Thor Eti Call Ec Note Dwl

HEP/123-qed

Topics and Methods in Theoretical Physics II (PHYS 3160)

Daniel B. Erenso and Victor J. Montemayor

Department of Physics & Astronomy, Middle Tennessee State University

1

Contents

I. Lecture 1 Introduction to the Calculus of Variations 3

A. Lecture 2 Applications of the Calculus of Variations 5

II. Lecture 3 Introduction to the Eigenvalue Problem 14

III. Lecture 4 Applications of the Eigenvalue Problem: Normal Modes of

Vibration 23

IV. Lecture 5: Special Integral Functions: The Gamma, Beta, and Error

Functions 31

V. Lecture 6 Stirling�s Formula and Elliptic Integrals 40

VI. Lecture 7 Power Series Solutions to Di¤erential Equations 47

VII. Lecture 8 Complete Sets of Functions and the Legendre Polynomials 54

A. Lecture 9 The Generating Function for Legendre Polynomials 62

VIII. Lecture 10 Legendre Series, Associated Legendre Functions, and

Spherical Harmonics 68

IX. Lecture 11: The Addition Theorem for Spherical Harmonics 79

X. Lecture 12: The Method of Frobenius and Bessel Functions 80

XI. Lecture 13 The Orthogonality and Normalization of Bessel Functions 89

XII. Lecture 14 Introduction to Partial Di¤erential Equations and the

Separation of Variables 103

XIII. Lecture 15: Laplaces�s Equation in Spherical Coordinates 111

A. Lecture 16 Laplace�s Equation in Cylindrical Coordinates 120

XIV. Lecture 17 Poisson�s Equation 127

XV. Functions of Complex Variables Lecture 18 135

2

A. Analytic Functions 135

XVI. Lecture 19 Contour Integration and Cauchy�s Theorem 141

A. Lecture 20 Residues and the Residue Theorem 150

B. Lecture 21 Applications of the Residue Theorem 158

C. Lecture 22 The Calculus of Residues Applied: The Kramers-Kronig Relations 168

XVII. Lecture 23 Introduction to Integral Transforms and the Laplace

Transform 179

A. Lecture 24 Applications of Laplace Transforms 189

XVIII. Lecture 25 Introduction to the Fourier Transform 197

XIX. Lecture 26 Applications of the Fourier Transform: The Heisenberg

Uncertainty Principle 209

XX. Lecture 27 Fourier Transforms and Convolution 216

I. LECTURE 1 INTRODUCTION TO THE CALCULUS OF VARIATIONS

� Geodesic: The curve along a surface which marks the shortest distance between twoneighboring points. Finding geodesics is one of the problems which we can solve using

the calculus of variation.

� Stationary point : A point on a give function f (x) is said to be stationary point when

df (x)

dx= 0:

Ex. 1 A ball of mass m is kicked from ground level with an initial speed vo at an angle �o

above the horizontal. Find the value of the time t after the ball is kicked that makes

the height function of the ball, y(t), stationary.

Sol: We recall that from kinematics of a projectile the motion of the ball along the y

direction is determined by Newton�s second law

mdvydt

= �mg ) dvydt

= �g ) vy (t) = v0y � gt

3

Noting that

vy (t) =dy (t)

dt

the value of the time after the ball is kicked that makes the height function of the

ball,y(t), stationary is given by

vy (t) =dy (t)

dt= 0) t =

v0yg=v0 sin (�)

g

� The Problem:

We consider some unknown function y(x). We assume that this function is known at

two �xed points, y(x1) and y(x2). We wish to �nd the function y(x) that makes the

integral

I =

Z x2

x1

F (x; y; y0) dx

stationary for some known function F (x; y; y0).

� The Euler-Lagrange Equation:

@

@x

�@F

@y0

�� @F

@y= 0

where F = F (x; y; y0)

Ex. 2 A geodesic in a given space (or on a given surface) is a curve drawn in that space

(or on that surface) that has the shortest length between two given points. Consider

two points in a 3-D Euclidean space. Prove that the shortest distance between the

two points is the distance measured along a straight line. In other words, show that

straight lines are geodesics in a 3-D Euclidean space.

Sol:

4

A. Lecture 2 Applications of the Calculus of Variations

� Recall : The function y(x) that makes, for two �xed points y(x1) and y(x2), the integral

I =

Z x2

x1

F (x; y; y0) dx

stationary is that same function y(x) that satis�es the Euler (Euler-Lagrange) di¤er-

ential equation:@

@x

�@F

@y0

�� @F

@y= 0

The Brachystochrone Problem: If two points A and B are given, at di¤erent heights

but not lying one above the other (as shown in the �gure below), it is required to �nd

among all possible curves connecting them, that one along which a material point slides

from A to B under the in�uence of gravity (neglecting friction) in the shortest possible

time. This curve is called a Brachistochrone curve (Gr. ��o&, brachistos - the

shortest,��o�o&, chronos - time), or curve of fastest descent.[From Wikipedia, the free

encyclopedia]

This problem occupied at the time the leading mathematicians in the whole of Europe:

Newton, Leibniz, Bernoulli, L�Hospital, and others. From then on, the calculus of

variations developed as a special mathematical discipline.

Ex. 3 Solve the brachystochrone problem, assuming the �material point�starts from rest.

Sol: We are given the two points (x1; y1) and (x2; y2);we chose axes through the point 1

with the y axis positive downward as shown in Figure below. We want to �nd the

curve joining the two points, down which a bead will slide (from rest) in the least time.

That means we want to minimize time t:

5

which means we want to �nd the stationary value of the integral

I =

Z 2

1

dt =

Z 2

1

ds

v:

The total energy of the bead is zero since it starts from rest assuming the zero energy

level is the origin. If there is no friction, then we can write the energy at any point

below the origin describe by the coordinates (x; y) is given by

1

2mv2 �mgy = 0) v =

p2gy:

Then

I =

Z 2

1

ds

v=

Z 2

1

dsp2gy

:

Noting that

ds =pdx2 + dy2

we have

I =

Z 2

1

pdx2 + dy2p2gy

=

Z 2

1

r1 +

�dxdy

�2p2gy

dy ) I =1p2g

Z y2

y1

p1 + x02py

dy;

so that the stationary value of this integral is determined from the Euler_Lagrange

equation@

@y

�@F

@x0

�� @F

@x= 0;

where

F (y; x; x0) =1p2g

p1 + x02py

:

where

x0 =dx

dy:

6

Noting that@F

@x= 0

and@F

@x0=

1p2g

x0p1 + x02

py

we have

@

@x

1p2g

x0p1 + x02

py

!= 0

) x0p1 + x02

py=pc:

where c is a constant. Solving for x0; we �nd

x0p1 + x02

py=pc) x02 = c

�1 + x02

�y

) x02 (1� cy) = cy ) dx

dy=

rcy

1� cy) x =

Z y

0

rcy

1� cydy:

Introducing the transformation de�ned by

cy = sin2��

2

�=1

2(1� cos (�))

) dy =1

csin

��

2

�cos

��

2

�d�

we have

x =

Z y

0

rcy

1� cydy =

1

c

Z �

0

ssin2

��2

�1� sin2

��2

� sin��2

�cos

��

2

�d�

) x =1

c

Z �

0

sin��2

�cos��2

� sin��2

�cos

��

2

�d�

=1

c

Z �

0

sin2��

2

�d� =

1

c

Z �

0

1

2(1� cos (�)) d� ) x =

1

2c(� � sin (�)) :

Therefore, the trajectory of the bead that takes a smallest possible time is given by

x =1

2c(� � sin (�)) ; y = 1

2c(1� cos (�)) :

Cycloid: Consider a circle of radius r rolling along the positive x-axis with a constant

angular velocity staring from the origin. If you mark the point on the circle coinciding

with the origin at the initial time and follow the trajectory of this point

7

[http://upload.wikimedia.org/wikipedia/commons/6/69/

Cycloid_f.gif], its x and y coordinates of this point are given by

x = r (� � sin (�)) ; y = r (1� cos (�)) :

where � is the angle that the circle (the point) rotated:For example the �gure below

shows this trajectory for a point on a circle of unit radius (r = 1).

For a given �, the circle�s centre lies at

x = r�; y = r:

On the other hand if the circles center is

x = r�; y = �r:

then the trajectory looks like the �gure shown below

8

� The Case of Multiple Dependent Functions: Suppose we are given a function, F , thatdepends on y; z; dy=dx; dz=dx, and x, and we want to �nd two curves y = y(x) and

z = z(x) which make

I =

Z x2

x1

F (x; y; z; y0; z0) dx

stationary, then we must solve the Euler-Lagrange equations

@

@x

�@F

@y0

�� @F

@y= 0

@

@x

�@F

@z0

�� @F

@z= 0

~ The Hamiltonian (H): The sum of the kinetic energy (T ) and potential energy (V )

H = T + V

~ The Lagrangian (L): The kinetic energy minus the potential energy

L = T � V

~ The classical action (S): the integral of the Lagrangian with respect to time over a given

period of time

S =

Z t2

t1

Ldt

9

~ Hamilton�s Principle: The motion of a given system from time t1 to time t2 is such that

the classical action

S =

Z t2

t1

Ldt

has a stationary value for the correct path of the motion. The path actually followed

by a system, as speci�ed in terms of the generalized coordinates qi, is that path that

makes the action integral stationary:

�I = �

Z t2

t1

L(qi; _qi; t) dt = 0

for i = 1; 2; 3:::n:This means that the Lagrangian must satisfy the set of equations

@

@t

�@L@ _qi

�� @L@qi

= 0

for i = 1; 2; 3:::n:

Ex. 4 Use Lagrange�s equations to �nd the equation of motion for a particle traveling along

the x-y plane under the in�uence of a potential energy function U(x).

Sol: The kinetic energy of a particle moving in the x-y plane is can be expressed as

T =1

2m�v2x + v2y

�=1

2m�_x2 + _y2

�Then the Lagrangian

L = T � U =1

2m�_x2 + _y2

�� U(x):

Since L = L (x; y; _x; _y; t) = 12m ( _x2 + _y2) � U(x) which means it is a function of two

variables and we must have two Euler-Lagrange equations

@

@t

�@L@ _x

�� @L@x

= 0

@

@t

�@L@ _y

�� @L@y

= 0:

Therefore, using the Lagrangian we �nd

@

@t

�@L@ _x

�� @L@x

= 0) @

@t(m _x) +

@U (x)

@x= 0) m�x = �@U (x)

@x

@

@t

�@L@ _y

�� @L@y

= 0) m�y = 0 (Zero acceleration)

10

Ex. 5 The Atwood�s Machine: A string passes over a frictionless pulley connecting two

masses, m1and m2. Find an expression for the acceleration of the masses in the

system.

Sol: Let�s de�ne the origin of the y axis at center of mass m2; m1 > m2; and the length of

the string is l: At a given time t let the position of m1 and m2 be y1 and y2; then the

kinetic energy of the system can be expressed as

T =1

2m1 _y

21 +

1

2m2 _y

22

and the gravitational potential energy

V = m1gy1 +m2gy2

Then the Lagrangian

L = T � V

of the system can be written as

L = 1

2m1 _y

21 +

1

2m2 _y

22 �m1gy1 �m2gy2:

This equation appears to be a function of two variables. However, because of the

constraint

y1 + y2 + l = C;

where C is a constant, we end up with a Lagrangian that depends on only one variable.

If we replace

y2 = C � y1 � l) _y2 = � _y1

11

we have

L = 1

2(m1 +m2) _y

21 �m1gy1 �m2g (C � y1 � l) :

which we may write as

L = 1

2(m1 +m2) _y

21 � (m1 �m2) gy1 � C1:

where we replaced C1 = m2g (C � l) : Now using

@

@t

�@L@ _qi

�� @L@qi

= 0

for qi = y1 and

@L@y1

= � (m1 �m2) g

@L@ _y1

= (m1 +m2) _y1

we �nd@

@t[(m1 +m2) _y1] = � (m1 �m2) g ) a1 = �y1 = �

m1 �m2

m1 +m2

g

Recalling that

_y2 = � _y1

the acceleration of the second mass becomes

a1 =m1 �m2

m1 +m2

g:

The minus sign indicates the �rst mass is accelerating in the negative y-direction.

Ex. 6 Central Forces: Describe the properties of the motion of a mass m moving under the

in�uence of a central force (that is, a force acting only along the radial direction) given

by

~F = f(r)r

for some function f(r). Assume that the motion is con�ned to a plane.

Sol: The kinetic energy

T =1

2mv2:

Using polar coordinates the magnitude of the velocity can be expressed as

v = _r2 + r2 _�2

12

and the kinetic energy becomes

T =1

2m�_r2 + r2 _�2

�:

The potential energy is related to the central force by

~F = �r � U (r)

where U (r) is the potential energy. Since the force is a central force it is directed

along the radial direction and it depends on r only. Therefore the potential energy

can be expressed as

U (r) = �Zf(r) dr:

Then the Lagrangian can be expressed as

L�t; r; _r; �; _�

�= T � U =

1

2m�_r2 + r2 _�2

�+

Zf(r) dr

Then using the Euler-Lagrange�s equation

@

@t

�@L@ _qi

�� @L@qi

= 0

we have

@

@t

�@L@ _�

�� @L@�

= 0

@

@t

�@L@ _r

�� @L@r

= 0

so that using

@L@�

= 0;@L@ _�

= mr2 _�;@L@r

= mr _�2 + f(r);

@L@ _r

= m _r

we �nd

@

@t

�@L@ _�

�= 0) @L

@ _�= const) mr2 _� = cont) I! = cons:

(Conservation of Ang. Mom.)

@

@t

�@L@ _r

�� @L@r

= 0) m�r = mr _�2 + f(r)

13

II. LECTURE 3 INTRODUCTION TO THE EIGENVALUE PROBLEM

� A Brief Matrix Review [Lecture 8 PHYS 3150]:

Matrix Arithmetic and Manipulation: Consider the following matrices:

A =

0@ 2 3 12 1 0

1A ; B =

0BBB@2 4

1 �13 �1

1CCCA

C =

0BBB@2 1 3

4 �1 �2�1 0 1

1CCCA ; D =

0BBB@�2 0 1

1 �1 23 1 0

1CCCA~ Multiplication by a Scalar : Any matrix can be multiplied by a scalar:

2A =

0@ 2� 2 3� 2 1� 2 �4� 22� 2 1� 2 0� 2 5� 2

1A2A =

0@ 4 6 2 �84 2 0 10

1A~ Addition and subtraction: Two matrices can be added or subtracted if and only if they

have the same dimensions. From matrices A;B;C, and D we can add/ subtract only

matrices C and D

C +D =

0BBB@2 1 3

4 �1 �2�1 0 1

1CCCA+0BBB@�2 0 1

1 �1 23 1 0

1CCCA =

0BBB@2� 2 1 + 0 3 + 1

4 + 1 �1� 1 �2 + 2�1 + 3 0 + 1 1 + 0

1CCCA

=

0BBB@0 1 4

5 �2 02 1 1

1CCCA~ Matrix Multiplication: two matrices can be multiplied if and only if the number of columns

of the �rst matrix is equal to the number of rows of the second matrix. If matrices

have the same dimension, then they can be multiplied. From the above matrices we

14

can make the multiplications:

AB =

0@ 2 3 12 1 0

1A0BBB@2 4

1 �13 �1

1CCCA =

0@ (ab)11 (ab)12(ab)21 (ab)22

1A

CD =

0BBB@2 1 3

4 �1 �2�1 0 1

1CCCA0BBB@�2 0 1

1 �1 23 1 0

1CCCA =

0BBB@(cd)11 (cd)12 (cd)13

(cd)21 (cd)22 (cd)23

(cd)31 (cd)32 (cd)33

1CCCAbut we can not make the matrix multiplications BC or BD

The element in row i and column j of the product matrix AB is equal to row i of A

times column j of B. In index notation

(ab)ij =nXk=1

aikbkj

where (ab)ij is the element of the product matrix AB:

~ Commutativities: For any two multiplyable matrices C and D;

CD 6= DC

~ Commutator : For square matrices C and D the Commutator [C;D] is de�ned as

[C;D] = CD �DC

~ For any three matrices, F;G;and H that can be multiplied we can write

The Associative Law :

F (GH) = (FG)H

The Distributive Law :

F (G+H) = FG+ FH

~ The Identity Matrix; I (Boas: The Unit Matrix, U )

IA = AI = A

15

~ Transpose of a Matrix : The transpose of the matrix A is denoted by AT :

A =

0@ 2 3 12 1 0

1A ; AT =

0BBB@2 2

3 1

1 0

1CCCA

B =

0BBB@2 4

1 �13 �1

1CCCA ; BT =

0@ 2 1 3

4 �1 �1

1A

~ Adjoint of a Matrix : The adjoint of a square matrix, A, is given by

adj(A) = [cof(A)]T

where cof(A) is the cofactor of the matrix A. We recall that the minor of matrix A

(Mij) is the determinant of the matrix formed from matrix A by removing the ith row

and jth column. For the cofactor matrix the elements are expressed as

[cof(A)]ij = (�1)i+jMij:

~ Inverse of a (Square) Matrix : A�1 (The inverse need not exist.)

A�1A = AA�1 = I

We can determine the inverse of an invertible matrix (det jAj 6= 0) using row reductionor the adjoint matrix.

a. Row reduction in this approach for the matrix, for example,

A =

26664a11 a12 a13

a21 a22 a23

a31 a32 a33

37775we start from 26664

a11 a12 a13

a21 a22 a23

a31 a32 a33

��1 0 0

0 1 0

0 0 1

37775

16

and do elementary row operation until we end up with266641 0 0

0 1 0

0 0 1

��b11 b12 b13

b21 b22 b23

b31 b32 b33

37775so that the inverse of the Matrix A is given by

A�1 =

0BBB@b11 b12 b13

b21 b22 b23

b31 b32 b33

1CCCA :

b. Using the adjoint matrix: Using the adjoint matrix the inverse can be expressed as

A�1 =[cof (A)]T

det jAj

Ex 9.1 Find the inverse of the matrix

A =

0BBB@�1 2 3

2 0 �4�1 �1 1

1CCCAusing

a. Row reduction approach

b. The adjoint matrix approach

Sol: a. In the row reduction approach we start from26664�1 2 3

2 0 �4�1 �1 1

��1 0 0

0 1 0

0 0 1

37775and try to get 26664

1 0 0

0 1 0

0 0 1

��a11 a12 a13

a21 a22 a23

a31 a32 a33

37775

17

so that we can get the inverse matrix

A�1 =

0BBB@a11 a12 a13

a21 a22 a23

a31 a32 a33

1CCCA :

b. Find the cofactor Cij of the element Aij in row i and column j which is equal

to (-1)i+j times the value of the determinant remaining when we cross o¤ row i

and column j. After you obtained all elements of the cofactor matrix write the

cofactor matrix C, transpose and divide the cofactor matrix with the determinant

of matrix A. The resulting matrix is A�1:

A�1 =1

jAjCT :

Ans:

A�1 = �12

0BBB@4 5 8

�2 �2 �22 3 4

1CCCA :

� Orthogonal Matrices: matrices that make an orthogonal transformation of vectors.In an orthogonal transformation of vectors the magnitude of the vectors remains the

same. For an orthogonal matrix

M�1 =MT

~ The Rotation Operator : For a counter-clockwise rotation about the z-axis by an

angle �, we denote the rotation matrix by R:Rz(�) = R. Then,

r0 = Rr

)

0BBB@x0

y0

z0

1CCCA =

0BBB@cos � sin � 0

� sin � cos � 00 0 1

1CCCA0BBB@x

y

z

1CCCA� Eigenvalues and Eigenvectors: In general for any linear transformation of the vector~r to a vector ~r0 by the transformation matrix M may be written as

~r0 =M~r

18

or 0BBB@x0

y0

z0

1CCCA =

0BBB@M11 M12 M13

M21 M22 M23

M31 M32 M33

1CCCA0BBB@x

y

z

1CCCA :

Under this transformation if there are new vectors ~r0 which are related to the vector

~r by

~r0 = �~r

where � is a constant, then the vector ~r is called the eigenvector (characteristic vector)

and � is the eigenvalue (characteristic value) of the Matrix M:Which means

M~r = �~r )

0BBB@M11 M12 M13

M21 M22 M23

M31 M32 M33

1CCCA0BBB@x

y

z

1CCCA = �

0BBB@x

y

z

1CCCAUsing the identity matrix I

I =

0BBB@1 0 0

0 1 0

0 0 1

1CCCAwe can put the above expression in the following form0BBB@

M11 M12 M13

M21 M22 M23

M31 M32 M33

1CCCA0BBB@x

y

z

1CCCA = �

0BBB@1 0 0

0 1 0

0 0 1

1CCCA0BBB@x

y

z

1CCCA =

0BBB@� 0 0

0 � 0

0 0 �

1CCCA0BBB@x

y

z

1CCCAwhich can be rewritten as26664

0BBB@M11 M12 M13

M21 M22 M23

M31 M32 M33

1CCCA�0BBB@� 0 0

0 � 0

0 0 �

1CCCA377750BBB@x

y

z

1CCCA = 0

)

0BBB@M11 � � M12 M13

M21 M22 � � M23

M31 M32 M33 � �

1CCCA0BBB@x

y

z

1CCCA = 0:

The eigenvalues are obtained from the condition��M11 � � M12 M13

M21 M22 � � M23

M31 M32 M33 � �

�� = 019

which is known as the eigenvalue equation (characteristic equation). To �nd the eigen-

vectors we substitute the eigenvalues and solve the resulting equations.

From The VNR Concise Encyclopedia of Mathematics (Van Nostrand Reinhold Co.,

publishers, 1977):

Eigenvalues:

Eigenvalue problems are important in many branches of physics. They make it possible to

�nd coordinate systems in which the transformations in question take on their simplest forms.

In mechanics for instance, the principal moments of a rigid body are found with the help

of the eigenvalues of the symmetric matrix representing the inertia tensor.... Eigenvalues

are of central importance in quantum mechanics, in which the measured values of physical

�observables�appear as the eigenvalues of certain operators. The term �transformation� is

used predominantly in pure mathematical (geometrical) context, whereas �operator�is more

customary in applications (physics, technology).

Ex. 7 Find the eigenvalues and the corresponding eigenvectors of the matrix

M =

0BBB@0 1 0

1 0 0

0 0 0

1CCCASol: The eigenvalue equation��

0� � 1 0

1 0� � 0

0 0 0� �

�� = 0) ��

�� 0

0 ��

�� 1 0

0 ��

�� = 0) ��3 + � = 0) �1 = 0; �2 = 1; �3 = �1:

The corresponding eigenvectors are determined from0BBB@M11 � �i M12 M13

M21 M22 � �i M23

M31 M32 M33 � �i

1CCCA0BBB@xi

yi

zi

1CCCA = 0

which gives, for �1 = 00BBB@0 1 0

1 0 0

0 0 0

1CCCA0BBB@x1

y1

z1

1CCCA = 0) x1 = 0; y1 = 0;

20

Then the eigenvector for �1 = 0 would be

j�1i =

0BBB@0

0

z1

1CCCA :

Normalizing this vector

h�1 j�1i = 1)�0 0 0

�0BBB@0

0

z1

1CCCA = 1) z1 = 1) j�1i =

0BBB@0

0

1

1CCCASimilarly for �2 = 1 0BBB@

�1 1 0

1 �1 0

0 0 �1

1CCCA0BBB@x2

y2

z2

1CCCA = 0

) �x2 + y2 = 0; x2 � y2 = 0; z2 = 0) x2 = y2; z2 = 0

) j�2i = x2

0BBB@1

1

0

1CCCA) j�2i =1p2

0BBB@1

1

0

1CCCAand for �3 = �1 0BBB@

1 1 0

1 1 0

0 0 1

1CCCA0BBB@x3

y3

z3

1CCCA = 0

) x3 + y3 = 0; x3 + y3 = 0; z3 = 0) x3 = �y3; z3 = 0

) j�3i = y3

0BBB@�11

0

1CCCA) j�3i =1p2

0BBB@�11

0

1CCCA :

Using Mathematica:

21

� Hermitian Matrices, their Eigenvalues, and their Eigenvectors: Hermitian Matricesare matrices which satisfy

(M�)T =M:

The eigenvalues area real and the eigenvectors are orthogonal.

� The Similarity Transformation: The similarity transformation of the matrix M is

given by

D � C�1MC:

where C is a matrix whose columns are the eigenvectors of the Eigenvalue equation

for matrixM . The matrix D is a diagonal matrix where the diagonal elements are the

eigenvalues.

22

III. LECTURE 4 APPLICATIONS OF THE EIGENVALUE PROBLEM: NOR-

MAL MODES OF VIBRATION

Ex. 7 Consider a system consisting of two equal masses m connected by three identical

springs of spring constant k.

The masses can slide on a horizontal, frictionless surface. The springs are at their

unstretched/uncompressed lengths when the masses are at their equilibrium positions.

At t = 0, the masses are displaced from their equilibrium positions by the amounts

x10 and x20 and released from rest, as shown in the �gure above. Completely describe

the resulting motion.

� The Equations of Motion:

� Similarity Transformation and the Eigenvalue Problem (The Eigenvalues and Eigen-

vectors):

� Solving the Decoupled Transformed Equations of Motion:

� The Propagator Matrix, U :

� The Normal Modes of Vibration:

Sol:

� The Equations of Motion: To �nd the equations of motion I will use Euler-Lagrangeequations. But you can use Newton�s second law. To use the Euler-Lagrange equation

we need to �nd the kinetic energy and the potential energy. Suppose at a given instant

of time the position of the �rst mass is x1 and the second mass is x2 as measured from

23

their corresponding equilibrium positions. Then the corresponding kinetic energy can

be expressed as

K1 =1

2m _x21; K2 =

1

2m _x22

so that the total kinetic energy be

K =1

2m�_x21 + _x22

�:

The potential energy (elastic) is due to the displacement of the springs from their

equilibrium position. If the �rst mass is displaced by x1from the equilibrium and the

second mass by x2 in the positive x direction, then the �rst spring will be stretched

by x1; the second spring will be compressed by x1 and at the same time it will be

stretched by x2; as a result the net displacement will be x2� x1: The third spring willbe compressed by x2: Therefore, the total potential energy will be

K =1

2kx21 +

1

2kx22 +

1

2k (x2 � x1)

2 :

Then the Lagrangian

L (t; x1; x2; _x1; _x2) = T � U

) L (t; x1; x2; _x1; _x2) =1

2m�_x21 + _x22

�� 12kx21 �

1

2kx22 �

1

2k (x2 � x1)

2

which leads to

@L@x1

= �kx1+k (x2 � x1) ;@L@x2

= �kx2�k (x2 � x1) ;@

@t

�@L@ _x1

�= m�x1;

@

@t

�@L@ _x2

�= m�x2:

Using the Euler-Lagrange�s equation

@

@t

�@L@ _xi

�� @L@xi

= 0 for i = 1; 2

we �nd

m�x1 � [�kx1 + k (x2 � x1)] = 0) m�x1 = �2kx1 + kx2

and

m�x2 � [�kx2 � k (x2 � x1)] = 0) m�x2 = �2kx2 + kx1:

If we introduce a constant

! =

rk

m;

24

then the above two equations can be put in the form

�x1 = �2!2x1 + !2x2

and

�x2 = !2x1 � 2!2x2:

These two equations describe the equations of motion for the two masses. As you can

see the two equations are coupled second order di¤erential equations.

� Similarity Transformation and the Eigenvalue Problem: In a matrix form the two

equations can be put in the form0@ �x1

�x2

1A =

0@ �2!2 !2

!2 �2!2

1A0@ x1

x2

1Aor

::

~r =M~r

where

~r = x1e1 + x2e2;

M =

0@ �2!2 !2

!2 �2!2

1A ;

and

e1 =

0@ 10

1A ; e2 =

0@ 01

1A :

We recall that for a matrix M the similarity transformation is given by

D = C�1MC;

where C is a matrix whose columns are the eigenvectors of the Eigenvalue equation

for matrixM . The matrix D is a diagonal matrix where the diagonal elements are the

eigenvalues. Suppose if we can �nd eigenvectors ~R such that

M ~R = �~R

then the eigenvalue equation can be written as

det

�� 2!2 � � !2

!2 �2!2 � �

�� = 0) �2!2 + �

�2 � �!2�2 = 0)�2!2 + �� !2

� �2!2 + �+ !2

�= 0) �1 = �!2; �2 = �3!2:

25

The corresponding eigenvectors are obtained from0@ �2!2 � �1 !2

!2 �2!2 � �1

1A0@ X1

X2

1A = 0

and 0@ �2!2 � �2 !2

!2 �2!2 � �2

1A0@ X1

X2

1A = 0:

Solving these equations for �1 = �!2 we �nd0@ �!2 !2

!2 �!2

1A0@ X1

X2

1A = 0) �X1 +X2 = 0) X2 = X1

for �2 = �3!2 0@ !2 !2

!2 !2

1A0@ X1

X2

1A = 0) X1 +X2 = 0) X2 = �X1:

Then the two eigenvectors will be

~R1 = X1

0@ 11

1A~R2 = X1

0@ 1

�1

1A :

If we normalize these vectors, we �nd

R1 =1p2

0@ 11

1Aand

R2 =1p2

0@ 1

�1

1A :

Note: using Mathematica you can determine the eigenvalues and unnormalized eigen-

vectors as follows:

26

� Solving the Decoupled Transformed Equations of Motion: The matrix C and its inverseC�1 can be expressed as

C =

0@ 1p2

1p2

1p2� 1p

2

1A ; C�1 =

0@ 1p2

1p2

1p2� 1p

2

1A :

N.B. You must use the procedure we studied to �nd the inverse of the matrix C. But,

here I am going to use mathematica to �nd the inverse

27

We note that

CC�1 =

0@ 1p2

1p2

1p2� 1p

2

1A0@ 1p2

1p2

1p2� 1p

2

1A) CC�1 =

0@ 1 00 1

1A = C�1C = I

so that the equation::

~r =M~r

can be written as

C�1::

~r = C�1M~r

) C�1::

~r = C�1MCC�1~r:

But we also know that

D = C�1MC =

0@ �1 0

0 �2

1A) D = C�1MC =

0@ �!2 0

0 �3!2

1A :

If we introduce the variables de�ned by

C�1~r = C�1

0@ x1

x2

1A =

0@ y1

y2

1A) C�1::

~r = C�1

0@ �x1

�x2

1A =

0@ �y1

�y2

1A ;

we can express the equations of motion as0@ �y1

�y2

1A =

0@ �!2 0

0 �3!2

1A0@ y1

y2

1A) �y1 = �!2y1; �y2 = �3!2y2

so that the solutions can be expressed as

y1 = A cos (!t) +B sin (!t) ; y2 = C cos�p3!t�+D sin

�p3!t�:

Now substituting these back into

C�1

0@ x1

x2

1A =

0@ y1

y2

1Awe �nd 0@ 1p

21p2

1p2� 1p

2

1A0@ x1

x2

1A =

0@ A cos (!t) +B sin (!t)

C cos�p3!t�+D sin

�p3!t�1A

28

Using the initial conditions x1 = x10; x2 = x20; we �nd0@ 1p2

1p2

1p2� 1p

2

1A0@ x10

x20

1A =

0@ A

C

1A) A =1p2(x10 + x20) ; C =

1p2(x10 � x20)

and using _x1 = _x2 = 0, since both masses released from rest, we have0@ 1p2

1p2

1p2� 1p

2

1A0@ _x1

_x2

1A =

0@ �!A sin (!t) + !B cos (!t)

�p3!C sin

�p3!t�+p3!D cos

�p3!t�1A

)

0@ 00

1A =

0@ !Bp3!D

1A) B = 0; D = 0

Therefore 0@ 1p2

1p2

1p2� 1p

2

1A0@ x1

x2

1A =

0@ 1p2(x10 + x20) cos (!t)

1p2(x10 + x20) cos

�p3!t�1A :

Once again recalling that

CC�1 =

0@ 1p2

1p2

1p2� 1p

2

1A0@ 1p2

1p2

1p2� 1p

2

1A) CC�1 =

0@ 1 00 1

1A = C�1C = I

we may write 0@ x1

x2

1A =

0@ 1p2

1p2

1p2� 1p

2

1A0@ 1p2(x10 + x20) cos (!t)

1p2(x10 � x20) cos

�p3!t�1A

x1 =1

2(x10 + x20) cos (!t) +

1

2(x10 � x20) cos

�p3!t�

x2 =1

2(x10 + x20) cos (!t)�

1

2(x10 � x20) cos

�p3!t�:

� The Propagator Matrix, U : If we simplify the above expressions, we �nd

x1 =1

2

�cos (!t) + cos

�p3!t��

x10 +1

2

�cos (!t)� cos

�p3!t��

x20

x2 =1

2

�cos (!t)� cos

�p3!t��

x10 +1

2

�cos (!t) + cos

�p3!t��

x20

which can be put in a matrix form as0@ x1

x2

1A =1

2

0@ cos (!t) + cos �p3!t� cos (!t)� cos �p3!t�cos (!t)� cos

�p3!t�cos (!t) + cos

�p3!t�1A0@ x10

x20

1A ;

29

or 0@ x1

x2

1A = U(t)

0@ x10

x20

1A) ~r = U(t)~r0;

where

U(t) =1

2



�p3!t�1A

is called the propagator.

� The Normal Modes of Vibration: Suppose the initial state of the two masses is de-scribed by the �rst eigenvector. That means

~r0 =

0@ x10

x20

1A =1p2

0@ 11

1Athen

~r = U(t)~r0

gives 0@ x1

x2

1A =1

2



�p3!t�1A 1p

2

0@ 11

1Awhich leads to 0@ x1

x2

1A =1p2

0@ cos (!t)cos (!t)

1A) x1 (t) = x2 (t) :

The two masses oscillate with a frequency ! in the same direction. On the other hand,

if initially state of the two masses is given by the second eigenvector

~r0 =

0@ x10

x20

1A =1p2

0@ 1

�1

1A :

then we have0@ x1

x2

1A =1

2



�p3!t�1A 1p

2

0@ 1

�1

1Awhich gives 0@ x1

x2

1A =1p2

0@ cos�p3!t�

� cos�p3!t�1A) x1 (t) = �x2 (t) :

The two masses oscillate with a frequencyp3! out of phase by 180�:

The two modes of vibrations we saw above are called Normal Modes of vibration.

30

IV. LECTURE 5: SPECIAL INTEGRAL FUNCTIONS: THE GAMMA, BETA,

AND ERROR FUNCTIONS

� The Factorial, n!: We recall the factorial function n! for a positive integer or zero isde�ned as

n! = n� (n� 1)� (n� 2)(n� 3)(n� 4):::3� 2� 1� 0!

where

0! = 1:

� The integral form of the Factorial function : Consider the integral function given by

F (p) =

Z 1

0

e��xdx:

For any real number � > 0, the value of this integral is

F (p) =

Z 1

0

e��xdx = �e��x

�

��10

=1

�:

Now let�s di¤erentiate this integral with respect to � as many as we can, say m time

(i.e. @n

@�n). For the �rst derivative n = 1Z 1

0

e��xdx =1

�Z 1

0

x0e��xdx =0!

�1) @

@�

�Z 1

0

e��xdx =1

�

�)Z 1

0

@

@�

�e��x

�dx =

@

@�

�1

�

�Z 1

0

�xe��xdx = � 1

�2

This can be put in the form Z 1

0

x1e��xdx =1!

�2:

For the second derivative (n = 2)

@2

@�2

�Z 1

0

e��xdx

�=

@2

@�2

�1

�

�) @

@�

�Z 1

0

�xe��xdx�=

@

@�

�� 1!�2

�)Z 1

0

x2e��xdx =2� 1!�3

=2!

�3:

For the third derivative (n = 3)

@3

@�3

�Z 1

0

e��xdx

�=

@3

@�3

�1

�

�) @

@�

�Z 1

0

x2e��xdx

�=

@

@�

�2!

�3

�)Z 1

0

�x3e��xdx = �3� 2� 1!�3

)Z 1

0

x3e��xdx =3!

�3:

31

Therefore it is not di¢ cult to see for the nth derivative we �ndZ 1

0

xne��xdx =n!

�n:

Now we set � = 1, we �nd

n! =

Z 1

0

xne�xdx

which is the integral form of the Factorial function which is valid for n zero or positive

integer.

� The Gamma function: If we replace n in the integral functionZ 1

0

xne�xdx

by p � 1 for any real number p > 0, we �nd a more generalized function know as theGamma Function given by

�(p) =

Z 1

0

xp�1e�xdx:

If we replace p by n+ 1; where n is zero or any positive integer, then we �nd

�(n+ 1) =

Z 1

0

xne�xdx:

which is the Factorial function we obtained earlier. Therefore, the factorial function

in terms of the Gamma function can be expressed as

n! = �(n+ 1) =

Z 1

0

xne�xdx:

� The Recursion Relation for the Gamma Function: If we replace p by p + 1 in theexpression for the Gamma function

�(p) =

Z 1

0

xp�1e�xdx:

we get

�(p+ 1) =

Z 1

0

xpe�xdx:

If we denote

u = xp; dv = e�xdx) du = pxp�1; v = �e�x

32

then using integration by parts Zudv = uv �

Zvdu

we �nd

�(p+ 1) = �e�xxp��10�Z 1

0

pxp�1��e�x

�dx:

In the above expression the �rst term is zero

�(p+ 1) = p

Z 1

0

xp�1e�xdx:

We know that

�(p) =

Z 1

0

xp�1e�xdx

hence

�(p+ 1) = p�(p):

Ex. 1 Show that

�

�1

2

�=p�

and Z 1

0

e�x2

dx =��12

�2

Using the de�nition of Gamma function, we can write

�

�1

2

�=

Z 1

0

1pxe�xdx .

if we introduce a new variable de�ned by

u2 = x) 2udu = dx

we can write

�

�1

2

�=

Z 1

0

1

ue�u

2

2udu = 2

Z 1

0

e�u2

du .

If we square both sides, we have

�2�1

2

�= 4

Z 1

0

Z 1

0

e�u2

due�v2

dv ) �2�1

2

�= 4

Z 1

0

Z 1

0

e�(u2+v2)dudv

Introducing polar coordinates de�ned by

u = r cos �; v = r sin � ) dudv = rdrd�; u2 + v2 = r2

33

the above integral can be put in the form

�2�1

2

�= 4

Z 1

0

e�r2

rdr

Z �=2

0

d�:

Here the limits of integration for � is �=2 since both u and v are positive we must integrate

only in the �rst quadrant.

Therefore, integrating with respect to r and � leads to

�2�1

2

�= 4 �e

�r2

2

��1

0

�

2= � ) �

�1

2

�=p�:

Now substituting this result into

�

�1

2

�= 2

Z 1

0

e�u2

du

we can see that Z 1

0

e�u2

du =

r�

2:

� The Beta Function: For p > 0 and q > 0, the beta function B(p; q) is de�ned by a

de�nite integral

B(p; q) =

Z 1

0

xp�1 (1� x)q�1 dx:

The are di¤erent forms of representations of the beta function. These includes the fol-

lowing

~ Replace x = y=a) dx = dy=a

B(p; q) =1

a

Z a

0

�ya

�p�1 �1� y

a

�q�1dy ) B(p; q) =

1

ap+q�1

Z a

0

yp�1 (a� y)q�1 dy:

~ Replace x = sin2 (�)) dx = 2 sin (�) cos (�) d�

B(p; q) =

Z �=2

0

sin2p�2 (�)�1� sin2 (�)

�q�12 sin (�) cos (�) d�

B(p; q) = 2

Z �=2

0

sin2p�1 (�) cos2q�1 (�) d�

~ Replace x = y= (1 + y) ) dx = dy= (1 + y) � ydy= (1 + y)2 = dy= (1 + y)2 and use the

limits of integration

x = 0) y = 0

x = 1) y

1 + y= 1, lim

y!1

y

1 + y= 1

34

and we �nd

B(p; q) =

Z 1

0

�y

1 + y

�p�1�1� y

1 + y

�q�1dy

(1 + y)2

) B(p; q) =

Z 1

0

�y

1 + y

�p�1�1

1 + y

�q�1dy

(1 + y)2) B(p; q) =

Z 1

0

yp�1dy

(1 + y)p+q

Ex. 2 Prove that the Gamma and the Beta Functions are related by

B(p; q) =�(p)�(q)

�(p+ q):

Proof: For the Gamma functions

�(p) =

Z 1

0

xp�1e�xdx ,�(q) =Z 1

0

yq�1e�ydy

if we use new variables de�ned by

u2 = x) 2udu = dx; v2 = y ) 2vdv = dy

we �nd

�(p) = 2

Z 1

0

u2p�1e�u2

du , �(q) = 2Z 1

0

v2q�1e�v2

dv .

Multiplying the two functions, we have

�(p)�(q) = 4

Z 1

0

Z 1

0

u2p�1v2q�1e�(u2+v2)du dv,

so that using the polar coordinates

v = r cos �; u = r sin � ) dudv = rdrd�; u2 + v2 = r2

we �nd

�(p)�(q) = 4

Z 1

0

Z �=2

0

(r sin �)2p�1 (r cos �)2q�1 e�r2

rdrd� .

This can be rewritten as

�(p)�(q) = 4

Z 1

0

r2(p+q�1)e�r2

rdr

Z �=2

0

(sin �)2p�1 (cos �)2q�1 d� .

The �rst integral

4

Z 1

0

r2(p+q�1)e�r2

rdr = 2

Z 1

0

R(p+q�1)e�RdR = 2� (p+ q)

35

and using

B(p; q) = 2

Z �=2

0

sin2p�1 (�) cos2q�1 (�) d�

we note that the second integralZ �=2

0

(sin �)2p�1 (cos �)2q�1 d� =B(p; q)

2:

Therefore

B(p; q) =�(p)�(q)

� (p+ q).

� The Error Function: the error function is de�ned as the area under

erf(x) =2p�

Z x

0

e�t2

dt

There are also other related integrals which sometimes referred as error functions.

These includes:

(a) The standard normal or Gaussian cumulative distribution function �(x)

�(x) =1p2�

Z x

�1e�t

2=2dt =1

2+1

2erf�x=p2�

Noting that

1p2�

Z x

�1e�t

2=2dt =1p2�

Z 0

�1e�t

2=2dt+1p2�

Z x

0

e�t2=2dt

) 1p2�

Z x

�1e�t

2=2dt =1p2�

Z 1

0

e�t2=2dt+

1p2�

Z x

0

e�t2=2dt

) 1p2�

Z x

�1e�t

2=2dt =1p2�

r�

2+

1p2�

Z x

0

e�t2=2dt

) 1p2�

Z x

�1e�t

2=2dt =1

2+

1p2�

Z x

0

e�t2=2dt

the error function is given by

�(x)� 12=1

2erf�x=p2�=

1p2�

Z x

0

e�t2=2dt:

(b) The complementary error function:

erf c(x) =2p�

Z 1

x

e�t2=2dt = 1 + erf

�x=p2�

erf c(xp2) =

r2

�

Z 1

x

e�t2=2dt

36

Ex. 3 Consider a criterion that either is or is not satis�ed. We look at a system that has

many elements, each of which satis�es or does not satisfy the criterion.

For example: Consider a test with many multiple-choice questions. Criterion: the

answer to a test question is correct. Each answer on the test is either correct (satis�es

criterion) or is incorrect (does not satisfy the criterion).

We then look at a large number of these systems (for example, a large number of tests

consisting of multiple-choice questions). We let x represent the number of elements

within a given system satisfying the criterion. We then de�ne the following:

x � The average number of elements satisfying the criterion

� � The standard deviation about the mean of the number of elements satisfying thecriterion.

The probability that any one system will have x to x + dx elements satisfying the

criterion is then given by the Gaussian distribution:

�(x)dx =1p2��

e�(x�x)2

2�2 dx

Find an expression in terms of the error function for the probability that the number

of elements of a given system satisfying the criterion, x, will be in the range

x� n� � x � x+ n�

for some real value of n (usually integral).

Sol: The probability that one system will have x to x+dx elements satisfying the criterion

is

�(x)dx =1p2��

e�(x�x)2

2�2 dx

then the probability that the number of elements in the range x � n� � x � x + n�

satisfying the criterion will be

Pn (x) =

Z x+n�

x�n��(x)dx =

Z x+n�

x�n�

1p2��

e�(x�x)2

2�2 dx:

Introducing a new variable de�ned by

x� xp2�

= y ) dx =p2�dy

37

and noting that for x1 = x� n� and x2 = x+ n�

y1 =x1 � xp2�

=x� n� � xp

2�=�np2

y2 =x2 � xp2�

=x+ n� � xp

2�=

np2

we can write

Pn (x) =

Z np2

�np2

1p2��

e�y2p2�dy

Pn (x) =1p�

Z np2

�np2

e�y2

dy:

If we split the integral in to two regions��np2; 0�and

�0; np

2

�; we have

Pn (x) =1p�

"Z 0

�np2

e�y2

dy +

Z np2

0

e�y2

dy

#

and noting that Z 0

�np2

e�y2

dy =

Z 0

np2

e�y2

d (�y) =Z np

2

0

e�y2

dy

we �nd

Pn (x) =2p�

Z np2

0

e�y2

dy:

Recalling the de�nition for the error function

erf(x) =2p�

Z x

0

e�t2

dt

we �nd that

Pn (x) = erf

�np2

�:

You can get the values of the error function for di¤erent values of n using, for example,

Mathematica. You will �nd the following results

38

n Pn (x)

1 68:26%

2 95:44%

3 99:74%

39

V. LECTURE 6 STIRLING�S FORMULA AND ELLIPTIC INTEGRALS

� We recall the Gamma Function

�(p) =

Z 1

0

xp�1e�xdx;

when p is zero or positive integer, gives the factorial function

p! = �(p+ 1) =

Z 1

0

xpe�xdx:

We will get an approximate formula when p is very large. This approximate formula

is known as Stirling�s formula and is given by

p! = �(p+ 1) �= ppe�pp2�p;

or if we take the natural logarithm of both sides

ln(p!) �= ln�ppe�p

p2�p�= ln (pp) + ln

�e�p�+ ln

�(2�p)

12

�= p ln (p)� p ln (e) +

1

2ln (2�p)

ln(p!) �= p ln p� p+1

2ln (2�p)

If p is very large, the last term is very small as compared to the �rst two terms and

the Stirling�s formula is given by

ln(p!) �= p ln p� p:

Proof: Introducing a new variable de�ned by

x = p+ ypp) dx =

ppdy;

x = 0) y = �pp; x!1) y !1;

the � function

p! = �(p+ 1) =

Z 1

0

xpe�xdx

can be put in the form

p! =

Z 1

�pp(p+ y

pp)p e�(p+y

pp)ppdy:

40

Noting that

(p+ ypp)p = eln[(p+y

pp)p] = ep ln(p+y

pp);

we can write

p! =

Z 1

�ppep ln(p+y

pp)e�(p+y

pp)ppdy ) p! =

pp

Z 1

�ppep ln(p+y

pp)�p�yppdy:

We recall that the Taylor series expansion for f(y) about y = 0 is given by

f(y) = f(0) +1

1!

df(y)

dy

��y=0

y +1

2!

d2f(y)

dy2

��y=0

y2 + :::

so that for f(y) = ln�p+ y

pp�; using

f(0) = ln (p) ;

df(y)

dy

��y=0

=

pp

p+ ypp

��y=0

=

pp

p;

d2f(y)

dy2

��y=0

=d

dy

� pp

p+ ypp

��y=0

= �pppp�

p+ ypp�2��y=0

= �1p;

we �nd an approximate expression

ln (p+ ypp) �= ln (p) +

ppy

p� y2

2p:

Then the approximate expression for the factorial becomes

p! �=pp

Z 1

�ppep ln(p+y

pp)�p�yppdy =

pp

Z 1

�ppe

�p

�ln(p)+

ppy

p� y2

2p

��p�ypp

�dy

=pp

Z 1

�ppe

�p ln(p)�p� y2

2

�dy =

ppep ln(p)�p

Z 1

�ppe�

y2

2 dy

) p! �=ppep ln(p)�p

Z 1

�ppe�

y2

2 dy:

Noting that

41

we may write

p2� =

Z 1

�1e�

y2

2 dy =

Z �pp

�1e�

y2

2 dy +

Z 1

�ppe�

y2

2 dy

)Z 1

�ppe�

y2

2 dy =

Z 1

�1e�

y2

2 dy �Z �pp

�1e�

y2

2 dy )Z 1

�ppe�

y2

2 dy =p2� �

Z �pp

�1e�

y2

2 dy;

we can write

p! �=ppep ln(p)�p

Z 1

�ppe�

y2

2 dy =p2p�ep ln(p)�p �ppep ln(p)�p

Z �pp

�1e�

y2

2 dy:

The second integral

limp!1

Z �pp

�1e�

y2

2 dy ! 0:

Hence, the Stirling�s approximation for the factorial function will be

p! �=p2p�ep ln(p)�p =

p2p�eln(p

p)e�p ) p! �=p2p�ppe�p:

or

p! �= p ln p� p:

Ex. 4 Consider a classroom full of gas molecules. There are approximately N = 5000NA

= 3� 1027 molecules in the room. From the Binomial Theorem, it can be shown that

the probability for n of the molecules to be in the front half and n0 = N�n molecules

to be in the back half of the room is given by

P (n) =

0@ N

n

1A pnqn0=

N !

n! (N � n)!pnqN�n

where p is the probability that a molecule will be found in the front half of the room,

and q is the probability that it will be found in the back half. From the symmetry of

the problem, we must have that

p =1

2; q = 1� p =

1

2= p

On the average, we would expect to �nd half of the molecules in the front half of

the room and the other half of the molecules in the back half of the room. Find the

probability that 0:1% of the molecules in the room have shifted from the front to the

back half of the room. That is, �nd the value of P (n) = P (0:499N).

42

Sol: Since q = p, we can write

P (n) =N !

n! (N � n)!pnpN�n =

N !

n! (N � n)!pN :

On the average there are nave = 0:5N of molecules in the front half of the room and

n0ave = N � nave = 0:5N in the back half of the room. Here we want to �nd the

probability that 0:1% of the molecules shifted to the back half of the room. In other

words we want to determine the probability that the number of molecules in the front

half (nave) is reduced by 0:1%. Which means we want to �nd P (n) for

n = 0:5N � (N � 0:1%)) n = 0:499N

which is given by

P (n) =N !

n! (N � n)!pN :

Obviously, both N and n are very large number and we can make Stirling�s approxi-

mation

lnn! �= n lnn� n

for the factorial. Thus

ln [P (n)] = ln

�N !

n! (N � n)!pN�= lnN !� lnn!� ln(N � n)! + ln pN

can be approximated as

ln [P (n)] �= N lnN �N � (n lnn� n)� ((N � n) ln (N � n)� (N � n)) +N ln p

) ln [P (n)] �= N lnN �N � n lnn+ n�N ln (N � n) + n ln (N � n) +N � n+N ln p

) ln [P (n)] �= N [lnN � ln (N � n) + ln p] + n [ln (N � n)� lnn]

) ln [P (n)] �= N ln

�N

N � np

�+ n ln

�N � n

n

�) ln [P (n)] �= n ln

�N � n

n

��N ln

�N � n

Np

�Now using

N = 3� 1027; n = 0:499N ) N � n = 0:501N; p = 0:5

we �nd

43

and

P (n) �= exp[�6� 1021]:

� The Complete Elliptic Integral of the First Kind

K (�) �Z �=2

0

d'p1� �2 sin2 '

Ex. 5 The Not-So Simple Pendulum...

Consider a simple pendulum with a mass m suspended from the end of a light rigid

rod of length L. We pull the pendulum to the side by an angle � and release it from

rest. Find an expression for the period of the pendulum, T , assuming that � = 0 and

d�=dt > 0 at t = 0, where � is the angle of the pendulum from the vertical. Then �nd

an approximate expression for the period of the pendulum for not-so-small and small

amplitudes of motion.

Sol: Using conservation of Mechanical energy

MEI =ME�

where MEI is the initial mechanical energy (when the pendulum is pulled to the side

by an angle �) which is just only the gravitational potential energy given by

MEI = mghmax = mgl(1� cos�)

and ME� is the mechanical energy at some instant of time (at an angle �) which is

the sum of kinetic and potential energy given by

ME� = mgh+1

2mv2 = mgl(1� cos �) + 1

2ml2

�d�

dt

�2:

Then

MEI =ME�

mgl(1� cos�) = mgl(1� cos �) + 12ml2

�d�

dt

�2) �g cos� = �g cos �) + 1

2l

�d�

dt

�2d�

dt=

r2g

l(cos � � cos�):

44

Noting that the period is the time for one complete oscillation which we can express

as

T = 2

Z t1=2

0

dt

where t = 0 is the time at which the pendulum at the maximum displacement from

the vertical (� = ��) and t = t1=2 is the time at which the pendulum reached to the

position (� = �). Therefore, using

d�

dt=

r2g

l(cos � � cos�)) dt =

d�q2gl(cos � � cos�)

we can write

T = 2

Z t1=2

0

dt = 2

Z �

��

d�q2gl(cos � � cos�)

Using the

cos � = 1� 2 sin2��

2

�;

cos� = 1� 2 sin2��2

�we have r

2g

l(cos � � cos�) =

s2g

l

�1� 2 sin2

��

2

�� 1 + 2 sin2

��2

��

=

s2g

l

�2 sin2

��2

�� 2 sin2

��

2

��so that

T = 2

Z �

��

d�q2gl

�2 sin2

��2

�� 2 sin2

��2

�� =sl

g

Z �

��

d�qsin2

��2

�� sin2

��2

� :Introducing the transformation de�ned by

sin

��

2

�= sin

��2

�sin') 1

2cos

��

2

�d� = sin

��2

�cos'd'

) d� =2 sin

��2

�cos'

cos��2

� d' =2 sin

��2

�cos'q

1� sin2��2

� ) d� =2 sin

��2

�cos'q

1� sin2��2

�sin2 '

d'

) � = ��) sin' = �1) ' = ��2

we have

45

T =

sl

g

Z �=2

��=2

d�qsin2

��2

�� sin2

��2

� =sl

g

Z �=2

��=2

2 sin(�2 ) cos'q1�sin2(�2 ) sin

2 'd'q

sin2��2

�� sin2

��2

�sin2 '

T =

sl

g

Z �=2

��=2


2 'd'

sin��2

�p1� sin2 '

T =

sl

g

Z �=2

��=2


2 'd'

sin��2

�cos'

T = 2

sl

g

Z �=2

��=2

d'q1� sin2

��2

�sin2 '

T = 4

sl

g

Z �=2

0

d'p1� �2 sin2 '

where

� = sin��2

�:

Therefore, the period is given by

T = 4

sl

gK (�)

where

K (�) =

Z �=2

0

d'p1� �2 sin2 '

is the elliptic integral of the �rst kind.

46

VI. LECTURE 7 POWER SERIES SOLUTIONS TO DIFFERENTIAL EQUA-

TIONS

� The di¤erential equations we are going to solve are linear di¤erential equation likethose we studied in PHYS 3150. However, unlike those equations the coe¢ cients in

the di¤erential we want to solve are functions of x instead of constants. Which means

we consider linear di¤erential equations of the form

d2y (x)

dx2+ f(x)

dy (x)

dx+ g(x)y(x) = 0:

Such kind of di¤erential equations can be solved using series substitution method. We

assume the solution of the di¤erential equation can be expressed as power series

y (x) =1Xn=0

anxn = a0 + a1x+ a2x

2 + a3x3 + :::anx

n + :::

We will illustrate this method by solving the following problem.

Ex. 13 A mass m is attached to a horizontal spring of spring constant k whose other end is

attached to a rigid vertical wall. The mass slides on a horizontal, frictionless surface.

At t = 0 the mass is stretched from its equilibrium position by a distance xmax and

released from rest. Find the equation for the position of the mass as a function of

time, x(t).

Sol: Using Newton�s second law the equation of motion for the mass m can be written as

Fnet = ma) md2x (t)

dt2= �kx

which put in the formd2x (t)

dt2+ !2x = 0;

where

! =

rk

m:

Let�s assume that the solution of this di¤erential equation can expressed as

x (t) =1Xn=0

antn = a0 + a1t+ a2t

2 + a3t3 + :::ant

n + :::

47

so that

dx (t)

dt=

1Xn=0

and (tn)

dt=

1Xn=0

nantn�1 = a1 + 2a2t+ 3a3t

2 + :::nantn�1 + :::

and

d2x (t)

dt2=

1Xn=0

and2 (tn)

dt2=

1Xn=0

n (n� 1) antn�2 = 2a2 + 3 � 2a3t+ 4 � 3a4t2 + :::

+n (n� 1) antn�2 + :::

The di¤erential equation can then be rewritten as

1Xn=0

n (n� 1) antn�2 + !21Xn=0

antn = 0:

Noting that

1Xn=0

n (n� 1) antn�2 = 2a2 + 3 � 2a3t+ 4 � 3a4t2 + :::n (n� 1) antn�2 + :::

=1Xn=0

(n+ 2) (n+ 1) an+2tn

we can write1Xn=0

(n+ 2) (n+ 1) an+2tn+!2

1Xn=0

antn = 0)

1Xn=0

�(n+ 2) (n+ 1) an+2 + !2an

�tn = 0:

From the above equation we �nd the following recursion relation

(n+ 2) (n+ 1) an+2 + !2an = 0) an+2 = �an

(n+ 2) (n+ 1)!2:

Now let�s examine the �rst few terms for this recursion relation. First we consider

when n is even

n = 0) a2 = �a02 � 1!

2 ) n = 0) a2�1 =a0 (�1)1

2!!2

n = 2) a4 = �a24 � 3!

2 =a0 (�1)2

4 � 3 � 2 � 1!2�2 ) n = 2) a2�2 =

a04!(�1)2 !2�n

n = 4) a6 = �a4!

2

6 � 5 =a0 (�1)3 !2�36 � 5 � 4 � 3 � 2 � 1 ) n = 4) a2�3 =

a0 (�1)3 !2�36!

Therefore for the even terms we �nd

a2n =a0 (�1)n !2n

(2n)!:

48

Now lets consider the odd terms

an+2 = �an

(n+ 2) (n+ 1)!2

n = 1) a3 =a1 (�1)1

3 � 2 !2 ) n = 1) a(2�2�1) =a1 (�1)1

(2� 2� 1)!!2�2�2

n = 3) a5 = �a35 � 4!

2 =a1 (�1)2 !2�3�25 � 4 � 3 � 2 � 1 ) n = 3) a(2�3�1) =

a1 (�1)2 !2�3�2(2� 3� 1)!

n = 5) a7 = �a5!

2

7 � 6 =a1 (�1)3 !2�4�27 � 6 � 5 � 4 � 3 � 2 � 1 ) n = 5) a(2�4�1) =

a1 (�1)3 !2�4�2(2� 4� 1)!

Then for the odd terms we can write

a(2n�1) =a1 (�1)n !2n�2(2n� 1)! :

where n = 1; 2; 3:::Therefore the general solution for the di¤erential equation

d2x (t)

dt2+ !2x = 0

given by

x (t) =1Xn=0

antn ) x (t) =

1Xn=0

a2nt2n +

1Xn=1

a2n�1t2n�1

using

a2n =a0 (�1)n !2n

(2n)!:

and

a(2n�1) =a1 (�1)n !2n�2(2n� 1)! :

can be written as

x (t) =

1Xn=0

antn ) x (t) = a0

1Xn=0

(�1)n !2n(2n)!

t2n

+a1

1Xn=1

(�1)n !2n�2(2n� 1)! t2n�1

) x (t) = a0

1Xn=0

(�1)n (!t)2n

(2n)!+ a1!

1Xn=1

(�1)n !2n�1t2n�1(2n� 1)!

x (t) = a0

1Xn=0

(!t)2n

(2n)!+ a1!

1Xn=1

(�1)n (!t)2n�1

(2n� 1)!

49

If we replace n by k + 1 in the second term of the series, we can write

x (t) = a0

1Xn=0

(�1)n (!t)2n

(2n)!+ a1!

1Xk=0

(�1)k (!t)2k+1

(2k + 1)!

where we included �1 into the constant a1: We recall that the Taylor series expansionsfor sin (t) and cos (t) are

sin (t) =1Xk=0

(�1)k (t)2k+1

(2k + 1)!

cos (t) =1Xn=0

(�1)n (t)2n

(2n)!

we can write the solution as

x (t) = a0 cos (!t) + a1! sin (!t)

or

x (t) = A cos (!t) +B sin (!t)

where A = a0 and B = a1! are constants determined by the initial conditions. Since

initially spring was stretched by xmax and the mass is released from rest, we have

x (t) = a0 cos (!t) + a1! sin (!t) = xmax for t = 0) a0 = xmax

and

dx (t)

dt= �a0 sin (!t) + a1!

2 cos (!t) = 0 for t = 0) a1!2 = 0) a1 = 0:

Therefore, the equation for the position of the mass as a function of time, x(t) is given

by

x (t) = xmax cos (!t)

where

! =

rk

m;

is the angular frequency.

� Orthogonal vectors and Dirac Notation: We recall that if the dot product of two realvectors ~A and ~B is zero, given by

~A � ~B =3Xi=1

AiBi = 0

50

the two vectors are said to be orthogonal. If these vectors are complex and orthogonal,

we must write

~A� � ~B =3Xi=1

A�iBi = 0:

where ~A� is the complex conjugate of ~A: Using Dirac notation any vector ~A is denoted

by a ket vector jAi and its complex conjugate ~A� by a bra vector hAj : Then the dotproduct of two vectors, using Dirac notation, can be expressed as

hA jBi =3Xi=1

A�iBi

and if they are orthogonal we have

hA jBi =3Xi=1

A�iBi = 0:

� Orthonormal Sets of Functions: two functions A(x) and B(x) are said to be orthogonalon (a; b) if and only if Z b

a

A�(x)B(x)dx = 0

where A�(x) is the complex conjugate of the function A(x): Suppose we have a whole

set of functions

A1 (x) ; A2 (x) ; A3 (x) :::An (x) :::

if these functions satisfy the condition

Z b

a

A�n(x)Am(x)dx =

8<: 0 n 6= m

constant n = m

then these functions are said to be orthogonal set of functions. Using Dirac notation

this can be expressed as

hAn (x) jAm (x)i =Z b

a

A�n(x)Am(x)dx

hAn (x) jAm (x)i =

8<: 0 n 6= m

constant n = m

Suppose for n = m

hAn (x) jAn (x)i = Cn

51

then we can write the set of normalized orthogonal functions de�ned by

Fn (x) =An (x)pCn

so that

hFn (x) jFm (x)i =Z b

a

F �n (x)Fm (x) dx

hFn (x) jFm (x)i = �nm:

where

�nm =

8<: 0 n 6= m

1 n = m:

If a set of functions Fn (x) satisfy the condition

hFn (x) jFm (x)i = �nm;

then these functions are said to form an orthonormal function set.

Ex. 14 Show that the set of functions de�ned by

Fn (x) =1p�sin (nx)

for n = 1; 2; 3::::form an orthonormal set of functions on (��; �).

Sol: Using Euler�s formula we can express

Fm (x) =1p�sin (mx) =

eimx � e�imx

2ip�

) F �n (x) =e�inx � einx

�2ip�

so that


a

F �n (x)Fm (x) dx

becomes

hFn (x) jFm (x)i

=

Z �

��

�e�inx � einx

�2ip�

��eimx � e�imx

2ip�

�dx

=1

4�

Z �

��

�ei(m�n)x � e�i(m+n)x

�ei(m+n)x + e�i(m�n)x�dx

1

4�i

�ei(m�n)x

m� n+e�i(m+n)x

m+ n

= �ei(m+n)x

m+ n� e�i(m�n)x

m� n

��

52

hFn (x) jFm (x)i =1

4�

�4 sin ((m� n)�)

m� n� 4 sin ((m+ n)�)

m+ n

�whether m = n or m 6= n the second term is always zero and we can write

hFn (x) jFm (x)i =sin ((m� n)�)

(m� n)�= �nm:

where we have applied applied LeHospital rule.

53

VII. LECTURE 8 COMPLETE SETS OF FUNCTIONS AND THE LEGENDRE

POLYNOMIALS

� Orthonormal Sets of Functions: We recall from previous lecture a set of functions,

Fn (x) ; form an orthonormal set on the interval (a; b) when


a

F �n (x)Fm (x) dx = �nm:

We now study when these sets of functions form a complete set on the interval (a; b)

in the function space. But before that we review the vector space and study complete

sets of vectors in a vector space.

� Complete sets of vectors in a vector space: Consider a three dimensional real space R3

in Cartesian coordinate system. We recall the unit vectors x; y, and z or x1; x2; and

x3 form an orthonormal set of vectors in 3-D vector space since

xn � xm = �nm

or using Dirac notation

hxnj xmi = �nm:

Any vector ~r � R3 can be expressed in terms of these three unit vectors

~r =3Xi=1

rixi

or using Dirac notation

jri =3Xi=1

ri jxii :

Multiplying both sides from the left by hxjj we �nd

hxj jri =3Xi=1

ri hxj jxii :

Since the vectors jxii form an orthonormal set of vectors

hxj jri =3Xi=1

ri�ji ) rj = hxj jri

If one of the vectors xi is missing, then we can not express any vector ~r � R3 in terms

of the remaining two vectors. But if we have all these three vectors any vector ~r �

R3 can be expressed in terms of these vectors. Then we say that the set of vectors

fxig = fx1; x2; x3g is a complete orthonormal set in R3:

54

� Complete Sets of Functions in a Function Space: We now consider the in�nite-

dimensional function space on the interval (a; b) : The in�nite set of orthonormal func-

tion fFn (x)g = fF0 (x) ; F1 (x) ; F2 (x) ; F3 (x) ; :::g are said to be complete set if afunction g (x) de�ned on the interval (a; b) can be expressed as a linear combination

of these orthonormal functions. That means

jg (x)i =1Xn=0

cn jFn (x)i ;

where the expansion coe¢ cient cn is determined using the orthonormality of the func-

tions Fn (x) : Multiplying both sides by the bra vector hFm (x)j from the left, we have

hFm (x) jg (x)i =1Xn=0

cn hFm (x) jFn (x)i ;

so that using


a

F �n (x)Fm (x) dx = �nm

hFm (x) jg (x)i =Z b

a

F �m (x) g (x) dx

we �nd

hFm (x) jg (x)i =1Xn=0

cn hFm (x) jFn (x)i )Z b

a

F �m (x) g (x) dx =1Xn=0

cn

Z b

a

F �m (x)Fn (x) dx

)Z b

a

F �m (x) g (x) dx =1Xn=0

cn�nm = cm

) cm =

Z b

a

F �m (x) g (x) dx:

Ex. 14 Any periodic function g(x) de�ned on the interval (��; �) can be expressed in termsof the set of orthonormal function�

Fn (x) =1p�sin (nx)

�=

�1p�sin (x) ;

1p�sin (2x) ;

1p�sin (3x) ; :::

�as

jg (x)i =1Xn=1

cn jFn (x)i ) g (x) =1Xn=1

cnp�sin (nx)

where

) cn =

Z �

��F �n (x) g (x) dx) cn =

1p�

Z �

��sin (nx) g (x) dx:

Thus the set of functionsnFn (x) =

1p�sin (nx)

ois a complete orthonormal set. We

will see that this is nothing but a Fourier series expansion of a periodic functions.

55

� The Legendre Di¤erential Equation: We have mentioned that Series substitutionmethod is useful to solve a linear di¤erential equation

an (x)dny

dxn+ an�1 (x)

dn�1y

dxn+ an�2 (x)

dn�2y

dxn�2+ :::a1 (x)

dy

dx+ a0 (x) = 0:

For the case where the coe¢ cients an (x) are independent of x, we have seen the

application of this method by solving the harmonic oscillator problem. Next we will

solve Legendre di¤erential equation which has coe¢ cients an (x) that dependent on

the variable x. The Legendre di¤erential equation is given by�1� x2

�y00 � 2xy0 + l (l + 1) y = 0;

where l is a constant. This equation arises in the solution of partial di¤erential equa-

tions in spherical coordinates. We will proceed to �nd the solution of the Legendre

di¤erential equation using series substitution. To this end, we assume a solution y(x)

expressed in power series as

y (x) =1Xn=0

anxn = a0 + a1x+ a2x

2 + a3x3 + :::anx

n + :::

The �rst derivative of this function

y0 (x) =1Xn=0

and

dx(xn) =

1Xn=0

annxn�1 ) xy0 (x) =

1Xn=0

annxn:

The second derivative

y00(x) =

1Xn=0

annd

dx

�xn�1

�=

1Xn=0

ann (n� 1)xn�2 ) x2y00(x) =

1Xn=0

ann (n� 1)xn:

If we expand the series for y00(x)

y00(x) =

1Xn=0

ann (n� 1)xn�2 = 0 + 0 + 2a2 + 3:2a3x1 + :::

we note that it can rewritten as

y00(x) =

1Xn=0

an+2 (n+ 2) (n+ 1) xn:

Now substituting the expressions for y; xy0; x2y00; and y00 into the Legendre di¤erential

equation�1� x2

�y00 � 2xy0 + l (l + 1) y = 0) y00 � x2y00 � 2xy0 + l (l + 1) y = 0

56

we �nd1Xn=0

an+2 (n+ 2) (n+ 1) xn �

1Xn=0

ann (n� 1)xn � 21Xn=0

annxn + l (l + 1)

1Xn=0

anxn = 0:

Simplifying this expression

1Xn=0

fan+2 (n+ 2) (n+ 1) +an (�n (n� 1)� 2n+ l (l + 1))gxn = 0

)1Xn=0

fan+2 (n+ 2) (n+ 1) +an (l (l + 1)� n (n+ 1))gxn = 0:

There follows that

an+2 (n+ 2) (n+ 1) + an (l (l + 1)� n (n+ 1)) = 0

an+2 (n+ 2) (n+ 1) + an�l2 + l � n2 � n

�= 0) an+2 = �

�l2 + l � n2 � n

(n+ 2) (n+ 1)

�an

Noting that

l2 + l � n2 � n = l2 � n2 + l � n = (l � n) (l + n) + (l � n)

) l2 + l � n2 � n = (l � n) (l + n+ 1)

we �nd

an+2 = �(l � n) (l + n+ 1)

(n+ 2) (n+ 1)an:

� Some details of applying the recursion relation:

Even n:

a2 = �(l � 0) (l + 0 + 1)(0 + 2) (0 + 1)

a0 ) a2 = �l (l + 1)

2!a0

a4 = �(l � 2) (l + 2 + 1)(2 + 2) (2 + 1)

a2 ) a4 = �(l � 2) (l + 3)

4 � 3 a2

) a4 =l (l + 1) (l � 2) (l + 3)

4!a0

a6 = �(l � 4) (l + 4 + 1)(4 + 2) (4 + 1)

a4 ) a6 = �(l � 4) (l + 5)

6 � 5 a4

) a6 = �l (l + 1) (l � 2) (l + 3) (l � 4) (l + 5)

6!a0

Odd n:

a3 = �(l � 1) (l + 1 + 1)(1 + 2) (1 + 1)

a1 ) a3 = �(l � 1) (l + 2)

3!a1

57

a5 = �(l � 3) (l + 3 + 1)(3 + 2) (3 + 1)

a3 ) a5 = �(l � 3) (l + 4)

5 � 4 a3

) a5 =(l � 1) (l + 2) (l � 3) (l + 4)

5!a1

a7 = �(l � 5) (l + 5 + 1)(5 + 2) (5 + 1)

a5 ) a7 = �(l � 5) (l + 6)

7 � 6 a5

a7 = �(l � 1) (l + 2) (l � 3) (l + 4) (l � 5) (l + 6)

7!a1:

Therefore, the solution of the Legendre equation is a sum of two series containing two

constants a0 and a1 determined by the initial conditions:

y (x) = a0

�1� l (l + 1)

2!x2 +

l (l + 1) (l � 2) (l + 3)4!

x4

� l (l + 1) (l � 2) (l + 3) (l � 4) (l + 5)6!

x6 + :::

�

+a1

�x� (l � 1) (l + 2)

3!x3 +

(l � 1) (l + 2) (l � 3) (l + 4)5!

x5

�(l � 1) (l + 2) (l � 3) (l + 4) (l � 5) (l + 6)7!

x7 + :::

�� The Legendre Polynomials: These are the polynomial functions that we generate fora convergent series for the function y(x) for di¤erent values of l: First we determine

the radius of convergence for the series

limn!1

��an+1anx

�� < 1:From the recursion relation we derived earlier, we have

an+2 = �(l � n) (l + n+ 1)

(n+ 2) (n+ 1)an

) limn!1

��an+2an

�� = limn!1

��(l � n) (l + n+ 1)

(n+ 2) (n+ 1)

��If we determine the interval of convergence for

limn!1

��an+2anx2�� < 1

it means we have determined the interval of convergence for the even series

ye (x) = a0

�1� l (l + 1)

2!x2 +

l (l + 1) (l � 2) (l + 3)4!

x4

� l (l + 1) (l � 2) (l + 3) (l � 4) (l + 5)6!

x6 + :::

�

58

and the interval of convergence for the odd series

yo (x) = a1

�x� (l � 1) (l + 2)

3!x3 +

(l � 1) (l + 2) (l � 3) (l + 4)5!

x5

�(l � 1) (l + 2) (l � 3) (l + 4) (l � 5) (l + 6)7!

x7 + :::

�Therefore, the interval of convergence for the function

y (x) = ye (x) + yo (x)

would be the intervals of convergence that is common to both series. Since each of the

coe¢ cients in both the even and odd series satisfy the condition

limn!1

��an+2anx2�� = lim

n!1

��(l � n) (l + n+ 1)

(n+ 2) (n+ 1)x2�� < 1

) limn!1

��(l � n) (l + n+ 1)

(n+ 2) (n+ 1)x2�� ' lim

n!1jxj < 1

we can conclude that the series

y (x) = ye (x) + yo (x)

is convergent in the interval �1 < x < 1: Since the ratio test fails at x = �1; weneed to examine the series at the boundaries x = �1:For x = �1; the even solutionbecomes

ye (x) = a0

�1� l (l + 1)

2!+l (l + 1) (l � 2) (l + 3)

4!

� l (l + 1) (l � 2) (l + 3) (l � 4) (l + 5)6!

+ :::

�and the odd solution

yo (x) = a1

�1� (l � 1) (l + 2)

3!+(l � 1) (l + 2) (l � 3) (l + 4)

5!

�(l � 1) (l + 2) (l � 3) (l + 4) (l � 5) (l + 6)7!

+ :::

�where we included the � in the constant a1:Now for l = 0; we have

ye (x) = a0

and

yo (x) = a1

�1 +

1� 23!

+1 (2) (3) (4)

5!+(1) (2) (3) (4) (5) (6)

7!+ :::

�

59

) yo (x) = a1

�1 +

1

3+1

5+1

7:::: +

1

2n+ 1+ :::

�= a1

1Xn=0

1

2n+ 1

Does this series convergent? Using the integral test, we haveZ 1

0

dn

2n+ 1=1

2ln (2n+ 1)

��10

=1

which means the series is divergent. Since we are looking for a well de�ned solution

for the di¤erential equation in the interval �1 � x � 1; we must have a1 = 0 so that

) yo (x) = a1

1Xn=0

1

2n+ 1= 0

and

y (x) = ye (x) = a0 for l = 0

For l = 1; we have

ye (�1) = a0

�1� 1 (1 + 1)

2!

+1 (1 + 1) (1� 2) (1 + 3)

4!

�1 (1 + 1) (1� 2) (1 + 3) (1� 4) (1 + 5)6!

+ :::

�

) ye (�1) = a0

�1� 2

2!� 1 (2) (4)

4!

�1 (2) (4) (3) (6)6!

+ :::

�

) ye (�1) = �a0�1

3+1

5+1

7

� 1

2n+ 3+ :::

�= a0

1Xn=0

1

2n+ 3

which is a divergent series and we must have a0 = 0 so that

) ye (�1) = 0:

On the other hand, for the odd term for all x (i.e. �1 � x � 1) at l = 1 we �nd

yo (x) = a1x:

60

which leads to

y (x) = yo (x) = a1x for l = 1:

For the same reason discussed above it can be shown that

y (x) =

8<: ye (x) l = even

yo (x) l = odd

which gives

l = 2) y (x) = a0�1� 3x2

�l = 3) y (x) = a1

�x� 5

3x3�

If we chose a0 and a1 such that y (x) = 1 when x = 1; we �nd

l = 0) y0 (x) = 1;

l = 1) y1 (x) = x

l = 2) y2 (x) =1

2

�3x2 � 1

�l = 3) y3 (x) =

1

2

�5x3 � 3x

�These polynomials are called Legendre Polynomials.

61

A. Lecture 9 The Generating Function for Legendre Polynomials

� The Legendre Polynomials

P0 (x) = 1; P1 (x) = x; P2 (x) =1

2

�3x2 � 1

�� The Generating function: Let�s �nd the Taylor series expansion for the function

� (x; h) =�1� 2xh+ h2

��1=2about h = 0. Which can be expressed as

� (x; h) = � (x; 0) +

�d

dh� (x; 0)

�h

1!

+

�d2

dh2� (x; 0)

�h2

2!+

�d3

dh3� (x; 0)

�h2

3!+ :::

Noting that

� (x; 0) = 1 = P0 (x)

d

dh� (x; h) = (x� h)

�1� 2xh+ h2

��3=2 ) d

dh� (x; 0) = x = P1 (x)

d2

dh2� (x; h) =

d

dh

h(x� h)

�1� 2xh+ h2

��3=2i= �

�1� 2xh+ h2

��3=2+ 3 (x� h)2

�1� 2xh+ h2

��5=2d2

dh2� (x; 0) = �1 + 3x2 ) d2

dh2� (x; 0) = 2!

3

2

�x2 � 1

�= 2!P2 (x)

d3

dh3� (x; h) =

d

dh

h��1� 2xh+ h2

��3=2= +3 (x� h)2

�1� 2xh+ h2

��5=2i

= �3 (x� h)�1� 2xh+ h2

��5=2 � 3� 2 (x� h)�1� 2xh+ h2

��5=2+3� 5 (x� h)3

�1� 2xh+ h2

��7=2) d3

dh3� (x; 0) = 15x3 � 3x) d3

dh3� (x; 0) = 3!

1

2

�5x3 � x

�= 3!P3 (x)

we can then conclude that

l!Pl (x) =dl

dhl

�1p

1� 2xh+ h2

��h=0

62

Therefore, the Taylor series expansion for the function

� (x; h) =�1� 2xh+ h2

��1=2about h = 0 can be written as

� (x; h) =

1Xl=0

Pl (x)hl:

Ex. 15 Consider a point charge Q located at the position ~r0 (the source point). Find an

expression for the electrostatic potential V (r) at the point P located at the position

~r (the �eld point).

Sol: For a point charge located at the position described by the vector ~r0 the electric

potential at the point at position ~r is inversely proportional to the distance between

the charge position and the point p (j~r � ~r0j) and directly proportional to the charge.It can be expressed as

V (r) =Q

4��0

1

j~r � ~r0j :

Noting that1

j~r � ~r0j =�r2 + r02 � 2rr0 cos �

��1=2For r0 < r we may write

1

j~r � ~r0j =1

r

1 +

�r0

r

�2� 2r

0

rcos �

!�1=2:

If we introduce the variables

h =r0

r; x = cos �

we have1

j~r � ~r0j =1

r

�1� 2xh+ h2

�1=2so that using �

1� 2xh+ h2��1=2

=

1Xl=0

Pl (x)hl

for

h =r0

r; x = cos �

63

the expression for the electric potential becomes

V (r) =Q

4��0

1

j~r � ~r0j

V (r) =Q

4��0r

1Xl=0

Pl (cos �)

�r0

r

�lwhen r0 < r: On the other hand when r0 > r; we must write

1

j~r � ~r0j =1

r0

�1 +

� rr0

�2� 2 r

r0cos �

��1=2:

or1

j~r � ~r0j =1

r0�1� 2xh+ h2

�1=2where

h =r0

r; x = cos �:

In this case the potential is given by

V (r) =Q

4��0

1

j~r � ~r0j

V (r) =Q

4��0r0

1Xl=0

Pl (cos �)� rr0

�l� The Determination of Legendre Polynomials: Rodrigues�Formula

Pl (x) =1

2ll!

dl

dxl�x2 � 1

�lEx. 16 Use Rodrigues�formula to �nd the equation for P2(x).

Sol: Using Rodrigues�Formula P2(x) can be expressed as

P2 (x) =1

222!

d2

dx2�x2 � 1

�2which can be simpli�ed as follows:

P2 (x) =1

23!

d2

dx2�x4 � 2x2 + 1

�) P2 (x) =

1

23!

d

dx

�4x3 � 4x

�) P2 (x) =

1

23!

�12x2 � 4

�) P2 (x) =

4

23!

�3x2 � 1

�) P2 (x) =

1

2

�3x2 � 1

�Leibniz�Rule: it is useful when we need to �nd a higher order derivatives of a product

(for example when we apply Rodrigues�Formula). It is given by

dN

dxN(uv) =

NXn=0

0@ N

n

1A�dnudxn

��dN�nv

dxN�n

�

64

Ex. 17 Use Leibniz�Rule to �ndd8

dx8�x2e2x

�:

Sol: For N = 8 Leibniz�Rule can be expressed as

d8

dx8(uv) =

8Xn=0

0@ 8

n

1A�dnudxn

��d8�nv

dx8�n

�:

For u = x2; we have

d0u

dx0= x2;

d1u

dx1= 2x;

d2u

dx2= 2;

dnu

dxn= 0 for n = 3

so that

d8

dx8(uv) =

3Xn=0

0@ 8

n

1A�dnudxn

��d8�nv

dx8�n

�

=

0@ 80

1A�d0udx0

��d8v

dx8

�+

0@ 81

1A�d1udx1

��d7v

dx7

�+

0@ 82

1A�d2udx2

��d6v

dx6

�

) d8

dx8(uv) =

�8!

8!0!

�x2�d8v

dx8

�+

�8!

7!1!

�2x

�d7v

dx7

�+

�8!

6!2!

�2

�d6v

dx6

�) d8

dx8(uv) = x2

�d8v

dx8

�+ 16x

�d7v

dx7

�+ 56

�d6v

dx6

�:

Noting that for v = e2x

dnv

dxn= 2ne2x

we �nd

d8

dx8(uv) = x2

�28e2x

�+ 16x

�27e2x

�+ 56

�26e2x

�d8

dx8(uv) = 28

�x2 + 8x+ 14

�e2x

d8

dx8(uv) =

�256x2 + 2048x+ 3584

�e2x

A Recursion Relation : We recall the generating function

�1� 2xh+ h2

��1=2=

1Xl=0

Pl (x)hl

65

so thatd

dh

�1� 2xh+ h2

��1=2=

d

dh

1Xl=0

Pl (x)hl

) (x� h)�1� 2xh+ h2

��3=2=

1Xl=0

lPl (x)hl�1

) (x� h)�1� 2xh+ h2

��1 �1� 2xh+ h2

��1=2=

1Xl=0

lPl (x)hl�1:

If we use �1� 2xh+ h2

��1=2=

1Xl=0

Pl (x)hl

we can write the above expression as

(x� h)�1� 2xh+ h2

��1 1Xl=0

Pl (x)hl =

1Xl=0

lPl (x)hl�1

) (x� h)

1Xl=0

Pl (x)hl =�1� 2xh+ h2

� 1Xl=0

lPl (x)hl�1

) x1Xl=0

Pl (x)hl�

1Xl=0

Pl (x)hl+1 =

1Xl=0

lPl (x)hl�1� 2x

1Xl=0

lPl (x)hl +

1Xl=0

lPl (x)hl+1

1Xl=0

lPl (x)hl�1�2x

1Xl=0

lPl (x)hl�x

1Xl=0

Pl (x)hl+

1Xl=0

lPl (x)hl+1+

1Xl=0

Pl (x)hl+1 = 0

1Xl=0

lPl (x)hl�1 �

1Xl=0

x (1 + 2l)Pl (x)hl +

1Xl=0

(1 + l)Pl (x)hl+1 = 0

Noting that

1Xl=0

lPl (x)hl�1 = 1P1 (x) + 2P2 (x)h

1 + 2P3 (x)h2 + 4P4 (x)h

3 + :::

)1Xl=0

lPl (x)hl�1 =

1Xl=0

(l + 1)Pl+1 (x)hl

and1Xl=0

(1 + l)Pl (x)hl+1 = P0 (x)h+ 2P1 (x)h

2

+3P2 (x)h3 + 4P3 (x)h

4 + 5P4 (x)h5 + ::: = 0� P�1 (x) + P0 (x)h+ 2P1 (x)h

2

+3P2 (x)h3 + 4P3 (x)h

4 )1Xl=0

(1 + l)Pl (x)hl+1 =

1Xl=0

lPl�1 (x)hl

66

we can write that

1Xl=0

(l + 1)Pl+1 (x)hl �

1Xl=0

x (1 + 2l)Pl (x)hl +

1Xl=0

lPl�1 (x)hl = 0

)1Xl=0

[(l + 1)Pl+1 (x)� x (1 + 2l)Pl (x) +lPl�1 (x)]hl = 0

There follows that

(l + 1)Pl+1 (x)� x (1 + 2l)Pl (x) + lPl�1 (x) = 0

) (l + 1)Pl+1 (x) = x (1 + 2l)Pl (x)� lPl�1 (x) :

67

VIII. LECTURE 10 LEGENDRE SERIES, ASSOCIATED LEGENDRE FUNC-

TIONS, AND SPHERICAL HARMONICS

� Orthonormality of the Legendre polynomials: Legendre polynomials form an orthogo-

nal set of functions. The orthogonality conditions is given by

hPl (x)j Pm (x)i =2

2l + 1�lm

for �1 � x � 1:

� Completeness of the Legendre Polynomials:The Legendre polynomials also form com-

plete set on the interval (�1; 1) : That means any function f (x) can be expressed asa linear combination of the Legendre polynomials

jf (x)i =1Xm=0

am jPm (x)i :

The expression for the expansion coe¢ cients is determined using the orthogonality

property. Multiplying both sides from the left by hPl (x)j ; we can write

hPl (x)j f (x)i =1Xm=0

am hPl (x) jPm (x)i

) hPl (x)j f (x)i =1Xm=0

am2

2l + 1�lm

) hPl (x)j f (x)i = al2

2l + 1

) al =2l + 1

2hPl (x)j f (x)i :

Recalling that

hPl (x)j f (x)i =1Z

�1

P �l (x) f (x) dx

and noting that Pl (x) is real, the expression for the expansion coe¢ cients is given by

al =2l + 1

2

1Z�1

Pl (x) f (x) dx:

Ex. 18 Expand the function f(x) in a Legendre series, for

f (x) =

8<: �1 �1 � x � 0+1 0 � x � 1

68

Sol: The Legendre series is given by

f (x) =1Xm=0

amPm (x)

where

al =2l + 1

2

1Z�1

P �l (x) f (x) dx:

The function has di¤erent values in the interval (�1; 0) and (0; 1). Hence the expansioncoe¢ cients, al can be expressed as

al =2l + 1

2

24 0Z�1

Pl (x) f (x) dx

+

1Z0

Pl (x) f (x) dx

35al =

2l + 1

2

24� 0Z�1

Pl (x) dx+

1Z0

Pl (x) dx

35

al =2l + 1

2

24� 0Z1

Pl (�x) d (�x) +1Z0

Pl (x) dx

35al =

2l + 1

2

24 0Z1

Pl (�x) dx+1Z0

Pl (x) dx

35al =

2l + 1

2

24� 1Z0

Pl (�x) dx+1Z0

Pl (x) dx

35al =

2l + 1

2

24 1Z0

[Pl (x)� Pl (�x)] dx

35Using the Rodrigues formula for the Legenedre polynomials

Pl (x) =1

2ll!

dl

dxl�x2 � 1

�lif we replace x by �x, we have

Pl (�x) =1

2ll!

dl

d (�x)l�x2 � 1

�l) Pl (�x) = (�1)l

1

2ll!

dl

dxl�x2 � 1

�l69

we leads to

Pl (�x) =

8<: Pl (x) l = even

�Pl (x) l = odd

Applying this result we �nd for al

al =

8>>><>>>:(2l + 1)

1Z0

Pl (x) dx l = odd

0 l = even

Using

P0 (x) = 1; P1 (x) = x

P2 (x) =1

2

�3x2 � 1

�; P3 (x) =

1

2

�5x3 � 3x

�we �nd

a1 = 3

1Z0

xdx =3

2

a3 =7

2

1Z0

�5x3 � 3x

�dx = �7

3

Therefore, the Legendre series is

f (x) =3

2P1 (x)�

7

3P2 (x) + :::

� Least-Squares Fit and Legendre Series: Suppose we are given a curve or a graph thatis described by a function f(x) in the interval (�1; 1) and we want to �nd a polynomial

gn (x) = anxn + an�1x

n�1 + :::a0

gn (x) =

nXk=0

akxk

that can best �t to the curve described by the function f(x):Then the polynomial

gn (x) can best �t the curve described by f(x) when the integralZ 1

�1[f(x)� gn (x)]

2 dx

a minimum. This polynomial is nothing but the nth order Legendre polynomial.

70

� The Associated Legendre Di¤erential Equation: We recall the Legendre di¤erentialequation �

1� x2�y00 � 2xy0 + l (l + 1) y = 0:

which is a special case of the Associated Legendre Di¤erential Equation given by

�1� x2

�y00 � 2xy0 +

�l (l + 1)� m2

1� x2

�y = 0:

where m2 � l2. If we set m = 0, we �nd the Legendre Di¤erential equation. We

can solve this DE using series substituting method. However, here we use a di¤erent

approach. To this end we introduce a transformation of variable de�ned by

y (x) =�1� x2

�m=2u (x)

) y0 (x) =�1� x2

�m=2u0 (x)

�m�1� x2

�(m�2)=2xu (x)

) y00(x) =

�1� x2

�m=2u00 (x)

�m�1� x2

�(m�2)=2xu0 (x)

�m�1� x2

�(m�2)=2u (x)

+m (m� 2)�1� x2

�(m�4)=2x2u (x)

�m�1� x2

�(m�2)=2xu0 (x)

) y00(x) =

�1� x2

�m=2u00 (x)

�2m�1� x2

�(m�2)=2xu0 (x)

�m�1� x2

�(m�2)=2u (x)

+m (m� 2)�1� x2

�(m�4)=2x2u (x)

)�1� x2

�y00(x) =

�1� x2

�m+2=2u00 (x)

�2m�1� x2

�m=2xu0 (x)

�m�1� x2

�m=2u (x)

+m (m� 2)�1� x2

�(m�2)=2x2u (x)

71

) 2xy0 (x) = 2�1� x2

�m=2xu0 (x)

�2m�1� x2

�(m�2)=2x2u (x)�

l (l + 1)� m2

1� x2

�y

= l (l + 1)�1� x2

�m=2u (x)

�m2�1� x2

��m=2u (x)

so that the DE �1� x2

�y00 � 2xy0 +

�l (l + 1)� m2

1� x2

�y = 0

becomes �1� x2

�m+2=2u00 (x)� 2m

�1� x2

�m=2xu0 (x)

�m�1� x2

�m=2u (x)

+m (m� 2)�1� x2

�(m�2)=2x2u (x)

� 2�1� x2

�m=2xu0 (x)

+ 2m�1� x2

�(m�2)=2x2u (x)

+l (l + 1)�1� x2

�m=2u (x)

�m2�1� x2

��m=2u (x) = 0:

Dividing the entire equation by (1� x2)m=2

; we �nd�1� x2

�u00 (x)� 2mxu0 (x)

�mu (x) +m (m� 2) x2

1� x2u (x)

�2xu0 (x) + 2m x2

1� x2u (x)

+l (l + 1)u (x)�m2 1

1� x2u (x) = 0

)�1� x2

�u00 (x)� 2mxu0 (x)� 2xu0 (x)

+l (l + 1)u (x)�mu (x) +m (m� 2)x21� x2

u (x)

+2mx2

1� x2u (x)� m2

1� x2u (x) = 0

72

)�1� x2

�u00 (x)� 2 (m+ 1) xu0 (x)

+

�l (l + 1)�m+

m2x2 � 2mx21� x2

+2mx2

1� x2� m2

1� x2

�u (x) = 0

)�1� x2

�u00 (x)� 2 (m+ 1) xu0 (x)

+

�l (l + 1)�m+

m2 (x2 � 1)1� x2

�u (x) = 0

)�1� x2

�u00 (x)� 2 (m+ 1) xu0 (x)

+ [l (l + 1)�m (m+ 1)]u (x) = 0:

For m = 0; this equation reduces to

�1� x2

�u00 (x)� 2xu0 (x) + l (l + 1)u (x) = 0:

which is the Legendre equation and the solution can be expressed as

u (x) = Pl (x) :

For m = 1; we have

�1� x2

�u00 (x)� 4xu0 (x)

+ [l (l + 1)� 2]u (x) = 0:

which has the same form as

d

dx

��1� x2

�u00 (x)� 2xu0 (x)

+l (l + 1)u (x) = 0g ;

)�1� x2

� hu0(x)i00� 2x [u0 (x)]0

�2x [u0 (x)]0 � 2u0 (x)

+l (l + 1)u0 (x) = 0:

73

)�1� x2

� hu0(x)i00� 4x [u0 (x)]0

+ [l (l + 1)� 2]u0 (x) = 0:

Since we know that the solution of the Legendre equation

��1� x2

�u00 (x)� 2xu0 (x)

+l (l + 1)u (x) = 0g ;

is

u (x) = Pl (x)

we can write the solution for

d

dx

��1� x2

�u00 (x)� 2xu0 (x)

+l (l + 1)u (x) = 0g

asdPl (x)

dx:

Therefore the solution of the di¤erential equation, for m = 1;

�1� x2

�u00 (x)� 4xu0 (x)

+ [l (l + 1)� 2]u (x) = 0:

can be expressed as

u (x) =dPl (x)

dx:

Following the same procedure it can be easily shown that for a given m; the solution

of the di¤erential equation

�1� x2

�u00 (x)� 2 (m+ 1) xu0 (x)

+ [l (l + 1)�m (m+ 1)]u (x) = 0:

can be expressed as

u (x) =dmPl (x)

dxm:

Substituting this into

y (x) =�1� x2

�m=2u (x)

74

the solution to the Associated Legendre Di¤erential equation�1� x2

�y00 � 2xy0 +

�l (l + 1)� m2

1� x2

�y = 0:

can be expressed as

y (x) =�1� x2

�m=2 dm

dxmPl (x)

� The Associated Legendre Functions: The solutions to the Associated Legendre Di¤er-ential equation are the Associated Legendre functions and are given by

Pml (x) =�1� x2

�m=2 dm

dxmPl (x) :

Note that when m = 0 we �nd the Legendre Polynomials

P 0l (x) = Pl (x) :

� Orthogonality of The Associated Legendre Functions:

hPml (x)j Pml0 (x)i =2

2l + 1

(n+m)!

(n�m)!�ll0

� The Laplace Equation in spherical coordinates: The Laplace equation

r2 = 0

in spherical coordinates can be written as

1

r2@

@r

�r2@

@r

�+

1

r2 sin �

@

@�

�sin �

@

@�

�+

1

r2 sin2 �

@2

@'2= 0

Using separation of variables

(r; �; ') = R (r)� (�) � (')

we may write

�(�) � (')

r2d

dr

�r2dR (r)

dr

�+R (r) � (')

r2 sin �

d

d�

�sin �

d�(�)

d�

�+R (r)� (�)

r2 sin2 �

d2� (')

d'2= 0:

75

Dividing the entire equation by R (r)� (�) � ('), we have

1

R (r) r2d

dr

�r2dR (r)

dr

�+

1

� (�) r2 sin �

d

d�

�sin �

d�(�)

d�

�+

1

� (') r2 sin2 �

d2� (')

d'2= 0;

and multiplying by r2 sin2 � gives

sin2 �

�1

R (r)

d

dr

�r2dR (r)

dr

�+

1

� (�) sin �

d

d�

�sin �

d�(�)

d�

��+

1

� (')

d2� (')

d'2= 0:

Since this equation consists of two independent terms

F (r; �) +G (') = 0

each of these terms must be a constant. If we intruduce the constant m2 for F (r; �) ;

then obviously we see that G (') must be �m2; which means

1

� (')

d2� (')

d'2= �m2

sin2 �

�1

R (r)

d

dr

�r2dR (r)

dr

�+

1

� (�) sin �

d

d�

�sin �

d�(�)

d�

��= m2:

The �rst equation can be rewritten as

d2� (')

d'2+m2� (') = 0

and the solution is given by

� (') = A (m) eim':

For the second equation we have

1

R (r)

d

dr

�r2dR (r)

dr

�+

1

� (�) sin �

d

d�

�sin �

d�(�)

d�

�� m2

sin2 �= 0

76

Again these equation involves two independent terms. These terms must be a constant.

We chose these constant to be l(l + 1) so that

1

R (r)

d

dr

�r2dR (r)

dr

�= l (l + 1)

1

� (�) sin �

d

d�

�sin �

d�(�)

d�

�� m2

sin2 �= �l (l + 1) :

If we simplify the second equation we �nd

1

sin �

d

d�

�sin �

d�(�)

d�

�+

�l (l + 1)� m2

sin2 �

��(�) = 0:

Noting that

�d (cos �) = sin �d�;

we may write

1

sin �

d

d�

�sin �

d�(�)

d�

�+

�l (l + 1)� m2

sin2 �

��(�) = 0:

as

d

d (cos �)

�sin2 �

d�(cos �)

d (cos �)

�+

�l (l + 1)� m2

sin2 �

��(cos �) = 0:

Introducing the variable

x = cos � ) sin2 � = 1� x2

we �nd

d

dx

��1� x2

� d�(x)dx

�+

�l (l + 1)� m2

1� x2

��(x) = 0

or �1� x2

� d2�(x)dx2

� 2xd�(x)dx

+

�l (l + 1)� m2

1� x2

��(x) = 0

77

which is the associated Legendre polynomials and solution is given by the associated

Legendry functions

�(x) = Pml (x) =�1� x2

�m=2 dm

dxmPl (x) :

Therefore the angular part of the Laplace equation in spherical coordinates has a

solution that can be expressed as

�(cos �) � (') = A (l;m)Pml (cos �) eim'

If we normalize this function we �nd what is known as Spherical harmonics, Ylm (�; ').

� The Spherical Harmonics:

Ylm (�; ') = (�1)ms2l + 1

4�

(l �m)!

(l +m)!Pml (cos �)

� eim'

� The Orthonormality of the Spherical Harmonics:

hYlm (�; ')j Yl0m0 (�; ')i

=

Z �

�=0

Z 2�

'=0

Y �lm (�; ')Yl0m0 (�; ') sin �d�d'

= �ll0�mm0

� Completeness of the Spherical Harmonics

f (�; ') =1Xl=0

lX�l

almYlm (�; ') ;

where

alm =

Z �

�=0

Z 2�

'=0

Y �lm (�; ') f (�; ') sin �d�d':

78

IX. LECTURE 11: THE ADDITION THEOREM FOR SPHERICAL HARMON-

ICS

� Recall: The Inverse-Distance Between Two Points

1

j~r � ~r0j =1Xl=0

rl<rl+1>

Pl (cos )

� The Addition Theorem for Spherical Harmonics:A mathematical result of consid-

erable interest and the use is called the addition theorem for spherical Harmon-

ics. Two coordinate vectors ~r and ~r0 with spherical coordinates (r; �; ') and

(r0; �0; '0) ;respectively,have an angle between them as shown in the �gure below.

The addition theorem expresses a Legendre polynomial of order l in the angle in

terms of the products of the spherical Harmonics of the angles �; ' and �0; '0 :

Pl (cos ) =4�

2l + 1

lXm=�l

Ylm (�; ')Y�lm (�

0; '0)

where

cos = cos � cos �0 + sin � sin �0 cos ('� '0)

Ex. 19 A solid sphere of radius R has a constant volume-charge density �. The sphere is

centered at the origin of coordinates. The point P is a distance d > R from the center

of the sphere. Find an expression for the electrostatic potential at the point P , �P ,

due to the charged sphere.

Some Useful Things to Recall :

Ylm (�; ') = (�1)ms2l + 1

4�

(l �m)!

(l +m)!

� Pml (cos �) gm (')

Pml (cos �) � Pml (x) =�1� x2

�m=2 dm

dxmPl (x)

gm (') = eim'

hPn (x)j Pl (x)i =2

2l + 1�l;n: For jxj � 1

Sol:

79

X. LECTURE 12: THE METHOD OF FROBENIUS AND BESSEL FUNCTIONS

� When the Standard Power Series Solutions fail : The power series solution

y (x) =

1Xn=0

anxn

we considered so far is applicable when the di¤erential equation involves no singular

point for 8 x"R: In such cases the solution of the di¤erential equation y (x) can beexpressed as power of x where all x have zero or positive exponents,

y (x) = a0 + a1x+ a2x2 + a3x

3:::

However, there are di¤erential equations which are not de�ned for 8 x"R. The solutionsof such di¤erential equations involves a factor with negative or fractions as exponent

for x. In such cases the standard power series solution fails and we need to use a

slightly modi�ed method known The Method of Frobenius which we will see shortly.

Before we do that we �rst consider some di¤erential equations whose solutions do not

satisfy the standard power series solution.

Ex. 20 Solve the di¤erential equation

y0 +y

x= 0

Sol: First we note that this di¤erential equation is not de�ned at x = 0. But it can be

solved easily as follows:

y0 +y

x= 0) dy

dx= �1

xy

) dy

y= �dx

x)Z y

y0

dy

y= �

Z x

x0

dx

x

) ln

�y

y0

�= � ln

�x

x0

�= ln

�x

x0

��1

) ln

�y

y0

�= ln

�x0x

�) y

y0=x0x) y = (x0y0)x

�1

Noting that y0x0 = a1 (constant), the solution can be written as

y = a1x�1;

80

which clearly show that x has a negative exponent and the standard power series

expansion does not work!


y0 � 3y2x= 0

Sol: Here also the di¤erential equation is singular at x = 0. The solution is given by

y0 � 3y2x= 0) dy

dx=3

2

1

xy

) dy

y=3

2

dx

x)Z y

y0

dy

y=3

2

Z x

x0

dx

x

) ln

�y

y0

�=3

2ln

�x

x0

�= ln

�x

x0

�3=2) y

y0=

�x

x0

�3=2) y =

�y0x

3=20

�x3=2

Noting that y0x3=20 = a1 (constant), the solution can be written as

y = a1x3=2 = x1=2 (a1x) :

which also involve x with a fraction exponent.

� The Method of Frobenius: When we face a di¤erential equation which are not de�nedfor all x"R like the examples we saw above, we use the method of Frobenius. We used

a generalized power series

y (x) = xs1Xn=0

anxn =

1Xn=0

anxn+s

where s is a number to be found to �t the problem. It may be positive, negative,

fraction, or even complex number although we do not consider the complex number

case here.

Ex. 22 The Bessel Di¤erential equation: As an application of the Method of Frobenius we

will solve the Bessel di¤erential equation

x2y00 + xy0 +�x2 � p2

�y = 0;

where p is a constant parameter characterizing the di¤erential equation. The Bessel

di¤erential equation, usually, derived from the Laplace equation in cylindrical coordi-

nates.

81

Sol: Using a generalized power series

y (x) = xs1Xn=0

anxn =

1Xn=0

anxn+s

we have �x2 � p2

�y =

1Xn=0

anxn+s+2 �

1Xn=0

p2anxn+s;

y0 =

1Xn=0

(n+ s) anxn+s�1

) xy0 =1Xn=0

(n+ s) anxn+s;

y00 =

1Xn=0

(n+ s) (n+ s� 1) anxn+s�2

) x2y00 =1Xn=0

(n+ s) (n+ s� 1) anxn+s;

so that substituting these expressions into the Bessel di¤erential equation

x2y00 + xy0 +�x2 � p2

�y = 0

we �nd

)1Xn=0

(n+ s) (n+ s� 1) anxn+s

+1Xn=0

(n+ s) anxn+s

+

1Xn=0

anxn+s+2 � p2

1Xn=0

anxn+s = 0

)1Xn=0

f(n+ s) (n+ s� 1) + (n+ s)

�p2anx

n+s +1Xn=0

anxn+s+2 = 0

)1Xn=0

�(n+ s)2 � p2

�anx

n+s

+

1Xn=0

anxn+s+2 = 0

82

Expanding the �rst series in the above expression

1Xn=0

�(n+ s)2 � p2

�anx

n+s

=�(0 + s)2 � p2

�a0x

0+s

+�(1 + s)2 � p2

�a1x

1+s

+1Xn=2

�(n+ s)2 � p2

�anx

n+s

)1Xn=0

�(n+ s)2 � p2

�anx

n+s

=�s2 � p2

�a0x

s +�(1 + s)2 � p2

�a1x

s+1

+

1Xn=2

�(n+ s)2 � p2

�anx

n+s

and replacing n by m� 2 and noting that n = 2) m = 0; we can write

1Xn=0

�(n+ s)2 � p2

�anx

n+s

=�s2 � p2

�a0x

s +�(1 + s)2 � p2

�a1x

s+1

+1Xm=0

�(m+ 2 + s)2 � p2

�am+2x

m+s+2:

Since m is a dummy variable we can replace it by n and rewrite the above expression

as

1Xn=0

�(n+ s)2 � p2

�anx

n+s

=�s2 � p2

�a0x

s +�(1 + s)2 � p2

�a1x

s+1

+

1Xn=0

�(n+ 2 + s)2 � p2

�an+2x

n+s+2:

Now substituting this expression into

1Xn=0

�(n+ s)2 � p2

�anx

n+s +

1Xn=0

anxn+s+2 = 0

83

we �nd �s2 � p2

�a0x

s +�(1 + s)2 � p2

�a1x

s+1

+

1Xn=0

�(n+ 2 + s)2 � p2

�an+2x

n+s+2

+1Xn=0

anxn+s+2 = 0

�s2 � p2

�a0x

s +�(1 + s)2 � p2

�a1x

s+1

+

1Xn=0

��(n+ 2 + s)2 � p2

�an+2 + an

xn+s+2

= 0:

There follows that �s2 � p2

�a0x

s = 0;�(1 + s)2 � p2

�a1x

s+1 = 0;1Xn=0

��(n+ 2 + s)2 � p2

�an+2 + an

xn+s+2 = 0;

which leads to �s2 � p2

�a0 = 0;�

(1 + s)2 � p2�a1 = 0;�

(n+ 2 + s)2 � p2�an+2 + an = 0

If we solve the �rst equation, we �nd

s = �p

and substituting this value into the second equation�(1 + s)2 � p2

�a1 = 0

we have �(1� p)2 � p2

�a1 = 0

) [(1� p)� p] [(1� p) + p] a1:

84

For the "+" case we �nd

(1� 2p) a1 = 0:

and for the "�" case(1 + 2p) a1 = 0:

For these two equations to hold true for any value of p, we must have

a1 = 0:

The third equation �(n+ 2 + s)2 � p2

�an+2 + an = 0

leads to the recursion relation

an+2 = �1

(n+ 2 + s)2 � p2an:

This recursion relation gives to two di¤erent functions for the two di¤erent values of

s (= �p).

I. First solution (s=p): For this case the recursion formula becomes

an+2 = �1

(n+ 2 + p)2 � p2an

= � 1

(n+ 2)2 + 2 (n+ 2) pan

) an+2 = �1

(n+ 2) (n+ 2 + 2p)an

Since we found a1 = 0 all the odd terms vanish and the recursion relation can be

expressed, for the even terms using n+ 2 = 2m) n = 2m� 2, as

a2m = �1

(2m) (2m+ 2p)a2m�2

) a2m = �a2m�2

22m (m+ p):

For the �rst few terms, we have

m = 1) a2 = �a0

221! (1 + p)

m = 2) a4 = �a2

222 (2 + p)

) a4 =a0

242! (1 + p) (2 + p)

85

m = 3) a6 = �a4

223 (3 + p)

) a6 = �a0

263! (1 + p) (2 + p) (3 + p)

Recalling that the Gamma function

� (p+ 1) = p� (p)

� (p+ 2) = (p+ 1)� (p+ 1)

= p (p+ 1)� (p) ;

� (p+ 3) = (p+ 2)� (p+ 2)

= p (p+ 1) (p+ 2)� (p) ;

we can rewrite

m = 1) a2 = �a0

221! (1 + p)

= � a0p� (p)

221!p (1 + p) � (p)

) a2 = �a0� (p+ 1)

221!� (p+ 2)

m = 2) a4 =a0

242! (1 + p) (2 + p)

=a0p� (p)

242!p (1 + p) (2 + p) � (p)

) a4 =a0� (p+ 1)

242!� (p+ 3)

m = 3) a6 = �a0

263! (1 + p) (2 + p) (3 + p)

= � a0p� (p)

263!p (1 + p) (2 + p) (3 + p) � (p)

) a6 = �a0� (p+ 1)

263!� (p+ 4):

Therefore, the solution

y (x) = xs1Xn=0

anxn =

1Xn=0

anxn+s

86

can be expressed as

y (x) = xp�a0 �

a0� (p+ 1)

221!� (p+ 2)x2

+a0� (p+ 1)

242!� (p+ 3)x4

� a0� (p+ 1)

263!� (p+ 4)x6 + :::

�which can be simpli�ed into

y (x) = a0xp� (p+ 1)

�1

� (p+ 1)

� 1

1!� (p+ 2)

�x2

�2+

1

2!� (p+ 3)

�x2

�4� 1

3!� (p+ 4)

�x2

�6+ :::

�:

Noting that

� (1) = � (2) = 1

n! = � (n+ 1)

the above expression can be rewritten as

y (x) = a02p�x2

�p� (p+ 1)

�1

� (1) � (p+ 1)

� 1

� (2) � (p+ 2)

�x2

�2+

1

� (3) � (p+ 3)

�x2

�4� 1

� (4) � (p+ 4)

�x2

�6+ :::

�:

If we divide this equation by 2pa0� (p+ 1), we have

Jp (x) =1

2pa0� (p+ 1)y (x)

) Jp (x) =�x2

�p� 1

� (1) � (p+ 1)

� 1

� (2) � (p+ 2)

�x2

�2+

1

� (3) � (p+ 3)

�x2

�4� 1

� (4) � (p+ 4)

�x2

�6+ :::

�:

or

Jp (x) =

1Xn=0

(�1)n

� (n+ 1)� (n+ 1 + p)

�x2

�2n+p:

This function Jp (x) is called the Bessel function of the �rst kind of order p.

87

II. Second Solution s=-p: The second solution, when s = �p can easily obtained fromthe �rst solution by replacing p by �p. It is given by

J�p (x) =

1Xn=0

(�1)n

� (n+ 1)� (n+ 1� p)

�x2

�2n�p:

If p is not an integer, Jp (x) is a series starting with xp and J�p (x) is a series starting

with x�p. Then Jp (x) and J�p (x) are two independent solutions and the linear combi-

nation of them is a general solution. The combination is is called either the Neumann

or Weber function and is denoted by either Np (x) or Yp (x) and is given by

Np (x) = Yp (x) =cos (�p) Jp (x)� J�p (x)

sin (�p):

However, if p is an integer, then the �rst few terms in J�p (x) are zero because

� (n+ 1� p) is a the gama function of a negative integer, which is in�nite.

J�p (x) = (�1)pJp (x)

and Jp (x) and J�p (x) are not independent solutions.

88

XI. LECTURE 13 THE ORTHOGONALITY AND NORMALIZATION OF

BESSEL FUNCTIONS

� Origin of the Bessel Equation: Consider the Laplace equation in cylindrical coordi-nates,

r2V =1

s

@

@s

�s@V

@s

�+1

s2@2V

@'2+@2V

@z2= 0


V (s; '; z) = R (s)� (')Z (z)

we have

� (')Z (z)

s

d

ds

�sdR (s)

ds

�+R (s)Z (z)

s2d2� (')

d'2

+R (s)� (')d2Z (z)

dz2= 0

so that multiplying by s2=R (s)� (')Z (z), we �nd

s2�

1

sR (s)

d

ds

�sdR (s)

ds

�+

1

Z (z)

d2Z (z)

dz2

�+

1

� (')

d2� (')

d'2= 0:

There follows that

s2�

1

sR (s)

d

ds

�sdR (s)

ds

�+

1

Z (z)

d2Z (z)

dz2

�= p2;

1

� (')

d2� (')

d'2= �p2:

We can write the di¤erential equation

s2�

1

sR (s)

d

ds

�sdR (s)

ds

�+

1

Z (z)

d2Z (z)

dz2

�= p2

as

1

sR (s)

d

ds

�sdR (s)

ds

�+

1

Z (z)

d2Z (z)

dz2=p2

s2

1

sR (s)

d

ds

�sdR (s)

ds

�� p2

s2+

1

Z (z)

d2Z (z)

dz2= 0

89

which leads to

1

sR (s)

d

ds

�sdR (s)

ds

�� p2

s2= �

�q

r0

�2;

1

Z (z)

d2Z (z)

dz2=

�q

r0

�2:

where q is a constant and r0 is the radius of the cylinder. The di¤erential equation

1

sR (s)

d

ds

�sdR (s)

ds

�� p2

s2= �

�q

r0

�2can be rewritten as

sd

ds

�sdR (s)

ds

�� p2R (s) = �

�q

r0

�2s2R (s)

) sd

ds

�sdR (s)

ds

�+

"�q

r0

�2s2 � p2

#R (s) = 0

) s2d2R (s)

ds2+ s

dR (s)

ds

+

"�q

r0

�2s2 � p2

#R (s) = 0:

Introducing the transformation of variable de�ned by

qs = r ) dr = qds

we have

r2d2R (r)

dr2+ r

dR (r)

dr+

�r2

r2o� p2

�R (r) = 0

If the radius of the cylinder is ro; then we can introduce a dimensionless variable

de�ned by

x =r

ro; dr = rodx;R (r) = y (x)

so that

x2d2y (x)

dx2+ x

dy (x)

dx+�x2 � p2

�y (x) = 0:

Since

0 � r � ro; and x =r

ro

the solution to the di¤erential equation

x2d2y (x)

dx2+ x

dy (x)

dx+�x2 � p2

�y (x) = 0:

exists only for 0 � x � 1:

90

� The Orthogonality and Normalization: In the previous lecture we did solve the BesselDi¤erential Equation

x2y00 + xy0 +�x2 � p2

�y = 0

and have found the solution is the Bessel function Jp (x) : We now determine the

orthogonality and orthonormality of these functions. To this end, noting that

x2y00 + xy0 = x (xy00 + y0) = xd

dx

�xdy

dx

�we can rewrite the Bessel Di¤erential Equation as

xd

dx

�xdy

dx

�+�x2 � p2

�y = 0

and its solutions is y(x) = Jp (x) : Suppose the variable x is u instead of x, the

di¤erential equation would be

ud

du

�udy

du

�+�u2 � p2

�y = 0

and the solution y(u) = Jp (u) = Jp (�x) : Let�s assume u = �x; where � is arbitrary

constant ) du = �dx, then

�xd

�dx

��x

dy

�dx

�+��2x2 � p2

�y = 0

) xd

dx

�xdy

dx

�+��2x2 � p2

�y = 0:

Since � is arbitrary constant we can consider two solutions given by y1(x) = Jp (�1x)

and y2(x) = Jp (�2x) and the corresponding di¤erential equations can be written as

xd

dx

�xdJp (�1x)

dx

�+��21x

2 � p2�Jp (�1x) = 0;

xd

dx

�xdJp (�2x)

dx

�+��22x

2 � p2�Jp (�2x) = 0

Multiplying the �rst equation by Jp (�2x) and the second by Jp (�1x), we have

xJp (�2x)d

dx

�xdJp (�1x)

dx

�+��21x

2 � p2�Jp (�1x) Jp (�2x) = 0;

91

xJp (�1x)d

dx

�xdJp (�2x)

dx

�+��22x

2 � p2�Jp (�1x) Jp (�2x) = 0;

so that subtracting these two equations leads to

xJp (�2x)d

dx

�xd

dxJp (�1x)

��xJp (�1x)

d

dx

�xd

dxJp (�2x)

�

+�21Jp (�2x) Jp (�1x)x2 � p2Jp (�2x) Jp (�1x)

��22Jp (�1x) Jp (�2x)x2

+p2Jp (�1x) Jp (�2x) = 0

xJp (�2x)d

dx

�xd

dxJp (�1x)

��xJp (�1x)

d

dx

�xd

dxJp (�2x)

�+��21 � �22

�Jp (�1x) Jp (�2x)x

2 = 0:

Dividing by x; we have

Jp (�2x)d

dx

�xd

dxJp (�1x)

�

�Jp (�1x)d

dx

�xd

dxJp (�2x)

�+��21 � �22

�Jp (�1x) Jp (�2x)x = 0:

and integrating with respect to x from 0 to 1Z 1

0

Jp (�2x)d

dx

�xd

dxJp (�1x)

�dx

�Z 1

0

Jp (�1x)d

dx

�xd

dxJp (�2x)

�dx

+��21 � �22

� Z 1

0

Jp (�1x) Jp (�2x)xdx = 0:

92

Using integration by parts Z 1

0

udv

dxdx = uvj10 �

Z 1

0

vdu

dxdx

we have Z 1

0

Jp (�2x)d

dx

�xd

dxJp (�1x)

�dx

= Jp (�2x)xd

dx(Jp (�1x))

��10

�Z 1

0

xd

dx(Jp (�1x))

d

dx(Jp (�2x)) dx

and Z 1

0

Jp (�1x)d

dx

�xd

dxJp (�2x)

�= Jp (�1x)x

d

dx(Jp (�2x))

��10

�Z 1

0

xd

dx(Jp (�2x))

d

dx(Jp (�1x)) dx

so that Z 1

0

Jp (�2x)d

dx

�xd

dxJp (�1x)

�dx

�Z 1

0

Jp (�1x)d

dx

�xd

dxJp (�2x)

�dx

= Jp (�2x)xd

dx(Jp (�1x))

��10

� Jp (�1x)xd

dx(Jp (�2x))

��10

�Z 1

0

xd

dx(Jp (�1x))

d

dx(Jp (�2x)) dx

+

Z 1

0

xd

dx(Jp (�2x))

d

dx(Jp (�1x)) dx

)Z 1

0

Jp (�2x)d

dx

�xd

dxJp (�1x)

�dx

�Z 1

0

Jp (�1x)d

dx

�xd

dxJp (�2x)

�dx

93

= Jp (�2x)xd

dx(Jp (�1x))

��10

� Jp (�1x)xd

dx(Jp (�2x))

��10

Since at x = 0, we �nd xJp (�x) = 0, we �nd

)Z 1

0

Jp (�2x)d

dx

�xd

dxJp (�1x)

�dx

�Z 1

0

Jp (�1x)d

dx

�xd

dxJp (�2x)

�dx

= Jp (�2)d

dx(Jp (�1x))

��x=1

� Jp (�1)d

dx(Jp (�2x))

��x=1

Substituting this result intoZ 1

0

Jp (�2x)d

dx

�xd

dxJp (�1x)

�dx

�Z 1

0

Jp (�1x)d

dx

�xd

dxJp (�2x)

�dx

+��21 � �22

� Z 1

0

Jp (�1x) Jp (�2x)xdx = 0

we �nd

Jp (�2)d

dx(Jp (�1x))

��x=1

� Jp (�1)d

dx(Jp (�2x))

��x=1

+��21 � �22

� Z 1

0

Jp (�1x) Jp (�2x)xdx = 0:

The zeroes of the Bessel function: The zeroes of the bessel function f�1; �2; �3; :::g arethe values of x at which

Jp (x) = 0:

in the �gure below the points on the x axis where the Jp (x) intersects (i.e. Jp (x) = 0)

are the zeroes of the Bessel function.

94

If we chose �1 and �2 to be the zero�s of the Bessel function (i.e. Jp (�1) = Jp (�2) = 0),

the we �nd ��21 � �22

� Z 1

0

Jp (�1x) Jp (�2x)xdx = 0:

For �1 6= �2; we must have Z 1

0

Jp (�1x)xJp (�2x) dx = 0:

This is the orthogonality condition for the Bessel function. It can be expressed, using

the Dirac notation as

hJp (�ix)j xJp (�jx)i =

8<: 0 i 6= j

cons: i = j:

To �nd the normalization condition (i.e. �1 = �2) let �2 = �; �1 = � + �, when � is

the zero�s of the Bessel function (i.e. Jp (�2) = Jp (�) = 0) then using

�21 � �22 = (�+ �)2 � �2 = 2��+ �2

Jp (�2x) = Jp (�x)

Jp (�1x) = Jp ((�+ �)x) = f (�)

the equation

Jp (�2)d

dx(Jp (�1x))

��x=1

� Jp (�1)d

dx(Jp (�2x))

��x=1

+��21 � �22

� Z 1

0

Jp (�1x) Jp (�2x)xdx = 0:

95

can be written as the equation

� Jp (�+ �)d

dx(Jp (�x))

��x=1

+�2��+ �2

� Z 1

0

Jp ((�+ �)x) Jp (�x)xdx = 0:

)�2��+ �2

� Z 1

0

Jp ((�+ �)x) Jp (�x)xdx

= Jp (�+ �)d

dx(Jp (�x))

��x=1

)�2��+ �2

� Z 1

0

Jp ((�+ �)x) Jp (�x)xdx

= Jp (�+ �)d

dx(Jp (�x))

��x=1

Noting thatd

dx(Jp (�x))

��x=1

= �dJp (x)

dx

��x=�

= �J 0p (�)

we can write

)�2��+ �2

� Z 1

0

Jp ((�+ �)x) Jp (�x)xdx

= �J 0p (�) Jp (�+ �)

)Z 1

0

Jp ((�+ �)x) Jp (�x)xdx

=�J 0p (�) Jp (�+ �)

(2��+ �2)

If we make a Taylor series expansion for f (�) about � = 0, we have

f (�) = f (0) + �df (�)

d�

��=0

+�2

2!

d2f (�)

d�2

��=0

+�3

3!

d3f (�)

d�3

��=0

+ :::

then for f (�) = Jp ((�+ �)) ; we have

f(�) = Jp (�+ �)) f(0) = Jp (�)

df (�)

d�

��=0

=dJp ((�+ �))

d�

��=0

= J 0p (�)

96

so that

Jp (�+ �) = Jp (�) + �J 0p (�)

+�2

2!J00

p�x+ :::

Since � is the zero�s of the Bessel function we know Jp (�) = 0:Hence

Jp (�+ �) = �J 0p (�) +�2

2!J00

p�x+ :::

Substituting this into Z 1

0

Jp ((�+ �)x) Jp (�x)xdx

=�J 0p (�) Jp (�+ �)

(2��+ �2)

we �nd Z 1

0

Jp ((�+ �)x) Jp (�x)xdx

=�J 0p (�)

h�J 0p (�) +

�2

2!J00p�x+ :::

i(2��+ �2)

)Z 1

0

Jp ((�+ �)x) Jp (�x)xdx

=�J 0p (�)

�J 0p (�) +

�2!J00p�x+ :::

�(2�+ �)

Now if we set � = 0, we �ndZ 1

0

Jp (�x) Jp (�x)xdx =�J 0p (�) J

0p (�)

2�

)Z 1

0

Jp (�x) Jp (�x)xdx =J 02p (�)

2:

Therefore, the orthonormality condition for the Bessel function can be written as

hJp (�ix)j xJp (�jx)i =J 0p (�i) J

0p (�j)

2�ij:

� Limiting (Asymptotic) Forms for the Bessel Functions: The Bessel functions takes thefollowing approximate expression for di¤erent limiting cases:

97

(a) For x << 1 : We recall the Bessel Function

Jp (x) =�x2

�p� 1

� (1) � (p+ 1)

� 1

� (2) � (p+ 2)

�x2

�2+

1

� (3) � (p+ 3)

�x2

�4� 1

� (4) � (p+ 4)

�x2

�6+ :::

�:

so that dropping all higher order terms beginning from�x2

�2for x << 1;we �nd

Jp (x) '�x2

�p 1

� (1) � (p+ 1)

Using

� (1) = 1;� (p+ 1) = p!

we �nd

Jp (x) '1

p!

�x2

�p:

Also using


sin (�p):

and

J�p (x) =�x2

��p� 1

� (1) � (�p+ 1)

� 1

� (2) � (�p+ 2)

�x2

�2+

1

� (3) � (�p+ 3)

�x2

�4� 1

� (4) � (�p+ 4)

�x2

�6+ :::

�:

we can show that

Np (x) � �(p� 1)!

�

�2

x

�p(b) For x >> 1

Jp (x) �r2

�xcos

�x� 2p+ 1

4�

�Np (x) �

r2

�xsin

�x� 2p+ 1

4�

�Note:Many di¤erential equations occur in practice that are not of the standard form

of the Bessel di¤erential equation we saw but whose solution can be written in terms

98

of the Bessel functions. For example the di¤erential equation

y00 +1� 2ax

y0 +

��bcxc�1

�2+a2 � p2c2

x2

�y = 0

has the solution

y = xaZp (bxc)

where Zp stands for Jp or Np or any linear combination of them, and a; b; c are con-

stants. See the lengthening pendulum example.

Fuch�s Theorem for Second-Order Di¤erential Equations: So far we have seen a second

order di¤erential equation of the form

y00 + f (x) y0 + g (x) y = 0

can be solved using the standard power series expansion method (e.g. the Legendre

di¤erential equation) or the more general Frobenius method (e.g. the Bessel equa-

tion). If you think that the Frobenius method can be used to solve any second order

di¤erential equation, then you are wrong. There is a necessary condition that must be

satis�ed if the di¤erential equation is solvable using Frobenius method. Fuch�s Theo-

rem determines the necessary and su¢ cient condition to apply the Frobenius Method.

It states that for the di¤erential equation

y00 + f (x) y0 + g (x) y = 0

the solution can be obtained using the Frobenius method if and only if the functions

de�ned by F (x) = f (x) y0 and G (x) = g (x) y can be expressed as power series

(i:e. a convergent series)

F (x) = f (x) y0 =Xn=0

anxn

G (x) = g (x) y =Xn=0

bnxn:

When this is the case the solutions to the di¤erential equation consist of either

1. two Frobenius series

y1 (x) =

1Xn=0

Anxn+s; y2 (x) =

1Xn=0

Bnxn+s

or

99

2. one solution y1 (x) which is a Frobenius series, and the second solution which is given

by

y2 (x) = y1 (x) ln x+ y (x)

where y (x) is a Frobenius series. That means

y1 (x) =1Xn=0

Anxn+s;

y2 (x) = y1 (x) ln x+1Xn=0

Bnxn+s

The second case occurs only when the roots of the indicial equation are equal or

di¤er by an integer.

Ex. 22 Consider the di¤erential equation

y00 +1

x2y0 � 2

x3y = 0

which has the solution

y(x) = e1=x:

Discuss the applicability of a Frobenius-type generalized power-series solution to the

solution of this di¤erential equation.

Sol: We note that the di¤erential equation has the form

y00 + f (x) y0 + g (x) y = 0

where

f (x) =1

x2; g (x) = � 2

x3

Using the the given solution

y(x) = e1=x

we see that

F (x) = f (x) y0 =1

x2d

dx

�e1=x

�= � 1

x4e1=x

and

G(x) = g (x) y = � 2x3e1=x

100

These two functions F (x) and G(x) have a singular point at x = 0 and can not be

expressed as Frobenius series and therefore the Frobenius Method can not be used to

�nd the solution of the di¤erential equation.

Ex. 23 Bessel Functions: An Application (The lengthening Pendulum): Consider a simple

pendulum that consists of a mass m attached to string of length l0 at the initial time

t = 0: The length of the pendulum is increasing at a steady rate v: Find the equation

of motion and determine the solution for small oscillation.

Sol: At a given time t let the length of the pendulum be l =px2 + y2. Then the gravita-

tional potential energy can be expressed by

U = �mgl cos (�)

and the kinetic energy

T =1

2m�l2 _�2 + _l2�2

�:

Then the Lagrangian becomes

L = T � U

=1

2m�l2 _�2 + _l2

�+mgl cos (�) :

Using Euler-Lagrange equations

d

dt

�@L

@ _�

�� @L

@�= 0

we �ndd

dt

�ml2 _�

�+mgl sin (�) = 0

) d

dt

�l2 _��+ gl sin (�) = 0

The length is increasing at a constant rate (v), which means

l = l0 + vt) _l = v ) dl

v= dt

then transforming the DE from t to l we �nd

vd

dl

�l2v

d�

dl

�+ gl sin (�) = 0

101

) l2v2d2�

dl2+ 2lv2

d�

dl+ gl sin (�) = 0

) d2�

dl2+2

l

d�

dl+

g

lv2sin (�) = 0:

For small angle �, we have sin (�) ' � and the DE becomes

) d2�

dl2+2

l

d�

dl+

g

lv2� = 0:

We now compare this with the DE

y00 +1� 2ax

y0 +

��bcxc�1

�2+a2 � p2c2

x2

�y = 0

whose solution is given by the Bessel function

y = xaZp (bxc) :

To this end, we note that

1� 2ax

=2

l) x = l; a =

�12

which gives �bclc�1

�2+

�12

�2 � p2c2

l2=

g

lv2

There follows that �1

2

�2� p2c2 = 0;

�bclc�1

�2=

g

lv2

) bclc�1 =

pg

vl�1=2 ) bc =

pg

v; lc�1 = l�1=2

) c =1

2) b = 2

pg

v;

�1

2

�2� p2c2 = 0) p = �1

Therefore, the solution can be expressed as

� (l) = l�1=2Z1

�2

pg

vx1=2

�:

102

XII. LECTURE 14 INTRODUCTION TO PARTIAL DIFFERENTIAL EQUA-

TIONS AND THE SEPARATION OF VARIABLES

� Review for ordinary di¤erential equation (ODE): An ODE is an equation in someunknown function, say f , involving possibly di¤erent orders of derivatives of that

function with respect to a single variable, say x. We solve the physical problem

described by the ODE by �rst obtaining the general solution, and then applying the

boundary conditions (BCs) imposed by the physical constraints to the system.


d2f (x)

dx2= �k2f (x) :

Sol: Assuming a solution of the form

f (x) = Ae�x

we �nd the indicial equation

�2 � k2 = 0:

For the plus case the solutions are

�1 = k; �2 = �k

and the general solution is given by

f (x) = A1e�kx + A2e

kx:

For the minus case the solution to the indicial equation are

�1 = ik; �2 = �ik

and the general solution becomes

f (x) = B1 cos (kx) +B2 sin (kx) :

� A partial di¤erential equation (PDE): A PDE is an equation in some unknown functionof more than one variable, say f = u(x; y; z; t), involving possibly di¤erent orders of

partial derivatives of that function. The following are some examples of PDEs

103

1. Gauss�Law for the Electric Field : The di¤erential form of Gauss�law states that the

electric �eld, ~E (x; y; z) of a volume charge distribution � (x; y; z) satis�es the PDE

r � ~E (x; y; z) = � (x; y; z)

�0:

If we use

~E (x; y; z) = Ex (x; y; z) x+ Ey (x; y; z) y

+ Ez (x; y; z) z

r = @

@x+

@

@y+

@

@z

we may write Gauss�Law as

@Ex (x; y; z)

@x+@Ey (x; y; z)

@y+@Ez (x; y; z)

@z

=� (x; y; z)

�0:

2. Poisson�s Equation: If we express the electric �eld, ~E (x; y; z) in terms of the electric

potential V (x; y; z) which is given by

~E (x; y; z) = �rV (x; y; z)

Gauss�s Law can be expressed as

r � [�rV (x; y; z)] = � (x; y; z)

�0

) r2V (x; y; z) = �� (x; y; z)�0

or

@2V (x; y; z)

@x2+@2V (x; y; z)

@y2+@2V (x; y; z)

@z2

= �� (x; y; z)�0

:

This PDE is known as Poisson�t equation.

3. Laplace�s Equation: If the charge density � (x; y; z) = 0, the Poisson�s equation be-

comes Laplace equation. It is given by

@2V (x; y; z)

@x2+@2V (x; y; z)

@y2

+@2V (x; y; z)

@z2= 0:

104

4. The di¤usion or heat �ow equation:

5. Wave equation:

6. Helmholtz Equation:

7. Schrödinger Equation:

We will study a technique commonly used to solve the PDEs discussed above. This

technique is called the separation of variables technique. We will learn this technique

by solving problems.

Ex. 24 Consider an in�nitely long, rectangular waveguide of width w and height h with the

top surface held at the constant potential V (y = h) = V o, and with all other sides

grounded. A vacuum pump has been connected to the waveguide to remove most

of the air within it. Find the electrostatic potential V (x; y; z) everywhere within the

waveguide.

Sol: Inside the waveguide there is no charge and the electric potential satis�es the Laplace

equation@2V (x; y; z)

@x2+@2V (x; y; z)

@y2+@2V (x; y; z)

@z2= 0:

Using separation of variables we express the electric potential as a product of three

independent functions A (x) ; B (y) ; and C (z) as

V (x; y; z) = A (x)B (y)C (z) :

105

We substituting this expression into the PDE

@2A (x)B (y)C (z)

@x2+@2A (x)B (y)C (z)

@y2

+@2A (x)B (y)C (z)

@z2= 0

we �nd

B (y)C (z)d2A (x)

dx2+ A (x)C (z)

d2B (y)

dy2

+A (x)B (y)d2C (z)

dz2= 0:

Now we divide the entire equation by V (x; y; z) = A (x)B (y)C (z)

B (y)C (z)

A (x)B (y)C (z)

d2A (x)

dx2

+A (x)C (z)

A (x)B (y)C (z)

d2B (y)

dy2

+A (x)B (y)

A (x)B (y)C (z)

d2C (z)

dz2= 0

so that

1

A (x)

d2A (x)

dx2+

1

B (y)

d2B (y)

dy2

+1

C (z)

d2C (z)

dz2= 0:

This equation consists of three independent terms and the sum of these terms must

be zero. This is possible if and only if these terms are constants. Therefore, we can

write

1

A (x)

d2A (x)

dx2= �k2x;

1

B (y)

d2B (y)

dy2= �k2y;

1

C (z)

d2C (z)

dz2= k2z ;

so that

k2x + k2y � k2z = 0:

These equations are ODE and can be solved using any of the techniques we have

studied so far. Considering the x dependent, we have

1

A (x)

d2A (x)

dx2= �k2x )

d2A (x)

dx2+ k2xA (x) = 0:

106

Similarly for the y dependence we �nd

d2B (y)

dy2+ k2yB (y) = 0:

Assuming kx and ky are real constants the solutions to the above ODEs, using the

results in Ex. 23, may be expressed as

Akx (x) = E cos (kxx) +H sin (kxx) ;

By (y) = C cos (kyy) +D sin (kyy) :

For the z dependentd2C (z)

dz2� k2zC (z) = 0

the solution, again applying the result in Ex. 23, can be expressed as

Ckxky (z) = Fekzz +Ge�kzz:

Then the general solution for the electric potential becomes

V (x; y; z) =Xkxky ;kz

AkxBkyCkxky

=Xkxky

[Ekx cos (kxx) +Hkx sin (kxx)]

��Cky cos (kyy) +Dky sin (kyy)

��Fekzz +Ge�kzz

�:

Since the cylinder is in�nite along the z-direction and the potential must be �nite, we

must have

Fekzz +Ge�kzz = cont

) kz = 0:

This result along with the equation

k2x + k2y � k2z = 0:

leads to

k2x + k2y = 0:

107

Then the potential can rewritten as

V (x; y; z) =Xkxky

Akx (x)Bky (y)Ckxky (z)

=Xkxky

[Ekx cos (kxx) +Hkx sin (kxx)]


�:

Now we apply the boundary conditions. We are told that the electric potential has

the following boundary conditions

Vkxky (x = 0; y; z) = 0;

Vkxky (x = w; y; z) = 0;

Vkxky (x; y = 0; z) = 0;

Vkxky (x; y = h; z) = V0:

Applying the �rst boundary conditions, we �nd

Vkxky (x = 0; y; z) =Xkxky

Ekx


�= 0:

) Ekx = 0

using this result and the second boundary condition, we obtain

Vkxky (x = w; y; z) =Xkxky

Hkx sin (kxw)


�= 0

) sin (kxw) = 0

) kx =n�

w;

where n = 0; 1; 2; 3:::.Substituting Ekx = 0, kx = n�wand using the third boundary

condition, we see that

Vkxky (x; y = 0; z) =Xkxky

Hkx sin (kxx)Cky = 0

) Cky = 0:

108

Before we apply the last boundary condition it is better to look at the behavior of

the electric potential for z. We know the electric potential must be �nite at all points

inside the waveguide. However, since the waveguide is in�nitely long the z dependent

part diverges for positive as well as negative z as z tends to in�nity. To eliminate the

divergence of the electric potential for large z we must have

limz!�1

Vnm (x; y; z) = Constant8 x and y:

) Fkxkyepk2x+k

2yz +Gkxkye

�pk2x+k

2yz = Constant:

Therefore, the general solution becomes

V (x; y; z) =Xn=1

H 0n;ky sin

�n�wx�sin (kyy) :

where we have de�ned a new constant

H 0kx = HkxDky :

Recalling that

k2x + k2y = 0) k2y = �k2x

) ky = �ikx =in�

w

sin (kyy) = sin

�in�

wy

�= i sinh

�n�wy�

Using this result the potential can be rewritten as

V (x; y; z) =Xn=1

H 0n sin

�n�wx�sinh

�n�wy�;

where we have included i into the constant H 0n:Now applying the fourth boundary

condition, we have

Vkxky (x; y = h; z) = V0

)Xn=1

H 0n sin

�n�wx�sinh

�n�

h

w

�= V0

where n = 1; 2:::.Multiplying both sides by sin�n�wx�and integrating with respect to

x over the width of the waveguide, we haveXn=1

H 0n sinh

�n�

h

w

�Z W

0

sin�n�wx�sin�m�wx�dx

= V0

Z W

0

sin�m�wx�dx

109

so that noting that Z W

0

sin�n�wx�sin�m�wx�dx =

w

�

Z W

0

sin�n�wx�sin�m�wx�d�x�w

�=w

�

Z �

0

sin (nu) sin (mu) d (u) =w

�

��2�mn

�)Z W

0

sin�n�wx�sin�m�wx�dx =

w

2�mn

and Z W

0

sin�m�wx�dx = � w

m�cos�m�wx��w0

)Z W

0

sin�m�wx�dx =

w

m�[1� (�1)m]

we �nd Xn=1

H 0n sinh

�n�

h

w

�w

2�mn = V0

w

m�[1� (�1)m]

) H 0m sinh

�m�

h

w

�= V0

2

m�[1� (�1)m]

H 0m =

8<: 0 m = even

4V0m� sinh(m� hw)

m = odd

Then the potential becomes

V (x; y; z) =4V0�

Xn=0

1

(2n+ 1) sinh�(2n+ 1) � h

w

�� sin

�(2n+ 1) �

wx

�sinh

�(2n+ 1) �

wy

�:

110

XIII. LECTURE 15: LAPLACES�S EQUATION IN SPHERICAL COORDINATES

� Laplace�s Equation: We have seen an example in the previous lecture illustrating howwe solve the Laplaces�s equation in Cartesian coordinates

@2

@x2V (x; y; z) +

@2

@y2V (x; y; z)

+@2

@z2V (x; y; z) = 0

We now proceed solving the Laplaces�s equation in spherical coordinates.

� The Laplacian in Spherical Coordinates: In spherical coordinates the Laplacian isgiven by

r2 =1

r2@

@r

�r2@

@r

�+

1

r2 sin �

@

@�

�sin �

@

@�

�+

1

r2 sin2 �

@2

@'2= 0:

we shall illustrate how to solve the Laplaces�s equation in spherical coordinates by

considering the following electrostatic example.

Ex. 3 All of space is initially �lled with a uniform electric �eld of magnitude Eo pointing in

the positive z-direction. A grounded conducting sphere of radius a is introduced into

the space with its center at the origin of coordinates. We wish to �nd the electrostatic

potential at all points outside of the sphere.

Sol: Since there is no free charge outside the sphere, the electrical potential V (~r) satis�es

the Laplaces�s equation. Because of the spherical symmetry, it is easy to solve the

Laplaces�s equation in spherical coordinates,

r2V (~r) = 0

) 1

r2@

@r

�r2@

@rV (r; �; ')

�+

1

r2 sin �

@

@�

�sin �

@

@�V (r; �; ')

�+

1

r2 sin2 �

@2

@'2V (r; �; ') = 0:

While solving this partial di¤erential equation I will outline the basic steps which we

used in the previous example. It can be used as a general guideline to solve partial

di¤erential equations.

111

(1) Identify and write down the boundary conditions for the given problem: For this prob-

lem we know that the sphere is grounded. Hence, the electric potential at any point

on the surface of the sphere must be zero. That means

V (r = a; �; ') = 0:

In addition we should expect that the presence of the conducting sphere far away from

the sphere (in�nity) is negligible. In other words at in�nity the electric potential is

just the electric potential of the constant electric �eld pointing along the z-direction.

Recalling that the electric potential V and electric �eld ~E are related by

~E = �rV;

for an electric �eld pointing along the positive z-direction, we have

~E = Eoz = �rV (x!1; y !1; z !1)

) V (x!1; y !1; z !1) = �Eoz:

We need to express this potential in spherical coordinates

V (r !1; �; ') = �Eor cos �:

(2) Introduce separation of variables: Write the function as a product of independent

functions (this may now work for all partial di¤erential equations). For Laplaces�s

equation in spherical coordinates, we may write

V (r; �; ') = R (r)� (�) � (') :

(3) Substitute the product functions, get independent terms, and write the resulting di¤er-

ential equations: using the product function in the partial di¤erential equation and

simplifying the resulting expression, �nd three (if possible) or two parts which depend

on single or two variables, respectively. For this example, we have

1

r2@

@r

�r2@

@r[R (r)� (�) � (')]

�+

1

r2 sin �

@

@�

�sin �

@

@�[R (r)� (�) � (')]

�+

1

r2 sin2 �

@2

@'2[R (r)� (�) � (')] = 0:

112

�(�) � (')

r2d

dr

�r2dR (r)

dr

�+R (r) � (')

r2 sin �

d

d�

�sin �

d�(�)

d�

�+R (r)� (�)

r2 sin2 �

d2� (')

d'2= 0:

Dividing the entire equation by R (r)� (�) � ('), we have

1

r2R (r)

d

dr

�r2dR (r)

dr

�+

1

r2 sin ��(�)

d

d�

�sin �

d�(�)

d�

�+

1

r2 sin2 �� (')

d2� (')

d'2= 0;

and multiplying by r2

1

R (r)

d

dr

�r2dR (r)

dr

�+

1

sin ��(�)

d

d�

�sin �

d�(�)

d�

�+

1

sin2 �� (')

d2� (')

d'2= 0:

This can be put in the form

F1 (r) + F2 (�; ') = 0

where

F1 (r) =1

R (r)

d

dr

�r2dR (r)

dr

�;

F2 (�; ') =1

sin ��(�)

d

d�

�sin �

d�(�)

d�

�+

1

sin2 �� (')

d2� (')

d'2= 0:

We note that F1 (r) depends on r and F2 (�; ') depends on � and ' which are inde-

pendent. Therefore, each of these function must be a constant,

F1 (r) = l (l + 1) ; F2 (�; ') = �l (l + 1) :

113

There follows that

1

R (r)

d

dr

�r2dR (r)

dr

�= l (l + 1)

) d

dr

�r2dR (r)

dr

�� l (l + 1)R (r) = 0

) r2d2R (r)

dr2+ 2r

dR (r)

dr� l (l + 1)R (r) = 0

and

1

sin ��(�)

d

d�

�sin �

d�(�)

d�

�+

1

sin2 �� (')

d2� (')

d'2= �l (l + 1)

) sin �

�(�)

d

d�

�sin �

d�(�)

d�

�+

1

� (')

d2� (')

d'2= �l (l + 1) sin2 �

) sin �

�(�)

d

d�

�sin �

d�(�)

d�

�+ l (l + 1) sin2 �

+1

� (')

d2� (')

d'2= 0:

We note that in the above expression the �rst two terms depend on � and the third

term depend on ': Therefore we must have

sin �

�(�)

d

d�

�sin �

d�(�)

d�

�+ l (l + 1) sin2 � = m2

1

� (')

d2� (')

d'2= �m2:

Simplifying these equations, we �nd

sin �d

d�

�sin �

d�(�)

d�

�+sin2 �

�l (l + 1)� m2

sin2 �

��(�) = 0

d2� (')

d'2+m2� (') = 0:

114

(4) Solve the resulting di¤erential equations: If we introduce transformation of variables

de�ned by

x = cos � ) dx = � sin �d�; sin � =p1� x2

) d� = � dx

sin �) d�

sin �= � dx

sin2 �= � dx

1� x2

we may write the di¤erential equation

sin �d

d�

�sin �

d�(�)

d�

�+sin2 �

�l (l + 1)� m2

sin2 �

��(�) = 0

as

��1� x2

� d

dx

��1� x2

� d�(x)dx

�+�1� x2

� �l (l + 1)� m2

1� x2

��(x) = 0

)�1� x2

� d2�(x)dx2

� 2xd�(x)dx

+

�l (l + 1)� m2

1� x2

��(x) = 0

This is the associated Legendry di¤erential equation, which we saw in the previous

chapter, whose solutions are given by the associated Legendry functions Pml (x) : There-

fore, for the � dependence, we �nd

�lm (�) = Pml (cos �) :

The di¤erential equation for the ' dependence is

d2� (')

d'2+m2� (') = 0

and its solution is given by

�m (') = Am cos (m') +Bm sin (m')

where �l � m � l from the periodicity of the function (i.e. � (') = � ('+ 2�)). Now

115

we need to �nd the solution for the di¤erential equation for the r dependent part

d

dr

�r2dR (r)

dr

�� l (l + 1)R (r) = 0

) r2d2R (r)

dr2+ 2r

dR (r)

dr� l (l + 1)R (r) = 0

) d2R (r)

dr2+2

r

dR (r)

dr� l (l + 1)

r2R (r) = 0

The above di¤erential equation is singular at r = 0. In such cases we know that the

solution is determined by using Frobenius method. Let�s assume a solution given by

R (r) =Xn=0

anrn+s

so that

dR (r)

dr=Xn=0

an (n+ s)xn+s�1;

d2R (r)

dr2=Xn=0

an (n+ s) (n+ s� 1)xn+s�2:

Then the di¤erential equation

d2R (r)

dr2+2

r

dR (r)

dr� l (l + 1)

r2R (r) = 0

becomes Xn=0

an (n+ s) (n+ s� 1)xn+s�2

+2Xn=0

an (n+ s)xn+s�2

�l (l + 1)Xn=0

anxn+s�2 = 0

Xn=0

an [(n+ s) (n+ s� 1)

+2 (n+ s)� l (l + 1)] xn+s�2

= 0:

Xn=0

an [(n+ s) (n+ s+ 1)� l (l + 1)] xn+s�2

= 0:

116

There follows that

(n+ s) (n+ s+ 1)� l (l + 1) = 0

) s2 + (2n+ 1)s+ n(n+ 1)� l(l + 1) = 0

and solving for s, we �nd

s1 = � (n+ l + 1) ; s2 = l � n

Therefore, the solution to the di¤erential equation becomes for s1 = � [n+ (l + 1)]

R1 (r) =Xn=0

anrn+s1 =

Pn=0 anrl+1

=Alrl+1

and for s2 = l � n

R2 (r) =Xn=0

anrn+s2 =

Xn=0

anrl = Alr

l:

Then the general solution is a linear combination of these two functions which we can

write as

Rl (r) = Blrl +

Clrl+1

:

Now we can write the general solution for the Laplace equation in spherical coordinates

(in this case the the electric potential) as

V (r; �; ') =1Xl=0

lXm=�l

Rl (r)�lm (�) �m (')

or

V (r; �; ') =1Xl=0

lXm=�l

�Blr

l +Clrl+1

�Pml (cos �)

� (Am cos (m') +Bm sin (m')) :

(4) Apply the boundary conditions: The sphere is grounded

V (r = a; �; ') = 0:

which means

V (r = a; �; ') =1Xl=0

lXm=�l

�Bla

l +Clal+1

�Pml (cos �)

� (Am cos (m') +Bm sin (m')) = 0:

117

Blal +

Clal+1

= 0) Cl = �Bla2l+1

Using this result we may want to rewrite the potential as

V (r; �; ') =1Xl=0

lXm=�l

�Blr

l � Bla2l+1

rl+1

�Pml (cos �)


) V (r; �; ') =1Xl=0

lXm=�l

Bl

�rl � a2l+1

rl+1

��Pml (cos �) (Am cos (m') +Bm sin (m')) :

Now using the second boundary condition

V (r !1; �; ') = �E0r cos �

we �nd

V (r !1; �; ') = limr!1

1Xl=0

lXm=�l

Blrl

�Pml (cos �) (Am cos (m') +Bm sin (m'))

= �E0r cos �

Since there is no ' dependence on the right hand side, we must have

m = 0

which leads to

limr!1

1Xl=0

A0BlrlP 0l (cos �) = �E0r cos �:

Expanding the series and including the constant A0 into Bl; we have

B0 +B1r cos � + limr!1

1Xl=2

BlrlP 0l (cos �) = �E0r cos �:

comparing the right and left hand side of this equation we �nd

B0 = 0; B1 = �E0; Bl = 0 for l > 1:

118

Substituting this result into the reduced expression for the potential

) V (r; �; ') =

1Xl=0

Bl

�rl � a2l+1

rl+1

�P 0l (cos �)

we �nd

V (r; �; ') =

�a3

r3� 1�E0r cos �:

119

A. Lecture 16 Laplace�s Equation in Cylindrical Coordinates

� In cylindrical coordinates the Laplace�s equation

r2V (~r) = 0

is given by

) 1

s

@

@s

�s@

@sV (s; '; z)

�+1

s2@2

@�2(V (s; '; '))

+@2

@z2V (s; �; z) = 0:

Ex. 4 A right, circular conducting cylindrical shell of radius r0 and length L has its axis

coincident with the z-axis and its ends at z = 0 and z = L. All sides of the cylinder

are grounded except for the face at z = 0, which is maintained at a position-dependent

potential speci�ed by

V (s; '; z = 0) = s cos'

Find the electrostatic potential everywhere inside the cylinder.

Sol: Using separation of variable

r2V =1

s

@

@s

�s@V

@s

�+1

s2@2V

@'2+@2V

@z2= 0


V (s; '; z) = R (s)� (')Z (z)

we have

� (')Z (z)

s

d

ds

�sdR (s)

ds

�+R (s)Z (z)

s2d2� (')

d'2

+R (s)� (')d2Z (z)

dz2= 0

so that multiplying by s2=R (s)� (')Z (z), we �nd

s2�

1

sR (s)

d

ds

�sdR (s)

ds

�+

1

Z (z)

d2Z (z)

dz2

�+

1

� (')

d2� (')

d'2= 0:

120

There follows that

s2�

1

sR (s)

d

ds

�sdR (s)

ds

�+

1

Z (z)

d2Z (z)

dz2

�= p2;

1

� (')

d2� (')

d'2= �p2:

We can write the di¤erential equation

s2�

1

sR (s)

d

ds

�sdR (s)

ds

�+

1

Z (z)

d2Z (z)

dz2

�= p2

as

1

sR (s)

d

ds

�sdR (s)

ds

�+

1

Z (z)

d2Z (z)

dz2=p2

s2

1

sR (s)

d

ds

�sdR (s)

ds

�� p2

s2+

1

Z (z)

d2Z (z)

dz2= 0

which leads to

1

sR (s)

d

ds

�sdR (s)

ds

�� p2

s2= �

�q

r0

�2;

1

Z (z)

d2Z (z)

dz2=

�q

r0

�2:

where q is a constant and r0 is the radius of the cylinder. The di¤erential equation

1

sR (s)

d

ds

�sdR (s)

ds

�� p2

s2= �

�q

r0

�2can be rewritten as

sd

ds

�sdR (s)

ds

�� p2R (s) = �

�q

r0

�2s2R (s)

) sd

ds

�sdR (s)

ds

�+

"�q

r0

�2s2 � p2

#R (s) = 0

) s2d2R (s)

ds2+ s

dR (s)

ds

+

"�q

r0

�2s2 � p2

#R (s) = 0:

Introducing the transformation of variable de�ned by

qs = r ) dr = qds

121

we have

r2d2R (r)

dr2+ r

dR (r)

dr+

�r2

r2o� p2

�R (r) = 0

If the radius of the cylinder is ro; then we can introduce a dimensionless variable

de�ned by

x =r

ro; dr = rodx;R (r) = y (x)

so that

x2d2y (x)

dx2+ x

dy (x)

dx+�x2 � p2

�y (x) = 0:

Since

0 � r � ro; and x =r

ro

the solution to the di¤erential equation

x2d2y (x)

dx2+ x

dy (x)

dx+�x2 � p2

�y (x) = 0:

exists only for 0 � x � 1:This is the Bessel DE and its solutions are given by the

Bessel functions

y (x) = apNp (x) + bpNp (x) :

Noting that the solution for the ' dependent of the di¤erential equation

1

� (')

d2� (')

d'2= �p2

the solution is given by

� (') = Ap cos (p') +Bp sin (p')

and for the z-dependence1

Z (z)

d2Z (z)

dz2=

�q

r0

�2the solution is given by

Z (z) = Cqeqr0z+Dqe

� qr0z:

We also recall the variable

x =r

ro; qs = r

122

Therefore the general solution for the Laplace equation is given by

V (s; '; z) =Xq;p

�Cqe

qr0z+Dqe

� qr0z�

� (Ap cos (p') +Bp sin (p'))

��apJp

�qs

ro

�+ bpNp

�qs

ro

��:

Now we write the boundary conditions

V (s; '; z = 0) = s cos'

V (s = r0; '; z) = 0:

V (s; '; z = L) = 0

and we also recall

Jp (x) =1Xn=0

(�1)n

� (n+ 1)� (n+ 1 + p)

�x2

�2n+p:

and


sin (�p):

If we apply the �rst boundary condition, we have

V (s; '; z = 0) =Xq;p

(Cq +Dq)

� (Ap cos (p') +Bp sin (p'))

��apJp

�qs

ro

�+ bpNp

�qs

ro

��= s cos':

which leads to

Bp = 0; p = 1

Since p is an integer the �rst few terms in J�p (x) are zero because � (n+ 1� p) is a

the gama function of a negative integer, which is in�nite

J�p (x) = (�1)pJp (x)

and Jp (x) and J�p (x) are not independent solutions. Therefore, we can rewrite the

potential as

V (s; '; z) =Xq

�Cqe

qr0z+Dqe

� qr0z�

� cos (') J1�qs

ro

�:

123

Using the third boundary condition

V (s; '; z = 0) =Xq

�Cqe

qr0L+Dqe

qr0L�

� cos (') J1�qs

ro

�= 0:

we �nd

Cqeqr0L+Dqe

� qr0L= 0) Dq = �Cqe2

qr0L:

Then the simpli�ed expression for the potential would be

V (s; '; z) =Xq

Cq

�eqr0z � e

2 qr0Le� qr0z�

� cos (') J1�qs

ro

�:

or

V (s; '; z) =Xq

2Cqeqr0L

e� qr0(L�z) � e

qr0(L�z)

2

!

� cos (') Jp�qs

ro

�:

V (s; '; z) =Xq

Fq sinh

�q

r0(L� z)

�� cos (') J1

�qs

ro

�:

Now we apply the second boundary condition

V (s = r0; '; z) =Xq

Fq sinh

�q

r0(L� z)

�� cos (') J1 (q) = 0:

this requires

J1 (q) = 0

which means

q = qii = 1; 2; 3:::

124

are the zeros of the Bessel function. Thus the potential can be written as

V (s; '; z) =Xi

Fi sinh

�qir0(L� z)

�� cos (') J1

�qis

ro

�:

To �nd Fi, we recall the orthonormality of the Bessel function

hJp (qix)j xJp (qjx)i =Z 1

0

Jp (qix) Jp (qjx)xdx

=J 0p (qi) J

0p (qj)

2�ij�ij:

and the boundary condition

V (s; '; z = 0) = s cos'

which leads to

V (s; '; z = 0) =Xi

Fi sinh

�qir0L

�� cos (') J1

�qis

ro

�= s cos':

)Xi

Fi sinh

�qir0L

�J1

�qis

ro

�= s:

Multiplying both sides by J1�qjs

ro

�srod�sro

�and integrating over s, we have

Xi

Fi sinh

�qir0L

��Z r0

0

J1

�qis

ro

�J1

�qjs

ro

�s

rod

�s

ro

�=

Z r0

0

sJ1

�qjs

ro

�s

rod

�s

ro

�:

and replacings

ro= x

Xi

Fi sinh

�qir0L

��Z 1

0

J1 (qix) J1 (qj)xd (x)

=

Z 1

0

r0xJ1 (qjx)xd (x) :

125

)Xi

Fi sinh

�qir0L

�J 0p (qi) J

0p (qj)

2�ij

=

Z 1

0

r0xJ1 (qjx)xdx:

) Fj sinh

�qjr0L

�J 02p (qj)

2=

Z 1

0

r0x2J1 (qjx) dx

) Fj =r0R 10x2J1 (qjx) dx

sinh�qjr0L�J 02p (qj)

2

:

Using the result

we may write

Fj =r0J2 (qj)

qj sinh�qjr0L�J 021 (qj)2

:

or applying

J 0p (qj) = JP+1 (qj)

we �nd

) Fj =2r0

qj sinh�qjr0L�J2 (qj)

:

Therefore the potential becomes

V (s; '; z) =Xj

2r0

qj sinh�qjr0L�J2 (qj)

� sinh�qjr0(L� z)

�cos (') J1

�qjs

ro

�:

where qj are the zeroes of the Bessel function

126

XIV. LECTURE 17 POISSON�S EQUATION

� In the previous three lectures we have seen several examples from electrostatics to

determine the electric potential of a given charge distribution in a region where there

is no charge with speci�c boundary conditions. In such cases the electric potential

satisfy the Laplace�s equation. We have seen how to solve that the Laplace�s equation

in Cartesian coordinates where it is given by

@2

@x2V (x; y; z) +

@2

@y2V (x; y; z)

+@2

@z2V (x; y; z) = 0;

in cylindrical coordinates

1

r

@

@r

�r@

@rV (r; '; z)

�+1

r2@2

@'2V (r; '; z)

+@2

@z2V (r; '; z) = 0;

and in spherical coordinates

1

r2@

@r

�r2@

@rV (r; �; ')

�+

1

r2 sin �

@

@�

�sin �

@

@�V (r; �; ')

�+

1

r2 sin2 �

@2

@'2V (r; �; ') = 0:

We know see how we determine the electric potential in a region where there is some

volume charge distribution, � (~r). In such cases the electric potential, V (x; y; z) satis-

�es the Poisson�s Equation given by

r2V (x; y; z) = �� (~r)�0

where �0 is the electrical permittivity of free space. I will discuss the basic procedures

for solving Poisson�s Equation and applying the boundary conditions using an example

in spherical coordinates.

� Approach to Solving Poisson�s Equation: when we studied how to solve inhomogenioussecond order linear di¤erential equations, for example

d2y

dx2+ a

dy

dx+ y = f (x)

127

we have seen that the solution is sum of the homogenous yh and the particular yp

solutions given by

y = yh + yp:

The Homogenous solution yh satis�es the equation

d2yhdx2

+ adyhdx

+ yh = 0

and the particular solution yp is determined using the di¤erent techniques we discussed.

The same principle is applied in solving Poisson�s equation. For the Poisson�s equation

r2V (x; y; z) = �� (~r)�0

the homogenous solutions Vh is basically is the solution to the Laplace equation

r2Vh (x; y; z) = 0

which we already know how to determine the solution. Suppose we determine the

particular solution Vp, then the solutions is given by

V (x; y; z) = Vh + Vp:

We can easily show that this solution satis�es the Poisson�s equation

r2V (x; y; z) = r2 (Vh + Vp)

= r2Vh +r2Vp

Since Vh is the solution to the Laplace�s equation, we

r2Vh = 0

and Vp is the particular solution

r2Vp = �� (~r)

�0

we can see that

r2V (x; y; z) = �� (~r)�0

:

Once we determine the homogenous and particular solutions we then write the general

solution and apply the boundary conditions.

128

Ex. 4 Point Charge and Grounded Conducting Sphere

Note: This is the same example as Boas works out (Ex. 13.8.1), but she uses Gaussian

units, so the equations look a bit di¤erent. We are using SI units, as usual.

A point charge Q is located a distance a from the center of a grounded, conducting

sphere of radius R, where R < a. Find the electrostatic potential everywhere outside

of the sphere. Assume that the center of the sphere is at the origin of coordinates,

and that the point charge is on the positive z axis.

Sol: We want to �nd the electric potential at some point outside the sphere r > R: Since

there is a point charge along the z axis at a distance a (> R) which is in the region

outside the sphere we obviously see that the charge density is not zero in this region

and the Laplace�s equation is not valid in all regions outside the sphere. Therefore,

to �nd the electric potential we need to �nd the solution to the Poisson�s Equation.

From the nature of the problem it is better to use spherical coordinates

r2V (r; �; ') = �� (~r)�0

) 1

r2@

@r

�r2@

@rV (r; �; ')

�+

1

r2 sin �

@

@�

�sin �

@

@�V (r; �; ')

�+

1

r2 sin2 �

@2

@'2V (r; �; ') = �� (~r)

�0:

The Homogenous solution for this equation is the general solution to the Laplace�s

equation in spherical coordinates

1

r2@

@r

�r2@

@rVh(r; �; ')

�+

1

r2 sin �

@

@�

�sin �

@

@�Vh(r; �; ')

�+

1

r2 sin2 �

@2

@'2Vh(r; �; ') = 0

and we recall that the general solution is given by

Vh (r; �; ') =

1Xl=0

lXm=�l

�Blr

l +Clrl+1

�Pml (cos �)


129

We recall that for a point charge Q located at a position ~r0 the electric potential at a

distance ~r is given by

VQ (r; �; ') =1

4��0

Q

j~r � ~r0j :

For ~r = r sin � cos'x+ r sin � sin'y + r cos �z and ~r0 = az, we have

j~r � ~r0j =�r2 + a2 � 2ra cos �

�1=2so that

VQ (r; �; ') =1

4��0

Q

(r2 + a2 � 2ra cos �)1=2:

To �nd the particular solution we note that the charge outside the sphere is zero except

the point charge at a point on the z axis a distance a from the origin. Therefore, the

particular solution for the Poisson�s equation

r2V (r; �; ') = �� (~r)�0

is that of the electrical potential of this point charge

Vp (r; �; ') = VQ (r; �; ')

=1

4��0

Q

(r2 + a2 � 2ra cos �)1=2:

The general solution to the Poisson�s equation can then be written as

V (r; �; ') = Vh (r; �; ') + Vp (r; �; ') :

V (r; �; ') =1Xl=0

lXm=�l

�Blr

l +Clrl+1

�Pml (cos �)

� (Am cos (m') +Bm sin (m'))

+1

4��0

Q

(r2 + a2 � 2ra cos �)1=2:

Now we will apply the boundary conditions.

1. The electric potential must be �nite as r !1

2. At any point on the surface of the sphere the electric potential must be zero since

the sphere is grounded. That means

130

V (r = R; �; ') = 0:

Applying the �rst boundary condition V (r !1; �; ')must be �nite (basically it must

be zero since there are no other charges) gives

limr!1

V (r; �; ')

= limn!1

1Xl=0

lXm=�l

�Blr

l +Clrl+1

�Pml (cos �)


+1

4��0

Q

(r2 + a2 � 2ra cos �)1=2:

) limr!1

V (r; �; ') = limn!1

1Xl=0

lXm=�l

Blrl

�Pml (cos �) (Am cos (m') +Bm sin (m'))

since this expression diverges, we must have

Bl = 0:

Therefore, we may rewrite the electric potential as

V (r; �; ') =

1Xl=0

lXm=�l

Clrl+1

Pml (cos �)


+1

4��0

Q

(r2 + a2 � 2ra cos �)1=2:

Now using the second boundary condition

V (r = R; �; ') = 0, we have

V (r = R; �; ') =

1Xl=0

lXm=�l

ClRl+1

Pml (cos �)


+1

4��0

Q

(R2 + a2 � 2Ra cos �)1=2= 0:

131

Recall that Generating Function for the Legendry Polynomials

1

j~r � ~r0j =1p

r2 + r02 � 2rr0 cos �

=

1Xl=0

rl<rl+1>

Pl (cos �)

we may write

1

4��0

Q

(R2 + a2 � 2Ra cos �)1=2

=Q

4��0

1Xl=0

Rl

al+1Pl (cos �)

for R < a and the electric potential on the surface of the sphere becomes

V (r = R; �; ') =1Xl=0

lXm=�l

ClRl+1

Pml (cos �)


+Q

4��0

1Xl=0

Rl

al+1Pl (cos �) = 0:

The above equation to be zero independent of '; we at least have the coe¢ cients for

Pml (cos �) must be independent of cos (m') and sin (m') : This requires

Am = 0; Bm = 0 for m 6= 0:

and the above expression becomes

V (r = R; �; ') =

1Xl=0

ClRl+1

Pl (cos �)

+Q

4��0

1Xl=0

Rl

al+1Pl (cos �) = 0:

V (r = R; �; ') = 0

)1Xl=0

�ClRl+1

+Q

4��0

Rl

al+1

�Pl (cos �) = 0:

There follows that

ClRl+1

+Q

4��0

Rl

al+1= 0

) Cl = �Q

4��0

R2l+1

al+1:

132

Therefore, the electric potential is given by

V (r; �; ') = � Q

4��0

1Xl=0

1

rl+1R2l+1

al+1Pl (cos �)

+1

4��0

Q

(r2 + a2 � 2ra cos �)1=2:

If we rewrite the series term as

� Q

4��0

1Xl=0

1

rl+1R2l+1

al+1Pl (cos �) =

�QRa

4��0

1Xl=0

1

rl+1R2l

alPl (cos �)

) � Q

4��0

1Xl=0

1

rl+1R2l+1

al+1Pl (cos �) =

q0

4��0

1Xl=0

r0l

rl+1Pl (cos �)

with r0 = R2

aand q0 = �QR

a; we can apply the generating function for Legendry

polynomials

1

j~r � ~r0j =1p

r2 + r02 � 2rr0 cos �

=1Xl=0

r<r>Pl (cos �)

and write

� Q

4��0

1Xl=0

1

rl+1R2l+1

al+1Pl (cos �) =

+q0

4��0

1pr2 + r02 � 2rr0 cos �

:

Then the electric potential can be expressed as

V (r; �; ') =q0

4��0

1pr2 + r02 � 2rr0 cos �

+Q

4��0

1pr2 + a2 � 2ra cos �

;

where r0 = R2

aand q0 = �QR

a:The above result shows that the electric potential due to

a point chargeQ placed outside the sphere at a distance a from the center of a grounded

133

conducting sphere can be imagined as sum of the potential due to the point charge

Q and a negative "image charge"�q0 = �QR

a

�located inside the sphere at a distance

r0 = R2

afrom the center of the sphere. In reality this image charge is not a charge

existing at this position. It is the total induced charge on the grounded conducting

sphere due to the electric �eld produced by the point charge Q. The method of

replacing this induced charge by an image charge to �nd the electric potential is called

the method of images. It is a useful method in determining the electric potential

of charges placed around conducting sphere, in�nitely long conducting cylinder, or

in�nitely wide conducting plate.

N.B. To learn more about Method of Images sign up for Electricity &

Magnetism (PHYS 4310).

134

XV. FUNCTIONS OF COMPLEX VARIABLES LECTURE 18

A. Analytic Functions

Complex Variables and Complex Functions [Summary Phys 3150]

� Real, Imaginary, and Complex Numbers: numbers that have the form

z = a+ ib

with a and b real numbers and

i =p�1) i2 = �1

are called complex numbers. In any complex number z = a+ ib; a is the real part and

b is the imaginary part

Re z = a; Im z = b

� Rectangular and Polar Representation of Complex Numbers: In the rectangular com-plex plane the real part a is the x coordinate and the imaginary part b is the y

coordinate

a = x; b = y

In polar coordinate the complex number z = a+ ib is represented by

a = r cos �; b = r sin �

where r is called the modulus or the magnitude and � is called the phase of the complex

number z

� Euler�s Equation and Exponential Representation of Complex Numbers : Any complexnumber z = x+ iy can be expressed in an exponential form

z = x+ iy = r exp (i�) = r cos � + ir sin �

where the magnitude and the phase are given by

jzj = r =px2 + y2; � = tan�1

�yx

�

135

� Complex Conjugate and Magnitude of a Complex Number : The complex conjugate ofa complex number z = x+ iy is denoted by z� and is given by

z� = x� iy

The magnitude of a complex number z is related to its complex conjugate by

jzj =pzz� =

px2 + y2

Ex. 1 Consider the complex function

f (z) =2z + 1

z � i

Find the real and imaginary parts of this function.

Sol: To �nd the real and imaginary part of a complex function for

z = x+ iy

we must be able to express the function f (z) as

f (x+ iy) = u (x; y) + iv (x; y) :

In order to do that we substitute z = x+ iy into the given function

f (x+ iy) =2 (x+ iy) + 1

(x+ iy)� i=2x+ 1 + i2y

x+ i (y � 1)

) f (x+ iy) =[2x+ 1 + i2y] [x� i (y � 1)][x+ i (y � 1)] [x� i (y � 1)]

) f (x+ iy) =[2x+ 1 + i2y] [x� i (y � 1)]

x2 + (y � 1)2

) f (x+ iy) = [(2x+ 1) x+ 2y (y � 1)

+i [2xy � (2x+ 1) (y � 1)]] =�x2 + (y � 1)2

�) f (x+ iy) = [(2x+ 1) x+ 2y (y � 1)

+i [�y + 2x+ 1]] =�x2 + (y � 1)2

�There follows that

u (x; y) =(2x+ 1) x+ 2y (y � 1)

x2 + (y � 1)2

v (x; y) =2x� y + 1

x2 + (y � 1)2

136

Ex. 2 Find the real and imaginary parts of the complex function

f(z) = z1=4:

Sol: Whenever we are given exponential complex functions it is better to use polar coordi-

nates. Let

z = r exp (i�)

where r and � are related to Cartesian coordinates x and y by

r =px2 + y2; � = tan�1

�yx

�:

The function can then be expressed as

f(z) = z1=4 ) f�rei��= [r exp (i�)]1=4

) f(z) = r1=4 exp

�i�

4

�= r1=4 cos

��

4

�+ ir1=4 sin

��

4

�so that the real and imaginary parts can be expressed as

u (r; �) = r1=4 cos

��

4

�; v (r; �) = r1=4 sin

��

4

�� Analytic Functions: a function f (x) is analytic (or regular or holomorphic or mono-genic) in a region of the complex plane if it has a (unique) derivative at every point of

the region. The statement "f (z) is analytic at a point z = z0" means that f (z) has

a derivative at every point inside some small circle about z = z0:

df

dz

��z=z0

= lim�z!0

f (z0 +�z)� f (z0)

�z

Example:

d (z2)

dz= 2z;

d (cos z)

dz= � sin z;

d

dz

�p1 + z2

�=

zp1 + z2

(chain rule)

Non-analytic function

f (z) = jzj2

How Can We Tell if a Given Function is Analytic?

137

� The Cauchy-Riemann Conditions: If f(z) = u (x; y) + iv (x; y) is analytic in a region,

then in that region@u

@x=@v

@y;@v

@x= �@u

@y

Proof : Noting that z = x+ iy, we can write

@f

@x=@f

@z

@z

@x) @f

@z=@f

@x;

@f

@y=@f

@z

@z

@y) @f

@y= i

@f

@z

) @f

@z= �i@f

@y:

which leads to@f

@x= �i@f

@y

And using f(z) = u (x; y) + iv (x; y) ; we may also write

@f

@x=@u

@x+ i

@v

@x@f

@y=@u

@y+ i

@v

@y

and substituting these expressions into

@f

@x= �i@f

@y

we get

@u

@x+ i

@v

@x= �i

�@u

@y+ i

@v

@y

�) @u

@x+ i

@v

@x=@v

@y� i

@u

@y

Two complex function are equal if and only if their real part are equal and their

imaginary part are equal. Hence

@u

@x=@v

@y;@v

@x= �@u

@y:

� Some Terminology:

Regular Point: The point zo is said to be a regular point of the function f(z) if the

function f(z) is analytic at that point.

138

Singularity: If f(z) is not analytic at a point zo, then zo is said to be a singular point,

or singularity of f(z).

Isolated Singularity: If zo is the only singularity of a function f(z) within an arbitrarily

small region surrounding zo, then zo is said to be an isolated singularity of the function

f(z) = u (x; y) + iv (x; y)

� Harmonic and Conjugate Harmonic Functions: Consider �rst condition in the TheCauchy-Riemann Conditions for analyticity of a complex function

@u

@x=@v

@y

and di¤erentiate with respect to x

@2u

@x2=

@2v

@x@y

and also the second condition@v

@x= �@u

@y

and di¤erentiate it with respect to y

@2v

@x@y= �@

2u

@y2) @2u

@y2= � @2v

@x@y:

If we add the two equations, we have

@2u

@x2+@2u

@y2=

@2v

@x@y� @2v

@x@y= 0

) r2u = 0

which is Laplace�s equation in two dimension. Functions satisfying Laplace�s equation

are called harmonic functions. Thus solutions of Laplace�s equation which are real

and imaginary parts of a function f (z) are called conjugate harmonic functions.

� An Important Theorem: Let f(z) be analytic within a region R, with one or moresingularities on the borders (and possibly outside) of R. Then f(z) has derivatives of

all orders that exist at any point zo inside R, so that f(z) can be expanded in a Taylor

series about the point zo. This series converges for all points within a circle centered

at zo that extends to the closest singularity of f(z).

139

Ex. 3 Expand the complex function given below in a Taylor series about the origin, and �nd

the circle of convergence for this series.

f (z) =pz + 2i (0 � � � 2�)

Sol: We recall that the Tailor series of a function f (z) is given by

f (z) =1Xn=0

1

n!

dnf (z0)

dzn(z � z0)

n :

For the complex function f (z) =pz + 2i, we have

f (z) =pz + 2i

df (z)

dz=1

2

1pz + 2i

d2f (z)

dz2= � 1

221

(z + 2i)3=2

d3f (z)

dz3=3

231

(z + 2i)5=2

Therefore

f (z) =1Xn=0

1

n!

dnf (z0)

dzn(z � z0)

n

) f (z) =pz0 + 2i+

1

2

1

(z0 + 2i)1=2(z � z0)

� 123

1

(z0 + 2i)3=2(z � z0)

2

+1

241

(z0 + 2i)5=2(z � z0)

3 :::

From the above expression we note that

limz0!�2i

1

z0 + 2i!1

which shows that the function f (z) is singular at z0 = �2i: Therefore, the radius ofconvergence is

jz0j = 2:

Which means the series is convergent for all z inside the circle of radius R = 2 centered

about the origin

140

XVI. LECTURE 19 CONTOUR INTEGRATION AND CAUCHY�S THEOREM

� Cauchy�s Theorem: Consider a function f(z) that is analytic inside a region R of thecomplex plane. Within the region R is a closed curve C that does not cross itself, and

that has a �nite number of sharp corners. Then Cauchy�s theorem states thatIC

f (z) dz = 0

Proof : Using

z = x+ iy

for a complex function

f (z) = u (x; y) + iv (x; y)

the closed integral can be expressed asIC

f (z) dz

=

IC

[(u(x; y) + iv (x; y)) (dx+ idy)]

=

IC

[(u(x; y)dx� v (x; y) dy)]

+ i

IC

[(v (x; y) dx+ u(x; y)dy)]

Applying Green�s theorem in the plane [PHYS 3150]IC

[(P (x; y) dx+Q (x; y) dy)]

=

ZZA

�@Q

@x� @P

@y

�dxdy:

we may write IC

[(u(x; y)dx� v (x; y) dy)]

=

ZZA

�@ [�v]@x

� @u

@y

�dxdy

+ i

ZZA

�@v

@x� @u

@y

�dxdy

141

)IC

f (z) dz = �ZZA

�@v

@x+@u

@y

�dxdy

+ i

ZZA

�@u

@x� @v

@y

�dxdy:

For analytic function f (z) = u (x; y)+ iv (x; y), the Cauchy-Riemann condition states

that@u

@x=@v

@y;@v

@x= �@u

@y

which leads to IC

f (z) dz = �ZZA

��@u@y+@u

@y

�dxdy

+i

ZZA

�@v

@y� @v

@y

�dxdy = 0:

� The Equation of a Circle in the Complex Plane: Consider a circle on a complex planeas shown in the �gure below.

Any point on this circle, z can be described by the equation

z � zo = jz � zoj ei�:

� Cauchy�s Integral Formula: if f (z) is analytic on and inside a simple closed curve C,the value of f (z) at a point z = a inside C is given by

f (a) =1

2�i

IC

f (z)

z � adz:

142

Proof : Consider the function

g (z) =f (z)

z � a

which is analytic everywhere inside the the closed curve C except z�a:We divide theregion closed by the curve C as shown in the �gure below so that the singular point

z = a can be excluded from the curve C.

In the region bounded by C1; C2; l1; and l2, the function

g (z) =f (z)

z � a

is analytic. Then according to Cauchy�s Theorem, we must haveIC0

g (z) dz = 0

where C 0 is the curve shown in the �gure above. This integral can be expressed asIC0

g (z) dz =

ZC1

g (z) dz +

Zl2

g (z) dz

+

ZC2

g (z) dz +

Zl1

g (z) dz = 0

Since we are integrating along l1 and l2 for the same length but in opposite direction

we have Zl1

g (z) dz = �Zl2

g (z) dz

143

so that ZC1

g (z) dz +

ZC2

g (z) dz = 0

or IC1

f (z)

z � adz = �

IC2

f (z)

z � adz:

Note that from the �gure the integration over C1 is counterclockwise where as the

integration over C1 is clockwise. We can make the integration over C1 counterclockwise

and drop the minus sign. IC1

f (z)

z � adz =

IC2

f (z)

z � adz:

Now using equation of a circle on a complex plane we saw earlier, we may write

z � a = �ei� ) z = a+ �ei�

) dz = i�ei�d�

so that IC2

f (z)

z � adz =

2�Z0

f�a+ �ei�

��ei�

i�ei�d�

= i

2�Z0

f�a+ �ei�

�d�

which leads to IC1

f (z)

z � adz =

IC2

f (z)

z � adz = i

2�Z0

f�a+ �ei�

�d�:

Now if we let �! 0, we note that C1 becomes C and f�a+ �ei�

�= f (a) and we �ndI

C1

f (z)

z � adz = i

2�Z0

f (a) d� = 2�if (a)

) f (a) =1

2�i

IC1

f (z)

z � adz:

Note: A line integral in the complex plane is called a contour integral.

Ex. 4 Evaluate the contour integral

I =

Z 2+4i

1+i

z2dz

along the paths indicated.

144

(a) A straight line joining the points z1 = 1 + i and z2 = 2 + 4i.

(b) Two straight lines: the �rst from the point z1 = 1 + i to zo = 2 + i, and the second

from the point zo to the point z2 = 2 + 4i.

(C) Find the integral

I =

IC

z2dz

for the closed triangular curve

Sol:

(a) The equation of the line joining the two points can be expressed as

y � y1x� x1

=y2 � y1x2 � x1

) y � 1x� 1 =

4� 12� 1 = 3

) y = 3x� 2:

Using

z = x+ iy

145

we may write Z 2+4i

1+i

z2dz

=

Z 2+4i

1+i

�x2 � y2 + 2ixy

�(dx+ idy)

=

Z 2+4i

1+i

��x2 � y2 + 2ixy

�dx

+��2xy + i

�x2 � y2

��dy�

=

Z 2+4i

1+i

��x2 � y2

�dx� 2xydy

�+2i

Z 2+4i

1+i

�xydx+

�x2 � y2

�dy�

so that replacing

y = 3x� 2) dy = 3dx

we �nd Z 2+4i

1+i

z2dz

=

Z 2

1

��x2 � (3x� 2)2

�dx� 2x (3x� 2) 3dx

�+i

Z 2

1

�2x (3x� 2) dx+

�x2 � (3x� 2)2

�3dx�

=

Z 2

1

��26x2 + 24x� 4

�dx

+i

Z 2

1

��18x2 + 32x2 � 12

�dx

= �863� 6i

(b) the integral along the �rst line from the point z1 = 1+ i to zo = 2+ i, and the second

line from the point zo to the point z2 = 2 + 4i can be expressed asZ 2+4i

1+i

z2dz =

Z 2

1

�x2 � 1 + 2ix

�dx

+

Z 4

1

�22 � y2 + 4iy

�idy

146

=

�x3

3+ ix2 � x

�21

+ i

��y

3

3+ 2iy2 + 4y

�41

=4

3+ 3i� 30� 9i

)Z 2+4i

1+i

z2dz = �863� 6i

which is the same result as in part a.

(c) Using Cauchy�s theorem IC

f (z) dz = 0

for the function

f (z) = z2

which is analytic everywhere inside and on the triangle, we can easily �ndIC

z2dz = 0:

Noting that IC

z2dz =

Z z0

z1

z2dz +

Z z2

z0

z2dz +

Z z1

z2

z2dz:

)IC

z2dz =

Z z0

z1

z2dz +

Z z2

z0

z2dz �Z z2

z1

z2dz:

and using the result we obtained in part a and b,IC

z2dz = �863� 6i�

��863� 6i

�= 0:

Ex. 4 Evaluate the contour integral

I =

Iez

z (z + 1)dz

for the closed path C given by jz � 1j = 3.

Sol: Noting that1

z (z + 1)=1

z� 1

z + 1

147

we may write

I =

IC

ez

z (z + 1)dz

=

IC

ez

zdz +

IC

ez

z + 1dz

For the contour de�ned by the curvejz � 1j = 3 is shown in the �gure below

Noting that the function

f(z) = ez

is analytic inside the curve bounded by C and z = 0 and z = �1 are inside the thiscurve, applying Cauchy�s integral formula we can writeI

C

ez

zdz = 2�if (0) = 2�iI

C

ez

z + 1dz = 2�if (�1) = 2�ie�1

so that

I =

IC

ez

z (z + 1)dz = 2�i

�1 + e�1

�:

� Laurent�s Theorem:Consider a region R between two circles C1 and C2 centered at thesame point z = zo, and let f(z) be analytic in R.

148

Then f(z) can be expanded in a series of the form

f(z) = a0 + a1 (z � z0) + a2 (z � z0)2

+ :::+b1

z � z0+

b2

(z � z0)2 + :::

=1Xn=0

an (z � z0)n +

1Xn=1

bn(z � z0)

n

which converges for any value of z within the region R.

� Some Terminology:

If bn = 0 for all values of n, then f(z) is analytic at z = zo, and zo is said to be a

regular point of f(z).

The coe¢ cient b1 is called the residue of f(z).

If the principal part of the series has terms only up to bn(i.e. bn 6= 0, but all the bn+1= 0); then f(z) is said to have a pole of order n at the point z = zo

If the principal part of the series has only the single term b1 (i.e. bn+1 > 0 for all

n > 1), then f(z) is said to have a simple pole at z = zo.

If there are an in�nite number of b0s di¤erent from zero, f (z) has an essential singu-

larity at z = z0

149

A. Lecture 20 Residues and the Residue Theorem

� The Residue Theorem: Let zk be an isolated singular point of f (z) inside a closedcurve de�ned by C; the residue theorem states thatI

C

f (z) dz = 2�iR (zk) ;

where the integral is in a counterclockwise direction and R (zk) is the residue of the

function f (z) at z = zk:

Proof: If we de�ne a small circle of radius r = �, as shown in the �gure below

we can write [applying the procedure we followed when we proof the Cauchy integral

formula] that IC

f (z) dz =

IC2

f (z) dz:

We can expand the function f (z) in Laurent series in the shaded region since the

function is analytic in this region

f(z) = a0 + a1 (z � zk) + a2 (z � zk)2

+ :::+b1

z � zk+

b2

(z � zk)2 + :::

=

1Xn=0

an (z � zk)n +

1Xn=1

bn(z � zk)

n

150

This leads to IC

f (z) dz = a0

IC2

dz + a1

IC2

(z � zk) dz

+a2

IC2

(z � zk)2 dz

+:::+

IC2

b1z � zk

dz +

IC2

b2

(z � zk)2dz

+

IC2

b2

(z � zk)3dz:::

Since the function (z � z0)n is analytic everywhere in the region bounded by the curve

C2 for n = 0, all the integrals IC2

(z � zk)n dz = 0

for all n = 0 and we �ndIC

f (z) dz =

IC2

b1z � zk

dz +

IC2

b2

(z � zk)2dz

+

IC2

b2

(z � zk)3dz:::

Now using

z � zk = �ei� ) z = zk + �ei�

) dz = i�ei�d�

we have IC

f (z) dz = b1

IC2

1

�ei�i�ei�d�

+

IC2

b2

(�ei�)2i�ei�d�

+

IC2

b2

(�ei�)3i�ei�d�:::

which may want to put in the formIC

f (z) dz = 2�ib1 +

1Xn=2

ibn�n�1

IC2

e�i(n�1)�d�

151

IC

f (z) dz = 2�ib1

+

1Xn=2

ibn�n�1

�Z 2�

0

fcos [(n� 1) �]� i sin [(n� 1) �]g d�

Since Z 2�

0

cos ((n� 1) �) d�

= �iZ 2�

0

sin ((n� 1) �) d� = 0

for all n > 2;we �nd IC

f (z) dz = 2�ib1 = 2�iR (zk)

where

R (zk) = b1

is the residue of the function f (z) at z = zk

Note: If there are more than one residue, the residue theorem states thatIf (z) dz = 2�i

nXk=1

R (zk) :

Methods of Finding Residues

� Method 1 : Expand f(z) about the point zo and simply read the value of the residueo¤ of the series.

Ex. 5 Find R(0) for the complex function

f (z) = z cos

�1

z

�Sol: We recall that the series expansion of the function

f (z) = cos (z)

is

cos (z) = 1� z2

2!+z4

4!� z6

6!:::

152

so that

cos

�1

z

�= 1� 1

z21

2!+1

z41

4!

� 1

z61

6!:::

and

f (z) = z cos

�1

z

�= z � 1

z

1

2!+1

z31

4!� 1

z51

6!:

There follows that for zk = 0

R (zk) = b1 = �1

2:

Ex. 6 Find R(1) for the complex function

f (z) =ez

(z � 1)2:

Sol: The function has isolated singular point at zk = 1 and we need to expand the function

g (z) = ez

about this point. The series expansion about this point is is given by

g (z) = e

�1 +

1

2!(z � 1)2 + 1

3!(z � 1)3

+1

4!(z � 1)4 ::::

�then

f (z) =ez

(z � 1)2:

becomes

f (z) =e

(z � 1)2�1 +

1

1!(z � 1) + 1

2!(z � 1)2

+1

3!(z � 1)3 + 1

4!(z � 1)4 ::::

�

f (z) =e

(z � 1)2+

e

(z � 1) +1

2!e+

1

3!(z � 1)

+1

4!(z � 1)2 ::::

�There follows that for zk = 1

R (zk) = b1 = e:

153

Ex. 7 Find R(�) for the complex function

f (z) =sin z

z � �

Sol: The function has an isolated singular point at

zk = �

and we need to expand the function sin (z) in series about this point

sin (z) = � (z � �) +(z � �)3

3!

� (z � �)5

5!+(z � �)7

7!:::

so that we can write

f (z) =sin z

z � �

as

f (z) = �1 + (z � �)2

3!

� (z � �)4

5!+(z � �)6

7!:::

There follows that for zk = �

R (zk) = b1 = 0:

� Method 2: If f(z) has a simple pole (a pole of order 1) at zo, then multiply f(z) by(z � zo) and evaluate the result at z = zo (or take the limit as z approaches zo):

R (zo) = limz!zo

[(z � z0) f (z)]

Ex. 8 Find R(�1) for the complex function

f (z) =z

(z + 1) (z + 2):

Sol: We note that f (z) has a simple pole (a order of 1) at zo = �1. Then we can use the

154

second method discussed above to �nd the residue zo = �1

R (�1) = limz!�1

[(z + 1) f (z)]

) R (�1) = limz!�1

�(z + 1)

z

(z + 1) (z + 2)

�) R (�1) = lim

z!�1

�z

(z + 2)

�) R (�1) = �1

(�1 + 2) = �1:

Ex. 9 Find R(�) for the complex function

f (z) = cot (z)

Sol: The cotangent function is given by

f (z) = cot (z) =cos (z)

sin (z):

Let�s assume that this function has a simple pole at z0 = � which is true as we will

see shortly. Then using the second method the residue at z0 = � can be expressed as

R (�) = limz!�

[(z � �) f (z)]

) R (�) = limz!�

�(z � �)

cos (z)

sin (z)

�As z ! �, the limit becomes 0=0 and we can apply Le�Hospital rule

R (�) = limz!�

[(z � �) f (z)]

) R (�) = limz!�

�(z � �)

cos (z)

sin (z)

�) R (�) = lim

z!�

"ddz[(z � �) cos (z)]

ddzsin (z)

#

) R (�) = limz!�

�cos (z)� (z � �) sin (z)

cos (z)

�) R (�) =

cos (�)� (� � �) sin (�)

cos (�)

R (�) = 1

Now if we make a series expansion for the function

f (z) = cot (z) =cos (z)

sin (z):

155

we will �nd

f (z) =1

z � �� z � �

3� 1

45(z � �)3

� 2

945(z � �)5 + ::::

which indeed shows the function has a simple pole at z0 = �:

� Method 3 (Generalization of method 2): To �nd the residue of f(z) at zo when f(z)has a pole of order n at zo, compute

R (zo) =

�1

(m� 1)!dm�1

dzm�1f(z � z0)

m f (z)g�z=zo

;

where m > n:

Ex. 10 Find R (3) for

f (z) =zezt

(z � 3)2

where t is a(possible complex) parameter.

Sol: We note that f (z) has a pole at zo = 3 of order n = 2. Thus we can apply the third

method to �nd the residue at zo which we write as

R (3) =

�1

(2� 1)!d2�1

dz2�1

�(z � 3)2 zezt

(z � 3)2��

z=3

) R (3) =

�d

dz

�zezt

�z=3

) R (3) =�(1 + zt) ezt

�z=3

) R (3) = (1 + 3t) e3t:

Ex. 11 Find R(�2) for the complex function

f (z) =e2z

z (z + 2)3

156

Sol: The function has a pole at zo = �2 of order n = 3. Using the third method for

m = n = 3, we have

R (�2)

=

�1

(3� 1)!d3�1

dz3�1

�(z + 2)3

e2z

z (z + 2)3

��z=�2

) R (�2) =�1

2!

d2

dz2

�e2z

z

��z=�2

) R (�2) =�1

2!

d

dz

�2e2z

z� e2z

z2

��z=�2

) R (�2) = 1

2!

�4e2z

z� 2e

2z

z2� 2e

2z

z2+2e2z

z3

�z=�2

) R (�2) =�2e2z

z� 2e

2z

z2+e2z

z3

�z=�2

) R (�2) =��1� 1

2� 18

�e4 = �13

8e�4:

157

B. Lecture 21 Applications of the Residue Theorem

� The Residue Theorem: we recall that if fz1; z2; ::zk:::zng are singular points of f (z)inside a closed curve de�ned by C; the residue theorem states thatI

C

f (z) dz = 2�i

nXk=1

R (zk) ;

where R (zk) is the residue of the function f (z) at z = zk: We are going to use this

theorem and the methods of �nding residues to evaluate several di¤erent types of

de�nite integrals. The methods are best shown by examples.

Ex. 12 Evaluate the real integral

I =

1Z0

dx

x4 + 1

Sol: Consider the a semicircle of radius R on the upper half complex plane and the complex

function

f (z) =1

z4 + 1:

Now let�s integrate the complex function over the semicircular contour in the upper

half plane in the counterclockwise direction as shown in the �gure below

which we write as Idz

z4 + 1=

Z R

�R

dx

x4 + 1+

ZC

dz

z4 + 1

For the curved part, we have

z = R exp[i�]) dz = iR exp[i�]d�;

158

where 0 � � � �; thenIdz

z4 + 1=

Z R

�R

dx

x4 + 1+

Z �

0

iR exp[i�]d�

R4 exp[4i�] + 1:

Now if we let R ! 1, we can see that the integral over the curved part (the secondintegral) approaches to zero and we �nd thatI

dz

z4 + 1=

Z 1

�1

dx

x4 + 1:

since x4 + 1 is an even function we can writeZ 1

�1

dx

x4 + 1= 2

Z 1

0

dx

x4 + 1

)Z 1

0

dx

x4 + 1=1

2

Idz

z4 + 1:

The integral of the complex function

f (z) =1

z4 + 1

on the closed curve de�ned by this semicircle of in�nite radius, using the Residue

theorem, can be expressed asIf (z) dz = 2�i

nXk=1

R (zk) :

In order to �nd the residues we �rst need to get the poles. Noting that

z4 + 1 = 0) z = (�1)1=4 =�ei(�+n(2�))

�1=4;

where n = 0; 1; 2; 3 [Phys 3150 Roots of a Complex number ], the poles for the function

f (z) are found to be

z1 = ei�4 = cos

��4

�+ i sin

��4

�) z1 =

p2

2(1 + i) ;

z2 = ei3�4 =

p2

2(�1 + i) ;

z3 = ei5�4 = �

p2

2(1 + i) ;

z4 = ei7�4 =

p2

2(1� i)

159

Remember, we are integrating on the upper half of the complex plane. Hence we are

interested in the poles whose phase is 0 � � � �: These poles are

z1 = ei�4 =

p2

2(1 + i) ;

z2 = ei3�4 =

p2

2(�1 + i) :

We now determine the residue of these poles. We recall that the residue of a pole

z = zk of order n is given by

R (zk) =

�1

(m� 1)!dm�1

dzm�1f(z � zk)

m f (z)g�z=zk

where m > n: We note that the complex function f (z) can then be expressed as

f (z) =1

z4 + 1

= 1 /(z � z1) (z � z2) (z � z3) (z � z4)

f (z)

= 1.�z � ei

�4

� �z � ei

3�4

��z � ei

5�4

��z � ei

7�4

�This shows that the two poles at z1 = ei

�4 and z2 = ei

3�4 are simple poles (of order

one) and we can use the second method to �nd the residues at these poles

R (zo) = limz!zo

[(z � z0) f (z)] :

For z1 = ei�4 , we have

R�ei

�4

�= lim

z!ei�4

��z � ei

�4

��z � ei

�4

� �z � ei

3�4

��z � ei

5�4

��z � ei

7�4

�o

R�ei

�4

�=

= limz!ei

�4

h1.�

z � ei3�4

��z � ei

5�4

��z � ei

7�4

�iR�ei

�4

�= 1

.�ei

�4 � ei

3�4

��ei

�4 � ei

5�4

��ei

�4 � ei

7�4

�:

160

so that using

ei�4 � ei

3�4 =

p2

ei�4 � ei

5�4 =

p2 (1 + i)

ei�4 � ei

7�4 = i

p2

the residue becomes

R�ei

�4

�= 1

.�p2��p

2 (1 + i)��

ip2�

) R�ei

�4

�= �1

.2p2 (1� i) :

Similarly, the residue at the second pole zo = ei3�4 is given by

R�ei

3�4

�= lim

z!ei3�4

h�z � ei

3�4

�1.�z � ei

�4

� �z � ei

3�4

��z � ei

5�4

��z � ei

7�4

�i

R�ei

3�4

�=

= limz!ei

3�4

h1.�z � ei

�4

� �z � ei

5�4

��z � ei

7�4

�i

R�ei

3�4

�=

= 1.�

ei3�4 � ei

�4

��ei

3�4 � ei

5�4

��ei

3�4 � ei

7�4

�:

and using

ei3�4 � ei

�4 = �

p2

ei3�4 � ei

5�4 = i

p2

ei3�4 � ei

7�4 =

p2 (�1 + i)

we �nd

R�ei

3�4

�= 1

.��p2��

ip2��p

2 (�1 + i)�

) R�ei

�4

�= 1

.2p2 (i+ 1) :

161

Now using the residue theoremIf (z) dz = 2�i

nXk=1

R (zk)

we can see that Idz

z4 + 1= 2�i

�� 1

2p2 (1� i)

+1

2p2 (i+ 1)

�)I

dz

z4 + 1= �i

��i� 1 + 1� i

2p2

�= �i

��2i2p2

�)I

dz

z4 + 1=

�p2:

Therefore Z 1

0

dx

x4 + 1=1

2

Idz

z4 + 1

)Z 1

0

dx

x4 + 1=

�

2p2:

� Mathematica Result :

Ex. 13 Evaluate the real integral

I =

2�Z0

cos (3�)

5� 4 cos (�)d�

Sol: Noting that

cos (�) =1

2(exp (i�) + exp (�i�))

=1

2

�exp (i�) +

1

exp (i�)

�=1

2

�z +

1

z

�cos (�) =

1

2

�z2 + 1

z

�

162

and

cos (3�) =1

2(exp (3i�) + exp (�3i�))

=1

2

�(exp (i�))3 +

1

(exp (i�))3

�=1

2

�z3 +

1

z3

�) cos (3�) =

1

2

�z6 + 1

z3

�;

where z = exp (i�) ; we can write

cos (3�)

5� 4 cos (�) =12

�z6+1z3

�5� 2(z2+1)

z

) cos (3�)

5� 4 cos (�) =z6 + 1

2z2 [5z � 2 (z2 + 1)]

) cos (3�)

5� 4 cos (�) =z6 + 1

�2z2 [2z2 � 5z + 2]

For z = exp (i�) ; noting that

dz = i exp (i�) d� ) d� = �i exp (�i�) dz

) d� = �idzz;

the integral becomes

I =

2�Z0

cos (3�)

5� 4 cos (�)d�

=

Iz6 + 1

�2z2 [2z2 � 5z + 2]

��idz

z

�=

Ii (z6 + 1) dz

2z3 [2z2 � 5z + 2]

)2�Z0

cos (3�)

5� 4 cos (�)d� =If (z) dz;

where

f (z) =i (z6 + 1)

2z3 (2z2 � 5z + 2)and the closed curve is a unit circle centered about the origin shown in the �gure below

163

We can then use the Residue theoremIf (z) dz = 2�i

nXk=1

R (zk)

to �nd the integral. To this end, we �rst �nd out the residues inside the unit circle.

Noting that

2z2 � 5z + 2 = (2z � 1) (z � 2)

) 2z2 � 5z + 2 = 2 (z � 1=2) (z � 2)

we may write

f (z) =i (z6 + 1)

4 (z � 0)3 (z � 1=2) (z � 2):

This expression shows that the function f (z) has poles z1 = 0; z2 = 1=2; z3 = 2:

However, only the �rst two poles (z1 = 0; z2 = 1=2) are inside the unit circle. Thus we

need only the residue of these two poles. The second pole z2 = 1=2 is a simple pole

164

since the order is one and the residue can be expressed as

R (1=2) = limz!1=2

[(z � 1=2) f (z)]

) R (1=2) = limz!1=2

�i (z6 + 1)

4z3 (z � 2)

�) R (1=2) =

i�164+ 1�

12

�12� 2� = i

65

64

��43

�) R (1=2) = �i65

48:

The �rst pole z0 = z1 = 0 is not a simple pole. It has order n = 3. Thus we use the

third method to �nd the residue for this pole

R (zo) =

�1

(m� 1)!dm�1

dzm�1f(z � z0)

m f (z)g�z=zo

;

where m > n: For m = 3 we may write

R (0)

=

�1

(3� 1)!d3�1

dz3�1

�z3i (z6 + 1)

2z3 (2z2 � 5z + 2)

��z=0

) R (0) =

�1

2

d2

dz2

�i (z6 + 1)

2 (2z2 � 5z + 2)

��z=0

) R (0) =

�1

2

d2

dz2

�i (z6 + 1)

2 (2z2 � 5z + 2)

��z=0

R (0) =1

2

�� 6iz5 (4z � 5)(2z2 � 5z + 2)2

+15iz4

2z2 � 5z + 2

+i (1 + z6) (4z � 5)2

(2z2 � 5z + 2)3� 2i (1 + z6)

(2z2 � 5z + 2)2

#z=0

R (0) =i

2

(�5)2

23� 2

22

!=i

2

�25

8� 12

�) R (0) =

21i

16

165

Therefore the integral givesIf (z) dz = 2�i

nXk=1

R (zk) = 2�i

��i6548+i21

16

�If (z) dz =

4�

48=

�

12

)2�Z0

cos (3�)

5� 4 cos (�)d� =If (z) dz =

�

12

� Mathematica result :

� Standard Methods of Integration using Contour Integrals:

Method I: A real integral over the entire x-axis

Method 2 : A real integral involving sines and cosines from 0 to 2�

166

167

C. Lecture 22 The Calculus of Residues Applied: The Kramers-Kronig Relations

� The Residue Theorem: we recall thatIf (z) dz = 2�iR (zk)

where R (z0) is the residue of the function f (z) at z = zk: If the are more than one

residue, the residue theorem states thatIf (z) dz = 2�i

nXk=1

R (zk)

Here the poles zk must be inside the closed curve over which the contour integral is

being carried on. What if there are poles exactly on the curve. Which means the

function f (z) is not analytic at z0k that is/are exactly located somewhere on the curve.

For example, let�s say we are interested to evaluate the real integral

I1 =

Z 1

�1

dx

x2 � 1 :

To solve this integral we need to consider the closed curve shown in the �gure below

and evaluate the contour integral

I2 =

If (z) dz;

where

f (z) =1

z2 � 1 :

168

But we we know that f (z) has a simple poles at z01 = �1 and z02 = 1 which are exactlyon the contour integral not inside the contour integral. So how can we evaluate I1 by

directly using the residue theorem. We can evaluate such kind of integral using two

di¤erent approaches. We need to either put the poles inside or outside the contour by

modifying the semicircular contour if we are going to use the residue theorem directly.

As we will see either of these approach lead to the same result which result in a

modi�cation of the residue theorem.

� The revised Residue Theorem: states thatIf (z) dz = 2�i

"nXk=1

R (zk) +1

2

nXk=1

R (z0k)

#:

where zk is the pole inside the contour and z0k is the simple pole on the contour.

Proof :: First Approach (Put the pole inside the curve): The contour shown in the

�gure below keeps the pole inside the contour. For now let�s assume there are no other

poles inside this contour except z0 = z01 and z0 = z02 so thatIf (z) dz = 2�i

2Xk=1

R (z0k) :

For the modi�ed contour, we can writeIf (z) dz =

Z z01��

�Rf (z) dz +

Z z01+�

z01��f (z) dz

+

Z z02��

z01+�

f (z) dz +

Z z02+�

z02��f (z) dz +

Z R

z02+�

f (z) dz

Ex. 15 Evaluate the integral

I =

Z 1

0

sin (x)

xdx

� Dispersion (The Frequency Dependence of permittivity) The electrons in a noncon-ducting medium are bound to speci�c molecules. These electrons oscillates about the

equilibrium position with a small amplitude. The electrons experiencing this kind

of motion can be modeled as a harmonic oscillator. Then the electron experiences a

spring force Fs given by

Fs = �kx:

169

In addition to this force, the electron can also experience some damping force Fd which

is proportional to the speed of the electron

Fd = � mdx

dt:

If the electron is exposed to an EM wave with frequency ! and polarized in the x-

direction

~E = E0 cos (!t) x

it will experience a driving force

Fd = qE = qE0 cos (!t)

Then using newtons second law, the net force acting on the electron can be written as

md2x

dt2= qE0 cos (!t)�

dx

dt� kx

) md2x

dt2+ m

dx

dt+ kx = qE0 cos (!t)

) d2x

dt2+

dx

dt+k

mx =

qE0mcos (!t)

ord2x

dt2+

dx

dt+ !20x =

qE0mcos (!t)

where

!0 =

rk

m

is the natural frequency of the electron. In terms of the complex variables ~x and ~E,

we may write this equation as

d2~x

dt2+

d~x

dt+ !20~x =

qE0mexp (�i!t) (1)

The Homogenous solution to this di¤erential equation is given by

~xH (t) = e� t=2 (A cos (�t) +B sin (�t)) :

Substituting a particular solution of the form

~xp (t) = ~x0 exp (�i!t)

into the di¤erential equation, we �nd

�m!2~x0 � i !m~x0 + k~x0 = qE0 (2)

170

which gives

�m!2~x0 � im !~x0 +m!20~x0 = qE0

) ~x0 =qE0=m

!20 � !2 � i !(3)

and the particular solution becomes

~xp (t) =qE0=m

!20 � !2 � i !exp (�i!t)

Therefore the complex displacement of the electron is given by

~x (t) = ~xH (t) + ~xp (t)

~x (t) = e� t=2 (A cos (�t) +B sin (�t))

+qE0=m

!20 � !2 � i !exp (�i!t)

But we are interested in the steady state of the electron which happen if we waited

long enough (i.e. t ! 1). Thus for steady state the complex displacement of theelectron becomes

~x (t) ' qE0=m

!20 � !2 � i !exp (�i!t)

which shows at steady state the electron begins to oscillates with frequency of the EM

�eld as we would expect. The complex dipole moment of the electron can then be

expressed as

~p (t) = q~x (t) =q2E0=m

!20 � !2 � i !exp (�i!t)

This can also be put in the form

~p (t) =q2E0=m (!

20 � !2 + i !)

(!20 � !2)2+ ( !)2

exp (�i!t)

) ~p (t) = j~p0j exp [� (i!t� ')]

where

j~p0j =q2E0=mq

(!20 � !2)2+ ( !)2

and

' = tan�1�

!

!20 � !2

�:

171

This means that p is out of phase by ' with respect to the electric �eld ~E. Lagging

behind by ' which is very small when ! � !0 and rises to � when ! � !0. If there

are N molecules per unit volume and fj electrons per molecule. If these electrons in

the jth molecule is oscillating with the natural frequency !j and damped by j; then

the total dipole moment of the electrons per unit volume (the polarization ~P ) can be

expressed as

~P =Nq2

m

Xj=1

fj!2j � !2 � i j!

E0 exp (�i!t)

~P =Nq2

m

Xj=1

fj!2j � !2 � i j!

~E

This can also be put in the form

~P = �0�e ~E

where

�e =Nq2

�0m

Xj=1

fj!2j � !2 � i !

:

Recalling that the complex permittivity

~� = �0 (1 + �e)

and the complex dielectric constant can be expressed as

~�r =~�

�0= 1 + �e = 1 +

Nq2

�0m

Xj=1

fj!2j � !2 � i j!

(4)

Eq. (4) shows the dielectric constant (the permittivity of the medium) depends on the

frequency of the EM wave-the medium is dispersive. Still wave equation is satis�ed by

the electric and magnetic �elds. For a dispersive medium it is given by

r2 ~E = ~��0@2 ~E

@t2

with a plane wave

~E = ~E0 exphi�~kz � !t

�i:

where the wave number ~k is complex and is given by

!~k=

1p~��0

) ~k =p~��0!

172

If we express it as

~k = kRe + ikIm

The electric �eld may be put in the form

~E = ~E0 exp (�kImz) exp [i (kRez � !t)] :

and the intensity which is proportional to the square of the amplitude of the electric

�eld becomes

I /�� ~E0�� exp (�2kImz)

which shows a damping in the �eld amplitude due to the absorption by the medium.

For that reason the absorption coe¢ cient of the medium is given by

� = 2kIm:

The wave velocity is given by

v =!

kRe

and the refractive index of the medium

n =c

v=ckRe!

The complex wave number

~k =p~��0! =

r~�

�0�0�0! =

p�r!

c

) ~k =!

c

"1 +

Nq2

�0m

Xj=1

fj!2j � !2 � i j!

#1=2For gases the second terms is very small and we can make a binomial expansion for

the square root which gives

~k ' !

c

1 +

Nq2

2�0m

Xj=1

fj!2j � !2 � i j!

!:

Then we can �nd the real and imaginary part of the wave number to be

kRe '!

c

1 +

Nq2

2�0m

Xj=1

fj�!2j � !2

��!2j � !2

�2+ ( j!)

2

!

kIm 'Nq2!2

2c�0m

Xj=1

fj j�!2j � !2

�2+ ( j!)

2

173

so that the absorption coe¢ cient and the refractive index of the medium can be ex-

pressed as

� =Nq2!2

�0m

Xj=1

fj j�!2j � !2

�2+ ( j!)

2

n = 1 +Nq2

2�0m

Xj=1

fj�!2j � !2

��!2j � !2

�2+ ( j!)

2

For just one particular molecule (jthmolecule) we may write that

� =Nq2!2

�0m

fj j�!2j � !2

�2+ ( j!)

2

n = 1 +Nq2

2�0m

fj�!2j � !2

��!2j � !2

�2+ ( j!)

2

Introducing the transformation de�ned by the dimensionless variables

� =!

!j; � =

j!j

we can write

� =Nq2!3j�0m!4j

�2�

(1� �2)2 + (��)2

n = 1 +Nq2!2j2�0m!4j

fj (1� �2)

(1� �2)2 + (��)2

� =Nq2

�0m!j

�2�

(1� �2)2 + (��)2

n = 1 +Nq2fj2�0m!2j

(1� �2)

(1� �2)2 + (��)2

� = �1�2�

(1� �2)2 + (��)2

n = 1 + �2(1� �2)

(1� �2)2 + (��)2

where

�1 =Nq2

�0m!j;�2 =

Nq2fj2�0m!2j

=�1fj2!j

The plot for such particular molecule the absorption coe¢ cient is shown in �gure below.

It shows that as the damping constant increases the range of the spectrum that would

be absorbed by the medium increases with a peak absorption at resonance when the

frequency of the EM wave is the same as the natural frequency of the electrons (i.e.

� = 1) ! = !j).

174

On the other hand the refractive index shown in the �gure below shows unusual behav-

ior near the resonant region. In this region the refractive index decreases in contrary

what we know in optics. The range of the spectrum where this unusual behavior is

observed increases with increase in damping e¤ect.

Comparison of the absorption coe¢ cient and the refractive index is shown in the �gure

below. The curves shown by the dotted lines are for the absorption and the solid line

are for the refractive index. In the immediate neighborhood of a resonance the index

of refraction drops sharply. The material may be practically opaque in this frequency

range since it coincides with the region of maximum absorption. This is because of

175

the electron are forced to oscillate with their favorite frequency and the amplitude

of oscillation is maximum and correspondingly high amount of energy is dissipated

by the damping mechanism. Because this behavior is atypical, it is called anomalous

dispersion (the region !1 < ! < !2).

Refractive index increases for ! < !1 and ! > !2 which is consistent with our expe-

rience from optics (dispersion). The most familiar example of dispersion is probably

a rainbow, in which dispersion causes the spatial separation of a white light into com-

ponents of di¤erent wavelengths (di¤erent colors).

Refractive index decreases for !1 < ! < !2, in the immediate neighborhood of a reso-

nance the index of refraction drops sharply. The material may be practically opaque

in this frequency range since it coincides with the region of maximum absorption. This

is because of the electron are forced to oscillate with their favorite frequency and the

amplitude of oscillation is maximum and correspondingly high amount of energy is

176

dissipated by the damping mechanism. Because this behavior is atypical, it is called

anomalous dispersion (the region !1 < ! < !2).

Refractive index less than one for ! > !2 which means the wave speed wave exceeds

c. This is not an alarm since energy does not travel at the wave velocity but rather at

a group velocity. Moreover we considered only one molecule.

Far away from resonance (! >> !j): For the case where we are far away from

resonance we can ignore the damping term and we can write�!2j � !2

�2+ ( j!)

2 '�!2j � !2

�2: Then index of refraction can be expressed as

n = 1 +Nq2

2�0m

Xj=1

fj�!2j � !2

�For transparent materials, the nearest signi�cant resonance typically lie in the ultra-

violet, then ! < !j: Taking this into account and noting that we are very far from

resonance we can make the approximation

1

!2j � !2=

1

!2j

�1� !2

!2j

� ' 1

!2j

�1 +

!2

!2j

�so that

n = 1 +Nq2

2�0m

Xj=1

1

!2j

�1 +

!2

!2j

�

) n = 1 +

Nq2

2�0m

Xj=1

fj!2j

!+ !2

Nq2

2�0m

Xj=1

fj!4j

!If we express the frequency in terms of the wave length in a vacuum

! = 2�c=�

we can write

n = 1 +

Nq2

2�0m

Xj=1

fj!2j

!+1

�2

2�2c2Nq2

�0m

Xj=1

fj!4j

!or

n = 1 + A

�1 +

B

�2

�This is known as Cauchy�s Formula: the constant A is called the coe¢ cient of Re-

fraction and B is called the coe¢ cient of Dispersion. It applies in most gases, in the

optical region.

177

Ex. 16 The Kramers-Kronig Relations: If we treat ! as a complex quantity, then k(!) is

analytic in the upper-half complex-! plane. We consider the integral

I =

Z 1

0

k (!)� 1! � !0

dx

where !0 is a general real angular frequency.

178

XVII. LECTURE 23 INTRODUCTION TO INTEGRAL TRANSFORMS AND

THE LAPLACE TRANSFORM

� Integral Transforms: Frequently in physics we encounter pairs of functions related byan integral of the form

F (p) =

Z b

a

K (p; t) f (t) dt

The function F (p) is called the (integral) transform of f (t) by the kernelK (p; t) : Here

we are interested in two fundamental transforms the Laplace and Fourier transforms.

Integral transform operation may also be described as mapping a function f (t) in

t-space into another function F (p) in p-space. This interpretation takes on physical

signi�cance in the time-frequency relation of Fourier transforms.

� Laplace Transforms: the Laplace transform F (p) or L of a function f (t) is de�nedby

F (p) = L [f (t)] = limb!1

Z b

0

e�ptf (t) dt:

Note:

(i) The Laplace transform exists even if the integral

I =

Z 1

0

f (t) dt

does not exist.

(ii) For a function f (t) its Laplace transform to exist there must be a positive constant

M such that ��e�s0tf (t)�� M

for su¢ ciently large t, t > t0:

f (t) = et2

:

(iii) The Laplace transform fails to exist if the function has a strong singularity as

t! 0

f (t) = tn for n � �1:

(iv Laplace Transform is linear. That means if the Laplace transform for the two

functions f (t) and g (t) exist, then we can write that

L [af (t) + dg (t)] = aL [f (t)] + dL [g (t)] :

179

Ex. 1 Show that the Laplace Transform is linear

Sol: Consider the function

h (t) = af (t) + dg (t)

so that the Laplace transform of this function can be expressed as

H (p) = L [h (t)]

= limb!1

Z b

0

e�pt [af (t) + dg (t)] dt:

) H (p) = limb!1

Z b

0

e�ptaf (t) dt

+ lima!1

Z a

0

e�ptdg (t) dt

) H (p) = a lima!1

Z a

0

e�ptf (t) dt

+ d lima!1

Z a

0

e�ptg (t) dt

) H (p) = aL [f (t)] + dL [g (t)]

) H (p) = L [h (t)]

L [af (t) + bg (t)] = aL [f (t)] + dL [g (t)]

Ex. 2 Find the Laplace transform of the following functions and specify the conditions [if

there is any speci�c condition that must be satis�ed to do the transform]

(a)

f (t) = 1

Sol: Recalling that

F (p) = L [f (t)] =Z 1

0

e�ptf (t) dt:

we �nd

L [1] =Z 1

0

e�ptdt =1

p:

for p > 0:

180

(b)

f (t) = e!t

where ! is a constant.

Sol:

F (p) = L�e!t�=

Z 1

0

e�(p�!)tdt:

) F (p) =1

p� !

for p > !

(c)

f1 (t) = cosh (!t) ; f2 (t) = sinh (!t)

Sol: Noting that

cosh (!t) =e!t + e�!t

2

sinh (!t) =e!t � e�!t

2

we may write

F (p) = L [cosh (!t)] = L�e!t + e�!t

2

�Recalling that Laplace transform is linear one can write

L [cosh (!t)] = 1

2

�L�e!t�+ L

�e�!t

��:

and applying the result in the previous example we �nd

L [cosh (!t)] = 1

2

�1

p� !+

1

p+ !

�) L [cosh (!t)] = p

p2 � !2

Similarly

F (p) = L [sinh (!t)] = L�e!t � e�!t

2

�Recalling that Laplace transform is linear one can write

L [sinh (!t)] = 1

2

�L�e!t�� L

�e�!t

��:

181

and applying the result in the previous example we �nd

L [sinh (!t)] = 1

2

�1

p� !� 1

p+ !

�) L [cosh (!t)] = !

p2 � !2

(d)

f (t) = cos (!t) ; f (t) = sin (!t)

Sol: Recalling that

cos (!t) = cosh (i!t) ; sin (!t) = �i sinh (i!t)

and applying the result above we can easily write that

L [cos (!t)] = L [cosh (i!t)] = p

p2 + !2

L [sin (!t)] = L [�i sinh (i!t)]

= �iL [sinh (i!t)]) L [sin (!t)] = !

p2 + !2

Ex. 3 For the unit step function, also called the Heaviside function, is de�ned such that

U (t� a) =

8<: 0 for t < a

1 for t > a:

Find

L [U (t� a)] :

Sol: We may write the Laplace transform

F (p) = L [f (t)] =Z 1

0

e�ptf (t) dt:

for the unit step function

L [f (t)] =Z a

0

e�ptf (t) dt+

Z 1

a

e�ptf (t) dt:

L [f (t)] =Z 1

a

e�ptdt =e�pt

�p

��1a

:

L [f (t)] = e�pa

p:

182

Ex. 4 Evaluate

L [f 0 (t)] = L�df (t)

dt

�;

L [f 00 (t)] = L�d2f (t)

dt2

�;

Lhf000(t)i= L

�d3f (t)

dt3

�;

and show that the general relation can be written as

L�f (n) (t)

�= L

�dnf (t)

dtn

�= pnL [f (t)]� pn�1f (0)� pn�2f 0 (0)

�pn�3f 00 (0)� :::� f (n�1) (0) :

Sol: Recalling that

L [f (t)] =Z 1

0

e�ptf (t) dt:

we have

L [f 0 (t)] =Z 1

0

e�pt�d

dtf (t)

�dt

so that applying integration by parts we may write

L [f 0 (t)] = e�ptf (t)��10+ p

Z 1

0

e�ptf (t) dt

) L [f 0 (t)] = �f (0) + p

Z 1

0

e�ptf (t) dt

) L [f 0 (t)] = pL [f (t)]� f (0)

Similarly for

L [f 00 (t)] =Z 1

0

e�pt�d2

dt2f (t)

�dt

we have

L [f 00 (t)] = e�ptd

dtf (t)

��10

+ p

Z 1

0

e�ptd

dtf (t) dt

) L [f 00 (t)] = �f 0 (0) + p

Z 1

0

e�ptd

dtf (t) dt

) L [f 00 (t)] = pL [f 0 (t)]� f 0 (0)

183

and using the result

L [f 0 (t)] = pL [f (t)]� f (0)

we �nd

L [f 00 (t)] = p2L [f (t)]� pf (0)� f 0 (0) :

For

L [f 000 (t)] =Z 1

0

e�pt�d3

dt3f (t)

�dt

we get

L [f 000 (t)] = e�ptd2

dt2f (t)

��10

+p

Z 1

0

e�ptd2

dt2f (t) dt

) L [f 000 (t)] = pL [f 00 (t)]� f 00 (0)

) L [f 000 (t)] = p3L [f (t)]� p2f 0 (0)

� pf (0)� f 00 (0)

) L [f 000 (t)] = p3L [f (t)]� p2f 0 (0)

� pf (0)� f 00 (0) :

From the above result we may write the general relation

L�f (n) (t)

�= L

�dnf (t)

dtn

�= pnL [f (t)]� pn�1f (0)� pn�2f 0 (0)

�pn�3f 00 (0)� :::� f (n�1) (0) :

� Inverse Laplace Transform: The inverse of a Laplace transform is represented by

L�1 [F (p)] = f (t) :

In order to perform inverse Laplace transform it is important to know the Laplace

transform of some basic functions we determined in the examples we considered. These

results are tabulated below

184

.f (t) 1 e!t sin (!t) cos (!t)

L [f (t)] 1p

1p�!

!p2+!2

pp2+!2

f (t) sinh (!t) cosh (!t) U (t� a)

L [f (t)] !p2�!2

pp2�!2

e�ap

p

Ex. 5 Evaluate the following inverse Laplace transforms

(a)

L�1�5

p+ 2

�(a)

L�1�4p� 3p2 + 4

�Sol:

(a) Noting that

L�1�5

p+ 2

�= 5L�1

�1

p� (�2)

�and recalling

L�e!t�=

1

p� !

we can write

L�1�5

p+ 2

�= 5e�2t:

(b) Here also we want to write

L�1�4p� 3p2 + 4

�= L�1

�4p� 3(p2 + 22)

�

) L�1�4p� 3p2 + 22

�= L�1

�4p

p2 + 22� 3

p2 + 22

�

) L�1�4p� 3p2 + 4

�= L�1

�4p

p2 + 22

�� L�1

�3

p2 + 22

�

185

) L�1�4p� 3p2 + 4

�= 4L�1

�p

p2 + 22

�� 32L�1

�2

p2 + 22

�We recall

L [cos (!t)] = p

p2 + !2

and

L [sin (!t)] = !

p2 + !2

so that

L�1�4p� 3p2 + 4

�= 4 cos (2t)� 3

2sin (2t)

� Mathematica Result

:

Ex. 6 Evaluate the integral

f (t) =

Z 1

0

sin (tx)

xdx

applying Laplace and inverse Laplace transform

Sol: We �rst apply Laplace transform

L [f (t)] =Z 1

0

e�pt�Z 1

0

sin (tx)

xdx

�dt

which can also be expressed, by interchanging the order of the integral, as

L [f (t)] =Z 1

0

1

x

�Z 1

0

sin (tx) e�ptdt

�dx

) L [f (t)]

=

Z 1

0

1

x

�Z 1

0

�eitx � e�itx

2i

�e�ptdt

�dx

) L [f (t)]

=

Z 1

0

1

x

�Z 1

0

�e�(p�ix)t � e�(p+itx)

2i

�dt

�dx

186

) L [f (t)]

=

Z 1

0

1

2ix

�e�(p�ix)t

� (p� ix)+e�(p+itx)

p+ ix

�10

dx

) L [f (t)] =Z 1

0

1

2ix

�1

p� ix� 1

p+ ix

�dx

) L [f (t)] =Z 1

0

1

2ix

�p+ ix� p+ ix

p2 + x2

�dx

) L [f (t)] =Z 1

0

dx

p2 + x2

Using the Residue theorem we can show that the integralZ 1

0

dx

p2 + x2=

�

2p

Now we take the inverse Laplace transform

L�1 [f (t)] = L�1�Z 1

0

dx

p2 + x2

�) L�1 [f (t)] = �

2L�1

�1

p

�

) L�1 [f (t)] = �

2

) f (t) =

Z 1

0

sin (tx)

xdx =

�

2for t > 0

Noting that for

f (t) = sin (tx)

we have

f (�t) = �f (t)

which leads to

) L�1 [f (t)] = �

2

) f (t) =

Z 1

0

sin (tx)

xdx = ��

2for t < 0:

For t = 0, obviously

f (t) =

Z 1

0

sin (tx)

xdx = 0:

187

Therefore we can write

Z 1

0

sin (tx)

xdx =

�

2[2U (t)� 1] =

8>>><>>>:�2

t > 0

0 t = 0

��2t < 0

where U (t) is the unit step function (Heaviside function) we saw in above, which can

be expressed as

U (t) =

8<: 0 for t < 0

1 for t > 0:

188

A. Lecture 24 Applications of Laplace Transforms

In this section we will see applications of Laplace Transforms by solving several di¤erential

equations. To this end we recall the following "Table of Laplace Transforms" we determined

in the previous class.

f (t) 1 e!t sin (!t) cos (!t)

L [f (t)] 1p

1p�!

!p2+!2

pp2+!2

f (t) sinh (!t) cos (!t) U (t� a)

L [f (t)] !p2�!2

pp2�!2

e�ap

p

and

L�f (n) (t)

�= L

�dnf (t)

dtn

�= pnL [f (t)]� pn�1f (0)� pn�2f

00(0)

�pn�3f 000 (0)� :::� f (n�1) (0) :

Ex. 7 Solve the following di¤erential equation given the initial conditions y (0) = 1 and

y0 (0) = 0d2y

dx2+ y (x) = 1

Sol: We note that using Laplace transform we have

L�d2y

dx2

�+ L [y (x)] = L [1]

so that with the help of the results given earlier, one can write

L [1] = 1

p;

L�d2y

dx2

�= p2L [y (x)]� py (0)� y0 (0)

Substituting the given initial conditions

y (0) = 1; y0 (0) = 0

we obtain

L�d2y

dx2

�= p2L [y (x)]� p:

189

Then the di¤erential equation becomes

p2L [y (x)]� p+ L [y (x)] = 1

p

) L [y (x)] = 1

p

There follows that

y (x) = L�1�1

p

�= 1

Ex. 8 Solve the following di¤erential equation given the initial conditions y (0) = 2 and

y0 (0) = �1d2y

dt2+ 3

dy

dt+ 2y (t) = 2e�t

Sol: Take the Laplace transform of each term and apply the given initial conditions

L�d2y

dt2

�= p2L [y (t)]� py (0)� y0 (0)

) L�d2y

dt2

�= p2L [y (t)]� 2p+ 1

L�dy

dt

�= pL [y (t)]� y (0)

) L�dy

dt

�= pL [y (t)]� 2

L�e�t�= L

�e�t�=

1

p+ 1:

The di¤erential equation can be expressed as

L�d2y

dt2

�+ 3L

�dy

dt

�+ 2L [y (t)] = 2L

�e�t�

) p2L [y (t)]� 2p+ 1

+3 (pL [y (t)]� 2) + 2L [y (t)] = 2 1

p+ 1

190

)�p2 + 3p+ 2

�L [y (t)]� 2p� 5 = 2

p+ 1

) (p+ 2) (p+ 1)L [y (t)] = 5 + 2p+ 2

p+ 1

) L [y (t)] = 5 + 2p

(p+ 2) (p+ 1)+

2

(p+ 2) (p+ 1)2

Simplifying this expression we �nd

L [y (t)] = 2p2 + 7p+ 7

(p+ 2) (p+ 1)2:

Noting that

2p2 + 7p+ 7

(p+ 2) (p+ 1)2=

A

p+ 2+

B

p+ 1

+C

(p+ 1)2

2p2 + 7p+ 7

(p+ 2) (p+ 1)2

=A (p2 + 2p+ 1) +B (p2 + 3p+ 2) + C(p+ 2)

(p+ 2) (p+ 1)2

2p2 + 7p+ 7

(p+ 2) (p+ 1)2

=(A+B) p2 + (2A+ 3B + C) p+ A+ 2B + 2C

(p+ 2) (p+ 1)2

There follows that

A+B = 2;

(2A+ 3B + C) = 7;

A+ 2B + 2C = 7

Solving these equations using Mathematica

191

we can write

2p2 + 7p+ 7

(p+ 2) (p+ 1)2=

1

p+ 2+

1

p+ 1

+2

(p+ 1)2

so that

L [y (t)] = 2p2 + 7p+ 7

(p+ 2) (p+ 1)2

=1

p+ 2+

1

p+ 1+

2

(p+ 1)2:

Using Mathematica:

There follows that

) y (t) = L�1�1

p+ 2

�+ L�1

�1

p+ 1

�+ L�1

�2

(p+ 1)2

�Noting that

L�1�1

p+ 2

�= e�2t;L�1

�1

p+ 1

�= e�t

and

2

(p+ 1)2=

��2 d

d�

�1

p+ �

��=1

) L�1�

2

(p+ 1)2

�=

��2 d

d�

�L�1

�1

p+ �

��=1

) L�1�

2

(p+ 1)2

�=

��2 d

d�e��t

��=1

) L�1�

2

(p+ 1)2

�=�2te��t

�=1

= 2te�t

192

the expression

) y (t) = L�1�1

p+ 2

�+ L�1

�1

p+ 1

�+ L�1

�2

(p+ 1)2

�becomes

y (t) = e�2t + e�t + 2te�t

) y (t) = e�2t�1 + et (1 + 2t)

�Using Mathematica:

By directly solving the di¤erential equation using mathematica:

Ex. 9 Consider the circuit shown in the �gure below. At t = 0 the switch S is closed to

position A. Find the resulting circuit response, i(t). Use R = 10; L = 2H; " (t) =

50 sin (5t)

193

Sol: Applying Kircho¤�s voltage rule we can write

" (t)� IR� LdI

dt= 0

) 50 sin (5t)� 10I � 2dIdt= 0:

We take the Laplace transform of this equation

) 25L [sin (5t)]� 5 [I]� L�dI

dt

�= 0

Using

L [sin (5t)] = 5

p2 + 52

L�dI

dt

�= pL [I (t)]� I (0)

we �nd

125

p2 + 52� 5L [I]� pL [I (t)] + I (0) = 0

) L [I (t)] = 1

p+ 5

�125

p2 + 52+ I (0)

�Noting that

125

(p2 + 52) (p+ 5)=

A

p+ 5+BP + C

p2 + 52

=Ap2 + 25A+BP 2 + 5Bp+ Cp+ 5C

p2 + 52

=(A+B) p2 + (5B + C) p+ 25A+ 5C

p2 + 52

) A+B = 0; 5B + C = 0; 25A+ 5C = 125

) A =5

2; B = �5

2; C =

25

2

Just to check using Mathematica:

194

we may write

L [I (t)] = 5

2

1

p+ 5� 52

P

p2 + 52

+25

2

1

p2 + 52+I (0)

p+ 5

so that

I (t) =5

2L�1

�1

p+ 5

�� 52L�1

�P

p2 + 52

�+5

2L�1

�5

p2 + 52

�+ I (0)L�1

�1

p+ 5

�

I (t) =5

2e�5t � 5

2cos (5t) +

5

2sin (5t) + I (0) e�5t

in RL circuit the inductor acts as an open circuit at the initial time and we can set

I (0) = 0

I (t) =5

2e�5t +

5

2(sin (5t)� cos (5t))

Using Mathematica:

Note that we can write

5

2e�5t + I (0) e�5t =

�5

2+ I (0)

�e�5t

= c[1]e�5t

195

Ex. 10 Damped Harmonic oscillator : Consider a mass m oscillating under the in�uence of

a spring, spring constant k and damped by a friction force which is proportional to

the velocity (Ff = �bv). Assuming that the particle starts from rest at x (0) = x0;

x0 (0) = 0. Find the equation of motion for the mass m and determine the position of

the mass as function of time

Ans.

x (t) = x0e(�b=2m)t

�cos (!1t) +

b

2m!1sin (!1t)

�where

!21 =k

m� b2

4m2

196

XVIII. LECTURE 25 INTRODUCTION TO THE FOURIER TRANSFORM

In PHYS 3150 we have been introduced to Fourier series. It is bene�cial to revise some

of the things we introduced to.

� Periodic Functions : A function f (x) is said to be a periodic function when the

function repeats itself after a period T; which means

f (t) = f (t� T )

for all t.

� Periodic Motion: Periodic motion is when the motion of an object continually repeatsitself. This can be repeatedly moving back and forth or it could be moving in a circular

orbit or rotation. Since the Law of Inertia states that an object moves in a straight

line unless acted upon by a force, periodic motion requires some sort of force to create

this special type of motion. Characteristics of periodic motion are the velocity of the

object, the period of motion and the amplitude of the motion. Example of Periodic

motion include circular motion ( e.g. when you swing an object around you that is

held on a rope or string, motion of the planets around the sun) and back-and-forth

motion (e.g. a bouncing ball, pendulum and spring).

� The Fourier Series Expansion of a Periodic Function: Any function of periodicity 2�can be expanded in a Fourier series of the form

f (t) =1

2a0 +

1Xn=1

[an cos (nt) + bn sin (nt)]

where the Fourier Coe¢ cients are given by

an =1

�

�Z��

f (t) cos (nt) dt; n = 0; 1; 2; 3:::

and

bn =1

�

�Z��

f (t) sin (nt) dt; n = 1; 2; 3:::

The conditions that must be satis�ed so that the Fourier Series expansion of the

function f (t) given by the above expression does exist is known asDirchlet Conditions.

197

Dirchlet conditions state that f (t) must have only a �nite number of discontinuities

and a �nite number of extreme values (i.e. maxima, minima in the interval), [��; �]and the integral

R �� jf (t) dtj must be �nite

� The Fourier Transform: the Fourier transform of a periodic or non periodicfunction,

f (t) ; that satis�es all the Dirchlet conditions does exist and is given by

F [f (t)] = 1p2�

Z 1

�1f (t) e�i!tdt = F (!)

� The Inverse Fourier Transform: We may develop the inverse Fourier Transform start-ing from the Fourier series expansion. Suppose we are considering the interval [�T; T ]instead of [��; �] the Fourier series expansion for the function f (t) can be expressedas

f (t) =1

2a0 +

1Xn=1

han cos

��nTt�+ bn sin

��nTt�i

where the Fourier Coe¢ cients are given by

an =1

T

TZ�T

f (x) cos��nTx�dx; n = 0; 1; 2; 3:::

and

bn =1

T

TZ�T

f (x) sin��nTx�dx; n = 1; 2; 3:::

The resulting Fourier series can be written as

f (t) =1

2T

TZ�T

f (x) dx+

+1

T

1Xn=1

cos��nTt� TZ�T

f (x) cos��nTx�dx

+1

T

1Xn=1

sin��nTt� TZ�T

f (x) sin��nTx�dx

198

) f (t) =1

2T

TZ�T

f (x) dx+

+1

T

1Xn=1

TZ�T

f (x)ncos��nTt�cos��nTx�

+sin��nTt�sin��nTx�o

dx

f (t) =1

2T

TZ�T

f (x) dx

+1Xn=1

1

T

TZ�T

f (x) cosh�nT(t� x)

idx:

We set

! =�n

T) �! =

�

T

so that

f (t) =1

2��!

TZ�T

f (x) dx

+1

T

1Xn=1

�!

TZ�T

f (x) cos (! (t� x)) dx

or

f (t) =1

�

1Xn=0

�!

TZ�T

f (x) cos (! (t� x)) dx

where we set

limT!1

�!

2�

TZ�T

f (x) dx

= lim�!!0

�!

2�

1Z�1

f (x) dx = 0

assuming that the integral for the function f (x) exists, that means

1Z�1

f (x) dx <1:

199

We now let the parameter T approach in�nity, transforming the interval [�T; T ] intothe in�nite interval (�1;1) : In this limit we have

�! ! d!

and if the function f satis�es the Dirchlet conditions, we can replace the summation

by integration

f (t) =1

�

Z 1

0

d!

1Z�1

f (x) cos [! (t� x)] dx

Noting that cos [! (t� x)] is an even function and sin (! (t� x)) is an odd function

of !; we have Z 1

0

d!

1Z�1

f (x) cos [! (t� x)] dx

=1

2

Z 1

�1d!

1Z�1

f (x) cos [! (t� x)] dx

andi

2

Z 1

�1d!

1Z�1

f (x) sin [! (t� x)] dx = 0

so that

f (t) =1

�

Z 1

0

d!

1Z�1

f (x) cos [! (t� x)] dx

=1

2�

Z 1

�1d!

1Z�1

f (x) cos [! (t� x)] dx

+i

2�

Z 1

�1d!

1Z�1

f (x) sin [! (t� x)] dx

f (t) =1p2�

1Z�1

d!ei!t1p2�

1Z�1

f (x) e�i!xdx

We recall

F (!) = F [f (t)] = 1p2�

Z 1

�1f (t) e�i!tdt

so that

f (t) =1p2�

1Z�1

F (!) ei!td! = F�1 [F (!)]

200

which the Inverse Fourier Transform.

Note: If f(t) has a discontinuity, then the inverse Fourier transform will return a

value for f(t) that is at the midpoint of the discontinuity at the x-value at which the

discontinuity occurs.

Ex. 1 Find and describe the Fourier transform of the function f(x) de�ned by

f (x) =

8<: 1 jxj < 10 jxj > 1

which is an even function of x. This is the single slit di¤raction problem of physical

optics. The slit is described by f (x) :The di¤raction pattern amplitude is given by the

Fourier transform F (!)

Sol: We recall

F [f (x)] = 1p2�

Z 1

�1f (x) e�i!xdx

so that for the given function we can write

F [f (x)] = 1p2�

Z 1

�1e�i!xdx

=1p2�

��e

�i!x

i!

�1�1=

1p2�

ei!x � e�i!x

i!

=2p2�

ei!x�e�i!x2i

!=

r2

�

sin (!)

!

F [f (x)] =r2

�

sin (!)

!

Note: The function

f (x) =sin (x)

x

is known as the sinc function commonly expressed as sinc (x) :

201

Ex. 3 Find the Fourier Transform of a Gaussian function

f (x) = e�a2x2

Sol: For the given function we can write

F [f (x)] = 1p2�

Z 1

�1e�(a

2x2+i!x)dx

202

and using

a2x2 + i!x = a2

"�x+

i!

2a2

�2+� !

2a2

�2#

= a2�x+

i!

2a2

�2+!2

4a2

we have

F [f (x)] = e�!2

4a2

p2�

Z 1

�1e�a

2(x+ i!2a2)2

dx

To evaluate the integral

I =

Z 1

�1e�a

2(x+ i!2a2)2

dx

consider the integral Ie�a

2z2dz

for the rectangular closed loop with vertices the rectangle with vertices�T; T; T +i!2a2;�T + i!

2a2. For this integral applying the residue theorem we can write

Ie�a

2z2dz =

TZ�T

e�a2x2dx+ i

!2a2Z0

e�a2(T+iy)2dy

+

�TZT

e�a2(x+ i!

2a2)2

dx+ i

0Z� !2a2

e�a2(�T+iy)2dy

)Ie�a

2z2dz =

TZ�T

e�a2x2dx

+i

!2a2Z0

e�a2(T+iy)2dy

+

�TZT

e�a2(x+ i!

2a2)2

dx+ i

0Z� !2a2

e�a2(T�iy)2dy

Noting that in the limit T !1;

!2a2Z0

e�a2(T+iy)2dy =

0Z� !2a2

e�a2(T�iy)2dy = 0

203

and there is no poles inside the curve, we may write

)Ie�a

2z2dz = 0

=

1Z�1

e�a2x2dx+

�1Z1

e�a2(x+ i!

2a2)2

dx = 0

)Ie�a

2z2dz = 0

=

1Z�1

e�a2x2dx�

1Z�1

e�a2(x+ i!

2a2)2

dx = 0

)Ie�a

2z2dz = 0

1Z�1

e�a2(x+ i!

2a2)2

dx =

1Z�1

e�a2x2dx

We may write1Z

�1

e�a2x2dx =

1

a

1Z�1

e�y2

dy

so that applying the result0@ 1Z�1

e�y2

dy

1A2

=

1Z�1

e�u2

du

1Z�1

e�v2

dv

)

0@ 1Z�1

e�y2

dy

1A2

=

1Z�1

e�(u2+v2)dudv

)

0@ 1Z�1

e�y2

dy

1A2

=

1Z�1

e�(u2+v2)dudv

)

0@ 1Z�1

e�y2

dy

1A2

=

1Z0

Z 2�

0

e�r2

rdrd�

)

0@ 1Z�1

e�y2

dy

1A2

= 2�

1Z0

e�r2

rdr

= ��e�r2��10= �

204

)1Z

�1

e�y2

dy =p�

we �nd1Z

�1

e�a2(x+ i!

2a2)2

dx =

p�

a:

Therefore the Fourier integral transform

F [f (x)] = e�!2

4a2

p2�

Z 1

�1e�a

2x2dx

becomes

F [f (x)] = e�!2

4a2

ap2= F (!) :

� The Dirac Delta Function: we recall the inverse Fourier transform

f (t) =1p2�

1Z�1

F (!) ei!td!

so that substituting

F (!) = F [f (t)] = 1p2�

Z 1

�1f (t0) e�i!t

0dt0

we have

f (t) =1p2�

1Z�1

24 1p2�

1Z�1

f (t0) e�i!t0dt0

35 ei!td!) f (t) =

1Z�1

f (t0)

24 12�

1Z�1

e�i!(t�t0)d!

35 dt0or

f (t) =

1Z�1

f (t0)� (t� t0) dt0

205

where

� (t� t0) =1

2�

1Z�1

e�i!(t�t0)d!

is known as the Dirac Delta function. The Dirac Delta function has di¤erent forms.

One of this form is given by

� (t� t0) =1

2�lima!1

aZ�a

e�i!(t�t0)d!

) � (t� t0) =1

2�lima!1

eia(t�t0) � e�ia(t�t

0)

i (t� t0)

) � (t� t0) = lima!1

sin [a (t� t0)]

� (t� t0):

Suppose the function

f (t) = 1) f (t0) = 1

for

f (t) =

1Z�1

f (t0)� (t� t0) dt0

we then �nd1Z

�1

� (t� t0) dt0 = 1

Now let�s examine the graph of the Dirac Delta function

) � (t� t0) = lima!1

sin [a (t� t0)]

� (t� t0):

for di¤erent value of a at t0 = 2: For a = 10

206

For a = 100

For a = 1000

For a = 1000000

207

From the above graphs it is easy to conclude that

� (t� t0) =

8<:1 t = t0

0 otherwise

which also leads to

bZa

f (t0)� (t� t0) dt0 =

8<: f (t) a < t < b

0 otherwise

208

XIX. LECTURE 26 APPLICATIONS OF THE FOURIER TRANSFORM: THE

HEISENBERG UNCERTAINTY PRINCIPLE

� The Time-Dependent Schroedinger Equation

� ~2

2mr2(~r; t) + U (~r) (~r; t) = i~

@(~r; t)

@t

� The Special Case of a Free Particle Traveling in the Positive-x Direction: For a freeparticle, we have

U (~r) = 0

and the time independent part of the S.E. for a particle traveling in the positive

x-direction can be written as

� ~2

2m

d2 (x)

dx2= E (x)

) d2 (x)

dx2= �2mE

~2 (x)

Using De Broglie wavelength of the particle, we can relate the momentum with the

wave vector

� =h

p) p =

h

�=2�

�

h

2�

) p = k~:

Then the total energy for a free particle

E =p2

2m

can be expressed as

E =k2~2

2m:

Using this result we may write

d2 (x)

dx2= �2mE

~2 (x)

asd2 (x)

dx2= �k2 (x)

209

The solution of which is given by

(x) = A (k) eikx:

Since for a free particle k takes continuous value, we may write the general solution as

(x) =

1Z�1

A (k) eikxdk:

What must be the condition that must be satis�ed by the function A (k) : To �nd this

condition we recall that the probability of �nding the function anywhere between �1and +1 is one. That means

1Z�1

j (x)j2 dx =1Z

�1

� (x) (x) dx = 1:

Substituting the integral expression for (x) we �nd

1Z�1

24 1Z�1

A� (k0) e�ik0xdk0

35�

24 1Z�1

A (k) eikxdk

35 dx = 1

)1Z

�1

A (k)

�1Z

�1

A� (k0)

0@ 1

2�

1Z�1

e�ix(k�k0)dx

1A dk0dk

=1

2�

Applying the de�nition of the Dirac Delta function

� (t� t0) =1

2�

1Z�1

e�i!(t�t0)d!

we may write1Z

�1

A (k)

24 1Z�1

A� (k0)� (k � k0) dk0

35 dk=1

2�

210

and using the property of the Dirac Delta function

f (t) =

1Z�1

f (t0)� (t� t0) dt0

we �nd

1Z�1

A (k)A� (k) dk =1

2�

)1Z

�1

jA (k)j2 dk = 1

2�:

This is the condition for A (k) :Therefore if we express

A (k) =1p2�� (k) ;

where � (k) is the momentum eigen function satisfying the normalization condition

1Z�1

j� (k)j2 dk = 1;

we may write a normalized function wave function for the free particle

� The (Inverse) Fourier Transform for a Wavepacket

(x) =1p2�

1Z�1

� (k) eikxdk

� Uncertainties

�x = �x =

qhx2i � hxi2

�k = �k =

qhk2i � hki2

� Expectation Values

hf (x)i =1Z

�1

� (x) f (x) (x) dx

hg (k)i =1Z

�1

�� (k) g (k)� (k) dk

211

Ex. 4 For a particle described by a Gaussian Wavepacket

(x) = Ae��x2

=

�2�

�

�1=4e��x

2

Some Useful Integrals

1Z�1

e�ay2

dy =

r�

a

1Z�1

y2e�y2

dy =

p�

2

(a) The Momentum eigen function is given by

� (k) =Ap2�e�k

2=4�

(b) Show that the expectation values

hki = 0;k2�= �:

(c) Find the Expectation values

hxi ;x2�:

(d) Show that the Heisenberg Uncertainty Principle

�x�px =~2

is satis�ed.

Sol:

(a) We recall

(x) =1p2�

1Z�1

� (k) eikxdk

which is the inverse Fourier transform of the momentum eign function � (k) ; we can

write

� (k) =1p2�

1Z�1

(x) e�ikxdx:

212

Then substituting

(x) = Ae��x2

=

�2�

�

�1=4e��x

2

we �nd

� (k) =Ap2�

1Z�1

e��x2

e�ikxdx

=Ap2�

1Z�1

e��(x2+ ik

� )dx:

Noting that

x2 +ik

�=

�x+

ik

2�

�2+

k2

4�2

we can write

� (k) =Ap2�e�

k2

4�

1Z�1

e��(x+ik2�)

2

dx:

Using the result in the previous example or Mathematica

we can easily show that

� (k) =Ap2�e�

k2

4� :

(b) In the momentum space we can express the expectation values

hki =1Z

�1

�� (k) k� (k) dk

k2�=

1Z�1

�� (k) k2� (k) dk

and using the momentum eigen function, we may write

hki =�

Ap2�

�2 1Z�1

ke�k2

2�dk = 0

213

k2�=

�Ap2�

�2 1Z�1

k2e�k2

2�dk:

Using the transformation of variable

x =kp2�) dk =

p2�dx

we have k2�=

�Ap2�

�2 1Z�1

2�x2e�x2p2�dx:

k2�=

�Ap2�

�2(2�)3=2

1Z�1

x2e�x2

dx:

so that using the relation1Z

�1

y2e�y2

dy =

p�

2

we �nd k2�=

�Ap2�

�2(2�)3=2

p�

2:

)k2�= A2 (2�)1=2

p�

2:

Substituting

A =

�2�

�

�1=4we �nd

k2�=

p2�p�

�p2�� p�2= �:

(c) It is easier to �nd these expectation values in position space rather than in momentum

space, which we may write as

hxi =1Z

�1

� (x)x (x) dx;

x2�=

1Z�1

� (x)x2 (x) dx:

Using the wave function

(x) = Ae��x2

=

�2�

�

�1=4e��x

2

214

the expectation values become

hxi = A21Z

�1

xe�2�x2

dx = 0;

x2�= A2

1Z�1

x2e�2�x2

dx:

Replacing

y =p2�x) dx =

dyp2�

we have x2�=

A2�p2��3=2

1Z�1

y2e�y2

dy

which gives x2�=

A2�p2��3=2p�2

or x2�=

1�p2��3=2

p2�p�

p�

2=1

4�

(d) The Uncertainty in position and momentum

�x = �x =

qhx2i � hxi2

�px = ~�k = ~�k = ~qhk2i � hki2

would then be

�x = �x =1

2p�;

�px = ~�k = ~�k = ~p�:

Then the uncertainty principle

�x =

�x�px =1

2p�~p� =

~2

is satis�ed. This means the Gaussian wave packet represent the minimum uncertainty

state.

215

XX. LECTURE 27 FOURIER TRANSFORMS AND CONVOLUTION

� Convolution: For two functions f (x) and g (x) with corresponding Fourier Transforms,F (!) and G (!) ; respectively, the convolution of these two functions over the interval

(�1;1), denoted by f � g; is de�ned as

f � g = 1p2�

Z 1

�1g (y) f (x� y) dy

=1p2�

Z 1

�1f (y) g (x� y) dy = g � f:

This form of integrals appears in probability theory in the determination of the proba-

bility density of two random, independent variables. One good example is the solution

of the Poisson�s equation

V (~r) =1

4��0

Z� (~r0)

1

j~r � ~r0jd3r0:

This may be interpreted as convolution of a charge distribution � (~r0) (as g (y))and

a weighting function (4��0 j~r � ~r0j)�1 (as f (x� y)). The operation of convolution

sometimes referred to as the Faltung, the German term for "folding"

� The Convolution theorem: the convolution theorem states that the convolution of twoor more functions is the same as the Fourier inverse transforms of a product of Fourier

Transforms. That means

f1 � f2 � f3:::fn

= F�1 [F1 (!)F3 (!)F3 (!) ::Fn (!)]

Proof: We shall proof this theorem by considering the case of two functions

f1 � f2 =1p2�

Z 1

�1f2 (y) f1 (x� y) dy

Let Fourier Transform of the function f1 (x� y) is F1 (!) ; then may write

f1 (x� y) = F�1 [F1 (!)]

=1p2�

Z 1

�1F1 (!) e

i!(x�y)d!

216

so that the convolution integral can be put in the form

f1 � f2 =1p2�

Z 1

�1f2 (y)

� 1p2�

�Z 1

�1F1 (!) e

i!(x�y)d!

�dy

f1 � f2 =1p2�

Z 1

�1F1 (!)

� 1p2�

�Z 1

�1f2 (y) e

�i!ydyei!x�d!

Noting that

F2 (!) =

Z 1

�1f2 (y) e

�i!ydy

we can write the above expression as

f1 � f2 =1p2�

�Z 1

�1F1 (!)F2 (!) e

i!xd!

�) f1 � f2 = F�1 [F1 (!)F2 (!)]

Ex. 4 Discuss, compute, and plot the convolution of the functions f(x) and g(x) given below.

Sol: We note that the two functions shown above can be expressed as

f (x) =

8<: 1 �10 < x < 10

0 otherwise

g (x) =

8<: 110x 0 < x < 10

0 otherwise

217

For g (x� y) ; we may write

g (x� y) =

8<: 110(x� y) 0 < x� y < 10

0 otherwise

or

g (x� y) =

8<: 110(x� y) x� 10 < y < x

0 otherwise

Applying the convolution formula

f � g = 1p2�

Z x

x�10f (y)

1

10(x� y) dy

) f � g = 1

10p2�

Z x

x�10f (y) (x� y) dy

this can be rewritten as

f � g = 1

10p2�

Z x

x�10f (y) (x� y) dy

we have to be careful in evaluating this integral because of the function f (y) :

Case 1 : x � �10

For this case the range of y would be (�1;�20) :In this interval we have f (y) = 0;and we �nd

f � g = 0

Case 2 : �10 < x < 0

For this case the range of y would be (�20; 0) :In this interval we have f (y) = 0 fory < �10 and f (y) = 1 for �10 < y < 0; and we �nd

f � g = 1

10p2�

Z x

�10(x� y) dy

) f � g = 1

10p2�

�xy � y2

2

��x�10

) f � g = 1

10p2�

�x2

2+ 10x+ 50

�

) f � g = 1p2�

�x2

20+ x+ 5

�Case 3 : 0 < x < 10

218

For this case the range of y would be (�10; 10) :In this interval we have f (y) = 1; andwe �nd

f � g = 1

10p2�

Z x

x�10(x� y) dy

) f � g = 1

10p2�

�xy � y2

2

��xx�10

) f � g = 1

10p2�

�x2

2� (x (x� 10)

�(x� 10)2

2

!#

) f � g = 1

10p2�

�x2

2��x2 � 100

2

��) f � g = 5p

2�

Case 4 : 10 < x < 20

For this case the range of y would be (0; 20) :In this interval we have f (y) = 0 for

10 < y < 20 and f (y) = 1 for 0 < y < 10; and we �nd

f � g = 1

10p2�

Z 10

x�10(x� y) dy

) f � g = 1

10p2�

�xy � y2

2

��10x�10

) f � g = 1

10p2�

�20x� 100� x2 + 100

2

�f � g = 1p

2�

�x� x2

20

�The graph of the convoluted functions is shown below

-20 -10 0 10 20 300.0

0.5

1.0

1.5

2.0

f*g

x

219

� Important Properties of Convolution

1. f(x) � g(x) = g(x) � f(x)

2. convolution tends to smooth out f(x)

3. convolution tends to spread out f(x)

220

Thor Eti Call Ec Note Dwl

Documents

correspondingly

jo cy22

taylor series

decoupled

tr10 12cos

refractive

fourier series

2dr i0