fourier series part - a...(or) explain half range sine series in the interval β€ β€π . solution:...
TRANSCRIPT
Fourier series
Part - A
1.Define Dirichletβs conditions
Ans:
A function defined in π β€ π₯ β€ π + 2π can be expanded as an infinite trigonometric series of
the form
π0
2+ ππ cos
πππ₯
π+ ππ sin
πππ₯
π , provided
β
π=1
i) π π₯ is single valued and periodic in π, π + 2π
ii) π π₯ is continuous or piecewise continuous with finite number of finite discontinuous in
π, π + 2π
iii) π π₯ has no or finite number of maxima or minima in π, π + 2π .
2. Write the Fourier series for the function f(x) in the interval π β€ π β€ π + ππ.
Solution:
π π₯ =π0
2+ ππ cos
πππ₯
π+ ππ sin
πππ₯
π
β
π=1
Where
π0 =1
π π π₯ ππ₯
π+2π
π
ππ =1
π π π₯ πππ
πππ₯
πππ₯
π+2π
π
ππ =1
π π π₯ π ππ
πππ₯
πππ₯
π+2π
π
3. Write the Fourier series for the function f(x) in the interval π β€ π β€ ππ.
Solution:
π π₯ =π0
2+ ππ cos
πππ₯
π+ ππ sin
πππ₯
π
β
π=1
Where
π0 =1
π π π₯ ππ₯
2π
0
ππ =1
π π π₯ πππ
πππ₯
πππ₯
2π
0
ππ =1
π π π₯ π ππ
πππ₯
πππ₯
2π
0
4. Write the Fourier series for the function f(x) in the interval βπ β€ π β€ π.
Solution:
π π₯ =π0
2+ ππ cos
πππ₯
π+ ππ sin
πππ₯
π
β
π=1
Where
π0 =1
π π π₯ ππ₯
π
βπ
ππ =1
π π π₯ πππ
πππ₯
πππ₯
π
βπ
ππ =1
π π π₯ π ππ
πππ₯
πππ₯
π
βπ
5. Write the Fourier series for the function f(x) in the interval π β€ π β€ ππ .
Solution:
π π₯ =π0
2+ ππ cos ππ₯ + ππ sin ππ₯
β
π=1
Where
π0 =1
π π π₯ ππ₯
2π
0
ππ =1
π π π₯ πππ ππ₯ππ₯
2π
0
ππ =1
π π π₯ π ππ ππ₯ππ₯
2π
0
6.Write the Fourier series for the function f(x) in the interval βπ β€ π β€ π .
Solution:
π π₯ =π0
2+ ππ cos ππ₯ + ππ sin ππ₯
β
π=1
Where
π0 =1
π π π₯ ππ₯
π
βπ
ππ =1
π π π₯ πππ ππ₯ππ₯
π
βπ
ππ =1
π π π₯ π ππ ππ₯ππ₯
π
βπ
7. Write the Fourier series for the even function f(x) in the interval βπ β€ π β€ π .
(or) Explain half range cosine series in the interval π β€ π β€ π .
Solution:
π0 =2
π π π₯ ππ₯
π
0
ππ =2
π π π₯ πππ ππ₯ππ₯
π
0
ππ = 0, Then the Fourier series is reduced to
π π₯ =π0
2+ ππ cos ππ₯
β
π=1
8. Write the Fourier series for the odd function f(x) in the intervalβπ β€ π β€ π .
(or) Explain half range sine series in the intervalπ β€ π β€ π .
Solution:
π0 = 0, ππ = 0, ππ =2
π π π₯ π ππ ππ₯ππ₯
π
0
,
Then the Fourier series is reduced to
π π₯ = ππ sin ππ₯
β
π=1
9. Write the Fourier series for the even function f(x) in the interval βπ β€ π β€ π.
(or) Explain half range Fourier cosine series in the interval π β€ π β€ π
Solution:
π0 =2
π π π₯ ππ₯
π
0
ππ =2
π π π₯ πππ
πππ₯
πππ₯
π
0
ππ = 0, Then the Fourier series is reduced to
π π₯ =π0
2+ ππ cos
πππ₯
π
β
π=1
10. Write the Fourier series for the odd function f(x) in the interval βπ β€ π β€ π.
(or) Explain half range Fourier sine series in the interval π β€ π β€ π
Solution:
π0 = 0, ππ = 0, ππ =2
π π π₯ π ππ
πππ₯
πππ₯
π
0
Then the Fourier series is reduced to
π π₯ = ππ sinπππ₯
π
β
π=1
11. Define periodic function and give examples.
Ans:
A function π π₯ is said to be periodic, if and only if π π₯ + π = π π₯ where T is called period
for the function π π₯ .
Eg: sin π₯ πππ cos π₯ are periodic functions with period 2π.
12. Write any two advantages of Fourier series.
Ans:
i) Discontinuous function can be represented by Fourier series. Although derivatives of the
discontinuous functions do not exist.
ii) The Fourier series is useful in expanding the periodic functions. Since outside the closed
interval, there exists a periodic extension of the function
iii) Fourier series of a discontinuous function is not uniformly convergent at all points.
iv) Expansion of an oscillation function by Fourier series gives all modes of oscillation which
is extremely useful in physics.
13. Define Parsevalβs identity for the Fourier series in the interval π, ππ
Solution:
1
π π π₯ 2
2π
0
ππ₯ =π0
2
2+ ππ
2 + ππ2
β
π=1
14. Define Parsevalβs identity for the Fourier series in the interval π, ππ
Solution:
1
π π π₯ 2
2π
0
ππ₯ =π0
2
2+ ππ
2 + ππ2
β
π=1
15. Define Parsevalβs identity for the Fourier series in the interval βπ , π
Solution:
1
π π π₯ 2
π
βπ
ππ₯ =π0
2
2+ ππ
2 + ππ2
β
π=1
16. Define Parsevalβs identity for the Fourier series in the interval βπ, π
Solution:
1
π π π₯ 2
π
βπ
ππ₯ =π0
2
2+ ππ
2 + ππ2
β
π=1
17. Define Parsevalβs identity for the half range Fourier cosine series in the interval π, π
Solution:
2
π π π₯ 2
π
0
ππ₯ =π0
2
2+ ππ
2
β
π=1
18. Define Parsevalβs identity for the half range Fourier sine series in the interval π, π
Solution:
2
π π π₯ 2
π
0
ππ₯ = ππ2
β
π=1
19. Define Parsevalβs identity for the half range Fourier cosine series in the interval π, π
Solution:
2
π π π₯ 2
π
0
ππ₯ =π0
2
2+ ππ
2
β
π=1
20. Define Parsevalβs identity for the half range Fourier sine series in the interval π, ππ
Solution:
2
π π π₯ 2
π
0
ππ₯ = ππ2
β
π=1
21. Define Root-Mean Square value of a function π π
Solution:
RMS value of a function π π₯ = π π₯ 2ππ₯π
π
πβπ
22. Find the value of ππin the cosine series expansion of π π = π in the interval π, ππ .
Solution:
ππ =2
π π π₯ πππ
πππ₯
πππ₯
π
0
=2
10 ππππ
πππ₯
10ππ₯
10
0
=π
5 πππ
πππ₯
10ππ₯
10
0
=π
5 π ππ
πππ₯10
ππ10
0
10
=π
5Γ
10
ππ π ππ
10ππ
10β sin 0
ππ = 0
23. Find the root mean square value of the function π π = π in the interval π, π
Solution:
RMS value of a function
π π₯ = π π₯ 2ππ₯π
π
π β π
= π₯2ππ₯π
0
π β 0
=
π₯3
3
0
π
π
=
π3
3π
=π
3
24. State the convergence theorem on Fourier series.
Ans:
i) The Fourier series of π π₯ converges to π π₯ at all points where π π₯ is continuous. Thus,
if π π₯ is continuous at π₯ = π₯0 , the sum of the Fourier series when π₯ = π₯0 is π π₯0 .
ii) At a point π₯ = π₯0 where π π₯ has a finite discontinuity, the sum of the Fourier series
=1
2 π¦1 + π¦2 where π¦1 and π¦2 are the two values of π π₯ at the point of finite
discontinuity.
25. The function π π =π
πβπ π β€ π β€ π cannot be expanded as Fourier series.
Explain why?
Ans:
Given π π₯ =1
π₯β2 0 β€ π₯ β€ 3
At π₯ = 2, π π₯ becomes infinity. So it does not satisfy one of the Direchletβs condition.
Hence it cannot be expanded as a Fourier series.
26. Can you expand π π = πβππ
π+ππ as a Fourier series in any interval.
Solution:
Let π π₯ = 1 β π₯2
1 + π₯2
This function is well defined in any finite interval in the range ββ, β it has no
discontinuous in the interval
Differentiate π β² π₯ = 1 + π₯2 β2π₯ β 1 β π₯2 2π₯
1 + π₯2 2
π β² π₯ =β4π₯
1 + π₯2 2
π π₯ is maximum or minimum when πβ² (π₯) = 0
(i.e.,) when 4π₯ = 0 π. π. , π₯ = 0.
So it has only one extreme value
(i.e.,) a finite number of maxima and minima in the interval ββ, β
Since it satisfies all the Diricheltβs conditions, it can be expanded in a Fourier series i
aspecified interval in the range ββ, β
27. Find ππ, π(π) = πππππ in (βπ < π₯ < π ). Solution: Since is π(π₯) = π₯π πππ₯ is an even function
π0 =2
π π π₯ ππ₯
π
0
=2
π π₯ π πππ₯ ππ₯
π
0
=2
π π₯(β cos π₯) β 1(β sin π₯) 0
π
=2
π π(β cos π) β 1(β sin π) β 0
= 2
29. Find ππ, π(π) = π β ππ in(π,π). Solution:
π0 =2
π π π₯ ππ₯
π
0
=2
π (π β π₯2)ππ₯
π
0
=2
π ππ₯ β
π₯3
3
0
π
=2
π π2 β
π3
3
30. Find ππ, π(π) = ππ in (βπ,π). Solution: Since π(π₯) = π₯3 is an odd function
ππ =2
π π₯3π ππ ππ₯ππ₯
π
0
ππ =2
π π₯3
βπππ ππ₯
π β 3π₯2
βπ ππ ππ₯
π2 + 6π₯
πππ ππ₯
π3 β 1
π ππ ππ₯
π4
0
π
ππ =2
π π3
βπππ ππ
π β 3π2
βπ ππ ππ
π2 + 6π πππ ππ
π3 β 1 π ππ ππ
π4 β 0
=2
π π3
β1 π
π + 6π
β1 π
π3
31. Find ππ in π π = πβπ in (βπ,π). Solution:
ππ =1
π π π₯ πππ ππ₯ππ₯
π
βπ
ππ =1
π πβπ₯πππ ππ₯ππ₯
π
βπ
=1
π
πβπ₯
1 + π2 βcos ππ₯ + ππ ππππ₯
βπ
π
=1
π β
πβπ
1 + π2 cos ππ + ππ ππππ +
ππ
1 + π2 cos(βππ) + ππ ππ(βππ)
=1
π β
πβπ
1 + π2 β1 π +
ππ
1 + π2 β1 π
= β1 π
π(1 + π2) ππ β πβπ
=2 β1 π
π(1 + π2)π πππ π₯
32. Find ππ , π(π) =( πβπ)
π in(π, ππ).
Solution:
ππ =1
π π π₯ π ππ ππ₯ππ₯
2π
0
ππ =1
π
( π β π₯)
2π ππ ππ₯ππ₯
2π
0
=1
π ( π β π₯)
2 β
πππ ππ₯
π β (β1) β
π ππ ππ₯
π
0
2π
=1
π ( π β 2π)
2 β
πππ 2ππ
π β β1 β
π ππ 2ππ
π β
( π β 0)
2 β
πππ 0
π + 0
=1
π π
2
1
π β 0 β
π
2 β
1
π + 0 =
1
π
Fourier Transform
Part-A
1. State the Fourier integral theorem.
Ans: If is a given function π(π₯) defined in βπ, π and satisfies Dirichletβs conditions, then
π π₯ =1
π π π‘ cos π π‘ β π₯ ππ₯ ππ
β
ββ
β
0
2. State the convolution theorem of the Fourier transform.
Ans: if πΉ(π ) and πΊ(π ) are the functions of π(π₯) and π(π₯) respectively then the Fourier
transform of the convolution of π(π₯) and π(π₯) is the product of their Fourier transform
πΉ (π β π π₯] = πΉ π . πΊ(π )
1
2π π β π π₯ πππ π₯
β
ββ
ππ₯ = 1
2π π π₯ πππ π₯
β
ββ
ππ₯ 1
2π π π₯ πππ π₯
β
ββ
ππ₯
3. Write the Fourier transform pairs.
Ans: πΉ π π₯ and πΉβ1 πΉ(π ) are Fourier transform pairs.
4. Find the Fourier sine transform of π(π) = πβππ(π > 0).
Ans: The Fourier sine transform of π(π₯) is
πΉπ π π₯ = 2
π π π₯ sin π π₯ ππ₯
β
0
= 2
π πβππ₯ sin π π₯ ππ₯
β
0
= 2
π
π
π 2 + π2 πππππ πβππ₯ sin ππ₯ ππ₯ =
π
π2 + π2
β
0
5. What is the Fourier transform of π(π β π) if the Fourier transform of π(π) is π(π).
Solution: πΉ π π₯ β π = ππππ πΉ(π )
6. State the Fourier transforms of the derivatives of a function.
Solution: πΉ ππ π(π₯)
ππ₯ π = βππ ππΉ π .
7. Find the Fourier sine transform of πβπ.
Solution: The Fourier sine transform of π(π₯) is
πΉπ π π₯ = 2
π π π₯ sin π π₯ ππ₯
β
0
= 2
π πβπ₯ sin π π₯ ππ₯
β
0
= 2
π
π
π 2 + 1 πππππ πβππ₯ sin ππ₯ ππ₯ =
π
π2 + π2
β
0
8. Prove that ππ π ππ =π
πππ
π
π , π > 0.
Solution:
πΉπ π ππ₯ = 2
π π ππ₯ cos π π₯ ππ₯
β
0
Put ππ₯ = π¦ when π₯ = 0, π¦ = 0.
ππ₯ =ππ¦
π When π₯ = β, π¦ = β
= 2
π π π¦ cos
π π¦
π
ππ¦
π
β
0
=1
π
2
π π π¦ cos
π
π π¦ ππ¦
β
0
=1
ππΉπ
π
π
9. Find Fourier sine transform of πππ.
Solution: The Fourier sine transform of π(π₯) is
The Fourier sine transform of π(π₯) is
πΉπ π π₯ = 2
π π π₯ sin π π₯ ππ₯
β
0
= 2
π πππ₯ sin π π₯ ππ₯
β
0
= 2
π
π
π 2 + π2 πππππ πππ₯ sin ππ₯ ππ₯ =
π
π2 + π2
β
0
10. if π π = π π π , then prove that π ππ π = βπ π [π π ]
π π
Solution: πΉ π =1
2π π π₯ πππ π₯ ππ₯
β
ββ
π[πΉ π ]
ππ =
π
ππ
1
2π π π₯ πππ π₯ ππ₯
β
ββ
=1
2π π π₯
π
ππ (πππ π₯ )ππ₯
β
ββ
=1
2π π π₯ ππ₯(πππ π₯ )ππ₯
β
ββ
= ππΉ π₯π π₯
πΉ π₯π π₯ = βπ π[πΉ π ]
ππ
11. Find Fourier sine transform of π
π
πΉπ π π₯ = 2
π π π₯ sin π π₯ ππ₯
β
0
= 2
π
1
π₯sin π π₯ ππ₯
β
0
= 2
π
π
2=
π
2
πππππ 1
π₯sin ππ₯ ππ₯
β
0=
π
2, π > 0 .
12. Find the Fourier cosine transform of πβπ.
Solution: The Fourier sine transform of π(π₯) is
πΉπ π π₯ = 2
π π π₯ cos π π₯ ππ₯
β
0
= 2
π πβπ₯ cos π π₯ ππ₯
β
0
= 2
π
1
π 2 + 1 πππππ πβππ₯ cos ππ₯ ππ₯ =
π
π2 + π2
β
0
13. If ππ π is the Fourier sine transform of π(π), show that
ππ π π ππ¨π¬ ππ = π
π ππ π + π + ππ π β π
Solution:
πΉπ π π₯ πππ ππ₯ = 2
π π π₯ cos ππ₯ sin π π₯ ππ₯
β
0
= 2
π.1
2 π π₯ sin π + π π₯ β sin π β π π₯
β
0
ππ₯
= 2
π.
1
2 π π₯ sin π + π π₯
β
0ππ₯ β
2
π.
1
2 π π₯ sin π β π π₯
β
0ππ₯
= 1
2 πΉπ π + π + πΉπ π β π
14. If π(π) is the Fourier transform of π(π), write the formula for the Fourier transform of
π(π)πππ ππ in terms of π.
Solution: πΉ π π₯ cos ππ₯ =1
2π π π₯ cos ππ₯ πππ π₯ ππ₯
β
ββ
=1
2π π π₯
ππππ₯ + πβπππ₯
2 πππ π₯ ππ₯
β
ββ
=1
2π π π₯
ππ(π +π)π₯
2ππ₯
β
ββ
+1
2π π π₯
ππ(π βπ)π₯
2ππ₯
β
ββ
πΉ π π₯ cos ππ₯ =1
2 πΉ π + π + πΉ(π β π)
15. Find the Fourier transform of π π = π ππ π < ππ ππ π > π
Solution: We know that the Fourier transform fo π(π₯) is given by
πΉ π(π₯) =1
2π π π₯ πππ π₯ ππ₯
β
ββ
=1
2π πππ π₯ ππ₯
π
βπ
=1
2π πππ π₯
ππ
βπ
π
=1
2π ππππ β πβπππ
ππ
= 2
π
sin ππ
π
16. State any two properties of Fourier transform
Solution: i) Linearity property: If π(π₯) and π(π₯)are any two functions then
πΉ ππ(π₯) + ππ(π₯) = ππΉ π(π₯) + ππΉ π(π₯) , where a and b are constants.
ii)Shifting property: If πΉ π(π₯) = πΉ π , then πΉ π(π₯ β π) = ππππ πΉ(π )
17. Define the infinite Fourier cosine transform and its inverse
Solution: The infinite Fourier cosine transform is given by
πΉπ π π₯ = πΉπ π = 2
π π π₯ cos π π₯ ππ₯
β
0
Then the inverse Fourier cosine transform is given by
πΉπβ1 πΉπ π = π(π₯) =
2
π πΉπ π cos π π₯ ππ
β
0
18. State Parsevalβs identity for Fourier transform.
Solution: If πΉ π is the Fourier transform of π(π₯), then
|π π₯ |2
β
ββ
ππ₯ = |πΉ π |2
β
ββ
ππ
19. Define the Fourier sine transform of π(π₯) in (0, π). Also give the inversion formula
Solution: The finite Fourier sine transform of a function π π₯ in 0, π is given by
ππ π = π π₯ π πππππ₯
π
π
0
ππ₯,
Then the inversion formula is given by
π π₯ =2
π ππ π
β
π=1
π πππππ₯
π.
20. Define the Fourier cosine transform of π(π) in (π, π). Also give the inversion formula
Solution: The finite Fourier cosine transform of a function π π₯ in 0, π is given by
ππ π = π π₯ ππππ₯
π
π
0
ππ₯,
Then the inversion formula is given by
π π₯ =1
πππ(0) +
2
π ππ π
β
π=1
πππ πππ₯
π.
21. Define the infinite Fourier sine transform and its inverse
Solution: The infinite Fourier sine transform is given by
πΉπ π π₯ = πΉπ π = 2
π π π₯ sin π π₯ ππ₯
β
0
Then the inverse Fourier cosine transform is given by
πΉπ β1 πΉπ π = π(π₯) =
2
π πΉπ π sin π π₯ ππ
β
0
22. Prove that if π(π) is an even function of π, its Fourier transform π(π) will also be an
even function of s.
Solution: By definition
πΉ π =1
2π π π₯ πππ π₯ ππ₯
β
ββ
β¦ 1
Changing s into βs in both sides of (1),
πΉ βπ =1
2π π π₯ πβππ π₯ ππ₯
β
ββ
β¦ 2
In the right hand side integral in (2), put π₯ = βπ’.
then ππ₯ = βππ’; when π₯ = β, π’ = ββ
and when π₯ = ββ, π’ = β.
So (2) becomes
πΉ βπ =1
2π π βπ’ πππ π’ . (βππ’)
ββ
β
=1
2π π βπ’ πππ π’ ππ’
β
ββ
=1
2π π βπ₯ πππ π₯ ππ₯
β
ββ [Changing the dummy variable u into x]
=1
2π π π₯ πππ π₯ ππ₯
β
ββ
= πΉ π ππ¦ 1 .
Here πΉ π is an even function of s.
23. Prove that if π(π) is an odd function of π, its Fourier transform π(π) will also be an
odd function of s.
Solution: By definition
πΉ π =1
2π π π₯ πππ π₯ ππ₯
β
ββ
β¦ 1
Changing s into βs in both sides of (1),
πΉ βπ =1
2π π π₯ πβππ π₯ ππ₯
β
ββ
β¦ 2
In the right hand side integral in (2), put π₯ = βπ’.
then ππ₯ = βππ’; when π₯ = β, π’ = ββ
and when π₯ = ββ, π’ = β.
So (2) becomes
πΉ βπ =1
2π π βπ’ πππ π’ . (βππ’)
ββ
β
=1
2π π βπ’ πππ π’ ππ’
β
ββ
=1
2π π βπ₯ πππ π₯ ππ₯
β
ββ [Changing the dummy variable u into x]
=1
2π βπ π₯ πππ π₯ ππ₯
β
ββ
= βπΉ π ππ¦ 1 .
Here πΉ π is an odd function of s.
24. Find the Fourier transform of the function defined by
π π = π, π < ππ, π < π₯ < ππ , π > π
Solution:
πΉ π(π₯) =1
2π π π₯ πππ π₯ ππ₯
β
ββ
=1
2π π π₯ πππ π₯ ππ₯
π
ββ
+ π π₯ πππ π₯ ππ₯ + π π₯ πππ π₯ ππ₯
β
π
π
π
=1
2π πππ π₯ ππ₯
π
π
=1
2π πππ π₯
ππ π
π
=1
2π
πππ π β πππ π
ππ
25. Show that π(π) = π, π < π₯ < β cannot be represented by a Fourier integral.
Solution: π π₯ ππ₯ = ππ₯β
0
β
0= π₯ 0
β = β β 0 = β
π. π. , π π₯ ππ₯
β
0
is not convergent
Hence π(π₯) = 1 cannot be represented by a Fourier integral.
26. Find the Fourier cosine transform of π π = π, 0 < π₯ < ππ, π₯ > π
Solution:
We know that
πΉπ π π₯ = πΉπ π = 2
π π π₯ cos π π₯ ππ₯
β
0
= 2
π 1 cos π π₯ ππ₯
π
0
= 2
π sin π π₯
π
0
π
= 2
π sin π π
π β 0
= 2
π sin π π
π
27. Find the Fourier transform of π π = π ππ π β€ ππ ππ π > 1
Solution: We know that the Fourier transform fo π(π₯) is given by
πΉ π(π₯) =1
2π π π₯ πππ π₯ ππ₯
β
ββ
=1
2π πππ π₯ ππ₯
1
β1
=1
2π πππ π₯
ππ
β1
1
=1
2π πππ β πβππ
ππ
= 2
π
sin π
π
28. Find the Fourier transform of πβπ|π|, π > 0.
Solution: We know that the Fourier transform fo π(π₯) is given by
πΉ π(π₯) =1
2π π π₯ πππ π₯ ππ₯
β
ββ
=1
2π π π₯ πππ π₯ ππ₯
0
ββ
+1
2π π π₯ πππ π₯ ππ₯
β
0
=1
2π π(ππ +π)π₯ππ₯
0
ββ
+1
2π π(ππ βπ)π₯ππ₯
β
0
=1
2π
π(ππ +π)π₯
ππ + π ββ
0
+ πβ(πβππ )π₯
ππ β π
0
β
=1
2π
1
ππ + πβ
1
ππ β π
=1
2π
ππ β π β ππ β π
ππ 2 β π2
= 2
π
π
π 2 + π2
29. State the Fourier transforms of derivates of a function.
Solution:
πΉ ππ(π₯)
ππ₯ =
1
2π πβ² π₯ πππ π₯ ππ₯
β
ββ
=1
2π πππ π₯ π π(π₯)
β
ββ
=1
2π πππ π₯ π π₯
ββ
ββ π π₯ ππ πππ π₯ ππ₯
β
ββ
= βππ
2π π π₯ πππ π₯ ππ₯
β
ββ
(assuming π(π₯) β 0 ππ π₯ β Β±β)
= βππ πΉ π .
πΉ π2π(π₯)
ππ₯2 =
1
2π πβ²β² π₯ πππ π₯ ππ₯
β
ββ
=1
2π πππ π₯ π πβ²(π₯)
β
ββ
=1
2π πππ π₯ πβ² π₯
ββ
ββ πβ² π₯ ππ πππ π₯ ππ₯
β
ββ
= βππ
2π πβ² π₯ πππ π₯ ππ₯
β
ββ
(assuming πβ²(π₯) β 0 ππ π₯ β Β±β)
= βππ
2π πππ π₯ π π(π₯)
β
ββ
= βππ
2π πππ π₯ π π₯
ββ
ββ π π₯ ππ πππ π₯ ππ₯
β
ββ
= βππ 2
2π π π₯ πππ π₯ ππ₯
β
ββ
(assuming πβ²(π₯) β 0 ππ π₯ β Β±β)
= βππ 2πΉ π .
In general,
πΉ πππ(π₯)
ππ₯π = βππ ππΉ π .
30. Find the finite Fourier cosine transform of π π = πβππin (π, π).
Solution: ππ π = π π₯ πππ πππ₯
π
π
0ππ₯, n being a positive integer
= πβππ₯ πππ πππ₯
π
π
0
ππ₯
= πβππ₯
π2 + πππ₯π
2 βππππ πππ₯
π+
πππ₯
ππ ππ
πππ₯
π
0
π
=π2πβππ
ππ 2 + ππ 2 βπ cos ππ β
π2
ππ 2 + ππ 2(βπ)
=ππ2
ππ 2 + ππ 2 1 + πβππ (β1)π+1
Partial Differential Equations
Part-A
1. Form a partial differential equation by eliminating arbitrary constants a and b from
π = π + π π + (π + π)π
Solution: Given
π§ = π₯ + π 2 + (π¦ + π)2 β¦ (1)
π =ππ§
ππ₯= 2 π₯ + π β¦ (2)
π =ππ§
ππ¦= 2 π¦ + π β¦ (3)
Substituting (2)& (3) in (1), we get π§ =π2
4+
π2
4
2. Solve (π«π β ππ«π«β + π«βπ)π = π
Solution: A.E is π2 β 2π + 1 = 0
π β 1 2 = 0
π = 1, 1
π¦ = π1 π¦ + π₯ + π₯π2(π¦ + π₯)
3. Form a partial differential equation by eliminating arbitrary constants a and b from
π β π π + (π β π)π = ππππππ πΆ
Solution: Given
π₯ β π 2 + π¦ β π 2 = π§2πππ‘2 πΌ β¦ 1
Partially differentiating w.r.t βπ₯β and βπ¦β we get
2 π₯ β π = 2π§ππππ‘2 πΌ β¦ 2
2 π¦ β π = 2π§ππππ‘2 πΌ β¦ 3
π₯ β π = π§ππππ‘2 πΌ β¦ 4
π¦ β π = π§ππππ‘2 πΌ β¦ 5
Substituting (4)& (5) in (1), we get π§2π2πππ‘4 πΌ + π§2π2πππ‘4 πΌ = π§2πππ‘2 πΌ
π2 + π2 = π‘ππ2πΌ
4. Find the complete solution of the differential equation
ππ + ππ β πππ = π
Solution: Given
π2 + π2 β 4ππ = 0 β¦ (1)
Let us assume that π§ = ππ₯ + ππ¦ + π β¦ (2) be the solution of (1).
Partially differentiating (2) w.r.t βπ₯β and βπ¦β we get
π =
ππ§
ππ₯= π
π =ππ§
ππ¦= π
β¦ 3
Substituting (3) in (1) we get
π2 + π2 β 4ππ = 0 β¦ (4)
From (4) we get π =4πΒ± 16π2β4π2π2
2=
4πΒ±2π 4βπ2
2= π Β± π 4 β π2 β¦ 5
Substituting (5) in (2) we get
π§ = π Β± π 4 β π2π₯ + ππ¦ + π
5.Find the solution of πππ + πππ = ππ
Solution: The S.E is
ππ₯
π₯2=
ππ¦
π¦2=
ππ§
π§2
Taking 1st two members, we get
ππ₯
π₯2=
ππ¦
π¦2
Integrating we get
β1
π₯= β
1
π¦+ π1
i.e., 1
π¦β
1
π₯ = π1
taking last two members, we get
ππ¦
π¦2=
ππ§
π§2
β1
π¦= β
1
π§+ π2
i.e., 1
π§β
1
π¦ = π2
The complete solution is π· 1
π¦β
1
π₯,
1
π§β
1
π¦ = 0
6. Find the partial differential equation of all planes having equal intercepts on the π and
π axis.
Solution:
The equation of the plane is
π₯
π+
π¦
π+
π§
π= 1 β¦ (1)
Partially differentiating (1) w.r.t βπ₯β and βπ¦β we get
1
π+
π
π= 0
π = βπ
πβ¦ (2)
1
π+
π
π= 0
π = βπ
πβ¦ (3)
From (2) and (3) we get π = π
7. Form the partial differential equation by eliminating the arbitrary function from
π± ππ β ππ,π
π = π
Solution: Given
π· π§2 β π₯π¦,π₯
π§ = 0 β¦ 1
Let π’ = π§2 β π₯π¦
π£ =π₯
π§
Then the given equation is of the form
π· π’, π£ = 0 β¦ (2)
The elimination of π· from the equation (2), we get,
ππ’
ππ₯
ππ£
ππ₯
ππ’
ππ¦
ππ£
ππ¦
= 0
i.e.,
2π§π β π¦π§βππ₯
π§2
2π§π β π₯βπ₯π
π§2
= 0
i.e., 2π§π β π¦ βπ₯π
π§2 β 2π§π β π₯ π§βππ₯
π§2 = 0
i.e.,ππ₯2 β π π₯π¦ β 2π§2 = π§π₯
8. Find the singular integral of the partial differential equation π = ππ + ππ + ππ β ππ
Solution: The complete integral is π§ = ππ₯ + ππ¦ + π2 β π2 β¦ (1)
Now,
ππ§
ππ= π₯ + 2π = 0 π =
βπ₯
2
ππ§
ππ= π¦ β 2π = 0 π =
π¦
2
β¦ 2
Substituting (2) in (1), we get
π§ = βπ₯2
2+
π¦2
2+
π₯2
4β
π¦2
4= β
π₯2
4+
π¦2
4
i.e., π¦2 β π₯2 = 4π§ which is the singular integral.
9. Find the complete solution of ππππ + ππππ = ππ
Solution: Given π₯2π2 + π¦2π2 = π§2 β¦ 1
Equation (1) can be written as π₯π 2 + π¦π 2 = π§2 β¦ 2
Put log π₯ = π and log π¦ = π
π₯π = π, π€ππππ π =ππ§
ππ
π¦π = π, π€ππππ π =ππ§
ππ
β¦ (3)
π2 + π2 = π§2 β¦ 4
This is of the form πΉ π§, π, π = 0
Let π§ = π π + ππ be the solution of (4).
Put π’ = π + ππ.Then π§ = π π’
π =ππ§
ππ’, π = π
ππ§
ππ’β¦ 5
Substituting (5) in (4), we get
ππ§
ππ’
2
1 + π2 = π§2
ππ§
ππ’=
π§
1 + π2
Separating the variables we get
ππ§
π§=
ππ’
1 + π2
Integrating we get
log π§ =π’
1 + π2+ π
log π§ =π + ππ
1 + π2+ π
Replacing π ππ¦ log π₯ πππ π ππ¦ log π¦, we get
log π§ =πππ π₯ + ππππ π¦
1 + π2+ π
Which gives the complete solution of (1).
10.Solve ππ + ππ = ππ
Solution: π2 + π2 = π2 β¦ 1
Let us assume that π§ = ππ₯ + ππ¦ + π be a solution of (1). ... (2)
Partially differentiating (2) w.r.t β²π₯ β²and β²π¦ β² , we get
ππ§
ππ₯= π = π,
ππ§
ππ¦= π = π β¦ (3)
Substituting (3) in (1) we get
π2 + π2 = π2 β¦ 4
Hence π§ = ππ₯ + ππ¦ + π is the solution of (1).
11. Form a partial differential equation by eliminating arbitrary constants a and b from
π = πππ + πππ
Solution: Given
π§ = ππ₯π + ππ¦π β¦ 1
π =ππ§
ππ₯= π. ππ₯πβ1 π =
π
ππ₯πβ1β¦ (2)
π =ππ§
ππ¦= π. ππ¦πβ1 π =
π
ππ¦πβ1β¦ (3)
Substituting (2) and (3) in (1), we get
π§ =π
ππ₯πβ1π₯π +
π
ππ¦πβ1π¦π
π§ =1
π ππ₯ + ππ¦
12. Solve π«π + ππ«ππ«β² + ππ«π«β² π + π«β² π π = π.
Solution:
A.E is π3 + 3π2 + 3π + 1 = 0
π + 1 3 = 0
i.e., π = β1, β1, β1.
π§ = π1 π¦ β π₯ + π₯π2 π¦ β π₯ + π₯2π3(π¦ β π₯)
13. Form a partial differential equation by eliminating arbitrary constants a and b from
π = ππ + ππ ππ + ππ
Solution:
Given
π§ = π₯2 + π2 π¦2 + π2 β¦ 1
π =ππ§
ππ₯= 2π₯ π¦2 + π2
π
2π₯= π¦2 + π2 β¦ (2)
π =ππ§
ππ¦= 2π¦ π₯2 + π2
π
2π¦= π₯2 + π2 β¦ (3)
Substituting (2) and (3) in (1) we get
π§ =π
2π₯
π
2π¦
ππ = 4π₯π¦π§
14. Solve π«π β π«π«β² + π«β² β π π = π.
Solution: The given equation can be written as
π· β 1 π· β π·β² + 1 π§ = 0.
We know that the complementary function corresponding to the factors
π· β π1π·β² β πΌ1 π· β π2π·
β² β πΌ2 π§ = 0 ππ
π§ = ππΌ1π₯π1 π¦ + π1π₯ + ππΌ2π₯π2(π¦ + π2π₯)
Here πΌ1 = 1, πΌ2 = β1, π1 = 0, π2 = 1.
π§ = ππ₯π1 π¦ + πβπ₯π2(π¦ + π₯)
15. Form the partial differential equation by eliminating the arbitrary function from
π = π ππ
π
Solution:
π§ = π π₯π¦
π§
π = πβ² π₯π¦
π§ π§π¦ β π₯π¦π
π§2β¦ (1)
π = πβ² π₯π¦
π§ π§π₯ β π₯π¦π
π§2β¦ (2)
From (1) we get
πβ² π₯π¦
π§ =
ππ§2
π§π¦ β π₯π¦πβ¦ (3)
Substituting (3) in (2), we get
π =ππ§2
π§π¦ β π₯π¦π
π§π₯ β π₯π¦π
π§2
16. Form a partial differential equation by eliminating arbitrary constants a and b from
π β π π + π β π π + ππ = π
Solution: Given
π₯ β π 2 + π¦ β π 2 + π§2 = 1 β¦ 1
Partially differentiating (1) w.r.t. β²π₯ β²and β²π¦ β² , we get
2 π₯ β π + 2π§π = 0
i.e., π₯ β π = βπ§π β¦ (2)
2 π¦ β π + 2π§π = 0
π¦ β π = βπ§π β¦ (3)
Substituting (2) and (3) in (1), we get
π§2π2 + π§2π2 + π§2 = 1
π2 + π2 + 1 =1
π§2
17. Find the complete integral of π + π = ππ
Solution: Given π + π = ππ β¦ (1)
Let us assume that π§ = ππ₯ + ππ¦ + π be a solution of (1). ... (2)
Partially differentiating (2) w.r.t β²π₯ β²and β²π¦ β² , we get
ππ§
ππ₯= π = π,
ππ§
ππ¦= π = π β¦ (3)
Substituting (3) in (1) we get
π + π = ππ β¦ 3
π =π
π β 1β¦ (4)
Substituting (4) in (2) we get
π§ = ππ₯ + π
π β 1 π¦ + π
which is the complete integral.
18. Solve π«π β ππ«π«β² π + ππ«β² π π = π.
Solution:
A.E. is π3 β 3π + 2 = 0
π = 1, 1 πππ β 2.
The solution is π§ = π1 π¦ + π₯ + π₯π2 π¦ + π₯ + π3(π¦ β 2π₯).
19. Find the general solution of ππππ
πππ β πππππ
ππππ+ π
πππ
πππ = π.
Solution:
A.E. is 4π2 β 12π + 9 = 0
π =12 Β± 144 β 144
8
π =3
2,3
2
The general solution is π§ = π1 π¦ +3
2π₯ + π₯π2 π¦ +
3
2π₯
20. Form the partial differential equation by eliminating the arbitrary function from
π = π π
π
Solution:
π§ = π π¦
π₯ β¦ (1)
Partially differentiating (1) w.r.t β²π₯ β²and β²π¦ β² , we get
π = πβ² π¦
π₯ βπ¦
π₯2β¦ (2)
π = πβ² π¦
π₯
1
π₯β¦ (3)
From (1) we get
πβ² π¦
π₯ = β
ππ₯2
π¦β¦ (4)
Substituting (4) in (3), we get
π = βππ₯2
π¦
1
π₯
ππ₯ + ππ¦ = 0 is the required partial differential equation.
21. Form the partial differential equation by eliminating the arbitrary function from
π§ = π₯ + π¦ + π π₯π¦ .
Solution: Given π§ = π₯ + π¦ + π π₯π¦ β¦ 1
Partially differentiating (1) w.r.t β²π₯ β²and β²π¦ β² , we get
π = 1 + πβ² π₯π¦ π¦ π β 1 = πβ² π₯π¦ π¦β¦ 2
π = 1 + πβ² π₯π¦ π₯ π β 1 = πβ² π₯π¦ π₯ β¦ 3
(2) Γ· 3 gives πβ1
πβ1=
π¦
π₯
ππ₯ β π₯ = ππ¦ β π¦ ππ ππ₯ β ππ¦ = π₯ β π¦ is the required partial differential equation.
22. Form the partial differential equation by eliminating the arbitrary function from
π§ = π π₯π¦ .
Solution: Given π§ = π π₯π¦ β¦ 1
Partially differentiating (1) w.r.t β²π₯ β²and β²π¦ β² , we get
π = πβ² π₯π¦ π¦β¦ 2
π = πβ² π₯π¦ π₯ β¦ 3
(2) Γ· 3 gives π
π=
π¦
π₯
ππ₯ = ππ¦ ππ ππ₯ β ππ¦ = 0 is the required partial differential equation.
23. Find the partial differential equation of all planes through the origin.
Solution:
The general equation of the plane is
ππ₯ + ππ¦ + ππ§ + π = 0 β¦ (1)
If (1) passes through the origin, then d=0.
So (1) becomes
ππ₯ + ππ¦ + ππ§ = 0 β¦ (2)
Partially differentiating (2) w.r.t βπ₯β and βπ¦β we get
π + ππ = 0 π = βππ β¦ 3
π + ππ = 0 π = βππ β¦ (4)
Substituting (3) and (4) in (1), we get
βπππ₯ β πππ¦ + ππ§ = 0 ππ₯ + ππ¦ = π§ is the required partial differential equation.
24. Write down the complete solution of π = ππ + ππ + π π + ππ + ππ.
Solution: The given equation is Clairautβs type.
So replacing p by a and q by b, the complete solution is
π§ = ππ₯ + ππ¦ + π 1 + π2 + π2
25. Find the differential equations of all spheres of radius c having their centers in the πππ
plane.
Solution: Let the center of the sphere be (π, π, 0), a point in the π₯ππ¦ plane. c is the given
radius.
The equation of the sphere is π₯ β π 2 + π¦ β π 2 + π§2 = π2 β¦ (1)
Partially differentiating (1) w.r.t βπ₯β and βπ¦β we get
2 π₯ β π + 2π§π = 0 π₯ β π = βπ§π β¦ (2)
2 π¦ β π + 2π§π = 0 π¦ β π = βπ§π β¦ (3)
Substituting (2) and (3) in (1), we get
π§2π2 + π§2π2 + π§2 = π2
π§2 π2 + π2 + 1 = π2 which is the required partial differential equation.
26. Eliminate π from πππ = π(π + π + π)
Solution: Given π₯π¦π§ = π π₯ + π¦ + π§ β¦ 1 .
Partially differentiating (1) w.r.t βπ₯β and βπ¦β we get
π¦ π§ + π₯π = πβ² π₯ + π¦ + π§ 1 + π β¦ (2)
π₯ π§ + π¦π = πβ² π₯ + π¦ + π§ 1 + π β¦ (3)
(2) Γ· 3 gives 1+π
1+π =
π¦ π§+π₯π
π₯ π§+π¦π
1 + π π₯ π§ + π¦π = 1 + π π¦ π§ + π₯π
π π₯π¦ β π₯π§ + π π¦π§ β π₯π¦ = π₯π§ β π¦π§
ππ₯ π¦ β π§ + ππ¦ π§ β π₯ = π§(π₯ β π¦) is the required partial differential equation.
27. Mention three types of partial differential equation.
Solution: i) A solution of a partial differential equation which contains the maximum
possible number of arbitrary constants is called complete integral.
ii) A solution obtained by giving particular values to the arbitrary constants in a complete
integral is called a particular integral.
iii) A solution of a partial differential equation which contains the maximum possible
number of arbitrary functions is called general integral.
28. Define singular integral.
Solution:
Let πΉ π₯, π¦, π§, π, π = 0 β¦ 1 be a partial differential equation and its complete integral be
π· π₯, π¦, π§, π, π = 0 β¦ 2
Differentiating (2) partially w.r.t. a and b in turn, we get
ππ·
ππ= 0 β¦ (3)
ππ·
ππ= 0 β¦ (4)
The eliminant of a and b from (2), (3) and (4), if it exits, is called singular integral.
29. Solve π + π = π
Solution: Given π + π = 1 β¦ 1
Let us assume that π§ = ππ₯ + ππ¦ + π be a solution of (1). ... (2)
Partially differentiating (2) w.r.t β²π₯ β²and β²π¦ β² , we get
ππ§
ππ₯= π = π,
ππ§
ππ¦= π = π β¦ (3)
Substituting (3) in (1), we get
π + π = 1 π = 1 β π 2β¦ (4)
Substituting (4) in (2), we get
π§ = ππ₯ + 1 β π 2π¦ + π is the complete integral of (1).
30. Find the particular integral of π«π β ππ«ππ«β² β ππ«π«β² π + πππ«β²π π = π¬π’π§(π + ππ)
Solution:
Particular integral =sin (π₯+2π¦ )
π·3β3π·2π·β² β4π·π·β² 2+12π·β²3
β¦ (1)
Replacing π·2 ππ¦ β 12 ππππ·β² 2ππ¦ β 22 in(1)
=sin(π₯ + 2π¦)
βπ· + 3π·β² + 16π· β 48π·β²
=sin(π₯ + 2π¦)
15π· β 45π·β²
=sin(π₯ + 2π¦)
15 π· β 3π·β²
= π· + 3π·β² sin(π₯ + 2π¦)
15 π·2 β 9π·β²2
=cos π₯ + 2π¦ + 6 cos(π₯ + 2π¦)
15 π·2 β 9π·β²2
=7cos π₯ + 2π¦
15 β1 β 9(β4)
=cos π₯ + 2π¦
75
Z βTransform
Part-A
1. Find π π .
Solution:
π π = ππ§βπ =
β
π=0
π
π§π=
1
π§
β
π=0
+2
π§2+
3
π§3+ β―
=1
π§ 1 +
2
π§+
3
π§2+
4
π§3+ β― =
1
π§ 1 β
1
π§
β2
=1
π§
π§2
π§ β 1 2
π π =π§
π§β1 2 [The region of convergence is 1
π§ < 1 ππ π§ > 1]
2. Form the difference equation from ππ = π + πππ
Solution: π¦π = π + π3π , π¦π+1 = π + π3π+1` , π¦π+2 = π + π3π+2
π¦π+2 β 4π¦π+1 + 3π¦π = 0
3. Find the value of π π(π) ππππ π π = πππ
Solution: π πππ = π(π) π§βπ§
π πππππ π πππ(π‘) = πΉ
π§
π
= π§
π§ β 1 2 π§β
π§π
=
π§π
π§π β 1
2 =ππ§
π§ β π 2. πππππ π π =
π§
π§ β 1 2
4. Find π ππ
π! in Z-transform.
Solution: π ππ
π ! = π
1
π!
π§βπ§
π
πππππ π πππ(π‘) = πΉ π§
π
= π1 π§ π§β
π§π
= ππ π§
5.Find π πβπππ using Z-transform.
Solution: π πβπππ‘ = π πβπππ‘ . 1
= π 1 π§βπ§π πππ πππππ π πβπππ‘ . π(π‘) = πΉ(π§ππππ )
= π§
π§ β 1 π§βπ§π πππ
=π§ππππ
π§ππππ β 1
6. Express π π π + π in terms of π π .
Solution: π π π + 1 = π§πΉ π§ β π§π 0 .
7. State and prove initial value theorem in Z-transform.
Solution: If π π π = πΉ π§ then lim
z β βπΉ π§ = π 0 =
limt β 0
π(π‘)
We know that
π π π = π π π§βπ
β
π=0
= π 0 + π 1 π§β1 + π 2 π§β2 + β―
πΉ π§ = π 0 +π 1
π§+
π 2
π§2+ β―
limz β β
πΉ π§ = π 0 πππππ lim
z β β
1
z= 0
limz β β
πΉ π§ = π 0 = lim
t β 0π(π‘)
8. Find the Z-transform of (π + π)(π + π)
Solution: π (π + 1)(π + 2) = π π2 + 3π + 2
= π π2 + 3π π + π 2
=π§(π§ + 1)
π§ β 1 3+ 3
π§
π§ β 1 2+ 2
π§
π§ β 1
9. Find the Z-transform of πππ ππ
Solution: π ππππ₯ = π πππ₯ π
=π§
π§ β πππ₯ πππππ π ππ =
π§
π§ β π
=π§
π§ β cos π₯ + π sin π₯ =
π§
π§ β cos π₯ + π sin π₯
π[cos ππ₯ + ππ ππ ππ₯] =π§ π§ β cos π₯ β π sin π₯
π§ β cos π₯ 2 + sin2 π₯
π[cos ππ₯] + ππ[π ππ ππ₯] =π§ π§ β cos π₯
π§2 β 2π§πππ π₯ + 1β
ππ ππ π₯
π§2 β 2π§πππ π₯ + 1
Equating real and imaginary parts we get
π(cos ππ₯) =π§ π§ β cos π₯
π§2 β 2π§πππ π₯ + 1
10. Find the Z-transform of ππ
Solution:
π ππ = πππ§βπ
β
π=0
= 1 + ππ§β1 + π2π§β2 + π3π§β3 + β―
= 1 + ππ§β1 + ππ§β1 2 + ππ§β1 3 + β― πππππ 1 + π₯ + π₯2 + π₯3 + β― = (1 β π₯)β1
=1
1 β ππ§β1=
π§
π§ β π
11. Let π π π = π π πππ π > 0then, Prove that
π π π β π = πβππ π
Solution:
π π π β π = π π β π π§βπ
β
π=0
= π π β π π§βπ
β
π=π
πππππ π β1 , π β2 , β¦πππ πππ‘ πππππππ
= π0π§βπ + π1π§
β π+1 + π2π§β π+2 + β―
= π§βπ π0 + π1π§β1 + π2π§β2 + β―
= π§βπ π π π§βπ
β
π=0
= π§βππΉ π§
12. Let π π π = π π πππ π > 0then, Prove that
π π π + π = ππ π π β π π β π π πβπ β β― π(π β π)πβ(πβπ)
Solution:
π π π + π = π π + π π§βπ
β
π=0
Multiply and divide by π§π in RHS, we get
= π π + π π§βπ
β
π=0
π§π
π§π
= π§π π π + π π§β(π+π)
β
π=0
= π§π π π π§βπ + π π + 1 π§β(π+1) + π π + 2 π§β(π+2) + β―
= π§π π π π§βπ
β
π=0
β π π β π π§βπ
πβ1
π=0
= π§π πΉ π§ β π π β π 1 π§β1 β β― π(π β 1)π§β(πβ1)
13. Find the Z-transform of βπ π
Solution:
π β1 π = β1 ππ§βπ
β
π=0
= 1 β π§β1 + π§β2 β π§β3 + β―
= 1 + π§β1 β1
=π§
π§ + 1
14. Find the Z-transform of ππ
Solution:
π π2 = π π. π
= βπ§ππ π
ππ§ πππππ π ππ π‘ = βπ§
ππΉ(π§)
ππ§
= βπ§π
ππ§
π§
π§ β 1 2
= βπ§ π§ β 1 2 β 2 π§ β 1 π§
π§ β 1 4
= βπ§ π§ β 1 β 2π§
π§ β 1 3
=π§ π§ + 1
π§ β 1 3
15. State convolution theorem for Z-transform
Solution: If π π πππ π(π) are two casual sequences,
π π(π) β π(π) = π π(π) . π π(π) = πΉ π§ . πΊ(π§)
16. If π π =π
πβπβπ», find π₯π’π¦
π β β π(π)
Solution: We know that lim
t β β π π‘ =
limz β 1
(π§ β 1)πΉ π§
=lim
z β 1 π§ β 1
π§
π§ β πβπ= 0
17. Find the Z-transform of ππ π ππ π π π = π(π)
Solution:
πΉ π§ = π π π‘ = π ππ π§βπ
β
π=0
Differentiate both sides w.r.t. π§β1.
ππΉ(π§)
ππ§= βππ ππ π§βπβ1
β
π=0
π§ππΉ(π§)
ππ§= β ππ ππ π§βπ
β
π=0
= βπ ππ π‘
π ππ π‘ = βπ§ππΉ(π§)
ππ§
18. If π π π = π(π) then prove that π πβπππ π = π(ππππ»)
Solution:
π πβππ‘π π‘ = πβπππ π ππ π§βπ
β
π=0
= π ππ π§πππ βπ
β
π=0
= πΉ π§πππ πππππ πΉ π§ = π ππ π§βπ
β
π=0
19. Find the Z-transform of πβππ πππ ππ
Solution:
π πβππ‘ cos ππ‘ = π[cos ππ‘] π§βπ§πππ πππππ π πβππ‘π π‘ = πΉ(π§πππ )
= π§ π§ β cos ππ
π§2 β 2π§πππ ππ + 1
π§βπ§πππ
=π§πππ π§πππ β cos ππ
π§2π2ππ β 2π§ππππππ ππ + 1
20. If π π =πππ
πβπ (πβπ) ππππ π π .
Solution:
π 0 =lim
z β βπΉ π§ =
limz β β
10π§
π§ β 1 π§ β 2
=lim
z β β
10π§
π§ 1 β1π§ π§ β 2
=lim
z β β
10
1 β1π§ π§ β 2
= 0
21. Find Z-transform of π
π, π β₯ π.
Solution:
π 1
π =
1
ππ§βπ
β
π=1
= 1
π§+
1
2π§2+
1
3π§3+ β―
= β log 1 β1
π§ ππ
1
π§ < 1
= log π§
π§ β 1 ππ π§ > 1
22. Find Z-transform of π
π!
Solution:
π 1
π! =
1
π!π§βπ =
β
π=0
1
π! π§β1 π
β
π=0
= 1 +π§β1
1!+
π§β1 2
2!+
π§β1 3
3!+ β― = ππ§β1
π 1
π! = π1 π§
23. Find the Z-transform of πππ
Solution:
π πππ = ππππ§βπ
β
π=0
= π π
π§
πβ
π=0
= π 1 +π
π§+
π
π§
2
+ β―
= π1
1 βππ§
ππ π
π§ < 1
=ππ§
π§ β π ππ π§ > |π|
24. Find the inverse Z-transform by using convolution theorem
ππ
π β π π
Solution:
πβ1 π§2
π§ β π 2 = πβ1
π§
π§ β π.
π§
π§ β π
= πβ1 π§
π§ β π β πβ1
π§
π§ β π = ππ β ππ
= ππβπ . ππ =
π
π=0
ππ
π
π=0
= π + 1 πππ’(π)
25. Define Z-transform
Solution:
The two sided Z-transform πΉ(π§) for a sequence π(π₯) is defined as
πΉ π§ = π π₯ π§βπ
β
π=ββ
The one sided Z-transform πΉ(π§) for a sequence π(π₯) is defined as
πΉ π§ = π π₯ π§βπ
β
π=0
26. Find the value of π πππ π
Solution:
π πππ π₯ = πππ π₯ π§βπ
β
π=0
= π π₯ π§
π
βπβ
π=0
= πΉ π§
π πππππ πΉ π§ = π π₯ π§βπ
β
π=0
27. Find the value of π π .
Solution:
π 1 = 1. π§βπ
β
π=0
= 1 +1
π§+
1
π§2+
1
π§3+ β―
= 1 β1
π§
β1
ππ 1
π§ < 1
=π§
π§ β 1 ππ π§ > 1.
28. Find the Z-transform of πΉ(π β π)
Solution:
π πΏ(π β π) = πΏ(π β π)π§βπ
β
π=0
=1
π§π, if k is positive integer
29. Find the Z-transform of π(π β π)
Solution:
π π’(π β 1) = 1. π§βπ
β
π=1
=1
π§+
1
π§2+
1
π§3+ β―
=1
π§ 1 β
1
π§
β1
ππ 1
π§ < 1
=1
π§ β 1 ππ π§ > 1
30. Find the Z-transform of πππΉ(π β π)
Solution:
π 3ππΏ(π β 1) = π πΏ(π β π) π§β
π§3
= 1
π§ π§β
π§3
=3
π§
Applications of partial differential equations
Part-A
1. Classify the partial differential equation ππππ + ππππ + πππ β πππ = π.
Solution: Given 3π’π₯π₯ + 4π’π₯π¦ + 3π’π¦ β 2π’π₯ = 0.
π΄ = 3, π΅ = 4, πΆ = 0
π΅2 β 4π΄πΆ = 16 > 0, Hyperbolic
2. The ends A and B of a rod of length ππ ππ long have their temperature kept at πππ πͺ
and πππ πͺ. Find the steady state temperature distribution on the rod.
Solution: When the steady state condition exists the heat flow equation is
π2π’
ππ₯2= 0
i.e., π’(π₯) = π1π₯ + π2 . . . (1)
The boundary conditions are (π) π’(0) = 20, (π) π’(10) = 70
Applying (a) in (1), we get π’ 0 = π2 = 20
Substituting π2 = 20 in (1), we get
π’ π₯ = π1π₯ + 20 β¦ (2)
Applying (b) in (2), we get π’ 10 = π110 + 20 = 70 π1 = 5
Substituting π1 = 5 in (2), we get
π’ π₯ = 5π₯ + 20
3.Write the one dimensional wave equation with initial and boundary conditions in which
the initial position of the string is π(π) and the initial velocity imparted at each point π is
π(π).
Solution: the one dimensional wave equation is
π2π¦
ππ‘2= πΌ2
π2π¦
ππ₯2
The boundary conditions are
π π¦ 0, π‘ = 0 ππ π¦ π, π‘ = 0, πππ π¦ π₯, 0 = π π₯ ππ£ ππ¦ (π₯, 0)
ππ‘= π(π₯)
4.In steady state conditions derive the solution of one dimensional heat flow equation.
Solution: when steady state conditions exist the heat flow equation is independent of time
t.
ππ’
ππ‘= 0
The heat flow equation becomes
π2π’
ππ₯2= 0
The solution of heat flow equation is π’ π₯ = π1π₯ + π2
5. What is the basic difference between the solutions of one dimensional wave equation
and one dimensional heat equation.
Solution: Solution of the one dimensional wave equation is of periodic in nature. But
solution of one dimensional heat equation is not of periodic in nature.
6. What are the possible solutions of one dimensional wave equation?
Solution:
π¦ π₯, π‘ = π1πππ₯ + π2π
βππ₯ π3ππππ‘ + π4π
βπππ‘ β¦ (1)
π¦ π₯, π‘ = π5 cos ππ₯ + π6 sin ππ₯ π7 cos πππ‘ + π8 sin πππ‘ β¦ (2)
π¦ π₯, π‘ = π9π₯ + π10 π11π‘ + π12 β¦ (3)
7. Write down possible solutions of the Laplace equation.
Solution:
π’ π₯, π¦ = π1πππ₯ + π2π
βππ₯ π3 cos ππ¦ + π4 sin ππ¦ β¦ (1)
π’ π₯, π¦ = π5 cos ππ₯ + π6 sin ππ₯ π7πππ¦ + π8π
βππ¦ β¦ (2)
π’ π₯, π¦ = π9π₯ + π10 π11π¦ + π12 β¦ (3)
8. State Fourier law of conduction.
Solution:
Fourier law of heat conduction:
The rate at which heat flows across an area A at a distance x from one end of a bar is given
by
π = βπΎπ΄ ππ’
ππ₯
π₯
where πΎ is the thermal conductivity and ππ’
ππ₯
π₯means the temperature gradient at x.
9. In the wave equation πππ
πππ= πΆπ πππ
πππ what does πΆπ stands for?
Solution:
πΌ2 = π =ππππ πππ
πππ π
10. State any two laws which are assumed to derive one dimensional heat equation.
Solution:
(i) The sides of the bar are insulated so that the loss or gain of heat from the sides by
conduction or radiation is negligible.
(ii) The same amount of heat is applied at all points of the face.
11. Classify the partial differential equation πππ + ππππ = π.
Solution:
Here π΄ = 1, π΅ = π₯, πΆ = 0
π΅2 β 4π΄πΆ = π₯2
π πππππππππ ππ π₯ = 0.
ππ π»π¦ππππππππ ππ π₯ > 0 πππ π₯ < 0.
12. A rod 30 cm long has its ends A and B kept at πππ πͺ and πππ πͺ respectively until
steady state conditions prevail. Find the steady state temperature in the rod.
Solution: Let π = 30 ππ
When the steady state condition prevail the heat flow equation is
π2π’
ππ₯2= 0
i.e., π’ π₯ = ππ₯ + π β¦ (1)
When the steady state condition exists the boundary conditions are
π’(0) = 20 , π’(π) = 80 . . . (2)
Applying (2) in (1) we get
π’(0) = π = 20. . . (3)
and π’(π) = ππ + 20 = 80
π =60
π=
60
30= 2 β¦ (4)
Substituting (3) and (4) in (1) we get
π’ π₯ = 2π₯ + 20
13. Classify the partial differential equation
π ππ πππ β ππππππ + ππ πππ + πππ β ππ = π
π ππ πππ + πππ + πππ + ππ
π + π = π
Solution:
a) π΄ = π¦2, π΅ = β2π₯π¦, πΆ = π₯2
π΅2 β 4π΄πΆ = 4π₯2π¦2 β 4π₯2π¦2 = 0
Parabolic
b) π΄ = π¦2, π΅ = 0, πΆ = 1
π΅2 β 4π΄πΆ = β4π¦2 < 0
Elliptic.
14. A rod 60 cm long has its ends A and B kept at πππ πͺ and πππ πͺ respectively until
steady state conditions prevail. Find the steady state temperature in the rod.
Solution: Let π = 60 ππ
When the steady state condition prevail the heat flow equation is
π2π’
ππ₯2= 0
i.e., π’ π₯ = ππ₯ + π β¦ (1)
When the steady state condition exists the boundary conditions are
π’(0) = 30 , π’(π) = 40 . . . (2)
Applying (2) in (1) we get
π’(0) = π = 30. . . (3)
and π’(π) = ππ + 30 = 40
π =10
π=
10
60=
1
6β¦ (4)
Substituting (3) and (4) in (1) we get
π’ π₯ =1
6π₯ + 20
15. Classify the partial differential equation
π) ππππ
πππ+ π
πππ
ππππ+
πππ
πππβ π
ππ
ππβ π
ππ
ππβ πππ = π
π) πππ
πππ+
πππ
πππ=
ππ
ππ
π
+ ππ
ππ
π
Solution:
a) π΄ = 4, π΅ = 4, πΆ = 1
π΅2 β 4π΄πΆ = 16 β 16 = 0
Parabolic.
b) π΄ = 1, π΅ = 0, πΆ = 1
π΅2 β 4π΄πΆ = β4 < 0
Elliptic.
16. Classify the partial differential equation
π) πππ
πππ=
πππ
πππ π)
πππ
ππππ=
ππ
ππ
ππ
ππ + ππ
Solution:
a) Here π΄ = 1, π΅ = 0, πΆ = β1
π΅2 β 4π΄πΆ = 4 > 0
Hyperbolic.
b) Here π΄ = 0, π΅ = 1, πΆ = 0
π΅2 β 4π΄πΆ = 1 > 0
Hyperbolic.
17. State one dimensional heat equation with the initial and boundary conditions.
Solution: The one dimensional heat equation is
πΌ2π2π’
ππ₯2=
ππ’
ππ‘
Initial condition means condition at π‘ = 0
Boundary condition means condition at π₯ = 0 πππ π₯ = π
18. Write the boundary conditions and initial conditions for solving the vibration of string
equation, if the string is subjected to initial displacement π(π) and initial velocity π(π).
Solution:
Initial displacement π(π₯)
π π¦ = 0 π€πππ π₯ = 0
ππ π¦ = 0 π€πππ π₯ = π
πππ ππ¦
ππ‘= 0 π€πππ π‘ = 0
ππ£ π¦ = π(π₯) π€πππ π‘ = 0
Initial velocity π(π₯)
π π¦ = 0 π€πππ π₯ = 0
ππ π¦ = 0 π€πππ π₯ = π
πππ ππ¦
ππ‘= π(π₯) π€πππ π‘ = 0
ππ£ π¦ = 0 π€πππ π‘ = 0
19. Write down the differential equation for two dimensional heat flow equation for the
unsteady state.
Solution: The two dimensional heat flow equation for the unsteady state is given by
πΌ2 π2π’
ππ₯2+
π2π’
ππ¦2 =
ππ’
ππ‘
20. What is the steady state heat flow equation in two dimensions in Cartesian form?
Solution: The two dimensional heat flow equation for the steady state is given by
π2π’
ππ₯2+
π2π’
ππ¦2= 0. πΏππππππ πππ’ππ‘πππ
21. Write down the polar form of two dimensional heat flow equation in steady state.
Solution:
π2π2π’
ππ2+ π
ππ’
ππ+
π2π’
ππ2= 0. πΏππππππ πππ’ππ‘πππ
22. Write all the solutions of Laplace equation in polar coordinates.
Solution:
π’ π, π = π1ππ + π2π
βπ π3 cos ππ + π4 sin ππ β¦ (1)
π’ π, π = π5 cos ππππ π + π6 ππππ π π7πππ + π8π
βππ β¦ (2)
π’ π, π = π9 log π + π10 π11π + π12 β¦ (3)
23. Explain the initial and boundary value problems.
Solution: The values of a required solution, on the boundary of some domain will be given.
These are called boundary conditions. In other cases, when time t is one of the variables,
the values of the solution at t=0 may be presented. These are called initial conditions.
The partial differential equation together with these conditions constitutes a boundary
value problem or an initial value problem, according to the nature of the condition.
24. What is meant by steady state condition in heat flow?
Solution:
Steady state condition in heat flow means that the temperature at any point in the body
does not vary with time. i.e., it is independent of time t.
25. Distinguish between steady and unsteady states in heat conduction problems.
Solution:
In unsteady state, the temperature at any point of the body depends on the position of the
point and also the time t. In steady state, the temperature at any point depends only on the
position of the point and is independent of the time t.
26. Solve using separation of variables method πππ + πππ = π.
Solution:
Given π¦π’π₯ + π₯π’π¦ = 0 β¦ (1)
Let π’ = π π₯ . π π¦ β¦ 2
be the solution of (1). From (2) we get
π’π₯ = πβ²π β¦ 3
π’π¦ = ππβ² β¦ 4
Substituting (3) and (4) in (1), we get
π¦. πβ²π + π₯. ππβ² = 0
π¦. πβ²π = βπ₯. ππβ²
πβ²
π₯π= β
πβ²
π¦π= π
πβ² = ππ₯π ππ πβ² = βππ¦π
ππ
ππ₯= ππ₯π ππ
ππ
ππ¦ = βπππ¦
ππ
π= ππ₯ ππ₯ ππ
ππ
π = βππ¦ ππ¦
log π = ππ₯2
2+ π1 ππ log π = βπ
π¦2
2+ π2
π = π1πππ₯2
2 ππ π = π2πβπ
π¦2
2 β¦ 5
Substituting (5) in (2) we get
π’ π₯, π¦ = π1π2ππ(π₯2βπ¦2)
2
27. By the method of separation of variables solve πππ + πππ = π
Solution: Given
π₯2ππ§
ππ¦+ π¦3
ππ§
ππ₯= 0 β¦ 1
Let π§ = π π₯ π π¦ β¦ (2) be the solution of (1).
Then
π§π₯ = πβ²π β¦ 3
π§π¦ = ππβ²β¦ 4
Substituting (3) and (4) in (1), we get
π¦3. πβ²π + π₯2. ππβ² = 0
βπ¦3. πβ²π = π₯2. ππβ²
πβ²
π₯2π= β
πβ²
π¦3π= π
πβ² = ππ₯2π ππ πβ² = βππ¦3π
ππ
ππ₯= ππ₯2π ππ
ππ
ππ¦ = βπππ¦3
ππ
π= ππ₯2 ππ₯ ππ
ππ
π = βππ¦3 ππ¦
log π = ππ₯3
3+ π1 ππ log π = βπ
π¦4
4+ π2
π = π1πππ₯3
3 ππ π = π2πβπ
π¦4
4 β¦ 5
Substituting (5) in (2) we get
π’ π₯, π¦ = π1π2πππ₯3
3 πβππ¦4
4
28. State the assumptions made in the derivation of one dimensional wave equation.
Solution:
(i) The mass of the string per unit length is constant
(ii) The string is perfectly elastic and does not offer any resistance to bending
(iii) The tension caused by stretching the string before fixing it at the end points is so large
that the action of the gravitational force on the string can be neglected.
(iv) The string performs a small transverse motion in a vertical plane, that is every particle of
the string moves strictly vertically so that the deflection and the slope at every point of the
string remain small in absolute value.
29. Write down the boundary condition for the following boundary value problem β If a
string of length βlβ initially at rest in its equilibrium position and each of its point is given
the velocity ππ
ππ
π=π= πππππ
π π π
π π < π₯ < π, determine the displacement function
π(π, π)β?
Solution: The boundary conditions are
π) π¦ 0, π‘ = 0, π‘ > 0 ππ) π¦ π, π‘ = 0, π‘ > 0
πππ) π¦ π₯, 0 = 0, 0 < π₯ < π ππ£) ππ¦(π₯, 0)
ππ‘= π£0π ππ
3ππ₯
π, 0 < π₯ < π
30. If the ends of a string of length βlβ are fixed and the midpoint of the string is drawn
aside through a height βhβ and the string is released from rest, write the initial conditions.
Solutions:
π π¦ 0, π‘ = 0
ππ π¦ π, π‘ = 0
πππ ππ¦(π₯, 0)
ππ‘= 0
ππ£ π¦ π₯, 0 =
2ππ₯
π, 0 < π₯ <
π
22π
π π β π₯ ,
π
2< π₯ < π