W1 = W2.

Note that Wu = 0 if and only if u = αw, α , since if u = αw then Wu = αw × w = 0, while conversely, if Wu = w × u = 0, then u∈is < a scalar multiple of w by Theorem 1.2.1. This result justiﬁes the use of the terminology ‘axial vector’ used for w. Also note that by virtue of the uniqueness of w, the vector αw, α , is a one-dimensional subspace of V. Similarly, W2u = 0 if and only if u = αw, α ∈, < since if W2u = 0, by the above result, Wu = w × u = βw, which is only possible when∈

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Introduction to Tensors 31

By choosing u = ej and taking the dot product of both sides with ei, Eqn. (1.87) can be expressed in component form as

W = e w , ij − ijk k 1 (1.90) w = e W . i − 2 ijk jk

More explicitly, if w = (w1, w2, w3), then   0 w w − 3 2 W =  w 0 w  .  3 − 1 w w 0 − 2 1 From Eqns. (1.89) or (1.90), it follows that W : W = tr (W2) = 2w w. Since tr W = − · tr W T = tr W and det W = det W T = det W, we have tr W = det W = 0. The second − 2 − 2 invariant is given by I2 = [(tr W) tr (W )]/2 = (W : W)/2 = w w. Thus, from the characteristic equation, we get the eigenvalues− of W as (0, i w , i w·), and the ‘spectral resolution’ as | | − | |

W = i w (n ⊗ nˆ nˆ ⊗ n), | | − where n is the eigenvector corresponding to i w , andn ˆ is its complex conjugate. Note that we have used the convention outlined at the| end| of Section 1.3.1 in writing the above result.

1.5 Orthogonal Tensors

T 1 A second-order tensor Q is said to be orthogonal if Q = Q− , or, alternatively by Eqn. (1.67), if

QT Q = QQT = I, (1.91)

where I is the identity tensor.

Theorem 1.5.1. A tensor Q is orthogonal if and only if it has any of the following properties of preserving inner products, lengths and distances:

(Qu, Qv) = (QT Qu, v) = (Iu, v) = (u, v) u, v V. ∀ ∈

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Conversely, if Eqn. (1.92a) holds, then

0 = (Qu, Qv) (u, v) = (u, QT Qv) (u, v) = (u, (QT Q I)v) u, v V, − − − ∀ ∈ which implies that QT Q = I (by Eqn. (1.24)), and hence Q is orthogonal. By choosing v = u in Eqn. (1.92a), we get Eqn. (1.92b). Conversely, if Eqn. (1.92b) holds, i.e., if (Qu, Qu) = (u, u) for all u V, then ∈ ((QT Q I)u, u) = 0 u V, − ∀ ∈ which, by virtue of Problem 29, leads us to the conclusion that Q Orth. By replacing u by (u v) in Eqn. (1.92b), we obtain Eqn. (1.92c),∈ and, conversely, by setting v to zero− in Eqn. (1.92c), we get Eqn. (1.92b).

As a corollary of the above results, it follows that the ‘angle’ between two vectors u 1 and v, deﬁned by θ := cos− (u v)/( u v ), is also preserved. Thus, physically speaking, multiplying the position vectors· of all| points| | | in a domain by Q corresponds to rigid body rotation of the domain about the origin. From Eqns. (1.49), (1.51) and (1.91), we have det Q = 1. Orthogonal tensors with determinant +1 are said to be proper orthogonal or rotations± (henceforth, this set is denoted by Orth+). Since det( R) = ( 1)3 det R = 1 for R Orth+, R is an orthogonal tensor that is not a rotation.− On the− other hand, if−Q is an∈ orthogonal− tensor that is not a rotation, Q is a rotation. Thus, when the underlying vector space dimension is odd (which includes− n = 3), every orthogonal tensor is either a rotation, or the product of a rotation with I: − Orth = R; R Orth+ . {± ∈ } 0 1 This result is not true when the dimension n is even, as the example Q = 1 0 , which cannot be written as R, where R Orth+, shows. For Q Orth+, we have − ∈ ∈ T cof Q = (det Q)Q− = Q, (1.93)

so that by Eqn. (1.54),

Q(u × v) = (Qu) × (Qv) u, v V. (1.94) ∀ ∈ A characterization of a rotation is as follows:

Theorem 1.5.2. Let e¯ ,e ¯ ,e ¯ and e , e , e be two orthonormal bases. Then { 1 2 3} { 1∗ 2∗ 3∗} Q = e¯1 ⊗ e1∗ + e¯2 ⊗ e2∗ + e¯3 ⊗ e3∗, is a proper orthogonal tensor. Conversely, every Q Orth+ can be represented in the above manner. ∈ Proof. If e¯ ,e ¯ ,e ¯ and e , e , e are two orthonormal bases, then { 1 2 3} { 1∗ 2∗ 3∗} T QQ = [e¯1 ⊗ e1∗ + e¯2 ⊗ e2∗ + e¯3 ⊗ e3∗][e1∗ ⊗ e¯1 + e2∗ ⊗ e¯2 + e3∗ ⊗ e¯3]

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= [e¯1 ⊗ e¯1 + e¯2 ⊗ e¯2 + e¯3 ⊗ e¯3] (by Eqn. (1.341)) = I, (by Eqn. (1.38))

and, by Eqn. (1.53),

det Q = [e¯1,e ¯2,e ¯3][e1∗, e2∗, e3∗] = 1. Conversely, if Q Orth+, and e , e , e is an orthonormal basis, then ∈ { 1∗ 2∗ 3∗} Q = QI

= Q [e1∗ ⊗ e1∗ + e2∗ ⊗ e2∗ + e3∗ ⊗ e3∗] (by Eqn. (1.38))

= (Qe1∗) ⊗ e1∗ + (Qe2∗) ⊗ e2∗ + (Qe3∗) ⊗ e3∗ (by Eqn. (1.343))

= e¯1 ⊗ e1∗ + e¯2 ⊗ e2∗ + e¯3 ⊗ e3∗, where e¯ ,e ¯ ,e ¯ = Qe , Qe , Qe is an orthonormal basis since e¯ = { 1 2 3} { 1∗ 2∗ 3∗} | i| Qei∗ = ei∗ = 1 for each i, and since, by virtue of Eqn. (1.94),

e¯ j × e¯k = (Qe∗j ) × (Qek∗) = Q(e∗j × ek∗) = eijkQei∗ = eijke¯i.

As a result of the above characterization, we have the following theorem.

Theorem 1.5.3. Every T Lin can be expressed as a linear combination of proper orthogonal tensors, i.e., ∈

Lin = span Orth+ . { } Proof. Let e , e , e denote the canonical Cartesian basis vectors. Then { 1 2 3} 1 e ⊗ e = [I + (e ⊗ e e ⊗ e e ⊗ e )] , 1 1 2 1 1 − 2 2 − 3 3 1 e ⊗ e = [(e ⊗ e + e ⊗ e + e ⊗ e ) + (e ⊗ e e ⊗ e e ⊗ e )] , 1 2 2 1 2 2 3 3 1 1 2 − 2 3 − 3 1 with similar expressions for the remaining combinations of dyadic products. Thus, the tensor T = T e ⊗ e can be expressed as ∑ φ Q , where Q ij i j i i i i ∈ Orth+.

If e and e¯ are two sets of orthonormal basis vectors, then they are related as { i} { i} T e¯i = Q ei, i = 1, 2, 3, (1.95)

where Q = ek ⊗ e¯k is a proper orthogonal tensor by virtue of Theorem 1.5.2. The components of Q with respect to the e basis are given by Q = e (e ⊗ e¯ )e = δ e¯ e = { i} ij i · k k j ik k · j e¯ e . Thus, ife ¯ and e are two unit vectors, we can always ﬁnd Q Orth+ (not necessarily i · j ∈

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unique), which rotatese ¯ to e, i.e., e = Qe¯. Let u and v be two vectors. Since u/ u and v/ v are unit vectors, there exists Q Orth+ such that | | | | ∈ v u = Q . v u | | | | Thus, if u and v have the same magnitude, i.e., if u = v , then there exists Q Orth+ such that u = Qv. | | | | ∈ We now study the transformation laws for the components of tensors under an orthogonal transformation of the basis vectors. Let ei ande ¯i represent the original and new orthonormal basis vectors, and let Q be the proper orthogonal tensor in Eqn. (1.95). From Eqn. (1.9), we have

e¯ = (e¯ e )e = Q e , (1.96a) i i · j j ij j e = (e e¯ )e¯ = Q e¯ . (1.96b) i i · j j ji j Using Eqn. (1.8) and Eqn. (1.96a), we get the transformation law for the components of a vector as

v¯ = v e¯ = v (Q e ) = Q v e = Q v . (1.97) i · i · ij j ij · j ij j In a similar fashion, using Eqn. (1.28), Eqn. (1.96a), and the fact that a tensor is a linear transformation, we get the transformation law for the components of a second-order tensor as

T¯ = e¯ Te¯ = Q Q T . (1.98) ij i · j im jn mn Conversely, if the components of a matrix transform according to Eqn. (1.98), then they all generate the same tensor. To see this, let T¯ = T¯ije¯i ⊗ e¯ j and T = Tmnem ⊗ en. Then

T¯ = T¯ije¯i ⊗ e¯ j

= QimQjnTmne¯i ⊗ e¯ j

= Tmn(Qime¯i) ⊗ (Qjne¯ j)

= Tmnem ⊗ en (by Eqn. (1.96b)) = T.

Due to this property, often, an alternate viewpoint is followed by many authors who take the transformation law given by Eqn. (1.98) as the deﬁnition of second-order tensors. We can write Eqns. (1.97) and (1.98) as

[v¯ ] = Q[v], (1.99) [T¯ ] = Q[T]QT. (1.100)

where [v¯ ] and [T¯ ] represent the components of the vector v and tensor T, respectively, with respect to thee ¯i coordinate system. Using the orthogonality property of Q, we can write the reverse transformations as

[v] = QT[v¯ ], (1.101)

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e¯1 e¯2

Fig. 1.1 Example of a coordinate system obtained from an existing one by a rotation about the 3-axis.

[T] = QT[T¯ ]Q. (1.102)

As an example, the Q matrix for the conﬁguration shown in Fig. 1.1 is     e¯1 cos θ sin θ 0 Q = e¯  =  sin θ cos θ 0 .  2 −  e¯3 0 0 1 The only real eigenvalues of Q Orth can be either +1 or 1, since if λ and6 n denote the eigenvalue and eigenvector of Q∈, i.e., Qn = λn, then −

(n, n) = (Qn, Qn) = λ2(n, n), 6 which implies that (n, n)(λ2 1) = 0. If λ and n are real, then (n, n) = 0 and λ = 1, while if λ is complex, then (n−, n) = 0. Let λˆ andn ˆ denote the complex conjugates6 of λ and± n, respectively. To see that the complex eigenvalues have a magnitude of unity observe that λλˆ (nˆ, n) = (Qnˆ, Qn) = (nˆ, n), which implies that λλˆ = 1 since (nˆ, n) = 0. Now, 6 using a similar argument as used in showing (n, n) = 0, we also have (n1, n2) = 0 for n1 and n2 corresponding to distinct complex eigenvalues of Q (complex conjugates not being considered6 as ‘distinct’–thus, this result is relevant only when the dimension n > 3), and (n, e) = (nˆ, e) = 0, where e is the eigenvector of Q corresponding to the eigenvalue of +1 or 1. −If R = I is a rotation, then the set of all vectors e such that 6 Re = e (1.103)

forms a one-dimensional subspace of V called the axis of R. To prove that such a vector always exists, we ﬁrst show that +1 is always an eigenvalue of R, and 1 is always an eigenvalue of an improper rotation. Since det R = 1, −

det(R I) = det(R RRT) = (det R) det(I RT) = det(I RT)T − − − − = det(I R) = det(R I), − − − which implies that det(R I) = 0, or that +1 is an eigenvalue. If e is the eigenvector corresponding to the eigenvalue− +1, then Re = e. Since every improper rotation is a product of I times a proper orthogonal tensor, 1 is an eigenvalue of an improper orthogonal tensor,− with the same eigenvector as the proper− orthogonal tensor from which it is generated.

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To show that the set of vectors e satisfying Eqn. (1.103) is a one-dimensional subspace of V, premultiply both sides of this equation with RT to get RTe = e. Combining this relation with Eqn. (1.103), we get (R RT)e = 0. There are two possibilities: (i) R is unsymmetric, or (ii) R is symmetric. If R is− unsymmetric, then R RT is a nonzero skew-symmetric tensor with e a scalar multiple of the axial vector of R −RT. Since we have shown that a scalar multiple of the axial vector of a skew-symmetric tensor− constitutes a one-dimensional subspace, αe, α is a one-dimensional subspace of V. Now consider the other possibility where R is symmetric.∈ < By Theorem 1.6.1 all the eigenvalues of R are real, and as shown above the only real eigenvalues can be 1. Taking into account that det R = 1, R = I and that one of the real eigenvalues is +±1, the set of eigenvalues in this case has to6 be +1, 1, 1 . Since the eigenvalue +1 is distinct from the other two, again invoking The- orem{ − 1.6.1,− it} follows that the corresponding eigenvector e is unique. Using Eqn.(1.123), it also follows that R Orth+ Sym I has to be of the form 2e ⊗ e I, where e is a unit vector. ∈ ∩ − { } − Conversely, given a vector w, there exists a proper orthogonal tensor R, and an improper one R0 such that Rw = w and R0w = w. To see this, consider the family of tensors − 1 1 R(w, α) = I + sin α W + (1 cos α)W2, (1.104) w w 2 − | | | | where W is the skew-symmetric tensor with w as its axial vector, i.e., Ww = 0. Using the Cayley–Hamilton theorem, we have W3 = w 2 W, from which it follows that W4 = − | | w 2 W2. Using this result, we get − | | " #" # sin α (1 cos α) sin α (1 cos α) RT R = I W + − W2 I + W + − W2 = I. − w w 2 w w 2 | | | | | | | | Since R has now been shown to be orthogonal, det R = 1. However, since det[R(w, 0)] = det I = 1, by continuity, we have det[R(w, α)] = 1 for any± α. Thus, R is a proper orthogonal tensor that satisﬁes Rw = w. Taking R0 = R, we get the required improper orthogonal tensor. It is easily seen that Rw = w. Essentially,− R rotates any vector in the plane perpendicular to w through an angle α. By Eqn. (1.89), W2 = w ⊗ w w 2 I. Thus, from Eqn. (1.104), we get R(w, 0) = I and R(w, π) = 2e ⊗ e I, where e =−w| /| w . Using this result, we have the following representation− theorem for proper| | orthogonal tensors.

Theorem 1.5.4. Let W be a skew-symmetric tensor with an axial vector of unit magnitude, and let e, q, r be an orthonormal set of vectors. Then tensors of the form { } R = I + sin α W + (1 cos α)W2, (1.105a) − R = e ⊗ e + cos α(I e ⊗ e) + sin α(r ⊗ q q ⊗ r). (1.105b) − − are proper orthogonal. Conversely, there exist W Skw, and orthonormal vectors e, q, r , such that every proper orthogonal tensor can∈ be expressed in the above forms. { }

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Proof. The tensor R in Eqn. (1.105a) is just a special case of the tensor R in Eqn. (1.104), which we have already shown to be proper orthogonal. Let e, q, r be an orthonormal set of vectors. Choose W = r ⊗ q q ⊗ r, so that W{ 2 = }e ⊗ e I. Substituting for W and W2 into Eqn. (1.105a), we− get the representation in− Eqn. (1.105b). Note that since Re = e, the vector e is the axis of R. Conversely, we now show that every proper orthogonal tensor can be represented as Eqns. (1.105a) and (1.105b). Let e be the axis of R, and let q, r be such that e, q, r form an orthonormal basis. Using the fact that Re = RTe = e, we have{(Re, q}) = (Re, r) = (e, Rq) = (e, Rr) = 0. Thus, Rq and Rr lie in the plane perpendicular to e. If (Rq, q) = (Rr, r) = cos α, then

R = RI = R(e ⊗ e + q ⊗ q + r ⊗ r) = (Re) ⊗ e + (Rq) ⊗ q + (Rr) ⊗ r = e ⊗ e + [(Rq, e)e + (Rq, q)q + (Rq, r)r] ⊗ q+ [(Rr, e)e + (Rr, q)q + (Rr, r)r] ⊗ r = e ⊗ e + cos α(q ⊗ q + r ⊗ r) + sin α(r ⊗ q q ⊗ r) − = e ⊗ e + cos α(I e ⊗ e) + sin α(r ⊗ q q ⊗ r), − − which is Eqn. (1.105b). By taking W = r ⊗ q q ⊗ r, we have W2 = e ⊗ e I, from which we get Eqn. (1.105a). − −

Using the results for the eigenvalues/eigenvectors of R, it is also possible to write the following “spectral resolutions” (using the convention outlined at the end of Section 1.3.1): I = e ⊗ e + n ⊗ nˆ + nˆ ⊗ n, R = e ⊗ e + λn ⊗ nˆ + λˆ nˆ ⊗ n, (1.106) R2 = e ⊗ e + λ2n ⊗ nˆ + λˆ 2nˆ ⊗ n,

where λ = λˆ = 1 in case all eigenvalues of R = I are real (so that R = 2e ⊗ e I), and n n = nˆ nˆ = e −n = e nˆ = 0 and n nˆ = 1 in case6 λ is complex. − · Note· from Eqn.· (1.105b)· that tr R· = 1 + 2 cos α. It is easily seen by multiplying the expression φ e ⊗ e + φ (q ⊗ q + r ⊗ r) + φ (r ⊗ q q ⊗ r) = 0 1 2 3 − successively by e, q and r that φ1 = φ2 = φ3 = 0, which implies that e ⊗ e, q ⊗ q + r ⊗ r, r ⊗ q q ⊗ r are linearly independent. Using the representation given{ by Eqn. (1.105b), we have− }       I 1 1 0 e ⊗ e        R  = 1 cos α sin α  q ⊗ q + r ⊗ r . R2 1 cos 2α sin 2α r ⊗ q q ⊗ r − The determinant of the square matrix in the above equation is 2 sin α(1 cos α). Thus, from Theorem 1.1.2 it follows that I, R, R2 are linearly independent for all− α except when (i) { }

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α = 0, in which case R = R2 = I and (ii) α = π, in which case R = 2e ⊗ e I Sym, and R2 = I (in this case I, R are linearly independent). − ∈ Using the representation{ } in Eqn. (1.104), let

1 1 2 R1(w1, α) = I + sin α W1 + 2 (1 cos α)W1, w1 w − | | | 1| 1 1 2 R2(w2, φ) = I + sin φ W2 + 2 (1 cos φ)W2, w2 w − | | | 2|

be two proper orthogonal tensors. If the axes of these two tensors coincide, i.e., if e1 = w / w = w / w = e , then we have W / w = W / w , and hence R and R 1 | 1| 2 | 2| 2 1 | 1| 2 | 2| 1 2 commute, i.e., R1R2 = R2R1. However, the converse is not true since, for example, R1 = diag[1, 1, 1] and R = diag[ 1, 1, 1] commute, but do not have the same axes. If e − − 2 − − 1 and e2 are the axes of R1 and R2, respectively, then we assume the (unnormalized) axis of R R to be of the form w = αˆ e + βê + γê × e (since e , e , e × e constitute 2 1 1 2 1 2 { 1 2 1 2} a basis when e1 and e2 are linearly independent) and find (αˆ , βˆ, γˆ) using the condition R R w = w. Using this procedure, we obtain the axis of R R as e = w/ w , where 2 1 2 1 | | w = (1 cos α)(1 + cos φ)e + sin α sin φ e (1 cos α) sin φ e × e . (1.107) − 1 2 − − 1 2 In the representation given by Eqn. (1.104), w is a fixed vector; however, in problems of rigid-body dynamics, we shall need to consider the case when the angular velocity vector is a function of time, and which, in general, will not coincide with the axis of Q. In such a situation, it is helpful to express Q Orth+ in terms of generalized coordinates. Problem 26 shows that 3 parameters are required∈ to characterize an orthogonal tensor in three-dimensional space, and presents one such characterization. However, the most com- monly used characterization is the one using the Euler angles θ, φ and ψ. The representation of Q in terms of these angles is       cos ψ sin ψ 0 1 0 0 cos φ sin φ 0 − − Q = sin ψ cos ψ 0 0 cos θ sin θ sin φ cos φ 0    −    0 0 1 0 sin θ cos θ 0 0 1   cos φ cos ψ cos θ sin ψ sin φ sin φ cos ψ cos θ sin ψ cos φ sin θ sin ψ − − − = cos φ sin ψ + cos θ cos ψ sin φ sin φ sin ψ + cos θ cos ψ cos φ sin θ cos ψ . (1.108)  − −  sin φ sin θ cos φ sin θ cos θ

The axis of Q is [0, 0, 1] when θ = 0, and   (1 cos θ)(sin φ + sin ψ) − e = (1 cos θ)(cos φ cos ψ) (θ = 0). (1.109)  − −  6 sin θ[1 cos(φ + ψ)] − We now present an expression for the eigenvalues of Q. Since from Eqn. (1.105b), tr Q = 1 + 2 cos α, and since we have shown that the magnitude of the eigenvalues is 1, the + iα iα eigenvalues of Q Orth are 1, e , e− . Since 1 cos α 1, we have 1 tr Q 3, with tr Q = 3 if and∈ only if Q ={ I (α = 0 in} Eqn. (1.105b)).− ≤ The≤ other bound tr−Q≤= 1 is≤ at- tained when α = π (or when θ or φ + ψ is equal to π in Eqn. (1.108)), and from Eqn. (1.105b),−

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we see that Q = 2e ⊗ e I Sym with eigenvalues 1, 1, 1 . When 1 < tr Q < 3, two eigenvalues and the− corresponding∈ eigenvectors are{ complex.− − } − + T Since for Q Orth , I2 = tr (cof Q) = tr [(det Q)Q− ] = tr Q, by the Cayley– Hamilton theorem,∈ we get

1 T 2 Q− = Q = Q (tr Q)Q + (tr Q)I. (1.110) − If Q , Q Orth+, which have the representations (see Theorem 1.5.2) 1 2 ∈

Q1 = a1 ⊗ b1 + a2 ⊗ b2 + a3 ⊗ b3, Q2 = c1 ⊗ d1 + c2 ⊗ d2 + c3 ⊗ d3,

where ai , bi , ci , di are orthonormal sets of vectors, then there exist proper orthog- { } { } { } { } T onal tensors R1 = ck ⊗ ak and R2 = dk ⊗ bk such that Q2 = R1Q1R2 . However, if the trace of Q1 and Q2 are the same (which implies that all their principal invariants are also the same), the following stronger result holds.

Theorem 1.5.5. Two proper orthogonal tensors Q1 and Q2 have the same trace, i.e., tr Q1 = tr Q2, if and only if there exists an orthogonal tensor Q0 (which means that either Q or Q is a rotation) such that Q = Q Q QT. 0 − 0 2 0 1 0 T Proof. If there exists Q0 Orth such that Q2 = Q0Q1Q0 , then using the same method of proof as used∈ to prove Eqn. (1.137), we see that all the principal invariants of Q1 and Q2 are the same. To prove the converse, note that if all the principal invariants of Q1 and Q2 are the same, then their characteristic equations, and hence their eigenvalues are the same. Thus, using Eqn. (1.106), we can write ˆ Q1 = e ⊗ e + λn ⊗ nˆ + λnˆ ⊗ n, ˆ Q2 = f ⊗ f + λg ⊗ gˆ + λgˆ ⊗ g. The tensor

Q0 = f ⊗ e + g ⊗ nˆ + gˆ ⊗ n T is orthogonal since Q0 Q0 = e ⊗ e + n ⊗ nˆ + nˆ ⊗ n = I, and using the relations T following Eqn. (1.106), we also see that Q2 = Q0Q1Q0 . In the above treatment, we have assumed that two eigenvalues are complex. If (1, 1, 1) are the eigenvalues of Q and Q , then they have the representations −Q −= 2e ⊗ e I and 1 2 1 − Q2 = 2 f ⊗ f I, so that any tensor that rotates e into f can be taken to be Q0. An alternative− proof can also be given using Eqn. (1.105b) (recall that tr R = 1 + 2 cos α) by writing

Q = e ⊗ e + cos α(I e ⊗ e) + sin α(r ⊗ q q ⊗ r), 1 − − Q = f ⊗ f + cos α(I f ⊗ f ) sin α(r˜ ⊗ q˜ q˜ ⊗ r˜), 2 − ± − where e, q, r and f,q ˜,r ˜ are orthonormal sets of axes, and then taking Q0 as f ⊗ e +{q˜ ⊗ q}+ r˜ ⊗{r. }

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1.6 Symmetric Tensors

In this section, we examine some properties of symmetric second-order tensors. We ﬁrst discuss the properties of the principal values (eigenvalues) and principal directions (eigenvectors) of a symmetric second-order tensor. 1.6.1 Principal values and principal directions We have the following result:

Theorem 1.6.1. Every symmetric tensor S has at least one principal frame, i.e., a right- handed triplet of orthogonal principal directions, and at most three distinct principal values. The principal values are always real. For the principal directions three possibilities exist: If all the three principal values are distinct, the principal axes are unique (modulo • sign reversal). If two eigenvalues are equal, then there is one unique principal direction, and the • remaining two principal directions can be chosen arbitrarily in the plane perpendicular to the ﬁrst one, and mutually perpendicular to each other. If all three eigenvalues are the same, then every right-handed frame is a principal • frame, and S is of the form S = λI. The components of the tensor in the principal frame are   λ1 0 0   S∗ =  0 λ2 0  . (1.111) 0 0 λ3

Proof. We seek λ and n such that (S λI)n = 0. (1.112) − But this is nothing but an eigenvalue problem. For a nontrivial solution, we need to satisfy the condition that det(S λI) = 0, − or, by Eqn. (1.79), λ3 I λ2 + I λ I = 0, (1.113) − 1 2 − 3 where I1, I2 and I3 are the principal invariants of S. We now show that the principal values given by the three roots of the cubic equation Eqn. (1.113) are real. Suppose that two roots, and hence the eigenvectors associated with them, are complex. Denoting the complex conjugates of λ and n by λˆ andn ˆ, we have Sn = λn, (1.114a)

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Snˆ = λˆ nˆ, (1.114b)

where Eqn. (1.114b) is obtained by taking the complex conjugate of Eqn. (1.114a) (S being a real matrix is not affected). Taking the dot product of both sides of Eqn. (1.114a) withn ˆ, and of both sides of Eqn. (1.114b) with n, we get

nˆ Sn = λn nˆ, (1.115) · · n Snˆ = λˆ nˆ n. (1.116) · · Using the deﬁnition of a transpose of a tensor, and subtracting the second relation from the ﬁrst, we get

STnˆ n n Snˆ = (λ λˆ )n nˆ. (1.117) · − · − · Since S is symmetric, ST = S, and we have

(λ λˆ )n nˆ = 0. − · Since n nˆ = 0, λ = λˆ , and hence the eigenvalues are real. · 6 The principal directions n1, n2 and n3, corresponding to distinct eigenvalues λ1, λ2 and λ3, are mutually orthogonal and unique (modulo sign reversal). We now prove this. Taking the dot product of

Sn1 = λ1n1, (1.118) Sn2 = λ2n2, (1.119)

with n2 and n1, respectively, and subtracting, we get

0 = (λ λ )n n , 1 − 2 1 · 2

where we have used the fact that S being symmetric, n2 Sn1 n1 Sn2 = 0. Thus, since we assumed that λ = λ , we get n n .· Similarly,− · we have 1 6 2 1 ⊥ 2 n2 n3 and n1 n3. If n1 satisfies Sn1 = λ1n1, then we see that n1 also satisfies⊥ the same⊥ equation. This is the only other choice possible that− satisfies Sn1 = λ1n1. To see this, let r1, r2 and r3 be another set of mutually perpendicular eigenvectors corresponding to the distinct eigenvalues λ1, λ2 and λ3. Then r1 has to be perpendicular to not only r2 and r3, but to n2 and n3 as well. Similar comments apply to r and r . This is only possible when r = n , r = n 2 3 1 ± 1 2 ± 2 and r3 = n3. Thus, the principal axes are unique modulo sign reversal. To prove± that the components of S in the principal frame are given by Eqn. (1.111), assume that n1, n2, n3 have been normalized to unit length, and then let e1∗ = n1, e2∗ = n2 and e3∗ = n3. Using Eqn. (1.28), and taking into account the orthonormality of e1∗ and e2∗, the components S11∗ and S12∗ are given by

S∗ = e∗ Se∗ = e∗ (λ e∗) = λ , 11 1 · 1 1 · 1 1 1 S∗ = e∗ Se∗ = e∗ (λ e∗) = 0. 12 1 · 2 1 · 2 2

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Similarly, on computing the other components, we see that the matrix representation of S with respect to e∗ is given by Eqn. (1.111). If there are two repeated roots, say, λ = λ , and the third root λ = λ , then 2 3 1 6 2 let e1∗ coincide with n1, so that Se1∗ = λ1e1∗. Choose e2∗ and e3∗ such that e1∗-e3∗ form a right-handed orthogonal coordinate system. The components of S with respect to this coordinate system are   λ1 0 0   S∗ =  0 S22∗ S23∗  . (1.120) 0 S23∗ S33∗

By Eqn. (1.80), we have

S22∗ + S33∗ = 2λ2, 2 2 λ S∗ + (S∗ S∗ (S∗ ) ) + λ S∗ = 2λ λ + λ , 1 22 22 33 − 23 1 33 1 2 2 (1.121) h 2i 2 λ S∗ S∗ (S∗ ) = λ λ . 1 22 33 − 23 1 2

Substituting for λ2 from the ﬁrst equation into the second, we get

2 2 (S∗ S∗ ) = 4(S∗ ) . 22 − 33 − 23

Since the components of S are real, the above equation implies that S23∗ = 0 and λ2 = S22∗ = S33∗ . This shows that Eqn. (1.120) reduces to Eqn. (1.111), and that Se2∗ = S12∗ e1∗ + S22∗ e2∗ + S32∗ e3∗ = λ2e2∗ and Se3∗ = λ2e3∗ (thus, e2∗ and e3∗ are eigenvectors corresponding to the eigenvalue λ2). However, in this case the choice of the principal frame e∗ is not unique, since any vector lying in the plane of e2∗ and e3∗, given by n∗ = c1e2∗ + c2e3∗ where c1 and c2 are arbitrary constants, is also an eigenvector. The choice of e1∗ is unique (modulo sign reversal), since it has to be perpendicular to e2∗ and e3∗. Though the choice of e2∗ and e3∗ is not unique, we can choose e2∗ and e3∗ arbitrarily in the plane perpendicular to e1∗, and such that e e . 2∗ ⊥ 3∗ Finally, if λ1 = λ2 = λ3 = λ, then the tensor S is of the form S = λI. To show this choose e1∗–e3∗, and follow a procedure analogous to that in the previous case. We now get S22∗ = S33∗ = λ and S23∗ = 0, so that [S∗] = λI. Using the transformation law for second-order tensors, we have

T T [S] = Q [S∗]Q = λQ Q = λI.

Thus, any arbitrary vector n is a solution of Sn = λn, and hence every right- handed frame is a principal frame.

As a result of Eqn. (1.111), we can write

S = λ1e1∗ ⊗ e1∗ + λ2e2∗ ⊗ e2∗ + λ3e3∗ ⊗ e3∗, (1.122)

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which is called as the spectral resolution of S. The spectral resolution of S is unique since If all the eigenvalues are distinct, then the eigenvectors are unique, and consequently • the representation given by Eqn. (1.122) is unique. If two eigenvalues are repeated, then, by virtue of Eqn. (1.38), Eqn. (1.122) reduces to •

S = λ1e1∗ ⊗ e1∗ + λ2e2∗ ⊗ e2∗ + λ2e3∗ ⊗ e3∗ = λ e∗ ⊗ e∗ + λ (I e∗ ⊗ e∗), (1.123) 1 1 1 2 − 1 1

from which the asserted uniqueness follows, since e1∗ is unique. If all the eigenvalues are the same then S = λI. • Although the eigenvalues of S Sym are real, its eigenvectors can be complex (note that in this case, real eigenvectors,∈ obtained by combining the complex eigenvectors, also exist). For example, if S = diag[1, 2, 3], then n = (1 + i)[1, 0, 0]/√2 andn ˆ = (1 i)[1, 0, 0]/ − √2 are eigenvectors corresponding to the eigenvalue 1. If S = diag[1, 1, 2], then n = [1, i, 0]/√2 andn ˆ = [1, i, 0]/√2 are eigenvectors corresponding to the repeated eigenvalue 1. Note that the eigenvectors− in such a case occur in complex conjugate pairs (which is proved simply by taking the complex conjugate of the relation Sn = λn, so that (n,n ˆ ) are the eigenvectors corresponding to the eigenvalue λ). Note, however, that although the eigenvectors can be complex, the eigenprojections Pi as given by Eqn. (J.4) are real.

Theorem 1.6.2. Let S be given by Eqn. (1.122), and let e denote the canonical basis { k} of V. Then Q = ek ⊗ ek∗ is a proper orthogonal tensor that diagonalizes S, i.e., the matrix representation of QSQT with respect to the canonical basis is diagonal. Conversely, if Q Orth+ diagonalizes a tensor S, i.e., QSQT = Λ where Λ is a diagonal matrix, then∈S is symmetric, the diagonal matrix Λ comprises the eigenvalues of S, and the rows of Q are the eigenvectors of S. Proof. By Theorem 1.5.2, Q = e ⊗ e Orth+, and since e = Qe , we have k k∗ ∈ i i∗ " # T T QSQ = Q ∑ λiei∗ ⊗ ei∗ Q i

= ∑ λi(Qei∗) ⊗ (Qei∗) i = ∑ λiei ⊗ ei, i

which is a diagonal matrix with λi along the diagonals. Conversely, if QSQT = Λ where Λ is a diagonal matrix, then S = QTΛQ is clearly a symmetric tensor. By Theorem 1.6.10, S and Λ have the same eigenvalues, and hence Λ has the eigenvalues of S along the diagonal. If follows that " # T T T T S = Q ΛQ = Q ∑ λiei ⊗ ei Q = ∑ λi(Q ei) ⊗ (Q ei), i i

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T T T which implies that S(Q ei) = λi(Q ei) (no sum on i); thus, Q ei are the eigenvectors of S. Since Q = IQ, we have

T Q = (ek ⊗ ek)Q = ek ⊗ (Q ek) = ek ⊗ ek∗, which shows that the rows of Q are the eigenvectors of S.

The Cayley–Hamilton theorem (Theorem 1.3.5) applied to S Sym yields ∈ S3 I S2 + I S I I = 0. (1.124) − 1 2 − 3 We have already proved this result for any arbitrary tensor. However, the following simpler proof can be given for symmetric tensors. Using Eqn. (1.341), the spectral resolutions of S, S2 and S3 are

S = λ1e1∗ ⊗ e1∗ + λ2e2∗ ⊗ e2∗ + λ3e3∗ ⊗ e3∗, 2 2 2 2 S = λ1e1∗ ⊗ e1∗ + λ2e2∗ ⊗ e2∗ + λ3e3∗ ⊗ e3∗, (1.125) 3 3 3 3 S = λ1e1∗ ⊗ e1∗ + λ2e2∗ ⊗ e2∗ + λ3e3∗ ⊗ e3∗. Substituting these expressions into the left-hand side of Eqn. (1.124), we get

3 2 3 2 LHS = (λ I λ + I λ I )(e∗ ⊗ e∗) + (λ I λ + I λ I )(e∗ ⊗ e∗) 1 − 1 1 2 1 − 3 1 1 2 − 1 2 2 2 − 3 2 2 3 2 + (λ I λ + I λ I )(e∗ ⊗ e∗) = 0, 3 − 1 3 2 3 − 3 3 3 since λ3 I λ2 + I λ I = 0 for i = 1, 2, 3. i − 1 i 2 i − 3 S I1 I/3 : S Sym constitutes the set of traceless symmetric tensors, while tensors of the{ form− (a ⊗ b∈+ b ⊗}a)/2, a, b V have determinant zero (a fact that can be veriﬁed by writing (a ⊗ b + b ⊗ a)/2 as (a∈⊗ b + b ⊗ a + 0 ⊗ 0)/2 and then using Eqn. (1.53)). However, not all symmetric tensors with determinant zero can be represented as (a ⊗ b + b ⊗ a)/2 since, from the Cauchy–Schwartz inequality it follows that the second invariant of (a ⊗ b + b ⊗ a)/2 computed using Eqn. (1.63) is always negative. The following explicit expressions can be given for the eigenvalues of S Sym when it is not of the form λI [298]: ∈ I λ = 1 + 2√p cos φ, 1 3 I λ = 1 √p(cos φ + √3 sin φ), 2 3 − I λ = 1 √p(cos φ √3 sin φ), 3 3 − − where 1 I I p = S 1 I : S 1 I , 6 − 3 − 3 1 I q = det S 1 I , 2 − 3

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p 3 2 1 1 p q φ = tan− − , 0 φ π. 3 q ≤ ≤

3 2 As can be seen from the above expressions, λ2 = λ3 when p = q . Alternative expressions for ﬁnding the eigenvalues are

I p λ = 1 2 Q cos α, 1 3 − I1 p 2π λ2 = 2 Q cos α + , (1.126) 3 − 3 I p 2π λ = 1 2 Q cos α , 3 3 − − 3

where I2 3I Q = 1 − 2 , 9 1 h i U = 9I I 2I3 27I , 54 1 2 − 1 − 3 ! 1 1 U α = cos− p . 3 Q3

2 If Q = 0 (i.e., if I1 = 3I2), then all the eigenvalues are repeated, λ1 = λ2 = λ3 = I1/3. If Q = 0 and Q3 = U2 (i.e., if I2 I2 4I3 4I3 I + 18I I I 27I2 = 0), then λ = I /3 2√Q 6 1 2 − 2 − 1 3 1 2 3 − 3 1 1 − and λ2 = λ3 = I1/3 + √Q if U > 0, and λ2 = I1/3 + 2√Q and λ1 = λ3 = I1/3 √Q if U < 0. If Q = 0 and Q3 = U2, then the eigenvalues are distinct. − We saw in6 Theorem 1.6.16 that the component form of a symmetric tensor is a diagonal matrix in a particular basis. The following theorem states the conditions under which the matrix representation has all the diagonal terms zero [18], [27], [117]:

Theorem 1.6.3. There exists an orthonormal basis n , i = 1, 2, 3, for which { i} n Sn = n Sn = n Sn = 0, (1.127) 1 · 1 2 · 2 3 · 3 if and only if tr S = 0. Proof. If Eqn. (1.127) holds then clearly tr S = n Sn + n Sn + n Sn = 0. 1 · 1 2 · 2 3 · 3 Thus, we just need to prove the converse. If tr S = λ1 + λ2 + λ3 = 0, then the spectral resolution of S can be written as

S = λ (e∗ ⊗ e∗ e∗ ⊗ e∗) + λ (e∗ ⊗ e∗ e∗ ⊗ e∗). 1 1 1 − 3 3 2 2 2 − 3 3

We have seen that ei∗ , i = 1, 2, 3 can always be chosen to be an orthonormal basis, and we assume{ that} such a choice has been made. Note that when tr S = 0, at most one eigenvalue can be zero, since if two or more are zero, the tensor S

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is the zero tensor. Thus, we can assume, without loss in generality, that both λ1 and λ2 are nonzero, and λ3 (λ1 + λ2) can be either zero or nonzero. If we choose ≡ − √ n1 = (e1∗ + e2∗ + e3∗)/ 3, (1.128) it is clear that n Sn = 0. We now try and ﬁnd n and n in the plane perpen- 1 · 1 2 3 dicular to n1 such that n2 Sn2 = n3 Sn3 = 0. Since n1 n2 = 0, n2 and n3 are of the form · · ·

γ1e1∗ + γ2e2∗ (γ1 + γ2)e3∗ n2 = q − , 2 2 2 γ1 + γ2 + (γ1 + γ2) (1.129) (γ1 + 2γ2)e1∗ + (2γ1 + γ2)e2∗ + (γ2 γ1)e3∗ n3 = n1 × n2 = − q − , √ 2 2 2 3 γ1 + γ2 + (γ1 + γ2)

where γ1, γ2 are constants to be determined. Both the conditions n2 Sn2 = 0 and n Sn = 0 yield the equation · 3 · 3 2 2 λ1γ2 + 2(λ1 + λ2)γ1γ2 + λ2γ1 = 0,

which on solving yields (recall that both λ1 and λ2 are nonzero)  q  (λ + λ ) λ2 + λ λ + λ2 − 1 2 ± 1 1 2 2 γ2 =   γ1. λ1

It sufﬁces to consider only the solution with the positive sign, since it can be shown that the solution obtained with the negative sign is simply a rearrange- ment of the one obtained with the positive one, i.e., if (n˜ 1, n˜ 2, n˜ 3) is the solution obtained with the positive sign, then (n˜ , n˜ , n˜ ) is the solution obtained with 1 − 3 2 the negative one. Substituting for γ2 into Eqns. (1.129), we get

g1e1∗ + g2e2∗ + g3e3∗ n2 = q , 2 2 2 g1 + g2 + g3 (1.130) h1e1∗ + h2e2∗ + h3e3∗ n3 = q , 2 2 2 h1 + h2 + h3

where

g1 = λ1, q g = λ2 + λ λ + λ2 (λ + λ ), 2 1 1 2 2 − 1 2 q g = λ λ2 + λ λ + λ2, 3 2 − 1 1 2 2

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q h = λ + 2λ 2 λ2 + λ λ + λ2, 1 1 2 − 1 1 2 2 q h = λ λ + λ2 + λ λ + λ2, 2 1 − 2 1 1 2 2 q h = 2λ λ + λ2 + λ λ + λ2. 3 − 1 − 2 1 1 2 2 Equations (1.128) and (1.130) are the desired solutions. The solution that has been presented above is only one choice in an infinite one-parameter family of solutions in terms of a parameter θ. To find this infinite one-parameter family, assume a unit vector in the principal basis ei∗ to be given by { }

n1 = (sin θ cos φ, sin θ sin φ, cos θ),

and also assume without loss in generality that λ1 and λ2 are chosen such that λ = λ . Using the fact that S in this coordinate system is given by 1 6 2 diag[λ1, λ2, (λ1 + λ2)], and eliminating φ by using the condition n1 Sn1 = 0 yields − ·

 q 2  (λ1+2λ2) cos θ λ2 λ λ − ± 1− 2  q ( + ) 2  n1(θ) =  λ2 2λ1 cos θ λ1  .  λ λ −  ± 2− 1 cos θ ± The second vector n also satisﬁes n Sn = 0, and hence has to be of the form 2 2 · 2 n1(α), where α is found in terms of θ by imposing the condition that n1 n2 = 0. Denoting t cos2 θ, we get · ≡ h √h cos2 α = 1 ± 2 , (1.131) D where

h = λ λ ( 2 + 7t 5t2) + (λ2 + λ2)( 1 + 3t 2t2), 1 1 2 − − 1 2 − − h = (λ4 + λ4)(1 2t)2t2 + (λ3λ + λ λ3)( 1 + 8t 22t2 + 20t3)t+ 2 1 2 − 1 2 1 2 − − λ2λ2( 2 + 15t 38t2 + 33t3)t, 1 2 − − D = 2(λ2 + λ2)( 1 + 2t) + 2λ λ ( 2 + 5t). 1 2 − 1 2 − Denoting the two roots for cos2 α corresponding to the plus and minus signs 2 2 in Eqn. (1.131) by cos α1 and cos α2, we get n2 = n1(α1) and n3 = n1(α2). The plus or minus signs for the components of n1 can be chosen arbitrarily; the corresponding signs for n are obtained by imposing the condition n n = 0, 2 1 · 2 and then n3 is found as n1 × n2.

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The result in Theorem 1.6.3 actually holds for an arbitrary traceless tensor T. This is seen by splitting T into its symmetric and antisymmetric parts, and since the diagonal elements of the antisymmetric part are zero in any basis, the basis found for the symmetric part works. As a simple example of the above theorem, let the matrix representation of S with respect to a basis be diag[1, 1, 0]. Then, by convention, we choose λ1 and λ2 to be the nonzero eigenvalues, i.e., λ −= 1, λ = 1. The corresponding eigenvectors are 1 2 − e1∗ = (1, 0, 0), e2∗ = (0, 1, 0) and e3∗ = (0, 0, 1). From Eqns. (1.128) and (1.130), we get n1 = (1, 1, 1)/√3, n2 = (1, 1, 2)/√6, and n3 = ( 1, 1, 0)/√2. A more complicated example is given in Section 3.7. − − The decomposition of a matrix into two symmetric components, i.e., T = S1S2, where S1, S2 Sym, and where one of the factors can be taken to be invertible [26], can be found ∈ 1 as follows. Assuming S2 to be invertible, we get S1 = TS2− . Since S1 Sym, we get 1 T 1 T ∈ S− T = TS− , or, alternatively, T S = S T. This results in at most n(n 1)/2 indepen- 2 2 2 2 − dent equations for the n(n + 1)/2 components of S2. For example, for dimension n = 2, we get     T12 T21 T11 T22 S11 − −     0 0 0  S22 = 0, (1.132) 0 0 0 S12

while for dimension n = 3, we get     T12 T21 0 T11 T22 T31 T32 S11 − − −     T 0 T T T T T  S   13 31 23 21 11 33  22 − − −     0 T T T T T T  S   23 32 13 22 33 12   33  − − −    = 0.  0 0 0 0 0 0  S     12      0 0 0 0 0 0  S23     0 0 0 0 0 0 S13

The decomposition is obviously not unique since, e.g., I can be factorized as (I)(I) or as ( I)( I). − − In general, it is not possible to express the factors S1 and S2 as a function of T. To see 0 w this, let T = w 0 , whose determinant is nonzero, so that both S1 and S2 are invertible. Using Eqn. (1.132),− we get the most general solution as " # " #" # 0 w 1 bw aw a b = − , a, b . w 0 a2 + b2 aw bw b a ∈ < − − − −

One can see that it is not possible to express either S1 or S2 as a function of T. 1.6.2 Positive deﬁnite tensors and the polar decomposition A second-order symmetric tensor S is positive deﬁnite if

(u, Su) 0 u V with (u, Su) = 0 if and only if u = 0. ≥ ∀ ∈

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We denote the set of symmetric, positive deﬁnite tensors by Psym. Since by virtue of Eqn. (1.33), all tensors T can be decomposed into a symmetric part Ts, and a skew-symmetric part Tss, we have

(u, Tu) = (u, Tsu) + (u, Tssu), = (u, Tsu),

because (u, Tssu) = 0 by Eqn. (1.86). Thus, the positive definiteness of a tensor is decided by the positive definiteness of its symmetric part. In Theorem 1.6.4, we show that a symmetric tensor is positive definite if and only if its eigenvalues are positive. Although the eigenvalues of the symmetric part of T should be positive in order for T to be positive definite, positiveness of the eigenvalues of T itself does not ensure its positive definiteness 1 10 as the following counterexample shows. If T = 0− 1 , then T is not positive definite since u Tu < 0 for u = (1, 1), but the eigenvalues of T are (1, 1). Conversely, if T is positive definite,· then by choosing u to be the real eigenvectors n of T, it follows that its real eigenvalues λ = (n Tn) are positive. · Theorem 1.6.4. Let S Sym. Then the following are equivalent: ∈ 1. S is positive definite. 2. The principal values of S are strictly positive. 3. The principal invariants of S are strictly positive. Proof. We first prove the equivalence of (1) and (2). Suppose S is positive definite. If λ and n denote the principal values and principal directions, respectively, of S, then Sn = λn, which implies that λ = (n, Sn) > 0 since n = 0. Conversely, suppose that the principal values of S are greater6 than 0. Assum- ing that e1∗, e2∗ and e3∗ denote the principal axes, the representation of S in the principal coordinate frame is (see Eqn. (1.122))

S = λ1e1∗ ⊗ e1∗ + λ2e2∗ ⊗ e2∗ + λ3e3∗ ⊗ e3∗. Then

Su = (λ1e1∗ ⊗ e1∗ + λ2e2∗ ⊗ e2∗ + λ3e3∗ ⊗ e3∗)u = λ (e∗ u)e∗ + λ (e∗ u)e∗ + λ (e∗ u)e∗ 1 1 · 1 2 2 · 2 3 3 · 3 = λ1u1∗e1∗ + λ2u2∗e2∗ + λ3u3∗e3∗, and

2 2 2 (u, Su) = u Su = λ (u∗) + λ (u∗) + λ (u∗) , (1.133) · 1 1 2 2 3 3 which is greater than or equal to zero since λi > 0. Suppose that (u, Su) = 0. Then by Eqn. (1.133), ui∗ = 0, which implies that u = 0. Thus, S is a positive deﬁnite tensor. To prove the equivalence of (2) and (3), note that, by Eqn. (1.80), if all the principal values are strictly positive, then the principal invariants are also

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strictly positive. Conversely, if all the principal invariants are positive, then I3 = λ1λ2λ3, is positive, so that all the λi are nonzero in addition to being real. Each λi has to satisfy the characteristic equation

λ3 I λ2 + I λ I = 0, i = 1, 2, 3. i − 1 i 2 i − 3

If λi is negative, then, since I1, I2, I3 are positive, the left-hand side of the above equation is negative, and hence the above equation cannot be satisﬁed. We have already mentioned that λi cannot be zero. Hence, each λi has to be positive.

Theorem 1.6.5. If S Psym, then there exists a unique H Psym, such that H2 := HH = S. The tensor∈H is called the positive definite square∈ root of S, and we write H = √S. Proof. Before we begin the proof, we note that a positive definite, symmetric tensor can have square roots that are not positive definite. For example, diag[1, 1, 1] is a non-positive definite square root of I. Here, we are interested only in− those square roots that are positive definite. The short proof of uniqueness given below is due to Stephenson [305]. Since 3 S = ∑ λiei∗ ⊗ ei∗, i=1

is positive definite, by Theorem 1.6.4, all λi > 0. Define 3 p H := ∑ λiei∗ ⊗ ei∗. (1.134) i=1 Since the e are orthonormal, it is easily seen that HH = S. Since the eigen- { i∗} values of H given by √λi are all positive, H is positive definite. Thus, we have shown that a positive definite square root tensor of S given by Eqn. (1.134) exists. We now prove uniqueness. With λ and n denoting the principal value and principal direction of S, we have 0 = (S λI)n − = (H2 λI)n − = (H + √λI)(H √λI)n. − Calling (H √λI)n = n¯, we have − (H + √λI)n¯ = 0. This implies thatn ¯ = 0. For, if not, √λ is a principal value of H, which contra- dicts the fact that H is positive definite.− Therefore, (H √λI)n = 0; −

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i.e., n is also a principal direction of H with associated principal values √λ. If H = √S, then it must have the form given by Eqn. (1.134) (since the spectral decomposition is unique), which establishes its uniqueness.

Now we prove the polar decomposition theorem, which plays a crucial role in the characterization of frame-indifferent constitutive relations.

Theorem 1.6.6 (Polar Decomposition Theorem). Let F be an invertible tensor. Then, it can be factored in a unique fashion as F = RU = VR, where R is an orthogonal tensor, and U, V are symmetric and positive definite tensors. One has p U = FT F p V = FFT. Proof. The tensor FT F is obviously symmetric. It is positive definite since (u, FT Fu) = (Fu, Fu) 0, ≥ with equality if and only if u = 0 (Fu = 0 implies that u = 0, since F is invertible). Let U = √FT F. U is unique, symmetric and positive definite by 1 Theorem 1.6.5. Define R = FU− , so that F = RU. The tensor R is orthogonal, because T 1 T 1 R R = (FU− ) (FU− ) T T 1 = U− F FU− 1 T 1 = U− (F F)U− (since U is symmetric) 1 1 = U− (UU)U− = I. 1 Since det U > 0, we have det U− > 0. Hence, det R and det F have the same sign. Usually, the polar decomposition theorem is applied to the deformation gradient F satisfying det F > 0. In such a case det R = 1, and R is a rotation. 1 1 1 T Next, let V = FUF− = FR− = RUR− = RUR . Thus, V is symmetric since U is symmetric. V is positive definite since (u, Vu) = (u, RURTu) = (RTu, URTu) 0, ≥ with equality if and only if u = 0 (again since RT is invertible). Note that FFT = VV = V 2, so that V = √FFT. Finally, to prove the uniqueness of the polar decomposition, we note that since 1 U is unique, R = FU− is unique, and hence so is V.

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Let (λi, ei∗) denote the eigenvalues/eigenvectors of U. Then, since V Rei∗ = RUei∗ = λi(Rei∗), the pairs (λi, f i∗), where f i Rei∗ are the eigenvalues/eigenvectors of V. Thus, F and R can be represented as ≡

3 3 3 F = RU = R ∑ λiei∗ ⊗ ei∗ = ∑ λi(Rei∗) ⊗ ei∗ = ∑ λi f i∗ ⊗ ei∗, i=1 i=1 i=1 (1.135) 3 3 R = RI = R ∑ ei∗ ⊗ ei∗ = ∑(Rei∗) ⊗ ei∗ = ∑ f i∗ ⊗ ei∗. i=1 i i=1 When F is invertible, the polar decomposition can be given in explicit form in terms of the eigenvalues λ1, λ2, λ3 of U, and powers of C or B. First, one determines the eigen- 2 2 {2 } T T T values λ1, λ2, λ3 of C = F F (which are also the eigenvalues of B = FF = RCR ), using Eqn.{ (1.126).} Then, depending on whether the eigenvalues are repeated or distinct, the relevant expressions for the component matrices in the polar decomposition are ([126], [153], [326]) All eigenvalues distinct (λ = λ = λ = λ ): 1 6 2 6 3 6 1 h a a a i R = F 1 I + 2 C + 3 C2 , a a a a a a U 1 = 1 I + 2 C + 3 C2, − a a a a a a V 1 = 1 I + 2 B + 3 B2, − a a a b b b U = 1 I + 2 C + 3 C2, b b b b b b V = 1 I + 2 B + 3 B2, b b b where

2 2 2 2 2 2 2 2 2 a1 = (λ1 + λ2 + λ3)(λ1λ2 + λ2λ3 + λ1λ3) + λ1λ2λ3(λ1 + λ2 + λ3 + λ1λ2 + λ2λ3 + λ1λ3), h i a = (λ + λ + λ )(λ2 + λ2 + λ2) + λ λ λ , 2 − 1 2 3 1 2 3 1 2 3 a3 = λ1 + λ2 + λ3, a = λ λ λ [(λ + λ + λ )(λ λ + λ λ + λ λ ) λ λ λ ] , 1 2 3 1 2 3 1 2 2 3 1 3 − 1 2 3 b1 = λ1λ2λ3(λ1 + λ2 + λ3), 2 2 2 b2 = λ1 + λ2 + λ3 + λ1λ2 + λ2λ3 + λ1λ3, b = 1, 3 − b = (λ + λ + λ )(λ λ + λ λ + λ λ ) λ λ λ . 1 2 3 1 2 2 3 1 3 − 1 2 3 Two distinct eigenvalues (λ = λ = λ ): 1 6 2 3 h a a i R = F 1 I + 2 C , a a a a U 1 = 1 I + 2 C, − a a

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a a V 1 = 1 I + 2 B, − a a b b U = 1 I + 2 C, b b b b V = 1 I + 2 B, b b where

2 2 a1 = λ1 + λ1λ2 + λ2, a = 1, 2 − a = λ1λ2(λ1 + λ2), b1 = λ1λ2, b2 = 1, b = λ1 + λ2.

All eigenvalues repeated (λ = λ = λ λ): 1 2 3 ≡ 1 R = F, U = V = λI. λ Rates of the stretch and rotation tensors are discussed in Section 1.9.4. 1.6.3 Isotropic functions We shall frequently make use of the following theorem:

Theorem 1.6.7. Let X, Y and Z be normed vector spaces, and let f : X Z and g : X Y be two mappings such that → → x, x0 X and g(x) = g(x0) = f (x) = f (x0). { ∈ } ⇒ Then, there exists a mapping h : g(X) Y Z such that ⊂ → f (x) = h(g(x)) x X. ∀ ∈

A scalar function φ : V is isotropic if φ(v) = φ(Qv) for all v V and Q Orth+. We have the following characterization:→ < ∈ ∈

Theorem 1.6.8. φ is isotropic if and only if there exist a function φˆ such that φ(v) = φˆ( v ) v V. | | ∀ ∈ Proof. If φˆ exists, then φ(Qv) = φˆ( Qv ) = φˆ( v ) = φ(v) v V. | | | | ∀ ∈ To prove the converse, we need to show that u = v implies φ(u) = φ(v), and then the result follows from Theorem 1.6.7.| | As| noted| after the proof of

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Theorem 1.5.2, if u = v , there exists a proper orthogonal tensor that rotates u into v, i.e., | | | |

v = Qu.

Now by the assumed isotropy, we have

φ(v) = φ(Qu) = φ(u).

A vector function g : V V is isotropic if Qg(v) = g(Qv) for all v V and Q Orth+. We have the following→ characterization: ∈ ∈

Theorem 1.6.9. g is isotropic if and only if there exists a function φ such that

g(v) = φ( v )v v V. | | ∀ ∈ Proof. If the function φ exists, then

g(Qv) = φ( Qv )Qv = φ( v )Qv = Qg(v). | | | | To prove the converse, let v be an arbitrary vector, and let Q = I be a proper orthogonal tensor with v/ v as its axis so that Qv = v. By the assumed6 isotropy, we have | |

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

Since the axis of Q is a one-dimensional subspace, g(v) is parallel to v, i.e., g(v) = α(v)v. Again by the assumed isotropy, α(v) = α(Qv), and the desired result follows by Theorem 1.6.8.

In what follows, we restrict the domain of the functions under consideration to a subset of Sym, and denote the list of principal invariants (I1, I2, I3) by . A scalar function φ : Psym is an isotropic function if it remains invariant under orthogonalI transformations, i.e.,→ <

φ(S) = φ(QSQT) S Psym, Q Orth+. (1.136) ∀ ∈ ∈ An isotropic tensor function G : Psym Sym is one that satisfies → QG(S)QT = G(QSQT) S Psym, Q Orth+. ∀ ∈ ∈ Although we have restricted S to be in Psym in the above definitions, all the following results will hold even if S lies in any subset of Sym that is invariant under Orth+, i.e., if A QAQT when A , Q Orth+. ∈ A ∈ A ∈ This condition is imposed in order that φ(QSQT) and G(QSQT) make sense. As is obvious from the above definition, Sym itself is invariant under Orth+. To show that Psym is invariant under Orth+, let S Psym. By Theorem 1.6.4, all the eigenvalues of S are strictly ∈

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positive. The following theorem shows that the eigenvalues of QSQT are the same as the eigenvalues of S. Then, again appealing to Theorem 1.6.4, we conclude that QSQT Psym. ∈ Theorem 1.6.10. The list of principal invariants is isotropic, i.e., I + = T S Psym, Q Orth . (1.137) IS IQSQ ∀ ∈ ∈ Proof. Note that

det(QSQT λI) = det(QSQT λQQT) − − = det[Q(S λI)QT] − = (det Q)(det QT) det(S λI) − = [det(QQT)] det(S λI) − = det(S λI). − Thus, the characteristic equation for QSQT and S is the same, leading to the same principal invariants (and hence, also to the same eigenvalues).

Theorem 1.6.11. The tensors A, B Sym have the same principal invariants if and ∈ only if there exists a proper orthogonal tensor Q such that A = QBQT.

Proof. If there exists Q Orth+ such that A = QBQT, then by Eqn. (1.137) (which also holds when ∈S Sym), A and B have the same principal invariants. Conversely, if A and B have∈ the same principal invariants, then by Eqn. (1.113) they have the same eigenvalues. By the spectral resolution theorem, we can write

A = ∑ λiei∗ ⊗ ei∗, i B = ∑ λie¯i ⊗ e¯i. i

Since e and e¯ are orthonormal sets of vectors, by Eqn. (1.95) there exists { i∗} { i} Q = e ⊗ e¯ Orth+ such that e = Qe¯ . Using Eqn. (1.344), we now have k∗ k ∈ i∗ i T T QBQ = ∑ λiQ(e¯i ⊗ e¯i)Q = ∑ λi(Qe¯i) ⊗ (Qe¯i) = ∑ λiei∗ ⊗ ei∗ = A. i i i

Theorem 1.6.12. A function φ : Psym is isotropic if and only if there exists a function φ˜ : (S) such that → < I → < φ(S) = φ˜( ) S Psym. (1.138) IS ∀ ∈

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Proof. Assuming Eqn. (1.138) holds, we conclude that φ is isotropic from Eqn. (1.137). To prove the converse, we need to show that

= = φ(A) = φ(B), IA IB ⇒ for A, B Psym, and then the desired result follows from Theorem 1.6.7. Since A and B∈have the same invariants, by Theorem 1.6.11 there exists Q Orth+ ∈ such that A = QBQT. Since φ is isotropic,

φ(A) = φ(QBQT) = φ(B).

The generalization of the above theorem to arbitrary tensors is given by the following [246, 294]:

Theorem 1.6.13. A function φ : Lin is isotropic if and only if there exists a function φ˜ such that → <

φ(T) = φ˜(tr T, tr T2, tr TTT), T Lin (n = 2), ∀ ∈ φ(T) = φ˜(tr T, tr T2, tr T3, tr TTT, tr T2TT, tr T2(TT)2, tr TTT T2(TT)2) T Lin (n = 3). ∀ ∈

Now we state and prove the Rivlin–Ericksen representation theorem for isotropic tensor functions (the generalization to an isotropic symmetric tensor-valued function of two symmetric tensors is stated in Eqn. (7.56)).

Theorem 1.6.14 (Rivlin–Ericksen Representation Theorem). A function G : Psym Sym is isotropic if and only if there exist scalar functions φ0, φ1, φ2 : (S) , such→ that I → < G(S) = φ ( )I + φ ( )S + φ ( )S2 S Psym. (1.139) 0 IS 1 IS 2 IS ∀ ∈ Proof. Assume Eqn. (1.139) holds. Choose S Psym and Q Orth+. By Eqn. (1.137), we have ∈ ∈

T T T T G(QSQ ) = φ ( T )I + φ ( T )QSQ + φ ( T )QSQ QSQ 0 IQSQ 1 IQSQ 2 IQSQ = φ ( )QQT + φ ( )QSQT + φ ( )QS2QT 0 IS 1 IS 2 IS = QG(S)QT,

which proves that G is isotropic. To prove the converse, assume that G is isotropic, i.e., G(QSQT) = QG(S)QT for all Q Orth+. We ﬁrst prove that every eigenvector of S is an eigenvector of G(S), or,∈ alternatively, any matrix that diagonalizes S also diagonalizes G(S). Let the spectral resolution of S be given by S = ∑i λiei∗ ⊗ ei∗. If we form a

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matrix Q with the eigenvectors of S along its rows, i.e., if Q = ek ⊗ ek∗, then by Theorem 1.6.2, Q diagonalizes S, and is proper orthogonal. Let

Q = e ⊗ e e ⊗ e e ⊗ e , 1 1 1 − 2 2 − 3 3 Q = e ⊗ e + e ⊗ e e ⊗ e . 2 − 1 1 2 2 − 3 3

Then, we see that Q1, Q2, Q1Q and Q2Q are all proper orthogonal tensors. In addition,

T T T QβQSQ Qβ = ∑ λiei ⊗ ei = QSQ , β = 1, 2 (no sum on β). i

By the assumed property of isotropy, we have

T T T T T T QβQG(S)Q Qβ = G(QβQSQ Qβ ) = G(QSQ ) = QG(S)Q , β = 1, 2. (1.140)

A straightforward computation shows that the above equations imply that QG(S)QT is diagonal, or, in other words, again appealing to Theorem 1.6.2, the eigenvectors of S and G(S) are the same. Hence

G(S) = µ1e1∗ ⊗ e1∗ + µ2e2∗ ⊗ e2∗ + µ3e3∗ ⊗ e3∗, (1.141)

where µi are the eigenvalues of G(S). Let us now establish that G(S) can be written as

G(S) = α (S)I + α (S)S + α (S)S2 S Psym, (1.142) 0 1 2 ∀ ∈

where α0, α1, α2 are real-valued functions of S. Assume ﬁrst that S has three dis- 2 tinct eigenvalues λi, with associated eigenvectors ei∗. Then the two sets I, S, S and e ⊗ e , e ⊗ e , e ⊗ e span the same subspace of the vector{ space of} { 1∗ 1∗ 2∗ 2∗ 3∗ 3∗} symmetric tensors. To see this, we ﬁrst observe that e1∗ ⊗ e1∗, e2∗ ⊗ e2∗, and e3∗ ⊗ e3∗ are symmetric tensors. They are also linearly independent, since if

αe1∗ ⊗ e1∗ + βe2∗ ⊗ e2∗ + γe3∗ ⊗ e3∗ = 0, then carrying out the operation e ()e , where () denotes the left- and right- 1∗ · 1∗ hand sides of the above equation, and using the orthogonality of e1∗, e2∗ and e3∗, we get α = 0; similarly, the operations e ()e and e ()e yield β = 0 and 2∗ · 2∗ 3∗ · 3∗ γ = 0. Thus, Lsp e1∗ ⊗ e1∗, e2∗ ⊗ e2∗, e3∗ ⊗ e3∗ is a three-dimensional subspace of the vector space of{ symmetric tensors, and,} hence, is a vector space itself. Since, by Eqns. (1.38) and (1.125),

I = e1∗ ⊗ e1∗ + e2∗ ⊗ e2∗ + e3∗ ⊗ e3∗, S = λ1e1∗ ⊗ e1∗ + λ2e2∗ ⊗ e2∗ + λ3e3∗ ⊗ e3∗, 2 2 2 2 S = λ1e1∗ ⊗ e1∗ + λ2e2∗ ⊗ e2∗ + λ3e3∗ ⊗ e3∗,

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and since the van der Monde determinant   1 1 1   det λ1 λ2 λ3 = (λ1 λ2)(λ2 λ3)(λ3 λ1) = 0, 2 2 2 − − − 6 λ1 λ2 λ3 by virtue of the eigenvalues being assumed to be distinct, Theorem 1.1.2 yields the result

2 Lsp e∗ ⊗ e∗, e∗ ⊗ e∗, e∗ ⊗ e∗ = Lsp I, S, S . { 1 1 2 2 3 3} { } Hence, Eqn. (1.141) can be written in the form given by Eqn. (1.142). Assume, next, that the tensor S has a double eigenvalue, say λ = λ = 2 3 6 λ1. Then, the two sets I, S and e1∗ ⊗ e1∗, e2∗ ⊗ e2∗ + e3∗ ⊗ e3∗ span the same subspace, since, in this case,{ we} can{ write }

I = e1∗ ⊗ e1∗ + (e2∗ ⊗ e2∗ + e3∗ ⊗ e3∗), S = λ1e1∗ ⊗ e1∗ + λ2(e2∗ ⊗ e2∗ + e3∗ ⊗ e3∗), and " # 1 1 det = λ2 λ1 = 0. λ1 λ2 − 6

All the vectors in the subspace spanned by e2∗ and e3∗ are eigenvectors of S, and hence those of G(S) as already proved. Thus,

G(S)e2∗ = µ2e2∗, G(S)e3∗ = µ3e3∗, G(S)(e2∗ + e3∗) = µ(e2∗ + e3∗),

which implies that µ = µ2 = µ3. Therefore, we can write

G(S) = µ1e1∗ ⊗ e1∗ + µ(e2∗ ⊗ e2∗ + e3∗ ⊗ e3∗), as

G(S) = α0(S)I + α1(S)S.

Assume, ﬁnally, that S has three repeated eigenvalues. Now, the eigenvalues of G(S) are also all repeated (proved in the same way as in the previous case), and all unit vectors are eigenvectors of S, and hence also of G(S). Thus, by using the spectral resolution of G(S), we conclude that it is a multiple I, i.e.,

G(S) = α0(S)I,

proving the assertion in all cases. Note from the above discussion that α2(S) = 0, when there are two repeated eigenvalues, and α1(S) = α2(S) = 0, when all eigenvalues are identical.

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It remains to show that the functions α , α and α are only functions of . 0 1 2 IS By virtue of Theorem 1.6.12, it sufﬁces to prove that α0, α1 and α2 are isotropic functions. This follows from the isotropy of G(S):

T h T T T 2i T G(QSQ ) = Q α0(QSQ )I + α1(QSQ )S + α2(QSQ )S Q

T h 2i = QG(S)Q = Q α0(S)I + α1(S)S + α2(S)S QT S Psym, Q Orth+, ∀ ∈ ∈ which implies that

α (QSQT) = α (S) S Psym, Q Orth+; for i = 0, 1, 2, i i ∀ ∈ ∈ again by the uniqueness of the expansion of G(S) in the spaces spanned by I, S, S2 , I, S or I according to which case is being considered. Thus, by Theorem{ } 1.6.12,{ } there{ exist} functions φ( ) such that IS α (S) = φ ( ) S Psym; i = 0, 1, 2. i i IS ∀ ∈

Note that the above theorem does not say that an isotropic function is a quadratic function of the elements of S, since the functions φi( S), i = 0, 1, 2, can be arbitrary. For linear functions, we have the following simplerI result:

Theorem 1.6.15. (Representation Theorem for Linear Isotropic Tensor Func- tions). A linear function G(S) : Sym Sym is isotropic if and only if there exist scalars µ and λ such that → G(S) = λ(tr S)I + 2µS S Sym. ∀ ∈ Proof. Although one could try and specialize the result from the previous theorem to the present case, it is easier to proceed directly; the following short proof is due to Gurtin [108]. Let V1 be the set of all unit vectors. For e V1, the tensor + ∈ e ⊗ e has eigenvalues (0, 0, 1) (since there exists Q Orth such that e = Qe1, T ∈ which implies that e ⊗ e = Q(e1 ⊗ e1)Q , so that by Eqn. (1.137), the principal invariants, and hence the eigenvalues, of e ⊗ e and e1 ⊗ e1 are the same). Since two eigenvalues of e ⊗ e are the same, analogous to the case of repeated eigenvalues in the previous theorem’s proof, G(e ⊗ e) also has a repeated eigenvalue, and thus by Eqn. (1.123) we have G(e ⊗ e) = λ(e)I + 2µ(e)e ⊗ e e V . (1.143) ∀ ∈ 1 Choose e, f V1, and let Q be a proper orthogonal tensor such that Qe = f. Since ∈ Q(e ⊗ e)QT = f ⊗ f, and since G is isotropic, 0 = QG(e ⊗ e)QT G( f ⊗ f ) = [λ(e) λ( f )]I + 2[µ(e) µ( f )] f ⊗ f. − − −

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By multiplying both sides of αI + β f ⊗ f = 0 by a vector g perpendicular to f, it follows that α = β = 0, implying in turn that I and f ⊗ f are linearly independent; thus,

λ(e) = λ( f ), µ(e) = µ( f ).

Therefore, λ and µ are scalar constants, and we conclude from Eqn. (1.143) that

G(e ⊗ e) = λI + 2µe ⊗ e. (1.144)

The spectral resolution of S Sym is ∈

S = ∑ αiei∗ ⊗ ei∗, i

with ei∗ orthonormal; therefore, in view of Eqn. (1.144), and the linearity of G,

G(S) = ∑ αiG(ei∗ ⊗ ei∗) = λ(α1 + α2 + α3)I + 2µS. i

Since tr S = α1 + α2 + α3, the above equation implies that

G(S) = λ(tr S)I + 2µS.

The converse assertion that the above representation is an isotropic function is left as an exercise (see Problem 37).

1.7 Higher-Order Tensors

A detailed discussion about the decomposition of higher-order tensors into irreducible tensors, and their eigenvalues and eigentensors can be found in [8], [150], [151], [260], [304] and [360]. Similar to the deﬁnition of a second-order tensor, a third-order tensor, which we represent by B, is deﬁned as a linear transformation that transforms a vector v to a second-order tensor T, i.e.,

Bv = T. (1.145)

In addition, the map is linear, i.e.,

B(av + bv ) = aBv + bBv a, b and v , v V. 1 2 1 2 ∀ ∈ < 1 2 ∈ In order to ﬁnd the representation for a third-order tensor in terms of its components, we deﬁne the dyadic or tensor product of three vectors a, b and c as

(a ⊗ b ⊗ c)d := (a ⊗ b)(c d) d V. (1.146) · ∀ ∈ Since a ⊗ b is a second-order tensor, and since the above transformation is linear, the tensor product of three vectors is a linear transformation that transforms vectors into

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second-order tensors, and hence is a third-order tensor. Just as a second-order tensor can be expressed as the sum of 9 dyads (Eqn. (1.37)), a third-order tensor can be expressed as the sum of 27 dyads as follows:

B = Bijkei ⊗ ej ⊗ ek, (1.147)

where

B = e (Be )e . (1.148) ijk i · k j In order to prove this, we consider the action of B on an arbitrary vector u:

Bu = (Bu)ij(ei ⊗ ej) = e (Bu)e (e ⊗ e ) i · j i j = e [B(u e )] e (e ⊗ e ) i · k k j i j = e [B((e u)e )]e (e ⊗ e ) i · k · k j i j = e (Be )e (e u)(e ⊗ e ) i · k j k · i j = e (Be )e (e ⊗ e ⊗ e )u . i · k j i j k Since u is arbitrary, we have proved that the tensor B can be represented by Eqn. (1.147), with the components Bijk of this tensor given by Eqn. (1.148). Note that

Bem = Bijmei ⊗ ej.

In [360], using the identity epjrepbr = 2δjb, the authors write

1 B = e e B , ijk 2 pjr pbr ibk then, assuming all eigenvalues to be distinct, use the spectral resolution of a fourth-order 9 tensor to write Bibkepbr in the form ∑m=1 λm(Pm)ipkr, from which it follows that

9 1 Bijk = ∑ λm epjr(Pm)ipkr . (1.149) m=1 2

Based on this representation, the authors claim that the λi’s (not all independent, since 9 Bibiepbp = ∑m=1 λm = 0) are eigenvalues of B. However, since the eigenvalue problem for a third-order tensor is never deﬁned in this work, it is not clear how the eigenvalues of the fourth-order tensor are also the eigenvalues of the third-order tensor. In general, an nth order tensor is deﬁned as a linear transformation that transforms a vector to an n 1th order tensor. It can be proved by induction that an nth order tensor can be represented− by the sum of 3n dyads as

T = T e ⊗ e ⊗ e ⊗ e , i1i2i3...in i1 i2 i3 ··· in where Ti1i2i3...in = ei1 (((Tein )ein 1 ) )ei2 . · − ···

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As in the case of second-order tensors, we deﬁne the sum of two nth order tensors S and T as

(S + T)u := Su + Tu u V, ∀ ∈ and the scalar multiple of T by α as ∈ < (αT)u := α(Tu) u V. ∀ ∈ It is easily seen from the deﬁnition of a tensor as a linear transformation that S + T and αT are nth order tensors. The transformation law for third order tensors can be deduced from Eqn. (1.148) as follows:

B¯ = Q e [B(Q e )]Q e ijk il l · kn n jm m = Q Q Q e (Be )e il jm kn l · n m = Qil QjmQknBlmn.

The alternate tensor E whose components are given by Eqn. (1.11) is an example of a third- order tensor. This can be proved by noting that

e¯ = e¯ (e¯ × e¯ ) ijk i · j k = Q Q Q e (e × e ) il jm kn l · m n = Qil QjmQknelmn,

which is nothing but the transformation law for the components of a third-order tensor. In fact, it is an isotropic third-order tensor, as we show in Section 1.8. Note that [[Ew]v]u = [u, v, w]. Similarly, the transformation law for a fourth-order tensor is

C¯ ijkl = QipQjqQkrQlsCpqrs. (1.150)

Generalizing the above arguments, the transformation law for an nth order tensor is given by

¯ Ti1i2i3...in = Qi1 j1 Qi2 j2 Qi3 j3 ... Qin jn Tj1 j2 j3...jn . (1.151)

We have seen how to construct fourth-order tensors as a dyadic product of vectors. However, another alternative way that is very widely used is to construct it as a dyadic product of two second-order tensors. This procedure is analogous to constructing second- order tensors using the dyadic product of vectors. We now discuss this methodology, which should be compared with the one for second-order tensors presented in Section 1.3. A fourth-order tensor is deﬁned as a linear transformation that maps second-order tensors to second-order tensors. Thus, if C is a fourth-order tensor that maps a second-order tensor A to a second-order tensor B, we write it as

B = CA. (1.152)

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C satisﬁes the property C(αA1 + βA2) = αCA1 + βCA2, for all α, β and A1, A2 Lin. The sum, scalar multiple and composition of fourth-order tensors is∈ deﬁned < as ∈

(C + D)A = CA + DA A Lin, ∀ ∈ (λC)A = λ(CA) A Lin, ∀ ∈ (CD)A = C(DA) A Lin. ∀ ∈ For presenting the component form of fourth-order tensors, it is convenient to introduce the notation E e ⊗ e . We have ij ≡ i j

T : Eij = tr (T(ej ⊗ ei)) = tr ((Tej) ⊗ ei) = Tkjtr (ek ⊗ ei) = Tkjδki = Tij. (1.153)

By letting T E , we get ≡ kl

Eij : Ekl = δikδjl. (1.154)

Analogous to the deﬁnition of the components of a second-order tensor (see Eqn. (1.27)), we now deﬁne the components via Eqn. (1.152) as

CEkl = Cijkl Eij. (1.155)

Taking the dot product on either side with Emn and using Eqn. (1.154), we get Cmnkl = Emn : CEkl, which, on replacing the indices (m, n) with (i, j) can be written as

Cijkl = Eij : CEkl. (1.156)

Two tensors C and D are said to be equal if

CB = DB B Lin, (1.157) ∀ ∈ or, alternatively, if

(A, CB) = (A, DB) A, B Lin. ∀ ∈

By choosing A Eij and B Ekl in the above equation, we see that equal tensors have equal components.≡ ≡ Using Eqn. (1.155) and the linearity of fourth-order tensors, Eqn. (1.152) can be written as

BijEij = C(Akl Ekl) = AklCEkl = Cijkl Akl Eij,

which by virtue of the linear independence of the E implies that { ij}

Bij = Cijkl Akl.

Similarly, we have

(CD)ijkl = (C)ijmn(D)mnkl.

We now give some examples of fourth-order tensors. The fourth-order identity tensor I is deﬁned as one that maps a second-order tensor into itself. The transposer T, symmetrizer

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S and skew-symmetrizer W are deﬁned as ones that map a second-order tensor into its transpose, symmetric and skew-symmetric part, respectively, i.e., IA = A A Lin, ∀ ∈ TA = AT A Lin, ∀ ∈ 1 SA = (A + AT) A Lin, 2 ∀ ∈ 1 WA = (A AT) A Lin. 2 − ∀ ∈ Note that S = (I + T)/2, I = S + W, (1.158) W = (I T)/2, T = S W. − − We have 1 1 1 ST = (I + T)T = (T + TT) = (T + I) = S = TS, (1.159) 2 2 2 1 1 1 WT = (I T)T = (T TT) = (T I) = W = TW, 2 − 2 − 2 − − 1 1 1 SS = (I + T)S = (S + TS) = (S + S) = S, 2 2 2 1 1 1 WW = (I T)W = (W TW) = (W + W) = W, 2 − 2 − 2 1 1 1 SW = (I + T)W = (W + TW) = (W W) = 0 = WS. (1.160) 2 2 2 − The components of I, T, S and W are obtained using Eqn. (1.156), and are given by

Iijkl = δikδjl,

Tijkl = δilδjk, 1 (1.161) S = (δ δ + δ δ ), ijkl 2 ik jl il jk 1 W = (δ δ δ δ ). ijkl 2 ik jl − il jk Using the above component form of S, we have 1 h i [SAS] = A + A + A + A , (1.162) ijkl 4 ijkl ijlk jikl jilk where A is any fourth-order tensor. The transpose of C, denoted by CT is deﬁned using the inner product as

(A, CT B) := (B, CA) A, B Lin. (1.163) ∀ ∈ T The components of C are obtained by taking A Eij and B Ekl in the above equation, and are given by ≡ ≡

T (C )ijkl = Cklij. (1.164)

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Similar to second-order tensors, one has

(CT)T = C, (1.165a) (CD)T = DTCT. (1.165b)

Analogous to the dyadic product of two vectors, the dyadic product of two tensors A and B can be deﬁned as

(A ⊗ B)C := (B : C)A C Lin. (1.166) ∀ ∈ The component form of this dyadic product is

(A ⊗ B)ijkl = Eij : (A ⊗ B)Ekl = (A : Eij)(B : Ekl) = AijBkl.

The transpose of A ⊗ B is B ⊗ A since

(C, (A ⊗ B)T D) = (D, (A ⊗ B)C) = (B : C)(A : D) = (C, (B ⊗ A)D).

By taking C u ⊗ v in Eqn. (1.166), we get ≡ (A ⊗ B)(u ⊗ v) = (u Bv)A. · A tensor C can be written as C E ⊗ E since, for an arbitrary A Lin, we have ijkl ij kl ∈

CA = (CA)ijEij

= [(CA) : Eij]Eij

= Akl[Eij : (CEkl)]Eij

= Cijkl(A : Ekl)Eij

= Cijkl[Eij ⊗ Ekl]A.

Also analogous to the results for dyadic products of vectors, we have

C(A ⊗ B) = (CA) ⊗ B, (A ⊗ B)C = A ⊗ (CT B).

Another tensor product that is of great use is the “square tensor product” deﬁned as

(A B)C := ACBT C Lin. ∀ ∈ The components of A B are given by

T (A B)ijkl = Eij : (A B)Ekl = Eij : (AEkl B ) = (ei ⊗ ej) : [(Aek) ⊗ (Bel)] = (e Ae )(e Be ) = A B . i · k j · l ik jl T Since (I I)A = IAI = A, for all A Lin, it is clear from Eqn. (1.157) that the fourth- order identity tensor can be written as ∈

I = I I, (1.167)

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with components (I)ijkl = δikδjl. T T The transpose of A B is given by A B since

T T T T T (C, (A B) D) = (D, (A B)C) = (D, ACB ) = (C, A DB) = (C, (A B )D). From the deﬁnition of the square tensor product, it follows that

(H I)A = HA, T (I H )A = AH. As an example of the use of these equations, note that the tensor equation HA + AH = 0 T can be written as (H I + I H )A = 0. The composition rule is given by

(A B)(C D) = (AC) (BD), (1.168) since

(A B)(C D)E = (A B)(CEDT) = ACEDT BT = [(AC) (BD)]E E Lin. ∀ ∈ Some additional properties are

T (A ⊗ B)(C D) = A ⊗ (C BD), (1.169) T (A B)(C ⊗ D) = (ACB ) ⊗ D, (1.170) (A B)(u ⊗ v) = (Au) ⊗ (Bv), (1.171) (a ⊗ b) (u ⊗ v) = (a ⊗ u) ⊗ (b ⊗ v). (1.172) As per our convention outlined at the end of Section 1.3.1, the above relations are valid even if the involved second-order tensors and vectors are complex-valued. The ﬁrst relation is proved as follows:

T T T (A ⊗ B)(C D)E = (A ⊗ B)(CED ) = [B : (CED )]A = [(C BD) : E]A = [A ⊗ (CT BD)]E E Lin. ∀ ∈ The remaining relations are proved in a similar manner. The transposer T commutes with the tensor product A B in the following way:

T(A B) = (B A)T, (1.173) since

T(A B)E = T(AEBT) = BET AT = (B A)ET = (B A)TE E Lin. ∀ ∈ In an identical fashion, one can show that

S(A B)S = S(B A)S, (1.174) S(A B + B A) = (A B + B A)S = S(A B + B A)S = 2S(A B)S, (1.175) S(A A) = (A A)S = S(A A)S, (1.176)

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where the second relation in Eqn. (1.175) follows from the ﬁrst by multiplying both sides by S, and using the fact that SS = S, while the third relation in Eqn. (1.175) follows from Eqn. (1.174). Equation (1.176) follows from Eqn. (1.175) by taking B = A. The component form of (A B)T is Ail Bjk while that of T(A B) is AjkBil. A summary of the component forms of the various dyadic products is

[A ⊗ B]ijkl = AijBkl,

[A B]ijkl = AikBjl, (1.177) [(A B)T]ijkl = Ail Bjk, [T(A B)]ijkl = AjkBil.

The tensor C is said to have the ﬁrst minor symmetry if C = TC, or, in indicial notation, if

Cijkl = TijmnCmnkl = δinδjmCmnkl = Cjikl,

and is said to have the second minor symmetry if C = CT, which in indicial notation reads

Cijkl = Cijlk.

Note that the ﬁrst minor symmetry condition can be written as (I T)C = 2WC = 0, while the second minor symmetry condition can be written as CW− = 0. Equivalently, using S = (I + T)/2, one can also write the two symmetry conditions as C = SC and C = CS, respectively. One has

C = SC and C = CS C = SCS, ⇐⇒ since given the conditions on the left, we directly get C = SCS, while, conversely, if C = SCS, then SC = SSCS = SCS = C, and CS = SCSS = SCS = C. Thus a tensor C has both the minor symmetries if and only if C = SCS. The tensor C is said to have major symmetry if C = CT, or, in indicial notation (using Eqn. (1.164)), if Cijkl = Cklij. The symmetric and skew-symmetric parts of C are given by (C + CT)/2 and (C CT)/2. Using Eqn. (1.161) or directly, it can be shown that I, T, S and W are all symmetric.− Assume that C has the ﬁrst minor symmetry, i.e., C = TC. By transposing this equation, and using Eqn. (1.165b) and the symmetry of T, we get

CT = CTT.

Thus, if C has the first minor symmetry, then CT has the second minor symmetry. It also follows that if C has major symmetry and one of the minor symmetries, it also has the other minor symmetry. The eigenvalue problem for a fourth-order tensor A that does not possess minor symmetries, involves finding the eigenvalue and eigentensor pair (λ, A), A Lin, which satisfies ∈

AA = λA. (1.178)

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By applying the Ψ-mapping as per Eqn. (I.1e) to the above equation, we get

[Ψ(A) λI]Ψ(A) = 0. − This equation has a nontrivial solution Ψ(A) = 0, if and only if the characteristic equation det[Ψ(A) λI] = 0 is satisﬁed. The characteristic6 equation can also be written in expanded form− as

λ9 I (A)λ8 + I (A)λ7 + I (A) = 0, − 1 2 · · · − 9 where (see Eqn. (J.17))

I (A) = tr Ψ(A) = λ + λ + + λ , 1 1 2 ··· 9 1 I (A) = [(tr Ψ(A))2 tr (Ψ(A))2] = λ λ + λ λ + + λ λ , 2 2 − 1 2 2 3 ··· 1 9 ...

I9(A) = det Ψ(A) = λ1λ2 ... λ9,

are a strict subset of the complete set of the principal invariants of A (see the references below for the case where A has minor symmetries, denoted by C there). Similar to the result for second-order tensors,

det AT = det Ψ(AT) = det[Ψ(A)T] = det Ψ(A) = det A. (1.179)

If A = A1A2, then, again, similar to the result for second-order tensors,

det A = det Ψ(A) = det[Ψ(A1)Ψ(A2)] = [det Ψ(A1)][det Ψ(A2)] = (det A1)(det A2). (1.180)

By virtue of Eqn. (1.171), we have

Theorem 1.7.1. If (λi, ui) and (γi, vi), i = 1, 2, . . . , n, where n is the underlying space dimension, are the (possibly complex-valued) eigenvalues/eigenvectors of diagonalizable tensors A, B, then (λiγj, ui ⊗ vj), i, j = 1, 2, . . . , n, are the eigenvalues/eigentensors of A B.

The theorem may not hold for non-diagonalizable tensors. For example, if A is non- T diagonalizable but invertible, and B = A− , then (1, I) is an eigenvalue/eigentensor of T A A− , but ui ⊗ vj (or their linear combinations) corresponding to λiγj = 1 may not equal I. An application of the above theorem is presented in Theorem 1.7.2. An orthogonal fourth-order tensor Q is deﬁned as one for which

QTQ = QQT = I. (1.181)

A fourth-order tensor is orthogonal if and only if it preserves inner products, i.e., Q is orthogonal if and only if

(QA, QB) = (A, B) A, B Lin, ∀ ∈

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the proof being identical to that for second-order orthogonal tensors. An example of an orthogonal tensor is Q Q , where Q , Q Orth. This follows since 1 2 1 2 ∈ T T T T T (Q1 Q2) (Q1 Q2) = (Q1 Q2 )(Q1 Q2) = (Q1 Q1) (Q2 Q2) = I I = I.

A special class of fourth-order orthogonal tensors are fourth-order rotations obtained by choosing Q = Q = Q, where Q Orth+. Equation (1.150) can be expressed as 1 2 ∈ [A¯ ] = [Q][A][Q]T, (1.182)

where Q := Q Q, Q Orth+, is a fourth-order rotation. ∈ Since Ψ(I) = I9 9, we have det I = 1. It follows from Eqns. (1.179)–(1.181) that × det Q = 1. ± From Eqns. (1.180)–(1.182), we get

det(A¯ λI) = det[Q(A λI)QT] = det(QQT) det(A λI) = (det I) det(A λI) − − − − = det(A λI), (1.183) − showing that the invariants that occur in the characteristic equation (and hence the eigenvalues) of A are independent (as they should be) of the choice of basis. The Cayley–Hamilton theorem reads

0 = A9 I A8 + I A7 + I I − 1 2 · · · − 9 9 = ∏(A λiI). i=1 −

The minimal polynomial of A is given by

k mi q(A) = ∏(A λiI) , i=1 −

where k is the number of distinct eigenvalues, and 1 mi ni, with ni denoting the ≤ ≤ T T algebraic multiplicity of each eigenvalue λi. If A is “normal”, i.e., if A A = AA , or, in particular, symmetric, then each mi = 1. 1 The tensor A is said to be invertible if there exists another tensor, denoted by A− , such that

1 1 A− A = AA− = I.

Similar to Eqns. (1.68) and (1.69), we have

1 1 1 (UV)− = V− U− , (1.184a)

T 1 1 T T (C )− = (C− ) =: C− , (1.184b)

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From Eqn. (1.184b), it follows that if C has major symmetric then so does its inverse. The inverse of A B (assuming that A and B are invertible) is given by 1 1 1 (A B)− = A− B− , (1.185) since, by Eqns. (1.167) and (1.168), we have

1 1 1 1 (A− B− )(A B) = (A− A) (B− B) = I I = I. 9 A fourth-order tensor A is invertible if and only if I9(A) = ∏i=1 λi is nonzero, in which case,

1 1 h 8 7 i A− = A I1A + + I8I . I9 − ··· By virtue of the relation (AT)T = A, we have TTA = A for all A Lin, which implies that ∈ TT = I. (1.186) Thus, the inverse of the transposer tensor T is T itself. If A has major symmetry (but not necessarily the minor symmetries), then [Ψ(A)]T = Ψ(AT) = Ψ(A), so that, similar to the case of second-order symmetric tensors, all the nine eigenvalues are real, and the corresponding eigentensors Ai, i = 1, 2, . . . , 9 are orthogonal in the sense that

(Ai, Aj) = Ai : Aj = δij. (1.187) Thus, we can write the spectral representation of A as

9 9 A = ∑ λi Ai ⊗ Ai, with ∑ Ai ⊗ Ai = I. i=1 i=1 If k is the number of distinct eigenvalues, then, similar to Eqn. (J.4), we have the following explicit formula:

 A λ I ∏k − j , k > 1  j=1 λi λj Ai ⊗ Ai = j=i − (1.188) 6 I, k = 1

1 When A is invertible, the spectral decomposition of A− is given by

9 1 1 A− = ∑ Ai ⊗ Ai. i=1 λi A is said to be positive definite if T : AT > 0 for all nonzero T Lin. Similar to symmetric second-order tensors (see Theorem 1.6.4), one can show that A∈ is positive definite if and only if all the eigenvalues λi (or all the invariants that occur in the characteristic equation) are strictly positive. If A is positive semi-definite, the square root of A defined via the relation HH = A is given by

9 p H = ∑ λi Ai ⊗ Ai. i=1

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Now consider the case when the fourth-order tensor C has both minor symmetries, i.e., C = SCS. The transpose is still deﬁned by Eqn. (1.163), except that A, B Sym. Using this deﬁnition, one can show that when C and D have both minor symmetries,∈ then Eqns. (1.165) still hold. The eigenvalue problem given by Eqn. (1.178) is now written as CS = λS = λSS, S Sym, or, alternatively, using Eqn. (I.14), as ∈ h 1 i χ(C)H− λI6 6 χ(S) = 0. − × Thus, C has only six eigenvalues, with the remaining three eigenvalues zero, and the corresponding eigentensors being linearly independent skew-symmetric tensors. The above 1 equation has a nontrivial solution χ(S) = 0, if and only if det[χ(C)H− λI] = 0. This characteristic equation can be written in6 expanded form as −

λ6 I (C)λ5 + I (C)λ4 + + I (C) = 0, − 1 2 ··· 6 where

1 I (C) = tr [χ(C)H− ] = λ + λ + + λ , 1 1 2 ··· 6

1 1 2 1 2 I (C) = [(tr [χ(C)H− ]) tr (χ(C)H− ) ] = λ λ + λ λ + + λ λ , 2 2 − 1 2 2 3 ··· 1 6 ...

1 I6(C) = det[χ(C)H− ] = 8 det χ(C) = λ1λ2 ... λ6, are a strict subset of the set of principal invariants of C [4, 20, 141, 238] (the results of Ahmad and Norris on the one hand, and Betten on the other regarding the number of quadratic principal invariants seem to be contradictory). As an example,

1 1 det S = det[χ(S)H− ] = det[HH− ] = 1. (1.189) Using Eqn. (I.11c), we now have, similar to Eqn. (1.179),

det CT = det C. (1.190)

If C = C1C2, then using Eqn. (I.11f), we again have similar to Eqn. (1.180),

det C = (det C1)(det C2). (1.191) If C has both the minor symmetries, then [C¯ ] = [S][C¯ ][S] and [C] = [S][C][S], so that Eqn. (1.182) can be written as

[C¯ ] = [Q][C][Q]T, (1.192)

where Q := [S][Q Q][S], Q Orth+. Now applying Eqns. (1.165b), (I.11c) and (I.11f), ∈ and using the symmetry of S and H, we get

1 1 T T χ([C¯ ]) = χ(Q)H− χ([C])H− χ(Q) = Aχ([C])A ,

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1 where A := χ(Q)H− is found by using Eqn. (I.5); the ﬁnal expression for A is given by Eqn. (I.12) with J QT. Using Eqns. (1.165b) and (1.176), we get ≡ T T QQ = Q Q = S(I I)S = SIS = S. (1.193) It follows from Eqns. (1.189)–(1.191) that

det Q = 1. ± From Eqns. (1.189) and (1.191)–(1.193), we get

det(C¯ λS) = det[Q(C λS)QT] = det(QQT) det(C λS) − − − = (det S) det(C λS) = det(C λS), (1.194) − − showing that the invariants that occur in the characteristic equation (and hence the eigenvalues) of C are independent of the choice of basis. The spectral decomposition of C reads

6 6 T C = ∑ λi Ai ⊗ Ai, with Ai = Ai and ∑ Ai ⊗ Ai = S. i=1 i=1 In place of Eqn. (1.188), we now have

 C λ S ∏k − j , k > 1  j=1 λi λj Ai ⊗ Ai = j=i − 6 S. k = 1

1 C is invertible if there exists another tensor, denoted by C− , such that

1 1 C− C = CC− = S.

The necessary and sufﬁcient condition for C to be invertible is that I6(C) be nonzero. If this condition is satisﬁed, then

6 1 1 C− = ∑ Ai ⊗ Ai. i=1 λi

1 As an example, since SS = S, we have S− = S. Let U, V and C be tensors that have minor symmetries. Then, since ST = S, both relations given by Eqns. (1.184) hold. From Eqn. (1.176), it follows (provided A is invertible) that

1 1 1 [S(A A)S]− = S(A− A− )S. 1 1 1 Note that, in general, [S(A B)S]− = S(A− B− )S. C is said to be positive semi-deﬁnite6 if S : CS 0 for all S Sym. If C is positive semi-deﬁnite, its square root H is given by ≥ ∈

6 p T H = ∑ λi Ai ⊗ Ai, with Ai = Ai . i=1

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C is positive deﬁnite if and only if all the λi, i = 1, 2, . . . , 6, are strictly positive. Examples of some spectral resolutions are

9 I = ∑ Ai ⊗ Ai, i=1 6 9 T = ∑ Ai ⊗ Ai ∑ Ai ⊗ Ai, i=1 − i=7 6 T S = ∑ Ai ⊗ Ai, Ai = Ai i=1 3 T W = ∑ Ai ⊗ Ai, Ai = Ai i=1 − I I I ⊗ I = 3 ⊗ . √3 √3

Let e be the axial vector of Q Orth+ Sym, and let (λ, n) and (λˆ ,n ˆ ) denote the remaining eigenvalue/eigenvector pairs,∈ with\ the “hat” denoting a complex conjugate. The following result, which follows directly from Theorem 1.7.1, holds:

Theorem 1.7.2. The eigenvalue/eigentensor pairs of Q := Q Q, where Q Orth+ Sym, are (1, I), (1, Q), (1, Q2), (λ, e ⊗ n), (λ, n ⊗ e), (λˆ , e ⊗ nˆ ), (λˆ ,n ˆ ⊗ e∈), (λ2, n ⊗\ n), and (λˆ 2,n ˆ ⊗ nˆ ).

If the dimension of the underlying space is four, and (λ1, n1), (λˆ 1,n ˆ 1), (λ2, n2), (λˆ 2,n ˆ 2) denote the eigenvalue/eigenvector pairs of Q, then the eigenvalue/eigentensor pairs of 2 3 2 ˆ 2 2 ˆ 2 Q are (1, I), (1, Q), (1, Q ), (1, Q ), (λ1, n1 ⊗ n1), (λ1,n ˆ 1 ⊗ nˆ 1), (λ2, n2 ⊗ n2), (λ2,n ˆ 2 ⊗ nˆ 2), (λ1λ2, n1 ⊗ n2), (λ1λ2, n2 ⊗ n1), (λ1λˆ 2, n1 ⊗ nˆ 2), (λ1λˆ 2,n ˆ 2 ⊗ n1), (λˆ 1λ2,n ˆ 1 ⊗ n2), (λˆ 1λ2, n2 ⊗ nˆ 1), (λˆ 1λˆ 2,n ˆ 1 ⊗ nˆ 2) and (λˆ 1λˆ 2,n ˆ 2 ⊗ nˆ 1). In general, when the underlying space dimension is n, the eigenvalue one is repeated n times, with the corresponding eigenten- n 1 sors given by I, Q,..., Q − , and the remaining n(n 1) eigenvalue/eigentensor pairs are formed by combinations of each complex eigenvalue/eigenvector− of Q with itself and with all the remaining eigenvalues/eigenvectors of Q barring the complex conjugate of that eigenvalue/eigenvector. We now discuss the solution of the tensor equation AXB + CXD = H, where A, B, C, D and H are given second-order tensors. We write this equation as

T T [A B + C D ]X = H.

In component form, the above equation reads

Tijkl Xkl = Hij,

where

Tijkl = AikBlj + CikDlj, (1.195)

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Applying Eqn. (I.1e), we get

    1   X11 T1111 T1122 T1133 T1112 T1123 T1131 T1121 T1132 T1113 − H11   T T T T T T T T T    X22  2211 2222 2233 2212 2223 2231 2221 2232 2213 H22       X33 T3311 T3322 T3333 T3312 T3323 T3331 T3321 T3332 T3313 H33       X  T T T T T T T T T  H   12  1211 1222 1233 1212 1223 1231 1221 1232 1213  12       X23 = T2311 T2322 T2333 T2312 T2323 T2331 T2321 T2332 T2313 H23 ,       X  T T T T T T T T T  H   31  3111 3122 3133 3112 3123 3131 3121 3132 3113  31   T T T T T T T T T    X21  2111 2122 2133 2112 2123 2131 2121 2132 2113 H21       X32 T3211 T3222 T3233 T3212 T3223 T3231 T3221 T3232 T3213 H32 X13 T1311 T1322 T1333 T1312 T1323 T1331 T1321 T1332 T1313 H13 (1.196)

with the components Tijkl given by Eqn. (1.195). It is clear that this method can be extended to solve a matrix equation of the type ∑i AiXBi = H of any dimension n by inverting an n2 n2 matrix. ×Of particular importance is the case when B = C = I and D = A S Sym. In this ≡ ∈ case, Tijkl = Sikδlj + δikSlj. Although the above approach is suitable for numerical purposes, a tensorial solution is more suitable for proving properties of the solution X. We now present a method, based on 1 [155], for determining (S I + I S)− , S Sym, which is applicable for any dimension n (see Rosati [273] and references therein for∈ other methods). T Consider the tensor T I + I T , T Lin. If (λi, ui) and (λi, vi) are the eigenvalue/eigenvectors of T and TT, respectively,∈ then

T (T I + I T )(ui ⊗ vj) = T(ui ⊗ vj) + (ui ⊗ vj)T

T = (Tui) ⊗ vj + ui ⊗ (T vj)

= (λi + λj)(ui ⊗ vj),

which shows that (λi + λj, ui ⊗ vj), i, j = 1, 2, . . . , n, are eigenvalue/eigentensor pairs of T T I + I T . However, unfortunately, these are not the only eigenvalue/eigentensor pairs, especially when the eigenvectors u are not linearly independent. For example, { i} when T = e1 ⊗ e2 + e3 ⊗ e4 + e4 ⊗ e5, with n = 5, all the 25 eigenvalues of T I + T I T are zero, and there are nine linearly independent eigentensors given by e1 ⊗ e2, e1 ⊗ e5, e3 ⊗ e2, e3 ⊗ e5, e3 ⊗ e3 e4 ⊗ e4 + e5 ⊗ e5, e3 ⊗ e4 + e4 ⊗ e5, e3 ⊗ e1 + e4 ⊗ e2, e ⊗ e + e ⊗ e and e ⊗ e−+ e ⊗ e . Although− the first four eigentensors are of the − 1 4 2 5 − 1 1 2 2 form ui ⊗ vj, i, j = 1, 2, the remaining five are not. A similar situation arises even in the case of the tensor T I + I T, which again has nine linearly independent eigentensors. Nevertheless, as we now show, λi + λj, i, j = 1, 2, . . . , n, are the eigenvalues of T I + I T T T (or T I + I T). Thus, the necessary and sufficient condition for T I + I T to be n invertible is that (det T) ∏ i=1 (λi + λj) = 0. For example, if n = 3, this condition is given j=i+1 6 by λ λ λ (λ + λ )(λ + λ )(λ + λ ) = 0. 1 2 3 1 2 2 3 1 3 6

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1 T 1 Let T = H1 J1 H1− and T = H2 J2 H2− , where J1 and J2 are matrices in Jordan normal form. Then

T 1 1 T I + I T = (H1 J1 H1− ) I + I (H2 J2 H2− )

1 1 = (H1 H2)(J1 I + I J2)(H1− H2− )

1 = (H1 H2)(J1 I + I J2)(H1 H2)− .

T Thus, the eigenvalues of T I + I T and J1 I + I J2 are the same. Since the matrix form of J1 I + I J2 is an upper triangular matrix with λi + λj, i, j = 1, 2, . . . , n, on the diagonals, we get the desired result. A similar proof can be devised for T I + I T. Let (λ , n ) be the eigenvalues/eigenvectors of S Sym. Then, the eigenvalue/eigentensor i i ∈ pairs of S(S I + I S)S when all the λi are distinct are λi + λj, ni ⊗ nj + nj ⊗ ni, i, j = 1, 2, . . . , n (the eigenvalue/eigentensors when the eigenvalues of S are repeated can be found similar to Eqns. (1.258) and (1.259)). Thus, the necessary and sufﬁcient condition n that S(S I + I S)S be invertible is that (det S) ∏ i=1 (λi + λj) = 0. j=i+1 6 Let λi, i = 1, 2, . . . , k, be the distinct eigenvalues of S Sym, and let Pi be the associated projections, which are given by Eqn. (J.4). We have ∈

" k ! k ! k ! k !# S(S I + I S)S = S ∑ λiPi ∑ Pj + ∑ Pi ∑ λjPj S i=1 j=1 i=1 j=1

" k k # = S ∑ ∑(λi + λj)Pi Pj S. i=1 j=1

Assuming that the necessary and sufﬁcient condition for invertibility is satisﬁed, the desired inverse is

" k k # 1 Pi Pj [S(S I + I S)S]− = S ∑ ∑ S, (1.197) i=1 j=1 λi + λj

since, on using Eqn. (1.175),

" k k # " k k # Pi Pj (PiS) Pj + Pi (PjS) S ∑ ∑ S [S I + I S] S = S ∑ ∑ S i=1 j=1 λi + λj i=1 j=1 λi + λj

" k k # = S ∑ ∑ Pi Pj S i=1 j=1

" k ! k !# = S ∑ Pi ∑ Pj S i=1 j=1

= S [I I] S = S,

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k where we have used the properties PmPi = δimPi (no sum on i), ∑i=1 Pi = I and SPi = k (∑m=1 λmPm)Pi = λiPi = PiS (no sum on i). Thus, the solution of the tensor equation S(S I + I S)SX = H is

k k T Pi(H + H )Pj X = ∑ ∑ . (1.198) i=1 j=1 2(λi + λj)

The above method can obviously be generalized to the case when T is diagonalizable. T Thus, if the tensor equation to be solved is (T I + I T)X = TX + XT = H, where k T = ∑i=1 λiPi, with Pi given by Eqn. (J.26), then assuming that T is such that the conditions for invertibility of T I + I T are satisﬁed, we have

1 X = (T I + I T)− H

" k k # Pi Pj = ∑ ∑ H i=1 j=1 λi + λj

k k T Pi HPj = ∑ ∑ . i=1 j=1 λi + λj

We now show how the solutions given by Eqns. (1.198) reduce to some of the solutions presented in the literature for the case H = W Skw. When n = 3, and when the ∈ eigenvalues are distinct, for a given i, we have PiWPi = (ei∗ Wei∗)Pi = 0. Multiplying both sides of Eqn. (1.198) by the determinant of ·

S˜ := (tr S)I S = (λ + λ )P + (λ + λ )P + (λ + λ )P , − 2 3 1 1 3 2 1 2 3

which is given by (λ1 + λ2)(λ2 + λ3)(λ1 + λ3), we get

(det S˜ )X = SW˜ S˜, (1.199)

which is the solution presented by Scheidler [282]. Since the solution has no explicit dependence on the eigenvalues λi, it holds even under the case when the eigenvalues 2 are repeated. Alternatively, by writing W as (P1 + P2 + P3)W(P1 + P2 + P3), S W as 2 2 2 (λ1P1 + λ2P2 + λ3P3)W(P1 + P2 + P3), etc. and noting that (λ1 + λ2)(λ2 + λ3)(λ1 + λ3) = I (S)I (S) I (S), one can also show that when H = W, Eqn. (1.198) reduces to 1 2 − 3 (I I I )X = (I2 I )W (S2W + WS2), 1 2 − 3 1 − 2 − which is the solution presented by Guo [104]. Finally, consider the case when the dimension of the underlying vector space n is two, and H is arbitrary. Using I1(S) = λ1 + λ2, I2(S) = λ1λ2, and writing H = (P1 + P2)H(P1 + P2), SH = (λ1P1 + λ2P2)H(P1 + P2), etc., one can show that Eqn. (1.198) can be written as

(2I I )X = (I2 + I )H I (SH + HS) + SHS, 1 2 1 2 − 1

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which is the solution presented by Hoger and Carlson [127]. Alternatively, if S is invertible, 1 writing S− = P1/λ1 + P2/λ2, we can also show that

1 1 2I1X = H + I2(S− HS− ), which is the solution presented by Ting [327].

1.8 Isotropic Tensors

A second-order tensor is said to be isotropic if it’s components are the same in all orthonor- T mal bases, i.e., if T = Tijei ⊗ ej = Tije¯i ⊗ e¯ j. Sincee ¯i = Q ei, this condition can also be written as

QTQT = T Q Orth+, (1.200) ∀ ∈ or, alternatively, as

QT = TQ Q Orth+. (1.201) ∀ ∈ We have the following characterization for isotropic tensors.

Theorem 1.8.1. A tensor T is isotropic if and only if it is of the form T = λI, where λ . ∈ < Proof. If T = λI, then obviously Eqn. (1.200) holds. To prove the converse, let v be an arbitrary vector, and let R be the proper orthogonal tensor for which v is along its axis, i.e., Rv = v (see Eqn. (1.104)). Since RT = TR, we have

RTv = TRv = Tv,

i.e., Tv also lies along the axis of R. Since the axis of R is a one-dimensional subspace, we have

Tv = λv = λIv.

Since the vector v is arbitrary, we have T = λI by the deﬁnition of equality of tensors.

Another proof can be given as follows. Express Eqn. (1.200) as

QT = T,

2 where Q = Q Q. By Theorem 1.7.2, I, Q and Q are the eigentensors of Q corresponding to the eigenvalue one. Hence, the above equation implies that

T = α0(Q)I + α1(Q)Q + α2(Q)Q,

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where α0, α1 and α2 are, in general, constants that depend on Q. Since T is independent of Q, we have

2 T = α0 I + α1Q0 + α2Q0,

where, now, α0, α1 and α2 are constants. Substituting this expression into Eqn. (1.201), we get α1 = α2 = 0, since, in general, rotations do not commute. This methodology can be extended to other dimensions as well. For example, if the dimension n is four, then we 2 3 get T = α0 I + α1Q0 + α2Q0 + α3Q0, but again, since rotations do not commute, we get all the constants other than α0 to be zero. Now consider the case when n is two. In this case, T = α0 I + α1Q0, but now since rotations do commute when n = 2, both α0 and α1 are nonzero. Using the representation of Q0 in terms of cos θ0 and sin θ0, we can write the α β most general form of an isotropic second-order tensor when n = 2 as T = β α . A third-order tensor is isotropic if and only if it is of the form λE, where−E is the alternate tensor with components eijk. To see the ‘if’ part, note that

¯ = e¯ (e¯ × e¯ ) = QTe [cof QT(e × e )] = QTe [QT(e × e )] = e (e × e ) = . Eijk i · j k i · j k i · j k i · j k Eijk

The reverse implication is proved by making special choices of Q (such as e1 ⊗ e1 + e2 ⊗ e e ⊗ e , e ⊗ e + e ⊗ e e ⊗ e , etc.) in the relation 3 − 3 2 2 2 3 1 − 1 3

Bijk = QipQjqQkr Bpqr,

and showing that Bijk has to be a scalar multiple of eijk. A fourth-order tensor is said to be isotropic if it has the same components in all orthonormal bases, i.e., if

C = QCQT, (1.202)

+ where Q := Q Q, with Q Orth . Alternatively, the above condition can also be written as ∈

Q(CA) = C(QAQT) A Lin. (1.203) ∀ ∈ We have the following characterization for isotropic fourth-order tensors:

Theorem 1.8.2. A fourth-order tensor C : Lin Lin is isotropic if and only if it is of the form → C = λI ⊗ I + (µ + γ)I + (µ γ)T (1.204) − = λI ⊗ I + 2µS + 2γW, or, in other words, a fourth-order isotropic tensor is a linear combination of tensors with components δijδkl, δikδjl and δilδjk. If C has the ﬁrst or second minor symmetry, i.e., if C = SC or C = CS, then using Eqn. (1.160), it follows as a corollary that C = λI ⊗ I + 2µS. (1.205)

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Proof. The proof is based on [157]. To see the “if” part, note that

T Q(I ⊗ I)Q = (QI) ⊗ (QI) = [(Q Q)I] ⊗ [(Q Q)I] = I ⊗ I,

QIQT = QQT = I,

T T T T QTQ = (Q Q)TQ = T(Q Q)Q = TQQ = T. (by Eqn. (1.173))

To prove the converse, take A to be Q Orth+ Sym in Eqn. (1.203) to get ∈ \ Q(CQ) = CQ.

Thus, CQ is a linear combination of the eigentensors of Q corresponding to the eigenvalue one, i.e., by virtue of Theorem 1.7.2,

CQ = α (Q)I + α (Q)Q + α (Q)Q2 Q Orth+ Sym. 0 1 2 ∀ ∈ \ In the limiting cases following Eqn. (1.105b), namely, (i) α 0, we have Q = I, so that α and α can be taken to be zero, and (ii) α π,→ we have Q2 = I, so 1 2 → that α2 can be taken to be zero in the above expression. Thus, by virtue of the continuous dependence of Q on α, we can write

CQ = α (Q)I + α (Q)Q + α (Q)Q2 Q Orth+, (1.206) 0 1 2 ∀ ∈ with α = α = 0 when Q = I, and α = 0, when Q = 2e ⊗ e I. 1 2 2 − We now show that the functions αi, i = 0, 1, 2, are in fact functions of the principal invariants of Q (which is just the trace, since, for proper orthogonal tensors, the determinant is equal to unity, and the second invariant is equal to the trace). By letting A R Orth+ in Eqn. (1.203), we get ≡ 0 ∈ CR = QTC(QR QT)Q, Q Orth+. 0 0 ∀ ∈ T Substituting the expressions for CR0 and C(QR0Q ) obtained from Eqn. (1.206) into the above equation, we get h i h i α (R ) α (QR QT) I + α (R ) α (QR QT) 0 0 − 0 0 1 0 − 1 0 h i R + α (R ) α (QR QT) R2 = 0. 0 2 0 − 2 0 0 2 Since, as shown in the discussion on page 38, I, R0, R0 , I, R0 and I are { } { +} { } linearly independent sets, depending on whether R0 Orth Sym, R0 Orth+ Sym I (α = 0), or R = I (α = α = 0), we∈ have \ ∈ ∩ − { } 2 0 1 2 α (R ) = α (QR QT), i = 0, 1, 2, Q Orth+, (1.207) i 0 i 0 ∀ ∈ i.e., the αi’s are isotropic scalar-valued functions of R0. To show that they are functions of I1(R0), we have to show that I (Q ) = I (Q ) = α (Q ) = α (Q ), i = 0, 1, 2, 1 1 1 2 ⇒ i 1 i 2

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and then the conclusion follows from Theorem 1.6.7. If I1(Q1) = I1(Q2), then + T by Theorem 1.5.5, there exists Q0 Orth such that Q2 = Q0Q1Q0 . Thus, ∈ T using Eqn. (1.207), we get αi(Q2) = αi(Q0Q1Q0 ) = αi(Q1). It follows from Theorem 1.6.7 that there exist functions αˆ i such that αi(Q) = αˆ i(I1(Q)). Equation (1.206) can now be written as

2 CQ = αˆ 0(I1)I + αˆ 1(I1)Q + αˆ 2(I1)Q . (1.208)

Eliminating Q2 from Eqn. (1.208) using Eqn. (1.110), we get

CQ = [αˆ (I ) αˆ (I )I ]I + [αˆ (I ) + αˆ (I )I ]Q + αˆ (I )QT. 0 1 − 2 1 1 1 1 2 1 1 2 1 Since C is independent of Q, and a linear transformation from Lin to Lin, the right-hand side of the above expression has to be linear in Q, i.e., the coefﬁcients of Q and QT are constants, and the coefﬁcient of I has to be a multiple of tr Q. Thus, set αˆ 0(I1) (λ + µ γ)I1, αˆ 1(I1) (µ + γ) (µ γ)I1, and αˆ 2(I1) µ γ, where λ, µ≡and γ are− constants. This≡ leads to − − ≡ − CQ = λ(tr Q)I + (µ + γ)Q + (µ γ)QT − = [λ(I ⊗ I) + (µ + γ)I + (µ γ)T] Q Q Orth+. (1.209) − ∀ ∈ The last step is to show that the above equation implies the expression given in Eqn. (1.204). Thus, by virtue of Eqn. (1.157), we have to show that Eqn. (1.209) holds for an arbitrary tensor T in place of Q. Let Ciso := [λ(I ⊗ I) + (µ + γ)I + (µ γ)T]. Since fourth-order tensors are a linear transformation from Lin to Lin,− Eqn. (1.209) implies that

+ [C Ciso] (∑ φiQi) = 0 φi , Qi Orth . − i ∀ ∈ < ∈

Since by Theorem 1.5.3, any T Lin can be expressed as a linear combination of Q Orth+, the proof is complete.∈ ∈

For various other proofs, see, e.g., [24], [102], [103], [105], [110], [142], [148], [175], [210], [250] and [251]. Note that this result does not hold when the space dimension n = 2, as C α β the counterexample = T T, where T = β α shows (the above proof fails in the last step for this case, since an arbitrary tensor cannot− be expressed as a linear combination of rotations). However, the result is applicable for any n if we assume C to be symmetric.

1.9 Diﬀerentiation of Tensors

In the subsequent chapters, we shall deal with tensor ﬁelds, where a tensor T is a function of a scalar quantity t, and a vector quantity x (e.g., t representing time, and x the position vector). In this section, we deﬁne the concept of the derivative of a tensor quantity. Let X

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and Y denote normed vector spaces, and let f : Ω X Y. We say that f (u) approaches zero faster than u, and write f (u) = o (u) if ⊂ →

f (u) lim k k = 0. u 0 u → k k 1.9.1 The directional derivative Let g be a function whose values are scalars, vectors or tensors, and whose domain Ω is an open interval of . The derivative Dg(t) is deﬁned by < d 1 Dg(t) := g(t) = lim [g(t + α) g(t)] . (1.210) dt α 0 α − → From Eqn. (1.210), it follows that

1 lim [g(t + α) g(t) αDg(t)] = 0, α 0 α − − → or, equivalently,

g(t + α) = g(t) + αDg(t) + o (α) . (1.211)

Thus, we see that the derivative of a vector function is a vector, or the derivative of a tensor function is a tensor. We now extend the above concept of a derivative to domains that lie in spaces of dimensions higher than one. Let L(X, Y) denote the vector space of all continuous linear mappings from X to Y, and let g : Ω X Y. We say that g is differentiable at x Ω if there exists a linear transformation Dg⊂(x) →L(X, Y) such that ∈ ∈ g(x + u) = g(x) + Dg(x)[u] + o (u) . (1.212)

The quantity Dg(x) L(X, Y) is called the Frechet derivative, while the quantity Dg(x)[u] Y is called∈ the directional derivative or Gateaux derivative. Note that the directional derivative∈ is computed by the action of the Frechet derivative on elements of X. If the directional derivative exists, it is unique, and can be written as

1 d Dg(x)[u] = lim [g(x + αu) g(x)] = g(x + αu) . (1.213) α 0 α − dα → α=0 When Ω , by comparing Eqns. (1.211) and (1.212), we see that ⊂ < dg Dg(t)[α] = α α . dt ∀ ∈ < Often, the easiest way of computing the directional derivative is to appeal directly to its deﬁnition. We now consider some examples. Consider the function φ(v) = v v. Then · φ(v + u) = (v + u) (v + u) · = v v + 2v u + u u · · · = φ(v) + 2v u + o (u) , ·

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so that

Dφ(v)[u] = 2v u. · Now consider the function G : Lin Lin deﬁned by G(T) = T2, where T is a second- order tensor. Then we have →

G(T + U) = (T + U)2 = T2 + TU + UT + U2 = G(T) + TU + UT + o (U) ,

which implies that DG(T)[U] = TU + UT. Similarly for G(T) = T3, we have

G(T + U) = G(T) + T2U + TUT + UT2 + o (U) ,

which implies that DG(T)[U] = T2U + TUT + UT2. For φ(T) = tr (T), we have φ(T + U) = φ(T) + tr U, which yields

D(tr T)[U] = tr U = I : U. (1.214)

If φ(T) = tr Tk, then (with T0 I) ≡ k k k k i i 1 (T + U) = T + ∑ T − UT − + o (U) , i=1

and on using the linearity of the trace operator, and the fact that tr AB = tr BA, we get

k k i i 1 φ(T + U) = φ(T) + ∑ tr [T − UT − ] + o (U) i=1 k k 1 = φ(T) + ∑ tr [T − U] + o (U) i=1 k 1 T = φ(T) + k(T − ) : U + o (U) ,

so that

k 1 T Dφ(T)[U] = k(T − ) : U. (1.215)

If φ(T) = det T, then from Problem 14, we have

det(T + U) = det T + cof T : U + T : cof U + det U = det T + cof T : U + o (U) ,

which yields

Dφ(T)[U] = cof T : U. (1.216)

T Recall that when T is invertible, cof T = (det T)T− . We now discuss the invariance properties of the derivative of a tensor function.

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Theorem 1.9.1. Let G : Sym Sym be an isotopic tensor function. Then the derivative is invariant in the following→ sense:

QDG(S)[U]QT = DG(QSQT)[QUQT] Q Orth+, S, U Sym. ∀ ∈ ∈ (1.217)

Proof. We have already remarked that Sym is invariant under Orth+. Thus, we simply need to prove Eqn. (1.217). Since G(S) is isotropic, we have G Q(S + U)QT = QG(S + U)QT

= QG(S)QT + QDG(S)[U]QT + o (U) = G(QSQT) + QDG(S)[U]QT + o (U) .

On the other hand, we also have G Q(S + U)QT = G(QSQT + QUQT)

= G(QSQT) + DG(QSQT)[QUQT] + o (U) .

From the above equations, and the uniqueness of the derivative, we get Eqn. (1.217).

1.9.2 Product rule Quite often, we will be required to compute the directional derivative of a bilinear map π( f, g). Examples of bilinear maps are [108] the product of a scalar and a vector • π(φ, v) = φv,

the inner product of two vectors • π(u, v) = u v, · the tensor product of two vectors • π(u, v) = u ⊗ v,

the action of a tensor on a vector • π(T, v) = Tv,

and so forth. Assuming that f and g are differentiable at a point x in the common domain of f and g, the product π( f (x), g(x)) is differentiable, and its directional derivative is given by (see [108] for the proof) Dπ( f (x), g(x))[u] = π( f (x), Dg(x)[u]) + π(D f (x)[u], g(x)) u. (1.218) ∀

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When x , then we simply have (with x replaced by t) ∈ < π˙ ( f (t), g(t)) = π ( f (t),g ˙ (t)) + π f˙ (t), g(t) .

Another result that is used frequently is the chain rule, which we now state. 1.9.3 Chain rule

Let f : Ω2 Ω3 and g : Ω1 Ω2, and let g(x) be differentiable at x Ω1. Further, let f be differentiable→ at y = g(x).→ Then the composition f g is differentiable∈ at x, and ◦ D( f g)(x)[u] = D f (g(x))[Dg(x)[u]]. (1.219) ◦ In case x , then writing t in place of x, we have ∈ < d D( f g)(t)[α] = α ( f g), ◦ dt ◦ dg Dg(t)[α] = α , dt and hence,

d dg f (g(t)) = D f (g(t)) . (1.220) dt dt

We have already shown how to ﬁnd the directional derivatives of the ﬁrst and third invariants. To illustrate the product and chain rules, we now compute the directional deriva- 1 tive of the second invariant of a second-order tensor I2(T) = (det T)(tr T− ) given by

h 1 1 1 1 i h Ti DI (T)[U] = det T tr (T− U)tr T− tr (T− UT− ) = (tr T)I T : U. (1.221) 2 − − 1 1 First we calculate the derivative of T− . Since TT− = I, we have

1 1 DT[U]T− + TDT− [U] = 0,

1 1 1 1 and since DT[U] = U, we get DT− [U] = T− UT− . Next, denoting tr T− by φ(G(T)), 1 − where G(T) = T− , we have

1 1 1 1 Dφ(T)[U] = Dφ(G)[DG(T)[U]] = Dφ(G)[T− UT− ] = tr (T− UT− ). − −

Finally, writing I2 as (det T)φ(T), and applying the product rule, we get

DI2(T)[U] = D(det T)(T)[U]φ(T) + det TDφ(T)[U] 1 1 1 1 = det Ttr (T− U)tr T− det Ttr (T− UT− ), − which proves the ﬁrst relation in Eqn. (1.221). To get the second relation, we note from Eqn. (1.80) that I2 is also given by 1 h i (tr T)2 tr (T2) . 2 −

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Hence, we have 1 DI [U] = [2(tr T)tr U 2tr (TU)] 2 2 − h i = (tr T)I TT : U. − The second relation can also be obtained directly from the ﬁrst using the Cayley–Hamilton theorem. One other useful result is the following:

d dT (det T) = cof T : . (1.222) dt dt To see this note that d dT (det T) = D det(T) (by Eqn. (1.220)) dt dt dT = cof T : (by Eqn. (1.216)). dt

T When T is invertible, then using cof T = (det T)T− , we can also write Eqn. (1.222) as

d dT (det T) = det T tr T 1 . (1.223) dt dt −

By taking the directional derivative of the Cayley–Hamilton equation, we get the following “Rivlin’s identity” [68]:

1 h i T2U + TUT + UT2 (tr U)T2 (tr T)(TU + UT) + (tr T)2 tr T2 U − − 2 − + [(tr T)(tr U) tr (TU)] T (cof T : U)I = 0. − − Similar identities can be derived in other space dimensions by taking the directional derivatives of the relevant Cayley–Hamilton equation. Identities involving three tensors can be derived by taking the directional derivative twice. Other identities can be derived by taking the directional derivative of the Cayley–Hamilton equation with a tensor function [68]. 1.9.4 Gradient, divergence and curl When X is a Hilbert space, the directional derivative of a real-valued function φ : Ω X can be identiﬁed with an element of the space X. The derivative Dφ(x) is an element⊂ → of < the dual space X0 = L(X, ) and thus, since the space X is a Hilbert space, there exists by the Riesz representation theorem< [63] a unique element ∇φ(x) in the space X that satisﬁes

(∇φ(x), u) = Dφ(x)[u], (1.224)

where (., .) denotes the inner product of the space X. The element ∇φ(x) is called the gradient of φ. Note that while Dφ(x) belongs to the dual space X0, the gradient ∇φ(x) belongs to the space X.

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As an example, consider the case when X = 3 equipped with the Euclidean inner product (u, v) = u v. Then we have < · φ(x + u) = φ(x) + ∇φ(x) u + o (u) . · To ﬁnd the component form of the gradient, we take u = ei to get 1 ∂φ(x) (∇φ)i = Dφ(x)[ei] = lim [φ(x + αei) φ(x)] = . α 0 α − ∂x → i Thus, we have ∂φ Dφ(x)[u] = ∇φ u = ui. · ∂xi As another example, consider the case when X = Lin, with the matrix inner product (A, B) = A : B. Then the gradient of φ(T) is the matrix ∂φ/∂Tij . By deﬁnition, we have ∂φ ∂φ Dφ(T)[U] = : U = Uij. ∂T ∂Tij

Similarly, if V(T) is a tensor-valued function of T, then the gradient and directional derivative of V are related as ∂V DV(T)[U] = U. (1.225) ∂T

The gradients of the invariants I1, I2 and I3 are obtained from Eqns. (1.214), (1.216) and (1.221), and are given by ∂I 1 = I, (1.226a) ∂T ∂I 2 = (tr T)I TT, (1.226b) ∂T − ∂I h iT 3 = cof T = I I I T + T2 . (1.226c) ∂T 2 − 1 The gradient of φ(T) = tr Tk is obtained using Eqn. (1.215), and is given by

∂ k k 1 T (tr T ) = k(T − ) . (1.227) ∂T

Since ∂Tij/∂Tkl = ∂Tji/∂Tlk = δikδjl and ∂Tji/∂Tkl = ∂Tij/∂Tlk = δjkδil, we have

∂T ∂TT = = I, ∂T T ∂T (1.228) ∂TT ∂T = = T. ∂T ∂TT

In general, if φ1(T) and φ2(T) are scalar-valued functions of T, then ∂(φ φ ) ∂φ ∂φ 1 2 = φ 2 + φ 1 . ∂T 1 ∂T 2 ∂T

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If φ(T) and U(T) are scalar and tensor-valued functions of T, then

∂φ ∂φ = T , (1.229) ∂TT ∂T ∂U ∂U = T. (1.230) ∂TT ∂T

Since (φU)ij = φUij, we have

∂(φU) ∂φ ∂U = U ⊗ + φ . (1.231) ∂T ∂T ∂T

If U(T) and V(T) are tensor-valued functions of T, then writing U : V as UmnVmn, we get

∂(U : V) ∂U T ∂V T = V + U. (1.232) ∂T ∂T ∂T

Similarly, using the relation (UV)ij = UimVmj, we get ∂(UV) ∂Vmj ∂Uim = Uim + Vmj ∂T ijkl ∂Tkl ∂Tkl ∂Vmn ∂Unm = Uimδnj + δinVmj ∂Tkl ∂Tkl ∂Vmn ∂Umn = Uimδnj + δimVnj, ∂Tkl ∂Tkl which in tensorial form reads ∂(UV) ∂V ∂U = (U I) + (I V T) . (1.233) ∂T ∂T ∂T Using Eqns. (1.158), (1.173), (1.228) and (1.233), we have the following results:

∂(T2) = T I + I TT, ∂T ∂(TT T) = (I TT)T + TT I = 2S(TT I), (1.234) ∂T ∂(TTT) = (I T) + (T I)T = 2S(I T), (1.235) ∂T 1 ∂(T− ) 1 T T 1 = T− T− = (T T )− , (1.236) ∂T − − T ∂(T− ) 1 T T 1 T 1 = T(T− T− ) = (T− T− )T = T(T T )− . (1.237) ∂T − − − 1 The fourth relation is obtained by taking U and V to be T− and T, respectively, and then 1 differentiating T− T = I. We get

1 1 T ∂T− 0 = T− I + (I T ) ∂T

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1 ∂T− T 1 1 = = (I T )− (T− I) ⇒ ∂T − T 1 = (I T− )(T− I) (by Eqn. (1.185)) − 1 T = T− T− . (by Eqn. (1.168)) − T 1 The ﬁfth relation is obtained by writing T− as TT− and then using Eqn. (1.236). If φ(G(T)) is a scalar-valued function of a tensor function G(T), then by the chain rule, we have ∂φ ∂φ ∂G = kl , ∂Tij ∂Gkl ∂Tij

which implies that

∂φ ∂G T ∂φ = . (1.238) ∂T ∂T ∂G

Similarly, if H(G(T)) is a tensor-valued function of a tensor function G(T), then

∂H ∂H ∂G = . (1.239) ∂T ∂G ∂T Special care has to be exercised in computing the gradients with respect to a symmetric tensor S. As a speciﬁc example, let us consider the computation of the ﬁrst and second- order derivatives of a scalar-valued function W˜ (C), where C Sym. Let S˜ and C denote these derivatives. We have ∈ 1 ∂W˜ ∂W˜ S˜ (C) = + 2 ∂C ∂CT 1 ∂W˜ ∂W˜ = + T (by Eqn. (1.229)) 2 ∂C ∂C 1 ∂W˜ = (I + T) 2 ∂C ∂W˜ = S . (by Eqn. (1.158)) ∂C In indicial notation, the above equation reads ! 1 ∂W˜ ∂W˜ S˜ij = + . 2 ∂Cij ∂Cji

The second-order gradient C = ∂S˜ /∂C is computed as follows:

1 ∂(SS˜ ) ∂(SS˜ ) C = + 2 ∂C ∂CT 1 ∂(SS˜ ) ∂(SS˜ ) = + T (by Eqn. (1.230)) 2 ∂C ∂C

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1 ∂S˜ = S (I + T) 2 ∂C ∂S˜ = S S (1.240) ∂C ∂2W˜ = S S. (1.241) ∂C∂C In other words, to ﬁnd ∂2W˜ /∂C∂C when C Sym, we compute it as if C is not symmetric, and then pre- and post-multiply the result with∈ the symmetrizer S. In indicial notation, the above equations read ! 1 ∂S˜ij ∂S˜ji ∂S˜ij ∂S˜ji Cijkl = + + + 4 ∂Ckl ∂Ckl ∂Clk ∂Clk ! 1 ∂2W˜ ∂2W˜ ∂2W˜ ∂2W˜ = + + + . 4 ∂Cij∂Ckl ∂Cji∂Ckl ∂Cij∂Clk ∂Cji∂Clk

Now consider the evaluation of the stretch and rotation tensors of F, and their deriva- 2 tives for arbitrary space dimension n. Let λi , i = 1, 2, . . . , k be the distinct eigenval- k 2 ues, and Pi be the corresponding projections of C Psym, so that C = ∑i=1 λi Pi and k ∈ U = ∑i=1 λiPi, with Pi given via Eqn. (J.4) as

 2 C λj I U λj I ∏k − = ∏k − , k > 1 j=1 λ2 λ2 j=1 λi λj Pi = j=i i − j j=i − (1.242) 6 6 I, k = 1.

The rotation tensor R is now obtained as FU 1 = F ∑k P /λ . Note that det F = − i=1 i i | | det U = √det C. Now we evaluate the derivatives of these tensors. Differentiating the relation C = UU using Eqn. (1.233), we get ∂U S = S [U I + I U] S S. ∂C

Since the eigenvalues λi, i = 1, 2, . . . , n, of U are positive, the necessary and sufﬁcient condition that [U I + I U] be invertible is satisﬁed1. Thus, using (1.197), we get the unique solution ∂U/∂C = S/(2λ) for k = 1, while for k > 1, we have

∂U 1 = S [U I + I U] S − ∂C { } " k k # Pi Pj = S ∑ ∑ S, (1.243) i=1 j=1 λi + λj

Now using Eqns. (1.175), (1.234) and (1.239), we get ∂U ∂U ∂C = S S ∂F ∂C ∂F 1If C is positive semi-deﬁnite instead of positive deﬁnite, then ∂U/∂C does not exist when C is singular.

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" k k # Pi Pj h T i = 2S ∑ ∑ S F I i=1 j=1 λi + λj " k k T # (Pi F ) Pj = 2S ∑ ∑ . (1.244) i=1 j=1 λi + λj By differentiating F = RU, and using Eqns. (1.158), (1.168), (1.185), (1.173), we get ∂R 1 ∂U = (I U)− I (R I) ∂F − ∂F   k k T 1 1 (Pi F ) Pj = I U− 2[R U− ]S ∑ ∑  − i=1 j=1 λi + λj

T 1 h 1 T i k k (RPi F ) (U− Pj) + T (U− Pi F ) (RPj) 1 = I U− ∑ ∑ − i=1 j=1 λi + λj k k T 1 T 1 T (λi + λj)(RPi R ) (U− Pj) λi(RPi R ) (U− Pj) [(RPi) (PjR )]T = ∑ ∑ − − i=1 j=1 λi + λj k k T T (RPi R ) Pj [(RPi) (PjR )]T = ∑ ∑ − . (1.245) i=1 j=1 λi + λj Note that we have interchanged i and j in the last term in the second-to-last step. Now using Eqn. (1.243), and the fact that U˙ = (∂U/∂C)C˙ (or, alternatively, Eqn. (1.244) and the fact that U˙ = (∂U/∂F)F˙ ), and in a similar manner using (1.245), we get U˙ = C˙ /(2λ) and T RT R˙ = (RT F˙ F˙ R)/(2λ) for k = 1, while for k > 1, − k k k k T T PiCP˙ j Pi(F F˙ + F˙ F)Pj U˙ = ∑ ∑ = ∑ ∑ . (1.246) i=1 j=1 λi + λj i=1 j=1 λi + λj k k T T Pi(R F˙ F˙ R)Pj R˙ = R ∑ ∑ − . (1.247) i=1 j=1 λi + λj

In a similar fashion, one can ﬁnd DFU(F)[Z] = (∂U/∂F)Z. If F is the deformation gradient, then see Eqns. (2.70) and (2.77). The corresponding results for the left stretch tensor V are

" k k # ∂V Gi Gj = S ∑ ∑ S, ∂B i=1 j=1 λi + λj " k k # ∂V Gi (GjF) = 2S ∑ ∑ , ∂F i=1 j=1 λi + λj k k k k T T GiBG˙ j Gi(FF˙ + FF˙ )Gj V˙ = ∑ ∑ = ∑ ∑ , (1.248) i=1 j=1 λi + λj i=1 j=1 λi + λj " k k T T # Gi(FR˙ RF˙ )Gj R˙ = ∑ ∑ − R, (1.249) i=1 j=1 λi + λj

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T where Gi = RPiR are the projections of V, which are obtained using Eqn. (1.242) by replacing C and U by B and V, respectively. If F is the deformation gradient, then see Eqns. (2.71) and (2.76). We now verify the following relations derived by Chen and Wheeler [47] and Wheeler [352] when n = 3:

∂U T U = F, ∂F ∂R T F = 0, ∂F ∂U 1 L = RT L U˜ (RT L LT R)UU˜ , ∂F − det U˜ − ∂R 1 L = RU˜ (RT L LT R)U˜ , ∂F det U˜ − where L Lin, and ∈ U˜ := (tr U)I U = (λ + λ )P + (λ + λ )P + (λ + λ )P . − 2 3 1 1 3 2 1 2 3

Note that det U˜ = (λ1 + λ2)(λ2 + λ3)(λ1 + λ3), and that

UU˜ = λ1(λ2 + λ3)P1 + λ2(λ1 + λ3)P2 + λ3(λ1 + λ2)P3. (1.250)

T T T T T T T T T Using the facts (A B) = A B , (C1C2) = C2 C1 , TA = A , S = S, T = T, T k h k i PiPj = δijPj (no sum on j), Pi = Pi, ∑i=1 Pi = I, UPj = ∑m=1 λmPm Pj = λjPj = PjU (no sum on j), we have

T k k ∂U FPjUPi + FPiUPj U = ∑ ∑ ∂F i=1 j=1 λi + λj

" k k # PjUPi + PiUPj = F ∑ ∑ i=1 j=1 λi + λj

" k k # λi(PjPi + PiPj) = F ∑ ∑ i=1 j=1 λi + λj

k λ P = F ∑ i i i=1 λi = F, T k k T T T ∂R RPiR FPj T(PiR FPjR ) F = ∑ ∑ − ∂F i=1 j=1 λi + λj

k k RPiUPj RPjUPi = ∑ ∑ − i=1 j=1 λi + λj

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k k λjR(PiPj PjPi) = ∑ ∑ − i=1 j=1 λi + λj

k λ R(P P ) = ∑ i i − i i=1 2λi = 0, k k T T ∂U Pi L FPj + Pi F LPj L = ∑ ∑ ∂F i=1 j=1 λi + λj

k k T T Pi L RUPj + PiUR LPj = ∑ ∑ i=1 j=1 λi + λj

k k T T λjPi L RPj + λiPiR LPj = ∑ ∑ i=1 j=1 λi + λj

k k k k T T T λjPi(R L L R)Pj = ∑ ∑ PiR LPj ∑ ∑ − i=1 j=1 − i=1 j=1 λi + λj

k ! k ! k k T T T λjPi(R L L R)Pj = ∑ Pi R L ∑ Pj ∑ ∑ − i=1 j=1 − i=1 j=1 λi + λj

k k T T λjPi(R L L R)Pj = RT L ∑ ∑ − − i=1 j=1 λi + λj 1 = RT L U˜ (RT L LT R)UU˜ , (by Eqn. (1.250)) − det U˜ − ∂R 1 1 T 1 T T L = (I U− )L (R U− ) R L U˜ (R L L R)UU˜ ∂F − − det U˜ −

1 T 1 1 T T = LU− RR LU− + RU˜ (R L L R)U˜ − det U˜ − 1 = RU˜ (RT L LT R)U˜ . det U˜ −

We now discuss the computation of the gradients of the eigenvalues λ1, λ2, λ3 of a diagonalizable tensor T for the case n = 3, and then specialize the results to the case of a symmetric tensor–a more comprehensive discussion of the derivatives of the eigenvalues of a symmetric tensor for the case n = 3 may be found in [344]. By Eqn. (J.25), T can be k expressed as ∑i=1 λiPi, where k is the number of distinct eigenvalues, Pi are projections k given by Eqn. (J.26), and ∑i=1 Pi = I. Consider the case when all eigenvalues are distinct, i.e., λ1 = λ2 = λ3 = λ1. It is known (see, e.g., the remark following (2.2.4) in [37]) that the6 eigenvalues6 and6 associated eigenprojections are differentiable in this case. Using Eqn. (1.60), we can show, for example, that cof (T λ I) = (T λ I)T(T λ I)T = PT. − 1 − 2 − 3 1 Since det(T γI (λi γ)I) = det(T λi I) = 0, i = 1, 2, 3 and γ , we see that λ γ, i = 1,− 2, 3− are the− eigenvalues of −T γI, γ . Thus, (λ λ∈), <(λ λ ) and i − − ∈ < 1 − 1 2 − 1

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(λ3 λ1) are the eigenvalues of (T λ1 I), which implies that (0, 0, (λ2 λ1)(λ3 λ1)) are the eigenvalues− of cof (T λ I), which− in turn leads to tr cof (T λ I−) = (λ −λ )(λ − 1 − 1 1 − 2 1 − λ3). By differentiating the relation det(T λi I) = 0, i = 1, 2, 3, and using the chain rule, we get − ∂λ (I 1 ⊗ I)cof (T λ I) = 0, i = 1, 2, 3, (1.251) − ∂T − 1 or, in other words, ∂λ cof (T λ I) (T λ I)T(T λ I)T 1 = − 1 = − 2 − 3 = PT. (1.252) ∂T tr cof (T λ I) (λ λ )(λ λ ) 1 − 1 1 − 2 1 − 3 The corresponding results for ∂λ2/∂T and ∂λ3/∂T are obtained by permuting the indices 1, 2 and 3. Equation (1.252) can also be directly obtained by differentiating the relation λ3 I λ2 + I λ I = 0 with respect to T, and using Eqns. (1.226a)–(1.226c); this shows 1 − 1 1 2 1 − 3 that the second expression for ∂λ1/∂T in Eqn. (1.252) is valid even for an arbitrary tensor, provided the denominator is nonzero. By differentiating the expression P1 = (T λ2 I)(T λ3 I)/[(λ1 λ2)(λ1 λ3)] using Eqns. (1.170), (1.231) and (1.233), we get − − − −

∂P1 1 h T = T I + I T (λ2 + λ3)I ∂T (λ1 λ2)(λ1 λ3) − − − h ii + (λ + λ 2λ )P ⊗ PT (λ λ ) P ⊗ PT P ⊗ PT . 2 3 − 1 1 1 − 2 − 3 2 2 − 3 3 3 3 T By using Eqn. (J.29) and the fact that I = ∑i=1 Pi = ∑i=1 Pi , the above relation simpliﬁes to T T T T ∂P1 P1 P + P2 P P1 P + P3 P = 2 1 + 3 1 . (1.253) ∂T λ λ λ λ 1 − 2 1 − 3 The gradients of P2 and P3 are obtained by cyclically permuting the indices 1, 2 and 3. The above expressions are valid for an arbitrary tensor with distinct eigenvalues (since such a tensor is diagonalizable). Now consider the case when two of the eigenvalues are repeated, i.e., λ = λ = λ . 1 6 2 3 Note that now, Eqn. (J.4) yields P1 = (T λ2 I)/(λ1 λ2). Since P1P1 = P1, and since λ = λ , (1.252) also holds. Using the fact− that λ + λ −= I λ , we get 2 3 2 3 1 − 1 ∂ λ I TT (λ + λ ) = I PT = 1 − . (1.254) ∂T 2 3 − 1 (λ λ ) 1 − 2 The gradient of P1 is obtained by setting λ2 = λ3 and using P2 + P3 = I P1 in Eqn. (1.253) as − T T ∂P1 P1 (I P1) + (I P1) P = − − 1 . ∂T (λ λ ) 1 − 2 Since P + P = I P , we get 2 3 − 1 ∂ ∂P (P + P ) = 1 . ∂T 2 3 − ∂T

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When all three eigenvalues are repeated, the gradient of the sum of the eigenvalues is I, but the individual gradients are undeﬁned. Since symmetric second-order tensors are T diagonalizable, all the above results can be applied to them by setting Pi = Pi . For a generalization to any underlying space dimension n, see Eqn. (J.27). The foregoing development can be used to ﬁnd the gradients of scalar or tensor-valued k functions of a diagonalizable tensor T = ∑i=1 λiPi, where (λi, Pi), i = 1, 2, . . . , k, are the distinct eigenvalues and corresponding eigenprojections of T (see, e.g., [37], [47], [104], [127], [244], [272], [352] and [355] for various other methods). For example, if G(T) = k ∑i=1 f (λi)Pi is a tensor-valued function of T, then using the above results, we get ∂G/∂T = f 0(λ)I for the case k = 1, while for k > 1,

k k k ∂G ∂ f f (λi) f (λj) = P PT + − P PT. (1.255) ∂T ∑ ∂λ i i ∑ ∑ λ λ i j i=1 i i=1 j=1 i − j j=i 6 The above formula is derived by ﬁrst considering the case of distinct eigenvalues (i.e., k = n), and then appropriate limits are taken (e.g., λ3 λ2 in case λ2 = λ3) to obtain the results for repeated eigenvalues. By virtue of Eqn.→ (J.27), the above result holds for any k arbitrary underlying space dimension n. In case G(S) = ∑i=1 f (λi)Pi is a symmetric- valued function of S Sym, then the above result reduces to ∂G/∂S = f 0(λ)S for the case k = 1, while for k > 1,∈

" k k k # ∂G ∂ f f (λi) f (λj) = S P P + − P P S. (1.256) ∂S ∑ ∂λ i i ∑ ∑ λ λ i j i=1 i i=1 j=1 i − j j=i 6 An application of Eqn. (1.256) can be found towards the end of Section 2.4. We now ﬁnd the eigenvalue/eigentensors of ∂G/∂S in the above equation. First consider the case when all the eigenvalues λi of S are distinct. Then, if ni represent the eigenvectors of S, each Pi can be represented as ni ⊗ ni (no sum on i). Using Eqns. (1.172) and (1.342), the eigenvalues/eigentensors (the eigentensors have not been normalized) are found to be

∂ f ∂ f ∂ f , n1 ⊗ n1 , , n2 ⊗ n2 , , n3 ⊗ n3 , ∂λ1 ∂λ2 ∂λ3 f (λ ) f (λ ) f (λ ) f (λ ) 1 − 2 , n ⊗ n + n ⊗ n , 2 − 3 , n ⊗ n + n ⊗ n , (1.257) λ λ 1 2 2 1 λ λ 2 3 3 2 1 − 2 2 − 3 f (λ ) f (λ ) 1 − 3 , n ⊗ n + n ⊗ n , λ = λ = λ = λ . λ λ 1 3 3 1 1 6 2 6 3 6 1 1 − 3

Next consider the case when two eigenvalues of S are repeated, say, λ1 = λ2 = λ3. We now have 6 ∂G ∂ f ∂ f = S P1 P1 + (I P1) (I P1) ∂S ∂λ1 ∂λ2 − − f (λ ) f (λ ) + 1 − 2 [P (I P ) + (I P ) P ] S. λ λ 1 − 1 − 1 1 1 − 2

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The distinct eigenvalue/eigentensor pairs of ∂G/∂S in this case are

∂ f ∂ f , n1 ⊗ n1 , , I n1 ⊗ n1 , ∂λ1 ∂λ2 − f (λ ) f (λ ) 1 − 2 , n ⊗ [n (n n )n ] + [n (n n )n ] ⊗ n , λ = λ = λ , λ λ 1 − · 1 1 − · 1 1 1 1 6 2 3 1 − 2 (1.258)

where n is such that n × n = 0, but is otherwise arbitrary. Finally, for the case when all 1 6 eigenvalues are repeated, i.e., λ1 = λ2 = λ3 λ, the distinct eigenvalue/eigentensor pair is ≡ ∂ f , A , λ = λ = λ λ, (1.259) ∂λ 1 2 3 ≡

where A Sym. The derivatives∈ of the eigenvectors e of S Sym are given by [244] { i∗} ∈ ∂e∗ e∗ ⊗ e∗ ⊗ e∗ + e∗ ⊗ e∗ ⊗ e∗ e∗ ⊗ e∗ ⊗ e∗ + e∗ ⊗ e∗ ⊗ e∗ 1 = 2 1 2 2 2 1 + 3 1 3 3 3 1 , ∂S 2(λ λ ) 2(λ λ ) 1 − 2 1 − 3 ∂e e∗ ⊗ e∗ ⊗ e∗ + e∗ ⊗ e∗ ⊗ e∗ e ⊗ e ⊗ e + e ⊗ e ⊗ e 2∗ = 1 1 2 1 2 1 + 3∗ 2∗ 3∗ 3∗ 3∗ 2∗ , ∂S 2(λ λ ) 2(λ λ ) 2 − 1 2 − 3 ∂e e∗ ⊗ e∗ ⊗ e∗ + e∗ ⊗ e∗ ⊗ e∗ e ⊗ e ⊗ e + e ⊗ e ⊗ e 3∗ = 1 1 3 1 3 1 + 2∗ 2∗ 3∗ 2∗ 3∗ 2∗ . ∂S 2(λ λ ) 2(λ λ ) 3 − 1 3 − 2 Similar to the gradient of a scalar field, we define the gradient of a vector field v as

∇v(x)u := Dv(x)[u]. (1.260)

The terminology ‘gradient’ used for the above quantity is misleading since the gradient was deﬁned in terms of the inner product, whereas the above quantity is not [58]. The quantity ∇v is in reality simply the matrix representation of the Frechet derivative of v, and hence a second-order tensor. Taking u = ej, and taking the dot product of the above equation with ei, we get the components of ∇v as

∂vi (∇v)ij = . ∂xj

Thus, we have

∂vi ∇v = ei ⊗ ej. ∂xj

The scalar ﬁeld

∂vi ∂vi ∂vi ∂vi ∇ v := tr ∇v = tr (ei ⊗ ej) = ei ej = δij = , (1.261) · ∂xj ∂xj · ∂xj ∂xi

is called the divergence of v.

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The gradient of a second-order tensor T is a third-order tensor deﬁned in a way similar to the gradient of a vector ﬁeld as

∇T(x)u := DT(x)[u]. (1.262)

∇T can be written in terms of its components as

∂Tij ∇T = ei ⊗ ej ⊗ ek. ∂xk The divergence of a second-order tensor T, denoted as ∇ T, is deﬁned as · (∇ T) u := ∇ (TTu) constant u V, (1.263) · · · ∀ ∈ leading to the component form

∂Tij ∇ T = ei. · ∂xj

The curl of a vector v, denoted as ∇ × v, is deﬁned by

(∇ × v) × u := [∇v (∇v)T]u u V. (1.264) − ∀ ∈ Thus, ∇ × v is the axial vector corresponding to the skew tensor [∇v (∇v)T]. In component form, we have −

∂vk ∂vk ∇ × v = eijk(∇v)kjei = eijk ei = ej × ek. ∂xj ∂xj

The curl of a tensor T, denoted by ∇ × T, is deﬁned by

(∇ × T)u := ∇ × (TTu) constant u V. (1.265) ∀ ∈ In component form,

∂Tjs ∂Tjs ∇ × T = eirs ei ⊗ ej = (er × es) ⊗ ej. (1.266) ∂xr ∂xr The Laplacian of a scalar function φ(x) is deﬁned by

∇2φ := ∇ (∇φ). (1.267) · In component form, the Laplacian is given by

∂2φ ∇2φ = . ∂xi∂xi

If ∇2φ = 0, then φ is said to be harmonic. The Laplacian of a tensor function T(x), denoted by ∇2T, is deﬁned by

(∇2T) : H := ∇2(T : H) constant H Lin. (1.268) ∀ ∈

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In component form,

2 2 ∂ Tij (∇ T)ij = . ∂xk∂xk In the remaining chapters, we shall assume that a function is sufﬁciently smooth (without explicitly stating what these smoothness properties are), so that all the equations presented make sense. For example, in the equation

∇ τ + ρb = 0, · if we are given that ρ and b are continuous functions of the position vector x, then we implicitly assume τ to be continuously differentiable. 1.9.5 Examples Although it is possible to derive tensor identities involving differentiation using the above definitions of the operators, the proofs can be quite cumbersome, and hence we prefer to ∂ use indicial notation instead. In what follows, u and v are vector fields, and ∇ ei (this ≡ ∂xi is to be interpreted as the ‘del’ operator acting on a scalar, vector or tensor-valued field, e.g., ∇φ = ∂φ e ): ∂xi i 1. Show that

∇ × ∇φ = 0. (1.269)

2. Show that 1 ∇(u u) = (∇u)Tu. (1.270) 2 ·

3. Show that ∇ [(∇u)v] = (∇u)T : ∇v + v [∇(∇ u)]. · · · 4. Show that

∇ (∇u)T = ∇(∇ u), (1.271a) · · ∇2u := ∇ (∇u) = ∇(∇ u) ∇ × (∇ × u). (1.271b) · · − From Eqns. (1.271a) and (1.271b), it follows that

∇ [(∇u) (∇u)T] = ∇ × (∇ × u). · − − From Eqn. (1.271b), it follows that if ∇ u = 0 and ∇ × u = 0, then ∇2u = 0, i.e., u is harmonic. · 5. Show that ∇ (u × v) = v (∇ × u) u (∇ × v). · · − · 6. Let W Skw, and let w be its axial vector. Then show that ∈ ∇ W = ∇ × w, · − ∇ × W = (∇ w)I ∇w. (1.272) · −

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Solution: 1. Consider the ith component of the left-hand side: ∂2φ (∇ × ∇φ)i = eijk ∂xj∂xk ∂2φ = eikj (interchanging j and k) ∂xk∂xj ∂2φ = eijk , − ∂xj∂xk which implies that ∇ × ∇φ = 0.

1 1 ∂(uiui) ∂ui T 2. ∇(u u) = ej = ui ej = (∇u) u. 2 · 2 ∂xj ∂xj 3. We have ∇ [(∇u)v] = ((∇u)v) · ∇j j ∂ ∂uj = vi ∂xj ∂xi 2 ∂vi ∂uj ∂ uj = + vi ∂xj ∂xi ∂xi∂xj = (∇u)T : ∇v + v ∇(∇ u). · · 4. The ﬁrst identity is proved as follows: T ∂ ∂uj [∇ (∇u) ]i = · ∂xj ∂xi ! ∂ ∂uj = ∂xi ∂xj = [∇(∇ u)] . · i To prove the second identity, consider the last term ∇ × (∇ × u) = e (∇ × u) e ijk∇j k i = e (e u )e ijk∇j kmn∇m n i 2 ∂ un = eijkemnk ei ∂xj∂xm 2 ∂ un = (δimδjn δinδjm) ei − ∂xj∂xm " 2 2 # ∂ uj ∂ ui = ei ∂xi∂xj − ∂xj∂xj = ∇(∇ u) ∇ (∇u). · − ·

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5. We have ∂(u × v) ∇ (u × v) = i · ∂xi ∂(ujvk) = eijk ∂xi ∂uj ∂vk = eijkvk + eijkuj ∂xi ∂xi ∂uj ∂vk = ekijvk ejikuj ∂xi − ∂xi = v (∇ × u) u (∇ × v). · − · 6. Using the relation W = e w , we have ij − ijk k ∂wk (∇ W) = eijk ei · − ∂xj = ∇ × w. − ∂Wjn (∇ × W)ij = eimn ∂xm ∂wr = eimnejnr − ∂xm ∂ur = (δijδmr δirδmj) − ∂xm ∂ur ∂ui = δij , ∂xr − ∂xj which is the indicial version of Eqn. (1.272).

1.10 The Exponential and Logarithmic Functions

The exponential of a tensor T(t) can be deﬁned either in terms of its series representation as 1 eT(t) := I + T(t) + [T(t)]2 + , (1.273) 2! ··· or in terms of a solution of the initial value problem X˙ (ξ) = T(t)X(ξ) = X(ξ)T(t), ξ > 0, (1.274) X(0) = I, (1.275) for the tensor function X(ξ), where the parameter ξ is independent of t. Note that the superposed dot in the above equation denotes differentiation with respect to ξ. The existence theorem for linear differential equations tells us that this problem has exactly one solution X : [0, ∞) Lin, which we write in the form → X(ξ) = eT(t)ξ.

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From Eqn. (1.273), it is immediately evident that

T e[T(t)] = (eT(t))T, (1.276)

1 and that if A Lin is invertible, then e(A− BA) = A 1eB A for all B Lin. ∈ − ∈ Theorem 1.10.1. For each t 0, eT(t) belongs to Lin+, and ≥ det(eT(t)) = etr T(t). (1.277)

Proof. If (λi, n) is an eigenvalue/eigenvector pair of T(t), then from Eqn. (1.273), it follows that (eλi(t), n) is an eigenvalue/eigenvector pair of eT(t). Hence, the determinant of eT(t), which is just the product of the eigenvalues, is given by

n T(t) n λi(t) ∑i=1 λi(t) tr T(t) det(e ) = Πi=1e = e = e .

Since etr T(t) > 0 for all t, eT(t) Lin+. ∈ From Eqn. (1.277), it directly follows that

det(eAeB) = det(eA) det(eB) = etr Aetr B = etr (A+B) = det(eA+B).

Theorem 1.10.2. Let A(t), B(t) Lin. If ∈ e[A(t)+B(t)]ξ = eA(t)ξ eB(t)ξ ξ (1.278) ∀ then A(t)B(t) = B(t)A(t). Conversely, if A(t)B(t) = B(t)A(t), then

e[A(t)+B(t)]ξ = eA(t)ξ eB(t)ξ = eB(t)ξ eA(t)ξ ξ (1.279) ∀ Proof. We shall write A(t) and B(t) simply as A and B for notational conve- Aξ Bξ (A+B)ξ nience. Let X A(ξ) := e , XB(ξ) := e , and X A+B := e . Then, we have

X˙ A+B = (A + B)X A+B, (1.280) ˙ X AXB = X˙ AXB + X AX˙ B = AX AXB + X ABXB, (1.281)

where, as before, the superposed dot denotes differentiation with respect to ξ. Let Z := X X X . From Eqns. (1.280) and (1.281), we get A+B − A B Z˙ = (A + B)X AX X X BX . (1.282) A+B − A B − A B

If AB = BA, then X AB = BX A, so that Eqn. (1.282) becomes

Z˙ = (A + B) [X X X ] = (A + B)Z. (1.283) A+B − A B

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The above differential equation is to be solved subject to the initial condition Z(0) = 0. The general solution of the differential equation is Z = e(A+B)ξ C, where the tensor C is a constant tensor. On using the initial condition, we get C = 0, from which it follows that Z = 0, thus proving Eqn. (1.279). In an analogous fashion, by taking Z := X X X , one can also show that if A+B − B A AB = BA, then e(A+B)ξ = eBξ eAξ. Thus, Eqn. (1.279) follows. Conversely, if Eqn. (1.278) holds, i.e., if X A+B = X AXB, then Eqn. (1.282) reduces to

BX X = X BX ξ. A B A B ∀ + 1 By Theorem 1.10.1, X Lin . Hence multiplying the above equation by X− , B ∈ B we get BX A = X AB for all ξ. Differentiating this relation with respect to ξ, and evaluating at ξ = 0, we get BA = AB.

As a corollary of the above theorem, it follows, by taking ξ = 1 in Eqn. (1.279), that

A(t)B(t) = B(t)A(t) = eA(t)+B(t) = eA(t)eB(t) = eB(t)eA(t). (1.284) ⇒ However, the converse of the above statement may not be true. Indeed, if AB = BA, one can have eA+B = eA = eB = eAeB, or eAeB = eA+B = eBeA or even eAeB = eBeA6 = eA+B as the following examples show: 6 6 1. Wood [354] presents the following example:

 √   √  0 0 3 0 0 3 2 − 2 A = 2π  0 0 1  , B = 2π  0 0 1  ,  − 2   − 2  √3 1 0 √3 1 0 − 2 2 2 2 so that eA+B = eA = eB = eAeB = I. The eigenvalues of A and B are (0, 2πi, 2πi). Another example presented by Horn and Johnson [136] is −

0 0 0 0  0 0 1 0  0 0 0 0  0 0 0 1  A =   , B =   , 0 0 0 2π 0 0 0 2π  −   −  0 0 2π 0 0 0 2π 0

for which, again, one has eA+B = eA = eB = eAeB = I. The eigenvalues of A and B are (0, 0, 2πi, 2πi). − 2. Wermuth [350] presents the following example where eAeB = eA+B = eBeA: if z = a + ib is a solution of ez z = 1, e.g., a = 2.088843 . . . and b = 7.4614896 . . ., then for − 0 0 0 0  0 0 1 0 0 0 0 0  0 0 0 1 A =   , B =   , 0 0 a b 0 0 0 0  −    0 0 b a 0 0 0 0

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we have

1 0 0 0  1 0 1 0 0 1 0 0  0 1 0 1 eA =   , eB =   , (1.285) 0 0 a + 1 b  0 0 1 0  −    0 0 b a + 1 0 0 0 1

so that

1 0 1 0  1 0 a + 1 b  − 0 1 0 1  0 1 b a + 1 eAeB = eA+B =   , eBeA =   . 0 0 a + 1 b  0 0 a + 1 b   −   −  0 0 b a + 1 0 0 b a + 1

3. An example for the case eAeB = eBeA = eA+B, again provided in [136], is 6 0 π 0 0 0 0 1 0 −     π 0 0 0 0 0 0 1 A =   , B =   , 0 0 0 π 0 0 0 0 0 0 π 0 0 0 0 0 − which yields

1 0 1 0   A A+B B 0 1 0 1 e = e = I, e =   , − 0 0 1 0 0 0 0 1

so that eAeB = eBeA = eA+B. 6 T(t) T(t) As an application of Eqn. (1.284), since T(t) and T(t) commute, we have e − = T(t) T(t) − I = e e− . Thus, T(t) 1 T(t) (e )− = e− . (1.286)

In fact, one can extend this result to get

(eT(t))n = enT(t) integer n. ∀ As an application of Eqn. (1.286), consider the solution of the tensorial differential equation

X˙ + AX + XBT = F(t), (1.287)

subject to the initial condition X(0) = C. The differentiation is with respect to t, and A and B are assumed to be independent of t. Multiplying Eqn. (1.287) by eAt eBt, we get

At Bt At BT t At T BT t At BT t (e e )X˙ + Ae Xe + e XB e = e F(t)e .

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The above equation can be written as

d h i T eAt eBt X = eAtF(t)eB t. dt Integrating the above equation between the limits [0, t], we get

Z t At Bt Aξ BT ξ (e e )X(t) = e F(ξ)e dξ + C, 0 which, by virtue of Eqn. (1.286), leads to

Z t A(t ξ) BT (t ξ) At BT t X(t) = e− − F(ξ)e− − dξ + e− Ce− 0 Z t Aξ BT ξ At BT t = e− F(t ξ)e− dξ + e− Ce− . (1.288) 0 − What we have effectively shown is that

e(AI+IB)t = eAt eBt t. ∀ The result given by Eqn. (1.288) also holds if B = 0, and X, F and C are vectors. This observation allows us to solve the tensorial differential equation

X˙ + AXB = F(t),

which can be written as

T X˙ + (A B )X = F(t),

Using the Ψˆ mapping of Appendix I, the above equation can be written as

T Ψˆ (X˙ ) + Ψˆ (A B )Ψˆ (X) = Ψˆ (F), whose solution is given by

Z t Ψˆ (A BT )ξ Ψˆ (A BT )t Ψˆ [X(t)] = e− Ψˆ [F(t ξ)] dξ + e− Ψˆ (C). 0 − For the exponential of a skew-symmetric tensor, we have the following theorem:

Theorem 1.10.3. Let W(t) Skw for all t. Then eW(t) is a rotation for each t 0. ∈ ≥ Conversely, for R(t) Orth+, there exists a W(t) Skw such that R(t) = eW(t). ∈ ∈ Proof. By Eqn. (1.286),

W(t) 1 W(t) W T (t) W(t) T (e )− = e− = e = (e ) , where the last step follows from Eqn. (1.276). Thus, eW(t) is a orthogonal tensor. By Theorem 1.10.1, det(eW(t)) = etr W(t) = e0 = 1, and hence eW(t) is a rotation.

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The converse is proved simply by taking W(t) = log R(t), which we show to be a skew-symmetric tensor later in this section.

In the three-dimensional case, by using the Cayley–Hamilton theorem, we get W3(t) = w(t) 2 W(t), where w(t) is the axial vector of W(t). Thus, W4(t) = w(t) 2 W2(t), − | | − | | W5(t) = w(t) 4 W(t), and so on. Substituting these terms into the series expansion of the exponential| function,| and using the representations of sine and cosine functions, we get2 sin( w(t) ) [1 cos( w(t) )] R(t) = eW(t) = I + | | W(t) + − | | W2(t). (1.289) w(t) w(t) 2 | | | | The directional derivative DR(W)[U] can be obtained from the above expression using the fact that w 2 = w w = W : W/2, so that 2 w D w (W)[U] = W : U. | | · | | | | Another proof of Eqn. (1.289) is obtained by assuming R(ξ) eW(t)ξ to be of the form 2 ≡ h1(ξ)I + h2(ξ)W + h3(ξ)W , and then determining the unknown functions h1(ξ), h2(ξ), and h3(ξ) by using the governing equation R˙ = RW (where the dot denotes differentiation with respect to ξ) subject to the initial conditions R(0) = I and R˙ (0) = W. By using the fact that W3(t) = w(t) 2 W(t), and that I, W, W2 is a linearly independent set (see Problem 4), we get − | | { }

h˙ 1(ξ) = 0, h˙ (ξ) = h w(t) 2 h (ξ), 2 1 − | | 3 h˙ 3(ξ) = h2(ξ), which are to be solved subject to the initial conditions

h1(0) = 1, h˙ 1(0) = 0,

h2(0) = 0, h˙ 2(0) = 1,

h3(0) = 0, h˙ 3(0) = 0. We get the solution as sin( w(t)ξ ) [1 cos( w(t)ξ )] h1(ξ) = 1, h2(ξ) = | | , h3(ξ) = − | | , w(t) w(t) 2 | | | | 2Similarly, in the two-dimensional case, if " # 0 γ(t) W(t) = , γ(t) 0 − where γ is a parameter which is a function of t, then " # sin γ(t) cos γ(t) sin γ(t) R(t) = eW(t) = cos γ(t)I + W(t) = . γ(t) sin γ(t) cos γ(t) − R t Let γ˜(t) = 0 γ(ξ) dξ. Then the solution of dz/dt = W(t)z in this two-dimensional case is " # cos γ˜(t) sin γ˜(t) z(t) = z(0). sin γ˜(t) cos γ˜(t) − This result is a special case of the result in Theorem 1.10.4.

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which, on setting ξ = 1, yields Eqn. (1.289). Yet another proof is obtained by using Eqn. (1.292) with T W(t), λ = 0, µ = i w(t) , ≡W(t) i w(t) |i w(t) | ν = i w(t) and ξ = 1 (which implies that the eigenvalues of e are (1, e | |, e− | |)). Not− | surprisingly,| Eqn. (1.289) has the same form as Eqn. (1.104) with α = w(t) . Equa- tion (1.289) is known as Rodrigues formula. If W(t) is a complex-valued skew-symmetric| | p tensor, replace w(t) in Eqn. (1.289) by w(t) w(t) if it is nonzero; if it is zero, then W | | · is nilpotent, and using the series expansion, we get eW(t) = I + W(t) + W2(t)/2. A generalization of Rodrigues formula to the case when n 4 was developed by Gallier and Xu [86]; we shall present this formula later in this section≥ (see Eqn. (1.295)), although we use a different method for deriving it than the one that they have used. Let a = sin( w(t) )/ w(t) , b = [1 cos( w(t) )]/ w(t) 2 and c = w(t) 2, and let | | | | − | | | T| | | R(t) be given by Eqn. (1.289). Using the facts RR˙ T + RR˙ = 0, 2b b2c a2 = 0, WWW˙ = c˙W/2, and WWW˙ 2 W2WW˙ = 0 (where now the superposed− dot− denotes differentiation− with respect to t), we− get

1 RR˙ T = (w w˙ )(1 a)W + aW˙ + b(WW˙ WW˙ ), (1.290a) c · − − 1 RT R˙ = (w w˙ )(1 a)W + aW˙ b(WW˙ WW˙ ). (1.290b) c · − − − The axial vectors of RR˙ T and RT R˙ are, thus, (w w˙ )(1 a)w/c + aw˙ + b(w × w˙ ) and (w w˙ )(1 a)w/c + aw˙ b(w × w˙ ), respectively.· Of course,− for the special case when · − − T T W(t) = W0t, where W0 is independent of t, the above formulae reduce to RR˙ = R R˙ = W0. In this connection, one has the following theorem [354]:

Theorem 1.10.4. For T(t) Lin, ∈ d(eT(t)) = T˙ eT(t) = eT(t)T˙ , (1.291) dt

R t T(ξ) dξ if and only if TT˙ = TT˙ . Thus, the solution of Z˙ = T(t)Z is Z(t) = e 0 Z(0) if and only if TT˙ = TT˙ .

Proof. If TT˙ = TT˙ , then T˙2 = TT˙ + TT˙ = 2TT˙ = 2TT˙ , and so on for derivatives of higher powers of T. Thus, by simply differentiating Eqn. (1.273), we get Eqn. (1.291). To prove the converse, note from Eqn. (1.273) that eT T = TeT . Differentiating both sides of this relation, and using the fact that Eqn. (1.291) holds, we get

T˙ eT T = TT˙ eT .

T T T 1 T Noting again that e T = Te , and multiplying both sides by (e )− = e− , we get the desired result. R t By replacing T(t) by 0 T(ξ) dξ in the ﬁrst relation in Eqn. (1.291), and post- multiplying by Z(0), we get the stated solution of the differential equation Z˙ = T(t)Z.

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In what follows T(t) is written simply as T. Explicit formulae for eTξ (where T can be complex-valued) when n = 3 and n = 4 are given by Cheng and Yau [52] in terms of the eigenvalues and powers of T.

Theorem 1.10.5. For n = 3, we have the following cases: 1. λ = λ = λ λ 1 2 3 ≡ (a) If q(T) = (T λI), then − eTξ = eλξ I.

(b) If q(T) = (T λI)2, then − eTξ = eλξ [ξ(T λI) + I] . − (c) If q(T) = (T λI)3, then − ξ2 eTξ = eλξ (T λI)2 + ξ(T λI) + I . 2 − −

2. λ µ = λ = λ λ 1 ≡ 6 2 3 ≡ (a) If q(T) = (T λI)(T µI), then − − eλξ eµξ λeµξ µeλξ eTξ = − T + − I. λ µ λ µ − − (b) If q(T) = (T µI)(T λI)2, then − − ξeλξ eλξ eµξ eTξ = − (T λI)2 + eλξ [ξ(T λI) + I] . λ µ − (λ µ)2 − − − −

3. λ1 λ = λ2 µ = λ3 ν = λ1. We have q(T) = (T λI)(T µI)(T νI)≡, and6 ≡ 6 ≡ 6 − − −

(T µI)(T νI) (T λI)(T νI) (T λI)(T µI) eTξ = eλξ − − + eµξ − − + eνξ − − . (λ µ)(λ ν) (µ λ)(µ ν) (ν λ)(ν µ) − − − − − − (1.292)

We now present two methods for the explicit determination of the exponential function for arbitrary n (for a detailed survey, see [225]). The first method is based on the Jordan canonical form given by Eqn. (J.37), while the second method is based on the definition given by Eqn. (1.274). The first method yields the formula

k " m 1 j !# i− ξ Tξ λiξ j e = ∑ e Pi + ∑ Ni , (1.293) i=1 j=1 j!

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where mi is as in Eqn. (J.24). To prove this result, we ﬁrst note that Pi and Ni commute since they are polynomials in T. Using Eqns. (J.37) and (1.279), and the properties PiPj = δijPi j (no sum on i), Ni N j = Pi N j = 0 for i = j, Pi Ni = Ni (no sum on i), and Ni = 0, j mi, we have 6 ≥

T k P k N e ξ = e∑i=1 λi iξ e∑i=1 iξ

" k #" k m 1 j # i− ξ λiξ j = ∑ e Pi I + ∑ ∑ Ni i=1 i=1 j=1 j! k " m 1 j !# i− ξ λiξ j = ∑ e Pi + ∑ Ni . i=1 j=1 j!

As an example, if T is given by Eqn. (J.39), we get

 α β  eα eβ e +(β α 1)e eα − − α−β (α β)2  −  eT = eαP + eβ(P + N ) =  β −β  . 1 2 2  0 e e  0 0 eβ

The exponential tensor eT for a diagonalizable (in particular, symmetric) tensor T is obtained either as a special case of Eqn. (1.293), or by substituting Eqn. (J.25) into Eqn. (1.273), and is given by

k Tξ λiξ e = ∑ e Pi, (1.294) i=1

with Pi given by Eqn. (J.26). The method proposed by Putzer [259] essentially yields the same result as that given by Eqn. (1.293), although more indirectly. Writing λm as (λr)m + i(λs)m, we can write Eqn. (1.293) as

k h i T (λr)m 2 e = ∑ e sin((λs)m)Bm cos((λs)m)Bm + [cos((λs)m) + i sin((λs)m)] Gm , m=1 −

nm 1 1 j 3 where Bm = iPm and Gm = ∑ − N . Note that B = iPm = Bm. The above j=1 j! m m − − form allows us to write a Rodrigues-type formula for any dimension n when T is a skew- symmetric tensor [86]. As discussed earlier, the eigenvalues of a real-valued W Skw ∈ are either zero or of the form iλm, m = 1, 2, . . . , k, where k is the number of distinct ± (1) (2) nonzero eigenvalues. Also, since W is normal, it is diagonalizable. Let Bm := i[Pm Pm ], (1) (2) − where Pm and Pm are the projections associated with the eigenvalues iλm and iλm, 2 (1) (2) k 2 − respectively. Since Bm = [Pm + Pm ], it follows that I = ∑m=1 Bm. From Eqn. (1.294), we get − −

k W h 2i e = I + ∑ sin(λi)Bi + [1 cos(λi)] Bi , (1.295) i=1 −

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k where W = ∑i=1 λiBi, with each Bi being a real-valued skew tensor. The λi are determined 2 by taking the square roots of the eigenvalues of W , while the matrices Bi are determined by solving the following system of k equations −

k W = ∑ λiBi, i=1 k 3 3 W = ∑ λi Bi, − i=1 , ··· k k+1 2k 1 2k 1 ( 1) W − = ∑ λi − Bi, − i=1 by inverting the van der Monde matrix [153]   λ1 λ2 ... λk  3 3 3   λ λ2 ... λ   1 k  ......  ,     ......  2k 1 2k 1 2k 1 λ1 − λ2 − ... λk − to obtain

 W2+λ2 I ( 1)k+1 W ∏k j , k > 1 λi j=1 λ2 λ2 Bi(W) = − j=i i − j  W 6 λ , k = 1. An identical procedure can be used to ﬁnd the exponential of a skew-Hermitian tensor (which is unitary) since, again, in this case, the real part of the eigenvalues and the Gi’s are zero. We have already noted that some complex-valued skew-symmetric tensors can be nilpotent, and hence, no Rodrigues-type formula can be derived for its exponential (which is an orthogonal tensor), in general. The second method that we discuss is the one devised by Fulmer [85]. Note that since dk/dξk(eTξ ) = TkeTξ, for k = 1, 2, . . . , n, we have " # dk eTξ = I and (eTξ ) = Tk. (1.296) ξ=0 dξk ξ=0

Also, by virtue of the Cayley–Hamilton theorem,

n n 1 n Tξ (T I T − + + ( 1) I I)e = 0, − 1 ··· − n or, alternatively,

n n 1 d d − n Tξ n I1 n 1 + + ( 1) In e = 0. (1.297) dξ − dξ − ··· −

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As a result of Eqns. (1.296) and (1.297), we see that eTξ is the unique solution of the nth order initial value problem

dkG(ξ) DG(ξ) = 0, G(0) = I, = Tk, k = 1, 2, . . . , n 1, (1.298) dξk − ξ=0

where D is the differential operator in parenthesis in Eqn. (1.297). If T has the characteristic polynomial given by Eqn. (J.23), then the general solution of DG(ξ) = 0 is

λ ξ n 1 λ ξ n 1 G(ξ) = e 1 (C + ξC + + ξ 1− C ) + + e k (C + ξC + + ξ k− C ), 11 12 ··· 1n1 ··· k1 k2 ··· knk (1.299)

where the Cij are n matrices to be determined from the n initial conditions

I = C + C + + C , 11 21 ··· k1 T = (λ C + C ) + + (λ C + C ), 1 11 12 ··· k k1 k2 T2 = (λ2C + 2λ C ) + + (λ2C + 2λ C ), 1 11 1 12 ··· k k1 k k2 ··· n 1 n 1 n 2 (n 1)! n n1 T − = λ1− C11 + (n 1)λ1− C12 + + − λ1− C1n1 + − ··· (n n1)! ··· − n 1 n 2 (n 1)! n nk + λ − C + (n 1)λ − C + + − λ − C . k k1 − k k2 ··· (n n )! k knk − k

The Cij are obtained as polynomials in T by inverting the n n coefficient matrix in the above equations; substituting these expressions into Eqn. (1.299),× we get the desired expression for eTξ. Note that the above method is based on the characteristic polynomial3, whereas the first method is based on the minimal polynomial. As an illustration, let us again find the expression for eT , with T given by Eqn. (J.39). Since the characteristic polynomial is given by (T αI)(T βI)2, the general solution is αξ βξ − − G(ξ) = e C11 + e (C21 + ξC22), where the Cij are obtained from the initial conditions

I = C11 + C21, T = αC11 + βC21 + C22, 2 2 2 T = α C11 + β C21 + 2βC22.

T α β Solving for Cij and substituting into e = e C11 + e (C21 + C22), we get the same expression as before. Now we discuss the differentiation of eT with respect to T, where T Lin, for any underlying space dimension n; the treatment is based on [155], alternative treatments∈ may be found in [143] and [144]. We shall ﬁrst give an alternate derivation of a formula that

3Actually, the method can be modiﬁed quite easily to work with the minimal polynomial as well–in this case k the number of Cij matrices to be determined is m = ∑i=1 mi instead of n. This is particularly advantageous, for example, if T is diagonalizable, since then m = 1 for i = 1, 2, . . . , k, so that C = 0 for i = 1, 2, . . . , k and j 2. i ij ≥

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appears in Mathias [211] and Ortiz et al. [241]. Let H ∂X/∂T, where X = eTξ. Differen- tiating the expression X˙ = TX with respect to T using≡ Eqn. (1.233), we get

T H˙ = (T I)H + I X .

Tξ Multiplying this equation by the “integrating factor” e− I, we get

d h Tξ i Tξ Tξ T (e− I)H = e− (e ) . dξ

Integrating this equation subject to the condition H ξ=0 = 0, and using Eqn. (1.286), we get |

Z ξ Tξ h Tη Tη Ti H(ξ) = (e I) e− (e ) dη 0 Z ξ h T(t η) Tη Ti = e − (e ) dη. 0

Similarly, by differentiating X˙ = XT, we also get

Z xi h Tη TT (t η)i H(ξ) = e e − dη. 0

Since ∂eT /∂T = H , we obtain |ξ=1 T Z 1 ∂(e ) h T(1 η) Tη Ti = e − (e ) dη ∂T 0 Z 1 T Tη T(1 η) = e e − dη. (1.300) 0

Explicit expressions for eTη such as those given by Eqn. (1.293) (or those presented in The- orem 1.10.5 for n = 3) should be substituted into the above expressions to get explicit formulae for ∂eT /∂T. Note that these formulae are valid for any T Lin and any n. Simpler expressions for a diagonalizable tensor T can be obtained∈ using Eqns. (1.294) λ and (1.300), or directly by using Eqn. (1.255) with f (λi) = e i . The ﬁnal result that we obtain for a diagonalizable tensor (for any n) is eλI for k = 1, and " # ∂(eT ) k k k eλi eλj = eλi P PT + − P PT , k > 1 ∂T ∑ i i ∑ ∑ λ λ i j i=1 i=1 j=1 i − j j=i 6

where k is the number of distinct eigenvalues of T, and Pi is given by Eqn. (J.26). When T T Sym, we obviously have Pi = Pi, and the above result should be pre- and post- multiplied∈ by S. In what follows, we shall continue to denote T(t) as T; the treatment is based on [158]. Recall that difficulties arise in defining a unique scalar-valued logarithmic function since eλ = eλ+2πin where n is an integer. Similar difficulties arise in defining the logarithm of a tensor. If one defines the solution X of eX = T as log T, then a non-singular T may have

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an inﬁnite number of real and complex logarithms. In order to avoid this nonuniqueness and ensure that the logarithm of a real tensor is real, we deﬁne the principal logarithmic function, denoted by log T, of a matrix T with no eigenvalues that are negative or zero (i.e. log T λi / ( ∞, 0], i = 1, 2, . . . , n), as the unique function such that e = T. The eigenvalues of log∈ −T have imaginary parts that lie strictly between π and π, since as we shall show below, if reiθ, π < θ < π, are the eigenvalues of T, then− log(reiθ) = log r + iθ, π < θ < − − π, are the eigenvalues of log T. Thus, if λi denote the eigenvalues of T, then 1. elog T = T if and only if λ / ( ∞, 0] i. i ∈ − ∀ 2. log eT = T if and only if π < Im(λ ) < π i. − i ∀ In series form the logarithmic function is given by

∞ ( 1)i 1 log(I + T) = ∑ − − (T)i, (1.301) i=1 i or, alternatively as

∞ ( 1)i 1 log(T) = ∑ − − (T I)i. (1.302) i=1 i −

The series given by Eqn. (1.301) is absolutely convergent only if T < 1. It is immediately evident from the above representation that log(I) = 0 and logk(TkT) = (log T)T. Since, A 1BA 1 B 1 for all invertible A Lin, e − = A− e A for all B Lin, we get log(A− BA) = 1 ∈ ∈ A− (log B)A. If H(ξ) : [0, 1] Lin is a tensor-valued function of ξ with no eigenvalues on the closed negative real axis→ for any ξ [0, 1], an alternative deﬁnition can be given as the solution of the differential equation ∈

dX dH H(ξ) = , X(0) = 0, (1.303) dξ dξ

with the solution denoted by X(ξ) = log H. For example, if H(ξ) = I + (T I)ξ, with no eigenvalues of T on the closed negative real axis, then, from the above deﬁnition,− we get

Z ξ 1 log[I + (T I)ξ] = [I + (T I)η]− (T I) dη. (1.304) − 0 − − If one violates the restriction on the eigenvalues of T, then, from the above deﬁnition, one can get a complex-valued logarithm even for a real tensor as is seen, for example, by taking T = diag[1, 1, 1]. Let (λ, n) denote an eigenvalues/eigenvector pair of T. Then, by using the fact that−

1 1 In = [I + (T I)η]− [I + (T I)η]n = (1 + (λ 1)η)[I + (T I)η]− n, − − − − we get from Eqn. (1.304),

Z ξ λ 1 log[I + (T I)ξ]n = − dη n = log[1 + (λ 1)ξ]n, − 0 1 + (λ 1)η − −

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so that (log[1 + (λ 1)ξ], n) are the corresponding eigenvalue/eigenvector of log[I + (T I)ξ]. In particular,− for ξ = 1, (log λ, n) are the eigenvalue/eigenvector pair of log T. Simi-− 1 larly, one can show that ( log λ, n) are the eigenvalue/eigenvector pair of log(T− ). Anal- ogous to Eqn. (1.277), we− get

n tr log T = ∑ log λi = log(det T). (1.305) i=1

We have the following analogue of Theorem 1.10.2:

Theorem 1.10.6. Let A, B be such that no eigenvalues of A, B and AB are on the closed negative real axis. If

log [I + (A I)ξ][I + (B I)ξ] = log[I + (A I)ξ] + log[I + (B I)ξ] { − − } − − ξ [0, 1] (1.306) ∀ ∈ then AB = BA. Conversely, if AB = BA (so that A, B and AB have a common set of eigenvec- a tors), and if π < Im(log αi + log βi) < π for each i, where αi and βi denote the eigenvalues of−A and B corresponding to the same eigenvector, then

log [I + (A I)ξ][I + (B I)ξ] = log [I + (B I)ξ][I + (A I)ξ] = { − − } { − − } log[I + (A I)ξ] + log[I + (B I)ξ] ξ [0, 1] (1.307) − − ∀ ∈ Proof. If Eqn. (1.306) holds, then by differentiating it with respect to ξ using Eqn. (1.303), we get

1 1 [I + (B I)ξ]− [I + (A I)ξ]− [A + B 2I + 2ξ(A I)(B I)] = − − − − − 1 1 [I + (A I)ξ]− (A I) + [I + (B I)ξ]− (B I). − − − − Differentiating the above relation once again with respect to ξ (using the fact that 1 1 1 d(T− )/dξ = T− [dT/dξ]T− ), and evaluating the resulting expression at ξ = 0, we get AB =−BA. Conversely, given that AB = BA, using Eqn. (1.303), we have log [I + (A I)ξ][I + (B I)ξ] { − − } Z ξ 1 1 = [I + (B I)η]− [I + (A I)η]− [A + B 2I + 2η(A I)(B I)] dη 0 − − − − − Z ξ 1 1 = [I + (B I)η]− [I + (A I)η]− [I + (A I)η](B I) + [I + (B I)η] 0 − − { − − − (A I) dη − } Z ξ Z ξ 1 1 = l [I + (B I)η]− (B I) dη + [I + (A I)η]− (A I) dη 0 − − 0 − − = log[I + (A I)ξ] + log[I + (B I)ξ]. − − In a similar manner, one gets the same expression for log [I + (B I)ξ][I + (A I)ξ] , leading to Eqn. (1.307). { − − }

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aThis condition is imposed since, as stated, the imaginary part of the eigenvalues of the log of any tensor has to lie between π and π. −

As a corollary, it follows by taking ξ = 1 in Eqn. (1.307), that

AB = BA and π < Im(log α + log β ) < π i = log(AB) − i i ∀ ⇒ = log(BA) = log A + log B. (1.308) The above implication need not be true if AB = BA, but the constraints on the eigenvalues of A and B are violated. Cheng et al. [53] have presented the following (complex-valued) (π e)i scalar example: if a = b = e − , where e is small and positive, then

log(ab) = 2ei = 2(π e)i = log(a) + log(b). − 6 − We present a (real-valued) matrix example that is a modiﬁcation of the example of Wood [354]. If

 √    0 0 3 0.499963 0.866004 0.00866011 2 − A = (π 0.01)  0 0 1  , H = eA  0.866004 0.500012 0.00499992 , −  − 2  ≈  −  √3 1 0 0.00866011 0.00499992 0.99995 − 2 2 − − then     0 0 2.71204 0 0 0.0173205 − log H  0 0 1.5658 , log H2  0 0 0.01  , ≈  −  ≈   2.71204 1.5658 0 0.0173205 0.01 0 − − so that log H2 = 2 log H. The eigenvalues of log H and log H2 are (approximately) (0, (π 0.01)i, (π 0.016 )i) and (0, 0.02i, 0.02i), respectively. − The− converse− assertion of Eqn.− (1.308) is true for dimension n = 2, as seen in [226, 227], and also for symmetric tensors (for any n), as is evident by taking the transpose. For n = 4, it may appear that by taking X = eA, Y = eB, where eA and eB are given by Eqn. (1.285), we have log(XY) = log X + log Y, but XY = YX. However, since both A and A + B do not 6 A A+B satisfy the constraint π < Im(λi) < π, we have log(e ) = A and log(e ) = A + B. Thus, the question of− whether the converse− assertion of Eqn.6 (1.308) is true for6 n > 2 remains unresolved. Let T be a tensor with no eigenvalues λi on the closed negative real axis. Since T and T 1 commute, and since log λ + log(λ ) 1 = 0, by taking A T and B T 1, we get − i i − ≡ ≡ − 1 log(T− ) = log T. (1.309) − If R Orth+ is such that it does not have any eigenvalues on the closed negative real axis, i.e., if∈R Orth+/Sym + I , then log R Skw. This is proved by noting that ∈ { } ∈ T T 1 [log R] = log(R ) = log(R− ) = log(R), − where the last step follows from Eqn. (1.309).

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We now discuss the explicit determination of the logarithm of a tensor; an alternative method may be found in [34]. Analogous to Eqn. (J.38), one can prove that

k " mi 1 j# 1 1 − ηNi [I + (T I)η]− = P + − . − ∑ [1 + (λ 1)η] i ∑ [1 + (λ 1)η] i=1 i − j=1 i − Substituting this relation into Eqn. (1.304), we get

k ( m 1 j 1 j 1 j ) Z ξ λ 1 i− ( 1) − η − N [I + (T I) ] = i P + − i d log ξ ∑ − i ∑ j+1 η − 0 [1 + (λi 1)η] [1 + (λ 1)η] i=1 − j=1 i − k ( mi 1 j 1 j j ) − ( 1) − ξ Ni = log[1 + (λi 1)ξ]Pi + − . (1.310) ∑ − ∑ j[1 + (λ 1)ξ]j i=1 j=1 i − In particular, for ξ = 1, we get

k ( m 1 j 1 j ) i− ( 1) N − − i log T = ∑ (log λi)Pi + ∑ j . (1.311) i=1 j=1 jλi An alternative derivation of Eqn. (1.310) can be given as follows. Since m 1 m 1 P , P ,..., P , N ,..., N 1− ,..., N ,..., N k− { 1 2 k 1 1 k k } constitutes a basis for any tensor, we write

k " mi 1 # − j log[I + (T I)ξ] = ∑ fi(ξ)Pi + ∑ fij(ξ)(Ni) , − i=1 j=1

where the fi(ξ) and fij(ξ) are functions to be determined. Substituting into the definition given by Eqn. (1.303), and using the linear independence of the basis, we get for each i the equations [1 + (λ 1)ξ] f˙ = (λ 1), i − i i − [1 + (λ 1)ξ] f˙ + ξ f˙ = 1, i − i1 i [1 + (λ 1)ξ] f˙ + ξ f˙ = 0, i − i2 i1 . . ˙ ˙ [1 + (λi 1)ξ] fi(m 1) + ξ fi(m 2) = 0, − i− i− which are to be solved subject to the condition that fi(0) = fij(0) = 0. We first solve for fi using the first equation, then for fi1 using the second one, and so on. The solution is given j 1 j j by fi = log[1 + (λi 1)ξ] and fij = ( 1) − ξ /[j(1 + (λi 1)ξ) ], which yields Eqn. (1.310). For a diagonalizable− tensor T, Eqn.− (1.311) simplifies− to

k log T = ∑(log λi)Pi, (1.312) i=1

with Pi given by Eqn. (J.26). By specializing Eqn. (1.311) to the case n = 3, we get the following analogue of Theorem 1.10.5:

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Theorem 1.10.7. For n = 3, we have the following cases: 1. λ = λ = λ λ 1 2 3 ≡ (a) If q(T) = (T λI), then − log T = (log λ)I.

(b) If q(T) = (T λI)2, then − T λI log T = (log λ)I + − . λ

2. λ µ = λ = λ λ 1 ≡ 6 2 3 ≡ (a) If q(T) = (T λI)(T µI), then − − T λI T µI log T = log µ − + log λ − . µ λ λ µ − − (b) If q(T) = (T µI)(T λI)2, then − − (T λI)2 (T µI)(T (2λ µ)I) (T λI)(T µI) log T = log µ − log λ − − − + − − . (λ µ)2 − (λ µ)2 λ(λ µ) − − −

3. λ1 λ = λ2 µ = λ3 ν = λ1. We have q(T) = (T λI)(T µI)(T νI)≡, and6 ≡ 6 ≡ 6 − − −

(T µI)(T νI) (T λI)(T νI) (T λI)(T µI) log T = log λ − − + log µ − − + log ν − − . (λ µ)(λ ν) (µ λ)(µ ν) (ν λ)(ν µ) − − − − − −

Some examples are    1  e 1 0 1 e 0     T = 0 e 0 , log T = 0 1 0 , (Case 1(b)), 0 0 e 0 0 1    1 2e 1  e 1 1 1 e 2−e2 T = 0 e 1 , log T =  1  , (Case 1(c)),   0 1 e  0 0 e 0 0 1    1 e3 e2 1  e 1 1 1 e(e 1) e3(−e 1−)2 − T = 0 e2 1  , log T =  1−  , (Case 2(b)).   0 2 e2  0 0 e2 0 0 2

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+ iα iα As an application, the logarithm of R Orth /Sym, with eigenvalues 1, e , e− (and hence tr R = 1 + 2 cos α) is obtained using∈ Case 3 in the above theorem as{ }

log R = iα(P P ) 2 − 3 iα iα iα(R I) R e− I R e I = iα − iα −iα + −iα e e− e 1 e− 1 α− − − = (R RT) 2 sin α − R RT = √2α − . R RT − By substituting W log R ( w = α) into the right-hand side of Eqn. (1.289), we can easily verify that elog R = ≡R. | | By differentiating Eqn. (1.304) with respect to T and using Eqns. (1.233) and (1.236), we get (for any underlying space dimension n)

Z ξ ∂ h 1 Ti log(I + (T I)ξ) = (I + (T I)η)− (I + (T I)η)− dη. ∂T − 0 − −

For a diagonalizable tensor, the above equation simpliﬁes to

∂ log(I + (T I)ξ) = ∂T − k k k ξ log[1 + (λi 1)ξ] log[1 + (λj 1)ξ] P PT + − − − P PT, ∑ 1 + (λ 1)ξ i i ∑ ∑ λ λ i j i=1 i − i=1 j=1 i − j j=i 6 a result that we could also have obtained directly by differentiating Eqn. (1.312) using Eqn. (1.255). By setting ξ = 1 in the above equation, we get

k k k ∂ 1 log λi log λj log T = P PT + − P PT. ∂T ∑ λ i i ∑ ∑ λ λ i j i=1 i i=1 j=1 i − j j=i 6 For A Sym, ∈

" k k k # ∂ 1 log λi log λj log A = S P P + − P P S, k > 1, ∂A ∑ λ i i ∑ ∑ λ λ i j i=1 i i=1 j=1 i − j j=i 6 1 = S λ = λ = λ λ, λ 1 2 3 ≡ (1.313)

with Pi given by Eqn. (J.4). An application is presented in Section 2.4.

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1.11 Divergence and Stokes’ Theorems

We state the divergence, Stokes’, potential and localization theorems that are used quite frequently in the following development. The divergence theorem relates a volume integral to a surface integral, while the Stokes’ theorem relates a contour integral to a surface integral. Let S represent the surface of a volume V, n represent the unit outward normal to the surface, φ a scalar field, u a vector field, and T a second-order tensor field. Then we have Divergence theorem (also known as the Gauss’ theorem)

Z Z ∇φ dV = φn dS. (1.314) V S

Applying Eqn. (1.314) to the components ui of a vector u, we get Z Z ∇ u dV = u n dS, (1.315) V · S · Z Z ∇ × u dV = n × u dS, V S Z Z ∇u dV = u ⊗ n dS. V S

Similarly, on applying Eqn. (1.314) to ∇ T, we get the vector equation · Z Z ∇ T dV = Tn dS. (1.316) V · S

Note that the divergence theorem is applicable even for multiply connected domains provided the surfaces are closed. Stokes’ theorem Let C be a contour, and S be the area of any arbitrary surface enclosed by the contour C. Then I Z u dx = (∇ × u) n dS, (1.317) C · S · I Z h i u × dx = (∇ u)n (∇u)Tn dS. (1.318) C S · −

In what follows, we assume n to be C0 continuous over the entire surface S (so that ∇ n makes sense). By taking u as w × n in Eqn. (1.317), using (w × n) λ = w (n × λ), where· λ = dx/ds is the unit tangent to C, and then applying Eqn. (1.354),· we get· I Z w (n × λ) ds = [(∇ n)(w n) ∇ w + n (∇w)n] dS. (1.319) C · S · · − · ·

Thus, for a closed surface S, we get Z Z (∇ n)(w n) dS = [I n ⊗ n] : ∇w dS. (1.320) S · · S −

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R By choosing w to be a constant vector c in the above equation, we get c S(∇ n)n dS = 0. Since the choice of c is arbitrary, we obtain · · Z κn dS = 0, (1.321) S where κ = ∇ n is the curvature. For a more detailed discussion of this result, see [22]. By choosing w =·x, w = c × x, w = (x x)c and w = (x ⊗ x)c in Eqn. (1.320), we get · Z κ(x n) dS = 2S, (1.322a) S · Z κ(x × n) dS = 0, (1.322b) S Z Z κ(x x)n dS = 2 [I n ⊗ n] x dS, (1.322c) S · S − Z Z κ(x n)x dS = [3I n ⊗ n] x dS. (1.322d) S · S − R R Note with relation to Eqns. (1.321) and (1.322b) that S n dS and S(x × n) dS are also zero, as can be seen by using the divergence theorem. Potential Theorems 1. Let V be a simply connected region, and let u = ∇φ be a vector field on V. Then ∇ × u = 0. Conversely, if ∇ × u = 0, then there exists a scalar potential φ such that u = ∇φ. We have already proved the forward assertion (see Eqn. (1.269)). To prove the con- x verse, we note from Eqn. (1.317) that H u dx = 0, and thus, the value of R u(y) dy C · x0 · evaluated along any curve in V joining a fixed point x0 and x, just depends on x. Let Z x φ(x) = u(y) dy. x0 · Then u = ∇φ. 2. If T is a tensor field such that ∇ × T = 0, then there exists a vector field u such that T = ∇u. To see this, using the definition of the curl of a tensor, we have ∇ × (TTw) = (∇ × T)w = 0, where w is a constant unit vector. By the above result this implies that there exists a scalar field φ such that TTw = ∇φ, or, alternatively, T = w ⊗ ∇φ = ∇(φw). Letting u = φw, we get the desired result. 3. If T is a tensor field such that ∇ × T = 0, and in addition, if tr T = 0, then there exists W Skw such that T = ∇ × W. To prove∈ this, we take the trace of the relation T = ∇u to get ∇ u = 0. Let W T = W be the skew-symmetric tensor whose axial vector is u. Then by· Eqn. (1.272), we get− T = ∇u = ∇ × W. 4. If v and T are vector and tensor fields such that Z v n dS = 0, S · Z TTn dS = 0, S

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for every closed surface S in V, then there exist vector and tensor fields u and U such that v = ∇ × u and T = ∇ × U, respectively. To prove the second statement from the first, take the dot product of the equation R TTn dS = 0 with a constant unit vector w. Then, by the definition of the transpose, S R we get S Tw n dS = 0, which by virtue of the first result implies the existence of a vector field u such· that Tw = ∇ × u, or, alternatively, T = (∇ × u) ⊗ w = ∇ × U, where U = w ⊗ u. If the region V does not contain any ‘holes’ in the interior (i.e., the boundary of V is comprised of only the outer surface), then by means of the divergence theorem, we R conclude that S v n dS = 0 for every closed surface S is equivalent to ∇ v = 0. Thus, for a vector field· v that satisfies ∇ v = 0 on such a body, there exists· a vector field u such that v = ∇ × u. · Now we state the localization theorem, which is used in Chapter 3 to obtain the differential equations from the integral form of the governing equations.

Theorem 1.11.1 (Localization theorem). Let φ be a continuous scalar, vector or tensor ﬁeld on V. Then for any given x V, 0 ∈ 1 Z φ(x0) = lim φ dV, r 0 V(Br) B → r where B is the closed ball of radius r > 0 centered at x , and V(B ) denotes its volume. r R 0 r It follows that if B φ dV = 0 for every closed ball B, then φ = 0. Proof. We have

1 Z 1 Z φ(x0) lim φ dV = [φ(x0) φ(x)] dV − r 0 V(Br) B V(Br) B − → r r 1 Z φ(x0) φ(x) dV ≤ V(Br) Br | − |

sup φ(x0) φ(x) , ≤ x B | − | ∈ r which tends to zero as r 0 since φ is continuous. → R Note that there is no localization theorem for surfaces, i.e., S φ dS = 0 for every closed surface S does not imply that φ = 0. To see this, take, for example, φ = a n, where a is a constant vector. ·

1.12 Groups

The treatment in this section is based on [35]. A group is a set, say , equipped with a function from , called combination and denoted by G G × G (a, b) a b, → ∗

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which satisﬁes the following axioms:

1. Associativity: For all a, b, c ∈ G (a b) c = a (b c). ∗ ∗ ∗ ∗ 2. Existence of a neutral element: There exists n such that ∈ G n a = a n = a a . ∗ ∗ ∀ ∈ G r 3. Existence of reverse elements: For each a , there exists a such that ∈ G ∈ G r r a a = a a = n. ∗ ∗ If, in addition, the commutative property

a b = b a a, b ∗ ∗ ∀ ∈ G is satisfied, then is said to be a commutative or Abelian group. Since the combinationG function in the definition of a group may be thought of as taking two elements from and producing an element also in , it is often referred to as a closed binary operation. TheGclosure property, i.e., the fact that theG function value a b is also in , is often called the fundamental closure property. ∗ G As an example, the set of real numbers is a group with respect to addition. We write a + b instead of a b, i.e., the combination function is defined on to by ∗ < × < < (a, b) a + b. → The three defining axioms are satisfied since

(a + b) + c = a + (b + c) a, b, c , ∀ ∈ < 0 + a = a + 0 a , ∀ ∈ < ( a) + a = a + ( a) = 0 a . − − ∀ ∈ < Thus, 0 is the neutral element, and reverse elements are the negatives. In fact, is a commutative group since <

a + b = b + a a, b . ∀ ∈ < We have the following useful results:

Theorem 1.12.1. Let be a group. Then for any a, b , there exists a unique x , G r ∈ G ∈ G such that a x = b. In fact, x = a b. Similarly, there exists a unique y such that ∗ r ∗ ∈ G y a = b. In fact, y = b a. ∗ ∗

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Proof. Suppose, tentatively, that there exists a x such that a x = b. Then r ∈ G ∗ combining on the left with a, we have

r r a b = a (a x) ∗ ∗ ∗ r = (a a) x (associativity) ∗ ∗ = n x (reverse) ∗ = x. (neutral)

r Thus, if x exists, it must be precisely a b. This establishes the uniqueness. To prove existence, one must show that this∗ x has the desired property. First of all, r we note that by closure a b . Next, consider ∗ ∈ G r r a (a b) = (a a) b (associativity) ∗ ∗ ∗ ∗ = n b (reverse) ∗ = b. (neutral)

r Hence, x = a b does the job. The result for y can be proved similarly. ∗

Theorem 1.12.2 (Cancellation Property). Let be a group. Then for a, b, c G ∈ G b a = c a = b = c. ∗ ∗ ⇒ a b = a c = b = c. ∗ ∗ ⇒ Proof. We have

r r (b a) a = (c a) a. ∗ ∗ ∗ ∗ Using the associativity, reverse and neutral properties, we get

b = c.

The second statement can be proved in a similar manner.

Theorem 1.12.3. A given group has only one neutral element.

Proof. Suppose n and n0 are both neutral elements for a group , i.e., n, n0 and for all a G ∈ G ∈ G a n = n a = a and n0 a = a n0 = a. ∗ ∗ ∗ ∗ Consider a n = a. We have ∗ 0 r n0 = a a = n. ∗

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Theorem 1.12.4. A given element of a group has only one reverse.

r r Proof. Let a . Suppose a and a0 are both reverses of a, i.e., ∈ G r r r r a a = a r = n and a0 a = a a0 = n. ∗ ∗ ∗ ∗ r Consider a a0 = n. We have ∗ r r r a0 = a n = a. ∗

Theorem 1.12.5. Let n be a neutral element of a group. Then

r n = n. r Proof. For every element a , the reverse axiom requires a a = n. Taking r ∈ G ∗ a = n, we get n n = n. Hence, ∗ r r n = n n = n. ∗

Theorem 1.12.6. Let a be any element of a group. Then

r r a = a. r Proof. We have a a = n. Hence, ∗ r r r r a = a n = a. ∗

Theorem 1.12.7. Let a and b be any two elements of a group. Then

r r r a b = b a. ∗ ∗ Proof. We have

r r r r (a b) (b a) = a (b (b a)) (associativity) ∗ ∗ ∗ ∗ ∗ ∗ r r = a ((b b) a) (associativity) ∗ ∗ ∗ r = a (n a) (reverse) ∗ ∗ r = a a (neutral) ∗ = n. (reverse)

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In the same way

r r (b a) (a b) = n. ∗ ∗ ∗ r r Thus, b a is a reverse element of a b. By the uniqueness of the inverse, it is the reverse.∗ ∗

In general, subsets of groups will not be groups, e.g., the subset n . We introduce the following terminology: G − { } Deﬁnition: Let be a nonempty [proper] subset of a group . Then, if is a group with respect to the combinationH function of , it is called a [proper]G subgroupH of .4 Since , the combinationG function for is the restriction ofG the combination functionH × for H ⊂ Gto × G . However, the function values,H which necessarily belong to , are not automaticallyG H× in H , i.e., for an arbitrary subset, closure could fail. An example ofG a subgroup of a group Hwith neutral element n is the singleton set n . The following theorem provides a sufﬁcientG condition for a subset to be a subgroup. { }

r Theorem 1.12.8. Let be a nonempty subset of a group . If for all a, b , a b , then is a subgroupH of . G ∈ H ∗ ∈ H H G Proof. As noted, the combination function for makes sense for . Hence, the associativity requirement is automatically satisﬁed.G We have toH show that the neutral and reverse elements of all a lie in , and that is closed under the combination operation. ∈ H H H Assume that for a, b ∈ H r a b . (1.323) ∗ ∈ H r r Choosing b = a, we get a a . But a a = n. Thus, n . ∗ ∈ H ∗ r ∈ Hr r Next, choose a = n. Then, we have n b . But n b = b. Thus, b r ∗ ∈ H ∗ ∈ H implies b . So contains the reverses of all its elements. The reverses necessarily∈ exist H becauseH of the properties of the ‘parent’ group . r r G Finally, let a, b . We have shown that b . Letting b play the role of b in Eqn. (1.323), we have∈ H ∈ H

r r a b . ∗ ∈ H r r But b = b. Thus, a b , and this proves that is closed under the combination operation. ∗ ∈ H H

4The bracket device above is used to make two statements simultaneously. To get the ﬁrst statement, include the bracketed material. To get the second leave out the bracketed material.

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We now come to a result that plays a key role in the classiﬁcation of materials. First, we deﬁne the set of unimodular tensors Unim, and the set of proper unimodular tensors Unim+ as Unim : = H : det H = 1 , { | | } Unim+ : = H : det H = 1 . { }

Theorem 1.12.9. Unim and Unim+ are groups with respect to tensor multiplication, and Unim+ is a proper subgroup of Unim. Hence, Unim is called the unimodular group of tensors, while Unim+ is called the proper unimodular group.

Proof. Closure holds since H , H Unim implies that det(H H ) = 1 2 ∈ | 1 2 | det H1 det H2 = 1. Associativity follows since tensor multiplication is as- sociative.| | | The reverse| element is simply the inverse, while the neutral element is the identity tensor. Similar arguments can be made for Unim+. Unim+ is a proper subgroup of Unim, since there are elements in Unim which are not in Unim+, e.g., diag[ 1, 1, 1]. −

Theorem 1.12.10. The orthogonal group Orth is a proper subgroup of the unimodular group Unim, and the proper orthogonal group Orth+ is a proper subgroup of the proper unimodular group Unim+.

Proof. The closure property of Orth follows since Q , Q Orth implies that 1 2 ∈ Q1Q2 Orth. The other properties are proved as in the previous proof. Orth is a proper∈ subgroup since there are unimodular tensors that are not orthogonal. e.g.,   1 0 0   1 1 0 . 0 0 1

A similar proof can be given for Orth+.

Before stating the next theorem, we prove the following lemma: + Lemma: Let S0 Psym Unim , and let it have at least two distinct eigenvalues, say λu ∈ ∩ h 2 2i λl λu + and λl, with λu > λl. Then, for every ξ , , there exists R Orth such ∈ λu λl ∈ that the symmetric and proper unimodular tensor

1 2 1 1 T = S0− RS0R− S0− has the spectrum ξ, ξ 1, 1 . { − } Proof. Since S Psym Unim+, all the eigenvalues of S are positive, and we can write 0 ∈ ∩ 0 the spectral resolution of S0 as 1 S0 = λue1∗ ⊗ e1∗ + λle2∗ ⊗ e2∗ + e3∗ ⊗ e3∗. λlλu

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Let R(θ) be the usual rotation matrix about the e3∗ axis, i.e.,   cos θ sin θ 0 R(θ) =  sin θ cos θ 0 . −  0 0 1

In terms of its action on the basis e , e , e , we have { 1∗ 2∗ 3∗}

R(θ)e1∗ = cos θe1∗ + sin θe2∗, R(θ)e∗ = sin θe∗ + cos θe∗, 2 − 1 2 R(θ)e3∗ = e3∗.

Note that R 1(θ) = RT(θ) = R( θ). Now, consider − − 1 2 1 1 T(θ) = S0− R(θ)S0R− (θ)S0− . (1.324) That T(θ) is symmetric and proper unimodular follows from straightforward calculations. Hence, its eigenvalues are all real. Moreover, +1 is an eigenvalue of T(θ) for all θ since

1 2 1 T(θ)e∗ = S− R(θ)S R( θ)S− e∗ 3 0 0 − 0 3 1 2 = λ λ S− R(θ)S R( θ)e∗ l u 0 0 − 3 1 2 = λlλuS0− R(θ)S0e3∗ 1 1 = S0− R(θ)e3∗ λlλu 1 1 = S0− e3∗ λlλu

= e3∗.

For θ = 0, we have R = I, which when substituted into Eqn. (1.324) yields T = I, i.e., the spectrum of T(0) is 1, 1, 1 . At θ = π/2, we have R(π/2)e = e , R(π/2)e = e , and { } 1∗ 2∗ 2∗ − 1∗ R(π/2)e3∗ = e3∗. Hence,

π 1 π 2 π 1 T e∗ = S− R S R S− e∗ 2 1 0 2 0 − 2 0 1 1 1 π 2 π = S0− R S0R e1∗ λu 2 − 2 1 1 π 2 = S0− R S0e2∗ − λu 2 2 λl 1 π = S0− R e2∗ − λu 2 2 λl 1 = S0− e1∗ λu 2 λl = 2 e1∗. λu

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2 n 2 2 o λu λl λu Similarly, T(π/2)e2∗ = 2 e2∗. Thus, the ordered spectrum of T(π/2) is 2 , 1, 2 . λl λu λl Since +1 is always an eigenvalue, and since T(θ) depends continuously on θ, we see that h 2 2 i λl λu given any ξ 2 , 2 , there exists θ [0, π/2] for which the lemma holds. ∈ λu λl ∈

Thus, if ei0 are the eigenvectors of T, then the spectral resolution of T is 1 T = ξe0 ⊗ e0 + e0 ⊗ e0 + e0 ⊗ e0 , (1.325) 1 1 ξ 2 2 3 3

h 2 2 i λl λu where ξ 2 , 2 . Now we are in a position to prove the main result of this ∈ λu λl section. The proof that we present is due to Noll (see [237], page 200).

Theorem 1.12.11. (Maximality of the Orthogonal Group in the Unimodular Group). If is a group with respect to tensor multiplication such that G Orth Unim, ⊂ G ⊂ then either = Orth or = Unim. In other words, there is no group between the orthogonal groupG and the unimodularG group. The corresponding result for Orth+ and Unim+ is that Orth+ is maximal in Unim+. Proof. There are two possibilities: 1. contains a tensor S Psym Unim with at least two distinct eigenval- G 0 ∈ ∩ ues, say λu > λl. 2. does not contain such a tensor, i.e., the only tensor from Psym Unim thatG it contains is the identity tensor. ∩ Consider the ﬁrst possibility. Using the polar decomposition, any tensor H Unim can be decomposed as H = QU, where Q is an orthogonal tensor and∈ U Psym Unim. Hence, ife ˆi are the eigenvectors of U, we can write the spectral∈ resolution∩ of U as

1 U = γ1eˆ1 ⊗ eˆ1 + γ2eˆ2 ⊗ eˆ2 + eˆ3 ⊗ eˆ3, γ1γ2

where γ1, γ2 > 0. We can decompose U as

U = U1U2,

where 1 U1 = γ1eˆ1 ⊗ eˆ1 + eˆ2 ⊗ eˆ2 + eˆ3 ⊗ eˆ3, γ1 1 U2 = eˆ1 ⊗ eˆ1 + γ1γ2eˆ2 ⊗ eˆ2 + eˆ3 ⊗ eˆ3. γ1γ2

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1/m1 Now choose integers m1, m2 large enough so that ξ1 := γ1 and ξ2 := 1/m (γ1γ2) 2 satisfy

2 ! 2 ! λl λu 2 ξi 2 . λu ≤ λl

(Choosing m ln γ /[2 ln(λu/λ )] and m ln(γ γ )/[2 ln(λu/λ )] does 1 ≥ | 1 l | 2 ≥ | 1 2 l | the job.) By the Lemma, there exist tensors T1 and T2 whose spectral resolution is given by

1 T1 = ξ1e¯1 ⊗ e¯1 + e¯2 ⊗ e¯2 + e¯3 ⊗ e¯3, ξ1 1 T2 = e10 ⊗ e10 + ξ2e20 ⊗ e20 + e30 ⊗ e30 . ξ2 Thus,

m1 1 T1 = γ1e¯1 ⊗ e¯1 + e¯2 ⊗ e¯2 + e¯3 ⊗ e¯3, γ1

m2 1 T2 = e10 ⊗ e10 + γ1γ2e20 ⊗ e20 + e30 ⊗ e30 . γ1γ2

Let R1 and R2 be the two proper orthogonal tensors that rotatee ¯i and ei0 intoe ˆi. Then, we have

m1 T 1 T R1T1 R1 = R1(γ1e¯1 ⊗ e¯1 + e¯2 ⊗ e¯2 + e¯3 ⊗ e¯3)R1 γ1 1 = γ1(R1e¯1) ⊗ (R1e¯1) + (Re¯2) ⊗ (R1e¯2) + (R1e¯3) ⊗ (R1e¯3) γ1 1 = γ1eˆ1 ⊗ eˆ1 + eˆ2 ⊗ eˆ2 + eˆ3 ⊗ eˆ3 γ1

= U1.

Similarly,

m2 T R2T2 R2 = U2. Hence,

H = QU

= QU1U2 m1 T m2 T = Q R1T1 R1 R2T2 R2

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m m h 1 2 1 1i 1 T h 1 2 1 1i 2 T = QR1 S0− R(θ1)S0R− (θ1)S0− R1 R2 S0− R(θ2)S0R− (θ2)S0− R2 , (1.326)

for some θ1, θ2 [0, π/2]. Thus, any H Unim can be expressed in terms of powers of S ∈Psym Unim, Q Orth,∈ and R(θ ), R(θ ), R , R Orth+. But 0 ∈ ∩ ∈ 1 2 1 2 ∈ Orth and S0 , and since is a group, any element generated by powers of S ⊂and G elements∈ Gof Orth has toG lie in . Thus, we have Unim . But, by 0 G ⊂ G hypothesis, Unim. Hence, for the case when S0 has at least two distinct eigenvalues G= ⊂ Unim. Now considerG the case when the only member from Psym Unim in is the ∩ G 1 identity tensor. For any H , we have H = QU, or, alternatively, U = Q− H, where Q Orth , and∈U G Psym Unim. Since is a group, and since ∈ ⊂ G 1 ∈ ∩ G H, Q , we have U = Q− H . However, since the only member from Psym ∈Unim G in is I, we have U ∈= GI. Therefore, H = Q Orth, i.e., Orth. But, by∩ hypothesis,G Orth . Hence, in this case, we have∈ = Orth. G ⊂ To prove that Orth+ is⊂ maximal G in Unim+, note that if HG Unim+, then Q in Eqn. (1.326) belongs to Orth+. Thus, any H Unim+ can be∈ expressed in terms of powers of S Psym Unim and tensors∈ in Orth+, and the result follows. 0 ∈ ∩

EXERCISES

1. Show that

(w × u) × (w × v) = [w ⊗ w](u × v). (1.327) [u × v, v × w, w × u] = [u, v, w]2 . (1.328)

From Eqn. (1.327), it follows that if n is a unit vector, then

[n, n × u, n × v] = [n, u, v] .

2. Which of the spaces Sym, Psym, Skw, Orth+ are linear subspaces of Lin? Justify. Evaluate if the matrices       0 1 1 0 0 1 0 1 0 − −  1 0 0  ,  0 0 1 , 1 0 1 , −   −    1 0 0 1 1 0 0 1 0 − − constitute a basis for Skw, and if the matrices       1 1 0 1 0 1 0 0 0       1 1 0 , 0 0 0 , 0 1 1 , 0 0 0 1 0 1 0 1 1

constitute a basis for Sym. If not, then give a ‘canonical’ basis for Skw and Sym. Also give a basis for deviatoric symmetric tensors.

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n 3. Let pi and qi , i = 1, 2, . . . , n, be sets of vectors in . Determine if the set pi ⊗ q ,{i =}1, 2, . .{ . , n}, is linearly dependent or independent< if { i} (a) p is linearly independent, and q is linearly independent; { i} { i} (b) p is linearly independent, but q is not; { i} { i} (c) q is linearly independent, but p is not; { i} { i} (d) p is linearly dependent, and q is also linearly dependent. { i} { i} 4. Let W Skw. Determine if the set I, W, W2 is linearly dependent or linearly ∈ { } independent. Deduce from this result and Eqn. (1.105a) if I, Q, QT , where Q Orth+, is linearly dependent or independent. { } ∈ 5. Show that the decomposition of a tensor into a symmetric and skew-symmetric part as given by Eqn. (1.33) is unique.

6. Show that tr RS = tr SR = tr RTST = tr ST RT. 7. If S and W are symmetric and skew-symmetric tensors, respectively, and T is an arbitrary second-order tensor prove that

1 S : T = S : TT = S : (TT + T) , (1.329) 2 1 W : T = W : TT = W : (T TT) , − 2 − S : W = 0. (1.330)

8. Prove the following: (i) If A : B = 0 for every symmetric tensor B, then A Skw. (ii) If A : B = 0 for every skew tensor B, then A Sym.∈ ∈ 9. Using Eqns. (1.35) and (1.52) prove that

cof (RS) = (cof R)(cof S), det(RS) = (det R)(det S).

10. Using indicial notation prove that 1 h i tr (cof T) = (tr T)2 tr (T2) , (1.331) 2 − [(cof T)u] × v = T(u × TTv), (1.332) cof (cof T) = (det T)T.

11. Show that an arbitrary tensor T cannot be represented as

αa ⊗ b + βb ⊗ c + γc ⊗ a,

by showing, in particular, that the identity tensor I cannot be represented in this way.

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12. Let T = u ⊗ p + v ⊗ q + w ⊗ r. Show using Eqn. (1.60) that

cof T = (u × v) ⊗ (p × q) + (v × w) ⊗ (q × r) + (w × u) ⊗ (r × p). (1.333)

Using Eqn. (1.64), deduce that

(u × v) ⊗ w + (v × w) ⊗ u + (w × u) ⊗ v = [u, v, w] I.

Using Eqn. (1.53), show that Eqn. (1.328) is recovered for the case [u, v, w] = 0 (Eqn. (1.328) holds even when [u, v, w] = 0). Note that the principal invariants6 of T are

I = tr T = u p + v q + w r, 1 · · · I = tr (cof T) = (u × v) (p × q) + (v × w) (q × r) + (w × u) (r × p), 2 · · · I3 = det T = [u, v, w][p, q, r] ,

1 T and that T− (when it exists) is given by (cof T) / det T.

13. Let λi, i = 1, 2, , n, denote the eigenvalues of T for space-dimension n. Using ··· n Eqn. (J.11), show that the eigenvalues of cof T are ∏j=1 λj, i = 1, 2, . . . , n. For the j=i 6 2 case n = 3, deduce Eqn. (1.331), I2(cof T) = tr T(det T) and I3(cof T) = (det T) . Now consider the case when T is diagonalizable with distinct eigenvalues, so that T = ∑n λ P (thus, T γI = ∑n (λ γ)P ; tr P = e e = 1). Using Eqn. (J.11), i=1 i i − i=1 i − i i ∗ · ∗ n n T n h n i T show that cof T = ∑i=1 ∏j=1 λj Pi , and cof (T γI) = ∑i=1 ∏j=1(λj γ) Pi . j=i − j=i − 6 6 Thus, for γ = λ say, cof (T λ I) = ∏n (λ λ )PT, and tr cof (T λ I) = 1 − 1 j=2 j − 1 1 − 1 ∏n (λ λ ). j=2 j − 1 14. If T, R and S are second-order tensors, then show that 1 1 det T = tr [(cof T)T T] = T : cof T, (1.334) 3 3 det(R + S) = det R + cof R : S + R : cof S + det S, (1.335) cof (R + S) = cof R + cof S + [(tr R)(tr S) tr (RS)] I (tr R)ST (tr S)RT − − − + (SR)T + (RS)T. (1.336)

By putting R = I in Eqn. (1.336), deduce that

cof (I + S) = (1 + tr S)I ST + cof S. (1.337) − By putting S = u ⊗ v in the above equation and using Eqn. (1.84), deduce that

cof (I + u ⊗ v) = (1 + u v)I v ⊗ u. · − Using Eqns. (1.335) and (1.337), show that for W Skw, ∈ 1 I W + w ⊗ w (I + W)− = − . 1 + w 2 | |

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15. Apply the Cayley–Hamilton theorem to W Skw, and ﬁnd the axial vector of W3. ∈ 16. If w is the axial vector of W Skw, show that w [u × Wu] 0 u, with equality if and only if (u, w) are linearly∈ dependent. · ≥ ∀ 17. Let W Skw, and let w be its axial vector. Using Eqn. (1.332), show that (cof T)w is the axial∈ vector of TWTT. If T Q Orth+, then deduce using Eqn. (1.93) that Qw ≡ ∈ is the axial vector of QWQT. Also, ﬁnd the axial vector of QW3QT. 18. Show that a (not necessarily symmetric) tensor T commutes with every skew tensor W if and only if T = λI. 19. Either using the result of the previous problem, or independently, show that a (not necessarily symmetric) tensor T commutes with every orthogonal tensor Q if and only if T = λI. 20. Show that a (not necessarily symmetric) tensor T commutes with every symmetric tensor S if and only if T = λI. 21. Show that (u, Tv) = 0 u, v that are mutually orthogonal, if an only if T = λI. ∀ 22. Let a, b be arbitrary vectors, and u, v be unit vectors such that u v = 1. · 6 − (a) Show that a × b is the axial vector of the skew tensor b ⊗ a a ⊗ b. It follows that if b ⊗ a = a ⊗ b, then a × b = 0, so that by Theorem− 1.2.1, a and b are linearly dependent. (b) Using (a) and Eqn. (1.104), show that the unique orthogonal tensor that rotates u into v about an axis perpendicular to u and v is given by

R = I + rv ⊗ u s(u ⊗ u + u ⊗ v + v ⊗ v). − where r = (1 + 2u v)/(1 + u v) and s = 1/(1 + u v). · · · 23. If w is the axial vector of W Skw, then show using Eqns. (1.54) and (1.327) (or Eqns. (1.60) and (1.90)) that cof∈W = w ⊗ w. Deduce using Eqn. (1.273) that

2 (e w 1) ecof W = I + | | − w ⊗ w. w 2 | | tr (cof W) w 2 The determinant of the above tensor is e = e| | .

24. Let W1 and W2 be skew-symmetric tensors with axial vectors w1 and w2. Show using Eqns. (1.21a) and (1.87) that

W W = w ⊗ w (w w )I, (1.338) 1 2 2 1 − 1 · 2 from which it follows that W W W W = w ⊗ w w ⊗ w . By using the 1 2 − 2 1 2 1 − 1 2 result of Problem 22a (or, independently, by means of Eqns. (1.21)), we see that w1 × w is the axial vector of W W W W . If w = αw , where α , then, by 2 1 2 − 2 1 1 2 ∈ < Eqn. (1.90), W1 = αW2, and hence W1W2 = W2W1. Conversely, if W1W2 = W2W1, then w × w , the axial vector of W W W W , is 0, and by Theorem 1.2.1, it 1 2 1 2 − 2 1

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follows that w1 and w2 are linearly dependent. Thus, two skew-symmetric tensors W1 and W2 commute if and only if their axial vectors are linearly dependent. It also follows from Eqn. (1.338) that

W W + W W = w ⊗ w + w ⊗ w 2(w w )I. 1 2 2 1 1 2 2 1 − 1 · 2 The eigenvalues/eigenvectors of the above symmetric tensor are

w1 w2 ( w1 w2 w1 w2), + , | | | | − · w1 w2 |w | |w | ( w w + w w ), 1 2 , − | 1| | 2| 1 · 2 w − w | 1| | 2| 2(w w ), w × w . − 1 · 2 1 2 One can now easily compute the principal invariants of W1W2 + W2W1 using these eigenvalues. 25. For W Skw, ﬁnd the polar decomposition of I + W in terms of W and its axial vector w∈. 26. Prove that there is a one-to-one correspondence between Skw and members of Orth+ with no eigenvalue equal to 1,5 by showing that − 1 + (a) For every W Skw, Q = (I W)− (I + W) Orth . If w is the axial vector of W, then for∈ n = 3, show− using Eqn. (1.66)∈ and Problem 14 that the above expression simpliﬁes to 1 h i Q = (1 + w w)I + 2W + 2W2 , (1.339) 1 + w w · · with w/ w as the axis. In expanded form, if | |   0 γ β − W =  γ 0 α ,  −  β α 0 − then   1 + α2 β2 γ2 2(αβ γ) 2(β + αγ) 1 − − − Q =  2(αβ + γ) 1 α2 + β2 γ2 2(α βγ)  . 1 + α2 + β2 + γ2  − − − −  2( β + αγ) 2(α + βγ) 1 α2 β2 + γ2 − − − 5For members of Orth+ with at least one eigenvalue equal to 1, the correspondence when the dimension n is three, is given by − Q = I + 2W2, where W is of the form   0 sin θ cos θ sin φ − W =  sin θ 0 cos θ cos φ .  −  cos θ sin φ cos θ cos φ 0 −

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By comparing Eqn. (1.104) with Eqn. (1.339), we see that sin α 2 w /(1 + ≡ | | w 2) and cos α = (1 w 2)/(1 + w )2, which corresponds to letting α [0, π) | | − | | | | ∈ in Eqn. (1.104). If w1/ w1 and w2/ w2 are the axes of R1 and R2, respectively, then, from Eqn. (1.107)| and| the above| expressions| for cos α and sin α, the axis of R R is e = w/ w , where 2 1 | | w = w + w w × w . 1 2 − 1 2 (b) For every Q Orth+ such that 1 is not an eigenvalue of Q, W = (Q I)(Q +∈I) 1 Skw. − − − ∈ Thus, the number of parameters required to deﬁne an orthogonal tensor is the number of independent components in a skew-symmetric matrix, namely n(n 1)/2; e.g., for n = 3, the number of parameters is 3. − 27. If u is an arbitrary vector, show that there exists W Skw and a vector v (both dependent on u) such that Wv = u. Use this result to show∈ that if WT = WT W 1 2 ∀ ∈ Skw, then T1 = T2. 28. Show that a scalar-valued function φ : Skw is isotropic if and only if there exists a function φ˜ : tr (W2) such that → < → < φ(W) = φ˜(tr W2) W Skw. ∀ ∈ The above result follows directly as a corollary of Theorem 1.6.13, or can be proved independently by using the representation given by Eqn. (1.89), and mimicking the proof of Theorem 1.6.12. 29. For S Sym, show using the spectral resolution of S or otherwise that ∈ (u, Su) = 0 u V, ∀ ∈ if and only if S = 0. It follows that (u, Tu) = 0 u V, if and only if T Skw. ∀ ∈ ∈ 30. A set of Cartesian axes is rotated about the origin to coincide with the unit vectors

√3 1 √3 e¯ = ( , , ), 1 4 4 2 3 √3 1 e¯ = ( , , ), 2 4 4 − 2 1 √3 e¯ = ( , , 0). 3 − 2 2 Write down the rotation matrix corresponding to this rotation, and transform the components of the tensor T with respect to e given by { i}   8 4 0 − [T] =  4 3 1 . − − −  0 1 2 − −

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31. Using direct notation show that (also verify using indicial notation)

(a ⊗ b)T = b ⊗ a, (1.340) (a ⊗ b)(c ⊗ d) = (b c)a ⊗ d, (1.341) · (a ⊗ b) : (u ⊗ v) = (a u)(b v), (1.342) · · T(a ⊗ b) = (Ta) ⊗ b, (1.343) (a ⊗ b)T = a ⊗ (TTb), (1.344) T : (a ⊗ b) = a Tb. (1.345) · Use Eqn. (1.341) to show that for a symmetric S, the spectral decompositions of Sn 1 and S− (when it exists) are n n n n S = λ1 e1∗ ⊗ e1∗ + λ2 e2∗ ⊗ e2∗ + λ3 e3∗ ⊗ e3∗, 1 1 1 1 S− = λ1− e1∗ ⊗ e1∗ + λ2− e2∗ ⊗ e2∗ + λ3− e3∗ ⊗ e3∗.

32. The Fibonacci numbers satisfy the recurrence relation Fn+1 = Fn + Fn 1 with F1 = − F2 = 1. In matrix form this recurrence relation can be written as " # " # F F n+1 = S n , Fn Fn 1 − where S = 1 1 . Use the above matrix form to ﬁrst express Fn+1 in terms of F2 , 1 0 Fn F1 and then use this to ﬁnd an explicit expression for Fn in terms of the eigenvalues λ1 and λ2 of S. 33. Let B be an invertible tensor. Show that if n is an eigenvector of A, then Bn is an eigenvector of BAB 1 corresponding to the same eigenvalue. It follows that if Q − ∈ Orth+ and if n is an eigenvector of a tensor T, then Qn is an eigenvector of QTQT corresponding to the same eigenvalue.

34. If the eigenvalues of a symmetric tensor S are ordered such that λ1 λ2 λ3, show using the spectral decomposition of S that ≤ ≤ u Su u Su λ1 = min · , λ3 = max · . u u u u u u · · 35. If T Lin and u V, show that ∈ ∈ (Tu, Tu) (T, T)(u, u). ≤ (Hint: TT T is a symmetric, positive semi-deﬁnite tensor.)

2 36. If R Lin, show that RT R R (Hint: RT R is a symmetric, positive semi-deﬁnite tensor).∈ Use this result and≤ the| Cauchy–Schwartz| inequality to show that if R, S Lin, then ∈

RS R S . | | ≤ | | | |

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As a result of the Cauchy–Schwartz inequality, we also have (R, S) R S . How- ever, among the two quantities (R, S) and RS , either can be| greater.| ≤ | For| | example,| 1 0 1 1 | | | | 1 1 let R = 1 2 ; if S = 0 1 , then (R, S) = 3 and RS = 2, while if S = −0 1 , then (R, S) =−1 and RS = 2. | | | | | | | | n 1 37. Show that S , S− and λ(tr S)I + 2µS are isotropic functions. 38. Show that for any orthogonal Q, the tensor QQ˙ T is skew at each t. 39. Show using Eqn. (1.336) that if G(T) = cof T, then

DG(T)[U] = [tr Ttr U tr (TU)] I + [TT (tr T)I]UT + [UT (tr U)I]TT. − − − (1.346)

Alternatively, using Eqn. (1.60), show that

[DG(T)[U]]ij = eimnejpqTmpUnq, (1.347)

∂Gij = eikmejlnTmn. (1.348) ∂Tkl Equation (1.348) in direct tensorial notation, obtained using Eqns. (1.55), (1.226a), (1.226b) and (1.233), is given by

∂ (cof T) = tr T[I ⊗ I T] (I ⊗ TT + TT ⊗ I) + (I T)T + (TT I)T. ∂T − − (1.349)

If T C Sym, then, by virtue of Eqn. (1.241), we just pre- and post-multiply the above≡ result∈ with S, and use Eqn. (1.175) to get ∂ (cof C) = tr C [I ⊗ I S] (I ⊗ C + C ⊗ I) + S [I C + C I] S. (1.350) ∂C − − The indicial notation form of the above equation, obtained using Eqns. (1.161) and (1.348), is (for the engineering form, see Eqn. (I.6))

∂ 1 h i (cof C) = eikrejls + eilrejks + ejkreils + ejlreiks Crs. (1.351) ∂C ijkl 4

1 Finally, consider the case when C Sym is invertible, i.e., cof C = (det C)C− . Show using Eqns. (1.236) and (1.240) that∈

∂ h 1 1 1 1 i (cof C) = det C C− ⊗ C− S(C− C− )S . (1.352) ∂C −

1 1 T 1 1 1 40. Let φ1(T) = tr (T− T− ) = T− : T− , φ2(T) = (det T)T− : T− , and φ3(T) = (cof T) : (cof T) = (T : T)2 (TT T) : (TT T) /2. Using either Eqns. (1.228)–(1.237) or by means of the directional− derivative, show that

∂φ1 T 3 = 2(T− ) , ∂T −

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136 Continuum Mechanics

∂φ2 1 1 T 1 T = (T− : T− )cof T 2(det T)T− T− T− , ∂T − ∂φ 3 = 2(T : T)T 2TTT T. ∂T −

41. Prove that

∇(φv) = φ∇v + v ⊗ ∇φ, ∇ (φv) = φ(∇ v) + v (∇φ), (1.353) · · · ∇ × (u × v) = (∇ v)u (∇v)u (∇ u)v + (∇u)v, (1.354) · − − · ∇(u v) = (∇u)Tv + (∇v)Tu, · ∇[(u v)w] = (u v)∇w + w ⊗ [(∇v)Tu] + w ⊗ [(∇u)Tv], · · ∇ (u ⊗ v) = u∇ v + (∇u)v, · · ∇ (TTv) = T : ∇v + v (∇ T), (1.355) · · · ∇ (φT) = φ∇ T + T∇φ, · · ∇(φT) = φ∇T + T ⊗ ∇φ, ∇(φTv) = φ∇(Tv) + (Tv) ⊗ ∇φ,

∇2(u v) = u ∇2v + v ∇2u + 2∇u : ∇v, · · · ∇ [a × (∇ × b)] = (∇ × a) (∇ × b) a [∇ × (∇ × b)]. (1.356) · · − · By integrating Eqn. (1.356) over a closed volume V with surface S show that Z Z (∇ × b) (a × n) dS = (∇ × a) (∇ × b) a [∇ × (∇ × b)] dV. − S · V { · − · }

42. Show that

∇ [(∇u)u] = ∇u : (∇u)T + u [∇(∇ u)], · · · ∇u : (∇u)T = ∇ [(∇u)u (∇ u)u] + (∇ u)2. · − · · 43. Show that Z h i Z u ⊗ (∇ TT) + (∇u)T dV = u ⊗ (TTn) dS. V · S

44. Show that 0 is a commutative group with respect to multiplication. < − { }

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