Chapter I Matrices, Vectors, & Vector Calculus

1-1, 1-9, 1-10, 1-11, 1-17, 1-18, 1-25, 1-27, 1-36, 1-37, 1-41.

Concept of a Scalar

Consider the array of particles shown in the figure.

The mass of the particle at

(x,y) can be expressed as.

Thus the mass of 4-grams can be expressed as gramM 4)3,2(

The mass is unchanged if the axes is transformed, i.e.,

gramMM 4)5.3,4()3,2(

Or in general we write )1.1(),(),( yxMyxM

The quantities that are invariant under coordinate transformation are called

scalars, otherwise they are called vectors.

),( yxM

Coordinate Transformation

)2.1(ˆˆˆ kAjAiAA zyx

A vector A can be expressed relative to the triad ijk as.

Relative to a new triad i`j`k` having a different orientation from that of the

triad ijk , the same vector can be expressed as

)3.1(ˆˆˆ kAjAiAA zyx

It is clear now that we can write

)4.1(

ˆˆˆˆˆˆˆ

ˆˆˆˆˆˆˆ

ˆˆˆˆˆˆˆ

zyxz

zyxy

zyxx

AkkAkjAkikAA

AjkAjjAjijAA

AikAijAiiiAA

In matrix notation, Eq.(1.4) can be expressed as

)5.1(ˆˆˆˆˆˆ

ˆˆˆˆˆˆ

ˆˆˆˆˆˆ

z

y

x

z

y

x

A

A

A

kkkjki

jkjjji

ikijii

A

A

A

The 3-by- 3 matrix in Eq.(1.5) is called the transformation matrix, or the rotation

matrix. In summation notation we write

3,2,13

1

iAAj

jiji

ij is called the direction cosines of the x`i-axis relative to the xj-axis.

Similarly, the unprimed components can be expressed as

zyxz

zyxy

zyxx

AkkAkjAkikAA

AjkAijAjijAA

AikAijAiiiAA

ˆˆˆˆˆˆˆ

ˆˆˆˆˆˆ

ˆˆˆˆˆˆˆ

Or in matrix form we can write

)6.1(ˆˆˆ

ˆˆˆ

ˆˆˆˆˆˆ

z

y

x

z

y

x

A

A

A

kkkjki

jkjjji

ikijii

A

A

A

Comparing the transformation matrices of equation 1.5 and 1.6 we note that the

columns and rows are interchanged.

Example 1.1 A point P is represented in the (x1, x2, x3) system by P(2,1,3). The same point is represented as

P(x`1, x`2, x`3) where x2 has been rotated toward x3 around x1-axis by 30o. Find the rotation matrix and

determine P(x`1, x`2, x`3)

Soution Form the figure, the rotation

matrix is given by

866.05.005.0866.00

001

30cos120cos90cos60cos30cos90cos90cos90cos0cos

ˆˆˆˆˆˆ

ˆˆˆˆˆˆ

ˆˆˆˆˆˆ

kkkjki

jkjjji

ikijii

Using Eq.(1.5) we find

1.2866.05.0

37.25.0866.0

2

zyz

zyy

xx

AAA

AAA

AA

Note that the length of the position vector is invariant under transformation, i.e.,

74.3232

22

123

22

21 xxxxxxr

Elementary Scalar & Vector Operations

The Scalar Product:

The scalar product of two vectors is a scalar defined as

)7.1(,cos BAABBABAi

ii

The scalar product obey the commutative and the distributive laws as

)8.1(ABBA

)9.1(CABACBA

The Unit Vectors:

They are vectors having a magnitude of unity and directed along a specific

coordinate axis. The unit vector along the radial direction is given by

)9.1(R

ReR

If any two unit vectors are orthogonal

)10.1(ijji ee

The Vector Product:

The vector product of two vectors is a third vector defined as

)11.1(BAC

The component of the vector C is given by

)12.1(,

kjkjijki BAC

The symbol ijk is called the permutation symbol (Levi-Civita density) with

)13.1(

npermutatio oddan formindecestheif1-

npermutatioeven an formindecestheif1

indexotheranytoequalisindexanyif0

ijk

Using Eq.(1.12), the components of C can be evaluated as

331332313213131

321232212212121

3111321112111111

BABABABABABA

BABABAC

The only nonvanishing terms are 123 & 132 , i.e.,

233223132321231 BABABABAC

31132 BABAC

Similarly, we find

12213 BABAC

The magnitude of the vector C is define also as

)14.1(,sin BABABAC

Geometrically, C is the area of the parallelogram defined by the two

vectors A&B

ecommutativnotABBA

ACBBACCBA

B

A

C

Acos

volume of parallelepiped

rulecabbac BACCABCBA

zyx

zyx

BBB

AAABAkji

Gradient Operator (Directional Derivative)

Let be a scalar function of 3-variables, i.e., ),,( zyx

dzz

dyy

dxx

d

This tells us how varies as we go a small distance (dx, dy, dz) away from the point (x, y, z). Let us rewrite the above equation as

kdzjdyidxkz

jy

ix

d ˆˆˆˆˆˆ

sddor

kz

jy

ix

with ˆˆˆ

is called the gradient of the function .

The symbol is called the del operator. It can be written

as k

zj

yi

xˆˆˆ

The gradient operator can operate on a scalar function, can be used in a scalar

product with a vector function (divergence), or can be used in a vector product

with a vector function (curl), i.e.,

)15.1(

ii

i

ex

)16.1(

ii

i

ex

grade

)17.1(

i i

i

x

AAAdiv

)18.1(iijk j

kijk e

x

AAAcurl

The gradient points in the direction of maximum increase of the function .

The magnitude gives the slope along this maximal direction.

is directed perpendicular to the surface =constant.

The successive operation of the gradient produces the Laplacian operator

)18.1(2

22

i ix

The Laplacian of a scalar function is written as

)19.1(2

22

i ix

Integral Calculus

Line Integral The line integral is expressed as b

aldA

where is a vector function and is an infinitesimal displacement vector

along a path from point a to point b. A

ld

If the path forms a closed loop, a circle is put on the integral sign, i.e., ldA

If the line integral is independent on the path followed, the vector is called

conservative A

Surface Integral The surface integral is expressed as

S SdA

where is a vector function and is an infinitesimal element of area. A

Sd

Again if the surface is closed we put a circle on the integral sign, that is

SdA

The direction of is perpendicular to the surface an directed outward for

closed surfaces and arbitrary for open surfaces. Sd

Volume Integral The volume integral is expressed as

V dT

where T is a vector function and d is an infinitesimal element of volume.

The Divergence Theorem (The Guass’s Theorem)

It states that the integral of a divergence over a volume is equal to the value of

the function at the boundary.

)20.1( Sv SdAdA

Stokes' Theorem

)21.1( ldASdAS

Since the boundary line for any closed surface shrink down to a point, then

)22.1(0 SdAS

Curvilinear Coordinates

Spherical coordinates (r, ,)

r: is the distance from the origin (from 0 to ∞)

: the polar angle, is the angle between r and the z-axis (from 0 to )

: the azimuthal angle is the angle between the projection of r to the x-y plane and the x-axis (from 0 to 2)

The relation between the Cartesian coordinates and

the spherical coordinates can be written as

)23.1(cos,sinsin,cossin rzryrx

The unit vectors associated with the spherical coordinates are related to the

corresponding unit vectors in the Cartesian coordinates as

)24.1(ˆcosˆsinsinˆcossinˆ kjier

)25.1(ˆsinˆsincosˆcoscosˆ kjie

)26.1(ˆcosˆsinˆ jie

The spherical coordinates can be considered as a result of two rotations: first

the rotation of the x-y axis an angle about the z-axis and then the x'-z‘ about

the y‘-axis an angle with x=, y=, and z=r. The rotational matrix of such

two rotations is

cossinsincossin0cossin

sinsincoscoscos

1000cossin0sincos

cos0sin010

sin0cos

Now the unit vectors associated with the spherical coordinates can be found

using Eq. 1.5 as

Which are identical to Equations 1.24-1.26

k

ji

e

e

e

rˆ

ˆˆ

cossinsincossin0cossin

sinsincoscoscos

ˆ

ˆ

ˆ

Equation 1.23 can be found using Equation 1.6 as

And the Cartesian unit vectors can be found in terms of their spherical unit

vectors as

rz

yx

00

cos0sinsinsincossincoscossinsincoscos

re

e

e

k

ji

ˆ

ˆ

ˆ

cos0sinsinsincossincoscossinsincoscos

ˆ

ˆˆ

eeei r ˆsinˆcoscosˆcossinˆ

eeej r ˆcosˆsincosˆsinsinˆ

eek r ˆsinˆcosˆ

The infinitesimal displacement vector in spherical coordinates is expressed as

)27.1(ˆsinˆˆ drrdrdrld

The volume element is expressed as

)28.1(sin2 drddrdldldld r

For the surface elements we have

constantisˆˆ1 drdrdldlSd r

constantisˆsinˆ2 drdrdldlSd r

constantˆsinˆ2

3 isrrddrrdldlSd

To find the volume of a sphere of radius R we have

3342

02

00sin RdddrrdV

R

To find the gradient in spherical coordinates let T=T (r, , ) so

)29.1(sin drTrdTdrTldTdT r

)30.1(

dT

dT

drr

TdT

Equating the above two equations we get

rr

TT r ˆ

ˆ1

r

T

rT

,

ˆsin

1

r

T

rT

ˆ

sin

1ˆ1ˆorr

T

rr

T

rr

r

TT

)31.1(ˆsin

1ˆ1ˆ

rrrr

rr

Similarly, one can find the divergence and the curl in spherical coordinates

)32.1(sin

1sin

sin

11 22

A

rA

rAr

rrA r

)33.1(

sin

ˆsinˆˆ

sin

12

ArrAA

r

rrr

rA

r

)34.1(sin

1sin

sin

112

2

222

2

2

2

T

r

T

rr

Tr

rrT

The Laplacian is defined as

Cylindrical coordinates (, , z)

: is the distance from the z-axis (from 0 to ∞)

: the azimuthal angle is the angle between and the x-axis (from 0 to 2)

z: the distance from the x-y plane (from -∞ to ∞)

The relation between the Cartesian coordinates

and the cylindrical coordinates can be written as

)35.1(,sin,cos zzyx

The unit vectors associated with the cylindrical coordinates are related to the

corresponding unit vectors in the Cartesian coordinates as

)36.1(ˆsinˆcosˆ ji

)37.1(ˆcosˆsinˆ ji

)38.1(ˆˆ kz

The cylindrical coordinates can be considered as a result a rotation of the x-y

axis an angle about the z-axis x'=, y'=, and z'=z. The rotational matrix of

such a rotation is

1000cossin0sincos

Now the unit vectors associated with the cylindrical coordinates are

Applying Eq.1.5 we recover Equations 1.36-1.38

k

ji

e

e

e

zˆ

ˆˆ

1000cossin0sincos

ˆ

ˆ

ˆ

Applying Eq.1.6, the Cartesian unit vectors can be found in terms of the

cylindrical unit vectors as

ze

e

e

k

ji

ˆ

ˆ

ˆ

1000cossin0sincos

ˆ

ˆˆ

The infinitesimal displacement vector in cylindrical coordinates is expressed as

)39.1(ˆˆˆ zdzddld

The volume element is expressed as

)40.1(dzdddldldld z

For the surface elements we have

constantiszˆˆ1 zddzdldlSd

constantisˆˆ2 dzddldlSd zr

constantisˆˆ3 dzddldlSd z

To recover Equation 1.35 we again apply Eq.1.6 as

zz

yx

01000cossin0sincos

)41.1(ˆˆ1

ˆ zz

The Del

The Divergence )42.1(11z

AAAA z

The Curl )43.1(

ˆsinˆˆ

1

zAAAz

A

)44.1(11

2

2

2

2

2

2

z

TTTT

The Laplacian

Position, Velocity, and Acceleration vectors:

In Cartesian coordinates, the position vector is written as

321 ˆˆˆ ezeyexr

While , the velocity, and the acceleration vectors are

ii

iexedt

dze

dt

dye

dt

dxr

dt

rdv ˆˆˆˆ 321

ii

iexrdt

rda ˆ

2

2

Note that the unit vectors in Cartesian coordinates are constant in time (both in

magnitude and direction)

In spherical coordinates, the position vector is written as

rerr ˆ

While , the velocity vector is

dt

edrerer

dt

dv rrr

ˆˆˆ

kjieBut rˆcosˆsinsinˆcossinˆ

kj

idt

ed r

ˆsinˆcossinsincos

ˆsinsincoscosˆ

cosˆsinsinˆsinˆsincosˆcoscosˆ ikjidt

ed r

eedt

ed r ˆsinˆˆ

)45.1(ˆsinˆˆ erererv r

and the acceleration vector is

dt

edrererer

dt

edrerer

dt

edrer

dt

vda rr

ˆsinˆsinˆcosˆsin

ˆˆˆ

ˆˆ

ee

dt

edbut r ˆcosˆ

ˆ

eedt

edand r ˆcosˆsin

ˆ

)46.1(ˆcos2sin2sin

ˆcossin2ˆsin2222

errr

errrrerrra r

In cylindrical coordinates, the position vector is written as

zezer ˆˆ

While , the velocity vector is

zr ezdt

ederv ˆ

ˆˆ

ejidt

edbut ˆˆcosˆsin

ˆ

)47.1(ˆˆˆ zr ezeerv

and the acceleration vector is

zezdt

edee

dt

ede

dt

vda ˆ

ˆˆˆ

ˆˆ

rejidt

edˆˆsinˆcos

ˆusing

)48.1(ˆˆ2ˆ2 zezeea

Second Order Linear Differential Equations:

A second order linear differential equation is any equation of the form

)49.1(xfbyyay

If f(x)= 0, Eq.(1.49) is called homogeneous differential equation, with the form

)50.1(0 byyay

Eq.(1.50) has the following important properties:

(1) If y1(x) is a solution of Eq.(1.50), then c1y1(x) is also a solution

(2) If y1(x) and y2(x) are solutions, then y1(x)+ y2(x) is also a solution

(3) If y1(x) and y2(x) are linearly independent solutions, then the general

solution of Eq.(1.50) is c1y1(x)+ c2y2(x).

The functions y1(x) and y2(x) are linearly independent if and only if the equation

021 xyxy

Is satisfied only by ==0

Eq.(1.50) has a solution of the form

)51.1(rxey

Substituting Eq.( 1.51) into Eq.(1.50) we obtain an algebraic equation called the

auxiliary equation:

)52.1(02 barr

)53.1(42

221 ba

arwith

If the two roots are not identical, the general solution of Eq.(1.50) is

If the two roots are identical, the general solution of Eq.(1.50) is

)54.1(21 21xrxr

ececy

)55.1(21rxrx xececy

If 4b > a2, then the roots are imaginary and can be written as

)56.1( ir

)57.1(4,2

221 aband

awith

xixixxrxr ececeececythen 2121 21

xccixcceyor x sincos 2121

)58.1(sincos xBxAeyor x

Eq.(1.58) can be written in a more convenient form as

)59.1(sin xAey x

)60.1(cos xAeyor x

Note that Eq.(1.59) is written

xAxAexxAey xx coscossincossincoscossin

Which is the same as Eq.(1.58). The same thing can be said about Eq.(1.60)

Example Solve the equation

Solution The auxiliary equation is

042 yyy

0422 rr 31 ir

3,1, andHence

The general solution is, from Eq.(58) or Eq.(1.59), respectively

xBxAey x 3sin3cos

xAeyor x 3sin

In Eq.(1.49) if f(x)≠ 0, it is called nonhomogeneous differential equation.

Theorem The general solution of the second order nonhomogeneous linear

Equation (Eq.149) can be expressed in the form

)61.1(pc yyy

With yc, called the complementary solution, is the general solution of the

corresponding homogeneous differential equation.

and yp, called the particular solution, is the solution of the inhomogeneous

differential equation.

The complementary solution, yc, is already known. One of the main two

methods to find yp, is the method of Undetermined coefficients.

The complementary solution, yc, is already known. One of the main two

methods to find yp, is the method of Undetermined coefficients.

The method is simply relies on guessing a solution depending on the type of

f(x) by multiplying the derivative of f(x) by some coefficients to be determined by substituting back into the main equation.

(I) If f(x) is an exponential function, then

)62.1(xAfy p

(II) If f(x) is a sine or cosine, then

)63.1()(cos)(sin xuBxuAy p

(III) If f(x) is a polynomial function of degree n, then

)64.1(11

21 onn

p AxAxAxAy

(IV) If f(x) is a product of two or more simple functions, then then our basic choice (before multiplying by x, if necessary) should be a product consists of the corresponding choices of the individual components of f(x).

Note: whenever your initial choice yp has any term in common with the

complementary solution, then you must alter it by multiplying your initial choice

of yp by x, as many times as necessary.

Example Solve the equation

Solution The auxiliary equation is

xxyyy 265 2

0230652 rrrr

The complementary solution is xx

c ececy3

22

1

The particular solution is predicted to be op AxAxAy 12

2

122 AxAyNow p 22Ayand p

Substituting back into the differential equation we get

xxAxAxAAxAA o 26252 2122122 xxAAAxAAxAor o 26526106

21212

22

Equating coefficients of like powers of x we get

0652

2610

16

12

12

2

oAAA

AA

A

108

11,

18

1,

6

112 oAAA

The particular solution is therefor

108

11

18

1

6

1 2 xxyp

The general solution is then

108

11

18

1

6

1 232

21

xxececyyy xxpc

Example Solve the equation

Solution The auxiliary equation is

xxyy cos34

2,02042 irr

The complementary solution is

xcxcxcxcey xc 2sin2cossincos 2121

The particular solution is predicted to be

xBxBxAxAy oop 2sin2cos 11

Substituting back into the original differential equation we get

0sin3cos13cos32sin23 110110 xxBxxAxABxAB

The coefficient of each term must vanish so we have

0,01,032,023 110110 BAABAB

0,3

2,1 11 BABA oo

The general solution is then

xxxxcxcyyy pc sin3

2cos2cos2sin 21