FOURIER TRANSFORMS

We have the Fourier integral

1
R +∞ R +∞
f (x) = 2π −∞ −∞
f (t)cosu(x − t)dtdu
1
R +∞ R +∞ R +∞
= 2π [
−∞ −∞
f (t)cosu(x − t)dt + i −∞ f (t)sinu(x − t)dt]du
1
R +∞ R +∞
= 2π [
−∞ −∞
f (t)eiu(x−t) dt]du
R +∞ R +∞
= 2π1
[
−∞ −∞
f (t)e−iut dt]eiux du
R +∞
where F (u) = √12π −∞ f (t)e−iut dt is the Fourier transform of f (x), denoted by F{f (x)}
and is defined by
Z +∞
1
F{f (x)} = F (u) = √ f (t)e−iut dt (1)
2π −∞

where
Z +∞
1
f (x) = √ F (u)eiux du (2)
2π −∞

and f (x) is called inverse Fourier transform of F (u), denoted by F−1 {F (u)} = f (x).
Example 1 Find the Fourier transform of f (x), where f (x) = 1 if |x| < a and f (x) = 0
if |x| > a.
Solution:
By definition,
R +∞
F{f (x)} = F (u) = √12π −∞ f (t)e−iut dt
R −a Ra R +∞
= √12π [ −∞ 0e−iut dt + −a 1e−iut dt + a 0e−iut dt]
Ra −iut 1 −iua
= √12π −a 1e−iut dt = − √12π [ e iu ]a−a = − √12π iu [e − eiua ]

2
p
= √12π iu
1
(2i)sin(au) = ( π2 ) u1 sin(au)

Example 2 Find the Fourier transform of f (x), where f (x) = 1 − x2 if |x| < 1 and
f (x) = 0 if |x| > 1.
Solution:
By definition,
R +∞
F{f (x)} = F (u) = √12π −∞ f (t)e−iut dt
R −1 R1 R +∞
= √12π −∞ 0e−iut dt + √12π −1 (1 − t2 )e−iut dt + √1
2π 1
0e−iut dt
R1 2 R1 −iut
= √1
2π −1
(1 − t2 )e−iut dt = − √12π [ (1−t
iu
) −iut 1
e ]−1 + √1
2π −1
(−2t) e iu dt

1
2

−iut R1 e−iut
= − √12π iu
2 te
[ −iu ]1−1 + √1 2
2π iu −1 −iu
dt

= − √12π { u22 [e−iu + eiu ] + 2

iu3
[e−iu − eiu ]}

= − √12π { u42 cosu + 2

iu3
(−2isinu)} = √4 [ 13 sinu
2π u
− 1
u2
cosu]

= √4 13 (sinu
2π u
− ucosu) 2

Fourier Sine and Cosine Transforms

The Fourier Sine Transform of f (x) is given by
r Z +∞
2
FS {f (x)} = F (λ) = f (t)sin(λt)dt (3)
π 0
and the Inverse Fourier Sine Transform of F (λ) is given by
r Z +∞
2
FS −1 {F (λ)} = f (x) = F (λ)sin(λx)dλ (4)
π 0

The Fourier Cosine Transform of f (x) is given by

r Z +∞
2
FC {f (x)} = F (λ) = f (t)cos(λt)dt (5)
π 0
and the Inverse Fourier Cosine Transform of F (λ) is given by
r Z +∞
−1 2
FC {F (λ)} = f (x) = F (λ)cos(λx)dλ (6)
π 0

Example 1 Find the Fourier Sine transform of f (x) = e−ax .

Solution: By definition,
q R
+∞
FS {e } = F (λ) = π2 0 e−at sin(λt)dt
−ax

q q
−at
= π2 [ ae2 +λ2 (−asin(λt) − λcos(λt)]+∞
0 = 2 λ
π a2 +λ2
2

Example 2 Find the Fourier Cosine transform of f (x) = e−ax .

Solution: By definition,
q R
+∞
FC {e−ax } = F (λ) = π2 0 e−at cos(λt)dt
q q
−at
= π2 [ ae2 +λ2 (−acos(λt) + λsin(λt)]+∞
0 = 2 a
π a2 +λ2
2

Example 3 Find the Fourier Cosine transform of f (x) = e−2x + 4e−3x .

Solution: By definition,
q R
−2x +∞
FC {e + 4e } = F (λ) = π2 0 (e−2t + 4e−3t )cos(λt)dt
−3x
 3

q q q q
= 2 2
π 4+λ2
+4 2 3
π 9+λ2
= 2 2
π 4+λ2
+ 2 12
π 9+λ2
2

Example 4 Find the Fourier Sine transform of f (x) = x1 .

Solution: By definition,
q R
+∞
1
FS { x } = F (λ) = π2 0 1t sin(λt)dt
1 λ
Let λt = s. Then t
= s
and dt = λ1 ds.
Thus, we get
q R q R q
+∞ +∞
F (λ) = π2 0 λs sin(s). λ1 ds = π2 0 sins 2 π
pπ
s
ds = ( )
π 2
= 2
R +∞
since 0 sins
s
ds = π2 . 2

Leibniz’s Rule: Let f (x, λ) and its derivative ∂f∂λ

be continuous functions of their vari-
R +∞
ables with −∞ < x < +∞ and −∞ < λ < +∞. Furthermore, let −∞ |f (x, λ)|dx be
R +∞
finite and | ∂f
∂λ
| ≤ h(x) piecewise continuous and that −∞ h(x)dx be finite. Then
Z +∞ Z +∞
d d
f (x, λ)dx = f (x, λ)dx (7)
dλ −∞ −∞ dλ

e−ax
Example 5 Find the Fourier Sine transform of f (x) = x
.
Solution: By definition,
q R
−ax +∞ −at
FS { e x } = F (λ) = π2 0 e t sin(λt)dt
Applying Leibniz’s rule, we get
q R q
d 2 +∞ e−at 2
dλ
F (λ) = π 0 t
. tcos(λt)dt = . a
π a2 +λ2
q R q
F (λ) = 2
π
a
a2 +λ2
dλ = 2
π
tan−1 λa +C
0 ⇒ C = 0.
If λ = 0, then F (λ) =q
Thus, we get F (λ) = 2
π
tan−1 λa . 2

Example 6 Find the Fourier Cosine transform of f (x) = x if 0 < x < 12 ; f (x) = 1 − x if
1
2
< x < 1; f (x) = 0 if x > 1.
Solution: By definition,
q R
+∞
FC {f (x)} = F (λ) = π2 0 f (t)cos(λt)dt
q R1 q R q R
2 2 1 2 +∞
= π
2
0
tcos(λt)dt + π 1 (1 − t)cos(λt)dt + π 1
0cos(λt)dt
2

1
q q R1 q q R
2 t 2 1 2 1−t 2 1 1
= [ sin(λt)]02
π λ
− π 0 λ
sin(λt)dt
2
+ [ sin(λt)]11
π λ
+ π 1
λ
sin(λt)dt
2 2

1 1
q q
2
= {[ t sin(λt)]02
π λ
+ [ λ12 cos(λt)]02 } + 2 1−t
[ sin(λt)
π λ
− 1
λ2
cos(λt)]11
2
4

q q
2 1
= [ sin λ2
π 2λ
+ 1
λ2
cos λ2 − 1
λ2
] + 2
π
[− λ12 cosλ − 1
2λ
sin λ2 + 1
λ2
cos λ2 ]
q
2 2
= [ cos λ2
π λ2
− 1
λ2
cosλ − 1
λ2
]
q
Thus, F (λ) = − 2 1
[ (1
π λ2
+ cosλ − 2cos λ2 ]. 2

Example 7 Find the Fourier Sine and Cosine transform of f (x) = e−x cosx.
Solution: By definition,
q R
+∞
FC {e−x cosx} = F (λ) = π2 0 e−t costcos(λt)dt
q R +∞
= 21
π2 0
e−t [cos(λ + 1)t + cos(λ − 1)t]dt
q
−t
= 21
[ e
π 2 1+(λ+1)2
{−cos(λ + 1)t + (λ + 1)sin(λ + 1)t}]∞
0
q
e−t ∞
+ π2 12 [ 1+(λ−1) 2 {−cos(λ − 1)t + (λ − 1)sin(λ − 1)t}]0

q
21
= { 1
π 2 1+(λ+1)2
+ 1
1+(λ−1)2
}
q q
2 1 2(λ2 +2) 2 λ2 +2
= π 2 λ4 +4
= π λ4 +4

By definition,
q R
2 +∞
FS {e−x cosx} = F (λ) = π 0
e−t costsin(λt)dt
q R +∞
= 21
π2 0
e−t {sin(λ + 1)t + sin(λ − 1)t}dt
q
−t
= 21
[ e
π 2 1+(λ+1)2
{−sin(λ + 1)t − (λ + 1)cos(λ + 1)t}]∞
0
q
e−t ∞
+ π2 12 [ 1+(λ−1) 2 {−sin(λ − 1)t − (λ − 1)cos(λ − 1)t}]0

q
21
= { λ+1
π 2 1+(λ+1)2
+ λ−1
1+(λ−1)2
}
q q
2 1 2λ3 2 λ3
= π 2 λ4 +4
= π λ4 +4
2

The Finite Fourier Sine and Cosine Transforms

The Finite Fourier Sine Transform of a function f defined in the interval (0, L) is denoted
and defined as
Z L
nπx
FS {f (x)} = fS (n) = f (x)sin dx, n ∈ Z (8)
0 L
In this case, the function f (x) is called the Inverse Finite Fourier Sine Transform of fS (n)
which is denoted and defined as
 5

∞
2X nπx
F−1
S {fS (n)} = f (x) = fS (n)sin , n∈Z (9)
L n=1 L

The Finite Fourier Cosine Transform of a function f defined in the interval (0, L) is
denoted and defined as
Z L
nπx
FC {f (x)} = fC (n) = f (x)cos dx, n ∈ Z (10)
0 L
In this case, the function f (x) is called the Inverse Finite Fourier Cosine Transform of
fC (n) which is denoted and defined as
∞
1 2X nπx
F−1
C {fC (n)} = f (x) = fC (0) + fC (n)cos , n∈Z (11)
L L n=1 L

Applications to Initial-Boundary Value Problems

Example 1 Solve the following model using Finite Fourier Sine Transform:

∂U ∂ 2U
= 3 2 , 0 < x < 2, t > 0 (12)
∂t ∂x
U (0, t) = U (2, t) = 0, t ≥ 0
U (x, 0) = x, 0 ≤ x ≤ 2
Solution: Taking Finite Fourier Sine Transform on both sides of (12), we get
Z 2 Z 2 2
∂U nπx ∂ U nπx
sin dx = 3 2
sin dx (13)
0 ∂t 2 0 ∂x 2
R2
Let u(n, t) = 0 U (x, t)sin nπx 2
dx. Then
d
R2 R2 2
dt
u(n, t) = 0 ∂U ∂t
sin nπx
2
dx = 3 0 ∂∂xU2 sin nπx 2
dx, by (13)
R2
= 3[sin nπx ∂U 2
2 ∂x 0
] − 3nπ
2 0
cos nπx
2 ∂x
∂U
dx
3n2 π 2 2
= −3 nπ nπx
sin nπx
2
R
2
[cos 2
U (x, t)] 0 − 4 0 2
U (x, t)dx
2 2 2
= 0 − 3n4π 0 U (x, t)sin nπx
R
2
dx, since given U (0, t) = U (2, t) = 0
2 2
= − 3n4π u(n, t)
du 2 2 2 2 2 2
Therefore, dt
= − 3n4π u ⇒ du
u
= − 3n4π dt ⇒ logu = − 3n4π t + logC
3n2 π 2
⇒ u(n, t) = Ce− 4
t

R2 R2
We have u(n, t) = 0 U (x, t)sin nπx
2
dx ⇒ u(n, 0) = 0 U (x, 0)sin nπx
2
dx
R2
= 0 xsin nπx
2
dx, since given that U (x, 0) = x
6

2x
R2
= − nπ cos nπx
2 0
]2 + 2
nπ 0
cos nπx
2
dx
4 4
= − nπ cosnπ + n2 π 2
[sin nπx
2 0
]2 4
= − nπ (−1)n
3n2 π 2
4
Hence C = u(n, 0) = − nπ 4
(−1)n and u(n, t) = − nπ (−1)n e− 4
t

Finally, taking Inverse Finite Fourier Sine Transform, we get

∞ 3n2 π 2
2 4
(−1)n e− t
sin nπx
P
U (x, t) = 2
− nπ 4
2
n=1
∞ 3n π 2 2
4
(−1)n+1 sin nπx e− 4 t 2
P
= nπ 2
n=1

Example 2 Solve the following model using Finite Fourier Sine Transform:

∂U ∂ 2U
= , 0 < x < π, t > 0 (14)
∂t ∂x2
U (0, t) = U (π, t) = 0, t ≥ 0
U (x, 0) = 2x, 0 ≤ x ≤ π
Solution: Taking Finite Fourier Sine Transform on both sides of (14), we get
Z π Z π 2
∂U ∂ U
sin(nx)dx = 2
sin(nx)dx (15)
0 ∂t 0 ∂x
Rπ
Let u(n, t) = 0 U (x, t)sin(nx)dx. Then
d
Rπ Rπ 2
dt
u(n, t) = 0 ∂U ∂t
sin(nx)dx = 0 ∂∂xU2 sin(nx)dx, by (15)
Rπ
= sin(nx) ∂U ]π − n 0 cos(nx) ∂U
∂x 0 ∂x
dx
R π ∂U
= 0 − n 0 ∂x cos(nx)dx
Rπ
= −n[cos(nx)U (x, t)]π0 − n2 0 U (x, t)sin(nx)dx
= 0 − n2 u(n, t), since it is given that U (0, t) = U (π, t) = 0
du du
Therefore, dt
= −n2 u ⇒ u
= −n2 dt ⇒ logu = −n2 t + logC
2t
⇒ u(n, t) = Ce−n
Rπ Rπ
We have u(n, t) = 0 U (x, t)sin(nx)dx ⇒ u(n, 0) = 0 U (x, 0)sin(nx)dx
Rπ
= 0 2xsin(nx)dx, since it is given that U (x, 0) = 2x
Rπ
= −2[ xsin(nx)
n
]π0 + n2 0 cos(nx)dx
2π 2
=0− n
cos(nπ) + n2
[sin(nx)]π0 = − 2π
n
cos(nπ)
2 2t
Hence C = u(n, 0) = − 2π
n
(−1)n and u(n, t) = Ce−n t = − 2π
n
(−1)n e−n
Finally, taking Inverse Finite Fourier Sine Transform, we get
 7

∞
2
2
− 2π (−1)n e−n t sin(nx)
P
U (x, t) = π n
n=1
∞
2t
(−1)n+1 n1 sin(nx)e−n
P
=4
n=1
2