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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