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SUBJECT:MATHEMATICS SUBJECTCODE:10MAT31

DEPT OF MATHEMATICS/SJBIT

Engineering Mathematics – III

Unit 1: Fourier Series ……………………………1-25

Unit 2: Fourier Transforms…………………………26-38

Unit 3: Applications Of Partial Differential Equations….39-48

Unit 4: Curve Fitting & Optimization ……………….49-66

Unit 5: Numerical Methods I………………………….67-83

Unit 6: Numerical Methods II …………….………….84-105

Unit 7: Numerical Methods III……………………….106-119

Unit 8: Difference Equations & Z – Transforms……120-136

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UNIT- I

FOURIER SERIES

CONTENTS:

Introduction

Periodic function

Trigonometric series and Euler’s formulae

Fourier series of period 2

Fourier series of even and odd functions

Fourier series of arbitrary period

Half range Fourier series

Complex form of Fourier series

Practical Harmonic Analysis0

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UNIT- I

FOURIER SERIES

DEFINITIONS :

A function y = f(x) is said to be even, if f(-x) = f(x). The graph of the even function is always symmetrical about the y-axis.

A function y=f(x) is said to be odd, if f(-x) = - f(x). The graph of the odd function is always symmetrical about the origin.

For example, the function f(x) = x in [-1,1] is even as f(-x) = xx = f(x) and the

function f(x) = x in [-1,1] is odd as f(-x) = -x = -f(x). The graphs of these functions are shown below :

Graph of f(x) = x Graph of f(x) = x

Note that the graph of f(x) = x is symmetrical about the y-axis and the graph of f(x) = x

is symmetrical about the origin.

1. If f(x) is even and g(x) is odd, then

h(x) = f(x) x g(x) is odd h(x) = f(x) x f(x) is even h(x) = g(x) x g(x) is even

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For example,1. h(x) = x2 cosx is even, since both x2 and cosx are even functions2. h(x) = xsinx is even, since x and sinx are odd functions3. h(x) = x2 sinx is odd, since x2 is even and sinx is odd.

2. If f(x) is even, then

a

a

a

dxxfdxxf0

)(2)(

3. If f(x) is odd, then

a

a

dxxf 0)(

For example,

a

a

a

xdxxdx0

,cos2cos as cosx is even

and

a

a

xdx ,0sin as sinx is odd

PERIODIC FUNCTIONS :-

A periodic function has a basic shape which is repeated over and over again. The fundamental range is the time (or sometimes distance) over which the basic shape is defined. The length of the fundamental range is called the period.

A general periodic function f(x) of period T satisfies the condition

f(x+T) = f(x)

Here f(x) is a real-valued function and T is a positive real number.

As a consequence, it follows thatf(x) = f(x+T) = f(x+2T) = f(x+3T) = ….. = f(x+nT)

Thus,f(x) = f(x+nT), n=1,2,3,…..

The function f(x) = sinx is periodic of period 2 sinceSin(x+2n) = sinx, n=1,2,3,……..

The graph of the function is shown below :

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Note that the graph of the function between 0 and 2 is the same as that between 2 and 4 and so on. It may be verified that a linear combination of periodic functions is also periodic.

FOURIER SERIES

A Fourier series of a periodic function consists of a sum of sine and cosine terms. Sines and cosines are the most fundamental periodic functions.

The Fourier series is named after the French Mathematician and Physicist Jacques Fourier (1768 – 1830). Fourier series has its application in problems pertaining to Heat conduction, acoustics, etc. The subject matter may be divided into the following sub topics.

FORMULA FOR FOURIER SERIES

Consider a real-valued function f(x) which obeys the following conditions called Dirichlet’s conditions :

FOURIER SERIES

Series with arbitrary period

Half-range series Complex series Harmonic Analysis

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1. f(x) is defined in an interval (a,a+2l), and f(x+2l) = f(x) so that f(x) is a periodic function of period 2l.

2. f(x) is continuous or has only a finite number of discontinuities in the interval (a,a+2l).

3. f(x) has no or only a finite number of maxima or minima in the interval (a,a+2l).

Also, let

la

a

dxxfl

a2

0 )(1

(1)

la

a

n xdxl

nxf

la

2

,cos)(1

,.....3,2,1n (2)

la

a

n xdxl

nxf

lb

2

,sin)(1

,......3,2,1n (3)

Then, the infinite series

1

0 sincos2 n

nn xl

nbx

l

na

a (4)

is called the Fourier series of f(x) in the interval (a,a+2l). Also, the real numbers a0, a1, a2, ….an, and b1, b2 , ….bn are called the Fourier coefficients of f(x). The formulae (1), (2) and (3) are called Euler’s formulae.

It can be proved that the sum of the series (4) is f(x) if f(x) is continuous at x. Thus we have

f(x) =

1

0 sincos2 n

nn xl

nbx

l

na

a ……. (5)

Suppose f(x) is discontinuous at x, then the sum of the series (4) would be

)()(2

1 xfxf

where f(x+) and f(x-) are the values of f(x) immediately to the right and to the left of f(x) respectively.Particular CasesCase (i) Suppose a=0. Then f(x) is defined over the interval (0,2l). Formulae (1), (2), (3) reduce to

2 2

0

0 0

1 1( ) ( )cos ,

l l

n

na f x dx a f x xdx

l l l

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2

0

1( )sin ,

l

n

nb f x xdx

l l

,......2,1n (6)

Then the right-hand side of (5) is the Fourier expansion of f(x) over the interval (0,2l).

If we set l=, then f(x) is defined over the interval (0,2). Formulae (6) reduce to

a0 =

2

0

)(1

dxxf

2

0

cos)(1

nxdxxfan , n=1,2,….. (7)

2

0

sin)(1

nxdxxfbn n=1,2,…..

Also, in this case, (5) becomes

f(x) =

1

0 sincos2 n

nn nxbnxaa

(8)

Case (ii)

Suppose a=-l. Then f(x) is defined over the interval (-l , l). Formulae (1), (2) (3) reduce to

(9)

n =1,2,……

Then the right-hand side of (5) is the Fourier expansion of f(x) over the interval (-l , l).

If we set l = , then f(x) is defined over the interval (-, ). Formulae (9) reduce to

a0 =

dxxf )(

1

l

l

n

l

l

xdxl

nxf

la

dxxfl

a

cos)(

1

)(1

0

l

l

n xdxl

nxf

lb ,sin)(

1

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nxdxxfan cos)(

1

, n=1,2,….. (10)

nxdxxfbn sin)(

1 n=1,2,…..

Putting l = in (5), we get f(x) =

1

0 sincos2 n

nn nxbnxaa

Some useful results :

1. The following rule called Bernoulli’s generalized rule of integration by parts is useful in evaluating the Fourier coefficients.

.......3''

2'

1 vuvuuvuvdx

Here uu , ,….. are the successive derivatives of u and

,......, 121 dxvvvdxv

We illustrate the rule, through the following examples :

166

86

43

2

cos2

sin2

cossin

2222

2323

3222

xxxxx ee

xe

xe

xdxex

n

nx

n

nxx

n

nxxnxdxx

2. The following integrals are also useful :

bxbbxaba

ebxdxe

bxbbxaba

ebxdxe

axax

axax

cossinsin

sincoscos

22

22

3. If ‘n’ is integer, thensin n = 0 , cosn = (-1)n , sin2n = 0, cos2n=1

Problems

1. Obtain the Fourier expansion of

f(x) = x2

1in - < x <

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We have,

dxxdxxfa )(2

11)(

10

=

22

1 2xx

nxdxxnxdxxfan cos)(2

11cos)(

1

Here we use integration by parts, so that

002

1

2cos

)1(sin

2

1

n

nx

n

nxxna

n

n

nx

n

nxx

nxdxxb

n

n

)1(

sin)1(

cos

2

1

sin)(2

11

2

Using the values of a0 , an and bn in the Fourier expansion

1 1

0 sincos2

)(n n

nn nxbnxaa

xf

we get,

1

sin)1(

2)(

n

n

nxn

xf

This is the required Fourier expansion of the given function.

2. Obtain the Fourier expansion of f(x)=e-ax in the interval (-, ). Deduce that

2

2 1

)1(2cos

n

n

nech

Here,

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0

1 1

2sinh

axax

a a

ea e dx

a

e e a

a a

2 2

2 2

1cos

1cos sin

2 ( 1) sinh

axn

ax

n

n

a e nxdx

ea a nx n nx

a n

a a

a n

bn =

nxdxe ax sin

1

=

nxnnxa

na

e ax

cossin1

22

=

22

sinh)1(2

na

an n

Thus,f(x) =

1

221

22sin

)1(sinh

2cos

)1(sinh2sinh

n

n

n

n

nxna

nanx

na

aa

a

a

For x=0, a=1, the series reduces to

f(0)=1 =

1

2 1

)1(sinh2sinh

n

n

n

or

1 =

22 1

)1(

2

1sinh2sinh

n

n

n

or 1 =

22 1

)1(sinh2

n

n

n

Thus,

2

2 1

)1(2cos

n

n

nech This is the desired deduction.

3. Obtain the Fourier expansion of f(x) = x2 over the interval (-, ). Deduce that

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......3

1

2

11

6 22

2

The function f(x) is even. Hence 0

0

1 2( ) ( )a f x dx f x dx

=

0 0

32

3

22 xdxx

or 3

2 2

0

a

nxdxxfan cos)(

1

=

0

,cos)(2

nxdxxf since f(x)cosnx is even

=

0

2 cos2

nxdxx

Integrating by parts, we get

Also,

0sin)(

1nxdxxfbn since f(x)sinnx is odd.

Thus2

21

( 1) cos( ) 4

3

n

n

nxf x

n

22

21

2

21

14

3

1

6

n n

n

Hence, .....3

1

2

11

6 22

2

2

032

2

)1(4

sin2

cos2

sin2

n

n

nx

n

nxx

n

nxxa

n

n

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2,2

0,)(

xx

xxxf

Deduce that ......5

1

3

11

8 22

2

Here, a0 =

dxxf )(

1=

0

)(2

dxxf

0

0

cos)(2

cos)(1

2

nxdxxfnxdxxfa

xdx

n

since f(x)cosnx is

even.

=

0

cos2

nxdxx

=

1)1(2

cos1

sin2

2

02

n

n

n

nx

n

nxx

Also,

0sin)(

1nxdxxfbn , since f(x)sinnx is odd

Thus the Fourier series of f(x) is

nxn

xfn

n cos1)1(12

2)(

12

For x= , we get

)(f

n

nn

n cos1)1(12

2 12

or

1

2)12(

)12cos(22

2 n n

n

Thus,

12

2

)12(

1

8 n n

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or ......5

1

3

11

8 22

2

This is the series as required.


f(x) =

xx

x

0,

0,

Deduce that ......5

1

3

11

8 22

2

Here,

2

1 0

0

0

xdxdxa

1)1(1

coscos1

2

0

0

n

n

n

nxdxxnxdxa

n

n

n

nxdxxnxdxb

)1(211

sinsin1 0

0

Fourier series is

f(x) =

112

sin)1(21

cos1)1(11

4 n

n

n

n nxn

nxn

Note that the point x=0 is a point of discontinuity of f(x). Here f(x+) =0, f(x-)=- at x=0.

Hence 2

02

1)]()([

2

1 xfxf

The Fourier expansion of f(x) at x=0 becomes

12

2

12

]1)1[(1

4

]1)1[(11

42

n

n

n

n

nor

n

Simplifying we get,

......5

1

3

11

8 22

2

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as f(x) cos(nx) is even

6. Obtain the Fourier series of f(x) = 1-x2 over the interval (-1,1).

The given function is even, as f(-x) = f(x). Also period of f(x) is 1-(-1)=2

Here a0 =

1

1

)(1

1dxxf =

1

0

)(2 dxxf

= 2

1

0

1

0

32

32)1(

xxdxx

3

4

1

0

1

1

)cos()(2

)cos()(1

1

dxxnxf

dxxnxfan

= 1

0

2 )cos()1(2 dxxnx


1

0

322

)(

sin)2(

)(

cos)2(

sin12

n

xn

n

xnx

n

xnxan

= 22

1)1(4

n

n

1

1

)sin()(1

1dxxnxfbn =0, since f(x)sin(nx) is odd.

The Fourier series of f(x) is

f(x) =

12

1

2)cos(

)1(4

3

2

n

n

xnn

7. Obtain the Fourier expansion of www.vtuc

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f(x) =

2

30

3

41

02

3

3

41

xinx

xinx

Deduce that ......5

1

3

11

8 22

2

The period of f(x) is 32

3

2

3

Also f(-x) = f(x). Hence f(x) is even

2/3

0

2

2/3

0

2/3

2/3

2/3

0

2/3

2/3

2/3

0

0

3

2

3

2cos

3

4

3

23

2sin

3

41

3

4

3

2cos)(

2/3

2

2/3cos)(

2/3

1

03

41

3

4

)(2/3

2)(

2/3

1

n

xn

n

xnx

dxxn

xf

dxxn

xfa

dxx

dxxfdxxfa

n

= n

n)1(1

422

Also,

23

23

0

23

sin)(3

1dx

xnxfbn

Thus

f(x) =

122 3

2cos)1(1

14

n

n xn

n

putting x=0, we get

f(0) =

1

22)1(1

14

n

n

n

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or 1 =

......

5

1

3

11

8222

Thus, ......5

1

3

11

8 22

2

HALF-RANGE FOURIER SERIES

The Fourier expansion of the periodic function f(x) of period 2l may contain both sine and cosine terms. Many a time it is required to obtain the Fourier expansion of f(x) in the interval (0,l) which is regarded as half interval. The definition can be extended to the other half in such a manner that the function becomes even or odd. This will result in cosine series or sine series only.

Sine series :Suppose f(x) = (x) is given in the interval (0,l). Then we define f(x) = -(-x) in (-l,0). Hence f(x) becomes an odd function in (-l , l). The Fourier series then is

1

sin)(n

n l

xnbxf

(11)

where

l

n dxl

xnxf

lb

0

sin)(2

The series (11) is called half-range sine series over (0,l).

Putting l= in (11), we obtain the half-range sine series of f(x) over (0,) given by

1

sin)(n

n nxbxf

0

sin)(2

nxdxxfbn

Cosine series :

Let us define

in (0,l) given..... in (-l,0) …..in order to make the function even.Then the Fourier series of f(x) is given by

1

0 cos2

)(n

n l

xna

axf

(12)

where,

)(

)()(

x

xxf

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l

n

l

dxl

xnxf

la

dxxfl

a

0

0

0

cos)(2

)(2

The series (12) is called half-range cosine series over (0,l)

Putting l = in (12), we get

Problems :

1. Expand f(x) = x(-x) as half-range sine series over the interval (0,).

We have,

0

2

0

sin)(2

sin)(2

nxdxxx

nxdxxfbn


n

n

n

n

nx

n

nxx

n

nxxxb

)1(14

cos)2(

sin2

cos2

3

032

2

The sine series of f(x) is

1

3sin)1(1

14)(

n

n nxn

xf

2. Obtain the cosine series of

xx

xxxf

2,

20,

)( ),0( over

..1,2,3,ncos)(2

)(2

cos2

)(

0

0

0

1

0

nxdxxfa

dxxfa

where

nxaa

xf

n

nn

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2

0

02

2

02

2: ( )

2

2cos ( ) cosn

Solution a xdx x dx

a x nxdx x nxdx

Performing integration by parts and simplifying, we get

,.....10,6,2,8

2cos2)1(1

2

2

2

nn

n

na n

n

Thus, the Fourier cosine series is

f(x) =

......

5

10cos

3

6cos

1

2cos2

4 222

xxx

3. Obtain the half-range cosine series of f(x) = c-x in 0<x<c

Here

c

cdxxcc

a0

0 )(2

c

n dxc

xnxc

ca

0

cos)(2

Integrating by parts and simplifying we get,

nn n

ca )1(1

222

The cosine series is given by

f(x) =

122

cos)1(112

2 n

n

c

xn

n

cc

COMPLEX FORM OF FOURIER SERIES

The standard form of Fourier series of f(x) over the interval (α, α+2l) is :

1

0 sincos2

)(n

nn l

xnb

l

xna

axf

If we use Euler formulae

sincos ie i

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then we obtain a complex exponential Fourier series

n

xl

ni

necxf

)( ………..(*)

where21

( )2

nl i xl

nc f x e dxl

n = 0, 1, 2, 3, …….

Note:-(1)

Putting l in (*), we get Fourier series valid in ( , )l l as

n

xl

ni

necxf

)( ………..(**)

where

1( )

2

nl i xl

nl

c f x e dxl

Note:-(2)

Putting α = 0 and l in (*), we get Fourier series valid in (0,2 ) as

( ) inx

nn

f x c e

where

1( )

2inx

nc f x e dx

Note:-(3)

Putting l in (**), we get Fourier series valid in ( , ) as

( ) inx

nn

f x c e

where

2

0

1( )

2inx

nc f x e dx

Problems:

1. Obtain the complex Fourier series of the function f(x) defined by f(x) = x over the interval (- , ).Here f(x) is defined over the interval (- , ). Hence complex Fourier series of f(x) is

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n

inxnecxf )( (1)

where

dxxe

dxexfc

inx

inxn

2

1

)(2

1

n = 0, 1, 2, 3, …….

Integrating by parts and substituting the limits, we get

n

ic

n

n

)1( , n 0

Using this in (1), we get

n

inxn

en

ixf)1(

)( , n 0

This is the complex form of the Fourier series of the given function

2. Obtain the complex Fourier series of the function f(x) = eax over the interval (- , ).

As the interval is again (- , ) we find n

c which is given by

( )

( )

1 1( )

2 2

1 1

2 2 ( )

inx ax inxn

a in xa in x

c f x e dx e e dx

ee dx

a in

2 2

( 1) ( )sinh

( )

n a in a

a n

Hence the complex Fourier series for f(x) is

n

inxnecxf )(

n

inxn

ena

inaa22

)()1(sinh

HARMONIC ANALYSIS

The Fourier series of a known function f(x) in a given interval may be found by finding the Fourier coefficients. The method described cannot be employed when f(x) is not known explicitly, but defined through the values of the function at some equidistant

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points. In such a case, the integrals in Euler’s formulae cannot be evaluated. Harmonic analysis is the process of finding the Fourier coefficients numerically.

To derive the relevant formulae for Fourier coefficients in Harmonic analysis, we employ the following result : The mean value of a continuous function f(x) over the interval (a,b) denoted by [f(x)] is defined as The Fourier coefficients defined through Euler’s formulae, (1), (2), (3) may be redefinedas

b

a

dxxfab

xf )(1

)( 2

0

12 ( ) 2[ ( )]

2

a l

a

a f x dx f xl

.

212 ( )cos 2 ( ) cos

2

a l

n

a

n x n xa f x dx f x

l l l

212 ( )sin 2 ( )sin

2

a l

n

a

n x n xb f x dx f x

l l l

Using these in (5), we obtain the Fourier series of f(x). The term a1cosx+b1sinx is called the first harmonic or fundamental harmonic, the term a2cos2x+b2sin2x is called the

second harmonic and so on. The amplitude of the first harmonic is 21

21 ba and that

of second harmonic is 22

22 ba and so on.

Problems:

1. Find the first two harmonics of the Fourier series of f(x) given the following table :

x 0 3

32 3

43

5 2f(x) 1.0 1.4 1.9 1.7 1.5 1.2 1.0

Note that the values of y = f(x) are spread over the interval 0 x 2 and f(0) = f(2) = 1.0. Hence the function is periodic and so we omit the last value f(2) = 0. We prepare the following table to compute the first two harmonics.

x0 y = f(x) cosx cos2x sinx sin2x ycosxycos2

xysinx ysin2x

0 1.0 1 1 0 0 1 1 0 0

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We have

]sin[2sin)(2

]cos[2cos)(2

nxyl

xnxfb

nxyl

xnxfa

n

n

as the length of interval= 2l = 2 or

l=

Putting, n=1,2, we get

1.06

)3.0(2

6

2cos2]2cos[2

367.06

)1.1(2

6

cos2]cos[2

2

1

xyxya

xyxya

0577.06

2sin2]2sin[

0392.16

sin2]sin[

2

1

xyxyb

xyxyb

The first two harmonics are a1cosx+b1sinx and a2cos2x+b2sin2x. That is (-0.367cosx + 1.0392 sinx) and (-0.1cos2x – 0.0577sin2x)

2. Express y as a Fourier series upto the third harmonic given the following values :

60 1.4 0.5 -0.5 0.866 0.866 0.7 -0.7 1.2124 1.2124

120 1.9 -0.5 -0.5 0.866 -0.866 -0.95 -0.95 1.6454 -1.6454

180 1.7 -1 1 0 0 -1.7 1.7 0 0

240 1.5 -0.5 -0.5 -0.866 0.866 -0.75 -0.75 1.299 1.299

300 1.2 0.5 -0.5 -0.866 -0.866 0.6 -0.6 -1.0392 -1.0392

Total -1.1 -0.3 3.1176 -0.1732

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x 0 1 2 3 4 5y 4 8 15 7 6 2

The values of y at x=0,1,2,3,4,5 are given and hence the interval of x should be 0 x < 6. The length of the interval = 6-0 = 6, so that 2l = 6 or l = 3.The Fourier series upto the third harmonic is

l

xb

l

xa

l

xb

l

xa

l

xb

l

xa

ay

3sin

3cos

2sin

2cossincos

2 3322110

or

3

3sin

3

3cos

3

2sin

3

2cos

3sin

3cos

2 3322110 x

bx

ax

bx

ax

bx

aa

y

Put 3

x , then

3sin3cos2sin2cossincos2 332211

0 bababaa

y ……… (1)

We prepare the following table using the given values :

x =3

xy ycos ycos2 ycos3 ysin ysin2 ysin3

0 0 04 4 4 4 0 0 0

1 600 08 4 -4 -8 6.928 6.928 0

2 1200 15 -7.5 -7.5 15 12.99 -12.99 0

3 1800 07 -7 7 -7 0 0 0

4 2400 06 -3 -3 6 -5.196 5.196 0

5 3000 02 1 -1 -2 -1.732 -1.732 0

Total 42 -8.5 -4.5 8 12.99 -2.598 0www.vtuc

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0]3sin[2

667.2)8(6

2]3cos[2

866.0)598.2(6

2]2sin[2

5.1)5.4(6

2]2cos[2

33.4)99.12(6

2]sin[2

833.2)5.8(6

2]cos[2

14)42(3

1

6

2][2)]([2

3

3

2

2

1

1

0

yb

ya

yb

ya

yb

ya

yyxfa

Usingthesein(1),we get

xxxxx

y cos667.2

3

2sin866.0

3

2cos5.1

3sin)33.4(

3cos833,27

This is the required Fourier series upto the third harmonic.

3. The following table gives the variations of a periodic current A over a period T :

t(secs) 0 T/6 T/3 T/2 2T/3 5T/6 T

A (amp) 1.98 1.30 1.05 1.30 -0.88 -0.25 1.98

Show that there is a constant part of 0.75amp. in the current A and obtain the amplitude of the first harmonic.

Note that the values of A at t=0 and t=T are the same. Hence A(t) is a periodic function

of period T. Let us denote tT

2

. We have

]sin[22

sin2

]cos[22

cos2

][2

1

1

0

AtT

Ab

AtT

Aa

Aa

(1)

We prepare the following table:

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tT

t 2 A cos sin Acos Asin

0 0 1.98 1 0 1.98 0

T/6 600 1.30 0.5 0.866 0.65 1.1258

T/3 1200 1.05 -0.5 0.866 -0.525 0.9093

T/2 1800 1.30 -1 0 -1.30 0

2T/3 2400 -0.88 -0.5 -0.866 0.44 0.7621

5T/6 3000 -0.25 0.5 -0.866 -0.125 0.2165

Total 4.5 1.12 3.0137

Using the values of the table in (1), we get

0046.13

0137.3

6

sin2

3733.03

12.1

6

cos2

5.13

5.4

6

2

1

1

0

Ab

Aa

Aa

The Fourier expansion upto the first harmonic is

T

t

T

t

T

tb

T

ta

aA

2sin0046.1

2cos3733.075.0

2sin

2cos

2 110

The expression shows that A has a constant part 0.75 in it. Also the amplitude of the first

harmonic is 21

21 ba = 1.0717.

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UNIT-II

FOURIER TRANSFORMS

CONTENTS:

Introduction

Finite Fourier transforms and Inverse finite Fourier transforms

Infinite Fourier transform (Complex Fourier transform) and

Inverse Fourier transforms

Properties [Linearity, Change of scale, Shifting and Modulation

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Fourier cosine and Fourier sine transforms & Inverse Fourier

cosine and sine transforms

FOURIER TRANSFORMS

Introduction

Fourier Transform is a technique employed to solve ODE’s, PDE’s,IVP’s, BVP’s and Integral equations.The subject matter is divided into the following sub topics :

FOURIER TRANSFORMS

Infinite Sine Cosine ConvolutionFourier Transform Transform Theorem &Transform Parseval’s

Identity

Infinite Fourier Transform

Let f(x) be a real valued, differentiable function that satisfies the following conditions:

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andinterval,finiteevery in itiesdiscontinusimple

ofnumber finiteaonly haveor ,continuousarexfderivativeitsandf(x)1)

bydefinedisf(x)ofTransformFourier infiniteThe parameter.realzero-nonbelet Also,

exists.xfintegral the2)-

dx

dxexfxfFf xiˆ

provided the integral exists.

bydefinedis f̂F

by denotedf̂ofTransformFourier inverseThe Transform.Fourier just theor TransformFourier complex calledalsoisTransformFourier infiniteThe

1-

defxffF xi

ˆ2

1ˆ1

Note : The function f(x) is said to be self reciprocal with respect to Fourier transform

ff ˆif .

Basic Properties

1. Linearity Property

For any two functions f(x) and (x) (whose Fourier Transforms exist) and any two constants a and b,

xbFxfaFxbxafF

Proof :

By definition, we have

dxexbxafxbxafF xi

xbFxfaF

dxexbdxexfa xixi

This is the desired property.

In particular, if a = b = 1, we get xFxfFxxfF

Again if a= -b = 1, we get xFxfFxxfF

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2. Change of Scale Property

have wea,constant zero-nonany for then ,xfFf̂If

aa

f̂

1xfF

Proof : By definition, we have

)1(dxeaxfaxfF xi

Suppose a > 0. let us set ax = u. Then expression (1) becomes

a

dueufaxfF

ua

i

)2(ˆ1

af

a

Suppose a < 0. If we set again ax = u, then (1) becomes

a

dueufaxfF a

ui

dueufa

ua

i

1

)3(ˆ1

af

a

Expressions (2) and (3) may be combined as

af

aaxfF

ˆ1

This is the desired property3. Shifting Properties

For any real constant ‘a’,

feaxfFi ai ˆ)(

afxfeFii iax ˆ)(Proof : (i) We have

dxexffxfF xi

ˆ

Hence, dxeaxfaxfF xi

Set x-a = t. Then dx = dt.Then,

dtetfaxfF ati )(

= aie dtetf ti

= aie f̂ii) We have

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dxexfaf xai

ˆ

dxeexf xiiax

iaxiax exfxgwheredxexg )()(,

xgF

xfeF iaxThis is the desired result.

4. Modulation Property

,ˆ fxfFIf

afafaxxfFthen ˆˆ2

1cos,

where ‘a’ is a real constant.

Proof : We have

2

cosiaxiax ee

ax

Hence

2cos

iaxiax eexfFaxxfF

property.desired theisThis

.propertiesshift andlinearity usingby ,ˆˆ2

1afaf

Note : Similarly

afafaxxfF ˆˆsin2

1

Problems:0whereef(x)function theofTransformFourier theFind1. x-a a

For the given function, we have

dxeexfF xixa

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dxeedxee xixaxixa

0

0

get we0, x -x,-xand x 0 x,xfact that theUsing

dxeedxeexfF xiaxxiax

0

0

dxedxexiaxia

0

0

0

0

ia

e

ia

e xiaxia

iaia

11

22

2

a

a

2. Find the Fourier Transform of the function

ax

ax

0,

1,f(x)

where ‘a’ is a positive constant. Hence evaluate

d

xai

cossin)(

dii

0

sin)(


dxexfxfF xi)(

dxexfdxexfdxexf xi

a

xia

a

xia )()()(

dxe

a

a

xi

asin

2

)1(sin

2ˆ

a

fxfFThus

get weformula,inversion employingby f̂Inverting

de

axf xi

sin

22

1

d

xixa sincossin1

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d

xaid

xa sinsincossin1

Here, the integrand in the first integral is even and the integrand in the second integral is odd. Hence using the relevant properties of integral here, we get

d

xaxf

cossin)( 1

or

)(

cossinxfd

xa

ax

ax

0,

,

d

sinyields this1,a0,For x

Since the integrand is even, we have

0

sin2 d

or

20

sin

d

Transform.Fourier respect to with reciprocalselfis22x

ef(x) that Deduce

constant.positiveaisa'' where2x2a-ef(x)ofTransformFourier theFind3.

Here

dxeexfF xixa 22

dxe xixa 22

dxeaa

iax

2

22

42

dxee a

iax

a

2

2

2

24

getwe,2a

i-ax t Setting

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a

dteexfF ta

22

2

4

dteea

ta

0

422

2

21

function.gammausing,1 2

2

4

aea

2

2

4ˆ aea

f

This is the desired Fourier Transform of f(x).

2

2x-

2

2

2

2f̂

hence,andef(x)get we

2x2a-ef(x)in 21aFor

e

getwe,ef(x)in xputtingAlso 2x2- 2

2-e)f(

.

Hence, )f( and f̂ are same but for constant multiplication by 2 .

f̂)f(Thus

reciprocalselfis e)f( that followsIt 22

-x

x

FOURIER SINE TRANSFORMS

Let f(x) be defined for all positive values of x.

ThusxfFor

xdxxf

s .f̂by

denotedisThis f(x).ofTransformSineFourier thecalledis sinintegralThe

s

0

dxxxfxfFs sinf̂0s

dxf s sinˆ2integralethrough th

definedisf̂ofTransformsineFourier inverseThe

0

s

dxff

Thusf

ss

s

sinˆ2Ff(x)

.For f(x)by denotedisThis

0

1-s

-1s

Properties

The following are the basic properties of Sine Transforms.

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(1) LINEARITY PROPERTYIf ‘a’ and ‘b’ are two constants, then for two functions f(x) and (x), we

have xgbFxfaFxbxafF sss


dxxxbxafxbxafFs sin0

xbFxfaF ss This is the desired result. In particular, we have

xFxfFxxfF sss and

xFxfFxxfF sss

(2) CHANGE OF SCALE PROPERTY

have we0,afor then ,f̂xfFIf s s

aa s

f̂

1axfFs

Proof : We have

dxxaxf sinaxfF0s

Setting ax = t , we get

a

dtt

atf

sinaxfF

0s

aa s

f̂

1

(3) MODULATION PROPERTY

have we0,afor then ,ˆxfFIf s sf

afafaxxfF sss ˆˆcos2

1

Proof : We have

0

sincoscos dxxaxxfaxxfFs

dxxaxaxf sinsin

2

10

property.Linearity usingby ,ˆˆ

2

1 afaf ss

Problems:

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1. Find the Fourier sine transform of

ax

ax

,0

0,1xf


a

adxxdxx sin0sinf̂

0s

ax

0

cos

acos1

x

ef(x)of transformsineFourier theFind2.

-ax

Here

0s

sinf̂

x

dxxe ax

Differentiating with respect to , we get

0s

sinf̂

x

dxxe

d

d

d

d ax

0

sin dxxx

axe

performing differentiation under the integral sign

0

cos dxxxx

e ax

022

sincos xxaa

e ax

22

a

a

Integrating with respect to , we get

ca

1s tanf̂

00f̂But s when

c=0

a

1s tanf̂

3. Find f(x) from the integral equation

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2,0

21,2

10,1

sinxf0

xdx

Let () be defined by

2,0

21,2

10,1

Given

0

ˆsin s

fxdxxf

Using this in the inversion formula, we get

dxx

sin2

xf0

dxdxdx sinsinsin

2

2

1

1

02

2

1

1

0

2 0sin2sin

dxdx

xxx

2cos2cos12

FOURIER COSINE TRANSFORMS

dxxxf cosintegralThe x.of valuespositivefor definedbef(x)Let 0

Thus.For f̂by denotedisandf(x)ofTransformCosineFourier thecalledis cc xf

xdxxfxf

cos2

Ff̂0cc

Thus .or xfby denotedisThis .2

integralthe

through definedisofTransformCosineFourier inverseThe

cf̂

1-cFcosˆ

cf̂

0

dxfc

0

1-c cosˆ2ˆFxf

dxff c

Basic Properties

The following are the basic properties of cosine transforms :

(1) Linearity property

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xxfxafFhave we(x),andf(x)functionsfor two then constants, twoareb''anda''If

c

cc bFaFxb

(2) Change of scale property

aa

cc

cc

f̂1

axfF

have we0,afor then ,f̂xfFIf

(3) Modulation property

aaax

ccc

cc

f̂f̂2

1cosxfF

have we0,afor then ,f̂xfFIf

The proofs of these properties are similar to the proofs of the corresponding properties of Fourier Sine Transforms.

Problems: 1) Find the cosine transform of the function

2,0

21,2

10,

x

xx

xx

xf

We have

xdxxdxxxdxx

xdxxf

cos0cos2cos

cosf̂

2

2

1

1

0

0c


2

12

1

0

2c

cos1

sin2

cossinf̂

xx

xxx

x

2

12coscos2

2) -ax Find the cosine transform of f(x) e , a 0. Hence evaluate 2 20

cos kx

x adx

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Here

22c

0

22

0c

f̂

sincosa

cosf̂

a

a

Thus

xxae

xdxe

ax

ax

Using the definition of inverse cosine transform, we get

xda

x cosa

2f

0 22

or

0 22

cos

2

da

xe

aax

Changing x to k, and to x, we get

a

edx

a

kx ax

2x

cos0 22

3) Solve the integral equation

aedxxxf cos

0

Let () be defined by() = e-a

c0

f̂cosGiven

dxxxf

Using this in the inversion formula, we get

22

00 22

0

0

2

sincos2

cos2

cos2

f

xa

a

xxaxa

e

xde

xdx

a

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UNIT III Applications of Partial Differential Equations

CONTENTS:

Introduction

Various possible solutions of the one dimensional wave equation

Various possible solutions of the one dimensional heat equation

Various possible solutions of the two dimensional Laplace’s equation

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D’Alembert’s solutions of the one dimensional wave equation

APPLICATION OF PARTIAL DIFFERENTIAL EQUATIONS

Introduction

A number of problems in science and engineering will lead us to partial differential equations. In this unit we focus our attention on one dimensional wave equation ,one dimensional heat equation and two dimensional Laplace’s equation.Later we discuss the solution of these equations subject to a given set of boundary conditions referred to as boundary value problems.Finally we discuss the D’Alembert’s solution of one dimensional wave equation.

Various possible solutions of standard p.d.es by the method of separation of variables.We need to obtain the solution of the ODEs by taking the constant k equal to i) Zero ii)positive:k=+p2 iii)negative:k=-p2

Thus we obtain three possible solutions for the associated p.d.e

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Various possible solutions of the one dimensional wave equation utt =c2uxx by the method of separation of variables.

Consider 2 2

22 2

u uc

t x

Let u= XT where X=X(x),T=T(t) be the solution of the PDEHence the PDE becomes

2 2 2 22 2

2 2 2 2

XT XT d T d Xc or X c

t x dt dx

Dividing by c2XT we have 2 2

2 2 2

1 1d T d X

c T dt X dx

Equating both sides to a common constant k we have2

2

1 d X

X dx=k and

2

2 2

1 d T

c T dt=k

2

20

d XkX

dx and

22

20

d Tc kT

dt

2 0D k X and 2 2 0D c k T

Where D2 =2

2

d

dxin the first equation and D2 =

2

2

d

dtin the second equation

Case(i) : let k=0AEs are m=0 amd m2=0 amd m=0,0 are the roots Solutions are given byT = 0 0

1 1 2 3 2 3t xc e c and X c x c e c x c

Hence the solution of the PDE is given byU= XT= 1 2 3c c x cOr u(x,t) =Ax+B where c1c2=A and c1c3=BCase(ii) let k be positive say k=+p2

AEs are m –c2p2=0 and m2-p2=0m= c2p2 and m=+psolutions are given by

2 2' ' '1 2 3

c p t px pxT c e andX c e c e Hence the solution of the PDE is given by

2 2'1

c p tu XT c e .( ' '2 3

px pxc e c e )

Or u(x,t) =2 2'

1c p tc e (A’ pxe +B pxe ) where c1’c2’=A’ and c1’c3’=B’

Case(iii): let k be negative say k=-p2

AEs are m+ c2p2=0 and m2+p2=0m=- c2p2 and m=+ipsolutions are given by

2 2'' '' ''1 2 3cos sinc p tT c e and X c px c px

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Hence the solution of the PDE is given by2 2'' '' ''

1 2 3.( cos sin )c p tu XT c e c px c px 2 2 '' ''( , ) ( cos sin )c p tu x t e A px B px

Various possible solutions of the two dimensional Laplace’s equation uxx+uyy=0 by the method of separation of variables

Consider 2 2

2 20

u u

x y

Let u=XY where X(x), Y=Y(y) e the solution of the PDEHence the PDE becomes

2 2

2 2

( )0

XY XY

x y

2 2

2 2

( )0

d X d YY X and dividing by XY we have

dx dy

2 2

2 2

1 1 ( )d X d Y

X dx Y dy

Equating both sides to a common constant k we have

2

2

1 d X

X dx=k and

2

2

1 ( )d Y

Y dy =k

Or (D2-k)X=0 and (D2+k)Y=0

Where D =d

dxin the first equation and D =

d

dyin the second equation

Case(i) Let k=0AE arem2=0 in respect of both the equationsM=0,0 and m=0,0Solutions are given byX= c1x+c2 and Y= c3y+c4

Hence the solution of the PDE is given byU=XY=( c1x+c2 ) (c3y+c4)

Case(ii) : let k be positive ,say k=+p2

m2-p2=0 and m2+p2=0m=+p and m=+ipsolutions are given by

' ' ' '1 2 3 4cos sinpx pxX c e c e and Y c py c py

Hence the solution of the PDE is given by

' ' ' '1 2 4( 3cos sin )px pxu XY c e c e c py c py

Case(iii) Let k be negative say k= -p2

AEs are m2+p2=0 and m2-p2=0

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M=+ip and m=+pSolutions are given by

'' ''1 2( cos sin )X c px c px and '' ''

3 4( )py pyY c e c e Hence the solution of the PDE is given by

'' '' '' ''1 2 3 4( cos sin )( )py pyu XY c px c px c e c e

EXAMPLES

1.Solve the wave equation utt=c2uxx subject to the conditions

u(t,0)=0 ,u(l,t)=0, ,0 0u

xt

and u(x,0) =u0sin3( x/l)

Soln: 1

, sin cosnn

n x n ctu x t b

l l

Consider u(x,0) =u0sin3( x/l)

1

,0 sinnn

n xu x b

l

30

1sin sinn

n

x n xu b

l l

30

1

3 1 3sin sin sin

4 4 nn

x x n xu b

l l l

comparing both sides we get

01

3

4

ub , 2 0b , 0

3 4 5, 0 04

ub b b

,

Thus by substituting these values in the expanded form we get

0 03 3 3( , ) sin cos sin cos

4 4

u ux ct x ctu x t

l l l l

2.Solve the wave equation utt=c2utt subject to the conditions

u(t,0)=0 ,u(l,t)=0, ,0 0u

xt

when t=0and u(x,0) =f(x)

Soln: : 1

, sin cosnn

n x n ctu x t b

l l

Consider u(x,0)=f(x) then we have

Consider u(x,0) = 1

sinnn

n xb

l

F(x) = 1

sinnn

n xb

l

The series in RHS is regarded as the sine half range Fourier series of f(x) in (0,l) and hence

0 01 2 3

3 3 2 3sin sin sin sin sin

4 4

u ux x x x xb b b

l l l l l

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0

2( )sin

l

n

n xb f x dx

l l

Thus we have the required solution in the form

1

, sin cosnn

n x n ctu x t b

l l

3. Solve the Heat equation 2

22

u uc

t x

given that u(0,t)=0,u(l,0)=0 and u(x,0)=

100x/l

Soln: 0

2 100sin

l

n

x n xb dx

l l l

=

20

200sin

l n xx dx

l l

22

0

. cos sin2001

/ /

l

n

n x n xx

l lbl n l n l

1

2

200 1 200 1200 1. cos .

n n

nb l nl n n n

The required solution is obtained by substituting this value of nb

Thus 2 2 21

21

200 1( , ) sin

n n c t

n

n xu x t e

n l l

4.Obtain the solution of the heat equation2

22

u uc

t x

given that u(0,t)=0,u(l,t)and

u(x,0) =f(x)where

20

2( )2

2

Tx lin x

lf xT l

l x in x ll

Soln: 0

2( )sin

l

n

n xb f x dx

l l

2

02

2 2 2sin ( )sin

l

l

nl

Tx n x Tx n xb dx l x dx

l l l l l

2

02

4sin ( )sin

l

l

l

T n x n xx dx l x dx

l l l

2 2

8sin

2n

T nb

n

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The required solution is obtained by substituting this value of bn

Thus 2 2 2

2 2 21

8 1( , ) sin sin

2

n c t

n

T n n xu x t e

n l l

5. Solve the heat equation 2

2

u u

t x

with the boundary conditions u(0,t)=0,u(l,t)and

u(x,0) =3sin x

Soln: 2

( , ) ( cos sin )............................(1)p tu x t e A px B px Consider u(0,t)=0 now 1 becomes

0=2p te (A) thus A=0

Consider u(1,t)=0 using A=0 (1) becomes

0=2p te (Bsinp)

Since B≠0,sinp=0or p=n2 2 2

( , ) ( sin )n c t

u x t e B n x

In general 2 2 2

1

( , ) sinn c t

nn

u x t b e n x

Consider u(x,0)= 3sin n x and we have

3 1 2 3sin sin sin 2 sin 3n x b x b x b x

Comparing both sides we get 1 2 33, 0, 0b b b We substitute these values in the expanded form and then get

2

( , ) 3 (sin )t

u x t e x

6.Solve 2 2

2 20

u u

x y

subject to the conditions u(0,y)=0,u( ,y)=0 and u(x,∞) =0 and

u(x,0)=ksin2xSoln: The befitting solution to solve the given problem is given by

( , ) ( cos sin ) py pyu x y A px B px Ce De

( , ) 0 ( sin ) 0,py pyu y gives B p Ce De

Since A=0,B≠0 and we must havesin p =0,therefore p=n where n is a integer

( , ) ( sin ) ny nyu x y B nx Ce De

The condition u(x,∞) =0 means that u→as y→∞

(0, ) 0 ( ) 0, 0py pyu y gives A Ce De A

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0=(Bsinnx)( nyCe )since 0nye asy Since B≠0 we must have C=0

We now have u(x,y) =BDsinnx nye

Taking n= 1,2,3,……… and BD =b1 ,b2 ,b3 ,b4…………………

We obtain a set of independent solutions satisfying the first three conditions Their sum also satisfy these conditions hence

1

( , ) sin nyn

n

u x y b n xe

Consider u(x,0)=ksin2x and we have

1

( ,0) sinnn

u x b nx

Ksin2x=b1sinx+b2sin2x+b3sin3x+…………….Comparing both sides we get b1=0, b2=0, b3=0Thus by substituting these values in the expanded form we get

2( , ) sin 2 yu x y k xe

D’Alembert’s solution of the one dimensional wave equation

We have one dimensional wave equation

2 2

22 2

u uc

t x

Let v=x+ct and w=x-ctWe treat u as a function of v and w which are functions of x and tBy chain rule we have,

. .u u v u w

x v x w x

Since v=x+ct and w=x-ct, 1 1v w

andx x

.(1) .(1)u u u u u

x v w v w

2

2

u u u u

x x x x v w

Again by applying chain rule we have,

2

2. .

u u u v u u w

x v v w x w v w x

2 2 2 2 2

2 2 2.(1) .(1)

u u u u u

x v v w w v w

But 2u

v w

=2u

w v

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But 2 2 2 2

2 2 22

u u u u

x v w v w

Similarly . .u u v u w

x v t w t

Since v=x+ct and w=x-ct, v w

c and ct t

.( ) .( ) ,u u u u u u

c c or ct v w t v w

Again applying chain rule we have,

2

2. .

u u u u v u u wc c

r t t v w v w t w v w t

2 2 2 2 2

2 2 2.( ) .( )

u u u u uc c c c

r v v w w v w

2 2 2 2 22 2

2 2 2

u u u u uc c

t v v w w v w

2 2 2 22

2 2 22

u u u uc

t v w v w

2 2 22

2 22

u u uc

v w v w

=2 2 2

22 2

2u u u

cv w v w

2

4u

w v

=0 or

2u

w v

=0

We solve this PDE by direct integration, writing it in the form,

0u

w v

Integrating w.r.t w treating v as constant we get ( )u

f vv

Now integrating w.r.t, we get ( ) ( )u f v dv G w U=F(v)+G(w) where F(v)= ( )f v dvBut v=x+ct and w=x-ctThus u=u(x,t)=F(x+ct)+G(x-ct)This is the D’Alembert’s solution of the one dimensional wave equation

EXAMPLES

1. Obtain the D’Alembert’s solution of the wave equation utt=c2uxx subject to the

conditions u(x,0)=f(x) and ( ,0) 0u

xt

Soln: D’Alembert’s solution of the wave equation is given by

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u(x,t)=F(x+ct0=G(x-ct)………………………………………..(1)Consider u(x,0)=f(x) (10 becomesu(x,0)=F(x)+G(x)f(x)=F(x)+G(x)…………………………………………………….(2)Differentiating partially wrt t we have

u

t

(x,t) =F’(x+ct).(c)+G’(x-ct).(-c)

u

t

(x,)=c[F’(x)-G’(x)]

0= c[F’(x)-G’(x)][F’(x)-G’(x)]=0 integrating wrt x we have[F(x)-G(x)]=kWhere k is the constant of integration……………………………(3)By solving simultaneously the equations,[F(x)+G(x)]=f(x)[F(x)-G(x)]=k

We obtain F(x)= 1 1( ) ( ) ( )

2 2f x k andG x f x k

Thus F(x+ct)= 1( )

2f x ct k and G(x-ct)= 1

( )2

f x ct k

Substituting these in(10 we have

U(x,t)= 1 1( ) ( )

2 2f x ct k f x ct k

Thus the required solution is

u(x,t)= 1( ) ( )

2f x ct f x ct

2. Obtain the D’Alembert’s solution of the wave equation utt=c2uxx given that u(x,0)=f(x)=l2-x2and ut(x,0)=0

Soln. Assuming the D’Alembert’s solution

u(x,t)= 1( ) ( )

2f x ct f x ct and f(x)=l2-x2

u(x,t)= 2 2 2 21( ) ( )

2l x ct l x ct

2 2 2 212 2 2

2l x c t

u(x,t)= 2 2 2 2l x c t

3. . Obtain the D’Alembert’s solution of the wave equation utt=c2uxx given that u(x,0)=asin2πx and

0 0u

when tt

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Soln: Assuming the D’Alembert’s solution

u(x,t)= 1( ) ( )

2f x ct f x ct

By data f(x)= asin2πx or f(x)= 1 cos 22

ax

u(x,t)= 1. 1 cos2 ( ) 1 cos 2 ( )

2 2

ax ct x ct

u(x,t)= 2 cos 2 2 cos 2 24

ax ct x ct

2 2cos2 cos 24

ax ct

u(x,t)= 1 1cos2 cos22

ax ct

UNIT IVCURVE FITTING AND OPTIMIZATION

CONTENTS:

Curve fitting by the method of least squares

Fitting of curves of the form

y ax b

2y ax bx c

bxy ae

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by ax

Optimization :

Linear programming

Graphical method

Simplex method

CURVE FITTING AND OPTIMIZATION

CURVE FITTING [BY THE METHOD OF LEAST SQUARE]:

We can plot ‘n’ points ( , )i ix y where i=0,1,2,3,………At the XY plane. It is difficult to draw a graph y=f(x) which passes through all these points but we can draw a graph which passes through maximum number of point. This curve is called the curve of best fit. The method of finding the curve of best fit is called the curve fitting.

FITTING A STRAIGHT LINE Y = AX + B

We have straight line that sounds as best approximate to the actual curve y=f(x)

passing through ‘n’ points ( , )i ix y , i=0,1.2………..n equation of a straight line is (1)y a bx

Then for ‘n’ points (2)i iY a bx

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Where a and b are parameters to be determined; iY is called the estimated value. The

given value iY corresponding to ix .

2

2

2

(3)

=

S = (4)

i i

i i

i i

Let S y Y

y a bx

y a bx

We determined a and b so that S is minimum (least). Two necessary conditions for this S

0 and 0b

S

a

.differentiate (4) w.r.t a and b partially

i

i

i

0

0 = y

0= y

y

or y

i i

i i

i

i

i

Sy a bx

a

y a bx

a b x

na b x

na b x

na b x

2

2

0 2

i i i

i i i i

Sy a bx x

b

x y ax bx

2

i0 = yi i ix a x b x

2i

2

y

or

where n = number of points or value.

i i ix n x b x

xy a x b x

FITTING A SECOND DEGREE PARABOLA Y=AX2 + BX + C

Let us take equation of parabola called parabola of best fit in the form2 (1)y a bx cx www.vt

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Where a, b, c are parameters to be determined. Let be the value of corresponding to the v

2

2

i

22

i

2i

(2)

Also

S= y (3)

y

y (4)

i i i

i

i i

i i

Y a bx cx

Y

a bx cx

S a bx cx

We determine a, b, c so that S is least (minimum).The necessary condition for this are

0, 0 & 0S S S

a b c

diff (4) w.r.t ‘a’ partially

2

2

2

2

2

0 2

0

0

i i i

i i i

i i i

i i i

Sy a bx cx

a

y a bx cx

y a bx cx

y a b x c x

2

2

2

i i i

i i i

y a b x c x

y na b x c x

or y na b x c x

diff (4) w.r.t ‘b’ partially

2

2 3

2

0 2

i i i i

i i i i i

Sy a bx cx x

b

x y ax bx cx

2 3

2 3

0

0

i i i i i

i i i i i

x y ax bx cx

x y a x b x c x

2 3

2 3

i i i i ix y a x b x c x

or xy a x b x c x

diff (4) w.r.t ‘c’ partiallywww.vtuc

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2 2

2 2 3 4

2 2 3 4

2 2 3 4

2 2 3 4

2 2 3 4

2

0 2

0

0

i i i i

i i i i i

i i i i i

i i i i i

i i i i i

Sy a bx cx x

c

x y ax bx cx

x y ax bx cx

x y a x b x c x

x y a x b x c x

or x y a x b x c x

Hence the normal equation for second degree parabola are 2y na b x c x

2 3xy a x b x c x 2 2 3 4x y a x b x c x

PROBLEMS:

1) Fit a straight line y a bx to the following data

: 5 10 15 20 25

: 16 19 23 26 30

x

y

Solution: Let (1)y a bx

Normal equation

2

(2)

(3)

y na b x

xy a x b x

2

5 16 25 80

10 19 100 190

15 23 225 345

20 26 400 520

25 30

x y x xy

2

625 750

75 =114 =1375 =1885

x y x xy

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(1) & (2)

114 15 75

1885 75 1375

12.3

0.7

(1) 12.3 0.7

a b

a b

a

b

y x

2) Fit equation of straight line of best fit to the following data

: 1 2 3 4 5

: 14 13 9 5 2

x

y

Solution:

Let (1)y a bx

Normal equation

2

2

(2)

(3)

1 14 1 14

2 13 4 26

3 9 9

y na b x

xy a x b x

x y x xy

2

27

4 5 16 20

5 2 25 10

15 =43 =55 =97

x y x xy

(2) & (3)

43 15 15

97 15 56

a b

a b

Solving above equations we get

18.2

3.2

(1) 18.2 3.2

a

b

y x

3) The equation of straight line of best fit find the equation of best fit

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: 0 1 2 3 4

: 1 1.8 3.3 4.5 6.3

x

y

Solution: let (1)y a bx

Normal equation

2

2

(2)

(3)

0 1 0 0

1 1.8 1 1.8

2 3.3

y na b x

xy a x b x

x y x xy

2

4 6.6

3 4.5 9 13.5

4 6.3 16 25.2

10 =16.9 =30 =47.1

(2) & (3)

16.9 5 10

47.1 10 30

0.72

1.33

(1)

x y x xy

a b

a b

a

b

0.72 1.33y x

4) If p is the pull required to lift a load by means of pulley block. Find a linear block of the form p=MW+C Connected p &w using following data

: 50 70 100 120

: 12 15 21 25p

Compute p when W=150. Solution: Given p=y & W=x equation of straight line is:

let (1 )y a bx

Normal equationswww.vtuc

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2

2

(2)

(3)

=

50 12 2500 600

70 15

y na b x

xy a x b x

x p y x xy

2

4900 1050

100 21 10000 2100

120 25 144000 3000

10 =16.9 =30 =x y x xy 47.1

(2) & (3)

73 4 340

6750 340 31800

2.27

0.187

(1) 2.27 0.187

put 150

30.32

a b

a b

a

b

y x

y

5) Fit a curve of the form xy ab

Solution: Consider

y

(1)

Take log on both side

logy=log

=loga+log

logy=loga+log

(2) ; logy=Y y=e

Corresponding normal equation loga=

x

x

x

x

y ab

ab

b

b

y A Bx

A

B

2

A a=e

(3) logb=B =e

(4)

Y nA B x b

xY A x B x

Solving the normal equation (3) & (4) for a & b . Substitute these values in (1) we get

curve of best fit of the form xy ab

6) Fit a curve of the curve xy ab for the data

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: 1 2 3 4 5 6 7 8

: 1.0 1.2 1.8 2.5 3.6 4.7 6.6 9.1

x

y

Solution:

2

2

(1)

Normal equations are

(2); A=loga

(3) Y=logy, B=logb

Y=logy

xlet y ab

Y nA B x

xY A x B x

x y x

1 1.0 0 1 0

2 1.2 0.182 4 0.364

3 1.8 0.587

xY

9 1.761

4 2.5 0.916 16 3.664

5 3.6 1.280 25 6.4

6 4.7 1.547 36 9.282

7 6.6 1.887 49 13.209

8 9.1 2.208 64 17.664

36 Y =8.607 x

2

A

B

=204 =52.34

(1) & (2)

8.607 8 36

52.34 36 203

0.382

0.324

then e 0.682

e 1.382

(1) 0.682 1.382x

x xY

A B

A B

A

B

a

b

y

7) Fit a curve of the form by ax for the following data

: 1 1.5 2 2.5

: 2.5 5.61 10.0 15.6

x

y Solution:: Consider

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2

2

(1)

Normal equations are

(2); Y=logy

(3) A=loga, X=logx

X= log

blet y ax

Y nA b x

XY A X B X

x y x X

log XY

1 2.5 0 0 0.916 0

1.5 5.62 0.405 0.164 1.7

Y y

26 0.699

2 10.0 0.693 0.480 2.302 1.595

2.5 15.6 0.916 0.839 2.747 2.5

2

A

2

16

X=2.014 X =1.483 Y=7.691 XY=4.81

(1) & (2)

7.691 4 2.014

4.81 2.014 1.483

A=0.916, e 2.499 2.5

1.999 2

(1) 2.5

A b

A b

a

b

y x

8) Fit a parabola 2y a bx cx for the following data

: 1 2 3 4

: 1.7 1.8 2.3 3.2

x

y Sol:

2

2

2 3

2 2 3 4

(1)

Normal equation

(2)

(3)

(4)

y a bx cx

y na b x c x

xy a x b x c x

x y a x b x c x

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2 3 4 2

1 1.7 1 1 1 1.7

x y x x x xy x y

1.7

2 1.8 4 8 16 3.6 7.2

3 2.3 9 27 81

2 3 4 2

6.9 20.7

4 3.2 16 64 256 12.8 51.2

9 30 100 354 25 80.8

(1), (3) &

x y x x x xy x y

2

(4)

9 4 10 30

25 10 30 100

80.8 30 100 354

2

-0.5

0.2

(1) 2 - 0.5 0.2

a b c

a b c

a b c

a

b

c

y x x

9) Fit a curve of the form bxy ae for the following

: 0 2 4

: 8.12 10 31.82

x

y Sol:

2

2

(1)

Normal equation

Y=nA+b (2); Y=logy

(3) A=loga

y Y=logy

0

bxy ae

x

xy A x b x

x x xY

8.12 2.094 0 0

2 10 2.302 4 4.604

4 31.82 3..46 16 2

0..341

13.84

6 Y=7.86 20 18.444

(2) & (3)

7.856=3A+6b

18.444=6A+20b

A=1.935, 6.924

b=0.341

(1) 6.924

A

x

x x xY

a e

y e

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10) Fit a II degree parabola 2ax bx c to the least square method & hence

find y when x=6

: 1 2 3 4 5

: 10 12 13 16 19

x

y

Sol:

2

2

2

2 3

2 2 3 4

(1)

(1)

Normal equation

(2)

(3)

(4)

y ax bx c

y c bx ax

y nc b x a x

xy c x b x a x

x y c x b x a x

2 3 4 2

1 10 1 1 1 10 10

2 12 4 8 16 24

x y x x x xy x y

14

3 13 9 27 81 39 117

4 16 16 64 256 64 256

5 19 25 125 625 95

2

475

15 70 55 225 979 232 906

70 5 15 55

232 15 55 225

906 55 225 979

0.285

0.485

9.4

(1) 0.285 0.485 9.4

at 6, 22.6

c b a

c b a

c b a

a

b

c

y x x

x y

OPTIMIZATION:Optimization is a technique of obtaining the best result under the given conditions. Optimization means maximization or minimization

LINEAR PROGAMMING:Linear Programming is a decision making technique under the given constraints on the condition that the relationship among the variables involved is linear. A general relationship among the variables involved is called objective function. The variables involved are called decision variables.

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The optimization of the objective function Z subject to the constraints is the mathematical formulation of a LPP.

A set real values ),........,( 21 nxxxX which satisfies the constraint BAX )( is called solution.

A set of real values xi which satisfies the constraints and also satisfies non negativity

constraints 0ix is called feasible solution.

A set of real values xi which satisfies the constraints along with non negativity restrictions and optimizes the objective function is called optimal solution.

An LPP can have many solutions.

If the optimal value of the objective function is infinity then the LPP is said to have unbounded solutions. Also an LPP may not possess any feasible solution.

GRAPHICAL METHOD OF SOLVING AN LPP

LPP involved with only two decision variables can be solved in this method. The method is illustrated step wise when the problem is mathematically formulated.

The constraints are considered in the form of 1

b

y

a

x

which graphically represents straight lines passing through the points (a,0) and (0,b) since there are only two decision variables.These lines along with the co-ordinate axes forms the boundary of the region known as the feasible region and the figure so formed by the vertices is called the convex polygon.

The value of the objective function nn xcxcxcZ ......2211 is found at all these vertices.The extreme values of Z among these values corresponding the values of the decision variables is required optimal solution of the LPP.

PROBLEMS:

1) Use the graphical method to maximize Z = 3x + 4y subject to the constraints .0,0,18052,402 yxyxyx

Solution: Let us consider the equations

)2.(..........1

3690)1...(..........1

4020

18052402

yxyx

yxyxwww.vtuc

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Let (1) and (2) represent the straight lines AB and CD respectively where we have A = (20, 0), B = (0, 40) ; C = (90 , 0), D = (0,36)We draw these lines in XOY plane.

Shaded portion is the feasible region and OAED is the convex polygon. The point E being the point of intersection of lines AB and CD is obtained by solving the equation:

)35,5.2(),(.18052,402 yxEyxyx

The value of the objective function at the corners of the convex polygin OAED are tabulated.

Corner

Value of Z = 3x + 4y

O(0,0) 0A(20, 0) 60E(2.5, 35) 147.5D(0,36) 144

35,5.25.147)( yxwhenZThus Max

2) Use the graphical method to maximize 21 53 xxZ subject to the constraints .0,,600,1500,20002 2122121 xxxxxxx


)3.........(600)2.(..........115001500

)1.....(110002000

600,1500,2000

22121

22121

xxxxx

xxxxx

Let (1) and (2) represent the straight lines AB and CD respectively where we have A = (2000, 0), B = (0, 1000) ; C = (1500 , 0), D = (0,1500)

.6002 x is a line parallel to the 1x axis.We draw these lines in XOY plane.www.vt

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On solving

)600,800(,600,20002

)600,900(,600,1500

)500,1000(,1500,2000

221

221

2121

Ggetwexxx

Fgetwexxx

Egetwexxxx

Also we have C = (1500 , 0), H= (0, 600)


Corner Value of 21 53 xxZ

O(0,0) 0C(1500, 0) 4500E(1000, 500) 5500F(900, 600) 5700G(800,600) 5400

600,9005700)( 21 xxwhenZThus Max

3) Use the graphical method to minimize 21 1020 xxZ subject to the constraints .0,,6034,303,402 21212121 xxxxxxxx


)3.........(12015

)2.(..........13010

)1.....(12040

60334,303,402

212121

212121

xxxxxx

xxxxxx

Let (1) ,(2) and (3) represent the straight lines AB ,CD and EF respectively where we have A = (40, 0), B = (0,20) ; C = (10 , 0), D = (0,30) ; E = (15,0), F = (0, 20)We draw these lines in XOY plane.www.vt

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Shaded portion is the feasible region and EAHG is the convex polygon. The point G being the point of intersection of lines EF and CD. The point H being the point of intersection of lines AB and CDis obtained by solving the equation:


Corner Value of 21 53 xxZ

A(15 , 0) 300A(40, 0) 800H(4 , 18) 260G (6, 12) 240

12,6240)( 21 xxwhenZThus MIN

4) Show that the following LPP does not have any feasible solution.Objective function for maximization:Z = 20x+30y.

Constraints: .0,0,6397,2443 yxyxyx


)2.(..........1

79)1...(..........1

68

63972443

yxyx

yxyx

Let (1) and (2) represent the straight lines AB and CD respectively where we have A = (8, 0), B = (0,6) ; C = (9 , 0), D = (0,7)We draw these lines in XOY plane.

It is evident that there is no feasible region.Thus we conclude that the LPP does not have any feasible solution.

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SIMPLEX METHOD

Simplex method is an efficient algebraic method to solve a LPP by systematic procedure and hence an algorithm can be evolved called the simplex algorithm.

In this method it is necessary that all the constraints in the inequality formn is converted into equality form thus arriving at a system of algebraic equations.

If the constraint is involved with we add a non zero ariables 0)(1 says to the LHS to make it an equality and the same variable is called slack variable.

LPP with all constraints being equalities is called a standard form of LPP.

A minimizing LPP is converted into an equivalent maximization problem. Minimizing the given objective function P is equivalent to maximizing –P under the same constraints and Min.P = -(Max value of –P )

PROBLEMS:

1) Use Simplex method to maximize z= 2x + 4y subject to the constraints 0,0,2432,223 yxyxyx

Solution: Let us introduce slack variables 1s and 2s to the inequalities to write them in

the following form.

1 2

1 2

1 2

3 1. 0. 22

2 3 0. 1. 24

2 4 0. 0. is the objective function

x y s s

x y s s

x y s s z

Solution by the simplex method is presented in the following table.

Thus the maximum value of Z is 32 at x=0 and y=8

NZV x y1s 2s Qty Ratio

1s 3 1 1 0 22 22/1=22 8 is least, 3 is pivot,

2s is replaced by y.

Also 1/3.R2 2s 2 3 0 1 24 24/3=8

Indicators(∆)

-2 -4 0 0 0

1s 3 1 1 0 221 2 1R R R

y 2/3 1 0 1/3 8∆ -2 -4 0 0 0

3 2 34R R R

1s 7/3 0 1 -1/3 14

y 2/3 0 0 1/3 8∆ 2/3 0 0 4/3 32 No negative indicators

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2) Use Simplex method to maximize Z =2x + 3y + z subject to the constraints 3 2 11, 2 5 19,3 4 25 0, 0, 0x y z x y z x y z x y z

Solution: Let us introduce slack variables 1s , 2s , 3s to the inequalities to write them in

the following form.

3

3

3

3

1 2

1 2

1 2

1 2

3 2 1. 0. 0. 11

2 5 0. 1. 0. 19

3 4 0. 0. 1. 25

2 3 0. 0. 0. is theobjectivefunction

x y z s s s

x y z s s s

x y z s s s

x y z s s s P


Thus the maximum value of P is 19 at x = 8 and y = 1, z = 0

NZV x y z1s 2s 3s Qty Ratio

1s 1 3 2 1 0 0 11 11/3=3.33 3.33 is least, 3 is pivot 1s is replaced

by y. Also 1/3.R1 2s 1 2 5 0 1 0 19 19/3=9.5

3s 3 1 4 0 0 1 35 25/1=25

Indicators(∆)

-2 -3 -1 0 0 0 0

y 1/3 1 2/3 1/3 0 0 11/3

2s 1 2 5 0 1 0 192 1 22R R R

3s 3 1 4 0 0 1 253 1 3R R R

∆ -2 -3 -1 0 0 0 04 1 43R R R

Y 1/3 1 2/3 1/3 0 0 11/3 11/3÷1/3=11 8 is least, 8/3 is pivot 3s is replaced

by x. Also 3/8.R3 2s 1/3 0 11/3 -2/3 1 0 35/3 35/3÷1/3=35

s3 8/3 0 10/3 -1/3 0 1 64/3 64/3÷8/3=8∆ -1 0 1 1 0 0 11y 1/3 1 2/3 1/3 0 0 11/3

1 3 11/ 3R R R

2s 1/3 0 11/3 -2/3 1 0 35/32 3 21/ 3R R R

x 1 0 10/8 -1/8 0 3/8 8∆ -1 0 1 1 0 0 1

4 3 4R R R y 0 1 1/4 3/8 0 -

1/81

2s 0 0 13/4 -5/8 1 -1/8

9

x 1 0 10/8 -1/8 0 3/8 9∆ 0 0 9/4 7/8 0 3/8 19 No negative

indicatorswww.vtuc

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3) Use Simplex method to minimize P =x - 3y + 2z subject to the constraints 3 2 7, 2 4 12, 4 3 8 10 0, 0, 0x y z x y x y z x y z

Solution: The given LPP is equivalent to maximizing the objective function –P subject to the same constraintsThat is –P=P’=-x + 3y - 2z is to be maximizedLet us introduce slack variables 1s , 2s , 3s to the inequalities to write them in the

following form.

3

3

3

3

1 2

1 2

1 2

1 2

3 2 1. 0. 0. 7

2 4 0 0. 1. 0. 12

4 3 8 0. 0. 1. 10

3 2 0. 0. 0. is theobjectivefunction

x y z s s s

x y z s s s

x y z s s s

x y z s s s P


Thus the maximum value of P is 19 at x = 8 and y = 1, z = 0

NZV x y z1s 2s 3s Qty Ratio

1s 3 -1 2 1 0 0 7 7/-1=-7 3 is least, 4 is pivot 2s is

replaced by y. Also 1/4.R2

2s -2 4 0 0 1 0 12 12/4=3

3s -4 3 8 0 0 1 10 10/3=3.3

Indicators(∆)

1 -3 2 0 0 0 0

s1 3 -1 2 1 0 0 71 1 2R R R

y -1/2 1 0 0 1/4 0 3

3s -4 3 8 0 0 1 103 2 33R R R

∆ 1 -3 2 0 0 0 04 2 43R R R

s1 5/2 0 2 1 1/4 0 10 10÷5/2=4 4 is least, 5/2 is pivot, 1s is

replaced by x. Also 2/5.R1

y -1/2 1 0 0 1/4 0 3 3÷-1/2=-6

3s -5/2 0 8 0 -3/4 1 1 1÷-5/2=-2/5

∆ -1/2 0 2 0 3/4 0 9x 1 0 4/5 2/5 1/10 0 4y -1/2 1 0 0 1/4 0 3

2 1 21/ 2R R R

3s -5/2 0 8 0 -3/4 1 1 3 1 35 / 2R R R

∆ -1/2 0 2 0 3/4 0 94 1 41/ 2R R R

x 1 0 4/5 2/5 1/10 0 4y 0 1 2/5 1/5 3/10 0 5

3s 0 0 10 1 -1/2 1 11

∆ 0 0 12/5 1/5 4/5 0 11 No negative indicators

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UNIT VNUMERICAL METHODS - 1

CONTENTS:

Introduction

Numerical solution of algebraic and transcendental equations

Regula –falsi method

Newton –Raphson method

Iterative methods of solution of a system of equation

Gauss –Seidel method

Relaxation method

Largest eigen value and the corresponding eigen vector by

Rayeigh’s power method

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NUMERICAL METHODS-1Introduction:Limitations of analytical methods led to the evolution of Numerical methods. Numerical Methods often are repetitive in nature i.e., these consist of the repeated execution of the same procedure where at each step the result of the proceeding step is used. This process known as iterative process is continued until a desired degree of accuracy of the result is obtained.

Solution of Algebraic and Transcendental Equations:

The equation f(x) = 0 said to be purely algebraic if f(x) is purely a polynomial in x.If f(x) contains some other functions like Trigonometric, Logarithmic, exponential etc. then f(x) = 0 is called a Transcendental equation.

Ex: (1) x4 - 7x3 + 3x + 5 = 0 is algebraic(2) ex - x tan x = 0 is transcendental

Method of false position or Regula-Falsi Method:This is a method of finding a real root of an equation f(x) = 0 and is slightly an improvisation of the bisection method.Let x0 and x1 be two points such that f(x0) and f(x1) are opposite in sign.

Let f(x0) > 0 and f(x1) < 0The graph of y = f(x) crosses the x-axis between x0 and x1

Root of f(x) = 0 lies between x0 and x1www.vtuc

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Now equation of the Chord AB is

)1...()xx(xx

)x(f)x(f)x(fy 0

01

010

When y =0 we get x = x2

)2...()x(f)x(f)x(f

xxxx.e.i 0

01

0102

Which is the first approximationIf f(x0) and f(x2) are opposite in sign then second approximation

)x(f)x(f)x(f

xxxx 0

02

0203

This procedure is continued till the root is found with desired accuracy.

Poblems:

1. Find a real root of x3 - 2x -5 = 0 by method of false position correct to three decimal places between 2 and 3.

Answer:Let f(x) = x3 - 2x - 5 = 0

f(2) = -1f(3) = 16

a root lies between 2 and 3Take x0 = 2, x1 = 3 x0 = 2, x1 = 3

)x(f)x(f)x(f

xxxxNow 0

01

0102

)1(116

232

= 2.0588 f(x2) = f(2.0588) = -0.3908

Root lies between 2.0588 and 3Taking x0 = 2.0588 and x1 = 3

f(x0) = -0.3908, f(x1) = 16

)x(f.)x(f)x(f

xxxxgetWe 0

01

0103

)3908.0(3908.16

9412.00588.2

= 2.0813f(x3) = f(2.0813) = -0.14680

Root lies between 2.0813 and 3

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Taking x0 = 2.0813 and x1 = 3f(x0) = 0.14680, f(x1) =16

0897.2)14680.0(1468.16

9187.00813.2x 4

Repeating the process the successive approximations arex5 = 2.0915, x6 = 2.0934, x7 = 2.0941, x8 = 2.0943

Hence the root is 2.094 correct to 3 decimal places.

2. Find the root of the equation xex = cos x using Regula falsi method correct to three decimal places.

Solution: Let f(x) = cosx - xex

Observe f(0) = 1 f(1) =cos1 - e = -2.17798

root lies between 0 and 1Taking x0 = 0, x1 = 1f(x0) = 1, f(x1) = -2.17798

)x(f.)x(f)x(f

xxxx 0

01

0102

31467.0)1(17798.3

10

f(x2) = f(0.31467) = 0.51987 +ve Root lies between 0.31467 and 1

x0 = 0.31467, x1 = 1f(x0) = 0.51987, f(x1) = -2.17798

44673.0)51987.0(51987.017798.2

31467.0131467.0x3

f(x3) = f(0.44673) = 0.20356 +ve Root lies between 0.44673 and 1

49402.020356.038154.2

55327.044673.0x 4

Repeating this process

x5 = 0.50995, x6 = 0.51520, x7 = 0.51692, x8 = 0.51748x9 = 0.51767, etc

Hence the root is 0.518 correct to 4 decimal places

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Newton Raphson Method

This method is used to find the isolated roots of an equation f(x) = 0, when the derivative of f(x) is a simple expression.

Let m be a root of f(x) = 0 near a. f(m) = 0We have by Taylor's series

.....)a(f!2

)ax()a(f)ax()a(f)x(f ''

2'

.....)a(f)am()a(f)m(f ' Ignoring higher order termsf(m) = f(a) + (m - a) f' (a) = 0

)a(f

)a(famor

'

)a(f

)a(famor

'

Let a = x0, m = x1

ionapproximatfirst theis)x(f

)x(fxxthen

0'

001

.

.

.

ionapproximatsecond theis)x(f

)x(fxx

1'

112

MethodRaphsonNewtonforformulaiterativetheis)x(f

)x(fxx

k'

kk1k

1. Using Newton's Raphson Method find the real root of x log10 x = 1.2 correct to four decimal places.Answer:Let f(x) = x log10 x - 1.2 f(1) = -1.2, f(2) = -0.59794, f(3) = 0.23136

10log

xlog1)x(f2.1

10log

xlogx)x(fhaveWe

e

e'

e

e

xlogelog 1010

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K1010

k10kk1k xlogelog

2.1xlogxxx

Let x0 = 2.5 (you may choose 2 or 3 also)

7465.25.2logelog

2.15.2log5.25.2x

1010

101

7406.27465.2logelog

2.17465.2log7465.27465.2x

10102

Repeating the procedure7406.2x3

equationgiventheofroottheis7406.2x

2.Using Newton's Method, find the real root of xex = 2. Correct to 3 decimal places.Answer:Let f(x) = xex - 2 f(0) = -2 f(1) = e - 2 = 0.7182Let x0 = 1 f' (x) = (x + 1) ex

We have

kxk

kxk

k1ke)1x(

2exxx

8678.0e2

2e1x1

8527.0e)8678.1(

2e)8678.0(8678.0x

8678.0

8678.0

2

8526.0e)8527.1(

2e)8527.0(8527.0x

8527.0

8527.0

3

placesdecimal3toCorrect.rootrequiredtheis,8526.0x

3. Find by Newton's Method the real root of 3x = cosx + 1 near 0.6, x is in radians. Correct for four decimal places.

Answer:Let f(x) = 3x - cosx - 1 f'(x) = 3 + sinx

k

1kkk1k xsin3

xcosx3xx

6071.0)6.0(sin3

1)6.0(cos)6.0(36.0x6.0xWhen 10

6071.0)6071.0(sin3

1)6071.0(cos)6071.0(36071.0x 2

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Since x1 = x2

The desired root is 0.6071

4. Obtain the iterative formula for finding the square root of N and find 41Answer:

NxLet or x2 - N = 0 f(x) = x2 - N f'(x) = 2xNow

k

2k

k1k x2

Nxxx

k

kk x2

N

2

xx

k

k1k x

Nx

2

1x.e.i

41findTo

4136thatObserve

6xChoose 0

4166.66

416

2

1x1

4031.64166.6

414166.6

2

1x 2

4031.64031.6

414031.6

2

1x3

Since x2 = x3 = 6.4031

4031.641ofvalueThe

5. Obtain an iterative formula for finding the p-th root of N and hence find (10)1/3

correct to 3 decimal places.Answer:Let xp = Nor xp - N = 0Let f(x) = xp - N

1p' px)x(f

1pk

pk

k1kpx

NxxxNow

Observe that 8 < 10

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3/13/1 108 3/1)10(2.e.i

Use x0 = 2, p = 3, N=10

1666.2)2(3

1022x

2

3

1

1545.2)1666.2(3

10)1666.2(1666.2x

2

3

2

1544.2)1545.2(3

10)1545.2(1545.2x

2

3

3

1544.2)10( 3/1

6. Obtain an iterative formula for finding the reciprocal of p-th root of N. Find (30)-1/5 correct to 3 decimal places.

Answer:Let x -p = Nor x -p - N = 0 f(x) = x -p - N f'(x) = -px -p - 1

Now

1p

k

pk

k1kxp

Nxxx

5.02

1)32(cesin 5/1

We use x0 = 0.5, p = 5, N = 30

processthepeatingRe,50625.0)5.0(5

30)5.0(5.0x

6

5

1

506495.0x,506495.0x 32

5065.0)30( 5/1

GAUSS-SEIDAL ITERATION METHOD (to solving the system linear simultaneous equations.)

Example 1.Use Gauss-Seidal iteration method to solve the following system of equations. 3x + 20y - 2z = -18 20x + y - 2z = 17 2x - 3y + 20z = 25

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Solution:. Rearranging thegiven system of equations 20x + y - 2z = 17 3x + 20y - 2z = -18 ………………….(1) 2x - 3y + 20z =25

The above system equations is arranged such that, diagonally dominant. System (1)

x =

201 [ 17- y +2z ]

y =

20

1 [-18-3x + z ] ………… (2)

z =

201 [25 –2x + 3y ]

Let the initial approximations to the solution of the system (A) be

,0)0( x ,0)0( y 0)0( zFirst Iteration:Using ( 2 )

)1(x201 [ 17- y

)0(

+2z)0(

]

)1(x201 [ 17- 0 + 2(0) ] = 8500.0

y)1(

=

20

1 [-18-3x)1(

+ z)0(

]

y)1(

=

201 [ -18-3(0.8500 ) + 0 ] = -1.0275

z)1(

=

201 [25 –2x

)1(

+ 3y)1(

]

z)1(

=

201 [25 –2(0.8500) + 3(- 1.0275)]

= 1.0109

Second Iteration: Usining ( 2 )

)2(x201 [ 17- y

)1(

+2z)1(

]

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)2(x201 [ 17- (-10275) + 2(1.0109 ) ]

= 1.0025

y)2(

=

201 [-18-3x

)2(

+ z)1(

]

y)2(

=

201 [ -18-3(1.0025 ) + 1.0109 ]

= - 0.99928

z)2(

=

20

1 [25 –2x)2(

+ 3y)2(

]

z)2(

=

201 [25 –2(1.0025) + 3(- 0.99928)]

= 0.9998

Third Iteration: Usining ( 2 )

)3(x201

[ 17- y)2(

+2z)2(

]

)3(x201 [ 17- (- 0.99928)+ 2(0.9998) ]

= 1.0000

y)3(

=

201 [-18-3x

)3(

+ z)2(

]

y)3(

=

20

1 [ -18-3(1.0000) +0.9998 ]

= -1.0000

z)3(

=

201 [25 –2x

)3(

+ 3y)3(

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z)3(

=

201 [25 –2(1.00) + 3(-1.0000)]

= 1.0000

Answer after three iterations x =1.000, y = -1.0000 and z = 1.000

2) Use Gauss-Seidal iteration method to solve the following system of equations. x + y +54z=110 27x +6 y - z = 85 …………(A) 6x+15y + 2z =72Solution :. Rearranging the system of equations (A)

27x +6 y - z = 85

6x+15y + 2z =72 (B) x + y +54z=110

The above system equations is arranged such that, the system is diagonally dominant.

System (B) x =

271 [ 85-6 y +z ]

y =

151 [72-6x -2z ] (C)

z =

541 [110 –x -y ]

Let the initial approximations to the solution of the system (A) be

,1)0( x ,0)0( y 0)0( zFirst Iteration:Using ( C )

)1(x271 [ 85-6 y

)0(

+z)0(

]

)1(x271

[ 85- 6 (0) +0] =3.148148

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y)1(

=

151

[72- 6x)1(

-2 z)0(

] y)1(

=

151 [72- 6(3.148148) -2 (0) ]

=3.54074

z)1(

=

541 [110 –x

)1(

-y)1(

]

z)1(

=

541 [110 –3.148148 -3.54074 ]=1.913168

Second Iteration: Usining ( C )

)2(x27

1 [ 85-6 y)1(

+z)1(

]

)2(x271 [ 85- 6 (3.54074) +1.913168]

=2.432175

y)2(

=

151 [72- 6x

)2(

-2 z)1(

]

y)2(

=

151 [72- 6(2.432175) -2( 1.913168) ]

= 3.57204

z)2(

=

541 [110 –x

)2(

-y)2(

]

z)2(

=

541 [110 –2.432175 -3.57204 ]

=1.925837

ert45t

Third Iteration: Usining ( C )

)3(x271 [ 85-6 y

)2(

+z)2(

]

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)3(x27

1 [ 85- 6 (3.57204) +2(1.925837)]

=2.425689

y)3(

=

151 [72- 6x

)3(

-2 z)2(

]

y)3(

=

15

1 [72- 6(2.425689) -2( 1.913168) ]

=3.57313

z)3(

=

541 [110 –x

)2(

-y)2(

]

z)3(

=

541 [110 –2.432175 -3.57204 ]

=1.925947

Answer after three iterations x =2.425689, y = 3.57313 and z =

RELAXATION METHOD:

The method is illustrated for the following diagonally dominant system of three independent equations in three unknowns.

11 1 12 2 13 3 1

21 1 22 2 23 3 2

31 1 32 2 33 3 3

a x a x a x b

a x a x a x b

a x a x a x b

From these equations the residuals 1 2 3, ,R R R are defined as follows

1 11 1 12 2 13 3 1

2 21 1 22 2 23 3 2

3 31 1 32 2 33 3 3

R a x a x a x b

R a x a x a x b

R a x a x a x b

The values of 1 2 3, ,x x x that make the residuals 1 2 3, ,R R R simultaneously zero

constitutes the exact solution of the system of equations. If this is not possible we need to make the residuals as close to zero as possible.

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The values of 1 2 3, ,x x x at that stage constitutes an approximate solution. The

relaxation method aims in reducing the values of residuals as close to zero as possible by modifying the value of the variables at every stage.

At every step the numerically largest residual is reduced to zero by choosing an appropriate integral value for the corresponding variable.

These integral values are called the increments in 1 2 3, ,x x x denoted by 1 2 3, ,x x x respectively. If the system possess exact solution then 1 2 3, ,R R R becomes zero

simultaneously and the required solution will be the sum of all these increments.i.e., 1 1 2 2 3 3, ,x x x x x x

When the system does not possess exact solution, we need to continue the process b y magnifying the prevailing residuals on multiplication with 10. Later we divide by 10 to obtain the solution correct to one decimal place.

The process will be contined to meet the requirement of the solution to the desired accuracy.

PROBLEMS:1) Solve the following system of equations by relaxation method.

1 2 3

1 2 3

1 2 3

12 31

2 8 24

3 4 10 58

x x x

x x x

x x x

Solution: The given system of equations are diagonally dominant.

1 1 2 3

2 1 2 3

3 1 2 3

12 31

2 8 24

3 4 10 58,

Let R x x x

R x x x

R x x x

be the residuals.

Let 1 2 3, ,x x x respectively represent the increments for 1 2 3, ,x x x in the following

relaxation table.

1x 2x 3x 1R 2R 3R

0 0 0 -31 -24 -580 0 6 -25 -30 20 4 0 -21 2 182 0 0 3 6 240 0 -2 1 8 40 -1 0 0 0 0

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1 2 3, ,R R R are all zero and the exact solution of the given system is

1 1 2 2 3 32, 4 1 3, 6 2 4x x x x x x Thus ( 1 2 3, ,x x x ) = (2, 3, 4 ) is the solution of the given system of equations.

2) Solve the following system of equations by relaxation method.

10 2 3 205

2 10 2 154

2 10 120

x y z

x y z

x y z

Solution: The given system of equations are diagonally dominant.

1

2

3

10 2 3 205

2 10 2 154

2 10 120,

Let R x y z

R x y z

R x y z

be the residuals.

Let , ,x y z respectively represent the increments for , ,x y z in the following relaxation table.

x y z 1R 2R 3R

0 0 0 -205 -154 -12021 0 0 5 -196 -1620 20 0 -35 4 -1820 0 18 -89 -32 -29 0 0 1 -50 -200 5 0 -9 0 250 0 3 -18 -6 52 0 0 2 -10 10 1 0 0 0 0

1 2 3, ,R R R are all zero and the exact solution of the given system is

32, 26, 21x x y y z z Thus ( , ,x y z ) = (32 , 26, 21 ) is the solution of the given system of equations.

Rayleigh’s power method:

Given square matrix A, if there exist a scalar λ and a non zero column matrix X, such that AX = λ X, then λ is called an eigen value of A and X is called an eigen vector of A corresponding to an eigen value λ.

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Rayleigh’s power method is an iterative method to determine the numerically largest eigen value and the corresponding eigen vector of a square matrix.

Working procedure: Suppose A is the given square matrix, we assume initially an eigen vector 0X in a

simple for like [1,0,0] [0,1,0] [0,0,1] [1,1,1]or or or and find the matrix

product 0AX which will also be a column matrix.

We take out the largest element as the common factor to obtain 110AX X .

We then find 1

AX and again put in the form 1 22AX X by normalization.

The iterative process is continued till two consecutive iterative values of λ and X are same upto a desired degree of accuracy.

The values so obtained are respectively the largest eigen value and the corresponding eigen vector of the given square matrix A.

Problems:

1) Using the Power method find the largest eigen value and the corresponding eigen vector starting with the given initial vector.

Tgiven 001

201

020

102

0 1 1

1 2 2

2 0 1 1 2 1

: 0 2 0 0 0 2 0

1 0 2 0 1 0.5

2 0 1 1 2.5 1

0 2 0 0 0 2.5 0

1 0 2 0.5 2 0.8

Solution AX X

AX X

2 3 3

2 0 1 1 2.8 1

0 2 0 0 0 2.8 0

1 0 2 0.8 2.6 0.93

AX X

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3 4 4

4 5 5

5

2 0 1 1 2.93 1

0 2 0 0 0 2.93 0

1 0 2 0.93 2.86 0.98

2 0 1 1 2.98 1

0 2 0 0 0 2.98 0

1 0 2 0.98 2.96 0.99

2 0 1 1

0 2 0 0

1 0 2 0.99

AX X

AX X

AX

6 6

6 7 7

2.99 1

0 2.99 0

2.98 0.997

2 0 1 1 2.997 1

0 2 0 0 0 2.997 0

1 0 2 0.997 2.994 0.999

X

AX X

Thus the largest eigen value is approximately 3 and the corresponding eigen vector is [1 ,0, 1]’

2) Using the Power method find the largest eigen value and the corresponding eigen vector starting with the given initial vector.

Tgiven 8.08.01

512

132

114

0 1 1

1

4 1 1 1 5.6 1

: 2 3 1 0.8 5.2 5.6 0.93

2 1 5 0.8 5.2 0.93

4 1 1 1 5.86 1

2 3 1 0.93 5.72 5.86 0.98

2 1 5 0.93 5.72 0.98

Solution AX X

AX

2 2X

2 3 3

4 1 1 1 5.96 1

2 3 1 0.98 5.92 5.96 0.99

2 1 5 0.98 5.92 0.99

AX X

3 4 4

4

4 1 1 1 5.98 1

2 3 1 0.99 5.96 5.98 0.997

2 1 5 0.99 5.96 0.997

4 1 1 1 5.994 1

2 3 1 0.997 5.988 5.994 0.999

2 1 5 0.997 5.988 0.999

AX X

AX

5 5X

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Thus after five iterations the numerically largest eigen value is 5.994 and corresponding eigen vector is [1, 0.999, -0,999]’

UNIT VI

NUMERICAL METHOD -2

CONTENTS:

Introduction

Finite difference:

Forward and Backward difference

Newtons forward and backward interpolation formula

Newtons divided difference formula

Lagranges interpolation formula

Numerical integration

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NUMERICAL METHODS-2

Finite Differences

Let y = f(x) be represented by a table

x : x0 x1 x2 x3 …. xn

y : y0 y1 y2 y3 … yn

where x0, x1,x2….xn are equidistant. (x1 - x0 = x2 - x1 = x3 - x2 =….=xn - xn-1 = h)

We now define the following operators called the difference operators.

Forward difference operator (∆)

)x(f)hx(f)x(f

1n,...,2,1,0r,yyy r1rr

sdifferenceforwardfirst

yyy

.

.

yyy

yyy

1nn1n

121

010

sdifferenceondsecthecalledare,....,y,y,y 22

12

02

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)yy()y(yNow 01002

)yy()yy(yy 011201

012 yy2y

12312ly yy2yy|||

r1r2rr2 yy2yy

012303 yy3y3yy:Note

rk

2kr2k

1kr1k

krrk C)1(....yCyCyy

Difference Tablex y y 2y 3y 4y 5y

x0 y0

y0

x1 y1 2y0

y1 3y0

x2 y2 2y1 4y0

y2 3y0

x3 y3 2y2

y3

x4 y4

.sdifferenceleadingthecalledare,....y,y,y 03

02

0

Ex: The following table gives a set of values of x and the corresponding values ofy = f(x)

x : 10 15 20 25 30 35

y : 19.97 21.51 22.47 23.52 24.65 25.89

Form the difference table and find )15(f),20(f),10(f),10(f 432

x y 2 3 4 5

10 19.97

1.54

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15 21.51 -0.58

0.96 0.67

20 22.47 0.09 -0.68

1.05 -0.01 0.72

25 23.52 0.08 0.04

1.13 0.03

30 24.65 0.11

1.24

35 25.89

04.0)15(f,03.0)20(f,58.0)10(f,54.1)10(f 432

Note: The nth differences of a polynomial of n the degree are constant.

Backward difference operator ()

Let y = f(x)

We define f(x) = f(x) - f(x - h)

i.e. y1 = y1 - y0 = ∆ y0

y2 = y2 - y1 = ∆ y1

y3 = y3 - y2 = ∆ y2

'

'

yn = yn - yn-1 = ∆ yn - 1

1r1rrr yyyy

Note:

1. f(x + h) = f(x + h) - f(x) = ∆ f(x)

2. 2 f(x + 2h) = (f(x + 2h))

= {f(x + 2h) - f(x + h)}

= f(x + 2h) - f(x + h)

= f(x + 2h) - f(x + h) - f(x + h) + f(x)

= f(x + 2h) -2f(x + h) + f(x)

= ∆2 f(x)

|||ly n f(x + nh) = ∆n f(x)

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Backward difference tablex y y 2y 3y

X0 y0

y1

X1 y1 2y2

y2 3y3

X2 y2 2y3

y3

X3 y3

1. Form the difference table for

x 40 50 60 70 80 90

y 184 204 226 250 276 304

and find y (30), 2y (70), 5y (90)

Soln:

x y y 2y 3y 4y 5y

40 184

20

50 204 2

22 0

60 226 2 0

24 0 0

70 250 2 0

26 0

80 276 2

28

90 304

y (80) = 26, 2y (70) = 2, 5y (90) = 0

2. Given

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x 0 1 2 3 4

f(x) 4 12 32 76 156

Construct the difference table and write the values of f (4), 2f (4), 3f (3)

x y y 2y 3y

0 4

8

1 12 12

20 12

2 32 24

44 12

3 76 36

80

4 156

3) Find the missing term from the table:

x 0 1 2 3 4

y 1 3 9 - 81

Explain why the value obtained is different by putting x = 3 in 3x.

Denoting the missing value as a, b, c ..etc. Construct a difference table and solve.

x y y 2y 3y 4y

0 1 2

1 3 6 4

2 9 a - 9 a - 15 a - 19 -4a + 124

3 a 81 - a 81 - a -3a +105

4 81

Put 4y = 0 (assuming f(x) its be a polynomial of degree 3)

i.e., -4a + 124 = 0

a = 31

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Since we have assumed f(x) to be a polynomial of degree 3 which is not 3x we obtained a

different value.

4) Given u1 = 8, u3 = 64, u5 = 216 find u2 and u4

x u u 2u 3u

x1 8

x2 a a - 8 -2a + 72 b + 3a - 200

x3 64 64 - a b + a - 128 -3b - a + 408

x4 b b - 64 -2b + 280

x5 216 216 -b

We carryout upto the stage where we get two entries ( 2 unknowns) and equate each of

those entries to zero. (Assuming) to be a polynomial of degree 2.

b + 3a - 200 = 0

-3b - a + 408 = 0 We get a = 24 b = 128

Interpolation:

The word interpolation denotes the method of computing the value of the function y =

f(x) for any given value of x when a set (x0, y0), (x1, y1),…(xn, yn) are given.

Note:

Since in most of the cases the exact form of the function is not known. In such cases the

function f(x) is replaced by a simpler function (x) which has the same values as f(x) for

x0, x1, x2….,xn.

....y!3

)2u()1u(uy

!2

)1u(uyuy)x( 0

30

200

0n y

!n

)1nu...()2u()1u(u

is called the Newton Gregory forward difference formula

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Note :

1. Newton forward interpolation is used to interpolate the values of y near the beginning

of a set of tabular values.

2. y0 may be taken as any point of the table but the formula contains those values of y

which come after the value chosen as y0.

Problems:

1) The table gives the distances in nautical miles of the visible horizon for the given

heights in feet above the earths surface.

x = height 100 150 200 250 300 350 400

y = distance 10.63 13.03 15.04 16.81 18.42 19.90 21.27

Find the values of y when i) x = 120, ii) y =218

Solution:

x y 2 3 4 5 6

100 10.632.40

150 13.03 -0.392.01 0.15

200 15.04 -0.24 -0.071.77 0.08 0.02

250 16.81 -0.16 -0.05 0.021.61 0.03 0.04

300 18.42 -0.13 -0.011.48 0.02

350 19.90 -0.111.37

400 21.27

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Choose x0 = 100

i) 4.050

100120u,120x

)39.0(!2

)14.0()4.0()40.2(

!1

4.063.10)120(f

)15.0(!3

)24.0()14.0()4.0(

)07.0(!4

)34.0()24.0()14.0()4.0(

)02.0(!5

)44.0()34.0()24.0()14.0()4.0(

649.11)02.0(!6

)54.0()44.0()34.0()24.0()14.0()4.0(

ii) Let x = 218, x0 = 200, 36.050

18

50

200218u

)16.0(2

)64.0(36.0)77.1(36.004.15)218(f

...)03.0(6

)64.1()64.0(36.0

= 15.7

2) Find the value of f(1.85).

x y y 2y 3y 4y 5y 6y1.7 5.474

0.5751.8 6.049 0.062

0.637 0.0041.9 6.686 0.066 0.004

0.703 0.008 -0.0042.0 7.389 0.074 0 0.004

0.777 0.008 02.1 8.166 0.082 0

0.859 0.0082.2 9.025 0.090

0.94923 9.974

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5.01.0

8.185.1

h

xxu85.1x,8.1xChoose 0

0

)066.0(2

)5.0()5.0()637.0()5.0(049.6)85.1(f

)008.0(6

)5.1()5.0)(5.0(

0.00050.0008-0.31856.049

= 6.359

3) Given sin 45o = 0.7071, sin 50o = 0.7660, sin 55o =0.8192, sin 60o = 0.8660.

Find sin 48o.

x y 2 3

45 0.7071

0.589

50 0.7660 -0.0057

0.0532 0.0007

55 0.8192 -0.0064

0.0468

60 0.8660

6.0h

xxu5h;45x,48x 0

0

)0589.0()6.0(7071.048sin o

)0057.0(2

)4.0)(6.0(

7431.0)0007.0(

6

)4.1()4.0()6.0(

4) From the following data find the number of students who have obtained 45

marks. Also find the number of students who have scored between 41 and 45 marks.

Marks 0 - 40 41 - 50 51 - 60 61 -70 71 - 80

No. of students 31 42 51 35 31

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x y 2 3 4

40 31

42

50 73 9

51 -25

60 124 -16 37

35 12

70 159 -4

31

80 190

2

9)5.0()5.0()42()5.0(31)45(f

!3

)25()5.1()5.0()5.0(

488672.47)37(!4

)5.2()5.1()5.0()5.0(

f(45) - f(40) = 70 = Number of students who have scored between 41 and 45.

5) Find the interpolating polynomial for the following data:

f(0) = 1, f(1) = 0, f(2) = 1, f(3) = 10. Hence evaluate f(0.5)

xy 2 3

0 1

-1

1 0 2

1 6

2 1 8

9

3 10

x1

0xu

)2(!2

)1x(x)1(x1)x(f

1x2x6

!3

)2x()1x(x 23

6) Find the interpolating polynomial for the following data:

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x: 0 1 2 3 4

f(x) : 3 6 11 18 27

x y 2 3 4

0 33

1 6 25 0

2 11 2 07 0

3 18 29

4 27

x1

0xu

)2(2

)1x(x)3(x3)x(f

)0(

!x

)1x(x 2xx23

Newton Gregory Backward Interpolation formula

....y!3

)2u()1u(uy

!2

)1u(uyuyy n

3n

2nn

h

xxuwhere n

1) The values of tan x are given for values of x in the following table. Estimate

tan (0.26)

x 0.10 0.15 0.20 0.25 0.30

y 0.1003 0.1511 0.2027 0.2553 0.3093

x y 2 3 4

0.10 0.10030.0508

0.15 0.1511 0.0008

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0.0516 0.00020.20 0.2027 0.0010 0.0002

0.0526 0.00040.25 0.2553 0.0014

0.05400.30 0.3093

8.005.0

3.026.0u

)054.0)(8.0(3093.0)26.0(f )0014.0()2.0(2

)8.0( 2659.0)0004.0(

6

)2.1()2.0()8.0(

2) The deflection d measured at various distances x from one end of a cantilever is

given by the following table. Find d when x = 0.95

95.0xwhen3308.0d25.02.0

195.0u

x d 2 3 4 5

0 00.0347

0.2 0.0347 0.04790.0826 -0.0318

0.4 0.1173 0.0161 0.00030.0987 -0.0321 -0.0003

0.6 0.2160 -0.016 00.0827 -0.032

0.8 0.2987 -0.04810.0346

1.0 0.3333

3) The area y of circles for different diameters x are given below:

x : 80 85 90 95 100

y : 5026 5674 6362 7088 7854

Calculate area when x = 98

x y y 2y 3y 4y80 5026

648

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85 5674 40688 -2

90 6362 38 4726 2

95 7088 40766

100 7854

Answer:

4.0n

xxu n

y = 7542

4) Find the interpolating polynomial which approximates the following data.

x 0 1 2 3 4

y -5 -10 -9 4 35

x y 2 3 4

0 -5-5

1 -10 61 6

2 -9 12 013 6

3 4 1831

4 35

1

4xu

!2

18)3x()4x()31()4x(35)x(f

!3

)6()2x()3x()4x(

f(x) 5-6x 2x x 23

Interpolation with unequal intervalsNewton backward and forward interpolation is applicable only when x0, x1,…,xn-1 are

equally spaced. Now we use two interpolation formulae for unequally spaced values of x.

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i) Lagranges formula for unequal intervals:

If y = f(x) takes the values y0, y1, y2,….,yn corresponding to x = x0, x1, x2,…,xn then

)x(f)xx)...(xx()xx(

)xx)...(xx()xx()x(f 0

n02010

n21

)x(f)xx)...(xx()xx()xx(

)xx)...(xx()xx()xx(1

n1312101

n320

....)x(f)xx)...(xx()xx()xx(

)xx)...(xx()xx()xx(2

n2321202

n310

)x(f)xx)...(xx()xx()xx(

)xx)...(xx()xx()xx(n

1nn2n1n0n

1n210

is known as the lagrange's

interpolation formula

ii) Divided differences ()

]x,x[xx

yyy)x(f 10

01

0100

]x,x[xx

yyy 12

12

121

]x,x[xx

yyy n1n

1nn

1nn1n

second divided difference

02

010

20

2

xx

yyy)x(f

]x,x,x[xx

]x,x[]x,x[210

02

0112

]x,x,x[xx

]x,x[]x,x[

xx

yyy|| 321

13

1223

13

121

2ly

similarly definedbecan,....y03

Newton's divided difference interpolation formula

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03

21002

10000 y)xx)(xx()xx(y)xx()xx(y)xx(y)x(fy

0n

n10 y)xx...()xx()xx(...

is called the Newton's divided difference formula.

Note:Lagrange's formula has the drawback that if another interpolation value were

inserted, then the interpolation coefficients need to be recalculated.

Inverse interpolation: Finding the value of y given the value of x is called interpolation

where as finding the value of x for a given y is called inverse interpolation.

Since Lagrange's formula is only a relation between x and y we can obtain the inverse

interpolation formula just by interchanging x and y.

0n02010

n21 x.)yy)...(yy()yy(

)yy)....(yy()yy(x

...x)yy...()yy()yy()yy(

)yy)...(yy()yy()yy()yy(1

n1312101

n3200

n1nn1n0n

1n10 x.)yy)...(yy()yy(

)yy)...(yy()yy(...

is the Lagranges formula for inverse interpolation

1) The following table gives the values of x and y

x : 1.2 2.1 2.8 4.1 4.9 6.2

y : 4.2 6.8 9.8 13.4 15.5 19.6

Find x when y = 12 using Lagranges inverse interpolation formula.

Using Langrages formula

05040302010

54321 x)yy()yy()yy()yy()yy(

)yy()yy()yy()yy()yy(x

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445251505

43210 x)yy)...(yy()yy()yy(

)yy()yy()yy()yy()yy(....

055.0964.0419.3252.1234.0022.0

= 3.55

2) Given the values

x : 5 7 11 13 17

f(x) : 150 392 1452 2366 5202

Evaluate f(9) using (i) Lagrange's formula (ii) Newton's divided difference formula.

i) Lagranges formula

392.)177()137()117()57(

)179()139()119()59()150(

)175()135()115()75(

)179()139()119()79()9(f

)2366()1713()1113()713()513(

)179()119()79()59()1452(

)1711()1311()711()511(

)179()139()79()59(

810)5202()1317()1117()717()517(

)139()119()79()59(

f(9) = 810

ii)

5 150

121

7 392 24

265 1

11 1452 32 0

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457 1

13 2366 42

709

17 5202

f(9) = 150 + 121 (9 - 5) + 24 (9 - 5) (9 - 7) + 1(9 - 5) (9 - 7) (9 - 11) = 810

3) Using i) Langranges interpolation and ii) divided difference formula. Find the

value of y when x = 10.

x : 5 6 9 11

y : 12 13 14 16

i) Lagranges formula

13)116)(96)(56(

)1110)(910)(510(12

)115()95()65(

)1110()910()610()10(fy

16)911)(611)(511(

)910)(610)(510(14

)119)(69)(59(

)1110)(610)(510(

3

44

ii)Divided differencex y 2 3

5 12

1

6 13

6

1

4

3/2

3

1

20

1

6

10/3

690

27

5116

1

15

2

9 14

15

2

5

3/2

12

2

11 16

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20

1)910)(610)(510(

6

1).610)(510()510(12)10(f

3

44

4) If y(1) = -3, y(3) = 9, y(4) =30, y(6) = 132 find the lagranges interpolating

polynomial that takes the same values as y at the given points.

Given:

x 1 3 4 6

y -3 9 30 132

9.)63)(43)(13(

)6x)(4x)(1x()3(.

)61)(41)(31(

)6x()4x()3x()x(f

132.)46)(36)(16(

)4x)(3x)(1x(30.

)64)(34)(14(

)6x)(3x)(1x(

= x3 - 3x2 + 5x - 6

5) Find the interpolating polynomial using Newton divided difference formula for

the following data:

x 0 1 2 5

y 2 3 12 147

x y 2 3

0 2

1

1 3 4

9 1

2 12 9

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45

5 147

F(x) = 2 + (x - 0)(1) + (x - 0) (x - 1) (4) + (x - 0) (x - 1) (x - 2) 1

= x3 + x2 - x + 2

Numerical Integration:-

To find the value of b

a

dxyI numerically given the set of values ),( ii yx ,

ni ,...,2,1,0 at regular intervals.(i) Simpson’s one third rule:-

2421310 .....2.....43 nnn yyyyyyyyh

I

when n is even.

(ii) Simpson’s three-eighth rule:-

363154210 .....2....38

3 nnn yyyyyyyyyy

hI

when n is a multiple of 3.

(iii) Weddle’s rule:-

......56510

36543210 yyyyyyy

hI

when n is a multiple of 6.

Problems:

1) Using Simpson’srd

3

1rule evaluate

1

021 x

dxby dividing the interval 1,0 into

4 equal sub intervals and hence find the value of correct to four decimal places.

Solution: Let us divide [0,1] into 4 equal strips (n = 4)1 0 1

length of each strip: 4 4

1 2 1 3 4The points of division are 0, , , , 1

4 4 2 4 4

h

x

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By data2

1

1y

x

Now we have the following table

x 0 1/4 1/2 3/4 1

2

1

1y

x

1 16/17 4/5 16/25 1/2

0y 1y 2y 3y 4y

Simpson’srd

3

1rule for n = 4 is given by

0 4 1 3 24 23

b

a

hydx y y y y y 1

20

1

20

1 1/ 4 1 16 16 41 4 2. 0.7854

1 3 2 17 25 5

10.7854

1

dxx

Thus dxx

To deduce the value of π: We perform theoretical integration and equate the resulting value to the numerical value obtained.

111 1 1

2 00

1tan tan (1) tan (0)

1 4

, 0.7854 4(0.7854) 3.14164

3.1416

dx xx

Wemust have

Thus

2) Given that x 4 4.2 4.4 4.6 4.8 5 5.2

xlog 1.3863 1.4351 1.4816 1.5261 1.5686 1.6094 1.6487

Evaluate 2.5

4

log dxx using Simpson’s th

8

3rule

Solution: Simpson’sth

8

3rule for n = 6 is given by

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0 6 1 2 4 5 3

5.2

4

5.2

4

33 2

8

3(0.2)log 1.3863 1.6487 3 1.4351 1.4816 1.5686 1.6094 2 1.5261

8

log 1.8279

b

a

e

e

hydx y y y y y y y

xdx

xdx

3) Using Weddle’s rule evaluate 1

20 1

xdx

x by taking seven ordinates and hence

find loge2Solution: Let us divide [0,1] into 6 equal strips ( since seven ordinates)

1 0 1 length of each strip:

6 61 2 1 3 1 4 2 5 6

The points of division are 0, , , , , , 16 6 3 6 2 6 3 6 6

h

x

By data2

1

1y

x

Now we have the following table

x 0 1/6 1/3 1/2 2/3 5/6 1

21

xy

x

0 6/37 3/10 2/5 6/13 30/61 1/2

0y 1y 2y 3y 4y 5y 6y

Weddle’s rule for n = 6 is given by

0 1 2 3 4 5 6

1

20

1

20

35 6 5

10

3(1/ 6)0 5(6 / 37) 3 /10 6(2 / 5) 6 /13 5(30 / 61) 1/ 2

1 10

0.34661

b

a

hydx y y y y y y y

xdx

x

xdx

x

To deduce the value of loge2: We perform theoretical integration and equate the resulting value to the numerical value obtained. www.vt

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om



112

200

1

20

1 1 1log (1 ) log 2 log 1

1 2 2 2

1log 2

1 2

1, log 2 0.3466 log 2 2(0.3466) 0.6932

2

log 2 0.6932

e e e

e

e e

e

xdx x

x

xHence dx

x

Wemust have

Thus

UNIT VIINUMERICAL METHODS -3

CONTENTS:

Numerical solution of PDE

Finite difference approximation to derivatives

Numerical solution of

One dimensional wave equation

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One dimensional heat equation

Two dimensional Laplace’s equation

UNIT-VIINUMERICAL METHODS-3

Classification of PDE s of second orderThe general second order linear PDE in two independent variables x,yis of the formAuxx+ Buxy+ Cuyy+ Dux+ Euy+ Fu=0Where A,B,C,D,E,F are in general functions of x,yThis equation is said to be

i) Parabolic if B2-4AC=0ii) Elliptic if B2-4AC<0iii) Hyperbolic if B2-4AC>0

Now let us examine the nature of three PDE which are under our discussionP.D.E A B C B24AC=0 Nature of

PDE1 One dimensional wave

equationc2uxx-utt=0C2 0 -1 4C2>0 Hyperbolic

2 One dimensional heat equationc2uxx-ut=0

C2 0 0 0 parabolic

3 two dimensional Laplace’sequationuxx+uyy=0

1 0 C2 -4<0 elliptic

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Numerical solution of a PDEConsider a rectangular region in the x—y plane. Let us divide this region into a network of rectangular of sides h and k. In other words , we draw lines x=ih,y=jk:I,j=1,2,3…….being parallel to the Y-axis and X-axis respectively resulting into a network of rectangles.the points of intersection of these lines are called mesh points or grid points.

We write u(x,y) =u(ih,jk) and the finite difference approximation for the the partial derivatives are put in the modified form

1, ,

1x i j i ju u u

h

, 1,

1y i j i ju u u

h

1, 1,

1

2x i j i ju u uh

, 1 ,

1y i j i ju u u

k

, , 1

1y i j i ju u u

k

, 1 , 1

1

2y i j i ju u uk

1, , 1,2

12xx i j i j i ju u u u

h

, 1 , , 12

12yy i j i j i ju u u u

k

The substitution of these finite diference approximation into the given PDE converts the PDE into a finite difference equation.Now we discuss numerical solution of the three important PDE namely

1) One dimensional wave equation2) One dimensional heat equation

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3) Two dimensional laplaceequation

Numerical solution of the one dimensional wave equationWe seek the numerical solution of the wave equation

2.................................(1)xx ttc u u

Subject to the boundary conditionsu(0,t)=0………………………(2)u(l,t)=0……………………….(3)and the initial conditionsu(x,0)=f(x)……………………(4)ut(x,0)=0……………………(5)

We shall substitute the finite difference approximation for the partial derivatives present in (1)

21, , 1, , 1 , , 12 2

1 1. 2 2i j i j i j i j i j i jc u u u u u uh k

22

1, , 1, , 1 , , 122 2i j i j i j i j i j i j

hc u u u u u u

k

Taking k/h= we have

2 21, , 1, , 1 , , 12 2i j i j i j i j i j i jc u u u u u u

2 2 2 2, 1 , 1, 1, , 12 1i j i j i j i j i ju c u c u u u ………………..(6)

For convenience let us choose such that 1- 2 2c =02 2

2 22 2

1 0k k h

c or k kh h c

Thus (6) reduces to the form

, 1i ju = 1, 1, , 1i j i j i ju u u

This is called the explicit formula for the solution of wave equation.

Examples

1.Solve the wave equation 2 2

2 24

u u

t x

subject to u=(0,t) =0,u(4,t)=0,ut(x,0)=0

and u(x,0)=x(4-x) by taking h=1,k=0.5 upto four steps

Soln: Below is the initial table wherein the first and last column are zero since (0,t)=0=u(4,t)

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Now consider u(x,0)=x(4-x)

1,0 2,0 3,0(1,0) 3; (2,0) 4; 3u u u u u

Next consider

,1 1,0 1,0

1,1 0,0 2,0

2,1 1,0 3,0

1,1 0,0 4,0

1

21 1

0 4 22 21 1

3 3 32 21 1

4 0 22 2

i i iu u u

u u u

u u u

u u u

We now consider the explicit formula to find the remaining values in the table

, 1 1, 1, , 1i j i j i j i ju u u u

1,2 0,1 2,1 1,0 0 3 3 0u u u u

2,2 1,1 3,1 2,0 2 2 4 0u u u u

3,2 2,1 4,1 3,0 3 0 3 0u u u u

The third row is completed

1,3 0,2 2,2 1,1 0 0 2 2u u u u

2,3 1,2 3,2 2,1 0 0 3 3u u u u

3,3 2,2 4,2 3,1 0 0 2 2u u u u

The fourth row is completed

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1,4 0,3 2,3 1,2 0 3 0 3u u u u

2,4 1,3 3,3 2,2 2 2 0 4u u u u

3,4 2,3 4,3 3,2 3 0 0 3u u u u The last row is completedThus the required values 0f ui,j are tabulated

t⁄ x

0 1 2 3 4

0 0 3 4 3 00.5 0 2 3 2 01 0 0 0 0 01.5 0 -2 -3 -2 02 0 -3 -4 -3 0

2. Solve 2 2

2 2

u u

t x

given that u(x,0) =0; u(0,t)=0 ut(x,0)=0 and u(l,t)=100sin(πt)

in the range 0≤t≤1 by taking h=1/4

Soln: Comparing the wave equation c2uxx = utt with the equation uxx= utt we have c2=1 or c=1By data h=1/4,k=h/c=1/4

We have seen that the condition ut(x,0)=0 will lead us to the formula

, 1,0 1,0

1

2i j i iu u u

1,1 0,0 2,0

1 10 0 0

2 2u u u

2,1 1,0 3,0

1 10 0 0

2 2u u u

3,1 2,0 4,0

1 10 0 0

2 2u u u

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4,1 3,0 5,0

1

2u u u

Hence

4,1u = 4 1, (1,1/ 4) 100sin( / 4) 0.707u x t u (Second row of the table is completed)We shall now conider the explicit formula to find the remaining values in the table

, 1 1, 1, , 1i j i j i j i ju u u u

1,2 0,1 2,1 1,0 0u u u u

2,2 1,1 3,1 2,0 0u u u u

3,2 2,1 4,1 3,0 70.7u u u u

4,2 3,1 5,1 4,0u u u u is inapplicable

4,2 4 2, (1,1/ 2) 100sin( / 2) 100u u x t u (Third row of the table is completed)

1,3 0,2 2,2 1,1 0u u u u

2,3 1,2 3,2 2,1 70.7u u u u

3,3 2,2 4,2 3,1 100u u u u

4,3 4 3, (1,3 / 4) 100sin(3 / 4) 70.7u u x t u (Fourth row of the table is completed)

1,4 0,3 2,3 1,2 70.7u u u u

2,4 1,3 3,3 2,2 100u u u u

3,4 2,3 4,3 3,2 70.7u u u u

4,4 4 4, (1,1) 100sin( ) 0u u x t u (Last row is completed)Thus the required values 0f ui,j are tabulated

t⁄ x

0 1/4 1/2 3/4 1

0 0 0 0 0 00.25 0 0 0 0 70.70.5 0 0 0 70.7 100.75 0 0 70.7 100 70.71 0 70.7 100 70.7 0

Numerical solution of the one dimensional heat equation

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We seek the numerical solution of heat equation ut=c2uxx..........................................(1)subject to the boundary conditions u(0,t)=0……………………………………………(2) u(l,t)=0…………………………………………….(3)and the initial condition u(x,0)=f(x)……………………………………….(4)we shall substitute the finite difference approximation for the partial derivatives present in (1)

2, 1 , 1, , 1,2

1 12i j i j i j i j i ju u c u u u

k h 2

, 1 , 1, , 1,22i j i j i j i j i j

kcu u u u u

h

Taking 2

2

kc

h=a the above equation becomes

, 1 , 1, , 1,2i j i j i j i j i ju u au au au

, 1 1, , 1,(1 2 )i j i j i j i ju au a u au …………………….(5)

This is called Schmidt explicit formula valid for 0<a<1/2For convenience let us set 1-2a=0 or a=1/2

That is 2

2

kc

h=1/2

Thus (5) becomes , 1 1, 1,

1

2i j i j i ju u u This is called Bendre Schmidt formula .

EXAMPLES

1) Find the numerical solution of the parabolic equation2

22

u u

x t

When u(0,t)=0=u(4,t) and u(x,00=x94-x0 by taking h=1

Soln: the standard form of the one dimensional heat equation is ut=c2uxx and the given equation can be put in the form ut=(1/2)uxx

c2=1/2 since h=1,k=2

21

2

h

c

the values of x=0 in 0<x<4 with h=1 are 0,1,2,3,4 and the values of t with k=1 are 0,1,2,3,4,5The initial table is given byConsider the initial condition u(x,0)=x(4-x)u1,0=u(1,0)=3; u2,0=4; u3,0=3

(First row in the table is completed)

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Now we consider the relation

, 1 1, 1,

1

2i j i j i ju u u

In particular

,1 1,0 1,0

1

2i i iu u u

1,1 0,0 2,0

1 10 4 2

2 2u u u

2,1 1,0 3,0

1 13 3 3

2 2u u u

3,1 2,0 4,0

1 14 0 2

2 2u u u

(The second row in the table is completed)

Again ,2 1,1 1,1

1

2i i iu u u

1,2 0,1 2,1

1 10 3 1.5

2 2u u u

2,2 1,1 3,1

1 12 2 2

2 2u u u

3,2 2,1 4,1

1 13 0 1.5

2 2u u u

(The third row in the table is completed)

Again ,3 1,3 1,2

1

2i i iu u u

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1,3 0,2 2,2

1 10 2 1

2 2u u u

2,3 1,2 3,2

1 11.5 1.5 1.5

2 2u u u

3,3 2,2 4,2

1 12 0 1

2 2u u u

(Fourth row in the table is completed)

Again ,4 1,3 1,3

1

2i i iu u u

1,4 0,3 2,3

1 10 1.5 0.75

2 2u u u

2,4 1,3 3,3

1 11 1 1

2 2u u u

3,4 2,3 4,3

1 11.5 0 0.75

2 2u u u

(Fifth row in the table is completed)

Again ,5 1,4 1,4

1

2i i iu u u

1,5 0,4 2,4

1 10 1 0.5

2 2u u u

2,5 1,4 3,4

1 10.75 0.75 0.75

2 2u u u

3,5 2,4 4,4

1 11 0 0.5

2 2u u u

He last row in the table is completed)Thus the required values 0f ui,j are tabulated

t⁄ x

0 1 2 3 4

0 0 3 4 3 01 0 2 03 2 02 0 1.5 02 1.5 03 0 1 701.5.7 1 04 0 0.75 1001 0.75 05 0 0.5 0.75 0.5 0

Numerical solution of the Laplace’s equation in two dimensions

Laplace’s equation in two dimensions is

2 2

2 20

u u

x y

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Here we consider a rectangular region R for which u(x,y) is know at the boundary. Let us suppose that the region is such that it can divided into a network of square mesh of side h.

We shall substitute the finite difference approximation for the partial derivatives ,hence we have

1, , 1, , 1 , , 12 2

1 12 2 0i j i j i j i j i j i ju u u u u u

h h

1, 1, , 1 , 1 ,4i j i j i j i j i ju u u u u

, 1, 1, , 1 , 1

1

4i j i j i j i j i ju u u u u ……………………(2)

This is called the standard five point formula. It may be observed that the value of ui,j at any interior mesh point is the average of its values at four neighboring points which is given in the figure below

, 1, _ 1 1, 1 1, 1 1, 1

1

4i j i j i j i j i ju u u u u ……………………………(3)

Equation (30 is called diagonal five point formula.

Examples1.solve Laplace’s uxx+uyy=0 for the following square mesh with boundary values as shown belowwww.vt

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u5 is located at the centre of the region.

5

10 21 17 12.1 12.525 .

4u by S F

Next we shall compute u7, u9, u1, u3by applying D.F

7

10 12.525 0 12.1 6.15625

4u

9

112.1 21 12.525 9 13.65625

4u

1

10 17 0 12.525 7.38125

4u

3

112.525 18.6 17 21 17.28125

4u

Finally we shall compute u2, u4, u6, u8 by applying S.F

2

17.38125 17.28125 17 12.525 13.546875

4u

4

10 12.525 7.38125 6.15625 6.515625

4u

6

112.525 21 17.28125 13.65625 16.115625

4u

8

16.15625 13.65625 12.525 12.1 11.109375

4u

Thus, u1=7.38, u2=13.55 , u3=17.28, u4=6.25, u5=12.53, u6=16.12, u7=6.16, u8=11.11, u9=13.66,

2.Solve uxx+uyy=0 in the following square region with the boundary conditions as indicated in the figure belowwww.vt

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Soln: we shall apply standard five point formula for u1, u2, u3, u4, to obtain a system of equations

1 2 3

120 100

4u u u

2 1 4

10 100

4u u u

3 4 1

130 0

4u u u

4 3 2

10 0

4u u u

Now we have a system of equations to be solved

1 2 34 120u u u ……………………………(1)

1 2 44 100u u u ………………………….(2)

1 3 44 30u u u ……………………………..(3)

2 3 44 0u u u ……………………………..(4)

Let us eliminate 1u from (1) and (2);(20 and (3)

That is 15 2u - 3u -4 4u =520………………………..(5)

4 2u -4 3u =70 …………..(6)

we shall now eliminate 4u from (4) and (5)

That is 14 2u -2 3u =520

Let us solve (6) and (7) : 2 2u -2 3u =35

14 2u -2 3u =520

Thus 2u =40.4, and 3u =22.9 hence from (1)

1u =45.825 and from(4) 4u =15.825

Thus the required values at th2 interior mesh points are

1u =45.825 , 2u =40.4, 3u =22.9 4u =15.825

3. solve the elliptic equation uxx+uyy=0 at the pivotal points for the following square mesh www.vt

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Soln: it is evident that the values at the boundary points on AB are respectively 200 and 100

Thus 1 2 3

1300 200

4u u u

2 1 4

1100 100

4u u u

3 4 1

1400 400

4u u u

4 3 2

1200 300

4u u u

1 2 34 500u u u ……………………………(1)

1 2 44 200u u u ………………………….(2)

1 3 44 800u u u ……………………………..(3)

2 3 44 500u u u ……………………………..(4)

Thus by solving simultaneously we obtain the values at the pivotal points

1u =250 , 2u =175, 3u =325 4u =250

4. Solve the elliptic partial differential equation for the following square mesh using the 5 point difference formula by setting up the linear equations at the unknown points P,Q,R,S

The linear equations for the unknownfor the 5 point difference formula are

Thus 11 1

4P Q S

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14 2

4Q P R

15 5

4R S Q

12 4

4S R P

4P-Q-S=2……………………………(1)

-P+4Q-R=6…………………………(2)

-Q+4R-S=10………………………(3)

-P-R+4S=6…………………………(4)

Thus by solving we get P=2,Q=3,S=3,R=4

UNIT VIIIDIFFERENCE EQUATIONS AND

Z- TRANSFORMSCONTENTS:

Introduction

Z-Transforms –Definition

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Damping Rule

Shifting RuleTheorem

Inverse Z-Transforms

Power Series Method

Applications Of Z-Transforms To Solve Difference Equations

DIFFERENCE EQUATIONS ANDZ- TRANSFORMS

Introduction

The Z-transform plays an important role plays in the study of communications, sample data control systems, discrete signal processing , solutions of difference equations etc.

Definition

Let un = f(n) be a real-valued function defined for n=0,1,2,3,….. and un = 0 for n<0. Then the Z-transform of un denoted by Z(un) is defined by

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0

)()(n

nnn zuuZzu (1)

The transform also is referred to as the one sided Z-transform or unilateral Z-transform. Next, we define un = f(n) for n=0, 1, 2, …… ∞.

The two-sided Z-transform is defined by

nnn zuuZ )( (2)

The region of the Z-plane in which the series (1) or (2) converges is called the region of convergence of the transform.

Properties of Z-transform

1. Linearity property

Consider the sequences {un} and {vn} and constants a and b. Then

Z[aun + bvn ] = aZ(un) + bZ(vn)


)()(

][][

0 0

0

nn

n n

nn

nn

n

nnnnn

vbZuaZ

zvbzua

zbvaubvauZ

In particular, for a=b=1, we get

Z[un+vn] = Z(un) + Z(vn)and for a=-b=1, we get

Z[un - vn] = Z(un) - Z(vn)

2. Damping property

Let Z(un) = )(zu . Then (i) azuuaZ n

n )( (ii) )()( azuuaZ nn

Proof : By definition, we havewww.vtuc

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a

zu

a

zuzuauaZ

n

n

nn

nn

nn

n

0 0

)()(

Thus

a

zuuaZ n

n )(

This is the result as desired. Here, we note that that if Z(un) = )(zu , then

][)( zuuaZ nn

aZZ

=

a

zu

Next,

)(

)()()(00

azu

azuzuauaZn

nn

n

nn

nn

n

Thus

)()( azuuaZ nn

This is the result as desired.

3. Shifting property

(a) Right shifting rule :

If Z(un) = )(zu , then Z(un-k) = z-k )(zu where k>0


0n

nknkn zuuZ

Since un = 0 for n<0, we have un-k = 0 for n=0,1,……(k-1)

Hence

)(

].......[

.......

0

110

)1(10

zuz

zuz

zuuz

zuzu

zuuZ

k

n

nn

k

k

kk

kn

nknkn

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Thus

Z(un-k) = z-k )(zu

(b)Left shifting rule :

]......)([)( )1(1

22

110

k

kk

kn zuzuzuuzuzuZ

Proof :

0

1

0

0

)(

0

)(

0

,

n

k

n

nn

nn

k

n

nkkn

k

n

nkkn

k

n

nknkn

zuzuz

knmwherezuzzuz

zuuZ

= ]......)([ )1(1

22

110

kk

k zuzuzuuzuz

Particular cases :

In particular, we have the following standard results :

1. ])([)( 01 uzuzuZ n

2. ])([)( 110

22

zuuzuzuZ n

3. ])([)( 22

110

33

zuzuuzuzuZ n etc.

Some Standard Z-Transforms :

1.Transform of an

By definition, we have

.....1

)(

0

2

0

n

n

n

nnn

z

a

z

a

z

a

zaaZwww.vtuc

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The series on the RHS is a Geometric series. Sum to infinity of the series is

az

zor

za 1

1Thus,

az

zaZ n

)(

In particular, when a=1, we get Z(1) = 1z

z

2. Transform of ean

Here)()( nan kZeZ where k = ea

aez

z

kz

z

Thus

aan

ez

zeZ

)(

3. Transform of np , p being a positive integer

We have,

0

)1(1

0

)(

n

np

n

npp

nznz

znnZ

Also, we have by defintion

0

11)(n

npp znnZ

Differentiating with respect to z, we get

0

)1(1

0

11

)(

)(

n

np

n

npp

znn

zndz

dnZ

dz

d

Using this in (1), we get

)]([)( 1 pp nZdz

dznZ

Particular cases of )( pnZ :-

1. For p = 1, we get

2)1(1

)1()(

z

z

z

z

dz

dzZ

dz

dznZ

(1)

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Thus,

2)1()(

z

znZ


3

2

22

)1()1()()(

z

zz

z

z

dz

dznZ

dz

dznZ

Thus,

3

22

)1()(

z

zznZ


4

233

)1(

4)(

z

zzznZ

4. Transform of nan

By damping property, we have

22

2

)(1

)1()()(

az

az

a

za

z

z

znZnaZ

aZZ

aZZ

n

, in view of damping

propertyThus,

2)()(

az

aznaZ n

5. Transform of n2an

We have,

a

ZZ

a

ZZ

n

z

zznZanZ

3

222

)1()()(

Thus,

3

222

)()(

az

zaazanZ n

6. Transforms of coshn and sinhn

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We have

)(2

1)(cosh

2cosh

nn

nn

eeZnZ

een

)()(2

1 nn eZeZ , by using the linearity property

1cosh2

cosh

1)(2

)(

1)(2

2

1

2

22

zz

zz

eezz

eez

zeezz

ezezz

ez

z

ez

z

Next,

1cosh2

sinh

1cosh22

11

2)(sinh

2sinh

2

2

zz

z

zz

eez

ezez

znZ

een

nn

7. Transforms of cosn and sinn

We have

)(2

1)(cos

2cos

inin

inin

eeZnZ

een

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1cos2

]cos[

1)(

)(2

2

2

1

2

2

zz

zz

eezz

eezz

ez

z

ez

z

inin

inin

ii

Next,

1cos2

sin

1cos22

2

1)(sin

2sin

2

2

zz

z

zz

ee

i

z

ez

z

ez

z

inZ

i

een

ii

ii

inin

Examples :

Find the Z-transforms of the following :

1. nn

nu

4

1

2

1

We have,

nn

n ZuZ4

1

2

1)(

)14)(12(

)38(2

14

4

12

2

4

1

2

1

4

1

2

1

zz

zz

z

z

z

z

z

z

z

z

ZZnn

2. !

1

nun

Here

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z

n

n

n

nn

ezz

nz

znn

ZuZ

1

2

00

........!2

1

!1

1

1

!

1

!

1

!

1)(

,

3. nau nn cos

We have

1cos2

)cos()(cos

2

zz

zznZ

By using the damping rule, we get

22

2

2

2

)cos(

1cos2

cos

1cos2

)cos()cos(

aazz

azz

a

z

a

z

a

z

a

z

zz

zznaZ

aZZ

n

4. un = )!2(

1

n

Let us denote vn = !

1

n, so that vn+2 =

)!2(

1

n= un

Here

zn evZ

1

)( Hence

z

vvvZzvZuZ nnn

10

22 )()()( , by left shifting rule

zez

zez

z

z

11

!1

1.

1

!0

1

12

12

List of standard inverse Z-transforms

, by exponential theorem

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1. 11

1

z

zZT 2. n

T kkz

zZ

1

3.

nz

zZT

21

14.

nk

kz

kzZ n

T

21

5.

23

21

1n

z

zzZT

6.

2

3

221 nk

kz

zkkzZ n

T

7.

34

231

1

4n

z

zzzZT

8.

3

4

32231 4

nkkz

zkzkkzZ n

T

9. )2sin(12

1 nz

zZT

10. )2cos(

12

21 n

z

zZT

Evaluation of inverse Z-transforms:We obtain the inverse Z-transforms using any of the following three methods:( I ) Power series Method:- This is the simplest of all the method of finding the inverse Z-transform. If U (z) is expressed as the ratio of two polynomials which cannot be factorized, we simply divide the numerator by the denominator and take the inverse Z-transform of each term in the quotient.Problems:

(1)Find the inverse transform of log 1z

z by power series method.

Putting

2 3

1 2 3

11 11 , ( ) log log(1 ) ...

1 2 31

1 1...

2 3

yz U z y y y yt

y

z z z

1

)(,.n

nn zuzUei

0 0

( 1)nn

for nThus u

otherwisen

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(2) Find the inverse Z-transform of 2( 1)z

z by division method.

)1(

12)(

2122

2

zzz

zzz

zzU

1

1

432

14111

21

212

)1(

14

13

12

1)(

)()(3)(211

11

)1(

32

n

nn

zzz

z

z

nZ

zzzzzU

z

z

z

z

Thus nu nn

1)1( ( II ) Partial fractions Method:- This method is similar to that of finding the inverse

Laplace transforms using partial fractions. The method consists of decomposing ( )U zz

into partial fractions, multiplying the resulting expansion by z and then inverting the same.

Problems:

(1) 22 3

( 2)( 4)

z z

z z

We write 22 3

( )( 2)( 4)

z zU z

z z

as

( ) 2 3 1 116 6( 2)( 4) 2 4

U z z A Bwhere A and B

z z z z z

Therefore

1 11( )

6 2 6 4

z zU z

z z

On inversion, we have

1 11( 2) (4)

6 6n n

nu nkz

z kz )(1

(2) 3

3

20

( 2) ( 4)

z z

z z

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We write 3

3

20( )

( 2) ( 4)

z zU z

z z

as

2 2

3 3

( ) 20

4( 2) ( 4) ( 2)

U z z A Bz Cz D

z zz z z

Multiplying throughout by

we get

Comparing the coefficients of z in the above equationwe get A = 6, B=0, C = ½ and D = -½ .Thus

On inversion, we get

(3)Find the inverse Z-transform of 32,)3)(2(

)5.65(22

2

zforzz

zz

Splitting into partial fractions, we obtain

2

22

2

)3(

1

3

1

2

1)(

1)3(32)3)(2(

)5.65(2)(

zzzzU

CBAwherez

C

z

B

z

A

zz

zzzU

22

3 3

1 16 0( ) 20 2 24( 2) ( 4) ( 2)

z zU z z

z zz z z

3( 2) ( 4)z z

2 2 320 ( )( 4) ( 2) .z A Bz Cz z D z

22

3 3

1 16 0( ) 20 2 24( 2) ( 4) ( 2)

z zU z z

z zz z z

3 2 2 3

3 3

3 2 22 2 3

3 3

2 2

3

1 12 1 1 12 4 4 1( ) .

2 2 4 2 2 4( 2) ( 2)

4 4 8 48 4 4 41 1 1 1

2 2 4 2 2 4( 2) ( 2)

1 ( 2) 4 8

2 ( 2)

z z z z z z z zU z

z zz z

z z z z zz z z z z z z

z zz z

z z z z

z

2

3

1 1 2 4 12

2 4 2 2 2 4( 2)

z z z z z

z z zz

2

1 2 2 1

1(2 2. 2 ) 4

2

2 2 2

n n nn

n n n

u n

n

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3227

4

9

3

3

21

9

1

27931

3

18421

1

113

19

1

31

3

121

2

1

3232

32

32

211

zwherezzzzzz

zzzz

andthatsozz

zz

z

.0,3)2(1,2,

)1(2

3

4

3

3

3

2

3

1

3333

1222

2

1

21

2

31

00

1

31

1

1

5

3

4

2

324

3

3

2

24

3

3

2

2

nnuandnugetweinversionon

znzz

zzzzzz

zzz

nn

nn

nn

n

n

n

n

n

nn

Difference Equations

INTRODUCTIONDifference equations arise in all situations in which sequential relation exists at various discrete values of the independent variable. The need to work with discrete functions arises because there are physical phenomena which are inherently of a discrete nature. In control engineering, it often happens that the input is in the form of discrete pulses of short duration. The radar tracking devices receive such discrete pulses from the target which is being tracked. As such difference equations arise in the study of electrical networks, in the theory of probability, in statistical problems and many other fields.

Just as the subject of difference equations grew out of differential calculus to become one of the must powerful instruments in the hands of a practical mathematician when dealing with continuous processes in nature, so the subject of difference equations is forcing its way to the fore for the treatment of discrete processes. Thus the difference equations may be thought of as the discrete counterparts of the differential equations.

DefinitionA difference equation is a relation between the differences of an unknown function at one or more general values of the argument.Eg: 1) 1 2n ny y

2) 2 5 6 0n n ny y y

3) 3 2 23 2n n n ny y y y x are difference equations.

An alternative way of writing a difference equation is as follows:Putting 1E , we get(1) may be written as,

1

1 1

2 1

( 1) 2

2 [sin ]

2 (4)

n n

rn n n n n r

n n n

E y y

Ey y y ce E y y

y y y

(2) may be written as,

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)5(023

052

06)55()12(

06)1(5)1(

12

112

2

2

nnn

nnnnn

nnn

nnn

yyy

yyyyy

yyEyEE

yyEyE

(3) may be written as

)6(5116

2236333

)1(2)12(3)133(

)1(2)1(3)1(

2123

2112123

2223

223

xyyyy

xyyyyyyyyyy

xyyEyEEyEEE

xyyEyEyE

nnnn

nnnnnnnnnn

nnnn

nnnn

The equations (4), (5) and (6) can also be written in terms of the operator E. i.e,

223

2

2

)5116(

0)23(

2)1(

xyEEE

yEE

yEE

n

n

n

Order of a difference equation:The order of a difference equation is the difference between the largest and the smallest arguments occurring in the difference equation divided by the unit of increment.

(interval)increment ofunit

argumentsmallest -argument largest .eqdifferenceaoforder . ei

Note: To find the order, the equation must be expressed in a form free of s . [because the highest power of does not give the order of difference equation]

therefore the order of the difference equation

(4) is = 21

)()2(

nn

(5) is = 21

)()2(

nn

(6) is = 31

)()3(

nn

The order of a difference equation can also be obtained by considering the highest power of the operator E involved in the equation.

Solution of a difference Equation is an expression for ( )ny which satisfies the given

difference equation.The general solution of a difference equation is that in which the number of arbitrary constants is equal to the order of the difference equation.

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A particular solution (or particular integral) is that solution which is obtained from the general solution by giving particular values to the constants.

APPLICATION OF Z-TRANSFORM TO SOLVE DIFFERENCE EQUATIONS

Working Procedure to solve a linear difference equation with constant coefficient by Z-transforms:1) Take the Z-transform of both sides of the difference equations using the Z-transforms formulae and the given conditions.2) Transpose all terms without )(zU to the right.3) Divide by the coefficient of )(zU , getting )(zU as a function of z.4) Express this function in terms of Z-transforms of known functions and take the inverse Z-transform of both sides. This gives nu as a function of n which is the desired solution.

Problems:(1) Using the Z-transform, solve 1,0334 1012 uuwithuuu n

nnn

)2()2(

)()(

,)()(),()(1

102

2

01

zzZAlso

zuuzUzuZ

uzUzuZthenzUuZIfthatnoteWe

n

n

nn

Taking the Z-transforms of both sides of the above difference equation, we get

)3()(3)(4)( 01

102

zzzUuzUzzuuzUz

Using the given conditions, it reduces to

,3

1

12

5

3

1

24

1

1

1

8

3

)3)(1)(3(

1

)3)(1(

1)(

)3()34)(( 2

zzz

zzzzzz

zU

zzzzzzU

on breaking into partial fractions, then

312

5

324

1

18

3)(

z

z

z

z

z

zzU

On inversion, we obtain

nnn

n z

zZ

z

zZ

z

zZu

)3(12

53

24

1)1(

8

3

312

5

324

1

18

3 111

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om



(2) Solve ,0296 1012 yywithyyy nnnn using Z-transform.

If 110

2201 )()(,)()(),()(

zyyzYzyZyzYzyZthenzYyZ nnn

Also )2()2( zzZ n

Taking Z-transforms of both sides, we get

)2()(9)(6)( 01

102

zzzYyzYzzyyzYz

Since )2()96)((,0,0 210 z

zzzzYhaveweyandy

Or ,)3(

5

3

1

2

1

25

1

)3)(2(

1)(22

zzzzzz

zYon splitting into partial

fractions.

Or

2)3(

53225

1)(

z

z

z

z

z

zzY

On taking inverse Z-transform of both sides, we obtain

nnnn

zz

zz

zz

n

naaz

azZn

ZZZy

21

35

)3(31

35

31

21

)()3()3(2

25

1

25

12

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