1. Vectors and Linear Spaces Jan 2013

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1. Vectors and Linear Spaces Jan 2013

  1. 1. SSG 805 Mechanics of Continua INSTRUCTOR: Fakinlede OA oafak@unilag.edu.ng oafak@hotmail.com ASSISTED BY: Akano T manthez@yahoo.com Oyelade AO oyelade.akintoye@gmail.com Department of Systems Engineering, University of Lagos
  2. 2. Purpose of the Course  Available to beginning graduate students in Engineering  Provides a background to several other courses such as Elasticity, Plasticity, Fluid Mechanics, Heat Transfer, Fracture Mechanics, Rheology, Dynamics, Acoustics, etc. These courses are taught in several of our departments. This present course may be viewed as an advanced introduction to the modern approach to these courses  It is taught so that related graduate courses can build on this modern approach.Department of Systems Engineering, University of Lagos 2 oafak@unilag.edu.ng 12/31/2012
  3. 3. What you will need  The slides here are quite extensive. They are meant to assist the serious learner. They are NO SUBSTITUTES for the course text which must be read and followed concurrently.  Preparation by reading ahead is ABSOLUTELY necessary to follow this course  Assignments are given at the end of each class and they are due (No excuses) exactly five days later.  Late submission carry zero grade.Department of Systems Engineering, University of Lagos 3 oafak@unilag.edu.ng 12/31/2012
  4. 4. Scope of Instructional Material Course Schedule: Slide Title Slides Weeks Text Read Pages Vectors and Linear Spaces 70 2 Gurtin 1-8 Tensor Algebra 110 3 Gurtin 9-37 Tensor Calculus 119 3 Gurtin 39-57 Kinematics: Deformation & Motion 106 3 Gurtin 59-123 Theory of Stress & Heat Flux 43 1 Holz 109-129 Balance Laws 58 2 Gurtin 125-205 The read-ahead materials are from Gurtin except the part marked red. There please read Holzapfel. Home work assignments will be drawn from the range of pages in the respective books.Department of Systems Engineering, University of Lagos 4 oafak@unilag.edu.ng 12/31/2012
  5. 5. Examination The only remedy for late submission is that you fight for the rest of your grade in the final exam if your excuse is considered to be genuine. Ordinarily, the following will hold: Evaluation Obtainable Home work 50 Midterm 25 Final Exam 25 Total 100Department of Systems Engineering, University of Lagos 5 oafak@unilag.edu.ng 12/31/2012
  6. 6. Course Texts  This course was prepared with several textbooks and papers. They will be listed below. However, the main course text is: Gurtin ME, Fried E & Anand L, The Mechanics and Thermodynamics of Continua, Cambridge University Press, www.cambridge.org 2010  The course will cover pp1-240 of the book. You can view the course as a way to assist your reading and understanding of this book  The specific pages to be read each week are given ahead of time. It is a waste of time to come to class without the preparation of reading ahead.  This preparation requires going through the slides and the area in the course text that will be covered.Department of Systems Engineering, University of Lagos 6 oafak@unilag.edu.ng 12/31/2012
  7. 7. Software  The software for the Course is Mathematica 9 by Wolfram Research. Each student is entitled to a licensed copy. Find out from the LG Laboratory  It your duty to learn to use it. Students will find some examples too laborious to execute by manual computation. It is a good idea to start learning Mathematica ahead of your need of it.  For later courses, commercial FEA Simulations package such as ANSYS, COMSOL or NASTRAN will be needed. Student editions of some of these are available. We have COMSOL in the LG LaboratoryDepartment of Systems Engineering, University of Lagos 7 oafak@unilag.edu.ng 12/31/2012
  8. 8. Texts  Gurtin, ME, Fried, E & Anand, L, The Mechanics and Thermodynamics of Continua, Cambridge University Press, www.cambridge.org 2010  Bertram, A, Elasticity and Plasticity of Large Deformations, Springer-Verlag Berlin Heidelberg, 2008  Tadmore, E, Miller, R & Elliott, R, Continuum Mechanics and Thermodynamics From Fundamental Concepts to Governing Equations, Cambridge University Press, www.cambridge.org , 2012  Nagahban, M, The Mechanical and Thermodynamical Theory of Plasticity, CRC Press, Taylor and Francis Group, June 2012  Heinbockel, JH, Introduction to Tensor Calculus and Continuum Mechanics, Trafford, 2003Department of Systems Engineering, University of Lagos 8 oafak@unilag.edu.ng 12/31/2012
  9. 9. Texts  Bower, AF, Applied Mechanics of Solids, CRC Press, 2010  Taber, LA, Nonlinear Theory of Elasticity, World Scientific, 2008  Ogden, RW, Nonlinear Elastic Deformations, Dover Publications, Inc. NY, 1997  Humphrey, JD, Cadiovascular Solid Mechanics: Cells, Tissues and Organs, Springer-Verlag, NY, 2002  Holzapfel, GA, Nonlinear Solid Mechanics, Wiley NY, 2007  McConnell, AJ, Applications of Tensor Analysis, Dover Publications, NY 1951  Gibbs, JW “A Method of Geometrical Representation of the Thermodynamic Properties of Substances by Means of Surfaces,” Transactions of the Connecticut Academy of Arts and Sciences 2, Dec. 1873, pp. 382-404.Department of Systems Engineering, University of Lagos 9 oafak@unilag.edu.ng 12/31/2012
  10. 10. Texts  Romano, A, Lancellotta, R, & Marasco A, Continuum Mechanics using Mathematica, Fundamentals, Applications and Scientific Computing, Modeling and Simulation in Science and Technology, Birkhauser, Boston 2006  Reddy, JN, Principles of Continuum Mechanics, Cambridge University Press, www.cambridge.org 2012  Brannon, RM, Functional and Structured Tensor Analysis for Engineers, UNM BOOK DRAFT, 2006, pp 177-184.  Atluri, SN, Alternative Stress and Conjugate Strain Measures, and Mixed Variational Formulations Involving Rigid Rotations, for Computational Analysis of Finitely Deformed Solids with Application to Plates and Shells, Computers and Structures, Vol. 18, No 1, 1984, pp 93-116Department of Systems Engineering, University of Lagos 10 oafak@unilag.edu.ng 12/31/2012
  11. 11. Texts  Wang, CC, A New Representation Theorem for Isotropic Functions: An Answer to Professor G. F. Smiths Criticism of my Papers on Representations for Isotropic Functions Part 1. Scalar-Valued Isotropic Functions, Archives of Rational Mechanics, 1969 pp  Dill, EH, Continuum Mechanics, Elasticity, Plasticity, Viscoelasticity, CRC Press, 2007  Bonet J & Wood, RD, Nonlinear Mechanics for Finite Element Analysis, Cambridge University Press, www.cambridge.org 2008  Wenger, J & Haddow, JB, Introduction to Continuum Mechanics & Thermodynamics, Cambridge University Press, www.cambridge.org 2010Department of Systems Engineering, University of Lagos 11 oafak@unilag.edu.ng 12/31/2012
  12. 12. Texts  Li, S & Wang G, Introduction to Micromechanics and Nanomechanics, World Scientific, 2008  Wolfram, S The Mathematica Book, 5th Edition Wolgram Media 2003  Trott, M, The Mathematica Guidebook, 4 volumes: Symbolics, Numerics, Graphics &Programming, Springer 2000  Sokolnikoff, IS, Tensor Analysis, Theory and Applications to Geometry and Mechanics of Continua, John Wiley, 1964Department of Systems Engineering, University of Lagos 12 oafak@unilag.edu.ng 12/31/2012
  13. 13. Linear Spaces Introduction
  14. 14. Unified Theory  Continuum Mechanics can be thought of as the grand unifying theory of engineering science.  Many of the courses taught in an engineering curriculum are closely related and can be obtained as special cases of the general framework of continuum mechanics.  This fact is easily lost on most undergraduate and even some graduate students.Department of Systems Engineering, University of Lagos 14 oafak@unilag.edu.ng 12/30/2012
  15. 15. Continuum MechanicsDepartment of Systems Engineering, University of Lagos 15 oafak@unilag.edu.ng 12/30/2012
  16. 16. Physical Quantities Continuum Mechanics views matter as continuously distributed in space. The physical quantities we are interested in can be • Scalars or reals, such as time, energy, power, • Vectors, for example, position vectors, velocities, or forces, • Tensors: deformation gradient, strain and stress measures.Department of Systems Engineering, University of Lagos 16 oafak@unilag.edu.ng 12/30/2012
  17. 17. Physical Quantities  Since we can also interpret scalars as zeroth-order tensors, and vectors as 1st-order tensors, all continuum mechanical quantities can generally be considered as tensors of different orders.  It is therefore clear that a thorough understanding of Tensors is essential to continuum mechanics. This is NOT always an easy requirement;  The notational representation of tensors is often inconsistent as different authors take the liberty to express themselves in several ways.Department of Systems Engineering, University of Lagos 17 oafak@unilag.edu.ng 12/30/2012
  18. 18. Physical Quantities  There are two major divisions common in the literature: Invariant or direct notation and the component notation.  Each has its own advantages and shortcomings. It is possible for a reader that is versatile in one to be handicapped in reading literature written from the other viewpoint. In fact, it has been alleged that  “Continuum Mechanics may appear as a fortress surrounded by the walls of tensor notation” It is our hope that the course helps every serious learner overcome these difficultiesDepartment of Systems Engineering, University of Lagos 18 oafak@unilag.edu.ng 12/30/2012
  19. 19. Real Numbers & Tuples  The set of real numbers is denoted by R  Let R 𝑛 be the set of n-tuples so that when 𝑛 = 2, R 2 we have the set of pairs of real numbers. For example, such can be used for the 𝑥 and 𝑦 coordinates of points on a Cartesian coordinate system.Department of Systems Engineering, University of Lagos 19 oafak@unilag.edu.ng 12/30/2012
  20. 20. Vector Space A real vector space V is a set of elements (called vectors) such that, 1. Addition operation is defined and it is commutative and associative underV : that is, 𝒖 + 𝐯 ∈V, 𝒖 + 𝐯 = 𝐯 + 𝒖, 𝒖 + 𝐯 + 𝒘 = 𝒖 + 𝐯 + 𝒘, ∀ 𝒖, 𝐯, 𝒘 ∈ V. Furthermore,V is closed under addition: That is, given that 𝒖, 𝐯 ∈ V, then 𝒘 = 𝒖 + 𝐯 = 𝐯 + 𝒖, ⇒ 𝒘 ∈ V. 2. V contains a zero element 𝒐 such that 𝒖 + 𝒐 = 𝒖 ∀ 𝒖 ∈V. For every 𝒖 ∈ V, ∃ − 𝒖: 𝒖 + −𝒖 = 𝒐. 3. Multiplication by a scalar. For 𝛼, 𝛽 ∈ R and 𝒖, 𝒗 ∈ V, 𝛼𝒖 ∈ V , 1𝒖 = 𝒖, 𝛼 𝛽𝒖 = 𝛼𝛽 𝒖, 𝛼 + 𝛽 𝒖 = 𝛼𝒖 + 𝛽𝒖, 𝛼 𝒖 + 𝐯 = 𝛼𝒖 + 𝛼𝐯.Department of Systems Engineering, University of Lagos 20 oafak@unilag.edu.ng 12/30/2012
  21. 21. Magnitude & Direction  We were previously told that a vector is something that has magnitude and direction. We often represent such objects as directed lines. Do such objects conform to our present definition?  To answer, we only need to see if the three conditions we previously stipulated are met:Department of Systems Engineering, University of Lagos 21 oafak@unilag.edu.ng 12/30/2012
  22. 22. Dimensionality From our definition of the Euclidean space, it is easy to see that, 𝐯 = 𝛼1 𝒖1 + 𝛼2 𝒖2 + ⋯ + 𝛼 𝑛 𝒖 𝑛 such that, 𝛼1 , 𝛼2 , ⋯ , 𝛼 𝑛 ∈ R and 𝒖1 , 𝒖2 , ⋯ , 𝒖 𝑛 ∈ E, is also a vector. The subset 𝒖1 , 𝒖2 , ⋯ , 𝒖 𝑛 ⊂ E is said to be linearly independent or free if for any choice of the subset 𝛼1 , 𝛼2 , ⋯ , 𝛼 𝑛 ⊂ R other than 0,0, ⋯ , 0 , 𝐯 ≠ 0. If it is possible to find linearly independent vector systems of order 𝑛 where 𝑛 is a finite integer, but there is no free system of order 𝑛 + 1, then the dimension ofE is 𝑛. In other words, the dimension of a space is the highest number of linearly independent members it can have.Department of Systems Engineering, University of Lagos 22 oafak@unilag.edu.ng 12/30/2012
  23. 23. Basis  Any linearly independent subset ofE is said to form a basis forE in the sense that any other vector inE can be expressed as a linear combination of members of that subset. In particular our familiar Cartesian vectors 𝒊, 𝒋 and 𝒌 is a famous such subset in three dimensional Euclidean space.  From the above definition, it is clear that a basis is not necessarily unique.Department of Systems Engineering, University of Lagos 23 oafak@unilag.edu.ng 12/30/2012
  24. 24. Directed Line 1. Addition operation for a directed line segment is defined by the parallelogram law for addition. 2. V contains a zero element 𝒐 in such a case is simply a point with zero length.. 3. Multiplication by a scalar 𝛼. Has the meaning that 0 < 𝛼 ≤ 1 → Line is shrunk along the same direction by 𝛼 𝛼 > 1 → Elongation by 𝛼 Negative value is same as the above with a change of direction.Department of Systems Engineering, University of Lagos 24 oafak@unilag.edu.ng 12/30/2012
  25. 25. Other Vectors  Now we have confirmed that our original notion of a vector is accommodated. It is not all that possess magnitude and direction that can be members of a vector space.  A book has a size and a direction but because we cannot define addition, multiplication by a scalar as we have done for the directed line segment, it is not a vector.Department of Systems Engineering, University of Lagos 25 oafak@unilag.edu.ng 12/30/2012
  26. 26. Other Vectors Complex Numbers. The set C of complex numbers is a real vector space or equivalently, a vector space over R. 2-D Coordinate Space. Another real vector space is the set of all pairs of 𝑥 𝑖 ∈ R forms a 2-dimensional vector space overR is the two dimensional coordinate space you have been graphing on! It satisfies each of the requirements: 𝒙 = {𝑥1 , 𝑥2 }, 𝑥1 , 𝑥2 ∈ R, 𝒚 = {𝑦1 , 𝑦2 }, 𝑦1 , 𝑦2 ∈ R.  Addition is easily defined as 𝒙 + 𝒚 = {𝑥1 + 𝑦1 , 𝑥2 + 𝑦2 } clearly 𝒙 + 𝒚 ∈ R 2 since 𝑥1 + 𝑦1 , 𝑥2 + 𝑦2 ∈ R. Addition operation creates other members for the vector space – Hence closure exists for the operation.  Multiplication by a scalar: 𝛼𝒙 = {𝛼𝑥1 , 𝛼𝑥2 }, ∀𝛼 ∈ R.  Zero element: 𝒐 = 0,0 . Additive Inverse: −𝒙 = −𝑥1 , −𝑥2 , 𝑥1 , 𝑥2 ∈ R Type equation here.A standard basis for this : 𝒆1 = 1,0 , 𝒆2 = 0,1 Type equation here.Any other member can be expressed in terms of this basis.Department of Systems Engineering, University of Lagos 26 oafak@unilag.edu.ng 12/30/2012
  27. 27. N-tuples 𝒏 −D Coordinate Space. For any positive number 𝑛, we may create 𝑛 −tuples such that, 𝒙 = {𝑥1 , 𝑥2 , … , 𝑥 𝑛 } where 𝑥1 , 𝑥2 , … , 𝑥 𝑛 ∈ R are members of R 𝑛 – a real vector space over theR. 𝒚 = {𝑦1 , 𝑦2 , … , 𝑦 𝑛 }, 𝑦1 , 𝑦2 , … , 𝑦 𝑛 ∈ R. 𝒙 + 𝒚 = {𝑥1 + 𝑦1 , 𝑥2 + 𝑦2 , … , 𝑥 𝑛 + 𝑦 𝑛 },. 𝒙 + 𝒚 ∈ R 𝑛 since 𝑥1 + 𝑦1 , 𝑥2 + 𝑦2 , … , 𝑥 𝑛 + 𝑦 𝑛 ∈ R Addition operation creates other members for the vector space – Hence closure exists for the operation.  Multiplication by a scalar: 𝛼𝒙 = {𝛼𝑥1 , 𝛼𝑥2 , … , 𝛼𝑥 𝑛 }, ∀𝛼 ∈ 𝑅.  Zero element 𝒐 = 0,0, … , 0 . Additive Inverse −𝒙 = −𝑥1 , −𝑥2 , … , −𝑥 𝑛 , 𝑥1 , 𝑥2 , … , 𝑥 𝑛 ∈ R There is also a standard basis which is easily proved to be linearly independent: 𝒆1 = 1,0, … , 0 , 𝒆2 = 0,1, … , 0 , … , 𝒆 𝑛 = {0,0, … , 0}Department of Systems Engineering, University of Lagos 27 oafak@unilag.edu.ng 12/30/2012
  28. 28. Matrices Let R 𝑚×𝑛 denote the set of matrices with entries that are real numbers (same thing as saying members of the real spaceR, Then, R 𝑚×𝑛 is a real vector space. Vector addition is just matrix addition and scalar multiplication is defined in the obvious way (by multiplying each entry by the same scalar). The zero vector here is just the zero matrix. The dimension of this space is 𝑚𝑛. For example, in R3×3 we can choose basis in the form, 1 0 0 0 1 0 0 0 0  0 0 0 , 0 0 0 , …, 0 0 0 0 0 0 0 0 0 0 0 1Department of Systems Engineering, University of Lagos 28 oafak@unilag.edu.ng 12/30/2012
  29. 29. The Polynomials Forming polynomials with a single variable 𝑥 to order 𝑛 when 𝑛 is a real number creates a vector space. It is left as an exercise to demonstrate that this satisfies all the three rules of what a vector space is.Department of Systems Engineering, University of Lagos 29 oafak@unilag.edu.ng 12/30/2012
  30. 30. Global Chart A mapping from the Euclidean Point spaceE to the product space R 3 is called a global chart. 𝜙: E → R 𝑛 So that ∀ 𝑥 ∈ 𝑉, ∃ {𝜙1 𝑥 , 𝜙 2 𝑥 , 𝜙 3 (𝑥)} which are the coordinates identified with each 𝑥.  A global chart is sometimes also called a coordinate system.Department of Systems Engineering, University of Lagos 30 oafak@unilag.edu.ng 12/30/2012
  31. 31. Euclidean Vector Space  An Inner-Product (also called a Euclidean Vector) SpaceE is a real vector space that defines the scalar product: for each pair 𝒖, 𝐯 ∈ E, ∃ 𝑙 ∈ R such that, 𝑙 = 𝒖 ⋅ 𝐯 = 𝐯 ⋅ 𝒖 . Further, 𝒖 ⋅ 𝒖 ≥ 0, the zero value occurring only when 𝒖 = 0. It is called “Euclidean” because the laws of Euclidean geometry hold in such a space.  The inner product also called a dot product, is the mapping " ⋅ ": V × V → R from the product space to the real space.Department of Systems Engineering, University of Lagos 31 oafak@unilag.edu.ng 12/30/2012
  32. 32. Co-vectors  A mapping from a vector space is also called a functional; a term that is more appropriate when we are looking at a function space.  A linear functional 𝐯 ∗: V → R on the vector spaceV is called a convector or a dual vector. For a finite dimensional vector space, the set of all convectors forms the dual spaceV ∗ ofV . If V is an Inner Product Space, then there is no distinction between the vector space and its dual.Department of Systems Engineering, University of Lagos 32 oafak@unilag.edu.ng 12/30/2012
  33. 33. Magnitude & Direction Again Magnitude The norm, length or magnitude of 𝒖, denoted 𝒖 is defined as the positive square root of 𝒖 ⋅ 𝒖 = 𝒖 2 . When 𝒖 = 1, 𝒖 is said to be a unit vector. When 𝒖 ⋅ 𝐯 = 0, 𝒖 and 𝐯 are said to be orthogonal. Direction Furthermore, for any two vectors 𝒖 and 𝐯, the angle between them is defined as, −1 𝒖⋅ 𝐯 cos 𝒖 𝐯 The scalar distance 𝑑 between two vectors 𝒖 and 𝐯 𝑑 = 𝒖− 𝐯Department of Systems Engineering, University of Lagos 33 oafak@unilag.edu.ng 12/30/2012
  34. 34. 3-D Euclidean Space  A 3-D Euclidean space is a Normed space because the inner product induces a norm on every member.  It is also a metric space because we can find distances and angles and therefore measure areas and volumes  Furthermore, in this space, we can define the cross product, a mapping from the product space " × ": V × V → V Which takes two vectors and produces a vector.Department of Systems Engineering, University of Lagos 34 oafak@unilag.edu.ng 12/30/2012
  35. 35. Cross Product Without any further ado, our definition of cross product is exactly the same as what you already know from elementary texts. We simply repeat a few of these for emphasis: 1. The magnitude 𝒂 × 𝒃 = 𝒂 𝒃 sin 𝜃 (0 ≤ 𝜃 ≤ 𝜋) of the cross product 𝒂 × 𝒃 is the area 𝐴(𝒂, 𝒃) spanned by the vectors 𝒂 and 𝒃. This is the area of the parallelogram defined by these vectors. This area is non-zero only when the two vectors are linearly independent. 2. 𝜃 is the angle between the two vectors. 3. The direction of 𝒂 × 𝒃 is orthogonal to both 𝒂 and 𝒃Department of Systems Engineering, University of Lagos 35 oafak@unilag.edu.ng 12/30/2012
  36. 36. Cross Product The area 𝐴(𝒂, 𝒃) spanned by the vectors 𝒂 and 𝒃. The unit vector 𝒏 in the direction of the cross product can be 𝒂×𝒃 obtained from the quotient, 𝒂×𝒃 .Department of Systems Engineering, University of Lagos 36 oafak@unilag.edu.ng 12/30/2012
  37. 37. Cross Product The cross product is bilinear and anti-commutative: Given 𝛼 ∈ R , ∀𝒂, 𝒃, 𝒄 ∈ V , 𝛼𝒂 + 𝒃 × 𝒄 = 𝛼 𝒂 × 𝒄 + 𝒃 × 𝒄 𝒂 × 𝛼𝒃 + 𝒄 = 𝛼 𝒂 × 𝒃 + 𝒂 × 𝒄 So that there is linearity in both arguments. Furthermore, ∀𝒂, 𝒃 ∈ V 𝒂 × 𝒃 = −𝒃 × 𝒂Department of Systems Engineering, University of Lagos 37 oafak@unilag.edu.ng 12/30/2012
  38. 38. Tripple Products The trilinear mapping, [ , ,] ∶ V × V × V → R From the product set V × V × V to real space is defined by: [ u,v,w] ≡ 𝒖 ⋅ 𝒗 × 𝒘 = 𝒖 × 𝒗 ⋅ 𝒘 Has the following properties: 1. [ a,b,c]=[b,c,a]=[c,a,b]=−[ b,a,c]=−[c,b,a]=−[a,c,b] HW: Prove this 1. Vanishes when a, b and c are linearly dependent. 2. It is the volume of the parallelepiped defined by a, b and cDepartment of Systems Engineering, University of Lagos 38 oafak@unilag.edu.ng 12/30/2012
  39. 39. Tripple product Parallelepiped defined by u, v and wDepartment of Systems Engineering, University of Lagos 39 oafak@unilag.edu.ng 12/30/2012
  40. 40. Summation Convention  We introduce an index notation to facilitate the expression of relationships in indexed objects. Whereas the components of a vector may be three different functions, indexing helps us to have a compact representation instead of using new symbols for each function, we simply index and achieve compactness in notation. As we deal with higher ranked objects, such notational conveniences become even more important. We shall often deal with coordinate transformations.Department of Systems Engineering, University of Lagos 40 oafak@unilag.edu.ng 12/30/2012
  41. 41. Summation Convention  When an index occurs twice on the same side of any equation, or term within an equation, it is understood to represent a summation on these repeated indices the summation being over the integer values specified by the range. A repeated index is called a summation index, while an unrepeated index is called a free index. The summation convention requires that one must never allow a summation index to appear more than twice in any given expression.Department of Systems Engineering, University of Lagos 41 oafak@unilag.edu.ng 12/30/2012
  42. 42. Summation Convention  Consider transformation equations such as, 𝑦1 = 𝑎11 𝑥1 + 𝑎12 𝑥2 + 𝑎13 𝑥3 𝑦2 = 𝑎21 𝑥1 + 𝑎22 𝑥2 + 𝑎23 𝑥3 𝑦3 = 𝑎31 𝑥1 + 𝑎32 𝑥2 + 𝑎33 𝑥3  We may write these equations using the summation symbols as: 𝑛 𝑦1 = 𝑎1𝑗 𝑥 𝑗 𝑗=1 𝑛 𝑦2 = 𝑎2𝑗 𝑥 𝑗 𝑗=1 𝑛 𝑦3 = 𝑎2𝑗 𝑥 𝑗 𝑗=1Department of Systems Engineering, University of Lagos 42 oafak@unilag.edu.ng 12/30/2012
  43. 43. Summation Convention  In each of these, we can invoke the Einstein summation convention, and write that, 𝑦1 = 𝑎1𝑗 𝑥 𝑗 𝑦2 = 𝑎2𝑗 𝑥 𝑗 𝑦3 = 𝑎2𝑗 𝑥 𝑗  Finally, we observe that 𝑦1 , 𝑦2 , and 𝑦3 can be represented as we have been doing by 𝑦 𝑖 , 𝑖 = 1,2,3 so that the three equations can be written more compactly as, 𝑦 𝑖 = 𝑎 𝑖𝑗 𝑥 𝑗 , 𝑖 = 1,2,3Department of Systems Engineering, University of Lagos 43 oafak@unilag.edu.ng 12/30/2012
  44. 44. Summation Convention Please note here that while 𝑗 in each equation is a dummy index, 𝑖 is not dummy as it occurs once on the left and in each expression on the right. We therefore cannot arbitrarily alter it on one side without matching that action on the other side. To do so will alter the equation. Again, if we are clear on the range of 𝑖, we may leave it out completely and write, 𝑦 𝑖 = 𝑎 𝑖𝑗 𝑥 𝑗 to represent compactly, the transformation equations above. It should be obvious there are as many equations as there are free indices.Department of Systems Engineering, University of Lagos 44 oafak@unilag.edu.ng 12/30/2012
  45. 45. Summation Convention If 𝑎 𝑖𝑗 represents the components of a 3 × 3 matrix 𝐀, we can show that, 𝑎 𝑖𝑗 𝑎 𝑗𝑘 = 𝑏 𝑖𝑘 Where 𝐁 is the product matrix 𝐀𝐀. To show this, apply summation convention and see that, for 𝑖 = 1, 𝑘 = 1, 𝑎11 𝑎11 + 𝑎12 𝑎21 + 𝑎13 𝑎31 = 𝑏11 for 𝑖 = 1, 𝑘 = 2, 𝑎11 𝑎12 + 𝑎12 𝑎22 + 𝑎13 𝑎32 = 𝑏12 for 𝑖 = 1, 𝑘 = 3, 𝑎11 𝑎13 + 𝑎12 𝑎23 + 𝑎13 𝑎33 = 𝑏13 for 𝑖 = 2, 𝑘 = 1, 𝑎21 𝑎11 + 𝑎22 𝑎21 + 𝑎23 𝑎31 = 𝑏21 for 𝑖 = 2, 𝑘 = 2, 𝑎21 𝑎12 + 𝑎22 𝑎22 + 𝑎23 𝑎32 = 𝑏22 for 𝑖 = 2, 𝑘 = 3, 𝑎21 𝑎13 + 𝑎22 𝑎23 + 𝑎23 𝑎33 = 𝑏23 for 𝑖 = 3, 𝑘 = 1, 𝑎31 𝑎11 + 𝑎32 𝑎21 + 𝑎33 𝑎31 = 𝑏31 for 𝑖 = 3, 𝑘 = 2, 𝑎31 𝑎12 + 𝑎32 𝑎22 + 𝑎33 𝑎32 = 𝑏32 for 𝑖 = 3, 𝑘 = 3, 𝑎31 𝑎13 + 𝑎32 𝑎23 + 𝑎33 𝑎33 = 𝑏33Department of Systems Engineering, University of Lagos 45 oafak@unilag.edu.ng 12/30/2012
  46. 46. Summation Convention The above can easily be verified in matrix notation as, 𝑎11 𝑎12 𝑎13 𝑎11 𝑎12 𝑎13 𝐀𝐀 = 𝑎21 𝑎22 𝑎23 𝑎21 𝑎22 𝑎23 𝑎31 𝑎32 𝑎33 𝑎31 𝑎32 𝑎33 𝑏11 𝑏12 𝑏13 = 𝑏21 𝑏22 𝑏23 = 𝐁 𝑏31 𝑏32 𝑏33 In this same way, we could have also proved that, 𝑎 𝑖𝑗 𝑎 𝑘𝑗 = 𝑏 𝑖𝑘  Where 𝐁 is the product matrix 𝐀𝐀T . Note the arrangements could sometimes be counter intuitive.Department of Systems Engineering, University of Lagos 46 oafak@unilag.edu.ng 12/30/2012
  47. 47. Vector Components Suppose our basis vectors 𝐠 𝑖 , 𝑖 = 1,2,3 are not only not unit in magnitude, but in addition are NOT orthogonal. The only assumption we are making is that 𝐠 𝑖 ∈ V , 𝑖 = 1,2,3 are linearly independent vectors. With respect to this basis, we can express vectors 𝐯, 𝐰 ∈ V in terms of the basis as, 𝐯 = 𝑣 𝑖 𝐠 𝑖, 𝐰 = 𝑤 𝑖 𝐠 𝑖 Where each 𝑣 𝑖 is called the contravariant component of 𝐯Department of Systems Engineering, University of Lagos 47 oafak@unilag.edu.ng 12/30/2012
  48. 48. Vector Components Clearly, addition and linearity of the vector space ⇒ 𝐯 + 𝐰 = (𝑣 𝑖 +𝑤 𝑖 )𝐠 𝑖 Multiplication by scalar rule implies that if 𝛼 ∈ R , ∀𝐯 ∈ V , 𝛼 𝐯 = (𝛼𝑣 𝑖 )𝐠 𝑖Department of Systems Engineering, University of Lagos 48 oafak@unilag.edu.ng 12/30/2012
  49. 49. Reciprocal Basis For any basis vectors 𝐠 𝑖 ∈ V , 𝑖 = 1,2,3 there is a reciprocal basis defined by the reciprocity relationship: 𝐠 𝑖 ⋅ 𝐠 𝑗 = 𝐠 𝑗 ⋅ 𝐠 𝑖 = 𝛿 𝑗𝑖 Where 𝛿 𝑗𝑖 is the Kronecker delta 𝑖 0 if 𝑖 ≠ 𝑗 𝛿𝑗 = 1 if 𝑖 = 𝑗 Let the fact that the above equations are actually nine equations each sink. Consider the full meaning:Department of Systems Engineering, University of Lagos 49 oafak@unilag.edu.ng 12/30/2012
  50. 50. Kronecker Delta Kronecker Delta: 𝛿 𝑖𝑗 , 𝛿 𝑖𝑗 or 𝛿 𝑗𝑖 has the following properties: 𝛿11 = 1, 𝛿12 = 0, 𝛿13 = 0 𝛿21 = 0, 𝛿22 = 1, 𝛿23 = 0 𝛿31 = 0, 𝛿32 = 0, 𝛿33 = 1 As is obvious, these are obtained by allowing the indices to attain all possible values in the range. The Kronecker delta is defined by the fact that when the indices explicit values are equal, it has the value of unity. Otherwise, it is zero. The above nine equations can be written more compactly as, 0 if 𝑖 ≠ 𝑗 𝛿 𝑗𝑖 = 1 if 𝑖 = 𝑗Department of Systems Engineering, University of Lagos 50 oafak@unilag.edu.ng 12/30/2012
  51. 51. Covariant components  For any ∀𝐯 ∈ V , 𝐯 = 𝑣 𝑖 𝐠 𝑖 = 𝑣𝑖 𝐠 𝑖 Are two related representations in the reciprocal bases. Taking the inner product of the above equation with the basis vector 𝐠 𝑗 , we have 𝐯 ⋅ 𝐠 𝑗 = 𝑣 𝑖 𝐠 𝑖 ⋅ 𝐠 𝑗 = 𝑣𝑖 𝐠 𝑖 ⋅ 𝐠 𝑗 Which gives us the covariant component, 𝐯 ⋅ 𝐠 𝑗 = 𝑣 𝑖 𝑔 𝑖𝑗 = 𝑣 𝑖 𝛿 𝑗𝑖 = 𝑣 𝑗Department of Systems Engineering, University of Lagos 51 oafak@unilag.edu.ng 12/30/2012
  52. 52. Contravariant Components In the same easy manner, we may evaluate the contravariant components of the same vector by taking the dot product of the same equation with the contravariant base vector 𝐠 𝑗 : 𝐯 ⋅ 𝐠 𝑗 = 𝑣 𝑖 𝐠 𝑖 ⋅ 𝐠 𝑗 = 𝑣𝑖 𝐠 𝑖 ⋅ 𝐠 𝑗 So that, 𝑗 𝐯 ⋅ 𝐠 𝑗 = 𝑣 𝑖 𝛿 𝑖 = 𝑣 𝑖 𝑔 𝑖𝑗 = 𝑣 𝑗Department of Systems Engineering, University of Lagos 52 oafak@unilag.edu.ng 12/30/2012
  53. 53. Metric coefficients The nine scalar quantities, 𝑔 𝑖𝑗 as well as the nine related quantities 𝑔 𝑖𝑗 play important roles in the coordinate system spanned by these arbitrary reciprocal set of basis vectors as we shall see. They are called metric coefficients because they metrize the space defined by these bases.Department of Systems Engineering, University of Lagos 53 oafak@unilag.edu.ng 12/30/2012
  54. 54. Levi Civita Symbol  The Levi-Civita Symbol: 𝑒 𝑖𝑗𝑘  𝑒111 = 0, 𝑒112 = 0, 𝑒113 = 0, 𝑒121 = 0, 𝑒122 = 0, 𝑒123 = 1, 𝑒131 = 0, 𝑒132 = −1, 𝑒133 = 0 𝑒211 = 0, 𝑒212 = 0, 𝑒213 = −1, 𝑒221 = 0, 𝑒222 = 0, 𝑒223 = 0, 𝑒231 = 1, 𝑒232 = 0, 𝑒233 = 0 𝑒311 = 0, 𝑒312 = 1, 𝑒313 = 0, 𝑒321 = −1, 𝑒322 = 0, 𝑒323 = 1, 𝑒331 = 0, 𝑒332 = 0, 𝑒333 = 0Department of Systems Engineering, University of Lagos 54 oafak@unilag.edu.ng 12/30/2012
  55. 55. Levi Civita Symbol  While the above equations might look arbitrary at first, a closer look shows there is a simple logic to it all. In fact, note that whenever the value of an index is repeated, the symbol has a value of zero. Furthermore, we can see that once the indices are an even arrangement (permutation) of 1,2, and 3, the symbols have the value of 1, When we have an odd arrangement, the value is -1. Again, we desire to avoid writing twenty seven equations to express this simple fact. Hence we use the index notation to define the Levi-Civita symbol as follows: 1 if 𝑖, 𝑗 and 𝑘 are an even permutation of 1,2 and 3  𝑒 𝑖𝑗𝑘 = −1 if 𝑖, 𝑗 and 𝑘 are an odd permutation of 1,2 and 3 0 In all other casesDepartment of Systems Engineering, University of Lagos 55 oafak@unilag.edu.ng 12/30/2012
  56. 56. Cross Product of Basis Vectors Given that 𝑔 = det 𝑔 𝑖𝑗 of the covariant metric coefficients, It is not difficult to prove that 𝐠 𝑖 ∙ 𝐠 𝑗 × 𝐠 𝑘 = 𝜖 𝑖𝑗𝑘 ≡ 𝑔𝑒 𝑖𝑗𝑘  This relationship immediately implies that, 𝐠 𝑖 × 𝐠 𝑗 = 𝜖 𝑖𝑗𝑘 𝐠 𝑘 .  The dual of the expression, the equivalent contravariant equivalent also follows from the fact that, 𝐠1 × 𝐠 2 ⋅ 𝐠 3 = 1/ 𝑔Department of Systems Engineering, University of Lagos 56 oafak@unilag.edu.ng 12/30/2012
  57. 57. Cross Product of Basis Vectors This leads in a similar way to the expression, 𝑒 𝑖𝑗𝑘 𝐠𝑖 × 𝐠𝑗 ⋅ 𝐠 𝑘 = = 𝜖 𝑖𝑗𝑘 𝑔 It follows immediately from this that, 𝐠 𝑖 × 𝐠 𝑗 = 𝜖 𝑖𝑗𝑘 𝐠 𝑘Department of Systems Engineering, University of Lagos 57 oafak@unilag.edu.ng 12/30/2012
  58. 58. Exercises  Given that, 𝐠 𝟏 , 𝐠 2 and 𝐠 3 are three linearly independent vectors and satisfy 𝐠 𝑖 ⋅ 𝐠 𝑗 = 𝛿 𝑗𝑖 , show 1 𝟏 1 that 𝐠1 = 𝐠 2 × 𝐠 3 , 𝐠 2 = 𝐠 3 × 𝐠1 , and 𝐠 3 = 𝐠1 × 𝑉 𝑽 𝑉 𝐠 2 , where 𝑉 = 𝐠1 ⋅ 𝐠 2 × 𝐠 3Department of Systems Engineering, University of Lagos 58 oafak@unilag.edu.ng 12/30/2012
  59. 59. It is clear, for example, that 𝐠1 is perpendicular to 𝐠 2 as well as to 𝐠 3 (an obvious fact because 𝐠1 ⋅ 𝐠 2 = 0 and 𝐠1 ⋅ 𝐠 3 = 0), we can say that the vector 𝐠1 must necessarily lie on the cross product 𝐠 2 × 𝐠 3 of 𝐠 2 and 𝐠 3 . It is therefore correct to write, 1 = 𝟏 𝐠 𝐠2 × 𝐠3 𝑽 Where 𝑉 −1 is a constant we will now determine. We can do this right away by taking the dot product of both sides of the equation (5) with 𝐠1 we immediately obtain, 𝐠1 ⋅ 𝐠1 = 𝑉 −1 𝐠1 ⋅ 𝐠 2 × 𝐠 3 = 1 So that, 𝑉 = 𝐠1 ⋅ 𝐠 2 × 𝐠 3 the volume of the parallelepiped formed by the three vectors 𝐠1 , 𝐠 2 , and 𝐠 3 when their origins are made to coincide.Department of Systems Engineering, University of Lagos 59 oafak@unilag.edu.ng 12/30/2012
  60. 60. Show that 𝑔 𝑖𝑗 𝑔 𝑗𝑘 = 𝛿 𝑖 𝑘 First note that 𝐠 𝑗 = 𝑔 𝑖𝑗 𝐠 𝑖 = 𝑔 𝑗𝑖 𝐠 𝑖 . Recall the reciprocity relationship: 𝐠 𝑖 ⋅ 𝐠 𝑘 = 𝛿 𝑖 𝑘 . Using the above, we can write 𝐠 𝑖 ⋅ 𝐠 𝑘 = 𝑔 𝑖𝛼 𝐠 𝛼 ⋅ 𝑔 𝑘𝛽 𝐠 𝛽 = 𝑔 𝑖𝛼 𝑔 𝑘𝛽 𝐠 𝛼 ⋅ 𝐠 𝛽 = 𝑔 𝑖𝛼 𝑔 𝑘𝛽 𝛿 𝛽𝛼 = 𝛿 𝑖 𝑘 which shows that 𝑔 𝑖𝛼 𝑔 𝑘𝛼 = 𝑔 𝑖𝑗 𝑔 𝑗𝑘 = 𝛿 𝑖 𝑘 As required. This shows that the tensor 𝑔 𝑖𝑗 and 𝑔 𝑖𝑗 are inverses of each other.Department of Systems Engineering, University of Lagos 60 oafak@unilag.edu.ng 12/30/2012
  61. 61. Show that the cross product of vectors 𝒂 and 𝒃 in general coordinates is 𝑎 𝑖 𝑏 𝑗 𝜖 𝑖𝑗𝑘 𝐠 𝑘 or 𝜖 𝑖𝑗𝑘 𝑎 𝑖 𝑏 𝑗 𝐠 𝑘 where 𝑎 𝑖 , 𝑏 𝑗 are the respective contravariant components and 𝑎 𝑖 , 𝑏 𝑗 the covariant. Express vectors 𝒂 and 𝒃 as contravariant components: 𝒂 = 𝑎 𝑖 𝐠 𝑖 , and 𝒃 = 𝑏 𝑖 𝐠 𝑖 . Using the above result, we can write that, 𝒂 × 𝒃 = 𝑎𝑖 𝐠𝑖 × 𝑏 𝑗 𝐠 𝑗 = 𝑎 𝑖 𝑏 𝑗 𝐠 𝑖 × 𝐠 𝑗 = 𝑎 𝑖 𝑏 𝑗 𝜖 𝑖𝑗𝑘 𝐠 𝑘 . Express vectors 𝒂 and 𝒃 as covariant components: 𝒂 = 𝑎 𝑖 𝐠 𝑖 and 𝒃 = 𝑏 𝑖 𝐠 𝑖 . Again, proceeding as before, we can write, 𝒂 × 𝒃 = 𝑎 𝑖 𝐠 𝑖 × 𝑏𝑗 𝐠 𝑗 = 𝜖 𝑖𝑗𝑘 𝑎 𝑖 𝑏 𝑗 𝐠 𝑘 Express vectors 𝒂 as a contravariant components: 𝒂 = 𝑎 𝑖 𝐠 𝑖 and 𝒃 as covariant components: 𝒃 = 𝑏 𝑖 𝐠 𝑖 𝒂 × 𝒃 = 𝑎 𝑖 𝐠 𝑖 × 𝑏𝑗 𝐠 𝑗 = 𝑎 𝑖 𝑏𝑗 𝐠 𝑖 × 𝐠 𝑗 = 𝒂⋅ 𝒃 𝐠𝑖 × 𝐠𝑗Department of Systems Engineering, University of Lagos 61 oafak@unilag.edu.ng 12/30/2012
  62. 62. Given that, 𝛿 𝑖𝑟 𝛿 𝑗 𝑟 𝛿 𝑘𝑟 𝑠 𝑟𝑠𝑡 𝛿 𝑖𝑗𝑘 ≡ 𝑒 𝑟𝑠𝑡 𝑒 𝑖𝑗𝑘 = 𝛿 𝑖 𝛿 𝑗𝑠 𝛿 𝑘𝑠 𝛿 𝑖𝑡 𝛿 𝑗𝑡 𝛿 𝑘𝑡 Show that 𝛿 𝑖𝑗𝑘 = 𝛿 𝑖𝑟 𝛿 𝑗𝑠 − 𝛿 𝑖𝑠 𝛿 𝑗𝑟 𝑟𝑠𝑘 Expanding the equation, we have: 𝛿 𝑗 𝑟 𝛿 𝑘𝑟 𝛿 𝑖𝑟 𝛿 𝑘𝑟 𝛿 𝑖𝑟 𝛿 𝑗𝑟 𝑒 𝑖𝑗𝑘 𝑒 𝑟𝑠𝑘 = 𝛿 𝑖𝑗𝑘 = 𝛿 𝑖 𝑘 𝑠 𝑟𝑠𝑘 𝑠 − 𝛿𝑗 𝑘 +3 𝑠 𝛿𝑗 𝛿𝑘 𝛿 𝑖𝑠 𝛿 𝑘𝑠 𝛿𝑖 𝛿 𝑗𝑠 = 𝛿 𝑖 𝑘 𝛿 𝑗𝑟 𝛿 𝑘𝑠 − 𝛿 𝑗𝑠 𝛿 𝑘𝑟 − 𝛿 𝑗 𝑘 𝛿 𝑖𝑟 𝛿 𝑘𝑠 − 𝛿 𝑖𝑠 𝛿 𝑘𝑟 + 3 𝛿 𝑖𝑟 𝛿 𝑗 𝑠 − 𝛿 𝑖𝑠 𝛿 𝑗𝑟 = 𝛿 𝑗𝑟 𝛿 𝑖 𝑠 − 𝛿 𝑗 𝑠 𝛿 𝑖𝑟 − 𝛿 𝑖𝑟 𝛿 𝑗𝑠 + 𝛿 𝑖𝑠 𝛿 𝑗 𝑟 +3(𝛿 𝑖𝑟 𝛿 𝑗 𝑠 − 𝛿 𝑖𝑠 𝛿 𝑗 𝑟 ) = −2(𝛿 𝑖𝑟 𝛿 𝑗𝑠 − 𝛿 𝑖𝑠 𝛿 𝑗𝑟 ) + 3(𝛿 𝑖𝑟 𝛿 𝑗 𝑠 − 𝛿 𝑖𝑠 𝛿 𝑗𝑟 ) = 𝛿 𝑖𝑟 𝛿 𝑗𝑠 − 𝛿 𝑖𝑠 𝛿 𝑗𝑟Department of Systems Engineering, University of Lagos 62 oafak@unilag.edu.ng 12/30/2012
  63. 63. 𝑟𝑗𝑘 Show that 𝛿 𝑖𝑗𝑘 = 2𝛿 𝑖𝑟 Contracting one more index, we have: 𝑟𝑗𝑘 𝑗 𝑗 𝑒 𝑖𝑗𝑘 𝑒 𝑟𝑗𝑘 = 𝛿 𝑖𝑗𝑘 = 𝛿 𝑖𝑟 𝛿 𝑗 − 𝛿 𝑖 𝛿 𝑗𝑟 = 3𝛿 𝑖𝑟 − 𝛿 𝑖𝑟 = 2𝛿 𝑖𝑟 These results are useful in several situations.Department of Systems Engineering, University of Lagos 63 oafak@unilag.edu.ng 12/30/2012
  64. 64. Show that 𝐮 ⋅ 𝐮 × 𝐯 = 0 𝐮 × 𝐯 = ϵ 𝑖𝑗𝑘 𝑢 𝑖 𝑣 𝑗 𝐠 𝑘 𝐮 ⋅ 𝐮 × 𝐯 = 𝑢 𝛼 𝐠 𝛼 ⋅ ϵ 𝑖𝑗𝑘 𝑢 𝑖 𝑣 𝑗 𝐠 𝑘 = 𝑢 𝛼 ϵ 𝑖𝑗𝑘 𝑢 𝑖 𝑣 𝑗 𝛿 𝑘𝛼 = ϵ 𝑖𝑗𝑘 𝑢 𝑖 𝑣 𝑗 𝑢 𝑘 =0 On account of the symmetry and antisymmetry in 𝑖 and 𝑘.Department of Systems Engineering, University of Lagos 64 oafak@unilag.edu.ng 12/30/2012
  65. 65. Show that 𝐮 × 𝐯 = − 𝐯 × 𝐮 𝐮 × 𝐯 = ϵ 𝑖𝑗𝑘 𝑢 𝑖 𝑣 𝑗 𝐠 𝑘 = −ϵ 𝑗𝑖𝑘 𝑢 𝑖 𝑣 𝑗 𝐠 𝑘 = −ϵ 𝑖𝑗𝑘 𝑣 𝑖 𝑢 𝑗 𝐠 𝑘 =− 𝐯 × 𝐮Department of Systems Engineering, University of Lagos 65 oafak@unilag.edu.ng 12/30/2012
  66. 66. Show that 𝐮 × 𝐯 × 𝐰 = 𝐮 ⋅ 𝐰 𝐯 − 𝐮 ⋅ 𝐯 𝐰. Let 𝐳 = 𝐯 × 𝐰 = ϵ 𝑖𝑗𝑘 𝑣 𝑖 𝑤 𝑗 𝐠 𝑘 𝐮 × 𝐯 × 𝐰 = 𝐮 × 𝐳 = 𝜖 𝛼𝛽𝛾 𝑢 𝛼 𝑧 𝛽 𝐠 𝛾 = 𝜖 𝛼𝛽𝛾 𝑢 𝛼 𝑧 𝛽 𝐠 𝛾 = 𝜖 𝛼𝛽𝛾 𝑢 𝛼 ϵ 𝑖𝑗𝛽 𝑣 𝑖 𝑤 𝑗 𝐠 𝛾 = ϵ 𝑖𝑗𝛽 𝜖 𝛾𝛼𝛽 𝑢 𝛼 𝑣 𝑖 𝑤 𝑗 𝐠 𝛾 𝑗 𝑗 = 𝛿 𝑖𝛾 𝛿 𝛼 − 𝛿 𝑖𝛼 𝛿 𝛾 𝑢 𝛼 𝑣 𝑖 𝑤 𝑗 𝐠 𝛾 = 𝑢 𝑗 𝑣 𝛾 𝑤𝑗 𝐠 𝛾 − 𝑢 𝑖 𝑣 𝑖 𝑤 𝛾 𝐠 𝛾 = 𝐮⋅ 𝐰 𝐯− 𝐮⋅ 𝐯 𝐰Department of Systems Engineering, University of Lagos 66 oafak@unilag.edu.ng 12/30/2012
  67. 67. Department of Systems Engineering, University of Lagos 67 oafak@unilag.edu.ng 12/30/2012

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