Madhya Pradesh Bhoj ( Open) University, Bhopalmpbou.edu.in/slm/mscphy1p1.pdf · 1.7 Let Us Sum Up 1.8 Check Your Progress: The Key 1. 0 INTRODUCTION A Vector is a directed „line

Madhya Pradesh Bhoj (Open) University,

Bhopal

BLOCK – I Vectors, Matrices, Tensors and Partial Differential Equations

UNIT-I Vectors and Matrices

UNIT-II Tensors and Partial Differential Equations

PAPER –I MATHEMATICAL PHYSICS

Madhya Pradesh Bhoj ( Open) University,

Bhopal

BLOCK-2 Group Theory and Functions of Complex variable UNIT-I Group Theory UNIT-II Functions of Complex variable


Madhya Pradesh Bhoj ( Open) University,

Bhopal

BLOCK-3 Special Functions and Integral Transform UNIT-I Special Functions and spherical harmonics UNIT-II Integral Transform


UNIT 1 VECTORS AND MATRICES

Structure

1.0 Introduction

1.1 Objectives

1.2 Curvilinear coordinates

1.2.1 Differential operator in terms of curvilinear coordinates.

1.3 Orthogonal curvilinear coordinate system

1.3.1 Derivation of Differential operators in spherical coordinates

1.4 Cylindrical coordinates

1.4.1 Derivations of gradients and curl in Polar, spherical and cylindrical

coordinate systems.

1.5 Eigen value problem

1.5.1 Example –Find the Eigen values

1. 6 Cayley- Hamilton theorem

1.6.1 Example –Find the Cayley- Hamilton Equation

1.6.2 Some important theorem of Eigen values and Eigen vectors

1.7 Let Us Sum Up

1.8 Check Your Progress: The Key

1. 0 INTRODUCTION

A Vector is a directed „line segment‟ having direction and magnitude both. Accordingly,

a vector A can be represented by an arrow of OP finite length directed from initial point

O to the terminal point P, the length of arrow represents the magnitude of vector A and

arrow head denotes the direction of vector A .In printed matter a vector quantity is

represented by the clarendon letter A. The magnitude or modulus of vector A is

represented by A or simply by A. The unit vector along vector A is represented by A.

The formulation of a law of Physics in terms of vector is independent of the choice of

axes of reference.

And a matrix may be defined as square or rectangular arrays of numbers or functions that

obey certain laws. The individual numbers (of functions) of the array are called the

element of the Matrix .It is convenient to think of every element of a matrix as belonging

to a certain row and the certain column of the matrix.

Vectors, Matrices, Tensors and Partial differential Equations

A matrix consisting of „m‟ rows and „n‟ column is said to be the matrix of order m x n

(read as m by n). A general matrix of order m x n can be conveniently written as,

Where the elements may be real or complex numbers or functions. This matrix may

briefly be written as,

A = [ aij]m x n

This means that A is matrix of order m x n whose ijth

element is aij . The first symbol ( i )

denotes row and the second symbol ( j ) denotes column to which the element aij

belongs. Obviously,1 i m and 1 j n.

1. 1 OBJECTIVES

The objective of this unit is to study vector and Matrix. After completing this unit we will

able to –

Define the vectors and matrix.

Represent the vector quantities in curvilinear coordinate system.

Derive the Gradient, divergence and curl in Polar, spherical and cylindrical

coordinate systems.

Solve Eigen value problem for the matrix.

Apply and solve the Cayley-Hamilton theorem.

Find the function of matrix.

Perform the Kronecker delta symbols.

1. 2 CURVILINEAR COORDINATES

In many problems of Physics and applied mathematics the Cartesian coordinates are not

very useful so it become necessary to rewrite the vector equations in terms of suitable

coordinate system called curvilinear coordinate system which is more

a11 a12 a13……..a1n

a21 a22 a23……a2n

…………………………...

…………………………...

am1 am2 am3…….amn

A=

Vectors and Matrices

sophisticated in use before the final solution of specific problem is obtained. The

curvilinear coordinates are so general in nature that it becomes very easy to transform

them in to any one of the several kinds of special coordinate systems, which have been

found useful in physical problems.

In Cartesian coordinate system the position of a point P (x,y,z) is determined by the

intersection of the mutually perpendicular planes x= constant, y= constant and

z= constant. Now if we superimpose three other families of surfaces on this system.

These surfaces may be parallel or may be plane. The three assumed families need not be

mutually perpendicular. These new families may be described by q1= constant,

q2= constant and q3= constant, intersect at the point P. The values of q1,q2 and q3 for the

three surfaces intersecting at the point P are called the curvilinear coordinates of P. The

three new surfaces are then called the coordinate surfaces or curvilinear surfaces.

If the coordinate surfaces are mutually perpendicular at every point P (x, y, z) then the

curvilinear coordinates (q1, q2, q3) are said to be orthogonal curvilinear coordinate of

point P. The coordinate axes are determined by the tangent to the coordinate lines at the

intersection of the three surfaces.

Obviously any given point P may be identified by curvilinear coordinates (q1, q2, q3) as

well as by Cartesian coordinates (x,y,z). This means that in principle we may write,

x= x (q1, q2, q3)

y = y (q1, q2, q3)

z= z (q1, q2, q3) (1.1)

(x,y,z)

Which inverses

q1= q (x,y,z)

q2 = q (x,y,z )

q3= q (x,y,z) (1.2)

Fig.-1.1

with each family of surface q1= constant, we can associate a unit vector ui normal to each

surface qi =constant and in the direction of increasing qi It may be noted that the three

curvilinear coordinates q1, q2,q3 need not be lengths. The scale factors hi may depend on

q,s and they have dimensions . The product hidqi must have the dimension of length.

o

x

y

z

P


1. 2.1 Differential operators in terms of curvilinear coordinates

Consider three mutually perpendicular coordinate surface described by q1 = constant,

q2=constant and q3= constant. Let (q1, q2, q3) be scalar function and V be a vector

function with components V1, V2, V3, in the three directions in which q1, q2, q3 increase.

If u1, u2, u3 are the unit vectors along the direction of increasing q1, q2, q3 respectively,

then vectors along V in terms of orthogonal curvilinear coordinates may be written as,

V = u1V1+ u2V2+ u3V3 (1.3)

(1) Gradient - The gradient of a scalar function is a vector whose magnitude and

direction give the maximum space rate of change of scalar function . From this

interpretation the component of ( q1, q2, q3) in the direction normal to the surface q1=

constant and hence in the direction of q1 is,

(1.4)

Where s1= h1q1 is the differential length in the direction of increasing q1 and

represents an increasing in on traveling a distance s1 in the limit s1 0.

By repeating equation (1.4) for q2 and again for q3, we get

(1.5)

(1.6)

Adding equations (1.4),(1.5) and (1.6) vectorially, the gradient of scalar function in

orthogonal curvilinear coordinates becomes ,

(1.7)

Thus the operator „grad‟ in orthogonal curvilinear coordinates is ,

(1.8)

1 =

Lim

s1 0

s1 =

s1 =

h1q1 =

1

h1

q1

2 = 1

h2

q2

3 = 1

h3

q3

grad = = u1

h1 q1 +

u2

h2 q2 +

u3

h3 q3

grad = = u1

h1 q1 +

u2

h2 q2 +

u3

h3 q3


(2) Divergence - The divergence of a vector V is written as,

div V = .V = . (u1V1 +u2V2+ u3V3) (1.9)

But the divergence

div. ( u1Vi) = . (u1Vi)= Vi .ui + ui . Vi (1.10)

Thus, in order to find .V, it is needed to obtain .uiVi which may be obtained as,

div V =.V =

(1.11)

This equation represents div V in orthogonal curvilinear coordinates.

(3) Laplacian - Laplacian may be obtained by combining equations (1.7) and (1.11).

Substituting V = (q1,q2,q3) in equation (11), we get,

Substituting components of from equation (1.7); we get

(1.12)

(4) Curl - The curl of vector V is written as ,

Curl V = x V = x (u1V1+ u2V2 + u3V3) = x (u1V1) + x (u2V2) + x (u3V3) (1.13)

Keeping in mind the relation Curl (A) = Curl A – A x grad , we can find the value

Curl V in form of curvilinear coordinates in form of determinant as ,

(1.14)

And, Curl V = (1.15)

1 (V1h2h3) (V2h3h1) (V3h1h2)

h1h2h3 q1 q2 q3 + +

1

h1h2h3 q1 q2 q3 + + h2h3(1) h3h1(2) h1h2(3) . =

2 =

1

h1h2h3 q1 q2 q3 + +

2 =

h2h3

h1

q1

h3h1

h2

q2

h1h2

h3

q3

Curl V = xV = 1

h1h2h3

h1u1 h2u2 h3u3

q1 q2 q3

V1h1 V2h2 V3h3

Vx Vz

z x

Vy Vz

x y Vz Vy

y z i + j + k


1. 3 ORTHOGONAL CURVILINEAR SPHERICAL POLAR

COORDINATES (r,, )

The spherical polar coordinate system consists of ,

(i) Concentric spheres about the origin O ( r= ( x2 + y

2 + z

2) = constant )

(ii) Right circular cones about z-axis with the vertices at the origin O

( = cos-1

(iii) Half planes through the Z-axis ( = tan-1

(y/x) = constant)

is the longitudinal or azimuthal angle, i.e. the angle between positive X-axis and X-Y

plane.

From Fig. (1.2) the transformation between rectangular coordinates (x, y, z) and spherical

coordinates (r,,) are given by,

x= r sin cos

y= r sin sin and

z= r cos (1.16)

On differentiating we get,

(1.17)

dy = sin sin dr + r cos sin d

+ r sin cos d (1.18)

Fig.-1.2

And,

dz= cos dr – r sin d (1.19)

The line element „ds‟ in Cartesian coordinates is given by,

ds2 = dx

2+ dy

2 +dz

2 (1.20)

Substituting the values of dx, dy and dz from above equations the expression for the line

element in spherical polar coordinates becomes

ds2 = dr

2 + r

2d

2+ r

2sin

2 d

2 (1.21)

Comparing this equation by specifying the scale factor notation equation i.e.,

z

r = Constant)

dx = dr + d + d x

r

x

x

= sin cosd + r cos cos d

- r sin sin d

x

P (x,y,z)

r

x

y

y

z

( x, y, 0)


ds2 = (h1dq1)

2 + (h2dq2)

2 +( h3dq3)

2 (1.22a)

We get the coordinates

q1=r, q2= , q3= and h1 =1, h2 =r and h3= r sin (1.22b)

1. 3.1 Derivation of differential operators in spherical polar coordinates

(1) Gradient - In curvilinear coordinates „grad ‟ is,

If ur, u,u are unit vectors along r,, axes respectively, then using (1.22), grad in

spherical polar coordinates may be expressed as,

(1.23)

or, grad = = (1.24)

(2) Divergence - In orthogonal curvilinear coordinates div V is,

Using equation (22) div V in spherical polar coordinates becomes,

(1.25)

(3) Laplacian - In orthogonal curvilinear coordinate 2 is,

Using equation (1.25) 2 in spherical polar coordinates becomes,

u1 u2 u3 h1 q1 h2 q2 h3 q3

+ + grad =

r

+

grad = ur u 1

r +

1

r sin u

r

+

ur u 1

r +

1

r sin u

1 (V1h2h3) (V1h2h3) (V1h2h3)

h1h2h3 q1 q2 q3 + + div V =

1 (r2

sin Vr) (r sin V) (r V)

r2

sin r + + div V =

1 (r2

Vr) (r sin V) (r V)

r2

r + + div V =

1

r sin

1

r sin

2 =

1 h2h3 h3h1 h2h1

h1h2h3 q1 h1 q1 q2 h2 q2 q3 h3 q3 + +


(1.26)

(4) Curl - In orthogonal curvilinear coordinates curl V is,

Using equation (1.25) we obtain the value of curl in form of spherical polar coordinates

(1.27)

1. 4 CYLINDRICAL COORDINATES

The cylindrical coordinates system consists of: -

(i) Right circular cylinders having z-axis as common axis, which form families of

concentric circles about the origin O in X-Y plane.

r = (x2 + y

2)1/2

= constant

(ii) Half planes through to z-axis, =tan-1

(y/x)=constant.

(iii) Planes parallel to X-Y plane, z=constant

Let P be a point in space such that its Cartesian coordinates are (x,y,z). Then to find the

position coordinates of P in cylindrical coordinates draw a radius vector r from O as

1 r2

sin sin

r2

sin r r + +

2 =

1

sin

1

r2 r

r2

r + =

1

r2

sin sin

+

1

r2

sin

2

2

Curl V = xV = 1

h1h2h3

h1u1 h2u2 h3u3

q1 q2 q3

V1h1 V2h2 V3h3

Curl V = xV = 1

r2sin

ur r u r sin u

r

Vr r V r sin V


shown in Fig.1.3 b . Then draw PM parallel to z-axis and finally OM parallel to OP .

Then angle MOX thus formed is .

Thus the position of point P is cylindrical coordinates is specified by (r,,z) where r is the

distance is the X-Y plane from the origin to the cylinder on which the point P lies, is

MoX in the X-Y plane and z is the distance from the XY plane to the point P.

From Fig.1.3 the transformations between rectangular coordinates (x,y,z) and cylindrical

coordinates (r,,z) are given by ,

Fig.1.3

From Fig.1.3 the transformations between rectangular coordinates (x,y,z) and cylindrical

coordinates (r,,z) are given by,

x= r cos

y = r sin

z=z (1.28)

Therefore, we have

(1.29)

dz =dz

As the line element is Cartesian coordinates is given by,

ds2 = dx

2 + dy

2 + dz

2

z

P

y

x

y

r

z

o

P

M

O

O

r

x

y

x

z

x

r dr

x

+ d = cos dr – r sin d dx =

y

r dr

y

+ d = sin dr + r cos d dy =


Substituting values of dx, dy and dz from equation (1.29), the line element ds in

cylindrical coordinates is,

ds2 = dr

2 + r

2d

2 +dz

2 (1.30)

Comparing this equation with, ds2 = (h1dq1)

2 + (h2dq2)

2 +( h3dq3)

2, we get

q1=r, q2= , q3=z and h1 =1, h2 =r and h3= 1 (1.31)

1. 4. 1 Differential operators in cylindrical coordinates

(1) Gradient -In orthogonal curvilinear coordinates grad is,

If ur,u and uz are the unit vector along r, ,z axis respectively, then using equation

(1.28), grad in cylindrical becomes,

(1.32)

(2) Divergence - In orthogonal curvilinear coordinate div V is given by ,

Using equation (1.31) div V in cylindrical coordinate becomes,

(1.33)

( 3) Laplacian - In orthogonal curvilinear coordinates 2

2 in spherical polar coordinates becomes,

grad = = u1

h1 q1 +

u2

h2 q2 +

u3

h3 q3

grad = =

r +

+

z

ur u

uz 1

r

1 (V1h2h3) (V2h3h1) (V3h1h2)

h1h2h3 q1 q2 q3 + + Div V = .V =

1 (r Vr) 1 V Vz

r r r z + + Div V =

2 =

1 h2h3 h3h1 h2h1

h1h2h3 q1 h1 q1 q2 h2 q2 q3 h3 q3 + +

2 =

1

r2 r

r2

r

1

r2 sin

+

Sin 1

r2 sin

+

2

2


And in cylindrical coordinates the value 2 becomes by using (1.31) equation

(1.34)

(4) Curl - In orthogonal curvilinear coordinates curl V is

Using (1.31), Curl V in cylindrical coordinate

(1.35)

2 =

2 1 1

2

r2 r r r

2

+ + +

2

z2

Curl V = xV = 1

h1h2h3

h1u1 h2u2 h3u3

q1 q2 q3

V1h1 V2h2 V3h3

Curl V = xV = 1

r

ur ru uz

r z

Vr Vr Vz

1 Vz V

r z ur

+ Vr Vz

z r u

+ 1 (rV) Vr

r r uz xV =

Check Your Progress 1

Note: a) Write your answers in the space given below,

b) Compare your answers with the ones given at the end of the unit.

i) Define curvilinear coordinates and write how these are differ from cartesian

coordinates?

ii) How cylindrical Coordinates are represented and how rectangular

coordinates can be transformed in to cylindrical Coordinates?

………………………………………………………………………………

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………………………………………………………………………………

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………………………………………………………………………………

………………………………………………………………………………

……………………………………………………………………..


1. 5 EIGENVALUES, EIGEN VECTORS:

CHARACTERISTICS EQUATION OF A MATRIX

Consider the linear transformation

AX =X (1.36)

Where is a scalar and A is the square matrix of order n. Obviously, the only effect of

the matrix A on the vector X is to multiply it simply by the constant scalar factor

A vector X defined by equation (1.36) is called an invariant vector under the linear

transformation.

Equation (1.36) may be written as,

(`1.37)

(A - I)X = 0

Where I is the unit matrix.

Any value of for which (1.36) or (1.37) has a non zero solution ( i.e. X 0) is called as

eigen value or characteristic root or latent root of the matrix A and the corresponding

non-zero solution X is called an eigen vector or characteristic vector of A corresponding

to that value of .

The matrix (A- .I) is called the characteristic matrix of A .

The determinant ()= (A - .I) is called the characteristic polynomials of A .

The system of homogeneous equation (1.36) and (1.37) has non –trivial equation if and

only if (A - .I) is singular i.e.,

()= (A - .I) =0 (1.38)

Or its equivalent;

()= a0 + a 1 + a 2

2 + ………..+ an

n (1.39)

Where a‟s are the function of the elements of A.

Equation (1.38) or (1.39) is called the characteristic equation or secular equation of A.

From these equations it follows that every characteristic root of A is a root of its


characteristic equation.. the n-root 1, 2, 3…………n of the characteristic equation are not

necessarily different.

1.5.1 Example -

Find the Eigen values and normalized Eigen vectors of the matrix

Solution.

Let A =

We have

A - I =

The characteristic equation of A is,

i.e. (1- )[ (1- )2 – 1] =0

i.e (1- ) ( 2

–2 ) = 0

i.e. (1-)(-2) =0 .

i.e. = 0,1,2 (1.40)

Thus the Eigen values of the matrix A are 0,1, 2

Eigen value equation is, (1.41)

(A - I) X =0

For =0, equation (1) reduces to

1 0 0

0 1 1

0 1 1

1 0 0

0 1 1

0 1 1

1 0 0

0 1 1

0 1 1

- 1 0 0

0 1 0

0 0 1

1- 0 0

0 1- 0

0 0 1-

=

A - I =

1- 0 0

0 1- 0

0 0 1- = 0


This is equivalent to the following equations

x1 =0

x2+x3 =0

x2+x3 =0 (1.42)

Solving these equations, we get

x1=0, x3=-x2

Within an arbitrary scale factor and an arbitrary sign factor eigen vector corresponding to

=0 is given by,

(1.43)

If the Eigen vectors be normalized to unity, then X1 = 1

i.e. [ 02 + k

2 + (-k)

2] =1 or, k = (1/2)

So that the normalized Eigen vector of matrix A corresponding to Eigen value =0 is

given by,

X1 = 0, (1/2), (-1/2)

This is equivalent to the following equations

x2=0 and x3=0.

So that X2 ={1,0,0}

For =2, equation (1) reduces to,

x1

x2

x3

1 0 0

0 1 1

0 1 1

= 0

0

0

X1 =

x1

x2

x3 = { x1,x2,x3} = { 0,k,-k}

x1

x2

x3

0 0 0

0 0 1

0 1 0

= 0

0

0

x1

x2

x3

-1 0 0

0 -1 1

0 1 -1

= 0

0

0

Vector and, Matrices

Which is equivalent to the following equations

-x1 =0

-x2 + x3 =0

x2 – x3 =0 (1.44)

Solving these equations, we get

x1 =0, x2 =x3

Within the arbitrary scale factor, the Eigen vector corresponding to =2 is given by,

X3 = {x1,x2,x3} = (0,k,k)

For X3 normalized to unity,

02 +k

2 + k

2

As the Eigen vectors are normalized to unity, therefore, the normalized Eigen vector of

matrix A corresponding to Eigen value =2 is given by ,

X3 = {0, (1/ 2 ), (1/ 2 )} (1.45)

Thus the normalized eigen vector of the given A corresponding to the eigen values 0,1,2

are,

{0, (1/ 2 ), (-1/ 2 )} , {1,0,0}, {0, (1/ 2 ), (1/ 2 )} respectively. (1.46)

1. 6 CAYLEY-HAMILTON THEOREM

Theorem: - Every square matrix satisfies its own characteristic equation i.e., if for a

square matrix A of order n, the characteristic polynomial is,

A -I

Then the matrix equation,

a0I+a1X + a2X2+……….+ anX

n = 0

is satisfied by X =A

=1 i.e. k = (1/ 2 )

=a0 +a1 +a22 + ……..+an

n


Proof

The characteristic polynomial is,

A -I

(1.47)

The characteristic equation of A is,

A -I

(1.48)

The matrix equation is,

(1.49)

If A satisfies this equation, then we have to show that

(1.50)

Since each element of characteristic matrix (A -I) is an ordinary polynomial of degree

„n‟ therefore the cofactor of every element of A -I is an ordinary polynomial of

degree (n-1) consequently, each element of,

B= adj (A -I )

is an ordinary polynomial of degree (n-1).

Therefore we can write,

B= adj (A -I) =B0 +B1+ B22 +…….+ Bn

n-1 (1.51)

Where B0, B1, B2,……. Bn-1,are all square matrices of the some order „n‟ whose elements

are polynomials in the elements of A .

By the theorem

(A-I) adj (A -I) = (A-I) (1.52)

Using equations (5) and (1) , we get ,

(A-I)[ B0 +B1+ B22 +…….+ Bn

n-1] = (a0 +a1+ a2

2 +…….+ an

n )I (1.53)

Comparing the coefficients of like power of on both the sides, we get

AB0 = a0I

AB1 – B0 = a1I

AB2 – B1 = a2I

=a0 +a1 +a22 + ……..+an

n

=a0 +a1 +a22 + ……..+an

n = 0

a0I + a1X + a2X2 +……..anX

n = 0

a0I + a1A + a2A2 +……..anA

n = 0


………………………….

………………………….

ABn-1 – Bn-2 = an-1I

-Bn-1 = anI

Now multiplying these equations respectively by,

I, A, A2, A

3…….A

n

And then adding, we get

(1.54)

Which is equation (1.3). This proves the theorem.

1.6.1 Example

Find the characteristic equation of the following matrix and verify the Cayley-

Hamilton theorem.

Solution: - Given matrix, is A =

A -I =

i.e.,

Hence the characteristic equation of given matrix A is,

Now, in order t verify Cayley –Hamilton theorem we have to show that,

Where, and , A =

0 = a0I + a1A+a2A2+…….+anA

n

1 2 3

2 -1 4

3 1 1

1 2 3

2 -1 4

3 1 1

1 2 3

2 -1 4

3 1 1

- 1 0 0

0 1 0

0 0 1 =

1- 2 3

2 -1- 4

3 1 1-

A -I = 1- 2 3

2 -1- 4

3 1 1- = -

3 +

2 +18 +30

-3 +

2 +18 +30 =0

-A3 + A

2 +18A +30I =0

1 0 0

0 1 0

0 0 1 I =

1 2 3

2 -1 4

3 1 1


A2 =

And A3 = A

2. A =

This shows that Cayley-Hamilton theorem is verified.

1 2 3

2 -1 4

3 1 1

1 2 3

2 -1 4

3 1 1 =

14 3 14

12 6 6

8 9 14

14 3 14

12 6 6

8 9 14

1 2 3

2 -1 4

3 1 1

=

62 39 68

48 21 78

62 24 62

-A3 + A

2 +18A +30I = =

62 39 68

48 21 78

62 24 62 +

14 3 14

12 6 6

8 9 14

+ 18

1 2 3

2 -1 4

3 1 1 + 30

1 0 0

0 1 0

0 0 1

-62+14+18+30 -39+3+36+0 -68+14+54+0

-48+12+36+0 -21+9-18+30 -78+6+72+0

-62+8+54+0 -24+6+18+0 -62+14+18+30 =

0 0 0

0 0 0

0 0 0 = = 0


1. 6. 2 Some Important theorems on Eigen values and Eigen vectors

Theorem: -I - The Eigen values of a hermitian matrix are real.

Proof: -

For a hermitian matrix A

= A (1.55)

Where is the transposed conjugate of A.

Let X be any Eigen vector of A corresponding to the eigen value.

Then AX = X (1.56)

Premultiplying equation (56) by, we get

(1.57)

Taking the transpose conjugate of both sides in (1.57)

i.e., ( ) =

(1.58)

(Using equation (1.55))

Now by using (1.56)

i.e., ( - * ) X = 0 (1.59)

As X is an Eigen vector X0, X 0 ,

Then equation (1.58) gives,

-* =0 or, =

*

This means that the conjugate of is equal to itself. This is only possible only when is

real.

A

A

X

X AX = X = X X X

(X AX ) = ( X ) X

X A X X * X

X A X = X * X

X A X = X * X

X X = X * X

X

X


Theorem – II - The Eigen values of a real symmetric matrix are all real.

Proof: -

For a real symmetric matrix A, we have (1.60)

A* =A

And, AT =A (1.61)

Taking complex conjugate of (1.61) and keeping in mind equation (1.55), we get

(AT)* =A, i.e. =A

Taking the complex conjugate of (1.61) and using equation (1.55), we get

(AT)* =A ,i.e. =A (1.62)

Thus real symmetric matrix is a hermitian matrix.

According to the Ist theorem the Eigen values of a hermitian matrix are all real.

Consequently, the Eigen values of a real symmetric matrix are all real. .

Theorem –III –The Eigen values (characteristic roots) of skew –hermitian matrix

are either zero or purely imaginary

Proof: -If A is a skew –Hermitian matrix, then we have

(1.63)

Let X be an Eigen vector A corresponding to eigen value .Then

AX=X (1.64)

(iA) X = (i)X (1.65)

We have (iA) = -i (-A) Using by equation (1.55)

(iA) = iA

This proves that iA is hermitian matrix.

According to equation (1.65) i is the Eigen value of hermitian matrix iA corresponding

to the eigen vector X. Therefore theorem Ist implies that i i.e. is zero or purely

imaginary number. This proves the theorem.

A

A

A = -A


Theorem: - IV The Eigen values of a real skew-symmetric matrix are either zero

or purely imaginary.

Proof

For a real skew symmetric matrix A, we have

A* =A (1.66)

And, AT = -A

* (1.67)

Taking complex conjugate of equation (1.67) we get,

(AT

)* = -A

* (1.68)

A = - A (Using equation (1.66))

This implies that the matrix A is skew –hermitian.

Hence according to theorem III, the Eigen values of a real skew symmetric matrix are

either zero or partial imaginary.

Theorem –V Any two eigen vectors corresponding to two distinct eigen values of a

Hermitian matrix are orthogonal.

Proof

Let X1, X2 be the two Eigen vectors corresponding to two distinct Eigen values 1, 2 of

a hermitian matrix A. Then

(1.69)

AX1 =1X1 (1.70)

AX2 =1X2 (1.71)

Here 1 and 2 are real numbers

Hence, (1.72)

1 = 1*, 2 =

*2

Premultiplying (1.70) and (1.71) by and respectively, we get

(1.73)

(1.74)

Taking transpose conjugate of (1.73), we get

A =A

X1 X2

X2 AX1 = 1X1 X2

X1 AX2 = 2X2 X1

(X2 AX1) = (1X1 X2) X1 A (X2 ) = *1 X1 (X2) Or,


(Using equation (1.72) )

(Using equation (1.69) ) (1.75)

Comparing equation (1.74) and (1.75), we get

i.e. (2 - 1) X1X2 = 0 (1.76)

As the eigen values are different, i.e. 2 1 or , (2 - 1) o, therefore we must have

X1X2 = 0 (1.77)

Which is the condition of Eigen vectors X1 and X2 to be orthogonal. Hence any two

Eigen vectors corresponding to two distinct eigen values of a Hermitian matrix are real.

X1 A (X2 ) = *1 X1 (X2)

X1 A (X2 ) = *1 X1 (X2)

1X1 X2 = *1 X1 (X2)




i) Find the eigen values of the following matrix:

3 1 4

0 2 6

0 0 5

ii) Define Cayley-Hamilton theorem and state its importance.

…………………………………………………………………………………………

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…………………………………………………………………………………………

…………………………………………………………………………………………

…………………………………………………………………………………………

…………………………………………………………………………………………

…………………………………………………………………………………………

…………………………………………………………………………………………

…………………………………………………………………………………………

…………………………………………………………………………………………

…………………………………………………………………………………………

…………………………………………………………………………………………

…………………………………………….


1. 7 LET US SUM UP

A vector is a directed „line segment‟ having direction and magnitude both.

The three assumed families which need not to be mutually perpendicular

described by q1 = constant, q2 = constant, q3 = constant ,intersecting at any point

„P‟ then the values of q1, q2, q3 for the three surfaces intersecting at the point „P‟

are called the “Curvilinear coordinates”.

“Curvilinear coordinates” q1, q2, q3 need not to be length, these coordinate are

expressed in terms of a scale factor hi .

Differential operators like Gradient, Divergence, Laplacian and Curl can be

expressed in terms of “Curvilinear coordinates”.

Differential operators like Gradient, Divergence, Laplacian and Curl can be

expressed in terms of “Spherical Polar coordinates( r,, )”

Where x=r sin cos, y= r sin sin and z = r cos

Differential operators in orthogonal curvilinear coordinates can be represented in

terms of Cylindrical coordinates where x=r cos , y=r sin and z=z.

The matrix (A -I) is called the characteristic matrix A, Where I is the unit matrix

and A is any given matrix.

The system of homogeneous equation (A -I)X =0 has non-trivial equation if and

only if (A -I) is singular, i.e.() =(A -I)=0.

Cayley -Hamilton Theorem states that Every square matrix equation statisfies its

own characteristic equation.

The Eigen values of a hermitian matrix are real.

The Eigen values of real symmetric matrix are real and the Eigen values

(characteristic roots) of skew –hermitian matrix are either zero or purely

imaginary.


1. 8 CHECK YOUR PROGRESS: THE KEY

1. i) The new families which may be described by q1= constant, q2= constant and

q3= constant, intersect at any point P on the three dimensionally surface and the

values of q1, q2 and q3 for the these three surfaces intersecting at the point P then

the coordinates q1 , q2 and q3 are called the curvilinear coordinates of P and the

three new surfaces are then called the coordinate surfaces or curvilinear surfaces.

In Cartesian coordinate system the position of a point P (x,y,z) is determined by

the intersection of the mutually perpendicular planes x= constant, y= constant and

z= constant while in case of curvilinear coordinates the families need not be

mutually perpendicular.

ii) The cylindrical coordinates system consists of:

(a) Right circular cylinders having z-axis as common axis, which form

families of concentric circles about the origin O in X-Y plane.

r = (x2 + y

2)1/2

= constant

(b) Half planes through to z-axis, =tan-1

(y/x)=constant.

(c ) Planes parallel to X-Y plane, z=constant

The transformations between rectangular coordinates (x,y,z) and cylindrical

coordinates (r,,z) are given by,

x= r cos , y = r sin , z=z

2. i) To find the eigen values of the given matrices follow just like example no.1.5.1

then the result will be (3-)((2-)(5-) = 0, then the values are = 2,3,5.

ii) The Cayley-Hamilton theorem is defined for square matrix as -

Every square matrix satisfies its own characteristic equation i.e., if for a square

matrix A of order n, the characteristic polynomial is,

A -I

Then the matrix equation

a0I+a1X + a2X2+……….+ anX

n = 0

is satisfied by X =A. This theorem is important uin order to find out the

characteristic of a square matrix.

=a0 +a1 +a22 + ……..+an

n

UNIT 2 TENSORS AND PRATIAL DIFFERENTAIL EQUATIONS

Structure

2.0 Introduction

2.1 Objectives

2.2 Definition of Tensor in three Dimensions

2.3 Coordinate Transformation

2.3.1 Indicial and summation convention

2.3.2 Dummy and real indices

2.3.3 Kronecker delta function

2.3.4 Properties of Kronecker delta function

2.3.5 Generalized Kronecker delta function

2.4 Types of Tensors

2.5 Rank of tensor

2.5.1 Tenor of higher ranks

2.6 Example

2.7 Algebraic operations of tensors

2.7.1 Addition and subtraction of tensors

2.7.2 Product of tensors

2.7.3 Contraction

2.7.4 Quotient law of tensor

2.8 Christofiels 3 index symbols.

2.8.1 Relationship between christoffels first and second kind symbols

2.8.2 Transformation laws for Christoffels symbol of first kind

2.9 Solved example

2.10 Partial differential equations

2.10.1 Solution of Laplace equation

2.10.2 Green function for Poisson equation and its solution

2.10.3 Solution of heat equation

2.10.4 Solution of wave equation

2.11 Solved example

2.11.1 Exercise

2.12 Let Us Sum Up



2. 0 INTRODUCTION

Tensor analysis is that part of study which is rather suitable for the mathematical

formulation of the natural laws in forms of which are invariant with respect to different

frames of reference. The tensor formulation was originated by G.Ricci and it become

popular when Albert Einstein used it as a natural tool for the description of his general

theory of relativity. Tensor analysis is the generalization of vector analysis.

The physical quantities scalars, vectors, dyadics, triadics…are collectively called tensors

of rank 0,1,2….In this way tensor analysis can be regarded as one possible extension of

vector analysis.

The partial differential equations are required to solve the special theoretical problems in

applied and theoretical physics. In this part of unit we shall discuss the solution of a few

partial differential equations which are frequently used in a wide variety of situations of

physical interest. The most frequently encountered partial differential equations are (1)

Laplace equation (ii) Poisson equation (iii) Heat flow equation (iv) Wave equation and

(v) Helmholtz differential equation.

2.1 OBJECTIVES

After completing this unit we are able are able to-

Define and classify the tensor by using coordinate transformation.

Perform the tensor addition, subtraction, multiplication and contraction

operations.

Define and establishes the Quotient law

Use and define Christofiels symbols

Define and solve Laplace, Poisson, wave and Heat conduction equations.

Apply partial differential equations to rectangular bar with finite and infinite

length.

2. 2 DEFINITION OF TENSOR IN THREE DIMENSIONS

The tensor is related with rotational motion and can be defined in terms of rotation. If we

consider the transformation of entities which are obtainable from the vectors. Let us

Tensors and Partial differential Equations

consider two vectors A and B. Let the linear relation connect their components.

(2.1)

After the rotation of the axes the new vector component will be connected by the

analogous relation

(2.2

With Tik =

Which is transformation equation for Tik when the axes are rotated. Such a set is called a

tensor of rank two. Similarly the tensors of the higher order (rank) in the three –

dimensional space may be defined,

And so on. (2.3)

2. 3 COORDINATE TRANSFORMATIONS

Tensor analysis is immediately connected with the subject transformations. Consider two

sets of variables (x1,x

2,x

3 …..x

n) and ( x

1, x

2, x

3 …..x

n) which determine the coordinates

of a point in an n- dimensional space in two different frames of reference. Let the two

sets of variables be related to each other by the transformations.

x1 =

1(x

1,x

2,x

3,x

4……….x

n)

x2 =

2(x

1,x

2,x

3,x

4……….x

n)

…………………………….

……………………………..

xn =

n (x

1,x

2,x

3,x

4……….x

n)

x =

(x

1,x

2,x

3,x

4……….x

) ( =1,2,3……n) (2.4)

Where function are single valued, continuous differentiable functions of coordinates. It

is essential that the n-functions of coordinates. It is essential that the n-function

be

independent.

Equation (2.4) can be solved for coordinates x as function of x

to yield

ij klTll i,j =1

3

ij kp l n Tjpn i,j =1

3

Tikl =

Bi = Tik Ak k =1

3


x =

(x

1,x

2,x

3,x

4……….x

) (2.5)

From equation (2.4) the differentials d x are transformed as

( = 1,2,3…..n) (2.6)

2.3.1 Indicial and Summation Conventions

Let us now introduce the following two conventions.

(1) Indicial Convention: - Any index, used either as subscript or superscript will

take all values from 1 to n unless the contrary is specified. Thus equation (2.3) can be

briefly written as,

(2.7)

The convention reminds us that there are „n‟ equations with = 1,2…..n and are the

functions of „n‟ coordinates x

with = 1,2,….n.

(2) Einstein‟s Summation convention: - If any index is repeated in a term, then a

summation with respect to that index over the range 1,2,…..n is implied. This convention

is called Einstein‟s summation convention.

According to this convention instead of expression , we merely write x

Using these two conventions equation (2.5) may be written as,

(2.8)

Thus the summation convention means the drop of sigma sign for the index appearing

twice in a given term. In other words, the summation convention implies the sum of the

term for the index appearing twice in that term over defined range.

d x

= dx1 + dx

2 +…………… dx

n

x

x1

x

x2

x

xn

n

=

=1

x

x

dx

x

=

( x

)

n

=1

x

dx = dx

x

x


2. 3. 2 Dummy and real indices-

Any index, which is repeated in a given term, so that the summation convention applies,

is called a dummy index, and may be replaced freely by any other index not already used

in that term. For example is the dummy in a

x.

Any index, which is not repeated in a given term, is called real index. For example, is a

real index in a

x. Any real index can not be replaced by another real index, e.g.

a

x a

x.

2.3.3 Kronecker delta Symbol -

Kronecker delta symbol is defined as,

(2.9)

2.3. 4 Properties of Kronecker delta function

(1) If x1, x

2, x

3,…..x

n are independent variables, then

(2.10)

(ii) An obvious property of Kronecker delta symbol is

(2.11)

(iii) If we are dealing with n-dimensions, then

(2.12)

(iv)

=

(2.13)

(v) (2.14)

2. 3. 5 Generalized Kronecker delta The generalized Kronecker delta is symbolized as,

(2.15)

= 1 if =

0 if

x

x =

A

= A

=

= n

x

x

x

x x

x = =

1 2 3………m

1 2 3………m


And are defined as follows: -

(i) The subscripts and superscripts can have any value from 1 to n .

(ii) If either at least two superscripts or t least any two subscripts have the same value

or the subscripts are not same set as superscripts, then the generalized Kronecker

delta is zero.

(2.1)

(iii) If all the subscripts are separately different and are the same set of numbers as the

subscripts, then the generalized Kronecker coordinates delta has the value +1 or –

1 according as whether it requires an even or odd number of permutations to

arrange the superscripts in the same order the subscripts. For example

(2.17a)

(2.17b)

2. 4 TYPES OF TENSORS

(a) Scalars: - Consider a function in a system of variables x and let this function

have the value in another system of variables x . Then if

=

The function is said to be scalar or invariant or tensor of order zero.

(b) Contravariant Vectors or Contravariant Tensor of first rank: -consider a set

„n‟ quantities A1,A

2, A

3…..A

n in a system of variables x

and these quantities has values

A1,A

2,A

3…..A

n in another system of variables x

. If these quantities obey the

transformation relation

(2.18)

= =

= 0

123

231 123

123 = =

1452

4125 = +1

123

132 123

213 = =

1452

4152 = -1

A = A

x

x


Then the quantities A are said to be the components of a contravariant vector or a

contravariant tensor of first rank. The components of a contravariant vector are actually

the components of a contravariant tensor of rank one.

Let us now consider a further change of variables from x to x

, then the new

components A must be given by,

(2.19)

This indicates that the transformations of contravariant vector form a group. Thus in

general, the tensors whose components transform like coordinate differential are called

contravariant tensor and is written as,

(2.20)

Where are the coordinate differentials.

The tensor of rank two contravariant in both indices transform as follows,

(2.21)

( c) Covariant Vectors : Consider a set of „n‟ quantities A1, A2, A3….An in a system

of variables x and let these quantities have values A1, A2, A3….An in another system of

variables x . If these quantities obey the transformation equations

(2.22)

Then the quantities A are said to be the components of a covariant vector or a covariant

tensor of rank one.

Let us now consider a further change of variables from x to x

. Then the new

component A must be given by,

[ Using equation (2.22)]

(2.23)

A = A

x

x

= x

x

x

x

A x

x

= A

Al = Ak

xi xk

K=0

3

xi xk

Tlk

= Tjl

j,l =0

3 xi xj

xk

xi

A = A xk

x

A = A = x

x

x

x

x

x

A

x

x

A =


This indicates that the transformations of covariant vectors from a group. The tensor

whose components transform like the partial derivatives of the coordinate are called

covariant tensors. The transformation law for a covariant vector is written as,

(2.24)

Where are the partial derivatives of the coordinates. Similarly, a covariant tensor

of rank two transforms as,

(2.25)

(d) Mixed Tensor: A mixed tensor is covariant in some indices and contravariant in

the others. It component transform according to,

(2.26)

This tensor Tkl is a mixed tensor, contravariant in one index I and covariant in the second

index k.

2. 5 RANK OF TENSOR

The rank of the tensor is equal to the number of suffixes or the indices attached to it .The

rank of the tensor when raised as power the number of the dimensions gives the number

of components of the tensor. Thus a tensor of rank N in three-dimensional space has 3N

components and in four-dimensional space it has 4N component. Therefore the rank of the

tensor gives the number of the mode of changes of a physical quantity when passing from

one system to another one, which is in rotation relative to the first. Thus a quantity that

does not change when the axes are rotated is a tensor of the zero rank and so on.

2. 5. 1 Tensors of higher ranks

The laws of transformation of vectors are,

Al = Ak xk

xl

k=1

3

xk

xl

Tlk

= Tjl

j,l =1

3 xj

xj

xi

xk

Tkl

= Tlj

j,l =1

3 xj xj

xi

xk


Contravariant:

(2.27)

Covariant:

(2.28)

(a) Contravariant tensors of second rank: - Let us consider (n) 2 quantities A

(here and take the values from 1 to n independently) in a system of variables x and

let these quantities have values A

in another system of variables x .If these quantities

obey the transformation equations,

(2.29)

Then the quantities A

are said to be the components of a contravariant tensor of second

rank.

(b) Covariant tensor of second rank: - If (n)2 quantities A in a system of variable

x are related to other (n)

2 quantities in another system of variables x

by the

transformation equations

(2.30)

Then the quantities A are said to be the components of a covariant tensor of second

rank

(c) Mixed Tensors of second rank :- - If (n)2 quantities A

in a system of variable

x are related to other (n)

2 quantities A

in another system of variables x

by the

transformation equation equations

(2.31)

Then the quantities A are said to be the components of a mixed tensor of second rank..

An important example of mixed tensor of second rank is the Kronecker delta .

A = A

x

x

A = A x

x

A

= A

x

x

x j

x

A = A x

x x

xj

A = A

x

x

x

x


2. 6 EXAMPLE

1. Show that the transformation of a tensor forms a Group

Solution:

Let A

be a mixed tensor in a system of variables

x . According to tensor

transformation law the components A

are related to the component A of the tensor in

the system of variables x as

(2.32)

Let us now consider a further change of variables from x and x, then the new

components A must be given by,

[ using (2.32)]

(2.33)

A = A

x

x

x

x

A = A

x

x

x

x

x

x

x

x

= x

xi

x j

x Aj

i

x

x

x

xi

= x

j

x

x

x Aj

i

= Aji

x

xi

x j

x




i) Define tensor and its types? What is the basic difference between tensor and

vector?

ii) What are the real and dummy indices used in tensor analysis?

…………………………………………………………………………………

…………………………………………………………………………………

…………………………………………………………………………………

…………………………………………………………………………………

…………………………………………………………………………………

…………………………………………………………………………………

…………………………………………………………………………………

…………………………………………………………………………………

…………………………………………………………………………………

………


This has the same form as equation (2.32). This means that if we make the direct

transformation from the system of variables x and x,, we get the same transformation

law .This proves that the transformations of tensors form a group

2. 7 ALGEBRAIC OPERRATIONS OF TENSORS

Addition, Subtraction, multiplication and Contraction operations with tensor. Quotient

law, Christofell‟s symbols.

2.7.1 Addition and Subtraction

The addition and subtraction of tensor is defined only in the case of tensors of same rank

and same type. Same type means the same number of contravariant and covariant indices.

The addition or subtraction of two tensors, like vector, involves the elements .To add or

subtract two tensors the corresponding elements are added or subtracted.

“ The sum or difference of two tensors of the same rank and same type is also a tensor of

the same rank and same type.”

For examples - If there are two tensors A

and B

of the same rank and same type,

then the laws of addition and subtraction are given by,

A

+ B

= C

(2.34)

A

- B

= D

(2.35)

Where C

and D

are the tensors of the same rank and same type as given tensors.

The transformation laws for the given tensors are,

(2.36)

(2.37)

Adding equation (2.36) and (2.37), we get

(2.38)

A

= A

x

x

x

x

x

x

B

= B

x

x

x

x

x

x

C

= C

x

x

x

x

x

x


Which is transformation law for the sum and is similar to transformation laws for A

and B

as given by equation (2.36) and (2.37).

Hence the sum C

(= A

+ B

) is itself a tensor of the same rank and same type as

given tensors.

Subtracting equation (2.37) from (2.36), we get

(2.39)

Which is transformation law for the difference and is similar to transformation laws for

A

and B

as given by equation (2.36) and (2.37).

Hence the difference D

(= A

- B

) is itself a tensor of the same rank and same

type as given tensors.

The addition of tensor is commutative and associative.

2.7. 2 Product of tensor

(i) Outer product: - The outer product of two tensors is a tensor whose rank is the

sum of the ranks of given tensors.

Thus if r and r are the ranks of two tensors, their outer product will be a tensor of rank

(r+ r).

For example if A

and B are the two tensors of rank 3 and 2 respectively, then

A

B = C

( say) is the tensor of rank (3+2=) 5.

Proof

Let us write the transformation equations of given tensors as,

(2.40)

(2.41)

Multiplying (2.40) and (2.41) we get

(2.42)

A

- B

= ( A

- B

) x

x

x

x

x

x

( A

- B

) x

x

x

x

x

x

D

=

B = B

x

x

x

x

A

= A

x

x

x

x

x

x

A

B = B A

x

x

x

x

x

x

x

x

x

x


(2.43)

Which is a transformation law for a tensor of rank 5. Hence the outer product of two

tensors A

and B of rank 3 and 2 is a tensor C

of rank 5. The outer product of

tensor is commutative and associative.

(ii) Inner product: - The product of two tensors followed by a contraction results in

a new tensor called an inner product of the two tensors and the process is called the inner

multiplication of two tensors.

Examples: -

(a) Consider two tensors A

and B

The outer product of these two tensors is

A

B = C

(2.44)

Applying contraction process by setting =, we obtain,

A

B = C

= D

( a new tensor )

(2.45)

The new tensor D

is the inner product of two tensors A

and B.

2.7. 3 Contraction of tensor

The algebraic operation by which the rank of a mixed tensor is lowered by 2 is known as

contraction. In the process of contraction one contravariant index and one covariant

index of a mixed tensor are set equal and the repeated index assumed over, the result is a

tensor of rank lower by two than the original tensor.

For example, consider a mixed tensor A

of rank 5 with contravariant indices,,

and covariant indices , .

The transformation law for the given tensor is,

(2.46)

To apply the process of contraction, we put = and obtain

A

= x

x

x

x

x

x x

x

x

x

A

A

= x

x

x

x

x

x x

x

x

x

A

A

= x

x

x

x

x

x x

x

x

x

A

x x

x

C

= B A

x

x

x

x

x

x

x

x

x

x


(2.47)

Which is also a transformation law for a mixed tensor of rank 3. Hence A

is a

mixed tensor of rank 3and may be denoted by A

. In this example we can further apply

the contraction process and the contravariant A

or A .Thus the process of

contraction enables us to obtain a tensor of rank (r-2) from a tensor of rank r.

2.7.4 Quotient law

In tensor analysis it is often necessary to ascertain whether a given entity is a tensor or

not. A direst method is used to test this and the method is known as Quotient law, which

stated as,

“Any entity whose inner product with an arbitrary tensor (Contravariant or

covariant) is a tensor is itself a tensor.”

2.8 CHRISTOFELL‟S 3 INDEX SYMBOLS

Christofell‟s 3 Index symbols are not tensor but formed fundamental tensors of first and

second kind namely,

(i) Christofell‟s symbol of first kind :-

[, ] = , = (2.48)

(ii) Christofell‟s symbol of second kind :-

(2.49)

From the symmetry property of g it follows that ,

[,, ] = [,, ] or, , = ,

(2.50)

and,

or,

(2.51)

There by indicating that Christofell‟s symbols, and ,

are symmetric with

respect to indices and .

g

x

g

x

+ - g

x

1

2

= = (1/2) g

g

x

g

x

+ - g

x

=

=

A

= x

x

x

x x

x

A


2. 8.1 Relationship between Christofell‟s symbols of first and second kind

From equation (2.48)

, =

Replacing by in above equation we get ,

(2.52)

Multiplying both sides of above equation by g

, we get

i.e. (2.53)

(ii) Interchanging and in equation (2.49), we get

Multiplying above equation by g , we get

(2.54)

i.e.

Equation (2.53) and (2.54) represent the relation between Christofell‟s symbols of first

and second kind.

2. 8. 2 Transformation Laws of Christofell‟s symbols

(i) The transformation law for Christofell‟s 3 index symbols of first kind :-

Using the definition of Christofell‟s symbol of first kind the transformation law for

Christofell‟s 3 index symbols of first kind can be represented as,

(2.55)

g

x

g

x

+ - g

x

1

2

g

x

g

x

+ - g

x

1

2 , =

g , =

g

x

g

x

g

x

+ - 1

2 g

=

= g

,

= (1/2) g

g

x

g

x

+ - g

x

g = (1/2) g

g

g

x

g

x

+ - g

x

= ,

, = g

,

+

2x

x

x

x x

g , = x

x

x

x

x

x


Obviously the Christofell‟s symbols of first kind do not transform like tensors unless the

second terms on R.H.S. is zero. This clearly indicates that the Christofell‟s symbols of

first kind are not the tensors.

(ii) The transformation law for Christofell‟s 3 index symbols of second kind: -

Transformation law for the contravariant fundamental transformation ism given as,

(2.56)

This equation represents the transformation law for the Christofell‟s 3 index symbols of

second kind. Obviously, the Christofell‟s 3 index symbols of second kind do not obey

the tensor transformation law unless the second term on the R.H.S. become zero in

equation (2.56). Thus from equation (2.55) and (2.56) it follows that the Christofell‟s 3

index symbols are not the tensors.

However, in a special case of linear transformation of coordinates of the type

x = a

x

+ b

n

(2.57)

Where a

and bn are constant so that then equation (2.55) and (2.56)

becomes

(2.58)

(2.59)

Equation (2.58) and (2.59) indicates that the Christofell‟s 3 index symbols transform like

tensors relative to the linear transformation of coordinates of type (2.57). In such cases

the Christofell‟s 3 Index symbols are called pseudo tensors.

= x

x

x

xn

x

x

n +

2x

x

x

x x

2x

x

x

= 0

, = x

x

x

x

x

x

,

= x

x

x

xn

x

x

n




i) Define the Quotient law and contraction for the tensor?

…………………………………………………………………………………………

…………………………………………………………………………………………

…………………………………………………………………………………………

…………………………………………………………………………………………

…………………………………………………………………………………………

…


2.9 SOLVED EXAMPLE

Show that

(i) [, ] + [ ,] =

(ii)

(iii)

Solution

(i) By definition of Christofell‟s 3 index symbols of first kind

(2.60)

(2.61)

Keeping the symmetry property in mind of g ; adding equation (2.60) and (2.61), we

get

(2.62)

(ii) Differentiating the determinant g = g , remembering that g

g is the

cofactor of g in this determinant, we obtain ,

(Since g

= g

)

i.e. =

(2.63)

i.e.,

g

x

x

= log g

gpq

xm

= - g

p - g

q

q

m

p

m

g

x

g

x

+ - g

x

1

2 [, ] = , =

g

x

g

x

+ - g

x

1

2 [,] = , =

[, ] + [ ,] = g

x

g

x = g

g

g

x

= g g

(, + , )

= g ( g

, + g

, )

= g ( g

, + g

, )

= g ( + g

) = 2g

1 g

2g x

= log g

x


If g is negative , then equation (2.64) becomes

(2.64)

(iii) From the properties of fundamental tensors

Differentiating this equation with respect to xm

, we get

Taking inner product of above equation with , we get

Using equation (2.62)

(since g

is symmetric )

Now replacing by q, and by , we get ,

Keeping in mind the symmetry property of g

, above equation may be rewritten as,

2.10 PARTIAL DIFFERENTIAL EQUATIONS

The most frequently encountered partial differential equations and the kinds of the

physical problems, which lead to each of them, are as given below –

= (log -g )

x

gg

= g

g + g

g

xm

g

xm

= 0

g p

g + g

g

xm

g

xm

= 0 g

p g

p

+ g

g

xm

= 0

p g

p ( , m + , m )

+ g

gp

xm

= 0 g

p , m + g

, m

+ gp

xm

= 0 g

p m

0 + g

m

p

= - gp

xm

g

p m

0 - g

m

p

= - gp

xm

g

p m

p - g

q m

p

= - gpq

xm

g

p - g

q

q

m p

m


(A) Laplaces equation: - The Laplace equation‟s differential is,

2 = 0 (2.65)

This is the most important not most commonly used differential equation and occurs in

studying the :-

(i) Gravitational potential in regions containing no matter.

(ii) The electrostatic potential in a uniform dielectric, in the theory of electrostatic

(iii) The magnetic potential in free space, in the theory of magnetostatic.

(iv) The electric potential in the theory of the steady of the steady flow of electric

currents in solid conductors .

(v) The temperature in the theory of thermal equilibrium of solids.

(vi) The velocity potential a points of a homogeneous liquid moving irrotationally

in hydrodynamic problems.

(B) Poisson‟s Equation: - The Poisson differential equation is ,

2 = (2.66)

Where is a function of position coordinates and is called the source density. The

function may represent the same physical quantities as for the Laplace equation

(Gravitational potential, electric potential, magnetic potential, the temperature) but in a

region containing matter or electric charge or magnetic source or heat source or fluid

source depending on physical situation.

(C) Heat flow Equation: - The time dependent heat flow equation is,

(2.67)

Where „h‟ is a constant and is called the diffusivity whereas may be the non-steady

state temperature with no heat source or it may be the concentration of diffusing material.

(D) The wave Equation: - The wave equation is,

(2.68)

2 =

1

h2 t

2 =

1 2

c2 t

2

2 - 1

2

c2 t

2

=0 Or,


2.10.1 Solution of Laplace‟s Equation

( i) In Cartesian coordinates

The Laplace equation in Cartesian coordinates system is

(2.69)

The solution of this equation may be obtained by using the method separation of

variables as,

(2.70)

Where N k1 k 2 k3 =ABC is an arbitrary constant and may be evaluated by initial

conditions of the specified problem.

(ii) In two dimensional cylindrical coordinates (r, )

The Laplace equation 2u = 0 in cylindrical coordinates is

(2.71)

The general solution of this equation based of circular harmonics is ,

(2.72)

Where a0,an,bn,dn and c0 are arbitrary constants.

Solution of Laplace equation (2.73) in general cylindrical coordinates (general cylindrical

Harmonics) is given by

(2.73)

This solution remain finite at r=0 and is especially useful in certain electrical problems

and the problems of steady state heat conduction. The constants in the solution may be

evaluated by using the boundary conditions of the specified problems

(iii) In spherical polar coordinates (special Harmonics)

The Lap lace equation 2u = 0 in spherical polar coordinates (r,, ) takes the form,

(2.74)

2u =

2u

2u

2u

x2 y

2 z

2

+ + = 0

u = N k k k e k1

x + k2

y + k3

z 1 2 3

1 u 1 2u

2u

r r r r2

2 z

2

r + + = 0

u= a0 loge r + rn ( an cos n + bn sin n ) + r-

n ( cn cos n + dn sin n) + c0

n =1

n =1

u= [ekz

( Am cos m + Bm sin n ) + e-kz

( Cn cos m + Dn sin m)] Jm(kr) m =1

m =1

u 1 u

r r sin

r2 + + = 0 sin

1

sin2

2u

2


The solution of this equation may be obtained by the method of separation of variables

and may be expressed as,

u= ( A rn + B r

-n-1) Sn (2.75)

This solution is called the spherical harmonic. The subscript „n‟ on Sn signifies that the

same value of „n‟ must be used in both terms of equation (2.75).

The general solution of Lapalce equation when Sn is independent of is given by –

(2.76)

Where An and Bn are constants to be determined by boundary conditions of specified

problem.

2. 10. 2 Green‟s Function for Poisson equation and it‟s solution

The Poisson equation in ( S.I.) system is given by,

2 ( r) = (2.77)

where (r) is electrostatic potential, (r) is the charge density at the point (r), and 0 is

the permittivity of the free space. The function satisfies certain boundary conditions and

if the charges qi, then the general solutions of Poisson‟s equation is the superposition of

single point charge solutions, given by

(2.78)

If the place of the discrete point charges the system consists of continuous charge

distribution with charge density , then above equation becomes ,

(2.79)

Where d is small element of volume.

If the charge is away from the origin at r = ri , then electrostatic potential at position r2 is

given by

(2.80)

u= ( An rn + Bn r

-n-1) Pn ( Cos )

n = 0

( r)

0 - -

(r) = ( 1/40) (qi / ri ) i

(r) = ( 1/40) (d / r ) i

( r2) = ( 1/40) ((r1)d /( r2 – r1 ) i V


2.10.3 Solution of Heat Flow equation

The three –dimensional heat flow equation is

(2.81)

Here u (x, y, z, t) = (x, y, z, t) (t) and the complete solution of above equation is given

as,

u (x, y, z, t) = (x, y, z, t) (t) = (x, y, z, t) e- k h t

(2.82)

Here the reason for choosing the constant (-k2) to be negative is that temperature of the

body decreases with increase of time. The value of the constant „k‟ satisfies the

conditions, 2 +k

2 =0 and ( / t) = -k

2h

2

2.10.4 Solution of Wave equation

The wave equation is given by

(2.83)

When these waves propagate along z-axis, we can assume that this equation satisfies the

given solution below.

(2.84)

2u =

1 u

h2 t

2 2

2 - 1

2

c2 t

2

=0

(r ,t) = (x, y) ei k z – i t

0




i) Define Laplace differential equation? Write its importance in applied Physics?

…………………………………………………………………………………………

…………………………………………………………………………………………

…………………………………………………………………………………………

…………………………………………………………………………………………

…………………………………………………………………………………………

…………………………………………………………………………………………

…………………………………………………………………………………………

…………………………………………………………………………………………

…………………………………………………………………………………………

…………………………………………………………………………………………

…


2.11 SOLVED EXAMPLES

Q.No.1: - Determine the steady state temperature distribution in a thin plate bounded by

the lines x =0,x=l ,y =0 and y= ; Assuming that heat can not escape from either surface

of the plate, the edges x=0,x=l and y= being kept at zero temperature; while the edge

y=0 being kept at steady temperature F (x).

Solution: - The diffusion equation is

(2.85)

Where „h2‟ is a constant and is called the diffusivity of the substance. Under steady state

(u / t) =0 , so that equation in two dimensions becomes ,

(2.86) BLOCK-1

This is two dimensional Laplace‟s equation in Cartesian coordinates. The boundary

conditions are

(i) u=0 at x=0 (ii) u=0 at x=l (iii) u=0 at y= and (iv) u= F(x) at y=0 (2.87)

In order to apply the method of separation of variables, let us substitute

u (x, y) =X(x) Y(y) (2.88)

Substituting u from equation (2.88) in equation (2.86) and diving by XY; we get

Or , (2.89)

In this equation left hand side is the function of „y‟ only, while right hand side is the

function of x only; therefore each side must be equal to the same constant.

(2.90)

and (2.91)

The solution of equations (2.91) are given by,

(2.92)

(2.93)

2u =

1 u

h2 t

2u =

2u

2u

x2 y

2

+ = 0

1 2X

X x2 +

1 2X

Y y2 = 0

1 2X

X x2 = -

1 2X

Y y2

1 2X

X x2 = -

1 2X

Y y2 = k

2

2X

x2 + k

2 X =0

2X

y2 - k

2 Y =0

X = C1 cos kx + C2 sin kx

Y = C3 ek y

+ C4 e

-k y

u = 0 u = 0

x

y

x= l


Substituting these values in equation (2.88); we get

(2.94)

The constant A,B,C,D are arbitrary constants and may be evaluated using boundary

conditions, we get A=B=C=0 so the equation (2.94) becomes,

(2.95)

For a non-trivial solution D 0, hence sin kl =0 or kl = r; r= 1,2,3.. (2.96)

( Here r0 since k 0)

And, k = and r=1,2,3….

Hence each value of r; there is a solution given by (2.95)

(2.97)

For all possible value of „r‟ the above solution may be written as,

Now applying the last boundary condition (iv) we obtain the value of Dr as

Substituting this value of Dr in equation (2.97); we

(2.98)

This is the required temperature in this plate.

Q.No.-2 A neutral conducting sphere is placed in a uniform electric field. Find the new-

perturbed electrostatic potential.

Solution: - If V is the electrostatic potential, the Laplace equation is 2V= 0, Due to

spherical shape, we choose spherical coordinates, with origin at the center of sphere and

z-axis parallel to applied field E0, then V is symmetric with respect to .

The solution of Laplace equation is,

(2.99)

As unperturbed (or applied) electric field is E0, then

u (x,y) = (C1 cos kx + C2 sin kx)( C3 ek y

+ C4 e

-k y )

= ek y

( A cos kx + B sin kx) + e-k y

( C cos kx + D sin kx)

u (x,y) = D e-k y

sin k x

r

l

ur = Dr e- r y / l

sin (r / l )

u = Dr e- r y / l

sin (r / l ) r =0

r =

Dr = ( 2 / l ) ( F (t) ( sin r t/l) dt l

r = 0

u = e

- r y / l sin (r / l ) ( 2 / l ) [ ( F (t) ( sin r t/l) dt ]

r =0 0

l

V = ( An rn + Bn r

-n-1) Pn (Cos )

n =0


V ( r ) = E0z = - E0 r cos = - E0 r P1 Cos (2.100)

Therefore for ( r ), equation (2.99) gives

(2.101)

Comparing coefficient of Pn(Cos ) on both sides, we get

An =0, n >1 and A1 = -E0 choosing conducting sphere and the plane =/2 at zero

potential, the equation (2.99), in view of condition (2.101) gives

(2.102)

This equation holds for all values of , hence coefficients of Pn( Cos ) must vanishes

separately .Thus A0 =B0 ,Bn = 0 for n 2 and B1 = E0 r

30

So the value of V, i.e. perturbed potential outside the sphere is,

(2.103)

2.11.1 Exercise

Q.No.-1 Determine the steady state temperature distribution of a thin rectangular

plate bounded by the lines x=0,x=l,y=0,y=b assuming that the edges

x=0,x=l,y=0 are at zero temperature and the edge y=b is maintained at

steady state temperature.

Q.No.-2 A long cylinder is made of two halves thermally insulated from each

other, the upper half is at temperature T1 and the lower at T2 . Find the

steady state temperature inside the cylinder inside the cylinder

[ Hint:- Boundary conditions u =T1 at r=R,0<< and u=T2 at r=R , < <

2 ]

2.12 LET US SUM UP

Tensor – The physical quantities scalars, vectors, dyadics, triadics…are collectively

called tensors and it is related with rotational motion. Indicial conventions and

Einstein‟s summation convention are used to represent a tensor.

- E0 r P1 Cos = ( An rn + Bn r

-n-1) Pn (Cos )

n =0

(V)r=r0 = 0 = A0 + ( B0 / r0) + ( B1/r2

0 – E0r0) P1 (Cos ) + [Bn Pn (Cos ) / r0n+1

] n=2

V = -E0r P1 (Cos ) + (E0r3

0 /r

2) P1 (Cos ) = - E0r P1 (Cos ) ( 1- r

20/r

2)


Rank of tensor – The rank of tensor is equal to the number of suffixes or the indices

attached to it. Rank of the tensor gives the number of the mode of changes of a physical

quantity when passing from one system to another one, which is in rotation relative to the

first.

Types of tensors - Tensors can be classified as (i) Contravariant (ii) Covariant (iii)

Mixed tensor

The sum or difference of two tensors of the same rank and same type is also a tensor of

same rank and same type.

Product of tensors – (i) Outer product (ii) Inner product. The outer product of two

tensors is a tensor whose rank is the sum of the ranks of given tensors. The product of

two tensors followed by a contraction results in a new tensor called an inner product of

the two tensors. Contraction is the algebraic operation by which the rank of a mixed

tensor is lowered by 2 is known as contraction

Quotient law - Any entity whose inner product with an arbitrary tensor (contravariant or

covariant) is a tensor is it self a tensor.

Partial differential equations –The most important partial differential equations are –

(i) Laplace‟s equation (ii) Poisson‟s equation (iii) Heat flow equation (iv) The wave

equation.

2.13 CHECK YOUR PROGRESS: THE KEY

1. i) The physical quantities scalars, vectors, dyadics, triadics…are collectively

called tensors of rank 0,1,2….In this way tensor analysis can be regarded as one

possible extension of vector analysis. In vector analysis is the analysis of vector

quantities is done only on the basis of triangle and polygons rules

Tensor analysis is the generalization of vector analysis.

ii) Dummy and real indices- Any index, which is repeated in a given term, so that

the summation convention applies, is called a dummy index, and may be replaced

freely by any other index not already used in that term. For example is the

dummy in a

x.


Any index, which is not repeated in a given term, is called real index. For

example, is a real index in a

x. Any real index can not be replaced by another

real index, e.g.

a

x a

x.

2. Quotient law - In tensor analysis it is often necessary to ascertain whether a given

entity is a tensor or not. A direst method is used to test this and the method is

known as Quotient law, which stated as,

“Any entity whose inner product with an arbitrary tensor (Contravariant or

covariant) is a tensor is itself a tensor.”

Contraction of a tensor- The algebraic operation by which the rank of a mixed

tensor is lowered by 2 is known as contraction. In the process of contraction one

contravariant index and one covariant index of a mixed tensor are set equal and

the repeated index assumed over, the result is a tensor of rank lower by two than

the original tensor.

For example, consider a mixed tensor A

of rank 5 with contravariant

indices,, and covariant indices , .

3. Laplaces equation - The Laplace equation‟s differential is, 2 = 0. This is the

most important not most commonly used differential equation and occurs in

studying the :-

Gravitational potential in regions containing no matter.

The electrostatic potential in a uniform dielectric, in the theory of

electrostatic

The magnetic potential in free space, in the theory of magnetostatic.

The electric potential in the theory of the steady of the steady flow of

electric currents in solid conductors .The temperature in the theory of

thermal equilibrium of solids. The velocity potential a points of a

homogeneous liquid moving irrotationally in hydrodynamic problems.

REFERENCES AND SUGGESTED TEXT BOOKS :

1 Group theory and Quantum mechanics by M.Tinkam

2 Mathematical methods for Physicist by G.Arfken

3 Mathematical Physics for Physicist & Engineers by L.Pipes.

4 Mathematical Physics by Satyaprakash

5 Mathematical Physics by B.D.Gupta.

UNIT 3 GROUP THEORY

Structure

3.0 Introduction

3.1 Objectives

3.2 Group

3.3 Sub-Group

3.3.1 Properties of sub group

3.3.2 The concept of normal sub group

3.3.3 Conjugate elements and classes

3.4 Factor group

3.4.1 Properties of factor group

3.5 complexes of Group

3.6 Isomorphism and Homomorphism groups

3.6.1 Isomorphism

3.6.2 Homomorphism

3.7 Reducible and irreducible representations

1.7.1 Theorems on representations (1) General Theorem (ii) Schur‟s Lemma.

1.7.2 Orthogonality theorem on representations

3.8 Character of representation and character Table

3.9 Direct product of group

3.10 Applications of Group theory in Physics .

3.11 Let Us Sum Up


3. 0 INTRODUCTION

The idea of a group has pervaded the whole of mathematics both pure and

applied. Now a day the group theory is developed in an abstract way so that it can

be applied in many different circumferences particularly in investigations of a

number of properties of symmetric molecules. The theory can also be used to

draw a number of deductions regarding the nature of vibrations of symmetric

molecules and the electronic properties and the electronic transitions of such

molecules.

3.1 OBJECTIVES

The main objective of this unit is to study group theory. After completing this unit you

will be able to-

Group Theory and Functions of Complex variable

Define various definitions of groups concepts particularly used in theoretical

physics.

Find out the way to solve particular problems relating to the physics using

group concept.

Find various relations, concepts and other related findings.

To solve the specific problem regarding to the theoretical physics.

3. 2 GROUP

A group is a set of distinct elements G ={ E,A,B,C,D…} finite or infinite in

number, endowed with the laws of compositions ( Such as addition,

multiplication, matrix multiplication) such that the following properties are

satisfied -

(i) Closure : The product of any two elements of the group is a unique element

which also belongs to that group.

AB G or BA G (1.1)

The two elements may be distinct or same this property is known as closure property of

the group.

(ii) Associative : When three or more elements are multiplied, the order of

multiplication makes no difference,

i.e. A(BC) = (A.B).C (1.2)

Associative law holds good for all elements they form is then combined with A.

(iii) Identity: Among the elements there is an identity element, denoted by E, with

the property of leaving the elements unchanged on multiplication i.e.

A.E =E.A =A ,B.E=E.B=B, C.E.=E.C=C (1.3)

(iv) Inverse: Each elements, A, in the group has an inverse (or reciprocal) A-1,

,such

that,

A.A-1

= A-1

A =E (1.4)

The number of elements in a group is known as its order. A group containing finite

number of elements is called a finite group : a group containing an infinite number of

elements is known as infinite group. A group which consists of a single element A and

Group Theory

its powers A2…A

p =E is called a cyclic group of order p; if p is the smallest positive

integer for which Ap =E, p is then the order of the cyclic group and is also called the

order of the element A. If the multiplication is commutative i.e. for every pair of element

A and B we have A.B= B. A.A group is said to be commutative or Abelian group,

however the product of group elements is not necessarily commutative.

3. 3 SUB GROUP

When a number of elements selected from a group do themselves form a group it is

known as a sub Group. The associated law for the subgroup is implied automatically by

the associated law for the group .The identity element for the group is also the identity

element for the sub group G. Every group G has two trivial subgroups – the identity

element E and group G itself . A subgroup H different from group G is called proper

subgroup i.e., subgroup A subgroup different from group G and identity E is called non

trivial subgroup. One easy way in which subgroups may be found is by forming the

powers of an element. In general, the symbols An for integral „n‟ can be shown similarly

to be uniquely defined .If the group is finite, all these powers of A must belong to the

group , and to prevent their number becoming infinite, they must begin to repeat after

some point. In particular there is some integer „m‟ such that,

Am

= E (1.5)

Thus the elements A,A2,A

3….A

m =E (1.6)

They form a subgroup known as cyclic subgroup generated by A and A is its generators;

because of the uniqueness Am

.

A.Am-1

= Am-1

.A =Am

= E (1.7)

So that the inverse of Am

is Am-1

and similarly the inverse of Ar is A

m-r. Thus the group

axioms are completely satisfied.

3. 3. 1 Important property of Subgroup

The subgroups of a finite group have the following property-

The order of a subgroup is a divisor of the group. This means that if G be a group of

order N and H a subgroup of order h, then N is an integral multiple of h. To prove this


important property of subgroups, let H=(E,H2,H3,…Hh) be the subgroup of a group G,

which is of order N. Suppose „x‟s any element of G: then

XH = (XE,XH2,XH3….XHh) (1.8)

It is quite obvious that H may be or may not be an element in the subgroup H. If X is a

number of H, then

XH = H if XH (1.9)

While contrary to this if X is not an element of a subgroup H, then no element of the set

XH belongs to H .

3. 3. 2 The concept of Normal Sub-Group

If the left and right cosets i.e. XH and HX of a subgroup G with respect to X, provided

XG, are the same, then H is called a normal subgroup. Mathematically it can be

expressed as

XH = HX or, X-1.

X .H =H for all X G.

A normal subgroup consists of complete classes of the bigger group i.e. of complete

classes of group G, and then H is a normal subgroup of G i.e. HG or G H.

Properties of Normal subgroups

(a) If l be the identity in G then the whole group „G‟ and {l } are normal subgroup

of G.

(b) Every subgroup „H‟ of a commutative five group „G‟ is normal since a left

coset „xH‟ is the same as the right coset‟ Hx‟ since.

xH x-1

Hx = H a, xG.

(C) The alternating group An is an invariant subgroup of the symmetric group Sn

( d) The interaction of any two normal subgroup of a group is a normal subgroup.

3.3.3 Conjugate Elements and Classes

Let A,B and X be any elements of a group G then if there exists a relation such that

X-1

A X =B (1.10)

Group Theory

„B‟ is said to be the transform of „A‟ by the element „X‟, the operation is said to be a

simplicity transformation of A by X and the two elements „A‟ and „B‟ are said to each

other. The conjugate elements have the properties.

(i) Every element is conjugate with itself.

(ii) If „A‟ is conjugate with „B‟, then „B‟ is conjugate with A.

(iii) If two elements „A‟ and „B‟ are conjugate to a third element C, then they( A

and B) are also conjugate one another.

The elements of a group which are conjugate to one another form a class of the Group. If

the group contains the elements A1(E),A2,A3 … Ag, the class of A may be found by

forming the sequence.

E-1

A E = A,A2-1

A A2 ----Ag-1

A Ag . (1.11)

All the elements of this sequence are conjugate to A and to one another, more over every

element conjugate to A in the sequence. The elements of a group can, therefore be

divided in to classes and every element appears in one and only one class. Except for the

class which consists only to the identity element E, no class is a sub group. Each class

consists of just one element in an abelian group.

All the elements of a class have the same order.

In a group of matrices , all the matrices which belongs to the same class have the

same trace

Product of Classes : Let Ci= (A1,A2,A3,…An) and Cj = ( B1,B2…Bn) be two classes

( Same or distinct) of a group containing „m‟ and „n‟ elements respectively . Then their

product is defined as a set containing all the elements obtained by taking the products of

each element of Ci with every element of Cj, keeping each element as man times as it

occurs in the product. Thus,

CiCj =(A1B1,A1B2…A1Bk ….Am Bn ) (1.12)

It may be easily seen that the set Ci Cj consists of complete classes .It would be enough to

show that if an element AlBk belongs to the set CiCj , then any element conjugate to Al Bk

with respect to some element „X‟ of the whole group is..

X-1

(Al Bk) X = (X-1

Al X)( X-1

Bl X) = ArBs (say) (1.13)

Group Theory and Functions of Complex variable.

Where, by definition of a class, Ar belongs Ci and Bs belongs to Cj. Hence ArBs belongs

to the set Ci Cj .

Therefore aijk are non –negative integers giving the number of times the class Ck is

contained in the product Ci Cj and the sum extends over all the classes of the group.

3. 4 FACTOR GROUP

If H be a normal subgroup of a group G then the group of all cosets of H in G is known as

factor group or quotient group of G by H and denoted by G/H.

The cosets of H combine among themselves under the operation O, H is the identity

element in the factor group. The associative law for the elements in the factor Group is a

consequence of the associative law in G. The inverse of any coset aH in the factor group

is the coset containing the element a-1

.

Since the group E, aa2 is the self conjugate subgroup the factor group G/H consists of two

elements H and bH and the multiplication table of G/H is given below

3. 4.1 Properties of Factor Group

(a) The order of a factor group G/H is equal to the index of H in G.

(b) Each quotient group of an abelian group is abelian but its converse is not true,

since.

And if S3 be a symmetric group and A3 an alternating group each of degree three then

S3/A3 is an abelian group of degree „3‟ whereas S3 is not abelian . The group S3/A3 is of

order 2 and so it is abelian group as every group of order 2, is abelian.

H H bH

BH bH H

H bH

5. 5 COMPLEXES OF GROUP

A non empty subset „H‟ of a group G is called as a complex of the group G. Hence a

complex is set, from a group, considered as the whole. If the complex contains A,B,C,

i.e. = [A,B,C], then the complex A = [A2,AB,AC].

Group Theory

Properties of complexes

(1) If „z‟ be a complex containing the elements a, b, c of a group „G‟ then z

={a,b,c}.

(2) If z ={a,b,c} be a complex then a z ={ a2, ab, bc } etc.

(3) If z1 and z2 be two complexes of a group G, then the product of z1 and z2 is

defined as,

z1z2 = { x: x =z1z2, z1z1, z2z2}

(4) The sub group „H‟ of a group „G‟ also gives a complex i.e. HH =H2=H

(5) A group can be expressed as a sum of complexes.

(6) The number of complexes in a group is equal to the index of a sub group „H‟ in

G and infact it is the order of the group divided by the order of the subgroup

„H‟.

(7) The product of complexes is associative.

3. 6 ISOMORPHISM AND HOMOMORPHISM GROUP REPRESENTATION

The group can be represented in two categories as Isomorphism and Homorphism which

may be defined as,

3. 6. 1 Isomorphism

Two groups G={E,A,B,C} and G = {E, A,B,C…} both of same order g, are said to be

isomorphism if there exists a unique one to one correspondence between their elements in

such a way that products correspond to products.

For example, if one to one correspondence is denoted by A A, B B,

C Cetc. then the product such as say, AB =C in the group G implies that AB= C,

Thus two isomorphic groups have similar group multiplication tables of the same

structure. The group multiplication table of G can be obtained from that of G simply by

replacing the elements of G by the corresponding elements of G. Hence two isomorphic

groups are essentially identical; the individual elements are merely labeled differently.

Clearly every group is isomorphic itself.

Mathematically, an isomorphic between two groups G1and G2 can be explained as

follows.


If G1 = (E1, A1,B1,C1…) and G2 = (E2,A2,B2,C2…) (1.14)

Both are of the same order N and if there exists unique correspondence between elements

G1 and G2 i.e., A1 corresponds to A2, B1corresponds to B2, C1 corresponds to C2 and so

on; then the product of two elements, say A1,B1=C1 in the group G1means that A2B2 =C2

in group G2.Thus the multiplication table of G2 can simply obtained from that of G1 by

the corresponding elements of G2.

3.6.2 Homomorphism

Between two groups resembles isomorphism except that the correspondence is not

required to be one to one but many to one. The idea of homomorphism in its complete

generality may be developed as follows. The group „G‟ is homomorphic on to group „H‟

if one and only one element of „H‟ correspond to every element of H and if the

correspondence is such that if AB=C in the group „H‟, then the product of Ai and Bj (1i,

j n) of G belongs to the set( Ci = C1,C2,…Cn) of G i.e. AiBj = Ck, 1 k n, for any „i‟

and „j‟ and some k.

Accordingly, Homomorphism is not a reciprocal property .If the group „H‟, and then we

have the following important results.

(a) The identity E1 of G corresponds to identity E of H.

(b) The set (Ei) is a normal subgroup of „G‟

(c) The group „H‟ is isomorphic to the factor group of the normal subgroup (Ei) of G.

Representation of Group

The representation of an abstract group generally refers to any group composed of

concrete mathematical entities which is homomorphic to the original group.

The simplest representation of group is obtained by associating unity with each element

of the group and the representation is then called the identity representation, this

representation provides the trivial homomorphism of any group on to the group which

contains only the identity and hence is unfaithful representation.

Consider a group G = { E,A,B,C…}of finite order „g‟ with E as the identity element. Let

= { T(E),T(A),T(B),T(C)…} be set of square matrices all of the same order having the

property .

T(A) T(B) =T(AB) Group Theory

i.e., if AB=C, in the group G, then

T(A)T(B) =T(C)

Such a set of the matrices is said to be representation of the group G and the number of

rows and column of the matrices of the set T is called the dimension of the

representation. Another representation of the some group of three elements may be

obtained by taking the determinant of each other each matrix, because of the fact that,

T(A)T(B)= T(A) T(B)=T(AB)

This operation reduces the matrices to ordinary number 1.Thus representation consists

of any two distinct matrices for six group elements and hence is unfaithful representation

and when two groups G and T are isomorphic to each other then such a representation is

called true or faithful representation.

3.7 REDUCIBLE AND IRREDUCIBLE REPRESENTATIONS

The reducible representations may be defined as the representations which may be

expressed in terms of representation of lower dimensionality by means of similarity

transformation. The representations for which this can not be done are said to be

irreducible. The super –matrix representations of the form indicated as,

(1.15a)

T(1)

E 0

0 T(2)

(E)

T(1)

A 0

0 T(2)

(A)

T(1)

B 0

0 T(2)

B

And those which may be brought to form (1.15a) by the similarity transformation are said

to be reducible. In above example, we would write T=T(1)

+T(2)

or more generally T =

ai T(i)

. (The notation term does not indicate the summation). The irreducible

representations have significant role in quantum mechanics, since each irreducible

representation displays the properties of a set of degenerate Eigen functions and hence

the numbers of irreducible representations gives the number of distinct energy levels.


3.7.1 Theorem on Representation

Theorem 1- Any representation by matrices with non vanishing determinant is

equivalent to a representation by unity matrices through a similarity transformation. The

physical interpretation of the similarity transformation of the representation T is that a

transformation from an oblique system of coordinates axis to an orthogonal one .This

implies that the basis vectors in now representation are orthogonal .By the use of this

theorem we may consider the representation by unitary matrices only without loss of

generality.

Theorem 2 - Schur‟s Lemma

Any matrix commute with all matrices of an irreducible representation must be a

constant matrix, i.e. ( a multiple of E).

Proof -

By the use of Ist theorem, we may restrict ourselves to unitary representations. Let „M‟ be

a matrix which commutes with all matrices of representation (Ai), i.e.

M (Ai) = (Ai)M, (I=1,2,3…h) (1.15b)

Taking adjoint of both sides, we get

M (Ai) = (Ai) M (1.16)

Pre and Post multiplying of equation (1.16) by (Ai), we get

(Ai)M (Ai) (Ai) = (Ai) (Ai) M (Ai)

i.e. (Ai)M = M (Ai) (1.17)

Thus if „M‟ commutes, then M also commutes with all matrices of the representation

(Ai).

Now we may define two Hermitian matrices H1 and H2 given by

H1=M+M and H2 = i(M- M ) (1.18)

Which commute with all (Ai). From equation (1.18) we have

M=H2 –iH2 (1.19)

Group Theory

Now we have to show that a commuting hermitian matrix is constant, then it follows that

„M‟ given by equation (1.19) is also a constant. Confining the attention to hermitian

commuting matrices we can always reduce them to diagonal form by unitary

transformation Md = U-1

MU.If (Al) is the transformed matrices of (Ai), i.e.

(Ai) = U-1(Ai) Ui , then

(Al) Md = Md (Ai) (1.20)

(Since matrix equations remains invariance now it remains to show that „Md‟ is not only

diagonal but a constant too. For this consider „‟ element of (1.20), i.e.

(Ai) (Md) = (Md) - (Ai)

or, (Ai)[(Md) -(Md) ] =0 for I=1,2,3…. (1.21)

Now, if (Md) (Md) , i.e. matrix is not constant, then (Ai) =0, for all Ai .This

means that the unitary transformation „‟ has brought (Ai) to block form, indicating

that, the representation was in fact, reducible. On the other hand, if we assume that

representation was irreducible, then it follows that (Md) = ( Md) , i.e. any

commutating matrix must be a constant matrix in the case of an irreducible

representation. This completes the proof. The importance of this theorem lies in fact that

its converse is also true .Thus, if we can show that if a non-constant matrix which

commute with all matrices of a representation exists, then the representation is reducible,

whereas if non exist or a constant commuting matrix exists, the representation is

irreducible.

Main features of Reducible and irreducible Representations

(i) The number of non-equivalent irreducible representation is the same as

the number of classes .

(ii) If there are „h‟ elements in a group then the number of times the jth

irreducible

representation is given by aj = XXj/h.

(iii) All the irreducible representations of an abelian group are one group are one

dimensional. Group Theory and Functions of Complex variable

(iv) A representation by nonsingular matrices can be transformed in to a

representation by unitary matrices through a similarity transformation.

(v) Any matrix commuting with all the matrices of an irreducible representation is a

constant matrix, i.e. a unit matrix multiplied by a constant scalar.

(vi) When a matrix „A” commutes with every matrix of a given representation of a

group, then either „A‟ is a scalar matrix or the representation is reducible & the

transformation used to diagonalize A, wholly or partially reduces the

representation.

(vii) The direct product of irreducible representations of two different group is also an

irreducible representation of the direct product of the group.

3.7. 2 Orthogonality theorem for an irreducible representation

Let (E), (A2), (A3)…. (Ah) and (E), (A2), (A3)…. (Ah) between two

nonequivalent irreducible representations of the same group G, then

(1.22)

Holds for all elements j, k, and pa, where the summation extends over all group elements

E, A2, A3…Ah , E being identity element.

Proof

Assuming the representation in unitary form as a similarity transformation always leads

multiples of the unit matrix unchanged, a matrix „M‟ commutes all the matrices (E=A2),

A2…Ah of the representation of the order „h‟, i.e.

Ai M = M Ai, i=1,2,3…h (1.23)

(R)j k (R)p q =0

But if there are two irreducible representations (A1), (A2)… (Ah) and (A1),

(A2)… (Ah) of dimensions d1and d2 respectively, then

M (Ai) =(Ai)M, i=1,2,…h , (1.24)

where M is the null matrix whereas for d1= d2; M is either a null matrix or a non –

singular matrix. Here equation (1.22) asserts that a matrix which satisfy equation (1.24),

must be a null matrix and one which satisfies (1.23) must be multiple of the identity

matrix. On account of group property of the representation all matrices of the form,

Group Theory

(1.25)

for arbitrary l rowed and m-columned matrix „x‟ satisfy (1.24). Also the group property

follows that ;

, ; since the same matrix appear on the left

and right except in different order.

Hence;

or, ( in concise form) (1.26)

so that by feature (viii) M must be a null matrix, i.e. for arbitrary Xir

; (1.27)

while on setting all matrix elements Xir =0, except one number Xqk=1 , the generalized

from (1.24) is

; where (R) and (R) must be irreducible, but not necessarily

unitary. In case (R-1

) c j(R) are unitary (R-1

) = [(R)]-1

and hence (R-1

) = (R) , so

that (M) reduces to unitary representation.

Geometrical Interpretation

The orthogonality theorem has a very elegant interpretation in the element group space.

This space may be considered as n-dimensional vector space in which the axes or

components are labeled by various group elements Ai =A1= E,A2,A3…An. Themselves

are labeled by the three indices .The representation index „i‟ and the subscript,

M= (R) x (R-1

)

(SR) x D{S(R-1

)} = (R) x D{S(R-1

)} = M

(S)M = (S) (R)x D(R-1

) = (SR)x (SR-1

) (S)

(S) M = M(S)

R

R

R R

R

M pj = (R)pi Xi r (R-1) r j R i r

(R)p q ( R) k j

indicating rows and columns with in the representation matrix. The theorem then states

that all these vectors are mutually orthogonal in this n-dimensional space. From this

result we may draw an important conclusion, since each irreducible representation of

finite group of dimensionality li provides us with l2i provides us with li

2 vector (Ap)

Where , =1,2,…li), the total number of orthogonal vectors is li2, where „i‟ run over

all the distinct irredecucible representation. If all the distinct irreducible representations

of the same group are considered, then equality sign holds, i.e. li2 =n

This equation is called the dimensionality theorem.


3. 8 CHARACTER OF REPRESENTATION AND CHRACTER TABLE

It is troublesome to find all the irreducible representation of a given group; since they are

not unique, but arbitrary. In most physical applications it is sufficient to know the trace of

the matrices of representation, a quantity called the „character‟ in group theory. Let be

the representation (reducible or irreducible) of a group G. Then the character of

representation is the set of the traces of all the matrices of representation, i.e.

(A p)= (A p), A p being a member of group (1.28)




i) Define a group and its type?

ii) What you mean by factor group? Write its main properties.

iii) Define the Isomorphism and homomorphism group representations?

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G {E=A1, A2….Ah}, i.e. character of representation, is the set of the traces of all the

matrices of representation;, is the set of „h‟ number { (E), (A2)… (Ah)}. The

characters of conjugate elements in a representation under a similarly transformation.

Thus, if A p and A q are the conjugate elements, then there exists an elements A r such

that

A p = Ar-1

Aq A r or (A p) = (Ar-1

) (A q) (A r)

Taking the trace of both sides; we get Group Theory

Trace { (A p)} = Trace { (A b)} or (A p) = (A q) (1.29)

since the matrix representations of all the elements in the same class are related by

similarity transformations (by definition of conjugate group elements); the invariance of

traces of traces show that all elements in the same class have the same character. This

enables us to specify the character of any given representation by simply giving the trace

of one matrix from each class of group elements. The character of Kth

class of

representation is denoted by (CK).

Orthogonality theorem : We can immediately transform for the character of irreducible

representations of a group. Setting = and = in orthogonality theorem (1.26) of

section above as stated that

(1.30)

We have,

(1.31)

Now summing over and and using definition given in (1.28) of character of

representation; we get

(1.32)

Thus the characters from a set of orthogonal vectors in group-element space .Now,

collecting the group elements according to classes ; then if N k is the number of elements

in the class Ck of the group, then equation (1.32 ) reduces to ,

(1.33)

(i) (A p)

( j )(A p) = ( h/l1 )ij

p

p

(i) (A p)

( i )(A p) = ( h/l1 )ij

( j )

(Ck ) ( l )

(A p) = (h/ l1)ij = (h/ l1)ij x li = hij

( j )

(Ck ) ( l )

(Ck).Nk = (h )ij

Here the sum over „k‟ runs for all classes .

Equation (1.31) may be expressed as

(1.34)

This equation indicates that the characters of various irreducible representations also

form an orthogonal vector system in the class –space where axes are labeled by classes

Ck rather then group elements Ap . Since the number mutually orthogonal vectors in a

space can not exceed its dimensionality hence the number of irreducible representations


can not exceed the number of classes. In fact they are always equal i.e. the number of

irreducible representation = number of classes.

Character Table

It is convenient to represent the characters of various representations in a character table

for any given group .The columns are labeled various classes, produced by the number of

Nk of elements in the class. The rows are labeled by the reducible representations and the

entries in the table are X(i)

(Ck). Thus, the character table for any group may be expressed

as,

Class

Irreducible

representation

N1C1 N2C2 N3C3……. ….. NKCK

(1) X1(C1) X

1(C2) X

1(C3) ….. X

1(Ck)

(2) X2(C1) X

2(C2) X

2(C3) …… X

2(Ck)

…… ….. ….. ….. …… …..

(k) (xi)(C1) (x

i)(C2) (x

i)(C3) …… (x

i)(Ck)

Main features of character

1. The character of the direct product is the product of the character, i.e. x(i)

(AB) =

x( i )

(A).x( j)

2. The characters form an orthogonal system.

3. If the representations (A) and (B) are the first degree then the direct product

(AB) is irreducible .In case both degree are higher then one, (AB) is reducible.

{(Nk/h)}(i)

().{(Nk/h)}(j)

(Ck ) = ij

4. Two irreducible representations are equivalent if and only if they have the same

character.

3. 9 DIRECT PRODUCT OF GROUP

If G1 and G2 are two groups with the same composition rule and if the elements of G1

commute with the elements of G2 and there being no common element between G1 and

G2 except the identity elements E, then the collection of all the elements formed by

multiplying the elements of G4 by the elements of G2 form a group G, which is called the

direct product of G1 & G2. The direct product of G may be symbolically written as,

Group Theory

G= G1xG2

Thus the direct product of two groups,

G1= (E1, X2, X3,X4 ….XN1…) of order N1

& G2 = (E1, Y2, Y3,Y4 ….YN1…) of order N2

is defined as group G of order N=N1N2 consisting of elements obtained by taking the

products of each element of G1 with every elements of G2 and is written as,

G =G1xG2 = ( E,EY2,EY3…EYN, X2Y2…X2YN….YN ) (1.35)

The concepts of direct product of group can be used in enlarging a group, further more, it

is also used in study of symmetry of physical systems such as atoms and elementary

particles.

N




i) Define a character in group representation and write its main properties?

ii) What you mean by direct product of a group?

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3.10 APPLICATIONS OF GROUP THEORY IN PHYSICS

[1] The Unitary Group- The collection of all non-singular matrices of order „n‟ with

matrix multiplication of the law of combination is called full linear group (FLG). The

order of this group is infinite since its elements are the infinite number of linear

transformations that change a given vector in to a new vector. If we impose certain


restrictions on the matrices of its transformation, we may obtain sub groups of the full

linear Group. A sub group of F.L.G. whose elements are square unitary matrices of order

„n‟ are called n-dimensional unitary group. Further, if the determinant of unitary group is

+1, the subgroup is called unitary special group (U.S.G.) of order „n‟ and is denoted by

SU(n). The special unitary group may be of any order such as SU(2), SU(3) etc.

Elementary particle symmetrical : The classification of elementary particles has met

with success in predicting new particles. The method developed for the arrangement is

based on group theory. It has been found that, in general the conservation law represents

an invariance which corresponds to an appropriate symmetry operation. The set of

operators that represents the symmetry constitutes the groups from which the theory gets

its name The irreducible representation of a group consists of a number of states,

quantities or objects to which the symmetry operations are applicable. Thus the

appropriate group operation can transform any one of these states in to other in the same

representation. The fundamental representation is the one containing the smallest number

of states for the particular group. If the system is invariant with respect to displacement in

space linear momentum is conserved, if it is w.r.t. time energy is conserved

The simplest unitary group U(1)contains transformation which add a phase factor only to

particle wave functions .The invariance under such transformations gives conservations

of Charge Q, Baryon B, lepton L and hypercharge Y.

Unitary Symmetry [SU(2) symmetry]

The particular symmetry group applicable to isospin conservation is a form of unitary

symmetry known as a U(2), which can be expressed by asset of 2x2 matrices. This group

may be reduced to a special unitary group SU(2), which is also written as SU2. It is

special because a restriction reduces by unity the number of operators in the group. The

two dimensions refer to basic states which make up fundamental representations in this

case . The restriction of special reduces the number of operators 2x2 =4 to three. The

group is then said to have to have three generators. By the use of algebra of SU(2) group

it can be shown that all irreducible representations of the symmetry consists of a

multiplets of 2T+1 states . All the members of multiplets have the same isospin T and are

Group Theory

essentially identical except for charge. If the symmetry was exact i.e. isospin is strictly

conserved, the components of multiplets would differ in charge and T3 . The SU(2)

symmetry is violated by the electromagnetic interaction for which conservation of

isospin is not applicable. The nucleon states p >, n>have anti nucleon state p >, n>

omitting> brackets for clarity and separating , trackless part, the combination of

nucleons with an anti nucleon may be represented as,

The first terms of r.h.s represents singlet ( -meason, T=0, I =0- ) and the second term

represents the triplet array ( pions T=1,I =0- ) The second term can be written as,

(1.36)

Eight fold way [SU (3) symmetry

The SU(3) theory is a generalization of the theory of isospin. This stands for special

unitary group in three dimensions. The term, three dimensions refers to the three basic

state which make up the fundamental representation in this case. In the three dimension

unitary group there are, in general 3x3=9 operators, but the restriction of “special”

reduces the number of eight.

p

n ( p n ) x

( p p + n n )

2 1 0

0 1 +

( pp – n n )1/2

pn

np ( pp – n n )1/2

0/2

+

-

0/2

Application of group algebra showed that the SU(3) symmetry should give rise to six

supermultiplets ,containing 1,8,8,10,10,27 members .The 10 multiplet is equivalent to the

10 but with hypercharge of opposite signs. In each of theses super multiplets the parity

and intrinsic spin of members are the same, while hypercharge and isotonic spin are not

same. Among above mentioned groups, 8 and 10 member groups are of particular

interest. IN case B=0, we may form particle anti particle states to fill a 3x3 array.


It can be identified with known zero spin zero spin .

(1.37)

The neutral meason is now written as [ = (pp + nn + 2 )/6]

There is in addition the symmetrical neutral combination or singlet

= (pp + nn + 2 )/3 (1.38)

Since these measons are formed from fermions particles –antiparticle pairs, hence have

odd parity. These eight particles with B=0 and Ip =0 – should be arranged.

Other application of group theory in nuclear physics –

(a) Charge conjugation

(b) Parity (Space –inversion invariance)

(c) Time reversal

(d) Combined inversion of charge and parity

(e) Combined inversion of charge, parity and time.

[2] Theory of point group

A large class of groups which one important in Physics and Chemistry are the symmetry

groups. The symmetry of a body is described by giving the set of all those

p

n

x p n

1/3 (2pp – nn - ) pn p

np 1/3 (-pp +2 np + ) n

p p 1/3 (-pp – nn + 2)

(0/2) +(/6)

+ K

+

- (-

0/2) +(/6) K

0

K- K

0 (- 2/6)

transformations which preserve the distances between all pairs of points of the body and

bring the body in to coincidence with it self . The distance preserving transformation can

all be built up from three fundamental operations-

(1) Rotation through a definite in a plane.

(2) Mirror reflection in a plane.

(3) Parallel displacement (Translation).

Group Theory

The least symmetry operation can occur if the body is of infinite extent ( i.e. infinite

crystal lattice).For a body of finite extension can exist first two symmetry operations . In

fact, all transformations of a symmetry group of a symmetry group of a finite body must

leave at least one point of the body fixed .In other words, all axes of rotation and all

planes of reflection must intersect must intersect at least one point, since successive

rotations about non intersecting axes of reflections in non-intersecting planes will result

in the introduction of translation and continual shift of the body. The symmetry group of

finite bodies, which leaves at least one point of the body fixed, are called point groups.

While the symmetry groups which include the translation symmetry are called space

groups. The examples of point groups are two dimensional rotation group R+(2), three

dimensional rotational group R+ (3) and the symmetry of a square C4v . The crystal

symmetry point groups are called the crystallographic point groups and together 32 in

number.

[3] The electronic structure of solids

The solution of Schroedinger equation gives the energies and the wave functions of an

electron in a crystal,

H (r) = (1.39)

Where V (r) is the potential experienced by the electron in the periodic lattice. The full

symmetry group of the crystal –Hamiltonian is the space group to which the crystal

h2

2

2m + V ( r) (r) = E(r)

belongs. The solution of this appears in the form of Bloch function k (r) with the

corresponding energy Eigen values E (k). The functional relation E = E ( k) is known as

electronic structure of the crystal . The knowledge of E (k) enables us to determine a

number of observable properties of the crystal, such as, in particular, the transport

properties and the optical properties. From the study of electronic structure we can infer

whether the crystal is a metal, a semiconductor, semimetal or a insulator. Since we are

concerned only with the group theoretical aspect of the electronic structure, the exact

analytical form of the potential V (r) =0 i.e., the case of a constant potential. This is the

elementary quantum mechanical problem of electrons in a box and the solution of the

equation above is just the plane waves,


With energies E (k) = h2k

2/2m

This is known as free electron approximation and this gives a considerable insight in to

the problem of the electronic band structure of crystals.

[ 4 ] The symmetry of the Reciprocal lattice

Let A be an any element of the point group of the direct lattice and G be the reciprocal

lattice vector for any transnational vector then from equation,

G. t = 2 (hn1+ kn2 + ln3) = 2 x integer

We have,

G. At = 2 x integer, for all„t‟ (1.40)

Since the operations of the point group are orthogonal transformations, the scalar product

of two vectors remains invariant under their action

i.e. A-1G.A-

1( At) =G. At.

Thus the equation G. At = 2 x integer, for all„t‟, gives (1.41)

A-1G. (t) = 2 x integer, for all„t‟

Since this is true for all lattice vectors„t‟ shows that A-1G must be of the form q1b1+

q2b2 + q3b3 with the integral qi . In other words A-1

G must be a reciprocal lattice vector

.As this holds for any element A of the point group, it is evident that the reciprocal lattice

has the same point group symmetry as the direct lattice.

( r) = eikr




i) What are the importances of group theory in applied and theoretical Physics?

…………………………………………………………………………………

…………………………………………………………………………………

…………………………………………………………………………………


Group Theory

3. 11 LET US SUM UP

Group – A group a set of distinct elements G = {E, A, B, C, D,…..} finite or infinite in

number, endowed with the laws of compositions (Such as addition, multiplication, matrix

multiplication) such that the properties of (i) Closure (ii) Associative (iii) Identity (iv)

Inverse.

Finite and infinite group – The number of elements in a group is known as its order. A

group containing finite number of elements is called a finite group and a group

containing an infinite number of elements is known as infinite group. A group which

consists of a single element A and its power A2,….A

P =E, is called a cyclic group. If the

multiplication is commutative i.e. for every pair of element A and B we A.B =B.A…A.

group is said to be commutative or Abelain group.

Sub Group –When a number of elements selected from a group do themselves form, it is

known as a sub group.

Factor group - If H be a normal subgroup of a group G then the group of all cosets

of H in g is known as factor group or quotient group of G by H and denoted by G/H.

Complexes of a group – An nonempty subset „H‟ of a group G is called as a complex of

the group G. Hence a complex is set, from a group, considered as a whole.

Isomorphism group – Two groups C= {E, A,B,C} and G= { E,A,B,C} both are of

same order g, are said to be Isomorphism group if there exists a unique one to one

correspondence between their elements in such away that products correspond to

products.


Homomorphism group – Between the two groups resembles isomorphism except that

the correspondence is not required to be one to one but many to one is called

Homomorphism group.

Reducible and irreducible representations – The reducible representations may be

defined as the representations which may be expressed in terms of lower dimensionally

by means of similarity transformation. The representations for which this can not be done

are said to be irreducible.

Schur‟s lemma – Any matrix commute with all matrices of an irreducible representation

must be a constant matrix.

Applications of group theory in Physics –The group theory is used in several branches

in applied and theoretical physics mainly in quantum and nuclear physics


1. i) Group : A group a set of distinct elements G = {E, A, B, C, D,…..} finite or

infinite in number, endowed with the laws of compositions (Such as addition,

multiplication, matrix multiplication) such that the properties of (i) Closure (ii)

Associative (iii) Identity (iv) Inverse.

Types of Group (i) Finite group (ii) Infinite group (iii) Cyclic group (iv)

commutative or abelian group.

ii) Factor group : If H be a normal subgroup of a group G then the group of

all cosets of H in g is known as factor group or quotient group of G by H and

denoted by G/H.

Properties of Factor Group:

The order of a factor group G/H is equal to the index of H in G.

Each quotient group of an abelian group is abelian but its converse is not

true, since.

and if S3 be a symmetric group and A3 an alternating group each of degree three

then S3/A3 is an abelian group of degree „3‟ whereas S3 is not abelian . The group

S3/A3 is of order 2 and so it is abelian group as every group of order 2, is abelian.

Group Theory

iii) Isomorphism group : Two groups C= {E, A,B,C} and G= { E,A,B,C} both

are of same order g, are said to be Isomorphism group if there exists a unique one

to one correspondence between their elements in such away that products

correspond to products.

Homomorphism group : Between the two groups resembles isomorphism except

that the correspondence is not required to be one to one but many to one is called

Homomorphism group.

2. Character in group representation: In most physical applications it is sufficient to

know the trace of the matrices of representation, a quantity called the „character‟

in group theory. Let be the representation (reducible or irreducible) of a group

G. Then the character of representation is the set of the traces of all the matrices

of representation, i.e.

(A p)= (A p), A p being a member of group

G {E=A1, A2….Ah}, i.e. character of representation, is the set of the traces of

all the matrices of representation;,

Main features of character

The character of the direct product is the product of the character, i.e. x(i)

(AB)

= x( i )

(A).x( j)

The characters form an orthogonal system.

If the representations (A) and (B) are the first degree then the direct

product (AB) is irreducible .In case both degree are higher then one, (AB)

is reducible.

Two irreducible representations are equivalent if and only if they have the

same character.

Direct Product of Group - If G1 and G2 are two groups with the same

composition rule and if the elements of G1 commute with the elements of G2 and

there being no common element between G1 and G2 except the identity elements

E, then the collection of all the elements formed by multiplying the elements of

G1 by the elements of G2 form a group G, which is called the direct product of G1

& G2. The direct product of G may be symbolically written as, G= G1xG2 .


3. i) Application of group theory in nuclear physics –

Charge conjugation

Parity (Space –inversion invariance)

Time reversal

Combined inversion of charge and parity

Combined inversion of charge, parity and time.

Application of group theory in solid state physics-

Theory of point group

The electronic structure of solids

The symmetry of the Reciprocal lattice

UNIT 4 FUNCTIONS OF COMPLEX VARAIBALE

Structure

4.1 Introduction

4.1 Objective

4.2 definition

4.3 Argand diagram

4.4 Function of complex variable

4.5 Analytic function

4.5.1 Condition for analyticity

4.5.2 Cauchy Riemann equation in Polar form

4.6 Complexes Integration

4.7 Cauchy‟s theorem

4.8 Cauchy‟s integral formula

4.8.1 Derivatives of an analytic function.

4.9 Residue & contour integration

4.9.1 Cauchy‟s Residue theorem.

4.9.2 Computational of residue.

4.10 Application of complex analysis.

4.11 Problem

4.12 Let us Sum up


4. 0 INTRODUCTION

Cantor, Dedikind and Weierstrass etc. extended the conception of rational

numbers to a large field known as real numbers which constitute rational as well as

irrational numbers. Evidently the system of real numbers is not sufficient for all

mathematical. It was therefore felt necessary by Euler Gauss, Hamilton, Cauchy, Reiman

and weierstrass etc. to extend the field of real numbers to the still larger field of complex

numbers. Euler for the first time introduced the symbol „i‟ with the property i2 =-1 and

then Gauss introduced a number of the form ( + i) which satisfies every algebraic

equation with real coefficients. Such a number ( + i) with i = -1, and and being

real is known as complex number.


4. 1 OBJECTIVES

After completing this unit we are able to-

Define the complex variable and use the Argand diagram of complex function.

Derive the Cauchy – Riemann condition of complex variables to be analytic

State and derive the Cauchy‟s theorem

Sate and derive the Cauchy‟s integral formula and prove sits theorem.

Find out the poles of function.

Sate and prove the residue theorem

Calculate the integration problems by contour integration.

4. 2 DEFINITION

Complex number

An ordered pair of real numbers such as(x, y) is termed as a complex number. If we write

z = (x, y) or x + i y, where i = (-1),then „x‟ is called the real part and „y‟ is the

imaginary part of the complex number „z‟ & denoted by,

x = Rz or

R(z) or Re(z )

y = Iz or I(z) or Im (z)

Equality of complex numbers

Two complex numbers (x,y) & (x,y) are equal if x=x & y=y.

Modulus of complex number

If z=x+iy be a complex number then its modulus ( or module) is denoted by z

and given by z = x+iy= + (x2+ y

2)1/2

.

Evidently z=0,if x=0, y=0.

4. 3 ARGAND DIAGRAM

The plane whose points are represented by complex numbers is known as Argand plane

or argand diagram or complex plane or Gaussian plane and complex number z

representing the point (x, y) is some times called as affix of point (x, y). Consider a point

Functions of Complex variable

„P‟ in x y plane. Let an ordered pair of values of x, y corresponds to the coordinates of

the point „P‟ then a complex number z may be made to correspond to the point P where

P= x + iy, here z is called the complex coordinate of the point P . In the adjoining fig 2.1.;

the x-axis is called the real axis and y-axis is called the imaginary axis or axis of

imaginaries.

If (r, ) be the polar coordinates of the point „P‟ the polar form of the complex number „z‟

is

Z= r ( cos + i sin ) = r ei

; where r= absolute value of complex z and Angle or

argument of „z‟ & usually written as arg z.

Application of Argand diagram

Sum, difference, product and quotient of

Complex numbers can be geometrically

Represented on the argand plane as follows

Taking z2 and z2 two complex numbers

represented by the points P and Q on Argand

Plane,

Fig.No.2.1

Z = (x2+y

2)1/2

Y( Imaginary

Axis)

X ( real axis)

o

P (r,)

x M

y r

o

Fig.No.2.2

Sum OP =z2,OQ =z2 and OR = OP + OQ= z2+ z2

(2.1)

( Sum)

Fig.No.2.3


Difference OP =z2,OQ =z2 and OR = OP - OQ= z2- z2

(2.2)

(Difference)

Fig.No.2.4

Product

The modulus of product of two complex numbers is equal to the product of their moduli

and argument of the product of two complex numbers is the sum of their arguments.

Geometrically represented on an argand plane the product of „n‟ complex quantities

z1,z2,z3….zn as shown in fig. When z1=z2 =…zn =z (say), the above results may be

summarized as follows,

o x

y

P

Q R

z2

z1

z1+z2

o x

y

R

Q P

z2

z1-z2

z1

x

y

z1

z2 zn

z1.z2.z3….zn

1 2

n

1+ 2,+3, --n

z2

z1

1

2

0 R

A x

y

z1

z2

P

Q

Fig.No.2.6

Fig.No.2.5

zn = r

n ( cos n + i sin n), under the assumptions

r1=r2 =…rn = r (say) and 1= 2 = 3……n (say).

i.e. zn = r

n = z

n and amplitude z

n = n = n ( amp z) (2.3)

Also if r=1, we get the DeMoivre,s theorem for positive integral exponents such as

zn =(cos n + i sin n)

Quotient

(2.4)


OR = ;

which shows that the radius

Vector of the point R is

4.4 FUNCTIONS OF A COMPLEX VARIABLE

If w = u + iv and z=x + i y are two complex numbers, then w is said to be the function of

z and written as w = f (z), if to every value of „z‟ in a certain domain D, there correspond

one or more values of w.

If „w‟ takes only one value for each value of „z‟ in the domain D, then „w‟ is said

to be uniform or single valued function of „z‟ and if it takes more than one values

for some or all valued of „z‟ in the domain D, then „w‟ is known as many valued

or multi valued function of „z‟.

Since U and v both are functions of „x‟ , „y‟; then

W = f(z) = u(x, y)+ iv(x, y) (2.5)

It is however notable that the path of a complex variable „z‟ is either a straight line or a

curve.

Continuity

= z1

z2 OR

OA =

OP

OQ =

OP

OQ

z1

z2

z1

z2

The function f (z) of a complex variable „z‟ in continuous at the point „z0‟ if given a

positive number >0, a number can be so found that,

f (z) – f (z0)< , for all points „z‟ of the domain „D‟ satisfying z-z0<, where

depends up on , and also in general depends upon z0 i.e. = (, z0).

If is independent of z0 or rather say that if a number h() can be found independent of

z0 such that f (z) – f (z0)< holds for every ( pairs of point) z,z0 of the domain D for

which z-z0<h, then the function f(z) is called uniformly continuous in Domain D.

“ It should be noted that if a function f is continuous at z=z0 i.e. if f = u+ iv is continuous

at z=z0 then it will be so if its real and imaginary parts are separately continuous function

of „x‟ and „y‟ at the point (x, y) = (x0, y0)”.

The continuity of u and v for the function f to be continuous at z=z0 is the necessary

conditions. Conversely, if u(x, y) and v(x, y) are continuous then


u (x, y) u(x0,y0) and v (x, y) v(x0,y0) as z z0

so that f(z) =u(x ,y)+ iv (x ,y) u(x0,y0)+ iv (x0,y0) = f(z0). (2.6)

So the condition is sufficient also.

Differentiability

If f(z) be single –valued function defined in a domain D of the Argand diagram , then

f(z) is

said to be differentiable at z=z0 at a point D if tends to a unique limit when

z z0 provided that z is also appoint z is also a point of D.

A function f(z) is said to be differentiable at a point z=z0, if Lim

Exists and is a finite quantity provided by whatever path z z0 .The finite limit when

exists is denoted by f (z0) and termed as the differential coefficient or derivative of f(z) at

z=z0.; i.e.

Lim (2.7)

z z0 f (z0) =

f (z) – f (z0)

(z-z0)

z z0

f (z+z) – f(z)

z

f (z) – f (z0)

(z - z0)

4. 5 ANALYTIC (OR REGULAR, HPMOMORPHIC OR MONOGESIC

FUNCTION

“ A function f (z) which is single valued and differentiable at every point of a domain D,

is said to regular in the given domain D”. A function may be differentiable in a Domain

D have possibly for a finite number of points. Such points are called singularities or

singular points of f(z).

4. 5. 1 The necessary and sufficient conditions for f (z) to be regular

Necessary Condition

If w=f (z), where w=u+ iv and z=x+ iy . As such u and v both are the functions of „x‟

and „y‟ and therefore we can write, w = f (z) = u (x ,y) +iv(x, y).Now if f(z) = u(x, y) + iv

(x, y) is

differentiable at a given point „z‟ then the ratio must from the relation

z=x +i y and z =x + iy,


If we take z to be wholly real, so that y= 0, then

Lim

must exist and tend to a definite limit,

(say) (2.8)

i.e. partial derivatives ux, vy must exist at the point (x, y)and the limiting value is ux+ ivx .

Similarly again if z be taken wholly imaginary, so that x =0. We find that the partial

derivatives uy, vy must exist at the point (x, y) and the limiting value is vy - iuy . Since the

function is differentiable, the two limits so obtained must be identical,

i.e. ux + ivx = vy - i uy

Equating real and imaginary parts, we get

ux = vy and vx = uy

and (2.9)

x0

w

x

= Lim x0

u (x+x,y) –u(x,y) v (x+x,y) –v(x,y)

x x

+ i

dw u v

dx x x = + i = ux + ivx

u v

x y

= v u

x y

= -

f (z+z) – f(z)

z

These two relations, which are necessary conditions for a function to be analytic and are

called Cauchy Reimann differential Equations.

Sufficient Condition

The sufficient conditions for the function f(z) = u(x, y) + iv(x, y) to be analytic in the

same domain D are that the real functions u(x, y) and v(x, y) of realm variables „x‟ and

„y‟ should have continuous first order partial derivatives that satisfy the cauchy Reimann

equations in domain D. The Cauchy Reimann equations are fundamental because they are

not only necessary but are also sufficient for a function to be analytic.

4. 5. 2 Cauchy – Reimann equations in Polar Form

The Cauchy –Reimann equations in polar form can be obtained by putting x = r cos , y

= r sin , r2 = (x

2 +y

2 )1/ 2

as

And,

(2.10)

Here it is to be noted that a function „w‟ ceases to be analytic when (dz /dw) = 0,; z=0

and a function is regular if it is independent of z and is the function of z.


4. 6 COMPLEX INTEGRATION

Riemann‟s definition of integration

Consider a function w = f (z) which is continuous ( But not necessarily analytic ) along a

curve „C‟ with end points „A‟ and „B‟. Divide the arc „C‟ in to „n‟ axis by the points z0 =

2,z2, z2..z r -2…zr , zn = with z0 being at „A‟ and zn at B. Taking points 2, 2….r, n

such that n lies on the arc zr-1,zr ; then the definite integral of f(z) along the curve „C‟ is

given by,

(2.11)

Note -1

If f(z) = u+ iv, z=x+ iy = (t) +i(t), so that

d z = d x+ i dy = { (t) +i (t) }dt, t .

Then we have,

= ( u+ iv) { (t) + i (t) }dt

u 1 v

r r

= u 1 u

r r

=

zn =

zn =

B

C zr

zr-1

z2

z1

y

x 0

f (z) dz = f (z) dz = f () (z r – zr-1 ) C AB

Lim

n n=1

n

f (z) dz

C

= ( u+ iv)( dx+idy)

Fig.2.7

= ( u dx – v dy) + i (v dx + u dy)

Note-2- If „C‟ consists of a number of arcs Cr,

(2.12)

4. 7 CAUCHY‟S THEOREM

“If f (z) is a regular function of „z‟ and if f(z) is continuous at each point within and on a

closed contour C,

then, (2.13)


i. e. the integral of the function around a closed contour is zero.”

[Contour – It is continuous Jordan curve consisting of a chain of finite number of regular

arcs and a continuous arc having no multiple points is called a Jordan Arc.]

Proof

Greens theorem states that if P(x, y),Q(x, y), are all continuous function of

„x‟ and „y‟ in Domain „D‟. Then,

(2.14)

Let us assume that f(z) = u+ iv, where z=x+ iy and so that, dz = dx+ idy

Substituting these values in , we get

(2.15)

[ using the results of Green‟s theorem]

A

C

f (z) dz = f(z) dz Cr

C

f(z) dz =0

Q

x

P

y

C

f(z) d z

C

D

Q P

x y

dx dy (Pdx +Qdy) =

C

f(z) dz = C

(u+ iv)( dx+idy) = C

(u dx-vdy) +i(vdx +u dy)

dx dy D

v u

x y

+ i D

u v

x y

dx dy =

= 0. as and [ Cauchy‟s Reimann equation ]

Goursat‟s Proof (Rigorous Proof)

Goursat proved Cauchy‟s integral theorem with out assuming that f(z) is continuous.

This progress is quite important. To prove this theorem let us first consider two lemmas.

Lemma –I :- If „C‟ is a closed contour, then

dz =0 and zdz =0 (2.16)

It follows from the definition of integral that,

f(z) dz = (2.17)

Taking f(z) =1, we have


Considering „z‟ point is at the zr-1.

= +

(2.18)

Lemma –II

Proof of the theorem

Suppose that we enclose „C‟ in a large square , of area „A‟ and apply the process

of repeated quardisection . It is obvious that the meshes obtained by the

subdivision of the interior of „C‟ will be squares and others will not be squares.

Let C1,C2,C3…Cn.. be the complete squares & D1,D2,D3….Dn be the partial

squares, then ,

u v

y x

= u v

x y

=

C

C

C

Lim n (zr –zr-1 ) f (z) = 0

r =1

n

C

Lim

n

(zr – zr-1).z = Lim n

r =1

{ zr-1 ( zr – zr-1 )}

1

2 Lim n

r=1

n { zr-1 ( zr – zr-1 )} Lim

n { zr-1 ( zr – zr-1 )}

r=1

n

Lim n

r=1

n { (zr+zr-1) ( zr – zr-1 )} 1

2 =

Lim n

r=1

n

{ ( z2

r – z2

r-1 )} 1

2 = = 0

C

f (z)dz = f(z) dz Cm

Dn

+ f(z) dz +……

zdz =

n

(2.19)

Also we have from the Goursat‟s lemma

f(z) = f(z0) + f(z0)(z-z0) + (z-z0); when < (2.20)

Then,

Fig.2.8

, As dz = z dz =0 from lemma 1.

So, ,sides of square being lni and Max

z-z0 =2 ln , where 2 ln is the

length of the diagonal of squares.


2 ln ds , s being the entire perimeter of the square

2 ln .4ln 42 ln.An‟ l2n =An; the area of the square (2.21)

Similarly,

Sn being length of arc forming the curved boundary of Dn.

(2.22)

The curved boundary of Dn. [Where An is the arc of square Dn of side ln ]

Hence equation (2) gives,

(2.23)

Cm

Dn

Sn [f(z0) + f(z0)(z-z0) + (z-z0)] dz

Cm

f (z)dz = Cm

= f (z0)dz – f (z0) z0 dz + f (z0) zdz + (z-z0) dz Cm

Cm

Cm

Cm

= (z-z0) dz Cm

Cm Cm

Cm

f (z)dz z-z0dz 2 ln dz Cm

Cm

Cn

Dn f (z)dz z-z0dz 2 ln dz

Dn Dn

2 ln ds Dn

2 ln (4ln+ sn)

42 An+ 2sn ln

C

f (z)dz 2 {4(An+ An) + snln }

n=1

2 {(4A+ SL) }

„S‟ being perimeter of contour C, L is the length of a side of some square enclosing C and

a is the total area 0, S and L both being finite so that

4.8 CAUCHY‟S INTEGRAL FORMULA

If the function f(z) is regular with in on a closed contour „C‟ and if z0 be point with in „C‟

then ,

(2.24)

Where the integral is taken in the counter clock-wise sense.

Proof

Although f(z) is assumed to be analytic, the function . It is not analytic at

point z=z0. Hence Cauchy‟s integral theorem is not applicable on the contour „C‟ for the

function (z). Let us describe a small circle of C0 radius about z0


which lies entirely with in C. Now the function (z) is analytic in the region between C

and C0. By making a cross cut joining any point C0 to any point C we form a closed

contour C with in which (z) is analytic; so that the Cauchy‟s integral theorem applies,

i.e.;

, i.e. (2.25)

In transversing the contour C in the counter clock sense the crosscut is transverse twice;

once in each sense, hence it follows:-

(2.26)

where all integrals are taken in clock wise sense,

Hence

C

f (z)dz =0 , this proves the theorem.

f (z0) = 1

2i

f (z) dz

(z-z0)

(z) = f (z)

(z-z0)

(z) dz =0 C

C

f (z)

(z-z0) dz =0

C

f (z)

(z-z0) dz =

C

f (z)

(z-z0) dz -

C0

f (z)

(z-z0) dz =0

C

f (z)

(z-z0) dz =

C0

f (z)

(z-z0) dz

C0 C

z0 . r

o x

y

(2.27)

writing f(z) = f(z0) + [ f(z) – f(z0)] on right

hand and resembling that a common may be

taken out from under the integral sigh, we get

(2.28) Fig.2.9

Now on C0 ; z-z0 =rei

; dz = ir ei

d ; so that,

, for every positive „r‟

(2.29)

Since f(z) is analytic and therefore continuous at z=z0; hence for every positive real

number , however small, there exists a positive number such that,

; so that

(2.30)


But in the limit r= z-z0, and since the other two integrals in equation (2.28)

are independent of , in view of equation (2.28), hence the value of the integral (2.30)

must be equal to zero. Then equation (2.27) yields.

(2.31)

This proves Cauchy‟s integral formula.

C

f (z)

(z-z0) dz =

C0

f (z0)

(z-z0) dz -

C0

f (z)-f(z0 )

(z-z0) dz =0

C0

f (z)

(z-z0) dz =

C0

ir ei d

rei = i d =2i

f(z) –f(z0)<

C0

f (z)-f(z0 )

(z-z0) dz

C0

f (z)-f(z0 )

(z-z0) dz <

C0

ir ei d

rei = d = 2

C

f (z)

(z-z0) dz = f(z0).2 i + 0 f(z0) =

1

2i C

f (z) dz

(z-z0)




i) Define analytic function and write the required conditions for a function to

be analytic?

ii) State the Cauchy‟s theorem and its importance?

iii) State the Cauchy‟s integral formula?

…………………………………………………………………………………………

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…………………………………………………………………………………………

4.8.1 Derivatives of an analytic function

Ist derivative


IInd

derivative

Hence in general (n=1,2,3…)

(2.32)

4. 9 RESIDUE AND COUNTOUR INTEGRATION

Definition of Residue

The residue of a function f(z) at the pole z=a, is defined to be the coefficient of (z-a)-2

in

the Laurent‟s expansion of the function f(z), i.e.,

(2.33)

Where z=a is a pole of order „m‟

If z=a be the pole of order one, then the residue is

f (z0) = 1

2i C

f (z) dz

(z-z0)2

f (z0) = 1

2i C

f (z) dz

(z-z0)3

f n (z0) =

!n

2i C

f (z) dz

(z-z0)n+1

f (z) = an (z-a)n + bn (z-a)

-n

n=1

n=1

Lim (z-a) f(z) za

b1= (2.34)

i.e. In case of simple pole at z=a,

Residue = (2.35)

4.9.1 Cauchy‟s Residue theorem

If the function f(z) be single valued, continuous and regular with in and on a closed

contour C, except a finite number of poles ( Singularities ) with C; then,

(2.36)

Where R represents the sum of the residues say R2,R2…..Rn. of the f(z) at the poles 2,

2, 3 …. n with in a.

Proof

Let us draw a set of circles 2, 2, …..n with centers 2, 2, …n and , such that they do

not intersect each other and lie entirely with in the closed curve C.

Then f(z) is regular with in the region enclosed between C and the small 2, 2, …..n. The

entire region „C‟ may be deformed to consist of these small circles and the polygon „P‟.

Now by the Cauchy‟s theorem we have,


(2.37)

But the integral round the polygon „P‟ vanishes since f(z) is regular with in and on the

closed contour P, therefore.

(2.38)

Let us now consider z=a, a pole of order m, then by Laurent expansion.

Where, is regular

with in and on r and no pole.

Lim (z-a) f(z) za

f (z) dz = 2iR C

f(z) dz = f(z) dz + f(z) dz C P r

r=1

n

f(z) dz = f(z) dz C r

r=1

n

z =a

1

2

3

4

5

6

y

C

f(z) = an (z-a)n + [bs /(z-a)s] n=0

s=0

m

= (z) + [bs /(z-a)s]

s=0

m

(z) {= an (z-a)n }

So,

[Where , by Cauchy‟s

Fundamental theorem]

So, Fig.2.10

, Putting (z-a) = ei

, where a varies from 0 to

2 then, dz= ei

d . As the point „z‟ makes a circuit which consists of the circle r,

therefore

Now,

So,

, where b2 is called the residue for the function.


Let the residues for r=1,2,3…n, be respectively R1,R2,R3….Rn , then

………………..

………………..

Hence, (2.39)

This proves the theorem.

4.9.2 Computation of Residue

(i) Residue of f(z) for a simple pole z=a, i.e. a pole of order one

Residue of f(z) at z=a is = Lim (z-a) f(z) (2.40)

f(z) dz = (z) dz + bs ( z-a)s dz

r

r

r

(z) dz = 0 r

f(z) dz = bs r

r

dz

(z-a)s 1

m

f(z) dz = bsi r

ei d

sei s S=1

m

0

2

= bs (1-s)

. i e

i(1-s) . d

0

2

0

2

ei(1-s)

. d = 0 if s 1. But if s=1 all the terms will be zero except one.

r f(z) dz = b.i d = 2 i b1

0

2

f(z) dz = 2iR1 r1

f(z) dz = 2iR2 r2

f(z) dz = 2iRn rn

f(z) dz = =2iR C

f(z) dz r1

1

m

z a

(z)

(z)

If we put f(z) = where (z) = (z-a) f (z), Provided that F(a) 0, but (a) =0

then

for a simple pole, residue at z=a is

(2.41)

(ii) Residue at z=a, a pole of „m‟

Let f(z)be of the form,

The residue of f(z) at z=a, a pole of order „m‟ is = (2.42)

(iii) Laurent‟s expansion is,

Putting z-a =t, i.e. z=a+t, we have

(2.43)

Where „b1‟is the residue of f(z) at z=a, the pole of any order is the coefficient of (1/t).

Hence the coefficient of (1/t) is the residue of f(z) at z=a, a pole of any order, in the

expansion of f(z) after putting z=(a + t).


(iv) Residue of f (z) at infinity

The residue of any function f(z) at infinity z is Lim {-z f (z) }. (2.44)

(v) Residue of f(z) at infinity is the negative of the coefficient of (1/z) in the expansion of

f(z) for values of „z‟ in the neighbourhood of z=, i.e.

Residue of f(z) at infinity = -a1= The negative of the coefficient of (1/z) in

the expansion of f(z) for values of „z‟ in

the neighbourhood of z=.

(a)

(a)

f(z) = (z)

(z-a)m

m-1(a)

!(m-1)

f(z)= an (z-a)n +{ (bm /(z-a)

m}

n=0

m=1

f(z)= an (z-a)n +

n=0

b1

(z-a)

b2

(z-a)2

+ bm,

(z-a)m

+…..

f(a+t)= an tn +

n=0

b1

t b2

t2

+ bm,

t m

+…..

z




i) Define residue and state the Cauchy‟s residue theorem formula.?

ii) Compute the residue of a function f(z) at (i) a simple pole and (ii) at infinity

…………………………………………………………………………………………

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4.10 APPLICATIONS OF COMPLEX ANALYSIS

(A) Evaluation of , let f(z) be a function such that,

(i) f(z) is analytic throughout the upper half plane except at certain points which

are its poles.

(ii) f(z) has no poles on the real axis ,i.e., if R ( R being the radius of semi

circle), then it will cover entire upper half plane.

(iii) z f (z) uniformly as z for 0 arg z.

(iv) and both converge , then = 2iR+, where


R+ denotes the sum of the residues of f(z) at its poles in the upper half plane.

Then

(2.45)

Where both the integrals on R.H.S. individually converge. if is sufficiently large to

enclose all the poles of f(z) in the upper half plane, Then

Here zf(z) 0 as z,

So, and (2.46)

f(x) dx

-

f(x) dx

0

f(x) dx

-

0

f(x) dx

-

f(x) dx =

-

f(x) dx + 0

f(x) dx

-

o

C

f(x) dx + - R

a-

f(z) dz +

f(x) dx + a+

R

f(z) dz =2iR+

f(z) dz

0 f(z) dz

= -iR0

f(z) dz = 2 iR+

Where „R0‟ is the residue at z=a, the simple pole on real axis. Hence, when R , we

have ,

, (2.47)

R0 denotes the sum of residue of f(z) at its simple poles on the real axis, for each pole

on the real axis can be treated similarly. The left hand side of this equation is known as

principal value for „dz‟ from - to and denotes as „P‟ so that,

P

(2.48)

Where,

(B) Jordan‟s Lemma to evaluate Infinite integrals

If be a semicircle with center as given as origin and radius R and let f(z) be a function

such that,

(i) f(z) is analytic in the upper half plane

(ii) f(z)0, uniformly asz for 0 arg .


(iii) „m‟ is positive

then, . Let us assume that f(z)=, when „z‟ is sufficiently

large and 0 as z,

Also let z= Rei

and dz = iR ei

d and dz= iR ei

d.

,

(2.49)

on solving the above integral by imposing certain boundary conditions we obtain the

result.

(/m)[ 2-e-mR

] 0 when z and 0.

Hence, (2.50)

f(x) dx + - R

a-

f(x) dx a+

R

= 2iR+ + iR0

f(x) dx = 2 i R+ + i R0

-

f(x) dx + - R

a-

f(x) dx a+

R

= f(x) dx

Lim -

0

eimz

f(z)0 as R

e

imz f(z) dz =

0 eimRe

f (Rei ) Rei d i

eimz

f(z)0 as R .

4. 11 SOLVED PROBLEM BASED ON COMPLEX ANALYSIS

Prob. (1):- Find the residue of at z=ia.

As given f(z) =

Here z=ia is = Lim (z-ia) f(z) = Lim (z-ia) = Lim

(2.51)

Prob. (2):- (a) Find the residue of at z=1.

(b) Find the residues of and at infinity.

(a) Here f(z) =

(2.52)

To find the residue at z=1, which is a pole of order 4.

f(z) = ,

where,

.


similarly going on we obtain,

Thus

Thus residue at z=2 is = (2.53)

(b) Both the functions are analytic at infinity.

(i) (2.54)

Residue of f(z) at z=0=Lim { -z f(z)} = Lim = Lim

= -1

(2.55)

z2

(z2+a

2)

z2

(z2+a

2)

z2

(z-ia)(z+ia) =

zia z

2

(z-ia)(z+ia) zia zia z

2

(z+ia)

= (ia)

2

2ai = (ia/2)

z4

(z-1)4(z-2)(z-3)

z

(z-a)(z-b)

(z2 – z

2 +1)

z2

z4

(z-1)4(z-2)(z-3)

1 z4

(z-1)4

(z-2)(z-3) . =

(z)

(z-1)4

z4

(z-2)(z-3) (z) = = (z

2 + 5z+19) -

16

(z-2) +

81

(z-3)

(Breaking in to partial fraction )

(z) = 2z+5 + - 16

(z-2)2

81

(z-3)2

(z) = - 96

(z-2)4

486

(z-3)4

(1) = - 96

(-1)4

486

(-2)2 = 96- (243/8) = (525/8)

(1)

3! = (175/16)

f(z) = z

(z-a)(z-b)

z z z

2

(z-a)(Z-b) -

z 1

(1-a/z) (1-b/z) -

(2.56)

(ii) f(z) =

Here, Lim { -z f(z)} does not exist

So we can write, f(z) = 1- + … Residue at f(z) at z=.

= Negative of the coefficient of (1/z) in the expansion of f(z) in the neighborhood of

z=. = -(-1) =1

Prob. (3)

Apply the calculus of residue to prove that,

(i)

(ii)

(iii) ,a>0.

Solution

(2.57)

(i) I =

Choosing the contour as a circle of unit radius (z=2),

z = ei

; so, dz= i ei

d= iz d Functions of Complex variable

so that, , where varies from 0 to 2.

Also when z =ei

=cos + i sin , and (2/z) = cos - sin.

Adding , z+ (2/z) = 2 cos, i.e. cos = ½{ z+ (2/z)}

Thus,

I = = (2/i) , where C denote the unit circle

z=2.Poles of the integrand will be given by z2+4z+2 =0

or, z =

(z2 – z

2 +1)

z2

z 1

z

1

z2

d

(2+cos) 0

2

= 2

3

ad

(a2+sin

2)

0

0

2 Sin

2 d

(a+bcos)

2

b2

= { a-(a2-b

2)}, if a>b>0

(1+a2)

=

d

(2+cos) 0

2

d = (dz /iz)

0 )

1

[iz { 2 +1/2 (z+1/z)}]

dz

C

C

dz

(z2+4z+1)

-4(16-4)

2 = -2 3

of which the pole z= -2- 3 lies outside the contour and therefore the only pole that lies

the contour is z= -2+3 ( which is of the order of one ).

So, the residue at z= -2+3 is,

Where f(z) =

Hence by Cauchy‟s residue theorem we have,

I =2 iR, where R represents the sum of the residues inside the contour.

I = 2 i x(2/i 3) = (2 / 3)

(2.58)


(ii) Suppose

(2.59)

Choosing the contour C as a circle (z=2) of unit radius.

z=ei

, so that (z+2/z) = 2cos

dz = iz d , (z- 2/z) = 2i sin .

so, I=

where, = and = are the roots of the quadratic

(z2+2b/a +2).

2

i (z2+4z+1)

z= (-2+3) Lim {z-( -2+3)} f(z)

2

i z= (-2+3) Lim {z-( -2+3)}

1

(z2+4z+1)

2

i z= (-2+3) Lim

{z-( -2+3)}

{z +2-3)} {z+2+3)} =

=

2

i z= (-2+3) Lim

1

{z+2+3)} = = (1/i 3)

0

2 Sin

2 d

(a+ b cos)

2

b2

= { a-(a2-b

2)}, if a>b>0 I =

I =

{1/2i) (z-1/z) } .dz

{a+b/2 (z+1/z)} z

{1/2i (z-1/z) } .dz

{a+b/2 (z+1/z)} z

1

i

1

i

C C

=

=

C C

-1

2ib

1

2ib

(z2 –1)

2 dz

z2 ( z-)(z-)

-a + (a2 –b

2)

b

-a - (a2 –b

2)

b

So, = (-2a/b), and . =2.

As such . =2, where >

So, = (2/)>2, i.e. <2 and so, z = , is the only simple pole with in the

contour C. z = lies outside the contour. Also z=0 is a pole of order two, which lies inside

the contour.

Residue at z = is, Lim (z - )

=

, where ( -)2 = ( +)

2 -4( .) = [(4a

2/b

2) – 4]

= [4(a2-b

2)/b

2] = (-2/2ib).(2/b) (a

2-b

2) =

Again residue at z=0 ( a pole of order two) is the coefficient of (2/z) in

, where z is a small quantity near the pole,


=

So, the residue at z=0 is = (a/ib2),

Hence by the Cauchy‟s residue theorem, I= 2iR = (2.60)

(iii) Suppose I =

(2.61)

Choosing the contour C as a semicircle of unit radius.

z=ei

, then dz =iz d , or d = (dz/iz) and sin = (2/2i)(z-2/z)

Here varies from 0 to . Then putting the values of d and sin from above

we obtain,

-1

2ib

1

2ib

(z2 –1)

2 dz

z2 ( z-)(z-)

z

-1

2ib

1

2ib

(z2 –1)

2 dz

z2 (z-)

Lim (z - ) z

-1

2ib

1

2ib

(z –1/z)2 dz

(z-) Lim (z - ) z

( -)2

-2ib( -) = =

As =1 so, = 1/

( -)

-2ib =

(a2- b

2)

-ib2

1 (z2 –1)

2

-2ib z2 ( z

2 +1+2az/b)

1 1 2a 1

-2ib z2 bz z

2

1- 1+ +

2 -1

2 { a-(a2-b

2)},

b2

ad

(a2+sin

2)

0

Poles are given by -z2 (4a

2+2)+z

4+2=0

Let us assume that, -z2 (4a

2+2)+z

4+2 (z

2 -

2)( z

2 -

2)

Thus, 2 +

2 = 4a

2+2 and

2

2 =2

If < 2, then >2, implies that the pole with in the semi –circle will be z=, which is the

only pole inside C and z= will lie outside the contour.

Thus the residue at z= is,

= Lim (z-)

Hence by Cauchy‟s Residue theorem ,

I = 2 iR = 2 i x = , where (2-

2) = [ ((

2+

2)2 - 4(

2

2)]

= 4a (2+a2)

= = .

(2.62)

Group Theory and Functions of Complex variable Prob.4

Prove by contour integration that;

(i)

(ii)

(ii

Solution

(i) (2.63)

Here f(z) =

I = -4a zdz

i {-z2 (4a

2+2)+z

4+1}

C

-4a zdz

i (z2 -

2)( z

2 -

2) z

= - 2a

i(2 -

2)

-2a

i(2 -

2)

4a

(2-

2)

4a

4a (1+a2)

(1+a2)

dx

(a+bx2)n 2

n b

= 0

0

0

0

xsin ax

(x2 + k

2)

dx =1/2( e-ak

) ,where a>0

0

0

dx

(1+x2) = /2

dx

(a+bx2)n

0

0

dz

(a+bz2)n

0

0

Choosing the closed contour C consisting of the real axis from –R to +R and the upper

half of a large circle z=R represented by , we have Cauchy‟s residue‟s theorem,

And since , , where R+ is the sum of residues,

So,

Now the poles f(z) are given by,

( a+bz2)n =0 or b

n { z+ i(a/b)}

n{ z- i(a/b)}

n =0

z = i(a/b) ( poles of order „n‟),

Only the pole z=i(a/b) of order „n‟ lies within the contour.

Residue at z=i(a/b) ( a pole of order „n‟) is =

Where (z) =

n-2

(z) =


Thus residue at z= (ia/b) (pole of order „n‟) is,

Now = , , which tends to zero as z.

Hence when R, the relation (2) yields.

On simplification we obtain ,

f(z) dz = f(x) dx + f(z) dz

C

-R

+R

f(z) dz C

=2iR+

2iR+ = f (x) dx + f(z) dz

-R

+R

n-1

(i(a/b)

(n-1)! 1

bn (z+i(a/b)n

(-1)n-1

.n(n+1)(n+2)…(2n-2)

bn (z+i(a/b)2n-1

n-1(ia/b)= (-1)

n-1.n(n+1)(n+2)…(2n-2)

bn (2i(a/b)2n-1

(-1)n-1

.n(n+1)(n+2)…(2n-2)

(n-1)! bn (2i(a/b)2n-1

f(z) dz dz

(a+bz2)n

dz

(a+bz2)n

dx

(a+bx2)n

-

= 2i x (-1)

n-1.n(n+1)(n+2)…(2n-2)

(n-1)! bn (2i(a/b)2n-1

= 2i x (-1)

n-1.1.3.5…(2n-3)x 2.4.6…(2n-2)

(n-1)! (n-1)! bn (2i(a/b)2n-1

dx

(a+bx2)n

0

= i x (-1)

n-1.1.3.5…(2n-3)x 2.4.6…(2n-2)

(n-1)! (n-1)! bn (2i(a/b)2n-1

(2.64)

(ii) suppose f(z) = , where is the imaginary part of f(z).

Choosing the closed C, which consists of real axis from –R to R and upper half of a large

circle

z=R, represented by , we have

(2.65)

Singularities of f(z) are given by z2 +k

2 =0, i.e. z = ik of which only the simple pole

z=+ik lies with in the contour.

Residue at z=+ik ( a simple pole) = Lim (z-ik)


So, Since modulus can not be negative .

Hence (2.65) yield when R.

And

Equating the imaginary parts of either sides, we get

(iii)

zeiax

z2+k

2

z sinaz

z2 +k

2


C

-R

+R

= 2iR+

zeiax

(z+ik)(z-ik) zik

e-ak

2 =

zeiaz

dz

(z2+k

2)

f(z) dz = ze

iaz dz

(z2+k

2)

0 As z

f(z) dz = 0,

xeiax

dx

x2+k

2

- = i e

-ak

xeiax

dx

x2+k

2

0 = i e

-ak /2

x sin ax dx

x2+k

2

0 = e-

ak /2

0

0

dx

(1+x2) = /2

dz

(1+z2) -R

+R 0

dx 1.3.5…..(2n-3) 1

(a+bx2)n 2

n b 1.2…….(n-1) a

1/2(2n-1) = 0

0

Let f(z) = , so that zf(z)0 as z.

Poles of f(z) are given by 2+z2 =0, i.e. z=i of which z=i lies in the upper half plane i.e.

in the contour chosen as consisting of a large semicircle along with the real axis from –

R to +R.

Residue of f(z) at z=i is Lim (z-i) = Lim

By Cauchy‟s residue theorem, we have,

Where the value of so the value of , then

From equation (2)

or, 2 , i.e.


4.11.1 Exercise

( A) Apply calculus of residue to prove that ,

(i)

(ii)

(B) Prove by contour integration that

(i)

(ii) (iii)

zi

1

(1+z2) zi

1

(z+i) =

1

2i


C

-R

+R

= (2i /2i) =

f(z) dz 0 as R f(z) dz =0

-

dx

(1+x2) =

0

dx

(1+x2) =

0

dx

(1+x2) = /2




i) Find the residue of [ (z2/ (z

2 + a

2)] at z= -ia.

…………………………………………………………………………………………

…………………………………………………………………………………………

…………………………………………………………………………………………

…….

0

2 d

(5-4cos) = (40 /27)

0

2 cosd

(5+4cos) = ( /6)

0

Log(1+x

2)

( 1+x2) = ( log 2)

= (32 /16) where a>0. 0

x

6 dx

( a4+x

4)

0

x

-1 dx

( 1+x) = cot ,0<<1

(C) Evaluate the following integrals by contour integration:-

(i) (ii) (iii)

(iv) (iv) (v)

[Ans. (i) (/2) (ii) (8/3), (iii) (/2)aea (iv) (/8a

3) (v) (/2) (vi) (/2).]

4.12 LET US SUM UP

Complex Number – An ordered pair of real numbers such as (x, y) is termed as a

complex number if we write z= (x, y) = x + i y , where i=(-1) ,Here x is called the real

part and „y‟ is the imaginary part of the complex number „z‟.

Argand diagram – The plane whose points are represented by complex numbers is

known as Argand palne or Argand diagram or Guassian palne.

Function of Complex variable – If w= u + iv and z=x + i y are two complex numbers,

then w is said to be the function of z and written as w = f ( z ),if to every value of „z‟ in a

certain domain d, there correspond one or more values of w.


Analytic or regular function – “A function f(z) which is single valued and differentiable

at every point of a domain D, is said to be regular is said to regular in the given domain

D.

Necessary condition of a function to be analytic – If w= f (z) = u (x, y) + i (x, y),then

(u/x) = ( v/ y) and ( v/x) = - ( u / y) are the necessary conditions for a function

to be analytic. In polar coordinates it is (u/r) = - ( v/r ) and (v/r) =- ( u /r ).

Cauchy‟s theorem – If f(z) is a regular function of „z‟ and if f(z) is continuous at each

point with in and on a closed contour C, Then c f(z) dz =0.

Cauchy‟s Integral formula – If the function f (z) is regular with in on a closed contour

„C‟

0

sin x dx

x 0

2 d

(5/4 + sin) 0

cos x dx

( a2+x

2)

-

x

2 dx

( a2+x

2)3

0

(1-cos x)dx

( x2)

0

x

2 dx

( 1+x4)

and z0 be point with in „C‟ then, .

Cauchy‟s Residue theorem – If the function f (z) be a single valued , continuous and

regular with in and on a closed contour C, expect a finite number of poles ( Singularities)

with C; then C f (z) dz =0.


1. i) Analytic or regular function : “A function f(z) which is single valued and

differentiable at every point of a domain D, is said to be regular is said to regular

in the given domain D. A function may be differentiable in a Domain D have

possibly for a finite number of points. Such points are called singularities or

singular points of f(z).

Conditions of analytically of a function f(z) :

And in Cartesian coordinate function .

And in Polar coordinates

ii) Cauchy„s Theorem : This theorem states that,

“If f (z) is a regular function of „z‟ and if f(z) is continuous at each point within

and on a closed contour C then,

i. e. the integral of the function around a closed contour is zero.”

This theorem is important in order to find out the integration around a closed

counter C of a given function f(z).


iii) Cauchy‟s Integral Formula: If the function f(z) is regular with in on a closed

contour „C‟ and if z0 be point with in „C‟ then ,

Where the integral is taken in the counter clock-wise sense.

2. i) Definition of Residue

The residue of a function f(z) at the pole z=a, is defined to be the coefficient of

(z-a)-2

in the Laurent‟s expansion of the function f(z), i.e.,

Where z=a is a pole of order „m‟

If z=a be the pole of order one, then the residue is

1 f(z)

2i (z-z0 )

u v

x y

= v u

x y

= -

u 1 v

r r

= u 1 u

r r

=

C

f(z) dz =0

f (z0) = 1

2i f (z) dz

(z-z0)

f (z) = an (z-a)n + bn (z-a)

-n

n=1

n=1

Lim (z-a) f(z) za

b1=

i.e. In case of simple pole at z=a,

Residue =

Cauchy‟s Residue theorem If the function f(z) be single valued, continuous and regular with in and on a

closed contour C, except a finite number of poles ( Singularities ) with C; then,

Where R represents the sum of the residues say R2,R2…..Rn. of the f(z) at the

poles 2, 2, 3 …. n with in a.

3 Attempts as solved problem no.1.

REFERENCES AND SUGGESTED TEXT BOOKS






Lim (z-a) f(z) za

f (z) dz = 2iR C

UNIT 5 SPECIAL FUNCTIONS AND SPHERICAL HARMINICS

Structure

5.0 Introduction

5.1 Objectives

5.2 Legendre Differential equations and Legendre functions

5.2.1 Legendre‟s Equation:

5.2.2 Generating Function of Legendre Polynomial

5.2.3 Rodrigue‟s formula for Legendre Polynomials

5.2.4 Orthogonal Properties of Legendre Polynomials

5.2.4 Recurrence Formulae for Legendre Pn(x).

5.3 Bessel‟s Differential Equation and Bessel‟s Polynomials

5.3.1 Bessel‟s Differential Equation

5.3.2 Generating Function for Bessel‟s function Jn(x)

5.3.3 Recurrence formulae for Jn(x)

5.3.4 Orthonormality of Bessel‟s functions

5.4 Hermite Differential Equation and Hermite Polynomials

5.4.1 Hermite Differential Equation

5.4.2 Hermite Polynomials

5.4.3 Generating Function for Hermite function Jn(x)

5.4.4 Recurrence formulae for Jn(x)

5.4.5 Rodrigue‟s formula for Hermite Polynomials

5.4.6 Orthonormality of Hermite functions

5.5 Laguerre‟s Differential Equation and Laguerre‟s Polynomials

5.5.1 Laguerre‟s Differential Equation

5.5.2 Generating Function for Laguerre‟s function Jn(x)

5.5.3 Rodrigue‟s formula for Laguerre‟s Polynomials

5.5.4 Recurrence relations for Laguerre polynomials

5.5.5 Orthonormality of Laguerre‟s functions

5.6 Representation of various functions in terms of hypergeometic

function

5.6.1 Legendre Polynomials

5.6.2 Laguerre Polynomials

5.6.3 Bessel‟s Function

5.7 Problems on polynomials and related functions

5.8 Let Us Sum Up

5.9 Check Your Progress: The key

Special Functions and Integral Transform

5. 0 INTRODUCTION

The differential equations may be divided in to two large classes, (i) Linear equations and

(ii) Non-Linear Equations .The non-linear equations of second and higher orders are

rather difficult to solve while linear equations are much simpler in many respect because

various properties of their solutions can be characterized in a general way and standard

methods are available for solving many of these equations. The linear differential

equations play an important role in theoretical physics in connection with mechanical

vibrations, elastic circuits and network, Quantum and nuclear physics.

5. 1 OBJECTIVES

The main purpose of this unit is to study special functions and Spherical harmonics in

respect of various differential equations. After completing this unit you will be able to :-

Define various linear differential equations particularly used in theoretical

physics.

Find out the way to solve these particular equations.

Find the generating functions, their orthogonal properties and important

recurrence relations.

To solve the specific problem regarding to the theoretical physics .

5. 2 LEGENDRE DIFFERENTIAL EQUATIONS AND LEGENDRE

FUNCTIONS

Legendre differential equation can be solved by using the Legendre function as given

below

5. 2.1 Legendre‟s Equation The differential equation

or, (1.1)

is called Legendre equation. This equation can be solved in series of ascending or

descending powers of „x‟; but the solutions in descending powers of „x‟ is more physical

importance.

Special Functions and spherical harmonics

d2y

dx2

(1-x2) - 2x dy

dx + n (n+1) y = 0

dy

dx (1-x

2 ) + n (n+1) y = 0

d dx

Solution of Legendre‟s equation in descending power of x - Let the solution of

Legendre‟s equation in descending powers of „x‟ may be put as,

y = xk

[ a0 + a1x-1

+ a2 x-2

+ a2 x-3

+ ….+ ar x-r +…..] (1.2)

This equation on differentiation yields,

(1.3)

And, (1.4)

Substituting these values of y, (dy / dx) , (d2y/dx

2) from equations (1.2),(1.3) and (1.4) in

equation 1.1, we get

Or,

(1.5)

This equation is an identity; therefore the coefficients of various powers of „x‟ must be

equal to zero. To obtain a general relation between coefficients of series; let us equate the

coefficients of xk-r-2

in equation (1.5) to zero; i.e.

(1.6)

This equation gives

(1.7)

Equating the coefficients of xk-1

to zero, by putting r=1 in equation (1.5) we obtain a1=0,

therefore equation (1.7) implies that

a1= a 3= a 5=a7 =0 (1.8)

i.e., all the coefficients of a‟s having odd suffixes are zero.

Now two cases are arises when k = n and k =-(n+1)


r = 0

[(k-r) (k-r-1) x k – r - 2 + { n(n+1)-(K-r)(K- r+1) }x k - r] ar = 0

ar +2 = ( k-r)(k-r-1)

{(k-r)+(n-1)} {(k-r) – ( n+2)} ar

= ar xk-r

d2y

dx2 a r (k-r) (k-r-1) x

k-r-2 =

dy

dx a r (k-r) x

k-r-1 =

r =0

r =0

r =0

( k-r)(k-r-1) ar + { n(n+1) –(k-r-2)(k-r-1)}ar + 2 = 0

Case (i) :- when k =n , we get from equation (1.7)

Substituting r = 0,2,4 ….we get.

Similarly,

And so on,

Also we have, a1= a 3= a 5=a7 =0 (1.9)

Substituting the values of various coefficients of a‟s in equation (1.2); we get the series

solution for k=n as,

(1.10)

where „a0‟ is an arbitrary constant and „n‟ is appositive integer . If

Then the above solution is called the Legendre polynomials or Legendre function of first

kind and is represented by Pn (x).

Thus,

(1.11)

This series is a terminating series and for different values of „n; we get Legendre

polynomials.


ar +2 = ( n-r)(n-r-1)

(2n-r-1) (-r – 2) ar

( n-r)(n-r-1)

(2n-r-1) (r + 2) ar = -

( n)(n-1)

(2n-1) ( 2) a0 , a2 =

( n-2)(n-3)

(2n-3) ( 4) a2 a4 =

( n-2)(n-3)

(2n-3) 2. 4 a4 =

( n)(n-1)

(2n-1) a0 ,

( n-2)(n-3) (n-4)(n-5)

(2n-3) (2n-5) 2. 4.6 a6 =

( n)(n-1)

(2n-1) a0 ,

y = a0 xn -

( n)(n-1)

(2n-1) ( 2) xn-2 +

( n-2)(n-3)

(2n-3) 2. 4

( n)(n-1)

(2n-1) x n-4 - ……..

1..3.5……(2n-1)

n! a0 =

1..3.5……(2n-1)

n! Pn(x) = x

n -

( n)(n-1)

(2n-1) ( 2) xn-2 +

( n-2)(n-3)

(2n-3) 2. 4

( n)(n-1)

(2n-1) x n-4 - ..

Case (ii) :- By putting k= -(n+1) in equation (1.7) we obtain another polynomial

called Legendre function of second kind and is denoted by Qn(x).

(1.12)

This is an infinite or non-terminating series, since „n‟ is a positive integer. As Pn(x) and

Qn(x) are two independent solutions of Legendre equations; therefore the most general

solution of Legendre equation may be expressed as,

y= APn(x) + B Qn(x)

(1.13)

Where A and B are arbitrary constants.

5.2.2 Generating Function of Legendre Polynomial

Legendre polynomials Pn(x) is the coefficient of Zn in the expansion of [ 1-

2xz+z2]

-1/2 in the ascending power of „z‟. i.e.,

(1.14)

[ 1-2xz+z2]

-1/ 2 =

5.2.3 Rodrigue‟s formula for Legendre Polynomials

It is given by (1.15)

5.2.4 Orthogonal Properties of Legendre Polynomials

To show that

(1.16)

(1.17)

The above two equations can be written equivalently in the form of a single equation

using Kronecker delta symbol mn „

[ if m n and ]

(1.18)

x - n- 5 + . n!

1..3.5…(2n+1) Qn(x) = x

-n-1 -

( n+1)(n+2)

2.(2n+3) x –n- 3 +

( n+1)(n+2)(n+3)(n+4)

2.4.(2n+3)(2n+5)

n=0

Pn(x)z n

Pn(x) = (x-2 –1)

n 1 d

n

2n n! dx

n

Pm(x) Pn(x) dx =0 for m n -1

+ 1

[Pn(x)]2 dx for m=n -1

+ 1 2

2n+1 =

Pm(x) Pn(x) dx = -1

+ 1 2

2n+1 mn mn

= 0 mn

= 1


5. 2. 5 Recurrence Formulae for Legendre Pn(x)

Formula (i) nPn = (2n-1) x Pn-1- (n- 1)Pn-2 (1.19)

Proof. From the generating function Pn(x); we have

(1.20)

Differentiating above equation with respect to „z‟ we get;

i.e. (1.21)

Multiplying both sides with (1-2xz+z2); we get

(x-z)( (1-2xz+z2)-1/2

= (1-2xz+z2 )

(1.22)

Equating coefficients of zn-1

from both sides, we get

(1.23)

This is first recurrence relation. This may be written in alternative form by substituting

(n+1) for n in this relation or equating the coefficients of zn from both sides of equation

(1.22) as;

(1.24)

Formula (ii) (1.25)

Where dashes denote differentiation with respect to „x‟ .

Proof . We have from generating function

(n+1) Pn+1 = (2n+1) xPn – nPn-1

nPn = xPn - Pn-1

(1-2xz+z2)-1/2

= Pn(x)z n

n =0

n=0

Pn(x)z n (1-2xz+z

2)-1/2

=

(-1/2) (1-2xz+z2)-3/2

(-2x+2z) = n z n-1

Pn(x)

n=o

(1-2xz+z2)-3/2

(x -z) = n z n-1

Pn(x)

n=o

n=o

n z n-1

Pn(x)

n=o

n=o

z n Pn(x) =

(1-2xz+ z2

) n z n-1

Pn(x)

(x-z)

xPn-1 –

Pn-2 = n Pn –2x (n-1) Pn-1 + (n+2) Pn-2

or, n Pn = (2n-1) xPn-1 – (n-1)Pn-2

(1.26)


Differentiating this equation with respect to „z‟,we get

Or, (1.27)

Now differentiate equation (1.26) with respect to „x‟, we get

BLOCK-3

Or, (1.28)

Dividing the equation (1.27) by (1.28); we get

i.e.,

(1.29)

Equating coefficients of zn from both sides of equation (1.28), we get

(1.30)

Formula (iii) -

(1.31)

Proof - From relation (1.24), we have

Differentiating with respect to „x‟; we get

(1.32)

From relation (1.30);

substituting this value of xPn in equation (1.32); we get

(1-2xz+z2)-1/2

(-2x+2z) = n z n-1

Pn(x)

-1

2

(1-2xz+z2)-3/2

( x - z) = n z n-1

Pn(x)

(1-2xz+z2)-1/2

(-2z) = n z n Pn(x)

-1

2

(1-2xz+z2)-3/2

( z) = n z n Pn(x)

n =0

n =0

n =0

n

nzn-1

Pn(x)

zn Pn

x-z

z =

n

(x-z) n

zn Pn = z

n nz

n-1 Pn(x)

x Pn – Pn-1 = nPn(x) or, nPn(x) = xPn(x) - Pn-1(x)

Pn +1(x) - Pn-1(x) = (2n+1)Pn(x)

(n+1) Pn+1 = (2n+1) xPn – nPn-1

(n+1) Pn+1 = (2n+1) Pn + (2 n+1)xPn - nPn-1

xPn(x) = nPn(x) + Pn-1(x)

n =0

(1.33)


Collecting coefficients of Pn and Pn-1 and rearranging; we get

i.e., (1.34)

This is third recurrence relation. This may be expressed in alternative form if we

substitute (n-1) for n in above relation, i.e.

(1.35)

Formula (iv) (1.36)

From relation (1.24); we have ,

Differentiating with respect to „x‟ we get

(1.37)

This is IV recurrence relation. This may be expressed in alternative form if we substitute

(n-1) for „n‟ in equation (1.37 ); i.e.,

(1.38)

Formula – (v ) (1.39)

From relation ( 1.38) we have;

From relation (1.30) ;

(n+1) Pn+1 = (2n+1) Pn + (2 n+1)[nPn+Pn-1] + nPn-1

(n+1)Pn+1 – (n+1) Pn –1 = (n+1)(2n+1)Pn

Pn+1 – Pn –1 = (2n+1)Pn

Pn – Pn – 2 = (2n-1)Pn

Pn+1 – x Pn = (n+1)Pn

(n+1) Pn+1 = (2n+1) xPn – nPn-1

(n+1) Pn+1 = (2n+1) Pn + (2 n+1)xPn – nPn-1

= (2n+1) Pn + (n+ n+1)xPn – nPn-1

= (2n+1) Pn + ( n+1)xPn + n (xPn - Pn-1)

= (2n+1) Pn + ( n+1)xPn + n (nPn )

[ Using relation (1.30)]

= ( n+1)xPn + Pn{ (2n+1) Pn +n2}

i.e., (n+1) Pn+1 = ( n+1)xPn + Pn(n+1)2

or, Pn+1 = xPn + Pn(n+1)

Pn+1 - xPn = Pn(n+1)

Pn - xPn-1 = n Pn-1

(1-x2) Pn = n (Pn-1 – x Pn )

n Pn-1 = Pn - xPn-1

n Pn(x) = xPn(x) - Pn-1(x)

Multiplying equation (1.30) by „x‟ and then subtracting from equation (1.39), we obtain

the relation

(1.40)


Formula –(vi)

(1.41)

From relation (1.24),

Or,

5. 3 BESSEL‟S DIFFRENTIAL EQUATION AND BESSEL‟S POLYNOMIALS

Solution of Bessel‟s differential equation can be obtained as follows-

5.3.1 Bessel‟s Differential Equation

The differential equation

(1.42)

is called Bessel‟s differential equation. This equation can also be put in the form

(1-x2) Pn = n (Pn-1 – x Pn )

(1-x2)Pn = (n+1) (x Pn – Pn+1 )

(n+1) Pn+1 = (2n+1) x Pn – nPn-1

= (n + n+1) x Pn – nPn-1

This can be expressed (n+1)[ Pn+1-xPn ] = n [ x Pn – Pn-1] = -n [ Pn-1 – xPn]

= - (1-x2)Pn Using relation (1.40)

(1-x2)Pn = (n+1) [ Pn+1-xPn ]

d2y

dx2

x2 + x dy

dx + (x2-n2) y = 0




i) State the Legendre differential equation with its general solution?

ii) Write the generating function of Legendre polynomials?

…………………………………………………………………………………………

…………………………………………………………………………………………

…………………………………………………………………………………………

…………………………………………………………………………………………

…………………………………………………………………………………………

…………………………………………………………………………………………

…………………………………………………………………………………………

…………………………………………………………………………………………

…………………………………………………………………………………………

………………………………………

(1.42a)

The solution of this equation are called Bessel‟s functions of order „n‟ Special Functions and Integral Transform

Solution of Bessel‟s equation in descending power of x

Let the series solution of Bessel‟s equation in ascending powers of „x‟ may be written as,

y = xk

[ a0 + a1x + a2 x

2 + a2 x

3 + ….+ ar x

r +…..] (1.43)

This equation on differentiation yields,

(1.44)

And, (1.45)

Substituting these values of y, (dy/dx) , (d2y/dx

2) from equations (1.43),(1.44) and (1.45)

in equation (1.42), we get

on simplification we get;

or, (1.46)

This equation is an identity; therefore the coefficients of various powers of „x‟ must be

equal to zero. To obtain a general relation between coefficients of series; let us equate the

coefficients of xk -2

in equation (1.46) to zero; we get,

But a0 0; since the first term of the series is non vanishing therefore;

(k2- n

2)= 0 or k = n (1.47)

Now equating to zero the coefficient of xk –1

in (1.46), we get;

a1[(k+1)2 – n

2] =0

But (k+1)2 –n

2 0 since k = n; hence we have a1= 0.

Again, equating to zero coefficients of general term i.e. x k+r

in (1.46) we get;

a r (k+r) x k+ r-1

r = 0

- (1/x)

r = 0

+ ( 1- n

2/x

2 ) ar xk+r = 0 a r (k + r) (k+r-1) x

k+r-2

r = 0

a0(k2- n2)= 0

ar +2 = -1

(k+ r+ n+2)} (k +r – n+2) ar

= ar xk+r

r =0

dy

dx a r (k + r) x

k+ r -1 =

r =0

d2y

dx2 a r (k + r) (k + r-1) x

k+ r-2 =

r =0

d2y

dx2 +

dy

dx + {(1- (n2/x2) y} = 0 1

x

r = 0

[ar {(k + r) 2 -n2} x k + r - 2 + ar x k + r] = 0

(1.48)


Equating the coefficients of xk-1

to zero, by putting r=1 in equation (1.46) we obtain a1=0,

therefore equation (1.48) implies that

a1= a 3= a 5=a7 =0 ( 1.49)

i.e., all the coefficients of a‟s having odd suffixes are zero.

Now two cases are arises when k =

Case (i) :- when k =n , we get from equation (1.48)

(1.50)

Substituting r = 0,2,4 ….we get.

Similarly,

And so on,

Also we have, a1= a 3= a 5=a7 =0 (1.51)

Substituting the values of various coefficients of a‟s in equation (1.43); we get the series

solution for k=n as,

(1.52)

where „a0‟ is an arbitrary constant and „n‟ is appositive integer . If

ar +2 = -1

(2n+r+2) (r +2) ar

(-1)

(2n+ 2).2 a0 = a2 =

(n+2) 24. 2! a4 =

(-1)2

(n+1) a0 ,

(2n+3) 26.3! a6 =

(-1)

(n+1)(n+2) a0 ,

y = a0 xn

(-1)

(n+1) 22.1! x2 + (-1)2

(n+ 1)

(n+2) 24.2! x 4 + ……+

(1)

2n (n+1)!

a0 =

(-a0)

(n+ 1).2.2

(-a0)

(n+ 1).2.2 = =

(-a0)

(n+ 1).22.1!

1+

(-1)r.

(n+1)(n+2)…(n+ r) 22r.r!

x 2r +

Then the above solution is called the Bessel‟s polynomials or Bessel‟s function of first

kind and is represented by Jn (x).


(1.53)

General Solution

When „n‟ is not an integer, J-n(x) is distinct from Jn(x); hence the most general solution of

Bessel‟s equation is,

Y = A Jn(x) + BJ-n(x) (1.54)

Where A and B are arbitrary constants.

However, when „n‟ is integer and since „n‟ appears in the differential equation .Only as

„n2‟, there is no loss of generality in taking „n‟ to be positive integer J-n(x) is not distinct

from Jn(x), in this case the denominator of the first „n‟ terms of series of J-n(x) for value

of r=0,1,2…(n-1) will have gamma function of negative numbers. As the gamma function

of negative numbers is always infinite; so

For r=0,1,2….(n-1). There by indicating that the first „n‟

Terms of the series for J-n(x) vanish. Therefore, we shall the terms left for r=n and

onwards, i.e.

Substituting s = r-n , i.e.; r = n+ s; we get

Jn(x) = x

n

2n (n+1)

1 + (-1)

(n+1) .22.1! x2 +

(-1)2

(n+1)(n+2) 24.2! x 4 + ..

(-1)r.

(n+1)(n+2)…(n+ r) 22r.r!

x 2r +

xn

2n (n+1)

=

r =0

(-1)r.

(n+1)(n+2)…(n+ r) 22r.r!

x 2r X =

r =0

(-1)r

.1 x –n+2r

r! (-n+r+1) 2

1

(-n +r +1) = 0

J-n(x) = r =n

(-1)r

.1 x –n+2r

r! (-n+r+1) 2

J-n(x) = s =0

(-1)n+s

.1 x n+2s

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

As (n+ s)! = (n+s+1) and (s+1) =s!; we have


i.e. J-n(x) = (-1)n Jn(x) (1.55)

Thus, in this case we no longer have two linearly independent solutions of Bessel‟s

equation.

Limiting values of Jn(x)

A precise analysis shows

That is, for large values of argument „x‟, the Bessel‟s functions behave like trigonometric

functions of decreasing amplitude.

Also,

5.3.2 Generating Function for Jn(x)

Jn(x), the Bessel‟s function of first kind of order „n‟ , is the coefficients of zn in the

expansion of the function e(x/2) (z-1/z)

. That is ;

(1.56)

5. 3. 3 Recurrence formulae for Jn(x)

(i) x . Jn(x) = n Jn (x) – x Jn+1(x) (1.57)

Proof

We know that Bessel‟s function of first kind is

Differentiating above equation with respect to „x‟, we get

BLOCK-3

Multiplying both sides by „x‟

J-n(x) = s =0

(-1)s

x n+2s

(n+s)!( s+1) 2 (-1)

n

X

Lim Jn (x) Cos ( x- /4 -n /2)

( x/2)

X0

Lim Jn (x) xn

2n n!

e(x/2) (z-1/z) = Jn(x)zn

-

= r =0

(-1)r

.1 x n+2r

r! (-n+r+1) 2 Jn(x)

= r =0

(-1)r

x n+2 r -1

r! (-n+r+1) 2 Jn(x) (n+2r) 1

2

= r =0

(-1)r

x n+2 r -1

r! (-n+r+1) 2 xJn(x) (n+2r)

x

2


[ Since (-1)! = ]

Now substituting r-1= s; we get

Hence

(ii) (1.58)

Proof :- We know that Bessel‟s function of first kind is

Differentiating above equation with respect to „x‟, we get

Multiplying both sides by „x‟

= r =0

n. (-1)r

x n+2 r

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

r =0

2r. (-1)r

x n+2 r -1

r! (n+r+1) 2 (n+2r) x

2

xJn (x) = nJn(x) +x r =0

(-1)r

x n+2 r -1

(r-1)! (n+r+1) 2

xJn (x) = nJn(x) +x r =1

(-1)r

x n+2 r -1

(r-1)! (n+r+1) 2

xJn (x) = nJn(x) +x s =0

(-1)s+1

x n+2 s+1

s! (n+s+2) 2

xJn (x) = nJn(x) -x r =1

(-1)s x

( n+1) +2s

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

xJn (x) = nJn(x) –x Jn+1

xJn (x) = -nJn(x) +x Jn-1

= r =0

(-1)r

.1 x n+2r

r! (-n+r+1) 2 Jn(x)

= r =0

(-1)r

x n+2 r -1

r! (-n+r+1) 2 Jn(x) (n+2r) 1

2

= r =0

(-1)

r x

n+2 r

r! (-n+r+1) 2 xJn(x) (n+2r)

= r =0

(-1)r

x n+2 r

r! (n+r+1) 2

(2n+2r-n)

(-1)r

x n+2 r -1

(r)! (n+r) 2 xJn (x) =- nJn(x) +x

r =0

xJn (x) = -nJn(x) +x r =1

(-1)r

x –n+2 r -1

(r )! (n-1+r+1) 2

= r =0

(-1)r

x n+2 r

r! (n+r+1) 2

(2n+2r-n)

Hence,


(iii) (1.59)

Proof

Recurrence relation (I) and (II) are

Adding above equations, we get

(iv) (1.60)

Proof


Subtracting relation (2) from (1) , we get

So,

(v) [ x -n

Jn(x)] = - x-n

Jn+1(x) (1.61)

Proof -

[ x -n

Jn(x)] = -nxn-1

Jn(x) + x-n

Jn(x) = xn-1

[ -n Jn(x) + x Jn(x)]

Using recurrence relation I, i.e.,

We have

[ x -n

Jn(x)] = x -n-1

[ -n Jn (x) + n Jn (x) –xJn+1(x) ] = x -n-1

[-xJn+1(x)]

[ x -n

Jn(x)] = x -n-1

[-xJn+1(x)]

xJn (x) = -nJn(x) +x Jn-1

2Jn(x) =Jn-1(x) – Jn+1(x)

x. Jn(x) = n Jn(x) – x Jn+1(x) x Jn (x) = -n Jn (x) + x Jn-1

2Jn(x) =Jn-1(x) – Jn+1(x)

2Jn(x) = x [Jn-1(x) + Jn+1(x)]

x. Jn(x) = n Jn(x) – x Jn+1(x) x Jn (x) = -n Jn (x) + x Jn-1

0 = 2n Jn(x) – xJn+1(x) –xJn-1(x)

2Jn(x) = x [Jn-1(x) + Jn+1(x)]

d

dx

d

dx

x. Jn(x) = n Jn(x) – x Jn+1(x) d

dx

d

dx

similarly using the second recurrence relation, we obtain the relation

[ x n Jn(x)] = x

n [Jn-1(x)]


5. 3. 4 Orthonormality of Bessel‟s functions

If and are the two roots of the equations Jn( ) = 0; then

The condition of orthogonality of Bessel‟s function over the interval (0, 1) with weight

function x is

With the condition of normalization is

Both the above equations represent the condition of orthogonality and may be written in

the form of a single equation as

(1.62)

Where is Kronecker delta symbol defined as =1 for = and 0 for .

5.4 HERMITE DIFFERNTIAL EQUATION AND HERMITE POLYNOMIALS

The Hermite differential equations can be solved as -

5. 4. 1 Hermite Differential Equation

d

dx

0

1

Jn(x) Jn(x) x dx = 0 for

0

1

x[Jn(x)]2 dx = (1/2) J

2n+1(x)

0

1

[Jn(x) Jn(x) xdx] = (1/2) J2n+1()




i) Discuss the orthogonality of Bessel‟s function?

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…………………………………………………………………………………………

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The differential equation is

(1.63)


Where „n‟ is a constant, is called Hermite differential equation. The series solution of the

Hermite equation (1.63) may be expressed as,

(1.64)

Differentiate equation (1.63) and putting the values of (dy/dx) and (d2y/dx

2) and equating

to zero the coefficients of lowest power of „x‟[obtained by putting r=0] i.e. coefficient of

xk-2

, we get k=0 and k=1 as a0 0.Now equating to zero the coefficient of general term

xk+r

; we get

(1.65)

Now there arise two cases when k=0 and K=1. When K =0 we obtain from equation

(1.64)

(1.66)

Case-II when k=1, we have equation (1.65)

(1.67)

Substituting r= 1,3,5,… we obtain the values of various coefficients a3 =a5 =a7= ..=(0)

each. Substituting r=0,2,4…etc. in equation (4), we obtain the values of a4=a6=…..Then

from equation(2).

(1.68)

In view of equations (1.66) and (1.67) a general solution may be written as,

y = Ay1+ By2, (1.69)

where A and B are arbitrary constants and y1 and y2 are given by the equation (1.66) and

(1.68).

1. 4. 2 Hermite Polynomial

The Hermite polynomial of degree n, for „n‟ being a positive integer is;

d2y

dx2

-2 x dy

dx + 2n y = 0

r=0

a r xk+r y =

a r+2 = 2 (k+ r) -2n

(k+r+2)(k+r+1) a r

y = a0 1- 2n

2! x

2 +2

2 n(n-2)

4! x

4 +…+ (-2)

m n(n-2)….(n-2m+2)2m

2m! x

2m = y1

a r+2 = 2 (1+ r) -2n

(r+3)(r+2) a r

y = a0 1- 2(n-1)

3! x

3 +2

2 (n-1)(n-3)

5! x

5 -…+ (-2)

m

(n-1)(n-3)…(n-2m+1)2m

(2m+1)! x

2m+1 +…= y2

Hn(n) = (-1)r

(2x) n-2r

n!

r!(n-2r)! r=0

p

(1.70)

From equation we observe that,


Hn(0) = (-1) n n!/{(n/2)!} if „n‟ is an integer.

Hn(0) = 0 if „n‟ is an odd integer . From equation (1.70) we can write values of Hermite

polynomials of different orders, even or odd. Some of them are written as,

H0(x) = 1,H1(2) = 2x,H4(x) = 4x2 –2, H3(x) = 8x

3 –12x, H(4) =16x

4 –48 x

2+12.

5.4. 3 Generating function of Hermite equation

The function is called the generating function of Hermite polynomials.

i.e. f(x,z) = (1.71)

5. 4. 4 Recurrence formula for Hermite Polynomials

(i) Hn(x) = 2nHn-1(x) (1.72)

Proof

We have

Differentiating above equation with respect to „x‟; we get

Equating the coefficients of zn on both sides, we get

i.e. Hn(x) = 2n.Hn-1(x) (1)

(ii) 2xHn(x) = 2nHn-1(x) + Hn+1(x) (1.73)

Proof

Where p = n/2 if „n‟ is even

½( n-1) if „n‟ is odd

e2zx –z

n=0

Hn(x)

n! z

n = e2zx –z

n=0

Hn(x)

n! z

n = e2zx –z

n=0

Hn(x)

n! z

n = e2zx –z 2 .2z = 2z

n=0

Hn(x)

n! z

n = e2zx –z 2

n=0

Hn(x)

n! z

n+1 2 =

n=0

Hn-1(x)

(n- 1)! z

n 2 =

Hn(x)

n!

Hn-1(x)

(n- 1)! 2 =

n=0

Hn(x)

n! z

n = e2zx –z

2

2

2

2

We have ; differentiating with respect to z;

We get,


(Since term on left hand side corresponding to n=0 is zero)

Rearranging we get

Equating coefficients of zn on the both the sides, we get

Or, 2x Hn(x) = 2nHn-1(x) +Hn+1(x) (1.74)

(iii) Hn(x) = 2xHn(x) – Hn+1(x) (1.75)

Proof


Hn(x) = 2n Hn-1(x)

2x Hn(x) = 2nHn-1(x) + Hn+1(x)

Subtracting (ii) from (i), we get

Hn(x) - 2x Hn(x) = 2n Hn-1(x) - 2nHn-1(x) - Hn+1(x)

Hn(x) = 2nHn(x) –Hn+1(x)

(iv) Hn(x) –2xHn + 2nHn(x) =0 (1.76)

Proof

Hermite differential equation is,

y -2xy +2ny =0

As Hn(x) is the solution of above equation, i.e., substituting Hn(x) for „y‟ we get,

Hn(x) –2xHn + 2nHn(x) =0

5. 4. 5 Rodrigue‟s formula for Hermite Polynomials

n=0

Hn(x)

n! nz

n-1 = e2zx –z .(2x-2z)

2

n=0

Hn(x)

(n-1)! z

n-1 = e2zx –z .2x- 2z. e 2zx -z

2 2

n=0

Hn(x)

(n-1)! z

n-1 = 2x - 2z.

n=0

Hn(x)

n! z

n

n=0

Hn(x)

n! z

n

n=0

Hn(x)

(n)! z

n = 2 +

n=1

Hn-1(x)

(n-1)! z

n

n=0

Hn+1(x)

n! z

n 2x

Hn(x) = (-1)n e

x (1.77)

The above equation represents differential form of Hermite polynomials and is called the

Rodrigue‟s formula for Hermite polynomials. Special Functions and Integral Transform

5. 4. 6 Orthogonality of Hermite Polynomials

The orthogonal property of Hermite polynomials is,

(1.78)

Where mn is Kronecker delta symbol such that mn=1 for m=n and 0 for mn.

The Hermite polynomials are specifically useful in analyzing the Quantum mechanical

problem of Hermite oscillator.

5. 5 LAGUERRE‟S DIFFRENTIAL EQUATION AND LAGUERRE‟S

POLYNOMIALS

The Laguerre‟s differential equation can be solved by using the series solution as-

5.5.1 Laguerre‟s Differential Equation

(1.79)

Let the series solutions of above equation be

(1.80)

so that we can obtain the values of (dy/dx )and (d2y/dx

2) from equation (1.80) and putting

the values in equation (1.79) we obtain;

(1.81)

Equation (1.81) is an identity; therefore the various powers of „x‟ must be zero. Equating

to zero the coefficients of lowest power of „x‟ i.e. xk-r

(putting r=0); we get a0k2 =0, as a0

0, being the coefficient of first term of the series; therefore k =0 . Again equating to

zero the coefficient of general term xk+r

; we get

2 dn

dxn (e

-x ) 2

0

e-x2

Hn(x) Hm(x) dx = 2n n! mn

d2y

dx2

x + (1- x) dy

dx + n y = 0

y = arxk+r

r =0

ar[ (k+r)2 x

k+r-2 – (k+ r-n)x

k+r ] = 0

r =0

ar+1 =

r-n

(r+1)2

ar

As k=0; we have

Substituting r=0,1,2,3…etc. in equation(1.81);we get


Therefore from equation (1.80); we have (k=0);

(1.82)

y=

In case „n‟ is a positive integer and a0 =1 the series terminates after nth

term, the solution

is said to be Laguerre polynomial of degree „n‟ and is denoted by;

(1.83)

Then the solution of Laguerre equation for „n‟ to be a positive integer is;

Y = ALn(x) (1.84)

5. 5. 2 Generating function for Laguerre polynomials

The generating function for Laguerre polynomials is

f( x,z) = (1.85)

This function generates all the Laguerre‟s polynomials and hence it is called the

generating function of Laguerre polynomials.

5. 5. 3 Rodrigue‟s formula for Laguerre Polynomials The Rodrigue‟s formula for Laguerre polynomials is,

(1.86)

This representation is specially useful in finding the Laguerre polynomials. Substituting

n=0,1,2,3…etc. in equation (2); we get,

L0(x) =1, L1(x) =[1-x] , L2(x) = [(2-4x+x2)/2!], L3(x) = (6-18x+9x

2)/3! (1.87)

5. 5. 4 Recurrence relations for Laguerre polynomials

ar+1 =

k + r-n

(k+r+1)2 ar

ar = (-1)r

n(n-1)(n-2)….(n-r+1)

(r!)2

a0

a0 (-1)

r n!

(r!)2(n-r)!

r =0

xr

Ln(x) = (-1)

r n

(r!)2(n-r)!

r =0

x

r =

(-1)r n

(r!)2(n-r)!

r =0

x

r

ex

dn

n! dxn

Ln(x) = (xn e

-x )

e-xz / (1-z)

(1-z) n=0

Ln(x)z

n for z<1 =

y = arxk+r

= r =0

arx

r

r =0

(i) (n+1)Ln+1(x) = (2n+1-x)Ln(x) -nLn-1(x) (1.88)

Proof

We have from the property of generating function of Laguerre polynomials,


Differentiating both sides with respect to „z‟; we get

Multiplying throughout by (1-z)2 and using (i); we get

Comparing coefficients of zn on either side, finally we get

(n+1) Ln+1(x) = (2n+1-x)Ln(x) -nLn-1(x)

(ii ) (1.89)

Proof –

We have

Differentiating with respect to „x‟, we get

Equating coefficients of zn on either side, we get

e-xz / (1-z)

(1-z) n=0

Ln(x)z

n for z<1 =

1

(1-z)2 e

-zx/(1-z)

1

(1-z) + e

-zx/(1-z)

-x

(1-z2)

n=0

Ln(x)nz

n-1 =

(1-z)

- x

n=0

Ln(x)nz

n-1 =

n=0

Ln(x)z

n

n=0

Ln(x)z

n (1-z

2 )

(1-x)

-

n=0

Ln(x)z

n

n=0

Ln(x)z

n+1 =

n=0

Ln(x)nz

n-1 n

n=0

Ln(x)z

n - 2 n

n=0

Ln(x)z

n + n

xLn(x) = nLn(x) – nLn-1(x)

e-xz / (1-z)

(1-z) n=0

Ln(x)z

n =

1

(1-z) e

-xz/(1-z)

-z

1-z n=0

Ln(x)z

n =

n=0

Ln(x)z

n e

-xz/(1-z)

-z

1-z = (1-z) Or,

Ln(x)zn+1

= Ln(x)zn - Ln(x)z

n+1

n=0

n=0

n=0

-

-Ln-1(x) = Ln(x) -Ln-1(x)

or, Ln(x) = Ln-1(x) – Ln-1(x) (ii)

Differentiating recurrence relation I with respect to x, we get

(n+1)Ln+1 = (2n+1-x)Ln(x) -Ln(x) –nLn-1(x) (1.90)


Replacing n by (n+1) in (ii); we get

Ln+1(x) =Ln(x) -Ln(x) (1.91)

Also from (ii)

Ln-1(x) = Ln(x) +Ln+1(x) (1.92)

Substituting values of Ln+1 and Ln –1 (iv) and (v); in (iii); we get,

(n+1)[Ln(x) -Ln(x)] = (2n+1-x)Ln(x) -Ln(x)-n[Ln(x) +Ln-1(x)]

Rearranging the coefficients of Ln(x),Ln(x) and Ln-1(x) and Ln-1(x); we get

xLn(x) = nLn(x) – nLn-1(x)

5. 5. 5 Orthogonal Property to Laguerre Polynomials

The Laguerre Polynomials do not themselves form an orthogonal set. However the

related set of functions

(x) =e-(x/2)

Ln(x)

form an orthogonal set for the interval 0 x , i.e.

i.e. (1.93)

0

m(x)n(x) dx = 0

e-x/2

Lm(x).e-x/2

Ln(x) dx = mn

0

e-x

Lm(x).e-x/2

Ln(x) dx = mn




i) State Hermite Polynomials and Generating function?

ii) Write the Rodrigue‟s formula for the Laguerre‟s polynomials?

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5.6 REPRESENTATION OF VARIOUS FUNCTIONS IN TERMS OF

HYPERGEOMETRIC FUNCTION

Various familiar function of mathematical physics may be expressed as the particular

cases of confluent hypergeometric functions corresponding to suitable choices of

parameters,, and variable x.

5.6.1 Legendre Polynomials-

(i) From Rodrigue‟s formula polynomials, we have

Expanding the above equation using the Leibneitz theorem we get,

So, Pn(x) = 2F1 (1.94)

(ii) in (1.94); we get

Pn (cos ) = (1.95)

Substituting x = -cos in equation(1.94); we get

(-1)nPn (cos) = [as, Pn(-x) = (-1)

n Pn(x)](1.96)

As, Pn (cos) = (1.97)

(Since in hypergeometric series and can be interchanged)

1

2nn!

dn

dxn (x2-1)n = Pn(x)

1

n!

dn

dxn [{(x-1)n.1/2(x+1)n}] =

1

n!

dn

dxn [(x-1)n.{1-(x+1)n}] =

= 1+ (1-x)

2 (-n).(n+1)

1.1!

(-n).(-n+1)(n+1)(n+2)

2.1.2! + (1-x)

2

2 +……

x=cos ,

-n, n+1, 1, 1-x

2

2

2F1 -n, n+1, 1, 2

sin 2

Pn ( -cos ) =

2F1 -n, n+1, 1, 2

cos 2

2F1 -n, n+1, 1, 2

cos 2

(-1)n

(-1)n

2F1 -n, n+1, 1, 2

cos 2

Equation (1.94),(1.95),(1.96) and (1.97) represent Legendre polynomials in terms of

hypergeometric function.

5.6.2 Laguerre Polynomials

The Laguerre polynomials are defined as,

(1.98)


But

Or,

5. 6. 3 Bessel‟s Function - Bessel‟s differential equation

(1.99)

If we substitute y=zn e

-iz u in this equation, we get

zu+ [(2n+1) –2iz]u -i(2n+1)u=0

which is a confluent hypergeometric function for =n+1/2, = 2n+1 and x=2iz.

Therefore, its solution is

(1.100)

U=1F1(n+1/2, 2n+1,2iz)

The solution of equation (8) which remains finite at z=0 is the Bessel function of

first kind .Therefore, we have

Jn = czn e

-iz 1F1(n+1/2,2n+1,2iz)

Substituting C = ; we get

Jn = zn e

-iz 1F1(n+1/2,2n+1,2iz) (1.101)

This is required expression for Bessel function in terms of confluent hypergeometric

function.

= (-1)

r n!

(r!)2(n-r)!

r =0

x

r Ln(x)

(-1)r (-1)

rn(n-1)(n-2)….(n-r+1)

(n-r)! n! =

(-n) (-n+1)(-n+2) …(n+r-1)

n! = =

(-n)r

n!

= (-n)

r n!

(n!) (r)

2!

r =0

x

r Ln(x) =

(-n)r

(r)

2!

r =0

x

r

= (-n)

r

(r)

2!

r =0

x

r = 1F1(-n,1,x)

d2y

dx2

x2 + x dy

dx + (x2-n2) y = 0

1

2n n!




i) State the Bessel‟s function in terms of Hypergeometric function?

…………………………………………………………………………………………

…………………………………………………………………………………………


5.7 PROBLEMS ON POLYNOMIALS AND RELATED FUNCTIONS

Prob-1. Express the electrostatic between two electric charges, at a distance„d‟ apart, as a

series of Legendre polynomials.

Solution - The electric force between two electric charges at the distance„d‟ is

proportional to 1/a2 and the associated electronic potential is proportional to 1/d, i.e.

(i)

Where K is an appropriate constant. Let the two charges at the distance r and R as shown

in fig.1 .Then the distance between them is

d = R- r= [R2 – 2Rr cos + r

2]

1/2 = R[1-2r/R cos + (r/R)

2] (ii)

Then equation (i) becomes

For r=r< R, we may make the change of variables.

Then equation (iii) becomes,

(iii)

(Using the generating function of Legendre polynomial)

In terms of r and this equation becomes

(iv)

K

d V=

V= K

R 1 - cos +

2r

R

r

R

2 -1/ 2

r

R h= ; x = cos

V= K

R 1 - 2hx + h

2

-1/ 2

= K

R Pn(x) h

n

n=0

V= K

R Pn (cos ) (r

n/R

n)

n=0

In many applications the distances R is much larger than r and the terms in the series

decreases rapidly in magnitude due to the (r/R)n. Thus the potential can be approximated

by using only a few terms of the series.

Prob.2:-Find the upper and lower bonds for Legendre polynomial of order n.

Solution :

The series solution of the Legendre polynomials may be written as


Pn(cos) = , where all the am are positive (i)

Thus Pn (cos) is numerically greatest when each cosine in this series is maximum i.e.

unity (i.e. for =0). But =0. Pn (cos ) =Pn(1) =1. (ii)

It follows that Pn(cos)1,n=0,1,2….or -1 Pn(cos), which gives upper and lower

bonds for Legendre polynomials.

Prob.-3 If (u, , z) and (r,,) be the cylindrical and spherical co-ordinates of the same

point and if = cos, Prove that

Pn() =

Solution :- In the given co-ordinate systems r2 = x

2+ y

2 +z

2 = u

2+z

2.

So,

(i)

Which is a function of u and z .

By Taylor theorem , we may have

(ii)

But by the equation (i), we have

= [r2 –2r cos k- k

2]

-1/2 since z=r cos

= r-1

[ 1-2cos (k/r) +(k/r)2]1/2

= (1/r) (k/r)n Pn cos (iii)

Thus, from equation (ii) and (iii), we have

m=0

am cos m

(-1)n r

n+1

n

n! zn

1

r

1

r = (u

2 + z

2)-1

= (u, z)

(u, z -k) =(u,z) –k (u,z) + k2/2! (u,z)….+ (-1)

n (k

n/n!) (u,z)

z

2

z2

n

zn

n=0

(-1)n

(kn/n!) (u,z)

n

zn

=

(u,z-k) = [ u2+ (z-k)

2]

-1/2 = [u

2 +z

2 – 2zk+k

2]

-1/2

n=0

n=0

(-1)n

(kn/n!) (u ,z)

n

zn

= (1/r) (k/r)n Pn cos = n=0

Equating the coefficients of kn both sides,

Pn(cos) = (iv)


Prob:-4 Find the solution of differential equation of cylindrical wave

Solution: Assuming its solution in the form u= ( r) eit

(i)

Then the equation becomes

Putting r= cx/, this equation reduces to

, which is Bessel‟s differential equation of zeroth

order .Its general solution then becomes 0 = ao J0(x).

Prob. 5. - When calculating the dependence of the current density upon the distance

from the axis, we come across a scalar equation of the following form in cylindrical co-

ordinates:

Find a solution for „u‟ which is a periodic function of the time.

Solution -

Suppose that u= ()eit

is a solution; then

And ;also

Substituting these values in the given equation, we get

(-1)n

rn+1

n!

zn (u,z) (-1)

n

rn+1

n!

zn

(1/r) =

2u

r2

+ u

r

1

r = 1

c2

2u

r2

2

r2

+

r

1

r =

2

c2

2

r2

2

r2

+

r

1

x + =0

2u

2

+ u

1

=

4

c2

u

t

du

d

d

d e

it =

2u

2

d2

d2

eit

= du

dt = i()e

it

d2

d2 +

1

d

d + k

2 =0 4i

c2

; where k2 =

Now in order to change independent variables, let us out x=k, so that (dx/d) =k.

And,

With these substitutions equations (1) reduce to

, which is the Bessel‟s differential equation of zeroth


order .Its solution is,

Hence u=a0J0(x).eit

where a0 is the amplitude factor.

Which is the required solution.

5. 8 LET US SUM UP

The differential equations are of two types (i) Linear and (ii) non-linear

equation. The non-linear equations of second and higher orders.

Some of the differential equations used in applied physics are (i) Legendre

differential equation (ii) Bessel‟s differential (iii) Hermite differential (iv)

Laguerre‟s Differential equation.


1. i) Legendre‟s Equation

The differential equation

or,

is called Legendre equation.

Solution of Legendre‟s equation in descending power of x - Let the solution of

Legendre‟s equation in descending powers of „x‟ may be put as,

d

d = k

d

dx

d2

d2 = k

2 d2

dx2

d2

dx2

+ 1

x d

dx + =0

= a0J0(x) =a0 1- x

2 x

4 x

6

22

22.4 2

2.4

2.6

2

+ - +…..

= a0J0(k).eit

= a0J0 {(-i )(4 )./c}.ei t

d2y

dx2

(1-x2) - 2x dy

dx + n (n+1) y = 0

dy

dx (1-x

2 ) + n (n+1) y = 0

d

dx

= ar xk-r

r =0

y = xk

[ a0 + a1x-1

+ a2 x-2

+ a2 x-3

+ ….+ ar x-r +…..]

ii) Generating Function of Legendre Polynomial

Legendre polynomials Pn(x) is the coefficient of Zn in the expansion

of [ 1-2xz+z2]

-1/2 in the ascending power of „z‟. i.e.,

[ 1-2xz+z2]

-1/ 2 =


2. i) Orthonormality of Bessel‟s functions

If and are the two roots of the equations Jn( ) = 0; then

The condition of orthogonality of Bessel‟s function over the interval (0, 1) with

weight function x is

With the condition of

normalization is both the above equations represent the condition of orthogonality

and may be written in the form of a single equation as

Where is Kronecker delta symbol defined as =1 for = and 0 for .

3. i) Hermite Polynomial

The Hermite polynomial of degree n, for „n‟ being a positive integer is;

From equation we observe that,

Hn(0) = (-1) n n!/{(n/2)!} if „n‟ is an integer.

Hn(0) = 0 if „n‟ is an odd integer .

Generating function of Hermite equation

The function is called the generating function of Hermite polynomials.

i.e. f(x,z) =

n=0

Pn(x)z n

0

1

Jn(x) Jn(x) x dx = 0 for

0

1

x[Jn(x)]2 dx = (1/2) J

2n+1(x)

0

1

[Jn(x) Jn(x) xdx] = (1/2) J2n+1()

Hn(n) = (-1)r

(2x) n-2r

n!

r!(n-2r)! r=0

p

Where p = n/2 if „n‟ is even

½( n-1) if „n‟ is odd

e2zx –z

n=0

Hn(x)

n! z

n = e2zx –z

2

2

ii) Rodrigue‟s formula for Laguerre Polynomials

The Rodrigue‟s formula for Laguerre polynomials is,


4.

i) Bessel‟s differential equation in terms of hypergeometric function

If we substitute y=zn e

-iz u in this equation, we get

zu+ [(2n+1) –2iz]u -i(2n+1)u=0

which is a confluent hypergeometric function for =n+1/2, = 2n+1 and x=2iz.

Therefore, its solution is

U=1F1(n+1/2, 2n+1,2iz)

The solution of Bessel‟s equation which remains finite at z=0 is the Bessel

function of first kind .Therefore, we have

Jn = czn e

-iz 1F1(n+1/2,2n+1,2iz)

Substituting C = ; we get

Jn = zn e

-iz 1F1(n+1/2,2n+1,2iz)

This is required expression for Bessel function in terms of confluent

hypergeometric function.

REFERENCES AND SUGGESTED TEXT BOOKS



ex

dn

n! dxn

Ln(x) = (xn e

-x )

d2y

dx2

x2 + x dy

dx + (x2-n2) y = 0

1

2n n!




UNIT6 INTEGRAL TRANSFORMS

Structure

6.0 Introduction

6.1 Objectives

6.2 Fourier Transform and its Properties

2.2.1 Infinite Fourier Sine and Cosine Transform

2.2.2 Properties of Fourier transform

6..3 Applications of Fourier transform

2.3.1 Solved Problems

6.4 Laplace Transform

2.4.1 Properties of Laplace Transform

2.4.2 Laplace Transform of some special transform

6.5 Solved problems

6.6 Let Us Sum Up


6. 0 INTRODUCTION

In mathematical physics we frequently use pairs of functions related by an expression of the

following form,

g() = f (t) k (,t) dt (2.1)

The function g() is called the integral transform of f (t), with in the kernel k (,t).The

integral Transform are useful in mathematical analysis and in physical applications.

There are different kinds of integrals transform depending on the choice of kernel k (,t)

and the range of integration. The transforms of the function f (t) for the kernels e-it

, e-t

,

t Jn (t), t -1

are called Fourier, Laplace Hankel and Melin‟s transform respectively; i.e.,

g() = f (t) e-it

dt (Fourier Transform)

(2.2)

g() = f (t) e- t

dt (Laplace Transform)

g() = f (t) t Jn (t) dt (Hankel Transform)

a

b

0

0

0

(2.3)


6. 1 OBJECTIVES

In this unit we shall discuss the Fourier and Laplace integral transforms since they are especially

useful in physical applications. After going through this unit you should be able to:

Understand the Fourier and Laplace integral transforms.

Apply these integral transform in various Physical and mathematical

problems.

Use Theses transforms in the theory of communication Engineering, solid

state physics etc. and evaluating certain integrals.

6. 2 FOURIER‟S TRANSFORM

If f (x) is the periodic function of x, then the Fourier integral of f (x) is defined as

(2.4)

This may be expressed as,

(2.5)

Where, g () =

(2.6)

The function g() is called the Fourier transform of f (t) and f (t) is called Fourier

inverse transform of g(). The integral in equation (2.6) transforms a time function

f (t) in to its equivalent frequency function g(); while integral process in equation

(2.5) is reverse one.

6.2.1 Infinite Fourier Sine and Cosine Transform

The Fourier transform of f(t) is given by

Integral Transform

f (x) = ei x d f (t) e- i xdt 1

2 - -

+ +

+ f (x) = ei x d g()d()

1

(2) -

f (t) e-i t dt

1

(2) -

+

f (t) e-i t dt 1

(2) -

+

g () =

(2.7)

( Replacing „ t‟ by „-t‟ in the first integral )

Now,

(2.8)

Thus Equation (3) gives,

Now using = Cos t and Sin t = ; we get

(2.9)

(2.10)

The integral in equation (5) is called infinite Fourier cosine transform and the integral

in equation (6) is called infinite Fourier sine transform and they are denoted by the

equations,

(2.11)

And

(2.12) Special Functions and Integral Transform

+

1

-

0 f (t) e-it dt +

o f(t) e-it dt =

+ 0

1

-

f (-t) e it dt + o f(t) e-it dt =

f (t) = f (-t) If function f (t) is even

-f (-t) If function f (t) is odd

+

1

f (t) (e it dt + e- i t) dt o

For even function f (t)

1

f (t) (e - it dt e i t) dt

o

+

For odd function f (t) g () =

(e it + e-it)

2

(e - it - eit)

2i

+

1

f (t) Cos t dt o


1

o

+ For odd function f (t)

g () =

f (t) Sin t dt .2

.2

gc() = (2/) f (t) Cos t dt

o

+

g s() = (2/) f (t) Sin t dt

o

+

The inverse Fourier transforms leads to functions;

(2.13)

And

(2.14)

Equations (2.11) and (2.13) form a pair of Fourier cosine transforms while equations

(2.12) and (2.14) form a pair of Fourier Sine transforms.

6.2.2 Properties of Fourier Transform

(1) Addition Theorem or Linearity Theorem

If f (t) = a1 f 1 (t) + a 2 f 2 (t)+…;then the Fourier transform of f (t) is given by,

g() = a1 g 1() + a 2 g 2() +… (2.15)

where g 1() , g 2()….are Fourier transforms of f 1(t), f 2(t),… and a1,a 2…are constants.

Proof The Fourier transform of function f (t) is given by ,

(2.16)

(2) Similarity Theorem or Change of Scale Property

If g() is the Fourier transform of function f (t), the Fourier transform of function f (at)

is,

Integral Transform

f (t) = (2/) gc() Cos t dt

o

+

f (t) = (2/) gs() Sin t dt

o

+

1

(2) -

+

f (t) e-i t dt g() =

1

(2) -

+

e- i t dt = [ a1 f 1(t) + a 2 f 2(t)+….]

a1 (2) -

+

e- i t dt

= f 1(t) + a2. f (t) e-i t dt +…… 1

(2) -

+

= a1 g 1() + a 2 g 2() +…

1

a a g

Proof

Denoting the Fourier transform of function f (t) by F.T.{ f (t) }, we have,

F.T. of { f (t) } =

Hence F.T. of { f (at) } =

Substituting y=at, in above integral, we get,

(2.17)

This theorem is well known in its application to wave forms and spectra, where

compression of time scale by given factor compresses the periods of all harmonics

components equally and therefore increases the frequency of every component by the

same factor.

(3) Shifting Property

If g() is the Fourier transform of function f (t), then the Fourier transform of

function f (t a) will be given by,e i a

g(); where „a‟ is any constant.

Proof:- By definition of Fourier transform,

F.T. of { f (t a)} =

Substituting (t a) =y; i.e. dt=d y; we have,

F.T. of { f (t a)} =

=

= e i a

g() (2.18) Special Functions and Integral Transform

f (t) e-i t dt 1

(2) -

+

= g()

1

(2) -

+ f (at) e-i t dt

1

(2) - f (y) e -i y/a dy

a 1

(2) -

+ f (y) e -i y/a dy

a = F.T. of { f (at) } =

1

(2) -

+

f (t a) e-i t dt

1

(2) -

+

f (y ) e-i (y a) dy

1

(2) -

+

f (y ) e-i y dy e i a

g ( -a) + g ( -a) 1

2

1

2 1

a g

According to this theorem if a given function be shifted in the positive or negative

direction by an amount „a‟, no Fourier component change in amplitude; but its Fourier

transform suffers phase changes.

(4) Modulation Theorem

If g() is the Fourier transform of function f (t), then the Fourier transform of

function f ( t) Cos at is given by;

Proof F.T. of { f (t) Cos at } =

(2.19)

( 5) Convolution Theorem The transform of a product of two functions is given by

a convolution integral.

Proof

Let f 1(t) and f 2(t) be the two given functions and their product functions f (t) i.e. f (t) = f1

(t) f2 (t) .

From definition F.T. of { f (t) } = (2.20)

If g1() is the Fourier transform of f1(t), then the Fourier inverse transverse g1() is,

(2.21)

Integral Transform

Substituting values of f1(t) from equation (2.21) in equation (2.20); we get

1

(2) -

+

f (t) Cos at e-i t dt

1

(2) -

+

f (t) e-i t dt = e+ iat + e- iat

2

1

2 = 1

-

+

e-i( -a)t f (t) dt+ e-i(+ a )t f (t) dt

1

2 = 1

-

+

e-i( -a)t f (t) dt+ e-i(+ a )t f (t) dt 1

-

+

= [ g (-a) + g ( + a)] 1

2

1

-

+

f1(t) f 2(t) e-it dt

1

-

+ e-i t d f1(t) = g1()

F.T. of { f (t) } = 1

-

+ f 2(t) e-it dt 1

-

+ e-i t d g1()

+

+ 1 f 2(t) e- i ( -)t dt d g1() =

(2.22)

Now the Fourier transform of f 2(t) is given by

Replacing by - in above equation; we get

(2.23)

Combining (2.22) and (2.23), the Fourier transform of f (t) becomes;

(2.24)

Thus the Fourier transform of a product of two functions f 1(t) and f2(t) is given by an

integral, known as convolution integral where the functions g1 () and g2 () are said to

convolve with each other .

( 6 ) Parsevals Theorem

The Fourier transform of a convolution integral is given by the product of transforms of

the convolving functions.

(2.25)

Where g () is the Fourier transform of a convolution and g1 () and g2 () are Fourier

transform of f 1(t) and f 2(t) respectively, we have,

Proof Let f (t) be given convolution integral, i.e.

(2.26)


The Fourier transform of f (t) is,

g2() = -

+

f 2(t) e-it dt

g2( - ) = -

+

f 2(t) e-i( - )t dt

1

-

+ g2( - ) d F.T. of { f (t) } = g1()

g () = g1 ()g2 ()

1

-

+ f1 (t) f 2(t - t) dt f (t) =

g () = F.T. of { f (t) } 1

f 2(t-t) e- i t dt dt f1(t) =

-

+

-

+

1 f 2(t-t) e- i (t -t) dt dt f1(t) =

+

+

(2.27)

If g1() and g2() are Fourier transforms of f1(t) and f2(t) respectively , we have,

(2.28)

(2.29)

BLOCK-3

Changing„t‟ by (t-t) in equation (2.29); we get

(2.30)

Hence from equations (2.28),(2.29) and (2.30) we have;

(2.31)

6. 3 APPLICATIONS OF FOURIER TRANSFORM

(i) Evaluation of integrals

Using Fourier transforms certain integrals may be evaluated.

Examples Using Fourier Sine and Cosine transforms evaluate,

And

Solutions Let us consider;

I1 = and (2.32) Integral Transform

(2.33)

1

-

+

e-i t dt g1() = f1(t)

1

-

+

e-i t dt g2() = f2(t)

1

-

+

e-i (t-t) dt g2() = f2(t-t)

g () = g1 ()g2 ()

0

+ cos nx a2 + n2

dn 0

+ n sin nx

a2 + n2

dn

0

+

e-ax cos nx dx

0

+

e-ax sin nx dx I 2 =

Integrating by parts ; we get .

I1= -

=

(2.34)

Similarly,

I2 = = (2.35)

Solving (2.34) and (2.35); we get

I1 = and (2.36)

Now choosing f (x) = e-ax

; the cosine and sine transform of f (x) are;

(2.37)

and ,

(2.38)

So that the Fourier inverse transformations yield ;

(2.39)

(2.40)

Equations (2.39) and (2.40) leads to the integral

(2.41)


1

a

0

+

e-ax sin nx dx e-ax cos nx

0

+ 1

a

1 n a

I2

e-ax sin nx

0

+ 1

a

-1

a + 0

+

e-ax cos nx dx n

a I1

a

a2

+ n2

I2 = n

a2

+ n2

2

0

+

e-ax cos nx dx gc(n) = = 2

a

a2 + n2

2

0

+

e-ax sin nx dx gs(n) = = 2

n

a2 + n2

f (x) = e-ax = 2

0

+ cos nx dn gc(n) =

2

a

a2 + n2 Cos nx dn

-

+

f (x) = e-ax = 2

0

+ sin nx dn gs(n) =

2

n

a2 + n2 sin nx dn

-

+

0

+ cos nx a2 + n2

dn 2a

= e-ax

0

+ n sin nx

a2 + n2

dn 2

= e-ax

(2.42)

(ii ) Solution of Boundary Value Problem :- The Fourier transforms may be applied

to solve certain boundary value problems arising in Physics and applied physical

conveniently .

Example

Solve , x > 0, t>0

Subject to the conditions

(i) u( 0,t) = 0

(ii) u(x,0) =

(iii) u(x, t) is bounded

BLOCK-3

Solutions:- The given differential equation is,

(2.43)

Taking Fourier transforms of both the sides,

or,

or,

Where, us = ; and as (2.44)

or, Integral Transform

( for as )

u (x, t) 2 u (x,t)

t x2

=

1; 0<x<1

0; x 1

u 2 u

t x2

=

0

+

sin x dx 2

u

t =

0

+

sin x dx 2

2 u

x2

0

+

u sin x dx 2

t =

0

+

sin x 2

u

x 0

u

x Cos x dx

us

t =

2

0

+

u

x Cos x dx

u sin x dx 2

0

+ u

x 0 x

us

t =

2

u Cos x dx

0 +

0

+ u sin x dx

= 0

+ u sin x dx

2

u (0,t) 2

x u o

(Using the boundary condition (i)

or,

(2.45)

This equation may also be written as,

(2.46)

Integrating equation (2.46) , we get

, , log A being the constant of integration.

or, (2.47)

At t=0, this gives , us (,0) =A (2.48)

Using (2.44), we have,

i.e.

(2.49)

Now equation (2.46) becomes,

(2.50)

Applying inverse Fourier sine transform,


= - 2 us,

u

t + 2 us = 0

u

us - 2 t =

log us = - 2 t + log A

us = A e - t 2

= 0

+

sin x dx 2

u ( x, 0) A = us (,0)

= 0

+ 1

1. sin x dx 2

+ 0

+

sin x dx 0.

= cos x

2

- 0

1

= 1- cos x

2

- uS e - t 2

= 0

+

sin x d 2

us us (x,t)

= 0

+

sin x d 2

1- cos x

2

e - t 2

0

= 1- cos x

2

sin x d

= 1- cos x

2

- A

(2.51)

This is required solution.

6. 3.1.Solved Examples

Ex. 1. Find the Fourier transform of,

(a)

(b) Use the results of (a) to evaluate

Solution.

(a) Fourier transform of function f (x) is ( by definition);

Integral Transform

For a given function;

f (x) = 1 for t< a

0 for t< a

-

+ Sin a Cos x

d

1

-

+

e-i t dt g() = f (x)

0

+

0

+

-a

+ a

a

+




i) Define Fourier‟s transform and state infinite sine and cosine transform.

ii) State the Linearity theorem of Fourier‟s transform?

…………………………………………………………………………………………

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Fourier transform of the given function is;

=

;

using k 0, we get g()

(b) From integral theorem, we have

Then from part (a) above, we have

d

(i)

R.H.S. of the above equation (i) is,

(ii) Special Functions and Integral Transform

( i.e. the second integral in (ii) vanishes as its integrand is odd function)

f (t) e-it dt = (o).e -it dt + 1.e-it dt + (0) .e-it dt

= 1. e-it dt -a

+ a

1

g() =

-a

+ a 1. e-it dt =

1

e-it

-i -a

+ a 1

(e ia - e - ia)

i

= 2

Sin a

k 0

= 2

limit K 0

a sin a

a 2

= .a

1

-

+

e-i t dt , then g() = f (t)

-

+ 1

ei t d , then f (t) = g ()

ei t d 1

f (t) =

-

+

2

Sin a

ei t d

1

f (t) =

-

+ Sin a

1

=

-

+ Sin a

(cos t + i sin t)

-

+

d 1

sina cos t

-

+

d i

sina sin t

+ =

-

+

d 1

sina cos t

=

f (t) =

1 for t< a

½ for t= a

0 for t> a

But,

Here we are taking f (t) = ½ for t = a, as in this case, ½ will be average value of to left

handed and right handed limits of the functions.

Integral Transform

6.4 LAPLACE TRANSFORM AND ITS PROPERTIES

The Laplace transform of a function F(t), denoted by L {F(t)} or by f (s)m is denoted as,

-

+ d 1

sina cos t

1 for t< a

½ for t= a

0 for t> a

=

-

+

d = sina cos t

for t< a

/2 for t= a

0 for t> a

0

a f (s) = L {F(t)} = Lim e-st F (t) dt = e-st F (t) dt

a 0




i) Deduce the value of

-

+ Sin

d

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(2.52)

Here the operator L is called the Laplace transformation operator. The parameter„s‟ may

be real or complex number; but generally it is taken to be real positive number. The

Laplace transform of a function F (t) exist only if the function satisfies the following

conditions.

(i) The function F(t) should be an arbitrary piecewise continuous function in

every finite interval and that F(t)= 0 for all negative values of „t.

(ii) The function F(t) should be exponential order.

This means that the finite integral F(t) dt need not to be exist i.e. F(t) may diverge

exponentially for large values of t; however if there is some constant s0 such that;

es0

t F(t) M , a is a positive constant. (2.53)

For sufficiently large t‟, then the function F(t) is said to be of exponential order s0 as

t.

6. 4.1 Properties of Laplace Transform

1. Linearity Property

If a1 and a2 are constants and the Laplace transforms of F1(t) and F2(t) are f1(s) and f2(s)

respectively, then the laplace transform of a1F1(t) + a 2F2(t) is given by a1f1(s) + a 2 f 2(s),

i.e.

L { a1 F1(t) + a 2F2(t)} = a1L { F1(t)} + a 2L { F2(t)} (2.54)

Proof:- Let L { F1(t)} = f1(s) =


And, Let L { F2 (t)} = f2(s) =

0

e-st F1 (t) dt 0

+

e-st [a1F1 (t) + a 2F2 (t)] 0

+

e-st F2 (t) dt 0

+

So, L { a1 F1(t) + a 2F2(t)} = dt

=

= a1{ f1(t)} + a1{ f1(t)} (2.55)

Generalizing this result

(2.56)

2. Change of Scale Property

If f (s) is the Laplace transform of F(t) then the Laplace transform of F(at) is

Proof :- We have

Therefore,

Substituting at=u, i.e. dt = du/a ; we get

(2.57)

4. First Translation (shifting) Property

If f (s) is the Laplace transform of F(t) then the Laplace transform of will be f ( s-

a).

(2.58)

Integral Transform

Proof :-

Then

=

e-st a1F1 (t) dt 0

+

+ e-st a 2F2 (t) dt 0

+

am Fm(t) m =1

n

L am L {Fm(t)} m =1

n

=

1 s

a a f

L { F (t) } = f (s) = e-st F (t) dt 0

+

L { F (at) } = f (s) = e-st F (at) dt 0

+

eat F(t)

L { F (t) } = f (s) = e-st F (t) dt 0

+

L {eat F (t) } = f (s) = e-st {e at F (t) }dt 0

+

F(t) e-( s - a) dt 0

+

= f ( s-a)

0

+ L { F (at) } = f (s) = e - (s/a) u F (u)

du

a

0

+

e - (s/a)u F (u) du = = 1 s

a f

(2.59)

Similarly it can be shown that

(2.60)

5. Second Transition Property (Heaviside shifting Theorem)

If L{F(t)} = f(s) and G(t) = ; then L {G(t)} = e-as

f (s).

BLOCK-3

Proof

= +

Substituting t=p + a , i.e. dt = dp ; we get ,

(2.61)

This result may be expressed as;

(2.62)

Where U(t- a) = is called Heaviside‟s unit step input function.

6. Derivatives of Laplace Transform

If f (s) is Laplace transform of F(t), then f (s)= and in general


Proof

We have f (s) =

f ( s+ a) = L { e -at F (t)

0 if t <a

F(t -a) if t >a

L { G (t) } = e-st G (t) dt 0

+

e-st G (t) dt 0

a e-st G (t) dt

a

+

= e-st .0. dt 0

a

+ e-st F (t-a) dt a

+

= 0 + e-st F (t-a) dt a

+

L { G (t) } = e-s( p+ a) F(p) dp a

+

= e-as

e-sp F(p) dp a

+

= e-as

f (s)

= e-as

f (s) L { F(t-a) U(t-a)}

1 if t >a

0 if t <a

d n f

d s n

L {t n F(t)} = (-1)

n f

n (s) = (-1)

n

e-st F (t) dt 0

+

d f

d s = L {-t F(t)}

Differentiating both sides with respect to s; we get,

=

i.e. (2.63)

Carrying out the process of differentiation „n‟ times ; we get ;

(2.64 )

6. 4. 2 Laplace Transform of Some Special functions

1. Gamma function

The Gamma function is defined as,

n = (2.65)

Integral Transform

and we have the Laplace transform

L {tn} = (2.66)

f (s) = L{ -t F(T)}

d f

d s (-t)e-st F (t) dt 0

+

= e-st [-t F (t)] dt 0

+

= f (s)

dn f

d sn = f n (s) = L{ (-1)

n tn F(T)} = (-1)

n L{t

n F(T)}

e-n xn-1 dx 0

+

e-st tn dt 0

+




i) State Laplace transform and the conditions of its existence ?

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Substituting s t=x, we get

L {tn} =

Or, L-1

, substituting n=1/2; we get

L-1

{ t-1/2

} [ since (1/2) = ] (2.67)

2. Dirac delta function

The dirac delta function (t) is given by

(2.68)

F(t) being a continuous function of variable „ t‟

Now, if F(t) =0 for t<a, then we have

(2.69)

Replacing F(t) by e-st

in equation (2); we get

(2.70)

L { (t-a)} = e-s a

(2.71)

If a=1, then L { (t)} = 1

6. 5 SOLVED EXAMPLES

Example-1:- Find the Laplace transform of;

(i) K (constant) (ii) t (iii) Kt (iv) tn, n0 (integer) (v) e

-at.

Solution:- From definition of Laplace transform,


(i) Here F(t) =k ( Constant)

f (s) = L{k} = e-st F(t) dt 0

+

e-st

-s t = 0

t =

= k

s

.

. . f (s) = L{k} = e-st k dt

0

+

= k

1

s n+1

e-x xn dx

0

+

= e-x 0

+ x s

n dx

s

(n+1)

s n+1

=

(n+1)

s n+1

= t

n

(1/2)

s = = (/s)

+

F (t) (t-a)dt = F(a) -

+

F (t) (t-a)dt = F(a) 0

+

e-st (t-a)dt = [ e-st]t = a = e-s a

0

(ii) Here F(t) = t

or, =

(iii) Here F(t) = k t

(iv) Here F(t) = tn ; where „n‟ is integer 0

(i)

From result (ii), we have;

(ii)

Differentiating with respect to s; we get

i.e. L {t2} =

(iii)

Differentiating (2) (n-1) times with respect to „s‟; we get,

(iv)

(v) Here f (t) = e-at

Integral Transform

Example-2. Find Laplace transform of

(i) Sin h (at) (ii) Cos h(at) (iii) eat Sin (t) (iv) e

at Cos (t)

e-st

-s t = 0

t =

.

. . L{t} = e-st t dt

0

+

= t +

0

e-st

-s dt

1 e-st

1

s -s s2

=

0

e-st

-s t = 0

t =

L{kt} = e-st k t dt 0

+

=k t +

0

e-st

-s dt 1 e

-st k

s -s s2

=

0

k =

L{tn} = e-st tn dt

0

+

e-st t dt

0

+

= 1

s2

e-st (–t2)dt 0

+

= - 2

s3

2

s3

L{tn} = e-st tn dt

0

+ n!

sn+1

=

L{e-at

} = e-st e - at dt 0

+

= e- (s+a)t dt 0

+ e

-(s+ a) t

-(s+a) = 0

= 1

(s+a)

Solution. We have

(i ) Here F(t) = Sin h (at)

(ii) Here F(t) = cos h (at)

If Re s>a (ii)

(iii) Here F(t) = eat sin (t)

, (iii)

Where Re s>a and „a‟ and are real.

Example-3. Find Laplace transform of the function f (t) = sin h at sin at

Solution. We have;


f (t) = sin h at sin at =

L{f(t)} = e-st F(t) dt 0

+

. . .

L{sin h (at)} = e-st sin h at dt 0

+ ea t

– e-a t

2 e-st dt

0

+

=

. . .

L{cos h (at)} = e-st cos h at dt

+ ea t

+ e-a t

2 e-st dt

0

+

=

[e-( s – a )t + e + ( s+ a) t] dt

+

0

1

2 = 1 1 1

2 (s-a) (s + a) =

s

(s2 – a

2)

=

+

. . .

L{sin h (at)} = e-st eat sin (t)dt 0

+

e-st dt 0

+

= (eat sin (t))

e -st dt 0

+

= eat (ei t

– e-i t

)

2i

[e -( s – a - i )t - e - ( s - a + i ) t] dt

+

0

1

2i =

= 1 1 1

2i (s-a- i) (s + a+ i) +

= 2i

2i{(s – a)2 + 2

=

{(s – a)2 + 2

ea t

– e-a t

2

ei a t

– e-i a t

2i .

1

4i

Substituting, a(1+i) = p and a(1-i) = q; we get,

L { sin h at sin at} =L

But L { e p i

} = and L {e-p i

} =

(i)

Now, p2 = a

2(1+ i)

2 = a

2 ( 1- 1+2i) = 2ia

2

q2 = a

2(1- i)

2 = a

2 ( 1- 1-2i) = -2ia

2

and - i( p2

– q2) = - i ( 2ia

2 + 2ia

2) = 4a

2

Also, (s2 – p

2) (s

2 – q

2) = (s

2 – 2ia

2) (s

2 + 2ia

2) = ( s

2 + 4a

2)

Substituting these values in (1), we get

Integral Transform

6. 6 LET US SUM UP

[e i a( 1+ i ) + e -i a( 1+ i ) - e - t a( 1- i ) - e t a( 1- i )] dt

=

1

4i [e i p + e - i p - e - t q - e t q] dt

1

4i [L {e i p} + L {e - i p} – L {e - t q} – L {e t q }] dt

=

1

(s-p)

1

(s + p)

1 1 1 1 1

4i (s-p) (s + p) (s + q) (s-q) + - - Thus, L { sin h (at) sin (at)} =

1 2s 2s

4i (s2-p

2 ) (s

2-q

2 )

- = = is ( p

2 – q

2 )

2s (s2 – p

2 )( s

2 – p

2 )

Thus, L {sin h (at) sin (at)} = 4a2s

2(s4+ 4a

2 )

2a2s

(s4+ 4a

2 )

=




i) Find the Laplace transform of Cosh (at)?

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Fourier Transform – If f(x) is the periodic function of x, then the function g ( )

defined as,

is called the Fourier transform of function f(t).

Infinite Fourier sine and cosine transform - If f(x) is the periodic function of x, then

the function g ( )

is called infinite Fourier cosine Transform and infinite Fourier sine Transform

respectively given in above equation.

Laplace Transform -The Laplace transform of a function F(t), denoted by L {F(t)} or by

f (s)m is denoted as,

Here the operator L is called the Laplace transformation operator.

6. 7 CHECK YOUR PROGRESS: THE KEY

1. i) This may be expressed as,

Where, g () =

The function g() is called the Fourier transform of f (t) and f (t) is called

Fourier inverse transform of g().


f (t) e-i t dt

1

(2) -

+ g ( ) =

+

1

f (t) Cos t dt o


1

o


g () =

f (t) Sin t dt .2

.2

0


a 0

+

f (x) = ei x d g()d() 1

(2) - f (t) e-i t dt

1

(2) -

+

and

(i)

(ii)

The integral in equation (i) is called infinite Fourier cosine transform and the

integral in equation (ii) is called infinite Fourier sine transform and they are

denoted by the equations,

ii) Addition Theorem or Linearity Theorem

If f (t) = a1 f 1 (t) + a 2 f 2 (t)+…;then the Fourier transform of f (t) is given by,

g() = a1 g 1() + a 2 g 2() +…

where g 1() , g 2()….are Fourier transforms of f 1(t), f 2(t),… and a1,a 2…are

constants

2. i) Put t=0 and a=1 in the result of solved problem 2.3.1( b ), we get

, as the integrand is „a‟ even function of „k‟, so

3.

i)The Laplace transform of a function F(t), denoted by L {F(t)} or by f (s)m is

denoted as,

Here the operator L is called the Laplace transformation operator. The parameter„s‟

may be real or complex number; but generally it is taken to be real positive number.

The Laplace transform of a function F (t) exist only if the function satisfies the

following conditions.

(i) The function F(t) should be an arbitrary piecewise continuous function in

every finite interval and that F(t)= 0 for all negative values of „t.

(ii) The function F(t) should be exponential order.

Integral Transform

+

1

f (t) Cos t dt o


1

o


g () =

f (t) Sin t dt .2

.2

-

+ sin

d =

0

+ sin

d = ; or 2 sin

d = /2 ;

0

+

0


a 0

0

4.

(ii) Here F(t) = cos h (at)

If Re s>a

REFERENCES AND SUEGGESTED TEXT BOOKS

1. Group theory and Quantum mechanics by M.Tinkam

2. Mathematical methods for Physicist by G.Arfken

3. Mathematical Physics for Physicist & Engineers by L.Pipes.

4. Mathematical Physics by Satyaprakash

5. Mathematical Physics by B.D.Gupta.

. . .

L{cos h (at)} = e-st cos h at dt

+ ea t

+ e-a t

2 e-st dt

0

+

=

[e-( s – a )t + e + ( s+ a) t] dt

+

0

1

2 = 1 1 1

2 (s-a) (s + a) =

s

(s2 – a

2)

=

+

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