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Physics Department, Yarmouk University, Irbid Jordan
Phys. 201 Methods of Theoretical Physics 1
Phys. 201 Methods of Theoretical Physics 1
Supplements
Dr. Nidal M. Ershaidat
Table of Contents
Vector Space ............................................................................................................ 1
Vector Identities...................................................................................................... 2
The Curl ................................................................................................................... 3
Complex Numbers ................................................................................................. 7
Complex Logarithms ........................................................................................... 10
Determinants......................................................................................................... 11
Ordinary Differential Equations (ODE’s) ......................................................... 16
ODE – An Example .............................................................................................. 19
Exact Differential .................................................................................................. 20
Method of Undetermined Coefficient or Guessing Method.......................... 21
Method of Variation of Parameters ................................................................... 24
Fourier Series......................................................................................................... 27
Curvilinear Coordinates...................................................................................... 35
Physics Department, Yarmouk University, Irbid Jordan
Phys. 201 Methods of Theoretical Physics 1
Dr. Nidal M. Ershaidat Doc. 1
1
Vector Space
A. The Concept
One of the fundamental concepts of linear algebra is the concept of vector
space. For example, many function sets studied in mathematical analysis are
with respect to their algebraic properties vector spaces. In analysis the
notion “linear space” is used instead of the notion “vector space”.
B. Definition
A set X is called a vector space over the number field KKKK if to every pair (x,y)
of elements of X there corresponds a sum x + y ∈ X, and to every pair (α,x)
where α ∈ KKKK and x ∈ X, there corresponds an element αx ∈ X, with the
properties 1-8:
1) x + y = y + x (commutability of addition);
2) x + (y + z) = (x + y) + z (associativity of addition);
3) ∃ 0 ∈ X such that 0 + x = x ∈ X (existence of null element);
4) ∀ x ∈ X ⇒∃ - x ∈ X: x + (-x) = 0 (existence of the inverse element);
5) 1 . x = x (unitarism);
6) α (βx) = (αβ) x (associativity with respect to number multiplication);
7) α (x + y) = α x + α y (distributivity with respect to vector addition);
8) (α + β ) x = α x + β x (distributivity with respect to number addition).
The properties 1-8 are called the vector space axioms. Axioms 1-4 show that
X is a commutative group or an Abelian group with respect to vector
addition. The second correspondence is called multiplication of the vector by
a number, and it satisfies axioms 5-8. Elements of a vector space are
called vectors. If KKKK = RRRR, then one speaks of a real vector space, and if KKKK =
CCCC, then of a complex vector space. Instead of the notion “vector space” we
shall use the abbreviative “space”.
Physics Department, Yarmouk University,
Irbid Jordan
Phys. 201 Mathematical Physics 1
Dr. Nidal M. Ershaidat Doc. A1
2
Vector Identities
In the following φ and ψ are scalar functions, →→→→A and
→→→→B are vectors.
→→→→→→→→→→→→→→→→→→→→→→→→→→→→
⋅⋅⋅⋅∇∇∇∇++++⋅⋅⋅⋅∇∇∇∇====
++++⋅⋅⋅⋅∇∇∇∇ BABA (1)
→→→→→→→→→→→→→→→→→→→→→→→→→→→→
××××∇∇∇∇++++××××∇∇∇∇====
++++××××∇∇∇∇ BABA (2)
××××∇∇∇∇φφφφ++++××××φφφφ∇∇∇∇====
φφφφ××××∇∇∇∇
→→→→→→→→→→→→→→→→→→→→→→→→AAA (3)
××××∇∇∇∇⋅⋅⋅⋅−−−−
××××∇∇∇∇⋅⋅⋅⋅====
××××⋅⋅⋅⋅∇∇∇∇
→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→BAABBA (4)
⋅⋅⋅⋅∇∇∇∇−−−−
⋅⋅⋅⋅∇∇∇∇++++
∇∇∇∇⋅⋅⋅⋅−−−−
∇∇∇∇⋅⋅⋅⋅====
××××××××∇∇∇∇
→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→ABBABAABBA (5)
0====
××××∇∇∇∇⋅⋅⋅⋅∇∇∇∇
→→→→→→→→→→→→A (6)
φφφφ∆∆∆∆≡≡≡≡φφφφ∇∇∇∇====
φφφφ∇∇∇∇⋅⋅⋅⋅∇∇∇∇
→→→→→→→→ 2 (Laplacian) (7)
0====
φφφφ∇∇∇∇××××∇∇∇∇
→→→→→→→→ (8)
→→→→→→→→→→→→→→→→→→→→→→→→→→→→
∇∇∇∇−−−−
⋅⋅⋅⋅∇∇∇∇∇∇∇∇====
××××∇∇∇∇××××∇∇∇∇ AAA
2 (9)
Physics Department, Yarmouk University, Irbid Jordan
Phys. 201 Methods of Theoretical Physics 1 Dr. Nidal M. Ershaidat Doc. 3
3
The Curl
A) Geometrical significance of the Curl
The divergence and curl of a vector field are two vector operators whose
basic properties can be understood geometrically by viewing a vector field as
the flow of a fluid or gas. Here we give an overview of basic properties of
curl than can be intuited from fluid flow. The curl of a vector field captures
the idea of how a fluid may rotate. Imagine that the below vector field F�
represents fluid flow (See Fig. 1). The vector field indicates that the fluid is
circulating around a central axis.
B) Measuring the Curl
In order to measure the density of some matter at a point, we measure the
mass (dm) of a small volume (dV) around the point, and then divide by the
volume. Mathematically, this can be expressed by:
dV
dm=ρ (1)
The smaller the volume, the better the approximation. Actually we define
the density ρ as being the limit:
dV
dm
dV 0lim
→=ρ (2)
A similar procedure is used to measure the strength of the rotation of a fluid.
If the vector field is interpreted as velocity of fluid flow, the fluid appears to
flow in circles.
This macroscopic circulation of fluid around circles actually is not what curl
measures. But, it turns out that this vector field also has curl, which we
might think of as “microscopic circulation”.
To test for curl, imagine that you immerse a small sphere into the fluid flow,
and you fix the center of the sphere at some point so that the sphere cannot
follow the fluid around (See Fig. 2).
4
(a) Time t (b) Time t’
Figure 1: The “rotating” vector field F�
at two different times.
Although you fix the center of the sphere, you allow the sphere to rotate in
any direction around its center point. The rotation of the sphere measures
the curl of the vector field F�
at the point in the center of the sphere. (The
sphere should actually be really really small, because, remember, the curl is
microscopic circulation.)
(a) Time t (b) Time t’
Figure 2: A small sphere immersed into the fluid flow.
The vector field F�
determines both in what direction the sphere rotates, and
the speed at which it rotates. We define the curl of F�
, denoted curl F�
, by a
vector that points along the axis of the rotation and whose length
corresponds to the speed of the rotation.
We can draw the vector corresponding to curl F�
as follows. We make the
length of the vector curl F�
proportional to the speed of the sphere's
rotation. The direction of curl F�
points along the axis of rotation, but we
need to specify in which direction along this axis the vector should point. We
will (arbitrarily?) set the direction of the curl vector by using the following
5
“right hand rule.” To see where curl F�
should point, curl the fingers of your
right hand in the direction the sphere is rotating; your thumb will point in
the direction of curl F�
. For our example, curl F�
is shown by the green
arrow.
Figure 3: The direction of curl F
�
is conventionally chosen using a right-hand rule.
The curl is a three-dimensional vector, and each of its three components
turns out to be a combination of derivatives of the vector field F�
. Once you
have the formula, calculating the curl of a vector field is a simple matter.
The curl is sometimes called the rotation, or "rot".
C) The Curl in different system of coordinates
The curl of a vector function is the vector product of the del operator with
this vector function:
The Curl in Cartesian coordinates
ky
F
x
Fj
x
F
z
Fi
z
F
y
FF xyzxyz ˆˆˆ
∂∂
−∂
∂+
∂∂
−∂
∂+
∂
∂−
∂∂
=×∇→→
(3)
where kji ˆ,ˆ,ˆ are unit vectors in the x, y, z directions. It can also be
expressed in determinant form:
zyx FFF
zyx
kji
∂∂
∂∂
∂∂
ˆˆˆ
(4)
The Curl in cylindrical polar coordinates
6
The curl in cylindrical polar coordinates, expressed in determinant form is:
zθr
θr
FFrF
zθr
r
k
r
∂∂
∂∂
∂∂
ˆ1
1
(5)
The Curl in spherical polar coordinates
The curl in spherical polar coordinates, expressed in determinant form is:
φθr FθrFrF
φθr
rθrθr
F
sin
1
sin
1
sin
12
∂∂
∂∂
∂∂
=×∇→→
(6)
Reference:
- http://hyperphysics.phy-astr.gsu.edu/hbase/curl.html
- (See the applets in) http://mathinsight.org/curl_idea
Physics Department, Yarmouk University, Irbid Jordan
Phys. 201 Methods for Theoretical Physics 1
Dr. Nidal M. Ershaidat Doc. 4
7
Complex Numbers
A. Definition of a complex number
A complex number z is defined by z = x + i y
where x and y are real numbers and 1−−−−====i .
x is called the real part of z and denoted x = Re(z)
y is called the imaginary part of z and denoted y = Im(z)
The form z = x + i y is called the rectangular form of the complex number z.
Note: In this document the letter z refers to a complex number and any
other letter, in particular, x and y, refer to real numbers or variables.
B. The Complex (Numbers) Space
Complex numbers define a space called the space of complex numbers or
simply the complex space. The eight axioms necessary to define a space in
linear analysis are verified, as follows:
1. Commutability of addition
For any two complex numbers z1 and z2 we define
z1 + z2 = z2 + z1 (∀∀∀∀ z1, z2)
2. Associativity of addition
For any set of complex numbers z1, z2 and z3 we define
z1 + (z2 + z3) = (z1 + z2) + z3
3. Existence of null element:
∃∃∃∃ 0 ∈∈∈∈ C such that 0 + z = z ∈ C
4. Existence of the inverse element:
∀ z ∈ C ⇒ ∃ - z ∈ C such that z + (-z) = 0
5. Unitarism:
∀ z ∈ C ⇒ 1 . z = z
And for (α, β and z ∈ C) we have:
6. Associativity with respect to number multiplication
α (β z) = (α β) z
7. Distributivity with respect to complex number addition
α (z1 + z2) = α z1 + α z2
8. Distributivity with respect to number addition
(α + β ) z = α z + β z
8
C. Complex conjugate of a complex number
The complex conjugate of a complex number z (z = x + i y) is defined by
z* = x - i y
D. Basic Operations in the Complex Space
Consider the complex numbers z1 = x1 + i y1 and z2 = x2 + i y2
- Addition
z = z1 + z2 = (x1 + x2) + i (y1 + y2) = z2 + z1
- Subtraction
z = z1 - z2 = (x1 - x2) + i (y1 - y2)
z' = z2 – z1 = (x2 – x1) + i (y2 – y1) # z1 - z2
-Multiplication
z = z1 . z2 = (x1 + i y1) . (x2 + i y2) = x1x2 + i x1 y1+ i x2 y2 - y1y2
-Division
22
11
yix
yix
2
1
z
zz
++++++++
========
(((( ))))22
22
21211221
22
22
22
11 .yx
yyyxiyxixx
yix
yix
yix
yix
++++
++++−−−−++++====
−−−−
−−−−
++++
++++====
22
22
2112
22
22
2121
yx
yxiyxi
yx
yyxx
++++
−−−−++++
++++
++++++++====
E. Graphical Representation of complex numbers
A complex number z = x + i y is represented in a 2D plane, called the Argand
Plane, by a point P whose coordinates are (x and y) (Fig. 1).
Figure 1: Graphical Representation of z
9
The vector →
OP represents the complex number z.
Fig. 1 representing z is the so-called Argand diagram of z.
The complex conjugate z* is represented by the vector →
*OP in Fig. 2.
Figure 2: Graphical Representation of z*
From Fig. 1 one can see that:
22 yxr +=
x
y1tan−=θ
r and θθθθ are called the polar coordinates of P and alternatively we have:
x = r cos θθθθ
y = r sin θθθθ
z can be written in the form:
z = r (cos θθθθ + i sin θθθθ)
Using Maclaurin’s series for eiθ, cosθ and sin θ we can write:
eiθ
= cos θ + i sin θ
z can be rewritten in the form:
z = r eiθ
Which we call the polar format of z. This form is very useful when defining
and using functions of complex variables.
F. Modulus (Magnitude) and Argument of z
r represents the magnitude or modulus of z and θ is called the argument of z.
And we write:
( ) 22 yxzModr +==
( )x
yz 1tanarg −==θ
© Nidal M. Ershaidat 2013
10
Complex Logarithms
The exponential function can be extended to a function which gives a
complex number as ex for any arbitrary complex number x; simply use the
infinite series with x complex. This exponential function can be inverted to
form a complex logarithm that exhibits most of the properties of the
ordinary logarithm. There are two difficulties involved: no x has ex = 0; and it
turns out that e2ππππi = 1 = e
0. Since the multiplicative property still works for the
complex exponential function, inzzee
ππππ++++==== 2 , for all complex z and integral n.
So the logarithm cannot be defined for the whole complex plane, and even
then it is multi-valued – any complex logarithm can be changed into an
"equivalent" logarithm by adding 2πi at will. The complex logarithm can only
be single-valued on the cut plane. For example, ln i = ½ πi or 5/2 π i or −3/2 π i,
etc.; and although i4 = 1, 4 log i can be defined as 2π i, or 10π i or −6 π i, and so
on.
z = Re(ln(x+iy)) z = |Im(ln(x+iy))| z = |ln(x+iy)|
Plots of the natural logarithm function on the complex plane (principal
branch)
Physics Department, Yarmouk University, Irbid Jordan
Phys. 201 Methods for Theoretical Physics 1
Dr. Nidal M. Ershaidat Doc. 5
11
Determinants
Consider the square matrix of order 2:
=
dcba A
The matrix A is invertible if and only if 0#cbda − . This number is called the
determinant of A. It is clear from this, that we would like to have a similar
result for bigger matrices (meaning higher orders). So is there a similar
notion of determinant for any square matrix, which determines whether a
square matrix is invertible or not?
In order to generalize such notion to higher orders, we will need to study the
determinant and see what kind of properties it satisfies. First let us use the
following notation for the determinant:
Determinant of cbdadcba
dcba
dcba −==
=
det
Properties of the Determinant
1. Any matrix A and its transpose have the same determinant, meaning
det A = det AT
This is interesting since it implies that whenever we use rows, a similar
behavior will result if we use columns. In particular we will see how row
elementary operations are helpful in finding the determinant. Therefore, we
have similar conclusions for elementary column operations.
2. The determinant of a triangular matrix is the product of the entries on the
diagonal, that is
dadc
a dba == 0
0.
3. If we interchange two rows, the determinant of the new matrix is the
opposite of the old one, that is
badc
dcba −= .
4. If we multiply one row with a constant, the determinant of the new matrix
is the determinant of the old one multiplied by the constant, that is
12
dλcλba
dcbaλ
dc
bλaλ == .
In particular, if all the entries in one row are zero, then the determinant is
zero.
5. If we add one row to another one multiplied by a constant, the
determinant of the new matrix is the same as the old one, that is
bλdaλcba
dcba
dc
dλbcλa++==++
.
Note that whenever you want to replace a row by something (through
elementary operations), do not multiply the row itself by a constant.
Otherwise, you will easily make errors (due to Property 4).
6. We have
det (AB) = det(A) det (B)
In particular, if A is invertible (which happens if and only if det (A) #0), then
( ) ( )AA
det1det 1- =
If A and B are similar, then det (A) = det(B).
Let us look at an example, to see how these properties work.
Example 1. Evaluate 3112
− .
Let us transform this matrix into a triangular one through elementary
operations. We will keep the first row and add to the second one the first
multiplied by ½. We get
2
70
12
3112 =− .
Using the Property 2, we get
72
72
2
70
12=⋅= .
Therefore, we have
3112
− = 7
which one may check easily.
13
Determinants of Matrices of Higher Order
As we said before, the idea is to assume that previous properties satisfied by
the determinant of matrices of order 2, are still valid in general.
So let us see how this works in case of a matrix of order 4.
Example 2. Evaluate
2113846287654321
.
We have
2113423187654321
2
2113846287654321
= .
If we subtract every row multiplied by the appropriate number from the first
row, we get
10850
0110
128404321
2113423187654321
−−−−
−−−=
We do not touch the first row and work with the other rows. We interchange
the second with the third to get
10850
12840
01104321
10850
0110
128404321
−−−−−−
−−=
−−−−
−−−.
If we subtract every row multiplied by the appropriate number from the
second row, we get
101300
121200
01104321
10850
12840
01104321
−−−−
−=
−−−−−−
−.
Using previous properties, we have
1013001100
01104321
12
101300
121200
01104321
−−
−−=
−−−−
−.
If we multiply the third row by 13 and add it to the fourth, we get
14
30001100
01104321
1013001100
01104321
−=
−−
−
which is equal to 3. Putting all the numbers together, we get
( ) ( ) .7231212
2113846287654321
=⋅−⋅−⋅=
These calculations seem to be rather lengthy. We will see later on that a
general formula for the determinant does exist.
Example 3. Evaluate 321
111021
− .
In this example, we will not give the details of the elementary operations.
We have
.9300130021
321111021
==−
Example 4. Evaluate 112
110211
−.
We have
.5500
010211
510110211
112110211
−=−
=−−
=−
General Formula for the Determinant
Let A be a square matrix of order n. Write A = (aij), where aij is the entry on
the row number i and the column number j, for i = 1, 2, 3, … n and j = 1, 2,
… n. For any i and j, set Cij (called the cofactors) to be the determinant of
the square matrix of order (n-1) obtained from A by removing the row
number i and the column number j multiplied by (-1)i+j. We have
( ) ∑=
=
=nj
jijijAdet
1
Ca
for any fixed i, and
( ) ∑=
=
=ni
i
ijij Ca1
Adet
15
for any fixed j. In other words, we have two types of formulas: along a row
(number i) or along a column (number j). Any row or any column will do.
The trick is to use a row or a column which has a lot of zeros.
In particular, we have along the rows
hgedc
kg
fdb
kh
fea
khg
fedcba
+−=
or
hgbaf
kgcae
khcbd
khg
fedcba
−+−= ,
or
edbak
fdcah
fecbg
khg
fedcba
+−= .
As an exercise write the formulas along the columns.
Example 5. Evaluate 104
312123
− .
We will use the general formula along the third row. We have
12231
32130
31124
104
312123
+−−−=− ( ) ( ) 29431164 −=−+−−=
Which technique to evaluate a determinant is easier? The answer depends
on the person who is evaluating the determinant. Some like the elementary
row operations and some like the general formula. All that matters is to get
the correct answer.
Note that all of the above properties are still valid in the general case. Also
you should remember that the concept of a determinant only exists for
square matrices.
http://www.sosmath.com/matrix/determ0/determ1.html Author: M.A. Khamsi
Physics Department, Yarmouk University, Irbid Jordan
Phys. 201 Methods of Theoretical Physics 1
Dr. Nidal M. Ershaidat Doc. 6
Author’s email: enidal@yu.edu.jo Address: Physics Department, Yarmouk University 21163 Irbid Jordan
16
Ordinary Differential Equations (ODE’s)
A) Definition
An ordinary differential equation is any equation which has exact
differentials or exact derivatives in both sides. The order of the highest
derivative is called the order of the differential equations. The solution of an
ODE is any relationship between the unknown function with its variable
which verifies the ODE.
B) First Order Differential Equations
Separable 1st order DE
A 1st order differential of the form:
( ) ( ) 0=+ dyyGdxxF (1)
Is called “separable 1st order DE”.
The general solution of Eq. 1 is obtained by integrating both sides, i.e.
( ) ( ) CdyyGdxxF =+ ∫∫ (2)
Linear 1st order DE
( ) ( )xQyxPdx
dy=+ (3)
If Q(x) = 0 then Eq. 3 can be written as:
( ) 0=+ dxxPy
dy (4)
which is a separable 1st order ODE (See Eq. 1), where ( ) ( )xPxF = and
( )y
yG1
= .
- Homogeneous Linear 1st order DE
Eq. 4 is also called homogeneous 1st order DE.
( ) 0=+ yxPdx
dy
The general solution of Eq. 4 is obtained by integrating both sides, i.e.
Author’s email: enidal@yu.edu.jo Address: Physics Department, Yarmouk University 21163 Irbid Jordan
17
( ) ( ){ }∫−= dxxPAxy exp (5)
- Inhomogeneous Linear 1st order DE
( ) ( )xQyxPdx
dy=+ (6)
The general solution of Eq. 6 is given by:
( ) ( ) ( ) ( ) ( )xIxIxI eCdxxQeexy −− += ∫ (7)
where
( ) ( ) ( )dxxPexI xI∫= (8)
Exact Equations
A first order ordinary differential is called exact equation if its lhs can be
expressed as an exact differential dU of a function U(x,y) such that:
( ) ( ) 0=+= dyy,xNdxy,xMdU (9)
This is verified if and only of:
yxx
N
y
M
∂∂
=
∂
∂ (10)
xy
M
∂
∂ is the partial derivative of the function M(x,y) with respect to y. This
means that we derive the function M(x,y) considering x as a constant (which
we symbolize by the subscript x.
yx
N
∂∂ is the partial derivative of the function N(x,y) with respect to x. This
means that we derive the function N(x,y) considering y as a constant (which
we symbolize by the subscript y.
Example: if ( ) 232, yxyxM = then 226 yxx
M
y
=
∂
∂ and yx
y
M
x
34=
∂
∂.
Homogeneous First Order DE
A first order ordinary differential is called exact equation if the lhs can be
expressed as an exact differential dU of a function U(x,y) such that:
=x
yF
dx
dy (11)
© Nidal M. Ershaidat 2013 18
The general solution is:
( )∫∫ +−
= CvvF
dv
x
dx (12)
Where
( )xvxy = (13)
Second Order Differential Equations
Linear 2nd order DE
A 2nd order DE equation is linear if and only of it can be written in the form:
( ) ( ) ( )xRyxQdx
dyxP
dx
yd=++
2
2
(14)
Where P(x), Q(x) and R(x) are general functions of the independent variable
x,
Various methods are used to solve such equations. These methods depend
on the nature of P(x), Q(x) and R(x).
Reference: Phys. 201 Methods of Theoretical Physics 1 – Textbook Introduction to Mathematical Physics, 2nd Edition 2004, Authors: N. Laham and N. Ayoub
Physics Department, Yarmouk University, Irbid Jordan
Phys. 201 Methods of Theoretical Physics 1
Dr. Nidal M. Ershaidat Doc. 8
Author’s email: enidal@yu.edu.jo, Address: Physics Department, Yarmouk University 21163 Irbid Jordan © Nidal M. Ershaidat 2013
19
ODE – An Example
Example 1:
Solve 054 =+−′′ yy
This is an inhomogeneous linear 2nd order differential equation.
( )xRybyay =+′+′′ (1)
with:
4,0 −== ba and ( ) constantxR =−= 5
Its general solution y(x) is the sum yC(x) + yP(x) where yC(x) is the
complementary solution i.e. the solution of the corresponding homogeneous
equation ( 04 =−′′ yy ) and yP(x) is the (unique) particular solution obtained
using the method of undetermined coefficients.
( ) ( ) ( )xyxyxy PC += (2)
with
( ) ( ) ( )dxxFeexy xxP ∫ β−αβ= (3)
and
( ) ( )dxxRexF x∫ α−= (4)
Complementary Solution
( ) xxC eCeCxy βα += 21 (5)
Where α and β are the roots of the auxiliary equation (r2 – 4 = 0), i.e.
2=α and 2−=β (6)
Particular Solution
( ) ( ) xx edxexF α−α−
α=−= ∫
55 (7)
⇒⇒⇒⇒( ) ( )
4
555
5
−=βα
−=α
=
α=
∫
∫β−β
α−β−αβ
dxee
dxeeexy
xx
xxxP
(8)
which gives:
( )4
522
21 ++= − xx eCeCxy (9)
Physics Department, Yarmouk University, Irbid Jordan
Phys. 201 Methods of Theoretical Physics 1
Dr. Nidal M. Ershaidat Doc. 8
Author’s email: enidal@yu.edu.jo, Address: Physics Department, Yarmouk University 21163 Irbid Jordan © Nidal M. Ershaidat 2013
20
Exact Differential A differential of the form
( ) ( )dyyxQdxyxPdf ,, += (1)
is exact (also called a total differential) if ∫df is path-independent. This will
be true if
.dyy
fdx
x
fdf
∂∂
+∂∂
= (2)
So P and Q must be of the form
( ) ( )y
fyxQand
x
fyxP
∂∂
=∂∂
= ,, (3)
But
,2
xy
f
y
P
∂∂∂
=∂∂
(4)
.2
yx
f
x
Q
∂∂∂
=∂∂
(5)
which yields
.x
Q
y
P
∂∂
=∂∂
(6)
The notion of exact differentials plays an important role in physics. In
thermodynamics, properties of a thermodynamic system (for example, the
pressure, volume and temperature for a gas) should be exact differentials.
Physics Department, Yarmouk University, Irbid Jordan
Phys. 201 Methods of Theoretical Physics 1
Dr. Nidal M. Ershaidat Doc. 9
21
Method of Undetermined Coefficient or Guessing Method
Principle
This method is based on a guessing technique. That is, we will guess the form
of yP and then plug it in the equation to find it. However, it works only under
the following two conditions:
Condition 1: the associated homogeneous equations has constant coefficients
Condition 2: the nonhomogeneous term R(x) is a special form
( ) ( ) ( )xexPxRx β= α
cos (1)
Or
( ) ( ) ( )xexLxRx β= α
sin (2)
where P(x) and L(x) are polynomial functions. Note that we may assume that
R(x) is a sum of such functions (see the remark below for more on this).
Assume that the two conditions are satisfied. Consider the equation
( )xRycybya =+′+′′ (3)
where a, b and c are constants and
( ) ( ) ( )xexPxRx
n β= αcos or ( ) ( ) ( )xexPxR
xn β= α
sin (4)
where Pn(x) is a polynomial function with degree n. Then a particular solution yP
is given by
( ) ( ) ( ) ( ) ( )( )xexUxexTxxyx
nx
ns
P β+β= ααsincos (5)
Where
( ) 22210 xAxAxAAxT nn ++++= … (6)
And
( ) 22210 xBxBxBBxU nn ++++= …
Where the constants Ak and Bk are to be determined.
The value of the power s depends on α + iβ in the following manner:
� if α + iβ is not a root of the characteristic equation then s=0.
� if α + iβ is a simple root of the characteristic equation then s=1.
� if α + iβ is a double root of the characteristic equation then s=2.
© Nidal M. Ershaidat 22
Remark: If the nonhomogeneous term R(x) satisfies the following
( ) ( )∑=
=
=Ni
i
i xRxR1
(7)
where Ri(x) are of the forms cited above, then we split the original equation
into N equations:
( ) ( )NixRycybya i ,,2,1 …==+′+′′ (8)
Then find a particular solution yi. A particular solution to the original
equation is given by
( ) ∑=
=
=Ni
i
iP yxy1
. (8)
Summary
Let us summarize the steps to follow in applying this method:
(1) First, check that the two conditions mentioned above are satisfied;
(2) If the equation is given as
( )∑=
=
=+′+′′Nk
k
k xRycybya1
, (9)
( ) ( ) ( )xexPxRx
n β= αcos or ( ) ( ) ( )xexPxR
xn β= α
sin (10)
where Pn is a polynomial function with degree n, then split this equation into
N equations
( )xRycybya k=+′+′′ , (11)
where
(3) Write down the characteristic equation a r2 + b r + c = 0 and find its roots;
(4) Write down the number αk+ i βk. Compare this number to the roots of the
characteristic equation found in the previous step.
(4.1) If αk+ iβk is not one of the roots, then set s = 0;
(4.2) If αk+ iβk is one of the two distinct roots, then set s = 1;
(4.3) If αk+iβk is equal to both roots (which means that the
characteristic equation has a double root, then set s = 2;
In other words, s measures how many times αk+ i βk is a root of the
characteristic equation;
(5) Write down the form of the particular solution
© Nidal M. Ershaidat 23
( ) ( ) ( ) ( ) ( )( )xexUxexTxxyx
nx
ns
P β+β= ααsincos , (12)
where
( ) 22210 xAxAxAAxT nn ++++= … (13)
and
( ) .22
210 xBxBxBBxU nn ++++= … (14)
(6) Find the constants Ai and Bi by plugging Rk into the equation
( )xRycybya k=+′+′′ ,
(7) Once all the particular solutions yk are found, then the particular solution
of the original equation is the series
( ) ∑=
=
=Nk
k
kP yxy1
(15)
Reference: http://www.sosmath.com/diffeq/second/guessing/guessing.html
Physics Department, Yarmouk University, Irbid Jordan
Phys. 201 Methods of Theoretical Physics 1
Dr. Nidal M. Ershaidat Doc. 10
24
Method of Variation of Parameters
Principle
This method has no prior conditions to be satisfied. Therefore, it may sound
more general than the method of undetermined coefficient or guessing
method. We will see that this method depends on integration while the other
one cited is purely algebraic which, for some at least, is an advantage.
Consider the equation
( ) ( ) ( )xRyxQyxPy =+′+′′ (1)
In order to use the method of variation of parameters we need to know that
{y1, y2} is a set of fundamental solutions of the associated homogeneous
equation ( ) ( ) 0=+′+′′ yxQyxPy . We know that, in this case, the general
solution of the associated homogeneous equation is 2211 ycycyh += . The
idea behind the method of variation of parameters is to look for a particular
solution such as
( ) ( ) ( ) ( ) ( )xxx 2211 yxuyxuyP += , (2)
where u1 and u2 are functions. From this, the method got its name.
The functions u1 and u2 are solutions to the system
( )
=′′+′′=′+′
xRyuyu
yuyu
2211
2211 0 (3)
which implies
( ) ( ) ( )( )( )
( ) ( ) ( )( )( )
=
−=
∫
∫
.,
,
21
12
21
21
dxxyyW
xRxyxu
dxxyyW
xRxyxu ,
(4)
Where (((( ))))(((( ))))xyyW 21 , is the wronskian of y1 and y2. Therefore, we have
( ) ( ) ( ) ( )( )( )
( ) ( ) ( )( )( )∫∫ +−= dx
xyyW
xRxyxydx
xyyW
xRxyxyxyP
21
12
21
21
,, (5)
25
Summary
Let us summarize the steps to follow in applying this method:
(1) Find {y1, y2} a set of fundamental solutions of the associated
homogeneous equation (((( )))) (((( )))) 0====++++′′′′++++′′′′′′′′ yxQyxPy ;
(2) Write down the form of the particular solution
(((( )))) (((( )))) (((( )))) (((( )))) (((( ))))xyxuxyxuxyP 2211 ++++==== (6)
(3) Write down the system:
(((( ))))
====′′′′++++′′′′
====′′′′++++′′′′
xRyuyu
yuyu
2211
2211 0 (7)
(4) Solve it. That is, find u1 and u2;
(5) Plug u1 and u2 into the equation giving the particular solution.
Example
Find the particular solution to ( )xyy tan1+=+′′ ; 22
π<<
π− x
Solution
Let us follow the steps in the summary:
(1) A set of fundamental solutions of the associated homogeneous equation
0=+′′ yy is {y1 = cos(x), y2 = sin(x)};
(2) The particular solution is given as
( ) ( ) ( ) ( ) ( ) .sincos 21 xxuxxuxyP +=
(3) We, thus, have the system:
( ) ( )( ) ( ) ( )
+=′+′−=′+′
xxuxu
xuxu
tan1cossin
0sincos
211
21
(4) We solve for 1u′ and 2u′ and get:
( ) ( )( )( ) ( ) ( )( )
+=′+−=′
xxxu
xxu
tan1cossin
tan1sin
2
1
Using Integration techniques, we get:
( ) ( ) ( ) ( ) ( )( ) ,tanseclnsincos1 xxxxxu +−+=
( ) ( ) ( ) .cossin2 xxxu −=
(5) The particular solution is
( ) ( ) ( ) ( ) ( )( )( ) ( ) ( ) ( )( )xxxxxxxxyP cossinsintanseclnsincoscos −++−+=
or
26
( ) ( ) ( )( ) .tanseclncos1 xxxyP +−=
Remark: Note that since the equation is linear, we may still split if
necessary. For example, we may split the equation
( )xyy tan1+=+′′
into the two equations 1=+′′ yy (R1) and ( )xyy tan=+′′ (R2)
then, find the particular solutions y1 for (R1) and y2 for (R2), to generate a
particular solution for the original equation by 21 yyyP +=
There are no restrictions on the method to be used to find y1 or y2. For
example, we can use the method of undetermined coefficients to find y1,
while for y2, we are only left with the variation of parameters.
Reference: http://www.sosmath.com/diffeq/second/variation/variation.html
Physics Department, Yarmouk University, Irbid Jordan
Phys. 201 Methods of Theoretical Physics 1
Dr. Nidal M. Ershaidat Doc. 11
27
Fourier Series
A) Definition
A Fourier series is an expansion of a periodic function f(x) in terms of an
infinite sum of sines and cosines. Fourier series make use of the
orthogonality relationships of the sine and cosine functions. The computation
and study of Fourier series is known as harmonic analysis and is extremely
useful as a way to break up an arbitrary periodic function into a set of simple
terms that can be plugged in, solved individually, and then recombined to
obtain the solution to the original problem or an approximation to it to
whatever accuracy is desired or practical.
B) Illustration
Examples of successive approximations to common functions using Fourier
series are illustrated below.
(a) (b)
(c) (d)
Figure 1: Using Fourier series in order to approximate some functions.
a) The square wave, b) The sawtooth wave, c) the triangle wave and d) the
semicircle.
28
C) Application - solution of ordinary differential equations
In particular, since the superposition principle holds for solutions of a linear
homogeneous ordinary differential equation, if such an equation can be
solved in the case of a single sinusoid, the solution for an arbitrary function
is immediately available by expressing the original function as a Fourier
series and then plugging in the solution for each sinusoidal component. In
some special cases where the Fourier series can be summed in closed form,
this technique can even yield analytic solutions.
D) Generalized Fourier Series
Any set of functions that form a complete orthogonal system have a
corresponding generalized Fourier series analogous to the Fourier series. For
example, using orthogonality of the roots of a Bessel function of the first
kind gives a so-called Fourier-Bessel series.
E) Computation of Fourier series
The computation of the (usual) Fourier series is based on the following
integral identities which represent the orthogonality relationships of the sine
and the cosine functions:
( ) ( )∫π+
π−
δπ= nmdxxnxm sinsin (1)
( ) ( )∫π+
π−
δπ= nmdxxnxm coscos (2)
( ) ( )∫π+
π−
= 0cossin dxxnxm (3)
( )∫π+
π−
= 0sin dxxm (4)
( )∫π+
π−
= 0cos dxxm (5)
For 0, ≠nm , where δmn is the Kronecker delta.
Using the method for a generalized Fourier series, the usual Fourier series
involving sines and cosines is obtained by taking f1(x) = cos x and f2(x) = sin x.
Since these functions form a complete orthogonal system over [- π, π], the
29
Fourier series of a function f(x) is given by
( ) ( ) ( )∑∑∞
=
∞
=
++=11
0 sincos2
1
n
n
n
n xnbxnaaxf (6)
a0, a1 …an are called the Fourier coefficients.
These coefficients are obtained using the following relations:
( )∫π+
π−π
= dxxfa1
0 (7)
( ) ( )∫π+
π−π
= dxxnxfan cos1
(8)
( ) ( )∫π+
π−π
= dxxnxfbn sin1
(9)
n = 1, 2, 3, …. Note that the coefficient of the constant term a0 has been
written in a special form compared to the general form for a generalized
Fourier series in order to preserve symmetry with the definitions of an and bn.
A Fourier series converges to the function f (equal to the original function at
points of continuity or to the average of the two limits at points of
discontinuity)
( ) ( )
( ) ( )
ππ−=
+
π<<π−
+
=
−+
+−
π→π→
→→
,limlim2
1
limlim2
1
0
000
xforxfxf
xforxfxf
f
xx
xxxx (10)
if the function satisfies so-called Dirichlet conditions.
As a result, near points of discontinuity, a "ringing" known as the Gibbs
phenomenon, illustrated above, can occur.
30
For a function f(x) periodic on an interval [- L, L] instead of [- π, π], a simple
change of variables can be used to transform the interval of integration from
[- π, π] to [- L, L].
Let
xL
x ′π= (11)
xdL
dx ′π= (12)
Solving for x′ gives xL
xπ
=′ , and plugging this in Eq. 6 gives
( ) ∑∑∞
=
∞
=
′π+
′π+=′
11
0 sincos2
1
n
n
n
n xL
nbx
L
naaxf (13)
Therefore,
( )∫+
−
′′=L
L
xdxfL
a1
0 (14)
( )∫+
−
′
′π′=L
L
n xdxL
nxf
La cos
1 (15)
( )∫+
−
′
′π′=L
L
n xdxL
nxf
Lb sin
1 (16)
Similarly, the function is instead defined on the interval [0,2L], the above
equations simply become
( )∫ ′′=L
xdxfL
a
2
0
0
1 (17)
( )∫ ′
′π′=L
n xdxL
nxf
La
2
0
cos1
(18)
( )∫ ′
′π′=L
n xdxL
nxf
Lb
2
0
sin1
(19)
In fact, for f(x) periodic with period 2 L, any interval [x0, x0+2L ] can be used,
with the choice being one of convenience or personal preference (Arfken
1985, p. 769).
The coefficients for Fourier series expansions of a few common functions are
31
given in Beyer (1987, pp. 411-412) and Byerly (1959, p. 51). One of the
most common functions usually analyzed by this technique is the square
wave. The Fourier series for a few common functions are summarized in the
table below.
F) Examples
Function f(x) Fourier series
sawtooth wave L
x
2 ∑
∞
=
π−
1
sin1
2
1
n
xL
n
n
square wave 112 −
−−
L
xH
L
xH ∑
∞
=
ππ
…,5,3,1
sin14
n
xL
n
n
triangle wave T(x) ( )∑
∞
=
−
π−
π…,5,3,1
2
21
2sin
18
n
n
xL
n
n
If a function is even so that f(x) = f(- x), then f(x) sin(nx) is odd. (This follows
since sin(nx) is odd and an even function times an odd function is an odd
function.) Therefore, bn=0 for all n. Similarly, if a function is odd so that f(x) = -
f(-x), then f(x) cos(nx) is odd. (This follows since cos(nx) is even and an even
function times an odd function is an odd function.) Therefore, an=0 for all n.
G) Complex Coefficients
The notion of a Fourier series can also be extended to complex coefficients.
Consider a real-valued function f(x). Write
( ) ∑∞
=
=0n
xnin eAxf (20)
Now examine
( ) ∫ ∑∫π+
π−
−∞
∞−=
π+
π−
−
= dxeeAdxexf xmi
n
xnin
xmi (21)
( )∫∑
π+
π−
−∞
∞−=
= dxeA xmni
n
n (22)
( ) ( )[ ]∫∑π+
π−
∞
∞−=
−+−= dxmnimnAn
n sincos (23)
32
nm
n
nA δπ= ∑∞
∞−=
2 (24)
mAπ= 2 , (25)
So
( )∫π+
π−
−
π= dxexfA xni
n2
1 (26)
The coefficients can be expressed in terms of those in the Fourier series
( ) ( ) ( )[ ]∫π+
π−
+π
= dxxnixnxfAn sincos2
1 (27)
( ) ( ) ( )[ ]
( )
( ) ( ) ( )[ ]
>−π
=π
<+π
=
∫
∫
∫
π+
π−
π+
π−
π+
π−
0sincos2
1
02
1
0sincos2
1
ndxxnixnxf
ndxxf
ndxxnixnxf
(28)
( )
( )
<−π
<π
<+π
=
02
1
02
1
02
1
0
nforbia
nfora
nforbia
nn
nn
(29)
For a function periodic in [- L/2, L/2], these become
( ) ( )∑∞
∞−=
π=n
Lxnin eAxf
2 (30)
( ) ( )∫−
π−=2
2
21L
L
Lxnin dxexf
LA (31)
These equations are the basis for the extremely important Fourier transform,
which is obtained by transforming An from a discrete variable to a continuous
one as the length L → ∞.
33
H) References
Arfken, G. "Fourier Series." Ch. 14 in Mathematical Methods for Physicists,
3rd ed. Orlando, FL: Academic Press, pp. 760-793, 1985.
Askey, R. and Haimo, D. T. "Similarities between Fourier and Power Series."
Amer. Math. Monthly 103, 297-304, 1996.
Beyer, W. H. (Ed.). CRC Standard Mathematical Tables, 28th ed. Boca
Raton, FL: CRC Press, 1987.
Brown, J. W. and Churchill, R. V. Fourier Series and Boundary Value
Problems, 5th ed. New York: McGraw-Hill, 1993.
Byerly, W. E. An Elementary Treatise on Fourier's Series, and Spherical,
Cylindrical, and Ellipsoidal Harmonics, with Applications to Problems in
Mathematical Physics. New York: Dover, 1959.
Carslaw, H. S. Introduction to the Theory of Fourier's Series and Integrals,
3rd ed., rev. and enl. New York: Dover, 1950.
Davis, H. F. Fourier Series and Orthogonal Functions. New York: Dover,
1963.
Dym, H. and McKean, H. P. Fourier Series and Integrals. New York:
Academic Press, 1972.
Folland, G. B. Fourier Analysis and Its Applications. Pacific Grove, CA:
Brooks/Cole, 1992.
Groemer, H. Geometric Applications of Fourier Series and Spherical
Harmonics. New York: Cambridge University Press, 1996.
Körner, T. W. Fourier Analysis. Cambridge, England: Cambridge University
Press, 1988.
Körner, T. W. Exercises for Fourier Analysis. New York: Cambridge
University Press, 1993.
Krantz, S. G. "Fourier Series." §15.1 in Handbook of Complex Variables.
Boston, MA: Birkhäuser, pp. 195-202, 1999.
Lighthill, M. J. Introduction to Fourier Analysis and Generalised Functions.
Cambridge, England: Cambridge University Press, 1958.
Morrison, N. Introduction to Fourier Analysis. New York: Wiley, 1994.
Sansone, G. "Expansions in Fourier Series." Ch. 2 in Orthogonal Functions,
rev. English ed. New York: Dover, pp. 39-168, 1991.
34
Weisstein, E. W. "Books about Fourier Transforms."
http://www.ericweisstein.com/encyclopedias/books/FourierTransforms.html.
Whittaker, E. T. and Robinson, G. "Practical Fourier Analysis." Ch. 10 in The
Calculus of Observations: A Treatise on Numerical Mathematics, 4th ed. New
York: Dover, pp. 260-284, 1967.
Source internet
Weisstein, Eric W. "Fourier Series." From MathWorld - A Wolfram Web
Resource. http://mathworld.wolfram.com/FourierSeries.html
© 1999 CRC Press LLC, © 1999-2007 Wolfram Research, Inc. | Terms of
Use
Physics Department, Yarmouk University, Irbid Jordan
Phys. 201 Methods of Theoretical Physics 1
Dr. Nidal M. Ershaidat Doc. 12
35
Curvilinear Coordinates
What are curvilinear coordinates?
A curvilinear coordinate system is composed of intersecting surfaces. If the intersections are all at right angles, then the curvilinear coordinates are said to form an orthogonal coordinate system. If not, they form a skew coordinate system.
Curvilinear q1 q2 q3 h1 h2 h3 →1a
→2a
→3a
Cartesian x y z 1 1 1 i j k
Spherical r θ φ 1 r r sin θ 0r θ φ
Cylindrical r (or ρ) φ z 1 r 1 0r or ρ φ k
→→→++= 333222111 adqhadqhadqhld
�
∑∑=
→
=
→→
∂φ∂
=∂
φ∂=φ∇
3
1
3
1 i iii
i ii
qha
la
( ) ( ) ( )
∂∂
+∂
∂+
∂∂
=⋅∇→→
3
213
2
312
1
321
321
1
q
hhA
q
hhA
q
hhA
hhhA
332211
321
332211
321
1
AhAhAh
qqq
hahaha
hhhA
∂∂
∂∂
∂∂
=×∇
→→→
→→
∂φ∂
∂∂
+
∂φ∂
∂∂
+
∂φ∂
∂∂
=φ∇=φ∇⋅∇→→
33
21
322
31
211
32
1321
2 1
qh
hh
qqh
hh
qqh
hh
qhhh
Physics Department, Yarmouk University, Irbid Jordan
Phys. 201 Methods of Theoretical Physics 1
Dr. Nidal M. Ershaidat Doc. 12
36
Cartesian to Cylindrical Cylindrical to Cartesian Cartesian to Spherical Spherical to Cartesian
φφ−φ= sinˆcosˆˆ0ri φ+φ= sinˆcosˆ
0 jir φφ−φθθ+φθ= sinˆcoscosˆcossinˆˆ0ri θ+φθ+φθ= cosˆsinsinˆcossinˆ
0 kjir
φφ+φ= cosˆsinˆˆ0rj φ+φ−=φ cosˆsinˆˆ ji φφ+φθθ+φθ= cosˆsincosˆsinsinˆˆ
0rj θ−φθ+φθ=θ sinˆsincosˆcoscosˆˆ kji
kk ˆˆ = kk ˆˆ = θθ−θ= sinˆcosˆˆ0rk φ+φ−=φ cosˆsinˆˆ ji
Cartesian Spherical Cylindrical
ld�
kdzjdyidxld ˆˆˆ ++=�
φφθ+θθ+= ˆsinˆ0 drdrrdrld
�
kdzdrrdrld ˆˆ0 +φφ+=
�
ψ∇→
zk
yj
xi
∂ψ∂
+∂ψ∂
+∂ψ∂ ˆˆˆ
φ∂ψ∂
φθ
+θ∂ψ∂
θ+∂ψ∂
=ψ∇→
ˆsin
11ˆ0
rrrr
zk
rrr
∂ψ∂
+φ∂ψ∂
φ+∂ψ∂
=ψ∇→
ˆ1ˆ0
→→⋅∇ A z
A
y
A
x
A zyx
∂∂
+∂
∂+
∂∂
( ) ( ) ( )
φ∂
∂+θ
θ∂∂
+θ∂∂
θ=⋅∇ φθ
→→ArArAr
rrA r sinsin
sin
1 2
2
( ) ( )
∂
∂+
φ∂
∂+
∂
∂=⋅∇ φ→→
z
ArA
r
Ar
rA zr1
→→×∇ A
zyx AAA
zyx
kji
∂∂
∂∂
∂∂
ˆˆˆ
φθ
→→
θφ∂∂
θ∂∂
∂∂
φθθ
θ=×∇
ArArAr
rrr
rA
r sin
ˆsinˆˆ
sin
10
2
zr AArAzr
krr
rA
φ
→→
∂∂
φ∂∂
∂∂
φ
=×∇
ˆˆˆ
10
ψ∇2 2
2
2
2
2
22
zyx ∂
ψ∂+
∂
ψ∂+
∂
ψ∂=ψ∇
φ∂
ψ∂
θ+
θ∂ψ∂
θθ∂
∂
θ+
∂ψ∂
∂∂
=ψ∇2
2
222
2
2
2
sin
1sin
sin
11
rrrr
rr
2
2
2
2
22
22 11
zrrrr ∂
ψ∂+
φ∂
ψ∂+
∂ψ∂
+∂
ψ∂=ψ∇
Or
φ∂
ψ∂θ
+
θ∂ψ∂
θθ∂
∂+
∂ψ∂
θ∂∂
θ=ψ∇
2
22
2
2
sin
1sinsin
sin
1
rr
rr
∂ψ∂
∂∂
+
φ∂ψ∂
φ∂∂
+
∂ψ∂
∂∂
=ψ∇z
rzrr
rrr
112