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Abrashina.Pdf УДК 514.742.4(075.8)+514.743(075.8) ББК 22.151.5я73-1 А16 Рецензенты: кафедра фундаментальной и прикладной математики Гродненского государственного университета имени Янки Купалы (заведующий кафедрой доктор физико-математических наук, профессор Е. А. Ровба); заведующий отделом вычислительной математики Института математики НАН Беларуси кандидат физико-математических наук Г. Ф. Громыко; доцент кафедры теории и практики перевода № 1 Минского государственного лингвистического университета кандидат филологических наук В. Г. Минина Абрашина-Жадаева, Н. Г. А16 Векторный и тензорный анализ в примерах и задачах = Vector and Tensor Analysis through Examples and Exercises : учеб. посо- бие / Н. Г. Абрашина-Жадаева, И. А. Тимощенко. – Минск : БГУ, 2019. – 250 с. ISBN 978-985-566-793-4. В учебном пособии излагаются основы векторного и тензорного анали- за. Приводятся базовые теоретические сведения, многочисленные упраж- нения для самостоятельного изучения материала. Основное внимание со- средоточено на методах решения задач. Предназначено для студентов высших учебных заведений по специ- альности «Компьютерная физика». УДК 514.742.4(075.8)+514.743(075.8) ББК 22.151.5я73-1 ISBN 978-985-566-793-4 © Абрашина-Жадаева Н. Г., Тимощенко И. А., 2019 © БГУ, 2019 Preface This textbook is written based on lecture courses and practical classes that were held at the Faculty of Physics of the Belarusian State Univer- sity under the curriculum “Mathematical Analysis” and “Fundamentals of Vector and Tensor Analysis”. It is intended for foreign students of physi- cal, mathematical, engineering-physical and engineering-technical special- ties and for those who independently study mathematics in English. The training of future engineers, teachers of the specialties “Physics”, “Chemistry”, “Biology” etc., is closely related to acquiring mathematical knowledge and practical skills. Therefore, the authors of the book sought to present the basics of mathematical information that a qualified natu- ral scientist must have in an accessible and convenient form. The goal is to assist with self-mastering important topics of higher mathematics for students, as well as in preparing classes on these topics for teachers. The book is also for those readers who want both to understand the basics of higher mathematics and to learn how to apply them. The main emphasis in the textbook is not on the theoretical aspect, but on explana- tions of the fundamentals of the subject with the help of examples. The authors, using illustrative examples, show the meaning of the most difficult concepts, methods of their application, usefulness, and significance. Plenty of exercises serves as a support for mastering skills. Everything valuable in the book belongs to the mathematical commu- nity, all errors, of course, belong to the authors. Comments and suggestions can be sent to [email protected] or [email protected]. Chapter 1 Introduction to tensor algebra 1.1. Index notation In tensor calculus according to the method of root letters and indices any tensor is specified by using a root letter and an ordered set of indices, i ki which take a certain range, e. g., p; v ; wkl; t l (i; k; l = 1; : : : ; n). The number, sequence and position (upper or lower) of indices define algebraic and transformational properties of objects. If the index is assigned a certain value, then this index is called fixed. To simplify formula manipulation the Einstein summation convention is assumed: 1) upper and lower identical indices are to be summed over their range. The sum sign P is omitted. Repeated indices are called dummy or sum- mation indices. Pair of dummy indices can occur in a formula only once, but can be easily renamed without changing the result of an expression: im km aib = akb ; 2) an index which is not a dummy index is called the free index. Free indices in the different parts of an expression are to be the same: ik k correct expression : a vi = b ; ik j incorrect expression : a vi = b : Kronecker delta: 1; if i = k; δ k = δk = i i 0; if i 6= k: k Example 1. Write the short expression aikx = bi; i; k = 1; 2; 3; in the full form. Solution. Let i = 1. Taking into account the summation convention we can expand k 1 2 3 b1 = a1kx = a11x + a12x + a13x : 4 For the index values i = 2 and i = 3 the expansion is similar. Thus we k obtain the following full form of aikx = bi: 8 1 2 3 a11x + a12x + a13x = b1; < 1 2 3 a21x + a22x + a23x = b2; 1 2 3 : a31x + a32x + a33x = b3: j Example 2. Factorize the expression ak − ck aj. Solution. Using Kronecker delta we can express the single element ak as j ak = δk aj: Therefore j j j j j ak − ck aj = δk aj − ck aj = (δk − ck )aj: Exercises Write expressions in the full form. k i k i 1.1. ai = bi ck; i; k = 1; 2; 3: 1.4. c k = a ; i; k = 1; 2; 3: i k i j i 1.2. d = aikb b ; i; k = 1; 2: 1.5. c kja = a bk: i; j; k = 1; 2: k 1.3. d = a k; k = 1; 2; 3: Simplify expressions. i k k k l ij l k m 1.6. a δi = b : 1.8. δi δj a = δm b c : l m 1.7. δi δk albm − aibk: Factorize expressions. i k i i j i r s 1.9. a kb − b : 1.11. a jkb − a rsb b ck. i k i i j k l i rs 1.10. a kbic − a i: 1.12. a jkb b − a rsclb d : k 1.13. Show that δk = n; k = 1; n: Find the value of expressions (i; j; : : : = 1; n). i j k l i j l k 1.14. δj δk δl δm . 1.15. δj δl δk δi . 5 1.2. Conjugate linear spaces Let V n be a linear space of the dimension n containing elements ~x;~y; etc. Any ordered linearly independent set of n elements (~e1;~e2; : : : ;~en); n n ~ei 2 V ; i = 1; n, is called the basis of the space V . Any element ~x in n V can be uniquely expressed as a linear combination of vectors ~ei: 1 2 n i ~x = x ~e1 + x ~e2 + ::: + x ~en = x ~ei: Coefficients xi of this linear combination are called components or coordi- nates of the element ~x in the basis (~ei). Let Vn be another linear space of the dimension n containing elements f; g; etc. Suppose there is a functional that maps each pair of elements n ~~x 2~V and f 2 Vn into a real number h~x; fi 2 R having the following properties: ~ ~ a) h~x; fi is linear with respect to each argument, that is 8α; β 2 R ~ hα~x + β~y; fi = αh~x; fi + βh~y; fi; ~ ~ ~ h~x; αf + βgi = αh~x; fi + βh~x; gi; ~ ~ ~ ~ b) h~x; fi is homogeneous, that is if h~x; fi = 0 for all f, then ~x is the null element~ of the space V n; and if h~x; fi ~= 0 for all ~x, then~ f is the null element of the space Vn; ~ ~ c) h~x; fi is symmetric, that is h~x; fi = hf; ~xi for all ~x and f. A real~ number h~x; fi is called the~bundle~. ~ n ~ Spaces V and Vn, for elements of which the bundle is defined, are called conjugate or dual spaces. 1 2 n If a set of elements (e ; e ;::: e ) is a basis in space Vn, then we can express any element f 2~V n~as ~ ~ i f = fi e : ~ ~ k Bases (~ei) and (e ) satisfying conditions ~ k k h~ei; e i = δ ~ i are called dual or reciprocal bases. In such bases we can calculate compo- i nents x and fi as follows: i i i x = h~x; e i; fi = hf;~eii; and h~x; fi = x fi: ~ ~ ~ 6 A function f : V n 7! R that maps an element of a linear space V n into a real number satisfying 8~x;~y 2 V n and 8α 2 R conditions f(~x + ~y) = f(~x) + f(~y) and f(α~x) = αf(~x) is called the linear function. A set V ∗ of all linear functions defined in the space V n is also a linear space of the dimension n; which is the conjugate space to V n, if the bundle is hf; ~xi = f(~x). ~ Example 1. Let V3 be a linear space of linear functions defined in a space V 3. Find the reciprocal basis to ~e1 = (1; 0; 1); ~e2 = (0; 1; 1); ~e3 = (1; 2; 0): Calculate coordinates of the element ~x = (2; 3; 2) in given basis. Solution. The coordinates of vectors ~ei; i = 1; 2; 3; are given in some ^ i i other basis (~ei). Consider the basis (e^ ); e^ 2 V3, that is reciprocal to ^ ~ ~ the basis (~ei). Let a linear function has coordinates f = (a1; a2; a3) in i ~ 3 reciprocal basis and f = ai e^ : Then the bundle of f 2 V3 and ~x 2 V ~ i ~ ~ i is equal to hf; ~xi = aix . We are to find three elements e that satisfy i i ~ ~ he ;~eki = δ k. Let ~ k ^ i i k ~ei = x ~ek; e = ak e^ : i ~ ~ k i i The coordinates x are known. Then we write the condition he ;~eki = δ k as i ~ 1 1 k 1 1 1 1 he ;~e1i = ak x = a1 · 1 + a2 · 0 + a3 · 1 = δ 1 = 1; ~ 1 1 1 k 1 1 1 1 he ;~e2i = ak x = a1 · 0 + a2 · 1 + a3 · 1 = δ 2 = 0; ~ 2 1 1 k 1 1 1 1 he ;~e3i = ak x = a1 · 1 + a2 · 2 + a3 · 0 = δ 3 = 0: ~ 3 1 1 1 This is a system of linear equations for variables a1; a2; a3, which solu- tion is 1 2 1 1 1 1 1 2 1 1 a1 = ; a2 = − ; a3 = or e = ; − ; : 3 3 3 ~ 3 3 3 Doing the same for e2 and e3 we obtain ~ ~ 2 1 2 1 1 1 e2 = − ; ; and e3 = ; ; − : ~ 3 3 3 ~ 3 3 3 7 i i To find coordinates of ~x in the basis (~ei) we use the formula x = h~x; e i: ~ 2 1 1 x1 = he1; ~xi = · 2 − · 3 + · 2 = 1; ~ 3 3 3 2 1 2 x2 = he2; ~xi = − · 2 + · 3 + · 2 = 1; ~ 3 3 3 1 1 1 x3 = he3; ~xi = · 2 + · 3 − · 2 = 1: ~ 3 3 3 Thus ~x = 1 · ~e1 + 1 · ~e2 + 1 · ~e3.
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