‫قســـــــــــــــــــــــــم الرياضيــــــــــــــــــــــــــات‬
‫‪Lectures in‬‬
‫‪Mathematics 3‬‬
‫‪For Computers and‬‬
‫‪Information Students‬‬
‫‪-1-‬‬
CONTENTS
Laplace Transforms
3
Inverse Laplace Transforms
13
Applications of Laplace Transforms
26
Fourier Series
37
Fourier Transform
53
Power Series Solutions of Linear Differential Equations
104
Gamma Function
154
Beta Function
156
Bessel Function
167
Bibliography
185
-2-
CHAPTER 1
Laplace Transforms
1. Introduction and Definition
The dynamics of physical phenomena are represented mathematically by
differential equations. Laplace transform is an efficient tool for solving both
ordinary and partial differential equations. Ordinary differential equations are
transformed into algebraic equations which simplifies the problem. Similarly
Laplace transform converts partial differential equations into ordinary
differential equations which are easier to solve. Laplace transform can be used
to analyze and
study the responses of electrical and mechanical systems
subjected to discontinuous forcing terms. This property facilitates studying
circuits where the input may be a voltage which remains on for a while, switches
off and switches on periodically. Thus Laplace transform finds applications in
signal analysis and is also a tool for investigating the effect of sudden impulsive
forces on dynamical systems like the spring mass coupled system .
Mathematically Laplace transform is a mapping which assigns to a function
𝑓(𝑡) another function 𝐹(𝑠) ,called the Laplace transform of function 𝑓(𝑡). It is
defined for all 𝑡 ≥ 0
∞
𝑭(𝒔) = 𝓛 {𝒇(𝒕)} = ∫𝟎 𝒆−𝒔𝒕 𝒇(𝒕)𝒅𝒕
(1)
for all values of 𝑠 for which this integral converges. This is an improper integral
over an infinite interval , it can be represented as a sum of improper integrals
within intervals which are bounded. As for example,
∞
𝑎
∫0 ℎ(𝑡)𝑑𝑡 = lim ∫0 ℎ(𝑡)𝑑𝑡
𝑎→∞
-3-
If this limit exists , then the given improper integral is said to converge. In
eq.(1) ,the integral will converge to a function of 𝑠 which may converge for
some range or values of 𝑠 and diverge for others. While deriving expressions for
Laplace transform of functions it is therefore essential to state the condition on 𝑠
for the transform to exist.
2. Conditions for the existence of Laplace transforms
Definition 2.1
A function 𝒇(𝒕) is piecewise continuous over an interval if it can be divided
into several sub intervals such that 𝒇(𝒕) is continuous within these
subintervals and has finite limits at the ends of every subinterval except
possibly at 𝒕 → ±∞.
Definition 2.2
A function 𝒇(𝒕) with 𝒕 > 0 is said to be of exponential order if |𝒇(𝒕)| ≤
𝒌𝒆𝒑𝒕
as 𝒕 → ∞ with 𝒌, 𝒑, 𝑻 all nonnegative for all 𝒕 ≥ 𝑻 .
This implies that the given function 𝑓(𝑡), for 𝑡 → ∞ , grows less slowly than a
positive multiple of some exponential function.
Existence Theorem: The Laplace transform of 𝑓(𝑡) defined as
ℒ {𝑓(𝑡) = 𝐹(𝑠) exists for 𝑠 > 𝑝 if
a) 𝒇(𝒕) is defined and is piecewise continuous on [𝟎, ∞]
b) 𝒇(𝒕) is of exponential order with respect to 𝒆𝒑𝒕 as 𝒕 → ∞.
-4-
𝑑
Given a function ℎ(𝑡),
∫𝑐 ℎ(𝑡)𝑑𝑡 exists if ℎ(𝑡) is piecewise continuous on
the bounded interval [c,d].
𝑑
Thus ∫0 𝑓(𝑡)𝑒 −𝑠𝑡 𝑑𝑡 exists for 𝑑 < ∞. To limit the growth of 𝑓(𝑡) as 𝑡 → ∞ ,
we put a constraint on 𝑓(𝑡).
Proof:
𝑒 𝑝𝑡 ≥ 1 if 𝑡 ≥ 0, |𝑓(𝑡)| ≤ 𝑘 𝑒 𝑝𝑡 ;
𝑎
Consider ∫0 |𝑒 −𝑠𝑡 𝑓(𝑡)| 𝑑𝑡 ; We let 𝑎 → ∞ , and check for the boundedness of
this integral
𝑎
𝑎
𝑎
∫|𝑒 −𝑠𝑡 𝑓(𝑡)| 𝑑𝑡 ≤ ∫|𝑒 −𝑠𝑡 𝑘𝑒 𝑝𝑡 | 𝑑𝑡 = 𝑘 ∫ 𝑒 −(𝑠−𝑝)𝑡 𝑑𝑡
0
0
0
𝑘
= (𝑠−𝑝) (1 − 𝑒 −𝑎(𝑠−𝑝) )
If 𝑠 > 𝑝 , 𝑎𝑛𝑑 𝑎 → ∞ , then the second term within the bracket → 0 and the
integral becomes
𝑘
𝑠−𝑝
.
𝑎
∞
𝑘 ∫0 𝑒 −(𝑠−𝑝)𝑡 𝑑𝑡 ≤ 𝑘 ∫0 𝑒 −(𝑠−𝑝)𝑡 𝑑𝑡
∞
𝐹(𝑠) = ∫0 𝑓(𝑡)𝑒 −𝑠𝑡 𝑑𝑡
∞
∞
𝑘
|𝐹(𝑠)| = |∫0 𝑓(𝑡)𝑒 −𝑠𝑡 𝑑𝑡| ≤ ∫0 |𝑒 −𝑠𝑡 𝑓(𝑡)| 𝑑𝑡 ≤
.
𝑠−𝑝
For a function 𝐹(𝑠) to be a Laplace transform of some function 𝑓(𝑡) the
condition is
𝐥𝐢𝐦, 𝑭(𝒔) = 𝟎
(2)
𝒔→∞
-5-
Example 2.1
𝐹(𝑠) =
𝑠
, lim
𝑠+5
𝑠
=1
𝑠→∞ 𝑠+5
lim 𝐹(𝑠) ≠ 0
𝑠→∞
∴ 𝐹(𝑠) cannot be the Laplace transform of any function.
Initial Value Theorem
𝐥𝐢𝐦 𝒇(𝒕) = 𝐥𝐢𝐦 𝒔𝑭(𝒔)
𝒕→𝟎
𝒔→∞
∞
ℒ{𝑓 ′ (𝑡)} = ∫0 𝑒 −𝑠𝑡 𝑓 ′ (𝑡)𝑑𝑡 = 𝑠𝐹(𝑠) − 𝑓(0)
Assuming the continuity of 𝑓(𝑡) at 𝑡 = 0, and assuming that the ℒ{𝑓 ′ (𝑡)} exists
and therefore is piecewise continuous and of exponential order,
∞
lim ∫0 𝑒 −𝑠𝑡 𝑓 ′ (𝑡)𝑑𝑡 = 0 = lim 𝑠𝐹(𝑠) − 𝑓(0) or
𝑠→∞
𝑠→∞
𝐥𝐢𝐦 𝒔𝑭(𝒔) = 𝒇(𝟎)
(3)
𝒔→∞
Final Value Theorem
𝐥𝐢𝐦 𝒇(𝒕) = 𝐥𝐢𝐦 𝒔𝑭(𝒔)
𝒕→∞
𝒔→𝟎
∞
ℒ{𝑓 ′ (𝑡)} = ∫0 𝑒 −𝑠𝑡 𝑓 ′ (𝑡)𝑑𝑡 = 𝑠𝐹(𝑠) − 𝑓(0)
∞
lim ∫0 𝑒 −𝑠𝑡 𝑓 ′ (𝑡)𝑑𝑡 = lim [f(t)]k0 = lim [𝑓(𝑘) − 𝑓(0)]
𝑠→0
RHS
k→∞
𝑘→∞
lim[𝑠𝐹(𝑠) − 𝑓(0)]
𝑠→0
comparing both ,we have
𝐥𝐢𝐦[𝒔𝑭(𝒔)] = 𝐥𝐢𝐦[𝒇(𝒕)]
𝒔→𝟎
(4)
𝒕→∞
-6-
3. Laplace Transform of functions
Example 3. 1:
Determine the Laplace transform of the following functions
a) 𝑓(𝑡) = 𝑡 𝑛
Definition 3.1
Gamma function 𝚪(𝒏) is defined by the following integral
∞
𝚪(𝒏) = ∫𝟎 𝒆−𝒙 𝒙𝒏−𝟏 𝒅𝒙
∞
𝐹(𝑠) = ∫0 𝑡 𝑛 𝑒 −𝑠𝑡 𝑑𝑡 = Γ(𝑛 + 1)/𝑠 𝑛+1
b) 𝑓(𝑡) = cos 𝜔𝑡
∞ 𝑒 −𝑡(𝑠+𝑖𝜔)
∞
𝐹(𝑠) = ∫0 𝑒 −𝑠𝑡 cos 𝜔𝑡 𝑑𝑡 = ∫0
2
∞ 𝑒 −𝑡(𝑠−𝑖𝜔)
𝑑𝑡 + ∫0
2
c) 𝑓(𝑡) = sin 𝜔𝑡
∞
𝜔
𝐹(𝑠) = ∫0 sin 𝜔𝑡 𝑒 −𝑠𝑡 𝑑𝑡 = 2 2
𝑠 +𝜔
d) 𝑓(𝑡) = cosh 𝜔𝑡
∞
𝑠
𝐹(𝑠) = ∫0 cosh 𝜔𝑡 𝑒 −𝑠𝑡 𝑑𝑡 = 2 2
𝑠 −𝜔
∞
𝜔
f ) 𝑓(𝑡) = sinh 𝜔𝑡 ; 𝐹(𝑠) = ∫0 sinh 𝜔𝑡 𝑒 −𝑠𝑡 𝑑𝑡 = 2 2
𝑠 −𝜔
∞
1
g) 𝑓(𝑡) = 𝑒 𝑎𝑡 ; 𝐹(𝑠) = ∫0 𝑒 −𝑡(𝑠−𝑎) 𝑑𝑡 =
𝑠−𝑎
-7-
𝑑𝑡 =
𝑠
𝑠 2 +𝜔2
4. Properties of Laplace transform
a) Linearity property
ℒ{𝑐 𝑓(𝑡) + 𝑑 𝑔(𝑡)} = 𝑐ℒ{𝑓(𝑡)} + 𝑑ℒ{𝑔(𝑡)} with 𝑐 and 𝑑 as scalars.
𝓛{𝒄 𝒇(𝒕) + 𝒅 𝒈(𝒕)} = 𝒄𝑭(𝒔) + 𝒅𝑮(𝒔)
(5)
Example 4.1:
Determine the Laplace transform of the following function
𝑓(𝑡) = 𝑒 2𝑡 + 𝑠𝑖𝑛4𝑡 − 𝑐𝑜𝑠2𝑡
ℒ{𝑒 2𝑡 + 𝑠𝑖𝑛4𝑡 − 𝑐𝑜𝑠2𝑡} = ℒ{𝑒 2𝑡 } + ℒ{𝑠𝑖𝑛4𝑡} − ℒ{cos 2𝑡}
=
1
𝑠−2
+
4
𝑠 2 +16
−
𝑠
𝑠 2 +4
b) Shifting theorem: First Translational Property
𝓛{𝒆𝒂𝒕 𝒇(𝒕)} = 𝑭(𝒔 − 𝒂)
(6)
Proof: Applying the definition of Laplace transform, we have
∞
𝑎𝑡
ℒ{𝑒 𝑓(𝑡)} = ∫ 𝑒
∞
−𝑠𝑡
𝑎𝑡
𝑒 𝑓(𝑡)𝑑𝑡 = ∫ 𝑒 −(𝑠−𝑎)𝑡 𝑓(𝑡)𝑑𝑡 = 𝐹(𝑠 − 𝑎)
0
0
Multiplying a function by 𝒆𝒂𝒕 results in the shifting of the Laplace
transform along the 𝒔 axis by 𝒂 .
Example 4.1:
Evaluate ℒ{𝑒 3𝑡 𝑠𝑖𝑛4𝑡}
Since ℒ{𝑠𝑖𝑛4𝑡} =
ℒ{𝑒 3𝑡 𝑠𝑖𝑛4𝑡} =
4
𝑠 2 +16
, Thus applying the shifting theorem , we obtain,
4
(𝑠−3)2 +16
-8-
Example 4.2:
Evaluate ℒ{𝑐𝑜𝑠 2 𝑡}
𝑐𝑜𝑠 2 𝑡 = (1 + 𝑐𝑜𝑠2𝑡)/2 ; using the linearity property
𝑠
ℒ{𝑐𝑜𝑠 2 𝑡} =
2
+
𝑠
(𝑠 2 +4 2 )2
c) Second translation property
In various problems in engineering ,the dependent variables may have jumps
which results in the functions having discontinuities at these points. These
discontinuities can be represented mathematically by a unit step function 𝑈(𝑡 −
𝑎) ,𝑡 = 𝑎 being a point of discontinuity for the given function.
A unit step function is defined as
𝑈(𝑡 − 𝑎) = 1 𝑓𝑜𝑟 𝑡 ≥ 𝑎
(7)
𝑈(𝑡 − 𝑎) = 0 𝑓𝑜𝑟 𝑡 < 𝑎
ℒ{𝑓(𝑡 − 𝑎)𝑈(𝑡 − 𝑎)} = 𝑒 −𝑎𝑠 𝐹(𝑠)
Proof: Using integration by parts and the basic definition of LaplaceTransform
𝑎
∞
ℒ{𝑓(𝑡 − 𝑎)𝑈(𝑡 − 𝑎)} = ∫ 𝑓(𝑡 − 𝑎)0𝑒
−𝑠𝑡
𝑑𝑡 + ∫ 𝑓(𝑡 − 𝑎) 𝑒 −𝑠𝑡 𝑑𝑡
0
substitute
𝑡−𝑎 = 𝑥,
𝑎
𝑑𝑡 = 𝑑𝑥; the limits transform to (0, ∞ )and
∞
expression becomes ∫0 𝑓(𝑥)𝑒 −(𝑥+𝑎)𝑠 𝑑𝑥 = 𝑒 −𝑎𝑠 𝐹(𝑠)
The unit step function is also expressed as 𝑈(𝑡 − 𝑎) = 𝑈𝑎 (𝑡)
𝓛 {𝑼𝒂 } =
𝒆−𝒂𝒔
(8)
𝒔
-9-
the
Within a given interval if there are several discontinuities which can be
expressed as step functions, a linear combination of unit step functions can be
invoked to express these discontinuities. For example
𝑓(𝑡) = 𝑎, 𝑓𝑜𝑟 𝑡 𝑖𝑛 (0, 𝑡1 )
𝑓(𝑡) = 𝑏 , 𝑓𝑜𝑟 𝑡 𝑖𝑛(𝑡1 , 𝑡2 )
𝑓(𝑡) = 𝑐 𝑓𝑜𝑟 𝑡 𝑖𝑛 (𝑡2 , 𝑡3 ) and so on. The complete function 𝑓(𝑡) is expressed as
𝑓(𝑡) = 𝑎 + (𝑏 − 𝑎)𝑈𝑡1 + (𝑐 − 𝑏)𝑈𝑡2 + ⋯
Example 4.3:
Evaluate ℒ{(𝑡 − 1)𝑈(𝑡 − 1)}
comparing the expression, we obtain 𝑎 = 1, 𝑓(𝑡) = 𝑡 , 𝑡ℎ𝑢𝑠 𝐹(𝑠) =
substitution the result is 𝑒 −𝑠 /𝑠 2
Example 4.4:
Find the LT of 𝑐𝑜𝑠𝑡 𝑈(𝑡 − 2𝜋)
since the periodicity of 𝑐𝑜𝑠 function is 2𝜋 , thus cos 𝑡 = cos(𝑡 − 2𝜋)
thus ℒ{𝑐𝑜𝑠𝑡 𝑈(𝑡 − 2𝜋)} = ℒ{𝑐𝑜𝑠(t-2π)𝑈(𝑡 − 2𝜋)}
using the second translation property, and linearity property we obtain
ℒ{𝑐𝑜𝑠 (𝑡 − 2𝜋)𝑈(𝑡 − 2𝜋) }= 𝑠
𝑒 −2𝜋𝑠
𝑠 2 +1
Example 4.5:
Evaluate ℒ{𝑓(𝑡)},
𝑓(𝑡) = 0 for 0 ≤ 𝑡 ≤ 3;
𝑓(𝑡) = 𝑝, 𝑓𝑜𝑟 𝑡 ≥ 3
- 10 -
1
𝑠2
and on
This function has a discontinuity at 𝑡 = 3 .
using the first principles we obtain the LT of the given function,
∞
3
∫0 𝑓(𝑡)𝑑𝑡𝑒 −𝑠𝑡 =∫0 0𝑑𝑡
∞
+ ∫3 𝑒 −𝑠𝑡 𝑝 𝑑𝑡 = 𝑝
𝑒 −3𝑠
𝑠
;𝑠 > 0
Example 4.6
A continuous electric pulse which is switched on only for 2 s say between 𝑡 = 3
and 𝑡 = 5 can be modelled using step functions. Assuming that the pulse is of 4
volts.
𝑓(𝑡) = 0 𝑖𝑛 (0,3) and (5, ∞)
𝑓(𝑡) = 4 𝑖𝑛 (3,5)
𝑓(𝑡) = 0 + (4 − 0)𝑈3 (𝑡) + (0 − 5)𝑈5 (𝑡)
ℒ {𝑓(𝑡)} = 4𝑈3 (𝑡) − 5𝑈5 (𝑡) 4
𝑒 −3𝑠
𝑠
−5
𝑒 −5𝑠
𝑠
5. Laplace transform of derivatives
∞
∞
−𝑠𝑡
ℒ [𝑓 ′ (𝑡)] = ∫0 𝑓 ′ (𝑡)𝑒 −𝑠𝑡 𝑑𝑡 = [ 𝑓(𝑡)𝑒 −𝑠𝑡 ]∞
𝑑𝑡
0 - ∫0 𝑓(𝑡)(−𝑠)𝑒
with 𝑓(𝑡) bounded , the upper limit → 0 in the first term.
𝓛 [𝒇′ (𝒕)] = 𝒔 𝑭(𝒔) − 𝒇(𝟎)
(9)
ℒ [𝑓 ′′ (𝑡)] = 𝑠 𝐺(𝑠) − 𝑔(0) = 𝑠[𝑠 𝐹(𝑠) − 𝑓(0)] − 𝑔(0) with
𝑔(𝑡) = 𝑓 ′ (𝑡) ; and 𝐺(𝑠) = ℒ [𝑔(𝑡)] . Thus
𝓛 [𝒇′′ (𝒕)] = 𝒔𝟐 𝑭(𝒔) − 𝒔 𝒇(𝟎) − (𝒇′ (𝒕))𝒕=𝟎
- 11 -
(10)
This property is particularly useful for converting ordinary differential equations
into algebraic equations.
6. Laplace transforms of integrals
𝐼𝑓 ℒ{𝑓(𝑡)} = 𝐹(𝑠),
𝒙
𝓛{∫𝟎 𝒇(𝒕)𝒅𝒕} =
𝑭(𝒔)
(11)
𝒔
The result of the integration will be some function of 𝑥 ,say 𝑝(𝑥)
𝑥
Let ∫0 𝑓(𝑡)𝑑𝑡 = 𝑝(𝑥); then 𝑝(0) = 0 𝑎𝑛𝑑 𝑝′ (𝑥) = 𝑓(𝑥)
Considering the Laplace transform ℒ {𝑝′ (𝑥)} = ℒ{𝑓(𝑥)}
or 𝑠𝑃(𝑠) − 𝑝(0) = 𝐹(𝑠) or
𝑥
𝑠ℒ{∫0 𝑓(𝑡)𝑑𝑡} = 𝐹(𝑠)
Example 6.1:
𝑥
Evaluate ℒ{∫0 𝑒 −2𝑡 𝑑𝑡}
Applying the above theorem , 𝑓(𝑡) = 𝑒 −2𝑡 , ℒ {𝑒 −2𝑡 } =
𝑥
Thus ℒ{∫0 𝑒 −2𝑡 𝑑𝑡} =
1
𝑠(𝑠+2)
- 12 -
1
𝑆+2
CHAPTER 2
Inverse Laplace transform
1. Definition
Inverse Laplace transform is an essential
operation for all applications of
Laplace transformation. This transformation is used for the retrieval of the
original function.
ℒ −1 𝐹(𝑠) = 𝑓(𝑡)
(1)
The inverses of functions can be obtained by inspection. We discuss the LT
inverse of functions which are not so obvious There are various methods of
determining the inverse Laplace transform of a function.
2. Properties of inverse Laplace transforms
. a) First Shifting theorem : 𝓛−𝟏 𝑭(𝒔 − 𝒂) = 𝒆𝒂𝒕 𝒇(𝒕)
∞
∞
𝐹(𝑠 − 𝑎) = ∫ 𝑓(𝑡)𝑒 −(𝑠−𝑎)𝑡 𝑑𝑡 = ∫ [𝑓(𝑡)𝑒 𝑎𝑡 ]𝑒 −𝑠𝑡 𝑑𝑡 = ℒ{𝑓(𝑡)𝑒 𝑎𝑡 }
0
0
𝓛−𝟏 {𝑭(𝒔 − 𝒂)} = 𝒆𝒂𝒕 𝒇(𝒕)
(2)
b) Second shifting theorem: 𝓛−𝟏 {𝒆−𝒂𝒔 𝑭(𝒔)} = 𝒇(𝒕 − 𝒂)𝑼(𝒕 − 𝒂)
∞
∞
𝑒 −𝑎𝑠 𝐹(𝑠) = 𝑒 −𝑎𝑠 ∫0 𝑓(𝑡)𝑒 −𝑠𝑡 𝑑𝑡 = ∫0 𝑓(𝑡)𝑒 −𝑠(𝑡+𝑎) 𝑑𝑡
We substitute 𝑡 + 𝑎 = 𝑥 ; 𝑑𝑡 = 𝑑𝑥 . The limit of 𝑥 is from 𝑎 to ∞.
∞
∴ 𝑒 −𝑎𝑠 𝐹(𝑠) = ∫𝑎 𝑓(𝑥 − 𝑎)𝑑𝑥
Using the property of the unit step function,
- 13 -
𝑈(𝑡 − 𝑎) = 1 for 𝑡 > 𝑎 , 𝑈(𝑡 − 𝑎) = 0 for 𝑡 < 𝑎
∞
∞
∫𝑎 𝑓(𝑥 − 𝑎)𝑑𝑥 = ∫0 𝑓(𝑥 − 𝑎) 𝑈(𝑥 − 𝑎)𝑑𝑥
𝑎
as ∫0 𝑓(𝑥 − 𝑎)𝑈(𝑥 − 𝑎) 𝑑𝑥 = 0
∞
𝒆−𝒂𝒔 𝑭(𝒔) = ∫𝟎 𝒇(𝒙 − 𝒂) 𝑼(𝒙 − 𝒂)𝒅𝒙
c)
𝓛−𝟏 {
𝑭(𝒔)
𝒔
(3)
𝒕
} = ∫𝟎 𝒇(𝒗)𝒅𝒗
Example: 2.1
Determine the Laplace inverse (ℒ −1 { 𝑒 −2𝑠 𝑠 −3 })
Using the second shifting property ,
𝑎 = 2; 𝐹(𝑠) =
1
𝑠
;𝑓(𝑡) =
3
1 2
𝑡
2
;
∴ Applying ℒ −1 {𝑒 −𝑎𝑠 𝐹(𝑠)} = 𝑓(𝑡 − 𝑎)𝑈(𝑡 − 𝑎)
1
ℒ −1 { 𝑒 −2𝑠 𝑠 −3 } = (𝑡 − 2)2 for 𝑡 ≥ 2
2
= 0 for 0 ≤ 𝑡 < 2
Example 2:
Determine ℒ −1 {
1−𝑒 −4𝑠
𝑠2
}
Applying the linearity property
ℒ −1 {
1−𝑒 −4𝑠
𝑠2
1
𝑒 −4𝑠
𝑠
𝑠2
} = ℒ −1 { 2} − ℒ −1 {
} = 𝑡 − (𝑡 − 4)𝑈(𝑡 − 4)
- 14 -
= 4 for 𝑡 ≥ 4 ; for 0 ≤ 𝑡 < 4, ℒ −1 {
1−𝑒 −4𝑠
𝑠2
}=𝑡
Example 3:
2
Determine ℒ −1 {(𝑠−4)2 }
+4
𝐹(𝑠) =
2
; 𝑓(𝑡) = 𝑠𝑖𝑛 2𝑡 ;
𝑠 2 +22
Applying the first shifting theorem , we obtain after substituting 𝑎 = 4,
2
ℒ −1 {(𝑠−4)2 } = 𝑒 4𝑡 sin 2𝑡
+4
Example 4:
𝑠−1
ℒ −1 {
𝑠 2 −4𝑠+8
=
𝑠
𝑠 2 −4𝑠+8
−
}
1
𝑠 2 −4𝑠+8
=
𝑠
(𝑠−2)2 +4
1
− (𝑠−2)2
+4
= 𝑒 2𝑡 cos 2𝑡 − 𝑒 2𝑡 sin 2𝑡
Example 5:
Determine ℒ −1 {
ℒ −1 {
𝐹(𝑠)
𝑠
1
𝑠(𝑠−4)
}
𝑡
𝑡
1
} = ∫0 𝑓(𝑣)𝑑𝑣 = ∫0 𝑒 4𝑣 𝑑𝑣 = [𝑒 4𝑡 − 1]
4
3. Methods for determination of inverse Laplace transforms
a) Partial fraction method
If the given function 𝐹(𝑠) is of the form 𝑁(𝑠)/𝑄(𝑠) with 𝑁(𝑠) and 𝑄(𝑠) as
polynomials in 𝑠 and with the degree of 𝑁(𝑠) < degree of 𝑄(𝑠) we have the
following forms for the partial fractions.
𝑁(𝑠)
=
(𝑠−𝑎)(𝑠−𝑏)
𝐴
(𝑠−𝑎)
+
𝐵
(4)
(𝑠−𝑏)
- 15 -
𝑁(𝑠)
(𝑠−𝑎)𝑛
=
𝑁 (𝑠)
(𝑠 2 +𝑏𝑠+𝑐)
𝐴
𝑠−𝑎
=
𝑁(𝑠)
(𝑠 2 +𝑏𝑠+𝑐)𝑚
+
𝐵
(𝑠−𝑎)2
+
𝐶
(𝑠−𝑎)3
+ ⋯+
𝐷
(5)
(𝑠−𝑎)𝑛
𝐴𝑠+𝐵
(6)
(𝑠 2 +𝑏𝑠+𝑐)
=
𝐴𝑠+𝐵
(𝑠 2 +𝑏𝑠+𝑐)
+
𝐷𝑠+𝐸
(𝑠 2 +𝑏𝑠+𝑐)2
+. . +
𝐹𝑠+𝐺
(𝑠 2 +𝑏𝑠+𝑐)𝑛
(7)
Our assumptions are that 𝑁(𝑠)and 𝑄(𝑠) do not have any common factors and
the degree of 𝑁(𝑠) > degree of 𝑁(𝑠).
𝑄(𝑠) can be written either in the linear factor form as (𝑠 − 𝑎)𝑚 with 𝑎 real.
The other possibility is to express 𝑑(𝑠) in terms of powers of an irreducible
quadratic factor (𝑠 2 + 𝑏𝑠 + 𝑐) with 𝑏 and 𝑐 real. In algebra , irreducible factor
implies that it cannot be written as (𝑠 − 𝑎)𝑚 with 𝑚 as a positive integer.
Example 3.1:
Determine the inverse LT of
3𝑠+2
(𝑠−1)(𝑠 2 +1)
=
𝐴𝑠+𝐵
(𝑠 2 +1)
put 𝑠 = 1 , 𝐶 =
5
2
+
𝐶
𝑠−1
3𝑠+2
(𝑠−1)(𝑠 2 +1)
⇒ (𝐴𝑠 + 𝐵)(𝑠 − 1) + 𝐶(𝑠 2 + 1) = 3𝑠 + 2
putting 𝑠 = 0,
−𝐵 +
5
2
= 2 𝑜𝑟 𝐵 =
1
2
1
comparing the coefficient of 𝑠 on both the sides we obtain − 𝐴 + = 3 or 𝐴 =
2
−5
2
;
On substitution of the values of the constants, we obtain
−1
−5𝑠 1
+
2
2
𝑠 2 +1
𝐿 {
}+
5
2
𝑠−1
}=
1
2
5
5
2
2
𝑠𝑖𝑛𝑡 − 𝑐𝑜𝑠𝑡 + 𝑒 𝑡
- 16 -
Example 3.2:
𝑠−4
Determine ℒ −1 {{𝑠2
{
𝑠−4
𝑠{𝑠 2 +2}
𝐴𝑠+𝐵
}=
𝑠 2 +2
+2}
+
𝐶
𝑠
} using partial fraction decomposition.
; 𝑠(𝐴𝑠 + 𝐵) + (𝑠 2 + 2)𝐶 = 𝑠 − 4 ; 𝑠 = 0 gives
comparing coefficients of various powers of 𝑠 results in
𝐵 = 1; 𝐴 = 2; 𝐶 = −2;
The above problem reduces to ℒ −1 {
2 cos √2 𝑡 +
1
√2
2𝑠
𝑠 2 +2
+
1
𝑠 2 +2
2
− }
𝑠
𝑠𝑖𝑛√2 𝑡 − 2
Example 3.3:
Evaluate using partial fraction decomposition
2𝑠+3
=
𝑠 2 +𝑠−2
𝐴
(𝑠+2)
+
𝐵
(𝑠−1)
; 𝐴(𝑠 − 1) + 𝐵(𝑠 + 2) = 2𝑠 + 3 ;
𝐴 + 𝐵 = 2; −𝐴 + 2𝐵 = 3; 𝐴 =
ℒ −1 {
2𝑠+3
𝑠 2 +𝑠−2
1
5
3
3
1
3
; 𝐵=
5
3
} = 𝑒 −2𝑡 + 𝑒 𝑡
b. Heaviside Expansion :
A quick method for inverse Laplace transform determination
In the partial fraction method, if an unrepeated factor of 𝑄(𝑠) is 𝑠 − 𝑎, then that
results in a term like
𝐴
𝑠−𝑎
. If 𝑃(𝑠) denotes the sum of all the fractions with all
factors of 𝑄(𝑠) except (𝑠 − 𝑎) then
𝑁(𝑠)
𝑄(𝑠)
=
𝐴
𝑠−𝑎
+ 𝑃(𝑠)
- 17 -
𝑁(𝑠)(𝑠−𝑎)
𝑄(𝑠)
= 𝐴 + (𝑠 − 𝑎)𝑃(𝑠) ;
We next consider lim[ 𝐴 + (𝑠 − 𝑎)𝑃(𝑠)] = 𝐴 .Therefore ,
𝑠→𝑎
𝐴 = lim {
𝑁(𝑠)
} ; Since (𝑠 − 𝑎) is a factor of 𝑄(𝑠), Thus lim
𝑠→𝑎 𝑄(𝑠)/(𝑠−𝑎)
𝑄(𝑠)
𝑠→𝑎 𝑠−𝑎
= 0/0
form. Using L'Hospital's rule this reduces to [𝑄′ (𝑠)]𝑠=𝑎
thus 𝐴 = {lim
𝑠→𝑎
where 𝑄′ (𝑠) =
𝑁(𝑠)
𝑄′ (𝑠)
𝑑
𝑑𝑠
} and ℒ −1 {
𝐴
𝑠−𝑎
}=
𝑁(𝑎)
𝑄′ (𝑎)
𝑄(𝑠) ; ℒ −1 {𝐹(𝑠)} =
𝑒 𝑎𝑡
𝑁(𝑎)
𝑄′ (𝑎)
𝑒 𝑎𝑡 + ℒ −1 {𝑃(𝑠)}
If 𝑄(𝑠) has another factor (𝑠 − 𝑏) , then ℒ −1 {𝐹(𝑠)} will have another term with
𝑁(𝑏)
𝑄′ (𝑏)
𝑒 𝑏𝑡 and in general if there are 𝑖 factors of denominator 𝑄(𝑠), factorized as
((𝑠 − 𝑎1 )(𝑠 − 𝑎2 )(𝑠 − 𝑎3 ). . (𝑠 − 𝑎𝑛 )
𝓛−𝟏 {𝑭(𝒔)} = ∑𝒏𝒊=𝟏
𝑵(𝒂𝒊 )
𝑸′ (𝒂𝒊 )
𝒆𝒂𝒊𝒕 + 𝓛−𝟏 {𝑷(𝒔)}
(8)
Example 3.4:
Determine 𝑓(𝑡), given 𝐹(𝑠) =
𝑠 2 +1
𝑠(𝑠+1)(𝑠+2)
The zeros of the denominator are , 𝑠 = 0, 𝑠 = −1, 𝑠 = −2 ;
𝑁(𝑠) = 𝑠 2 + 1;
𝑄(𝑠) = 𝑠 3 + 3𝑠 2 + 2𝑠 ; 𝑄′ (𝑠) = 3𝑠 2 + 6𝑠 + 2
We now determine the values of 𝑁(𝑠) 𝑎𝑛𝑑 𝑄′ (𝑠) at these values.
𝑁(0) = 1; 𝑁(−1) = 2; 𝑁(−2) = 5
𝑄′ (0) = 2; 𝑄′(−1) = −1; 𝑄′ (−2) = 2
- 18 -
𝑓(𝑡) = ∑𝑘𝑖=1
𝑁(𝑎𝑖 )
𝑄′ (𝑎
𝑖)
1
2
2
−1
𝑒 𝑎𝑖 𝑡 = +
5
𝑒 −𝑡 + 𝑒 −2𝑡
2
Example 3.5:
Evaluate using Heaviside formula ℒ −1 {
𝑠−3
𝑠−3
(𝑠 2 +4)(𝑠+7)
}
1
(𝑠 2 +4)(𝑠+7)
= 𝐴ℒ −1 {(𝑠+7)} + 𝑃(𝑠)
𝑁(𝑠) = 𝑠 − 3; 𝑁(𝑠 = −7) = −10 ; 𝐷(𝑠) = 𝑠 3 + 7𝑠 2 + 4𝑠 + 28
𝐷′ (𝑠) = 3𝑠 2 + 14𝑠 + 4 ; 𝐷′ (𝑠) 𝑎𝑡 𝑠 = −7 is 53
Thus one term will be
−10
53
𝑒 −7𝑡 ;
The other roots of 𝐷(𝑠) are 𝑠 = 2𝑖, 𝑠 = −2𝑖
at 𝑠 = 2𝑖, 𝐷′ (𝑠) = −8 + 28 i , 𝑁(𝑠) = 2𝑖 − 3
−3+2𝑖
−8+28𝑖
𝑒 2𝑖𝑡
is another term.
The third term is at the root 𝑠 = −2𝑖
𝐷′ (𝑠) = −8 − 2𝑖
ℒ −1 {
𝑠−3
}=
;
−10
(𝑠 2 +4)(𝑠+7)
53
𝑒 −7𝑡 +
𝑁(𝑠) = −2𝑖 − 3 ; the
−3+2𝑖
−8+28𝑖
𝑒 2𝑖𝑡 +
−3−2𝑖
−8−28𝑖
term is
−3−2𝑖
−8−2𝑖
𝑒 −2𝑖𝑡
𝑒 −2𝑖𝑡
The 2nd and 3rd terms are the complex conjugates of each other. If 𝑍 is a
complex number with its complex conjugate as 𝑍 ∗ ,then
(𝑍 + 𝑍 ∗ )/2 = real part of 𝑍 .
−3+2𝑖
−8+28𝑖
−10
53
𝑒 2𝑖𝑡 +
−3−2𝑖
−8−28𝑖
𝑒 −7𝑡 + 𝑅𝑒 {
𝑒 −2𝑖𝑡 = Re {
40−34𝑖
212
𝑒 2𝑖𝑡 } =
−3+2𝑖
−4+14𝑖
−10
53
𝑒 2𝑖𝑡 }
𝑒 −7𝑡 +
- 19 -
10
53
cos 2𝑡 −
17
106
sin 2𝑡
If the denominator has terms like (𝑠 − 𝑎)𝑘 . We invoke residue theorem to
determine the inverse Laplace transform.
Without getting into the complexities we just state that the formula for getting
the inverse laplace transforms is
If 𝑭(𝒔) =
𝑵(𝒔)
;
(𝒔−𝒂)𝒌
𝒇(𝒕) = 𝐥𝐢𝐦
𝒔→𝒂
𝒅𝒌−𝟏
𝟏
𝑵(𝒔)
(𝒌−𝟏)! 𝒅𝒔𝒌−𝟏
{(𝒔 − 𝒂)𝒌 [𝒆𝒔𝒕 (𝒔−𝒂)𝒌]}
(9)
This is an extremely efficient method of determining the inverse Laplace
transform.
Example 3.5:
−3𝑠−2
Evaluate ℒ −1 {(𝑠+4)2 } =𝑓(𝑡) using the residue theorem formula as above.
Comparing with eq.(3.5) , we obtain 𝑘 = 2; 𝑎 = −4
𝑓(𝑡) = lim
1 𝑑
𝑠→−4 1! 𝑑𝑠
[𝑒 𝑠𝑡
−3𝑠−2
1
] = lim [𝑡𝑒 𝑠𝑡 (−3𝑠 − 2) − 3𝑒 𝑠𝑡 ]
𝑠→−4
10𝑡𝑒 −4𝑡 − 3𝑒 −4𝑡
4. More inverse Laplace transforms
Example 4.1. Determine the inverse Laplace transform of 𝑒 −√𝑠 .
Expressing 𝑒 −√𝑠 as an infinite series , we have
𝑒 −√𝑠 = {1 −
𝑠 1/2
Since ℒ{𝑡 𝑛 } =
1
+
(𝑠 1/2 )2
Γ(𝑛+1)
𝑠 𝑛+1
2!
−
(𝑠 1/2 )3
3!
+
(𝑠 1/2 )4
4!
𝑜𝑟 ℒ −1 (𝑠 𝑛+1 )−1 =
- 20 -
− ⋯}
𝑡𝑛
Γ(𝑛+1)
(10)
1
1
1
1
1
Γ ( ) = − Γ (− ) and Γ ( ) = √𝜋 Thus Γ (− ) = −2√𝜋
2
2
2
2
2
1
3
3
3
−2
Γ (− ) = − Γ (− ) or Γ (− ) =
2
2
2
2
5
5
3
−2
− Γ (− ) = Γ (− ) ,
2
2
2
5
−8
Γ (− ) =
2
15
5
3
4
(−2)√𝜋 = √𝜋
3
3
5
Γ (− ) = Γ (− ) or
2
2
√𝜋
Now Γ(1) = 0Γ(0) but by the basic definition of the gamma function , we have
Γ(1) = 1 and ∴ Γ(0) is infinite.
Similarly Γ(0) = −1. Γ(−1) , thus Γ(−1) is infinite. Similarly we can prove
that Γ(−2) is infinite and for all positive integers 𝑛 , Γ(−𝑛) is infinite.
1
22 2 2
2
2
13 5 7
2𝑛+1
Γ(−n − ) = (−1)𝑛+1 . . . …
√𝜋
(11)
3
We need to determine ℒ
ℒ{𝑡 𝑛 } =
or ℒ
−1
Γ(n+1)
𝑠 𝑛+1
{𝑠
𝑝+
1
2
−1
{1 − √𝑠 +
𝑠
2!
−
𝑠2
3!
+
𝑠2
4!
, putting 𝑛 = −𝑝 − 3/2 , or ℒ {
}=
5
𝑠2
−
𝑡
5!
+⋯}
3
−𝑝−
2
1
Γ(−p−2)
𝑡
3
−𝑝−
2
1
Γ(−p−2)
1 35
= . . …
2 22
2𝑝+1 (−1)𝑝+1
2
√𝜋
𝑡
−𝑝−
} = s p+1/2
3
2
(12)
.𝑡 −5/2
(13)
Using eqn(10) and eqn(11) , we obtain
3
𝑝 = 0, ℒ
−1
√𝑠 =
−
𝑡 2
1
Γ(−2)
3
=
−
𝑡 2
(−2√𝜋)
3
1 3
, ℒ −1 {𝑠 2 } = . .
1
2 2 √𝜋
For all positive 𝑝 𝑎𝑛𝑑 𝑝 = 0 𝑤𝑖𝑡ℎ 𝑝 ∈ 𝐼 , ℒ −1 {𝑠 𝑝 } = 0
ℒ −1 {𝑠} = ℒ −1 {𝑠 2 } = ℒ −1 {𝑠 3 } = ℒ −1 {𝑠 4 } = 0 ;
- 21 -
Substituting the values from eqn (3) , we get
3
ℒ −1 {𝑒 −√𝑠 } =
−
𝑡 2
(2√𝜋)
1 3
− . .
1
2 2 √𝜋
.𝑡 −5/2
1
3!
1 3 5
+ . . .
1
2 2 2 √𝜋
𝑡 −7/2
𝟑 −𝟏
−𝟏
𝓛
{𝒆−√𝒔 }
=
−
𝒕 𝟐 𝒆 𝟒𝒕
(14)
𝟐√𝝅
5. Convolution theorem for Laplace transforms
Statement: The Laplace transform of the convolution of 2 functions is the
product of their Laplace transforms.
𝒕
𝑭(𝒔)𝑮(𝒔) = 𝓛 {∫𝟎 𝒇(𝒗)𝒈(𝒕 − 𝒗)𝒅𝒗}
Proof:
∞
LHS 𝐹(𝑠)𝐺(𝑠) = 𝐹(𝑠) ∫0 𝑔(𝑣)𝑒 −𝑠𝑣 𝑑𝑣 , 𝐹(𝑠)𝑒 −𝑠𝑣 is the Lapalce transform of
𝑓(𝑡 − 𝑣)𝐻(𝑡 − 𝑣) where 𝐻(𝑡 − 𝑣) is the unit Heaviside function which is
defined in eq ( 4.3 ) chapter 1.
∞
∞
LHS is ∫0 𝑓(𝑡 − 𝑣)𝑈(𝑡 − 𝑣)𝑒 −𝑠𝑡 𝑑𝑡 ∫0 𝑔(𝑣)𝑑𝑣
Changing the order of integration, we obtain
∞
∞
∫0 𝑒 −𝑠𝑡 𝑑𝑡 ∫0 𝑓(𝑡 − 𝑣)𝑈(𝑡 − 𝑣)𝑔(𝑣)𝑑𝑣
𝐻(𝑡 − 𝑣) = 1 for 𝑡 > 𝑣 or 𝑣 < 𝑡 and 𝑈(𝑡 − 𝑣) = 0 for 𝑣 > 𝑡 , we thus change
our limit of integration from 0 to ∞ to 0 to 𝑡.
∞
𝑡
𝑡
LHS is ∫0 𝑒 −𝑠𝑡 𝑑𝑡[∫0 𝑓(𝑡 − 𝑣)𝑔(𝑣)𝑑𝑣] =ℒ ∫0 𝑓(𝑡 − 𝑣)𝑔(𝑣)𝑑𝑣 = RHS
Example 1:
𝑡
Determine the Laplace transform of ∫0 𝑔(𝑢)𝑑𝑢 using the convolution theorem.
- 22 -
Applying the convolution theorem and putting 𝑓(𝑡 − 𝑢) = 1
1
thus 𝐹(𝑠) = and 𝐺(𝑠) = ℒ{𝑔(𝑢)} we obtain
𝑠
𝑡
ℒ ∫0 𝑔(𝑢)𝑑𝑢 = 𝐹(𝑠) 𝐺(𝑠) =
1
𝑠
𝐺(𝑠)
(DU)
Example 2:
Determine the inverse Laplace transform of
𝐹(𝑠) =
1
𝑠
; 𝐺(𝑠) =
𝑎2
𝑠 2 +𝑎2
𝑎2
𝑠(𝑠 2 +𝑎2 )
; Thus 𝑓(𝑢) = 1; 𝑔(𝑢) = 𝑎 𝑠𝑖𝑛𝑎𝑢
Applying Convolution theorem , we obtain
𝑎2
ℒ −1 {
𝑡
} = ∫0 𝑎 𝑠𝑖𝑛𝑎𝑢 𝑑𝑢 = (1 − 𝑐𝑜𝑠𝑎𝑡)
(DU)
𝑠(𝑠 2 +𝑎2 )
Example 3:
Show that the inverse Laplace transform of
𝑒 −2𝑠
𝑠2
is (𝑡 − 2)𝑈(𝑡 − 2) with 𝑈 as
the unit step function.
ℒ −1 {𝐹(𝑠)𝑒 −𝑎𝑠 } = 𝑓(𝑡 − 𝑎)𝑈(𝑡 − 𝑎) eq.(2.2) Chapter 2
Comparing this with the given problem, 𝐹(𝑠) =
1
𝑠2
;𝑎 = 2
𝑓(𝑡) = 𝑡 and ℒ −1 {𝐹(𝑠)𝑒 −𝑎𝑠 } = (𝑡 − 2)𝑈(𝑡 − 2)
Example 4:
Evaluate ℒ −1 {(𝑠2
Let (𝑠2
𝑠
+𝑎2 )
𝑠2
} using Convolution theorem
+𝑎2 )(𝑠 2 −𝑐 2 )
= 𝐹(𝑠);
𝑠
(𝑠 2 −𝑐 2 )
= 𝐺(𝑠)
- 23 -
ℒ −1 {(𝑠2
𝑠2
𝑡
+𝑎2 )(𝑠 2 −𝑐 2 )
} = ∫0 cosh 𝑐𝑣 cos (𝑡 − 𝑣) 𝑑𝑣
expressing cosh 𝑐𝑣 =
𝑒 𝑐𝑣+𝑒 −𝑐𝑣
and cos(𝑡 − 𝑣) =
2
The above integration reduces to
1
𝑎2 +𝑐 2
𝑒 𝑖(𝑡−𝑣)+𝑒 −𝑖(𝑡−𝑣)
2
[asin 𝑎𝑡 + 𝑐 sinh 𝑐𝑡]
Example 5:
Using convolution theorem , prove ℒ −1 {
𝐹(𝑆)
𝑆
𝑡
} = ∫0 𝑓(𝑣)𝑑𝑣
1
In the convolution theorem let 𝐺(𝑠) = ; Thus 𝑔(𝑣) = 1, 𝑔(𝑡 − 𝑣) = 1
𝑠
Interchanging 𝐹(𝑆) and 𝐺(𝑠) ,convolution theorem being commutative we
𝑡
𝑡
have ℒ −1 {𝐹(𝑠)𝐺(𝑠)} = ∫0 𝑓(𝑣)𝑔(𝑡 − 𝑣)𝑑𝑣 = ∫0 𝑓(𝑣)𝑑𝑣
Exercises Chapter 2
1
𝑒 −2𝑠
𝑠
𝑠(1−𝑒 −𝑠 )
1. Determine ℒ −1 { 2 +
𝑠 𝑒 −4𝑠
2. Determine ℒ −1 {
𝑠 2 −9
3. Determine ℒ −1 {
}
};𝑠>3
𝑠+6
}
𝑠 2 +2𝑠−8
𝑠
4. Determine ℒ −1 { (𝑠+1)2}
5. Determine the inverse Laplace transform of 𝐹(𝑠).
𝐹(𝑠) = ln ( 1 +
6. Evaluate ℒ −1 {
𝑘2
𝑠2
)
3
𝑠 2 (𝑠 2 +𝑎2 )
}
- 24 -
7. Find ℒ −1 {
𝑎2 𝑠
𝑠 4 −𝑎4
}
8. Determine the convolution 𝑓(𝑦) ∗ 𝑔(𝑦),
given 𝑓(𝑦) = 𝑦 2 𝑎𝑛𝑑 𝑔(𝑦) = 𝑦
9. Using convolution theorem find ℒ −1 {
1
2
𝑠
(𝑠 2 +16)
}
𝑠
10. Using convolution theorem , determine ℒ −1 {(𝑠+1)(𝑠2
11. Evaluate ℒ −1 {
2
𝑠 3 (𝑠 2 +2𝑠)
}
- 25 -
+1)
}
CHAPTER 3
Applications of Laplace transforms
1. Solving initial value problems: solutions of linear differential equations
with constant coefficients
step 1: Perform Laplace transform on both sides of the differential equation.
step 2: The initial value problem is converted to an algebraic equation.
step3: Algebraic equation is solved.
step4: Consider the inverse Laplace transform of this solution
step 5: This is the solution of the initial value problem.
Example 1:
Solve the initial value problem which is a first order differential equation
𝑦 ′ (t) + 5 y(t) = t ; y(t = 0) = 2;
Considering Laplace transform on both the sides and applying the formula for
Laplace transform of a derivative , we obtain
𝑠𝑌(𝑠) − 2 + 5𝑌(𝑠) =
(𝑠 + 5)𝑌(𝑠) =
𝑌(𝑠) =
1
𝑠2
1
(𝑠+5)𝑠 2
1
𝑠2
....... step 1
+2
+
2
𝑠+5
....... step 2
We solve for 𝑌(𝑠) using partial fractions.
1
(𝑠+5)𝑠 2
𝐴
𝐵
𝑠
𝑠2
= +
+
1
𝐴 𝑠 (𝑠 + 5) + 𝐵(𝑠 + 5) + 𝐶𝑠 2 = 1 ; putting 𝑠 = 0 , 𝐵 = ;
5
- 26 -
𝐶
𝑠+5
putting 𝑠 = −5, 𝐶 =
or 𝐴 =
𝑌(𝑠) =
−1
25
1
6
1
5
25
; putting 𝑠 = 1, 6𝐴 + +
25
=1
; Substituting the values of 𝐴, 𝐵, 𝐶
−1
25𝑠
+
1
5𝑠 2
+
1
25(𝑠+1)
........ step3
𝑦(𝑡) = ℒ −1 [𝑌(𝑠)]
𝑦(𝑡) = −
1
25
1
1
5
25
+ 𝑡+
𝑒 −5𝑡
........step4
Example 2:
We now solve 𝑦 " (𝑡) + 2 𝑦 ′ (𝑡) + 6 = 𝑡 2
𝑦(𝑡 = 0) = 2; 𝑦 ′ (𝑡 = 0) = 0
ℒ {𝑦 " (𝑡) + 2𝑦 ′ (𝑡) + 6} = ℒ{𝑡 2 }
6
2
𝑠
𝑠3
𝑠 2 𝑌(𝑠) − 𝑠𝑦(𝑡 = 0) − 𝑦 ′ (𝑡 = 0) + 2𝑠𝑌(𝑠) − 4 + =
6 2
𝑌(𝑠)(𝑠 2 + 2𝑠) = 2𝑠 + 4 − + 3
𝑠 𝑠
𝑌(𝑠) =
𝑦(𝑡) = 2 − (3𝑡 +
3𝑒 −2𝑡
2
2𝑠 + 4
6
2
−
+
𝑠 2 + 2𝑠 𝑠(𝑠 2 + 2𝑠) 𝑠 3 (𝑠 2 + 2𝑠)
3
1
𝑡
𝑡2
2
8
4
4
− ) + 𝑒 −2𝑡 + −
+
𝑡3
6
−
1
8
The significance of Laplace transform for solving differential equations is that it
can be used
a) For discontinuous forcing functions while the conventional method cannot be
used for solving such differential equations.
- 27 -
b) Initial conditions are used as a part of the transformation from differential to
algebraic equation unlike the conventional method which uses these initial
conditions for evaluating the constants for complete solution of the differential
equations. For example consider the conventional simple harmonic motion with
damping. The differential equation representing it is
𝑚
𝑑2𝑥
𝑑𝑡 2
+𝑏
𝑑𝑥
𝑑𝑡
+ 𝑘𝑥 = 𝑓(𝑡)
With 𝑚, 𝑏, 𝑘 respectively as the mass ,damping constant and the force constant.
If the forcing term 𝑓(𝑡) is not a continuous function, it is still possible to solve
this differential equation using Laplace transforms.
Example 3:
Solve the following initial value problem using Laplace transform.
𝑑2𝑥
𝑑𝑡 2
ℒ{
+ 4𝑥 = 𝑠𝑖𝑛3𝑡, 𝑥(0) = 𝑥 ′ (0) = 0 ; 𝑥 ′ (0) =
𝑑2𝑥
𝑑𝑡 2
𝑑𝑥
𝑑𝑡
+ 4𝑥} = ℒ {𝑠𝑖𝑛3𝑡}
𝑠 2 𝑋(𝑠) + 4𝑋(𝑠) − 𝑠𝑥(0) − 𝑥 ′ (0) =
Thus 𝑥(𝑡) = ℒ −1 {(𝑠2
𝑥(𝑡) =
3
10
2
+4)(𝑠 2 +9)
3
𝑠 2 +9
; 𝑋(𝑠) = (𝑠2
3
+4)(𝑠 2 +9)
;
}; Using partial fraction decomposition we obtain
1
𝑠𝑖𝑛2𝑡 − sin 3𝑡
5
Physical significance: The above differential equation represents the motion of a
simple harmonic motion with an external driving force. It could represent a
spring mass system which is subjected to an external force 𝑠𝑖𝑛 3𝑡 . The
resultant displacement is represented by 𝑥(𝑡) which represents a motion with
natural frequency (𝜔 = 2) and the forcing term (𝜔 = 3).
- 28 -
Example 4:
We next a differential equation for a spring mass system with damping.
𝑥 ′′ − 3𝑥 ′ + 2𝑥 = 0 ; 𝑥(0) = 0; 𝑥 ′ (0) = 0;
ℒ{𝑥 ′′ − 3𝑥 ′ + 2𝑥} = 𝑠 2 𝑋(𝑠) − 1 − 3𝑠𝑋(𝑠) + 2𝑋(𝑠) = 0
𝑋(𝑠) =
1
(𝑠 2 −3𝑠+2)
=
1
(𝑠−2)
−
1
𝑠−1
; 𝑥(𝑡) = 𝑒 2𝑡 − 𝑒 𝑡
Example 5:
This example is that of a series LCR circuit with an external source 𝑉(𝑡) and 𝐿,
𝑅, 𝐶
as the inductance , resistance and capacitance respectively. The
instantaneous charge and the current are 𝑞(𝑡) and 𝑖(𝑡) respectively. At 𝑡 =
0, charge on the capacitor and the current in the circuit is 0.
𝐿
𝑑𝑖
𝑑𝑡
+𝑅𝑖+
since
𝑑𝑞
𝑑𝑡
𝑞
𝐶
=𝑉
= 𝑖 , we obtain 𝐿
𝑑2𝑞
𝑑𝑡 2
+𝑅
𝑑𝑞
𝑑𝑡
+
𝑞
𝐶
=𝑉
Considering the Laplace transform on both the sides we obtain ,
𝐿 [𝑠 2 𝑄(𝑠) − 𝑠𝑞(0) −
𝑑𝑞
𝑑𝑡 𝑡=0
] + 𝑅[𝑠𝑄(𝑠) − 𝑞(0)] +
𝑄(𝑠)
𝐶
=
𝑉
𝑠
Assigning values to the constants, 𝑅 = 10 𝑜ℎ𝑚𝑠, 𝐿 = 1𝐻, 𝐶 = .01 𝐹 , 𝑉 =
220𝑉 , we obtain
𝑠 2 𝑄(𝑠) + 10𝑠 𝑄(𝑠) + 100 𝑄(𝑠) = 220/𝑠 ; Solving for 𝑄(𝑠) ,
𝑄(𝑠) =
𝑞(𝑡)
220
𝑠(𝑠 2 +10𝑠+100)
;
= 11/5 − (11 ∗ (𝑐𝑜𝑠(5 ∗ 3^(1/2) ∗ 𝑡) + (3^(1/2) ∗ 𝑠𝑖𝑛(5 ∗
3^(1/2) ∗ 𝑡))/3))/(5 ∗ 𝑒𝑥𝑝(5 ∗ 𝑡))
- 29 -
2. Application of Laplace transform for solving simultaneous differential
equations.
This finds applications in coupled circuits like those of two inductive coils with
mutual inductance between them. Determination of currents in such circuits
involves solving simultaneous differential equations.The second application is
coupled oscillators like coupled spring mass systems.
𝑥 ′ = 𝑦 + 3𝑥
(1)
𝑦 ′ = 4𝑦 − 𝑥
(2)
𝑥(𝑡 = 0) = 1;
𝑦(𝑡 = 0) =0
Taking Laplace transform of (1) , we obtain
ℒ{𝑥 ′ (𝑡)} = ℒ{ 𝑦 + 3𝑥 } or 𝑠𝑋(𝑠) − 1 = 𝑌(𝑠) + 3𝑋(𝑠) or
𝑋(𝑠)[𝑠 − 3] − 𝑌(𝑠) = 1
(3)
ℒ{𝑦 ′ (𝑡)} = ℒ{4𝑦 − 𝑥} or 𝑠𝑌(𝑠) − 0 = 4𝑌(𝑠) − 𝑋(𝑠)
𝑋(𝑠) + 𝑌(𝑠){𝑠 − 4} = 0
(4)
substituting for 𝑌(𝑠) from eq. (3) in eq. (4) ,we obtain
𝑋(𝑠) + {𝑠 − 4}{ 𝑋(𝑠)[𝑠 − 3]-1} = 0
𝑋(𝑠){𝑠 2 − 7𝑠 + 13} = 𝑠 − 4 or
𝑋(𝑠) =
𝑠−4
𝑠 2 −7𝑠+13
;
𝑥(𝑡) = exp((7t)/2)(cos((3^(1/2)t)/2) - (3^(1/2)sin((3^(1/2)t)/2))/3)
- 30 -
2. We apply Laplace transform for solving the family of differential equations
for successive disintegration in radioactivity. Let 𝑁𝐴 (𝑡) be the parent nucleus
which disintegrates into the daughter nucleus 𝑁𝐵 (𝑡) which further disintegrates
into the stable nucleus 𝑁𝑐 (𝑡).
𝑁𝐴 → 𝑁𝐵 → 𝑁𝐶
. Determine the number of nuclides 𝑁𝐴 , 𝑁𝐵 and 𝑁𝐶 at any
instant of time 𝑡, given the initial number of nuclides are
𝑁𝐴 = 𝑁0 ; 𝑁𝐵 = 0; 𝑁𝐶 = 0
𝑑𝑁𝐴
𝑑𝑡
= −𝜆𝐴 𝑁𝐴
(5)
= −𝜆𝐵 𝑁𝐵 + 𝜆𝐴 𝑁𝐴
(6)
𝑑𝑁𝐵
𝑑𝑡
𝑑𝑁𝐶
𝑑𝑡
= 𝜆𝐵 𝑁𝐵
Taking Laplace transform of eq(5) , we obtain
𝑠 ℒ{𝑁𝐴 } − 𝑁0 = −ℒ{𝜆𝐴 𝑁𝐴 } ; (𝑠 + 𝜆)ℒ{𝑁𝐴 } = 𝑁0 ;
ℒ{𝑁𝐴 } =
𝑁0
(𝑠+𝜆)
𝑁𝐴 = 𝑁0 𝑒 −𝜆𝐴𝑡
Considering the Laplace transform of eq.(6) we obtain
𝜆 𝑁
0
𝑠ℒ{𝑁𝐵 } − 0 = 𝜆𝐴 ℒ{𝑁𝐴 } − 𝜆𝐵 ℒ{𝑁𝐵 } ; ℒ{𝑁𝐵 } = (𝑆+𝜆 𝐴)(𝑆+𝜆
𝐴)
𝐵
ℒ{𝑁𝐵 } = 𝜆𝐴 𝑁0
1
[
1
𝜆𝐴 −𝜆𝐵 𝑆+𝜆𝐴
−
1
𝑆+𝜆𝐵
] ; 𝑁𝐵 =
𝜆𝐴 𝑁0
𝜆𝐴 −𝜆𝐵
[𝑒 −𝜆𝐵𝑡 − 𝑒 −𝜆𝐴𝑡 ]
We repeat the process for eq(3) and obtain 𝑁𝐶 .
𝑠ℒ{𝑁𝐶 } − 0 = 𝜆𝐴 𝑁0
ℒ{𝑁𝐶 } =
𝑘
𝑠
𝜆𝐵
𝜆𝐴 −𝜆𝐵
[𝑒 −𝜆𝐵𝑡 − 𝑒 −𝜆𝐴𝑡 ] ; Let 𝑘 = 𝜆𝐴 𝑁0 𝜆
𝜆𝐵
𝐴 −𝜆𝐵
𝑡
𝑡
[𝑒 −𝜆𝐵𝑡 − 𝑒 −𝜆𝐴𝑡 ]; 𝑁𝐶 = 𝑘 [∫0 𝑒 −𝜆𝐵𝑣 𝑑𝑣 + ∫0 𝑒 −𝜆𝐴 𝑣 𝑑𝑣]
- 31 -
;
Using ℒ −1 {
𝐹(𝑠)
𝑠
𝑡
} = ∫0 𝑓(𝑣)𝑑𝑣 ;
Thus 𝑁𝐶 = 𝑘 [
1
𝜆𝐵
−
𝑒 −𝜆𝐵 𝑡
𝜆𝐵
+
𝑒 −𝜆𝐴𝑡
𝜆𝐴
−
1
𝜆𝐴
]
3. Application of Laplace transform for solving partial differential
equations
a) Solution of heat flow along semi infinite bar
The heat diffusion equation is
∇2 𝑢(r, 𝑡) =
𝑐𝜌 𝜕𝑢(𝒓,𝑡)
𝑘
𝜕𝑡
with 𝑢(r, 𝑡), 𝑐, 𝜌, 𝑘 respectively as the temperature
distribution in the material, specific heat capacity, density of the material and its
thermal conductivity. This equation assumes that there are no external heat
sources.
For a semi infinite bar , 𝑥 > 0 , we consider the 1-d form of the heat diffusion
equation. Let 𝑢(𝑥, 𝑡) represent the temperature at a distance 𝑥 from one end of
the bar at time 𝑡. The boundary condition is that 𝑢(0, 𝑡) = 𝑢0 and the initial
condition is 𝑢(𝑥, 0) = 0
As it is a semi infinite bar, we solve this problem using Laplace transform (𝑥 >
0). For an infinite bar , we employ Fourier transform.
Let
𝑘
𝑐𝜌
= 𝛼, the thermal diffusivity , We consider the Laplace transform on both
the sides of the heat diffusion equation
ℒ {𝛼
𝜕2 𝑢(𝑥,𝑡)
𝜕𝑢(𝑥,𝑡)
𝜕𝑥 2
𝜕𝑡
} = ℒ{
∞ 𝜕2 𝑢(𝑥,𝑡)
𝛼 ∫0
𝜕𝑥 2
} or
𝑒 −𝑠𝑡 𝑑𝑡 = 𝑠𝑈(𝑥, 𝑠) − 𝑢(𝑥, 0)
transform of 𝑢(𝑥, 𝑡)
- 32 -
with
𝑈(𝑥, 𝑠)
as
the
Laplace
The LHS integration is with respect to 𝑡 while the double differentiation is with
respect to 𝑥.
∞ 𝜕2 𝑢(𝑥,𝑡)
𝛼 ∫0
𝜕𝑥 2
𝑒 −𝑠𝑡 𝑑𝑡 = 𝛼
𝜕2
∞
∫ 𝑢(𝑥, 𝑡)𝑒 −𝑠𝑡 𝑑𝑡 = 𝛼
𝜕𝑥 2 0
𝜕2 𝑈(𝑥,𝑠)
𝜕𝑥 2
Using the initial condition 𝑢(𝑥, 0) = 0, we obtain
𝛼
𝜕2 𝑈(𝑥,𝑠)
𝜕𝑥 2
= 𝑠𝑈(𝑥, 𝑠), let
𝑠
𝛼
= 𝑝2
𝑈(𝑥, 𝑠) = 𝐴𝑒 𝑝𝑥 + 𝐵𝑒 −𝑝𝑥 with 𝐴 and 𝐵 as constants to be evaluated.
𝑢(𝑥, 𝑡) = 0 as 𝑥 → ∞ for the boundedness of the temperature. Thus 𝑈(𝑥, 𝑠)= 0
as 𝑥 → ∞ which implies that 𝐴 = 0
Since 𝑢(0, 𝑡) = 𝑢0 therefore 𝑈(0, 𝑠) = ℒ{𝑢0 } =
𝑢0
𝑠
with 𝑢0 as the constant
temperature at 𝑥 = 0.
𝑢0
=𝐵
𝑠
and the complete solution is 𝑈(𝑥, 𝑠) =
𝑢0
𝑠
𝑒 −𝑝𝑥 =
𝑢0
𝑠
𝑒 −𝑥√𝑠/∝
The solution 𝑢(𝑥, 𝑡) is the Laplace inverse of 𝑈(𝑥, 𝑠).
𝑢0
ℒ −1 {
𝑠
𝑒 −𝑝𝑥 } = 𝑢0 [1 − erf (
𝑥
√4𝛼𝑡
]
We first discuss the Laplace inverse of the function 𝑒 −√𝑠/∝ /𝑠
Using
the
property
𝑡
of
𝐼𝑓 ℒ𝑓(𝑣) = 𝐹(𝑠), ℒ {∫0 𝑓(𝑣)𝑑𝑣} =
ℒ −1 {
𝐹(𝑠)
𝑠
𝑡
} = ∫0 𝑓(𝑣)𝑑𝑣 ,
𝐹(𝑠)
𝑠
𝐹(𝑠)
𝑠
the
Laplace
transform
or
= 𝑒 −√𝑠/∝ /𝑠 , we obtain from eqn(4)
- 33 -
3
−1
−
𝑡 𝑣 2 𝑒 4𝑣
∫0 2√𝜋
∞
𝑑𝑣 , substituting 𝑣 = 1/4𝑦 2 we get
2
∫1/2√𝑡
2
√𝜋
𝑒 −𝑦 𝑑𝑦 = erfc (1/2√𝑡)
ℒ −1 {
We next consider
𝑒 −𝑥√𝑠
𝑠
}
𝑡
Using the property that ℒ { 𝑓 ( )} = 𝑎𝐹(𝑎𝑠) and putting 𝑎 = 𝑥 2
𝑎
−√𝑥2 𝑠
−1 𝑒
ℒ { 2 }
𝑠𝑥
𝑒 −𝑥√𝑠
ℒ −1 {
𝑠
=
1
𝑥2
1
𝑒𝑟𝑓𝑐 (
𝑡
)
2√ 2
𝑥
𝑥
} = 𝑒𝑟𝑓𝑐 (2 𝑡)
√
Example 2:
Consider the 1-d wave equation,
𝜕2 𝑢(𝑥,𝑡)
𝜕𝑥 2
=
1 𝜕2 𝑢(𝑥,𝑡)
𝑣2
𝜕𝑡 2
𝑢(𝑥, 𝑡)is the displacement of the string form the equilibrium position at
position 𝑥 at time 𝑡, 𝑣 represents the velocity of the wave.
We consider the Laplace transform of the wave equation to obtain
∞ 𝜕2 𝑢(𝑥,𝑡)
∫0
𝜕𝑥 2
𝜕2
𝜕𝑥 2
𝑒 −𝑠𝑡 𝑑𝑡 =
∞
∫ 𝑢(𝑥, 𝑡)
𝜕𝑥 2 0
𝑒 −𝑠𝑡 𝑑𝑡 as the integration is only over 𝑡 so
can be outside the integral. Secondly after integration, the dependence on 𝑡
vanishes and
𝓛{
𝜕2
𝝏𝟐 𝒖(𝒙,𝒕)
𝝏𝒙𝟐
𝜕2
𝜕𝑥 2
}=
can be converted into an ordinary differential.
𝒅𝟐 𝑼(𝒙,𝒔)
𝒅𝒙𝟐
- 34 -
Example 3:
Determine the ℒ {
𝜕𝑢(𝑥,𝑡)
By definition , ℒ {
𝜕𝑡
}
𝜕𝑢(𝑥,𝑡)
𝜕𝑡
∞ 𝜕𝑢(𝑥,𝑡)
} = ∫0
𝑒 −𝑠𝑡 𝑑𝑡
𝜕𝑡
∞
−𝑠𝑡
integrating by parts [𝑢(𝑥, 𝑡)𝑒 −𝑠𝑡 ]∞
𝑑𝑡
0 + 𝑠 ∫0 𝑢(𝑥, 𝑡)𝑒
Assuming that 𝑢(𝑥, 𝑡) is bounded at 𝑡 → ∞ ,
𝓛{
𝓛{
𝓛{
𝓛{
𝝏𝒖(𝒙,𝒕)
𝝏𝒕
𝝏𝟐 𝒖
𝝏𝒕𝟐
} = 𝒔𝑼(𝒙, 𝒔) − 𝒖(𝒙, 𝟎)
}
( 7)
can be determined by considering
𝜕𝑢
𝜕𝑡
= 𝑓(𝑡) and using eq.( )
𝝏𝟐 𝒖
𝜕𝑢
𝝏𝒕
𝜕𝑡 𝑡=0
} = 𝑠 𝐹(𝑠) − 𝑓(𝑥, 0) = 𝑠[𝑠𝑈(𝑥, 𝑠) − 𝑢(𝑥, 0)] −
𝟐
𝝏𝟐 𝒖
𝝏𝒕𝟐
} = 𝒔𝟐 𝑼(𝒙, 𝒔) − 𝒔𝒖(𝒙, 𝟎) −
𝝏𝒖
𝝏𝒕 𝒕=𝟎
Exercises
1. Solve the wave equation for a string fixed at both ends
𝑥 = 0 𝑎𝑛𝑑 𝑥 = 1.
with 𝑢𝑡 =
𝑢(0, 𝑡) = 0 , 𝑢(1, 𝑡) = 0; 𝑢(𝑥, 0) = 𝑠𝑖𝑛4𝜋𝑥, 𝑢𝑡 (𝑥, 0) = 0
𝜕𝑢
𝜕𝑡
2. Consider 2 masses 𝑚1 and 𝑚2 which are coupled by a massless spring, of
spring constant 𝑘. Solve the simultaneous differential equations and hence
determine the displacement of the masses at any instant of time 𝑡. Initial
conditions are 𝑥1 = 𝑥2 = 0;
𝑑𝑥1
𝑑𝑡
= 𝑢;
𝑑𝑥2
𝑑𝑡
=0
3. Solve the following simultaneous differential equation.
- 35 -
𝑑𝑥
= 4𝑥 − 𝑦;
𝑑𝑡
𝑑𝑦
𝑑𝑡
= 𝑥 + 4𝑦
4. In a series L C R circuit , at 𝑡 = 0 a voltage 𝑉 = 20 sin 400𝑡volts is suddenly
switched on. Assume the values of L = 1𝐻, R= 1000ohms and C =
6.25 𝑚𝑖𝑐𝑟𝑜 𝐹𝑎𝑟𝑎𝑑𝑠 . Determine the value of current 𝑖(𝑡) and charge 𝑞(𝑡) at
any instant of time.
5. In an open circuit with L = 10 𝑚𝐻; R= 250 𝛺 , C = 1𝜇𝐹 ,at 𝑡 = 0,
the capacitor has a charge of 10−5 𝑐𝑜𝑢𝑙𝑜𝑚𝑏𝑠 . At this instant the circuit is
closed. Determine the charge 𝑄(t) .
6. Solve the second order differential equation
𝑑2𝑦
𝑑𝑡 2
+ 9𝑦 = 9𝑈(𝑡 − 3) , 𝑦(0) =
𝑑𝑦
𝑑𝑡 𝑡=0
=0
7. Using Laplace transform , solve
𝜕𝑢(𝑥,𝑡)
𝜕𝑡
=
𝜕2 𝑢(𝑥,𝑡)
𝜕𝑥 2
,𝑢(𝑥, 0) = 3 sin 2𝜋𝑥, 𝑢(0, 𝑡) = 0, 𝑢(1, 𝑡) = 0
(D.U.)
0 < 𝑥 < 1; 𝑡 > 0
8. Solve the simultaneous differential equations using Laplace transforms.
𝑑𝑥
𝑑𝑡
+𝑦 =0;
𝑑𝑦
𝑑𝑡
− 𝑥 = 0;
𝑥(0) = 1, 𝑦(0) = 0
9. Solve using Laplace transform
𝑑𝑥
𝑑𝑡
= −𝑥 − 𝑦;
𝑑𝑦
𝑑𝑡
= −4𝑥 − 𝑦 ; 𝑥(𝑡 = 0) = 1; 𝑦(𝑡 = 0) = 0
- 36 -
(D.U.)
CHAPTER 4
Fourier Series
The Fourier series provide a method of analysis periodic function in to their
constituent component.
J.B. Fourier (1768-1830) was among the first to investigate this problem. In his
book ``Théorie Analytique de la Chaleur'', written in 1822, he introduced the
concept of Fourier series which he used extensively. Recall that a Fourier series
is any expression of the form
Periodic Function:
A Function f(x) is said to be periodic if F( x+p)=F(x) for all values of x, where p
is same positive number P is the interval between two successive repetitions and
is called the period of the function.
Example: f (x ) = sin 4x
 

) = sin  4(x + ) 
2
2 

= sin ( 4x + 2 )
f (x +

= sin 4x cos 2 + cos 4x sin 2
= sin 4x
= f (x )
hence sin nx is periad

2
, observe that 2 is also a period of sin 4x
Example: f (x ) = cos x
f (x + 2 ) = cos(x + 2 )
= cos x cos 2 − sin x sin 2
= cos x
= f (x )
Hence cos x is periodic of period 2
- 37 -
Example:
Of periodic functions are sin x and cos x of period equal to 2π.
Example:
➢ The function sin(x) has a period 2π, since sin(x+2π)=sin(x)
➢ The period of sin (nx) or cos (nx), where n is a positive integer, is 2π/n.
- 38 -
Example
 3
f (x ) = 
−3
0x 5
-5  x  0
Period = 10
0 x 
  x  2
Period = 2
Example
sinx
f (x ) = 
0
Example
0

f (x ) = 1
0

0x 2
2x 4
4x 6
Period = 6
Finding the smallest positive period of the following function:
 = 2 f
,f = 1/ T
2
2
=
 T=
T

1)
2)
3)
4)
5)
6)
7)
cos x
sinx
cos2x
sin2x
cos  x
sin2 x
cosnx
2 x
8) sin
k
Fourier series:
The basic of the Fourier series is that all function of practical significance which
are defined in the Interval
−  x  
convergent trigonometric series of the form:
- 39 -
can be expressed in terms of the
f ( )
f ( )
0
0


T
T
F (x ) = a0 + a1 cos x + a2 cos 2x + a3 cos3x + .........
+b1 sin x + b2 sin 2x + b3 sin 3x + .........

F (x ) = a0 +  (an cos nx +b n sin nx )
n =1


n =1
n =1
F (x ) = a0 +  (an cos nx ) +  (b n sin nx )
where the Fourier Coefficients a n and b n
Fourier coefficients of f ( x ) , given by the Euler formulas
We need to work out the Fourier coefficients (a0 , an and b n ) for given functions f (x ).
This process is broken down into three steps
T /2
STEP ONE
1
a 0 =  f (x )dx
T −T /2
STEP TWO
2
a n =  f (x )cos(n0 x )dx
T −T /2
ST EP THREE
2
bn =  f (x )sin(n0 x )dx
T −T /2
T /2
n=0,1,2,....
T /2
n=0,1,2,....
Problem: Find the Fourier series corresponding to the function
0
f (x ) = 
3
−5  x  0
0x 5
Period = 10
- 40 -
Solution :T = 10, 0 =
STEP ONE
2 2 
=
=
T
10 5
5
5
5
1
1
3
3
3
− f (x )dx = a0 = 10 -5 f (x )dx = 10 0 3dx = 10 = 10 (5 - 0) = 2
0
2
an =
T
2

3 5


f ( x ) cos( n 0 x )dx =  3cos( nx )dx = 
sin  nx 

10 0
5
5 n
5
0
-T /2
an =
S TEP TWO

1
a0 =
2
bn
T /2
3
n
2
=
T
5
5


 3
sin( 5  5n ) - sin(0)  = n  sin(n  ) = 0
1

f (x ) sin(n 0 x )dx =  3sin( nx )dx

5 -5
5
-T /2
T /2
5
3 5

3 5

3 


bn = - 
cos( nx ) = 
cos( nx ) =
1- cos(  5n ) 

5 n
5
5 n
5
n 
5

0
5
3
bn =
1- cos(n )
n
n even
n even
+1
0
cos( n  ) = 
 bn = 
n odd
 −1
 6 / n n odd
5
the corresponding Fourier Series is


n =1
n =1
F (x ) = a0 +  (an cos nx ) +  (b n sin nx )
F (x ) =
3 6

1
3
1

+  sin( x ) + sin( x ) + sin( x + ......... 
2 
5
3
5
5

- 41 -
0
Example
−k
f (x ) = 
 k
k is cons tan t
STEP ONE
-  x  0
0 x 
,T = 2 , 0 =
1
a0 =
T
a0 =
2
=1
T
T /2

f ( x )dx
−T /2
1
2

 kdx =
−
0

 1
1
kdx
+
kdx  =  − +   = 0


  −
0
 
T /2
STEP TWO
an =
2
f (x ) cos(n 0 x )dx
T −T/ 2

0


1
k
cos
(
nx
)
dx
+
k cos(nx )dx 


 −
  −
0

0


1  −k
k
an =  sin(nx ) + sin(nx ) 
 n
n
−
0
an = 0 (odd Function symetric about origin )
an =
1
bn =
2
f ( x )sin( n0 x )dx
T −T/2
 k cos(nx )dx =
T /2
STEP THREE
bn
bn
0


1
=  f (x )sin(nx )dx =   −k sin(nx )dx +  k sin(nx )dx 
 −
  −
0

1
0

=  −k sin(nx ) − −k sin(nx ) 0 

bn =
N ote :
1

k
 2 − 2 cos n 
nk
When n is Odd
bn =
k
4k
(2 + 2) =
nx
n


n =1
n =1
F (x ) = a0 +  (an cos nx ) +  (b n sin nx )
F (x ) =
4k

4k
F (x ) =

4k
F (x ) =


4k
F( ) =
2


1
 n sin nx
n =1
 1
 1

  1

sin( 2 ) + 3 sin( 2 ) + 5 sin( 2 ) + 7 sin( 2 ) + ............
1 1 1
 1 1 1

1 + 3 + 5 + 7 + + 7 + 9 + 11 + ...........
(0.7543) = 0.964 k
- 42 -
Fourier Coefficients of Even Functions:
f (t ) = f (−t )
f (t ) =
a0 =
a0 
+  an cos n0t
2 n =1
T /2
1
T

T /2
2
a0 =
T
 f (x )dx
0
2
an =
T
an =
4
T
f (x )dx
−T /2
T /2

f (x ) cos(n0 x )dx
−T /2

T /2
0
f (t ) cos(n 0t )dt
Fourier Coefficients of Odd Functions:
f (t ) = −f (−t )

f (t ) =  b n sin n 0t
n =1
bn =
4
T

T /2
0
f (t ) sin(n 0t )dt
Some useful properties of even and odd functions are as follows.
➢ The product of two odd functions is an even function.
➢ The product of two even functions is an even function.
➢ The product of an odd function and an even function is an odd function.
- 43 -
Exercise 1
Classify each of the following function as even, odd or neither even nor odd.
x
1) − f (x ) = 
-x
x
2) − f (x ) = 
x
2
1) − f (x ) = 
-2
cosx
2) − f (x ) = 
0
0  x 1
-1  x  0
0  x 1
-1  x  0
0x 3
-3  x  0
0 x 
  x  2
➢ The Fourier series of an even function f(x) is expressed in terms of a
cosine series

f ( x ) = a0 +  an cos nx
, bn=0
n =1
➢ The Fourier series of an odd function f(x) is expressed in terms of a
sine series

f ( x ) =  b n sin nx
n =1
, a0=0 , an=0
Example:
Find the Fourier series of the following periodic function.
f ( x ) = x 2 when
−  x  
- 44 -
f(x)
−
0

3
5
7
9
x
Exercise 2
Determine the Fourier series for the periodic function:
 −2
1)- f(x) 
+2
when
when

 0 when


2)- f(x)  1 when


−1 when
0
1)- f(x) 
sin( )
- < x< 0
0<x<
- < x< 0
0 < x<

2
when
when

2
< x< 
- <  < 0
0< <
Expansion of non-periodic functions
If a function f(x) is not periodic then it cannot be expanded in a Fourier series.
Given a non-periodic function, a new function may be constructed by taking the
values of f(x) in the given range and then repeating them outside of the given
range at intervals of 2π For example, the function
- 45 -
f(x)=x
is not a periodic function
However, if a Fourier series for f(x)=x is required then the function is
constructed outside of this range so that it is periodic with period 2π .
For determining a Fourier series of a non-periodic function over a range 2π,
exactly the same formulae for the Fourier coefficients are used


n =1
n =1
F (x ) = a0 +  (an cos nx ) +  (b n sin nx )
it is periodic with period T= 2 ⎯⎯
→ 0 =
STEP ONE
a0 =
STEP T WO
an =
STEP THREE
bn =

1
2
1

1

2
2
⎯⎯
→ 0 =
⎯⎯
→ 0 = 1
T
2
 f (x )dx
−

 f (x ) cos(nx )dx
n=0,1,2,....
−

 f (x )sin(nx )dx
n=0,1,2,....
−
Example;
Determine the Fourier series to represent the function f(x) =2x
in the range −π
to+π.
The function f(x) =2x is not periodic. The function is shown in the range −π to
π of period 2π
Solution:


n =1
n =1
F (x ) = a0 +  (an cos nx ) +  (b n sin nx )
- 46 -
STEP ONE

1
a0 =
2
 2xdx
-

1
1
 x 2  =
 2 - (- )2 ) = 0
(

2
2

1
=  f (x ) cos(nx )dx
n =0,1, 2,....
a0 =
STEP TWO
an

an =
-
1


 2x cos(nx )dx
-

2  x sin nx
sin nx

an = 
-
dx 
 n
n
 -

2  x sin nx cos nx 
an = 
+
 n
n 2  -
an =
STEP THREE
2 
cos n
0+

 
n2
bn =
bn =
bn

1
 f (x ) sin(nx )dx

1

cos n (- )  
 
 -0 +
 = 0
n2
 

n =0,1, 2,....
-

 2x sin(nx )dx
-
2  -x cos n  - cos n 
= 
-
dx
 
n
n
bn =
2  -x cos n  sin nx
+
 
n
n2
  -(- ) cos n (- ) sin n (- )  
+
-

n
n2
 

   cos(-n )   -4
-
  = cos n
n
 
 
4
4
4
where n is odd , bn = , thus b1 = 4 , b 3 = b 5 = , and so on .
n
3
5
thus
4
4
4
f (x ) = 2x = 4sin x - sin 2x + sin 3x - sin 4x + ..........
2
3
4
bn =
2  - cos n
 
n



  -
- 47 -
Example: Obtain a Fourier series for the function defined as:
when 0  x  
 x
f (x ) = 
  x  2
0 when
Problems
- 48 -
Half range Fourier series
When a function is defined over the range 0 to π instead of from 0 to 2π it may
be expanded in a series of sine terms only or of cosine terms only. The series
produced is called a half-range Fourier series.
(a) If a half-range cosine series is required for the function
f(x)=x
in the range 0 to π then an even periodic function is required.
f(x)=x is shown plotted from x=0 to x=π.
Since an even function is symmetrical about the y- axis the line AB is
constructed as shown. When a half-range cosine series is required then the
Fourier coefficients a0 and an are calculated

F (x ) = a0 +  (an cos nx )
n =1
STEP ONE
ST EP TWO
a0 =
an =
1

2


 f (x )dx
0

 f (x ) cos(nx )dx
0
(b) If a half-range sine series is required for the function
f(x)=x in the range 0 to π then an odd periodic function is required.
f(x)=x is shown plotted from x=0 to x=π.
- 49 -
Since an odd function is symmetrical about the origin the line CD is
constructed as shown. When a half-range sine series is required then the
Fourier coefficient bn is calculated.

F (x ) =  b n sin nx
n =1
bn =
2


 f (x )sin(nx )dx
−
Example:
1.
Determine the half-range Fourier cosine series to represent the
function f(x) =3x in the range 0 ≤ x ≤ π.
2.
Determine the half-range Fourier sine series to represent the function
f(x) =3x in the range 0 ≤ x ≤ π.
- 50 -
1.
Fourier cosine series

f (x ) = a0 +  (an cos nx )
n =1
f (x ) = 3x

STEP ONE
a0 =

1
1
f
(
x
)
dx
=
3xdx
2 0
2 0

3  x 2  3
a0 =   =
  2 0 2
STEP TWO
an =
2


2

 f (x ) cos(nx )dx =   3x cos(nx )dx
0
0

6   sin nx cos nx 
an = 
+
 n
n 2  0
an =
6   sin n cos n
+
  n
n2
cos 0  
 
 −  0 + 2 
n 
 
6
cos n cos 0 
− 2 
0 +

n2
n 
6
a n = 2 (cos n − 1)
n
when n is EVEN , an -0
6
−12
when n is ODD, an= 2 (−1 − 1) =
n
n2
−12
−12
−12
Hence a1 =
, a3 2 , a5
, and So on

3
n2
3 12
1
1
f(x)=3x=
− (cos x + 2 cos 3x + 2 cos 5x .......)
2 
3
5
an =
- 51 -
2.
Fourier sine series

F (x ) =  b n sin nx
n =1
where
f (x ) = 3x
bn =
2



2
 f (x ) sin(nx )dx =   3x sin(nx )dx
−
−

bn =
6  −x cos nx sin nx 
+
 
n
n 2  0
6  −x cos n  sin n 
+
 
n
n2
6
bn =- cos n
bn =


 − (0 + 0) 



6
when n is ODD , bn = .
n
6
6
6
Hence b1= , b3= , b5= , So on.
1
3
5
6
when n is EVEN , bn =- .
n
6
6
6
Hence b2=- , b4=- , b6=- , So on.
2
4
6
Hence the Half-range Fourier Sine series is given by:
1
1
1
1


f(x)=3x=6  sinx- sin 2x + sin 3x − sin 4x + sin 5x + ........ 
2
3
4
5


- 52 -
CHAPTER 5
FOURIER TRANSFORMS
5.1 Introduction
The Fourier series expresses any periodic function into a sum of sinusoids. The
Fourier transform is the extension of this idea to non-periodic functions by
taking the limiting form of Fourier series when the fundamental period is made
very large (infinite).
Fourier transform finds its applications in astronomy,
signal processing, linear time invariant (LTI) systems etc.
Some useful results in computation of the Fourier transforms:
∞
𝜆
∞
𝑎
1. ∫0 𝑒 −𝑎𝑥 sin 𝜆𝑥 𝑑𝑥 = 2 2
𝑎 +𝜆
2. ∫0 𝑒 −𝑎𝑥 cos 𝜆𝑥 𝑑𝑥 = 2 2
𝑎 +𝜆
∞ sin 𝜆𝑥
3. ∫0
𝑥
𝜋
𝑑𝑥 = , 𝜆 > 0
2
∞ 𝑠𝑖𝑛𝑥
When 𝜆 = 1, ∫0
4. sin 𝑎𝑥 =
5. cos 𝑎𝑥 =
∞
6. ∫0 𝑒 −𝑎
𝑥
𝑑𝑥 =
𝜋
2
𝑒 𝑖𝑎𝑥 −𝑒 −𝑖𝑎𝑥
2𝑖
𝑒 𝑖𝑎𝑥 +𝑒 −𝑖𝑎𝑥
2𝑥2
2
𝑑𝑥 =
√𝜋
2𝑎
∞
2
When 𝑎 = 1, ∫0 𝑒 −𝑥 𝑑𝑥 =
√𝜋
2
7. Heaviside Step Function or Unit step function 𝐻(𝑡)or 𝑈(𝑡) =
0, when 𝑡 < 0
{
1, when 𝑡 ≥ 0
At 𝑡 = 0, 𝐻(𝑡) is sometimes taken as 0.5 or it may not have any
specific value.
- 53 -
Shifting at 𝑡 = 𝑎
0, when 𝑡 < 𝑎
𝐻(𝑡 − 𝑎) or 𝑈(𝑡 − 𝑎) = {
1, when 𝑡 ≥ 𝑎
8. Dirac Delta Function or Unit Impulse Function is defined as 𝛿(𝑡 − 𝑎) = 0,
∞
t≠a such that ∫0 𝛿(𝑡 − 𝑎)𝑑𝑡 = 1, 𝑎 ≥ 0. It is zero everywhere except
one point 'a'. Delta function in sometimes thought of having infinite value
at 𝑡 = 𝑎. The delta function can be viewed as the derivative of
the Heaviside step function
Dirichlet’s Conditions for Existence of Fourier Transform
Fourier transform can be applied to any function 𝑓(𝑥) if it satisfies the following
conditions:
∞
1. 𝑓(𝑥) is absolutely integrable i.e. ∫−∞|𝑓(𝑥)|𝑑𝑥 is convergent.
2. The function 𝑓(𝑥) has a finite number of maxima and minima.
3. 𝑓(𝑥) has only a finite number of discontinuities in any finite
5.2 Fourier Transform, Inverse Fourier Transform and Fourier Integral
̅
The Fourier transform of 𝑓(𝑥), −∞ < 𝑥 < ∞, denoted by 𝑓(λ)
where λ ∈ N , is
given by
̅
𝐹{𝑓(𝑥)} ≡ 𝑓(λ)
=
1
∞
𝑒 𝑖𝜆𝑥 𝑓(𝑥)𝑑𝑥 …①
∫
−∞
2𝜋
√
̅
Also inverse Fourier transform of 𝑓(λ)
gives 𝑓(𝑥) as:
𝑓(𝑥) =
∞ −𝑖𝜆𝑥
̅
𝑒
𝑓(λ)𝑑𝜆
…②
∫
√2𝜋 −∞
1
- 54 -
̅
Rewriting ① as 𝑓 (λ)
=
1
√2𝜋
∞
∫−∞ 𝑒 𝑖𝜆𝑡 𝑓(𝑡)𝑑𝑡 and using in ②, Fourier integral
representation of 𝑓(𝑥) is given by:
𝑓(𝑥) =
∞
1
∞
∫ ∫ 𝑒 𝑖𝜆(𝑡−𝑥) 𝑓(𝑡)𝑑𝑡 𝑑𝜆
2𝜋 −∞ −∞
2.2.1 Fourier Sine Transform (F.S.T.)
Fourier Sine transform of 𝑓(𝑥), 0 < 𝑥 < ∞, denoted by 𝑓𝑠̅ (λ), is given by
2 ∞
𝐹𝑠 {𝑓(𝑥)} ≡ 𝑓𝑠̅ (λ) = √ ∫0 𝑓(𝑥) sin 𝜆𝑥 𝑑𝑥…③
𝜋
Also inverse Fourier Sine transform of 𝑓𝑠̅ (λ) gives 𝑓(𝑥) as:
∞
2
𝑓(𝑥) = √ ∫0 𝑓𝑠̅ (λ) sin 𝜆𝑥 𝑑𝜆 … ④
𝜋
2 ∞
Rewriting ③ as 𝑓𝑠̅ (λ) = √ ∫0 𝑓(𝑡) sin 𝜆𝑡 𝑑𝑡 and using in ④, Fourier sine
𝜋
integral representation of 𝑓(𝑥) is given by:
∞
2
∞
𝑓(𝑥) = ∫0 ∫0 𝑓(𝑡) sin 𝜆𝑡 sin 𝜆𝑥 𝑑𝑡𝑑𝜆
𝜋
2.2.2 Fourier Cosine Transform (F.C.T.)
Fourier Cosine transform of 𝑓(𝑥), 0 < 𝑥 < ∞, denoted by 𝑓𝑐̅ (λ), is given by
2 ∞
𝐹𝑐 {𝑓(𝑥)} ≡ 𝑓𝑐̅ (λ) = √ ∫0 𝑓(𝑥) cos 𝜆𝑥 𝑑𝑥 …⑤
𝜋
Also inverse Fourier Cosine transform of 𝑓𝑐̅ (λ) gives 𝑓(𝑥) as:
2
∞
𝑓(𝑥) = √ ∫0 𝑓𝑐̅ (λ) cos 𝜆𝑥 𝑑𝜆 … ⑥
𝜋
- 55 -
2
∞
Rewriting ⑤ as 𝑓𝑐̅ (λ) = √ ∫0 𝑓(𝑡) cos 𝜆𝑡 𝑑𝑡 and using in ⑥, Fourier cosine
𝜋
integral representation of 𝑓(𝑥) is given by:
2
∞
∞
𝑓(𝑥) = ∫0 ∫0 𝑓(𝑡) cos 𝜆𝑡 cos 𝜆𝑥 𝑑𝑡𝑑𝜆
𝜋
Remark:
• Parameter λ may be taken as p, s or ω as per usual notations.
̅
• Fourier transform of 𝑓(𝑥) may be given by 𝑓 (λ)
=
1
∞
∫ 𝑒 −𝑖𝜆𝑥 𝑓(𝑥)𝑑𝑥
√2𝜋 −∞
,
then Inverse Fourier transform of 𝑓 (̅ λ)is given by 𝑓(𝑥) =
1
√2𝜋
∞
̅
∫−∞ 𝑒 𝑖𝜆𝑥 𝑓(λ)𝑑𝜆
• Sometimes Fourier transform of 𝑓(𝑥) is taken as
̅
𝑓(λ)
=
∞
∫−∞ 𝑒 𝑖𝜆𝑥 𝑓(𝑥)𝑑𝑥,
thereby
Inverse
Fourier
transform
is
given
𝑓(𝑥) =
by
1
2𝜋
∞
̅
∫−∞ 𝑒 −𝑖𝜆𝑥 𝑓(λ)𝑑𝜆
Similarly
if
Fourier
Sine
transform
is
taken
as
𝑓𝑠̅ (λ) =
∞
∫0 𝑓(𝑥) sin 𝜆𝑥 𝑑𝑥,
2
∞
then Inverse Sine transform is given by 𝑓(𝑥) = ∫0 𝑓𝑠̅ (λ) sin 𝜆𝑥 𝑑𝜆
𝜋
Similar is the case with Fourier Cosine transform.
- 56 -
Example 1 State giving reasons whether the Fourier transforms of the following
functions exist:
i. sin
1
ii.𝑒 𝑥
𝑥
iii. 𝑓(𝑥) =
1, if 𝑥 is rational
{
0, if 𝑥 is irrational
Solution: i. The graph of sin
1
𝑥
oscillates infinite number of times at 𝑥 =
𝑛𝜋, 𝑛 ∈ Z
∴ 𝑓(𝑥) sin
1
𝑥
is having infinite
(−∞, ∞). Hence
number of maxima and minima in the interval
Fourier transform of 𝑓(𝑥) = sin
1
𝑥
does not exist.
∞
ii. For 𝑓(𝑥) = 𝑒 𝑥 , ∫−∞|𝑒 𝑥 |𝑑𝑥 is not convergent. Hence Fourier
transform of
𝑒 𝑥 does not exist.
1, if 𝑥 is rational
iii. 𝑓(𝑥) = {
is having infinite number of maxima
0, if 𝑥 is irrational
minima in the interval (−∞, ∞). Hence Fourier
and
transform of 𝑓(𝑥) does not exist.
Example 2 Find Fourier Sine transform of
i.
1
ii. 2𝑒 −3𝑥 + 3𝑒 −2𝑥
𝑥
2
∞
Solution: i. By definition, we have 𝐹𝑠 {𝑓(𝑥)} ≡ 𝑓𝑠̅ (λ) = √ ∫0 𝑓(𝑥) sin 𝜆𝑥 𝑑𝑥
𝜋
2 ∞1
2 𝜋
𝜋
∴ 𝑓𝑠̅ (λ) = √ ∫0 sin 𝜆𝑥 𝑑𝑥 = √ . = √
𝜋
𝑥
𝜋 2
2
2 ∞
ii. By definition, 𝐹𝑠 {𝑓(𝑥)} ≡ 𝑓𝑠̅ (λ) = √ ∫0 𝑓(𝑥) sin 𝜆𝑥 𝑑𝑥
𝜋
- 57 -
2
∞
2
∞
∴ 𝑓𝑠̅ (λ) = √ ∫0 (2𝑒 −3𝑥 + 3𝑒 −2𝑥 ) sin 𝜆𝑥 𝑑𝑥
𝜋
2
∞
= √ ∫0 2𝑒 −3𝑥 sin 𝜆𝑥 𝑑𝑥 + √ ∫0 3𝑒 −2𝑥 sin 𝜆𝑥 𝑑𝑥
𝜋
𝜋
2 2𝑒 −3𝑥
=√ [
𝜋 9+𝜆2
∞
2 3𝑒 −2𝑥
(−3 sin 𝜆𝑥 − 𝜆 cos 𝜆𝑥] + √ [
𝜋 4+𝜆2
0
(−2 sin 𝜆𝑥 −
∞
𝜆 cos 𝜆𝑥]
0
2
2𝜆
𝜋
9+𝜆2
= √ [0 +
2
3𝜆
𝜋
4+𝜆2
] + √ [0 +
2
2𝜆
𝜋
9+𝜆2
]=√ [
+
3𝜆
4+𝜆2
]=
5𝜆3 +35𝜆
2
√ [(9+𝜆2 )(4+𝜆2 )]
𝜋
Example 3 Find Fourier transform of Delta function 𝛿(𝑥 − 𝑎)
Solution: 𝐹{𝛿(𝑥 − 𝑎)} =
1
∞
1
𝑒 𝑖𝜆𝑎
∫ 𝑒 𝑖𝜆𝑥 . 𝛿(𝑥 − 𝑎) 𝑑𝑥
√2𝜋 −∞
=
√2𝜋
∞
∵ ∫−∞ 𝑓(𝑡) 𝛿(𝑡 − 𝑎)𝑑𝑡 = 𝑓(𝑎) by virtue of fundamental property of Delta
function where 𝑓(𝑡) is any differentiable function.
Example 4 Show that Fourier sine and cosine transforms of 𝑥 𝑛−1 are
⌈𝑛
𝜆𝑛
sin
𝑛𝜋
2
and
⌈𝑛
𝜆𝑛
cos
𝑛𝜋
2
respectively.
∞
Solution: By definition, ∫0 𝑒 −𝑡 𝑡 𝑛−1 𝑑𝑡 = ⌈𝑛
Putting 𝑡 = 𝑖𝜆𝑥 so that 𝑑𝑡 = 𝑖𝜆𝑑𝑥
- 58 -
∞
⇒ ∫0 𝑒 −𝑖𝜆𝑥 (𝑖𝜆𝑥)𝑛−1 𝑖𝜆 𝑑𝑥 = ⌈𝑛
∞
⇒ ∫0 𝑥 𝑛−1 𝑒 −𝑖𝜆𝑥 𝑑𝑥 =
⌈𝑛 𝑖 −𝑛
𝜆𝑛
⌈𝑛
∞
⇒ ∫0 𝑥 𝑛−1 (cos 𝜆𝑥 − 𝑖 sin 𝜆𝑥) 𝑑𝑥 =
𝜆𝑛
(cos
𝑛𝜋
− 𝑖 sin
2
∵ 𝑖 −𝑛 = (cos
∞
⌈𝑛
∞
⇒ ∫0 𝑥 𝑛−1 cos 𝜆𝑥 𝑑𝑥 − 𝑖 ∫0 𝑥 𝑛−1 sin 𝜆𝑥 𝑑𝑥 =
𝜆𝑛
𝑛𝜋
2
𝑛𝜋
2
cos
)
− 𝑖 sin
𝑛𝜋
2
−𝑖
𝑛𝜋
2
⌈𝑛
𝜆𝑛
)
sin
𝑛𝜋
2
Equating real and imaginary parts, we get
⌈𝑛
∞
∫0 𝑥 𝑛−1 cos 𝜆𝑥 𝑑𝑥 = 𝜆𝑛 cos
⇒ 𝑓𝑐̅ (λ) =
⌈𝑛
𝜆𝑛
cos
𝑛𝜋
2
𝑛𝜋
2
∞
and ∫0 𝑥 𝑛−1 sin 𝜆𝑥 𝑑𝑥 =
and 𝑓𝑠̅ (λ) =
⌈𝑛
𝜆𝑛
sin
⌈𝑛
𝜆𝑛
sin
𝑛𝜋
2
𝑛𝜋
2
𝑥, 0 < 𝑥 < 1
Example 5 Find Fourier Cosine transform of 𝑓(𝑥) = {2 − 𝑥, 1 < 𝑥 < 2
0,
𝑥>2
2 ∞
Solution: By definition, we have 𝐹𝑐 {𝑓(𝑥)} ≡ 𝑓𝑐̅ (λ) = √ ∫0 𝑓(𝑥) cos 𝜆𝑥 𝑑𝑥
𝜋
2 ∞
∴ 𝑓𝑐̅ (λ) = √ ∫0 𝑓(𝑥) cos 𝜆𝑥 𝑑𝑥
𝜋
2
1
2
= √ [∫0 𝑥 cos 𝜆𝑥 𝑑𝑥 + ∫1 (2 − 𝑥) cos 𝜆𝑥 𝑑𝑥 +
𝜋
∞
∫2 0. cos 𝜆𝑥 𝑑𝑥 ]
2
= √ [[(𝑥) (
𝜋
(−1) (−
cos 𝜆𝑥
𝜆2
sin 𝜆𝑥
𝜆
) − (1) (−
cos 𝜆𝑥
𝜆2
1
sin 𝜆𝑥
0
𝜆
)] + [(2 − 𝑥) (
2
)] ]
1
- 59 -
)−
2 sin 𝜆
=√ [
𝜋
𝜆
+
cos 𝜆
𝜆2
1
−
𝜆2
−
cos 2𝜆
𝜆2
−
sin 𝜆
𝜆
+
cos 𝜆
𝜆2
2 2 cos 𝜆−cos 2𝜆−1
]=√ [
𝜆2
𝜋
]
Example 6 Find Fourier Sine and Cosine transform of 𝑓(𝑥) = 𝑒 −𝑥 and hence
show that
∞ cos 𝑚𝑥
∫0
1+𝑥 2
𝜋
𝑑𝑥 =
2
∞ 𝑥 sin 𝑚𝑥
𝑒 −𝑚 = ∫0
1+𝑥 2
𝑑𝑥
Solution: To find Fourier Sine transform
2 ∞
𝐹𝑠 {𝑓(𝑥)} ≡ 𝑓𝑠̅ (λ) = √ ∫0 𝑓(𝑥) sin 𝜆𝑥 𝑑𝑥
𝜋
2 ∞
2
⇒ 𝑓𝑠̅ (λ) = √ ∫0 𝑒 −𝑥 sin 𝜆𝑥 𝑑𝑥 = √ (
𝜆
𝜋 1+𝜆2
𝜋
)……①
Taking inverse Fourier Sine transform of ①
∞
2
𝑓(𝑥) = √ ∫0 𝑓𝑠̅ (λ) sin 𝜆𝑥 𝑑𝜆
𝜋
∞
2
𝜆
⇒ 𝑓(𝑥) = ∫0
𝜋
1+𝜆2
𝑠𝑖𝑛𝜆𝑥𝑑𝜆…..②
Substituting 𝑓(𝑥) = 𝑒 −𝑥 in ②
⇒ 𝑒 −𝑥 =
2
∞ 𝜆𝑠𝑖𝑛𝜆𝑥
∫
𝜋 0
1+𝜆2
𝑑𝜆
Replacing 𝑥 by 𝑚 on both sides
2
∞ 𝜆𝑠𝑖𝑛𝜆𝑚
⇒ 𝑒 −𝑚 = ∫0
𝜋
1+𝜆2
𝑑𝜆
𝑏
𝑏
Now by property of definite integrals ∫𝑎 𝑓(𝑥)𝑑𝑥 = ∫𝑎 𝑓(𝑦)𝑑𝑦
∴
𝜋
∞ 𝑥 sin 𝑚𝑥
𝑒 −𝑚 = ∫0
2
1+𝑥 2
𝑑𝑥 ….③
- 60 -
Similarly taking Fourier Cosine transform of 𝑓(𝑥) = 𝑒 −𝑥
2
∞
𝐹𝑐 {𝑓(𝑥)} ≡ 𝑓𝑐̅ (λ) = √ ∫0 𝑓(𝑥) cos 𝜆𝑥 𝑑𝑥
𝜋
2 ∞
2
⇒ 𝑓𝑐̅ (λ) = √ ∫0 𝑒 −𝑥 cos 𝜆𝑥 𝑑𝑥 = √ (
1
𝜋 1+𝜆2
𝜋
)……④
Taking inverse Fourier Cosine transform of ④
∞
2
𝑓(𝑥) = √ ∫0 𝑓𝑐̅ (λ) cos 𝜆𝑥 𝑑𝜆
𝜋
∞
2
1
⇒ 𝑓(𝑥) = ∫0
𝜋
1+𝜆2
cos 𝜆𝑥 𝑑𝜆….. ⑤
Substituting 𝑓(𝑥) = 𝑒 −𝑥 in ⑤
⇒ 𝑒 −𝑥 =
2
∞ cos 𝜆𝑥
∫
𝜋 0
1+𝜆2
𝑑𝜆
Replacing 𝑥 by 𝑚 on both sides
∞ cos 𝜆𝑚
2
⇒ 𝑒 −𝑚 = ∫0
𝜋
1+𝜆2
𝑑𝜆
𝑏
𝑏
Again by property of definite integrals ∫𝑎 𝑓(𝑥)𝑑𝑥 = ∫𝑎 𝑓(𝑦)𝑑𝑦
∴
𝜋
2
∞ cos 𝑚𝑥
𝑒 −𝑚 = ∫0
1+𝑥 2
𝑑𝑥 …. ⑥
From ③and ⑥, we get
∞ cos 𝑚𝑥
∫0
1+𝑥 2
𝑑𝑥 =
𝜋
2
∞ 𝑥 sin 𝑚𝑥
𝑒 −𝑚 = ∫0
1+𝑥 2
𝑑𝑥
Example 7 Find Fourier transform of 𝑓(𝑥) = {
∞ 𝑥cos 𝑥−sin 𝑥
and hence evaluate ∫0 (
𝑥3
- 61 -
1 − 𝑥 2 , |𝑥| < 1
|𝑥| > 1
0,
𝑥
) cos 2 𝑑𝑥
Solution: Fourier transform of 𝑓(𝑥) is given by
=
=
=
1
√2𝜋
√2𝜋 𝑖 𝜆
1
∫ (1 − 𝑥 2 )𝑒 𝑖𝜆𝑥 𝑑𝑥
√2𝜋 −1
[(1 −
[2
∫ 𝑒 𝑖𝜆𝑥 𝑓(𝑥)𝑑𝑥
√2𝜋 −∞
1
2𝑒 𝑖𝜆
1
∞
1
̅
𝐹{𝑓(𝑥)} ≡ 𝑓(λ)
=
𝑒 𝑖𝜆𝑥
𝑥 2) ( )
𝑖𝜆
−
2
2𝑒 −𝑖𝜆
𝑖 3𝜆
𝑖 2 𝜆2
+
3
𝑒 𝑖𝜆 +𝑒 −𝑖𝜆
𝜋
𝜆2
2
2 cos 𝜆
𝜋
𝜆2
= √ [−
∴ 𝑓 (̅ λ) =
+
+
+
𝑒 𝑖𝜆 −𝑒 −𝑖𝜆
𝑖𝜆3
2 sin 𝜆
𝜆3
𝑒 𝑖𝜆𝑥
1
− (−2𝑥) ( 2 2 ) + (−2) ( 3 3 )]
𝑖 𝜆
𝑖 𝜆
2𝑒 𝑖𝜆
2
= √ [−
𝑒 𝑖𝜆𝑥
2𝑒 −𝑖𝜆
𝑖 3 𝜆3
−1
]
∵ 𝑖 2 = −1 and 𝑖 3 = −𝑖
]
]
2√2 sin 𝜆−𝜆cos 𝜆
(
)
𝜆3
√𝜋
…..①
Taking inverse Fourier transform of ①
𝑓(𝑥) =
2
∞ −𝑖𝜆𝑥
̅
𝑒
𝑓(λ)𝑑𝜆
∫
−∞
√2𝜋
∞
1
sin 𝜆−𝜆cos 𝜆
⇒ 𝑓(𝑥) = ∫−∞ 𝑒 −𝑖𝜆𝑥 (
𝜋
2
𝜆3
) 𝑑𝜆
∞
𝜆cos 𝜆−sin 𝜆
⇒ 𝑓(𝑥) = ∫−∞(cos 𝜆𝑥 − 𝑖 sin 𝜆𝑥) (
𝜋
𝜆3
∵ 𝑒 −𝑖𝜆𝑥 =
) 𝑑𝜆
cos 𝜆𝑥 − 𝑖 sin 𝜆𝑥
2
∞
sin 𝜆−𝜆cos 𝜆
⇒ 𝑓(𝑥) = ∫−∞ [cos 𝜆𝑥 (
𝜋
𝜆3
1 − 𝑥 2 , |𝑥| < 1
Substituting 𝑓(𝑥) = {
in ②
|𝑥| > 1
0,
- 62 -
sin 𝜆−𝜆cos 𝜆
) − 𝑖 sin 𝜆𝑥 (
𝜆3
)] 𝑑𝜆….②
⇒{
sin 𝜆−𝜆cos 𝜆
1 − 𝑥 2 , |𝑥| < 1 2 ∞
= ∫−∞ [cos 𝜆𝑥 (
)−
𝜋
𝜆3
|𝑥| > 1
0,
sin 𝜆−𝜆cos 𝜆
𝑖 sin 𝜆𝑥 (
𝜆3
)] 𝑑𝜆
Equating real parts on both sides, we get
𝜋
∞
sin 𝜆−𝜆cos 𝜆
) 𝑑𝜆
∫−∞ cos 𝜆𝑥 (
𝜆3
= {2
(1 − 𝑥 2 ), |𝑥| < 1
|𝑥| > 1
0,
1
Putting 𝑥 = on both sides
2
∞
𝜆 sin 𝜆−𝜆cos 𝜆
∫−∞ cos 2 (
∞
𝜆3
𝜋
𝜆 sin 𝜆−𝜆cos 𝜆
⇒ 2 ∫0 cos (
2
1
) 𝑑𝜆 = 2 (1 − 4)
𝜆3
) 𝑑𝜆 =
3𝜋
𝜆 sin 𝜆−𝜆cos 𝜆
∵ cos (
2
8
𝜆3
) is an even
function of 𝜆
𝑏
𝑏
Now by property of definite integrals ∫𝑎 𝑓(𝑥)𝑑𝑥 = ∫𝑎 𝑓(𝑦)𝑑𝑦
∞ 𝑥cos 𝑥−sin 𝑥
∴ ∫0 (
𝑥3
𝑥
3𝜋
) cos 2 𝑑𝑥 = − 16
Example 8 Find the Fourier cosine transform of 𝑓(𝑥) =
1
1+𝑥 2
Solution: To find Fourier cosine transform
2 ∞
𝐹𝑐 {𝑓(𝑥)} ≡ 𝑓𝑐̅ (λ) = √ ∫0 𝑓(𝑥) cos 𝜆𝑥 𝑑𝑥
𝜋
2
∞
⇒ 𝑓𝑐̅ (λ) = √ ∫0
𝜋
1
1+𝑥 2
cos 𝜆𝑥 𝑑𝑥 …..①
To evaluate the integral given by ①
Let 𝑔(𝑥) = 𝑒 −𝑥 …… ②
- 63 -
∞
2
𝐹𝑐 {𝑔(𝑥)} ≡ 𝑔̅𝑐 (λ) = √ ∫0 𝑔(𝑥) cos 𝜆𝑥 𝑑𝑥
𝜋
2
∞
⇒ 𝑔̅𝑐 (λ) = √ ∫0 𝑒 −𝑥 cos 𝜆𝑥 𝑑𝑥
𝜋
2
=√ [
𝑒 −𝑥
𝜋 1+𝜆2
⇒ 𝑔̅𝑐 (λ) = √
2
(− cos 𝜆𝑥 + 𝜆 sin 𝜆𝑥)]
∞
0
1
𝜋 1+𝜆2
Again taking Inverse Fourier cosine transform
2
∞
𝑔(𝑥) = ∫0
𝜋
2
1+𝜆2
∞
⇒ 𝑔(𝜆) = ∫0
𝜋
∞
⇒ ∫0
1
1+𝑥 2
1
1
1+𝑥 2
cos 𝜆𝑥 𝑑𝑥 =
𝜋
2
cos 𝜆𝑥 𝑑𝜆
cos 𝜆𝑥 𝑑𝑥
𝑔(𝜆) ….. ③
Using ② in ③, we get
∞
⇒ ∫0
1
𝜋
1+𝑥 2
cos 𝜆𝑥 𝑑𝑥 = 𝑒 −𝜆 ……④
2
Using ④ in ①, we get
2
∞
𝑓𝑐̅ (λ) = √ ∫0
𝜋
1
2 𝜋
1+𝑥 2
cos 𝜆𝑥 𝑑𝑥 = √ .
𝜋 2
Example 9 Find the Fourier sine transform of
evaluate
∞
𝑥
∫0 𝑡𝑎𝑛−1 (𝑎) 𝑠𝑖𝑛𝑥𝑑𝑥
Solution: To find Fourier sine transform
- 64 -
𝜋
𝑒 −𝜆 = √ 𝑒 −𝜆
2
𝑓(𝑥) =
𝑒 −𝑎𝑥
𝑥
and use it to
∞
2
𝐹𝑠 {𝑓(𝑥)} ≡ 𝑓𝑠̅ (λ) = √ ∫0 𝑓(𝑥) sin 𝜆𝑥 𝑑𝑥
𝜋
2 ∞𝑒
⇒ 𝑓𝑠̅ (λ) = √ ∫0
𝜋
−𝑎𝑥
𝑥
sin 𝜆𝑥 𝑑𝑥
To evaluate the integral, differentiating both sides with respect to 𝜆
𝑑
2
∞ 𝑒 −𝑎𝑥
𝑓 ̅ (λ) = √ ∫0
𝑑𝜆 𝑠
𝜋
2
𝑥
(cos 𝜆𝑥)𝑥𝑑𝑥
∞
2
𝑎
= √ ∫0 𝑒 −𝑎𝑥 𝑐𝑜𝑠𝜆𝑥𝑑𝑥 = √ 2 2
𝜋
𝜋 𝑎 +𝜆
Now integrating both sides with respect to 𝜆
2
𝑎
𝑓𝑠̅ (λ) = √ ∫ 2 2 𝑑𝜆
𝜋 𝑎 +𝜆
2
𝜆
⇒ 𝑓𝑠̅ (λ) = √ 𝑡𝑎𝑛−1 ( ) + 𝑐
𝜋
𝑎
when 𝜆 = 0, 𝑓𝑠̅ (λ) = 0, ⇒ c = 0
2
𝜆
∴ 𝑓𝑠̅ (λ) = √ 𝑡𝑎𝑛−1 ( )
𝜋
𝑎
Again taking Inverse Fourier Sine transform
2
∞
𝜆
𝑓(𝑥) = ∫0 𝑡𝑎𝑛−1 ( ) sin 𝜆𝑥 𝑑𝜆
𝜋
𝑎
Substituting 𝑓(𝑥) =
𝑒 −𝑎𝑥
𝑥
2
∞
𝑒 −𝑎𝑥
𝑥
on both sides
𝜆
= ∫0 𝑡𝑎𝑛−1 ( ) sin 𝜆𝑥 𝑑𝜆
𝜋
𝑎
Putting 𝑥 =1 on both sides
- 65 -
𝜋
2
∞
𝑒 −𝑎 = ∫0 𝑡𝑎𝑛−1
∞
𝜆
𝑎
sin 𝜆 𝑑𝜆
𝜆
𝜋
𝑎
2
⇒ ∫0 𝑡𝑎𝑛−1 sin 𝑥 𝑑𝑥 = 𝑒 −𝑎
∞ cos 𝜆𝑡
Example 10 If 𝑡 > 0 Show that i. ∫0
𝜆2 + 𝑎 2
∞ 𝜆sin 𝜆𝑡
ii. ∫0
Solution: i. Let 𝑓(𝑡) =
𝜋
2𝑎
𝜆2 + 𝑎 2
𝑑𝜆 =
𝑑𝜆 =
𝜋
2𝑎
𝜋
2
𝑒 −𝑎𝑡 , 𝑎 > 0
𝑒 𝑎𝑡 , 𝑎 ≤ 0
𝑒 −𝑎𝑡 , 𝑎 > 0, 𝑡 > 0
Taking Fourier cosine transform of 𝑓(𝑡), we get
2 ∞
𝐹𝑐 {𝑓(𝑡)} ≡ 𝑓𝑐̅ (λ) = √ ∫0 𝑓(𝑡) cos 𝜆𝑡 𝑑𝑡
𝜋
=
𝜋
∞
2
√ ∫ 𝑒 −𝑎𝑡 cos 𝜆𝑡 𝑑𝑡
2𝑎 𝜋 0
1
𝜋
𝑎
= √ 2 2
𝑎 2 𝑎 +𝜆
Also inverse Fourier cosine transform of 𝑓𝑐̅ (λ) gives 𝑓(𝑡) as:
∞
2
𝑓(𝑡) = √ ∫0 𝑓𝑐̅ (λ) cos 𝜆𝑡 𝑑𝜆
𝜋
1
𝜋
2
∞
= √ √ ∫0
𝑎 2 𝜋
∞ cos 𝜆𝑡
⇒ 𝑓(𝑡) = ∫0
∞ cos 𝜆𝑡
∴ ∫0
𝜆2 + 𝑎 2
𝑑𝜆 =
𝜆2 + 𝑎 2
𝜋
2𝑎
𝑎
𝑎2 +𝜆2
cos 𝜆𝑡 𝑑𝜆
𝑑𝜆
𝑒 −𝑎𝑡 , 𝑎 > 0
𝜋
ii. Again let 𝑔(𝑡) = 𝑒 𝑎𝑡 , 𝑎 ≤ 0, 𝑡 > 0
2
- 66 -
Taking Fourier sine transform of 𝑔(𝑡), we get
∞
2
𝐹𝑠 {𝑔(𝑡)} ≡ 𝑔̅𝑠 (λ) = √ ∫0 𝑔(𝑡) sin 𝜆𝑡 𝑑𝑡
𝜋
𝜋
∞
2
= √ ∫0 𝑒 𝑎𝑡 sin 𝜆𝑡 𝑑𝑡, 𝑎 ≤ 0
2 𝜋
∞
𝜋
= √ ∫0 𝑒 −𝑎𝑡 sin 𝜆𝑡 𝑑𝑡 , 𝑎 > 0
2
=√
𝜋
𝜆
2
𝑎2 +𝜆2
Also inverse Fourier sine transform of 𝑔̅𝑠 (λ) gives 𝑔(𝑡) as:
2
∞
𝜋
2
𝑔(𝑡) = √ ∫0 𝑔̅𝑠 (λ) sin 𝜆𝑡 𝑑𝜆
𝜋
∞
= √ √ ∫0
2 𝜋
𝜆
𝑎2 +𝜆2
∞ 𝜆sin 𝜆𝑡
⇒ 𝑔(𝑡) = ∫0
𝜆2 + 𝑎 2
∞ 𝜆sin 𝜆𝑡
𝜋
∴ ∫0
𝜆2 + 𝑎 2
𝑑𝜆 =
2
sin 𝜆𝑡 𝑑𝜆
𝑑𝜆
𝑒 𝑎𝑡 , 𝑎 ≤ 0
Example 11 Prove that Fourier transform of 𝑒
−𝑥2
2
Solution: Fourier transform of 𝑓(𝑥) is given by
̅
𝐹{𝑓(𝑥)} ≡ 𝑓 (λ)
=
∴ 𝐹 {𝑒
−𝑥2
2
} = 𝑓(𝜆) =
∞ 𝑖𝜆𝑥
1
𝑒 𝑓(𝑥)𝑑𝑥
∫
−∞
2𝜋
√
2
∞ −𝑥
∫ 𝑒 2
√2𝜋 −∞
1
𝑒 𝑖𝜆𝑥 𝑑𝑥
- 67 -
is self reciprocal.
=
2
∞ −𝑥 + 𝑖𝜆𝑥
𝑒 2
∫
√2𝜋 −∞
1
∞
1
𝑑𝑥 =
(𝑥 2 −2𝑖𝜆𝑥 + (𝑖𝜆)2 − (𝑖𝜆)2 )
∫ 𝑒2
√2𝜋 −∞
=
2 2
∞ −1(𝑥−𝑖𝜆)2 +𝑖 𝜆
2
2
∫ 𝑒
√2𝜋 −∞
=
=
1
√2𝜋
−1
∞
∫−∞ 𝑒 2
−𝜆2
𝑒 2
(𝑥−𝑖𝜆)2
(𝑥 2 − 2 𝑖𝜆𝑥)
𝑑𝑥
𝑑𝑥
𝑑𝑥
2
∞ −𝑧
𝑒 2
∫
√2𝜋 −∞
−𝜆2
=
−1
∞
∫ 𝑒2
√2𝜋 −∞
−1
=
−𝜆2
𝑒 2
1
∞ −𝑧
𝑒 2
∫
√2𝜋 0
2𝑒 2
By putting 𝑧 = (𝑥 − 𝑖𝜆)
𝑑𝑧
2
𝑑𝑧
𝑒
−𝑧2
2
being even function of
𝑧
−𝜆2
=
Put
𝑧
√2
𝑧 2
∞ −( )
𝑒 √2
∫
√2𝜋 0
2𝑒 2
𝑑𝑧
= 𝑡 ⇒ 𝑑𝑧 = √2 𝑑𝑡
−𝜆2
∴ 𝑓(𝜆) =
∞
2√2𝑒 2
√2𝜋
−𝜆2
=
∴ 𝐹 {𝑒
−𝑥2
2
2𝑒 2
√𝜋
}=𝑒
2
∫0 𝑒 −𝑡 𝑑𝑡
√𝜋
.
2
=𝑒
−𝜆2
2
∞
2
∵ ∫0 𝑒 −𝑡 𝑑𝑡 =
√𝜋
2
−𝜆2
2
Hence we see that Fourier transform of 𝑒
−𝑥2
2
is given by 𝑒
transformed to 𝜆. ∴ We can say that Fourier transform of 𝑒
- 68 -
−𝜆2
2
−𝑥2
2
. Variable 𝑥 is
is self reciprocal.
2
Example 12 Find Fourier Cosine transform of 𝑒 −𝑥 .
2 ∞
Solution: By definition, 𝐹𝑐 {𝑓(𝑥)} ≡ 𝑓𝑐̅ (λ) = √ ∫0 𝑓(𝑥) cos 𝜆𝑥 𝑑𝑥
𝜋
2
2 ∞
⇒ 𝑓𝑐̅ (λ) = √ ∫0 𝑒 −𝑥 cos 𝜆𝑥 𝑑𝑥
𝜋
∞
2
=
=
∞
2
1
∞
2 +𝑖𝜆𝑥
∫ (𝑒 −𝑥
√2𝜋 0
𝑖𝜆 2
=
=
𝑖𝜆
𝑖 𝜆
∞
−(𝑥− ) +
2
4
(𝑒
∫
√2𝜋 0
=
2
1
−𝜆2
𝑒 4
∞ −(𝑥−𝑖𝜆)
2
[∫0 𝑒
√2𝜋
2 2
𝑑𝑥 +
−𝜆2
𝑒 4
√𝜋
=
[
√2𝜋 2
+𝑒
2
) 𝑑𝑥
2
∫ ( 𝑒 −𝑥 𝑒 𝑖𝜆𝑥 + 𝑒 −𝑥 𝑒 −𝑖𝜆𝑥 )𝑑𝑥
√2𝜋 0
𝑖𝜆 2
𝑖𝜆
2
1
−(𝑥 2 −2 ( )𝑥 + ( ) − ( ) )
∞
2
2
2
(𝑒
∫
√2𝜋 0
1
𝑒 𝑖𝜆𝑥 +𝑒 −𝑖𝜆𝑥
2
𝑒 −𝑥 (
= √ ∫0
𝜋
+𝑒
∞ −(𝑥+𝑖𝜆)
2
∫0 𝑒
2 −𝑖𝜆𝑥
𝑖𝜆
2
) 𝑑𝑥
𝑖𝜆 2
2
𝑖𝜆 2
2
−(𝑥 2 +2 ( )𝑥 + ( ) − ( ) )
𝑖𝜆 2 𝑖2 𝜆2
2
4
−(𝑥+ ) +
+ 𝑒 −𝑥
) 𝑑𝑥
) 𝑑𝑥
2
𝑑𝑥]
−𝜆2
+
√𝜋
]
2
⇒ 𝑓𝑐̅ (λ) =
=
√𝜋𝑒 4
√2𝜋
1
−
√2
e
𝜆2
4
Or
2
Fourier Cosine transform of 𝑒 −𝑥 can also be found using the method given
below:
- 69 -
∞
2
2
𝑓𝑐̅ (λ) = √ ∫0 𝑒 −𝑥 cos 𝜆𝑥 𝑑𝑥 ….①
𝜋
Differentiating both sides with respect to λ
⇒
𝑑
∞
2
𝑓 ̅ (λ) = −√ ∫0
𝑑𝜆 𝑐
𝜋
2
𝑥𝑒 −𝑥 sin 𝜆𝑥 𝑑𝑥
2
2
= √ [sin 𝜆𝑥.
𝜋
𝜆
2
𝑒 −𝑥
2
∞
] −
0
∞
2
= 0 − √ ∫0 𝑒 −𝑥 cos 𝜆𝑥 𝑑𝑥
2 𝜋
⇒
⇒
𝑑 ̅
𝑓 (λ)
𝑑𝜆 𝑐
𝑓𝑐̅ (λ)
𝑑
𝑑𝜆
=−
𝜆
𝑓𝑐̅ (λ) = − 𝑓𝑐̅ (λ)
2
∞
𝑒 −𝑥
√ ∫0 𝜆cos 𝜆𝑥.
𝜋
2
2
𝑑𝑥
….②
using ① in ②
2
𝜆
2
Integrating both sides with respect to λ
2
𝜆
⇒ log 𝑓𝑐̅ (λ) = − + log 𝑘 , where log 𝑘 is the constant of
4
integration
𝜆2
⇒ 𝑓𝑐̅ (λ) = 𝐾e− 4 …..③
⇒
∞
√ ∫0
𝜋
2
𝑒
−𝑥 2
−
cos 𝜆𝑥 𝑑𝑥 = 𝑘e
Putting λ = 0 on both sides
2
∞
2
√ ∫0 𝑒 −𝑥 𝑑𝑥 = 𝑘
𝜋
2 √𝜋
𝜋 2
⇒𝑘=√ .
=
1
√2
….. ④
- 70 -
𝜆2
4
Using ④ in ③, we get
𝑓𝑐̅ (λ) =
1
√2
−
e
𝜆2
4
2
Example 13 Find Fourier transform of 𝑥𝑒 −𝑎𝑥 , 𝑎 > 0
2
Solution: By definition, 𝐹{ 𝑥𝑒 −𝑎𝑥 } = 𝑓(𝜆) =
=
=
=
=
∞
1
∫ 𝑥𝑒 −𝑎𝑥
√2𝜋 −∞
2 +𝑖𝜆𝑥
∞
1
2
𝑥𝑒 −𝑎𝑥 𝑒 𝑖𝜆𝑥 𝑑𝑥
∫
√2𝜋 −∞
𝑑𝑥
𝑖𝜆 2
𝑖𝜆
𝑖𝜆 2
−𝑎(𝑥 2 −2 ( )𝑥 + ( ) − ( ) )
∞
2𝑎
2𝑎
2𝑎
𝑥𝑒
𝑑𝑥
∫
√2𝜋 −∞
1
2
2 2
𝑖𝜆
𝑖 𝜆
∞
−𝑎(𝑥− ) +
2𝑎
4𝑎
∫ 𝑥𝑒
√2𝜋 −∞
1
−𝜆2
𝑒 4𝑎
∞
[∫−∞ (𝑥
2𝜋
√
−
𝑖𝜆
2𝑎
)𝑒
𝑑𝑥
−𝑎(𝑥−
𝑖𝜆 2
)
2𝑎
𝑑𝑥 +
∞ −𝑎(𝑥− 𝑖𝜆 )
2𝑎
∫ 𝑒
2𝑎 −∞
𝑖𝜆
2
𝑑𝑥]
−𝜆2
=
𝑒 4𝑎
√2𝜋
∞
2
[∫−∞ 𝑡𝑒 −𝑎𝑡 𝑑𝑡 +
𝑖𝜆
∞
2
−𝜆2
=
𝑒 4𝑎
√2𝜋
[0 +
∞
𝑖𝜆
2
∫ 𝑒 −𝑎𝑡 𝑑𝑡]
𝑎 0
∵
2
2
𝑡𝑒 −𝑎𝑡 is odd function and 𝑒 −𝑎𝑡 is even function in 𝑡
−𝜆2
=
𝑒 4𝑎
√2𝜋
.
𝑖𝜆
𝑎
∞
2
∫0 𝑒 −(√𝑎 𝑡) 𝑑𝑡
−𝜆2
=
𝑒 4𝑎
.
𝑖𝜆
√2𝜋 𝑎√𝑎
∞
𝑖𝜆
∫ 𝑒 −𝑎𝑡 𝑑𝑡] , Putting (𝑥 − 2𝑎) = 𝑡
2𝑎 −∞
2
∫0 𝑒 −𝑧 𝑑𝑧 , Putting √𝑎 𝑡 = 𝑧
- 71 -
−𝜆2
=
𝑒 4𝑎
.
𝑖𝜆
√2𝜋 𝑎√𝑎
14
2
∵ ∫0 𝑒 −𝑧 𝑑𝑧 =
√𝜋
2
−𝜆2
𝑖𝜆 𝑒 4𝑎
⇒ 𝑓(𝜆) =
Example
∞
√𝜋
2
.
2𝑎√2𝑎
Find
Fourier
cosine
integral
representation
of 𝑓(𝑥) =
𝑥 2, 0 < 𝑥 < 𝑎
{
0 , 𝑥>𝑎
Solution: Taking Fourier Cosine transform of 𝑓(𝑥) = {
𝑥 2, 0 < 𝑥 < 𝑎
0 , 𝑥>𝑎
2 ∞
𝐹𝑐 {𝑓(𝑥)} ≡ 𝑓𝑐̅ (λ) = √ ∫0 𝑓(𝑥) cos 𝜆𝑥 𝑑𝑥
𝜋
2 𝑎
⇒ 𝑓𝑐̅ (λ) = √ ∫0 𝑥 2 cos 𝜆𝑥 𝑑𝑥
𝜋
2
𝑠𝑖𝑛𝜆𝑥
𝜋
𝜆
= √ [(𝑥 2 ) (
2
𝑎2
⇒ 𝑓𝑐̅ (λ) = √ [(
𝜋
−𝑐𝑜𝑠𝜆𝑥
) − (2𝑥) (
𝜆
−
2
𝜆3
𝜆2
) + (2) (
−𝑠𝑖𝑛𝜆𝑥
𝜆3
)]
𝑎
0
2𝑎
) 𝑠𝑖𝑛𝜆𝑎 + 𝜆2 𝑐𝑜𝑠𝜆𝑎]
Now taking Inverse Fourier Cosine transform
2
∞
𝑎2
2
2𝑎
𝑓(𝑥) = ∫0 [( − 3 ) 𝑠𝑖𝑛𝜆𝑎 + 2 𝑐𝑜𝑠𝜆𝑎] cos 𝜆𝑥 𝑑𝜆
𝜋
𝜆
𝜆
𝜆
This is the required Fourier cosine integral representation of 𝑓(𝑥) =
𝑥 2, 0 < 𝑥 < 𝑎
{
0 , 𝑥>𝑎
Example 15 If 𝑓(𝑥) = {
sin 𝑥 , 0 < 𝑥 < 𝜋
,
0,
𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒
- 72 -
prove that
𝑓(𝑥) =
∞ 2cos 𝜆𝑥 + cos(𝜋+𝑥)𝜆+cos(𝜋−𝑥)𝜆
∫
𝜋 0
1−𝜆2
1
𝜋𝑡
𝑑𝜆 . Hence evaluate
∞ cos
∫0 1−𝑡22
𝑑𝑡
sin 𝑥 , 0 < 𝑥 < 𝜋
Solution: Given 𝑓(𝑥) = {
0,
𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒
To find Fourier cosine integral representation of 𝑓(𝑥), taking Fourier Cosine
transform of 𝑓(𝑥)
∞
2
𝜋
2
𝑓𝑐̅ (λ) = √ ∫0 𝑓(𝑥)𝑐𝑜𝑠𝜆𝑥𝑑𝑥 = √ ∫0 sin 𝑥 𝑐𝑜𝑠𝜆𝑥𝑑𝑥
𝜋
𝜋
=√
=√
=√
=√
=√
1
2𝜋
1
2𝜋
1
2𝜋
1
2𝜋
1
2𝜋
⇒ 𝑓𝑐̅ (λ) = √
1
2𝜋
Taking
2
𝜋
∫0 (sin(𝜆 + 1)𝑥 − sin(𝜆 − 1)𝑥) 𝑑𝑥
[−
[−
cos(𝜆+1)𝑥
(𝜆+1)
cos(𝜆+1)𝜋
(𝜆+1)
cos 𝜆𝜋
[ (𝜆+1) −
[
+
+
cos 𝜆𝜋
(𝜆−1)
cos(𝜆−1)𝑥 𝜋
(𝜆−1)
]
0
cos(𝜆−1)𝜋
(𝜆−1)
1
1
(𝜆+1)(𝜆−1)
−2 cos 𝜆𝜋−2
𝜆2 −1
Inverse
2 1+cos 𝜆𝜋
]=√ [
𝜋
1−𝜆2
Fourier
∞ 1+cos 𝜆𝜋
1
1−𝜆2
] cos 𝜆𝑥 𝑑𝜆
∞ 2cos 𝜆𝑥 + 2 cos 𝜆𝜋 cos 𝜆𝑥
= ∫0
𝜋
1−𝜆2
- 73 -
]
Cosine
∞
2
𝜆−1−𝜆−1
+ (𝜆+1)(𝜆−1)]
√ ∫0 𝑓𝑐̅ (λ) cos 𝜆𝑥 𝑑𝜆
𝜋
⇒ 𝑓(𝑥) = ∫0 [
𝜋
1
+ (𝜆+1) − (𝜆−1)]
(𝜆−1) cos 𝜆𝜋 − (𝜆+1) cos 𝜆𝜋
[
1
+ (𝜆+1) − (𝜆−1)]
𝑑𝜆
transform,
𝑓(𝑥) =
∞ 2cos 𝜆𝑥 + cos(𝜋+𝑥)𝜆+cos(𝜋−𝑥)𝜆
1
= ∫0
𝜋
1−𝜆2
∞ 2cos 𝜆𝑥 + cos(𝜋+𝑥)𝜆+cos(𝜋−𝑥)𝜆
1
⇒ 𝑓(𝑥) = ∫0
𝜋
1−𝜆2
𝑑𝜆
𝑑𝜆
sin 𝑥 , 0 < 𝑥 < 𝜋
Putting 𝑓(𝑥) = {
0,
𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒
1
1−𝜆2
Putting 𝑥 =
⇒
𝜋
2
on both sides
𝜋𝜆
𝜋
𝜋
∞ 2cos 2 + cos(𝜋+ 2 )𝜆+cos(𝜋− 2 )𝜆
∫
𝜋 0
1−𝜆2
1
𝜋𝜆
⇒
sin 𝑥 , 0 < 𝑥 < 𝜋
𝑑𝜆 = {
0,
𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒
∞ 2cos 𝜆𝑥 + cos(𝜋+𝑥)𝜆+cos(𝜋−𝑥)𝜆
⇒ ∫0
𝜋
∞ cos
∫0 1−𝜆22
𝜋
𝑑𝜆 =
2
𝜋𝑡
⇒
∞ cos
∫0 1−𝑡22
𝑑𝑡 =
𝑑𝜆 = 1
𝜋
2
∞
1 − 𝜆, 0 ≤ 𝜆 ≤ 1
Example 16 Solve the integral equation ∫0 𝑓(𝑥)𝑐𝑜𝑠𝜆𝑥𝑑𝑥 = {
0,
𝜆>1
∞ 𝑠𝑖𝑛2 𝑡
Hence deduce that ∫0
𝑡2
𝑑𝑡 =
𝜋
2
∞
1 − 𝜆, 0 ≤ 𝜆 ≤ 1
Solution: Given that ∫0 𝑓(𝑥)𝑐𝑜𝑠𝜆𝑥𝑑𝑥 = {
……①
0,
𝜆>1
⇒
∞
√ ∫0 𝑓(𝑥)𝑐𝑜𝑠𝜆𝑥𝑑𝑥
𝜋
2
2
={
√ (1 − 𝜆), 0 ≤ 𝜆 ≤ 1
𝜋
0, 𝜆 > 1.
2
√ (1
⇒ 𝑓𝑐̅ (λ) = { 𝜋 − 𝜆), 0 ≤ 𝜆 ≤ 1
0, 𝜆 > 1.
Taking Inverse Fourier Cosine transform
2
∞
𝑓(𝑥) = √ ∫0 𝑓𝑐̅ (λ) cos 𝜆𝑥 𝑑𝜆
𝜋
- 74 -
⇒ 𝑓(𝑥) =
1
2
∫ (1 − 𝜆) cos 𝜆𝑥 𝑑𝜆
𝜋 0
2
𝑠𝑖𝑛𝜆𝑥
𝜋
𝑥
= [(1 − 𝜆) (
2
𝑐𝑜𝑠𝑥
𝜋
𝑥2
= [−
⇒ 𝑓(𝑥) =
+
1
𝑥2
4𝑠𝑖𝑛2
𝜋𝑥 2
−𝑐𝑜𝑠𝜆𝑥
) − (−1) (
2 1−𝑐𝑜𝑠𝑥
]= [
𝑥
2
𝜋
𝑥2
𝑥2
1
)]
0
𝑥
]=
2
2 2𝑠𝑖𝑛 2
𝜋 𝑥2
….②
Using ②in ①, we get
2𝑥
∞ 4𝑠𝑖𝑛
∫0 𝜋𝑥 2 2
1 − 𝜆, 0 ≤ 𝜆 ≤ 1
𝑐𝑜𝑠𝜆𝑥𝑑𝑥 = {
0,
𝜆>1
Putting 𝜆 = 0 on both sides
2𝑥
⇒
∞ 𝑠𝑖𝑛 2
𝑑𝑥
∫
𝜋 0
𝑥2
⇒
∞ 𝑠𝑖𝑛
∫0 𝑥 2 2 𝑑𝑥
4
=1
2𝑥
Putting
𝑥
2
∞ 𝑠𝑖𝑛2 𝑡
∴ ∫0
4𝑡 2
=
𝜋
4
= 𝑡 , 𝑑𝑥 = 2𝑑𝑡
2𝑑𝑡 =
𝜋
4
∞ 𝑠𝑖𝑛2 𝑡
⇒ ∫0
𝑡2
𝑑𝑡 =
𝜋
2
Example 17 Find the function 𝑓(𝑥) if its Cosine transform is given by:
(i)
sin 𝑎𝜆
𝜆
1
(ii) {√
𝜆
(𝑎 − 2) , 𝜆 < 2𝑎
2𝜋
0
,
𝜆 ≥ 2𝑎
sin 𝑎𝜆
Solution: (i) Given that 𝑓𝑐̅ (λ) =
𝜆
Taking Inverse Fourier Cosine transform
- 75 -
∞
2
𝑓(𝑥) = ∫0 𝑓𝑐̅ (λ) cos 𝜆𝑥 𝑑𝜆
𝜋
∞ sin 𝑎𝜆
2
⇒ 𝑓(𝑥) = ∫0
𝜋
𝜆
cos 𝜆𝑥 𝑑𝜆
∞ 2sin 𝑎𝜆 cos 𝜆𝑥
1 2
= . ∫0
2 𝜋
𝜆
∞ sin(𝑎+𝑥)𝜆
1
= ∫0
𝜋
𝜆
𝑑𝜆
∞ sin (𝑎−𝑥) 𝜆
1
𝑑𝜆 + ∫0
𝜋
𝜆
𝑑𝜆
Now 0 < 𝑥 < ∞ ∴ 𝑎 + 𝑥 > 0
⇒ 𝑓(𝑥) =
1 𝜋
𝜋
{𝜋1 2𝜋
2
𝜋
𝜋 2
2
[ + ] , 𝑎 − 𝑥 > 0 𝑖. 𝑒. 𝑥 < 𝑎
[ − ] , 𝑎 − 𝑥 < 0 𝑖. 𝑒. 𝑥 > 𝑎
∞ sin 𝜆𝑥
∵ ∫0
𝑥
2
0
1 ,𝑥 < 𝑎
⇒ 𝑓(𝑥) = {
0, 𝑥 > 𝑎
1
𝜆
(𝑎 − 2) , 𝜆 < 2𝑎
(ii) Given that 𝑓𝑐̅ (λ) = {√2𝜋
0 ,
𝜆 ≥ 2𝑎
Taking Inverse Fourier Cosine transform
∞
2
𝑓(𝑥) = √ ∫0 𝑓𝑐̅ (λ) cos 𝜆𝑥 𝑑𝜆
𝜋
2𝑎 1
2
⇒ 𝑓(𝑥) = √ ∫0
𝜋
1
√
𝜆
(𝑎 − 2) cos 𝜆𝑥 𝑑𝜆
2𝜋
2𝑎
𝜆
= ∫0 (𝑎 − ) cos 𝜆𝑥 𝑑𝜆
𝜋
2
1
𝜆
sin 𝜆𝑥
= [(𝑎 − ) (
𝜋
2
1
cos 2𝑎𝑥
𝜋
2𝑥 2
= [−
+
𝑥
1
2𝑥
1
) − (− 2) (−
]=
2
1
2𝜋𝑥
- 76 -
cos 𝜆𝑥
𝑥2
)]
2𝑎
0
[1 − cos 2𝑎𝑥] =
2
𝜋
𝑑𝑥 = , 𝜆 >
sin2 𝑎𝑥
𝜋𝑥 2
Example 18 Find the function 𝑓(𝑥) if its Sine transform is given by:
(i) 𝑒 −𝑎𝜆
(ii)
𝜆
1+𝜆2
Solution: (i) Given that 𝑓𝑠̅ (λ) = 𝑒 −𝑎𝜆
Taking Inverse Fourier Sine transform
2
∞
2
∞
𝑓(𝑥) = ∫0 𝑓𝑠̅ (λ) sin 𝜆𝑥 𝑑𝜆
𝜋
2
𝑥
⇒ 𝑓(𝑥) = ∫0 𝑒 −𝑎𝜆 sin 𝜆𝑥 𝑑𝜆 = . 2 2
𝜋
𝜋 𝑎 +𝑥
(ii) Given that 𝑓𝑠̅ (λ) =
𝜆
1+𝜆2
Taking Inverse Fourier Sine transform
2
∞
2
∞
𝑓(𝑥) = ∫0 𝑓𝑠̅ (λ) sin 𝜆𝑥 𝑑𝜆
𝜋
⇒ 𝑓(𝑥) = ∫0
𝜋
2
∞
= ∫0
𝜋
2
𝜆
1+𝜆2
𝜆2
2
∞ (1+𝜆2 )−1
sin 𝜆𝑥 𝑑𝜆 = ∫0
𝜆(1+𝜆2 )
𝜋
∞ sin 𝜆𝑥
= ∫0
𝜋
sin 𝜆𝑥 𝑑𝜆
𝜆
2
∞ sin 𝜆𝑥
2
𝑑𝜆 − ∫0
𝜋
∞ sin 𝜆𝑥
⇒ 𝑓(𝑥) = 1 − ∫0
𝜋
𝜆(1+𝜆2 )
𝜆(1+𝜆2 )
𝜆(1+𝜆2 )
𝑑𝜆
……①
𝑑𝜆
∞ sin 𝜆𝑥
∵ ∫0
Differentiating with respect to 𝑥
2
∞ 𝜆 cos 𝜆𝑥
⇒ 𝑓′(𝑥) = 0 − ∫0
𝜋
𝜆(1+𝜆2 )
sin 𝜆𝑥 𝑑𝜆
𝑑𝜆
- 77 -
𝜆
𝜋
𝑑𝜆 = , 𝑥 > 0
2
2
∞ cos 𝜆𝑥
⇒ 𝑓′(𝑥) = − ∫0
𝜋
2
(1+𝜆2 )
𝑑𝜆 ……②
∞ 𝜆 sin 𝜆𝑥
Also 𝑓′′(𝑥) = ∫0
𝜋
(1+𝜆2 )
𝑑𝜆 = 𝑓(𝑥)
⇒ 𝑓 ′′ (𝑥) − 𝑓(𝑥) = 0….. ③
This is a linear differential equation with constant coefficients
③may be written as (𝐷2 − 1)𝑓(𝑥) = 0
Auxiliary equation is 𝑚2 − 1 = 0
⇒ m = ±1
Solution of ③ is given by
𝑓(𝑥) = 𝑐1 𝑒 𝑥 + 𝑐2 𝑒 −𝑥 ….. ④
⇒ 𝑓 ′ (𝑥) = 𝑐1 𝑒 𝑥 − 𝑐2 𝑒 −𝑥 …..⑤
Now from ①, 𝑓(𝑥) = 1, at 𝑥 = 0
Using in ④, we get 𝑐1 + 𝑐2 = 1 ….⑥
2
∞
Again from ②, 𝑓′(𝑥) = − ∫0
𝜋
1
(1+𝜆2 )
𝑑𝜆, at 𝑥 = 0
2
⇒ 𝑓′(𝑥) = − [tan−1 𝜆]∞
0 = −1 at 𝑥 = 0
𝜋
Using in ⑤, we get 𝑐1 − 𝑐2 = −1 …. ⑦
Solving ⑥ and ⑦, we get 𝑐1 = 0, 𝑐2 = 1
Using in ④, we get 𝑓(𝑥) = 𝑒 −𝑥
- 78 -
Note: Solution of the differential equation 𝑓 ′′ (𝑥) − 𝑓(𝑥) = 0 may be written
directly
as 𝑓(𝑥) = 𝑒 −𝑥
Example 19 Find the Fourier transform of the function 𝑓(𝑥) = 𝑒 −𝑎|𝑥| , −∞ <
𝑥<∞
Solution: 𝑓(𝑥) = {
Fourier
𝑒 𝑎𝑥 , 𝑥 < 0
𝑒 −𝑎𝑥 𝑥 ≥ 0
of 𝑓(𝑥)
transform
is
given
by
∞
∫−∞ 𝑒 𝑖𝜆𝑥 𝑓(𝑥)𝑑𝑥
0
∞
⇒ 𝑓 (̅ λ) = ∫−∞ 𝑒 𝑎𝑥 𝑒 𝑖𝜆𝑥 𝑑𝑥 + ∫0 𝑒 −𝑎𝑥 𝑒 𝑖𝜆𝑥 𝑑𝑥
0
∞
= ∫−∞ 𝑒 𝑥(𝑎+𝑖𝜆) 𝑑𝑥 + ∫0 𝑒 −𝑥(𝑎−𝑖𝜆) 𝑑𝑥
=[
⇒ 𝑓 (̅ λ) =
𝑒 𝑥(𝑎+𝑖𝜆)
1
𝑎+𝑖𝜆
𝑭{𝒆−𝒂|𝒙| } =
Example 20 Find 𝐹 −1 [
Solution: 𝐹 −1 [
]
(𝑎+𝑖𝜆) −∞
+
∴ 𝐹{𝑒 −𝑎|𝑥| } =
Result:
0
1
𝑎−𝑖𝜆
− [
=
𝑒 −𝑥(𝑎−𝑖𝜆)
(𝑎−𝑖𝜆)
∞
]
0
2𝑎
𝑎2 +𝜆2
2𝑎
𝑎2 +𝜆2
𝟐𝒂
𝟐𝒂
−𝟏
[
] = 𝒆−𝒂|𝒙|
⇒
𝑭
𝟐
𝟐
𝟐
𝟐
𝒂 +𝝀
𝒂 +𝝀
1
(9+𝜆2 )(4+𝜆2 )
1
(9+𝜆2 )(4+𝜆2 )
]
1
1
5
9+𝜆2
] = 𝐹 −1 [−
+
1
1
5
32 +𝜆2
= 𝐹 −1 [−
- 79 -
1
4+𝜆2
+
]
1
22 +𝜆2
]
̅
𝐹{𝑓(𝑥)} ≡ 𝑓(λ)
=
−1
=
30
−1
=
30
𝐹 −1 [
6
9+𝜆2
𝑒 −3|𝑥| +
]+
1
20
1
20
𝐹 −1 [
𝑒 −2|𝑥|
4
4+𝜆2
]
∵ 𝐹 −1 [
2𝑎
𝑎2 +𝜆2
] = 𝑒 −𝑎|𝑥|
Example 21 Find the Fourier transform of the function 𝑓(𝑥) = 𝑒 −𝑎𝑥 𝑈(𝑥), 𝑎 >
0
where 𝑈(𝑥) represents unit step function
0, 𝑥 <0
0, 𝑥 <0
Solution: 𝑓(𝑥) = 𝑒 −𝑎𝑥 {
= { −𝑎𝑥
1, 𝑥 ≥ 0
𝑒 , 𝑥≥0
Fourier
transform
of 𝑓(𝑥)
is
given
by
̅
𝐹{𝑓(𝑥)} ≡ 𝑓(λ)
∞
∫−∞ 𝑒 𝑖𝜆𝑥 𝑓(𝑥)𝑑𝑥
∞
⇒ 𝑓 (̅ λ) = ∫0 𝑒 −𝑎𝑥 𝑒 𝑖𝜆𝑥 𝑑𝑥
∞
= ∫0 𝑒 −𝑥(𝑎−𝑖𝜆) 𝑑𝑥
=− [
⇒ 𝑓 (̅ λ) =
∴ 𝐹{𝑓(𝑥)} =
𝑒 −𝑥(𝑎−𝑖𝜆)
(𝑎−𝑖𝜆)
]
0
1
𝑎−𝑖𝜆
1
𝑎−𝑖𝜆
or 𝐹{𝑒 −𝑎𝑥 𝑈(𝑥)} =
Result:
∞
𝑭{𝒆−𝒂𝒙 𝑼(𝒙)} =
1
𝑎−𝑖𝜆
𝟏
𝟏
] = 𝒆−𝒂𝒙 𝑼(𝒙) = 𝒆−𝒂𝒙 𝑯(𝒙)
⇒ 𝑭−𝟏 [
𝒂 − 𝒊𝝀
𝒂 − 𝒊𝝀
- 80 -
=
If Fourier transform of 𝒇(𝒙) = 𝒆−𝒂𝒙 𝑼(𝒙) is taken as
Note:
∞
∫−∞ 𝒆−𝒊𝝀𝒙 𝒆−𝒂𝒙 𝑼(𝒙)𝒅𝒙,
𝑭−𝟏 [
then
𝟏
𝒂+𝒊𝝀
] = 𝒆−𝒂𝒙 𝑼(𝒙) =
𝒆−𝒂𝒙 𝑯(𝒙)
Example 22 Find the inverse transform of the following functions:
i.
1
ii.
2−3𝑖𝜆− 𝜆2
Solution: i. 𝐹 −1 [
1
2−3𝑖𝜆− 𝜆2
1
iii.
8+6𝑖𝜆− 𝜆2
5
6−5𝑖𝜆− 𝜆2
1
1
] = 𝐹 −1 [(1−𝑖𝜆)(2−𝑖𝜆)] = 𝐹 −1 [(1−𝑖𝜆) −
1
1
(2−𝑖𝜆)
]
1
= 𝐹 −1 [(1−𝑖𝜆)] − 𝐹 −1 [(2−𝑖𝜆)]
= 𝑒 −𝑥 𝐻(𝑥) − 𝑒 −2𝑥 𝐻(𝑥)
∵ 𝐹 −1 [
1
𝑎−𝑖𝜆
]=
𝑒 −𝑎𝑥 𝐻(𝑥)
⇒ 𝐹 −1 [
ii. 𝐹 −1 [
1
(𝑒 −𝑥 − 𝑒 −2𝑥 ), 𝑥 ≥ 0
]
=
{
2−3𝑖𝜆− 𝜆2
0 , 𝑥<0
1
1
8+6𝑖𝜆− 𝜆2
1
] = 𝐹 −1 [(4+𝑖𝜆)(2+𝑖𝜆)] = 𝐹 −1 [(4+𝑖𝜆) −
1
1
]
(2+𝑖𝜆)
1
= 𝐹 −1 [(4+𝑖𝜆)] − 𝐹 −1 [(2+𝑖𝜆)]
= 𝑒 −4𝑥 𝐻(𝑥) − 𝑒 −2𝑥 𝐻(𝑥)
∵ 𝐹 −1 [
𝑒 −𝑎𝑥 𝐻(𝑥)
⇒ 𝐹 −1 [
iii. 𝐹 −1 [
5
6−5𝑖𝜆− 𝜆2
(𝑒 −𝑥 − 𝑒 −2𝑥 ), 𝑥 ≥ 0
]
=
{
8+6𝑖𝜆− 𝜆2
0 , 𝑥<0
1
1
1
] = 5 𝐹 −1 [(2−𝑖𝜆)(3−𝑖𝜆)] = 5𝐹 −1 [(2−𝑖𝜆) −
1
1
1
(3−𝑖𝜆)
= 5𝐹 −1 [(2−𝑖𝜆)] − 5𝐹 −1 [(3−𝑖𝜆)]
- 81 -
]
1
𝑎+𝑖𝜆
]=
= 5𝑒 −2𝑥 𝐻(𝑥) − 5𝑒 −3𝑥 𝐻(𝑥)
∵ 𝐹 −1 [
1
𝑎−𝑖𝜆
]=
𝑒 −𝑎𝑥 𝐻(𝑥)
⇒ 𝐹 −1 [
5(𝑒 −2𝑥 − 𝑒 −3𝑥 ), 𝑥 ≥ 0
]
=
{
6−5𝑖𝜆− 𝜆2
0 , 𝑥<0
5
Example 23 Find the Fourier transform of 𝑓(𝑥) =
Solution: We know 𝐹 −1 [
1
𝑎−𝑖𝜆
⇒ 𝐹 −1 [
⇒
2−𝑖𝑥
] = 𝑒 −𝑎𝑥 𝐻(𝑥)
1
2−𝑖𝜆
∞
1
1
] = 𝑒 −2𝑥 H(𝑥)
1
𝑒 −𝑖𝜆𝑥 𝑑𝜆 = 𝑒 −2𝑥 H(𝑥)
∫
2𝜋 −∞ 2−𝑖𝜆
Interchanging 𝑥 and 𝜆, we get
1
∞
1
𝑒 −𝑖𝜆𝑥 𝑑𝑥 = 𝑒 −2𝜆 H(𝜆)
∫
2𝜋 −∞ 2−𝑖𝑥
0, 𝜆<0
= { −2𝜆
𝑒 , 𝜆≥0
∞
⇒ ∫−∞
⇒ F{
0,
𝜆<0
𝑒 −𝑖𝜆𝑥 𝑑𝑥 = {
−2𝜆
2−𝑖𝑥
2𝜋𝑒 , 𝜆 ≥ 0
1
0,
𝜆<0
}={
−2𝜆
2−𝑖𝑥
2𝜋𝑒 , 𝜆 ≥ 0
1
5.3 Properties of Fourier Transforms
Linearity:
If 𝑓(𝜆) and 𝑔(𝜆) are Fourier transforms of 𝑓(𝑥) and 𝑔(𝑥)
respectively, then
𝐹{𝑎𝑓(𝑥) + 𝑏𝑔(𝑥)} = 𝑎𝑓(𝜆) + 𝑏𝑔(𝜆)
- 82 -
Proof: 𝐹{𝑎𝑓(𝑥) + 𝑏𝑔(𝑥)} =
=𝑎
1
∞
1
[𝑎𝑓(𝑥)
∫
−∞
2𝜋
√
∞
+ 𝑏𝑔(𝑥)] 𝑒 𝑖𝜆𝑥 𝑑𝑥
∞
1
∫ 𝑓(𝑥)𝑒 𝑖𝜆𝑥 𝑑𝑥 + 𝑏 √2𝜋 ∫−∞ 𝑔(𝑥)𝑒 𝑖𝜆𝑥 𝑑𝑥
√2𝜋 −∞
= 𝑎𝑓(𝜆) + 𝑏𝑔(𝜆)
Change of scale: If 𝑓(𝜆) is Fourier transforms of 𝑓(𝑥), then 𝐹{𝑓(𝑎𝑥)} =
1
𝑎
𝜆
𝑓( )
𝑎
Proof: 𝐹{𝑓(𝑎𝑥)} =
∞
1
∫ 𝑓(𝑎𝑥). 𝑒 𝑖𝜆𝑥
√2𝜋 −∞
Putting 𝑎𝑥 = 𝑡 ⇒ 𝑎𝑑𝑥 = 𝑑𝑡
∴ 𝐹{𝑓(𝑎𝑥)} =
𝑡
∞
𝑖𝜆 𝑑𝑡
𝑎
∫ 𝑓(𝑡). 𝑒 . 𝑎
√2𝜋 −∞
1
1
=
𝜆
∞
𝑖( )𝑡
𝑎
.
𝑑𝑡
∫ 𝑓(𝑡). 𝑒
𝑎 √2𝜋 −∞
1
1
𝜆
= 𝑓( )
𝑎
𝑎
Shifting Property: If 𝑓(𝜆) is Fourier transforms of 𝑓(𝑥), then 𝐹{𝑓(𝑥 − 𝑎)} =
𝑒 𝑖𝜆𝑎 𝑓(𝜆)
Proof: 𝐹{𝑓(𝑥 − 𝑎)} =
∞
1
∫ 𝑓(𝑥 − 𝑎). 𝑒 𝑖𝜆𝑥
√2𝜋 −∞
Putting (𝑥 − 𝑎) = 𝑡 ⇒ 𝑑𝑥 = 𝑑𝑡
∴ 𝐹{𝑓(𝑥 − 𝑎)} =
= 𝑒 𝑖𝜆𝑎
1
1
√2𝜋
∞
∫−∞ 𝑓(𝑡). 𝑒 𝑖𝜆(𝑡+𝑎) 𝑑𝑡
∞
∫ 𝑓(𝑡). 𝑒 𝑖𝜆𝑡 𝑑𝑡 = 𝑒 𝑖𝜆𝑎 𝑓(𝜆)
√2𝜋 −∞
Modulation Theorem: If 𝑓(𝜆) is Fourier transforms of 𝑓(𝑥), then
i.
1
𝐹{𝑓(𝑥) cos 𝑎𝑥} = {𝑓(𝜆 + 𝑎) + 𝑓(𝜆 − 𝑎)}
2
- 83 -
1
ii.
Fs[𝑓(𝑥) cos 𝑎𝑥] = {𝑓𝑠 (𝜆 + 𝑎) + 𝑓𝑠 (𝜆 − 𝑎)}
2
iii.
Fc[𝑓(𝑥) sin 𝑎𝑥] = {𝑓𝑠 (𝜆 + 𝑎) − 𝑓𝑠 (𝜆 − 𝑎)}
2
iv.
Fc[𝑓(𝑥) cos 𝑎𝑥] = {𝑓𝑐 (𝜆 + 𝑎) + 𝑓𝑐 (𝜆 − 𝑎)}
2
v.
Fs[𝑓(𝑥) sin 𝑎𝑥] = {𝑓𝑐 (𝜆 − 𝑎) − 𝑓𝑐 (𝜆 + 𝑎)}
2
1
1
1
Proof: i. 𝐹{𝑓(𝑥) cos 𝑎𝑥} =
∫ 𝑓(𝑥) cos 𝑎𝑥 . 𝑒 𝑖𝜆𝑥
√2𝜋 −∞
=
1
= [
1
2 √2𝜋
1
∞
1
∞
𝑒 𝑖𝑎𝑥 +𝑒 −𝑖𝑎𝑥 𝑖𝜆𝑥
𝑓(𝑥)
𝑒 𝑑𝑥
∫
2
√2𝜋 −∞
1
∞
∫−∞ 𝑓(𝑥)𝑒 𝑖(𝜆+𝑎)𝑥 𝑑𝑥 +
∞
∫ 𝑓(𝑥)𝑒 𝑖(𝜆−𝑎)𝑥 𝑑𝑥 ]
√2𝜋 −∞
1
= {𝑓 (𝜆 + 𝑎) + 𝑓(𝜆 − 𝑎)}
2
2
∞
ii. Fs[𝑓(𝑥) cos 𝑎𝑥] = √ ∫0 𝑓(𝑥) cos 𝑎𝑥 sin 𝜆𝑥 𝑑𝑥
𝜋
1
2
∞
= √ ∫0 𝑓(𝑥) [sin(𝜆 + 𝑎)𝑥 +
2 𝜋
sin(𝜆 − 𝑎)𝑥 ]𝑑𝑥
1
2
∞
2
∞
= [√ ∫0 𝑓(𝑥) sin(𝜆 + 𝑎)𝑥 𝑑𝑥 + √ ∫0 𝑓(𝑥) sin(𝜆 −
2
𝜋
𝜋
𝑎)𝑥 𝑑𝑥]
1
= {𝑓𝑠 (𝜆 + 𝑎) + 𝑓𝑠 (𝜆 − 𝑎)}
2
2
∞
iii. Fc[𝑓(𝑥) sin 𝑎𝑥] = √ ∫0 𝑓(𝑥) sin 𝑎𝑥 cos 𝜆𝑥 𝑑𝑥
𝜋
- 84 -
1
2
∞
= √ ∫0 𝑓(𝑥) [sin(𝜆 + 𝑎)𝑥 −
2 𝜋
sin(𝜆 − 𝑎)𝑥 ]𝑑𝑥
1
2
∞
2
∞
= [√ ∫0 𝑓(𝑥) sin(𝜆 + 𝑎)𝑥 𝑑𝑥 − √ ∫0 𝑓(𝑥) sin(𝜆 −
2
𝜋
𝜋
𝑎)𝑥 𝑑𝑥]
1
= {𝑓𝑠 (𝜆 + 𝑎) − 𝑓𝑠 (𝜆 − 𝑎)}
2
2
∞
2
∞
iv. Fc[𝑓(𝑥) cos 𝑎𝑥] = √ ∫0 𝑓(𝑥) cos 𝑎𝑥 cos 𝜆𝑥 𝑑𝑥
𝜋
1
= √ ∫0 𝑓(𝑥) [cos(𝜆 + 𝑎)𝑥 +
2 𝜋
cos(𝜆 − 𝑎)𝑥 ]𝑑𝑥
1
2
∞
2
∞
= [√ ∫0 𝑓(𝑥) cos(𝜆 + 𝑎)𝑥 𝑑𝑥 + √ ∫0 𝑓(𝑥) cos(𝜆 −
2
𝜋
𝜋
𝑎)𝑥 𝑑𝑥]
1
= {𝑓𝑐 (𝜆 + 𝑎) + 𝑓𝑐 (𝜆 − 𝑎)}
2
∞
2
v. Fs[𝑓(𝑥) sin 𝑎𝑥] = √ ∫0 𝑓(𝑥) sin 𝑎𝑥 sin 𝜆𝑥 𝑑𝑥
𝜋
1
2
∞
= √ ∫0 𝑓(𝑥) [cos(𝜆 − 𝑎)𝑥 −
2 𝜋
cos(𝜆 + 𝑎)𝑥 ]𝑑𝑥
- 85 -
1
2
∞
2
∞
= [√ ∫0 𝑓(𝑥) cos(𝜆 − 𝑎)𝑥 𝑑𝑥 − √ ∫0 𝑓(𝑥) cos(𝜆 +
2
𝜋
𝜋
𝑎)𝑥 𝑑𝑥]
1
= {𝑓𝑐 (𝜆 − 𝑎) − 𝑓𝑐 (𝜆 + 𝑎)}
2
Convolution theorem: Convolution of two functions 𝑓(𝑥) and 𝑔(𝑥) is defined
as
∞
𝑓(𝑥) ∗ 𝑔(𝑥) = ∫−∞ 𝑓(𝑢)𝑔(𝑥 − 𝑢)𝑑𝑢
If 𝑓(𝜆) and 𝑔(𝜆) are Fourier transforms of 𝑓(𝑥) and 𝑔(𝑥) respectively,
then
Convolution theorem for Fourier transforms states that
𝐹{𝑓(𝑥) ∗ 𝑔(𝑥)} = 𝐹{𝑓(𝑥)} . 𝐹{𝑓𝑔(𝑥)} ≡ 𝑓(𝜆). 𝑔(𝜆)
̅
Proof: By definition 𝑓(λ)
=
∞ 𝑖𝜆𝑥
1
𝑒 𝑓(𝑥)𝑑𝑥
∫
−∞
2𝜋
√
∞
and 𝑔̅ (λ) = ∫−∞ 𝑒 𝑖𝜆𝑥 𝑔(𝑥)𝑑𝑥
∞
Now 𝑓(𝑥) ∗ 𝑔(𝑥) = ∫−∞ 𝑓(𝑢)𝑔(𝑥 − 𝑢)𝑑𝑢
∞
∞
∴ 𝐹{𝑓(𝑥) ∗ 𝑔(𝑥)} = ∫−∞ 𝑒 𝑖𝜆𝑥 [∫−∞ 𝑓(𝑢)𝑔(𝑥 − 𝑢)𝑑𝑢]𝑑𝑥
Changing the order of integration, we get
∞
∞
∴ 𝐹{𝑓 ∗ 𝑔} = ∫−∞ 𝑓(𝑢)[∫−∞ 𝑒 𝑖𝜆𝑥 𝑔(𝑥 − 𝑢)𝑑𝑥 ]𝑑𝑢
Putting 𝑥 − 𝑢 = 𝑡 ⇒ 𝑑𝑥 = 𝑑𝑡 in the inner integral, we get
∞
∞
𝐹{𝑓 ∗ 𝑔} = ∫−∞ 𝑓(𝑢)[∫−∞ 𝑒 𝑖𝜆(𝑢+𝑡) 𝑔(𝑡)𝑑𝑡]𝑑𝑢
∞
∞
= ∫−∞ 𝑒 𝑖𝜆𝑢 𝑓(𝑢)[∫−∞ 𝑒 𝑖𝜆𝑡 𝑔(𝑡)𝑑𝑡]𝑑𝑢
∞
= ∫−∞ 𝑒 𝑖𝜆𝑢 𝑓(𝑢)𝑔(𝜆) 𝑑𝑢
- 86 -
∞
= 𝑔(𝜆) ∫−∞ 𝑒 𝑖𝜆𝑢 𝑓(𝑢)𝑑𝑢
= 𝑓(𝜆). 𝑔(𝜆)
Example 24
2
Find the Fourier transform of 𝑒 −𝑥 . Hence find Fourier
transforms of
i. 𝑒
−𝑎𝑥 2
, 𝑎 > 0 ii. 𝑒
−𝑥2
𝟐
𝟐
∞
1
2
Solution: F {𝑒 −𝑥 } = 𝑓(𝜆) =
2
∫ 𝑒 −𝑥 𝑒 𝑖𝜆𝑥 𝑑𝑥
√2𝜋 −∞
∞
1
2
iii. 𝑒 2(𝑥−3) iv. 𝑒 −𝑥 cos 2𝑥
2 +𝑖𝜆𝑥
=
∫ 𝑒 −𝑥
√2𝜋 −∞
=
∞ −(𝑥 2 −2 ( 2 )𝑥 + ( 2 ) − ( 2 ) )
∫ 𝑒
√2𝜋 −∞
𝑖𝜆 2
𝑖𝜆
1
=
𝑑𝑥
2
−𝜆2
=
=
=
∞ −(𝑥−𝑖𝜆)
2
∫ 𝑒
√2𝜋 −∞
𝑒 4
−𝜆2
𝑒 4
√2𝜋
∞
𝑑𝑥
𝑖𝜆
2
√2𝜋
∞
𝑑𝑥
2
∫−∞ 𝑒 −𝑧 𝑑𝑧
𝜆2
−
2𝑒 4
𝑑𝑥
2 2
∞ −(𝑥−𝑖𝜆) +𝑖 𝜆
2
4
∫ 𝑒
√2𝜋 −∞
1
𝑖𝜆 2
By putting 𝑧 = (𝑥 − )
2
2
2
∫0 𝑒 −𝑧 𝑑𝑧
𝑒 −𝑧 being even function of
𝑧
−𝜆2
∴ 𝑓(𝜆) =
2𝑒 4
√2𝜋
.
√𝜋
2
∴ We have 𝐹{𝑓(𝑥)} = 𝑓(𝜆) =
=
1
√2
1
√2
𝑒
𝑒
−𝜆2
4
−𝜆2
4
- 87 -
…..①
if 𝑓(𝑥) = 𝑒 −𝑥
2
i.
2
2
Now 𝐹{𝑒 −𝑎𝑥 } = 𝐹 {𝑒 (√𝑎 √𝑥) }
=
∴ 𝐹{𝑒
ii.
−𝑎𝑥 2
}=
Putting 𝑎 =
𝐹 {𝑒
𝑥2
−
2
1
2
1
1
𝜆
√
√
1
.
√𝑎 √2
𝑒
1 𝜆 2
)
4 √𝑎
− (
}=
.𝑒
1
2
=
1
√2𝑎
𝑒
𝜆2
4𝑎
−
Using ①in ②
in i.
𝜆2 .
1
By change of scale property…. ②
𝑓( )
𝑎
𝑎
− 4
.
= 𝑒
2
𝜆2
2
−
√2.
3
iii. To find 𝐹{𝑒 −2(𝑥−3) } , Put 𝑎 = 2 in i.
𝐹{𝑒
−2𝑥 2
∴ 𝐹{𝑒
1
} = 2𝑒
−2(𝑥−3)2
𝜆2
8
−
2
}=
𝜆
1
𝑒 3𝑖𝜆 . 𝑒 − 8
2
∵ By shifting property 𝐹{𝑓(𝑥 − 𝑘)} =
𝑒 𝑖𝜆𝑘 𝑓(𝜆)
2
iv. To find Fourier transform of 𝐹{ 𝑒 −𝑥 𝑐𝑜𝑠2𝑥}
1
𝐹{𝑓(𝑥) cos 𝑎𝑥} = 𝑓(𝜆 + 𝑎) + 𝑓(𝜆 − 𝑎) By modulation theorem
2
Now 𝐹{𝑒
∴ 𝐹{𝑒
−𝑥 2
−𝑥 2
} ≡ 𝑓(𝜆) =
1
𝑐𝑜𝑠2𝑥} = [
2
1
√2
𝑒
1
√2
𝑒
𝜆2
4
−
(𝜆+2)2
4
−
.
+
1
√2
𝑒
(𝜆−2)2
4
−
Example 25 Using Convolution theorem, find 𝐹 −1 [
Solution: 𝐹 −1 [
1
12−7𝑖𝜆− 𝜆2
1
]
1
12−7𝑖𝜆− 𝜆2
1
]
] = 𝐹 −1 [(4−𝑖𝜆)(3−𝑖𝜆)] = 𝐹 −1 [(4−𝑖𝜆) .
- 88 -
1
]
(3−𝑖𝜆)
Now by Convolution theorem
𝐹{𝑓(𝑥) ∗ 𝑔(𝑥)} = 𝑓(𝜆). 𝑔(𝜆) ⇒ 𝐹 −1 [𝑓(𝜆). 𝑔(𝜆) ] = 𝑓(𝑥) ∗ 𝑔(𝑥)
1
∴ 𝐹 −1 [(4−𝑖𝜆) .
1
1
] = 𝐹 −1 [(4−𝑖𝜆)] ∗ 𝐹 −1 [
(3−𝑖𝜆)
1
(3−𝑖𝜆)
]
= 𝑒 −4𝑥 𝐻(𝑥) ∗ 𝑒 −3𝑥 𝐻(𝑥)
∵ 𝐹 −1 [
1
𝑎−𝑖𝜆
]=
𝑒 −𝑎𝑥 𝐻(𝑥)
∞
= ∫−∞ 𝑒 −4𝑢 𝐻(𝑢)𝑒 −3(𝑥−𝑢) 𝐻(𝑥 − 𝑢)𝑑𝑢
∞
∵ 𝑓(𝑥) ∗ 𝑔(𝑥) = ∫−∞ 𝑓(𝑢)𝑔(𝑥 −
𝑢)𝑑𝑢
∞
=𝑒 −3𝑥 ∫−∞ 𝑒 −𝑢 𝐻(𝑢)𝐻(𝑥 − 𝑢)𝑑𝑢
𝐻(𝑢)𝐻(𝑥 − 𝑢) =
Now
1,
𝑢 ≥ 0, 𝑥 − 𝑢 ≥ 0, 𝑖. 𝑒. 0 ≤ 𝑢 ≤ 𝑥
{
0, 𝑢 < 0, 𝑥 − 𝑢 < 0, 𝑖. 𝑒. 𝑢 < 0 and 𝑢 > 𝑥
∴ 𝐹 −1 [
𝑥
1
12−7𝑖𝜆− 𝜆2
] = 𝑒 −3𝑥 ∫0 𝑒 −𝑢 𝑑𝑢 = −𝑒 −3𝑥 [𝑒 −𝑢 ]0𝑥 = −𝑒 −3𝑥 [𝑒 −𝑥 −
1], 𝑥 ≥ 0
= 𝑒 −3𝑥 − 𝑒 −4𝑥 , 𝑥 ≥ 0
⇒ 𝐹 −1 [
1
12−7𝑖𝜆− 𝜆
]={
2
𝑒 −3𝑥 − 𝑒 −4𝑥 , 𝑥 ≥ 0
0 , 𝑥<0
Example 26 Find the inverse Fourier transforms of
Solution: i. We know that 𝐹 −1 [
∴ 𝐹 −1 [
1
2−𝑖𝜆
1
𝑎−𝑖𝜆
] = 𝑒 −𝑎𝑥 𝐻(𝑥)
] = 𝑒 −2𝑥 𝐻(𝑥)
- 89 -
𝑒 3𝑖𝜆
2−𝑖𝜆
Now By shifting property 𝐹{𝑓(𝑥 − 𝑘)} = 𝑒 𝑖𝜆𝑘 𝑓(𝜆)
⇒ 𝐹 −1 [𝑒 𝑖𝜆𝑘 𝑓(𝜆)] = 𝑓(𝑥 − 𝑘)
∴ 𝐹 −1 [
𝑒 3𝑖𝜆
2−𝑖𝜆
] = 𝑒 −2(𝑥−3) 𝐻(𝑥 − 3)
5.4 Fourier Transforms of Derivatives
Let 𝑢(𝑥, 𝑡) be a function of two independent variables 𝑥 and 𝑡, such that
Fourier
of 𝑢(𝑥, 𝑡) is
transform
denoted
𝑢(𝜆, 𝑡)
by
i.e
𝑢(𝜆, 𝑡) =
∞
∫−∞ 𝑒 𝑖𝜆𝑥 𝑢(𝑥, 𝑡)𝑑𝑥
Again let 𝑢,
𝜕𝑢
𝜕𝑥
,
𝜕2 𝑢
𝜕𝑥 2
, … → 0 as 𝑥 → ±∞,
Then Fourier transforms of
∞
𝜕𝑢
𝜕𝑢 𝜕2 𝑢
,
𝜕𝑥 𝜕𝑥 2
, … with respect to 𝑥 are given by:
∞
𝜕𝑢
∞
1. 𝐹 { } = ∫−∞ 𝑒 𝑖𝜆𝑥 𝑑𝑥 = [𝑒 𝑖𝜆𝑥 𝑢]−∞ − 𝑖𝜆 ∫−∞ 𝑒 𝑖𝜆𝑥 𝑢𝑑𝑥 = −𝑖𝜆 𝑢(𝜆, 𝑡)
𝜕𝑥
𝜕𝑥
𝐹{
𝜕2 𝑢
𝜕𝑢 ∞
𝜕2 𝑢
∞
} = ∫−∞ 𝑒 𝑖𝜆𝑥 2 𝑑𝑥 = [𝑒 𝑖𝜆𝑥 ]
𝜕𝑥 2
𝜕𝑥
𝜕𝑥
−∞
∞
− 𝑖𝜆 ∫−∞ 𝑒 𝑖𝜆𝑥
𝜕𝑢
𝜕𝑥
𝑑𝑥 =
(−𝑖𝜆)2 𝑢(𝜆, 𝑡)
⋮
𝑭{
𝝏𝒏 𝒖
𝝏𝒙𝒏
} = (−𝒊𝝀)𝒏 𝒖(𝝀, 𝒕)
2. Fourier sine transform of
𝐹𝑠 {
𝜕2 𝑢
𝜕2 𝑢
𝜕𝑥 2
is given by:
∞ 𝜕2 𝑢
} = ∫0
𝜕𝑥 2
𝜕𝑥
sin 𝜆𝑥 𝑑𝑥 = [sin 𝜆𝑥
2
=
∞
𝜆2 ∫0 sin 𝜆𝑥
𝜕2 𝑢
𝜕𝑥 2
𝜕𝑢 ∞
- 90 -
𝜕𝑢
0
0
𝑑𝑥
∞
] − 𝜆 ∫0 cos 𝜆𝑥 𝑑𝑥
𝜕𝑥
𝜕𝑥
−𝜆[cos 𝜆𝑥 . 𝑢(𝑥, 𝑡)]∞
0 −
∴ 𝑭𝒔 {
𝝏𝟐 𝒖
𝝏𝒙𝟐
} = 𝝀𝒖(𝟎, 𝒕) − 𝝀𝟐 𝒖𝒔 (𝝀, 𝒕)
3. Fourier cosine transform of
𝐹𝑐 {
𝜕2 𝑢
𝜕2 𝑢
𝜕𝑥 2
is given by:
∞ 𝜕2 𝑢
} = ∫0
𝜕𝑥 2
𝜕𝑥
cos 𝜆𝑥 𝑑𝑥 = [cos 𝜆𝑥
2
∞
∴ 𝑭𝒄 {
𝝏𝟐 𝒖
𝝏𝒙𝟐
𝜕2 𝑢
𝜕𝑥 2
𝜕𝑢
𝜕𝑥 𝑥=0
𝝏𝒖
} = −[ ]
𝝏𝒙 𝒙=𝟎
∞
𝐹 { } = ∫−∞ 𝑒 𝑖𝜆𝑥
𝜕𝑡
𝜕𝑢
𝜕𝑥
𝑑𝑥
+ 𝜆[sin 𝜆𝑥 . 𝑢(𝑥, 𝑡)]∞
0 −
𝑑𝑥
4. Fourier transforms of
𝜕𝑢
∞
] + 𝜆 ∫0 sin 𝜆𝑥
𝜕𝑥 0
−[ ]
=
𝜆2 ∫0 cos 𝜆𝑥
𝜕𝑢 ∞
𝜕𝑢
𝜕𝑡
𝜕𝑢
𝜕𝑡
𝒅
𝝏𝒕
𝒅𝒕
𝝏𝒖
𝒅
𝝏𝒕
𝒅𝒕
𝝏𝒖
𝒅
𝝏𝒕
𝒅𝒕
Similarly 𝑭𝒔 { } =
𝑭𝒄 { } =
with respect to 𝑥 are given by:
𝑑𝑥 =
𝝏𝒖
∴ 𝑭{ } =
− 𝝀𝟐 𝒖𝒄 (𝝀, 𝒕)
𝑑
∞
∫ 𝑒 𝑖𝜆𝑥 𝑢(𝑥, 𝑡)𝑑𝑥
𝑑𝑡 −∞
𝒖(𝝀, 𝒕)
𝒖𝒔 (𝝀, 𝒕)
𝒖𝒄 (𝝀, 𝒕)
5.5 Applications of Fourier Transforms to boundary value problems
Partial differential equation together with boundary and initial conditions can be
easily solved using Fourier transforms. In one dimensional boundary value
problems, the partial differential equations can easily be transformed into an
ordinary differential equation by applying a suitable transform and solution to
boundary value problem is obtained by applying inverse transform. In two
dimensional problems, it is sometimes required to apply the transforms twice
and the desired solution is obtained by double inversion.
- 91 -
Algorithm to solve partial differential equations with boundary values:
1. Apply the suitable transform to given partial differential equation. For this
check the range of 𝑥
i.
If −∞ < 𝑥 < ∞, then apply Fourier transform.
ii.
If 0 < 𝑥 < ∞, then check initial value conditions
a) If value of 𝑢(0, 𝑡) is given, then apply Fourier sine transform
𝜕𝑢
b) If value of [ ]
𝜕𝑥 𝑥=0
is given, then apply Fourier cosine
transform
An ordinary differential equation will be formed after applying the
transform.
2. Solve the differential equation using usual methods.
3. Apply Boundary value conditions to evaluate arbitrary constants.
4. Apply inverse transform to get the required expression for 𝑢(𝑥, 𝑡).
Example 27 The temperature 𝑢(𝑥, 𝑡) at any point of an infinite bar satisfies the
𝜕𝑢
equation
𝜕𝑡
=
𝜕2 𝑢
𝜕𝑥 2
, −∞ < 𝑥 < ∞, 𝑡 > 0 and the initial
temperature along the length
of
1 𝑓𝑜𝑟 |𝑥| < 1
the bar is given by 𝑢(𝑥, 0) = {
0 𝑓𝑜𝑟 |𝑥| > 1
Determine the expression for 𝑢(𝑥, 𝑡).
Solution: As range of 𝑥 is (−∞, ∞), applying Fourier transform to both sides of
the
given equation :
𝜕𝑢
𝜕2 𝑢
𝜕𝑡
𝜕𝑥 2
𝐹{ } = 𝐹{
}
- 92 -
𝑑
⇒
𝑑𝑡
𝑢(𝜆, 𝑡) = −𝜆2 𝑢(𝜆, 𝑡)
𝜕𝑢
𝑑
𝜕𝑡
𝑑𝑡
∵ 𝐹{ }=
𝑢(𝜆, 𝑡) and
𝐹{
𝜕2 𝑢
𝜕𝑥 2
}=
(−𝑖𝜆)2 𝑢(𝜆, 𝑡)
Rearranging the ordinary differential equation in variable separable
form:
⇒
𝑑𝑢
𝑢
= −𝜆2 𝑑𝑡 … ①
where 𝑢 ≈ 𝑢(𝜆, 𝑡)
Solving ① using usual methods of variable separable differential
equations
log 𝑢 = −𝜆2 𝑡 + log 𝐴
𝑢
⇒ log = −𝜆2 𝑡
𝐴
2
⇒ 𝑢(𝜆, 𝑡) = 𝐴 𝑒 −𝜆 𝑡 … ②
Putting 𝑡 = 0 on both sides
⇒ 𝑢(𝜆, 0) = 𝐴 … ③
1 𝑓𝑜𝑟 |𝑥| < 1
Now given that 𝑢(𝑥, 0) = {
0 𝑓𝑜𝑟 |𝑥| > 1
Taking Fourier transform on both sides, we get
⇒ 𝑢 (𝜆, 0) =
=
=
=
1
∞
1
1
∫ 𝑢(𝑥, 0) 𝑒 𝑖𝜆𝑥 𝑑𝑥
√2𝜋 −∞
∫ 𝑒 𝑖𝜆𝑥 𝑑𝑥
√2𝜋 −1
1
√2𝜋
1
√2𝜋
1
1
[𝑒 𝑖𝜆𝑥 ]−1
𝑖𝜆
1
[𝑒 𝑖𝜆 − 𝑒 −𝑖𝜆 ] =
𝑖𝜆
- 93 -
1
2𝑖 𝑒 𝑖𝜆 −𝑒 −𝑖𝜆
√2𝜋 𝑖𝜆
[
2𝑖
]
⇒ 𝑢 (𝜆, 0) =
2
𝑠𝑖𝑛𝜆
√2𝜋
𝜆
… ④
From ③and ④, we get
𝐴=
2
𝑠𝑖𝑛𝜆
√2𝜋
𝜆
…⑤
Using ⑤ in ②, we get
𝑢(𝜆, 𝑡) =
2
𝑠𝑖𝑛𝜆
√2𝜋
𝜆
2𝑡
𝑒 −𝜆
Taking Inverse Fourier transform
𝑢(𝑥, 𝑡) =
⇒ 𝑢(𝑥, 𝑡) =
∞
1
∫ 𝑒 −𝑖𝜆𝑥 𝑢(𝜆, 𝑡)𝑑𝜆
√2𝜋 −∞
2
∞ 𝑠𝑖𝑛𝜆
∫
2𝜋 −∞
1
∞ 𝑠𝑖𝑛𝜆
⇒ 𝑢(𝑥, 𝑡) = ∫−∞
𝜋
2
𝜆
∞
𝜆
2
2
𝑒 −𝜆 𝑡 𝑒 −𝑖𝜆𝑥 𝑑𝜆
2
𝑒 −𝜆 𝑡 (cos 𝜆𝑥 − 𝑖 sin 𝜆𝑥)𝑑𝜆
sin 𝜆 cos 𝜆𝑥
⇒ 𝑢(𝑥, 𝑡) = ∫0 𝑒 −𝜆 𝑡 (
𝜋
𝜆
) 𝑑𝜆
sin 𝜆 sin 𝜆𝑥
∵(
𝜆
) is odd function
of 𝜆 Example 28 Using Fourier transform, solve the equation
𝑥 < ∞, 𝑡 > 0
𝜕𝑢
𝜕𝑡
=𝑘
𝜕2 𝑢
𝜕𝑥 2
,0 <
subject to
conditions:
i.
𝑢(0, 𝑡) = 0, 𝑡 > 0
ii.
𝑢(𝑥, 0) = 𝑒 −𝑥 , 𝑥 > 0
iii.
𝑢 and
𝜕𝑢
𝜕𝑥
both tend to zero as 𝑥 → ±∞
Solution: As range of 𝑥 is (0, ∞), and also value of 𝑢(0, 𝑡) is given in initial
value
conditions, applying Fourier sine transform to both sides of
the given equation:
- 94 -
𝜕𝑢
𝜕2 𝑢
𝜕𝑡
𝜕𝑥 2
𝐹𝑠 { } = 𝑘𝐹𝑠 {
𝑑
⇒
𝑑𝑡
}
𝑢𝑠 (𝜆, 𝑡) = 𝑘𝜆𝑢(0, 𝑡) − 𝑘𝜆2 𝑢𝑠 (𝜆, 𝑡)
𝜕𝑢
𝑑
𝜕𝑡
𝑑𝑡
∵ 𝐹𝑠 { } =
𝑢𝑠 (𝜆, 𝑡) and
𝐹𝑠 {
𝜕2 𝑢
𝜕𝑥 2
} = 𝜆𝑢(0, 𝑡) −
𝜆2 𝑢𝑠 (𝜆, 𝑡)
𝑑
⇒
𝑑𝑡
𝑢𝑠 (𝜆, 𝑡) = −𝑘𝜆2 𝑢𝑠 (𝜆, 𝑡)
∵ 𝑢(0, 𝑡) = 0
Rearranging the ordinary differential equation in variable separable form:
⇒
𝑑𝑢
𝑢
= −𝑘𝜆2 𝑑𝑡 … ①
where 𝑢 ≈ 𝑢𝑠 (𝜆, 𝑡)
Solving ① using usual methods of variable separable differential
equations
log 𝑢 = −𝑘𝜆2 𝑡 + log 𝐴
𝑢
⇒ log = −𝑘𝜆2 𝑡
𝐴
2
⇒ 𝑢𝑠 (𝜆, 𝑡) = 𝐴 𝑒 −k𝜆 𝑡 … ②
Putting 𝑡 = 0 on both sides
⇒ 𝑢𝑠 (𝜆, 0) = 𝐴 … ③
Now given that 𝑢(𝑥, 0) = 𝑒 −𝑥
Taking Fourier sine transform on both sides, we get
2
∞
⇒ 𝑢𝑠 (𝜆, 0) = √ ∫0 𝑢(𝑥, 0) sin 𝜆𝑥 𝑑𝑥
𝜋
- 95 -
∞
2
= √ ∫0 𝑒 −𝑥 sin 𝜆𝑥 𝑑𝑥
𝜋
⇒ 𝑢𝑠 (𝜆, 0) = √
2
𝜆
𝜋 1+𝜆2
… ④
From ③and ④, we get
𝐴=√
2
𝜆
𝜋 1+𝜆2
…⑤
Using ⑤ in ②, we get
𝑢𝑠 (𝜆, 𝑡) = √
2
𝜆
𝜋
1+𝜆2
2𝑡
𝑒 −𝑘𝜆
Taking Inverse Fourier sine transform
∞
2
𝑢(𝑥, 𝑡) = √ ∫0 𝑢𝑠 (𝜆, 𝑡) sin 𝜆𝑥 𝑑𝜆
𝜋
2
∞
⇒ 𝑢(𝑥, 𝑡) = ∫0
𝜋
Example 29
𝜆
1+𝜆2
2
𝑒 −𝑘𝜆 𝑡 sin 𝜆𝑥 𝑑𝜆
The temperature 𝑢(𝑥, 𝑡) in a semi-infinite rod 0 < 𝑥 < ∞ is
determined
by the differential equation
𝜕𝑢
𝜕𝑡
=2
𝜕2 𝑢
𝜕𝑥 2
subject to
conditions:
i.
ii.
𝑢 = 0, when 𝑡 = 0 , 𝑥 ≥ 0
𝜕𝑢
𝜕𝑥
= −𝑘 (𝑎 𝑐𝑜𝑠𝑡𝑎𝑛𝑡), when 𝑥 = 0, 𝑡 > 0
𝜕𝑢
Solution: As range of 𝑥 is (0, ∞), and also value of [ ]
𝜕𝑥 𝑥=0
value
is given in initial
conditions, applying Fourier cosine transform to both sides
of the equation:
- 96 -
𝜕𝑢
𝜕2 𝑢
𝜕𝑡
𝜕𝑥 2
𝐹𝑐 { } = 2𝐹𝑐 {
𝑑
⇒
𝑑𝑡
𝜕𝑢
𝑢𝑐 (𝜆, 𝑡) = −2 [ ]
𝜕𝑥 𝑥=0
− 2𝜆2 𝑢𝑐 (𝜆, 𝑡)
𝜕𝑢
𝑑
𝜕𝑡
𝑑𝑡
∵ 𝐹𝑐 { } =
}
𝑢𝑐 (𝜆, 𝑡) and
𝐹𝑐 {
𝜕2 𝑢
𝜕𝑥 2
𝜕𝑢
} = −[ ]
𝜕𝑥 𝑥=0
𝜆2 𝑢𝑐 (𝜆, 𝑡)
⇒
⇒
𝑑
𝑢𝑐 (𝜆, 𝑡) = 2𝑘 − 2𝜆2 𝑢𝑐 (𝜆, 𝑡)
𝑑𝑡
𝑑𝑢
𝑑𝑡
+ 2𝜆2 𝑢 = 2𝑘 … ①
where 𝑢 ≈ 𝑢𝑐 (𝜆, 𝑡)
This is a linear differential equation of the form
𝑑𝑦
𝑑𝑥
+ 𝑃𝑦 = 𝑄
where 𝑃 = 2𝜆2 , 𝑄 = 2𝑘
2 𝑑𝑡
Integrating Factor (IF) = 𝑒 ∫ 𝑃𝑑𝑡 = 𝑒 ∫ 2𝜆
2𝑡
= 𝑒 2𝜆
Solution of ① is given by
2
2
𝑢.𝑒 2𝜆 𝑡 = ∫ 2𝑘. 𝑒 2𝜆 𝑡 𝑑𝑡 + 𝐴
2
⇒ 𝑢.𝑒 2𝜆 𝑡 =
2
2𝑘𝑒 2𝜆 𝑡
2𝜆2
⇒ 𝑢𝑐 (𝜆, 𝑡) =
𝑘
𝜆2
+𝐴
2
+ 𝐴𝑒 −2𝜆 𝑡 … ②
Putting 𝑡 = 0 on both sides
⇒ 𝑢𝑐 (𝜆, 0) =
𝑘
𝜆2
+𝐴 … ③
Now given that 𝑢(𝑥, 0) = 0
Taking Fourier cosine transform on both sides, we get
- 97 -
−
∞
⇒ 𝑢𝑐 (𝜆, 0) = ∫0 𝑢(𝑥, 0) cos 𝜆𝑥 𝑑𝑥 = 0
⇒ 𝑢𝑐 (𝜆, 0) = 0 … ④
From ③and ④, we get
𝐴=−
𝑘
𝜆2
…⑤
Using ⑤ in ②, we get
𝑢𝑐 (𝜆, 𝑡) =
𝑘
2
(1 − 𝑒 −2𝜆 𝑡 )
𝜆2
Taking Inverse Fourier cosine transform
2
∞
𝑢(𝑥, 𝑡) = ∫0 𝑢𝑐 (𝜆, 𝑡) cos 𝜆𝑥 𝑑𝜆
𝜋
⇒ 𝑢(𝑥, 𝑡) =
2
∞ 1−𝑒 −2𝜆 𝑡
∫ ( 𝜆2 ) cos 𝜆𝑥 𝑑𝜆
𝜋 0
2𝑘
Example 30 Using Fourier transforms, solve the equation
𝑡>0
𝜕𝑦
𝜕𝑡
=𝑘
𝜕2 𝑦
𝜕𝑥 2
,𝑥 > 0 ,
subject to conditions:
i.
𝑦 = 𝛼, 𝑤ℎ𝑒𝑛 𝑥 = 0, 𝑡 > 0
ii.
𝑦 = 0, 𝑤ℎ𝑒𝑛 𝑡 = 0, 𝑥 > 0
Solution: As range of 𝑥 is (0, ∞), and also value of 𝑦(0, 𝑡) is given in initial
value
conditions, applying Fourier sine transform to both sides of
the given equation:
𝜕𝑦
𝜕2 𝑦
𝜕𝑡
𝜕𝑥 2
𝐹𝑠 { } = 𝑘𝐹𝑠 {
⇒
𝑑
𝑑𝑡
}
𝑦𝑠 (𝜆, 𝑡) = 𝑘𝜆𝑦(0, 𝑡) − 𝑘𝜆2 𝑦𝑠 (𝜆, 𝑡)
- 98 -
𝜕𝑦
𝑑
𝜕𝑡
𝑑𝑡
∵ 𝐹𝑠 { } =
𝑦𝑠 (𝜆, 𝑡) and
𝐹𝑠 {
𝜕2 𝑦
𝜕𝑥 2
} = 𝜆𝑦(0, 𝑡) −
𝜆2 𝑦𝑠 (𝜆, 𝑡)
𝑑
⇒
𝑑𝑡
⇒
𝑦𝑠 (𝜆, 𝑡) = 𝑘𝛼𝜆 − 𝑘𝜆2 𝑦𝑠 (𝜆, 𝑡)
𝑑𝑦
𝑑𝑡
∵ 𝑦(0, 𝑡) = 𝛼
+ 𝑘𝜆2 𝑦 = 𝑘𝛼𝜆 … ①
where 𝑦 ≈ 𝑦𝑠 (𝜆, 𝑡)
This is a linear differential equation of the form
𝑑𝑦
𝑑𝑥
where 𝑃 = 𝑘𝜆2 , 𝑄 = 𝑘𝛼𝜆
2 𝑑𝑡
Integrating Factor (IF) = 𝑒 ∫ 𝑃𝑑𝑡 = 𝑒 ∫ 𝑘𝜆
2𝑡
= 𝑒 𝑘𝜆
Solution of ① is given by
2
2
𝑦.𝑒 𝑘𝜆 𝑡 = ∫ 𝑘𝛼𝜆. 𝑒 𝑘𝜆 𝑡 𝑑𝑡 + 𝐴
⇒ 𝑦.𝑒
𝑘𝜆2 𝑡
=
2
𝑘𝛼𝜆𝑒 𝑘𝜆 𝑡
𝑘𝜆2
+𝐴
𝛼
2
⇒ 𝑦𝑠 (𝜆, 𝑡) = + 𝐴𝑒 −𝑘𝜆 𝑡 … ②
𝜆
Putting 𝑡 = 0 on both sides
𝛼
⇒ 𝑦𝑐 (𝜆, 0) = + 𝐴 … ③
𝜆
Now given that 𝑦(𝑥, 0) = 0
Taking Fourier sine transform on both sides, we get
∞
⇒ 𝑦𝑠 (𝜆, 0) = ∫0 𝑦(𝑥, 0) sin 𝜆𝑥 𝑑𝑥 = 0
⇒ 𝑦𝑠 (𝜆, 0) = 0 … ④
- 99 -
+ 𝑃𝑦 = 𝑄
From ③and ④, we get
𝛼
𝐴=− …⑤
𝜆
Using ⑤ in ②, we get
𝛼
2
𝑦𝑠 (𝜆, 𝑡) = (1 − 𝑒 −𝑘𝜆 𝑡 )
𝜆
Taking Inverse Fourier sine transform
∞
2
𝑦(𝑥, 𝑡) = ∫0 𝑦𝑠 (𝜆, 𝑡) sin 𝜆𝑥 𝑑𝜆
𝜋
⇒ 𝑦(𝑥, 𝑡) =
2
∞ 1−𝑒 −𝑘𝜆 𝑡
∫ ( 𝜆 ) sin 𝜆𝑥 𝑑𝜆
𝜋 0
2𝛼
Example 31 An infinite string is initially at rest and its initial displacement is
by 𝑓(𝑥), −∞ < 𝑥 < ∞. Determine the displacement
given
𝑦(𝑥, 𝑡) of the string.
Solution: The equation of the vibrating string is given by
𝜕2 𝑦
𝜕𝑡 2
= 𝑐2
𝜕2 𝑦
𝜕𝑥 2
Initial conditions are
i.
ii.
𝜕𝑦
]
𝜕𝑡 𝑡=0
=0
𝑦(𝑥, 0) = 𝑓(𝑥)
Taking Fourier transform on both sides
𝐹{
⇒
𝜕2 𝑦
𝜕2 𝑦
𝜕𝑡
𝜕𝑥 2
} = 𝑐2𝐹 {
2
𝑑2
𝑑𝑡 2
}
𝑦(𝜆, 𝑡) = −𝑐 2 𝜆2 𝑦(𝜆, 𝑡) where 𝐹{𝑦(𝑥, 𝑡)} ≡ 𝑦(𝜆, 𝑡)
- 100 -
⇒
𝑑2𝑦
𝑑𝑡 2
+ 𝑐 2 𝜆2 𝑦 = 0 … ①
Solution of ① is given by
𝑦(𝜆, 𝑡) = 𝐴 cos 𝑐𝑝𝑡 + 𝐵 sin 𝑐𝑝𝑡 …②
Putting 𝑡 = 0 on both sides
𝑦(𝜆, 0 ) = 𝐴 … ③
Given that 𝑦(𝑥, 0) = 𝑓(𝑥)
⇒ 𝑦(𝜆, 0 ) = 𝑓(𝜆) … ④
From ③ and ④
𝐴 = 𝑓(𝜆) … ⑤
Using ⑤ in ②
𝑦(𝜆, 𝑡) = 𝑓(𝜆)cos 𝑐𝑝𝑡 + 𝐵 sin 𝑐𝑝𝑡 …⑥
⇒
⇒
𝜕𝑦
𝜕𝑡
𝜕𝑦
= −𝑐𝑝𝑓(𝜆)sin 𝑐𝑝𝑡 + 𝑐𝑝𝐵 cos 𝑐𝑝𝑡
]
𝜕𝑡 𝑡=0
= 𝑐𝑝𝐵 … ⑦
Also given that
𝜕𝑦
]
𝜕𝑡 𝑡=0
= 0… ⑧
From ⑦ and ⑧, we get 𝐵 = 0 … ⑨
Using ⑨ in ⑥, we get
𝑦(𝜆, 𝑡) = 𝑓(𝜆) cos 𝑐𝑝𝑡
Taking inverse Fourier transform
- 101 -
where 𝑦 ≈ 𝑦(𝜆, 𝑡)
𝑦(𝑥, 𝑡) =
∞
1
𝑦(𝜆, 𝑡)𝑒 −𝑖𝜆𝑥 𝑑𝜆
∫
−∞
2𝜋
√
⇒ 𝑦(𝑥, 𝑡) =
1
∞
∫ 𝑓(𝜆) cos 𝑐𝑝𝑡 𝑒 −𝑖𝜆𝑥 𝑑𝜆
√2𝜋 −∞
Exercise 2A
𝑎 − |𝑥|, |𝑥| ≤ 𝑎
1. Find the Fourier transform of 𝑓(𝑥) = {
0, |𝑥| ≤ 𝑎
∞ sin2 𝑥
Hence prove that ∫0
𝑥2
𝑑𝑥 =
𝜋
2
∞
2. Solve the integral equation ∫0 𝑓(𝑥) cos 𝜆𝑥 𝑑𝑥 = 𝑒 −𝜆 , 𝜆 > 0
𝑥, 0<𝑥<1
3. Obtain Fourier sine integral of the function 𝑓(𝑥) = {2 − 𝑥, 1 < 𝑥 < 2
0, 𝑥 >2
|𝑥| ≤ 1
is given
𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒
1,
4. Prove that Fourier integral of the function 𝑓(𝑥) = {
0,
by
5. Find the Fourier sine and cosine transforms of 𝑥𝑒 −𝑎𝑥
2
∞ sin 𝜆 cos 𝜆𝑥
𝑓(𝑥) = ∫0
𝜋
𝜆
∞ sin 𝑥
𝑑𝜆.
Hence show that ∫0
𝑥
𝑑𝑥 =
𝜋
2
6. The temperature 𝑢(𝑥, 𝑡) at any point of a semi infinite bar satisfies the
𝜕𝑢
equation
𝜕𝑡
=
𝜕2 𝑢
𝜕𝑥 2
, 0 < 𝑥 < ∞, 𝑡 > 0 , subject to conditions
i. 𝑢(0, 𝑡) = 0, t > 0
ii.
1,
. 𝑢(𝑥, 0) = {
0,
0<𝑥<1
Determine the expression for
𝑥>1
𝑢(𝑥, 𝑡)
- 102 -
7. Determine the distribution of temperature in the semi infinite medium, 𝑥 ≥ 0,
end at 𝑥 = 0 is maintained at zero temperature and initial
when the
distribution of temperature is
𝑓(𝑥).
Answers
1.
2
2(1−cos 𝑎𝜆)
∞ (2 sin 𝜆−sin 2𝜆)
∫
𝜋 0
5.
2. 𝑓(𝑥) =
𝜆2
𝜆2
(𝑎2 +𝜆
,
2 )2
(𝑎2 +𝜆2 )2
2
∞
3. 𝑓(𝑥) =
𝜋(1+𝑥 2 )
sin 𝜆𝑥 𝑑𝜆
𝑎2 −𝜆2
2𝑎𝜆
2
2
∞ 1−cos 𝜆
6. 𝑢(𝑥, 𝑡) = ∫0
𝜋
7. 𝑢(𝑥, 𝑡) = ∫0 𝑓𝑠̅ (𝜆)𝑒 −𝑐
𝜋
2 𝜆2 𝑡
sin 𝜆𝑥 𝑑𝜆
- 103 -
𝜆
2
𝑒 −𝜆 𝑡 sin 𝜆𝑥 𝑑𝜆
Chapter 6
Power Series Solutions of Linear Differential Equations
6.1
Review of Properties of Power Series
6.2
Solutions about Ordinary Points
6.3
Solutions about Regular Singular Points - The Method of Frobenius
6.4
Bessel’s Equations and Functions
6.5
Legendre’s Equations and Polynomials
6.6
Orthogonality of Functions
6.7
Sturm – Liouville Theory
6.8
Exercises
We have seen in chapter 5 that we can solve linear differential equations
of order two or more with constant coefficients. The Cauchy-Euler equation is
exception. In fact most linear differential equations of higher order with variable
coefficients cannot be solved in terms of elementary functions. The usual
strategy for solving such type of equations is to assume a solution in the form of
an infinite series and proceed in a manner similar to the method of undetermined
coefficients (Section 5.6). Since these series solutions often turn out to be power
series, it is appropriate to summarise properties of power series in the first
section of this chapter. We conclude this chapter with the Sturm-Liouville
theory dealing with eigenvalues and eigenfunctions. Strum-Liouville’s
differential equation includes Bessel’s and Legendre’s equations as special
cases. Examples of Strum-Liouville problems are presented.
- 104 -
6.1 Review of Properties of Power Series
A power series in (x-a) is an infinite series of the form
c0+ c1 (x-a) + c2 (x-a)2 +- - - - =


n =o
c n ( x − a)n
(6.1)
Series of (6.1) is also called a power series centered at a. The power
series centered at a=0 is often referred as the power series, that is, the series


n=o
c n x n A power series centered at a is called convergent at a specified value
N
of x if its sequence of partial sums SN(x) =  c n ( x − a)n , that is, {SN (x)} is
n =o
convergent. In other words the limit of {SN (x)} exists. If the limit does not exist
the power series is called divergent. The set of points x at which the power
series is convergent is called the interval of convergence of the power series.

For R >o, a power series

n =o
c n ( x − a)n converges if x − a <R and diverges if
x − a >R. If the series converges only at a then R=0, and if it converges for all x
then R=. x − a <R is equivalent to a-R<x<a+R. A power series may or may not
converge at the end points a-R and a+R of this interval.

A power series is called absolutely convergent if the series
c
n =o
n
( x − a)n
converges. A power series converges absolutely within its interval of
convergence. By the Ratio test a power series centered at a, series given in (6.1)
is absolutely convergent if L= x-a nlim
→
c n+1
cn
is less than 1, that is, L <1, the
series diverges if L>1, and test fails if L=1. A power series defines a function

f(x)=

n =o
c n ( x − a)n whose domain is the interval of convergence of the series. If
the radius of convergence R>o, then f is continuous, differentiable and
- 105 -
integrable on the interval (a-R, a+R). Moreover f’(x) and f(x)dx can be found
by term by term differentiation and integration. Convergence at an endpoint
may be either lost by differentiation or gained through integration.

Let y =

n=o

y' =

n =o
y” =


n =o
cnxn
nc n x n−1
n(n − 1)c n x n−2
We observe that the first term in y' and first two terms in y' are zero.
Keeping this in mind we can write

y' =

n=1

y'' =

n=2

Identity property:
If

n =o
(6.2)
nc n x n−1
n(n − 1)c n x n−2
c n ( x − a)n =0, R>o for all x in the interval of
convergence, then cn=0 for all n.
Analytic at a point. A function f is analytic at a point a if it can be represented
by a power series in x-a with a positive or infinite radius of convergence. A
power series where cn=
f ( n ) (a )
, that is, the series of the type
n!
- 106 -


n =o
cn
f ( n ) (a )
( x − a) n
n!
is called the Taylor series. If a=o then Taylor series is called Maclaurin series. In
calculus it is shown that ex, cos x, sin x, ln (x-1) can be written in the form of a
power series more precisely in the form of Maclaurin series. For example
x2
+---2!
x3 x5
sin x = x −
+
---3! 5!
x2 x4 x6
cos x = 1 −
+
−
+---2! 4! 6!
for | x |  .
e x = 1+ x +
Arithmetic of Power Series: Two power series can be combined through the
operation of addition, multiplication, and division. The procedures for power
series are similar to those by which two polynomials are added, multiplied, and
divided. For example:



x2 x3 x4
x3
x5
x7
e x sin x = 1 + x +
+
+
+ - - - -  x −
+
−
+ - - - - 
2
6 24
6 120 5040



1
1  5
 1 1
 1 1
 1
= (1)x + x 2 +  − +  x 3 +  − +  x 4 + - - - - 
−
+
x + - - -  6 2
 6 6
 120 12 24 
= x + x2 +
x3 x5
−
−---3 30
Since the power series for ex and sin x converge for x<, the product
series converges on the same interval.
Shifting the Summation Index: In order to discuss power series solutions of
differential equations it is advisable to learn combining two or more summations
as a single summation.

Example 6.1 Express

n=2
n(n − 1)c n x n−2 +


n =0
- 107 -
c n x n+1 as one power series.
Solution:
In order to add the two given series, it is necessary that both
summation indices start with the same number and the powers of x in each series
be such that if one series starts with a multiple of x to the first power, then we
want that the other series to start with the same power. In this problem the first
series starts with xo where as the second series starts with x1. By writing the first
term of the first series outside the summation notation,


n=2


n =0
n =3
n(n − 1)c n x n +  c n x n+1 =2.1c2x0+  n(n − 1)c n x n +
−2
−2


n =0
c n x n+1
Both series on the right hand side start with the same power of x, namely
x1. Let k=n-2 and k=n+1 respectively in first and second series. Then the right
hand becomes


k =1
k =1
2 c2+  (k + 2)(k + 1)c k +2 x k +  c k −1x k
(6.3)
Keeping in mind that it is the value of the summation index that is
important not the summation index which is a dummy variable say k=n -1 or
k=n+1. Now we are in position to add the series in (6.3) term by term and we
have


n=2

n(n − 1)c n x n−2 +  c n x n+1
n =0

=2c2+  [(k + 2)(k + 1)c k +2 x + c k −1 ]x k .
k =1
6.2 Solution about Ordinary Point
We look for power series solution of linear second-order differential
equation about a special point:
a 2 ( x)
d2 y
dy
+ a1 ( x )
+ a 0 ( x )y = 0
2
dx
dx
(6.4)
- 108 -
where a2 (x)  0.
This can be put into the standard form
d 2 y a1 ( x ) dy a 0 ( x )
+
+
y=0
dx 2 a 2 ( x ) dx a 2 ( x )
or
d2 y
dy
+ P( x )
+ Q( x )y = 0
2
dx
dx
(6.5)
A point xo is said to be an ordinary point of the differential equation
(6.4) if P(x) and Q (x) of (6.5) are analytic at x o, that is, P(x) and are Q(x)
represented by a power series. A point that is not an ordinary point is called a
singular point.

A solution of the form y =  c n ( x − x o )n is said to be a solution about
n=0
the ordinary point x0.
Remark 6.1 It has been proved that if x=x0 is an ordinary point of (6.4)
then there exist two linearly independent solutions in the form of a power series

centered at x0, that is, y =

n=0
some interval defined by
c n ( x − x o )n . A series solution converges at least on
x − x o <R, where R is the distance from xo to the
closest singular point.
Power series solution about an ordinary point:

Let
y=

n=0
c n x n and substitute values of y,
dy
dy 2
= y', 2 = y" in
dx
dx
(6.5)
Combine series as in Example 6.1, and then equate all coefficients to the
right hand side of the equation to determine the coefficients c n. We illustrate the
method by the following examples. We also see through these examples how the
- 109 -

single assumption that y=

n =0
c n x n leads to two sets of coefficients, so we have
two distinct power series y1 (x) and y2(x) both expanded about the ordinary point
x=0. The general solution of the differential equation is y=C 1y1(x)+C2y2(x),
infact it can been shown that C1=c o and C2=c1.
The differential equation
d2 y
+ xy = 0 is known as Airy’s equation and
dx 2
used in the study of diffraction of light, diffraction of radio waves around the
surface of the earth, aerodynamics etc. We discuss here power series solution of
this equation around its ordinary point x=0.
Example 6.2 Write the general solution of Airy’s equation y'+xy=0.
Solution: In view of the remark, two power series solutions centred at 0,


n=o
n=2
convergent for x < exist. By substituting y=  c n x n , y =  n(n − 1)c n x n−2 into
Airy’s differential equation we get


n=2
n =0
y''+xy=  c nn (n − 1)x n−2 + x  c n x n ,

=

n=2

c n n (n − 1)x n−2 +  c n x n+1
(6.6)
n =0
As seen in the solution of Example 6.1, (6.6) can be written as

y''+xy=2c2+  [(k+1) (k+2)ck+2+ck-1]xk=0
(6.7)
k =1
Since (6.7) is identically zero, it is necessary that coefficient of each
power of x be set equal to zero, that is,
2c2=0 (It is the coefficient y x0) and
(k+1)(k+2) ck+2+ck-1=0, k=1,2,3 - - - -- - - -..
- 110 -
(6.8)
The above holds in view of the identity property. It is clear that c2=0. The
expression in (6.8) is called a recurrence relation and it determines the ck in
such a manner that we can choose a certain subset of the set of coefficients to be
non-zero. Since (k+1)(k+2)0 for all values of k, we can solve (6.8) for ck+2 in
terms of ck-1.
ck+2= -
c k −1
, k = 1,2,3, - - - (k + 1)(k + 2)
For k=1, c3 = -
co
2.3
For k = 2, c4 = -
co
3.4
For k= 3, c5 = -
c2
= 0 as c2=0
4 .5
For k= 4, c6 = -
c3
1
c0
=
2.3.5.6.
5.6
For k= 5, c7 = -
− c4
1
=
c1
6.7 3.4.6.7
For k= 6. c8 = -
− c5
= 0 as c5=0
7.8
For k= 7. c9 = -
c6
1
=−
c0
8 .9
2.3.5.6.8.9.
For k = 8, c10 = -
For k = 9, c11= -
c7
1
=−
c1
9.10
3.4.6.7.8.10
c8
= 0 as c8=0
10.11
and so on,
- 111 -
(6.9)

Substituting the coefficients just obtained into y=  c n x n
n=0
=c0+c1x+c2x2+c3x3+c4x4+c5x5+c6x6+c7x7+c8x8+c9x9+c10x10- - - we get
y=c0+c1x+0
−
co 3 c1 4
c0
c1
c0
c1
x −
x +0+
x6 +
x7 + 0 −
x9 −
x10 + 0 + - - - 2.3
3.4
2.3.5.6
3.4.6.7
2.3.5.6.8.9.
3.4.6.7.9.10
After grouping the terms containing co and the terms containing c1, we
obtain y=c0y1(x)+c1y2(x), where
y1(x)=1
= 1+ 
k =1
1 3
1
1
x +
x6 −
x9 + - - - 2.3
2.3.5.6
2.3.5.6.8.9
( −1)k
x 3k
2.3 - - - - (3k − 1)(3k )
y2(x) = x 
= x+ 
k =1
1 4
1
1
x +
x7 −
x 10 + - - - 3.4
3.4.6.7
3.4.6.7.9.10
+1
( −1)k
x 3k
3.4 - - - - (3k )(3k + 1)
Since the recursive use of (6.9) leaves c0 and c1 completely undetermined,
they can be chosen arbitrarily.
y=c0y1(x)+c1y2(x) is the general solution of the Airy’s equation.
Example 6.3 : Find two power series solutions of the differential equation y"xy=0 about the ordinary point x=0.

Solution: Substituting y =  c n x n into the differential equation we get
n =0
- 112 -

y"-xy=

n=2

n(n − 1)c n x n−2 −  c n x n+1
n =0


k =0
k =1
=  (k + 2)(k + 1)c k + 2 xk −  c k −1 xk

= 2c2 +  [(k + 2)(k + 1)c k +2 − c k −1 ]x k
k =1
Thus c2 = 0,
(k+2)(k+1)ck+2 –ck-1= 0
and
c k +2 =
1
c k −1,k = 1,2,3....
(k + 2)(k + 1)
Choosing co= 1 and c1=0 we find
c3 =
1
1
, c 4 = c 5 = 0, c 6 =
and so on.
6
180
For c0=0 and c1=1 we obtain
c 3 = 0, c 4 =
1
6
1
1
, c 5 = c 6 = 0,, c 7 =
and so on. Thus two solutions are
12
504
y1 = 1 + x 3 +
y2 = x +
1 6
x + - - - - and
180
1 4
1 7
x +
x +---12
504
6.3 Solutions about Regular Singular Points – The Method of Frobenius
A singular point x0 of (6.4) is called a regular singular point of this
equation if the functions p(x) = (x-xo) P(x) and q(x)=(x-xo)2Q(x) are both
analytic at x0. A singular point that is not regular is said to be on irregular
- 113 -
singular point of the equation. This means that one or both of the functions
p(x)=(x-x0) P(x) and q(x) = (x-x0)2Q(x) fail to be analytic at x0.
In order to solve a differential equation given by (6.4) about a regular
singular point we employ the following theorem due to Frobenius.
Theorem 6.1 (Frobenius Theorem) If x=x0 is a regular singular point of the
differential equation (6.4), then there exists at least one solution of the form
y=(x-xo)r


n =o

c n ( x − x o )n =  c n ( x − x o )n +r
n=0
where r is constant to be determined. The series will converge at least on
some interval 0<x-x0<R.
The method of Frobenius: Finding series solutions about a regular singular
point x0, is similar to the method of previous section in which we substitute y=


n =o
c n ( x − x o )n+r into the given differential equation and determine the unknown
coefficients cn by a recurrence relation. However, we have an additional task in
this procedure. Before
determining coefficients we must find unknown
exponent r. Equate to 0 the coefficient of the lowest power of x. This equation is
called the indicial equation and determines the value(s) of the index r.
If r is found to be number that is not a non negative integer, then the

corresponding solution y=

n =o
c n ( x − x o )n+r is not a power series. For the sake of
simplicity we assume that the regular singular point is x=0.
Example 6.4 Apply the Method of Frobenius to solve the differential equation
2x y"+3y’-y=0 about the regular singular point x=0.
Solution: Let us assume that the solution is of the form
- 114 -

y=

c n x n+r then
n=o

y' =

n =o

y"=

n =o
c n (n + r )x n+r −1
c n (n + r )(n + r − 1)x n+r −2 ,
Substituting these values of y', y' and y'' into 2x y''+3 y'-y=0, we get

2

n =o
c n (n + r )(n + r − 1)x n+r −1 + 3


n=o
c n (n + r )x n+r −1 −


n =o
c n x n + r = 0.
Shifting the index in the third series and combing the first two yields


n =o

c n (n + r ) (2n + 2r + 1)x n+r −1 -  c n−1x n+r −1 =0
n=o
Writing the term corresponding to n=0 and combining the terms for n/
into one series,

cor(2r+1)xr-1+  [c n (n + r ) (2n+2r+1)-cn-1]xn+r-1 = 0
n =1
Equating the coefficients of xr-1 to zero yields the indicial equation
c0r(2r+1)=0
Since c0 0, either r=0 or = -
1
2
Hence two linearly independent solutions of the given differential
equation have the form

y1 = F0 (x) =  c n x n and
n =o
- 115 -
y2 = F −1/ 2 (x) =x
-1/2


c *n xn
n =o
Since cn(n+r) (2n+2r+1) -cn-1=0 for all n  1, we have the following
information on the coefficients for the two series:
1
c n−1
n(2n + 1)
(i)
co is arbitrary, and for n1, cn=
(ii)
c*o is arbitrary, and for n1,cn*=
1
*
c n−1
n(2n − 1)
Iteration of the formula for cn yields
n=1, c1 =
2c
1
2
c0 =
c0 = 0
1.3
1.2.3
3!
22 c 0
1
1
n= 2, c2=
c1 =
c0 =
2.5
2.3.5
5!
n= 3, c3 =
1
1 22 c 0 23 c 0
c2 =
=
3.7
3.7 5!
7!
Each term of cn was multiplied by
2
to make the denominator (2n+1)!. The
2
general form of cn is then
cn =
2n c 0
(2n + 1)!
2n c 0
Similarly, the general form of cn is found to be cn =
.
(2n)!
*
*
The two solutions are

2n
xn,y2= co*x-1/2
n = o ( 2n + 1)!
y1=co 


n =o
2n n
x
(2n)!
- 116 -
y2 is not a power series.
Example 6.5 Apply the method of Frobenius to obtain two linearly independent
series solution of the differential equation
2x y" – y'+2y= 0
about a regular singular point x=0 of the differential equation.

Solution: Substituting y =  c n x n+r ,
n=o

y' =

n =o
c n (n + r )x n+r −1
and
n+r −1

y" =
c
n =o
n
(n + r )(n + r − 1)x
into the differential equation and collecting terms, we obtain

2x y''- y'+2y=(2r2-3r)c0xr-1+  [2(k+r-1)(k+r)ck -(k+r)ck+2ck-1]xk+r-1=0,
k =1
which implies that
2r2-3r=r(2r-3)=0
and
(k+r)(2k+2r-3)ck+2ck-1=0.
3
2
The indicial roots are r=0 and r= .For r=0 the recurrence relation is
ck = -
2c k −1
, k= 1,2,3, - - - k( 2k − 3)
- 117 -
and
c1 = 2c0, c2= - 2c0, c3=
4
c0
9
For r=
3
the recurrence relation is
2
ck = -
2c k −1
, k=1,2,3,- - - ( 2k + 3)k
and
c1 = -
2
2
4
c 0 ,c 2 =
c 0 ,c 3 = −
c 0.
5
35
945
The general solution is
4
9
y = C1 (1+2x-2x2+ x3+- - - - )
2
5
+C2 x3/2 (1- +
2 2 4 3
xx +- - - -)
35
945
6.4 Bessel's equation
x2 y''+x y'+(x2-v2)y=0
(6.10)
(6.10) is called Bessel's equation.
Solution of Bessel's Equation:
Because x=0 is a regular singular point of Bessel's equation we know that there

exists at least one solution of the form y=  c n x n+r .
n=o
expression into (6.10) gives
- 118 -
Substituting the last


x y"+x y'+(x -v )y=  c n (n + r )(n + r − 1)x + 
2
2
2
n +r
n =o
n =o

c n (n + r )x n+r +  c n x n+r + 2
n =o

-v2  c n x n+r = c0(r2-r+r-v2)xr
n=o


+xr
n =1

c n [(n + r )(n + r − 1) + (n + r ) − v 2 ]x n + x r  c n x n+ 2
n =o


n =1
n=o
r
= c 0 (r 2 − v 2 )x + xr  c n [(n + r )2 ] − v 2 ]xn + xr  c n xn + 2
(6.11)
From (6.11) we see that the indicial equation is r2-v2=0, so the indicial roots are
r1=v and r2 = -v. When r1=v, (6.11) becomes


n =1
n =o
xv  c nn(n + 2v )x n + x v  c n x n+2





n=2
n =0





k =0

=xv (1 + 2v )c 1x +  c nn(n + 2v )x n +  c n x n+2 
=xv (1 + 2v )c 1x +  [(k + 2)(k + 2 + 2v )c k +2 + c k ]x k +2  = 0
Therefore by the usual argument we can write (1+2v)c1=0 and
(k+2) (k+2+2v)ck+2+ck=0
or ck+2=
− ck
, k = 0,1,2,- - - (k + 2)(k + 2 + 2v )
(6.12)
The choice c1=0 in (6.12) implies c3=c5=c7= - - - - = 0, so for k=0,2,4, - - - - we
find, after letting k +2 = 2n, n = 1,2,3, - - - - that
c2n = -
c 2n−2
(6.13)
2 n(n + v )
2
- 119 -
Thus c2 = -
c0
2 .1(1 + v )
2
c4 = -
c0
c2
= 4
2 2(2 + v ) 2 .2.1(1 + v )(2 + v )
c6 = -
c0
c4
=− 6
2 .3(3 + v )
2 .1.2.3(1 + v )(2 + v )(3 + v )
:
:
c2n =
( −1)n c 0
, n = 1,2,3,- - - 2 2 n! (1 + v )(2 + v )...(n + v )
2
2
(6.14)
It is standard practice to choose c0 to be specific value – namely.
c0 =
1
2 (1 + v )
v
where  (1+v) is the gamma function. (See Appendix) Since this latter
function possesses the convenient property  (1+) = (), we can reduce the
indicated product in the denominator of (6.14) to one term.
For example:
 (1+v+1)= (1+v)  (1+v)
 (1+v+2)= (2+v)  (2+v)= (2+v)(1+v)(1+v).
Hence we can write (6.14) as
c 2n =
( −1)n
( −1)n
=
2 2n+ v n! (1 + v )(2 + v )...(n + v )(1 + v ) 2 2n+ v n! (1 + v + n)
for n=0,1,2, - - - -
- 120 -
Bessel Function of the First Kind: Using the coefficients c2n just obtained and

r=v, a series solution of (6.10) is y=  c 2n x 2n+ v This solution is usually denoted
n =0
by Jv ( x ) :

( −1)n
x
 
n=0 n! (1 + v + n)  2 
Jv ( x ) = 
2n+ v
(6.15)
.
If v0, the series converges at least on the interval [o,  ). Also, for the second
exponent r2= -v we obtain, in exactly the same manner,

( −1)n
x
 
n=0 n! (1 − v + n)  2 
J−v ( x ) = 
2n− v
(6.16)
.
The functions Jv(x) and J-v(x) are called Bessel functions of the first kind of
order v and –v, respectively. Depending on the value of v, (6.16) may contain
negative powers of x and hence converge on (0,  ).*
6.5 Legendre's Equation
(1-x2) y"-2x y'+n(n+1)y = 0
(6.17)
Equation (6.17) is known as Legendre's equation.
Solution of Legendre's Equation:
Since x=0 is an ordinary point of the

equation, we substitute the power series y=  c n x n , , shift summation indices, and
n =0
combine series to get
(1-x2) y"-2x y'+n(n+1)y=[n(n+1)c0+2c2]+[(n-1)(n+2)c1+6c3]x

+  [ j + 2)( j + 1)c j+2 + (n − j)(n + j + 1)c j ]x j = 0,
j= 2
*
When we replace x by x , the series given in (6.15) and (6.16) converge for 0< x <  .
- 121 -
which implies that n(n+1)c0+2c2=0
(n-1)(n+2)c1+6c3=0
(j+2)(j+1)cj+2+(n-j)(n+j+1)cj=0
or
c2 = -
c3 = -
c j+ 2 −
n(n + 1)
c0
2!
(n − 1)(n + 2)
c1
3!
(n − j)(n + j + 1)
c j , j = 2,3,4,- - - ( j + 2)( j + 1)
(6.18)
If we let j take on the values 2,3,4, - - - -, the recurrence relation (6.18) yields
c4= −
(n − 2)(n + 3)
(n − 2)n(n + 1)(n + 3)
c2 =
c0
4.3
4!
c5 = −
(n − 3)(n + 4)
(n − 3)(n − 1)(n + 2)(n + 4)
c3 =
c1
5. 4
5!
c6 = −
(n − 4)(n + 5)
(n − 4)(n − 2)n(n + 1)(n + 3)(n + 5)
c4 = −
c0
6. 5
6!
C7 = −
=−
(n − 5)(n + 6)
c5
7. 6
(n − 5)(n − 3)(n − 1)(n + 2)(n + 4)(n + 6)
c1
7!
and so on. Thus for at least |x| <1 we obtain two linearly independent power
series solutions:
 n(n + 1) 2 (n − 2)n(n + 1)(n + 3) 4
y 1( x ) = c 0 1 −
x +
x
2!
4!

−
(6.19)
(n − 4)(n − 2)n(n + 1)(n + 3)(n + 5) 6

x + - - - -
6!

- 122 -

(n − 1)(n + 2) 3 (n − 3)(n − 1)(n + 2)(n + 4) 5
y 2 ( x) = c 1 x −
x +
x
3!
5!

−
(n − 5)(n − 3)(n − 1)(n + 2)(n + 4)(n + 6) 7

x + - - - -
7!

Notice that if n is an even integer, the first series terminates, whereas
y2(x) is an infinite series. For example, if n=4, then
35 4 
 4 .5 2 2 .4 .5 .7 4 

y1( x ) = c 0 1 −
x +
x  = c 0 1 − 10 x 2 +
x .
2!
4!
3




Similarly, when n is an odd integer, the series for y2(x) terminates with xn;
that is, when n is a nonnegative integer, we obtain an nth-degree polynomial
solution of Legendre's equation.
Since we know that a constant multiple of a solution of
Legendre's
equation is also a solution, it is traditional to choose specific values for c0 or c1,
depending on whether n is an even or odd positive integer, respectively. For n=0
we choose c0=1, and for n = 2,4,6, - - - -,
c 0 = ( −1)n / 2
1.3 - - - - (n − 1)
;
2 .4 - - - - n
where as for n=1 we choose c1 = 1, and for n=3,5,7, - - - -,
c 1 = ( −1)(n−1) / 2
1.3 - - - - n
.
2.4 - - - - (n − 1)
For example, when n=4, we have
y1( x ) = ( −1)4 / 2
(
)
1 .3 
35 4  1
1 − 10 x 2 +
x  = 35 x 4 − 30 x 2 + 3 .

2 .4 
3
 8
Legendre Polynomials These specific nth-degree polynomial solutions are
called Legendre polynomials and are denoted by Pn(x). From the series for
- 123 -
y1(x) and y2(x) and from the above choices of c0 and c1 we find that the first
several Legendre polynomials are
P0(x) =1
P1(x) = x
P2 ( x ) =
1
(3 x 2 − 1)
2
P4 ( x ) =
1
(35 x 4 − 30 x 2 + 3)
8
P3 ( x ) =
1
(5 x 3 − 3 x )
2
P5 ( x ) =
(6.20)
1
(63 x 5 − 70 x 3 + 15 x ).
8
Remember, P0(x), P1(x), P2(x), P3(x), - - - - are, in turn, particular solutions of
the differential equations
(1- x 2 )y"-2xy' = 0
n = 0:
n = 1:
(1− x 2 )y"−2xy'+2y = 0
n = 2:
(1− x2 )y"−2xy'+6y = 0
n = 3:
(1− x 2 )y"−2xy'+12y = 0
(6.21)
Properties You are encouraged to verify the following properties using the
Legendre polynomials in (6.20)
(i)
Pn(-x)=(-1)nPn(x)
(ii)
Pn(1)=1
(iii)
Pn(-1)=(-1)n
(iv)
Pn(0)=0, n odd
(v)
P' n (0) = 0 , n even
6.6 Orthogonality of Functions
The concept of orthogonality of functions is the generalization of the
notion of orthogonality or perpendicularity of two vectors in the plane.
- 124 -
Deprition 6.1 (i) (Orthogonal Function) Two functions 1 and 2 defined on an
interval (a,b) into R are said to be orthogonal if
b

1(x) 2 (x) dx = 0, 12
a
 0, 1= 2
(ii) A set of real-valued functions {1(x), 2 (x)- - - -} is said to be
orthonormal if
b

m(x) n (x) dx = 0, mn
a
0, m=n
(iii) A set of real-valued functions {0(x), 1 (x),2 (x)- - - -} is said to be
orthonormal if
b

m(x) n (x) dx = 0, mn
a
=1, m=n
In other words if {n(x)} is an orthogonal set of functions on the interval
b
[a,b] with the property that

|n(x)2dx = 1 for n= 0,1,2,3- - - -then {n(x)} is
a
orthonormal set on the interval.
(iv) A set of functions {0,1,2,
b
respect to weight function p(x), if

- - - -
} is said to be orthogonal with
m (x) n(x) p (x) dx = 0, mn
a
 0, m=n
- 125 -
Example 6.6 The set {1, cos x, cos 2x,- - - -} is orthogonal on the interval [,]
Verification: If we make the identification o(x)=1 and n(x) = cos nx, we must

then show that

0 ( x)n ( x)dx = 0,if n0, and
−


−
m ( x )n ( x )dx = 0,m  n
We have, in the first place,


−

o ( x )n ( x )dx =  cos n x d x

1

=  sin nx 
n
 −
=
1
sin n − sin( −n) = 0, n  0.
n
In the second place


−

m
( x ) n ( x )dx =  cos mx cos nx dx
−

1
=  cos (m + n)x + cos (m − n)x dx, by using a well known
2 −
trigonometric identity,

=
1  sin( m + n)x sin( m − n)x 
+
2  m + n
m − n  − 
m  n.
= 0,

Example 6.7 (i) Compute  | n ( x ) |2 dx where
−
- 126 -
o ( x) = 1, 1( x) = cos x, 2 ( x) = cos 2x - - -  1

cos x cos 2x
,
,- - - -


 2

Show that the set 
,
is orthonormal on the interval [-,]. Solution: (i)

have
| 
−
For o(x) =1 we

0
( x ) |2 dx =  dx
−
= x− =  − (−) = 2



−
−
For n (x) = cosnx,n>0,  | n ( x ) |2 dx =  | cos nx |2 dx

 cos
=
2
nxdx
−
=
1
2

 [ + cos 2 nx ]dx
−
=

Thus for n>0,  | n ( x ) |2 dx = 
−
1

2
or   | n ( x ) |2 dx  = 
 −

We are required to show that

(a)  |
−
o ( x)
2
|2 dx = 1
- 127 -

(b)  |
n ( x )

−
|2 dx = 1

Verification of (a) :  |
0 ( x)
−
=
1
2
|2 dx =
2
1
2

| 
0
( x ) |2 dx
−

 1dx = 1
−

Verification of (b):  |
n ( x )

−

|2 dx =
1
|  n ( x ) |2 dx
 −

1
=  | cos nx |2 dx
 −

=  cos 2 nx dx
−

1
[1+ cos 2 nx ]dx
2 −
=
2
=1
2
Orthogonal Series Expansion
Let {n (x)} be an infinite orthonormal set of functions on interval
[a,b]. and f(x) be a function defined on [a,b]. Then f(x) can be written as
f(x)=coo(x)+c12(x)+c22(x)+- - - - +cnn(x)+ - - - b
where c n
 f ( x) ( x)dx = 1
=
 |  ( x) | dx
n
a
b
a
(6.23)
2
n
n=0, 1,2,3 …
- 128 -
(6.22)
The series on the right hand side of (6.22) is called orthogonal expansion
of f(x) defined on [a,b] in terms of the orthonormal set of functions {n(x)}
defined on [a,b]. cn's given by (6.23) are called coefficients of orthogonal
expansion of f. If orthonormal set of Example 6.6 is considered we get cosine
Fourier expansion of f(x), that is, (6.22) will be cosine Fourier series and (6.23)
will give cosine Fourier coefficients. One can consider expansion of a function
in terms of Bessel's orthonormal set of functions and Legendre's orthonormal set
of functions.
6.7
Sturm-Liouville Theory
Consider the linear differential equation of order two
y"+R(x) y'+(Q(x)+ P(x)) y=0
(6.24)
Given an interval on which the coefficients R(x) and (Q(x)+ P(x) are
continuous we seek values of  for which (6.24) has non-trivial solutions.
We can also seek values of  when (6.24) is given with boundary
conditions. Let us put this differential equation in a more convenient form.
Multiply (6.24) by r=e  R( x )dx to get
y''e  R( x )dx + R(x) y'e  R( x )dx +(Q(x)+  P (x)) ye  R( x )dx = 0
(6.25)
Since r(x)0, equation (6.25) has the same solutions as (6.24). This
equation can be written as
(r y')' + (q+ p) y=0
(6.26)
where q(x)=Q(x) e  R( x )dx , p(x) = P(x) e  R( x )dx
Equation (6.26) is called the Sturm-Liouville differential equation, or
the Sturm-Liouville form of equation (6.24). Through out this section we
- 129 -
assume that p.q, and r and r’ are continuous on [a,b] or at least on (a,b), and p(x)
>0 and r(x) >0 on (a,b).
Remark 6.2 Bessel’s equation given by (6.10)
and Legendre’s equation given by (6.17) are special cases of the SturmLiouville differential equation (6.26).
For Bessel’s equation we can choose r(x)=
x2
, q(x)=x2,
2
p(x)=1, = -v2 in (6.26).
For Legendre’s equation we take r(x) = 1-x2
q(x)=0, p(x)=1, and  = n(n+1) in (6.26)
The Regular Sturm-Liouville Problem: Find numbers  for which there are
non-trivial solutions of (6.26) subject to the regular boundary conditions having
the following form
A1y(a) + A2 y' (a)=0, B1y(b) + B2 y' (b)=0
where A1 and A2 are given constants, at least one must be non-zero. Similarly B1
and B2 are given constants, at least one must be non-zero.
The Periodic Sturm-Liouville Problem:
Find numbers  for which there are non-trivial solutions of (6.26) on an
interval [a,b] where r(a)=r(b) and subject to the periodic boundary conditions
y(a)=y(b), y' (a)= y' (b)
- 130 -
The Singular Sturm-Liouville Problem:
Find numbers  for which there are non-trivial solutions of the SturmLiouville equation on (a,b), subject to one of the following three kinds of
boundary conditions:
Case I. r(a)=0 and there is no boundary condition at a, while at b the boundary
condition is
B1y(b)+B2 y' (b)=0,
where B1 and B2 are not both zero.
Case 2. r(b)=0 and there is no boundary conditions at b, while at a the condition
is
A1y(a)+A2 y' (a)=0,
with A1 and A2 not both zero.
Case 3. r(a)=r(b)=0, and no boundary condition is specified at a or b. We seek
solutions that are bounded functions on [a,b].
Definition 6.2 A number  for which Sturm-Liouville differential equation,
(6.26), subject to boundary conditions of one of these three problems, has
nontrivial solution is called an eigenvalue of the problem. A corresponding
nontrivial solution is called an eigenfunction associated with this eigenvalue.
Remark 6.3 (i) The zero function cannot be an eigenfunctiion. Any nonzero
constant multiple of an eigenfunction is an eigenfunction.
(ii) In mathematical models of real systems, eigenvalues have some
physical meaning. For example in the study of wave motion the eigenvalues are
fundamental frequencies of vibration of the system.
- 131 -
The fundamental properties of Sturm-Liouville problems are described by
the following theorem which is considered as the heart of Sturm-Liouville
theory.
Theorem 6.2 (a) Each regular and each periodic Sturm-Liouville problem has
an infinite number of distinct real eigenvalues. If these are labeled 1, 2. - - -,
so that n<n+1, then lim  n=.
n→ .
(b) If n and m are distinct eigenvalues of any of the three kinds of SturmLiouville problems defined on an interval (a,b) and n and m are corresponding
eigenfunctions, then
b
 p( x)
n
( x )m ( x )dx = 0
a
(c) All eigenvalues of a Sturm-Liouville problem are real numbers.
(d) For a regular Sturm-Liouville problem, any two eigenfunctions
corresponding to a single eigenvalue are constant multiples of each other.
Interested readers will find the proof of this theorem in references [.,.] or
at website no. [
].
Remark 6.4 (i) Part (a) assures existence of eigenvalues, at least for regular and
periodic problems. A singular problem may also have an infinite sequence of
eigenvalues say for example, for Bessel’s equation. This part also asserts that the
eigenvalues spread out so that if arranged in increasing order, they increase
without bound. For example, numbers 1-
1
n
could not be eigenvalues of a
Sturm-Liouville problem, since these numbers approach 1 as n→
- 132 -
(ii) Part (b) can be stated as “Eigenfunctions associated with distinct eigenvalues
are orthogonal on [a,b], with weight function p(x). The weight function p is the
coefficient of  in the Sturm-Liouville equation.
This orthogonality provides the possibility of expansion of functions in
series of eigenfunctions of a Sturm-Liouville problem, analogue of equation
(6.22) is possible for eigenfunctions.
(iii) Part (c) states that a Sturm Liouville problem can have no complex
eigenvalue.
(iv) Part (d) applies only to regular Sturm-Liouville problems.
Example 6.8 Discuss solutions of regular Sturm-Liouville problem:
y''+ y=0, y(0)=y(l)=0
on an interval [0,l] in cases (i)  = 0,
(ii)  is negative number, and (iii)  is positive number
Case (i) Let =0, then y''=0 and integrating it twice we get y(x)=cx+d for some
constants c and d. Now y(0)=d=0, and y(l)=cl=0 implies c=0. This means that
y(x)=cx+d must be the trivial solution. In the absence of a non-trivial solution,
=0 is not an eigenvalue of this problem.
Case (ii) Suppose that  is negative, say =-k2 for k>0
Now y"-k2y=0. This is homogeneous linear differential equation with
constant coefficients. The auxiliary equation is m2-k2=0. Roots are m1=k,m2= -k.
The general solution is
y(x)=c1ekx+c2e-kx
- 133 -
Since y(0)=c1+c2=0, then
c2=-c1, so y=c1(ekx-e-kx). since
sin hkx =
ekx − ekx
, we have
2
y=2c1 sinh kx. But then
y(l)=2c1 sinh kl=0
Since kl>0, sinh kl>0, so c1, = 0
This case also leads to the trivial solution, so this Sturm-Liouville
problem has no negative eigenvalue.
Case (iii)  is positive, say =k2
Now y''+k2y=0. The auxiliary equation of this homogeneous linear
differential equation with constant coefficients is
m2+k2=0. Roots are m1=ik, m2=-ik.
As discussed in Section 5.5, equation (5.18) the general solution is
y(x)=c1cos (kx) +c2 sin(kx)
Now
y(o)=c11 +c2.0=0 or c1=0
y(x)=c2 sin (kx). Finally, we need
y(l)=c2 sin kl=0
To avoid trivial solution, we need c20.
- 134 -
Then we must choose k so that sin kl=0, which means that kl must be a
positive multiple of , say kl = n.Then
n =
n2 2
for n=1,2,3,- - - -- - - -.
l2
Each of these numbers is an eigenvalue of this Sturm-Liouville problem.
Corresponding to each n, the eigenfunctions are
 nx 
,
 l 
yn(x) = c sin 
where c is any non-zero real number.
Example 6.9 Discuss solution of periodic Sturm-Liouville problem:
y''+y=0, y(-l)=y(l), y'(-l)= y'(l)
on an interval [-l,l] for cases
(i)
 = 0 (ii)  <0 (ii) >0
Solution: Case (i) =0 Then y=cx+d. (See example 6.8) Now
y(-l) = - cl+d=y(l)=cl+d implies c=0. The constant function y=d satisfies both
boundary conditions. Thus =0 is an eigenvalue with nonzero constant
eigenfunctions.
Case (ii) <0, say = -k2
Solving as in case (ii) of Example 6.8
y(x)=c1ekx+c2e-kx is the general solution.
Since y(-l)=y(l), then
- 135 -
c1e-kl+c2ekl=c1ekl+c2e-kl
(6.27)
And y' (-l)= y' (l) gives us after dividing out the common factor k
c1e-kl-c2ekl=c1ekl-c2e-kl
(6.28)
Rewrite equation (6.27), as
c1 (e-kl-ekl)=c2 (e-kl-ekl)
This implies that c1=c2. The equation (6.28) becomes
c1(e-kl-ekl) = c1(ekl-e-kl)
But this implies that c1=-c1, hence c1=0. This solution is therefore trivial, hence
this problem has no negative eigenvalue.
Case (iii)  >0, say =k2
Now as in Example 6.8 (case iii) the general solution is
y(x)=c1cos (kx)+c2 sin (kx)
Now
y(-l)=c1cos kl-c2 sin (kl)=y(l)=c1cos (kl)+c2sin (kl)
But this implies that
-c2sin (kl)=c2 sin (kl)
or –c2=c2 implying c2=0
Also y' (-l)=kc1sin (kl) + kc2 cos (kl)
= y' (l)=-kc1 sin (kl) +kc2 cos (kl).
Then
- 136 -
kc1 sin (kl)=0
If sin (kl)0, then c1=c2=0, leaving the trivial solution. Thus we assume
that sin kl=0 which requires that kl=n for some positive integer n. Therefore,
the numbers
n=
n22
ll
2
are eigonvalues for n=1,2,- - - - , with corresponding eigenfunctions
yn(x)=c1 cos (
nx
nx
) +c2 sin (
)
l
l
where c1 and c2 are not both zero
6.8
Exercises:
Review of Power Series
1.
Write excos x in the form of a power series. Examine whether this power
series is convergent.
Solution About Ordinary Points:
Find the general solution of the following differential equations about an
ordinary point in terms of two power series
2.
y''-(1+x)y=0
3.
y''+(cos x) y = 0
4.
y''+x2y=0
5.
y''+y=ex
6.
y'+xy=x2-2x
- 137 -
7.
(x2-1) y'+y=0
8.
y"-(x+1) y'-y=0
Use the power series method to solve the following initial value problems
9.
(x-1) y''-x y'+y=0
y(0)=-2, y' (0)=6
10.
(x2+1) y''+2x y'=0, y(0)=0, y' (0)=1
11.
y"+xy=0, y(0)=1, y' (0)=1
12.
xy"+y+x=0, y(1)=1, y' (1)=1
Solution About Regular Singular Point: The method of Frobenius
Use the method of Frobenius to solve the following differential equations
13.
x y''-x y'+y=0
14.
y''+
15.
x y''+ y'+y=0
16.
2x y'-3 y'-
17.
4x2y''+(3x+1)y=0
18.
x y''-(x+5) y'+3y=0
19.
x y''+(x-5) y'+3y=0
20.
x y''+ y'+xy=0
3
y'-2y=0
x
3+x
y=0
x
Bessel’s Equation
- 138 -
Find the general solution of the following equations
21.
x2 y''+x y'+(x2-1)y=0
22.
x y''+x y'+xy=0
23.
Verify that y=xnJn(x) is a particular solution of x y"+(1-2n) y'+xy=0, x>0
Legendre’s Equation
Solve the following equations
24.
(1-x2) y''-2x y'=0
25.
(1-x2) y''-2x y'+12y=0 subject to initial conditions
y(0)=0, y' (0)=1.
Sturm Liouville Theory
In each of problems 26 through 35, classify the Sturm-Liouville problem
as regular, periodic, or singular; state the relevant interval, find the eigenvalues;
corresponding to each eigenvalue, find an eigenfunction.
26.
y''+ y=0;
y' (0)= y' (l)=0
27.
y''+ y=0;
y(0)=0, 3y(1)+ y' (1)=0
28.
y''+ y=0;
y(0)=0, y' (l)=0
29.
y''+ y=0;
y' (0)=0, y' (l)=0
30.
y''+ y=0;
y' (0)=y(4)=0
31.
y''+ y=0;
y(0)=y(),y' (0)= y' ()
32.
y''+ y=0;
y(-3)=y(3), y' (-3)= y' (3)
- 139 -
33.
y''+ y=0;
y(0)=0, y()+2 y' ()=0
34.
y'+ y=0;
y(0)-2 y' (0)=0, y' (1)=0
35.
y''+2 y'+(1+)y=0,
y(0)=y(1)=0.
- 140 -
Solved Problems
1- PROPERTIES OF LAPLACE TRANSFORMS
EXERCISE 1
1. Determine the Laplace transforms: (a) 2te2t
(a) ℒ 2te 2t  = 2 ℒ t1e2t  = (2)
(b) ℒ t 2et  =
2!
( s − 1)
2+1
=
1!
(s − 2)
1+1
(b) t2et
2
=
( s − 2)
2
2
( s − 1)
3
2. Determine the Laplace transforms: ( a) 4t3e −2 t

(b)
1 4 −3 t
te
2

24
=
4
3+1
 (s + 2) 
( s + 2)


(a) ℒ 4t 3e−2t  = 4 
3!
12
1
4! 
 =
5
4 +1
2  ( s + 3) 
( s + 3)
(b) ℒ  t 4e−3t  = 
1
2

3. Determine the Laplace transforms: (a) e t cos t

(b) 3e2tsin 2t

s −1
s −1
= 2
= 2
2
2
 ( s − 1) + 1  s − 2s + 1 + 1
s − 2s + 2


(a) ℒ et cos t = 
s −1


6
6
= 2
=
 ( s − 2 )2 + 22  s 2 − 4s + 4 + 4
s − 4s + 8


(b) ℒ 3e2t sin 2t = 3 
2
4. Determine the Laplace transforms: (a) 5e −2 t cos 3t
- 141 -
(b) 4e −5 t sin t

5 ( s + 2)
5 (s + 2)
=
=

2
 ( s + 2 ) + 32  s 2 + 4s + 4 + 9
s + 4s + 13



s+2
(a) ℒ 5e−2t cos3t = 5 
2


4
4
= 2
= 2
2
2
 ( s + 5) + 1  s + 10s + 25 + 1
s + 10s + 26


(b) ℒ 4e−5t sin t = 4 
1
5. Determine the Laplace transforms: (a) 2e t sin2 t
(b)
1 3t
e cos2 t
2
1
s −1
1 − cos 2t  
t
t
−
  = ℒ e  - ℒ e cos 2t =
2
s − 1 ( s − 1)2 + 22



(a) ℒ 2e t sin 2 t = ℒ 2e t 



=
1
1
s −1
− 2
s − 1 s − 2s + 5
1 + cos 2t  

2


(b) ℒ  e3t cos 2 t  = ℒ  e3t 
2
2
1



= ℒ  e3t  + ℒ  e3t cos 2t  = 
1
4
1
4


1
4
1
s −3
+
s − 3 ( s − 3)2 + 22

s−3 
s −3
 1 1
+ 2
= 
4  s − 3 s − 6s + 9 + 4  4  s − 3 s − 6s + 13 
1 1
= 
+
2
6. Determine the Laplace transforms: (a) et sinh t

(b) 3e2t cosh 4t

1
1
1
=
=
=

2
2
 ( s − 1) − 12  s − 2s + 1 − 1 s − 2s
s(s − 2)


(a) ℒ et sinh t = 
1
2

3 ( s − 2)
3(s − 2)
= 2
= 2
2
2
 ( s − 2 ) − 4  s − 4s + 4 − 16
s − 4s − 12



(b) ℒ 3e2t cosh 4t = 3 
s−2
7. Determine the Laplace transforms: (a) 2e − t sinh 3t
- 142 -
(b)
1 −3t
e cosh 2t
4


6
6
= 2
=
 ( s + 1) − 32  s2 + 2s + 1 − 9
s + 2s − 8


(a) ℒ 2e− t sinh 3t = 2 
3
2
 1
s+3
1
s+3
s+3

=
=  2

2
2
2
4  ( s + 3) − 2  4  s + 6s + 9 − 4 
4 s + 6s + 5
(b) ℒ  e−3t cosh 2t  = 
1
4

(
8. Determine the Laplace transforms: (a) 2e t (cos 3t - 3 sin 3t)
)
(b) 3e − 2t (sinh
2t - 2 cosh 2t)
ℒ 2et ( cos 3t − 3sin 3t )
(a)

ℒ 2e t cos 3t
-
 

3
−
6



2
2 
 ( s − 1)2 + 32 
s
−
1
+
3
(
)

 

6e sin 3t = 2 
t
=
s −1
=
2 ( s − 1)
18
2s − 2 − 18
2s − 20
− 2
= 2
= 2
2
s − 2s + 10 s − 2s + 10 s − 2s + 10 s − 2s + 10
=
2 ( s − 10 )
s 2 − 2s + 10
(b) ℒ 3e−2t ( sinh 2t − 2 cosh 2t ) = ℒ 3e−2t sinh 2t - ℒ 6e−2t cosh 2t

 

s+2
−
6



2
 ( s + 2 ) − 22   ( s + 2 ) − 22 

 

= 3
2
2
=
6 (s + 2)
6 (s + 2)
6
6
− 2
= 2
− 2
s + 4s + 4 − 4 s + 4s + 4 − 4 s + 4s s + 4s
=
−6 ( s + 1 )
6 − 6s − 12 −6s − 6
=
=
s ( s + 4)
s (s + 4)
s (s + 4)
2
- 143 -
ℒ
EXERCISE 2
1. Derive the Laplace transform of the first derivative from the definition of a
Laplace transform.
Hence derive the transform ℒ {1} =
1
s
Let f(t) = 1 then f (t) = 0 and f(0) = 1
From equation (3), page 1051 of textbook, ℒ f '(t) = s ℒ f (t) − f (0)
ℒ{ 0 } = sℒ 1 - 1
Hence,
1 = s ℒ 1
i.e.
ℒ 1 =
and
1
s
2. Use the Laplace transform of the first derivative to derive the transforms:
(a) ℒ{e at } =
1
s−a
(b) ℒ{3t2} =
6
s3
(a) Let f(t) = eat then f (t) = a eat and f(0) = 1
From equation (3), page 1051 of textbook, ℒ f '(t) = s ℒ f (t) − f (0)
ℒ{ a eat } = sℒ eat  - 1
Hence,
1 = (s – a)ℒ eat 
i.e.
ℒ eat  =
and
- 144 -
1
s−a
(b) Let f(t) = 3t 2 then f (t) = 6t and f(0) = 0
Since
ℒ f '(t) = s ℒ f (t) − f (0)
then,
ℒ{ 6t } = sℒ 3t 2  + 0
i.e.
6
= sℒ 3t 2 
2
s
ℒ 3t 2  =
and
6
s3
3. Derive the Laplace transform of the second derivative from the definition of
a Laplace
transform. Hence derive the transform ℒ{sin at} =
a
s + a2
2
Let f(t) = sin at, then f (t) = a cos at and f (t) = −a 2 sin at , f(0) = 0 and f
(0) = a
From equation (4), page 1051 of textbook,
ℒ f ''(t) = s 2 ℒ f (t) − sf (0) − f '(0)
Hence,
ℒ −a 2 sin at = s2 ℒ sin at - s(0) - a
i.e.
−a 2 ℒ sin at = s 2 ℒ sin at - a
a = ( s 2 + a 2 ) ℒ sin at
Hence,
ℒ sin at =
and
a
s + a2
2
4. Use the Laplace transform of the second derivative to derive the transforms:
(a) ℒ{sinh at} =
a
s − a2
2
(b) ℒ{cosh at} =
- 145 -
s
s − a2
2
(a) Let f(t) = sinh at then f (t) = a cosh at and f (t) = a 2 sinh at , f(0) = 0
and f (0) = a
ℒ f ''(t) = s 2 ℒ f (t) − sf (0) − f '(0)
Hence,
ℒ a 2 sinh at = s 2 ℒ sinh at - s(0) - a
i.e.
a 2 ℒ sinh at = s 2 ℒ sinh at - a
i.e.
and
a = ( s 2 − a 2 ) ℒ sinh at
ℒ sinh at =
a
s − a2
2
(b) Let f(t) = cosh at then f (t) = a sinh at and f (t) = a 2 cosh at , f(0) = 1
and f (0) = 0
ℒ f ''(t) = s 2 ℒ f (t) − sf (0) − f '(0)
Hence,
ℒ a 2 cosh at = s 2 ℒ cosh at - s(1) - 0
i.e.
a 2 ℒ cosh at = s 2 ℒ cosh at - s
i.e.
and
s = ( s 2 − a 2 ) ℒ cosh at
ℒ cosh at =
s
s − a2
2
- 146 -
EXERCISE 3
1. State the initial value theorem. Verify the theorem for the functions
(b) (t - 4) 2 and state their initial values.
(a) 3 - 4 sin t
The initial value theorem states: limit f(t) = limit  s ℒ f (t) 
t →0
s→
(a) Let f(t) = 3 – 4 sin t
Hence,
then ℒ{f(t)} = ℒ{3 – 4 sin t} =
3
4
− 2
s s +1
 3
4 
lim it 3 − 4sin t  = lim it s  − 2  
t →0
s →
  s s + 1 


= lim
it 3 − 2 
s →
 s + 1
4s
3 – 4 sin 0 = 3 -
i.e.
i.e.
3=3

 +1
2
which verifies the theorem.
The initial value is 3
2
(b) Let f(t) = ( t − 4 ) = t 2 − 8t + 16
then
Hence,
ℒ t 2 − 8t + 16 =
2 8 16
− +
s3 s 2 s
  2 8 16  
lim it  t 2 − 8t + 16  = lim it s  3 − 2 +  
t →0
s →
s 
 s s


it
− + 16 
= lim
s →  s 2
s


2
i.e.
8
16 = 16 which verifies the theorem
- 147 -
The initial value is 16
2. Verify the initial value theorem for the voltage functions:
(a) 4 + 2 cos t
(b) t - cos 3t and state their initial values.
The initial value theorem states: limit f(t) = limit  s ℒ f (t) 
t →0
s→
then ℒ{f(t)} = ℒ{4 + 2 cos t} =
(a) Let f(t) = 4 + 2 cos t
4
2s
+ 2
s s +1
 4
2s  
lim it  4 + 2 cos t  = lim it s  + 2  
t →0
s →
  s s + 1 
Hence,

2s 2 
= lim it  4 + 2 
s →
 s + 1
i.e.
4 + 2 cos 0 = 4 +
2 2
2 + 1
i.e.
4+2=4+2
i.e.
6=6
which verifies the theorem.
The initial value is 6
(b) Let f(t) = t - cos 3t
then
Hence,
ℒ t − cos 3t =
1
s
− 2 2
2
s s +3
 1
s 
lim it  t − cos 3t  = lim it s  2 − 2

t →0
s →
  s s + 9 
- 148 -
1
s2 
= limit  − 2 
s →
s s + 9 
0 − cos 0 =
i.e.
i.e.
1
2
− 2
  +9
- 1 = - 1 which verifies the theorem
The initial value is - 1
3. State the final value theorem and state a practical application where it is of
use. Verify the
theorem for the function 4 + e −2 t (sin t + cos t) representing a displacement
and state its final
value.
limit f(t) = limit  s ℒ f (t)
The final value theorem states:
t →
s →0

The final value theorem is used in investigating the stability of systems such as
in automatic aircraft-landing systems.
Let f(t) = 4 + e−2t (sin t + cos t) = 4 + e−2t sin t + e−2t cos t
then
Hence,
4
s
ℒ f (t) = +
1
(s + 2)
2
+1
+
s+2
(s + 2)
2
+1
4
1
s+2 
limit  4 + e−2t sin t + e−2t cos t  = limit  s  +
+

t →
s →0
 s ( s + 2 )2 + 1 ( s + 2 )2 + 1 



= limit 4 +
s →0
- 149 -

s (s + 2) 

2
2
( s + 2 ) + 1 ( s + 2 ) + 1
s
+
i.e.
4+0+0=4+0+0
i.e.
4 = 4 which verifies the theorem
The final value is 4
4. Verify the final value theorem for the function 3t 2e-
4t
and determine its
steady state value.
The final value theorem states:
limit f(t) = limit  s ℒ f (t)
t →
s →0

Let f(t) = 3t2e- 4t
then

2! 
6
=
2 +1 
3
 (s + 4)  (s + 4)
ℒ f (t) = 3 
Hence,
i.e.
i.e.
 6 
limit 3t 2 e− 4t  = limit  s 

t →
s →0
 ( s + 4 )3 


3t  e− 4 = 0
0 = 0 which verifies the theorem
The final value is 0
- 150 -
Problems
(Laplace Transform)
1. If f (t ) is not continuous at t = a, 0  a   but f (a + 0) and f (a − 0) exist,
then find L{ f ' (t )} .
2. Show that L(
sin at
cos at
) exists, but L(
) does not exist.
t
t
(ii) L{1 − erf (t )}
(i) L(5t )
3. Find :
(iii) L(
(v) L(sin at − at cos at )
(iv) L{t 2u(t − 3)}
sin t
)
t
(vi)
L(cos t.log t. (t −  ))
4. Find L(
1
t
) given L(2
t

t
sin t
dt , t → s}
(i) L{
t
0
5. Find

6. (i)
)=
e− at − e−bt
0 t dt
1
s3/2
.

(ii)  sin t 2dt
0
t
(ii) Evaluate  e−2t t sin 3 tdt
0


(iii) Find  sin(tx )dx and hence evaluate  sin 3x 2 dx
2
0
7. Evaluate the following:
L−1{
5
s
(i) L−1{ 2 + (
s −1 2
7
) −
}
s
3s + 2
(ii)
2s 2 − 6s + 5
}
s 3 − 6s 2 + 11s − 6
(v)
s
}
s + s2 + 1
4
(iii) L−1 (
L−1{
0
1
), n  0
s−n
(iv) L−1{
s 2 + 2s + 3
}
( s 2 + 2s + 2)( s 2 + 2s + 5)

p2 
2
2
 ( p + 4) 
(vi) L−1 
- 151 -
8. Show that:
(i)
L−1{
1
eat erf ( at )
}=
( s − a) s
a
(ii)
1
s+2
e− x − e−2 x
L−1{ log
}= 
dx
s
s +1
x
0
t
 2t
 T , 0  t  T / 2
f (t + T ) = f (t )
9. Find L{ f (t )} (i) f (t ) = 
2(
T
−
t
)

, T /2t T
 T
7

cos(t − ), t  7 / 3
(ii) f (t ) = 
3

0, t  7 / 3
(iii) f (t ) = k sin
10.Prove that
t
T
[u (t − 2T ) − u (t − 3T )] .
1 2 3
n
n n −1
* * *...*
=
t , where* is convolution.
2 3 4
n + 1 n + 1!
11. Express the following in terms of unit step function and obtain the
Laplace transform
(i)
t , 0  t  2
f (t ) = 
0, 2  t
4, 0  t  1
(ii) f (t ) = −2, 1  t  3
5, t  3

Answers
3. (i)
1
, s  log 5
s − log 5
(v)
4.
2a 3
(s 2 + a 2 )2
(ii)
2 6 9
1
(iv) e−3s [ 3 + 2 + ]
s s s
s + 1( s + 1 + 1)
(vi) −e− s log 
1
s 3/2
1
s
5. (i) cot −1 s (ii)
1 
2 2
- 152 -
6. (i) log
b
a
(ii)

3( s + 2) 
1
1
−

2
2
2
2
2s  [( s + 2) + 1] [( s + 2) + 9] 
t
7
3
−2
7. (i) 6t + 1 − 4( )1/2 − e 3

t
(ii)
2
t
3
sinh sin( t )
2
2
3
(iii)
1  1 
,
2 2t 2 6
(iii) Does not exists
1 t 2 t 5 3t
e +e + e
2
2
1
3
(v) e−t (sin t + sin 2t )
9. (i)
(vi)
t cos 2t sin 2t
+
2
4
2
sT
s
k 2
−7 s /3
tanh
e
,
s

0
(e−2 sT − e−3sT )
(ii)
(iii)
Ts 2
2
s2 + 1
s 2T 2 +  2
11. (i)
1 − (1 + 2s)e−2 s
s2
(ii)
4 − 6e− s + 7e−3s
s
- 153 -
(iv)
Gamma & Beta Functions
I. Gamma Function
Definition
Γ(n) =
∞
0
x n-1 e -x dx ; n > 0
& Γ(n) = Γ(n+1) / n
; n Є R - Z0
Results:
(1) Γ(n+1) = n Γ(n) ; n > 0 , where Γ(1) = 1
(2) Γ(n+1) = n!
; nЄ N
(3) Γ(n) Γ(1- n) = π /sin(nπ)
( convention: 0! = 1)
;0<n<1
In Particular;
Γ(1/2) = π
Examples:
Example(1)
Evaluate
∞
0
x 4 e -x dx
Solution
∞
0
x 4 e -x dx =
∞
0
x 5-1 e -x dx = Γ(5)
Γ(5) = Γ(4+1) = 4! = 4(3)(2)(1) = 24
Exercise
Evaluate
∞
0
x 5 e -x dx
Example(2)
- 154 -
Evaluate
∞
0
∞
0
x 1/2 e -x dx
x 1/2 e -x dx =
∞
0
x 3/2-1 e -x dx = Γ(3/2)
3/2 = ½ + 1
Γ(3/2) = Γ(½+ 1) = ½ Γ(½ ) = ½ π
Exercise
Evaluate
∞
0
x 3/2 e -x dx
Example(3)
Evaluate
∞
0
∞
0
x 3/2 e -x dx
x 3/2 e -x dx =
∞
0
x 5/2-1 e -x dx = Γ(5/2)
5/2 = 3/2 + 1
Γ(5/2) = Γ(3/2+ 1) = 3/2 Γ(3/2 ) = 3/2 . ½ Γ(½ ) = 3/2 . ½ . π =
¾ π
Exercise
Evaluate
∞
0
x 5/2 e -x dx
Example(4)
Find Γ(-½)
(-½) + 1 = ½
Γ(-1/2) = Γ(-½ + 1) / (-½) = - 2 Γ(1/2 ) = - 2 π
Example(5)
Find Γ(-3/2)
- 155 -
(-3/2) + 1 = - ½
Γ(-3/2) = Γ(-3/2 + 1) / (-3/2) = Γ(-1/2 ) / (-2/3) = ( - 2 π ) / (-2/3) = 4
π /3
Exercise
Evaluate Γ(-5/2)
- 156 -
II. Beta Function
Definition
B(m,n) =
1
0
x m-1 (1 – x ) n-1 dx ; m > 0 & n > 0
Results:
(1) B(m,n) = Γ(m) Γ(n) / Γ(m+ n)
(2) B(m,n) = B(n,m)
(3)
π/2
0
(4)
∞
0
sin 2m-1x . cos 2n-1x dx = Γ(m) Γ(n) / 2 Γ(m+ n) ; m>0 & n>0
x q-1 / (1+x) . dx = Γq) Γ(1-q) = Π / sin(qπ) ; 0<q<1
Examples:
Example(1)
Evaluate
1
0
x4 (1 – x ) 3 dx
Solution
1
0
x 4 (1 – x ) 3 dx = x 5-1 (1 – x ) 4-1 dx
= B(5,4) = Γ(5) Γ(4) / Γ(9) = 4! . 3! / 8! = 3!/(8.7.6.5) = 1/ (8.7.5) = 1/280
Exercise
Evaluate
1
0
x2 (1 – x ) 6 dx
Example(2)
Evaluate I =
1
0
[ 1 /
3
[x2 (1 – x )] ] dx
Solution
I =
1
0
x -2/3 (1 – x ) -1/3 dx =
1
0
x 1/3 - 1 (1 – x ) 2/3 - 1 dx
- 157 -
= B(1/3,2/3) = Γ(1/3) Γ(2/3) / Γ(1)
Γ(1/3) Γ(2/3) = Γ(1/3) Γ(1- 1/3) = π /sin(π/3) = π / ( 3/2) = 2π / 3
Exercise
Evaluate I =
1
0
[ 1 /
4
[x3 (1 – x )] ] dx
Example(3)
Evaluate I =
1
0
x . (1 – x ) dx
Solution
I =
1
0
x 1/2 (1 – x ) dx =
1
0
x 3/2 - 1 (1 – x ) 2 - 1 dx
= B(3/2 , 2) = Γ(3/2) Γ(2) / Γ(7/2)
Γ(3/2) = ½ π
Γ(5/2) = Γ(3/2+ 1) = (3/2) Γ(3/2 ) = (3/2) . ½ π = 3π / 4
Γ(7/2) = Γ(5/2+ 1) = (5/2) Γ(5/2 ) = (5/2) . (3π / 4) = 15 π / 8
Thus,
I = (½ π ) . 1! / (15 π / 8) = 4/15
Exercise
Evaluate I =
1
0
x5 . (1 – x ) dx
II. Using Gamma Function to Evaluate Integrals
Example(1)
Evaluate: I =
∞
0
x 6 e -2x dx
Solution:
- 158 -
Letting y = 2x, we get
I = (1/128) 0∞ y 6 e -y dy = (1/128) Γ(7) = (1/128) 6! = 45/8
Example(2)
Evaluate: I =
∞
0
x e –x^3 dx
Solution:
Letting y = x3 , we get
I = (1/3) 0∞ y -1/2 e -y dy = (1/3) Γ(1/2) = π / 3
Example(3)
Evaluate: I =
∞
0
xm e – k x^n dx
Solution:
Letting y = k xn , we get
I = [ 1 / ( n . k (m+1)/n) ] 0∞ y [(m+1)/n – 1] e -y dy = [ 1 / ( n . k (m+1)/n) ]
Γ[(m+1)/n ]
II. Using Beta Function to Evaluate Integrals
Formulas
(1) 01 x m-1 (1 – x ) n-1 dx = B(m,n) = Γ(m) Γ(n) / 2 Γ(m+ n) ; m > 0 &
n>0
(3)
π/2
0
(4)
∞
0
sin 2m-1x . cos 2n-1x dx = (1/2) B(m,n)
; m>0 & n>0
x q-1 / (1+x) . dx = Γ(q) Γ(1-q) = Π / sin(qπ) ; 0 < q < 1
Using Formula (1)
- 159 -
Example(1)
Evaluate: I = 02 x2 / (2 – x ) . dx
Solution:
Letting x = 2y, we get
I = (8/2) 01 y 2 (1 – y ) -1/2 dy = (8/2) . B(3 , 1/2 ) = 642 /15
Example(2)
Evaluate: I = 0a x4  (a2 – x2 ) . dx
Solution:
Letting x2 = a2 y , we get
I = (a6 / 2) 01 y 3/2 (1 – y )1/2 dy = (a6 / 2) . B(5/2 , 3/2 ) = a6 /3 2
Exercise
Evaluate: I = 02 x  (8 – x3 ) . dx
Hint
Lett x3 = 8y
Answer
I = (8/3) 01 y-1/3
(1 – y ) 1/3 . dy = (8/3) B(2/3 , 4/3 ) = 16 π / ( 9 3 )
Using Formula (3)
Example(3)
Evaluate: I = 0∞ dx / ( 1+x4 )
- 160 -
Solution:
Letting x4 = y , we get
I = (1 / 4) 0∞ y -3/4 dy / (1 + y ) = (1 / 4) . Γ (1/4) . Γ (1 - 1/4 )
= (1/4) . [ π / sin ( ¼ . π ) ] = π 2 / 4
Using Formula (2)
Example(4)
a. Evaluate: I = 0π/2 sin 3 . cos 2x dx
b. Evaluate: I = 0π/2 sin 4 . cos 5x dx
Solution:
a. Notice that: 2m - 1 = 3 → m = 2 & 2n - 1 = 2 → m = 3/ 2
I = (1 / 2) B( 2 , 3/2 ) = 8/15
b. I = (1 / 2) B( 5/2 , 3 ) = 8 /315
Example(5)
a. Evaluate: I = 0π/2 sin6 dx
b. Evaluate: I = 0π/2 cos6x dx
Solution:
a. Notice that: 2m - 1 = 6 → m = 7/2 & 2n - 1 = 0 → m = 1/ 2
I = (1 / 2) B( 7/2 , 1/2 ) = 5π /32
b. I = (1 / 2) B( 1/2 , 7/2 ) = 5π /32
- 161 -
Example(6)
a. Evaluate: I = 0π cos4x dx
b. Evaluate: I = 02π sin8 dx
Solution:
a. I = 0π cos4x = 2 0π/2 cos4x = 2 (1/2) B (1/2 , 5/2 ) = 3π / 8
b. I = I = 0π sin8x = 4 0π/2 sin8x = 4 (1/2) B (9/2 , 1/2 ) = 35π / 64
Details
I.
Example(1)
Evaluate: I =
∞
0
x 6 e -2x dx
x = y/2
x 6 = y 6 /64
dx = (1/2)dy
x 6 e -2x dx = y 6 /64 e –y . (1/2)dy
Example(2)
I=
∞
0
x e –x^3 dx x=y1/3
x= y1/6
dx=(1/3)y-2/3 dy
x e –x^3 dx = y1/6 e –y . (1/3)y-2/3 dy
- 162 -
Example(3)
Evaluate: I =
∞
0
xm e – k x^n dx
y = k xn
x = y1/n / k1/n
xm = ym/n / km/n
dx = (1/n) y(1/n-1) / k1/n dy
xm e – k x^n dx = ( ym/n / km/n ) . e – y . (1/n) y(1/n-1) / k1/n dy
m/n + 1/n – 1 = (m+1)/n - 1
-m/n – 1/n = - (m+1)/n
I = [ 1 / ( n . k (m+1)/n) ] 0∞ y [(m+1)/n – 1] e -y dy
II.
Example(1)
I = 02 x2 / (2 – x ) . dx
x = 2y
dx=2dy
x2 = 4 y2
(2 – x ) = (2 – 2y ) =2 (1 – y )
x2 / (2 – x ) . dx = 4 y2 / 2 (1 – y ) 2dy
y=0 when x=0
y=1 when x=2
- 163 -
Example(2)
Evaluate: I = 0a x4  (a2 – x2 ) . dx
x2 = a2 y , we get
x4 = a4 y2
x= a y1/2
dx= (1/2)a y-1/2 dy
 (a2 – x2 ) =  (a2 – a2 y ) = a (1 – y )1/2
x4  (a2 – x2 ) . dx = a4 y2 a (1 – y )1/2 (1/2)a y-1/2 dy
y=0 when x=0
y=1 when x=a
Example(3)
I = 0∞ dx / ( 1+x4 )
x4 = y
x=y1/4
dy= (1/4) y-3/4 dy
dx / ( 1+x4 ) = (1 / 4) y -3/4 dy / (1 + y )
Proofs of formulas (2) & (3)
Formula (2)
We have,
B(m,n) =
1
0
x m-1 (1 – x ) n-1 dx
Let x = sin2y
Then dy = 2 sinx cox dx
- 164 -
&
x m-1 (1 – x ) n-1 dx = (sin2y) m-1 ( cos2y ) n-1 ( dy / 2 sinx cox )
= 2 sin 2m-1y . cos 2n-1y dy
When x=0 , we have y = 0
When x=1, we hae y = π/2
Thus,
I = 2 0π/2 sin 2m-1y . cos 2n-1y dy
I = 0π/2 sin 2m-1y . cos 2n-1y dy = B(m,n) / 2
Formula (3)
We have,
I = 0∞ x q-1 / (1+x) dx
Let
y = x / (1+x)
Hence, x = y / 1-y
, 1 + x = 1 + (y / 1-y) = 1/(1-y)
& dx = - [ (1- y) – y(-1)] / (1-y)2 . dy= 1 / (1-y)2 . dy
whn x = 0 , we have y = 0
when x→∞ , we have y = lim x→∞ x / (1+x) = 1
Thus,
I = 0∞ [ x q-1 / (1+x) ] dx = 0∞ [ ( y / 1-y ) q-1 / (1/(1-y)) ] . 1 / (1-y)2 . dy
- 165 -
= 01 [ y
q-1
/ (1-y) -q ] dy
= B(q , 1-q) = Γ(q) Γ(1-q)
Proving that
Γ(1/2) =
∞
0
Γ(1/2
) = π
x 1/2-1 e -x dx = 0∞ x -1/2 e -x dx
Let y = x ½
x = y2
dx = 2y dy
Γ(1/2) =
∞
0
y -1 e –y^2 2y dy
= 2 0∞ e –y^2 dy
=2 (π / 2 ) = π
- 166 -
Bessel Functions and their Applications
1. Bessel’s Equation:
As we have pointed above that the equation
𝑑2𝑦
𝑑𝑥 2
+
1 𝑑𝑦
𝑥 𝑑𝑥
+ (1 −
𝜈2
𝑥2
)𝑦 = 0
…………(1)
In which 𝜈 is a constant, is known as Bessel’s equation and its solutions are
called as Bessel functions. These functions were used by Bessel in the
nineteenth century in a problem of dynamical Astronomy but since an
orthonormal set of function can be defined with Bessel functions it has
enormous use in different branches of mathematical physics and Engineering.
We have shown the two solutions of (1) may be taken as 𝐽𝜈 (𝑥) and 𝐽−𝜈 (𝑥)
[𝜈
is not an integer] when
𝐽𝜈 (𝑥) = ∑∞
𝑟=0
𝑥 𝜈+2𝑟
2
(−1)𝑟 ( )
𝑟 ! Γ(𝜈+𝑟+1)
……….…….(2)
If 𝜈 is an integer then
𝐽−𝑛 (𝑥) = ∑∞
𝑟=0
𝑥 −𝑛+2𝑟
2
(−1)𝑟 ( )
𝑟 ! Γ(−𝑛+𝑟+1)
…………..(3)
Now, using Γ(𝑛 − 𝑟) Γ(1 − 𝑛 + 𝑟) = 𝜋 𝑒𝑎𝑠𝑒 (𝑛 − 𝑟)𝜋 and hence Γ(1 − 𝑛 + 𝑟)
is infinite when r = 0, 1, 2….(n-1). Therefore
𝐽−𝑛 (𝑥) = ∑∞
𝑟=𝑛
𝑥 −𝑛+2𝑟
2
(−1)𝑟 ( )
𝑟 ! Γ(−𝑛+𝑟+1)
………….(4)
Writing r = s + n, this become
𝐽−𝑛 (𝑥) = ∑∞
𝑠=0
𝑥 𝑛+2𝑠
2
(−1)𝑠+𝑛 ( )
(𝑠+𝑟) ! Γ(𝑠+1)
= (−1)𝑛 ∑∞
𝑠=0
𝑥 𝑛+2𝑠
2
(−1)𝑠 ( )
Γ(𝑛+𝑠+1) 𝑠!
= (−1)𝑛 𝐽𝑛 (𝑥)
……….(5)
- 167 -
In this case the other solution is taken to Yn(x) [given in (8) in 1.2]
Example. show that
𝐽1 (𝑥) = √
2
2
2
sin 𝑥, 𝐽−1 (𝑥) = √
cos 𝑥
𝜋𝑥
𝜋𝑥
2
1
Putting 𝜈 = in (2)
2
1
𝑥 2
𝑥 2
𝑥 4
(2)
(2)
(2)
𝐽1 (𝑥) =
+
……… ]
[1 −
3
3
3 5
2
Γ( )
1.
2
2 1.2. 2 . 2
=√
2
𝑥3 𝑥5
2
sin 𝑥
[𝑥 − + … … … ] = √
𝜋𝑥
3! 5!
𝜋𝑥
1
For the second relation put 𝜈 = − in (3)
2
The results are special cases of an important general theorem which states that
𝐽𝜈 (𝑥) is expressible in finite terms by means of algebraic and trigonometric
function of x whenever 𝜈 is half of an odd integer. The functions 𝐽𝑛+1 (𝑥) and
2
𝐽−(𝑛+1) (𝑥) when n is a positive integer or zero are called spherical Bessel
2
functions and have important application in wave motion in which spherical
polar coordinates are appropriate.
2 Bessel function of order zero
When 𝜈 = 0, the first solution of Bessel’s equation is given by
𝑥 2𝑟
𝐽𝑜 (𝑥) =
(−1)𝑟 ( )
2
∞
∑𝑟=𝑜
𝑟 ! Γ(𝑟+1)
=1−
𝑥2
𝑥4
2
22 4
+
2
−
2
𝑥6
22 .4 2 .62
+ ⋯(1)
Let Yo (x) is the second independent solution and we take it as
- 168 -
2
𝑥
𝑌𝑜 (𝑥) = (𝑙𝑜𝑔𝑒 + 𝛾) 𝐽𝑜 (𝑥) − 𝑉𝑜 (𝑥) ………….(2)
𝜋
2
𝑥
when Y is the Euler’s constant and 𝑉𝑜 ( ) is a series is ascending powers of x,
𝜋
Now
𝑌𝑜 ′(𝑥) =
𝑌𝑜 ′′(𝑥) =
2
𝑥
2
𝐽 (𝑥) − 𝑉𝑜 ′(𝑥)
(𝑙𝑜𝑔𝑒 + 𝛾) 𝐽𝑜 ′(𝑥) +
𝜋
2
𝜋𝑥 𝑜
2
𝑥
4
2
𝐽𝑜 ′(𝑥) − 2 𝐽𝑜 (𝑥) − 𝑉𝑜 ′′(𝑥)
(𝑙𝑜𝑔𝑒 + 𝛾) 𝐽𝑜 ′′(𝑥) +
𝜋
2
𝜋𝑥
𝜋𝑥
Substituting in Bessel equation with 𝜈 = 0 i.e.
𝑑2𝑦
𝑑𝑥 2
+
1 𝑑𝑦
𝑥 𝑑𝑥
𝑦=0
………….(3)
We find
2
𝑥
1
4 ′
𝐽 (𝑥)
(𝑙𝑜𝑔𝑒 + 𝛾) {𝐽𝑜′′ (𝑥) + 𝐽𝑜′ (𝑥) + 𝐽𝑜 (𝑥)} +
𝜋
2
𝑥
𝜋𝑥 𝑜
1
− {𝑉𝑜′′ (𝑥) + 𝑉𝑜 (𝑥) + 𝑉𝑜 (𝑥)} = 0
𝑥
i.e.
𝑉𝑜′′ (𝑥)
1
+ 𝑉𝑜 ′(𝑥) + 𝑉𝑜 (𝑥) =
𝑥
4
𝐽′ (𝑥)
𝜋𝑥 𝑜
=
2
∑∞
𝜋 𝑟=1
𝑥 2𝑟−2
2
(−1)𝑟 ( )
𝑟 ! (𝑟−1)!
𝑥 2𝑟
2
To solve (4) let 𝑉𝑜 (𝑥) = ∑∞
𝑏 ( )
𝜋 𝑟=0 𝑟 2
1
2
1
𝑥 2𝑟−2
i.e. 𝑉𝑜 ′(𝑥) = ∑∞
𝑟 𝑏𝑟 ( )
𝑥
𝜋 𝑟=1 2
2
∞
2
1
𝑥 2𝑟−2
′′ (𝑥)
𝑉𝑜
= ∑ 𝑟(𝑟 − )𝑏𝑟 ( )
𝜋
2
2
𝑟=1
1
𝑥 2𝑟−2
2
i.e. 𝑉𝑜′′ (𝑥) + 𝑉𝑜 ′(𝑥) + 𝑉𝑜 (𝑥) = ∑∞
(𝑟 2 𝑏𝑟 + 𝑏𝑟−1 ) ( )
𝑟=1
𝑥
𝜋
2
comparing with (4) we get
- 169 -
………(4)
2
𝑟 𝑏𝑟 + 𝑏𝑟−1
(−1)𝑟
=
, 𝑟 = 1,2,3, … … … …
𝑟! (𝑟 − 1)!
taking 𝑏𝑜 = 0 the above recurrence relation gives 𝑏1 = −1 and
4𝑏2 + 𝑏1 =
1
1
1
, 9𝑏3 + 𝑏2 = −
, 16 𝑏4 + 𝑏3 =
, … ….
2!
2! 3!
3! 4!
these gives successively
𝑏2 =
1
1
1
1 1
1
1 1 1
(1 + ) , 𝑏3 = −
(1 + + ) , 𝑏4 = (1 + + + ) , … …
2!
2
(3!)
2 3
4!
2 3 4
and in general
(−1)𝑟
1 1
1
𝑏𝑟 =
(1 + + + ⋯ + )
𝑟!
2 3
𝑟
substitution in (2) we get
2
𝑥
2
𝑌𝑜 (𝑥) = (𝑙𝑜𝑔𝑒 + 𝛾) 𝐽𝑜 (𝑥) − ∑∞
𝜋
2
𝜋 𝑟=1
𝑥 2𝑟
2
(−1)𝑟 ( )
𝑟!𝑟!
1
1
……….(5)
(1 + 2 + ⋯ + 𝑟 )
Yo(x) is called the Bessel function of the second kind of order zero and the
complete solution of Bessel’s equation of order zero (3) is
𝑦 = 𝐴 𝐽𝑂 (𝑥) + 𝐵 𝑌𝑜(𝑥) …………….(6)
Both Jo(x) and Yo(x) are oscillatory functions and their graphs are shown below.
It should be noted that both Jo(x) and Yo(x) vanish at an infinite sequence of
values of x. In this respect they
behave
similarly
to
the
trigonometrical functions cos x
and
sin
x
which
vanish
1
respectively when 𝑥 = (𝑟 + )𝜋
2
and
- 170 -
𝑥 = 𝑟𝜋
.
In
practical
application the positive values of x for which the Bessel function vanish are of
great importance and these values are known as positive zeroes of the functions.
If we denote the rth zero of Jo(x), Yo(x) by 𝛼𝑟 and 𝛽𝑟 respectively, It can be
shown that these zeroes interlace i.e. 𝛽𝑟 < 𝛼𝑟 < 𝛽𝑟+1 for all r. Unlike those of
the trigonometrical functions, the positive zeroes of the Bessel function are not
equally spaced along the x-axis, but it can be shown that 𝛼𝑟 and 𝛽𝑟 approximate
1
3
4
4
respectively to(𝑟 − )𝜋 and (𝑟 − )𝜋 for large values of r. Numerical values of
the zeroes can be found in many hand book and numerical Tables. It is easily
seen that for small values of x
𝐽𝑂 (𝑥) ≃ 1 and 𝑌𝑜 (𝑥) ≃
2
𝑥
[𝑙𝑜𝑔𝑒 + 𝛾]
𝜋
2
Thus 𝐽𝑜 (𝑥) → 1 and 𝑌𝑜 (𝑥) → −∞ as 𝑥 → 0 and also it can be shown that both
the function tends to zero as 𝑥 → ∞
Note: 𝑌𝑛 (𝑥)can also be developed as above. In fact from (6) and (8) of 1.2 we
can write
𝑥 2𝑟+1
(−1)𝑟 ( )
𝑥
𝑥3
𝑥5
2
𝐽1 (𝑥) = ∑
= − 3
+
− ⋯ 𝑎𝑛𝑑
𝑟 ! (𝑟 + 1)!
2 2 1! 2! 25 2! 3!
∞
𝑟=0
2
𝑥
2
1
𝑌1 (𝑥) = (𝑙𝑜𝑔𝑒 + 𝛾) 𝐽1 (𝑥) − − ∑∞
𝜋
2
𝜋𝑥
𝜋 𝑟=0
𝑥 2𝑟+1
2
(−1)𝑟 ( )
𝑟 !(𝑟+1)!
1
𝑟+1
}
1
1
{2 (1 + + ⋯ + ) +
2
𝑟
……….(6)
the first term on the right being x/2.
Below we give the rough graphs of
J1(x) and Y1(x). Note the interlacing
of their positive zeroes. For small x,
- 171 -
J1(x) and Y1(x) behaves like x/2 and
−2
𝜋𝑥
respectively. Both the function tends to
zero as 𝑥 → ∞. These positive zeroes are extensively tabulated. To avoid the
necessity of having to write the general solution of Bessel’s equation in the two
different forms.
i.e. 𝑦 = 𝐴 𝐽𝜈 (𝑥) + 𝐵 𝐽−𝜈 (𝑥), 𝜈 not zero or positive integer
𝑦 = 𝐴 𝐽𝑛 (𝑥) + 𝐵 𝑌𝑛 (𝑥), n zero or a positive integer. It is possible to define
the second solution as
𝑌𝜈 (𝑥) =
𝐽𝜈 (𝑥) 𝑐𝑜𝑠 𝜈𝜋−𝐽−𝜈 (𝑥)
sin 𝜈𝜋
………….(7)
and then for 𝜈 → 𝑛 (n zero or integer) the expression on r,h,s, of (7) tends to
Yn(x) [Use L’ Hospital’s theorem]. Thus with the form (7), the general solution
of Bessel’s equation of order 𝜈 for all values of 𝜈 is 𝑦 = 𝐴 𝐽𝜈 (𝑥) +
𝐵 𝑌𝜈 (𝑥)………(8)
3.3 Hankel functions.
Although 𝐽𝜈 (𝑥) and 𝑌𝜈 (𝑥) are independent solutions of Bessel’s equation it is
sometimes convenient to take the fundamental solution in a slightly different
form. Henkel functions often called Bessel functions of the third kind, are often
defined by
(1)
(2)
𝐻𝜈 (𝑥) = 𝐽𝜈 (𝑥) + 𝑖 𝑌𝜈 (𝑥), 𝐻𝜈 (𝑥) = 𝐽𝜈 (𝑥) − 𝑖 𝑌𝜈 (𝑥) ………..(9)
And there are independent solutions of Bessel’s equation. In terms of these
function, the general solution of the equation is
(1)
(2)
𝑦 = 𝐴 𝐻𝜈 (𝑥) + 𝐵 𝐻𝜈 (𝑥) ………..(10)
- 172 -
The Hankel functions bear the same relation to the Bessel function of the first
and third kind as the function 𝑒 ± 𝑖 𝜈 𝑥 bear to 𝑐𝑜𝑠 𝜈𝑥 and 𝑠𝑖𝑛 𝜈𝑥 and they are
convenient in mathematical analysis for similar reasons. Note that by (7) and (9)
(1)
𝐻𝜈 (𝑥) =
(2)
𝐻𝜈 (𝑥) =
Use of
−𝑖
𝑠𝑖𝑛 𝜈𝜋
𝑖
𝑠𝑖𝑛 𝜈𝜋
[𝐽𝜈 (𝑥)𝑒 𝑖 𝜈 𝜋 − 𝐽−𝜈 (𝑥)] ………(11)
(1)
(2)
𝐻𝜈 (𝑘𝑥)as 𝐻𝜈 (𝑘𝑥)
(1),(2)
(𝑥)𝑎𝑟𝑒
𝐻𝜈
√
2
𝜋𝑘𝑥
𝑒
[𝐽𝜈 (𝑥)𝑒 − 𝑖 𝜈 𝜋 − 𝐽−𝜈 (𝑥)]
𝜋 𝜈𝜋
)
4 4
± 𝑖 (𝑘𝑥− −
in the propagation of waves with
and
if
the
time
dependent
part
is
(1)
(2)
𝑒 − 𝑖 𝜔 𝑡 , (𝜔 > 0)then 𝐻𝜈 (𝑘𝑥) generates forward wave and 𝐻𝜈 (𝑘𝑥) generates
backward wave.
3.4 Some properties of Bessel functions.
Bessel function possesses properties of which great use can be made in the
discussion of physical problems. A few of the more important of these are given
in this section but for an extensive treatment one may refer to the standard text
by Watson.
(a) Recurrence formulae
We have seen that 𝐽𝜈 (𝑥) = ∑∞
𝑟=0
𝑥 𝜈+2𝑟
2
(−1)𝑟 ( )
𝑟 ! Γ(𝜈+𝑟+1)
𝑥 2𝑟
i.e. 𝑥
−𝜈
𝐽𝜈 (𝑥) = 2
−𝜈
(−1)𝑟 ( )
2
∑∞
𝑟=0 𝑟 ! Γ(𝜈+𝑟+1) , ℎ𝑒𝑛𝑐𝑒
𝑥 2𝑟−1
𝑟
(−1)
𝑟
(
)
𝑑
2
−𝜈 (𝑥)
−𝜈
(𝑥 𝐽𝜈 ) = 2 ∑
𝑑𝑥
𝑟 ! Γ(𝜈 + 𝑟 + 1)
∞
𝑟=1
𝑥 2𝑟+1
= −2
−𝜈
(−1)𝑟 ( )
2
∑∞
𝑟=0 𝑟 ! Γ(𝜈+𝑟+2)
= −𝑥 −𝜈 𝐽𝜈+1 (𝑥)
- 173 -
……..(1)
This can be written in the form
𝑥 −𝜈 𝐽′ 𝜈 (𝑥) − 𝜈𝑥 −𝜈−1 𝐽𝜈 (𝑥) = −𝑥 −𝜈 𝐽𝜈 (𝑥)
And it follows that
𝜈
𝐽 (𝑥) −
𝑥 𝜈
𝐽′ 𝜈 (𝑥) = 𝐽𝜈+1 (𝑥)
………….(2)
This is the first of the required recurrence formulae. A useful particular case of
this is 𝐽′𝑜 (𝑥) = −𝐽1 (𝑥)[puting 𝜈 = 0 in (2)]
𝑥 2𝜈+2𝑟
𝜈
Again, 𝑥 𝐽𝜈 (𝑥) = 2
𝜈
(−1)𝑟 ( )
2
∞
∑𝑟=0
𝑟 ! Γ(𝜈+𝑟+1)
We have
𝑥 2𝜈+2𝑟−1
𝑟 (𝜈
(−1)
+
𝑟)
(
)
𝑑
2
𝜈 (𝑥)
𝜈
(𝑥 𝐽𝜈 ) = 2 ∑
𝑑𝑥
𝑟 ! Γ(𝜈 + 𝑟 + 1)
∞
𝑟=0
= 𝑥 𝜈 ∑∞
𝑟=0
𝑥 𝜈−1+2𝑟
2
(−1)𝑟 ( )
𝑟 ! Γ(𝜈+𝑟)
= 𝑥 𝜈 𝐽𝜈−1 (𝑥)[𝜈 + 𝑟 = 𝜈 − 1 + 𝑟 + 1] ……(3)
This can be written
𝑥 𝜈 𝐽′ 𝜈 (𝑥) + 𝜈𝑥 𝜈−1 𝐽𝜈 (𝑥) = 𝑥 𝜈 𝐽𝜈−1 (𝑥)
i.e.
𝜈
𝐽 (𝑥) +
𝑥 𝜈
𝐽′ 𝜈 (𝑥) = 𝐽𝜈−1 (𝑥) …………(4)
Two other useful recurrence formulae followed by addition and subtraction of
(2) and (4). Thus
- 174 -
2𝜈
𝐽𝜈 (𝑥) = 𝐽𝜈−1 (𝑥) + 𝐽𝜈+1 (𝑥)
.…….…(5)
2 𝐽′ 𝜈 (𝑥) = 𝐽𝜈−1 (𝑥) − 𝐽𝜈+1 (𝑥)
………..(6)
𝑥
The function of the second kind 𝑌𝜈 (𝑥) and the Hankel functions
(1)
(2)
𝐻𝜈 (𝑥), 𝐻𝜈 (𝑥) satisfy the same recurrence formulae. For example
𝑌𝜈−1 (𝑥) =
𝐽𝜈−1 (𝑥) cos(𝜈 − 1)𝜋 − 𝐽−𝜈+1 (𝑥) 𝐽𝜈−1 (𝑥) cos 𝜈𝜋 + 𝐽−𝜈+1 (𝑥)
=
sin(𝜈 − 1)𝜋
sin 𝜈𝜋
𝑌𝜈+1 (𝑥) =
𝐽𝜈+1 (𝑥) cos(𝜈 + 1)𝜋 − 𝐽−𝜈−1 (𝑥) 𝐽𝜈+1 (𝑥) cos 𝜈𝜋 + 𝐽−𝜈−1 (𝑥)
=
sin(𝜈 + 1)𝜋
sin 𝜈𝜋
Thus adding and using (5), we have
2𝜈
2𝜈
𝐽𝜈 cos 𝜈𝜋 + (− ) 𝐽−𝜈 (𝑥)
2𝜈
𝑥
𝑌𝜈−1 (𝑥) + 𝑌𝜈+1 (𝑥) = 𝑥
=
𝑌 (𝑥) … … … (7)
sin 𝜈𝜋
𝑥 𝜈
The other recurrence formulae can be proved in the same way.
(b) The generating function for the Bessel coefficients: Provided t is not zero,
1𝑥
1
the functions 𝑒 2 𝑥 𝑡 and 𝑒 −2 𝑡 can be expanded in powers of t, and product of
these expansion gives
∞
∞
𝑟=0
𝑠=0
∞
∞
1
1
1 𝑥𝑡 𝑟
1 −𝑥 𝑠
(−1)𝑠 𝑥 𝑟+𝑠
exp 𝑥 (𝑡 − ) = ∑ ( ) ∑ ( ) = ∑ ∑
+𝑟−𝑠
( )
2
𝑡
𝑟! 2
𝑠! 2𝑡
𝑟! 𝑠! 2
𝑟=0 𝑠=0
Term by term multiplication of the series being permissible because of the
absolute convergence of the separate series. The coefficients of tn(n a positive
integer or zero) is found by taking n = r-s i.e. r = n+s and taking s very form 0
to infinity. Thus the coefficients of tn is
- 175 -
∞
(−1)𝑛
𝑥 𝑛+2𝑠
= 𝐽𝑛 (𝑥)
∑
( )
(𝑛 + 𝑠)! 𝑠! 2
𝑠=0
The coefficients of t
taking
s
very
(−1)𝑠
-n
(n a positive integer) is found by taking r = -n+s and
from
𝑥 −𝑛+2𝑠
=∑∞
𝑠=𝑛 (−𝑛+𝑠)! 𝑠! (2)
𝑥
n
to
infinity.
Thus
the
coefficient
of
t
-n
= 𝐽−𝑛 (𝑥) [see equ. (4) of 3.1]
1
𝑛
Hence exp { (𝑡 − )} = ∑+∞
−∞ 𝑡 𝐽𝑛 (𝑥)
2
𝑡
……….(8)
And the exponential function on the left can be regarded as the generating
function of Jn(x). Because of the form of (8), the functions Jn(x) when n=0, 1, 2,
3… are often called the Bessel coefficients.
Example: Show that
∞
∞
exp(−𝑖 𝑥 sin 𝜃) = 𝐽𝑜 (𝑥) + 2 ∑ 𝐽2𝑛 (𝑥) cos 2𝑛𝜃 − 2𝑖 ∑ 𝐽2𝑛+1 (𝑥)
𝑛=1
𝑛=0
sin(2𝑛 + 1)𝜃
Writing 𝑡 = 𝑒 −𝑖𝜃 in (8)
+∞
1
exp {− 𝑥(𝑒 𝑖𝜃 − 𝑒 −𝑖𝜃 )} = ∑(−1)𝑛 𝑒 𝑖𝑛𝜃 𝐽𝑛 (𝑥)
2
−∞
∞
= 𝐽𝑜 (𝑥) + ∑(−1)𝑛 {𝑒 𝑖𝑛𝜃 + (−1)𝑛 𝑒 −𝑖𝑛𝜃 } 𝐽𝑛 (𝑥) . . … . (9)
𝑛=1
Since J-n(x) =(-1)n Jn(x). The required result follows when we note that 𝑒 𝑖𝜃 −
𝑒 −𝑖𝜃 = 2 𝑖 sin 𝜃 and
(−1)𝑛 𝑒 𝑖𝑛𝜃 + 𝑒 −𝑖𝑛𝜃 = 2 cos 𝑛𝜃 , 𝑛 𝑒𝑣𝑒𝑛,
= −2𝑖 sin 𝑛𝜃, 𝑛 𝑜𝑑𝑑
- 176 -
(c) Bessel Integral
Equation (9) can be written as
𝑛 𝑖𝑛𝜃
exp(−𝑖 𝑥 sin 𝜃) = 𝐽𝑜 (𝑥) + ∑∞
+ 𝑒 −𝑖𝑛𝜃 } 𝐽𝑛 (𝑥)
𝑛=0{(−1) 𝑒
………(10)
2𝜋
Since ∫𝑜 𝑒 𝑖𝑛𝜃 𝑑𝜃 = 2𝜋, when r=0 and vanishes when r is an integer,
multiplication of (10) by 𝑒 𝑖𝑛𝜃 and integration w.r.t. 𝜃 between o and 2𝜋 gives
formally
2𝜋
∫𝑜 exp{𝑖 (𝑛𝜃 − 𝑥 sin 𝜃)} 𝑑𝜃 = 2𝜋 𝐽𝑛 (𝑥),
………(11)
n = 0, 1, 2, 3,
A rigorous proof however exists else where [Lebedeo –Special function and
their applications]
since Jn(x) is real, we can show by equating real and
imaginary parts that
1 2𝜋
𝐽𝑛 (𝑥) =
∫ cos (𝑛𝜃 − 𝑥 sin 𝜃) 𝑑𝜃
2𝜋 𝑜
1 𝜋
= ∫ cos (𝑛𝜃 − 𝑥 sin 𝜃) 𝑑𝜃 … … (12)
𝜋 𝑜
and the expression (11) gives Bessel’s integral for Jn(x).
(d) Some integrals involving Bessel functions.
Using the result (3)
𝑎
∫ 𝑥 𝜈 𝐽𝜈−1 (𝑥)𝑑𝑥 = [𝑥 𝜈 𝐽𝜈 (𝑥)]𝑎0
𝑜
If 𝜈 is real and positive then as 𝑥 → 0, 𝑥 𝜈 𝐽𝜈 (𝑥) → 0 and so
𝑎
∫𝑜 𝑥 𝜈 𝐽𝜈−1 (𝑥)𝑑𝑥 = 𝑎𝜈 𝐽𝜈 (𝑎), (𝜈 > 0) ………(13)
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and a similar integral can do obraiaed from the result (1). Other integrals of this
type can be evaluated by such devices as integration by parts and use of
recurrence formulae. Below we discuss an important integral
𝑎
𝐼 = ∫𝑜 𝑥 𝐽𝜈 (𝜆𝑥)𝐽𝜈 (𝜇𝑥) 𝑑𝑥
……….(14)
Let 𝑢 = 𝐽𝜈 (𝜆𝑥) then u satisfies Bessel’s equation when u is written in place of
y and 𝜆 in place of x. This can be written in the form,
𝑥2
𝑑2𝑢
𝑑𝑢
𝑑𝑥
𝑑𝑥
+𝑥
2
+ (𝜆2 𝑥 2 − 𝜈 2 )𝑢 = 0 ………(15)
Similarly 𝜈 = 𝐽𝜈 (𝜇𝑥) satisfies the equation
𝑥2
𝑑2𝜈
𝑑𝑥 2
+𝑥
𝜈
𝑢
𝑥
𝑥
𝑑𝜈
𝑑𝑥
+ (𝜇2 𝑥 2 − 𝜈 2 )𝑢 = 0
……..(16)
Multiplying by and respectively and subtracting
(𝜆2 −𝜇2 )𝑥 𝑢𝜈 = 𝑥 (𝑢
𝑑2𝜈
𝑑2𝑢
𝑑𝑥
𝑑𝑥 2
−𝜈
2
𝑑𝜈
𝑑𝑢
𝑑
𝑑𝜈
𝑑𝑢
) + (𝑢 𝑑𝑥 − 𝜈 𝑑𝑥 ) = 𝑑𝑥 {𝑥 (𝑢 𝑑𝑥 − 𝜈 𝑑𝑥 )}
….(17)
Substituting for u,v and integrating w.r.t. x between o and a
𝑎
(𝜆2
−𝜇
2)
∫ 𝑥 𝐽𝜈 (𝜆𝑥)𝐽𝜈 (𝜇𝑥) 𝑑𝑥 = [𝑥{ 𝐽𝜈 (𝜆𝑥)𝐽′ 𝜈 (𝜇𝑥) − 𝐽𝜈 (𝜇𝑥)𝐽′ 𝜈 (𝜆𝑥)}]𝑎𝑜
𝑜
= 𝑎[𝜇 𝐽𝜈 (𝜆𝑎)𝐽′ 𝜈 (𝜇𝑎) − 𝜆 𝐽𝜈 (𝜇𝑎)𝐽′ 𝜈 (𝜆𝑎)]
And we have obtained the value of I in (14) when 𝜆2 ≠ 𝑢2 i.e.
𝜆 ≠ ± 𝜇 . If 𝜆 = 𝜇, the integral (14) becomes
𝑎
𝐼 = ∫0 𝑥 J𝜈2 (𝜆𝑥)𝑑𝑥
and from (18)
- 178 -
… … … (18)
𝑎
∫ 𝑥
𝐽𝜈2 (𝜆𝑥)𝑑𝑥
0
= lim [−
𝜇→𝜆
𝑎. 𝜇 𝐽𝜈 (𝜆𝑎)𝐽𝜈′ (𝜇𝑎) − 𝑎𝜆 𝐽𝜈 (𝜇𝑎)𝐽𝜈′ (𝜆𝑎)
= lim
𝜇→𝜆
𝜆2 − 𝜇2
𝑎 𝛿
{𝜇 𝐽𝜈′ (𝜇𝑎)𝐽𝜈 (𝜆𝑎) − 𝜆 𝐽𝜈 (𝜇𝑎)𝐽𝜈 𝐽𝜈′ (𝜆𝑎)}]
2𝜇 𝛿𝜇
[L’ Hospital’s theorem]
=−
=
𝑎2
2
𝑎
lim {𝐽𝜈 (𝜆𝑎) 𝐽𝜈′ (𝜇𝑎) + 𝑎𝜇 𝐽𝜈 (𝜆𝑎) 𝐽𝜈′′ (𝜇𝑎) − 𝜆𝑎 𝐽𝜈′ (𝜆𝑎)𝐽𝜈′ (𝜇𝑎)}
2𝜆 𝜇→𝜆
[{𝐽𝜈′ (𝜆𝑎)}2 − 𝐽𝜈 (𝜆𝑎)𝐽𝜈′′ (𝜆𝑎) −
1
𝐽
𝜆𝑎 𝜈
(𝜆𝑎)𝐽𝜈′ (𝜆𝑎)] ……….(19)
Since 𝐽𝜈 (𝜆𝑎) satisfies the Bessel’s equation
𝐽𝜈′′ (𝜆𝑎) +
1
𝜆𝑎
𝐽𝜈′ (𝜆𝑎) + (1 −
𝜈2
𝜆2 𝑎 2
) 𝐽𝜈 (𝜆𝑎)=0 ………….....(20)
We get using (19) and (20)
𝑎
∫0 𝑥 𝐽𝜈2 (𝜆𝑥)𝑑𝑥 =
𝑎2
2
𝜈2
[{𝐽𝜈′ (𝜆𝑎)}2 + (1 − 2 2) 𝐽𝜈2 (𝜆𝑎)] …….(21)
𝜆 𝑎
Interesting results comes out when 𝜆 is such that 𝐽𝜈 (𝜆𝑎) = 0
This case we shall discuss below.
(e) Fourier Bessel series
Taking
𝜆 = 𝛼𝑟 , 𝜇 = 𝛼𝑠
when 𝛼𝑟
and 𝛼𝑠
are two positive roots of the
equation 𝐽𝜈 (𝛼𝑎) = 0 and substituting in (18) and (21) we have
𝑎
∫0 𝑥 𝐽𝜈 (𝛼𝑟 𝑥) 𝐽𝜈 (𝛼𝑠 𝑥)𝑑𝑥 = 0, 𝑟 ≠ 𝑠 …………………(22)
𝑎
∫0 𝑥 𝐽𝜈2 (𝛼𝑟 𝑥) 𝑑𝑥
1
= 𝑎2 {𝐽𝜈′ (𝛼𝑟 𝑎)}2 , 𝑟 = 1, 2, 3…………….(23)
2
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1
Hence the functions
𝑥 2 𝐽𝜈 (𝛼𝑟 𝑥), 𝑟 = 1, 2, 3 … … where 𝛼𝑟 are the roots of
𝐽𝜈 (𝛼𝑎) =0 are orthogonal in the interval (0, a), and the possibility of expanding
an arbitry functions 𝑓(𝑥) in the form
𝑓(𝑥) = 𝑏1 𝐽𝜈 (𝛼1 𝑥) + 𝑏2 𝐽𝜈 (𝛼2 𝑥) + … . . +𝑏𝑟 𝐽𝜈 (𝛼𝑟 𝑥) + … , (0 < 𝑥 < 𝑎)
……(24)
Assuming this to be possible [for a certain class of function 𝑓(𝑥)]
multiplication of (24) by 𝑥 𝐽𝜈 (𝛼𝑟 𝑥) and the use of (22) and (23) leads to
𝑎
1
∫0 𝑥 𝑓(𝑥) 𝐽𝜈 (𝛼𝑟 𝑥) 𝑑𝑥𝑏𝑟 . 2 𝑎2 {𝐽𝜈′ (𝛼𝑟 𝑎)}2 …………………(25)
the other term on the r.h.s. vanishes by the virtue of (22). The recurrence
relation (2) gives
𝜈
𝐽𝜈 (𝛼𝑟 𝑎) − 𝐽𝜈′ (𝛼𝑟 𝑎) = 𝐽𝜈+1 (𝛼𝑟 𝑎)
𝛼𝑟 𝑎
and since 𝐽𝜈 (𝛼𝑟 𝑎) = 0 it follows that 𝐽𝜈′ (𝛼𝑟 𝑎) = − 𝐽𝜈+1 (𝛼𝑟 𝑎)
Thus (25) gives
𝑏𝑟 =
𝑎
2
2
𝑎2 𝐽𝜈+1
(𝛼𝑟
∫ 𝑥 𝑓(𝑥)𝐽𝜈 (𝛼𝑟 𝑥) 𝑑𝑥……………………(26)
𝑎) 0
Equation (26) gives a formulae from which the coefficient in the series on right
of (24) can be calculated and, from the analogy with the Fourier series, this
series is called Fourier Bessel series.
The analysis given above is, of course, purely formal and no attempt has been
made to discuss the condition under which such an expansion of an arbitrary
function is legitimate. A full discussion of these will be found in Watson’s
standard treaties. By choosing 𝛼𝑟 to be the roots of equations other than
𝐽𝜈 (𝛼𝑟 𝑎) = 0, other expansions can be obtained, an example will be found in
- 180 -
Exercise III [usually needed in Acoustics or Electro-Magnetic diffraction in a
mixed boundary value problem].
𝟑. 𝟓 Asymptotic Values
In physical problems it is often describe to be able to approximate to the Bessel
functions 𝐽𝜈 (𝑥) and 𝑌𝜈 (𝑥) when x is large For such values of 𝑥, the series
defining these function converge slowly but useful approximation can be
obtained as follows. Writing 𝑦 = 𝑢 𝑒 𝑖𝑥 in Bessel’s equation we obtain
𝑑2𝑢
(𝑥 2 𝑑𝑥 2 = 𝑥
𝑑𝑢
𝑑𝑥
𝑑𝑢
− 𝜈 2 𝑢) + 𝑖 (2𝑥 2 + 𝑥 𝑢) = 0 …(1)
𝑑𝑥
𝜌+n
and we seek a solution of the form 𝑢 = ∑∞
substituting in (1), the term
𝑟=0 𝑎𝑛 𝑥
with the highest power of 𝑥 is 𝑖𝑎0 (2𝜌 + 1)𝑥 𝜌+1 and equating its coefficient to
zero, 𝜌 = −
1
2
with this value of 𝜌 the recurrence relation between the
coefficients 𝑎𝑟+1 and 𝑎𝑟 is found in the usual way to be
𝑎𝑟+1 = 𝑖 {
4𝜈2 − (2𝑟+1)2
23 (𝑟+1)
} 𝑎𝑟
……………..(2)
The resulting series is not convergent but it can, nevertheless, be used to find
numerical values of the solution. For large values of 𝑥, we can take a first
approximation to the solution of Bessel’s equation to be 𝑦 = 𝑢𝑒 𝑖𝑥
i.e. 𝑦 =
1
𝑎0 𝑥 − 2 𝑒 𝑖𝑥
𝑏0 𝑥
−
1
2
and another approximate solution can similarly be found to be
𝑒 − 𝑖𝑥 .
The standard solution 𝐽𝜈 (𝑥), 𝑌𝜈 (𝑥)can be identified with
combinations of these approximate solutions by writing
𝐽𝜈 (𝑥) ~ √
𝑌𝜈 (𝑥) ~ √
2
𝜋𝑥
2
𝜋𝑥
𝑐𝑜𝑠 (𝑥 −
𝑠𝑖𝑛 (𝑥 −
𝜋
4
𝜋
4
−
−
𝜈𝜋
2
𝜈𝜋
2
)
)…………………………….(3)
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1
It is easy to verify that when 𝜈 = , these approximation yield the exact solution
2
𝐽1 (𝑥) = √
2
2
𝜋𝑥
sin 𝑥 and 𝑌1 (𝑥) = − √
2
2
𝜋𝑥
𝑐𝑜𝑠 𝑥. The result (3) can of course be
improved by using the recurrence relation (2) to calculate the coefficients further
turns in the series giving 𝑢 in descending powers of 𝑥. Such series are called
asymptotic series and the reader is referred to more advanced treaties for full
discussion.
3.6 Modified Bessel function.
Writing 𝑖𝑥 in place of 𝑥, Bessel’s equation becomes
𝑑2𝑦
1 𝑑𝑦
𝑑𝑥
𝑥 𝑑𝑥
+
2
− (1 +
𝜈2
𝑥2
)𝑦 = 0
………………(1)
And this ‘modified’ equation plays a significant role in science and engineering.
Working exactly in the same way as in example 2, 1.1, one solution of this
equation is given by the function
𝑥 𝜈+2𝑟
( )
2
𝐼𝜈 (𝑥) ∑∞
𝑟=0 𝑟! Γ(ν+r+1)
= (𝑖)−𝜈 𝐽𝜈 (𝑖𝑥)
……………..(2)
The introduction of 𝐽𝜈 (𝑖𝑥) in the second part of Equation (2) is useful but it is of
course, only formal as we have not defined the Bessel function for imaginary
arguments. However this procedure can be justified by appealing to works
discussing Bessel function of the complex variable. The function 𝐼𝜈 (𝑥) is called
the modified Bessel function of the first kind of order 𝜈 and it is a solution of
Bessel’s modified equation for all values of 𝜈.
As was the case with Bessel’s equation, certain difficulties arise in finding a
second
solution
of
the
modified
equation
when
𝜈=𝑛
(𝑛 zero or positive integer). It is conventional to denote the second solution by
𝑘𝜈 (𝑥) and to define it by the relation.
- 182 -
2
𝜋
……………(3)
𝑘𝜈 (𝑥) = 𝑐𝑜𝑠𝑒𝑐 𝜈𝜋 {𝐼−𝜈 (𝑥) − 𝐼𝜈 (𝑥)}
or by the limit of the expression on the right 𝜈 = 𝑛. The function 𝐾𝜈 (𝑥) so
defined is known as the modified Bessel function of the second kind of order 𝜈
and it can be shown that
𝑥
𝑥 2𝑟
𝐾0 (𝑥) = − (𝑙𝑜𝑔𝑒 + 𝛾)
2
( )
2
𝐼0 (𝑥)+∑∞
𝑟=1 𝑟! 𝑟!
(1 +
1
1
1
+ + ⋯ + ) ………..(4)
2
3
𝑟
While for 𝑛 = 1, 2, 3 ….
𝑥
1
(−1)𝑟 (𝑛−𝑟−1)!
2
2
𝑟!
𝐾𝑛 (𝑥) = (−1)𝑛+1 (𝑙𝑜𝑔𝑒 + 𝛾) 𝐼𝑛 (𝑥)+ ∑𝑛−1
𝑟=0
𝑥 2𝑟+𝑛
+
( )
1
2
(−1)𝑛 ∑∞
𝑟=0
2
𝑟! (𝑛+𝑟)!
{1 +
1
2
1
1
3
𝑟
+ +⋯ +
𝑥 2𝑟−𝑛
(2 )
1
1
2
𝑛+𝑟
+1+ +⋯ +
} ……(5)
The term corresponding to 𝑟 = 0 in the second series being
𝑥 𝑛 1
1 1
1
( ) .
{1 + + + ⋯ + }
2
𝑛!
2 3
𝑛
In terms of these functions, the general solution of the modified equation (1) is, for
all values of
𝑦 = 𝐴 𝐼𝜈 (𝑥) + 𝐵𝐾𝜈 (𝑥)………………………………(6)
Where A, B are arbitrary constants.
The modified function which occur most frequently in practical applications are
those of order zero and unity. Rough graph are shown
- 183 -
It is more convenient to plot 𝑒 −𝑥 𝐼𝑛 (𝑥) and 𝑒 𝑥 𝐾𝑛 (𝑥), than the function
themselves because of the behavior of the function for large value of 𝑥. This can
be shown to be
𝐼𝜈 (𝑥)~
𝑒𝑥
√2𝜋𝑥
, 𝐾𝜈 (𝑥)~ √
𝜋
2𝑥
𝑒 −𝑥
……………(7)
The modified function 𝐼𝜈 (𝑥), 𝐾𝜈 (𝑥) bear to exponential function 𝑒 ±𝑥 .
Similar relations to those which the function 𝐽𝜈 (𝑥), 𝑌𝜈 (𝑥)
bear to the
trigonometrical function, and the modified functions have no zeroes for real values
of 𝑥. Some useful properties are given in the exercises. Numerical values for
various 𝜈 and 𝑥 may be found in Tables.
- 184 -
References:
1-
A.M. Matha Hans J. Haubold: Special Functions for Applied Scientists.
ISBN: 978-0-387-75893-0 e-ISBN: 978-0-387-75894-7. Springer
Science+Business Media, LLC. (2008)
2-
Mathai, A.M. (1993). A Handbook of Generalized Special Functions for
Statistical and Physical Sciences, Oxford University Press, Oxford.
3-
Zafar Ahsan: Mathematical Methods. Edition: First. Publisher: Paths
International Ltd. United Kingdom. ISBN: ISBN-10: 1844640434,
ISBN-13: 978-1844640430.
4-
Riley, K. F., M. P. Hobson, and S. J. Bence, Mathematical Methods for
Physics and Engineering: A Comprehensive Guide, Cambridge
University Press, second edition, 2002.
- 185 -