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SRM Institute of Science and Technology
Ramapuram Campus
Department of Mathematics
Continuous Assessment I
18MAB201T/ Transforms and Boundary Value Problems
Part-A
The order and degree of a PDE
1
. a) 2,1
b) 1,2
is
c) 2,2
(CLO-1
Ans: (a)
d) 1,1
Remember)
The order and degree of a PDE
2
.
is
a)1,2
b) 2,1
c) 1,3
(CLO-1
Remember)
Ans: (b)
d) 3,1
While forming the PDE, if the number of arbitrary
3 constants to be eliminated is equal to the number of
.
independent variables,then the resulting PDE will be
of _________ order.
a)1st
b) 2nd
c)
3rd
(CLO-1
Remember)
Ans: (a)
d) >1
The complete integral of
is
4
. a)
b)
Ans: (a)
c)
d)
(CLO-1 Apply)
+c
5. The solution of
is
Ans: (a)
a)
(CLO-1 Apply)
b)
c)
d) a
6.
The P.I of
a)
7
is
b)
c)
The P.I of
a)
c)
d) 0
Ans: (a)
(CLO-1 Apply)
is
b)
Ans: (b)
d) 0
(CLO-1 Apply)
8.
9.
The complementary function of
is
a)
b)
c)
d)2x
The P.I of
a)
Ans: (a)
(CLO-1
Apply)
is
b)
c)
d)
Ans: (b)
(CLO-1
Apply)
c)
d)
Ans: (a)
(CLO-1 Apply)
Ans: (a)
(CLO-1 Apply)
x
The P.I of
10.
a)
b)
0
11.
12
The P.I of
is
a)
b)
c)
d) 0
The complete integral of
is
(a)
c)
13
14.
15.
16.
(CLO-1 Apply)
b)
d)
Ans: (a)
+2
The complete integral of F(p,q) =0 is
a) 0
b) px + qy + c
c) Z = ax + f(a)y + c d)1
While forming the PDE, if the number of
arbitrary constants to be eliminated is more than
the number of independent variables, then the
resulting PDE will be of _________ order.
a) 1st
b) 2nd and higher
c) only 3rd
nd
d) only 2
(CLO-1 Apply)
Ans: (c)
(CLO-1
Remember)
Ans: (b)
The complete integral of z= px+qy+p+q is
a)z=ax+by+c
b) z= ax+by+a+b
c)z=ax+by+b
d) z= ax+by+a
The P.I of
Ans: (b)
is
(CLO-1 Apply)
(CLO-1 Apply)
a)
17.
b)
c) 8
Ans: (b)
d) 0
The P.I of
is
(CLO-1 Apply)
a)
b)
c) 0
Ans: (a)
d) 1
18.
The solution of
is
Ans: (a)
a)
b)
(CLO-1 Apply)
c)
d) xy
19.
The P.I of
is
(CLO-1
a)
b)
c)
Ans: (b)
Apply)
d) 0
20.
The solution of r − 4s + 4t = 0 is
a)
Ans: (b)
b)
(CLO-1 Apply)
c)
21.
d) z=1
The complete integral of
is
Ans: (a)
a)
(CLO-1 Apply)
b)
c)
d)
22.
23.
The complete integral of
is
a)
b)
c)
d) z=ay+bx
The solution of
Ans: (a)
is
(CLO-1 Apply)
(CLO-1 Apply)
Ans: (a)
a)
b) z = ax
c)
24.
25.
d)
Equation of the form Pp+Qq=R is called --------a) Clairaut’s type
b) Lagrange’s Linear
equations
c) Euler form
The P.I of
is
(CLO-1
Remember)
d) Laurent’s Form
a)
Ans: (a)
b)
c)
Ans: (b)
(CLO-1 Apply)
d) 0
The solution of r − 4s + 4t = 0 is
(CLO-1 Apply)
26
a)
c)
27
Ans: (b)
b)
d) z=6
While forming the PDE, if the number of
arbitrary constants to be eliminated is more than
the number of independent variables, then the
resulting PDE will be of _________ order.
a) 1st
b) 2nd and higher
c) 3rd and
higher d) only 2nd
(CLO-1
Remember)
Ans: (b)
The order and degree of a PDE
is
(CLO-1 Apply)
28
a)1,2
b) 2,1
c) 1,3
The solution of
a)
Ans: (b)
is
Ans: (a)
(CLO-1 Apply)
29
30
d) 3,1
The complete solution of
is
(CLO-1 Apply)
a)
b)
Ans: (c)
c)
d)
The solution of
is
a)
Ans: (d)
b)
(CLO-1 Apply)
31
c)
d)
The complete Integral of
is
a)
Ans: (a)
b)
32
(CLO-1 Apply)
c)
d)
The Particular Integral of
is
a)
Ans: (c)
b)
(CLO-1 Apply)
33
c)
d)
The Particular Integral of
Ans: (a)
is
34
a)
b)
(CLO-1 Apply)
c)
d)
The Particular Integral of
is
a)
35.
Ans: (a)
b)
(CLO-1 Apply)
c)
d)
The complete integral of z=px+qy+p-q is
(CLO-1 Apply)
36
(a)z= ax+by+a-b
c)
+c
b) z=ax+by
d)
Ans: (a)
+2
18MAB201T – TRANSFORMS AND BOUNDARY VALUE PROBLEMS
Partial Differential Equations
SRM INSTITUTE OF SCIENCE AND TECHNOLOGY
RAMAPURAM CAMPUS
DEPARTMENT OF MATHEMATICS
Year/Sem : II/III
Branch: Common to All branches
Unit I – Partial Differential Equations
1.Form PDE of 𝑧 = 𝑎𝑥 + 𝑎2 𝑦 2 + 𝑏
(a) 𝒒 = 𝟐𝒑𝟐 𝒚
by eliminating arbitrary constants
(b) 𝑞 = 2𝑝𝑦
(c) 𝑝 = 2𝑞2 𝑦
(d) 𝑝 = 2𝑞𝑦
Solution:
Given 𝑧 = 𝑎𝑥 + 𝑎2 𝑦 2 + 𝑏 ------(1)
Differentiate (1) partially with respect to 𝑥 𝑎𝑛𝑑 𝑦 ,
𝑝=𝑎
− −−→ (2)
𝑞 = 2𝑎2 𝑦 − −→ (3)
Substituting (2) in (3) we get 𝑞 = 2𝑝2 𝑦 which is the required PDE.
2. Form PDE of 2𝑧 = (𝑎𝑥 + 𝑦)2 + 𝑏 , by eliminating arbitrary constants
(a) 𝑝𝑥 + 𝑞𝑦 = 𝑝2 (b) 𝑝𝑦 + 𝑞𝑥 = 𝑞 2 (c) 𝒑𝒙 + 𝒒𝒚 = 𝒒𝟐 (d) 𝑝2 𝑥 + 𝑞 2 𝑦 = 𝑞 2
Solution:
Given 2𝑧 = (𝑎𝑥 + 𝑦)2 + 𝑏 ------(1)
Differentiate (1) partially with respect to 𝑥 𝑎𝑛𝑑 𝑦 ,
2𝑝 = 2𝑎(𝑎𝑥 + 𝑦)
− −−→ (2)
2𝑞 = 2(𝑎𝑥 + 𝑦)
− −→ (3)
Dividing (2) by (3) we get
𝑝
𝑞
= 𝑎 𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑖𝑛𝑔 𝑖𝑛 (3) 𝑤𝑒 𝑔𝑒𝑡
𝑝𝑥 + 𝑞𝑦 = 𝑞 2 which is the required PDE.
SRMIST, Ramapuram
1
Department of Mathematics
18MAB201T – TRANSFORMS AND BOUNDARY VALUE PROBLEMS
2 2𝑦
3. Form PDE of 𝑧 = 𝑎𝑥𝑒 𝑦 + 𝑎 2𝑒
(a) 𝑝 = 𝑥𝑞 + 𝑝2
Partial Differential Equations
+ 𝑏 , by eliminating arbitrary constants
(b) 𝒒 = 𝒙𝒑 + 𝒑𝟐 (c) 𝑝 = 𝑥𝑞 + 𝑞 2
(d) 𝑞 = 𝑥 + 𝑝2
Solution
Given 𝑧 = 𝑎𝑥𝑒 𝑦 +
𝑎2 𝑒 2𝑦
2
+ 𝑏 ------(1)
Differentiate (1) partially with respect to 𝑥 𝑎𝑛𝑑 𝑦 ,
𝑝 = 𝑎𝑒 𝑦
− −−→ (2)
𝑦
𝑞 = 𝑎𝑥𝑒 + 𝑎2 𝑒 2𝑦
− −→ (3)
Substituting (2) in (3) we get 𝑞 = 𝑥𝑝 + 𝑝2 which is the required PDE.
4. Form PDE of 𝑧 = 𝑓(𝑥 2 − 𝑦 2 ) by eliminating arbitrary function
(a) 𝒚𝒑 + 𝒙𝒒 = 𝟎 (b) 𝑦𝑝 − 𝑥𝑞 = 0
(c) 𝑦𝑞 + 𝑥𝑝 = 0 (d) 𝑦𝑞 − 𝑥𝑝 = 0
Solution:
Given 𝑧 = 𝑓(𝑥 2 − 𝑦 2 ) − −→ (1)
Differentiate (1) partially with respect to 𝑥 𝑎𝑛𝑑 𝑦 ,
𝑝 = 2𝑥𝑓 ` (𝑥 2 − 𝑦 2 ) − −→ (2)
𝑞 = −2𝑦𝑓 ` (𝑥 2 − 𝑦 2 ) − −→ (3)
𝑝
𝑞
From (2) & (3) 𝑓 ` (𝑥 2 − 𝑦 2 ) =
& 𝑓 ` (𝑥 2 − 𝑦 2 ) =
So,
𝑝
2𝑥
=
2𝑥
𝑞
−2𝑦
⇒ 𝑦𝑝 + 𝑥𝑞 = 0 𝑤ℎ𝑖𝑐ℎ 𝑖𝑠 𝑡ℎ𝑒 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑 𝑃𝐷𝐸.
−2𝑦
5. Form PDE of 𝜑(𝑥 2 + 𝑦 2 + 𝑧 2 , 𝑙𝑥 + 𝑚𝑦 + 𝑛𝑧) = 0
(a)
(c)
𝟐𝒙+𝟐𝒛𝒑
𝟐𝒚+𝟐𝒛𝒒
2𝑥+2𝑧𝑝
2𝑦+2𝑧𝑞
=
=
𝒍+𝒏𝒑
𝒎+𝒏𝒒
𝑚+𝑛𝑞
𝑙+𝑛𝑝
(b)
(d)
2𝑦+2𝑧𝑞
2𝑥+2𝑧𝑝
2𝑥−2𝑧𝑝
2𝑦−2𝑧𝑞
=
=
𝑙+𝑛𝑝
𝑚+𝑛𝑞
𝑙+𝑛𝑝
𝑚+𝑛𝑞
Solution:
𝑥 2 + 𝑦 2 + 𝑧 2 = 𝜑( 𝑙𝑥 + 𝑚𝑦 + 𝑛𝑧) − −→ (1)
Differentiate Partially (1) with respect to 𝑥 𝑎𝑛𝑑 𝑦
2𝑥 + 2𝑧𝑝 = (𝑙 + 𝑛𝑝)𝜑 ` (𝑙𝑥 + 𝑚𝑦 + 𝑛𝑧) − −→ (2)
2𝑦 + 2𝑧𝑞 = (𝑚 + 𝑛𝑞)𝜑 ` (𝑙𝑥 + 𝑚𝑦 + 𝑛𝑧) − −→ (3)
(2)
(3)
⇒
2𝑥+2𝑧𝑝
2𝑦+2𝑧𝑞
=
𝑙+𝑛𝑝
𝑚+𝑛𝑞
SRMIST, Ramapuram
which is the required PDE
2
Department of Mathematics
18MAB201T – TRANSFORMS AND BOUNDARY VALUE PROBLEMS
Partial Differential Equations
6.Form a PDE of 𝑧 = 𝑓(𝑥 2 + 𝑦 2 ) 𝑏𝑦 𝑒𝑙𝑖𝑚𝑖𝑛𝑎𝑡𝑖𝑛𝑔 𝑡ℎ𝑒 𝑎𝑟𝑏𝑖𝑡𝑟𝑎𝑟𝑦 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛
(a) 𝑝𝑥 = 𝑦𝑞
(b) 𝑝 + 𝑦 = 𝑥𝑞
(c) 𝑝𝑦 = 𝑥 + 𝑞
(d) 𝒑𝒚 = 𝒙𝒒
Solution:
Given 𝑧 = 𝑓(𝑥 2 + 𝑦 2 ) − −→ (1)
Differentiate partially (1) with respect to 𝑥 𝑎𝑛𝑑 𝑦 𝑤𝑒 𝑔𝑒𝑡
𝑝 = 2𝑥 𝑓 ` (𝑥 2 + 𝑦 2 )
𝑞 = 2𝑦 𝑓 ` (𝑥 2 + 𝑦 2 )
𝑝
𝑥
Therefore, = ⇒ 𝑝𝑦 = 𝑥𝑞 𝑤ℎ𝑖𝑐ℎ 𝑖𝑠 𝑡ℎ𝑒 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑 𝑃𝐷𝐸
𝑞
𝑦
7. Solve 𝑝2 + 𝑞2 = 4
(a) 𝒛 = 𝒂𝒙 ± √𝟒 − 𝒂𝟐 + 𝒄
(c) 𝑧 = 𝑎𝑥 ± √4 − 𝑎 + 𝑐
(b) 𝑧 = 𝑎𝑥 ± √4 + 𝑎2 + 𝑐
(d) 𝑧 = 𝑎𝑥 ± √4 + 𝑎 + 𝑐
Solution:
Given 𝑝2 + 𝑞 2 = 4 − − → (1)
Let us assume that 𝑧 = 𝑎𝑥 + 𝑏𝑦 + 𝑐 − −→ (2) be a solution of (1)
Partially differentiating (2) with respect to 𝑥 𝑎𝑛𝑑 𝑦 we get
𝑝 = 𝑎, 𝑞 = 𝑏
Substituting above in (1) we get
𝑎2 + 𝑏 2 = 4 − −→ (3)
From (3) we get 𝑏 = ±√4 − 𝑎2
Substituting in (2) we get 𝑧 = 𝑎𝑥 ± √4 − 𝑎2 + 𝑐 which is the complete integral
of (1). There is no singular integral of this type 𝑓(𝑝, 𝑞) = 0
8. Find the complete integral of 𝑝 = 𝑞
(a) 𝑧 = 𝑎(𝑥 − 𝑦) + 𝑐
(c) 𝑧 = 𝑎𝑥 + 𝑦 + 𝑐
(b) 𝑧 = 2𝑎𝑥 + 𝑐
(d) 𝒛 = 𝒂(𝒙 + 𝒚) + 𝒄
Solution:
Given 𝑝 = 𝑞 − −→ (1)
Let us assume that 𝑧 = 𝑎𝑥 + 𝑏𝑦 + 𝑐 − −→ (2) be a solution of (1)
Partially differentiating (2) with respect to 𝑥 𝑎𝑛𝑑 𝑦 we get
𝑝 = 𝑎, 𝑞 = 𝑏
Substituting above in (1) we get
𝑎 = 𝑏 − −→ (3)
SRMIST, Ramapuram
3
Department of Mathematics
18MAB201T – TRANSFORMS AND BOUNDARY VALUE PROBLEMS
Partial Differential Equations
Substituting in (2) we get 𝑧 = 𝑎(𝑥 + 𝑦) + 𝑐 which is the complete integral of
(1).
9.Find the complete solution of √𝑝 + √𝑞 = 1
1
2
𝟏
𝟐
(a) 𝑧 = 𝑎𝑥 + (1 + √𝑎) 𝑦 + 𝑐
(b) 𝒛 = 𝒂𝒙 + (𝟏 − √𝒂) 𝒚 + 𝒄
1
2
1
2
(c) 𝑧 = 𝑎𝑥 − (1 − √𝑎) 𝑦 + 𝑐
(d) 𝑧 = 𝑎𝑥 − (1 + √𝑎) 𝑦 + 𝑐
Solution:
Given √𝑝 + √𝑞 − 1 = 0
− −→ (1)
The Complete Solution is given by 𝑧 = 𝑎𝑥 + 𝑏𝑦 + 𝑐 − −→ (2)
Replace 𝑝 𝑏𝑦 𝑎 and 𝑞 𝑏𝑦 𝑏 𝑖𝑛 (1), 𝑤𝑒 𝑔𝑒𝑡 √𝑎 + √𝑏 − 1 = 0
√𝑏 = 1 − √𝑎 ⇒ 𝑏 = (1 − √𝑎)
1
2
1
2
Substituting in (2), we get 𝑧 = 𝑎𝑥 + (1 − √𝑎) 𝑦 + 𝑐 which is required complete
solution.
10. Find the Singular integral of 𝑧 = 𝑝𝑥 + 𝑞𝑦 + 𝑝𝑞
(a) 𝑧 = 𝑥𝑦
(c) 𝒛 = −𝒙𝒚
(b) 𝑧 = −
(d) 𝑧 =
𝑥
𝑥
𝑦
𝑦
Solution:
The Complete Integral is 𝑧 = 𝑎𝑥 + 𝑏𝑦 + 𝑎𝑏 − −→ (1)
Partially differentiating (1) with respect to `𝑎` 𝑎𝑛𝑑 `𝑏` 𝑎𝑛𝑑 𝑒𝑞𝑢𝑎𝑡𝑖𝑛𝑔 𝑡𝑜 0
𝜕𝑧
= 𝑥 + 𝑏 = 0 − −→ (2)
𝜕𝑎
𝜕𝑧
= 𝑦 + 𝑎 = 0 − −→ (3)
𝜕𝑏
From (2) and (3) 𝑎 = −𝑦, 𝑏 = −𝑥 𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑖𝑛𝑔 𝑖𝑛 (1)
𝑧 = −𝑥𝑦 − 𝑥𝑦 + 𝑥𝑦
So, 𝑧 = −𝑥𝑦 𝑖𝑠 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑 𝑠𝑖𝑛𝑔𝑢𝑙𝑎𝑟 𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑙.
SRMIST, Ramapuram
4
Department of Mathematics
18MAB201T – TRANSFORMS AND BOUNDARY VALUE PROBLEMS
Partial Differential Equations
11.Solve 𝑝 + 𝑞 = 1
(a) 𝑓(𝑥 + 𝑦, 𝑦 − 𝑧) = 0
(c) 𝒇(𝒙 − 𝒚, 𝒚 − 𝒛) = 𝟎
(b) 𝑓(𝑥 − 𝑦, 𝑦 + 𝑧) = 0
(d) 𝑓(𝑥 + 𝑦, 𝑦 + 𝑧) = 0
Solution:
𝑝 + 𝑞 = 1 𝑖𝑛 𝑡ℎ𝑒 𝑓𝑜𝑟𝑚 𝑜𝑓 𝑝𝑃 + 𝑞𝑄 = 𝑅
Comparing 1st and 2nd & 2nd
𝑑𝑥 𝑑𝑦 𝑑𝑧
=
=
1
1
1
rd
and 3 term and integrating
∫ 𝑑𝑥 = ∫ 𝑑𝑦
& ∫ 𝑑𝑦 = ∫ 𝑑𝑧
𝑥 − 𝑦 = 𝑐1 ,
Solution is 𝑓(𝑥 − 𝑦, 𝑦 − 𝑧) = 0
𝑦 − 𝑧 = 𝑐2
12.Find the complete integral of 𝑝2 = 𝑞𝑧
(a) 𝑧 = 𝑘𝑒 𝑎(𝑥−𝑎𝑦)
(c) 𝑧 = 𝑘𝑒 𝑥−𝑎𝑦
(b) 𝑧 = 𝑘𝑒 𝑥+𝑎𝑦
(d) 𝒛 = 𝒌𝒆𝒂(𝒙+𝒂𝒚)
Solution:
Given 𝑝2 = 𝑞𝑧 − −→ (1) which of the form 𝑓(𝑝, 𝑞, 𝑧) = 0
Let 𝑢 = 𝑥 + 𝑎𝑦 , 𝑤ℎ𝑒𝑟𝑒 `𝑎` 𝑖𝑠 𝑎𝑟𝑏𝑖𝑡𝑎𝑟𝑦 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡.
Replace 𝑝 𝑏𝑦
𝑑𝑧
𝑑𝑢
𝑑𝑧
𝑎𝑛𝑑 𝑞 𝑏𝑦 𝑎 ( ) 𝑖𝑛 (1), we get
𝑑𝑢
𝑑𝑧
= 𝑎𝑧
𝑑𝑢
𝑑𝑧
= 𝑎𝑑𝑢
𝑧
Integrating both sides , 𝑙𝑜𝑔𝑧 = 𝑎𝑢 + 𝑐 (𝑜𝑟) 𝑧 = 𝑘𝑒 𝑎𝑢
The complete integral is given by 𝑧 = 𝑘𝑒 𝑎(𝑥+𝑎𝑦)
13.Find the complete integral of 𝑧 = 𝑝𝑞
(a) 𝟒𝒂𝒛 = (𝒙 + 𝒂𝒚 + 𝒃)𝟐
(c) 4𝑎𝑧 = (𝑥 − 𝑎𝑦 + 𝑏)2
(b) 4𝑎𝑧 = (𝑥 + 𝑎𝑦 + 𝑏)3
(d) 4𝑎𝑧 = (𝑥 − 𝑎𝑦 + 𝑏)3
Solution:
Given 𝑧 = 𝑝𝑞 − −→ (1) which of the form 𝑓(𝑝, 𝑞, 𝑧) = 0
Let 𝑢 = 𝑥 + 𝑎𝑦 , 𝑤ℎ𝑒𝑟𝑒 `𝑎` 𝑖𝑠 𝑎𝑟𝑏𝑖𝑡𝑎𝑟𝑦 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡.
Replace 𝑝 𝑏𝑦
𝑑𝑧
𝑑𝑢
𝑑𝑧
𝑎𝑛𝑑 𝑞 𝑏𝑦 𝑎 ( ) 𝑖𝑛 (1), we get
𝑑𝑢
SRMIST, Ramapuram
5
Department of Mathematics
18MAB201T – TRANSFORMS AND BOUNDARY VALUE PROBLEMS
Partial Differential Equations
𝑑𝑧 2
𝑧 = 𝑎( )
𝑑𝑢
𝑑𝑧
𝑑𝑢
𝑑𝑧
√𝑧
= ±√
= ±
𝑧
𝑎
𝑑𝑢
√𝑎
Integrating on both sides, ±2√𝑎𝑧 = 𝑢 + 𝑘
Squaring on both sides and substituting 𝑢 = 𝑥 + 𝑎𝑦 𝑤𝑒 𝑔𝑒𝑡
4𝑎𝑧 = (𝑥 + 𝑎𝑦 + 𝑏)2 which is the required
complete integral.
14.
Solve 𝑝2 + 𝑞 2 = 𝑥 + 𝑦
(a) 𝑧 =
(c) 𝒛 =
2
3
𝟐
𝟑
3
(𝑥 + 𝑘)2 +
𝟑
𝟐
(𝒙 + 𝒌) +
3
2
𝟐
𝟑
3
(𝑦 − 𝑘)2
(𝒚 − 𝒌)
(b) 𝑧 =
𝟑
𝟐
(d) 𝑧 =
3
2
3
2
3
(𝑥 + 𝑘)2 +
3
2
(𝑥 + 𝑘) +
2
3
3
2
3
(𝑦 − 𝑘)2
3
(𝑦 − 𝑘)2
Solution:
The given problem can be written as
𝑝2 − 𝑥 = 𝑦 − 𝑞 2 − −→ (1)
This is of the form 𝑓1 (𝑥, 𝑝) = 𝑓2 (𝑦, 𝑞) and there is no singular integral for this
type. We will find the complete integral.
Let 𝑝2 − 𝑥 = 𝑦 − 𝑞 2 = 𝑘 (𝑠𝑎𝑦)
Then 𝑝 = √𝑥 + 𝑘 , 𝑞 = √𝑦 − 𝑘
We know that 𝑑𝑧 = 𝑝𝑑𝑥 + 𝑞𝑑𝑦
Integrating both sides
𝑧 = ∫ √𝑥 + 𝑘 𝑑𝑥 + ∫ √𝑦 − 𝑘 𝑑𝑦
3
2
2
3
𝑧 = (𝑥 + 𝑘)2 + (𝑦 − 𝑘)2
3
3
which is the required solution
15.Solve 𝑦𝑝 = 2𝑦𝑥 + log 𝑞
(a) 𝒛 = (𝒙𝟐 + 𝒌𝒙) +
(c) 𝑧 = (𝑥 2 + 𝑘𝑥) −
𝒆𝒌𝒚
𝒌
𝑒 𝑘𝑦
𝑘
+𝑪
(b)
𝑧 = (𝑥 2 − 𝑘𝑥) −
+𝐶
(d)
𝑧 = (𝑥 2 − 𝑘𝑥) +
𝑒 𝑘𝑦
𝑘
𝑒 𝑘𝑦
𝑘
+𝐶
+𝐶
Solution:
The given problem can be written as
𝑝 − 2𝑥 =
SRMIST, Ramapuram
𝑙𝑜𝑔𝑞
−→ (1)
𝑦
6
Department of Mathematics
18MAB201T – TRANSFORMS AND BOUNDARY VALUE PROBLEMS
Partial Differential Equations
This is of the form 𝑓1 (𝑥, 𝑝) = 𝑓2 (𝑦, 𝑞) and there is no singular integral for this
type. We will find the complete integral.
𝑙𝑜𝑔𝑞
𝑝 − 2𝑥 =
= 𝑘 (𝑠𝑎𝑦)
𝑦
𝑙𝑜𝑔𝑞
i.e., 𝑝 − 2𝑥 = 𝑘 ,
=𝑘
𝑦
𝑝 = 2𝑥 + 𝑘,
𝑙𝑜𝑔𝑞 − 𝑘𝑦 = 0
𝑞 = 𝑒 𝑘𝑦
𝑝 = 2𝑥 + 𝑘,
𝑧 = ∫ 𝑝𝑑𝑥 + ∫ 𝑞𝑑𝑦
𝑧 = ∫(2𝑥 + 𝑘)𝑑𝑥 + ∫ 𝑒 𝑘𝑦 𝑑𝑦
𝑒 𝑘𝑦
𝑧 = (𝑥 2 + 𝑘𝑥) +
+𝐶
𝑘
𝑤ℎ𝑖𝑐ℎ 𝑖𝑠 𝑡ℎ𝑒 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
16.Solve 𝑥𝑝 + 𝑦𝑞 = 𝑥
𝑦
(a) 𝜑 ( , 𝑥 − 𝑧) = 0
(b) 𝜑(𝑥𝑦, 𝑥 − 𝑧) = 0
(c) 𝝋 ( , 𝒙 − 𝒛) = 𝟎
(d) 𝜑 ( , ) = 0
𝑦 𝑥
𝑥
𝒙
𝑥 𝑧
𝒚
Solution:
This is of Lagrange`s type of PDE where 𝑃 = 𝑥, 𝑄 = 𝑦, 𝑅 = 𝑥
𝑑𝑥
𝑑𝑦
𝑑𝑧
The subsidiary equations are = =
𝑥
𝑑𝑥
𝑥
𝑎𝑛𝑑 𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑡𝑖𝑛𝑔 𝑤𝑒 𝑔𝑒𝑡 𝑙𝑜𝑔𝑥 = 𝑙𝑜𝑔𝑦 + 𝑙𝑜𝑔𝑐1
𝑥
𝑥
𝑖𝑒. , = 𝑐1 𝑆𝑜, 𝑢 =
𝑦
𝑦
𝑑𝑥
𝑑𝑧
Taking first and last, =
⇒ 𝑑𝑥 = 𝑑𝑧 𝑎𝑛𝑑 𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑡𝑖𝑛𝑔 𝑤𝑒 𝑔𝑒𝑡 𝑥 = 𝑧 + 𝑐2
𝑥
𝑥
So 𝑣 = 𝑥 − 𝑧
Taking first two
𝑥
=
𝑦
𝑑𝑦
𝑦
𝑥
The Solution is given by 𝜑(𝑢, 𝑣) = 0 𝑖𝑒. , 𝜑 ( , 𝑥 − 𝑧) = 0
𝑦
SRMIST, Ramapuram
7
Department of Mathematics
18MAB201T – TRANSFORMS AND BOUNDARY VALUE PROBLEMS
Partial Differential Equations
17.Solve (𝐷2 − 4𝐷𝐷` − 5𝐷`2 )𝑧 = 0
(a)𝑧 = 𝑓1 (𝑦 − 𝑥) + 𝑓2 (𝑦 − 5𝑥)
(b)𝒛 = 𝒇𝟏 (𝒚 − 𝒙) + 𝒇𝟐 (𝒚 + 𝟓𝒙)
(c)𝑧 = 𝑓1 (𝑦 + 𝑥) + 𝑓2 (𝑦 + 5𝑥)
(d) 𝑧 = 𝑓1 (𝑦 − 𝑥) + 𝑓2 (𝑦 − 5𝑥)
Solution:
Replace 𝐷 𝑏𝑦 𝑚 𝑎𝑛𝑑 𝐷` 𝑏𝑦 1
The auxiliary equation is given by 𝑚2 − 4𝑚 − 5 = 0
⇒ 𝑚 = 5, −1
𝑇ℎ𝑒 𝑔𝑒𝑛𝑒𝑟𝑎𝑙 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑖𝑠 given by
𝑧 = 𝑓1 (𝑦 − 𝑥) + 𝑓2 (𝑦 + 5𝑥)
18.Solve 25𝑟 − 40𝑠 + 16𝑡 = 0
(a)𝒛 = 𝒇𝟏 (𝟓𝒚 + 𝟒𝒙) + 𝒇𝟐 (𝟓𝒚 + 𝟒𝒙) (b) 𝑧 = 𝑓1 (5𝑦 + 4𝑥) + 𝑓2 (4𝑦 + 5𝑥)
(c)𝑧 = 𝑓1 (4𝑦 + 5𝑥) + 𝑓2 (5𝑦 + 4𝑥)
(d) 𝑧 = 𝑓1 (𝑦 + 4𝑥) + 𝑓2 (5𝑦 + 𝑥)
Solution
Since 𝑟 =
𝜕2 𝑧
𝜕𝑥 2
= 𝐷2 𝑧 , 𝑡 =
𝜕2 𝑧
𝜕𝑦 2
2
= 𝐷` 𝑧 𝑠 =
𝜕2 𝑧
𝜕𝑥𝜕𝑦
= 𝐷𝐷` 𝑧
The auxiliary equation is given by 25𝑚2 − 40𝑚 + 16 = 0 ⇒ (5𝑚 − 4)2 = 0
4 4
So, 𝑚 = ,
5 5
𝑇ℎ𝑒 𝑔𝑒𝑛𝑒𝑟𝑎𝑙 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑖𝑠 given by
𝑧 = 𝑓1 (5𝑦 + 4𝑥) + 𝑓2 (5𝑦 + 4𝑥)
SRMIST, Ramapuram
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Department of Mathematics
18MAB201T – TRANSFORMS AND BOUNDARY VALUE PROBLEMS
Partial Differential Equations
3
𝜕 𝑧
19. Solve 𝜕𝑧
=0
3
(a) 𝑧 = 𝑓1 (𝑦) + 𝑥 2 𝑓2 (𝑦) + 𝑥𝑓3 (𝑦)
(b) 𝑧 = 𝑓1 (𝑦) − 𝑥𝑓2 (𝑦) + 𝑥 2 𝑓3 (𝑦)
(c)𝒛 = 𝒇𝟏 (𝒚) + 𝒙𝒇𝟐 (𝒚) + 𝒙𝟐 𝒇𝟑 (𝒚)
(d)𝑧 = 𝑓1 (𝑦) + 𝑥𝑓2 (𝑦) − 𝑥 2 𝑓3 (𝑦)
Solution:
The auxiliary equation is given by 𝑚3 = 0
𝑚 = 0,0,0
Hence the general solution is given by
𝑧 = 𝑓1 (𝑦) + 𝑥𝑓2 (𝑦) + 𝑥 2 𝑓3 (𝑦)
20.Find the particular integral of
(a) −𝒄𝒐𝒔𝒙 (b) 𝑐𝑜𝑠𝑥
𝜕𝑧
𝜕𝑥
+
𝜕𝑧
= 𝑠𝑖𝑛𝑥
𝑑𝑦
(c) −𝑠𝑖𝑛𝑥
(d) 𝑠𝑖𝑛𝑥
Solution:
𝑃. 𝐼 =
1
𝐷+𝐷`
𝑠𝑖𝑛𝑥 = ∫ 𝑠𝑖𝑛𝑥 𝑑𝑥 = −𝑐𝑜𝑠𝑥
𝑃. 𝐼 = −𝑐𝑜𝑠𝑥
21.
(a)
(c)
Find the Particular Integral of (𝐷2 − 2𝐷𝐷` + 𝐷`2 )𝑧 = cos(𝑥 − 3𝑦)
−𝟏
𝟏𝟔
−1
32
𝐜𝐨𝐬 (𝒙 − 𝟑𝒚)
(b)
cos(𝑥 − 3𝑦)
(d)
1
16
𝑥
16
cos (𝑥 − 3𝑦)
cos (𝑥 − 3𝑦)
Solution:
𝑃. 𝐼 =
1
cos(𝑥 − 3𝑦)
𝐷2 − 2𝐷𝐷` + 𝐷`2
Replace 𝐷2 𝑏𝑦 − 1, 𝐷`2 𝑏𝑦 − 9, 𝐷𝐷` 𝑏𝑦 3
𝑃. 𝐼 =
SRMIST, Ramapuram
1
−1−6−9
cos(𝑥 − 3𝑦)
9
=
−1
16
cos (𝑥 − 3𝑦)
Department of Mathematics
18MAB201T – TRANSFORMS AND BOUNDARY VALUE PROBLEMS
22.
(a)
𝟏
𝟐
Find the Particular Integral of
𝒆𝒙+𝒚
(b)
1
2
𝑒 𝑥−𝑦
(c)
1
2
𝜕2 𝑧
𝜕2 𝑧
𝜕𝑥
𝜕𝑥𝜕𝑦
−5
2
𝑒 2𝑥+𝑦
(d)
1
2
+6
Partial Differential Equations
𝜕2 𝑧
𝜕𝑦 2
= 𝑒 𝑥+𝑦
𝑒 2𝑥−𝑦
Solution:
The given equation can be written as
(𝐷2 − 5𝐷𝐷` + 6𝐷`2 )𝑧 = 𝑒 𝑥+𝑦
𝑃. 𝐼 =
=
1
𝑒 𝑥+𝑦
𝐷2 − 5𝐷𝐷` + 6𝐷`2
1
1−5+6
𝑒 𝑥+𝑦 =
1
2
𝑒 𝑥+𝑦
23. Solve (𝐷3 − 6𝐷2 𝐷` + 11𝐷𝐷`2 − 6𝐷`3 )𝑧 = 0
(a) 𝑧 = 𝑓1 (𝑦 − 𝑥) + 𝑓2 (𝑦 + 2𝑥) + 𝑓3 (𝑦 + 3𝑥)
(b) 𝒛 = 𝒇𝟏 (𝒚 + 𝒙) + 𝒇𝟐 (𝒚 + 𝟐𝒙) + 𝒇𝟑 (𝒚 + 𝟑𝒙)
(c) 𝑧 = 𝑓1 (𝑦 + 𝑥) + 𝑓2 (𝑦 − 2𝑥) + 𝑓3 (𝑦 + 3𝑥)
(d) 𝑧 = 𝑓1 (𝑦 − 𝑥) + 𝑓2 (𝑦 + 2𝑥) + 𝑓3 (𝑦 − 3𝑥)
Solution:
The auxiliary Equation is given by 𝑚3 − 6𝑚2 + 11𝑚 − 6 = 0
Solving 𝑚 = 1,2,3
C.F is 𝑓1 (𝑦 + 𝑥) + 𝑓2 (𝑦 + 2𝑥) + 𝑓3 (𝑦 + 3𝑥)
𝑤ℎ𝑒𝑟𝑒 𝑓1 , 𝑓2 , 𝑓3 𝑎𝑟𝑒 𝑎𝑟𝑏𝑖𝑡𝑟𝑎𝑟𝑦 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛𝑠
Solution is 𝑧 = 𝑓1 (𝑦 + 𝑥) + 𝑓2 (𝑦 + 2𝑥) + 𝑓3 (𝑦 + 3𝑥)
SRMIST, Ramapuram
10
Department of Mathematics
18MAB201T – TRANSFORMS AND BOUNDARY VALUE PROBLEMS
Partial Differential Equations
24.Find the Particular Integral of (𝐷2 + 3𝐷𝐷` − 4𝐷`2 )𝑧 = 𝑠𝑖𝑛𝑦
1
1
(a) − 𝑠𝑖𝑛𝑦
(b) 𝑐𝑜𝑠𝑦
4
4
1
𝟏
4
𝟒
(c)- 𝑐𝑜𝑠𝑦 (d) 𝒔𝒊𝒏𝒚
Solution:
𝑃. 𝐼 =
1
𝐷2 +3𝐷𝐷` −4𝐷`2
𝑠𝑖𝑛𝑦
Replace 𝐷2 𝑏𝑦 0, 𝐷`2 𝑏𝑦 − 1, 𝐷𝐷` 𝑏𝑦 0
𝑃. 𝐼 =
25. Solve (𝐷3 − 3𝐷2 𝐷`
1
1
𝑠𝑖𝑛𝑦 = 𝑠𝑖𝑛𝑦
0 + 0 − 4(−1)
4
+ 4𝐷`3 )𝑧 = 0
(a) 𝑧 = 𝑓1 (𝑦 + 𝑥) + 𝑓2 (𝑦 + 2𝑥) + 𝑥𝑓3 (𝑦 − 2𝑥)
(b) 𝑧 = 𝑓1 (𝑦 + 𝑥) + 𝑓2 (𝑦 + 2𝑥) + 𝑥𝑓3 (𝑦 + 2𝑥)
(c) 𝑧 = 𝑓1 (𝑦 + 𝑥) + 𝑓2 (𝑦 − 2𝑥) + 𝑥𝑓3 (𝑦 − 2𝑥)
(d) 𝒛 = 𝒇𝟏 (𝒚 − 𝒙) + 𝒇𝟐 (𝒚 + 𝟐𝒙) + 𝒙𝒇𝟑 (𝒚 + 𝟐𝒙)
Solution:
The auxiliary equation is 𝑚3 − 3𝑚2 + 4 = 0
The roots are 𝑚 = −1,2,2
C.F = 𝑓1 (𝑦 − 𝑥) + 𝑓2 (𝑦 + 2𝑥) + 𝑥𝑓3 (𝑦 + 2𝑥)
Solution is 𝑧 = 𝑓1 (𝑦 − 𝑥) + 𝑓2 (𝑦 + 2𝑥) + 𝑥𝑓3 (𝑦 + 2𝑥)
SRMIST, Ramapuram
11
Department of Mathematics
SRM Institute of Science and Technology
Ramapuram Campus
Department of Mathematics
Question Bank of Module-2(Fourier Series)
(2020–2021-ODD)
Subject.Code: 18MAB201T
Subject.Name: Transforms and Boundary Value Problems
Part-A(1*20=20)
Year/Sem: II/III
Branch: Common to All branches
Module-2( Fourier Series)
1.
sin x is
a periodic function with period
(a)  (b)
2.
1*20=20

(c) 2 (d) 4
2
ANS
-
(CLO-2,
Remember)
c
Which one of the following function is an even function
(a) (a) sin x (b) x (c) ex (d) x2
(CLO-2,
ANS
-
Remember)
a
(CLO-2,
d
3.
a
 f(x)dx  0 if f(x) is
a
ANS-
(a)odd (b) even
4.
a

a
Remember)
(c) periodic (iv) zero
a
f(x)dx  2 f(x)dx if f(x) is
0
(a) even
(CLO-2,
(b) odd (c) neither even nor odd (iv) periodic
ANS
-
Remember)
a
5.

 | x | dx is equal to
ANS



 /2
0
0
0
(a) 2 xdx (b) 0 (c) 2 (x)dx (iv) 4  xdx
a
-
(CLO-2,
Remember)
6.
tan x
is a periodic function with period
(CLO-2,
(b) 2 (c) 3 (d)  / 2
(a) 
Remember)
ANS
-
a
7.
The constant a0 of the Fourier series for the function f(x) = x is
0  x  2
ANS
(a) 2 (b)  (c) 3
8.
(d) 0
ANS
0  x  2
9.
(d)
Apply)
2
f(x) cosnx dx (b) 0
l 0
(c)
2
f(x) sinnx dx
l 0
ANS
(d)
1
x dx
l l
If f(x) is an even function in (, ) then the value of bn in the Fourier
series expansion of f(x) is
(CLO-2,
Remember)
ANS
–
c


1
2
(a)  f(x) cosnx dx (b)  f(x) cosnx dx (c) 0
 
0
–
b
l
l
l
10.
(CLO-2,
k
2
If f(x) is an odd function in (l,l) then value of an in the Fourier series
expansion of f(x) is
(a)
-
b
(b) 2k (c) 0
(d)
(CLO-2,

2
 f(x) sinnx dx
 
11.
Remember)
The RMS value of f(x) in a  x  b is
b
(a) 0
12.
(b)
 [f(x)] dx
b
2
a
ba
(c)
 [f(x)] dx
a
ba
ANS
b
 f(x)dx
2
(d)
–
(CLO-2,
b
Remember)
a
ba
The RMS value of f(x) = x in 1  x  1 is
(a) 1 (b) 0 (c)
13.
(CLO-2,
Apply)
b
The constant a0 of the Fourier series for the function f(x) = k,
(a) k
-
1
3
ANS
(d) -1
If y is the RMS value of f(x) in (0,2l) then
–
Apply)
c
a20 1  2
  (an  bn2 )
4 2 n1
(CLO-2,
(CLO-2,
is
ANS
–
Remember)
(a)
14.
y2
2
(b) y (c)
y
2
Half range cosine series for f(x) in (0, ) is

a
(a) 0   an cos nx
2 n1

(c)
b
n1
15.
d
2
(d) y
n
(b)
(d)
(CLO-2,
Remember)
a
n1
–
a
a20 1  2
  (an  bn2 )
4 2 n1

sinnx
ANS
n
cosnx
Half range sine series for f(x) in (0, ) is
a
(a)  an cosnx (b) 0   an cos nx (c)
2
n1

ANS

b
n1
n
-
c
sinnx (d)
(CLO-2,
a0
  an cos nx
2
16.
Remember)
x,
x,
The function defined by f(x)  
(a) odd
  x  0
is
0x
(b) neither odd nor even
(CLO-2,
(c) periodic
(d) even
ANS-
Remember)
d
17.
g(x), 0  x  
is
g(x),   x  0
ANS
The function f(x)  
-
b
(CLO-2,
18.
(a) even function
(b) odd function
(c) increasing function
(d) periodic function
The value of Fourier series of f(x) in 0  x  2 at x = 0 is
(a) f(0) (b) f(2)
19.
Remember)
(c)
f(0)  f(2)
(d) 0
2
ANS-
c
(CLO-2,
Remember)
A function f(x) with period T if
(a) f(x  T)  f(T)
(CLO-2,
(b) f(x  T)  f(x)
Remember)
ANS-
(c) f(x  T)  f(x) (d) f(x  T).f(x)  0
b
20.
An example for a function which neither even nor odd
(a) x sin x
(b) eax
(c) x2 sin x
(d) x cos x
(CLO-2,
ANS-
b
21.
Write the formula for finding Euler’s constant of a0 Fourier series in (0,2π)
Apply)
a ) a0 
c) a0 
22.
1

l

2

2

f ( x)dx

f ( x)dx
(CLO-2
Ans
(a)
0
2

(b) 1
Remember)
f ( x)dx
0
x 0  x  1
at x  0
2 1  x  2
(d)0
Sum the Fourier series for f ( x)  
(c) 3
x 0  x  1
at x  1
2 1  x  2
1
3
(c)
(d)
4
2
Ans
(b)
(CLO-2
Remember)
Sum the Fourier series for f ( x)  
1
(b)
6
1
(a)
3
24.

2
1
d ) a0 
2
0
(a) 2
23.
b) a0 
f ( x)dx
0
2
(CLO-2
Ans
(c)
Remember)
What is the constant term a0 and the coefficient of cosnx, an in the Fourier
series expansion of f(x) = x - x3 in (-π, π)?
(a)

,0
3
(b) 0, 
(c) 0,

(d) 0,0
2
Ans
(d)
(CLO-2
Remember)
Write the formula for finding Euler’s constant of an Fourier series in (0,2π)
25
a ) an 
c) an 
26.
2
1
2
1


f ( x) cos nxdx
b ) an 
0
1

2

d ) an 
f ( x) sin nxdx
0
1

2

f ( x) cos nxdx
0

(CLO-2
Ans
(b)
Remember)
Ans
(a)
Remember)
 f ( x) cos nxdx
0
State Parseval’s Identity for full range expression of f(x) as Fourier series in
(0,2l)
1
a)
2l
2l
  f ( x) 
2
0
dx 
a0 2 1 
  (an 2  bn 2 )
4 2 1
b)
a0 2 1 
1
2
f
(
x
)
dx

  (an 2  bn 2 )
 

l0
4 2 1
c)
a0 2 1 
1
2
f
(
x
)
dx

  (an 2  bn 2 )


l 0
2 2 1
l
2l
1
d)
2l
a0 2 1 
2
2
0  f ( x) dx  4  4 0 (an  bn )
2l
2
What is the constant term a0 and the coefficient of cosnx, an in the Fourier
series expansion of f(x) = x3 in (-π, π)?
27
(a)0,0
(b)  ,1
(c)

2
,0
(d)

3
,0
(CLO-2
Write the formula for finding Euler’s constant of bn Fourier series in (0,2π)
a)bn 
28
c) bn 
1
2


b) bn 
f ( x) sin nxdx
0
2
1


f ( x) sin nxdx d ) bn 
0
1
2
1


 f ( x) sin nxdx
0


(CLO-2
Ans
(c)
f ( x) sin nxdx
0
Find a Fourier sine series for the function f(x)=1; 0<x< π.
29.
a)
4

sin nx
 n1,3,5 n

b)
4

sin nx
4
c)
n
3
n 1


Remember)

(CLO-2

sin nx
2
sin nx
d)

 n1,3,5 n
n 1,3,5 2n

Ans
(a)
Remember)
Find an in expanding e x as Fourier series in (-π, π)
30
(1) n 1
(1) n
a)
2 cosh  b)
cosh 
(1  n 2 )
(1  n 2 )
(1)
2 cosh 
 (1  n 2 )
n 1
n
c)
(CLO-2
Ans
(c)
d)
(1)
cosh 
 (1  n 2 )
Find the value of an for f(x)= c in (0,10) in cosine series expansion
31
(a) 10
(b) c
(c) c/10
(d)0
If f(x) is an odd function defined in (-l , l ) what are the values of a0 and an.?
32
(a) 0,0
(b) 0,2l
(c) 2l,0
(d)1,1
(CLO-2
Ans
(d)
33

3
(b) 
(c)

(d) 0
2
Remember)
(CLO-2
Ans
(a)
Find bn in the expansion of cosx as a Fourier series in (-π, π)
(a)
Remember)
Remember)
(CLO-2
Ans
(d)
Remember)
  ,  x  0
 x,0  x  
Find a0 in the expansion of f(x) = 
34

(a)
2
(b) 
(c) 

2
finding Euler’s constant of a0 for f(x) =
35
36
(a)

3
(b) 
(c)

2
(CLO-2
(d) 
1
  x  in - < x < 
2
(d) 0
Find the fourier constant an of periodicity 3 for f(x) = 2x-x2 in 0<x<3
Ans
(c)
Remember)
(CLO-2
Ans
(b)
Remember)
Ans
(CLO-2
(c)
Remember)
18MAB201T – TRANSFORMS AND BOUNDARY VALUE PROBLEMS
Fourier Series
SRM INSTITUTE OF SCIENCE AND TECHNOLOGY
RAMAPURAM CAMPUS
DEPARTMENT OF MATHEMATICS
Year/Sem : II/III
Branch: Common to All branches
Unit 2 – Fourier Series
1. Write the formula for finding Euler’s constants of a Fourier series in 0  x  2 .
Solution:
Euler’s constants of a Fourier series in 0  x  2 is given by
a0 

2
an 
1

2
bn 
1
1
0
0

2
0
f ( x) dx
f ( x) cos nx dx
f ( x) sin nx dx
2. Write the formula for finding Euler’s constants of a Fourier series in 0  x  2l .
Solution:
Euler’s constants of a Fourier series in 0  x  l is given by
1 2l
f ( x) dx
l 0
1 2l
a n   f ( x) cos nx dx
l 0
1 2l
bn   f ( x) sin nx dx
l 0
a0 
3. Write the formula for Fourier constants for f(x) in the interval    x   .
Solution:
1

a0 
f ( x) dx
 
an 
1
f ( x) cos nx dx
 
bn 
1




f ( x) sin nx dx
 

4. Write the formula for Fourier constants for f(x) in the interval  l  x  l .
Solution:
1 l
f ( x) dx
l l
1 l
a n   f ( x) cos nx dx
l l
1 l
bn   f ( x) sin nx dx
l l
a0 
SRMIST, Ramapuram
1
Deparment of Mathematics
18MAB201T – TRANSFORMS AND BOUNDARY VALUE PROBLEMS
Fourier Series
5. Find the constant value a0 of the Fourier series for the function f(x) = k, 0  x  2 .
Solution:
a0 
1

2

f ( x) dx
0
a) K
b)
1

2
0
2k

k dx
k
(2 )

c) 0
 2k
d) k/2
6. Find the constant value a0 of the Fourier series for the function f(x) = x, 0  x   .
Solution:
a0 

2


a) 
7.

f ( x) dx
0
b)2 
2



0

x dx
2 ( ) 2
 2

d)  /2
c)0
If f ( x)  e x in    x   , find an
Solution:
an 
c)
8.
a)


  f ( x) cos nx dx

 1n

a)
1
 1  n
2
 e
 1n
 e
 1n
 e
 1  n 2
 1  n 2

 e 

1



e
x


1  ex
cos nx  n sin nx 
 
2
 1 n
 

cos nx dx



 e 

b)

 e 

d)
 1n
 1  n 2 
e

 e 

 1n e   e  

If f ( x)  e in    x   , find bn
x
Solution:
bn 
1


  f ( x) sin nx dx


1



e

x
sin nx dx

1  ex
sin nx  n cos nx 
 
2
 1 n
 


n  1
e   e 
2
 1 n
n 1


n  1
e  e 
a)
 1  n 
2
e  e 
c)
 1  n 


n 1


2

2

b)
d)

1
e   e 
2
 1 n


 1n e   e  

b)
SRMIST, Ramapuram
2
Deparment of Mathematics
18MAB201T – TRANSFORMS AND BOUNDARY VALUE PROBLEMS
Fourier Series
 2x
1  l ,  l  x  0
9. Check whether the function is even or odd, where f ( x)  
1  2 x , 0  x  l

l
Solution:
2( x)
2x
 1
 f ( x), where 0  x  l
l
l
 the given function is even function
for  l  x  0, f ( x)  1 
a)
b)
c)
d)
Even function
odd function
constant function
neither even nor odd
10. Find the constant value an of the Fourier series for the function f(x) = x, 0  x   .
Solution:
2 
2
an   f ( x) cos nx dx 





0
0
x cos nx dx

n
2  sin nx cos nx 
2   1  1 


x



 
n
n 2  0
  n 2

n
2   1  1 


a)

  n 2

b)2 
c)
2 1
n
d)  /2

11. Find the constant value bn of the Fourier series for the function f(x) = x,    x   .
Solution:
2 
2
bn   f ( x) sin nx dx 


0


0
x sin nx dx


a)
2   cos nx sin nx 
 2(1) n
x


 
n
n 2  0
n
n
2   1  1 



  n 2

b)
 1n

c)
 2(1) n
n
d)  /2
12. Find the constant term of the Fourier series for the function f(x) = |x|,    x   .
Solution:
a0 
a)
1

f ( x) dx
 



b)2 
2


0
x dx  
d)  /2
c)0
13. Find the Fourier coefficient bn of the Fourier series for the function f(x) = x, 0  x  2 .
Solution:
SRMIST, Ramapuram
3
Deparment of Mathematics
18MAB201T – TRANSFORMS AND BOUNDARY VALUE PROBLEMS
bn 
1


2
f ( x) sin nx dx 
0


  2(1) n / n
a)


1


2
0
Fourier Series
2
1   cos nx sin nx 
x sin nx dx 
x

 
n
n 2  0

b)  2(1) n / n
d)3 
c)0
14. Half-range cosine series for f(x) in 0 ,   is
a)
d)
a0 
2
b)
4

b
n 1
n 1
a0 2



  a n cos nx
n
cos nx
a
c)
n 1
n
cos nx
1 
an 2  bn 2

2 n 1
15. If f(x) = x2 in   ,   then the value of bn is?
Solution:
Since the given function is even in the given interval, the Fourier coefficient bn value is
zero in this case.
a) 1
b)0
c)-1
d)2
16. In the Fourier series expansion of f(x) = sin x in   ,   . What is the value of an?
Solution:
The function 𝑓(−𝑥) = sin(−𝑥) = −𝑠𝑖𝑛𝑥 = −𝑓(−𝑥) 𝑠𝑜 𝑓 (𝑥)𝑖𝑠 𝑜𝑑𝑑 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛
So an = 0
a) 1
b)0
c) 
d)  
17. Find the constant value a0 from the following table:

X
0
2 / 3
 /3
F(x) = y
1
1.4
1.9
1.7
Solution:
a0 
2 y
n
a) 1.9

4 / 3
5 / 3
2
1.5
1.2
1
21  1.4  1.9  1.7  1.5  1.2
 2.9
6
b)2.9
c)4.9
d)6.9
18. What is the sum of the Fourier series at a point x = 1 where the function has a finite
discontinuity.
Solution:
𝑓 (𝑥 + 𝑥0 ) + 𝑓(𝑥 − 𝑥0 )
𝑓 (𝑥) =
2
Here 𝑥0 = 1 , 𝑓(𝑥) =
a) f(x) =
𝑓(𝑥+1)+𝑓(𝑥−1)
f(x+1)+f(x−1)
2
2
b)𝑓 (𝑥) =
𝑓(𝑥+1)
2
c)𝑓 (𝑥) =
𝑓(𝑥−1)
2
d) 𝑓 (𝑥) = 𝑓 (𝑥 + 1) + 𝑓 (𝑥 − 1)
19. In the expansion of 𝑓(𝑥) = 𝑠𝑖𝑛ℎ𝑥 𝑖𝑛 (−𝜋, 𝜋) as a Fourier Series, find the coefficient of 𝑎𝑛 .
Solution:
e x  ex
f ( x)  sin h x 
2
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f (  x) 
Fourier Series
ex  e x
 (e x  e  x )

  sin h x   f ( x)
2
2
So, 𝑓 (𝑥)𝑖𝑠 𝑎𝑛 𝑜𝑑𝑑 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛, 𝑡ℎ𝑒 𝑓𝑜𝑢𝑟𝑖𝑒𝑟 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑎𝑛 𝑖𝑠 0
a) 0
b)1
c)2
d)3
20. To what value, the Fourier series corresponding to f(x) = x2 in (0, 2  ) converges at x = 0?
Solution:
The Fourier series converges to
f (0)  f (2 )
 2 2
2
a)π
c)π2
b)2π
d)𝟐𝛑𝟐
1
21. Examine whether the function 𝑓 (𝑥) = 1−𝑥 , can be expanded in Fourier series in any
interval including 𝑥 = 1
Solution:
1
At 𝑥 = 1, 𝑡ℎ𝑒 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 𝑓(𝑥) = 1−𝑥 𝑖𝑠 𝑛𝑜𝑡 𝑐𝑜𝑛𝑡𝑖𝑛𝑢𝑜𝑢𝑠
By Dirichlet`s condition, we cannot expand 𝑓(𝑥) as a Fourier series.
a) Can be expanded since f(x) is not continuous
b) Cannot be expanded since f(x) does not satisfies Dirichlet`s condition
c) Can be expanded since f(x) satisfies Dirichlet`s Condition
d) Cannot be expanded since f(x) is continuous
22. Find the constant term in the Fourier series corresponding to 𝑓 (𝑥) = 𝑥 − 𝑥 3 𝑖𝑛 (−𝜋, 𝜋)
Solution:
𝑓 (𝑥) = 𝑥 − 𝑥 3
𝑓 (−𝑥) = −𝑥 + 𝑥 3 = −(𝑥 − 𝑥 3 ) = −𝑓(𝑥)
𝑓 (𝑥)𝑖𝑠 𝑎𝑛 𝑜𝑑𝑑 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 (−𝜋, 𝜋)
Hence Constant term 𝑎0 = 0
a) 0
b)1
c)2
d)3
23. Find the R.M.S Value of the function 𝑓 (𝑥) = 𝑥 𝑖𝑛 (0, 𝑙).
Solution:
𝑙
3
R. M. S =
a)
𝑙
√∫ x 2 dx
0
𝒍
√𝟑
𝑙
b)
l
√2
√(x )
3
=
𝑙3
𝑙
= √
=
3𝑙
√3
0
𝑙
c)
l2
d)
√3
l2
√2
24. Find the value of an in the cosine series expansion of f(x) = k in (0,10).
Solution:
𝑛𝜋𝑥 10
1 10
𝑛𝜋𝑥
𝑘 𝑠𝑖𝑛 10
𝑎𝑛 = ∫ 𝑘 𝑐𝑜𝑠
𝑑𝑥 = [ 𝑛𝜋 ]
=0
5 0
10
5
10
0
a) 0
b)10
c)20
d)30
25. Find the R.M.S Value of the function 𝑓 (𝑥) = 𝑘 𝑖𝑛 (−𝑙, 𝑙).
Solution:
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Deparment of Mathematics
18MAB201T – TRANSFORMS AND BOUNDARY VALUE PROBLEMS
𝑙
R. M. S =
a)
26.
√∫−𝑙 k 2 dx
2𝑙
b)𝐤
𝐥
√𝟑
==
√k 2 (x)−𝑙 𝑙
2𝑙
l2
c)
Fourier Series
=𝑘
d)
√3
l2
√2
In the Fourier series expansion of f(x) = |sin x| in   ,   . What is the value of bn?
Solution:
𝑓 (−𝑥) = |sin(−𝑥)| = |−𝑠𝑖𝑛𝑥| = |𝑠𝑖𝑛𝑥| = 𝑓(𝑥)
𝑠𝑖𝑛𝑐𝑒 𝑓(𝑥) 𝑖𝑠 𝑎𝑛 𝑜𝑑𝑑 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 𝑏𝑛 = 0
a) 1
27.
c) 
b)0
d)  
Find 𝑏1, if f(x) = k in 0 < x <  .
Solution:
2 𝜋
2 𝜋
∫ 𝑓(𝑥)𝑠𝑖𝑛𝑥 𝑑𝑥 = ∫ 𝑘𝑠𝑖𝑛𝑥 𝑑𝑥
𝜋 0
𝜋 0
2
2
4
𝜋
= 𝑘(−𝑐𝑜𝑠𝑥)0 = (−(−1) + 1) =
𝜋
𝜋
𝜋
𝑏1 =
a)
28.
2
b)
𝜋
𝟒
c)
𝝅
d)
1
4𝜋
Find the R.M.S Value of the function 𝑓 (𝑥) = 2𝑥 𝑖𝑛 (0,3).
Solution:
3
3
R.M .S 
 4x
2
dx
0
a) 2
29.
1
2𝜋
3

b)6
 x3 
4 
6
 3 0

 2
3
3
c)9
d)0
F(x) = x + x2 in   ,   is ____________
Solution:
𝐹 (−𝑥) = −𝑥 + (−𝑥)2 = −𝑥 + 𝑥 2 ≠ −𝑓(𝑥)so it is neither ever nor odd function
a) Odd function
c) Constant function
30.
Find the series value
b)Even function
d)Neither odd nor even
1
1
1
 2  2  ... , if f(x) = x2 in the interval (-  ,  ), a0 =2  2/3 ,
2
1
2
3
an =4(-1)n+1/n2
Solution:
The Fourier cosine series is
f ( x) 
a0

2

2
n 1
3
 an cos nx 

4 1

n2
n 1

n 1
cos nx
When x = 0,
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f (0) 
2
3

4 1

n2
n 1

n 1
f ( )  f ( )  2


2
3
2 
2
3

4 1

n2
n 1
2
12
a)
4 1

n2
n 1

n 1

b)
Fourier Series
2

2
8
n 1
12
c)

 1n1
n 1
n2

2
2
d)0
Answers
5
b
6
a
7
a
8
a
9
a
10
a
11
c
12
a
13
b
14
a
15
b
16
b
17
b
18
a
19
a
20
d
21
b
22
a
23
a
24
a
25
b
26
b
27
b
28
a
29
d
30
a
SRMIST, Ramapuram
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Deparment of Mathematics
SRM Institute of Science and Technology
Ramapuram Campus
Department of Mathematics
Question Bank of Module-3(Application of PDE)
(2020–2021-ODD)
Subject.Code: 18MAB201T
Subject.Name: Transforms and Boundary Value Problems
Part-A (1*20=20)
Year/Sem: II/III
1.
Branch: Common to All branches
1 mark
The proper solution of the problems on vibration of string is
(a)
y(x, t)  ( Aex  Bex )(Ceat  Deat )
(b)
y(x, t)  ( Ax  B)(Ct  D)
(CLO-3
Ans (c)
(c) y(x, t)  ( A cosx  B sin x)(C cosat  D sin at)
(d) y(x, t)  ( Ax  B)
2.
1 mark
The one dimensional wave equation is
u
(a)
 2
u
2
(b)
In wave equation
2 y
t
T
(a)
(b)
m
4.
5.
2

2
2
a
2 y
x
2
, a2
k
(b)

u
t
 y
(CLO-3
k

(c)
m
T
2
2 u
x2
Ans (b)
stands for
c
In heat equation
(a)

2
a
t 2
x 2
2
x
y2 a t2 2y
(d)

t
x2
a 2 y2
y
(c) t  x
3.
 y
2
(c)
m
k
c
(d)
k
m
(CLO-3
Ans (a)
u
t
0
2 u
(c)
x
2

u
t
(b)
0
2 u
0
t 2
2 u
(d)
x
2

0
Remember)
1 mark
(d)
k
c
(CLO-3
Ans (c)
Remember)
1 mark
The one dimensional heat equation in steady state is
(a)
Remember)
1 mark
, 2 stands for
T
Remember)
(CLO-3
Ans (d)
Remember)
6.
The proper solution of ut  2 u is
1 mark
xx
(a) u  ( Ax  B)C
(b)
(CLO-3
2 2
u  ( A cosx  B sin x)e   t
x
x
(c) u  ( Ae  Be )e
7.
 t
Ans (b)
Remember)
(d) u  At  B
The proper solution in steady state heat flow problems is
(a) u  ( Aex  Bex )e  t
1 mark
(b) u  Ax  B
(CLO-3
2 2
(c) u  ( A cosx  B sin x)e  t
x
(d) u  ( Ae  Be
 x
8.
)(Ce
 at
 De
Ans (b)
 at
)
1 mark
The one dimensional heat equation is
u
2
(a)
u
2


0
(b)
x2 y 2
2u
2u
(c)
 2
a
t 2
x 2
(d)
u
t
u
x
Remember)
 2
u
2
(CLO-3
x2
2
2 u
Ans (b)

t 2
How many initial and boundary conditions are required to
9.
solve
u
t
1 mark

 2u
2
Remember)
x2
(CLO-3
(a) Four
(b) Two
(c) Three
(d) Five
Ans (c)
Remember)
How many initial and boundary conditions are required to
10. solve
2 y

t 2
2
a
2 y
1 mark
x 2
(CLO-3
(a) Two
(b) Three
(c) Five
(d) Four
Ans (d)
1 mark
11. One dimensional wave equation is used to find
(a) Temperature
(CLO-3
(b) Displacement
Ans (b)
(c) Time
12.
13.
(d) Mass
(b) Temperature distribution
(d) Displacement
Heat flows from
(a) Higher to Lower
(c) Lower to higher
(CLO-3
Ans (b)
14. The tension T caused by stretching the string before fixing it
at the end points is
Remember)
1 mark
temperature
(b) Uniform
(d) Stable
Remember)
1 mark
One dimensional heat equation is used to find
(a) Density
(c) Time
Remember)
(CLO-3
Ans (a)
Remember)
1 mark
(a) Increasing
(c) Constant
(CLO-3
(b) Decreasing
(d) Zero
Ans (c)
A string is stretched between two fixed points x = 0 and x =
15. l. The initial conditions are
(a) y(0, t)  0, y(x, t)  0
(b) y(x, 0)  0,
(c) y(0, t)  0, y(l, t)  0
(d)
y
t
Remember)
1 mark
(x, 0)  0
(CLO-3
Ans (c)
Apply)
 y 
 y 
 x   0,  x   0
 (0,t )
 (l ,t )
16. The amount of heat required to produce a given temperature
change in a body is proportional to
(a) Weight of the body
(b) Mass of the body
(c) Density of the body
(d) Tension of the body
1 mark
(CLO-3
Ans (b)
The general solution for the displacement y(x,t) of the string
17. of length l vibrating between fixed end points with initial
velocity zero and initial displacement f(x) is
nx
 nat 
(a)  Bn sin   cos
 

 l 
(b)


18.
1 mark

l
 nx 
 nat 
sin



 l   l 
 B sin 
n
(CLO-3
 nx 
(c)
 nat 
sin
n
 

 l   l 
 nx 
Bn sin 

 l 
 B cos 
Ans (a)
(d)
The steady state temperature of a rod of length l whose ends
are kept at 30 and 40 is
(a) u 
10x
 30
(b) u 
20x
l
(c) u 
Remember)
10x
Remember)
1 mark
 30
(CLO-3
l
 20
(d) u 
l
Ans (a)
10x
Apply)
l
When the ends of a rod is non-zero for one dimensional heat
flow equation, the temperature function u(x, t) is modified as
19.
the sum of steady state and transient state temperatures. The
transient part of the solution which
(a) Increases with increase of time
(b) Decreases with increase of time
(c) Increases with decrease of time
(d) Decreases with decrease of time
1 mark
(CLO-3
Ans (b)
A rod of length l has its ends A and B kept at 0 and 100
20. respectively, until steady state conditions prevail. Then the
initial condition is given by
1 mark
100x
(a) u(x, 0)  ax  b 100l
(b) u(x, 0) 
(c) u(x, 0)  100xl
l
u(x,
0)

(x
 l)100
(d)
Remember)
(CLO-3
Ans (b)
Apply)
21.
In wave equation
2 y
t
Tension
Mass
Time
(c)
Mass

2
(a)
22.
In heat equation
a
2 y
x
2
, a2
1 mark
Temperature
Mass
Mass
(d)
Time
Ans (a)
u
 2
 stands
for
2
1 mark
2u
t
(CLO-3
,
x2
In heat equation
u
 2
2u
Ans (a)
k
 = 2 , here k stands for
,
x2
c
t
In heat equation
u
 2
1 mark
Ans (a)
k
 = 2 , here  stands for
,
x2
c
t
Remember)
(CLO-3
(b)time
(d) mass
2u
Remember)
(CLO-3
(b)time
(d) mass
(a)Thermal conductivity
(c) zero
24.
stands for
(b)
(a) diffusivity
(c) tension
23.
2
Remember)
1 mark
(CLO-3
(a) density
25.
(b) tension
In heat equation
u
 2
2u
(d) zero
k
 = 2 , here c stands for
,
x2
c
t
(a) specific heat
(c)mass
(b) tension
(d) zero
u
u
(a)
t
2u
(c)
x
2
2 u
 0

u
t
2
(b)

t 2
u
(d)
0
u
In heat equation
t
solution is
 2
Ans (a)
Remember)
1 mark
(CLO-3
Ans (a)
Remember)
2 u
In heat equation  
, the steady state condition
t
x2
is
26.
27.
(c)mass
x
1 mark
0
(CLO-3
Ans (a)
Remember)
0
2 u
, the unsteady state
x2
1 mark
(a) u  ( Ax  B)C
(b) u  ( A cosx  B sin x)e  t
(c) u  ( Aex  Bex )e  t
(CLO-3
Ans (b)
(d) u  At  B
If B  4 AC  0 , then the 2nd order
partial differential equation is classified
as
2
28.
1 mark
Remember)
29.
(a) Elliptic
(b) Hyperbolic
(c)parabolic
(d) Laplace equation
33.
(b) Hyperbolic
(c)parabolic
(d) Laplace equation
Ans (b)
x 2
(d) Laplace equation
The partial differential equation
xf xx  yf yy  0 , x  0, y  0 is classified as
1 mark
(CLO-3
Ans (c)
(d) Laplace equation
The partial differential equation
xf xx  yf yy  0 , x  0, y  0 is classified as
(b) Hyperbolic
(c)parabolic
(d) Laplace equation
The partial differential equation
u xx  4u xy  4u yy  0 is classified as
Remember)
1 mark
(CLO-3
Ans (b)
Remember)
1 mark
(CLO-3
Ans (c)
Remember)
1 mark
(b) Hyperbolic
(a) Elliptic
Remember)
 2u
The partial differential equation u xx  2u xy  u yy  0
is classified as
(a) Elliptic
(b) Hyperbolic
(c)parabolic
36.
(CLO-3
2
2y
2 y


t 2
x 2
is classified as
(a) Elliptic
(a) Elliptic
35.
2
Remember)
1 mark
(d) Laplace equation
The one dimensional wave equation
(c)parabolic
34.
Ans (a)
(d) Laplace equation
The one dimensional heat equation
t
is classified as
(a) Elliptic
(b) Hyperbolic
(c)parabolic
32.
(CLO-3
(d) Laplace equation
u
Remember)
1 mark
If B 2  4 AC  0 , then the 2nd order
partial differential equation is classified
as
(a) Elliptic
(b) Hyperbolic
(c)parabolic
31.
Ans (c)
If B 2  4 AC  0 , then the 2nd order
partial differential equation is classified
as
(a) Elliptic
(b) Hyperbolic
(c)parabolic
30.
(CLO-3
(CLO-3
Ans (a)
Remember)
1 mark
(CLO-3
Ans (b)
1 mark
Remember)
37.
38.
39.
(a) Elliptic
(b) Hyperbolic
(c)parabolic
(d) Laplace equation
The partial differential equation
2u xx  3u xy  4u yy  0 is classified as
(a) Elliptic
(b) Hyperbolic
(c)parabolic
(d) Laplace equation
The partial differential equation
u xx  3u xy  2u yy  0 is classified as
(a) Elliptic
(b) Hyperbolic
(c)parabolic
(d) Laplace equation
The partial differential equation
f xx  f xy  f yy  f y  0 is classified as
(a) Elliptic
Ans (c)
(CLO-3
Ans (a)
(CLO-3
Ans (b)
(CLO-3
The partial differential equation
2 f xx  f xy  f yy  2 f y  0 is classified as
1 mark
(c)parabolic
(d) Laplace equation
Remember)
1 mark
Ans (a)
(b) Hyperbolic
Remember)
1 mark
(b) Hyperbolic
(a) Elliptic
Remember)
1 mark
(d) Laplace equation
(c)parabolic
40.
(CLO-3
Remember)
(CLO-3
Ans (b)
Remember)
18MAB201T – TRANSFORMS AND BOUNDARY VALUE PROBLEMS
Applications of PDE
SRM INSTITUTE OF SCIENCE AND TECHNOLOGY
RAMAPURAM CAMPUS
DEPARTMENT OF MATHEMATICS
Year/Sem : II/III
Branch: Common to All branches
Unit III – Applications of Partial Differential Equations
1.Write the possible solutions of one dimensional wave equation
(a) 𝑦(𝒙, 𝒕) = (𝒄𝟏 𝒆𝒑𝒙 + 𝒄𝟐 𝒆−𝒑𝒙 )(𝒄𝟑 𝒆𝒑𝒂𝒕 + 𝒄𝟒 𝒆−𝒑𝒂𝒕)
𝒚(𝒙, 𝒕) = (𝒄𝟓 𝒄𝒐𝒔𝒑𝒙 + 𝒄𝟔 𝒔𝒊𝒏𝒑𝒙)(𝒄𝟕 𝒄𝒐𝒔𝒑𝒂𝒕 + 𝒄𝟖 𝒔𝒊𝒏𝒑𝒂𝒕)
𝒚(𝒙, 𝒕) = (𝒄𝟗 𝒙 + 𝒄𝟏𝟎 )(𝒄𝟏𝟏 𝒕 + 𝒄𝟏𝟐 )
(b) 𝑦(𝑥, 𝑡) = (𝑐1𝑒 𝑝𝑥 + 𝑐2𝑒 −𝑝𝑥 )(𝑐3𝑒 𝑝𝑎𝑡 + 𝑐4𝑒 −𝑝𝑎𝑡 )
𝑦(𝑥, 𝑡) = (𝑐5𝑐𝑜𝑠𝑝𝑥 − 𝑐6𝑠𝑖𝑛𝑝𝑥)(𝑐7𝑐𝑜𝑠𝑝𝑎𝑡 + 𝑐8 𝑠𝑖𝑛𝑝𝑎𝑡)
𝑦(𝑥, 𝑡) = (𝑐9 𝑥 + 𝑐10)(𝑐11𝑡 + 𝑐12𝑡 2 )
(c) 𝑦(𝑥, 𝑡) = (𝑐1𝑒 𝑝𝑥 + 𝑐2𝑒 −𝑝𝑥 )(𝑐3𝑒 𝑎𝑡 + 𝑐4𝑒 −𝑎𝑡 )
𝑦(𝑥, 𝑡) = (𝑐5𝑐𝑜𝑠𝑝𝑥 + 𝑐6𝑠𝑖𝑛𝑝𝑥)(𝑐7𝑐𝑜𝑠𝑝𝑎𝑡 + 𝑐8 𝑠𝑖𝑛𝑝𝑎𝑡)
(d) 𝑦(𝑥, 𝑡) = (𝑐1𝑒 𝑝𝑥 + 𝑐2𝑒 −𝑝𝑥 )(𝑐3𝑒 𝑝𝑎𝑡 + 𝑐4𝑒 −𝑝𝑎𝑡 )
𝑦(𝑥, 𝑡) = (𝑐9 𝑥 + 𝑐10𝑥 2)(𝑐11𝑡 + 𝑐12 )
Solution:
The possible solution of one dimensional wave equation
𝝏𝟐 𝒚
𝝏𝒕𝟐
= 𝒂𝟐
𝝏𝟐 𝒚
𝝏𝒙𝟐
𝒂𝒓𝒆
 𝑦(𝑥, 𝑡) = (𝑐1𝑒 𝑝𝑥 + 𝑐2𝑒 −𝑝𝑥 )(𝑐3 𝑒 𝑝𝑎𝑡 + 𝑐4𝑒 −𝑝𝑎𝑡 )
 𝑦(𝑥, 𝑡) = (𝑐5𝑐𝑜𝑠𝑝𝑥 + 𝑐6𝑠𝑖𝑛𝑝𝑥)(𝑐7𝑐𝑜𝑠𝑝𝑎𝑡 + 𝑐8𝑠𝑖𝑛𝑝𝑎𝑡)
 𝑦(𝑥, 𝑡) = (𝑐9 𝑥 + 𝑐10 )(𝑐11𝑡 + 𝑐12)
SRMIST, Ramapuram
1
Deparment of Mathematics
18MAB201T – TRANSFORMS AND BOUNDARY VALUE PROBLEMS
Applications of PDE
2. Write the all the possible solutions of one dimensional heat flow equation
(a) 𝑢(𝑥, 𝑡) = 𝑐1 (𝑐2𝑥 2 + 𝑐3 ) ,
𝑢 (𝑥, 𝑡) = 𝑐3 𝑒 𝛼
2 𝑝2 𝑡
𝑢 (𝑥, 𝑡) = 𝑐6 𝑒 −𝛼
(𝑐4𝑒 𝑝𝑥 + 𝑐5𝑒 −𝑝𝑥 ) ,
2 𝑝2 𝑡
(𝑐7 𝑐𝑜𝑠𝑝𝑥 + 𝑐8𝑠𝑖𝑛𝑝𝑥 )
(b) 𝒖(𝒙, 𝒕) = 𝒄𝟏 (𝒄𝟐 𝒙 + 𝒄𝟑 ) ,
𝟐 𝒑𝟐 𝒕
𝒖(𝒙, 𝒕) = 𝒄𝟑 𝒆𝜶
(𝒄𝟒 𝒆𝒑𝒙 + 𝒄𝟓 𝒆−𝒑𝒙 ) ,
𝟐 𝒑𝟐 𝒕
𝒖(𝒙, 𝒕) = 𝒄𝟔 𝒆−𝜶
(𝒄𝟕 𝒄𝒐𝒔𝒑𝒙 + 𝒄𝟖 𝒔𝒊𝒏𝒑𝒙)
(c) 𝑢 (𝑥, 𝑡) = 𝑐1 (𝑐2𝑥 + 𝑐3) ,
𝑢 (𝑥, 𝑡) = 𝑐3 𝑒 −𝛼
𝑢 (𝑥, 𝑡) = 𝑐6 𝑒 𝛼
2 𝑝2 𝑡
2 𝑝2 𝑡
(𝑐4 𝑒 𝑝𝑥 + 𝑐5𝑒 −𝑝𝑥 ) ,
(𝑐7𝑐𝑜𝑠𝑝𝑥 + 𝑐8𝑠𝑖𝑛𝑝𝑥 )
(d) 𝑢 (𝑥, 𝑡) = 𝑐1 (𝑐2𝑥 + 𝑐3) ,
2
𝑢 (𝑥, 𝑡) = 𝑐3 𝑒 𝑝 𝑡 (𝑐4𝑒 𝑝𝑥 + 𝑐5𝑒 −𝑝𝑥 ) ,
2
𝑢 (𝑥, 𝑡) = 𝑐6 𝑒 −𝑝 𝑡 (𝑐7𝑐𝑜𝑠𝑝𝑥 + 𝑐8𝑠𝑖𝑛𝑝𝑥 )
Solution:
The possible solutions of one dimensional heat flow equation
𝜕𝑢
𝜕𝑡
= 𝜶𝟐
𝝏𝟐 𝒖
𝝏𝒙𝟐
are
 𝑢(𝑥, 𝑡) = 𝑐1(𝑐2𝑥 + 𝑐3)
 𝑢(𝑥, 𝑡) = 𝑐3𝑒 𝛼
2 𝑝2 𝑡
 𝑢(𝑥, 𝑡) = 𝑐6𝑒 −𝛼
SRMIST, Ramapuram
(𝑐4𝑒 𝑝𝑥 + 𝑐5 𝑒 −𝑝𝑥 )
2 𝑝2 𝑡
(𝑐7𝑐𝑜𝑠𝑝𝑥 + 𝑐8𝑠𝑖𝑛𝑝𝑥 )
2
Deparment of Mathematics
18MAB201T – TRANSFORMS AND BOUNDARY VALUE PROBLEMS
Applications of PDE
3. Write all boundary conditions for one dimensional wave equation with zero
initial velocity
(a) 𝑦(0, 𝑡) = 0 , 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑡 > 0
𝑦(𝑙, 𝑡) = 𝑙 , 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑡 > 0
𝜕𝑦(𝑥,0)
𝜕𝑡
=0
𝑦(𝑥, 0) = 𝑓(𝑥)
(b) 𝑦(0, 𝑡) = 𝑙 , 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑡 > 0
𝑦(𝑙, 𝑡) = 0 , 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑡 > 0
𝑦(𝑥, 0) = 0
𝜕𝑦(𝑥,0)
𝜕𝑡
= 𝑓(𝑥)
(c) 𝑦(0, 𝑡) = 0 , 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑡 > 0
𝑦(𝑙, 𝑡) = 0 , 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑡 > 0
𝑦(𝑥, 0) = 0
𝜕𝑦(𝑥,0)
𝜕𝑡
= 𝑓(𝑥)
(d) 𝒚(𝟎, 𝒕) = 𝟎 , 𝒇𝒐𝒓 𝒂𝒍𝒍 𝒕 > 𝟎
𝒚(𝒍, 𝒕) = 𝟎 , 𝒇𝒐𝒓 𝒂𝒍𝒍 𝒕 > 𝟎
𝝏𝒚(𝒙,𝟎)
𝝏𝒕
=𝟎
𝒚(𝒙, 𝟎) = 𝒇(𝒙)
Solution
One dimensional wave equation is
𝝏𝟐 𝒚
𝝏𝒕𝟐
= 𝒂𝟐
𝝏𝟐 𝒚
𝝏𝒙𝟐
, boundary conditions are
 𝑦(0, 𝑡) = 0 , 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑡 > 0
 𝑦(𝑙, 𝑡) = 0 , 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑡 > 0

𝜕𝑦(𝑥,0)
𝜕𝑡
= 0 (𝐼𝑛𝑖𝑡𝑖𝑎𝑙 𝑉𝑒𝑙𝑜𝑐𝑖𝑡𝑦 = 0)
 𝑦(𝑥, 0) = 𝑓(𝑥)
SRMIST, Ramapuram
3
Deparment of Mathematics
18MAB201T – TRANSFORMS AND BOUNDARY VALUE PROBLEMS
Applications of PDE
4. Write all boundary conditions for one dimensional heat flow equation
(a) 𝑢(0, 𝑡) = 𝑙 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑡 > 0
𝑢 (𝑙, 𝑡) = 0 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑡 > 0
𝑢 (𝑥, 0) = 𝑓(𝑥) 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑥 𝑖𝑛 (0, 𝑙)
(b) 𝑢 (0, 𝑡) = 0 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑡 > 0
𝑢 (𝑙, 𝑡) = 0 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑡 > 0
𝑢 (𝑥, 0) = 0 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑥 𝑖𝑛 (0, 𝑙)
(c) 𝒖(𝟎, 𝒕) = 𝟎 𝒇𝒐𝒓 𝒂𝒍𝒍 𝒕 > 𝟎
𝒖(𝒍, 𝒕) = 𝟎 𝒇𝒐𝒓 𝒂𝒍𝒍 𝒕 > 𝟎
𝒖(𝒙, 𝟎) = 𝒇(𝒙) 𝒇𝒐𝒓 𝒂𝒍𝒍 𝒙 𝒊𝒏 (𝟎, 𝒍)
(d) 𝑢 (0, 𝑡) = 0 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑡 > 0
𝑢 (𝑙, 𝑡) = 𝑙 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑡 > 0
𝑢 (𝑥, 0) = 𝑓(𝑥) 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑥 𝑖𝑛 (0, 𝑙)
Solution:
One dimensional heat flow equation
𝜕𝑢
𝜕𝑡
= 𝜶𝟐
𝝏𝟐 𝒖
𝝏𝒙𝟐
𝒂𝒓𝒆
 𝑢(0, 𝑡) = 0 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑡 > 0
 𝑢(𝑙, 𝑡) = 0 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑡 > 0
 𝑢(𝑥, 0) = 𝑓(𝑥) 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑥 𝑖𝑛 (0, 𝑙)
SRMIST, Ramapuram
4
Deparment of Mathematics
18MAB201T – TRANSFORMS AND BOUNDARY VALUE PROBLEMS
Applications of PDE
5. A string is stretched between two fixed points at a distance 2𝑙 apart and the
points of the string are given velocity 𝑓 (𝑥 ), 𝑥 being the distance from end
point. Formulate the problem to find the displacement of the string at any time
(a) 𝒚(𝟎, 𝒕) = 𝟎 , 𝒇𝒐𝒓 𝒂𝒍𝒍 𝒕 > 𝟎
𝒚(𝟐𝒍, 𝒕) = 𝟎 , 𝒇𝒐𝒓 𝒂𝒍𝒍 𝒕 > 𝟎
𝒚(𝒙, 𝟎) = 𝟎 𝒇𝒐𝒓 𝒂𝒍𝒍 𝒙 𝒊𝒏 (𝟎, 𝟐𝒍)
𝝏𝒚(𝒙,𝟎)
𝝏𝒕
= 𝒇(𝒙)𝒇𝒐𝒓𝒂𝒍𝒍 𝒙 𝒊𝒏 (𝟎, 𝟐𝒍)
(a) 𝑦(0, 𝑡) = 0 , 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑡 > 0
𝑦(2𝑙, 𝑡) = 0 , 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑡 > 0
𝜕𝑦(𝑥,0)
𝜕𝑡
= 0𝑓𝑜𝑟𝑎𝑙𝑙 𝑥 𝑖𝑛 (0,2𝑙)
𝑦(𝑥, 0) = 𝑓(𝑥) 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑥 𝑖𝑛 (0,2𝑙)
(a) 𝑦(0, 𝑡) = 2𝑙 , 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑡 > 0
𝑦(2𝑙, 𝑡) = 0 , 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑡 > 0
𝑦(𝑥, 0) = 0 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑥 𝑖𝑛 (0,2𝑙)
𝜕𝑦(𝑥,0)
𝜕𝑡
= 𝑓 (𝑥 )𝑓𝑜𝑟𝑎𝑙𝑙 𝑥 𝑖𝑛 (0,2𝑙)
(a) 𝑦(0, 𝑡) = 0 , 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑡 > 0
𝑦(2𝑙, 𝑡) = 2𝑙 , 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑡 > 0
𝑦(𝑥, 0) = 0 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑥 𝑖𝑛 (0,2𝑙)
𝜕𝑦(𝑥,0)
𝜕𝑡
= 𝑓 (𝑥 )𝑓𝑜𝑟𝑎𝑙𝑙 𝑥 𝑖𝑛 (0,2𝑙)
SRMIST, Ramapuram
5
Deparment of Mathematics
18MAB201T – TRANSFORMS AND BOUNDARY VALUE PROBLEMS
Applications of PDE
Solution:
One dimensional wave equation is
𝝏𝟐 𝒚
𝝏𝒕𝟐
= 𝒂𝟐
𝝏𝟐 𝒚
𝝏𝒙𝟐
, boundary conditions are
 𝑦(0, 𝑡) = 0 , 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑡 > 0
 𝑦(2𝑙, 𝑡) = 0 , 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑡 > 0
 𝑦(𝑥, 0) = 0 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑥 𝑖𝑛 (0,2𝑙)

𝜕𝑦(𝑥,0)
𝜕𝑡
= 𝑓 (𝑥 )𝑓𝑜𝑟𝑎𝑙𝑙 𝑥 𝑖𝑛 (0,2𝑙)
6.A string of length 2𝑙 is stretched to a constant tension T, is fastened at both
the ends and hence fixed. The mid points of the string is taken to a height ‘𝑏′
and then released from rest in that position. Find the equation of the string in
its initial position
𝑏𝑥
(a) 𝑦(𝑥, 0) = {𝑏
𝑙
𝑙
(2𝑙 + 𝑥 )
𝒃𝒙
(b) 𝒚(𝒙, 𝟎) = {𝒃
𝒍
(c) 𝑦(𝑥, 0) = {𝑏
𝑙
(d) 𝑦(𝑥, 0) = {𝑏
𝑙
𝒍
(𝟐𝒍 − 𝒙)
−
𝑏𝑥
𝑙
(2𝑙 − 𝑥 )
−
𝑏𝑥
𝑙
(2𝑙 + 𝑥 )
𝑤ℎ𝑒𝑛 0 < 𝑥 < 𝑙
𝑤ℎ𝑒𝑛 𝑙 < 𝑥 < 2𝑙
𝒘𝒉𝒆𝒏 𝟎 < 𝒙 < 𝒍
𝒘𝒉𝒆𝒏 𝒍 < 𝒙 < 𝟐𝒍
𝑤ℎ𝑒𝑛 0 < 𝑥 < 𝑙
𝑤ℎ𝑒𝑛 𝑙 < 𝑥 < 2𝑙
𝑤ℎ𝑒𝑛 0 < 𝑥 < 𝑙
𝑤ℎ𝑒𝑛 𝑙 < 𝑥 < 2𝑙
Solution:
The initial displacement of the string is in the form
𝑏𝑥
𝑤ℎ𝑒𝑛 0 < 𝑥 < 𝑙
𝑙
𝑦(𝑥, 0) = {
𝑏
(2𝑙 − 𝑥 ) 𝑤ℎ𝑒𝑛 𝑙 < 𝑥 < 2𝑙
𝑙
SRMIST, Ramapuram
6
Deparment of Mathematics
18MAB201T – TRANSFORMS AND BOUNDARY VALUE PROBLEMS
Applications of PDE
7.Classify the equation 𝑈𝑥𝑥 − 𝑦 4 𝑈𝑦𝑦 = 2𝑦 3 𝑈𝑦
(a) Elliptic
(b) Hyperbolic
(c)Parabolic
(d) Concentric
Solution:
Here 𝐴 = 1, 𝐵 = 0, 𝐶 = −𝑦 4
𝐵 2 − 4𝐴𝐶 = 0 − 4(1)(−𝑦 4 ) = 4𝑦 4 > 0
The equation is hyperbolic
8. Classify the equation 𝑥 2 𝑓𝑥𝑥 + (1 − 𝑦 2 )𝑓𝑦𝑦 = 0 𝑤ℎ𝑒𝑛 𝑥 = 0
(a) Elliptic
(b) Hyperbolic
(c)Parabolic
(d) Concentric
Solution:
Here 𝐴 = 𝑥 2 , 𝐵 = 0, 𝐶 = 1 − 𝑦 2
𝐵 2 − 4𝐴𝐶 = 0 − 4(𝑥 2 )(1 − 𝑦 2 ) = 0 (𝑠𝑖𝑛𝑐𝑒 𝑥 = 0)
The equation is parabolic
9. Classify the equation 4𝑈𝑥𝑥 + 4𝑈𝑥𝑦 + 𝑈𝑦𝑦 + 2𝑈𝑥 − 𝑈𝑦 = 0
(a) Elliptic (b) Hyperbolic
(c)Parabolic
(d) Concentric
Solution:
Here 𝐴 = 4, 𝐵 = 4, 𝐶 = 1
𝐵 2 − 4𝐴𝐶 = 16 − 4(4)(1) = 0
The equation is parabolic
10. Classify 𝑈𝑥𝑥 + 𝑈𝑦𝑦 = 0
(a) Elliptic
(b) Hyperbolic
(c)Parabolic
(d) Concentric
Solution:
Here 𝐴 = 1, 𝐵 = 0, 𝐶 = 1
𝐵 2 − 4𝐴𝐶 = −1 < 0
The equation is elliptic
SRMIST, Ramapuram
7
Deparment of Mathematics
18MAB201T – TRANSFORMS AND BOUNDARY VALUE PROBLEMS
Applications of PDE
11.Classify 𝑈𝑥𝑥 + 5𝑈𝑥𝑦 + 4𝑈𝑦𝑦 + 𝑈𝑥 + 𝑈𝑦 = 0
(a) Elliptic
(b) Hyperbolic
(c)Parabolic
(d) Concentric
Solution:
Here 𝐴 = 1, 𝐵 = 5, 𝐶 = 4
𝐵 2 − 4𝐴𝐶 = 25 − 16 > 0
The equation is hyperbolic
2
2
12.Classify 𝜕𝜕𝑡𝑢2 = 𝑐 2 𝜕𝜕𝑥𝑢2
(a) Elliptic
(b) Hyperbolic
(c)Parabolic
(d) Concentric
Solution:
Here 𝐴 = 1, 𝐵 = 0, 𝐶 = −𝑐 2
𝐵 2 − 4𝐴𝐶 = 0 − 4(1)(−𝑐 2) = 4𝑐 2 > 0
The equation is hyperbolic
2
𝜕 𝑢
13. Classify 𝜕𝑢
= 𝑐2 2
𝜕𝑡
𝜕𝑥
(a) Elliptic
(b) Hyperbolic
(c)Parabolic
(d) Concentric
Solution:
Here 𝐴 = 𝑐 2 , 𝐵 = 0, 𝐶 = 0
𝐵 2 − 4𝐴𝐶 = 0 − 4(0)(𝑐 2) = 0
The equation is parabolic
14.
Write the conditions for classification of PDE to be hyperbolic, parabolic
and hyperbolic
(a) 𝑩𝟐 − 𝟒𝑨𝑪 < 𝟎 [𝑬𝒍𝒍𝒊𝒑𝒕𝒊𝒄 𝑬𝒒𝒖𝒂𝒕𝒊𝒐𝒏]
𝑩𝟐 − 𝟒𝑨𝑪 = 𝟎 [𝑷𝒂𝒓𝒂𝒃𝒐𝒍𝒊𝒄 𝑬𝒒𝒖𝒂𝒕𝒊𝒐𝒏]
𝑩𝟐 − 𝟒𝑨𝑪 > 𝟎 [𝑯𝒚𝒑𝒆𝒓𝒃𝒐𝒍𝒊𝒄 𝑬𝒒𝒖𝒂𝒕𝒊𝒐𝒏]
(a) 𝐵 2 − 4𝐴𝐶 > 0 [𝐸𝑙𝑙𝑖𝑝𝑡𝑖𝑐 𝐸𝑞𝑢𝑎𝑡𝑖𝑜𝑛]
𝐵 2 − 4𝐴𝐶 < 0 [𝑃𝑎𝑟𝑎𝑏𝑜𝑙𝑖𝑐 𝐸𝑞𝑢𝑎𝑡𝑖𝑜𝑛]
𝐵 2 − 4𝐴𝐶 = 0 [𝐻𝑦𝑝𝑒𝑟𝑏𝑜𝑙𝑖𝑐 𝐸𝑞𝑢𝑎𝑡𝑖𝑜𝑛]
(a) 𝐵 2 − 4𝐴𝐶 = 0 [𝐸𝑙𝑙𝑖𝑝𝑡𝑖𝑐 𝐸𝑞𝑢𝑎𝑡𝑖𝑜𝑛]
𝐵 2 − 4𝐴𝐶 > 0 [𝑃𝑎𝑟𝑎𝑏𝑜𝑙𝑖𝑐 𝐸𝑞𝑢𝑎𝑡𝑖𝑜𝑛]
SRMIST, Ramapuram
8
Deparment of Mathematics
18MAB201T – TRANSFORMS AND BOUNDARY VALUE PROBLEMS
Applications of PDE
𝐵 2 − 4𝐴𝐶 < 0 [𝐻𝑦𝑝𝑒𝑟𝑏𝑜𝑙𝑖𝑐 𝐸𝑞𝑢𝑎𝑡𝑖𝑜𝑛]
(a) 𝐵 2 − 4𝐴𝐶 = 0 [𝐸𝑙𝑙𝑖𝑝𝑡𝑖𝑐 𝐸𝑞𝑢𝑎𝑡𝑖𝑜𝑛]
𝐵 2 − 4𝐴𝐶 > 0 [𝑃𝑎𝑟𝑎𝑏𝑜𝑙𝑖𝑐 𝐸𝑞𝑢𝑎𝑡𝑖𝑜𝑛]
𝐵 2 − 4𝐴𝐶 < 0 [𝐻𝑦𝑝𝑒𝑟𝑏𝑜𝑙𝑖𝑐 𝐸𝑞𝑢𝑎𝑡𝑖𝑜𝑛]
Solution:
Let a second order PDE in the function 𝑢 of the two independent variables 𝑥, 𝑦
be of the form
𝐴(𝑥, 𝑦)
𝜕2 𝑢
𝜕2 𝑢
𝜕2𝑢
𝜕𝑢 𝜕𝑢
(
)
(
)
+
𝐵
𝑥,
𝑦
+
𝐶
𝑥,
𝑦
+
𝑓
(𝑥,
𝑦,
𝑢,
, ) = 0 − −→ (1)
𝜕𝑥 2
𝜕𝑥𝜕𝑦
𝜕𝑦 2
𝜕𝑥 𝜕𝑦
Equation (1) is classified as elliptic, parabolic or hyperbolic at the points of a
given region R depending on whether
𝐵 2 − 4𝐴𝐶 < 0 [𝐸𝑙𝑙𝑖𝑝𝑡𝑖𝑐 𝐸𝑞𝑢𝑎𝑡𝑖𝑜𝑛]
𝐵 2 − 4𝐴𝐶 = 0 [𝑃𝑎𝑟𝑎𝑏𝑜𝑙𝑖𝑐 𝐸𝑞𝑢𝑎𝑡𝑖𝑜𝑛]
𝐵 2 − 4𝐴𝐶 > 0 [𝐻𝑦𝑝𝑒𝑟𝑏𝑜𝑙𝑖𝑐 𝐸𝑞𝑢𝑎𝑡𝑖𝑜𝑛]
15.Classify 𝑥 2𝑈𝑥𝑥 + 2𝑥𝑦𝑈𝑥𝑦 + (1 + 𝑦 2 )𝑈𝑦𝑦 = 0
(a) Elliptic
(b) Hyperbolic
(c)Parabolic
(d) Concentric
Solution:
Here 𝐴 = 𝑥 2 , 𝐵 = 2𝑥𝑦, 𝐶 = (1 + 𝑦 2 )
𝐵 2 − 4𝐴𝐶 = 2𝑥𝑦 − 4(𝑥 2 )(1 + 𝑦 2 ) = −4𝑥 2 < 0
The equation is elliptic
2
𝜕 𝑢
𝜕𝑢
𝜕𝑢
16.Classify 𝜕𝑥𝜕𝑦
= ( ) ( ) + 𝑥𝑦
𝜕𝑥
𝜕𝑦
(a) Elliptic
(b) Hyperbolic
(c)Parabolic
(d) Concentric
Solution:
Here 𝐴 = 0, 𝐵 = 1, 𝐶 = 0
𝐵 2 − 4𝐴𝐶 = 1 − 4(0)(0) = 1 > 0
The equation is hyperbolic
SRMIST, Ramapuram
9
Deparment of Mathematics
18MAB201T – TRANSFORMS AND BOUNDARY VALUE PROBLEMS
Applications of PDE
17.A rod of length 𝑙 𝑐𝑚 whose one side is kept at 20°𝐶 and the other end is
kept at 50°𝐶 is maintained until steady state prevails. Find the steady state
temperature.
(a) 𝑢(𝑥 ) =
20
(b) 𝑢 (𝑥 ) =
20
(c) 𝒖(𝒙) =
𝟑𝟎
(d) 𝑢 (𝑥 ) =
30
𝑙
𝑥 − 20
𝑙
𝒍
𝑙
𝑥 + 30
𝒙 + 𝟐𝟎
𝑥 − 20
Solution:
𝑢(𝑥 ) = 𝑎𝑥 + 𝑏
When 𝑥 = 0, 𝑢 (0) = 𝑏 ⇒ 𝑏 = 20𝐶
When 𝑥 = 𝑙, 𝑢(𝑙) = 𝑎𝑙 + 20 ⇒ 50𝐶 = 𝑎𝑙 + 20
𝑎𝑙 = 30 ⇒ 𝑎 =
𝑠𝑜, 𝑢(𝑥 ) =
30
𝑙
30
𝑥 + 20
𝑙
18. A rod of length 𝑙 𝑐𝑚 whose one side is kept at 0°𝐶 and the other end is
kept at 100°𝐶 is maintained until steady state prevails. Find the steady state
temperature.
(a) 𝒖(𝒙) =
𝟏𝟎𝟎
(b) 𝑢 (𝑥 ) =
10
(c) 𝑢 (𝑥 ) =
− 100
(d) 𝑢 (𝑥 ) =
100
𝒍
𝑙
𝑥
𝑙
𝑙
𝒙
𝑥
𝑥2
Solution:
𝑢(𝑥 ) = 𝑎𝑥 + 𝑏
When 𝑥 = 0, 𝑢 (0) = 𝑏 ⇒ 𝑏 = 0
SRMIST, Ramapuram
10
Deparment of Mathematics
18MAB201T – TRANSFORMS AND BOUNDARY VALUE PROBLEMS
Applications of PDE
When 𝑥 = 𝑙, 𝑢(𝑙) = 𝑎𝑙 + 0 ⇒ 100 = 𝑎𝑙
100
𝑙
100
𝑠𝑜, 𝑢(𝑥 ) =
𝑥
𝑙
𝑎𝑙 = 100 ⇒ 𝑎 =
19. A rod of length 𝑙 𝑐𝑚 whose one side is kept at 0°𝐶 and the other end is
kept at 120°𝐶 is maintained until steady state prevails. Find the steady state
temperature.
(a) 𝑢(𝑥 ) = −
120
𝑙
(b) 𝒖(𝒙) =
𝟏𝟐𝟎
(c) 𝑢 (𝑥 ) =
120
(d) 𝑢 (𝑥 ) =
120
𝒍
𝑙
𝑥
𝒙
𝑥2
𝑙
Solution:
𝑢(𝑥 ) = 𝑎𝑥 + 𝑏
When 𝑥 = 0, 𝑢 (0) = 𝑏 ⇒ 𝑏 = 0
When 𝑥 = 𝑙, 𝑢(𝑙) = 𝑎𝑙 + 0 ⇒ 120 = 𝑎𝑙
120
𝑙
120
𝑠𝑜, 𝑢(𝑥 ) =
𝑥
𝑙
𝑎𝑙 = 120 ⇒ 𝑎 =
20. A rod of length 20 𝑐𝑚 whose one side is kept at 30°𝐶 and the other end is
kept at 70°𝐶 is maintained until steady state prevails. Find the steady state
temperature.
(a) 𝑢(𝑥 ) = 3𝑥 − 20
(b) 𝑢 (𝑥 ) = 2𝑥 − 30
(c) 𝒖(𝒙) = 𝟐𝒙 + 𝟑𝟎
(d) 𝑢 (𝑥 ) = 3𝑥 + 30
SRMIST, Ramapuram
11
Deparment of Mathematics
18MAB201T – TRANSFORMS AND BOUNDARY VALUE PROBLEMS
Applications of PDE
Solution:
𝑢(𝑥 ) = 𝑎𝑥 + 𝑏
When 𝑥 = 0, 𝑢 (0) = 𝑏 ⇒ 𝑏 = 30
When 𝑥 = 20, 𝑢(20) = 70 = 20𝑎 + 30
⇒ 40 = 20𝑎
⇒𝑎=2
𝑠𝑜, 𝑢(𝑥 ) = 2𝑥 + 30
21.
A Uniform string of length `l` is struck in such a way that an initial velocity
of 𝑉0 is imparted to the portion of the string between
𝑙
4
𝑎𝑛𝑑
3𝑙
4
while the
string is in its equilibrium position. Write the wave equation and its boundary
conditions
(a)
𝝏𝟐 𝒚
𝝏𝟐 𝒚
𝝏𝒕
𝝏𝒙𝟐
= 𝒂𝟐
𝟐
,
𝒚(𝟎, 𝒕) = 𝟎, 𝒕 > 𝟎
𝒚(𝒍, 𝒕) = 𝟎, 𝒕 > 𝟎
𝒚(𝒙, 𝟎) = 𝟎
𝟎
𝝏𝒚(𝒙, 𝟎)
=
𝝏𝒕
𝒗𝟎
{𝟎
(b)
𝜕2 𝑦
𝜕𝑥
= 𝑎2
𝜕2 𝑦
𝜕𝑡 2
𝒍
𝟒
𝒍
𝟑𝒍
≤𝒙≤
𝟒
𝟒
𝟑𝒍
<𝒙≤𝒍
𝟒
𝟎<𝒙<
,
𝑦(0, 𝑡) = 0, 𝑡 > 0
𝑦(𝑙, 𝑡) = 0, 𝑡 > 0
𝑦(𝑥, 0) = 𝑥
𝑙
4
𝜕𝑦(𝑥, 0)
𝑙
3𝑙
= 1
≤𝑥≤
𝜕𝑡
4
4
3𝑙
{0 4 < 𝑥 ≤ 𝑙
𝑣0
SRMIST, Ramapuram
12
0<𝑥<
Deparment of Mathematics
18MAB201T – TRANSFORMS AND BOUNDARY VALUE PROBLEMS
(c)
𝜕2 𝑦
𝜕𝑡 2
= 𝑎2
𝜕2 𝑦
𝜕𝑥 2
Applications of PDE
,
𝑦(0, 𝑡) = 0, 𝑡 > 0
𝑦(𝑙, 𝑡) = 0, 𝑡 > 0
𝜕𝑦(𝑥,0)
𝜕𝑡
=0
𝑙
4
𝑙
3𝑙
≤𝑥≤
4
4
3𝑙
<𝑥≤𝑙
4
0
𝑦(𝑥, 0) =
0<𝑥<
𝑣0
{0
(d)
𝜕2 𝑦
𝜕𝑡 2
= 𝑎2
𝜕2 𝑦
𝜕𝑥 2
,
𝑦(0, 𝑡) = 0, 𝑡 > 0
𝑦(𝑙, 𝑡) = 0, 𝑡 > 0
𝜕𝑦(𝑥,0)
𝜕𝑡
=0
𝑙
4
𝑙
3𝑙
𝑦(𝑥, 0) = −𝑣0
≤𝑥≤
4
4
3𝑙
{ 1 4 <𝑥≤𝑙
0
0<𝑥<
Solution:
The wave equation is
𝜕2 𝑦
𝜕𝑡 2
= 𝑎2
𝜕2 𝑦
𝜕𝑥 2
The boundary conditions are
𝑦(0, 𝑡) = 0, 𝑡 > 0
𝑦(𝑙, 𝑡) = 0, 𝑡 > 0
𝑦(𝑥, 0) = 0
0
𝜕𝑦(𝑥, 0)
=
𝜕𝑡
𝑣0
{0
SRMIST, Ramapuram
13
𝑙
4
𝑙
3𝑙
≤𝑥≤
4
4
3𝑙
<𝑥≤𝑙
4
0<𝑥<
Deparment of Mathematics
18MAB201T – TRANSFORMS AND BOUNDARY VALUE PROBLEMS
Applications of PDE
22.
Write the one dimensional wave equation and also general solution for
the displacement 𝑦(𝑥, 𝑡) of the string 𝑙 vibrating between fixed end points with
initial zero and initial displacement 𝑓 (𝑥 ).
(a)
(a)
(a)
(a)
𝜕2 𝑦
𝜕𝑡 2
𝝏𝟐 𝒚
𝝏𝒕𝟐
𝜕2 𝑦
𝜕𝑡 2
𝜕2 𝑦
𝜕𝑥 2
𝜕2 𝑦
= 𝑎2
𝜕𝑥 2
𝝏𝟐 𝒚
= 𝒂𝟐
𝝏𝒙𝟐
𝜕2 𝑦
= 𝑎2
𝜕𝑥 2
= 𝑎2
𝜕2 𝑦
𝜕𝑡 2
, ∑ 𝐵𝑛 cos (
𝑛𝜋𝑥
𝑙
𝑛𝜋𝑥
𝑙
, ∑ 𝐵𝑛 sin (
𝑛𝜋𝑎𝑡
𝑙
)
𝒏𝝅𝒙
𝒏𝝅𝒂𝒕
𝒍
𝒍
, ∑ 𝑩𝒏 𝐬𝐢𝐧 (
, ∑ sin (
) sin (
) 𝐜𝐨𝐬 (
) cos (
𝑛𝜋𝑥
𝑙
𝑛𝜋𝑎𝑡
𝑙
) cos (
)
)
𝑛𝜋𝑎𝑡
𝑙
)
Solution:
The one dimensional wave equation is given by
𝝏𝟐 𝒚
𝝏𝒕𝟐
= 𝒂𝟐
𝝏𝟐 𝒚
𝝏𝒙𝟐
The general solution for the displacement 𝑦(𝑥, 𝑡)𝑖𝑠 𝑔𝑖𝑣𝑒𝑛 𝑏𝑦
∑ 𝐵𝑛 sin (
𝑛𝜋𝑥
𝑛𝜋𝑎𝑡
) cos (
)
𝑙
𝑙
23. Classify 𝑥 2 𝑓𝑥𝑥 + (1 − 𝑦 2 )𝑓𝑦𝑦 ) = 0 𝑓𝑜𝑟 − 1 < 𝑦 < 1, −∞ < 𝑥 < ∞
𝑎𝑛𝑑 𝑥 ≠ 0
(a) Elliptic
(b) Hyperbolic
(c)Parabolic
(d) Concentric
Solution:
Here 𝐴 = 𝑥 2 , 𝐵 = 0, 𝐶 = 1 − 𝑦 2
𝐵 2 − 4𝐴𝐶 = 0 − 4(𝑥 2)(1 − 𝑦 2 ) = 4𝑥 2 (𝑦 2 − 1) < 0
(Since 𝑥 2 𝑖𝑠 𝑎𝑙𝑤𝑎𝑦𝑠 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒 𝑖𝑛 − ∞ < 𝑥 < ∞ &
𝑖𝑛 − 1 < 𝑦 < 1, 𝑦 2 − 1 𝑖𝑠 𝑛𝑒𝑔𝑎𝑡𝑖𝑣𝑒)
The equation is elliptic
SRMIST, Ramapuram
14
Deparment of Mathematics
18MAB201T – TRANSFORMS AND BOUNDARY VALUE PROBLEMS
Applications of PDE
24.Classify the one dimensional wave equation
(a) Elliptic
(b) Hyperbolic
(c)Parabolic
(d) Concentric
Solution:
The one dimensional wave equation is given by
𝝏𝟐 𝒚
𝝏𝒕𝟐
= 𝒂𝟐
𝝏𝟐 𝒚
𝝏𝒙𝟐
Here 𝐴 = 𝑎2 , 𝐵 = −1, 𝐶 = 0
𝐵 2 − 4𝐴𝐶 = (−1)2 − 4(𝑎2 )(0) = 1 > 0
The equation is hyperbolic.
25. Write the one dimensional heat flow equation and also general solution
for 𝑢(𝑥, 𝑡) where 𝑢(𝑥, 0) = 𝑓 (𝑥 ).
(a)
(b)
(c)
(d)
𝝏𝒖
𝝏𝒕
𝜕𝑢
𝜕𝑡
𝜕𝑢
𝜕𝑥
𝜕𝑢
𝜕𝑥
= 𝜶𝟐
= 𝛼2
= 𝛼2
= 𝛼2
𝝏𝟐 𝒖
𝝏𝒙𝟐
𝜕2 𝑢
𝜕𝑥 2
𝜕2 𝑢
𝜕𝑡 2
, ∑∞
𝒏=𝟏 𝑩𝒏 𝐬𝐢𝐧
𝒏𝝅𝒙
, ∑∞
𝑛=1 𝐵𝑛 cos
𝑛𝜋𝑥
, ∑∞
𝑛=1 𝐵𝑛 sin
𝜕2 𝑢
𝜕𝑡 2
𝒍
𝑙
𝑛𝜋𝑥
𝑙
, ∑∞
𝑛=1 𝐵𝑛 cos
𝑛𝜋𝑥
𝑙
Solution:
One dimensional heat flow equation
Solution is 𝑢(𝑥, 𝑡) =
∑∞
𝑛=1 𝐵𝑛
sin
SRMIST, Ramapuram
15
𝑛𝜋𝑥
2
𝑛𝜋𝑥
𝑙
𝑙
where 𝐵𝑛 = 𝑏𝑛 = ∫ 𝑓 (𝑥 ) sin
𝜕𝑢
𝜕𝑡
= 𝜶𝟐
𝝏𝟐 𝒖
𝝏𝒙𝟐
𝑙
𝑑𝑥
Deparment of Mathematics
SRM Institute of Science and Technology
Ramapuram Campus
Department of Mathematics
Question Bank of Module-IV(Fourier Transforms)
(2020–2021-ODD)
Subject.Code: 18MAB201T
Subject.Name: Transforms and Boundary Value Problems
Part-A (1*20=20)
Year/Sem: II/III
1.
Branch: Common to All branches
1 mark
The Fourier transform of a function f(x) is
a)
b)
(CLO-4
c)
Ans (b)
Remember)
d)
1 mark
2.
The Fourier transform of f(x)=
a)
is
b)
(CLO-4
Ans (a)
c)
3.
Remember)
d)
The Fourier cosine transform of
1 mark
is
a)
b)
(CLO-4
Ans (d)
Remember)
c)
d)
4.
Under Fourier cosine transform f ( x)  e  a
2
x2
is----
1
mark
--function
a) self-reciprocal
b) cosine
c) inverse function
d) sine
(CLO-4
Ans (a)
5.
The Fourier sine transform of x
a) 0
b)
c)
d) 1
Remember)
1
is
mark
(CLO-4
Ans (b)
6.
Remember)
1 mark
a)
b)
(CLO-4
c)
7.
d)
Ans (c)
Remember)
1 mark
The
a)
b)
(CLO-4
c)
d)
Ans (d)
8.
Remember)
1 mark
9.
a)
F(s+a)
b) F(s-a)
c)
F(sa)
d) F(s/a)
(CLO-4
Ans (a)
F[ f(x) cosax ] =
a) [ f(a)+f(s-a)]/ 2
Remember)
1 mark
b) [ f(sa)+f(s+a)]/
2
(CLO-4
c) [ f(s+a)+f(s-a)]/ 2
d) [ f(s+a)-f(s-
Ans (c)
Remember)
a)]/ 2
10.
F[ f(x) *g(x) ] =
1 mark
a) F(s) + G(s)
b) F(s) - G(s)
c) F(s)G(s)
d) F(s)|G(s)
(CLO-4
Ans (c)
Remember)
11.
1 mark
If F(s) = F [f(x)] then
a)
b)
(CLO-4
Ans (b)
c)
Remember)
d)
12.
1 mark
a)
b)
c)
(CLO-4
Ans (c)
Remember)
d)
1 mark
13.
a)
b)
(CLO-4
c)
Ans (a)
Remember)
d)
1 mark
14.
a)
b)
c)
d)
(CLO-4
Ans (d)
Remember)
15.
The relation between Fourier transform and Laplace
transform is
1 mark
a)
b)
(CLO-4
Ans (a)
c)
Remember)
d)
16.
1 mark
The Fourier cosine transform of
a)
b)
(CLO-4
c)
d)
Ans (a)
The Fourier transform of an odd function of x is
Remember)
1 mark
17.
a) an odd function of s
b) even function of s
(CLO-4
c) an odd function of x
Ans (a)
Remember)
d) even function of x
The Fourier transform of an even function of x is
1 mark
18.
a) an odd function of s
(CLO-4
b) even function of s
c) an odd function of x
Ans (b)
Remember)
d) even function of x
19.
1 mark
The Fourier sine transform of
a)
b)
(CLO-4
Ans (c)
c)
d)
Remember)
20.
21
F [e ibx f ( x)] 
1 mark
a) F(s/b)
b) F(s+b)
c)
d) F(s-b)
F(bs)
(CLO-4
Ans (b)
If f(x) is a function in(-l,l) and satisfies dirichlets
conditions then
1 mark
1
∞
∞
(CLO-4
1
∞
∞
Remember)
a) 𝑓 (𝑥) = 𝜋 ∫0 ∫−∞ 𝑓(𝑡) 𝑐𝑜𝑠𝛌(𝐭 − 𝐱)𝑑𝑡𝑑𝛌
b) 𝑓 (𝑥) = 𝜋 ∫0 ∫−∞ 𝑓(𝑡) 𝑐𝑜𝑠𝐱(𝐭 − 𝐱)𝑑𝑥𝑑𝛌
1
∞
∞
c) 𝑓 (𝑥) = 2𝜋 ∫0 ∫−∞ 𝑓 (𝑡) 𝑐𝑜𝑠𝛌(𝐭 − 𝐱)𝑑𝑡𝑑𝛌
d) 𝑓 (𝑥) =
Remember)
∞
∫
𝜋 0
2
∞
∫−∞ 𝑓(𝑡) 𝑐𝑜𝑠𝛌(𝐭 −
Ans (c )
𝐱)𝑑𝑡𝑑𝛌
1
Under Fourier cosine transform 𝑓(𝑥) = √𝑥
1 mark
22
a)
b)
c)
d)
Self-reciprocal function
Cosine function
Inverse function
Complex function
(CLO-4
Ans ( a)
Remember)
If F(f(x)) = F(s) and f(x)→0 as x→±∞ then F(f’(x)) is
1 mark
23
a) –isF(s)
b) is F(s)
c) sF(s)
d) –F(s)
(CLO-4
Ans (a )
Find the Fourier sine transform of
Remember)
, a>0
1 mark
24
2
𝑠
a)
𝐹𝑠 [𝑒 −𝑎𝑥 ] = √𝜋 [𝑠2 +𝑎2 ]
b)
𝐹𝑐 [𝑒 𝑎𝑥 ] = √𝜋 [𝑠2 +𝑎2]
c)
𝐹𝑠 [𝑒 −𝑎𝑥 ] = √𝜋 [𝑠2 +𝑎2 ]
d)
𝐹𝑠 [𝑒 𝑎𝑥 ] = √𝜋 [𝑠2 +𝑎2]
2
𝑎
1
2
𝑠
𝑠
(CLO-4
Ans(a)
Remember)
Find the Fourier Cosine transform of
, a>0
1 mark
25
2
𝑎
a)
𝐹𝑐 [𝑒 −𝑎𝑥 ] = √𝜋 [𝑠 2+𝑎2 ]
b)
𝐹𝑐 [𝑒 𝑎𝑥 ] = √𝜋 [𝑠2+𝑎2 ]
c)
𝐹𝑠 [𝑒 −𝑎𝑥 ] = √𝜋 [𝑠 2+𝑎2 ]
d)
𝐹𝑠 [𝑒 𝑎𝑥 ] = √𝜋 [𝑠 2+𝑎2 ]
2
𝑎
2
2
𝑠
(CLO-4
An(a)
Remember)
𝑠
Modulation theorem F[f(x)cosax]
1 mark
26
a)
b)
c)
d)
1
2
1
2
1
4
1
4
[𝐹(𝑠 + 𝑎) + 𝑓(𝑠 − 𝑎)]
[𝐹(𝑠 + 𝑎) − 𝑓(𝑠 − 𝑎)]
[𝐹(𝑠 + 𝑎) + 𝑓(𝑠 − 𝑎)]
(CLO-4
Ans (a)
Remember)
[𝐹(𝑠 + 𝑎) + 𝑓(𝑠 − 𝑎)]
Find the fourier transform of
1 mark
27
2 𝑠𝑖𝑛𝑠𝑎−𝑎𝑠𝑐𝑜𝑠𝑠𝑎
a)
𝑖√𝜋 ⌈
b)
√𝜋 ⌈
c)
𝑖√𝜋 ⌈
d)
𝑖√𝜋 ⌈
𝑠2
1 𝑠𝑖𝑛𝑠𝑎−𝑎𝑠𝑐𝑜𝑠𝑠𝑎
𝑠2
⌉
2 𝑠𝑖𝑛𝑠𝑎−𝑎𝑠𝑐𝑜𝑠𝑠𝑎
𝑠
2 𝑠𝑖𝑛𝑠𝑎−𝑎𝑠𝑐𝑜𝑠𝑠𝑎
𝑠3
⌉
⌉
(CLO-4
Ans (a)
Remember)
⌉
Find the fourier transform of
1 mark
28
a)
2 𝑠𝑖𝑛𝑠𝑎
√𝜋 ⌈
𝑠
(CLO-4
1 𝑠𝑖𝑛𝑠𝑎
b)
√𝜋 ⌈
c)
𝑖√𝜋 ⌈
⌉
𝑠2
⌉
2 𝑎𝑠𝑐𝑜𝑠𝑠𝑎
𝑠
Ans (a)
⌉
Remember)
d)
2 𝑠𝑖𝑛𝑠𝑎
𝑖√ ⌈
𝜋
𝑠3
⌉
Find the fourier cosine transform of
1 mark
29
a)
b)
c)
d)
𝜋
2
𝜋
4
𝜋
2
1
2
𝑒 −|𝑥|
𝑒 −|𝑥|
(CLO-4
𝑒 |𝑥|
Ans (a)
Remember)
𝑒 −|𝑥|
Find the Fourier Cosine transform of
1 mark
30
2
15
2
15
10
a)
𝐹𝑐 [𝑒 −𝑎𝑥 ] = √𝜋 [𝑠 2+25 + 𝑠 2 +4]
b)
𝐹𝑠 [𝑒 −𝑎𝑥 ] = √ [
c)
𝐹𝑐 [𝑒 −𝑎𝑥 ] = √𝜋 [𝑠 2+25 + 𝑠 2 +4]
d)
𝐹𝑠 [𝑒 −𝑎𝑥 ] = √𝜋 [𝑠 2+25 + 𝑠 2+4]
𝜋 𝑠 2+25
+
10
𝑠 2+4
1
15
10
1
15
10
]
(CLO-4
Ans (a)
Remember)
Find the Fourier sine transform of
1 mark
31
2
𝑠
a)
𝐹𝑠 [𝑒 −𝑎𝑥 ] = √ [
b)
𝐹𝑐 [𝑒 𝑎𝑥 ] = √𝜋 [𝑠2+32 ]
c)
𝐹𝑠 [𝑒 −𝑎𝑥 ] = √𝜋 [𝑠 2+32 ]
d)
𝐹𝑠 [𝑒 𝑎𝑥 ] = √𝜋 [𝑠 2+32 ]
𝜋 𝑠 2+32
2
]
𝑎
1
2
𝑠
(CLO-4
Ans (a)
Remember)
𝑠
Find the Fourier sine transform of
32
1 mark
𝜋
a)
√2
b)
√4
c)
√2
d)
√𝜋
𝜋
(CLO-4
Ans(b)
1
Remember)
1
Find the Fourier transform of
.
1 mark
33
(a) 𝐹(𝑆) =
1
𝑒
√2
(b) F (𝑆) =
−𝑠2
1
√
(c) 𝐹(𝑆) = 𝑎
d) 𝐹 (𝑆) = 𝑎
−𝑠2
4𝑎2
1
√
1
√2
(CLO-4
𝑒 4𝑎2
3
−𝑠2
Ans (d)
𝑒 4𝑎2
4
𝑒
Remember)
−𝑠2
4𝑎2
Find the Fourier cosine transform of
1 mark
34
(a)
1
(b)
√3
1
(c)
𝑎 √4
1
(d)
𝑎 √2
𝑒
(CLO-4
−𝑠2
4𝑎2
𝑒
−𝑠2
4𝑎2
𝑒
−𝑠2
4𝑎2
Ans (a)

evaluate
35
 (x
0
2
dx
 a2 )2
Remember)
1 mark
𝜋
a)3𝑎2
b)
(CLO-4
3𝜋
4𝑎 3
Ans( c )
𝜋
c)4𝑎3
𝜋
d)𝑎3
Find the Fourier cosine transform of
36
e 3x
1 mark
Remember)
2
3
𝜋
32 +𝑠 2
a)√ [
]
(b)
(c) √𝜋 [𝑏+𝑠 3]
(𝑑)
2
𝑏
2
2
𝜋
4+𝑠 3
√ [
2
1
]
√𝜋 [1+𝑠 2 ]
(CLO-4
Ans (a)
Remember)
18MAB201T – TRANSFORMS AND BOUNDARY VALUE PROBLEMS
Fourier Transforms
SRM INSTITUTE OF SCIENCE AND TECHNOLOGY
RAMAPURAM CAMPUS
DEPARTMENT OF MATHEMATICS
Year/Sem : II/III
Branch: Common to All branches
Unit 4 – Fourier Transforms
1. The Fourier Transforms of 𝑓(𝑥 ) = 1 𝑖𝑛 𝑎 < 𝑥 < 𝑏 𝑖𝑠
𝟏
(a) 𝑭[𝒇(𝒙)] =
[𝒆𝒊𝒃𝒔 − 𝒆𝒊𝒂𝒔]
(b) 𝐹 [𝑓(𝑥 )] =
𝒊𝒔√𝟐𝝅
1
𝑖𝑏𝑠
√2𝜋
1
𝐹 [𝑓(𝑥 )] =
(c)
(d) 𝐹 [𝑓(𝑥 )] =
[𝑒
𝑖 √𝜋
1
𝑖𝑠√2𝜋
− 𝑒 𝑖𝑎𝑠 ]
[𝑒 𝑖𝑏𝑠 − 𝑒 𝑖𝑎𝑠 ]
[𝑒 𝑖𝑎𝑠 − 𝑒 𝑖𝑏𝑠 ]
𝐹 [𝑓(𝑥 )] =
Solution:
1
∞
∫ 𝑓 (𝑥 )𝑒 𝑖𝑠𝑥 𝑑𝑥
√2𝜋 −∞
𝑏
𝑒 𝑖𝑠𝑥
1
𝑖𝑠𝑥
=
∫ 1. 𝑒 𝑑𝑥 =
[
] =
[𝑒 𝑖𝑏𝑠 − 𝑒 𝑖𝑎𝑠 ]
√2𝜋 𝑎
√2𝜋 𝑖𝑠 𝑎 𝑖𝑠 √2𝜋
𝑏
1
1
Ans: a
2. The Fourier Transforms of 𝑓(𝑥 ) = 𝑒 −𝑎|𝑥|, 𝑎 > 0 𝑖𝑠
(a) √
1
2𝜋
[
𝑎
𝑎 2+𝑠2
𝟐
𝒂
𝝅
𝒂𝟐+𝒔𝟐
] (b) √ [
Solution: 𝐹[𝑒 −𝑎|𝑥| ] =
1
]
1
1
𝜋
𝑎 2+𝑠2
(c) √ [
]
(d) √
2𝑎
𝜋
[
𝑠
𝑎 2+𝑠2
]
∞
∫ 𝑒 −𝑎|𝑥|𝑒 𝑖𝑠𝑥 𝑑𝑥
√2𝜋 −∞
𝑎
1
=
∫ (𝑐𝑜𝑠 𝑠𝑥 + 𝑖𝑠𝑖𝑛𝑠𝑥)𝑒 −𝑎|𝑥|𝑑𝑥
√2𝜋 0
=
𝑎
𝟐
𝒂
∫ 𝑐𝑜𝑠 𝑠𝑥 𝑒 −𝑎𝑥 𝑑𝑥 = √ [ 𝟐
]
𝝅 𝒂 + 𝒔𝟐
√2𝜋 0
2
Ans: b
3. 𝐼𝑓 𝐹 [𝑓(𝑥 )] =
(a)
1
𝑖𝑠
𝐹 (𝑠 )
1
∞
∫−∞ 𝑓 (𝑥 )𝑒 𝑖𝑠𝑥 𝑑𝑥
2𝜋
√
(b) −
1
2𝑖𝑠
𝐹 (𝑠 )
𝑥
= 𝐹 (𝑠), 𝑡ℎ𝑒𝑛 𝐹[∫𝑎 𝑓(𝑥)𝑑𝑥] =
𝟏
(c) − 𝑭(𝒔)
𝒊𝒔
2
(d) − 𝐹(𝑠)
𝑖𝑠
Solution:
𝑥
Let ∫𝑎 𝑓(𝑥)𝑑𝑥 = 𝑔(𝑥) 𝑖. 𝑒. , 𝑓(𝑥)𝑑𝑥 = 𝑔′ (𝑥)
If 𝐹[𝑔(𝑥 )] = 𝐺 (𝑠), 𝐹 [𝑔′ (𝑥)] = −𝑖𝑠𝐺(𝑠)
SRMIST, Ramapuram
1
Deparment of Mathematics
18MAB201T – TRANSFORMS AND BOUNDARY VALUE PROBLEMS
Fourier Transforms
𝟏
𝐹 [𝑓(𝑥 )] = −𝑖𝑠𝐹 [𝑔(𝑥 )] = − 𝑭(𝒔)
𝒊𝒔
Ans: c
4. The Fourier Transforms of 𝑓(𝑥 ) = 𝑒 𝑖𝑘𝑥 , 𝑎 < 𝑥 < 𝑏 𝑖𝑠
(a)
(b)
(c)
(d)
1
𝑖(𝑘𝑠)√2𝜋
1
[𝑒 𝑖(𝑘𝑠)𝑏 − 𝑒 𝑖(𝑘𝑠)𝑎 ]
[𝑒 𝑖(𝑘−𝑠)𝑏 − 𝑒 𝑖(𝑘−𝑠)𝑎 ]
𝑖(𝑘−𝑠)√2𝜋
2
[𝑒 𝑖(𝑘+𝑠)𝑏 − 𝑒 𝑖(𝑘+𝑠)𝑎 ]
𝑖(𝑘+𝑠)√𝜋
𝟏
𝒊(𝒌+𝒔)𝒃
𝒊(𝒌+𝒔)𝒂
[𝒆
𝒊(𝒌+𝒔)√𝟐𝝅
−𝒆
]
𝐹 [𝑓(𝑥 )] =
Solution:
1
∞
∫ 𝑓 (𝑥 )𝑒 𝑖𝑠𝑥 𝑑𝑥
√2𝜋 −∞
𝐹[𝑒 𝑖𝑘𝑥 ] =
=
𝑏
1
√2𝜋
𝑒 𝑖(𝑘+𝑠)𝑥 𝑑𝑥
∫
𝑎
𝑏
1
√2𝜋
∫ 𝑒 𝑖𝑘𝑥 𝑒 𝑖𝑠𝑥 𝑑𝑥
𝑎
𝑏
𝑒 𝑖(𝑘+𝑠)𝑥
1
=
[
] =
[𝑒 𝑖(𝑘+𝑠)𝑏 − 𝑒 𝑖(𝑘+𝑠)𝑎 ]
𝑖(𝑘
+
𝑠)
2𝜋
𝑖(𝑘
+
𝑠)
2𝜋
√
√
𝑎
1
Ans:d
5. The Fourier Transforms of 𝑓(𝑥 ) = 1, |𝑥 | ≤ 𝑎 𝑖𝑠
𝟐 𝐬𝐢𝐧 𝒂𝒔
1 sin𝑎𝑠
𝝅
𝜋
(a) √ (
) (b) √ (
𝒔
𝑠
𝐹 [𝑓(𝑥 )] =
Solution:
=
1
√2𝜋
) (c) √
1
√
𝑠
2 cos 𝑎𝑠
) (d) √ (
𝜋
𝑠
)
∞
1. 𝑒 𝑖𝑠𝑥 𝑑𝑥
−𝑎
=
cos 𝑎𝑠
(
∫ 𝑓 (𝑥 )𝑒 𝑖𝑠𝑥 𝑑𝑥
2𝜋 −∞
𝑎
∫
1
2𝜋
𝑎
𝑒 𝑖𝑠𝑥
1
=
[
] =
[𝑒 𝑖𝑠𝑎 − 𝑒 −𝑖𝑎𝑠 ]
𝑖𝑠
2𝜋
𝑖𝑠
2𝜋
√
√
−𝑎
1
1
2 sin 𝑎𝑠
[2𝑖 sin 𝑎𝑠] = √ (
)
𝜋
𝑠
𝑖𝑠√2𝜋
Ans:a
6. If 𝑓(𝑥 ) = 1 𝑎𝑛𝑑 𝐹 [𝑓(𝑥 )] =
2 sin 𝑎𝑠
√ (
𝜋
𝑠
∞
) 𝑡ℎ𝑒𝑛 𝑏𝑦 𝑃𝑎𝑟𝑠𝑒𝑣𝑎𝑙′ 𝑠 𝐼𝑑𝑒𝑛𝑡𝑖𝑡𝑦, ∫−∞ (
(a) 2𝜋𝑎
(b) 𝝅𝒂
Solution:
SRMIST, Ramapuram
(c) 𝜋/𝑎
(d) 2𝜋
𝑎
∞
𝑠
) 𝑑𝑠 𝑖𝑠
2 sin 𝑎𝑠 2
∫−𝑎 1𝑑𝑥 = ∫−∞ √𝜋 (
2
sin 𝑎𝑠 2
𝑠
) 𝑑𝑠
Deparment of Mathematics
18MAB201T – TRANSFORMS AND BOUNDARY VALUE PROBLEMS
2
∞
2𝑎 = ∫−∞ (
𝜋
sin 𝑎𝑠 2
𝑠
∞
) 𝑑𝑠 = ∫−∞ (
Fourier Transforms
sin 𝑎𝑠 2
) 𝑑𝑠 = 𝜋𝑎
𝑠
Ans: b
7. 𝐼𝑓 𝐹𝑠 [𝑥 𝑓(𝑥 )] = −
𝟐
𝒂𝒔
(a) √ [
𝝅 (𝒂𝟐+𝒔𝟐) 𝟐
𝑑
𝐹[
𝑑𝑠 𝑐
𝑓(𝑥 )], 𝑡ℎ𝑒𝑛 𝐹𝑠 [𝑥 𝑒 −𝑎𝑥 ] 𝑖𝑠
𝟏
]
𝟐𝒂𝒔
(b) √ [
𝝅 (𝒂𝟐+𝒔𝟐) 𝟐
𝟐
]
(c) √ [
]
𝟐
(d) √ [
𝟐𝝅
𝝅 (𝒂𝟐+𝒔𝟐) 𝟐
]
𝑑
𝐹 [𝑒 −𝑎𝑥 ]
𝑑𝑠 𝑐
𝐹𝑠 [𝑥 𝑒 −𝑎𝑥 ] = −
Solution:
𝟐𝒂𝒔
𝝅 (𝒂𝟐+𝒔𝟐) 𝟐
𝑑
=−
𝑑𝑠
𝟐
𝒂
√ [
𝝅 (𝒂𝟐+𝒔𝟐)
𝟐
]= √ [
𝟐𝒂𝒔
𝝅 (𝒂𝟐+𝒔𝟐) 𝟐
]
Ans:c
8. 𝐼𝑓 𝐹𝑐 [𝑥 𝑓(𝑥 )] =
𝒂𝟐
𝟐
(a) √ [
𝟐
𝟐
(b) √ [
]
𝟐) 𝟐
(d) √ [
𝒔𝟐
𝝅 (𝒂𝟐+𝒔𝟐) 𝟐
𝒂𝟐−𝒔𝟐
𝟐
]
𝒂𝟐−𝒔𝟐
𝝅 (𝒂𝟐+𝒔𝟐) 𝟐
𝝅 (𝒂𝟐−𝒔
Solution:
𝑓(𝑥 )], 𝑡ℎ𝑒𝑛 𝐹𝑐 [𝑥 𝑒 −𝑎𝑥 ] 𝑖𝑠
]
𝟐) 𝟐
𝝅 (𝒂𝟐+𝒔
(c) √ [
𝑑
𝐹[
𝑑𝑠 𝑠
𝑑
𝐹𝑐 [𝑥 𝑒 −𝑎𝑥 ] =
𝐹 [𝑒
𝑑𝑠 𝑠
=
𝑑
𝑑𝑠
]
−𝑎𝑥 ]
𝟐
𝒔
√ [
𝟐
𝝅 (𝒂𝟐+𝒔𝟐)
]=√ [
𝒂𝟐 −𝒔𝟐
𝝅 (𝒂𝟐+𝒔𝟐) 𝟐
].
Ans: d
𝟐
𝒂
𝝅
(𝒂𝟐+𝒔𝟐)
9. 𝐼𝑓 𝑓 (𝑥 ) = 𝑒 −𝑎𝑥 𝑎𝑛𝑑 𝐹𝑐 [𝑥 𝑒 −𝑎𝑥 ] = √ [
∞
𝑡ℎ𝑒𝑛 𝑏𝑦 𝑃𝑎𝑟𝑠𝑒𝑣𝑎𝑙′𝑠 𝐼𝑑𝑒𝑛𝑡𝑖𝑡𝑦 ∫0
(a)
𝝅
𝟒𝒂𝟑
(b)
𝝅
𝒂𝟑
(c)
𝟏
(𝒂𝟐+𝒔𝟐)
𝝅
𝟐
(d)
𝟐𝒂𝟐
]
𝑑𝑠 𝑖𝑠
𝟑𝝅
𝟒𝒂𝟑
Solution:
By Parseval’s Identity,
∞
∞ |𝑓 (𝑥 )|2 𝑑𝑥
∫0 |𝐹𝑐 [ 𝑠]|2 𝑑𝑠 = ∫0
∞
𝟐𝒂𝟐 ∞
𝟏
∫
𝑑𝑠 = ∫ 𝑒 −2𝑎𝑥 𝑑𝑥
𝝅 0 (𝒂𝟐 + 𝒔𝟐) 𝟐
0
∞
𝟏
𝝅
∫
𝑑𝑠 =
𝟐
𝟒𝒂𝟑
0 (𝒂𝟐 + 𝒔𝟐)
Ans: a
𝟐
𝒔
𝝅
(𝒂𝟐+𝒔𝟐)
10. 𝐼𝑓 𝑓 (𝑥 ) = 𝑒 −𝑎𝑥 𝑎𝑛𝑑 𝐹𝑠 [ 𝑒 −𝑎𝑥 ] = √ [
SRMIST, Ramapuram
3
]
Deparment of Mathematics
18MAB201T – TRANSFORMS AND BOUNDARY VALUE PROBLEMS
𝒔𝟐
∞
𝑡ℎ𝑒𝑛 𝑏𝑦 𝑃𝑎𝑟𝑠𝑒𝑣𝑎𝑙′𝑠 𝐼𝑑𝑒𝑛𝑡𝑖𝑡𝑦 ∫0
(a)
𝜋
(b)
4𝑎 3
𝝅
(c)
𝟒𝒂
(𝒂𝟐+𝒔𝟐)
𝝅
𝑑𝑠 𝑖𝑠
𝟐
(d)
𝟐𝒂𝟐
Fourier Transforms
𝟑𝝅
𝟒𝒂𝟑
By Parseval’s Identity,
Solution:
∞
∞
| ( )|2
∫ |𝐹𝑐 [ 𝑠]|2 𝑑𝑠 = ∫ 𝑓 𝑥 𝑑𝑥
0
0
∞
𝟐 ∞
𝒔𝟐
∫
𝑑𝑠
=
∫
𝑒 −2𝑎𝑥 𝑑𝑥
𝝅 0 (𝒂𝟐 + 𝒔𝟐) 𝟐
0
2
∞
𝑠
𝜋
∫
𝑑𝑠 =
2
4𝑎
0 (𝑎2 + 𝑠 2)
Ans: b
11. If
𝟐
∞
∫
𝝅 0
𝟏𝟎
(𝒔𝟐+𝟒)(𝒔𝟐+𝟐𝟓)
∞
then ∫0
(a)
𝒅𝒙
(𝒙𝟐+𝟒)(𝒙𝟐+𝟐𝟓)
𝜋
(b)
120
Solution:
∞
∫
0
∞ −7𝑥 𝑑𝑥 ,
𝑑𝑠 = ∫0 𝑒
=
𝝅
(c)
𝟏𝟔𝟎
𝝅
𝟑𝝅
(d)
𝟏𝟒𝟎
𝟏𝟖𝟎
𝜋
𝑑𝑥
𝜋 ∞ −7𝑥
𝑒
𝑑𝑥 =
=
∫
140
(𝑠 2 + 4)(𝑠 2 + 25)
20 0
Ans: c
𝒂
∞ −𝑎𝑥 cos 𝑏𝑥 𝑑𝑥 =
,
−5𝑥 + 5𝑒 −2𝑥 is
12. If ∫0 𝑒
(𝒂𝟐+𝒃𝟐) then FCT of 𝑓 (𝑥 ) = 3𝑒
2
(a) √ [
25
𝜋 (25+𝑠2)
2
(c) √ [
10
𝜋 (25+𝑠2)
+
+
10
(4+𝑠2 )
15
(4+𝑠2 )
2
]
(b) √ [
]
(d) √ [
5
𝜋 (25+𝑠2 )
𝟐
𝟏𝟓
𝝅 (𝟐𝟓+𝒔𝟐)
+
+
20
(4+𝑠2)
]
𝟏𝟎
(𝟒+𝒔𝟐 )
]
𝐹𝑐 [𝑓(𝑥 )] = 3𝐹𝑐 [𝑒 −5𝑥 ] + 5𝐹𝑐 [𝑒 −2𝑥 ]
Solution:
𝟐
𝟓
𝝅
(𝟓𝟐+𝒔𝟐 )
=√ {𝟑 [
𝟐
=√ [
𝟏𝟓
𝝅 (𝟐𝟓+𝒔𝟐)
+
] + 𝟓[
𝟐
(𝟐𝟐 +𝒔𝟐 )
𝟏𝟎
(𝟒+𝒔𝟐 )
]}
]
Ans: d
13. By Fourier Transforms identity, if
∞ −(𝑎+𝑏)𝑥
2𝑎𝑏 ∞
1
𝑑𝑥 ,
∫
𝑑𝑠 = ∫ 𝑒
2
2
2
2
𝜋 0 (𝑠 + 𝑎 )(𝑠 + 𝑏 )
0
SRMIST, Ramapuram
4
Deparment of Mathematics
18MAB201T – TRANSFORMS AND BOUNDARY VALUE PROBLEMS
∞
Then ∫0
𝟏
(𝒙𝟐+𝒂𝟐)(𝒙𝟐+𝒃𝟐)
𝝅
(a)
𝑑𝑥 is
𝝅
(b)
𝟐𝒂𝒃(𝒂+𝒃)
Fourier Transforms
(c)
𝟐𝒃(𝒂+𝒃)
𝝅
(d)
𝟐𝒂(𝒂+𝒃)
𝝅
𝟐𝒂𝒃(𝒂−𝒃)
Solution:
∞
∞ −(𝑎+𝑏)𝑥
𝟏
𝝅
𝑑𝑥 ,
∫
𝑑𝑠 =
∫ 𝑒
𝟐
𝟐
𝟐
𝟐
𝟐𝒂𝒃 0
0 (𝒔 + 𝒂 )(𝒔 + 𝒃 )
𝝅
=
𝟐𝒂𝒃(𝒂+𝒃)
Ans: a
𝑒 −𝑎𝑥
14. If 𝐹𝑐 [
𝑥
1
(a)
√2𝜋
1
(c)
√2𝜋
]= −
√2𝜋
𝑏2
log (
𝑒 −𝑎𝑥 −𝑒 −𝑏𝑥
1
𝑎 2+𝑠2
𝑎2
log (
𝑎 2+𝑠2
log(𝑎2 + 𝑠 2 ) then 𝐹𝑐 [
)
(b)
)
(d)
𝑥
𝒃𝟐+𝒔𝟐
𝟏
√𝟐𝝅
1
√2𝜋
𝐥𝐨𝐠 (
)
log (
)
𝒂𝟐+𝒔𝟐
𝑏2−𝑠2
𝑎 2+𝑠2
] is
Solution:
𝑒 −𝑎𝑥 −𝑒 −𝑏𝑥
𝐹𝑐 [
𝑥
]=−
1
√2𝜋
log(𝑎2 + 𝑠 2 ) +
1
√2𝜋
log(𝑏 2 + 𝑠 2 )
𝑏2 + 𝑠 2
=
log ( 2
)
𝑎 + 𝑠2
√2𝜋
1
Ans:b
15. If 𝐹𝑠 [
𝑒 −𝑎𝑥
𝑥
2
∞ 𝑒 −𝑎𝑥 −𝑒 −𝑏𝑥
𝑎
] = −√ 𝑡𝑎𝑛 −1 ( ) then ∫0 (
𝜋
𝑠
𝑥
) sin 𝑠𝑥 𝑑𝑥 is
𝑎
𝑏
𝑏
𝑎
𝑠
𝑠
𝑠
𝑠
(a) (𝑡𝑎𝑛 −1 ( ) −𝑡𝑎𝑛 −1 ( )) (b) (𝑡𝑎𝑛 −1 ( ) +𝑡𝑎𝑛 −1 ( ))
𝒃
𝒂
(c) (𝒕𝒂𝒏−𝟏 ( ) −𝒕𝒂𝒏−𝟏 ( ))
𝒔
𝒔
(d)
𝑎
𝑏
(𝑡𝑎𝑛 −1 ( ) +𝑡𝑎𝑛 −1 ( ))
𝑠
𝑠
Solution:
𝑒 −𝑎𝑥 −𝑒 −𝑏𝑥
𝐹𝑐 [
𝑥
∞ 𝑒 −𝑎𝑥 −𝑒 −𝑏𝑥
2
√ ∫0 (
𝜋
∞
𝑥
𝑒 −𝑎𝑥 −𝑒 −𝑏𝑥
∫0 (
𝑥
2
𝑏
𝑎
𝜋
𝑠
𝑠
] = √ (𝑡𝑎𝑛 −1 ( ) −𝑡𝑎𝑛 −1 ( ))
2
𝑏
𝑎
𝜋
𝑠
𝑠
) sin 𝑠𝑥 𝑑𝑥 = √ (𝑡𝑎𝑛 −1 ( ) −𝑡𝑎𝑛 −1 ( ))
𝑏
𝑎
𝑠
𝑠
) sin 𝑠𝑥 𝑑𝑥 = (𝑡𝑎𝑛 −1 ( ) −𝑡𝑎𝑛 −1 ( ))
Ans: c
16. If 𝑓(𝑥 ) =
𝟐𝒂
∞ 𝒄𝒐𝒔 𝒔𝒙
𝒅𝒔
∫
𝝅 𝟎 𝒔𝟐+𝒂𝟐
SRMIST, Ramapuram
(where 𝑓(𝑥 ) = 𝑒 −𝑎𝑥 , 𝑎 > 0)
5
Deparment of Mathematics
18MAB201T – TRANSFORMS AND BOUNDARY VALUE PROBLEMS
∞ 𝒄𝒐𝒔 𝒎𝒙
then by Inverse Fourier Transform ∫𝟎
(a)
𝝅
𝟐𝒂
𝑒 −𝑚𝑠
(b)
𝝅
𝟐𝒎
𝑒 −𝑚𝑎
𝝅
(c)
𝟒𝒂
𝒙𝟐+𝒂𝟐
Fourier Transforms
𝒅𝒙 is
𝑒 −𝑚𝑎
(d)
𝝅
𝟐𝒂
𝒆−𝒎𝒂
Solution:
∞
𝒄𝒐𝒔 𝒔𝒙
𝝅 −𝑎𝑥
𝒅𝒔 =
𝑒
𝟐
𝟐
𝟐𝒂
𝟎 𝒙 +𝒂
By changing the dummy variables of integration s=x and x=m then
∞ 𝒄𝒐𝒔 𝒎𝒙
𝝅 −𝑚𝑎
∫ 𝟐
𝒅𝒙
=
𝑒
𝟐
𝟐𝒂
𝟎 𝒙 +𝒂
Ans:d
∫
𝟐
∞ 𝒔 𝒔𝒊𝒏 𝒔𝒙
17. If 𝑓(𝑥 ) = ∫𝟎
𝝅
𝒔𝟐+𝒂𝟐
𝒅𝒔 (where 𝑓 (𝑥 ) = 𝑒 −𝑎𝑥 , 𝑎 > 0)
∞ 𝒙 𝒔𝒊𝒏 𝒎𝒙
then by Inverse Fourier Transform ∫𝟎
𝝅
(a) 𝒆−𝒂𝒎
𝟐
(b)
𝝅
𝑒 −𝑚𝑎
𝟐𝒎
(c)
𝝅
𝟒𝒂
𝒙𝟐+𝒂𝟐
𝒅𝒙 is
𝑒 −𝑚𝑎
(d)
𝜋
2𝑎
𝑒 −𝑚𝑎
Solution:
∞𝒔
𝒔𝒊𝒏 𝒔𝒙
𝝅 −𝑎𝑥
𝒅𝒔
=
𝑒
𝟐
𝟐
𝟐
𝟎 𝒙 +𝒂
By changing the dummy variables of integration s=x and x=m then
∞ 𝑥 𝑠𝑖𝑛 𝑚𝑥
𝜋
∫
𝑑𝑥 = 𝑒 −𝑎𝑚
2
2
2
0 𝑥 +𝑎
Ans: a
∫
∞ 𝒔𝒊𝒏 𝒕
18. If ∫𝟎
𝒕
(a) √
𝒅𝒕 =
2
𝝅
𝟐
𝟏
, then 𝐹𝑠 [ ] is
𝒙
(b) √
𝜋
𝝅
(c) √
𝟐
1
2
𝑥
𝜋
𝐹𝑠 [ ] = √
Solution:
∞1
∫0
𝑥
𝑃𝑢𝑡 𝑠𝑥 = 𝜃 𝑡ℎ𝑒𝑛
𝑑𝑥 =
𝑑𝜃
𝑠
1
(d) √
𝜋
1
2𝜋
𝑠𝑖𝑛 𝑠𝑥 𝑑𝑥
1
𝑠
=
𝑥
𝜃
1
𝜋
𝑥
2
, then 𝐹𝑠 [ ] = √
Ans: b
∞ −𝑎𝑥 cos 𝑏𝑥 𝑑𝑥 =
19. If ∫0 𝑒
2
(a) √ [
2
𝜋 (24+𝑠2)
SRMIST, Ramapuram
+
𝒂
(𝒂𝟐+𝒃𝟐)
3
(9+𝑠2 )
]
, then FCT of 𝑓(𝑥 ) = 𝑒 −2𝑥 + 3𝑒 −𝑥 is
2
(b) √ [
1
𝜋 (25+𝑠2)
6
+
2
(4+𝑠2)
]
Deparment of Mathematics
18MAB201T – TRANSFORMS AND BOUNDARY VALUE PROBLEMS
𝟐
𝟐
(c) √ [
𝝅 (𝟒+𝒔𝟐)
+
𝟑
(𝟏+𝒔𝟐 )
2
]
5
(d) √ [
𝜋 (25+𝑠2 )
+
Fourier Transforms
1
(4+𝑠2)
]
𝐹𝑐 [𝑓(𝑥 )] = 𝐹𝑐 [𝑒 −2𝑥 ] + 3𝐹𝑐 [𝑒 −𝑥 ]
Solution:
𝟐
𝟐
𝟏
√ {[
]
+
𝟑
[
]}
𝝅 (𝟒 + 𝒔𝟐)
(𝟏 + 𝒔𝟐)
2
=√ [
2
𝜋 (4+𝑠2)
3
+
(1+𝑠2 )
]
Ans:c
20. The Fourier sine Transforms of 𝑓(𝑥 ) = 1, 0 < 𝑥 < 1 𝑖𝑠
2 cos 𝑠
(a) √ [
𝜋
𝑠
2 1+cos 𝑠
]
(b) √ [
𝜋
𝑠
1 2−cos 𝑠
] (c) √ [
𝜋
2
𝑠
𝟐 𝟏−𝐜𝐨𝐬 𝒔
] (d) √ [
𝝅
∞
2
𝒔
]
1
𝐹𝑠 [𝑓(𝑥 )] = √ ∫0 𝑓 (𝑥 ) sin 𝑠𝑥 𝑑𝑥 = √ ∫0 sin 𝑠𝑥 𝑑𝑥 =
𝜋
𝜋
Solution:
2
cos 𝑠𝑥 1
2 1 − cos 𝑠
√ [−
] =√ [
]
𝜋
𝑠 0
𝜋
𝑠
=
1
2 sin 𝑎𝑠
[2𝑖 sin 𝑎𝑠] = √ (
)
𝜋
𝑠
𝑖𝑠√2𝜋
Ans:d
𝒃
∞ −𝑎𝑥 sin 𝑏𝑥 𝑑𝑥 =
, then FST of 𝑓 (𝑥 ) = 𝑒 −2𝑥 + 3𝑒 −𝑥 is
21. If ∫0 𝑒
(𝒂𝟐+𝒃𝟐)
𝟐
𝒔
𝝅
(𝟒+𝒔𝟐 )
2
2
(a) √ [
(c) √ [
𝜋 (4+𝑠2 )
Solution:
+
−
𝟑𝒔
3
(1+𝑠2 )
2
𝑠
𝜋
(25+𝑠2)
2
5𝑠
]
(b) √ [
]
(d) √ [
(𝟏+𝒔𝟐)
𝜋 (25+𝑠2)
+
+
2𝑠
(4+𝑠2 )
𝑠
(4+𝑠2)
]
]
𝐹𝑠 [𝑓(𝑥 )] = 𝐹𝑐 [𝑒 −2𝑥 ] + 3𝐹𝑐 [𝑒 −𝑥 ]
𝟐
𝒔
𝒔
√ {[
]+ 𝟑[
]}
𝟐
𝝅 (𝟒 + 𝒔 )
(𝟏 + 𝒔𝟐)
2
=√ [
𝑠
𝜋 (4+𝑠2)
+
3𝑠
(1+𝑠2 )
]
Ans:a
SRMIST, Ramapuram
7
Deparment of Mathematics
18MAB201T – TRANSFORMS AND BOUNDARY VALUE PROBLEMS
Fourier Transforms
22. The Fourier cosine Transforms of 𝑓(𝑥 ) = 1, 0 < 𝑥 < 1 𝑖𝑠
2 cos 𝑠
(a) √ [
𝜋 𝑠−𝑎
]
𝟐 𝐬𝐢𝐧 𝒔
1 cos 𝑠
𝝅
𝜋
(b) √ [
𝒔
] (c) √ [
2
𝑠
𝟐 𝟏−𝐬𝐢𝐧 𝒔
] (d) √ [
𝝅
𝒔
∞
2
]
1
𝐹𝑐 [𝑓(𝑥 )] = √ ∫0 𝑓(𝑥 ) cos 𝑠𝑥 𝑑𝑥 = √ ∫0 cos 𝑠𝑥 𝑑𝑥
𝜋
𝜋
Solution:
2 sin 𝑠𝑥 1
2 sin 𝑠
𝜋
𝜋
=√ [
] =√ [
𝑠
0
𝑠
]
Ans:b
23. If
𝟐
∞
∫
𝝅 0
(𝒔𝟐+𝟒)(𝒔𝟐+𝟐𝟓)
∞
then ∫0
(a)
∞ −4𝑥
𝑑𝑠 = ∫0 𝑒 𝑑𝑥 ,
𝒔
𝜋
𝒙 𝒅𝒙
(𝒙𝟐+𝟒)(𝒙𝟐+𝟐𝟓)
(b)
12
Solution:
=
𝜋
(c)
16
𝝅
(d)
𝟏𝟒
𝝅
𝟖
𝜋
𝑠𝑑𝑠
𝜋 ∞ −4𝑥
𝑒
𝑑𝑥
=
∫
= ∫
2
2
8
2 0
0 (𝑠 + 4)(𝑠 + 25)
∞
Put s=x.
Ans: d
24. By Fourier Transforms identity, if
∞
Then ∫0
(a)
𝟏
𝜋
𝜋
(b)
∞ −(𝑎+𝑏)𝑥 𝑑𝑥 ,
= ∫0 𝑒
𝑑𝑥 is
(𝒙𝟐+𝟗)(𝒙𝟐+𝟏𝟔
162
2𝑎𝑏 ∞
1
𝑑𝑠
∫
2
2
0
𝜋
(𝑠 +𝑎 )(𝑠2+𝑏2 )
186
(c)
𝝅
(d)
𝟏𝟔𝟖
𝝅
𝟖𝟏𝟔
Solution:
∞
∫
0
∞
𝟏
𝝅
𝝅
−7𝑥 𝑑𝑥 =
𝑑𝑠
=
∫
𝑒
(𝒔𝟐 + 𝟑𝟐)(𝒔𝟐 + 𝟒𝟐)
𝟐(𝟑)(𝟒) 0
𝟏𝟔𝟖
Ans: c𝑒 −7𝑥 𝑑𝑥
25. The Fourier Transforms of 𝑓(𝑥 ) = 1 𝑖𝑛 1 < 𝑥 < 2 𝑖𝑠
(a) 𝑭[𝒇(𝒙)] =
(b) 𝐹 [𝑓(𝑥 )] =
1
√2𝜋
(c) 𝐹 [𝑓(𝑥 )] =
(d) 𝐹 [𝑓(𝑥 )] =
SRMIST, Ramapuram
1
𝑖𝑠√2𝜋
[𝑒 𝑖2𝑠 − 𝑒 𝑖𝑠 ]
[𝑒 𝑖𝑠 + 𝑒 𝑖2𝑠 ]
1
𝑖 √𝜋
[𝑒 𝑖𝑏 − 𝑒 𝑖𝑎 ]
1
𝑖𝑠 √2𝜋
[𝑒 𝑖𝑠 − 𝑒 𝑖2𝑠 ]
8
Deparment of Mathematics
18MAB201T – TRANSFORMS AND BOUNDARY VALUE PROBLEMS
Solution:
𝐹 [𝑓(𝑥 )] =
=
1
√
∞
∫ 𝑓 (𝑥 )𝑒 𝑖𝑠𝑥 𝑑𝑥
2𝜋 −∞
1
√
Fourier Transforms
2
∫ 1. 𝑒 𝑖𝑠𝑥 𝑑𝑥 =
2𝜋 1
1
√2𝜋
𝑒 𝑖𝑠𝑥
[
𝑖𝑠
2
] =
1
1
𝑖𝑠√2𝜋
[𝑒 𝑖2𝑠 − 𝑒 𝑖𝑠 ]
Ans: a
SRMIST, Ramapuram
9
Deparment of Mathematics
18MAB201T – TRANSFORMS AND BOUNDARY VALUE PROBLEMS
Z Transforms
SRM INSTITUTE OF SCIENCE AND TECHNOLOGY
RAMAPURAM CAMPUS
DEPARTMENT OF MATHEMATICS
Year/Sem : II/III
Branch: Common to All branches
Unit V – Z Transforms
1.Find 𝑍(𝑛)
𝐳
𝑧
(a) (𝐳−𝟏)𝟐
(b) (𝑧+1)2
(c)
𝑧
(d)
𝑧−1
𝑧
𝑧+1
Solution:
∞
𝑍(𝑛) = ∑ 𝑛𝑧 −𝑛
0
1
2
𝑧
1
𝑧2
= +
+
2. Find Z-Transform of
𝑧
(a) (𝑧+𝑎)2
𝑧
(b) (𝑧−𝑎)2
+⋯
𝑧3
1 −2
= (1 − )
𝑧
3
𝑧
𝑧
= (𝑧−1)2
𝑛𝑎𝑛
𝒂𝒛
𝑎𝑧
(c) (𝒛−𝒂)𝟐 (d) (𝑧+𝑎)2
Solution:
𝑑
𝑍(𝑎𝑛 )
𝑑𝑧
𝑑
𝑧
= −𝑧 ( )
𝑑𝑧 𝑧−𝑎
𝑎
= −𝑧 (−
)
(𝑧 − 𝑎 ) 2
𝑎𝑧
= (𝑧−𝑎)2
𝑍(𝑛𝑓(𝑛)) = −𝑧
SRMIST, Ramapuram
1
Department of Mathematics
18MAB201T – TRANSFORMS AND BOUNDARY VALUE PROBLEMS
3. Find Z-Transform of
(a) 𝑒
𝑎
−𝑧
𝑎𝑛
(b) 𝒆
Z Transforms
1
𝑛!
𝒂
𝒛
1
1
(d) 𝑒 −𝑧
(c) 𝑒 𝑧
Solution
𝑍 ( 𝑎𝑛
1
1
) = 𝑍( ) 𝑧
𝑛!
𝑛! 𝑧→
𝑎
1
𝑧
= (𝑒 )
𝑧
𝑧→𝑎
𝑎
= 𝑒𝑧
4. Find Z-Transform of
(a)
𝒛
(b)
𝒛−𝒓𝒆𝒊𝜽
𝑟 𝑛 𝑒 𝑖𝑛𝜃 by eliminating arbitrary function
𝑧
(c)
𝑧+𝑟𝑒 𝑖𝜃
1
𝑧−𝑟𝑒 𝑖𝜃
(d)
−1
𝑧−𝑟𝑒 𝑖𝜃
Solution:
We know that 𝑍(𝑎𝑛 ) =
Put 𝑎 = 𝑟𝑒
𝑧
𝑧−𝑎
𝑖𝜃
𝑍( 𝑟 𝑛 𝑒 𝑖𝑛𝜃 ) =
𝑧
𝑧 − 𝑟𝑒 𝑖𝜃
5. Find 𝑍(𝑛2 )
1+𝑧
1+𝑧
(a) −𝑧 [(𝑧−1)3 ]
(b) 𝑧 [(𝑧−1)3 ]
1+𝑧
1+𝑧
(c) −𝑧 [(𝑧+1)3 ]
(d𝑧 [(𝑧+1)3 ]
Solution:
𝑍(𝑛2 ) = −𝑧
= −𝑧
𝑑
𝑑𝑧
𝑑
𝑑𝑧
𝑍(𝑛)
𝑧
((𝑧−1)2 )
1+𝑧
= −𝑧 [(𝑧−1)3 ]
SRMIST, Ramapuram
2
Department of Mathematics
18MAB201T – TRANSFORMS AND BOUNDARY VALUE PROBLEMS
Z Transforms
6.Find 𝑍(𝑛(𝑛 − 1))
𝑧
𝑧
(a) (𝑧+1)3
2𝑧
(b) (𝑧−1)3
(c) (𝑧+1)3
𝟐𝒛
(d) (𝒛−𝟏)𝟑
Solution:
𝑍(𝑛(𝑛 − 1)) = 𝑍(𝑛2 − 𝑛)
= 𝑍(𝑛2 ) − 𝑍(𝑛)
𝑧(𝑧+1)
𝑧
= (𝑧−1)3 − (𝑧−1)2
2𝑧
= (𝑧−1)3
7. Find 𝑍(𝑛(1 + 5𝑛 ))
𝒛
𝟓𝒛
𝑧
5𝑧
(a)(𝒛−𝟏)𝟐 + (𝒛−𝟓)𝟐
𝑧
5𝑧
𝑧
5𝑧
(b) (𝑧+1)2 + (𝑧+5)2
(c) (𝑧−1)2 + (𝑧+5)2
(d) (𝑧+1)2 + (𝑧−5)2
Solutionz9:
𝑍(𝑛 (1 + 5𝑛 )) = 𝑍(𝑛) + 𝑍(𝑛. 5𝑛 )
𝑧
5𝑧
= (𝑧−1)2 + (𝑧−5)2
𝑧
7𝑧
8. Find the inverse Z-Transform of 𝑧+1
+
, 𝑓𝑜𝑟 𝑛 > 0
𝑧−3
(a) (−1)𝑛 + 7(−3)𝑛
(c) (−1)𝑛 − 7(−3)𝑛
(b) (−1)𝑛 − 7(3)𝑛
(d) (−𝟏)𝒏 + 𝟕(𝟑)𝒏
Solution:
𝑍 −1 (
9.Find 𝑍−1 (
1 𝑛
2𝑧
𝑧
7𝑧
+
) = (−1)𝑛 + 7(3)𝑛
𝑧+1 𝑧−3
)
2𝑧−1
𝟏 𝒏
(a) ( )
(b) ( )
(c) ( )
(d) ( )
2
1 𝑛
𝟐
1 𝑛
2
2
Solution:
𝑍
SRMIST, Ramapuram
−1
2𝑧
2 −1
𝑧
1 𝑛
(
)= 𝑍 (
)=( )
1
2𝑧 − 1
2
2
𝑧−
2
3
Department of Mathematics
18MAB201T – TRANSFORMS AND BOUNDARY VALUE PROBLEMS
10. Find
𝑧
𝑍 −1 (
Z Transforms
)
4𝑧+1
𝑛+1
1
(a) (−1)𝑛 ( )
3
𝒏+𝟏
𝒏 𝟏
1 𝑛+1
(b) (−2)𝑛 ( )
(c) (−𝟏) ( )
𝟒
Solution:
(d)
3
1 𝑛+1
(−2)𝑛 ( )
4
𝑧
1
𝑧
𝑍 −1 (
) = 𝑍 −1 (
)
1
4𝑧 + 1
4
𝑧+
4
1
1 𝑛
1 𝑛+1
= (− ) = (−1)𝑛 ( )
4
4
4
11.Find 𝒁(3𝑛 (1 + 𝑛))
(a)
(c)
𝑧
3𝑧
𝑧+3
𝒛
𝒛−𝟑
+ (𝑧+3)2
(b)
𝟑𝒛
(𝒛−𝟑)𝟐
(d)
+
𝑧
3𝑧
𝑧+3
𝑧
𝑧−3
+ (𝑧−3)2
3𝑧
+ (𝑧+3)2
Solution:
𝒁(3𝑛 (1 + 𝑛)) = 𝑍(3𝑛 ) + 𝑍(𝑛. 3𝑛 )
=
12.If 𝐹 (𝑧)𝑧
𝑛−1
=
(a) 2−𝑛
(c) 3𝑛
𝑧𝑛
(𝑧−1)(𝑧−2)
𝑧
𝑧−3
3𝑧
+ (𝑧−3)2
𝑡ℎ𝑒𝑛 𝑓𝑖𝑛𝑑 𝑟𝑒𝑠𝑖𝑑𝑢𝑒 𝑎𝑡 𝑙𝑎𝑟𝑔𝑒𝑠𝑡 𝑝𝑜𝑙𝑒
(b) 3−𝑛
(d)𝟐𝒏
Solution:
𝑅𝑒𝑠𝑧=2𝐹(𝑧)𝑧 𝑛−1 = lim (𝑧 − 2)
=
𝑛
𝑧
13. If 𝐹 (𝑧)𝑧 𝑛−1 = (𝑧−1)(𝑧−2)
𝑧→2
2𝑛
2−1
𝑧𝑛
(𝑧 − 1)(𝑧 − 2)
= 2𝑛
𝑡ℎ𝑒𝑛 𝑓𝑖𝑛𝑑 𝑟𝑒𝑠𝑖𝑑𝑢𝑒 𝑎𝑡 𝑠𝑚𝑎𝑙𝑙𝑒𝑠𝑡 𝑝𝑜𝑙𝑒
(a) −𝟏
(c) 1
(b) (−1)𝑛
(d) (−2)𝑛
SRMIST, Ramapuram
4
Department of Mathematics
18MAB201T – TRANSFORMS AND BOUNDARY VALUE PROBLEMS
Solution:
𝑅𝑒𝑠𝑧=1𝐹(𝑧)𝑧
𝑛−1
Z Transforms
𝑧𝑛
= lim (𝑧 − 1)
𝑧→2
(𝑧 − 1)(𝑧 − 2)
1
=
= −1
−1
14.
Find 𝑦(𝑧)𝑓𝑜𝑟 𝑡ℎ𝑒 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 𝐸𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝑦𝑛+1 − 𝑦𝑛 = 0, 𝑦0 = 1
𝑧
𝑧
(a)𝑦(𝑧) =
(b) 𝑦(𝑧) =
𝑧+1
𝒛
(c) 𝒚(𝒛) =
(d) 𝑦(𝑧) =
𝒛−𝟏
1−𝑧
−𝑧
𝑧+1
Solution:
𝑍(𝑦𝑛 ) − 𝑧𝑦0
𝑧𝑦(𝑧) − 𝑧𝑦0 − 𝑦(𝑧) = 0
(𝑧 − 1)𝑦(𝑧) = 𝑧
𝑧
𝑦(𝑧) =
𝑧−1
15.Find 𝑍[(𝑒
(a)
(c)
𝒛
𝒛−𝒆𝟏𝟎𝟎
𝑧
𝑧−𝑒 100
+
+
𝑛 )100
+ (𝑒
𝑛 )200
)]
𝒛
(b)
𝒛−𝒆𝟐𝟎𝟎
𝑧
(d)
𝑧+𝑒 200
𝑧
𝑧+𝑒 100
𝑧
𝑧+𝑒 100
+
+
𝑧
𝑧−𝑒 200
𝑧
𝑧+𝑒 200
Solution:
𝑍[(𝑒 𝑛 )100 + (𝑒 𝑛 )200 )] = 𝑍 ((𝑒 100)𝑛 ) + 𝑍(((𝑒 100 )𝑛 )
𝑧
𝑧
+
𝑧 − 𝑒 100 𝑧 − 𝑒 200
1+𝑧 −1
16.Using Final value theorem evaluate 𝑓(𝑧) = 1−0.25𝑧
−1
(a) 𝟎
(b) 1
(c) 2
(d) 3
=
Solution:
Let 𝐹 (𝑧) =
𝑧+1
𝑧−0.25
lim(𝑧 − 1)𝐹 (𝑧) = lim (𝑧 − 1)
𝑧→1
SRMIST, Ramapuram
𝑧→1
5
𝑧+1
=0
𝑧 − 0.25
Department of Mathematics
18MAB201T – TRANSFORMS AND BOUNDARY VALUE PROBLEMS
17. Find the initial value of
(a) 0
(b) 1
𝐹 (𝑧) =
Z Transforms
𝑧
2𝑧 2 −3𝑧+1
(c)-1 (d)2
Solution:
𝑓 (0) = 𝐥𝐢𝐦 𝑭(𝒛)
𝒛→∞
0=
18.Evaluate
1
𝑧
=
3
3 1
𝑧 (2 − 𝑧 + 𝑧 2 ) 𝑧 2 (2 − 𝑧 + 2 )
𝑧
𝑍[(𝑘 − 1)𝑎𝑘−1]
𝒂
(a)(𝒛−𝒂)𝟐
𝑎
(b) (𝑧+𝑎)2
1
(c)(𝑧−𝑎)2
1
(d) (𝑧+𝑎)2
Solution
𝑍[(𝑘 − 1)𝑎𝑘−1 ] = 𝑧 −1𝑍[𝑘𝑎𝑘 ]
𝑎𝑧
𝑎
= 𝑧 −1 ((𝑧−𝑎)2 ) = (𝑧−𝑎)2
19. Find 𝑌(𝑧)
(a) 𝑌(𝑧) =
(c)𝒀(𝒛) =
𝑓𝑜𝑟 𝑦𝑛+2 + 4𝑦𝑛 = 0 , 𝑦0 = 0, 𝑦1 = −2
𝑧
𝑧 2 +4
𝟐𝒛
𝒛𝟐 +𝟒
(b)𝑌 (𝑧) =
(d)𝑌(𝑧) =
2𝑧
𝑧 2 −4
−2𝑧
𝑧 2 +4
Solution:
𝑧 2 𝑌(𝑧) − 𝑧 2 𝑦(0) − 𝑧𝑦(1) − 4𝑌 (𝑧) = 0
𝑧 2 𝑌 (𝑧) − 𝑧(2) + 4𝑌 (𝑧) = 0
SRMIST, Ramapuram
6
Department of Mathematics
18MAB201T – TRANSFORMS AND BOUNDARY VALUE PROBLEMS
𝑌 (𝑧) =
20.Find
Z Transforms
2𝑧
𝑧2 + 4
𝑍{𝑒 3𝑛 }
(a) −𝒄𝒐𝒔𝒙 (b) 𝑐𝑜𝑠𝑥
(c) −𝑠𝑖𝑛𝑥
(d) 𝑠𝑖𝑛𝑥
Solution:
𝑍 {𝑒 3𝑛 } = 𝑍{(𝑒 3 )𝑛 } =
21.
(a)
𝑧
𝑧 − 𝑒3
Find 𝑍{(−3)𝑛 }
𝒛
(b)
𝒛+𝟑
−𝑧
(c)
𝑧+3
𝑧
𝑧−3
(d)
2𝑧
𝑧+3
Solution:
𝑧
𝑧−𝑎
𝑧
𝑧
𝑍{(−3)𝑛 } =
=
𝑧 − (−3) 𝑧 + 3
𝑍 {𝑎𝑛 } =
22.
(a)
Find 𝑍{2𝑛 + (−5)𝑛 }
𝑧
𝑧+2
+
𝑧
𝑧−5
(b)
𝑧
𝑧−2
+
𝑧
𝑧−5
(c)
𝒛
𝒛−𝟐
+
𝒛
𝒛+𝟓
(d)
𝑧
𝑧+2
+
𝑧
𝑧−5
Solution:
𝑍{2𝑛 + (−5)𝑛 } = 𝑍{2𝑛 } + 𝑍{(−5)𝑛 }
=
𝑧
𝑧
+
𝑧−2 𝑧+5
23. Find 𝑍{4.8𝑛 + (−4)(−6)𝑛 }
(a)
(b)
(c)
(d)
4𝑧
𝑧−8
𝟒𝒛
𝒛−𝟖
4𝑧
𝑧+8
4𝑧
𝑧+8
+
−
−
+
4𝑧
𝑧+6
𝟒𝒛
𝒛+𝟔
4𝑧
𝑧−6
4𝑧
𝑧−6
SRMIST, Ramapuram
7
Department of Mathematics
18MAB201T – TRANSFORMS AND BOUNDARY VALUE PROBLEMS
Z Transforms
Solution:1
1𝑍 {4.8𝑛 + (−4)(−6)𝑛 } = 4𝑍{8𝑛 } + (−4)𝑍{(−6)𝑛 }
=
4𝑧
𝑧−8
−
4𝑧
𝑧+6
𝑧
𝑧
24.Find 𝑍 −1 {𝑧+7
+ 2.
}
𝑧−3
(a) (7)n + 2(−3)𝑛
(b) (−7)n − 2(3)𝑛
(c)(7)n + 2(3)𝑛
(d) (−𝟕)𝐧 + 𝟐(𝟑)𝒏
Solution:
𝑍 −1 {
𝑧
𝑧
𝑧
2
+ 2.
} = 𝑍 −1 {
} + 2 𝑍 −1 {
}
𝑧+7
𝑧−3
𝑧+7
𝑧−3
= (−7)n + 2(3)𝑛
𝑧
𝑧
25.Find 𝑍 −1 {𝑧−8
+ 3.
}
𝑧+3
(a) (8)n − 3(−3)𝑛
(c)(−8)n + 3(−3)𝑛
(b) (−8)n + 3(3)𝑛
(d)(𝟖)𝐧 + 𝟑(−𝟑)𝒏
Solution:
𝑍 −1 {
𝑧
𝑧
𝑧
2
+ 3.
} = 𝑍 −1 {
} + 3 𝑍 −1 {
}
𝑧−8
𝑧+3
𝑧−8
𝑧+3
= (8)n + 3(−3)𝑛
SRMIST, Ramapuram
8
Department of Mathematics
SRM Institute of Science and Technology
Ramapuram Campus
Department of Mathematics
Question Bank of Module-V(Z-Transform)
(2020–2021-ODD)
Subject.Code: 18MAB201T
Subject.Name: Transforms and Boundary Value Problems
Part-A (1*20=20)
Year/Sem: II/III
1.
Branch: Common to All branches
1 mark
What is z(5)?
a)
z
z −1
z
c)
5( z − 1)
b)
5z
z −1
(CLO-5
Ans (b)
z −1
z
d)
z[( −1) n ] = _______
2.
z +1
z
z
c)
( z + 1)
a)
3.
1 mark
z
2z − 1
−z
d)
z +1
b)
Ans (c)
(CLO-5
b)
Ans (c)
What is z[( −2) n ] ?
z+2
z
−z
c)
( z + 2)
a)
z
( z − 1) 2
2z
c)
( z − 1) 2
Remember)
1
z
z−2
z
d)
z+2
b)
Ans (d)
Remember)
1
−z
( z + 1) 2
z
d)
( z + 1) 2
mark
(CLO-5
What is z[n ] ?
a)
Remember)
1 mark
z
if z  a
z+a
z
d)
if z  a
z+a
z
if z  a
z−a
z
c)
if z  a
z−a
5.
(CLO-5
z[( a) n u(n)] = _________
a)
4.
Remember)
b)
mark
(CLO-5
Ans (a)
Remember)
6.
z[e −5n ] = _______
z
z + e −5
z
c)
z + e5
z
z − e −5
2z
d)
z + e5
What is Z-Transform of nan ?
a)
7.
az
( z + a) 2
z
c)
( z − a) 2
a)
8.
az
( z + a) 2
z
c)
( z − a) 2
If z[f(t)]=F(z) then
a) f (0)
(CLO-5
b)
az
( z − a) 2
z
d)
( z + a) 2
Ans (b)
Remember)
1 mark
b)
(CLO-5
Ans (b)
Remember)
1 mark
What is z[n 2 ] ?
a)
9.
1 mark
az
( z − a) 2
z
d)
( z + a) 2
b)
(CLO-5
Ans (a)
lim
F ( z) = ?
n→
Remember)
1 mark
b) f (1)
(CLO-5
c ) f ( )
d) lim f (t )
t →
Ans (a)
 1 
?
n ! 
10.
Remember)
1 mark
What is Z-Transform of z 
a) e1 / z
c) e −1 / z
11.
d ) e−z
Ans (a)
 n 
z sin
= __________
2 

z
( z + 1)
z
c) 2
( z − 1)
a)
12.
(CLO-5
b) e z
2
1 mark
z
( z − 4)
2z
d) 2
( z + 1)
b)
(CLO-5
2
Ans (a)
n 

z cos
= __________
2 

z2
a) 2
( z + 1)
z
c) 2
( z + 1)
b)
Remember)
1 mark
z
( z − 1)
z2
d) 2
( z − 1)
Remember)
(CLO-5
Ans (a)
Remember)
13.
−1 
z 
z 
 = __________
z − a
a n +1
a)
14.
c)
na n +1
na
b) na
n
d ) na
c)
−1
a n +1
a
Ans (d)
n
d) a
(CLO-5
n −1
Ans (d)
 
1
n +1
b)
c)
n
1
(n + 1)!
1
d)
n!
n −1
b)
(CLO-5
Ans (d)
n +1
d)
1
n
c)
na n +1
na n
(CLO-5
Ans (a)
Remember)
1 mark
b) na
(CLO-5
d ) na n−1
Ans (c)
19.
Remember)
1 mark
−1 
az 
z 
= __________
2 
 ( z − 1) 
a)
Remember)
1 mark
−1 
z 
z 
= __________
2 
 ( z − 1) 
a)
Remember)
1 mark
b) a
1
c)
(n − 1)!
18.
(CLO-5
z e1 / z = __________
a)
17.
n −1
Remember)
1 mark
−1 
1 
z 
 = __________
 ( z − a) 
a)
16.
Ans (c)
−1 
z 
z 
= __________
2 
 ( z − a) 
a)
15.
(CLO-5
b) a
d ) a n −1
an
c)
1 mark
−1 
z 
z 
 = __________
 ( z + 1) 
Remember)
1 mark
(−1) n
a)
c)
b)
(−1) n −1
(−1) n +1
n(−1) n
a)
(CLO-5
Ans (a)
20.
1 mark
What is z[ f (n) * g (n) ]
a) F ( z ).G −1 ( z )
b) F −1 ( z ).G −1 ( z )
c) F −1 ( z ).G( z )
d ) F ( z ).G( z )
(CLO-5
Ans (d)
(
Z a n .n
21.
az
( z + a) 2
az
(c)
( z − a) 2

Z 5.3 n − 2(−1) n
z
( z − a) 2
z
(d)
a( z − a) 2

 z 
 z 
(b) 4
 + 2

 z − 3
 z − 1
 z   z 
(d) 5
+

 z − 3  z − 2
Ans (c)
Remember)
(CLO-5
Ans (a)
Remember)
1
Z 
n
 z 
(a) z log 

 z − 1
 z − 1
(b) z log 

 z 
 z −1
(c) log 

 z 
 z 
(d) log 

 z − 1
24.
(CLO-5
(b)
 z   z 
(a) 5
 − 2

 z − 3   z + 1
 z 
 z 
(c) 5
 + 2

 z − 4
 z + 1
23.
Remember)
)
(a)
22.
Remember)




1
 residue at z = 1
Z −1 
1
1 
2

 ( z − )( z − ) 
2
3 

(CLO-5
Remember)
Ans (d)
6
(a)
2
(d) −
25.
6
(b)
n +1
2
(c)
6
2
n+2
(CLO-5
6
Ans (b)
2 n +1
Remember)


8z 2

Z −1 
 (2 z − 1)( 4 z − 1) 
n
1 1
(a)     
2 4
1
n
(b) (1)   
4
n
1
(d)  
2
n −1
n
(CLO-5
n
Ans (a)
1 1
(c)     
8  4
26.
n −1
Remember)
n
1
 
4
n
y (n + 2) − 4 y (n + 1) + 4 y (n) = 0 where y (0) = 1, y (1) = 0
(a) Y ( z )( z 2 − 4 z + 4) = z 2 + 4 z
(b) Y ( z )( z 2 + 4 z + 4) = z 2 − 4 z
(CLO-5
(c) Y ( z )( z 2 − 4 z + 4) = z 2 − 4 z
Ans (c)
(d) Y ( z )( z 2 − 4 z − 4) = z 2 + 4 z
27.
Remember)
Z [t] is
n =
(a) Z t  = T  n z
n =1
n =
(b) Z t  = T  n z −n
(CLO-5
Ans (b)
n =1
n =
Remember)
1
(c) Z t  = T  z − n
n =1 n
n =
(d) Z t  = T  z − n
n =1
28.


Z − n 2 is
(a) −
z ( z + 1)
(z − 1)
Tz
(d)
(z − 1)2
3
(b)
z
(z − 1)2
(c)
z ( z + 1)
(z − 1)3
(CLO-5
Ans (a)
Remember)
29.
(
)
Z e 3t −7 is
(a)
e7 z
z + e − 3t
(b)
e −7 z
z − e − 3t
e −7
z − e −t
(c)
(d)
e7
z + e −t
30.
(CLO-5
Ans (b)
Remember)
Z- Transform formula
n =
(a) Z  f (n) =  f (n) z −n
n =1
n =
(b) Z  f (n) =  ( f (n) z )
−n
(CLO-5
Ans (a)
n =1
n =
(c) Z  f (n) =  z
Remember)
n
n =1
n =
(d) Z  f (n) =  f (n) z −n
n =1
Inverse Z transform of
31.
z2
is
1

 z −  (z − 4)
2

  1 n  1 n 
(b) 2  +   
  2   4  
 1  n  1  n 
(d)   +   
 2   4  
  1 n  1 n 
(a) 2  −   
  2   4  
 1  n  1  n 
(c)   −   
 2   4  
(CLO-5
Ans (a)
Remember)
( )
Z e t +7 is
32.
(a)
e10 z
z + e −t
(b)
e7 z
z − et
e10
z − e −t
(c)
(d)
e10
z + e −t
(CLO-5
Ans (b)
Remember)
Z n(2n − 1) is
33.
(a)
2z
(z + 1)2
(b)
z ( z + 3)
(z − 1)
3
(c)
2z
(z − 1)
3
(d)
z
(z − 1)
(CLO-5
Ans (b)
Z n − 1 is
34.
Remember)
z
( z − 1) 2
2 z (1 − z )
c)
( z − 1) 2
a)
b)
− 2z
( z + 1) 2
z
d)
( z + 1) 2
(CLO-5
Ans (c)
Remember)
z
inverse
Ztransform
of
2
35.
z − 7 z + 10
1
1
+
3( z − 5) 3( z + 2)
1
1
(c)
−
3( z − 5) 3( z − 2)
(a) −
36.
1
1
−
3( z − 5) 3( z − 2)
1
1
(d) −
+
( z − 5) ( z − 2)
(b) −
(CLO-5
Ans (c)
Remember)
Z (sin t ) is
(a)
z sin T
z − 2 z cos T − 1
(b)
z sin T
z − 2 z cos T + 1
(c)
z 2 sin T
z 2 − 2 z cos T + 1
(d)
z sin T
2
z + 2 z cos T + 1
2
2
(CLO-5
Ans (b)
Remember)
SRM Institute of Science and Technology
DEPARTMENT OF MATHEMATICS
Transforms and Boundary Value Problems
1. The order and degree of the PDE
(A) 2, 1
(B) 1, 2
∂2z
∂x2
∂z 2
+ 2xy( ∂x
) +
∂z
∂y
(C) 1, 1
= 5, respectively
(D) 2, 2
ANSWER: A
2. The Partial Differential Equation corresponding to Z = (x + a)(y + b) is
(A) p2 + q 2 = z
(B) pq = z
(C) p2 − q 2 = z
(D) z = p2 q 2
ANSWER: B
√
√
3. The complete solution of p + q = 1 is
√
√
(A) z = ax + (1 + a)2 y + c
(B) z = ax + (1 − a)2 y
√
√
(D) z = ax − (1 − a)2 y + c
(C) z = ax + (1 − a)2 y + c
ANSWER: C
4. The general solution of px + qy = z is
(A) f (x, y) = 0
(B) f ( xy , yz ) = 0
(C) f (xy, yz) = 0 (D) f (x2 + y 2 ) = 0
ANSWER: B
5. The general solution of (y − z)p + (z − x)q = x − y is
(A) f (x + y + z) = x2 + y 2 + z 2
(B) f (xyz) = x2 + y 2 + z 2
(C) f (x + y + z) = xyz
(D) f (x2 + y 2 + z 2 ) = x2 y 2 z 2
ANSWER: A
6. The complete solution of z = px + qy + p2 + q 2 is
(A) z = (x + a)(y + b)
(B) z = ax + by + c
(C) z = ax + by + c2 + d2
(D) z = ax + by + a2 + b2
ANSWER: D
7. The solution of the linear PDE (D2 + 4DD0 − 5D02 )z = 0 is
(A) z = f1 (y + x) + f2 (y + 5x)
(B) z = f1 (y − x) + f2 (y − 5x)
(C) z = f1 (y + x) + f2 (y − 5x)
(D) z = f1 (y − x) + f2 (y + 5x)
ANSWER: C
8. The solution of
∂3z
∂x3 =
2
0 is
(A) z = (1 + x + x )f (y)
(B) z = (1 + y + y 2 )f (x)
(C) z = f1 (y) + xf2 (y) + x2 f3 (y)
(D) z = f1 (x) + yf2 (x) + y 2 f3 (x)
ANSWER: C
9. The solution of p + q = z is
(A) f (x + y, y + logz)
(B) f (xy, ylogz)
(C) f (x − y, y − logz)
(D) f (xy, y − logz)
ANSWER: C
10. The particular solution of (D2 − 2DD0 + D02 )z = sinx
(A) −sinx
(B) sinx
(C) cosx
(D) −cosx
ANSWER: A
11. The period of sin5x is
(A)
8π
5
(B)
6π
5
(C)
4π
5
(D)
2π
5
ANSWER: D
12. If f (x) = xsinx in (−π, π) then the value of bn in Fourier series expansion is
(A) 0
(B) 1
(C) 2
(D) 3
ANSWER: A
13. Fourier coefficient a0 in the Fourier series expansion of a function represents
the
(A) maximum value of the function
(B) 2 mean value of the function
(C) minimum value of the function
(D) mean value of the function
ANSWER: B
14. If the Fourier series of the function f (x) in (−`, `) has only cosine terms then
f (x) must be
(A) odd function
(B) even function
(C) neither even nor odd function
(D) multi-valued function
ANSWER: B
15. If f (x) = x2 + x in (0, `) then the even extension in (−`, 0) is
(A) −x2 − x
(B) −x2 + x
(C) x2 + x
(D) x2 − x
ANSWER: D
16. Compute the constant term
lowing data:
x
f(x)
(A) 8.7
ANSWER: A
a0
2
of the Fourier series of f (x) given by the fol-
π
2π
4π
5π
0
π
2π
3
3
3
3
1.0 1.4 1.9 1.7 1.5 1.2 1.0
(B) 9.7
(C) 2.9
(D) 1.45
x
f(x)
0 1 2
9 18 24
3 4 5
28 26 20
17. Compute a1 of the Fourier series of f (x) given by the table:
(A) -8.33
(B) -25
(C) -1.155
(D) 0.519
ANSWER: A
18. The root mean square value of the function f (x) over the interval (a, b) then
ȳ =
qR
q
Rb
b
1
2
2
(A)
|
f
(x)
|
dx
(B)
a
b−a a | f (x) | dx
q
q
Rb
Rb
1
1
2
(C) a−b a | f (x) | dx
(D) b−a
a | f (x) | dx
ANSWER: B
19. The period of tan2x is
(A)
2π
n
(B)
π
n
(C)
π
2
(D)
2
π
ANSWER: C
20. The Fourier cosine series of the function f (t) = sin( πt` ), 0 < t < ` then the
value of a0 is
(A)
1
π
(B)
2
π
(C)
3
π
(D)
4
π
ANSWER: D
21. In one dimensional wave equation
(A)
T
m
(B)
k
c
∂2y
∂t2
2
∂ y
2
= a2 ∂x
stands for
2, a
(C)
m
T
(D)
k
m
ANSWER: A
22. One dimensional wave equation is used to find the
(A) time
(B) displacement (C) heat flow
(D) mass
ANSWER: B
23. Heat flows from
(A) higher to lower temperature
(B) lower to higher temperature
(C) constant temperature
(D) uniform temperature
ANSWER: A
24. The steady state temperature of the rod of length 20cm whose ends are kept at
30C and 80C is
(A) 30 − 52 x
ANSWER: D
(B) 30 + 25 x
(C) 10 + 25 x
(D) 30 + 52 x
25. The tension T caused by stretching the string before fixing it at the end points
is
(A) decreasing
(B) increasing
(C) zero
(D) constant
ANSWER: D
26. The amount of heat required to produce a given temperature change in a body
is proportional to the
(A) mass of the body
(B) weight of the body
(C) density of the body
(D) Tension of the body
ANSWER: A
27. The one dimensional heat equation is of the form
(A)
(C)
∂2u
∂t2
∂2u
∂x2
2
= a2 ∂∂xu2
(B)
∂2u
∂y 2
(D)
+
=0
∂u
2 ∂2u
=
α
∂t
∂x2
∂2u
∂2u
∂x2 + ∂y 2 =
f (x, y)
ANSWER: B
28. The proper solution of one dimensional heat flow equation is u(x, t)
(B) (Aeλx + Be−λx )eα
(A) Ax + B
(C) (A cos px + B sin px)e−α
2 2
λ t
2 2
λ t
(D) (Aeλx + Be−λx )(Ceλat + De−λat )
ANSWER: C
29. The slope of the deflection curve in vibrating string is assumed to be
(A) small at all points and at all times
(B) large at all points and at all times
(C) small at all points but not at all times
(D) large at all points but not at all times
ANSWER: A
30. How many initial and boundary conditions are required to solve the equation
∂2u
2 ∂2u
∂t2 = a ∂x2
(A) 1
(B) 2
(C) 3
(D) 4
ANSWER: D
31. If F [f (x)] = F (s), then F [f (x − a)] =
(A) eisa F (s)
(B) e−isa F (s)
(C) eisx F (s)
(D) eis(x−a) F (s)
ANSWER: A
32. If F [f (x)] = F (s), then F [f (ax)] =
(A) F ( as )
ANSWER: D
(B) F ( as )
(C)
1
a
|a| F ( s )
(D)
1
s
|a| F ( a )
33. If F [f (x)] = F (s), then F [eiax f (x)] =
(A) F (s − a)
(B) F(s+a)
(C) eisa F (s)
(D) e−isa F (s)
ANSWER: B
34. If f (x) = e−ax , then Fourier sine transform of f (x) is
q
q
p a
2 a
s
(A) π s2 +a2
(B) π2 s2 +a
(C) π2 s2 +a
2
2
(D)
pπ
s
2 s2 +a2
ANSWER: B
35. If f (x) = e−ax , then Fourier cosine transform of f (x) is
q
q
p a
p s
s
2 a
(A) π s2 +a2
(B) π2 s2 +a
(C) π2 s2 +a
(D) π2 s2 +a
2
2
2
ANSWER: A
36. If f (x) = x1 , then Fourier sine transform of f (x) is
q
pπ
(A) 2
(B) π2
(C) π2
(D)
2
π
ANSWER: A
37. Under Fourier cosine transform f (x) =
√1
x
is
(A) cosine function
(B) sine function
(C) self reciprocal function
(D) complex function
ANSWER: C
38. If F [f (x)] = F (s), then
R∞
(A) −∞ | Fs (s) |2 ds
R∞
(C) 0 | F (s) |2 ds
R∞
−∞
| f (x) |2 dx =
R∞
(B) −∞ | Fc (s) |2 ds
R∞
(D) −∞ | F (s) |2 ds
ANSWER: D
39. The Fourier transform of a function f (x) is
R∞
R∞
(A) √12π −∞ f (x)eisx dx
(B) √12π −∞ F (s)e−isx ds
R∞
R∞
(D) √12π 0 f (x)eisx dx
(C) √1π −∞ f (x)eisx dx
ANSWER: A
40. The Fourier cosine transform of 5e−2x is
q
q
q
2 10
2 2
(A) π s2 +4
(B) π s2 +4
(C) π2 s2s+4
(D)
q
2 5s
π s2 +4
ANSWER: A
41. If Z[f (n)] = F (z), then Z( 31n ) =
(A)
z
z−1
ANSWER: B
(B)
3z
3z−1
(C)
z
z−3
(D)
z
z−3n
42. If Z[f (n)] = F (z), then Z(3n sin nπ
2 ) =
(A)
z2
(B)
z 2 +9
z
z 2 +9
(C)
z
3
z 2
( 3 ) +1
(D)
z
z−3n
(C)
az
(z−a)2
(D)
z 2 +z
(z−1)3
ANSWER: C
43. If Z[f (n)] = F (z), then Z[an n] =
(A)
z
(z−a)2
(B)
az 2 +a2 z
(z−a)3
ANSWER: C
z
44. If Z[f (n)] = F (z), then Z −1 [ (z−1)(z−2)
]=
(A) 1 − 2n
(B) 2n + 1
(C) −2n − 1
(D) 2n − 1
(C) n2 an
(D) (−a)n
ANSWER: D
z
45. If Z[f (n)] = F (z), then Z −1 [ z−a
]=
(A) an
(B) nan
ANSWER: A
46. If Z[f (n)] = F (z), then the poles of F (z) =
z
(z−1)(z−2)
are
(A) z = −1, z = 2
(B) z = 1, z = 2
(C) z = 1, z = −2
(D) z = −1, z = −2
ANSWER: B
47. If F (z)z n−1 =
zn
(z−1)(z−2) ,
(A) 1, 2n
then the residue of F (z)z n−1 at each pole, respectively
(B) −1, 2n
(C) 1, (−2)n
(D) −1, −2
ANSWER: B
48. If Z[f (n)] = F (z), then Z[(−3)n ] =
z
z
(A) (z−3)
(B) z+3
(C)
2
z
(z+3)2
(D)
z
z−3
(C)
z
z+1
(D)
z
z−1
(C)
z
z−e−5
(D)
z
z+e5
ANSWER: B
49. If Z[f (n)] = F (z), then Z[K] =
(A)
Kz
z−1
(B)
Kz
z+1
ANSWER:A
50. If Z[f (n)] = F (z), then Z[e−5n ] =
(A)
z
z+e−5
(B)
z
z−e5
ANSWER:C
51. If Z[f (n)] = F (z), then Z[ n!1 ] =
1
(A) e− z
ANSWER: C
(B) ez
1
(C) e z
(D) e−z
2
z
52. If Z[f (n)] = F (z), then Z −1 [ (z−a)
2] =
(A) (n + 1)(−a)n (B) (n − 1)(−a)n (C) (n + 1)(a)n
(D) (n − 1)(a)n
ANSWER: C
53. If Z[f (n)] = F (z), then Z[an cos nπ
2 ] =
(A)
az 2
z 2 +a2
ANSWER: B
(B)
z2
z 2 +a2
(C)
az 2
z 2 −a2
(D)
az
z 2 +a2
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UNIT– I
PARTIAL DIFFERENTIAL EQUATIONS
This unit covers topics that explain the formation of partial differential equations
and the solutions of special types of partial differential equations.
1.1 INTRODUCTION
A partial differential equation is one which involves one or more partial
derivatives. The order of the highest derivative is called the order of the equation. A
partial differential equation contains more than one independent variable. But, here we
shall consider partial differential equations involving one dependent variable „z‟ and only
two independent variables x and y so that z = f(x,y). We shall denote
z
------- = p,
x
z
2z
2z
2z
----------- = q, ---------- = r, ---------- = s, ---------- = t.
xy
y2
y
x2
A partial differential equation is linear if it is of the first degree in the dependent
variable and its partial derivatives. If each term of such an equation contains either the
dependent variable or one of its derivatives, the equation is said to be homogeneous,
otherwise it is non homogeneous.
1.2 Formation of Partial Differential Equations
Partial differential equations can be obtained by the elimination of arbitrary constants or
by the elimination of arbitrary functions.
By the elimination of arbitrary constants
Let us consider the function
 ( x, y, z, a, b ) = 0 -------------(1)
where a & b are arbitrary constants
Differentiating equation (1) partially w.r.t x & y, we get
∂
∂
+
p
∂x
∂
+ q
∂y
= 0
_________________ (2)
= 0
_________________ (3)
∂z
∂
∂z
Eliminating a and b from equations (1), (2) and (3), we get a partial differential
equation of the first order of the form f (x,y,z, p, q) = 0
Example 1
Eliminate the arbitrary constants a & b from z = ax + by + ab
Consider
z = ax + by + ab
____________ (1)
Differentiating (1) partially w.r.t x & y, we get
∂z
= a
i.e, p= a
= b
i.e, q = b
__________ (2)
∂x
∂z
________
(3)
∂y
Using (2) & (3) in (1), we get
z = px +qy+ pq
which is the required partial differential equation.
Example 2
Form the partial differential equation by eliminating the arbitrary constants a and b
from
z = ( x2 +a2 ) ( y2 + b2)
Given z = ( x2 +a2 ) ( y2 + b2)
__________ (1)
Differentiating (1) partially w.r.t x & y , we get
p = 2x (y2 + b2 )
q = 2y (x + a )
Substituting the values of p and q in (1), we get
4xyz = pq
which is the required partial differential equation.
Example 3
Find the partial differential equation of the family of spheres of radius one whose centre
lie in the xy - plane.
The equation of the sphere is given by
( x – a )2 + ( y- b) 2 + z2
=1
_____________ (1)
Differentiating (1) partially w.r.t x & y , we get
2 (x-a ) + 2 zp =
2 ( y-b ) + 2 zq =
0
0
From these equations we obtain
x-a = -zp _________ (2)
y -b = -zq _________ (3)
Using (2) and (3) in (1), we get
z2p2 + z2q2 + z2 = 1
or z2 ( p2 + q2 + 1) = 1
Example 4
Eliminate the arbitrary constants a, b & c from
x2
y2
+
2
z2
+
2
a
= 1 and form the partial differential equation.
2
b
c
The given equation is
x2
y2
+
2
a
z2
+
b
2
=1
2
c
_____________________________ (1)
Differentiating (1) partially w.r.t x & y, we get
2x
2zp
+
= 0
a2
c2
2y
2zq
+
b2
= 0
c2
Therefore we get
x
zp
+
a2
y
= 0 _________________ (2)
c2
zq
+
= 0
2
____________________ (3)
2
b
c
Again differentiating (2) partially w.r.t „x‟, we set
(1/a2 ) + (1/ c2 ) ( zr + p2 ) = 0
_______________ (4)
Multiplying ( 4) by x, we get
x
p2x
xz r
+
a2
+
c2
=0
c2
From (2) , we have
zp
+
c2
p2x
xzr
+
c2
=0
c2
or -zp + xzr + p2x = 0
By the elimination of arbitrary functions
Let u and v be any two functions of x, y, z and Φ(u, v ) = 0, where Φ is an
arbitrary function. This relation can be expressed as
u = f(v) ______________ (1)
Differentiating (1) partially w.r.t x & y and eliminating the arbitrary
functions from these relations, we get a partial differential equation of the first order
of the form
f(x, y, z, p, q ) = 0.
Example 5
Obtain the partial differential equation by eliminating „f „ from z = ( x+y ) f ( x2 - y2 )
Let us now consider the equation
z = (x+y ) f(x2- y2) _____________ (1)
Differentiating (1) partially w.r.t x & y , we get
p = ( x + y ) f ' ( x2 - y2 ) . 2x + f ( x2 - y2 )
q = ( x + y ) f ' ( x2 - y2 ) . (-2y) + f ( x2 - y2 )
These equations can be written as
p - f ( x2 - y2 ) = ( x + y ) f '( x2 - y2 ) . 2x ____________ (2)
q - f ( x2 - y2 ) = ( x + y ) f '( x2 - y2 ) .(-2y) ____________ (3)
Hence, we get
p - f ( x2 - y2 )
x
= -
q - f ( x2 - y2 )
i.e,
i.e,
y
py - yf( x2 - y2 ) = -qx +xf ( x2 - y2 )
py +qx
= ( x+y ) f ( x2 - y2 )
Therefore, we have by(1),
py +qx = z
Example 6
Form the partial differential equation by eliminating the arbitrary function f
from
z = ey f (x + y)
Consider z = ey f ( x +y ) ___________ ( 1)
Differentiating (1) partially w .r. t x & y, we get
p = ey f ' (x + y)
q = ey f '(x + y) + f(x + y). ey
Hence, we have
q=p+z
Example 7
Form the PDE by eliminating f & Φ from z = f (x +ay ) + Φ ( x – ay)
Consider
z = f (x +ay ) + Φ ( x – ay) _______________ (1)
Differentiating (1) partially w.r.t x &y , we get
p = f '(x +ay ) + Φ' (x – ay)
___________________ (2)
q = f ' (x +ay ) .a + Φ' (x – ay) ( -a) _______________ (3)
Differentiating (2) & (3) again partially w.r.t x & y, we get
r = f "( x+ay) + Φ "( x – ay)
t = f "( x+ay) .a2 + Φ"( x – ay) (-a)2
i.e,
t = a2 { f"( x + ay) + Φ"( x – ay)}
or
t = a2r
Exercises:
1.
„b‟
Form the partial differential equation by eliminating the arbitrary constants „a‟ &
from the following equations.
(i)
z = ax + by
(ii)
x2 + y2
z2
+
=1
a2
b2
(iii) z = ax + by + a2 + b2
(iv)
ax2 + by2 + cz2 = 1
(v)
z = a2x + b2y + ab
2.
Find the PDE of the family of spheres of radius 1 having their centres lie on the
xy plane{Hint: (x – a)2 + (y – b)2 + z2 = 1}
3.
Find the PDE of all spheres whose centre lie on the (i) z axis (ii) x-axis
4.
Form the partial differential equations by eliminating the arbitrary functions in the
following cases.
(i)
z = f (x + y)
(ii)
z = f (x2 – y2)
(iii) z = f (x2 + y2 + z2)
(iv)
 (xyz, x + y + z) = 0
(v)
(vi)
(vii)
z = x + y + f(xy)
z = xy + f (x2 + y2)
z = f xy
z
(viii) F (xy + z2, x + y + z) = 0
(ix)
z = f (x + iy) +f (x – iy)
(x)
z = f(x3 + 2y) +g(x3 – 2y)
1.3 SOLUTIONS OF A PARTIAL DIFFERENTIAL EQUATION
A solution or integral of a partial differential equation is a relation connecting the
dependent and the independent variables which satisfies the given differential equation. A
partial differential equation can result both from elimination of arbitrary constants and
from elimination of arbitrary functions as explained in section 1.2. But, there is a basic
difference in the two forms of solutions.
A solution containing as many arbitrary
constants as there are independent variables is called a complete integral. Here, the partial
differential equations contain only two independent variables so that the complete
integral will include two constants.A solution obtained by giving particular values to the
arbitrary constants in a complete integral is called a particular integral.
Singular Integral
Let f (x,y,z,p,q) = 0 ---------- (1)
be the partial differential equation whose complete integral is
 (x,y,z,a,b) = 0
----------- (2)
where „a‟ and „b‟ are arbitrary constants.
Differentiating (2) partially w.r.t. a and b, we obtain

-------- = 0
a

and
--------- = 0
b
----------- (3)
----------- (4)
The eliminant of „a‟ and „b‟ from the equations (2), (3) and (4), when it exists, is
called the singular integral of (1).
General Integral
In the complete integral (2), put b = F(a), we get
 (x,y,z,a, F(a) ) = 0
---------- (5)
Differentiating (2), partially w.r.t.a, we get


------- + -------- F'(a) = 0
a
b
-------- (6)
The eliminant of „a‟ between (5) and (6), if it exists, is called the general integral of (1).
SOLUTION OF STANDARD TYPES OF FIRST ORDER PARTIAL
DIFFERENTIAL EQUATIONS.
The first order partial differential equation can be written as
f(x,y,z, p,q) = 0,
where p = z/x and q = z / y. In this section, we shall solve some standard forms
of equations by special methods.
Standard I : f (p,q) = 0. i.e, equations containing p and q only.
Suppose that z = ax + by +c is a solution of the equation f(p,q) = 0, where f (a,b)
= 0.
Solving this for b, we get b = F (a).
Hence the complete integral is z = ax + F(a) y +c
------------ (1)
Now, the singular integral is obtained by eliminating a & c between
z = ax + y F(a) + c
0 = x + y F'(a)
0 = 1.
The last equation being absurd, the singular integral does not exist in this case.
To obtain the general integral, let us take c =  (a).
z = ax + F(a) y +  (a)
Then,
-------------- (2)
Differentiating (2) partially w.r.t. a, we get
0 = x + F'(a). y + '(a)
--------------- (3)
Eliminating „a‟ between (2) and (3), we get the general integral
Example 8
Solve pq = 2
The given equation is of the form f (p,q) = 0
The solution is z = ax + by +c, where ab = 2.
2
Solving,
b = ------.
a
The complete integral is
2
Z = ax + ------ y + c
a
---------- (1)
Differentiating (1) partially w.r.t „c‟, we get
0 = 1,
which is absurd. Hence, there is no singular integral.
To find the general integral, put c =  (a) in (1), we get
2
Z = ax + ------ y +  (a)
a
Differentiating partially w.r.t „a‟, we get
2
0 = x – ------ y + (a)
a2
Eliminating „a‟ between these equations gives the general integral.
Example 9
Solve pq + p +q = 0
The given equation is of the form f (p,q) = 0.
The solution is z = ax + by +c, where ab + a + b = 0.
Solving, we get
b= –
a
-------1+a
a
Hence the complete Integral is z = ax – ------1+a
y+c
------ (1)
Differentiating (1) partially w.r.t. „c‟, we get
0 = 1.
The above equation being absurd, there is no singular integral for the given partial
differential equation.
To find the general integral, put c =  (a) in (1), we have
a
z = ax – --------1 +a
y +  (a)
------(2)
Differentiating (2) partially w.r.t a, we get
1
0 = x – -------- y + (a)
(1 + a)2
1
----- (3)
Eliminating „a‟ between (2) and (3) gives the general integral.
Example 10
Solve p2 + q2 = npq
The solution of this equation is z = ax + by + c, where a2 + b2 = nab.
Solving, we get
b= a
n + (n2 – 4)
------------------
2
Hence the complete integral is
n + n2 – 4
z =ax +a -----------------2
y+c
-----------(1)
Differentiating (1) partially w.r.t c, we get 0 = 1, which is absurd. Therefore, there is no
singular integral for the given equation.
To find the general Integral, put C =  (a), we get
z = ax +
a
n + n2 – 4
------------------- y +  (a)
2
Differentiating partially w.r.t „a‟, we have
n + n2 – 4
0 = x + ------------------2
y +  (a)
The eliminant of „a‟ between these equations gives the general integral
Standard II : Equations of the form f (x,p,q) = 0, f (y,p,q) = 0 and f (z,p,q) = 0.
i.e, one of the variables x,y,z occurs explicitly.
(i)
Let us consider the equation f (x,p,q) = 0.
Since z is a function of x and y, we have
z
z
dz = ------- dx + -------- dy
x
y
or
dz = pdx + qdy
Assume that q = a.
Then the given equation takes the form f (x, p,a ) = 0
Solving, we get
Therefore,
p = (x,a).
dz = (x,a) dx + a dy.
Integrating, z =  (x,a) dx + ay + b which is a complete Integral.
(ii) Let us consider the equation f(y,p,q) = 0.
Assume that p = a.
Then the equation becomes f (y,a, q) = 0
Solving, we get q =  (y,a).
Therefore, dz = adx + (y,a) dy.
Integrating, z = ax + (y,a) dy + b, which is a complete Integral.
(iii) Let us consider the equation f(z, p, q) = 0.
Assume that q = ap.
Then the equation becomes f (z, p, ap) = 0
Solving, we get p = (z,a). Hence dz = (z,a) dx + a (z, a) dy.
dz
ie, ----------- = dx + ady.
 (z,a)
Integrating,
dz
 ----------- = x + ay + b, which is a complete Integral.
 (z,a)
Example 11
Solve q = xp + p2
Given
q = xp + p2 -------------(1)
This is of the form f (x,p,q) = 0.
Put q = a in (1), we get
a = xp + p2
i.e, p2 + xp – a = 0.
Therefore,
-x +(x2 + 4a)
p = -------------------2
Integrating ,
– x ± x2 + 4a
z =  -------------------2
Thus,
x2
x
x
z = – ------ ± ------ (4a + x2)+ a sin h–1 ----4
4
2a
Example 12
Solve q = yp2
This is of the form f (y,p,q) = 0
Then, put p = a.
Therfore, the given equation becomes q = a2y.
Since dz = pdx + qdy, we have
dz = adx + a2y dy
a2y2
Integrating, we get z = ax + ------- + b
2
Example 13
Solve 9 (p2z + q2) = 4
This is of the form f (z,p,q) = 0
Then, putting q = ap, the given equation becomes
9 (p2z + a2p2) = 4
Therefore,
2
p = ± ---------3 (z + a2)
and
2a
q = ± ---------3 (z + a2)
Since dz = pdx + qdy,
dx + ay + b
+ ay + b
2
dz = ± -----3
1
2
1
---------- dx ± -------- a ---------- dy
z + a2
3
z + a2
Multiplying both sides by z + a2, we get
2
2
z + a dz = ------ dx + ------ a dy , which on integration gives,
3
3
2
(z+a2)3/2
2
2
------------- = ------ x + ------ ay + b.
3/2
3
3
(z + a2)3/2 = x + ay + b.
or
Standard III : f1(x,p) = f2 (y,q). ie, equations in which ‘z’ is absent and the variables
are
separable.
Let us assume as a trivial solution that
f(x,p) = g(y,q) = a (say).
Solving for p and q, we get p = F(x,a) and q = G(y,a).
But
z
z
dz = -------- dx + ------- dy
x
y
Hence dz = pdx + qdy = F(x,a) dx + G(y,a) dy
Therefore, z = F(x,a) dx +  G(y,a) dy + b , which is the complete integral of the given
equation containing two constants a and b. The singular and general integrals are found in
the usual way.
Example 14
Solve pq = xy
The given equation can be written as
p
y
----- = ------ = a (say)
x
q
Therefore,
and
p
----- = a
x
y
------ = a
q
implies
p = ax
y
implies q = ----a
Since dz = pdx + qdy, we have
y
dz = axdx + ------ dy, which on integration gives.
a
ax2
y2
z = ------- + ------- + b
2
2a
Example 15
Solve p2 + q2 = x2 + y2
The given equation can be written as
p2 – x2 = y2 – q2 = a2 (say)
and
But
p2 – x2 = a2
implies p = (a2 + x2)
y2 – q2 = a2
implies q = (y2 – a2)
dz = pdx + qdy
ie, dz =  a2 + x2 dx + y2 – a2 dy
Integrating, we get
x
a2
x
y
a2
y
2
2
–1
2
2
-1
z = ----x + a + ----- sinh
----- + ----y – a – ----- cosh ----- + b
2
2
a
2
2
a
Standard IV (Clairaut’s form)
Equation of the type z = px + qy + f (p,q) ------(1) is known as Clairaut‟s form.
Differentiating (1) partially w.r.t x and y, we get
p = a and
q = b.
Therefore, the complete integral is given by
z = ax + by + f (a,b).
Example 16
Solve z = px + qy +pq
The given equation is in Clairaut‟s form.
Putting p = a and q = b, we have
z = ax + by + ab
-------- (1)
which is the complete integral.
To find the singular integral, differentiating (1) partially w.r.t a and b, we get
0=x+b
0=y+a
Therefore we have, a = -y and b= -x.
Substituting the values of a & b in (1), we get
z = -xy – xy + xy
or
z + xy = 0, which is the singular integral.
To get the general integral, put b = (a) in (1).
Then z = ax + (a)y + a (a)
---------- (2)
Differentiating (2) partially w.r.t a, we have
0 = x + '(a) y + a'(a) + (a)
---------- (3)
Eliminating „a‟ between (2) and (3), we get the general integral.
Example 17
Find the complete and singular solutions of z = px + qy +  1+ p2 + q2
The complete integral is given by
z = ax + by +  1+ a2 + b2
-------- (1)
To obtain the singular integral, differentiating (1) partially w.r.t a & b. Then,
a
0 = x + --------------------1 + a2 + b2
b
0 = y + --------------------1 + a2 + b2
Therefore,
–a
x = ----------------- -----------(2)
(1 + a2 + b2)
–b
y = ------------------------ (3)
2
2
(1 + a + b )
and
Squaring (2) & (3) and adding, we get
a2 + b2
x2 + y2 = -----------------1 + a2 + b2
Now,
i.e,
1
1 – x – y = ----------------1 + a2 + b2
1
1 +a2 + b2 = ---------------1 – x2 – y2
2
2
Therefore,
1
(1 +a2 + b2) = ---------------1 – x2 – y2
Using (4) in (2) & (3), we get
-----------(4)
x = – a 1 – x2 – y2
y = – b 1 – x2 – y2
and
-x
a = ------------1–x2–y2
Hence,
and
-y
b = ---------------1–x2–y2
Substituting the values of a & b in (1) , we get
- x2
y2
1
z = ------------- – -------------- + --------------1–x2–y2
1–x2–y2
1–x2–y2
which on simplification gives
z = 1 – x2 – y2
or
x2 + y2 + z2 = 1,
which is the singular integral.
Exercises
Solve the following Equations
1. pq = k
2. p + q = pq
3. p +q = x
4. p = y2q2
5. z = p2 + q2
6. p + q = x + y
7. p2z2 + q2 = 1
8. z = px + qy - 2pq
9. {z – (px + qy)}2 = c2 + p2 + q2
10. z = px + qy + p2q2
EQUATIONS REDUCIBLE TO THE STANDARD FORMS
Sometimes, it is possible to have non – linear partial differential equations of the
first order which do not belong to any of the four standard forms discussed earlier. By
changing the variables suitably, we will reduce them into any one of the four standard
forms.
Type (i) : Equations of the form F(xm p, ynq) = 0 (or) F (z, xmp, ynq) = 0.
Case(i) : If m  1 and n  1, then put x1-m = X and y1-n = Y.
z
z
X
z
Now, p = ----- = ------ . ------- = ------ (1-m) x -m
x
X
x
X
z
z
Therefore, x p = ------ (1-m) = (1 – m) P, where P = ------X
X
z
Similarly, ynq = (1-n)Q, where Q = -----Y
Hence, the given equation takes the form F(P,Q) = 0 (or) F(z,P,Q) = 0.
Case(ii) : If m = 1 and n = 1, then put log x = X and log y = Y.
m
z
z
X
z
1
Now, p = ----- = ------- . ------- = ------- -----x
X
x
X
x
z
Therefore, xp = ------ = P.
X
Similarly, yq =Q.
Example 18
Solve x4p2 + y2zq = 2z2
The given equation can be expressed as
(x2p)2 + (y2q)z = 2z2
Here m = 2, n = 2
Put X = x1-m = x -1 and Y = y 1-n = y -1.
We have xmp = (1-m) P and ynq = (1-n)Q
i.e,
x2p = -P and y2q = -Q.
Hence the given equation becomes
P2 – Qz = 2z2
----------(1)
This equation is of the form f (z,P,Q) = 0.
Let us take Q = aP.
Then equation (1) reduces to
P2 – aPz =2z2
a  (a2 + 8)
P = ----------------2
Hence,
z
a  (a2 + 8)
and
Q = a ---------------- z
2
Since dz = PdX + QdY, we have
a  (a2 + 8)
a  (a2 + 8)
dz = ---------------- z dX + a --------------- z dY
2
2
i.e,
dz
a  (a2 + 8)
------ = ---------------- (dX + a dY)
z
2
Integrating, we get
a   a2 + 8
log z = ---------------- (X + aY) +b
2
Therefore,
a  (a2 + 8)
log z = ---------------2
1
a
---- + ----- + b which is the complete solution.
x
y
Example 19
Solve x2p2 + y2q2 = z2
The given equation can be written as
(xp)2 + (yq)2 = z2
Here m = 1, n = 1.
Put X = log x and Y = log y.
Then xp = P and yq = Q.
Hence the given equation becomes
P2 + Q2 = z2 -------------(1)
This equation is of the form F(z,P,Q) = 0.
Therefore, let us assume that Q = aP.
Now, equation (1) becomes,
Hence
and
P2 + a2 P2 = z2
z
P = -------(1+a2)
az
Q = -------(1+a2)
Since dz = PdX + QdY, we have
z
az
dz = ----------dX + ---------- dY.
(1+a2)
(1+a2)
dz
2
i.e, (1+a ) ------ = dX + a dY.
z
Integrating, we get
(1+a2) log z = X + aY + b.
Therefore, (1+a2) log z = logx + alogy + b, which is the complete solution.
Type (ii) : Equations of the form F(zkp, zkq) = 0 (or) F(x, zkp) = G(y,zkq).
Case (i) : If k  -1, put Z = zk+1,
Z
Z
z
z
k
Now ------- = -------- ------- = (k+1)z . ------- = (k+1) zkp.
x
z
x
x
1
Z
Therefore, z p = ----- ------k+1
x
k
1
Z
Similarly, z q = ------- -----k+1
y
k
Case (ii) : If k = -1, put Z = log z.
Z
Z z
1
Now, ------- = -------- ------- = ----- p
x
z
x
z
Z
1
Similarly,
----- = ----- q.
y
z
Example 20
Solve z4q2 – z2p = 1
The given equation can also be written as
(z2q)2 – (z2p) =1
Here k = 2. Putting Z = z k+1 = z3, we get
1
Z
Z p = ------ -----k+1 x
k
1
Z
i.e, Z p = ------ -----3
x
2
and
1
Z
Z q = ------ -----k+1 y
and
1
Z
Z q = ------ -----3
y
k
2
Hence the given equation reduces to
Q 2
P
------  ------ = 1
3
3
i.e,
Q2 – 3P – 9 = 0,
which is of the form F(P,Q) = 0.
Hence its solution is Z = ax + by + c, where b2 – 3a – 9 = 0.
Solving for b, b = ± (3a +9)
Hence the complete solution is
Z = ax + (3a +9) . y + c
or
z3 = ax + (3a +9) y + c
Exercises
Solve the following equations.
1. x2p2 + y2p2 = z2
2. z2 (p2+q2) = x2 + y2
3. z2 (p2x2 + q2) = 1
4. 2x4p2 – yzq – 3z2 = 0
5. p2 + x2y2q2 = x2 z2
6. x2p + y2q = z2
7. x2/p + y2/q = z
8. z2 (p2 – q2) = 1
9. z2 (p2/x2 + q2/y2) = 1
10. p2x + q2y = z.
1.4 Lagrange’s Linear Equation
Equations of the form Pp + Qq = R ________ (1), where P, Q and R are
functions of x, y, z, are known as Lagrange‟s equations and are linear in „p‟ and „q‟.To
solve this equation, let us consider the equations u = a and v = b, where a, b are arbitrary
constants and u, v are functions of x, y, z.
Since „u ‟ is a constant, we have du = 0 -----------(2).
But „u‟ as a function of x, y, z,
∂u
du =
∂u
dx +
∂x
Comparing (2) and (3), we have
∂u
dx +
∂x
Similarly,
∂v
∂y
dz
∂z
∂u
∂u
dy +
∂y
= 0 ___________ (3)
dz
= 0 ___________ (4)
∂v
dy +
∂y
dz
∂z
∂v
dx +
∂x
∂u
dy +
∂z
By cross-multiplication, we have
dx
dy
dz
=
∂u
∂v
∂u
∂v
=
∂u
∂v
∂z
∂u
∂v
∂u
∂v
-
∂y
∂y
∂z
∂x
∂z
∂u
∂v
∂x
∂y
∂z
∂x
∂y
∂x
(or)
dx
dy
=
P
dz
=
______________ (5)
Q
R
Equations (5) represent a pair of simultaneous equations which are of the first
order and of first degree.Therefore, the two solutions of (5) are u = a and v = b. Thus,
( u, v ) = 0 is the required solution of (1).
Note :
To solve the Lagrange‟s equation,we have to form the subsidiary or auxiliary
equations
dx
dy
dz
=
=
P
Q
R
which can be solved either by the method of grouping or by the method of
multipliers.
Example 21
Find the general solution of px + qy = z.
Here, the subsidiary equations are
dx
dy
=
x
Taking the first two ratios ,
dz
=
y
dx
x
Integrating, log x = log y + log c1
z
=
dy
y
or
x = c1 y
i.e,
c1 = x / y
From the last two ratios,
dy
y
=
dz
z
Integrating, log y = log z + log c2
or
y = c2 z
i.e,
c2 = y / z
Hence the required general solution is
Φ( x/y, y/z) = 0, where Φ is arbitrary
Example 22
Solve p tan x + q tan y = tan z
The subsidiary equations are
dx
dy
=
dz
=
tanx
tany
tanz
Taking the first two ratios ,
ie,
dx
dy
=
tanx
tany
cotx dx = coty dy
Integrating, log sinx = log siny + log c1
ie, sinx = c1 siny
Therefore,
c1 = sinx / siny
Similarly, from the last two ratios, we get
siny = c2 sinz
i.e,
Hence the
c2 = siny / sinz
general solution is
sinx
Φ
siny
,
siny
= 0, where Φ is arbitrary.
sinz
Example 23
Solve (y-z) p + (z-x) q = x-y
Here the subsidiary equations are
dx
dy
=
y-z
dz
=
x –y
z- x
Using multipliers 1,1,1,
dx + dy + dz
each ratio =
0
Therefore, dx + dy + dz =0.
Integrating, x + y + z = c1 ____________ (1)
Again using multipliers x, y and z,
xdx + ydy + zdz
each ratio =
0
Therefore,
Integrating,
or
xdx + ydy + zdz = 0.
x2/2 + y2/2 +z2/2 = constant
x2 + y2 + z2 = c2 __________ (2)
Hence from (1) and (2), the general solution is
Φ ( x + y + z, x2 + y2 + z2) = 0
Example 24
Find the general solution of (mz - ny) p + (nx- lz)q = ly - mx.
Here the subsidiary equations are
dx
dy
dz
=
mz- ny
=
nx - lz
ly - mx
Using the multipliers x, y and z, we get
xdx + ydy + zdz
each fraction =
0
`  xdx + ydy + zdz = 0, which on integration gives
x2/2 + y2/2 +z2/2 = constant
x2 + y2 + z2 = c1 __________ (1)
or
Again using the multipliers l, m and n, we have
ldx + mdy + ndz
each fraction =
0
`  ldx + mdy + ndz = 0, which on integration gives
lx + my + nz = c2 __________ (2)
Hence, the required general solution is
Φ(x2 + y2 + z2 , lx + my + nz ) = 0
Example 25
Solve (x2 - y2 - z2 ) p + 2xy q = 2xz.
The subsidiary equations are
dx
x2-y2-z2
dy
dz
=
=
2xy
2xz
Taking the last two ratios,
dx
dz
=
2xy
2xz
2xy
2xz
dy
dz
ie,
=
y
Integrating, we get
or
z
log y = log z + log c1
y = c1z
i.e, c1 = y/z __________ (1)
Using multipliers x, y and z, we get
xdx + y dy + zdz
each fraction =
xdx + y dy + zdz
x (x2-y2-z2 )+2xy2+2xz2
=
x ( x2+ y2 + z2 )
Comparing with the last ratio, we get
xdx + y dy + zdz
dz
=
x ( x2+ y2 + z2 )
2xz
2xdx + 2ydy + 2zdz
dz
i.e,
=
x2+ y2 + z2
Integrating,
z
log ( x2+ y2 + z2 ) = log z + log c2
x2+ y2 + z2
or
= c2 z
x2+ y2 + z2
i.e,
c2 =
___________ (2)
z
From (1) and (2), the general solution is Φ(c1 , c2) = 0.
x2+ y2 + z2
i.e,
Φ (y/z) ,
= 0
z
Exercises
Solve the following equations
1. px2 + qy2 = z2
2. pyz + qzx = xy
3. xp – yq = y2 – x2
4. y2zp + x2zq = y2x
5. z (x – y) = px2 – qy2
6. (a – x) p + (b – y) q = c – z
7. (y2z p) /x + xzq = y2
8. (y2 + z2) p – xyq + xz = 0
9. x2p + y2q = (x + y) z
10. p – q = log (x+y)
11. (xz + yz)p + (xz – yz)q = x2 + y2
12. (y – z)p – (2x + y)q = 2x + z
1.5 PARTIAL DIFFERENTIAL EQUATIONS OF HIGHER ORDER WITH
CONSTANT COEFFICIENTS.
Homogeneous Linear Equations with constant Coefficients.
A homogeneous linear partial differential equation of the nth order is of the form
nz
nz
nz
c0 ------ + c1 ----------- + . . . . . . + cn -------- = F (x,y)
xn
xn-1y
yn
--------- (1)
where c0, c1,---------, cn are constants and F is a function of „x‟ and „y‟. It is
homogeneous because all its terms contain derivatives of the same order.
Equation (1) can be expressed as
or
(c0Dn + c1Dn-1 D' + ….. + cn D'n ) z = F (x,y)
f (D,D') z = F (x,y)
---------(2),


where, -----  D and -----  D' .
x
y
As in the case of ordinary linear equations with constant coefficients the complete
solution of (1) consists of two parts, namely, the complementary function and the
particular integral.
The complementary function is the complete solution of f (D,D') z = 0-------(3),
which must contain n arbitrary functions as the degree of the polynomial f(D,D'). The
particular integral is the particular solution of equation (2).
Finding the complementary function
Let us now consider the equation f(D,D') z = F (x,y)
The auxiliary equation of (3) is obtained by replacing D by m and D' by 1.
i.e, c0 mn + c1 mn-1 + ….. + cn = 0
---------(4)
Solving equation (4) for „m‟, we get „n‟ roots. Depending upon the nature of the roots,
the Complementary function is written as given below:
Roots of the auxiliary
Nature of the
equation
roots
m1,m2,m3 ……. ,mn
distinct roots
m1 = m2 = m, m3 ,m4,….,mn two equal roots
m1 = m2 = …….= mn = m
all equal roots
Complementary function(C.F)
f1 (y+m1x)+f2(y+m2x) + …….+fn(y+mnx).
f1(y+m1x)+xf2(y+m1x) + f3(y+m3x) + ….+
fn(y+mnx).
f1(y+mx)+xf2(y+mx) + x2f3(y+mx)+…..
+ …+xn-1 fn (y+mx)
Finding the particular Integral
Consider the equation f(D,D') z = F (x,y).
1
Now, the P.I is given by --------- F (x,y)
f(D,D')
Case (i) : When F(x,y) = eax +by
1
P.I = ----------- eax+by
f (D,D')
Replacing D by „a‟ and D' by „b‟, we have
1
P.I = ----------- eax+by,
where f (a,b) ≠ 0.
f (a,b)
Case (ii) : When F(x,y) = sin(ax + by) (or) cos (ax +by)
1
P.I = ----------------- sin (ax+by) or cos (ax+by)
f(D2,DD',D'2)
Replacing D2 = -a2, DD' 2 = -ab and D' = -b2, we get
1
P.I = ----------------- sin (ax+by) or cos (ax+by) , where f(-a2, - ab, -b2) ≠ 0.
f(-a2, - ab, -b2)
Case (iii) : When F(x,y) = xm yn,
1
P.I = ---------- xm yn = [f (D, D')]-1 xm yn
f(D,D')
Expand [f (D,D')]-1 in ascending powers of D or D' and operate on xm yn term by term.
Case (iv) : When F(x,y) is any function of x and y.
1
P.I = ---------- F (x,y).
f (D,D')
1
Resolve----------- into partial fractions considering f (D,D') as a function of D alone.
f (D,D')
Then operate each partial fraction on F(x,y) in such a way that
1
--------- F (x,y) =  F(x,c-mx) dx ,
D–mD'
where c is replaced by y+mx after integration
Example 26
Solve(D3 – 3D2D' + 4D'3) z = ex+2y
The auxillary equation is m=m3 – 3m2 + 4 = 0
The roots are m = -1,2,2
Therefore the C.F is f1(y-x) + f2 (y+ 2x) + xf3 (y+2x).
ex+2y
P.I.= ---------------------- (Replace D by 1 and D' by 2)
D3–3D2D'+4D'3
ex+2y
= ------------------1-3 (1)(2) + 4(2)3
ex+2y
= --------27
Hence, the solution is z = C.F. + P.I
ex+2y
ie,
z = f1 (y-x) + f2(y+2x) + x f3(y+2x) + ---------27
Example 27
Solve (D2 – 4DD' +4 D' 2) z = cos (x – 2y)
The auxiliary equation is m2 – 4m + 4 = 0
Solving, we get m = 2,2.
Therefore the C.F is f1(y+2x) + xf2(y+2x).
1
 P.I = ---------------------2 cos (x-2y)
D2 – 4DD' + 4D'
Replacing D2 by – 1, DD' by 2 and D' 2 by –4, we have
P.I
1
= ----------------------- cos (x-2y)
(-1) – 4 (2) + 4(-4)
cos (x-2y)
= – -------------25
 Solution is z = f1(y+2x) + xf2(y+2x) – --------------- .
25
Example 28
Solve (D2 – 2DD') z = x3y + e5x
The auxiliary equation is m2 – 2m = 0.
Solving, we get m = 0,2.
Hence the C.F is f1 (y) + f2 (y+2x).
x3y
P.I1 = --------------D2 – 2DD'
1
= --------------2D'
2
D 1– ------D
1
2D'
= ------- 1– ------D2
D
(x3y)
–1
(x3y)
2
1
2D'
4D'
= ----- 1 + ------- + ---------- + . . . . . (x3y)
D2
D
D2
1
2
4
2
3
'
3
= ------ (x y) + ----- D (x y) + ------ D' (x3y) + . . . . .
D2
D
D2
1
2
4
3
3
= ------- (x y) + ----- (x ) + ------ (0) + . . . . .
D2
D
D2
P.I1
1
2
= ------- (x3y) + ------ (x3)
D2
D3
P.I1
x5y
x6
= ------- + -----20
60
P.I2
e5x
= -------------- (Replace D by 5 and D' by 0)
D2 – 2DD'
e5x
= -----25
x5y
x6
e5x
Solution is Z = f1(y) + f2 (y+2x) + ------- + ------ + -----20
60
25
Example 29
2
Solve (D2 + DD' – 6 D‟) z = y cosx.
The auxiliary equation is m2 + m – 6 = 0.
Therefore, m = –3, 2.
Hence the C.F is f1(y-3x) + f2(y + 2x).
y cosx
P.I = -----------------------2
D2 + DD' – 6D'
y cosx
= --------------------------(D + 3D') (D – 2D')
1
= ------------(D+3D')
1
--------------- y cosx
(D – 2D')
1
= ------------(D+3D')
 (c – 2x) cosx dx, where y = c – 2x
1
= -------------  (c – 2x) d (sinx)
(D+3D')
1
= ------------- [(c – 2x) (sinx) – (-2) ( - cosx)]
(D+3D')
1
= ------------- [ y sin x – 2 cos x)]
(D+3D')
=  [(c + 3x) sinx – 2 cosx] dx , where y = c + 3x
=  (c + 3x) d(– cosx) – 2 cosx dx
= (c + 3x) (– cosx) – (3) ( - sinx) – 2 sinx
= – y cosx + sinx
Hence the complete solution is
z = f1(y – 3x) + f2(y + 2x) – y cosx + sinx
Example 30
Solve r – 4s + 4t = e 2x +y
2z
2z
2z
Given equation is -------- – 4 ---------- + 4 ----------- = e2x + y
xy
y2
x2
i.e,
(D2 – 4DD' + 4D' 2 ) z = e2x + y
The auxiliary equation is m2 – 4m + 4 = 0.
Therefore,
m = 2,2
Hence the C.F is f1(y + 2x) + x f2(y + 2x).
e2x+y
P.I. = ------------------------2
D2 – 4DD'+4D'
Since D2 – 4DD'+4D'2 = 0 for D = 2 and D' = 1, we have to apply the general rule.
e2x+y
P.I. = --------------------------(D – 2D') (D – 2D')
1
1
= ------------- -------------- e2x+y
(D – 2D') (D – 2D')
1
= -------------  e2x+c – 2x dx , where y = c – 2x.
(D – 2D')
1
= -------------  ec dx
(D – 2D')
1
= ------------- ec .x
(D – 2D')
1
= ------------- xe y+2x
D – 2D'
=  xec-2x + 2x dx ,
where y = c – 2x.
=  xec dx
= ec. x2/2
x2ey+2x
= ------------2
Hence the complete solution is
1
z = f1(y+2x) + f2(y+2x) + ----- x2e2x+y
2
1.6 Non – Homogeneous Linear Equations
Let us consider the partial differential equation
f (D,D') z = F (x,y)
------- (1)
If f (D,D') is not homogeneous, then (1) is a non–homogeneous linear partial differential
equation. Here also, the complete solution = C.F + P.I.
The methods for finding the Particular Integrals are the same as those for
homogeneous linear equations.
But for finding the C.F, we have to factorize f (D,D') into factors of the form D – mD' – c.
Consider now the equation
(D – mD' – c) z = 0 -----------(2).
This equation can be expressed as
p – mq = cz ---------(3),
which is in Lagrangian form.
The subsidiary equations are
dx
dy
dz
--------- = --------- = ----------- -----------(4)
1
–m
cz
The solutions of (4) are y + mx = a and z = becx.
Taking b = f (a), we get z = ecx f (y+mx) as the solution of (2).
Note:
1. If (D-m1D' – C1) (D – m2D'-C2) …… (D – mnD'-Cn) z = 0 is the partial
differential equation, then its complete solution is
z = ec1x f1(y +m1x) + ec2x f2(y+m2x) + . . . . . + ecnx fn(y+mnx)
2. In the case of repeated factors, the equation (D-mD' – C)nz = 0 has a complete
solution z = ecx f1(y +mx) + x ecx f2(y+mx) + . . . . . +x n-1 ecx fn(y+mx).
Example 31
Solve (D-D'-1) (D-D' – 2)z = e 2x – y
Here m1 = 1, m2 = 1, c1 = 1, c2 = 2.
Therefore, the C.F is ex f1 (y+x) + e2x f2 (y+x).
e2x-y
P.I. = ------------------------------ Put D = 2, D' = – 1.
(D – D' – 1) (D-D' – 2)
=
e2x-y
------------------------------------(2 – ( – 1) – 1) (2 – ( – 1) – 2)
e2x-y
= -------------2
Hence the solution is z = ex f1 (y+x) + e2x f2 (y+x) +
e 2x-y
----------.
2
Example 32
Solve (D2 – DD' + D' – 1) z = cos (x + 2y)
The given equation can be rewritten as
(D-D'+1) (D-1) z = cos (x + 2y)
Here m1 = 1, m2 = 0, c1 = -1, c2 = 1.
Therefore, the C.F = e–x f1(y+x) + ex f2 (y)
1
2
2
'
'
P.I = --------------------------- cos (x+2y) [Put D = – 1,DD = - 2 ,D = – 4]
(D2 – DD' + D' – 1)
1
= --------------------------- cos (x+2y)
– 1 – (– 2) + D' – 1
1
= ------- cos (x+2y)
D'
sin (x+2y)
= ---------------2
sin(x+2y)
-x
x
Hence the solution is z = e f1(y+x) e f2(y) + ---------------- .
2
Example 33
Solve [(D + D'– 1) (D + 2D' – 3)] z = ex+2y + 4 + 3x +6y
Here m1 = – 1, m2 = – 2 , c1 = 1, c2 = 3.
Hence the C.F is z = ex f1(y – x) + e3x f2(y – 2x).
ex+2y
P.I1 = ---------------------------------(D+D' – 1) (D + 2D' – 3)
[Put D = 1, D'= 2]
ex+2y
= -------------------------(1+2 – 1) (1+4 – 3)
ex+2y
= -----------4
1
P.I2 = -------------------------------- (4 + 3x + 6y)
(D+D' – 1) (D + 2D' – 3)
1
= --------------------------------------- (4 + 3x + 6y)
D + 2D'
3 [1 – (D+D')] 1 – -----------3
-1
1
D + 2D'
' -1
= ------ [1 – (D + D )] 1 – -------------- (4 +3x+6y)
3
3
1
D + 2D'
1
'
' 2
= ----[1 + (D + D )+ (D+D ) + . . .] 1+ ------------- + ----- (D+2D')2 + ….. .]
3
3
9
. (4 + 3x +6y)
1
4
5
= ---- 1 + ----- D + ------D' + . . . . . (4 + 3x + 6y)
3
3
3
1
4
5
= ---- 4 +3x + 6y + ----- (3) + -----(6)
3
3
3
= x + 2y + 6
Hence the complete solution is
ex+2y
z = exf1 (y-x) + e3x f2 (y – 2x) + ------- + x + 2y + 6.
4
Exercises
(a) Solve the following homogeneous Equations.
1.
2z
2z
2z
---------- + --------- – 6 --------- = cos (2x + y)
x2
xy
y2
2.
2z
2z
---------- – 2 --------- = sin x.cos 2y
x2
xy
3. (D2 + 3DD' + 2D'2) z = x + y
4. (D2 – DD'+ 2D' 2) z = xy + ex. coshy
ey + e–y
ex+y + e x-y
Hint: e . coshy = e . ------------- = -----------------2
2
3
'2
'3
2x+y
5. (D – 7DD – 6D ) z = sin (x+2y) + e
x
x
6. (D2 + 4DD' – 5D'2) z = 3e2x-y + sin (x – 2y)
7. (D2 – DD' – 30D'2) z = xy + e6x+y
8. (D2 – 4D' 2) z = cos2x. cos3y
9. (D2 – DD' – 2D'2) z = (y – 1)ex
10. 4r + 12s + 9t = e3x – 2y
(b) Solve the following non – homogeneous equations.
1. (2DD' + D' 2 – 3D') z = 3 cos(3x – 2y)
2. (D2 + DD' + D' – 1) z = e-x
3. r – s + p = x2 + y2
4. (D2 – 2DD' + D'2 – 3D + 3D' + 2)z = (e3x + 2e-2y)2
5. (D2 – D'2 – 3D + 3D') z = xy + 7.
UNIT–II
FOURIER SERIES
2.1 INTRODUCTION
The concept of Fourier series was first introduced by Jacques Fourier (1768–
1830), French Physicist and Mathematician. These series became a most important tool
in Mathematical physics and had deep influence on the further development of
mathematics it self.Fourier series are series of cosines and sines and arise in representing
general periodic functions that occurs in many Science and Engineering problems. Since
the periodic functions are often complicated, it is necessary to express these in terms of
the simple periodic functions of sine and cosine. They play an important role in solving
ordinary and partial differential equations.
2.2 PERIODIC FUNCTIONS
A function f (x) is called periodic if it is defined for all real „x‟ and if there is
some positive number „p‟ such that
f (x + p ) = f (x) for all x.
This number „p‟ is called a period of f(x).
If a periodic function f (x) has a smallest period p (>0), this is often called the
fundamental period of f(x). For example, the functions cosx and sinx have fundamental
period 2.
DIRICHLET CONDITIONS
Any function f(x), defined in the interval c  x  c + 2, can be developed as
a0

a Fourier series of the form ------- +  (an cosnx + bn sinnx) provided the following
n=1
2
conditions are satisfied.
f (x) is periodic, single– valued and finite in [ c, c + 2 ].
f (x) has a finite number of discontinuities in [ c, c + 2].
f (x) has at the most a finite number of maxima and minima in [ c,c+ 2].
These conditions are known as Dirichlet conditions. When these conditions are satisfied,
the Fourier series converges to f(x) at every point of continuity. At a point of
discontinuity x = c, the sum of the series is given by
f(x) = (1/2) [ f (c-0) + f (c+0)] ,
where f (c-0) is the limit on the left and f (c+0) is the limit on the right.
EULER’S FORMULAE
The Fourier series for the function f(x) in the interval c < x < c + 2 is given by
a0

f (x) = ------- +  (an cosnx + bn sinnx), where
n=1
2
1 C + 2
a0 = -----  f (x) dx.
 C
1 C + 2
an = -----  f(x) cosnx dx.
 C
1 C + 2
bn = -----  f (x) sinnx dx.
 C
These values of a0, an, bn are known as Euler‟s formulae. The coefficients a0, an, bn are
also termed as Fourier coefficients.
Example 1
Expand f(x) = x as Fourier Series (Fs) in the interval [ -π, π]
Let
f(x)
=
∞
+ ∑
n=1
ao
---2
π
∫ f (x) dx
π -π
1 π
=
∫ x dx
π -π
1
Here ao =
π
=
1
π
2
x
2
-π
1
π2
π
2
=
π2
-
ao = 0
= 0
2
[ an cos nx + bn sin nx ] ----------(1)
1 π
an = --- ∫ f(x) cosnx dx
π -π
π
∫ x
-π
1
=
π
sin nx
-------n
d
π
=
1
(x) sin nx
π
=
- (1)
-cos nx
n2
n
1
π
cos nπ
n2
-π
cos nπ
n2
= 0
π
∫ f(x) sin nx dx
π -π
1
π
-cos nx
∫ x d
n
π
-π
1
=
bn
=
1
=
-cosnx
(x)
π
π
-sin nx
- (1)
n
n2
-π
=
- πcos nπ
n
1
π
πcosnπ
n
= -2π cosnπ
nπ
bn
= 2 (-1)n+1
n
[
cos nπ = (-1)n]
Substituting the values of ao, an & bn in equation (1), we get

f(x)
x
= 
n=1
=
2(-1)n+1
n
2
sinx
1
sin nx
- 1 sin2x + 1 sin 3x -……
2
3
Example 2
Expand
that
1.
1 - 1 + 1 - 1 + ………… = 2
12
2.
f(x) = x2 as a Fourier Series in the interval ( -  x   ) and hence deduce
22
32
42
12
1 + 1 + 1 + 1 + ………… = 2
12 22 32 42
6
1 + 1 + 1 + 1 + ………… = 2
12 32 52 72
8

Let f(x) = a0 +  [ an cosnx + bn sinnx ]
2
n=1
Here

a0 = 1
 f(x) dx
 -
3.
= 1



-
3
x2 dx

= 1

x
3 -
= 1

3
3
3 + 3
ao = 22
3

an = 1 
 -
=
f(x) cosnx dx

1  x2 cosnx dx
 -
=

1  x2 d
 -
=
sinnx
n

(x2) sinnx – (2x)-cosnx + (2) – sinnx
n
n2
n3
1

-
1
=
2 cosn + 2 cosn
n2
n2

4
an =
(-1)n
n2

bn = 1 
 -
=
=
=
f(x) sinnx dx

1  x2 d
 -
-cosnx
n
1

(x2) –cosnx
n
– (2x) -sinnx + (2) cosnx
n2
n3
1
-2 cosn
2 cosn
+

n
2 cosn
+
n
2cosn
-
n3
bn = 0
Substituting the values of a0, an & bn in equation (1) we get

f(x) = 22 + 
6
n= 1
i.e,
i.e,
x2 =
2
x =
4
n2
(-1)n cosnx

2 +  4 (-1)n cosnx
3 n=1 n2

 +4 
2
(-1)n cosnx
n3

-
3
2 + 4
3
=
x2
n=1 n2
-cosx + cos2x – cos3x + …..
12
22
32
2 - 4 cosx + cos2x + cos3x - …..
3
12
22
32
Put x = 0 in equation (2) we get
=
0
i.e,
=
2 - 4
3
-----------(2)
1 – 1 + 1 – 1 + …..
12 22 32 42
1 – 1 + 1 - …… = 2
12 22 32
12
-------------(3)
Put x =  in equation (2) we get
2 = 2 - 4
3
-1 - 1
12
22
i.e,
2 - 2 = 4
3
1 + 1 + 1 + ………
12
22
32
i.e,
1 + 1 + 1 + ……
12
22
32
-
1 - ………
32
= 2
6
-------------(4)
Adding equations (3) & (4) we get
1 - 1 + 1 - ….. + 1 + 1 + 1 +…. = 2 + 2
12
22 32
12
22 32
12
6
i.e,
2
i.e,
12 + 12 + 12 + .......
1 3
5
= 32
12
1 + 1 + 1 + 1 …..
12 32
52
72
= 2
8
Example 3
Obtain the Fourier Series of periodicity 2π for f(x) = ex in [-π, π]
Let
a0
f(x) = ---2
∞
+ ∑( an cosnx + bn sinnx)
n=1
-------------(1)
π
a0 = 1 ∫ f(x) dx
π -π
π
= 1 ∫ ex dx
π -π
π
x
= 1 [e ]
π
-π
= 2 {eπ – eπ}
2π
a0 = 2 sin hπ
π
π
an = 1 ∫ f(x) cos nx dx
π-π
π
= 1 ∫ ex cos nx dx
π -π
π
x
=1 e
[cosnx + n sin nx]
π (1+n2)
-π
=1
e π (-1)n 
π 1+n2
=
an =
e π (-1)n
1+n2
(-1)n
( e π - eπ )
(1+n2) π
2 ( -1)n sin hπ
π(1+n2)
1
π
bn = ----- ∫ f(x) sin nx dx
π -π
π
= 1 ∫ ex sin nx dx
π -π
π
x
=1 e
(sinnx – n cosnx)
π (1+n2)
-π
π
n

- π
=1
e {-n(-1) }
e
{-n(-1)n}
π 1+n2
1+n2
=
bn =
( e π - e π )
n(-1)n+1
π(1+n2)
2n(-1)n+1 sin hπ
π(1+n2)

f(x) = 1 sin hπ + 
π
n=1
ex =
ie, ex=
2(-1)n sinhπ cosnx + 2(-n)(-1)n sinhπ sinnx
π(1+n2)
π(1+n2)

1 sin hπ + 2sin hπ  (-1)n2
π
π
n=1 1+n

sin hπ 1 + 2  (-1)n
π
n=1 1+n2
Example 4
(cos nx – n sin nx)
(cos nx – n sin nx)
x
in
(O, )
(2 - x)
in
(, 2)
Let f (x) =
1
2
Find the FS for f (x) and hence deduce that  ---------- = ------n=1
(2n-1)2
8
a0

Let f (x) = ------- +  an cosnx + bn sin nx
--------- (1)
n=1
2

1 
2
Here a0 = ------  f(x) dx +  f (x) dx

 o
1 
2
= ------  x dx+  (2 - x) dx

 o
1
= ----
1
= -----
i.e, ao = 
x2 
– (2 - x)2 2
------- + -------------2 0
2

2
2
------ + -----=
2
2
1
an = ----

2
 x cos nx dx +  (2 - x) cos nx dx
o

1
sin nx
sin nx

2
= ----  x d --------- +  (2 - x) d --------
 0
n
n
1
sin nx
-cos nx
= ---- (x) --------- – (1) ---------
n
n2

sin nx
– cosnx
+ (2-x) --------- – (–1) ---------n
n2
o
1
cos n
1
cos2n
cosn
= ---- ----------- – ------- – ------------ + ---------
n2
n2
n2
n2
1 2cos n
2
= ---- ------------ – ------
n2
n2
an
2
= -------- [( –1)n -1]
n2
1

2
bn = -----  f(x) sin nx dx +  f(x) sin nx dx

 o
1
= ---
–cos nx
–cos nx

2
 x d --------- +  (2 - x) d --------o

n
n
1
–cos nx
–sinnx
= ---- (x) --------- – (1) ----------

–cos nx
– sinnx
+ (2-x) --------- – (–1) ----------
2

n
n2
o
1 –  cos n
 cosn
= ---- -------------- + --------------
n
n
i.e, bn = 0.
n
n2

=0

2

f (x) = ----- +  ------- [ (– 1)n – 1] cos nx
n=1
2
n2

4
cosx
cos3x
cos5x
= ----- – ------ -------------- +----------- +------------ + . . . . .
2

12
32
52
-----(2)
Putting x = 0 in equation(2), we get

4
1
1
1
0 = ------- – ------- ------ + ------ + ------ + . . . . .
2

12
32
52
1
1
1
i.e, ------ + ------ + ------ + . . . . . =
12
32
52
2
------8
1
2
i.e,  ------------- = ------n=1
(2n – 1)2
8

Example 5
Find the Fourier series for f (x) = (x + x2) in (- < x < ) of percodicity 2 and hence

deduce that
 (1/ n2) = 2 /6.
n=1
a0

Let f(x) = ------- +  ( an cosnx + bn sinnx)
n=1
2
Here,
a0
1

= ------  (x + x2) dx
 –
1
x2
x3 
= ----- ----- +----2
2
3
0
1
2
3
2
3
= ------ ------ + ------ – -------- + ------
2
3
2
3
22
ao = -------3
1

an = -----  f (x) cos nx dx
 –
1
sin nx

2
= -----  (x + x ) d --------- –
n
1
= ----
sin nx
– cosnx
(x + x2) ---------- – (1+2x) ---------n
n2

– sinnx
+ (2) ---------n3
–
1
(– 1)n
(– 1)n
= ------- (1+ 2) --------- – (1 – 2)---------
n2
n2
an
4 (– 1)n
= ------------n2
1

bn = -----  f (x) sin nx dx
 –
1
– cos nx

2
= -----  (x + x ) d --------- –
n
1
– cosnx
–sinnx
cosnx
2
= ----- (x + x ) ---------- – (1+2x) ---------- + (2) ---------
n
n2
n3

–
1
= ------
– 2(– 1)n
 (–1)n
 (– 1)n
2 (–1)n
---------------- – ------------- – ----------- + -----------n
n
n
n2
2 (– 1)n+1
bn = ------------n
2
4 (– 1)n
2(– 1)n+1

f(x)
= ------ +  ----------- cos nx + -------------- sinnx
3 n=1
n2
n
2
cosx
cos2x
cos3x
= -----  4 ----------  ---------- + ----------  …..
sin2x
+ 2 sin x  ---------+ . . .
.
12
3
22
32
2
Here x = - and x =  are the end points of the range.  The value of FS at x =  is the
average of the values of f(x) at x =  and x = -.
f ( - ) + f ()
 f(x) = -------------------2
-  + 2 +  + 2
= -----------------------2
= 2
Putting x = , we get
2
1
1
1
2 = ------ + 4 ------ + ------ + ------- + . . . .
3
12
22
32
2
i.e, ------ =
6
1
1
1
------ + ------ + ------ +. . . . . . . .
12
22
32
1
2
Hence,  ----- = -------.
n=1
n2
6

Exercises:
Determine the Fourier expressions of the following functions in the given interval
1.f(x) = ( - x)2, 0 < x < 2
2.f(x) = 0 in - < x < 0
=  in 0 < x < 
3.f(x) = x – x2 in [-,]
4.f(x) = x(2-x) in (0,2)
5.f(x) = sinh ax in [-, ]
6.f(x) = cosh ax in [-, ]
7.f(x) = 1 in 0 < x < 
= 2 in  < x < 2
8.f(x) = -/4 when - < x < 0
= /4 when 0 < x < 
9.f(x) = cosx, in - < x < , where „‟ is not an integer
10.Obtain a fourier series to represent e-ax from x = - to x = . Hence derive the series
for  / sinh
2.3 Even and Odd functions
A function f(x) is said to be even if f (-x) = f (x). For example x2, cosx, x sinx, secx are
even functions. A function f (x) is said to be odd if f (-x) = - f (x). For example, x3, sin x,
x cos x,. are odd functions.
(1) The Euler‟s formula for even function is
a0

f (x) = ------ +  an consnx
n=1
2
2 
2 
where ao = ----  f(x) dx ; an = -----  f (x) cosnx dx
 0
 0
(2) The Euler‟s formula for odd function is

f (x) =  bn sin nx
n=1
2 
where bn = -------  f(x) sin nx dx
 0
Example 6
Find the Fourier Series for f (x) = x in ( - , )
Here, f(x) = x is an odd function.

 f(x) =  bn sin nx
------- (1)
n=1
2 
bn = -----  f (x) sin nx dx
 0
2 
- cos nx
= -----  x d ------------- 0
n
2
= ----
–cos nx
– sin nx
(x) ----------- - (1) -----------n
n2
2
= ----
–  cos n
-------------n
2 (– 1) n+1
bn = ----------------n
2 ( – 1)n+1
f (x)=  --------------- sin nx
n=1
n

2 ( – 1)n+1
x =  ------------- sin nx
n=1
n

i.e,
Example 7

0
Expand f (x) = |x| in (-, ) as FS and hence deduce that
1
1
1
2
------ + ----- + ------ . . . . . =. -------12
32
52
8
Solution
Here f(x) = |x| is an even function.
ao

 f (x) = -------- +  an cos nx
n=1
2
2 
ao = -----  f(x) dx
 0
2 
= -----  x dx
 0
------- (1)

2
x2
= ----- ----=

2
0
2 
an = -----  f (x) cos nx dx
 0
2 
sin nx
= -----  x d --------- 0
n
2
= ----
sin nx
– cos nx
(x) ---------- – (1) ----------n
n2

0
2
cos n
1
= ----- ---------- – -----
n2
n2
2
an = ----- [(– 1) n – 1]
n2

2

f (x)= -------+  ------- [(–1)n – 1] cos nx
n=1
2
n2

4
cos x
cos3x
cos5x
i.e, |x| = ----- – ------ ----------- + ---------- + --------- + . . . . . .
2

12
32
52
----------(2)
Putting x = 0 in equation (2), we get

4
0 = ------ – ------2

1
1
1
------ + ------ + ------- + . . . . . . .
12
32
52
1
1
1
Hence, ------ + ------ + ------+ ….
12
32
52
2
= ------8
Example 8
2x
If f (x) = 1 + ----- in ( - , 0)

2x
= 1 – ----- in ( 0,  )


Then find the FS for f(x) and hence show that  (2n-1)-2 = 2/8
n=1
Here f (-x) in (-,0) = f (x) in (0,)
f (-x) in (0,) = f (x) in (-,0)
 f(x) is a even function
ao

Let f (x) = -------- +  an cos nx
n=1
2
2 
ao = ----- 
 0
2x
1 – ------- dx

2
2x2
= ----- x – ------
2
a0 = 0

0
------- (1).
2 
2x
an = -----  1 – ------- cos nx dx
 0

2 
2x
sin nx
= -----  1 – -------- d ---------- 0

n
2
= ----
2x
1 – -----
sin nx
--------- –
n
–2
– cosnx
------- ---------
n2

0
4
an = ------- [(1– (– 1)n]
2n2
4
f (x)=  -------- [1– (– 1)n]cos nx
n=1
2n2

4
2cos x
2cos3x
2cos5x
= ----- ----------- + ---------- + ------------ + . . . . . .
2
12
32
52
-----------(2)
Put x = 0 in equation (2) we get
2
------- = 2
4
1
1
1
------ + ------- + --------+ . . . . . . .
12
32
52
1
1
1
2
==> ------ + ------ + ------ + … = ------12
32
52
8
1
2
 ----------- = -------n=1
(2n–1)2
8

or
Example 9
Obtain the FS expansion of f(x) = x sinx in (- < x<) and hence deduce that
1
1
1
-2
------ – ------- + ------- – . . . . . = --------1.3
3.5
5.7
4.
Here f (x) = xsinx is an even function.
ao

Let f (x) = -------- +  an cos nx
------------- (1)
n=1
2
2 
Now, ao = -----  xsin x dx
 0
2 
= -----  x d ( - cosx)
 0
2
= ----- (x) (- cosx) – (1) (- sin x)


0
a0 = 2
2 
an = -----  f (x) cos nx dx
 0
2 
= -----  x sin x cosnx dx
 0
1 
= -----  x [ sin (1+n)x + sin (1 – n)x] dx
 0
1 
– cos (1+n)x
cos (1 – n) x
= ------  x d --------------- – ------------------ 0
1+n
1–n

1
– cos (1+n)x cos (1 – n) x
– sin (1+n)x
sin (1 – n) x
= ------ (x) --------------- – --------------- – (1) --------------- – ----------------
1+n
1–n
(1 + n)2
(1 – n)2
0
1
– cos (1+n)
 cos (1 – n) 
= ------ ------------------- – ------------------
1+n
1–n
- [cos cosn - sin  sinn]
[cos cosn - sin  sin n ]
=
---------------------------------- – ---------------------------------1+n
1–n
=
(1+n) ( – 1) n + (1 – n ) ( – 1)n
------------------------------------1 – n2
2(–1)n
an = --------1 – n2
When n = 1
, Provided n  1
2 
a1 = -----  x sinx cos x dx
 0
1 
= -----  x sin2x dx
 0
1 
- cos2x
= -----  x d --------- 0
2
1
– cos 2x
-sin 2x
= ------ (x) ----------- – (1) -----------
2
4

0
Therefore, a1 = -1/2
a0

f (x)= -------+ a1 cosx +  ancos nx
n=2
2
1
 2( -1)n
= 1 – ------ cosx +  ----------- cosnx
n=2
2
1-n2
1
cos2x
cos3x
cos4x
ie, x sinx = 1 – ------ cos x – 2 ----------- - ------------ + ----------- - . . . .
2
3
8
15
Putting x = /2 in the above equation, we get

1
1
1
----- – 1 = 2 ------ – ------- + -------- – . . . . . . .
2
1.3
3.5
5.7
1
1
1
-2
Hence, ------ – ------ + ------- – . . . . . . = ---------1.3
1.5
5.7
4
Exercises:
Determine Fourier expressions of the following functions in the given interval:
i. f(x) = /2 + x, - < x < 0
/2 - x, 0 < x < 
ii. f(x) = -x+1 for + - < x < 0
x+1 for 0 < x < 
iii. f(x) = |sinx|, - < x < 
iv. f(x) =x3 in - < x < 
v. f(x) = xcosx, - < x < 
vi. f(x) = |cosx|, - < x < 
2sin a sin x
2sin 2x
3sin3x
vii. Show that for - < x < , sin ax = --------- -------- - ------ + -------- - …….

12 - 2 22 - 2
32 - 2
2.4 HALF RANGE SERIES
It is often necessary to obtain a Fourier expansion of a function for the range
(0, ) which is half the period of the Fourier series, the Fourier expansion of such a
function consists a cosine or sine terms only.
(i) Half Range Cosine Series
The Fourier cosine series for f(x) in the interval (0,) is given by
a0

f(x) = ---- +  an cosn x
2
n=1
2

 f(x)
where a0 = ------
dx
and
0
2 
an = -------  f(x) cosnx
 0
(ii) Half Range Sine Series
dx
The Fourier sine series for f(x) in the interval (0,) is given by

f(x) =  bn sinnx
n=1
2

where bn = ------ f(x) sinnx dx
0

Example 10
If c is the constant in ( 0 < x <  ) then show that
c = (4c / ) { sinx + (sin3x /3) + sin5x / 5) + ... ... ... }
Given
f(x) = c in (0,).

Let f(x) =  bn sinnx
 (1)
n=1
bn
2
= ------

2
= ------

 f(x) sin nx dx
0
 c sin nx dx
0
- cosnx
------------n
2c
= ---
-(-1)n
1
------ + ---n
n
bn = (2c/n) [ 1 – (-1)n ]


2c
= ------
0
 f(x) =  (2c / n) (1-(-1)n ) sinnx
n=1
4c
= --
i.e, c
sin3x
sinx + --------3
sin5x
+ ---------- + … … …
5
Example 11
Find the Fourier Half Range Sine Series and Cosine Series for f(x) = x in the interval
(0,).
Sine Series

f(x) =  bn
n=1
Let
sinnx
2
bn = ------
Here
=
2 
 f(x) sinnxdx = ------  x d ( -cosnx / n)
0
 0

- cosnx
- sinnx

(x) ----------- - (1) -------------n
n2
0
2
---
=
-------(1)
-  (-1) n
-------------n
2
----
2(-1) n+1
bn = ---------n

 f(x) = 
n=1
2
---- (-1)n+1 sin nx
n
Cosine Series
a0 
Let f(x) = ---- +  an cosnx
2 n=1
2 
Here a0 = ------  f(x)dx
 0
---------(2)
2 
= -------  xdx
 0
2
= ------
an
f(x)
x2
--2
2
an = ------

2
an = ------


=
0
 f(x) cosnx dx
0
 x d (sinnx / n )
0
sinnx
- cosnx

(x) ----------- - (1) -------------n
n2
0
=
2
---
=
2
------ (-1)n -1
n2
=

2

--- +  ----- [ (-1)n - 1] cosnx
2
n=1 n2
=> x
 4
= --- + --2

cosx
cos3x
---------- + ---------12
32
cos5x
+ --------- …… …
52
Example 12
Find the sine and cosine half-range series for the function function .
f(x) = x , 0 < x  π/2
= π-x, π/2x< 
Sine series

Let f (x) =  bn sin nx.
n=1
π
bn= (2/ )  f (x) sin nx dx
0
/2

=(2/ )  x sin nx dx +  (-x) sin nx dx
0
/2
/2
= (2/ )  x .d
0
-cos nx
n

+  (-x) d
/2
-cos nx
n
/2
-cos nx
= (2/ ) x
-sin nx
-(1)
n2
n
0

cos nx
+ (-x) - n
= (2/)
sin nx
-(-1) n2
-(/2)cos n(/2)
+
n
/2
sin n(/2)
-(/2)cosn(/2)

n2
2sinn(/2)
n2
4
=
sin (n/2)
n
2
 sin(n/2)
Therefore , f(x)= ( 4/ ) 
sin nx
n=1
n2
sin3x
ie, f (x)= ( 4/ ) sinx 
sin5x
+

n
= (2/)
 --------
sin (/2 )
n2
32
52
Cosine series

.Let f(x) = (ao /2) + an cosnx., where
n=1

ao = (2/)  f(x) dx
0
/2

=(2/)  x dx+  (-x) dx
0
/2
=(2/)
/2

(x2/2) + (x –x2/2)
0
/2
= /2

an = (2/)  f(x) cosnx dx
0
/2

=(2/ )  x cosnx dx +  (-x) cosnx dx
0
/2
/2 sinnx

=(2/ )  x d
+  (-x) d
0
n
/2
sinnx
n
/2
sinnx
= (2/ ) x
-cosnx
-(1)
n2
n
0

sinnx
+ (-x)
cosnx
-(-1) -
n
= (2/)
( /2) sinn(/2)
+
n2
/2
cos n(/2)

1
n2
n
( /2) sinn(/2)
cosnx
+ 

n2
n2
cos n(/2)
+
n2
n
2cosn ( /2) - {1+(-1)n}
=(2/)
n2
 2 cos n( /2)- {1+(-1)n }
Therefore, f(x)= ( /4)+(2/) 
cosnx .
2
n=1
n
cos6x
= ( /4)-(2/) cos2x+
+------------32
Exercises
1.Obtain cosine and sine series for f(x) = x in the interval 0< x < . Hence show that 1/12
+ 1/32 + 1/52 + … = 2/8.
2.Find the half range cosine and sine series for f(x) = x2 in the range 0 < x < 
3.Obtain the half-range cosine series for the function f(x) = xsinx in (0,)..
4.Obtain cosine and sine series for f(x) = x (-x) in 0< x < 
5.Find the half-range cosine series for the function
6.f(x) = (x) / 4 , 0<x< (/2)
= (/4)(-x), /2 < x < .
7.Find half range sine series and cosine series for
f(x) = x in 0<x< (/2)
= 0 in /2 < x < .
8.Find half range sine series and cosine series for the function f(x) ==  - x in the interval
0 < x < .
9.Find the half range sine series of f(x) = x cosx in (0,)
10.Obtain cosine series for
f(x) = cos x ,
0<x< (/2)
= 0, /2 < x < .
2.5 Parseval’s Theorem
Root Mean square value of the function f(x) over an interval (a, b) is defined as
[f (x)] r m s =
b
∫ [f(x)]2 dx
a
b–a
The use of r.m.s value of a periodic function is frequently made in the
theory of mechanical vibrations and in electric circuit theory. The r.m.s value is
also known as the effective value of the function.
Parseval’s Theorem
If f(x) defined in the interval (c, c+2π ), then the Parseval‟s Identity is given by
c+2π
∫ [f (x)]2 dx = (Range)
c
ao2
1
4
or2
ao2
= ( 2π)
∑ ( an2 + bn2 )
+
1
∑ ( an2 + bn2 )
+
4
2
Example 13
Obtain the Fourier series for f(x) = x2 in – π < x < π
Hence show that
we have ao =
π4
90
1 + 1 + 1 +. . .=
14
24
34
2π2
3 ,
an =
4 (-1)n
n2
,
bn = 0, for all n (Refer Example 2).
By Parseval‟s Theorem, we have
π
ao2
∞
∫ [ f(x)]2 dx = 2π
-π
36
π4
x5
5 -π
π4
9
=>
9
Hence
n4
∞
1
n=1
n4
1
∞
n=1
n4
π4
1
∑
n=1
n=1
+8 ∑
=
5
∞
16(-1) 2n
+8 ∑
= 2π
π4
∞
+ 1/2 ∑
= 2π
π
i.e,
n=1
4
4π4
π
∫ x4 dx
-π
i.e,
+ ½ ∑ (an2 + bn2)
=
4
n
90
1 + 1
14
24
+ 1 + . . .=
34
π4
90
2.6 CHANGE OF INTERVAL
In most of the Engineering applications, we require an expansion of a given
function over an interval 2l other than 2.
Suppose f(x) is a function defined in the interval c< x < c+2l. The Fourier
expansion for f(x) in the interval c<x<c+2l is given by
f(x)
=
a0 
----- + 
2
n=1
an
nx
nx
cos ---- + bn sin ---l
l
1
where a0 = ----l
c+2 l

c
f(x)dx
1
an = ----l
c+2 l

c
f(x) cos (nx / l ) dx
1
bn = ----l
c+2 l

c
f(x) sin (nx / l ) dx
&
Even and Odd Function
f(x)
If f(x) is an even function and is defined in the interval ( c, c+2 l ), then
a0 
nx
=
----- + 
an cos ---2
n=1
l
2
where a0 = ----l
l
2
an = ----l
l

0
 f(x)dx
0
f(x) cos (nx / l ) dx
If f(x) is an odd function and is defined in the interval ( c, c+2 l ), then
f(x)
=


n=1
nx
bn sin ---l
where
l

0
2
bn = ----l
f(x) sin (nx / l ) dx
Half Range Series
Sine Series
f(x)
=


n=1
nx
bn sin ---l
where
2
bn = ----l
l
 f(x) sin (nx / l ) dx
0
Cosine series
f(x)
=
a0

----- + 
2
n=1
nx
an cos ---l
l

0
2
where a0 = ----l
f(x)dx
2
l
an = ---- f(x) cos (nx / l ) dx
l
0
Example 14
Find the Fourier series expansion for the function
f(x) = (c/ℓ)x in 0<x<ℓ
= (c/ℓ)(2ℓ- x) in ℓ<x<2ℓ
Let
a0
nx
nx

f (x) = ------ +  an cos ------ + bn sin -----n=1
2
ℓ
ℓ
1
Now, a0 = ----l
2l
 f(x)dx
0
1
ℓ
2ℓ
= ------ (c/ℓ)  x dx + (c/ℓ)  (2ℓ - x) dx
o
ℓ
ℓ
1
ℓ
2ℓ
2
2
= ----- (c/ℓ) (x / 2) + (c/ℓ) (2ℓx - x /2)
0
ℓ
ℓ
c
= ---- ℓ2 = c
ℓ2
1
an = ----ℓ
2ℓ
 f(x) cos (nx / ℓ ) dx
0
1
=
ℓ
c
=
ℓ2
ℓ
nx
 (c/ℓ)x cos
0
ℓ
ℓ
 x d
0
2ℓ
nx
dx +  (c/ℓ)(2ℓ- x) cos
ℓ
ℓ
sin(nx /ℓ)
n /ℓ
2ℓ
+  (2ℓ- x) d
ℓ
dx
sin(nx /ℓ)
n /ℓ
ℓ
nx
nx
sin
-cos
c
ℓ
=
ℓ
 (1)
(x)
ℓ2
n
ℓ
n22
ℓ2
nx
0
2ℓ
nx
-cos
ℓ
sin
ℓ
(-1)
+ (2ℓ- x)
n22
ℓ2
n
ℓ
ℓ
c
ℓ2 cosn
2
n
=
ℓ
2 2
c
ℓ2
2
n
=
ℓ
ℓ2

n
2 2
ℓ2 cos2n
+ 
n
2 2
ℓ2 cosn
+
n22
{ 2 cosn 2 }
2 2
2c
=
n22
1
bn =
ℓ
1
=
ℓ
c
=
ℓ2
{ (-1)n 1}
2ℓ
nx
 f(x) . sin
dx
0
ℓ
ℓ
nx
2ℓ
nx
 (c/ℓ)x sin
dx +  (c/ℓ)(2ℓ- x) sin
dx
0
ℓ
ℓ
ℓ
ℓ
cos(nx /ℓ)
 x d 0
n /ℓ
2ℓ
cos(nx /ℓ)
+  (2ℓ- x) d ℓ
n /ℓ
ℓ
nx
nx
cos
sin
c
ℓ
=

(x)
ℓ
ℓ
(1) 
2
n
ℓ
n22
ℓ2
0
nx
nx
cos
ℓ
+
(2ℓ- x)
2ℓ
sin

ℓ
 (-1) 
n22
ℓ2
n
ℓ
ℓ
ℓ2 cosn
c
ℓ2 cosn

=
ℓ
2
+
n
n
= 0.
Therefore, f(x) =
c
2c
--- + ---2
2


n=1
{ (-1)n 1}
------------- cos (nx /ℓ)
n2
Example 15
Find the Fourier series of periodicity 3 for f(x) = 2x – x2 , in 0  x  3.
Here 2ℓ = 3.
 ℓ = 3 / 2.
Let
a0
2nx
2nx

f (x) = ------ +  an cos ------ + bn sin -----n=1
2
3
3
3
where
ao = (2 / 3)  (2x - x2) dx
0
= (2 / 3) 2 (x2/2) – (x3/3) dx
3
0
= 0.
2nx
an = (2 / 3)  (2x - x ) cos ------ dx
0
3
sin(2nx /3)
3
= (2 / 3)  (2x - x2) d
0
(2n/3)
3
2
3
sin(2nx /3)
(2x - x2)
= (2 / 3)
(2n/3)
cos(2nx /3)
sin(2nx/3)
– (2 -2x) + (-2) (4n22/9)
(8n33/27)
0
= (2 / 3) - ( 9 / n22) – ( 9 / 2n22)
= - 9 / n22
2nx
bn = (2 / 3)  (2x - x ) sin ------ dx
0
3
3
2
cos(2nx /3)
3
= (2 / 3)  (2x - x2) d –
0
(2n/3)
3
cos(2nx /3)
= (2 / 3) (2x - x2) –
(2n/3)
sin(2nx /3)
– (2 -2x) -
(4n22/9)
cos(2nx/3)
+ (-2)
(8n33/27)
0
= (2 / 3) ( 9 /2n) – ( 27/ 4n33) + ( 27/ 4n33)
= 3 / n

Therefore,
f (x) = 
n=1
2nx
2nx
- ( 9 / n22) cos ------ + (3 / n) sin -----3
3
Exercises
1.Obtain the Fourier series for f(x) = x in 0 < x < 2.
2.Find the Fourier series to represent x2 in the interval (-l, l ).
3.Find a Fourier series in (-2, 2), if
f(x) = 0, -2 < x < 0
= 1, 0 < x < 2.
4.Obtain the Fourier series for
f(x) = 1-x in 0 < x < l
= 0 in l < x < 2 l. Hence deduce that
1- (1/3 ) +(1/5) – (1/7) + … = /4 &
(1/12) + (1/32) + (1/52) + … = (2/8)
5.If f(x) = x,
0<x<1
= (2-x), 1 < x < 2,
Show that in the interval (0,2),
cos x cos3x
cos 5x
f(x) = (/2) – (4/) --------- + -------- + ------ + …….
12
32
52
6.Obtain the Fourier series for
f(x) = x in 0 < x < 1
= 0 in 1 < x < 2
7.Obtain the Fourier series for
f(x) = (cx /l ) in 0 < x < l
= (c/l ) (2 l - x ) in l < x < 2 l .
8.Obtain the Fourier series for
f(x) = (l + x ), - l < x < 0.
= (l - x ), 0 < x < l.
Deduce that

1
2
 -------= -----1 (2n – 1)2
8
9.Obtain half-range sine series for the function
f(x) = cx
in 0 < x < ( l /2)
= c (l – x) in (l/2) < x < l
10.Express f(x) = x as a half – range sine series in 0 < x < 2
11.Obtain the half-range sine series for ex in 0 < x < 1.
12.Find the half –range cosine series for the function f(x) = (x-2)2 in the interval
0 < x < 2.
Deduce that

1
 -------1 (2n – 1)2
2
= ----8
2.7 Harmonic Analysis
The process of finding the Fourier series for a function given by numerical values
is known as harmonic analysis.
a0

f (x) = ------- +  (an cosnx + bn sinnx), where
n=1
2
ie, f(x) = (a0/2) + (a1 cosx + b1 sinx) + (a2 cos2x + b2 sin2x) +
(a3cos3x + b3sin3x)+… -------------(1)
2 ∑ f(x)
Here a0 = 2 [mean values of f(x)] = ----------n
2 ∑ f(x) cosnx
an = 2 [mean values of f(x) cosnx] = --------------------n
2 ∑ f(x) sinnx
bn = 2 [mean values of f(x) sinnx] = ------------------n
In (1), the term (a1cosx + b1 sinx) is called the fundamental or first harmonic,
the term (a2cos2x + b2sin2x) is called the second harmonic and so on.
&
Example 16
Compute the first three harmonics of the Fourier series of f(x) given by the
following table.
x:
0
π/3
2π/3
π
4π/3
5π/3
2π
f(x):
1.0
1.4
1.9
1.7
1.5
1.2
1.0
We exclude the last point x = 2π.
Let f(x) = (a0/2) + (a1 cosx + b1 sinx) + (a2 cos2x + b2 sin2x) + …………
To evaluate the coefficients, we form the following table.
x
0
π/3
2π/3
π
4π/3
5π/3
f(x)
1.0
1.4
1.9
1.7
1.5
1.2
cosx
1
0.5
-0.5
-1
-0.5
0.5
sinx
0
0.866
0.866
0
-0.866
-0.866
2 ∑f(x)
Now, a0 =
cos2x
1
-0.5
-0.5
1
-0.5
-0.5
sin2x
0
0.866
-0.866
0
0.866
-0.866
cos3x
1
-1
1
-1
1
-1
sin3x
0
0
0
0
0
0
2 (1.0 + 1.4 + 1.9 + 1.7 + 1.5 + 1.2)
=
= 2.9
6
6
2 ∑f(x) cosx
a1 = ---------------- = -0.37
6
2 ∑f(x) cos2x
a2 = -------------------- = -0.1
6
2 ∑f(x) cos3x
a3 = -------------------- = 0.033
6
2 ∑f(x) sinx
b1 = ---------------- = 0.17
6
2 ∑f(x) sin2x
b2 = -------------------- = -0.06
6
2 ∑f(x) sin3x
b3 = -------------------- = 0
6
 f(x) = 1.45 – 0.37cosx + 0.17 sinx – 0.1cos2x – 0.06 sin2x + 0.033 cos3x+…
Example 17
Obtain the first three coefficients in the Fourier cosine series for y, where y is
given in the following table:
x:
0
1
2
3
4
5
y:
4
8
15
7
6
2
Taking the interval as 60o, we have
0o
60o
120o
180o
240o
300o
:
x:
0
1
2
3
4
5
y:
4
8
15
7
6
2
 Fourier cosine series in the interval (0, 2π) is
y = (a0 /2) + a1cos + a2cos2 + a3cos3 + …..
To evaluate the coefficients, we form the following table.
o
0o
60o
120o
180o
240o
300o
cos
1
0.5
-0.5
-1
-0.5
0.5
cos2
1
-0.5
-0.5
1
-0.5
-0.5
cos3
1
-1
1
-1
1
-1
Total
Now, a0 = 2 (42/6) = 14
y
4
8
15
7
6
2
42
y cos
4
4
-7.5
-7
-3
1
-8.5
y cos2
4
-4
-7.5
7
-3
-1
-4.5
y cos3
4
-8
15
-7
6
-2
8
a1 = 2 ( -8.5/6) = - 2.8
a2 = 2 (-4.5/6) = - 1.5
a3 = 2 (8/6) = 2.7
y = 7 – 2.8 cos - 1.5 cos2 + 2.7 cos3 + …..
Example 18
The values of x and the corresponding values of f(x) over a period T are given
below. Show that f(x) = 0.75 + 0.37 cos + 1.004 sin,where  = (2πx )/T
x:
y:
0
1.98
T/6
1.30
T/3
1.05
T/2
1.30
2T/3
-0.88
5T/6
-0.25
T
1.98
We omit the last value since f(x) at x = 0 is known.
Here  = 2πx
T
When x varies from 0 to T,  varies from 0 to 2π with an increase of 2π /6.
Let f(x) = F( ) = (a0/2) + a1 cos + b1 sin.

0
π/3
2π/3
Π
4π/3
5π/3
To evaluate the coefficients, we form the following table.
y
cos
sin
y cos
1.98
1.0
0
1.98
1.30
0.5
0.866
0.65
1.05
-0.5
0.866
-0.525
1.30
-1
0
-1.3
-0.88
-0.5
-0.866
0.44
-0.25
0.5
-0.866
-0.125
4.6
1.12
Now, a0 = 2 ( ∑ f(x) / 6)= 1.5
y sin
0
1.1258
0.9093
0
0.762
0.2165
3.013
a1 = 2 (1.12 /6) = 0.37
a2 = 2 (3.013/6) = 1.004
Therefore, f(x) = 0.75 + 0.37 cos + 1.004 sin
Exercises
1.The following table gives the variations of periodic current over a period.
t (seconds) : 0
T/6
T/3
T/2
2T/3 5T/6
T
A (amplitude): 1.98 1.30 1.05 1.30 -0.88 -0.25 1.98
Show that there is a direct current part of 0.75 amp in the variable current and obtain the
amplitude of the first harmonic.
2.The turning moment T is given for a series of values of the crank angle  = 75

:
0
30
60
90
120
150
180
T
:
0
5224 8097 7850 5499 2626 0
Obtain the first four terms in a series of sines to represent T and calculate
T for  = 75
3. Obtain the constant term and the co-efficient of the first sine and cosine terms in the
Fourier expansion of „y‟ as given in the following table.
X
:
0
1
2
3
4
5
Y
:
9
18
24
28
26
20
4. Find the first three harmonics of Fourier series of y = f(x) from the following data.
X : 0  30
60 90
120 150 180 210 240 270 300 330
Y : 298 356
373 337
254
155
80
51
60
93 147
221
2.8 Complex Form of Fourier Series
The series for f(x) defined in the interval (c, c+2π) and satisfying
∞
Dirichlet‟s conditions can be given in the form of f(x) = ∑ cn e-inx ,
n = -∞
where ,
c+2π
1 ∫ f(x) e – i nx dx
2π c
In the interval (c, c+2ℓ), the complex form of Fourier series is given by
cn =
where,
∞
inπx
f(x) = ∑ cn e ℓ
n=-∞
1
c+2ℓ
-inπx
cn = ------- ∫ f(x) e ℓ dx
c
2ℓ
Example 19
Find the complex form of the Fourier series f(x) = e –x in -1 ≤ x ≤ 1.
∞
inπx
f(x) = ∑ cn e
n=-∞
We have
l
where
cn =
1
2
cn =
1
2
∫
=
=
dx
-1
1
=
-inπx
e –x e
- (1+ i n π) x
∫ e
dx
-1
1
2
e - (1+i nπ) x
- (1+inπ)
1
-1
1 .
-2 ( 1+inπ ) e - (1 + i n π) x -e (1+ i nπ)
(1-inπ) [ e-1 ( cos nπ – isin nπ) - e (cos nπ + i sin nπ) ]
-2 ( 1+n2π2)
= (1-inπ) cos nπ ( e-1 - e )
-2 ( 1+n2π2)
(1-inπ)
Cn = ----------- (-1)n sinh1
(1+n2π2)
∞
(1-inπ)
inπx
 f(x) = ∑
----------- (-1)n sinh1 e
n= - ∞ (1+n2π2)
Example 20
Find the complex form of the Fourier series f(x) = ex in - π < x < π.
∞
We have f(x) = ∑ Cn e i nx
n=-∞
1
π
where Cn = ------ ∫ f(x) e – i nx dx
2π - π
=
=
1
π
------ ∫ ex e –i nx dx
2π - π
1
π
------- ∫ e (1-i n) x dx
2π - π
π
1
2π
=
(1-in)x
e
( 1-in)
-π
=
1
[ e(1-in)π -e - (1-i n) π]
2π(1-in)
(1+in)
= ----------- [ eπ ( cos nπ – i sin nπ ) –e -π ( cosn π + i sin nπ)]
2π(1+n)2
(1+in) (-1)n . e π – e-π
= ----------------------( 1+n2)
2π
(-1)n(1+in) sin h π
= -------------------------( 1+n2) π
 f(x) =
∞ (-1)n(1+in) sin h π
∑ ----------------------n= - ∞
( 1+n2) π
e i nx
Exercises
Find the complex form of the Fourier series of the following functions.
1.f(x) = eax, -l < x < l.
2.f(x) = cosax, - < x < .
3.f(x) = sinx, 0 < x < .
4.f(x) = e-x, -1 < x < 1.
5.f(x) = sinax, a is not an integer in (-,  ).
2.9 SUMMARY(FOURIER SERIES)
A Fourier series of a periodic function consists of a sum of sine and cosine terms.
Sines and cosines are the most fundamental periodic functions.The Fourier series is
named after the French Mathematician and Physicist Jacques Fourier (1768 – 1830).
Fourier series has its application in problems pertaining to Heat conduction, acoustics,
etc. The subject matter may be divided into the following sub topics.
FOURIER SERIES
Series with
arbitrary period
Half-range series
Complex series
Harmonic Analysis
FORMULA FOR FOURIER SERIES
Consider a real-valued function f(x) which obeys the following conditions called
Dirichlet‟s conditions :
1. f(x) is defined in an interval (a,a+2l), and f(x+2l) = f(x) so that f(x) is a periodic
function of period 2l.
2. f(x) is continuous or has only a finite number of discontinuities in the interval
(a,a+2l).
3. f(x) has no or only a finite number of maxima or minima in the interval (a,a+2l).
Also, let
1
a0 
l
an 
1
l
1
bn 
l
Then, the infinite series
a  2l
 f ( x)dx
(1)
a
a  2l

a
a  2l

a
 n 
f ( x) cos
 xdx,
 l 
n  1,2,3,..... (2)
 n 
f ( x) sin
 xdx,
 l 
n  1,2,3,...... (3)
a0 
 n 
 n 
  an cos
 x  bn sin
x
2 n 1
 l 
 l 
(4)
is called the Fourier series of f(x) in the interval (a,a+2l). Also, the real numbers a0, a1,
a2, ….an, and b1, b2 , ….bn are called the Fourier coefficients of f(x). The formulae (1),
(2) and (3) are called Euler‟s formulae.
It can be proved that the sum of the series (4) is f(x) if f(x) is continuous at x. Thus we
have

a
 n 
 n 
f(x) = 0   an cos
(5)
 x  bn sin
 x …….
2 n 1
 l 
 l 
Suppose f(x) is discontinuous at x, then the sum of the series (4) would be
1
f ( x )  f ( x )
2
where f(x+) and f(x-) are the values of f(x) immediately to the right and to the left of f(x)
respectively.


Particular Cases
Case (i)
Suppose a=0. Then f(x) is defined over the interval (0,2l). Formulae (1), (2), (3) reduce
to
2l
1
a0   f ( x)dx
l0
1
 n 
an   f ( x) cos
 xdx,
l0
 l 
2l
n  1,2,......
(6)
1
 n 
f ( x) sin 
 xdx,

l0
 l 
2l
bn 
Then the right-hand side of (5) is the Fourier expansion of f(x) over the interval (0,2l).
If we set l=, then f(x) is defined over the interval (0,2). Formulae (6) reduce to
a0 =
an 
bn 
1

1

 f ( x)dx
0
1

2
2
 f ( x) cos nxdx ,
0
n=1,2,….. 
(7)
2
 f ( x) sin nxdx
n=1,2,….. 
0
Also, in this case, (5) becomes
f(x) =
a0 
  an cos nx  bn sin nx
2 n 1
(8)
Case (ii)
Suppose a=-l. Then f(x) is defined over the interval (-l , l). Formulae (1), (2) (3) reduce
to
l
a0 
1
f ( x)dx
l l
1
 n 
an   f ( x)cos
 xdx
l l
 l 
l
n =1,2,…… 
1
 n
f ( x)sin

l l
 l
l
bn 
(9)

 xdx,

Then the right-hand side of (5) is the Fourier expansion of f(x) over the interval (-l , l).
If we set l = , then f(x) is defined over the interval (-, ). Formulae (9) reduce to
a0 =
an 
bn 
1

1

 f (x)dx

1



 f (x)cos nxdx ,

n=1,2,….. 
(10)

 f (x)sin nxdx
n=1,2,….. 

Putting l =  in (5), we get
f(x) =
a0 
  an cos nx  bn sin nx
2 n 1
Some useful results :
1. The following rule called Bernoulli‟s generalized rule of integration by parts is useful
in evaluating the Fourier coefficients.
'
''
 uvdx  uv1  u v2  u v3  .......
Here u , u  ,….. are the successive derivatives of u and
v1   vdx,v2   v1dx,......
We illustrate the rule, through the following
examples
:
  sin nx   cos nx 
2
2   cos nx 
sin


2
x
x
nxdx
x



  2 3 
2

n 

 n
  n 
2x
2x


 e2 x   e2 x 
3 2x
3 e
2 e





  6

x
e
dx

x

3
x

6
x

 2 
 4 




 8   16 
2.
The following integrals are also useful :
e ax
 e cos bxdx  a 2  b2 a cos bx  b sin bx
e ax
ax
a sin bx  b cos bx
e
sin
bxdx


a 2  b2
ax
3.
If „n‟ is integer, then
sin n = 0 ,
cosn = (-1)n ,
sin2n = 0,
cos2n=1
ASSIGNMENT
1. The displacement y of a part of a mechanism is tabulated with corresponding angular
movement x0 of the crank. Express y as a Fourier series upto the third harmonic.
x0
0
30
60
90
120
150
180
210
240
270
300
330
y
1.80
1.10
0.30
0.16
1.50
1.30
2.16
1.25
1.30
1.52
1.76
2.00
2. Obtain the Fourier series of y upto the second harmonic using the following table :
x0
45
90
135
180
225
270
315
360
y
4.0
3.8
2.4
2.0
-1.5
0
2.8
3.4
3. Obtain the constant term and the coefficients of the first sine and cosine terms in the
Fourier expansion of y as given in the following table :
x
0
1
2
3
4
5
y
9
18
24
28
26
20
4. Find the Fourier series of y upto the second harmonic from the following table :
x
0
2
4
6
8
10
12
Y
9.0
18.2
24.4
27.8
27.5
22.0
9.0
5. Obtain the first 3 coefficients in the Fourier cosine series for y, where y is given below
x
0
1
2
3
4
5
y
4
8
15
7
6
2
UNIT – III
APPLICATIONS OF PARTIAL DIFFERENTIAL
EQUATIONS
3.1 INTRODUCTION
In Science and Engineering problems, we always seek a solution of the
differential equation which satisfies some specified conditions known as the boundary
conditions. The differential equation together with the boundary conditions constitutes a
boundary value problem. In the case of ordinary differential equations, we may first find
the general solution and then determine the arbitrary constants from the initial values. But
the same method is not applicable to partial differential equations because the general
solution contains arbitrary constants or arbitrary functions. Hence it is difficult to adjust
these constants and functions so as to satisfy the given boundary conditions. Fortunately,
most of the boundary value problems involving linear partial differential equations can be
solved by a simple method known as the method of separation of variables which
furnishes particular solutions of the given differential equation directly and then these
solutions can be suitably combined to give the solution of the physical problems.
3.2Solution of the wave equation
The wave equation is
2y
2y
2
t
= a
2
x
2
-----------(1) .
Let y = X(x) . T(t) be the solution of (1), where „X‟ is a function of „x‟ only and „T‟ is a
function of „t‟ only.
2y
2y
Then
= X T′′
and
= X′′ T.
2
2
t
x
Substituting these in (1), we get
X T′′ = a2 X′′ T.
X′′
i.e,
T′′
=
X
---------------(2).
a2T
Now the left side of (2) is a function of „x‟ only and the right side is a function of „t‟ only.
Since „x‟ and „t‟ are independent variables, (2) can hold good only if each side is equal to
a constant.
X′′
T′′
Therefore,
=
= k (say).
2
X
aT
Hence, we get X′′  kX = 0 and T′′  a2 kT = 0. --------------(3).
Solving equations (3), we get
(i) when „k‟ is positive and k = 2, say
X = c1 ex + c2 e - x
T = c3 eat + c4 e - at
(ii) when „k‟ is negative and k = 2, say
X = c5 cosx + c6 sin x
T = c7 cosat + c8 sin at
(iii) when „k‟ is zero.
X = c9 x + c10
T = c11 t + c12
Thus the various possible solutions of the wave equation are
y = (c1 ex + c2 e - x) (c3 eat + c4 e - at)
------------(4)
y = (c5 cosx + c6 sin x) (c7 cosat + c8 sin at) -----------(5)
y = (c9 x + c10) (c11 t + c12)
------------(6)
Of these three solutions, we have to select that particular solution which suits the
physical nature of the problem and the given boundary conditions. Since we are dealing
with problems on vibrations of strings, „y‟ must be a periodic function of „x‟ and „t‟.
Hence the solution must involve trigonometric terms.
Therefore, the solution given by (5),
i.e,
y = (c5 cosx + c6 sin x) (c7 cosat + c8 sin at)
is the only suitable solution of the wave equation.
llustrative Examples.
Example 1
If a string of length ℓ is initially at rest in equilibrium position and each of its points is
given
y
x
the velocity
t
= vo sin
t=0
, 0  x  ℓ. Determine the displacement y(x,t).
ℓ
Solution
The displacement y(x,t) is given by the equation
2y
2 y
= a2
t
2
-----------(1)
x
2
The boundary conditions are
i. y(0,t) = 0, for t  0.
ii. y(ℓ,t) = 0, for t  0.
iii. y(x,0) = 0, for 0 x  ℓ.
y
x
iv.
t
= vo sin
t=0
, for 0 x  ℓ.
ℓ
Since the vibration of a string is periodic, therefore, the solution of (1) is of the form
y(x,t) = (Acosx + Bsinx)(Ccosat + Dsinat) ------------(2)
Using (i) in (2) , we get
0 = A(Ccosat + Dsinat) , for all t  0.
Therefore,
A=0
Hence equation (2) becomes
y(x,t) = B sinx(Ccosat + Dsinat) ------------(3)
Using (ii) in (3), we get
0 = Bsinℓ (Ccosat + Dsinat), for all t  0, which gives ℓ = n.
n
=
Hence,
, n being an integer.
ℓ
nx
Thus , y(x,t) = Bsin
nat
Ccos
ℓ
nat
+ Dsin
ℓ
ℓ
------------------(4)
Using (iii) in (4), we get
nx
0 = Bsin
.C
ℓ
which implies

C = 0.
y(x,t) = Bsin
nx
.
ℓ
nat
Dsin
ℓ
nx
= B1sin
nat
. sin
ℓ
, where B1= BD.
ℓ
The most general solution is

nx
y(x,t) =  Bn sin
sin
n=1
ℓ
nat
----------------(5)
ℓ
Differentiating (5) partially w.r.t t, we get
y
=
t

nx
nat
 Bn sin
.cos
.
n=1
ℓ
ℓ
na
ℓ
Using condition (iv) in the above equation, we get
x
vo sin

= 
n=1
ℓ
na
Bn .
x
nx
. sin
ℓ
ℓ
a
i.e, vo sin
· · · · ·
=
x
B1 .
2a
. sin
ℓ
ℓ
+
B2 .
ℓ
2x
. sin
ℓ
+ · ·
ℓ
Equating like coefficients on both sides, we get
a
B1
2a
= vo , B2 .
ℓ
3a
= 0, B3
ℓ
= 0, · · · · · · · ·
ℓ
voℓ
i.e, B1 =
a
,
B2 = B3 = B4 = B5 = · · · · · · · · = 0.
Substituting these values in (5), we get the required solution.
x
voℓ
i.e,
y(x,t) =
sin
a
at
.
sin
ℓ
ℓ
Example 2
A tightly stretched string with fixed end points x = 0 & x = ℓ is initially at
rest in its equilibrium position . If it is set vibrating by giving to each of its points a
velocity
y/t = kx(ℓ-x) at t = 0. Find the displacement y(x,t).
Solution
The displacement y(x,t) is given by the equation
2y
2
2 y
=a
t2
-----------(1)
x2
The boundary conditions are
i.
ii.
iii.
iv.
y(0,t) = 0, for t  0.
y(ℓ,t) = 0, for t  0.
y(x,0) = 0, for 0 x  ℓ.
y
= kx(ℓ – x), for 0 x  ℓ.
t t = 0
Since the vibration of a string is periodic, therefore, the solution of (1) is of the form
y(x,t) = (Acosx + Bsinx)(Ccosat + Dsinat) ------------(2)
Using (i) in (2) , we get
0 = A(Ccosat + Dsinat) , for all t  0.
which gives
A = 0.
Hence equation (2) becomes
y(x,t) = B sinx(Ccosat + Dsinat) ------------(3)
Using (ii) in (3), we get
0 = Bsinℓ(Ccosat + Dsinat), for all t  0.
which implies ℓ = n.
n
Hence,
=
, n being an integer.
ℓ
nx
nat
nat
Thus , y(x,t) = Bsin
Ccos
+ Dsin
------------------(4)
ℓ
ℓ
ℓ
Using (iii) in (4), we get
nx
0 = Bsin
.C
ℓ
Therefore, C = 0 .
nx
nat
Hence, y(x,t) = Bsin
. Dsin
ℓ
ℓ
nx
= B1sin
nat
. sin
ℓ
ℓ
, where B1= BD.
The most general solution is

nx
y(x,t) =  Bn sin
sin
n=1
ℓ
nat
----------------(5)
ℓ
Differentiating (5) partially w.r.t t, we get
y

nx
nat
=  Bn sin
.cos
.
t
n=1
ℓ
ℓ
Using (iv), we get

na
kx(ℓ-x) =  Bn.
n=0
ℓ
2
na
i.e, Bn .
=
ℓ
ℓ
na
ℓ
nx
. sin
ℓ
ℓ
nx
 f(x). sin
dx
0
ℓ
2
i.e,
Bn
=
ℓ
nx
 f(x). sin
dx
na 0
ℓ
2
ℓ
nx
 kx(ℓ – x) sin
dx
na 0
ℓ
nx
ℓ
– cos
2k 
ℓ
2
=
 (ℓx – x ) d
na 0
nx
ℓ
=
nx
nx
-cos
=
2k
na
(ℓx- x2) d
-sin
ℓ
n
ℓ
- (ℓ-2x)
ℓ
n22
ℓ2
2k
-2cosn
2
=
+
na
n33
n33
ℓ3
ℓ3
2ℓ3
2k
=
.
n
{1 - cosn}
3 3
na
4 kℓ3
i.e,
Bn =
{1 – (-1)n}
n a
4 4
8kℓ3
or
Bn =
,
if n is odd
n a
4 4
0,
if n is even
Substituting in (4), we get

8kℓ3
nat
nx
y(x,t) = 
.
sin
sin
n=1,3,5,…… n44 a
ℓ
ℓ
Therefore the solution is
8kℓ3

l
(2n-1)at
(2n-1)x
y(x,t) =

sin
sin
4 a n=1 (2n-1)4
ℓ
ℓ
Example 3
A tightly stretched string with fixed end points x = 0 & x = ℓ is initially in a
position given by y(x,0) = y0sin3(x/ℓ). If it is released from rest from this position, find
the displacement y at any time and at any distance from the end x = 0 .
Solution
The displacement y(x,t) is given by the equation
2y
2 y
2
=a
t
2
-----------(1)
x
2
The boundary conditions are
(i) y(0,t) = 0,  t  0.
(ii) y(ℓ,t) = 0,  t  0.
(iii) y
= 0, for 0 < x < ℓ.
t t = 0
(iv) y(x,0) = y0 sin3((x/ℓ), for 0 < x < ℓ.
The suitable solution of (1) is given by
y(x,t) = (Acosx + Bsinx)(Ccosat + Dsinat) ------------(2)
Using (i) and (ii) in (2) , we get
n
A=0 & =
ℓ
nat
nx
 y(x,t) = B sin
(Ccos
ℓ
y
Now,
nat
+ Dsin
) -----------(3)
ℓ
ℓ
nx
nat
= B sin
- Csin
t
na
.
ℓ
ℓ
nat
+ Dcos
ℓ
na
.
ℓ
Using (iii) in the above equation , we get
nx
0 = B sin
na
D
ℓ
ℓ
Here, B can not be zero . Therefore D = 0.
Hence equation (3) becomes
nx
y(x,t) = B sin
= B1sin
nat
. Ccos
ℓ
ℓ
nx
nat
. cos
ℓ
ℓ
, where B1 = BC
The most general solution is

nx
nat
ℓ
y(x,t) =  Bn sin
n=1
ℓ
cos
---------------- (4)
ℓ
Using (iv), we get

nx
=  Bnsin
n=1
ℓ
n
y0 sin3
ℓ

nx
i.e,  Bnsin
n=1
ℓ
x
i.e, B1sin
x
3
= y0
4
ℓ
2x
+ B2 sin
ℓ

sin
4
+ ….
ℓ
x
3y0
=
3x
sin
ℓ
3x
+B3 sin
ℓ
1
sin
4
y0
-
ℓ
3x
sin
4
ℓ
Equating the like coefficients on both sides, we get
3y0
-y0
B1 =
, B2 = B4 = … = 0 .
, B3 =
4
4
Substituting in (4), we get
x
3y0
y(x,t) =
sin
4
at
. cos
ℓ
y0
-
ℓ
3x
sin
4
3at
. cos
ℓ
ℓ
Example 4
A string is stretched & fastened to two points x = 0 and x = ℓ apart.
Motion is
started by displacing the string into the form y(x,0) = k(ℓx-x2) from which it is
released at
time t = 0. Find the displacement y(x,t).
Solution
The displacement y(x,t) is given by the equation
2 y
2
=a
-----------(1)
t2
x2
2y
The boundary conditions are
(i) y(0,t) = 0,  t  0.
(ii) y(ℓ,t) = 0,  t  0.
(iii)
y
t
= 0, for 0 < x < ℓ.
t=0
(iv) y(x,0) = k(ℓx – x2), for 0 < x < ℓ.
The suitable solution of (1) is given by
y(x,t) = (Acosx + Bsinx)(Ccosat + Dsinat) ------------(2)
Using (i) and (ii) in (2) , we get
n
A=0 & =
.
ℓ
 y(x,t) = B sin
nx
nat
+ Dsin
ℓ
) -----------(3)
ℓ
y
Now,
nat
(Ccos
ℓ
nx
nat
= B sin
- Csin
t
na
.
ℓ
ℓ
nat
+ Dcos
ℓ
na
.
ℓ
ℓ
Using (iii) in the above equation , we get
nx
0 = B sin
na
D
ℓ
ℓ
Here, B can not be zero
D=0
Hence equation (3) becomes
nx
y(x,t) = B sin
nat
. Ccos
ℓ
nx
= B1sin
ℓ
nat
.
cos
ℓ
ℓ
, where B1 = BC
The most general solution is

nx
y(x,t) =  Bnsin
nat
cos
---------------- (4)
n=1
ℓ
ℓ

nx
Using (iv), we get
kx(ℓx – x2) =  Bnsin
---------------- (5)
n=1
ℓ
The RHS of (5) is the half range Fourier sine series of the LHS function .
2
ℓ
nx
 Bn =
 f(x) . sin
dx
ℓ 
ℓ
0
nx
ℓ
-cos
2k  (ℓx- x2) d
ℓ
=

n
ℓ 0
ℓ
nx
nx
-cos
2k
-sin
ℓ
2
=
(ℓx- x ) d
ℓ
- (ℓ-2x)
ℓ
n22
ℓ2
n
ℓ
nx
ℓ
cos
ℓ
+ (-2)
n33
ℓ3
2k
=
-2cos n
n33
ℓ3
ℓ
.
ℓ
{1- cos n}
n33
4kℓ2
i.e, Bn
{1- (-1)n }
=
n33
+
2ℓ3
2k
=
2
n33
ℓ3
0
or Bn
=
8kℓ2
n33 , if n is odd
0,
if n is even

8kℓ2
nat
nx
y(x,t) = 
cos
.sin
n=odd n33
ℓ
ℓ
or
y(x,t) =
8k 
1
(2n-1)at
(2n-1)x

cos
.sin
3
3
 n=1 (2n-1)
ℓ
ℓ
Example 5
A uniform elastic string of length 2ℓ is fastened at both ends. The
midpoint of the string is taken to the height „b‟ and then released from rest in
that position . Find the displacement of the string.
Solution
The displacement y(x,t) is given by the equation
2y
2 y
= a2
t
2
-----------(1)
x
2
The suitable solution of (1) is given by
y(x,t) = (Acosx + Bsinx)(Ccosat + Dsinat) ------------(2)
The boundary conditions are
(i) y(0,t) = 0,  t  0.
(ii) y(ℓ,t) = 0,  t  0.
(iii) y
t
= 0, for 0 < x < 2ℓ.
t=0
b
O(0,0)
ℓ
(b/ℓ)x ,
B(2ℓ,0)
x
0<x<ℓ
(iv) y(x,0) =
-(b/ℓ)(x-2ℓ), ℓ<x<2ℓ
[Since, equation of OA is y = (b/ℓ)x and equation of AB is (y-b)/(o-b) = (x-ℓ)/(2ℓ-ℓ)]
Using conditions (i) and (ii) in (2), we get
n
A=0 & =
2ℓ
 y(x,t) = B sin
nx
nat
(Ccos
2ℓ
2ℓ
nx
nat
= B sin
t
) -----------(3)
2ℓ
y
Now,
nat
+ Dsin
- Csin
2ℓ
na
.
2ℓ
+ Dcos
2ℓ
Using (iii) in the above equation , we get
nx
0 = B sin
na
D
2ℓ
2ℓ
Here B can not be zero, therefore D = 0.
Hence equation (3) becomes
nx
y(x,t) = B sin
nat
. Ccos
2ℓ
nx
nat
2ℓ
nat
na
.
2ℓ
2ℓ
= B1sin
.
cos
2ℓ
, where B1 = BC
2ℓ
The most general solution is

nx
y(x,t) =  Bnsin
n=1
2ℓ
nat
cos
---------------- (4)
2ℓ
Using (iv), We get

nx
y(x,0) =  Bn .sin
---------------- (5)
n=1
2ℓ
The RHS of equation (5) is the half range Fourier sine series of the LHS function .
2 2ℓ
nx
 Bn =
 f(x) . sin
dx
2ℓ 
2ℓ
0
1
ℓ
nx
 f(x) . sin
dx

2ℓ
0
=
ℓ
1
=
ℓ
1
=
ℓ
2ℓ
nx
+  f(x) . sin
dx

2ℓ
ℓ
ℓ b
nx
2ℓ -b
nx

x sin
dx + 
(x-2ℓ) sin
dx
 ℓ
2ℓ
 ℓ
2ℓ
0
ℓ
nx
nx
ℓ
-cos
2ℓ
-cos
b 
2ℓ
b  (x-2ℓ) d
ℓ
 xd

ℓ 0
n
ℓ ℓ
n
2ℓ
2ℓ
ℓ
cos
1
b
=
nx
2ℓ
 (1)
(x)
ℓ
ℓ
nx
sin
2ℓ
n
2ℓ

n
4ℓ2
2 2
0
n
n
b
n
sin
-ℓcos
2
=
2
n
2ℓ
8b sin (n/2)
sin
2
+
ℓ2
n
ℓcos
+
2
+
n22
4ℓ2
n
2ℓ
n22
4ℓ2
=
n22
Therefore the solution is

nat
y(x,t) =  8bsin(n/2) cos
n=1
n22
2ℓ
nx
sin
2ℓ
Example 6
A tightly stretched string with fixed end points x = 0 & x = ℓ is
initially in
the position y(x,0) = f(x). It is set vibrating by giving to each of its points a
velocity
y
= g(x) at t = 0 . Find the displacement y(x,t) in the form of Fourier series.
t
Solution
The displacement y(x,t) is given by the equation
2y
2 y
= a2
t
2
-----------(1)
x
2
The boundary conditions are
(i) y(0,t) = 0,  t  0.
(ii) y(ℓ,t) = 0,  t  0.
(iii) y(x,0) = f(x) , for 0  x  ℓ.
(iv)  u
t
= g(x), for 0  x  ℓ.
t=0
The solution of equation .(1) is given by
y(x,t) = (Acosx + Bsinx)(Ccosat + Dsinat) ------------(2)
where A, B, C, D are constants.
Applying conditions (i) and (ii) in (2), we have
n
A = 0 and  =
.
ℓ
Substituting in (2), we get
nx
y(x,t) = B sin
nat
(Ccos
ℓ
nx
y(x,t) = sin
ℓ
nat
+ Dsin
ℓ
nat
(B1cos
ℓ
+ D1 sin
)
ℓ
nat
) where B1 = BC and D1 = BD.
ℓ
The most general solution. is

nat
y(x,t) =  Bn cos
n=1
ℓ
nat
+ Dn .sin
nx
.sin
ℓ
--------------(3)
ℓ
Using (iii), we get

nx
f(x) =  Bn .sin
n=1
ℓ
---------------- (4)
The RHS of equation (4) is the Fourier sine series of the LHS function.
2
ℓ
nx
 Bn =
 f(x) . sin
dx
ℓ 
ℓ
0
Differentiating (3) partially w.r.t „t‟, we get
y
t

nat
=  -Bn sin
n=1
ℓ
na
nat
na
+ Dn .cos
ℓ
nx
.sin
ℓ
ℓ
ℓ
Using condition (iv) , we get

na
g(x) =  Dn
n=1
ℓ
nx
. sin
-----------------(5)
ℓ
The RHS of equation (5) is the Fourier sine series of the LHS function.
ℓ
nx
 Dn .
=
 g(x) . sin
dx
ℓ
ℓ 
ℓ
0
2
ℓ
nx
 Dn =
 g(x) . sin
dx
na 
ℓ
0
Substituting the values of Bn and Dn in (3), we get the required solution of the
given equation.
na
2
Exercises
(1) Find the solution of the equation of a vibrating string of length „ℓ‟, satisfying the
conditions
y(0,t) = y(ℓ,t) = 0 and y = f(x), y/ t = 0 at t = 0.
(2) A taut string of length 20 cms. fastened at both ends is displaced from its position of
equilibrium, by imparting to each of its points an initial velocity given by
v=x
in 0  x  10
= 20  x in 10  x  20,
„x‟ being the distance from one end. Determine the displacement at any subsequent time.
(3) Find the solution of the wave equation
2u
2u
2
=c
,
t2
x2
corresponding to the triangular initial deflection f(x ) = (2k/ ℓ) x
when 0 x ℓ/ 2
= (2k/ ℓ) (ℓ – x) when ℓ/ 2 x ℓ,
and initial velocity zero.
(4) A tightly stretched string with fixed end points x = 0 and x = ℓ is initially at rest in its
equilibrium position. If it is set vibrating by giving to each of its points a velocity y/ t
= f(x)
at t = 0. Find the displacement y(x,t).
(5) Solve the following boundary value problem of vibration of string
i.
ii.
iii.
iv.
y(0,t) = 0
y(ℓ,t) = 0
y
(x,0) = x (x – ℓ), 0 x ℓ.
t
y(x,0) = x
in 0 x ℓ/ 2
= ℓ – x in ℓ/ 2 x ℓ.
(6) A tightly stretched string with fixed end points x = 0 and x = ℓ is initially in a
position given by y(x,0) = k( sin(x/ ℓ) – sin( 2x/ ℓ)). If it is released from rest, find the
displacement of „y‟ at any distance „x‟ from one end at any time „t‟.
3.3 Solution of the heat equation
The heat equation is
u
t
= 2
2u
----------------(1).
x
2
Let u = X(x) . T(t) be the solution of (1), where „X‟ is a function of „x‟ alone and „T‟ is a
function of „t‟ alone.
Substituting these in (1), we get
X T′ = 2 X′′ T.
X′′
i.e,
T′
=
---------------(2).
X
T
Now the left side of (2) is a function of „x‟ alone and the right side is a function of „t‟
alone. Since „x‟ and „t‟ are independent variables, (2) can be true only if each side is
equal to a constant.
2
X′′
Therefore,
T′
=
X
= k (say).
2T
Hence, we get X′′  kX = 0 and T′  2 kT = 0. --------------(3).
Solving equations (3), we get
(i) when „k‟ is positive and k = 2, say
X = c1 ex + c2 e - x
2 2
 t
T = c3 e 
(ii) when „k‟ is negative and k = 2, say
X = c4 cosx + c5 sin x
2 2
 t
T = c6 e  
(iii) when „k‟ is zero.
X = c7 x + c8
T = c9
Thus the various possible solutions of the heat equation (1) are
u = (c1 ex + c2 e - x) c3 e 
2 2
 t
u = (c4 cosx + c5 sin x) c6 e
u = (c7 x + c8) c9
2 2
  t
-----------(4)
----------(5)
------------(6)
Of these three solutions, we have to choose that solution which suits the physical
nature of the problem and the given boundary conditions. As we are dealing with
problems on heat flow, u(x,t) must be a transient solution such that „u‟ is to decrease with
the increase of time „t‟.
Therefore, the solution given by (5),
u = (c4 cosx + c5 sin x) c6 e
2 2
  t
is the only suitable solution of the heat equation.
Illustrative Examples
Example 7
A rod „ℓ‟ cm with insulated lateral surface is initially at temperature f(x) at an
inner point of distance x cm from one end. If both the ends are kept at zero temperature,
find the temperature at any point of the rod at any subsequent time.
Let the equation for the conduction of heat be
u
2u
--------- = 2 ---------t
x2
The boundary conditions are
(i)
u (0,t) = 0,
 t≥0
(ii)
u (ℓ,t) = 0,
t>0
(iii) u (x,0) = f (x), 0 < x < ℓ
------------- (1)
The solution of equation (1) is given by
2 2
 t
u (x,t) = (A cosx + B sinx) e –
--------------- (2)
Applying condition (i) in (2), we have
0 = A.e - 
2 2
t
which gives A = 0
2  2t
 u (x,t) = B sinx e –
-------------- (3)
2 2
Applying condition (ii) in the above equation, we get 0 = Bsinℓ e -  t
n
i.e, ℓ = n or  = --------- (n is an integer)
ℓ
-n222
nx
------------- t
 u (x,t) = B sin --------e
ℓ2
ℓ
Thus the most general solution is
-n222

nx
------------- t
u (x,t) =  Bn sin --------e
ℓ2
------------- (4)
n=1
ℓ
By condition (iii),

nx
u (x,0) =  Bn sin ----------- = f (x).
n=1
ℓ
The LHS series is the half range Fourier sine series of the RHS function.
2
nx
ℓ
 Bn = ------  f (x) sin -------- dx
0
ℓ
ℓ
Substituting in (4), we get the temperature function

2
nx
ℓ
u (x,t) =  ------  f (x) sin -------- dx
n=1
ℓ 0
ℓ
Example 8
nx
sin --------ℓ
-n222
------------ t
e
ℓ2
u
2u
The equation for the conduction of heat along a bar of length ℓ is ------ = 2 -------
--,
t
x2
neglecting radiation. Find an expression for u, if the ends of the bar are maintained at
zero temperature and if, initially, the temperature is T at the centre of the bar and falls
uniformly to
zero at its ends.
x
P
A
B
Let u be the temperature at P, at a distance x from the end A at time t.
u
2u
The temperature function u (x,t) is given by the equation ------ = 2 ---------, ----------(1)
t
x2
The boundary conditions are
(i)
u (0,t) = 0,  t > 0.
(ii)
u (ℓ,t) = 0,  t > 0.
u(x,0)
A(ℓ/2,T)
T
B(ℓ,0)
O(0,0)
L
ℓ
X
2Tx
ℓ
u(x,0) = ----------, for 0 < x < ----ℓ
2
2T
ℓ
= ----------(ℓ - x), for ----- < x < ℓ
ℓ
2
The solution of (1) is of the form
2 2
 t
u (x,t) = (A cosx + B sinx) e -
Applying conditions (i) and (ii) in (2), we get
----------(2)
n
A = 0 &  = ------ℓ
nx
 u (x,t) = B sin --------ℓ
-n222
------------- t
e
ℓ2
Thus the most general solution is

nx
 u (x,t) =  Bn sin --------n=1
ℓ
-n222
------------- t
e
ℓ2
------------- (3)
Using condition (iii) in (3), we have

nx
u (x,0) =  Bn sin ---------------------- (4)
n=1
ℓ
We now expand u (x,0) given by (iii) in a half – range sine series in (0,ℓ)
2
nx
ℓ
Here Bn = ------  u (x,0) sin -------- dx
ℓ 0
ℓ
2
ie, Bn = -----ℓ
2Tx
nx
2T
nx
ℓ
 --------- sin -------- dx +  -------- (ℓ-x) sin -------- dx
0
ℓ/2
ℓ
ℓ
ℓ
ℓ
ℓ/2
nx
nx
- cos --------- cos ----------4T ℓ/2
ℓ
ℓ
ℓ
= ------  x d
---------------- + (ℓ-x) d ------------------------ℓ2
4T
= -----ℓ2
0
n/ℓ
nx
- cos --------ℓ
(x) -----------------n/ℓ
ℓ/2
– (1)
n/ℓ
nx
- sin ----------ℓ
------------------------
ℓ/2
+
n22/ℓ2
o
- cos --------- sin ---------ℓ
ℓ
(ℓ - x) ------------------ – (-1) ---------------------n/ℓ
n22/ℓ2
ℓ
ℓ/2
4T
- ℓ2
n
ℓ2
n
ℓ2
n
ℓ2
n
= -------- --------- cos ------- + ---------- sin ------- + ------- cos ----- +-------- sin -----ℓ2
2n
2
n22
2
2n
2
n22
2
4T
= ------ℓ2
2ℓ2
n
-------- sin ------n22
2
8T
n
Bn = --------- sin------n22
2
Hence the solution is
-n222
8T
n
nx

------------- t
u (x,t) =  ---------- sin -------- sin --------- e
ℓ2
n=1
2 2
n
2
ℓ
or
-n222
8T
n
nx
------------- t
u (x,t) = 
---------- sin -------- sin --------- e ℓ2
n=1,3,5…
n22
2
ℓ
or
-2 (2n-1)22
----------------- t
8T 
(-1) n+1
(2n-1)x
ℓ2
u (x,t) = ---------  -------------- sin --------------- e
2 n=1 (2n-1)2
ℓ

Steady - state conditions and zero boundary conditions
Example 9
A rod of length „ℓ‟ has its ends A and B kept at 0C and 100C until steady state
conditions prevails. If the temperature at B is reduced suddenly to 0C and kept so while
that of A is maintained, find the temperature u(x,t) at a distance x from A and at time „t‟.
The heat-equation is given by
u
2u
2
--------- =  -------------------- (1)
t
x2
Prior to the temperature change at the end B, when t = 0, the heat flow was
independent of time (steady state condition).
When the temperature u depends only on x, equation(1) reduces to
2u
-------- = 0
x2
Its general solution is u = ax + b
------------- (2)
100
Since u = 0 for x = 0 & u = 100 for x = ℓ, therefore (2) gives b = 0 & a = --------ℓ
100
u (x,0) = ------- x, for 0 < x < ℓ
ℓ
Hence the boundary conditions are
(i) u (0,t)
(ii) u (ℓ,t)
(iii) u (x,0)
= 0,
 t0
= 0,
 t0
100x
= ----------- , for 0 < x < ℓ
ℓ
The solution of (1) is of the form
22 t
u (x,t) = (A cos x + B sin x) e -
Using, conditions (i) and (ii) in (3), we get
n
A = 0 &  = -------ℓ
-n222
nx
------------- t
 u (x,t) = B sin --------e
ℓ2
ℓ
Thus the most general solution is
-n222
--------------- (3)

nx
 u (x,t) =  Bn sin --------n=1
ℓ
Applying (iii) in (4), we get
------------- t
e
ℓ2
------------- (4)
nx

u (x,0) =  Bn sin --------n=1
ℓ
100x
nx

ie, ------- =  Bn sin --------n=1
ℓ
ℓ
==> Bn
2 ℓ 100x
nx
= -----  -------- sin ------- dx
ℓ 0
ℓ
ℓ
200 ℓ
= --------  x
ℓ2
0
200
= -------ℓ2
nx
- cos -------ℓ
d ------------------n
-------ℓ
nx
- cos ---------ℓ
(x) ------------------n
-------ℓ
200
–ℓ2
= --------- ------- cos n
ℓ2
n
200 (-1) n+1
Bn = -----------------n
Hence the solution is
nx
- sin ---------ℓ
– (1) ---------------------n22
----------ℓ2
ℓ
0
-n222 t
e
ℓ2
200 (-1) n+1
nx
 -------------- sin ---------n=1
n
ℓ

u (x,t) =
Example 10
A rod, 30 c.m long, has its ends A and B kept at 20C and 80C respectively, until
steady state conditions prevail. The temperature at each end is then suddenly reduced to
0C and kept so. Find the resulting temperature function u (x,t) taking x = 0 at A.
The one dimensional heat flow equation is given by
u
2u
------- = 2 --------t
------------ (1)
x2
u
In steady-state, ------ = 0.
t
2u
Now, equation (1) reduces to --------- = 0
x2
Solving (2), we get u = ax + b
------------- (2)
------------- (3)
The initial conditions, in steady – state, are
u = 20, when x = 0
u = 80, when x = 30
Therefore, (3) gives b = 20, a = 2.
u (x) = 2x + 20
------------- (4)
Hence the boundary conditions are
(i)
u (0,t) = 0,
 t>0
(ii)
u (30,t) = 0,  t > 0
(iii) u (x,0) = 2x + 20, for 0 < x < 30
The solution of equation (1) is given by
2 2
 t
u (x,t) = (A cos x + Bsinx) e -
Applying conditions (i) and (ii), we get
----------------- (5)
n
A = 0,  = --------, where „n‟ is an integer
30
-2n22
nx---------- t
 u (x,t) = B sin --- e ---- 900
------------- (6)
30
The most general solution is

 u (x,t) =  Bn sin
n=1
-2n22
nx ----------- t
e ----- 900
30
------------- (7)
Applying (iii) in (7), we get
nx
u (x,0) =  Bn sin ------- = 2x +20, 0 < x < 30.
n=1
30

 Bn
2 30
= -----  (2x + 20) sin
30 0
1 30
= --------  (2x+ 20) d
15 0
nx
------- dx
30
nx
- cos ---------30
------------------n
-------30
nx
1
= -----15
- cos ---------30
(2x+20) ------------------n
-------30
nx
-----30
– (2) ---------------------n22
-------900
30
- sin
0
1
= ------15
Bn
–2400 cosn
600
----------------- + -------n
n
40
= ------- {1 – 4 ( - 1)n }
n
Hence, the required solution is
-2n22
----------- t
 40
nx
n
u (x,t) =  ------- {1 – 4 ( -1) } sin ---e ---- 900
30
n=1 n
Steady–state conditions and non–zero boundary conditions
Example 11
The ends A and B of a rod 30cm. long have their temperatures kept at 20C and
80C, until steady–state conditions prevail. The temperature of the end B is suddenly
reduced to 60C and kept so while the end A is raised to 40C. Find the temperature
distribution in the rod after time t.
Let the equation for the heat- flow be
u
2u
------- = 2 --------t
x2
----------- (1)
2u
In steady–state, equation (1) reduces to -------- = 0.
x2
Solving, we get
u = ax + b
-------------- (2)
The initial conditions, in steady–state, are
u = 20,
u = 80,
when x = 0
when x = 30
From (2), b = 20 & a = 2.
Thus the temperature function in steady–state is
u (x) = 2x + 20
-------------- (3)
Hence the boundary conditions in the transient–state are
(i)
u (0,t) = 40,  t > 0
(ii)
u (30,t) = 60,  t > 0
(iii) u (x,0) = 2x + 20, for 0 < x < 30
we break up the required funciton u (x,t) into two parts and write
u (x,t) = us (x) + ut (x,t)
--------------- (4)
where us (x) is a solution of (1), involving x only and satisfying the boundary
condition (i) and (ii). ut (x,t) is then a function defined by (4) satisfying (1).
Thus us(x) is a steady state solution of (1) and ut(x,t) may therefore be regarded
as a transient solution which decreases with increase of t.
To find us(x)
2u
we have to solve the equation --------- = 0
x2
Solving, we get us(x) = ax + b
------------- (5)
Here us(0) = 40, us(30) = 60.
Using the above conditions, we get b = 40, a = 2/3.
2
us(x) = ------ x + 40
3
To find ut(x,t)
-------------- (6)
ut ( x,t) = u (x,t) – us (x)
Now putting x = 0 and x = 30 in (4), we have
ut (0,t) = u (0,t) – us (0) = 40–40 = 0
and
ut (30,t) = u (30,t) – us (30) = 60–60 = 0
Also ut (x,0) = u (x,0) – us (x)
2
= 2x + 20 – ------ x– 40
3
4
= ------- x – 20
3
Hence the boundary conditions relative to the transient solution ut (x,t) are
ut (0,t) = 0
--------------(iv)
ut (30,t) = 0
--------------(v)
ut (x,0) = (4/3) x – 20 -------------(vi)
and
-22t
We have
e Bsinx)
ut(x,t) = (Acosx +
--------------(7)
---e
Using condition (iv) and (v) in (7), we get
n
A = 0 &  = --------30
Hence equation (7) becomes
-2n22
nx
----------- t
ut (x,t) = B sin ----e ---- 900
e
30
The most general solution of (1) is
-2n22
nx

----------- t
ut(x,t) =  Bnsin ---- e --- 900
-----------------(8)
n=1
30
Using condition (vi) ,
nx

ut (x,0) =  Bn sin ------- = (4/3) x–20, 0 < x < 30.
n=1
30
 Bn
2 30
nx
= -----  {(4/3) x–20} sin -------- dx
30 0
30
1 30 4
= ------  ----- x–20 d
15 0 3
nx
- cos ---------30
------------------n
-------30
nx
nx
- cos ---------1
= -----15
4
----- x–20
3
1
= ------15
30
--------------n
-------30
- sin ---------4
– ----3
30
------------------n22
-------900
–600 cosn
600
----------------- – -------n
n
– 40
= ------- {1 + cos n }
n
– 40 { 1 + (-1)n }
Bn = ------------------------n
or Bn
= 0 , when n is odd
-80
-------, when n is even
n

ut (x,t)
-80
=
------n=2,4,6, . . . n
-2n22
- ---------- t
nx 900
sin -- e ---30
 u (x,t) = us (x) + ut (x,t)
-2n22
2
80 
1
nx
----------- t
ie, u (x,t) = ------x + 40 – ------ 
----- sin -- e ----- 900
3
n=2,4,6,..
n
30
Exercises
(1) Solve u/ t = 2 (2u / x2) subject to the boundary conditions u(0,t) = 0,
u(l,t) = 0, u(x,0) = x, 0 x l.
(2) Find the solution to the equation u/ t = 2 (2u / x2) that satisfies the conditions
i.
u(0,t) = 0,
ii. u(l,t) = 0,  t  0,
iii. u(x,0) = x for 0 x l/ 2.
= l – x for l/ 2 x l.
(3) Solve the equation u/ t = 2 (2u / x2) subject to the boundary conditions
i.
u(0,t) = 0,
ii. u(l,t) = 0,  t  0,
iii. u(x,0) = kx(l – x), k  0, 0  x  l.
(4) A rod of length „l‟ has its ends A and B kept at 0o C and 120o C respectively until
steady state conditions prevail. If the temperature at Bis reduced to 0o C and kept so while
that of A is maintained, find the temperature distribution in the rod.
(5) A rod of length „l‟ has its ends A and B kept at 0o C and 120o C respectively until
steady state conditions prevail. If the temperature at Bis reduced to 0o C and kept so while
10o C and at the same instant that at A is suddenly raised to 50o C. Find the temperature
distribution in the rod after time „t‟.
(6) A rod of length „l‟ has its ends A and B kept at 0o C and 100o C respectively until
steady state conditions prevail. If the temperature of A is suddenly raised to 50o C and
that of B to
150o C, find the temperature distribution at the point of the rod and at any time.
(7) A rod of length 10 cm. has the ends A and B kept at temperatures 30o C and 100o C,
respectively until the steady state conditions prevail. After some time, the temperature at
A is lowered to 20o C and that of B to 40o C, and then these temperatures are maintained.
Find the subsequent temperature distribution.
(8) The two ends A and B of a rod of length 20 cm. have the temperature at 30o C and
80o C respectively until th steady state conditions prevail. Then the temperatures at the
ends A and B are changed to 40o C and 60o C respectively. Find u(x,t).
(9) A bar 100 cm. long, with insulated sides has its ends kept at 0o C and 100o C until
steady state condition prevail. The two ends are then suddenly insulated and kept so. Find
the temperature distribution
(10) Solve the equation u/ t = 2 (2u / x2) subject to the conditions (i) „u‟ is not
infinite
as t  (ii) u = 0 for x = 0 and x = ,  t (iii) u = x  x2 for t = 0 in (0, ).
3.4 Solution of Laplace’s equation(Two dimentional heat equation)
The Laplace equation is
2u
2u
+
x
= 0
y
2
2
Let u = X(x) . Y(y) be the solution of (1), where „X‟ is a function of „x‟ alone and „Y‟ is
a function of „y‟ alone.
2u
2u
Then
= X′′ Y
and
= . X Y′′
x2
y2
Substituting in (1), we have
X′′ Y + X Y′′ = 0
X′′
Y′′
= 
i.e,
X
---------------(2).
Y
Now the left side of (2) is a function of „x‟ alone and the right side is a function of „t‟
alone. Since „x‟ and „t‟ are independent variables, (2) can be true only if each side is
equal to a constant.
X′′
Y′′
=
= k (say).
X
Y
Hence, we get X′′  kX = 0 and Y′′ + kY = 0. --------------(3).
Therefore,
Solving equations (3), we get
(i) when „k‟ is positive and k = 2, say
X = c1 ex + c2 e - x
Y = c3 cosy + c4 sin y
(ii) when „k‟ is negative and k = 2, say
X = c5 cosx + c6 sin x
Y = c7 ey + c8 e - y
(iii) when „k‟ is zero.
X = c9 x + c10
Y = c11 x + c12
Thus the various possible solutions of (1) are
u = (c1 ex + c2 e - x) (c3 cosy + c4 sin y) ------------(4)
u = (c5 cosx + c6 sin x) (c7 ey + c8 e - y) ----------(5)
u = (c9 x + c10) (c11 x + c12)
------------(6)
Of these three solutions, we have to choose that solution which suits the physical
nature of the problem and the given boundary conditions.
Example 12
An infinitely long uniform plate is bounded by two parallel edges x = 0 & x = ℓ
and an end at right angles to them. The breadth of this edge y = 0 is ℓ and this edge is
maintained at a temperature f (x). All the other 3 edges are at temperature zero. Find the
steady state temperature at any interior point of the plate.
Solution
Let u (x,y) be the temperature at any point x,y of the plate.
2u
2u
Also u (x,y) satisfies the equation -------- + --------- = 0
---------- (1)
x2
y2
Let the solution of equation (1) be
u(x,y) = (A cos x + B sinx) (Cey + De–y)
Y
y=
x=0
x=ℓ
0<x<ℓ
0<y<
ℓ
y=0
f (x)
The boundary conditions are
0
(i) u (0, y) = 0,
(ii) u (ℓ, y) = 0,
(iii) u (x, ) = 0,
(iv) u (x, 0) = f(x),
X
for 0 < y < 
for 0 < y < 
for 0 < x < ℓ
for 0 < x < ℓ
------------ (2)
Using condition (i), we get
0 = A (Cey + De-y)
i.e, A = 0
 Equation (2) becomes,
u (x,y) = B sinx ( Cey + De -y)
------------------- (3)
Using cndition (ii), we get
n
 = --------ℓ
Therefore,
nx
(ny/ℓ ) (-ny/ℓ)
u (x,y) = B sin --------- { Ce
+ De
}
----------------- (4)
ℓ
Using condition (iii), we get C = 0.
nx
(- ny/ℓ)
u (x,y) = B sin -------- De
ℓ
nx
(- ny/ℓ)
i.e, u (x,y) = B1 sin -------- e
, where B1 = BD.
ℓ
The most general solution is

nx
(- ny/ℓ)
u (x,y) =  Bn Sin ------- e
n =1
ℓ
-------------- (5)
Using condition (iv), we get

nx
f (x) =  Bn Sin ------------------------- (6)
n=1
ℓ
The RHS of equation (6) is a half – range Fourier sine series of the LHS function.
2
ℓ
nx
Bn = --------  f (x). Sin -------- dx
ℓ 0
ℓ
-------------- (7)
Using (7) in (5), we get the required solution.
Example 13
A rectangular plate with an insulated surface is 8 cm. wide and so long compared
to its width that it may be considered as an infinite plate. If the temperature along short
edge y = 0 is u(x,0) = 100 sin (x/8), 0 < x < 8, while two long edges x = 0 & x = 8 as
well as the other short edges are kept at 0C. Find the steady state temperature at any
point of the plate.
Solution
The two dimensional heat equation is given by
2u
2u
--------- + ---------- = 0
-------------- (1)
x
y
The solution of equation (1) be
2
2
u (x,y) = (A cosx + B sinx) (Cey + De-y)
----------------- (2)
The boundary conditions are
(i) u (0, y) = 0,
for 0 < y < 
(ii) u (8, y) = 0,
for 0 < y < 
(iii) u (x, ) = 0,
for 0 < x < 8
(iv) u (x, 0) = 100 Sin (x/8,) for 0 < x < 8
Using conditions (i), & (ii), we get
n
A = 0 ,  -------8
nx
(ny / 8)
(-ny / 8)
u (x,y) = B sin -------- Ce
+ De
8
(ny / 8)
(-ny / 8)
nx
= B1e
+ D1e
sin ----------, where B1 = BC
8
D1 = BD
The most general soln is

(ny / 8)
(-ny / 8)
nx
u (x,y) =  Bne
+ Dne
sin ----------------------- (3)
n=1
8
Using condition (iii), we get Bn = 0.

(- ny / 8)
nx
Hence, u (x,y) =  Dne
sin ---------n=1
8
Using condition (iv), we get
--------------- (4)
x

nx
100 sin --------- =  Dn sin --------8
n=1
8
x
x
2x
3x
i.e, 100 sin --------- = D1 sin ------- + D2 sin -------- + D3 sin --------- + . . . . .
8
8
8
8
Comparing like coefficients on both sides, we get
D1 = 100, D2 = D3 = . . . . = 0
Substituting in (4), we get
(-y / 8)
u (x,y) = 100 e
sin (x / 8)
Example 14
A rectangular plate with an insulated surface 10 c.m wide & so long compared to
its width that it may considered as an infinite plate. If the temperature at the short edge y
= 0 is given by
u (x,0) = 20 x,
0<x < 5
20 (10-x), 5 < x < 10
and all the other 3 edges are kept at temperature 0C. Find the steady state temperature at
any point of the plate.
Solution
The temperature function u (x,y) is given by the equation
2u
2u
---------- + ---------- = 0
--------------- (1)
2
2
x
y
The solution is
u (x,y) = (A cosx + B sinx) (Cey + De-y)
---------------- (2)
The boundary conditions are
= 0,
for 0 < y < 
(ii) u (10, y) = 0,
for 0 < y < 
(iii) u (x, ) = 0,
for 0 < x < 10
(iv) u (x, 0) = 20 x,
if
0<x < 5
if
5 < x < 10
(i) u (0, y)
20 (10-x),
Using conditions (i), (ii), we get
n
A = 0 &  = --------10
Equation (2) becomes
nx
(ny / 10)
u (x,y) = B sin ------ Ce
10
+ De
(ny / 10)
= B1e
(- ny/10)
(- ny/10)
+ D1 e
nx
where B1 = BC,
sin --------10
D1 = BD
The most general solution is

(ny / 10)
(- ny/10)
u (x,y) =  Bne
+ Dn e
n =1
Using condition (iii), we get Bn= 0.
nx
sin --------10
------------ (3)
 Equation (3) becomes

(- ny/10)
u (x,y) =  Dne
n =1
Using condition (iv), we get
nx
sin --------10
------------ (4)

nx
u (x,0) =  Dn. sin --------n =1
10
------------ (5)
The RHS of equation (5) is a half range Fourier sine series of the LHS function
2 10
nx
Dn = --------  f (x) sin -------- dx
10 0
10
2
= -----10
nx
- cos ---------10
nx
- sin ---------10
(20x) ------------------- – (20) ---------------------n
n22
-------------10
100
nx
nx
- cos ---------- sin ---------10
10
+ [20 (10–x)] ------------------- – (-20) ---------------------n
n22
-------------10
100
n
800 sin -------2
i.e,
Dn = ------------------------n22
Substituting in (4) we get,
n
800 sin -------2
(-ny / 10)

u (x,y) =  ------------------------- e
n =1
n22
nx
sin ---------10
Example 15
A rectangular plate is bounded by the lines x = 0, x = a, y = 0 & y = b.
The edge temperatures are u (0,y) = 0, u (x,b) = 0, u (a,y) = 0 &
5
0
10
5
u (x,0) = 5 sin (5x / a) + 3 sin (3x / a). Find the steady state temperature distribution at
any point of the plate.
The temperature function u (x,y) satisfies the equation
2u
---------- +
x2
2u
------------ = 0
y2
---------- (1)
Let the solution of equation (1) be
u (x,y) = (A cosx + Bsinx) (Cey + De - y)
------------ (2)
The boundary conditions are
(i) u (0,y) = 0,
for 0 < y < b
(ii) u (a,y) = 0,
for 0 < y < b
(iii) u (x, b) = 0,
for 0 < x < a
(iv) u (x,0) = 5 sin (5x / a) + 3 sin (3x / a), for 0 < x < a.
y y=b
x=0
O
x=a
y =0
x
Using conditions (i), (ii), we get
n
A = 0,  = --------a
nx
(ny / a)
(-ny / a)
u (x,y) = B sin -------- Ce
+ De
a
nx
(ny / a)
(-ny / a)
= sin -------B1e
+ D1e
a
The most general solution is
(ny / a)
(-ny / a)
nx
u (x,y) =  Bne
+ Dn e
sin ----------- -------(3)
n=1
a

Using condition (iii) we get
(nb / a)
(-nb / a)
 Bne
+ Dn e

0
=
n=1
(nb / a)
==> Bne
+

Dn = Bn
nx
sin ----------a
(-nb / a)
Dn e
=0
e (nb / a)
---------------- = - Bne(2nb / a)
-e (-nb / a)
Substituting in (3), we get
nx

u (x,y) =  Bne (ny / a) - Bne (2nb / a) e (-ny / a) sin ------------n=1
a
Bn
nx

=  ----------- e(ny / a) e(-nb / a)  e (2nb / a) e (-ny / a) e(-nb / a) sin ------n=1 (-nb)/a
e
a
=
2 Bn
e(n (y-b) / a) - e(-n (y-b) / a)
nx
sin
e(-nb / a)
2
a
2Bn
n (y–b)
nx
=  --------------- sin h --------------- sin ----------e(-nb / a)
a
a
n (y –b)
nx

i.e, u (x,y) =  Cn sin h ---------- sin -----n=1
a
a
----------- (4)
Using condition (iv), we get
5x
3x
n (-b)
nx

5 sin -------- + 3 sin --------- =  Cn sin h ------- sin -----n=1
a
a
a
a
5x
3x
nb
nx

ie, 5 sin -------- + 3 sin --------- =  - Cn sin h ------ sin ------a
a
n=1
a
a
5x
3x
b
x
2b
2x
ie, 5 sin ------ + 3 sin ------- = - C1 sinh ------ sin ------ - C2 sin h------ sin ------ - …
a
a
a
a
a
a
Comparing the like coefficients on both sides, we get
3b
- C3 sinh ------------ = 3
&
a
5b
- C5 sinh ------------ = 5,
a
C1 = C2 = C4 = C6 = . . . = 0
-3
-5
==> C3 = ---------------- & C5 = ---------------sinh (3b /a)
sinh(5b/ a )
Substituting in (4), we get
u (x,y) = -
3
3 (y-b)
------------- sin h ----------sinh(3b / a)
a
3x
sin ---------a
5
5 (y-b)
5x
 ------------ sin h ---------- sin ----------sinh(5b / a)
a
a
3
3 (b-y)
3x
u (x,y) = ------------- sin h ----------- sin ---------sinh(3b / a)
a
a
i.e,
+
5
5 (b-y)
5x
--------------- sin h ----------- sin -----sinh(5b / a)
a
a
Exercises
2u
(1) Solve the Laplace equation
+
x
i.
u(0,y) = 0 for 0  y  b
2u
2
= 0 , subject to the conditions
y
2
ii.
u(a,y) = 0 for 0  y  b
iii. u(x,b) = 0 for 0  x  a
iv.
u(x,0) = sin3(x/ a) ,0  x  a.
(2) Find the steady temperature distribution at points in a rectangular plate with insulated
faces and the edges of the plate being the lines x = 0, x = a, y = 0 and y = b. When three
of the edges are kept at temperature zero and the fourth at a fixed temperature o C.
2u
2u
(3) Solve the Laplace equation
+
= 0 , which satisfies the conditions
x2
y2
u(0,y) = u(l,y) = u(x,0) = 0 and u(x,a) = sin(nx/ l).
2u
(4) Solve the Laplace equation
2u
+
= 0 , which satisfies the conditions
x
y
u(0,y) = u(a,y) = u(x,b) = 0 and u(x,0) = x (a – x ).
2u
2u
(5) Solve the Laplace equation
+
= 0 , subject to the conditions
2
2
x
y
i.
u(0,y) = 0, 0  y  l ii. u(l,y) = 0, 0  y  l
iii. u(x,0) = 0, 0  x  l
iv. u(x,l) = f(x), 0  x  l
2
2
(6) A square plate is bounded by the lines x = 0, y = 0, x = 20 and y = 20. Its faces are
insulated.
The temperature along the upper horizontal edge is given by u(x,0) = x (20 – x), when 0 
x 20,
while other three edges are kept at 0o C. Find the steady state temperature in the plate.
(7) An infinite long plate is bounded plate by two parallel edges and an end at right
angles to them.The breadth is . This end is maintained at a constant temperature „u0‟ at
all points and the other edges are at zero temperature. Find the steady state temperature at
any point (x,y) of the plate.
(8) An infinitely long uniform plate is bounded by two parallel edges x = 0 and x = l, and
an end at right angles to them. The breadth of this edge y = 0 is „l‟ and is maintained at a
temperature f(x). All the other three edges are at temperature zero. Find the steady state
temperature at any interior point of the plate.
(9) A rectangular plate with insulated surface is 8 cm. wide and so long compared to its
width that it may be considered infinite in length without introducing an appreciable
error. If the temperature along one short edge y = 0 is given by u(x,0) = 100 sin(x/ 8), 0
 x  8, while the two long edges x = 0 and x = 8 as well as the other short edge are kept
at 0o C, show that the steady state temperature at any point of the plane is given by u(x,y)
= 100 e-y/ 8 sin x/ 8 .
(10) A rectangular plate with insulated surface is 10 cm. wide and so long compared to
its width that it may be considered infinite length. If the temperature along short edge y =
0 is given
u(x,0) = 8 sin(x/ 10) when 0  x  10, while the two long edges x = 0 and x = 10 as
well as the other short edge are kept at 0o C, find the steady state temperature distribution
u(x,y).
UNIT-IV
FOURIER TRANSFORMS
4.1 Introduction
This unit starts with integral transforms and presents three well-known integral
transforms, namely, Complex Fourier transform, Fourier sine transform, Fourier cosine
transform and their inverse transforms. The concept of Fourier transforms will be
introduced after deriving the Fourier Integral Theorem. The various properties of these
transforms and many solved examples are provided in this chapter. Moreover, the
applications of Fourier Transforms in partial differential equations are many and are not
included here because it is a wide area and beyond the scope of the book.
4.2 Integral Transforms
~
The integral transform f(s) of a function f(x) is defined by
~
b
f(s) =  f(x) K(s,x) dx,
a
if the integral exists and is denoted by I{f(x)}. Here, K(s,x) is called the kernel of the
transform. The kernel is a known function of „s‟ and „x‟. The function f(x) is called the
inverse transform
~
of f(s). By properly selecting the kernel in the definition of general integral transform,
we get various integral transforms.
The following are some of the well-known transforms:
(i) Laplace Transform
L{f(x)} =  f(x) e – sx dx
0
(ii) Fourier Transform
1
F{f(x)} =
2

 f(x) eisx dx
-
(iii) Mellin Transform

 f(x) x s-1 dx
M{f(x)} =
0
(iv) Hankel Transform

Hn{f(x)} =
0
 f(x) x Jn(sx) dx,
where Jn(sx) is the Bessel function of the first kind and order „n‟.
4.3 FOURIER INTEGRAL THEOREM
If f(x) is defined in the interval (-ℓ,ℓ), and the following conditions
(i) f(x) satisfies the Dirichlet‟s conditions in every interval (-ℓ,ℓ),

(ii)   f(x)  dx converges, i.e. f(x) is absolutely integrable in (-,)
-
 
are true, then f(x) = (1∕)   f(t) cos(t-x) dt d.
0
-
Consider a function f(x) which satisfies the Dirichlet‟s conditions in every interval (-ℓ,ℓ)
so that, we have
a0
nx
nx

f(x) = ----- + 
an cos ---- + bn sin ----------(1)
2
n=1
ℓ
ℓ
where a0
1
ℓ
= -----  f(t) dt
ℓ -ℓ
1
ℓ
an = -----  f(t) cos (nt / ℓ ) dt
ℓ -ℓ
1
ℓ
bn = -----  f(t) sin (nt / ℓ ) dt
ℓ -ℓ
Substituting the values of a0, an and bn in (1), we get
and
f(x)
=
1
----2ℓ
ℓ
1 
ℓ
n(t – x)
 f(t) dt + --- 
 f(t) cos ----------- dt
-ℓ
ℓ n=1 -ℓ
ℓ
-------(2)
1
ℓ
1
ℓ
-----  f(t) dt
 -----   f(t) dt ,
2ℓ -ℓ
2ℓ -ℓ
then by assumption (ii), the first term on the right side of (2) approaches zero as ℓ  .
As ℓ  , the second term on the right side of (2) becomes
Since,
1


n(t – x)
ℓim
--- 
 f(t) cos ----------- dt
ℓ   ℓ n=1 -
ℓ
=
.
ℓim
0
1 

---    f(t) cos { n  (t – x) } dt ,on taking (∕ℓ) =
 n=1
-
By the definition of integral as the limit of sum and (n∕ℓ ) =  as ℓ   , the second
term of (2) takes the form
1  
---   f(t) cos  (t – x) dt d ,
 0 -
Hence as ℓ  , (2) becomes
1  
f(x) = ---   f(t) cos  (t – x) dt d
 0 -
which is known as the Fourier integral of f(x).
---------(3)
Note:
When f(x) satisfies the conditions stated above, equation (3) holds good at a point
of continuity. But at a point of discontinuity, the value of the integral is (1/ 2) [f(x+0) +
f(x-0)] as in the case of Fourier series.
Fourier sine and cosine Integrals
The Fourier integral of f(x) is given by
f(x) =
=
1  
---   f(t) cos (t – x) dt d
 0 -
1  
---   f(t) { cost . cosx + sint . sinx } dt d
 0 -
1


1


= ---  cosx  f(t) cost dt d +
 sinx  f(t) sint dt d ----(4)
-
-
0

 0
When f(x) is an odd function, f(t) cost is odd while f(t) sint is even. Then the first
integral of (4) vanishes and, we get
2
f(x)



 sinx  f(t) sint dt d
=
-
0
-------(5)
which is known as the Fourier sine integral.
Similarly, when f(x) is an even function, (4) takes the form
2
f(x)



 cosx  f(t) cost dt d
=
-
0
-------(6)
which is known as the Fourier cosine integral.
Complex form of Fourier Integrals
The Fourier integral of f(x) is given by
f(x) =
=
1  
---   f(t) cos (t – x) dt d
 0 -
1
--

 f(t)
-

 cos (t – x) d dt
0
Since cos (t – x) is an even function of , we have by the property of definite integrals
f(x) =
1 

---  f(t) (1/ 2)  cos (t – x) d dt
-
 -
1
 
i.e., f(x) = ---   f(t) cos (t – x) dt d ---------(7)
2 - -
Similarly, since sin (t – x) is an odd function of , we have
 
---   f(t) sin (t – x) dt d
2 - -
Multiplying (8) by „i ‟ and adding to (7), we get
0
=
f(x) =
1
--2
---------(8)
 
  f(t) ei(t – x) dt d
- -
---------(9)
which is the complex form of the Fourier integral.
4.4 Fourier Transforms and its properties
Fourier Transform
We know that the complex form of Fourier integral is
1


f(x) =
2
Replacing  by s, we get
-
-
1

 e- isx ds
f(x) =
2

 f(t) ei(t-x) dt d.
-

 f(t) eist dt .
-
It follows that if
1
F(s) =
2
1
Then,

 f(t) eist dt --------------- (1)
-

 F(s) e-isx ds --------------- (2)
f(x) =
2
-
The function F(s), defined by (1), is called the Fourier Transform of f(x). The function
f(x), as given by (2), is called the inverse Fourier Transform of F(s). The equation (2)
is also referred to as the inversion formula.
Properties of Fourier Transforms
(1) Linearity Property
If F(s) and G(s) are Fourier Transforms of f(x) and g(x) respectively, then
F{a f(x) + bg(x)} = a F(s) + bG(s),
where a and b are constants.
1

We have
F(s) =
 eisx f(x) dx
2 - 
1

G(s) =
 eisx g(x) dx
2 - 
Therefore,
1

F{a f(x) + b g(x)} =
 eisx {a f(x) + bg(x)}dx
2 - 
1

1

isx
= a
 e f(x) dx + b
 eisx g(x) dx
2 - 
2 - 
= a F(s) + bG(s)
i.e, F{a f(x) + bg(x)} = a F(s) + bG(s)
(2) Shifting Property
(i)
If F(s) is the complex Fourier Transform of f(x), then
F{f(x-a)} = eisa F(s).

F(s) =
 eisx f(x) dx ----------------( i )
2 - 
1
We have

Now,
F{f(x-a)} =
 eisx f(x-a) dx
2 - 
Putting x-a = t, we have
1

F{f(x-a)} =
 eis(t+a) f(t) dt .
2 - 
1

ias
=e
 eist f(t) dt .
2 - 
1
= eias . F(s).
( by (i) ).
(ii) If F(s) is the complex Fourier Transform of f(x), then
F{eiax f(x) } = F(s+a).
1

We have
F(s) =
 eisx f(x) dx ----------------( i )
2 - 

f(x)} =
 eisx .eiax f(x) dx.
2 - 
1
Now,
iax
F{e

 ei(s+a)x. f(x) dx .
2 - 
1
=
= F(s+a)
by (i) .
(3) Change of scale property
If F(s) is the complex Fourier transform of f(x), then
F{f(ax)} =1/a F(s/a), a  0.

F(s) =
 eisx f(x) dx ----------------( i )
2 - 
1
We have

F{f(ax)} =
 eisx f(ax) dx.
2 - 
1
Now,
Put ax = t, so that dx = dt/a.

 e ist/a .f(t) dt/a .
2 - 
1
F{f(ax)} =
1
=
a
1

.  ei(s/a)t f(t) dt .
2 - 
1
=
. F(s/a).
( by (i) ).
a
(4) Modulation theorem.
If F(s) is the complex Fourier transform of f(x),
Then
F{f(x) cosax} = ½{F(s+a) + F(s-a)}.

F(s) =
 eisx f(x) dx
2 - 
1
We have

 eisx .f(x) cosax. dx.
2 - 
1
Now,
F{f(x) cosax} =

eiax + e-iax
=
 eisx. f(x)
dx .
2 - 
2
1
1
=
2

1 
i(s+a)x
 e
.f(x) dx +
 ei(s-a)x f(x) dx
2 - 
2 -
1
1
=
{ F(s+a) + F(s-a)}
2
(5) nth derivative of the Fourier Transform
If F(s) is the complex Fourier Transform of f(x),
F{xn f(x)} = (-i)n dn/dsn .F(s).
1

We have
F(s) =
 eisx f(x) dx ------------------- (i)
2 - 
Then
Differentiating (i) „n‟ times w.r.t „s‟, we get
dn F(s)
dsn

=
 (ix)n. eisx f(x) dx
2 - 
1

 eisx {xn f(x)} dx
2 -
(i)n
=
= ( i )n F{xn f(x)}.
dn F(s)
1
 F{x f(x)} =
n
.
(i)n
dsn
dn
n
n
i.e, F{x f(x)} = (-i)
F(s).
ds
n
(6) Fourier Transform of the derivatives of a function.
If F(s) is the complex Fourier Transform of f(x),
Then, F{f „(x)} = -is F(s) if f(x)  0 as x   .

 eisx f(x) dx .
2 - 
1
We have
F(s) =

F{f „(x)} =
 eisx f „(x) dx.
2 - 
1
Now,

=
 eisx d{f (x)}.
2 - 
1
1


e .f(x) - is  f(x). eisx dx.
-
-
isx
=
2

= - is
 eisx f(x) dx , provided f(x) = 0
2 - 
as x    .
1
= - is F(s).
i.e, F{f ‟(x)} = - is F(s) ---------------------( i )
Then the Fourier Transform of f  (x),

 eisx f (x) dx.
2 - 
1
i.e, F{f (x)} =

=
 eisx d{f ‟(x)}.
2 - 
1


e .f „(x) -  f „(x). eisx .(is)dx.
- -
1
=
isx
2

 eisx f „(x) dx , provided f „(x) = 0
2 - 
as x    .
1
= - is
= - is F{f „(x).}
= (-is).(-is)F(s).
by( i ).
= (-is)2 . F(s).
i.e, F{f “(x)} = (- is)2 .F(s) , Provided f , f‟ 0
as x   .
In general, the Fourier transform of the nth derivative of f(x) is given by
F{f n(x)} = (-is)n F(s),
provided the first „n-1‟ derivatives vanish as x  .
Property (7)
x
If F(s) is the complex Fourier Transform of f(x), then F  f(x)dx
a
F(s)
=
(-is)
Let
x
g(x) =  f(x) dx .
a
Then, g‟(x) = f(x).
------------( i )
f [g„(x)] = (-is) G(s), by property (6).
Now
= (-is). F{g(x)}
x
= (-is). F  f(x) dx .
a
x
i.e, F{g‟(x)} = (-is). F
 f(x) dx .
a
x
1
i.e, F  f(x) dx =
. F{g‟(x)}.
a
(-is)
1
=
F{f (x)}.
[ by ( i )]
(-is)
x
Thus, F  f(x) dx
a
F(s)
=
.
(-is)
Property (8)
If F(s) is the complex Fourier transform of f(x),

Then, F{f(-x)} = F(s), where bar denotes complex conjugate.
Proof
 
F(s) =
 f(x) e-isx dx .
2 -

1
Putting x = -t, we get
 
F(s) =
 f(-t) eisx dt .
2 -

1

= F{f(-x)} .
Note: If F{f(x)} = F(s), then
(i)
F{f(-x)} = F(-s).
(ii)
F{f(x)} = F(-s).

Example 1
Find the F.T of f(x) defined by
f(x) = 0 x<a
= 1 a<x<b
=0
x>b.
The F.T of f(x) is given by
1

F{f (x)} =
 eisx f (x) dx.
2 - 
1
=
2 a
b
 eisx .dx .
1
eisx
b
is
a
=
2
eibs – eias
1
=
.
2
is
Example 2
Find the F.T of f(x) = x for  x   a
= 0 for  x  > a.

F{f (x)} =
 eisx f (x) dx.
2 - 
1
1
a
=
 eisx .x.dx.
2 -a
eisx
1
a
=
 x .d
2 -a
is
a
xeisx
1
=
eisx
-
2
(is)2
is
-a
aeisa
1
=
2
is
1
a
(is)
is
1
-2ai
-isa
s2
2i
sinsa
s
2
[sinsa - as cossa].
2
i [sinsa - as cossa]
(2/)
s2
2
Example 3
f(x) = eiax , 0 < x < 1
= 0
otherwise
The F.T of f(x) is given by

F{f (x)} =
 eisx f (x) dx.
2 - 
1
1
(eisa - e-isa )
)+
1
.
s
(is)2
1
s
2i
=
+
is
cossa +
2
=
2
(e + e
2
e-isa
+
isa
=
Find the F.T of
ae-isa
-
=
=
eisa
1
 eisx . eiax dx.
2
0
1
1
=
 ei(s+a)x .dx .
2 0
1
ei(s+a)x 1
2
i(s+a) 0
=
1
{ei(s+a)x -1}
=
i2.(s+a)
i
{1- ei(s+a)}
=
2.(s+a)
Example 4
2 2
-a x
Find the F.T of e
,
a>0 and hence deduce that the F.T of e
2
-x / 2
2
-s / 2
is e
.
The F.T of f(x) is given by

F{f (x)} =
 eisx f (x) dx.
2 - 
1
F e-a
2x 2
 22
 e –a x . eisx .dx.
2 -
1
=
2
2


-
-s / 4a
e
=
2
2
2
e-[ax – (is/2a)] dx .
2
 2
=
 e-t dt, by putting ax –(is/2a) = t
a2 -
e-s / 4a
e-s
2 2
/ 4a
=
a2
 2
.  , since  e-t dt =  (using Gamma functions).
-
1
2
2
e-s / 4a .
=
----------------(i)
2.a
2
To find F{e-x
/2
}
Putting a = 1/ 2 in (1), we get
2
F{e-x
/2
2
} = e-s
/2
.
Note:
If the F.T of f(x) is f(s), the function f(x) is called self-reciprocal. In the above
example e -x
2
/2
is self-reciprocal under F.T.
Example 5
Find the F.T of
f(x) = 1 for x<1.
= 0 for x>1.

 sinx
x
Hence evaluate
dx.
0
The F.T of f(x),

 eisx f (x) dx.
2 - 
1
i.e.,
F{f (x)} =
1
1
=
 eisx .(1).dx .
2 -1
1
1
eisx
=
2
is
-1
eis – e -is
1
=
.
2
is
sins
, s≠0
=(2/)
s
sins
Thus, F{f(x)}= F(s) =(2/).
,
s≠0
s
Now by the inversion formula , we get

 f(s). e-isx .ds.
2 - 
1
f(x) =

sins
=  (2/)
-
s
or
1 for x<1
. e-isx .ds.=
0
 sins
i.e,

e-isx . ds.=
 -
s
1
for x>1.
1 for x<1
0
for x>1.
Putting x = 0, we get
 sins

ds = 1
 -
s
1
 sins
i.e,

ds = 1, since the integrand is even.
 0
s
2



0

sins
ds =
s
 sinx

Hence, 
dx =
0
x
2
Exercises
(1) Find the Fourier transform of
1
for x<a
f(x) =
0
for x>a.
(2) Find the Fourier transform of
x2
for xa
f(x) =
0
for x>a.
2
(3) Find the Fourier transform of
a2 – x2 , x<a
f(x) =
0,
x >a>0.
 sint - tcost


dt =
-
t3
4
Hence deduce that
(4) Find the Fourier transform of e-ax and x e-ax. Also deduce that
 cosxt


dt =
e-ax
- a2 + t2
2a
d
-ax
{Hint : F{x. e
F{ e-ax}}
}= -i
ds
4.5 Convolution Theorem and Parseval’s identity.
The convolution of two functions f(x) and g(x) is defined as
1 
f(x) * g(x) =
 f(t). g(x-t). dt.
2 -
Convolution Theorem for Fourier Transforms.
The Fourier Transform of the convolution of f(x) and g(x) is the product of their
Fourier Transforms,
i.e, F{f(x) * g(x)} = F{f(x).F{g(x)}.
Proof:
F{f(x) * g(x)} = F{(f*g)x)}

=
 (f *g)(x). eisx . dx.
2 - 
1

=

2 -
1

 f(t). g(x-t). dt eisx dx .
2 -
1

1

=
 f(t)
 g(x-t). eisx dx . dt.
2 -
2 -
(by changing the order of integration).
1

=
 f (t).F{g(x-t)}. dt.
2 - 
1

=
 f(t). eits .G(s). dt. (by shifting property)
2 - 
1

= G(s).
 f(t). eist dt.
2 - 
1
= F(s).G(s).
Hence, F{f(x) * g(x)} = F{f(x).F{g(x)}.
Parseval’s identity for Fourier Transforms
If F(s) is the F.T of f(x), then


2
 f(x) dx =  F(s)2 ds.
-
-
Proof:
By convolution theorem, we have
F{f(x) * g(x)} = F(s).G(s).
Therefore, (f*g) (x) = F-1{F(s).G(s)}.

1 
 f(t). g(x-t). dt =
 F(s).G(s).e-isx ds. ----------(1)
2 -
2 -
(by using the inversion formula)
Putting x = 0 in (1) , we get
1
i.e,


 f(t). g(-t). dt =  F(s).G(s).ds. ----------(2)
-
-


Since (2) is true for all g(t), take g(t) = f(-t) and hence g(-t) = f(t) ---------(3)
Also, G(s) = F{g(t)}

= F{f(-t)}

= F(s) ----------------(4) (by the property of F.T).
Using (3) & (4) in (2), we have
 


 f(t).f(t). dt =  F(s).F(s).ds.
-
-


2
  f(t) dt =  F(s)2 ds.
-
-


i.e,  f(x)2 dx =  F(s)2 ds.
-
-
Example 6
Find the F.T of f (x) = 1-x for x   1.
=0
for x > 1

and hence find the value  sin4t dt.
0 t4
1
Here, F{f(x)}=
1
 (1- x )eisx dx.
2 -1
1
1
=
 (1- x) (cossx + i sinsx) dx.
2 -1
1
1
i
1
=
 (1- x) cossx dx.+
(1- x) sinsx dx.
2 -1
2 -1
=
2
1
1
2  (1- x) cossx dx. by the property of definite integral.
0
1
sinsx
= (2/)  (1-x) d
0
s
1
sinsx
= (2/)
(1-x)
cossx
-(-1) s2
s
0
1- coss
= (2/)
s2
Using Parseval‟s identity, we get

1
2
 (1-coss) ds. = (1- x)2 dx.
 -
s4
-1
2

1
 (1-coss)2 ds. = 2 (1- x)2 dx = 2/3.
0
s4
0
4


16
i.e,


 sin4(s/2) ds. = 2/3.
0
s4
Setting s/2 = x , we get
 sin4 x

0 16x4
2.dx. = 2/3.
 sin4 x
 
0
x4
dx. = /3.
16

Example 7
Find the F.T of f(x) if
1
for x<a
0
for x>a>0.
f(x) =

Using Parseval‟s identity, prove  sint
0 t
Here,
2
dt. = /2.
1
a
F{f(x)} =
 eisx .(1) .dx .
2 -a
1
eisx
a
2
is
-a
=
1
eisa – eisa
2
is
=
sinas
= (2/)
s
sinas
i.e.,
F(s) = (2/)
.
s
Using Parseval‟s identity


  f (x)  2 dx =   F(s)  2 ds,
-
-
we have
a

 1 . dx =  (2/)
-a
-
sinas
ds.
s
 sinas
2a = (2/) 
-
s
Setting as = t, we get
2
2
ds.

(2/) 
-


-
i.e.,
sint
dt/a = 2a
( t/a)
sint


0
Hence,
2
dt
= 
dt
= 
t
 sint
2 
0
t

2
sint
2
2
dt
=  / 2.
t
4.6 Fourier sine and cosine transforms:
Fourier sine Transform
We know that the Fourier sine integral is
2


 sin x .  f(t) sint dt.d.
f(x) =

0
0
Replacing  by s, we get
2


 sinsx .  f(t) sinst dt. ds.
f(x) =

0
0
It follows that if

Fs(s) = (2/ )  f(t) sinst dt..
0

i.e.,
Fs(s) = (2/ )  f(x) sinsx dx.
------------(1)
0

then f(x)
= (2/ )  Fs(s) sinsx ds.
0
------------(2)
The function Fs(s), as defined by (1), is known as the Fourier sine transform of f(x).
Also the function f(x), as given by (2),is called the Inverse Fourier sine transform of
Fs(s) .
Fourier cosine transform
Similarly, it follows from the Fourier cosine integral
2


f(x) =
 cos x .  f(t) cost dt.d.
0
0


Fc(s) = (2/ )  f(x) cossx dx.
that if
------------(3)
0

then
f(x)
= (2/ )  Fc(s) cossx ds.
------------(4)
0
The function Fc(s), as defined by (3), is known as the Fourier cosine transform of
f(x). Also the function f(x), as given by (4),is called the Inverse Fourier cosine
transform of Fc(s) .
Properties of Fourier sine and cosine Transforms
If Fs(s) and Fc(s) are the Fourier sine and cosine transforms of f(x) respectively, the
following properties and identities are true.
(1) Linearity property
Fs [a f(x) + b g(x) ] = a Fs { f(x) } + b Fs { g(x) }.
and Fc [a f(x) + b g(x) ] = a Fc { f(x) } + b Fc { g(x) }.
(2) Change of scale property
Fs [ f(ax) ] = (1/a) Fs [ s/a ].
and
Fc [ f(ax) ] = (1/a) Fc [ s/a ].
(3) Modulation Theorem
i.
Fs [ f(x) sinax ] = (1/2) [ Fc (s-a) - Fc (s+a)].
ii.
Fs [ f(x) cosax ] = (1/2) [ Fs (s+a) + Fs (s-a)].
iii.
Fc[ f(x) cosax ] = (1/2) [ Fc (s+a) + Fc (s-a) ].
iv.
Fc[ f(x) sinax ] = (1/2) [ Fs (s+a) - Fs (s-a) ].
Proof
The Fourier sine transform of
f(x)sinax is given by

Fs [ f(x) sinax ] =(2/ )  (f(x) sinax) sinsx dx.
0

= (1/2) (2/ )  f(x) [cos(s-a)x – cos(s+a)x] dx.
0
= (1/2) [ Fc (s-a) – Fc (s+a) ].
Similarly, we can prove the results (ii), (iii) & (iv).
(4) Parseval’s identity


0

Fc(s) Gc(s) ds =  f(x) g(x) dx .
0


0

Fs(s) Gs(s) ds =  f(x) g(x) dx .
0


0

Fc(s)
2
Fs(s)
2
0
2
dx .
2
dx .
0


ds =  f(x)

ds =  f(x)
0
Proof


0


0
0
Fc(s) Gc(s) ds =  Fc(s) [(2/ )  g(t) cosst dt] ds


=  g(t) [(2/ )  Fc(s) cosst ds] dt
0
0

=  g(t) f(t) dt

i.e.,
0

 Fc(s) Gc(s) ds =  f(x) g(x) dx .
0
Similarly, we can prove the second identity and the other identities follow by setting
g(x) = f(x) in the first identity.
Property (5)
If Fs(s) and Fc(s) are the Fourier sine and cosine transforms of f(x) respectively, then
d
(i) Fs{ x f(x) } = -
Fc(s) .
ds
d
(ii) Fc{ x f(x) } = -
Fs(s) .
ds
Proof
The Fourier cosine transform of f(x),

i.e.,
Fc(s) = (2/ )  f(x) cossx dx.
0
Differentiating w.r.t s, we get
d

[ Fc(s) ] = (2/ )  f(x) {- x sin sx } dx.
0
ds

= - (2/ )  ( x f(x)) sin sx dx.
0
= - Fs{x f(x)}
d
i.e., Fs{x f(x)} = { Fc(s) }
ds
Similarly, we can prove
d
Fc{x f(x)} = { Fs(s) }
ds
Example 8
Find the Fourier sine and cosine transforms of e-ax and hence deduce the inversion
formula.
The Fourier sine transform of f(x) is given by

Fs { f(x) } = (2/ )  f(x) sinsx dx.
0

Now ,
-ax
Fs { e
} = (2/ )  e-ax sinsx dx.
0
e-ax ( - a sinsx – s cossx)

= (2/ )
a2 + s2
0
s
= (2/ )
, if a>0
2
a +s
The Fourier cosine transform of f(x) is given by
2

Fc { f(x) } = (2/ )  f(x) cossx dx.
0

Now ,
-ax
Fc { e
} = (2/ )  e-ax cossx dx.
0

-ax
e
( - a cossx + s sinsx)
= (2/ )
a2 + s2
0
a
= (2/ )
, if a>0
2
a +s
2
Example 9
x,
for 0<x<1
2 – x, for 1<x<2
0,
for x>2
Find the Fourier cosine transform of f(x) =
The Fourier cosine transform of f(x),
1
2
i.e., Fc { f(x) } = (2/ )0  x cossx dx. + (2/ )1  (2 - x ) cossx dx.
sinsx
1
= (2/ )  x d
0
sinsx
2
+ (2/ )  ( 2 – x) d
1
s
s
1
sinsx
= (2/ )
cossx

x
(1)
s2
s
0
2
sinsx
+ (2/ )
cossx
 ( - 1) +
(2 – x)
s
s2
sins
coss
= (2/ )
1
+
s2
sins
s
cos2s
+ -
s2
coss
-
+
s2
s2
s
2 coss
cos2s
= (2/ )
-
1
-
s2
s2
s2
Example 10

Find the Fourier sine transform of e- x . Hence show that
m>0.
The Fourier sine transform of f(x) is given by

0
dx =
1+x
2

Fs { f(x) } = (2/ )  f(x) sinsx dx.
0

= (2/ )  e-x sinsx dx.
0
e-x ( - sinsx – s cossx)

= (2/ )
1 + s2
0
s
= (2/ )
.
1 + s2
Using inversion formula for Fourier sine transforms, we get
s

(2/ )  (2/ )
0
sin sx ds. = e-x
1+s
2
Replacing x by m,
e
-m

s sinms
0
1 + s2

x sinmx
= (2/ ) 
ds
e-m
x sinmx
,
2
= (2/ ) 
dx
0


Hence,
1 + x2
e-m
x sinmx
dx
0
1+x
=
2
2
Example 11
x
1
Find the Fourier sine transform of
and the Fourier cosine transform of
.
a2+x2
a2+x2
x
To find the Fourier sine transform of
,
2
2
a +x
We have to find Fs { e-ax }.

Fs { e-ax } = (2/ )  e-ax sin sx dx.
Consider,
0
s
= (2/ )
.
a2 + s2
Using inversion formula for Fourier sine transforms, we get
s

-ax
e
= (2/ )  (2/ )
sinsx ds.
0
a2 + s2


i.e.,
0
e-ax
s sinsx
ds
=
, a>0
s2 + a2
2
x sinsx
e-as
Changing x by s, we get


0
dx =
2
x +a
2
x
Now
Fs
= (2/ ) 
2
------------(1)
2
2
x

sinsx dx
0
x +a
2
2
x +a
e-as
.
= (2/ )
,
2
using (1)
= (/2) e-as
1
Similarly,for finding the Fourier cosine transform of
2
a +x
, we have to findFc{e-ax}.
2

Fc{ e-ax } = (2/ )  e-ax cossx dx.
Consider ,
0
a
= (2/ )
.
a2 + s2
Using inversion formula for Fourier cosine transforms, we get
a

e-ax = (2/ )  (2/ )
cossx ds.
0
2
2
a +s
cossx
0
s2 + a2
2a
cossx
e-as

i.e.,
e-ax

ds
=
Changing x by s, we get


0
dx =
x2 + a2
1
Now,
Fc
------------(2)
2a
1

= (2/ ) 
x2 + a2
cossx dx
0
x2 + a2
e-as
= (2/ )
.
,
using (2)
2a
e-as
= (/2)
a
Example 12
Find the Fourier cosine transform of e-a
of
2 2
-a x
xe
.
2 2
2 2
x
and hence evaluate the Fourier sine transform
The Fourier cosine transform of e-a
2 2
-a x
Fc{e

x
2 2
-a x
} = (2/ )  e
is given by
cossx dx
0

= Real part of (2/ )  e-a
2 2
x
e isx dx
0
1
e -s
= Real part of
2
2
/ 4a
.
(Refer example (4) of section 4.4)
a .2.
1
e -s
=
2
2
/ 4a
.
----------------(i)
a .2.
d
But ,
Fs {x f(x)} = -
Fc (s)
ds
 Fs {x e-a
d
2 2
x
1
2
e -s
} = -
2
/ 4a
, by (1)
a 2
ds
1
e -s
= -
2
2
/ 4a
( - s / 2a2).
a 2
s
=
2
e -s
2
/ 4a
.
2 2. a3
Fc [ 1 / x ] = 1 / s
and
Fs [ 1 / x ] = 1 / s
This shows that 1 / x is self-reciprocal.
Example 13

Evaluate
dx

0
using transform methods.
(a2 + x2)(b2 + x2)
Let f(x) = e –ax ,
g(x) = e- bx

Then
= (2/ )  e-ax cossx dx.
Fc{ s }
0
a
= (2/ )
.
a2 + s2
b
Gc{ s } = (2/ )
Similarly,
.
2
2
b +s
Now using Parseval‟s identity for Fourier cosine transforms,


0
0
 Fc(s) . Gc(s) ds =  f(x) g(x)dx.
i.e.,
2
we have,
2ab
or


0


=  e–(a+b)x dx
ds
(a2 + s2)(b2 +s2)
ds

0
ab


0
e–(a+b)x

–(a+b)
0
=
(a2 + s2)(b2 +s2)
= 1 / ( a+b )


dx

Thus,
0
=
(a2 + x2)(b2 + x2)
2ab(a+b)
Example 14
Using Parseval‟s identity, evaluate the integrals

0

dx

and
(a2 + x2)2
0

(a2 + x2)2
Let f(x) = e-ax
s
Then Fs(s) = (2/ )
Fc(s) = (2/ )
,
a2 + s2
a
a2 + s2
x2
dx
if a > 0
Now, Using Parseval‟s identity for sine transforms,


i.e.,
0
we get, (2/ )

Fs(s)
(2/ )
ds =  f(x)

dx .
s2

ds
=
(a2 + s2)2
 e-2ax dx
0
s2

e-2ax

0
ds
=
0
-2a

x2

0
1

=
(a2 + s2)2

Thus
2
0

0
or
2
dx
=
(a2 + x2)2
, if a > 0
4a
Now, Using Parseval‟s identity for cosine transforms,


i.e.,
0
we get, (2/ )

Fc(s)
or
(2a / )
ds =  f(x)

ds
1


 e-2ax dx
0
ds

=
(a2 + s2)2
dx

0
=
(a2 + s2)2
0
dx .
a2

Thus ,
2
0

0
2
2
2a

=
2
2 2
(a + x )
, if a > 0
4a
3
2a
Exercises
1. Find the Fourier sine transform of the function
sin x , 0  x < a.
0
, x>a
f(x) =
2. Find the Fourier cosine transform of e-x and hence deduce by using the inversion
formula

 cos x dx

=
e -
0
2
(1 + x )
2
3. Find the Fourier cosine transform of e-axsin ax.
4. Find the Fourier cosine transform of e-2x + 3 e-x
5. Find the Fourier cosine transform of
(i)
e-ax / x
(ii) ( e-ax - e-bx ) / x
6. Find, when n > 0
(i) Fs[xn-1] and

(ii) Fc[xn-1]
(n)
Hint: e-ax xn-1dx =
0
,n>0,a>0
n
a
7. Find Fc[xe-ax] and Fs[xe-ax]
8. Show that the Fourier sine transform of 1 / (1 + x2) is (/2) e-s.
9. Show that the Fourier sine transform of x / (1 + x2) is (/2) e-s.
2
10. Show that x e-x
/ 2
is self reciprocal with respect to Fourier sine transform.
11. Using transform methods to evaluate

(i)
dx

0
and
(x2+1)( x2+4)
UNIT-V
Z – Transforms AND DIFFERENCE EQUATIONS
5.1 Introduction
The Z-transform plays a vital role in the field of communication Engineering and
control Engineering, especially in digital signal processing. Laplace transform and
Fourier transform are the most effective tools in the study of continuous time signals,
where as Z – transform is used in discrete time signal analysis. The application of Z –
transform in discrete analysis is similar to that of the Laplace transform in continuous
systems. Moreover, Z-transform has many properties similar to those of the Laplace
transform. But, the main difference is Z-transform operates only on sequences of the
discrete integer-valued arguments. This chapter gives concrete ideas about Z-transforms
and their properties. The last section applies Z-transforms to the solution of difference
equations.
Difference Equations
Difference equations arise naturally in all situations in which sequential relation
exists at various discrete values of the independent variables. These equations may be
thought of as the discrete counterparts of the differential equations. Z-transform is a very
useful tool to solve these equations.
A difference equation is a relation between the independent variable, the
dependent variable and the successive differences of the dependent variable.
For example, 2yn + 7yn + 12yn = n2
and
3yn - 3yn - 2yn = cos n
are difference equations.
----------- (i)
---------- (ii)
The differences yn, 2yn, etc can also be expressed as.
yn = yn+1 - yn,
2yn = yn+2 - 2yn+1 + yn.
3yn = yn+3 - 3yn+2 + 3yn+1 - yn and so on.
Substituting these in (i) and (ii), the equations take the form
2
----------- (iii)
yn+2 + 5yn+1 +6yn = n
and
yn+3 - 3yn+2 = cos n
----------- (iv)
Note that the above equations are free of s.
If a difference equation is written in the form free of s, then the order of the
difference equation is the difference between the highest and lowest subscripts of y‟s
occurring in it. For example, the order of equation (iii) is 2 and equation (iv) is 1.
The highest power of the ys in a difference equation is defined as its degree when
it is written in a form free of s. For example, the degree of the equations
yn+3 + 5yn+2 + yn = n2 + n + 1 is 3 and y3n+3 + 2yn+1 yn = 5 is 2.
5.2 Linear Difference Equations
A linear difference equation with constant coefficients is of the form
a0 yn+r + a1 yn+r -1 + a2 yn+r -2 + . . . . +aryn = (n).
i.e., (a0Er + a1Er-1 + a2 Er-2 + . . . . + ar)yn = (n) ------(1)
where a0,a1, a2, . . . . . ar are constants and (n) are known functions of n.
The equation (1) can be expressed in symbolic form as
f(E) yn = (n)
If (n) is zero, then equation (2) reduces to
----------(2)
f (E) yn = 0
----------(3)
which is known as the homogeneous difference equation corresponding to (2).The
solution
of (2) consists of two parts, namely, the complementary function and the particular
integral.
The solution of equation (3) which involves as many arbitrary constants as the order of
the equation is called the complementary function. The particular integral is a
particular solution of equation(1) and it is a function of „n‟ without any arbitrary
constants.
Thus the complete solution of (1) is given by yn = C.F + P.I.
Example 1
Form the difference equation for the Fibonacci sequence .
The integers 0,1,1,2,3,5,8,13,21, . . . are said to form a Fibonacci sequence.
If yn be the nth term of this sequence, then
yn = yn-1 + yn-2 for n > 2
or yn+2 - yn+1 - yn = 0 for n > 0
5.3 Z - Transforms and its Properties
Definition
Let {fn} be a sequence defined for n = 0,1,2,…….,then its Z-transform F(z) is defined as

F(z) = Z{fn} =  fn z –n ,
n=0
whenever the series converges and it depends on the sequence {fn}.
The inverse Z-transform of F(z) is given by Z-1{F(z)} = {fn}.
Note: If {fn} is defined for n = 0, ± 1, ± 2, ……., then

F(z) = Z{fn} =  fn z –n , which is known as the two – sided Z- transform.
n=-
Properties of Z-Transforms
1. The Z-transform is linear.
i.e, if F(z) = Z{fn} and G(z) = Z{gn}, then
Z{afn + bgn} = aF(z) + bG(z).
Proof:

Z{ afn + bgn} =  { afn + bgn} z-n
(by definition)
n=0


= a  fn z + b gn z-n
-n
n=0
n=0
= aF(z) + b G(z)
2. If Z{fn} = F(z), then Z{anfn} = F (z/a)
Proof: By definition, we have

Z { anfn} =  an fn z-n
n=0

=  fn (z/a)-n = F(z/a)
n=0
Corollary:
If Z{fn} = F (z), then Z{ a-nfn} = F(az).
dF (z)
3. Z{nfn} = -z --------dz
Proof

We have
F(z)=  fn z-n
n=0
Differentiating, we get
dF(z)

------ =  fn (-n) z-n -1
n=0
dz
1 
= - -----  nfn z-n
z n=0
1
= - --- Z{nfn}
z
dF (z)
Hence, Z{nfn} = -z --------dz
4. If Z{fn} = F(z), then
Z{fn+k}= zk{ F(z) – f0 – (f 1 / z ) - … - ( fk-1 / zk-1) } (k > 0)
Proof

Z { fn+k} =  fn+k z-n , by definition.
n=0

=  fn+k z-n zk z-k
n=0

= z  fn+k z - (n+k)
k
n=0

= zk  fm z-m , where m = n+k .
m=k
= zk {F(z) – f0 – (f1/z) - .. .. .. – ( fk-1 / z k-1) }
In Particular,
(i) Z{f n+1} = z {F(z) - f0}
(ii) Z{f n+2}= z2 { F(z) – f0 – (f1/z) }
Corollary
If Z{fn} = F(z), then Z{fn–k} = z-k F(z).
(5) Initial value Theorem
If Z {fn} = F (z), then fo = ℓt
F(z)
z-
Proof
We know that F (z) = f0 + f1 z-1 + f2z-2 + . . .
Taking limits as z   on both sides, we get
ℓt F(z) = f0
z 
Similarly, we can find
f1 = ℓt { z [F(z) – f0]}; f2 = ℓt { z2 [F(z) – f0- f1z-1]} and so on.
z 
z 
(6) Final value Theorem
If Z{fn} = F(z), then ℓt fn = ℓt (z-1) F(z)
n 
z 1
Proof
By definition, we have

Z {fn+1 – fn} =  {fn+1 – fn} z-n
n=0

Z{fn+1} – Z{fn} =  {fn+1 – fn} z-n
n=0

ie, z {F(z) – f0} – F(z) =  {fn+1 – fn} z-n
n=0

(z –1) F(z) – f0z
=  {fn+1 – fn} z-n
n=0
Taking, limits as z  1 on both sides, we get

ℓt {(z –1) F(z)} – f0
z 1
= ℓt
 {fn+1 – fn} z-n
z 1 n=0

=  (fn+1 – fn) = (f1 – f0) + (f2 –f1) + . . . + (fn+1 – fn)
n=0
= ℓt
n 
i.e,
ℓt {(z –1) F(z)} – f0 = f - f0
z 1
f = ℓt [(z-1) F(z)]
Hence,
i.e,
z 1
ℓt
n 
fn = ℓt [(z-1) F(z)]
z 1
fn+1 – f0
SOME STANDARD RESULTS
Z{an} = z / (z-a), for |z| > |a|.
1.
Proof
By definition, we have

Z{a } =  an z-n
n
n=0

=  (a/z)n
n=0
1
= --------1-(a/z)
= z / (z-a), for |z| > |a|
In particular, we have
Z{1} = z / (z-1), (taking a = 1).
and
Z{(-1)n} = z / (z +1), (taking a = -1).
2. Z{nan} = az /(z-a)2
Proof: By property, we have
dF(z)
Z{nfn} = -z -------dz
d
= -z -------- Z{an}
dz
d
z
az
Z{nan} = -z ----- ------- = --------dz
z-a
(z-a)2
Similarly, we can prove
Z{n2an} = {az(z+a)}/ (z-a)3
d
(3) Z{n } = -z ------ Z{nm-1}, where m is a positive integer.
dz
m
Proof

Z{nm} =  nm z-n
n=0

= z  nm – 1 n z-(n+1) ----------------(1)
n=0
Replacing m by m-1, we get

Z{nm-1}= z  nm – 2 n z-(n+1)
n=0

Z{nm-1}=  nm – 1 z –n.
i.e,
n=0
Differentiating with respect to z, we obtain
d

dz
n=0
------ Z{nm-1} =  nm – 1 (-n) z-(n+1) ----------(2)
Using (2) in (1), we get
d
Z{nm} = -z ------ Z{nm-1}, which is the recurrence formula.
dz
In particular, we have
d
Z{n} = -z ----- Z{1}
dz
d
z
z
= -z ----- ------- = --------dz
z-1
(z-1)2
Similarly,
d
Z{n2} = -z ----- Z{n}
dz
d
z
= -z ----- ------dz (z-1)2
z (z+1)
= -------------.
(z-1)3
z (z - cos)
4. Z {cosn } = -------------------- and
z2 - 2z cos +1
z sin
Z {sinn } = -------------------z2 - 2z cos +1
We know that
Z{an} = z /(z-a), if |z| > |a|
Letting a = e i, we have
z
z
Z{e } = ---------- = -------------------z-ei
z–(cos  + isin)
z
Z{cosn + isinn} = --------------------------(z–cos ) - isin
in
z {(z–cos ) + isin}
= -----------------------------------------------{(z–cos ) - isin} {(z–cos ) + isin}
z (z–cos ) + izsin
= ---------------------------z2 - 2z cos +1
Equating the real & imaginary parts, we get
z (z - cos)
Z (cosn ) = -------------------z2 - 2z cos +1
z sin
Z (sinn ) = -------------------z2 - 2z cos +1
z (z - rcos)
5 . Z{r cosn } = --------------------z2 – 2rz cos +r2
n
and
and
zr sin
Z{r sinn} = --------------------z2 – 2rz cos +r2
We know that
n
if |z|>|r|
Z{an} = z /(z-a), if |z|>|a|
Letting a = rei , we have
Z{rnein } = z /(z -re i) .
z
i.e, Z{r (cosn + isinn) } = ------------z - rei
z
= --------------------------z – r(cos + isin)
n
z {(z - rcos) + i rsin}
= ---------------------------------------------------------{(z – rcos) – i rsin}{(z – rcos) + i rsin}
z (z - rcos) + i rzsin
= --------------------------------------(z – rcos)2 +r2 sin2
z (z - rcos) + i rzsin
= -----------------------------------z2 – 2rz cos +r2
Equating the Real and Imaginary parts, we get
z (z- rcos)
Z{r cosn} = ------------------------ and
z2 – 2zrcos + r2
n
zrsin
Z{rn sinn} = ------------------------; if | z | > | r |
z2-2zrcos + r2
Table of Z – Transforms
1.
fn
F(z)
1
z
--------z–1
2.
(-1)n
3.
an
4.
n
2
z
--------z+1
z
--------z–a
z
----------(z–1)2
z2 + z
------------(z–1)3
5.
n
6.
n(n-1)
2z
---------(z–1)3
7.
n(k)
k!z
-----------(z–1)k+1
8.
nan
az
----------(z–1)2
9.
cosn
z (z-cos)
-------------------z2 – 2zcos+1
10.
sinn
z sin
-------------------z2–2zcos + 1
11.
rn cosn
z (z-rcos)
---------------------z2–2rz cos + r2
n
12.
r sinn
13.
cos(n/2)
rz sin
---------------------z2–2rzcos +r2
z2
-----------z2 + 1
14.
sin(n/2)
z
-----------z2 + 1
Tz
15
t
----------(z–1)2
T2 z(z + 1)
-----------------(z–1)3
t2
16
17
e
z
----------z – eaT
18
e-at
z
----------z – e-aT
19
Z{cost}
z (z - cosT)
---------------------z2 - 2z cosT +1
Z{sint }
z sinT
---------------------z2 - 2z cosT +1
20
21
22
at
Z{e –at cos bt}
Z{e
–at
sin bt}
zeaT (zeaT – cos bT)
-------------------------------z2e2aT – 2zeaT cos bT +1
zeaT sin bT
-------------------------------z2e2aT – 2zeaT cos bT +1
2 (z – 1)3 3
2 (z – 1)
Example 2
Find the Z– transform of
(i)
n(n-1)
(ii)
n2 + 7n + 4
(iii) (1/2)( n+1)(n+2)
(i) Z { n(n-1)}= Z {n2} – Z {n}
z (z+1)
z
= -------------- - ---------(z-1)3
(z-1)2
z (z+1) – z (z-1)
= ----------------------(z – 1)3
2z
= -----------(z-1)3
(iii)
Z{ n2 + 7n + 4}= Z{n2} + 7 Z{n}+ 4 Z{1}
z (z+1)
z
z
= -------------- + 7 ---------- + 4 ----------(z-1)3
(z-1)2
z-1
z {(z+1) + 7(z-1) + 4(z-1)2}
= -----------------------------------(z – 1)3
2
2z(z -2)
= --------------(z-1)3
(n+1) (n+2)
1
(iii) Z ---------------- = ------{ Z{n2} + 3Z{n}+2Z{1}}
2
2
1
z(z+1)
3z
2z
= ------ ----------- + -------- + -------- if |z | > 1
2
(z-1)3
(z-1)2
(z-1)
z3
= ------------(z-1)3
Example 3
Find the Z- transforms of 1/n and 1/n(n+1)
.
1
1

(i ) Z ----=  ------- z-n
n=1
n
n
1
1
1
= ----- + ------ + -------- + . . . .
z
2z2
3z3
= - log (1 – 1/z ) if |1/z| < 1
= - log (z-1 / z)
= log (z/z-1), if | z | >1.
1
1
1
(ii) Z --------- = Z ------ - ------n(n+1)
n
n+1
1
1


=  ----- z-n   ------ z-n
n=1
n=0
n
n+1
z
= log ------ 
z-1
z
= log ------  z
z-1
1
1
1+ ------- + ------- + . . .
2z
3z2
1
1
1 2
1
1 3
---- + ----- ------ + ------ ----- + . . .
z
2
z
3
z
z
= log ------  z { - log (1 – 1/z)}
z-1
z
= log ------  z log (z/z-1)
z-1
= (1- z) log {z/(z-1)}
Example 4
Find the Z- transforms of
(i) cos n/2
(ii) sin n/2
n
(i) Z{cos n/2} =  cos ------ z-n
n=0
2
1
1
= 1  ------ + ------  ……………..
z2
z4

1
= 1 + -----z2
z2 + 1
= ---------z2
-1
-1
z2
= ---------- , if | z | > |
z2 + 1
n

(ii) Z{sin n/2} =  sin ------ z-n
n=0
2
1
1
1
= -----  ------ + ------  . . . . . . . . . . .
z
z3
z5
1
1
1
= ------ 1 - ------ + ------- - . . .
z
z2
z4
1
1
= ---- 1 + -----z
z2
-1
1
z2 +1
= ---- ---------z
z2
-1
1
z2
z
= ----- ---------- = --------z
z2 + 1
z2 + 1
Example 5
Show that Z{1/ n!} = e1/z and hence find Z{1/ (n+1)!} and Z{1/ (n+2)!}
1
Z ------n!
1
=  ------ z-n
n=0
n!

(z-1)n
=  --------n=0
n!
z-1
(z-1)2
= 1 + ------- + --------- + . . .
1!
2!

-1
= e z = e1/z
To find
1
Z -------(n+1)!
We know that Z{fn+1} = z { F(z) – f0}
Therefore,
1
Z -------(n+1)!
1
= z Z ----- – 1
n!
= z { e1/z -1}
Similarly,
1
Z -------(n+2)!
= z2 { e1/z -1 – (1/z)}.
Example 6
Find the Z- transforms of the following
(i) f(n) =
n, n > 0
0, n< 0
(ii) f (n) = 0, if n > 0
1, if n < 0
(iii)f(n) = an / n!, n > 0
0, otherwise

Z{f(n)} =  f (n) z –n
(i )
n=0

=  n z –n
n=0
= (1 / z) + (2/ z2) + ( 3/ z3) + . . .
= (1 / z) {1+ (2/z) + (3/z2) + . . .}
= (1/z){ 1 – (1/z)}-2
z-1 -2
= 1/z -----z
= z / (z-1)2, if |z |> |

Z{f(n)}=  f (n) z –n
(ii)
n=-

=  z –n
n=-

=  zn
n=0
= (1/1 – z), if | z | < 1.

Z{f(n)} =  f (n) z
(iii)
–n
n= 0
an
=  ------ z –n
n=0
n!

(az-1)n
=  -----------n=0
n!

= eaz
-1
= e a/z
Example 7
2z2 + 3z +12
If F(z) = --------------------- , find the value of „f2‟ and „f3‟.
(z-1)4
2z2 + 3z +12
Given that
F(z) = ---------------------.
(z-1)4
This can be expressed as
1
2 + 3z-1 +12z-2
F(z) = ----- ---------------------.
z2
(1- z-1)4
By the initial value theorem, we have
fo = ℓt
F(z) = 0.
z-
f1 = ℓt {z[F(z) - fo]} = 0.
Also,
z-
f2 = ℓt {z2 [F(z) - fo – (f1 /z)]}
Now,
z-
= ℓt
z-
2 + 3z-1 +12z-2
--------------------- - 0 – 0.
(1- z-1)4
= 2.
and
f3 = ℓt {z3 [F(z) - fo – (f1 /z) – (f2/ z2)]}
z-
= ℓt
3
z
z-
Given that
= ℓt
z-
2 + 3z-1 +12z-2
2
--------------------- – ------(1- z-1)4
z2
11z3 + 8z -2
z ---------------------. = 11.
z2 (z-1)4
3
5.4 Inverse Z – Transforms
The inverse Z – transforms can be obtained by using any one of the following
methods.They are
I.
II.
III.
IV.
Power series method
Partial fraction method
Inversion Integral method
Long division method
I.
Power series method
This is the simplest method of finding the inverse Z –transform. Here F(z) can
be expanded in a series of ascending powers of z -1 and the coefficient of z –n will be the
desired inverse Z- transform.
Example 8
Find the inverse Z – transform of log {z /(z+1)} by power series method.
1
1/y
Putting z = -------, F (z) = log -------------y
(1 / y) + 1
1
= log --------1+y
= - log (1+y)
y2
y3
= - y + -------- - --------- + . . . . . .
2
3
1
1
(-1)n
= -z-1 + ------ z –2 - ------ z-3 + . . . . . +-------- z-n
2
3
n
0,
Thus, fn =
for n = 0
(-1)n / n , otherwise
II. Partial Fraction Method
Here, F(z) is resolved into partial fractions and the inverse transform can be taken
directly.
Example 9
z
Find the inverse Z – transform of -----------------z2 + 7z + 10
z
Let F (z)
= ------------------z2 + 7z + 10
F(z)
1
1
Then -------- = ------------------ = -------------------z
z2 + 7z + 10
(z+2) (z+5)
1
A
B
Now , consider ------------------ = --------- + --------(z+2) (z+5)
z+2
z+5
1
1
1
= ------- -------- - ------3
z +2
3
1
--------z +5
1
z
1
z
F (z) = ------- -------- - ------- ---------3
z +2
3
z+5
Therefore,
Inverting, we get
1
1
= ------- (-2)n - ------ (-5)n
3
3
Example 10
Find the inverse Z – transform of
8z2
--------------------(2z–1) (4z–1)
8z2
z2
Let F (z) = --------------------- = ---------------------(2z–1) (4z–1)
(z– ½ ) (z – ¼)
F (z)
z
Then ---------- = --------------------------z
(z– ½ ) (z – ¼)
Now,
We get,
z
A
B
-------------------- = --------- + ---------(z– ½ ) (z – ¼)
z– ½
z–¼
F (z)
2
1
---------- = -----------  ------------z
z– ½
z–¼
z
z
F (z) = 2 -----------  ----------z– ½
z–¼
Therefore,
Inverting, we get
z
z
fn = Z –1{F(z)}= 2 Z-1 -----------  Z-1 ----------z– ½
z–¼
fn = 2 (1 / 2)n – (1 / 4)n , n = 0, 1, 2, . . . . . .
i.e,
Example 11
Find
Let
-1
Z
4- 8z-1 + 6z-2
------------------(1+z-1) (1-2z-1)2
by the method of partial fractions.
4- 8z-1 + 6z-2
F(z) = ------------------(1+z-1) (1-2z-1)2
4z3 - 8z2 + 6z
= --------------------(z + 1) (z - 2)2
F(z)
Then ----- =
z
4z2 - 8z + 6
A
B
C
-------------------- = ------- + --------- + --------, where A = B = C = 2.
(z + 1) (z - 2)2
z+1
z-2
(z-2)2
F(z)
2
2
2
So that ------- = ------- + --------- + ---------z
z+1
z-2
(z -2)2
2z
2z
2z
Hence, F(z) = ------- + --------- + ---------z+1
z-2
(z -2)2
Inverting, we get
fn =
i.e,
2(-1)n + 2(2)n + n.2n
fn = 2(-1)n + (n+2)2n
Inversion Integral Method or Residue Method
The inverse Z-transform of F (z) is given by the formula
1
fn = ------  F(z) z n-1 dz
2i C
= Sum of residues of F(z).zn-1 at the poles of F(z) inside the contour C which is
drawn according to the given Region of convergence.
Example 12
Using the inversion integral method, find the inverse Z-transform of
3z
------------(z-1) (z-2)
3z
Let F(z) = ---------------.
(z-1) (z-2)
Its poles are z = 1,2 which are simple poles.
By inversion integral method, we have
1
fn = ------  F(z). z n-1 dz = sum of resides of F(z). z n-1 at the poles of F(z).
2i C
1
3z
1
3zn
i.e, fn = ------  -------------. zn-1 dz = ------  ------------- dz = sum of residues
2i C (z-1)(z-2)
2i C (z-1)(z-2)
---------(1).
Now,
3zn
Residue (at z =1) = ℓt (z-1). ------------- = -3
z-1
(z-1)(z-2)
Residue (at z =2) = ℓt
z2
3zn
(z-2). -------------- = 3.2 n
(z-1)(z-2)
Sum of Residues = -3 + 3.2n = 3 (2n-1).
Thus the required inverse Z-transform is
fn = 3(2n-1), n = 0, 1, 2, …
Example 13
z(z+1)
Find the inverse z-transform of ----------- by residue method
(z-1)3
z(z+1)
Let F(z) = ----------(z-1)3
The pole of F(z) is z = 1, which is a pole of order 3.
By Residue method, we have
1
fn = --------  F(z). z n-1 dz = sum of residues of F(z).z n-1 at the poles of F(z)
2i C
1
z+1
n
i.e., fn = --------  z . -------- dz = sum of residues .
2i C
(z-1)3
1
d2
Now, Residue (at z = 1) = ---- ℓt ------ ( z -1)3
2! z1 dz2
1
= ---- ℓt
2! z1
Hence,
zn(z+1)
----------(z-1)3
d2
----- { zn ( z +1)}
dz2
1
d2
= ---- ℓt ----- { zn+1 + zn)}
2! z1 dz2
1
= ---- ℓt { n(n+1) z n-1 + n(n-1) z n-2 }
2 z1
1
= ---- { n(n+1) + n (n-1)} = n2
2
fn = n2, n=0,1,2,…..
IV. Long Division Method
If F(z) is expressed as a ratio of two polynomials, namely, F(z) = g(z-1) / h(z-1),
which can not be factorized, then divide the numerator by the denominator and the
inverse transform can be taken term by term in the quotient.
Example 14
1+2z-1
Find the inverse Z-transform of --------------------, by long division method
1-z-1
1+2z-1
Let F (z) = ---------------1-z-1
By actual division,
1+3z-1 +3z-2+3z-3
1 – z -1
1 + 2z-1
1 – z -1
+3z -1
3z-1 – 3z-2
+ 3z -2
3z -2 – 3z-3
+ 3z -3
3z -3 – 3z -4
+3z -4
Thus F(z) = 1 + 3z-1 + 3z-2 + 3z-3 + . . . . . .
Now, Comparing the quotient with

 fnz-n = fo + f1z-1 + f2z-2 + f3z-3 + . . . . . .
n=0
We get the sequence fn as f0 = 1, f1 = f2 = f3 = . . . . . . = 3.
Hence fn =
1,
for n = 0
3,
for n > 1
Example 15
Find the inverse Z-transform of
z
-----------z2 – 3z +2
By actual division
z -1 + 3z -2 + 7z –3 + ... ... ...
1-3z-1 + 2z-2
z-1
z -1 - 3z -2 + 2z –3
3z-2 – 2z -3
3z-2 – 9z –3 + 6z -4
7z -3 – 6z –4
7z -3 – 21z –4 + 14z-5
+15 z -4 – 14z –5
 F(z) = z -1 + 3z -2 + 7z –3 + ... ... ...
Now comparing the quotient with

 f n z –n = f0 + f1 z-1 + f2 z-2 + f3 z –3 + .. ..
n=0
We get the sequence fn as f0 = 0, f1 = 1, f2 = 3, f3 = 7, .... ... ....
Hence, fn = 2n-1, n = 0, 1, 2, 3, ...
Exercises
1. Find Z-1 {4z / (z-1)3} by the long division method
2. Find Z-1
z (z2 – z + 2)
------------------- by using Residue theorem
(z+1) (z-1)2
-1
z2
------------------- by using Residue theorem
(z+2) (z2+4)
3. Find Z
4. Find Z-1 (z/z-a) by power series method
5. Find Z-1 (e-2/z) by power series method
z3 – 20z
---------------- by using Partial fraction method
(z-4) (z-2)3
6. Find Z-1
5.5 CONVOLUTION THEOREM
If Z-1{F (z)} = fn and Z-1{G(z)} = gn, then
n
Z-1{F(z). G(z)} =  fm. gn-m = fn* gn, where the symbol * denotes the operation of
m=0
convolution.
Proof


We have F (z) =  fn z-n, G (z) =  gnz-n
n=0
n=0
 F(z) .G (z) = (f0 + f1z-1 + f2z-2 + ... + fnz-n + ... ). (go + g1z-1 + g2z-2 + ...+ gnz-n+
...)

=  (fogn+f1gn-1+f2gn-2+ . . .+ fngo)z-n
n=0
= Z (fogn+f1gn-1+f2gn-2+ . . .+ fngo)
n
= Z  fm gn-m
m=0
= Z {fn * gn}
Hence,
Z-1 { F(z) . G(z)} = fn * gn
Example 16
Use convolution theorem to evaluate
z2
----------------(z-a) (z-b)
-1
Z
We know that Z-1 {F(z). G(z)} = fn*gn.
z
z
Let F (z) = ---------- and G (z) ----------z-a
z-b
-1
Then fn = Z
z
z
n
-1
---------- = a & gn = Z ----------- = bn
z-a
z-b
Now,
Z-1 {F(z). G (z)} = fn * gn = an * bn
n
=  am bn-m
m=0
a
m
= b  ------which is a G.P.
m=0
b
n
n
n
=b
-1
ie, Z
z2
---------------(z-a) (z-b)
(a/b) n+1 - 1
----------------------(a/b) – 1
an+1 – bn+1
= --------------------a–b
Example 17
Find z-1
z 3
-------- by using convolution theorem
(z-1)
z2
z
Let F (z) = ---------- and G (z) ----------(z-1)2
(z-1)
Then fn = n+1 & gn = 1
By convolution Theorem, we have
n
Z-1 { F(z). G (z) } = fn * gn = (n+1) * 1 =  (m+1) . 1
m=0
(n+1) (n+2)
= -------------------2
Example 18
Use convolution theorem to find the inverse Z- transform of
1
-------------------------------[1 – (1/2)z –1] [1- (1/4)z-1]
Given
-1
Z
1
z2
-1
------------------------------ - = Z --------------------------[1 – (1/2)z –1] [1- (1/4)z-1]
[z-(1/2)] [z – (1/4)]
z
z
Let F (z) = -------------- & G (z) = ------------z – (1/2)
z – (1/4)
Then fn = (1/2)n
& gn = (1/4)n.
We know that Z-1{ F(z). G(z)} = fn * gn
= (1/2)n * (1/4)n
1 m 1
=  ------- ------m=0
2
4
n
n-m
1 n n
1 m 1
= ------  ------- ------m=0
4
2
4
-m
1 n n
= ------  2 m
m=0
4
1 n
= ------ { 1+2+22+ . . . + 2n} which is a G.P
4
1
= -----4
1
= -----4
n
n
2n+1 - 1
-------------2-1
{2n+1 – 1}
1
1 n
= ---------- - ------2n-1
4
1
1
 Z-1 --------------------------------- = ------ [1 – (1/2)z –1] [1- (1/4)z-1]
2 n-1
1
------4n
5.6 Application of Z - transform to Difference equations
As we know, the Laplace transforms method is quite effective in solving linear
differential equations, the Z - transform is
useful tool in solving linear difference
equations.
To solve a difference equation, we have to take the Z - transform of both sides of
the difference equation using the property
Z{fn+k}= zk{ F(z) – f0 – (f 1 / z ) - … - ( fk-1 / zk-1) } (k > 0)
Using the initial conditions, we get an algebraic equation of the form F(z) = (z).
By taking the inverse Z-transform, we get the required solution fn of the given difference
equation.
Exmaple 19
Solve the difference equation yn+1 + yn = 1, y0 = 0, by Z - transform method.
Given equation is yn+1 + yn = 1
---------- (1)
Let Y(z) be the Z -transform of {yn}.
Taking the Z - transforms of both sides of (1), we get
Z{yn+1} + Z{yn} = Z{1}.
ie, z {Y(z) - y0} + Y(z) = z /(z-1).
Using the given condition, it reduces to
z
(z+1) Y(z) = -------z-1
z
i.e, Y(z) = --------------------(z - 1) (z + 1)
or
1
z
z
Y(z) = ----- --------- - -------2
z-1
z+1
On taking inverse Z-transforms, we obtain
yn = (1/2){1 - (-1)n}
Example 20
Solve yn+2 + yn = 1, y0 = y1 = 0 , using Z-transforms.
Consider
yn+2 + yn = 1
------------- (1)
Taking Z- transforms on both sides, we get
Z{yn+2}+ Z{yn} = Z{1}
y1
z
z2 {Y(z) - y0 - ------ } + Y(z) = --------z
z-1
z
(z2 + 1) Y(z) = ----------z-1
z
or Y(z) = -------------------(z - 1) (z2 + 1)
Y(z)
1
A
Bz + C
------ = --------------- = ------- + ----------z
(z-1)(z2+1)
z-1
z2+1
Now,
Therefore,
1
= ----2
1
z
1
------- - --------  ---------z-1
z2 + 1
z2 + 1
1
Y(z) = ----2
z
z2
z
------- - --------  ---------z-1
z2 + 1
z2 + 1
Using Inverse Z-transform, we get
yn =(½){1 - cos (n / 2) - sin (n / 2)}.
Example 21
Solve yn+2 + 6yn+1 + 9yn = 2n, y0 = y1 = 0, using Z-transforms.
Consider yn+2 + 6yn+1 + 9yn = 2n
-------- (1)
Taking the Z-transform of both sides, we get
Z{yn+2} + 6Z{yn+1} + 9Z{yn} = Z {2n}
y1
i.e,
z
z2 Y(z)-y0 - ------ + 6z {Y(z) - y0} + 9Y(z) = -------z
z–2
z
(z + 6z + 9) Y(z) = -------z-2
2
i.e,
Y(z)
Therefore,
z
= ---------------(z-2) (z+3)2
Y(z)
1
-------- = ---------------z
(z-2)(z+3)2
Y(z)
1
1
1
1
1
1
ie, ------ = ------ ------- - ------- ------- - ----- -------- ,
z
25
z-2
25
z+3
5 (z+3)2
using partial fractions.
Or
1
z
z
5z
Y(z) = ------ -------  -------  -------25
z-2
z+3
(z+3)2
On taking Inverse Z-transforms, we get
yn = (1/ 25){ 2n - (-3)n + (5/3) n (-3)n}.
Example 22
Solve the simultaneous equations
xn+1 - yn = 1; yn+1 - xn = 1 with x (0) = 0; y (0) = 0.
The given equations are
xn+1 - yn = 1,
yn+1 - xn =1,
x0 = 0
y0 = 0
------------- (1)
-------------- (2)
Taking Z-transforms, we get
z
z {X(z) - x0} – Y(z) = -------z-1
z
z {Y(z) - y0} – X(z) = -------z-1
Using the initial conditions, we have
z
z X(z) – Y(z) = -------z-1
z
z Y(z) – X(z) = -------z-1
Solving the above equations, we get
z
X(z) = -------- and
(z-1)2
z
Y(z) = --------.
(z-1)2
On taking the inverse Z-transform of both sides, we have xn = n and yn = n ,
which is the required solution of the simultaneous difference equations.
Example 23
Solve xn+1 = 7xn + 10yn ; yn+1 = xn + 4yn, with x0 = 3, y0 = 2
Given
xn+1 = 7xn + 10yn
yn+1 = xn + 4yn
------------- (1)
------------- (2)
Taking Z- transforms of equation(1), we get
z { X(z) - x0} = 7 X(z) + 10 Y(z)
(z - 7) X(z) – 10 Y(z) = 3z ----------(3)
Again taking Z- transforms of equation(2), we get
z {Y(z) - y0} = X(z) + 4Y(z)
-X(z) + (z - 4)Y(z) = 2z ---------- (4)
Eliminating „x‟ from (3) & (4), we get
2z2 - 11z
2z2 - 11z
Y(z) = ---------------- = ------------------z2 - 11z+8
(z-9) (z-2)
Y(z)
2z - 11
A
B
---- = ---------------- = ------- + ------- ,where A =1 and B = 1.
z
(z-9) (z-2)
z-9
z-2
so that
Y(z)
1
1
ie, ----- = ---------- + --------z
z-9
z–2
z
z
ie, Y(z) = ---------- + --------z-9
z–2
Taking Inverse Z-transforms, we get yn = 9n + 2n.
From (2),
xn = yn+1 - 4yn = 9n+1 + 2n+1 - 4 (9n + 2n)
= 9.9n + 2.2n - 4.9n - 4.2n
Therfore,
xn = 5.9n - 2.2n
Hence the solution is xn = 5.9n - 2.2n and yn = 9n + 2n.
Exercises
Solve the following difference equations by Z – transform method
1. yn+2 + 2yn+1 + yn = n, y0 = y1 = 0
2. yn+2 – yn = 2n, y0 = 0, y1 = 1
3. un+2 – 2cos un+1+ un=0, u0 = 1, u1 = cos
4. un+2 = un+1 + un, u0 = 0, u1 = 1
5. yn+2 – 5yn+1+ 6yn = n (n-1), y0 = 0, y1 = 0
6. yn+3 – 6yn+2 + 12yn+1 – 8yn = 0, y0 = -1, y1 = 0, y2 = 1
5.7 FORMATION OF DIFFERENCE EQUATIONS
Example
Form the difference equation
yn  a 2n  b(2) n
yn 1  a 2n 1  b(2) n 1
 2a2n  2b(2)n
yn2  a2n2  b(2)n1
 4a2n  4b(2)n
Eliminating a and b weget,
yn 1 1
yn 1
2 2 =0
yn  2
4
4
yn (8  8) 1(4 yn1  2 yn2 )  1(4 yn1  2 yn2 )  0
16 yn  4 yn  2  0
4( yn  2  4 yn )  0
yn  2  4 yn  0
Exercise:
1. Derive the difference equation form yn  ( A  Bn)(3)n
2. Derive the difference equation form U n  A2n  Bn
BIBLIOGRAPHY
1. Higher Engineering Mathematics – Dr.B.S. Grewal
2. Engineering Mathematics – Vol III – P. Kandasamy
3. Engineering Mathematics-II – T.Veerarajan
4. Higher Engineering Mathematics – N.P.Bali & others
4. Advanced Mathematics For Engineering- Narayanan
5. Advanced Engineering Mathematics- C.Ray & C.Barrett
6. Advanced Engineering Mathematics- Erwin Kreyszig
UNIT –I
PARTIAL DIFFERENTIAL EQUATIONS
1. Explain how PDE are formed?
PDE can be obtained
(i)
By eliminating the arbitrary constants that occur in the functional relation
between the dependent and independent variables.
(ii)
By eliminating arbitrary functions from a given relation between the dependent
and independent variables.
2. From the PDE by eliminating the arbitrary constants a & b from z  ax  by .
Given z  ax  by
Diff. p.w.r. to x we get,
z
a
x
i.e.,
pa
Diff. p.w.r. to y we get,
z
b
y
i.e.,
Substituting in (1) we get
q b
z  px  qy .
3. From the PDE by eliminating the arbitrary constants a & b from z  ( x2  a 2 )( y 2  b2 ) .
Given z  ( x2  a 2 )( y 2  b2 ) ---------(1)
p
z
 (2 x)( y 2  b2 )
x
p
 y 2  b2
2x
q
---------(2)
z
 (2 y)( x 2  a 2 )
y
q
 x2  a2
2y
---------(3)
Substituting (2) & (3) in (1) we get the required p.d.e.
 q   p  pq
i.e., z     
 2 y   2 x  4 xy
4xyz  pq .
4. Eliminate f from z  f ( x 2  y 2 ) .
Given z  f ( x 2  y 2 )
…(1)
Diff (1) p.w.r. to x and y we get,
z
 f ( x 2  y 2 )  2 x 
x
i.e.,
p  f ( x 2  y 2 )  2 x
...(2)
z
 f ( x 2  y 2 )  2 y 
y
i.e.,
q  f ( x2  y 2 )  2 y 
(2)
p x


 3 q y
py  qx  0 .
...(3)
5. Obtain PDE from z  f (sin x  cos y) .
Given
z  f (sin x  cos y)
p 
z
 f  (sin x  cos y) cos x 
x
...(2)
q 
z
 f  (sin x  cos y)   sin y 
y
...(3)
 2 
 3
p
cos x

q
 sin y
p sin y   q sin y
p sin y  q sin y  0 .
6. Solve
z
 sin x .
x
Given
…(1)
z
 sin x
x
Integrating w.r to x on both sides
z   cos x  c
But z is a function of x and y
 z   cos x  f ( y)
Hence c  f ( y) .
7. Mention three types of solution of a p.d.e (or) Define general and complete integrals of a
p.d.e.
(i) A solution which contains as many arbitrary constants as there are independent
variables is called a complete integral (or) complete solution.
(ii) A solution obtained by giving particular values to the arbitrary constants in a
complete integral is called a particular integral (or) particular solution.
(iii)A solution of a p.d.e which contains the maximum possible number of arbitrary
functions is called a general integral (or) general solution.
8. Solve
p  q 1
p  q 1
Given
This is of the form F(p,q) = 0.
Hence the complete integral is z  ax  by  cz .
where,
a  b  1, b  1  a
b  (1  a )2
Therefore the complete solution is
z  ax  (1  a )2 y  c
--------- (1)
Diff.p.w.r. to c we get,
0 1
There is no singular integral.
Taking c  f  a  when f is arbitrary.
z  ax  (1  a )2 y  f (a)
--------- (2)
Diff. p.w.r.to ' a '
 1 
0  x  2(1  a ) 
 y  f (a)
2 a 
--------- (3)
Eliminating ' a ' between (2) & (3) we get the general solution.
9. Find the complete integral of z  px  qy  p 2  q 2 .
Given z  px  qy  p 2  q 2 .
This equation is of the form z  px  qy  f ( p, q) .
By Clairaut’s type,put p  a, q  b .
Therefore the complete integral is z  ax  by  a 2  b2 .
10. Find the complete integral of q  2 px .
Given
q  2 px .
This equation of the form
Let q  a , then
But dz 
p
f ( x, p, q)  0 .
a
2x .
a
dx  ady .
2x
Integrating on both sides,
a
 dz   2 xdx   ady .
a
z  log x  ay  b .
2
11. Find the complete integral of pq  xy .
Given
Hence
pq  xy .
p y
 .
x q
It is of the form f ( x, p)   ( y, q) .
Let
p y
 a.
x q
 p  ax
and
q
y
.
a
Hence dz  pdx  qdy .
dz  axdx 
y
dy .
a
Integrating on both sides,
za
x2 y 2

c.
2 2a
2az = a 2 x2  y 2  b is the required complete integral.
12. Solve px  qy  z .
Given
px  qy  z
--------- (1)
This equation is of the form Pp  Qq  R
when P  x, Q  y, R  z
The subsidiary equations are
ie.,
Take
dx dy dZ


P Q
R
dx dy dZ


x
y
z
dx dy

x
y
dx
dy
 x  y
log x  log y  log c1
Take
log x  log( zc2 )
log x  log( yc1 )
i.e.,
dx dz

x
z
dx
dz
 x  z
log x  log z  log c2
x  yc1
x  zc2
x
 c1
y
x
u
y
x
 c2
z
x
v
z
 x x
Therefore the solution of the given p.d.e is   ,   0 .
y z
13. Solve ( D2  4DD  3D2 ) z  0 .
Given ( D2  4DD  3D2 ) z  0
The auxiliary equation is m2  4m  3  0
m  m  3   1 m  3   0
m  3 , m 1
The roots are distinct.
Hence C.F  1  y  x   2  y  3x  .
 z  C.F .
i.e.,
z  1  y  x   2  y  3x  .
14. Solve 2r  5s  3t  0 .
Given 2r  5s  3t  0 .
The given differential equation can be written as,
2
2 z
2 z
2 z

5

3
 0.
x 2
xy
y 2
 2D
i.e.,
2
 5DD  3D2  z  0 .
The auxiliary equation is, 2m2  5m  3
0 .
2m 2  6m  m  3  0
2m  m  3   1  m  3   0
 m  3 2m  1
0
m3 , m
1
2
1 

C.F  1  y  3x   f  y  x 
2 

 z  1  y  3x   2  2 y  x  .
15. Find the P.I of  D2  DD  z  e x y .
Given
D
2
 DD  z  e x y
P.I 
1
e x y
D  DD
2
P.I 
1 x y
1
e
 ex y .
1 1
0
If we replace D by 1 and D’ by -1 we get dr  0 .
P.I 
x
x
e x y 
e x y
2 D  D
2 1  1
.
x
 e x y  x e x y
1
16. Find the P.I of  D2  2DD  D2  z  cos  x  3 y  .
Given  D2  2DD  D2  z  cos  x  3 y 
1
cos  x  3 y 
D  2 DD  D2
cos  x  3 y 
.

1  2(3)  9
1
 cos  x  3 y 
16
P.I 
2
PART-B
1.Solve z  px  qy  1  p 2  q 2 .
Soln:
Given: z  px  qy  1  p 2  q 2
This is of the form z=px+qy+f(p,q)
Hence, the complete integral is z  ax  by  1  a 2  b 2 ------------->(1)
Where a & b are arbitrary constant.
To Find The Singular integral:
Diff (1) p.w.r.to a,
We get, 0  x  0 
1
2 1  a2  b2
( 2a )
a   x 1  a 2  b 2 ---------(2)
Diff (1) p.w.r.to b,
We get, 0  y  0 
1
2 1  a2  b2
(2b)
b   y 1  a 2  b 2 ---------(3)
(1)=> z   x 2 1  a 2  b 2  y 2 1  a 2  b 2  1  a 2  b 2
z  (1  x 2  y 2 ) 1  a 2  b 2 ----------(4)
(4)=> z  (1  x 2  y 2 )
1
1  x2  y2
z2 1 x2  y2
x2  y2  z 2 1
Which is the singular solution.
To Get the general integral:
Put b   (a) in (1) , we get
z  ax   (a) y  1  a 2  [ (a)]2 ---------------(5)
Diff (5) p.w.r.to a, we get
0  x   ' (a) y 
2a  2 (a) ' (a)
2 1  a 2  [ (a)] 2
-----------------------(6)
Eliminate a between (5) abd (6) to get the general solution.
2.Solve y2p-xyq=x(z-2y)
Soln:
Given y2p-xyq=x(z-2y)
This equation of the form Pp+Qq=R
Here, P=y2 ,Q=-xy , R= x(z-2y)
The Lagrange’s subsidiary equation are
\i.e,
dx dy dz


P Q R
dy
dx
dz


y 2  xy x( z  2 y )
Take ,
dx
dy

2
y
 xy
dy
dz

 xy x( z  2 y )
dx dy

y x
dy
dz

 y ( z  2 y)
xdx=-ydy
(z-2y)dy=-ydz
 x dx   y dy
z dy-2y dy=-ydz
y 2 c1
x2


2
2
2
ydz+zdy=2ydy
x2+y2=c1
 d ( yz )   2 y dy
u=x2+y2
yz=y2+c2
v=yz-y2
Hence the general solution is f(x2+y2 , yz-y2)=0.
3.Solve:(3z-4y)p+(4x-2z)q=2y-3x
Soln:
Given: (3z-4y)p+(4x-2z)q=2y-3x
This equation of the form Pp+q=R
Here, P= (3z-4y)
,Q=(4x-2z) , R= 2y-3x
The Lagrange’s subsidiary equation are
\i.e,
dx
(3z - 4y)

dx dy dz


P Q R
dy
dz

--------------------(1)
(4x - 2z) 2y - 3x
Use Lagrangian multipliers x,y,z,
We get the ratio in (1)
=
xdx  ydy  zdz
xdx  ydy  zdz
=
0
(3z - 4y)x + (4x - 2z)y  (2y - 3x)z
Xdx+ydy+zdz=0
Integrating we get
 x dx   y dy   z dz  0
x2 y2 z2 a



2
2
2 2
i.e, x2+y2+z2=a.
Again use Lagrangian multipliers 2,3,4,
We get the ratio in (1)
=
2dx  3dy  4dz
2dx  3dy  4dz
=
(6z - 8y - 12x - 6z  8y - 12x
0
2dx  3dy  4dz =0
Integrating, we get
 2 dx   3dy   4dz  0
2x+3y+4z=b.
Hence the general solution is,
F(x2+y2+z2 , 2x+3y+4z)=0.
4.Find the general solution of x(y2-z2)p+y(z2-x2)q=z(x2-y2)
Soln;
Given; x(y2-z2)p+y(z2-x2)q=z(x2-y2)
This equation of the form Pp+q=R
Here, P= x(y2-z2)
,Q= y(z2-x2) , R= z(x2-y2)
The Lagrange’s subsidiary equation are
\i.e,
dx
x(y - z 2 )
2

dx dy dz


P Q R
dy
dz

--------------------(1)
2
2
y(z - x )
z(x - y 2 )
2
Use Lagrangian multipliers x,y,z,
We get the ratio in (1)
=
xdx  ydy  zdz
xdx  ydy  zdz
=
2
2
2
2
0
x(y - z ) + y(z - x )  z(x - y )
2
2
xdx+ydy+zdz=0
Integrating we get
 x dx   y dy   z dz  0
x2 y2 z2 a



2
2
2 2
i.e, x2+y2+z2=a.
Again use Lagrangian multipliers
1 1 1
, , ,
x y z
We get the ratio in (1)
1
1
1
1
1
1
dx  dy  dz
dx  dy  dz
x
y
z
x
y
z
= 2
=
0
y  z2  z2  x2  x2  y2
1
1
1
dx  dy  dz =0
x
y
z
Integrating, we get
1
1
1
 x dx   y dy   z dz  0
logx +logy+logz=log b
Hence the general solution is,
F(x2+y2+z2 , logx +logy+logz)=0.
5.Solve:[D3-2D2D’]z=ex+2y+4sin (x+y)
Soln:
Given: [D3-2D2D’]z=ex+2y+4sin (x+y)
The auxiliary equation is m3-2m2=0
Replace D by m and D’ by 1
m2(m-2)=0
m=0,0 and m=2
C.F= 1 ( y)  x 2 ( y)  3 ( y  2 x)
1
e x + 2y
2
D - 2D D'
1

e x + 2y
3
2
(1) - 2(1) (2)
Replace D by 1 and D' by 2
1 x  2y

e
3
P.I1 
3
1
4 sin( x  y )
D - 2D 2 D'
1
 I .P 4 3
e i( x y)
2
'
D  2D D
1
 I .P 4 3
e i( x y)
2
(i )  2(i ) (i )
Replace D by i and D' by i
P.I 2 
3
1
e i( x y)
 i  2i
1
 4 I .P e i ( x  y )
i
 4 IP (i (cos( x  y )  i sin ( x  y ))
 4 I .P
 4 cos ( x  y )
Hence the general solution is
Z= 1 ( y)  x 2 ( y)  3 ( y  2 x) 
1 x  2y
 4 cos ( x  y)
e
3
UNIT II
FOURIER SERIES
PART – A
1. Explain periodic function with examples.
A function f  x  is said to have a period T if for all x , f  x  T   f  x , where T is a
positive constant. The least value of T  0 is called the period of f  x  .
Example : f  x   sin x ; f  x  2   sin  x  2   sin x .
Here f  x   f  x  2  . sin x is a periodic function with period 2 .
2. State Dirichlet’s conditions for a function to be expanded as a Fourier series.
Let a function f ( x) be defined in the interval c  x  c  2 with period 2 and
satisfies the following conditions can be expanded as a Fourier series in (c, c  2 ) .
(i)
f ( x) is a well defined function.
(ii)
f ( x) is finite or bounded.
(iii)
f ( x) has only a finite number of discontinuous point.
(iv)
f ( x) has only a finite number of maxima and minima.
3. State whether y  tan x can be expressed as a Fourier series. If so how?. If not why?
tan x cannot be expanded as a Fourier series. Since tan x not satisfies Dirichlet’s
condition.
4. State the convergence condition on Fourier series.
(i) The Fourier series of f ( x) converges to f ( x) at all points where f ( x) is
continuous.
(ii) At a point of discontinuity x0 , the series converges to the average of the left limit and
right limit
of f ( x) at x0
f  x0  
1
lim f  x0  h   lim f  x0  h   .
h 0

2  h 0
5. To what value does the sum of Fourier series of f ( x) converge at the point of continuity
x  a?
The sum of Fourier series of f ( x) converges to the value f (a) at the continuous point
x  a.
6. To what value does the sum of Fourier series of f ( x) converge at the point of
discontinuity x  a ?
At the discontinuous point x  a , the sum of Fourier series of f ( x) converges to
 f  x0  h   f  x0  h  
f  x0   lim 
.
h 0
2


7. If f ( x)  x 2  x is expressed as a Fourier series in  2, 2  , to which value this series
converges at
x  2 ?.
f ( x)  x2  x,  2  x  2
The value to which the Fourier series of f ( x) converges at x  2 which is an end points
is given by

f  2   f  2 
2

 4  2   4  2
2
 2.
 The Fourier series converges at x  2 to the value 4.
cos x
8. If f  x   
50
if 0  x  
and f  x   f  x  2  for all x, find the sum of the
if   x  2
Fourier series
of f  x  at x   .
Sum of the Fourier series of the function f  x  at x   .
 f   
f     f   
2

cos   50
2

1  50
49

.
2
2
9. If f  x   sinh x is defined in   x   , write the value of a0 , an .
Given f  x   sinh x
f   x   sinh   x    sinh x
  f  x .
 sinh x is an odd function.
 a0  0, an  0 .
10. Write the formulae for Fourier constants for f ( x) in the interval ( ,  ) .
The Fourier constants for f ( x) in the interval ( ,  ) are given by
a0 
1



an 
f ( x).dx

bn 
1

1



f ( x)cos nx.dx



f ( x)sin nx.dx .

11. Find the constant a0 of the Fourier series for function
a0 
i.e.,
1

2

f ( x)dx 
0
1

2

0
f x   x
2
xdx
in
0  x  2 .

1  x2 
1  4 2
    
 0  2
  2 0   2

a0  2 .
12. If f  x  = x expanded as a Fourier series in (  π , π ) , find a0.
The given function f ( x)  x is an even function.
a0 

1



1
f ( x)dx 




2  x2 
x dx   xdx      .
0
  2 0
2


13. Find the Fourier coefficients a0 of f  x   e x in    x   .
a0 
1


 e dx
x


1

 e x 



e

1

 e   
2sinh 
14. Find bn in the expansion of x2 as a Fourier series in

.
  ,   .
Since f ( x)  x 2 is an even function, the value of bn  0 .
15. Find the constant term a0 in the Fourier series corresponding to f  x  = x  x 3 in
( π, π ) .
3
Given f x   x  x


f  x    x  x 3   x  x 3   f x 
i.e,
f  x    f x 

f ( x) is an odd function in   ,  
Hence a 0  0 .
16. If f  x   x 2  x 4 is expanded as a Fourier series in  l , l  , find the value of bn .
Given f  x   x2  x4 ,  l  x  l
f  x    x    x   x 2  x 4  f x 
2

4
f  x  is an even function in  l , l  .
Hence bn  0 .
 2x
1 + π , -π < x < 0
17. In the Fourier expansion of f ( x ) = 
in  π,π  , find the value of
1  2 x , 0 < x < π

π
bn the
coefficient of sin nx .
 2x
1  π , 0 < x < π
f ( x ) = 
1  2 x ,  π < x < 0

π
 f  x  is an even function of x in   ,  
The coefficient of sin nx , bn  0 . Since the Fourier series of f ( x) consists of cosine
terms only.
18. Find the constant a0 of the Fourier series for the function
π < x < π .
f ( x)  x cos x
f  x  = x cos x in
f ( x)   x cos x   f ( x)
 f ( x) is an odd function. Hence a0  0 .
19. Write the Fourier sine series of k in
0,   .

The Fourier sine series is given by f ( x)   bn sin nx
n 1
where
bn
1
=





f ( x)sin nx.dx  2  k sin nxdx

0

 4k
 , if n is even
i.e., bn   n
0, if n is odd
2k   cos nx
2k
1  (1)n 




  n  0 n

f ( x) 
4k
4k 
1
sin
nx
sin[(2n  1) x] .
=
 n

 n1 (2n  1)
n is odd
20. Obtain the sine series for unity in (0, π ) .
Here f x   1; f ( x) 

 bn sin nx
n 1
where
bn =
1





f ( x)sin nx.dx  2 1.sin nx. dx


0
2   cos nx
2
1  (1)n 
 


  n  0 n
 4
 , if n is even
i.e., bn   n
0, if n is odd

f ( x) 
4 
1
4
=
sin[(2n  1) x] .
sin
nx

 n
 n1 (2n  1)
n is odd
21. Find the value of an , in the cosine series expansion of f  x   k in the interval  0,10  .
n x 

sin

k
10 
 

5  n 
 10  0
10
2
n x
k cos
dx

10 0
10
10
an 

k  10
 sin n  0   0 .

5  n

22. If f  x  is defined in 3  x  3 what is the value of Fourier coefficients.
3
a0 
1
f  x  dx
3 3
1
n x
f  x  cos
dx

3 3
3
3
;
an 
1
n x
f  x  sin
dx .

3 3
3
3
;
bn 
23. Define Root Mean Square value of a function.
The root mean square value of y  f ( x) in (a, b) is denoted by y . It is defined as
b
 y dx
2
R.M.S., y 
a
(b  a)
.
24. Find the R.M.S value of y  x2 in   ,   .
1
y 
2
2

 x


2 2

1  x5 
1


5 5 



2  5  
10
dx
y2 
4
y 
5
2
5
.
25. Find the R.M.S value if f ( x ) = x 2 in π  x  π .

b
 y dx
2
Since R.M.S
y

a
(b  a )

 x dx



[  ( )]

4


2
 x 2  dx
2 x dx

2
0
2


4
 x dx
4

0


 x5 
 
 5 0

5 2
.

5
5
26. State the Parseval’s Identity (or) theorem on Fourier series.
If f ( x) is a periodic function of period 2 in (c, c  2 ) with Fourier coefficients
a0 , an and bn , then
1

c  2

c
a02  2
 f  x   dx 
  (an  bn2 ) .
2 n1
2
27. Write the complex form of Fourier series for f(x) defined in the interval (c, c+2l).
The series for f ( x) defined in the interval (c, c  2 ) and satisfying Dirichlet’s
conditions can be given in the form of
f ( x) 

 cne
n 
inx
1
, where cn 
2
c  2

f ( x)einx dx .
c
28. What do you mean by Harmonic analysis?
The process of finding the Fourier series of the periodic function y  f  x  of period
2l  or  2 using the numerical values of x and y is known as Harmonic analysis.
PART B
1)
Express f(x)=
as a Fourier series with period
Hence deduce the value of the series
Solution:
We know that the Fourier series be
to be valid in the interval 0 to
.
Sub in (1) we get
Put
2)
is a point continuity
Obtain Fourier series for f(x) of period
Hence deduce that
Solution:
Given
We know that the Fourier series is
Where
and defined as follows
and
Substituting the values in equation (1) we get
This is the required Fourier series
i)
Put
is a point of continuity
ii)
Put
is a point of continuity
Hence proved
ODD AND EVEN FUNCTION
3. Find the Fourier series of
Solution:
Given
Therefore f(x) is neither even nor odd function
We know that the Fourier series is
Where
Substituting the values in equation (1) we get
This is the required Fourier series.
FOURIER SINE SERIES
3)
Expand
in a Fourier sine series in the interval
Solution:
Given
We know that the half range fourier sine series is
Where
Substituting
4)
value in equation (1) we get
This is the required half range Fourier sine series.
HALF RANGE COSINE SERIES
Obtain the half range cosine series for
in the interval (0,2).
olution:
Given
We know that the Fourier half range cosine series is
Where
Here
Substituting these values in equation (1) we get
This is the required Fourier series
COMPLEX FORM OF FOURIER SERIES
6) Find the complex form of the Fourier series of
=
Solution:
Given
=
in -1<x≤1
We know that the Fourier series is
where
Here = 1 ,
in -1 < x ≤ 1
HARMONIC ANALYSIS
7)Computeupto first harmonics of the Fourier series of f(x) given by the following table
X
0
T/6
T/3
T/2
2T/3
5T/6
T
F(x)
1.98
1.3
1.05
1.3
-0.88
-0.25
1.98
Solution:
First and last value are same. Hence we omit the last value.
When x varies from 0 to T
varies from 0 to
We know that the Fourier series is
…..(1)
x
y
0
0
1.98
1.0
0
1.98
0
T/6
1.30
0.5
0.866
0.65
1.1258
T/3
1.05
-0.5
0.866
-0.525
0.9093
T/2
1.30
-1
0
-1.3
0
2T/3
-0.88
-0.5
-0.866
0.44
0.762
5T/6
-0.25
0.5
-0.866
-0.125
0.2165
Sum
4.5
1.12
3.013
Substituting the above value in equation (1) we get
This is the required Fourier series
UNIT III
APPLICATIONS OF PARTIAL DIFFERENTIAL EQUATIONS
PART – A
1. What conditions are assumed in deriving the one dimensional wave equation?
The wave equation is
2
2 y
2  y

a
.
t 2
x 2
In deriving this equation we make the following assumptions.
(i) The motion takes place entirely in one plane i.e., XY plane.
(ii) We consider only transverse vibrations the horizontal displacement of the
particles of the string is negligible.
(iii)The tension T is constant at all times and at all points of the deflected string.
(iv) T is considered to be so large compared with the weight of the string and hence
the force of gravity is negligible.
(v) The effect of friction is negligible.
(vi) The string is perfectly flexible.
2. State the wave equation and give the various solutions of it?
2
2 y
2  y
The wave equation is
.
a
t 2
x 2
The various possible solutions of this equation are
(i) y( x, t )  ( A1e px  A2 e  px )( A3 e pat  A4 e  pat ) .
(ii) y( x, t )  ( A5 cos px  A6 sin px)( A7 cos pat  A8 sin pat ) .
(iii) y( x, t )  ( A9 x  A10 )( A11t  A12 ) .
3. Find the nature of PDE 4u xx  4uxy  u yy  2ux  u y  0 .
This is of the form Au xx  Buxy  cu yy  f  x, y, u, ux, uy   0 .
Here A  4, B  4, C  1 .
B2-4AC=16-4(4)(1)=0.
Therefore the equation is Parabolic.
1. Classify the equation uxx-y4uyy=2y3uy.
Solution:
This is of the form Auxx+Buxy+Cuyy+f(x,y,u,ux,uy)=0.
Here A=1, B=0, C=-1.
B2-4AC=0-4(1)(-1)=4>0.
Therefore the equation is Hyperbolic.
2. Classify: x2uxx+2xyuxy+(1+y2) uyy-2ux=0.
Solution:
This is of the form Auxx+Buxy+Cuyy+f(x,y,u,ux,uy)=0.
Here A=x2, B=2xy, C=1+y2.
B2-4AC=4x2y2-4(x2)(1+y2)
= 4x2y2-4 x2-4(x2 y2)
=-4x2<0.
Therefore the equation is Elliptic.
3. A string is stretched and fastened to two point l apart. Motion is started by
displacing the string into the form y  y 0 sin
x
l
from which it is released at time
t=0. Formulate this problem as the boundary value problem.
Solution:
The displacement y(x,t) is the solution of the wave equation.
2
2 y
2  y

a
t 2
x 2
The boundary conditions are:
i) y(0, t )  0 for all t  0 .
ii) y(l , t )  0 for all t  0 .
iii)
y
x,0  0 .
t
iv) y ( x,0)  f ( x)  y 0 sin
x
l
.
4. What is the constant a2 in the wave equation
2
2 y
2  y

a
t 2
x 2
(or)
2
2 y
2  y
In the wave equation 2  c
what does c2 stand for?
2
t
x
Solution:
a 2 or c 2 
T
Tension

M Mass per unit length of the string
5. State the suitable solution of one dimensional heat equation
u
 2u
 a2 2 .
t
x
Solution:
u( x, t )  ( A cos px  B sin px)e c
2
p 2t
.
6. State the governing equation for one dimensional heat equation and necessary
conditions to solve the problem.
Solution:
The one dimensional heat equation is
u
 2u
 a 2 2 where u(x,t) is the
t
x
temperature at time t at a point distance x from the left end of the rod.
The boundary conditions are
i) u(0, t )  k10 C for all t  0
ii) u(l , t )  k 20 C for all t  0
iii) the initial condition is u( x,0)  f ( x), 0  x  l .
7. Write all variable separable solutions of the one dimensional heat equation
u
 2u
 a2 2 .
t
x
Solution:
i) u( x, t )  ( A1e x  B2 e x )C1e
t
2 2
ii) u( x, t )  ( A2 cos x  B2 sin x) C 2 e 
t
2 2
iii) u( x, t )  ( A3 x  B3 )C3 .
8. Write down the diffusion problem in one dimension as a boundary value problem in
two different forms.
Solution:
u
 2u
 a 2 2 is the one dimensional heat flow.
t
x
Here a 2 
k
is called the diffusivity.
pc
In the steady state
d 2u
 0.
dx 2
9. State any two laws which are assumed to derive one dimensional heat equation.
Solution:
i)
Heat flows from higher to lower temperature
ii)
The rate at which the heat flows across any area is proportional to the area and
to the temperature gradient normal to the curve. This constant is
proportionality is known as the thermal conductivity (k) of the material. It is
known as Fourier law of heat conduction.
10. Write any two solutions of the Laplace equation Uxx+Uyy=0 involving exponential
terms in x or y.
Solution:
i) u( x, y)  ( A1e px  A2 e  px )( A3 cos py  A4 sin py ) .
ii) u( x, y)  ( A1 cos px  A2 sin px)( A3 e py  A4 e  py ) .
11. In steady state conditions derive the solution of one dimensional heat flow equation.
Solution:
The PDE of unsteady one dimensional heat flow is
2
u
2  u
a
t
x 2
….. (1)
In steady state condition, the temperature u depends only
on x and not on t.
Hence
u
0
t
Therefore equation (1) reduces to
 2u
 0.
x 2
The general solution is u=ax+b, where a, b are arbitary.
12. Write the boundary condition and initial conditions for solving the vibration of
string equation, if the string is subjected to initial displacement f(x) and initial
velocity g(x).
Solution:
The wave equation is
2
2 y
2  y
.

a
t 2
x 2
The initial and boundary conditions are
i) y(0, t )  0 .
ii) y(l , t )  0 .
iii)
y
x,0  g ( x) .
t
iv) y( x,0)  f ( x)
13. Write down the governing equation of two dimensional steady state heat equation.
Solution:
 2u  2u
The required equation is

 0.
x 2 y 2
14. The ends A and B of a rod of length 10cm long have their temperature distribution
kept at 20oC and 70oC. Find the steady state temperature distribution of the rod.
Solution:
The steady state equation of one dimensional heat flow is
d 2u
0
dx 2
….. (1)
The general solution of equation (1) is u(x)=ax+b ….. (2)
The boundary conditions are u(0)=20, u(l)=70.
Put x=0 in (2) we get u(0)=a(0)+b
 b=20
Put x=l in (2) we get u(l)=al+b
70= al+20
al=50
a= 50/l
Therefore equation (2)  u(x)= 50x/l+20
Here l=10 cm
Therefore u(x)= 50x/10+20
u(x)=5x+20.
15. Write down the different solutions of Laplace equation in polar coordinates.
 2r
r  2 u
r
r
 2  0.
 2


2
Solution:
i) u(r,  )  (C1 r p  C2 r  p )(C3 cos p  C4 sin p )
ii) u(r, )  (C5 cos( p log r )  C6 sin( p log r )(C7 e p  C8 e  p )
iii) u(r, )  (C9 log r  C10 )(C11  C12 ) .
16. What is the general solution of a string of length l whose end points are fixed and
which starts from rest?
Solution:

y ( x, t )   Bn sin
n 1
nx
nat
cos
.
l
l
17. How many boundary conditions and initial conditions are required to solve the one
dimensional wave equation?
Solution:
Two boundary conditions and two initial conditions are required.
PART B
1.A string is stretched and fastened to two points x = 0 and x= l apart. Motion is started by
displacing the string into the form y = k (l x – x2 ) from which it is released at time t=0. Find
the displacement of any point on the sting at a distance of x from one end at time t.
Solution: The ODWE ytt  c 2 ycc
Solution :
y(x,t)= (Acos px +Bsin px)(Ccos pct +Dsin pct)
Boundary and initial conditions are (i) y(0,t) = 0 (ii) y(l,t) = 0
(iii) y t (x,0)=0 (iv) y(x,0)=f(x), 0< x < l .
Using Boundary and initial conditions:
i)
y(0,t) = 0, put x=0
A(Ccos pct +Dsin pct)=0  A=0
 Suitable solution
ii)
y(x,t)= Bsin px (Ccos pct +Dsin pct)
y(l,t) = 0 , put x=l
Bsin pl (Ccos pct +Dsin pct)=0  B  0 Bsin pl =0
 pl = n 
 Suitable solution
iii)
 p=
y(x,t)= Bsin
n
l
nx
nct
nct
(Ccos
+Dsin
)
l
l
l
y t (x,0)=0
Bsin
nx l
nct
nct
(C(-sin
) +Dcos
)
l
l
l nc
Put t=0  Bsin
nx l
(Dcos0)=0
l nc
 Suitable solution
y(x,t)= Bsin

 D=0
nx
nct
Ccos
l
l
General solution: y(x,t)=  Bn sin
n 1
iv)
nx
nct
cos
l
l
y(x,0)=f(x), 0< x < l .
Here t=0 

B
n 1
n
sin
nx
= f(x)= k lx  x 2
l


By Half range sine series:
2
nx
f x sin
dx

l 0
l
l
Bn =
2
nx
k lx  x 2 sin
dx

l 0
l
l
=


l
2
3
2k 
l
nx
nx  nx 
nx 
 l 
2


(
lx

x
)
cos

(
l

2
x
)
sin

2



 cos


l 
n
l
l
l 
 n 
 l 
0
3
3
2k 
 l  
n l 

  2
 
 2(1) 
l 
 n 
 n  
3

2k   l 
n

 [(1)  1]
2
l   n 

=

4kl 2
n
1   1
3 3
n

 8kl 2

=  n 3 3 , n  odd
 0, n  even
Required Solution:
8kl 2
nx
nct
sin
cos

3 3
l
l
n 1, 3, 5 n 

y(x,t)=

8kl 2

3

1
 (2n  1)
n 0
3
sin
(2n  1)x
(2n  1)ct
cos
l
l
2.A taut string of length 2l is fastened at both ends . The midpoint of the string is taken to a
height b and then released from rest in that position. Find the displacement of the string at
any time.
Solution: let L=2l
0 x
Equation of AC:
L
2
By two point formula: at (0,0) and (
x
y
2bx
 y

L/2 b
L
Equation of CB:
L
<x<L
2
L
,b)
2
By two point formula, at (
y
L
,b)and (L,0)
2
2b( L  x)
L
L
 2bx
,
0

x


L
2
 y(x,0)= 
2b( L  x) L

, xL
L
2

The ODWE ytt  c 2 ycc
Suitable solution
y(x,t)= (Acos px +Bsin px)(Ccos pct +Dsin pct)
Boundary and initial conditions are (i) y(0,t) = 0
(iii) y t (x,0)=0
(ii) y(l,t) = 0
(iv) y(x,0)=f(x), 0< x < l .
Using Boundary and initial conditions:
i) y(0,t) = 0, put x=0
A(Ccos pct +Dsin pct)=0  A=0
 Suitable solution
y(x,t)= Bsin px (Ccos pct +Dsin pct)
ii) y(l,t) = 0 , put x=l
Bsin pl (Ccos pct +Dsin pct)=0  B  0 Bsin pl =0
 pl = n 
 p=
 Suitable solution
v)
n
l
y(x,t)= Bsin
nx
nct
nct
(Ccos
+Dsin
)
l
l
l
y t (x,0)=0
Bsin
nx l
nct
nct
(C(-sin
) +Dcos
)
l
l
l nc
Put t=0  Bsin
nx l
(Dcos0)=0  D=0
l nc
 Suitable solution
y(x,t)= Bsin

General solution: y(x,t)=
B
n 1
vi)
n
sin
nx
nct
Ccos
l
l
nx
nct
cos
l
l
y(x,0)=f(x), 0< x < l .
L
 2bx
,0  x 

nx
L
2
Here t=0   Bn sin
= f(x)= 
2
b
(
L

x
)
L
l
n 1

, xL
L
2


Half range sine series:
nx
Bn = 2
f x sin
dx

l 0
l
l
 L2

L
2  2bx
nx
2b( L  x)
nx 
sin
dx  
sin
dx 
= 
L 0 L
L
L
L
L


2
L
2


nx  
  cos nx 


sin
2 2b  
L   (1) 
L 
=
( x) 
2
n 

L L  
 n   

 
 
 
L 

  L    0
L


nx 
nx  




cos
sin


2 2b 
L
L



( L  x) 
 (1) 

2 

n 
  n  
L L


 
 
L 


  L    L
2
2
2
4b  L2
n  L 
n
L2
n  L 
n 
 2 
cos


cos

 sin
 sin

L  2n
2  n 
2 2n
2  n 
2 
=
4b  L2
n 
2 2 2 sin

2 
L  n
2 
=
8b
n
sin
2 2
n
2
Required Solution:Put L=2l

8b
n 
y(x,t)=
2
n 1

=
2
sin
8b
n 
n 1
2
2
n
nx
nct
sin
cos
2
L
L
sin
n
nx
nct
sin
cos
2
2l
2l
3.If a string of length ' l ' is initially at rest in its equilibrium position and each of its points
 y 
3x
, 0  x  l . Determine the displacement y(x, t).
  v0 Sin
l
 t t 0
is given the velocity 
Solution :
Let l=20
The ODWE ytt  c 2 ycc
Suitable solution
y(x,t)= (Acos px +Bsin px)(Ccos pct +Dsin pct)
Boundary and initial conditions are (i) y(0,t) = 0
(ii) y(l,t) = 0
(iii) y (x,0)=0 (iv) y t (x,0)=f(x)  v0 sin 3
x
l
Using Boundary and initial conditions:
i) y(0,t) = 0, put x=0
A(Ccos pct +Dsin pct)=0  A=0
 Suitable solution
ii) y(l,t) = 0 , put x=l
y(x,t)= Bsin px (Ccos pct +Dsin pct)
Bsin pl (Ccos pct +Dsin pct)=0  B  0 Bsin pl =0
 pl = n 
 p=
n
l
 Suitable solution
y(x,t)= Bsin
nx
nct
nct
(Ccos
+Dsin
)
l
l
l
iii) y (x,0)=0
Bsin
nx
.C=0
l
 C=0
 Suitable solution
y(x,t)= Bsin

General solution: y(x,t)=
B
n 1
n
sin
nx
nct
Dsin
l
l
nx
nct
sin
l
l
iv) y t (x,0)=f(x), 0< x < l .

y t (x,t)=
B
y t (x,0)=
B
c
x
n 1
n
sin
nx nc
nct
cos
l
l
l
sin
x
nx nc
 v0 sin 3
l
l
l

l
B1 sin
n 1
l

v
2c
2x 3c
3x
B2 sin

B3 sin
 ...  0
l
l
l
l
4
v
3c
B3 = - 0
4
l
3v
c
B1 = 0
4
l
B1 
n
l 3v0
c 4
B3  
x
3x 

3 sin l  sin l 
B2  B4  B5  ...  0
lv 0
12c
Required Solution:
y(x,t)= B1 sin

x
l
sin
ct
l
 B3 sin
3x
3ct
sin
l
l
3v0
lv
x
ct
3x
3ct
 0 sin
sin sin
sin
4c
12c
l
l
l
l
4.A rod 30 cm long has its ends A and B kept at 200C and 800C respectively until steady
state conditions prevail the temperature at each end is then suddenly reduced to 0 0 c
and kept so. Find the resulting temperature function u(x,t) taking x=0 at A.
Solution:
Let l= 30
In steady state u xx  0
ba
In initial temperature u(x) = 
x + a
 l 
u(x,0)=
60 x
 20
l
After change ODHE: ut   u xx
2
Suitable Solutions:
u(x,t)= (Acospx + Bsin px) e 
2
p 2t
Boundary and Initial Conditions:
i) u(0,t)=0
ii) u(l,t)=0
Using Boundary and initial conditions:
i) u(0,t)=0
Here x=0  u(0,t)= Ae 
2
p 2t
 A=0
Suitable solution: u(x,t)= Bsin px
ii) u(l,t)=0
e 
2 2
p t
iii) u(x,0)=
60 x
 20
l
Here x=l  B sin ple 
Sinpl==0  pl=nπ
2
 p=

iii) u(x,0)=f(x)=

B
n 1
n
sin
n
l
nx
e
l
Suitable solution: u(x,t)= Bsin
General Solution: u(x,t)=
0
p 2t
B
n 1
n
sin
 2 n 2 2t
2
l2
nx
e
l
 2 n 2 2t
2
l2
60 x
 20
l
nx 0 60 x
e =
 20
l
l
By Half range sine series,
2
nx
f x sin
dx

l 0
l
l
Bn =
2 60 x
nx
(
 20) sin
dx

l 0 l
l
l

l


nx 
nx  


 cos l  60  sin l  
2 60 x
 (
 20) 
  ( ) 
2
n

l l
l
 n   



 
 
l



  l    0

2
n
 (80) cos
l 
l



 l

 n

 60  n
  ( ) sin
l 
l


2 l
40
 80(1) n  20 
1  4(1) n
l n
n

2
 l  

 
 n   
Required Solution:
40
: u(x,t)=

40
=


 (1  4(1)
n
nx
e
l
) sin
n 1

 (1  4(1)
n
nx
30
) sin
n 1
e
 2 n 2 2t
2
l2
 2 n 2 2t
900
2
5. An infinitely long rectangular plate with insulated surface 10 cm wide. The two long
edges and one short edge are kept at 00 temperature, while the other short edge x=0 is kept
at temperature given by u=20y, 0  y  5, u=20(10-y),5  y  10. Find the steady state
temperature in the plate.
Solution : Steady state two dimensional heat equation:
u xx  u yy  0
Infinite plate extended in x-direction :
Let l=10
I Boundary Conditions
i) u(x,0) = 0
ii) u(x,l) = 0
l

 20 y,0  y  2
iii) u(∞,y) = 0 iv) u(0,y) = f(y)= 
l
20(l  y ),  y  l
2

II Suitable Solution:


u(x,y)= Ae  px  Be px (Ccospy + Dsinpy)
Using boundary conditions:


 px
px
i) u(x,0) = Ae  Be C  0
C=0
Suitable Solution:


u(x,y)= Ae  px  Be px Dsinpy
ii) u(x,l) = 0
Dsinpl=0
Sinpl==0  pl=nπ
 p=
n
l
Suitable Solution:
nx
  nx

ny
u(x,y)=  Ae l  Be l  D sin
l


iii) u(∞,y) = 0
B e  =0  B=0

General Solution: u(x,y)=  Bn e

nx
l
sin
n 1
l

 20 y,0  y  2
(iv) u(0,y) = f(y)= 
l
20(l  y ),  y  l
2

l

20
y
,
0

y


nx
2
Bn sin
= f(x)= 

l
l
n 1
20(l  y ),  y  l
2


Half range sine series:
nx
Bn 2
=  f x sin
dx
l 0
l
l
 2l

l
40 
nx
nx 
dx   (l  x) sin
dx 
=   x sin
l 0
l
l
l


2
ny
l
l
2


nx  
  cos nx 


sin

40  
l
l


=
( x) 
  (1) 
2 

n

l
 n  

 


 
l

 
  l    0
l


nx 
nx  




cos
sin

40 
l   (1) 
l 

(l  x) 
n 
  n  2  
l 



 
 
l



  l    l
2
2
2
40  l 2
n  l 
n
l2
n  l 
n 
 
cos


cos

 sin
 sin

l  2n
2  n 
2 2n
2  n 
2 
=
40  l 2
n 
2 2 2 sin

l  n
2 
=
80l
n
sin
2 2
2
n
Required Solution:Putl=10
nx
800
n  10
ny
e
u(x,y)=  2 2 sin
sin
2
10
n 1 n 

UNIT IV
FOURIER TRANSFORMS
PART – A
1. State Fourier integral theorem.
If f(x) is piece-wise continuously differentiable and absolutely integrable in (-  ,  )
then
f  x 
f  x 
1

1
2
 

  f t  e
is x t 
dtds
(or) equivalently
 
 
  f t  cos  t  x  dt d  .
0 
This is known as Fourier integral theorem or Fourier integral formula.
2. Define Fourier transform pair (or) Define Fourier transform and its inverse transform.
The complex (or infinite) Fourier transform of f(x) is given by
F  f  x    F  s  
1
2

 f  x e
isx
dx

Then the function f(x) is the inverse Fourier Transform of F(s) and is given by
f  x 
1
2

 F s e
 isx
dx .

F  f  x  and F 1  F  s  its also called Fourier Transform Pairs.
3. Show that f(x) = 1, 0 < x <  cannot be represented by a Fourier integral.


0
0
 f  x  dx 1dx   x

0
  and this value tends to  as x   .

i.e.,  1 f  x  dx is not convergent. Hence f  x   1 cannot be represented by a Fourier
0
integral.
4. State and prove the linear property of FT.
Stt:
F a f  x   b g  x   a F  s   bG  s 
Proof:
F  f  x    F  s  
1
2
F  a f  x   b g  x   
1
2

1
2

 a F  s   bG  s  .
a
2

 f  x e
isx
dx


 a f  x   b g  x  e
isx
dx



a f  x  eisx dx 




f  x  eisx dx 
1
2
b
2

 b g  x e
isx


 g  x e

isx
dx
dx
5. State and prove the Shifting property of FT.
Stt:
F  f  x  a    eias F  s  .
Proof:
1
2
F  f  x    F  s  
F  f  x  a   
1
2

 f  x e

 f  x  ae
isx
when x   , y  
when x  , y  
1
2

 f  ye
is  y  a 
dy
 f  ye
1
2



isy
dx

dx  dy
eisa

2
dx

Put x  a  y  x  y  a

isx
eisa

2
dy

6. State and prove the Change of scale property of FT.
Stt:
1 s
F  f  ax    F   , a  0 .
a a
Proof:
F  f  x    F  s  
1
2

 f  x e

isx
dx

 f  ye
isy isa
e dy


 f  x e

isx
dx
 eisa F  s  .
1
2
F  f  ax   
Put ax  y  x 

 f  ax  e
1 1
 
a  2

 f  ye
1
2
s
i  y
a




dx

y
a
when x   , y  
a dx  dy i.e., dx 

isx
f  ye
dy
a
 y
is  
a

when x  , y  
dy
a

1
a 2


s
i  y
f  y  e  a  dy


dy 

1 s
F
.
a  a 
7. If F  f  x   F  s  , prove that F  f  x  eiax   F  s  a  .
Proof:
1
2
F  f  x    F  s  
F eiax f  x   
1
2

1
2

e
iax

 f  x e
isx
dx

f  x  eisx dx



 f  x e
i sax
dx
 F s  a .

8. State and prove the Modulation property of FT. (OR) If Fourier transform of f(x) is
F(s).
1
F (s  a)  F (s  a) .
2
Prove that the Fourier transform of f  x  cos ax is
Stt:
F  f  x  cos ax  
1
 f  s  a   f  s  a  where f  s   F  f  x  .
2
Proof:

1  1

2  2
F  f  x  cos ax  
1
2

1
2

 f  x e e
isx

iax

 f  x  cos ax e
isx
dx




 eiax  eiax
f  x  eisx 
2


 dx


 eiax  dx 



1 1
2 2


f  x e 
i s a x
dx 

1
1
f s  a  f s  a
2
2
1 1
2 2



 f  x e
i s a  x
dx

1
 f  s  a   f  s  a  .
2
9. What is meant by self-reciprocal with respect to FT?
If the Fourier transform of f  x  is obtained just by replacing x by s, then f  x  is called
self-reciprocal with respect to FT.
Example:
f  x  e

x2
2
F  f  x   F  s   e

s2
2
.
1
10. Prove that Fc  f  x  cos ax    Fc (s  a)  Fc ( s  a) where Fc denotes the Fourier
2
cosine
transform f  x  .
2
The F.C.T is, Fc  f  x   

Fc  f  x  cos ax  




 f ( x) cos sx dx
0

2
 f ( x) cos ax cos sxdx

0
2


 f ( x) cos sx cos axdx
0
f ( x) cos(s  a) x  cos(s  a) xdx

2
2
1
0


 2 
1  2 




 
f
(
x
)
cos(
s

a
)
xdx

f
(
x
)
cos(
s

a
)
xdx

  
2   0

0

1
 [ Fc ( s  a)  Fc ( s  a)] .
2
11. Prove that Fc  x f  x   
2
W.k.t Fs  f  x  

d Fs  f  x  
.
ds

 f ( x)sin sx dx
0
d Fs  f  x   d  2


ds
ds  


0

 f ( x)sin sx dx 


2 
d
  f ( x)  sin sx  dx 
 0
ds


2



  x f ( x) cos sx dx



2 
  f ( x) cos sx . x dx 
 0

 Fc  x f  x  .
0
12. Define Fourier cosine transform (FCT) pair.
The infinite Fourier cosine transform of f(x) is defined by

2
Fc  f  x   
f  x  cos sx dx
 0
The inverse Fourier cosine transform Fc  f  x   is defined by
f  x 
2


 F  f  x  cos sx dx .
c
0
Fc  f  x   and Fc 1  Fc  f  x   are called Fourier Cosine Transform Pairs.
cos x if 0  x  a
13. Find the Fourier Cosine transform of f(x) = 
.
if x  a
0
We know that

2
Fc  f  x   
f  x  cos sx dx
 0

2

a
 cos x cos sx dx
0
a

2 1
cos 1  s  x  cos 1  s  x  dx
 2 0 
1  sin(1  s) x sin(1  s) x 


1  s  0
2  1  s
a

1  sin(1  s)a sin(1  s)a 

 1  s  1  s    0  0  
2 



1  sin(1  s)a sin(1  s)a  



1  s  
2  1  s
provided s  1 ; s  1 .
14. Find the Fourier Cosine transform of e ax , a > 0.
Given f  x   e ax

2
We know that F.C.T is, Fc  f  x   
f  x  cos sx dx
 0

2


e
 ax

cos sx dx
0
Here a  a, b  s
Fc e ax  
2  a 
, a  0.
  a 2  s 2 
15. Find the Fourier Cosine transform of e  x .
We know that
But  e ax cos bx dx 
0
a
a  b2
2

2
Fc  f  x   
f  x  cos ax dx
 0
2

Fc  e  
e x cos ax dx =

0
x
2 1 
.
 1  a 2 
16. Define Fourier sine transform (FST) pair.
The infinite Fourier sine transform of f(x) is defined by

2
Fs  f  x   
f  x  sin sx dx .
 0
The inverse Fourier sine transform of Fs  f  x  is defined by
f  x 
2


 F  f  x  sin sx dx .
s
0
Fs  f  x  and Fs 1  Fs  f  x   are called Fourier Sine Transform Pairs.
17. Find the Fourier Sine transform of e 3 x .

2
The FST is, Fs  f  x   
f  x  sin sx dx . Here f  x   e3 x .
 0
Fs e
3 x
2

 
e3 x sin sx dx

0
2
s

2
2 

 s  3 


2  ax
Formula Fs e ax  
e sin sx dx .
 0
18. Find the Fourier Sine transform of f(x)= e x .
We know that Fs [ f ( x)] 
2
Fs [e x ] 


2



 f ( x)sin sx dx
e
0
x
sin sx dx
0
   ax

b
] .
   e sin bx dx  2
2 
a b 
 0
2 s 
 1  s 2 
19. Find the Fourier Sine transform of 3e 2 x .
Let f  x   3e2 x

2
W.k.t Fs  f  x   
f  x  sin sx dx
 0
3
2

e

2 x

2


 3e
2 x
sin sx dx
0
sin sx dx
0


2  e 2 x
 2 sin sx  s cos sx
3

2
 4  s
0
3
2 s 
  s 2  4 
20. Find the Fourier Sine transform of
We know that

1
.
x
2  3s 
.
  s 2  4 
3
2
0   1 2  s  

 
4  s


2
Fs  f  x   
f  x  sin sx dx
 0

1
Fs   =
x
2 1
sin sxdx
 0 x
x 0   0
Let sx = 
x 0   0
sdx = d 

=
2 s
d
  sin 

 0  
ds

=
2 sin 

0

d
2  
=
  2 
=

2
.
21. State the Convolution theorem on Fourier transform.
If F  s  and G  s  are the Fourier transform of f  x  and g  x  respectively. Then the
Fourier transforms of the convolution of f  x  and g  x  is the product of their Fourier
transforms.
F  f  x  * g  x   F  s  G  s 
 F  f  x  F  g  x  .
22.State the Parseval’s formula or identity on Fourier Transform.
If F  s  is the Fourier transform of f  x  , then




f ( x) dx 
2


PART B
1. State and prove the convolution theorem for Fourier Transforms.
2
F ( s) ds .
Statement:
If F  s  and G  s  are the Fourier transform of f  x  and g  x  respectively. Then the
Fourier transforms of the convolution of f  x  and g  x  is the product of their Fourier
transforms.
F  f  x  * g  x   F  s  G  s 
 F  f  x  F  g  x  .
Where  f  g x 
1
2

 f (t ) g ( x  t )dt

PROOF: By convolution of two functions:
 f  g x 

1
2
 f (t ) g ( x  t )dt

The Fourier transform of f  g is
F[ f  g ] 
1

2

1
2



1
2
 ( f  g )e
 1

 2

 isx
{
f
(
t
)
g
(
x

t
)
dt
e dx






f (t )dt  g ( x  t )e isx dx

F[ f  g ] 
1
2
dx

x= 
Put u=x-t du=dx

isx
1
2




u=  and x=-  u=- 

f (t )dt  g (u )e is (u t ) du



f (t )e ist dt

1
2

 g (u)e

isu
du
F[( f  g )(x)]=F(s)G(s)
2
2

a  x , in x  a
2.Find the Fourier transform of f ( x)  
in x  a

0,

Hence evaluate
2
2

a  x , in x  a
Given: f ( x)  
in x  a

0,
Solution:
1
2
F(s) 

1
2

 sin t  t cos t 
dt 
3
4
t

0
 

 f ( x )e
isx
dx

a
 (a
2
 x 2 )e isx dx
a
a
1
2
2

 (a  x ) cos sxdx
2 0

2
2
  2a cos as 2 sin as  

0
 
s2
s 3  
 

4
2
 sin as  as cos as 


s3


2
2  sin as  as cos as 



s3

3. Show that e

x2
2
Solution:
Fourier transform:
is reciprocal with respect to Fourier transforms

1
2
F[f(x)] 
 f ( x)e
1

2
1

2
x  is
2
e
e
F ( s)  e
f(x)= e




x2
2
e

x2
 isx
2

 ( x is ) 


2 

e


dx

e
2
e

s2
2
dx


1
dx
2
e
 y2
e

s2
2
x    y   and x    y  
2dy

s2
2

eisx dx

dy 
1
F ( s) 
2
dx

1

2
y
isx
s2
2
1

2

2 e  y dy
2
0

 2

where  e  y dy 
2
0

2
s2
2
x2
2
is self reciprocal with respect to Fourier transform.
1  x if x  1
4. Find the Fourier transform of f ( x)  
. Hence deduce that
 0 if x  1

4

 sin t 
0  t  dt  3
Solution:
Fourier transform:
F[f(x)] 


1
2
 f ( x)e
isx
dx

1
2
1
 (1  x )e
isx
dx
1
1
2

(1  x ) cos sxdx
2 0
1
2
sin sx
  cos sx  

 (1  x)
 (1)
 
2

s
 s
 0

2   cos sx 1 
 2

  s2
s 
2  1  cos sx 


  s2 
F ( s) 

By parseval’s identity,




f ( x) dx =


2

 1  x  dx 2 1  x  dx
2
1

2
1
1
2

F ( s) ds   f ( x) dx
2
0


1
2
1  x 3 0 = 2 .
3
3
8  sin 4 ( s / 2) 
2 1  cos s 
F ( s)  


  s 2 
 
s4

2
2



 sin 4 ( s / 2) 
16  sin 4 ( s / 2) 
 F (s) ds    s 4 ds   0  s 4 ds

2
8
Put t=s/2 2t=s 2dt=ds
s  0  t  0 ands    t  



16  sin 4 ( s / 2) 
2  sin 4 t 
2
dt




dt
 0  2t 4 
 0  t 4 


1
2
1  x 3 0
3

2  sin 4 t 
2
 4 dt 

 0  t  
3

 sin 4 t 

0  t 4 dt  3
 x for 0  x  1

5. Find the Fourier cosine transform of f ( x)  2  x for 1  x  2
0
for x  2

Solution: Fc [ f ( x)] 
2


 f ( x) cos sxdx
0
1
2

2
Fc [ f ( x)] 
 x cos sxdx   (2  x) cos sxdx
 0
1


1
2
2  sin sx cos sx  
sin sx
cos sx  
x


(
2

x
)

(

1
)


 
s
s 2  0 
s
s 2 1 

2  sin s cos s 1 cos 2s sin s cos s  
 2  2 2 
 2 

  s
s
s
s
s
s 

2  2 cos s cos 2s 1  
 2  2 

  s 2
s
s 
6. Find the Fourier cosine transform of e  a x
2 2
Solution:
Fc [ f ( x)] 
Fc [ f ( x)] 
2

2

 f ( x) cos sxdx
0

e

0
a 2 x 2
cos sxdx

2 1 a 2 x 2 isx
e
e dx
 2 


2 2
2 1
R.P  e a x isx dx
 2




is 
2
   ax 
1
2a 


R.P  e 
2

1
e
2

 s2 
 2 
 4 a 

R.P  e

s2 

4 a 2 
dx
2

is 
s2 
   ax   2 
2 a  4 a 
 
dx

Put t  ax 
is
2a
dt=adx
x    t   andx    t  
 s2 
 s2 

2 dt
1   4 a 2  1
1   4 a 2 

e
R.P 

e
R.P  e t
a
a
2

2

Fc [ f ( x)] 
1
a 2
e
 s2 
 2 
 4 a 
7. Find Fourier sine transform of e
 ax

, a  0 and deduce that
s
0
Solution:
2
Fs [ f ( x)] 



 f ( x) sin sxdx
0
2 s 
  s 2  a 2 
By inversion formula,
f ( x) 
2


 F [ f ( x)]sin sxds
s
0
2
s

sin sxdx  e ax
2
a
2
2

2 s 
sin sxds  f ( x)
  s 2  a 2 

0


  s
2
0
s 


sin sxds  f ( x)  e ax ,a>0
2
2
2
a 

8. Evaluate
 x
2
0
dx
using Fourier Cosine Transform.
 a x2  b2
2
Solution: Fc [e ax ] 


2 a 
  s 2  a 2 
By Parseval’s identity


0
0
 Fc (s)Gc (s)ds  f ( x) g ( x)dx


2 a  2 b 
ds   e axe bxdx
  s 2  a 2    s 2  b 2 
0

0
2ab


 (s
0

 (x
2
2
0
ds
1

2
2
 a )(s  b )
 ( a  b)
2
dx

1

2
2
 a )( x  b ) 2ab (a  b)
2

9. Evaluate
 (x
0
2
dx
 1)( x 2  4)
Solution:

Proving
 (x
0
2
dx

1

2
 1)( x  4) 2ab (a  b)
Put a=1 and b=2

 (x
0
2
dx




2
 1)( x  4) (2)(1)(2)(3) 12
put s=x

10. Using Parseval’s identity evaluate:
 x
0
x 2 dx
2
 a2

2
Solution;
Consider the function f ( x)  e  ax
2 s 
  s 2  a 2 
Fs [e ax ] 


By parseval’s identity,  [ Fc ( s)] ds   [ f ( x)]2 dx
2
0
0
2

 2  s 
 ax 2
ds

0    s 2  a 2   0 [e ] dx




2 
s2

 0  s 2  a 2


x2

0  x 2  a 2





 e 2 ax 
ds



2

  2a  0

put s=x

1  
dx 

2
2a 2 4a


UNIT V
Z -TRANSFORMS AND DIFFERENCE EQUATIONS
PART – A
   Z z a is
1. Prove that Z a
n
z a

We know that
.
Z  x  n    x  n  z  n
n 0
   a z

Za
n
n
n
n 0

a
  
n 0  z 
 a
 1  
 z
z
z  12
2
a a
 1      ....
z z
 1  x
1
z  a


 z 
2. Prove that Z n  
n
1

1

 1  x  x 2  ...
z
,z a.
za
.

We know that
Z xn    xn z n
n 0

Z n   nz n
n 0

n
n
n 0 z

 0
1 2
3
 2  2  ...
z z
z
2

1
1 1
 1  2   3   ....
z 
z z

1  1 
 1  
z  z 
2



 1  x
2

 1  2 x  3x 2  ....
1  z  1 
 

z  z 
3. Find
2



1 z 
 
z  z  1
2

z
z  12
 
za
n
.
We know that

Z xn  
 xnz
n
n  
  a
Za

n
n
z n
n  

1
a
n

z   a n z n
n n
n0

  
 .......  a 3 z 3  a 2 z 2  az  z a n
 
z 

n

z
a


z  a 


az
z

1  az z  a
z  a2z

1  az z  a  .
4. Find
 .
Z e  an
a 

 G.P.  1  r 


.
We know that
 
Z an 
z
za
 

 
z
za
Z e  an  Z e  a
5. Find
 
n
z
z  e a
Here
a  e a .
 .
Z a n 1
We know that
Z an 
  
Z a n 1  Z a n a 1

 
 a 1Z a n
 z 
 a 1 
 z  a 
 1 
6. Find Z 
.
 nn 1
1
A
B
 
nn  1 n n  1
1  An  1  Bn
Put n  0 we get,
1 A
Put n  1 we get,
1   B (i.e) B  1
1
1
1
 
nn  1 n n  1

1  z 
a  z  a  .
We know that
z
1
Z    log
z 1
n
z
 1 
Z

z
log
z 1
 n  1

 1 
1
 1 
Z

z

Z

 n 
 n  1
 nn  1
 log
z
z
 z log
z 1
z 1
 1  z  log
7. Find

Z a n cos n
z
.
z 1
.


z
We know that Z a n f n   F  
a


Z a n cos n  Z cos n zz / a
 z z  cos   
 2
 z  2 z cos   1 z  z / a
z z


cos


a  a

 2
z
z

2
cos   1
a
a2

by linearity
zz  a cos  
z 2  2az cos   a 2
an 
.
n
!
 
8. Find Z 
Sol:
We know that


z
Z a n f n   F  
a
 1    1 
Z a n    Z   
 n!   n!  z  z / a
 
 e1 / z
zz / a
1
e
z / a
e
9. Prove that Z nf n    z
Give.,
a
z
d
F Z  .
dz
F Z   Z  f n

F Z    f n z  n
n 0

d
F Z     n f nz n1
dz
n 0
z n
  nf n 
z
n 0


 1 
1/ z
 Z  n!   e
 

z

d
F Z    nf n z n
dz
n 0
 Z nf n
Z nf x    z
10. Find
d
F Z 
dz
 .
Z n2
We know that
Z nf n   Z
d
F Z 
dz
 
Z n 2  Z nn   z
 z
d
Z n
dz
d  z 


dz   z  12 
 z  12 1  z2z  1
 z

z  12


 z  1  2z 
 z
2 


z

1


 1 z 
 z
2
 z  1 

z  1
z2  z
z

z  13 z  13
11. Find the Z-transform of nC k  .
 1  nC1 z 1  nC 2 z 2  ..............  nC n z  n
This is the expansion of binominal theorem.
 1  z 1 
n
12. Find

Z e t t 2
.
We know that


Z e  at f t   Z  f t z  zeaT

   
Z e t t 2  Z t 2

z  zeT


T 2 ze T ze T  1
ze
T

1
3
13. Define Unit Sample sequence.
The unit sample sequence
1
0
 n   
 n  is defined the sequence with values
for
n0
for
n0
14. Define Unit step sequence.
The unit step sequence u n  has values.
1
u n   
0
for
n0
for
n0
15. Find


Z 2 n  n  2 .


Z 2n  n  2  Z  n  2z z / 2
1
4
1
 2


2
z
 z  zz / 2  z 
 
2
f 0  lim F z  .
16. If Z  f n  F z , , then
z 

Z  f n    f n z  n
n 0
 f 0 
f 1 f 2
 2  .......
z
z
f 1 f 2


lim Z  f n   lim  f 0 
 2  .....
x 
z 
z
z


lim F z   f 0 .
x 
17. Find the Z-transform of


Z na n u n   z 1
 z 1
na n un .
d
dz 1
 z 
 z  a  by def . of u (n)



d
1  az 1
1
dz

1
 z 1  11  az 1   a
2
 az 1 1  az 1 
2

az 1
1  az 
1 2
.
18. Define convolution of sequences.
i) The convolution of two sequences xn  and yn  is defined as

a. xn * y n    f K g n  K  if the sequences are non – causal
K  
b. xn  * y n  
and
n
 f K g n  K  if the sequences are causal.
K 0
ii) The convolution of two functions f(t) and g (t) is defined as
n
 f t  * g t    f KT g n  K T ,
K 0
where is T is the sampling period.
PART B
1. Using the Z transforms, Solve
Solution:
Given
[
(
) U (z)-
-2z-3z = 0
[
]
U (z) =
U (z) =
=
=
+
…………… (1)
Then
Put z = -1, we get
(1)
4 =A
3 = -B
A=4
B = -3
=

Put z = -2, we get
U (z) =
-[
-3[
Z [u(n)] =
-3[
u(n) =
-3
= 4(
- 3(
= [4-3(
)] (
[
2. Solve the difference equation
Solution:
Given
[
[
[
[
[
[
Y (z) =
=
=
=
=
……….(1)
]
Put z=1 , we get
Put z =-2 , we get
8 = 3A
-4 = -3B
A=8/3
B = 4/3
(1)
Z[y(n)] =
y(n) =
=
3. Using Z transforms, Solve
,
Solution:
Given
]
[
(
) U (z)-z =
(
) U (z) = z+
,
=
=
=
]
=
=
=
=
…………… (1)
+
Put z = 2, we get
(2)
Put z = 4, we get
1 =-2A
1= 2B
A=-
B=
=

+[
U (z) = -
+[
Z [u(n)] = -
+ [
u(n) = -
[
= -
4.Using Z transforms, Solve
,
Solution:
Given
[
(
(
) Y (z) =
) Y (z) =
Y (z) =
,
=
=
=
+
Put z = 2, we get
+
…………… (1)
Put z = -3, we get
1 =25A
1= - 5C
A=
C=
Equating
co-eff.
on both sides, we get
0=A+B
B = -A, B = -
(1)
=
=
ie,


z2
5. Find Z 
.
 z  a z  b 
1
Solution:


z2
z
z 
1 
Z 1 
.
Z 

 z  a z  b
 z  a z  b 
 z 
 z 
 Z 1 
* Z 1 .


z  a
 z  b
 a n *b n
n
 a b
m
nm
n 0
a
 b  
m0  b 
n
m
n
n 1
a
  1
b
 bn
being a G.P
a
1
b
a n 1  b n 1

a b

a n  1
2
n 1
1  a  a  ...  a  a  1 


Note:


z2
6. Find Z 
.
 z  1z  3
1
Solution:


z2
z 
 z
Z 
 Z 1 
.

 z  1 z  3 
 z  1z  3
1
 z 
 z 
 Z 1 
* Z 1 .

 z  1
 z  3 
 1n * 3n
n
 1 3
n 0
m nm
1
 3  
m 0  3 
n
n
m
n 1
1
  1
n  3
3
being a G.P
1
1
3
3n1  1

2
Note:

a n  1
2
n 1
1  a  a  ...  a  a  1 


SRM Institute of Science and Technology
DEPARTMENT OF MATHEMATICS
Transforms and Boundary Value Problems
1. The order and degree of the PDE
(A) 2, 1
(B) 1, 2
∂2z
∂x2
∂z 2
+ 2xy( ∂x
) +
∂z
∂y
(C) 1, 1
= 5, respectively
(D) 2, 2
ANSWER: A
2. The Partial Differential Equation corresponding to Z = (x + a)(y + b) is
(A) p2 + q 2 = z
(B) pq = z
(C) p2 − q 2 = z
(D) z = p2 q 2
ANSWER: B
√
√
3. The complete solution of p + q = 1 is
√
√
(A) z = ax + (1 + a)2 y + c
(B) z = ax + (1 − a)2 y
√
√
(D) z = ax − (1 − a)2 y + c
(C) z = ax + (1 − a)2 y + c
ANSWER: C
4. The general solution of px + qy = z is
(A) f (x, y) = 0
(B) f ( xy , yz ) = 0
(C) f (xy, yz) = 0 (D) f (x2 + y 2 ) = 0
ANSWER: B
5. The general solution of (y − z)p + (z − x)q = x − y is
(A) f (x + y + z) = x2 + y 2 + z 2
(B) f (xyz) = x2 + y 2 + z 2
(C) f (x + y + z) = xyz
(D) f (x2 + y 2 + z 2 ) = x2 y 2 z 2
ANSWER: A
6. The complete solution of z = px + qy + p2 + q 2 is
(A) z = (x + a)(y + b)
(B) z = ax + by + c
(C) z = ax + by + c2 + d2
(D) z = ax + by + a2 + b2
ANSWER: D
7. The solution of the linear PDE (D2 + 4DD0 − 5D02 )z = 0 is
(A) z = f1 (y + x) + f2 (y + 5x)
(B) z = f1 (y − x) + f2 (y − 5x)
(C) z = f1 (y + x) + f2 (y − 5x)
(D) z = f1 (y − x) + f2 (y + 5x)
ANSWER: C
8. The solution of
∂3z
∂x3 =
2
0 is
(A) z = (1 + x + x )f (y)
(B) z = (1 + y + y 2 )f (x)
(C) z = f1 (y) + xf2 (y) + x2 f3 (y)
(D) z = f1 (x) + yf2 (x) + y 2 f3 (x)
ANSWER: C
9. The solution of p + q = z is
(A) f (x + y, y + logz)
(B) f (xy, ylogz)
(C) f (x − y, y − logz)
(D) f (xy, y − logz)
ANSWER: C
10. The particular solution of (D2 − 2DD0 + D02 )z = sinx
(A) −sinx
(B) sinx
(C) cosx
(D) −cosx
ANSWER: A
11. The period of sin5x is
(A)
8π
5
(B)
6π
5
(C)
4π
5
(D)
2π
5
ANSWER: D
12. If f (x) = xsinx in (−π, π) then the value of bn in Fourier series expansion is
(A) 0
(B) 1
(C) 2
(D) 3
ANSWER: A
13. Fourier coefficient a0 in the Fourier series expansion of a function represents
the
(A) maximum value of the function
(B) 2 mean value of the function
(C) minimum value of the function
(D) mean value of the function
ANSWER: B
14. If the Fourier series of the function f (x) in (−`, `) has only cosine terms then
f (x) must be
(A) odd function
(B) even function
(C) neither even nor odd function
(D) multi-valued function
ANSWER: B
15. If f (x) = x2 + x in (0, `) then the even extension in (−`, 0) is
(A) −x2 − x
(B) −x2 + x
(C) x2 + x
(D) x2 − x
ANSWER: D
16. Compute the constant term
lowing data:
x
f(x)
(A) 8.7
ANSWER: A
a0
2
of the Fourier series of f (x) given by the fol-
π
2π
4π
5π
0
π
2π
3
3
3
3
1.0 1.4 1.9 1.7 1.5 1.2 1.0
(B) 9.7
(C) 2.9
(D) 1.45
x
f(x)
0 1 2
9 18 24
3 4 5
28 26 20
17. Compute a1 of the Fourier series of f (x) given by the table:
(A) -8.33
(B) -25
(C) -1.155
(D) 0.519
ANSWER: A
18. The root mean square value of the function f (x) over the interval (a, b) then
ȳ =
qR
q
Rb
b
1
2
2
(A)
|
f
(x)
|
dx
(B)
a
b−a a | f (x) | dx
q
q
Rb
Rb
1
1
2
(C) a−b a | f (x) | dx
(D) b−a
a | f (x) | dx
ANSWER: B
19. The period of tan2x is
(A)
2π
n
(B)
π
n
(C)
π
2
(D)
2
π
ANSWER: C
20. The Fourier cosine series of the function f (t) = sin( πt` ), 0 < t < ` then the
value of a0 is
(A)
1
π
(B)
2
π
(C)
3
π
(D)
4
π
ANSWER: D
21. In one dimensional wave equation
(A)
T
m
(B)
k
c
∂2y
∂t2
2
∂ y
2
= a2 ∂x
stands for
2, a
(C)
m
T
(D)
k
m
ANSWER: A
22. One dimensional wave equation is used to find the
(A) time
(B) displacement (C) heat flow
(D) mass
ANSWER: B
23. Heat flows from
(A) higher to lower temperature
(B) lower to higher temperature
(C) constant temperature
(D) uniform temperature
ANSWER: A
24. The steady state temperature of the rod of length 20cm whose ends are kept at
30C and 80C is
(A) 30 − 52 x
ANSWER: D
(B) 30 + 25 x
(C) 10 + 25 x
(D) 30 + 52 x
25. The tension T caused by stretching the string before fixing it at the end points
is
(A) decreasing
(B) increasing
(C) zero
(D) constant
ANSWER: D
26. The amount of heat required to produce a given temperature change in a body
is proportional to the
(A) mass of the body
(B) weight of the body
(C) density of the body
(D) Tension of the body
ANSWER: A
27. The one dimensional heat equation is of the form
(A)
(C)
∂2u
∂t2
∂2u
∂x2
2
= a2 ∂∂xu2
(B)
∂2u
∂y 2
(D)
+
=0
∂u
2 ∂2u
=
α
∂t
∂x2
∂2u
∂2u
∂x2 + ∂y 2 =
f (x, y)
ANSWER: B
28. The proper solution of one dimensional heat flow equation is u(x, t)
(B) (Aeλx + Be−λx )eα
(A) Ax + B
(C) (A cos px + B sin px)e−α
2 2
λ t
2 2
λ t
(D) (Aeλx + Be−λx )(Ceλat + De−λat )
ANSWER: C
29. The slope of the deflection curve in vibrating string is assumed to be
(A) small at all points and at all times
(B) large at all points and at all times
(C) small at all points but not at all times
(D) large at all points but not at all times
ANSWER: A
30. How many initial and boundary conditions are required to solve the equation
∂2u
2 ∂2u
∂t2 = a ∂x2
(A) 1
(B) 2
(C) 3
(D) 4
ANSWER: D
31. If F [f (x)] = F (s), then F [f (x − a)] =
(A) eisa F (s)
(B) e−isa F (s)
(C) eisx F (s)
(D) eis(x−a) F (s)
ANSWER: A
32. If F [f (x)] = F (s), then F [f (ax)] =
(A) F ( as )
ANSWER: D
(B) F ( as )
(C)
1
a
|a| F ( s )
(D)
1
s
|a| F ( a )
33. If F [f (x)] = F (s), then F [eiax f (x)] =
(A) F (s − a)
(B) F(s+a)
(C) eisa F (s)
(D) e−isa F (s)
ANSWER: B
34. If f (x) = e−ax , then Fourier sine transform of f (x) is
q
q
p a
2 a
s
(A) π s2 +a2
(B) π2 s2 +a
(C) π2 s2 +a
2
2
(D)
pπ
s
2 s2 +a2
ANSWER: B
35. If f (x) = e−ax , then Fourier cosine transform of f (x) is
q
q
p a
p s
s
2 a
(A) π s2 +a2
(B) π2 s2 +a
(C) π2 s2 +a
(D) π2 s2 +a
2
2
2
ANSWER: A
36. If f (x) = x1 , then Fourier sine transform of f (x) is
q
pπ
(A) 2
(B) π2
(C) π2
(D)
2
π
ANSWER: A
37. Under Fourier cosine transform f (x) =
√1
x
is
(A) cosine function
(B) sine function
(C) self reciprocal function
(D) complex function
ANSWER: C
38. If F [f (x)] = F (s), then
R∞
(A) −∞ | Fs (s) |2 ds
R∞
(C) 0 | F (s) |2 ds
R∞
−∞
| f (x) |2 dx =
R∞
(B) −∞ | Fc (s) |2 ds
R∞
(D) −∞ | F (s) |2 ds
ANSWER: D
39. The Fourier transform of a function f (x) is
R∞
R∞
(A) √12π −∞ f (x)eisx dx
(B) √12π −∞ F (s)e−isx ds
R∞
R∞
(D) √12π 0 f (x)eisx dx
(C) √1π −∞ f (x)eisx dx
ANSWER: A
40. The Fourier cosine transform of 5e−2x is
q
q
q
2 10
2 2
(A) π s2 +4
(B) π s2 +4
(C) π2 s2s+4
(D)
q
2 5s
π s2 +4
ANSWER: A
41. If Z[f (n)] = F (z), then Z( 31n ) =
(A)
z
z−1
ANSWER: B
(B)
3z
3z−1
(C)
z
z−3
(D)
z
z−3n
42. If Z[f (n)] = F (z), then Z(3n sin nπ
2 ) =
(A)
z2
(B)
z 2 +9
z
z 2 +9
(C)
z
3
z 2
( 3 ) +1
(D)
z
z−3n
(C)
az
(z−a)2
(D)
z 2 +z
(z−1)3
ANSWER: C
43. If Z[f (n)] = F (z), then Z[an n] =
(A)
z
(z−a)2
(B)
az 2 +a2 z
(z−a)3
ANSWER: C
z
44. If Z[f (n)] = F (z), then Z −1 [ (z−1)(z−2)
]=
(A) 1 − 2n
(B) 2n + 1
(C) −2n − 1
(D) 2n − 1
(C) n2 an
(D) (−a)n
ANSWER: D
z
45. If Z[f (n)] = F (z), then Z −1 [ z−a
]=
(A) an
(B) nan
ANSWER: A
46. If Z[f (n)] = F (z), then the poles of F (z) =
z
(z−1)(z−2)
are
(A) z = −1, z = 2
(B) z = 1, z = 2
(C) z = 1, z = −2
(D) z = −1, z = −2
ANSWER: B
47. If F (z)z n−1 =
zn
(z−1)(z−2) ,
(A) 1, 2n
then the residue of F (z)z n−1 at each pole, respectively
(B) −1, 2n
(C) 1, (−2)n
(D) −1, −2
ANSWER: B
48. If Z[f (n)] = F (z), then Z[(−3)n ] =
z
z
(A) (z−3)
(B) z+3
(C)
2
z
(z+3)2
(D)
z
z−3
(C)
z
z+1
(D)
z
z−1
(C)
z
z−e−5
(D)
z
z+e5
ANSWER: B
49. If Z[f (n)] = F (z), then Z[K] =
(A)
Kz
z−1
(B)
Kz
z+1
ANSWER:A
50. If Z[f (n)] = F (z), then Z[e−5n ] =
(A)
z
z+e−5
(B)
z
z−e5
ANSWER:C
51. If Z[f (n)] = F (z), then Z[ n!1 ] =
1
(A) e− z
ANSWER: C
(B) ez
1
(C) e z
(D) e−z
2
z
52. If Z[f (n)] = F (z), then Z −1 [ (z−a)
2] =
(A) (n + 1)(−a)n (B) (n − 1)(−a)n (C) (n + 1)(a)n
(D) (n − 1)(a)n
ANSWER: C
53. If Z[f (n)] = F (z), then Z[an cos nπ
2 ] =
(A)
az 2
z 2 +a2
ANSWER: B
(B)
z2
z 2 +a2
(C)
az 2
z 2 −a2
(D)
az
z 2 +a2
SRM Institute of Science and Technology
Kattankulathur
DEPARTMENT OF MEATHEMATICS
18MAB201T- TRANSFORMS AND
BOUNDARY VALUE PROBLEMS
UNIT - I Partial Differential Equations
Tutorial Sheet - 1
Questions
Part - A
Sl. No.
1
Form the PDE by eliminating arbitrary constants ‘a’ and ‘b’ from
2
2
2
( x − a ) + ( x − b) + z = c
2
2
Form the PDE by eliminating arbitrary constants ‘a’ and ‘b’ from
log( az − 1) = x + ay + b
3
2
Solve
p + q =1
p = q( z − p)
z = ax + (1 − a ) 2 y + c
Solve the equation pq + p + q = 0
5
( p 2 + q 2 + 1) z 2 = c 2
py = qx
2
Eliminate the arbitrary function ‘f’ from z = f ( x + y )
4
Answer
z = ax −
a
y+c
a +1
Part - B
6
2
2
2
Form the PDE by eliminating ‘f’ from f ( x + y + z , xyz ) = 0
x ( y 2 − z 2 ) p + y ( z 2 − x 2 )q
= z( x2 − y 2 )
7
8
2
2
2
Form the PDE by eliminating ‘f’ from z = xy + f ( x + y + z )
Form the PDE by eliminating ‘f’ from xyz = f ( x + y + z )
p ( y + xz ) − q ( x + yz )
= y2 − x2
x( y − z ) p + y ( z − x)q
= z( x − y)
9
Form the PDE by eliminating ‘f’ and ‘g’ from
z = f ( x + ct ) + g ( x − ct )
10
Obtain the PDE by eliminating ‘a’ , ‘b’ and ‘c’ from
2
2
2
x
y
z
+ 2 + 2 =1
2
a
b
c
q2 = c2 p2
zs + pq = 0
SRM Institute of Science and Technology
Kattankulathur
DEPARTMENT OF MEATHEMATICS
18MAB201T- TRANSFORMS AND
BOUNDARY VALUE PROBLEMS
UNIT - I Partial Differential Equations
Tutorial Sheet - 2
Sl. No.
Questions
Part - A
Find the singular integral of the PDE
1
z = px + qy + p 2 − q 2
2 Find the complete integral of the PDE
p+ q= x
Answer
4 z = y 2 − x2
x2
4 a 32
z = + ax −
x
2
3
+ ay + c
z = x 2 + ax +
3 Solve yp = 2 xy + log q
4 Find the complete integral of p + q = sin x + sin y
eay
+c
a
z = ax − cos x − cos y − ay + c
 sin x sin y 
φ
,
=0
 sin y sin z 
5 Solve p tan x + q tan y = tan z
Part - B
6 Solve (3 z − 4 y ) p + (4 x − 2 z )q = 2 y − 3x
φ ( x 2 + y 2 + z 2 ,2 x + 3 y + 4 z ) = 0
7 Solve ( x 2 − yz ) p + ( y 2 − zx )q = z 2 − xy
x− y

φ
, xy + yz + zx  = 0
 y−z

8 Solve (2 z − y ) p + ( x + z )q + 2 x + y = 0
φ ( x2 + y 2 + z 2 , z + 2 y − x ) = 0
9 Solve ( y + z ) p + ( z + x)q = x + y
10 Solve x 2 ( y − z ) p + y 2 ( z − x)q = z 2 ( x − y )
 x− y

φ
, ( x + y + z )( x − y ) 2  = 0
 y−z

1 1 1

φ  + + , xyz  = 0
x y z

SRM Institute of Science and Technology
Kattankulathur
DEPARTMENT OF MEATHEMATICS
18MAB201T- TRANSFORMS AND
BOUNDARY VALUE PROBLEMS
UNIT - I Partial Differential Equations
Tutorial Sheet - 3
Sl. No.
1
Questions
Part - A
Answer
Solve ( D 2 − 3DD ' + 2 D '2 ) z = 0.
z = φ1 ( y + x) + φ2 ( y + 2 x )
Solve ( D 2 − 4 DD ' + 4 D '2 ) z = e 2 x+ y .
x 2 2 x+ y
z = φ1 ( y + 2 x) + xφ2 ( y + 2 x) + e
2
3
Solve ( D 3 − 2 D 2 D ' ) z = 4sin( x + y ).
z = φ1 ( y ) + xφ2 ( y ) + φ3 ( y + 2 x) − 4cos( x + y )
4
Solve ( D 2 − 6 DD ' + 5D ' 2 ) z = xy.
5
Solve ( D 2 − DD ' ) z = sin x sin 2 y.
2
z = φ1 ( y + x) + φ2 ( y + 5 x) +
z = φ1 ( y ) + φ2 ( y + x)
−
6
x3 y x 4
+
6
4
1
( 2 cos x cos 2 y − sin x sin 2 y )
3
Part - B
z = φ1 ( y − x) + xφ2 ( y − x)
Solve ( D + 2 DD + D ) z = 2cos y − x sin y.
2
'
'2
+ x sin y + 2cos y
7
Solve ( D 3 + D 2 D ' − DD '2 − D '3 ) z = e x cos(2 y ).
z = φ1 ( y − x) + xφ2 ( y − x) + φ3 ( y + x)
Solve ( D 3 − 2 D 2 D ' ) z = sin( x + 2 y ) + 3x 2 y.
ex
( cos 2 y + 2sin 2 y )
25
z = φ1 ( y ) + xφ2 ( y ) + φ3 ( y + 2 x)
+
8
1
x5 y x 6
− cos( x + 2 y ) +
+
3
20 60
9
Solve ( D 2 + DD ' − 6 D '2 ) z = y cos x.
z = φ1 ( y + 2 x) + φ2 ( y − 3 x)
+ sin x − y cos x
10 Solve ( D 2 − 3DD ' + 2 D '2 ) z = (2 + 4 x )e x +2 y .
z = φ1 ( y + x ) + φ2 ( y + 2 x )
2
+ e x + 2 y (11 + 6 x)
9
SRM INSTITUTE OF SCIENCE AND TECHNOLOGY
DEPARTMENT OF MATHEMATICS
18MAB201T/Transforms and Boundary value problems
UNIT III - APPLICATIONS OF PARTIAL DIFFERENTIAL EQUATIONS
ANSWERS-TUTORIAL SHEET -1,2 AND 3
PART-B QUESTIONS
1. The equation is parabolic at all points.
2. The equation is hyperbolic for all x, y.
3. The equation is elliptic.
4.
• The motion takes place entirely in one plane. This plane is chosen as the xy plane.
• In this plane, each particle of the string moves in a direction perpendicular to the
equilibrium position of the string.
• The tension T caused by stretching the string before fixing it at the end points is constant at
all times at all points of the deflected string.
• The tension T is very large compared with the wight of the string and hence the
gravitational force may be neglected.
• The effect of friction is negligible.
• The string is perfectly flexible. It can transmit only tension but not bending or shearing
forces.
• The slope of the deflection curve is small at all points and at all times.
PART-C QUESTIONS
4. y(x, t) =
5. y(x, t) =
6. y(x, t) =
3y0
4
k
4
sin
πx
l
cos
πat
l
−
y0
4
sin
3πx
l
cos
3πat
l
[sin x cos t + sin 3x cos 3t]
∞
∑
n=1
8k
n2 π 2
sin
nπ
2
sin
nπx
l
cos
nπct
l
TUTORIAL SHEET -2
PART-B QUESTIONS
1. The possible solutions are
(i) y(x, t) = (A1 eλx + B1 e−λx )(C1 eλat + D1 e−λat )
(ii) y(x, t) = (A2 cos λx + B2 sin λx)(C2 cos λat + D2 sin λat)
(iii) y(x, t) = (A3 x + B3 )(C3 t + D3 )
The correct solution is
y(x, t) = (A2 cos λx + B2 sin λx)(C2 cos λat + D2 sin λat).
2. The initial and boundary conditions are
(i) y(0, t) = 0 for
t≥0
(ii) y(l,
t)
=
0
for
t≥0
)
(
∂y
at (x,0)= 0 for
0≤x≤l
(iii)
∂t




2b
l


x
if 0 ≤ x ≤

l
2
(iv) y(x, 0) =



2b
1


(l − x) if
≤x≤l

l
2
3. The initial and boundary conditions are
(i) y(0, t) = 0 for t ≥ 0
(ii) y(l, t) = 0 for t ≥ 0
(iii) (
y(x, 0)
) = 0 for 0 ≤ x ≤ l
∂y
(iv)
at t=0 = 3x(l − x) for 0 ≤ x ≤ l
∂t
PART-C QUESTIONS
4. y(x, t) =
3lv0
πx
πat
sin
v0 l
−
sin
3πx
sin
3πat
l
l
12πa
l
l
]
∞ [
4l3 ∑ 1 + 2(−1)n
nπx
nπct
5. y(x, t) = − 3
sin
. cos
3
π n=1
n
l
l
6. y(x, t) =
4πa
sin
∞
16cl ∑ 1
aπ 3
n=1
n3
sin
nπ
2
sin
nπx
2l
sin
nπat
2l
TUTORIAL SHEET -3
PART-B QUESTIONS
1.
• Heat flows from a higher to lower temperature
• The amount of heat required to produce a given temperature change in a body is
proportional to the mass of the body and to the temperature change. This constant of
proportionality is known as the specific heat (c) of the conducting material.
• The rate at which heat flows through an area is proportional to the area and to the
temperature gradient normal to the area. This constant of proportionality is known as the
thermal conductivity k of the material.
2. The possible solutions are
2 2
(i) u(x, t) = (A1 eλx + B1 e−λx )(C1 eα λ t )
2 2
(ii) u(x, t) = (A2 cos λx + B2 sin λx)(C2 e−α λ t )
(iii) u(x, t) = (A3 x + B3 )C3
The correct solution is
2 2
u(x, t) = (A2 cos λx + B2 sin λx)(C2 e−α λ t ).
3. The initial and boundary conditions are
(i) u(0, t) = 0 for all t
(ii) u(l, t) = 0 for all t



2T x
l


if 0 < x <

l
2
(iii) u(x, 0) =



2T
1


(l − x) if
<x<l

l
2
4. u(x) = 2x + 20
PART-C QUESTIONS
5. u(x, t) =
6. u(x, t) =
∞
4l ∑ 1
π2
n=1
n2
∞
∑
200
n=1
nπ
sin
nπ
2
sin
(−1)n+1 sin
nπx
l
nπx
l
−α2 n2 π 2 t
e
l2
−α2 n2 π 2 t
e
7. u(x, t) = us (x) + ut (x, t) = 50 − 4x −
l2
∞
60 ∑ 1
π
n=1
n
sin
nπx
5
−α2 n2 π 2 t
e
25
SRM INSTITUTE OF SCIENCE AND TECHNOLOGY
DEPARTMENT OF MATHEMATICS
18MAB201T/Transforms and Boundary value problems
UNIT II – FOURIER SERIES
TUTORIAL SHEET -1
PART B Questions
1. State Dirichlet condition’s for a given function to expand in Fourier series.
sin x, 0  x  
2.Find a1 for the periodic function f ( x) = 
0,   x  2
3.Find a0 for the periodic function f ( x) = e− x , 0  x  2 .
4.Find an for the Fourier series of periodicity 3 for f ( x) = 2 x − x 2 in 0 < x <3
5.Find half –range cosine series for f ( x) = x, 0  x  
PART C Questions
6.Find the Fourier series to represent ( x − x 2 ) in the interval [-  ,  ].Deduce the value of
1 1 1
− + − ........
12 22 32
7.Obtain the Fourier series expansion for f ( x) = x 2in −   x   and hence the sum of the
series
1 1 1
+ + + ........
14 24 34


sin x, 0  x  4
8. If f ( x) = 
.Express f(x) in a series of sines.
cos x,   x  

4
2
9.Find the Fourier series for f ( x) = cos x in −   x   of periodicity 2  .
10.Find the Fourier series for f ( x) = sin x in −   x   of periodicity 2  .
SRM INSTITUTE OF SCIENCE AND TECHNOLOGY
DEPARTMENT OF MATHEMATICS
18MAB201T/Transforms and Boundary value problems
UNIT II – FOURIER SERIES
TUTORIAL SHEET -2
PART B Questions
1. Expand f ( x) = ( x − 1) 2 in 0 < x <1 in a Fourier series of sine series only.
2.The Fourier series of the function f ( x) = x + x 2 , −  x   is

. Then deduce that 
1
1 2
=
n2
6
sin x, 0  x  
3. Find b1 for the function f ( x) = 
.
0,   x  2
4.The Fourier series of the function f ( x) = ( − x)2 ,0  x  2 is

. Then deduce the sum 
1
1
n2
5. .Express f ( x) = x( − x), 0  x   , as a Fourier series of periodicity 2 containing sine terms
only.
PART C Questions
6.Find the Fourier series of f ( x) = x + x 2 , −2  x  2 .Hence find the sum of the series
1 1 1
+ + + ........
12 22 32
7.Find the half –range cosine series for the function f ( x) = ( x − 1)2 , in0  x  1 .Hence show that
 2 = 6
1 1 1
+ + + ........
12 22 32
.
x
 l , 0  x  l
8.If f ( x) = 
.Express f(x) as a Fourier series of periodicity 2 l .
 2l − x , l  x  2l
 l
 x, −1  x  0
and deduce the sum of
 x + 2, 0  x  1
9.Find the Fourier series of periodicity 2 for f ( x) = 
1 1 1
+ − + ........
3 5 7
10.Express f(x) = x in half-range cosine series and sine series of periodicity 2 l in the range
1 1 1
0  x  l and deduce the value of 4 + 4 + 4 + ........ .
1 3 5
1−
SRM INSTITUTE OF SCIENCE AND TECHNOLOGY
DEPARTMENT OF MATHEMATICS
18MAB201T/Transforms and Boundary value problems
UNIT II – FOURIER SERIES
TUTORIAL SHEET -3
PART B Questions
1.Find R.M.S value of f ( x) = x − x 2 , −1  x  1
2.Find R.M.S value of f ( x) = x 2 , −  x   .
3.Define Root Mean Square and find the RMS value of f ( x) = 1 − x, 0  x  1
4.Find the half-range Fourier sine series for f ( x) = x, 0  x  
5.Obtain the half –range cosine series for f ( x) = x( − x), 0  x  
PART C Questions
6.Compute the first two harmonic of the Fourier series of f(x) given by the following table:
5
2
4


3
3
3
3
f(x)
1
1.4
1.9
1.7
1.5
1.2
7.Compute first three harmonics of the half-range cosine series of y = f(x) from
x
2
0
1
x
0
1
2
3
4
5
f(x)
4
8
15
7
6
2
8. Compute the first two harmonic of the Fourier series of f(x) given by the following table:
x
f(x
)
0
6.82
4
30
7.97
6
60
8.02
6
90
7.20
4
120
5.67
6
150
3.67
4
180
1.76
4
210
0.55
2
240
0.26
2
270
0.90
4
300
2.49
2
330
4.73
6
9.The values of x and the corresponding values of f(x) over period T are given below .Show that
2 x
f(x) = 0.75 + 0.37cos  +1.004 sin  where  =
.
T
x
0
5T
2T
T
T
T
T
6
3
6
3
2
f(x)
1.98
1.30
1.05
1.30
-0.88
-0.25
1.98
2
10.Expand f ( x) = x − x as a Fourier series in −1  x  1 and using this series find the RMS value
of f(x) in the interval.
Tutorial Sheet-1
Answers
Part-A
1. State any three condition
2. a 1 = 0
3.
4.
5.
a 0=
1 − e−2
.

−9
a n= 2 2
n
 −4
; if n is odd

a n =  n2
0
if n is even
Part - B
−
−4(−1)
−2(−1) n
1 1 1
2
+
cos
nx
+
sin
nx
and
−
+
......
=

3
n2
n
12 22 32
3
n =1
n =1

2
6. f(x)=
7. f(x)=
8. f(x)=
9. f(x)=
10. f(x)=
2
3

n
(−1)n
1 1 1
4
cos
nx
and
+
+
+
......

=
2
14 24 34
90
n =1 n

+ 4
4 2  sin 2 x sin 6 x sin10 x

−
+
− ...

  1.3
5.7
9.11

2

2

−
4

n

n=2
−
4

1
 n
cos 
−1
 2
2
 (4n

n =1
1
2
− 1)

 cos nx

cos 2nx
Tutorial Sheet-2
Answers
Part-A
 4(−1)n 2
4 
+
− 3 3  sin n x
3 3
n n  
n =1  n 

1. f(x)=  
2. Take x =  and
3.
1 2
=

2
6
n =1 n

b1 = 1/ 2
4. Take x = 0 and
5. f(x)=
8



1
1 2
=

2
6
n =1 n

sin x
n3
Part- B
4
3

6. f(x)= + 
1
we get
1
3

1
3

16(−1)n+1
 n
cos 
2 2
n
 2
1
8. f(x) = + 
n =1
2
n +1
sin
n
x & take x = 2,
2
1 1 1
2
+
+
+
...
=
12 22 32
4
7. f(x) = + 
9. f(x) = 1 +

4(−1)

x
+


n

1

−4
1 2
cos
n

x
and
take
x
=
0,
we
get
1 n2 = 6
n2 2
4
n 2

2
cos n x
1
 n 1 − 2(−1)

n =1
n

 sin n x & take x = 1 , we get1 − 1 + 1 − 1 + ... =
2
3
5
7
4
 n
cos 
l 4l
10. cosine series: f(x) = − 2   2l
2  n =1
n


n
 l
sine series f(x) =  bn sin 
1

x

1 1 1
2

x
and
+
+
+
...
=
.

14 34 54
96

Tutorial Sheet-3
Answers
Part -A
1.
2.
3.
R.M.S = 815
R.M.S=  2
R.M.S: The root mean square value of a function
b 2 
  y dx 
−
−


Y=f(x) over a given interval (a,b) is defined as y =  a
and
y
=

 b−a 


1
(−1)n −1
sin nx.
n
n =1

4.
f(x)= 2
5.
f(x) =
2
6
+
−4
cos nx
2
n is even n

Part-B
6.
7.
8.
f(x) = 1.45 − 0.33cos x − 0.1cos 2 x + 0.03cos 3x + .. + 0.17 sin x − 0.06sin 2 x + ..

2
3 x
x − 1.66 cos
f(x) = 7 + 4.565cos x − 2.833cos
6
6
6
f(x) = 4.174 + 2.450 cos x + 0.120 cos 2 x + 0.08cos 3 x + 3.160sin x + 0.034sin 2 x +
0.010sin 3x
9.
f(x)= 0.75 + 0.37 cos  + 1.005sin 

2
 4

(−1) n +1 cos n x +
(−1) n +1 sin n x  & R.M .S = 8
2 2
15
n

n =1  n 
10. f(x) = − 1 3 +  
3
SRM INSTITUTE OF SCIENCE AND TECHNOLOGY
DEPARTMENT OF MATHEMATICS
18MAB201T/Transforms and Boundary value problems
UNIT III - APPLICATIONS OF PARTIAL DIFFERENTIAL EQUATIONS
TUTORIAL SHEET -1
PART-B QUESTIONS
1. Classify the equation x2 fxx + (1 − y 2 )fyy = 0.
2. Classify the equation (1 + x2 )fxx + (5 + 2x2 )fxy + (4 + x2 )fyy = 2 sin(x + y).
3. Classify the equation
∂ 2u
∂x2
+
∂ 2u
∂y 2
= f (x, y).
4. Write down the assumptions made in deriving one-dimensional wave equations.
PART-C QUESTIONS
4. A tightly stretched(string
) with fixed end points x = 0 and x = l is initially in a position given by
πx
y(x, 0) = y0 sin
. If it is released from rest from this position, find the displacement y
l
at any time and at any distance from the end x = 0.
5. An elastic string is stretched between two fixed points at a distance π apart. In its initial position
the string is in the shape of the curve f (x) = k(sin x − sin3 x). Obtain y(x, t) the vertical
2
∂ 2y
2∂ y
displacement if y satisfies the equation
=
c
.
∂t2
∂x2
2
∂ 2y
2∂ y
6. Find the solution of the wave equation
=
c
, corresponding to the triangular initial
∂t2
∂x2




2kx
l



,0 < x <

l
2
deflection f (x) =
and the initial velocity is zero.




2k
l


(l − x), < x < l

l
2
SRM INSTITUTE OF SCIENCE AND TECHNOLOGY
DEPARTMENT OF MATHEMATICS
18MAB201T/Transforms and Boundary value problems
UNIT III - APPLICATIONS OF PARTIAL DIFFERENTIAL EQUATIONS
PART-B QUESTIONS
TUTORIAL SHEET -2
1. Write the possible solutions and correct solution of the one dimensional wave equations.
2. A string is tightly stretched and its ends are fastened at two points x = 0 and x = l. The mid
point of the string is displaced transversely through a small distance ′ b′ and the string is released
from rest in that position. Write down the initial and boundary conditions.
3. A tightly stretched string with fixed end points x = 0 and x = l is initially at rest in its
equilibrium position. If it is set vibrating giving each point a velocity 3x(l − x), write down the
initial and boundary conditions.
PART-C QUESTIONS
′ ′
4. If a string of(length
) l is initially at rest in its equilibrium position and each point of it is given
∂y
πx
the velocity
= v0 sin3
, 0 < x < l. Determine the transverse displacement
∂t t=0
l
y(x, t).
5. A tightly stretched string has its ends fixed at x = 0 and x = l. Initially the string is in the form
y = kx2 (l − x), where k is a constant, and then released from rest. Find the displacement at
any point x and any time t > 0.
6. A string is stretched between two fixed points at a distance 2l apart and the points of the string
are given initial velocities

cx



,
in 0 < x < l
l
v=

c

 (2l − x), in l < x < 2l
l
x being the distance form an end point. Find the displacement of the string at any subsequent
time.
SRM INSTITUTE OF SCIENCE AND TECHNOLOGY
DEPARTMENT OF MATHEMATICS
18MAB201T/Transforms and Boundary value problems
UNIT III - APPLICATIONS OF PARTIAL DIFFERENTIAL EQUATIONS
PART-B QUESTIONS
TUTORIAL SHEET -3
1. Write down the assumptions made in deriving one-dimensional heat equation.
2. Write down the possible solutions and correct solution of one dimensional heat equations.
3. A homogeneous rod of conducting material of length l units has ends kept at zero temperature
and the temperature at the centre is T and falls uniformly to zero at the two ends. Write down
the initial and boundary conditions.
4. A rod 30cm long has its ends A and B kept at 20◦ C and 80◦ C respectively until steady state
conditions prevail. Find the steady state temperature in the rod.
PART-C QUESTIONS
∂ 2u
= α2
subject to (i) u(0, t) = 0, for t ≥ 0 (ii) u(l, t) = 0, for t ≥ 0
∂t
∂x2
l


 x,
for 0 ≤ x ≤
2
(iii) u(x, 0) =

l

 l − x, for
≤ x ≤ l.
2
5. Solve
∂u
6. A rod of length l has its ends A and B kept at 0◦ C and 100◦ C until steady state condition
prevail. If the temperature at B is reduced suddenly to 0◦ C and kept so while that of A is
maintained, find the temperature u(x, t) at a distance x from A and at time t.
7. A bar, 10 cm long, with insulated sides, has its ends A and B kept at 20◦ C and 40◦ C
respectively until steady state conditions prevail. The temperature at A is then suddenly raised
to 50◦ C and at the same instant that at B is lowered to 10◦ C. Find the subsequent temperature
at any point of the bar at any time.
SRM INSTITUTE OF SCIENCE AND TECHNOLOGY
DEPARTMENT OF MATHEMATICS
18MAB201T/Transforms and Boundary value problems
UNIT IV-FOURIER TRANSFORMS
TUTORIAL SHEET -1
PART-B QUESTIONS
1. If F {f (x)} = F (s), then F {f (ax)} =
1
|a|
F
( )
s
.
a
2. State and Prove Modulation theorem.
3. If F {f (x)} = F (s), then F {xn f (x)} = (−i)n
{
4. Find the complex Fourier transform of f (x) =
PART-C QUESTIONS
dn
dsn
F (s).
x, for|x| ≤ a
0, for|x| > a
{
a2 − x2 , |x| < a
5. Show that the Fourier transform of f (x) =
is
0,
|x| > a > 0
√ (
)
∫ ∞
sin t − t cos t
2 sin as − as cos as
π
. Hence deduce that
dt = . Using
2
3
3
π
s
t
4
)0
∫ ∞(
sin t − t cos t 2
π
Parseval’s identity show that
.
dt =
t3
15
0
{
1 − x2 , |x| < 1
6. Find the Fourier transform of f (x) =
and hence evaluate
0,
|x| > 1
)
∫ ∞(
x cos x − sin x
x
cos
dx
x3
2
0
{
1, |x| < a
7. Find the Fourier transform of f (x) given by f (x) =
and hence evaluate
0, |x| > a > 0
∫ ∞
∫ ∞
sin x
sin as cos sx
dx and
ds.
x
s
0
−∞
{
1, |x| < a
8. Find the Fourier transform of f (x) given by f (x) =
and using
0, |x| > a > 0
)
∫ ∞(
sin t 2
π
Parseval’s identity, prove
dt = .
t
2
0


9. Show that the transform of e
2 2
e−a x , a > 0.
−x2
2




is e
−s2
2


by finding the Fourier transform of
{
10. Find the Fourier transform of f (x) given by f (x) =
∫
∞
value of
0
sin4 t
t4
1 − |x|, |x| < 1
and hence find the
0,
|x| > 1
dt.
∗ ∗ ∗ ∗ ∗ ∗ ALL THE BEST ∗ ∗ ∗ ∗ ∗ ∗
UNIT IV TUTORIAL 2
Answer all the questions
PART B
1. Find
(
sin2x)
2. Find
(
cos2x)
3. Find (
sin2x)
4. Find
cos 2x)
(
5. Find (x
and
(x
6. State convolution of two functions in Fourier transforms.
7. If (
)=
,Find
(
) using change of scale property.
PART C
8. Find
9. Find
10. Find
(
(
and
(
and hence derive the inversion formula.
) and hence find If
(
(
).
) and use it to evaluate
) sinx dx.
11.State and prove convolution theorem in fourier transforms.
12.Find the function if its sine transform is
13. Find
(
14.Prove that
.
).
(
))= -
( ( )) and
(
))=
( ( ))
UNIT IV TUTORIAL 3
Answer all the questions
PART B
1.
Prove that F(
2. Prove that F(
f(x)) =
)=
(F(s)).
F(s).
3. StateParseval’s identity for both sine and cosine transforms.
4. If
(
5. If f(x)=
)=
,find
(
(x)) and
, find F(x
(
(x)) .
F(
.
PART C
6. f(x)=
, find the value of
7. UsingParseval’s identity evaluate
dx and
8. Solve the integral equation
=
9. Solve the integral equation
10. Solve for f(x) if
11 .Prove that
dt.
dx
.
=
=
.
is self reciprocal under Fourier transforms.
12. Find the Fourier sine and cosine transforms of
and hence show that
is self
reciprocal under both the transforms.
13. Find the Fourier transform of
and hence find F(
and F(
.
ANSWERS FOR THE QUESTIONS IN TUTORIAL 2.
1.
(
sin2x) =
=
2.
(
cos2x) =
=
3.
(
sin2x) =
=
4.
(
cos 2x) =
=
7.
(
9.
(
10.
(
12.
) = F(
.
and
)=
(
13.
=
(
(x
)=
and
)=
sin x dx=
.
=
ANSWERS FOR THE QUESTIONS IN TUTORIAL 3
(
(f’(x))= -s
5. If f(x)=
F(
6.
7.
(f’’(x))= -
and
,then F(
=
(s)-
=-
=
and F(
=
dt=
dx =
and
dx=
8. f(x)=
9. f(x)=
10 .f(x)=
13. F(
F(
)=
.
)=
and F(
)=
.
SRM INSTITUTE OF SCIENCE AND TECHNOLOGY
DEPARTMENT OF MATHEMATICS
18MAB201T/Transforms and Boundary value problems
UNIT IV-FOURIER TRANSFORMS
ANSWERS-TUTORIAL SHEET -1
PART-B QUESTIONS
5. F {f (x)} =
2i
1
[sin sa − as cos sa]
√
2π
s2
PART-C QUESTIONS
[
]
)
∫ ∞(
−4 s cos s − sin s
x cos x − sin x
x
3π
7. F {f (x)} = √
and
cos dx = −
3
3
s
x
2
16
2π
0
√
8. F {f (x)} =
∫
∞
−∞
∫
∞
0
2 sin as
s 

π


for |x| < a
sin as cos sx
2
ds =

s

 0 for |x| > a
sin x
x
dx =
π
π
2
√
12. F {f (x)} =
2
π
(
1 − cos s
s2
)
∫
∞
and
0
sin4 t
t4
dt =
π
3
SRM Institute of Science and Technology
Kattankulathur
DEPARTMENT OF MEATHEMATICS
18MAB201T Transforms and Boundary Value
Problems
1
UNIT - V : Z Transforms
Tutorial Sheet - 13
Sl.No.
Questions
Part – B
n
Find the z -transforms of cos .
2
2
Find the z -transforms of cos 2 t.
3
If z[f (n)] = F(z), then prove that z[a − n f (n)] = F(az).
4
z
If z[f (n)] = F(z), then prove that z[a n f (n)] = F  .
a
Find z[n 2 + a n+3 ].
5
6
7
8
9
10
Part – C
f z[f (n)] = F(z), then prove that z[f (n − k)] = z − k F(z) for
k  0.
Find z[(n + 1)(n + 2)].
 2n + 3 
Find z 
.
(n
+
1)(n
+
2)


 1 
Find z 
.
n(n
−
1)


 n  
+ .
Find the z -transforms of cos 
 2 4
Answer
n 
z2

z cos  = 2
2  z +1

z
z(z − cos 2T)
z cos t =
+
2(z − 1) 2(z − 2zcos 2T + 1)
2
2
z(z + 1) a 3z
+
(z − 1)3 z − a
z(z + 1)
3z
2z
+
+
3
2
(z − 1) (z − 1) z − 1
 z 
(z 2 + z)log 
−z
 z −1 
 z −1   z −1 

 log 

 z   z 
1 z(z − 1)
2
2 (z + 1)
SRM Institute of Science and Technology
Kattankulathur
DEPARTMENT OF MEATHEMATICS
18MAB201T Transforms and Boundary Value
Problems
1
2
UNIT – V : Z Transforms
Tutorial Sheet - 14
Sl.No.
Questions
Part – B
n
Find the z -transforms of sin .
2
 n 
Find the z -transforms of sin 3   .
 6 
3
 n 
Find the z -transforms of sin 2   .
 4 
4
Use
initial
value
theorem
zeaT (zeaT cos bT)
f (z) = 2 2aT
.
z e - 2zeaT cos bT + 1
Answer
z
 n 
z sin  = 2
2  z +1


3z
 n  
z sin 3    =
2
 6   4(z − z 3 + 1)

z
−
4(z 2 + 1)

z
 n  
z sin 2    =
 4   2(z − 1)

2
z
−
2(z 2 + 1)
to
find
f (0)
when
f (0) = 1
f () = 0
TzeaT
.
Use final value theorem to find f () when f (z) =
(zeaT − 1) 2
Part – C
(i) f (n) = 2n + 1
6
2z 2 + 4z
z2 + z
Find the inverse z -transforms of (i)
(ii)
by
2 n
(ii) f (n) = n 2
(z − 1) 2
(z − 2)3
long division division method.
f (n) = 1 + 2u(n − 1)
7
1 + 2z −1
Find the inverse z -transform of
by
long
division
1 − z −1
division method.
8
5z
1
Find the inverse z -transform of
by partial fraction
f (n) = 3n − n
2
(2z − 1)(z − 3)
method.
9
3
5
z 2 + 2z
f (n) = − 4.2n + .3n
Find the inverse z -transform of
by partial
2
2
(z − 1)(z − 2)(z − 3)
fraction method.
10
4z 2 − 12z
Find the inverse z -transform of 3
by partial fraction
20 8
z − 3z + 2
f (n) =
− n
9
3
method.
20
− (−2) n
9
5
SRM INSTITUTE OF SCIENCE AND TECHNOLOGY
DEPARTMENT OF MATHEMATICS
18MAB201T - Transforms and Boundary value problems
UNIT V – Z-Transform
TUTORIAL SHEET -15
PART-B
Sl.No
Questions
1
Answer
 2 z 2  3z 

 by partial fraction method.
 ( z  2)( z  4) 
(2) n 11 (4) n

6
6
1
Find Z
2
If F ( z ) 
3z
, find the residue of F ( z ) z n1 at z  2 .
( z  1) ( z  2)
3.(2) n
3
If F ( z ) 
z3
, find the residue of F ( z ) z n1 at z  1 .
( z  1) ( z  2)
2
(1) n
3
1


z2

.
 ( z  1)( z  3) 
4
Using convolution Theorem evaluate Z
5
Solve: y n1  2 y n  0 given y 0  3 using Z transforms.
1 n 1
(3  1)
2
3.(2) n
PART-C
6
8z 2
Find the inverse Z transform of
by using
(2 z  1)(4 z  1)
convolution theorem.
7
Find by Residue method if Z
8
Find Z
9
10
1
1
 2z 2  4z 
.

3 
 ( z  2) 


z3
by using method of partial fraction.

2 
(
z

2
)(
z

1
)


Solve: y n 2  6 y n1  9 y n  2 n , given y0  y1  0 , using Z
transforms.
Solve y n 2  7 y n1  12 y n  2 n , given y 0  0 , y1  0 , using Z
transform method.
 1  n  1  2 n 1 
2     
 2   2  
n 2 .(2) n
 3.1n  n  4.2 n
1 n 1
1
.2  (3) n  n(3) n
25
25
15
1 n
1
. 2  3n  . 4 n
2
2