Mathematical Preliminaries

In the introduction, we saw that some of the quantities like force is a vector or first order tensor, stress is a second order tensor or simply a tensor. The algebra and calculus associated with these quantities differs, in some aspects from that of scalar quantities. In this chapter, we shall study on how to manipulate equations involving vectors and tensors and how to take derivatives of scalar valued function of a tensor or tensor valued function of a tensor.

A physical quantity, completely described by a single real number, such as temperature, density or mass is called a scalar. A vector is a directed line element in space used to model physical quantities such as force, velocity, acceleration which have both direction and magnitude. The vector could be denoted as v or v. Here we denote vectors as v. The length of the directed line segment is called as the magnitude of the vector and is denoted by |v|. Two vectors are said to be equal if they have the same direction and magnitude. The point that vector is a geometric object, namely a directed line segment cannot be overemphasized. Thus, for us a set of numbers is not a vector.

The sum of two vectors yields a new vector, based on the parallelogram law of addition and has the following properties:

Then the scalar multiplication αu produces a new vector with the same direction as u if α > 0 or with the opposite direction to u if α < 0 with the following properties:

The dot (or scalar or inner) product of two vectors u and v denoted by u ⋅ v assigns a real number to a given pair of vectors such that the following properties hold:

| (2.12) |

A vector e is called a unit vector if |e| = 1. A nonzero vector u is said to be orthogonal (or perpendicular) to a nonzero vector v if: u ⋅ v = 0. Then, the projection of a vector u along the direction of a vector e whose length is unity is given by: u ⋅ e

So far algebra has been presented in symbolic (or direct or absolute) notation. It represents a very convenient and concise tool to manipulate most of the relations used in continuum mechanics. However, particularly in computational mechanics, it is essential to refer vector (and tensor) quantities to a basis. Also, for carrying out mathematical operations more intuitively it is often helpful to refer to components.

We introduce a fixed set of three basis vectors e_{1}, e_{2}, e_{3} (sometimes introduced
as i, j, k) called a Cartesian basis, with properties:

| (2.13) |

so that any vector in three dimensional space can be written in terms of these three basis vectors with ease. However, in general, it is not required for the basis vectors to be fixed or to satisfy (2.13). Basis vectors that satisfy (2.13) are called as orthonormal basis vectors.

Any vector u in the three dimensional space is represented uniquely by a linear
combination of the basis vectors e_{1}, e_{2}, e_{3}, i.e.

| (2.14) |

where the three real numbers u_{1}, u_{2}, u_{3} are the uniquely determined Cartesian
components of vector u along the given directions e_{1}, e_{2}, e_{3}, respectively. In other
words, what we are doing here is representing any directed line segment as a linear
combination of three directed line segments. This is akin to representing any real
number using the ten arabic numerals.

If an orthonormal basis is used to represent the vector, then the components of the vector along the basis directions is nothing but the projection of the vector on to the basis directions. Thus,

| (2.15) |

Using index (or subscript or suffix) notation relation (2.14) can be written as u
= ∑
_{i=1}^{3} u_{
i}e_{i} or, in an abbreviated form by leaving out the summation symbol,
simply as

| (2.16) |

where we have adopted the summation convention, invented by Einstein. The
summation convention says that whenever an index is repeated (only
once) in the same term, then, a summation over the range of this index is
implied unless otherwise indicated. The index i that is summed over is
said to be a dummy (or summation) index, since a replacement by any
other symbol does not affect the value of the sum. An index that is not
summed over in a given term is called a free (or live) index. Note that in the
same equation an index is either dummy or free. Here we consider only
the three dimensional space and denote the basis vectors by {e_{i}}_{i∈{1,2,3}}
collectively.

In light of the above, relations (2.13) can be written in a more convenient form as

| (2.17) |

where δ_{ij} is called the Kronecker delta. It is easy to deduce the following
identities:

| (2.18) |

The projection of a vector u onto the Cartesian basis vectors, e_{i} yields the i^{th}
component of u. Thus, in index notation u ⋅ e_{i} = u_{i}.

As already mentioned a set of numbers is not a vector. However, for ease of computations we represent the components of a vector, u obtained with respect to some basis as,

| (2.19) |

instead of writing as u = u_{i}e_{i} using the summation notation introduced above.
The numbers u_{1}, u_{2} and u_{3} have no meaning without the basis vectors which are
present even though they are not mentioned explicitly.

If u_{i} and v_{j} represents the Cartesian components of vectors u and v
respectively, then,

The cross (or vector) product of u and v, denoted by u ∧ v produces a new vector satisfying the following properties:

The cross product characterizes the area of a parallelogram spanned by the vectors u and v given that

| (2.26) |

where θ is the angle between the vectors u and v and n is a unit vector normal to the plane spanned by u and v, as shown in figure 2.1.

In order to express the cross product in terms of components we introduce the
permutation (or alternating or Levi-Civita) symbol ϵ_{ijk} which is defined
as

| (2.27) |

with the property ϵ_{ijk} = ϵ_{jki} = ϵ_{kij}, ϵ_{ijk} = -ϵ_{ikj} and ϵ_{ijk} = -ϵ_{jik}, respectively.
Thus, for an orthonormal Cartesian basis, {e_{1},e_{2},e_{3}}, e_{i} ∧ e_{j} = ϵ_{ijk}e_{k}

It could be verified that ϵ_{ijk} may be expressed as:

| (2.28) |

It could also be verified that the product of the permutation symbols ϵ_{ijk}ϵ_{pqr} is
related to the Kronecker delta by the relation

| (2.29) |

We deduce from the above equation (2.29) the important relations:

| (2.30) |

The triple scalar (or box) product: [u,v,w] represents the volume V of a parallelepiped spanned by u, v, w forming a right handed triad. Thus, in index notation:

| (2.31) |

Note that the vectors u, v, w are linearly dependent if and only if their scalar triple product vanishes, i.e., [u,v,w] = 0

The product (u ∧ v) ∧ w is called the vector triple product and it may be verified that

| (2.36) |

Thus triple product, in general, is not associative, i.e., u ∧ (v ∧ w) ≠ (u ∧ v) ∧ w.

A second order tensor A, for our purposes here, may be thought of as a linear function that maps a directed line segment to another directed line segment. This we write as, v = Au where A is the linear function that assigns a vector v to each vector u. Since A is a linear function,

| (2.37) |

for all vectors u, v and all scalars α, β.

If A and B are two second order tensors, we can define the sum A + B, the difference A - B and the scalar multiplication αA by the rules

If the relation v ⋅ Av ≥ 0 holds for all vectors, then A is said to be a positive semi-definite tensor. If the condition v ⋅ Av > 0 holds for all nonzero vectors v, then A is said to be positive definite tensor. Tensor A is called negative semi-definite if v ⋅ Av ≤ 0 and negative definite if v ⋅ Av < 0 for all vectors, v ≠ o, respectively.

The tensor (or direct or matrix) product or the dyad of the vectors u and v is denoted by u ⊗ v. It is a second order tensor which linearly transforms a vector w into a vector with the direction of u following the rule

| (2.40) |

The dyad satisfies the linearity property

| (2.41) |

The following relations are easy to establish:

A dyadic is a linear combination of dyads with scalar coefficients, for example,
α(u ⊗ v) + β(w ⊗ x). Any second-order tensor can be expressed as a dyadic. As
an example, the second order tensor A may be represented by a linear
combination of dyads formed by the Cartesian basis {e_{i}}, i.e., A = A_{ij}e_{i} ⊗ e_{j}.
The nine Cartesian components of A with respect to {e_{i}}, represented by A_{ij} can
be expressed as a matrix [A], i.e.,

| (2.45) |

This is known as the matrix notation of tensor A. We call A, which is resolved
along basis vectors that are orthonormal and fixed, a Cartesian tensor of order
two. Then, the components of A with respect to a fixed, orthonormal basis
vectors e_{i} is obtained as:

| (2.46) |

The Cartesian components of the unit tensor 1 form the Kronecker delta
symbol, thus 1 = δ_{ij}e_{i} ⊗ e_{j} = e_{i} ⊗ e_{i} and in matrix form

| (2.47) |

Next we would like to derive the components of u ⊗ v along an orthonormal
basis {e_{i}}. Using the representation (2.46) we find that

| (2.49) |

The product of two second order tensors A and B, denoted by AB, is again a second order tensor. It follows from the requirement (AB)u = A(Bu), for all vectors u.

Further, the product of second order tensors is not commutative, i.e., AB ≠
BA. The components of the product AB along an orthonormal basis {e_{i}} is
found to be:

The following properties hold for second order tensors:

Note that the relations AB = 0 and Au = o does not imply that A or B is 0 or u = o.

The unique transpose of a second order tensor A denoted by A^{t} is governed by
the identity:

| (2.60) |

for all vectors u and v.

Some useful properties of the transpose are

From identity (2.60) we obtain e_{i} ⋅ A^{t}e_{
j} = e_{j} ⋅ Ae_{i}, which gives, in regard to
equation (2.46), the relation (A^{t})_{
ij} = A_{ji}.

The trace of a tensor A is a scalar denoted by tr(A) and is defined as:

| (2.65) |

where u, v, w are any vectors such that [u,v,w] ≠ 0, i.e., these vectors span the entire 3D vector space. Thus, tr(m ⊗ n) = m ⋅ n. Let us see how: Without loss of generality we can assume the vectors u, v and w to be m, n and (m ∧ n) respectively. Then, (2.65) becomes

The dot product between two tensors denoted by A ⋅ B, just like the dot product of two vectors yields a real value and is defined as

| (2.71) |

Next, we record some useful properties of the dot operator:

The norm of a tensor A is denoted by |A| (or ∥A∥). It is a non-negative real number and is defined by the square root of A ⋅ A, i.e.,

| (2.76) |

The determinant of a tensor A is defined as:

| (2.77) |

where u, v, w are any vectors such that [u,v,w] ≠ 0, i.e., these vectors span the entire 3D vector space. In index notation:

| (2.78) |

Then, it could be shown that

A tensor A is said to be singular if and only if det(A) = 0. If A is a
non-singular tensor i.e., det(A) ≠ 0, then there exist a unique tensor A^{-1}, called
the inverse of A satisfying the relation

| (2.81) |

If tensors A and B are invertible (i.e., they are non-singular), then the properties

Corresponding to an arbitrary tensor A there is a unique tensor A^{*}, called the
adjugate of A, such that

| (2.88) |

for any arbitrary vectors a and b in the vector space. Suppose that A is invertible and that a, b, c are arbitrary vectors, then

| (2.90) |

between the inverse of A and the adjugate of A^{t}.

An orthogonal tensor Q is a linear function satisfying the condition

| (2.91) |

for all vectors u and v. As can be seen, the dot product is invariant (its value does not change) due to the transformation of the vectors by the orthogonal tensor. The dot product being invariant means that both the angle, θ between the vectors u and v and the magnitude of the vectors |u|, |v| are preserved. Consequently, the following properties of orthogonal tensor can be inferred:

It follows from (2.92) that Q Figure 2.2 shows what happens to two directed line segments, u_{1} and v_{1} under
the action the two kinds of orthogonal tensors. A proper orthogonal tensor
corresponding to a rotation about the e_{3} basis, whose Cartesian coordinate
components are given by

| (2.94) |

transforms, u_{1} to u_{1}^{+} and v_{
1} to v_{1}^{+}, maintaining their lengths and the angle
between these line segments. Here α = Φ - Θ. An improper orthogonal tensor
corresponding to reflection about e_{1} axis, whose Cartesian coordinate components
are given by

| (2.95) |

transforms, u_{1} to u_{1}^{-} and v_{
1} to v_{1}^{-}, still maintaining their lengths and the angle
between these line segments a constant.

A symmetric tensor, S and a skew symmetric tensor, W are such:

| (2.96) |

therefore the matrix components of these tensor reads

| (2.97) |

thus, there are only six independent components in symmetric tensors and three independent components in skew tensors. Since, there are only three independent scalar quantities that in a skew tensor, it behaves like a vector with three components. Indeed, the relation holds:

| (2.98) |

where u is any vector and ω characterizes the axial (or dual) vector of the skew tensor W, with the property |ω| = |W|∕ (proof is omitted). The relation between the Cartesian components of W and ω is obtained as:

Any tensor A can always be uniquely decomposed into a symmetric tensor, denoted by S (or sym(A)), and a skew (or antisymmetric) tensor, denoted by W (or skew(A)). Hence, A = S + W, where

| (2.103) |

Next, we shall look at some properties of symmetric and skew tensors:

where B denotes any second order tensor. The first of these equalities in the above equations is due to the property of the dot and trace operator, namely A ⋅ B = A

Consider any vector u and a unit vector e. Then, we write u = u_{∥} + u_{⊥}, with u_{∥}
and u_{⊥} characterizing the projection of u onto the line spanned by e and onto the
plane normal to e respectively. Using the definition of a tensor product (2.40) we
deduce that

Any tensor of the form α1, with α denoting a scalar is known as a spherical tensor.

Every tensor A can be decomposed into its so called spherical part and its deviatoric part, i.e.,

| (2.115) |

where α = tr(A)∕3 = A_{ii}∕3 and dev(A) is known as a deviator of A or a
deviatoric tensor and is defined as dev(A) = A - (1∕3)tr(A)1, or [dev(A)]_{ij} =
A_{ij} - (1∕3)A_{kk}δ_{ij}. It then can be easily verified that tr(dev(A)) = 0, for any
second order tensor A.

Above we saw two additive decompositions of an arbitrary tensor A. Of comparable significance in continuum mechanics are the multiplicative decompositions afforded by the polar decomposition theorem. This result states that an arbitrary invertible tensor A can be uniquely expressed in the forms

| (2.116) |

where Q is an orthogonal tensor and U, V are positive definite symmetric tensors. It should be noted that Q is proper or improper orthogonal according as det(A) is positive or negative. (See, for example, Chadwick [1] for proof of the theorem.)

A fourth order tensor A may be thought of as a linear function that maps second
order tensor A into another second order tensor B. While this is too narrow a
viewpoint^{1} ,
it suffices for the study of mechanics. We write B = A : A which defines a linear
transformation that assigns a second order tensor B to each second order tensor
A.

We can express any fourth order tensor, A in terms of the three Cartesian basis vectors as

| (2.117) |

where A_{ijkl} are the Cartesian components of A. Thus, the fourth order tensor A
has 3^{4} = 81 components. Remember that any repeated index has to be summed
from one through three.

One example for a fourth order tensor is the tensor product of the four vectors u, v, w, x, denoted by u ⊗ v ⊗ w ⊗ x. We have the useful property:

| (2.118) |

If A = u ⊗ v ⊗ w ⊗ x, then it is easy to see that the Cartesian components of A,
A_{ijkl} = u_{i}v_{j}w_{k}x_{l}, where u_{i}, v_{j}, w_{k}, x_{l} are the Cartesian components of the vectors
u, v, w and x respectively.

Another example of a fourth order tensor, is the tensor obtained from the
tensor product of two second order tensors, i.e., D = A ⊗ B, where A, B are
second order tensors and D is the fourth order tensor. In index notation we may
write: D_{ijkl} = A_{ij}B_{kl} where D_{ijkl}, A_{ij}, B_{kl} are the Cartesian components of the
respective tensors.

The double contraction of a fourth order tensor A with a second order tensor B results in a second order tensor, denoted by A : B with the property that:

| (2.121) |

Next, we compute the Cartesian components of A : B to be:

Then, the following can be established:

The unique transpose of a fourth order tensor A denoted by A^{t} is governed by
the identity

| (2.125) |

for all the second order tensors B and C. From the above identity we deduce the
index relation (A^{t})_{
ijkl} = A_{klij}. The following properties of fourth order tensors can
be established:

Next, we define fourth order unit tensors I and so that

| (2.128) |

for any second order tensor A. These fourth order unit tensors may be represented by

The deviatoric part of a second order tensor A may be described by means of a fourth order projection tensor, ℙ where

| (2.131) |

Thus the components of dev(A) and A are related through the expression
[dev(A)]_{ij} = P_{ijkl}A_{kl}, with P_{ijkl} = δ_{ik}δ_{jl} - (1∕3)δ_{ij}δ_{kl}.

Similarly, the fourth order tensors S and W given by

| (2.132) |

are such that for any second order tensor A, they assign symmetric and skew part of A respectively, i.e.,

| (2.133) |

Till now, we represented second order tensor components in a matrix form, for the advantages that it offers in computing other quantities and defining other operators. However, when we want to study mapping of second order tensor on to another second order tensor, this representation seem to be inconvenient. For this purpose, we introduce this alternate representation. Now, we view the second order tensor as a column vector of nine components instead of a 3 by 3 matrix as introduced in (2.45). The order of these components is subjective. Keeping in mind the application of this is to study elasticity, we order the components of a general second order tensor, A, as,

| (2.134) |

In view of this, the fourth order tensor, which for us is a linear function that maps a second order tensor to another second order tensor, can be represented as a 9 by 9 matrix as,

| (2.135) |

where A and B are second order tensors and ℂ is a fourth order tensor. Note that as before the fourth order tensor has 81 (=9*9) components. Thus, now the fourth order tensor is a matrix which is the reason for representing the second order tensor as vector.

The scalar λ_{i} characterize eigenvalues (or principal values) of a tensor A if there
exist corresponding nonzero normalized eigenvectors _{i} (or principal directions or
principal axes) of A, so that

| (2.136) |

To identify the eigenvectors of a tensor, we use subsequently a hat on the vector quantity concerned, for example, .

Thus, a set of homogeneous algebraic equations for the unknown eigenvalues
λ_{i}, i = 1,2,3, and the unknown eigenvectors _{i}, i = 1,2,3, is

| (2.137) |

For the above system to have a solution _{i} ≠ o the determinant of the system
must vanish. Thus,

| (2.138) |

where

| (2.139) |

This requires that we solve a cubic equation in λ, usually written as

| (2.140) |

called the characteristic polynomial (or equation) for A, the solutions of which are
the eigenvalues λ_{i}, i = 1,2,3. Here, I_{i}, i = 1,2,3, are the so-called principal scalar
invariants of A and are given by

A repeated application of tensor A to equation (2.136) yields A^{α}_{
i} = λ_{i}^{α}_{
i}, i
= 1,2,3, for any positive integer α. (If A is invertible then α can be any integer;
not necessarily positive.) Using this relation and (2.140) multiplied by _{i}, we
obtain the well-known Cayley-Hamilton equation:

| (2.142) |

It states that every second order tensor A satisfies its own characteristic equation.
As a consequence of Cayley-Hamilton equation, we can express A^{α} in terms of
A^{2}, A, 1 and principal invariants for positive integer, α > 2. (If A is invertible,
the above holds for any integer value of α positive or negative provided α ≠
{0, 1, 2}.)

For a symmetric tensor S the characteristic equation (2.140) always has three
real solutions and the set of eigenvectors form a orthonormal basis {} (the proof
of this statement is omitted). Hence, for a positive definite symmetric tensor A,
all eigenvalues λ_{i} are (real and) positive since, using (2.136), we have λ_{i} = _{i} ⋅A_{i}
> 0, i = 1,2,3.

Any symmetric tensor S may be represented by its eigenvalues λ_{i}, i = 1,2,3,
and the corresponding eigenvectors of S forming an orthonormal basis {_{i}}. Thus,
S can be expressed as

| (2.143) |

called the spectral representation (or spectral decomposition) of S. Thus, when orthonormal eigenvectors are used as the Cartesian basis to represent S then

| (2.144) |

The above holds when all the three eigenvalues are distinct. On the other hand, if
there exists a pair of equal roots, i.e., λ_{1} = λ_{2} = λ ≠ λ_{3}, with an unique
eigenvector _{3} associated with λ_{3}, we deduce that

| (2.145) |

where P_{3}^{∥} and P_{3}^{⊥} denote projection tensors introduced in (2.109) and (2.110)
respectively. Finally, if all the three eigenvalues are equal, i.e., λ_{1} = λ_{2} = λ_{3} = λ,
then

| (2.146) |

where every direction is a principal direction and every set of mutually orthogonal basis denotes principal axes.

It is important to recognize that eigenvalues characterize the physical nature of the tensor and that they do not depend on the coordinates chosen.

Let C be symmetric and positive definite tensor. Then there is a unique positive definite, symmetric tensor U such that

| (2.147) |

We write for U. If spectral representation of C is:

| (2.148) |

then the spectral representation for U is:

| (2.149) |

where we have assumed that the eigenvalues of C, λ_{i} are distinct. On the other
hand if λ_{1} = λ_{2} = λ_{3} = λ then

| (2.150) |

If λ_{1} = λ_{2} = λ ≠ λ_{3}, with an unique eigenvector _{3} associated with λ_{3}
then

| (2.151) |

(For proof of this theorem, see for example, Gurtin [2].)

Consider two sets of mutually orthogonal basis vectors which share a common
origin. They correspond to a ‘new’ and an ‘old’ (original) Cartesian coordinate
system which we assume to be right-handed characterized by two sets of basis
vectors {_{i}} and {e_{i}}, respectively. Hence, the new coordinate system could
be obtained from the original one by a rotation of the basis vectors e_{i}
about their origin. We then define the directional cosine matrix, Q_{ij},
as,

| (2.152) |

Note that the first index on Q_{ij} indicates the ‘old’ components whereas the second
index holds for the ‘new’ components.

It is worthwhile to mention that vectors and tensors themselves remain invariant upon a change of basis - they are said to be independent of the coordinate system. However, their respective components do depend upon the coordinate system introduced. This is the reason why a set of numbers arranged as a 3 by 1 or 3 by 3 matrix is not a vector or a tensor.

Consider some vector u represented using the two sets of basis vectors {e_{i}} and
{_{i}}, i.e.,

| (2.154) |

from the definition of the directional cosine matrix, (2.152). We assume that the
relation between the basis vectors e_{i} and _{j} is known and hence given the
components of a vector in a basis, its components in another basis can be found
using equation (2.154).

In an analogous manner, we find that

| (2.155) |

The results of equations (2.154) and (2.155) could be cast in matrix notation as

| (2.156) |

respectively. It is important to emphasize that the above equations are not
identical to = Q^{t}u and u = Q, respectively. In (2.156) [] and [u]
are column vectors characterizing components of the same vector in two
different coordinate systems, whereas and u are different vectors, in the
later. Similarly, [Q] is a matrix of directional cosines, it is not a tensor
even though it has the attributes of an orthogonal tensor as we will see
next.

Combining equations (2.154) and (2.155), we obtain

| (2.157) |

Hence, (Q_{ai}Q_{ji} -δ_{ja})u_{a} = 0 and (Q_{ij}Q_{ia} -δ_{ja})ũ_{a} = 0 for any vector u. Therefore,

To determine the transformation laws for the Cartesian components of any second-order tensor A, we proceed along the lines similar to that done for vectors. Since, we seek the components of the same tensor in two different basis,

| (2.160) |

Then it follows from (2.46) that,

In matrix notation, [] = [Q]
| (2.162) |

We emphasize again that these transformations relates the different matrices []
and [A], which have the components of the same tensor A and the equations []
= [Q]^{t}[A][Q] and [A] = [Q][][Q]^{t} differ from the tensor equations =
Q^{t}AQ and A = QQ^{t}, relating two different tensors, namely A and
.

Finally, the 3^{n} components A_{
j1j2…jn} of a tensor of order n (with n indices j_{1},
j_{2}, …, j_{n}) transform as

| (2.163) |

This tensorial transformation law relates the different components Ã_{i1i2…in} (along
the directions _{1}, _{2}, _{3}) and A_{j1j2…jn} (along the directions e_{1}, e_{2}, e_{3}) of the same
tensor of order n.

We note that, in general a second order tensor, A will be represented as
A_{ij}e_{i} ⊗ E_{j}, where {e_{i}} and {E_{j}} are different basis vectors spanning the same
space. It is not necessary that the directional cosines Q_{ij} = E_{i} ⋅_{i} and q_{ij} = e_{i} ⋅_{i}
be the same, where _{i} and _{j} are the ‘new’ basis vectors with respect to which
the matrix components of A is sought. Thus, generalizing the above is
straightforward; each directional cosine matrices can be different, contrary to the
assumption made.

A tensor A is said to be isotropic if its components are the same under arbitrary rotations of the basis vectors. The requirement is deduced from equation (2.161) as

| (2.164) |

Of course here we assume that the components of the tensor are with respect to a single basis and not two or more independent basis.

Note that all scalars, zeroth order tensors are isotropic tensors. Also, zero tensors and unit tensors of all orders are isotropic. It can be easily verified that for second order tensors spherical tensor is also isotropic. The most general isotropic tensor of order four is of the form

| (2.165) |

where α, β, γ are scalars. The same in component form is given by: αδ_{ij}δ_{kl} +
βδ_{ik}δ_{jl} + γδ_{il}δ_{jk}.

Having got an introduction to algebra of tensors we next focus on tensor calculus. Here we shall understand the meaning of a function and define what we mean by continuity and derivative of a function.

A function is a mathematical correspondence that assigns exactly one element
of one set to each element of the same or another set. Thus, if we consider scalar
functions of one scalar variable, then for each element (value) in the subset of
the real line one associates an element in the real line or in the space of
vectors or in the space of second order tensors. For example, Φ = (t),
u = (t) = u_{i}(t)e_{i}, A = (t) = A_{ij}(t)e_{i} ⊗ e_{j} are scalar valued, vector
valued, second order tensor valued scalar functions with a set of Cartesian
basis vectors assumed to be fixed. The components u_{i}(t) and A_{ij}(t) are
assumed to be real valued smooth functions of t varying over a certain
interval.

The first derivative of the scalar function Φ is simply = dΦ∕dt. Recollecting from a first course in calculus, dΦ∕dt stands for that unique value of the limit

| (2.166) |

The first derivative of u and A with respect to t (rate of change) denoted by
= du∕dt and = dA∕dt, is given by the first derivative of their associated
components. Since, de_{i}∕dt = o, i = 1, 2, 3, we have

| (2.167) |

In general, the n^{th} derivative of u and A (for any desired n) denoted by
d^{n}u∕dt^{n} and d^{n}A∕dt^{n}, is a vector-valued and tensor-valued function whose
components are d^{n}u_{
i}∕dt^{n} and d^{n}A_{
ij}∕dt^{n}, respectively. Again we have assumed that
de_{i}∕dt = o.

By applying the rules of differentiation, we obtain the identities:

Since, AA^{-1} = 1, = 0. Hence, we compute:

| (2.176) |

A tensor function is a function whose arguments are one or more tensor variables and whose values are scalars, vectors or tensors. The functions Φ(B), u(B), and A(B) are examples of so-called scalar-valued, vector-valued and second order tensor-valued functions of one second order tensor variable B, respectively. In an analogous manner, Φ(v), u(v), and A(v) are scalar-valued, vector-valued and second order tensor-valued functions of one vector variable v, respectively.

Let 𝔇 denote the region of the vector space that is of interest. Then a scalar field Φ, defined on a domain 𝔇, is said to be continuous if

| (2.177) |

where 𝔙 denote the set of all vectors in the vector space. The properties of continuity are attributed to a vector field u and a tensor field A defined on 𝔇, if they apply to the scalar field u ⋅ a and a ⋅ Ab ∀ a and b ∈𝔙.

Consider a scalar valued function of one second order tensor variable A, Φ(A). Assuming the function Φ is continuous in A, the derivative of Φ with respect to A, denoted by , is the second order tensor that satisfies

| (2.178) |

for any second order tensor U.

Example: If Φ(A) = det(A), find both when A is invertible and when it is not.

Taking trace of the Cayley-Hamilton equation (2.142) and rearranging we get

| (2.179) |

Hence,

| (2.180) |

Using the properties of trace and dot product we find that

from which, we deduce that using equation (2.178). Substituting equations (2.184) through (2.186) in (2.180) we get
| (2.187) |

If A is invertible, then multiplying (2.142) by A^{-t} and rearranging we
get

| (2.188) |

In light of the above equation, (2.187) reduces to

| (2.189) |

Next, we consider a smooth tensor valued function of one second order tensor variable A, f(A). As before, the derivative of f with respect to A, denoted by , is the fourth order tensor that satisfies

| (2.190) |

for any second order tensor U.

It is straight forward to see that when f(A) = A, where A is any second order tensor, then (2.190) reduces to

| (2.191) |

hence, = I, the fourth order unit tensor where we have used equation (2.128a).

If f(A) = A^{t} where A is any second order tensor, then using (2.190) it can be
shown that = obtained using equation (2.128b).

Now, say the tensor valued function f is defined only for symmetric tensors, and that f(S) = S, where S is a symmetric second order tensor. Compute the derivative of f with respect to S, i.e., .

First, since the function f is defined only for symmetric tensors, we have to
generalize the function for any tensor. This is required because U is any second
order tensor. Hence, f(S) = f([A + A^{t}]) = (A), where A is any second order
tensor. Now, find the derivative of with respect to A and then rewrite the result
in terms of S. The resulting expression is the derivative of f with respect to S.
Hence,

| (2.192) |

Substituting the above in equation (2.190) we get

| (2.193) |

Thus,

| (2.194) |

Consequently,

| (2.195) |

Example - 1: Assume that A is an invertible tensor and f(A) = A^{-1}. Then
show that

| (2.196) |

Recalling that all invertible tensors satisfy the relation AA^{-1} = 1, we
obtain

| (2.197) |

Differentiating the above equation with respect to s

| (2.198) |

Evaluating the above equation at s = 0, and rearranging

| (2.199) |

Hence proved.

Example - 2: Assume that S is an invertible symmetrical tensor and f(S) =
S^{-1}. Find .

As before generalizing the given function, we have A) = {[A + A^{t}]}^{-1}.
Following the same steps as in the previous example we obtain:

| (2.200) |

Differentiating the above equation with respect to s and evaluating at s = 0 and rearranging we get

| (2.201) |

Therefore,

| (2.202) |

Let Φ be a smooth scalar-valued function and A, B smooth tensor-valued functions of a tensor variable C. Then, it can be shown that

In this section, we consider scalar and vector-valued functions that assign a scalar and vector to each element in the subset of the set of position vectors for the points in 3D space. If x denotes the position vector of points in the 3D space, then (x) is a function that assigns a scalar Φ to each point in the 3D space of interest and Φ is called the scalar field. A few examples of scalar fields are temperature, density, energy. Similarly, the vector field (x) assigns a vector to each point in the 3D space of interest. Displacement, velocity, acceleration are a few examples of vector fields.

Let 𝔇 denote the region of the 3D space that is of interest, that is the set of position vectors of points that is of interest in the 3D space. Then a scalar field Φ, defined on a domain 𝔇, is said to be continuous if

| (2.205) |

where 𝔈 denote the set of all position vectors of points in the 3D space and a is a constant vector. The scalar field Φ is said to be differentiable if there exist a vector field w such that

| (2.206) |

It can be shown that there is at most one vector field, w satisfying the above equation (proof omitted). This unique vector field w is called the gradient of Φ and denoted by grad(Φ).

The properties of continuity and differentiability are attributed to a vector field u and a tensor field T defined on 𝔇, if they apply to the scalar field u ⋅ a and a ⋅ Tb ∀ a and b ∈𝔈. Given that the vector field u, is differentiable, the gradient of u denoted by grad(u), is the tensor field defined by

| (2.207) |

The divergence and the curl of a vector u denoted as div(u) and curl(u), are respectively scalar valued and vector valued and are defined by

When the tensor field T is differentiable, its divergence denoted by div(T), is the vector field defined by
| (2.210) |

If grad(Φ) exist and is continuous, Φ is said to be continuously differentiable and this property extends to u and T if grad(u⋅a) and grad(a⋅Tb) exist and are continuous for all a, b ∈𝔈.

An important identity: For a continuously differentiable vector field, u

| (2.211) |

(Proof omitted)

Let Φ and u be some scalar and vector field respectively. Then, the Laplacian
operator, denoted by Δ (or by ∇^{2}), is defined as

| (2.212) |

The Hessian operator, denoted by ∇∇ is defined as

| (2.213) |

If a vector field u is divergence free (i.e. div(u) = 0) then it is called solenoidal. It is called irrotational if curl(u) = o. It can be established that

| (2.214) |

Consequently, if a vector field, u is both solenoidal and irrotational, then Δ(u) = o (follows from the above equation) and such a vector field is said to be harmonic. If a scalar field Φ satisfies Δ(Φ) = 0, then Φ is said to be harmonic.

Potential theorem: Let u be a smooth point field on an open or closed simply connected region and let curl(u) = o. Then there is a continuously differentiable scalar field on such that

| (2.215) |

From the definition of grad (2.207), it can be seen that it is a linear operator. That is,

| (2.216) |

Consequently, all other operators div, Δ are also linear operators.

Before concluding this section we collect a few identities that are useful subsequently. Here Φ, Ψ are scalar fields, u, v are vector fields and A is second order tensor field.

Till now, in this section we defined all the quantities in closed form but we
require them in component form to perform the actual calculations. This
would be the focus in the reminder of this section. In the following, let
e_{i} denote the Cartesian basis vectors and x_{i} i = 1, 2, 3, the Cartesian
coordinates.

Let Φ be differentiable scalar field, then it follows from equation (2.206), on
replacing a by the base vectors e_{i} in turn, that the partial derivatives exist in
𝔇 and that, moreover, w_{i} = . Hence

| (2.228) |

Let u be differentiable vector field in 𝔇 and a some constant vector in 𝔈. Then, using (2.228) we compute

| (2.229) |

where a_{p} and u_{p} are the Cartesian components of a and u respectively.
Consequently, appealing to definition (2.207) we obtain

| (2.230) |

Then, according to the definition for divergence, (2.208)

| (2.231) |

Next, we compute div(u ∧ a) to be

| (2.232) |

using (2.231). Comparing the above result with the definition of curl in (2.209) we infer

| (2.233) |

Finally, let T be a differentiable tensor field in 𝔇, then we compute

| (2.234) |

using equation (2.231). Then, we infer

| (2.235) |

from the definition (2.210).

The component form for the other operators, namely Laplacian and Hessian can be obtained as

Above we obtained the components of the different operators in Cartesian coordinate system, we shall now proceed to obtain the same with respect to other coordinate systems. While using the method illustrated here we could obtain the components in any orthogonal curvilinear system, we choose the cylindrical polar coordinate system for illustration.

First, we outline a general procedure to find the basis vectors for a given
orthogonal curvilinear coordinates. A differential vector dp in the 3D vector
space is written as dp = dx_{i}e_{i} relative to Cartesian coordinates. The same
coordinate independent vector in orthogonal curvilinear coordinates is written
as

| (2.238) |

where z_{i} denotes curvilinear coordinates. But

| (2.239) |

where g_{i} is the basis vectors in the orthogonal curvilinear coordinates.
Thus,

| (2.240) |

Comparing equations (2.239) and (2.240) we obtain the desired transformation relation between the bases to be:

| (2.241) |

The basis g_{i} so obtained will vary from point to point in the Euclidean space and
will be orthogonal (by the choice of curvilinear coordinates) at each point but not
orthonormal. Hence we define

| (2.242) |

and use these as the basis vectors for the orthogonal curvilinear coordinates.

Let (x,y,z) denote the coordinates of a typical point in Cartesian coordinate system and (r,θ,Z) the coordinates of a typical point in cylindrical polar coordinate system. Then, the coordinate transformation from cylindrical polar to Cartesian and vice versa is given by:

Consequently, the orthonormal cylindrical polar basis vectors (e_{r},e_{θ},e_{Z})
and Cartesian basis vectors (e_{1},e_{2},e_{3}) are related through the equations

Now, we can compute the components of grad(Φ) in cylindrical polar coordinates. Towards this, we obtain

To obtain the components of grad(u) one can follow the same procedure outlined above or can compute the same using the above result and the definition of grad (2.207) as detailed below:

where we have used the following identities:
| (2.259) |

Using the techniques illustrated above the following identities can be established:

Suppose u(x) is a vector field defined on some convex three dimensional region,
in physical space with volume v and on a closed surface a bounding this
volume^{2}
and u(x) is continuous in and continuously differentiable in the interior of .
Then,

| (2.262) |

where, n is the outward unit normal field acting along the surface a, dv and da are infinitesimal volume and surface element, respectively. Proof of this theorem is beyond the scope of these notes.

Using (2.262) the following identities can be established:

where Φ, u, A are continuously differentiable scalar, vector and tensor fields defined in v and continuous in v and n is the outward unit normal field acting along the surface a.Stokes theorem relates a surface integral, which is valid over any open surface a, to a line integral around the bounding single simple closed curve c in three dimensional space. Let dx denote the tangent vector to the curve, c and n denote the outward unit vector field n normal to s. The curve c has positive orientation relative to n in the sense shown in figure 2.3. The indicated circuit with the adjacent points 1, 2, 3 (1, 2 on curve c and 3 an interior point of the surface s) induced by the orientation of c is related to the direction of n (i.e., a unit vector normal to s at point 3) by right-hand screw rule. For a continuously differentiable vector field u defined on some region containing a, we have

| (2.268) |

Let Φ and u be continuously differentiable scalar and vector fields defined on a and c and n is the outward drawn normal to a. Then, the following can be shown:

Finally, we record Greens’s theorem. We do this first for simply connected domain and then for multiply connected domain. If any curve within the domain can be shrunk to a point in the domain without leaving the domain, then such a domain is said to be simply connected. A domain that is not simply connected is said to be multiply connected.

Applying Stoke’s theorem, (2.268) to a planar simply connected domain with
the vector field given as u = f(x,y)e_{1} + g(x,y)e_{2},

| (2.271) |

The above equation (2.271) when f(x,y) = 0 reduces to

| (2.272) |

where θ is the angle made by the tangent to the curve c with e_{1} and n_{x} is the
component of the unit normal n along e_{1} direction, refer to figure 2.4. Then, when
g(x,y) = 0 equation (2.271) yields,

| (2.273) |

where n_{y} is the component of the unit normal along the e_{2} direction.

Similarly, applying Stoke’s theorem (2.268) to a vector field, u = (y,z)e_{2} +
ĝ(y,z)e_{3} defined over a planar simply connected domain,

| (2.274) |

The above equation (2.274) when (y,z) = 0 reduces to

| (2.275) |

where n_{y} is the component of the normal to the curve along the e_{2} direction and
when ĝ(y,z) = 0 yields

| (2.276) |

where n_{z} is the component of the normal to the curve along the e_{3} direction.

In a simply connected domain, the region of interest has a single simple closed curve. In a multiply connected domain, the region of interest has several simple closed curves, like the one shown in figure 2.5. If the line integrals are computed by traversing in the direction shown in figure 2.5,

| (2.277) |

where the functions f and g are continuously differentiable in the domain, C_{0} is
the outer boundary of the domain, C_{i}’s are the boundary of the voids in the
interior of the domain and we have assumed that there are N such voids in the
domain. We emphasize again our assumption that we transverse the interior voids
in the same sense as the outer boundary.

The above equation (2.277) when f(x,y) = 0 reduces to

| (2.278) |

where n_{x}^{i} is the component of the unit normal n along e_{
1} direction of the i^{th}
curve. Then, when g(x,y) = 0 equation (2.277) yields,

| (2.279) |

where n_{y}^{i} is the component of the unit normal along the e_{
2} direction of the i^{th}
curve.

In this chapter, we looked at the mathematical tools required to embark on a study of mechanics. In particular, we began with a study on algebra of tensors and then graduated to study tensor calculus. Most of the mathematical results that we would be using is summarized in this chapter.

- Give reasons for the following: (a) δ
_{ii}= 3, (b) δ_{ij}δ_{ij}= 3, (c) δ_{ij}a_{i}= a_{j}, (d) δ_{ij}a_{jk}= a_{ik}, (e) δ_{ij}a_{kj}= a_{ki}, (f) δ_{ij}a_{ij}= a_{ii}, (g) δ_{ij}δ_{jk}δ_{ik}= 3, (h) δ_{ik}δ_{jm}δ_{ij}= δ_{km}(i) ϵ_{kki}= 0, (j) δ_{ij}ϵ_{ijk}= 0, (k) ϵ_{ijk}ϵ_{ijk}= 6, where δ_{ij}denotes the Kronecker delta and ϵ_{ijk}the permutation symbol - Show that a
_{ij}= a_{ji}when ϵ_{ijk}a_{jk}= 0. ϵ_{ijk}is the permutation symbol - Expand a
_{ij}b_{ij} - Expand D
_{ij}x_{i}x_{j}. If possible simplify the expression D_{ij}x_{i}x_{j}when (a) D_{ij}= D_{ji}(b) D_{ij}= -D_{ji} - Express the expression c
_{i}= A_{ij}b_{j}- b_{i}as c_{i}= B_{ij}b_{j}and thus obtain B_{ij} - Consider two vectors a = e
_{1}+ 4e_{2}+ 6e_{3}and b = 2e_{1}+ 4e_{2}, where {e_{i}} are orthonormal Cartesian basis vectors. For these vectors compute,- Norm of a
- The angle between vectors a and b
- The area of the parallelogram bounded by the vectors a and b
- b ∧ a
- The component of vector a in the direction z = e
_{1}+ e_{2}.

- Show that y = mx where y and x are vectors and m is a scalar is a linear transformation. Find the second order tensor that maps the given vector x on to y.
- Show that y = mx + c where y, x and c are vectors and m is a scalar is not a linear transformation.
- Establish the following ϵ
_{ijk}T_{jk}e_{i}= (T_{23}-T_{32})e_{1}+ (T_{31}-T_{13})e_{2}+ (T_{12}-T_{21})e_{3}and use this to show that δ_{ij}ϵ_{ijk}e_{k}= o. Here {e_{i}} denotes the orthonormal Cartesian basis, δ_{ij}the Kronecker delta and ϵ_{ijk}the permutation symbol - Show that if tensor A is positive definite then det(A) > 0
- Using Cayley-Hamilton theorem, deduce that
for an arbitrary tensor A.

- Let A be an arbitrary tensor. Show that A
^{t}A and AA^{t}are positive semi-definite symmetric tensors. If A is invertible, prove that these tensors - A^{t}A and AA^{t}- are positive definite. - Let u, v, w, x be arbitrary vectors and A an arbitrary tensor. Show
that
- (u ⊗ v)(w ⊗ x) = (v ⋅ w)(u ⊗ x)
- A(u ⊗ v) = (Au) ⊗ v
- (u ⊗ v)A = u ⊗ (A
^{t}v)

- Let Q be a proper orthogonal tensor. If {p,q,r} form an
orthonormal basis then a general representation for Q is: Q =
p ⊗ p + (q ⊗ q + r ⊗ r) cos(θ) + (r ⊗ q - q ⊗ r) sin(θ). Now, show
that
- I
_{1}= I_{2}= 1 + 2 cos(θ) - I
_{3}= 1 - Q has only one real eigen value if θ ≠ 0,π

- I
- Let A be an invertible tensor which depends upon a real parameter, t.
Assuming that exists, prove that
- Let ϕ(A) = det(A) and A be an invertible tensor. Then, show that =
I
_{3}A^{-1} - Let ϕ, u and T be differentiable scalar, vector and tensor fields.
Using index notation verify the following for Cartesian coordinate
system:
- grad(ϕu) = u ⊗ grad(ϕ) + ϕgrad(u)
- div = div(u) -u ⋅ grad(ϕ)
- div(ϕT) = T
^{t}grad(ϕ) + ϕdiv(T) - curl(grad(ϕ)) = o
- grad(u ⋅ u) = 2grad(u)
^{t}u - div(T
^{t}u) = T ⋅ grad(u) + u ⋅ div(T)

- If A is an invertible tensor, then it can be decomposed uniquely as A = RU
= VR where R is an orthogonal tensor, and U, V are positive definite
symmetric tensors. Then, show that AA
^{t}= V^{2}and A^{t}A = U^{2}. Using these results prove that the eigenvalues of A^{t}A and AA^{t}are the same and that their eigenvectors are related as: p_{i}= Rq_{i}, where p_{i}are the eigenvectors of AA^{t}and q_{ i}are the eigenvectors of A^{t}A. - The matrix components of a tensor A when represented using Cartesian
coordinate basis are
For this tensor find the following:

- The spherical and deviatoric parts of A
- The eigenvalues of A
- The eigenvectors of A

- Let a new right-handed Cartesian coordinate system be represented by the
set {
_{i}} of basis vectors with transformation law_{2}= - sin(θ)e_{1}+ cos(θ)e_{2},_{3}= e_{3}. The origin of the new coordinate system coincides with the old origin.- Find
_{1}in terms of the old set {e_{i}} of basis vectors - Find the orthogonal matrix [Q]
_{ij}that expresses the directional cosine of the new basis with respect to the old one - Express the vector u = -6e
_{1}- 3e_{2}+ e_{3}in terms of the new set {_{i}} of basis vectors - Express the position vector, r of an arbitrary point using both
{e
_{i}} and {_{i}} basis and establish the relationship between the components of r represented using {e_{i}} basis and {_{i}} basis.

- Find
- For the following components of a second order tensor:
relative to a Cartesian orthonormal basis, {e

_{i}}, determine- The principal invariants
- The principal values, p
_{1}, p_{2}, p_{3} - The orientation of the principal directions with respect to the Cartesian basis, {e}
- The orientation of the orthonormal basis, {
_{i}} with respect to the basis, {e_{i}}, so that the stress tensor has a matrix representation of the form, T = diag[p_{1},p_{2},p_{3}] when expressed with respect to the {_{i}} basis - The transformation matrix, [Q]
_{ij}which transforms the set of orthonormal basis {e_{i}} to {_{i}}. - The components of the same tensor T with respect to a basis {
_{i}} when_{1}= (e_{1}+ e_{2})∕2,_{2}= (-e_{1}+ e_{2})∕2,_{3}= e_{3}.

- Let a tensor A be given by: A = α[1-e
_{1}⊗e_{1}] + β[e_{1}⊗e_{2}+ e_{2}⊗e_{1}], where α and β are scalars and e_{1}and e_{2}are orthogonal unit vectors. - Show that u ∧ v is the axial vector for v ⊗ u - u ⊗ v. Using this establish the identity a∧ (b∧c) = (b⊗c-c⊗b)a, ∀ a,b,c ∈𝔙. Use could be made of the identity [(u ∧ v) ∧ u] ⋅ v = (u ∧ v) ⋅ (u ∧ v).
- Let be a regular region and T a tensor field which is continuous in and
continuously differentiable in the interior of . Show that,
where x is the position of a representative point of and τ is the axial vector of T - T

^{t}. (Hint: Make use of the identity obtained in problem 23.) - Let (r,θ,ϕ) denote the spherical coordinates of a typical point related to
the Cartesian coordinates (x,y,z) through x = r cos(θ) cos(ϕ), y =
r cos(θ) sin(ϕ), z = r sin(θ). Then express the spherical coordinate basis in
terms of the Cartesian basis and show that

where α is a scalar field, v is a vector field represented in spherical coordinates and T is a tensor field also represented using spherical coordinate basis vectors.