Векторный и тензорный анализ в примерах и задачах = Vector and Tensor Analysis through Examples and Exercises

252 Pages·5.636 MB·Russian

by  Абрашина-Жадаева, Наталья Григорьевна

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,
which take a certain range, e. g., p, vi, w , tki (i,k,l = 1,...,n). The
kl l
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.
(cid:80)
The sum sign 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:
a bim = a bkm;
i k
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:
correct expression : aikv = bk;
i
incorrect expression : aikv = bj.
i
Kronecker delta:
(cid:26)
1, if i = k,
δ k = δk =
i i 0, if i (cid:54)= k.
Example 1. Write the short expression a xk = b , i,k = 1,2,3, in
ik i
the full form.
Solution. Let i = 1. Taking into account the summation convention we
can expand
b = a xk = a x1 + a x2 + a x3.
1 1k 11 12 13
4
For the index values i = 2 and i = 3 the expansion is similar. Thus we
obtain the following full form of a xk = b :
ik i

a x1 + a x2 + a x3 = b ,
 11 12 13 1
a x1 + a x2 + a x3 = b ,
21 22 23 2
 a x1 + a x2 + a x3 = b .
31 32 33 3
j
Example 2. Factorize the expression a − c a .
k k j
Solution. Using Kronecker delta we can express the single element a as
k
j
a = δ a . Therefore
k k j
j j j j j
a − c a = δ a − c a = (δ − c )a .
k k j k j k j k k j
Exercises
Write expressions in the full form.
1.1. a = b kc , i,k = 1,2,3. 1.4. ci k = ai, i,k = 1,2,3.
i i k k
1.2. d = a bibk, i,k = 1,2. 1.5. ci aj = aib . i,j,k = 1,2.
ik kj k
1.3. d = ak , k = 1,2,3.
k
Simplify expressions.
1.6. aiδ k = bk. 1.8. δ kδ laij = δ lbkcm.
i i j m
1.7. δ lδ ma b − a b .
i k l m i k
Factorize expressions.
1.9. ai bk − bi. 1.11. ai bj − ai brbsc .
k jk rs k
1.10. ai b ck − ai . 1.12. ai bjbk − al c bidrs.
k i i jk rs l
1.13. Show that δ k = n, k = 1,n.
k
Find the value of expressions (i,j,... = 1,n).
1.14. δ iδ jδ kδ l. 1.15. δ iδ jδ lδ k.
j k l m j l k i
5
1.2. Conjugate linear spaces
Let V n be a linear space of the dimension n containing elements (cid:126)x, (cid:126)y,
etc. Any ordered linearly independent set of n elements ((cid:126)e ,(cid:126)e ,...,(cid:126)e ),
1 2 n
(cid:126)e ∈ V n, i = 1,n, is called the basis of the space V n. Any element (cid:126)x in
i
V n can be uniquely expressed as a linear combination of vectors (cid:126)e :
i
(cid:126)x = x1(cid:126)e + x2(cid:126)e + ... + xn(cid:126)e = xi(cid:126)e .
1 2 n i
Coefficients xi of this linear combination are called components or coordi-
nates of the element (cid:126)x in the basis ((cid:126)e ).
i
Let V be another linear space of the dimension n containing elements
n
f, g, etc. Suppose there is a functional that maps each pair of elements
(cid:126)(cid:126)x ∈(cid:126)V n and f ∈ V into a real number (cid:104)(cid:126)x,f(cid:105) ∈ R having the following
n
properties: (cid:126) (cid:126)
a) (cid:104)(cid:126)x,f(cid:105) is linear with respect to each argument, that is ∀α,β ∈ R
(cid:126)
(cid:104)α(cid:126)x + β(cid:126)y,f(cid:105) = α(cid:104)(cid:126)x,f(cid:105) + β(cid:104)(cid:126)y,f(cid:105);
(cid:126) (cid:126) (cid:126)
(cid:104)(cid:126)x,αf + βg(cid:105) = α(cid:104)(cid:126)x,f(cid:105) + β(cid:104)(cid:126)x,g(cid:105);
(cid:126) (cid:126) (cid:126) (cid:126)
b) (cid:104)(cid:126)x,f(cid:105) is homogeneous, that is if (cid:104)(cid:126)x,f(cid:105) = 0 for all f, then (cid:126)x is the
(cid:126) (cid:126) (cid:126)
null element of the space V n; and if (cid:104)(cid:126)x,f(cid:105) = 0 for all (cid:126)x, then f is the null
element of the space V ; (cid:126) (cid:126)
n
c) (cid:104)(cid:126)x,f(cid:105) is symmetric, that is (cid:104)(cid:126)x,f(cid:105) = (cid:104)f,(cid:126)x(cid:105) for all (cid:126)x and f.
(cid:126) (cid:126) (cid:126) (cid:126)
A real number (cid:104)(cid:126)x,f(cid:105) is called the bundle.
Spaces V n and V(cid:126), for elements of which the bundle is defined, are
n
called conjugate or dual spaces.
If a set of elements (e1,e2,...en) is a basis in space V , then we can
n
(cid:126) (cid:126) (cid:126)
express any element f ∈ V n as
(cid:126)
f = f ei.
i
(cid:126) (cid:126)
Bases ((cid:126)e ) and (ek) satisfying conditions
i
(cid:126)
(cid:104)(cid:126)e ,ek(cid:105) = δ k
i i
(cid:126)
are called dual or reciprocal bases. In such bases we can calculate compo-
nents xi and f as follows:
i
xi = (cid:104)(cid:126)x,ei(cid:105), f = (cid:104)f,(cid:126)e (cid:105), and (cid:104)(cid:126)x,f(cid:105) = xif .
i i i
(cid:126) (cid:126) (cid:126)
6
A function f : V n (cid:55)→ R that maps an element of a linear space V n into
a real number satisfying ∀(cid:126)x,(cid:126)y ∈ V n and ∀α ∈ R conditions
f((cid:126)x + (cid:126)y) = f((cid:126)x) + f((cid:126)y) and f(α(cid:126)x) = αf((cid:126)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 (cid:104)f,(cid:126)x(cid:105) = f((cid:126)x).
(cid:126)
Example 1. Let V be a linear space of linear functions defined in a
3
space V 3. Find the reciprocal basis to
(cid:126)e = (1,0,1), (cid:126)e = (0,1,1), (cid:126)e = (1,2,0).
1 2 3
Calculate coordinates of the element (cid:126)x = (2,3,2) in given basis.
Solution. The coordinates of vectors (cid:126)e ,i = 1,2,3, are given in some
i
other basis ((cid:126)eˆ). Consider the basis (eˆi), eˆi ∈ V , that is reciprocal to
i 3
ˆ (cid:126) (cid:126)
the basis ((cid:126)e ). Let a linear function has coordinates f = (a ,a ,a ) in
i 1 2 3
reciprocal basis and f = a eˆi. Then the bundle of f (cid:126)∈ V and (cid:126)x ∈ V 3
i 3
is equal to (cid:104)f,(cid:126)x(cid:105) =(cid:126)a xi. W(cid:126)e are to find three elem(cid:126) ents ei that satisfy
i
(cid:104)ei,(cid:126)e (cid:105) = δi (cid:126). Let (cid:126)
k k
(cid:126)
(cid:126)e = xk(cid:126)eˆ , ei = ai eˆk.
i k k
i (cid:126) (cid:126)
The coordinates xk are known. Then we write the condition (cid:104)ei,(cid:126)e (cid:105) = δi
k k
i (cid:126)
as
1 1 1 1
(cid:104)e1,(cid:126)e (cid:105) = a xk = a · 1 + a · 0 + a · 1 = δ1 = 1;
1 k 1 2 3 1
(cid:126) 1
1 1 1 1
(cid:104)e1,(cid:126)e (cid:105) = a xk = a · 0 + a · 1 + a · 1 = δ1 = 0;
2 k 1 2 3 2
(cid:126) 2
1 1 1 1
(cid:104)e1,(cid:126)e (cid:105) = a xk = a · 1 + a · 2 + a · 0 = δ1 = 0.
3 k 1 2 3 3
(cid:126) 3
1 1 1
This is a system of linear equations for variables a ,a ,a , which solu-
1 2 3
tion is
(cid:18) (cid:19)
2 1 1 2 1 1
1 1 1
a = , a = − , a = or e1 = ,− , .
1 2 3
3 3 3 (cid:126) 3 3 3
Doing the same for e2 and e3 we obtain
(cid:126) (cid:126)
(cid:18) (cid:19) (cid:18) (cid:19)
2 1 2 1 1 1
e2 = − , , and e3 = , ,− .
(cid:126) 3 3 3 (cid:126) 3 3 3
7
To find coordinates of (cid:126)x in the basis ((cid:126)e ) we use the formula xi = (cid:104)(cid:126)x,ei(cid:105):
i
(cid:126)
2 1 1
x1 = (cid:104)e1,(cid:126)x(cid:105) = · 2 − · 3 + · 2 = 1;
(cid:126) 3 3 3
2 1 2
x2 = (cid:104)e2,(cid:126)x(cid:105) = − · 2 + · 3 + · 2 = 1;
(cid:126) 3 3 3
1 1 1
x3 = (cid:104)e3,(cid:126)x(cid:105) = · 2 + · 3 − · 2 = 1.
(cid:126) 3 3 3
Thus (cid:126)x = 1 ·(cid:126)e + 1 ·(cid:126)e + 1 ·(cid:126)e .
1 2 3
Exercises
Let V be a linear space of linear functions defined in a space V 3. Find
3
the reciprocal basis to the basis ((cid:126)e ,(cid:126)e ,(cid:126)e ) and calculate coordinates of the
1 2 3
element (cid:126)x = (2,3,2) in given basis.
1.16. (cid:126)e = (1,2,−3), (cid:126)e = (2,−2,0), (cid:126)e = (−2,0,1), (cid:126)x = (1,2,3).
1 2 3
1.17. (cid:126)e = (2,−1,0), (cid:126)e = (−1,0,1), (cid:126)e = (0,2,−1), (cid:126)x = (3,−2,1).
1 2 3
1.18. (cid:126)e = (0,−1,1), (cid:126)e = (2,0,1), (cid:126)e = (−1,−1,2), (cid:126)x = (1,1,1).
1 2 3
1.19. (cid:126)e = (1,1,1), (cid:126)e = (1,0,0), (cid:126)e = (0,1,0), (cid:126)x = (−1,1,−1).
1 2 3
1.20. Let (cid:126)x = (x1,x2) ∈ V 2 and y = (y ,y ) ∈ V . Define the bundle
1 2 2
as (cid:126)
(cid:104)(cid:126)x,y(cid:105) = ai xky ,
k i
(cid:126)
where (ai ) is a matrix 2 × 2. Prove that spaces V 2 and V are dual.
k 2
Let a bundle for elements of spaces V 2 and V be defined as it done in
2
problem 1.20. Find the reciprocal basis to ((cid:126)e ,(cid:126)e ). Calculate coordinates
1 2
of elements (cid:126)x and y in given bases and their bundle.
(cid:126) (cid:18) (cid:19)
1 1
1.21. (cid:126)e = (1,2), (cid:126)e = (1,1), (ai ) = , (cid:126)x = (2,3),
1 2 k 2 1
y = (−1,2).
(cid:126) (cid:18) (cid:19)
3 −4
1.22. (cid:126)e = (1,0), (cid:126)e = (0,1), (ai ) = , (cid:126)x = (1,1),
1 2 k 3 −2
y = (−2,1).
(cid:126) (cid:18) (cid:19)
1 2
1.23. (cid:126)e = (−2,3), (cid:126)e = (1,−1), (ai ) = , (cid:126)x = (−1,0),
1 2 k 3 4
y = (1,−2).
(cid:126)
8
(cid:18) (cid:19)
1 −1
1.24. (cid:126)e = (1,−1), (cid:126)e = (−2,3), (ai ) = , (cid:126)x = (−2,1),
1 2 k 1 1
y = (1,−3).
(cid:126)
1.25. Consider a space of polar vectors (cid:126)x,(cid:126)y,... and a space of lamellar
vectors f,g,.... Polar vector is defined by an ordered pair of points O and
M, the(cid:126)po(cid:126)int O being common point for all vectors. A lamellar vector f
is defined by an ordered pair of parallel planes ω and µ (or by a pair o(cid:126)f
parallel lines in the two dimensional case) and the plane ω contains the
point O. Define the bundle of polar and lamellar vectors by following rule:
a) if O ∈ ω and M ∈ ω, then (cid:104)(cid:126)x,f(cid:105) = (cid:104)f,(cid:126)x(cid:105) = 0;
b) if M ∈/ ω, then (cid:104)(cid:126)x,f(cid:105) = (cid:104)f,(cid:126)x(cid:105)(cid:126)is the(cid:126)coefficient of proportionality
(cid:126) (cid:126)
between line segments OM and ON (fig. 1.1).
Fig. 1.1
The point N ∈ µ and the points M and N are on the same straight
line OM. In addition, (cid:104)f,(cid:126)x(cid:105) > 0 if the points M and N are on the same
side of the plane ω and(cid:126)(cid:104)f,(cid:126)x(cid:105) > 0 otherwise. Prove that spaces of polar
and lamellar vectors are d(cid:126)ual.
1.26. Let P be some point in a three dimensional space and C1 be a
P
set of all differentiable functions at P. Consider a set
T (P) = {df(P)|f ∈ C1}.
1 P
The set T (P) is a linear space. Define a bundle of an element (cid:126)l ∈ V 3 and
1
df(P) ∈ T (P) as1:
1
∂f
(cid:126) (cid:126)
(cid:104)l,df(P)(cid:105) = |l| .
∂l
Prove that spaces T (P) and V 3 are dual.
1
1The definition of directional derivative can be found in section 3.1.
9
1.3. Changing of bases
Suppose that in a linear space V n there are an arbitrary “old” basis
((cid:126)e ),i = 1,...,n, and some “new” basis ((cid:126)e ), i(cid:48) = 1(cid:48),...,n(cid:48) = n. In the
i i(cid:48)
tensor calculus the new basis elements are denoted by the same root letter
“e” but with primed indices. Since ((cid:126)e ) is a basis, we can expand its
i(cid:48)
elements (cid:126)e as
i
(cid:126)e = Ai (cid:126)e ,
i(cid:48) i(cid:48) i
where n2 numbers Ai form a matrix, called the transformation matrix
i(cid:48)
from the “old” basis to the “new” one. The element of the ith row and the
i(cid:48)th column of the matrix (Ai ) is the ith coordinate of the element (cid:126)e in
i(cid:48) i(cid:48)
the basis (cid:126)e . The matrix (Ai ) is nondegenerate that is det(Ai ) (cid:54)= 0.
i i(cid:48) i(cid:48)
An “old” basis can be expressed through a “new” basis by inverse trans-
formation matrix (Ai(cid:48) ):
i
(cid:126)e = Ai(cid:48) (cid:126)e ,
where i i i(cid:48)
Ak(cid:48) Ai = δk(cid:48) , Ak(cid:48) Aj = δ j.
i j(cid:48) j(cid:48) i k(cid:48) i
Coordinates of a vector (cid:126)x ∈ V n in an “old” and a “new” bases are
connected by the relation
xi(cid:48) = Ai(cid:48) xi, xi = Ai xi(cid:48).
i i(cid:48)
Let ei and ei(cid:48) be an “old” and a “new” reciprocal bases in a dual space
(cid:126) (cid:126)
V correspondingly. Then
n
ei(cid:48) = A i(cid:48)ei, ei = A iei(cid:48);
i i(cid:48)
(cid:126) (cid:126) (cid:126) (cid:126)
f = A if , f = A i(cid:48)f ,
i(cid:48) i(cid:48) i i i i(cid:48)
where (A i) is the inverse transposed transformation matrix and (A i(cid:48)) is
i(cid:48) i
the transposed transformation matrix:
A k(cid:48)A i = δk(cid:48) , A k(cid:48)A j = δ j and Ai A j(cid:48) = δj(cid:48) .
i j(cid:48) j(cid:48) i k(cid:48) i k(cid:48) i k(cid:48)
You can see that coordinates of elements of a dual space V are changing
n
in the same way as basis elements of space V n, and vise versa, coordinates
of elements of an initial space V n are changing in the same way as basis
elements of the dual space V . Arithmetic vectors (xi) and (f ) (i. e.
n i
ordered set of numbers) are called contravariant vectors and covariant
vectors or briefly vectors and covectors.
10