Mathematics (Математика). Учебное пособие

151

(

)

rrr

r

However

(

)

()()()(

)

rrr

r

(

)

r

(

)

r

rrr

r

( )

.2kyjxzizMa

++==

xz

y

¶

y

+

¶

z

¶

=

Madiv

¶

z

+

¶

x

22

.0

=

Fig.14

Fig.14 – graphic representation of the surface integral of the 2nd order

2

òò

s

,

ydxdyxzdydzdydzzI ++=

where closed surface σ consists of an

outer side of a part of a paraboloid’s surface

s

1

Therefore,

22

òòò

V

,0,4:

³-=+ zzyx

and also of a part of a plane .0:

==

dVI .00

s

=z

2

Example. Figure out whether vector field

kzxjyziyxMa

++-++= 2 is conservative.

Solution. Vector field

the domain

( )

rr

[ ]

aMarot

=Ñ=

=Marot Calculating

¶

-Ma

is conservative if at every point М of

.0

kji

rr

æ

R

¶

ç

=

ç

zyx

¶

è

RQP

ö

Q

¶

-

z

y

¶

P

¶

æ

÷

i

÷ ø

-

+

ç

z

¶

è

æ

R

x

Q

¶

ö

ç

j

÷ ø

-

+

ç

x

¶

è

To remind ourselves how to calculate the curl in this formula let’s use the formal Hamilton's operator "nabla" (or, as it is also known, “Del”)

r

ö

P

÷

,, k

÷

y

ø

152

rrr

¶

+

¶

()()()(

)

r

¶

(

)

r

(

)

r

(

)

r

¶

()()()(

)

rrrrrrr

r

(

)

r

3

6

p

+

and

Q

x

j

+

M

z

¶

zyxRzyxQzyxPa ,,,,,,,,

in every point М

R

+

¶

z

Q

=

¶

z

¶

0=

.

0=Madiv

.

R

=

x

R

.0,2,1,0,0,1 =

¶

y

,02102010

¹---=-+-+-= kjikjiMarot

y

¶

Q

¶

y

=

i

=Ñ

x

¶

that works the same as finding a vector product in Cartesian coordinates.

For other types of fields, which we will work with in exercise 8, let’s establish what they mean:

Solenoidal field of domain V meets the condition

r

( ) ( ) ( ) ( )

Madiv

Harmonic field

and conservative, i.e.

P

In our case

Then

therefore, the field

Try to solve a contour integral of the 1st kind yourself первого рода

=

¶

y

x

ò

y

L

P

= M

Ma

¶

Ma

d

l

M

¶

x

is at every point of domain V both solenoidal

0

=Marot

P

=

z

is not conservative.

, where L is the arc of a circle

¶

,

k

,cos2,sin2

££== ttytx

.

153

Section 6. Differential equations

22

ydyxdx

yx

-=+-

,ln(1)0.5ln10.5ln

yxc

11

=±--

()()0

++-=

(1)(1)

yxdyxydx

22

ydyxdx

yx

-=+-

,ln(1)0.5ln10.5ln

yxc

11

=±--

()(

)

()(

)

Topic 19. Basic differential equations

Introduction. The solution of geometrical, physical and engineering problems often lead to equations that relate the independent variables to any function of these variables and the derivatives of this function.

Example. Find the general integral of a differential equation

.

11

Solution. Because the variables in this equationвare already separat- ed, we only need to integrate its left and right sides.

We get

Answer:

ydyxdx

=-+=-+

22

òò

11

yx

+-

ycx

2

Example. Find the general integral of a differential equation

22

xyxdxyxydy

Solution. Let’s transform this equation to a form of

22

-=-+

. This is an equation with separating varia-

22

.

bles. Let’s separate them:

ydyxdx

on the equation:

òò

=-+=-+

22

11

yx

+-

On the basis of the properties of logarithms and methods of potentia-

tion, we get an answer

ycx

Example. Find the general solution for differential equations.

а)

Solution. Let’s separate the variables. For that, let's factor out the common multiplier:

11

2

.

0=+++ dyxxydxyxy

. Now let’s integrate both parts

22

.

and put y to the left and

011 =+++ dyyxdxxy

.

154

()(

)

11

yxdxxydy

dy

dx

fxfy

dx

dxdy

æö

ç÷

èø

Þ

x

xyx

dy

dx

y

x

dyy

dxx

æö

ç÷

èø

x

+=+

¢

=

¢

xdxU

sin

xdxU

sin

+=-+

1

ö

æ ç

è

yx=+

Cxye

.

y

.

dx

+

1

÷

x

ø

y +=

; when expressing

dy

= 

() ()

12

11

æö

11

+=-+

ç÷

xy

èø

ö

æ

1

÷

ç

òò

è

Cyyxx lnlnln +--=+

,

yx=+

Cxye

is the general integral of the

sin

y

¢

dy

+-=

1

÷

y

ø

.

yx

+

. We need to make sure

lnlnln =+

.

Cexy

.

out of

which

x to the right:

this equation, we need to make sure that

would mean that our equation is a differential equation with dividable

variables. Now let’s separate them.

Then, integrate this form by the corresponding variables:

We get

Thus, we made sure that given equation.

Answer:

¢

б)

Solution. Let’s make sure that it is impossible to separate the varia-

bles by dividing both parts by x:

xyx +=

that derivative

F

=

equation of the first kind. We will solve it by replacing the corresponding variables.

Let’s introduce a new variable

dU

in the equation depends only on the ratio of

which would mean that this is a homogenous differential

=

sin

òò

;

UUUxU

tg lnln

ln +=

;

y

= ;

dU

;

UxU sin

U

2

Cx

;

¢

=

U

tg =

2

¢

UxUyU

+

.

; let’s integrate it

tg =

;

y

2

Cx

, i.e.

Cx

x

;

155

arctgxCxy

×

=

×

=

112

yxyxCyxCxC

=Þ=+Þ=++

12

yxCxC

=++

=-¢+¢

¢

2 is the general solution to this equation.

Answer: arctgxCxy

2 .

Exercise 2. Find the general solution for differential equation y'' = x2.

Solution. Because the derivative in our case is a function that de-

pends only on x, it is solvable by serial integration:

234

Answer:

¢¢¢

4

/12

/3/12

.

Do this yourself and reference the answers after doing so:

Find the general solution (general integral of the differential equation.

1. ex+3ydy = xdx. (Answer: е3y = 3(C — хе-x – е--x).)

2. y'sin х = у In у. (Answer: In у = С tg (х/2).)

3. у' = (2х— I) ctg у. (Answer: In |cos y| =x — x2+C.)

4. sec2 х tg ydy+ sec2 у tg xdy = 0. (Answer: С = = tg у tg x.)

Topic 20. Liner differential equations

Introduction. In mathematics uses many types of linear differential equations. In this workshop we will focus on the study of linear homogeneous differential equations with constant coefficients.

Example. Find the specific solution of the differential equation, that

¢

¢¢

meets these initial conditions

¢

Solution.

¢¢

+

32 =-

+

x

2

eyyy

is a non-homogeneous differential

equation with constant coefficients of the 2nd kind. The solution is actually the sum of other solutions: of the general solution of a homogenous equation y and the specific solution of a non-homogeneous equation y*,

which we will find via the right side. But first, let’s find y.

yy

032

. Let’s make a defining equation:

3;1

-== kk

21

Therefore, the general solution of the homogenous equation is:

+=

2

x

xx

321-

eCeCy

.

¢

1)0(;1)0(;32

==-

=

yyeyyy

.

2

;032

=-+ kk

156

y* we will find as

*2*2

xx

yAeyAe

¢²

=¢=

320

¢¢¢¢¢¢

++=

xeA2

. y* is a particular solution, so it will turn it

×

into a true numerical identity. Let’s substitute it into our equation and find А.

2;4

==

Therefore,

*

( ) ( )

2222

xxxx

xey2*

2,0=

. Thus, the general solution to the equation is

xxx

-

21

23

eeCeCyyy

2,0++=+=

. To calculate the particular solution

.

2,0344

=Þ=-+ AeAeAeAe

.

let’s define the values of constants from the initial conditions:

xxx

23

eeCeCy

2,0++=

;

yy

0

2

=

C

ì

1

í

=

C

î

2

xxx

23

.

1)0(;1)0(

;

-

0*20*3

;4,031

eeCeC +-=

75.0

05.0

Answer:

-

21

23

¢

0

1

-

2

ì í î

-

21

2,01 eeCeC ++=

CC

21

CC

xxx

;4,03

eeCeCy +-=

0*20*3

;

++=

1

2,01

®

+-=

4,031

21

-

eeey

2,005.075.0 ++=

Try to do these yourself:

¢¢

¢¢¢

1)

yyy

2)

+

¢

2

123 xyyy -=

157

Section 7. Elements of field theory

(

)

(

)

l

¶

l

¶

r

rrr

¶

x

¶

z

(

)

rrr

Topic 21. Field theory

Introduction. This workshop is devoted to the discussion of the following concepts: vector potential, the gradient of the function f, rotor of vector, the divergence of a vector, Stokes formula.

Example 1.

а) Find the magnitude and direction of the maximum field change

222

575 xyzzxyyzxMU +-=

at point

M

Solution. It is proven that the scalar field U(M) has at point М0 the

U

( )

maximum derivative in the direction

M

¶

gradient’s modulus of field U at point:

U

( ) ( )

=

if we take vector

¶

, which points at the differentiation direction, for

l

the direction of vector gradU(M0). That is why we actually need to find vector

( ) ( ) ( ) ( )

MgradU

U

=

x

¶

U

iM

+

y

¶

Here are the relevant calculations:

U

( ) ( )

22

+-=

yzzyxyzM

¶

U

( ) ( )

¶

y

U

( ) ( )

¶

0

22

+-=

22

+-=

xzxyzzxM

+-=

xyzxyyxM

848

kjiMgradU

U

M

0

¶

U

M

0

¶

y

U

M

0

¶

.1,1,1

0

, which is equal to the

0

,max

MgradUM

00

U

jM

+

z

¶

.

kM

0000

,

85710,5710

=+-=

45145,5145

-=+-=

,

81075,1075

=+-=

¶

gradU

¶

U

( ) ( ) ( ) ( )

=

M

U

¶

l

MgradUM

000

2

12848max

=+-+==

158

б) Find out, whether the vector field

()()()(

)

rrr

r

(

)

r

(

)

r

rrr

¶

()()()(

)

r

¶

(

)

r

(

)

r

(

)

r

¶

()()()(

)

rrrrrrr

r

(

)

r

is conservative.

kzxjyziyxMa

++-++= 2

Solution. Vector field

the domain of field

( )

rr

[ ]

aMarot

=Ñ=

¶

-Ma

is conservative, is at every point М of

.0

=Marot

Let’s find

kji

æ

R

¶

ç

=

ç

zyx

¶

è

RQP

ö

Q

¶

-

z

y

¶

P

¶

æ

÷

i

+

ç

÷

z

¶

è

ø

rr

R

¶

ö

j

-

+

÷

x

¶

ø

æ

Q

¶

ç

¶

è

ö

P

¶

÷

-

÷

y

x

¶

ø

In this formula, to simplify memorizing the method of calculating the rotor we used the formal Hamilton's operator "nabla":

k

i

=Ñ

x

¶

j

+

y

¶

,

z

¶

according to the rule of finding the vector product in rectangular Cartesian coordinates.

For other field types, explored in exercise 8, here are their definitions:

Solenoid field

zyxRzyxQzyxPa ,,,,,,,,

at every point М of

domain V has to meet the condition:

r

( ) ( ) ( ) ( )

= M

Madiv

¶

Harmonic field

Ma

is both conservative and solenoid at point of

domain V at the same time, so

In our case

P

¶

y

P

=

¶

z

So

fore, field

Ma

is not conservative.

M

Q

+

¶

y

=Marot

Q

=

¶

x

P

x

M

0

+

and

Q

¶

z

R

¶

z

=

0=

.

0=Madiv

.

¹---=-+-+-= kjikjiMarot

R

¶

y

,02102010

.0,2,1,0,0,1 =

there-

R

=

¶

x

Example 2: Find the derivative of the scalar field

2222

at point

)4;1;3(M

at the direction

)1ln(),,( zxyxzyxU +-++=

r

,, k

a. Of the vector kjis 232 --= .

159

b. Of the normal to the surface σ:

¶

222

ing an acute angle with the positive direction of the axis Oz.

c. Of the perpendicular to the level surfaces of the function

),,( zyxU

, going through point

)1,0,0(1M

.

Solution:

We can find the derivative of the direction via formula:

U

¶

l

U

x

¶

U

=

y

¶

U

z

¶

a) Find

2

=

1

2yxy

1

++

-=

0

Bacause 17)2()3(2||

¶

Thus

U

¶

l

M

b) Let’s find

0

=×=

lgradU

x

-

22

yx

++

,

22

z

,

22

zx

+

2

3

55

17

0

U

¶

x

22

zx

+

2

ì

0

l

í

17

î

2

æ

-×+×-=

ç

11

è

U

+×

¶

y

¶

,

3

17

U

¶

x

¶

U

¶

y

¶

U

¶

z

kjil ×+×+×=

gba

coscoscos

222

=-+-+=s , we get

3

,

17

4

æ

ö

-×-

ç

÷

5

è

ø

M

, i.e.

--=

2

17

coscoscos

U

coscoscos)(

×

+×

¶

z

3

6

11

2

=

11

4

-=

5

2

ü ý

17

þ

ö ÷

ø

kjil ×+×+×=

gba

l =

.

, т.е.

5

52

1755

-=-=

s

||0s

==

l =

where N is the vector of a normal to the surface σ.

Then

ì

¶

F

=

N ,,

í

¶

x

î

222

ü

¶

F

¶

y

,

ý

¶

z

þ

0496),,(

=-+++= zzyxxzyxF

55

,

zzyxx 496

=+++

, form-

gba

3

1752

.

935

N

,

||0N

160

¶

{

}

>

g

F

¶

x

¶

F

¶

y

¶

F

¶

z

x

M

y

M

z

We get

Notice that if

12)62(

=+=

M

1818

==

4)42(

=-=

M

6

4,18,12=N

,

0cos

, then the found vector forms an acute angle

222

2241812||

=++=N

ì

0

,

=

l

í

11

î

with the axis Oz, therefore, the requirement of the task is met.

Thus,

16

11

254

-=×-×+×-=

.

605

9

2

6

55

3

11

¶

U

¶

l

M

c) Find the surface of level of function ),,( zyxU , going through point M1:

110)001ln(),,( -=+-++== CzyxU

.

2222

,

+

2222

=++-++= zxyxzyxF

2

ö

æ

÷

ç

11

ø

è

.

1)1ln(

-=+-++ zxyx

l =

01)1ln(),,(

¶

F

¶

z

M

222

4

ö

æ ç

è

=

-+

÷

5

ø

We get surface σ1:

Similar to the previous paragraph we find the vector

is a vector, perpendicular to surface of a level σ1

N

3

¶

F

¶

x

N

0

l

6

11

M

3

ì

,

í

55

î

3

ì í

2045

î

11

-=-=

55

5

4

2

ü

--=

,

ý

5

þ

,

10

,

2045

¶

F

¶

,

y

||

44

--=

,

2045

2

=

11

M

3

ö

æ

-=N

÷

ç

55

ø

è

ü ý þ

,

N

9

11

||0N

-=

2045

55

2

ü

,

ý

11

þ

, where

4

5

,

.