1-Engineering-Mathematics-III.pdf

Transforms and Partial differential equations UNIT I

I year / I sem

TRANSFORMS AND PARTIAL DIFFERENTIAL EQUATIONS FOURIER SERIES

9

Fourier series – Odd and even functions – Half range sine series – Half range cosine series – Complex form of Fourier Series – Parseval’s identify – Harmonic Analysis. UNIT II

FOURIER TRANSFORM

9

Fourier integral theorem (without proof) – Sine and Cosine transforms – Properties (without Proof) – Transforms of simple functions – Convolution theorem – Parseval’s identity – Finite Fourier transform – Sine and Cosine transform. UNIT III

Z -TRANSFORM AND DIFFERENCE EQUATIONS

9

Z-transform - Elementary properties (without proof) – Inverse Z – transform – Convolution theorem Formation of difference equations – Solution of difference equations using Z - transform. UNIT IV

PARTIAL DIFFERENTIAL EQUATIONS

9

Solution of First order partial differential equation reducible to standard forms – Lagrange’s linear equation – Linear partial differential equations of second order and higher order with constant coefficients. UNIT V

BOUNDARY VALUE PROBLEMS

9

Solutions of one dimensional wave equation – One dimensional heat equation – Steady state solution of two-dimensional heat equation (Insulated edges excluded) – Fourier series solutions in Cartesian coordinates. TUTORIAL :15 TOTAL: 60 TEXT BOOKS

1. Andrews, L.A., and Shivamoggi B.K., “Integral Transforms for Engineers and Applied Mathematicians”, Macmillen , New York ,1988. 2. Grewal, B.S., “Higher Engineering Mathematics”, Thirty Sixth Edition, Khanna Publishers, Delhi, 2001. 3. Kandasamy, P., Thilagavathy, K., and Gunavathy, K., “Engineering Mathematics Volume III”, S. Chand & Company ltd., New Delhi, 1996. REFERENCES

1. Narayanan, S., Manicavachagom Pillay, T.K. and Ramaniah, G., “Advanced Mathematics for Engineering Students”, Volumes II and III, S. Viswanathan (Printers and Publishers) Pvt. Ltd. Chennai, 2002. 2. Churchill, R.V. and Brown, J.W., “Fourier Series and Boundary Value Problems”, Fourth Edition, McGraw-Hill Book Co., Singapore, 1987.

Prepared by : P.Padma

WWW.Vidyarthiplus.Com

Transforms and Partial differential equations

I year / I sem

CHAPTER 1 FOURIER SERIES 1.1 PERIODIC FUNCTIONS A function is said to have a period T all if for x,

,

positive constant. The least value of T>0 is called the period of

where T is a

.

EXAMPLES 1.1

We know that 4

,6

=

sin x = sin (x + 4

)=

…

Therefore the function has period 2

,

, etc. However, 2 is the least value and therefore is the period of f(x). Similarly cos x is a periodic function with the period 2

and tan x has period

.

1.2 DIRICHLET’S CONDITIONS

A function series of the form

defined in c

x

c+2l can be expanded as an infinite trigonometric

+

provided

1.

is single- valued and finite in (c , c+2l)

2.

is continuous or piecewise continuous with finite number of finite discontinuities in (c , c+2l). has no or finite number of maxima or minima in (c , c+2l).

3.

1.3 EULER’S FORMULAS

If a function series

defined in (c , c+2l) can be expanded as the infinite trigonometric

+

[ Formulas given above for

then

and

are called Euler’s formulas for Fourier coefficients]

1




I year / I sem

1.4 DEFINITION OF FOURIER SERIES

The infinite trigonometric series Fourier series of

in the interval c

+ x

is called the c+2l, provided the coefficients are given by the

Euler’s formulas. EVEN FUNCTION

If

=

in (-l , l) such that

=

, then

is said to be an even

function of x in (-l , l).

If Such that

=

or

=

, then

is said to be an even function of x in

(-l , l). EXAMPLE y = cosx , y =

a re even functions.

ODD FUNCTION

If

=

in (-l , l) such that

= -

, then

is said to be an odd

function of x in (-l , l).

If Such that

= -

or

=-

, then

is said to be an odd function of x in

(-l , l). EXAMPLE y = sin x , y = x are odd functions. 1.5 FOURIER SERIES OF EVEN AND ODD FUNCTIONS

1. The Fourier series of an even function

in (-l , l) contains only cosine terms

(constant term included), i.e. the Fourier series of an even function

in (-l , l) is

given by 2



Transforms and Partial differential equations =

I year / I sem

+

where in (-l , l) contains only sine terms, i.e.

2. The Fourier series of an odd function the Fourier series of an odd function =

in (-l , l) is given by

,

where 1.6 PROBLEMS 1. Find the Fourier series of period 2l for the function the sum of

= x(2l

–

x) in (0 , 2l). Deduce

=

Solution:

Let

=

in (0 , 2l)

+

…………(1)

using Bernoulli’s formula. =

=0 3 Prepared by : P.Padma



I year / I sem

Using these values in (1), we have = x (2l - x)

(0, 2l)

in

The required series

…

……………..(2)

can be obtained by putting x = l in the Fourier

series in (2). x = l lies in (0 , 2l) and is a point of continuity of the function Sum the Fourier series in (2)

= x(2l – x).

= f(l)

= l(2l - l)

i.e.

i.e.. -

=

…

2. Find the Fourier series of period 2

for the function

= x cos x in 0 < x < 2

.

Solution:

Let

=

.……..…………(1)

+

if n =0,

if n

1

1

=0

4 Prepared by : P.Padma



I year / I sem

if n =

,

if n

1 1

= Using these values in (1), we get f(x) =

3. Find the Fourier series expansion of

= sin ax in (-l , l).

Solution:

is defined in a range of length 2l, we can expand

Since

in Fourier series of

period 2l. Also

= sin[a(-x)] = -sin ax = is an odd function of x in (-l , l).

Hence Fourier series of Let

will not contain cosine terms. ………………….(1)

=




I year / I sem

Using these values in (1), we get

4. Find the Fourier series expansion of

=

. Hence obtain a series for

cosec Solution:

Though the range

is symmetric about the origin,

is neither an even function

nor an odd function. Let

=

+

..…..…………(1)

in




I year / I sem

Using these values in (1), we get =

in

[Since x=0 is a point of continuity of f(x)] i.e.,

i.e.,

i.e.,




I year / I sem

1.7 HALF-RANGE FOURIER SERIES AND PARSEVAL’S THEOREM

(i) The half range cosine series in (0 , l) is =

+

where

(ii) The half range sine series in (0 , l) is =

,

where (iii) The half range cosine series in (0 ,

=

) is given by

+

where

(iv) The half range sine series in (0 ,

=

) is given by

,

where




I year / I sem

1.7.1 ROOT-MEAN SQUARE VALUE OF A FUNCTION Definition

If a function y =

is defined in (c , c+2l), then

is called the root mean-

square(R.M.S.) value of y in (c , c+2l) and is denoted by

Thus 1.7.2 PARSEVAL’S THEOREM

If y=

can be expanded a s a Fourier series of the form in (c , c+2l), then the root-mean square value

+ in

of y =

(c , c+2l) is given by

PROOF

=

in (c , c+2l)

+

....……………….(1)

By Euler’s formulas for the Fourier coefficients, ..…………………(2)

…....……………..(3) Now, by definition, =

=

using (1)

= =

,

by using (2) and (3)




I year / I sem

= 1.7.3 EXAMPLES 1. Find the half-range (i) cosine series and (ii) sine series for Solution:

(i) To get the half-range cosine series for extension for i.e. put Now

in (

, 0).

=

=

is even in (

,

in (

in (0 ,

………………….(1)

is given by

(ii) To get the half-range sine series of

in (0 ,

). ), we should give an odd extension

, 0). Put

==-

Now

), we should give an even

+

in (0 ,

i.e.

)

, 0)

The Fourier half-range cosine series of

in (-

in (0 ,

). =

for

=

is odd in (-

,

in (in (-

, 0)

, 0)

).




I year / I sem ……………….(2)

=

Using this value in(2), we get the half-range sine series of

2. Find the half-range sine series of

in (0 ,

).

= sin ax in (0 , l).

Solution:

We give an odd extension for i.e. we put

in (-l , 0).

= -sin[a(-x)] = sin ax in (-l , 0) is odd in (-l , l)

Let

=




I year / I sem

Using this values in (1), we get the half-range sine series as

3. Find the half-range cosine series of

= a in (0 , l). Deduce the sum of

. Solution:

in (-l , 0),

Giving an odd extension for Let

is made an odd function in (-l , l). ..……………(1)

f(x) =

Using this value in (1), we get a= Since the series whose sum is required contains constant multiples of squares of

, we apply

Parseval’s theorem.




4. Expand

=

-

r.m.s. value of

I year / I sem

as a Fourier series in -1 < x < 1 and using this series find the in the interval.

Solution:

The Fourier series of =

in (-1 , -1) is given by .………………(1)

+

……………….(3)




I year / I sem

Substituting (2), (3), (4) in (1) we get = We know that r.m.s. value of f(x) in (-l , l) is ……………….(5) From (2) we get .………………..(6) From (3) we get ………………..(7) From (4) we get ..………………(8) Substituting (6), (7) and (8) in (5) we get

5. Find the Fourier series for

=

in

Hence show that

Solution:

The Fourier series of

in (-1 , 1) is given by =

+ 14



Transforms and Partial differential equations The co-efficients

I year / I sem

are

Parseval’s theorem is

i.e.,

=

1.8 HARMONIC ANALYSIS The process of finding the Fourier series for a function given by numerical value is known as harmonic analysis. In harmonic analysis the Fourier coefficients of the

function y =

in (0 , 2

) are given by

= 2[mean value of y in (0 , 2

)]

= 2[mean value of y cos nx in (0 , 2 )] = 2[mean value of y sin nx in (0 , 2 )] is defined in the interval (0 , 2l), then its Fourier series is,

(i) Suppose the function = and now,

+

= 2[mean value of y in (0 , 2l)] =




I year / I sem

= in (0 , l) is,

(ii) If the half range Fourier sine series of =

, then

= (iii) If the half range Fourier sine series of =

in (0 ,

) is,

, then

= in (0 , l) is,

(iv) If the half range Fourier cosine series of =

+

, then

= 2[mean value of y in (0 , l)] = (v) If the half range Fourier cosine series of =

+

in (0 ,

) is,

, then

= 2[mean value of y in (0 ,

)]

=

.

1.8.1 EXAMPLES 1. The following table gives the variations of a periodic function over a period T. x

1.98

Show that

= 0.75 + 0.37

1.3

+1.004

1.05

,

1.3

-0.88

-0.25

1.98

where

Solution:




I year / I sem

Here the last value is a mere repetition of the first therefore we omit that value and consider the remaining 6 values.

n = 6. ..………………..(1)

Given

when x takes the values0,of

,

,

,

,

,

,

take

s the values 0,

,

,

. (By using (1))

Let the Fourier series be of the form ………………(2)

where

n=6 cos

sin

1.98 1.30

1.0 0.500

0 0.866

1.98 0.65

0 1.1258

1.05

-0,500

0.866

-0.525

0.9093

1.30

-1

0

-1.3

0

-0.88

-0.500

-0.866

0.44

0.762

-0.25

0.500

-0.866

-0.125

0.2165

1.12

3.013

y

4.6

Substituting these values of

y cos

y sin

in (2), we get = 0.75 + 0.37 cos + 1.004 sin 17




I year / I sem

2. Find the Fourier series upto the third harmonic for the function y = (0 ,

defined in

) from the table x

0

2.34

2.2

1.6

0.83

0.51

0.88

1.19

Solution: We can express the given data in a half range Fourier sine series. ..………………...(1)

x

y = f(0)

sin x

sin 2x

sin 3x

y sin x

y sin 2x

y sin 3x

0

2.34

0

0

0

0

0

0

30

2.2

0.5

0.87

1

1.1

1.91

2.2

60

1.6

0.87

0.87

0

1.392

1.392

0

90

0.83

1

0

-1

0.83

0

-0.83

120

0.51

0.87

-0.87

0

0.44

-0.44

0

150

0.88

0.5

-0.87

1

0.44

0.76

0.88

180

1.19

0

0

0

0

0

0

4.202

3.622

2.25

Now

Substituting these values in (1), we get

= 1.4 sin x + 1.21 sin 2x + 0.75 sin 3x 3. Compute the first two harmonics of the Fourier series for f(x) from the following data x

0

30

60

90

120

150

180

0

5224

8097

7850

5499

2626

0




I year / I sem

Solution:

Here the length of the interval is

we can express the given data in a half range

Fourier sine series ………………………(1)

i.e., x

y

sin x

sin 2x

0

0

0

0

30

5224

.5

0.87

60

8097

0.87

0.87

90

7850

1

0

120

5499

0.87

-0.87

150

2626

0.5

-0.87

Now

= 7867.84 sin x + 1506.84 sin 2x 4. Find the Fourier series as far as the second harmonic to represent the function given in the following data. x

0

1

2

3

4

5

9

18

24

28

26

20

Solution:

Here the length of the interval is 6 (not 2 ) i.e., 2l = 6 or l = 3 The Fourier series is …………………..(1)

y 0

0

0

9

9

0

9

0




I year / I sem

1

18 9

-9

15.6

2

24

-12

20.9

-24

0

3

28

-28

0

28

0

4

26

-13

-22.6

-13

22.6

5

20

10

-17.4

-10

-17.4

125

-25

-3.4

-19

20.8

Substituting these values of

15.7

in (1), we get

1.9 COMPLEX FORM OF FOURIER SERIES

The equation of the form is called the complex form or exponential form of the Fourier series of coefficient

When l =

in (c , c+2l). The

is given by

, the complex form of Fourier series of

in (c , c+2

) takes the form

where




I year / I sem

1.9.1 PROBLEMS 1. Find the complex form of the Fourier series of

=

in (0 , 2).

Solution:

Since 2l = 2 or l = 1, the complex form of the Fourier series is

Using this value in (1), we get

2. Find the complex form of the Fourier series of

= sin x in (0 ,

).

Solution:

Here 2l =

or l =

.

The complex form of Fourier series is …………………..(1)




I year / I sem

Using this value in (1), we get in (0 , 3. Find the complex form of the Fourier series of

=

) in (-l , l).

Solution:

Let the complex form of the Fourier series be

Using this value in (1), we have




I year / I sem in (-l , l)

4. Find the complex form of the Fourier series of

= cos ax in (-

), where a is

,

neither zero nor an integer. Solution:

Here 2l = 2

or l =

.

The complex form of Fourier series is ………………….(1)

Using this value in (1), we get in (-

,

).

2mark PART – A

1. Determine the value of Ans:

in the Fourier series expansion of

is an odd function.

2. Find the root mean square value of

in the interval

.

Ans:

RMS Vale of

in

is




3. Find the coefficient

of

I year / I sem

in the Fourier cosine series of the function

in

the interval Ans: Here

Fourier cosine series is =

+

, where

4. If

and

series of Ans:

at Here

for all x, find the sum of the Fourier

. is a point of discontinuity.

The sum of the Fourier series is equal to the average of right hand and left hand limit of the given function at

.

i.e.,

5. Find Ans:

Since 6. If

in the expansion of

as a Fourier serinies

.

=0 is an even function in

.

is an odd function defined in (-l , l) what are the values of 24



Transforms and Partial differential equations Ans:

I year / I sem

=0 since

is an odd function.

7. Find the Fourier constants Ans:

for

in

.

=0

Since

is an even function in

.

8. State Parseval’s identity for the half-range cosine expansion of

in (0 , 1).

Ans:

where

9. Find the constant term in the Fourier series expansion of

in

.

Ans:

= 0 since

is an odd function in

.

10. State Dirichlet’s conditions for Fourier series. Ans:

(i)

is defined and single valued except possibly at a finite number of points in

(ii)

is periodic with period 2

(iii)

and

(a)

.

are piecewise continuous in

Then the Fourier series of

.

.

converges to

if x is a point of continuity

(b)

if x is a point of discontinuity.

11. What you mean by Harmonic Analysis? Ans:




I year / I sem

The process of finding the Fourier series for a function given by numerical value is known as harmonic analysis. In harmonic analysis the Fourier coefficients function y =

in (0 , 2

of the

) are given by

= 2[mean value of y in (0 , 2

)]

= 2[mean value of y cos nx in (0 , 2 )] = 2[mean value of y sin nx in (0 , 2 )]

12. In the Fourier expansion of

in

. Find the value of

,

the coefficient of sin nx. Ans:

Since

is an even function the value of

= 0.

13. What is the constant term and the coefficient of

in the Fourier expansion of

in (-7 , 7)? Ans:

Given

The given function is an odd function. Hence

14. State Parseval’s identity for full range expansion of

are zero.

as Fourier series in (0 , 2l).

Ans:

= where




15. Find a Fourier sine series for the function

I year / I sem

= 1; 0 < x <

.

Ans:

…………………….(1)

The Fourier sine series of

16. If the Fourier series for the function

is

Deduce that Ans:

Putting

we get




I year / I sem

17. Define Root mean square value of a function? Ans:

If a function y =

is defined in (c , c+2l), then

is called the root mean-

square(R.M.S.) value of y in (c , c+2l) and is denoted by Thus

18. If

is expressed as a Fourier series in the interval (-2 , 2), to which value this

series converges at x = 2. Ans:

Since x = 2 is a point of continuity, the Fourier series converges to the arithmetic mean of at x = -2 and x = 2 i.e., 19. If the Fourier series corresponding to

in the interval

without finding the values of

is find the value of

Ans:

By using Parseval’s identity,

20. Find the constant term in the Fourier series corresponding to interval

expressed in the

.

Ans:

Given Now




I year / I sem

CHAPTER 2 FOURIER TRANSFORMS 2.1 INTEGRAL TRANS FORM

The integral transform of a function

is defined by

where

k(s , x) is a known function of s and x and it is called the kernel of the transform. When k(s , x) is a sine or cosine function, we get transforms called Fourier sine or cosine transforms. 2.2 FOURIER INTEGRAL THEOREM

If

is a given function defined in (-l , l) and satisfies Dirichlet’s conditions, then

At a point of discontinuity the value of the integral on the left of above equation is

2.3 EXAMPLES

1. Express the function

as a Fourier Integral. Hence evaluate

and find the value of

Solution: We know that the Fourier Integral formula for

is ……………….(1)

Here

= 1 for

i.e.,

f(t) = 1 in -1 < t < 1

= 0 for = 0 in

and

Equation (1)




I year / I sem

.………………(2)

[Using sin (A+B) + sin (A-B) = 2 sin A cos B] This is Fourier Integral of the given function. From (2) we get

=

……………….(3

………………..(4)

But Substituting (4) in (3) we get

=

Putting x = 0 we get

2. Find the Fourier Integral of the function

Verify the representation directly at the point x = 0. Solution: The Fourier integral of is ……………….(1)

……….………(2)




I year / I sem

Putting x = 0 in (2), we get

The value of the given function at x = 0 is

. Hence verified.

2.4 FOURIER SINE AND COSINE INTEGRALS

The integral of the form

is known as Fourier sine integral. The integral of the form

is known as Fourier cosine integral. 2.4.1 PROBLEMS 1. Using Fourier integral formula, prove that

Solution: The presence of in the integral suggests that the Fourier sine integral formula has been used. Fourier sine integral representation is given by




I year / I sem

2. Using Fourier integral formula, prove that

Solution: The presence of in the integral suggests that the Fourier cosine integral formula for has been used. Fourier cosine integral representation is given by




I year / I sem

2.5 COMPLEX FORM OF FOURIER INTEGRALS

The integral of the form

is known as Complex form of Fourier Integral. 2.5.1 FOURIER TRANS FORMS COMPLEX FOURIER TRANSFORMS

The function of

is called the Complex Fourier transform

.

INVERSION FORMULA FOR THE COMPLEX FOURIER TRANS FORM

The function

is called the inversion formula for the

Complex Fourier transform of

and it is denoted by

FOURIER SIN E TRANS FORMS

The function the function

is called the Fourier Sine Transform of

.

The function


Fourier sine transform and it is denoted by FOURIER COSIN E TRANS FORMS

The function Transform of

is called the Fourier Cosine .

The function


Fourier Cosine Transform and it is denoted by 2.5.2 PROBLEMS 1. Find the Fourier Transform of




I year / I sem

Hence prove that

Solution: We know that the Fourier transform of

is given by

By using inverse Fourier Transform we get




I year / I sem

The second integral is odd and hence its values is zero.

i.e., Putting

, we get

2. Find the Fourier sine transform of

,

(or)

, x > 0. Hence evaluate

Solution: The Fourier sine transform of f(x) is given by

Here

=

for x > 0

Using inverse Fourier sine transform we get




I year / I sem

Replacing x by m we get

[since s is dummy variable, we can replace it by x]

3. Find the Fourier cosine transform of Solution:

We know that Here

Let

………………(1)

Then

Differentiating on both sides w.r.t. ‘s’ we get,

Integrating w.r.t. ‘s’ we get 36 Prepared by : P.Padma



I year / I sem

4. Find the Fourier cosine transform of Solution: We know that the Fourier cosine transform of f(x) is

Here

5. Find

, if its sine transform is

Hence deduce that th e inverse sine

transform of Solution: We know that the inverse Fourier sine transform of

is given by

Here 37 Prepared by : P.Padma



I year / I sem

Differentiating w.r.t. ‘x’ on both sides, we get,

To find the inverse Fourier sine transform of Put a = 0, in (1), we get

2.5.3 PROPERTIES 1. Linearity Property

Proof:




I year / I sem

2. Change of Scale Property

If F(s) is the Fourier transform of

then

Proof:

Put

ax = y a dx = dy

i.e., dx =

When

3. Shifting Property ( Shifting in x )


then

Proof:

Put

x-a = y dx = dy

When

4. Shifting in respect of s


then

Proof:




I year / I sem

5. Modulation Theorem


then

Proof:

COROLLARIES

6. Conjugate Symmetry Property


then

Proof:

We know that

Taking complex conjugate on both sides we get 40 Prepared by : P.Padma



Put

I year / I sem

x = -y dx = -dy

When

7. Transform of Derivatives

If F(s) is the Fourier transform of continuously differentiable,

and

and if

is continuous,

are absolutely integrable in

is piecewise and

, then

8. Derivatives of the Transform


then

Proof:

Extending, we get,

41




I year / I sem

DEFINITION

is called the convolution product or simply the convolution of the functions

and

anddenoted is by

.

9. Convolution Theorem

If F(s) and G(s) are the Fourier transform of and respectively then the Fourier transform of the convolution of f(x) and g(x) is the product of their Fourier transforms. i.e., Proof :

Inverting, we get

10. Parseval’s Identity (or) Energy Theorem

If

is a given function defined in

where F(s) is the Fourier transform of Proof: We know that

then it satisfy the identity,

.

Putting x = 0, we get

………………..(1) .……………….(2) ………………..(3)

Let i.e.,

by property (9) ………………..(4)

i.e.,




I year / I sem

Substituting (2) and (4) in (1) we get

11. If

and

are given functions x and of

Fourier cosine transf and orms

and

and a

a

re their

re their Fourier sine transforms then

(i) (ii)

,

which is Parseval’s identity for Fourier cosine and sine transforms. Proof:

(i)

Changing the order of integration

Similarly we can prove the other part of the result. Replacing (ii) in and(i)noting that

and

, we get

i.e.,

12. If (i)

, then and

(ii) Proof:




I year / I sem

Similarly the result (ii) follows.




I year / I sem

2.5.4 PROBLEMS

1. Show that the Fourier transform of

. Hence deduce that

is

Using Parseval’s

identity show that

Solution: We know that

………………..(A)

When a = 1, Using inverse Fourier Transform, we get




I year / I sem

[The second integral is odd and hence its value is zero]

[since the integrand is an even function of s] Putting a = 1, we get

Putting x = 0, in the given function we get

Using Parseval’s identity,

[Using (A)]




2. Find the Fourier Transform of

I year / I sem

if

Hence deduce that Solution: We know that

Since

The second integral becomes zero since it is an odd function.




I year / I sem

Using Parseval’s identity

3. Evaluate

using transforms.

Solution:

We know that the Fourier cosine transform of Similarly the Fourier cosine transform of 48 Prepared by : P.Padma



I year / I sem

We know that

4. Find the Fourier transform of

and hence deduce that

( i)

(ii)

Ans :




I year / I sem

Using inversion formula, we get

Putting a = 1, we get,

2.6 FINITE FOURIER TRANSFORMS

If transform of

is a function defined in the interval (0 , l) then the finite Fourier sine in 0



I year / I sem

where ‘n’ is an integer

The inverse finite Fourier sine transform of

The finite Fourier cosine transform of

is

and is given by

in 0 < x < l is defined as where ‘n’ is an integer

The inverse finite Fourier cosine transform of

is

and is given by

2.6.1 PROBLEMS 1. Find the finite Fourier sine and cosine transforms of

in 0 < x < l.

Solution: The finite Fourier sine transform is

Here

The finite Fourier cosine transform is 51 Prepared by : P.Padma



I year / I sem

Here

2. Find the finite Fourier sine and cosine transforms of Solution: The finite Fourier sine transform of

.

is

Here

The finite Fourier cosine transform of

is

Here




3. Find

if its finite sine transform is given by

integer and

I year / I sem

where p is positive

.

Solution: We know that the inverse Fourier sine transform is given by

………………..(1)

Here

………………..(2)

=

Substituting (2) in (1), we get

4. If

find

if 0 < x <1.

Solution:

We know that

Here

Let




I year / I sem

2 mark PART A

1. State the Fourier integral theorem. Ans: If is a given function defined in (-l , l) and satisfies Dirichlet’s conditions, then

2. State the convolution theorem of the Fourier transform. Ans:

If F(s) and G(s) are the Fourier transform of and respectively then the Fourier transform of the convolution of f(x) and g(x) is the product of their Fourier transforms. i.e., 3. Write the Fourier transform pair. Ans: and are Fourier transform pairs. 4. Find the Fourier sine transform of Ans:

5. If the Fourier transform of Ans:

Put

(a > 0).

is F(s) then prove that .

x-a = y dx = dy

When

6. State the Fourier transforms of the derivatives of a function. 54 Prepared by : P.Padma



I year / I sem

Ans:

7. Find the Fourier sine transform of Ans:

Here

.

for x > 0

8. Prove that Proof:

Put

ax = y a dx = dy

i.e., dx =

When

9. If F(s) is the Fourier transform of

then prove that

Proof:




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10. Find the Fourier sine transform of Ans:

11. Find Fourier sine transform of Ans:

12. Find Fourier cosine transform of Ans:


then

Proof:





I year / I sem

then

Proof:


then

Proof:

Put

ax = y a dx = dy

i.e., dx =

When




I year / I sem

Chapter 3 Z-Transforms 3.1 Introduction: In Communication Engineering, two basic types of signals are encountered. They are (1) Continuous time signals. (2) Discrete time signals. Continuous time signals are defined for continuous values of the independent variable, namely time and are denoted by a function .

Discrete time signals are defined only at discrete set of values of the independent variable and are denoted by a sequence . Z-transform plays an important role in analysis of linear discrete time signals. 3.2 Definition of z-transform:

If

is a sequence defined for

.,then

the two-sided or bilateral Z-transform of

and denoted by

is called or

,where z is a complex variable in general. If

is a casual sequence, i.e if ,

for n<0,then the Z-transform is

called one-sided or unilateral Z-transform of

and i

s defined as

We shall mostly deal with one sided Z-transform which will be hereafter referred to as Z-transform. 3.3 Properties of Ztransforms: (1) Linearity:

The Z-transform is linear . Proof:

similarly,

.

(2)Time Shifting:

58




I year / I sem

(i) (ii) Proof:

Extending this result, we get

(3)Frequency Shifting:

(i) (ii) Proof:

Similarly (ii) can be proved. Corollary: If

, then 59




The result follows, if we replace

I year / I sem

in (ii).

(4)Time Reversal for Bilateral Z-Transform:

If Proof:

(5) Differentiation in the Z-Domain:

(i) (ii) Proof: (i)

Similarly, (ii) can be proved. (6) Initial Value Theorem: (i)

(ii) Proof:




I year / I sem

Similarly, (ii) can be proved. (7) Final value Theorem: (i)

(ii) Proof:

Taking limits as z tends to 1,

Similarly, (ii) can be proved, starting with property 2(ii). (8) Convolution Theorem: Definitions: The convolution of the two sequences

is defined as

(i)

, if the sequence are non causal and

(ii)

, if the sequences are causal.

The convolution of two functions

is defined as

where T is the sampling period.




I year / I sem

Statement of the theorem: (i)

(ii) Proof: (For the bilateral z=transform) (i)

By changing the order of summation, , by putting n-r=m

(ii) ……………(1) Say, where

……………(2) Using (2) in (1), we get




I year / I sem

Z-Transforms of some basic functions: (1) is the unit impulse sequ ence defined by

(2) defined by

Where k is a constant and

is the unit step sequence

(i)

Where the region of convergence (ROC) is

.

(ii) In particular, and

(3)

(i)

, where the ROC is

(ii)

, where the ROC is

(iii)

.

(iv)

. .

.

(v)

(4)

.

(i)




(ii)

I year / I sem

Where the ROC is

.

(iii) (iv) (v)

(5)

.

(i) (ii)

(6)

.

(i)

.

(ii)

.

(7)

(i)

. .

(ii) Putting a=1, we get

.

(8)

.

(i)

.

(ii)

.

(iii)

.

In particular, 64 Prepared by : P.Padma



I year / I sem

.

(iv)

.

In particular, .

(9)

.

(i)

.

(ii)

.

(iii)

.

(iv)

.

Problems: (1) Find the bilateral Z-transforms of (i) (ii) (iii) Solution: (i)

By property 3,




I year / I sem

(ii)

By property 3, which is true for bilateral Z-transform also. s

(iii)

by property5.

(2) Find the Z-transforms of (i) and (ii) Solution: (i)

(ii)




I year / I sem

(3) Find Z-transforms of (i)

, and

(ii) Solution: (i)

[Refer to basic transform (6)]

(ii)

, by partial fractions.





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(4) Find the Z-transforms of (i) (ii)

, and

(iii) Solution: (i)

Let


(ii) Let




I year / I sem

By basic transform (8)

(iii)

Let

(5) (i) Use initial value theorem to find

(ii) Use final value theorem to find

, when

, when

Solution: (i) By initial value theorem,

(ii)

By final value theorem,




I year / I sem

(6) Use convolution theorem to find the sum of the first n natural numb ers. Solution:

By convolution theorem,

Taking inverse Z-transforms,

(7) Use convolution theorem to find the inverse Z-transform of (i )

and

(ii) Solution: (i)




I year / I sem

(ii)

Inverse Z-transforms:

The inverse of Z-transform of , when

has been already defined as .

can be found out by any one of the following methods. Method 1 (Expansion method)



Transforms and Partial differential equations If

I year / I sem

can be expanded in a series of ascending powers of

form

, i.e in the

, by binomial, exponential and logarithmic theorems, the coefficient of

in the expansion gives

.

Method 2 (Long division method)

When the usual methods of expansion of then

isdivided by

fail and if

,

in the cla ssical manner and hence the expansion

is obtained in the quotient.

Method 3 (partial fraction Method) When is a rational function in which the denominator can be factorised,

is resolved in to partial fraction and then inverse Z-transforms of the partial fractions.

is derived as the sum of the

Method 4 (By Cauchy’s Residue Theorem) By using the relation between the Z-transform and Fourier transform of a sequence, it can be proved that

Where C is a circle whose centre is the origin and radius is sufficiently large to include all the isolated singularities of . By Cauchy’s residue theorem, x sum of the residues of

at the isolated singularities.

Sum of the residues of

at the isolated singularities.

Use of Z-transforms to solve Finite Difference equations: Z-transforms can be used to solve finite difference equation of the form with given values of y(0) and y(1). Taking Z-transforms on both sides of the given difference equation and using the values of y(0) and y(1), we will get . Then

will give To express (i) (ii)

and

.

in terms of

.

. .




I year / I sem

Problems: (1) Find the inverse Z-transform of

, by the long division method.

Solution:

Thus

,

(2) Find the inverse Z-transform of

, by the long division method.

Solution:

Thus

,




(3) Find

I year / I sem

by the method of partial fractions.

Solution:

Let

(4) Find

, by using Residue theorem.

Solution: , Where C is the circle whose

centre is the origin and which includes the singularities

. ,by Cauchy’s residue

theorem. are simple poles.




I year / I sem

UNIT 3 2 mark

(1)Form the difference equation from Ans:

(2)Express

in terms of

Ans:

(3)Find the value of Ans:

when

(4)Define bilateral Z-transform. Ans : If

is a sequence defined for

two-sided or bilateral Z-transform of a complex variable in general.

.,then and denoted by

(5)Find the z-transform of Ans:

(6)Find Ans:

using z-transform.



iscalled the or

,where z is


(7)Define unilateral Z-transform. Ans: If is a casual sequence, i.e if , one-sided or unilateral Z-transform of

(8)Find

I year / I sem

for n<0,then the Z-transform is called and is defined as

using z-transform.

Ans:

(9) State and prove initial value theorem in z-transform. Ans: (i) (ii) (i)

Similarly, (ii) can be proved. (10)Find the z-transform of n. Ans:




I year / I sem

(11) Find the Z-transforms of Ans:




I year / I sem

CHAPTER 4 4.1 PARTIAL DIFFERENTIAL EQUATIONS

A partial differential equation is an equation involving a function of two or more variables and some of its partial derivatives. Therefore a partial differential equation contains one dependent variable and one independent variable. Here z will be taken as the dependent variable and x and y the independent variable so that . We will use the following standard notations to denote the partial derivatives.

The order of partial differential equation is that of the highest order derivative occurring in it.

Formation of partial differential equation :

There are two methods to form a partial differential equation. (i) By elimination of arbitrary constants. (ii) By elimination of arbitrary functions.

4.2 Problems Formation of partial differential equation by elimination of arbitrary constants : (1)Form the partial differential equation by eliminating the arbitrary constants from

.

Solution:

……………... (1)

Given




I year / I sem

Here we have two arbitrary constants a & b. Differentiating equation (1) partially with respect to x and y respectively we get ……………… (2) ………………. (3) Substitute (2) and (3) in (1) we get , which is the required partial differential equation.

(2) Form the partial differential equation by eliminating the arbitrary constants a, b, c from

.

Solution:

We note that the number of constants is more than the number of independent variable. Hence the order of the resulting equation will be more than 1. .................. (1) Differentiating (1) partially with respect to x and then with respect to y, we get

Differentiating (2) partially with respect to x, ……………..(4)

Where

.

From (2) and (4) ,

.

From (5) and (6), we get 78 Prepared by : P.Padma



I year / I sem , which is the required partial differential

equation.

(3) Find the differential equation of all spheres of the same radius c having their center on the yoz-plane.

.

Solution:

The equation of a sphere having its centre at

, that lies on the

-plane

and having its radius equal to c is ……………. (1)

If a and b are treated as arbitrary constants, (1) represents the family of spheres having the given property. Differentiating (1) partially with respect to x and then with respect to y, we have …………… (2) …………….(3)

and

…………….(4)

From (2),

……………..(5)

Using (4) in (3), Using (4) and (5) in (1), we get

. i.e.

, which is the required partial differential

equation.

Problems Formation of partial differential equation by elimination of arbitrary functions:




I year / I sem

(1)Form the partial differential equation by eliminating the arbitrary function

‘f’

from

solution: Given

……………(1)

i.e.

Differentiating (1) partially with respect to x and then with respect to y, we get ………….(2) …………….(3) where Eliminating f’(u) from (2) and (3), we get

i.e.

(2) Form the partial differential equation by eliminating the arbitrary function ‘

………………(1)

Solution: Given

Let

,

Then the given equation is of the form The elimination of

from equation (2), we get,



.

’


I year / I sem

i.e.

i.e i.e

‘f’

(3) Form the partial differential equation by eliminating the arbitrary function from

.…………(1)

Solution: Given

Differentiating (1) partially with respect to x, ………….(2) Where

and

Differentiating (1) partially with respect to y, …………. (3) Differentiating (2) partially with respect to x and then with respect to y, …………. (4) ………….. (5)

and Differentiating (3) partially with respect to y,

………….. (6) Eliminating

and

from (4), (5) and (6) using determinants, we

have

=0

i.e. or




I year / I sem

(4) Form the partial differential equation by eliminating the arbitrary function ‘

’

from .

…………...(1)

Solution: Given

Where Differentiating partially with respect to x and y, we get …………(2) ………….(3) ………..(4) …………(5) …………..(6) From (4) and (6), we get

=

=

i.e.

Solutions of partial differential equations

Consider the following two equations

………..(1)




I year / I sem ………..(2)

and

Equation (1) contains arbitrary constants a and b, but equation (2) contains only one arbitrary function f. If we eliminate the arbitrary constants a and b from (1) we get a partial differential equation of the form

. If we eliminate the arbitrary function f from (2) we get

a partial differential equation of the form

.

Therefore for a given partial differential equation we may have more than one type of solutions. Types of solutions:

(a) A solution in which the number of arbitrary constants is equal to the number of independent variables is called Complete Integral (or) Complete solution. (b) In complete integral if we give particular values to the arbitrary constants we get Particular Integral.

(c) The equation which does not have any arbitrary constants is known as Singular Integral.

To find the general integral:

………....(1)

Suppose that

is a first order partial differential equation whose complete solution is ………..(2) Where a and b are arbitrary constants. Let

, where ‘f’is an arbitrary function.

Then (2) becomes ……….(3) Differentiating (3) partially with respect to ‘a’, we get

……….(4) 83 The eliminant of ‘a’ between the two equations (3) and (4), when it exists, is called the Prepared by : P.Padma



I year / I sem

general integral of (1).

Methods to solve the first order partial differential equation: Type 1:

………...(1)

Equation of the form

i.e the equation contains p and q only. …….....(2)

Suppose that is a solution of the equation

substitute the above in (1), we get

on solving this we can get

, where

is a known function.

Using this value of b in (2), the complete solution of the given partial differential equation is …………(3) is a complete solution, To find the singular solution, we have to eliminate ‘a’ and ‘c’ from

Differentiating the above with respect to ‘a’ and ‘c’, we get , and 0=1. The last equation is absurd. Hence there is no singular solution for the equation of Type 1.

Problems:

(1) Solve

. 84




I year / I sem

Solution:

………….(1)

Given: Equation (1) is of the form

. ………….(2)

Assume be the solution of equation (1). From (2) we get

.

(1) ……….(3) Substitute (3) in (2) we get ……....(4) This is a complete solution.

To find the general solution: in (4),where ‘f’ is an arbitrary function.

We put

…………(5)

i.e.

Differentiating (5) partially with respect to ‘a’, we get

……………(6) Eliminating ‘a’ between equations (5) and (6), we get the required general solution.

To find the singular solution: Differentiate (4) partially with respect to ‘a’ and ‘c’, we get , 0=1.(which is absurd) so there is no singular solution.

(2) Solve




I year / I sem

Solution:

………..(1)

Given: Equation (1) is of the form

………(2)

Assume be the solution of eq uation (1). From (2) we get (1)

…….....(3) Substituting (3) in (2), we get ………(4) This is a complete solution.

To find the general solution: We put

in (4), we get ……..(5)

Differentiating (5) partially with respect to ‘a’, we get

………..(6) Eliminating ‘a’ between equations (5) and (6), we get the required general solution

To find the singular solution: Differentiating (4) with respect to ‘a’ and ‘c’.

and 0=1 (which is absurd). So there is no singular solution.




I year / I sem

Type 2: (Clairaut’s type) The equation of the form

………(1) is known as Clairaut’s equation.

…………(2)

Assume be a solution of (1).

Substitute the above in (1), we get ………..(3) which is the complete solution.

Problem:

(1) Solve Solution:

……….(1)

Given: Equation (1) is a Clairaut’s equation

………..(2)

Let be the solution of (1). Put

in (1), we get ……….(3)

which is a complete solution.


in (3), we get ………(4)

Differentiate (4) partially with respect to ‘a’, we get 87 Prepared by : P.Padma



I year / I sem ………..(5)

Eliminating ‘a’ between equations (4) and (5), we get the required general solution

To find singular solution, Differentiate (3) partially with respect to ‘a’, we get

………..(6) Differentiate (3) partially with respect to ‘b’, we get

………..(7) Multiplying equation (6) and (7),we get

(2) Solve Solution:

……….(1)

Given: Equation (1) is a Clairaut’s equation

………...(2)

Let be the solution of (1). Put

in (1), we get ……….(3) 88




I year / I sem

which is the complete solution.


in (3), we get ……..(4)

Differentiate (4) partially with respect to ‘a’, we get ……..(5) Eliminating ‘a’ between equations (4) and (5), we get the required general solution

To find the singular solution: Differentiate (3) partially with respect to ‘a’,

.............. (4) Differentiate (3) partially with respect to ‘b’,

………….(5) Substituting equation (4) and (5) in equation (3), we get

Type 3: Equations not containing x and y explicitly, i.e. equations of the form

……….(1)




I year / I sem

For equations of this type ,it is known that a solution will be of the form ……….(2) Where ‘a’ is the arbitrary constant and Putting

is a specific function to be found out.

, (2) becomes

and If (2) is to be a solution of (1), the values of p and q obtained should satisfy (1). ……..(3)

i.e. From (3), we get

……….(4) Now (4) is a ordinary differential equation, which can be solved by variable separable method. The solution of (4), which will be of the form

, is the

complete solution of (1). The general and singular solution of (1) can be found out by usual method. Problems: (1)Solve

.

Solution:

…………(1)

iven: Equation (1) is of the form Assume

where,

be a solution of (1). …….(2) ……(3)

Substituting equation (2) & (3) in (1), we get 90 Prepared by : P.Padma



I year / I sem

By variable separable method,

By integrating, we get

……….(4) This is the complete solution. To find the general solution: We put

in (4), we get ……..(5)

Differentiate (5) partially with respect to ‘a’, we get ………..(6) 91 Prepared by : P.Padma



I year / I sem

Eliminating ‘a’ between equations (4) and (5), we get the required general solution.

To find the singular solution: Differentiate (4) partially with respect to ‘a’ and ‘k’, we get ………..(7)

and

(which is absurd)

So there is no singular solution.

(2)Solve

.

Solution:

………(1)

Given: Equation (1) is of the form Assume

where ,

be a solution of (1). …….(2)

……(3)




I year / I sem

Substituting equation (2) & (3) in (1), we get

Integrating the above, we get

………..(4) This is the complete solution.


in (4), we get ……..(5)

Differentiate (5) partially with respect to ‘a’, we get ……..(6) Eliminating ‘a’ between equations (4) and (5), we get the required general solution. To find the singular solution: Differentiate (4) partially with respect to ‘a’ and ‘k’, we get 93 Prepared by : P.Padma



I year / I sem ………..(7)

and

(which is absurd)

So there is no singular solution.

Type 4: Equations of the form

………..(1) i.e. equation which do not contain z explicitly and in which terms containing p and x can be separated from those containing q and y. To find the complete solution of (1), .where ‘a’ is an arbitrary constant.

We assume that Solving

,we can get

and solving

,we can get

. Now

i.e. Integrating with respect to the concerned variables, we get ……….(2) The complete solution of (1) is given by (2), which contains two arbitrary constants ‘a’ and ‘b’. The general and singular solution of (1) can be found out by usual method.

Problems: (1)Solve

.

Solution:

Given: ………..(1) Equation (1) is of the form 94 Prepared by : P.Padma


Transforms and Partial differential equations Let

I year / I sem

(say)

………….(2) …………(3)

Similarly, Assume

be a solution of (1)

Substitute equation (2) and (3) to the above, we get

Integrating the above we get,

………..(4) This is the complete solution. The general and singular solution of (1) can be found out by usual method.

(2) Solve

.

Solution:

Given: ………….. (1) Equation (1) is of the form Let

(say) …………. (2) ……………(3)

Similarly, Assume

be a solution of (1)

Substitute equation (2) and (3) to the above, we get

Integrating the above we get, 95 Prepared by : P.Padma



I year / I sem

……………(4) This is the complete solution. The general and singular solution of (1) can be found out by usual method.

Equations reducible to standard types-transformations: Type A: Equations of the form

.

Where m and n are constants, each not equal to 1. We make the transformations

.

Then

and

Therefore the equation

reduces to

.which is a

type 1 equation. The equation

.which is a type 3

reduces to

equation.

Problem: (1)Solve

.

Solution:

Given: This can be written as . Which is of the form

, where m=2,n=2.




I year / I sem

Put

Substituting in the given equation, . This is of the form Let

, where

Equation becomes,

Solving for

.

.

,

is a complete solution. The general and singular solution can be found out by usual method.

Type B: Equations of the form

.

Where k is a constant, which is not equal to -1. We make the transformations Then

. and




Therefore the equation

I year / I sem

reduces to

which is a type 1

equation. The equation

reduces to

which is a

type 4 equation.

Problems: (1)Solve: Solution:

Given: The equation can be rewritten as Which contains

………(1)

.

Hence we make the transformation

Similarly Using these values in (1), we get ………..(2) As (2) is an equation containing P and Q only, a solution of (2) will be of the form ………….(3) Now

obtained from (3) satisfy equation (2)

i.e. Therefore the complete solution of (2) is i.e complete solution of (1) is 98 Prepared by : P.Padma



I year / I sem

Singular solution does not exist. General solution is found out as usual. Type C: Equations of the form

, where

We make the transformations

Then

and Therefore the given equation reduces to

This is of type 1 equation.

Problem: (1)Solve Solution:

Given:

………..(1)

It can be rewritten as which is of the form we make the transformations

i.e.

Then , 99 Prepared by : P.Padma


Transforms and Partial differential equations Similarly,

I year / I sem

,

Using these in (1),it becomes …………(2) As (2) contains only P and Q explicitly, a solution of the equation will be of the form ………….(3) Therefore


i.e. Therefore the complete solution of (2) is

Therefore the complete solution of (1) is

Singular solution does not exist. General solution is found out as usual.

Type D: Equation of the form

By putting

the equation reduces

to where

.

Problems: (1)Solve

.

Solution:

Given:

…………….(1)

.

Rewriting (1), …………….(2)

.



Transforms and Partial differential equations As (2) contains

I year / I sem

, we make the substitutions

Then i.e. Similarly, Using these in (2), it becomes …………..(3) which contains only P and Q explicitly. A solution of (3) is of the form …………(4) Therefore


i.e. Therefore the complete solution of (3) is Therefore the complete solution of (1) is

……..(5)

General solution of (1) is obtained as usual.

General solution of partial differential equations:

Partial differential equations, for which the general solution can be obtained directly, can be divided in to the following three categories. (1) Equations that can be solved by direct (partial) integration. (2) Lagrange’s linear equation of the first order. (3) Linear partial differential equations of higher order with constant coefficients.

Equations that can be solved by direct (partial) integration : Problems:



Transforms and Partial differential equations (1)Solve the equation Also show that

if when

I year / I sem when

when

.

Solution:

…………….(1)

Given: Integrating (1) partially with respect to x,

…………….(2) When

in (2), we get

. …………….(3)

Equation (2) becomes Integrating (3) partially with respect to t, we get

……………(4) Using the given condition, namely

when

we get

Using this value in (4), the required particular solution of (1) is

Now

i.e. when

.

(2) Solve the equation

simultaneously.

Solution: Given

Integrating (1) partially with respect to x, 102 Prepared by : P.Padma



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Differentiating (1) partially with respect to y, …………...(4) Comparing (2) and (4), we get

Therefore the required solution is , where c is an arbitrary constant.

Lagran ge’s linear equation of the first order:

A linear partial differential equation of the first order , which is of the form

where

are functions of

is called Lagrange’s linear equation.

working rule to solve

(1)To solve

, we form the corresponding subsidiary simultaneous equations

(2)Solving these equations, we get two independent solutions (3)Then the required general solution is

. .

Solution of the simultaneous equations Methods of grouping:

By grouping any two of three ratios, it may be possible to get an ordinary differential equation containing only two variables, eventhough P;Q;R are in general, functions of x,y,z. By solving this equation, we can get a solution of the simultaneous equations. By this method, we may be able to get two independent solutions, by using different groupings.




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Methods of multipliers:

If we can find a set of three quantities l,m,n which may be constants or functions of the variables x,y,z, such that

, then the solution of the

simultaneous equation is found out as follows.

Since

If

differential of some function

, then we get

is an exact Integrating this, we get

, which is a solution of Similarly, if we can find another set of independent multipliers another independent solution

we can get

.

Problems: (1)Solve

.

Solution:

Given:

.

This is of Lagrange’s type of PDE where

.

The subsidiary equations are

Taking first two members Integrating we get

………….(1)

i.e.

Taking first and last members i.e.

. . ………......(2)

Integrating we get




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Therefore the solution of the given PDE is

.

(2)Solve the equation Solution:

Given: This is of Lagrange’s type of PDE where

.

The subsidiary equations are

……….(1)

Using the multipliers 1,1,1, each ratio in (1)=

.

.

……………(2)

Integrating, we get Using the multipliers y,x,2z, each ratio in (1)=

.

.

……………(3)

Integrating, we get Therefore the general solution of the given equation is

.

(3)Show that the integral surface of the PDE

.

Which contains the straight line

.

Solution:

The subsidiary equations of the given Lagrange ‘s equation are ……………(1)

Using the multipliers

. each ratio in (1)=

.

.

……………(2)

Integrating, we get




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Using the multipliers y,x,-1, each ratio in (1)=

.

.

……………(3)

Integrating, we get The required surface has to pass through

……………(4) Using (4) in (2) and (3), we have ……………(5) Eliminating x in (5) we get, …………….(6) Substituting for a and b from (2) and (3) in (6), we get , which is the equation of the required

surface.

Linear P.D.E.S of higher order with constant coefficients :

The standard form of a homogeneous linear partial differential equation of the order with constant coefficients is …………….(1) where a’s are constants. If we use the operators

, we can symbolically write equation (1)

as …………….(2) …………….(3) where

is a homogeneous polynomial of the

deg ree in

.

The method of solving (3) is similar to that of solving ordinary linear differential equations with constant coefficients.




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The general solution of (3) is of the form z = (complementary function)+(particular integral),where the complementary function is the R.H.S of the general solution of and the particular integral is given symbolically by

.

:

Complementary fun ction of

C.F of the solution of

is the R.H.S of the solution of …………(1)

.

,then we get an equation which is called the

In this equation, we put auxiliary equation.

Hence the auxiliary equation of (1) is ………….(2) Let the roots ofequation this be

.

Case 1:

The roots of (2) are real and distinct. The general solution is given by

Case 2:

Two of the roots of (2) are equal and others are distinct. The general solution is given by

Case 3:

‘r’ of the roots of (2) are equal and others distinct. The general solution is given by

To find particular integral: Rule (1): If the R.H.S of a given PDE is

, then

Put




If

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refer to Rule (4).

Rule (2): If the R.H.S of a given PDE is

, then

provided the

Replace denominator is not equal to zero. If the denominato r is zero, refer to Rule (4). Rule (3): If the R.H.S of a given PDE is

Expand

by using Binomial Theorem and then operate on

Rule (4): If the R.H.S of a given PDE

Rule(1),(2) and(3)] resolve

, then

.

is any other function [other than

into linear factors say

etc. then the

Note: If the denominator is zero in Rule (1) and (2) then apply Rule (4). Working rule to find P.I when denominator is zero in Rule (1) and Rule (2).

If the R.H.S of a given PDE is of the form Then This rule can be applied only for equal roots.

Problems: (1) Solve Solution:

Given: 108 Prepared by : P.Padma



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The auxiliary equation is

The general solution of the given equation is

(2)Solve Solution:

Given: The auxiliary equation is




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Therefore the general solution is

(3)Solve Solution:

Given: The auxiliary equation is




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Therefore the general solution is




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UNIT 4

2 mark

(1)Form a partial differential equation by eliminating arbitrary constants a and b from Ans: Given

……. (1)

Substituting (2) & (3) in (1), we get (2) Solve: Ans: Auxiliary equation

(3)Form a partial differential equation by eliminating the arbitrary constants a and b from the equation . Ans: Given: ….. (1) Partially differentiating with respect to ‘x’ and ‘y’ we get ….. (2) …... (3) (2)

…… (4)

(3)

….. (5)

Substituting (4) and (5) in (1) we get . . (4)Find the complete solution of the partial differential equation Ans: Given: …….. (1) Let us assume that ……… (2) be the solution of (1) Partially differentiating with respect to ‘x’ and ‘y’ we get 112 Prepared by : P.Padma



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…….. (3)

Substituting (3) in (1) we get From the above equation we get,

………. (4)

Substituting (5) in (2) we get

(5)Find the PDE of all planes having equal intercepts on the x and y axis. Ans: The equation of such plane is ……….

(1)

Partially differentiating (1) with respect to ‘x’ and ‘y’ we get ……….. (2) ……….. (3) From (2) and (3), we get (6)Find the solution of Ans: The S.E is

.

Taking first two members, we get

Integrating we get

i.e Taking last two members, we get

Integrating we get 113 Prepared by : P.Padma



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i.e The complete solution is

(7)Find the singular integral of the partial differential equation Ans: The complete integral is

Therefore

(8)Solve: Ans: …….. (1) Let us assume that …… (2) be the solution of (1) Partially differentiating with respect to ‘x’ and ‘y’ we get …….. (3)

Substituting (3) in (1) we get This is the required solution. (9)Form a partial differential equation by eliminating the arbitrary constants a and b from

Ans:

……… (1) Partially differentiating with respect to ‘x’ and ‘y’ we get 114




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…….. (2)


This is the required PDE. (10)Solve: Ans: Auxiliary equation

(11)Form a partial differential equation by eliminating the arbitrary constants a and b from Ans:

……. (1)

Given

Substituting (2) & (3) in (1), we get

(12)Solve: Ans: The given equation can be written as

We know that the C.F corresponding to the factors

In our problem




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(13)Form a partial differential equation by eliminate the arbitrary function f from

Ans:

From (1), we get

Substituting (3) in(2), we get

(14)Solve: Ans: Auxiliary equation

(15)Obtain partial differential equation by eliminating arbitrary constants a and b from Ans: Given

……. (1)

Substituting (2) & (3) in (1), we get

(16)Find the general solution of

Ans: Auxiliary equation is




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General solution is

(17)Find the complete integral of

Ans: Let us assume that

……… (1) be the solution of the given equation. Partially differentiating with respect to ‘x’ and ‘y’ we get

…….. (2)


Substituting the above in (1) we get

This gives the complete integral. (18)Solve: Ans: Auxiliary equation

(19)Find the PDE of the family of spheres having their centers on the line x=y=z. Ans: The equation of such sphere is Partially differentiating with respect to ‘x’ and ‘y’ we get

From (1),

From (2),




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From (3) and (4), we get

This is the required PDE.




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Chapter 5 Boundary value problems




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Solution :



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1-Engineering-Mathematics-III.pdf

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