AEM ADV03 Introductory Mathematics

Transcription

1 AEM ADV03 Introductory Mathematics 2017/2018

2 Table of Contents: Introductory Mathematics Function Expansion & Transforms Infinite series Convergence Power Series Taylor Series Fourier Series Complex Fourier series Termwise Integration and Differentiation Fourier series of Odd and Even functions Integral Transform Fourier Transform Laplace Transform Vector Spaces, vector Fields & Operators Scalar (inner) product of vector fields Lp norms Vector product of vector fields Vector operators Gradient of a scalar field Divergence of a vector field Curl of a vector field Repeated Vector Operations The Laplacian Linear Algebra, Matrices & Eigenvectors Basic definitions and notation Multiplication of matrices and multiplication of vectors and matrices Matrix multiplication Traces and determinants of square Cayley products The Kronecker product Matrix Rank and the Inverse of a full rank matrix Full Rank matrices

3 3.3.2 Solutions of linear equations Preservation of positive definiteness A lower bound on the rank of a matrix product Inverse of products and sums of matrices Eigensystems Diagonalisation of symmetric matrices Matrix Factorization Similarity Transform LU decomposition QR decomposition Singular Value decomposition Solution of linear systems Direct methods Iterative methods

4 Introductory Mathematics What is Mathematics? Different schools of thought, particularly in philosophy, have put forth radically different definitions of mathematics. All are controversial and there is no consensus. Leading definitions 1. Aristotle defined mathematics as: The science of quantity. In Aristotle's classification of the sciences, discrete quantities were studied by arithmetic, continuous quantities by geometry. 2. Auguste Comte's definition tried to explain the role of mathematics in coordinating phenomena in all other fields: The science of indirect measurement, The ``indirectness'' in Comte's definition refers to determining quantities that cannot be measured directly, such as the distance to planets or the size of atoms, by means of their relations to quantities that can be measured directly. 3. Benjamin Peirce: Mathematics is the science that draws necessary conclusions, Bertrand Russell: All Mathematics is Symbolic Logic, Walter Warwick Sawyer: Mathematics is the classification and study of all possible patterns, Most contemporary reference works define mathematics mainly by summarizing its main topics and methods: 6. Oxford English Dictionary: The abstract science which investigates deductively the conclusions implicit in the elementary conceptions of spatial and numerical relations, and which includes as its main divisions geometry, arithmetic, and algebra, American Heritage Dictionary: The study of the measurement, properties, and relationships of quantities and sets, using numbers and symbols, Other playful, metaphorical, and poetic definitions: 8. Bertrand Russell: The subject in which we never know what we are talking about, nor whether what we are saying is true, Charles Darwin: A mathematician is a blind man in a dark room looking for a black cat which isn't there. 10. G. H. Hardy: A mathematician, like a painter or poet, is a maker of patterns. If his patterns are more permanent than theirs, it is because they are made with ideas, Field of Mathematics Mathematics can, broadly speaking, be subdivided into the study of quantity, structure, space, and change (i.e. arithmetic, algebra, geometry, and analysis). In addition to these main concerns, there are also 4

5 subdivisions dedicated to exploring links from the heart of mathematics to other fields: to logic, to set theory (foundations), to the empirical mathematics of the various sciences (applied mathematics), and more recently to the rigorous study of uncertainty. Mathematical awards Arguably the most prestigious award in mathematics is the Fields Medal, established in 1936 and now awarded every four years. The Fields Medal is often considered a mathematical equivalent to the Nobel Prize. The Wolf Prize in Mathematics, instituted in 1978, recognizes lifetime achievement, and another major international award, the Abel Prize, was introduced in The Chern Medal was introduced in 2010 to recognize lifetime achievement. These accolades are awarded in recognition of a particular body of work, which may be innovational, or provide a solution to an outstanding problem in an established field. A famous list of 23 open problems, called Hilbert's problem, was compiled in 1900 by German mathematician David Hilbert. This list achieved great celebrity among mathematicians, and at least nine of the problems have now been solved. A new list of seven important problems, titled the Millennium Prize Problems, was published in A solution to each of these problems carries a $1 million reward, and only one (the Riemann hypothesis) is duplicated in Hilbert's problems. Mathematics in Aeronautics Mathematics in aeronautics includes calculus, differential equations, and linear algebra, etc. Calculus 1 Calculus has been an integral part of man's intellectual training and heritage for the last twenty-five hundred years. Calculus is the mathematical study of change, in the same way that geometry is the study of shape and algebra is the study of operations and their application to solving equations. It has two major branches, differential calculus (concerning rates of change and slopes of curves), and integral calculus (concerning accumulation of quantities and the areas under and between curves); these two branches are related to each other by the fundamental theorem of calculus. Both branches make use of the fundamental notions of convergence of infinite sequences and infinite series to a well-defined limit. Generally, modern calculus is considered to have been developed in the 17th century by Isaac Newton and Gottfried Leibniz, today calculus has widespread uses in science, engineering and economics and can solve many problems that algebra alone cannot. 1 Extracted from: Boyer, Carl Benjamin. The history of the calculus and its conceptual development. Courier Dover Publications,

6 Differential and integral calculus is one of the great achievements of the human mind. The fundamental definitions of the calculus, those of the derivative and the integral, are now so clearly stated in textbooks on the subject, and the operations involving them are so readily mastered, that it is easy to forget the difficulty with which these basic concepts have been developed. Frequently a clear and adequate understanding of the fundamental notions underlying a branch of knowledge has been achieved comparatively late in its development. This has never been more aptly demonstrated than in the rise of the calculus. The precision of statement and the facility of application which the rules of the calculus early afforded were in a measure responsible for the fact that mathematicians were insensible to the delicate subtleties required in the logical development of the discipline. They sought to establish the calculus in terms of the conceptions found in the traditional geometry and algebra which had been developed from spatial intuition. During the eighteenth century, however, the inherent difficulty of formulating the underlying concepts became increasingly evident, and it then became customary to speak of the metaphysics of the calculus, thus implying the inadequacy of mathematics to give a satisfactory exposition of the bases. With the clarification of the basic notions --which, in the nineteenth century, was given in terms of precise mathematical terminology-- a safe course was steered between the intuition of the concrete in nature (which may lurk in geometry and algebra) and the mysticism of imaginative speculation (which may thrive on transcendental metaphysics). The derivative has throughout its development been thus precariously situated between the scientific phenomenon of velocity and the philosophical noumenon of motion. The history of integration is similar. On the one hand, it had offered ample opportunity for interpretations by positivistic thought in terms either of approximations or of the compensation of errors, views based on the admitted approximate nature of scientific measurements and on the accepted doctrine of superimposed effects. On the other hand, it has at the same time been regarded by idealistic metaphysics as a manifestation that beyond the finitism of sensory percipiency there is a transcendent infinite which can be but asymptotically approached by human experience and reason. Only the precision of their mathematical definition --the work of the nineteenth century-- enables the derivative and the integral to maintain their autonomous position as abstract concepts, perhaps derived from, but nevertheless independent of, both physical description and metaphysical explanation. 6

7 1 Function Expansion & Transforms A series expansion is a representation of a particular function of a sum of powers in one of its variables, or by a sum of powers of another function ff(xx). There are many areas in engineering, such as the motion of fluids, the transfer of hear or processing of signals where the application of certain quantities involves functions as independent variables. Therefore, it is important for us to understand how to solve each function in the equations. In this chapter, we will cover infinite series, convergence and power series. Furthermore, in engineering, transforms in one form to another plays a major role in analysis and design. An area of continuing importance is the use of Laplace, Fourier, and other transforms in fields such as communication, control and signal processing. These will be covered later in this chapter. 1.1 Infinite series Many important differential equations appearing in practice cannot be solved in terms of these functions. Solutions can often only be expressed as infinite series, that is, infinite sums of simpler functions such as polynomials, or trig functions. A polynomial series may be written as: PP(xx) = aa kk xx kk kk=0 (1) where {aa kk, kk = 0, 1, } are the expansion coefficients of the power series. If {aa kk }, kk = 0, 1, is a sequence of numbers, the ordered sum of all its terms, namely: aa kk = aa oo + aa 1 + (2) kk=0 is called an infinite series. There are a few infinite series that you ll need to know as well as you know the multiplication table. These are: 1. The geometric series 1 + aa + aa 2 + aa 3 + aa 4 + aa 5 + = aa kk = kk=0 1 1 aa (3) We can also write geometric series as: 7

8 aarr kk kk=0 (4) where aa is the first term and rr is the common ratio. This series will converge if rr < 1 and diverges if otherwise. 2. The pp series or hyperharmonic series 1 kk pp kk=1 (5) This series will converge if pp > 1. When pp = 1, the pp series becomes a harmonic series which diverges. Other series you need to know include the following which are known as the power series: ee aa = 1 + aa + aa2 2! aakk + = kk! sin aa = aa aa3 3! + = ( 1)kk cos aa = 1 aa2 2! + = ( 1)kk aa 2kk+1 (2kk + 1)! aa 2kk (2kk)! (6) (7) (8) ln(1 + aa) = 1 aa2 2 + aa3 3 aakk = ( 1)kk+1 kk 0 ( aa < 1) (9) It is better to understand their derivation instead of just memorizing the series. For example, the cosine series is the derivative of the sine. Therefore, by knowing the sine series, you can differentiate it term by term to obtain the cosine series. The logarithm of (1 + aa) is an integral f 1/(1 + aa) so you can obtain its series from that of the geometric series. 8

9 1.2 Convergence Does an infinite series converges? Does the limit as NN of the sum below exist? NN uu kk 1 (10) There are a few methods used to prove convergence of a series. These methods are: 1. The nn th term rule If you have a series with a sum of aa nn < nn=0 (11) Where the lim nn aa nn = 0, then the series converges. Example 1: Determine whether this infinite series below converges or diverges using the nn th term rule: 4nn2 nn 3 7 3nn 3 nn=1 Solution: The first step is to rewrite the equation above as: 4nn 2 nn 3 lim nn 7 3nn 3 Since we have both nn 3 in the numerator and denominator, so we can divide the series by nn 3, and this gives us lim nn 4 nn 1 7 nn 3 3 As nn, we can then rewrite the equation above as: 9

10 lim nn 4 nn 1 = 7 nn = 1 3 Hence, 4nn 2 nn 3 lim nn 7 3nn 3 = Therefore, the series diverges. Example 2: Determine whether this infinite series below converges or diverges using the nn th term rule: 1 nn 2 Solution: By applying the n-th term rule, the equation above can be re-written as: Therefore, this series converges. nn=1 1 lim nn 2 = 0 nn 2. The ratio test To apply this comparison test, you will need a stable of known convergent series. Assuming you have a series of aa kk kk=0 (12) Then, let PP = lim kk aa kk+1 aa kk (13) If PP < 1 the series converges PP > 1 the series diverges PP = 1 then it is inconclusive 10

11 Example 3: Determine whether this infinite series below converges or diverges using the ratio test. ( 10)kk 4 2kk+1 (kk + 1) kk=0 Solution: The first thing to do is to identify the series above for aa kk, aa kk = ( 10)kk 4 2kk+1 (kk + 1) To compute aa kk+1, we need to substitute kk + 1 for all kk in the equation of aa kk above, therefore: aa kk+1 = ( 10) kk+1 4 2(kk+1)+1 ((kk + 1) + 1) = ( 10) kk+1 4 2kk+3 (kk + 2) Now, we can define P as: PP = lim kk aa kk+1 1 aa kk Substituting the values of aa kk+1 and aa kk into the equation above, we get: PP = lim kk ( 10) kk+1 4 2kk+3 (kk + 2) 42kk+1 (kk + 1) ( 10) kk 10 (kk + 1) PP = lim kk 4 2 (kk + 2) PP = lim kk + 1 kk kk + 2 When kk, the ratio of (kk + 1)/(kk + 2) becomes 1, and therefore: Since PP < 1, therefore the series converges. PP = < 1 11

12 3. The Root test The root test is similar to that of the ratio test. Assuming you have a series of Then, let aa kk kk=0 (14) nn rr = lim aa kk kk = lim kk aa kk 1 kk (15) If rr < 1 the series converges rr > 1 the series diverges rr = 1 then it is inconclusive Example 4: Determine whether this infinite series below converges or diverges using the root test. kkkk 3 1+2kk kk=1 Solution: Using the root test rule, we can re-write the equation above as: rr = lim kkkk kk 3 1+2kk 1 kk kk rr = lim kk 3 1 = kk = Therefore, since rr = > 1, this series diverges. Example 5: Determine whether this infinite series below converges or diverges using the root test. 5kk 3kk3 7kk Solution: Using the root test rule, we can re-write the equation above as: kk=0 kk 12

13 kk 5kk 3kk3 rr = lim kk 7kk kk 5kk 3kk3 rr = lim kk 7kk Therefore, solving for rr, we get rr = 2 9 Since rr < 1, this series converges. 4. Absolute Convergence An absolutely convergent series is convergent. Any absolute series that converges, then the original series converges too. If: aa kk kk=1 Converges Then: aa kk kk=1 Converges too! 5. Leibniz Theorem Leibniz Theorem is used for alternating series. For example, we have a series as below: ( 1) kk kk=0 aa kk (16) When aa kk > 0, aa kk > aa kk+1 and aa kk 0, then the series converges. Also, if aa kk is a decreasing sequence and that the 13

14 lim aa kk = 0 kk Then the series is convergent. Example 6: Determine whether this infinite series below converges using Leibniz Theorem. ( 1)kk+1 kk Solution: First, let s look at aa kk, the equation above can be re-written as: kk=1 Therefore, ( 1) kk+1 1 kk kk=1 aa kk = 1 kk Now, if we apply a limit to aa kk, lim aa 1 kk = lim kk kk kk = 0 And we can see that aa kk > aa kk+1, therefore this series converges. Of all the methods discussed above, it can be concluded that root test is more powerful because it can be shown that If: PP = lim kk aa KK+1 aa kk exist Then: rr = lim kk (aa kk ) 1 kk exist too! Therefore, PP = rr. However, rr may exist even when PP does not! 14

15 1.3 Power Series We must therefore give meaning to an infinite sum of constants, using this to give meaning to an infinite sum of functions. When the functions being added are the simple powers (xx xx oo ) kk, the sum is called a Taylor (power) series and if xx oo = 0, a Maclaurin series. When the functions are trig terms such as ssssss(kkkk) or cccccc(kkkk), the series might be a Fourier series, certain infinite sums of trig functions that can be made to represent arbitrary functions, even functions with discontinuities. This type of infinite series is also generalized to sums of other functions such as Legendre polynomials. Eventually, solutions of differential equations will be given in terms of infinite sums of Bessel functions, themselves infinite series Taylor Series Having understood sequences, series and power series, now we will focus to one of the main topic: Taylor polynomials. The Taylor polynomial approximation is given by: ff(xx) = pp nn (xx) + 1 xx nn! (xx tt)nn ff (nn+1) (tt)dddd aa (17) Where the nn-th degree Taylor polynomial pp nn (xx) is given by: pp nn (xx) = ff(aa) + ff (aa) 1! When aa = 0, the series is also called Maclaurin series. There are 2 conditions apply: (xx aa) + + ff(nn) (aa) (xx aa) nn (18) nn! 1. ff (xx), ff (1) (xx),, ff (nn+1) (xx) are continuous in a closed interval containing xx = aa. 2. xx is any point in the interval. A Taylor series represents a function for a given value as an infinite sum of terms that are calculated from the value of the function s derivatives. Therefore, the Taylor series of a function ff (xx) for a value aa is the power series, and can be written as: ff(xx) = ffnn (aa) (xx aa) nn (19) nn! nn=0 15

16 Example 7: Find the Maclaurin series of a function ff (xx) = ee xx and its radius of convergence. Solution: So, if ff (xx) = ee xx, then ff (nn) (xx) = ee xx, so ff (nn) (0) = ee 0 = 1 for all nn. Therefore, the Taylor series for ff at 0 (which is the Maclaurin series), so: ff(xx) = ffnn (0) (xx) kk = xxnn nn! nn! nn=0 nn=0 = 1 + xx 1! + xx2 2! + xx3 3! + To find the radius of convergence, let aa nn = xx nn /nn!. Then, xx nn+1 aa nn+1 = aa nn (nn + 1)! nn! xx nn = xx nn < 1 So, by Ratio Test, the series converges for all xx and the radius of convergence is RR = The conclusion we can draw from example 7 is that if ee xx has a power series expansion at 0, then: ee xx = xxnn nn! nn=0 So now, under what circumstances is a function equal to the sum of its Taylor series? Or if ff has derivatives of all orders, when is it that equation (19) is true? With any convergent series, this means that ff(xx) is the limit of the sequence of partial sums. In the case of Taylor series, the partial sums can be written as in as equation (18), where: pp nn (xx) = ff(aa) + ff (aa) 1! (xx aa) + ff (aa) 2! (xx aa) 2 + ff(nn) (aa) (xx aa) nn nn! For the example of the exponential function ff (xx) = ee xx, the results from example 7 shows that the Taylor polynomials at 0 (or Maclaurin polynomials) with nn = 1, 2 and 3 are: pp 1 (xx) = 1 + xx pp 2 (xx) = 1 + xx + xx2 2! 16

17 In general, ff (xx) is the sum of its Taylor series if If we let pp 3 (xx) = 1 + xx + xx2 2! + xx3 3! ff(xx) = lim nn pp nn (xx) (20) RR nn (xx) = ff(xx) pp nn (xx) so that ff(xx) = pp nn (xx) + RR nn (xx) (21) Then, RR nn (xx) is called the remainder of the Taylor series. If we show that lim nn RR nn (xx) = 0, then it follows that from equation (21): lim pp nn(xx) = lim [ff(xx) RR nn (xx)] = ff (xx) lim RR(xx) = ff(xx) nn nn nn We have therefore proved the following: If ff(xx) = pp nn (xx) + RR nn (xx), where pp nn is the nnth degree Taylor Polynomial of ff at aa and lim nn RR nn(xx) = 0 (22) for xx aa < RR, then ff is equals to the sum of its Taylor series on the interval xx aa < RR. Therefore, if ff has nn + 1 derivatives in an interval II that contains the number aa, then for xx in II there is a number zz strictly between xx and aa such that the remainder term in the Taylor series can be expressed as RR nn (xx) = ff(nn+1) (zz) (nn + 1)! (xx aa)nn+1 (23) Example 8: Find the Maclaurin series for sin xx and prove that it represents sin xx for all xx. Solution: First, we arrange our computation in two columns as follows: ff(xx) = sin xx ff(0) = 0 ff (1) (xx) = cos xx ff (1) (0) = 1 ff (2) (xx) = sin xx ff (2) (0) = 0 17

18 ff (3) (xx) = cos xx ff (3) (0) = 1 ff (4) (xx) = cos xx ff (4) (0) = 0 Since the derivatives repeat nn a cycle of four, we can write the Maclaurin series as follow: ff(0) + ff(1) (0) 1! xx + ff(2) (0) 2! xx 2 + ff(3) (0) 3! xx 3 + ff(4) (0) xx 4 + 4! = ! xx + 0 2! xx ! xx ! xx4 + = xx xx3 3! + xx5 5! xx7 7! + = ( 1) kk kk=0 xx 2kk+1 (2kk + 1)! You can try with different types of functions, and you will get a Maclaurin series table that looks like this: 1 1 xx = xxnn = 1 + xx + xx 2 + xx 3 + RR = 1 nn=0 ee xx = xxnn nn! nn=0 = 1 + xx 1! + xx2 2! + xx3 3! + RR = xx(2nn+1) sin xx = ( 1) nn (2nn + 1)! nn=0 xx2nn cos xx = ( 1) nn (2nn)! nn=0 tan 1 xx2nn+1 nn xx = ( 1) 2nn + 1 nn=0 xxnn nn 1 ln(1 + xx) = ( 1) nn nn=0 = xx xx3 3! + xx5 5! xx7 7! + RR = = 1 xx2 2! + xx4 4! xx6 6! + RR = = xx xx3 3 + xx5 5 xx7 7 + RR = 1 = xx xx2 2 + xx3 3 xx4 4 + RR = 1 18

19 Example 9: Find the first 3 terms of the Taylor series for the function sin ππππ centered at aa = 0.5. Use your answer to find an approximate value to sin ππ 2 + ππ 10 Solution: Let us first do the derivatives for the function given: ff(xx) = sin ππxx. Therefore, ff (1) xx = ππ cos ππππ, ff (2) xx = ππ 2 sin ππππ, ff (3) xx = ππ 3 cos ππππ, ff (4) xx = ππ 4 sin ππππ And so, Substituting this back into equation (17), we get: Therefore, sin ππππ = sin ππ 2 + xx ! = 1 ππ xx ! 2 sin ππ = 1 ππ ! ( ππ) 2 + xx ππ 4 + 4! + ππ xx ! + ππ = = ! + 19

20 1.3.2 Fourier Series As mentioned previously, a Fourier series decomposes periodic functions into a sum of sines and cosines (trigonometric terms or complex exponentials). For a periodic function ff(xx), periodic on [ LL, LL], its Fourier series representation is: ff(xx) = 0.5aa 0 + aa nn cos nnnnnn LL + bb nn sin nnnnnn LL nn=1 (24) where aa 0, aa nn and bb nn are the Fourier coefficients and they can be written as: aa 0 = 1 2LL LL LL ff(xx)dddd LL (25) aa nn = 1 ff(xx) cos nnnnnn dddd LL LL LL (26) LL bb nn = 1 ff(xx) sin nnnnnn dddd LL LL LL (27) where period, pp = 2LL. Equation (24) is also called Real Fourier series. There are 2 conditions apply: 1. ff(xx) is a piecewise continuous is piecewise continuous on the closed interval [ LL, LL]. A function is said to be piecewise continuous on the closed interval [aa, bb] provided that it is continuous there, with at most a finite number of exceptions where, at worst, we would find a removable or jump discontinuity. At both a removable and a jump discontinuity, the one-sided limits ff(tt + ) = lim ff(xx) and xx tt + ff(tt ) = lim ff(xx) exist and are finite. xx tt 2. A sum of continuous and periodic functions converges pointwise to a possibly discontinuous and non-periodic function. This was a startling realisation for mathematicians of the early nineteenth century. 20

21 Example 10: Find the Fourier series of (xx) = xx 2, 1 < xx < 1 Solution: In this example, period, pp = 2, but we know that pp = 2LL, therefore, LL = 1. First, let us find aa 0. From equation (25), aa 0 = 1 2LL LL ff(xx)dddd LL 1 aa 0 = 1 2 xx2 dddd = Next, let us find bb nn. From equation (27), LL bb nn = 1 ff(xx) sin nnnnnn dddd LL LL LL bb nn = xx2 sin nnnnnn dddd 1 = 0 Finally, we will find aa nn. From equation (26), LL aa nn = 1 ff(xx) cos nnnnnn dddd LL LL LL aa nn = xx2 cos nnnnnn dddd 1 Solving using integration by parts, we get: aa nn = 2xxxxxxxxxxxxxx nn 2 ππ aa nn = nn 2 ππ 2 [( 1)nn + ( 1) nn ] aa nn = 4( 1)nn nn 2 ππ 2 1 Therefore, the Fourier series can be written as: 21

22 ff(xx 2 ) = 1 4( 1)nn + 3 nn 2 ππ 2 cos(nnnnnn) nn= Complex Fourier series A function ff(xx) can also be expressed as a Complex Fourier series and can be defined to be: + ff(xx) = cc nn ee iiiiiiii/ll (28) where cc nn = 1 2ππ ππ ff(xx)ee iiiiii ππ (29) We know that: ee iiii = cos xx + ii sin xx ee iiii = cos xx ii sin xx ee iiii ee iiii = 2ii sin xx (30) ee iiii + ee iiii = 2 cos xx Therefore, from equation (29), cc nn = 1 2ππ cc nn = ππ ππ ff(xx)ee iiiiii ππ ππ ff(xx) cos nnnn dddd ii ππ Hence, we can write: 1 ππ ππ ff(xx) sin nnnn dddd ππ cc nn = 1 2 (aa nn iibb nn ), nn > 0 22

23 cc nn = 1 2 (aa nn + iibb nn ), nn < 0 cc nn = aa 0, nn = 0 Example 11: Write the complex Fourier transform of ff(xx) = 2 sin xx cos 10xx Solution: Here, we can expand the function by substituting the sin and cos functions from equation (30), we get: ff(xx) = 2 eeiiii ee iiii 2ii ee10iiii + ee 10iiii 2 ff(xx) = 1 ii eeiiii 1 ii ee iiii 1 2 ee10iiii 1 2 ee 10iiii Therefore: cc 1 = 1 ii, cc 10 = 1 2, cc 1 = 1 ii, cc 10 = Termwise Integration and Differentiation Parseval s Identity Consider a Fourier series below and expand it ff(xx) = aa 0 + {aa nn cos nnnn + bb nn sin nnnn} = aa 0 + aa 1 cos xx + bb 1 sin xx + aa 2 cos 2xx + bb 2 sin 2xx + nn=1 Square it, we get: NN NN ff 2 (xx) = aa aa 2 nn cos 2 nnnn + bb 2 nn sin 2 nnnn + 2aa 0 ( aa nn cos nnnn + bb nn sin nnnn) nn=1 nn=1 NN + 2aa 1 cos xx bb 1 sin xx + 2aa 1 cos xx ( aa nn cos nnnn + bb nn sin nnnn) + + 2aa NN cos NNNN bb NN sin NNNN nn=1 23

24 Integrate both sides, we get: ππ ff 2 (xx) dddd ππ ππ NN = aa aa 2 nn cos 2 nnnn + bb 2 nn sin 2 nnnn + ππ nn=1 dddd ππ ff 2 (xx) dddd ππ NN = 2ππaa ππππ nn 2 + ππππ nn nn=1 Parseval s Identity can be written as: LL 1 LL ff(xx) 2 dddd = 2 aa ( aa nn 2 + bb nn 2 ) LL nn=1 (31) If: a) ff(xx) is continuous, and ff (xx) is a piecewise continuous on [ LL, LL] b) ff(ll) = ff( LL) c) ff (xx) exist at xx in ( LL, LL), Therefore: ff (xx) = ππ nn LL nn=1 aa nn sin nnnnnn LL Example 12: From Example 10, we found that the Fourier series is: ff(xx 2 ) = 1 4( 1)nn + 3 nn 2 cos(nnnnnn), xx2 ππ2 nn=1 Solution: Applying Parseval s to the equation above, we get: nn 4 ππ 4 nn=1 1 = xx 4 1 dddd = bb nn cos nnnnnn (32) LL 24

25 16 nn 4 ππ 4 nn=1 1 nn 4 nn=1 = = 8 45 = ππ Fourier series of Odd and Even functions A function ff(xx) is called an eeeeeeee or ssssssssssssssssss function if it has the property A function ff(xx) is called an oooooo or aannnnnnnnnnnnnnnnnnnnnnnn function if ff( xx) = ff(xx) (33) ff( xx) = ff(xx) (34) A function that is neither even nor odd can be represented as the sum of an even and an odd function. Cosine waves are even, so any Fourier series representation of a periodic function must have an even symmetry. A function ff(xx) defined on [0, LL] can be extended as an even periodic function (bb nn = 0). Therefore, the Fourier series representation of an even function is: ff(xx) = 0.5aa 0 + aa nn cos nnnnnn LL nn=1 LL, aa nn = 2 ff(xx) cos nnnnnn LL LL 0 dddd (35) Similarly sine waves are odd, so any Fourier sine series representation of a periodic function must have odd symmetry. Therefore a function ff(xx) defined on [0, LL] can be extended as an odd periodic function (aa nn = 0) and the Fourier series representation of an even function is: ff(xx) = 0.5aa 0 + bb nn sin nnnnnn LL nn=1 Example 13: If ff(xx) is even, show that LL, bb nn = 2 ff(xx) sin nnnnnn LL LL 0 dddd (36) LL (a) aa nn = 2 ff(xx) cos nnnnnn LL 0 LL dddd 25

26 (b) bb nn = 0 Solution: For an even function, we can write the equation as: LL aa nn = 1 nnnnnn ff(xx) cos LL LL LL 0 dddd = 1 nnnnnn ff(xx) cos LL LL LL LL dddd + 1 nnnnnn ff(xx) cos LL LL 0 ddxx Let x=-u, we can re-write: 1 LL 0 nnnnnn ff(xx) cos LL LL LL dddd = 1 ff( uu) cos nnnnnn LL LL 0 LL dddd = 1 ff(uu) cos nnnnnn LL LL 0 dddd Since by definition of an even function f(-u) = f(u). Then: LL aa nn = 1 ff(uu) cos nnnnnn LL LL 0 LL dddd + 1 nnnnnn ff(xx) cos LL LL 0 LL dddd = 2 nnnnnn ff(xx) cos LL LL 0 dddd To show that bb nn = 0, we can write the expression as LL bb nn = 1 ff(xx) sin nnnnnn LL LL LL 0 dddd = 1 ff(xx) sin nnnnnn LL LL LL LL dddd + 1 ff(xx) sin nnnnnn LL LL 0 dddd If we make the transformation x=-u in the first integral on the right of the equation above, we obtain: 1 LL 0 ff(xx) sin nnnnnn LL LL dddd LL = 1 nnnnnn ff( uu) sin LL LL 0 LL dddd = 1 ff(uu) sin nnnnnn LL LL 0 dddd LL = 1 ff(uu) sin nnnnnn LL LL 0 LL dddd = 1 ff(xx) sin nnnnnn LL LL 0 dddd Therefore, substituting this into the equation for bb nn, we get LL bb nn = 1 ff(xx) sin nnnnnn LL LL 0 LL dddd + 1 ff(xx) sin nnnnnn LL LL dddd =

27 1.4 Integral Transform An integral transform is any transform of the following form With the following inverse transform xx 2 FF(ww) = KK(ww, xx)ff(xx) dddd (37) xx Fourier Transform ww 2 ff(xx) = KK 1 (ww, xx)ff(ww) dddd (38) ww 1 A Fourier series expansion of a function ff(xx) of a real variable xx with a period of 2LL is defined over a finite interval LL xx LL. If the interval becomes infinite and we sum over infinitesimals, we then obtain the Fourier integral ff(xx) = 1 2ππ FF(ww)eeiiiiii dddd (39) with the coefficients FF(ww) = ff(xx)ee iiiiii dddd (40) Equation (40) is the Fourier transform of ff(xx). The Fourier integral is also known as the inverse Fourier transform of FF(ww). In this example, xx 1 = ww 1 =, xx 2 = ww 2 = and KK(ww, xx) = ee iiiiii. The Fourier transform transforms a function of one variable (e.g. time in seconds) which lives in the time domain to a second function which lives in the frequency domain and changes the basis of the function to cosines and sines. Example 14: Find the Fourier transform of 1 2 < xx < 2 ff(xx) = 0 ooooheeeeeeeeeeee Solution: We can write the Fourier transform as in equation (40): 27

28 FF(ww) = 1 2ππ 2 ff(xx)ee iiiiii 2 dddd = 1 2ππ 2 ee iiiiii 2 dddd = 1 ee iiiiii 2ππ iiii 2 = 1 iiii 2ππ ee 2iiii ee 2iiii 2 = 1 [(cos 2ww ii sin 2ww) (cos 2ww + ii sin 2ww)] iiii 2ππ = 1 [ 2ii sin 2ww] iiii 2ππ = 2 ππ 2ww sin ww Example 15: Find the Fourier transform of tt 1 tt 1 ff(tt) = 0 eeeeeeeeeeheeeeee Solution: Recalling the Fourier transform in equation (40), we can write FF(ww) = ff(xx)ee iiiiii dddd 1 = tt ee iiiiii 1 dddd By applying integration by parts, we get: = tt 1 iiii eeiiiiii iiii ee iiiiii 1 dddd 28

29 We can also rewrite 1 ii = ii, therefore: = iiii 1 ww eeiiiiii iiii 1 iiii eeiiiiii 1 = iiii 1 ww eeiiiiii ww 2 eeiiiiii 1 = ii ww ee iiii + ee iiii + 1 ww 2 ee iiii + ee iiii = 2 ii 1 ww 2 ee iiii + ee iiww + 2ii ww 2 1 2ii eeiiii ee iiii = 2ii ww 2ii cos ww sin ww ww2 = 2ii ww cos ww 1 sin ww ww Laplace Transform The Laplace transform is an example of an integral transform that will convert a differential equation into an algebraic equation. The Laplace transform of a function ff(xx) of a variable xx is defined as the integral FF(ss) = L{ff(tt)} = ff(tt)ee ssss 0 dddd (41) Where s is a real, positive parameter that serves as a supplementary variable. The conditions are: if ff(tt) is piecewise continuous on (0, ), and of exponential order ( ff(tt) KKee αααα for some KK and αα > 0), then FF(ss) exists for ss > αα. Several Laplace transforms are given in the table below, where aa is a constant and nn is an integer. Example 16: Find the Laplace transforms of the following functions: ff(tt) = 3 0 < tt < 5 0 tt > 0 29

30 Solution: L{ff(tt)} = ff(tt) ee ssss 0 5 dddd = 3 ee ssss dddd + 0 ee ssss dddd 0 5 = 3 ee ssss 5 ss = 3 ee 5ss ss 1 ss = 3 ss (1 ee 5ss ) Example 17: Find the Laplace transforms of the following functions: tt 0 < tt < aa ff(tt) = bb tt > aa Solution: L{ff(tt)} = ff(tt) ee ssss 0 aa ddtt = tt ee ssss dddd + bb ee ssss dddd 0 aa = ee ssss ss aa ee ssss tt ss bb ee ssss ss 0 aa = ee aaaa aa ss 1 ss 2 ee0 0 1 ss 2 bb ss (0 ee aaaa ) = 1 aa + bb 1 ee aaaa ss2 ss ss2 Example 17: Determine the Laplace transform of the function below: ff(tt) = 5 3tt + 4 sin 2tt 6ee 4tt Solution: First, let s break the equation one by one, we get: 30

31 L{5} = 5 ss, RRRR (ss) > 0 L{tt} = 1, RRRR (ss) > 0 ss2 L{sin 2tt} = 2 ss 2 + 4, RRRR (ss) > 0 L{ee 4tt } = 1 ss 4, RRRR (ss) > 4 Therefore, by linearity property, L{ff(tt)} = L{5 3tt + 4 sin 2tt 6ee 4tt } = L{5} 3L{tt} + 4L{sin 2tt} 6L{ee 4tt } = 5 ss 3 ss ss ss 4 ff(xx) = LL 11 {FF(ss)} aa tt xx nn ee aaaa sin aaaa cos aaaa sinh aaaa cosh aaaa LAPLACE TRANSFORMS FF(ss) = LL{ff(ss)} aa ss 1 ss 2 (nn!) ss nn+1 1 (ss aa) aa (ss 2 + aa 2 ) ss (ss 2 + aa 2 ) aa (ss 2 aa 2 ) ss (ss 2 aa 2 ) 31

32 2. Vector Spaces, vector Fields & Operators In the context of physics we are often interested in a quantity or property which varies in a smooth and continuous way over some one-, two-, or three-dimensional region of space. This constitutes either a scalar field or a vector field, depending on the nature of property. In this chapter, we consider the relationship between a scalar field involving a variable potential and a vector field involving field, where this means force per unit mass or change. The properties of scalar and vector fields are described and how they lead to important concepts, such as that of a conservative field, and the important and useful Gauss and Stokes theorems. Finally examples will be given to demonstrate the ideas of vector analysis. There are basically four types of functions involving scalars and vectors: Scalar functions of a scalar, ff(xx) Vector function of a scalar, rr(tt) Scalar function of a vector, φφ(rr) Vector function of a vector, AA(rr) 1. The vector x is normalised if x T x = 1 2. The vectors x and y are orthogonal if x T y = 0 3. The vectors x 1, x 2,, x nn are linearly independent if the only numbers which satisfy the equation aa 1 x 1 + aa 2 x aa nn x nn = 0 are aa 1 = aa 2 =... = aa nn = 0 4. The vectors x 1, x 2,, x nn form a basis for a nn dimensional vector-space if any vector x in the vectorspace can be written as a linear combination of vectors in the basis thus x = aa 1 x 1 + aa 2 x aa nn x nn where aa 1, aa 2,, aa nn are scalars. Figure 1: Components of a vector 32

33 For example, a vector A from the origin in the figure above to a point P in the 3-dimensions takes the form AA = aa xx ıı + aa yy ȷȷ + aa zz kk (42) Where ıı, ȷȷ, kk are unit vectors along the {xx, yy, zz} axes, respectively. The vector components aa xx, aa yy, aa zz, are the corresponding distances along the axes. The length or magnitude of Vector AA is AA = aa xx 2 + aa yy 2 + aa zz 2 (43) 2.1 Scalar (inner) product of vector fields The scalar product of vector fields is also called as the dot product. For example, if we have 2 vectors as AA = (AA 1, AA 2, AA 3 ) and BB = (BB 1, BB 2, BB 3 ), therefore, AA, BB = AA BB = AA TT BB = AA 1 BB 1 + AA 2 BB 2 + AA 3 BB 3 (44) We can also write AA BB = AA BB cos θθ (45) where θθ is the angle between AA and BB satisfying 0 θθ ππ. The inner product of vectors is a scalar. The scalar product obeys the product laws which are listed below: Product laws: 1. Commutative: AA BB = BB AA 2. Associative: mmaa nnbb = mmmmaa BB 3. Distributive: AA (BB + CC) = AA BB + AA CC 4. Cauchy-Schwarz inequality: AA BB (AA AA) 1 2(BB BB) 1 2 Note that a relation such as AA BB = AA CC does not imply that BB = CC, as AA BB AA CC = AA (BB CC) = 0 (46) Therefore, the correct conclusion is that AA is perpendicular to the vector BB CC. Example 1: Determine the angle between AA = 1,3, 2 and BB = 2, 4, 1. 33

34 Solution: All we need to do here is to rewrite equation (45) as: Therefore we know that: cos θθ = cos θθ = We ll first have to compute the individual parameters AA BB AA BB AA BB AA BB AA BB = 12 AA = 14 BB = 21 Hence, the angle between the vectors is: Lp norms cos θθ = = θθ = cos 1 ( ) = There are many norms that could be defined for vectors. One type of norms is called the LL pp norm, often denoted as pp. For pp 1, it is defined as the pp nnnnnnnn and can be written as: nn xx pp = xx ii pp ii=1 1 pp, xx = [xx 1, xx nn ] TT (47) There are a few types of norms such as the following: 1. xx 1 = ii xx ii, also called the Manhattan norm because it corresponds to sums of distances along coordinate axes, as one would travel along the rectangular street plan of Manhattan. 2. xx 2 = xx ii 2 ii, also called the Euclidean norm, the Euclidean length, or just the length of the vector. 3. xx = mmaaaa ii xx ii, also called the max norm or the Chebyshev norm. 34

35 Some relationships of norms are as below: xx xx 2 xx 1 xx xx 2 nn xx (48) xx 2 xx 1 nn xx 2 If we define the inner product induced norm xx = xx, xx. Then, ( xx + yy ) 2 xx + yy 2, xx + yy 2 = xx 2 + yy xx, yy (49) Example 2: Given a vector vv = ıı 4ȷȷ + 5kk, determine the Manhattan norm, Euclidean length and Chebyshev norm. Solution: So, if we re-write the vector vv as vv = (1, 4,5), then we can calculate the norms easily. A. Manhattan norm (One norm): vv 1 = vv ii ii = = 10 B. Euclidean norm (Two norm) vv 2 = xx ii 2 ii = = 42 C. Chebyshev norm (Infinity norm) vv = mmmmmm ii xx ii = mmmmmm ii { 1, 4, 5 } = 5 35

36 Therefore, we can see that 2.2 Vector product of vector fields xx xx 2 xx The vector product of vector fields is also called as the cross product. For example, if we have 2 vectors as AA = (AA 1, AA 2, AA 3 ) and BB = (BB 1, BB 2, BB 3 ), therefore, AA BB = (AA 2 BB 3 AA 3 BB 2, AA 1 BB 3 AA 3 BB 1, AA 1 BB 2 AA 2 BB 1 ) (50) The cross product of the vectors AA and BB, is orthogonal to both AA and BB, forms a right-handed systems with AA and BB, and has length given by: AA BB = AA BB sin θθ (51) where θθ is the angle between AA and BB satisfying 0 θθ ππ. The vector product of a vector is a vector. A few additional properties of the cross product are listed below: 1. Scalar multiplication (aaaa) (bbbb) = aaaa(aa BB) 2. Distribution laws AA (BB + CC) = AA BB + AA CC 3. Anticommuttaion BB AA = AA BB 4. Nonassociativity AA (BB CC) = (AA CC)BB (AA BB)CC If we breakdown equation (9), we ca rewrite the cross product of vectors AA and BB as: AA BB = AA 2 AA 3 ıı AA 1 AA 3 ȷȷ + AA 1 AA 2 kk BB 2 BB 3 BB 1 BB 3 BB 2 BB 1 ıı ȷȷ kk = AA 1 AA 2 AA 3 BB 1 BB 2 BB 3 Example 3: If AA = (3, 2, 2) and BB = ( 1, 0, 5), compute AA BB and find the angle between the two vectors. Solution: It s a very simple solution here, all we have to do is the compute the cross product first, so AA BB = ıı ȷȷ kk 36

37 = 10ıı 13ȷȷ 2kk Angle between the two vectors are given as: AA BB = AA BB sin θθ. Rearranging equation (51), we get: AA BB sin θθ = AA BB = = ( 10) 2 + ( 13) 2 + ( 2) 2 (3) 2 + ( 2) 2 + ( 2) 2 ( 1) 2 + (0) 2 + (5) θθ = Vector operators Certain differential operations may be done on a scalar and vector fields. This may have a wide range of applications in physical sciences. The most important tasks are those of finding the gradient of a scalar field and the divergence and curl of a vector field. In the following topics, we will discuss the mathematical and geometrical definitions of these, which will rely on concepts of integrating vector quantities along lines and over surfaces. In the midst of these differential operations is the vector operator, which is called as del (or nabla) and in Cartesian coordinates, is defined as: ıı + ȷȷ + kk (52) Gradient of a scalar field The gradient of a scalar field φφ(xx, yy, zz) is defined as grad φ = φ = ıı + ȷȷ + kk (53) Clearly, φ is a vector field whose xx, yy and zz components are the first partial derivatives of φφ(xx, yy, zz) with respect to xx, yy and zz. 37

38 Example 4: Find the gradient of the scalar field φφ = xxyy 2 zz 3. Solution: We can easily solve this problem by using equation (12), so the gradient of the scalar field φφ = xxyy 2 zz 3 is grad φ = ıı + ȷȷ + kk = yy 2 zz 3 ıı + 2xxxxzz 3 ȷȷ + 3xxyy 2 zz 2 kk If we consider a surface in 3D space with φφ(rr) = cccccccccccccccc then the direction normal (i.e. perpendicular) to the surface at the point rr is the direction of grad φφ. The magnitude of the greater rate of change of φφ(rr) is the magnitude of grad φφ. φ φφ = constant Figure 2. Direction of gradient In a physical situations, we may have a potential, φφ, which varies over a particular region and this constitutes a field EE, satisfying: EE = φ = ıı + ȷȷ + kk Example 5: Calculate the electric field at point (xx, yy, zz) due to a charge qq 1 at (2, 0, 0) and a charge qq 2 at (-2, 0, 0) where charges are in coulombs and distances are in metres. Solution: We need to understand the equation for Electric field which is given by: 38

39 EE = kk cc qq rr where rr is the magnitude or position and kk cc is the Coulomb constant and is given by: Therefore, the potential at the point (xx, yy, zz) is kk cc = 1 4ππεε 0 φ(xx, yy, zz) = qq 1 As a result, the components of the fields are 4ππεε 0 (2 xx) 2 + yy 2 + zz 2 + qq 2 4ππεε 0 (2 + xx) 2 + yy 2 + zz 2 qq 1 (2 xx) EE xx = 4ππεε 0 {(2 xx) 2 + yy 2 + zz 2 } 3/2 + qq 2 (2 + xx) 4ππεε 0 {(2 + xx) 2 + yy 2 + zz 2 } 3/2 qq 1 yy EE yy = 4ππεε 0 {(2 xx) 2 + yy 2 + zz 2 } 3/2 + qq 2 yy 4ππεε 0 {(2 + xx) 2 + yy 2 + zz 2 } 3/2 qq 1 zz EE zz = 4ππεε 0 {(2 xx) 2 + yy 2 + zz 2 } 3/2 + qq 2 zz 4ππεε 0 {(2 + xx) 2 + yy 2 + zz 2 } 3/2 Example 6: The function that describes the temperature at any point in the room is given by: TT(xx, yy, zz) = 100 cos xx 10 sin yy cos zz 10 Find the gradient of TT, the direction of greatest change in temperature in the room at point (10ππ, 10ππ, ππ) and the rate of change of temperature at this point. Solution: First, let s find the gradient of the function TT, which is given by equation (53): TT = ıı + ȷȷ + kk = 10 sin xx 10 sin yy 10 cos zz ıı + 10 cos xx 10 cos yy cos zz ȷȷ cos xx 10 sin yy 10 sin zz kk 39

40 Therefore, at the point (10ππ, 10ππ, ππ) in the room, the direction of the greatest change in temperature is: TT = 0ıı 10ȷȷ + 0kk And the rate of change of temperature at this point is the magnitude of the gradient, which is TT = ( 10) 2 = Divergence of a vector field The divergence of a vector field AA(xx, yy, zz) is defined as the dot product of the operator and AA: div AA = AA = AA 1 xx + AA 2 + AA 3 (54) where AA 1, AA 2 and AA 3 are the xx, yy and zz components of AA. Clearly, AA is a scalar field. Any vector field AA for which AA = 0 is said to be solenoidal. Example 7: Find the divergence of a vector field AA = xx 2 yy 2 ıı + yy 2 zz 2 ȷȷ + xx 2 zz 2 kk Solution: This is a straight forward example, using equation (54) we can solve this easily: AA = AA 1 + AA 2 + AA 3 = 2(xxyy 2 + yyzz 2 + xx 2 zz) Example 8: Find the divergence of a vector field FF = (yyyyee xxxx, xxxxee xxxx, ee xxxx + 3 cos 3zz) Solution: Again, using equation (54) we can solve this easily: FF = FF 1 + FF 2 + FF 3 = yy 2 zzee xxxx + xx 2 zzee xxxx 9 sin 3zz 40

41 The value of the scalar div AA at point rr gives the rate at which the material is expanding or flowing away from the point rr (outward flux per unit volume) Theorem involving Divergence Divergence theorem, also known as Gauss theorem relates a volume integral and a surface integral within a vector field. Let FF be a vector field, SS be a closed surface and R be the region inside of SS, then: FF ddaa SS = FFdddd R (55) Example 9: Evaluate the following (3xxıı + 2yyȷȷ ) ddaa SS where SS is the sphere xx 2 + yy 2 + zz 2 = 9. Solution: We could parameterize the surface and evaluate the surface integral, but it is much faster to use the divergence theorem. Since: div (3xxıı + 2yyȷȷ ) = (3xx) + yy (2yy) + (0) = 5 The divergence theorem gives: (3xxıı + 2yyȷȷ ) ddaa SS = 5 dddd R = 5 (Volume of sphere) = 180π Example 10: Evaluate the following (yy 2 zzıı + yy 3 ȷȷ + xxxxkk ) ddaa SS 41

42 where SS is the boundary of the cube defined by 1 xx 1, 1 yy 1, aaaaaa 0 zz 2. Solution: First let s solve the divergence of the equation given: div yy 2 zzıı + yy 3 ȷȷ + xxxxkk = (yy2 zz) + (yy3 ) + (xxxx) The divergence theorem gives: = 3yy 2 + xx yy 2 zzıı + yy 3 ȷȷ + xxxxkk ddaa SS = (3yy 2 + xx) dddd R = (3yy 2 + xx) dddd dddd dddd = 2 6yy 2 dddd 1 = Curl of a vector field The vector product (cross product) of operator and the vector A is known as the curl or rotation of A. Thus in Cartesian coordinates, we can write: Therefore: ıı ȷȷ kk curl AA = AA = (56) AA 1 AA 2 AA 3 curl AA = AA = AA 3 AA 2 ıı, AA 1 AA 3 ȷȷ, AA 2 AA 1 kk, (57) 42

43 where AA = (AA 1, AA 2, AA 3 ). The vector curl AA at point r gives the local rotation (or vorticity) of the material at point r. The direction of curl AA is the axis of rotation and half the magnitude of curl AA is the rate of rotation or angular frequency of the rotation. Example 11: Find the curl of a vector field aa = xx 2 yy 2 zz 2 ıı + yy 2 zz 2 ȷȷ + xx 2 zz 2 kk Solution: This is a straight forward question. All we have to do is to put the equation in the form of equation (56), we get: ıı ȷȷ kk aa = xx 2 yy 2 zz 2 yy 2 zz 2 xx 2 zz 2 = (xx2 zz 2 ) (yy2 zz 2 ) ıı (xx2 zz 2 ) (xx2 yy 2 zz 2 ) ȷȷ + (yy2 zz 2 ) (xx2 yy 2 zz 2 ) kk = 2 yy 2 zzıı + (xxxx 2 xx 2 yy 2 zz)ȷȷ + xx 2 yyzz 2 kk Theorem involving Curl The theorem involving curl of vectors is better known as Stoke s theorem. If we consider a surface SS in R 3 that has a closed non-intersecting boundary, CC, the topology of, say, one half of a tennis ball. That is, if we move along C and fall to our left, we hit the side of the surface where the normal vectors are sticking out. Stoke s theorem states that for a vector field FF within which the surface is situated is given by: FF ddrr CC = ( FF) nn dddd SS (58) The theorem can be useful in either direction: sometimes the line integral is easier than the surface integral, and sometimes vice-versa. Example 12: Evaluate the line integral of the function FF(xx, yy, zz) = xx 2 yy 3, ee xxxx+zz, xx + zz 2 around a circle xx 2 + yy 2 = 1 in the plane yy = 0, oriented counterclock-wise as viewed from the positive yy direction. Solution: Whenever we want to integrate a vector field around a closed curve, and it looks like the computation might be messy, think of applying Stoke s Theorem. The circle CC in question is the positively- 43

44 oriented boundary of the disc SS given by xx 2 + yy 2 1, yy = 0, with the unit normal vector nn pointing in the positive yy direction. That is nn = ȷȷ = 0, 1, 0. Stoke s Theorem tells us that: Evaluating the curl of FF, we get: FF ddrr CC ıı ȷȷ kk FF = xx 2 yy 3 ee xxxx+zz xx + zz 2 = ( FF) nn dddd SS = ee xxxx+zz ıı ȷȷ + (yyee xxxx+zz 3xx 2 yy 2 )kk ( FF) nn = ee xxxx+zz ıı ȷȷ + (yyee xxxx+zz 3xx 2 yy 2 )kk (0, 1, 0) = 1 FF ddrr CC = ( FF) nn dddd SS = 1 dddd SS = aaaaaaaa(ss) = ππ 2.4 Repeated Vector Operations The Laplacian So far, note the following: i. grad must operate on a scalar field and gives a vector field in return ii. div operates on a vector field and gives a scalar field in return, and, iii. curl operates on a vector field and gives a vector field in return 44

45 In addition to the vector relations involving del ( ) mentioned above, there are six other combinations in which del appears twice. The most important one which involves a scalar is: dddddd gggggggg φφ = φ = 2 φφ (59) where φφ(xx, yy, zz) that is a scalar point function. The operator 2 =, is also known as the Laplacian, takes a particularly simple form in Cartesian coordinates, which are: 2 = 2 xx yy zz 2 (60) When applied to a vector, it yields a vector, which is given in Cartesian coordinates: The cross product of two dels operating on a scalar function yields 2 AA = 2 AA xx AA yy AA zz 2 (61) ıı ȷȷ kk φ = cccccccc gggggggg φ = = 0 (62) If AA = 0 for any vector AA, then AA = φφ. In this case, AA is irrotational. Similarly, AA = dddddd cccccccc AA = 0 (63) Finally, a useful expansion is given by: ( AA) = cccccccc cccccccc AA = ( AA) 2 AA (64) Other forms for other coordinate systems for 2 are as follows: 1. Spherical polar coordinates: 45

46 2 = 1 rr 2 rr2 + 1 rr 2 sin θθ sin θθ rr 2 sin 2 θθ φφ (65) 2 2. Two-dimensional polar coordinates: 3. Cylindrical coordinates: 2 = 2 rr rr (66) rr 2 θθ 2 2 = 2 rr rr rr 2 θθ (67) zz 2 Several other useful relations are summarised below: DEL OPERATOR RELATIONS Let φφ and ψψ be scalar fields and AA and BB be vector fields Sum of fields (φφ + ψψ) = φφ + ψψ (AA + BB) = AA + BB (AA + BB) = AA + BB Product of fields (φφφφ) = φφ( ψψ) + ψψ( φφ) (φφaa) = φφ( AA) + ( φφ) AA (φφaa) = φφ( AA) + ( φφ) AA (AA BB) = BB ( AA) AA ( BB) (AA BB) = AA ( BB) + (BB )AA BB( AA) (AA )BB (AA BB) = AA ( BB) BB( AA) + (BB )AA (AA )BB Laplacian ( φφ) = 2 φφ ( AA) = ( AA) 2 AA 46

47 (a) Example 13: If AA = 2yyyyıı xx 2 yyȷȷ + xxzz 2 kk, BB = xx 2 ıı + yyyyȷȷ xxxxkk and φφ = 2xx 2 yyzz 3, find (a) (AA )φφ (b) AA φφ (c) BB φφ (d) 2 φφ Solution: (AA )φφ = 2yyyyıı xx 2 yyȷȷ + xxzz 2 kk ıı + ȷȷ + kk φφ = 2yyyy xx2 yy + xxzz2 2xx2 yyzz 3 = 2yyyy (2xx2 yyzz 3 ) xx 2 yy (2xx2 yyzz 3 ) + xxzz 2 (2xx2 yyzz 3 ) = 2yyyy(4xxyyzz 3 ) xx 2 yy(2xx 2 zz 3 ) + xxzz 2 (6xx 2 yyzz 2 ) = 8xxyy 2 zz 4 2xx 4 yyzz 3 + 6xx 3 yyzz 4 (b) φφ = (2xx2 yyzz 3 )ıı + (2xx2 yyzz 3 )ȷȷ + (2xx2 yyzz 3 )kk Therefore = 4xxxxzz 3 ıı + 2xx 2 zz 3 ȷȷ + 6xx 2 yyzz 2 kk AA φφ = 2yyyyıı xx 2 yyȷȷ + xxzz 2 kk 4xxxxzz 3 ıı + 2xx 2 zz 3 ȷȷ + 6xx 2 yyzz 2 kk = 8xxyy 2 zz 4 2xx 4 yyzz 3 + 6xx 3 yyzz 4 (c) φφ = 4xxxxzz 3 ıı + 2xx 2 zz 3 ȷȷ + 6xx 2 yyzz 2 kk, therefore: 47

48 ıı ȷȷ kk BB φφ = xx 2 yyyy xxxx 4xxxxzz 3 2xx 2 zz 3 6xx 2 yyzz 2 = (6xx 2 yy 2 zz 3 + 2xx 3 yyzz 3 )ıı + ( 4xx 2 yy 2 zz 3 6xx 4 yyzz 2 )ȷȷ + (2xx 4 zz 3 4xxyy 2 zz 4 )kk (d) 2 φφ = 2 2 (2xx2 yyzz 3 ) (2xx2 yyzz 3 ) (2xx2 yyzz 3 ) = 4yyzz xx 2 yyyy 48

49 3. Linear Algebra, Matrices & Eigenvectors In many practical systems, there naturally arises a set of quantities that can conveniently be represented as a certain dimensional array, referred to as matrix. If matrices were simply a way of representing array of numbers, then they would have only a marginal utility as a means of visualising data. However, a whole branch of mathematics has evolved, involving manipulation of matrices, which has become a powerful tool for the solution f many problems. For example, consider the set of nn linear equations with nn unknowns aa 11 YY 1 + aa 12 YY aa 1nn YY nn = 0 aa 21 YY 1 + aa 22 YY aa 2nn YY nn = 0 aa nn1 YY 1 + aa nn2 YY aa nnnn YY nn = 0 (68) The necessary and sufficient condition for the set to have a non-trivial solution (other than YY 1 = YY 2 = = YY nn = 0) is that the determinant of the array of coefficients is zero: dddddd(aa) = Basic definitions and notation A matrix is an array of numbers with mm rows and nn columns. The (i, j) th element is the element found in row ii and column jj. For example, have a look at the matrix below. Tis matrix has mm = 2 rows, nn = 3 column, and therefore the matrix order is 2 3. The (i, j) th element is aa iiii AA = aa 11 aa 12 aa 13 aa 21 aa 22 aa 23 (69) Matrices may be categorized based on the properties of its elements. Some basic definitions include: 1. The transpose of matrix AA (or AA TT ) is formed by interchanging element aa iiii with element aa jjjj. Therefore: AA TT = aa jjjj, (AA + BB) TT = AA TT + BB TT, (AAAA) TT = AA TT BB TT (70) A symmetric matrix is equals to its transpose, AA = AA TT. 49