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.
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CHAPTER 1
FOURIER SERIES
1.1 PERIODIC FUNCTIONS
A function
is said to have a period T if for all x,
positive constant. The least value of T>0 is called the period of
, where T is a
.
EXAMPLES 1.1
= sin x = sin (x + 4 ) = … Therefore the function has period 2 ,
We know that
4 , 6 , 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
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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 = cos x , y =
are 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
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=
I year / I sem
+
where
2. The Fourier series of an odd function
the Fourier series of an odd function
=
in (-l , l) contains only sine terms, i.e.
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
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Using these values in (1), we have
x (2l - x) =
……………..(2)
in (0, 2l)
…
The required series
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)
i.e.
= x(2l – x).
= f(l)
= l(2l - l)
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 1
=0,
if n 1
=0
4
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if n 1
=
,
if n 1
=
Using these values in (1), we get
f(x) =
3. Find the Fourier series expansion of
= sin ax in (-l , l).
Solution:
Since
is defined in a range of length 2l, we can expand
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)
=
5
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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
6
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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.,
7
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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
8
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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 as a Fourier series of the form
+
in
in (c , c+2l), then the root-mean square value
of y =
(c , c+2l) is given by
PROOF
=
....……………….(1)
in (c , c+2l)
+
By Euler’s formulas for the Fourier coefficients,
..…………………(2)
…....……………..(3)
Now, by definition,
=
=
using (1)
=
=
,
by using (2) and (3)
9
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=
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)
).
10
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……………….(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
=
11
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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:
Giving an odd extension for
Let
in (-l , 0),
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.
12
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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)
13
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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
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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 )]
(i) Suppose the function
=
and now,
is defined in the interval (0 , 2l), then its Fourier series is,
+
= 2[mean value of y in (0 , 2l)]
=
15
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=
(ii) If the half range Fourier sine series of
=
in (0 , l) is,
, then
=
(iii) If the half range Fourier sine series of
=
in (0 ,
) is,
, then
=
(iv) If the half range Fourier cosine series of
=
+
in (0 , l) is,
, 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:
16
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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 values of 0,
,
,
,
,
,
,
takes the values 0,
,
,
. (By using (1))
Let the Fourier series be of the form
………………(2)
where
n=6
y
cos
sin
y cos
y sin
1.98
1.0
0
1.98
0
1.30
0.500
0.866
0.65
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
4.6
Substituting these values of
in (2), we get
= 0.75 + 0.37 cos + 1.004 sin
17
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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
18
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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
19
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1
18
9
15.7
-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
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
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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)
21
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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
22
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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
23
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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 series in
.
=0
is an even function in
.
is an odd function defined in (-l , l) what are the values of
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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
are piecewise continuous in
Then the Fourier series of
(a)
.
.
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:
25
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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 =
of the
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 )]
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
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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
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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
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CHAPTER 2
FOURIER TRANSFORMS
2.1 INTEGRAL TRANSFORM
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
= 0 for
= 0 in
i.e.,
f(t) = 1 in -1 < t < 1
and
Equation (1)
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.………………(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)
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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
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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
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2.5 COMPLEX FORM OF FOURIER INTEGRALS
The integral of the form
is known as Complex form of Fourier Integral.
2.5.1 FOURIER TRANSFORMS
COMPLEX FOURIER TRANSFORMS
The function
of
is called the Complex Fourier transform
.
INVERSION FORMULA FOR THE COMPLEX FOURIER TRANSFORM
is called the inversion formula for the
The function
Complex Fourier transform of
and it is denoted by
FOURIER SINE TRANSFORMS
The function
the function
is called the Fourier Sine Transform of
.
The function
is called the inversion formula for the
Fourier sine transform and it is denoted by
FOURIER COSINE TRANSFORMS
The function
Transform of
is called the Fourier Cosine
.
The function
is called the inversion formula for the
Fourier Cosine Transform and it is denoted by
2.5.2 PROBLEMS
1. Find the Fourier Transform of
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Hence prove that
Solution:
We know that the Fourier transform of
is given by
By using inverse Fourier Transform we get
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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
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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
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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 the inverse sine
transform of
Solution:
We know that the inverse Fourier sine transform of
is given by
Here
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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:
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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 )
If F(s) is the Fourier transform of
then
Proof:
Put
x-a = y
dx = dy
When
4. Shifting in respect of s
If F(s) is the Fourier transform of
then
Proof:
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5. Modulation Theorem
If F(s) is the Fourier transform of
then
Proof:
COROLLARIES
6. Conjugate Symmetry Property
If F(s) is the Fourier transform of
then
Proof:
We know that
Taking complex conjugate on both sides we get
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Put
I year / I sem
x = -y
dx = -dy
When
7. Transform of Derivatives
If F(s) is the Fourier transform of
and
continuously differentiable,
, then
and if
is continuous,
are absolutely integrable in
is piecewise
and
8. Derivatives of the Transform
If F(s) is the Fourier transform of
then
Proof:
Extending, we get,
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DEFINITION
is called the convolution product or simply the convolution
of the functions
and
and is denoted 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)
Let
.……………….(2)
i.e.,
………………..(3)
by property (9)
………………..(4)
i.e.,
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Substituting (2) and (4) in (1) we get
and
are given functions of x and
11. If
Fourier cosine transforms and
and
and
are their
are 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.
(ii) Replacing
in (i) and noting that
and
, we get
i.e.,
12. If
(i)
, then
and
(ii)
Proof:
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Similarly the result (ii) follows.
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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
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[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)]
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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.
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Using Parseval’s identity
3. Evaluate
using transforms.
Solution:
We know that the Fourier cosine transform of
Similarly the Fourier cosine transform of
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We know that
4. Find the Fourier transform of
and hence deduce that
(i)
(ii)
Ans :
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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 <x < l is defined as
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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
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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
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if its finite sine transform is given by
3. Find
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
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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.
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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:
13. If F(s) is the Fourier transform of
then
Proof:
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14. If F(s) is the Fourier transform of
I year / I sem
then
Proof:
15. If F(s) is the Fourier transform of
then
Proof:
Put
ax = y
a dx = dy
i.e., dx =
When
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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
,where z is a complex variable in general.
If
is a casual sequence, i.e if ,
called one-sided or unilateral Z-transform of
and denoted by
is called
or
for n<0,then the Z-transform is
and is 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:
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(i)
(ii)
Proof:
Extending this result, we get
(3)Frequency Shifting:
(i)
(ii)
Proof:
Similarly (ii) can be proved.
Corollary:
If
, then
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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:
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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.
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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
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Z-Transforms of some basic functions:
(1)
is the unit impulse sequence 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)
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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,
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.
(iv)
.
In particular,
.
(9)
.
(i)
.
(ii)
.
(iii)
.
(iv)
.
Problems:
(1) Find the bilateral Z-transforms of
(i)
(ii)
(iii)
Solution:
(i)
By property 3,
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(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)
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(3) Find Z-transforms of
(i)
, and
(ii)
Solution:
(i)
[Refer to basic transform (6)]
(ii)
, by partial fractions.
[Refer to basic transform (6)]
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(4) Find the Z-transforms of
(i)
(ii)
, and
(iii)
Solution:
(i)
Let
[Refer to basic transform (8)]
(ii) Let
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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,
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(6) Use convolution theorem to find the sum of the first n natural numbers.
Solution:
By convolution theorem,
Taking inverse Z-transforms,
(7) Use convolution theorem to find the inverse Z-transform of
(i)
and
(ii)
Solution:
(i)
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(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)
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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
is divided by
fail and if
,
in the classical 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
is derived as the sum of the
inverse Z-transforms of the partial fractions.
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.
at the isolated singularities.
Sum of the residues of
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
.
.
.
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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
,
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(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.
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UNIT 3
2 mark
(1)Form the difference equation from
Ans:
(2)Express
Ans:
in terms of
(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.
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is called the
or
,where z is
Transforms and Partial differential equations
(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:
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(11) Find the Z-transforms of
Ans:
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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
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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
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, 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:
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(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,
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.
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i.e.
i.e
i.e
(3) Form the partial differential equation by eliminating the arbitrary function ‘f’
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
from (4), (5) and (6) using determinants, we
and
have
=0
i.e.
or
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(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)
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………..(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)
The eliminant of ‘a’ between the two equations (3) and (4), when it exists, is called the
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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
.
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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
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Solution:
………..(1)
Given:
Equation (1) is of the form
………(2)
Assume
be the solution of equation (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.
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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.
To find the general solution:
We put
in (3), we get
………(4)
Differentiate (4) partially with respect to ‘a’, we get
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………..(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)
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which is the complete solution.
To find the general solution:
We put
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)
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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
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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)
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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)
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Substituting equation (2) & (3) in (1), we get
Integrating the above, 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)
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
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………..(7)
(which is absurd)
and
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
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Let
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(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,
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……………(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.
.which is a type 3
reduces to
The equation
equation.
Problem:
(1)Solve
.
Solution:
Given:
This can be written as
.
Which is of the form
, where m=2,n=2.
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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
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Therefore the equation
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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
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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
,
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Similarly,
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,
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
obtained from (3) satisfy equation (2)
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)
.
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As (2) contains
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, 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
obtained from (4) satisfy equation (3)
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:
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(1)Solve the equation
Also show that
if
when
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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,
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…………..(3)
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.
Lagrange’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
.
……….(1)
The subsidiary equations are
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
degree 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 function 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 of this equation 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
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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 denominator 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:
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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)
…… (4)
….. (5)
(2)
(3)
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
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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
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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
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…….. (2)
Substituting (2) in (1) we get
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 (2) in (1) we get
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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Transforms and Partial differential equations
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Transforms and Partial differential equations
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